Modeling and Control in Vibrational and Structural Dynamics: A Differential Geometric Approach (Chapman & Hall CRC Applied Mathematics & Nonlinear Science)

CHAPMAN & HALL/CRC APPLIED MATHEMATICS AND NONLINEAR SCIENCE SERIES MODELING AND CONTROL IN VIBRATIONAL AND STRUCTURAL...

Author: Peng-Fei Yao

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CHAPMAN & HALL/CRC APPLIED MATHEMATICS AND NONLINEAR SCIENCE SERIES

MODELING AND CONTROL IN VIBRATIONAL AND STRUCTURAL DYNAMICS A Differential Geometric Approach

Peng-Fei Yao Chinese Academy of Sciences Beijing, China

© 2011 by Taylor & Francis Group, LLC

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2011 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper Version Date: 20110506 International Standard Book Number: 978-1-4398-3455-8 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com


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To Professor John E. Lagnese


Contents

List of Symbols

ix

Preface

xi

1 Preliminaries from Differential Geometry 1.1 Linear Connections, Differential of Tensor Fields and Curvature 1.2 Distance Functions . . . . . . . . . . . . . . . . . . . . . . . 1.3 A Basic Computational Technique . . . . . . . . . . . . . . . 1.4 Sobolev Spaces of Tensor Field and Some Basic Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 21 38 49

2 Control of the Wave Equation with Variable Coefficients in Space 57 2.1 How to Understand Riemannian Geometry as a Necessary Tool for Control of the Wave Equation with Variable Coefficients 57 2.2 Geometric Multiplier Identities . . . . . . . . . . . . . . . . . 60 2.3 Escape Vector Fields and Escape Regions for Metrics . . . . 63 2.4 Exact Controllability. Dirichlet/Neumann Action . . . . . . . 73 2.5 Smooth Controls . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.6 A Counterexample without Exact Controllability . . . . . . . 108 2.7 Stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 2.8 Transmission Stabilization . . . . . . . . . . . . . . . . . . . 120 2.9 Notes and References . . . . . . . . . . . . . . . . . . . . . . 125 3 Control of the Plate with Variable Coefficients in Space 3.1 Multiplier Identities . . . . . . . . . . . . . . . . . . . . . . 3.2 Escape Vector Fields for Plate . . . . . . . . . . . . . . . . 3.3 Exact Controllability from Boundary . . . . . . . . . . . . 3.4 Controllability for Transmission of Plate . . . . . . . . . . 3.5 Stabilization from Boundary for the Plate with a Curved Middle Surface . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Notes and References . . . . . . . . . . . . . . . . . . . . .

. . . .

127 127 132 135 144

. .

150 158

4 Linear Shallow Shells. Modeling, and Control 4.1 Equations in Equilibrium. Green’s Formulas . . . . . . . . . 4.2 Ellipticity of the Strain Energy of Shallow Shells . . . . . . .

161 161 171 vii


viii 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . Multiplier Identities . . . . . . . . . . . . . . . . . . . . . . . Escape Vector Field and Escape Region for the Shallow Shell Observability Inequalities. Exact Controllability . . . . . . . Exact Controllability for Transmission . . . . . . . . . . . . . Stabilization by Linear Boundary Feedback . . . . . . . . . . Stabilization by Nonlinear Boundary Feedback . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . .

175 181 188 199 216 229 239 246

5 Naghdi’s Shells. Modeling and Control 5.1 Equations of Equilibrium. Green’s Formulas. Ellipticity of Strain Energy. Equations of Motion . . . . . . . . . . . . 5.2 Observability Estimates from Boundary . . . . . . . . . . 5.3 Stabilization by Boundary Feedback . . . . . . . . . . . . 5.4 Stabilization of Transmission . . . . . . . . . . . . . . . . 5.5 Notes and References . . . . . . . . . . . . . . . . . . . .

the . . . . . . . . . .

249 249 261 270 277 286

6 Koiter’s Shells. Modeling and Controllability 6.1 Equations of Equilibria. Equations of Motion . 6.2 Uniqueness for the Koiter Shell . . . . . . . . . 6.3 Multiplier Identities . . . . . . . . . . . . . . . 6.4 Observability Estimates from Boundary . . . . 6.5 Notes and References . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

287 287 301 311 315 328

7 Control of the Quasilinear Wave Equation in Higher Dimensions 7.1 Boundary Traces and Energy Estimates . . . . . . . 7.2 Locally and Globally Boundary Exact Controllability 7.3 Boundary Feedback Stabilization . . . . . . . . . . . . 7.4 Structure of Control Regions for Internal Feedbacks . 7.5 Notes and References . . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

329 330 343 355 372 385

. . . . .

. . . . .

. . . . .

Bibliography

387

Index

403


List of Symbols

R (M, g) C ∞ (M ) = T 0 (M ) X (M ) = T (M ) = Λ(M ) T k (M ) D [·, ·] Γkij ⊗ D2 f tr S RXY R(·, ·, ·, ·) Φ∗ κ(Ξ) N Π(·, ·) Ric (·, ·) L(γ) d(x, y) Σ(x0 ) expx0 Σ(x0 ) ∆ ∆g ∆ Λk (M ) ∧ i (X)T div X d Rn Ω Γ = ∂Ω L2 (Ω, T k ) L2 (Ω, Λk ) H k (Ω, Λ)

set of real numbers Riemannian manifold with metric g all C ∞ functions on M all vector fields on M all tensor fields of rank k on M Levi–Civita connection Lie bracket Christoffel symbol tensor product Hessian of f in metric g trace of tensor S of rank 2 curvature operator curvature tensor differential of map Φ sectional curvature of subspace Ξ of dimension 2 normal vector field of hypersurface second fundamental form Ricci tensor length of curve γ distance between x and y in metric g interior of tangent cut locus of x0 interior of cut locus of x0 Laplacian in metric Laplacian in metric g Hodge–Laplacian all k-forms exterior product interior product of tensor field T by vector field X divergence of vector field X exterior derivative Euclidean space of dimension n bounded open set in Rn , or middle surface of plate or shell boundary of Ω all square integrable tensor fields of rank k on Ω all square integrable k-forms on Ω all 1-forms with all its i-th differential square integrable ix


x Q (·, ·) and k · k ν 1 Υ = (g − g) 2 ρ=Π−Π µ

lo (ζ), L(ζ) ℵ Bm D (Σ0 ) Bm N (Σ0 )

for 0 ≤ i ≤ k on Ω formal adjoint of covariant differential operator inner product and norm of L2 (Ω), respectively outside normal of boundary in Euclidean metric strain tensor of middle surface change of curvature tensor of middle surface Poisson’s coefficient L2 (Ω) = (L2 (Ω, Λ))2 × (L2 (Ω))2 H1 (Ω) = (H 1 (Ω, Λ))2 × (H 1 (Ω))2 HΓ1ˆ (Ω) = (HΓˆ1 (Ω, Λ))2 × (HΓˆ1 (Ω))2 lower order terms canonical isomorphism space of boundary control in Dirichlet action space of boundary control in Neumann action


Preface

Control theory of partial differential equations is a many-faceted subject. Created to describe the control behavior of mechanical objects such as wave equations, plates, and shells, it has developed into a body of material that interacts with many branches of mathematics and mechanics, such as differential geometry and elasticity. The differential geometrical approach was introduced more than a decade ago. Since then, many important advances in modeling and control in vibrational and structural dynamics have been made. This book is intended to present a systematic, updated account of this direction of research. The original motivation for the differential geometrical approach may be said to have arisen from the need to cope with the following three situations: (i) The case where the coefficients of the partial differential equations are variable in space (waves/plates); (ii) The case where the partial differential equations themselves are defined on curved surfaces (shells); (iii) The case where the systems have quasilinear principal parts. Regarding the first need (i), the differential geometric energy methods described in this book may be viewed as far-reaching generalizations of the classical energy methods of the early/mid-1980s. The first reason this approach is necessary is that the curvature theory in Riemannian geometry provides us with checkable assumptions for controllability/stabilization for the systems with variable coefficients. The classical analysis was originally successful in dealing with the canonical (constant coefficients) wave/plate equations, but proved inadequate for treating variably coefficient cases mainly because controllability/stabilization is a global property while the classical case corresponds to the zero curvature. Although the curvature is locally defined at each point, it interprets global information on almost everything on manifolds, including controllability/stabilization. That is one of the reasons Riemannian geometry is called global geometry. As to the second need (ii), it would seem self-explanatory that if an equation is defined on a manifold (as in the case of shells), the natural setting that is necessary for its analysis should be that of differential geometry. In particular, modeling of shells in the form of free coordinates is the key step in analysis and control. Classically, the topic of static shells is covered by many books. They all assume the middle surface of a shell to be described by one coordinate patch: this is the image in R3 of a smooth function defined on a conxi © 2011 by Taylor & Francis Group, LLC

xii nected domain of R2 , rooted in classical differential geometry. This view has geometrical limitations, as it forces the exclusion of interesting objects such as a sphere. Moreover, the classical models use traditional geometry and end up with highly complicated resultant equations. In these, the explicit presence of the Christoffel symbols makes them unsuitable for energy method computations of the type needed for continuous observability/stabilization estimates. We view a middle surface of a shell as a Riemannian manifold of dimension 2 that is not subject to any particular coordinate patch. A basic computational technique, first used by Bochner, allows us to have mathematical models for shells in the form of free coordinates and to carry out multiplier schemes for controls. The basic ideas are as follows: In order to verify an identity or a pointwise estimate, it suffices to do so at each point p relative to a coordinate system or frame field that offers the greatest computational simplification. What prevents a computation from being as simple as the corresponding classical (i.e., Euclidean) situation is the presence of the Christoffel symbols. Thus, we use a coordinate system or a frame field relative to which the Christoffel symbols vanish at the given point p. Such a coordinate system varies with point p of the middle surface. Clearly, once the middle surface of a shell is fixed by one particular coordinate patch, this technique does not work. More importantly, it seems clear that, without this differential geometric tool, many of these more sophisticated theorems would probably not have been discovered or, at least, their discoveries would otherwise have been delayed. As the third need (iii), the quasilinearity of models comes from nonlinear materials, and their controllability/stabilization depends not only on their variable coefficients but also on their equilibria. Similarly, as a necessary tool, the curvature theory yields checkable assumptions. This book will cover the following topics: (i) control of the wave equation with variable coefficients (Chapter 2); (ii) control of plates with variable coefficients (Chapter 3); (iii) modeling and control of shallow shells (Chapter 4); (iv) modeling and control of Naghdi’s shells (Chapter 5); (v) modeling and control of Koiter’s shells (Chapter 6); (vi) control of the quasilinear wave equation (Chapter 7), with particular emphasis on research results using the differential geometrical approach. In order to make this book self-contained, I have condensed the main content of the book, Introduction to Riemannian Geometry (in Chinese), by H. Wu, C. L. Shen, and Y. L. Yu, into Chapter 1. I hope this chapter will help readers who are not familiar with Riemannian geometry to obtain some necessary knowledge for our problems. I thank Goong Chen for inviting me to write this book. I have greatly profited from the comments and thoughtful suggestions of many of my colleagues, friends and students, in particular: X. Cao, M. M. Cavalcanti, S. G. Chai, G. Chen, M. C. Delfour, L. Deng, S. Feng, D. X. Feng, H. Gao, B. Z. Guo, Y. X. Guo, V. Komornik, M. Krstic, I. Lasiecka, T. T. Li,


xiii S. J. Li, S. Li, B. Miara, B. P. Rao, Z. C. Shao, R. Triggiani, M. Tucsnak, G. Weiss, H. G. Wu, J. Q. Wu, B. Y. Zhang, and Z. F. Zhang. I am extremely grateful for the suggestions and lists of mistakes from drafts of this book sent to me by S. G. Chai, L. Deng, Y. X. Guo, S. Li, Z. C. Shao, J. Q. Wu, and Z. F. Zhang. I have been supported by the National Science Foundation of China during the writing, most recently under grants no. 60821091, no. 60334040, no. 60221301, no. 60774025, no. 10831007, and no. YB20098000101(the excellent PhD adviser program of Beijing).

Peng-Fei Yao Key Laboratory of Systems and Control Institute of Systems Science Academy of Mathematics and Systems Science Chinese Academy of Sciences Beijing 100190, P. R. China E-mail: [email protected]


Chapter 1 Preliminaries from Differential Geometry

1.1 1.2 1.3 1.4

Linear Connections, Differential of Tensor Fields and Curvature . . . . . . . . Exercise 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Distance Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A Basic Computational Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sobolev Spaces of Tensor Field and Some Basic Differential Operators . Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 19 21 37 38 48 48 56

This chapter will cover some necessary knowledge of Riemann geometry which will be needed in the subsequent chapters to make this book self-contained as far as possible. The main reference is book [206] in Chinese.

1.1

Linear Connections, Differential of Tensor Fields and Curvature

Connections Let M be a C ∞ manifold of dimension n. We say that g is a Riemannian metric on M which means that for each tangential space Mx (x ∈ M ), there is an inner product gx (·, ·) (sometimes denoted by h·, ·i) on Mx and this relation is C ∞ . Here the C ∞ means: For any local coordinate (x1 , · · · , xn ), if we let gij (x) = gx (∂xi , ∂xj ), ∂ where ∂xi = ∂x , then they are C ∞ functions of the coordinate (x1 , · · · , xn ). i Note that the inner products gx (·, ·) are equal to that the matrices (gij (x))n×n are positive, symmetrical. In the terminology of tensors, g is actually a C ∞ covariant tensor field of rank 2 such that, for any vector fields X, Y on M,

g(X, Y ) = g(Y, X),

g(X, X) ≥ 0,

and gx (X, X) = 0 if and only if X(x) = 0. In this chapter (M, g) will always denote a Riemannian manifold of dimension n with metric g. 1 © 2011 by Taylor & Francis Group, LLC

2

Modeling and Control in Vibrational and Structural Dynamics

The following examples are two of the most important Riemannian manifolds in our book. Example 1.1 Let G(x) be a symmetrical, positive matrix for each x ∈ Rn . We introduce an inner product on each Rnx = Rn by g(X, Y ) = hG(x)X, Y i

for

X, Y ∈ Rxn ,

where h·, ·i is the Euclidean metric in Rn . Then (Rn , g) is a Riemannian manifold. g is the Euclidean metric if and only if G(x) = I is the unit matrix on Rn . Example 1.2 Let M be a hypersurface in Rn+1 . For each x ∈ M, Mx is an n-dimensional tangential plane of M at x. We define an inner product on each Mx by g(X, Y ) = hX, Y i for X, Y ∈ Mx , (1.1) where h·, ·i is the Euclidean metric in Rn+1 . This metric is called the induced metric from Rn+1 . For example, let M = { x = (x1 , · · · , xn+1 ) ∈ R

n+1

|

n+1 X i=1

x2i = 1 },

be the unit sphere in Rn+1 . Let x ∈ M be given. Then α ∈ Mx if and only if there is a curve γ: (−ε, ε) → M such that γ(0) = x and γ(0) ˙ = α. Let γ(t) = (γ1 (t), · · · , γn+1 (t)). Then γ(t) ∈ M implies that n+1 X

γi2 (t) = 1

for

i=1

t ∈ (−ε, ε).

We differentiate the above identity in the variable t at t = 0 to obtain Mx = { α ∈ Rn+1 | hx, αi = 0 }. If xn+1 6= 0, then the induced metric g of (1.1) on Mx is g(α, β) = hα, βi = =

X

n+1 X i=1

αi βi =

n X

αi βi +

i=1

1 x2n+1

αi βj (δij + xi xj /x2n+1 )

n X

αi βj xi xj

ij=1

ij

for any α = (α1 , · · · , αn+1 ), β = (β1 , · · · , βn+1 ) ∈ Mx . Before any study of Riemannian metrics, let us introduce the concept of “connections.” Roughly speaking, a “connection” is a means to differentiate tensor fields. Let us recall how we differentiate a vector field in Rn . If v is a


Preliminaries from Differential Geometry

3

vector at p ∈ Rn and f is a function differentiable at p, then the directional derivative Dv f can be defined, as a real number, by Dv f = lim

t→0

f (p + tv) − f (p) . t

Let X be a vector field at p. Then X = (X1 , · · · , Xn ) where Xi are given by the equation n X X= Xi ∂xi , i=1

and (x1 , · · · , xn ) is the natural coordinates in Rn . Then the derivative of the vector field X along the vector v is defined by Dv X = (Dv X1 , · · · , Dv Xn ), or Dv X =

n X

(Dv Xi )∂xi .

i=1

Clearly, Dv X satisfies the following properties: (a) Dαv X = αDv X for any α ∈ R; (b) Dv (f X) = (Dv f )X + f Dv X for any function f ; (c) Dv (X1 + X2 ) = Dv X1 + Dv X2 for any vector fields X1 and X2 ; (d) Dv1 +v2 X = Dv1 X + Dv2 X for any vectors v1 and v2 . The above properties are exactly ones of the directional derivatives. Let V be a vector field such that V (p) = v. We define a vector field DV X by (DV X)(p) = Dv X. Then properties (a)-(d) become (C1) Df V +gW X = f DV X + gDW X; (C2) DV (f X) = (DV f )X + f DV X; (C3) DV (X + Y ) = DV X + DV Y, where V, W, X, and Y are vector fields on Rn and f, g are functions on Rn , respectively. Let us summarize the points of the above arguments: Given vector fields V and X in Rn , we define a directional derivative DV X which satisfies the properties (C1)-(C3). Now we come back to manifolds. Denote by C ∞ (M ) and X (M ) all C ∞ functions and vector fields, respectively, on manifold M. For X ∈ X (M ) and f ∈ C ∞ (M ), V (f ) denotes the directional derivative of f along the vector field V. Then V (f ) ∈ C ∞ (M ). Definition 1.1 Let M be a C ∞ manifold. A connection is a map D: X (M )× X (M ) → X (M ) which satisfies properties (C1)−(C3). In (C1)−(C3), DV f = V (f ).


4


Remark 1.1 Given a connection, DV X is called the covariant derivative of the vector field X along the vector field V. Then D is also called a covariant differential. Traditionally, “covariant” means that something varies with local coordinates. In the classical differential geometry, everything was done by local coordinates. However, what we can do is actually something which is independent of local coordinates. Here the definition of a connection is free of local coordinates. We do not use local coordinates except when they are necessary. Remark 1.2 Property (C1) implies that, if V and W are vector fields on M such that for some x ∈ M, V (x) = W (x), then (DV X)(x) = (DW X)(x)

(1.2)

(Exercise 1.1.1). Let D be a connection. By Remark 1.2, for any v ∈ Mx , we may define Dv X as follows: Taking any vector field V ∈ X (M ) with V (x) = v, let Dv X = (DV X)(x). Remark 1.3 Unlike that in Remark 1.2, for v ∈ Mx and V ∈ X (M ), we are not able to define DV v by extending v to a vector field (Exercise 1.1.2). Nevertheless, the following things are true and very useful: Let D be a connection on manifold M. Let v ∈ Mx and let γ: (−ε, ε) → M be a curve with γ(0) ˙ = v. We further assume that X, Y ∈ X (M ) are such that X(γ(t)) = Y (γ(t))

for

t ∈ (−ε, ε).

Then Dv X = Dv Y

(1.3)

(Exercise 1.1.3). There is a consequence of the relation (1.3): If γ: [0, a] → M is a curve and X ∈ X (M ) is a vector field, then Dγ(0) X is determined by X ◦ γ completely ˙ where X ◦ γ(t) = X(γ(t)). Remark 1.4 There are many connections on a manifold. For example, we can construct one for each local coordinate and then put them together by a partition of unity. The significant one among all connections is the so-called Levi-Civita connection which is the one uniquely given by the metric in Theorem 1.1 below. Theorem 1.1 Let M be a Riemannian manifold with the metric g = h·, ·i. Then there is a unique connection D that is determined by the following formulas: For vector fields X, Y and Z, (L1) XhY, Zi = hDX Y, Zi + hY, DX Zi; (L2) DX Y = DY X + [X, Y ] where [X, Y ] = XY − Y X is the Lie bracket.



5

Proof. First, we prove the uniqueness. Consider a local coordinate (x1 , · · · , xn ). Let functions Γkij be given by the following formulas D∂xi ∂xj =

n X

Γkij ∂xk

for

k=1

1 ≤ i, j ≤ n.

(1.4)

For our purposes, it suffices to prove that functions Γkij are uniquely determined by the metric g. As usual, let gij = h∂xi , ∂xj i. Then the formula (L1) implies that gjkxi = ∂xi gjk = ∂xi h∂xj , ∂xk i = hD∂xi ∂xj , ∂xk i + h∂xj , D∂xi ∂xk i n X (1.5) = (Γlij glk + Γlik gjl ). l=1

On the other hand, it follows from the relations (L2) and (1.4) that 0 = D∂xi ∂xj − D∂xj ∂xi − [∂xi , ∂xj ] =

n X

k=1

[Γkij − Γkji ]∂xk ,

that is, Γkij = Γkji

for 1 ≤ i, j, k ≤ n.

(1.6)

We take indexes i, j, and k in the formula (1.5) in turn to give gkixj =

n X

(Γljk gli + Γlji gkl ),

(1.7)

l=1

gijxk =

n X (Γlki glj + Γlkj gil ).

(1.8)

l=1

Summing up the formulas (1.5) and (1.7) minus (1.8) and using the formula (1.6), we obtain n X 2 Γlij gkl = gjkxi + gkixj − gijxk , l=1

or

n

Γkij =

1 X kl g (gjlxi + glixj − gijxl ) 2 l=1

for 1 ≤ i, j, k ≤ n,

(1.9)

where (g ij ) = (gij )−1 , which show that Γkij are uniquely determined by the metric g. To complete the proof, it remains to show that there exists a connection D which satisfies the relations (L1) and (L2). We do it as follows. Let U be any chart with a coordinate function (x1 , · · · , xn ). Let Γkij be given on U by the formulas (1.9). Let X and Y be vector fields on M which are given by X=

n X

fi ∂xi ,

i=1


Y =

n X i=1

hi ∂xi

on U.

6


We let DX Y =

n X

fi (hjxi +

ij=1

n X l=1

hl Γjil )∂xj

on U.

It is easy to check, as an exercise, that (a) If V is another chart, then DX Y is the same on V ∩ U; (b) D is a connection satisfying the relations (L1) and (L2).

Definition 1.2 The connection D, determined by Theorem 1.1 is called the Levi-Civita connection of M. In the sequel, D will always denote the Levi-Civita connection. Definition 1.3 In the formulas (1.9), Γkij are called the Christoffel symbols. The formula (1.9) can be written as a form of coordinates free. For any vector fields X, Y, and Z, hDX Y, Zi =

1 {XhY, Zi + Y hZ, Xi − ZhX, Y i + hZ, [X, Y ]i 2 +hY, [Z, X]i − hX, [Y, Z]i}.

(1.10)

If we take X, Y, and Z as ∂xi , ∂xj , and ∂xk , respectively, in the formula (1.10), then we have the formula (1.9). Conversely, it is easy to verify (1.10) by (1.9) (Exercise 1.1.5). A useful means for computation is the following: Definition 1.4 n fields E1 , · · · , En , defined on some open set O of M, is said to be an orthonormal frame (basis) if hEi , Ej i = δij for x ∈ O and 1 ≤ i, j ≤ n. Let (x1 , · · · , xn ) be a local coordinate. We orthogonalize ∂x1 , · · · , ∂xn to have Theorem 1.2 Let M be a Riemannian manifold of dimension n. Let U be a chart. Then there exists an orthonormal frame on U. Let E1 , · · · , En be an orthonormal frame on M. It follows from the formula (1.10) that hDEi Ej , Ek i =

1 {hEk , [Ei , Ej ]i + hEj , [Ek , Ei ]i − hXi , [Ej , Ek ]i}. 2

The above equations determine the Levi-Civita connection D completely. Parallel Transport To understand what a connection connects, we need the concept of parallel transports. Let γ: [a, b] → M be an embedded curve. Let X be a vector field along γ, that is, for each t ∈ [a, b], X(t) ∈ Mγ(t) such



7

that X(t) is C ∞ with respect to the variable t. Since γ is embedded, X can ˜ on M, that is, be extended to be a vector field X ˜ X(γ(t)) = X(t) for t ∈ [a, b]. ˜ is independent of the extension X, ˜ so we write as By Remark 1.3, Dγ(t) X ˙ Dγ(t) X but not mentioning the extension. ˙ Question When does Dγ(t) X = 0 for all t ∈ [a, b]? ˙ Pn Let us write D X = 0 in detail. Denote γ(t) ˙ = i=1 γ˙ i (t)∂xi (γ(t)) and γ(t) ˙ Pn X(t) = i=1 Xi (t)∂xi (γ(t)). Using the formula (1.4), we have X Dγ(t) X = Dγ(t) (Xj (t)∂xj (γ(t))) ˙ ˙ j

=

X

X˙ j ∂xj (γ(t)) +

j

=

X

X

Xj (t)Dγ(t) ∂xj ˙

j

[X˙ k (t) +

k

X

Γkij γ˙ i (t)Xj ]∂xk .

ij

Then Dγ(t) X = 0 is equivalent to ˙ X X˙ k (t) + Γkij γ˙ i (t)Xj = 0 ij

for 1 ≤ k ≤ n,

(1.11)

which is a system of linear ordinary differential equations. Its unknown is X = (X1 (t), · · · , Xn (t))τ where the superscript τ denotes the transpose of a vector. Thus, for any v ∈ Mγ(a) given, there is a unique solution X(t) such that X(a) = v, Dγ(t) X = 0. ˙ Remark 1.5 The linearity of the problem (1.11) guarantees its solutions to exist on the whole interval [a, b]. Otherwise, we only know that its solutions exist on a small interval [a, a + ε). Definition 1.5 A vector field X is said to be parallel along γ if Dγ(t) X = 0. ˙ A vector w ∈ Mγ(b) is called the parallel transport of a vector v ∈ Mγ(a) if there is a parallel vector field X along γ such that X(a) = v and X(b) = w. Remark 1.6 In the above argument, we assumed that γ is contained in a local chart. Clearly, it is not necessary. We could divide the interval [a, b] into many small intervals [a, a1 ], · · · , and [ak , b] such that each is in one chart. For any v ∈ Mγ(a) , we parallelize it to X(a1 ) ∈ Mγ(a1 ) along [a, a1 ], and then to X(a2 ) ∈ Mγ(a2 ) along [a1 , a2 ]. Repeating this procedure, we get X(b) = w. We have proved Theorem 1.3 Let γ(t): [a, b] → M be an embedded curve. For each v ∈ Mγ(a) , there is a unique w ∈ Mγ(b) such that it is the parallel transport of v along γ.


8


From Theorem 1.3, for each embedded curve γ: [a, b] → M, we are able to define a map Pγ : Mγ(a) → Mγ(b) such that Pγ v is the parallel transport of v along γ. Then Pγ is a linear isomorphism from Mγ(a) to Mγ(b) . Definition 1.6 Pγ is called a parallel isomorphism, given by the connection D. Now we explain what a connection connects. Let D be a connection of M. If x, y ∈ M are connected by a curve γ, then D yields a parallel isomorphism Pγ : Mx → My . In this sense, D connects any two tangential spaces Mx and My if there is a curve which connects x with y. For a curve, if its tangential vector field is parallel along itself, it will be especially interesting, that is, Definition 1.7 A curve γ is called a geodesic of the connection D if Dγ(t) γ˙ = ˙ 0. A geodesic exists at least locally. Let (x1 , · · · , xn ) be a local coordinate. Denote γ = (γ1 , · · · , γn ) where γi = xi ◦ γ. Then X γ(t) ˙ = γ˙ i (t)∂xi (γ(t)), i

i where γ˙ i (t) = dγ ˙ and the formulas (1.11) that γ is dt (t). It follows from X = γ a geodesic if and only if X γ¨k + (Γkij ◦ γ)γ˙ i γ˙ j = 0 for 1 ≤ k ≤ n. (1.12)

ij

This is a system of nonlinear ordinary differential equations of rank 2. We only have a unique local solution. Since (γ1 (0), · · · , γn (0)) and (γ˙ 1 (0), · · · , γ˙ n (0)) can be given arbitrarily, we have Theorem 1.4 Given v ∈ Mx (x ∈ M ) and a connection D of M, there exists a unique geodesic γ such that γ(0) = x,

γ(0) ˙ = v.

Differential of Tensor Field Let W be a linear space with an inner product h·, ·i and let k ≥ 0 be an integer. A k-linear functional α on W is a functional from W k = W ⊗ · · · ⊗ W → R such that, for any X1 , · · · , Xi−1 , Xi+1 , · · · , Xk fixed in W, α(X1 , · · · , Xi−1 , Xi , Xi+1 , · · · , Xk ) is a linear functional with respect to the variable Xi ∈ W. In particular, α is a 0-linear functional on W if and only if α is a constant; α is a 1-linear functional on W if and only if α ∈ W and α(X) = hα, Xi for


X ∈ W.


9

Definition 1.8 Let M be a Riemannian manifold. A C ∞ function is called a 0-rank tensor field on M ; a vector field is called a 1-rank tensor field on M. Let k ≥ 2 be an integer. T is said to be a tensor field of rank k if, for each x ∈ M, T is a k-linear functional on Mx such that T (X1 , · · · , Xk ) ∈ C ∞ (M ) where X1 , · · · , Xk are k vector fields on M. Remark 1.7 Let W be a linear space. In the literature tensors are defined as multiple linear functionals on W ⊗ · · · ⊗ W ⊗ W ∗ ⊗ · · · ⊗ W ∗ where W ∗ is the adjoint space of W. For convenience, we here assume that W has an inner product so that W = W ∗ in the duality sense. We denote by T k (M ) all k-rank tensor fields on M. Then T 0 (M ) = C (M ) and T 1 (M ) = X (M ). Let v ∈ Mx and T ∈ T k (M ) be a tensor field. In order to define Dv T, we consider parallel isometries. Let γ; [0, a] → M be a curve such that γ(0) = x and γ(0) ˙ = v. Let ∞

P(t) : Mx → Mγ(t)

for

t ∈ [0, a]

be the parallel isometry of the connection D along γ. We define Dv T as follows. For v1 , · · · , vk ∈ Mx ,

d [T (γ(t))(P(t)v1 , · · · , P(t)vk )]t=0 . (1.13) dt Then Dv T is again a k-linear functional on Mx . It is easy to check that the formula (1.13) is independent of the choice of the curve γ (Exercise 1.1.8). Dv T is called the covariant derivative of T in the direction v. Since Dv T is still linear in v ∈ Mx , we have Dv T (v1 , · · · , vk ) =

Definition 1.9 Let M be a Riemannian manifold with the Levi-Civita connection D and let k ≥ 0 be an integer. Let T ∈ T k (M ) be a tensor field of rank k. The differential of the tensor field T is a tensor field of rank k + 1 , denoted by DT, to be defined by the formula DT (X1 , · · · , Xk , X) = DX T (X1 , · · · , Xk )

(1.14)

for X, X1 , · · · , Xk ∈ X (M ), where, for each x ∈ M, the right hand side of (1.14) is defined by the formula (1.13). We have Theorem 1.5 (1) Let T1 , T2 be tensor fields. Then D(T1 ⊗ T2 ) = DT1 ⊗ T2 + T1 ⊗ DT2 . (2) Let X1 , · · · , Xk+1 ∈ X (M ) be vector fields and let T ∈ T k (M ) be a tensor field. Then DT (X1 , · · · , Xk+1 ) = Xk+1 (T (X1 , · · · , Xk )) −


k X i=1

T (X1 , · · · , DXk+1 Xi , · · · , Xk ).

(1.15)

10


Proof. Let x ∈ M be given. We assume that T1 ∈ T k (M ) and T2 ∈ T l (M ). Let X ∈ X (M ) and let γ: [0, a] → M be a curve such that γ(0) = x,

γ(0) ˙ = X(x).

For vector fields X1 , · · · , Xk , Y1 , · · · , Yl ∈ X (M ), we have, via the formula (1.13) at x, D(T1 ⊗ T2 )(X1 , · · · , Xk , Y1 , · · · , Yl , X)

= DX (T1 ⊗ T2 )(X1 , · · · , Xk , Y1 , · · · , Yl ) d = [T1 (P(t)X1 (x), · · · , P(t)Xk (x))T2 (P(t)Y1 (x), · · · , P(t)Yl (x))]|t=0 dt = [DX(x) T1 (X1 , · · · , Xk )]T2 (Y1 , · · · , Yl ) +T1 (X1 , · · · , Xk )DX(x) T2 (Y1 , · · · , Yl ) = [(DT1 ) ⊗ T2 ](X1 , · · · , Xk , X, Y1 , · · · , Yl ) +T1 ⊗ (DT2 )(X1 , · · · , Xk , Y1 , · · · , Yl , X).

We assume k = 2. Similar arguments will prove the formula (2) for the other k. Let E1 , · · · , En be a local basis on M. Then Ei ⊗ Ej (1 ≤ i, j ≤ n) forms a local basis in T 2 (M ). Let T =

n X

ij=1

Tij Ei ⊗ Ej .

It follows from the formulas (1) and (L1) that X DX T = [X(Tij )Ei ⊗ Ej + Tij DX Ei ⊗ Ej + Tij Ei ⊗ DX Ej ]. ij

We obtain X X(T (X1 , X2 )) = X[ Tij hX1 , Ei ihX2 , Ej i] ij

=

X ij

[X(Tij )hX1 , Ei ihX2 , Ej i + Tij hX1 , DX Ei ihX2 , Ej i

+Tij hX1 , Ei ihX2 , DX Ej i + Tij hDX X1 , Ei ihX2 , Ej i

+Tij hX1 , Ei ihDX X2 , Ej i]

= DX T (X1 , X2 ) + T (DX X1 , X2 ) + T (X1 , DX X2 ), that is, the formula (1.15) is true for k = 2.

Let f ∈ C ∞ (M ) be a C ∞ function. The differential Df is a vector field which is also called the gradient of the connection D. Sometimes we denote Df by df. For X ∈ X (M ), Df (X) = X(f ) = hDf, Xi for x ∈ M.



11

The differential D2 f of Df is a tensor field of rank 2. For a tensor field T, we denote D2 T = D(DT ), D3 T = D(D2 T ), · · · . Note that in general D2 T (· · · , X, Y ) 6= D2 T (· · · , Y, X)

(1.16)

where “ · · · ” denotes the number of the variables. Definition 1.10 Let f ∈ C ∞ (M ). D2 f is called the Hessian of f in the connection D. In general we cannot change the order of the differential on a tensor field as in (1.16) but if T is a 0 rank tensor field, we have Theorem 1.6 Let f ∈ C ∞ (M ). Then D2 f (X, Y ) = D2 f (Y, X)

(1.17)

for any X, Y ∈ X (M ). Proof. Using the formula (1.15) and (L2), we have D2 f (X, Y ) = Y (Df (X)) − Df (DY X) = Y X(f ) − DY X(f ) = Y X(f ) − DX Y (f ) − [Y, X]f = XY (f ) − DX Y (f ) = D2 f (Y, X). If S is a symmetric 2-rank tensor field, we define its trace tr S to be a function on M by tr S(x) =

n X i=1

S(ei , ei ) for x ∈ M,

(1.18)

where e1 , · · · , en is an orthonormal basis of Mx . It is easy to check that the above definition of tr S is independent of choices of the orthonormal basis (Exercise 1.1.11). For f ∈ C ∞ (M ), the formula (1.17) means that D2 f is a symmetrical tensor field of rank 2. We define ∆f = tr D2 f, and call ∆ the Laplacian acting on functions. It is also called the LaplaceBeltrami operator. If Rn has the Euclidean metric, a simple computation P yields ∆f = i fxi xi , the classical Laplacian. Exercise 1.1.12 presents some formulas for ∆ in a general metric g. The second operator which is important to us is the curvature operator that is defined as follows.


12


Definition 1.11 Let (M, g) be a Riemannian manifold. Let X, Y ∈ X (M ) be vector fields and let k ≥ 0 be an integer. We define a map RXY : T k (M ) → T k (M ) by RXY = −DX DY + DY DX + D[X,Y ] .

RXY is called the curvature operator.

As an exercise (Exercise 1.1.13) for the readers, we have Theorem 1.7 RXY has the following properties: (1) For T1 ∈ T k (M ) and T2 ∈ T l (M ), RXY (T1 ⊗ T2 ) = RXY T1 ⊗ T2 + T1 ⊗ RXY T2 ; (2) For any f ∈ C ∞ (M ) and T ∈ T k (M ), R(f X)Y T = RX(f Y ) T = RXY (f T ) = f RXY T ; (3) For any f ∈ C ∞ (M ), RXY f = 0. Let Z, W ∈ X (M ) be vector fields. Then RXY Z is a vector field. By (2), RXY Z is linear in each variable X, Y, and Z, respectively. We define R(X, Y, Z, W ) = hRXY Z, W i

(1.19)

for vector fields X, Y, Z, W ∈ X (M ). It is a tensor field of rank 4. Definition 1.12 The tensor field R of 4 rank, given by (1.19), is called the curvature tensor. Remark 1.8 Although it looks not to be very complicated, the curvature tensor controls almost everything in (M, g); see comments in [190], II. This tensor field is determined by the second derivatives of the metric g. In a local coordinate, we have the formulas ([190], II, 4D − 7 page) R(∂xi , ∂xj , ∂xk , ∂xl ) =

1 (gilxj xk + gjkxi xl − gikxj xl − gjlxi xk ) 2 X + (grs Γrjk Γsil + grs Γrjl Γsik ) (1.20) rs

where gij = g(∂xi , ∂xj ) and Γkij are the Christoffel symbols. The curvature tensor has the following properties (Exercise 16). Theorem 1.8 For any vector fields X, Y, Z, and W, we have (1) RXY = −RY X ; (2) RXY Z + RY Z X + RZX Y = 0 (the first Bianchi identity); (3) R(X, Y, Z, W ) = −R(X, Y, W, Z); (4) R(X, Y, Z, W ) = R(Z, W, X, Y ).



13

We introduce a quadratic form on Mx ⊗ Mx by Q(X, Y ) = R(X, Y, X, Y ), which determines the curvature tensor completely. Theorem 1.9 Let R and R′ be tensor fields which satisfy the properties (1)(4) in Theorem 1.8. If Q = Q′ on Mx ⊗ Mx , then R = R′ . Proof. Exercise 17.

Now we introduce the concept of sectional curvatures. Definition 1.13 Let Ξ be a 2-dimensional linear subspace of Mx . For any basis v1 , v2 of Ξ, we define the sectional curvature of Ξ by κ(Ξ) =

R(v1 , v2 , v1 , v2 ) |v1 ∧ v2 |2

(1.21)

2

where |v1 ∧ v2 |2 = |v1 |2 |v2 |2 − hv1 , v2 i . It follows from (1)-(4) in Theorem 1.8 that the right hand side of (1.21) is independent of choice of the basis of Π (Exercise 1.1.16). If e1 , e2 is an orthonormal basis of Ξ, then κ(Ξ) = R(e1 , e2 , e1 , e2 ). Remark 1.9 κ is a function which is defined on all 2-dimensional tangential spaces. It follows from Theorem 1.9 that the sectional curvature determines the whole curvature tensor. Remark 1.10 Although sectional curvatures are defined locally at point x ∈ M (we may compute it by the formula (1.20)), they will yield all the global information about the Riemannian manifold (M, g). We will explain this point more later. We have noted that in general it is not possible to change the order of the differentials, see (1.16). However, the following theorem, named the Ricci identity, tells us that their difference is just a curvature term. Theorem 1.10 Let T be a tensor field and let X, Y be vector fields. Then D2 T (· · · , X, Y ) = D2 T (· · · , Y, X) + (RXY T )(· · · ).

(1.22)

Proof. By the formula (1.15), we have D2 T (· · · , X, Y ) = DY (DT )(· · · , X) = DY DX T (· · · ) − DDY X T (· · · ) = DX DY T (· · · ) − DDX Y T (· · · ) − DX DY T (· · · ) +DY DX T (· · · ) − DDY X T (· · · ) + DDX Y T (· · · ) = D2 T (· · · , Y, X) + (RXY T )(· · · ),


14


since DX Y = DY X + [X, Y ].

Finally, we introduce the Ricci tensor and the Ricci curvature function. The Ricci tensor is a 2 rank tensor field which is defined by Ric (X, Y ) =

n X

R(X, ei , Y, ei )

i=1

at each x ∈ M

where X and Y are vector fields and e1 , · · · , en is an orthonormal basis of Mx . It is easy to check that Ric is independent of choices of bases. By (4) in Theorem 1.8, Ric is symmetric, that is, Ric (X, Y ) = Ric (Y, X). Let x ∈ M and X ∈ Mx with |X| = 1. We define Ric (X, X) to be the Ricci curvature of the vector X. Let e1 , · · · , en be an orthonormal basis of Mx such that e1 = X. Then Ric (X, X) =

n X

R(e1 , ei , e1 , ei ),

i=2

since R(e1 , e1 , e1 , e1 ) = 0 by (3) in Theorem 1.8. Then the Ricci curvature is the sum of n − 1 sectional curvatures. Isometry Map Let M and N be two Riemannian manifolds. Let Φ: M → N be a C ∞ map. Let x ∈ M be given. We define a linear map Φ∗ : Mx → NΦ(x) as follows. Let v ∈ Mx . Let γ: (−ε, ε) → M be a curve such that γ(0) = x, We define Φ∗ v ∈ NΦ(x) by

γ(0) ˙ = v.

˙ Φ∗ v = β(0)

where β(t) = Φ(γ(t)) for

t ∈ (−ε, ε)

is a curve on N with β(0) = Φ(x). Sometimes, we denote dΦ = Φ∗ and call it the differential of the map Φ. Then for any f ∈ C ∞ (N ), (Φ∗ v)(f ) = [f ◦ Φ ◦ γ(t)]′t=0 = v(f ◦ Φ). Definition 1.14 A C ∞ map Φ: M → N is called a local isometry between two Riemannian manifolds M and N, if, for each x ∈ M, Φ is a local isomorphism at x such that Φ∗ : Mx → NΦ(x) is an isometry between the two inner product spaces. Further, Φ is said to be an isometry from M to N, if Φ is a local isometry and, at the same time, a diffeomorphism. Remark 1.11 If there is a local isometry between M and N, then the two manifolds have the same dimension. A local isometry keeps the connections the same in the following sense.



15

Theorem 1.11 Let M and N be two Riemannian manifolds. Let D and D′ be the Levi-Civita connections of M and N, respectively. Let Φ: M → N be a local isometry. Let X, Y be vector fields on M. Then ′ Φ∗ D X Y = D Φ Φ∗ Y. ∗X

Proof. Let p ∈ M be given and let q = Φ(p). Let ϕ = (x1 , · · · , xn ) be a local coordinate system near p on M. Since Φ is a local isomorphism, y = ϕ ◦ Φ−1 is a local coordinate system around q on N. For f ∈ C ∞ (N ), we have ∂ ∂ [f ◦ (ϕ ◦ Φ−1 )−1 (y)] = [f ◦ Φ ◦ ϕ−1 (y)] ∂yi ∂yi = ∂xi (f ◦ Φ) = Φ∗ ∂xi (f ),

∂yi (f ) =

that is, ∂yi = Φ∗ ∂xi

for all i.

(1.23)

Let ′ gij = h∂yi , ∂yj iN

gij = h∂xi , ∂xj iM ,

for all i, j.

(1.24)

k

Let Γkij and Γ′ ij be the Christoffel symbols, respectively for D and D′ . Since Φ is a local isometry, the relations (1.23) imply ′ gij = gij ◦Φ

for all i,

(1.25)

k

Γkij = Γ′ ij ◦ Φ. for all i, j, k. Let X=

X

Xi ∂xi ,

Y =

i

Then by (1.23) Φ∗ X =

X i

X

(1.26)

Yi ∂xi .

i

(Xi ◦ Φ−1 )∂yi ,

Φ∗ Y =

X i

(Yi ◦ Φ−1 )∂yi .

Since X(Yi ) = X(Yi ◦ Φ−1 ◦ Φ) = Φ∗ X(Yi ◦ Φ−1 ) for all i, we obtain, via (1.23)-(1.26), n n X X Φ∗ DX Y = Φ∗ [ (X(Yk ) + Xi Yj Γkij )∂xk ] k=1

= =

n X

ij=1

[Φ∗ X(Yk ◦ Φ−1 ) +

k=1 ′ DΦ Φ∗ Y. ∗X

n X

ij=1

k

(Xi ◦ Φ−1 )(Yj ◦ Φ−1 )Γ′ ij ]∂yk


16


Theorem 1.12 Let Φ: M → N be a local isometry between the two Riemannian manifolds. Let R and R′ be the curvature tensors on M and N, respectively. Then, for any vector fields Xi ∈ X (M ) for 1 ≤ i ≤ 4, we have R(X1 , X2 , X3 , X4 ) = R′ (Φ∗ X1 , Φ∗ X2 , Φ∗ X3 , Φ∗ X4 ) ◦ Φ.

(1.27)

′ Proof. Let gij and gij be defined in (1.24) for all i, j. It follows from (1.23) and (1.26) that ′ ′ ′ gijxk = ∂xk (gij ◦ Φ) = ∂yk gij ◦ Φ = gijy ◦ Φ, k ′ ′ ′ gijxk xl = ∂xl (gijy ◦ Φ) = ∂yl (gijy ) ◦ Φ = gijy ◦ Φ, k k k yl

(1.28)

for all i, j, k, l. From the relations (1.20), (1.25), (1.26), and (1.28), we have the formula (1.27). Second Fundamental Form of Hypersurface In general it is not easy to compute sectional curvatures. We will study the second fundamental forms of hypersurfaces which will provide us with a useful tool to compute sectional curvatures. ˜ be an (n + 1)-dimensional Riemannian manifold. An n-dimensional Let M ˜ , denoted by M, is called a hypersurface. A vector field submanifold of M ˜ N ∈ X (M ) is called the normal vector field of M if for each p ∈ M, N ⊥ Mp and |N | = 1. We introduce a Riemannian metric on a hypersurface M as follows, which ˜: is called the induced metric from M g(X, Y ) = hX, Y i for

X, Y ∈ X (M ),

(1.29)

˜ . Let D ˜ and D be the Levi-Civita connections where h·, ·i is the metric of M ˜ of M and M. Then Lemma 1.1 For any X, Y ∈ X (M ), ˜ X Y − hD ˜ X Y, N iN. DX Y = D

(1.30)

Proof. We define a map P : X (M ) × X (M ) → X (M ) by ˜ X Y − hD ˜ X Y, N iN P (X, Y ) = D

for

X, Y ∈ X (M ).

Clearly P satisfies (C1)-(C3). Then P is a connection on M. It suffices to show that P satisfies (L1) and (L2) in Theorem 1.1. Indeed, for X, Y, Z ∈ X (M ), since N ⊥ X (M ), ˜ X Y, Zi + hY, D ˜ X Zi = hP (X, Y ), Zi + hY, P (X, Z)i, XhY, Zi = hD ˜XY − D ˜ Y X = [X, Y ]. P (X, Y ) − P (Y, X) = D Then DX Y = P (X, Y ).



17

Definition 1.15 The second fundamental form of a hypersurface M is a 2rank tensor field, given by ˜ X N, Y i Π(X, Y ) = hD

for

X, Y ∈ X (M ).

(1.31)

Lemma 1.2 The second fundamental form Π of M is symmetric, that is, Π(X, Y ) = Π(Y, X)

for

X, Y ∈ X (M ).

(1.32)

Proof. Since [X, Y ] ∈ X (M ), we have ˜ X N, Y i = XhN, Y i − hN, D ˜ X Y i = −hN, D ˜XY i Π(X, Y ) = hD ˜ ˜ = −hN, DY X + [X, Y ]i = −hN, DY Xi = Π(Y, X). Remark 1.12 From Lemmas 1.1 and 1.2, the following formula is true: ˜ X Y + Π(X, Y )N. DX Y = D

(1.33)

Theorem 1.13 For X, Y ∈ X (M ), ˜ R(X, Y, X, Y ) = R(X, Y, X, Y ) − Π(X, X)Π(Y, Y ) + Π2 (X, Y )

(1.34)

˜ and R are, the curvature tensors of M ˜ and M, respectively. where R Proof. The proof consists of a straight computation: ˜ ˜XD ˜ Y X, Y i + hD ˜Y D ˜ X X, Y i + hD ˜ [X,Y ] X, Y i, R(X, Y, X, Y ) = −hD ˜XD ˜ Y X, Y i = −XhD ˜ Y X, Y i + hD ˜ Y X, D ˜XY i −hD = −XhDY X, Y i + hDY X, DX Y i + Π2 (X, Y )(by (1.33)) = −hDX DY X, Y i + Π2 (X, Y ),

˜Y D ˜ X X, Y i = Y hD ˜ X X, Y i − hD ˜ X X, D ˜Y Y i hD = hDY DX X, Y i − Π(X, X)Π(Y, Y )(by (1.33)), ˜ [X,Y ] X, Y i = hD[X,Y ] X, Y i(by (1.33)). hD Summing up the above yields the formula (1.34).

Codazzi’s Equation We introduce the Codazzi equation which will be needed in the modeling of shells in this book. ˜ be the curvature tensor of M ˜ . Then Theorem 1.14 Let R ˜ DΠ(Z, X, Y ) = DΠ(Z, Y, X) + R(X, Y, N, Z) for all X, Y, Z ∈ X (M ).


(1.35)

18

Modeling and Control in Vibrational and Structural Dynamics Proof. For any X ∈ X (M ), ˜ X N, N i = X(|N |2 )/2 = 0. hD

(1.36)

From (1.32), (1.33), and (1.36), we have ˜ X N, Zi − XhD ˜ Y N, Zi Y (Π(Z, X)) − X(Π(Z, Y )) = Y hD ˜Y D ˜ X N, Zi − hD ˜XD ˜ Y N, Zi + hD ˜ X N, D ˜ Y Zi − hD ˜ Y N, D ˜ X Zi = hD ˜ XY N, Zi − hD ˜ [X,Y ] N, Zi + hD ˜ X N, DY Zi − hD ˜ Y N, DX Zi = hR ˜ = R(X, Y, N, Z) − Π([X, Y ], Z) + Π(X, DY Z) − Π(Y, DX Z)

(1.37)

for all X, Y, Z ∈ X (M ). On the other hand, the left hand side of (1.37) = DΠ(Z, X, Y ) − DΠ(Z, Y, X)

+Π(DY Z, X) + Π(Z, DY X) − Π(DX Z, Y ) − Π(Z, DX Y ) = DΠ(Z, X, Y ) − DΠ(Z, Y, X) − Π(Z, [X, Y ]) +Π(DY Z, X) − Π(DX Z, Y ).

Then the formula (1.35) follows from (1.37) and (1.38).

(1.38)

˜ = Rn+1 . Then M is a hypersurface of Rn+1 . Let X, Y, Z ∈ X (M ˜) Let M with Z = (Z1 , · · · , Zn+1 ). Then ˜XD ˜Y Z = D

n+1 X

XY (Zi )∂xi

i=1

which yields ˜ XY Z = 0. In particular, R(X, ˜ R Y, N, Z) = 0. It follows from Theorem 1.14 that Corollary 1.1 Let M be a hypersurface of Rn+1 and let Π be its second fundamental form. Then DΠ is symmetric in its variables. Definition 1.16 Let M be a hypersurface of Rn+1 . The induced metric g of M is called the first fundamental form of M. Remark 1.13 Let M and M ′ be two hypersurfaces of Rn+1 . If there is an isometry Φ: M → M ′ , then g(X, Y ) = g ′ (Φ∗ X, Φ∗ Y )

(1.39)

for all X,Y ∈ X (M ), where g and g ′ are the induced metrics of M and M ′ , respectively. We say that two isometrical hypersurfaces have the same first fundamental form in the sense (1.39).



19

In general, two isometrical hypersurfaces may have different shapes. In order that they have the same shape, the two isometrical hypersurfaces must have the same second fundamental form in the following sense. Theorem 1.15 Let M and M ′ be two hypersurfaces of Rn+1 . Let N and N ′ be the normals of M and M ′ , respectively. Let φ: M → M ′ be an isometry such that Π(X, Y ) = Π′ (φ∗ X, φ∗ Y ) for all X, Y ∈ X (M ), where Π and Π′ are the second fundamental forms of M and M ′ , respectively. Then there is an isometry Φ: Rn+1 → Rn+1 such that Φ|M = φ. For a proof of the above theorem, see [190], IV. Let us give an example to end this section. Example 1.3 Let M = { x = (x1 , · · · , xn+1 ) ∈ Rn+1 |

n+1 X i=1

x2i = 1 }

be the n-dimensional unit sphere in Rn+1 . Then (M, g) is a Riemannian manifold where g is the induced metric from Rn+1 . The normal field of M is N = x. Let X = (X1 , · · · , Xn+1 ), Y = (Y1 , · · · , Yn+1 ) ∈ X (M ). Then the second fundamental form of M is given by ˜ X N, Y i = h Π(X, Y ) = hD

n+1 X i=1

X(xi )∂xi , Y i =

n+1 X i=1

Xi Yi = hX, Y i,

that is, Π = g.

(1.40)

Let x ∈ M be given. Let e1 , e2 be an orthonormal basis of a 2-dimensional ˜ 1 , e2 , e1 , e2 ) = 0, by the formulas (1.34) and subspace Ξ of Mx . Since R(e (1.40), we obtain the sectional curvature of Ξ = R(e1 , e2 , e1 , e2 ) = Π(e1 , e1 )Π(e2 , e2 ) −Π2 (e1 , e2 ) = 1.


20


Exercise 1.1 1.1.1 Prove the relation (1.2). 1.1.2 Find out vector fields V, T1 and T2 on Rn such that, for some p ∈ Rn , T1 (p) = T2 (p) but (DV T1 )(p) 6= (DV T2 )(p). 1.1.3 Prove the relation (1.3). 1.1.4 Prove the relations (a) and (b) in the proof of Theorem 1.1. 1.1.5 Let M be a Riemannian manifold and let D be the Levi-Civita connection. Prove the formula (1.10). 1.1.6 Let M be a Riemannian manifold and let D be its Levi-Civita connection. Then D satisfies the relation (L1) in Theorem 1.1 if and only if all parallel transports, given by D, are isometries between the tangential spaces. 1.1.7 Let M be a Riemannian manifold and let D be its Levi-Civita connection. Let γ: [a, b] → M be a geodesic. We define the length of γ by L(γ) =

Z

a

where |γ(t)| ˙ = hγ(t), ˙ γ(t)i ˙

1/2

b

|γ(t)|dt ˙

. Prove L(γ) = (b − a)|γ(0)|. ˙

1.1.8 Prove that the formula (1.13) is independent of the choice of the curve γ. 1.1.9 Let M be a Riemannian manifold and let D be its Levi-Civita connection. A tensor field T is called parallel if DT = 0. Let A: Mx → My be a linear operator and α be a k-linear functional on My . We define A∗ α to be a k-linear functional on Mx by A∗ α(v1 , · · · , vk ) = α(Av1 , · · · , Avk )

for

vi ∈ Mx

(1.41)

with 1 ≤ i ≤ k. Prove that T is parallel if and only if, for each curve γ: [0, 1] → M, P∗ (1)T = T |γ(0) . 1.1.10 Let M be a Riemannian manifold with the metric g and let D be its Levi-Civita connection. Then g is parallel. 1.1.11 Let S be a tensor field of rank 2. Prove that the definition of tr S by (1.18) is free from choices of the orthonormal basis.



21

1.1.12 Let M be a Riemannian manifold with the metric g. Let (x1 , · · · , xn ) be a local coordinate. Let (g ij ) = (gij )−1 where gij = g(∂xi , ∂xj ) for 1 ≤ i, j ≤ n. Let f ∈ C ∞P (M ). Then (1) ∆f = ij g ij D2 f (∂xi , ∂xj ); √ P (2) ∆f = √1G ij (g ij Gfxi )xj where G = det(gij ). 1.1.13 Prove Theorem 1.7.

1.1.14 Prove Theorem 1.8([190], II). 1.1.15 Prove Theorem 1.9 (The proof of Lemma 1.6 later; [72]; [101], I, page 199). 1.1.16 Prove that the definition of (1.21) is independent of choice of bases of Ξ.

1.2

Distance Functions

Distance functions will play an important role in the application of Riemannian geometry. Let (M, g) be a complete Riemannian manifold. Let γ: [a, b] → M be a curve. The length of γ is defined as Z b L(γ) = |γ(t)|dt. ˙ a

If γ is contained in a local coordinate (x1 , · · · , xn ), then v Z b uX u n t L(γ) = gij (γ(t))γ˙ i (t)γ˙ j (t)dt, a

ij=1

P where gij = g(∂xi , ∂xj ) and γ(t) ˙ = ˙ i (t)∂xi (γ(t)). We remark that the iγ length of a (continuous and) piecewise smooth curve may be defined as the sum of the lengths of the smooth pieces. For x, y ∈ M, set curve (x, y) = { γ | γ : [a, b] → M piecewise smooth curve with γ(a) = x, γ(b) = y }.

The distance between two points x, y ∈ M can be defined: d(x, y) =


inf

γ∈ curve (x,y)

L(γ).

22


Remark 1.14 It is easy to show that any two points x, y ∈ M can be connected by a piecewise smooth curve and therefore, d(x, y) is always defined (Exercise 1.2.1). The distance function satisfies the usual axioms as in Exercise 1.2.2. We consider whether a geodesic γ L(γ). Formula for the first variation curve. A C ∞ map α(·, ·): [a, b] × [0, ε] curve γ if α(t, 0) = γ(t)

∈ curve (x, y) satisfies that d(x, y) = of arc length Let γ: [a, b] → M be a → M is called a variation of the base for

t ∈ [a, b].

If, in addition, α(a, s) = γ(a) and α(b, s) = γ(b) for all s ∈ [0, ε], then α is called a proper variation of the base curve γ. Let L(s) = L(α(·, s)) for s ∈ [0, ε]. We will compute L′ (0). For convenience, we assume that γ is parameterized by arc length, that is, |γ(t)| ˙ = 1 for all t ∈ [a, b]. For s ∈ [0, ε], let T = α(t, ˙ s) be the tangential vector field of the one Rb parameter family α of curves. Since L(s) = a |T (t, s)|dt, we have Z b d ′ L (s) = |T (t, s)|dt. ds a d Let us compute ds (|T |2 ). For each t fixed, α(t, ·): [0, ε] → M is a curve in the parameter s. Let U = αs (t, s) be the tangential vector field of the curve α(t, ·), which is called the transversal vector field of the one parameter family α of curves. Then

d (|T |2 ) = [hT, T i(α(t, s))]s = U hT, T i = 2hDU T, T i. ds In addition, it is easy to check that (Exercise 1.2.5) [T, U ] = 0. Then DU T = DT U and we have ′

L (s) =

Z

a

b

hT, DT U i dt. |T |

(1.42)

In particular, at s = 0, hT, DT U i hT, DT U i |s=0 = = T hT, U i − hDT T, U i |T | |γ| ˙ d = hT, U i − hDγ(t) γ, ˙ U i. ˙ dt We have the formula for the first variation of arc length Z b L′ (0) = hγ, ˙ U i|ba − hDγ(t) γ, ˙ U idt, ˙ a


(1.43)


23

where U (t) = U |s=0 . As an application of the above formula, we have Theorem 1.16 Let x, y ∈ M be given. Let γ: [a, b] → M be a curve with the arc length parameter such that γ(a) = x, γ(b) = y, and d(x, y) = L(γ). Then γ is a geodesic. Proof. Let α: [a, b] × [0, ε] → M be a proper variation of the base curve γ. Then α(a, s) = x and α(b, s) = y imply that U (a) = U (b) = 0. Then L′ (0) = 0 and the formula (1.43) yield Dγ˙ γ˙ = 0 for all t ∈ [a, b]. The above theorem tells us that, if a curve γ is shortest in length, it must be a geodesic. Conversely, this is not true, that is, a geodesic may not be shortest in length globally. Formula for the second variation of arc length A curve is called a normal geodesic if it is a geodesic with the arc length parameter. Let γ be a normal geodesic connecting x with y. Then L′ (0) = 0 for a proper variation of the base curve γ. From the formula (1.42), we obtain ′′

L (s) =

Z

a

b

d hT, DT U i dt, ds |T |

(1.44)

where the integrand is U(

hT, DT U i 1 1 2 ) = − 3 hT, DT U i + (hDT U, DT U i + hT, DU DT U i) |T | |T | |T | 1 |DT U |2 = − 3 (T hT, U i − hDT T, U i)2 + |T | |T | 1 + (−hT, RUT U i + hT, DT DU U i), |T |

where the relation [T, U ] = 0 is used again. Substitute it into the formula (1.44) and compute at s = 0. Noting that |T | = |γ| ˙ = 1,

DT T = Dγ˙ γ˙ = 0,

hT, DT DU U i = T hT, DU U i,

at s = 0, we obtain L′′ (0) = hγ, ˙ DU U i|ba Z b ′ + [U˙ (t)|2 − hγ, ˙ RU γ˙ U i − (hγ, ˙ U (t)i )2 ]dt, a

which is called the formula for the second variation of arc length.


(1.45)

24


Remark 1.15 In the formula (1.45) the crucial term is the curvature one which relates L′′ (0) to the sectional curvatures. hγ, ˙ DU U i|ba are called the boundary terms. If α is a proper variation, then hγ, ˙ DU U i|ba = 0. We decompose U (t) = U ⊥ (t) + hγ, ˙ U iγ˙

for

t ∈ [a, b],

where hU ⊥ (t), γ(t)i ˙ = 0. Since U˙ (t) = Dγ˙ [U ⊥ (t) + hγ, ˙ U iγ] ˙ = U˙ ⊥ (t) + ′ hγ(t), ˙ U (t)i γ(t), ˙ it follows from the formula (1.45) that ′′

L (0) =

hγ, ˙ DU U i|ba

+

Z

a

b

[U˙ ⊥ (t)|2 − hγ, ˙ RU ⊥ γ˙ U ⊥ i]dt,

(1.46)

The above formula will be needed when we compute the Hessian of the distance later. Exponential Map In order to show that a geodesic is locally shortest in length, we need the concept of the exponential map. Let (M, g) be a complete Riemannian manifold. Let x ∈ M be given. We define a map expx : Mx → M as follows: For any v ∈ Mx , let γ be a geodesic starting from x with the initial tangential vector v. Take the point y on γ such that the arc length of γ between x and y is |v|. Then we set expx v = y. Definition 1.17 The map expx : Mx → M is called the exponential map of M at x. Remark 1.16 By the theory of ordinary differential equations and from the equations (1.12), the exponential map is locally well-defined: Given x ∈ M, there are ε > 0 and a neighborhood U of x on M such that, for any y ∈ U and v ∈ My , if |v| < ε, then there is a unique geodesic γ satisfying γ(0) = y and γ(0) ˙ = v. Here we assume that M is complete so that expx has the domain Mx . For any v ∈ Mx , the exponential map maps a straight line tv in Mx into a geodesic γ(t) = expx tv satisfying γ(0) = x and γ(0) ˙ = v. Then the exponential map is a local diffeomorphism at the origin of Mx , that is, Lemma 1.3 Let x ∈ M be given. Then there is ε > 0 such that expx : Bx (ε) → M is a diffeomorphism, where Bx (ε) = { v ∈ Mx | |v| < ε }. Proof. Since Mx is an n-dimensional linear space, its tangential spaces are itself. It will suffice to prove that (d expx )|0 : Mx → Mx is invertible, where 0 is the origin of Mx . For any v ∈ Mx , c(t) = tv is a curve in Mx such that c(0) = 0 and c(0) ˙ = v. By the definition of d, (d expx )|0 v = [expx c(t)]′t=0 = v, that is, (d expx )|0 = I, the identical map from Mx to itself. Fix ε > 0 small such that expx on Bx (ε) = { v ∈ Mx | |v| < ε } is a



25

diffeomorphism. Let B(x, ε) = expx Bx (ε), which is called the geodesic ball centered at x with radius ε. We define a map by for t ∈ R, v ∈ Mx .

F (t, v) = expx tv

(1.47)

Then F˙ (t, v) = d expx v ∈ Mexpx tv is called the radial vector field on M starting from x. Since, for any v ∈ Mx fixed, F (t, v) is a geodesic, we have |d expx v|expx tv = |F˙ (t, v)|F (t,v) = |F˙ (0, v)|x = |v|

(1.48)

for all t ∈ R and v ∈ Mx . For 0 < s < ε, let Sx (s) = { v ∈ Mx | |v| = s } and S(x, s) = expx Sx (s). S(x, s) is called the geodesic sphere centered at x with radius s. The following is called the Gaussian lemma which states that the radial vector field orthogonalizes the tangential space of the sphere on Riemannian manifolds. Lemma 1.4 We have hd expx v, Xi(expx v) = 0

for all

X ∈ [S(x, |v|)]expx v .

Proof. Let y = expx v ∈ S(x, |v|) and let X ∈ [S(x, |v|)]y be given. Since expx is a diffeomorphism, there is a curve σ ˜ on Sx (|v|) such that σ ˜ (0) = v and σ ˜˙ (0) = (d expx )−1 X. In addition |˜ σ (s)| = |v|

for all s.

(1.49)

Then the curve σ(t) = expx σ ˜ (t) on S(x, |v|) satisfies that σ(0) = y and ˜˙ (0) = X. σ(0) ˙ = d expx σ Consider one parameter family of curves α(t, s) = expx t˜ σ (s)

for (t, s) ∈ [0, 1] × [0, a]

with the base curve γ(t) = α(t, 0) = expx tv for t ∈ [0, 1]. Let T (t, s) = α(t, ˙ s) = d expx σ ˜ (s) be the tangential vector field and U (t, s) = αs (t, s) = td expx σ ˜˙ (s) be the transversal vector field of α(t, s), respectively. Then [T, U ] = 0

for (t, s) ∈ [0, 1] × [0, a],

T (1, 0) = γ(1) ˙ = d expx v ∈ My , and U (1, 0) = X ∈ [S(x, |v|)]y . Moreover, it follows from (1.49) that |T (t, s)| = |α(t, ˙ s)| = |˜ σ (s)| = |v|

for (t, s) ∈ [0, 1] × [0, a].

(1.50)

In addition, U (0, 0) = 0, the origin of Mx . ˙ 0) = Dγ(t) We need to prove hγ(1), ˙ U (1, 0)i(y) = 0. Denote U(t, U. By the ˙ formula (L2) in Theorem 1.1, U˙ (t, 0) = DU(t,0) T + [T, U ](t, 0) = DU T. Let


26


f (t) = hγ(t), ˙ U (t, 0)i for t ∈ [0, 1]. Since γ is a geodesic, we have, via the formula (1.50), f˙(t) = γh ˙ γ, ˙ U (t, 0)i = hDγ(t) γ, ˙ U (t, 0)i + hγ, ˙ U˙ (t, 0)i ˙ 1 1 = hT, DU T i = U (|T |2 ) = U (|v|2 ) = 0 for t ∈ [0, 1], 2 2 which yields hd expx v, Xi = hγ(1), ˙ U (1, 0)i = hγ(0), ˙ U (0, 0)i = hv, 0i = 0. The following theorem shows that a geodesic is locally shortest in length. Theorem 1.17 Let ε > 0 be such that expx : Bx (ε) → B(x, ε) is a diffeomorphism. Let v ∈ Bx (ε). Let γ(t) = expx tv for t ∈ [0, 1]. Then d(x, y) = L(γ) = |v| where y = γ(1). Proof. Let σ: [0, 1] → M be a piecewise smooth curve such that σ(0) = x and σ(1) = y. We will prove L(σ) ≥ |v|. We assume that σ is C ∞ ; otherwise, we confine it to several small segments. Case 1 Let σ(s) ∈ B(x, ε) for all s ∈ [0, 1]. Let σ ˜ (s) = exp−1 x σ(s). Then σ ˜ (s) is a curve on Mx such that σ ˜ (0) = 0 and σ ˜ (1) = v. Consider one parameter family of curves α(t, s) = expx t

σ ˜ (s) . |˜ σ (s)|

Then σ(s) = α(|˜ σ (s)|, s). Let T = α(t, ˙ s) and U = αs (t, s). By the formula (1.48), |T | = 1 for all (t, s) ∈ [0, a] × (0, 1]. By the Gaussian lemma, hT, U i = 0

for (t, s) ∈ [0, a] × (0, 1].

Thus, we obtain |σ(s)| ˙ = |(

d d d |˜ σ (s)|)T + U | ≥ | |˜ σ (s)|| ≥ |˜ σ (s)| ds ds ds

which gives L(σ) =

Z

0

1

|σ(s)|ds ˙ ≥

Z

0

1

d |˜ σ (s)|ds = |˜ σ (1)| = |v|. ds

Case 2 We assume that σ starts from σ(0) = x, goes out of B(x, ε), and then comes back to arrive at σ(1) = y. Take |v| < ε0 < ε. Then the curve σ must intersect S(x, ε0 ), for example as the first time, at some point



27

y0 ∈ S(x, ε0 ) at s = s0 after it starts from x. Since σ: [0, s0 ] → B(x, ε), Case 1 implies that L(σ) ≥ L(σ|[0,s0 ] ) ≥ | exp−1 x y0 | = ε0 > |v|. How large ε in Theorem 1.17 can be taken is closely related to sectional curvatures. The following result, named as the Cartan-Hadamard theorem, is of globalness. In 1898, Hadamard proved that, for a complete, simply connected surface M in R3 , if its Gaussian curvature is nonpositive, then for any x ∈ M, expx : Mx → M is a diffeomorphism. In 1925, E. Cartan generalized this result to the n-dimensional case. Its proof involves much more knowledge in geometry and we omit it because of the limited extent of this chapter. For a proof, we refer to [38]. Theorem 1.18 (Cartan-Hadamard) Let (M, g) be a complete, simply connected Riemannian manifold. If (M, g) has non-positive sectional curvatures, then for any x ∈ M, expx : Mx → M is a diffeomorphism. It follows from Theorems 1.18 and 1.17 that Corollary 1.2 Let (M, g) be a complete, simply connected Riemannian manifold with non-positive sectional curvatures. Then all geodesics on (M, g) are shortest in length. Theorem 1.17 tells us that a geodesic is locally shortest in length without curvature information. However, it is actually shortest in a relatively large region which is confined by a cut point. Let γ = expx0 tv with v ∈ Mx0 and |v| = 1. Then that γ is shortest in the interval [0, t0 ] is equivalent to d(γ(0), γ(t0 )) = t0 = L(γ|[0,t0 ] ). Moreover, if γ|[0,t0 ] is shortest, then for any t < t0 , γ|[0,t] does. By Theorem 1.17, there is t0 > 0 (or t0 = ∞) such that, for any t < t0 , γ|[0,t] is shortest, but for t > t0 , γ|[0,t] is not. γ(t0 ) is called the cut point of γ with regard to γ(0) = x0 . Then t0 v ∈ Mx0 is called the tangent cut point. We define τ : Sx0 (1) → R by τ (v) = t0 for v ∈ Sx0 (1). Let C(x0 ) = { τ (v)v | v ∈ Sx0 (1) }. C(x0 ) is called the tangent cut locus of x0 . The set expx0 C(x0 ) ⊂ M is called the cut locus of x0 . Let M be a complete Riemannian manifold. Let x0 ∈ M. Let Σ(x0 ) = { tv | v ∈ Sx0 (1), 0 ≤ t < τ (v) },

(1.51)

which is called the interior of the tangent cut locus of x0 . Then expx0 : Σ(x0 ) → expx0 Σ(x0 ) is a diffeomorphism ([72]).


28


Definition 1.18 expx0 Σ(x0 ) is called the interior of the cut locus of x0 . Clearly, any geodesic γ(t) = expx0 tv is shortest on expx0 Σ(x0 ), that is, if γ(b) ∈ expx0 Σ(x0 ), then ρ(γ(b)) = b|v|, where ρ(x) = d(x, x0 ) is the distance function. Furthermore, expx0 Σ(x0 ) is a large region on M in the following sense. We have M = expx0 Σ(x0 ) ∪ expx0 C(x0 ).

(1.52)

It is easy to check that C(x0 ) is a zero measure set on Mx0 . Then expx0 C(x0 ) is a zero measure set on M since it is the image of the zero measure set C(x0 ), that is, expx0 Σ(x0 ) is M minus a zero measure set. Corollary 1.3 Let M be a complete Riemannian manifold and let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function from x ∈ M to x0 in the metric g. Let γ(t) = expx0 tv for t ∈ R where v ∈ Mx0 with |v| = 1. Then for x = γ(ρ(x)) ∈ expx0 Σ(x0 ) Dρ|γ(t) = γ(t) ˙

for

t ∈ [0, ρ(x)].

(1.53)

Proof. That x ∈ expx0 Σ(x0 ) implies that ρ(γ(t)) = t for t ∈ [0, ρ(x)]. Then hDρ, γ(t)i ˙ = 1 for t ∈ [0, ρ(x)]. (1.54) In addition, since Dρ ⊥ [S(x0 , t)]γ(t) , the relation (1.53) follows from (1.54).

In order to compute the Hessian D2 ρ of the distance function ρ, we introduce the concept of Jacobi field. Definition 1.19 Let γ: [a, b] → M be a geodesic. A vector field J along γ is called a Jacobi field if it satisfies the equation ¨ + RγJ J(t) ˙ =0 ˙ γ

for

t ∈ [a, b],

(1.55)

where J¨ = Dγ˙ Dγ˙ J and RγJ is the curvature operator. A Jacobi field J is ˙ called normal if hγ(t), ˙ J(t)i = 0 for all t ∈ [a, b]. The following lemma plays a basic role in the structure of D2 ρ. Lemma 1.5 Let M be a complete Riemannian manifold and let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function from x ∈ M to x0 in the metric g. Let γ(t) = expx0 tv for t ∈ R where v ∈ Mx0 with |v| = 1. Let x = expx0 bv ∈ expx0 Σ(x0 ). Then, for any X ∈ [S(x0 , b)]x , there is a normal Jacobi field J along γ such that J(0) = 0,

J(b) = X;

˙ D2 ρ(X, X) = hJ(b), J(b)i.


(1.56) (1.57)


29

Proof. Let σ: [0, ε] → M be a geodesic such that σ(0) = x and σ(0) ˙ = X. Denote σ ˜ (s) = exp−1 x0 σ(s) for s ∈ [0, ε]. Set α(t, s) = expx0 t

σ ˜ (s) b

for (t, s) ∈ [0, b] × [0, ε].

(1.58)

Then the base curve of the one parameter family α(·, s) of curves is α(t, 0) = γ(t). Let T (t, s) = α(t, ˙ s), U (t, s) = αs (t, s) be the tangential vector field and the transversal vector field of α, respectively. Then [T, U ] = 0 for (t, s) ∈ [0, b] × [0, ε]. (1.59) Since α(·, s) is a geodesic for each s ∈ [0, ε] fixed, DT T = 0

for (t, s) ∈ [0, b] × [0, ε].

(1.60)

Let J(t) = U (t, 0). We will prove that J is a Jacobi field along γ to satisfy the relations (1.56) and (1.57). The relation (1.59) implies DT U = DJ T. Then it follows from the relations (1.59) and (1.60) that ¨ J(t) = Dγ˙ Dγ˙ J = DT DT U = DT DU T = −[−DT DU T + DU DT T + D[T,U] T ] = −RT U T = −RγJ ˙ for t ∈ [0, b]. ˙ γ

(1.61)

In addition, since α(0, s) = x0 and α(b, s) = σ(s), we have J(b) = U (b, 0) = σ(0) ˙ = X, DU U |(0,s) = 0,

J(0) = U (0, 0) = 0,

(1.62)

DU U |(b,s) = Dσ(s) σ˙ = 0. ˙

Moreover, using the relations (1.61) and (3) of Theorem 1.8 , we have d2 ¨ hγ(t), ˙ J(t)i = hγ(t), ˙ J(t)i = −R(γ(t), ˙ J(t), γ(t), ˙ γ(t)) ˙ = 0, dt2 and, therefore, we obtain hγ(t), ˙ J(t)i = σ0 t + σ1 and, via (1.62), hγ(t), ˙ J(t)i = hγ(0), ˙ J(0)i = 0

for t ∈ [0, b],

that is, J is a normal Jacobi field with the relation (1.56). Note that expx0 Σ(x0 ) is star-shaped open set. Then that x ∈ expx0 Σ(x0 ) implies that, for ε > 0 small, α(t, s) ∈ expx0 Σ(x0 ) for (t, s) ∈ [0, b] × [0, ε], that is, ρ(α(b, s)) = L(α(·, s)|[0,b] ) for s ∈ [0, ε]. (1.63) On one hand

d ρ(α(b, s)) = hDρ, U (b, s)i. ds


30


Then d2 ρ(α(b, s)) = D2 ρ(J(b), J(b)) + Dρ(DU U )|(b,0) = D2 ρ(X, X). 2 ds s=0

On the other hand, by the formulas (1.46), (1.63) and (1.55), we obtain Z b d2 ′′ ˙ 2 − R(γ(t), ρ(α(b, s)) = L (0) = [|J(t)| ˙ J(t), γ(t), ˙ J(t)]dt ds2 s=0 0 Z b d ˙ ˙ = hJ(t), J(t)idt = hJ(b), J(b)i. 0 dt

Remark 1.17 On expx0 Σ(x0 ) a Jacobi field satisfying the conditions (1.56) is unique ([206]). Let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function from x ∈ M to x0 in the metric g. Denote by CD ρ2 all the points x on M such that D2 ρ2 (X, X) > 0 for all X ∈ Mx , X 6= 0.

The set CD ρ2 will play an important role in application; see the following chapters in this book. Consider the Euclidean space Rn , the distance function is ρ(x) = |x − x0 |. It is easy to check that (Exercise 1.2.6) D2 ρ2 (X, X) = 2|X|2

for all X ∈ Rnx , x ∈ Rn

where D is the classical connection, given by the Euclidean metric. Then CD ρ2 = Rn . For a general metric without any information on curvature, we will only know that CD ρ2 is just a neighborhood of x0 , see Theorem 1.20 later. However, the sectional curvatures will yield the global information on the set CD ρ2 , see Theorem 1.19 below. As an application of Lemma 1.5 and Theorem 1.18, we will prove Theorem 1.19 Let (M, g) be a simply connected, complete Riemannian manifold with non-positive sectional curvatures. Then CD ρ2 = M. Proof. By Theorem 1.18, expx0 Σ(x0 ) = M. Let x ∈ M be given. Let x = expx0 v. Let γ(t) = expx0 tv/|v| be the normal geodesic connecting x0 and x. Let X ∈ Mx be given with |X| 6= 0. We will prove D2 ρ2 (X, X) > 0. We decompose X as X = hX, γ(|v|)i ˙ γ(|v|) ˙ + X0 where hγ(|v|), ˙ X0 i = 0. It follows from the formula (1.53) that D2 ρ2 (X, X) = 2(Dρ ⊗ Dρ + ρD2 ρ)(X, X) 2

= 2hX, γ(|v|)i ˙ + 2|v|D2 ρ(X0 , X0 )


(1.64)


31

where the relation D2 ρ(X0 , γ) ˙ = hDγ˙ γ, ˙ X0 i = 0 is used. If X0 = 0, then by (1.64), D2 ρ2 (X, X) = 2|X|2 > 0. Let X0 6= 0. By Lemma 1.5, there is a normal Jacobi field J along γ such that J(0) = 0,

J(|v|) = X0 ,

and |v|

d ˙ hJ(t), J(t)idt dt 0 Z |v| Z |v| 2 ˙ ˙ 2 dt > 0 = [|J(t)| − R(γ, ˙ J, γ, ˙ J)]dt ≥ |J(t)|

˙ D ρ(X0 , X0 ) = hJ(|v|), J(|v|)i = 2

Z

0

0

˙ ˙ =0 because R(γ, ˙ J, γ, ˙ J) ≤ 0 and J(t) is not identically zero. Indeed, if J(t) for all t ∈ [0, |v|], then J is parallel along γ and therefore, |X0 | = |J(|v|)| = |J(0)| = 0, which is a contradiction. Without curvature information, CD ρ2 is just a neighborhood of x0 , that is, Theorem 1.20 Let M be a Riemannian manifold. Let x0 ∈ M and let ρ(x) = d(x, x0 ) be the distance function. Then D2 ρ2 is positive in a neighborhood of x0 . Proof. Let ε > 0 be such that expx0 : Bx0 (ε) → B(x0 , ε) is a diffeomorphism. For any v ∈ Mx0 , let γ(t) = expx0 tv/|v|. Then ρ2 (γ(t)) = t2

for t ∈ [0, ε).

We differentiate the above identity twice in t at t = 0 to have D2 ρ2 (v, v) = 2|v|2

(1.65)

since γ(0) ˙ = v/|v|. Then there is a neighborhood of x0 where D2 ρ2 > 0.

In Theorem 1.20, we have checked the convexity of the function ρ2 . We introduce Definition 1.20 A differentiable function f : M → R is called (strictly) convex if its Hessian D2 f is positive semidefinite (definite). Proposition 1.1 Let G(x) = (gij ) be a symmetric, positive, and smooth matrix for each x ∈ Rn . Consider (Rn , g) as a Riemannian manifold with the metric g = G(x). Let Ω ⊂ Rn be an open set. Then f is strictly convex function on Ω if and only if f satisfies that X X [2ail (akj fxk )xl − aijxl akl fxk ]ξi ξj ≥ c aij ξi ξj (1.66) ijkl

ij

for all (x, ξ) ∈ Ω × Rn , where c > 0 and (aij (x)) = G−1 (x) for x ∈ Rn .


32


Proof. Let x = (x1 , · · · , xn ) be the natural coordinate system. Then x ∈ Rn

gij (x) = h∂xi , ∂xj ig = hG(x)∂xi , ∂xj i for

for all 1 ≤ i, j ≤ n where h·, ·i denotes the Euclidean metric of Rn . The Christoffel symbols are given by X 2Γkpq = akl (gplxq + gqlxp − gpqxl ) l

=

X l

=−

[(akl gpl )xq + (akl gql )xp − akl gpqxl − aklxq gpl − aklxp gql ]

X [akl gpqxl + aklxq gpl + aklxp gql ]

(1.67)

l

for all 1 ≤ p, q, k ≤ n. It follows from (1.67) that 2

X

aip ajq Γkpq =

pq

X l

akl aijxl −

X l

aikxl ajl −

X

ajkxl ail

(1.68)

l

for all 1 ≤ i, j, k ≤ n. For ξ = (ξ1 , · · · , ξn ) ∈ Rn given, let X=

X

ξi ∂xi

x ∈ Ω.

for

i

Let D be the Levi-Civita connection of the metric g. It follows from (1.68) that X aip ajq ξi ξj D2 f (∂xp , ∂xq ) 2D2 f (G−1 X, G−1 X) = 2 ijpq

=2

X

ξi ξj

pq

ij

=

X

aip ajq [fxp xq − D∂xp ∂xq f ]

X X X [2 aik ajl fxk xl − 2 aip ajq Γkpq fxk ]ξi ξj ij

kl

pqk

X X X = [2 aik ajl fxk xl − akl aijxl fxk ]ξi ξj ij

kl

kl

XX X XX X + ( ξi aik )xl ( ξj ajl )fxk + ( ξj ajk )xl ( ξi ail )fxk kl

=

X ijkl

i

j

kl

j

[2ail (akj fxk )xl − aijxl akl fxk ]ξi ξj .

Then the inequality (1.66) is equivalent to 2D2 f (G−1 X, G−1 X) ≥ chG−1 X, Xi


i

(1.69)

Preliminaries from Differential Geometry for all X ∈ Rnx with x ∈ Ω, that is, D2 f (X, X) ≥

33

c |X|2g 2 1/2

for all X ∈ Rnx with x ∈ Ω where |X|g = hG(x)X, Xi the metric g.

is the norm of X in

D2 ρ in Spaces of Constant Curvature Let M be a Riemannian manifold. M is called a space of constant curvature κ, or a space form if all its sectional curvatures are the same number κ. Lemma 1.6 Let M be a space of constant curvature κ. Let x ∈ M and let R be the curvature tensor. Then R(X1 , X2 , X3 , X4 ) = κ[hX1 , X3 ihX2 , X4 i − hX2 , X3 ihX1 , X4 i]

(1.70)

for Xi ∈ Mx with 1 ≤ i ≤ 4. Proof. Let X1 ∈ Mx be given. Consider a bilinear form on Mx by b(X2 , X4 ) = R(X1 , X2 , X1 , X4 )

for X2 , X4 ∈ Mx .

By (1.21), 2

b(X2 , X2 ) = κ(|X1 |2 |X2 |2 − hX1 , X2 i ). By (4) of Theorem 1.8, b(X2 , X4 ) = b(X4 , X2 ). Then 4R(X1 , X2 , X1 , X4 ) = 4b(X2 , X4 ) = b(X2 + X4 , X2 + X4 ) − b(X2 − X4 , X2 − X4 ) 2

= κ(|X1 |2 |X2 + X4 |2 − hX1 , X2 + X4 i )

2

−κ(|X1 |2 |X2 − X4 |2 − hX1 , X2 − X4 i ) = 4κ(|X1 |2 hX2 , X4 i − hX1 , X2 ihX1 , X4 i), that is, R(X1 , X2 , X1 , X4 ) = κ(|X1 |2 hX2 , X4 i − hX1 , X2 ihX1 , X4 i)

(1.71)

for all X1 , X2 , X4 ∈ Mx . For Xi ∈ Mx with 1 ≤ i ≤ 4, set P (X1 , X2 , X3 , X4 ) = R(X1 , X2 , X3 , X4 ) −κ[hX1 , X3 ihX2 , X4 i − hX2 , X3 ihX1 , X4 i]. Then P ∈ T 4 (Mx ). We will prove that P (X1 , X2 , X3 , X4 ) = 0 for all Xi ∈ Mx with 1 ≤ i ≤ 4. It follows from (1.71) that P (X1 , X2 , X1 , X3 ) = 0


(1.72)

34


for all Xi ∈ Mx with 1 ≤ i ≤ 3. By (1)-(4) of Theorem 1.8 it is easy to check that P satisfies the following properties: P (X1 , X2 , X3 , X4 ) = −P (X2 , X1 , X3 , X4 ); (1.73) P (X1 , X2 , X3 , X4 ) + P (X2 , X3 , X1 , X4 ) + P (X3 , X1 , X2 , X4 ) = 0;

(1.74)

P (X1 , X2 , X3 , X4 ) = −P (X1 , X2 , X4 , X3 );

(1.75)

P (X1 , X2 , X3 , X4 ) = P (X3 , X4 , X1 , X2 ).

(1.76)

It follows from (1.73), (1.75), and (1.72) that P (X1 , X2 , X3 , X2 ) = P (X2 , X1 , X2 , X3 ) = 0.

(1.77)

Using (1.72) and (1.73), we obtain 0 = P (X1 + X2 , X3 , X1 + X2 , X4 ) = P (X1 , X3 , X1 + X2 , X4 ) + P (X2 , X3 , X1 + X2 , X4 ) = P (X1 , X3 , X1 , X4 ) + P (X1 , X3 , X2 , X4 ) + P (X2 , X3 , X1 , X4 ) +P (X2 , X3 , X2 , X4 ) = P (X1 , X3 , X2 , X4 ) + P (X2 , X3 , X1 , X4 ) = −P (X3 , X1 , X2 , X4 ) + P (X2 , X3 , X1 , X4 ).

(1.78)

Similarly, we have 0 = P (X1 , X2 + X4 , X3 , X2 + X4 ) = P (X1 , X2 , X3 , X4 ) −P (X2 , X3 , X1 , X4 ).

(1.79)

Summing up (1.78) with (1.79) yields P (X1 , X2 , X3 , X4 ) = P (X3 , X1 , X2 , X4 ).

(1.80)

In addition, the relation 0 = P (X1 + X3 , X2 , X1 + X3 , X4 ) implies that P (X1 , X2 , X3 , X4 ) = P (X2 , X3 , X1 , X4 ).

(1.81)

Finally, from (1.80), (1.81), and (1.74), we obtain 3P (X1 , X2 , X3 , X4 ) = P (X1 , X2 , X3 , X4 ) + P (X2 , X3 , X1 , X4 ) +P (X3 , X1 , X2 , X4 ) = 0. Remark 1.18 If Xi ∈ X (M ) are vector fields for 1 ≤ i ≤ 3, it follows from (1.70) that RX1 X2 X3 = κ[hX1 , X3 iX2 − hX2 , X3 iX1 ]. (1.82)



35

Let M be a space of constant curvature. Let x0 ∈ M be given. We will compute the Hessian D2 ρ of the distance function ρ on the interior of the cut locus of x0 , expx0 Σ(x0 ). For this purpose, we need to compute Jacobi fields. Lemma 1.7 Let M be a space of constant curvature κ. Let x ∈ expx0 Σ(x0 ) be given. Let γ(t) = expx0 tv with |v| = 1 such that x = γ(b). Let X ∈ [S(x0 , b)]x . Let e ∈ Mx0 be such that the parallel transport e(t) ∈ Mγ(t) of e along γ satisfies e(b) = X. Let J be a Jacobi field along γ such that J(0) = 0 and J(b) = X. Then J(t) = f0 (t)e(t)

t ∈ [0, b]

for

(1.83)

where √  1  sin √κb sin κt κ > 0; f0 (t) = 1b t, κ = 0; √  1 √ sinh −κt, κ < 0, sinh −κb

for

t ∈ [0, b].

Proof. Since X = e(b) is the parallel transport of e along γ, |e(t)| = |X| for all t ∈ [0, b]. Let e1 , e2 , · · · , en be an orthonormal basis of Mx0 with e1 = γ(0) ˙ and e2 = e/|X|. For 1 ≤ i ≤ n, we parallelize ei to ei (t) ∈ Mγ(t) along γ|[0,t] for 0 < t ≤ b. Then for each t ∈ [0, b], e1 (t), · · · , en (t) is an orthonormal basis of Mγ(t) such that e˙ i (t) = Dγ(t) ei = 0 for 1 ≤ i ≤ n where e1 (t) = γ(t) ˙ and ˙ e2 (t) = e(t)/|X|. Let n X J(t) = Ji (t)ei (t) for t ∈ [0, b]. i=1

Using the formula (1.70), we have RγJ ˙ = ˙ γ =

n X

i=1 n X

Ji (t)Rγ(t)e γ˙ = ˙ i (t)

n X

Ji (t)R(γ(t), ˙ ei (t), γ, ˙ ej (t))ej (t)

ij=1

Ji (t)R(γ(t), ˙ ei (t), γ, ˙ ej (t))ej (t)

ij=2 n X

=κ

Ji (t)ei (t)

i=1

for t ∈ [0, b].

Then J, being a Jacobi field, is equivalent to Jï (t) + κJi (t) = 0 for

t ∈ [0, b],

1 ≤ i ≤ n.

(1.84)

In addition, the conditions J(0) = 0 and J(b) = X yield J1 (0) = 0,

J1 (b) = |X|;


Ji (0) = Ji (b) = 0

(1.85)

36


for 2 ≤ i ≤ n. We solve the problem (1.84)-(1.85) to give the formula (1.83). Combining Lemmas 1.7 with 1.5, we have Theorem 1.21 Let M be a space of constant curvature κ. Let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function from x ∈ M to x0 in the metric g. Then for x ∈ expx0 Σ(x0 ) D2 ρ(X, X) = f (ρ)|X|2 where

for

X ∈ Mx , hDρ, Xi = 0,

√ √ κ cot κρ κ > 0;  1 , κ = 0; f (ρ) =  ρ √ √ −κ coth −κρ, κ < 0.

(1.86)

(1.87)

Hessian Comparison Theorem For a general Riemannian manifold with variable curvature, we can study the Hessian of the distance by comparing it with the one in a space of constant curvature. Let M be a Riemannian manifold and let γ(t) be a normal geodesic on M. Let R be the curvature tensor. Then for any X ∈ Mγ(t) with hX, γ(t)i ˙ = 0 and |X| = 1, the sectional curvature R(γ(t), ˙ X, γ(t), ˙ X) is called a radial curvature at γ(t). Let M1 and M2 be complete Riemannian manifolds. Let xi0 ∈ Mi be given for i = 1, 2. Denote by ∂Mi the radial vector fields on Mi starting from xi0 , respectively, for i = 1, 2. Let ρi be the distance functions on Mi starting from xi0 , respectively, for i = 1, 2. If µi are symmetric tensors of rank 2 on Mi /{ xi0 }, respectively, for i = 1, 2, then for xi ∈ Mi , µ1 (x1 ) ≺ µ2 (x2 )

(1.88)

means: for all Xi ∈ Mxi with |X1 | = |X2 | and hX1 , ∂M1 i = hX2 , ∂M2 i, the inequality µ1 (X1 , X1 ) ≤ µ2 (X2 , X2 ) is valid. We introduce the following theorem without proofs. For a proof, see [71]. Theorem 1.22 (Hessian comparison theorem) Let γi : [0, b] → Mi be normal geodesics with γi (0) = xi0 and γi (b) ∈ expxi0 Σ(xi0 ), respectively, for i = 1 and 2. If each radial curvature at γ2 (t) ≥ every radial curvature at γ1 (t), then D 2 ρ2 ≺ D 2 ρ1 for every t ∈ [0, b]. It follows from Theorems 1.21 and 1.22 that



37

Corollary 1.4 Let x0 ∈ M be given. Let γ: [0, b] → M be a normal geodesic with γ(0) = x0 and γ(b) ∈ expx0 Σ(x0 ). Set κ0 = inf {all radial curvatures at γ(t)}, t∈[0,b]

κ1 = sup {all radial curvatures at γ(t) }. t∈[0,b]

Then f0 (t)|X|2 ≥ D2 ρ(X, X) ≥ f1 (t)|X|2

(1.89)

for all X ∈ Mγ(t) with hX, γ(t)i ˙ = 0 for all t ∈ [0, b], where ρ = d(x0 , x) and f0 (ρ) = f (ρ) and f1 (ρ) = f (ρ) are given by the formula (1.87) with κ replaced by κ0 and by κ1 , respectively.

Exercise 1.2 1.2.1 Let (M, g) be a complete Riemannian manifold. Then for any two points x, y ∈ M, there is a piecewise smooth curve on M that connects x with y. 1.2.2 The distance function on the Riemannian manifold (M, g) satisfies the usual axioms: (a) d(x, y) ≥ 0 for all x, y, and d(x, y) > 0 for all x 6= y; (b) d(x, y) = d(y, x); (c) d(x, y) ≤ d(x, z) + d(z, y) (triangle inequality) for all x, y, z ∈ M. 1.2.3 Let (M, g) be a Riemannian manifold. Then the topology on M induced by the distance function d coincides with the original manifold topology of M. 1.2.4 If γ: [a, b] → M is a smooth curve, and ψ: [α, β] → [a, b] is a change of parameter, then L(γ ◦ ψ) = L(γ). 1.2.5 Consider a C ∞ map α(·, ·): [a, b] × [0, ε] → M. Let T = α(·, ˙ s) and U = αs (t, ·) be the tangential vector field and the transversal vector field of one parameter curve family α(t, ·), respectively. Then [T, U ] = 0. 1.2.6 Consider the Euclidean space Rn with the classical dot metric. Let ρ(x) = |x − x0 | be the distance function. Prove D2 ρ2 = 2I for all x ∈ Rn .


38


1.2.7 Let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function in the metric g. For x ∈ M given, let γ: [0, ρ(x)] → M be a geodesic such that γ(0) = x0 and γ(ρ(x)) = x. Then γ is a normal geodesic. 1.2.8 Let x0 ∈ M be given. Let ρ(x) = d(x, x0 ) be the distance function in the metric g. Let x ∈ expx0 Σ(x0 ). Then x = expx0 ρ(x)v for some v ∈ Mx0 with |v| = 1 , and D2 ρ(γ(ρ(x)), ˙ X) = 0

for

X ∈ Mx

where γ(t) = expx0 tv. 1.2.9 Let M be a Riemannian manifold. A differentiable function f : M → R is (strictly) convex if and only if, for any geodesic γ, ((f ◦γ)′′ > 0) (f ◦γ)′′ ≥ 0.

1.3

A Basic Computational Technique

This section presents a basic computational technique that is useful not only in differential geometry (particularly in the Bochner technique) but in many other mathematical problems as well. We will need it as a basic necessary computational tool to obtain models of shells and to carry out multiplier schemes for control problems in this book. The basic ideas are as follows: In order to verify an identity or a pointwise estimate on a Riemannian manifold, it suffices to do so at each point p relative to a coordinate system or frame field that offers the greatest computational simplification. What prevents a computation from being as simple as the corresponding classical (i.e., Euclidean) situation is the presence of the Christoffel symbols Γkij . Thus we hope to find a coordinate system or a frame field relative to which the Γkij ’s vanish at the given point p. Such a coordinate system or frame field is then generically referred to as being normal at p. The possibility of finding such a coordinate system or frame field under various circumstances is the content of the present section. Definition 1.21 Let M be a Riemannian manifold and let p ∈ M be given. Let (x1 , · · · , xn ) be a coordinate system around p. If

gij (p) = δij , D∂xi ∂xj (p) = 0

(1.90)

for all i, j = 1, · · · , n, then the coordinate system { xi } is said to be normal at p.



39

A well known fact is that a coordinate system normal at p can be obtained from geodesic coordinate systems: Let expp denote the exponential map defined on an open neighborhood of 0 in the tangent space Mp . Let e1 , · · · , en be an orthonormal basis of Mp . We define a coordinate system ϕ = (x1 , · · · , xn ) at p by n X ϕ(q) = (x1 , · · · , xn ) for q = expp xj ej ∈ M. (1.91) j=1

The above coordinate system is called a geodesic coordinate system.

Theorem 1.23 Let M be a Riemannian manifold and let p ∈ M be given. Then a geodesic coordinate system is normal at p. Proof. For any f ∈ C ∞ (M ), by definition ∂xi f =

n X ∂ f (expp xj ej ) = hDf, d expp ei i, ∂xi j=1

that is , for 1 ≤ i ≤ n.

∂xi = d expp ei

Since d expp is the identical map from Mp to itself at p, gij (p) = hei , ej i(p) = δij . We define a bilinear map η: Mp ⊗ Mp → Mp by η(v, w) =

n X

αi βj (D∂xi ∂xj )(p)

ij=1

P P P where v = i αi ei and w = i βi ei . For any v = i αi ei ∈ Mp fixed, let γ(t) = expp tv. Then ϕ(γ(t)) = t(α1 , · · · , αn ) which yields γ(t) ˙ =

n X

αi ∂xi (γ(t)).

i=1

Then that γ is a geodesic implies that η(v, v) = Dγ(0) γ˙ = 0. ˙ By the bilinearity of η, we have η(v, w) = 0 for any v, w ∈ Mp . In particular, the second system of equations in (1.90) hold true. Definition 1.22 Let M be a Riemannian manifold and let p ∈ M be given. Let there be a frame field {E1 , · · · , En } near p, i.e., E1 ,· · · , En are locally defined vector fields around p which satisfy hEi , Ej i = δij for all i, j. The frame field {Ei } is said to be normal at p if (DEi Ej )(p) = 0


for all

i, j.

(1.92)

40


Theorem 1.24 Given p ∈ M and an orthonormal basis {e1 , · · · , en } of Mp , there exists a frame field {E1 , · · · , En } normal at p such that Ei (p) = ei for 1 ≤ i ≤ n. Proof. Let ε > 0 be given such that expp : Bp (ε) ⊂ Mp → M is a diffeomorphism. Let U = expp Bp (ε) which is a neighborhood of p. Then a frame field {E1 , · · · , En } can be defined in U by: If q ∈ U, Ei (q) = the parallel transport of ei from p to q along the unique minimizing geodesic in U jointing p and q. Consider the geodesic γi (t) = expx tei . Since Ej is a parallel transport along γ(t) for 0 < t < ε, we have DEi Ej = Dγ(0) Ej = 0 ˙

at x

for all i, j.

Remark 1.19 For a frame field normal at p [Ei , Ej ](p) = 0

(1.93)

for all i, j because [Ei , Ej ](p) = (DEi Ej )(p) − (DEj Ei )(p) = 0. As the first example to show how the computational principle works, we have Lemma 1.8 Let {xi } be a local coordinate system on M normal at p ∈ M. Then for f ∈ C 2 (M ) n X ∂2f (p). (1.94) ∆f (p) = ∂x2i i=1

If E1 , · · · , En is a frame field normal at p, then ∆f (p) =

n X

Ei Ei (f ).

(1.95)

i=1

Proof. The conditions gij (p) = h∂xi , ∂xj i(p) = δij mean that ∂x1 , · · · , ∂xn is an orthonormal basis of Mp . Then ∆f = tr D2 f =

n X

D2 f (∂xi , ∂xi ) =

i=1

X ∂2 X f− D∂xi ∂xi f 2 ∂xi i i

which yields the formula (1.94) since (D∂xi ∂xi )(p) = 0 for all i. Let {Ei } be a frame field normal at p. Then the relation (DEi Ei )(p) = 0 yields, at p, X X X ∆f = D2 f (Ei , Ei ) = [Ei Ei (f ) − DEi Ei (f )] = Ei Ei (f ). i

i

i

In order to compute the Hodge-Laplacian ∆, we need



41

Definition 1.23 A vector field on M is called a 1-form. Let k ≥ 2. A k-rank tensor field ω ∈ T k (M ) is called a k-form if ω is antisymmetric, i.e., for X1 , · · · , Xk ∈ X (M ), ω(· · · , Xi , · · · , Xj , · · · ) = −ω(· · · , Xj , · · · , Xi , · · · ) for all i, j where the variables in the position “· · · ” are kept the same. All the k-forms on M is denoted by Λk (M ). In particular, Λ(M ) = X (M ) = T (M ). We have two important operations: First, the exterior product by a vector field X ∈ X (M ): Λk (M ) → Λk+1 (M ), ω 7−→ e(X)ω = X ∧ ω where X ∧ ω is defined by X ∧ ω(X1 , · · · , Xk+1 ) =

k+1 X i=1

ˆ i , · · · , Xk+1 ), (−1)i+1 hX, Xi iω(X1 , · · · , X

(1.96)

ˆ i denotes that the variable Xi does not for X1 , · · · , Xk+1 ∈ X (M ), where X appear. Second, the interior product by a vector field X ∈ X (M ): T k+1 (M ) → T k (M ), T 7−→ i (X)T, [ i (X)T ](X1 , · · · , Xk ) = T (X, X1 , · · · , Xk ) for all X1 , · · · , Xk ∈ X (M ). In particular, if X, Y ∈ X (M ) are vector fields, then i (X)Y = hX, Y i (1.97) is a function on M. Let M be an oriented n-dimensional Riemannian manifold. An n-form ω0 on M is called the volume element if for any frame field {E1 , · · · , En }, |ω0 (E1 , · · · , En )| = 1. Let X ∈ X (M ) be a vector field and ω0 ∈ Λn (M ) be a volume element. Then there is a unique function on M, denoted by div X, such that ( div X)ω0 = d( i (X)ω0 ) where d is the exterior derivative. The function div X is called the divergence of X. Lemma 1.9 Let E1 , · · · , En be a frame field. Then d=

n X i=1


Ei ∧ DEi .

(1.98)

42

Modeling and Control in Vibrational and Structural Dynamics Pn Proof. Let d0 = i=1 Ei ∧ DEi . We must show d = d0 . Note that d0 is independent of the choice of {Ei }: Let Y1 , · · · , Yn be P another frame field with the same domain of definition as {Ei } and let Yi = nj=1 αij Ej . Then (αij ) Pn is an orthogonal matrix and k=1 αki αkj = δij for all i, j. Thus n X

k=1

Yk ∧ DYk =

n n X X

ij=1 k=1

αki αkj Ei ∧ DEj =

n X i=1

Ei ∧ DEi .

P Let f ∈ C ∞ (M ). Clearly, d0 f = ni=1 Ei (f )Ei = Df = df. Now fix a point p ∈ M and let a coordinate system (x1 , · · · , xn ) be normal at p. It suffices to show that relative to this particular choice of {Ei }, d = d0 holds at the point p. Since both d and d0 are linear operators, by renumbering the indices if necessary, it suffices to show that, at p, d = d0 on the k-form f dx1 ∧· · ·∧dxk , where f is a C ∞ function defined near p. Since {xi } is normal at p, (DX ∂xi )(p) = 0 for all X ∈ X (M ). Then the fact that dxi (∂xj ) = δij implies that 0 = Ek [dxi (∂xj )] = (DEk dxi )(∂xj ) + dxi (DEk ∂xj ) = (DEk dxi )(∂xj ) at p for all i, j, k. Thus d0 (f dx1 ∧ · · · ∧ dxk ) = d0 f ∧ dx1 ∧ · · · ∧ dxk +f

n X j=1

k X Ej ∧ ( dx1 ∧ · · · ∧ DEj dxi ∧ · · · ∧ dxk ) i=1

= d(f dx1 ∧ · · · ∧ dxk ) at p.

(1.99)

Lemma 1.10 Let M be an oriented n-dimensional Riemannian manifold. Let X ∈ X (M ). Then div X = tr DX. (1.100) Let p ∈ M be given. If {Ei } is a frame field normal at p, then div X =

n X i=1

Ei hX, Ei i

at p.

(1.101)

Proof. Let p ∈ M be given. Let ω0 be the volume element. Let E1 , · · · , En be a frame field normal at p with the positive orientation. Then hEi , Ej i = δij around p, (DEi Ej )(p) = 0 for all i, j, and ω0 = E1 ∧ · · · ∧ En . It follows from (1.98) that X dEk = Ei ∧ DEi Ek = 0 at p (1.102) i

for all k. P Let X = i Xi Ei . We have

î , · · · , En ) = ω0 (X, E1 , · · · , E î , · · · , En ) i (X)ω0 (E1 , · · · , E i−1 i−1 = (−1) ω0 (E1 , · · · , X, · · · , En ) = (−1) Xi



43

for all i. Then i (X)ω0 =

n X i=1

(−1)i−1 Xi E1 ∧ · · · ∧ Eî ∧ · · · ∧ En .

Using the formula (1.102), we obtain, at p, d[ i (X)ω0 ] = =

n X

(−1)i−1 d(Xi ) ∧ E1 ∧ · · · ∧ Eî ∧ · · · ∧ En

i=1 n X

ij=1

(−1)i−1 Ej (Xi )Ej ∧ E1 ∧ · · · ∧ Eî ∧ · · · ∧ En =

n X

Ei (Xi )ω0

i=1

which gives div X =

n X

Ei (Xi ) =

i=1

X i

Ei hX, Ei i =

X i

hDEi X, Ei i = tr DX.

In particular, (1.101) holds true.

Remark 1.20 To define the divergence for a vector field it is not necessary to assume the Riemannian manifold to be orientable. The right hand side of the formula (1.100) can be regarded as the definition of the divergence of the vector field X for a Riemannian manifold without an orientation. Let p ∈ M be given. Let T k (Mp ) be all k-rank tensors on Mp . We will introduce an inner product, still denoted by h·, ·i, on T k (Mp ) as follows. For any α, β ∈ T k (Mp ), hα, βi =

n X

i1 ···ik =1

α(ei1 , · · · , eik )β(ei1 , · · · , eik )

(1.103)

where {ei } is an orthonormal basis. It is easy to check that the definition (1.103) is independent of the choice of {ei }. Let Λk (Mp ) be all skew-symmetric tensors of rank k on Mp . Then a similar inner product is defined by X hα, βi = α(ei1 , · · · , eik )β(ei1 , · · · , eik ) (1.104) i1 0 be given. Consider the problem   utt = div A(x)∇u in (0, T ) × Ω, u = 0 on (0, T ) × Γ1 , u = ϕ on (0, T ) × Γ0 , (2.1)  u(0, x) = u0 (x), ut (0, x) = u1 (x) on Ω. We say the problem (2.1) is exactly controllable on the time interval [0, T ] if we can find T > 0 such that, for any (u0 , u1 ), (ˆ u0 , u ˆ1 ) given, there is a boundary function ϕ on (0, T ) × Γ0 such that the solution u to the problem (2.1) satisfies u(T, x) = u ˆ0 (x),

ut (T, x) = uˆ1 (x)

for

x ∈ Ω.

(2.2)

The problem (2.1) is called of constant coefficient if A(x) is a constant matrix for all x ∈ Rn ; otherwise, it is said to be a variable coefficient problem. Extension Method Russell [180], [181], and [182] obtained exact controllability in the case of constant coefficients by extending the problem (2.1) to be a Cauchy problem of the wave equation on the whole space Rn . This method was also used in [148], [149]. They constructed boundary controls in terms of the solution formula of the Cauchy problem. However, this method does not work for the variable coefficient problem because the solution formula of the Cauchy problem is no longer useful as in the case of constant coefficients. Duality Method The concept of duality between control and observation was first noted by Kalman who gave a duality theorem in [92] (also see [102]) for linear systems and was generalized to distributed parameter systems by [58] and the references there: Exact controllability is equivalent to an observability inequality in an operator theory form (see Theorem 2.10). However, they did not apply it to the particular problem (2.1)(now we can apply their theorems to obtain the particular observability estimate of the problem (2.1); see Theorem 2.11). Lions [146], [147] clarified the problem (2.1), worked out the particular observability inequality, and obtained constructive boundary controls (see Theorem 2.14). Since then, instead of looking for a boundary control directly, one has struggled on proving the observability estimate. However, it was not easy to prove. It was first proved by [86] that the observability inequality is true in the case of constant coefficients by the multiplier methods. The multiplier methods have been used by [39], [40], [41], [104], [105], [99],[117]-[122], [146], [155], [156], and many others. But in the classical analysis the multiplier methods do not work for the variable coefficient problems. At that time the


Control of the Wave Equation with Variable Coefficients in Space

59

observability inequality for the wave equation with variable coefficients was not checkable and, thus, Lions [146] put forward his open question: Is the observability estimate true for the variable coefficient problem? Geometric Optics Method A systematic attempt to understand the wave equation controllability and non-controllability properties can be found in [174], [175] where the author proved that to have exact controllability the so-called Geometric Control Condition (GCC) needs to be satisfied. Later on it was proved by [8] that in fact GCC is also sufficient to assure exact controllability. GCC says that the problem (2.1) is exactly controllable if and only if any geodesic initiated from the domain must hit the control portion Γ0 after a finite time. This condition provided us with a visible description of controllability. Although it is useful to help us understand the problem, whether GCC is an answer to our question depends on whether it is checkable. In the case of constant coefficients, a geodesic is a straight line. Since the domain is bounded, a straight line from the domain must hit Γ0 if Γ0 is choosen appropriately large so that GCC is true. But in the variable coefficient case, a geodesic is a solution to a second order system of nonlinear ordinary differential equations (see (1.12)). It is almost impossible to verify GCC from its definition. Then in this sense GCC just changed a problem of linear PDEs into the one of nonlinear ODEs. Thus Lions’ question was still open after GCC. Operator Theory Method [207] introduced an assumption for the stabilization of the wave equation with variable coefficients. By Russell’s principle this assumption is sufficient to guarantee the exact controllability. However, there was no way to check this assumption until [64] proved that it is equivalent to the one given by [208]. Pseudodifferential Multiplier Method General pseudodifferential multipliers were first derived from pseudoconvex functions [87] where these techniques were applied to solutions with compact support. However, the pseudodifferential multiplier methods in [87] did not account for boundary traces (which are critical for continuous observability inequalities). The techniques with pseudodifferential Carleman multipliers which did account for the boundary traces were first proposed in [192]. However, [192] requires the existence of a pseudoconvex function, a property which essentially can be verified mostly, if not exclusively, in the case of constant coefficients. Differential Geometric Method Differential Geometric Methods were introduced by [208] where the original motivation was to give checkable conditions to the exact controllability of the wave equation with variable coefficients to answer Lions’ open question [146]. They were extended by [25], [30], [130], [131], [201], [209]-[211], [216], [215], [218], and many others. There are two great ingredients in Riemannian geometry that play the key role in extending the multipliers in [86] to the case of variable coefficients: One is the Bochner technique(Section 1.3) which provides us with a great computational tool to carry out the multiplier scheme. The other one is the curvature theory which yields the global information for our assumptions.


60


The basic ideas of Bochner technique(Section 1.3) are as follows: In order to verify an identity or a pointwise estimate, it suffices to do so at each point p relative to a coordinate system or frame field that offers the greatest computational simplification. What prevents a computation from being as simple as the corresponding classical (i.e., Euclidean) situation is the presence of the Christoffel symbols Γkij . Thus we use a coordinate system or a frame field relative to which the Γkij vanish at the given point p. The second great ingredient, the curvature theory, yields the global information for our assumptions: The multiplier techniques lead to exact controllability to assumption (A): There exists an escape vector field on the domain for the metric g = A−1 (x). One of the useful consequences is that a convex function yields an escape vector field. Now the key question is whether this assumption is checkable. The classical analysis can only tell us that assumption (A) is locally true whenever the domain of the wave equation is small enough. However, the Riemannian curvature theory provides us with an effective tool to check assumption (A)(see Theorem 2.6 and examples in Section 2.3). Although the curvature is locally defined at each point, it interprets global information on almost everything in the metric, including assumption (A). That is one of the reasons why Riemannian geometry is called global geometry. Moreover, as far as the author is aware, up to now, the curvature theory has been the only way to check exact controllability for the variable coefficient problems. Recently, observability estimates were derived by [60] under an assumption (H): There are a function d and a constant c > 0 such that X X [2ail (akj dxk )xl − aijxl akl dxk ]ξi ξj ≥ c aij ξi ξj for (x, ξ) ∈ Ω × Rn , ij

ijkl

where A(x) = (aij (x)) is the coefficient matrix. One would expect that assumption (H) represented a new assumption. However, it was not the case because this assumption is equivalent to that d is a convex function(see Proposition 1.1). Therefore it is a consequence of assumption (A) for which, again, only the curvature theory is capable of checkability (Section 2.3).

2.2

Geometric Multiplier Identities

Let Ω ⊂ Rn be a bounded, open set with smooth boundary Γ. We consider u as a regular solution to the problem utt = div A(x)∇u + f

in (0, ∞) × Ω

(2.3)

to establish some multiplier identities where A(x) are symmetric, positive, smooth n × n matrices for all x ∈ Rn which are given by materials in appli-



61

cation. The main multipliers are 2H(u) and 2hu where H and h are a vector field and a function on Ω, respectively. We introduce g = A−1 (x) for x ∈ Rn n

(2.4)

n

as a Riemannian metric on R and consider the couple (R , g) as a Riemannian manifold. We denote by g = h·, ·ig the inner product and by D the covariant differential of the metric g, respectively. Denote by h·, ·i = · the Euclidean product of Rn . Then hX, Y ig = hA−1 (x)X, Y i for X, Y ∈ Rnx , x ∈ Rn .

(2.5)

If applied to functions, sometimes, we denote by ∇g and ∇ the gradients of the metric g and the Euclidean metric, respectively. Similarly, div g and div are the divergences of the metric g and the Euclidean metric, respectively. With two metrics on Rn in mind, one the Euclidean metric and the other the Riemannian metric g, we have to deal with various notation carefully. First, let us present some basic relations between the two metrics on Rn . It follows from the relation (2.5) that Lemma 2.1 Let x = (x1 , x2 , · · · , xn ) be the natural coordinate system on Rn . Let H, X ∈ X (Rn ) be vector fields and let h, f be functions. Then i) hA(x)H(x), X(x)i g = hH(x), X(x)i = H(x) · X(x) for x ∈ Rn ; ii) ∇g f = A(x)∇f for x ∈ Rn ; iii) h∇g f, ∇g hig = ∇g f (h) = hA(x)∇f, ∇hi for x ∈ Rn where A(x) is the coefficient matrix, given in (2.3). The following identity plays a crucial role in establishing multiplier identities in a version of the metric g. Lemma 2.2 Let f, h be functions on Rn and let H be a vector field on Rn . Then h∇g f, ∇g H(h) ig + h∇g h, ∇g H(f ) ig = DH(∇g h, ∇g f ) + DH(∇g f, ∇g h) + div h∇g f, ∇g hig H −h∇g f, ∇g hig div H,

x ∈ Rn

(2.6)

where div H is the divergence of the vector field H in the Euclidean metric. Proof. We compute the identity (2.6) at each point by the Bochner technique. Let x ∈ Rn be given. Let E1 , · · · , En be a frame field normal at x in the metric g. Then hEi , Ej ig = δij


in a neighborhood of x,

(2.7)

62

Modeling and Control in Vibrational and Structural Dynamics DEi Ej = 0

Let H =

Pn

i=1

at the point x.

(2.8)

hi Ei . The relations (2.7) and (2.8) imply that ∇g h =

n X

Ei (h)Ei

in a neighborhood of x,

(2.9)

i=1

DH(∇g h, ∇g f ) =

n X

Ei (h)Ej (hi )Ej (f ),

(2.10)

ij=1

Ej Ei (f ) = D2 f (Ei , Ej ) = Ei Ej (f )

at x,

(2.11)

2

where D f denotes the Hessian of f in the metric g. Using the relations (2.9)(2.11), we obtain n X h∇g f, ∇g H(h) ig = Ej (f )Ej H(h) j=1

=

n X

ij=1

h i Ej (f ) Ej (hi )Ei (h) + hi Ej Ei (h)

= DH(∇g h, ∇g f ) +

n X j=1

H Ej (h) Ej (f )

(2.12)

at x. We change f and h into each other in the formula (2.12) and add them together. The identity (2.6) follows from the following formula n X j=1

n X H Ej (h) Ej (f ) + H Ej (f ) Ej (h) = H(h∇g f, ∇g hig ) j=1

= div h∇g f, ∇g hig H − h∇g f, ∇g hig div H.

Now we are ready to present our main multiplier identities. Theorem 2.1 Let u solve the problem (2.3). Suppose that H is a vector field. Then div {2H(u)A(x)∇u − (|∇g u|2g − u2t )H} + 2f H(u)

= 2[ut H(u)]t + 2DH(∇g u, ∇g u) + (u2t − |∇g u|2g ) div H.

(2.13)

Proof. Multiply the equation (2.3) by 2H(u). Using the formulas (2.6) and iii) of Lemma 2.1 with f = h = u, we obtain 2f H(u) = 2(utt − div A(x)∇u)H(u) h i h i = 2ut H(u) − div 2H(u)A(x)∇u − H(u2t ) + 2hA(x)∇u, ∇ H(u) i h it h i = 2 ut H(u) + div (|∇g u|2g − u2t )H − 2H(u)A(x)∇u t

+2DH(∇g u, ∇g u) + (u2t − |∇g u|2g ) div H,


Control of the Wave Equation with Variable Coefficients in Space that is, the identity (2.13).

63

Theorem 2.2 Let u solve the problem (2.3) and let p be a function on Ω. Then div [2puA(x)∇u − u2 A(x)∇p] + 2f pu = 2p(uut )t + 2p(|∇g u|2g − u2t ) − u2 div A(x)∇p.

(2.14)

Proof. Multiply the equation (2.3) by 2pu and we have, by the formulas ii) and iii) of Lemma 2.1, 2f pu = 2(utt − div A(x)∇u)pu

= 2p(uut )t − 2 div [puA(x)∇u] − 2pu2t + 2hA∇u, ∇(pu)i = 2p(uut )t + div [u2 A∇p − 2puA(x)∇u] +2p(|∇g u|2g − u2t ) − u2 div A∇p.

2.3

Escape Vector Fields and Escape Regions for Metrics

The duality method will transfer controllability/stabilization of the wave equation to an assumption which is: Definition 2.1 Let g be the metric given by (2.4) and let Ω ⊂ Rn be a bounded, open set. A vector field H is said to be an escape vector field for the metric g on Ω if the covariant differential DH of H in the metric g is a positive tensor field on Ω, i.e., there is a constant ̺0 > 0 such that DH(x) ≥ ̺0 g(x)

for all

x ∈ Ω.

(2.15)

Remark 2.1 Escape vector fields were introduced by [208] as a checkable assumption for the exact controllability of the wave equation with variable coefficients. In [156] a convex function was called an escape function. Remark 2.2 If A(x) = I, the unit matrix, then for any x0 ∈ Rn , H = x − x0 is an escape vector field for the Euclidean metric on the whole space Rn since DH = g for all x ∈ Rn . Remark 2.3 Roughly speaking, if there is an escape vector field, then all waves will go to the boundary along this vector field and boundary controllability is true.


64


In general it is very hard to verify this assumption. That is one of the reasons why the Riemannian geometric approach is necessary to us. The goal of this section is to show the checkability of the above assumption by the Riemannian curvature theory: It is locally true (when Ω is small in some sense); the sectional curvature provides the global information on its existence; see Theorems 2.4-2.6 below. First, we give a necessary condition for escape vector fields Theorem 2.3 A necessary condition for escape vector fields for the metric g on Ω is: Ω does not include any closed geodesic. Proof. By contradiction. Let there be a closed geodesic γ(t) ∈ Ω for all t ∈ [0, b]. Then γ(0) = γ(b) and γ(0) ˙ = γ(b). ˙ (2.16) Assume that H is an escape vector field on Ω. Let f (t) = hH(γ(t)), γ(t)i ˙ g

for t ∈ [0, b].

By (2.16), we get f (0) = f (b).

(2.17)

Since H is escaping, we have f˙(t) = DH(γ(t), ˙ γ(t)) ˙ > 0 for

t ∈ [0, b]

which implies that f (b) > f (0), that contradicts (2.17).

If h is a strictly convex function in the metric g on Ω, then the Hessian D2 h of h in the metric g is positive on Rnx × Rnx for all x ∈ Ω. Then the first useful criterion is Theorem 2.4 If h is a strictly convex function in the metric g on Ω, then H = Dh is an escape vector field for the metric g on Ω. Remark 2.4 The class of escape vector fields for the metric is larger than that which is given by all gradients of strictly convex functions. There is an escape vector field which is not a gradient of any strictly convex function; see Example 2.6 later. For x ∈ Rn , let Ξ ⊂ Rnx be a two-dimensional subspace. Denote the sectional curvature of Ξ by κ(x, Ξ) in the metric g. Let x0 ∈ Rn be given. Let Bg (x0 , γ) be a geodesic ball in (Rn , g) centered at x0 with radius γ > 0. Set κ(Ω) =

sup

κ(x, Ξ).

(2.18)

x∈Ω, Ξ⊂Rn x

Denote by ρ(x) = dg (x, x0 ) the distance function of the metric g from x ∈ Rn to x0 .



65

Theorem 2.5 If γ > 0 is such that 4γ 2 κ(Ω) < π 2 and Ω ⊂ Bg (x0 , γ), then H = ρDρ is an escape vector field for the metric g on Ω. Proof. Since the inequality (2.15) is only required to be true for each point in Ω, it is independent of points in Bg (x0 , γ)/Ω. So we may change the metric g(x) for x ∈ Bg (x0 , γ)/Ω such that 4γ 2 κ(Bg ) < π 2

(2.19)

where κ(Bg ) =

sup

κ(x, Ξ).

x∈Bg (x0 ,γ), Ξ⊂Rn x

Since (Rn , g) is simply connected, by Theorem 1.22, we have D2 ρ(X, X) ≥ f (ρ)|X|2g

for X ∈ Rnx ,

hX, Dρig = 0

for x ∈ Bg (x0 , γ) where  κ(Bg )p = 0;  1/ρ, p f (ρ) = pκ(Bg ) cot( κ(B p g )ρ), κ(Bg ) > 0;  −κ(Bg ) coth( −κ(Bg )ρ), κ(Bg ) < 0.

(2.20)

(2.21)

It follows from (2.20) and (2.21) that DH = ρD2 ρ + Dρ ⊗ Dρ ≥ ̺0 g

for x ∈ Ω ⊂ Bg (x0 , γ)

where ̺0 = minx∈Ω {ρf (ρ), 1} > 0.

(2.22)

Remark 2.5 Theorem 2.5 shows that an escape vector field for the metric g locally exists on Ω: When Ω is included in a neighborhood of some point in Rn , it exists. Remark 2.6 For convenience, we may change the geodesic ball Bg (x0 , γ) in the metric g in Theorem 2.5 into the one in the Euclidean metric as in Corollary 2.1 below if κ(Ω) > 0. Corollary 2.1 Let κ(Ω) > 0. Let σ0 =

sup x∈Ω, X∈Rn x , |X|=1

hA−1 (x)X, Xi.

If there is x0 ∈ Rn such that Ω ⊂ B(x0 , σ1 ), then p there is an escape vector field for the metric g on Ω, where σ1 = π/(2 σ0 κ(Ω)) and B(x0 , σ1 ) = { x ∈ Rn | |x − x0 | < σ1 }.


66

Modeling and Control in Vibrational and Structural Dynamics Proof. By a similar reason as for (2.19), we assume that hA−1 (x)X, Xi ≤ σ0

for X ∈ Rnx , |X| = 1,

for all x ∈ B0 (x0 , σ1 ). Let x ∈ B(x0 , σ1 ) be given. Let r(t) = tx0 + (1 − t)x. Then r(·) is a curve connecting x0 and x. Then Z 1 Z 1 1/2 dg (x0 , x) ≤ |r(t)| ˙ g dt = hA−1 (r(t))(x0 − x), x0 − xi dt 0 0 p 1/2 ≤ σ0 |x − x0 | < π/(2 κ(Ω)). p Then there is 0 < γ < π/(2 κ(Ω) such that Ω ⊂ Bg (x0 , γ).

The corollary follows from Theorem 2.5.

Theorem 2.6 Let metric g be given by (2.4). Then (a) If (Rn , g) has non-positive sectional curvature, then there exists an escape vector field for the metric g on the whole space Rn . (b) If (Rn , g) is noncompact, complete, and its sectional curvature is positive everywhere on Rn , then there exists an escape vector field in the metric g on the whole space Rn . Proof. (a) is a straight consequence of Theorem 2.5 and (b) is given by [71]. Examples We give some examples which verify existence or non-existence of escape vector fields for a metric. Example 2.1 Let a coefficient matrix A(x) = aij be constant, symmetric, positive. Then (Rn , g) is of zero sectional curvature. By Theorem 2.6, (a), for any Ω ⊂ Rn bounded, there is an escape vector field on Ω. Example 2.2 Let ai > 0 be constants for 1 ≤ i ≤ n. Let α(x) = (a1 x1 , · · · , an xn )τ and A(x) =

1 [(1 + |α(x)|2 )I − α(x) ⊗ α(x)] 1 + |α(x)|2

for

x ∈ Rn ,

where I is the n × n unit matrix. Then the metric on Rn is g = A−1 (x) = I + α(x) ⊗ α(x)

for

x ∈ Rn

and (Rn , g) is a complete noncompact Riemannian manifold. We claim: There is an escape vector field for the metric g on the whole space Rn .



67

Proof. Step 1 Let M = { (x1 , · · · , xn , xn+1 ) ∈ Rn+1 | 2xn+1 =

n X i=1

ai x2i }

n+1

be a hypersurface of R . Then M is a Riemannian manifold with the induced metric from Rn+1 . Let p = (x1 , · · · , xn+1 ) ∈ M. The tangential space of M at p is given by Mp = { β ∈ Rn+1 | βn+1 =

n X i=1

ai xi βi }.

Then the normal of M is N (p) =

1 (a1 x1 , · · · , an xn , −1) for p ∈ M. (1 + |α(x)|2 )1/2

(2.23)

Denote by h·, ·i the Euclidean metric of Rn+1 . Let D0 be the connection of the metric h·, ·i on Rn . Then the second fundamental form of M is given by 0 Π(X, Y ) = −hDX Y, N i for

Let Xi =

∂ ∂ + ai xi ∂xi ∂xn+1

for

X, Y ∈ X (M ). i = 1, · · · , n.

Then X1 , · · · , Xn is a vector field basis of M. Since D0 ∂

∂xi

i, j ≤ n + 1, we obtain

0 DX Xj = Xi (aj xj ) i

for 1 ≤ i, j ≤ n. Let X=

n X i=1

αi Xi ,

∂ ∂ = aj δij ∂xn+1 ∂xn+1

Y =

(2.24)

n X

βi X i .

∂ ∂xj

= 0 for 1 ≤ (2.25)

(2.26)

i=1

Let R be the curvature tensor. Then by Theorem 1.13 and (2.23)-(2.25), we have R(X, Y, X, Y ) = Π(X, X)Π(Y, Y ) − |Π(X, Y )|2 n n n X X X 1 2 2 = [( a α )( a β ) − ( ai αi βi )2 ] > 0 i i i i 1 + |α(x)|2 i=1 i=1 i=1 if and only if X, Y are linearly independent. Thus M has positive sectional curvature everywhere. Step 2 Consider a map Φ: M → (Rn , g) given by Φ(p) = x


for p = (x, xn+1 ) ∈ M.

68


Let p = (x, xn+1 ) ∈ M and β = (β1 , · · · , βn+1 ) ∈ Mp be given. Assume that γ: (−ε, ε) → M be a curve such that γ(0) = p and γ(0) ˙ = β. Let γ(t) = (γ1 (t), · · · , γn (t), γn+1 (t)). Then Φ(γ(t)) = (γ1 (t), · · · , γn (t)) imply that Φ∗ β = (β1 , · · · , βn ). We obtain g(Φ∗ β, Φ∗ β) = hA−1 (x)Φ∗ β, Φ∗ βi =

n X i=1

n X βi2 + ( ai xi βi )2 = gˆ(β, β) i=1

n+1

where gˆ denotes the induced metric of M from R . Thus Φ is an isometry between M and (Rn , g). Therefore (Rn , g) is of positive curvature everywhere, noncompact complete by Step 1. Then what we need follows from (b) in Theorem 2.6. Example 2.3 Set A(x) =

1 1 + x21 + x62

1 + x62 x1 x32 x1 x32 1 + x21

for

x = (x1 , x2 ) ∈ R2 .

Then the metric is −1

g=A

(x) =

1 + x21 −x1 x32 −x1 x32 1 + x62

for

x = (x1 , x2 ) ∈ R2 .

It is easy to check that (R2 , g) is isometric to a surface M in R3 , given by M = { (x1 , x2 , x3 ) ∈ R3 | 4x3 = 2x21 − x42 }.

We have the Gauss curvature of (R2 , g) at x = (x1 , x2 ):

κ(x) = the Gauss curvature of M at p = (x, x3 ) −3x22 = ≤ 0. (1 + x21 + x62 )2 By (a) in Theorem 2.6, there is an escape vector field for the metric g on the whole space R2 . For convenience to compute sectional curvature, we give a lemma. Lemma 2.3 Let R2 have a metric g = g1 dx1 dx1 + g2 dx2 dx2 . Then the Gaussian curvature is 1 2 2 κ = 2 2 [g2 g1x1 g2x1 + g1 g1x2 g2x2 + g1 g2x + g2 g1x − 2g1 g2 (g1x2 x2 + g2x1 x1 )]. 1 2 4g1 g2 In particular, if g1 = g2 = h, then κ=−

1 ∆ log h 2h

where ∆ is the Laplacian in the Euclidean metric of R2 .


(2.27)

Control of the Wave Equation with Variable Coefficients in Space Proof. Set

1 Xi = √ ∂xi gi

69

for i = 1, 2.

Then X1 , X2 forms an orthonormal frame on (R2 , g). We have κ = hRX1 X2 X1 , X2 ig =

1 hR∂x1 ∂x2 ∂x1 , ∂x2 ig g1 g2

(2.28)

where R∂x1 ∂x2 ∂x1 = −D∂x1 D∂x2 ∂x1 + D∂x2 D∂x1 ∂x1 is the curvature operator. Let Γkij be the coefficients of the connection D in the metric g. From the formula (1.9), we obtain Γ112 =

1 g1x , 2g1 2

Γ211 = −

1 g1x , 2g2 2

Γ212 =

1 g2x1 , 2g2

1 1 g1x , Γ222 = g2x . 2g1 1 2g2 2 It follows from the properties of the connection that Γ111 =

D∂x1 D∂x2 ∂x1 = D∂x1 (Γ121 ∂x1 + Γ221 ∂x2 ) = (· · · )∂x1 + [Γ112 Γ211 + (Γ212 )x1 + (Γ212 )2 ]∂x2 . Thus hD∂x1 D∂x2 ∂x1 , ∂x2 ig = [Γ112 Γ211 + (Γ212 )x1 + (Γ212 )2 ]g2 1 1 2 1 2 = g2x1 x1 − g1x2 − g . 2 4g1 4g2 2x1

(2.29)

A similar computation yields 1 1 1 hD∂x2 D∂x1 ∂x1 , ∂x2 ig = − g1x2 x2 + g1x g2x + g1x g2x . 2 4g1 1 1 4g2 2 2

(2.30)

From the relations (2.28), (2.29) and (2.30), we obtain the formula of κ. Example 2.4 Let a(x) be a smooth, positive function on R2 . Set a(x) 0 A(x) = for x ∈ R2 . 0 a(x) Then the metric is g=

1/a(x) 0 0 1/a(x)

for

x ∈ R2 .

By Lemma 2.3, the Gauss curvature of (R2 , g) is κ=

a∆ log a 2


for

x ∈ R2

70


where ∆ is the Laplacian of the Euclidean metric of R2 . By (a) in Theorem 2.6, if ∆ log a ≤ 0 for x ∈ R2 , then there is an escape vector field for the metric on the whole space R2 . In particular, let a(x) = ex1 +x2 for x ∈ R2 . Then κ = 0 for x ∈ R2 . There is an escape vector field for this metric on the whole space R2 . Example 2.5 Consider a coefficient matrix, given by x +x e 1 2 0 A(x) = for x ∈ R2 . 0 e−x1 −x2 The metric on R2 is g=

e−x1 −x2 0 0 ex1 +x2

.

By Lemma 2.3, we have 1 κ(x) = − (e−x1 −x2 + ex1 +x2 ) < 0 2

for

x ∈ R2 .

Then there is an escape vector field for the metric g on the whole space R2 from Theorem 2.6. Example 2.6 Set 3

3

A(x) = ex1 +x2 I

x ∈ R2

for

where I is the 2 × 2 unit matrix. Then the metric is g = hI

where

3

3

h = e−x1 −x2

for

x ∈ R2 .

Then the coefficients of the connection are 3 Γ111 = Γ212 = −Γ122 = − x21 , 2

3 Γ112 = Γ222 = −Γ211 = − x22 . 2

Let H = −e−x1 ∂x1 − e−x2 ∂x2

for

x ∈ R2 .

(2.31)

Let X = X1 ∂x1 + X2 ∂x2 . We have DX H = e−x1 X1 ∂x1 − e−x1 DX ∂x1 + e−x2 X2 ∂x2 − e−x2 DX ∂x2 = [(e−x1 − e−x1 Γ111 − e−x2 Γ112 )X1 − (e−x1 Γ112 + e−x2 Γ122 )X2 ]∂x1

+[−(e−x1 Γ211 + e−x2 Γ212 )X1 + (e−x2 − e−x1 Γ212 − e−x2 Γ222 )X2 ]∂x2

which yields DH(X, X) = hDX H, Xig =


2 X

ij=1

hij Xi Xj

(2.32)


71

where −x 3 2 −x2 e 1 + 32 (x21 e−x1 + x22 e−x2 ) − x22 e−x1 ) 2 (x1 e hij = h . 3 2 −x1 − x21 e−x2 ) e−x2 + 32 (x21 e−x1 + x22 e−x2 ) 2 (x2 e (2.33) It follows from (2.32) and (2.33) that

DH(X, X) ≥ ̺|X|2g

for

x ∈ R2

where ̺ = min(e−x1 , e−x2 ) > 0

for

x ∈ R2 ,

that is, the vector field H, given by (2.31), is an escape vector field for the metric g on the whole space R2 . However, H is not a gradient of any function because the matrix (hij ) is not symmetric. Next, we introduce geometric conditions on interior feedback regions. Let J be an integer. Let Ωi ⊆ Ω be open sets with C ∞ boundary ∂Ωi for 1 ≤ i ≤ J such that Ωi ∩ Ωj = ∅ for 1 ≤ i < j ≤ J. (2.34) Let H i be vector fields on Ωi for 1 ≤ i ≤ J such that DH i (X, X) ≥ ̺0 |X|2g

for

X ∈ Rxn , x ∈ Ωi

(2.35)

for all i where ̺0 > 0. Definition 2.2 An subset ω ⊂ Ω is said to be an escape region for the metric g on Ω if there is ε > 0 small such that ω ⊇ Ω ∩ Nε [ ∪Ji=1 Γi0 ∪ (Ω\ ∪Ji=1 Ωi ) ]

(2.36)

where Nε (S) = ∪x∈S { y ∈ Rn |y − x| < ε} ,

S ⊂ Rn ,

Γi0 = { x ∈ ∂Ωi | H i (x) · ν i (x) > 0 } ,

and ν i (x) is the unit normal of ∂Ωi in the Euclidean of Rn , pointing outside of Ωi . Remark 2.7 Escape regions were introduced by [151] for interior feedback stabilization of the wave equation with constant coefficients. They were given in the above form by [62] for the variable coefficient problem. Now we consider the structure of an escape region for the metric. Since the set in the right-hand side of (7.207) is always a subset of Ω, ω = Ω is an escape region for the metric. However, we are particularly interested in the case ω 6= Ω where ω is small as far as possible in some sense.


72


If there is an escape vector field H on Ω, then the inequality (7.206) holds for all x ∈ Ω. We can take J = 1 and Ω1 = Ω. In this case, an escape region ω for the metric only needs to be supported in a neighborhood of Γ0 , where Γ0 = { x ∈ Γ | H(x) · ν(x) > 0 } .

(2.37)

This roughly means that we can dig out almost the whole Ω from the domain Ω and that escape regions only need to be supported in a neighborhood of a subset of the boundary. Suppose that A(x) is the unit matrix for all x ∈ Rn . Then the metric is the Euclidean metric of Rn , and D = ∇ is the gradient of the standard metric. Let x0 ∈ Rn be given. Then the vector field H(x) = x − x0 on Ω meets condition (7.206) with ρ0 = 1. Thus, escape regions for the metric needs only to be supported in a neighborhood of Γ0 = { x ∈ Γ | (x − x0 ) · ν(x) > 0 } . Such regions were used in [196]. In general, whether there exists one escape vector field H satisfying estimate (7.206) on the whole domain Ω largely depends on the sectional curvature of the Riemannian metric g; see Theorems 2.5 and 2.6. Let x0 ∈ Rn be given. Let ρ(x) = ρ(x, x0 ) be the distance function from x ∈ Ω to x0 in the metric g. We define a vector field H(x, x0 ) on Ω by H(x, x0 ) = ρ(x)Dρ(x)

for

x∈Ω

(2.38)

If g is the Euclidean metric, then D = ∇ and H(x, x0 ) = x − x0 . It follows from Theorem 2.5 that Theorem 2.7 If γ > 0 is such that 4γ 2 κ(B) < π 2 and Ω ⊂ B(x0 , γ), an escape region for the metric on Ω can be supported in a neighborhood of Γ0 = { x ∈ Γ | H(x, x0 ) · ν(x) > 0 }

(2.39)

where κ(B) is given by (2.18) and H(x, x0 ) is given by (2.38). Remark 2.8 If (Rn , g) has nonpositive curvature, then κ(B) ≤ 0 for any γ > 0 and the results in Theorem 2.7 hold for any Ω ⊂ Rn bounded. If κ(B) > 0, one escape vector field on the whole Ω does not exist in general when Ω is large. A counterexample will be given by Example 2.7 later. √ But we can dig out a finite number of geodesic balls with radius ≤ π/2 κ from Ω and let an escape region be supported in a neighborhood of the remaining part of Ω. This is the following. Theorem 2.8 Let κ > 0. Suppose that there are points xi ∈ Ω for 1 ≤ i ≤ J such that π ρ(xi , xj ) > √ , 1 ≤ i, j ≤ J, i 6= j . κ



73

Let B(xi , r0 ) and S(xi , r0 ) be the geodesic ball and the geodesic sphere √ centered at xi with radius r0 in the metric g, respectively, where r0 < π/2 κ. Let Γi0 = { x | x ∈ S(xi , r0 ), H(x, xi ) · ν(x) > 0 }. If ω ⊇ Ω ∩ Nε [ ∪Ji=1 Γi0 ∪ ( Ω\ ∪Ji=1 B(xi , r0 )) ] , then ω is an escape region on Ω for the metric. Since we can dig out geodesic balls with radii small from Ω as many as we want, the following results are immediate. Theorem 2.9 For ε > 0 given, there is an escape region ω ⊂ Ω with meas (ω) < ε where meas (ω) is the n-dimensional Lebesgue measure of ω.

2.4

Exact Controllability. Dirichlet/Neumann Action

The concept of duality between control and observation was first noted by Kalman who gave a duality theorem in [92] (also see [102]) for linear systems and was generalized to distributed parameter systems by [58] and the references there. Duality Method Let X and Y stand for Banach spaces of general type. Let A be an operator defined on a subspace of one of these spaces. Its domain will be denoted by D(A). Its range is denoted by R(A). Consider an abstract linear system A : D(A) ⊂ X −→ Y

(2.40)

where A is a linear operator with dense domain. Along with (2.40), we consider a dual system A∗ : D(A∗ ) ⊂ Y ∗ −→ X ∗ (2.41) where X ∗ and Y ∗ are the conjugate spaces of X and Y, respectively, and A∗ is the adjoint operator for A. Definition 2.3 The system (2.40) is observable if there is c > 0 such that kAxkY ≥ ckxkX

for

x ∈ D(A).

(2.42)

Definition 2.4 The system (2.41) is exactly controllable if X ∗ ⊆ R(A∗ ).


(2.43)

74


The following is a duality theorem relating observability and controllability ([58] or [69]). Theorem 2.10 The system (2.40) is observable if and only if the system (2.41) is exactly controllable. Proof. Let (2.42) hold true. Let x∗ ∈ X ∗ be given. We define a linear functional y ∗ on the subspace R(A) of Y by y ∗ (Ax) = x∗ (x)

for all Ax ∈ R(A).

(2.44)

Then by (2.42) |y ∗ (Ax)| ≤ kx∗ kX ∗ kxkX ≤ kx∗ kX ∗ kAxkY /c

for all

Ax ∈ R(A).

By the Hahn-Banach Theorem, y ∗ can be extended from R(A) to Y to belong Y ∗ , still denoted by y ∗ . Then the relation (2.44) means that y ∗ ∈ D(A∗ ) and x∗ = A∗ y ∗ ∈ R(A∗ ). Conversely, let (2.43) be true. For x ∈ D(A), x 6= 0, we define fx ∈ X ∗∗ by x∗ (x) fx (x∗ ) = for all x∗ ∈ X ∗ . kAxkY

Then for each x∗ ∈ X ∗ fixed, the condition (2.43) implies that there is y∗ ∈ Y ∗ such that x∗ = A∗ y ∗ . For this x∗ , we have |fx (x∗ )| = |y ∗ (

Ax )| ≤ ky ∗ kY ∗ kAxkY

for all x ∈ D(A), x 6= 0.

By the resonance theorem, we have kxkX = sup kfx kX ∗∗ < ∞ x∈D(A), kxkX =1 kAxkY x∈D(A), kxkX =1 sup

that is equivalent to (2.42) being true.

Control in Dirichlet Boundary Condition Let Γ0 be a portion of the boundary Γ which is relatively open. Set Γ1 = Γ/Γ0 . Let T > 0 be given. Set Q = (0, T ) × Ω,

Σ = (0, T ) × Γ,

Σi = (0, T ) × Γi ,

We consider the control problem  utt = div A(x)∇u in Q,    u = 0 on Σ1 , u = ϕ on Σ0 ,    u(0) = u0 , ut (0) = u1 on Ω with control ϕ on Σ0 . Following [116], we have


for i = 0, 1.

(2.45)


75

Proposition 2.1 Let T > 0 given. For any (u0 , u1 ) ∈ L2 (Ω) × H −1 (Ω) and ϕ ∈ L2 (Σ0 ), the problem (2.45) has the unique solution u ∈ C[0, T ; L2(Ω)] ∩ C 1 [0, T ; H −1 (Ω)]. Definition 2.5 We say that the problem (2.45) is exactly L2 (Ω) × H −1 (Ω) controllable by L2 (Σ0 ) control at time T if, for any (u0 , u1 ), (ˆ u0 , u ˆ1 ) ∈ L2 (Ω)× −1 2 H (Ω), there is ϕ ∈ L (Σ0 ) such that the solution u to the problem (2.45) satisfies u(T ) = u ˆ0 , ut (T ) = uˆ1 on Ω. Now we formulize the controllability of the problem (2.45) in order to apply Theorem 2.10. Let X = H01 (Ω) × L2 (Ω),

Y = L2 (Σ0 ).

Then X ∗ = H −1 (Ω) × L2 (Ω),

Y ∗ = L2 (Σ0 ).

We define an operator A: X → Y as follows. For any (w0 , w1 ) ∈ X, we solve the problem   wtt = div A(x)∇w in Q, w = 0 on Σ, (2.46)  w(0) = w0 , wt (0) = w1 on Ω to define

A(w0 , w1 ) = wνA

on Σ0

where wνA = hA(x)∇w, νi and ν is the outside normal of Ω on Γ in the Euclidean metric of Rn . Next, we define an operator B: Y ∗ → X ∗ as follows. For any ϕ ∈ L2 (Σ0 ), we solve the problem  utt = div A(x)∇u in Q,    u(T ) = ut (T ) = 0 on Ω, (2.47)  u = 0 on Σ1 ,   u = ϕ on Σ0 to define

Bϕ = (ut (0), −u(0)) on Ω. Proposition 2.1 implies that B is a bounded, linear operator. R(B) is all initial data which are exactly controllable to the rest. Since the system (2.45) is time-reversible, the exact controllability for the problem (2.45) on X ∗ = H −1 (Ω) × L2 (Ω) is equivalent to the relation R(B) = X ∗ . For any ϕ ∈ Y ∗ and (w0 , w1 ) ∈ X, using the systems (2.46) and (2.47), we


76


obtain (Bϕ, (w0 , w1 ))L2 (Ω)×L2 (Ω) = (ut (0), w(0)) − (u(0), wt (0)) Z T = [(u(τ ), wt (τ )) − (ut (τ ), w(τ ))]t dτ 0

=

Z

0

T

[(u(τ ), wtt (τ )) − (utt (τ ), w(τ ))]dτ

= (ϕ, A(w0 , w1 ))L2 (Σ0 ) = A∗ ϕ((w0 , w1 )) = (A∗ ϕ, (w0 , w1 ))L2 (Ω)×L2 (Ω) where (·, ·) is the inner product of L2 (Ω), which implies that B = A∗ . In particular, A: X → Y is bounded since B is. It follows from Theorem 2.10 that Theorem 2.11 The system (2.45) is exactly L2 (Ω) × H −1 (Ω) controllable at time T > 0 by L2 (Σ0 ) control if and only if there is c > 0 such that kwνA k2L2 (Σ0 ) ≥ c(kw0 k2H 1 (Ω) + kw1 k2L2 (Ω) ) 0

(2.48)

for all (w0 , w1 ) ∈ H01 (Ω) × L2 (Ω). Remark 2.9 Proposition 2.1 implies that A: H01 (Ω) × L2 (Ω) → L2 (Σ0 ) is bounded. Then Theorem 2.11 means that k(w0 , w1 )k⋆ = kwνA kL2 (Σ0 ) defines an equivalent norm on H01 (Ω) × L2 (Ω). Definition 2.6 The system (2.46) and the inequality (2.48) are said to be the duality system and the observability estimate of the problem (2.45), respectively. Lions’ Construction [146], [147] First, we introduce a theorem which is capable of defining a space which is exactly controllable, given by J. L. Lions by the dual method. Let H and L be Hilbert spaces with H ⊂ L and H being dense in L such that kykL ≤ CkykH

for all y ∈ H.

Let H ∗ denote the completion of the following space {x ∈ L|

sup |(x, y)L | < ∞ }

kykH =1

with a norm kxkH ∗ = supkykH =1 |(x, y)L |. Clearly, H ⊂ L ⊂ H ∗. Definition 2.7 H ∗ is said to be the dual space of H with respect to L.



77

For each x ∈ H ∗ , (x, ·)L is a linear, bounded functional on H. Then, by the Riesz representation theorem, there is a unique element in H, denoted by P x, such that (x, y)L = (P x, y)H for all y ∈ H. (2.49) Then P : H ∗ → H is a linear operator. The following results are needed when we construct boundary control for the Neumann action. Theorem 2.12 The operator P : H ∗ → H is an isometry. Proof. It follows from (2.49) that kP xkH = sup (P x, y)H = sup (x, y)L = kxkH ∗ kykH =1

kykH =1

for all x ∈ H. Then R(P ) is a closed subspace of H. To complete the proof, it remains to show R(P ) = H. If it is not true, then there is x0 ∈ H, x0 6= 0, such that (P x, x0 )H = 0. The condition (2.49) yields (x, x0 )L = 0

for all x ∈ L,

that is, x0 = 0. This contradiction shows R(P ) = H. Definition 2.8 The operator ℵ = P isomorphism from H to H ∗ .

−1

: H → H

∗

is called the canonical

Remark 2.10 It follows from (2.49) that (x, y)H = (ℵx, y)L

for all

x, y ∈ H.

(2.50)

Suppose that there is a linear operator Λ: H → L with dense domain D(Λ) ⊂ H, such that (Λx, y)L = (x, Λy)L

for x, y ∈ D(Λ).

We have Theorem 2.13 Let there be a constant c > 0 such that (Λx, x)L ≥ ckxk2H

for all

x ∈ H.

(2.51)

Let R = { Λx | (Λx, x)L < ∞ }

with an inner product (Λx, Λy)R = (Λx, y)L . Then R ⊃ H ∗.

If, in addition to the condition (2.51), there is a constant C such that (Λx, x)L ≤ Ckxk2H

for all

then R = H ∗.


x ∈ H,

(2.52)

78

Modeling and Control in Vibrational and Structural Dynamics Proof. We introduce a space H1 = { x | (Λx, x)L < ∞ }

with an inner product (x, y)H1 = (Λx, y)L . It is easy to check that R = H1∗ . Let the inequality (2.51) hold. The inequality (2.51) implies H1 ⊂ H. Thus R ⊃ H ∗. Moreover, let the inequality (2.52) be true. Then H = H1 . Thus R = H1∗ = H ∗ . Remark 2.11 In application R represents a space which is exactly controllable. In addition, the condition (Λx, x)L < ∞ stands for the regularity of control action; see Remark 2.15 later. We are now back to the controllability problem (2.45) again. Let (w0 , w1 ) ∈ H01 (Ω) × L2 (Ω). We solve the duality problem (2.46) to obtain the solution w. Then we solve the problem  utt = div A(x)∇u in Q,    u(T ) = ut (T ) = 0 on Ω, (2.53) u = 0 on Σ1 ,    u = wνA on Σ0 where νA = A(x)ν. We define an operator Λ: H01 (Ω) × L2 (Ω) → H −1 (Ω) × L2 (Ω) by Λ(w0 , w1 ) = (ut (0), −u(0)) on Ω.

We take H = H01 (Ω) × L2 (Ω) and L = L2 (Ω) × L2 (Ω). Let w and w ˆ be the solutions to the problem (2.46) for the initial data (w0 , w1 ) and (w ˆ0 , w ˆ1 ), respectively. Suppose that u and uˆ are the solutions to the problem (2.53) with the boundary data wνA and w ˆνA , respectively, on Σ0 . Using (2.46) and (2.53), we have (Λ(w0 , w1 ), (w ˆ0 , w ˆ1 ))L2 (Ω)×L2 (Ω) = (ut (0), w(0)) ˆ − (u(0), w ˆt (0)) Z T = [(u, w ˆt ) − (ut , w)] ˆ t dτ = (wνA , w ˆνA )L2 (Σ0 ) . (2.54) 0

Then H ∗ represents the space which is exactly controllable by L2 (Σ0 ) control. Applying Theorem 2.13, we have Theorem 2.14 The system (2.45) is exactly L2 (Ω) × H −1 (Ω) controllable at time T > 0 by L2 (Σ0 ) control if and only if k(w0 , w1 )k⋆ = kwνA kL2 (Σ0 ) defines an equivalent norm on H01 (Ω) × L2 (Ω). Remark 2.12 In general the problem (2.53) only has distributed solutions whether the boundary control u = wνA is smooth or not. To have smooth controls, we need to smooth the boundary control u = wνA as in (2.100).



79

Let w be a solution to the duality problem (2.46). We introduce its energy by 2E(t) =

Z

Ω

(wt2 + hA(x)∇w, ∇wi)dx

for t ≥ 0

(2.55)

where h·, ·i is the Euclidean metric of Rn . Then E(t) = E(0) for all t ≥ 0. Then the exact controllability of the problem (2.45) on L2 (Ω) × H −1 (Ω) follows from Lemma 2.4 and Theorem 2.15 below. Lemma 2.4 For any T > 0, there is CT > 0 such that kwνA k2L2 (Σ) ≤ CT E(0)

(2.56)

for all solutions to the problem (2.46). ˆ be a vector field on Ω such that Proof. Let H ˆ = νA H

for

x∈Γ

where νA (x) = A(x)ν(x). For X ∈ X (Γ), by (i) in Lemma 2.1, we have h∇g w, Xig = h∇w, Xi = X(w) = 0

for x ∈ Γ

since w|Γ = 0. Then ∇g w = h∇g w,

νA νA νA i = wνA |νA |g g |νA |g |νA |2g

which yields |∇g w|2g = ∇g w(w) = wν2A /|νA |2g

for x ∈ Γ.

(2.57)

ˆ u = w, and f = 0 in the identity (2.13), we integrate it After taking H = H, over Q to have the estimate (2.56). Next, we need a uniqueness result which is introduced from [5] or [170] without proofs. Proposition 2.2 Let X be a vector field and let q be a smooth function on Ω, respectively. Then the problem div A(x)∇v = X(v) + qv has a unique zero solution. We have


on

Ω,

v|Γ0 = vνA |Γ0 = 0

(2.58)

80


Theorem 2.15 Let H be an escape vector field for the metric g on Ω, where g is given by (2.4). Then for any T > T0 , there is a constant cT > 0 such that kwνA k2L2 (Σ0 ) ≥ cT E(0)

(2.59)

for all solution w to the problem (2.46), where 2 sup |H|g (x), ̺0 x∈Ω

T0 =

Γ0 = { x ∈ Γ | hH, νi > 0 },

(2.60)

and ̺0 > 0 is given by (2.15). Proof. We decompose H into the direct sum in (Rnx , g(x)) as H = hH,

νA νA hH, νi i +Z = νA + Z |νA |g g |νA |g |νA |2g

for x ∈ Γ

where hZ, νA ig = hZ, νi = 0, which gives H(w) = hH, νiwνA /|νA |2g

for x ∈ Γ.

(2.61)

Taking f = 0 and u = w in the identity (2.13) and using the assumption (2.15), we obtain div X1 ≥ 2̺0 |∇g w|2g + (wt2 − |∇g w|2g ) div H + 2[wt H(w)]t

= ̺0 (wt2 + |∇g w|2g ) + (wt2 − |∇g w|2g )( div H − ̺0 ) + 2[wt H(w)]t

(2.62)

where X1 = 2H(w)A(x)∇w − (|∇g w|2g − wt2 )H. Letting f = 0, u = w, and p = ( div H − ̺0 )/2 in the identity (2.14), and then adding it to the inequality (2.62), we get div X2 ≥ ̺0 (wt2 + |∇g w|2g ) − w2 div A(x)∇p +2p(wwt )t + 2[wt H(w)]t

(2.63)

where X2 = X1 + 2pwA(x)∇w − w2 A(x)∇p. Moreover, it follows from the relations (2.61) and (2.57) and the boundary conditions in the problem (2.46) that hX2 , νi = hH, νiwν2A /|νA |2g

for x ∈ Γ.

We integrate the inequality (2.63) over Q and obtain, by (2.64), ckwνA k2L2 (Σ0 ) ≥ 2(̺0 T − ε − 2 sup |H|g )E(0) − Cε L(w) x∈Ω


(2.64)


81

for ε > 0 small where L(w) = (kwk2L2 (Q) + kw(0)k2 + kw(T )k2 ) is the lower order term. Since T > T0 , there are constants ci > 0 for 1 ≤ i ≤ 3 such that c1 kwνA k2L2 (Σ0 ) + c2 L(w) ≥ c3 E(0) (2.65) for all solutions w to the problem (2.46). Then the inequality (2.59) follows by Lemma 2.5 below. Remark 2.13 T0 in (2.60) represents a control time which can be shown to be the shortest in some cases of constant coefficients. Lemma 2.5 Let the inequality (2.65) hold true for all solutions w to the problem (2.46). Then there is c > 0, independent of solutions, such that kwνA k2L2 (Σ0 ) ≥ cE(0).

(2.66)

Proof. Step 1 Let Y = { w ∈ H 1 (Q) | w is a solution to (2.46) satisfying wνA |Σ0 = 0 }. Then Y = { 0 }.

(2.67)

Indeed, from the inequality (2.65), we have c2 L(w) ≥ c3 E(0)

for all w ∈ Y

which implies that any bounded, closed set in Y ∩H 1 (Q) is compact in H 1 (Q). Then Y is a finitely dimensional, linear space. For any w ∈ Y, wt ∈ Y. Then ∂t : Y → Y is a linear operator. Let Y 6= { 0 }. Then ∂t has at least one eigenvalue λ. Assume that v ∈ Y, v 6= 0, is one of its eigenfunctions. Then vt = λv. Thus v is a nonzero solution to the problem (2.58) with X = 0 and q = λ2 . This contradiction shows that (2.67) is true. Step 2 By contradiction. Suppose that the estimate (2.66) is not true. Then there are (w0k , w1k ) ∈ H01 (Ω) × L2 (Ω), whose solutions are denoted by wk , such that E(wk , 0) = 1,

kwνkA k2L2 (Σ0 ) ≤

1 k

for k ≥ 1.

(2.68)

Then kwk k2H 1 (Q) = 2T for all k ≥ 1. Thus there is a subsequence, still denoted by { wk }, such that { wk } converges in L2 (Ω) for each t ∈ [0, T ] and

(2.69)

{ wk } converges in L2 (Q).

(2.70)


82


It follows from the relations (2.65), (2.68), (2.69), and (2.70) that { wk } converges in H 1 (Q). Then there is a solution w0 to the problem (2.46) such that wk → w0

as k goes to ∞ in H 1 (Q).

Then E(w0 , 0) = 1,

wν0A |Σ0 = 0,

which contradicts the relation (2.67).

Remark 2.14 The proof of the above lemma is referred to as a compactnessuniqueness argument in the literature. Control in Neumann Condition (Γ, g) is an (n − 1)-dimensional Riemannian manifold where g is the induced metric from (Rn , g). Denote by ∇Γg the gradient of (Γ, g). Then for any v ∈ H 1 (Q), |∇g v|2g = vν2A /|νA |2g + |∇Γg v|2g

for

x ∈ Γ.

We consider the control problem   utt = div A(x)∇u in Q, uν = ϕ on Σ,  A u(0) = u0 , ut (0) = u1 on Ω,

with a Neumann action ϕ on Σ. The duality problem is   wtt = div A(x)∇w in Q, wν = 0 on Σ,  A w(0) = w0 , wt (0) = w1 on Ω.

(2.71)

(2.72)

For T > 0, let

L0 = L2 (Σ0 ),

H0 = H 1 ([0, T ]; L2 (Γ0 )),

L1 = L2 (Σ1 ),

H1 = L2 ([0, T ]; H 1 (Γ1 )).

Denote by ℵi : Hi → Hi∗ the canonical isomorphisms with respect to Li for i = 0, 1. Then it follows from (2.49) that Z (ℵ0 φ, ϕ)L2 (Σ0 ) = (φt ϕt + φϕ)dΣ (2.73) Σ0

for all φ, ϕ ∈ H 1 ([0, T ]; L2 (Γ0 )), and Z (ℵ1 φ, ϕ)L2 (Σ1 ) = (h∇Γg φ, ∇Γg ϕig + φϕ)dΣ Σ1

for all φ, ϕ ∈ L2 ([0, T ]; H 1 (Γ1 )).


(2.74)


83

For (w0 , w1 ) ∈ H 1 (Ω) × L2 (Ω), we solve the problem (2.72) to obtain w. Then we solve the problem  utt = div A(x)∇u in Q,    u(T ) =ut (T ) = 0 on Ω, (2.75) −ℵ0 w on Σ0 ,    uνA = −ℵ1 w on Σ1 . We define an operator Λ: H 1 (Ω) × L2 (Ω) → (H 1 (Ω))∗ × L2 (Ω) by Λ(w0 , w1 ) = (ut (0), −u(0)) on Ω. We take H = H 1 (Ω) × L2 (Ω) and L = L2 (Ω) × L2 (Ω). Let w and w ˆ be the solutions to the problem (2.72) for the initial data (w0 , w1 ) and (w ˆ0 , w ˆ1 ), respectively. Suppose that u and uˆ are the solutions to the problem (2.75) with the corresponding boundary data from w and w, ˆ respectively. Using (2.72), (2.75), (2.73) and (2.74), we have (Λ(w0 , w1 ), (w ˆ0 , w ˆ1 ))L2 (Ω)×L2 (Ω) = (ut (0), w(0)) ˆ − (u(0), w ˆt (0)) Z T = [(u, w ˆt ) − (ut , w)] ˆ t dτ = −(w, ˆ uνA )L2 (Σ) Z0 Z = (wt w ˆt + ww)dΣ ˆ + (h∇Γg w, ∇Γg wi ˆ g + ww)dΣ. ˆ (2.76) Σ0

Σ1

Remark 2.15 If a solution w to the duality problem (2.72) such that Z Z (wt2 + w2 )dΣ + (|∇Γg w|2g + w2 )dΣ Σ0

Σ1

= (Λ(w0 , w1 ), (w0 , w1 ))L2 (Ω)×L2 (Ω) < ∞,

that is, w|Σ0 ∈ H 1 ([0, T ]; L2(Γ0 )) and w|Σ1 ∈ L2 ([0, T ]; H 1 (Γ1 )), then the boundary control in the problem (2.75) satisfies ℵ0 w ∈ [H 1 ([0, T ]; L2(Γ0 ))]∗ ;

ℵ1 w ∈ [L2 ([0, T ]; H 1 (Γ1 ))]∗ .

Then in Theorem 2.13, (Λx, x)L < ∞ describes the regularity of control action. Applying Theorem 2.13, we have Theorem 2.16 If there is cT > 0 such that Z Z (wt2 + w2 )dΣ + (|∇Γg w|2g + w2 )dΣ ≥ cT E(0) Σ0

(2.77)

Σ1

for all solutions w to the problem (2.72), then the problem (2.71) is exactly L2 (Ω) × (H 1 (Ω))∗ controllable by the Neumann action. Next, we establish the observability inequality (2.77).


84


Theorem 2.17 Let H be an escape vector field for the metric g on Ω. Then the inequality (2.77) holds true for any T > T0 where T0 and Γ0 are given in (2.60). Proof. We take u = w and f = 0 in the identities (2.13) and (2.14). Let p = ( div H − ̺0 )/2 where ̺0 is given in (2.15). Then we add the identity (2.14) to the identity (2.13) and obtain, as in (2.63), div X ≥ ̺0 (wt2 + |∇g w|2g ) + 2[wt H(w) + pwwt ]t − w2 div A(x)∇p

(2.78)

where X = 2H(w)A(x)∇w − (|∇g w|2g − wt2 )H + 2pwA(x)∇w − w2 A(x)∇p. Using the boundary conditions wνA = 0, we have Z Z Z hX, νidΣ = [(wt2 − |∇g w|2g )hH, νi + w2 ]dΣ − w2 (1 + pνA )dΣ Σ

Σ

Σ

≤ CI(w) + Ckwk2H 1/2 (Q)

(2.79)

where I(w) denotes the left hand side of the inequality (2.77). We integrate the inequality (2.78) over Q and obtain, by (2.79), CI(w) + Cε L(w) ≥ 2(̺0 T − 2 sup |H|g − ε)E(0)

(2.80)

x∈Ω

for ε > 0 small enough, where L(w) are lower order terms with respect to the norm of H 1 (Q). The inequality (2.77) follows from (2.80) by a compactness-uniqueness argument as for Lemma 2.5. Remark 2.16 In Theorem 2.16 controls are taken in “large” spaces where ℵ0 w ∈ (H 1 ([0, T ]; L2 (Γ0 )))∗ and ℵ1 w ∈ (L2 ([0, T ]; H 1 (Γ1 )))∗ . In general, even if w is smooth, ℵi (w) may be not in L2 (Σi ). However, if the boundary Γ satisfies the following assumption (2.81), then controls can be taken in L2 (Σ), see Theorem 2.18 below. Let H be an escape vector field for the metric g such that hH, νi ≥ 0 for all x ∈ Γ.

(2.81)

Let T0 be given in (2.60). Then an argument as in the proof of Theorem 2.17 yields the estimate Z Σ

(wt2 + w2 )dΣ ≥ cT E(0)

where T > T0 is given and w are solutions to the problem (2.72). We use a trick in [146]. Let w1 satisfy Z w1 dx = 0. Ω


(2.82)


85

Assume that ω solves div A(x)∇ω = w1

on Ω,

Let

Z

v=

ωνA = 0

on Γ.

t

wdτ + ω.

0

Then v solves the problem (2.72) with initial data v(0) = ω and vt (0) = w0 . Applying v to the inequality (2.82) gives Z w2 dΣ ≥ cT (kw0 k2 + kw1 k2(H 1 (Ω))∗ ). (2.83) Σ

It follows from Theorem 2.13 that

Theorem 2.18 Let H be an escape vector field for the metric g on Ω and let T0 be given in (2.60). Then the problem (2.71) is exactly H 1 (Ω) × L2 (Ω) controllable by L2 (Σ) control by the Neumann action. Finally, we consider the exact controllability problem   utt = div A(x)∇u in Q, u|Σ1 = 0, uνA |Σ0 = ϕ,  u(0) = u0 , ut (0) = u1 on Ω

where Γ0 ∪ Γ1 = Γ. The duality problem of the problem (2.84) is   wtt = div A(x)∇w in Q, w|Σ1 = wνA |Σ0 = 0,  w(0) = w0 , wt (0) = w1 on Ω.

(2.84)

(2.85)

It is easy to check that the exact controllability of the problem (2.84) on the space L2 (Ω) × (HΓ11 (Ω))∗ leads to the following observability estimate: There is a constant cT > 0 such that kwt k2L2 (Σ0 ) ≥ cT E(0)

(2.86)

for all solutions w to the problem (2.85). We have Theorem 2.19 Let Γ1 6= ∅. Let H be an escape vector field for the metric g on Ω such that hH, νi ≤ 0 for x ∈ Γ1 . (2.87)

Then for T > T0 , the estimate (2.86) holds true.

Proof. Let Γ0+ = { x | hH, νi ≥ 0, x ∈ Γ0 }. Using the boundary conditions w|Σ1 = wνA |Σ0 = 0, we obtain, via (2.87), Z Z hX, νidΣ = hH, νiwν2A /|νA |2g dΣ ≤ 0 (2.88) Σ1

Σ1

where X is given in (2.78). Similar arguments as in the proof of Theorem 2.17 complete the proof.


86

2.5


Smooth Controls

We consider smooth controls which are needed for the control of the quasilinear wave equation in Chapter 7. Control with the Dirichlet Action Let F and q be a vector field and a function on Ω, respectively. Let T > 0 be given. We consider the problem  utt = div A(x)∇u + F (u) + qu in Q,    u = 0 on Σ1 , (2.89) u = ϕ on Σ0 ,    u(0) = u0 , ut (0) = u1 on Ω.

The duality system is   wtt = Aw in Q w = 0 on Σ  w(0) = w0 , wt (0) = w1

(2.90) on Ω

where the operator A is defined by

Aw = div A(x)∇w − F (w) + (q − div F )w. We have the following Green formula Z ⋆ (v, Au) = (A v, u) + [vuνA − uvνA − uvhF, νi] dΓ

(2.91)

(2.92)

Γ

for u, v ∈ H 2 (Ω), where A⋆ u = div A(x)∇u + F (u) + qu.

(2.93)

Let m ≥ 0 be a given integer. We shall study the exact controllability of the problem (2.89) on [H m+1 (Ω)∩HΓ11 (Ω)]×[H m (Ω)∩HΓ11 (Ω)] by appropriate boundary controls where HΓ11 (Ω) = { v ∈ H 1 (Ω) | v|Γ1 = 0 }. To this end, we make some preparation as follows. Space of Boundary Control Let B 0D (Σ0 ) = L2 (Σ0 ). For m ≥ 1, let m B D (Σ0 ) consist of all the functions k ϕ ∈ ∩m−1 [0, T ], H m−1/2−k (Γ0 ) , ϕ(k) ∈ H 1 (Σ0 ), (2.94) k=0 C for 0 ≤ k ≤ m − 1 with the norm kϕk2B m = D

m−1 X k=0

kϕ(k) k2C([0,T ],H m−k−1/2 (Γ0 )) +

m−1 X k=0

kϕ(k) k2H 1 (Σ0 )

where the superscript (k) denotes the k-th derivative with respect to time



87

variable t. Let Ξ00 (Ω) = L2 (Ω). For k ≥ 1 an integer, let Ξk0 (Ω) consist of the functions u in H k (Ω) with the boundary conditions i A u|Γ = 0, 0 ≤ i ≤ l − 1, if k = 2l; (2.95) Ai u|Γ = 0, 0 ≤ i ≤ l if k = 2l + 1 and with the norms of H k (Ω) where A is given by (2.91). Then Ξ10 (Ω) = H01 (Ω),

Ξ20 (Ω) = H 2 (Ω) ∩ H01 (Ω),

Ξ30 (Ω) = { w | w ∈ H 3 (Ω), w|Γ = 0, Aw|Γ = 0 }.

We consider the operator A on L2 (Ω), given by (2.91) with a domain D(A) = H 2 (Ω) ∩ H01 (Ω).

(2.96)

It follows from (2.91) and (2.92) that the adjoint operator of A is given by A∗ w = A⋆ w,

D(A∗ ) = H 2 (Ω) ∩ H01 (Ω)

(2.97)

where A⋆ is given by (2.93). We have Lemma 2.6 Let l be a positive integer. Consider the operator Al : H 2l (Ω) ∩ 2 Ξ2l 0 (Ω) → L (Ω). Then L2 (Ω) = R(Al ) ⊕ N (A∗ l ) where

(2.98)

N (A∗ l ) = { w | A∗ l w = 0, w ∈ H 2l (Ω) ∩ Ξl0 (Ω) }

is a finitely dimensional subspace of L2 (Ω).

Proof. It will suffice to prove that R(Al ) is a closed subspace of L2 (Ω) and N (A∗ l ) is finitely dimensional. Set σ0 > max{sup (q − div F ), 0}. x∈Ω

Then Aw = Aσ0 w + σ0 w

for w ∈ D(A)

where the operator Aσ0 is defined by Aσ0 w = div A(x)∇w − F (w) + (q − div F − σ0 )w, D(Aσ0 ) = H 2 (Ω) ∩ H01 (Ω). Since q − div F − σ0 ≤ 0, by the regularity of elliptic operators (see Theorem


88


2 2 8.12 in [68]), 0 belongs to the resolvent set of Aσ0 and A−1 σ0 : L (Ω) → H (Ω) ∩ 1 2 H0 (Ω) is a compact operator on L (Ω). Let l C = (I + σ0 A−1 σ0 ) − I

where I is the identity operator on L2 (Ω). Then C is a compact operator on L2 (Ω). Thus R(I + C) is closed in L2 (Ω). Then R(Al ) = R(Alσ0 (I + C)) is closed in L2 (Ω). Similarly, we can take a constant σ1 > max{q, 0}. Then q − σ1 ≤ 0. Thus the operator A∗σ1 = A∗ − σ1 I has the compact inverse. Then A∗ l = A∗ lσ1 +

l X i=1

Cil σ1i A∗ l−i σ1 .

Then A∗ l w = 0 if and only if w+

l X i=1

Cil σ1i A∗ −i σ1 w = 0.

∗l

Then N (A ) is finitely dimensional since the operator compact.

Pl

i=1

Cil σ1i A∗ −i σ1 is

Let T > T1 > 0 be given. We assume that z ∈ C ∞ (−∞, ∞) is such that 0 ≤ z(t) ≤ 1 with 0 t ≥ T, z(t) = (2.99) 1 t ≤ T1 .

For (w0 , w1 ) ∈ Ξm+1 (Ω) × Ξm 0 (Ω) given, we solve the duality problem 0 (2.90) and, then, we solve the following problem   utt = A⋆ u in Q, u(T ) = ut (T ) = 0 on Ω, (2.100)  u|Σ1 = 0, u|Σ0 = zwνA

where A⋆ is defined by (2.93). An operator Λ on Ξm+1 (Ω) × Ξm 0 (Ω) is defined by 0 Λ(w0 , w1 ) = (ut (0), −u(0)).

(2.101)

It is easy to check that, for any (w0 , w1 ), (w ˆ0 , w ˆ1 ) ∈ H01 (Ω) × L2 (Ω), Z (Λ(w0 , w1 ), (wˆ0 , w ˆ1 ))L2 (Ω)×L2 (Ω) = z(t)wνA w ˆνA dΣ (2.102) Σ0

where w and w ˆ are solutions to the duality problem (2.90) with initial data (w0 , w1 ) and (w ˆ0 , w ˆ1 ), respectively. A similar argument as for Lemma 2.4 yields



89

Lemma 2.7 There is a constant C > 0 such that kwνA k2L2 (Σ) ≤ Ck(w0 , w1 )k2H 1 (Ω)×L2 (Ω) 0

(2.103)

for all solutions w to the problem (2.90). The formula (2.102) implies the regularity of the operator Λ, that is, Proposition 2.3 below. For convenience, we denote by k · ki,j the norms of H i (Ω) × H j (Ω) for i, j ≥ −1. Proposition 2.3 Let m ≥ 0 be an integer. Then the operator Λ, given by m−1 (2.101), is bounded from Ξm+1 (Ω) × Ξm (Ω) × H m (Ω). 0 (Ω) to H 0 Proof. We use induction on m. (1) Let m = 0. It follows from the relations (2.102) and (2.103) that kΛ(w0 , w1 )k−1,0 =

|(zwνA , w ˆνA )L2 (Σ) | ≤ Ck(w0 , w1 )k1,0 ,

sup k(w ˆ 0 ,w ˆ 1 )k1,0 =1

that is, the claim is true for m = 0. (2) Assume that the claim holds for some m ≥ 0. We shall prove that the claim is true for m + 1. It will suffice if we establish the estimate kut (0)km + ku(0))km+1 ≤ C(kw0 km+2 + kw1 km+1 )

(2.104)

for all (w0 , w1 ) ∈ Ξm+2 (Ω) × Ξm+1 (Ω). 0 0 Case 1 Let m = 2l with l ≥ 0. Then m + 1 = 2l + 1. Let w ˆ be the solution to the duality problem (2.90) with the initial data (w ˆ0 , w ˆ1 ). Then w ˆ(2i) and w ˆ(2i+1) are the solutions to the problem (2.90) corresponding to the initial data (Ai w ˆ0 , Ai w ˆ1 ) and (Ai w ˆ1 , Ai+1 w ˆ0 ), respectively, for i ≥ 0, where A is given by (2.91). Using initial data (w0 , w1 ) and (Am+1 w ˆ0 , Am+1 w ˆ1 ) in the formula (2.102), we have Z (ut (0), Am+1 w ˆ0 ) − (u(0), Am+1 w ˆ1 ) = z(t)wνA w ˆν(2m+2) dΣ. (2.105) A Σ0

Noting that z(0) = 1 and z (i−1) (T ) = z (i) (0) = 0 for i ≥ 1, we obtain, by integration by parts with respect to the variable t over [0, T ], Z

Σ0

z(t)wνA w ˆν(2m+2) dΣ A

j=1

+(−1)m+1

Z

Σ0

=

l−1 Z X j=0

=

m+1 XZ

Γ0

(−1)j wν(j−1) (0)w ˆν(2m+2−j) (0)dΓ A A

[z(t)wνA ](m+1) w ˆν(m+1) dΣ A

j

m−j

(A w1 )νA (A Γ0

w ˆ0 )νA dΓ −

+(−1)m+1 I(w, w) ˆ


l Z X j=0

(Aj w0 )νA (Am−j w ˆ1 )νA dΓ

Γ0

(2.106)

90


where I(w, w) ˆ = +

Z

z(t)wν(m+1) w ˆν(m+1) dΣ A A

Σ0 m+1 XZ j=1

Σ0

w ˆν(m+1) dΣ. Cjm+1 z (j) (t)wν(m+1−j) A A

(2.107)

We assume that (w ˆ0 , w ˆ1 ) ∈ C0∞ (Ω) × C0∞ (Ω). Noting that (Ai w ˆ0 , Aj w ˆ1 ) ∈ ∞ × C0 (Ω) for all i, j, using the formula (2.92), and the boundary in (2.95), we obtain C0∞ (Ω)

(u(0), Am+1 w ˆ1 ) = (A⋆ l u(0), Al+1 w ˆ1 ) + = −(A∇(A⋆ l u(0)), ∇(Al w ˆ1 )) +

l Z X j=0

l−1 Z X j=0

Γ

A⋆ j u(0)(Am−j w ˆ1 )νA dΓ

u(2j) (0)(Am−j w ˆ1 )νA dΓ

Γ0

ˆ1 ) + Al w ˆ1 ( div F − q)) −(A⋆ l u(0), F (Al w

(2.108)

since u solves the problem (2.100). Similarly, we have (ut (0), Am+1 w ˆ0 ) = (A⋆ l ut (0), Al+1 w ˆ0 ) Z l−1 X + u(2j+1) (0)(Am−j w ˆ0 )νA dΓ. j=0

(2.109)

Γ0

(2j+1)

(2j)

(0) = (Aj w1 )νA and u(2j) (0) = wνA (0) = Since u(2j+1) (0) = wνA (A w0 )νA on Γ0 , it follows from the relations (2.105)-(2.108) that j

(A⋆ l ut (0), Al+1 w ˆ0 ) + (A∇(A⋆ l u(0)), ∇(Al w ˆ1 ))

= −I(w, w) ˆ − (A⋆ l u(0), F (Al w ˆ1 ) + Al w ˆ1 ( div F − q))

(2.110)

for all (w0 , w1 ) ∈ Ξm+2 (Ω) × Ξm+1 (Ω) and (wˆ0 , w ˆ1 ) ∈ C0∞ (Ω) × C0∞ (Ω). 0 0 Letting w ˆ0 = 0 in the identity (2.110), we obtain (A0 (A⋆ l u(0)), Al w ˆ1 )

= I(w, w) ˆ + (Al u(0), F (Al w ˆ1 ) + Al w ˆ1 ( div F − q))

for all w ˆ1 ∈ C0∞ (Ω), where A0 = div A(x)∇.


(2.111)


91

From (2.107) and (2.103), we have |I(w, w)| ˆ ≤ +C

l X j=0

≤C ≤C

|

l X j=0

l X j=0

Z

Z l X | j=0

Σ0

Σ0

m+1 (2j) C2j z (t)wν(2(l−j)+1) w ˆν(m+1) dΣ| A A

m+1 (2j+1) C2j+1 z (t)wν(2(l−j)) w ˆν(m+1) dΣ| A A

(kwν(2(l−j)+1) kL2 (Σ0 ) + kwν(2(l−j)) kL2 (Σ0 ) )kw ˆν(m+1) kL2 (Σ0 ) A A A (k(Al−j w1 , Al−j+1 w0 )k1,0

+k(Al−j w0 , Al−j w1 )k1,0 )k(Al w ˆ1 , 0)k1,0

≤ C(kw0 k2H m+2 (Ω) + kw1 k2H m+1 (Ω) )1/2 kAl w ˆ1 kH 1 (Ω) .

Next, we estimate

(2.112)

kA0 (A⋆ l u(0))kH −1 (Ω)

through the identity (2.111). For this purpose, Lemma 2.6 is needed. We decompose A0 (A⋆ l u(0)) as a direct sum A0 (A⋆ l u(0)) = [A0 (A⋆ l u(0))]0 + [A0 (A⋆ l u(0))]1 where [A0 (A⋆ l u(0))]0 ∈ IN (A∗ l ). By Lemma 2.8 later, we have k[A0 (A⋆ l u(0))]0 k ≤ Ckw0 kH m+3/2 (Ω) + Cku(0)kH m (Ω) .

(2.113)

Moreover, using the relations (2.98) and (2.112) in (2.111), we obtain, via (2.113), kA0 (A⋆ l u(0))kH −1 (Ω) ≤ Ck(w0 , w1 )km+2,m+1 + Cku(0)kH m (Ω) . (2.114) Moreover, on the boundary Γ the problem (2.100) implies that kA⋆ i u(0)kH m+1/2−2i (Γ) = ku(2i) (0)kH m+1/2−2i (Γ0 ) = kwν(2i) (0)kH m+1/2−2i (Γ0 ) A

= k(Ai w0 )νA kH m+1/2−2i (Γ) ≤ ckw0 kH m+2 (Ω)

for 0 ≤ i ≤ l.

(2.115)

Now, using the relations (2.114) and (2.115) and the ellipticity of the operator A⋆ and A0 , we obtain ku(0)kH m+1 (Ω) ≤ c(kA⋆ u(0)kH m−1 (Ω) + ku(0)kH m+1/2 (Γ) + ku(0)kH m (Ω) ) ≤ C(kA⋆ l u(0)kH 1 (Ω) + ku(0)kH m+1/2 (Γ) + ku(0)kH m (Ω) )

≤ C(kA0 A⋆ l u(0)kH −1 (Ω) + kA⋆ l u(0)kH 1/2 (Γ) + k(w0 , w1 )km+2,m+1 ) ≤ Ck(w0 , w1 )km+2,m+1


92


where the induction assumption ku(0)kH m (Ω) ≤ ck(w0 , w1 )km+1,m is used. A similar argument yields kut (0)km ≤ C(kw0 km+2 + kw1 km+1 ).

(2.116)

Case 2 Let m = 2l + 1. Similar arguments as in Case 1 yield the estimate (2.104). . Lemma 2.8 The estimate (2.113) holds true. Proof. Since IN (A∗ l ) is finitely dimensional, it suffices to estimate (A0 (A⋆ l u(0)), θ) for θ ∈ IN (A∗ l ) given. In fact, it follows from the Green formula that Z |(A0 (A⋆ l u(0)), θ)| = | − A∗ l u(0)θνA dΓ + (A∗ l u(0), A0 θ)| Γ

≤ Cku(m) (0)kL2 (Γ0 ) + CkA∗ l u(0)k

≤ CkAl w0 kH 3/2 (Ω) + Cku(0)kH m (Ω) ≤ Ckw0 kH m+3/2 (Ω) + Cku(0)kH m (Ω) . We need to indicate the control function space, that is, Lemma 2.9 Let m ≥ 0 be given. Let w solve the problem (2.90) with initial m data (w0 , w1 ) ∈ Ξm+1 (Ω) × Ξm 0 (Ω). Then wνA ∈ B D (Σ0 ). 0 Proof. We only prove the case m = 1. The proof may be completed by induction but we omit it. Let (w0 , w1 ) ∈ Ξ20 (Ω) × Ξ10 (Ω). We shall prove wνA ∈ C([0, T ], H 1/2 (Γ0 )) ∩ H 1 (Σ0 ).

(2.117)

Clearly, for any T > 0 given, there is cT > 0 such that kw(t)k2H 2 (Ω) ≤ cT (kw0 k2H 2 (Ω) + kw1 k2H 1 (Ω) )

∀ t ∈ [0, T ],

which implies wνA ∈ C([0, T ], H 1/2 (Γ)). Since wt is the solution of the problem (2.90) for the initial data (w1 , Aw0 ) ∈ H01 (Ω) × L2 (Ω), Lemma 2.7 implies wtνA ∈ L2 (Σ0 ). To complete the proof, it remains to show that wνA ∈ L2 (0, T ), H 1 (Γ0 ) . Let X be a vector field of the manifold Γ, that is, X(x) ∈ Γx for each x ∈ Γ. We extend X to the whole Ω to be a vector field on the manifold (Ω, g) where g = A−1 (x). Let v = X(w) for (t, x) ∈ Q = (0, T ) × Ω.



93

Then v solves the problem   v¨ = Av + [X, A]w in Q, v|Γ = 0, t ∈ (0, T ),  v(0) = X(w0 ) ∈ H01 (Ω), v(0) ˙ = X(w1 ) ∈ L2 (Ω), where [X, A]w = X(Aw) − AX(w) with the estimate

k[X, A]w(t)k2 ≤ cT (kw0 k2H 2 (Ω) + kw1 k2H 1 (Ω) ) for t ∈ [0, T ].

(2.118)

Let H be a vector field on Ω with H(x) = νA

for x ∈ Γ.

Letting u = v and f = [X, A]w − F (v) + v(q − div F ) in the identity (2.13), we integrate it over Q to obtain Z Z 2 T vνA dΣ = 2(v, ˙ H(v))|0 + [2DH(∇g v, ∇g v) + (v˙ 2 − |∇g v|2g ) div H]dQ Σ

Σ

+2(F (v) + v( div F − q) − [X, A]w, H(v))

which yields, via (2.118) Z vν2A dΣ ≤ cT (kw0 k2H 2 (Ω) + kw1 k2H 1 (Ω) ).

(2.119)

Σ

Since vνA = νA (X(w)) = X(wνA ) + [νA , X]w

for x ∈ Γ,

by (2.119), we have Z |X(wνA )|2 dΣ ≤ cT,X kw0 k2H 2 (Ω) + kw1 k2H 1 (Ω) , Σ

for any vector field X of the manifold Γ, that is, wνA ∈ L2 (0, T ), H 1 (Γ0 ) . We shall show that the problem (2.100) provides smooth controls by the following theorem.

Theorem 2.20 Let m ≥ 0 be an integer and let T > 0 be given. Then the problem (2.89) is exactly [H m (Ω)∩HΓ11 (Ω)]×[H m−1 (Ω)∩HΓ11 (Ω)] controllable by B m D (Σ0 ) controls through the relations (2.90), (2.100) and (2.101) if and only if there is c1T > 0 and c2T > 0 such that c1T k(w0 , w1 )km+1,m ≤ kΛ(w0 , w1 )km−1,m ≤ c2T k(w0 , w1 )km+1,m for all (w0 , w1 ) ∈ Ξm+1 (Ω) × Ξm 0 (Ω). 0


(2.120)

94


Proof. By Theorem 2.13 the claim is true for m = 0. We consider the case of m = 1. If Λ is surjective, then the inverse operator theorem implies that the estimate (2.120) holds true for m = 1. Conversely, we assume that the inequality (2.120) is true. If Λ is not surjective, we derive a contradiction as follows. Let there be (v0 , v1 ) ∈ L2 (Ω) × HΓ11 (Ω), (v0 , v1 ) 6= 0, such that (ut (0), v0 ) − (u(0), v1 )H 1 (Ω) = (Λ(w0 , w1 ), (v0 , v1 ))0,1 = 0

(2.121)

for all (w0 , w1 ) ∈ C0∞ (Ω) × C0∞ (Ω). Noting that u(0)|Γ1 = 0 and u(0)|Γ0 = (w0 )νA |Γ0 = 0, we have (u(0), v1 )H 1 (Ω) = (u(0), −∆v1 ) where ∆ is the Laplacion in the Euclidean metric. Using this formula in (2.121), we obtain, via (2.102), 0 = (wνA , w ˆνA )L2 (Σ0 ) = ((w0 , w1 ), Λ(v0 , −∆v1 ))L2 (Ω)×L2 (Ω) where w ˆ solves the problem (2.90) with the initial data (v0 , −∆v1 ), for all (w0 , w1 ) ∈ C0∞ (Ω) × C0∞ (Ω), that is, (v0 , −∆v1 ) = 0, which is a contradiction. We repeat this procedure to complete the proof. . Next, we assume that the observability inequality is true for distribution controls to derive smooth controls . Assumption (A): Let T1 > 0 be given such that there is cT1 > 0 such that kwνA k2L2 ((0,T1 )×Γ0 ) ≥ cT1 k(w0 , w1 )k21,0 (2.122) for all solutions w to the problem (2.90). Remark 2.17 Let F = 0 and q = 0. If there is an escape vector field for the metric g on Ω, then Assumption (A) is true, see Theorem 2.15. For any F and q, the inequality (2.122) was established in [130] under the assumption that there is a strictly convex function on Ω in the metric g. [154] proved that the assumption (A) is also true if there is an escape vector field for the metric g on Ω. We have Theorem 2.21 Let m ≥ 0 be given and let T > T1 be given where T1 > 0 is such that the assumption (A) holds true. Then the problem (2.89) is exactly [H m (Ω) ∩ HΓ11 (Ω)] × [H m−1 (Ω) ∩ HΓ11 (Ω)] controllable on [0, T ] by B m D (Σ0 ) controls. Proof. By Proposition 2.3, it will suffice to prove that the left hand side of (2.120) is true. We prove it by induction. If m = 0, it is just the assumption (A). We assume that the claim is true for some m ≥ 0. We will prove that it holds true for m + 1. Let us assume m = 2l for some l ≥ 0. Similar arguments can show the case m = 2l + 1.


Control of the Wave Equation with Variable Coefficients in Space Let w solves the problem (2.90) with initial data (w0 , w1 ) ∈ (2i) Ξm+1 (Ω). Then w(2i) and wt solve the same problem with 0 i i i i+1 (A w0 , A w1 ) and (A w1 , A w0 ), respectively, for 1 ≤ i ≤ l.

95

Ξm+2 (Ω) × 0 initial data

By taking (w ˆ0 , wˆ1 ) = (w0 , w1 ) in the relations (2.110) and (2.107), we obtain, via the inequalities (2.122) and (2.103), ckΛ(w0 , w1 )km,m+1 k(w0 , w1 )km+2,m+1

≥ |I(w, w)| − ckΛ(w0 , w1 )km−1,m k(w0 , w1 )km+2,m+1 ≥ c1 k(Al w1 , Al+1 w0 )k21,0 − ck(Al w1 , Al+1 w0 )k1,0 k(w0 , w1 )km+1,m −ckΛ(w0 , w1 )km−1,m k(w0 , w1 )km+2,m+1 .

(2.123)

On the other hand, (w0 , w1 ) ∈ Ξm+2 (Ω) × Ξm+1 (Ω) implies, by the ellip0 0 ticity of the operator A, k(w0 , w1 )km+2,m+1 ≤ c(k(Al w1 , Al+1 w0 )k1,0 + k(w0 , w1 )km+1,m ).

(2.124) Using (2.124) in (2.123), we obtain, by the induction assumption, that the claim holds true for m + 1. Control with the Neumann Action We now turn to the smooth control problem  utt = div A(x)∇u in Q,    u = 0 on Σ1 , (2.125) uν = ϕ on Σ0 ,    A u(0) = u0 , ut (0) = u1 on Ω.

The duality problem of the system (2.125) is   wtt = div A(x)∇w in Q, w|Σ1 = wνA |Σ0 = 0,  w(0) = w0 , wt (0) = w1 on Ω.

(2.126)

The problem (2.125) is exactly (HΓ1 (Ω))∗ × L2 (Ω) controllable if and only if the inequality (2.86) holds true. However, to find the smooth control, the one-side observability estimate (2.86) is insufficient. We have to seek boundary estimates of another type controlled by the initial energy on both sides from below and also from above, as in Theorem 2.14. Let ε > 0 be given small. Let ηε ∈ C ∞ (R) be such that 0 ≤ ηε ≤ 1 and ηε (t) = 1 t ≤ −ε;

ηε (t) = 0

t ≥ 0.

For any T > ε, let zε (t) = ηε (t − T ).

(2.127)

Then zε (t) = 1

0 ≤ t ≤ T − ε;


zε (t) = 0

t ≥ T.

96

Modeling and Control in Vibrational and Structural Dynamics Let H be a vector field. We introduce a bilinear form by Z Ψ(w, w) ˆ = zε (t)[wt w ˆt − h∇Γg w, ∇Γg wi ˆ g ]hH, νidΣ

(2.128)

Σ0

where w and w ˆ are solutions to the problem (2.126) with initial data (w0 , w1 ) and (wˆ0 , w ˆ1 ), respectively. Let H be an escape vector field for the metric g on Ω with ̺0 > 0 such that the inequality (2.15) is true. For ε > 0 fixed small, set T0 = [(1 + sup ε|ηε′ (t)|) sup |H|g . −ε≤t≤0

(2.129)

x∈Ω

We further assume that hH, νi ≤ 0 for

x ∈ Γ1 .

(2.130)

Another observability estimate we need is the following. Theorem 2.22 Let Γ1 6= ∅ and Γ0 ∩ Γ1 = ∅. Let H be an escape vector field for the metric g on Ω such that the assumption (2.130) holds. Let T0 be given in (2.129). For any T > 0, there are constant c0 > 0 and cT > 0 such that cT k(w0 , w1 )k21,0 ≥ Ψ(w, w) + c0 (zε w, w)L2 (Σ0 ) + c0 kw0 k2 + c0 kwk2L2 (Q) Z T ≥ [̺0 zε (t)dt − T0 − ε]k(w0 , w1 )k21,0 (2.131) 0

for all solutions w to the problem (2.126). Proof. We use the multipliers 2zε (t)H(w) and 2zε (t)pw for the equation in (2.126) as in Theorems 2.1 and 2.2, to obtain Z zε [2H(w)wνA + (wt2 − |∇g w|2g )hH, νi]dΣ Σ

Z

T

= −2(w1 , H(w0 )) − 2 zεt (t)(wt , H(w))dt T −ε Z + zε [2Dg H(∇g w, ∇g w) + (wt2 − |∇g w|2g ) div H]dQ,

(2.132)

Q

and 2

Z

zε (t)p(wt2

Q

−2

Z

−

|∇g w|2g )dQ

T

zεt (t)(wt , pw)dt +

T −ε

Z

respectively, where p is a function.


= −2(w1 , pw0 ) − Σ0

zε w2 pνA dΣ,

Z

zǫ w2 div A∇pdQ Q

(2.133)


97

Let ̺0 > 0 be given in (2.15). We take p = div H − ̺0 in the identity (2.133) and obtain the estimate Z 2 zε p(wt2 − |∇g w|2g )dQ ≥ −εk(w0 , w1 )k1,0 Q

−cε (kw0 k2 + kwk2L2 (Q) ) − c0 kwk2L2 (Q) − c0 (zε w, w)L2 (Σ0 ) .

(2.134)

On Σ1 , the boundary conditions w = 0 implies that H(w) = hH, νiwνA /|νA |2g and |∇g w|2g = wν2A /|νA |2g . Then Z Z 2 2 zε wν2A hH, νi/|νA |2g dΣ ≤ 0. zε [2H(w)wνA + (wt − |∇g w|g )hH, νi]dΣ = Σ1

Σ1

On Σ0 , wνA = 0 implies ∇g w = ∇Γg w. Using those relations in the identity (2.132), we have, via (2.134), Z Z 2 2 Ψ(w, w) ≥ ̺0 zε (wt + |∇g w|g dQ) + zε (wt2 − |∇g w|2g )pdQ Q

Q

−(1 + sup ε|ηε′ (t)|) sup |H|g k(w0 , w1 )k21,0 −ε≤t≤0

Z

x∈Ω

T

zε (t)dt − T0 − ε}k(w0 , w1 )k21,0 0 −c0 kwk2L2 (Q) − c0 (zε w, w)L2 (Σ0 ) − c0 kw0 k2

≥ {̺0

which proves that the right hand side of the inequality (2.131) holds. On the other hand, since Γ1 ∩ Γ0 = ∅, we take two open sets ω0 and ω1 in Rn such that ω0 ∩ ω1 = ∅ and Γi ⊂ ωi for i = 0, 1, respectively. Let h ∈ C ∞ (Rn ) be such that h(x) = 1 x ∈ ω0 ;

h(x) = 0 x ∈ ω1 .

Replacing H with hH in the identity (2.132) yields Ψ(w, w) ≤ cε T k(w0 , w1 )k21,0 . Let all the assumptions in Theorem 2.22 hold. Let 1/2 > ε > 0. Using a boundary trace estimate, Lemma 7.2 in [122], we have an estimate Z (2.135) |∇Γg w|2g dΣ ≤ cT ε (kwt k2L2 (Σ0 ) + kwk2H 1/2+ε (Q) ) (ε,T −ε)×Γ0

for all solutions to the problem (2.126). Let T > 2ε. We use the right hand side of the inequality (2.131) where the interval [0, T ] is replaced with [ε, T −ε] to have cT (zε wt , wt )L2 (Σ0 ) + cT L(w) ≥ [̺0 (T − 2ε) − T0 − ε)k(w0 , w1 )k21,0


(2.136)

98


for all solutions w to the problem (2.126), where L(w) are some lower order terms related to the norm kwk2H 1 (Q) = k(w0 , w1 )k21,0 . A compactnessuniqueness argument, as in Lemma 2.5, shows that the lower order terms in (2.136) can be absorbed if the following uniqueness problem holds true: The problem   wtt = div A(x)∇w in Q, w|Σ1 = 0, (2.137)  wt |Σ0 = wνA |Σ0 = 0, has the unique zero solution. However, the uniqueness problem (2.137) is true because the problem (2.137) implies that v = wt to solve the problem vtt = div A(x)∇v in Q, v|Σ0 = vνA |Σ0 = 0, which yields wt = 0 on Ω, that is, w = 0 on Q since Γ1 6= ∅. We have Corollary 2.2 Let all the assumptions in Theorem 2.22 hold. Then for T > (T0 + ε)/̺0 + 2ε, there is cT > 0 such that cT (zε wt , wt )L2 (Σ0 ) ≥ k(w0 , w1 )k21,0

(2.138)

for all solutions w to the problem (2.126). We introduce an operator by A0 v = div A(x)∇v,

D(A0 ) = { w | w ∈ H 2 (Ω), w|Γ1 = 0, wνA |Γ0 = 0 }.

Let (w0 , w1 ) ∈ D(A0 ) × HΓ11 (Ω). Then (w1 , A0 w0 ) ∈ HΓ11 (Ω) × L2 (Ω). Since wt solves the problem (2.126) with the initial data (w1 , A0 w0 ), the inequality (2.131) implies c2T k(w1 , A0 w0 )k21,0 ≥ Ψ(wt , wt ) + c0 (zε wt , wt )L2 (Σ0 ) + c0 kw1 k2 + c0 kwt k2L2 (Q) ≥ c1T k(w1 , A0 w0 )k21,0 ,

(2.139)

for T > (T0 + ε)/̺0 + 2ε. Moreover, by the inequality (2.138), we have kw1 k2 ≤ k(w0 , w1 )k21,0 ≤ cT (zε wt , wt )L2 (Σ0 ) ,

(2.140)

kwt k2L2 (Q) ≤ T k(w0 , w1 )k21,0 ≤ T cT (zε wt , wt )L2 (Σ0 ) .

(2.141)

We introduce a bilinear form by Ψ⋆ (w, w) ˆ = Ψ(w, w) ˆ + c0 cT (1 + T )(zε w, w) ˆ L2 (Σ0 )

(2.142)

where the positive constants c0 and cT are defined in (2.131) and (2.138), respectively. RT Noting that 0 zε (t)dt ≥ (T −ε), from the inequalities (2.131) and (2.141), we have



99

Lemma 2.10 Let all the assumptions in Theorem 2.22 hold. Then for T > (T0 + ε)/̺0 + 2ε given, there is constant cT > 0 such that cT k(w1 , A0 w0 )k21,0 ≥ Ψ⋆ (wt , wt )

≥ [̺0 (T − 2ε) − T0 − ε]k(w1 , A0 w0 )k21,0 ,

(2.143)

for all (w0 , w1 ) ∈ D(A0 ) × HΓ11 (Ω). Space of Boundary Control Let T > 0 be given. This time, we introduce a Banach space B m N (Σ0 ) as follows. We consider the space H 1 ([0, T ], L2(Γ0 )) which is the completion of C0∞ (Σ0 ) with an inner product Z (ϕ, φ)H 1 ([0,T ],L2 (Γ0 )) =

(ϕt φt + ϕφ)dΣ

Σ0

and set H01 ([0, T ], L2 (Γ0 )) = { ϕ | ϕ ∈ H 1 ([0, T ], L2 (Γ0 )), ϕ(0) = 0 }. For m = 0, we define B 0N (Σ0 ) = (H01 ([0, T ], L2 (Γ0 )))∗ that is the duality space of H01 ([0, T ], L2(Γ0 )) with respect to the space L2 (Σ0 ). For m = 1, let B 1N (Σ0 ) = L2 (Σ0 ). For m ≥ 2, let B m N (Σ0 ) consist of all the functions k m−k−3/2 ϕ ∈ ∩m−2 (Γ0 )), k=0 C ([0, T ], H

ϕ(k) ∈ H 1 (Σ0 ),

(2.144)

with the norm kϕk2B m (Σ0 ) = N

m−2 X k=0

kϕ(k) k2C([0,T ],H m−k−3/2 (Γ0 )) +

m−2 X k=0

kϕ(k) k2H 1 (Σ0 ) .

Let Ξ10N (Ω) = HΓ11 (Ω). For m ≥ 2, let Ξm 0N (Ω) consist of functions u in H (Ω) with the boundary conditions m

Ai0 u|Γ1 = (Ai0 u)νA |Γ0 = 0, Ai0 u|Γ1 = (Aj0 u)νA |Γ0 = 0,

0 ≤ i ≤ l − 1, if m = 2l; 0 ≤ i ≤ l, 0 ≤ j ≤ l − 1, if m = 2l + 1,

with the norms of H m (Ω). In particular, Ξ20N (Ω) = D(A0 ). Clearly, if Γ1 6= ∅ l and Γ1 ∩ Γ0 = ∅, Ξm 0N (Ω) has an equivalent norm: kA0 uk for m = 2l or kAl0 wkH 1 (Ω) for m = 2l + 1.


100


We now go back to the problem (2.125). Let w be a solution to the problem (2.126). We solve the problem   utt = div A(x)∇u in Q, u(T ) = ut (T ) = 0 on Ω, (2.145)  u|Σ1 = 0, uνA |Σ0 = zε [(w(3) − ∆Γg wt )h0 − λT wt ], where h0 = hH, νi and λT = c0 cT (1 + T ) + supx∈Γ0 |∆Γg h0 |/2. We define ΛN : m m−2 Ξm+1 (Ω) × H m−1 (Ω) by 0N (Ω) × Ξ0N (Ω) → H ΛN (w0 , w1 ) = (ut (0), −u(0)).

(2.146)

It is easy to check that (ΛN (w0 , w1 ), (w ˆ0 , w ˆ1 ))L2 (Ω)×L2 (Ω) = −

Z

wu ˆ νA dΣ

(2.147)

Σ0

for all (w0 , w1 ), (w ˆ0 , w ˆ1 ) ∈ HΓ11 (Ω) × L2 (Ω). We have the following observability estimates for the Neumann action. Theorem 2.23 Let Γ1 6= ∅ and Γ0 ∩ Γ1 = ∅. Let H be an escape vector field for the metric g on Ω such that the assumption (2.130) holds. Let m ≥ 1 be given. Then, for T > 0 suitable large, there are c1 > 0 and c2 > 0 satisfying c1 k(w0 , w1 )km+1,m ≤ kΛN (w0 , w1 )km−2,m−1 ≤ c2 k(w0 , w1 )km+1,m (2.148) m 1 for all (w0 , w1 ) ∈ Ξm+1 0N (Ω) × Ξ0N (Ω), where k · k−1,0 = k · k(HΓ

1

(Ω))∗ ×L2 (Ω) .

Proof. By induction on m. Step 1 Let m = 1. Let (w0 , w1 ) ∈ Ξ20N (Ω) × Ξ10N (Ω) be given. For any (w ˆ0 , w ˆ1 ) ∈ Ξ20N (Ω)× Ξ10N (Ω) given, suppose that w ˆ solves the problem (2.126) with the initial data (w ˆ0 , w ˆ1 ). Then w ˆt solves the problem (2.126) with the initial (wˆ1 , A0 w ˆ0 ) ∈ HΓ11 (Ω) × L2 (Ω). By the formulas (2.147) and (2.145), we obtain Z (ΛN (w0 , w1 ), (w ˆ1 , A0 w ˆ0 ))L2 (Ω)×L2 (Ω) = − w ˆt uνA dΣ Σ0

Z

Z

T

Z

= Ψ⋆ (wt , w ˆt ) + h0 w ˆ1 div A(x)∇w0 dΓ + zεt h0 w ˆt wtt dΓdt Γ0 T −ε Γ0 Z Z 1 − zε w ˆt ∇Γg h0 (wt )dΣ + sup |∆Γg h0 | zε wt w ˆt dΣ (2.149) 2 x∈Γ0 Σ0 Σ0 1/2

1/2

≤ Ψ⋆ (wt , wt )Ψ⋆ (w ˆt , w ˆt ) + ck(w1 , A0 w0 )k1,0 k(w ˆ1 , A0 w ˆ0 )k1,0 Z T +c kw ˆt kH 1/2 (Γ0 ) (k|∇Γg wt kH −1/2 (Γ0 ) + kwt kH 1/2 (Γ0 ) )dt 0

≤ cT k(w1 , A0 w0 )k1,0 k(w ˆ1 , A0 w ˆ0 )k1,0 ,



101

for all (w0 , w1 ) ∈ Ξ20N (Ω) × Ξ10N (Ω), which yields kΛN (w0 , w1 )k(HΓ1

1

(Ω))∗ ×L2 (Ω)

≤ cˆT k(w1 , A0 w0 )k1,0 ,

for (w0 , w1 ) ∈ Ξ20N (Ω) × Ξ10N (Ω). Since Γ0 ∩ Γ1 = ∅, Γ0 is a closed surface in Rn and, then, Z Z Z 2 2 zε wt ∇Γg h0 (wt )dΣ = zε ∇Γg h0 (wt )dΣ = − zε wt2 ∆Γg h0 dΣ, Σ0

Σ0

Σ0

that is, 2

Z

Σ0

zε wt ∇Γg h0 (wt )dΣ + sup |∆Γg h0 | x∈Γ0

Z

Σ0

zε wt2 dΣ ≥ 0.

(2.150)

Using the relations (2.149), (2.143) and (2.150), we have (ΛN (w0 , w1 ), (w1 , A0 w0 ))L2 (Ω)×L2 (Ω)

≥ [̺0 (T − 2ε) − T0 − T1 c(Γ0 ) − ε]k(w1 , A0 w0 )k21,0

(2.151)

for ε > 0 small, where T1 = [(1 + sup ε|ηε′ (t)|) sup |h0 | −ε≤t≤0

x∈Ω

and a constant c(Γ0 ) > 0 satisfies |(u, v)L2 (Γ0 ) | ≤ c(Γ0 )k(u, v)k1,0

for all (u, v) ∈ H 1 (Ω) × L2 (Ω).

Then there is a constant c1 > 0 such that kΛN (w0 , w1 )k(HΓ1

1

(Ω))∗ ×L2 (Ω)

≥ c1 k(w1 , A0 w0 )k1,0

when T > [T0 + T1 c(Γ0 ) + ε]/̺0 + 2ε. Then the inequality (2.148) is true for m = 1. Step 2 Suppose that the inequality (2.148) is true for some m ≥ 1. We shall prove that it is true for m + 1. Case I Let m = 2l for some l ≥ 1. Firstly, we assume that (w0 , w1 ) ∈ C0∞ (Ω) × C0∞ (Ω).

(2.152)

Suppose that w ˆ solves the problem (2.126) with an initial data (wˆ0 , w ˆ1 ) ∈ C0∞ (Ω) × C0∞ (Ω). Then w ˆ(2i) and w ˆ(2j+1) solve the problem (2.126) with the initial data (Ai0 w ˆ0 , Ai0 w ˆ1 ) and (Aj0 w ˆ1 , Aj+1 ˆ0 ), respectively, for 0 ≤ i ≤ l + 1 0 w and 0 ≤ j ≤ l.


102


Claim 1 The following identity is true. l −(A∇(Al−1 ˆ1 )) − (Al0 u(0), Al+1 w ˆ0 ) 0 ut (0)), ∇(A0 w Z 0 1 = Ψ⋆ (w(m+1) , w ˆ(m+1) ) + sup |∆Γg h0 | zε w(m+1) w ˆ(m+1) dΣ 2 x∈Γ0 Σ0 Z (m+1) (m+1) − zε w ˆ ∇Γg h0 (w )dΣ

+ +

Σ0 m−1 XZ T

Z

j=1 T −ε Γ0 m Z T Z X T −ε

j=1

Γ0

Cjm−1 zε(j) w(m+2−j) w ˆ(m+2) h0 dΓdt

Cjm zε(j) w ˆ(m+1) (∆Γg w(m+1−j) h0 + λT w(m+1−j) )dΓdt. (2.153)

Proof of (2.153) Using w ˆ(2m+1) in place of w ˆ in the formula (2.147), we obtain (ut (0), Am ˆ1 ) 0 w =− +λT

Z

−

z ε h0 w

ZΣ0

(u(0), Am+1 w ˆ0 ) 0

(3)

w ˆ

(2m+1)

dΣ +

=−

Z

Z

uνA w ˆ(2m+1) dΣ

Σ0

z ε h0 w ˆ(2m+1) ∆Γg wt dΣ

Σ0

zε wt w ˆ

(2m+1)

dΣ

Σ0

= Term 1 + Term 2 + λT Term 3.

(2.154)

We compute the terms in the right-hand side of (2.154) by integrating by parts over Σ0 = (0, T ) × Γ0 , respectively, as follows. Z

Term 1 =

zε w(m+2) w ˆ(m+2) h0 dΣ +

Σ0

+

m−1 XZ T

T −ε

j=1

=

− +

Z

Z

Γ0

j=2

l Z X j=1

w(j+3) (0)w ˆ(2m−j) (0)h0 dΓ

Γ0

Γ0

Aj0 w1 Am+1−j w ˆ0 h0 dΓ 0

Aj0 w0 Am+1−j w ˆ1 h0 dΓ 0

m−1 XZ T j=1

Z

Cjm−1 zε(j) w(m+2−j) w ˆ(m+2) h0 dΓdt

Σ0

Γ0

(−1)j

j=0

zε w(m+2) w ˆ(m+2) h0 dΣ +

l Z X

m−2 X

T −ε

Z

Γ0

Cjm−1 zε(j) w(m+2−j) w ˆ(m+2) h0 dΓdt;


(2.155)

Control of the Wave Equation with Variable Coefficients in Space Z Term 2 = z ε h0 w ˆ(m+1) ∆Γg w(m+1) dΣ

103

Σ0

+

m Z X

Z

T −ε

j=1

+

T

Γ0

m−1 X

Z

(−1)j+1

+

Z

Σ0

l Z X j=1

+

zε h∇Γg w(m+1) , ∇Γg w ˆ(m+1) ig h0 dΣ −

T

Z

T −ε

Term 3 =

Γ0

Z

j=0

zε w(m+1) w ˆ(m+1) dΣ +

j=1

m Z X j=1

l Z X

Γ0

Z

Σ0

zε w ˆ(m+1) ∇Γg h0 (w(m+1) )dΣ

h0 Am−j w ˆ0 ∆Γg Aj0 w1 dΓ 0

ˆ(m+1) ∆Γg w(m+1−j) h0 dΓdt; Cjm zε(j) w

Σ0

+

l−1 Z X

h0 Am−j w ˆ1 ∆Γg Aj0 w0 dΓ − 0

Γ0

m Z X j=1

h0 w ˆ(2m−j) (0)∆Γg w(j+1) (0)dΓ

Γ0

j=0

=−

Cjm zε(j) w ˆ(m+1) ∆Γg w(m+1−j) h0 dΓdt

Γ0

Aj0 w0 Am−j w ˆ1 dΓ − 0

T

T −ε

l−1 Z X j=0

Γ0

Z

Γ0

(2.156)

Cjm zε(j) w(m+1−j) w ˆ(m+1) dΓdt

Aj0 w1 Am−j w ˆ0 dΓ. 0

(2.157)

Moreover, by the problem (2.145), we have, on Γ0 for j ≥ 0, j+1 j+1 (Aj0 ut (0))νA = u(2j+1) (0) = (Aj+2 νA 0 w0 − ∆Γg A0 w0 )h0 − λT A0 w0 ; (2.158) j j (Aj0 u(0))νA = (Aj+1 0 w1 − ∆Γg A0 w1 )h0 − λT A0 w1 .

(2.159)

We substitute (2.155)-(2.159) into (2.154) to yield (ut (0), Am ˆ1 ) − (u(0), Am+1 w ˆ0 ) 0 w 0

= the right hand side of the identity (2.153) l−1 Z l−2 Z X X + (Aj0 u(0))νA A0m−j w ˆ0 dΓ − (Aj0 ut (0))νA Am−1−j w ˆ1 dΓ. (2.160) 0 j=0

Γ0

j=0

Γ0

m+1 On the other hand, for (w0 , w1 ), (w ˆ0 , w ˆ1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω), we obtain, via the Green formula, l (ut (0), Am ˆ1 ) = −(A∇(Al−1 ˆ1 )) 0 w 0 ut (0)), ∇(A0 w Z l−2 X (Aj0 ut (0))νA Am−1−j − w ˆ1 dΓ; 0 j=0


Γ0

(2.161)

104

Modeling and Control in Vibrational and Structural Dynamics −(u(0), Am+1 w ˆ0 ) = −(Al0 u(0), Al+1 ˆ0 ) 0 0 w Z l−1 X + (Aj0 u(0))νA Am−j w ˆ0 dΓ. 0 j=0

(2.162)

Γ0

After substituting (2.161) and (2.162) into the left-hand side of the identity (2.160) and eliminating the same terms from the both sides, we obtain the identity (2.153). Claim 2 There is Tm > 0 such that, for T > Tm given, there exists c1 > 0 such that (2.163) kΛN (w0 , w1 )k2m−1,m ≥ c1 k(w0 , w1 )k2m+2,m+1 m+1 for all (w0 , w1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω). Proof of (2.163) Replace w with w(m) in the inequality (2.143) and obtain 2 (m+1) cT k(Al0 w1 , Al+1 , w(m+1) ) 0 w0 )k1,0 ≥ Ψ⋆ (w 2 ≥ [ρ0 (T − 2ε) − c]k(Al0 w1 , Al+1 0 w0 )k1,0 ,

(2.164)

m+1 for all (w0 , w1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω), where c is a constant, independent of T > 0. We let w ˆ = w in the identity (2.153) and observe that Z Z 1 (m+1) (m+1) zε w ∇Γg h0 (w )dΣ = − zε [w(m+1) ]2 ∆Γg h0 dΣ; (2.165) 2 Σ0 Σ0

|

m−1 XZ T j=1

≤c

T −ε

m−1 X

Z

Γ0

Cjm−1 zε(j) w(m+2−j) w(m+2) h0 dΓdt|

sup

j=1 T −ε≤t≤T

kw(m+2−j) kH 1/2 (Γ0 ) kw(m+2) kH −1/2 (Γ0 )

2 ≤ ck(Al0 w1 , Al+1 0 w0 )k1,0 ;

|

m Z X j=1

≤c

T

T −ε

m X

Z

Γ0

Cjm zε(j) w(m+1) (∆Γg w(m+1−j) h0 + λT w(m+1−j) )dΓdt|

sup

j=1 T −ε≤t≤T

+λT c

m X

(2.166)

kw(m+1−j) kH 3/2 (Γ0 ) kw(m+1) kH 1/2 (Γ0 )

sup

j=1 T −ε≤t≤T

kw(m+1−j) kH 1/2 (Γ0 ) kw(m+1) kH −1/2 (Γ0 )

2 ≤ ck(Al0 w1 , Al+1 0 w0 )k1,0 .

(2.167)

Setting w ˆ = w in the identity (2.153) and via (2.143) and (2.165)-(2.167) l+1 2 l 2 l 2 kA∇(Al−1 0 ut (0))k + kA0 u(0)k ≥ [ρ0 (T − 2ε) − c]k(A0 w1 , A0 w0 )k1,0 ,



105

which yields Claim 2 by the ellipticity of the operator A0 . Claim 3 There is c2 > 0 such that kut (0)k2H m−1 (Ω) + ku(0)k2H m (Ω) ≤ c2 k(w0 , w1 )k2m+2,m+1

(2.168)

m+1 for all (w0 , w1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω). m+1 Proof of (2.168) We assume that (w0 , w1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω). We take w ˆ1 = 0 in the identity (2.153) and use the inequality (2.143). We obtain l+1 |(Al0 u(0), Al+1 ˆ0 )| ≤ ck(Al0 w1 , Al+1 ˆ0 k, 0 w 0 w0 )k1,0 kA0 w

that is,

kAl0 u(0)k ≤ ck(Al0 w1 , Al+1 0 w0 )k1,0 .

(2.169)

Using the ellipticity of the operator A0 and the equation in (2.145), we have ku(0)k2H m (Ω) ≤ ckA0 u(0)k2H m−2 (Ω) + ckuνA (0)k2H m−3/2 (Γ0 ) + cku(0)k2H m−1 (Ω) 2 ≤ ckA20 u(0)k2H m−4 (Ω) + cku(2) νA (0)kH m−7/2 (Γ0 )

+ckuνA (0)k2H m−3/2 (Γ0 ) + cku(0)k2H m−1 (Ω) .

Repeating this process gives ku(0)k2H m (Ω) ≤ ckAl0 u(0)k2 +c

l−1 X j=0

2 2 ku(2j) νA (0)kH m−2j−3/2 (Γ0 ) + cku(0)kH m−1 (Ω) .

(2.170)

We use the boundary control in (2.145) and the equation in the duality problem (2.126) to obtain l−1 X j=0

+c

2 ku(2j) νA (0)kH m−2j−3/2 (Γ0 ) ≤ c

l−1 X j=0

≤c

l−1 X j=0

kw(2j+1) (0)k2H m−2j−3/2 (Γ0 )

(kw(2j+3) (0)k2H m−2j−3/2 (Γ0 ) + k∆Γg w(2j+1) (0)k2H m−2j−3/2 (Γ0 ) )

l−1 X j=0

(kw(2j+3) (0)k2H m−2j−1 (Ω) + kA0 w(2j+1) (0)k2H m−2j−1 (Ω) )

≤ ckw(2l+1) (0)k21 ≤ ck(Al0 w1 , Al+1 0 w0 )k1,0 .

(2.171)

We combine (2.170) and (2.171) and use the induction assumption ku(0)k2H m−1 (Ω) ≤ ck(w0 , w1 )k2H m+1 (Ω)×H m (Ω) , to have ku(0)k2H m (Ω) ≤ ck(w0 , w1 )k2H m+2 (Ω)×H m+1 (Ω)


(2.172)

106


m+1 for all (w0 , w1 ) ∈ Ξm+2 0N (Ω) × Ξ0N (Ω). A similar argument establishes the estimate for ut (0). Case II Let m = 2l + 1 for some l ≥ 0. A similar argument shows that the inequality (2.148) holds with m replaced by m + 1 if it is true for m. Finally, the theorem follows by induction.

For the exact controllability of the problem (2.125), we have Theorem 2.24 Let Γ1 6= ∅ and Γ0 ∩ Γ1 = ∅. Let H be an escape vector field for the metric g on Ω such that hH, νi ≤ 0

for

x ∈ Γ1 .

Then, for m ≥ 0, there is Tm > 0 such that, for T > Tm , the problem (2.125) is exactly L2 (Ω)×(HΓ11 (Ω))∗ controllable for the case m=0 and exactly (H m (Ω)∩ HΓ11 (Ω)) × (H m−1 (Ω) ∩ HΓ11 (Ω)) controllable for the case m ≥ 1 on [0, T ] by Bm N (Σ0 ) controls. Proof. Step 1 Let the operator ΛN be given by (2.146). We shall m+1 m−1 prove that ΛN are isomorphisms from Ξm+2 (Ω) ∩ 0N (Ω) × Ξ0N (Ω) onto (H 1 m 1 HΓ1 (Ω)) × (H (Ω) ∩ HΓ1 (Ω)). By Theorem 2.23 it will suffice to prove that ΛN is surjective. Let m = 0. Let R(ΛN ) 6= (HΓ11 (Ω))∗ × L2 (Ω). Then there is a nonzero point (v0 , v1 ) ∈ (HΓ11 (Ω))∗ × L2 (Ω) such that (ut (0), v0 )(HΓ1

1

(Ω))∗

− (u(0), v1 ) = 0

(2.173)

for all (w0 , w1 ) ∈ D(A0 ) × HΓ11 (Ω). Recall that −A0 : HΓ11 (Ω) → (HΓ11 (Ω))∗ is the canonical isomorphism and the inner product of (HΓ11 (Ω))∗ is defined by (z1 , z2 )(HΓ1

1

(Ω))∗

−1 1 = (A−1 0 z1 , A0 z2 )HΓ

1

(Ω) .

Then, by (2.50), (ut (0), v0 )(HΓ1

1

(Ω))∗

−1 1 = (A−1 0 ut (0), A0 v0 )HΓ

1

(Ω)

= (ut (0), −A−1 0 v0 ).

Then it follows from the relation (2.173) that (ΛN (w0 , w1 ), (−A−1 0 v0 , v1 ))L2 (Ω)×L2 (Ω) = 0

(2.174)

for all (w0 , w1 ) ∈ D(A0 ) × HΓ11 (Ω). In particular, we take w0 = A−1 0 v1 and w1 = −A−1 v in (2.174) and obtain, via the inequality (2.151), that 0 0 k(−A−1 v , v )k = 0, that is, (v , v ) = 0. This contradiction shows that 0 1 1,0 0 1 0 ΛN is surjective. Repeating the above procedure, we may prove that ΛN are surjective when m ≥ 1. We omit it here. Step 2 We consider the regularity of the boundary control function in the



107

problem (2.145). Since wt is a lower order term in the the boundary control, zε (w(3) − ∆Γg wt )h0 is the principal part of the control. It will suffice to prove m+2 m+1 that zε (w(3) − ∆Γg wt )h0 ∈ B m N (Σ0 ) for (w0 , w1 ) ∈ Ξ0N (Ω) × Ξ0N (Ω) with m ≥ 0. Let m = 0. Let (w0 , w1 ) ∈ Ξ20N (Ω) × Ξ10N (Ω). For any ϕ ∈ 1 H ([0, T ], L2(Γ0 )), by Lemma 2.11 below, we have (zε (w(3) − ∆Γg wt )h0 , ϕ)L2 (Σ0 )

= −(zε (wtt − ∆Γg w)h0 , ϕt )L2 (Σ0 ) − (zεt (wtt − ∆Γg w)h0 , ϕ)L2 (Σ0 ) ≤ ckwtt − ∆Γg wkL2 (Σ0 ) kϕkH 1 ([0,T ],L2 (Γ0 )) for all ϕ ∈ H01 ([0, T ], L2(Γ0 )), which implies that zε (w(3) − ∆Γg wt )h0 ∈ B 0N (Σ0 ) Let m = 1 and (w0 , w1 ) ∈ Ξ30N (Ω) × Ξ20N (Ω). Then (wt (0), wtt (0)) ∈ 2 Ξ0N (Ω) × Ξ10N (Ω). By Lemma 2.11 below, w(3) − ∆Γg wt ∈ L2 (Σ0 ), that is, zε (w(3) − ∆Γg wt )h0 ∈ B 1N (Σ0 ). m+1 (m−1) Let m ≥ 2 and (w0 , w1 ) ∈ Ξm+2 (0), w(m) (0)) ∈ 0N (Ω)×Ξ0N (Ω). Then (w 2 × Ξ0N (Ω). Lemma 2.11 yields

Ξ30N (Ω)

w(m+1) − ∆Γg w(m−1) ∈ C([0, T ], H 1/2 (Γ0 )) ∩ H 1 (Σ0 ), that is, [zε (w(3) − ∆Γg wt )h0 ](m−2) ∈ C([0, T ], H 1/2 (Γ0 )) ∩ H 1 (Σ0 ). Then zε (w(3) − ∆Γg wt )h0 ∈ B m N (Σ0 ).

Remark 2.18 Unlike the control with the Dirichlet action, here ΛN : H m+1 (Ω) × H m (Ω) → H m−2 (Ω) × H m−1 (Ω). This is because the Neumann action loses a regularity of 1 order (actually, order 1/2 ). For this point, see [117]. Lemma 2.11 Let w solve the duality problem (2.126) with the initial data (w0 , w1 ) ∈ D(A0 ) × HΓ11 (Ω). Then wtt − ∆Γg w ∈ L2 (Σ0 ).

(2.175)

Furthermore, if (w0 , w1 ) ∈ Ξ30N (Ω) × Ξ20N (Ω), then wtt − ∆Γg w ∈ C([0, T ], H 1/2 (Γ0 )) ∩ H 1 (Σ0 ).

(2.176)

Proof. Let the Riemann metric g be given by (2.4). Then div A(x)∇w = ∆g w + F (w)


for x ∈ Ω,

(2.177)

108


where ∆g is the Laplacian of the metric g and F is a vector field on Ω give by F =−

1 A(x)∇ det A(x). 2 det A(x)

Using the boundary condition wνA = 0 on Σ0 and the relations (2.126) and (2.177), we obtain wtt − ∆Γg w =

1 D2 w(νA , νA ) + hF, ∇Γg wig |νA |2g

on Σ0

where D2 w(·, ·) is the Hessian of w in the metric g. Since k∇Γg wk2L2 (Σ0 ) ≤ cT k(w1 , A0 w0 )k21,0 , to get the relation (2.175) it will suffice to prove D2 w(νA , νA ) ∈ L2 (Σ0 ).

(2.178)

Let H be a vector field on Ω such that H = 0 for x ∈ Γ1 ;

H = νA

for x ∈ Γ0 .

We set v = H(w)

for

x ∈ Ω.

It is easy to check that this v solves the problem with the Dirichlet boundary conditions   vtt = div A∇v + [H, div A∇]w in Q, v = 0 on Σ, (2.179)  v(0) = H(w0 ), vt (0) = H(w1 ).

In addition, (w0 , w1 ) ∈ D(A0 )×HΓ11 (Ω) implies (v(0), vt (0)) ∈ H01 (Ω)×L2 (Ω). We integrate the identity (2.13) where u = v and f = [H, div A∇]w to obtain vνA = D2 w(νA , νA ) + h∇Γg w, DνA Hig ∈ L2 (Σ0 ),

which gives the relation (2.178). Next, we assume (w0 , w1 ) ∈ Ξ30N (Ω) × Ξ20N (Ω). Then (v(0), vt (0)) ∈ that 2 1 1 H (Ω) ∩ H0 (Ω) × H0 (Ω), where v solves the problem (2.179). A similar argument, as for the relation (2.117), shows that vνA ∈ C([0, T ], H 1/2 (Γ0 )) ∩ H 1 (Σ0 ), which implies that the relation (2.176) is true.

2.6

A Counterexample without Exact Controllability

A systematic attempt to understand the wave equation controllability and non-controllability properties can be found in the classical articles [174] and



109

[175] where the author proved that to have exact controllability the so-called geometric optics condition (GOC) needs to be satisfied. Later on it was proved by [8] that in fact GOC is also sufficient to assure exact controllability. Although it is hard to check as a sufficient condition, GOC is useful as a necessary one to construct counterexamples by the geometrical approach; see Theorem 2.25 later. Given the differential operator P u = utt − div A(x)∇u acting on functions, one sets fα = eiα(hx,ξi+tτ ) , and p(x, t, ξ, τ ) = lim f α P fα = hA(x)ξ, ξi − τ 2 α→∞

for (x, t, ξ, τ ) ∈ Rn × R × Rn × R. p(x, t, ξ, τ ) is called the principal symbol of the operator P. Definition 2.9 Given (x0 , t0 , ξ0 ) ∈ Rn × R × Rn \{0}, (x(s), t(s), ξ(s), τ (s)) is said to be a null bicharacteristic curve through (x0 , t0 , ξ0 ) if it satisfies the Hamiltonian system of ordinary differential equations x(s) ˙ =

1 ∇ξ p, 2

˙ = 1 ∂p , t(s) 2 ∂τ

1 ˙ ξ(s) = − ∇p, 2

τ˙ (s) = −

1 ∂p , 2 ∂t

(2.180)

with (x(0), t(0), ξ(0)) = (x0 , t0 , ξ0 ) and τ (0) chosen so that p(x0 , t0 , ξ0 , τ (0)) = 0, where ∇ξ denotes the gradient of the Euclidean metric with respect to the variable ξ ∈ Rn . The projection, (x(s), t(s)), of a bicharacteristic curve on (x, t)-space is called a ray. Note that there are two choices for τ (0) and p(x(s), t(s), ξ(s), τ (s)) = 0 for all s since the matrix A(x) is positive for x ∈ Rn . The following proposition shows that a ray is actually a geodesic in the metric g = A−1 (x). Proposition 2.4 Let x(t) be a geodesic on the Riemannian manifold (Rn , g) with x(0) = x0 parameterized by arc length. Then (x(t), ±t, ξ(t), ∓1) are bicharacteristic curves through (x0 , 0, ξ0 ), where ξ(t) = A−1 (x(t))x(t) ˙ for all t ∈ R. Proof. It will suffice to verify the relation (2.180). The relation ξ(t) = A−1 (x(t))x(t) ˙ yields x(t) ˙ = A(x(t))ξ(t) =


1 ∇ξ p 2

for t ∈ R.

(2.181)

110


Let A(x) = (aij (x)) and A−1 (x) = (gij (x)). Then n X l=1

glpxi alk = −

n X

1 ≤ i, p ≤ n.

(2.182)

Γjlp x˙ l (t)x˙ p (t) = 0 for 1 ≤ j ≤ n

(2.183)

glp alkxi

l=1

for

Since x(t) is a geodesic, we have x¨j (t) +

n X

lp=1

where Γjlp are the coefficients of the Levi-Civita connection, given by n

Γjlp =

1X ajk (gklxp + gkpxl − glpxk ), 2 k=1

from which we further have n X

gij Γjlp =

j=1

1 (gilxp + gipxl − glpxi ). 2

(2.184)

Using the relations (2.183), (2.184), (2.181) and (2.182), we obtain n n n X X X ˙ξi (t) = [ gil (x(t))x˙ l (t)]t = gilxp (x(t))x˙ l (t)x˙ p (t) + gij (x(t))¨ xj (t)

=

l=1 n X

1 2

=−

lp=1

1 2

j=1

lp=1

n 1 X glpxi x˙ l (t)x˙ p (t) = glpxi akl ajp ξk ξj 2

n X

lpkj=1

lpkj=1

glp ajp aklxi ξk ξj = −

1 ∂p 2 ∂xi

(2.185)

for 1 ≤ i ≤ j. Finally, it follows that p(x(t), ±t, ξ(t), ∓1) = =

X ijlp

n X

ij=1

aij (x(t))ξi (t)ξj (t) − 1

aij gil gjp x˙ l (t)x˙ p (t) − 1 =

X il

gli x˙ l (t)x˙ i (t) − 1 = hx(t), ˙ x(t)i ˙ g −1= 0

for all t ∈ R, since x(t) is parameterized by arc length. This completes our proof. From [8] and Proposition 2.4, the following result is immediate. Theorem 2.25 If there is a closed geodesic that is contained in Ω, then for any T > 0, the problem (2.45) has no exact controllability where Γ0 = Γ.



111

We give a counterexample to end this section. Example 2.7 Let (1 + |x|2 )2 0 0 (1 + |x|2 )2

(1 + |x|2 )−2 0 0 (1 + |x|2 )−2

A(x) =

,

x = (x1 , x2 ) ∈ R2 .

,

x = (x1 , x2 ) ∈ R2 .

The metric is g=

Let B = { x | |x|2 < 1 },

S = { x | |x| = 1 }.

We have the following conclusions. (i) If Ω ⊂ B, or Ω ⊂ R2 \B, then the problem (2.45) is exactly controllable; (ii) If S ⊂ Ω, then the problem (2.45) has no exact controllability even when controls act on the whole boundary Γ. Proof. Let k(x) be the Gaussian curvature function. By Lemma 2.3, we have k(x) = 4

for x ∈ R2 .

Let M be the sphere of radius 1/2 and centered at (0, 0, 1/2) in R3 , given by M = { (x1 , x2 , x3 ) | x21 + x22 + x23 = x3 }, with the induced Riemannian metric from R3 . Then (R2 ∪ {∞}, g) is isometric to M with an isometry Φ: M → (R2 ∪ {∞}, g), defined by Φ(x1 , x2 , x3 ) = (

x1 x2 , ) for (x1 , x2 , x3 ) ∈ M. 1 − x3 1 − x3

It is easy to check that sup dg (0, x) = π/2,

x∈R2

sup dg (0, x) = π/4,

x∈S

where dg is the distance function in the metric g. If Ω ⊂ { x | |x| < 1 }, or Ω ⊂ { x | |x| > 1 }, the exact controllability follows from Theorem 2.5. Let S ⊂ Ω. Since the big circle C = { (x1 x2 , 1/2) | x21 + x22 = 1/4} is a closed geodesic on the sphere M, S = Φ(C) is a geodesic in (R2 , g). Then (ii) follows from Theorem 2.25.


112


2.7

Stabilization

We will study stabilization from boundary and interior, respectively. Stabilization from Boundary Consider the stabilization problem  utt = div A(x)∇u in (0, ∞) × Ω,    u = 0 on (0, ∞) × Γ1 (2.186) uν + aut = 0 on (0, ∞) × Γ0 ,    A u(0) = u0 , ut (0) = u1 on Ω,

where a is a positive function on Γ0 . Let Z 2E(t) = (u2t + hA(x)∇u, ∇ui)dx Ω

be the energy. It follows that E(t) = E(0) −

Z tZ 0

Γ0

au2t dΓ

for t ≥ 0.

(2.187)

The well-posedness of the problem (2.186) can be obtained by the same arguments as in the case of constant coefficients but with a few modifications, for example, by the theory of semigroups ([39]-[41]). We omit the details. Our purposes are to seek geometrical assumptions on the metric g = A−1 (x) such that the energy of the problem (2.186) decays exponentially. Remark 2.19 In the case of constant coefficients the exponential decay of energy for the problem (2.186) from boundary was obtained under certain geometric conditions by [39], [40], [41], [99], [100], [122], [207], and many other authors. Trace Estimate on Boundary In order to remove some unnecessary restrictions on the control portion Γ0 , a trace estimate of the wave equation has to be introduced from [122]. Its proof needs a pseudodifferential analysis and we omit it. Let u solve the wave equation utt = div A(x)∇u + X(u) + pu + qut in Q, (2.188) u = 0 on Σ0 , where X is a vector field and p, q are functions, respectively, on Ω. Then Lemma 2.12 ([122]) Let T > 0 be given. let T /2 > ε > 0 be given small. Then there is cT ε > 0 such that Z T −ε Z Z |∇Γg u|2g dΓdt ≤ cT ε (u2νA + u2t )dΣ + cT ε kuk2H ε+1/2 (Q) (2.189) ε

Γ0


Σ0


113

for all solutions u to the problem (2.188) where ∇Γg u = ∇g u −

uνA Aν |Aν|2g

for

x ∈ Γ.

(2.190)

We have Theorem 2.26 Let H be an escape vector field on Ω for the metric g = A−1 (x) such that hH, νi ≤ 0 for x ∈ Γ1 . (2.191) Then there are C > 0 and σ > 0 such that E(t) ≤ Ce−σt

for

t≥0

(2.192)

for all solutions u to the problem (2.186). Proof. For convenience, we assume DH(X, X) ≥ |X|2g

for

X ∈ Rnx , x ∈ Ω.

(2.193)

Then div H ≥ n ≥ 2 for

Set Φ(t) = 2

Z

ut P u dx,

x ∈ Ω.

P u = H(u) + pu,

Ω

(2.194)

p = ( div H − 1)/2.

Then |Φ(t)| ≤ C(E(t) + ku(t)k2 ) for

t ≥ 0.

(2.195)

Noting that hH, Aνig = hH, νi for x ∈ Γ, we have a direct decomposition H=

hH, νi Aν + HΓg |Aν|2g

for x ∈ Γ

where hAν, HΓg ig = 0 for x ∈ Γ. It follows that P u|Σ1 = hH, νiuνA /|Aν|2g ,

(2.196)

P u|Σ0 = hH, νiuνA /|Aν|2g + HΓg (u) + pu.

(2.197)

Using the equation in (2.186) and the formula (2.6) and noting that div H − 2p = 1, we have 2utt P u = 2P u div A(x)∇u = 2 div (P uA(x)∇u) − 2hA(x)∇u, ∇P ui = div (2P uA(x)∇u − |∇g u|2g H) − 2DH(∇g u, ∇g u) +|∇g u|2g div H − 2p|∇g u|2g − A(x)∇p(u2 )

= div (2P uA(x)∇u − |∇g u|2g H − u2 A∇p)

−2DH(∇g u, ∇g u) + |∇g u|2g + u2 div A(x)∇p


114


which yields, via the formulas (2.196), (2.197), (2.193), and (2.194), Z Φt (t) = 2 [utt P u + ut P ut ] dx Z Ω = [2P uuνA + (u2t − |∇g u|2g )hH, νi]dΓ ZΓ − [2DH(∇g u, ∇g u) + (u2t − |∇g u|2g ) − u2 div A(x)∇p]dx ZΩ Z ≤ hH, νiu2νA /|Aν|2g dΓ + C (|∇Γg u|2g + u2νA + u2t )dΓ Γ1

Γ0

−E(t) +

Ckuk2L2 (Ω)

≤ −E(t) + C

Z

Γ0

(|∇Γg u|2g + u2νA + u2t )dΓ + Ckuk2L2 (Ω) .

(2.198)

Using (2.189), (2.198), and (2.195), we obtain Z T −ε E(τ )dτ ≤ |Φ(T − ε)| + |Φ(ε)| + Ckuk2L2 (Q) ε

+C

Z

T −ε

ε

Z

Γ0

(|∇Γg u|2g + u2νA + u2t )dΓdt

≤ C[E(T − ε) + E(ε) + ku(T − ε)k2 + ku(ε)k2 ] + Ckuk2L2 (Q)

+cT ε (kuνA k2L2 (Σ0 ) + kut k2L2 (Σ0 ) ) + cT ε kuk2H ε+1/2 (Ω)

(2.199)

where k · k denotes the norm of L2 (Ω). Using the estimate E(T ) ≤ E(t) for ε ≤ t ≤ T − ε in the left hand side of (2.199), using the relation (2.187) in the right hand side of (2.199), and using uνA = −aut for (t, x) ∈ Σ0 , we have (T − 2ε − 2C)E(T ) ≤ cT ε kut k2L2 (Σ0 ) + CT L(u), which yields, after we take T > 2ε + 2C, a constant cT > 0 such that E(T ) ≤ cT kut k2L2 (Σ0 ) + cT L(u)

(2.200)

where L(u) denote some lower order terms with respect to the norm of H 1 (Q). Then a compactness-uniqueness argument as in Lemma 2.5 gives a constant cT > 0 such that E(T ) ≤ cT kut k2L2 (Σ0 ) . (2.201) Let a ≥ a0 > 0 for x ∈ Γ0 . The relation (2.187) implies that Z a0 u2t dΣ ≤ E(0) − E(T ). Σ0

Using the above inequality in (2.201), we obtain E(T ) ≤


cT E(0) a 0 + cT


115

for all solutions u to the problem (2.186) which yields the formula (2.192) (Exercise 2.7). Stabilization from Interior Consider the stabilization problem   utt − div A(x)∇u + a(x)ut = 0 in (0, ∞) × Ω, u = 0 on (0, ∞) × Γ, (2.202)  u(0) = u0 , ut (0) = u1 on Ω

where a(x) is a nonnegative smooth function on Ω. Let E(t) be the energy of (2.202), given by (2.55). A simple computation gives Z E(T ) = E(0) − au2t dQ, Q = (0, T ) × Ω, T > 0. (2.203) Q

Remark 2.20 For the constant coefficient problem the earlier work concerning a local control was due to [103]; [151] used the piecewise multiplier method to study the local feedback; [196] studied the nonlinear feedback acting on a local region which was a neighborhood of a suitable subset of the boundary. We need ˆ ⊂ Ω be an open set and let νˆ be the normal of ∂ Ω ˆ in the Lemma 2.13 Let Ω ˆ Let H be a vector field on Ω. ˆ Euclidean metric of Rn pointing outside of Ω. Let ˆ = (0, T ) × Ω, ˆ ˆ = (0, T ) × ∂ Ω. ˆ Q Σ Then 2

Z

Z DH(∇g u, ∇g u)dQ = −2 ut [H(u) + pu]dx|T0 ˆ ˆ Q Ω Z Z ˆ ˆ + Φ(H, Ω)dQ + Φ(H, ∂ Ω)dΣ, ˆ Q

ˆ Σ

(2.204)

where u is a solution to the problem (2.202), 2p = div H, ˆ = u2 div A(x)∇p − 2aut [H(u) + pu], Φ(H, Ω) ˆ = 2H(u)uνÂ + (u2t − |∇g u|2g )hH, νî Φ(H, ∂ Ω) +2puuνÂ − u2 pνÂ ,

(2.205)

(2.206)

and νÂ = Aˆ ν. Proof. We take f = −a(x)ut in the identity (2.13) to have 2DH(∇g u, ∇g u) = div {2H(u)A(x)∇u − (|∇g u|2g − u2t )H} −2aut H(u) + 2p(|∇g u|2g − u2t ) − 2[ut H(u)]t .


(2.207)

116


Moreover, letting f = −aut in the identity (2.14) yields 2p(|∇g u|2g − u2t ) = div [2puA(x)∇u − u2 A(x)∇p] − 2paut u +u2 div A(x)∇p − 2p(uut )t .

(2.208)

ˆ we obtain the idenInserting (2.208) into (2.207), and integrating over Q, tity (2.204). Let ω = { x ∈ Ω | a(x) > 0 }.

(2.209)

We hope that the region ω can be small in some sense so this problem is also referred as to locally distributed control. We have Theorem 2.27 If ω is an escape region on Ω for the metric g = A−1 (x), then there are constants C > 0 and σ > 0 such that E(t) ≤ Ce−σt E(0)

(2.210)

for all solutions u to the problem (2.202). Proof. It suffices to prove that there are T > 0 and c > 0, independent of solutions to the problem (2.202), such that Z E(T ) ≤ c au2t dQ, Q = (0, T ) × Ω. (2.211) Q

Indeed, if (2.211) is true, then it follows from (2.203) that E(T ) ≤

c E(0), 1+c

which implies (2.210). Next, we prove (2.211). Let Ωi and H i be given in (2.34) and (7.206), respectively, for 1 ≤ i ≤ J. Let ε > 0 be such that ω ⊇ Ω ∩ Nε [∪Ji=1 Γi0 ∪ (Ω\ ∪Ji=1 Ωi )],

(2.212)


S ⊂ Rn ,

Γi0 = { x ∈ ∂Ωi | H i (x) · ν i (x) > 0 } ,

and ν i (x) is the unit normal of ∂Ωi in the Euclidean metric, pointing towards the exterior of Ωi for all i. For 0 < ε2 < ε1 < ε0 < ε, set Vj = Nεj [∪Ji=1 Γi0 ∪ (Ω\ ∪Ji=1 Ωi )] for


j = 0, 1, 2.

(2.213)


117

Then Ω\Vj ⊂ ∪Ji=1 Ωi

for j = 0, 1, 2,

V2 ⊂ V1 ⊂ V0 ⊂ V 0 ,

V 0 ∩ Ω ⊂ ω.

(2.214) (2.215)

Clearly, there is a0 > 0 such that a(x) ≥ a0

for x ∈ V 0 ∩ Ω.

Let φi satisfy φi ∈ C0∞ (Rn ), 0 ≤ φi ≤ 1, and 1 on Ωi \V1 , i φ = 0 on V2 ,

(2.216)

(2.217)

for 1 ≤ j ≤ J. Step 1 We estimate the integral Z

T

0

Z

Ω\V1

|∇g u|2g dxdt.

ˆ = Ωi in the identity We take H = φi H i , 2pi = div (φi H i ), and Ω (2.204), respectively, to have Z Z i i 2 D(φ H )(∇g u, ∇g u)dQ = −2 ut [φi H i (u) + pi u]dx|T0 Qi Ωi Z Z i i + Φ(φ H , Ωi )dQ + Φ(φi H i , ∂Ωi )dΣ (2.218) Qi

Σi

where Qi = (0, T ) × Ωi , Σi = (0, T ) × ∂Ωi , Φ(φi H i , Ωi ) = u2 div A(x)∇pi − 2aut [φi H i (u) + pi u],

(2.219)

and Φ(φi H i , ∂Ωi ) = 2φi H i (u)uνiA + φi (u2t − |∇g u|2g )hH i , νi i +2pi uuνi A − u2 piνiA

(2.220)

for all i. Since φi (x) = 1 for x ∈ Ωi \V1 , it follows from (2.217) that D(φi H i )(∇g u, ∇g u) ≥ ρ0 |∇g u|2g

for x ∈ Ωi \V1

(2.221)

for all i. Since ∂Ωi \Γ ⊂ Ω\ ∪Jj=1 Ωj ⊂ V2 and Γi0 ⊂ V2 , we have φi = 0 for x ∈ i Γ0 ∪ (∂Ωi \Γ) and Φ(φi H i , ∂Ωi ) = 0 for x ∈ Γi0 ∪ (∂Ωi \Γ)


118


for all i. In addition, if x ∈ ∂Ωi ∩Γ, then the boundary condition u = 0 implies Φ(φi H i , ∂Ωi ) = φi hH i , νi i

u2νiA |Aνi |2

for

x ∈ ∂Ωi ∩ Γ

for all i. Noting that ∂Ωi = [Γi0 ∪ (∂Ωi \Γ)] ∪ [(∂Ωi \Γi0 ) ∩ Γ], we have Φ(φi H i , ∂Ωi ) ≤ 0 for all x ∈ ∂Ωi

(2.222)

for all i. Using the estimates (2.221) and (2.222) in the identity (2.218) yields Z

T

Z

Z

T

Z

|∇g u|2g dxdt Z +C[E(T ) + E(0) + ku(0)k2 + ku(T )k2) + Cβ au2t dQ Qi Z +β |∇g u|2g dQ + Cβ kuk2L2(Q) 0

Ωi \V1

|∇g u|2g dxdt

≤C

0

Ωi ∩V1

Qi

where β > 0 can be chosen small, which gives, via the relation Ω ⊂ (∪Ji=1 Ωi ) ∪ V1 , that Z

T

0

≤C

Z

|∇g u|2g dxdt

Ω\V1 T

Z

0

Z

Ω∩V1

≤

XZ i

|∇g u|2g dxdt

+C[E(T ) + E(0) + L(u)].

T

0

+C

Z

Ωi

Z

Q

|∇g u|2g dxdt

au2t dQ (2.223)

Step 2 We estimate Z

T

0

Z

Ω∩V1

|∇g u|2g dxdt.

Let ξ ∈ C0∞ (Rn ) be such that 0 ≤ ξ ≤ 1 and ξ=

0 1

x ∈ Rn /V0 x ∈ V1 .

We take p = ξ and f = −aut in the identity (2.14), respectively, and integrate it over Q = (0, T ) × Ω to obtain 2

Z

Q

ξ|∇g u|2g dQ =

Z

Q

[2ξu2t − 2aut u + u2 div A(x)∇ξ]dQ − 2


Z

Ω

ξuut dx|T0


119

which yields, via (2.216), 2

Z

T

0

Z

|∇g u|2g dxdt ≤ 2

Ω∩V1 +Cβ kuk2L2(Q)

2 ≤ a0

Z

Q

T

Z

0

Z

V0 ∩Ω

u2t dxdt + β

Z

Q

u2t dQ

+ E(T ) + E(0) Z au2t dQ + β u2t dQ + Cβ kuk2L2(Q) + E(T ) + E(0) (2.224) Q

where β > 0 can be chosen small. Step 3 It follows from (2.223) and (2.224) that Z Z Z 2 2 |∇g u|g dQ ≤ C aut d + β u2t dQ Q

Q

Q

+Cβ [E(T ) + E(0) + L(u)]

(2.225)

where β > 0 can be chosen small. On the other hand, we take p = 1/2 in the identity (2.14) and then integrate it over Q = (0, T ) × Ω. We have Z Z Z aut udQ + uut dx|T0 (u2t − |∇g u|2g )dQ = Q Ω Q Z ≤ au2t dQ + C[E(T ) + E(0) + L(u)]. (2.226) Q

From (2.225) and (2.226), we obtain Z

T

E(t)dt = 0

≤C

Z

Q

Z

Q

|∇g u|2g dQ

au2t dQ + β

Z

T

1 + 2

Z

Q

(u2t − |∇g u|2g )dQ

E(t)dt + C[E(T ) + E(0) + L(u)].

(2.227)

0

From (2.203), we have E(t) ≥ E(T ) for all t ∈ [0, T ] and E(0) = E(T ) + 2 au t dQ. Using those facts and letting β = 1/2 in the inequality (2.227), we Q obtain Z (T /2 − 2C)E(T ) ≤ 2C au2t dQ + CL(u) R

Q

where constant C > 0 is given in the inequality (2.227). Then for T > 4C, there is a constant cT > 0 such that Z E(T ) ≤ cT au2t dQ + cT L(u) (2.228) Q

for all solutions u to the problem (2.202). By an argument as in Lemma 2.5, the lower order terms can be absorbed and the inequality (2.211) follows from (2.228).


120

2.8


Transmission Stabilization

We consider the stabilization of transmission of the wave equation with variable coefficients where the material is assumed to be composed of two portions with connecting conditions between them. Feedback controls will be acting on the boundary and the connecting area, respectively. Let Ω be a bounded domain in Rn with smooth boundary Γ. Let Ω be divided into two regions, Ω1 and Ω2 , by a smooth hypersurface Γ3 on Ω. Set Γi = ∂Ω ∩ ∂Ωi for i = 1 and 2. Denote by νi the normal of ∂Ωi pointing outside Ωi . Let ω be a domain near Γ3 such that Γ3 ⊂ ω ⊂ Ω. Let m > 0, a1 > 0, and a2 > 0 be constants. Definition 2.10 A function w on Ω is said to satisfy the connecting conditions for the wave equation if w1 = w2 ,

a1 w1ν1A + a2 w2ν2A = 0

for

x ∈ Γ3

(2.229)

where wi = w|Ωi for i = 1, 2. Let a(x) = a1

for x ∈ Ω1 ;

a(x) = a2

for x ∈ Ω2 .

We consider the following system  utt − a(x) div A(x)∇u + mχω ut = 0 in (0, ∞) × Ω\Γ3      u(x, 0) = u0 (x), ut (x, 0) = u1 (x) on Ω, u = 0 on (0, ∞) × Γ1 ,   uν + l(x)ut = 0 on (0, ∞) × Γ2    A u satisfies the connecting conditions on (0, ∞) × Γ3

(2.230)

where χω denotes the characteristic function of ω and l ∈ L∞ (Γ2 ) is such that l(x) ≥ l0 > 0 for all x ∈ Γ2 . Let V = {u ∈ HΓ11 (Ω)| uk ∈ H 2 (Ωk ), k = 1, 2, a1 u1ν1A + a2 u2ν2A = 0 on Γ3 }. The well-posedness for the problem (2.230) can be established by the semigroup theory: For all given initial data (u0 , u1 ) ∈ HΓ11 (Ω)×L2 (Ω), the problem (2.230) admits a unique global weak solution u ∈ C(R+ , HΓ11 (Ω)) ∩ C 1 (R+ , L2 (Ω)). Furthermore, if (u0 , u1 ) ∈ (V ∩ HΓ11 (Ω)) × HΓ11 (Ω)), then u ∈ C(R+ , V ) ∩ C 1 (R+ , HΓ11 (Ω)).



121

Let u be a solution to the problem (2.230). We define the energy of the problem (2.230) by Z 2E(t) = [u2t + a(x)hA(x)∇u, ∇ui]dx. (2.231) Ω

Lemma 2.14 For T > 0, E(T ) = E(0) − a2

Z

0

T

Z

Γ2

l(x)|u2t |2 dΓdt − m

T

Z

0

Z

ω

|ut |2 dx dt (2.232)

where u2 = u|Ω2 . Proof. Let ui = u|Ωi for i = 1, 2. Differentiating E(t) and using the connecting conditions, we have 2

X d E(t) = dt =

Z

k=1 Ωk 2 Z X

Γk ∪Γ3

k=1

=

Z

[uktt ukt + ak hA(x)∇uk , ∇ukt i]dx ak ukt ukνkA dΓ − m

Z

ω

|ut |2 dxdt

u1t (a1 u1ν1A + a2 u2ν2A )dΓdt + a2

Γ3

Z

u2t u2ν2A dΓdt

Γ2

Z

|ut |2 dxdt Z Z 2 = −a2 l(x)|u2t | dΓ − m |ut |2 dx −m

ω

Γ2

ω

which yields the identity (2.232).

The stabilization of the problem (2.230) is Theorem 2.28 Let Γ1 6= ∅ and Γ1 ∩ Γ2 = ∅. Let ω ⊃ Γ3 be such that ω ∩ Γi = ∅

for

i = 1, 2.

(2.233)

If there is an escape vector field H on Ω such that hH, ν1 i ≤ 0

for all

x ∈ Γ1 ,

(2.234)

then there are positive constants C and σ such that E(t) ≤ Ce−σt E(0)

for

t≥0

(2.235)

for all solutions of the problem (2.230) with (u0 , u1 ) ∈ HΓ11 (Ω) × L2 (Ω).


122


Proof. It will suffice to prove that there are T > 0 and CT > 0 such that Z E(T ) ≤ CT [

0

T

Z

m|ut |2 dxdt + C

ω

Z

Σ2

a2 l(x)|u2t |2 dΣ]

(2.236)

for all solutions u to the problem (2.230). Multiplying the wave equation in (2.232) by 2H(uk ) and integrating on Qk = (0, T ) × Ωk , we obtain Z [2ak H(uk )ukνkA + (u2kt − ak |∇g uk |2g )hH, νk idΣ ∂Ωk

=2 Z +

Z

Ωk

Qk

ukt H(uk )dx|T0

+ 2m

Z

0

T

Z

ukt H(uk )dxdt ωk

[2ak DH(∇g uk , ∇g uk ) + (u2kt − ak |∇g uk |2g ) div HdQ (2.237)

where Σk = (0, T ) × Γk and ωk = ω ∩ Ωk for k = 1, 2. Summing up the identities (2.237) for k = 1, 2 yields, via the connecting conditions in (2.230), Z [2a(x)H(u)uνA + (u2t − a(x)|∇g u|2g )hH, νi]dΣ + F (u) Σ

Z Z TZ =2 ut H(u)dx|T0 + 2m ut H(u)dxdt 0 ω Z Ω + [2a(x)DH(∇g u, ∇g u) + ̺0 (u2t − a(x)|∇g u|2g )]dQ Q Z + (u2t − a(x)|∇g u|2g )( div H − ̺0 )dQ

(2.238)

Q

where F (u) =

Z

Σ3

{2a1 [H(u1 ) − H(u2 )]u1ν1A

+(a2 |∇g u2 |2g − a1 |∇g u1 |2g )hH, ν1 i}dΣ

(2.239)

where Σ3 = (0, T ) × Γ3 and ̺0 > 0 is given in (2.15). By multiplying the wave equation in (2.232) by 2pu, a similar computation gives Z Z Z 2 p[u2t − a(x)|∇g u|2g ]dQ = 2 put udx|T0 − 2 a(x)u2 div A∇pdQ Q

Ω

Z

T

Z

Q

Z

+2m put udxdt + a(x)[2u2 pνA − 2puuνA ]dΣ Σ Z 0 ω 2 + 2(a1 − a2 )u1 hA∇p, ν1 idΣ Σ3


(2.240)


123

where p is a function on Ω. Letting p = ( div H − ̺0 )/2 in (2.240) and then inserting it into (2.238), we obtain, via the estimate (2.15), Z [2a(x)H(u)uνA + (u2t − a(x)|∇g u|2g )hH, νi]dΣ + F (u) Σ

≥ ̺0

Z

T

0

E(t)dt − 2

Z

Z

a(x)u2 div A(x)∇pdQ

Q

Z

pu]dx|T0

T

+2 ut [H(u) + + 2m 0 Z Ω Z 2 + a(x)[2u pνA − 2puuνA ]dΣ + Σ

Z

ut [H(u) + pu]dxdt

ω

Σ3

2(a1 − a2 )u21 hA∇p, ν1 idΣ. (2.241)

As in (2.57) and (2.61), the boundary condition u1 = 0 on Γ1 implies that H(u1 )u1ν1A = hH, ν1 i|∇g u1 |2g

for x ∈ Γ1 .

Noting that u1t = 0 and hH, ν1 i ≤ 0 on Γ1 , we have Z [2a1 H(u1 )u1ν1A + (u21t − a1 |∇g u1 |2g )hH, ν1 i]dΣ Σ1 Z = a1 |∇g u1 |2g hH, ν1 idΣ ≤ 0.

(2.242)

Σ1

Using (2.242) in (2.241), we obtain Z T Z ̺0 E(t)dt ≤ F (u) + C (|u2t |2 + |u2ν2A |2 + |∇Γg u2 |2g )dΣ + Cε L(u) 0

Σ2

+C[E(T ) + E(0)] + ε

Z

T

E(t)dt + Cε m

0

Z

0

T

Z

ω

u2t dxdt

(2.243)

where ∇Γg is defined in (2.190), ε > 0 can be chosen small, and L(u) = ku(0)k2 + ku(T )k2 + kuk2L2(Q) + kuk2H 1/2 (Q) . Moreover, the trace estimate in Lemma 2.12 and the feedback relation on Γ2 yield, for 1/2 > α > 0 small, Z T −α Z (|u2t |2 + |u2ν2A |2 + |∇Γg u2 |2g )dΓdt α Γ2 Z ≤ CT α a2 l(x)|u2t |2 dΣ + CT α kuk2H α+1/2 (Q) . (2.244) Σ2

Next, we estimate the transmission term F (u) in (2.239). Let us choose ˆ to be such that open subsets θ1 , θ2 and a vector field H ˆ =H H

on Γ3 ,

ˆ ⊂ θ1 , supp H


and Γ3 ⊂ θ1 ⊂ θ1 ⊂ θ2 ⊂ θ2 ⊂ ω.

124


ˆ = 0 for x ∈ Γ. Then it follows from (2.238) The conditions (2.233) implies H ˆ with H replaced by H that Z TZ F (u) ≤ C[E(T ) + E(0) + (m|ut |2 + a(x)|∇g u|2g )dxdt. (2.245) 0

θ1

C0∞ (Ω)

Let φ ∈ be such that 0 ≤ φ ≤ 1, φ = 1 for x ∈ θ1 , and φ = 0 for x∈ / θ2 . Letting p = φ in (2.240), we obtain Z TZ Z TZ a(x)|∇g u|2g dxdt ≤ C |ut |2 dxdt 0

θ1

0

+C[E(T ) + E(0) +

T

Z

Z

0

≤ C[E(T ) + E(0) +

Z

ω

T

Z

0

θ2

|ut |2 dxdt] + CL(u)

ω

|ut |2 dxdt] + CL(u) (by θ2 ⊂ ω). (2.246)

Using the estimates (2.246) and (2.245) in (2.243) and by taking 0 < ε < ̺0 /2, we obtain Z T Z E(t)dt ≤ C (|u2t |2 + |u2ν2A |2 + |∇Γg u2 |2g )dΣ 0

Σ2

+C[E(T ) + E(0)] + Cm

Z

T

0

Z

ω

u2t dxdt + CL(u).

(2.247)

Replacing the integral interval [0, T ] by [α, T − α] in (2.247) gives, via (2.244), Z T −α Z TZ Z 2 E(t)dt ≤ C m|ut | dxd + C a2 l(x)|u2t |2 dΣ α

0

ω

Σ2

+C[E(T − α) + E(α)] + CL(u) + kuk2H α+1/2 (Q) .

Furthermore, the relation (2.232) yields Z T −α E(t)dt ≥ (T − 2α)E(T ),

(2.248)

α

E(T − α) = E(T ) + a2 E(α) = E(T ) + a2

Z

Z

T

T −α T Z

α

Z

Γ2

2

Γ2

l(x)|u2t | dΓdt + m

l(x)|u2t |2 dΓdt + m

Z

Z T

α

T

T −α

Z

ω

Z

ω

|ut |2 dx dt,

|ut |2 dx dt.

Using those relations in (2.248), for T > 2α+2C, we obtain a constant CT > 0 such that Z TZ Z E(T ) ≤ CT [ m|ut |2 dxd + C a2 l(x)|u2t |2 dΣ] + CT L(u) (2.249) 0

ω

Σ2

where L(u) are lower order terms. Then (2.236) follows from (2.249) after the lower order terms are absorbed by a compactness-uniqueness argument as in Lemma 2.5.



125

Exercises 2.1 Prove the Green formula (2.92). 2.2 Let F and q be a vector field and a function on Ω, respectively. Consider an operator on L2 (Ω), given by Au = div A(x)∇u − F (u) + (q − div F )u,

D(A) = H 2 (Ω) ∩ H01 (Ω).

Then the adjoint operator of A is given by A∗ u = div A(x)∇u + F (u) + qu,

D(A∗ ) = H 2 (Ω) ∩ H01 (Ω).

2.3 Prove Lemma 2.7. 2.4 Let the operator Λ be defined by (2.101). Prove the estimate (2.116). 2.5 Let L = L2 (Ω) and H = H01 (Ω). Prove that (H01 (Ω))∗ = H −1 (Ω) and the canonical isomorphism ℵ: H01 (Ω) → H −1 (Ω) is given by ℵ = −∆ where ∆ is the classical Laplacian in Rn . 2.6 Let the inequality (2.200) hold true. Prove the inequality (2.201) by the compactness-uniqueness argument as in Lemma 2.5. 2.7 Let u be a solution to the problem (2.186) and let E(t) be its energy, defined by (2.55) where w is replaced with u. Let there be constants T > 0 and 0 < λ < 1 such that E(T ) ≤ λE(0) for all solutions u to the problem (2.186). Prove that there are constant C > 0 and σ > 0 such that E(t) ≤ Ce−σt E(0)

for

t≥0

for all solutions u.

2.9

Notes and References

Sections 2.2, 2.3, 2.4, and 2.6 are from [208]; Section 2.5 is from [216]; Section 2.7 is from [62], [63], [65], [66], and [80]; Section 2.8 is from [27].


126


The differential geometrical approach was first introduced by [208] for the controllability of the wave equation with variable coefficients. It was extended to cope with lower order terms by [130] and [201]. Then this method was used on various problems of the wave equation and the Schr¨ odinger with variable coefficients by [20], [21], [22], [23], [31], [55], [56], [79], [83], [84], [162], [163], [164], [178], and many others. Those results are not included in this chapter.


Chapter 3 Control of the Plate with Variable Coefficients in Space

3.1 3.2 3.3 3.4 3.5 3.6

Multiplier Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Escape Vector Fields for Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exact Controllability from Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Controllability for Transmission of Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stabilization from Boundary for the Plate with a Curved Middle Surface Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

127 131 135 144 150 158 158

We will establish Riemannian multiplier identities by the Bochner technique for the plate with variable coefficients to derive exact controllability/stabilization. Those multipliers are a geometric version of the classical ones for the plate with constant coefficients. Then we introduce a checkable assumption which guarantees exact controllability/stabilization, which is called escape vector fields for plate. It is different from the one for the wave equation. In particular, Section 3.2 is devoted to study the existence of escape vector fields for plate by curvature and many examples will be given. Finally, in Section 3.5, we will derive a plate model which is defined on a curved surface by the variational principle. In this model only the deformation along its normal is considered while the deformation in the tangential is neglected so it may serve as a curved plate. Stabilization by boundary feedbacks is presented for this model.

3.1

Multiplier Identities

Let Ω ⊂ Rn be a bounded, open set with smooth boundary Γ. We shall consider regular solutions u to the problem utt + A2 u = f

in (0, ∞) × Ω

(3.1)

and establish some multiplier identities where function f is given. In (3.1) we have defined Au = div A(x)∇u (3.2) n where A(x) = aij (x) is symmetrical, positive for each x ∈ R .

127


128


We introduce g = A−1 (x)

for

x ∈ Rn

(3.3)

as a Riemannian metric on Rn and consider the couple (Rn , g) as a Riemannian manifold. We denote by g = h·, ·ig the inner product and by D the covariant differential of the metric g, respectively. Denote by · = h·, ·i the Euclidean product of Rn . If applied to functions, we denote by ∇g and ∇ the gradients of the metric g and the Euclidean metric, respectively. Similarly, div g and div are the divergences of the metric g and the Euclidean metric, respectively. With two metrics on Rn in mind, one the Euclidean metric and the other the Riemannian metric g, we have to deal with various notations carefully. Let x ∈ Rn be given. We recall that E1 , · · · , En is a frame field normal at x on (Rn , g) if and only if it is a local basis for vector fields such that hEi , Ej ig = δij DEi Ej (x) = 0

in a neighborhood of x, for all 1 ≤ i, j ≤ n.

Tx2

Denote by all tensors of rank 2 on Rnx . Then Tx2 is an inner product space of dimension n2 with the inner product hF, Gig =

n X

F (ei , ej )G(ei , ej ) for

ij=1

F, G ∈ Tx2

(3.4)

where e1 , · · · , en is an orthonormal basis of (Rnx , g(x)). It is easy to check that h·, ·ig is independent of the choice of the orthnoromal base {ei } of (Rnx , g(x)). Let ∆g be the Laplacian in the metric g. Then in the natural coordinate systems x = (x1 , · · · , xn ) ∆g w =

n X p det A(x) ([det A(x)]−1/2 aij (x)wxi )xj = Aw − hDw, Dυg ig ij=1

for w ∈ C 2 (Rn ) (Exercise 1.1.12), that is, A = ∆g + Dυg ,

(3.5)

where 2υg = log det A(x)

for x ∈ Rn .

(3.6)

We denote by X (Rn ) all vector fields on Rn . Denote by ∆g : X (Rn ) → X (Rn ) the Hodge-Laplace operator in the metric g. We define an operator A: X (Rn ) → X (Rn ) by AH = −∆g H + DDυg H We have


for H ∈ X (Rn ).

(3.7)

Control of the Plate with Variable Coefficients in Space

129

Lemma 3.1 Let H be a vector field and w be a function on Rn . Then the following Weitzenbock formula holds true ∆g (H(w)) = −(∆g H)(w) + 2hDH, D2 wig +H(∆g w) + 2 Ric (H, Dw) for

x ∈ Rn

(3.8)

where Ric is the Ricci tensor of the metric g. Proof. Since ∆g Dw = (δd + dδ)dw = dδdw = −D(∆g w), the formula (3.8) follows from Theorem 1.27 by letting X = H and Y = Dw in the formula (1.120) where (M, g) = (Rn , g). The main multipliers we need are hu,

2hAu,

2H(u),

and 2H(Au)

where h is a function. Let v be the normal vector field of Γ = ∂Ω, pointing outside of Ω. For T > 0, let Q = (0, T ) × Ω, Σ = (0, T ) × Γ. Theorem 3.1 Let H be a vector field on Ω and let υg be given by (3.6). Suppose that u is a solution to the problem (3.1). Then Z {2Au(H(u))νA − 2H(u)(Au)νA + [u2t − (Au)2 ]hH, νi}dΣ Σ Z = 2 ut , H(u) |T0 + {[u2t − (Au)2 ] div H Q

2

+4AuhDH, D uig + 2AuP (Du) − 2f H(u)}dQ

(3.9)

where P (Du) = (AH)(u) + (2 Ric − D2 υg )(H, Du). Proof. First, the divergence formula yields 2utt H(u) = 2[ut H(u)]t − div (u2t H) + u2t div H.

(3.10)

Next, we compute term 2A2 uH(u). To this end, we need some formulas further. By (3.5), we obtain H(Au) = H(∆g u) + H(hDu, Dυg ig )

= H(∆g u) + D2 u(Dυg , H) + D2 υg (Du, H).


(3.11)

130


In addition, we have hDυg , D(H(u))ig = Dυg (hH, Duig )

= DH(Du, Dυg ) + D2 u(H, Dυg ).

(3.12)

It follows from the formulas (3.11), (3.12), (3.5), and (3.8) that A(H(u)) = ∆g (H(u)) + Dυg (hH, Duig ) = −(∆g H)(u) + 2hDH, D2 uig

+H(∆g u) + 2 Ric (H, Du) + DH(Du, Dυg ) + D2 u(H, Dυg ) = H(Au) + 2hDH, D2 uig + P (Du). (3.13)

From the formula (3.13), we have 2A2 uH(u) = 2 div H(u)A(x)∇(Au) − 2hD(Au), D(H(u))ig

= 2 div [H(u)A(x)∇(Au) − AuA(x)∇(H(u))] + 2AuA(H(u)) = div {2[H(u)A(x)∇(Au) − AuA(x)∇(H(u))] + (Au)2 H} +4AuhDH, D2 uig − (Au)2 div H + 2AuP (Du).

(3.14)

We multiply the equation (3.1) by 2H(u), integrate it over Q by parts, and obtain the formula (3.9) from the formulas (3.1), (3.10), and (3.14). Remark 3.1 In the right hand side of the identity (3.9), div H is the divergence of the vector field H in the Euclidean metric in Rn , not in the metric g. Theorem 3.2 Let H be a vector field on Ω and let u be a solution to the problem (3.1). Set p = div H. Then Z {2[put + H(ut )](ut )νA + 2H(Au)(Au)νA Σ

−(2ut Aut + |∇g ut |2g + |∇g (Au)|2g )hH, νi − u2t h∇g p, νi}dΣ Z = −2 ut , H(Au) |T0 + [2DH(∇g ut , ∇g ut ) + 2DH(∇g (Au), ∇g (Au)) Q

+(|∇g ut |2g − |∇g (Au)|2g )p − u2t Ap + 2f H(Au)]dQ.

(3.15)

Proof. We use the multiplier 2H(Au). Using the identity in Lemma 2.2, we have 2Aut H(ut ) = 2 div H(ut )A(x)∇ut − 2h∇g ut , ∇g (H(ut ))ig = div [2H(ut )A(x)∇ut − |∇g ut |2g H] −2DH(∇g ut , ∇g ut ) + |∇g ut |2g p.

(3.16)

In addition, 2put Aut = div [2put A∇ut − u2t A∇p] + u2t Ap − 2|∇g ut |2g p.


(3.17)


131

From the formulas (3.16) and (3.17), we have 2utt H(Au) = 2[ut H(Au)]t − 2 div (ut Aut )H + 2put Aut + 2H(ut )Aut = 2[ut H(Au)]t + div {2[put + H(ut )]A∇ut − u2t A∇p −(2ut Aut + |∇g ut |2g )H} + u2t Ap − |∇g ut |2g p −2DH(∇g ut , ∇g ut ).

(3.18)

On the other hand, by the identity in Lemma 2.2 again, 2A2 uH(Au) = div [2H(Au)A(x)∇(Au) − |∇g (Au)|2g H]

+|∇g (Au)|2g p − 2DH(∇g (Au), ∇g (Au)).

(3.19)

We multiply the equation (3.1) by 2H(Au), integrate it over Q, and obtain the identity (3.15) from (3.18) and (3.19). Theorem 3.3 Let p be a function on Ω. Suppose that u is a solution to the problem (3.1). Then Z Z (i) p[u2t − (Au)2 ]dQ = {Au[2Dp(u) + uAp] − f pu}dQ Q Q Z (3.20) +(ut , pu)|T0 + [pu(Au)νA − Au(pu)νA ]dΣ. Σ Z (ii) 2 p[|∇g ut |2g − |∇g (Au)|2g ]dQ Q Z = {2p[ut (ut )νA − Au(Au)νA ] + [(Au)2 − u2t ]pνA }dΣ Σ Z (3.21) −2(ut , pAu)|T0 + {Ap[u2t − (Au)2 ] + 2f pAu}dQ. Q

Proof. The divergence formula yields A(puAu) = A(pu)Au + 2h∇g (pu), ∇g (Au)ig + puA2 u = 2 div puA∇(Au) + A(pu)Au − puA2 u.

(3.22)

Using the formula (3.22) and the equation (3.1), we compute that Z T Z (ut , pu)|T0 = [(utt , pu) + (ut , put )]dt = (f pu − puA2 u + pu2t )dQ 0 Q Z Z = [Au(pu)νA − pu(Au)νA ]dΣ + [pu2t − A(pu)Au + f pu]dQ Σ

Q

which gives the identity (3.20). A similar computation gives the identity (3.21).


132

3.2


Escape Vector Fields for Plate

To have a real solution to the control problems for the systems with variable coefficients, it is necessary to obtain checkable geometric conditions. If the geometric conditions we have are uncheckable, we just turn one problem into another. Then the questions under consideration are still open. One of the main contributions of Riemann geometry is that it provides the control problems with the geometric conditions verifiable by the Riemann curvature theory. Now we introduce such a geometric condition for the control problems of the Euler-Bernoulli plate with variable coefficients. Let coefficient matrix A(x) be symmetric, positive for each x ∈ Rn . Consider the Riemannian manifold (Rn , g) where the metric is given by g = hA−1 (x)·, ·i

(3.23)

where h·, ·i is the Euclidean metric of Rn . Definition 3.1 Let g be the metric, given by (3.23), on Rn and let H be a vector field on Ω. Vector field H is said to be an escape vector field for plate if there is a function ϑ on Ω such that DH(X, X) = ϑ(x)|X|2g

for

X ∈ Rnx ,

x ∈ Ω,

(3.24)

and ϑ0 = min ϑ(x) > 0.

(3.25)

x∈Ω

Remark 3.2 Clearly, the conditions (3.24) and (3.25) imply that H is an escape vector field for the metric g on Ω. The converse is not true. In the case of constant coefficients where aij (x) = δij , for any x0 ∈ Rn fixed, the radial field H = x − x0 is an escape vector field for plate with ϑ = 1. The condition (3.24) has the following useful properties which can help us simplify computation: Theorem 3.4 Let H be a vector field such that the condition (3.24) holds. Then the condition (3.24) is equivalent to DH + D∗ H = 2ϑ(x)g

for

x ∈ Ω.

(3.26)

Moreover, if T ∈ T 2 (Ω) is a symmetric tensor field of rank 2 in the metric g, then hT, T (·, D·H)ig = ϑ(x)|T |2g for x ∈ Ω; (3.27) trg T (·, D· H) = ϑ(x)trg T

for

x∈Ω

(3.28)

where trg is the trace in the metric g and “·” denotes the position of variables.



133

Proof. Let x ∈ Ω be given. The condition (3.24) implies that DH(X + Y, X + Y ) − DH(X − Y, X − Y ) = 4ϑ(x)hX, Y ig which yields the identity (3.26). The converse is clearly true. Let x ∈ Ω be given. The symmetry of the tensor field T implies that there is an orthonormal basis e1 , · · · , en of the tangential space (Rnx , g) such that T (ei , ej ) = 0

at x for

i 6= j.

It follows from the formulas (3.29) and (3.24) that T (ei , Dei H) = T ei , DH(ei , ei )ei = ϑ(x)T (ei , ei )

(3.29)

(3.30)

for 1 ≤ i ≤ n. By the relations (3.30) and (3.29), we have hT, T (·, D· H)ig = =

n X

T (ei , ej )T (ei , Dej H)

ij=1 n X i=1

and trg T (·, D· H) =

T (ei , ei )T (ei , Dei H) = ϑ(x)|T |2g

n X

T (ei , Dei H) = ϑ(x)trg T.

i=1

Remark 3.3 If DH is symmetric, then DH = ϑg. Remark 3.4 In the case of dimension 2, a real plate, the condition (3.24) or equivalently (3.26) is always true; see Theorem 4.8 in Chapter 4. Moreover, Theorem 4.8 will show that an escape vector field for plate in the case n = 2 always exists locally. In general, it is not easy to find a vector field verifying the conditions (3.24) and (3.25). The following theorem shows that they have a close relationship with the curvature. For κ ∈ R, we define  √  sin κt for κ > 0; h(t) = t for for t ∈ R. √ κ = 0;  sinh −κt for κ < 0,

Theorem 3.5 Let g be the metric given by (3.23). Let there be a function ω such that the Riemann manifold (Rn , e2ω g) has constant curvature κ. Given x0 ∈ Rn . Denote by ρˆ(x) the distance function from x0 to x in the metric gˆ = e2ω g. Set ˆ ρˆ H = h(ˆ ρ)D (3.31)


134


ˆ is the Levi-Civita connection of the Riemann manifold (Rn , gˆ). Then where D DH(X, X) = [h′ (ˆ ρ) − H(ω)]|X|2g

for

X ∈ Rnx , x ∈ Rn .

(3.32)

Proof. Let x ∈ Rn be given and let X ∈ Rnx . We decompose X as a direct ˆ ρî D ˆ ˆ + X⊥ X = hX, D g ˆ ρ ˆ ρˆ, X ⊥ i = 0. Noting that D ˆ 2 ρˆ(D ˆ ρˆ, X ⊥ ) = 0 and (Rn , gˆ) has constant with hD g ˆ curvature κ, from Theorem 1.21, we have ˆ ˆ ρˆ ⊗ D ˆ ρˆ + h(ˆ ˆ 2 ρˆ](X, X) DH(X, X) = [h′ (ˆ ρ)D ρ)D = h′ (ˆ ρ)|X|2gˆ .

(3.33)

ˆ k the connection coefficients of the metrics g and gˆ, Denote by Γkij and Γ ij ∂ , ∂ i ˆ . Then gij = e−2ω gîj . We have respectively. Set gîj = h ∂x i ∂xj g n

Γkij =

1 X lk ∂gil ∂gjl ∂gij g ( + − ) 2 ∂xj ∂xi ∂xl l=1

ˆ kij − ωxi δkj − ωxj δki + gîj =Γ

n X

gˆkl ωxl .

(3.34)

l=1

Pn Pn ∂ ∂ . Let X = i=1 Xi ∂x be a vector field on Rn . It Denote H = i=1 hi ∂x i i follows from (3.34) that DX H =

n X

[X(hk ) +

k=1

n X

Xi hj Γkij ]

ij=1

∂ ∂xk

ˆ X H − X(ω)H − H(ω)X + hX, Hi Dω. ˆ =D g ˆ

(3.35)

Noting e2ω g = gˆ, from (3.35) and (3.33), we obtain ˆ DH(X, X) = e−2ω hDX H, Xigˆ = e−2ω [DH(X, X) − H(ω)|X|2gˆ ] = [h′ (ˆ ρ) − H(ω)]|X|2g .

Remark 3.5 The ideas in Theorem 3.5 are from Chapter 5 of the book [187]. Let us see several examples. It follows from Theorem 3.5 with ω = 0 that Example 3.1 Let the metric g have constant curvature κ. Given x0 ∈ Rn . Denote by√ρ the distance function from x0 to x in the metric g. If κ ≤ 0, then n H = sinh −κρDρ is an escape vector field for plate √ for any Ω ⊂ R . Let √ κ> 0. If there is x0 ∈ Rn such that Ω ⊂ B(x0 , π/(2 κ)), then H = sin κρDρ √ is an escape vector field for plate on Ω where √ B(x0 , π/(2 κ)) is the geodesic ball in (Rn , g) centered at x0 with radius π/(2 κ).



135

Example 3.2 Consider an operator Aw =

n X

(a(x)wxi )xi

i=1

where a(x) is a positive function on Rn . Then g = a−1 (x)I. Let gˆ = I be the Euclidean metric. Then 2ω = log a. Let x0 ∈ Rn be given. Then ρˆ = |x − x0 | ˆ ρˆ = x − x0 . It follows from Theorem 3.5 that and H = ρˆD DH(X, X) = [1 − If n

Ω ⊂ {x ∈ R |

n X i=1

H(a) ]|X|2g . 2a

(xi − x0i )axi (x) < 2a(x) },

then H = x − x0 is an escape vector field for plate on Ω. (a) Let a(x) = 1 + |x|2 and H = x. Since H(a) = 2|x|2 < 2a(x) for all x ∈ Rn , for any Ω ⊂ Rn , the vector field H is an escape vector field for plate. 2 (b) Let a(x) = e−|x| and H = x. Then H(a) = −2|x|2 a < 2a for all x ∈ Rn . Then H is an escape vector field for plate for any Ω ⊂ Rn . 2 (c) Let a(x) = e|x| and H = x. Then H(a) = 2|x|2 a. If Ω ⊂ { x | |x| < 1, x ∈ Rn }, then H is an escape vector field for plate on Ω.

3.3

Exact Controllability from Boundary

Let Γ0 be a portion of the boundary Γ which is relatively open. Set Γ1 = Γ \ Γ0 . Control in Fixed Boundary Condition Let T > 0 be given. We consider the control problem  utt + A2 u = 0 in Q,    u(0) = u0 , ut (0) = u1 on Ω, (3.36)  u = uνA = 0 on Σ1   u = ϕ, uνA = ψ on Σ0 ,

with controls ϕ and ψ, where Au = div A(x)∇u and Q = (0, T ) × Ω,

Σ1 = (0, T ) × Γ1 ,

The dual version for the above problem is in w   wtt + A2 w = 0 in Q, w(0) = w0 , wt (0) = w1  w = wνA = 0 on Σ

where Σ = (0, T ) × Γ.


Σ0 = (0, T ) × Γ0 .

on Ω,

(3.37)

136


Remark 3.6 In the case of constant coefficients where A = ∆, one control, uνA = ψ is enough, see [109]. We here add another control function u = ϕ in order to avoid the following uniqueness assumption: The problem 2 A w = λw in Ω, (3.38) w = wνA = Aw = 0 on Γ0 has the unique zero solution where λ is complex number, see [109]. Another uniqueness problem is considered by [119], that is, the conditions, w = wν = (∆w)ν = 0

on

Γ0 ,

imply that w = 0 if A = ∆. We do not know if the uniqueness problem (3.38) is true or not. We introduce the energy of the system (3.37) by Z 2E(t) = [wt2 + (Aw)2 ]dx.

(3.39)

Ω

Consider the exact controllability of the system (3.36) on the space L2 (Ω) × H −2 (Ω). Let w be the solution to the problem (3.37) for initial data (w0 , w1 ) ∈ H02 (Ω) × L2 (Ω). For T > 0, we solve the following time-reverse problem  2   utt + A u = 0 inQ,  u(T ) = ut (T ) = 0 on Ω, (3.40)  u = uνA = 0 on Σ1 ,   u = −(Aw)νA , uνA = Aw on Σ0 .

Then for the initial data (u(0), ut (0)), we have found control actions u = −(Aw)νA and uνA = Aw on Σ0 which steer the problem (3.36) to rest at a time T . Enlightened by [109], we define a map by Λ(w0 , w1 ) = ut (0), −u(0) (3.41) Denote by w ˆ and by uˆ the solutions to the problem (3.37) with initial data (w ˆ0 , w ˆ1 ) and to the problem (3.40) with boundary data uˆ = −(Aw) ˆ νA , u ˆνA = Aw ˆ on Σ0 , respectively. Using the equations (3.37) and (3.40), we obtain Λ(w0 , w1 ), (wˆ0 , w ˆ1 ) 2 = (ut (0), w ˆ0 ) − (u(0), w ˆ1 ) L (Ω)×L2 (Ω) Z = [(w ˆt , u) − (ut , w)]| ˆ T0 = (w ˆtt u − utt w)dQ ˆ Q Z = (wA ˆ 2 u − uA2 w)dQ ˆ Q Z = [w(Au) ˆ ˆνA − u(Aw) ˆ νA + Awu ˆ νA ]dΣ (3.42) νA − Auw ZΣ = [AwAwˆ + (Aw)νA (Aw) ˆ νA ]dΣ. (3.43) Σ0



137

Let H = H02 (Ω) × L2 (Ω),

L = L2 (Ω) × L2 (Ω),

R = { Λ(w0 , w1 ) | (Λ(w0 , w1 ), (w0 , w1 ))L2 (Ω)×L2 (Ω) < ∞ }.

Using Theorem 2.13, the formula (3.43), and Theorem 3.7 below, we obtain that H −2 (Ω) × L2 (Ω) = H ∗ ⊂ R. Since the system (3.36) is time-reverse, we have Theorem 3.6 Let T > 0 be given. Let H be an escape vector field on Ω for plate. Then the control problem (3.36) is exactly controllable on the space L2 (Ω) × H −2 (Ω) with controls (ϕ, ψ) ∈ L2 (Σ0 ) × L2 (Σ0 ), where Γ0 = { x | hH, νi > 0, x ∈ Γ },

Γ1 = Γ \ Γ0 .

We need Lemma 3.2 Let H be a vector field on Ω. Let w ∈ H 2 (Ω) satisfy w = wυA = ˆ where Γ ˆ is a relatively open subset of Γ. Then 0 on Γ ˆ (i) Dw = 0, in particular, H(w) = 0 on Γ. ˆ (ii) (H(w))νA = hH, νiAw on Γ. Proof. Let e1 = νA /|νA |g . Then e1 is the unit normal along Γ in the metric g where g is given by (3.23). For 2 ≤ i ≤ n, let ei be the tangential vector fields along Γ such that e1 , e2 , · · · , en forms an orthonormal basis in (Rn , g). Then the conditions, wνA |Γ = 0 and w|Γ = 0, imply that ei (w)|Γ = 0 for 1 ≤ i ≤ n. Thus Dw|Γ = 0 and, therefore, Dei Dw|Γ = 0

for 2 ≤ i ≤ n.

(3.44)

It follows from the relations (3.44) and (3.5) that (H(w))νA /|νA |g = e1 (H(w)) = hDw, De1 Hig + hDe1 Dw, Hi

= hH, e1 ig D2 w(e1 , e1 ) = hH, νi∆g w/|νA |g = hH, νiAw/|νA |g , that is, the formula (ii). The following lemma is from [170], also see [188]. We omit the proof here. ˆ be a relatively open subset of Γ. If w solves the problem Lemma 3.3 Let Γ 2 A w = F (w, Dw, D2 w, D3 w) on Ω, ˆ w = wνA = Aw = (Aw)νA = 0 on Γ, then w = 0 on Ω. Theorem 3.7 Let T > 0 be given. Let H be an escape vector field on Ω for plate. Then there is a constant cT > 0 such that Z [(Aw)2 + (Aw)2νA ]dΣ ≥ cT E(0) (3.45) Σ0

for all solutions w of the problem (3.37).


138


Proof. Let x ∈ Ω be given. Let e1 , · · · , en be an orthonormal basis of (Rnx , g). Using the symmetry of D2 w as a tensor of rank 2 on Rnx , and the formulas (3.26) and (3.5), we obtain 2

hDH, D wig = =

n X

DH(ei , ej )D2 w(ei , ej )

ij=1

X

[DH(ei , ej ) + DH(ej , ei )]D2 w(ei , ej )/2

ij

= ϑ(x)∆g w = ϑ(x)Aw − ϑ(x)Dυg (w).

(3.46)

Let w solve the problem (3.37). Then E(t) = E(0) for all t ≥ 0. Letting p = div H − 2ϑ0 , f = 0, and using the boundary conditions in the identity (3.20), we obtain Z [wt2 − (Aw)2 ]( div H − 2ϑ0 ) dQ Q Z = (wt , pw)|T0 + Aw[2Dp(w) + wAp]dQ Q

≤ ε[E(0) + E(T ) +

Z

T

E(τ )dτ ] + Cε L(w)

0

= (2 + T )εE(0) + Cε L(w)

(3.47)

for ε > 0 given small, where ϑ0 is given in (3.25) and L(w) = kw(0)k2 + k|Dw|g (0)k2 + kw(T )k2 + k|Dw|g (T )k2 +kwk2L2 (Q) + k|Dw|g k2L2 (Q)

for

t ≥ 0.

(3.48)

Next, using the estimate (3.47), the formula (3.46), the boundary conditions of the problem (3.37), and the inequality (3.25) in the identity (3.9) yields Z Z Z 2 2 c (Aw) dΣ ≥ (Aw) hH, νidΣ ≥ (Aw)2 hH, νidΣ Σ0 Σ0 Σ Z Z 2 2 2 ≥ 4ϑ0 (Aw) dQ + [wt − (Aw) ] div H dQ Q

Q

−(2 + T )εE(0) − Cε L(w) Z Z = 2ϑ0 [wt2 + (Aw)2 ]dQ + [wt2 − (Aw)2 ]( div H − 2ϑ0 ) dQ Q

Q

−(2 + T )εE(0) − Cε L(w)

≥ 2[2ϑ0 T − (2 + T )ε]E(0) − Cε L(w) which gives cT L(w) + c

Z

Σ0

[(Aw)2 + (Aw)2νA ]dΣ ≥ 2ϑ0 T E(0)



139

where we have set ε = ϑ0 T /(2 + T ). By Lemma 3.3, the lower terms in the right hand side of the above inequality can be absorbed as in the proof of Lemma 2.5. Then the inequality (3.45) holds true. Control in Simply Supported Boundary Condition We consider the exact controllability problem  utt + A2 u = 0 in (0, T ) × Ω,    u(0) = u0 , ut (0) = u1 on Ω, (3.49) u = Au = 0 on (0, T ) × Γ1 ,    u = ϕ, Au = ψ on (0, T ) × Γ0 ,

with controls ϕ and ψ. The dual version for the above problem is in w   wtt + A2 w = 0 in (0, T ) × Ω, w(0) = w0 , wt (0) = w1 on Ω, (3.50)  w = Aw = 0 on (0, T ) × Γ. This time, we define an energy of the system (3.50) by Z 2E1/4 (t) = (|∇g wt |2g + |∇g (Aw)|2g )dx.

(3.51)

Ω

Then E1/4 (t) = E1/4 (0) for all t ≥ 0. We introduce a selfadjoint operator A on L2 (Ω) by Aw = div A(x)∇w,

D(A) = H 2 (Ω) ∩ H01 (Ω).

(3.52)

The eigenvalues of −A are 0 < λ1 ≤ · · · ≤ λk ≤ · · · ,

with

lim λk = ∞.

k→∞

There is an orthonormal basis { ϕk } of L2 (Ω) such that −Aϕk = λk ϕk for all k ≥ 1. Moreover, By [87], Corollary 17.5.8, there are c1 , c2 > 0 such that |N (λ) − c1 λn/2 | ≤ c2 λ(n−1)/2 log λ for any λ > 0 where N (λ) denotes the number of the eigenvalues ≤ λ of −A. Letting λ = λk in the above inequality yields a constant c > 0 such that, for k large enough, λk ≥ ck 2/n , (3.53) from which we obtain ∞ X

k=1

λ−β k < ∞ for β > n/2.


(3.54)

140


Remark 3.7 A finer estimate than (3.53) was given by [187] for the eigenvalues of the Laplacian on Riemannian manifolds. For the constant coefficient problem with A(x) = I, a theorem due to H. Weyl (see [2]) states λk = (c + o(1))k 4/n

as

k → ∞.

First, we have Lemma 3.4 Let w be a complex solution to the problem (3.50). Then w can be discomposed as w = w+ + w− (3.55) which satisfies wt = i Aw+ − i Aw− .

(3.56)

Proof. It is easy to check that a solution w to the problem (3.50) can be expressed by ∞ ∞ X X w= α+,k e− i λk t ϕk + α−,k e i λk t ϕk (3.57) k=1

k=1

where 2α±,k = (w0 , ϕk ) ± i

1 (w1 , ϕk ) for k ≥ 1 λk

(3.58)

where (·, ·) is the complex product on L2 (Ω). The formula (3.56) is true if we let ∞ X w± = α±,k e∓ i λk t ϕk . (3.59) k=1

Lemma 3.5 Let H be an escape vector field for the metric g where g is given by (3.23). Let T > 0 be given. Then there are constants c1T > 0, c2T > 0 such that Z c1T E1/4 (0) ≤

Σ0

2 [wtν + (Aw)2νA ]dΣ ≤ c2T E1/4 (0) A

(3.60)

for all solutions w of the problem (3.50), Γ0 = { x | x ∈ Γ, hH, νi > 0 } and Σ0 = (0, T ) × Γ0 . Proof. Since H is escaping on Ω, there is ̺0 > 0 such that DH(X, X) ≥ ̺0 |X|2g

for

X ∈ Rnx , x ∈ Ω.

(3.61)

By the boundary conditions, w = Aw = 0 on Γ, we have |νA |2g H(wt ) = hH, νiwtνA , |νA |2g H(Aw) = hH, νi(Aw)νA ,


|νA |2g ∇g wt = wtνA νA ,

|νA |2g ∇g (Aw) = (Aw)νA νA .

(3.62) (3.63)


141

Letting p = div H/2 and f = 0, and using the boundary conditions of the problem (3.50) in the identity (3.21), we obtain Z | [|∇g wt |2g − |∇g (Aw)|2g ] div H dQ| ≤ CL(w) (3.64) Q

where L(w) = kwt (0)k2 + kwt (T )k2 + k|D2 w|g (0)k2 + k|D2 w|g (T )k2 +kwt k2L2 (Q) + k|D2 w|g k2L2 (Q) (3.65) are the lower order terms relatively to the energy E1/4 (0). Using the boundary conditions of the problem (3.50) and the relations (3.61)-(3.64) in the identity (3.15) gives Z Z hH, νi 2 2 2 2c [wtνA + (Aw)νA ]dΣ ≥ 2 [wtν + (Aw)2νA ] dΣ A |νA |2g Σ0 Σ ≥ 2(̺0 T − 2ε)E1/4 (0) − Cε L(w).

Letting ε = ̺0 T /4 and using Lemma 3.3, as in Lemma 2.5, to absorb the lower order terms, we obtain the left hand side of the inequality (3.60). ˆ on Ω such that Next, we choose a vector field H ˆ = A(x)ν H

for x ∈ Γ.

ˆ in the identity (3.15), and the relations (3.62) and Using the vector field H (3.63), we obtain the right hand side of the inequality (3.60). The following estimates are observability inequalities in the simply supported boundary conditions. Theorem 3.8 Let H be an escape vector field for the metric g. Let T > 0 be given. Then there is a constant cT > 0 such that Z [wν2A + (Aw)2νA ]dΣ ≥ cT E1/4 (0) (3.66) Σ0

for all solutions w of the problem (3.50) where Γ0 is given in Lemma 3.5. Proof. Let T > ε > 0 be given small. We consider complex solutions of the problem (3.50). Denote by Z± all the solutions to the problem (3.50) which are given by the formulas (3.58) and (3.59), respectively. For w± ∈ Z± , let Z 2E1/4 (w± ) = (|∇g (w± t (t))|2g + |∇g (Aw± (t))|2g )dx. Ω

Then E1/4 (w± )(t) = E1/4 (w± (0)). Since w± t = ± i Aw± for x ∈ Ω, we apply Lemma 3.5 with the interval [0, T ] replaced by [ε, T − ε] to obtain Z T −ε Z c1T ε E1/4 (w± )(0) ≤ |(Aw± )νA |2 dΣ ≤ c2T ε E1/4 (w± )(0). (3.67) ε


Γ0

142


In particular, letting w+ = e− i λk t ϕk in (3.67) yields Z c1 λk ≤ |ϕkνA |2 dΓ ≤ c2 λk for all k ≥ 1.

(3.68)

Γ0

Let φ ∈ C0∞ (R) be such that 0 ≤ φ ≤ 1 and φ(t) = 1

for t ∈ [ε, T − ε];

φ(t) = 0

for t ∈ / (0, T ).

Let φˆ denote the Fourier transform of φ. Since φ has a compact support, for β > n/2 given there is cβ > 0 such that ˆ |φ(ζ)| ≤ cβ ζ −β Let ηk =

∞ X l=1

for all ζ ≥ λ1 .

1 (λk + λl )β

(3.69)

for k ≥ 1.

By (3.53) it is easy to prove that ηk → 0

as k → ∞.

(3.70)

Let w and v be complex solutions to the problem (3.50). Set Z pΓ (w, v) = φ(t)(Aw)νA (Av)νA dΣ, Σ0

p1 (w, v) =

Z

Q

p2 (w, v) =

φ(t)h∇g (Aw), ∇g (Av)ig dQ, Z

Q

φ(t)h∇g wt , ∇g vt ig dQ.

By Lemma 3.4, w = w+ + w− where w± ∈ Z± . Then p2 (w± , w± ) = p1 (w± , w± ),

p2 (w+ , w− ) = −p1 (w+ , w− ),

(3.71)

and the conservation of the energy implies 2p1 (w± , w± ) = p1 (w± , w± )+p2 (w± , w± ) = 2E1/4 (w± )(0) p1 (w+ , w+ ) + p1 (w− , w− ) = E1/4 (w)(0)

Z

Z

T

φ(τ )dτ, (3.72)

0

T

φ(τ )dτ.

(3.73)

c1 pi (w± , w± ) ≤ pΓ (w± , w± ) ≤ c2 pi (w± , w± ) for i = 1, 2.

(3.74)

0

Then it follows from (3.67) and (3.72) that



143

Using the relations (3.59), (3.58), (3.68), (3.74), (3.69), and (3.70), we obtain Z T ∞ Z X φ(t)e− i (λk +λl )τ dτ | |pΓ (w+ , w− )| = | λk λl α+,k α−,l (ϕk )νA (ϕl )νA dΓ kl=1

=

√

2π|

∞ Z X

kl=1

Γ0

Γ0

0

ˆ k + λl )| λk λl α+,k α−,l (ϕk )νA (ϕl )νA dΓφ(λ

∞ ∞ 3/2 3/2 X X λk λl 1 |α ||α | ≤ C (λ3 |α+,k |2 + λ3l |α−,l |2 ) ≤C +,k −,l (λk + λl )β (λk + λl )β k

=C

kl=1 ∞ X k=1

kl=1

ηk λ3k |α+,k |2 + C

∞ X l=1

ηl λ3l |α−,l |2

≤ Cm (kw0 k2 + kw1 k2 ) + Cηm [p1 (w+ , w+ ) + p1 (w− , w− )] ≤ Cm (kw0 k2 + kw1 k2 ) + Cηm E1/4 (w)(0)(by (3.73))

(3.75)

for m being a positive integer. Finally, using the relations (3.74), (3.75), and (3.71), we obtain Z |(Aw)νA |2 dΣ ≥ pΓ (w, w) Σ0

≥ pΓ (w+ , w+ ) + pΓ (w− , w− ) − 2|pΓ (w+ , w− )|

≥ c1 [p1 (w+ , w+ ) + p1 (w− , w− )] − 2|pΓ (w+ , w− )| ≥ [c1 − 2Cηm ][p1 (w+ , w+ ) + p1 (w− , w− )] − Cm (kw0 k2 + kw1 k2 ) ≥ σm E1/4 (0) − 2Cm (kw0 k2 + kw1 k2 ).

where σm = c1 − 2Cηm , which gives Z Cm (kw0 k2 + kw1 k2 ) + (|wνA |2 + |(Aw)νA |2 )dΣ ≥ σm E1/4 (0). Σ0

We fixed m large enough such that σm > 0 to have the inequality (3.66) by absorbing the lower order terms as in Lemma 2.5. Let A−1 denote the inverse of the selfadjoint operator in (3.52). Then there are constants c1 > 0, c2 > 0 such that c1 (kw0 k2H 1 (Ω) + kw1 k2H −1 (Ω) ) 0

≤ k|∇g (A(A−1 w0 ))|g k2 + k|∇g (A−1 w1 )|g k2 ≤ c2 (kw0 k2H 1 (Ω) + kw1 k2H −1 (Ω) ) 0

for all (w0 , w1 ) ∈ H01 (Ω) × H −1 (Ω). We define k(u, v)k2B(Ω) = k|∇g (Au)|g k2 + k|∇g v|g k2 ,


(3.76)

144

Modeling and Control in Vibrational and Structural Dynamics B(Ω) = { (u, v) | u, v ∈ L2 (Ω), k(u, v)kB(Ω) < ∞ }.

The inequality (3.76) shows that A−1 : H01 (Ω) × H −1 (Ω) → B(Ω) is an isomorphism. Let (v0 , v1 ) ∈ H01 (Ω) × H −1 (Ω) be given. Let w be the solution of the problem (3.50) with initial data (w(0), wt (0)) = (A−1 v0 , A−1 v1 ) ∈ B(Ω). For T > 0, we solve the following time-reverse problem  utt + A2 u = 0 in Q,    u(T ) = ut (T ) = 0 on Ω, (3.77)  u = Au = 0 on Σ1 ,   u = (Aw)νA , Au = wνA on Σ0 . Then for the initial data (u(0), ut (0)), we have found control actions u = (Aw)νA and Au = wνA on the portion Γ0 of the boundary which steer the problem (3.36) to rest at a time T . We define a map by Λ(v0 , v1 ) = − A−1 ut (0), A−1 u(0) . Then

Λ(v0 , v1 ), (v0 , v1 )

L2 (Ω)×L2 (Ω)

=

Z

Σ0

[wν2A + (Aw)2νA ]dΣ.

(3.78)

By Proposition 3.8 and the inequality (3.76), there is cT > 0 such that Λ(v0 , v1 ), (v0 , v1 ) ≥ cT k(v0 , v1 )k2H 1 (Ω)×H −1 (Ω) L2 (Ω)×L2 (Ω)

0

for (v0 , v1 ) ∈ H01 (Ω) × H −1 (Ω). It follows from Theorem 2.13 that

Theorem 3.9 Let H be an escape vector field for the metric g. For each T > 0, the problem (3.49) is exactly controllable on the space H −1 (Ω) × AH −1 (Ω) with controls (ϕ, ψ) ∈ [L2 (Σ0 )]2 . Remark 3.8 For the Euler-Bernoulli plate with the simply supported boundary conditions, the system behaves like a wave equation so that the existence of an escape vector field in the metric g can guarantee the exact controllability. However, for the fixed boundary conditions, an escape vector field for plate is needed.

3.4

Controllability for Transmission of Plate

We consider exact boundary controllability of transmission of the plate with variable coefficients where the middle surface is assumed to be composed of two materials with connecting conditions between them.



145

Let Ω be a bounded domain in Rn with smooth boundary Γ. Let Ω be divided into two regions, Ω1 and Ω2 , by a smooth hypersurface Γ3 on Ω. Set Γi = ∂Ω∩∂Ωi for i = 1 and 2. Denote by νi the normal of ∂Ωi in the Euclidean metric of Rn pointing outside Ωi . Then ν1 + ν2 = 0 We define a(x) = where ai > 0 are given constants.

for all x ∈ Γ3 . a1 a2

on Ω1 , on Ω2 ,

Definition 3.2 A function w on Ω is said to satisfy connecting conditions on Γ3 for plate if w1 = w2 , w1ν1A + w2ν2A = 0 on Γ3 , (3.79) a1 Aw1 = a2 Aw2 , a1 (Aw1 )ν1A + a2 (Aw2 )ν2A = 0 on Γ3 , where wi = w|Ωi for i = 1, 2. Let T > 0 be given and let Γ0 ⊂ Γ be a subset. We consider controllability of the problem  utt + a(x)A2 u = 0 in (0, T ) × (Ω/Γ3 ),      u(0) = u0 (x), ut (0) = u1 (x) on Ω, u = uνA = 0 on Σ/Σ0 , (3.80)   u satisfies the connecting conditions for plate on Σ ,  3   u = ϕ, uνA = ψ on Σ0 , where ϕ, ψ are boundary controls, Σ3 = (0, T ) × Γ3 , and Σ0 = (0, T ) × Γ0 . The duality problem of the system (3.80) is  wtt + a(x)A2 w = 0 in (0, T ) × (Ω/Γ3 ),    w(0) = w0 (x), wt (0) = w1 (x) on Ω, (3.81) w = wνA = 0 on Σ,    w satisfies the connecting conditions for plate on Σ3 . First, we consider regularity of the system (3.81). Let

H 4 (Ω, Γ3 ) = { w ∈ H02 (Ω), wi ∈ H 4 (Ωi ), i = 1, 2, w satisfies the connecting conditions for plate on Γ3 }. Define an operator Aw = div A(x)∇w,

D(A) = H 2 (Ω) ∩ H02 (Ω).

Let A be given by A=

0 I −aA2 0


,

D(A) = H 4 (Ω, Γ3 ) × H01 (Ω).

(3.82)

146


It is readily shown that A is skew-adjoint on the space H02 (Ω) × L2 (Ω), i.e. A∗ = −A, where the inner product on H02 (Ω) is given by Z (w, v)H02 (Ω) = a(x)Aw(x)Av(x) dx. (3.83) Ω

So A generates a C0 -group eAt on H02 (Ω) × L2 (Ω). Then Proposition 3.1 For (w0 , w1 ) ∈ H02 (Ω) × L2 (Ω), the system (3.81) admits a unique weak solution w ∈ C([0, T ]; H02 (Ω)) ∩ C 1 ([0, T ]; L2(Ω)). Furthermore, if (w0 , w1 ) ∈ H 4 (Ω, Γ3 ) × H02 (Ω), then w ∈ C([0, T ]; H 4 (Ω, Γ3 )) ∩ C 1 ([0, T ]; H02 (Ω)). We need Lemma 3.6 Let H be a vector field on Ω. Let a function w on Ω satisfy the connecting conditions for plate on Γ3 . Then (H(w1 ))ν1A + (H(w2 ))ν2A = hH, ν1 i(Aw1 − Aw2 )

for

x ∈ Γ3

(3.84)

where wi = w|Ωi for i = 1. 2. Proof. Let x ∈ Γ3 be given. Let E1 , · · · , En be a frame field normal at x where E1 (x) = ν1A (x)/|Aν|2g . Then the connecting conditions imply that Dw1 = Dw2 for x ∈ Γ3 . Thus n X

2

D w1 (Ei , Ei ) =

i=2

n X

D2 w2 (Ei , Ei ) at x.

(3.85)

i=2

In addition, by (3.5), we have Aw1 =

n X 1 2 D w (ν , ν ) + D2 w1 (Ei , Ei ) + Dw1 (υg ) 1 1A 1A |Aν|2g i=2

(3.86)

at x. Using (3.85) and (3.86), we obtain (H(w1 ))ν1A + (H(w2 ))ν2A = D2 w1 (H, ν1A ) + D2 w2 (H, ν2A ) + Dw1 (Dν1A H) + Dw2 (Dν2A H) = hH, ν1 i(Aw1 − Aw2 ) at x.



147

Lemma 3.7 Let H be a vector field on Ω. Let w solve the problem (3.81). Then Z Z a(x)(Aw)2 hH, νidΣ + [a1 (Aw1 )2 − a2 (Aw2 )2 ]hH, ν1 idΣ Σ Σ3 Z = 2(wt , H(w))|T0 + [wt2 − a(x)(Aw)2 ] div HdQ Q Z a(x)Aw[2hDH, D2 wig + P (Dw)]dQ (3.87) +2 Q

where P (Dw) = AH(w) + Ric (H, Dw) − D2 υg (H, Dw),

υg = log det A(x)/2, and AH is given by (3.7).

Proof. We multiply the equation in (3.81) by 2H(wi ) and integrate it over Qi = (0, T )×Ωi for i = 1, 2. A similar computation as in the proof of Theorem 3.1 yields Z Z Z 2 − ai (Awi )2 ] div H ψ(wi )dΣ = 2 wit H(wi )dx|T0 + {[wit Σi ∪Σ3

Ωi

Qi

+4ai Awi hDH, D2 wi ig + 2ai Awi P (Dwi )}dQ

(3.88)

where Σi = (0, T ) × Γi and 2 ψ(wi ) = 2ai Awi (H(wi ))νiA − 2ai H(wi )(Awi )νiA + [wit − ai (Awi )2 ]hH, νi i,

for i = 1, 2. By Lemma 3.2, the boundary conditions wi = wiνiA = 0 on Γi imply that ψ(wi ) = ai (Awi )2 hH, νi i for

x ∈ Γi , i = 1, 2.

(3.89)

Since ν1 = −ν2 for x ∈ Γ3 , the connecting conditions for plate yield H(w1 ) = H(w2 ) for x ∈ Γ3 and ψ(w1 ) + ψ(w2 ) = 2a1 Aw1 [(H(w1 ))ν1A + (H(w2 ))ν2A ] + 2H(w1 )(a1 Aw1 − a2 Aw2 )

−2H(w1 )[a1 (Aw1 )ν1A + a2 (Aw2 )ν2A ] + [a2 (Aw2 )2 − a1 (Aw1 )2 ]hH, ν1 i = [a1 (Aw1 )2 − a2 (Aw2 )2 ]hH, ν1 i. (3.90)

Summing up the identities (3.88) for i = 1, 2, and using the formulas (3.89) and (3.90), we obtain the identity (3.87). Lemma 3.8 Let p be a function. Let w solve the problem (3.81). Then Z p[wt2 − a(x)(Aw)2 ]dQ = (wt , pw)|T0 Q Z + a(x)Aw[2Dp(w) + wAp]dQ. (3.91) Q


148


Proof. A similar computation as in the proof of Theorem 3.3 gives Z Z 2 p[wit − ai (Awi )2 ]dQ = ai Awi [2Dp(wi ) + wi Ap]dQ Qi Qi Z Z + pwit wi dx|T0 + ai [pwi (Awi )νiA − Awi (pwi )νiA ]dΣ (3.92) Ωi

Σi ∪Σ3

for i = 1, 2. In addition, the boundary conditions and the connecting conditions imply X ai [pwi (Awi )νiA − Awi (pwi )νiA ] = 0 for x ∈ Γ ∪ Γ3 . i

Then (3.91) follows by summing up (3.92) with i = 1, 2. We define the energy of the problem (3.81) by Z 2E(t) = [wt2 + a(x)(Aw)2 ]dx. Ω

Then it follows from (3.81) that 2

X d E(t) = dt i=1 + =

2 X

i=1 2 X i=1

Z

Ωi

ai ai

Z

(wit witt + ai Awi Awit )dx =

Ωi

Z

2 X

ai

i=1

Z

wit (Awi )νiA dΓ Γi ∪Γ3

(hA∇(Awi ), ∇wit i + Awi Awit )dx

Γi ∪Γ3

[(Awi )witνiA − wit (Awi )νiA ]dΓ = 0,

that is, E(t) = E(0) for

t ≥ 0.

The observability estimate is given by Theorem 3.10 Let H be an escape vector field for plate such that (a1 − a2 )hH, ν1 i ≥ 0

for

x ∈ Γ3 .

Let T > 0 be given. Then there is cT > 0 satisfying Z cT {[(Aw)2 + [(Aw)νA ]2 }dΣ ≥ E(0) Σ0

for all solutions w to the problem (3.81) where Σ0 = (0, T ) × Γ0 and Γ0 = { x | x ∈ Γ, hH, νi > 0 }.


(3.93)

(3.94)


149

Proof. Let ϑ0 > 0 be given by (3.25). By the formula (3.46), we have Aw[2hDH, D2 wig + P (Dw)] ≥ ϑ1 (Aw)2 − C|Dw|2g

(3.95)

where 0 < ϑ1 < 2ϑ0 is a constant and ϑ0 is given in (3.25). Moreover, the connecting conditions and the assumption (3.93) on Σ3 yield Z [a1 (Aw1 )2 − a2 (Aw2 )2 ]hH, ν1 idΣ Σ3

=

Z

Σ3

a2 (Aw2 )2 (a2 − a1 )hH, ν1 idΣ ≤ 0. a1

(3.96)

Using the estimates (3.95) and (3.96) in the identity (3.87), we obtain Z a(x)(Aw)2 hH, νidΣ Σ0 Z Z ≥ a(x)(Aw)2 hH, νidΣ + [a1 (Aw1 )2 − a2 (Aw2 )2 ]hH, ν1 idΣ Σ Σ3 Z Z ≥ 2ϑ1 a(x)(Aw)2 dQ + [wt2 − a(x)(Aw)2 ] div HdQ Q Q Z +2(wt , H(w))|T0 − C |Dw|2g dQ Q Z Z ≥ ϑ1 [wt2 + a(x)(Aw)2 ]dQ + p[wt2 − a(x)(Aw)2 ]dQ Q

2

Q 2

−Cε (k|Dw(T )|g k + k|Dw(0)|g k + k|Dw|g k2L2 (Q) ) −ε[E(T ) + E(0)]

(3.97)

where p = div H − ϑ1 and ε > 0 can be chosen small. Using the identity (3.91) in the inequality (3.97) gives Z c {(Aw)2 + [(Aw)νA ]2 }dΣ ≥ (T ϑ1 − 4ε)E(0) Σ0

−Cε [k|Dw(T )|g k2 + k|Dw(0)|g k2 + k|Dw|g k2L2 (Q) ] −Cε [kw(0)k2 + kw(T )k2 + kwk2L2 (Q) ],

which, by letting 0 < ε < T ϑ1 /4, yields Z {(Aw)2 + [(Aw)νA ]2 }dΣ + L(w) ≥ cT E(0) Σ0

for all solutions w to the problem (3.81), where L(w) = kw(0)k2 + kw(T )k2 + kwk2L2 (Q)

+k|Dw(T )|g k2 + k|Dw(0)|g k2 + k|Dw|g k2L2 (Q) .


(3.98)

150


Finally, the lower order terms can be absorbed as in Lemma 2.5 and the inequality (3.94) follows. It follows from Theorem 3.10 that Theorem 3.11 Let H be an escape vector field for plate such that the condition (3.93) holds true. Then for any T > 0, the transmission problem (3.80) is exactly L2 (Ω) × H −2 (Ω) controllable by L2 (Σ0 ) × L2 (Σ0 ) controls.

3.5

Stabilization from Boundary for the Plate with a Curved Middle Surface

Model Throughout this section M is a surface of R3 with the induced metric g from the Euclidean metric of R3 . We shall consider a curvy plate whose middle surface, Ω, is part of the surface M and where the extension effects along the tangential direction are neglected. Let boundary Γ of the middle surface Ω consist of two disjoint parts Γ0 ∪ Γ1 = Γ. We assume that the material undergoing obeys Hooke’s law. The potential energy is defined by Z P (u) = [(1 − µ)|D2 u|2 + µ(tr D2 u)2 ]dx Ω

where u is the displacement of the plate along the normal, 0 < µ < 1/2 is the Poisson coefficient, D denotes the covariant differential of the metric g, and tr is the trace in the metric g. If there is no external force, then the equations of motion for u are obtained by setting to zero the first variation of the Lagrangian: Z T [kut k2 − P (u)]dτ = 0 0

2

where k · k is the norm of L (Ω). The above variation is taken with respect to kinematically admissible displacements. We obtain, as a result of calculation by the above variation, the following system:  utt + ∆2 u + (1 − µ) div (κ∇u) = 0 in (0, ∞) × Ω,      u = uν = 0 on (0, ∞) × Γ1 , ∆u + (1 − µ)B1 u = 0 on (0, ∞) × Γ0 , (3.99)   (∆u) + (1 − µ)B u = 0 on (0, ∞) × Γ ,  ν 2 0   u(0) = u0 , ut (0) = u1 on Ω,

where ∆, div , ∇, κ, and ν are the Laplacian, the divergence, the gradient, the



151

Gauss curvature function of surface M , and the outside normal along Γ in the induced metric g, respectively. In the system (3.99), B1 and B2 are boundary operators, defined by B1 u = −D2 u(τ, τ ),

and B2 u =

∂ (D2 u(τ, ν)) + κuν , ∂τ

respectively, where τ is a unit tangential vector field on Γ such that ν, τ forms an orthonormal basis (Mx , g(x)) for each x ∈ Γ. Remark 3.9 The term div (κ∇u) in the system (3.99) comes from the curvedness of middle surface Ω. If M = R2 , then κ = 0 and the system is the same as in [109]. Lemma 3.9 Let M be a surface of R3 . Let Ric be the Ricci tensor of M . Then Ric = κg. (3.100) Proof. Let x ∈ M be given. Let e1 , e2 be an orthonormal basis of Mx . For any X, Y ∈ X (M ), by Theorem 1.8, we have R(ei , X, ei , Y ) = hX, ej ihY, ej iR(ei , ej , ei , ej ) = κhX, ej ihY, ej i for i 6= j, 1 ≤ i, j ≤ 2, where R is the curvature tensor. Then Ric (X, Y ) = R(e1 , X, e1 , Y ) + R(e2 , X, e2 , Y ) = κhX, Y i. We introduce a(u, v) = (1 − µ)hD2 u, D2 vig + µ tr D2 u tr D2 v,

(3.101)

Au = ∆2 u + (1 − µ) div (κ∇u),

(3.102)

Γ1 (u) = ∆u + (1 − µ)B1 u,

Γ2 (u) = (∆u)ν + (1 − µ)B2 u.

(3.103)

The following formula plays an important role in our problems, which presents the relationship between the interior and the boundary. Lemma 3.10 Let the boundary Γ of the middle surface Ω be a closed curve. Then Z Z a(u, v)dg = (Au, v) + [Γ1 (u)vν − Γ2 (u)v]dΓ (3.104) Ω

for u, v ∈ H 4 (Ω).


Γ

152


Proof. Using the formulas (1.155), (3.100), and (1.137), we have Z hD2 u, D2 vig dg = (D2 u, D2 v)L2 (Ω,T 2 ) Ω Z = (∆Du − κDu, Dv)L2 (Ω,Λ) + hDν Du, DvidΓ Γ Z = (δ∆du − δ(κ∇u), v) + [vh∆Du − κDu, νi + hDν Du, Dvi]dΓ Γ

= (∆2 u + div (κ∇u), v) Z + {D2 u(ν, ν)vν + D2 u(ν, τ )vτ − v[(∆u)ν + κuν ]}dΓ.

(3.105)

Γ

Since Γ is a closed curve, Z Z D2 u(ν, τ )vτ dΓ = − [D2 u(ν, τ )]τ vdΓ. Γ

(3.106)

Γ

Moreover, the Green formula yields Z tr D2 u tr D2 vdg = (∆u, ∆v) = (∆2 u, v) Ω Z + [∆uvν − v(∆u)ν ]dΓ.

(3.107)

Γ

The formula (3.111) follows from the relations (3.105)-(3.107).

We define the energy of the system by Z 2E(t) = [u2t + a(u, u)]dx. Ω

Let u satisfy the equation in (3.99) and the fixed boundary conditions, u = uν = 0 on Γ1 . Using Lemma 3.10, we obtain Z E ′ (t) = [Γ1 (u)utν − Γ2 (u)ut ]dΓ. Γ

If we act on Σ0 = (0, ∞) × Γ0 by

Γ1 (u) = −utν ,

Γ2 (u) = ut ,

then E(t) = E(s) −

Z tZ s

Γ0

(u2t + u2tν )dΓdτ

for

t ≥ 0, s ≥ 0.

We have a closed loop system  utt + ∆2 u + (1 − µ) div (κ∇u) = 0 in (0, ∞) × Ω,      u = uν = 0 on (0, ∞) × Γ1 , ∆u + (1 − µ)B1 u = −utν on (0, ∞) × Γ0 ,   (∆u)ν + (1 − µ)B2 u = ut on (0, ∞) × Γ0 ,    u(0) = u0 , ut (0) = u1 on Ω. © 2011 by Taylor & Francis Group, LLC

(3.108)

(3.109)


153

We shall establish the exponential decay of the energy for the above system. Well Posedness of the System (3.109) We introduce bilinear forms Z Z B(u, v) = a(u, v) dx, α(u, v) = (uv + uν vν )dΓ. Ω

Γ0

It follows from Lemma 3.10 that an appropriate variational formulation of the system (3.109) is as follows: Find w ∈ C([0, ∞), HΓ21 (Ω)) ∩ C 1 ([0, ∞), HΓ11 (Ω)) such that [(wt , v) + α(w, v)]t + B(w, v) = 0 for all v ∈ HΓ21 (Ω), (3.110) w(0) = w0 ∈ HΓ21 (Ω), wt (0) = w1 ∈ L2 (Ω). We assume that Γ1 6= ∅,

Γ0 ∩ Γ1 = ∅.

(3.111)

Then the condition (3.111) implies that B(·, ·) is an equivalent norm of HΓ21 (Ω). Let HΓ−2 (Ω) denote the dual space of HΓ21 (Ω) with respect to the L2 1 topology. Let ℵ be the canonical isomorphism from HΓ21 (Ω) endowed with the inner product B(·, ·) to HΓ−2 (Ω). Thus 1 B(u, v) = (ℵu, v) for u, v ∈ HΓ21 (Ω). Moreover, since 0 ≤ α(u, u) ≤ ckuk2H 2 HΓ21 (Ω)

→

HΓ−2 (Ω) 1

Γ1 (Ω)

(3.112)

, there is a positive operator ℵ0 :

such that

α(u, v) = (ℵ0 u, v) for u, v ∈ HΓ21 (Ω).

(3.113)

Then the variational problem (3.110) can be written as (wt + ℵ0 w)t + ℵw = 0

in HΓ−2 (Ω). 1

(3.114)

Let us formally rewrite the system (3.114) as CYt + IN Y = 0 where C=

ℵ0 0I

,

IN =

0 −ℵ ℵ ℵ0

(3.115)

,

Y =

w wt

.

We shall solve the problem (3.115) in HΓ21 (Ω) × L2 (Ω). Let D(IN ) = { (u, v) | u, v ∈ HΓ21 (Ω), ℵu + ℵ0 v ∈ L2 (Ω) }. (Ω) × Then IN : D(IN ) → HΓ−2 (Ω) × L2 (Ω). Since C: HΓ21 (Ω) × L2 (Ω) → HΓ−2 1 1 2 L (Ω) is the canonical isomorphism, the system (3.115) is equivalent to Yt + C−1 IN Y = 0.


(3.116)

154


Theorem 3.12 Let the condition (3.111) be true. Then −C−1 IN is the infinitestimal generator of a C0 -semigroup of contraction on HΓ21 (Ω) × L2 (Ω). Proof. Step 1 D(IN ) is dense in HΓ21 (Ω) × L2 (Ω). By Lemma 3.10, we have (ℵu + ℵ0 v, w) = B(u, w) + α(v, w) = (Au, w) Z + {[Γ1 (u) + vν ]wν + [v − Γ2 (u)]w} dΓ Γ0

which implies D(IN ) ⊃ D0 where D0 = { (u, v) ∈ [H 4 (Ω) ∩ HΓ21 (Ω)] × HΓ21 (Ω), Γ1 (u) + vν = 0, v − Γ2 (u) = 0 on Γ0 }. D(IN ) is dense in HΓ21 (Ω) × L2 (Ω) since D0 is. Step 2 −C−1 IN is dissipative and Range (λI +C−1 IN ) = HΓ21 (Ω)×L2 (Ω) for λ > 0. We need prove that for arbitrary (ϕ, ψ) ∈ HΓ21 (Ω)×L2 (Ω), there is (u, v) ∈ D(IN ), such that (I + C−1 IN )(u, v) = (ϕ, ψ). (Ω) is onto. It suffices to show that the map I + ℵ + ℵ0 : HΓ21 (Ω) → HΓ−2 1 Indeed, it is enough if there exists v ∈ HΓ21 satisfying (I + ℵ + ℵ0 )v = ψ − ℵϕ. Let the above relation be true. Choosing v in this way and setting u = v + ϕ, we will have (u, v) ∈ D(IN ), and (I + C−1 IN )(u, v) = (u − v, ℵu + ℵ0 v + v) = (ϕ, ψ). (Ω). Now we prove the surjection of the map I + ℵ + ℵ0 : HΓ21 (Ω) → HΓ−2 1 −2 2 For arbitrary f ∈ HΓ1 (Ω), consider the map F : HΓ1 (Ω) → R defined by Z 1 1 1 2 2 F (u) = kukH 2 (Ω) + kukL2(Ω) + (u2 + u2ν ) dΓ − (f, u)(H −2 (Ω), H 2 (Ω)) . Γ1 Γ1 Γ1 2 2 2 Γ It can be verified that F is well-defined, continuously differentiable and that Z F ′ (u)v = (u, v)HΓ2 (Ω) + (u, v)L2 (Ω) + uv + uν vν dΓ − (f, v)(H −2 (Ω), H 2 (Ω)) 1

Γ1

Γ

= B(u, v) + (u, v)L2 (Ω) + α(u, v) − (f, v)(H −2 (Ω), H 2 Γ1

= ((I + ℵ + ℵ0 )u − f, v)(H −2 (Ω), H 2 Γ1


Γ1 (Ω))

.

Γ1 (Ω))

Γ1


155

The convexity of F is easily verified. Finally F is coercive: F (u) → ∞ if kukHΓ2 (Ω) → ∞. This follows from the inequality 1

1 F (u) ≥ ( kukHΓ2 (Ω) − kf kH −2 (Ω) )kukHΓ2 (Ω) . Γ1 1 1 2 It follows that the infimum of F is attained at some point u ∈ HΓ21 (Ω). Then F ′ (u) = 0 i.e. (I + ℵ + ℵ0 )u = f . Boundary Trace In order to save space, we introduce an important result of Theorem 2.3 in [123] without proofs, which allows us to get rid of some geometrical conditions on the shape of the boundary. It reads as follows. Lemma 3.11 Let u solve the problem utt + ∆2 u + (1 − µ) div (κ∇u) = 0 u = uν = 0 on (0, ∞) × Γ1 .

in

(0, ∞) × Ω,

Let 0 < α < T /2 be given. Then there is cT α > 0 such that Z |D2 u|2 dΓdτ ≤ cT α (kΓ1 (u)k2L2 (Σ0 ) + kΓ2 (u)k2L2 ((0,T );H −1 (Γ0 )) (α,T −α)×Γ0

+kut k2L2 (Σ0 ) + kutν k2L2 ((0,T );H −1 (Γ0 )) + L(u))

where L(u) are some lower order terms relative to the energy

(3.117) RT 0

E(t)dt.

Stabilization of the Closed-Loop System (3.109) Let u ∈ H m (Ω) be given. We say that L(u) is a lower order term related to the norm of u in H m (Ω) if for any ε > 0 there is Cε > 0 such that |L(u)| ≤ εkuk2H m (Ω) + Cε kuk2. We need Lemma 3.12 Let H be a vector field satisfying the condition (3.24). Then Z Z Z 2 a(u, H(u))dx = a(u, u)hH, νidΓ + 2 ϑa(u, u)dx + L(u) (3.118) Ω

Γ

Ω

where L(u) denotes some lower order terms relative to u in the norm of H 2 (Ω). Proof. Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x. We compute at x, via the identity (1.22), D2 (H(u))(Ei , Ej ) = Ej Ei (Du(H)) = Ej (D2 u(H, Ei ) + Du(DEi H)) = D3 u(Ei , H, Ej ) + D2 u(DEj H, Ei ) + D2 u(DEi H, Ej ) + Du(DEj DEi H) = H(D2 u(Ei , Ej )) + D2 u(DEj H, Ei ) + D2 u(DEi H, Ej ) + lo (u),


(3.119)

156


where lo (u) denotes some terms of order ≤ 1. Since D2 u(·, ·) is symmetrical, it follows from the formulas (3.119), (3.27) and (3.28) that 2hD2 u, D2 (H(u))i = H(|D2 u|2 ) + 4ϑ|D2 u|2 + lo (u)

(3.120)

2 tr D2 u tr D2 (H(u)) = H((tr D2 u)2 ) + 4ϑ(tr D2 u)2 + lo (u).

(3.121)

and

Noting that the relation H(a(u, u)) = div a(u, u)H − 2ϑa(u, u), we obtain the identity (3.118) from the formulas (3.120) and (3.121). Lemma 3.13 Let H be a vector field satisfying the condition (3.24). Let u solve the problem utt + Au = 0 in (0, ∞) × Ω (3.122) where Au is given by (3.102). Then Z Z 2 ϑ[u2t + a(u, u)]dQ = SB dΣ − 2(ut , H(u))|T0 + L(u) Q

(3.123)

Σ

where SB = [u2t − a(u, u)]hH, νi + 2Γ1 (u)(H(u))ν − 2Γ2 (u)H(u) (3.124) RT and L(u) denotes some lower order terms related to the energy 0 E(τ )dτ .

Proof. Multiplying by 2H(u) the equation (3.122) and integrating over Q = (0, T ) × Ω, we obtain Z Z Z 2 utt H(u)dQ = 2(ut , H(u))|T0 + 2 ϑu2t dQ − u2t hH, νidΣ. (3.125) Q

Q

Σ

On the other hand, by using Lemmas 3.10 and 3.12, we have Z Z Z 2 AuH(u)dQ = a(u, u)hH, νidΣ + 2 ϑa(u, u)dQ Q Σ Q Z +2 [Γ2 (u)H(u) − Γ1 (u)(H(u))ν ]dΣ + L(u).

(3.126)

Σ

The identity (3.123) follows from the identities (3.125) and (3.126).

We have Theorem 3.13 Assume that the conditions (3.111) holds. Let H be an escape vector field on Ω for plate such that hH, νi ≤ 0

on

Γ1 .

(3.127)

Then there are constants c1 , c2 > 0 such that E(t) ≤ c1 e−c2 t E(0) for all solutions u to the system (3.109).


for

t≥0

(3.128)


157

Proof. To get the uniform stabilization, it will suffice to prove that there are a time T and a constant cT > 0, which are independent of solution u such that Z E(T ) ≤ cT (|ut |2 + |utν |2 )dΣ. (3.129) Σ0

Indeed, if the above inequality is true, then the identity (3.108) implies E(T ) ≤ cT (E(0) − E(T )), that is, cT E(0). 1 + cT

E(T ) ≤

We have the uniform stabilization. In the following, we prove the inequality (3.129). Step 1 On Γ1 , the boundary conditions u = uν = 0 imply that Du = 0,

D2 u(τ, τ ) = D2 u(τ, ν) = 0.

(3.130)

It follows from (3.130) that (H(u))ν = D2 u(ν, ν)hH, νi,

B1 u = B2 u = 0,

a(u, u) = [D2 u(ν, ν)]2 = (∆u)2

for x ∈ Γ1 ,

for x ∈ Γ1 .

(3.131) (3.132)

Let SB be given by (3.124). Then the condition (3.127) and the relations (3.130)-(3.132) yield SB = (∆u)2 hH, νi ≤ 0 on Γ1 .

(3.133)

Next, consider SB on Γ0 . Clearly, |(H(u))ν | ≤ c(|D2 u| + |Du|) and |H(u)| ≤ c|Du| for x ∈ Γ0 . From the formula (3.124), we obtain SB ≤ c(u2t + |D2 u|2 + |Γ1 (u)|2 + |Γ2 (u)|2 ) + L(u) on Γ0 .

(3.134)

Step 2 Let 0 < α < T /2 be given. Change the integral domain (0, T ) into (α, T − α) in the identity (3.123) with respect to time variable and use the inequalities (3.133), (3.134), and (3.117) to give 2ϑ0

Z

T −α

α

E(τ )dτ ≤ c

Z

T −α

α

Z

Γ0

(u2t + u2tν + |D2 u|2 )dΣ

+c[E(α) + E(T − α)] + L(u) Z ≤ cT (u2t + u2tν )dΣ + c[E(α) + E(T − α)] + L(u)

(3.135)

Σ0

where ϑ0 = minx∈Ω ϑ. On the other hand, using the identity (3.108), we have Z

T −α

α

E(t)dt ≥ (T − 2α)E(T ) − T


Z

Σ0

(u2t + u2tν )dΣ,

(3.136)

158

Modeling and Control in Vibrational and Structural Dynamics Z T −α Z 1 E(α) ≤ E(t)dt + (u2t + u2tν )dΣ. (3.137) T − 2α α Σ0

Noting that E(T − α) ≤ E(α), inserting (3.136) and (3.137) into (3.135) yields Z E(T ) ≤ cT (u2t + u2tν )dΣ + L(u) (3.138) Σ0

when T is large enough. The inequality (3.129) follows from the inequality (3.138) by the uniqueness argument as in Lemma 2.5, see Exercise 3.4.

Exercises 3.1 Prove the formula (3.78). 3.2 Let the operator A be given by (3.82). Prove that A is skew-adjoint on the space H02 (Ω) × L2 (Ω) where the inner product on H02 (Ω) is given by (3.83). 3.3 Let conditions (3.111) hold true. Then Z B 1/2 (u, u) = ( a(u, u)dx)1/2 Ω

defines an equivalent norm of HΓ21 (Ω). 3.4 Prove that the inequality (3.138) implies that the problem utt + ∆2 u + (1 − µ) div (κ∇u) = 0 in (0, T ) × Ω, u = uν = ∆u = (∆u)ν = 0 on Γ0 has a finite dimensional space Ξ of solutions. Then it follows from Lemma 3.3 that Ξ = ∅.

3.6


Sections 3.1, 3.2, and 3.3 are from [209]; Section 3.4 is from [78]; Section 3.5 is from [77]. For the control problems of the Euler-Bernoulli plate with constant coefficients, we refer to [9], [10], [89], [99], [127], [120], [146], and many other authors.



159

[209] was the first paper to introduce the Riemannian geometrical approach to establish observability estimates for the Euler-Bernoulli plate with variable coefficients. The content of this chapter again shows the virtues of Riemannian geometry: The Bochner technique was used to simplify computation to obtain multiplier identities (Sections 3.1) and plate models ( Section 3.5); the curvature theory provides us the global information on the existence of an escape vector field for plate which guarantees the exact controllability and the stabilization. An excellent survey in this direction is included in [75]. Recently, some further results in this direction have been obtained but not included in this chapter: [28] studied thermoelastic plates with variable coefficients; [81] studied the stabilization of elastic plates with dynamical boundary controls; [131] extended Carleman estimates to the Euler-Bernoulli plate with variable coefficients and lower order terms. Behavior of the Euler-Bernoulli plate varies with its boundary conditions. For instance, an escape vector field for the metric is enough to guarantee controllability/stabilization for the simply supported boundary conditions. But the fixed (or free) boundary conditions need something more, that is, an escape vector field for plate. Normally those assumptions are locally true, that is, when the middle surface of the plate is small in some sense. However, as the author knows, only the sectional curvature theory of Riemannian geometry has provided a useful tool to yield the global information on those assumptions in the case of variable coefficients. [179], based on the operator theory, established the exponential decay of the energy in the simply supported boundary conditions for the EulerBernoulli plate with variable coefficients. The stabilization here came down to verifying an assumption given by [207] for the wave equation. This assumption was not checkable until [64] proved that it is equivalent to an escape vector field for the metric. Thus its checkability is due to the curvature theory again.


Chapter 4 Linear Shallow Shells. Modeling, and Control

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

Equations in Equilibrium. Green’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ellipticity of the Strain Energy of Shallow Shells . . . . . . . . . . . . . . . . . . . . . . . . Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multiplier Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Escape Vector Field and Escape Region for the Shallow Shell . . . . . . . . . . . Observability Inequalities. Exact Controllability . . . . . . . . . . . . . . . . . . . . . . . . . Exact Controllability for Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stabilization by Linear Boundary Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stabilization by Nonlinear Boundary Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161 171 175 180 188 199 216 228 239 246

We view the middle surface of a shallow shell as a Riemannian manifold of dimension 2 to derive its mathematical models in the form of coordinates free with the help of the Bochner technique. Then analysis of controllability/stabilization is carried out.

4.1

Equations in Equilibrium. Green’s Formulas

We keep all the notations as in Chapter 1. Let M be a surface in R3 with the normal field N. Suppose that g is the induced metric of the surface M from the standard metric of R3 . We work on the standard Riemannian manifold (M, g) for the shell problem. Let us assume that the middle surface of a shell occupies a bounded region Ω of the surface M. The shell, a body in R3 , is defined by S = { p | p = x + zN (x), x ∈ Ω, −h/2 < z < h/2 }

(4.1)

where h is the thickness of the shell, small. Let the shell occupy a region F ( S ) in R3 after deformation, where F : S → R3 is a differentiable mapping. After deformation, the middle surface becomes 161 © 2011 by Taylor & Francis Group, LLC

162


another surface, given by F (Ω) =

n

o F (x) | x ∈ Ω .

(4.2)

Suppose that the deformation of a shell obeys the basic kinematical assumption of the classical Kirchhoff-Love theory: The normals to the undeformed middle surface move to the normals of the deformed middle surface without any change in length. Under the above Kirchhoff-Love assumption, the modeling of a shell is to find out the middle surface F (Ω) of the deformed shell by physics laws where the middle surface Ω of the undeformed shell is given. To this end, by Theorem 1.15, it suffices to work out the first and the second fundamental forms of the surface F (Ω). They are determined, respectively, by the strain tensor and the curvature tensor below. The strain tensor of the middle surface is defined by Υ=

1 (g − g), 2

(4.3)

where g is the induced metric of F (Ω) in R3 . The change of curvature tensor of the middle surface is defined by ρ = Π − Π,

(4.4)

where Π and Π are the second fundamental forms of the surfaces M and F (Ω), respectively. We recall that the classical Kirchhoff-Love assumption is contradicted by the statical assumption of approximate plane stress. For an isotropic material both assumptions agree in the statement that Υ(X, N ) = 0 for x ∈ Ω and X ∈ X (Ω), but they disagree in their predictions of the direct normal strain component Υ(N, N ). The assumption of plane stress implies that for a material obeying Hooke’s law µ Υ(N, N ) = − tr Υ for x ∈ Ω 1−µ where µ is Poisson’s coefficient, but the length of the normal keeping unchanged forces Υ(N, N ) = 0. However, we keep the classical Kirchhoff-Love assumption in this chapter for the shallow shell. We will use a complementary hypothesis of Koiter in Chapter 6 for the Koiter model. Let ζ(x) = (ζ1 , ζ2 , ζ3 ) denote the displacement vector which carries a generic material point x = (x1 , x2 , x3 ) on the undeformed middle surface Ω to a new position in space in a deformed configuration F (Ω) of the middle surface. The new position is thus given by F (x) = x + ζ(x)

∀ x ∈ Ω.

(4.5)

We decompose the displacement vector ζ into the sum ζ(x) = W (x) + w(x)N (x)


x ∈ Ω, W (x) ∈ Mx ,

(4.6)

Linear Shallow Shells. Modeling, and Control

163

where W and w are components of ζ on the tangent plane and on the normal of the undeformed middle surface Ω, respectively. Then W ∈ X (Ω) is a vector field and w is a function on Ω, respectively. Given x ∈ Ω. Let E1 , E2 be a frame field normal at x. Then E1 , E2 , N form a frame field in R3 at x. By (4.2), we have g(Ei , Ej ) = hF∗ Ei , F∗ Ej i for

1 ≤ i, j ≤ 2,

(4.7)

where F∗ is the differential of the deformation mapping F . The second fundamental form Π of the surface M is defined by ˜ X N, Y i, Π(X, Y ) = hD

∀ X, Y ∈ X (M ),

(4.8)

˜ is the covariant differential of R3 in the Euclidean metric. where D For any vector fields X and Y on M, since the manifold M is a submanifold of R3 , we obtain from the formula (4.8) ˜ X Y = DX Y + hD ˜ X Y, N iN = DX Y − Π(X, Y )N D

(4.9)

where D is the covariant differential of the surface M in the induced metric g. It follows from (4.6) and (4.9) that ˜ Ei ζ = D ˜ Ei W + Ei (w)N + wD ˜ Ei N D h i ˜ Ei N. = DEi W + Ei (w) − Π(Ei , W ) N + wD

(4.10)

We therefore obtain, by the formulas (4.5), (4.7), and (4.10),

˜ Ei ζ, Ej + D ˜ Ej ζi g(Ei , Ej ) = hEi + D ˜ Ej ζi + hEj , D ˜ Ei ζi + hD ˜ Ei ζ, D ˜ Ej ζi = δij + hEi , D = g(Ei , Ej ) + DW (Ei , Ej ) + DW (Ej , Ei ) ˜ Ei ζ, D ˜ Ej ζi +2wΠ(Ei , Ej ) + hD

(4.11) (4.12)

for i = 1, 2, since Π is symmetric. The relations (4.3) and (4.12) then yield the following displacement-strain relation of the middle surface Υ(ζ) =

1 (DW + D∗ W ) + wΠ + T (ζ), 2

(4.13)

where T (ζ) is a tensor field of rank 2 on Ω, defined by T (ζ)(X, Y ) =

1 ˜ ˜ Y ζi hDX ζ, D 2

for X, Y ∈ X (Ω).

(4.14)

After linearization with respect to the displacement, the formula (4.13) yields 1 Υ(ζ) = (DW + D∗ W ) + wΠ. (4.15) 2


164


In addition, we define the change of curvature tensor of the middle surface Ω as ρ(ζ) = −D2 w (4.16) in a form where D2 w is the Hessian of w. The choice of the tensor field (4.16) is justified for a shallow shell, see [94], p. 27, or [165], p. 355. Note that the formulas (4.15) and (4.16) are in a form coordinates free. Let the material of the shell be isotropic and obey Hooke’s law. The shell strain energy associated to a displacement vector field ζ of the middle surface Ω can be written as Z B1 (ζ, ζ) = hb B(ζ, ζ) dx, (4.17) Ω

where B(ζ, ζ) = a(Υ(ζ), Υ(ζ)) + γa(ρ(ζ), ρ(ζ)), b = E/(1 − µ2 ),

γ = h2 /12,

a(Υ(ζ), Υ(ζ)) = (1 − µ)|Υ(ζ)|2 + µ(tr Υ(ζ))2

(4.18) (4.19) (4.20)

for x ∈ Ω, where E, µ respectively denote Young’s modulus and Poisson’s coefficient of the material. In the formula (4.20) the tensor product hΥ(ζ), Υ(ζ)i and the trace tr Υ are defined in Chapter 1 by (1.103) and (1.18), respectively, for each x ∈ Ω. The expression (4.17) is an approximation to the shell strain energy. Its derivation from the three-dimensional elasticity theory is carried out by integration on the thickness of the shell, following the methods of asymptotic expansions. For those methods, we refer to [95], [158], or [183]. Traditionally, the formula (4.20) is expressed in a coordinate form. Lemma 4.1 below shows that they are the same. Let aα ∈ X (Ω) be a vector field basis on Ω where α = 1, 2. Then aαβ = aτα · aβ = haα , aβ i = g(α, β) is the first fundamental form of the middle surface Ω and γαβ = Υ(aα , aβ ) is in the traditional symbols. Let

and let

−1 aαβ = aαβ

γβα =

2 X

λ=1

We have


aαλ γλβ .

Linear Shallow Shells. Modeling, and Control Lemma 4.1

2 X

αβ=1

γβα γαβ = hΥ, Υi

2 X

γαα = tr Υ

for

α=1

for

165

x ∈ Ω;

x ∈ Ω.

Proof. Let x ∈ Ω be fixed. Let E1 ,E2 be an orthonormal basis of Mx . Let X C = Cij where Cαβ is such that aα = Ciα Ei at x. i

Then

aαβ = C τ C,

aαβ = C −1 C −τ .

We obtain X X (aαλ Cj1 α )(aβλ1 Cjβ )Ciλ Ci1 λ1 Υ(Ei , Ej )Υ(Ei1 , Ej1 ) γβα γαβ = X = cλj1 cλ1 j Ciλ Ci1 λ1 Υ(Ei , Ej )Υ(Ei1 , Ej1 ) X = δij1 δi1 j Υ(Ei , Ej )Υ(Ei1 , Ej1 ) i2 Xh = Υ(Ei , Ej ) = hΥ, Υi at x

−1 since the tensor field Υ is symmetric, where cαβ = Cαβ . Similarly, it follows that 2 X

α=1

γαα =

X

aαλ Cjα Ciλ Υ(Ei , Ej ) =

X

Υ(Ei , Ei ) = tr Υ

at x.

Now, with the expression (4.17), we are able to associate the following symmetric bilinear form, directly defined on the middle surface Ω: Z B(ζ, η) = B(ζ, η) dx, (4.21) Ω

where ζ is given by the formula (4.6) and η = U + uN

for U (x) ∈ Mx , x ∈ Ω.

(4.22)

Next, we consider the expression of the work done by the external load. For simplification, we assume that the external load P is given by P (p) = F (x) + f (x)N (x)


for

p = x + zN (x) ∈ S ,

x∈Ω

(4.23)

166


where F ∈ X (Ω) and f are a vector field and a function on Ω, respectively. Then, the work done by the external load associated with a displacement vector η of the middle surface can be written as Z L(η) = [hF, U i + f u]dv. S

By the formula (4.8), for each x ∈ Ω, there is a symmetric matrix A(x): Mx → Mx such that Π(X, Y ) = hA(x)X, Y i

for all X, Y ∈ Ωx .

Let λ1 and λ2 be the two eigenvalues of the matrix A(x). Then tr Π = λ1 + λ2

and κ = λ1 λ2

(4.24)

are the mean curvature and the Gaussian curvature, respectively. The smallest principal radius of the curvature of the undeformed middle surface is given by 1/ max{|λ1 |, |λ2 |} λ21 + λ22 6= 0, R= for x ∈ Ω. (4.25) ∞ λ1 = λ2 = 0, The following lemma is ready. Lemma 4.2 |κ| ≤

1 , R2

| tr Π| ≤

2 , R

and

|Π| ≤

√ 2 R

(4.26)

for all x ∈ Ω. We have L(η) =

Z

h/2

−h/2

Z

Ω

[hF, U i + f u](1 + tr Πz + κz 2 ) dxdz.

(4.27)

For a thin shell, it is assumed that h/R ≪ 1,

(4.28)

see [94], p.18. By the formula (4.27), L(η) can further be simplified as Z L(η) = h [hF, U i + f u] dx (4.29) Ω

since | tr Πz| ≤ h/R and |κz 2 | ≤ h2 /(4R2 ). Let the boundary Γ of the middle surface Ω be nonempty. We assume that the shell is clamped along a portion Γ0 of Γ and free on Γ1 , where Γ0 ∪Γ1 = Γ. Set HΓ10 (Ω, Λ) = { W | W ∈ H 1 (Ω, Λ), W |Γ0 = 0 },



167

∂w |Γ = 0 }, ∂ν 0 where H 1 (Ω, Λ) is the Sobolev space defined in Section 1.4 of Chapter 1. Reset ζ, η, and P as HΓ20 (Ω) = { w | w ∈ H 2 (Ω), w|Γ0 =

ζ = (W, w),

η = (U, u),

and P = (F, f ),

respectively. We then derive the following variational problem: For P ∈ L2 (Ω, Λ) × L2 (Ω) find ζ ∈ HΓ10 (Ω, Λ) × HΓ20 (Ω) such that ∀ η ∈ HΓ10 (Ω, Λ) × HΓ20 (Ω).

B1 (ζ, η) = L(η),

(4.30)

2

We recall some notation in Chapter 1. Let T (Ω) be all tensor fields of rank 2 on Ω and let Λ(Ω) be all 1-forms on Ω. Let X (Ω) be all vector fields on Ω. Let X ∈ X (Ω). Then X ∈ Λ(Ω) in the following sense: X(Y ) = hX, Y i for all Y ∈ X (Ω). In the above sense X (Ω) = Λ(Ω). Let f be a function. Then ∇f = df where ∇f and df denote the gradient of f and the covariant differential of f (which is a 1-form), respectively. We need the following: Proposition 4.1 We have QΠ = d tr Π,

2

(4.31)

where Q: T (Ω) → X (Ω) is an operator defined by the formula (1.144) of Chapter 1. Proof. By Corollary 1.1, DΠ is symmetric, that is, DΠ(X, Y, Z) = DΠ(Y, X, Z) = DΠ(Y, Z, X)

(4.32)

for X, Y, Z ∈ X (Ω). Given x ∈ Ω. Let E1 , E2 be a frame field normal at x. By the formulas (1.144) and (4.32), we have hQΠ, Ei i = tr lEi DΠ = =

2 X

2 X

DΠ(Ei , Ej , Ej ) =

j=1

Ei (Π(Ej , Ej )) = Ei (tr Π)

2 X

DΠ(Ej , Ej , Ei )

j=1

at x

(4.33)

j=1

since DEi Ej = 0 at x for i = 1, 2. The formula (4.31) follows from (4.33).

The following formula is the Green formula which expresses the relationship between the boundary and the interior of a displacement vector field. This formula plays an important role in the boundary-valued problems.


168


Theorem 4.1 Let the bilinear form B(·, ·) be given in (4.21). For ζ = (W, w), η = (U, u) ∈ H 1 (Ω, Λ) × H 2 (Ω), we have Z B(ζ, η) = (Aζ, η)L2 (Ω,Λ)×L2 (Ω) + ∂(Aζ, η) dΓ (4.34) Γ

where ∂(Aζ, η) = B1 (W, w)hU, νi + B2 (W, w)hU, τ i ∂u +γ ∆w + (1 − µ)B3 w ∂ν ∂∆w + (1 − µ)B4 w u, −γ ∂ν

(4.35)

ν, τ are the normal and the tangential along curve Γ, respectively, −∆µ W − (1 − µ)κW − H(w) Aζ = , γ[∆2 w − (1 − µ)δ(κdw)] + (tr2 Π − 2(1 − µ)κ)w + G(W )

(4.36) ∆µ is of the Hodge-Laplacian type, applied to 1-forms (or equivalently vector fields), defined by 1−µ ∆µ = −( δd + dδ), (4.37) 2 d the exterior differential, δ the formal adjoint of d, ∆ the Laplacian on manifold M, H(w) = (1 − µ) i (dw)Π + µ tr Πdw + wd tr Π, (4.38) G(W ) = (1 − µ)hDW, Πi − µ tr ΠδW,  B1 (W, w) = (1 − µ)Υ(ζ)(ν, ν) + µ(wH − δW ),    B (W, w) = (1 − µ)Υ(ζ)(ν, τ ),  2 2 (4.39) B 3 w = −D w(τ, τ ),   ∂ ∂w   B4 w = (D2 w(τ, ν)) + κ(x) , ∂τ ∂ν i (dw)Π is the interior product of Π by dw, and Π and κ are the second fundamental form and the Gaussian curvature of the surface M, respectively. Proof. Let Q be defined by (1.144). Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x, that is, DEi Ej = 0 at x

(4.40)

for i = 1, 2. By the formulas (1.144) and (4.40) hQ(wΠ), Ek i = D(wΠ)(Ek , E1 , E1 ) + D(wΠ)(Ek , E2 , E2 ) = E1 wΠ(Ek , E1 ) + E2 wΠ(Ek , E2 ) = Π(Ek , ∇w) + w tr i (Ek )DΠ


at x


169

which yields, via Proposition 4.1, Q(wΠ) = i (dw)Π + wd tr Π

(4.41)

since Dw = dw. In addition, by (4.24), we have 2

|Π|2 = tr Π − 2κ,

x ∈ M.

(4.42)

Using the formula (4.15), we obatin 1 hDW, DU + D∗ U i + hwΠ, DU i 2 +(hΠ, DW i + w|Π|2 )u.

hΥ(ζ), Υ(η)i =

(4.43)

We integrate the identity (4.43) over Ω. It follows from the relations (1.155), (1.156), (3.100), (1.145), (4.41), and (4.42) that Z hΥ(ζ), Υ(η)i dx Ω

1 (DW, DU + D∗ U )L2 (Ω,T 2 ) + (wΠ, DU )L2 (Ω,T 2 ) 2 +(hΠ, DW i + w|Π|2 , u) Z 1 1 = (∆W + dδW − 2κW, U )L2 (Ω,Λ) + [hDν W, U i + hDU W, νi] dΓ 2 2 Γ Z +(−Q(wΠ), U )L2 (Ω,Λ) + h i (ν)(wΠ), U i dΓ + (hΠ, DW i + w|Π|2 , u) =

Γ

1 = (( δd + dδ)W − κW − i (dw)Π − wd tr Π, U )L2 (Ω,Λ) 2 Z 2

+(hΠ, DW i + w(tr Π − 2κ), u) +

Υ(ζ)(ν, U ) dΓ

(4.44)

Γ

where the following formula is used i 1h Υ(ν, U ) = DW (ν, U ) + DW (U, ν) + wΠ(ν, U ) 2 i 1h = hDU W, νi + hDν W, U i + h i (ν)(wΠ), U i. 2 On the other hand, the formulas (1.111) and (1.100) yield tr DW = −δW.

(4.45)

By the equations (4.15), (4.45), and (1.137), we obtain the following Z Z tr Υ(ζ) tr Υ(η) dx = (−δW + w tr Π)(−δU + u tr Π) dx Ω

Ω

2

= (δW − w tr Π, δU ) + (− tr ΠδW + w tr Π, u)

2

= (dδW − d(w tr Π), U )L2 (Ω,Λ) + (− tr ΠδW + w tr Π, u) Z + hU, νi tr Υ(ζ) dΓ. (4.46) Γ


170


Using the formulas (4.46), (4.44), (4.18), and (4.21), we have Z a(Υ(ζ), Υ(η)) dx Ω

= −(∆µ W + (1 − µ)κW + H(w), U )L2 (Ω,Λ) 2

+(G(W ) + (tr Π − 2(1 − µ)κ)w, u) Z + [B1 (W, w)hU, νi + B2 (W, w)hU, τ i] dΓ

(4.47)

Γ

where the following formula is used Υ(ν, U ) = Υ(ν, ν)hU, νi + Υ(ν, τ )hU, τ i for x ∈ Γ. R Next, we compute the integration Ω a(ρ(ζ), ρ(η)) dx. Since δw = 0, the relations ∆ = dδ + δd and d2 = 0 imply δ∆dw = δdδdw = ∆2 w.

(4.48)

Noting that dw = Dw, we obtain from (4.16), (1.155), (1.137), and (4.48) that Z hρ(ζ), ρ(η)i dx = (Ddw, Ddu)L2 (Ω,T 2 ) Ω Z = (∆dw − κdw, du)L2 (Ω,Λ) + hDν dw, dui dΓ Γ Z Z ∂w 2 = (∆ w − δ(κdw), u) + u[hν, ∆dwi − κ ] dΓ + D2 w(ν, du) dΓ ∂ν Γ Γ Z ∂u ∂u 2 2 2 = (∆ w − δ(κdw), u) + [D w(ν, ν) + D w(ν, τ ) ] dΓ ∂ν ∂τ Γ Z ∂∆w ∂w − u[ +κ ] dΓ, (4.49) ∂ν ∂ν Γ since ∆ = −∆ if it is applied to functions and ∆dw = −d∆w. In addition, since trρ(ζ) = −∆w, by Green’s formula, we have Z tr ρ(ζ) tr ρ(η) dx = (∆w, ∆u) = (∆2 w, u) Ω Z ∂u ∂∆w + [∆w −u ] dΓ. (4.50) ∂ν ∂ν Γ It is generally assumed that Γ is a closed curve so that Z Z ∂u ∂ 2 D w(ν, τ ) dΓ = − u (D2 w(ν, τ )) dΓ. ∂τ ∂τ Γ Γ

(4.51)

Furthermore, we have ∆w = D2 w(ν, ν) + D2 w(τ, τ ),


on Γ.

(4.52)

Linear Shallow Shells. Modeling, and Control By the formulas (4.49)-(4.52), and (4.20), we obtain Z a(ρ(ζ), ρ(η)) dx = (∆2 w − (1 − µ)δ(κdw), u) Ω Z ∂u ∂∆w −( + (1 − µ)B4 w)u] dΓ. + [(∆w + (1 − µ)B3 w) ∂ν ∂ν Γ

171

(4.53)

Finally, the identity (4.34) follows from the formulas (4.21), (4.18), (4.47), and (4.53). By Theorem 4.1 the following proposition is immediate. Proposition 4.2 The variational problem (4.30) is equivalent to solving the following boundary value problem in unknown ζ = (W, w) bAζ = (F, f )

(4.54)

subject to the boundary conditions ( W |Γ0 = 0, (4.55) ∂w w|Γ0 = |Γ = 0, ∂ν 0    B1 (W, w) = B2 (W, w) = 0 on Γ1 , ∆w + (1 − µ)B3 w = 0 on Γ1 , (4.56)   ∂∆w + (1 − µ)B4 w = 0 on Γ1 , ∂ν where A and Bi (1 ≤ i ≤ 4) are defined in (4.36) and in (4.39), respectively.

4.2

Ellipticity of the Strain Energy of Shallow Shells

The ellipticity of the strain energy of shells is an important issue necessary to all the control problems. In this section we derive the ellipticity results for shallow shells. Let Γ0 ⊂ Γ be a portion of the boundary of the middle surface Ω with a positive length. Consider the Sobolev spaces HΓ10 (Ω, Λ) = { W | W ∈ H 1 (Ω, Λ), W |Γ0 = 0 } and

(4.57)

∂w |Γ = 0 } (4.58) ∂ν 0 with the norms of H 1 (Ω, Λ) and H 2 (Ω), respectively. In the following, we will show that the bilinear form B(·, ·), given in (4.21), induces an equivalent norm on the space HΓ10 (Ω, Λ) × HΓ20 (Ω), i.e., the energy norm for a shallow shell. We need several lemmas first. HΓ20 (Ω) = { w | w ∈ H 2 (Ω), w|Γ0 =


172


Lemma 4.3 Let W be a vector field (or equivalently a 1-form). Set U = DW + D∗ W.

(4.59)

Then 1 [DU (X, Y, Z) + DU (Z, X, Y ) − DU (Y, Z, X)] 2 +R(X, Y, Z, W ), ∀ X, Y, Z ∈ X (M ). (4.60)

D2 W (X, Y, Z) =

Proof. Given x ∈ M. Let E1 , E2 be a frame field normal at x. By (4.59), we have DU (Ei , Ej , Ek ) = Ek (U (Ei , Ej )) = D2 W (Ei , Ej , Ek ) + D2 W (Ej , Ei , Ek )

(4.61)

at x since (DEi Ej )(x) = 0. Changing i, j, and k into k, i, and j, respectively, in the formula (4.61), we obtain from the formula (1.22) DU (Ek , Ei , Ej ) = D2 W (Ek , Ei , Ej ) + D2 W (Ei , Ek , Ej ) = D2 W (Ek , Ei , Ej ) + D2 W (Ei , Ej , Ek ) + R(Ek , Ej , W, Ei )

(4.62)

at x where R(·, ·, ·, ·) is the curvature tensor. A similar argument gives DU (Ej , Ek , Ei ) = D2 W (Ej , Ei , Ek ) + D2 W (Ek , Ei , Ej ) +R(Ek , Ei , W, Ej ) + R(Ej , Ei , W, Ek )

(4.63)

at x. By the equations (4.61), (4.62), and (4.63), we obtain 2D2 W (Ei , Ej , Ek ) = [DU (Ei , Ej , Ek ) + DU (Ek , Ei , Ej ) − DU (Ej , Ek , Ei )] −R(Ek , Ej , W, Ei ) + R(Ek , Ei , W, Ej ) + R(Ej , Ei , W, Ek )

(4.64)

at x. By the first identity of Bianchi, (2) in Theorem 1.8, R(X, Y, Z, W ) + R(Y, Z, X, W ) + R(Z, X, Y, W ) = 0 and the properties of the curvature tensor R(X, Y, Z, W ) = −R(Y, X, Z, W ) = R(Y, X, W, Z) = R(W, Z, Y, X), we obtain −R(Ek , Ej , W, Ei ) + R(Ek , Ei , W, Ej ) + R(Ej , Ei , W, Ek )

= −R(Ej , Ek , Ei , W ) − R(Ek , Ei , Ej , W ) + R(Ei , Ej , Ek , W ) = 2R(Ei , Ej , Ek , W ) at x. (4.65) Inserting the formula (4.65) into the formula (4.64), we obtain the identity (4.60).



173

Lemma 4.4 Let W ∈ X (Ω) such that DW + D∗ W = 0

on

Ω.

(4.66)

Then ∆W = 2κW

on

Ω

(4.67)

where ∆ is the Hodge-Laplace operator and κ is the Gauss curvature. Proof. Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x such that DEi Ej (x) = 0 for 1 ≤ i, j ≤ 2. It follows from (4.60), (4.66), and (3.100) that 2 X

D2 W (Ej , Ei , Ei ) =

i=1

2 X

R(Ej , Ei , Ei , W )

i=1

= − Ric (Ej , W ) = −κhW, Ej i.

(4.68)

By (1.114) and (4.68) , we obtain at x ∆W = − =−

2 X

2 DE W+ i Ei

i=1

2 X

2 X

R(Ei , Ej , W, Ej )Ei

ij=1

D2 W (Ej , Ei , Ei )Ej + κW = 2κW.

ij=1

Remark 4.1 In the flat case where κ = 0, if W = (w1 , w2 ), then ∂w ∂wj i + . DW + D∗ W = ∂xj ∂xi Therefore the identity (4.67) is equivalent to 2 X ∂ 2 w1 i=1

∂x2i

=

2 X ∂ 2 w2 i=1

∂x2i

= 0.

Lemma 4.5 There is constant c > 0 such that kDW + D∗ W kL2 (Ω,T 2 ) ≥ ckW kH 1 (Ω,Λ)

for

W ∈ HΓ10 (Ω, Λ).

(4.69)

Proof. Set X = { W | W ∈ L2 (Ω, Λ), DW + D∗ W ∈ L2 (Ω, T 2 ) }.

(4.70)

Given W ∈ X . By the formula (4.60), D2 W ∈ H −1 (Ω, T 3 ) since DW + D W ∈ L2 (Ω, T 2 ), where ∗ H −1 (Ω, T 3 ) = HΓ1 (Ω, T 3 ) . (4.71) ∗


174


By a lemma of J. L. Lions (see [18]) in a local coordinate, we obtain W ∈ H 1 (Ω, Λ). Then X = H 1 (Ω, Λ). (4.72) In order to get the inequality (4.69), it suffices by the open mapping theorem to prove the following uniqueness: DW + D∗ W = 0 implies

W =0

for W ∈ HΓ10 (Ω, Λ).

(4.73)

Suppose that DW + D∗ W = 0 on Ω.

(4.74)

Then W satisfies the equation (4.67) by Lemma 4.4. Let ν and τ be the normal and the tangential along Γ, respectively. The condition W |Γ0 = 0 implies that Dτ W |Γ0 = 0, which gives, by combining the formula (4.74), DW |Γ0 = W |Γ0 = 0. (4.75) Applying [5], or [170] to the system of the equation (4.67) and the boundary condition (4.75), we obtain that W = 0 on Ω, since, in the standard coordinate system x = (x1 , x2 ), ∆W = (−∆w1 + first order terms, −∆w2 + first order terms) if W = (w1 , w2 ). We now assume that the shell is clamped on a portion Γ0 of boundary Γ with length Γ0 > 0. We have the following: Theorem 4.2 There is a constant c > 0 such that B(ζ, ζ) ≥ ckζk2H 1

2 Γ0 (Ω,Λ)×HΓ0 (Ω)

(4.76)

for all ζ = (W, w) ∈ HΓ10 (Ω, Λ) × HΓ20 (Ω). Hence B(·, ·) induces an equivalent norm over space HΓ10 (Ω, Λ) × HΓ20 (Ω). Proof. By means of Schwartz’s inequality, it follows from the formula (4.15) that 1 kDW + D∗ W k2L2 (Ω,T 2 ) 4 +(wΠ, DW + D∗ W )L2 (Ω,T 2 ) + kw|Π|k2 1 ≥ kDW + D∗ W k2L2 (Ω,T 2 ) − 2 max |Π|2 kwk2 . (4.77) x∈Ω 8

kΥ(ζ)k2L2 (Ω,T 2 ) =

By the formulas (4.20) and (4.77), we obtain Z a(Υ(ζ), Υ(ζ)) dx Ω

≥

1−µ [kDW + D∗ W k2L2 (Ω,T 2 ) − 16 max |Π|2 kwk2 ]. x∈Ω 8


(4.78)


175

In addition, by the relations (4.16) and (4.20), we obtain Z a(ρ(ζ), ρ(ζ)) dx ≥ (1 − µ)kD2 wk2L2 (Ω,T 2 ) .

(4.79)

Ω

From the inequalities (4.78) and (4.79) and Lemma 4.5, it is obvious that there are C1 , C2 > 0 such that B(ζ, ζ) + C1 kwk2 ≥ C2 kζk2H 1

2 Γ0 (Ω,Λ)×HΓ0 (Ω)

.

(4.80)

From the inequality (4.80), to get the inequality (4.76), it suffices by the compactness-uniqueness argument to prove the following uniqueness: B(ζ, ζ) = 0

implies

ζ = 0.

However, this is almost immediate. If B(ζ, ζ) = 0, the inequality (4.79) yields w = 0 so that we obtain from the inequality (4.78) that DW + D∗ W = 0. Finally, W = 0 follows from the inequality (4.69), i.e., ζ = 0. By Theorem 4.2 the following result is ready. Theorem 4.3 For (F, f ) ∈ L2 (Ω, Λ) × L2 (Ω), the boundary value problem (4.54)-(4.56) has a unique solution ζ ∈ HΓ10 (Ω, Λ) × HΓ20 (Ω).

4.3

Equations of Motion

Let us compute the kinetic energy for a shallow shell. Suppose that F (Ω) is the middle surface of the deformed shell which is given by (4.2). Given x ∈ Ω. Denote by N (F (x)) the normal field of the surface F (Ω) at F (x). Let E1 , E2 be a frame field normal at x, positively orientated, i.e., E1 , E2 , N is a frame field of R3 at x such that N = E1 × E2 . Then we have N (F (x)) = Set

F∗ E1 × F∗ E2 . |F∗ E1 × F∗ E2 |

˜ Ei ζ, Ej i, ζij = hD

1 ≤ i, j ≤ 3,

(4.81)

(4.82) (4.83)

where ζ is the displacement vector of the undeformed middle surface Ω, given in (4.6), and E3 = N. Using the formula (4.10), we have ζ13 E1 + ζ23 E2 = dw − i (W )Π


(4.84)

176


where i (W )Π is the interior product of tensor field Π by vector field W . It follows from the formulas (4.5), (4.81), (4.82), and (4.84) that ˜ E1 ζ) × (E2 + D ˜ E2 ζ) F∗ E1 × F∗ E2 = (E1 + D ˜ E2 ζ + D ˜ E1 ζ × D ˜ E2 ζ ˜ E1 ζ × E2 + E1 × D =N +D = (1 + ζ11 + ζ22 )N − dw + i (W )Π + N(ζ)

(4.85)

where N(ζ) denote all the nonlinear terms with respect to displacement field ζ. The equations (4.11) and (4.83) give |F∗ E1 × F∗ E2 |2 = |F∗ E1 |2 |F∗ E2 |2 − hF∗ E1 , F∗ E2 i2 ˜ E1 ζ|2 )(1 + 2ζ22 + |D ˜ E2 ζ|2 ) = (1 + 2ζ11 + |D 2 ˜ E1 ζ, D ˜ E2 ζi) −(ζ12 + ζ21 + hD = 1 + 2ζ11 + 2ζ22 + N(ζ),

that is, |F∗ E1 × F∗ E2 | = 1 + ζ11 + ζ22 + N(ζ).

(4.86)

Let us substitute the expressions (4.85) and (4.86) into the formula (4.82). After linearization, we obtain N (F (x)) = −dw + i (W )Π + N.

(4.87)

Let S be given in the formula (4.1). Given p = x+zN (x) ∈ S where x ∈ Ω and −h/2 < z < h/2. It follows from the classical Love-Kirchhoff assumptions that F (p) = F (x) + zN (F (x)) for p = x + zN (x). (4.88) Denote ζ(p) the displacement vector of the point p. Then, the relations (4.87) and (4.88) yield ζ(p) = ζ(x) − z(dw − i (W )Π)

for

p = x + zN (x), x ∈ Ω.

(4.89)

Now, we assume that displacement vector ζ depends on time t and, for simplicity, the mass density per unit of volume is unit. Then the kinetic energy per unit area of the undeformed middle surface is obtained by integration with respect to z Z h/2 H(t) = |ζ t |2 (1 + tr Πz + κz 2 ) dz. (4.90) −h/2

Set Y = dw − i (W )Π.

(4.91)

By the formula (4.89), we obtain |ζ t |2 (1 + tr Πz + κz 2 ) = |ζt |2 − (− tr Π|ζt |2 + 2hζt , Yt i)z

+(κ|ζt |2 − 2 tr Πhζt , Yt i + |Yt |2 )z 2 −(2κhζt , Yt i − tr Π|Yt |2 )z 3 + κ|Yt |2 z 4 .


(4.92)


177

Integrating both the sides of the equation (4.92) over (−h/2, h/2) with respect to z yields h3 h5 (κ|ζt |2 − 2 tr Πhζt , Yt i + |Yt |2 ) + κ|Yt |2 12 80 3 3 3 h h h = h|Wt |2 + hwt2 + |Dwt |2 + κ|Wt |2 κwt2 12 12 12 h3 h3 h3 h3 2 tr ΠhWt , Dwt i − Π(Wt , Dwt ) + | i (Wt )Π| + Π(Wt , Wt ) tr Π − 12 6 6 6 5 5 h5 h h + κ|Dwt |2 − κΠ(Wt , Dwt ) + κ| i (Wt )Π|2 (4.93) 80 40 80

H(t) = h|ζt |2 +

since dw = Dw. By using some ideas in [94], however, we can further simplify H(t) as H(t) = h|Wt |2 + hwt2 +

h3 |Dwt |2 . 12

(4.94)

In order to justify the approximation (4.94) it will be shown that the order of magnitude of the terms, which have been omitted in the formula (4.93), is indeed negligible compared with the expression (4.94). All the neglected terms are listed below: h3 κ|Wt |2 ; (4.95) 12 h3 2 κw ; 12 t

(4.96)

h3 | i (Wt )Π|2 ; 12

(4.97)

h3 Π(Wt , Wt ) tr Π; 6

(4.98)

−

h3 tr ΠhWt , Dwt i; 6

(4.99)

h3 Π(Wt , Dwt ); 6

(4.100)

−

h5 κ|Dwt |2 ; 80 −

(4.101)

h5 κΠ(Wt , Dwt ); 40

(4.102)

h5 κ| i (Wt )Π|2 . 80

(4.103)

It is almost obvious that the neglected terms (4.95), (4.96), and (4.101) are indeed negligible in a thin shell. In fact, if R is the smallest principal radius


178


of curvature of the undeformed middle surface given by (4.25), we have the estimates |(4.95)| ≤

h2 (h|Wt |2 ); 12R2

|(4.101)| ≤

3 20

|(4.96)| ≤

h R

2

h2 (hwt2 ); 12R2

h3 |Dwt |2 . 12

(4.104)

In addition, by Lemma 4.2, we have 1 |(4.97)| ≤ 6

h R

2

2

(h|Wt | );

1 |(4.103)| ≤ 5

√ 2 2 h |(4.98)| ≤ (h|Wt |2 ); 3 R

h R

4

(h|Wt |2 ).

(4.105)

Finally, by means of Schwartz’s inequality and Lemma 4.2, we obtain estimates √ 3 h h3 |(4.99)| ≤ (h|Wt |2 + |Dwt |2 ); (4.106) 3 R 12 √ 6 h h3 |(4.100)| ≤ (h|Wt |2 + |Dwt |2 ); 3 R 12 √ 3 6 h h3 |(4.102)| ≤ (h|Wt |2 + |Dwt |2 ). 20 R 12

(4.107)

(4.108)

By the assumption (4.28) for a thin shell and by the estimates (4.104)(4.108), the approximation (4.94) is justified so that we obtain the kinetic energy of the shell P=

Z

Ω

H(t) dx =

Z

Ω

(h|Wt |2 + hwt2 +

h3 |Dwt |2 ) dx. 12

(4.109)

The equations of motion for ζ are obtained by setting to zero the first variation of the Lagrangian Z

0

T

[P(t) + L(ζ) − B1 (ζ, ζ)] dt

(4.110)

(the “Principle of Virtual Work”), where L(·) and B1 (·, ·) are given by the formulas (4.29) and (4.17), respectively. We assume that there is no external loading on the shell. Then the variation of (4.110) is taken with respect to kinematically admissible displacement fields. We obtain, as a result of the calculation by the formulas (4.34) and (4.110), the following



179

Theorem 4.4 We assume that there are no external loads on the shell and that the shell is clamped along a portion Γ0 of Γ and free on Γ1 , where Γ0 ∪Γ1 = Γ. Then the displacement vector ζ = (W, w) satisfies the following boundary value problem:   Wtt − b[∆µ W + (1 − µ)κW + F (w)] = 0, wtt − γ∆wtt + b[γ ∆2 w − (1 − µ)δ(κdw) in Q∞ , (4.111)  2 +(tr Π − 2(1 − µ)κ)w + G(W )] = 0, (

where

W = 0, ∂w w= = 0, ∂ν

on

Σ∞ 0

   B1 (W, w) = B2 (W, w) = 0, ∆w + (1 − µ)B3 w = 0,   bγ[ ∂∆w + (1 − µ)B4 w] − γ ∂wtt = 0, ∂ν ∂ν

Q∞ = (0, ∞) × Ω,

Σ∞ 0 = (0, ∞) × Γ0 ,

(4.112)

on

Σ∞ 1 ,

Σ∞ 1 = (0, ∞) × Γ1 .

(4.113)

(4.114)

If the shell is flat, we have Corollary 4.1 Let the shell be flat, a plate. The above equations in (4.111) are uncoupled. In this case M = R2 with the Euclidean metric in R2 . In the natural coordinate, setting W = (w1 , w2 ), we then have  Wtt − b∆µ W = 0 in Q∞ ,    ∞    W = 0, on Σ0 , ∂w2 ∂w1 ∂w2 ∂w1 (4.115) (1 − µ)( ν1 + ν2 ) + µ( + ) = 0 on Σ∞ 1 ,  ∂ν ∂ν ∂x ∂y      ( ∂w2 + ∂w1 )ν2 + ( ∂w1 − ∂w2 )ν1 on Σ∞ , 1 ∂τ ∂ν ∂τ ∂ν where ν = (ν1 , ν2 ), τ = (ν2 , −ν1 ),  2 ∂ w1 1 − µ ∂ 2 w1 1 + µ ∂ 2 w2 + +  ∂x2 2 ∂y 2 2 ∂x∂y ∆µ W =   1 − µ ∂ 2 w2 ∂ 2 w2 1 + µ ∂ 2 w1 + + 2 ∂x2 ∂y 2 2 ∂x∂y and

 wtt − γ∆wtt + bγ∆2 w = 0, in Q∞ ,       w = ∂w = 0 on Σ∞ 0 , ∂ν ∆w + (1 − µ)B3 w = 0 on Σ∞  1    ∂∆w ∂wtt   bγ[ + (1 − µ)B4 w] − γ =0 ∂ν ∂ν




 , 

(4.116)

(4.117) on

Σ∞ 1 ,

180


where

∂2w ∂2w + . (4.118) ∂x2 ∂y 2 The equations (4.117) are the same as in [105], pp. 15-16, a Kirchhoff plate. ∆w =

Proof. Since M = R2 , we have Π=0

and

tr Π = κ = 0.

(4.119)

It follows from the formulas (4.119) and (4.43) that H(w) = 0

and G(W ) = 0.

(4.120)

It is easily checked that −dδW = (

∂ 2 w1 ∂ 2 w2 ∂ ∂ 2 w1 ∂ 2 w2 ∂ + ) + ( + ) ; ∂x2 ∂x∂y ∂x ∂x∂y ∂y 2 ∂y

(4.121)

∂ 2 w1 ∂ 2 w2 ∂ ∂ 2 w2 ∂ 2 w1 ∂ − ) +( − ) , (4.122) 2 2 ∂y ∂x∂y ∂x ∂x ∂x∂y ∂y ∂ ∂ since W = w1 + w2 . ∂x ∂y Inserting the formulas (4.121) and (4.122) into the formula (4.37) yields the formula (4.116). By the formulas (4.39) and (4.119), we obtain −δdW = (

B1 (W, w) = (1 − µ)DW (n, n) − µW ∂w1 ∂w2 ∂w1 ∂w2 = (1 − µ)( ν1 + ν2 )( + ); ∂ν ∂ν ∂x ∂y

(4.123)

1−µ (hDν W, τ i + hDτ W, νi) 2 ∂w2 ∂w1 ∂w2 1 − µ ∂w1 = [( − )ν1 + ( + )ν2 ]. (4.124) 2 ∂τ ∂ν ∂ν ∂τ By the formulas (4.120), (4.123), and (4.124), we obtain the equations (4.115) and (4.117) by Theorem 4.4. We check B3 and B4 on the boundary Γ in order to show that the equations (4.117) are the same as in [105], pp. 15-16. In fact, we have  2  ∂ w ∂2w  ∂x2 ∂x∂y  ν2  B3 w = −D2 w(τ, τ ) = −(ν2 , −ν1 )   ∂ 2 w ∂ 2 w  −ν1 ∂x∂y ∂y 2 ∂2w ∂2w ∂2w = 2ν1 ν2 − ν1 2 − ν2 2 ∂x∂y ∂y ∂x B2 (W, w) =

and B4 w =

∂ 2 ∂ ∂2w ∂2w ∂2w D w(τ, n) = [(ν22 − ν12 ) + ν1 ν2 ( 2 − )]. ∂τ ∂τ ∂x∂y ∂x ∂y 2



4.4

181


We will establish some multiplier identities by the Bochnner technique which are useful to control problems. Let M be a surface in R3 with the induced metric g and let Ω ⊂ M be the middle surface of a shell. Let ζ = (W, w) be a displacement vector field and let V be a vector field on the middle surface Ω. The main multipliers we will use are m(ζ) = DV W, V (w) and qζ where q is a function on Ω. For ζ = (W, w), let

e(ζ) = |Wt |2 + |wt |2 + γ|Dwt |2 + B(ζ, ζ)

(4.125)

for each x ∈ Ω where γ = h2 /12 and B(ζ, ζ) is given by (4.18). If ζ = (W, w) solves the shallow shell problem (4.111), the energy of the displacement vector field ζ is given by Z 1 E(t) = e(ζ) dx. (4.126) 2 Ω

In the sequel, lo (ζ) denotes all the lower order terms with respect to the energy to which for any ε > 0 there exists Cε > 0 such that | lo (ζ)|2 ≤ εe(ζ) + Cε l(ζ)

for all x ∈ Ω

(4.127)

where l(ζ) = |W |2 + |w|2 + |Dw|2 + γwt2 .

(4.128)

So lo (ζ) may be different from line to line and page to page. We need several lemmas for our computation. Lemma 4.6 Let V, W ∈ X (Ω) be vector fields on the middle surface Ω. Then D DV W = DV (DW ) + R(V, ·, W, ·) + DW (·, D· V ) (4.129)

where R(·, ·, ·, ·) is the curvature tensor of the middle surface Ω and “·” denotes the position of variables.

Proof. Given x ∈ Ω. Let E1 , E2 be a frame field normal at x. Using the relations DEi Ej (x) = 0 and (1.22), we have D(DV W )(Ei , Ej ) = Ej DV W (Ei ) = Ej DW (Ei , V ) = D2 W (Ei , V, Ej ) + DW (Ei , DEj V )

= D2 W (Ei , Ej , V ) + RV Ej W (Ei ) + DW (Ei , DEj V ) = DV (DW )(Ei , Ej ) + R(V, Ej , W, Ei ) +DW (Ei , DEj V ) at x


(4.130)

182


for 1 ≤ i, j ≤ 2. The above identity (4.130) means that D(DV W )(X, Y ) = DV (DW )(X, Y ) + R(V, X, W, Y ) + DW (X, DY V ) for all X, Y ∈ X (Ω) since R(V, ·, W, ·) is symmetric.

Let V ∈ X (Ω) be a vector field and let T ∈ T 2 (Ω) be a tensor field of rank 2. We define a tensor field of rank 2 on Ω by ∗ i 1h G(V, T ) = T (·, D· V ) + T (·, D· V ) (4.131) 2 ∗ Let where T (·, D· V ) denotes the transpose of the tensor field T (·, D· V ).

Lemma 4.7 Let V ∈ X (Ω) and ζ = (W, w) be given. Let m(ζ) = DV W, V (w) . Then Υ m(ζ) = DV Υ(ζ) + G(V, DW ) + lo (ζ) (4.132) where Υ is the strain tensor of the middle surface, given by (4.15). Moreover, ρ m(ζ) = DV ρ(ζ) + 2G(V, ρ(ζ)) + lo (ζ) (4.133) where ρ is the change of curvature tensor of the middle surface, given by (4.16).

Proof. We make the covariant differential on the formula (4.15) with respect to the vector field V, use the formula (4.129), and have i 1h DV Υ(ζ) = DV (DW ) + DV (D∗ W ) + V (w)Π + wDV Π. (4.134) 2 On the other hand, we make a transpose on the formula (4.129) to obtain D∗ DV W = DV (D∗ W ) + R(V, ·, W, ·) + DW (D· V, ·). (4.135)

Inserting the formulas (4.129) and (4.135) into the formula (4.134), we obtain the identity (4.132) where lo (ζ) = R(V, ·, W, ·) − wDV Π

since R(V, ·, W, ·) is symmetric. Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x. Using the relations DEi Ej (x) = 0 and the relation (1.22) , we have D2 V (w) (Ei , Ej ) = Ej Ei Dw(V ) = Ej D2 w(Ei , V ) + Dw(DEi V ) = D3 w(Ei , V, Ej ) + D2 w(Ei , DEj V )

+D2 w(DEi V, Ej ) + Dw(DEj DEi V ) = D2 w(Ei , Ej , V ) + R(V, Ej , Dw, Ei ) +D2 w(Ei , DEj V ) + D2 w(DEi V, Ej ) +Dw(DEj DEi V ).


(4.136)


183

The last term in the right hand side of (4.136) can be computed as follows: Dw(DEj DEi V ) = hDEj DEi V, Dwi = Ej DV (Dw, Ei ) − hDEi V, DEj Dwi

= D2 V (Dw, Ei , Ej ) + DV (DEj Dw, Ei ) −hDEi V, DEj Dwi

= i (Dw)D2 V (Ei , Ej ).

(4.137)

Inserting the formula (4.137) into the formula (4.136) yields the identity ∗ D2 V (w) = DV D2 w + D2 w(·, D· V ) + D2 w(·, D· V ) + i (Dw)Π + R(V, ·, W, ·).

(4.138)

Noting that ρ(ζ) = −D2 w and ρ m(ζ) = −D2 V (w) , we obtain the

identity (4.133) from the identity (4.138) where

lo (ζ) = − i (Dw)Π − R(V, ·, W, ·). Lemma4.8 Let V ∈X (Ω) be a vector field and let ζ = (W, w) be given. Let m(ζ) = DV W, V (w) . Suppose that the bilinear forms B(·, ·) and a(·, ·) are given by (4.18) and (4.20), respectively. Then

h i 2B ζ, m(ζ) = div B(ζ, ζ)V − B(ζ, ζ) div V +2a Υ(ζ), G(V, DW ) + 4γa ρ(ζ), G(V, ρ(ζ)) + lo (ζ)

(4.139)

where G(V, DW ) and G(V, ρ(ζ)) are defined by the formula (4.131). Proof. By the expression (4.132), we have 2hΥ(ζ), Υ m(ζ) i = V |Υ(ζ)|2 + 2hΥ(ζ), G(V, DW )i + lo (ζ) = div (|Υ(ζ)|2 V ) − |Υ(ζ)|2 div V +2hΥ(ζ), G(V, DW )i + lo (ζ).

(4.140)

Let x ∈ Ω be given and let E1 , E2 be a frame field normal at x. Then 2 2 X X V tr Υ(ζ) = V [ Υ(ζ)(Ei , Ei )] = DV Υ(Ei , Ei ) = tr DV Υ i=1


i=1

(4.141)

184


at x. From the formulas (4.141) and (4.132), we obtain 2 tr Υ(ζ) tr Υ m(ζ) = 2 tr Υ(ζ)V tr Υ(ζ) + 2 tr Υ(ζ) tr G(V, DW ) + lo (ζ) h 2 i 2 = div tr Υ(ζ) V − tr Υ(ζ) div V +2 tr Υ(ζ) tr G(V, DW ) + lo (ζ).

(4.142)

Using the formulas (4.140) and (4.142), we have h i 2a Υ(ζ), Υ m(ζ) = div a Υ(ζ), Υ(ζ) V − a Υ(ζ), Υ(ζ) div V +2a Υ(ζ), G(V, DW ) + lo (ζ). (4.143)

Next, we compute a ρ(ζ), ρ m(ζ) . It follows from the expression (4.133) that 2hρ(ζ), ρ m(ζ) i = div (|ρ(ζ)|2 V ) − |ρ(ζ)|2 div V 2

+4hρ(ζ), G(V, ρ(ζ))i + lo (ζ);

h i 2 tr ρ(ζ) tr ρ m(ζ) = div (tr ρ(ζ))2 V − (tr ρ(ζ))2 div V +4 tr ρ(ζ) tr G(V, ρ(ζ)) + lo (ζ).

(4.144)

(4.145)

The formulas (4.144) and (4.145) yield h i 2a ρ(ζ), ρ(m(ζ)) = div a ρ(ζ), ρ(ζ) V − a ρ(ζ), ρ(ζ) div V +4a ρ(ζ), G(V, ρ(ζ) + lo (ζ). (4.146) Finally, the formula (4.139) follows from the identities (4.143) and (4.146).

Now we consider multiplier identities for a displacement vector field to the equations of motion for the shallow shell. Let ζ = (W, w) be a solution to the following problem ζtt − γ(0, ∆wtt ) + Aζ = ϕ

in Ω

(4.147)

where A is the shallow shell operator, given by (4.36), and γ is given in (4.19), respectively and ϕ = (F, f ) denotes an external force acting on the middle surface of the shallow shell. In the case of zero external load, the equation (4.147) is different from the equations (4.111) by a coefficient b of the operator A but by a change of variable they are the same. For simplicity, we study the problem (4.147).



185

Let V be a vector field and let u be a function on Ω. By (2.6), 2hDu, D(V (u))i = 2DV (Du, Du) + div (|Du|2 V ) −|Du|2 div V.

(4.148)

In order to simplify computation in obtaining multiplier identities we have to choose a special vector field V for the multiplier m(ζ). We make the following: Vector field V on Ω satisfies DV (X, X) = ϑ(x)|X|2

for

X ∈ Mx ,

x∈Ω

(4.149)

where ϑ is a function on Ω. In the following L(ζ) denotes the lower order term with respect to the energy level, that is, for any ε > 0 there is Cε > 0 such that 0 ≤ L(ζ) ≤ ε

Z

0

T

E(t)dt + Cε [kl(ζ)(0)k2 + kl(ζ)(T )k2 + kl(ζ)k2L2 (Q) ] (4.150)

where l(ζ) is given by (4.128). Let Q = (0, T ) × Ω,

Σ = (0, T ) × Γ.

Theorem 4.5 Let ζ = (W, w) solve the problem (4.147) and let a vector field V satisfy the assumption (4.149). Then Z Z ∂wtt [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ + 2 [∂(Aζ, m(ζ)) + γV (w) ] dΣ ∂ν Σ Σ Z = 2Z(t)|T0 + 2 [a(Υ(ζ), G(V, DW )) + 2γa(ρ(ζ), G(V, ρ(ζ)))] dQ Q Z + {[|ζt |2 + γ|Dwt |2 − B(ζ, ζ)] div V − 2γDV (Dwt , Dwt )} dQ Q Z − hϕ, m(ζ)i dQ + L(ζ) (4.151) Q

where Z(t) = (ζt , m(ζ))L2 (Ω,Λ)×L2 (Ω) + γ(Dwt , D(V (w)))

(4.152)

and the tensor field G(V, DW ) of rank 2 is defined by (4.131). Proof. We multiply the equation (4.147) by 2m(ζ) and integrate over Q by parts. Letting u = wt in the formula (4.148) yields 2h∇wt , ∇V (wt )i = 2DV (Dwt , Dwt ) − |Dwt |2 div V + div (|Dwt |2 V ).


(4.153)

186


Let P = ζ − γ(0, ∆w). By the formula (4.153), we have 2hPtt , m(ζ)i = 2hζtt , m(ζ)i − 2γ∆wtt V (w) = −2γ div [V (w)∇wtt ] + 2hζtt , m(ζ)i + 2γh∇wtt , ∇V (w)i = −2γ div [V (w)∇wtt ] + 2[hζt , m(ζ)i + γh∇wt , ∇V (w)i]t −2[hζt , m(ζt )i + γh∇wt , ∇V (wt )i] = − div [(|ζt |2 + γ|Dwt |2 )V + 2γV (w)Dwtt ] +(|ζt |2 + γ|Dwt |2 ) div V − 2γDV (Dwt , Dwt )

+2[hζt , m(ζ)i + γh∇wt , ∇V (w)i]t .

Integrating the identity (4.154) over Q = (0, T ) × Ω gives Z 2 hPtt , m(ζ)i dQ = 2Z(t)|T0 Q Z + [(|ζt |2 + γ|Dwt |2 ) div V − 2γDV (Dwt , Dwt )] dQ Q Z − [2γV (w)hDwtt , νi + (|ζt |2 + γ|Dwt |2 )hV, νi] dΣ.

(4.154)

(4.155)

Σ

On the other hand, by the Green formula (4.34) and the formula (4.139) Z Z T 2 hAζ, m(ζ)i dQ = 2 (Aζ, m(ζ))L2 (Ω,Λ)×L2 (Ω) dt Q 0 Z Z =2 B(ζ, m(ζ)) dQ − 2 ∂(Aζ, m(ζ)) dΣ Q Σ Z Z = [B(ζ, ζ)hV, νi − 2∂(Aζ, m(ζ))] dΣ + 2 a(Υ(ζ), G(V, DW )) dQ Σ Q Z + [4γa(ρ(ζ), G(V, ρ(ζ))) − B(ζ, ζ) div V ] dQ + L(ζ). (4.156) Q

Finally, using the formulas (4.147), (4.155), and (4.156), we obtain the identity (4.151). Theorem 4.6 Let ζ = (W, w) solve the problem (4.147) and let q be a function on Ω. Then Z Z q[|Wt |2 − a(Υ(ζ), Υ(ζ))] dQ + hϕ, q(W, 0)i dQ Q Q Z =− ∂(Aζ, q(W, 0)) dΣ + L(ζ) (4.157) Σ

and

Z

Z q[wt2 + γ|Dwt |2 − γa(ρ(ζ), ρ(ζ))] dQ + hϕ, q(0, w)i dQ Q Q Z ∂wtt = − [∂(Aζ, q(0, w)) + γqw ] dΣ + L(ζ). (4.158) ∂ν Σ



187

Proof. Set η1 = (W, 0). We multiply the equation (4.147) by qη1 and integrate it over Q. Let P = ζ − γ(0, ∆w). Then hPtt , qη1 i = qhPt , η1 it − qhPt , η1t i = qhWt , W it − q|Wt |2 .

(4.159)

On the other hand, it is easy to check that D(qW ) = qDW + W ⊗ Dq.

(4.160)

Then 1 Υ(qη1 ) = qΥ(η1 ) + (W ⊗ Dq + Dq ⊗ W ) = qΥ(ζ) + lo (ζ). 2 Thus B(ζ, qη1 ) = a(Υ(ζ), Υ(qη1 )) = qa(Υ(ζ), Υ(ζ)) + lo (ζ). Using the formula (4.162) and the Green formula (4.34), we obtain Z Z hAζ, qη1 i dQ = qa(Υ(ζ), Υ(ζ)) dQ Q Q Z − ∂(Aζ, q(W, 0)) dΣ + L(ζ).

(4.161)

(4.162)

(4.163)

Σ

Then the identity (4.157) follows from the relations (4.163), (4.159), and (4.147). Next, we use a multiplier qη2 where η2 = (0, w). First, we have hPtt , qη2 i = −γ div (qw∇wtt ) + γh∇wtt , ∇(qw)i + qwtt w

= −γ div (qw∇wtt ) + γh∇wt , ∇(qw)it + q(wt w)t −qwt2 + lo (ζ).

(4.164)

It is easy to check that D2 (qw) = qD2 w + Dw ⊗ Dq + Dq ⊗ Dw + wD2 q.

(4.165)

Then ρ(qη2 ) = qρ(ζ) + lo (ζ).

(4.166)

Since Υ(qη2 ) = lo (ζ), the formula (4.166) yields B(ζ, qη2 ) = qγa(ρ(ζ), ρ(ζ)) + lo (ζ)

(4.167)

where the bilinear form B(·, ·) is given by (4.18). By the Green formula (4.34) and the relation (4.167), we obtain Z Z hAζ, qζi dQ = qγa(ρ(ζ), ρ(ζ)) dQ Q Q Z − (Aζ, q(0, w)) dΣ + L(ζ). (4.168) Σ

Finally, the identity (4.158) follows from the formulas (4.164) and (4.168).


188


Theorem 4.7 Let ζ = (W, w) solve the problem (4.147) and let a vector field V satisfy the assumption (4.149). Then Z [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ Σ Z ∂wtt ] dΣ +2 [∂(Aζ, m(ζ)) + γV (w) ∂ν Σ Z = 2Z(t)|T0 + 2 a(Υ(ζ), G(V, DW )) dQ Q Z +2 ϑ(x)[|ζt |2 + γa(ρ(ζ), ρ(ζ)) − a(Υ(ζ), Υ(ζ))] dQ Q Z − hϕ, m(ζ)i dQ + lo (ζ) (4.169) Q

where Z(t) is given by (4.152) and the tensor field G(V, DW ) is defined by (4.131). Proof. Since ρ(ζ) is symmetric, we have from (3.27) and (3.28) that hρ(ζ), G(V, ρ(ζ))i = hρ(ζ), ρ(ζ)(·, D· V )i = ϑ(x)|ρ(ζ)|2 ; tr ρ(ζ) tr G(V, ρ(ζ)) = tr ρ(ζ) tr ρ(ζ)(·, D· V ) = ϑ(tr ρ(ζ))2 , which yields a(ρ(ζ), G(V, ρ(ζ))) = ϑa(ρ(ζ), ρ(ζ)).

(4.170)

Moreover, the assumption (4.149) yields DV (Dwt , Dwt ) = ϑ|Dwt |2

and

div V = 2ϑ.

(4.171)

Using the relations (4.170) and (4.171) in the identity (4.151), we obtain the identity (4.169).

4.5

Escape Vector Field and Escape Region for the Shallow Shell

Some geometric conditions on the middle surface are necessary to obtain continuous observability estimates for a shallow shell. In this section we will introduce such conditions, study their existence, and present some examples. We assume that the ellipticity of the strain energy of the shallow shell holds, that is, there is a λ0 > 0 such that λ0 B(ζ, ζ) ≥ kDW k2L2 (Ω,T 2 ) + γkD2 wk2L2 (Ω,T 2 )


(4.172)


189

for all ζ = (W, w) ∈ H 1 (Ω, Λ) × H 2 (Ω) where the bilinear form B(·, ·) is defined by (4.21). The above inequality is established by Theorem 4.2 in Section 4.2 if displacement vector fields of the shallow shell are fixed along a portion Γ0 of the boundary with a positive length. Geometric Condition for Boundary Control Definition 4.1 A vector field V on Ω is said to be an escape vector field for the shallow shell if the following assumptions hold: (i) There is a function ϑ on Ω such that DV (X, X) = ϑ(x)|X|2

for all

X ∈ Mx ,

x ∈ Ω.

(4.173)

(ii) Let ι(x) = hDV, Ei/2

for

x∈Ω

(4.174)

where E denotes the volume element of the surface M. Suppose that the functions ϑ(x) and ι(x) satisfy the inequality 2 min ϑ(x) > λ0 (1 + µ) max |ι(x)| x∈Ω

x∈Ω

(4.175)

where λ0 is given in (4.172) and µ is Poisson’s coefficient of the material. Remark 4.2 Clearly, if a vector field is escaping for the shallow shell, then it is also escaping for plate, see Section 3.2. Remark 4.3 In Section 4.6 we shall show that existence of an escape vector field for the shallow shell can guarantee controllability/stabilization. Remark 4.4 We shall prove that on the 2-dimensional middle surface Ω there always exists a vector field V satisfying the assumption (4.173), see Theorem 4.8 later. Then the key to being an escape vector field for the shallow shell is the assumption (4.175). By Theorem 2.3, a necessary condition to the assumption (4.175) is that there is no closed geodesics inside the middle surface Ω. Remark 4.5 The function ι(x), given by (4.174), depicts the symmetry of the covariant differential DV of the vector field V in some sense. Clearly, if DV is symmetric, then ι(x) = 0 for all x ∈ Ω. If the shell is flat, a plate, then M = R2 . For any x0 ∈ R2 , set V (x) = x−x0 . It is easy to check that DV = g

and ι(x) = 0.

Then V = x − x0 is an escape vector field for the flat shell where ϑ(x) = 1 and ι(x) = 0. We shall show that if M is a surface of R3 that is of constant curvature or of revolution, then there exists a vector field V on the whole M with ι(x) = 0


190


for all x ∈ M, see Theorems 4.9 and 4.10 below. We will also present in such cases curvature conditions for existence of an escape vector field for the shallow shell, see Corollaries 4.3 and 4.4 below. First, we have Lemma 4.9 Let V ∈ X (Ω) satisfy the assumption (4.173). Then the twoorder tensor field DV can be decomposed as DV = ϑ(x)g + ι(x)E

for

x ∈ Ω.

(4.176)

Proof. We decompose DV into a sum of the symmetric part and the skew part as 1 1 DV = (DV + D∗ V ) + (DV − D∗ V ). (4.177) 2 2 Since the space Λ2 (M ) of 2-forms on the surface M is one-dimensional there is a function q(x) on Ω such that (DV − D∗ V )/2 = q(x)E

for x ∈ Ω.

(4.178)

Using the formula hD∗ V, Ei = hD, E ∗ i = −hDV, Ei and the formulas (4.174) and (4.178), we obtain q(x) = q(x)hE, Ei/2 = (hDV, Ei − hD∗ V, Ei)/2 = hDV, Ei/2 = ι(x)

(4.179)

since hE, Ei = 2. Finally the formula (4.176) follows from (3.26), (4.179), (4.178) and (4.177). Consider a set Θ which consists of all the C ∞ functions q such that there is a region ℵ ⊂ M satisfying that Ω ⊂ ℵ and ∆q = κ(x)

for all x ∈ ℵ

(4.180)

where κ is the Gaussian curvature function of the surface M. It is easily checked that Θ is nonempty by an elliptic boundary value problem for the Laplacian since Ω ⊂ M is a bounded region with a nonempty boundary Γ. For a given q ∈ Θ, consider a metric, given by gˆ = e2q g

(4.181)

where g is the induced metric of M in R3 . Denote by (ℵ, gˆ) the Riemannian manifold with the metric (4.181) and by ρ(x) ˆ the distance function on (ℵ, gˆ) ˆ be the covariant differential of (ℵ, gˆ) from x0 ∈ ℵ to x ∈ ℵ, respectively. Let D in the metric gˆ. The following theorem shows that there always exists a vector field satisfying the assumption (4.173) on the whole middle surface Ω and this vector field is locally escaping for the shallow shell.



191

Theorem 4.8 Let q ∈ Θ and x0 ∈ ℵ be given. Let ˆ V (x) = ρˆ(x)Dρ(x)

for

x ∈ ℵ.

(4.182)

Then the vector field V satisfies the assumption (4.173) where ϑ = 1 − V (q) and ι(x) = hDq ⊗ V, Ei such that lim ϑ(x) = 1

x→x0

and

lim ι(x) = 0.

x→x0

(4.183)

Locally, V is an escape vector field for the shallow shell. Proof. We compute in a coordinate. Let (x1 , x2 ) be a coordinate on ℵ. Set gij = h

∂ ∂ ∂ ∂ , i and gîj = gˆ( , ). ∂xi ∂xj ∂xi ∂xj

(4.184)

for 1 ≤ i, j ≤ 2. It follows from (4.181) that gîj = e2q gij . ˆ k the coefficients of connections D and D, ˆ reDenote by Γkij and by Γ ij spectively. As a computation in (3.34) in Chapter 3, we have the following formulas 2 X ˆ k − δik qxj − δjk qxi + gij g kl qxl , (4.185) Γkij = Γ ij l=1

ˆ X V − V (q)X − X(q)V + hX, V iDq. DX V = D

(4.186)

From the formulas (4.185), we obtain ∆q − κ + κ ê2q = 0 in ℵ

(4.187)

where κ ˆ is the Gaussian curvature function of (ℵ, gˆ) in the metric gˆ and −1 g ij = gij . The relations (4.181) and (4.187) imply that κ ˆ=0

for x ∈ ℵ,

(4.188)

that is, the Riemannian manifold (ℵ, gˆ) has zero curvature. It follows from Theorem 3.5 that DV (X, X) = hDX V, Xi = [1 − V (q)]|X|2 .

(4.189)

The assumption (4.173) is true for the vector field V where ϑ(x) = 1 − V (q). Let x ∈ ℵ be given. Let e1 , e2 be an orthonormal basis of Mx with the positive orientation. From the formulas (4.186) and (4.181), we have ˆ e2 V, e1 i + e1 (q)hV, e2 i − e2 (q)hV, e1 i DV (e1 , e2 ) = hDe2 V, e1 i = hD = e−2q gˆ(e1 , e2 ) + e1 (q)hV, e2 i − e2 (q)hV, e1 i = e1 (q)hV, e2 i − e2 (q)hV, e1 i;


(4.190)

192

Modeling and Control in Vibrational and Structural Dynamics DV (e2 , e1 ) = e2 (q)hV, e1 i − e1 (q)hV, e2 i.

(4.191)

It follows from the formulas (4.190) and (4.191) that ι(x) = hDV, Ei/2 = DV (e1 , e2 ) − DV (e2 , e1 ) /2 = hDq ⊗ V, Ei. Finally, from the relations (4.182) and (4.181), we have |V | = e−q |V |gˆ = ρˆ(x)e−q so that the limits in (4.183) follow.

Remark 4.6 For a detailed computation for the formula (4.187), we refer to Chapter 5 of the book [187]. Theorem 4.9 Let a surface M of R3 be of constant curvature κ. For x0 ∈ M given, let ρ be the distance function from x ∈ M to x0 ∈ M on (M, g), i.e., ρ(x) = dg (x0 , x). Set V = h(ρ)Dρ where h is defined by  √  sin κρ, κ > 0, κ=0 h(ρ) = ρ, √ for x ∈ expx0 (Σ(x0 )) (4.192)  sinh −κρ, κ < 0, where expx0 (Σ(x0 )) is the interior of the cut locus of the point x0 . Then DV = ϑ(x)g where

and

ι(x) = 0

for

x ∈ expx0 (E(x0 ))

√ √  κ cos κρ(x), κ > 0, ϑ(x) = 1, κ = 0,√ √ −κ cosh −κρ(x), κ < 0.

Proof. It follows from Theorem 1.21 that √ √  κ cot κρ(x), κ > 0, 1 , κ = 0, D2 ρ(τ, τ ) = ρ(x) √ √ −κ coth −κρ(x), κ < 0,

(4.193)

(4.194)

(4.195)

where τ ∈ Mx is such that τ, Dρ forms an orthonormal basis of Mx and D2 ρ is the Hessian of the distance ρ. Since D2 ρ(Dρ, X) = 0 for any X ∈ Mx , it follows from the formula (4.195) that DV (X, Y ) = h′ (ρ)X(ρ)Y (ρ) + h(ρ)D2 ρ(X, Y ) = h′ (ρ)hX, DρihY, Dρi + h(ρ)D2 ρ(τ, τ )hX, τ ihY, τ i = ϑ(x)hX, Y i. Moreover, the symmetry of DV implies ι(x) = 0.



193

Remark 4.7 It should be noted that, even if the surface M has zero Gaussian curvature, the vector field V, defined in Theorem 4.9, is only escaping on the open set expx0 Σ(x0 ). It may not be escaping on the whole surface M. For example, let M be a cylindrical surface. Then M has zero Gaussian curvature. Since a big circle on the cylinder is a closed geodesic for the cylinder, a necessary condition, Theorem 2.3, implies that there is no escape vector field on the middle surface Ω if Ω contains a big circle. In fact it is easy to show that for any x0 ∈ M, expx0 Σ(x0 ) 6= M. The following corollaries are immediate from Theorem 4.9. Corollary 4.2 Let M be a cylinder and let Ω ⊂ M be a middle surface. If there is a generator of the cylinder which does not intersect Ω, then there exists an escape vector field for the shallow shell. Corollary 4.3 Let M be a surface in R3 with constant curvature κ which is simply connected. If κ < 0, then for any middle surface Ω ⊂ M there exists an escape vector field for the shallow shell. Let κ > 0. If there is a geodesic ball √B(x0 , δ) ⊂ M centered at some point x0 ∈ M with radius 0 < δ < π/(2 κ) such that the middle surface Ω ⊂ B(x0 , δ), then an escape vector field for the shallow shell exists. Theorem 4.10 Let f be a function on R. Consider a surface of revolution, given by p M = { (x, y, z) | z = f (r), r = x2 + y 2 , (x, y) ∈ R2 }. (4.196)

The Gaussian curvature is f ′ (r)f ′′ (r) κ(p) = r(1 + f ′2 )2

for

p = (x, y, z) ∈ M.

(4.197)

Then there exist a vector field V and a function ϑ such that DV = ϑ(p)g

and

ι(p) = 0

for

p∈M

where g is the induced metric of M in R3 . Furthermore, if Z tκ(t) dt ≤ 1,

(4.198)

(4.199)

κ(t)>0

then ϑ(p) > 0 where

for all

p ∈ M,

f ′ (r(t))f ′′ (r(t)) for t > 0, r(t)(1 + f ′2 (r(t)))2 and the function r(t) is given by the equation Z r(t) p t= 1 + f ′2 (s) ds for t ≥ 0. κ(t) =

0


(4.200) (4.201)

(4.202)

194


Proof. Denote p0 = (0, 0, f (0)) ∈ M and denote by ρ(p) = dg (p, p0 ) the distance function in the induced metric g from p ∈ M to p0 . Let p = (x, y, f (r)) ∈ M where r2 = x2 + y 2 . Since M is of revolution, it is easily checked that a curve : [0, ρ(p)] → M, given by x y σ(t) = r(t) , r(t) , f (r(t) (4.203) r r

is a unique minimizing geodesic, parameterized by arc length, which joins p0 to p. Let τ0 ∈ M0 be such that σ(0), ˙ τ0 forms an orthonormal basis of M0 . Let τ = τ (t) ∈ Mσ(t) be the parallel transport of τ0 along the geodesic σ(t) for t > 0. By Lemma 1.5, there is a normal Jacobi field J(t) = h(t)τ (t) along σ such that ′′ h (t) + κ(t)h(t) = 0 t > 0, (4.204) h(0) = 0, h′ (0) = 1, and D2 ρ(τ, τ ) =

h′ (ρ) h(ρ)

for

ρ > 0.

(4.205)

Set V = h(ρ)Dρ. We shall prove that DV = h′ (ρ)g for all p ∈ M. By (4.205), we obtain DV (X, Y ) = h′ (ρ)X(ρ)Y (ρ) + h(ρ)D2 ρ(X, Y ) = h′ (ρ)hX, DρihY, Dρi + h(ρ)D2 ρ(τ, τ )hX, τ ihY, τ i = h′ (ρ)hX, Y i

where the formula D2 ρ(Dρ, X) = 0 for X ∈ Mp is used. Let the assumption (4.199) hold. We will prove that h′ (t) > 0 for all t ≥ 0. Set κ ˜(t) = max(κ(t), 0). Then the assumption (4.199) means that Z ∞ t˜ κ(t) dt ≤ 1. (4.206) 0

Let φ be the solution to the problem ′′ φ (t) + κ ˜ (t)φ(t) = 0 t > 0, φ(0) = 0, φ′ (0) = 1.

(4.207)

It follows from the above equations that φ(t) > 0

and φ′ (t) > 0 on (0, ∞).

(4.208)

Then Sturm’s theorem implies that h(t) ≥ φ(t) > 0 since κ ˜ (t) ≥ κ(t) for t > 0.


on (0, ∞)

(4.209)


195

Set u(t) = h′ (t)φ(t) − h(t)φ′ (t). By the formulas (4.204), (4.207), and (4.209), we obtain u′ (t) = (˜ κ − κ)φh ≥ 0

on (0, ∞)

which yields u(t) ≥ u(0) = 0, that is, by (4.208) and (4.209), h′ (t) ≥ h(t)

φ′ (t) >0 φ(t)

on (0, ∞).

From Theorem 4.10, we have Corollary 4.4 Let a surface M of revolution be given by (4.196). If the curvature condition (4.199) holds true, then for any middle surface Ω ⊂ M, there exists an escape vector field for the shallow shell. Geometric Condition for Internal Control Next, we consider an open region of the middle surface where a control action will take place. This control action may be for exact controllability in the open-loop case, or for stabilization by an interior feedback. We seek geometric conditions on such regions G. Clearly, one of the choices is G = Ω but this is trivial. We are particularly interested in the case where G is not the whole middle surface and is as small as possible in some sense. The goal of the following treatment is to try to choose a control region small. Definition 4.2 A region G ⊂ Ω is said to be an escape region for the shallow shell if the following assumptions hold: There are finite subregions Ωi ⊂ Ω with boundaries Γi for 1 ≤ i ≤ J where J is some natural number such that (i) Ωi ∩ Ωj = ∅ for all 1 ≤ i < j ≤ J; (4.210) (ii) for each Ωi there is an escape vector field Vi such that DVi (X, X) = ϑi (x)|X|2

Ωi ,

(4.211)

2 min ϑi (x) > λ0 (1 + µ) max ιi (x),

(4.212)

x∈Ωi

on x∈Ωi

where ιi (x) = hDVi , Ei/2 for all 1 ≤ i ≤ J; (iii) h i G ⊃ Ω ∩ Nε ∪Ji=1 Γi0 ∪ Ω/ ∪Ji=1 Ωj

(4.213)

where ε > 0 small and

Nε (S) = ∪x∈S { y ∈ Ω | dg (y, x) < ε} νi

for

S ⊂ M,

Γi0 = { x ∈ Γi hVi (x), νi (x)i > 0 },

are normals of Ωi pointing outside of Ωi ,

for all 1 ≤ i ≤ J


(4.214)

196


Remark 4.8 If there exists an escape vector field V on the whole middle surface Ω for the shallow shell, then the relations (4.211) and (4.212) hold true for V1 = V. In this case, we can take J = 1 and Ω1 = Ω. An escape region G needs only to be supported in a neighborhood of Γ0 , where Γ0 = { x ∈ ∂Ω | hV (x), n(x)i > 0 }.

(4.215)

This roughly means that we can dig out almost the whole Ω from the domain Ω and that the escape region is supported only in a neighborhood of a subset of the boundary. The following theorem is clearly true. Theorem 4.11 If there exists an escape vector field V for the shallow shell, then an escape region can be supported in a neighborhood of Γ0 where Γ0 is given by (4.215). In general an escape vector field for the shallow shell does not exist on the whole middle surface Ω. However, by Theorem 4.8, on a geodesic ball with a small radius an escape vector field exists. Then we can dig out from Ω as many as possible geodesic balls with radii small and let an escape region be supported in a neighborhood of the remaining. Therefore the following results are immediate. Theorem 4.12 For ε > 0 given, we can choose an escape region G ⊂ Ω with meas (G) < ε where meas (G) is the 2-dimensional Lebesgue measure of G. Here we give some examples verifying existence of an escape vector field for the shallow shell and showing the structure of an escape region. Example 4.1 Let a surface M be a cylinder given by M = { (x, y, z) | x2 + y 2 = 1, z ∈ R }. Let Φ(x, y) = { (x, y, z) ∈ M | z ∈ R }

be a generatrix of M for (x, y) fixed with x2 + y 2 = 1. (a) Let a middle surface Ω ⊂ M be such that there is (x0 , y0 ) ∈ R2 with x20 + y02 = 1 satisfying Ω ⊂ M/Φ(x0 , y0 ). Let p0 = (−x0 , −y0 , 0) ∈ M and let ρ(p) = dg (p, p0 ) be the distance function from p ∈ M to p0 . Clearly, for any p ∈ M/Φ(x0 , y0 ) there is a unique minimizing geodesic on M/Φ(x0 , y0 ) connecting p and p0 . Then M/Φ(x0 , y0 ) ⊂ expp0 Σ(p0 ). Let V = ρDρ. By Theorem 4.9, the relations (4.173) and (4.175) hold true. This means that there exists an escape vector field for the shallow shell.



197

(b) Let a middle surface Ω be a segment of the cylinder, given by Ω = { (x, y, z) | x2 + y 2 = 1, z ∈ (−1, 1) } with boundary Γ = { (x, y, −1) | x2 + y 2 = 1 } ∪ { (x, y, 1) | x2 + y 2 = 1 } Since big circles ω(z0 ) = { (x, y, z0 ) | x2 + y 2 = 1 } ⊂ Ω are closed geodesics for all z0 ∈ (−1, 1), an escape vector field on the whole middle surface Ω does not exit, by Theorem 2.3. We consider the structure of an escape region for this middle surface. Let p0 = (0, −1, 0) ∈ Ω. Let Ω1 = Ω/Φ(0, 1) and V1 = ρDρ. By Theorem 4.9, the relations (4.211) and (4.212) hold true with J = 1. Thus an escape region of the middle surface can be supported in a neighborhood of Γ ∪ Φ(0, 1), see Figure 4.1.

G

FIGURE 4.1: An escape region for a cylinder. Example 4.2 Let M be the unit sphere, given by M = { (x, y, z) | x2 + y 2 + z 2 = 1 }. For p ∈ M and δ > 0, let B(p, δ) = { q | q ∈ M, dg (p, q) < δ } be the geodesic ball centered at p with radius δ. (a) Let a middle surface Ω ⊂ M be a spherical cap such that there is p ∈ M satisfying Ω ⊂ B(p, π/2).


198


By Theorem 4.9, there exists an escape vector field for the shallow shell. (b) Let Ω = M be a middle surface without boundary. Then by Theorem 2.3 there is no escape vector field on the whole Ω. We consider the structure of an escape region of Ω. Let ω = { (x, y, 0) | x2 + y 2 = 1 }. Then ω is a big circle on Ω. For ε > 0 given small, let Ω1 = B (0, 0, 1), π/2 − ε and Ω2 = B (0, 0, −1), π/2 − ε .

By Theorem 4.9, there are escape vector fields Vi satisfying the relations (4.211) and (4.212) with i = 1, 2. Thus an escape region of Ω can be supported in a neighborhood of the big circle ω, see Figure 4.2.

O

G

FIGURE 4.2: An escape region for a closed spherical shell. Example 4.3 Let a middle surface Ω be a portion of a surface M of revolution, given by M = { (x, y, z) | z = log(1 + x2 + y 2 ), (x, y) ∈ R2 }.

(4.216)

Set f (r) = log(1 + r2 ). It is easy to check that the Gaussian curvature is κ(p) = −

2(1 + r2 ) 0 } is empty. By Corollary 4.4, an escape vector field for the shallow shell exists for any middle surface Ω ⊂ M. Example 4.4 Consider a helicoid, defined by M = { α(t, s) | (t, s) ∈ R2 , t > 0 }


(4.218)


199

where α(t, s) = (t cos s, t sin s, c0 s), −c20 /(t2

c0 > 0.

(4.219)

c20 )2 .

The Gaussian curvature is + We set E1 = αt = (cos s, sin s, 0); q q E2 = αs / t2 + c20 = (−t sin s, cos s, c0 )/ t2 + c20 .

Then E1 , E2 forms a frame field on the whole surface M. We obtain DE1 E1 = 0, DE1 E2 = 0,

DE2 E1 = tE2 /(t2 + c20 ), DE2 E2 = −tE1 /(t2 + c20 ),

where D is the Levi-Civita connection of the surface M. For any c > 0, set Vc = fc E1 + hE2 where fc =

q Z t t2 + c20 (t2 + c20 )−1/2 dt + c ,

h=

0

q t2 + c20 s.

A computation yields DVc = ϑc (x)g + ι(x)E where ϑc = 1 + tfc /(t2 + c20 )

and

for

c>0

q ι(x) = −st/ t2 + c20 .

Clearly, for any bounded middle surface Ω ⊂ M with Ω ⊂ M and for any c1 > 0, there is c > 0 large enough such that min ϑc (x) ≥ c1 max ι(x). x∈Ω

x∈Ω

Then an escape vector field for the shallow shell exists.

4.6

Observability Inequalities. Exact Controllability

In this section we shall derive some observability estimates of the shallow shell from boundary and also from interior. Those estimates are necessary both for exact controllability and stabilization of the shallow shell. Suppose that the ellipticity (4.172) of the strain energy of the shallow shell


200


holds. Observability Estimate from Boundary Let V be an escape vector field for the shallow shell with functions ϑ and ι such that the assumptions (4.173) and (4.175) hold. Set σ0 = max |V |, x∈Ω

σ1 = min ϑ(x) − λ0 (1 + µ) max |ι(x)|/2. x∈Ω

x∈Ω

(4.220)

For T > 0 given, set Q = Ω × (0, T ), Σ0 = Γ0 × (0, T ), 1

2

Σ = Γ × (0, T ), Σ1 = Γ1 × (0, T ).

Let ζ = (W, ζ) ∈ H (Ω, Λ) × L (Ω) be a solution to the following problem ζtt − γ(0, ∆wtt ) + Aζ = 0 in Ω

(4.221)

where A is the shallow shell operator, given by (4.36), and γ is given in (4.19), respectively. We define, for (t, x) ∈ Σ, SB = [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi ∂w tt +∂ Aζ, 2m(ζ) + (̺W, −ϑw) + γ 2V (w) − ϑw ∂ν

(4.222)

̺ ≥ λ0 (1 + µ) max |ι(x)|/2.

(4.223)

where m(ζ) = (DV W, V (w)) and ̺ = 2ϑ − σ1 . By (4.220) x∈Ω

We need the following two lemmas in order to obtain observability estimates. Lemma 4.10 Let Z(t) be given by (4.152). Then |Z(t)| ≤ σ0 λ0 E(t) + L(ζ)

(4.224)

where L(ζ) are the lower terms satisfying the estimate (4.150). Proof. It is easy to check that |D(V (w))| ≤ |D2 w||V | + c|Dw| ≤ σ0 |D2 w| + c|Dw|. We have 2hζt , m(ζ)i + 2γhDwt , D(V (w)i ≤ 2σ0 |Wt ||DW | + 2σ0 |wt ||Dw| + 2γ|Dwt ||D(V (w))| ≤ σ0 e(ζ) + lo (ζ)

(4.225)

for ε > 0 small. The inequality (4.224) follows from the ellipticity (4.172) and the formula (4.225). In the following lemma the assumption (4.175) plays a key role.



201

Lemma 4.11 Let V be an escape vector field for the shallow shell. Then Z Z σ1 a Υ(ζ), Υ(ζ) dx ≤ a Υ(ζ), G(V, DW ) dx + CL(ζ) (4.226) Ω

Ω

where Υ(ζ) is the strain tensor of the middle surface, given by (4.15), a(·, ·) is a bilinear form, given by (4.20), the tensor field G(V, DW ) of rank 2 is defined by (4.131) and σ1 > 0 is given in (4.220). Proof. Given x ∈ Ω. We compute at x. Since DW + D∗ W is symmetric, there is an orthnonrmal basis e1 , e2 of Mx with the positive orientation such that DW (e1 , e2 ) + DW (e2 , e1 ) = 0 at x

(4.227)

which yields Υ(ζ)(e1 , e2 ) = lo (ζ)

at x.

(4.228)

On the other hand, the formula (4.176) in Lemma 4.9 implies that De1 V = hDe1 V, e1 ie1 + hDe1 V, e2 ie2 = DV (e1 , e1 )e1 + DV (e2 , e1 )e2 = ϑ(x)e1 − ι(x)e2

at x

(4.229)

since E(e2 , e1 ) = −1. Similarly, we have De2 V = ι(x)e1 + ϑ(x)e2

at x.

(4.230)

Set Wij = DW (ei , ej ) at x for 1 ≤ i, j ≤ 2. Using the formulas (4.131), (4.229) and (4.230), we obtain G(V, DW )(e1 , e1 ) = DW (e1 , De1 V ) = ϑ(x)W11 − ι(x)W12 ; G(V, DW )(e2 , e2 ) = ι(x)W21 + ϑ(x)W22

at x.

(4.231) (4.232)

It follows from the formulas (4.228), (4.231), (4.232) and (4.227) that hΥ(ζ), G(V, DW )i = Υ(e1 , e1 )G(V, DW )(e1 , e1 ) + Υ(e2 , e2 )G(V, DW )(e2 , e2 ) + lo (ζ) 2 2 = ϑ(x)(W11 + W22 ) + ι(x)(W11 + W22 )W21 + lo (ζ) 2 = ϑ(x)|Υ(ζ)| + ι(x)(W11 + W22 )W21 + lo (ζ)

(4.233)

at x. Similarly, we obtain 2 tr Υ(ζ) tr G(V, DW ) = ϑ(x) tr Υ(ζ) +2ι(x)(W11 + W22 )W21 + lo (ζ)


at x.

(4.234)

202


From the relations (4.233), (4.234) and (4.20), we have a Υ(ζ), G(V, DW ) = ϑ(x)a Υ(ζ), Υ(ζ) + lo (ζ)

+(1 + µ)ι(x)(W11 + W22 )W21 ≥ ϑ(x)a Υ(ζ), Υ(ζ) − (1 + µ)|ι(x)||DW |2 /2 + lo (ζ) ≥ min |ϑ(x)|a Υ(ζ), Υ(ζ) x∈Ω

1+µ − max |ι(x)||DW |2 + lo (ζ). 2 x∈Ω

(4.235)

Furthermore, letting ζ = (W, 0) in the ellipticity assumption (4.172) yields Z Z |DW | dx ≤ λ0 a Υ(ζ), Υ(ζ) dx (4.236) Ω

Ω

Finally, using the escaping assumption (4.175) and the inequality (4.236), we integrate the inequality (4.235) over Ω to have the estimate (4.226). First, we have the estimate from boundary Theorem 4.13 Let V be an escape vector field for the shallow shell. Let ζ = (W, w) ∈ H 1 (Ω, Λ) × L2 (Ω) solve the problem (4.221) such that SB |Σ is finite where SB |Σ is defined by (4.222). Let T > 0 be given. Then 2σ1

Z

0

T

E(t) dt ≤

Z

SB dΣ + σ0 λ0 [E(0) + E(T )] + CT L(ζ)

(4.237)

Σ

where λ0 > 0 is the ellipticity number given by (4.172), σ0 > 0 and σ1 > 0 are given in (4.220), and SB is given by (4.222). Proof. Set Ψ = 2[σ1 − ϑ(x)]a(Υ(ζ), Υ(ζ)) + 2ϑ(x)[|ζt |2 + γa(ρ(ζ), ρ(ζ))] where σ1 is given in (4.220). Then Ψ = σ1 [|ζt |2 + γ|Dwt |2 + B(ζ, ζ)] +(2ϑ − σ1 )[|ζt |2 + γa(ρ(ζ), ρ(ζ))]

+(σ1 − 2ϑ)a(Υ(ζ), Υ(ζ)) − σ1 γ|Dwt |2 = σ1 e(ζ) + (2ϑ − σ1 )[|Wt |2 − a(Υ(ζ), Υ(ζ))] +(ϑ − σ1 )[wt2 + γ|Dwt |2 + γa(ρ(ζ), ρ(ζ))]

+ϑ(x)[γa(ρ(ζ), ρ(ζ)) − γ|Dwt |2 − wt2 ] + 2ϑwt2 where e(ζ) is defined by (4.125).


(4.238)


203

We use the identity (4.169) where ϕ is set zero and obtain, via the formula (4.226) and the relation (4.238), Z Z ∂wtt [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ + 2 [∂(Aζ, m(ζ)) + γV (w) ] dΣ ∂ν Σ Σ Z = 2Z(t)|T0 + 2 a(Υ(ζ), G(V, DW )) dQ Q Z +2 ϑ(x)[|ζt |2 + γa(ρ(ζ), ρ(ζ)) − a(Υ(ζ), Υ(ζ))] dQ + lo (ζ) Q Z ≥ 2Z(t)|T0 + Ψ dQ + lo (ζ) Q

≥

2Z(t)|T0

+

Z

Q

+ 2σ1

Z

T

E(t) dt +

0

Z

Q

(2ϑ − σ1 )[|Wt |2 − a(Υ(ζ), Υ(ζ))] dQ

ϑ(x)[γa(ρ(ζ), ρ(ζ)) − γ|Dwt |2 − wt2 ] dQ + lo (ζ)

(4.239)

since minx∈Ω ϑ(x) > σ1 . Next, we let q = 2ϑ(x) − σ1 in the identity (4.157) and let q = ϑ(x) in the identity (4.158), respectively, and have Z (2ϑ − σ1 )[|Wt |2 − a(Υ(ζ), Υ(ζ))] dQ Q Z =− ∂(Aζ, ̺(W, 0)) dΣ + lo (ζ) (4.240) Σ

and

Z

ϑ[wt2 + γ|Dwt |2 − γa(ρ(ζ), ρ(ζ))] dQ Z ∂wtt = − [∂(Aζ, ϑ(0, w)) + γϑw ] dΣ + lo (ζ). ∂ν Σ Q

(4.241)

Finally, we insert the identities (4.240) and (4.241) into the inequality (4.239), use the estimate (4.224), and obtain the inequality (4.237). In order to absorb the lower order terms in the estimate (4.237) through a compactness-uniqueness argument as in Lemma 2.5, we need the following uniqueness result on a the shallow shell. ˆ ⊂ Γ be relatively open. Proposition 4.3 Let λ be a complex number and let Γ Suppose that ζ = (W, w) solves the problem λ2 ζ − λ2 γ(0, ∆w) + Aζ = 0 subject to boundary conditions  ˆ  W = DW = 0 on Γ, ∂w ∂∆w w = = ∆w = =0 ∂ν ∂ν © 2011 by Taylor & Francis Group, LLC

on

on

Ω

ˆ Γ.

(4.242)

(4.243)

204


Then W =w=0

on

Ω.

(4.244)

Proof. It will suffice to show that the equation (4.242) is equivalent to a system of three equations of the fourth order differential operators with the same principal part ∆2 where ∆ is the Laplacian on the surface M in the induced metric g and the boundary conditions (4.243) yield the zero Cauchy ˆ for this system. Therefore, this proposition is covered by [188]. data on Γ Denote by LF (ζ) all the terms for which there is a constant C > 0 such that 3 X | LF (ζ)| ≤ C (|Di W | + |Di w|) on Ω. (4.245) i=1

From the formulas (4.36) and (4.37) the first equation of the system (4.242)

is

1 − µ

δd + dδ W = (κ − µκ + λ2 )W + H(w)

on Ω (4.246) 2 where H(w) is given in (4.38). Applying the operator dδ to both the sides of the equation (4.246), we have, via Theorem 1.25, dδdδW = LF (ζ)

on Ω.

(4.247)

δdδdW = LF (ζ)

on Ω.

(4.248)

on Ω

(4.249)

Similarly, Then the relations (4.247) and (4.248) yield ∆2 W = LF (ζ)

where ∆ = dδ + δd is the Hodge-Laplacian. Let (x1 , x2 ) be a coordinate on M and let W = (w1 , w2 ) in this coordinate. It is easy to check that ∆2 W = (∆2 w1 , ∆2 w2 ) + LF (w1 , w2 ). Then in this coordinate the system (4.242) is equivalent to a problem  2  ∆ w1 = LF (w1 , w2 , w), ∆2 w2 = LF (w1 , w2 , w), (4.250)  2 ∆ w = LF (w1 , w2 , w). To complete the proof, we have to show that the boundary conditions in ˆ From (4.243) mean that the system (4.250) has the zero Cauchy data on Γ. the boundary conditions in (4.243) it will suffice to prove D2 W |Γˆ = D3 W |Γˆ = 0.

(4.251)

ˆ be given. Let E1 , E2 be a frame field normal at x such that Let x ∈ Γ



205

E1 (x) = ν(x) and E2 (x) = τ (x) where ν(x) and τ (x) are respectively the normal and the tangential of the boundary point x. Then Dν(x) Ei = Dτ (x) Ei = 0

at x.

(4.252)

Since the vector τ is in a tangential direction along boundary, the boundary conditions (4.243) imply τ DW (Ei , Ej ) = 0 at x. (4.253) By (4.253) and (4.252), we obtain at x D2 W (Ei , Ej , τ ) = τ DW (Ei , Ej ) − DW (Dτ Ei , Ej ) −DW (Ei , Dτ Ej ) = 0.

(4.254)

Next, we shall prove D2 W (Ei , Ej , ν) = 0

at x.

(4.255)

If Ej = τ, then the formula (4.254) and Theorem 1.10 yield D2 W (Ei , τ, ν) = D2 W (Ei , ν, τ ) + Rτ ν W = 0

at x.

(4.256)

Therefore, to get (4.255) true, what remains is to prove D2 W (Ei , ν, ν) = 0

at x.

(4.257)

Using the formulas (1.117), (1.119), (4.254), and (4.256), we have at x X X δdW = − DEi DEi W + hDEj DEi W, Ej iEi i

=− =−

X

ij

2

D W (Ej , Ei , Ei )Ej +

ij

X

D2 W (Ej , Ei , Ej )Ei

ij

X

D W (Ej , ν, ν)Ej +

X

D2 W (Ej , Ej , Ei )Ei = −

2

j

X

D2 W (ν, Ei , ν)Ei

i

2

= −D W (τ, ν, ν)τ

(4.258)

and dδW = −

ij

2

= −D W (ν, ν, ν)ν.

X

D2 W (Ej , Ej , ν)ν

j

(4.259)

Using the relations (4.258), (4.259), and (4.246), one has 1 − µ 1−µ 2 − D W (τ, ν, ν)τ − D2 W (ν, ν, ν)ν = δd + dδ W = 0 2 2


206


at x, that is, the formula (4.257) holds true. ˆ A similar computation produces D3 W = 0 on Γ.

We now go back to boundary control. Control in Fixed Boundary Condition Let Γ = ∂Ω 6= ∅ with Γ = Γ0 ∪ Γ1 and Γ0 ∩ Γ1 = ∅. Consider the Dirichlet control problem in unknown ̟ = (Φ, φ)   ̟tt − γ(0, ∆φtt ) + A̟ = 0 on Q,     ̟(0) = ̟0 , ̟t (0) = ̟1 on Ω,    Φ|Γ = 0, Φ|Γ = U 0 < t < T, 1 0 (4.260) ∂φ  |Γ1 = 0 0 < t < T, φ|Γ1 =   ∂ν    ∂φ   φ|Γ0 = u, |Γ = v 0 < t < T, ∂ν 0 with a control action (U, u, v) on the portion Γ0 of the boundary. The question is to find some constant T0 > 0 such that for T > T0 , the following steering property of (4.260) holds true: for all initial data in an appropriate Hilbert space which will be specified later, there exists a suitable control function (U, u, v) such that the corresponding solution of (4.260) satisfies ̟(x, T ) ≡ 0,

̟t (x, T ) ≡ 0.

Exact controllability of (4.260) will be carried out by the dual method. The dual version of (4.260) in unknown ζ = (W, w) follows as  ζtt − γ(0, ∆wtt ) + Aζ = 0 on Q     ζ(0) = ζ0 , ζt (0) = ζ1 on Ω, (4.261) W = 0 on Σ,   ∂w   w= = 0 on Σ. ∂ν Remark 4.9 In the flat case, for the normal component, one control function ∂φ ∂ν = v is enough; see [109]. We here add another control function φ = u for the problem (4.260) in order to avoid the following uniqueness assumption: The problem  2 2   λ ζ − λ γ(0, ∆w) + Aζ = 0 on Ω, W = Dν W = 0 on Γ, (4.262)   w = ∂w = ∆w = 0 on Γ ∂ν admits the unique zero solution. The above uniqueness result does not fall into a class of systems to which the Holmgren theorem may be applied even if the coefficients are analytic since it is not the Cauchy problem for component w (it does for component W ). For the flat case, it has been proved in [109]. Lemma 4.12 Let ζ = (W, w) be a solution to the Dirichlet problem (4.261). Then E(t) = E(0) for t ≥ 0 (4.263)


Linear Shallow Shells. Modeling, and Control where E(t) is defined by (4.126); Z Z (|ζt |2 + γ|Dwt |2 ) dQ = B(ζ, ζ) dQ + lo (ζ); Q

207

(4.264)

Q

T E(0) = 2

Z

B(ζ, ζ) dxdt + lo (ζ);

(4.265)

Q

where B(·, ·) is the bilinear from, given in (4.18). Moreover,    B1 (W, w) = DW (ν, ν),  B2 (W, w) = (1 − µ)DW (τ, ν)/2,  B3 w = 0,   B4 w = 0

on

Γ

(4.266)

where Bi are given by (4.39) and

B(ζ, ζ) = (DW (ν, ν))2 + (1 − µ)(DW (τ, ν))2 /2 + γ(∆w)2 , SB = B(ζ, ζ)hV, νi

(4.267) (4.268)

for x ∈ Γ where SB is defined by (4.222). Proof. It is easy to check that E ′ (t) = 0, that is, E(t) = E(0). We let q = 1 and ϕ = 0 in the identities (4.157) and (4.158), respectively, and then add them up to obtain the formula (4.264), via the boundary conditions (4.261). The formula (4.265) follows from the formulas (4.263) and (4.264). ∂w Moreover, the conditions W |Γ = 0 and w|Γ = |Γ = 0 imply that on Γ, ∂ν Dτ W = 0, and δW = −DW (ν, ν), Υ(ζ)(ν, τ ) = DW (τ, ν)/2, D2 w(ν, τ ) = 0,

Υ(ζ)(ν, τ ) = DW (ν, ν), Dw = 0,

D2 w(τ, τ ) = 0,

∆w = D2 w(ν, ν),

∂V (w) = D2 w(V, ν) = hV, νi∆w, ∂ν which yield the formula (4.266) from the formula (4.39). Using the above formulas, we obtain by (4.20) and (4.35), respectively, the formula (4.267) and ∂(Aζ, 2m(ζ)) = 2B(ζ, ζ)hV, νi on Γ. (4.269) and

Finally, using the formulas (4.269) and (4.267) in the formula (4.222), we obtain the formula (4.268). For convenience, set ECSD = [H01 (Ω, Λ) × H02 (Ω)] × [L2 (Ω, Λ) × H01 (Ω)].


(4.270)

208


Let ζ = (W, w) be the solution to the problem (4.261) for the initial data (ζ0 , ζ1 ) ∈ ECSD . For T > 0 we solve the following time-reverse problem in unknown η = (Ψ, ψ)  ηtt − γ(0, ∆ψtt ) + Aη = 0 on Q,     η(T ) = ηt (T ) = 0 on Ω,     Ψ|Γ = 0, Ψ|Γ = −Dν W 0 < t < T, 1 0 (4.271) ∂ψ  = ψ| | = 0 0 < t < T,  Γ1 Γ1   ∂ν   ∂∆w ∂ψ   ψ|Γ0 = , |Γ = −∆w 0 < t < T. ∂ν ∂ν 0

Then for the initial data (η(0), ηt (0)), we have found a control action ∂∆w (−Dν W, , −∆w) on the portion Γ0 of the boundary which steers the ∂ν problem (4.260) to rest at a time T. Enlightened by [109], we define a map by Λ(ζ0 , ζ1 ) = ηt (0) − (0, γ∆ψt (0)), −η(0) + (0, γ∆ψ(0)) . (4.272) We multiply the equation (4.271) by ζ, multiply the equation (4.261) by η, integrate over Q = Ω × (0, T ), and obtain, via the Green formula (4.34) and Lemma 4.12, Λ(ζ0 , ζ1 ), (ζ0 , ζ1 ) 2 [L (Ω,Λ)×L2 (Ω)]2 = ηt (0), ζ0 − η(0), ζ1 L2 (Ω,Λ)×L2 (Ω)

L2 (Ω,Λ)×L2 (Ω)

+γ(∆ψ(0), w1 ) − γ(∆ψt (0), w0 ) Z Th i = (Aη, ζ)L2 (Ω,Λ)×L2 (Ω) − (Aζ, η)L2 (Ω,Λ)×L2 (Ω) dt L2 (Ω)

L2 (Ω)

0

= =

Z

T

Z0

Σ

Z0

+γ Z =

h i ∂(Aη, ζ) − ∂(Aζ, η) dt

[(DW (n, n))2 + (1 − µ)(DW (τ, n))2 /2] dΣ

∂∆w 2 ) ] dΣ ∂ν h ∂∆w 2 i B(ζ, ζ) + γ( ) dΣ ∂ν Σ0 [(∆w)2 + (

Σ0

where Σ0 = Γ0 × (0, T ). Since −∆: H01 (Ω) → H −1 (Ω) and −∆: L2 (Ω) → H −2 (Ω) are isomorphisms, respectively, it follows from Theorem 2.13 that Theorem 4.14 The problem (4.260) is exactly [L2 (Ω, Λ) × H01 (Ω)] × [H −1 (Ω, Λ) × L2 (Ω)]



209

controllable by L2 (Σ0 , Λ) × L2 (Σ0 ) × L2 (Σ0 ) controls if the following observability inequality is true: There is a T0 > 0 such that, for any T > T0 , there exists c > 0 satisfying Z h ∂∆w 2 i B(ζ, ζ) + γ( ) dΣ ≥ cE(0) (4.273) ∂ν Σ0 where ζ = (W, w) are solutions to the problem (4.261) with initial data (ζ0 , ζ1 ) ∈ ECSD where the space ECSD is given by (4.270). We now establish continuous observability inequality in the fixed boundary conditions. Theorem 4.15 Suppose that there is an escape vector field V for the shallow shell. Then for any T > T0 there is c > 0 such that the observability estimate (4.273) holds where T0 = λ0 σ0 /σ1 , (4.274) Γ0 = { x ∈ Γ | hV, ν(x)i > 0 },

(4.275)

and constants λ0 , σ0 and σ1 are given in (4.220). Proof. Let T > T0 be given. Let ε > 0 be given such that 2ε < σ1 (T − T0 ). Using Lemma 4.12 in Theorem 4.13, we obtain Z max hV, νi B(ζ, ζ) dΣ + CL(ζ) x∈Γ0

Σ0

≥ [2σ1 (T − T0 ) − 2ε]E(0).

(4.276)

Next, we shall use Proposition 4.3 to absorb the lower order terms in (4.276) to obtain (4.273) by a compactness-uniqueness argument as in Lemma 2.5. Control in Free Boundary Condition We let Γ1 6= ∅, Γ1 ∩ Γ0 = ∅, and consider the control problem in unknown ̟ = (Φ, φ) ̟tt − γ(0, ∆φtt ) + A̟ = 0 on Q, (4.277) ̟(0) = ̟0 , ̟t (0) = ̟1 on Ω. The unknown ̟ is subject to boundary condition on Σ1 = Γ1 × (0, T ) Φ=φ=

∂φ = 0 on Σ1 . ∂ν

We act on Σ0 = Γ0 × (0, T ) by    B1 (Φ, φ) = u1 , B2 (Φ, φ) = u2 , ∆φ + (1 − µ)B3 φ = v1 ,   ∂∆φ + (1 − µ)B4 φ − ∂φtt = v2 ∂ν ∂ν © 2011 by Taylor & Francis Group, LLC

(4.278)

on Σ0 .

(4.279)

210


The dual problem for the above is the following in ζ = (W, w). ζtt − γ(0, ∆wtt ) + Aζ = 0 on Q, ζ(0) = ζ0 , ζt (0) = ζ1 on Ω,

(4.280)

subject to boundary condition on Σ1 = Γ1 × (0, T ) W =w=

∂w =0 ∂ν

on Σ1 ,

and on Σ0 = Γ0 × (0, T )    B1 (W, w) = B2 (W, w) = 0, ∆w + (1 − µ)B3 w = 0,   ∂∆w + (1 − µ)B4 w − ∂wtt = 0 ∂ν ∂ν

(4.281)

on Σ0 .

(4.282)

We use the boundary conditions (4.281) and (4.282) in the formula (4.222) and obtain Lemma 4.13 Let V be an escape vector field for the shallow shell. Let ζ solve the problem (4.280)-(4.282). Then Z Z Z 2 2 SB dΣ = [|ζt | + γ|Dwt | − B(ζ, ζ)] dΣ + B(ζ, ζ)hV, νi dΣ. (4.283) Σ

Σ0

Σ1

By an argument as in the fixed boundary conditions the exact controllability of the problem (4.277)-(4.279) leads to the following continuous observability inequality in the free boundary conditions: Seek T0 > 0 such that for any T > T0 , there is c > 0 satisfying Z [|ζt |2 + γ|Dwt |2 ] dΣ ≥ cE(0) (4.284) Σ0

for all initial data (ζ0 , ζ1 ) ∈ ECSN Γ1 for which the left hand side of (4.284) is finite where ECSN Γ1 = [HΓ11 (Ω, Λ) × HΓ21 (Ω)] × [L2 (Ω, Λ) × L2 (Ω)].

(4.285)

Using an argument as in the proof of Theorem 4.15, we have Theorem 4.16 Let V be an escape vector field for the shallow shell. Then for any T > T0 , there is c > 0 such that the inequality (4.284) holds where T0 and Γ0 are defined by (4.274) and (4.275), respectively. Remark 4.10 Let the shell be flat, that is, M = R2 . Then the control system (4.260) or (4.277)-(4.279) becomes two systems, where one is a wave equation on the component Φ and the other is a plate equation on the component φ. We take λ0 = 1. If we set V = x − x0 , x0 a fixed point in R2 , then inequalities (4.273) and (4.284) on the component w are exactly the same as in [109]. In this case, σ1 = 1. It follows that T0 = 2diameter(Ω), which is the best for wave component W ; see [99]. In this sense, T0 , given by (4.274), is the best.



211

Finally, we study the internal exact controllability. Consider a control problem in unknown ̟ = (Φ, φ)    ̟tt − γ(0, ∆φtt + A̟ = b(x)(F, f ) on Ω × (0, T ), ̟(0) = ̟0 , ̟t (0) = ̟1 on Ω,   Φ = φ = ∂φ = 0 on Γ × (0, T ), ∂ν

(4.286)

where b ≥ 0 and b ∈ C 2 (Ω). In this case, the boundary can be empty. If Γ = ∅, there will be no boundary term in the system (4.286). The question is to find some constant T0 > 0 such that for T > T0 , the following steering property of (4.286) holds true: for all initial data ̟0 ∈ L2 (Ω, Λ) × H01 (Ω), ̟1 ∈ H −1 (Ω, Λ) × L2 (Ω), there exists a suitable control function (F, f ) ∈ L2 (0, T ; L2(Ω, Λ) × H −1 (Ω)) such that the corresponding solution of (4.286) satisfies ̟(x, T ) ≡ 0,

̟t (x, T ) ≡ 0.

In this case, we say that the dynamics (4.286) are exactly controllable in the interval [0, T ] on (L2 (Ω, Λ) × H01 (Ω)) × (H −1 (Ω, Λ) × L2 (Ω)) by means of the internal control b(F, f ). Let G = { x ∈ Ω | b(x) > 0 } be the region where the control acts. In the design of the controller, an important issue is how to choose G to be small in some sense out of all the open subsets of the middle surface Ω for which the exact controllability of (4.286) can be achieved. In fact, the structure of such a G has been studied in Section 4.5. In this section, we always assume that G is an escape region for the shallow shell with the structure (4.210)-(4.213). Set σi0 = max |Vi |, x∈Ωi

σi1 = min ϑi (x) − λ0 (1 + µ) max |ιi (x)|/2,

(4.287)

maxi σi0 . mini σi1

(4.288)

x∈Ωi

T0 = 2λ0

x∈Ωi

We will obtain the internal exact controllability by the dual method through the following observability inequality from the interior. Theorem 4.17 Let the middle surface Ω have a boundary or no boundary. Let G be an escape region for the shallow shell with the structure (4.210)(4.213). Then for any T > T0 there is c > 0 such that Z

0

T

Z

G

[B(ζ, ζ) + |ζ|2 ] dxdt ≥ cE(0)

(4.289)

for all solutions ζ to the problem (4.261) where T0 is given by (4.288).


212


Proof. For ε > ε0 > ε1 > ε2 > 0, set h i Θi = Nεi ∪Ji=1 Γi0 ∪ Ω/ ∪Ji=1 Ωj

(4.290)

for i = 0, 1, 2, where a set Nε is given by (4.214). Clearly, we have Θ2 ⊂ Θ1 ⊂ Θ0 ⊂ G. For each 1 ≤ i ≤ J, let hi ∈ C ∞ (M ) be such that   0 ≤ hi ≤ 1, h = 1 in Ωi /Θ1 ;  i hi = 0 in Θ2 .

(4.291)

(4.292)

For each i, we apply the identity (4.151) with Ω := Ωi , V := hi Vi , and ϕ = 0, to obtain Z TZ [Q1 (ζ)hhi Vi , νi i + 2Σ(ζ, hi Vi )] dΓdt = 2Zi (t)|T0 0

+

Z

0

∂Ωi T Z

[Q1 (ζ) div (hi Vi ) + Q2 (ζ, hi Vi )] dxdt

(4.293)

Q1 (ζ) = |ζt |2 + γ|Dwt |2 − B(ζ, ζ);

(4.294)

Ωi

where ∂wtt ; ∂ν Zi (t) = (ζt , mi (ζ))L2 (Ωi ,Λ)×L2 (Ωi ) + γ(Dwt , D(hi Vi (w)))L2 (Ωi ) ; Σ(ζ, hi Vi ) = ∂(Aζ, mi (ζ)) + γhi Vi (w)

mi (ζ) = hi (DVi W, Vi (w));

(4.295) (4.296) (4.297)

Q2 (ζ, hi Vi ) = 2[a(Υ(ζ), G(hi Vi , DW )) + 2γa(ρ(ζ), G(hi Vi , ρ(ζ)))] −2γD(hi Vi )(Dwt , Dwt ) + lo (ζ) (4.298) and the tensor field G(V, DW ) is defined by (4.131). Set ∂Ωi = I1 ∪ I2

where I1 = Γi0 ∪(∂Ωi /Γ) and I2 = (∂Ωi /Γi0 )∩Γ. Since ∂Ωi /Γ ⊂ Ω/∪Jj=1 Ωj ⊂ Θ2 and Γi0 ⊂ Θ2 , we have hi = 0 for x ∈ I1 and then obtain Z TZ [Q1 (ζ)hhi Vi , νi i + 2Σ(ζ, hi Vi )] d∂Ωi dt = 0. (4.299) 0

I1

In addition, since I2 ⊂ Γ ∩ { x ∈ ∂Ωi | hVi , νi i ≤ 0 } by the definition of Γi0 , the fixed boundary conditions in (4.261) and the formula (4.269) imply that Q1 (ζ)hhi Vi , νi i + 2Σ(ζ, hi Vi ) = hi B(ζ, ζ)hVi , νi i ≤ 0


(4.300)


213

for x ∈ I2 . It follows from the relations (4.299) and (4.300) that Z

T

Z

0

∂Ωi

[Q1 (ζ)hhi Vi , νi i + 2Σ(ζ, hi Vi )] d∂Ωi dt ≤ 0

(4.301)

for 1 ≤ i ≤ J. We obtain, from the identity (4.293) and the inequality (4.301), 2Zi (t)|T0 + +

Z

T

0

Z

0

Z

Q1 (ζ) div (hi Vi ) dxdt

Ωi

T

Z

Ωi

Q2 (ζ, hi Vi ) dxdt ≤ 0.

(4.302)

Now, we deal with the first integral in the left hand side of (4.302). We take Ω := Ωi , q = div (hi Vi ), and ϕ = 0 in the identities (4.157) and (4.158), respectively, and then add them up to have T

Z

0

Z

=−

Q1 (ζ) div (hi Vi ) dxdt + lo (ζ)

Ωi Z T 0

Z

[∂(Aζ, div (hi Vi )ζ) + γw

∂Ωi

∂wtt div (hi Vi )] d∂Ωi dt. ∂ν

(4.303)

Let ∂Ωi = (∂Ωi /Ω) ∪ (∂Ωi ∩ Ω). Since ∂Ωi /Ω ⊂ Γ, the integrand in the right hand side of (4.303) is zero by the Dirichlet boundary conditions of ζ on Γ. In addition, ∂Ωi ∩ Ω ⊂ Θ2 implies that the integrand is also zero by hi = 0 on Θ2 . Therefore Z TZ Q1 (ζ) div (hi Vi ) dxdt = L(ζ). (4.304) 0

Ωi

Next, we deal with the second integral in the left hand side of (4.302). On Ωi /Θ1 , since hi = 1 and the vector field Vi is escaping with the structure (4.210)-(4.213), using the formulas (4.170) and (4.171) in the relation (4.298) we obtain Q2 (ζ, hi Vi ) = 2a(Υ(ζ), G(Vi , DW )) + 4γϑi a(ρ(ζ), ρ(ζ)) −2γϑi |Dwt |2 + lo (ζ)

(4.305)

for x ∈ Ωi /Θ1 . We integrate (4.305) over Ωi /Θ1 and, by Lemma 4.11, have 2σi1 ≤

Z

Z

0

T

0 T Z

Z

B(ζ, ζ) dxdt + 2γ

Ωi /Θ1

Z

T

0

Q2 (ζ, hi Vi ) dxdt + L(ζ) Ωi /Θ1

where σi1 is given in (4.287).


Z

Ωi /Θ1

ϑi [a(ρ(ζ), ρ(ζ)) − |Dwt |2 ] dxdt (4.306)

214


Using the relations (4.306) and (4.304) in the inequality (4.302), we obtain T

Z

2σi1

Z

0

+2γ Z −

T

Z

B(ζ, ζ) dxdt ≤ 2(|Zi (0)| + |Zi (T )|)

Ωi /Θ1

Z

0

Ωi /Θ1

ϑi [|Dwt |2 − a(ρ(ζ), ρ(ζ))] dxdt

Q2 (ζ, hi Vi ) dxdt + L(ζ).

(4.307)

Ωi ∩Θ1

We estimate the first integral in the right hand side of the inequality (4.307) as follows. Applying the identity (4.158) with Ω := Ωi , q = div (hi Vi ), and ϕ = 0 yields γ

Z

T

0

=−

Z

Z

0

q[|Dwt |2 − a(ρ(ζ), ρ(ζ))] dxdt

Ωi T Z

[∂(Aζ, q(0, w)) + γqw

∂Ωi

∂wtt ] d∂Ωi dt + L(ζ) ∂ν

(4.308)

where the term qwt2 in (4.158) is eliminated as a lower order term. The same argument as that for (4.303) shows that the boundary integral in the right hand side of (4.308) is zero. Therefore 2γ

T

Z

0

=γ

Z

≤ cγ ≤c

Z

Z

T 0

Z

0

0 T

Ωi /Θ1

ϑi [|Dwt |2 − a(ρ(ζ), ρ(ζ))] dxdt

Z

Ωi ∩Θ1

T

Z

Z

q[|Dwt |2 − a(ρ(ζ), ρ(ζ))] dxdt + L(ζ)

Ωi ∩Θ1

Ωi ∩Θ1

[|Dwt |2 + a(ρ(ζ), ρ(ζ))] dxdt + L(ζ)

[γ|Dwt |2 + B(ζ, ζ)] dxdt + L(ζ).

(4.309)

We consider the second integral in the right hand side of (4.307). Clearly, it follows from (4.298) that |Q2 (ζ, hi Vi )| ≤ Cε [γ|Dwt |2 + B(ζ, ζ)] + ε(|DW |2 + |D2 w|2 ) for x ∈ Ωi . We apply Lemma 4.10 with Ω := Ωi and V := hi Vi to obtain Z 2|Zi (t)| ≤ (σi0 λ0 + ε) e(ζ) dx + lo (ζ).

(4.310)

(4.311)

Ωi

Using the formulas (4.310), (4.309), (4.311), and (4.172) in the inequality


Linear Shallow Shells. Modeling, and Control (4.307), we obtain Z TZ 2σi1 0

+c

Z

T

0

Ωi /Θ1

Z

Ωi ∩Θ1

B(ζ, ζ) dxdt ≤ 2σi0 λ0

Z

215

e(ζ) dxdt + εT E(0) Ωi

[γ|Dwt |2 + B(ζ, ζ)] dxdt + L(ζ).

(4.312)

Since Ω = (∪Ji=1 Ωi /Θ1 ) ∪ Θ1 , we add (4.312) from i = 1 to i = J to have Z TZ 2 min σi1 B(ζ, ζ) dxdt ≤ 2λ0 max σi0 E(0) + εT E(0) i

+c

0

Z

0

T

Z

i

Ω

B(ζ, ζ) dxdt + cγ

Θ1

Z

T

0

Z

Θ1

|Dwt |2 dxdt + L(ζ).

(4.313)

To finish the proof it remains to estimate the last integral in the right hand side of (4.313). Let q ∈ C ∞ (Ω) be 0 ≤ q ≤ 1 such that q=1

in Θ1 ;

q = 0 in G/Θ1 .

We add the identity (4.157) and the identity (4.158) up with the above q and ϕ = 0, and then obtain, by the Dirichlet boundary conditions of ζ, Z TZ Z TZ 2 γ |Dwt | dx ≤ q(|ζt |2 + γ|Dwt |2 ) dxdt 0

Θ1

0

=

Z

Ω

T

0

≤c

Z

qB(ζ, ζ) dxdt + L(ζ)

Ω

Z

0

T

Z

B(ζ, ζ) dxdt + L(ζ).

(4.314)

G

Finally, using the relations (4.314), (4.263), (4.265), and Lemma 4.14 below in the inequality (4.313), we obtain the inequality (4.289). By a compactness-uniqueness argument as in Lemma 2.5, we have Lemma 4.14 Let T > 0 and C1 > 0 be such that Z TZ C1 E(0) ≤ [B(ζ, ζ) + |ζ|2 ] dxdt + L(ζ) 0

(4.315)

G

for all solutions ζ to the problem (4.261). Then there is c > 0 such that the inequality (4.289) holds. Internal Exact Controllability Let ζ = (W, w) be a solution to the Dirichlet problem (4.261). We solve the following problem for η = (Ψ, ψ)    ηtt − γ(0, ∆ψtt ) + Aη = −b(x)(Aζ + ζ) on Ω × (0, T ), η(T ) = ηt (T ) = 0 on Ω, (4.316)   Ψ = ψ = ∂ψ = 0 on Γ × (0, T ). ∂ν © 2011 by Taylor & Francis Group, LLC

216


Therefore, we define a map Λ: ECSD → ( ECSD )′ by Λ(ζ0 , ζ1 ) = ηt (0) − γ(0, ∆ψt (0)), −η(0) + γ(0, ∆ψ(0))

(4.317)

where the space ECSD is given by (4.270). We have Proposition 4.4 Let (ζ0 , ζ1 ) ∈ ECSD . Then Λ(ζ0 , ζ1 ), (ζ0 , ζ1 )

[L2 (Ω,Λ)×L2 (Ω)]2

=

Z

T

[B(ζ, b(x)ζ) + b(x)|ζ|2 ] dxdt.

(4.318)

0

Proof. By the problems (4.316), (4.261), and the Green formula (4.34), the left hand side of (4.318) equals ηt (0), ζ0 − η(0), ζ1 L2 (Ω,Λ)×L2 (Ω)

L2 (Ω,Λ)×L2 (Ω)

+γ(∆ψ(0), w1 )L2 (Ω) − γ(∆ψt (0), w0 )L2 (Ω) Z T = (bAζ + bζ, ζ)L2 (Ω,Λ)×L2 (Ω) dt 0

=

Z

T

0

Z

[B(ζ, bζ) + b|ζ|2 ] dxdt.

(4.319)

Ω

Clearly, we have Z

0

T

Z

G

[B(ζ, ζ) + |ζ|2 ] dxdt ≤ CT E(0)

for all solutions ζ to the Dirichlet problem (4.261). It follows from Theorem 4.17 that Theorem 4.18 Let the middle surface Ω have a boundary or no boundary. Let G be an escape region for the shallow shell with the structure (4.210)(4.213). Let T0 > 0 be given by (4.288). Then for any T > T0 the internal control problem (4.286) is exactly {[H01 (Ω, Λ) × H02 (Ω)] × [L2 (Ω, Λ) × H01 (Ω)]}∗ controllable.

4.7

Exact Controllability for Transmission

Let a middle surface Ω be divided into two regions, Ω1 and Ω2 , by a smooth curve Γ3 on Ω. Set Γi = ∂Ω ∩ ∂Ωi for i = 1 and 2. Denote by νi the normal of



217

∂Ωi (pointing outside Ωi ) and by τi the tangential on ∂Ωi , respectively. Then νi = ν,

τi = τ

ν1 (x) = −ν2 (x),

for x ∈ Γ1 ∪ Γ2 = ∂Ω,

τ1 (x) = −τ2 (x)

for x ∈ Γ3 .

(4.320)

Suppose that the type of material in Ω1 is different from that in Ω2 , where Young’s modulus and Poisson’s coefficient have a jump across the separatrix Γ3 . First, let us clarify transmission conditions for the shallow shell. We assume that the notation here is the same as in Theorem 4.1. Let Ei and µi , respectively, denote Young’s modulus and Poisson’s coefficient of the material corresponding to the region Ωi , i = 1, 2. Let Ai denote the shallow shell operator, given by the formula (4.36) in which E and µ have been replaced with Ei and µi , respectively, for i = 1, 2. Let ζi = (Wi , wi ) be a displacement vector on Ωi for i = 1 and 2, respectively. Set on Γ3  TC i1 (ζi ) = (1 − µi )Υ(ζi )(νi , νi ) + µi (wi H − δWi ),    TC (ζ ) = (1 − µ )Υ(ζ )(ν , τ )  i2 i i i i i    TC (ζ ) = γ[∆w − (1 − µ )D2 w (τ , τ )]  i3 i i i i i i (4.321) ∂∆wi i  TC i4 (ζi ) = −(−1) γ{   ∂νi    ∂wi  ∂  +(1 − µi )[ ∂τi D2 wi (τi , νi ) + κ ]}. ∂νi

Vector fields ζi = (Wi , wi ) on Ωi are said to satisfy the transmission conditions for the shallow shell on Γ3 if the following relations hold:  W = W2 ,   1 ∂w1 ∂w2 on Γ3 . (4.322) w1 = w2 , + =0  ∂ν1 ∂ν2  b1 TC 1j (ζ1 ) = b2 TC 2j (ζ2 ) for j = 1, 2, 3, 4, It is easy to check that

Lemma 4.15 Let ζi = (Wi , wi ) satisfy the transmission (4.322) and let ηi = (Ψi , ψi ) satisfy the transmission conditions (4.322) on Γ3 , respectively. Let ∂(Ai ·, ·) be given by (4.35) in which µ is replaced by µi , for i = 1, 2. Then b1 ∂(A1 η1 , ζ1 ) + b2 ∂(A2 η2 , ζ2 ) = 0

on

Γ3 .

(4.323)

Let ̟i = (Φi , φi ) denote a displacement vector field on Ωi , for i = 1, 2. Let ω be a region such that Γ3 ⊂ ω ⊂ Ω. Let ωi = Ωi ∩ ω. We consider the control problem of transmission  ̟itt − γ(0, ∆φitt ) + bi Ai ̟i = χωi (Fi , fi ) on Ωi × (0, T ),     ̟i (0) = ̟i0 , ̟it (0) = ̟i1 on Ω,     ∂φ1 Φ1 = φ1 = = 0 on Γ2 × (0, T ), (4.324) ∂ν    ̟i satisfy the transmission conditions (4.322) on Γ3 ,     Φ = U, φ = u, ∂φ1 = v on Γ × (0, T ),  1 1 1 ∂ν © 2011 by Taylor & Francis Group, LLC

218


where χωi are the characteristic functions of ωi and bi = Ei /(1 − µ2i ) for i = 1 and 2. In the system (4.324) an internal control acts on a neighborhood of the transmission part of the two portions of the middle surface and another control (U, u, v) acts on the boundary Γ1 . The dual version of the system (4.324) in unknown ζ = (W, w) is given by  ζ − γ(0, ∆witt ) + bi Ai ζi = 0 on Ωi × (0, T ),    itt  ζ (0) = ζi0 , ζit (0) = ζi1 on Ωi ,    i  ∂w W =w= = 0 on Γ2 × (0, T ), (4.325) ∂ν   ζ satisfy the transmission conditions (4.322) on Γ ,  i 3    ∂w  W = w = = 0 on Γ1 × (0, T ). ∂ν

Let ζ be a solution to the transmission problem (4.325) such that ζi = ζ|Ωi . The total energy of the system (4.325) is defined by E(t) =

2 Z X

Ωi

i=1

[|ζit |2 + γ|Dwit |2 + bi Bi (ζi , ζi )] dx

(4.326)

where Bi (·, ·) are bilinear forms, given by (4.17) in which µ is replaced by µi for i = 1, 2. Let ω be a region such that Γ3 ⊂ ω. Let ζ = (W, w) be a solution to the problem (4.325) with an initial data (ζ0 , ζ1 ) satisfying E(0) < ∞. Let κ(x) ∈ C ∞ (Ω) be given such that supp κ ⊂ ω. We then solve the problem in η = (Ψ, ψ)  ηitt − γ(0, ∆ψitt ) + bi Aηi = −κ(x)Ai ζi on Ωi × (0, T ),     on Ωi ,  ηi (T ) = ζit (T ) = 0    ∂ψ Ψ=ψ= = 0 on Γ2 × (0, T ), (4.327) ∂ν   η satisfy the transmission conditions (4.322) on Γ ,  i 3    ∂∆w ∂ψ   Ψ = −Dν W, ψ = , = −∆w on Γ1 × (0, T ). ∂ν ∂ν

Therefore, we define a map Λ by Λ(ζ0 , ζ1 ) = ηt (0) − γ(0, ∆ψt (0)), −η(0) + γ(0, ∆ψ(0)) . Let

b(x) = bi = Ei /(1 − µ2i ), and Aζ = Ai ζi Then

We have

1 E(t) = 2

Z

Ω

B(ζ, ζ) = Bi (ζi , ζi ),

for x ∈ Ωi ,

i = 1, 2.

[|ζt |2 + γ|Dwt |2 + b(x)B(ζ, ζ)] dx.


(4.328)


219

Proposition 4.5 Let ζ be a solution to the problem (4.325). Then Λ(ζ0 , ζ1 ), (ζ0 , ζ1 ) +

Z

T

0

Z

[L2 (Ω,Λ)×L2 (Ω)]2

[b(x)B(ζ, ζ) + γ(

Γ1

=

Z

0

∂∆w 2 ) ] dΓdt, ∂ν

T

Z

b(x)B(ζ, κζ) dxdt

Ω

(4.329)

where B(·, ·)|Ωi = Bi (·, ·) for i = 1 and 2. Proof. Using the relations (4.325), (4.327) and the Green formula (4.34), we obtain Λ(ζ0 , ζ1 ), (ζ0 , ζ1 ) [L2 (Ω,Λ)×L2 (Ω)]2

=

2 Z X

T

i=1 0 2 X

+γ = +

i=1 2 X

[(ηi , ζitt )L2 (Ωi ,Λ)×L2 (Ωi ) − (ηitt , ζi )L2 (Ωi ,Λ)×L2 (Ωi ) ] dt

[(∆ψi (0), wit (0))L2 (Ωi ) − (∆ψit (0), wi (0))L2 (Ωi ) ]

bi

Z

T

0 i=1 Z 2 T X i=1

0

Z

∂Ωi

[∂(Ai ηi , ζi ) − ∂(Ai ζi , ηi )] d∂Ωi dt

(Ai ζi , κζi )L2 (Ωi ,Λ)×L2 (Ωi ) dt.

(4.330)

Let us compute the integrals in the right hand side of (4.330) on Ω1 and Ω2 , separately. It follows from the formulas (4.266) and (4.267) with µ = µ1 that Z [∂(A1 η1 , ζ1 ) − ∂(A1 ζ1 , η1 )] d∂Ω1 ∂Ω1 Z = [∂(A1 η1 , ζ1 ) − ∂(A1 ζ1 , η1 )] dΓ3 Γ3 Z ∂∆w 2 + [B1 (ζ1 , ζ1 ) + γ( ) ] dΓ (4.331) ∂ν Γ1 and b2

Z

∂Ω2

= b2

Z

[∂(A2 η2 , ζ2 ) − ∂(A2 ζ2 , η2 )] d∂Ω2

Γ3

= −b1

Z

[∂(A2 η2 , ζ2 ) − ∂(A2 ζ2 , η2 )] dΓ3

Γ3

[∂(A1 η1 , ζ1 ) − ∂(A1 ζ1 , η1 )] dΓ.


(4.332)

220


In the above computation, we have used the relation (4.320) and Lemma 4.15 on the common boundary Γ3 . Using the formulas (4.332) and (4.331) in the formula (4.330), we have the formula (4.329). We have proven Theorem 4.19 If there is λ0 ≥ 1 such that λ0

2 X i=1

bi

Z

Ωi

Bi (ζ, ζ) dx ≥ kDW k2L2 (Ω,T 2 ) + γkD2 wkL2 (Ω,T 2 )

(4.333)

for all ζ = (W, w) ∈ H 1 (Ω, Λ) × H 2 (Ω), then the problem (4.327) is exactly ECSD ∗ controllable if there is T0 > 0 such that for any T > T0 , there exists c > 0 satisfying Z Z ∂∆w 2 b(x)B(ζ, κζ) dQ + [b(x)B(ζ, ζ) + γ( ) ] dΓdt ≥ cE(0) (4.334) ∂ν Q Σ1 for all solutions ζ to the problem (4.325), where the space ECSD is given by (4.270). Next, we present appropriate geometric conditions on the material of the shallow shell for the observability estimate (4.334). Definition 4.3 A vector field V on Ω is said to be an escape vector field for transmission of the shallow shell if the relation (4.173) is true such that 2 min ϑ(x) > λ0 b0 (1 + µ0 ) max ι(x) x∈Ω

x∈Ω

(4.335)

where µ0 = max(µ1 , µ2 ) and b0 = max(b1 , b2 ). Existence of such vector fields has been studied and a number of examples are given in Section 4.5. Assumption (H1) There is λ0 ≥ 1 such that the ellipticity condition (4.333) holds true. Assumption (H2) Let V be an escape vector field for transmission of the shallow shell such that Γ1 = { x | x ∈ Γ, hV, νi ≥ 0 }.

(4.336)

Set σ0 = max |V |, x∈Ω

σ1 = min ϑ(x) − λ0 b0 (1 + µ0 ) max ι(x)/2. x∈Ω

x∈Ω

(4.337)

To prove the inequality (4.334), we need to clarify some multiplier identities for the transmission problem as follows.



221

Lemma 4.16 Let the assumptions (H1) and (H2) hold. Let ζ = (W, w) be a solution to the problem ζtt − γ(0, ∆wtt ) + bAζ = 0

in

(Ω/Γ3 ) × (0, T )

(4.338)

and ζi = ζ|Ωi satisfy the transmission conditions (4.322) on Γ3 × (0, T ). Let T > 0 be given. Then Z T Z 2σ1 E(t), dt ≤ σ0 λ0 [E(0) + E(T )] + Σ(ζ, b)dΣ 0 Σ Z + Σ3 (ζ, b) dΣ + CL(ζ) (4.339) Σ3

where h i Σ(ζ, b) = |ζt |2 + γ|Dwt |2 − b(x)B(ζ, ζ) hV, νi ∂w tt +∂ b(x)Aζ, 2m(ζ) + (̺W, −ϑw) + γ 2V (w) − ϑw , ∂ν Σ3 (ζ, b) = 2b1 ∂(A1 ζ1 , m(ζ1 )) + 2b2 ∂(A2 ζ2 , m(ζ2 )) h i + b2 B2 (ζ2 , ζ2 ) − b1 B1 (ζ1 , ζ1 ) hV, ν1 i, m(ζ) = (DV W, V (w)), Σ = (0, T ) × Γ,

(4.340)

(4.341)

̺ = 2ϑ − σ1 ,

Σ3 = (0, T ) × Γ3 .

Proof. Multiply (4.335) by 2m(ζ) and integrate by parts. Let P = ζ − γ(0, ∆w). Using the identity (4.153), the relations (4.320) and (4.322), we obtain Z

Q

= −

hPtt , 2m(ζ)idQ =

2Z|T0

+2

Q

XZ i

Z

Σi

2 Z X i=1

Qi

hPtt , 2m(ζ)idQ

ϑ|ζt |2 dQ

[(|ζit |2 + γ|Dwit |2 )hV, νi i + 2γV (wi )hDwitt , νi i]dΣ

Z = 2Z|T0 + 2 ϑ|ζt |2 dQ Q Z 2 − [(|ζt | + γ|Dwt |2 )hV, νi i + 2γV (w)hDwtt , νi]dΣ, Σ

where Σi = ∂Ωi × (0, T ) and Z Z = [hζt , m(ζ)i + γh∇wt , ∇V (w)i] dx. Ω


(4.342)

222


In addition it follows, similar to Lemma 4.10, that |Z| ≤ σ0 λ0 E(t) + CL(ζ).

(4.343)

On the other hand, using the relation (4.170) in the identity (4.156) in which Ω and A are replaced by Ωi and Ai , respectively, we have Z 2 hAi ζi , m(ζi )i dQ Qi Z Z = [Bi (ζi , ζi )hV, νi i − 2∂(Ai ζi , m(ζi ))] dΣ + 2 a(Υ(ζi ), G(V, DWi )) dQ Σ Qi Zi +2 ϑ[γai (ρ(ζi ), ρ(ζi )) − ai (Υ(ζi ), Υ(ζi ))] dQ + L(ζ), (4.344) Qi

where Qi = Ωi × (0, T ). Moreover, a similar argument as in Lemma 4.11 yields Z Z σ1 b(x)a(Υ(ζ), Υ(ζ)) dx ≤ b(x)a(Υ(ζ), G(V, DW )) dx + CL(ζ), (4.345) Ω

Ω

where a(Υ(ζ), Υ(ζ))|Ωi = ai (Υ(ζi ), Υ(ζi )) and a(Υ(ζ), G(V, DW ))|Ωi = ai (Υ(ζi ), G(V, DWi )). It follows from (4.344) and (4.345) that Z hb(x)Aζ, 2m(ζ)i dQ Q Z Z ≥ [b(x)B(ζ, ζ) − ∂(b(x)Aζ, 2m(ζ))] dΣ − Σ3 (ζ, b) dΣ Σ3 ZΣ +2 [(σ1 − ϑ)b(x)a(Υ(ζ), Υ(ζ)) + ϑb(x)γa(ρ(ζ), ρ(ζ))]dQ. (4.346) Q

From the relations (4.338), (4.342), and (4.346), we obtain Z [|ζt |2 + γ|Dwt |2 − b(x)B(ζ, ζ)]hV, νi dΣ Σ Z Z +2 [∂(b(x)Aζ, m(ζ)) + γV (w)hDwtt , νi]dΣ + Σ3 (ζ, b) dΣ Σ Σ3 Z ≥ Z|T0 + Ψb dQ + L(ζ) (4.347) Q

where Ψb = 2[σ1 − ϑ(x)]b(x)a(Υ(ζ), Υ(ζ)) + 2ϑ(x)[|ζt |2 + γb(x)a(ρ(ζ), ρ(ζ))]. A similar argument as in (4.238) yields Z Z Z T Ψb dQ ≥ σ1 E(t) dt + ̺[|Wt |2 − b(x)a(Υ(ζ), Υ(ζ))] dQ Q 0 Q Z + ϑ[γb(x)a(ρ(ζ), ρ(ζ)) − γ|Dwt |2 − wt2 ] dQ. (4.348) Q



223

Next, we multiply the equation (4.338) by ̺(W, 0) and ϑ(0, w), respectively, and use the transmission conditions on Γ3 to obtain Z ̺[|Wt |2 − b(x)a(Υ(ζ), Υ(ζ))] dQ Q Z =− ∂(b(x)Aζ, ̺(W, 0))dΣ + L(ζ), (4.349) Σ

Z

ϑ[γb(x)a(ρ(ζ), ρ(ζ)) − γ|Dwt |2 − wt2 ] dQ Q Z ∂wtt = [∂(b(x)Aζ, ϑ(0, w)) + γϑw ] dΣ + L(ζ). ∂ν Σ

(4.350)

Finally, we use the relations (4.343), (4.348), (4.349), and (4.350) in the equation (4.347) to have the estimate (4.339). Let ωi be regions such that Γ3 ⊂ ω1 ⊂ ω 1 ⊂ ω2 ⊂ ω 2 ⊂ ω. Let θ ∈ C0∞ (ω1 ) be such that 0 ≤ θ(x) ≤ 1;

θ(x) = 1 for

x ∈ Γ3 .

(4.351)

Lemma 4.17 Let the assumptions (H1) and (H2) hold and let the transmission conditions (4.322) be true. Let ζ solve the problem (4.325). Then Z Z Σ3 (ζ, b) dΣ + θb(x)B(ζ, ζ)hV, νi dΣ ≤ σ0 λ0 [E(0) + E(T )] Σ3

+c

Σ

Z

0

T

Z

b(x)B(ζ, ζ) dxdt + L(ζ).

(4.352)

ω

Proof. Using the multiplier θ(x)m(ζ) to the equation (4.338) and by integration by parts, we obtain, via the boundary conditions and the transmission conditions (4.322), Z Z Σ3 (ζ, b) dΣ + θ[∂(b(x)Aζ, 2m(ζ)) − b(x)B(ζ, ζ)]hV, νi dΣ Σ3 Σ Z = 2Z1 |T0 + [|ζt |2 + γ|Dwt |2 − b(x)B(ζ, ζ)] div (θV ) dQ Q Z +2 [b(x)a(Υ(ζ), G(θV, DW )) + γb(x)a(ρ(ζ), G(θV, ρ(ζ)))] dQ Q Z − γD(θV )(Dwt , Dwt ) dQ + L(ζ) (4.353) Q


224


where Z1 =

Z

θ[hζt , m(ζ)i + γhDwt , D(V (w))i] dx.

Ω

The condition 0 ≤ θ ≤ 1 implies that 2|Z1 | ≤ σ0 λ0 E(t) + L(ζ).

(4.354)

In addition, a similar computation as in (4.269) yields ∂(Ai ζi , m(ζi )) = Bi (ζi , ζi )hV, νi for x ∈ Γi

i = 1, 2.

(4.355)

On the other hand, using the multiplier (W, w) div (θV ) to the problem (4.325), we obtain that the second integral in the right hand side of (4.353) is a lower order term. Thus the relation supp θ ⊂ ω1 implies that the right hand side of (4.353) Z TZ ≤ Z1 |T0 + C [b(x)B(ζ, ζ) + γ|Dwt |2 ] dxdt + L(ζ). 0

(4.356)

Ω∩ω1

RT R Next, we estimate the term γ 0 ω1 |Dwt |2 dxdt. Let q ∈ C ∞ (M ) be such that 0 ≤ q ≤ 1; q = 1 on ω1 ; q = 0 on ω2 . Using the multiplier q(0, w) to the problem (4.325), we have Z TZ Z 2 γ |Dwt | dxdt ≤ q[wt2 + γ|Dwt |2 − γb(x)a(ρ(ζ), ρ(ζ))] dQ 0

+γ

Ω∩ω1

Z

Q

qb(x)B(ζ, ζ)dQ ≤ c

Z

Q T

0

Z

b(x)B(ζ, ζ) dxdt + L(ζ).

(4.357)

ω

Finally, the inequality (4.352) follows after inserting (4.357), (4.356), (4.355), and (4.354) into (4.353). We shall prove Theorem 4.20 Let the assumptions (H1) and (H2) hold. Let V be an escape vector field for transmission of the shallow shell. Let ω ⊂ M be an open set such that Γ3 ⊂ ω. Let θ ∈ C0∞ (ω) be such that the condition (4.351) holds. Then for any T > 2T0 there is c > 0 such that the observability estimate (4.334) holds where T0 = 2λ0 σ0 /σ1 . (4.358) Proof. Let ζ be a solution to the problem (4.325). Then it follows from the Dirichlet boundary conditions in the problem (4.325) and the formulas (4.340) and (4.355) that Z Z Σ(ζ, b) dΣ = b(x)B(ζ, ζ) dΣ. (4.359) Σ


Σ


225

Let 0 < ε0 < 1 be given. Set ω1 = { x | x ∈ ω, θ(x) > ε0 }. Clearly, ω1 is an open set such that Γ3 ⊂ ω. Let θˆ ∈ C0∞ (ω1 ) be such that 0 ≤ θˆ ≤ 1 and θˆ = 1 for x ∈ Γ3 . We apply Lemma 4.17 with θ := θˆ and ω := ω1 , and obtain Z Z ˆ Σ3 (ζ, b) dΣ + θb(x)B(ζ, ζ)hV, νi dΣ ≤ σ0 λ0 [E(0) + E(T )] Σ3

+

c ε0

Σ

Z

0

T

Z

b(x)B(ζ, θζ) dxdt + L(ζ).

(4.360)

Ω

Using the relations (4.360), (4.359), and (4.336) in the inequality (4.339), we obtain the observability (4.334) by a compactness-uniqueness argument as in Proposition 4.3. If the materials have a special property and the division curve Γ3 has some special shape as follows, then boundary control may be enough for the exact controllability of the transmission problem. We make the following Assumption (H3) b2 (1 − µ2 ) < b1 (1 − µ1 ), b2 (1 + µ2 ) < b1 (1 + µ1 ), and hV, ν1 i ≥ 0 for all x ∈ Γ3 . Remark 4.11 The assumption (H3) is a condition on the Young’s modulus and Poisson’s coefficient of the material of the shell. It may be necessary to establish the observability inequality if only the boundary controls are allowed. In the case of wave equations, [152] gave an example to show that the similar condition as the assumption (H3) is necessary for the stabilization and controllability by the boundary control. Consider the boundary control problem  ηtt − γ(0, ∆ψtt ) + b(x)Aη = 0 on Q,     on Ω,  η(T ) = ηt (T ) = 0    ∂ψ Ψ=ψ= = 0 on Σ2 , ∂ν   ηi satisfy the transmission conditions on Γ3 ,     ∂∆w ∂ψ   Ψ = −Dν W, ψ = , = −∆w on Σ1 . ∂ν ∂ν

(4.361)

We have

Theorem 4.21 Let the assumptions (H1), (H2) and (H3) hold. Then for any T > T0 , the problem (4.22) is exactly ECSD ∗ controllable by L2 (Σ, Λ) × L2 (Σ) × L2 (Σ) where T0 and ECSD are given by (4.358) and (4.270), respectively. The proof of the above theorem follows from the following observability estimate


226


Theorem 4.22 Let the assumptions (H1), (H2) and (H3) hold. Then for any T > T0 there is c > 0 such that Z

0

T

Z

[b(x)B(ζ, ζ) + γ(

Γ1

∂∆w 2 ) ] dΓdt ≥ cE(0) ∂ν

(4.362)

for all solutions ζ to the problem (4.325).

Proof. By Theorem 4.16 it will suffice to prove that Σ3 (ζ, b) ≤ lo (ζ)

(4.363)

where Σ3 (ζ, b) is given by (4.341) and lo (ζ) is the lower order terms given by (4.127). The transmission conditions (4.322) imply that on Γ3

and

 DW1 (τ1 , τ1 ) = DW2 (τ2 , τ2 ),      DW1 (ν1 , τ1 ) = DW2 (ν2 , τ2 ); Dw1 = Dw2 , V (w1 ) = V (w2 ),   D2 w1 (τ1 , τ1 ) = D2 w2 (τ2 , τ2 ),    2 D w(ν1 , τ1 ) = D2 w2 (ν2 , τ2 ),    b1 DW1 (ν1 , ν1 ) + b1 µ1 DW1 (τ1 , τ1 ) + lo (ζ1 )    = b2 DW2 (ν2 , ν2 ) + b2 µ2 DW2 (τ2 , τ2 ) + lo (ζ2 ), b1 (1 − µ1 )Υ(ζ1 )(ν1 , τ1 ) = b2 (1 − µ2 )Υ(ζ2 )(ν2 , τ2 ),   b1 D2 w1 (ν1 , ν1 ) + b1 µ1 D2 w1 (τ1 , τ1 )    = b2 D2 w2 (ν2 , ν2 ) + b2 µ2 D2 w2 (τ2 , τ2 ).

(4.364)

(4.365)

Furthermore the relations (4.365) and (4.364) reach on Γ3

 b2 DW2 (ν2 , ν2 ) = b1 DW1 (ν1 , ν1 )     +(b1 µ1 − b2 µ2 )DW1 (τ1 , τ1 ) + lo (ζ),    b2 (1 − µ2 )DW2 (τ2 , ν2 ) = b1 (1 − µ1 )DW1 (τ1 , ν1 ) +[b1 (1 − µ1 ) − b2 (1 − µ2 )]DW1 (ν1 , τ1 ) + lo (ζ),    2  b D w2 (ν2 , ν2 ) = b1 D2 w1 (ν1 , ν1 )  2   +(b1 µ1 − b2 µ2 )D2 w1 (τ1 , τ1 ).

(4.366)

Using the relations (4.320), (4.322) and (4.364) in the formula (4.35), we



227

obtain

b1 ∂(A1 ζ1 , m(ζ1 )) + b2 ∂(A2 ζ2 , m(ζ2 )) = [b1 TC 11 (ζ1 )DW1 (ν1 , ν1 ) − b2 TC 21 (ζ2 )DW2 (ν2 , ν2 )]hV, ν1 i +[b1 TC 12 (ζ1 )DW1 (τ1 , ν1 ) − b2 TC 22 (ζ2 )DW2 (τ2 , ν2 )]hV, ν1 i

+[b1 TC 13 (ζ1 )D2 w1 (ν1 , ν1 ) − b2 TC 23 (ζ2 )D2 w2 (ν2 , ν2 )]hV, ν1 i = [b1 (DW1 (ν1 , ν2 ))2 − b2 (DW2 (ν2 , ν2 ))2 ]hV, ν1 i +[b1 µ1 DW1 (ν1 , ν1 ) − b2 µ2 DW2 (ν2 , ν2 )]DW1 (τ1 , τ1 )hV, ν1 i

+2[b1 (1 − µ1 )(Υ(ζ1 )(ν1 , τ1 ))2 − b2 (1 − µ2 )(Υ(ζ2 )(ν2 , τ2 ))2 ]hV, ν1 i +γ[b1 (D2 w1 (ν1 , ν1 ))2 − b2 (D2 w2 (ν2 , ν2 ))2 ]hV, ν1 i

+γ[b1 µ1 D2 w1 (ν1 , ν1 ) − b2 µ2 D2 w2 (ν2 , ν2 )]D2 w1 (τ1 , τ1 )hV, ν1 i.

(4.367)

Moreover, by a similar computation, we have

b2 B2 (ζ2 , ζ2 ) − b1 B1 (ζ1 , ζ1 )

= b2 (DW2 (ν2 , ν2 ))2 − b1 (DW1 (ν1 , ν1 ))2 + (b2 − b1 )(DW1 (τ1 , τ1 ))2 +2b2 (1 − µ2 )(Υ(ζ2 )(ν2 , τ2 ))2 − 2b1 (1 − µ1 )(Υ(ζ1 )(ν1 , τ1 ))2

+2b2 µ2 DW2 (ν2 , ν2 )DW2 (τ2 , τ2 ) − 2b1 µ1 DW1 (ν1 , ν1 )DW1 (τ1 , τ1 ) +γ[b2 (D2 w2 (ν2 , ν2 ))2 − b1 (D2 w1 (ν1 , ν1 ))2 + (b2 − b1 )(D2 w1 (τ1 , τ1 ))2 ]

+2γ[b2 µ2 D2 w2 (ν2 , ν2 )D2 w2 (τ2 , τ2 ) − b1 µ1 D2 w1 (ν1 , ν1 )D2 w1 (τ1 , τ1 )] +2γ[b2 (1 − µ2 ) − b1 (1 − µ1 )](D2 w1 (ν1 , τ1 ))2 . (4.368)

It follows from the relations (4.368), (4.367), and (4.366) that

2b1 ∂(A1 ζ1 , m(ζ1 )) + 2b2 ∂(A2 ζ2 , m(ζ2 )) +[b2 B2 (ζ2 , ζ2 ) − b1 B1 (ζ1 , ζ1 )]hV, ν1 i

= [b1 (DW1 (ν1 , ν2 ))2 − b2 (DW2 (ν2 , ν2 ))2 + (b2 − b1 )(DW1 (τ1 , τ1 ))2 ]hV, ν1 i

+2[b1 (1 − µ1 )(Υ(ζ1 )(ν1 , τ1 ))2 − b2 (1 − µ2 )(Υ(ζ2 )(ν2 , τ2 ))2 ]hV, ν1 i +γ[b1 (D2 w1 (ν1 , ν1 ))2 − b2 (D2 w2 (ν2 , ν2 ))2 ]hV, ν1 i

+2γ[b2 (1 − µ2 ) − b1 (1 − µ1 )](D2 w1 (ν1 , τ1 ))2 ]hV, ν1 i +γ(b2 − b1 )(D2 w1 (τ1 , τ1 ))2 ]hV, ν1 i + lo (ζ)


228

Modeling and Control in Vibrational and Structural Dynamics n1 [b1 (b2 − b1 )(DW1 (ν1 , ν1 ))2 = b2 −2b1 (b1 µ1 − b2 µ2 )DW1 (ν1 , ν1 )DW1 (τ1 , τ1 ) +(b2 (b2 − b1 ) − (b1 µ1 − b2 µ2 )2 )(DW1 (τ1 , τ1 ))2 ] b1 (1 − µ1 ) +2 [b2 (1 − µ2 ) − b1 (1 − µ1 )](Υ(ζ1 )(ν1 , τ1 ))2 b2 (1 − µ2 ) 2γ + [b1 (b2 − b1 )(D2 w1 (ν1 , ν1 ))2 b2 −2b1 (b1 µ1 − b2 µ2 )D2 w1 (ν1 , ν1 )D2 w1 (τ1 , τ1 ) +(b2 (b2 − b1 ) − (b1 µ1 − b2 µ2 )2 )(D2 w1 (τ1 , τ1 ))2 ] o +2γ[b2 (1 − µ2 ) − b1 (1 − µ1 )](D2 w1 (ν1 , τ1 ))2 hV, ν1 i

+ lo (ζ).

(4.369)

Let us estimate the terms in (4.369) separately. The assumption (H3) implies that b2 − b1 < b2 µ2 − b1 µ1

and b2 − b1 < −(b2 µ2 − b1 µ1 ),

that is, b2 − b1 < 0. Thus b1 (b2 − b1 )(DW1 (ν1 , ν1 ))2

−2b1 (b1 µ1 − b2 µ2 )DW1 (ν1 , ν1 )DW1 (τ1 , τ1 ) +(b2 (b2 − b1 ) − (b1 µ1 − b2 µ2 )2 )(DW1 (τ1 , τ1 ))2 < −ε0 [(DW1 (ν1 , ν1 ))2 + (DW1 (τ1 , τ1 ))2 ]

(4.370)

for some ε0 > 0; [b2 (1 − µ2 ) − b1 (1 − µ1 )](Υ(ζ1 )(ν1 , τ1 ))2 < −ε0 (Υ(ζ1 )(ν1 , τ1 ))2 ;

(4.371)

b1 (b2 − b1 )(D2 w1 (ν1 , ν1 ))2

−2b1 (b1 µ1 − b2 µ2 )D2 w1 (ν1 , ν1 )D2 w1 (τ1 , τ1 ) +(b2 (b2 − b1 ) − (b1 µ1 − b2 µ2 )2 )(D2 w1 (τ1 , τ1 ))2 < −ε0 [(D2 w1 (ν1 , ν1 ))2 + (D2 w1 (τ1 , τ1 ))2 ];

[b2 (1 − µ2 ) − b1 (1 − µ1 )](D2 w1 (ν1 , τ1 ))2 < −ε0 (D2 w1 (ν1 , τ1 ))2 . Finally, the estimate (4.363) follows from (4.369)-(4.373).


(4.372) (4.373)


4.8

229

Stabilization by Linear Boundary Feedback

The goal of this section is to introduce some uniform stabilization results for the shallow shell model with suitable, natural, linear or nonlinear dissipative boundary feedback in the form of moments and shears applied to an edge of the shell. More explicitly, what this means is the following. First, with homogeneous boundary conditions, the shell model is conservative (energy preserving). Next, we impose suitable linear dissipative terms (tractions/shears/moments) in physical boundary conditions exercised only on a portion Γ0 of the boundary Γ of the shell and then seek to force the energy of the new corresponding closed loop, well-posed (Theorem 4.23) dissipative problem to decay to zero at a certain rate. The rate depends explicitly on pre-assigned growth properties of the dissipative terms. This is the content of Theorem 4.26. Consider the shallow shell in unknown ζ = (W, w) ζtt − γ(0, ∆wtt ) + Aζ = 0 in

Q∞ = Ω × (0, ∞)

(4.374)

and define the total energy of the shell by 2E(t) = kWt k2L2 (Ω,Λ) + kwt k2L2 (Ω) + γkDwt k2L2 (Ω,Λ) + B(ζ, ζ)

(4.375)

for t ≥ 0 where A and B(·, ·) are given by (4.36) and (4.21), respectively. By the Green formula (4.34) and the equation (4.374), we obtain d E(t) = 2(Wtt , Wt )L2 (Ω,Λ) + 2(wtt , wt )L2 (Ω) dt +2γ(Dwtt , Dwt )L2 (Ω,Λ) + 2B(ζ, ζt ) Z h i ∂wtt =2 wt + ∂(Aζ, ζt ) dΓ ∂ν ZΓ h =2 v1 (ζ)hWt , νi + v2 (ζ)hWt , τ i Γ

+v3 (ζ) where

i ∂wt ∂wtt + (v4 (ζ) + )wt dΓ ∂ν ∂ν

 v1 (ζ) = (1 − µ)Υ(ζ)(ν, ν) + µ(wH − δW ),     v2 (ζ) = (1 − µ)Υ(ζ)(ν, τ ), v3 (ζ) = γ[∆w − (1 − µ)D2 w(τ, τ )],     v4 (ζ) = −γ{ ∂∆w + (1 − µ)[ ∂ D2 w(τ, ν) + θ ∂w ]}. ∂ν ∂τ ∂ν

(4.376)

(4.377)

We assume that Γ = Γ0 ∪ Γ1 such that

Γ0 ∩ Γ1 = ∅.


(4.378)

230


Then the condition (4.378) implies that Γi are closed curves for i = 0 and 1. We assume that a solution to (4.374) is subject to the boundary conditions W = 0,

w=

∂w = 0 on Σ∞ 1 = Γ1 × (0, ∞). ∂ν

(4.379)

We will act on Σ∞ 0 = Γ0 × (0, ∞) by a feedback as follows. For ζ = (W, w), we set ζˆ = (hW, νi, hW, τ i, hDw, νi, hDw, τ i, w). (4.380) We consider feedback laws to be defined by vi (ζ) = Ji (ζt ) i = 1, 2, 3, on Σ∞ 0 = Γ0 × (0, ∞) v4 (ζ) + γhDwtt , νi = J4 (ζt ) where the feedback operators are given by ˆ Fi i 5 i = 1, 2, 3, Ji (ζ) = −hζ, R ˆ F5 i 5 + ∂ hζ, ˆ F4 i 5 J4 (ζ) = −hζ, R R ∂τ

(4.381)

(4.382)

where Fi = Fi (x) ∈ R5 for x ∈ Γ0 . If the 5 × 5 matrix F := (F1 , F2 , F3 , F4 , F5 ) satisfies that F is symmetric and positive semidefinite on Γ0 ,

(4.383)

then using the relations (4.377)-(4.382) in the formula (4.376), we have Z d E(t) = −2 hF ζˆt , ζˆt iR5 dΓ ≤ 0. (4.384) dt Γ0 Therefore, the following closed-loop system under the feedback laws of (4.381) and (4.382) is dissipative in the sense that E(t) is nonincreasing if the condition (4.383) is true:  ζtt − γ(0, ∆wtt ) + Aζ = 0 in Q∞ ,    ∂w   = 0 on Σ∞ W = w = 1 , ∂ν (4.385) ˆ vi (ζ) + hζt , Fi i = 0 i = 1, 2, 3 on Σ∞  0 ,   ∂ ∞   v4 (ζ) + γhDwtt , νi + hζˆt , F4 i = ∂τ hζˆt , F5 i on Σ0 ,  ζ(0) = ζ0 , ζt (0) = ζ1 on Ω.

Remark 4.12 When the tangent component W of the shell is zero, the feedback laws of the system (4.385) are what the paper [105] presented for the uniform stabilization of the Kirchhoff plate. We have proven

Lemma 4.18 Let ζ be a solution to the problem (4.385) with finite energy. Then, for any s ≤ t, the following identity holds true for the energy E(t) defined by (4.375): Z tZ E(t) + hF ζˆt , ζˆt iR5 dΓdt = E(s). (4.386) s


Γ0


231

The above lemma roughly tells us that a solution to the closed-loop system (4.385) can be obtained by a C0 semigroup of contraction on the space [HΓ1 (Ω, Λ)×HΓ21 (Ω)]×[L2 (Ω, Λ)×HΓ11 (Ω)] if the condition (4.383) holds true. Following some ideas in [105] for the Kirchhoff plate, we shall deal with this point in detail. Then the regularity of solutions we need for the stabilization is worked out by [3]. Variational formulation Set W = HΓ11 (Ω, Λ) × HΓ21 (Ω),

V = L2 (Ω, Λ) × HΓ11 (Ω),

L = L2 (Ω, Λ) × L2 (Ω). Introduce bilinear forms α(ζ, η) =

Z

[hζ, ηi + γhDw, Dui] dx

(4.387)

Ω

and α0 (ζ, η) =

Z

Γ0

ˆ ηî 5 dΓ hF ζ, R

(4.388)

for ζ = (W, w) and η = (U, u). It follows from Green’s formula (4.34) that an appropriate variational formulation of the system (4.385) is as follows: Find a vector field ζ ∈ C([0, ∞); W) ∩ C 1 ([0, ∞); V) such that [α(ζt , η) + α0 (ζ, η)]t + B(ζ, η) = 0 for all η ∈ W, (4.389) ζ(0) = ζ0 ∈ W, ζt (0) = ζ1 ∈ V. Well-posedness of the system (4.385) Let the ellipticity (4.172) of the shallow shell hold. Suppose that the condition (4.383) holds true. Then B(·, ·) and α(·, ·) are equivalent inner products on W and V, respectively. We identify L with its dual L∗ so that we have the dense and continuous embeddings W ⊂ V ⊂ L ⊂ V ∗ ⊂ W ∗.

(4.390)

Let A (respectively, P ) denote the canonical isomorphism of V (respectively, W) endowed with the inner product α(·, ·) (respectively, B(·, ·)) onto V ∗ (respectively, W ∗ ). Thus α(ζ, η) = (Aζ, η)L

for all ζ, η ∈ V,

(4.391)

B(ζ, η) = (P ζ, η)L

for all ζ, η ∈ W,

(4.392)

where (·, ·)L denotes the inner product of L. Moreover, clearly 0 ≤ α0 (ζ, ζ)| ≤ ckζk2W . Then there is a nonnegative continuous operator A0 : W → W ∗ such that α0 (ζ, η) = (A0 ζ, η)L for all ζ, η ∈ W. (4.393) Then the variational problem (4.389) can be written as (Aζt + A0 ζ)t + P ζ = 0 in W ∗ .


(4.394)

232


Let us formally rewrite (4.394) as the system P 0 ζ 0 −P ζ + =0 0 A ζt t P A0 ζt or CYt + IN Y = 0 where C=

P 0 0 A

,

IN =

0 −P P A0

,

(4.395) and Y =

ζ ζt

.

We wish to solve the problem (4.395) in the space W × V. In order to let the problem (4.395) make sense in that space it is natural to introduce D(IN ) = { (ζ, η) | ζ, η ∈ W, P ζ + A0 η ∈ V ∗ }.

(4.396)

Then IN : D(IN ) → W ∗ × V ∗ . Since C is the canonical isomorphism of W × V onto W ∗ × V ∗ , we rewrite the system (4.395) in the form Yt + C−1 IN Y = 0

in W × V.

(4.397)

Solutions of the system (4.385) are therefore defined via (4.397). Theorem 4.23 Let the ellipticity (4.172) of the shallow shell hold. Suppose that the condition (4.383) holds true. Then −C−1 IN is the infinitesimal generator of a C0 -semigroup of contraction on W × V. Proof. (i) We claim that D(IN ) is dense in W × V. By the definitions of P and A0 , for ς = (U, u) ∈ W, we obtain by the Green formula (4.34) and the relation (4.382) (P ζ + A0 η, ς)L = B(ζ, ς) + α0 (η, ς) Z Z = hAζ, ςi dx + [(v1 (ζ) − J1 (η))hU, νi + (v2 (ζ) − J2 (η))hU, τ i Ω

Γ0

+(v3 (ζ) − J3 (η))hDu, νi + (v4 (ζ) − J4 (η))u] dΓ

(4.398)

where Ji (·) are given by the formula (4.382) for 1 ≤ i ≤ 4. The expression on the right-hand side of the formula (4.398) implies the relation D(IN ) ⊃ D0 (4.399) where

D0 = {(ζ, η) | ζ ∈ W ∩ (H 2 (Ω, Λ) × H 4 (Ω)), η ∈ W, vi (ζ) = Ji (η), i = 1, 2, 3 }.

Indeed, if (ζ, η) ∈ D0 , it follows from the relations (4.398) and (4.399) that |(P ζ + A0 η, ς)L | ≤ kAζkL kςkL + ckv4 (ζ) − J4 (η)kH −1/2 (Γ0 ) kukH 1/2 (Γ0 ) ≤ ckζkH 2 (Ω,Λ)×H 4 (Ω) kςkL + c(kwkH 4 (Ω) + kηkW )kukH 1 (Ω) ≤ c(kζkH 2 (Ω,Λ)×H 4 (Ω) + kηkW )kςkV for all ς ∈ V,


(4.400)


233

that is, P ζ + A0 η ∈ V ∗ . Then D(IN ) is dense in W × V since D0 is dense in W × V. (ii) −C−1 IN is dissipative. Indeed, using the relations (4.391)-(4.393), we obtain C−1 IN (ζ, η), (ζ, η) = (−η, A−1 (P ζ + A0 η), (ζ, η) W×V

W×V

= −B(η, ζ) + α(A−1 (P ζ + A0 η), η) = −B(η, ζ) + (P ζ + A0 η, η)L = α0 (η, η) ≥ 0

for (ζ, η) ∈ D(IN ) where the inner products of W and V are B(·, ·) and α(·, ·), respectively. (iii) We also have Range (λI + C−1 IN ) = W × V for λ > 0. In fact, this is equivalent to Range (λ2 A + λA0 + P ) = V ∗ . But, by the Lax-Milgram theorem, it is actually true.

As a consequence of Theorem 4.23, we have the following result. Theorem 4.24 Let the ellipticity (4.172) of the shallow shell hold. Suppose that the condition (4.383) holds true. If initial data ζ0 ∈ W and ζ1 ∈ W are such that P ζ0 + A0 ζ1 ∈ V ∗ , (4.401) then the problem (4.385) admits a unique solution satisfying ζ ∈ C 1 ([0, ∞); W) ∩ C 2 ([0, ∞); V), Aζtt + A0 ζt + P ζ = 0, ζ(0) = ζ0 ,

ζtt ∈ C([0, ∞); V), t ≥ 0,

(4.402)

ζt (0) = ζ1 .

Trace Estimate on Boundary In order to remove some unnecessary restrictions on the control portion Γ0 , some trace estimates of the displacement vector field ζ = (W, w) of the shell have to be introduced. This will be done for the W -component and the w-component, separately. Consider the problem Wtt − ∆µ W = F in Q, (4.403) W |Γ1 = 0 on Σ1 where ∆µ is given by (4.37) and Q = (0, T ) × Ω, Set

Σ1 = (0, T ) × Γ1 .

   v1 (W, 0) = 1 − µ (DW + D∗ W )(ν, ν) − µδW, 2 (4.404) 1−µ   v2 (W, 0) = (DW + D∗ W )(ν, τ ). 2 We localize the problem (4.403) and use Proposition 3.2.2 in [126] to obtain


234


Proposition 4.6 Let 0 < α < T /2. Let W be a solution to the problem (4.403). Then there is CαT > 0 such that Z T −α Z Z T |Dτ W |2 dΓdt ≤ CαT kF k2H −1/2 (Ω,Λ) dt α 0 Z Γ0 2 2 +CαT (|Wt | + |v1 (W, 0)| + |v2 (W, 0)|2 ) dΣ + L(W ) (4.405) Σ0

where Σ0 = (0, T ) × Γ0 and L(W ) denotes some lower order terms related to the energy level kW k2H 1 (Q) . Remark 4.13 The proof of Proposition 4.6 in [126] was given by the microlocal arguments. To avoid introducing the microlocal analysis theory we refer the readers to [126] for the proof of Proposition 4.6 in detail. For the same reason we refer the readers to [123] for a detailed proof of Proposition 4.7 below. Consider the problem ( wtt − γ∆wtt + ∆2 w = f ∂w w= = 0 on Σ1 . ∂ν

in Q,

Let on Σ0 ( v3 (0, w) = γ[∆w − (1 − µ)D2 w(τ, τ )], ∂∆w ∂w ∂ v4 (0, w) = −γ{ + (1 − µ)[ ∂τ D2 w(τ, ν) + θ ]}. ∂ν ∂ν

(4.406)

(4.407)

Applying Theorem 2.1 in [123] in the case of dimension 2 to the problem (4.406), we obtain Proposition 4.7 Let 0 < α < T /2. Let 0 < ε < 1/2 and 0 < s0 < 1/2. Then the following inequality holds true for solutions w of the problem (4.406): Z T −α Z |D2 w|2 dΓdt ≤ Cα,ε,s0 ,T [kf k2H −s0 (Q) + kv3 (0, w)k2L2 (Σ0 ) α

Γ0

+kv4 (0, w) + γhDwtt , νik2H −1 (Σ0 ) + kDwt k2L2 (Σ0 ,Λ)

+kwt k2L2 (Σ0 ) + kwk2L2 (0,T ;H 3/2+ε (Ω)) ].

(4.408)

Now we are ready to prove the trace estimates: Theorem 4.25 Let T /2 > α > 0 be given. Then Z T −α Z (|DW |2 + |D2 w|2 ) dΓdt α Γ0 Z ≤ CT,α (|ζt |2 + |Dwt |2 ) dΣ + L(ζ) Σ0

for all solutions to the problem (4.385).


(4.409)


235

Proof. By the formulas (4.36) and (4.38), the W -component of ζ solves a problem (4.403) with F satisfying the estimate |F | ≤ c(|W | + |Dw| + |w|) which implies that Z

0

T

in Q,

kF k2H −1/2 (Ω,Λ) dt = L(ζ).

Noting that vi (W, 0) = vi (ζ) + lo (ζ), applying Proposition 4.6 to the component W yields, via the feedback laws in (4.385), T −α

Z

Z

Z |Dτ W |2 dΓdt ≤ CαT (|Wt |2 + |v1 (ζ)|2 + |v2 (ζ)|2 ) dΣ α Γ0 Σ0 Z 2 + lo (W ) ≤ CαT (|ζt | + |Dwt |2 ) dΣ + L(ζ). (4.410) Σ0

Moreover, by the formula (4.377) and the feedback laws (4.381), we obtain   DW (ν, ν) = F1 (ζt ) − [(1 − µ)Π(ν, ν)/2 + µH]w − µDW (τ, τ ) 2 F2 (ζt ) − DW (ν, τ )  DW (τ, ν) = 1−µ

which implies T −α

Z

α

Z

Γ0

|DνW |2 dΓdt ≤ C

Z

Σ0

(|ζt |2 + |Dτ W |2 + |w|2 )dΣ.

(4.411)

Then the estimates (4.410) and (4.411) together give Z

T −α

α

Z

Γ0

2

|DW | dΓdt ≤ CαT

Z

Σ0

(|ζt |2 + |Dwt |2 ) dΣ + L(ζ).

(4.412)

Next, we estimate the component w of ζ. By the formulas (4.36) and (4.38), the w-component of ζ solves a problem (4.406) with f satisfying the estimate |f | ≤ C(|DW | + |Dw| + |w|)

in Q

which implies that kf k2H −s0 (QT ) ≤ CkDW k2H −s0 (Ω,T 2 ) .

(4.413)

Let us prove kDW k2H −s0 (Ω,T 2 )

≤C

Z

0

T

kW k2H 1−s0 (Ω,Λ) dt.

(4.414)

For simplicity, we assume that Ω is a coordinate patch U with the coordinate


236


system x = (x1 , x2 ). Then Q = (0, T ) × U and W = (w1 , w2 ). Denote the Fourier transform variable of (t, x) by (s, y). By definition ∗ H −s0 (Q) = H0s0 (Q) . For u ∈ H −s0 (Q) given, we have

kuk2H −s0 (Q) = k(1 + s2 + |y|2 )−s0 /2 u ˆk2L2 (R3 ) ≤ k(1 + |y|2 )−s0 /2 u ˆk2L2 (R3 ) Z Z = (1 + |y|2 )−s0 |ˆ u|2 dsdy R2 R Z Z 2 −s0 = (1 + |y| ) |ˆ ux |2 dtdy R2 T

Z

=

0

0

kuk2H −s0 (Ω) dt

(4.415)

where u ˆx denotes the Fourier transform on the variable x. It follows from (4.415) that kDW k2H −s0 (Q,Λ) ≤ C ≤C ≤C ≤C

2 X i=1

(kw1xi k2H −s0 (Q) + kw2xi k2H −s0 (Q) )

2 Z X i=1 T

Z

0

Z

T

0

0

T

(kw1xi k2H −s0 (Ω) + kw2xi k2H −s0 (Ω) ) dt

(kw1 k2H 1−s0 (Ω) + kw2 k2H 1−s0 (Ω) ) dt kW k2H 1−s0 (Ω,Λ) dt.

By the inequalities (4.414) and (4.413), the term kf k2H −s0 (Q) is a lower order term. Noting that vi (0, w) = vi (ζ) for i = 1, 2, we apply Proposition 4.7 to the component w and obtain, via the feedback laws in (4.385), Z T −α Z |D2 w|2 dΓdt α

Γ0

∂ ˆ hζt , F5 i − hζˆt , F4 ik2H −1 (Σ0 ) ∂τ +kDwt k2L2 (Σ0 ,Λ) + kwt k2L2 (Σ0 ) + kwk2L2 (0,T ;H 3/2+ε (Ω)) ] + L(ζ) Z ≤ Cα,T (|ζt |2 + |Dwt |2 ) dΣ + L(ζ). (4.416) ≤ Cα,T [khζˆt , F3 ik2L2 (Σ0 ) + k

Σ0

The inequality (4.409) follows from (4.412) and (4.416).



237

Stabilization of the Closed-Loop System (4.385) For stabilization some geometric conditions are necessary. We make the following assumptions: Assumption H1 The ellipticity (4.172) holds; Assumption H2 There exists an escape vector field V for the shallow shell such that the conditions (4.173) and (4.175) hold; Assumption H3 Γ0 and Γ1 satisfy the following conditions Γ1 6= ∅,

Γ1 ∩ Γ0 = ∅,

and hV, νi ≤ 0

for x ∈ Γ1 ;

(4.417)

Assumption H4 F, given in (4.382), is symmetric and positive on Γ0 . Remark 4.14 The assumptions H1-H3 are geometric conditions on the middle surface of the shell, while the assumption H4 is on the feedback. For a plate the assumptions H1-H2 are automatically satisfied, if we set V = x − x0 . For the general case, the assumptions H1-H2 can be verified by the geometric method; see Section 4.5. Thanks to [123] and [126], the geometric assumption H3 is, generally, considered to be much weaker than the following: hV, νi ≤ 0

in

Γ1

and

hV, νi > 0

in

Γ0

(4.418)

which is used to avoid the complex trace estimates. We have Theorem 4.26 Let the assumptions H1-H4 hold. Then the closed-loop system (4.385) is exponentially stable: There are constants Ci > 0 such that E(t) ≤ C1 e−C2 t E(0)

t≥0

(4.419)

for all initial data ζ0 ∈ HΓ11 (Ω, Λ) × HΓ21 (Ω) and ζ1 ∈ L2 (Ω, Λ) × HΓ11 (Ω). Proof. By Theorem 4.13, we have 2σ1

Z

0

T

E(t)dt ≤

Z

SB dΣ + σ0 λ0 [E(0) + E(T )] + L(ζ)

(4.420)

Σ

where SB is given by (4.222) and σ0 , σ1 are given in (4.172). On Σ1 the boundary conditions in (4.385) and the assumption H3 imply that Z Z SB dΣ = B(ζ, ζ)hV, νidΣ ≤ 0 (4.421) Σ1

Σ1

(see the proof of Lemma 4.12). Let η = (U, u) := (2DV W + ̺W, 2V (w) − ϑw)


(4.422)

238


where ̺ is given in (4.222). Then by the formulas (4.377), (4.35) and the feedback laws in (4.385) on Σ0 ∂(Aζ, 2m(ζ) + (̺W, −ϑw)) + γ(2V (w) − ϑw)

∂wtt ∂ν

= ∂(Aζ, η) + γhDwtt , νiu ∂u = −hζˆt , F1 ihU, νi − hζˆt , F2 ihU, τ i − hζˆt , F3 i ∂ν ∂ ˆ +( hζt , F5 i − hζˆt , F4 i)u ∂τ which yields, via (4.388), Z ∂wtt ]dΓ [∂(Aζ, 2m(ζ) + (̺W, −ϑw)) + γ(2V (w) − ϑw) ∂ν Γ0 Z =− hζˆt , F ηîdΓ0 = −α0 (ζt , η). (4.423) Γ0

Using the formulas (4.423) and (4.222), we obtain Z Z SB dΣ = [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ Σ0 Σ0 Z − α0 (ζt , η)dt.

(4.424)

0

We estimate SB on Σ0 . Let si > 0 be given such that s1 |X|2 ≤ hF X, Xi ≤ s2 |X|2

for

X ∈ R5 , x ∈ Γ 0 .

Then the definition of (4.388) yields Z Z s1 (|ζ|2 + |Dw|2 ) dΓ ≤ α0 (ζ, ζ) ≤ s2 (|ζ|2 + |Dw|2 ) dΓ. Γ0

(4.425)

(4.426)

Γ0

It follows from the inequality (4.426) that |α0 (ζt , η)| ≤ [α0 (ζt , ζt )]1/2 [α0 (η, η)]1/2 Z ≤c (|ζt |2 + |Dwt |2 + |DW |2 + |D2 w|2 )dΓ + L(ζ).

(4.427)

Γ0

Using (4.427) in (4.424), we have Z Z SB dΣ ≤ (|ζt |2 + |Dwt |2 + |DW |2 + |D2 w|2 )dΓ + L(ζ). Σ0

(4.428)

Σ0

Next, change the integral domain (0, T ) into (α, T − α) in both sides of the inequality (4.420) and use Theorem 4.25 to give Z T −α σ1 E(t)dt ≤ CT [E(α) + E(T − α)] α Z + (|ζt |2 + |Dwt |2 )dΣ + L(ζ). (4.429) Σ0



239

d Note the relation E(t) = −α0 (ζt , ζt ) and the inequality (4.426) and we dt find, for any T > t > 0, E(t) = E(T ) +

Z

T

α0 (ζt , ζt )dt

t

≤ E(T ) + s2 ≥ E(T ) + s1

Z

Σ0

Z

Σ0

(|ζt |2 + |Dwt |2 )dΣ;

(4.430)

(|ζt |2 + |Dwt |2 )dΣ.

(4.431)

Using the inequalities (4.430) and (4.431) in the inequality (4.429), we obtain for T > 0 large enough Z E(T ) ≤ CT (|ζt |2 + |Dwt |2 )dΣ + L(ζ). Σ0

By the compactness and uniqueness (Proposition 4.3) approach, we now have Z E(T ) ≤ CT (|ζt |2 + |Dwt |2 )dΣ. (4.432) Σ0

Finally, using the inequality (4.432) and the left-hand side of the inequality (4.426) Z E(T ) ≤ CT

0

α0 (ζt , ζt )dt = CT (E(0) − E(T )),

that is, E(T ) ≤

CT E(0). 1 + CT

(4.433)

The estimate (4.419) follows from the inequality (4.433).

4.9

Stabilization by Nonlinear Boundary Feedback

We consider the stabilization problem of nonlinear feedbacks  ζtt − γ(0, ∆wtt ) + Aζ = 0 in Q∞ ,    ∂w  ∞    W = w = ∂ν = 0 on Σ1 ,  v1 (ζ) = −g1 (hWt , νi), v2 (ζ) = −g2 (hWt , τ i)  t  on Σ∞ v3 (ζ) = −g3 ( ∂w  0 , ∂ν ),   ∂wt ∂ ∞  v (ζ) + γhDw , νi = g ( 4 tt 4  ∂τ ∂τ ) on Σ0 ,  ζ(0) = ζ0 , ζt (0) = ζ1 on Ω © 2011 by Taylor & Francis Group, LLC

on Σ∞ 0 ,

(4.434)

240


where A is given by the formula (4.36) and vi (ζ) are given in (4.377) for 1 ≤ i ≤ 4. In the above feedback laws gi are nonlinear functions for 1 ≤ i ≤ 4. A starting point is, as usual, an equality which states how the energy E(t) in (4.375) of the entire system (4.434) is affected by feedback laws hi . This fact will tell us how to make assumptions on nonlinear functions gi such that the energy E(t) is decreasing. Using the boundary conditions of (4.434) in the formula (4.376), we obtain Lemma 4.19 Let ζ be a finite energy solution of system (4.434). Then, for any 0 ≤ s ≤ t , the following identity holds true for the energy E(t) defined by (4.375): Z tZ E(t) + i(ζt , ζt ) dΓdt = E(s) (4.435) s

Γ0

where

i(ζt , ζt ) =

4 X

gi (ζît )ζît ,

(4.436)

i=1

ζˆ1 = hW, νi,

ζˆ2 = hW, τ i,

ζˆ3 = hDw, νi,

ζˆ4 = hDw, τ i.

(4.437)

We make the following assumption on nonlinear function gi : ˜ Let gi ∈ C 1 (R) be such that gi (0) = 0 and Assumption H4 gi (s)s > 0

for s ∈ R, s 6= 0

(4.438)

for 1 ≤ i ≤ 4. Moreover, there exist positive constants mi > 0 for 1 ≤ i ≤ 3, such that for all s ∈ R with |s| ≥ m3 , we have m1 |s|2 ≤ gi (s)s ≤ m2 |s|2

for 1 ≤ i ≤ 4.

(4.439)

The well-posedness of the system (4.434) can be established by the nonlinear semigroup theory (see [6]). We only introduce the regularity results on this issue in Proposition 4.8 below. For the proofs, we refer the readers to [122] and [115]. ˜ hold. Proposition 4.8 Let the ellipticity (4.172) and the assumption H4 Then there exists a unique, global solution of finite energy to the problem (4.434). This is to say that for any initial data ζ0 ∈ HΓ11 (Ω, Λ) × HΓ21 (Ω) and ζ1 ∈ L2 (Ω, Λ) × HΓ11 (Ω), there exists a unique solution ζ ∈ C([0, T ]; HΓ11 (Ω, Λ) × HΓ21 (Ω)), ζt ∈ C([0, T ]; L2 (Ω, Λ) × HΓ11 (Ω)),

where T > 0 is arbitrary.

In addition, as in the case of the linear feedbacks, to avoid some unnecessary geometric assumptions on the uncontrolled part of the boundary, the trace estimates in [126] below are crucial, which can be proven by the similar arguments as in the proof of Theorem 4.25.



241

Theorem 4.27 Let 0 < α < T /2 be given. Then the inequality (4.409) holds true for all solutions ζ = (W, w) to the problem (4.434). ˜ hold. Following [115] Uniform Stabilization Let the assumption H4 we construct a function h below. Let a function h be concave and strictly increasing, and vanishing at the origin: h(0) = 0 such that the following inequalities are satisfied: h(gi (s)s) ≥ |s|2 + |gi (s)|2

|s| ≤ m3 ,

for

1≤i≤4

(4.440)

where m3 > 0 is given in (4.439). We define first the function h0 (·) by h0 (·) = C1 h(

· ) 8 meas Σ0

(4.441)

for some C1 > 0 which will be specified later. Since h0 is monotone increasing, for C > 0, CI + h0 is invertible. We next define the function p(s) = (C2 I + h0 )−1 (s)

(4.442)

for some C2 > 0 which will be specified later. Then p is a positive, continuous, strictly increasing function with p(0) = 0. Since p is positive, increasing, q(x) is. Finally, let q = I − (I + p)−1 . (4.443) We need Lemma 4.20 Consider a sequence { sk | k = 0, 1, · · · } of positive numbers which satisfies sk+1 + p(sk+1 ) ≤ sk for k ≥ 0 (4.444) where the function p is given by (4.442). Then sk ≤ S(k)

for all

k≥0

(4.445)

S(0) = s0 .

(4.446)

where S(t) is the solution to the problem d S(t) + q(S(t)) = 0, dt Moreover, limt→∞ S(t) = 0. Proof. Use induction. Assume that sk ≤ S(k) and prove that sk+1 ≤ S(k + 1). The inequality (4.444) is equivalent to sk+1 ≤ (I + p)−1 (sk ). On the other hand, the equation (4.446) gives Z S(t) ≤ S(t) + q(S(t))dt = S(s) for 0 ≤ s ≤ t, s


(4.447)

(4.448)

242


since q is positive. Since (I +p)−1 is monotone increasing, from (4.446)-(4.448), we have Z k+1 S(k + 1) = S(k) − q(S(t)) dt ≥ S(k) − q(S(k)) k

= (I − q)(S(k)) = (I + p)−1 (S(k)) ≥ (I + p)−1 (sk ) ≥ sk+1 .

Finally, it is obvious that the limit c := limt→∞ S(t) exists. Then the fact R∞ that the integral 0 q(S(t))dt converges implies 0 = limt→∞ q(S(t)) = q(c), that is, c = 0. We are ready to prove Theorem 4.28 Assume that the assumptions H1-H3 in Section 4.8 hold. Let ˜ in this section hold. Then there exists a constant T0 > 0 the assumption H4 such that the following estimate holds true: E(t) ≤ S(

t − 1) T0

for

t > T0

(4.449)

where S(t) is the solution of the ordinary differential equation d S(t) + q(S(t)) = 0 dt

S(0) = E(0),

(4.450)

where the function q(x) is given by (4.443). Proof. As in the proof of Theorem 4.26, we will use the observability inequality (4.237) in Theorem 4.13 as a base. Step 1 Let T /2 > α > 0 be given. Applying the observability inequality (4.237) in Theorem 4.13 with Q replaced by Ω × [α, T − α] yields Z T −α Z 2σ1 E(t) dt ≤ SB dΓdt + σ0 λ0 [E(α) + E(T − α)] α

(α,T −α)×Γ

+L(ζ)

(4.451)

where SB is given by (4.222). The key is to deal with the boundary integral in the right hand side of (4.451). Clearly, the inequality (4.421) still holds true for solution ζ of the problem (4.434) in which Σ1 is replaced by Γ1 × (α, T − α), that is, Z SB dΓdt ≤ 0. (4.452) (α,T −α)×Γ1

Therefore we need to estimate SB |Γ0 ×(α,T −α) where the feedback laws in (4.434) will enter the formula (4.222). Let η = (U, u) be given by (4.422): U = 2DV W + βW,


u = 2V (w) − ϑw.


243

Let ηˆ1 = hU, νi,

ηˆ2 = hU, τ i,

ηˆ3 = hDu, νi,

ηˆ4 = hDu, τ i.

(4.453)

(x, t) ∈ Σ0 .

(4.454)

It follows that 4 X i=1

|ˆ ηi |2 ≤ c(|DW |2 + |D2 w|2 ) + lo (ζ)

for

A similar computation as in (4.424) yields Z Z SB dΣ = [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ Γ0 ×(α,T −α) Γ0 ×(α,T −α) Z − i(ζt , η)dΣ (4.455) Γ0 ×(α,T −α)

where i(·, ·) is given in (4.436) and ζî , ηî are given in (4.437) and (4.453), respectively. ˜ for gi and the formulas (4.436) and (4.454), we Using the assumption H4 have 4 4 4 X X X |i(ζt , η)|2 ≤ ( |gi (ζît )|2 |ˆ ηi |2 )1/2 ≤ (|gi (ζît )|2 + |ˆ ηi |2 ) i=1 4 X

≤ c(

i=1

i=1

i=1

|gi (ζît )|2 + |DW |2 + |D2 w|2 ) + lo (ζ).

(4.456)

It follows from the relations (4.456), (4.455), and the trace estimates (4.409) that Z SB dΣ Γ0 ×(α,T −α)

≤ Cα,T

Z

= Cα,T

4 Z X

Γ0 ×(α,T −α)

i=1

Σ0

(|ζt |2 + |Dwt |2 +

4 X i=1

|gi (ζît )|2 )dΣ + L(ζ)

(|gi (ζît )|2 + |ζît |2 ) dΣ + L(ζ).

(4.457)

On the other hand, we use the identity (4.435) where t = T and s = T − α to obtain Z Z E(T − α) = E(T ) + i(ζt , ζt )gdΓdt T −α

≤ E(T ) + c


Γ0

4 Z X i=1

Σ0

(|gi (ζît )|2 + |ζît |2 ) dΣ.

(4.458)

244


Similarly, E(α) ≤ E(T ) + c

4 Z X

Σ0

i=1

(|gi (ζît )|2 + |ζît |2 ) dΣ.

(4.459)

Using the inequalities (4.459), (4.458), (4.457), and (4.452) in the inequality (4.451), we obtain 2σ1

Z

T −α

α

E(t) dt ≤ 2σ0 λ0 E(T ) + L(ζ) 4 Z X

+CT

i=1

(|gi (ζît )|2 + |ζît |2 ) dΣ.

Σ0

(4.460)

Moreover, we let t = T in the identity (4.435), integrate it with respect to the variable s over (α, T − α), and have, via (4.460) (T − 2α)E(T ) = Z

≤

T −α

Z

T −α α

4 Z X

E(s)ds + CT

α

Σ0

i=1

2σ0 λ0 E(T ) + CT σ1

≤

E(s)ds −

4 Z X i=1

Σ0

Z

T −α

α

Z Z s

i(ζt , ζt )dΓdtds Γ0

(|gi (ζît )|2 + |ζît |2 ) dΣ

(|gi (ζît )|2 + |ζît |2 ) dΣ + lo (ζ).

(4.461)

Combining the inequality (4.461) and the compactness/uniqueness argument by Proposition 4.3 yields, for T large enough, E(T ) ≤ CT

4 Z X i=1

Σ0

(|gi (ζît )|2 + |ζît |2 ) dΣ.

(4.462)

Step 2 Set ΣiA = { ζît ∈ L2 (Σ0 ) | |ζît | ≥ m3 },

ΣiB = Σ0 − ΣiA ,

where m3 > 0 is given in (4.439). By the assumption (4.439), we have m2 + 1 (|gi (ζît )|2 + |ζît |2 ) dΣ ≤ 2 m1 ΣiA

Z

Z

gi (ζît )ζît dΣ

for 1 ≤ i ≤ 4. On the other hand, it follows from the assumption (4.440) that Z Z (|gi (ζît )|2 + |ζît |2 ) dΣ ≤ h(gi (ζît )ζît )dΣ ΣiB


(4.463)

Σ0

Σ0

(4.464)


245

for 1 ≤ i ≤ 4. Since the function h is concave, the Jensen inequality states that 4 Z X i=1

h(gi (ζît )ζît )dΣ ≤

Σ0

4 Z X ≤ h0 2 i=1

Σ0

4 X

( meas Σ0 )h

i=1

1 meas Σ0

Z

gi (ζît )ζît dΣ Σ0

gi (ζît )ζît dΣ /CT

(4.465)

where h0 is given in (4.441) with C1 = 4CT meas Σ0 where CT is given in (4.462). Using the inequalities (4.465), (4.464), and (4.463) in the inequality (4.462), we obtain 4 Z m2 + 1 X E(T ) ≤ CT 2 I + h0 (2 gi (ζît )ζît dΣ , 2m1 i=1 Σ0

that is, via (4.442) and (4.435), p(E(T )) ≤ 2

4 Z X i=1

Σ0

gi (ζît )ζît dΣ = E(0) − E(T ),

or E(T ) + p(E(T )) ≤ E(0),

(4.466)

where C2 = CT (m22 + 1)/(2m1 ) in (4.442). Step 3 Applying the inequality (4.466) yields E((k + 1)T ) + p(E((k + 1)T )) ≤ E(kT ) for

k ≥ 0.

(4.467)

We now take sk = E(kT ) for k ≥ 0. Thus Lemma 4.20 gives E(kT ) ≤ S(k) for

k ≥ 0.

(4.468)

Setting t = kT + τ where 0 ≤ τ < T, the decreasing of E(t) and S(t) yields E(t) ≤ E(kT ) ≤ S(k) = S( for t ≥ T.

t−τ t ) ≤ S( − 1) T T

If we, in addition, assume that the functions gi are of a polynomial growth at the origin, then the following explicit decay rates are ready:


246


Corollary 4.5 Assume that the assumptions H1-H3 in Section 4.8 hold. Let ˜ in this section hold. Moreover, suppose that there are σ1 > the assumption H4 0 and σ2 > 0 such that gi (s)s ≤ σ1 s2 gi (s)s ≥ σ2 |s|p+1

for

for all

|s| ≤ 1,

s ∈ R,

1 ≤ i ≤ 4,

Then E(t) ≤ Ce−σ3 t

if p = 1;

some p ≥ 1.

2

E(t) ≤ Ct 1−p

if p > 1

where both constants C > 0 and σ3 > 0 depend in general on E(0). Proof. Let us construct a function h to meet the assumption (4.440). Set 1 + σ12

h(s) =

2/(p+1)

σ2

s2/(p+1) .

Then the inequalities (4.440) hold true with m3 = 1. Thus p(s) = (C2 I + h0 )−1 (s) and C2 p(s) + C3 pm (s) = s for m = 2/(1 + p) and a suitable constant C3 > 0. Near s = 0, we have a asymptotic behavior p(s) ∼ Cs1/m

and q(x) ∼ Cs1/m .

Then S(t) behaves like S(t) ∼

2

Ct 1−p p > 1, Ce−σ3 t p = 1

near ∞.

4.10


Sections 4.1, 4.2, and 4.3 are from [210]; Sections 4.4, 4.5, and 4.6 are from [211] and [67]; Section 4.7 is from [34]; Section 4.8 is from [32]; Section 4.9 is from [126]. Control problems of shallow shells were studied by several papers where the middles have special shapes. For circular cylindrical shells, the exponential decay of the energy by boundary feedbacks was obtained by [42]. For spherical shells uniform stabilization by boundary dissipation was given by [129] (for the linear case) and by [113] (for the nonlinear case). Classically, the topics of thin shells were covered by many books, for example, see [59], [94], and [95]. In addition, [19], [49]-[54], by constructing the geometric properties of the middle surface through the oriented boundary distance function in Rn and by using tangential differential operators on a



247

neighborhood of a submanifold of Rn , presented the modeling and analysis of thin shells. In classical shell theory the middle surface of a shell is described by one coordinate patch: A map from a domain of R2 to R3 . Therefore shell models end up with highly complicated resultant equations where the strain tensor and the change of curvature tensor of the middle surface are as unknowns. In these formulas, the explicit presence of the Christoffel symbols makes them unsuitable for energy method computations of the type needed for the control problems, such as continuous observability/stabilization estimates. For example, in [95], p. 33, Koiter commented on this point, “Expressing the strain measures in terms of the displacement components and their derivatives, we obtain three equations in terms of the three displacement components as unknowns. This approach is not very attractive because the resulting so-called displacement equations of equilibrium are undoubtedly extremely complicated. As far as we are aware, such general displacement equations have never appeared in print, not even in the linear theory of shells.” However, the above points of Koiter were just right in the context of classical geometry, that is, those were just the limits of classical geometry. We view the middle surface of a shell as a Riemannian manifold of dimension 2 to derive its mathematical models in the form of coordinates free. Then the Bochner technique helps us overcome the complicated computation which is necessary in the modeling and analysis of controllability/stabilization. More importantly, it seems clear that, without this differential geometric tool, many of these more sophisticated theorems would probably not have been discovered or, at least, their discoveries would otherwise have been much delayed. The shell model in the present form of free coordinates (Theorems 4.1, 4.4) was worked out by the author more than 10 years ago but those equations were just published in [210] recently. Based on those equations and with the help of the Bochner technique a series of results on control of shallow shells have been obtained. The ellipticity of the shallow shell was first obtained by [14] under the “shallowness” assumptions: Π and DΠ are small enough. Then the ellipticity was given by [210] which was based on some curvature assumptions. The results of Theorem 4.2 in Section 4.2 are new, first published here where we have removed the “shallowness” assumptions in [14] and the curvature assumptions in [210] due to Lemma 4.5. Observability estimates from boundary (Theorems 4.15, 4.16) were given in [211] where the assumptions (4.173) and (4.175) of an escape vector field for the shallow shell were introduced and a number of examples were given. The internal exact controllability of Theorem 4.17 was obtained by [67] where the geometric conditions (4.210)-(4.214) of an escape region for the shallow shell were introduced. As we can see in Section 4.7, the excellent work [34] made the important


248


contribution: Established new geometric conditions (H3) which reflect material properties and shapes of the division curve if we have exact controllability only from boundary for the transmission problem. Uniform stabilization by linear boundary feedbacks was given by [32] which closely followed [105] in seeking stabilization results. We use the trace estimates (Proposition 4.6) in [126] directly to clarify some unsatisfactory reasoning in the proofs of those in [32] which are crucial for avoiding geometric assumptions on uncontrolled portion. Uniform stabilization by nonlinear boundary feedbacks came from an excellent work [126] in which the nonlinear feedback laws are taken from [115]. One of the main contributions was a sharp trace theory of the elastic W component (Proposition 4.6 or Section 3.2 of [126]) by extending the microlocal argument which was originated in [122] for the wave equation.


Chapter 5 Naghdi’s Shells. Modeling and Control

5.1 5.2 5.3 5.4 5.5

Equations of Equilibrium. Green’s Formulas. Ellipticity of the Strain Energy. Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Observability Estimates from Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stabilization by Boundary Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stabilization of Transmission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

249 261 270 277 285 285

The strain tensors for Naghdi’s shells are from [157] and [158]. By viewing the middle surface of the shell as a Riemannian manifold with the induced metric in R3 , Naghdi’s shells will be formulated mathematically as in the case of shallow shells. Then the multiplier scheme is carried out with help of the Bochner technique. Boundary exact controllability and boundary stabilization are derived. Stabilization of transmission is also presented.

5.1

Equations of Equilibrium. Green’s Formulas. Ellipticity of the Strain Energy. Equations of Motion

Green’s Formulas and Equations of Equilibrium We keep all notations as before. Let us assume that the middle surface of the shell occupies a bounded region Ω of a surface M in R3 with a normal field N. The shell, a body in R3 , is defined by S = { p | p = x + zN (x), x ∈ Ω, −h/2 < z < h/2 }

(5.1)

where h is the thickness of the shell, small. The second fundamental form Π of M is given by ˜ X N, Y i ∀ X, Y ∈ X (M ) Π(X, Y ) = hD ˜ is the covariant differential of the standard metric of R3 . For each where D x ∈ Ω, Π is a symmetric, bilinear form over Mx × Mx so that there is a symmetric, linear operator S: Mx → Mx such that Π(X, Y ) = hSX, Y i

∀ X, Y ∈ Mx , x ∈ Ω.

(5.2) 249


250


The third fundamental form of the surface M is defined by ˜ X N, D ˜ Y N i ∀ X, Y ∈ X (M ). c(X, Y ) = hD

(5.3)

In the Naghdi model, the displacement vector ζ(p) of point p = x+zN (x) ∈ S of the three-dimensional shell can be approximated as follows: ζ(p) = ζ1 (x) + z ⊓ (x)

x∈Ω

(5.4)

(see (7.67) in [157]) where ζ1 (x) ∈ R3 is the displacement vector field of the middle surface and ⊓(x) ∈ R3 represents deformation of the normal N (x) for each x ∈ Ω. In the literature, ⊓(x) are called the director displacement vector which is capable of rotation and stretches independently of the deformation of material points. We decompose vector fields ζ1 and ⊓ into sums ζ1 (x) = W1 (x) + w1 (x)N (x)

and

⊓ (x) = V (x) + w2 (x)N (x),

(5.5)

respectively, where W1 , V ∈ X (Ω). [157] ([7.59] and [7.55] in [157]) yields the following tensor fields, directly defined on the middle surface, Υ(ζ) = X0 (ζ) =

1 (DW1 + D∗ W1 ) + w1 Π, 2

1 [DV + D∗ V + Π(·, D· W1 ) + Π(D· W1 , ·)] + w2 Π + w1 c, 2

(5.6) (5.7)

and

1 [Dw1 + V − i(W1 )Π], (5.8) 2 where Υ(ζ), X0 (ζ), and ϕ0 (ζ) denote respectively the strain tensor of the middle surface, the change of curvature tensor of the middle surface, and the rotation of the normal N . In the above expressions we have made the use of the canonical isomorphism: Mx = Mx∗ as Mx are inner product spaces of dimension 2 for x ∈ M. i(W1 )Π is the interior product of the tensor field Π by the vector field W1 . The formulas (5.6) − (5.8) are also called the strain tensors of Naghdi. We make a change of variables by ϕ0 (ζ) =

W2 = V + i(W1 )Π

for x ∈ Ω.

(5.9)

Let x ∈ Ω be given and let E1 , E2 be a frame field normal at x. Using the relations DEi Ej (x) = 0, we obtain DW2 (Ei , Ej ) = Ej hW2 , Ei i = Ej hV, Ei i + Ej (Π(W1 , Ei ))

= DV (Ei , Ej ) + Π(Ei , DEj W1 ) + DΠ(W1 , Ei , Ej ),

that is, DW2 = DV + Π(·, D· W1 ) + i(W1 )DΠ


for x ∈ Ω.

(5.10)

Naghdi’s Shells. Modeling and Control

251

Inserting (5.10) into (5.7) and (5.9) into (5.8), respectively, we have X0 (ζ) =

1 (DW2 + D∗ W2 ) + X0L (ζ) 2

(5.11)

1 Dw1 + ϕ0L (ζ), 2

(5.12)

and ϕ0 (ζ) = where

X0L (ζ) = −i(W1 )DΠ + w1 c + w2 Π,

1 ϕ0L (ζ) = −i(W1 )Π + W2 2

(5.13)

are zero order terms. Denote by Eˆ the strain tensor of the three-dimensional shell. By taking the first approximation from (7.61), (7.59) and (7.55) in [157], we obtain Naghdi’s displacement-strain relations (7.65) in [157] as follows:  ˆ   E|Ω (p) = Υ(ζ) + zX0z(ζ), ˆ i(N )E(p) = ϕ0 (ζ) + Dw2 , for p = x + xN (x) ∈ S (5.14)  2 ˆ E(N, N )(p) = w2 ,

ˆ Ω denotes the component of the three-dimensional shell strain Eˆ on where E| the middle surface. We assume that the material of the shell is homogenous and isotropic. Let x ∈ Ω be given. Let E1 , E2 be an orthonormal frame at x in the induced metric g of the surface M from the Euclidean metric of R3 . Then E1 , E2 and E3 forms an orthonormal frame in the Euclidean metric of R3 where E3 = N. The stress-strain relations of the three-dimensional shell on the middle surface are E ˆ µ ˆ ij ] for x ∈ Ω σij = [Eij + tr Eδ (5.15) 1+µ 1 − 2µ ˆ i , Ej ), E is Young’s modulus, for 1 ≤ i, j ≤ 3 where σij = σ(Ei , Ej ), Eîj = E(E and µ is Poisson’s ratio. It follows from the formula (5.15) that X ij

σij Eîj = =

E ˆ 2 + µ (tr E) ˆ 2] [|E| 1+µ 1 − 2µ 2 2 2 X X E X ˆ2 µ 2 2 [ Eij + 2 Eî3 + Eˆ33 + ( Eîi + Eˆ33 )2 ] 1 + µ ij=1 1 − 2µ i=1 i=1

E ˆ Ω |2 + 2|i(N )E| ˆ 2 + Eˆ2 (N, N ) [|E| 1+µ µ ˆ Ω + E(N, ˆ + (tr E| N ))2 ] 1 − 2µ

=

(5.16)

for x ∈ Ω. Then the strain energy of the three-dimensional shell is obtained


252


by integrating (5.16) over S : I(ζ) = α

Z Z Ω

h/2

J(x, z) dzdx

(5.17)

−h/2

where ˆ Ω |2 + 2|i(N )E| ˆ 2 + Eˆ2 (N, N ) J(x, z) = [|E| ˆ Ω + E(N, ˆ +β(tr E| N ))2 ](1 + tr Πz + κz 2 ), α = E/(1 + µ),

β = µ/(1 − 2µ),

(5.18) (5.19)

where κ is the Gauss curvature of the middle surface. By inserting the formula (5.14) into the formula (5.18), we have J(x, z) = {|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 +[2hΥ(ζ), X0 (ζ)i + hϕ0 (ζ), Dw2 i + 2β(tr Υ(ζ) + w2 ) tr X0 (ζ)]z 2

+[|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)]z 2 }(1 + tr Πz + κz 2 ) = |Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 +{[|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 ] tr Π

+2hΥ(ζ), X0 (ζ)i + hϕ0 (ζ), Dw2 i + 2β(tr Υ(ζ) + w2 ) tr X0 (ζ)}z 2

+{|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)

+[|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 ]κ +[2hΥ(ζ), X0 (ζ)i + hϕ0 (ζ), Dw2 i + 2β(tr Υ(ζ) + w2 ) tr X0 (ζ)] tr Π}z 2 +{[2hΥ(ζ), X0 (ζ)i + hϕ0 (ζ), Dw2 i + 2β(tr Υ(ζ) + w2 ) tr X0 (ζ)]κ 2

+[|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)] tr Π}z 3 2

+[|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)]κz 4 .

(5.20)

Then it follows from the formula (5.20) that Z

h/2

J(x, z) dz

−h/2

= [|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 ]h 2

+{|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ) +

+[|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 ]κ + [2hΥ(ζ), X0 (ζ)i +hϕ0 (ζ), Dw2 i + 2β(tr Υ(ζ) + w2 ) tr X0 (ζ)] tr Π}h3 /12 2

+[|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)]κ h5 /80.

(5.21)

As usual in the thin shell theory, we assume that h/R ≪ 1


(5.22)


253

where R is the smallest principal radius of the curvature of the undeformed middle surface is given by (4.25). By Lemma 4.2, the assumption (5.22), and the formula (5.21), the following approximation to the strain energy of the shell is justified Z I(ζ) = αh {|Υ(ζ)|2 + 2|ϕ0 (ζ)|2 + w22 + β(tr Υ(ζ) + w2 )2 Ω

2

+γ[|X0 (ζ)|2 + |Dw2 |2 /2 + β tr X0 (ζ)]} dx

(5.23)

where γ = h2 /12. Now we are able to associate the following symmetric, bilinear form with the strain energy, directly defined on space (H 1 (Ω, Λ))2 × (H 1 (Ω))2 , Z αh B0 (ζ, η) = B0 (ζ, η) dx (5.24) 2 Ω where B0 (ζ, η) = 2hΥ(ζ), Υ(η)i + 4hϕ0 (ζ), ϕ0 (η)i + 2w2 u2 +2β(tr Υ(ζ) + w2 )(tr Υ(η) + u2 ) + 2γhX0 (ζ), X0 (η)i +γhDw2 , Du2 i + 2γβ tr X0 (ζ) tr X0 (η)

(5.25)

where ζ, η ∈ (H 1 (Ω, Λ))2 × (H 1 (Ω))2 are given by ζ = (W1 , W2 , w1 , w2 ),

η = (U1 , U2 , u1 , u2 ).

ˆ be a relatively open portion of boundary Γ of the middle surface. Let Γ Set L2 (Ω) = (L2 (Ω, Λ))2 × (L2 (Ω))2 , 1

1

2

1

2

(5.26)

H (Ω) = (H (Ω, Λ)) × (H (Ω)) ,

(5.27)

HΓ1ˆ (Ω) = (HΓˆ1 (Ω, Λ))2 × (HΓˆ1 (Ω))2 .

(5.28)

Suppose that Load (·) is a linear functional which is associated with the external load, given by Load (ζ) = (ζ, F )L2 (Ω)

for all ζ ∈ L2 (Ω)

(5.29)

for some F ∈ L2 (Ω). Then, for a Naghdi’s shell which is clamped along Γ0 and free on Γ1 , where Γ0 ∪ Γ1 = Γ, we derive the following variational problem: For F ∈ L2 (Ω), find ζ ∈ HΓ1 0 (Ω) such that B(ζ, η) = Load (η)

(5.30)

for all η ∈ HΓ1 0 (Ω). Similarly, we may propose our problems for other kinds of boundary conditions by choosing different Sobolev spaces.


254


Let ∆β be an operator of the Hodge-Laplace type, given by ∆β = −[δd + 2(1 + β)dδ]

(5.31)

which will be applied to vector fields or equivalently to 1-forms, where d is the exterior derivative, δ is its formal adjoint, given by (1.109), and β is given in (5.19). For x ∈ Ω and G ∈ T 2 (Ω) given, hG, i(X)DΠi is a linear functional with respect to X ∈ Mx . Then there is a vector in Mx , denoted by PG, such that hG, i(X)DΠi = hPG, Xi for X ∈ Mx .

(5.32)

Therefore P: T 2 (Ω) → X (Ω) is a linear operator. The following Green formulas are the key to the boundary control problems which present the relations between the interior and the boundary of a displacement vector field. Theorem 5.1 Let the bilinear form B0 (·, ·) be given by (5.24). For ζ = (W1 , W2 , w1 , w2 ), η = (U1 , U2 , u1 , u2 ) ∈ H1 (Ω), we have Z αh αh B0 (ζ, η) = (A0 ζ, η)L2 (Ω) + ∂(A0 ζ, η) dΓ (5.33) 2 2 Γ where ∂(A0 ζ, η) = hB01 (ζ), U1 i + γhB02 (ζ), U2 i + 2hϕ0 (ζ), νiu1 + γ

∂w2 u2 , (5.34) ∂ν

ν, τ are the normal and the tangential along the boundary Γ, L2 (Ω), H1 (Ω) are the Sobolev spaces given by (5.26) and (5.27), respectively, B01 (ζ) = 2 i (ν)Υ(ζ) + 2β(tr Υ(ζ) + w2 )ν, (5.35) B02 (ζ) = 2 i (ν)X0 (ζ) + 2β(tr X0 (ζ))ν, A0 ζ = −(∆β W1 + F1 (ζ),

γ∆β W2 + F2 (ζ), ∆w1 + f1 (ζ), γ∆w2 + f2 (ζ)),

(5.36)

Fi (ζ), fi (ζ) are first order terms (≤ 1) for i = 1, 2, ∆β is given by (5.31), and ∆ is the Laplacian of the Riemann manifold M with the induced metric from R3 . Proof. Denote by F (ζ) lower order terms (≤ 1) which may be different from line to line and from term to term. It follows from the formula (5.25) that Z B0 (ζ, η) dx = 2(Υ(ζ), Υ(η))L2 (Ω,T 2 ) + 4(ϕ0 (ζ), ϕ0 (η))L2 (Ω,Λ) Ω

+2(w2 , u2 ) + 2β(tr Υ(ζ) + w2 , tr Υ(η) + u2 )

+2γ(X0 (ζ), X0 (η))L2 (Ω,T 2 ) + γ(Dw2 , Du2 )L2 (Ω,Λ) +2γβ(tr X0 (ζ), tr X0 (η)).


(5.37)


255

Let us compute each term in the right hand side of the formula (5.37), separately, as follows. By the formula (5.6) and the formula (4.44) in Chapter 4, we obtain 2(Υ(ζ), Υ(η))L2 (Ω,T 2 ) = ((δd + 2dδ)W1 + F (ζ), U1 )L2 (Ω,T 2 ) Z +(F (ζ), u1 ) + 2 Υ(ζ)(ν, U1 ) dΓ.

(5.38)

Γ

It follows from the divergence formula and the formulas (5.12) and (5.13) that 4(ϕ0 (ζ), ϕ0 (η))L2 (Ω,Λ) = (2ϕ0 (ζ), Du1 + 2ϕ0L (η))L2 (Ω,Λ) Z = −(∆w1 + F (ζ), u1 ) + 2 hϕ0 (ζ), νiu1 dΓ Γ

+(F (ζ), U1 )L2 (Ω,Λ) + (F (ζ), U2 )L2 (Ω,Λ)

(5.39)

where the following computation is used (by (5.2)) (2ϕ0 (ζ), −2 i (U1 )Π)L2 (Ω,Λ) Z Z = −4 hϕ0 (ζ), i (U1 )Πi dx = −4 Π(ϕ0 (ζ), U1 ) dx Ω

Ω

= −4(Sϕ0 (ζ), U1 )L2 (Ω,Λ) = (F (ζ), U1 )L2 (Ω,Λ) .

Using the formula (4.46), we have 2β(tr Υ(ζ) + w2 , tr Υ(η) + u2 )L2 (Ω) = 2β(dδW1 + F (ζ), U1 )L2 (Ω,Λ) + 2β

Z

(tr Υ(ζ) + w2 )hU1 , νi dΓ

Γ

+(F (ζ), u1 ) + (F (ζ), u2 ).

(5.40)

Using the formulas (5.11), (1.155) and (1.156), we obtain 2γ(X0 (ζ), X0 (η))L2 (Ω,T 2 )

1 (DU2 + D∗ U2 ) + X0L (η))L2 (Ω,Λ) 2 = γ(DW2 , DU2 )L2 (Ω,Λ) + γ(D∗ W2 , DU2 )L2 (Ω,Λ) + (F (ζ), U2 )L2 (Ω,Λ) +(F (ζ), U1 )L2 (Ω,Λ) + (F (ζ), u1 ) + (F (ζ), u2 ) Z = γ((δd + 2dδ)W2 , U2 )L2 (Ω,Λ) + γ (hDν W2 , U2 i + hDU2 W2 , νi) dΓ = γ(DW2 + D∗ W2 + X0L (ζ),

Γ

+(F (ζ), U2 )L2 (Ω,Λ) + (F (ζ), U1 )L2 (Ω,Λ) +(F (ζ), u1 ) + (F (ζ), u2 ).

(5.41)

In addition, Z

hDν W2 , U2 i + hDU2 W2 , νi) dΓ Z = 2γ X0 (ζ)(ν, U2 )dΓ + (F (ζ), U2 )L2 (Ω,Λ) γ

Γ

Γ


(5.42)

256


and the Green formula for the Laplacian states that Z ∂w2 γ(Dw2 , Du2 )L2 (Ω,Λ) = γ(−∆w2 , u2 ) + γ u2 dΓ. ∂ν Γ

(5.43)

Since tr X0 (η) = −δU2 − tr i (U1 )DΠ + u1 tr c + u2 tr Π, by (1.137) in Chapter 1, we have 2γβ(tr X0 (ζ), tr X0 (η)) = 2γβ(dδW2 + F (ζ), U2 )L2 (Ω,Λ) + 2γβ

Z

Γ

hU2 , νi tr X0 (ζ)dΓ

+(F (ζ), U1 )L2 (Ω,Λ) + (F (ζ), u1 ) + (F (ζ), u2 ).

(5.44)

Finally, we add up all the formulas from (5.38) to (5.44) to obtain the formula (5.33). If the middle surface is flat, then Π = c = 0. The formulas (5.6)-(5.8) become   2Υ(ζ) = DW1 + D∗ W1 , 2X0 (ζ) = DW2 + D∗ W2 , (5.45)  2ϕ0 (ζ) = Dw1 + W2 , where W2 = V. From the formulas (5.45) and the proof of Theorem 5.1, the following corollary and theorem are immediate. Corollary 5.1 For a plate, we have   ∆β W1 + 2βdw2   ∆β W2 − W2 − dw1  A0 ζ = −    ∆w1 − δW2 γ∆w2 − 2(1 + β)w2 + 2βδW1

(5.46)

where ζ = (W1 , W2 , w1 , w2 ). If Φ = (φ1 , φ2 ) in the usual coordinate system (x, y) of R2 , then ∂ 2 φ1 ∂ 2 φ1 ∆φ1 = + , ∂x2 ∂y 2   ∂ 2 φ1 ∂ 2 φ1 ∂ 2 φ2  2(1 + β) ∂x2 + ∂y 2 + 2(1 + β) ∂x∂y  . ∆β Φ =   ∂ 2 φ2 ∂ 2 φ2 ∂ 2 φ1  + + 2(1 + β) 2(1 + β) ∂y 2 ∂x2 ∂x∂y

Theorem 5.2 The variational problem (5.30) is equivalent to solving the following boundary value problem in unknown ζ = (W1 , W2 , w1 , w2 ) αh A0 ζ = (F1 , F2 , f1 , f2 ) 2


(5.47)


257

subject to the boundary conditions W1 |Γ0 = W2 |Γ0 = w1 |Γ0 = w2 |Γ0 = 0,

(5.48)

αh αhγ αhγ ∂w2 B01 (ζ)|Γ1 = B02 (ζ)|Γ1 = αhϕ0 (ζ)|Γ1 = =0 2 2 2 ∂ν where A0 and B0i are defined in (5.36) and (5.35), respectively, and

(5.49)

(F1 , F2 , f1 , f2 ) = F . Ellipticity of the Strain Energy Now we consider ellipticity of the strain energy (5.24). A key issue is the following uniqueness results. Lemma 5.1 Let ζ = (W1 , W2 , w1 , w2 ) ∈ HΓ1 0 (Ω) be such that Υ(ζ) = 0,

X0 (ζ) = 0,

ϕ0 (ζ) = 0,

and

w2 = 0,

(5.50)

then ζ = 0 for all x ∈ Ω where HΓ1 0 (Ω) is given by (5.28) and Γ0 has a positive length. Proof. By the formulas (5.6), (5.11), and (5.12), the assumptions (5.50) state that DW1 + D∗ W1 = −2w1 Π, (5.51) DW2 + D∗ W2 = −2w1 c + 2 i (W1 )DΠ,

(5.52)

Dw1 = 2 i (W1 )Π − W2 .

(5.53)

Let U = −2w1 Π. It is easy to check that DU = −2Π ⊗ Dw1 − 2w1 DΠ.

(5.54)

Using the formula (4.60) in Chapter 4 and the formula (5.54), we have D2 W1 (X, Y, Z) = −Π(X, Y )Z(w1 ) − Π(Z, X)Y (w1 ) + Π(Y, Z)X(w1 ) +R(X, Y, Z, W1 ) − w1 DΠ(X, Y, Z) (5.55) for X, Y, Z ∈ X0 (Ω), since DΠ is symmetric by Corollary 1.1, where R(·, ·, ·, ·) is the curvature tensor. Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x such that DEi Ej (x) = 0 for 1 ≤ i, j ≤ 2. By the formulas (5.55) and (1.114), we obtain at x ∆W1 = − =−

2 X

2 DE W1 + i Ei

i=1

2 X

2 X

R(Ei , Ej , W1 , Ej )Ei

ij=1

D2 W1 (Ej , Ei , Ei )Ej + κ W1

ij=1

= 2 i (Dw1 )Π − (tr Π)Dw1 + w1 D(tr Π) + 2κ W1


(5.56)

258


where ∆ is the Hodge-Laplace operator and κ is the Gauss curvature. Moreover, it follows from the formulas (5.53) and (5.52) that ∆w1 =

2 X

D(2 i (W1 )Π − W2 )(Ei , Ei )

i=1 2 X

=2

i=1

[DΠ(W1 , Ei , Ei ) + Π(DEi W1 , Ei )] − div W2

= 2hW1 , D(tr Π)i + 2 tr Π(D· W1 , ·) − div W2 = w1 tr c + hW1 , D(tr Π)i + 2 tr Π(D· W1 , ·).

(5.57)

Next, we check the values of DW1 and Dw1 on Γ0 . Since w1 |Γ0 = W1 |Γ0 = W2 |Γ0 = 0, it follows from the formulas (5.51)-(5.53) that DW1 + D∗ W1 = 0,

DW2 + D∗ W2 = 0 and Dw1 = 0

(5.58)

for all x ∈ Γ0 . Moreover, the conditions (DWi + D∗ Wi )|Γ0 = 0 and Wi |Γ0 = 0 for i = 1, 2 imply that DWi = 0 on Γ0 (5.59) Applying [5] to the elliptic system of the equations (5.56) and (5.57) yields that W1 = 0 and w1 = 0 on Ω. It follows from the formula (5.52) that DW2 + D∗ W2 = 0

on Ω

which yields W2 = 0 on Ω by Lemma 4.5.

Theorem 5.3 Let B0 (·, ·) be the bilinear form given by (5.24). Then there is c > 0 such that B0 (ζ, ζ) ≥ ckζk2H1

Γ0 (Ω)

for

ζ ∈ HΓ1 0 (Ω)

(5.60)

where HΓ1 0 (Ω) is given by (5.28) with Γ0 ⊂ Γ having a positive length. Proof. Using the formulas (5.6), (5.11), and (5.12) in the expression (5.25), we obtain B0 (ζ, ζ) ≥ c1 (|DW1 + D∗ W1 |2 + |DW2 + D∗ W2 |2 + |Dw1 |2 + |Dw2 |2 ) −c2 (|W1 |2 + |W2 |2 + w12 + w22 ) which yields B0 (ζ, ζ) + c2 kζk2L2 (Ω) ≥ c1 kζk2H1

Γ0 (Ω)

.

Then the inequality (5.60) follows from the above inequality and the uniqueness result, Lemma 5.1, by an argument as in Lemma 2.5. By Theorem 5.3, the following results are immediate.



259

Theorem 5.4 For F ∈ L2 (Ω) the problem (5.47)-(5.49) has a unique solution in HΓ1 0 (Ω). Equations of Motion In the last part of this section, we consider the equations of motion of the Naghdi model. For this purpose we need to compute the kinetic energy. The relations (5.4), (5.5), and (5.9) yield the following displacement expression h i ζ(p) = W1 (x) + w1 (x)N + z W2 (x) + w2 (x)N − i (W1 )Π (5.61) for p = x + zN ∈ S . Now we assume that the displacement vector field ζ depends on time t, and for simplicity, the mass density per unit of volume is one. Then the kinetic energy per unit area of the undeformed middle surface is obtained by integration with respect to thickness z Z

h/2

−h/2

|ζt |2 (1 + tr Πz + κ z 2 ) dz

(5.62)

where κ is the Gauss curvature of the middle surface Ω. As for the formula (4.94), a computation in the thin shell theory with a relative error h/R yields that the total kinetic energy of the middle surface Ω is Z 2 2 K(ζt , ζt ) = h [|W1t |2 + w1t + γ(|W2t |2 + w2t )] dx (5.63) Ω

where the term i (W1 )Π disappears because its error is h/R. The equations of motion for ζ are obtained by setting to zero the first variation of the Lagranian Z

0

T

h i K(ζt , ζt ) + Load (ζ) − B0 (ζ, ζ) dt

(5.64)

(the “Principle of Virtual Work”) where Load (·) and B0 (·, ·) are given by (5.29) and (5.24), respectively. We assume that there is no external loading on the shell. Then the variation (5.64) is taken with respect to kinematically admissible displacements. We obtain, as a result of the calculation by (5.33) and (5.64), Theorem 5.5 Suppose that there are no external loads on the shell and the shell is clamped along a portion Γ0 of the boundary Γ of the middle surface and free on Γ1 where Γ0 ∪ Γ1 = Γ. Then the displacement vector field ζ = (W1 , W2 , w1 , w2 ) satisfies the following (

αh A0 ζ = 0 on Q∞ 2 ζ(0, x) = ζ0 (x), ζt (0, x) = ζ1 (x)

hCζtt +


(5.65) on

Ω,

260


subject to the boundary conditions

where

ζ = 0 on Σ0∞ ,    αh B01 (ζ) = αh B02 (ζ) = 0, 2 2 αh αh ∂w2   hϕ0 (ζ), νi = = 0, 2 2 ∂ν   1000 0 γ 0 0  C= 0 0 1 0, 000γ

(5.66)

on

Σ1∞ ,

(5.67)

(5.68)

A0 ζ, B0i (ζ) and ϕ0 (ζ) are given by Theorem 5.1, and Q∞ = (0, ∞) × Ω,

Σ0∞ = (0, ∞) × Γ0 ,

Σ1∞ = (0, ∞) × Γ1 .

By a change of variables by η = C1/2 ζ and after changing t to t/c with c α = 2 in the problem (5.65), we obtain the following system ηtt + Aη = 0 on Q∞ (5.69) η(0) = η0 , ηt (0) = η1 on Ω, 2

subject to the boundary conditions η=0 (

on Σ0∞ ,

B1 (η) = B2 (η) = 0, ∂w2 = 0, hϕ(ζ), νi = ∂ν

on Σ1∞ ,

(5.70) (5.71)

where A = C−1/2 A0 C−1/2 .

(5.72)

The bilinear form associated with A becomes B(ζ, η) = B0 (C−1/2 ζ, C−1/2 η) = 2hΥ(ζ), Υ(η)i + 4hϕ(ζ), ϕ(η)iγ √ √ +2β(tr Υ(ζ) + w2 / γ)(tr Υ(η) + u2 / γ) + 2hX (ζ), X (η)i

+2β tr X (ζ) tr X (η) + hDw2 , Du2 i + 2w2 u2 /γ,  1  Υ(ζ) = (DW1 + D∗ W1 ) + w1 Π,    2  1 √ X (ζ) = (DW2 + D∗ W2 ) + w2 Π − γ( i (W1 )DΠ − w1 c), 2   1 1    ϕ(ζ) = Dw1 − i (W1 )Π + √ W2 , 2 γ

and this time the Green formula (5.33) reads Z B(ζ, η) = (Aζ, η)L2 (Ω) + ∂(Aζ, η) dΓ Γ


(5.73)

(5.74)

(5.75)

Naghdi’s Shells. Modeling and Control where B(ζ, η) =

Z

261

B(ζ, η) dx,

(5.76)

Ω

∂(Aζ, η) = hB1 (ζ), U1 i + hB2 (ζ), U2 i + 2hϕ(ζ), νiu1 +

√ B1 (ζ) = 2 i (ν)Υ(ζ) + 2β(tr Υ(ζ) + w2 / γ)ν, B2 (ζ) = 2 i (ν)X (ζ) + 2β(tr X (ζ))ν,

ζ = (W1 , W2 , w1 , w2 ),

5.2

∂w2 u2 , ∂ν

(5.77) (5.78)

and η = (U1 , U2 , u1 , u2 ).

Observability Estimates from Boundary

In this section it is assumed that all the regularities we need hold since we are mainly concerned with inequalities for controllability/stabilization. Let b(·, ·) be a bilinear form on T 2 (Ω) × T 2 (Ω), defined by b(T1 , T2 ) = hT1 , T2 i + β tr T1 tr T2

for

T1 , T2 ∈ T 2 (Ω)

(5.79)

where β > 0 is a constant, given in (5.19). For W ∈ H 1 (Ω, Λ), set S(W ) =

1 (DW + D∗ W ). 2

(5.80)

It is easy to check from Theorem 5.3 that there is a λ0 ≥ 1 such that Z λ0 [b(S(W ), S(W )) + |W |2 ] dx ≥ kDW k2L2 (Ω,T 2 ) (5.81) Ω

for W ∈ H 1 (Ω, Λ). Multiplier Identities For W ∈ X (Ω) and T ∈ T 2 (Ω), let G(W, T ) ∈ T 2 (Ω) be given by the formula (4.131) in Chapter 4. Let V be a vector field on Ω such that there is a function ϑ on Ω that satisfies the relation DV (X, X) = ϑ(x)|X|2

for all X ∈ Mx ,

x ∈ Ω.

(5.82)

For ζ = (W1 , W2 , w1 , w2 ) ∈ H1 (Ω), set m(ζ) = (DV W1 , DV W2 , V (w1 ), V (w2 )).


(5.83)

262


Lemma 5.2 We have 2B(ζ, m(ζ)) =

Z

+2

ZΓ

B(ζ, ζ)hV, νi dΓ − 2

Z

ϑB(ζ, ζ)dx

Ω

e(ζ, ζ) dx + lo (ζ)

(5.84)

Ω

where B(·, ·) is given in (5.76) and e(ζ, ζ) = 2b(S(W1 ), G(V, DW1 )) + 2b(S(W2 ), G(V, DW2 )) +4ϑ|ϕ(ζ)|2 + ϑ|Dw2 |2 . Proof. Let us compute Υ(m(ζ)), X (m(ζ)), ϕ(m(ζ)), and hDw2 , D(V (w2 ))i, separately. From the identity (4.132) in Chapter 4 it follows that Υ m(ζ) = DV Υ(ζ) + G(V, DW1 ) + lo (ζ).

(5.85)

(5.86)

Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x. We then obtain DV ( i (W1 )DΠ)(Ei , Ej ) = V (DΠ(W1 , Ei , Ej )) = DV (DΠ)(W1 , Ei , Ej ) + DΠ(DV W1 , Ei , Ej ) at x which yields DV ( i (W1 )DΠ) = i (DV W1 )DΠ + i (W1 )(DV DΠ).

(5.87)

Similarly, we have DV ( i (W1 )Π) = i (DV W1 )Π + i (W1 )DV Π

(5.88)

D(V (w1 )) = DV Dw1 + Dw1 (D· V ).

(5.89)

and Using the identity (4.132) in Lemma 4.7 with (W, w) = (W2 , w2 ) and the formula (5.87), we obtain X (m(ζ)) = DV X (ζ) + G(V, DW2 ) + lo (ζ).

(5.90)

It follows from the formulas (5.88) and (5.89) that ϕ(m(ζ)) = DV ϕ(ζ) + ϕ(ζ)(D· V ) + lo (ζ) which yields, via the relation (5.82), 2hϕ(ζ), ϕ(m(ζ))i = V (|ϕ(ζ)|2 ) + 2ϑ|ϕ(ζ)|2 + lo (ζ).


(5.91)


263

Using the formula (2.6) in Chapter 2 and the relation (5.82), we obtain 2hDw2 , D(V (w2 ))i = div (|Dw2 |2 V ) = V (|Dw2 |2 ) + 2ϑ|Dw2 |2 . (5.92) Now we compute B(ζ, m(ζ)). Noting that hΥ(ζ), G(V, DW1 )i + β tr Υ(ζ) tr G(V, DW1 ) = b(S(W1 ), G(V, DW1 )) + lo (ζ)

and hX (ζ), G(V, DW1 )i + β tr X (ζ) tr G(V, DW1 ) = b(S(W2 ), G(V, DW2 )) + lo (ζ), it follows from the relations (5.86), (5.90), and (5.91) that Z 2B(ζ, m(ζ)) = [V (B(ζ, ζ)) + 4b(S(W1 ), G(V, DW1 )) Ω

+4b(S(W2 ), G(V, DW2 )) + 2ϑ|Dw2 |2

+8ϑ|ϕ(ζ)|2 + lo (ζ)] dx Z Z = B(ζ, ζ)hV, νi dΓ − 2 ϑB(ζ, ζ) dx Ω ZΓ +2 e(ζ, ζ) dx + lo (ζ). Ω

Let ζ be a displacement vector field of the shell. The total energy of the shell is defined by 2E(t) = kζt k2L2 (Ω) + B(ζ, ζ) (5.93) where the bilinear form B(·, ·) is given by (5.76). Theorem 5.6 Let ζ = (W1 , W2 , w1 , w2 ) ∈ H1 (Ω) solve the problem ζtt + Aζ = 0

for

x ∈ Q = (0, T ) × Ω.

(5.94)

Then Z

[2∂(Aζ, m(ζ)) + (|ζt |2 − B(ζ, ζ))hV, νi]dΣ Z = 2(ζt , m(ζ))L2 (Ω) |T0 + 2 ϑ[|ζt |2 − B(ζ, ζ)] dQ Q Z +2 e(ζ, ζ) dQ + L(ζ) Σ

(5.95)

Q

where Σ = (0, T ) × Γ and e(ζ, ζ) is given by (5.85) and L(ζ) denotes lower


264


order terms relative to the energy (5.93). Moreover, if p is a function on Ω, then Z Z ∂(Aζ, pζ) dQ = p[B(ζ, ζ) − |ζt |2 ] dQ + L(ζ). (5.96) Σ

Q

Proof. We multiply the equation (5.94) by 2m(ζ) and integrate over Q by parts. By the relation (5.82), we have (ζtt , 2m(ζ))L2 (Ω) = 2[(ζt , m(ζ))L2 (Ω) ]t − 2(ζt , m(ζt ))L2 (Ω) Z = 2[(ζt , m(ζ))L2 (Ω) ]t + 2 ϑ|ζt |2 dx Ω Z 2 − |ζt | hV, νi dΓ.

(5.97)

Γ

Using the formulas (5.75) and (5.84), we obtain Z (Aζ, 2m(ζ))L2 (Ω) = [B(ζ, ζ)hV, νi − 2∂(Aζ, m(ζ))] dΓ ZΓ Z −2 ϑB(ζ, ζ) dx + 2 e(ζ, ζ) dx + lo (ζ). Ω

(5.98)

Ω

The identity follows from the formulas (5.97), (5.98), and the equation (5.94). We multiply the equation (5.94) by pζ and integrate over Q by parts to obtain the identity (5.96). Let us introduce the geometric assumptions for controllability/stabilization of the Naghdi shell. Definition 5.1 A vector field V on Ω is said to be an escape vector field for the Naghdi shell if the following assumptions hold: (i) There is a function ϑ on Ω such that the formula (5.82) holds true. (ii) Let ι(x) = hDV, Ei/2 for x ∈ Ω (5.99)

where E denotes the volume element of the middle surface Ω. Suppose that the functions ϑ(x) and ι(x) satisfy the inequality 2 min ϑ(x) > λ0 (1 + 2β) max |ι(x)| x∈Ω

x∈Ω

(5.100)

where λ0 ≥ 1 is given in (5.81) and β is given in (5.19). Existence of such a vector field has been studied in Section 4.5 in Chapter 4 and many examples are given there. Let V be an escape vector field for the Naghdi shell. Set σ0 = max |V |, x∈Ω

σ1 = min ϑ(x) − (1 + 2β) max |ι(x)|/2, x∈Ω


x∈Ω

(5.101)

Naghdi’s Shells. Modeling and Control Γ0 = { x | hV (x), ν(x)i > 0, x ∈ Γ },

Γ1 = Γ/Γ0 .

265 (5.102)

Now we have the following observability estimates from boundary. Theorem 5.7 Let V be an escape vector field for the Naghdi shell. Let ζ solve the problem (5.94). Then for T > 0 Z Z T SB dΣ + λ0 σ0 [E(0) + E(T )] ≥ 2σ1 E(t) dt + L(ζ) (5.103) Σ

0

where SB = ∂(Aζ, 2m(ζ) + ̺ζ) + [|ζt |2 − B(ζ, ζ)]hV, νi,

m(ζ) = (DV W1 , DV W2 , V (w1 ), V (w2 )),

̺ = 2ϑ − σ1 .

(5.104)

Proof. We let p = ̺ in the identity (5.96) and add it to the identity (5.95) to have Z Z SB dΣ = 2(ζt , m(ζ))L2 (Ω) |T0 + σ1 [|ζt |2 − B(ζ, ζ)]dQ Q Σ Z +2 e(ζ, ζ) dQ + L(ζ). (5.105) Q

On the other hand, by the formulas (5.73), (5.79), and (5.80), it is easy to check that B(ζ, ζ) = 2b(S(W1 ), S(W1 )) + 2βb(S(W2 ), S(W2 )) +4|ϕ(ζ)|2 + |Dw2 |2 + lo (ζ). By Lemma 5.3 below, we have Z Z e(ζ, ζ) dQ ≥ σ1 B(ζ, ζ) dQ + L(ζ). Q

(5.106)

(5.107)

Q

Using the formula (5.106) and the inequality (5.81), we obtain 2|(ζt , m(ζ))L2 (Ω) | ≤ σ0 [kζt k2L2 (Ω) +

2 X i=1

(kDWi k2L2 (Ω,T 2 ) + kDwi k2L2 (Ω,Λ) )]

≤ 2λ0 σ0 E(t) + L(ζ).

(5.108)

Now inserting the inequalities (5.108) and (5.107) into the formula (5.105) gives the inequality (5.103). A similar argument as in Lemma 4.11 in Chapter 4 yields Lemma 5.3 Let V be an escape vector field for the Naghdi model and let G(V, DW ) ∈ T 2 (Ω) be given by (4.131) for W ∈ H 1 (Ω, Λ). Then for W ∈ H 1 (Ω, Λ) Z Z σ1 b(S(W ), S(W )) dx ≤ b(S(W ), G(V, DW )) dx + lo (ζ). (5.109) Ω


Ω

266


Lemma 5.4 Let Γ0 ⊂ Γ be a relative open set and let X be a vector field on ˆ 1, W ˆ 2, w Ω. Let ζ = (W1 , W2 , w1 , w2 ), ζˆ = (W ˆ1 , w ˆ2 ) ∈ H1 (Ω) satisfy ζ = ζˆ = 0 on Γ0 . Then ˆ = B(ζ, ζ)

2 X

ˆ i (ν, ν) [2(1 + β)DWi (ν, ν)DW

i=1

î ˆ i (τ, ν) + ∂wi ∂ w +DWi (τ, ν)DW ] ∂ν ∂ν and

ˆ = B(ζ, ζ)hX, ˆ ∂(Aζ, DX ζ) νi

on

on

Γ0

Γ0

(5.110)

(5.111)

ˆ 1 , DX W ˆ 2 , X(w where ∂(A·, ·) is given by (5.77) and DX ζˆ = (DX W ˆ1 ), X(w ˆ2 )). Proof. The boundary conditions ζ|Γ0 = 0 yield Wi = Dτ Wi = 0,

ˆ wi = τ (wi ) = 0 on Γ

(5.112)

for i = 1, 2. By the formulas (5.74) and (5.112), we have Υ(ζ)(ν, ν) = DW1 (ν, ν),

Υ(ζ)(ν, τ ) =

1 DW1 (τ, ν), 2

Υ(ζ)(τ, τ ) = 0,

X (ζ)(ν, ν) = DW2 (ν, ν),

X (ζ)(ν, τ ) =

1 DW2 (τ, ν), 2

X (ζ)(τ, τ ) = 0,

1 ∂w1 ˆ ν on Γ. 2 ∂ν Moreover, the relations in (5.112) give  DX W1 = hX, νi(DW1 (ν, ν)ν + DW1 (τ, ν)τ ),       DX W2 = hX, νi(DW2 (ν, ν)ν + DW2 (τ, ν)τ ), ∂w1 X(w1 ) = hX, νi ,   ∂ν   ∂w   X(w2 ) = hX, νi 2 ∂ν and ϕ(ζ) =

(5.113)

on Γ0 .

(5.114)

In particular,

 DW1 (ν, X) = hX, νiDW1 (ν, ν),    DW1 (τ, X) = hX, νiDW1 (τ, ν), DW2 (ν, X) = hX, νiDW2 (ν, ν),    DW2 (τ, X) = hX, νiDW2 (τ, ν),

on Γ0 .

(5.115)

On the other hand, applying the relations in (5.112) and (5.113) to the formula (5.78), we obtain B1 (ζ) = 2(1 + β)DW1 (ν, ν)ν + DW1 (τ, ν)τ, (5.116) B2 (ζ) = 2(1 + β)DW2 (ν, ν)ν + DW2 (τ, ν)τ.



267

ˆ Note that all the relations (5.112)-(5.116) are true if we replace ζ by ζ. The formula (5.110) follows by using the relations in (5.113) in the formula ˆ respectively. (5.73) for ζ and ζ, ˆ Applying the relations (5.113)-(5.116) to the formula (5.77) for ζ and ζ, respectively, we obtain (5.111). By a similar argument as in Proposition 4.3, we have ˆ ⊂ Γ be relatively open. Proposition 5.1 Let λ be a complex number and Γ Then the problem    λη + Aη = 0 on Ω, Wi |Γˆ = Dν Wi |Γˆ = 0, (5.117)   wi | ˆ = ∂wi | ˆ = 0, i = 1, 2 Γ ∂ν Γ has the unique zero solution.

Dirichlet Control First, we consider the Dirichlet mixed problem in unknown ξ = (Φ1 , Φ2 , φ1 , φ2 ) for T > 0   ξtt + Aξ = 0 in Q, ξ(0) = ξ0 , ξt (0) = ξ1 , on Ω, (5.118)  ξ|Σ1 = 0, ξ|Σ0 = ς

with control functions ς ∈ L2 (Σ0 ) where

L2 (Σ0 ) = (L2 (Σ0 , Λ))2 × (L2 (Σ0 ))2 . Its dual version in ζ = (W1 , W2 , w1 , w2 ) is   ζtt + Aζ = 0 in Q, ζ = 0 on Σ,  ζ(0) = ζ0 , ζt (0) = ζ1

(5.119) on Ω.

Let ζ solve the Dirichlet problem (5.119). We solve the problem   ηtt + Aη = 0 in Q, η(T ) = ηt (T ) = 0 on Ω,  η|Σ1 = 0, η|Σ0 = Dν ζ

(5.120)

where Dν ζ = (Dν W1 , Dν W2 , w1ν , w2ν ) and define

Λ(ζ0 , ζ1 ) = (−ηt (0), η(0)). Let ζ and ζˆ solve the duality problem (5.119) with initial data (ζ0 , ζ1 ) and (ζˆ0 , ζˆ1 ), respectively. It follows from (5.120), (5.119), and (5.111) that, for


268


(ζ0 , ζ1 ), (ζˆ0 , ζˆ1 ) ∈ H01 (Ω) × L2 (Ω), (Λ(ζ0 , ζ1 ), (ζˆ0 , ζˆ1 ))L2 (Ω)×L2 (Ω) = −(ηt (0), ζˆ0 )L2 (Ω) + (η(0), ζˆ1 )L2 (Ω) Z T ˆ L2 (Ω) − (η, ζˆt )L2 (Ω) ]t dt = [(ηt , ζ) 0

Z

T

ˆ L2 (Ω) + (η, Aζ) ˆ L2 (Ω) ]dt [−(Aη, ζ) Z Z ˆ + (Aζ, ˆ η)]dΣ = ˆ Dν ζ)dΣ = [−∂(Aη, ζ) ∂(Aζ, Σ Σ0 Z ˆ = B(ζ, ζ)dΣ. =

0

(5.121)

Σ0

Remark 5.1 By (5.121) and (5.110), if a solution ζ to the duality problem (5.119) satisfies (Λ(ζ0 , ζ1 ), (ζ0 , ζ1 ))L2 (Ω)×L2 (Ω) < ∞, then the boundary control in (5.120) has a regularity Dν ζ ∈ L2 (Σ0 ). Let L = L2 (Ω) × L2 (Ω) and H = H01 (Ω) × L2 (Ω). Applying Theorem 2.13, we obtain Theorem 5.8 The problem (5.118) is exactly [L2 (Ω) × HΓ1 1 (Ω)]∗ controllable by (L2 (Σ0 , Λ))2 × (L2 (Σ0 ))2 controls on [0, T ] if and only if there are c1 > 0 and c2 > 0 such that Z c1 E(0) ≤ B(ζ, ζ) dΣ ≤ c2 E(0) (5.122) Σ0

for all solutions ζ of the problem (5.119). Moreover, we have Theorem 5.9 If there is an escape vector field for the Naghdi shell on the middle surface, then the observability inequality (5.122) holds true for any T > T0 where T0 = λ0 σ0 /σ1 . (5.123) Proof. Let SB be given in (5.104). By (5.111), SB = B(ζ, ζ)hV, νi for

(t, x) ∈ Σ.

Noting that E(t) = E(0) for t ≥ 0 we obtain by Theorem 5.7 Z SB dΣ ≥ 2(σ1 T − λ0 σ0 )E(0) + L(ζ). Σ


(5.124)

(5.125)


269

In addition, by the definition Γ0 in (5.102) and the formula (5.124), we have Z Z SB dΣ ≤ σ0 B(ζ, ζ) dΣ. (5.126) Σ

Σ0

Thus the left hand side of the inequality (5.122) follows from the estimates (5.125) and (5.126) by a compactness/uniqueness argument in terms of Proposition 5.1, as in Lemma 2.5. Finally, as usual, we take an vector field H ∈ X (Ω) such that H|Γ = ν and multiply the equation in (5.119) by (DH W1 , DH W2 , H(w1 ), H(w2 )). As a result of the calculation, we obtain the right hand side of the inequality (5.122). Neumann Control Let Γ1 6= ∅ and Γ0 ∩Γ1 = ∅. We consider the problem in unknown ξ = (Φ1 , Φ2 , φ1 , φ2 )  ξtt + Aξ = 0 in Q,    ξ = 0 on Σ1 (5.127) ∂Aξ = ς on Σ0 ,    ξ(0) = ξ0 , ξt (0) = ξ1 on Ω, where

∂Aξ = (B1 (ξ), B2 (ξ), ϕ(ξ), φ2 )

(5.128)

and ς is a boundary control. The dual problem for the above is the following in ζ = (W1 , W2 , w1 , w2 )  ζtt + Aζ = 0 in Q,    ζ = 0 on Σ1 (5.129) ∂Aζ = 0 on Σ0 ,    ζ(0) = ζ0 , ζt (0) = ζ1 on Ω. Let

ℵ:

H 1 ([0, T ], L2 (Σ0 )) → [H 1 ([0, T ], L2 (Σ0 ))]∗

be the canonical map with respect to L2 (Σ0 ) where

L2 (Σ0 ) = (L2 (Σ0 , Λ))2 × (L2 (Σ0 ))2 . Then, for any ψ ∈ H 1 ([0, T ], L2 (Σ0 )), ℵψ ∈ [H 1 ([0, T ], L2 (Σ0 ))]∗ satisfies Z (ℵψ, χ)L2 (Σ0 ) = (hψt , χt i + hψ, χi)dΣ (5.130) Σ0

for all χ ∈ L2 (Σ0 ). Let ζ solve the problem (5.129). We solve the problem   ηtt + Aη = 0 in Q, η(T ) = ηt (T ) = 0 on Ω,  η|Σ1 = 0, ∂Aη|Σ0 = −ℵζ © 2011 by Taylor & Francis Group, LLC

(5.131)

270


and define Λ(ζ0 , ζ1 ) = (−ηt (0), η(0)). ˆ Let ζ and ζ solve the duality problem (5.129) with initial data ˆ Σ are in the space (ζ0 , ζ1 ) and (ζˆ0 , ζˆ1 ), respectively, such that ζ|Σ0 and ζ| 0 H 1 ([0, T ], L2 (Σ0 )). As for (5.121), by (5.130), we have Z ˆ ∂(Aη, ζ)dΣ (Λ(ζ0 , ζ1 ), (ζˆ0 , ζˆ1 ))L2 (Ω)×L2 (Ω) = − Σ0 Z ˆ dΣ = (ℵζ, ζ) ˆ L2 (Σ ) =− h∂Aη, ζi 0 Σ0 Z ˆ = (hζt , ζˆt i + hζ, ζi)dΣ. Σ0

Then the estimate (Λ(ζ0 , ζ1 ), (ζ0 , ζ1 ))L2 (Ω)×L2 (Ω) < ∞ implies that the boundary control in (5.131) has a regularity −ℵζ ∈ [H 1 ([0, T ], L2 (Σ0 ))]∗ . Let L = L2 (Ω) × L2 (Ω) and H = HΓ1 1 (Ω) × L2 (Ω). Applying Theorem 2.13, we obtain, via Theorem 5.11 below, Theorem 5.10 Let V be an escape vector field for the Naghdi shell. Let T0 be given by (5.123). Then the problem (5.127) is exactly [L2 (Ω) × HΓ1 1 (Ω)]∗ controllable by [H 1 ([0, T ], L2 (Σ0 ))]∗ controls on [0, T ] for T > T0 . A similar argument as for Theorem 5.9 yields Theorem 5.11 Let V be an escape vector field for the Naghdi shell. Let T0 be given by (5.123). Then, for T > T0 , there is c > 0 such that Z (|ζt |2 + |ζ|2 ) dx ≥ cE(0) (5.132) Σ0

for all solutions to the problem (5.129) for which the left hand side of (5.132) is finite.

5.3

Stabilization by Boundary Feedback

Consider the following boundary feedback control problem in unknown ζ = (W1 , W2 , w1 , w2 ) ζtt + Aζ = 0 in Q∞ , (5.133) ζ(0) = ζ0 , ζt (0) = ζ1 on Ω,



271

subject to the boundary conditions ζ =0

on Σ1∞ = (0, ∞) × Γ1

and a boundary feedback  B1 (ζ) = J1 (ζt )ν + J2 (ζt )τ,     B2 (ζ) = J3 (ζt )ν + J4 (ζt )τ, 2ϕ(ζ) = J5 (ζt ),     ∂w2 = J6 (ζt ), ∂ν

on Σ0∞

(5.134)

(5.135)

where where Ji , i = 1, · · · , 6 are feedback operators. For simplicity, we denote ζ˘ = (hW1 , νi, hW1 , τ i, hW2 , νi, hW2 , τ i, w1 , w2 ) for any ζ = (W1 , W2 , w1 , w2 ). We will consider feedback laws to be defined by ˘ J(ζ) = −ζF (5.136) where J(ζ) = (J1 (ζ), · · · , J6 (ζ)) and F = fij (x) are 6 × 6 matrices. We assume that F is symmetric, positive semidefinite on Γ0 . Define the total energy of the system (5.133)-(5.135) by Z 2E(t) = [|ζt |2 + B(ζ, ζ)] dx

(5.137)

(5.138)

Ω

where the bilinear form B(·, ·) is given by the formula (5.73). Using the Green formula (5.75) and the relations (5.133)-(5.136), we have Z Z d E(t) = ∂(Aζ, ζt ) dΓ = − hζ˘t , F ζ˘t iR6 dΓ. ≤ 0 (5.139) dt Γ Γ0 Then the resulting closed-loop system under the feedback laws of (5.135) and (5.137) is dissipative in the sense that E(t) is decreasing. Variational Formula of the Closed-Loop System (5.133)-(5.137) Let the Sobolev spaces HΓ1 1 (Ω) and L(Ω) be given by (5.28) and (5.26), respectively. Introduce a bilinear form by Z ˘ F η˘i 6 dΓ hζ, (5.140) α(ζ, η) = R Γ0

where ζ = (W1 , W2 , w1 , w2 ), η = (U1 , U2 , u1 , u2 ), and F is the feedback matrix. Then it follows from the Green formula (5.75) that an appropriate variational


272


formulation for the system (5.133)-(5.137) is: Find ζ ∈ C([0, ∞); HΓ1 1 (Ω)) ∩ C 1 ([0, ∞), L2 (Ω)) such that d 1 2 dt [(ζt , η)L (Ω) + α(ζ, η)] + B(ζ, η) = 0 ∀ η ∈ HΓ1 (Ω), (5.141) 1 2 ζ(0) = ζ0 ∈ HΓ1 (Ω), ζt (0) = ζ1 ∈ L (Ω). Well-Posedness of the Closed-Loop System (5.133)-(5.137) Let the following assumption hold. Assumption (H1) Γ0 and Γ1 satisfy the following conditions Γ1 6= 0,

Γ0 ∩ Γ1 = ∅,

and hV, νi ≤ 0

on Γ1 .

(5.142)

By Theorem 5.3, the bilinear form B(·, ·) introduces an equivalent inner product on HΓ1 1 (Ω). By the assumption (5.137), α(·, ·) is continuous, symmet∗ ric, and non-negative. We identify L2 (Ω) with its dual L2 (Ω) so that we have the dense and continuous embeddings ∗

HΓ1 1 (Ω) ⊂ L2 (Ω) ⊂ HΓ1 1 (Ω) .

(5.143)

Let ℵ denote the canonical isomorphism of HΓ1 1 (Ω) endowed with the scalar product B(·, ·) onto HΓ1 1 (Ω). Then B(ζ, η) = (ℵζ, η)L2 (Ω)

ζ, η ∈ HΓ1 1 (Ω).

for

(5.144) ∗

Furthermore, there is a non-negative operator A ∈ L(HΓ1 1 (Ω), HΓ1 1 (Ω) ) such that α(ζ, η) = (Aζ, η)L2 (Ω) for ζ, η ∈ HΓ1 1 (Ω). (5.145)

Then the equation in (5.141) can be written as d (ζt + Aζ) + ℵζ = 0 dt

∗

in HΓ1 1 (Ω) .

(5.146)

Let us formally rewrite (5.146) as the following system: CYt + IN Y = 0 where C=

ℵ0 0I

,

t≥0 IN =

∗

in HΓ1 1 (Ω) × L2 (Ω) 0 −ℵ ℵ A

,

Y =

∗

ζ ζt

(5.147)

.

(5.148)

We shall solve the problem (5.147) in the space HΓ1 1 (Ω) × L2 (Ω). In order to make sense of (5.147) in that space it is natural to introduce D(IN ) = { (ζ, η) ∈ HΓ1 1 (Ω) × HΓ1 1 (Ω) | ℵζ + Aη ∈ L2 (Ω) }.

∗

Since C is the canonical isomorphism of HΓ1 1 (Ω)×L2 (Ω) onto HΓ1 1 (Ω) ×L2 (Ω), we rewrite (5.147) in the form Yt + C−1 IN Y = 0 in HΓ1 1 (Ω) × L2 (Ω). Then solutions of the system (5.133)-(5.137) are defined by (5.149). We have


(5.149)


273

Theorem 5.12 −C−1 IN is the infinitemal generator of a C0 -semigroup of contraction on HΓ1 1 (Ω) × L2 (Ω). Proof. Claim 1 D(IN ) is dense in HΓ1 1 (Ω) × L2 (Ω). By the definitions of the operators ℵ and A, we have (ℵζ + Aη, ς)L2 (Ω) = B(ζ, ς) + α(η, ς) Z = (Aζ, ς) + [∂(Aζ, ς) + h˘ η , F ς˘iR6 ] dΓ

(5.150)

Γ0

for ς ∈ HΓ1 1 (Ω). The expression on the right-hand side of the formula (5.150) implies the relation D(IN ) ⊃ D0 = { (ζ, η) | ζ ∈ HΓ1 1 (Ω) ∩ H2 (Ω), η ∈ HΓ1 1 (Ω), ∂(Aζ, ς) + h˘ η , F ς˘iR6 = 0 on Γ0 }, where H2 (Ω) = (H 2 (Ω, Λ))2 × (H 2 (Ω))2 .

Indeed, if (ζ, η) ∈ D0 , then

|(ℵζ + Aη, ς)L2 (Ω) | ≤ ckζkH2 (Ω) kςkL2 (Ω) which yields ℵζ + Aη ∈ L2 (Ω). Since D0 is dense in HΓ1 1 (Ω) × L2 (Ω), Claim 1 follows. Claim 2 −C−1 IN is dissipative. Noting that the inner product on HΓ1 1 (Ω) is B(·, ·), it is shown by (C−1 IN (ζ, η), (ζ, η))H1Γ

1

(Ω)×L2 (Ω)

= ((−η, ℵζ + Aη), (ζ, η))H1Γ

1

(Ω)×L2 (Ω)

= −B(ζ, η) + B(ζ, η) + α(η, η) = α(η, η) ≥ 0 for (ζ, η) ∈ D(IN ). Claim 3 Range(λI + C−1 IN ) = HΓ1 1 (Ω) × L2 (Ω) for λ > 0. In fact, this is equivalent to Range(λ2 I + λA + ℵ) = L2 (Ω). But, by the Lax-Milgram theorem, it is actually true.

As a consequence of Theorem 5.12, we have the following result. Theorem 5.13 Let the assumption (H1) in (5.142) hold. Then the problem (5.133)-(5.137) admits a unique solution with ζ ∈ C([0, ∞); HΓ1 1 (Ω)),

ζt ∈ C([0, ∞); L2 (Ω)).

Furthermore, if (ζ0 , ζ1 ) ∈ D(Q), then the solution ζ satisfies  ζ ∈ C 1 ([0, ∞); HΓ1 1 (Ω)) ∩ C 2 ([0, ∞); L2 (Ω)),    ℵζ + Aζt ∈ C([0, ∞); L2 (Ω)),  ζtt + Aζt + ℵζ = 0 t > 0,   ζ(0) = ζ0 , ζt (0) = ζ1 .


(5.151)

(5.152)

274


Solution ζ is called a weak solution if the conditions (5.151) are true. It is called a strong solution if the conditions (5.152) are true. Trace Estimates on Boundary Temporarily, we forget the definition of β in (5.19) and assume that µ and β are just positive constants, independent of each other. By comparing the formula (5.31) in this chapter with the formula β . Because of (4.37) in Chapter 4 it follows that ∆β = 2(1 + β)∆µ if µ = 1+β this relation the results in Proposition 4.6 in Chapter 4 hold true if replaced ∆µ by ∆β . Consider the problem

Wtt − ∆β W = F in Q, W |Γ1 = 0 on Σ1

(5.153)

where ∆β is given by (5.31). Then Proposition 5.2 Let 0 < α < T /2. Let W be a solution to the problem (5.153). Then there is cαT > 0 such that T −α

Z

α

+cαT

Z

Z

2

Γ0

ΣT 0

|Dτ W | dΓdt ≤ cαT

Z

T

0

kF k2H −1/2 (Ω,Λ) dt

(|Wt |2 + |S(W )(ν, ν) − βδW |2

+|S(W )(ν, τ )|2 ) dΣ + L(W )

(5.154)

where S(W ) is given by the formula (5.80). Now we have Theorem 5.14 Let ζ = (W1 , W2 , w1 , w2 ) be a solution of the closed-loop system (5.133)-(5.137). Let T /2 > α > 0 be given. Then 2 Z X i=1

T −α

α

Z

Γ0

(|DWi |2 + |Dwi |2 ) dΓdt ≤ cT α

Z

Σ0

|ζt |2 dΣ + L(ζ).

(5.155)

Proof. By Propositions 5.2 and Lemma 2.12, we need to estimate Dν Wi i and ∂w ∂ν . However, they are controlled by the feedback laws. It follows from the formulas (5.136), (5.135), (5.78), and (5.74) that    hB1 (ζ), νi = J1 (ζt ), hB2 (ζ), νi = J3 (ζt ),   2ϕ(ζ) = J5 (ζt ), © 2011 by Taylor & Francis Group, LLC

hB1 (ζ), τ i = J2 (ζt ), hB2 (ζ), τ i = J4 (ζt ), ∂w2 = J6 (ζt ) ∂ν

Naghdi’s Shells. Modeling and Control which yield  1   DW1 (ν, ν) = [J1 (ζt ) − 2βDW1 (τ, τ )] + lo 0 (ζ),   2(1 + β)    DW1 (τ, ν) = J2 (ζt ) − DW1 (ν, τ ) + lo 0 (ζ),     1   DW2 (ν, ν) = [J3 (ζt ) − 2βDW2 (τ, τ )] + lo 0 (ζ), 2(1 + β)  DW2 (τ, ν) = J4 (ζt ) − DW2 (ν, τ ) + lo 0 (ζ),    ∂w1    = J5 (ζt ) + lo 0 (ζ),   ∂ν   ∂w  2  = J6 (ζt ) ∂ν

where lo 0 (ζ) denote some zero order terms on ζ. By the relations (5.156), we obtain ( |DWi |2 ≤ c(|ζt |2 + |Dτ Wi |2 + |ζ|2 ), ∂wi 2 | + |ζ|2 ) |Dwi | ≤ c(|ζt |2 + | ∂τ

275

(5.156)

(5.157)

for i = 1, 2. Finally, the inequality (5.155) follows the estimates (5.157) and Propositions 5.2 and Lemma 2.12. Stabilization of the Closed-Loop System (5.133)-(5.137) For stabilization, we need a further assumption: Assumption (H2) The 6 × 6 matrix F is positive on Γ0 . We have Theorem 5.15 Suppose that there exists an escape vector field for the Naghdi shell such that relations (5.99) and (5.100) hold. Let the assumptions (H1) and (H2) be true where the assumption (H1) is given by (5.142). Then there are c1 , c2 > 0 such that E(t) ≤ c1 E(0)e−c2 t

for

t≥0

(5.158)

where E(t) is the total energy of the closed-loop system (5.133)-(5.137), given by (5.138). Proof. Let ζ = (W1 , W2 , w1 , w2 ) be a solution of the closed loop system (5.133)-(5.137). Let the boundary terms SB be given by the formula in (5.104). It follows from Lemma 5.4 and the boundary conditions (5.134) that Z Z SB dΣ = B(ζ, ζ)hV, νi dΣ ≤ 0, (5.159) Σ1

Σ1

via the assumption (H1). Moreover, using the formulas (5.140), (5.136), (5.135), and (5.77), we obtain Z ∂(Aζ, m(ζ) + ̺ζ) dΓ = −α(ζt , m(ζ) + ̺ζ). (5.160) Γ0


276


Let s1 , s2 > 0 be such that s1 |X|2 ≤ hX, F XiR6 ≤ s2 |X|2

X ∈ R6 , x ∈ Γ 0 .

for

Then the definition (5.140) of α(·, ·) yields Z Z 2 s1 |ζ| dΓ ≤ α(ζ, ζ) ≤ s2 Γ0

Γ0

|ζ|2 dΓ

(5.161)

which gives |α(ζt , 2m(ζ) + ̺ζ)| ≤ α1/2 (ζt , ζt )α1/2 (2m(ζ) + ̺ζ, 2m(ζ) + ̺ζ) 2 X 2 ≤ c(|ζt | + (|DWi |2 + |Dwi |2 ) + |ζ|2 ). (5.162) i=1

Using the relations (5.162), (5.160), and (5.104), we have Z Z T Z SB dΣ = − α(ζt , 2m(ζ) + ̺ζ) dt + [|ζt |2 − B(ζ, ζ)]hV, νidΣ Σ0

0

≤c

Z

Σ0

Σ0

[|ζt |2 +

2 X i=1

(|DWi |2 + |Dwi |2 )]dΣ + L(ζ).

(5.163)

Then applying the inequalities (5.163) and (5.159) to the inequality (5.103) in Theorem 5.7 yields σ1

Z

T

E(t)dt ≤ c

0

Z

Σ0

[|ζt |2 +

2 X i=1

(|DWi |2 + |Dwi |2 )]dΣ

+c[E(0) + E(T )] + L(ζ).

(5.164)

Next, change the integral domain Σ0 into [α, T − α] × Γ0 on the both sides of the inequalities (5.164) and use the inequality (5.155) to give Z T −α Z σ1 E(t) dt ≤ c[E(α) + E(T − α) + |ζt |2 dΣ] + L(ζ). (5.165) α

Σ0

Using the relations (5.139) and (5.140), we have for any T ≥ s ≥ 0 Z T Z E(s) = E(T ) + α(ζt , ζt ) dt ≤ E(T ) + c |ζt |2 dΣ s

which yields E(α) ≤ E(T ) + and Z

T −α α

Σ0

Z

Σ0

|ζt |2 dΣ

E(t)dt ≥ (T − 2α)E(T ) − cT


(5.166)

Z

Σ0

|ζt |2 dΣ.

(5.167)


277

Using the inequalities (5.166) and (5.167) in the inequality (5.165) gives Z E(T ) ≤ cT |ζt |2 dΣ (5.168) Σ0

for some T > 0 large where the lower order terms have been absorbed by Proposition 5.1 as in Lemma 2.5. Using the left hand side inequality in (5.161), the inequality (5.168), and the equations (5.139), we now have E(T ) ≤ cT

Z

T

0

α(ζt , ζt ) dt = cT (E(0) − E(T )),

that is, E(T ) ≤

cT E(0) 1 + cT

from which the exponential decay follows.

5.4

Stabilization of Transmission

Let a middle surface Ω be divided into two regions, Ω1 and Ω2 , by a smooth curve Γ3 on Ω. Set Γi = ∂Ω ∩ ∂Ωi for i = 1 and 2. Denote by νi and τi the normal and the tangential on ∂Ωi pointing outside Ωi , respectively. Then νi = ν,

τi = τ

ν1 (x) = −ν2 (x),

for x ∈ Γ1 ∪ Γ2 = ∂Ω,

τ1 (x) = −τ2 (x)

for x ∈ Γ3 .

(5.169)

Suppose that the type of material in Ω1 is different from that in Ω2 , where Young’s modulus and Poisson’s coefficient have a jump across the separatrix Γ3 . First, let us clarify transmission conditions for the Naghdi shell. Let Ei and µi , respectively, denote Young’s modulus and Poisson’s coefficient of the material corresponding to the region Ωi , i = 1, 2. Set αi = Ei /(1 + µi ),

βi = µi /(1 − 2µi ).

(5.170)

Let Ai be the Naghdi shell operator A0 on the portion Ωi , given by the formula (5.36) in which α and β have been replaced with αi and βi , respectively, for i = 1, 2. Let ζi = (Wi1 , Wi2 , wi1 , wi2 ) be a displacement vector on Ωi for i = 1 and 2, respectively. Similarly, let Bi1 (ζi ) = 2 i (νi )Υ(ζi ) + 2βi (tr Υ(ζi ) + wi2 )νi , (5.171) Bi2 (ζi ) = 2 i (νi )X0 (ζi ) + 2βi (tr X0 (ζi ))νi , where i (νi ) denotes the interior product.


278


Definition 5.2 A displacement vector field ζ is said to satisfy the transmission conditions for the Naghdi shell on the separatrix Γ3 if  ζ1 = ζ2 ,     α B (ζ ) + α2 B21 (ζ2 ) = 0,   1 11 1 α1 B12 (ζ1 ) + α2 B22 (ζ2 ) = 0, on Γ3 . (5.172)   α1 hϕ0 (ζ1 ), ν1 i + α2 hϕ0 (ζ2 ), ν2 i = 0,     α1 ∂w12 + α2 ∂w22 = 0 ∂ν1 ∂ν2

where ζi = ζ|Ωi for i = 1, 2.

Consider the following boundary feedback problem in unknown ζi = (Wi1 , Wi2 , wi1 , wi2 ) for x ∈ Ωi ζitt + αi Ai ζi = 0 in (0, ∞) × Ωi , (5.173) ζi (0) = ζi0 , ζit (0) = ζi1 on Ωi , i = 1, 2, which is subject to the boundary conditions ζ1 = 0 on (0, ∞) × Γ1 ,

(5.174)

the transmission conditions (5.172) on (0, ∞) × Γ3 , and a boundary feedback on (0, ∞) × Γ2    B21 (ζ2 ) = J1 (ζ2t )ν + J2 (ζ2t )τ,   B22 (ζ2 ) = J3 (ζ2t )ν + J4 (ζ2t )τ, 2ϕ(ζ2 ) = J5 (ζ2t ),     ∂w22 = J6 (ζ2t ) ∂ν

on (0, ∞) × Γ2

(5.175)

(5.176)

where Ji , i = 1, · · · , 6 are feedback operators. We introduce feedback laws by J1 (ζ), · · · , J6 (ζ) = −F ζ˘ on (0, ∞) × Γ2 (5.177) where F are symmetric, nonnegative 6 × 6 matrices for x ∈ Γ2 and ζ˘ = (hW1 , νi, hW1 , τ i, hW2 , νi, hW2 , τ i, w1 , w2 ) for ζ = (W1 , W2 , w1 , w2 ). For convenience, set α(x) = αi ,

β(x) = βi ,

Ax = Ai ,

and ζ = ζi

as x ∈ Ωi for i = 1, 2. Define the total energy of the closed loop system (5.173)-(5.177) by Z 2E(t) = [|ζt |2 + α(x)Bx (ζ, ζ)] dx (5.178) Ω

where Bx (·, ·) = Bi (·, ·) for x ∈ Ωi where Bi (·, ·) = B0 (·, ·) is given by the



279

formula (5.25) in which E and µ are replaced by Ei and µi for i = 1 and 2, respectively. It follows from the Green formula (5.33) on Ωi that Z Z Z Bi (ζ, η) dx = hAi ζ, ηi dx + ∂(Ai ζ, η) dΓ (5.179) Ωi

Ωi

Γi ∪Γ3

for i = 1 and 2. By the formulas (5.172)-(5.179), we obtain 2

X d E(t) = dt i=1 =

2 X

Z

Ωi

Z

αi

∂(Ai ζi , ζit ) dΓ = α2

Γi ∪Γ3

i=1

= −α2

[hζit , ζitt i + αi Bi (ζi , ζit )] dx

Z

Z

∂(A2 ζ2 , ζ2t ) dΓ

Γ2

hF ζ˘2t , ζ˘2t iR6 dΓ ≤ 0

Γ2

which implies that the closed-loop system (5.173)-(5.177) is dissipative. By similar arguments as in Section 5.3 , we obtain the existence, uniqueness and regularity of solutions to the closed-loop system (5.173)-(5.177) as follows. Set L2 (Ω) = (L2 (Ω, Λ))2 × (L2 (Ω))2 ,

HΓ1 1 (Ω) = (HΓ11 (Ω, Λ))2 × (HΓ11 (Ω))2 ,

W = { ζ | ζ ∈ HΓ1 1 (Ω), ζi = ζ|Ωi ∈ (HΓ21 (Ωi , Λ))2 × (HΓ21 (Ωi ))2 , i = 1, 2; ζ satisfies the transmission conditions (5.172) }.

Theorem 5.16 For (ζ 0 , ζ 1 ) ∈ HΓ1 1 (Ω) × L2 (Ω), the closed-loop system (5.173) − (5.177) admits a unique solution with ζ ∈ C([0, ∞), HΓ1 1 (Ω)) and ζt ∈ C([0, ∞), L2 (Ω)). Moreover, if the initial data satisfy ζ 0 ∈ W, ζ1 ∈ HΓ1 1 (Ω), and ∂(A2 ζ20 , η) = −hF ζ˘20 , η˘iR6 on Γ2 for all η ∈ HΓ1 1 (Ω), then ζ ∈ C([0, ∞); W ∩ HΓ1 1 (Ω)). Next, let us clarify some assumptions for the stabilization of the system (5.173)-(5.177), which will be based on the geometric properties and the materials of the Naghdi shell. Assumption (H1) Let V be an escape vector field V for the Naghdi shell. Let ϑ(x) and ι(x) be defined by the formulas (5.82) and (5.99), respectively. Moreover, suppose that ϑ and ι meet the inequality 2 min ϑ(x) > λ0 (1 + 2β0 ) max |ι(x)| x∈Ω

x∈Ω

(5.180)

where λ0 ≥ 1 is the elliptic number, given by the inequality (5.81) and β0 = max{β1 , β2 }. Set σ0 = max |V |, x∈Ω

σ1 = min ϑ(x) − λ0 (1 + 2β0 ) max |ι(x)|/2.


x∈Ω

x∈Ω

280


Assumption (H2) Γ1 6= ∅ and hV, νi ≤ 0 for x ∈ Γ1 . Assumption (H3) α1 6= α2 , (α2 − α1 )hV, ν1 i ≤ 0, and (α2 β2 − α1 β1 )hV, ν1 i ≤ 0 for x ∈ Γ3 . Assumption (H4) F ∈ C 1 (Γ2 ) is positive on Γ2 . For existence of an escape vector field with the relations (5.82) and (5.180), see Section 4.5 of Chapter 4. Then the main results of this section are Theorem 5.17 Let the assumptions (H1) − (H4) hold. Then there are c1 > 0 and c2 > 0 such that E(t) ≤ c1 e−c2 t

∀ t ≥ 0.

(5.181)

The proof of this theorem will be given at the end of this section. For this stabilization, we need the following observability estimate, Theorem 5.18 and the boundary trace estimate, Theorem 5.19. Using Theorem 5.7 on Ωi with A replaced by αi Ai and B(·, ·) replaced by αi Bi (·, ·), respectively for i = 1, 2, and the transmission condition (5.172), we obtain Theorem 5.18 Let the assumption (H1) hold. Suppose that ζ = (W1 , W2 , w1 , w2 ) solve the problem ζtt + α(x)Ax ζ = 0

(5.182)

such that the right hand side of the estimate (5.183) below is well defined. Let T > 0 be given. Then there is c > 0 independent of ζ such that Z T Z Z 2σ1 E(t) dt ≤ SB |Σ dΣ + SB |Σ3 dΣ3 0

Σ

Σ3

+σ0 λ0 [E(0) + E(T )] + L(ζ)

(5.183)

where SB |Σ = [|ζt |2 − α(x)Bx (ζ, ζ)]hV, νi + α(x)∂(Ax ζ, 2m(ζ) + ̺ζ), m(ζ) = (DV W1 , DV W2 , v(w1 ), V (w2 )),

̺ = 2ϑ − σ1 ,

(5.184)

SB Σ3 = 2[α1 ∂(A1 ζ1 , m(ζ1 )) + α2 ∂(A2 ζ2 , m(ζ2 ))] +[α2 B2 (ζ2 , ζ2 ) − α1 B1 (ζ1 , ζ1 )]hV, ν1 i.

(5.185)

In addition, by similar methods in Theorem 5.14, we have Theorem 5.19 Let ζ solve the system (5.173) − (5.176). Then for T /2 > α > 0 there is cT α > 0 such that 2 Z X i=1

T −α

α

Z

Γ2

(|DW2i |2 + |Dw2i |2 ) dΣ ≤ cT α


Z

Σ2

|ζt |2 dΣ + L(ζ). (5.186)


281

The following lemma is a key to the transmission problem. Lemma 5.5 The transmission conditions (5.172) and the material assumption (H3) imply that Z Σ3

SB |Σ3 ≤ L(ζ).

Proof. We compute on the division Γ3 . The transmission conditions (5.172) yield that on Γ3   Dτ1 W1i + Dτ2 W2i = 0, ∂w1i ∂w2i i = 1, 2, + =0  ∂τ1 ∂τ2

 α (1 + β1 )DW1i (ν1 , ν1 ) + α1 β1 DW1i (τ1 , τ1 )   1   = α2 (1 + β2 )DW2i (ν2 , ν2 ) + α2 β2 DW2i (τ2 , τ2 ) + lo (ζ),     α1 Υ(ζ1 )(ν1 , τ1 ) = α2 Υ(ζ2 )(ν2 , τ2 ), α1 X0 (ζ1 )(ν1 , τ1 ) = α2 X0 (ζ2 )(ν2 , τ2 ) + lo (ζ),   ∂w1i ∂w2i   α1 + α2 = 0,    ∂ν ∂ν2 1  i = 1, 2.

Moreover, the relations (5.187) and (5.188) reach  α1 (1 + β1 )DW1i (ν1 , ν1 ) = α2 (1 + β2 )DW2i (ν2 , ν2 )      +(α2 β2 − α1 β1 )DW1i (τ1 , τ1 ) + lo (ζ), α1 DW1i (τ, ν1 ) = α2 DW2i (τ2 , ν2 )   +(α2 − α1 )DW1i (ν1 , τ1 ) + lo (ζ),    i = 1, 2.

(5.187)

(5.188)

(5.189)

Note that m(ζi ) = DV ζi = hV, νi iDνi ζi +hV, τi iDτi ζi for i = 1 and 2. Then the transmission conditions (5.172) imply that α1 ∂(A1 ζ1 , m(ζ1 )) + α2 ∂(A2 ζ2 , m(ζ2 )) = [α1 ∂(A1 ζ1 , Dν1 ζ1 ) − α2 ∂(A2 ζ2 , Dν2 ζ2 )]hV, ν1 i. Then SB |Σ3 = {2α1 ∂(A1 ζ1 , Dν1 ζ1 ) − 2α2 ∂(A2 ζ2 , Dν2 ζ2 ) +α2 B2 (ζ2 , ζ2 ) − α1 B1 (ζ1 , ζ1 )}hV, ν1 i.

(5.190)

We shall compute the right hand side of (5.190). The definitions of ∂(Ai ζi , Dνi ζi ) and Bi (ζi , ζi ) read that ∂(Ai ζi , Dνi ζi ) = hBi1 (ζi ), Dνi Wi1 i + γhBi2 (ζi ), Dνi Wi2 i ∂wi1 ∂wi2 2 +2hϕ0 (ζi ), νi i + γ( ) ∂νi ∂νi


(5.191)

282


and Bi (ζi , ζi ) = 2|Υ(ζi )|2 + 2βi [tr Υ(ζi )]2 + 2γ|X0 (ζi )|2 + 2γβi [tr X0 (ζi )]2 +γ|Dwi2 |2 + 4|ϕ0 (ζi )|2 + lo (ζ), (5.192) respectively, on Σ3 for i = 1, 2. From the formulas (5.171), we obtain 2αi hBi1 (ζi ), Dνi Wi1 i − 2αi {|Υ(ζi )|2 + βi [tr Υ(ζi )]2 } = 2αi (1 + βi )[DWi1 (νi , νi )]2 − 2αi (1 + βi )[DWi1 (τi , τi )]2

+4αi [Υ(ζi )(νi , τi )]2 − 4αi Υ(ζi )(νi , τi )DWi1 (νi , τi ).

(5.193)

Let qi = αi (1 + βi ) for i = 1, 2,

p = q2 − q1 .

It follows from the formulas (5.193) and (5.187)-(5.189) that

2α1 hB11 (ζ1 ), Dν1 W11 i − 2α1 {|Υ(ζ1 )|2 + β1 [tr Υ(ζ1 )]2 } −2α2 hB21 (ζ2 ), Dν2 W21 i + 2α2 {|Υ(ζ2 )|2 + β2 [tr Υ(ζ2 )]2 }

= 2q1 [DW11 (ν1 , ν1 )]2 − 2q2 [DW21 (ν2 , ν2 )]2 + 2p[DW11 (τ1 , τ1 )]2 4α1 + (α2 − α1 )[Υ(ζ1 )(ν1 , τ1 )]2 + lo (ζ) α2 2q1 p 2 = [DW11 (ν1 , ν1 )]2 + [q2 p − (α1 β1 − α2 β2 )2 ][DW11 (τ1 , τ1 )]2 q2 q2 4q1 + (α2 β2 − α1 β1 )DW11 (ν1 , ν1 )DW11 (τ1 , τ1 ) q2 4α1 + (α2 − α1 )[Υ(ζ1 )(ν1 , τ1 )]2 + lo (ζ). (5.194) α2 Since q1 p = q1 (α2 − α1 ) + q1 (α2 β2 − α1 β1 ), we have

q2 p − (α1 β1 − α2 β2 )2 = q2 (α2 − α1 ) + q1 (α2 β2 − α1 β1 ),

the right hand side of (5.194) 2n = q1 (α2 − α1 )[DW11 (ν1 , ν1 )]2 + q2 (α2 − α1 )[DW11 (τ1 , τ1 )]2 q2 o +q1 (α2 β2 − α1 β1 )[DW11 (ν1 , ν1 ) + DW11 (τ1 , τ1 )]2 +

4α1 (α2 − α1 )[Υ(ζ1 )(ν1 , τ1 )]2 + lo (ζ). α2

(5.195)

From the relations (5.194) and (5.195) and the assumption (H3), we obtain n 2α1 hB11 (ζ1 ), Dν1 W11 i − 2α1 {|Υ(ζ1 )|2 + β1 [tr Υ(ζ1 )]2 } o −2α2 hB21 (ζ2 ), Dν2 W21 i + 2α2 {|Υ(ζ2 )|2 + β2 [tr Υ(ζ2 )]2 } hV, ν1 i ≤ lo (ζ).


(5.196)


283

A similar computation gives n 2α1 hB12 (ζ1 ), Dν1 W21 i − 2α1 {|X0 (ζ1 )|2 + β1 [tr X0 (ζ1 )]2 }

o −2α2 hB22 (ζ2 ), Dν2 W22 i + 2α2 {|X0 (ζ2 )|2 + β2 [tr X0 (ζ2 )]2 } hV, ν1 i ≤ lo (ζ).

(5.197)

Moreover, we have ∂w11 ∂w21 − 4α2 hϕ0 (ζ2 ), ν2 i ∂ν1 ∂ν2 +4α2 |ϕ0 (ζ2 )|2 − 4α1 |ϕ0 (ζ1 )|2 = lo (ζ) 4α1 hϕ0 (ζ1 ), ν1 i

(5.198)

and ∂w12 2 ∂w22 2 ) − 2α2 ( ) + α2 |Dw22 |2 − α1 |Dw12 |2 ∂ν1 ∂ν2 ∂w12 2 ∂w12 2 ∂w22 2 ∂w22 2 = α1 ( ) − α1 ( ) − α1 ( ) + α2 ( ) ∂ν1 ∂τ1 ∂ν2 ∂τ2 ∂w12 2 ∂w12 2 α1 = (α2 − α1 )( ) + (α2 − α1 )( ) α2 ∂ν1 ∂τ1 2α1 (

which yields n o ∂w12 2 ∂w22 2 2α1 ( ) − 2α2 ( ) + α2 |Dw22 |2 − α1 |Dw12 |2 hV, ν1 i ≤ 0. (5.199) ∂ν1 ∂ν2 Finally, we obtain the estimate Z SB dΣ ≤ lo (ζ) Σ3

by combining (5.190), (5.191), (5.192), (5.196), (5.197), (5.198), and (5.199). Proof of Theorem 5.17 Using Lemma 5.5 in Theorem 5.18 yields 2σ1

Z

T 0

E(t) dt ≤

Z

( SB 1 + SB 2 )dΣ

Σ

+σ0 λ0 [E(0) + E(T )] + lo (ζ)

(5.200)

where SB 1 = [|ζt |2 − α(x)Bx (ζ, ζ)]hV, νi,

(5.201)

SB 2 = α(x)∂(Ax ζ, 2m(ζ) + ̺ζ).

(5.202)

By the boundary condition (5.174) and the identity (5.111), we have SB 1 |Σ1 = −α1 B1 (ζ, ζ)hV, νi


284


and SB 2 |Σ1 = 2α1 B1 (ζ, ζ)hV, νi. It follows from the assumption (H2) that Z Z ( SB 1 + SB 2 )dΣ = α1 B1 (ζ, ζ)hV, νidΣ ≤ 0. Σ1

(5.203)

Σ1

Next, set a1 (ζ, η) =

Z

Γ2

˘ η˘i 6 dΓ. α2 hF ζ, R

Then the assumption (H4) implies that there are s1 , s2 > 0 such that for η = (U1 , U2 , u1 , u2 ) Z Z s1 |η|2 dΓ ≤ a1 (η, η) ≤ s2 |η|2 dΓ. (5.204) Γ2

Γ2

′

Noting that the relation E (t) = −a1 (ζ2t , ζ2t ), we have, for any T ≥ α ≥ 0, Z T E(α) = E(T ) + a1 (ζ2t , ζ2t ) dt. (5.205) α

From the feedback laws (5.176) and (5.177), we obtain Z Z T SB 2 dΣ = − a1 (ζ2t , 2m(ζ2 ) + ̺ζ) dt Σ2

0

Z

≤c

2

Σ2

[|ζ2t | +

2 X i=1

(|DW2i |2 + |Dw2i |2 )] dΣ + L(ζ). (5.206)

Moreover, clearly, the following estimate is true, Z

Σ2

SB 1 dΣ ≤ c

Z

Σ2

[|ζ2t |2 +

2 X i=1

(|DW2i |2 + |Dw2i |2 )] dΣ + L(ζ).

(5.207)

Substituting the relations (5.203), (5.206), and (5.207) in the inequality (5.200) yields 2σ1

Z

T 0

E(t) dt ≤ c

Z

Σ2

[|ζ2t |2 +

2 X i=1

(|DW2i |2 + |Dw2i |2 )] dΣ

+σ0 λ0 [E(0) + E(T )] + L(ζ).

(5.208)

Now, change the integral domain Σ2 into [α, T − α] × Γ2 in both sides of the inequalities (5.208) and use the inequality (5.186) and the relation (5.205) to get Z T −α Z n o 2σ1 E(t) dt ≤ c E(α) + E(T − α) + |ζ2t |2 dΣ + L(ζ). (5.209) α


Σ2

Naghdi’s Shells. Modeling and Control By the relation (5.205), we find, for any T ≥ s ≥ 0, Z Z E(T ) − c |ζ2t |2 dΣ ≤ E(s) ≤ E(T ) + c |ζ2t |2 dΣ. Σ2

285

(5.210)

Σ2

Using the inequality (5.210) in the inequality (5.209) and the uniqueness in Proposition 5.1, we obtain, for some T suitably large, Z E(T ) ≤ cT |ζ2t |2 dΣ. (5.211) Σ2

Finally, using the left hand side of the inequality (5.204) in the estimate (5.211), we have Z T E(T ) ≤ cT a1 (ζ2t , ζ2t ) dt = cT (E(0) − E(T )), 0

that is, E(T ) ≤

cT E(0). 1 + cT

Then the estimate (5.181) follows from the inequality (5.212).

(5.212)

Exercises 5.1 Let the Sobolev space H1 (Ω) be given by (5.27) and let the bilinear form B(·, ·) be given by (5.76). Prove that there is c > 0 such that B(ζ, ζ) ≤ ckζk2H1 (Ω)

for all

ζ ∈ H1 (Ω).

5.2 Let the operator A of the Naghdi shell be given by the formula (5.72). Then for ζ = (W1 , W2 , w1 , w2 ), Aζ = (∆β W1 , ∆β W2 , ∆w1 , ∆w2 ) + lo (ζ).

(5.213)

5.3 Write out the proof of Lemma 5.3 in detail by following the proof of Lemma 4.11 in Chapter 4. 5.4 Prove Proposition 5.1. 5.5 Prove Theorem 5.18. 5.6 Prove Theorem 5.19. 5.7 Prove that the transmission conditions (5.172) imply the relations (5.188) and (5.189).


286

5.5



Sections 5.1 and 5.2 are from [213]; Section 5.3 is from [25]; Section 5.4 is from [35]. The displacement-strain relations (5.14) come from [157] which is the foundation stone of the Naghdi shell. The Green formulas in Theorem 5.1 and the equations of motion in Theorem 5.5 are given in the form of coordinates free. Modeling of the Naghdi shell in the form of coordinates free is the key for a series of advances on this model. In this direction, [35] is an excellent work. One of its main contributions clarifies the transmission assumption (H3), by a complicated computation as in Lemma 5.5, which guarantees the transmission stabilization only by feedbacks on the portion Γ2 of the boundary. A recent advance on the Naghdi model is made by [30] which is not included in this chapter. [30] presents some well-posedness results and an explicit, simple formula of the feedthrough operator ([202], [203]), respectively, for the input-output system of the Naghdi shell. The well-posedness results in [30] can guarantee the equivalence between exact controllability and stabilization by Russell’s principle. Moreover, the feedthrough operator of the Naghdi model p in [30] shows that the speeds of propagation of the Naghdi shell are 1 and 1/ 2(1 + β) along the tangential direction and the normal, respectively, where β is given in (5.19).


Chapter 6 Koiter’s Shells. Modeling and Controllability

6.1 6.2 6.3 6.4 6.5

Equations of Equilibria. Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniqueness for the Koiter Shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multiplier Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Observability Estimates from Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

287 301 311 314 327 328

The Koiter shell was introduced by [98]. In this model the second fundamental form of the middle surface goes into the change of curvature tensor which makes its analysis much more complicated than that of the shallow shell. We are then faced with new tasks: to derive the dynamic system starting from the strain tensor and the curvature tensor and to set up appropriate geometric conditions for controllability/stabilization. Multiplier identities are established and exact controllability from boundary is presented.

6.1

Equations of Equilibria. Equations of Motion

Let us assume that the middle surface of the shell occupies a bounded region Ω of the surface M in R3 . The shell, a body in R3 , is defined by S = { p | p = x + zN (x), x ∈ Ω, −h/2 < z < h/2 } where h is the thickness of the shell, small. As usual, denote by ζ(x) the displacement vector of point x of the middle surface. We decompose the displacement vector ζ(x) into a direct sum ζ(x) = W (x) + wN (x)

W (x) ∈ Mx ,

x ∈ Ω,

that is, W and w are the components of ζ on the tangential plane and on the normal of the undeformed middle surface Ω, respectively. For convenience, we let ζ = (W, w). For the Koiter model, the linearized strain tensor and the change of curvature tensor of the middle surface Ω are given by Υ(ζ) =

1 (DW + D∗ W ) + wΠ 2

(6.1) 287


288


and ρ(ζ) = D2 w − Π(·, D· W ) − Π(D· W, ·) − i (W )DΠ − w c ,

(6.2)

respectively, where Π is the second fundamental form, c the third fundamental form of the surface M given by (5.3), D2 w the Hessian of w, and · denotes the position of the variable. In (6.2), i (W )DΠ is the interior product of the tensor field DΠ by the vector field W . The relations (6.1) and (6.2) are given by [98], also see [12] or [134]. Remark 6.1 If we describe the strain tensors (6.1) and (6.2) by a coordinate, they look as follows. Let the middle surface be given by a coordinate ϕ(x1 , x2 ) = (ϕ1 (x1 , x2 ), ϕ2 (x1 , x2 ), ϕ3 (x1 , x2 )) Set aα = (

∂ϕ1 ∂ϕ2 ∂ϕ3 , , ), ∂xα ∂xα ∂xα

for

(x1 , x2 ) ∈ R2 .

1 ≤ α ≤ 2.

Define aα by haβ , aα i = δαβ and let W = w1 a1 + w2 a2 . Then in the classical notation the tensors (6.1) and (6.2) become Υαβ =

1 (wα|β + wβ|α ) − bαβ w, 2

ραβ = w|αβ − cαβ w + bλα wλ|β + bλβ wλ|α + bλα|β wλ where bαβ = −Π(aα , aβ ),

w|αβ = D2 w(aα , aβ ),

cαβ = c(aα , aβ ),

wα|β = hDaβ W, aα i,

∂α N = bλα aλ , bλα|β = −DΠ(aλ , aα , aβ ).

The shell strain energy associated to the displacement vector field ζ of the middle surface Ω can be written as Z Eh B1 (ζ, ζ) = B(ζ, ζ) dx (6.3) 1 − µ2 Ω where B(ζ, ζ) = a(Υ(ζ), Υ(ζ)) + γa(ρ(ζ), ρ(ζ)), a(T, T ) = (1 − µ)hT, T i + µ(tr T )2 ,

γ = h2 /12,

(6.4)

T ∈ T 2 (Ω),

for x ∈ Ω, where E, µ respectively denote Young’s modulus and Poisson’s coefficient of the material. The expression (6.3) is an approximation to the shell strain energy. Its derivation from the three-dimensional elasticity theory is carried out by integration on the thickness of the shell, following the methods


Koiter’s Shells. Modeling and Controllability

289

of asymptotic expressions. For a formal utilization of these methods in shell theory, we refer to [95], [157], [183], and the bibliographies of these papers. For an isotropic material, with the expression (6.3), we are able to associate the following symmetric bilinear form, directly defined on the middle surface Ω: Z B(ζ, η) = B(ζ, η) dx (6.5) Ω

where η = (U, u).

The Ellipticity of the Strain Energy for the Koiter Shell The ellipticity is an indispensable result for all control problems. To establish it, we need Lemma 6.1 Let Γ0 ⊂ Γ have a positive length and let a displacement vector field ζ = (W, w) be such that W |Γ0 = 0,

w|Γ0 =

∂w |Γ = 0. ∂ν 0

(6.6)

If ρ(ζ) = Υ(ζ) = 0 for all x ∈ Ω, then ζ = 0 for x ∈ Ω. Proof. Let U = DW + D∗ W. Then Υ = 0 gives U = −2wΠ. It is easy to check that DU = −2Π ⊗ Dw − 2wDΠ. (6.7) Let x ∈ Ω be given. Let E1 , E2 be a frame field normal at x such that DEi Ej (x) = 0 for 1 ≤ i, j ≤ 2. By the formula (4.60) in Chapter 4, we have 2 X i=1

D2 W (Ej , Ei , Ei ) = −2Π(Dw, Ej ) + tr ΠhDw, Ej i −κhW, Ej i − whDH, Ej i at x

(6.8)

where κ is the Gauss curvature. It follows from the formulas (6.8) and (1.114) that at x ∆W = − =−

2 X

2 DE W+ i Ei

i=1

2 X

2 X

R(Ei , Ej , W, Ej )Ei

ij=1

D2 W (Ej , Ei , Ei )Ej + κW

ij=1

= 2 i (Dw)Π − HDw − wDH + 2κW.

(6.9)

On the other hand, by (6.2), the assumption ρ(ζ) = 0 implies that ∆w = tr D2 w = 2 tr Π(·, D· W ) + tr i (W )DΠ + w tr c for x ∈ Ω.


(6.10)

290


Moreover, the conditions in (6.6) and the relation DW + D∗ W = −2wΠ yield that DW |Γ0 = 0. Finally, applying the uniqueness theorem of [170] to the system (6.9) and (6.10), we obtain that ζ = 0. Theorem 6.1 Let Γ0 ⊂ Γ have a positive length. Then there is a constant c > 0 such that, for all ζ ∈ HΓ10 (Ω, Λ) × HΓ20 (Ω), B(ζ, ζ) ≥ ckζk2H 1

2 Γ0 (Ω,Λ)×HΓ0 (Ω)

.

(6.11)

Proof. By Lemma 4.5 and the inequality (4.78) in Chapter 4, we have constants c1 , c2 > 0 such that Z a(Υ(ζ), Υ(ζ)) dx ≥ c1 kW k2H 1 (Ω,Λ) − c2 kwk2L2 (Ω) . (6.12) Γ0

Ω

In addition, from (6.2), there is c3 > 0 such that |ρ(ζ)| ≥ |D2 w|2 − c3 (|DW |2 + |w|2 ) for

x ∈ Ω.

(6.13)

Let 0 < ε < 1 be such that ε(1 − µ)c3 < c1 . Then Z Z a(ρ(ζ), ρ(ζ)) dx ≥ ε(1 − µ) |ρ(ζ)|2 dx Ω

Ω

≥ ε(1 − µ)kwk2H 2

Γ0 (Ω)

−[ε(1 − µ) +

− ε(1 − µ)c3 kW k2H 1

c3 ]kwk2H 1 (Ω) .

Γ0 (Ω,Λ)

(6.14)

It follows from the inequalities (6.12) and (6.14) that there are constants c4 , c5 > 0 such that B(ζ, ζ) + c5 kwk2H 1 (Ω) ≥ c4 kζk2H 1 (Ω,Λ)×H 2 (Ω) .

(6.15)

Finally the inequality (6.11) follows from the inequality (6.15) and Lemma 6.1 by a compactness-uniqueness argument as in Lemma 2.5. We recall the following operators which are needed to describe the Koiter shell operator. Operator S Let the operator S: L2 (Ω, Λ) → L2 (Ω, Λ) be defined by (5.2), that is, Π(X, Y ) = hSX, Y i for X, Y ∈ Mx , x ∈ Ω. (6.16) Then for each x ∈ M, S: Mx → Mx is a symmetric, linear operator. It is clear that for X ∈ X (M ) ˜XN SX = D (6.17) ˜ is the covariant differential of the Euclidean metric of R3 and N is where D the normal of the surface M.



291

Operator Q Let the operator Q: L2 (Ω, T 2 ) → L2 (Ω, Λ) be defined by hX, QT i = tr i (X)DT

for

X ∈ X (M ),

T ∈ T 2 (Ω).

(6.18)

−Q is the formal adjoint of the covariant differential operator D. For the further properties of Q, see Section 1.4. Operator P Let the operator P be defined by (5.32). Then hT, i (X)DΠi = hPT, Xi for

T ∈ T 2 (M ),

X ∈ Mx , x ∈ M.

(6.19)

Lemma 6.2 We have Π(·, D· W ) + Π(D· W, ·) + i (W )DΠ = D(SW ) + D∗ (SW ) − i (W )DΠ. (6.20) Moreover, let δ be, given by (1.109), the formal adjoint of the exterior derivative d. Then Qρ = (δd + 2dδ)SW + d∆w − 2κSW + Q( i (W )DΠ) +κ dw − i (Dw) c − wQ c

(6.21)

where κ is the Gauss curvature function. Proof. Let x ∈ Ω be given and let E1 , E2 be a frame field normal at x. Using the relations DEi Ej (x) = 0 and the formula (6.16), we obtain at x D(SW )(Ei , Ej ) = Ej (hSW, Ei i) = Ej [Π(Ei , W )] = DΠ(W, Ei , Ej ) + Π(Ei , DEj W ) which yield the formula (6.20) since DΠ is symmetric by Corollary 1.1. Using the formulas (1.150) and (3.100), we have QD2 w = −∆dw + κdw = d∆w + κdw.

(6.22)

In addition, it follows from the identity (6.20) and the formulas (1.150) and (1.151) that Q[Π(·, D· W ) + Π(D· W, ·) + i (W )DΠ] = −(δd + 2dδ)SW + 2κSW − Q( i (W )DΠ).

(6.23)

Moreover, it is easy to check from (1.144) that Q(w c ) = i (Dw) c + wQ c .

(6.24)

Finally, we obtain the formula (6.21) by the formulas (6.2), (6.22), (6.23), and (6.24). Denote by Oj (ζ) some terms which contain at most the j-th order derivatives of the vector field ζ with respect to the spacial variables for 0 ≤ j ≤ 3. We have the following Green formula for the Koiter shell.


292


Theorem 6.2 Let the bilinear form B(·, ·) be given by (6.5). Then, for ζ = (W, w), η = (U, u) ∈ H 1 (Ω, Λ) × H 2 (Ω), Z B(ζ, η) = (Aη, ζ)L2 (Ω,Λ)×L2 (Ω) + ∂(Aη, ζ)dΓ, (6.25) Γ

where

∂(Aζ, η) = hV1 (ζ), U i + v2 (ζ)

∂u + v3 (ζ)u, ∂ν

(6.26)

ν is the normal along the curve Γ, Aζ =

(Aζ)1 (Aζ)2

(6.27)

where (Aζ)1 = −(∆µ + 4γS∆µ S)W − 2γSdδdw + γ(1 − µ)PD2 w +γµ(∆w)QΠ + O1 (ζ), (Aζ)2 = γ(∆2 w − 2δdδSW ) + γµ∆hW, QΠi +γ(1 − µ){δ[ i (Dw) c − Q( i (W )Π)] − hD2 w, c i} +γ[(1 − µ)κ − 2µ tr c]∆w + O1 (ζ),

(6.28)

(6.29)

∆µ is of the Hodge-Laplacian type, applied to 1-forms (or equivalent vector fields), defined by 1−µ ∆µ = −( δd + dδ), (6.30) 2 d the exterior differential, δ the formal adjoint of d, ∆ the Laplacian on the manifold M (∆ = −δd),    V1 (ζ) = B1 (ζ) − 2γSB2 (ζ), v2 (ζ) = γhB2 (ζ), νi, (6.31)   v3 (ζ) = −γ[hB3 (ζ), νi + (1 − µ) ∂ ρ(ζ)(ν, τ )], ∂τ τ is the tangential along the curve Γ, and   B1 (ζ) = (1 − µ) i (ν)Υ(ζ) + µ(tr Υ(ζ))ν, B2 (ζ) = (1 − µ) i (ν)ρ(ζ) + µ(tr ρ(ζ))ν, (6.32)  B3 (ζ) = (1 − µ)Qρ(ζ) + µd(tr ρ(ζ)). R Proof. We need to compute the integral Ω a(ρ(ζ), ρ(η)) dx. To this end, R R let us compute Ω hρ(ζ), ρ(η))i dx and Ω tr ρ(ζ) tr ρ(η) dx, separately. Using the formulas (1.145) and (6.21), we have Z (ρ(ζ), D2 u)L2 (Ω,T 2 ) = −(Qρ, du)L2 (Ω,Λ) + ρ(ν, du) dΓ Γ = ∆2 w − 2δdδSW − δQ( i (W )DΠ) + κ∆w + δ( i (Dw) c ) + O1 (ζ), u Z + [ρ(ν, du) − uhν, Qρi] dΓ. (6.33) Γ



293

Similarly, we obtain −

ρ(ζ), Π(·, D· U ) + Π(D· U, ·) + i (U )DΠ + u c 2 L (Ω,T 2 ) = − ρ(ζ), 2D(SU ) − i (U )DΠ + u c L2 (Ω,T 2 ) Z = 2(Qρ, SU )L2 (Ω,Λ) − 2 ρ(ν, SU ) dΓ + (PD2 w + O1 (ζ), U )L2 (Ω,Λ) Γ

+(hD2 w, c i + O1 (ζ), u) = − 2Sdδdw + 2S(δd + 2dδ)SW + PD2 w + O1 (ζ), U L2 (Ω,Λ) Z +(−hD2 w, c i + O1 (ζ), u) − 2 ρ(ν, SU ) dΓ. (6.34) Γ

It follows from the formulas (6.33) and (6.34) that Z

hρ(ζ), ρ(η)i dx = − 2Sdδdw + 2S(δd + 2dδ)SW + PD2 w + O1 (ζ), U L2 (Ω,Λ) + ∆2 w − 2δdδSW − δQ( i (W )DΠ) + κ∆w + δ( i (Dw) c ) −hD2 w, c i + O1 (ζ), u Z + {−2ρ(ν, SU ) + [ρ(ν, du) − uhν, Qρi]} dΓ. (6.35) Ω

Γ

R Next, we compute the integral Ω tr ρ(ζ) tr ρ(η) dx. Noting that tr DSU = −δSU, we have from the formulas (6.2) and (6.20) tr ρ(ζ) = ∆w + 2δSW + tr i (W )DΠ − w tr c ;

(6.36)

tr ρ(η) = ∆u + 2δSU + tr lU DΠ − u tr c .

(6.37)

By Green’s formula for the Laplacian, ∂u ∂ tr ρ(ζ) −u ] dΓ ∂ν ∂ν Γ = (∆2 w − 2δdδSW + ∆hW, QΠi − (tr c )∆w + O1 (ζ), u) Z ∂u ∂ tr ρ(ζ) + [tr ρ(ζ) −u ] dΓ (6.38) ∂ν ∂ν Γ (tr ρ(ζ), ∆u) = (∆ tr ρ(ζ), u) +


Z

[tr ρ(ζ)

294


since ∆ = −δd when it is applied to a function. Moreover, (tr ρ(ζ), 2δSU + tr i(U )DΠ − u tr c ) Z = 2(d tr ρ(ζ), SU )L2 (Ω,Λ) − 2 hSU, νi tr ρ(ζ) dΓ Γ

+((∆w)QΠ + O1 (ζ), U )L2 (Ω,Λ) + (−(tr c )∆w + O1 (ζ), u) = (−2Sdδdw + 4SdδSW + (∆w)QΠ + O1 (ζ), U )L2 (Ω,Λ) Z +(−(tr c )∆w + O1 (ζ), u) − 2 hSU, νi tr ρ(ζ) dΓ.

(6.39)

Γ

It follows from the formulas (6.36)-(6.39) that Z tr ρ(ζ) tr ρ(η) dx Ω

= (−2Sdδdw + 4SdδSW + (∆w)QΠ + O1 (ζ), U )L2 (Ω,Λ) + ∆2 w − 2δdδSW + ∆hW, QΠi − 2(tr c )∆w + O1 (ζ), u Z ∂u ∂ tr ρ(ζ) + [tr ρ(ζ) −u − 2hSU, νi tr ρ(ζ)] dΓ. ∂ν ∂ν Γ

(6.40)

Finally, using the formula (4.47), (6.35), and (6.40), we obtain the formula (6.25). Remark 6.2 There are higher order coupling terms −2γSdδdw and −2γδdδW in the principal parts of the Koiter shell (6.28) and (6.29), respectively, because the second fundamental form goes into the change of the curvature tensor (see (6.2)). It is a coupling relationship that makes analysis on the Koiter shell much more difficult than that on the shallow shell. If we set S = 0 in the formulas (6.28) and (6.29), then the formulas (6.27) become shallow shell ones where only the lower order terms are coupled. The detailed expressions (6.28) and (6.29) are necessary when we consider a uniqueness problem in Section 6.2 (see the proof of Theorem 6.5). If the shell is flat, a plate, the equations in the formulas (6.25) are uncoupled. The formulas on the components W and w are the same as in [105], [109], a Kirchhoff plate. Definition 6.1 The operator A, given by (6.27), is called the Koiter operator. The Kinetic Energy of the Koiter Shell We will consider the kinetic energies of the Koiter shell under the classical Kirchhoff-Love assumption and the Koiter complementary hypotheses, respectively. The Kinetic Energy of the Koiter Shell under the Classical Kirchhoff-Love Assumption In this case, the kinetic energy is the same as for the shallow shell (see Section 4.3) Z Z h3 P= H(t) dx = (h|Wt |2 + hwt2 + |Dwt |2 ) dx. (6.41) 12 Ω Ω



295

By the “Principle of Virtual Work” and the Green formula (6.25), we obtain the following displacement equations for the Koiter shell after changing t to λt with λ2 E/(1 − µ2 ) = 1. Theorem 6.3 We assume that there are no external loads on the shell and the shell is clamped along a portion Γ0 of Γ and free on Γ1 , where Γ0 ∪ Γ1 = Γ and Γ0 ∩ Γ1 = ∅. Then the displacement vector ζ = (W, w) satisfies the following boundary value problem:   Wtt + (Aζ)1 = 0, wtt − γ∆wtt + (Aζ)2 = 0, (6.42)  ζ(0) = ζ 0 , ζt (0) = ζ 1 in Q∞ , (

W = 0, ∂w w= = 0, ∂ν

on Σ0∞ ,

   V1 (ζ) = 0, v2 (ζ) = 0,   γ ∂wtt + v3 (ζ) = 0, ∂ν

on Σ1∞ ,

(6.43)

(6.44)

where (Aζ)1 and (Aζ)2 are given by the formulas (6.28) and (6.29), respectively, and Q∞ = Ω × (0, ∞),

Σ0∞ = Γ0 × (0, ∞),

Σ1∞ = Γ1 × (0, ∞).

(6.45)

The Kinetic Energies of the Koiter Shell in the Koiter Hypotheses In order to eliminate a contradiction in the classical Kirchhoff-Love assumptions, Koiter introduced complementary hypotheses which permit us to derive a satisfactory approximation of the displacement field of the shell from only the knowledge of the displacement field of the middle surface. These hypotheses are: Koiter’s Hypotheses ([95], pages 15-16) (i) The normal to the undeformed middle surface remains normal to the deformed middle surface after the deformation; (ii) During the deformation, the stresses are approximatively plane and parallel to the tangent plane to the middle surface. Let σ(·, ·) be the stress tensor of the three-dimensional shell. Then the Koiter assumption (ii) means i (N )σ = 0, that is, σ(X, N )(p) = 0

for X ∈ X (M )

(6.46)

and σ(N, N )(p) = 0 for p = x + zN ∈ S .


(6.47)

296


Let Υ(·, ·) be the strain tensor of the three-dimensional shell. For an isotropic material, we have σ(·, ·) =

E µE ˜ g (·, ·) Υ(·, ·) + (trΥ)˜ 1+µ (1 + µ)(1 − 2µ)

(6.48)

where E, µ respectively denote Young’s modulus and Poisson’s coefficient of ˜ and g˜(·, ·) are the trace and the Euclidean the material, respectively, and tr 3 metric in R , respectively. It is easy to check that the relations (6.46)-(6.48) yield the following formulas Υ(X, N ) = 0 for

X ∈ X (M )

(6.49)

and (1 − µ)Υ(N, N ) + µ tr Υ = 0 for

p = x + zN ∈ S

(6.50)

where tr is the trace of M in the induced metric from R3 . Let the shell occupy a region F ( S ) in R3 after deformation where F ( S ) = { F(p) | p = x + zN (x) ∈ S } and F : S → R3 is the deformation map. The Koiter assumption (i) implies F (p) = F (x) + zN (F (x))

for

p = x + zN (x) ∈ S

(6.51)

where N is the normal to the deformed middle surface Ω = { F(x) | x ∈ Ω } and z = z(p) is the distance from F (p) to the surface Ω. Lemma 6.3 Under the Koiter assumptions (i) and (ii), we have the following approximation z(p) = z[1 − α tr Υ(ζ)] −

z2 α tr D( i (W )Π − Dw) 2

(6.52)

for p = x + zN ∈ S where α = µ/(1 − µ) and Υ(ζ) is the strain tensor of the undeformed middle surface Ω and ζ = (W, w). Proof. Let ζ = F (p) − p be the displacement vector of the material point p = x + zN. Then ζ(p) = W + (w − z)N + zN

for

p = x + zN.

Using the formula (4.87), we have ˜ N ζ = dz N − N = dz ( i (W )Π − Dw) + ( dz − 1)N, D dz dz dz

(6.53)

˜Xζ = D ˜ X W + zD ˜ X ( i (W )Π − Dw) + X(z)( i (W )Π − Dw) D ˜ X N + [X(w) + X(z)]]N +(w + z − z)D (6.54)



297

˜ is the covariant differential of the Euclidean metric where X ∈ X (Ω) and D 3 of R . It follows from the formulas (6.53), (6.54), and the formula (4.87) that dz = 1 + Υ(N, N ), dz ˜ N ζ, Xi = hD

dz [Π(W, X) − X(w)], dz

˜ X ζ, Xi = DW (X, X) + zD( i (W )Π − Dw)(X, X) hD +X(z)[Π(W, X) − X(w)] + (w + z − z)Π(X, X)

(6.55) (6.56)

(6.57)

for X ∈ X (M ). Using the relations (6.49), (6.54), and (6.56), we obtain ˜ X ζ, N i + hD ˜ N ζ, Xi 0 = 2Υ(N, X) = hD = −Π(W, X) − zΠ( i (W )Π − Dw, X) + X(w) + X(z) dz + [Π(W, X) − X(w)] for X ∈ X (M ) dz which yields Dz = (

dz − 1)(Dw − i (W )Π) + z i ( i (W )Π − Dw)Π. dz

(6.58)

Let R be the smallest principal radius of curvature of the undeformed middle surface given by (4.25). Then the estimate |z i ( i (W )Π − Dw)Π| ≤ c

h | i (W )Π − Dw| R

for some c > 0 implies that the formula (6.58) can be approximated by Dz = (

dz − 1)(Dw − i (W )Π). dz

(6.59)

Combining the formulas (6.57) and (6.59) gives tr Υ = tr DW + z tr D( i (W )Π − Dw) + hDz, i (W )Π − Dwi +(w + z − z) tr Π dz = tr Υ(ζ) + z tr D( i (W )Π − Dw) − ( − 1)| i (W )Π − Dw|2 dz +(z − z) tr Π from which we have, by dropping the nonlinear terms, tr Υ = tr Υ(ζ) + z[tr D( i (W )Π − Dw) + tr Π] − z tr Π.


(6.60)

298


Using the formulas (6.60) and (6.55) in the formula (6.50), we have dz + α[tr D( i (W )Π − Dw) + tr Π]z = α tr Πz + 1 − α tr Υ(ζ) dz

(6.61)

with z(0) = 0. Since

h h , α| tr Πz| ≤ c R R for some c > 0, the terms α tr Πz and α tr Πz can be omitted compared with 1 from the equation (6.61) by the assumption (4.28). Thus α| tr Πz| ≤ 2α

dz + α tr D( i (W )Π − Dw)z = 1 − α tr Υ(ζ), dz

z(0) = 0.

(6.62)

We solve the equation (6.62), then linearize the solution with respect to the displacement vector, and obtain the formula (6.52). Using the formulas (6.52) and (4.87), we have, after a linearization, ζ(p) = ζ + zY1 + z 2 Y2

(6.63)

where Y1 = i (W )Π−Dw−α[tr Υ(ζ)]N,

1 Y2 = − α[tr D( i (W )Π−Dw)]N. (6.64) 2

Then the kinetic energy per unit area of the unformed middle surface is obtained by integration with respect to z Z h/2 (6.65) H(t) = |ζ t |2 (1 + tr Πz + κz 2 ) dz. −h/2

Since

h 1 h h |ζ |2 + ( )2 |ζ t |2 for |z| ≤ , R t 4 R 2 we have, by the formula (6.63), Z h/2 h3 H(t) = |ζ t |2 dz = (|Wt |2 + wt2 )h − αwt tr D( i (Wt )Π − Dwt ) 12 −h/2 |ζ t |2 | tr Πz + κz 2 | ≤

+|Y1t |2

h3 h5 + |Y2t |2 . 12 80

(6.66)

By a similar argument, the term |Y1t |2 h3 /12 in the right hand side of the formula (6.66) can be replaced with [|Dwt |2 + α2 (tr DWt )2 ]h3 /12. We then obtain h2 α tr D( i (Wt )Π − Dwt )]2 }h + |Dwt |2 h3 /12 24 +α2 (tr DWt )2 h3 /12 + α2 [tr D( i (Wt )Π − Dwt )]2 h5 /720. (6.67)

H(t) = {|Wt |2 + [wt −



299

Let the operator S be given by (6.16). We introduce a bilinear form on L2 (Ω, Λ) × L2 (Ω) by H(ζ, η) =

Z

Ω

{hW, U i + [w + ασ0 δ( i (W )Π − Dw)][u + ασ0 δ( i (U )Π − Du)]

+2σ0 hDw, Dui + 2α2 σ0 δW δU +α2 σ1 δ(SW − Dw)δ(SU − Du)} dx

(6.68)

where ζ = (W, w),

η = (U, u),

σ0 = h2 /24,

σ1 = h4 /720.

(6.69)

We need Lemma 6.4 We have H(ζ, η) = Φ(ζ), U + Ψ(ζ), u L2 (Ω,Λ) Z ∂u + [hΓ1 (ζ), U i + Γ2 (ζ) + Γ3 (ζ)u] dΓ ∂ν Γ

(6.70)

where Φ(ζ) = W + 2α2 σ0 dδW + α2 (σ02 + σ1 )SdδSW +ασ0 Sdw + α2 (σ02 + σ1 )Sd∆w, Ψ(ζ) = w − 2(1 − α)σ0 ∆w + α2 (σ02 + σ1 )∆2 w +ασ0 δSW − α2 (σ02 + σ1 )δdδSW,

Γ1 (ζ) = −2α2 σ0 δW ν − α[σ0 w + α(σ02 + σ1 )(δSW + ∆w)]Sν, Γ2 (ζ) = ασ0 w + α2 (σ02 + σ1 )[δ(SW ) + ∆w],

Γ3 (ζ) = σ0 (2 − α)

∂w ∂δSW ∂∆w − α2 (σ02 + σ1 )( + ). ∂ν ∂ν ∂ν

(6.71)

(6.72) (6.73) (6.74)

(6.75)

Proof. We compute each term in the right hand side of (6.70) where the formulas (1.136) and (1.137) have been used many times.


300


Noting that i (W )Π = SW and D = d if applied to a function, we have w + ασ0 δ( i (W )Π − Dw), u + ασ0 δ( i (U )Π − Du) = w + ασ0 δ(SW − Dw), u +ασ0 dw + ασ0 dδ(SW − Dw), SU − Du 2 L (Ω,Λ) Z −ασ0 [w + ασ0 δ(SW − Dw)]hSU − Du, νi dΓ Γ = w + 2ασ0 ∆w + α2 σ02 ∆2 w + ασ0 δSW − α2 σ02 δdδSW, u + ασ0 Sdw + α2 σ02 Sdδ(SW − Dw), U 2 L (Ω,Λ) Z −ασ0 [w + ασ0 δ(SW − Dw)]hSν, U i dΓ ZΓ ∂u +ασ0 [w + ασ0 δ(SW − Dw)] dΓ ∂ν Γ Z −ασ0 hDw + ασ0 dδ(SW − Dw), νiu dΓ; (6.76) Γ

Dw, Du

L2 (Ω,Λ)

Z = δdw, u + hDw, νiu dΓ;

δW, δU = dδW, U

(6.77)

Γ

L2 (Ω,Λ)

−

Z

(δW )hU, νi dΓ;

(6.78)

Γ

δ(SW − Dw), δ(SU − Du) = Sdδ(SW − Dw), U 2 L (Ω,Λ) Z ∂u − δdδ(SW − Dw), u − δ(SW − Dw)[hSν, U i − ] dΓ ∂ν Γ Z − hdδ(SW − Dw), νiu dΓ. (6.79) Γ

Inserting the formulas (6.76)-(6.79) in the formula (6.68), we obtain the formula (6.70). Equations of Motion of the Koiter Shell in Koiter’s Hypotheses By the “Principle of Virtual Work” and the Green formula (6.25), we obtain the following displacement equations for the Koiter shell after changing t to λt with λ2 E/(1 − µ2 ) = 1 : Theorem 6.4 We assume that there are no external loads on the shell and the shell is clamped along a portion Γ0 of Γ and free on Γ1 , where Γ0 ∪ Γ1 = Γ and Γ0 ∩ Γ1 = ∅. Let Φ(ζ), Ψ(ζ), Γ1 (ζ), Γ2 (ζ) and Γ3 (ζ) be defined by Lemma



301

6.4. Then the displacement vector ζ = (W, w) satisfies the following boundary value problem:   Φ(ζtt ) + (Aζ)1 = 0, Ψ(ζtt ) + (Aζ)2 = 0, (6.80)  ζ(0) = ζ 0 , ζt (0) = ζ 1 in Q∞ , ( W = 0, on Σ0∞ , (6.81) ∂w w= = 0, ∂ν   Γ1 (ζtt ) + V1 (ζ) = 0, Γ2 (ζtt ) + v2 (ζ) = 0, on Σ1∞ , (6.82)  Γ3 (ζtt ) + v3 (ζ) = 0, where (Aζ)1 and (Aζ)2 are given by the formulas (6.28) and (6.29), respectively, and Q∞ = Ω × (0, ∞),

6.2

Σ0∞ = Γ0 × (0, ∞),

Σ1∞ = Γ1 × (0, ∞).

(6.83)

Uniqueness for the Koiter Shell

We consider the uniqueness of the Cauchy problem for the Koiter shell which is needed to absorb the lower order terms in obtaining controllability/stabilization for the equations of motion in Theorem 6.3. Let M be a smooth surface in R3 and let a middle surface of the shell occupy a bounded region Ω of the surface M whose boundary Γ is smooth. ˆ ⊂ Γ have a positive length. Consider the following Cauchy problem in Let Γ unknown ζ = (W, w): O1 (ζ) Aζ = on Ω (6.84) O1 (ζ) + O2 (w) subject to the boundary conditions  ˆ  W = DW = 0 on Γ, ∂w ∂∆w ˆ w = = ∆w = = 0 on Γ ∂ν ∂ν

(6.85)

where A is the Koiter operator, given by the formula (6.27), and Oj denote some terms with the orders of the derivatives less than j. Let R be the smallest principal radius of curvature of the undeformed middle surface given by (4.25). In general the condition h/R ≪ 1 is assumed in the thin shell. Here we need √ h/R < 3. (6.86) In this section we will establish the following uniqueness results.


302


Theorem 6.5 Let the condition (6.86) be true. If one of the following assumptions holds: (a) inf x∈Ω (|Π|2 − 2κ) > 0; or (b) |Π|2 = 2κ for every x ∈ Ω, where κ is the Gauss curvature of the surface M, then the problem (6.84) − (6.85) has a unique zero solution. Remark 6.3 The assumptions (a) and (b) are easy to check. Clearly, (a) is true when Π 6= 0 and κ ≤ 0. Moreover, if λ1 and λ2 are the two eigenvalues of the second fundamental form Π, then |Π|2 − 2κ = (λ1 − λ2 )2 . Then the assumption (a) is equivalent to λ1 6= λ2

∀x ∈ Ω

and the assumption (b) means λ1 = λ2

∀ x ∈ Ω.

Let us make a preparation for the proof of Theorem 6.5 which will be given at the end of this section. We introduce some basic Carleman estimates of the first order differential operators from [223] where some of them were given by [191]. Those estimates will play an important role in our proof of Theorem 6.5. Consider a differential operator ∂ on R2 , given by ∂ = ∂x1 + λ∂x2

(6.87)

∂ and λ ∈ C 1 with Im λ 6= 0 at the origin (0, 0). From ∂xi Propositions 1.2 ([223, p. 3]) and 1.4 ([223, p. 45]), we have

where ∂xi =

Lemma 6.5 There exist positive constants C, k0 , T0 , and r such that, for all k ≥ k0 and 0 < T ≤ T0 , Z Z 2 C k(x1 −T )2 2 ek(x1 −T ) |∂w|2 dx1 dx2 (6.88) e |w| dx1 dx2 ≤ k F (T,r) F (T,r) and Z

2

ek(x1 −T ) (|∂x1 w|2 + |∂x2 w|2 ) dx1 dx2 F (T,r) Z 2 2 ≤ C(1 + kT ) ek(x1 −T ) |∂w|2 dx1 dx2

(6.89)

F (T,r)

for w ∈ C ∞ with supp w ⊂ F (T, r) where F (T, r) = { (x1 , x2 ) | 0 ≤ x1 ≤ T, |x2 | ≤ r }.


(6.90)


303

Next, we move the estimates in Lemma 6.5 to the Riemannian manifold M with the induced metric from the Euclidean metric of R3 . Let x0 ∈ M be given and let E1 , E2 be a frame at x0 . Consider a differential operator Φ, given by Φ = E1 + λE2 (6.91) where λ ∈ C 1 with Im λ 6= 0. Let ϕ(x) = (y1 , y2 ) be a coordinate system on M at x0 with ϕ(x0 ) = (0, 0). We have Lemma 6.6 There exist positive constants C, k0 , T0 , and r such that, for all k ≥ k0 and 0 < T ≤ T0 , Z Z 2 C k(y1 (x)−T )2 2 e |w| dg ≤ ek(y1 (x)−T ) |Φw|2 dg (6.92) k −1 −1 ϕ (F (T,r)) ϕ (F (T,r)) where dg is the volume element of the induced metric g of M from R3 and Z 2 ek(y1 (x)−T ) (|E1 w|2 + |E2 w|2 ) dg ϕ−1 (F (T,r)) Z 2 ≤ C(1 + kT 2 ) ek(y1 (x)−T ) |Φw|2 dg (6.93) ϕ−1 (F (T,r))

for w ∈ C ∞ with supp w ⊂ ϕ−1 (F (T, r)). Proof. Let Ei = αi1 ∂y1 + αi2 ∂y2 ∂ and (αij )2×2 are real matrices with det(αij ) 6= 0. Then ∂yi Im λ 6= 0 and det(αij ) 6= 0 together imply that α + λα 12 22 α11 + λα21 6= 0, Im 6= 0. α11 + λα21

where ∂yi =

Then Lemma 6.6 follows by applying Lemma 6.5 to Φ since α12 + λα22 Φ = (α11 + λα21 ) ∂y1 + ∂y2 . α11 + λα21

If we let Φ = E1 + i E2 , then |Φw|2 ≤ 2|Dw|2 . Applying Lemma 6.6 repeatedly, we have Lemma 6.7 For a positive integer m given, there exist positive constants C, k0 , T0 , and r such that, for all k ≥ k0 and 0 < T ≤ T0 , m−1 XZ 2 ek(y1 (x)−T ) |Dj w|2 dg j=0

≤

C k

ϕ−1 (F (T,r))

Z

ϕ−1 (F (T,r))


2

ek(y1 (x)−T ) |Dm w|2 dg

(6.94)

304


for w ∈ C ∞ with supp w ⊂ ϕ−1 (F (T, r)). Let the operators ∆µ and S be given by the formula (6.30) and (6.17), respectively. We introduce a fourth order differential operator A on H 4 (M, Λ) by AW = ∆∆µ W − βS(δd)2 SW (6.95) where ∆ = δd + dδ and β is a positive constant. We need Lemma 6.8 We have [S, ∆]W = O1 (W )

for

W ∈ H 2 (Ω, Λ).

(6.96)

Furthermore, if |Π|2 = 2κ for all x ∈ Ω, then [S, dδ]W = O1 (W ),

[S, δd]W = O1 (W )

(6.97)

for W ∈ H 2 (Ω, Λ). Proof. Let x ∈ Ω be given. Since S: Mx → Mx is a symmetric, linear operator, there exists an orthonormal basis e1 , e2 of Mx and real numbers λ1 , λ2 such that Sei = λi ei for i = 1, 2. (6.98) Let E1 , E2 be a frame normal at x such that Ei (x) = ei ,

DEi Ej (x) = 0

(6.99)

for 1 ≤ i, j ≤ 2. Then in a neighborhood of x, we have D(SW )(Ei , Ej ) = hDEj (SW ), Ei i = Ej (Π(W, Ei )) − Π(W, DEj Ei ) = DΠ(W, Ei , Ej ) + Π(DEj W, Ei ). (6.100) Using the formulas (6.98)-(6.100), we obtain at x D2 (SW )(Ei , Ej , Ek ) = Ek (D(SW )(Ei , Ej )) = D2 Π(W, Ei , Ej , Ek ) + DΠ(DEk W, Ei , Ej ) + DΠ(DEj W, Ei , Ek ) +Π(DEk DEj W, Ei ) = λi hDEk DEj W, Ei i + O1 (W ) = λi D2 W (Ei , Ej , Ek ) + O1 (W )

(6.101)

for 1 ≤ i, j, k ≤ 2. It follows from the formulas (6.98), (6.99), (6.101), and (1.114), that at x X X ∆SW (x) = − DEj DEj (SW ) + R(Ei , Ej , SW, Ej )Ei j

=−

X

ij

2

D (SW )(Ei , Ej , Ej )Ei + κSW

ij

= S∆W + O1 (W ).



305

Next, let |Π|2 = 2κ for all x ∈ Ω. Then λ1 = λ2 = λ for all x ∈ Ω. By the formulas (6.101) and (1.119), we have X X hDEi DEj (SW ), Ej iEi = − D2 (SW )(Ej , Ej , Ei )Ei dδSW = − ij

ij

= SdδW + O1 (W ).

Similarly, we have δdSW = SδdW + O1 (W ).

We consider the structure of the operator A in a frame field. To this end, we assume that x0 ∈ M is given such that |Π|2 (x0 ) > 2κ(x0 ).

(6.102)

Then there are two distinct eigenvalues λ1 and λ2 of the operators S for x in a neighborhood of x0 . Then there is a frame E1 , E2 at x0 such that SEi = λi Ei for x in a neighborhood of x0 . We have the formula Π = λ1 E1 ⊗ E1 + λ2 E2 ⊗ E2 .

(6.103)

Lemma 6.9 Let x0 ∈ M be such that the assumption (6.102) holds true. Let a frame E1 , E2 at x0 be such that the formula (6.103) is true. If W = w1 E1 + w2 E2 ,

(6.104)

−2AW = ∆f1 E1 + ∆f2 E2 + O3 (W )

(6.105)

f1 = 2E12 w1 + a1 E22 w1 + ςE1 E2 w2 ,

(6.106)

f2 = ςE1 E2 w1 + a2 E12 w2 + 2E22 w2 , ai = 1 − µ + 2βλ2i , for i = 1, 2, ς = 1 + µ − 2βκ,

(6.107)

then where

(6.108)

where κ is the Gauss curvature. Proof. Since ∆δd = (δd)2 , we have by the formula (6.96) AW = ∆(∆µ W − βSδdSW ) + O3 (W ). For the expression (6.105) it suffices to compute ∆µ W − βSδdSW. Using the formulas (6.104), (1.117), and (1.119), we have X X δdW = − DEi DEi W + hDEi DEj W, Ei iEj + O1 (W ) i

ij

= (E1 E2 w2 − E22 w1 )E1 + (E1 E2 w1 − E12 w2 )E2 + O1 (W )


(6.109)

(6.110)

306


and dδW = −

X ij

hDEi DEj W, Ej iEi + O1 (W )

= −(E12 w1 + E1 E2 w2 )E1 − (E2 E1 w1 + E22 w2 )E2 + O1 (W ).

(6.111)

It follows from the formulas (6.30), (6.110) and (6.111) that 2∆µ W = [2E12 w1 + (1 − µ)E22 w1 + (1 + µ)E1 E2 w2 ]E1

+[(1 − µ)E12 w2 + 2E22 w2 + (1 + µ)E1 E2 w1 ]E2 + O1 (W ).

(6.112)

In addition, replacing W with SW in the formula (6.110) and then applying the operator S to it, we obtain SδdSW = λ1 (λ2 E1 E2 w2 − λ1 E22 w1 )E1

+λ2 (λ1 E1 E2 w1 − λ2 E12 w2 )E2 + O1 (W ).

(6.113)

Then it follows from the formulas (6.112) and (6.113) that 2∆µ W − 2βSδdSW = f1 E1 + f2 E2 .

(6.114)

Finally, the formula (6.105) follows from the formulas (6.109) and (6.114). A simple computation yields Lemma 6.10 Let x0 ∈ Ω be such that the condition (6.102) holds. Let β > 0 satisfy β|Π|2 (x0 ) < 1 − µ. (6.115)

Then

(2 + ς)2 > a1 a2 > (2 − ς)2

(6.116)

where ai and ς are given in the formulas (6.108).

Now we establish a Carleman estimate for the operator A. Theorem 6.6 Let x0 ∈ Ω be such that the condition (6.102) holds. Let β > 0 satisfy the inequality (6.115). Let ϕ(x) = (y1 , y2 ) be a coordinate system on M at x0 with ϕ(x0 ) = (0, 0). Then there exist positive constants C, k0 , T0 , and r such that, for all k ≥ k0 and 0 < T ≤ T0 , 3 Z X j=0

≤

C k

2

ϕ−1 (F (T,r))

Z

ek(y1 (x)−T ) |Dj W |2 dg 2

ϕ−1 (F (T,r))

ek(y1 (x)−T ) |AW |2 dg

(6.117) (6.118)

for every W ∈ C ∞ (M, Λ) such that supp W ⊂ ϕ−1 (F (T, r)) where F (T, r) is given by the formula (6.90).



307

Proof. Step 1 Denote W = (w1 , w2 )τ . We define an operator Ψ on L (Ω) × L2 (Ω) by 2

ΨW = G0 E14 W + G1 E24 W + G2 E23 E1 W +G3 E22 E12 W + G4 E2 E13 W

(6.119)

where E2j E14−j W = (E2j E14−j w1 , E2j E14−j w2 )τ , 0 ≤ j ≤ 4, 2 0 a1 0 G0 = , G1 = , 0 a2 0 2 0ς a1 + 2 0 G2 = G4 = , G3 = ς 0 0 a2 + 2

(6.120) (6.121) (6.122)

where ai and ς are given in the formulas (6.108). Let f1 and f2 be given by the formulas (6.106) and (6.107), respectively. It is easy to check that ∆f1 ΨW = + O3 (W ). (6.123) ∆f2 Let Z = (z1 , z2 , z3 , z4 )τ where

on

8 L2 (Ω)

zj = (E24−j E1j−1 w1 , E24−j E1j−1 w2 )τ ,

It is easy to check that the following relation holds

1 ≤ j ≤ 4.

E1 Z + JE2 Z = O3 (W ), O3 (W ), O3 (W ), G−1 0 ΨW where



0  0 J =  0 G−1 0 G1

−I2 0 0 G−1 0 G2

0 −I2 0 G−1 0 G3

(6.124)

τ

 0 0   −I2  G−1 0 G4

(6.125)

(6.126)

and I2 is the unit matrix of 2 × 2. Step 2 We diagonalize the matrix J. Noting that

2λ4 + (a1 + 2)λ2 + a1 = (λ2 + 1)(2λ2 + a1 ) and a2 λ4 + (a2 + 2)λ2 + 2 = (λ2 + 1)(a2 λ2 + 2), we have 1 det(G0 λ4 − G4 λ3 + G3 λ2 − G2 λ + G1 ) 2a2 1 = {[2λ4 + (a1 + 2)λ2 + a1 ][a2 λ4 + (a2 + 2)λ2 + 2] 2a2 −ς 2 (λ2 + 1)2 λ2 } 1 = (λ2 + 1)2 f (λ2 ) (6.127) 2a2

det(λI − J) =


308


where f (λ2 ) = 2a2 λ4 + (4 + a1 a2 − ς 2 )λ2 + 2a1 .

(6.128)

2

Consider solutions to the equation f (λ ) = 0. Let 2a2 h2 + (4 + a1 a2 − ς 2 )h + 2a1 = 0.

(6.129)

The inequality (6.116) implies that (4 + a1 a2 − ς 2 )2 − 16a1 a2 √ √ √ = (2 − ς − a1 a2 )(2 + ς − a1 a2 )[(2 + a1 a2 )2 − ς 2 ] < 0. Then for a solution of the equation (6.129) and by (6.116) again Re h = −

4−ς 4 + a1 a2 − ς 2 0 in a neighborhood of x0 . We may assume that λ1 6= 0. Applying the operator Ej to both the sides of the equation (6.135) yields Ej E1 (∆w) = O3 (ζ)

for j = 1, 2.

(6.136)

In addition, using the equation (6.134), we have E2 E2 (∆w) = ∆2 w − E1 E1 (∆w) = O3 (ζ).

(6.137)

It follows from the relations (6.135) and (6.137) that D2 (∆w) = O3 (ζ).

(6.138)

Moreover, the equation (6.138) yields ∆(D2 w(Ei , Ej )) = ∆Ei Ej (w) + O3 (ζ) = −Ei Ej (∆w) + O3 (ζ) = O3 (ζ) for 1 ≤ i, j ≤ 2, that is,

∆D2 w = O3 (ζ).

(6.139)

γd∆2 w = 2γ(dδ)2 SW + O3 (ζ).

(6.140)

Applying the operator d to the equation (6.134) gives

Using the relations (6.96), (6.138), and (6.139), we obtain ∆Sdδdw = −∆Sd∆w = −S∆d∆w − [∆, S]d∆w = −S∆d∆w + O3 (ζ) = Sd(∆2 w) + O3 (ζ).


(6.141)

310


Applying the operator ∆ = dδ + δd to both the sides of the equation (6.133), we obtain, via the formulas (6.96), (6.30), (6.138), (6.139), (6.140), and (6.141) AW = ∆∆µ W − βS(δd)2 SW

= ∆[∆µ W + 4γS∆µ SW + 2γSdδdw] +4γS(dδ)2 SW − 2γ∆Sdδdw + O3 (ζ)

= 2γ[Sd(∆2 w) − ∆Sdδdw] + O3 (ζ) = O3 (ζ)

(6.142)

where β = 2γ(1 − µ). On the other hand, a similar argument as in the proof of Proposition 4.3 yields ( W |Γˆ = DW |Γˆ = D2 W |Γˆ = D3 W |Γˆ = 0, (6.143) ∂w ∂∆w w|Γˆ = |Γˆ = ∆w|Γˆ = |Γˆ = 0. ∂ν ∂ν Moreover, by the condition (6.86), we have β|Π|2 (x0 ) =

h2 h2 (1 − µ) 2 (λ1 + λ22 ) ≤ (1 − µ) < 1 − µ. 6 3R2

Then the Carleman estimate in Theorem 6.6 holds true for the operator A. By the conditions (6.143), we have that ζ = 0 on Ω after we apply Theorem 6.6 of this section to the component AW and Theorem 1.1 of Chapter Two of [223] to the component ∆2 w, respectively. (ii) Next, we assume that |Π|2 = 2κ for all x ∈ Ω. It is easy to check that there is a function ϕ ∈ C ∞ such that Π = ϕg,

SW = ϕW

for

x∈Ω

(6.144)

where g is the induced metric of the surface M from the Euclidean metric of R3 . Moveover, we have DΠ = g ⊗ Dϕ. Then the symmetry of DΠ (Corollary 1.1) implies that ϕ = λ is a constant. Then DΠ = 0, P = 0, and QΠ = 0. Using those results in the formulas (6.28) and (6.29), the system (6.84) becomes (1 + 4γλ2 )∆µ W − 2γλd(∆w) = O1 (ζ);

(6.145)

γ(∆2 w − 2λδdδW ) = O2 (ζ).

(6.146)

Applying the differential operator d to the equation (6.146), we have d(∆2 w) = 2λ(dδ)2 W + O2 (ζ).

(6.147)

Applying the differential operator dδ to the equation (6.145) and using the formula (6.147), we obtain (dδ)2 W =

4γλ2 2γλ d(∆2 w) + O3 (ζ) = (dδ)2 W + O3 (ζ) 2 1 + 4γλ 1 + 4γλ2



311

which yields (dδ)2 W = O3 (ζ).

(6.148)

Furthermore, applying the differential operator δd to the equation (6.145) gives (δd)2 W = O3 (ζ). (6.149) It follows from the formulas (6.148) and (6.149) that ∆2 W = O3 (ζ).

(6.150)

Finally, we apply [170] to the system (6.146), (6.150) and (6.143) to have ζ = 0.

6.3


In this section we shall establish some multiplier identities for the Koiter model. The multipliers we will need are f ζ1 ,

f ζ2 ,

and m(ζ) = (DV W, V (w))

where ζ = (W, w), ζ1 = (W, 0), ζ2 = (0, w), f is a function, and V is a vector field which satisfies the following assumption (6.151). Let V be a vector field on Ω such that there is a function ϑ on Ω that satisfies the relation DV (X, X) = ϑ(x)|X|2

for all X ∈ Ωx

x ∈ Ω.

(6.151)

For the existence of such a vector field, see Section 4.5. For W ∈ H 1 (Ω, Λ), we define F (V, W )(X, Y ) = Π(DX W, DY V ) + Π(DX V, DY W ) +DV Π(DX W, Y ) + DV Π(X, DY W )

(6.152)

for all X, Y ∈ Mx , x ∈ Ω. We need Lemma 6.11 Let a vector field V be such that the relation (6.151) holds and let m(ζ) = (DV W, V (w)). Then ρ(m(ζ)) = DV ρ(ζ) + ρ(ζ)(D· V, ·) + ρ(ζ)(·, D· V ) +F (V, W ) + O0 (W ) + O1 (w)

(6.153)

where the change of the curvature tensor ρ is defined by the formula (6.2).


312


Proof. It follows from the formula (4.133) that D2 (V (w)) = DV D2 w + 2G(V, D2 w) + O1 (w)

(6.154)

where the tensor G(V, D2 w) is defined by the formula (4.131) in Chapter 4. Let x ∈ Ω be given and let E1 , E2 be a frame normal at x. Then [V, Ej ] = DV Ej − DEj V = −DEj V

at x

for j = 1,2. Then DEj DV W = DEj DV W + DDEj V W + RV Ej W

at x

(6.155)

where RV Ej is the curvature operator. By the formulas (6.16) and (6.155), we have at x Π(Ei , DEj DV W ) = Π(Ei , DV DEj W ) + Π(Ei , DDEj V W ) + Π(Ei , RV Ej W ) = V (Π(Ei , DEj W )) − DV Π(Ei , DEj W ) + Π(Ei , DDEj V W )

+R(V, Ej , W, SEi )

(6.156)

for 1 ≤ i, j ≤ 2. Using the formulas (6.154) and (6.156), we obtain at x ρ(m(ζ))(Ei , Ej ) = D2 (V (w))(Ei , Ej ) − Π(Ei , DEj DV W )

−Π(Ej , DEi DV W ) − i (DV W )DΠ(Ei , Ej ) − V (w) c (Ei , Ej ) = V (ρ(Ei , Ej )) + ρ(Ei , DEj V ) + ρ(Ej , DEi V ) +F (V, W )(Ei , Ej ) + O0 (W ) + O1 (w)

which yields the formula (6.153).

Lemma 6.12 We have 2B(ζ, m(ζ)) =

Z

B(ζ, ζ)hV, νi dΓ + 2℘(V, ζ) + lo (ζ)

Γ

+2

Z

Ω

ϑ[γa(ρ(ζ), ρ(ζ)) − a(Υ(ζ), Υ(ζ))] dx

(6.157)

where B(·, ·) and B(·, ·) are given by the formulas (6.5) and (6.4), respectively, and Z ℘(V, ζ) = [a(Υ(ζ), G(V, DW )) + γa(ρ(ζ), F (V, W ))] dx. (6.158) Ω

Proof. The expression for a(Υ(ζ), Υ(m(ζ))) has been given in the formula (4.143) in Chapter 4. Here we need to compute a(ρ(ζ), ρ(m(ζ))). Using Theorem 3.4, we have a(ρ(ζ), ρ(ζ)(·, D· V )) = ϑ(x)a(ρ(ζ), ρ(ζ)).


(6.159)


313

It follows from the formulas (6.153), (6.159), and (6.151) that 2a(ρ(ζ), ρ(m(ζ))) = V (a(ρ(ζ), ρ(ζ))) + 4ϑa(ρ(ζ), ρ(ζ)) +2a(ρ(ζ), F (V, W )) + lo (ζ) = div [a(ρ(ζ), ρ(ζ))V ] + 2ϑa(ρ(ζ), ρ(ζ)) +2a(ρ(ζ), F (V, W )) + lo (ζ)

(6.160)

Using the formula (4.143) in Chapter 4 and the formula (6.160), we obtain 2B(ζ, m(ζ)) = div [B(ζ, ζ)V ] + 2ϑ[γa(ρ(ζ), ρ(ζ)) − a(Υ(ζ), Υ(ζ))] +2a(Υ(ζ), G(V, DW )) + 2γa(ρ(ζ), F (V, W )) + lo (ζ). (6.161) Integrating the identity (6.161) over Ω yields the formula (6.157).

Let ζ = (W, w) solve the problem ζtt + γ(0, −∆wtt ) + Aζ = 0

(6.162)

where A is the Koiter operator, given by the formula (6.27), and γ = h2 /12. We have Theorem 6.7 Let ζ = (W, w) solve the problem (6.162) and let p be a function. Then Z Z ∂(Aζ, pζ1 ) dΣ = p[B(ζ, ζ1 ) − |Wt |2 ]dQ + L(ζ); (6.163) Σ

Q

Z h ∂wtt i ∂(Aζ, pζ2 ) + γpw dΣ ∂ν Σ Z = p[B(ζ, ζ2 ) − wt2 − γ|Dwt |2 ] dQ + L(ζ)

(6.164)

Q

where ζ1 = (W, 0), ζ2 = (0, w), Σ = (0, T ) × Γ and Q = (0, T ) × Ω. Proof. Clearly, we have Υ(pζi ) = pΥ(ζi ) + lo (ζ),

ρ(pζi ) = pρ(ζi ) + lo (ζ)

for i = 1, 2, which yield, via the formula (6.4), B(ζ, pζi ) = pB(ζ, ζi ) + lo (ζ).

(6.165)

We multiply the equation (6.162) by pζ1 , integrate by parts, and obtain, by the Green formula (6.25) and the formula (6.165), Z T 0 = ζtt + γ(0, −∆wtt ) + Aζ, pζ1 2 dt L (Ω,Λ)×L2 (Ω) 0 Z Z = [hWtt , pW i + B(ζ, pζ1 )] dQ − ∂(Aζ, pζ1 ) dΣ Q Σ Z Z = p[B(ζ, ζ1 ) − |Wt |2 ] dQ − ∂(Aζ, pζ1 ) dΣ + L(ζ) Q


Σ

314


which is the formula (6.163). Similarly, we multiply the equation (6.162) by pζ2 and obtain the identity (6.164). Theorem 6.8 Let ζ = (W, w) solve the problem (6.162). Then Z

[|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νidΣ Z ∂wtt +2 [∂(Aζ, m(ζ)) + γV (w) ]dΣ ∂ν Σ Z T T = 2Z|0 + 2 ℘(V, ζ) dt + L(ζ) 0 Z +2 ϑ[|ζt |2 + γa(ρ(ζ), ρ(ζ)) − a(Υ(ζ), Υ(ζ))] dQ Σ

(6.166)

Q

where Z = (ζt , m(ζ))L2 (Ω,Λ)×L2 (Ω) + γ(Dwt , D(V (w)))L2 (Ω,Λ) .

(6.167)

Proof. We multiply the equation (6.162) by 2m(ζ) = 2(DV W, V (w)) and integrate by parts. We have T hζtt , 2m(ζ)i dQ = (ζt , 2m(ζ))L2 (Ω,Λ)×L2 (Ω) 0 Q Z Z +2 ϑ|ζt |2 dQ − |ζt |2 hV, νi dΣ;

Z

Q

(6.168)

Σ

Z ∂wtt h(0, ∆wtt ), 2m(ζ)i dQ − 2 V (w) dΣ = −2 hDwtt , D(V (w))i dQ ∂ν Q Σ Q Z = −2(Dwt , D(V (w))L2 (Ω,Λ) |T0 + 2 hDwt , D(V (wt ))i dQ Q Z = −2(Dwt , D(V (w))L2 (Ω,Λ) |T0 + |Dwt |2 hV, νi dΣ (6.169) Z

Z

Σ

where the following formula is used: 2hDwt , D(V (wt ))i = 2DV (Dwt , Dwt ) + div (|Dwt |2 V ) − |Dwt |2 div V = div (|Dwt |2 V ). By using the formulas (6.168), (6.169), and the Green formula (6.25) in the equation (6.162), we obtain the identity (6.166).



6.4

315

Observability Estimates from Boundary

We consider boundary controllability for the Koiter model. For that purpose, we introduce some assumptions. Let Γ1 ⊂ Γ and Γ1 6= ∅. By the ellipticity of the Koiter shell, Theorem 6.1, there is λ0 > 0 such that λ0 B(ζ, ζ) ≥ kζk2H 1

2 Γ1 (Ω,Λ)×HΓ1 (Ω)

(6.170)

for all ζ ∈ HΓ11 (Ω, Λ) × HΓ21 (Ω). Definition 6.2 Let a vector field V on Ω be such that the relation (6.151) holds. V is said to be an escape vector field for the Koiter shell if the following inequality holds: min ϑ(x) > max υ(x) (6.171) x∈Ω

x∈Ω

where υ(x) = 1 + [2 + λ0 (1 + µ)(1 + 2γ|Π|2 )]|ι(x)| + 16λ0 (1 + µ)γ|DV Π|2 , and the functions ϑ(x) and ι(x) are given in the formulas (6.151) and (5.99), respectively. Remark 6.4 By Theorem 4.8, there always exists a vector field V satisfying the assumption (6.151). Then the key to being an escape vector field for the Koiter shell is the assumption (6.171). For M being of constant curvature or of revolution, then there exists a vector field V on the whole M with ι(x) = 0 for all x ∈ M, see Theorems 4.9 and 4.10 in Chapter 4. In those cases the condition (6.171) becomes min ϑ(x) > 1 + 16λ0 (1 + µ)γ max |DV Π|2 . x∈Ω

x∈Ω

Let us see several examples. Example 6.1 (Circular Cylindrical Shells) Consider a circular cylindroid to be defined by M = { (x1 , x2 , x3 ) | x22 + x23 = 1 }.

Then M is of zero curvature with the induced metric g in R3 . Given x0 = (x01 , x02 , x03 ) ∈ M. Let A(x0 ) = { (x1 , −x02 , −x03 ) | x1 ∈ (−∞, ∞) }. Let ρ(x) = d(x0 , x) be the distance function on M from x0 to x ∈ M. Let V = 2ρDρ for x ∈ M/A(x0 ). (6.172) It is easy to check that DX V = 2X


∀ X ∈ Mx , x ∈ M/A(x0 );

(6.173)

316

Modeling and Control in Vibrational and Structural Dynamics ϑ(x) = 2,

ι(x) = 0,

DV Π = 0

M/A(x0 ).

on

(6.174)

From the relations (6.173) and (6.174) we have reached the following conclusion: For any Ω ⊂ M/A(x0 ), V, given by (6.172), is an escape vector field for the Koiter shell. Example 6.2 (Spherical Shells) Let M be the sphere of the constant curvature α = c2 . Then Π = cg, DΠ = cDg = 0 where g is the induced metric of M in R3 . Given x0 ∈ M. Let ρ(x) = d(x0 , x) be the distance function on M from x0 to x ∈ M. Set h(x) = − cos(cρ(x)) and define H = Dh = c sin cρDρ

x ∈ M/{ −x0 }.

for

We obtain DH = c2 cos cρ(x) g

for

x ∈ M/{ −x0 }.

We then arrive at the following conclusion: Let Ω ⊂ { x | x ∈ M, d(x0 , x)
0},

where α(t, s) = (t cos s, t sin s, c0 s),

c0 > 0.

Set E1 =

∂ = (cos s, sin s, 0), ∂t

1 1 ∂ E2 = p = p (−t sin s, t cos s, c0 ), 2 2 2 t + c0 ∂s t + c20

1 N=p (c0 sin s, −c0 cos s, t). 2 t + c20

Then E1 , E2 makes up a frame field and N is unit normal vector on the whole surface M.



317

Let Ω ⊂ M be a bounded open set such that Ω ⊂ M. Then c1 =

inf

α(t,s)∈Ω

We take

t p > 0, 2 t + c20

c2 =

t|s| p < ∞. t2 + c20 α(t,s)∈Ω sup

V = f1 E1 + f2 E2

(6.175)

where f1 =

Z q t2 + c20 (

t

0

We obtain

dt 1 + 2c2 + λ0 (1 + µ)c2 p ), + c1 t2 + c20

f2 =

q t2 + c20 s.

DV = ϑg + ιE, ϑ=1+

t f1 , t2 + c20

It is easy to check that

st ι = −p . 2 t + c20

min ϑ(x) ≥ 2 + 2c2 + λ0 (1 + µ)c2 ; x∈Ω

υ(x) ≤ 1 + 2c2 + λ0 (1 + µ)c2 + c3 γ|Π|2 where c3 = 2λ0 (1 + µ)

sup [c2 + α(t,s)∈Ω

32t2 2 t + c20

s2 +

1 2 f ]. t2 + c20 1

To make the error of the thin shell small, h/R has to be small. If h/R is so small that c3 h c3 γ|Π|2 ≤ ( )2 < 1, (6.176) 6 R we then have min ϑ(x) > max υ(x), x∈Ω

x∈Ω

that is, under the assumption (6.176), V, given by (6.175), is an escape vector field for the Koiter shell. Lemma 6.13 Let V be an escape vector field for the Koiter shell. Then for ζ = (W, w) ∈ H 1 (Ω, Λ) × H 2 (Ω) Z ℘(V, ζ) ≥ ϑ(x)[a(Υ(ζ), Υ(ζ)) − γa(ρ(ζ), ρ(ζ1 ))]dx Ω Z maxx∈Ω υ(x) − B(ζ, ζ) dx + lo (ζ) (6.177) 2 Ω where ℘(V, ζ) is defined by the formula (6.158) and ζ1 = (W, 0).


318


Proof. Note that a(Υ(ζ1 ), Υ(ζ1 )) = a(Υ(ζ), Υ(ζ))+ lo (ζ) and the inequality (4.235) in Chapter 4 yields a(Υ(ζ), G(V, DW )) ≥ ϑ(x)a(Υ(ζ), Υ(ζ)) 1+µ |ι(x)||DW |2 + lo (ζ). − 2

(6.178)

Let us estimate a(ρ(ζ), F (V, W )). Set æ(·, ·) = Π(D· V, D· W ) + Π(D· W, D· V );

(6.179)

Im (·, ·) = DV Π(·, D· W ) + DV Π(D· W, ·).

(6.180)

Let x ∈ Ω be given. Since Π is symmetric, we may take e1 , e2 to be an orthonormal basis of Mx , with the positive orientation such that Π(ei , ei ) = λi ,

i = 1, 2,

Π(e1 , e2 ) = Π(e2 , e1 ) = 0.

(6.181)

Then the formula (4.176) of Lemma 4.9 in Chapter 4 implies De1 V = ϑ(x)e1 − ι(x)e2

and De2 V = ι(x)e1 + ϑ(x)e2 .

(6.182)

Denote Wij = DW (ei , ej ) for 1 ≤ i, j ≤ 2. Then use of the relations (6.181) and (6.182) gives æ(e1 , e1 ) = Π(De1 V, De1 W ) + Π(De1 W, De1 V ) = ϑ(x)[Π(e1 , De1 W ) + Π(De1 W, e1 )] − 2λ2 ι(x)W21

= −ϑ(x)ρ(ζ1 )(e1 , e1 ) − 2λ2 ι(x)W21 + lo (ζ); æ(e1 , e2 ) = −ϑ(x)ρ(ζ1 )(e1 , e2 ) + ι(x)[λ1 W11 − λ2 W22 ] + lo (ζ); æ(e2 , e2 ) = −ϑ(x)ρ(ζ1 )(e2 , e2 ) + 2λ1 ι(x)W12 + lo (ζ).

(6.183) (6.184) (6.185)

From the formulas (6.183)-(6.185), one obtains a(ρ(ζ), æ) = −ϑ(x)a(ρ(ζ), ρ(ζ1 )) + 2ι(x){(1 − µ)[−λ2 W21 ρ(ζ)(e1 , e1 ) +λ1 W12 ρ(e2 , e2 ) + (λ1 W11 − λ2 W22 )ρ(ζ)(e2 , e1 )] +µ tr ρ(ζ)[λ1 W12 − λ2 W21 ]} + lo (ζ).

(6.186)

We apply Schwartz’s inequality to the second term in the right hand side of the inequality (6.186) to give a(ρ(ζ), æ) ≥ −ϑ(x)a(ρ(ζ), ρ(ζ1 )) − |ι(x)|a(ρ(ζ), ρ(ζ)) −|ι(x)||Π|2 |DW |2 + lo (ζ).

(6.187)

On the other hand, we apply Schwartz’s inequality again to obtain 2a(ρ(ζ), Im ) ≥ −a(ρ(ζ), ρ(ζ)) − a( Im , Im ) ≥ −a(ρ(ζ), ρ(ζ)) − 16(1 + µ)|DV Π|2 |DW |2 .


(6.188)


319

Finally, we combine the inequalities (6.178), (6.187), and (6.188) to get the inequality (6.177). Let ζ = (W, w) solve the problem (6.162). Define the total energy of the shell by E(t) =

1 [kWt k2L2 (Ω,Λ) + kwt k2 + γkDwt k2L2 (Ω,Λ) + B(ζ, ζ)], 2

(6.189)

σ1 = min ϑ(x) − max υ(x),

(6.190)

and we set x∈Ω

Q = Ω × (0, T ),

x∈Ω

Σ = Γ × (0, T ).

We say that L(ζ) are lower order terms if, any ε > 0, there is Cε > 0 such that Z T L(ζ) ≤ ε E(τ )dτ + Cε {kW (0)k2L2(Ω,Λ) + kw(0)k2 + γkwt (0)k2 0

+kDw(0)k2L2 (Ω,Λ) + kW (T )k2L2 (Ω,Λ) + kw(T )k2

+γkwt (T )k2 + kDw(T )k2L2 (Ω,Λ) Z T + [kW k2L2 (Ω,Λ) + kwk2 + γkwt k2 + kDwk2L2 (Ω,Λ) ]dt}.

(6.191)

0

First, we have the following observability estimates. Theorem 6.9 Let V be an escape vector field for the Koiter shell and let ζ = (W, w) solve the problem (6.162). Given T > 0. Then there is CT > 0, independent of ζ, such that Z T Z 2σ1 E(t) dt ≤ SB dΣ + σ0 λ0 [E(0) + E(T )] + CT L(ζ) (6.192) 0

Σ

where SB = [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi + ∂(Aζ, 2m(ζ)) ∂wtt +ϑ∂(Aζ, ζ1 − ζ2 ) + γ(2V (w) − ϑw) ∂ν

(6.193)

where m(ζ) = (DV W, V (w)),

ζ1 = (W, 0),

ζ2 = (0, w).

Proof. We take p = ϑ(x) in the identity (6.163) and p = −ϑ(x) in the identity (6.164), respectively, and then add them up to Z h Z ∂wtt i ∂ Aζ, ϑ(ζ1 − ζ2 ) − γϑw dΣ = ϑB(ζ, ζ1 − ζ2 ) dQ ∂ν Σ Q Z + ϑ(wt2 + γ|Dwt |2 − |Wt |2 ) dQ + L(ζ). (6.194) Q


320


Adding the identities (6.194) and (6.166) up and using the inequality (6.177), we further have Z [|ζt |2 + γ|Dwt |2 − B(ζ, ζ)]hV, νi dΣ Σ Z ∂w tt + [∂ Aζ, 2m(ζ) + ϑ(ζ1 − ζ2 ) + γ 2V (w) − ϑw ] dΣ ∂ν ZΣ = ϑ[|ζt |2 + γ|Dwt |2 + B(ζ, ζ)] dQ Q Z Z +2 ϑwt2 dQ + 2 ϑ[γa(ρ(ζ), ρ(ζ1 )) − a(Υ(ζ), Υ(ζ))]dQ Q

Q

+2Z|T0 + 2 ≥ 2σ1

Z

0

T

Z

T

℘(V, ζ) dt + L(ζ)

0

E(t) dt + 2Z|T0 + L(ζ)

(6.195)

where the following relations have been used B(ζ, ζ2 ) = γa(ρ(ζ), ρ(ζ2 )) + lo (ζ), γa(ρ(ζ), ρ(ζ)) − B(ζ, ζ2 ) = γa(ρ(ζ), ρ(ζ1 )) + lo (ζ).

Finally, we obtain the inequality (6.192) by Lemma 4.10 for the term 2Z|T0 . Dirichlet Action We consider the boundary control problem in unknown ς = (U, u)  ςtt − γ(0, ∆utt ) + Aς = 0 in Q,     ς(0) = ς 0 , ςt (0) = ς 1 in Ω,     U |Γ = 0, U |Γ = Φ for 0 < t < T, 1 0 (6.196) ∂u  u| = 0, |  Γ1 Γ1 = 0 for 0 < t < T,   ∂ν   ∂u   u|Γ0 = φ, |Γ = ϕ for 0 < t < T ∂ν 0 where Φ, φ and ϕ are boundary controls. It’s dual version in ζ = (W, w) is given by  ζtt − γ(0, ∆wtt ) + Aζ = 0 in Q,     ζ(0) = ζ 0 , ζt (0) = ζ 1 in Ω, (6.197) W |Γ = 0 for 0 < t < T,   ∂w   w|Γ = |Γ = 0 for 0 < t < T. ∂ν We have ˆ ⊂ Γ be given and let ζi = (Wi , wi ) ∈ H 1 (Ω, Λ) × H 2 (Ω) Lemma 6.14 Let Γ be such that ∂wi ˆ Wi = 0, wi = = 0 for x ∈ Γ (6.198) ∂ν



321

ˆ for i = 1, 2. Let X ∈ X (Ω) be a vector field. Then on Γ B(ζ1 , ζ2 ) = DW1 (ν, ν)DW2 (ν, ν) + (1 − µ)DW1 (τ, ν)DW2 (τ, ν)/2 +γ{ρ(ζ1 )(ν, ν)ρ(ζ2 )(ν, ν) + 2(1 − µ)DW1 (Sτ, ν)DW2 (Sτ, ν)}; (6.199) ∂(Aζ1 , DX ζ2 ) = B(ζ1 , ζ2 )hX, νi. Proof. The proof will be complete by a careful computation. It follows from the boundary conditions (6.198) that ( Dτ Wi = Dwi = Dτ Dwi = 0, ˆ for x ∈ Γ ∂X(wi ) = hX, νiD2 wi (ν, ν) ∂ν where τ is the unit tangential along Γ for i = 1, 2. Then the formulas ˆ imply that on Γ Υ(ζi )(ν, ν) = DWi (ν, ν), Υ(ζi )(τ, τ ) = 0, Υ(ζi )(ν, τ ) = DWi (τ, ν)/2, ρ(ζi )(ν, ν) = D2 wi (ν, ν) − 2DWi (Sν, ν), ρ(ζi )(τ, τ ) = 0, ρ(ζi )(ν, τ ) = −DWi (Sτ, ν), hB1 (ζi ), νi = DWi (ν, ν), hB1 (ζi ), τ i = (1 − µ)DWi (τ, ν)/2, hB2 (ζi ), νi = ρ(ζi )(ν, ν), hB2 (ζi ), τ i = −(1 − µ)DWi (Sτ, ν),

(6.200)

(6.201) (6.201)

(6.202) (6.203) (6.204)

for i = 1, 2. Inserting the formulas (6.202) and (6.203) into the formula (6.4), we obtain the formula (6.199). On the other hand, using (6.201)-(6.204) in the formulas (6.26), we obtain ∂X(w2 ) ∂ν = hX, νi[hB1 (ζ1 ), Dν W2 i − 2γhB2 (ζ1 ), SDν W2 i + γhB2 (ζ1 ), νiD2 w2 (ν, ν)] = hX, νi{hB1 (ζ1 ), νiDW2 (ν, ν) + hB1 (ζ1 ), τ iDW2 (τ, ν) ∂(Aζ1 , DX ζ2 ) = hV1 (ζ1 ), DX W2 i + v2 (ζ1 )

+γhB2 (ζ1 ), νi[D2 w2 (ν, ν) − 2DW2 (Sν, ν)] − 2γhB2 (ζ1 ), τ iDW2 (Sτ, ν)} = hX, νi{DW1 (ν, ν) + (1 − µ)DW1 (τ, ν)DW2 (τ, ν)/2 +γ[ρ(ζ1 )(ν, ν)ρ(ζ2 )(ν, ν) + 2(1 − µ)DW1 (Sτ, ν)DW2 (Sτ, ν)]} ˆ = hX, νiB(ζ1 , ζ2 ) for x ∈ Γ.

Let ζ = (W, w) solve the problem (6.197). Then we solve the problem in η = (Φ, φ)  ηtt − γ(0, ∆φtt ) + Aη = 0 in Q,     η(T ) = ηt (T ) = 0 in Ω,   ∂φ (6.205) Φ=φ= = 0 on Σ1 ,   ∂ν   ∂φ   Φ = −Dν W, φ = −v3 (ζ), = −D2 w(ν, ν) on Σ0 ∂ν © 2011 by Taylor & Francis Group, LLC

322


where v3 (ζ) is given in (6.31). We define Λ(ζ0 , ζ1 ) = (−ηt (0) + γ(0, ∆φt (0)),

η(0) − γ(0, ∆φ(0))).

Let ζ and ζˆ solve the problem (6.197) with initial data (ζ0 , ζ1 ) and (ζˆ0 , ζˆ1 ), respectively. It follows from (6.197), (6.225), and (6.25) that ((−ηt (0), η(0)), (ζˆ0 , ζˆ1 ))[L2 (Ω,Λ)×L2 (Ω)]2 = −(ηt (0), ζˆ0 )L2 (Ω,Λ)×L2 (Ω) + (η(0), ζˆ1 )L2 (Ω,Λ)×L2 (Ω) Z T d ˆ L2 (Ω,Λ)×L2 (Ω) − (η, ζˆt )L2 (Ω,Λ)×L2 (Ω) ]dt = [(ηt , ζ) dt 0 Z T =γ [(∆φtt , w) ˆ − (φ, ∆wˆtt )]dt Z

0 T

ˆ L2 (Ω,Λ)×L2 (Ω) − (η, Aζ) ˆ L2 (Ω,Λ)×L2 (Ω) ]dt [(Aη, ζ) Z ˆ η)dΣ. = γ[(∆φ(0), w ˆ1 ) − (∆φt (0), w ˆ0 )] − ∂(Aζ,

+

0

(6.206)

Σ0

From Lemma 6.14, we have (Λ(ζ0 , ζ1 ), (ζˆ0 , ζˆ1 ))[L2 (Ω,Λ)×L2 (Ω)]2 Z ˆ + v3 (ζ)v3 (ζ)]hV, ˆ = [B(ζ, ζ) νidΣ. Σ0

Let L = [L2 (Ω, Λ) × L2 (Ω)]2 and H = [H01 (Ω, Λ) × H02 (Ω)] × [L2 (Ω, Λ) × Since −∆: H01 (Ω) → H −1 (Ω) and −∆: L2 (Ω) → H −2 (Ω) are isomorphisms, respectively, applying Theorem 2.13 yields H01 (Ω)].

Theorem 6.10 The problem (6.196) is exactly [L2 (Ω, Λ) × H01 (Ω)] × [H −1 (Ω, Λ) × L2 (Ω)] controllable by L2 (Σ0 , Λ) × L2 (Σ0 ) × L2 (Σ0 ) controls on [0, T ] if the following observability inequality is true: There is cT > 0 satisfying Z [B(ζ, ζ) + v32 (ζ)]dΣ ≥ cT E(0) (6.207) Σ0

for all solutions ζ to the problem (6.197). We have taken one more control φ = v1 (ζ) in the problem (6.225) in order to have Lemma 6.15 Let ζ = (W, w) solve the problem (6.197). If B(ζ, ζ) = v3 (ζ) = 0

for

x ∈ Γ0 ,

(6.208)

then W = DW = 0,

w=

∂w ∂∆w = ∆w = =0 ∂ν ∂ν


for

x ∈ Γ0 .

(6.209)


323

Proof. It follows from (6.199) and (6.201) that DW = 0,

ρ(ζ)(ν, ν) = 0

for x ∈ Γ0 ,

which yields, via (6.203), ρ(ζ) = 0,

∆w = D2 w(ν, ν) = tr ρ(ζ) = 0

for x ∈ Γ0 .

In particular, D2 w = 0 for x ∈ Γ0 . ∂∆w It remains to prove = 0 for x ∈ Γ0 . Let x ∈ Γ0 be given. Let E1 , E2 ∂ν be a frame field on M normal at x such that E1 (x) = τ,

E2 (x) = ν,

and DEi Ej (x) = 0,

[Ei , Ej ](x) = 0

for all i, j. Then DEj DEi W = DEi DEj W + R Ei Ej W

at x

for all i, j, where REi Ej is the curvature operator. By W (x) = Dw(x) = 0, we have Π(DEj DEi W, Ek ) = hDEj DEi W, SEk i

= Π(DEi DEj W, Ek ) + R(Ei , Ej , W, SEk ) = Π(DEi DEj W, Ek ) at x and, by (1.22), D3 w(Ei , ν, Ei ) = D3 w(Ei , Ei , ν) + R(ν, Ei , Dw, Ei ) = D3 w(Ei , Ei , ν) at x, for all i, j, k. We obtain, at x, Dρ(ζ)(ν, Ei , Ei ) = Dρ(Ei , ν, Ei ) = Ei (ρ(ζ)(Ei , E2 )) = D3 w(Ei , ν, Ei ) − Π(DEi DE2 W, Ei ) − Π(Ei , DEi DE2 W )

= Dρ(ζ)(Ei , Ei , ν) at x, for i = 1, 2, which give tr i (ν)Dρ = Since

∂ tr ρ(ζ) ∂ν

∂ρ(ζ) =0 ∂τ


for x ∈ Γ0 .

for x ∈ Γ0 ,

(6.210)

324


we obtain, via (6.30), (6.31), (6.18), and (6.210), 0 = v3 (ζ) = −γhB3 (ζ), νi = −γ

∂ tr ρ ∂∆w = −γ ∂ν ∂ν

for

x ∈ Γ0 .

Now, let us consider the inequality (6.207). We make the following assumption for a uniqueness result: Assumption (H) Let one of the following conditions hold inf (|Π|2 − 2κ) > 0,

or |Π|2 = 2κ

x∈Ω

for all x ∈ Ω

(6.211)

where κ is the Gauss curvature of the surface M. See Remark 6.3. We have Theorem 6.11 Let V be an escape vector field for the Koiter shell such that the inequality (6.171) holds. Moreover, we assume that the assumption (H) is true. Then for any T > T0 , there exists cT > 0 such that the inequality (6.207) holds where T0 = λ0 σ0 /σ1 ,

Γ0 = { x | x ∈ Γ, hV (x), ν(x)i > 0 }

(6.212)

where σ1 is given by (6.190) and σ0 = max |V (x)|. x∈Ω

(6.213)

Proof. It follows from the boundary conditions in (6.197), Lemma 6.14, and the formula (6.193) that SB = B(ζ, ζ)hV, νi for

(t, x) ∈ Σ.

(6.214)

It follows from the estimate (6.192) and the formula (6.214) that for any ε > 0 small and for T > 0 Z 2(σ1 T − σ0 λ0 )E(0) ≤ B(ζ, ζ) dΣ + L(ζ) Σ0

where ζ solves the problem (6.197). Then the inequality (6.207) holds for T > T0 which differs by a lower order term. To get the inequality (6.207) by absorbing the lower order terms, as in Lemma 2.5, we need the following uniqueness: If η = (U, u) solves the problem λ2 (U, u − ∆u) + Aη = 0 where λ is a complex number and η satisfies the Dirichlet boundary conditions on Γ0 such that B(η, η) + v32 (η) = 0, (6.215)



325

then η=0

on Ω.

(6.216)

However, by Lemma 6.15, the above result is given by Theorem 6.5.

Neumann Control Let Γ1 6= ∅ and Γ0 ∩ Γ1 = ∅. We turn to the control problem in ς = (U, u) ςtt − γ(0, ∆u) + Aς = 0 in Q, (6.217) ς(0) = ς 0 , ςt (0) = ς 1 on Ω, where we act on Σ1 = (0, T ) × Γ1 by U = 0,

u=

∂u =0 ∂ν

(6.218)

and on Σ0 = (0, T ) × Γ0 by V1 (ς) = Φ,

v2 (ς) = φ,

γ

∂utt + v3 (ς) = ϕ, ∂ν

(6.219)

where Φ, φ and ϕ are control functions. The dual problem for the above is the following in ζ = (W, w) ζtt − γ(0, ∆wtt ) + Aζ = 0 in Q, (6.220) ζ(0) = ζ 0 , ζt (0) = ζ 1 on Ω, subject to the boundary condition

V1 (ζ) = 0,

W = 0,

w=

v2 (ζ) = 0,

γ

∂w = 0 on Σ1 , ∂ν

∂wtt + v3 (ζ) = 0 ∂ν

on Σ0 .

(6.221) (6.222)

We consider the spaces L0 = L2 (Σ0 , Λ) × L2 (Σ0 ) × L2 (Σ0 ), H 0 = H 1 ([0, T ], L2 (Γ0 , Λ) × L2 (Γ0 ) × H 1 (Γ0 )).

∗

Denote by H 0 the conjugate space of H 0 with respect to the space L0 . Let ∗ ℵ: H 0 → H 0 be the canonical map. Let ζ = (W, w) solve the problem (6.220)-(6.222) such that Z ∂wt 2 ∂wt 2 [|Wt |2 + γ( ) + |wt |2 + γ( ) ]dΣ < ∞. (6.223) ∂ν ∂τ Σ0 We define Γζ = (W,


∂w , w) ∂ν

for x ∈ Γ.

326

Modeling and Control in Vibrational and Structural Dynamics ∗

By Theorem 2.12, the condition (6.223) means that ℵΓζ ∈ H 0 satisfies (ℵΓζ, ̟)L0 = (Γζ, ̟)H 0 Z ∂wt ∂wt bt = (hWt , At i + γ at + γ )dΣ ∂ν ∂τ ∂τ Σ0

(6.224)

for all ̟ = (A, a, b) ∈ H 0 . Then we solve the problem in η = (Φ, φ)  ηtt − γ(0, ∆φtt ) + Aη = 0 in Q,     η(T ) = ηt (T ) = 0 in Ω,   ∂φ Φ=φ= = 0 on Σ1 ,   ∂ν   φ   (V1 (η), v2 (η), γ tt + v3 (η)) = ℵΓζ. on Σ0 . ∂ν

(6.225)

We define Λ by

Λ(ζ0 , ζ1 ) = (−ηt (0), η(0)).

(6.226)

This time, we use the space

L = L2 (Ω, Λ) × HΓ11 (Ω), and consider an equivalent inner product on L, given by

((W, w), (U, u))L =

Z

Ω

(hW, U i + wu + γhDw, Dui)dx.

Let ζ and ζˆ solve the problem (6.220)-(6.222) with initial data (ζ0 , ζ1 ) and (ζˆ0 , ζˆ1 ) such that (6.223) are true, respectively. Using (6.220)-(6.222),(6.225),



327

(6.25), and (6.224), we obtain (Λ(ζ0 , ζ1 ), (ζˆ0 , ζˆ1 ))L×L = −(ηt (0), ζˆ0 )L + (η(0), ζˆ1 )L Z T ˆ L − (η, ζˆtt )L ]dt = [(ηtt , ζ) 0 Z ˆ i + φtt w = [hΦtt , W ˆ + γhDφtt , DwidQ ˆ Q Z ˆ tt + φw − [hΦ, W ˆtt + γhDφ, Dwˆtt i]dQ Q Z ˆ − hη, ζˆtt − γ(0, ∆w = [hηtt − γ(0, ∆φtt ), ζi ˆtt )i]dQ Q Z φtt ∂w ˆtt +γ [w ˆ −φ ]dΣ ∂ν ∂ν Z Σ φtt ∂w ˆtt ˆ − ∂(Aζ, ˆ η) + γ[w = {∂(Aη, ζ) ˆ −φ ]}dΣ ∂ν ∂ν Σ Z ˆ ∂φtt ˆtt ˆ + γ ∂w ˆ i + v2 (η) ∂ w = {[hV1 (η), W + (v3 (η) + γ )w] ˆ − [v3 (ζ) ]φ}dΣ ∂ν ∂ν ∂ν Σ0 ˆ L2 (Σ ,Λ)×L2 (Σ )×L2 (Σ ) = (ℵΓ(ζ), Γ(ζ)) 0 0 0 Z ˆ = (hWt , Wt i + wt w ˆt + γhDwt , Dw ˆt i)dΣ. (6.227) Σ0

Let H = [HΓ11 (Ω) × HΓ21 (Ω)] × L2 (Ω, Λ) × HΓ11 (Ω).

Denote by H ∗ the conjugate space of H with respect to L. Then applying Theorem 2.13 gives ∗

Theorem 6.12 The problem (6.217)−(6.219) is exact H ∗ controllable by H 0 controls on [0, T ] if the following inequality is true: There is cT > 0 such that Z [|Wt |2 + |wt |2 + γ|Dwt |2 ]dΣ ≥ cT E(0) (6.228) Σ0

for all solutions ζ to the problem (6.220) − (6.222) for which the left hand side of (6.228) is finite. Furthermore, we have Theorem 6.13 Let V be an escape vector field for the Koiter shell and let the assumption (H) in (6.211) hold. Then, for any T > T0 , there is cT > 0 such that the inequality (6.228) holds where T0 and Γ0 are given in (6.212). Proof. It follows from the boundary conditions in (6.221)-(6.222) and Lemma 6.14 that 2 |ζt | + γ|Dwt |2 − B(ζ, ζ)] on Σ0 , SB = hV, νi (6.229) B(ζ, ζ)hV, νi on Σ1 . Applying Theorem 6.9, we obtain (6.228).


328


Exercises 6.1 Let M be a surface of R3 with the induced metric g from R3 and let Π be the second fundamental form of M. If |Π|2 = 2κ for all x ∈ M, then there is a function ϕ such that Π = ϕg,

SW = ϕW

for

x ∈ M, W ∈ X (M )

where the operator S is defined by the formula (6.17). 6.2 Prove Lemma 6.10. 6.3 Let ζ = (W, w) be a smooth solution to the problem (6.84)-(6.85). Prove that the formula (6.143) holds.

6.5


Sections 6.1, 6.3, and 6.4 are from [37]; Section 6.2 is from [26]. The motion equations for the Koiter shell in Theorem 6.3 were given by [37] where the kinetic energy of the shell was based on to the classical KirchhoffLove Assumptions. However, under Koiter’s Hypotheses [95] the kinetic energy of the displacement vector field for the Koiter shell is very complicated, see the formula (6.67), which leads to different motion equations in Theorem 6.4. The equations in Theorem 6.4 are new and first published here. The ellipticity of the strain energy for the Koiter shell was proved by [13] where the authors used the classical differential geometry where the middle surface of the shell was given by one coordinate path. Here we present a proof of the ellipticity (Theorem 6.1) that is very simple due to the Bochner technique. In the excellent work [40] the Bochner technique again plays an important role in establishing the corresponding Carleman estimates to obtain the uniqueness for the Koiter shell.


Chapter 7 Control of the Quasilinear Wave Equation in Higher Dimensions

7.1 7.2 7.3 7.4 7.5

Boundary Traces and Energy Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Locally and Globally Boundary Exact Controllability . . . . . . . . . . . . . . . . . . . Boundary Feedback Stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Structure of Control Regions for Internal Feedbacks . . . . . . . . . . . . . . . . . . . . Notes and References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

329 342 355 372 385

This chapter introduces the latest advances on control of the quasilinear wave equation in higher dimensions by combining the Riemannian geometrical approach with some knowledge from nonlinear partial differential equations where the quasilinearity arises in the principal part of the systems. Section 7.1 presents some necessary estimates from the nonlinear partial differential equations. In Section 7.2 we prove the locally exact controllability around the equilibrium under existence of an escape vector field for a metric. We then establish the globally exact controllability in such a way that the state of the quasilinear wave equation moves from an equilibrium in one location to an equilibrium in another location under some geometrical conditions. The Dirichlet action and the Neumann action are studied, respectively. Some examples are presented to verify the globally exact controllability. In Section 7.3 we consider the stabilization of the quasilinear wave equation with a structure of an input-output in the boundary when initial data and boundary inputs are near a given equilibrium of the system. We show that the stabilization of solutions depends not only on this dissipation structure but also on existence of an escape vector field for the Riemannian metric, given by the coefficients and the equilibrium of the system. In particular, we prove that the norm of the state of the system decays exponentially if the input stops after a finite time and if there exists an escape vector field in the metric, which implies the exponential stabilization of the system by boundary feedback. In Section 7.4 we study the stabilization of smooth solutions for the quasilinear wave equation by an internal local damping when initial data are close to a given equilibrium. We show that an escape region for the metric can guarantee the global solutions and their exponential stabilization. These escape regions depend not only on the sectional curvature of a Riemannian metric but also on the equilibrium of the system. 329 © 2011 by Taylor & Francis Group, LLC

330

7.1


Boundary Traces and Energy Estimates

We shall treat boundary traces and energy estimates of the quasilinear wave equation in three types of boundary data: Dirichlet data, Neumann data, and Robin data, respectively. Let Ω ⊂ Rn be an open, bounded set with the smooth boundary Γ. Suppose that Γ consists of two parts, Γ0 and Γ1 . If Γ0 6= ∅, we further assume that Γ1 ∩ Γ0 = ∅. Let A(x, y) = (aij (x, y)) be symmetric, positive on Ω × Rn where aij are smooth functions. Let b be a smooth function on Ω × Rn × R. Definition 7.1 w ∈ H 2 (Ω) is said to be an equilibrium if n X

ij=1

aij (x, ∇w)wxi xj + b(x, ∇w, 0) = 0

for

x ∈ Ω.

(7.1)

Let w be an equilibrium and let T > 0 be given. We consider the problem Pn utt = ij=1 aij (x, ∇u)uxi xj + b(x, ∇u, ut ) for (t, x) ∈ Q, (7.2) u = w for (t, x) ∈ Σ1 , where Q = (0, T ) × Ω,

Σ1 = Σ1 .

In this section we establish some energy and boundary trace estimates of the problem (7.2) in preparation for controllability/stabilization. The target space is H m (Ω) × H m−1 (Ω) where we need the integer m to satisfy m ≥ [n/2] + 3.

(7.3)

Remark 7.1 We assume the condition (7.3) in order to have the existence of small time solutions, for example, by [91] or [48], if initial data on Ω and boundary value on the portion Γ0 are specified, respectively. The target space will be larger if m is smaller. In the linear case, the number m can be any real number in (−∞, ∞), see [117], [192], and [193]. In [114], the authors were able to establish global solutions in H m (Ω) × H m−1 (Ω) for m ≥ 1 where m may be much smaller than in the condition (7.3) because they had a lower regularity for their model. Here we have to assume that the condition (7.3) holds since we are short of the corresponding local regularities when m < [n/2] + 3. In general, solutions of the problem (7.2) may blow up in a finite time even if the initial data and the boundary value are smooth. We here assume that the system (7.2) has small time solutions and study it’s energy estimates and boundary traces. k Let u ∈ ∩m [0, T ], H m−k (Ω) be a solution of the problem (7.2) for k=0 C


Control of the Quasilinear Wave Equation in Higher Dimensions

331

some T > 0. Let w ∈ H m (Ω) be an equilibrium of the problem (7.2). We can make a transform by u = w + φ and consider the φ-problem P φtt = ij aij (x, ∇w + ∇φ)φxi xj + b0 (x, ∇φ, φt ) for (t, x) ∈ Q, (7.4) φ = 0 for (t, x) ∈ Σ1 where b0 (x, y, s) = b(x, ∇w + y, s) for (x, y, s) ∈ Ω × Rn × R. We introduce a linear operator A(t) by A(t)v =

n X

ij=1

for v ∈ H 2 (Ω).

aij (x, ∇w + ∇φ)vxi xj

(7.5)

Then (−A(t)v1 , v2 ) = −

Z

Γ

v2 hA∇v1 , νidΓ + (A∇v1 , ∇v2 ) + (Cv1 , v2 )

(7.6)

for v1 , v2 ∈ H 1 (Ω) where A = (aij (x, ∇w + ∇φ)) , ν is the outside normal of Γ in the Euclidean metric, and Cv1 =

n X

ij=1

(aij (x, ∇w + ∇φ))xj v1xi .

Then the problem (7.4) can be rewritten as φtt = A(t)φ(t) + b0 (x, ∇φ, φt ), φ = 0 for (t, x) ∈ Σ1 .

for (t, x) ∈ Q,

(7.7)

Boundary Traces and EnergyEstimates with Dirichlet Data on k Γ0 Let φ ∈ ∩m [0, T ], H m−k (Ω) be a solution of the problem (7.7) for k=0 C some T > 0. We introduce E(t) =

m X

k=0

Q(t) =

kφ(k) (t)k2m−k ,

EΓ0 ,D (t) =

m−1 X k=0

kφ(k) (t)k2m−k−1/2,Γ0 ,

m X kφ(k) (t)k2 + k∇φ(k−1) (t)k2 ,

k=1

QΓ0 ,D (t) =

m X k=1

L(t) =

m X

k=2

E k (t),

kφ(k) (t)k2Γ0 + kφ(k−1) (t)k21,Γ0 ,

(7.8)

(7.9)

(7.10)

where k · k, k · kj , k · kΓ0 , and k · kj,Γ0 are norms of L2 (Ω), H j (Ω), L2 (Γ0 ), and H j (Γ0 ), respectively, for 1 ≤ j ≤ m. We shall establish


332


Theorem 7.1 Let γ > 0 be given. Let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that sup kφ(t)km ≤ γ

˙ sup kφ(t)k m−1 ≤ γ.

and

0≤t≤T

(7.11)

0≤t≤T

Then there is cγ > 0, which depends on the number γ but is independent of solution φ, such that Q(t) ≤ E(t) ≤ cγ Q(t) + cγ EΓ0 ,D (t) + cγ L(t)

for

0≤t≤T

(7.12)

and Q(t) ≤ cγ Q(0) + cγ

Z

0

t

[(1 + E 1/2 (t))Q(t) + QΓ0 ,D (t) + L(t)]dt

(7.13)

for t ∈ [0, T ]. Remark 7.2 We observe that the conditions (7.3) and (7.11) imply that φt , ∂xi φ ∈ C 1 (Ω)

for

1≤i≤n

(7.14)

which is also one of the reasons why we need the condition (7.3). The proof of Theorem 7.1 will be given after Lemma 7.5. First, we collect here a few basic properties of Sobolev spaces in the following lemma which play an important role in our estimates in the sequel. Let k · ks be the norm of H s (Ω) for s ≥ 0. In particular, denote by k · k = k · k0 the norm of L2 (Ω). Lemma 7.1 Let Ω ⊂ Rn be a bounded, open set. Then (i) Let s1 > s2 be given. For any ε > 0 there is cε > 0 such that kwk2s2 ≤ εkwk2s1 + cε kwk2

for

w ∈ H s1 (Ω).

(7.15)

(ii) Let si ≥ 0 be given for i = 1, 2. If 0 ≤ r ≤ min{s1 , s2 , s1 + s2 − [n/2] − 1}, then there is a constant c > 0 such that kf gkr ≤ ckf ks1 kgks2

for

f ∈ H s1 (Ω), g ∈ H s2 (Ω).

(7.16)

(iii) Let sj ≥ 0 be given for j = 1, · · · , k. Let 0 ≤ r ≤ min

min {sj1 + · · · + sji − (i − 1)([n/2] + 1)}.

1≤i≤k j1 ≤···≤ji

Then there is a constant c > 0 such that kf1 · · · fk kr ≤ ckf1 ks1 · · · kfk ksk

for

fj ∈ H sj (Ω), 1 ≤ j ≤ k.

(7.17)

For the proofs of (ii) and (iii) in the above lemma, for example, see books [1], or [195].



333

Lemma 7.2 (i) Let f (x, y, s) be a smooth function on Ω×Rn ×R. Set F (x) = f (x, ∇φ, φt ). For 0 ≤ k ≤ m − 1, there is c = c(supx∈Ω (|∇φ|, |φt |)) > 0 such that k X kF kk ≤ c (1 + kφkm + kφt km−1 )j . (7.18) j=0

(ii) Let φ be a solution of the problem (7.7) and let γ > 0 be given. Suppose that the condition (7.11) holds true. Then there is cγ > 0, which depends on the γ, such that kvk2k+1 ≤ cγ kA(t)vk2k−1 + kvk2k+1/2, Γ0 + kvk2k (7.19) for v ∈ H k (Ω) ∩ HΓ11 (Ω) and 0 ≤ k ≤ m − 1.

Proof. (i) By induction. The inequality (7.18) is clearly true for k = 0. Suppose that it holds for 0 ≤ k < m − 1. Since Fxi = fxi (x, ∇φ, φt ) +

n X j=1

fyj (x, ∇φ, φt )φxi xj + fs (x, ∇φ, φt )φtxi

for 1 ≤ i ≤ n, by using the inequality (7.16) and the induction assumption for fxi (x, ∇φ, φt ) and for fyj (x, ∇φ, φt ), respectively, we obtain kF kk+1 = (kF k2 + ≤c+c

n X i=1

n X i=1

kFxi k2k )1/2

kfxi (·, ∇φ, φt )kk + c

+ckfs (·, ∇φ, φt )kk kφtxi km−2 ≤c+c ≤c

k+1 X j=0

n X

ij=1

kfyj (·, ∇φ, φt )kk kφxi xj km−2

k X ˙ m−1 )j (1 + kφkm + kφk j=0

˙ m−1 )j . (1 + kφkm + kφk

(ii) A standard method as to the linearly elliptic problem can give the inequality (7.19); for example see [68]. Lemma 7.3 Let b be a smooth function on Ω × Rn × R. Let γ > 0 be given and let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that the condition (7.11) holds true. Then there is cγ > 0, which depends on the number γ, such that kb(k) (x, ∇φ, φt )k2m−k−2 ≤ cγ kφ(k) (t)k2m−k−1

+cγ kφ(k+1) (t)k2m−k−2 + cγ pk


(7.20)

334


where p1 = 0;

pk =

k X i=2

for 1 ≤ k ≤ m − 2 and

E i (t),

kA(j) (t)φ(k−j) (t)k2m−k−2 ≤ cγ

k≥2

j X i=1

E 1+i (t)

(7.21)

for 1 ≤ j ≤ k ≤ m − 2 where the operator A(t) is defined by (7.5). ˆ = (D, ∂) the covariant differential on the space Proof. Denote by D Rn × R with the Euclidean metric. Then it is easy to check that b(k) (x, ∇φ, φt ) =

k X

X

i=1 r1 +···+ri =k

ˆ i b((∇φ(r1 ) , φ(r1 ) ), · · · , (∇φ(ri ) , φ(ri ) )) D t t

ˆ i b0 denotes the covariant differential of the function b0 (x, y, s) of order where D i with respect to the variable (y, s) in the Euclidean metric of Rn × R. (k) (k) ˆ Let us estimate the term Db((∇φ , φt )) first. We have (k) (k) ˆ Db((∇φ , φt )) =

n X l=1

(k)

byl (x, ∇φ, φt )φ(k) xl (t) + bs (x, ∇φ, φt )φt .

It follows from the inequalities (7.16), (7.18) and (7.11) that X (k) (k) ˆ kDb((∇φ , φt ))km−k−2 ≤ c kbyl km−1 kφ(k) (t)km−k−1 l

(k+1)

+kbs km−1 kφ

km−k−2 ≤ cγ (kφ(k) (t)km−k−1 + kφ(k+1) (t)km−k−2 ). (7.22)

Denote x0 = t. For 2 ≤ i ≤ k, we observe that ˆ i b((∇φ(r1 ) , φ(r1 ) ), · · · , (∇φ(ri ) , φ(ri ) )) D t t are sums of terms such as (ri ) 1) f (x, ∇φ, φt )φ(r xl1 (t) · · · φxl (t) i

where r1 + · · · + ri = k for 0 ≤ lj ≤ n and 1 ≤ j ≤ i. Using the inequalities (7.17) and (7.18), we have (ri ) (r1 ) 1) kf φ(r (t)km−r1 · · · kφ(ri ) (t)km−ri xl · · · φxl km−k−2 ≤ cγ kφ 1

i

≤ cγ E i/2 (t).

(7.23)

The inequality (7.20) follows from the inequalities (7.22) and (7.23).



335

Moreover, by the inequalities (7.16) and (7.20), we obtain (k−j) ka(j) pq (·, ∇w + ∇φ)φxp xq km−k−2 (k−j) ≤ cka(j) pq (·, ∇w + ∇φ)km−j−2 kφxp xq km+j−k−2 ≤ cγ

j X i=1

E (i+1)/2 (t)

for 1 ≤ p, q ≤ n which implies that the inequality (7.21) is true.

Lemma 7.4 Let γ > 0 be given. Let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that the condition (7.11) holds true. Then there is cγ > 0 such that the inequality (7.12) is true. Proof. It is clear that kφ(m) (t)k2 + kφ(m−1) (t)k21 = kφ(m) (t)k2 + k∇φ(m−1) (t)k2 +kφ(m−1) (t)k2 ≤ Q(t).

(7.24)

Proceeding by induction, we assume that for some 1 ≤ j ≤ m − 1 kφ(j) (t)k2m−j ≤ cγ Q(t) + cγ EΓ0 ,D (t) + cγ L(t)

(7.25)

which, as shown above, is true for j = m and j = m − 1. For m ≥ j ≥ 3, we assume that the inequality (7.25) is true for j and j − 1. We need to prove that it is true for j − 2. Formal differentiation of the first equation in (7.7) by j − 2 times with respect to t yields (j−2)

φ(j) (t) = A(t)φ(j−2) (t) + b0 +

j−2 X i=1

(x, ∇φ, φt )

Cij−2 A(i) (t)φ(j−2−i) (t)

(7.26)

where Ckj−2 are the coefficients of Leibniz’s rule for differentiating a product. Using the inequalities (7.19) and (7.20), we obtain kφ(j−2) (t)k2m−j+2 ≤ cγ kA(t)φ(j−2) (t)k2m−j + cγ kφ(j−2) (t)k2m−j+3/2, Γ0

+cγ kφ(j−2) (t)k2m−j+1

≤ cγ kφ(j) (t)k2m−j + cγ kφ(j−1) (t)k2m−j+1 + cγ kφ(j−2) (t)k2m−j+1 +cγ EΓ0 ,D (t) + cγ

j−1 X i=2

E i (t)

(7.27)

where the relation kφ(j−1) (t)k2m−j ≤ kφ(j−1) (t)k2m−j+1 has been used. Then the inequality (7.12) follows by induction where the following inequality has also been used to absorb the term kφ(j−2) (t)k2m−j+1 in the right hand side of the inequality (7.27), kφ(j−2) (t)k2m−j+1 ≤ εkφ(j−2) (t)k2m−j+2 + cγ,ε kφ(j−2) (t)k2 for ε > 0 small, since kφ(j−2) (t)k2 ≤ Q(t) for m ≥ j ≥ 3.


336


Lemma 7.5 Let γ > 0 be given and let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that the condition (7.11) holds true. Let ϕ ∈ H 1 (Q) solve the linear problem ϕtt (t) = A(t)ϕ(t) + F (t) for (t, x) ∈ Q, (7.28) ϕ = 0 for (t, x) ∈ Σ1 . Set 2 Υ(t) = kϕ(t)k ˙ + k∇ϕ(t)k2 ,

Then there is cγ > 0 such that

2 2 ΥΓ0 ,D (t) = kϕ(t)k ˙ Γ0 + kϕk1,Γ0 .

Υ(t) ≤ cγ Υ(0) Z th i ˙ )km−1 )Υ(τ ) + ΥΓ ,D (τ ) + kF (τ )k2 dτ +cγ (1 + kφ(τ 0

(7.29)

0

for 0 ≤ t ≤ T.

Proof. Let 2 P (t) = kϕ(t)k ˙ + (A∇ϕ, ∇ϕ) .

(7.30)

Using the formula (7.6), we obtain

˙ P˙ (t) = 2 (ϕ(t), ¨ ϕ(t)) ˙ + 2 (A∇ϕ(t), ∇ϕ(t)) ˙ + A∇ϕ, ∇ϕ Z ˙ ϕϕ ˙ νA dΓ = 2 (F + Cϕ, ϕ) ˙ + A∇ϕ, ∇ϕ + 2 Γ0

where ϕνA = hA(x, ∇φ)∇ϕ, νi. It follows from the positiveness of A(x, ∇φ) that Υ(t) ≤ cγ P (t) ≤ cγ P (0) Z th i ˙ m−1 k∇ϕk2 dt +cγ (k∇ϕk + kF (t)k)kϕk ˙ + kφk 0 Z tZ Z tZ +ε ϕ2νA dΓdt + cγ,ε ϕ˙ 2 dΓdt 0

Γ0

0

(7.31)

Γ0

for 0 ≤ t ≤ T where ε > 0 is given small. To obtain the inequality (7.29) from the inequality (7.31), we have to RtR estimate the term 0 Γ ϕ2νA dΓdt. We now introduce a Riemannian metric g = A−1 (x, ∇φ)

on Ω for 0 ≤ t ≤ T so that the couple (Ω, g) are Riemannian manifolds for t ∈ [0, T ]. If Γ0 = ∅, the inequality (7.29) is clearly true due to the inequality (7.31). We assume that Γ0 6= ∅. Since Γ1 ∩ Γ0 = ∅, there is a vector field H on Ω such that H|Γ1 = 0 in a neigborhood of Γ1 ; H|Γ0 = νA . (7.32)



337

Applying the identity (2.13) in Theorem 2.1 to the problem (7.28) yields div {2H(ϕ)A(x, ∇φ)∇ϕ − (|∇g ϕ|2g − u2t )H} + 2[Cϕ + F (t)]H(ϕ)

= 2[ϕt H(ϕ)]t + 2DH(∇g ϕ, ∇g ϕ) + (ϕ2t − |∇g ϕ|2g ) div H.

We integrate the above equation over Ω to obtain Z 1 ϕ2νA + (ϕ˙ 2 − |∇g ϕ|2g )|νA |2g dΓ 2 Γ0 Z d 1 (ϕ, ˙ H(ϕ)) + Dg H(∇g ϕ, ∇g ϕ) + (ϕ˙ 2 − |∇g ϕ|2g ) div H dx = dt 2 Ω Z − (Cϕ + F (t)) H(ϕ)dx. (7.33) Ω

Using the relation |∇g ϕ|2g =

ϕ2νA + |∇Γg ϕ|2g |νA |2g

for

x∈Γ

in the formula (7.33) where ∇Γg is the gradient in the induced metric on Γ from the Riemannian metric g, we have Z tZ ϕ2νA dΓdt ≤ cγ [Υ(t) + Υ(0)] 0

+cγ

Γ0

Z tZ 0

[Υ(t) + ΥΓ0 ,D (t)] dxdt.

(7.34)

Ω

Finally, we insert the inequality (7.34) into the inequality (7.31), and choose ε > 0 so small that the term εcγ Υ(t) can be absorbed by the left hand side of the inequality (7.31) to obtain the inequality (7.29). Proof of Theorem 7.1 The inequality (7.12) has been proved in Lemma 7.4. Now we prove the inequality (7.13). We let ϕ = φ(j−2) (t) for 2 ≤ j ≤ m + 1 in the equation (7.28) and apply Lemma 7.5 to obtain kφ(j−1) (t)k2 + k∇φ(j−2) (t)k2 Z t h i ≤ cγ Q(0) + cγ 1 + E 1/2 (t) Q(t) + QΓ0 ,D (t) dt 0

+cγ

Z

t

0

(j−2) 2

(kb0

k +

j−2 X

k=1

kA(k) (t)φ(j−2−k) (t)k2 )dt.

(7.35)

In addition, a similar computation as in Lemma 7.3 yields (j−2) 2

kb0

k

≤ cγ k∇φ(j−2) k2 + cγ L(t) ≤ cγ Q(t) + cγ L(t),


(7.36)

338


and kA(k) (t)φ(j−2−k) (t)k2 ≤ cγ L(t)

for 1 ≤ k ≤ j − 2.

(7.37)

The inequality (7.13) follows from (7.35)-(7.37).

Boundary Traces and Energy Estimates with Neumann data on k Γ0 Let φ ∈ ∩m [0, T ], H m−k (Ω) be a solution of the problem (7.7) for k=0 C some T > 0. We define vνA = hA(x, ∇φ)∇v, νi

for v ∈ H 2 (Ω)

and let EΓ0 ,N (t) =

m−2 X k=0

2 kφ(k) νA (t)km−k−3/2,Γ0 ,

QΓ0 ,N =

m−1 X k=0

2 kφ(k) νA k1/2,Γ0 .

We have Theorem 7.2 Let γ > 0 be given and let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that the conditions (7.11) hold. Then there is cγ > 0, which only depends on the γ, such that Q(t) ≤ E(t) ≤ cγ Q(t) + cγ EΓ0 ,N (t) + cγ L(t)

(7.38)

for 0 ≤ t ≤ T and Q(t) ≤ cγ Q(0) + cγ

Z

0

t

[(1 + E 1/2 (t))Q(t) + QΓ0 ,N (t) + L(t)]dt

(7.39)

for t ∈ [0, T ] where E(t) and Q(t) are given in (7.8) and (7.9), respectively. Proof. It will suffice to make some revisions in the proofs of Lemmas 7.4 and 7.5, respectively. Using the ellipticity that there is cγ > 0 such that kvk2k+1 ≤ cγ kA(t)vk2k−1 + kvνA k2k−1/2, Γ0 + kvk2k (7.40) for v ∈ H m (Ω) ∩ HΓ11 (Ω) and 0 ≤ k ≤ m − 1 in the proof of Lemma 7.4 yields the inequality (7.38). Moreover, the second inequality (7.39) is based on the following

Lemma 7.6 Let γ > 0 be given and let φ be a solution of the problem (7.7) on the interval [0, T ] for some T > 0 such that the conditions (7.11) hold true. Let ϕ ∈ H 1 (Q) solve the linear problem ϕ(t) ¨ = A(t)ϕ + F (t) for (t, x) ∈ Q, (7.41) ϕ = 0 for (t, x) ∈ Σ1 .



339

Set 2 Υ(t) = kϕ(t)k ˙ + k∇ϕ(t)k2 ,

ΥΓ0 ,N (t) = kϕνA k2H 1/2 (Γ0 ) .

Then there is cγ > 0 such that Υ(t) ≤ cγ Υ(0) Z th i 2 ˙ +cγ (1 + kφ(t)k )Υ(τ ) + Υ (τ ) + kF (τ )k dτ m−1 Γ0 ,N

(7.42)

0

for 0 ≤ t ≤ T. Proof. Let P (t) = kϕk ˙ 2 + (A∇ϕ, ∇ϕ) . Then

Z ˙ ˙ P (t) = 2 (F, ϕ) ˙ + A∇ϕ, ∇ϕ + 2

ϕνA ϕdΓ. ˙

(7.43)

Γ0

Using the estimate

|(ϕνA , ϕ) ˙ L2 (Γ0 ) | ≤ kϕk ˙ H −1/2 (Γ0 ) kϕνA kH 1/2 (Γ0 ) ≤ cΥ(t) + cΥΓ0 ,N (t) in (7.43) gives the inequality (7.42).

Boundary Traces and Energy Estimates with Robin Data on Γ0 This time, we consider the wave equation in a divergence form: φtt = div a(x, ∇w + ∇φ) for (t, x) ∈ Q, (7.44) φ = 0 for (t, x) ∈ Σ1 where a(·,·) = (a1 (·,·), · · · , an (·, ·)): Ω × Rn → Rn is smooth nonlinear map such that aiyj (x, y) are symmetric and positive for all (x, y) ∈ Ω × Rn and w ∈ H m (Ω) is an equilibrium which is defined by div a(x, ∇w) = 0 for x ∈ Ω. k Let φ ∈ ∩m [0, T ], H m−k (Ω) be a solution to the problem (7.44) for k=0 C some T > 0. Let λ > 0 be given. We define I(t) = φt + λha(x, ∇w + ∇φ), νi, (7.45) O(t) = φt − λha(x, ∇w + ∇φ), νi. Let EΓ0 ,R (t) =

m−2 X

QΓ0 ,R,I =

m−1 X

k=0

k=0

kI(k) (t)k2m−k−3/2,Γ0 ,

kI(k) (t)k2Γ0 ,


L(t) =

QΓ0 ,R,O =

m−1 X k=0

2m X

k=3

E k/2 (t),

kO(k) (t)k2Γ0 ,

340

Modeling and Control in Vibrational and Structural Dynamics Qλ (t) = 2λ[kφt (t)k2 + (Bφ (t)∇φ, ∇φ)] +2λ

m−1 X k=1

[kφ(k+1) (t)k2 + (Aφ (t)∇φ(k) , ∇φ(k) )],

P(t) = 4λ(kφt k2 + (Bφ (t)∇φ, ∇φ) + kφk2 ) where Bφ (t) =

Z

1

0

aiyj (x, ∇w + s∇φ)ds ,

(7.47)

(7.48)

Aφ (t) = aiyj (x, ∇w + ∇φ) .

We have

(7.46)

(7.49)

Theorem 7.3 Let γ > 0 be given and let φ be a solution of the problem (7.44) on the interval [0, T ] for some T > 0 such that the conditions (7.11) hold. Then there are c0γ , cγ > 0, which only depend on the γ, such that c0,γ Qλ (t) ≤ E(t) ≤ cγ Qλ (t) + cγ EΓ0 ,R (t) + cγ kφ(t)k2 + cγ L(t)

(7.50)

for 0 ≤ t ≤ T, Q˙ λ (t) ≤ 4QΓ0 ,R,I (t) − QΓ0 ,R,0 (t) + cγ L(t),

(7.51)

−Q˙ λ (t) ≤ 2QΓ0 ,R,I (t) + 2QΓ0 ,R,0 (t) + cγ L(t),

(7.52)

P(t) ≤ {P(0) + cγ

Z

0

T

[EΓ0 ,R (τ ) + L(τ )] dτ }et

(7.53)

for 0 ≤ s ≤ t ≤ T where E(t) is given in (7.8). To prove Theorem 7.3, we need Lemma 7.7 Let γ > 0 be given and let φ solve the system (7.44). Let rj (t) =

j X

k=1

(k)

Ckj div Aφ ∇φ(j−k) ,

rΓ0 j (t) =

j X

k=1

(k)

Ckj hAφ ∇φ(j−k) , νi

for 1 ≤ j ≤ m − 2. Then krj k2m−2−j ,

krΓ0 ,j k2m−3/2−j ≤ cγ

for 1 ≤ k ≤ m − 2.


m−1 X k=2

E k (t)

(7.54)


341

Proof. Let b be a smooth function on Ω × Rn . We have b(k) (x, ∇φ) =

k X

X

i=1 r1 +···+ri =k

Dyi b(∇φ(r1 ) , · · · , ∇φ(ri ) )

(7.55)

for 1 ≤ k ≤ j where Dyi b are the differentials of b of order i with respect to variable y. Thus, rj (t) is a sum of some functions in the form (ri ) (j−k) 1) (f (x, ∇φ)φ(r )xq xj1 · · · φxji φxp

with r1 + · · · + ri = k for 1 ≤ i ≤ k. Using the inequalities (7.17) and (7.18), we have (ri ) (j−k) 1) k(f (x, ∇φ)φ(r )xq k2m−2−j xj1 · · · φxji φxp

(ri ) (j−k) 2 1) ≤ kf (x, ∇φ)φ(r km−1−j ≤ cγ E i+1 (t) xj1 · · · φxj φxp i

(7.56)

which yield the inequality (7.54) on rj (t) for all 1 ≤ j ≤ m − 2. An similar argument gives the inequality (7.54) on rΓ0 j (t) for all j. Proof of Theorem 7.3. The left-hand side of the inequality (7.50) is clearly true. We prove the right-hand side of it. We have kφ(m) (t)k2 + kφ(m−1) (t)k21 ≤ kφ(m) (t)k2 + cγ (Aφ (t)∇φ(m−1) , ∇φ(m−1) ) + kφ(m−1) (t)k2 ≤ cγ Qλ (t).

(7.57)

Proceeding by induction, we assume that for 3 ≤ l ≤ m, when j = l and j = l − 1, kφ(j) (t)k2m−j ≤ cγ Q(t) + cγ EΓ0 ,R (t) + cγ kφ(t)k2 + cγ L(t)

(7.58)

which, as shown in (7.57), is true for l = m. Formal differentiation of the equations in (7.44) and (7.45) for l − 2 times with respect to t yields   φ(l) = div Aφ (t)∇φ(l−2) + rl−3 (t) for (t, x) ∈ Q,    φ(l−2) = 0 for (t, x) ∈ Σ 1 (7.59) (l−2) (l−1) + λr = I(l−2) (t) for (t, x) ∈ Σ0 , φ + λφ  ν Γ ,l−3 A 0    φ(l−1) − λφ(l−2) − λr (l−2) (t) for (t, x) ∈ Σ0 νA Γ0 ,l−3 = O where rl−3 and rΓ0 ,l−3 are given in Lemma 7.7 if l ≥ 4 and they are zero when l = 3. Using the ellipticity (7.40), the equations in (7.59), and the estimates in Lemma 7.7, we obtain kφ(l−2) (t)k2m−l+2

2 (l−2) ≤ cγ (k div Aφ (t)∇φ(l−2) k2m−l + kφ(l−2) (t)k2m−l+1 ) νA kΓ0 ,m−l+1/2 + kφ

≤ cγ kφ(l) (t)k2m−l + cγ kI(l−2) (t)k2Γ0 ,m−l+1/2 + cγ kφ(l−1) (t)k2m−l +εkφ(l−2) (t)k2m−l+2 + cγ,ε kφ(l−2) (t)k2 + cγ L(t)


(7.60)

342


which implies that the inequality (7.58) is also true for l − 2. Then the inequality (7.58) holds for 1 ≤ j ≤ m by induction. To get the estimate (7.50), we need to prove that the inequality (7.58) is true for j = 0. The system (7.44)-(7.45) can be rewritten as  φtt = div Bφ (t)∇φ for (t, x) ∈ Q,    φ = 0 for (t, x) ∈ Σ1 , (7.61) φt + λφνB = I(t) (t, x) ∈ Σ0 ,    φt − λφνB = O(t) (t, x) ∈ Σ0 . Then similar estimates as in (7.60) yield the inequality (7.58) for j = 0 where this time instead of the formulas in (7.59), we use the formulas in (7.61). Let ϑ0 (t) = kφt (t)k2 + (Bφ (t)∇φ, ∇φ) (7.62) and, for 1 ≤ j ≤ m − 1, let ϑj (t) = kφ(j+1) (t)k2 + (Aφ (t)∇φ(j) (t), ∇φ(j) (t)).

(7.63)

For 1 ≤ j ≤ m − 1 using the formulas (7.59) for l = j + 2 and the Green formula, we obtain ϑ˙ j (t) = 2(rj−1 (t), φ(j+1) (t)) + 2(φ(j+1) (t), (φ(j) )νA (t)) +(A˙ φ (t)∇φ(j) (t), ∇φ(j) (t))

= (2λ)−1 (kI(j) (t) − λrΓ0 ,j−1 (t)k2Γ0 − kO(j) (t) + λrΓ0 ,j−1 (t)k2Γ0 ) +2(rj−1 (t), φ(j+1) (t)) + (A˙ φ (t)∇φ(j) (t), ∇φ(j) (t)). (7.64)

In the case j = 0 we use the formula (7.61) to obtain ϑ˙ 0 (t) = (2λ)−1 (kI(t)k2Γ0 − kO(t)k2Γ0 ).

(7.65)

The inequalities (7.51) and (7.52) follow from (7.64) and (7.65) since Qλ (t) = Pm−1 4λ j=0 ϑj (t) where the following estimates have been used |(A˙ φ (t)∇φ(j) (t), ∇φ(j) (t))| ≤ cγ E 3/2 (t)

for 1 ≤ j ≤ m − 1.

Similarly, we have ˙ P(t) ≤ kI(t)k2Γ0 + cγ E 3/2 (t) + P(t),

0 ≤ t ≤ T,

which yields P(t) ≤ P(0) + cγ

Z

9

T

[EΓ0 ,R (τ ) + E

3/2

(τ )] dτ ] +

Z

0

t

P(τ ) dτ

(7.66)

for 0 ≤ t ≤ T. The inequality (7.53) follows from (7.66) by Gronwall’s inequality.



7.2

343

Locally and Globally Boundary Exact Controllability

In this section we study the boundary exact controllability for the quasilinear wave equation. We derive the existence of long time solutions near an equilibrium, prove the local exact controllability around the equilibrium under existence of an escape vector field for the metric. We then establish the globally exact controllability in such a way that the state of the quasilinear wave equation moves from an equilibrium in one location to an equilibrium in another location under some geometrical conditions. The Dirichlet action and the Neumann action are studied, respectively. Escape vector fields here are determined by a Riemannian metric, given by the coefficients and equilibria of the quasilinear wave equation. Then the Riemmannian curvature theory provides a unique verifiable tool for this assumption. Some examples are presented to verify the globally exact controllability. Boundary Exact Controllability with Dirichlet Action Let Ω ⊂ Rn be an open, bounded set with the smooth boundary Γ. Suppose that Γ consists of two disjoint parts, Γ0 and Γ1 , where Γ0 6= ∅. Let w be an equilibrium, given by (7.1), and let T > 0 be given. We consider a controllability problem around the equilibrium w  Pn utt = ij=1 aij (x, ∇u)uxi xj + b(x, ∇u) for (t, x) ∈ Q,    u = w for (t, x) ∈ Σ1 , (7.67) u = w + ϕ for (t, x) ∈ Σ0 ,    u(0) = u0 , ut (0) = u1

where aij (x, y), b(x, y) are smooth functions on Ω × Rn such that A(x, y) = (aij (x, y)) > 0 for (x, y) ∈ Ω × Rn and b(x, 0) = 0 for x ∈ Ω. Let m ≥ [n/2] + 3 be a given positive integer. Definition 7.2 We say that (u0 , u1 ) ∈ H m (Ω) × H m−1 (Ω) and k ϕ ∈ ∩m−2 [0, T ], H m−k−1/2 (Γ0 ) k=0 C

satisfy the compatibility conditions of order m with the Dirichlet data if u0 = w,

uk ∈ H m−k (Ω),

ϕ = 0,

ϕ(k) (0) = uk ,

uk = 0 for

for

x ∈ Γ1 ,

x ∈ Γ0 ,

(7.68)

for 1 ≤ k ≤ m − 1 where for k ≥ 2, uk = u(k) (0), as computed formally (and recursively) in terms of u0 and u1 , using the equation in (7.67). Definition 7.3 Let u0 , u1 , uˆ0 , and u ˆ1 be given functions on Ω and T > 0 be


344


given. If there is a boundary function ϕ on Σ0 such that the solution of the problem (7.67) satisfies u(T ) = u ˆ0 ,

u(T ˙ ) = uˆ1

for

x ∈ Ω,

we say the system (7.67) is exactly controllable from (u0 , u1 ) to (ˆ u0 , u ˆ1 ) at time T by boundary with the Dirichlet action. We assume that short time solutions to the problem (7.67) exist for (u0 , u1 ) ∈ H m (Ω)×H m−1 (Ω) and study the possibility of moving it to another state in H m (Ω) × H m−1 (Ω) at time T via a boundary control k ϕ ∈ ∩m−2 [0, T ], H m−k−1/2 (Γ0 ) . k=0 C Solutions in Long Time In general, solutions of the system (7.67) may blow up in a finite time even if the initial data and the boundary control are smooth. However, in order to move one state to another, the control time must be larger than the wave length of the system. For those purposes, we shall establish the existence of long time solutions of the system around an equilibrium. Theorem 7.4 Let w ∈ H m (Ω) be an equilibrium of the problem (7.67). Let T > 0 be arbitrarily given. Then there is εT > 0, which depends on the time T, such that, if (u0 , u1 ) ∈ H m (Ω) × H m−1 (Ω) satisfy ku0 − wkm < εT ,

ku1 km−1 < εT , k and if ϕ ∈ ∩m−2 [0, T ], H m−k−1/2 (Γ0 ) with ϕ(k) ∈ H 1 (Σ0 ) for 0 ≤ k ≤ k=0 C m − 1 satisfies the compatibility conditions with (u0 , u1 ) of order m such that m−1 X k=0

kϕ(k) k2C ([0,T ],H m−1/2−k (Γ )) + 0

m−1 X k=0

kϕ(k) k2H 1 (Σ0 ) < εT ,

k then the system (7.67) has a solution u ∈ ∩m [0, T ], H m−k (Ω) . k=0 C

Proof. Let u be a short time solution to the problem (7.67) and let φ = u − w. Then φ(0) = u0 − w, φt (0) = u1 , φ|Γ0 = ϕ,

and φ is a short time solution to the problem (7.7) with b0 (x, ∇φ, φt ) = b(x, ∇φ) for x ∈ Ω. We take γ = 1. Let c1 = cγ ≥ 1

(7.69)

be fixed such that the corresponding inequalities (7.12) and (7.13) in Theorem 7.1 hold for t in the existence interval of the solution u, respectively.



345

Let T1 > 0 be arbitrarily given. We shall prove that, if initial data (u0 , u1 ) and boundary value ϕ are compatible of order m such that E(0) + max EΓ0 ,D (t) + 0≤t≤T1

T1

Z

QΓ0 ,D (t)dt ≤

0

1 −6c21 T1 e , 48c31

(7.70)

then solutions to the problem (7.67) exist at least on the interval [0, T1 ]. We set 1 1 1 η= ≤ < . (7.71) 4c1 4 2 Since E(0) ≤ η/4, the solution of short time must satisfy E(t) ≤ η ≤ 1/2 < 1

(7.72)

for some interval [0, δ]. Let δ0 be the largest number such that (7.72) is true for t ∈ [0, δ0 ). It follows from the inequalities (7.71) and (7.72) that m X i=2

1 E(t) 2c1

E i (t) ≤ 2E 2 (t) ≤

for t ∈ [0, δ0 ).

(7.73)

We shall prove δ0 ≥ T1 by contradiction. Suppose that δ0 < T1 . Then in this interval [0, δ0 ] the conditions (7.11) hold true. We apply Theorem 7.1. Using the inequalities (7.12) and (7.73), we obtain E(t) ≤ 2c1 [Q(t) + EΓ0 ,D (t)] for t ∈ [0, δ0 ). (7.74) Moreover, it follows from the inequalities (7.13), (7.12), (7.72), (7.73), and (7.69) that Q(t) ≤ c1 E(0) + 3c1

Z

0

t

E(τ ) dτ + c1

Z

t

QΓ0 ,D (τ ) dτ

(7.75)

0

for t ∈ [0, δ0 ). Inserting the inequality (7.75) into the inequality (7.74), we have Z T1 Z t E(t) ≤ 6c21 [E(0) + QΓ0 ,N (τ ) dτ + max EΓ0 ,N (t)] + 6c21 E(τ ) dτ 0≤t≤T1

0

0

which yields, via the condition (7.70) and the Gronwall inequality, E(δ0 ) ≤ η/2 < η. This contradicts the definition of δ0 .

In order to move one state to another in H m (Ω) × H m−1 (Ω), we recall


346


a boundary control space in (2.94) as follows, which is a Banach space. Let Bm D (Σ0 ) consist of all the functions k m−1/2−k ϕ ∈ ∩m−1 C [0, T ], H (Γ ) , 0 k=0 ϕ(k) ∈ H 1 (Σ0 )) ,

ϕ(k) (0)|Γ0 = 0

(7.76) (7.77)

for 0 ≤ k ≤ m − 1 with the norm kϕk2B m = D

m−1 X k=0

kϕ(k) k2C ([0,T ],H m−k−1/2 (Γ )) + 0

m−1 X k=0

kϕ(k) k2H 1 (Σ0 ) .

Clearly, the controllability of the problem (7.67) depends on the geometrical properties of not only the coefficients but also the equilibrium. We introduce the metric g = A−1 (x, ∇w) (7.78) on Ω and consider the couple (Ω, g) as a Riemannian manifold with a boundary Γ. Definition 7.4 An equilibrium w is said to be exactly controllable if there is an escape vector field on Ω for the metric (7.78). Remark 7.3 Existence of escape vector fields has been studied in Section 2.3. Some more examples will be given in the end of this section. Let w ∈ H m (Ω) be exactly controllable and let H be the strictly convex function on Ω in the metric g. Then there is a ̺0 > 0 such that DH(X, X) ≥ ̺0 |X|2g

for X ∈ Rnx ,

x ∈ Ω.

(7.79)

We have Theorem 7.5 Let w ∈ H m (Ω) be an equilibrium which is exactly controllable. Let 2 max |H|g . (7.80) T0 = ̺0 x∈Ω Furthermore, if Γ1 6= ∅, we assume that hH, νi ≤ 0

for

x ∈ Γ1 .

(7.81)

Then, for T > T0 given, there is a neighborhood of the point (w, 0) in H m (Ω)× H m−1 (Ω) where we can move one state of the problem (7.67) to another by a boundary control function in the space B m D (Σ0 ).



347

Proof. We invoke Theorem 7.4 to define a map for ϕ ∈ B m D (Σ0 ) by setting Φ(ϕ) = (u(T ), ut (T )) where u is the solution of the following problem  Pn utt = ij=1 aij (x, ∇u)uxi xj + b(x, ∇u) for    u = w for (t, x) ∈ Σ1 ,  u = w + ϕ for (t, x) ∈ Σ0 ,   u(0) = w, ut (0) = 0.

(7.82)

(t, x) ∈ Q, (7.83)

Let εT > 0 be given by Theorem 7.4. Then

Φ : B(0, εT ) → H m (Ω) × H m−1 (Ω) where B(0, εT ) ⊂ B m D (Σ0 ) is the ball with the radius εT centered at 0. We observe that Φ(0) = (w, 0). If we can prove that there is a neighborhood U of the point (w, 0) in H m (Ω) × H m−1 (Ω) such that U ⊂ the image Φ(B(0, εT )), then for each state in U there will be a boundary control ϕ ∈ B(0, εT ) which steers the state to (w, 0) by the time reversibility of the problem (7.83) and the proof will be complete. For the above purpose, we need to evaluate ∂ Φ(σϕ)|σ=0 ∂σ

for

ϕ ∈ Bm D (Σ0 ).

Φ′ (0)ϕ = (φ(T ), φt (T ))

for

ϕ ∈ Bm D (Σ0 )

Φ′ (0)ϕ = It is easy to check that

(7.84)

where φ(t, x) is the solution of the linear system with variable coefficients in the space variable    φtt = Aφ + F (φ) for (t, x) ∈ Q,  φ = 0 for (t, x) ∈ Σ1 , (7.85) φ = ϕ for (t, x) ∈ Σ0 ,    φ(0) = φt (0) = 0 for x ∈ Ω where

Aφ =

n X

ij=1

Fi =

aij (x, ∇w)φxi xj ,

X lj

F = (F1 , · · · , Fn ),

aljyi (x, ∇w)wxl xj + byi (x, ∇w).

From [154], an escape vector field for the metric (7.78) guarantees that the observability estimate (2.122) is true for T > T0 . By Theorem 2.21, the linear problem (7.85) is exactly H m (Ω) × H m−1 (Ω) controllable on [0, T ] with boundary control in B m D (Σ0 ). Then the linear map, given by (7.84), m m−1 Φ′ (0) : B m (Ω) D (Σ0 ) → H (Ω) × H


348


is surjective. Then the nonlinear map Φ is locally surjective at ϕ = 0 by Lemma 7.8 below. The proof is complete. The proof of the following lemma is classical; for example, see Theorem 3.1.19 in [11]. Lemma 7.8 Let X and Y be Banach spaces. Let a map Φ: O → Y, where O is an open subset of X, be Frech´ et differentiable. Let x0 ∈ O. If Φ′ (x0 ): X → Y is surjective, then there is an open neighborhood of y0 = Φ(x0 ) contained in the image Φ(O). Remark 7.4 The results in Theorem 7.5 are local. However, if we have enough equilibria exactly controllable, we can move the quasilinear wave state along a curve of equilibria, moving in successive small steps from one equilibrium to another nearby equilibrium until the target equilibrium is reached, see Theorem 7.6 below. This procedure can be done by the theorem of finite covering through finite steps. This idea was used by [186] for the quasilinear string. Globally Exact Controllability from One Equilibrium to Another with the Dirichlet Action Let w ∈ H m (Ω) be a given equilibrium. For α ∈ [0, 1], we assume that wα ∈ H m (Ω) are the solutions of the Dirichlet problem Pn ij=1 aij (x, ∇wα )wαxi xj + b(x, ∇wα ) = 0 for x ∈ Ω, (7.86) wα |Γ = αw|Γ such that sup kwα km < ∞.

(7.87)

α∈[0,1]

For the existence of the classical solution to the Dirchlet problem (7.86), for example, see [68]. We have Theorem 7.6 Let w ∈ H m (Ω) be an equilibrium which is exactly controllable. Let wα ∈ H m (Ω), given by (7.86), be also exactly controllable for all α ∈ [0, 1] such that the condition (7.87) holds. Then, there are T > 0 and ϕ ∈ B m D (Σ0 ) that is compatible with the initial data (w, 0) such that the solution of the system (7.67) with (u0 , u1 ) = (w, 0) satisfies u(T ) = u(T ˙ ) = 0. Proof. By Theorem 7.5 and the theorem of finite covering it will suffice to prove that wα : [0, 1] → H m (Ω) is continuous in α ∈ [0, 1] because the interval [0, 1] is compact.



349

∂ It is readily seen that vα = ∂α wα is the solution of the following linear, elliptic problem  Pn  ij=1 aij (x, ∇wα )vαxi xj +   n X n X [ aijyl (x, ∇wα )wαxi xj + byl (x, ∇wα )]vαxl = 0, (7.88)   ij l   vα |Γ = w|Γ

for each α ∈ [0, 1]. Moreover, by the maximum principle for the above problem (7.88), ∂ sup | wα | ≤ sup |w|. (7.89) x∈Ω ∂α x∈Γ Let A(α)v =

n X

ij=1

aij (x, ∇wα )vxi xj

for v ∈ H 2 (Ω), α ∈ [0, 1].

(7.90)

Using the uniform bound (7.87), the ellipticity of the operator A(α0 ), and the estimate (7.89), we have kwα − wα0 km ≤ ckA(α0 )(wα − wα0 )km−2 + c|α − α0 |kwkm−1/2, Γ + ckwα − wα0 k ≤ ckA(α0 )(wα − wα0 )km−2 + c|α − α0 |(kwkm−1/2, Γ + sup |w|).

(7.91)

x∈Γ

Next, let us estimate kA(α0 )(wα − wα0 )km−2 . [A(α0 ) − A(α)]wα and b(x, ∇wα ) − b(x, ∇wα0 ) can be written as sums of some terms, respectively, in the form f (x, ∇wα , ∇wα0 )(wα0 xl − wαxl )wαxi xj . Applying the inequality (7.17) to the above products gives, via the bound (7.87) and the estimate (7.89), kA(α0 )(wα − wα0 )km−2 ≤ k(A(α0 ) − A(α))wα km−2 + kb(x, ∇wa ) − b(x, ∇wα0 )km−2 ≤ ckwα − wα0 km−1 ≤ εkwα − wα0 km + cε |α − α0 | sup |w|.

(7.92)

x∈Γ

We obtain the desired result after substituting (7.92) into (7.91).

Remark 7.5 Since the quasilinear wave equation is time-reversible, an equilibrium can be moved to another if they can both be moved to zero. However, this result only gives the existence of the control time T. We do not know how large the T is because it is given by the theorem of finite covering.


350


Boundary Exact Controllability with the Neumann Action Next, we turn to the boundary control with the Neumann action. Let Γ = Γ0 ∪ Γ1 and Γ0 ∩ Γ1 = ∅ with Γ1 being nonempty. Let a(·, ·) = (a1 (·, ·), · · · , an (·, ·)) where ai (·, ·) are smooth functions on Ω × Rn such that a(x, 0) = 0 for x ∈ Ω and A(x, y) = aiyj (x, y) > 0 for (x, y) ∈ Ω × Rn . This time we assume that the quasilinear part of the system is in the divergence form. Let T > 0 be given. We consider a controllability problem  u ¨ = div a(x, ∇u) on Q,    u = w on Σ1 , (7.93) ha(x, ∇u), νi = ha(x, ∇w), νi + ϕ on Σ0 ,    u(0) = u0 , u(0) ˙ = u1 where ν is the normal of the boundary Γ in the Euclidean metric of Rn and w is an equilibrium, defined by div a(x, ∇w) = 0

on Ω.

(7.94)

We say that (u0 , u1 ) ∈ H m (Ω) × H m−1 (Ω) and k [0, T ], H m−k−3/2 (Γ0 ) ϕ ∈ ∩m−2 k=0 C

satisfy the compatibility conditions of order m with the Neumann boundary data on Γ0 and the Dirichlet data on Γ1 if the relations (7.68) hold and 0 for x ∈ Γ0 , k = 0, (k) ϕ (0) = hA(x, ∇w)∇uk , νi x ∈ Γ0 1 ≤ k ≤ m − 1, where uk = u(k) (0) are given by the equation in (7.93) for k ≥ 2. Solutions in Long Time Using the estimates in Theorem 7.2, similar arguments as in the proof of Theorem 7.4 yield Theorem 7.7 Let w ∈ H m (Ω) be an equilibrium of the problem (7.93). Let T > 0 be arbitrarily given. Then there is εT > 0, which depends on the time T, such that, if (u0 , u1 ) ∈ H m (Ω) × H m−1 (Ω) satisfy ku0 − wkm < εT ,

ku1 km−1 < εT , k and ϕ ∈ ∩m−2 [0, T ], H m−k−3/2 (Γ0 ) is such that ϕ(k) ∈ H 1 (Σ0 ) for 0 ≤ k=0 C k ≤ m − 1, which satisfies the compatibility conditions with (u0 , u1 ) of order m and m−2 X k=0

kϕ(k) k2C ([0,T ],H m−k−3/2 (Γ )) + 0

m−2 X k=0

kϕ(k) k2H 1 (Σ0 ) < εT ,

k then the system (7.93) has a unique solution u ∈ ∩m [0, T ], H m−k (Ω) . k=0 C



351

Let T > 0 be given. We recall a Banach space B m N (Σ0 ) by (2.144) as follows. Let B m N (Σ0 ) consist of all the functions k m−k−3/2 ϕ ∈ ∩m−2 (Γ0 )), k=0 C ([0, T ], H

ϕ(k) (0) = 0,

x ∈ Γ0 ,

ϕ(k) ∈ H 1 (Σ0 ),

0≤k ≤m−1

with the norm kϕk2B m (Σ0 ) = N

m−2 X k=0

kϕ(k) k2C ([0,T ],H m−3/2−k (Γ )) + 0

m−2 X k=0

kϕ(k) k2H 1 (Σ0 ) .

Geometry Condition for Local Controllability with the Neumann Action We introduce a Riemannian metric on Ω by g = A−1 (x, ∇w) and consider the couple (Ω, g) as a Riemannian manifold with a boundary Γ where w is an equilibrium, given by the problem (7.94). We have Theorem 7.8 Let Γ1 6= ∅. Let w ∈ H m+1 (Ω) be an equilibrium which is exactly controllable. Let H be an escape vector field on Ω such that hH, νi ≤ 0 for x ∈ Γ1 . Then there exists a time T0 > 0 such that for T > T0 given, there is a neighborhood of the point (w, 0) in H m+1 (Ω)× H m (Ω) where we can move one state of the problem (7.93) to another by a boundary control function in the space B m N (Σ0 ). Proof. We use Theorem 7.7 to define a map for ϕ ∈ B m N (Σ0 ) by setting ΦN (ϕ) = (u(T ), u(T ˙ )) where u is the solution of the following problem  utt = div a(x, ∇u) for (t, x) ∈ Q,    u|Γ1 = w for t ∈ (0, T ), ha(x, ∇u), νi|Γ0 = ha(x, ∇w), νi + ϕ for    u(0) = w, ut (0) = 0.

(7.95)

t ∈ (0, T ),

(7.96)

Let εT > 0 be given by Theorem 7.7. Then the map ΦN : B(0, εT ) → H m (Ω) × H m−1 (Ω) is well defined where B(0, εT ) ⊂ B m N (Σ0 ) is the ball with the radius εT centered at 0. We observe that, since w ∈ H m+1 (Ω), ΦN (0) = (w, 0) ∈ H m+1 (Ω) × H m (Ω). Then Theorem 7.8 is equivalent to the following claim: For some T > 0 there are ε1 > 0 and ε2 > 0 with εT ≥ ε2 such that BH m+1 (Ω)×H m (Ω) ((w, 0), ε1 ) ⊂ ΦN (B(0, ε2 )) (7.97) where BH m+1 (Ω)×H m (Ω) ((w, 0), ε1 ) is the ball with the radius ε1 centered at 0 in the space H m+1 (Ω) × H m (Ω).


352


It is easy to check that the map ΦN is F r´ echet differentiable on B(0, εT ). In particular, Φ′N (0)ϕ = (v(T ), v(T ˙ ))

for ϕ ∈ B m N (Σ0 )

(7.98)

where v(t, x) is the solution of the linear system with variable coefficients in the space variable  vtt = div A(x, ∇w)∇v for (t, x) ∈ Q,    v|Γ1 = 0 for t ∈ (0, T ), (7.99) vν |Γ = ϕ for t ∈ (0, T ),    A 0 v(0) = v(0) ˙ =0 where vνA = hA(x, ∇w)∇v, νi. For i = 1, 2, let Xi = B m+i−1 (Σ0 ), N

Yi = H m+i−1 (Ω) × H m+i−2 (Ω).

It is easy to check by Theorem 7.2 that the mappings ΦN , given by (7.95), are of C 1 from B Xi (0, r) to Yi for i = 1, 2, and for some r > 0. By Lemma 7.9 below, to prove the relation (7.97) is to establish the exact controllability of the linear system (7.99) on the space [H m+1 (Ω)∩HΓ11 (Ω)]×[H m (Ω)∩HΓ11 (Ω)], which has been given by Theorem 2.24. A similar argument as in the proof of Theorem (3.1.19) in [11] gives Lemma 7.9 Let X1 , X2 , Y1 , and Y2 be Banach spaces with X2 ⊂ X1 , Y2 ⊂ Y1 , X2 = X1 , and Y2 = Y1 which satisfy the relations kxkX1 ≤ kxkX2

and

kykY1 ≤ kykY2 .

Suppose that Φ : B Xi (0, r) → Yi are mappings of C 1 for i = 1, 2 such that Y2 ⊂ Φ′ (0)X1 .

(7.100)

B Y2 (Φ(0), ε) ⊂ Φ ( B X1 (0, r)) .

(7.101)

Then there is ε > 0 such that

Remark 7.6 Here we lose an explicit formula of T0 . Remark 7.7 Unlike the control with the Dirichlet action, we only have the exact controllability results in the space H m+1 (Ω) × H m (Ω) by a control ϕ ∈ Bm N (Σ0 ). This is because the Neumann action loses a regularity of order 1(actually, order 1/2). For this point, we may see [117]. In addition, although we can move one state to another in the space H m+1 (Ω) × H m (Ω), we can not guarantee the solution (u(t), u(t)) ˙ of the problem (7.93) always stays in H m+1 (Ω) × H m (Ω) in the process of the control for 0 ≤ t ≤ T where they are actually in the space H m (Ω) × H m−1 (Ω) for all t ∈ [0, T ] by Theorem 7.7. The same things happen to the globally exact controllability results in Theorem 7.9 below.



353

Globally Exact Controllability from One Equilibrium to Another with the Neumann Action Let w ∈ H m+1 (Ω) be given an equilibrium. For α ∈ [0, 1], we assume that wα ∈ H m+1 (Ω) are solutions to the problem div a(x, ∇wα ) = 0 for x ∈ Ω, (7.102) wα = αw for x ∈ Γ with, this time, a uniform bound sup kwα km+1 < ∞.

(7.103)

α∈[0,1]

Similar arguments, as in the proof of Theorem 7.6, yield Theorem 7.9 Let an equilibrium w ∈ H m+1 (Ω) be exactly controllable. Let wα ∈ H m+1 (Ω), given by (7.102), be also exactly controllable for all α ∈ [0, 1] such that the condition (7.103) holds. Then, there are T > 0 and ϕ ∈ B m N (Σ0 ) which is compatible with the initial data (w, 0) of order m such that the solution of the system (7.93) with (u0 , u1 ) = (w, 0) satisfies u(T ) = u(T ˙ ) = 0. Finally, let us see some examples to verify the globally exact controllability in Theorem 7.6. Example 7.1 Let n = 2 and m = 4. Consider the control problem   utt = (|∇u|2 + 1)∆u for (t, x) ∈ (0, T ) × Γ, u|Γ = ϕ for 0 ≤ t ≤ T,  u(0) = w ut (0) = 0 for x ∈ Ω,

(7.104)

∂2 ∂2 + 2. 2 ∂x ∂y Then w ∈ H 4 (Ω) is an equilibrium if and only if w is harmonic on Ω, i.e.,

where ∆ =

∆w = 0

for

x ∈ Ω.

(7.105)

Let w ∈ H 4 (Ω) be an equilibrium. Then metric (7.78) is given by |∇w|2 + 1 0 g = A−1 (x, ∇w), A(x, ∇w) = . 0 |∇w|2 + 1 Then the condition (7.105) implies that |∇(|∇w|2 + 1)|2 = 2|∇w|2 |D2 w|2 where D2 w is the Hessian of w in the Euclidean metric of R2 . By the formulas (2.27) and (7.105), the Gauss curvature of the Riemannian manifold (Ω, g) is |D2 w|2 κ(x) = for x ∈ Ω. (7.106) |∇w|2 + 1


354


Then the zero equilibrium, w = 0, is exactly controllable for any Ω ⊂ R2 . In addition, by Corollary 2.1 and Theorem 7.6, we have the conclusion: If two equilibria wi 6= 0 in H 4 (Ω) are such that there are xi ∈ Ω satisfying Ω ⊂ { x ∈ R2 | |x − xi | < γi } where γi =

(7.107)

π(1 + inf x∈Ω |∇wi |2 ) , 2 supx∈Ω |D2 wi |

(7.108)

for i = 1, 2, then there are a control time T > 0 and a control function ϕ ∈ B 4D (Σ0 ) such that the solution of the problem (7.104) with the initial (w1 , 0) satisfies u(T ) = w2 , u(T ˙ ) = 0. In particular, let w1 = a(x21 − x22 ), w2 = bx1 x2 , Ω = the unit disc, (7.109) √ √ where 0 < a < π/(4 2) and 0 < b < π/(2 2). It is easy to check that wi meet the conditions (7.107) for i = 1, 2. Then the state of the system (7.104) can be moved from (w1 , 0) to (w2 , 0) at some time T > 0. Example 7.2 Let n = 2. Consider the control problem   utt = (|∇u|2 + 1)−1 ∆u for (t, x) ∈ (0, T ) × Γ, u|Γ = ϕ for 0 ≤ t ≤ T,  u(0) = w ut (0) = 0 for x ∈ Ω.

(7.110)

Let w ∈ H 4 (Ω) be an equilibrium. The metric is |∇w|2 + 1 0 g= . 0 |∇w|2 + 1

By (2.27), the Gauss curvature of (Ω, g) is k(x) = −

|D2 w|2 ≤ 0, (|∇w|2 + 1)3

∀ x ∈ Ω.

By Theorems 2.6 and 7.6, for any two equilibria w1 , w2 ∈ H 4 (Ω) and any Ω ⊂ Rn , there are a control time T > 0 and a control function ϕ ∈ B 4D (Σ0 ) such that the state of the system (7.110) is moved from (w1 , 0) to (w2 , 0). Example 7.3 Let n = 2 and Ω ⊂ R2 be a bounded, open set. Consider a quasilinear wave equation utt (t, x) =

ux1 x1 (t, x) ux2 x2 (t, x) + 2 1 + ux1 (t, x) 1 + u2x1 (t, x)


on

Q.

(7.111)


355

Consider an equilibrium given by w(x) = λx1 x2

for

x = (x1 , x2 ) ∈ R2

(7.112)

where λ 6= 0 is constant. Let wα (x) = αw(x) Then

for

1  1 + α2 λ2 x22 A(x, ∇wα ) =  0 

α ∈ [0, 1].

(7.113) 

0

  1 1 + α2 λ2 x21

and the metrics are 1 + α2 λ2 x22 0 gα = 0 1 + α2 λ2 x21

for

α ∈ [0, 1].

By Lemma 2.3, the Gauss curvature is kα (x) = −

2 + α2 λ2 |x|2 ≤0 (1 + α2 λ2 x22 )(1 + α2 λ2 x21 )

for

x ∈ R2

and for α ∈ [0, 1]. Then wa (x) is exactly controllable for each α ∈ [0, 1]. By Theorem 7.6, the equilibrium w, given by (7.112), can be steered to rest at a time T by a boundary control.

7.3

Boundary Feedback Stabilization

We consider the existence of global solutions of the quasilinear wave equation in the divergence form with a boundary dissipation structure of an inputoutput when initial data and boundary inputs are near a given equilibrium of the system in a certain sense. We show that the existence of global solutions depends not only on this dissipation structure but also on a Riemannian metric, given by the coefficients and the equilibrium of the system, where an escape vector field for the metric will guarantee global solutions in time. In particular, we have exponential decays of the norm of the state of the system if the input stops after a finite time, which implies the exponential stabilization of the system by boundary feedback. Let Ω ⊂ Rn be a bounded, open set with the smooth boundary Γ = Γ0 ∪Γ1 and Γ0 ∩ Γ1 = ∅. Let a(x, y) = (a1 (x, y), · · · , an (x, y)) :


Ω × Rn → Rn

356


be a smooth mapping with a(x, 0) = 0 for x ∈ Ω such that A(x, y) = aiyj (x, y) are symmetrical and positive for (x, y) ∈ Ω × Rn . Let w be an equilibrium, defined by div a(x, ∇w) = 0 for

x ∈ Ω.

Let Q∞ = (0, ∞) × Ω,

Σ∞ 1 = (0, ∞) × Γ1 ,

Σ∞ 0 = (0, ∞) × Γ0 .

We consider the following problem   utt = div a(x, ∇w + ∇u) for (t, x) ∈ Q∞ , u = 0 for (t, x) ∈ Σ∞ 1 ,  u(0, x) = u0 , ut (0, x) = u1 for x ∈ Ω

(7.114)

I(t) = ut + λha(x, ∇w + ∇u), νi for

(t, x) ∈ Σ∞ 0 ,

(7.115)

O(t) = ut − λha(x, ∇w + ∇u), νi for

(t, x) ∈ Σ∞ 0 ,

(7.116)

with an input I(t) and an output O(t) on the portion Γ0 of the boundary Γ

n

where h·, ·i is the Euclidean metric of R , ν is the outside unit normal of Γ, and λ > 0 is a constant. Remark 7.8 If σ(x, y) is a smooth function on Ω × Rn such that a(x, y) = (σy1 (x, y), · · · , σyn (x, y)) , we define the total energy of the quasilinear system (7.114) − (7.116) at time t in the zero equilibrium by Z E(t) = 4λ [u2t /2 + σ(x, ∇u)] dx (7.117) Ω

and obtain by a simple computation Z Z dE(t) = |I(t)|2 dΓ − |O(t)|2 dΓ. dt Γ0 Γ0

(7.118)

This formula expresses that the rate of change of the energy is equal to the power supplied to the system by the input minus the power taken out by the output on Γ0 . The balance equation (7.118) means that the boundary structure (7.115) is dissipative. Remark 7.9 It is well known that smooth solutions of quasilinear hyperbolic systems usually develop singularities after some time, see [93] and [133]. Since the structure of the boundary dissipation (7.115) makes the energy dissipative, we expect that the introduction of the boundary structure (7.115) assures the existence of a globally smooth solution. The aim of this section is to seek



357

general geometrical conditions for a quasilinear part as in the problem (7.114) and for a general equilibrium w to assure the system (7.114) − (7.115) to have global solutions when initial data and inputs are small. The results in Theorems 7.10, 7.11 later show that an escape vector field for a metric, given by (7.119) below, is one of such geometrical conditions. Let m ≥ [n/2] + 3 be a given positive integer. We introduce a Banach space for inputs. Let I m (Σ∞ 0 ) consist of all the functions I(t) = I(t, x) on Σ∞ = (0, ∞) × Γ0 such that I(k) (t) ∈ L2 ((0, ∞), H m−k−3/2 (Γ0 )) ∩ C[0, ∞; H m−k−3/2 (Γ0 )] for 0 ≤ k ≤ m − 2 and I(m−1) (t) ∈ L2 ((0, ∞), L2 (Γ0 )) with a norm Z ∞ kIk2I m((0,∞),Γ0 ) = max EΓ0 ,R (t) + EΓ0 ,R (τ ) dτ 0≤t 0 is a constant. (H2) The boundary portions Γ1 and Γ0 satisfy hH, νi ≤ 0 for


x ∈ Γ1 ,

(7.121)

358

Modeling and Control in Vibrational and Structural Dynamics hH, νi ≥ 0 for

x ∈ Γ0 .

(7.122)

(H3) Let Γ1 6= ∅. If Γ1 = ∅, we further assume that the initial value u0 satisfies Z u0 dΓ = 0. (7.123) Γ

We assume that solutions of short time exist for the system (7.114)-(7.115) with appropriate initial data and boundary input. In fact, there are standard approaches to obtain solutions of short time, for example, see [91] and [48]. We shall establish the following Theorem 7.10 Let an equilibrium w ∈ H m (Ω) be such that the assumptions (H1), (H2), and (H3) hold. Let (u0 , u1 ) and I satisfy the compatibility conditions of m order. Then for any (u0 , u1 ) in H m (Ω) × H m−1 (Ω) small and any I in I m ((0, ∞), Γ0 ) small, the system (7.114)-(7.115) has a global solution u k m−k in ∩m (Ω)). k=0 C ((0, ∞), H If the input I(t) = 0 after a finite time T0 > 0, the energy will decay exponentially, which is the stabilization by feedback from boundary. Theorem 7.11 Let all the assumptions of Theorem 7.10 hold. If there is T0 > 0 such that I(t) = 0 for t ≥ T0 , then there are c1 > 0, c2 > 0, and Tˆ ≥ T0 such that E(t) ≤ c1 e−c2 t for t ≥ Tˆ (7.124) Pm (k) 2 where E(t) = k=0 ku km−k . The proofs of Theorems 7.10 and 7.11 will be given at the end of this section.

Remark 7.10 In the linear case the structure of dissipation, as above, has been studied thoroughly, for example, see [202] and references there. Remark 7.11 The assumption (7.122) can be removed if an estimate of boundary trace can be established for the quasilinear wave equation as that for the linear wave equation in [122]. Remark 7.12 We need the assumption (H3) for a uniqueness result to get rid of some lower order terms, see Lemma 7.11. Instead of (7.123) there are other options. Example 7.4 Let a(y) = (a1 (y1 ), a2 (y2 )) : R2 → R2 where ai are smooth functions on R for i = 1, 2 such that a′i (s) > 0 for s ∈ R. Let w(x) = x1 x2 for x = (x1 , x2 ) ∈ R2 .



359

Then div a(∇w) = 0

x ∈ R2 .

for

The metric is g = A−1 (x, ∇w) =

1/a′1 (x2 ) 0 0 1/a′2 (x1 )

.

Let a1 (s) = a2 (s) = arctan s for s ∈ R. Then 1 + x22 0 g= . 0 1 + x21 By Lemma 2.3, the Gauss curvature of (R2 , g) is κ(x) = −

2 + |x|2 ≤0 (1 + x21 )2 (1 + x22 )2

for

x = (x1 , x2 ) ∈ R2 .

By Theorem 2.6, the assumption (H1) holds true for any Ω ⊂ R2 .

We shall make preparations for the proofs of Theorems 7.10 and 7.11. Let γ > 0 be given and let u satisfy the problem (7.114) on the interval [0, T ] for some T > 0 such that sup kukm ≤ γ,

0≤t≤T

sup kuk ˙ m−1 ≤ γ.

(7.125)

0≤t≤T

For t ∈ [0, T ], we introduce a metric gu on Ω by

−1 gu = A−1 (x, ∇w + ∇u). u (t) = A

(7.126)

Consider the couple (Ω, gu ) as a Riemannian manifold for each fixed t. Let X, Y be vector fields on Ω and let f be a function. Then hX, Y igu = hA−1 u (t)X, Y i,

∇gu f = Au (t)∇f

(7.127)

where h·, ·i and h·, ·igu are the inner products of the Euclidean metric and the metric gu , respectively, and ∇ and ∇gu are the gradients of the Euclidean metric and the metric gu , respectively. By (7.119) and (7.126) g0 = A−1 (x, ∇w) = g.

(7.128)

In addition, it is easy to check from the assumption (7.125) that there are c0,γ > 0 and cγ > 0 such that c0,γ |∇g f |2g ≤ |∇gu f |2gu ≤ cγ |∇g f |2g

for

t ∈ [0, T ],

f ∈ C ∞ (Ω).

(7.129)

Let Dgu and D be the Levi-Civita connections of the Riemannian metrics gu and g, respectively. Let H be a vector field on Ω. We denote by Dgu H and by DH the covariant differentials of the metric gu and g, respectively. They are tensor fields of rank 2 on Ω. We define η = Dgu H − DH.


(7.130)

360


Lemma 7.10 Let H be a vector field on Ω. Suppose that the tensor field of rank 2 η = η(·, ·) is given by the formula (7.130). Let γ > 0 be given and u be such that supx∈Ω |∇u| ≤ γ. Then there is cγ > 0 such that |η(X, Y )| ≤ cγ (|∇u| + |∇2 u|)|X||Y |

for

X, Y ∈ Rnx , x ∈ Ω

(7.131)

where ∇ is the covariant differential of the Euclidean metric of Rn . Proof. Using the relations (7.128) and (7.127), we have Dgu H(

∂ ∂ ∂ ∂ ∂ , ) = hH, i − hH, (Dgu ) ∂ igu ∂xj ∂x ∂xi ∂xj ∂xj ∂xi gu i n X ∂ ∂ ∂ = hA−1 (t)H, i − Γkgu ij hA−1 i u u (t)H, ∂xj ∂xi ∂xk k=1

∂ ∂ ∂ ∂ = DH( , ) + η( , ) ∂xi ∂xj ∂xi ∂xj

(7.132)

where η(

∂ ∂ ∂ ∂ −1 , ) = h[A−1 (x, ∇w)]H, i u (t) − A ∂xi ∂xj ∂xj ∂xi n X ∂ − Γkgu ij hA−1 i u (t)H, ∂xk +

k=1 n X k=1

Γkgij hA−1 (x, ∇w)H,

∂ i, ∂xk

(7.133)

Γkgu ij and Γkgij are the coefficients of the connections Dgu and D, respectively. ij Let A−1 u (t) = (a (x, ∇w + ∇u)). Then we have n

1X = alk (x, ∇w + ∇u)[(alj (x, ∇w + ∇u))∂xi 2

Γkgu ij

l=1

+(ali (x, ∇w + ∇u))∂xj − (aij (x, ∇w + ∇u))∂xl ] = fijk (x, ∇u) + pijk (u)

(7.134)

where n

fijk (x, ∇u) =

1X jl alk (x, ∇w + ∇u)[ail xj (x, ∇w + ∇u) + axi (x, ∇w + ∇u) 2

l=1 ij −axl (x, ∇w

pijk (u) =

+ ∇u)],

n 1 X alk (x, ∇w + ∇u)[ail yh (x, ∇w + ∇u)uxh xj 2

(7.135)

lh=1

ij +ajl yh (x, ∇w + ∇u)uxh xi − ayh (x, ∇w + ∇u)uxh xl ].


(7.136)


361

The formula (7.136) yields |pijk (u)| ≤ cγ |∇2 u|.

(7.137)

In addition, we have, by (7.135), Γkgij = fijk (x, 0).

(7.138)

Thus by (7.133)-(7.138), n X

k=1

n

Γkgu ij hA−1 u (t)H,

X ∂ ∂ i = Γkgij hA−1 (x, ∇w)H, i ∂xk ∂xk k=1

+ + +

n X

[fijk (x, ∇u) − fijk (x, 0)]hA−1 u (t)H,

k=1 n X k=1 n X

−1 Γkgij h[A−1 (x, ∇w)]H, u (t) − A

pijk (u)hA−1 u (t)H,

k=1

∂ i ∂xk

∂ i ∂xk

∂ i. ∂xk

(7.139)

Moreover, the relations A−1 u (t)

−1

−A

(x, ∇w) =

n Z X k=1

fijk (x, ∇u) − fijk (x, 0) = imply that

1

0

aij yk (x, ∇w

n Z X

h=1

,

1

0

−1 |h[A−1 (x, ∇w)]H, u (t) − A

|

+ τ ∇u)dτ uxk

!

fijkxh (x, τ ∇u)dτ uxh , ∂ i| ≤ cγ |∇u|, ∂xk

∂ ∂ −1 h[A−1 (x, ∇w)]H, i| ≤ cγ (|∇u| + |∇2 u|), u (t) − A ∂xj ∂xi |fijk (x, ∇u) − fijk (x, 0)| ≤ cγ |∇u|.

(7.140) (7.141) (7.142)

The inequality (7.131) follows from the formulas (7.133) and the estimates in (7.140)-(7.142). Theorem 7.12 Let the assumptions (H1), (H2), and (H3) hold. Let γ > 0 be given. Let u be a solution to the problem (7.114) on the interval [0, T ] for some T > 0 such that the estimates in (7.125) hold true. Then there are cγ > 0 and Tγ > 3 supx∈Ω |H|g /̺0 , where ̺0 > 0 is given by (7.120), such that, if 0 ≤ s ≤ t ≤ T are such that t − s ≥ Tγ , then Z t Z t Qλ (τ ) dτ ≤ cγ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ (7.143) s

s


362


where Qλ (t), QΓ0 ,R,I (t), QΓ0 ,R,O (t), and L(t) are defined in Theorem 7.3, respectively. Proof. Let H be a vector field and let f be a function. We denote by H(f ) the directional derivative of the function f along the vector field H. Then H(f ) = hH, ∇f i = hH, ∇gu f igu .

(7.144)

We assume that solutions to the problem (7.114)-(7.116) exist on [0, T ] for some T > 0. Let 0 ≤ s ≤ t ≤ T be given. Step 1 We assume 1 ≤ j ≤ m − 1. We differentiate the equations in (7.114)-(7.116) by j times with respect to the variable t (see (7.59)) to obtain  (j)  utt = div Au (t)∇u(j) + rj−1 for (t, x) ∈ Q,    u(j) = 0 for (t, x) ∈ Σ 1 (7.145) (j) (j+1)  u + λu + λr = I(j) (t) for (t, x) ∈ Σ0 , ν Γ ,j−1 A 0    (j+1) (j) u − λuνA − λrΓ0 ,j−1 = O(j) (t) for (t, x) ∈ Σ0

where rj−1 (t) and rΓ0 ,j−1 (t) are given in Lemma 7.7 for j ≥ 2 and they are zero if j = 1. For 0 ≤ j ≤ m − 1, let ϕ = u(j) . We multiply the equation in (7.145) by 2H(ϕ), integrate by parts over Qts = (s, t) × Ω, and obtain (see the identity (2.13)) Z [2H(ϕ)ϕνA + (ϕ˙ 2 − |∇gu ϕ|2gu )hH, νi]dΣ Σts

Z t = 2 (ϕ, ˙ H(ϕ)) −2 s

+

Z

Qts

˙ [ϕ˙ H(ϕ) + rj−1 (t)H(ϕ)] dQ

Qts

[2Dgu H(∇gu ϕ, ∇gu ϕ) + (ϕ˙ 2 − |∇gu ϕ|2gu ) div H]dQ

(7.146)

where Σts = (s, t) × Γ. Let f ∈ C 2 (Ω). We multiply the equation in (7.145) by 2f ϕ, integrate by parts over Qts , and obtain Z t Z [ϕ2 fνA − 2f ϕϕνA ]dΣ 2 f (ϕ˙ 2 − |∇gu ϕ|2gu )dQ = 2 (ϕ, ˙ f ϕ)L2 (Ω) + s

Qts

−

Z

[ϕ2 div Au (t)∇f + 2rj−1 (t)f ϕ]dQ.

Σts

(7.147)

Qts

Next, we assume that the vector field H on Ω satisfies the assumptions (H1) and (H2). Noting that H does not depend on time t, we have from the identity (7.146), ⊎Σts = ⊎Qts (7.148)



363

where ⊎Σts =

Z

Σts

[2H(ϕ)ϕνA + (ϕ˙ 2 − |∇gu ϕ|2gu )hH, νi]dΣ

Z t ⊎Qts = 2 (ϕ, ˙ H(ϕ))L2 (Ω) −2 rj−1 (t)H(ϕ) dQ s Qts Z + [2Dgu H(∇gu ϕ, ∇gu ϕ) + (ϕ˙ 2 − |∇gu ϕ|2gu ) div H]dQ.

(7.149)

(7.150)

Qts

To obtain the inequality (7.143), we shall estimate ⊎Σts and ⊎Qts , respectively. Estimate on ⊎Σts Decompose ⊎Σts as ⊎Σts = ⊎Σ1 ts + ⊎Σ0 ts where Σ1 ts = (s, t) × Γ1 and Σ0 ts = (s, t) × Γ0 . On Σts = (s, t) × Γ we have a decomposition of the direct sum H = HΓgu + hH,

Au (t)ν Au (t)ν i . |Au (t)ν|gu gu |Au (t)ν|gu

(7.151)

Moreover, since the boundary condition ϕ = u(j) = 0 on Γ1 implies H(ϕ) =

ϕνA hH, νi, hAu (t)ν, νi

∇gu ϕ = ϕνA

Au (t)ν hAu (t)ν, νi

(7.152)

for x ∈ Γ1 , we obtain ⊎Σ1 ts =

Z

Σ1 ts

(ϕνA )2 hH, νi dΣ ≤ 0, hAu (t)ν, νi

(7.153)

via the assumption (7.121). On Σ0 ts = (s, t) × Γ0 , the assumption (7.122) and the boundary formulas in Theorem 7.3 yield, via the estimates in Lemma 7.7, respectively, Z ⊎Σ0 ts ≤ [2H(ϕ)ϕνA + ϕ˙ 2 hH, νi] dΣ Σ0 ts

≤ εkϕk2L2 ((s,t),H 1 (Γ0 )) + cγ,ε (kϕk ˙ 2L2 (Σ0 ts ) + kϕνA k2L2 (Σ0 ts ) ) Z t ≤ cγ,ε [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ )] dτ s Z t 2 +εkϕkL2 ((s,t),H 1 (Γ0 )) + cγ,ε L(τ ) dτ (7.154) s

where ε > 0 is small.


364


We now estimate the term kϕk2L2 ((s,t),H 1 (Γ0 )) . Let H1 be a vector field on Ω such that H1 = 0

for x ∈ Γ1 ;

H1 = Au (t)ν

for

x ∈ Γ0 .

(7.155)

1/2

Since the vector field Au (t)ν/hAu (t)ν, νi is the unit normal of the metric gu along the boundary Γ , we have a directional decomposition ∇gu ϕ = ∇Γgu ϕ + ϕνA which yields

Au (t)ν hAu (t)ν, νi

|∇gu ϕ|2gu = |∇Γgu ϕ|2gu + ϕ2νA /hAu (t)ν, νi.

(7.156)

We replace the vector field H in the identity (7.146) with the above −H1 , and, by (7.156), we obtain the left hand side of (7.146) Z = [(|∇Γgu ϕ|2gu hAu (t)ν, νi − (ϕ˙ 2 hAu (t)ν, νi + ϕ2νA )]dΣ

(7.157)

Σ0ts

and the right hand side of (7.146) Z t ≤ cγ Qλ (t) + cγ Qλ (s) + cγ [Qλ (τ ) + L(τ )]dτ.

(7.158)

s

Since Z

Σ0ts

(ϕ˙ 2 hAu (t)ν, νi + ϕ2νA )]dΣ ≤ cγ

Z

s

t

[QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ )] dτ,

it follows from (7.157) and (7.158) that kϕk2L2 ((s,t),H 1 (Γ0 )) ≤ cγ Qλ (s) + cγ Qλ (t) Z t +cγ [Qλ (τ ) + QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ.

(7.159)

s

Moreover, the inequalities (7.51) and (7.52) in Theorem 7.3 imply that Z t 1 max{Qλ (t), Qλ (s)} ≤ Qλ (τ ) dτ t−s s Z t +cγ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ. (7.160) s

Inserting (7.160) into (7.159), and then inserting (7.159) into (7.154) yield Z t ⊎Σts ≤ εcγ [1 + 1/(t − s)] Qλ (τ ) dτ s Z t +cγ,ε [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ (7.161) s



365

where ε > 0 can be small and 0 ≤ s ≤ t ≤ T. Estimate on ⊎Qts It follows from the identities (7.120) and (7.131) that Dgu H(∇gu ϕ, ∇gu ϕ) = DH(∇gu ϕ, ∇gu ϕ) + η(∇gu ϕ, ∇gu ϕ) ≥ ̺0 c0,γ |∇gu ϕ|2gu − cγ L(t).

(7.162)

Noting that Z

Ω

|∇gu ϕ|2gu dx = (Au (t)∇ϕ, ∇ϕ),

from (7.150) and (7.162), we have Z Z 2 ⊎Qts ≥ 2̺0 c0,γ |∇gu ϕ|gu dQ + Qts

Qts

(ϕ˙ 2 − |∇gu ϕ|2gu ) div HdQ

Z t −cγ Qλ (s) − cγ Qλ (t) − cγ L(τ ) dτ s Z t Z ≥ ̺0 c0,γ ϑj dτ + 2 f (ϕ˙ 2 − |∇gu ϕ|2gu )dQ s

Qts

−cγ Qλ (s) − cγ Qλ (t) − cγ

t

Z

s

L(τ ) dτ

(7.163)

where 2 ϑj (t) = kϕ(t)k ˙ L2 (Ω) + (Au (t)∇ϕ, ∇ϕ),

f=

div H − ̺0 c0,γ . 2

On the other hand, using the identity (7.147), we obtain Z 2 f (ϕ˙ 2 − |∇gu ϕ|2gu ) dQ Qts

2

2

≤ cγ [Qλ (s) + Qλ (t)] + cγ (kϕ(s)k + kϕ(t)k ) + cγ Z t Z t 2 2 +cγ kϕkΓ0 + kϕνA kΓ0 dτ + cγ L(τ ) dτ. s

Z

s

t

kϕ(τ )k2 dτ (7.164)

s

Inserting (7.164) into (7.163) and using the relations (7.148) and (7.161) give Z

s

t

Z t ϑj (τ ) dτ ≤ εcγ [1 + 1/(t − s)] Qλ (τ )dτ s Z t +cε,γ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )]dτ s Z t +cε,γ Υ(s) + cε,γ Υ(t) + cε,γ Υ(τ )dτ s


(7.165)

366


for 1 ≤ j ≤ m − 1 where Υ(t) =

m−1 X j=0

ku(j) (t)k2

(7.166)

is a lower order term with respect to E(t). Furthermore, from (7.166), we have m−1 ˙ X (k+1) (k) (u , u ) ≤ εQλ (t) + cε Υ(t), Υ(t) = 2 k=0

which implies

max{Υ(s), Υ(t)} ≤ ε

Z

s

t

Qλ (τ ) dτ + [cε + 1/(t − s)]

Z

t

Υ(τ ) dτ

(7.167)

s

for 0 ≤ s < t ≤ T. Inserting (7.167) into (7.165), we obtain constants cγ > 0 and Tγ > 3 supx∈Ω |H|g /̺0 such that, if 0 ≤ s < t ≤ T and t − s ≥ Tγ , then Z t ϑj (τ ) dτ ≤ εcγ Qλ (τ ) dτ s s Z t +cγ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ ) + Υ(τ )] dτ

Z

t

(7.168)

s

for 1 ≤ j ≤ m − 1 and ε > 0 small. Step 2 This time, we consider a metric hu = Bu−1 (t) =

Z

0

1

aiyj (x, ∇w + τ ∇u)dτ

to replace the metric in (7.126), and use the system  utt = div Bu (t)∇u (t, x) ∈ Q,      u|Γ1 = 0 t ∈ (0, T ), ut + λuυB = I(t) (t, x) ∈ Σ0 ,   u  t − λuυB = O(t) (t, x) ∈ Σ0 ,   u(0, x) = u0 , ut (0, x) = u1 x ∈ Ω

−1

instead of the system (7.145). By repeating the procedure of Step 1, we obtain that the inequality (7.168) holds when ϑj (t) is replaced with ϑ0 (t) where ϑ0 (t) = kut (t)k2 + (Bu (t)∇u, ∇u). Finally, the inequality (7.143) follows from (7.168) and Lemma 7.11 below.



367

Lemma 7.11 Let all the assumptions of Theorem 7.10 hold. Suppose that cγ > 0 and Tγ > 3 supx∈Ω |H|g /̺0 such that the inequality (7.168) holds. If 0 ≤ s < t ≤ T and t − s ≥ Tγ , then Z t Z t Υ(τ ) dτ ≤ cγ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ (7.169) s

s

where Υ(t) is given by (7.166). Proof. We use an idea from [66]. It suffices to prove that for any solution u to the problem (7.114)-(7.115), Z Tγ Z Tγ [QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ. (7.170) Υ(τ ) dτ ≤ cγ 0

0

In fact, if the inequality (7.170) is satisfied, then for any h > 0, we have Z Tγ +h Z Tγ Υ(τ ) dτ = Υ(s + h) ds h

0

≤ cγ ≤ cγ

Tγ

Z

[QΓ0 ,R,I (s + h) + QΓ0 ,R,O (s + h) + L(s + h)] ds

0

Tγ +h

Z

h

[QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ.

(7.171)

So for s ≥ 0, s + mTγ ≤ t ≤ s + (m + 1)Tγ where m is an integer, we have Z t Z s+mTγ Z t Z s+Tγ Υ(τ ) dτ + Υ(τ ) dτ Υ(τ ) dτ + · · · + Υ(τ ) dτ = ≤ cγ

s+(m−1)Tγ

s

s

Z

s+Tγ

s Z t

≤ 2cγ

s

+···+

Z

s+mTγ

+

s+(m−1)Tγ

Z

t

t−Tγ

s+mTγ

[QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ

QΓ0 ,R,I (τ ) + QΓ0 ,R,O (τ ) + L(τ )] dτ.

We prove the inequality (7.170) by contradiction. Suppose that (7.170) does not hold for some γ0 > 0. Then, there exist initial data (φk0 , φk1 ), inputs Ik , and the corresponding solutions φk of the problem (7.114)-(7.115) over [0, T ] such that sup kφk kH m (Ω) ≤ γ0 , 0≤t≤T

Z

0

Tγ

Υk (τ ) dτ ≥ k

where Υk (t) =

Pm−1 j=0

Z

Tγ

0

[QIk ,Γ0 (τ ) + QOk ,Γ0 (τ ) + Lk (τ )] dτ

k(φk )(j) (t)k2L2 (Ω) , Lk (t) =

Ek (t) =

m X j=0

k(φk )(j) k2H m−j (Ω)


P2m

j/2

j=3

for

Ej (t), and

k ≥ 1.

(7.172)

368


Set c2k

=

Z

Tγ

ψk = φk /ck ,

Υk (τ ) dτ,

ϑk = Ok /ck .

θk = Ik /ck ,

0

Then

m−1 X Z Tγ 0

j=0

(j)

kψk (τ )k2L2 (Ω) dτ = 1

(7.173)

(7.174)

and by (7.172) m−1 X Z Tγ j=0

(j) (kθk k2L2 (Γ0 )

0

as k → ∞. Since Z

+

Tγ

0

Υk (τ ) dτ ≤

(j) kϑk k2L2 (Γ0 ) ) dτ

Z

Tγ

0

Ek (τ ) dτ ≤

the inequality (7.172) implies that infinity, which, in turn, shows that

R Tγ 0

2m Z 1 X Tγ j/2 Ek (τ ) dτ → 0 + 2 ck j=3 0

Z

Tγ1/2 (

Tγ 0

Lk (τ ) dτ )1/2 ,

Ek (τ ) dτ goes to zero as k goes to

ck → 0.

(7.175)

We divide both sides of the inequality (7.168) by c2k where we have set s = 0, t = Tγ , and ϕ = (φk )(j) to see that m−1 X Z Tγ j=0

0

(j+1)

[kψk

(j)

(j)

(τ )k2L2 (Ω) + (Aφk (τ )∇ψk (τ ), ∇ψk (τ ))L2 (Ω) ]dτ

are bounded for all k ≥ 1. Then there is p ∈ H 1 ((0, Tγ ) × Ω) such that (j)

weakly in

H 1 ((0, Tγ ) × Ω) ,

(7.176)

(j)

strongly in L2 ((0, Tγ ) × Ω)

(7.177)

ψk ⇀ p(j) ψk → p(j)

for 0 ≤ j ≤ m − 1 as k goes to infinity. Next, we divide the system (7.145) by c2k where we have set u = φk to show that p(j) ∈ H 1 ((0, Tγ ) × Γ) satisfy  ¨ = Ap(j) , (t, x) ∈ (0, T ) × Ω,  p(j) γ    p(j) = 0 (t, x) ∈ (0, T ) × Γ , γ 1 (7.178) ˙ + λp(j) (j)  p = 0 (t, x) ∈ (0, T γ ) × Γ0 , νA    (j) (j) p ˙ − λpνA = 0 (t, x) ∈ (0, Tγ ) × Γ0 , © 2011 by Taylor & Francis Group, LLC


369

for 0 ≤ j ≤ m − 1 where Ap(j) = div A(x, ∇w)∇p(j) ,

(j) p(j) , νi. νA = hA(x, ∇w)∇p

By the observability inequality for the linear wave equation with variable coefficients, Theorem 2.15, the formulas (7.178) imply that p(j) = 0 for

1 ≤ j ≤ m − 1.

Then p is a constant. Moreover, the assumption H3 implies that p = 0. These contradict the relations (7.174), (7.176), and (7.177). By a similar argument as in Lemma 7.11 the right hand side of the inequality (7.50) of Theorem 7.3 can be improved into Lemma 7.12 Let all the assumptions in Theorem 7.10 hold. Let γ > 0 be given and φ satisfy the problem (7.114)-(7.115) on the interval [0, T ] for T > 3 supx∈Ω |H|g /̺0 such that (7.125) are true. Then there is cγ > 0 such that E(t) ≤ cγ Qλ (t) + cγ EΓ0 ,R (t) + cγ L(t)

(7.179)

for T ≥ t ≥ 3 supx∈Ω |H|g /̺0 . Now we are ready to give proofs for Theorems 7.10 and 7.11. Proof of Theorem 7.10 Let (φ0 , φ1 ) ∈ H m (Ω) × H m−1 (Ω) and I ∈ m I ((0, ∞), Γ0 ) be given such that the problem (7.114)-(7.115) has a short time solution. We shall look for a constant η0 > 0 such that, if E(0) + kIk2I m ((0,∞),Γ0 ) ≤ η0 , then the solution of the problem (7.114)-(7.115) is global in time. In the rest of this section we take γ = 1. Let c0,1 > 0,

c1 ≥ 1,

T1 > 3 sup |H|g /̺0 x∈Ω

be given such that the estimates in Theorems 7.3 and 7.12 hold true. Let Ξ(η) =

2m−3 X

2 2 Θ(η) = 2eη max{c1 c−1 0,1 + 2c1 λ, 5c1 , 3c1 Ξ(1)η}

η k/2 ,

k=0

for η ∈ (0, ∞). Let 0 < η < min{1, Θ−2(2T1 )/4}

(7.180)

be given. We assume that E(0) + kIk2I m((0,∞),Γ0 ) ≤ η 3/2 < η.


(7.181)

370


Then there is some δ > 0 such that E(t) < η

(7.182)

for t ∈ [0, δ). Let δ0 > 0 be the largest number such that the estimate (7.182) holds true for t ∈ [0, δ0 ). We shall prove that δ0 = ∞. Step 1 We have δ0 > 2T1 . (7.183) Indeed, letting (7.183) be not true, i.e., δ0 ≤ 2T1 , we shall have a contradiction as follows. Using (7.47), (7.46), and (7.50), we have P(0) ≤ 2(c−1 0,1 + 2λ)E(0),

(7.184)

which gives, by (7.53), 2 3/2 kφ(t)k2 ≤ 2{(c−1 Ξ(1)}e2T1 0,1 + 2λ)E(0) + c1 kIkI m ((0,∞),Γ0 ) + 2c1 T1 η

2 3/2 ≤ 2e2T1 max{c−1 ] 0,1 + 2λ, c1 , 2c1 T1 Ξ(1)}[E(0) + kIkI m + η

(7.185)

since L(t) ≤ η 3/2 Ξ(η) ≤ η 3/2 Ξ(1) for 0 ∈ [0, δ0 ) by (7.182). In addition, we multiply (7.51) by 2, add it to (7.52), then integrate it over (0, t), and obtain 2 Qλ (t) ≤ max{c−1 0,1 , 10, 6c1 T1 Ξ(1)}[E(0) + kIkI m

((0,∞),Γ0 )

+ η 3/2 ].

(7.186)

Inserting the estimates (7.185) and (7.186) into the right hand side of the inequality (7.50) yields, via (7.180) and (7.181), E(t) ≤ Θ(2T1 )[E(0) + kIk2I m

((0,∞),Γ0 )

+ η 3/2 ] ≤ 2Θ(2T1)η 1/2 η < η

(7.187)

for 0 ≤ t ≤ δ0 , which contradicts the definition of δ0 . Step 2 Let T1 ≤ s < t < δ0 with t − s > T1 . Integrating (7.179) over (s, t) yields Z t Z t Z t E(τ )dτ ≤ c1 Qλ (τ )dτ + c1 kIk2I m ((s,t),Γ0 ) + c1 Ξ(1)η 1/2 E(τ )dτ, s

s

that is, Z

s

t

s

Z t Qλ (τ )dτ + kIk2I m((s,t),Γ0 ) ), E(τ )dτ ≤ c1 [1 − c1 Ξ(1)η 1/2 ]−1 ( s

which, in turn, gives Z t Z t 1/2 L(τ )dτ ≤ f (η )( Qλ (τ )dτ + kIk2I m ((s,t),Γ0 ) ) s

s

where f (x) = c1 Ξ(1)x[1 − c1 Ξ(1)x]−1


for 0 < x < 1,

(7.188)

Control of the Quasilinear Wave Equation in Higher Dimensions 371 Z t Z t kIk2I m((s,t),Γ0 ) = max EΓ0 ,R (τ ) + EΓ0 ,R (τ ) dτ + kI(m−1) (τ )k2Γ0 dτ. s≤τ 0 be given. We multiply (7.189) by κ, then add it to (7.190), and have Z t Qλ (t) + [κ − c1 f (η 1/2 )] Qλ (τ )dτ ≤ Qλ (s) s

+{κc1 [1 − c1 f (η 1/2 )]−1 [1 + f (η 1/2 )] + 4 + c1 f (η 1/2 )}kIk2I m((s,t),Γ0 ) Z t +{κc1 [1 − c1 f (η 1/2 )]−1 − 1} QΓ0 ,R,O (τ )dτ (7.191) s

for T1 ≤ s ≤ t < δ0 with t − s ≥ T1 . Letting κ = [1 − c1 f (η 1/2 )]/c1 in (7.191) gives Z t Qλ (t) + ω(η) Qλ (τ )dτ ≤ Qλ (s) s

+ [5 + (1 + c1 )f (η 1/2 )]kIk2I m ((s,t),Γ0 )

(7.192)

1/2 for T1 ≤ s ≤ t < δ0 with t − s ≥ T1 , where ω(η) = c−1 ). 1 − (1 + c1 )f (η Let τ0 = max{c1 c−1 0,1 Θ(2T1 ), c1 (7 + c1 ), c1 Ξ(1)}.

We fix η > 0 such that η 1/2
0.

(7.193)

Then when the condition (7.181) holds, the inequalities (7.179), (7.187), and (7.192) imply that the estimate (7.182) holds with δ0 = ∞. Indeed, if δ0 < ∞, then we will have a contradiction as follows. The condition 2c1 Ξ(1)η < 1 implies f (η 1/2 ) ≤ 1. Then using (7.192) with s = T1 in (7.179), we have 2 3/2 E(t) ≤ c1 c−1 0,1 E(T1 ) + c1 (7 + c1 )kIkI m ((s,t),Γ0 ) + c1 Ξ(1)η


(7.194)

372


for 0 ≤ t ≤ δ0 . Finally, we use the inequality (7.187) with t = T1 in (7.194) and obtain, by (7.181) and (7.193), E(t) ≤ τ0 [E(0) + kIk2I m ((s,t),Γ0 ) + η 3/2 ] ≤ 2τ0 η 1/2 η < η for all T1 ≤ t ≤ δ0 , which contradicts the definition of δ0 again. Proof of Theorem 7.11 Let T0 > 0 be such that I(t) = 0

for x ∈ Γ0 , t ≥ T0 .

(7.195)

By (7.192), we have Qλ (t) + ω(η)

Z

t

s

Qλ (τ )dτ ≤ Qλ (s)

(7.196)

for max{T1 , T0 } ≤ s < t < ∞ with t − s > T1 . Let ω0 = max{T1 , T0 } and t > ω0 . We multiply (7.196) by sk , integrate it from 0 to t − ω0 , and obtain Z t−ω0 Z (t − ω0 )k+1 ω(η) t−ω0 k+1 k s Qλ (s)ds ≥ Qλ (t) + s Qλ (s)ds k+1 k+1 0 0 for all k ≥ 0 which yields Q(t) ≤ Q(0)e−ω(η)(t−ω0 )

for t ≥ ω0 .

(7.197)

The estimate (7.124) follows from (7.179) and (7.197) since the condition (7.195) implies EΓ0 ,R (t) = 0 for t ≥ T0 .

7.4

Structure of Control Regions for Internal Feedbacks

After an internal local damping act on an escape region for the metric, we study the existence of global smooth solutions and the exponential decay of the energy for the quasi-linear wave equation when initial data are close to a given equilibrium. Introduction and Main Results Let n ≥ 2 be an integer, Ω ⊂ Rn be a bounded open set with smooth boundary Γ, and a(x, y) = (a1 (x, y), a2 (x, y), . . . , an (x, y)) be a smooth mapping from Ω × Rn to Rn with a(x, 0) = 0 for

x∈Ω

(7.198)

such that (aij (x, y)) is symmetrical and (aij (x, y)) > 0


for (x, y) ∈ Ω × Rn

(7.199)


373

where aij = aiyj are the partial derivatives of ai with respect to the variable y. Let f (x, s): Ω × R → R be a smooth function such that f (x, 0) = 0 for

x ∈ Ω.

(7.200)

We consider the problem:  ¨ x) − div a(x, ∇w + ∇φ) + f (x, φ) ˙ = 0 in (0, ∞) × Ω,  φ(t, φ = 0 on (0, ∞) × Γ,  ˙ x) = φ1 (x) on Ω φ(0, x) = φ0 (x), φ(0,

(7.201)

div a(x, ∇w) = 0 for

(7.202)

where w is an equilibrium solution, which satisfies

x ∈ Ω.

It is well known that solutions to the problem (7.201) usually develop singularities after some time even if initial data (φ0 , φ1 ) are small ([93]). Here we study what condition on f can guarantee that the problem (7.201) admits global smooth solutions when initial data (φ0 , φ1 ) are small. For this purpose, we assume that f satisfies fs (x, s) ≥ 0 where fs (x, s) = Let

for x ∈ Ω, |s| ≤ 1

(7.203)

∂f (x, s) . ∂s G = { x ∈ Ω | fs (x, 0) > 0}.

(7.204)

Definition 7.5 G is called the damping region of the system (7.201). The structure of G reflects the effect of the internal dissipation f. We seek geometric conditions on G such that the problem (7.201) has global smooth solutions in time and its energy decays. We note that G = Ω is one of the choices for such purposes (see Theorem 7.13). However, we are particularly interested in the case where G is not the whole domain Ω and is as small as possible in a technical sense. One of such conditions is an escape region for a metric, given by (7.33) below, see Corollaries (2.7)-(2.8). Remark 7.13 Problems like (7.201) have been extensively studied, and a wealth of results on this subject is available in the literature. Let m ≥ [n/2] + 3 be a given positive integer. We say that (φ0 , φ1 ) ∈ H m (Ω) × H m−1 (Ω) satisfies the compatibility conditions of order m if φ0 |Γ = 0 ,

φk |Γ = 0 ,

1≤k ≤m−1

where for k ≥ 2, φk = φ(k) (0) as computed formally (and recursively) in terms of φ0 and φ1 , using (7.201).


374


We seek geometric conditions on G such that solutions φ(t, x) to the problem (7.201) are in m \ C k ((0, +∞), H m−k (Ω)), k=0

if (u0 , u1 ) ∈ H m (Ω) × H m−1 (Ω) are small in (w, 0) ∈ H m (Ω) × H m−1 (Ω), and satisfies the compatibility conditions of order m with w. Let A(x, y) = (aij (x, y)) be the n × n matrix, given in (7.199), for each (x, y) ∈ Ω × Rn . We define g = A−1 (x, ∇w(x))

for x ∈ Ω

(7.205)

as a Riemannian metric on Ω and consider the couple (Ω, g) as a Riemannian manifold with the boundary Γ. Here the metric g depends not only on the coefficients aij (·, ·) but also on the equilibrium w. We denote the covariant differential of the metric g by D. If H is a vector field on Ω, then the covariant differential DH of H is a tensor field of rank 2 on Ω. We recall that G ⊂ Ω is said to be an escape region for the metric g if the following things are true: There exist ε > 0, ̺0 > 0, Ωi ⊆ Ω with C ∞ boundary ∂Ωi and vector fields H i , i = 1, 2, . . . , J, such that Ωi ∩ Ωj = ∅ for 1 ≤ i < j ≤ J and DH i (X, X) ≥ ̺0 |X|2g

for X ∈ Rxn , x ∈ Ωi ,

G ⊇ Ω ∩ Nε [ ∪Ji=1 Γi0 ∪ (Ω\ ∪Ji=1 Ωi ) ]

(7.206) (7.207)


S ⊂ Rn ,

Γi0 = { x ∈ ∂Ωi | H i (x) · ν i (x) > 0 },

(7.208)

and ν i (x) is the unit normal of ∂Ωi at x in the Euclidean metric of Rn , pointing towards the exterior of Ωi . For the structure of an escape region, see Section 2.3. We assume that short time solutions to the system (7.201) exist to consider the global smooth solutions in time and the decay of the energy. Theorem 7.13 Let w ∈ H m (Ω) be an equilibrium and let G be an escape region for the metric g. Suppose that (φ0 , φ1 ) ∈ H m (Ω) × H m−1 (Ω) satisfies the compatibility conditions of order m. Then for (φ0 , φ1 ) small in H m (Ω) × H m−1 (Ω), the system (7.201) has a global solution φ in m \

C k ((0, +∞), H m−k (Ω)) .

k=0



375

Moreover, there exist positive constants C > 0, σ > 0 such that E(t) ≤ Ce−σt where E(t) =

Pm

k=0

t≥0

for

(7.209)

kφ(k) k2H m−k (Ω) .

The proof of Theorem 7.13 will be given at the end of the following Subsection. Let us see an example. Example 7.5 Let a, w, and g be given by Example 7.4. Then the Gauss curvature κ ≤ 0. Consider the problem  φx1 x1 φx2 x2  ˙ = 0 in (0, ∞) × Ω,  − + f (x, φ)  φ¨ − 2 1 + (x2 + φx1 ) 1 + (x1 + φx2 )2  φ = 0 on (0, ∞) × Γ,   ˙ φ(0) = φ0 , φ(0) = φ1 on Ω.

Let H be an escape vector field on Ω. By Theorem 2.7, the damping region of the above problem can be supported on a neighborhood of Γ0 = { x ∈ Γ | hH, νi > 0 }.

Proofs of the Main Results We assume that solution φ to the system (7.201) exists for some T > 0 where initial data (φ0 , φ1 ) are in H m (Ω) × H m−1 (Ω) satisfying the compatibility conditions of order m. Denote Bφ (t) = aij (x, ∇w + ∇φ) for (t, x) ∈ Q; (7.210) then

˙ [a(x, ∇w + ∇φ)]′ = Bφ (t)∇φ,

(7.211)

and for j ≥ 2, [a(x, ∇w + ∇φ)](j) = Bφ (t)∇φ(j) +

j−1 X

(k)

Ck Bφ (t)∇φ(j−k) .

(7.212)

k=1

˙ with respect to t and obtain We compute the j-th derivatives of f (x, φ) ˙ (j) = (f (x, φ))

j X

X

i=1 l1 +···+li =j

˙ φ˙ (l1 ) φ˙ (l2 ) · · · φ˙ (li ) fs(i) (x, φ)

˙ φ˙ (j) + = fs (x, φ)

j X

X

i=2 l1 +···+li =j

(7.213)

˙ φ˙ (l1 ) φ˙ (l2 ) · · · φ˙ (li ) . fy(i) (x, φ)

We define Bφ (t)v = div Bφ (t)∇v


for

v ∈ H 2 (Ω)

(7.214)

376


and for v ∈ H 2 (Ω), x ∈ Γ.

vνB = hBφ (t)∇v, νi

(7.215)

Then (Bφ (t)v, φ) = −(Bφ (t)∇v, ∇φ)

for

v, φ ∈ H 2 (Ω) ∩ H01 (Ω).

(7.216)

For T > 0, differentiating the system (7.201) j times with respect to t, we get

˙ + r (t) = 0 in Q∞ , ˙ φ(j) φ¨(j) − Bφ (t)φ(j) + fs (x, φ) j (j) ∞ φ = 0 on Σ

(7.217)

where rj (t) is defined by rj (t) =

j X

X

i=2 l1 +···+li =j

−

j−1 X

˙ φ˙ (l1 ) φ˙ (l2 ) · · · φ˙ (li ) fy(i) (x, φ)

(k)

div Bφ (t)∇φ(j−k)

(7.218)

k=1

for 2 ≤ j ≤ m − 1 and r1 (t) = 0. Similar arguments as in the proof of Lemma 7.7 yield Lemma 7.13 Let γ > 0 be given and let φ be a solution to the problem (7.201) on [0, T ) such that ˙ sup kφ(t)k H m−1 (Ω) ≤ γ.

sup kφ(t)kH m (Ω) ≤ γ,

0≤t≤T

(7.219)

0≤t≤T

Then for 2 ≤ j ≤ m − 1, there exists a constant cγ > 0, depending on γ, such that m−1 X 2 krj (t)kH m−1−j (Ω) ≤ cγ E k (t) (7.220) k=2

where

E(t) =

m X

k=0

kφ(k) (t)k2H m−k (Ω) .

(7.221)

For the system (7.217) with 1 ≤ j ≤ m − 1, we define the corresponding energy as ˙ (t)k2 + (B (t)∇φ(j) (t), ∇φ(j) (t)). ϑj (t) = kφ(j) (7.222) φ Moreover, we introduce an operator Nφ (t)v = div Nφ (t)∇v where Nφ (t) =

Z

0


1

for

v ∈ H 2 (Ω)

aij (x, ∇w + s∇φ) ds.

(7.223)

Control of the Quasilinear Wave Equation in Higher Dimensions Then the system (7.201) can be rewritten as ˙ = 0 in Q∞ , φ¨ − Nφ (t)φ + f (x, φ) ∞ φ = 0 on Σ

377

(7.224)

with the energy defined as 2 ˙ ϑ0 (t) = kφ(t)k + (Nφ (t)∇φ(t), ∇φ(t)).

(7.225)

Next, we define ϑ(t) =

m−1 X j=0

U(f, φ) =

ϑj (t),

L(t) =

2m X

k=3

E k/2 (t),

m−1 X

˙ (j+1) , φ(j+1) ) + (f (x, φ), ˙ φ), ˙ (fs (x, φ)φ

(7.226)

j=1

2 ˙ P(t) = kφ(t)k2 + kφ(t)k + (Nφ (t)∇φ(t), ∇φ(t)).

Theorem 7.14 Let γ > 0 be given and let φ be a solution to the problem (7.201) on the interval [0, T ] for some T > 0 such that the conditions (7.219) hold true. Then there are constants c0,γ > 0 and cγ > 0, which depend only on γ, such that for 0 ≤ t ≤ T, c0,γ ϑ(t) ≤ E(t) ≤ cγ ϑ(t) + cγ L(t),

(7.227)

˙ ≤ cγ L(t) + 2U(f, φ), −ϑ(t)

(7.228)

˙ + cγ L(t) ≥ 2c0,γ U(f, φ), −ϑ(t) ˙ ≤ cγ L(t), ϑ(t)

(7.229) (7.230)

Proof. Clearly, the conditions (7.219) imply that c0,γ Q(t) ≤ ϑ(t) ≤ c0,γ Q(t) for 0 ≤ t ≤ T where Q(t) is defined in (7.9). Then the estimate (7.227) follows from the inequality (7.12) in Theorem 7.1 since L(t) ≤ L(t) and EΓ0 ,D (t) = 0 for the problem (7.201). We differentiate ϑj (t) and have ϑ˙ j (t) = 2(φ(j+2) , φ(j+1) ) + 2(Bφ (t)∇φ(j) , ∇φ(j+1) ) +(B˙ φ (t)∇φ(j) , ∇φ(j) ).

(7.231)

Using the formulas (7.216) and (7.217) in (7.231), and by Lemma 7.13, we obtain −ϑ˙ j (t) = −(B˙ φ (t)∇φ(j) , ∇φ(j) ) + 2(rj (t), φ(j+1) (t)) ˙ (j+1) , φ(j+1) ) +2(fs (x, φ)φ ˙ (j+1) , φ(j+1) ) ≤ cγ L(t) + 2(fs (x, φ)φ


(7.232) (7.233)

378


for 1 ≤ j ≤ m − 1. In addition, the identity (7.232) also gives ˙ k2 + 2(f (x, φ)φ ˙ (j+1) , φ(j+1) ) −ϑ˙ j (t) ≥ −cε,γ L(t) − εkφ(j) s

(7.234)

for ε > 0 small. We have, by (7.224), ˙ φ). ˙ −ϑ˙ 0 (t) = −(N˙ φ (t)∇φ, ∇φ) + 2(f (x, φ), The inequalities (7.228)-(7.230) follow from (7.233)-(7.235).

(7.235)

Let φ satisfy the problem (7.217) on the interval [0, T ] for some T > 0. For t ∈ [0, T ], let gφ be the metric on Ω given by gφ = Bφ−1 (t)

(7.236)

where the matrix Bφ (t) is defined by (7.210). Consider the pair (Ω, gφ ) as a Riemannian manifold for fixed t ∈ [0, T ). ˆ ⊂ Ω be given. We apply the identities (2.13) and (2.14) to the Let Ω ˆ to have problem (7.217) and integrate them over (s, t) × Ω Lemma 7.14 Let φ(j) be a solution to (7.217) for 2 ≤ j ≤ m − 1 and let ˆ ⊆ Ω be a subset. Suppose that H and h are a vector field and a function on Ω ˆ respectively. Then Ω, Z tZ ˆ Φ(H, h, ∂ Ω) = 2 Dgφ H(∇gφ φ(j) , ∇gφ φ(j) )dxdτ s

ˆ Ω

ˆ +Φ(H, h, Ω)

(7.237)

where Z tZ

˙ )2 − |∇ φ(j) |2 hH, νi dxdτ (j) 2H(φ(j) )φ(j) + ( φ gφ νB gφ ˆ s ∂Ω Z tZ − (φ(j) )2 hνB − 2hφ(j) φ(j) (7.238) νB dxdτ,

ˆ = Φ(H, h, ∂ Ω)

s

ˆ ∂Ω

Z tZ ˙ , M(φ(j) ) t − ˆ = 2 φ(j) Φ(H, h, Ω) (φ(j) )2 Bφ (τ )h dxdτ s ˆ s Ω Z tZ i ˙ M(φ(j) )dxdτ ˙ φ(j) +2 rj (t) + fs (x, φ) ˆ s Ω Z tZ ˙ 2 − |∇ φ(j) |2 p(H, h)dxdτ, + φ(j) (7.239) gφ gφ s

ˆ Ω

M(φ(j) ) = H(φ(j) ) + hφ(j) ,


p(H, h) = ( div H − 2h).


379

Lemma 7.15 Let G, given by (7.204), be an escape region for the metric g. Let γ > 0 be given and let φ be a solution to the problem (7.201) on the interval [0, T ] for some T > 0 such that the conditions (7.219) are true. Then there exist constants cγ > 0 and Tγ > 3 supx∈Ω¯ |H|g /̺0 such that, if 0 ≤ s ≤ t ≤ T with t − s ≥ Tγ , then Z

t s

ϑ(τ )dτ ≤ cγ

where R(t) =

Pm−1 j=0

Z

t

L(τ )dτ + cγ

s

Z

t

R(τ )dτ + cγ

s

Z

s

t

U(f, φ)dxdτ (7.240)

kφ(j) (t)k2 and U(f, φ) is given by (7.226).

Proof. For 0 < ε2 < ε1 < ε0 < ε, set Qk = Nεk [∪Ji=1 Γi0 ∪ (Ω\ ∪Jj=1 Ωi )]

for k = 0, 1, 2 .

(7.241)

Obviously we have Q2 ⊂ Q1 ⊂ Q0 ⊂ G .

(7.242)

Let β i , i = 1, . . . , J satisfy i

β ∈

C0∞ (Rn )

for

i

0 ≤ β ≤ 1,

i

β =

1 on Ωi \ Q1 , 0 on Q2 .

(7.243)

For each i, 1 ≤ i ≤ J, set ˆ := Ωi , H := β i H i , h := 1 div (β i H i ) = hi . Ω 2

(7.244)

Step 1 For 1 ≤ j ≤ m − 1, we estimate ϑj (t). (a) We estimate the integral Z tZ s

Ω\Q1

|∇gφ φ(j) |2gφ dxdτ

where |∇gφ φ(j) |2gφ = Bφ (t)∇φ(j) , ∇φ(j) . It follows from (7.237) that 2

Z tZ s

Dgφ (β i H i )(∇gφ φ(j) , ∇gφ φ(j) )dxdτ

Ωi i i

= Φ(β H , hi , ∂Ωi ) − Φ(β i H i , hi , Ωi ).

(7.245)

We show that Φ(β i H i , hi , ∂Ωi ) ≤ 0. In fact, we notice that ∂Ωi = [Γi0 ∪ ∂Ωi \ Γ ] ∪ (∂Ωi \ Γi0 ) ∩ Γ ≡ I1 ∪ I2 .


(7.246)

380


Since ∂Ωi \ Γ ⊆ Ω \ ∪Ji=1 Ωi , we have β i = 0 in I1 ⊆ Q2 . Since I2 ⊆ Γ and φ(j) = 0 on I2 , (j) 2 i i i β i H i (φ(j) )φ(j) νB = |∇gφ φ |gφ hβ H , ν i

for x ∈ I2 .

We get Z tZ s

=

Z

s

∂Ωi tZ

i

i

(j)

2β H (φ

(∂Ωi \Γi0 )∩ Γ

)φ(j) νB

+

˙ (φ(j)

2

−

|∇gφ φ(j) |2gφ )hβ i H i , νi

∇gφ φ(j) 2 hβ i H i , νidxdτ ≤ 0 g

dxdτ (7.247)

φ

i i since, by (7.208), hβ ∈ ∂Ωi \ Γi0 . Furthermore, we decompose S H , νi T ≤ 0 for x S ∂Ωi = (∂Ωi \ Ω) (∂Ωi Ω) ≡ J1 J2 . Since J1 ⊆ Γ, by the boundary condition, we get Z tZ 2 (7.248) φ(j) (divβ i H i )νB − 2div(β i H i )φ(j) φ(j) νB dxdτ = 0. s

J1

It is easy to verify that J2 = ∂Ωi ∩ Ω ⊆ Ω \ ∪Ii=1 Ωi ⊆ Q2 and Z tZ (j) 2 i i i i (j) (j) φ (divβ H )νB − 2div(β H )φ φνB dxdτ = 0. s

(7.249)

J2

The estimate (7.246) is from (7.247), (7.248), (7.249) and (7.238). ˙ ≥ 0, we obtain, from Lemma Next, noting that p(H i , hi ) = 0 and fs (x, φ) 7.13 and the formula (7.239), that |Φ(β i H i , hi , Ωi )| ≤ cγ [ϑj (s) + ϑj (t)] Z t Z t Z t +ε ϑj (τ )dτ + cγ kφ(j) k2 dτ + cε,γ L(τ )dτ s s s Z tZ ˙ φ˙ (j) ]2 dxdτ +cε,γ fs (x, φ)[ (7.250) s

Ωi

for ε > 0 small and 1 ≤ j ≤ m − 1 where the estimate kφ(j) k2 ≤ cγ k∇gφ φ(j) k2 ≤ cγ ϑj has been used. Moreover, it follows from Lemma 7.10, the relations (7.245), (7.246), and (7.250) that Z tZ 2̺0 |∇gφ φ(j) |2gφ dxdτ s

≤

Ω\Q1

J Z t X i=1

s

Z

Ωi

2Dgφ (β i H i )(∇gφ φ(j) , ∇gφ φ(j) )dxdτ + cγ

Z t Z t ≤ cγ [ϑj (s) + ϑj (t)] + ε ϑj (τ )dτ + cγ kφ(j) k2 dτ s s Z tZ Z t (j) 2 ˙ φ˙ ] dxdτ + cε,γ +cε,γ fs (x, φ)[ L(τ )dτ s

Ω


s

Z

s

t

L(τ )dτ

(7.251)


381

since Ω/Q1 ⊂ ∪Ji=1 Ωi . (b) We estimate Z tZ s

Ω∩Q1

|∇gφ φ(j) |2gφ dxdτ .

Let ξ ∈ C0∞ (Rn ) such that 0 ≤ ξ ≤ 1 and ξ = 0 for

x ∈ R n \ Q0 ;

x ∈ Q1 .

ξ = 1 for

We multiply the equation in (7.217) by ξφ(j) , integrate it by parts, and obtain Z tZ

Z tZ ˙ ]2 dxdτ ξ|∇gφ φ(j) |2gφ dxdτ = −(φ˙ (j) , ξφ(j) )|ts + ξ[φ(j) s Ω s Ω Z tZ Z tZ (j) (j) ˙ ξφ(j) dxdτ ˙ φ(j) − φ h∇gφ φ , ∇gφ ξigφ dxdτ − fs (x, φ) s Ω s Ω Z tZ − rj (τ )ξφ(j) dxdτ s

Ω

which yields Z tZ s

+cγ

Ω∩Q1 Z tZ s

+cγ

˙ ]2 dxdτ + ε [φ(j)

Q0

Z tZ s

|∇gφ φ(j) |2gφ dxdτ ≤ cγ [ϑj (s) + ϑj (t)] Z

t

ϑj (τ )dτ + cε,γ

s

Z

t

s

kφ(j) k2 dτ

˙ ]2 dxdτ. ˙ φ(j) fs (x, φ)[

(7.252)

Ω

(c) Furthermore, we need to estimate Z tZ s

2

˙ ) dxdτ. (φ(j)

Q0

By (7.204) and (7.242), we see that there exists c0 > 0 such that fs (x, 0) ≥ c0 > 0

for x ∈ Q0 .

(7.253)

We can take γ small in (7.219), which can guarantee that ˙ ≥ c0 2fs (x, φ) so that we have Z tZ c0 s

Q0

2

˙ ) dxdτ ≤ 2 (φ(j)


for x ∈ Q0 ,

Z tZ s

Ω

(7.254)

2

˙ ) dxdτ. ˙ φ(j) fs (x, φ)(

(7.255)

382


(d) From the relations (7.251), (7.252), and (7.255), we obtain 2̺0

m−1 XZ t s

j=1

Z

Ω

|∇gφ φ(j) |2gφ dxdτ

≤ cγ [ϑ(s) + ϑ(t)] + ε +cε,γ

m−1 XZ t i=1

s

Z

Z

t

ϑ(τ )dτ + cγ

s

Z

t

R(τ )dτ

s

˙ φ˙ (j) ]2 dxdτ + cε,γ fs (x, φ)[

Ω

Z

t s

L(τ ).

(7.256)

Step 2 We shall estimate ϑ0 (t). We define gˆφ = Nφ−1 (t) and replace gφ in Step 1 by gˆφ . By a similar process, we obtain Z tZ Z t Z t 2 |∇gˆφ φ|gˆφ dxdτ ≤ cγ [ϑ0 (s) + ϑ0 (t)] + ε ϑ0 (τ )dτ + cγ kφk2 dτ s Ω s s Z tZ Z t ˙ φdxdτ ˙ +cε,γ f (x, φ) + cε,γ L(τ ). (7.257) s

Ω

s

Step 3 We multiply the equation in (7.217) by φ(j) , integrate it over (s, t) × Ω, and obtain for 1 ≤ j ≤ m − 1 Z t Z tZ ˙ k2 dτ ≤ kφ(j) |∇gφ φ(j) |2gφ dxdτ + ϑj (s) + ϑj (t) s s Ω Z tZ Z t ˙ ]2 dxdτ + c ˙ φ(j) + fs (x, φ)[ [L(τ ) + kφ(j) k2 ]dτ. (7.258) γ s

Ω

s

Similarly, we have Z t Z tZ ˙ 2 dτ ≤ kφk |∇gˆφ φ|2gˆφ dxdτ + ϑ0 (s) + ϑ0 (t) s s Ω Z tZ Z t ˙ φdxdτ ˙ + f (x, φ) + cγ [L(τ ) + kφk2 ]dτ. s

Ω

(7.259)

s

Now combining the relations (7.256)-(7.259) yields Z t Z t ϑ(τ )dτ ≤ cγ [ϑ(s) + ϑ(t)] + cγ [R(τ ) + L(τ )]dτ s s Z t +cγ U(f, φ)dτ.

(7.260)

s

Step 4 From the inequalities (7.228) and (7.230), we have Z t Z t 1 max{ϑ(t), ϑ(s)} ≤ ϑ(τ )dτ + cγ L(τ )dτ t−s s s Z t + 2 U(f, φ))dτ. s


(7.261)


383

Let Tγ > max{2cγ , 3 supx∈Ω¯ |H|g /̺0 } be given. Inserting the inequality (7.261) into the inequality (7.260), we obtain the inequality (7.240). An argument as in the proof of Lemma 7.11 yields Lemma 7.16 Let all assumptions in Lemma 7.15 hold. Then we have Z t Z t Z t R(τ )dτ ≤ cγ L(τ )dτ + cγ U(f, φ)dτ (7.262) s

s

s

for t − s ≥ T0 , where T0 ≥ 3 supx∈Ω¯ |H|g /̺0 and U(f, φ) is given in (7.226). Next, we shall establish the following Theorem 7.15 Let G be an escape region for the metric g. Let γ > 0 be given and let φ be a solution to the problem (7.201) on the interval [0, T ] for some T > 0 such that the conditions (7.219) are true. Then there are constants cγ > 0 and Tγ > 3 supx∈Ω¯ |H|g /̺0 such that for 0 ≤ s ≤ t ≤ T with t − s ≥ Tγ , Z t Z t ϑ(τ )dτ + cγ ϑ(t) ≤ cγ ϑ(s) + cγ L(τ )dτ (7.263) s

s

where L(t) and ̺0 are as given in Theorem 7.14 and (7.206), respectively. Proof. Using the inequality (7.262) in the inequality (7.240) gives Z t Z t Z t ϑ(τ )dτ ≤ cγ L(τ )dτ + cγ U(f, φ)dτ. (7.264) s

s

s

On the other hand, from the inequality (7.229), we obtain Z t Z t U(f, φ)dxdτ ≤ cγ [ϑ(s) − ϑ(t)] + cγ L(τ )dτ. s

(7.265)

s

Finally inserting the inequality (7.265) into the inequality (7.264), we obtain the inequality (7.263). Proof of Theorem 7.13 We assume that all the notation remains the same as before. Let (φ0 , φ1 ) ∈ H m (Ω) × H m−1 (Ω) be given such that the problem (7.201) has a short time solution. We look for a constant η0 such that, if E(0) ≤ η0 , (7.266) then solutions of the system (7.201) are global. Throughout this section we take γ = 1. Let 0 < c0,1 ≤ 1, c1 ≥ 1, and T1 >3 supx∈Ω¯ |H|g /̺0 be given according to γ = 1 in Theorems 7.14 and 7.15. Let 2m−3 X Ξ(η) = η k/2 , (7.267) k=0


384

Modeling and Control in Vibrational and Structural Dynamics 2 Θ(η) = 2 max{c1 c−1 0,1 , 2c1 Ξ(1), c1 Ξ(1)η}

(7.268)

0 < η < min{1, Θ−2(2T1 )/4}

(7.269)

for η ∈ (0, ∞). Let be given. We assume that E(0) ≤ η 3/2 < η.

(7.270)

Then there is some δ > 0 such that E(t) < η

(7.271)

for t ∈ [0, δ). Let δ0 > 0 be the largest number such that inequality (7.271) holds for t ∈ [0, δ0 ). We shall show that δ0 = +∞. Step 1 We claim δ0 > 2T1 . Suppose that this is not true, i.e., δ0 ≤ 2T1 . We have a contradiction as follows. Using the inequalities (7.227) and (7.230), we obtain ϑ(0) ≤ c−1 0,1 E(0),

3/2 ϑ(t) ≤ c−1 Ξ(1), 0,1 E(0) + 2c1 T1 η

for 0 ≤ t ≤ δ0 ≤ 2T1 , where the relations (7.270) and L(t) =

2m X

k=3

E k/2 (t) ≤ η 3/2 Ξ(η) ≤ η 3/2 Ξ(1)

(7.272)

have been used. From (7.227) and (7.270)-(7.272), we obtain 3/2 E(t) ≤ c1 c−1 0,1 E(0) + c1 (2c1 T1 + 1)Ξ(1)η

2 3/2 ≤ 2 max{c1 c−1 ] 0,1 , 2c1 T1 Ξ(1), 2c1 Ξ(1)}[E(0) + η

≤ Θ(2T1 )[E(0) + η 3/2 ] ≤ 2Θ(2T1 )η 1/2 η < η

(7.273)

for all t ∈ [0, δ0 ]. This contradicts the definition of δ0 . Step 2 Next, we prove that δ0 = +∞ by contradiction again. Let T1 ≤ s < t < δ0 with t − s ≥ T1 . Integrating the inequality (7.227) over interval (s, t) yields Z t Z t Z t E(τ )dτ ≤ c1 ϑ(τ )dτ + c1 Ξ(1)η 1/2 E(τ )dτ, s

s

s

that is, Z

s

t

E(τ )dτ ≤ c1 [1 − c1 Ξ(1)η 1/2 ]−1

Z

t

ϑ(τ )dτ,

s

which, in turn, shows that Z t Z t Z t E(τ )dτ ≤ h(η 1/2 ) ϑ(τ )dτ L(τ )dτ ≤ Ξ(1)η 1/2 s


s

s

(7.274)


385

where h(x) = c1 Ξ(1)x[1 − c1 Ξ(1)x]−1

0 ≤ x ≤ 1.

for

Furthermore, combining the inequalities (7.263) and (7.274), we obtain Z t Z t c1 ϑ(t) + ϑ(τ )dτ ≤ c1 ϑ(s) + c1 h(η 1/2 ) ϑ(τ )dτ , s

s

that is, ϑ(t) + w(η)

Z

t s

ϑ(τ )dτ ≤ ϑ(s)

(7.275)

1/2 where w(η) = c−1 ). 1 − h(η We fix η > 0 small such that −1 η 1/2 ≤ 2c1 c−1 , 0,1 Θ(2T1 ) + c1 Ξ(1)]

w(η) > 0.

If δ0 < ∞, we find a contradiction as follows. First, the inequality (7.275) implies that ϑ(t) ≤ ϑ(T1 ) for 2T1 ≤ t ≤ δ0 . (7.276) From (7.227), (7.276) and (7.273), we obtain 3/2 E(t) ≤ c1 ϑ(T1 ) + c1 Ξ(1)η 3/2 ≤ c1 c−1 0,1 E(T1 ) + c1 Ξ(1)η 1/2 ≤ [2c1 c−1 η

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