Kernel Determination Problems in Hyperbolic Integro-Differential Equations 9819922593, 9789819922598

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Infosys Science Foundation Series in Mathematical Sciences

Durdimurod K. Durdiev Zhanna D. Totieva

Kernel Determination Problems in Hyperbolic Integro-Differential Equations

Infosys Science Foundation Series

Infosys Science Foundation Series in Mathematical Sciences Series Editors Gopal Prasad, University of Michigan, Ann Arbor, USA Irene Fonseca, Carnegie Mellon University, Pittsburgh, PA, USA Editorial Board Chandrashekhar Khare, University of California, Los Angeles, USA Mahan Mj, Tata Institute of Fundamental Research, Mumbai, India Manindra Agrawal, Indian Institute of Technology Kanpur, Kanpur, India Ritabrata Munshi, Tata Institute of Fundamental Research, Mumbai, India S. R. S. Varadhan, New York University, New York, USA Weinan E, Princeton University, Princeton, USA

The Infosys Science Foundation Series in Mathematical Sciences, a Scopusindexed book series, focuses on high-quality content in the domain of pure and applied mathematics, biomathematics, financial mathematics, operations research and theoretical computer science. With this series, Springer hopes to provide readers with monographs, textbooks, edited volume, handbooks and professional books of the highest academic quality on current topics in relevant disciplines. Literature in this series will appeal to a wide audience of researchers, students, educators and professionals across the world.

Durdimurod K. Durdiev · Zhanna D. Totieva

Kernel Determination Problems in Hyperbolic Integro-Differential Equations

Durdimurod K. Durdiev The Institute of Mathematics Academy of Sciences of the Republic of Uzbekistan Tashkent, Uzbekistan

Zhanna D. Totieva Southern Mathematical Institute Vladikavkaz Scientific Center Russian Academy of Sciences Vladikavkaz, Russia

ISSN 2363-6149 ISSN 2363-6157 (electronic) Infosys Science Foundation Series ISSN 2364-4036 ISSN 2364-4044 (electronic) Infosys Science Foundation Series in Mathematical Sciences ISBN 978-981-99-2259-8 ISBN 978-981-99-2260-4 (eBook) https://doi.org/10.1007/978-981-99-2260-4 Mathematics Subject Classification: 35L20, 35R30, 35Q99 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

To the memory of our parents

Preface

Mathematical physics usually studies well-posed problems that have solutions, which are unique and stable to small changes in the data in suitable functional spaces. Such problems, as a rule, are called direct problems. In this case, it is assumed that a differential equation is specified, as well as certain initial and boundary conditions. However, in applications there are interesting problems, where the differential (or integro-differential) equation is only partially specified, namely some functions that are part of the differential equation (either in the right-hand side, integrant in integrodifferential equations or the initial and boundary conditions) remain unknown. The problems, in which these unknowns are to be determined on some information about the solutions of direct problems for differential equations, are called inverse problems. According to the current accepted terminology, the inverse problem is called one-dimensional if unknown functions depend only on one variable, and multidimensional if they are functions of several variables. Inverse problems of mathematical physics can be divided into groups depending on the additional information that is given; which functions are unknown; what differential equations are considered. The most frequently used is the classification of additional information. According to the type of additional information asked about the solution of the direct problem, the inverse problems for hyperbolic equations can be divided into three main groups: kinematic, spectral, and dynamic [1, 2]. In this book, the questions of constructing methods for solving one-dimensional and multidimensional inverse dynamical problems for hyperbolic equations with memory are studied, and furthermore, the theorems of uniqueness, stability, and existence of solutions of these inverse problems are obtained. The problems of memory restoration in hyperbolic equations are a relatively new direction in the theory of inverse problems, which arose at the end of the 80s of the last century. It is mainly represented by the works of Italian mathematicians A. Lorenzi, M. Grasselli, E. Paparoni, and others [3–7]. After 2000, this trend has been developed in the works of V. G. Romanov, A. L. Bukhgeym, S. I. Kabanikhin, V. G. Yakhno, J. Janno, L. von Wolfersdorf, and others [8–18, 6, 19–31]. Under certain conditions, convolutiontype operators that describe “prehistory” or “memory phenomena” are added to the equations for the spread of elastic and electromagnetic waves. vii

viii

Preface

Let us explain the example of the spread of electromagnetic waves. In rapidly varying electromagnetic fields, whose frequencies are not limited by the smallness condition as compared with the frequencies characteristic of establishing the electric and magnetic polarization of a substance, the unique dependence of D and B (induction of electric and magnetic fields, respectively) on the meanings of E and H (intensity of the corresponding fields ) at the same time. It turns out that the meanings of D and B at a given time depend not only on E and H, but also on the whole background of the action of these fields (such a medium is called an environment with aftereffect) [32]. This circumstance is an expression of the fact that the establishment of the electric and magnetic polarization of a substance does not manage to follow the change in the electromagnetic field. The most general view of the linear relationship between D(x, t), B(x, t) and the corresponding meanings of the functions E(x, t), H(x, t) at all previous points in time can be written as integral relations [32]: 

t

D(x, t) = εE +

ϕ(t − τ )E(x, τ )dτ

0



t

B(x, t) = μH +

ψ(t − τ )H (x, τ )dτ,

(1)

0

E = (E 1 , E 2 , E 3 ), H = (H1 , H2 , H3 ), D = (D1 , D2 , D3 ), B = (B1 , B1 , B3 ), x = (x1 , x2 , x3 ) ϕ(t) = diag (ϕ1 , ϕ2 , ϕ3 ), ψ(t) = diag (ψ1 , ψ2 , ψ3 ) are diagonal matrix functions representing memory. These functions are finite for all values of their argument and tend to zero at t → ∞. The latter circumstance is an expression of the fact that the values of D(x, t), B(x, t) at a given time cannot be noticeably influenced by the values of E(x, t), H(x, t) at very old moments. The physical mechanism underlying the integral dependencies of the form (1) is in the process of establishing electromagnetic polarization. Therefore, the interval of values in which the functions ϕ(t), ψ(t) differ noticeably from zero, the magnitude of the relaxation time characterizing the rate of these processes. Let, in accordance with equalities (1), the set of vectors E, H be the solution of the Cauchy problem for the system of Maxwell equations with zero initial data: ∇ × H = (∂/∂t)D(x, t) + σ E + j, ∇ × E = −(∂/∂t)B(x, t), (x, t) ∈ R4 , (E, H )|t 0.

(15)

i=0

Additionally, it is assumed that the coefficients aij are sufficiently smooth and are such that for each pair of points x, ξ ; ξ = (ξ1 , ξ2 , ξ3 ) a unique geodesic (x, ξ ), is corresponded, connecting them in a metric in which an element of length ds is found by the formula ds =



bi j (x)d xi d x j ,

(16)

i, j=1

  where bij are the matrix elements, which are the inverse of the matrix A = ai j . We denote by τ (x, ξ ) the distance in the metric (16) between the points x, ξ  τ (x, ξ ) =



(x, ξ )

bi j (x)d xi d x j .

i, j=1

The physical meaning of the function τ (x, ξ )is the travel time of the perturbation produced at the point ξ to the point x. Obviously τ (x, ξ ) = τ (ξ, x). S. L. Sobolev established the existence of the function σ (x, ξ ), which satisfies the equation in the first argument

2

3  i, j=1

⎡ ai j τxi σx j + σ ⎣

3 

i, j=1

ai j τxi x j +

3  i=1

⎛ ⎝2

3  ∂ai j j=1

∂x j

⎞

⎤

− b j ⎠τxi ⎦ = 0

(17)

xiv

Preface

possessing a number of remarkable properties [101] (p. 63), due to which the problem (13), (15) is reduced to an integral equation ¨  1 σ (ξ, x)Nξ ϕ(ξ ) − ϕ(ξ )Nξ σ (ξ, x) + σ (ξ, x) 4π S(x, t)   × ψ(ξ )Nξ τ (ξ, x) + ϕ(ξ )Rξ d Sξ ˚ 1 + σ (ξ, x) f (ξ, t − τ (ξ, x))dvξ 4π D(x, t) ˚ 1 + L˜ ∗ξ σ (ξ, x)u(ξ, t − τ (ξ, x))dvξ . 4π D(x, t)

u(x, t) =

Here ξ is the variable point of integration, S(x, t) is the quasisphere τ (x, ξ ) = t, D(x, t)) is the volume bounded by this quasisphere, dS, dv are the elements of surface area and the volume respectively, L˜ ∗ is an adjoint operator, according to Lagrange, with an operator ˜ = Lu

3 

ai j (x)u xi x j +

i, j=1

3 

bi (x)u xi + c(x)u,

(18)

i=1

N is the linear differential operator N ϕ(x) =

3 

ai j

i, j=1

3 

ai j ϕxi cos(n, xi ),

i, j=1

The index ξ of the operators L˜ ∗ , N shows that they are applied with respect to the variable ξ , R=

3  i=1

⎛ ⎝bi −

3  ∂ai j j=1

∂ xi

⎞ ⎠ cos(n, xi ),

n is the unit vector of the outer normal to the surface S(x, t). In the case when L˜ = L that is ai j = δi j (δi j is the Kronecker symbol), to all the conditions listed in [101] (p. 63) and the equation (19) satisfies the function    3 1 1 1 exp σ (x, ξ ) = bi (ξ + γ (x − ξ ))(xi − ξi )dγ . |x − ξ | 2 0 i=1

Preface

xv

Based on the method of S. L. Sobolev in Sect. 1.2, it has been constructed Volterratype integral equation which is an equivalent to problem (11), (12). In this case, it is shown that the constructed equation has a unique solution, and it can be obtained using the method of successive approximations. In the inverse problem, the kernel k(t), t > 0 is found for a given mode of oscillation of the point x = 0. Here, there will also be investigated three one-dimensional inverse problems for a system of thermoelasticity equations in a vertically inhomogeneous disconnected medium, an integro-differential equation of acoustics and a two-dimensional system of Maxwell equations with memory. The memory of the medium is determined by the concentrated source of disturbances for the first problem and with the distribution for the second. These problems are of both practical and theoretical interest, and the results of their research constitute local unique solvability theorems and estimates of the continuous dependence of a change in a solution on a change in information. Chapter 2 is devoted to the study of multidimensional inverse problems. We consider the problem of finding the function k(x, t), t ≥ 0, x ∈ Rn from the equations  u tt − u zz − u =

t

k(x, τ )u(x, z, t − τ )dτ, (x, t) ∈ Rn+1 , z > 0,

(19)

0

u|t 0 is a fixed number, occurring in the system of integro-differential thermoviscoelasticity equations. The medium density and the Lame parameters are assumed to be function of one variable. The integrand h(t), t ∈ [0; T ] is known. The inverse problem is replaced by the equivalent integral equation for unknown functions. The local theorem of unique solvability is proved, and the stability estimate of solving the inverse problem is obtained. In the third section, it is investigated the problem of determining the moduli of elasticity c11 (x3 ), c12 (x3 ), c44 (x3 ), x3 > 0, occurring in the system of integro-differential viscoelasticity equations for homogenous anisotropic medium. It is assumed that density of medium is constant and matrix kernel k(t) = diag(k1 , k2 , k3 )(t), t ∈ [0, T ] is known. As additional information is the Fourier transform of the first and third component of the displacements vector for x 3 = 0. The results are the local theorems on the existence of a unique solution of the inverse problems and the theorems of stability. Chapter 7 studies one and two-dimensional problems of determining the memory kernel from a system of viscoelasticity and thermoviscoelasticity equations. The first two chapters define one- and two-dimensional kernels of the system of viscoelastic equations. Wherein, under the assumption that the coefficients of the system depend on one spatial variable and the applied concentrated force is directed along one of the coordinate axes, the system is reduced to one second-order equation. The research of problems is based on the methods used in Chaps. 2 and 3. In the third section, it is studied the problem of determining the one-dimensional matrix kernel of the system of viscoelasticity. This problem reduces to the sequential determination of the diagonal elements of the matrix kernel from second-order hyperbolic equations. In the fourth section, the problem of recovering the memory kernel of thermoviscoelasticity equation is considered. Based on the methods of Chap. 1, a local existence and uniqueness theorem is proved and an estimate of the stability of the solution is obtained. Chapter 8 devoted to the problems of reconstructing the coefficients of differential equations and the convolution kernel of hyperbolic integro-differential equations by the method of Dirichlet-to-Neumann operator. The rise of interest to this kind of problems is generally related to Calderdon who posed [107] the question whether the conductivity γ (x) in the elliptic equation ∇(γ (x)∇u) = 0

(22)

over a bounded domain  ⊂ Rn is uniquely determined by the (stationary) Dirichletto-Neumann operator. The positive answer to this question for functions γ (x) ∈

xviii

Preface

C ∞ () was given in the classical article [108] by Sylvester and Uhlmann who proved existence of special solutions to (22). Using the results of [108], Alessandrini [109] further proved conditional stability of a solution to the problem. Much information on this topic is presented in the survey [110] (see also the bibliography therein). Uniqueness and conditional stability were proven in [111, 110] respectively for one nonstationary problem, namely the problem of reconstructing the coefficient q(x) in the hyperbolic equation u tt −  u + q(x)u = 0 over the domain ×(0, T ), with  a bounded domain in Rn , n ≥ 2, and T > 0. This chapter deals with one problem of determining the potential and two problems of restoring memory from the integro-differential hyperbolic equation using the method of Dirichlet-to-Neumann operator. Chapter 9 contains some opening problems. Tashkent, Uzbekistan Vladikavkaz, Russia October 2022

Durdimurod K. Durdiev Zhanna D. Totieva

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78. Colombo, F. 2007. A inverse problem for a parabolic integro-differential model in the theory of combustion. Physica D: Nonlinear Phenomena 236: 81–89. 79. Favaron, A. 2006. An identification problem arising in the theory of heat conduction for materials with memory. Journal of Inverse and Ill-Posed Problems 14 (3): 235–249. 80. Grasselli, M. 1992. An identification problem for a linear integro-differential equation occurring in heat flow. Mathematical Methods in the Applied Sciences 15: 167–186. 81. Karuppiah, K., J.K. Kim, and K. Balachandran. 2015. Parameter identification of an integrodifferential equation. Nonlinear Functional Analysis and Applications 20 (2): 169–185. 82. Janno, J. 2004. Determination of a time- and space-dependent heat flux relaxation function by means of a restricted Dirichlet-to-Neumann operator. Mathematical Methods in the Applied Sciences 27 (11): 1241–1260. 83. Beilina, L., and M.V. Klibanov. 2012. Approximate global convergence and adaptivity for coefficient inverse problems. Springer. 84. Bellassoued, M., and M. Yamamoto. 2017. Carleman estimates and applications to inverse problems for hyperbolic systems. Springer. 85. Bukhgeim, A.L., and M.V. Klibanov. 1981. Uniqueness in the whole of one class of multidimensional inverse problems. Doklady Akademii Nauk SSSR 260 (2): 269–272. 86. Klibanov, M.V., and A.A. Timanov. 2004. Carleman estimates for coefficient inverse problems and numerical applicationz. Utrecht-Boston. 87. Anikonov, Yu.E. 1995. Multidimensional inverse and Ill-posed problems for differential equations. Utrecht: VSP. 88. Anikonov, Yu.E. 1997. Formulas in inverse and Ill-posed problems. Utrecht: VSP. 89. Bukhgeim, A.L. 2000. Introduction to the theory of inverse problems. Walter de Gruyter. 90. Cannon, J.R., and U. Hornung. 1986. Inverse problems. Basel-Boston-Stuttgart: Birkhauser. 91. Kabanikhin, S.I. 1988. Projection—Difference methods for determining the coefficients of hyperbolic equations. Novosibirsk: Nauka (Russian). 92. Kabanikhin, S.I., and A. Lorenzi. 1999. Identification problems of wave phenomena: Theory and numerics. Utrecht: VSP. 93. Newton, R.G. 1989. Inverse Schrodinger scattering in Tree dimensions, texts and monographs in physics. Berlin: Springer. 94. Prilepko, A.I., D.G. Orlovsky, and I.A. Vasin. 1999. Methods for solving inverse problems in mathematical physics. New York: Marcel Dekker, Inc. 95. Ramm, A.G. 1994. Multidimensional inverse scattering problems. Moswcow: Mir (Russian). 96. Romanov, V.G. 2005. Stability in inverse problems. M.: Nauchniy Mir (Russian). 97. Vladimirov, V.S. 1971. Equations of mathematical physics. New York: Marcel Dekker, INC. 98. Sobolev, S.L. 1930. Wave equation for an inhomogenuous medium. Trudy Seismologicheskogo Instituta 6: 1–57 (Russian). 99. Sobolev, S.L. 1934. To the question on integration of wave equation in an inhomogenuous medium. Trudy Seismologicheskogo Instituta 1 (42) (Russian). 100. Sobolev, S.L. 1936. New method of solving Cauchy’s problem for partial differential equations of normal hyperbolic form. Matematicheskii Sbornik 1 (43) (Russian). 101. Romanov, V.G. 1972. Some inverse problems for hyperbolic–type equations. Novosibirsk: Nauka (Russian). 102. Ovsyannikov, L.V. 1965. Singular operator in the scale of the Banach spaces. Doklady Akademii Nauk SSSR 163 (4): 819–822 (Russian). 103. Ovsyannikov, L.V. 1971. Nonlinear Cauchy problem in the scales of the Banach spaces. Doklady Akademii Nauk SSSR 200 (4): 789–792 (Russian). 104. Nirenberg, L. 1974. Topics in nonlinear functional analysis. New York: Courant Institute Mathematical Sciences, New York University. 105. Durdiev, U.D. 2019. A problem of identification of a special 2D memory kernel in an integro-differential hyperbolic equation. Eurasian Journal of Mathematical and Computer Applications 7 (2): 4–19. 106. Durdiev, U.D., and Z.D. Totieva. 2019. A problem of determining a special spatial part of 3D memory kernel in an integro-differential hyperbolic equation. Mathematical Methods in Applied Sciences 42 (18): 7440–7451.

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107. Calderon, A.P. 1980. On an inverse boundary value problem. Seminar on Numerical Analysis and Its Applications to Continuum Physics (Rio de Janeiro, 1980), Soc. Brasil. Mat., Rio de Janeiro, (1980), 65–73. 108. Sylvester, J., and G. Uhlmann. 1987. A global uniqueness theorem for an inverse boundary value problem. Annals of Mathematics (2) 251 (1): 153–169. 109. Alessandrini, G. 1988. Stable determination of conductivity by boundary measurements. Applicable Analysis 1–3 (27): 153–172. 110. Nachman, A. 1988. Reconstructions from boundary measurements. Annals of Mathematics (2) 128 (3): 531–576. 111. Rakesh, and W.W. Symes. 1988. Uniqueness for an inverse problem for the wave equation. Communications in Partial Differential Equations 13 (1): 87–96.

Contents

1 Local Solvability of One-Dimensional Kernel Determination Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Inverse Problem for the Wave Equation in the Medium with Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 The Inverse Problem for Hyperbolic Second-Order Integro-differential Equation with Variable Coefficients for Lower Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Formulation of the Problem. The Derivation of Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 The Existence and Uniqueness Theorem . . . . . . . . . . . . . . . . . 1.2.3 The Stability Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Inverse Problem for the System of Thermoelasticity Equations for a Vertically Inhomogeneous Cohesionless Medium with Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Inverse Problem for an Integro-differential Equation of Acoustics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Problem of Determining the Memory in a Two-Dimensional System of Maxwell Equations with Distributed Data . . . . . . . . . . . . . 1.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The Solvability of Multidimensional Inverse Problems of a Memory Kernel Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Definition of a Multidimensional Memory Kernel in the Integro-differential Wave Equation . . . . . . . . . . . . . . . . . . . . . . 2.2 Definition of the Memory Kernel Standing at Some Derivatives of a Direct Problem Solution . . . . . . . . . . . . . . . . . . . . . . . 2.3 Definition of Memory Kernel in a Two-Dimensional System Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 2

12 12 20 29

34 44 55 62 63 65 66 81 94

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2.4 Differential Properties of the Solution of the Memory Kernel Determination Problem in a Two-Dimensional Case . . . . . . . . . . . . . 98 2.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 3 Global Solvability of Memory Reconstruction Problems . . . . . . . . . . . 3.1 The Problem of Determining the Memory Kernel from the Integro-Differential Equation of String Vibrations . . . . . . . 3.2 Memory Identification Problem from Integro-Differential Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Inverse Problem for an Integro-Differential Equation of Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Global Solvability of the Problem of Determining Two Unknowns in the Inverse Problem for the Integro-Differential Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 About Global Solvability of a Multidimensional Inverse Problem for an Equation with Memory . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Stability in Inverse Problems for Determining of Two Unknowns . . . 4.1 The Problem of Determining the Speed of Sound and Memory of Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Statement of the Problem and Preliminary Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Properties of the Direct Problem Solution . . . . . . . . . . . . . . . . 4.1.3 Investigation of the Inverse Problem . . . . . . . . . . . . . . . . . . . . 4.2 The Problem of Determining the Memory of Medium and Regular Part of Impulsive Source . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Statement of the Problem and Preliminary Constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Properties of the Direct Problem Solution. Statement and Proof of the Main Results . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Two-Dimensional Special Kernel Determination Problems . . . . . . . . . 5.1 A Problem of Identification of a Two-Dimensional Kernel . . . . . . . . 5.1.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Investigation of the Direct Problem . . . . . . . . . . . . . . . . . . . . . 5.1.3 Inverse Problem—The Existence and Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 A Problem of Determining a Spatial Part of a Memory Kernel . . . . 5.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

111 112 124 125

134 139 155 155 157 157 157 161 165 175 175 176 184 184 185 185 185 187 198 204 209 209

Contents

6 Some Inverse Problems of Viscoelasticity System with the Known Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 The Multidimensional Linearized Problem of Determining the Density Function for a Viscoelasticity System . . . . . . . . . . . . . . . 6.1.1 Statement of Problems and Main Results . . . . . . . . . . . . . . . . 6.1.2 Direct Problem of Isotropic Viscoelasticity and Properties of Its Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 The Fundamental Solution of the Cauchy Problem for a Hyperbolic Operator with Memory . . . . . . . . . . . . . . . . . 6.1.4 Direct and Inverse Linearized Problem . . . . . . . . . . . . . . . . . . 6.2 The Problem of Determining the Coefficient of Thermal Enlargement of Thermoviscoelasticity Equation . . . . . . . . . . . . . . . . . 6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Kernel Identification Problems in a Viscoelasticity System . . . . . . . . . 7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Problem of Determining the One-Dimensional Matrix Kernel of the System of Viscoelasticity . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Problem of Determining the One-Dimensional Kernel Thermoviscoelasticity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 The Dirichlet-to-Neumann Maps Method for a Half Space . . . . . . . . 8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 The Problem Statement and Results . . . . . . . . . . . . . . . . . . . . . 8.2.2 The Direct Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.3 Auxiliary Assertions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.4 Proof of the Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Recovering a Time- and Space-Dependent Kernel in a Hyperbolic Integro-Differential Equation from a Restricted Dirichlet-to-Neymann Operator . . . . . . . . . . . . . . . 8.3.1 Problem Formulation and Results . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.3 Direct Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.4 Uniqueness on the Boundary . . . . . . . . . . . . . . . . . . . . . . . . . .

xxvii

211 213 213 217 223 228 232 243 271 271 273 273 284 307 319 325 325 327 327 331 332 333 335 341

345 345 349 350 355

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8.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 9 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368

Chapter 1

Local Solvability of One-Dimensional Kernel Determination Problems

This chapter discusses one-dimensional inverse problems of determining the memory of a medium entering into an integral term of convolution type in hyperbolic equations. In Sect. 1.1, the problem of finding a kernel from  an integro-differential   wave equation in a half-space R3+ = (x1 , x2 , x3 ) ∈ R3  x3 > 0 is investigated. Further, in Sect. 1.2, the results obtained in the previous section are generalized to the case of a second-order hyperbolic-type equation with variable coefficients at lower derivatives with respect to x = (x1 , x2 , x3 ), the main part of which coincides with the wave operator. Vibration of the medium comes from a point source located in the point x = 0 at the moment t = 0. In both sections, the property of direct problems solution, which are well-posed boundary–initial value problems, is investigated. The mode of oscillation of point x = 0 is set to solve the inverse problem. In Sect. 1.3, the one-dimensional kernel memory is determined from the system of thermoelasticity equations for a vertically inhomogeneous cohesionless medium. In Sect. 1.4, it is investigated the local solvability of inverse problem for an integro-differential equation of acoustics. Section 1.5 studies a problem of determining of two functions representing the electrical and magnetic prehistory of the medium, respectively. These unknown functions are found from the two-dimensional Maxwell system of integro-differential equations with a distributed source of perturbation. The unique solvability theorems are proved, and the conditional stability estimates for the solutions of the direct and the inverse problems are obtained.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_1

1

2

1 Local Solvability of One-Dimensional Kernel …

1.1 The Inverse Problem for the Wave Equation in the Medium with Memory For x3 > 0 we consider the equation t u tt = u +

k(τ )u(x, t − τ )dτ + δ(x, t)

(1.1)

0

with zero initial and boundary conditions   u t 0, that is in the integral term of Eq. (1.1) given at the point x = 0, a generalized solution of problem (1.1) and (1.2):  (1.3) u x=0 = f (t), t > 0.

Definition 1.1 The solution of the inverse problem will be called function k(t), such that the corresponding solution to problem (1.1) and (1.2) satisfies condition (1.3). We continue the function u(x, t) to the half-space x3 < 0 as an even function with respect x3 . The solution of the problem (1.1) and (1.2) we present in the following form: u(x, t) = −2ε(x, t) + v(x, t), where ε(x, t) =

 θ (t)  2 δ t − |x|2 2π

is the fundamental solution of the wave equation. Then, inverse problem (1.1)–(1.3) for a scattered wave v(x, t) is reduced to the form θ (t − |x|) vtt − v = k(t − |x|) + 2π |x|

t k(τ )v(x, t − τ )dτ, (x, t) ∈ R4 ,

(1.4)

0

v|t 0.

(1.6)

1.1 The Inverse Problem for the Wave Equation in the Medium with Memory

3

Consider the Cauchy problem (1.4) and (1.5). Obviously, v|t 0 and for sufficiently large n the inequality  C

kT 4

n

(t − |x|)2n+1 < (2n + 1)!

is valid. From (1.19) we get that v(x ¯ 0 , t0 ) < , where  is an arbitrarily small quantity. For instance, choosing 0 <  < |v(x ¯ 0 , t0 )| /2, we get a contradiction. Therefore, v(x, ¯ t) = 0 and the solution of Eq. (1.12) is unique. We now turn to Eq. (1.13). For this equation we use the process of successive approximations according to the following scheme: k(t − |x|) , 4π wn (x, t) = Ak wn−1 (x, t) , t ≥ |x|, n = 1, 2, 3, . . . w0 (x, t) =

(1.20)

1.1 The Inverse Problem for the Wave Equation in the Medium with Memory

We show that the series

∞

7

wn (x, t)

n=0

converges uniformly in the domain D(T ). In this domain the following estimates take place 2n √ k kT (t − |x|) wn  ≤ , n = 1, 2, 3, . . . . 4π (2n)!22n Hence the series

∞ n=0

wn (x, t) is majorized by series ∞

k

4π n=0

√ 2n kT (t − |x|) , (2n)!22n

which, in turn, for all (x, t) ∈ D(T ) is majorized by the convergent numerical series ∞

k

4π n=0



kT 3

2n

(2n)!22n

=

  1 k ch kT 3 . 4π 2

The series ∞ n=0 wn (x, t) converges to a unique continuous solution of Eq. (1.13). The following estimate is valid:   1 k 3 ch kT . w ≤ 4π 2

(1.21)

Now we pass to the consideration of the inverse problem and suppose, as before, that k(t) is a continuous function. It is clear from this that the function f (t) for t → +0 must satisfy the additional condition f (+0) = 0, and there must be a continuous differentiable at t > 0. Putting in (1.12) x = 0 and using condition (1.6), we obtain 1 f (t) = 4π

t k(α)dα + Ak [v(x, t)]|x=0 , t > 0.

(1.22)

0

Differentiating equality (1.22) with respect to the variable t and introducing the notation F(t) = 4π f (t), we find k(t) = F(t) − 4π Ak [w(x, t)] |x=0 . Let F = maxt∈[0,T ] |F(t)| . Fix  > 0 and consider the ball

(1.23)

8

1 Local Solvability of One-Dimensional Kernel …

T (F, ) = {r (t) : r ∈ C[0, T ], F − r  ≤ ]} . We introduce the operator by formula U [k(t)] = F(t) − 4π Ak [w(x, t)] |x=0 .

(1.24)

Then, Eq. (1.23) is written in the more convenient form k(t) = U [k(t)] .

(1.25)

Let T ∗ = min (T1 , T2 ) , where  T1 =

3

4γ12 , T2 = F + 

 3

4γ22 , F + 

(1.26)

and γ1 is the positive root of the equation γ 2 chγ =

2 , F + 

(1.27)

γ2 is the positive root of the equation γ 2 chγ

 γ2 chγ = 1. 1+ 4

(1.28)

Theorem 1.1 Let f (t) ∈ C 1 [0, T ] and f (+0) = 0. Then, for ∗ t ∈ [0, T ], T ∈ (0, T ), there exists a unique continuous solution of Eq. (1.25) belonging to a ball T (F, ). We prove this theorem using contraction mapping principle. The operator U is a contraction operator if the following two conditions are fulfilled: (1) If ϕ ∈ T (F, ) , then U ϕ ∈ T (F, ) , (2) If ϕ 1 and ϕ 2 are arbitrary elements in T (F, ) , then the following inequality holds true: ||U ϕ 1 − U ϕ 2 || ≤ ρ||ϕ 1 − ϕ 2 ||, where 0 < ρ < 1. First, we show that for any T ∈ (0, T ∗ ) the operator U maps the ball T (F, ) into itself, i.e., from the condition k(t) ∈ T (F, ) it follows U [k(t)] ∈ T (F, ) . Indeed, for any continuous function k(t) the function U [k(t)], calculated by the formula (1.24), will be continuous function. In addition, estimating the norm of differences, we find

1.1 The Inverse Problem for the Wave Equation in the Medium with Memory

U [k(t)] − F(t) ≤ 4π kwA1 [1]|x=0

9

  T3 1 2 3 k ch ≤ kT . 8 2

Here we used the estimate (1.21) and easily verifiable equality |A1 [1]|x=0 = t 3 /8. We note that the function standing in the right side of this inequality, monotonically increases with increasing T, and if the function k(t) belongs to the ball T (F, ), then k ≤  + F. (1.29) Therefore, for all T ∈ (0, T ∗ ) this inequality will only intensify if instead of T we substitute T1 , and instead of k we substitute the expression  + F. Carrying out this and using formulas (1.26) and (1.27), we get U [k(t)] − F(t) ≤

 1 ( + F) γ 2 chγ γ =γ1 = . 2

Thus, we have proved that U [k(t)] ∈ T (F, ). Now show the fulfillment of the second condition of the contraction mapping ˜ principle. Consider two functions k(t), k(t), belonging to the ball T (F, ), and ˜ estimate the difference of their images U [k(t)], U [k(t)]. The solution of Eq. (1.13), ˜ corresponding to function k(t), we denoted by w(x, ˜ t). Thus, w(x, ˜ t) satisfies the equation 1 ˜ ˜ t)], t ≥ |x|. (1.30) w(x, ˜ t) = k(t − |x|) + Ak˜ [w(x, 4π By using this equation and (1.13), we estimate the norm of difference        ˜ ˜ ˜ + kw − w) ˜ A1 [1]|x=0  ≤ 4π k − kw U [k(t)] − U k(t)   T 3π ˜ ˜ k − kw + kw − w ˜ . (1.31) ≤ 2 ˜ Using Eqs. (1.13) and (1.30), we derive an estimate for w − w ˜ through k − k. From equalities (1.13) and (1.30) we make an integral equation for w − w: ˜ w(x, t) − w(x, ˜ t) =

 1  ˜ − |x|) + Ak−k˜ [w(x, t)] k(t − |x|) − k(t 4π  ˜ t) . (1.32) +Ak˜ w(x, t) − w(x,

Considering |w(x, t) − w(x, ˜ t)|, from Eq. (1.32) we get the following linear inequality with respect to |w(x, t) − w(x, ˜ t)| :

10

1 Local Solvability of One-Dimensional Kernel …

    T 3 k 1 1 3 ˜ + ch |w(x, t) − w(x, ˜ t)| ≤ k − k kT 4π 32π 2  ˜ 1 |w − w| ˜ . +kA Applying to this inequality the method of successive approximations, we find w(x, t) − w(x, ˜ t)       1  1 T 3 k 1   ≤ k − k˜  kT 3 ch kT 3 . + ch 4π 32π 2 2

(1.33)

Substituting the estimates (1.21) and (1.33) into inequality (1.31), we get      3   ˜ T k ch 1 kT 3 ˜  ≤ k − k U [k(t)] − U k(t) 8 2       3 3 ˜ T k 1 1 T k 3 3 ˜ ˜ +k − k ch ch kT kT 1+ . 8 2 8 2 (1.34) ˜ belong to the ball T (F, ), and, consequently, for each The functions k(t), k(t) of them the inequality (1.29) holds. We notice that the function on the right-hand ˜ and T . side of inequality (1.34) increases monotonically with increasing k, k ˜ Therefore, if k, k, in inequality (1.34), we replace by  + F, and instead of T we set T2 , the inequality will only intensify. Having done this, we find      ˜ ˜  ≤ ρ(T )k − k, U [k(t)] − U k(t)   1 ρ(T ) = γ 2 chγ 1 + γ 2 chγ √ 4 γ = 21 (+F)T 3   1 < γ 2 chγ 1 + γ 2 chγ = 1. 4 γ =γ2 From this inequality it follows that the distance in the functional space C between ˜ ˜ functions U [k(t)], U [k(t)] is less than the distance between the functions k(t), k(t) T in the same space. Therefore, the operator U in the ball (F, ) is a contraction operator. Then, according to Banach theorem, the operator U in the ball T (F, ) has a unique fixed point, that is, there is only one solution of Eq. (1.25). Theorem 1.1 is proven. Now we give a stability estimate of the solution of inverse problem and prove the uniqueness theorem for an arbitrary fixed T > 0. Denote by K (k0 ) the set of functions k(t) ∈ C[0, T ], satisfying the inequality |k(t)| ≤ k0 t ∈ [0, T ] with fixed positive constant k0 .

1.1 The Inverse Problem for the Wave Equation in the Medium with Memory

11

˜ ∈ K (k0 ) be solutions of the inverse problem Theorem 1.2 Let k(t) ∈ K (k0 ), k(t) (1.4)–(1.6) with the data f (t), f˜(t), respectively. Then, there exists a positive number C = C(k0 , T ), such that the following stability estimate is valid: ˜ ˜ k(t) − k(t) C[0,T ] ≤ C f (t) − f (t)C 1 [0,T ] .

(1.35)

˜ Proof Using equality (1.23), we compose an equation for difference k(t) − k(t)   ˜ = 4π f (t) − f˜ (t) k(t) − k(t)    −4π Ak−k˜ [w(x, t)]x=0 + Ak˜ w(x, t) − w(x, ˜ t) x=0 , (x, t) ∈ DT . (1.36) Here the function w˜ satisfies equality (1.30). We note that the domain DT allows an equivalent description:     T T  DT = (x, t) : 0 ≤ |x| ≤ − t −  , 0 ≤ t ≤ T . 2 2 Let     ˜  , ψ(t) = max max k(t) − k(t)

max

0≤|x|≤T /2−|t−T /2|

 |w(x, t) − w(x, ˜ t)| ,

where t ∈ [0, T ]. Using Eqs. (1.32) and (1.36), we find that for 0 ≤ t ≤ |x| ≤ T /2 − |t − T /2| inequalities t  T2 1   ˜ |w(x, t) − w(x, ˜ t)| ≤ (μ0 + k0 ) ψ(α)dα, k(t) − k(t) + 4π 4

(1.37)

0

t

      ˜  ≤ 4π   f (t) − f˜(t) k(t) − k(t)

+ π T (μ0 + k0 )

ψ(α)dα

2

C 1 [0,T ]

(1.38)

0

are valid, where μ0 := (k0 /4π )ch we get

  1 2

 kT 3 . From (1.37), using inequality (1.38),

    |w(x, t) − w(x, ˜ t)| ≤  f (t) − f˜(t)

T2 (μ0 + k0 ) + C 1 [0,T ] 2

t ψ(α)dα. (1.39) 0

It follows from the relations (1.38) and (1.39) that ψ(t) satisfies the following integral inequality

12

1 Local Solvability of One-Dimensional Kernel …

t

    ψ(t) ≤ 4π  f (t) − f˜(t)

+ π T (μ0 + k0 )

ψ(α)dα, t ∈ [0, T ].

2

C 1 [0,T ]

0

Using the Gronwall inequality, we obtain the estimate      |ψ(t)| ≤ 4π exp π T 2 (μ0 + k0 )  f (t) − f˜(t)

C 1 [0,T ]

, t ∈ [0, T ].

(1.40)

Inequality (1.35) is equivalent to (1.40) if   C = 4π exp π T 2 (μ0 + k0 ) . The uniqueness theorem for any fixed T > 0 follows from the stability Theorem 1.2. ˜ Theorem 1.3 Suppose that the functions {k(t), k(t)}∈C[0, T ]) and { f (t), f˜(t)} ∈ 1 C [0, T ] have the same meaning as in Theorem 1.2. If moreover f (t) = f˜(t), t ∈ ˜ for t ∈ [0, T ]. [0, T ], then k(t) = k(t)

1.2 The Inverse Problem for Hyperbolic Second-Order Integro-differential Equation with Variable Coefficients for Lower Derivatives The present section deals with a memory determination problem of a hyperbolic second-order integro-differential equation with constant principal part and variable coefficients at lower derivatives. The research method is based on reducing the problem to nonlinear system of integral equations of the second kind and uses the fundamental solution for a hyperbolic equation with variable coefficients, constructed by Sobolev [1–3]. The theorem of uniqueness and stability in global and the existence theorem in local are proved.

1.2.1 Formulation of the Problem. The Derivation of Integral Equations Consider the equation t u tt − Lu =

k(τ )u(x, t − τ )dτ + δ(x, t) 0

(1.41)

1.2 The Inverse Problem for Hyperbolic Second-Order …

13

at the condition u|t 0.

(1.44)

We investigate the problem (1.41), (1.42), and (1.44) by reducing it to a system of nonlinear integral equations with respect to unknown functions. Using the results of [1] (p. 64), we reverse the operator on the left side of (1.41) on the Cauchy data (1.42): 1 1 σ (x)δ (t − |x|) + u(x, t) = 4π 4π





L ∗ξ σ (x, ξ )u(ξ, t − |ξ − x|)

|ξ −x|≤t

 k(ν)u(ξ, t − |ξ − x| − ν)dν dξ,

t−|ξ  −x|

+ σ (x, ξ ) 0

where ⎧ 1 ⎫ 3 ⎨1 

⎬ 1 exp bi [ξ + α(x − ξ )](xi − ξi )dα , σ (x, ξ ) = ⎩2 ⎭ |ξ − x| i=1

(1.45)

0

σ (x) = σ (x, 0). L ∗ξ denotes an operator that is adjoint in the sense of Lagrange with the operator L. Equality holds true for it L ∗ξ [|ξ − x|σ (x, t)] = |ξ − x|L ∗ξ σ (x, ξ ), which is easily verified directly. Introducing a new function v(x, t) by formula

(1.46)

14

1 Local Solvability of One-Dimensional Kernel …

u(x, t) = v(x, t) +

1 σ (x)δ (t − |x|) , 4π

we get equation for it v(x, t) =



1 16π 2

σ (ξ )L ∗ξ σ (x, ξ )δ(t − |ξ | − |ξ − x|)dξ

D(x,t)



1 + 16π 2

t−|ξ  −x|

σ (ξ )σ (x, ξ )

k(ν)δ(t − |ξ | − |ξ − x| − ν)dν dξ 0

D(x,t)

1 + 4π

 

L ∗ξ σ (x, ξ )v(ξ, t − |ξ − x|)

D(x,t)



t−|ξ  −x|

+ σ (x, t)

k(ν)v(ξ, (t − |ξ − x| − ν)dν dξ, 0

in the space of variables ξ bounded by surface S(x, t) =  D(x, t) is the domain  where x  |ξ − x| + |ξ = t| . Using property of δ-function, we rewrite the last equality in the equivalent form  v(x, t) = S(x,t)

ds 0 (x, ξ )    |ξ ||ξ − x| ∇ξ (|ξ | + |ξ − x|) 

t +

ds dτ 1 (x, ξ )   |ξ ||ξ − x| ∇ξ (|ξ | + |ξ − x|)

k(t − τ ) |x|

S(x,t)

  v(ξ, t − |ξ − x|) + 2 (x, ξ ) |ξ − x| D(x,t)

3 (x, ξ ) + |ξ − x|

 k(ν)v(t − |ξ − x| − ν)dν dξ.

t−|ξ  −x|

(1.47)

0

Here ds is the element of the surface area S(x, t); 1 |ξ |σ (ξ )|ξ − x|L ∗ξ σ (x, ξ ), 16π 2 1 |ξ |σ (ξ )|ξ − x|σ (x, ξ ), (1.48) 1 (x, ξ ) = 16π 2 1 1 |ξ − x|L ∗ξ σ (x, ξ ), 3 (x, ξ ) = |ξ − x|σ (x, ξ ). 2 (x, ξ ) = 4π 4π

0 (x, ξ ) =

1.2 The Inverse Problem for Hyperbolic Second-Order …

15

Let us transform the integrals entering into the formula (1.47). We introduce, as in the previous section, at the same time with the coordinate system ξ, the Cartesian system η, for which the center is placed at the point 0, and the axis η3 is directed along the straight line connecting the points 0 and x toward the point x. We use a system in which ξ and η are related by the formula conversions [4]: ξ = Qη,

(1.49)

⎞ cos θ0 cos ϕ0 − sin θ0 sin ϕ0 sin θ0 cos ϕ0 Q = ⎝ cos θ0 sin ϕ0 sin θ0 cos ϕ0 sin θ0 sin ϕ0 ⎠ . − sin θ0 0 cos θ0 ⎛

Here θ0 , ϕ0 are the angular coordinates of the unit vector x/|x| at spherical coordinate system connected with the Cartesian system coordinates in the usual way. In addition, with this coordinate system, we associate in the usual way a spherical coordinate system r, θ, ϕ. Then, denoting the first term in (1.47) by J, we find 1 J (x, t) = |x|

t+|x|

2π

2

0 (x, ξ )dr.

dϕ 0

t−|x| 2

In this formula the variable point ξ is connected with η by formulas (1.49). Further, η = r (sin θ cos ϕ, sin θ sin ϕ, cos θ ), where r = (t 2 − |x 2 |)/2(t − |x| cos θ ). In the inner integral of J, we make the change of the integration variable r with the variable z as follows: r= Then 1 J (x, t) = 2

1 (t + |x|z). 2 2π

1 0 (x, ξ )dz,

dϕ 0

−1

and η is expressed in terms of the integration variables ϕ, z by the formulas   1 2 1 2 t − |x|2 1 − z 2 cos ϕ, η2 = t − |x|2 1 − z 2 sin ϕ, 2 2 1 η3 = (t z + |x|) . 2

η1 =

16

1 Local Solvability of One-Dimensional Kernel …

It is easy to see that the integral over S(x, τ ) in the second term of (1.47) differs from J only by the integrand. Therefore, the transformation of this integral is quite analogous to the transformation of J (x, t). In order to transform the remaining integral terms of (1.47), we use the equation of the ellipsoid S(x, τ ) in the form r (θ, τ ) =

τ 2 − |x|2 . 2(τ − |x| cos θ )

As a result, it is easy to get equality [4]: 2r 3 (θ, τ ) dξ = 2 dω dτ, |ξ − x| τ − |x|2 where dω = sin θ dθ dϕ is the element of the solid angle with the center at the point 0. In view of the above, Eq. (1.47) takes the form 1 v(x, t) = 2

2π

1 dϕ

0

t +2 |x|

−1

1 0 (x, ξ )dz + 2

dτ 2 τ − |x|2



t

2π k(t − τ )

|x|

1 (x, ξ )dz dτ

dϕ 0

−1

 r 3 (θ, τ ) v(ξ, t − τ + r )2 (x, ξ )

S(x,τ )



t−τ +3 (x, ξ )

1

k(ν)v(ξ, t − τ + r − ν)dν dω.

(1.50)

0

Thus, the regular part of the solution of the problem (1.41) and (1.42) satisfies the integral equation (1.50). We denote G T = {x : |x| ≤ T }, T is an arbitrary positive number. Let bi (x) ∈ C 2 (G T ) , i = 1, 2, 3, c(x) ∈ C (G T ) , k(t) ∈ C (0, T ] , where C 2 (G T ), C (G T ) are classes of twice continuously differentiable and continuous functions in the domain G T , respectively. Then, we show that the function v(x, t) is continuous with respect to the variables x, t in the domain DT =: {(x, t) : |x| ≤ t ≤ T − |x|} and it can be found from Eq. (1.50) by the method of successive approximation. In [1] (p. 63) it was received the estimate   ∗  L σ (x, ξ ) ≤ ξ

M0 , |x − ξ |

1.2 The Inverse Problem for Hyperbolic Second-Order …

17

where M0 is a positive constant depending only on the global properties of the coefficients of the operator L ∗ . Using the definitions of the functions i , i = 0, 1, 2, 3, through formulas (1.48) and the last inequality, we find |i (x, ξ )| ≤ M for all i = 0, 1, 2, 3, where a constant M depends only on T and the norms of the functions bi , i = 1, 2, 3, c in the spaces C 2 (G T ), C (G T ) , respectively. We apply to Eq. (1.50) the method of successive approximation according to the following scheme: 1 v (x, t) = 2

2π

0

1 dϕ

0

−1

1 0 (x, ξ )dz + 2 t

v n (x, t) = v 0 (x, t) + 2 |x|

t−τ +3 (x, ξ )

τ2

dτ − |x|2

t

2π k(t − τ )

|x|



1 1 (x, ξ )dz dτ,

dϕ 0

−1

 3 r (θ, τ ) v n−1 (ξ, t − τ + r )2 (x, ξ )

S(x,τ )

 k(ν)v n−1 (ξ, t − τ + r − ν)dν dω,

0

n = 1, 2, . . . . (1.51) It is obvious that each of the functions v n (x, t) is continuous in the domain DT . Denoting k0 = maxt∈[0,T ] |k(t)| , we estimate functions v n (x, t) for (x, t) ∈ DT :   0 v (x, t) ≤ 2π M [1 + k0 (t − |x|)] ,   1 v (x, t) ≤ 2π M [1 + k0 (t − |x|)]   (t − |x|)3 + 4π 2 M 2 T t − |x| + k0 (t − |x|)2 + k02 . 2! Thus

n m+1 m+ j



 n  j (t − |x|) v (x, t) ≤ 2π M . (2π M T )m k0 m! j! m=0 j=0

For Eq. (1.50), we write the Neumann series corresponding to the scheme (1.51) v 0 (x, t) +

∞

 n  v (x, t) − v n−1 (x, t) . n=1

18

1 Local Solvability of One-Dimensional Kernel …

Its partial sum coincides with the function v n (x, t), and therefore, this series can be majorized by ∞

n+1 n+ j

j (t − |x|) , 2π M (2π M T )n k0 n! j! n=0 j=0 which, in turn, for all (x, t) ∈ DT is majorized by convergent numerical series ∞

n+1

j

(2π M)n+1 k0

n=0 j=0

T 2n+ j . n! j!

Thus, the Neumann series converges uniformly, and therefore, its sum is a continuous function in the domain DT . As usual, it is not difficult to prove that the sum of a series is a solution of Eq. (1.50) and it is unique. Differentiating equality (1.50) with respect to the variable t, we find vt (x, t) 2π

t

=  4 t 2 − |x|2

1 dϕ

' ∇ξ 0 , Qη (t, x, ϕ, z) dz

−1

0



1 + π k(t − |x|)

&

1

−1

z+1 x, 0, 0, |x| dz 2

2π 1 & ' k(t − τ )τ dτ  ∇ξ 1 , Qη (τ, x, ϕ, z) dz dϕ τ 2 − |x|2 0 −1 |x|  2 + 2 r 3 (θ, t)v(ξ, r )2 (x, ξ )dω t − |x|2 1 + 4

t

S(x,t)

t +2 |x|

dτ τ 2 − |x|2 ⎡



 r 3 (θ, τ ) vt (ξ, t − τ + r )2 (x, ξ )

S(x,τ )

+ 3 (x, ξ ) ⎣k(t − τ )v(ξ, r ) +

t−τ

⎤

 ⎦ dω. k(ν)vt (ξ, t − τ + r − ν)dν

0

Here

  ∇ξ  j (x, ξ ) =  jξ1 ,  jξ2 ,  jξ3 (x, ξ ),

j = 0, 1,

(1.52)

1.2 The Inverse Problem for Hyperbolic Second-Order …

19

·, · is a scalar product in R3 ;      z 2 η (t, x, z, ϕ) = η1 , η2 , η3 = 1 − z 2 cos ϕ, 1 − z 2 sin ϕ, t − |x|2 . t We introduce linear integral operators by the formulas t A[k(t), v(x, t)] = 2 |x|

 r 3 (θ, τ ) v(ξ, t − τ + r )2 (x, ξ )



dτ τ 2 − |x|2

S(x,τ )

t−τ + 3 (x, ξ )

 k(ν)v(ξ, t − τ + r − ν)dν dω,

0

B[k(t), v(x, t)] t   k(t − τ )dτ = 8 t 2 − |x|2 τ 2 − |x|2 |x|

r 3 (θ, τ )v(ξ, r )3 (x, ξ )dω,

(1.53)

S(x,τ )

E[v(x, t)] = 



8 t2

−

|x|2

r 3 (θ, t)v(ξ, r )2 (x, ξ )dω. S(x,τ )

Denote 1 v0 (x, t) = 2

2π

1 0 (x, ξ )dz,

dϕ 0

−1

2π

1

w0 (x, t) = t

dϕ

&

' ∇ξ 0 , Qη dz.

(1.54)

−1

0

We also introduce a new function w(x, t) by the formula  w(x, t) = 4 t 2 − |x|2 vt (x, t). Then, Eqs. (1.50) and (1.52) we write in the form 1 v(x, t) = v0 (x, t) + 2

t

2π k(t − τ )

|x|

+ A[k(t), v(x, t)],

1 1 (x, ξ )dz dτ

dϕ 0

−1

(1.55)

20

1 Local Solvability of One-Dimensional Kernel …

w(x, t) = w0 (x, t) + 4π





1 t2

−

|x|2 k(t

− |x|)

1

−1

z+1 x, 0, 0, |x| dz 2

√2 2 t −|x|   2π   + t 2 − |x|2 k t − β 2 − |x|2 dβ dϕ 0

1 ×

0

&

' ∇ξ 1 , Qη |τ =√β 2 −|x|2 dz + B[k(t), v(x, t)]

−1

+ E[v(x, t)] +





 w(x, t) t 2 − |x|2 A k(t),  . t 2 − |x|2

(1.56)

Passing in the last equality to the limit at x → 0, and taking into account the conditions (1.44) we obtain the equation for the unknown function k(t) t k(t) = F(t) + w¯ 0 (t) − 2π 

2π k(t − β)dβ

0

1 dϕ

&

' ∇ξ 1 , Qη |τ =β,x=0 dz

−1

0

2π B[k(t), v(x, t)] + E[v(x, t)] t    w(x, t) 2 2 + t − |x| A k(t),  , t 2 − |x|2 x=0

−

where F(t) = 8π f (t), w¯ 0 (t) = −

(1.57)

2π w0 (0, t). t

The order of quantity, included in braces when x = 0, is equal to t 2 . Because of this, in particular, it follows from (1.57) the following equality k(+0) = F(+0).

(1.58)

1.2.2 The Existence and Uniqueness Theorem The system of Eqs. (1.55)–(1.57) is a closed system of integral equations with respect to functions v(x, t), w(x, t), k(t). We write it in the form of an operator equation ψ = U ψ,

(1.59)

1.2 The Inverse Problem for Hyperbolic Second-Order …

21

where ψ is vector function of variables x, t with components ψi , i = 1, 2, 3 at that ψ1 (x, t) = v(x, t), ψ2 (x, t) = w(x, t), ψ3 (x, t) ≡ ψ3 (t) = k(t). The operator U is defined on a set of functions ψ ∈ C [DT ] , and in accordance with equalities (1.55)– (1.57) has the form U = (U1 , U2 , U3 ) , 1 U1 ψ = v0 (x, t) + 2

t

2π ψ3 (t − τ )

|x|

1 1 (x, ξ )dz dτ + A [ψ3 (t), ψ1 (x, t)] ,

dϕ −1

0

1



U2 ψ = w0 (x, t) + 4π t 2 − |x|2 ψ3 (t − |x|) −1

 z+1 1 x, 0, 0, |x| dz 2

√2 2 t −|x|   2π   2 2 2 2 ψ3 t − β − |x| dβ dϕ + t − |x| 0

1 ×

0

&

' ∇ξ 1 , Qη |τ =√β 2 −|x|2 dz + B[ψ3 (t), ψ1 (x, t)]

−1

   ψ2 (x, t) 2 2 +E[ψ1 (x, t)] + t − |x| A ψ1 (t),  , t 2 − |x|2 t U3 ψ = F(t) + w¯ 0 (t) − 2π

2π ψ3 (t − β)dβ

0

1 dϕ

0

 2π B[ψ3 (t), ψ1 (x, t)] + E[ψ( x, t)] − t    (x, t) ψ 2 + t 2 − |x|2 A ψ3 (t),  . t 2 − |x|2 x=0

(1.60)

&

' ∇ξ 1 , Qη |τ =β,x=0 dz

−1

We make the following assumptions regarding the functions bi (x), i = 1, 2, 3, c(x): bi (x) ∈ C 3 [G T ] , c(x) ∈ C [G T ] . 

Introduce notations ψ = max

1≤i≤3

 max |ψi (x, t)| ,

(x,t)∈DT

ψ0 (x, t) = (ψ01 , ψ02 , ψ03 ) = (v0 (x, t), w0 (x, t), w¯ 0 (t)),

22

1 Local Solvability of One-Dimensional Kernel …

  b0 = max bi C 3 [G T ] , c0 = cC 1 [G T ] . 1≤i≤3

In the future, we need estimates for the functions & ' i (x, ξ ), i = 0, 1, 2, 3, ∇ξ  j (x, ξ ), Qη (t, x, ϕ, z) , j = 0, 1 for (x, ξ ) ∈ G T × G T , t ∈ [0, T ], ϕ ∈ [0, 2π ], z ∈ [−1, 1]. Therefore, we will carry out some calculations related to the definition of the dependence of the functions i (x, ξ ), i = 0, 1, 2, 3 on bi , i = 0, 1, 2, c. Using (1.45), (1.46), and (1.48), we find |ξ − x|L ∗ξ σ (x, ξ ) = L ∗ξ exp

= ξ exp

⎧ 1 3 ⎨1 

⎧ 1 3 ⎨1 

⎩2 0

bi [ξ + α(x − ξ )] (xi − ξi )dα

i=1

⎫ ⎬ ⎭

⎫ ⎬

bi [ξ + α(x − ξ )](xi − ξi )dα ⎩2 ⎭ i=1 0 ⎧ 1 ⎫⎫ ⎧ 3 3 ⎨1 

⎬⎬

∂ ⎨ − bi [ξ + α(x − ξ )](xi − ξi )dα bi (ξ ) exp ⎩2 ⎭⎭ ∂ξi ⎩ i=1 0 i=1 ⎧ 1 ⎫ 3 ⎨1 

⎬ + c(ξ ) exp bi [ξ + α(x − ξ )](xi − ξi )dα ⎩2 ⎭ 0 i=1 ⎧ 1 ⎫ 3 ⎨1 

⎬ := exp bi [ξ + α(x − ξ )] (xi − ξi )dα (x, ξ ), ⎩2 ⎭ 0

i=1

where (x, ξ ) ⎤ ⎡  

3 1

3 1 ⎣ b jξi x + α(ξ − x)⎦ (x j − ξ j )α = 4 i=1 j=1 0

 2 − bi [x + α(ξ − x)] dα

1 + 2

1 

3 0

i=1

 ξ bi [x + α(ξ − x)] (xi − ξi )α 2 − 2divξ b [x + α(ξ − x)] dα

1.2 The Inverse Problem for Hyperbolic Second-Order …

1

− bi (ξ ) 2 i=1 3

−

3

1 

3 0

23

 b jξi [x + α(ξ − x)] (x j − ξ j )α − bi [x + α(ξ − x)] dα

j=1

b jξi (ξ ) + c(ξ ).

(1.61)

i=1

Here the low index ξ in the operators , div indicates that they are applied by variable ξ. Consequently, 0 (x, ξ ) = 1 (x, ξ )(x, θ ), 2 (x, ξ ) = 3 (x, ξ )(x, ξ ).

(1.62)

Further, ∂ (x, ξ ) ∂ξν 1 = 0 (x, ξ ) 2

1 

3 0

 biξν (αξ )αξi + biξν [x + α(ξ − x)] α(xi − ξi )

i=1

 ∂ +bν (αξ ) − bν [x + α(ξ − x)] dα + 1 (x, ξ ) (x, ξ ), ν = 1, 2, 3, ∂ξν ∂ (x, ξ ) ∂ξν  3  1 

3 1

b jξi [x + α(ξ − x)] (x j − ξ j ) − bi [x + α(ξ − x)] dα = 4 i=1 j=1 0

×

3 1 

3

i=1 0

b jξi ξν [x + α(ξ − x)] (x j − ξ j ) − bνξi [x + α(ξ − x)] α

j=1

 −biξν [x + α(ξ − x)] α dα   1 ξ biξν [x + α(ξ − x)] α 3 − ξ bν [x + α(ξ − x)] α 2 + 2  − 2divξ bξν [x + α(ξ − x)] dα 1

− biξ (ξ ) 2 i=1 ν 3

1 

3 0

b jξi ξν [x + α(ξ − x)] (x j − ξ j )α 2

j=1

 − bνξi [x + α(ξ − x)] α − biξν [x + α(ξ − x)] α dα

−

3

i=1

b jξi ξν (ξ ) + cξν (ξ ).

(1.63)

24

1 Local Solvability of One-Dimensional Kernel …

 1 

3  biξν (αξ )αξi + biξν [x + α(ξ − x)] α(xi − ξi ) 1ξν (x, ξ ) = 1 (x, ξ ) 0

i=1



+bν (αξ ) + bν [x + α(ξ − x)] dα. From formulae (1.61) and (1.63) we obtain |(x, ξ )| ≤

3b02 4



 3 3 T +1 T +3 2 2

3b0 (T + 4) + c0 := χ0 , (x, ξ ) ∈ G T × G T , 2    3b02  ∂ 13b0 3b02 ≤  (T + 2) + + c0 := χ1 , (x, ξ ) + 2) (9T + 4) + (3T   ∂ξ 4 2 3 ν ν = 1, 2, 3, (x, ξ ) ∈ G T × G T . +

In view of these inequalities, we find the following estimates: |1 (x, ξ )| ≤

1 exp (3b0 T ) := α1 , 16π 2

|0 (x, ξ )| ≤ |1 (x, ξ )| |(x, t)| ≤ α1 χ0 := α0 , |3 (x, ξ )| ≤

 3 1 exp b0 T := α3 , 4π 2

|2 (x, ξ )| ≤ |3 (x, ξ )| |(x, t)| ≤ α3 χ0 := α2 .

(1.64)

From relations (1.63) we get  & '      ∇ξ  j (x, ξ ), Qη  ≤ max  jξ  max 1, η  i i 1≤i≤3

1≤i≤3

3 ≤ α0 b0 T + α1 χ1 := β0 , 2 j = 0, 1, (x, ξ ) ∈ G T × G T , t ∈ [0, T ], ϕ ∈ [0, 2π ], z ∈ [−1, 1]. Fix the number M > 0 and consider the ball    (ψ0 , M) = g(x, t) ∈ C [DT ] ψ0 − g ≤ M . It is easy to see that for ψ ∈  (ψ0 , M) it takes place the estimate ψ ≤ ψ0  + M ≡ N .

(1.65)

1.2 The Inverse Problem for Hyperbolic Second-Order …

Let

25

T ∗ = min {γi } , 1≤i≤6

where γi , i = 1, 2, 3, 4, 5, 6 are positive roots of the following equations, respectively:  4α1 2α2 M 2 γ+ , γ γ + = α3 N α3 N 2π α3 N 2  γ

4 ln γ + 3 6α3 

γ

8α2 + 6β0 4α1 γ + γ+ 3α3 N α3 N 2

8α2 β0 4 2β0 γ+ + + 2 3 3α3 N π α3 N α3 N γ

γ



 =

=

M , 2π α3 N 2 M

4π 2 α3 N 2

(1.66)

,

 3α1 3 3α2 γ+ = , γ2 + 4α3 N α3 N 2π α3 N

  N N α1 1 α2 + β0 1+ + γ+ = , ln γ γ 2 + 3 6α3 α3 N α3 N 4π α3 N

(1.67)

 4α2 + 3β0 3 γ 4γ + = . α3 N 8π 2 α2 N Theorem 1.4 Let all above-mentioned conditions about functions bi (x), i = 1, 2, 3, c(x) be fulfilled. Besides, f (t) ∈ C 1 [0, T ], and the following matching condition is satisfied: c(0) +

3 

3 i=1

4

 bi2 (0)

− 2bi xi (0) = 8π f (0).

(1.68)

Then in the domain DT , T ∈ (0, T ∗ ) , there exists a unique continuous solution of Eq. (1.59) belonging to a ball (ψ0 , M). For proof of Theorem 1.4, we use the following inequalities: |A[1, 1]| ≤

π (t − |x|) [3α2 (t + |x|) + α3 (t − |x|)(t + 2|x|)] 6  3  |B[1, 1]| ≤ 2π α3 t 2 − |x|2 ,  |E[1]| ≤ 4π α2 t t 2 − |x|2 ,

(1.69)

26

1 Local Solvability of One-Dimensional Kernel …

     3π   1 α3 ψ3  t   t 2 − |x|2 α2 + ≤  A ψ3 (t),   4 2 t 2 − |x|2   2 ψ3  x | |x|  , ln + 2 t + t 2 − |x|2 which easily follow from formulas (1.53) for all (x, t) ∈ DT . Let us prove, for example, the validity of the first of inequalities (1.69). By definition t A[1, 1] = 2 |x|

dτ 2 τ − |x|2

  r 3 (θ, τ ) 2 (x, ξ ) + 3 (x, ξ )(t − τ ) dω,

 S(x,τ )

we use the formula r 2 (θ, τ ) sin θ dθ = −

τ 2 − |x|2 dr, 2|x|

which follows from the equation of the family of confocal ellipsoids S(x, τ ) in the polar coordinate system: r (θ, τ ) =

τ 2 − |x|2 . 2(τ − |x| cos θ )

We proceed to the integration variable r and, taking into account the estimates from (1.64) for 2 , 3 , obtain 2π |A[1, 1]| ≤ |x|

τ +|x|

t

2 (α2 + α3 (t − τ )) dτ

|x|

r dr τ −|x| 2

π = (t − |x|) [3α2 (t + |x|) + α3 (t − |x|)(t + 2|x|)] . 6 We show that for any T ∈ (0, T ∗ ) the operator U maps the ball (ψ0 , M) into itself. Obviously, the operator U transforms ψ(x, t) ∈ C [DT ] into functions also belonging to space C [DT ]. In addition, composing the norm of differences and using formulas (1.60), inequalities (1.64), (1.65), and (1.69), we find U1 ψ − ψ01  ≤ 2π α1 ψ3  (t − |x|) + ψ1  |A [ψ3  , 1]| π α3 2 3 ≤ N T + π α2 N T 2 + 2π α1 N T, 2   t 2 − |x|2 U2 ψ − ψ02  ≤ 8π α1 ψ3  4πβ0 ψ3  t 2 + |x|2 + + ψ1  ψ3  B[1, 1]| + ψ1  |E[1]|

1.2 The Inverse Problem for Hyperbolic Second-Order …

27

     1  2  + ψ2   A ψ3  ,   t − |x|2  t 2 − |x|2     NT2 4π N T α3 N T 2 3 T α2 + + ln T ≤ 2π α3 N T + 3 2 4 +4π α2 N T 2 + 4πβ0 N T 2 + 8π α1 N T,

 2π U3 ψ − ψ03  ≤ 4πβ0 t + 8π β0 ψ3  t + ψ1  ψ3  B[1, 1]| t       1  2  + ψ1  |E[1]| + ψ2   A ψ3  ,   t − |x|2  t 2 − |x|2  x=0  2 8π N T α3 N α2 + + 8π 2 α2 N T ≤ 4π 2 α3 N 2 T 2 + 3 2 2

+8π 2 β0 N T + 4πβ0 T. For all T ∈ (0, T ∗ ) from these inequalities in view of (1.66) it follows U ψ − ψ0  ≤ M, i.e., U ψ ∈ (ψ0 , M). Now suppose that ψ (1) , ψ (2) are any two elements of (ψ0 , M), T < T ∗ . Then, using auxiliary inequalities of the following type       (1)   (1)  (1) (1) (2) (2)  (2)  ψi ψ j − ψi ψ j  ≤ ψi  ψ j − ψ j      (1)     (1) (2)   + ψ (2) − ψ − ψ (2)  , ψ  j i i  ≤ 2N ψ we obtain     (1) (2)  U1 ψ (1) − U1 ψ (2)  ≤ 2π α1 (t − |x|)  ψ3 − ψ3         +  A ψ3(1) , ψ1(1) − A ψ3(2) , ψ1(2)     π 2π ≤ 2π α1 T + α2 T 2 + α3 N T 3 ψ (1) − ψ (2)  , 2 3      (1) (2)  U2 ψ (1) − U2 ψ (2)  ≤ 8π α1 t 2 − |x|2  ψ3 − ψ3          +4πβ0 (t 2 − |x|2 ) ψ3(1) − ψ3(2)  + ψ1(1) ψ3(1) − ψ1(2) ψ3(2)  |B[1, 1]|        ψ2(1)  (1) (2)  (1) 2 2  + ψ1 − ψ1  |E[1]| + t − |x| ||  A ψ3 ,  t 2 − |x|2     ψ (2)  ≤ 8π α1 T + 4πβ0 T 2 + 4π α2 T 2 −A ψ3(2) ,  2  2 2 t − |x|     (1) 8π N 2 T 2 α3 T T ψ − ψ (2)  , α2 + + 4π α3 N T 3 + + ln T 3 2 4

28

1 Local Solvability of One-Dimensional Kernel …

    (1) (2)  U3 ψ (1) − U3 ψ (2)  ≤ 8π 2 β0 t  ψ3 − ψ3      2π  (1) (1)  (1) (2) (2)  (2)  + ψ1 ψ3 − ψ1 ψ3  |B[1, 1]| + ψ1 − ψ1  |E[1]| t         ψ2(1) ψ2(2) (1) (2) 2 2   + t − |x| ||  A ψ3 ,  − A ψ3 ,   t 2 − |x|2 t 2 − |x|2 x=0  2 T 8π ≤ 8π 2 α2 T + 8π 2 β0 T + 8π 2 α3 N T 2 + (α2 + α3 N T ) 3   × ψ (1) − ψ (2)  . If we take into account (1.67), then from the last inequalities one gets     U ψ (1) − U ψ (2)  ≤ T ψ (1) − ψ (2)  . ∗ T This implies that for any T ∈ (0, T ∗ ) the operator U implements a contraction mapping of the ball (ψ0 , M) onto itself. Then, by the Banach theorem in the ball (ψ0 , M), there exists a unique continuous solution of Eq. (1.59). Therefore, solving the system of Eqs. (1.55)–(1.57), for example, by the method of successive approximations (from the Banach theorem it follows its convergence to the solution), we uniquely construct in the domain DT for T ∈ (0, T ∗ ) the functions v(x, t), w(x, t), k(t). By that it is determined the solution of the inverse problem, that is, the function k(t) on the interval [0, T ]. Theorem 1.5 Under the conditions of Theorem 1.4, the function k(t) ∈ C[0, T ] is uniquely determined by the information (1.44). The proof of this theorem is carried out by contradiction. Then, by virtue of Theorem 1.4, there exists T ∈ (0, T ∗ ) , such that k1 (t) = k2 (t), t ∈ [0, T ].

(1.70)

We denote by T0 the exact upper bound of all possible numbers T ∈ (0, T ∗ ) , for which equality (1.67) holds. Now we consider a similar with (1.41), (1.42), and (1.44) problem, in which the Cauchy data are given on the hyperplane t = T0 . Since     u 1 t 0 is constant and that ρ = ρ(x3 ), μ = μ(x3 ), and λ = λ(x3 ) are functions of one variable x3 and satisfy the conditions ρ(x3 ) > 0, μ(x3 ) > 0, and λ(x3 ) + 2μ(x3 ) > 0. Consider system (1.85) and (1.86) with the following initial and boundary conditions: (1.88) u i |t 0; θ1 (t) = tθ (t), θ (t) = 1 for t ≥ 0, θ (t) = 0 for t < 0, and δ (t) is the derivative of the Dirac delta function. The boundary condition (1.91) simulates a heat shock on the surface of the halfspace x3 > 0. Whereby the temperature on the boundary increases from the initial temperature T0 to the value T1 . After that, there is convective heat exchange between the surface x3 = 0 and the medium filling the half-space x3 > 0. The problem of finding the displacement vector u(x, t) and the temperature increment H (x, t), satisfying relations (1.85)–(1.91) (in the generalized sense) for given functions α(y), h(t), ρ(x3 ), λ(x3 ), and μ(x3 ) and given constants k, γ , T0 , and T1 will be called the direct problem. Let us state the inverse problem: find the function h(t), t ∈ [0, T ], T > 0, occurring in Eq. (1.85) from the known additional condition u 3 (0, t) = g(t), t ∈ [0, T ]

(1.92)

of the solution of the direct problem. Thus, the functions α(y), ρ(x3 ), λ(x3 ), and μ(x3 ) and the constants k, γ , T0 , and T1 are assumed to be given. A study of various statements of inverse problems of determining the functions ρ(x3 ), λ(x3 ), μ(x3 ), and α(y) from the system of elasticity equations can be found in the monograph [6]. In the following, we assume that the functions α, ρ, λ, and μ are sufficiently smooth. The order of smoothness will be specified below. Problem (1.86), (1.89), and (1.91) can be viewed as a separate initial–boundary value problem for the heat equation. One can readily verify that it has the solution H (z, t) = θ (t)(T1 − T0 )    t  √ z z − exp(γ z + γ 2 kτ )erfc √ + γ kτ erfc √ dτ, × 2 kτ 2 kτ 0

where √ erfc(z) = 1 − erf(z), erf(z) = (2/ π )

z exp(−ξ 2 )dξ, z = x3 . 0

Therefore, Eqs. (1.85), (1.88), and (1.90) for the function H (z, t) are equivalent to the relations   ∂ 2u3 ∂u 3 ∂ ρ 2 = (λ(z) + 2μ(z)) − (3λ(z) + 2μ(z))R(H (z, t)) ∂t ∂z ∂z t + h(τ )u 3 (z, t − τ )dτ, (1.93) 0

36

1 Local Solvability of One-Dimensional Kernel …

  u 3 

≡ 0,

(1.94)

t y > 0.

0

Here (y, t) := {(τ, ξ )|0 ≤ ξ ≤ y, t − y + ξ ≤ τ ≤ t + y − ξ } ,

(1.106)

38

1 Local Solvability of One-Dimensional Kernel …

w0 (y, t) 1 = [g0 (t + y) + g0 (t − y)] 2 t+y t+y 1 a (0) + a(0)(3λ(0) + 2μ(0)) R (H (0, β)) dβ − g0 (β)dβ 2 4a(0) t−y t−y     1 ∂  (3λ˜ (ξ ) + 2μ(ξ ˜ ))R( H˜ (ξ, τ )) dτ dξ. (1.107) + a(ξ )a(0) 2 ∂ξ (y,t)

In relation (1.106), we pass to the limit as t → y + 0 and use relation (1.105):  y 2y−ξ   ξ

0

a(0) 1 h(τ − ξ ) − q(ξ )w(ξ, τ ) − ρ(ξ ˜ ) ρ(ξ ˜ )

= a (0) − a(0)

τ −ξ

 h(β)w(ξ, τ − β)dβ dτ dξ

0

y q(s)ds − 2w0 (y, y). 0

In particular, this implies one of the matching conditions between problem data: 2g0 (0) = a (0).

(1.108)

By differentiating the previous equation with respect to y, we obtain y  0

a(0) 1 h(2(y − ξ )) − q(ξ )w(ξ, 2y − ξ ) − ρ(ξ ˜ ) ρ(ξ ˜ )

d a(0)q(y) − w0 (y, y). =− 2 dy

 h(β)w(ξ, 2y − ξ − β)dβ dξ

2(y−ξ  )

0

(1.109)

Since d a (0) w0 (y, y) = g0 (2y) + a(0) [3λ(0) + 2μ(0)] R [H (0, 2y)] − g0 (2y) dy 2a(0) y   ∂  (3λ˜ (ξ ) + 2μ(ξ ˜ ))R( H˜ (ξ, 2y − ξ )) dξ, (1.110) a(ξ )a(0) + ∂ξ 0

1.3 Inverse Problem for the System of Thermoelasticity Equations …

39

from (1.109) we readily obtain the second matching condition  2g0 (0)a(0) − a g0 (0) = −a 2 (0) S (0) − 2(S (0))2 .

(1.111)

Recall that the functions a(y) and S(y) occurring in (1.108) and (1.111) are defined in (1.96). To obtain an integral equation for h, we differentiate relation (1.109) with respect to y after the replacement of the variable 2(y − ξ ) in the second integral by ξ we have: a(0) h(2y) − q(y)w(y, y) ρ(0) y  a(0)ρ˜ (ξ ) 2 h(2(y − ξ )) − h(2(y − ξ ))w(ξ, ξ ) + 2ρ˜ 2 (ξ ) ρ(ξ ˜ )

−

0

1 −2q(ξ )wt (ξ, 2y − ξ ) − ρ(ξ ˜ )

2(y−ξ  )

h(β)wt (ξ, 2y − ξ − β)dβ]dξ 0

a(0)q (y) d2 =− − 2 w0 (y, y). 2 dy By using relation (1.105) for the computation of w(y, y) and by solving the last relation for h, we obtain the equation y/2 h(y) = h 0 (y) + 0

2ρ(0) h(y − 2ξ )w(ξ, ξ ) a(0)ρ(ξ ˜ )

ρ(0)ρ˜ (ξ ) 2ρ(0) − h(y − 2ξ ) + q(ξ )wt (ξ, y − ξ ) 2 2ρ˜ (ξ ) a(0) ⎤ y−2ξ  ρ(0) + h(β)wt (ξ, y − ξ − β)dβ ⎦ dξ, a(0)ρ(ξ ˜ )

(1.112)

0

where ⎡

⎤ y/2 ⎣ ρ(0)a (0) − ρ(0) q(s)ds ⎦ h 0 (y) = q 2 2a(0) 2 0   ρ(0) d2 q (t) + − w (t, t) . 0 2ρ(0) a(0) dt 2 t=y/2 y



(1.113)

40

1 Local Solvability of One-Dimensional Kernel …

To find the function (d2 /dt 2 )w0 (t, t)|t=y/2 from the definition of the function h 0 (y), one should integrate by parts in (1.110), differentiate both sides of the resulting relation with respect to y, and replace y by y/2. Then we obtain d2 a (0) g (y) w0 (t, t)|t=y/2 = 2g0 (y) − 2 dt a(0) 0 1       y y  y a(0) (3λ˜ (t) + 2μ˜ (t)) , R H˜ + a t=y/2 2 2 2 1   y   y    y y  y + 2μ˜ α H˜ , ψ(y) + a( )a(0) 3λ˜ 2 2 2 2 2 y/2  a(0)  a (ξ ) (3λ˜ (ξ ) + 2μ(ξ − ˜ )) α( H˜ (ξ, y − ξ )) H˜ y (ξ, y − ξ )dξ, (1.114) a(ξ ) 0

where d d ˜ y y y , = Hz (z, t)|z=τ −1 (y/2),t=y/2 τ −1 ( ) H dy 2 2 dy 2 1 1 + Ht (z, t)|z=τ −1 (y/2),t=y/2 = − γ (T1 − T0 ) 2 2   y/2 √ z × exp(γ z + γ 2 kτ )erfc √ + γ kτ v(z) dτ 2 kτ z=τ −1 (y/2) 0   1 z + (T1 − T0 ) erfc √ 2 2 kt  √ z − exp(γ z + γ 2 kt)erfc √ + γ kt z=τ −1 (y/2),t=y/2 , 2 kt ˜ Hy (ξ, y − ξ ) = (T1 − T0 )   z − exp(γ z + γ 2 k(y − ξ )) × erfc √ 2 k(y − ξ )    z ×erfc √ . + γ k(y − ξ ) 2 k(y − ξ ) z=τ −1 (ξ )

ψ(y) :=

To close the system (1.106) and (1.112) with respect to the unknown functions, one needs an equation for wt (y, t). We use relation (1.106) for the computation of wt (y, t). By using the equivalent description of the domain (y, t) in the form (y, t) = {(ξ, τ )|t − y ≤ τ ≤ t + y, 0 ≤ ξ ≤ y − |t − τ |} ,

1.3 Inverse Problem for the System of Thermoelasticity Equations …

41

we differentiate equation (1.106) with respect to t: 1 wt (y, t) = w0t (y, t) + 2

t+y t−y

a(0) h(τ + |t − τ | − y) ρ(y ˜ − |t − τ |)

1 ρ(y ˜ − |t − τ |) τ +|t−τ   |−y h(β)w(y − |t − τ |, τ − β)dβ sgn(t − τ )dτ,

− q(y − |t − τ w(y − |t − τ |, τ ) −

(1.115)

0

where, according to (1.107), w0t has the form w0t (y, t) 1 1 g (t + y) + g0 (t − y) + a(0)(3λ(0) + 2μ(0)) = 2 0 2 a (0) × [R (H (0, t + y)) − R (H (0, t − y))] − [g0 (t + y) − g0 (t − y)] 4a(0)   y ∂ 1  ˜ (3λ(ξ ) + 2μ(ξ ˜ )) R( H˜ (ξ, t + y − ξ )) a(ξ )a(0) + 2 ∂ξ 0  −R( H˜ (ξ, t − y + ξ )) dξ a (0) 1 g0 (t + y) + g0 (t − y) − [g0 (t + y) − g0 (t − y)] 2 4a(0)  y  a(0) 1 ˜ )) R( H˜ (ξ, t + y − ξ )) a (ξ )(3λ˜ (ξ ) + 2μ(ξ − 4 a(ξ ) 0  −R( H˜ (ξ, t − y + ξ )) dξ.

=

Therefore, from relations (1.102)–(1.105), we obtain the system of integral equations (1.106), (1.112), and (1.115). Conversely, under fulfillment of the matching conditions (1.108) and (1.91), from these equations, one can readily derive the differential relations of the inverse problem (1.102)–(1.105). The above argument justifies the following assertion. Lemma 1.1 Let   g0 (t) ∈ C 2 [0, T ] , ρ(z) ∈ C 3 0, τ −1 (T /2) , λ(z) ∈ C 3 0, τ −1 (T /2) ,  μ(z) ∈ C 3 0, τ −1 (T /2) , α(y) ∈ C [0, Y ] ,

42

1 Local Solvability of One-Dimensional Kernel …

Y := H˜ (T /2, T /2) and let the matching conditions (1.108) and (1.111) be satisfied. Then solving the inverse problem (1.102)–(1.105) in the class of functions h(t) ∈ C [0, T ] and (w, wt ) ∈ C (DT ) , DT := {(y, t)|0 ≤ y ≤ t ≤ T − y} , is equivalent to solving system of Eqs. (1.106), (1.112), and (1.115). For system (1.106), (1.112), and (1.115), we have the following local unique solvability theorem. Theorem 1.7 Let the assumptions of the Lemma 1.1 be valid, and let T, T0 , T1 , k, and γ be given positive numbers; moreover, let T1 − T0 > 0. Then there exists T ∗ ∈ (0, T ] such that the system of integral equations (1.106), (1.112), and (1.115) has a unique solution in the domain DT ∗ . Proof The function H (z, t) occurring on the right-hand sides in the integral equations (1.112) and (1.115) is a regular solution of the heat equation (1.86) in the domain z > 0, t > 0 with the conditions (∂ H/∂z − γ H ) z=+0 = −γ (T1 − T0 )t and H |t=+0 = 0. Therefore, the functions H (z, t), Hz (z, t), and Ht (z, t) are continuous in the domain DT \{z = 0}. From the closed-form expressions for these functions, one can readily see that they are also continuous for z = 0. Under the assumptions of Theorem 1.7, this implies that relations (1.106), (1.112), and (1.115) form a closed system of Volterra integral equations of the second kind with continuous free terms and kernels for the unknown functions w(z, t), h(t), and wt (z, t). The small parameter in this system is the integration interval, whose length does not exceed T . Therefore, the Banach contraction mapping principle can be used for small T , which provides the existence of a unique solution of system (1.106), (1.112), and (1.115). This solution can be found by the successive approximation method. Each iteration leads to continuous functions. Hence the limit functions are continuous solutions of Eqs. (1.106), (1.112), and (1.115). Remark 1.1 The number T ∗ in Theorem 1.7 will be chosen from the condition that the integral operators in Eqs. (1.106), (1.112), and (1.115) are contraction operators on some set of continuous functions and satisfy the inequality H˜ (T ∗ /2, T ∗ /2) < T ∗ . Now let us obtain a stability estimate for the solution and prove the uniqueness theorem for arbitrary T > 0. Let K (h 0 ) be the set of functions h(t) ∈ C[0, T ] satisfying the inequality |h(t)| ≤ h 0 with a given constant h 0 for t ∈ [0, T ], and let R0 = max{ρ(z)C 3 [0,τ −1 (T /2] , λ(z)C 3 [0,τ −1 (T /2] , μ(z)C 3 [0,τ −1 (T /2] , α(y)C[0,Y ] , g0 (t)C 2 [0,T ] , ρ ∗ (z)C 3 [0,τ −1 (T /2] , λ∗ (z)C 3 [0,τ −1 (T /2] , μ∗ (z)C 3 [0,τ −1 (T /2] , α ∗ (y)C[0,Y ] , g0∗ (t)C 2 [0,T ] }, Theorem 1.8 Let h(t) ∈ K (h 0 ) and h ∗ (t) ∈ K (h 0 ) be the solutions of the inverse problem (1.102)–(1.105) with data

1.3 Inverse Problem for the System of Thermoelasticity Equations …

43

{ρ(z), λ(z), μ(z), α(y), g0 (t)} , and



 ρ ∗ (z), λ∗ (z), μ∗ (z), α ∗ (y), g0∗ (t) ,

respectively. Then there exists a positive number C = C(h 0 , R0 , γ , T, T0 , T1 , h 0 ) such that the following stability estimate holds:  h(t) − h ∗ (t)C[0,T ] ≤ C ρ − ρ ∗ C 3 [0,τ −1 (T /2] + λ − λ∗ C 3 [0,τ −1 (T /2]  (1.116) +μ − μ∗ C 3 [0,τ −1 (T /2] + α − α ∗ C[0,Y ] + g0 − g0∗ C 2 [0,T ] . Proof First, by considering Eq. (1.106) for h(t) ∈ K (h 0 ) as an integral equation for w(y, t) and by forming a series of successive approximations by the usual scheme, we note that the absolute value of the function w(y, t) in the domain DT is bounded by some constant c0 = c0 (h 0 , R0 , γ , T, T0 , T1 , h 0 ), i.e., |w(y, t)| ≤ c0 , (y, t) ∈ DT .

(1.117)

Let us introduce the differences w¯ = w − w∗ , h¯ = h − h ∗ , w¯ t = wt − wt∗ , ρ¯ = ρ − ρ ∗ , λ¯ = λ − λ∗ , μ¯ = μ − μ∗ , α¯ = α − α ∗ , g¯0 = g0 − g0∗ . The domain DT admits the equivalent description     T  T  DT = (y, t)|0 ≤ y ≤ − t −  , 0 ≤ t ≤ T . 2 2 Let 

  ¯ , ϕ(t) = max h(t)

 max

0≤y≤T /2−|t−T /2|

|w(y, ¯ t)|,

max

0≤y≤T /2−|t−T /2|

|w¯ t (y, t)| ,

¯ and w¯ t with the use of relations (1.106), t ∈ [0, T ] . By forming equations for w, ¯ h, (1.112), and (1.115) and by making simple estimates, we obtain the integral inequality  ϕ(t) ≤ c∗ ρ − ρ ∗ C 3 [0,τ −1 (T /2] + λ − λ∗ C 3 [0,τ −1 (T /2] + μ − μ∗ C 3 [0,τ −1 (T /2] ∗

+ α − α C[0,Y ] + g0 −

g0∗ C 2 [0,T ]



t + c∗∗

ϕ(τ )dτ, 0

(1.118)

44

1 Local Solvability of One-Dimensional Kernel …

for (y, t) ∈ DT , where c∗ and c∗∗ are positive constants depending on h 0 , R0 , k, γ , T, T0 , and T1 . By using the Gronwall inequality, from (1.118) we obtain the estimate (1.116) with some constant C. The uniqueness theorem for each T > 0 is a consequence of Theorem 1.8. Theorem 1.9 Let the functions (h(t), h ∗ (t)) ∈ C [0, T ] , ρ(z), λ(z), μ(z), α(y), g0 (t), and ρ ∗ (z), λ∗ (z), μ∗ (z), α ∗ (y), and g0∗ (t) have the same meaning as in Theorem 1.7. If, in addition ρ(z) = ρ ∗ (z), λ(z) = λ∗ (z), μ(z) = μ∗ (z) for z ∈  −1 0, τ (T /2 , α(y) = α ∗ (y) for y ∈ [0, Y ] , and g0 (t) = g0∗ (t) for t ∈ [0, T ] , then h(t) = h ∗ (t) on the segment [0, T ].

1.4 Inverse Problem for an Integro-differential Equation of Acoustics We consider the initial boundary-value problem for the one-dimensional integrodifferential equation of acoustics 1 c2 (z)

∂ 2v ∂ ln ρ(z) , z > 0, t > 0, = z − ∂t 2 ∂z v |t≤0 ≡ 0, (+0, t) = δ (t),

(1.119) (1.120)

where c(z) > 0 is the wave velocity, ρ(z) is the medium density, v(z, t) is the acoustic pressure; (z, t) is the stress; and the functions (z, t) and v(z, t) are related as ∂v (z, t) + (z, t) = ∂z

t k(t − τ ) 0

∂v (z, τ )dτ ; ∂z

(1.121)

δ (t) is as before the derivative of the Dirac delta function δ(t). Note that the solution to problem (1.119)–(1.121) is the limit (in the sense of generalized functions) of the solutions to problem (1.119)–(1.121), corresponding to δ (t − t0 ) instead of δ (t) in the boundary condition (1.120) at t0 tending to + 0. In this section the inverse problem is posed as follows: to determine the kernel k(t), t > 0, that enters Eq. (1.119) via relation (1.121), given that for the solution of the problem we have v(+0, t) = g(t), t > 0. (1.122) Equation (1.119), which allows for absorption by an imperfectly elastic medium by means of relation (1.121), is encountered in geophysics when the properties of the medium are studied by seismic waves. In fact, the system of Boltzmann equations (one of the most common for linear inelastic media) reduces to Eq. (1.119) in the one-dimensional case under appropriate assumptions about smoothness.

1.4 Inverse Problem for an Integro-differential Equation of Acoustics

45

Let us introduce a new variable x by the formula z x = ψ(z) = 0

dξ , c(ξ )

and adopt the notation 2 v (x, t) := v(ψ −1 (x), t), a(x) := c(ψ −1 (x)), b(x) := ρ(ψ −1 (x)). where ψ −1 (x) is an inverse of the function ψ(z). In what follows, we assume that c(z) > 0 and ρ(z) > 0. The main result in the section is the following theorem on the local unique solvability of the inverse problem. Theorem 1.10 Assume that a function g(t) can be represented as g(t) = −c(+0)δ(t) + θ (t)g0 (t), 2 where the function g0 (t) belongs to the space C [0, T ] , the functions c(z) and ρ(z) 3 −1 belong to the space C 0, ψ (T ) . Then, for sufficiently small T0 ∈ (0, T ), the solution of the inverse problem in Eqs. (1.119)–(1.122) exists and is unique in the class of kernels k(t) belonging to the space C 2 [0, T0 ] .

Relations (1.119)–(1.122) for the new functions 2 v , λ and new variable x become ⎤ ⎡  2 t ∂ 22 v ∂ λ (x) ∂ ⎣ = − v (x, τ )dτ ⎦ , 2 v (x, t) + k(t − τ )2 ∂t 2 ∂x2 λ(x) ∂ x 0

x > 0, t > 0, ∂2 v (+0, t) + 2 v |t≤0 ≡ 0, ∂x

(1.123)

t k(t − τ ) 0

∂2 v (+0, τ )dτ = c(+0)δ (t), ∂x

2 v (+0, t) = g(t), t > 0,

(1.124)

(1.125)

where λ(x) := a(x)b(x). Further, c(0), h(0), λ(0), etc. are understood to be the respective values of these functions for x tending to zero from the right. We transform the integro-differential equation (1.123) so that, first, the integrand does not contain the derivatives of the function 2 v with respect to x and, second, the coefficient in front of ∂∂2vx in the terms outside the integral sign is zero. These requirements can be fulfilled by introducing a new function u with the formula ⎡ ⎣2 v (x, t) +

t 0

⎤

  λ(0) k(0)t = u(x, t). k(t − τ )2 v (x, τ )dτ ⎦ exp − 2 λ(x)

46

1 Local Solvability of One-Dimensional Kernel …

Then it is easy to verify by direct calculations that the function 2 v can be expressed in terms of the function u using the formula ⎡ 2 v (x, t) = ⎣exp (k(0)t/2) u(x, t) +

t

⎤ h(t − τ ) exp (k(0)τ/2) u(x, τ )dτ ⎦

0

where

λ(x) , λ(0)

t h(t) = −k(t) −

k(t − τ )h(τ )dτ.

(1.126)

0

Let us introduce the notation 2λ(x)λ (x) − 3(λ (x))2 h 2 (0) − h (0) + , 4 4λ2 (x)   ˜ := h (t) exp h(0)t , g˜ 0 (t) := g(t) exp h(0)t , k(t) 2 2  h(0)t . h 0 (t) := k(t) exp 2 (x) :=

Then Eqs. (1.123)–(1.125) in the new functions u(x, t) and h(t) become ∂ 2u ∂ 2u = 2 + (x)u + 2 ∂t ∂x

t

˜ − τ )u(x, τ )dτ, x > 0, t > 0, k(t

(1.127)

0

u|t≤0 ≡ 0, ∂u  λ (0) 1 u(0, t), = c(0)δ (t) + c(0)k(0)δ(t) −  ∂ x x=+0 2 2λ(0)

(1.128) (1.129)

t u|x=+0 = g˜ 0 (t) +

h 0 (t − τ )g˜ 0 (τ )dτ,

(1.130)

0

It follows from the theory of hyperbolic equations that the function u(x, t) as a solution to the direct problem in Eqs. (1.127)–(1.130) possesses the property u ≡ 0, t < x, x > 0 and, in a neighborhood of the characteristic line t = x, has the structure: u(x, t) = α(x)δ(t − x) + θ (t − x)u(x, ˜ t), (1.131) where u(x, ˜ t) is a regular function.

1.4 Inverse Problem for an Integro-differential Equation of Acoustics

47

Let us denote β(x) := u(x, ˜ x + 0). Substituting the function (1.131) into Eqs. (1.127)–(1.130) and using the method of separation of singularities [7], we find λ (0) = 21 α (x) = 0, α(0) = −c(0), 2β (x) − (x)α(x) = 0, α (0) − β(0) + α(0) 2λ(0) c(0)k(0). We solve these ordinary differential equations to obtain ⎞ ⎛ x c(0) ⎝ λ (0) α(x) = −c(0), β(x) = − + (ξ )dξ ⎠ . k(0) + 2 λ(0) 0

It follows from the above that the function u(x, ˜ t) in the domain D := {(x, t) : t > x > 0} satisfies the equations

∂ 2 u˜ ∂ 2 u˜ ˜ − x) + = + (x)u˜ − c(+0)k(t ∂t 2 ∂x2

t

˜ − τ )u(x, k(t ˜ τ )dτ,

x

u| ˜ t=x+0 = β(x), 

∂ u˜ λ (0) + u(x, ˜ t) ∂x 2λ(0)

(1.132)

(1.133)

 = 0,

(1.134)

x=0

t u| ˜ x=0 = g˜ 0 (t) − c(0)h 0 (t) +

h 0 (τ )g˜ 0 (t − τ )dτ, t > 0.

(1.135)

0

By requiring that the functions u(x, ˜ t) and (∂ u/∂ ˜ x)(x, t) for x = t = 0, from relations (1.133)–(1.135) we obtain the following expressions for the values h(0) and h (0) in terms of the known values: h(0) = − h (0) =

2 λ (0) g˜ 0 (0) − , c(0) λ(0)

1 1 (2h 2 (0) − (0)) − (g˜ (0) + g˜ 0 (0)h(0)). 2 c(0) 0

When deriving the last expression, we used the following relations, which follow from Eq. (1.126):



t

h (t) = −k (t) − k(0)h(t) −

k (t − τ )h(τ )dτ,

0

h (0) = −k (0) + k 2 (0).

48

1 Local Solvability of One-Dimensional Kernel …

In what follows, we assume that, in the relations for (x), h(0) and h (0) have been replaced by their already established values. The proof of the theorem is based on the following lemma. Lemma 1.2 If the conditions of Theorem 1.10 are satisfied, then problem (1.132)– (1.135) for (x, t) ∈ DT , DT = ((x, t)|0 ≤ x ≤ t ≤ T − x) is equivalent to the prob˜ lem of determining functions u(x, ˜ t), (∂ u/∂t)(x, ˜ t), k(t), h 0 (t), h 0 (t), h 0 (t) from the system of equations: t u(x, ˜ t) = β(x) + x

∂ u˜ (x, τ )dτ, ∂τ

(1.136)

 ∂ u˜ 1 ˜ − x)x (x, t) = g˜ (t − x) − c(0)h 0 (t − x) − h(0)g˜ 0 (t − x) − c(0)k(t ∂t 2 0  t−x x +t λ (0) c(0) 1 h 0 (t − y − τ )g˜ 0 (τ )dτ + u(0, ˜ t − x) −  + 2λ(0) 4 2 2 0 ⎡ ⎤ x t−x 1 ⎣ ˜ )u(ξ, + (ξ )u(ξ, ˜ t − x + ξ) − k(τ ˜ t − x + ξ − τ )dτ ⎦ dξ 2 0

0

t+x

2  1 ˜ + x − 2ξ ) (ξ )u(ξ, ˜ t + x − ξ ) − c(0)k(t + 2 x t+x−2ξ 

+

 ˜ )u(ξ, k(τ ˜ t + x − ξ − τ )dτ dξ,

(1.137)

0

  ˜ = − 1  t − 2 g˜ 0 (t) − c(0)h 0 (t) + g˜ 0 (0)h 0 (t) − λ (0) ∂ u˜ (0, t) k(t) 2 2 c(0) 2λ(0) ∂t     t t 1 t t 2 2 t −τ ˜ −  β − dτ g˜ 0 (t − τ )h 0 (τ )dτ − k(τ )β 2 2 2 c(0) c(0) 2 t

2 + c(0)

2 0

0

⎡

⎣(ξ ) ∂ u˜ (ξ, t − ξ ) + ∂t

h 0 (t) = −h(0) +



0

t−2ξ 

0

⎤ ∂ u ˜ ˜ ) (ξ, t − ξ − τ )dτ ⎦ dξ, k(τ ∂t

t h (0) − h (0) t + (t − τ )h 0 (τ )dτ, 2

(1.138)

2

0

(1.139)

1.4 Inverse Problem for an Integro-differential Equation of Acoustics

h 0 (t)

h 2 (0) − h (0) + = 2

t

49

h 0 (τ )dτ,

(1.140)

0

h 0 (t)

˜ + = −k(t)



t h 2 (0) ˜ − τ )h 0 (τ )dτ. − h (0) h 0 (t) − k(t 4

(1.141)

0

Proof Note that the following factorization identities hold: ∂ 2 u˜ ∂ 2 u˜ − = ∂t 2 ∂x2



∂ ∂ − ∂t ∂x



∂ ∂ + ∂t ∂x



 u˜ =

∂ ∂ + ∂t ∂x



∂ ∂ − ∂t ∂x

u. ˜

Taking these identities into account, we integrate Eq. (1.132) along the corresponding characteristics of the first-order differential operators in the domain (x, t) ∈ DT . We integrate along the characteristic of the operator ∂/∂t − ∂/∂ x from the point (x, t) to the point ((x + t)/2, (x + t)/2) in the plane of variables  x+t  (ξ, τ ). Using the relation , which follows from condi ˜ + t)/2, (x + t)/2) = − c(0) (∂/∂t + ∂/∂ x) u((x 2 2 tion (1.133) after differentiation with respect to x, we obtain 

∂ ∂ + ∂t ∂x



 (x+t)/2   c(0) x +t (ξ )u(ξ, ˜ t + x − ξ) u(x, ˜ t) = −  + 2 2 x

˜ + x − 2ξ ) − −c(0)k(t

t+x−2ξ 

 ˜k(τ )u(ξ, ˜ t + x − ξ − τ )dτ dξ.

(1.142)

0

Let us perform integration along the characteristic of the operator ∂/∂t + ∂/∂ x from the point (0, t − x) to the point (x, t). We use formulas (1.134) and (1.135) to obtain 

∂ ∂ − ∂t ∂x



−c(0)h 0 (t x + 0

˜ − x)x u(x, ˜ t) = g˜ 0 (t − x) − h(0)g˜ 0 (t − x) − c(+0)k(t

λ (0) u(t ˜ − x, 0) + − x) + 2λ(0)

t−x 0

⎡

h 0 (t − x − τ )g˜ 0 (τ )dτ

⎤ t−x ˜ )u(ξ, ⎣(ξ )u(ξ, ˜ t − x + ξ) − k(τ ˜ t − x + ξ − τ )dτ ⎦ dξ.(1.143) 0

50

1 Local Solvability of One-Dimensional Kernel …

Summing (1.142) and (1.143), we arrive at Eq. (1.137). By setting x = 0 in Eq. (1.142) and using conditions (1.134) and (1.135), we find g˜ 0 (t)

+ g˜ 0 (0)h 0 (t) −

c(0)h 0 (t)

λ (0) u(0, ˜ t) + − 2λ(0)

t

g˜ 0 (t − τ )h 0 (τ )dτ

0

 t/2  t c(0) ˜ − 2ξ )  + =− (ξ )u(ξ, ˜ t − ξ ) − c(0)k(t 2 2 0 t−2ξ 

−

˜ )u(ξ, k(τ ˜ t − ξ − τ )dτ dξ.

0

Differentiating this relation with respect to t, after simple transformations, we obtain Eq. (1.138). To complete the system of integral equations (1.136)–(1.138), we use formulas (1.139)–(1.141), which follow from the definition of the function h 0 (t) and relation (1.126). Under the condition of Theorem 1.10, the equivalence of the system of integral equations (1.136)–(1.141) and the inverse problem in Eqs. (1.132)–(1.135) can be proved in the standard way [8]. Lemma 1.2 is proved. Let us introduce the notation h 2 (0) − h (0) =: r00 , 4

λ (0) =: λ0 , 2λ(0)

c(0) =: c0 , 2

 2 λ (0) λ (0)g˜ 0 (0) 1 g˜ 0 (0) + =: L 0 , =: M0 . 3 c(0) λ(0) 3λ(0)c(0) We write the system of Eqs. (1.136)–(1.141) as the operator equation ϕ = Aϕ,

(1.144)

where   ϕ = ϕ1 (x, t), ϕ2 (x, t), ϕ3 (t), ϕ4 (t), ϕ5 (t), ϕ6 (t)  ∂ u˜ ˜ − x)x + h 0 (t − x) + λ0 h 0 (t − x)), k(t) ˜ − 2h 0 (t) (x, t) + c0 (k(t = u(x, ˜ t), ∂t  ˜ +3L 0 h 0 (t) − 3M0 h 0 (t), h 0 (t), h 0 (t), h 0 (t) + k(t) − r00 h 0 (t) is a vector function with its first two components belonging to the space C(DT ) and the other components to the space C[0, T ] (we denote the linear space of such vector functions by C6 ), while the operator A acts in the space C6 and has the form

1.4 Inverse Problem for an Integro-differential Equation of Acoustics

51

A = (A1 , A2 , A3 , A4 , A5 , A6 )T . If we introduce the vector function ϕ0 (x, t) = (ϕ01 , ϕ02 , ϕ03 , ϕ04 , ϕ05 , ϕ06 )   x +t 1 c0 , := β(x), (g˜ 0 (t − x) + (λ0 − h(0))g˜ 0 (t − x)) −  2 2 2      t 1 t t 1 1 g˜ 0 (t) − λ0 g˜ 0 (t) −  − β , −  2 2 c0 2 2 2  2  h (0) h 2 (0) −h(0) + − h (0) t, − h (0), 0 , 2 2 then, according to Eqs. (1.136)–(1.141), the components of the operator A in Eq. (1.144) become t 



1 (2ϕ6 (τ − x) + ϕ3 (τ − x)) 3 x  2r00 )ϕ4 (τ − x) −λ0 c0 ϕ4 (τ − x) dτ, −L 0 ϕ5 (τ − x) + (M0 + 3 ϕ2 (x, τ ) − c0 ϕ5 (τ − x) − c0 x

A1 ϕ = ϕ01 +

A2 ϕ = ϕ02 +

λ0 2

t−x t−x 1 ϕ4 (τ )g˜ 0 (t − x − τ )dτ + ϕ5 (t − x − τ )g˜ 0 (τ )dτ 2 0

+

1 2

0

x  (ξ )ϕ1 (ξ, t − x + ξ ) 0

t−x −

1 2r00 (2ϕ6 (τ ) + ϕ3 (τ )) − L 0 ϕ5 (τ ) + (M0 + )ϕ4 (τ ) 3 3



0 t+x

2  1 (ξ )ϕ1 (ξ, t + x − ξ ) ×ϕ1 (ξ, t − x + ξ − τ )dτ dξ + 2 x  1 −2c0 (2ϕ6 (t + x − 2ξ ) + ϕ3 (t + x − 2ξ )) − L 0 ϕ5 (t + x − 2ξ ) 3 2r00 )ϕ4 (t + x − 2ξ ) +(M0 + 3 t+x−2ξ   1 2r00 (2ϕ6 (τ ) + ϕ3 (τ )) − L 0 ϕ5 (τ ) + (M0 + )ϕ4 (τ ) + 3 3 0  ×ϕ1 (ξ, t + x − ξ − τ )dτ dξ, 

52

1 Local Solvability of One-Dimensional Kernel …

λ0 A3 ϕ = ϕ03 + c0

t

ϕ4 (τ )g˜ 0 (t

1 − τ )dτ − c0

0

1 − c0

t 

t

ϕ4 (τ )g˜ 0 (t − τ )dτ

0

1 2r00 (2ϕ6 (τ ) + ϕ3 (τ )) − L 0 ϕ5 (τ ) + (M0 + )ϕ4 (τ ) 3 3

0

  t/2 1 (ξ ) ϕ2 (ξ, t − ξ ) + c0 ξ (2ϕ6 (t − ξ ) ×β 3 0 2r00 +ϕ3 (t − ξ )) − L 0 ϕ5 (τ ) + (M0 + )ϕ4 (t − ξ ) −c0 ϕ5 (t − ξ ) 3 t−2ξ   1 2r00 (2ϕ6 (τ ) + ϕ3 (τ )) − L 0 ϕ5 (τ ) + (M0 + )ϕ4 (τ ) + 3 3 0   1 (2ϕ6 (t − ξ − τ ) + ϕ3 (t − ξ − τ )) × ϕ2 (ξ, t − ξ − τ ) + c0 ξ 3 2r00 )ϕ4 (t − ξ − τ ) −L 0 ϕ5 (t − ξ − τ ) + (M0 + 3  −c0 ϕ5 (t − ξ − τ ) dτ dξ, 

t −τ 2

t A4 ϕ = ϕ04 +



1 dτ + c0



1 1 (t − τ ) (ϕ6 (τ ) − ϕ3 (τ )) + L 0 ϕ5 (τ ) − (M0 − r00 )ϕ4 (τ ) dτ, 3 3

0

t  A5 ϕ = ϕ05 +

1 1 (ϕ6 (τ ) − ϕ3 (τ )) + L 0 ϕ5 (τ ) − (M0 − r00 )ϕ4 (τ ) dτ, 3 3

0

t 

1 (2ϕ6 (t − τ ) + ϕ3 (t − τ )) − L 0 ϕ5 (t − τ ) 3 0 2 +(M0 + r00 )ϕ4 (t − τ ) ϕ4 (τ )dτ, 3

A6 ϕ = ϕ06 −

Let us consider the above-introduced linear space C6 of vector functions ϕ. In this space, we introduce the norm by the formula: ϕ = max ϕi , 1≤i≤6

where ϕi  = max(x,t)∈DT |ϕi (x, t)|, i = 1, 2; ϕi  = maxt∈[0,T ] |ϕi (t)|, i = 3, 4, 5, 6.

1.4 Inverse Problem for an Integro-differential Equation of Acoustics

53

By B(ϕ0 ) we denote the set of vector functions ϕ(x, t) that satisfy the inequality ||ϕ − ϕ0 || ≤ ||ϕ0 ||. Obviously, ϕ ≤ 2ϕ0  for ϕ(x, t) ∈ B(ϕ0 ). Let us prove that A is a contraction operator in the metric space B(ϕ0 ) provided the number T is sufficiently small. Let us verify the first condition of contraction operator. To simplify calculations, we denote 2r00 , 0 := max |(x)| , B0 x∈[0,T /2] 3    G 0 := max |g˜ 0 (t)| , G 1 := max g˜ 0 (t) , G 2 := max P0 := 1 + L 0 + M0 + t∈[0,T ]

t∈[0,T ]

t∈[0,T ]

:= max |β(x)| , x∈[0,T /2]   g˜ (t) . 0

Let ϕ ∈ B(ϕ0 ). Then, for (x, t) ∈ DT we have the estimates   t   1 (2ϕ6 (τ − x) + ϕ3 (τ − x)) |A1 ϕ − ϕ01 | ≤  ϕ2 (x, τ ) − c0 ϕ5 (τ − x) − c0 x 3 x    2r00 )ϕ4 (τ − x) −λ0 c0 ϕ4 (τ − x) dτ  −L 0 ϕ5 (τ − x) + (M0 + 3    P0 T ≤ ||ϕ0 ||T 2 + c0 + 2(1 + λ0 ) =: α1 ϕ0 , 2    3 |A2 ϕ − ϕ02 | ≤ ||ϕ0 ||T (λ0 + 1)G 0 + 0 + P0 ||ϕ0 ||T + c0 =: α2 ϕ0 , 2   2 1 1 λ0 G 1 + G 2 + P0 B0 + 0 + P0 T ||ϕ0 || |A3 ϕ − ϕ03 | ≤ ||ϕ0 ||T c0 2 2   P0 T ||ϕ0 || 0 + =: α3 ϕ0 , +0 + P0 T 4 6   r00 2 2 + L 0 + M0 + =: α4 ϕ0 , |A4 ϕ − ϕ04 | ≤ ||ϕ0 ||T 3 3   4 2r00 + 2L 0 + 2M0 + =: α5 ϕ0 , |A5 ϕ − ϕ05 | ≤ ||ϕ0 ||T 3 3 |A6 ϕ − ϕ06 | ≤ 4||ϕ0 ||2 T P0 =: α6 ϕ0 , It follows from the above estimates that the operator A maps the set B(ϕ0 ) to itself if the number T satisfies the condition max αi < 1.

1≤i≤6

(1.145)

54

1 Local Solvability of One-Dimensional Kernel …

Now we verify the second condition of contraction operator. Let ϕ k := (ϕ1k , ϕ2k , ϕ3k , ϕ4k , ϕ5k , ϕ6k ), and let ϕ k ∈ B(ϕ0 ), k = 1, 2. We use the obvious inequalities |ϕk1 ϕs1 − ϕk2 ϕs2 | ≤ |ϕk1 − ϕk2 ||ϕs1 | + |ϕk2 ||ϕs1 − ϕs2 | ≤ 4ϕ0 ϕ 1 − ϕ 2  to obtain t  ϕ21 (x, τ ) − ϕ22 (x, τ ) − c0 ϕ51 (τ − x) − ϕ52 (τ − x)

||(Aϕ − Aϕ )1 || = 1

2

x



1 (2ϕ61 (τ − x) − ϕ62 (τ − x) + ϕ31 (τ − x) − ϕ32 (τ − x)) −c0 x 3 −L 0 ϕ51 (τ − x) − ϕ52 (τ − x) + M0 ϕ41 (τ − x) − ϕ42 (τ − x)  +λ0 c0 ϕ41 (τ − x) − ϕ42 (τ − x) dτ    P0 T + 1 + λ0 =: ||ϕ 1 − ϕ 2 ||β0 , ≤ ||ϕ 1 − ϕ 2 ||T 1 + c0 4  ||(Aϕ 1 − Aϕ 2 )2 || ≤ ||ϕ 1 − ϕ 2 ||T =: ||ϕ 1 − ϕ 2 ||β2 ,

  3 1 1 1 (λ0 + 1)G 0 + 0 + P0 ||ϕ0 ||T + c0 2 2 2 2

 1 1 λ0 G 1 + G 2 + P0 B0 + 0 + P0 T ||ϕ0 || ||(Aϕ − Aϕ )3 || ≤ ||ϕ − ϕ ||T c0 2   P0 T ||ϕ0 || 0 1 + =: ||ϕ 1 − ϕ 2 ||β3 , + 0 + P0 T 2 8 6   1 r00 1 2 1 2 2 1 + (L 0 + M0 ) + =: ||ϕ 1 − ϕ 2 ||β4 , ||(Aϕ − Aϕ )4 || ≤ ||ϕ − ϕ ||T 3 2 6   2 r00 1 2 1 2 + L 0 + M0 + =: ||ϕ 1 − ϕ 2 ||β5 , ||(Aϕ − Aϕ )5 || ≤ ||ϕ − ϕ ||T 3 3 1

2



1

2

||(Aϕ 1 − Aϕ 2 )6 || ≤ 4||ϕ 1 − ϕ 2 ||T P0 ||ϕ0 || =: ||ϕ 1 − ϕ 2 ||β6 . Therefore, ||Aϕ 1 − Aϕ 2 || ≤ ρ||ϕ 1 − ϕ 2 ||, where ρ < 1, if T > 0 satisfies condition (1.145) Thus, if the number T is small enough for condition (1.145) to be satisfied, then A is a contraction operator in the metric space B(ϕ0 ). Then, according to the Banach principle, there exists a unique solution of Eq. (1.144) in the space B(ϕ0 ) [9] (pp. 87–97). Since h 0 (t) = k(t) exp (h(0)t/2), the obtained function h 0 (t) can be used to determine the function k(t) using the formula k(t) = h 0 (t) exp (−h(0)t/2) .

1.5 Problem of Determining the Memory in a Two-Dimensional …

55

1.5 Problem of Determining the Memory in a Two-Dimensional System of Maxwell Equations with Distributed Data The problems of finding the functions (x), μ(x), σ (x) which describe electrodynamic characteristics of the medium from the system of Maxwell equations are the subject of a large number of works (see, e.g., the bibliography in [10]). In electromagnetic fields with a greater frequency compared to the frequencies characteristic of establishing electrical and magnetic polarization of medium, the most general form of the dependence of D and B (inductions of electric and magnetic fields, respectively) on the values of E and H (stresses of the corresponding fields) can be written in the form of integral relations [11]: t ϕ(t − τ )E(x, τ )dτ,

D(x, t) = εE + 0

t ψ(t − τ )H (x, τ )dτ,

B(x, t) = μH +

(1.146)

0

E = (E 1 , E 2 , E 3 ), H = (H1 , H2 , H3 ), D = (D1 , D2 , D3 ), B = (B1 , B1 , B3 ), x = (x1 , x2 , x3 ), ϕ(t) = diag(ϕ1 , ϕ2 , ϕ3 ), ψ(t) = diag(ψ1 , ψ2 , ψ3 ) are diagonal matrices functions representing memory of the medium. In this section, the problem of determining the functions ϕ(t), ψ(t) is investigated. Wherein , μ, σ are considered as constants. Also assume that ϕ1 = ϕ2 = ϕ3 , ψ1 = ψ2 = ψ3 . In view of (1.146), we consider two-dimensional system of Maxwell equations [10] 

∂ H1 ∂ H3 ∂ E2 − + + (σ + ϕ0 )E 2 + ∂t ∂z ∂x

μ

∂ E2 ∂ H1 − + ψ0 H1 + ∂t ∂z

∂ E2 ∂ H3 + + ψ0 H3 + μ ∂t ∂x

t

t

ϕ (t − τ )E 2 (τ, x, z)dτ = 0,

0

ψ (t − τ )H1 (τ, x, z)dτ = 0,

(1.147)

0

t

ψ (t − τ )H3 (τ, x, z)dτ = 0,

0

where ϕ0 = ϕ(0), ψ0 = ψ(0). Further in (1.147) for simplicity of calculations, we put  = μ = σ = 1 and introduce the new functions

56

1 Local Solvability of One-Dimensional Kernel …

U1 = E 2 + H1 , U2 = E 2 − H1 , U3 = H3 . Then, the system of Eq. (1.147) is written in the following canonical form: 

t ∂ ∂ ∂ +E + N U + (τ )U (t − τ, x, z)dτ = 0. I3 + K ∂t ∂z ∂x

(1.148)

0

Here I3 is the unit three-dimensional matrix, K = diag(−1, 1, 0), U = (U1 , U2 , U3 )∗ , ∗ is the symbol of transposition, ⎞ ⎞ ⎛ 001 1 + ϕ0 + ψ0 1 + ϕ0 − ψ0 0 1 E = ⎝ 0 0 1 ⎠ , N = ⎝ 1 + ϕ0 − ψ0 1 + ϕ0 + ψ0 0 ⎠ , 2 1 1 0 0 ψ0 0 2 2 ⎞ ⎛ ϕ +ψ ϕ −ψ 0 1 = ⎝ ϕ − ψ ϕ + ψ 0 ⎠ (t). 2 0 0 ψ ⎛

Consider for the system (1.148) in the domain    G = (t, x, z) t > 0, −∞ < x < ∞, 0 < z < H initial–boundary value problem  U t=0 = f (x, z), −∞ < x < ∞, 0 < z < H,  U1 z=H = g1 (t, x), t > 0, −∞ < x < ∞,  U2 z=0 = g2 (t, x), t > 0, −∞ < x < ∞.

(1.149)

We set the inverse problem: determine the functions ϕ(t), ψ(t), if the following conditions with respect to the solutions of problem (1.148) and (1.149) are known: ∞ h 1 (t) =

U1 (t, x, 0) exp(iλx)dx, −∞

∞ h 2 (t) =

U2 (t, x, H ) exp(iλx)dx, t > 0

(1.150)

−∞

for some fixed λ. First, we assume that the functions f (x, z), g1 (t, x), g2 (t, x) are finite ones in x for fixed (t, z), consequently, this condition guarantees the existence of Fourier transforms:

1.5 Problem of Determining the Memory in a Two-Dimensional …

57

∞

f˜(λ, z) =

f (x, z) exp(iλx)dx, −∞ ∞

g˜ i (λ, t) =

gi (t, x) exp(iλx)dx, i = 1, 2. −∞

Then, due to hyperbolicity of the system (1.148) the solution of initial–boundary problem (1.148) and (1.149) has the same property. We write relations (1.148) and (1.149) in terms of the Fourier transformation with respect to variable x 

t ∂ ∂ ˜ I3 + K + N V + (τ )V (t − τ, λ, z)dτ = 0, ∂t ∂z

(1.151)

0

 V t=0 = f˜(λ, z), 0 < z < H,   V1 z=H = g˜ 1 (λ, t), V2 z=0 = g˜ 2 (λ, t), t > 0, where

(1.152)

∞ U (t, x, z) exp(iλx)dx,

V (t, λ, z) = −∞

⎞ 1 + ϕ0 + ψ0 1 + ϕ0 − ψ0 −iλ 1 N = ⎝ 1 + ϕ0 − ψ0 1 + ϕ0 + ψ0 −iλ ⎠ . 2 − iλ2 − iλ2 ψ0 ⎛

We introduce a new function W by the formula W (t, λ, z) = Vt (t, λ, z). To obtain relations for W , we differentiate with respect to t Eq. (1.151) and the boundary conditions in (1.152), and the value of W at t = 0 will be found using Eq. (1.151) and initial conditions. Then, the function W is the solution of the following problem: 

t ∂ ∂ I3 + K + N˜ W + (t) f˜(λ, x) + (τ )W (t − τ, x, z)dτ = 0, ∂t ∂z 0   ∂ W t=0 = − K + N˜ f˜(λ, z), 0 < z < H, (1.153) ∂z   = g˜ 1t (λ, t), W2  = g˜ 2t (λ, t), t > 0. W1  z=H

z=0

Let (t, z) be an arbitrary point of the domain G H = {(t, z)| t > 0, 0 < z < H }. On the plane of variables (τ, ξ ) we draw the characteristics of differential operators

58

1 Local Solvability of One-Dimensional Kernel …

through this point

∂ ∂ ∂ ∂ ∂ − , + , , ∂t ∂z ∂t ∂z ∂t

and continue them to the intersection in the τ ≤ t with the boundary of   domain G H . The intersection point is denoted by t0i , z 0i , i = 1, 2, 3. Integrating the ith component of equality (1.153) along the corresponding characteristic from t0i , z 0i to the point (t, z). Then one gets relations   Wi (t, λ, z) = W t0i , λ, z 0i −

t

3 t0i

−

 t 

3 t0i

+

n i j (τ )W j (τ, λ, ξ )

j=1

τ

3 0

ϕi j f˜j (λ, z + λi (τ − t))dτ

j=1

 ϕi j (τ )W j (τ − γ , λ, ξ )dγ

j=1

dτ, (1.154) ξ =z+λi (τ −t)

i = 1, 2, 3. Here ⎧ ⎪ ⎪ t + z − H, i = 1, t > H − z ⎨ 0, i = 1, t ≤ H − z 1 t0 (t, z) = t − z, i = 2, t > z ⎪ ⎪ ⎩ 0, i = 2, t ≤ 0, i = 3, (t, z) ∈ G H , ⎧ ⎫ g˜ 1t (t + z − H, λ), i = 1, t > H − z, ⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎪ ⎪ ⎪ f˜1z (λ, t + z) − 3j=1 n 1 j f˜j (λ, t + z), i = 1, t ≤ H − z, ⎪ ⎨ ⎬ i  2, t > z, , Wi t0 , λ, z 0i = g˜ 2t (t − z, λ), i = ⎪ ⎪ − f˜2z (λ, t + z) − 3 n 2 j f˜j (λ, t − z), i = 2, t ≤ z, ⎪ ⎪ ⎪ ⎪ j=1 ⎪ ⎪ ⎪ ⎪ ⎩ − 3 n f˜ (λ, z), i = 3, (t, z) ∈ G , ⎭ j=1

3j

j

H

n i j , ϕi j (i, j = 1, 2, 3) denote the elements matrices N˜ , , respectively. Consider now an arbitrary point (0, z) ∈ 0 = ((t, z) t = 0, 0 ≤ z ≤ H ), and through this point, we draw the characteristic of differential operators ∂ ∂ ∂ ∂ − , + ∂t ∂z ∂t ∂z up to their crossing the points of the lateral boundary of domain G H . Integrating the first two equations of the system (1.153) and using the initial data and the data (1.150), we find

1.5 Problem of Determining the Memory in a Two-Dimensional …

λi f˜i z (λ, z) −

3

n i j f˜j (λ, z) − h it (ti (z)) +

j=1

ti (z)

3 0

59

ϕi j (τ ) f˜j (λ, z + λi (τ − t))dτ

j=1

⎤ ⎡ ti (z)

τ

3 3 ⎣ + n i j W j (τ, λ, ξ ) + ϕi j (τ )W j (τ − γ , λ, ξ )dγ ⎦ 0

j=1

0

j=1

dτ,

ξ =z+λi τ

i = 1, 2,

(1.155)

Assuming in equality (1.156) z = 0 for i = 1, and z = H for i = 2 we obtain n 11 f˜1 (λ, 0) + n 12 f˜2 (λ, 0) = − f˜1z (λ, 0) − h 1 (0) + iλ f˜3 (λ, 0), n 21 f˜1 (λ, H ) + n 22 f˜2 (λ, H ) = − f˜2z (λ, H ) − h 2 (0) + iλ f˜3 (λ, H ). Since n 11 = n 11 , n 12 = n 21 , from the last equalities with the fulfilling the condition    f˜1 (λ, 0) f˜2 (λ, 0)   = 0,  (1.156) F(λ) =  ˜ f 2 (λ, H ) f˜1 (λ, H )  we uniquely find n 11

n 12

   1 ˜ ˜ ˜ = f 1 (λ, H ) − f 1z (λ, 0) − h 1 (0) + iλ f 3 (λ, 0) F(λ)   ˜ ˜ ˜ − f 2 (λ, 0) − f 2z (λ, H ) − h 2 (0) + iλ f 3 (λ, H ) ,    1 ˜ ˜ ˜ = f 1 (λ, 0) − f 2z (λ, H ) − h 2 (0) + iλ f 3 (λ, H ) F(λ)   ˜ ˜ ˜ − f 2 (λ, H ) − f 1z (λ, H ) − h 1 (0) + iλ f 3 (λ, H ) .

From these equalities, one can find numbers ϕ0 = n 11 + n 12 − 1, ψ0 = n 11 − n 12 .

(1.157)

Thus, the matrix N˜ becomes known. Lemma 1.3 Let ϕ ∈ C 1 [0, H ], ψ ∈ C 1 [0, H ], f˜i ∈ C 2 [0, H ], i = 1, 2, 3; g˜ i ∈ Wi (i = 1, 2) C 1 [0, H ], i = 1, 2 and λ is some fixed number. Then, the functions ∂∂z ∂ W3 are piecewise-continuous, and the function ∂z is continuous in the domain G H , W2 W1 moreover, ∂∂z has a finite discontinuity along the characteristic z = H − t and ∂∂z has a finite discontinuity along the characteristic z = t.

60

1 Local Solvability of One-Dimensional Kernel …

To prove Lemma 1.3, we differentiate the equalities (1.154) with respect to z: 3  i 

  i ϕi j (t0i ) f˜j λ, z + λi (t0i − t) Wi z (t, λ, z) = Wz t0 , λ, z 0 + j=1

  +n i j W j t0i , λ, z + λi (t0i − t) t0

3 i

+

0

t − t0i

τ + 0

  ϕi j (γ )W j t0i − γ , λ, z + λi (t0i − t) dγ



j=1 3

ϕi j (τ ) f˜j z (λ, z + λi (τ − t))dτ −

j=1 3

j=1

 t 

3 t0i

d i t dz 0 n i j (τ )W j z (τ, λ, ξ )

j=1

 ϕi j (γ )W j z (τ − γ , λ, ξ )dγ

ξ =z+λi (τ −t)

dτ, i = 1, 2, 3.

(1.158)

It is easy to see that the system of integral equations (1.158) has continuous kernels and piecewise continuous free members (first, second, third terms). Therefore, this equation has a unique solution and it can be obtained via the method of successive approximations. It is not difficult to believe that each iteration leads to W1 ∂ W2 piecewise-continuous functions ∂∂z , ∂z , with discontinuities along the correspondW3 ing characteristics, outgoing from the point (0, 0), (H, 0) and function ∂∂z which is continuous in the domain G H . Note that for the solvability of the inverse problem (1.148)–(1.150) when the conditions of the lemma 1.3 are satisfied, it is necessary that the functions h i (t) ∈ C 1 [0, H ], i = 1, 2, satisfies the matching conditions with the initial data: h 1 (0) = f˜1 (λ, 0), h 2 (0) = f˜2 (λ, H ), had piecewise-continuous second derivatives on [0, H ], moreover, for h i (t) finite discontinuities are admissible only at the point t = H. Theorem 1.11 Let the conditions of the lemma 1.3, inequalities (1.156) and necessary conditions on the data of the inverse problem be satisfied. Then there exists such number H ∗ , that for H ∈ (0, H ∗ ) solution of inverse problem (1.148)–(1.150) exists, is unique and is determined by setting h i (t) for t ∈ (0, H ), i = 1, 2. For each i, differentiating (1.155) with respect to z, and taking into account the conditions from (1.153), we obtain

1.5 Problem of Determining the Memory in a Two-Dimensional …

61

ϕ11 (z) f˜1 (λ, 0) + ϕ12 (z) f˜2 (λ, 0) = f˜1zz (λ, z) + h 1 (λ, z) +

3

n 1 j f˜j z (λ, z)

j=1

−

z

3 j=1

0

−

3

n 1 j W j (z, λ, 0) −

j=1

−

ϕ1 j (τ ) f˜j z (λ, z − τ )dτ

z

3

z

3 0

⎡

⎣n 1 j w j z (τ, λ, ξ ) +

τ

j=1

0

ϕ1 j (τ )W j (z − τ, λ, 0)dτ

j=1

⎤ ϕ1 j (ν)W j z (τ − ν, λ, ξ )dν ⎦

dτ ξ =z−τ

0

= 0,

(1.159)

ϕ11 (z) f˜1 (λ, H ) + ϕ12 (z) f˜2 (λ, H ) 3

= f˜2zz (λ, H + z) + h 2 (λ, z) − n 2 j f˜j z (λ, H + z) j=1

−

z

3 0

+

j=1

z

3 0

⎡

ϕ1 j (τ ) f˜j z (λ, H + z − τ )dτ −

3

n 2 j w j (z, λ, H )

j=1

ϕ1 j (τ )W j (z − τ, λ, H )dτ +

j=1

z

3 0

× ⎣n 2 j W j z (τ, λ, ξ ) +

τ

j=1

⎤

ϕ1 j (ν)W j z (τ − ν, λ, ξ )dν ⎦

dτ ξ =H +z+τ

0

= 0.

(1.160)

The following equalities are also used here: ϕ11 = ϕ22 , ϕ12 = ϕ21 , ϕ13 = ϕ23 = 0. The system of Eqs. (1.159) and (1.160) by virtue of condition (1.156) can be solved with respect to the functions ϕ11 , ϕ12 . Then, ϕ (t) = ϕ11 + ϕ12 , ψ (t) = ϕ11 − ϕ12 . We add to relations (1.154), (1.158)–(1.160) the obvious equalities z ϕ(z) = ϕ0 +



z

ϕ (τ )dτ, ψ(z) = ψ0 + 0

0

ψ (τ )dτ,

(1.161)

62

1 Local Solvability of One-Dimensional Kernel …

where the numbers ϕ0 , ψ0 are determined by formulas (1.157). Consider the square G H,H = {(t, z)| 0 ≤ t ≤ H, 0 ≤ z ≤ H } . The system of Eqs. (1.154), (1.158)– (1.161) in this square is closed with respect to functions W Wi , Wi z , i = 1, 2, 3, ϕ(z), ϕ (z), ψ(z), ψ (z). To obtain from this system the normal form of nonlinear integral equations of the second kind, it is sufficient for beginning, to express the nonintegral terms Wi (z, λ, 0), Wi (z, λ, H ), i = 1, 2, in (1.159) and (1.160) using equalities (1.150), the boundary conditions from (1.153) and W3 (z, λ, 0), W3 (z, λ, H ) with the corresponding values using Eq. (1.154) for i = 3. Then, the nonintegral terms ϕ1i , i = 1, 2 in (1.158) can be expressed using the solved system of Eqs. (1.159) and (1.160) with respect to ϕ1i , i = 1, 2. The system of Eqs. (1.154), (1.158)–(1.160) contains the integration interval as a small parameter, which does not exceed H . Therefore, for small H, to this system is applicable the contracted mapping principle. It is easy to check that each iteration leads to continuous functions W3 , W3z , ϕ(z), ϕ (z), ψ(z), ψ (z) in the domain G H,H and to piecewise-continuous functions Wi , Wi z , i = 1, 2 with discontinuities along the corresponding characteristics outgoing points (0, 0), (H, 0). Theorem 1.11 is proved.

1.6 Conclusions In this chapter, the one-dimensional inverse problems of determining the memory of a medium entering into an integral term of convolution type in the second-order hyperbolic integro-differential equations are investigated. Section 1.1 discussed the problem  from an  of finding a kernel  integro-differential wave equation in a half-space R3+ = (x1 , x2 , x3 ) ∈ R3  x3 > 0 . Further, in Sect. 1.2, the results obtained in the previous section are generalized to the case of a second-order hyperbolic-type equation with variable coefficients at lower derivatives with respect to x = (x1 , x2 , x3 ), the main part of which coincides with the wave operator. Based on the method of S. L. Sobolev in this section there have been constructed Volterra type integral equations which are equivalent to the inverse problem. The fixed point theorem (Banach’s theorem) was applied to these integral equations. In both sections, the properties of direct problems solution, which are well-posed boundary–initial value problems, are also studied. In subsequent sections, by the method of the previous sections there have also been investigated three one-dimensional inverse problems for a system of thermoelasticity equations in a vertically inhomogeneous disconnected medium, an integro-differential equation of acoustics, and a two-dimensional system of Maxwell equations with memory. The main results of this chapter are local existence theorems and conditional stability estimates for the solution of inverse problems.

References

63

References 1. Romanov, V.G. 1972. Some inverse problems for hyperbolic-type equations. Novosibirsk: Nauka (Russian). 2. Sobolev, S.L. 1930. Wave equation for an inhomogeneous medium, Tr. Seysm. ins. 6: 1–57 (in Russian). 3. Sobolev S.L. 1934. To the question on integration of wave equation in an inhomogeneous medium, Tr. Seysm. ins. 1 (42) (in Russian). 4. Lavrent’ev, M.M., V.G. Romanov, and S.P. Shishatskii. 1980. Ill-posed problems of mathematical physics and analyses. Moscow: Nauka (in Russian). 5. Kovalenko, A.D. 1975. Thermoelasticity. Kiev: Vish. shk (in Russian). 6. Yakhno, V.G. 1990. Inverse problems for differential equations of elasticity. Novosibirsk: Nauka (in Russian). 7. Courant, R., and D. Hilbert. 1962. Methods of mathematical physics, II, 830. New York/London: Interscience Publication. 8. Durdiev, D.K. 2009. Global solvability of one inverse problem for integro-differential equation of electrodynamics, Uzb. Mat. J. 3: 36–45 (in Russian). 9. Karchevsky, A.L., and A.G. Fat’yanova. 2001. Numerical solution of the inverse problem for an elasticity system with after effect for an upright nonuniform medium. Sibir. Zh. Vychisl. Mat. 4 (3): 259–268 (Russian). 10. Romanov, V.G. 1994. Kabanikhin Inverse problems for Maxwell’s equations. Utrecht: VSP. 11. Landau, L.D., and E.M. Lifshits. 1959. Electrodynamics of continuous media. Moscow: Nauka (Russian).

Chapter 2

The Solvability of Multidimensional Inverse Problems of a Memory Kernel Determination

For multidimensional inverse problems there are only special cases for which solvability is established. One of such classes of functions in which solvability takes place is the class of analytic functions. The technique used here is based on the scale method of Banach spaces of analytic functions, developed in the works of Ovsyannikov [1, 2] and Nirenberg [3]. This method was first applied to the problem of solvability of multidimensional inverse problems by Romanov [4] (Sect. 3). In this chapter, we consider inverse problems for the case when spatial variables belong to the half space, i.e., := Rn × R+ , R+ = {z ∈ R|z>0 } , (x, z) ∈ Rm+1 + and the inverse problem is to determine the memory kernel k(x, t), which is included in the integral term of the hyperbolic equation. In this case, it is assumed that the initial data in all the considered problems are equal to zero, and the Neumann boundary condition is given on   × [0, T ], ∂Rm+1 := (x, z) ∈ Rm+1 ∂Rm+1 + + + |z = 0 , T > 0, and is a smooth function in t and analytical in x ∈ Rn . The inverse problems are investigated under the condition that solutions of the corresponding initial–boundary value × [0, T ], and they are analytic functions problems are given for (x, z, t) ∈ ∂Rm+1 + of a real variable x. A scale of functions, which are analytic in x, is introduced, and local existence theorems for the solution of the inverse problems are established.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_2

65

66

2 The Solvability of Multidimensional Inverse …

2.1 Definition of a Multidimensional Memory Kernel in the Integro-differential Wave Equation Consider the problem of determining a pair of functions u and k satisfying the equations t u tt − u zz − u =

k(x, τ )u(x, z, t − τ )dτ, (x, t) ∈ Rm+1 , z > 0,

(2.1)

0

 u

t 0, of functions ϕ(x), x ∈ Rm , which are analytic in the neighborhood of the origin and satisfy the relation ∞  s |α| α |D ϕ(x)| < ∞. |x| 0, s > 0 and D α :=

∂ |α|

∂ x1α1 ...∂ xmαm

, α := (α1 , . . . , αm ),

|α| := α1 + · · · + αm , α! := (α1 )! . . . (αm )!. In the following, the parameter r will be considered fixed, while a parameter s is variable. Then, it is formed a scale of Banach spaces As (r ), s > 0 of analytic functions. The following property is obvious: if ϕ(x) ∈ As (r ), then ϕ(x) ∈ As  (r ) for all s  ∈ (0, s), consequently, As (r ) ⊂ As  (r ), if s  ∈ (0, s) and the following inequality is valid: ϕs (r ) D α ϕs  ≤ cα (2.4) (s − s  )|α| for any α with constant cα , which depends only on α.

2.1 Definition of a Multidimensional Memory …

67

The estimate (2.4) can be shown as follows [4] (p. 93): ∞   (s  )|β|  α+β D ϕ(x) |x| 0, x ∈ Rm ,  u t=z+0 ≡ 0, x ∈ Rm .

(2.9) (2.10)

Starting with these equalities, we now construct a system of integro-differential equations for the functions u and k. With the help of d’Alembert’s formula, from equalities (2.8) and (2.9) we obtain the equation

68

2 The Solvability of Multidimensional Inverse …

u(x, z, t) = u 0 (x, z, t) + τ −ξ +

1 2

  u(x, ξ, τ ) − k(x, τ − ξ ) (z,t)

k(x, γ )u(x, ξ, τ − γ )dγ dτ dξ, (x, z, t) ∈ D.

(2.11)

0

Here (z, t) = {(τ, ξ )| 0 ≤ ξ ≤ z, t − z + ξ ≤ τ ≤ t + z − ξ } , t+z h(x, τ )dτ.

1 1 u 0 (x, z, t) = [ f (x, t + z) + f (x, t − z)] + 2 2

(2.12)

t−z

Passing to the limit in the formula (2.11) as t → z + 0, taking conditions (2.7) and (2.10) into account we come to equality  z 2z−ξ   u(x, ξ, τ ) − k(x, τ − ξ ) ξ

0

τ −ξ +

k(x, γ )u(x, ξ, τ − γ )dγ dτ dξ = −2u 0 (x, z, z + 0).

0

Differentiating this equality with respect to z, we obtain z  u(x, ξ, 2z − ξ ) − k [x, 2(z − ξ )] 0

z−ξ +

k(x, γ )u(x, ξ, 2z − ξ − γ )dγ dξ = −

d u 0 (x, z, z + 0). dz

(2.13)

0

From formulas (2.12) and (2.13), it is easy to derive the following matching condition: f t (x, +0) = −h(x, +0), which, together with condition (2.7), is necessary for the solvability of the problem (2.1)–(2.4). The equation for k can be obtained by differentiating the equality (2.13) with respect to z after replacing the variable 2z − ξ in the second integral by ξ  :

2.1 Definition of a Multidimensional Memory …

69

k(x, z) = k0 (x, z) z−2ξ z/2  +2 v(x, ξ, z − ξ ) + k(x, γ )v(x, ξ, z − ξ − γ )dγ dξ, (2.14) 0

0

where k0 (x, z) = 8 [ f tt (x, t) + h t (x, t)]t=2z , v(x, z, t) ≡ u t (x, z, t). Now we use equality (2.11) for calculating v(x, z, t). Applying the equivalent description of the domain (z, t) in the form (z, t) = {(ξ, τ )| t − z ≤ τ ≤ t + z, 0 ≤ ξ ≤ z − |t − τ |} , we differentiate (2.11) with respect to t. Making the change of the integration variable τ − t = ξ  , we find v(x, z, t) = v0 (x, z, t) −z 1 u(x, z − |ξ |, ξ + t) − k(x, ξ + |ξ | + t − z) + 2 z

k(x, γ )u(x, z − |ξ |, t + ξ − γ )dγ sgn ξ dξ,

ξ +|ξ |+t−z

+

(2.15)

0

where v0 (x, z, t) = (∂/∂t)u 0 (x, z, t). The system of Eqs. (2.11), (2.14), and (2.15) is a closed integro-differential equations for the functions u, k, v. Note that the operator  in the functions u and v appears in the system only under the integral sign. We will henceforth consider system (2.11), (2.14), and (2.15) in the domain DT = G T × Rm , G T = {(z, t)| 0 ≤ z ≤ t ≤ T − z} , T > 0.

 Let C(z,t) G T ; As0 denote the class of functions with values in As0 which are continuous in the variables (z, t) in the domain G T . For fixed (z, t), the norm of a function

g(x, z,t) in As0 will be denoted by gs0 (z, t). The norm of a function g in C(z,t) G T ; As0 is defined by the equality gC(z,t) (G T ;

As0 )

= sup gs0 (z, t). (z,t)∈G T

(2.16)

Theorem 2.1 Suppose that the above assumptions on the functions f (x, t) and h(x, t) are satisfied. Moreover, assume

 ( f (x, t), h(x, t), f t (x, t), h t (x, t), f tt (x, t)) ∈ Ct [0, T ]; As0 ,

70

2 The Solvability of Multidimensional Inverse …

   T R hs0 (t),  f t s0 (t) ≤ , max  f s0 (t), max 1, 2 2   R max  f tt s0 (t), h t s0 (t) ≤ , t ∈ [0, T ], R > 0. 16   Then, there exists such a ∈ 0, 2sT0 that for every s ∈ (0, s0 ) in the domain sT =  DT {(x, z, t)| 0 ≤ z ≤ a(s0 − s)} there is a unique solution to the system of Eqs. (2.11), (2.14), and (2.15) in the class of functions

 (u(x, z, t), v(x, z, t)) ∈ C(z,t) PsT ; As0

 k(x, t) ∈ Ct [0, a(s0 − s)]; As0 , PsT = G T



{(z, t)| 0 ≤ z ≤ a(s0 − s)}

exists and it is unique, for which the estimates u − u 0 s (z, t) ≤ R, k − k0 s (z) ≤ v − v0 s (z, t) ≤

R , (s0 − s)2

R , (z, t) ∈ PsT s0 − s

are valid. Proof Under the conditions of Theorem 2.1 we have

 (u 0 , k0 , v0 ) ∈ C(z,t) G T ; As0 , u 0 s (z, t) ≤ R, k0 s (z) ≤ R, v0 s (z, t) ≤ R, (z, t) ∈ G T , 0 < s < s0 . Let an denote the members of the monotone decreasing sequence that is defined by the equalities an an+1 = , n = 0, 1, 2, . . . (2.17) 1 + 1/(n + 1)2 Denote a = lim an = a0 n→∞

∞ 

−1 . 1 + 1/(n + 1)2

(2.18)

n=0

  The number a0 ∈ 0, 2sT0 will be chosen in an appropriate way. For the system of Eqs. (2.11), (2.14), and (2.15) we construct the process of successive approximations according to the following scheme:

2.1 Definition of a Multidimensional Memory …

u n+1 (x, z, t) = u 0 (x, z, t) + τ −ξ +

1 2

71

  u n (x, ξ, τ ) − kn (x, τ − ξ ) (z,t)

kn (x, γ )u n (x, ξ, τ − γ )dγ dτ dξ,

0

kn+1 (x, z) = k0 (x, z) z−2ξ z/2  vn (x, ξ, z − ξ ) + +2 kn (x, γ )vn (x, ξ, z − ξ − γ )dγ dξ, 0

(2.19)

0

vn+1 (x, z, t) = v0 (x, z, t) −z 1 u n (x, z − |ξ |, ξ + t) − kn (x, ξ + |ξ | + t − z) + 2 z

kn (x, γ )u n (x, z − |ξ |, t + ξ − γ )dγ sgn ξ dξ.

ξ +|ξ |+t−z

+ 0

Define the function s  (z) by the formula s  (z) =

s + ν n (z) z , ν n (z) = s0 − . 2 an

(2.20)

Introduce the notations pn = u n+1 − u n , qn = kn+1 − kn , rn = vn+1 − vn , n = 0, 1, 2, . . . . Then, pn , qn , rn satisfy the relations 1 p0 (x, z, t) = 2

  u 0 (x, ξ, τ ) − k0 (x, τ − ξ ) (z,t)

τ −ξ +

k0 (x, γ )u 0 (x, ξ, τ − γ )dγ dτ dξ,

0

q0 (x, z) z−2ξ z/2  v0 (x, ξ, z − ξ ) + =2 k0 (x, γ )v0 (x, ξ, z − ξ − γ )dγ dξ, 0

0

72

2 The Solvability of Multidimensional Inverse …

r0 (x, z, t) −z 1 u 0 (x, z − |ξ |, ξ + t) − k0 (x, ξ + |ξ | + t − z) = 2 z

k0 (x, γ )u 0 (x, z − |ξ |, t + ξ − γ )dγ sgn ξ dξ,

ξ +|ξ |+t−z

+ 0

pn+1 (x, z, t) = u 0 (x, z, t)   1  pn (x, ξ, τ ) − qn (x, τ − ξ ) + 2 (z,t)

τ −ξ qn (x, γ )u n+1 (x, ξ, τ − γ ) + 0

 + kn (x, γ ) pn (x, ξ, τ − γ ) dγ dτ dξ, qn+1 (x, z) z/2 =2 rn (x, ξ, z − ξ ) 0 z−2ξ  

+

qn (x, γ )rn+1 (x, ξ, z − ξ − γ ) 0

 + kn (x, γ )rn (x, ξ, z − ξ − γ ) dγ dξ, rn+1 (x, z, t) −z 1 =  pn (x, z − |ξ |, ξ + t) − qn (x, ξ + |ξ | + t − z) 2 z ξ +|ξ |+t−z

qn (x, γ )u n+1 (x, z − |ξ |, t + ξ − γ )

+ 0

 + kn (x, γ ) pn (x, z − |ξ |, t + ξ − γ ) dγ sgn ξ dξ for n = 0, 1, 2, . . . .

2.1 Definition of a Multidimensional Memory …

73

Show that a0 ∈ (0, 2sT0 ) can be chosen so that the following inequalities be valid for all n = 0, 1, 2, . . .:   ν n (z) − s  pn s (z, t) , λn = max sup z (z,t,s)∈Fn  sup

(z,t,s)∈Fn

qn s (z)

(ν n (z) − s)3 , z

  (ν n (z) − s)2 sup rn s (z, t) < ∞, z (z,t,s)∈Fn u n+1 − u 0 s (z, t) ≤ R, kn+1 − k0 s (z) ≤ vn+1 − v0 s (z, t) ≤

R , (s0 − s)2

R , (z, t, s) ∈ Fn+1 , s0 − s

where Fn = {(z, t, s)| (z, t) ∈ G T , 0 ≤ z ≤ an (s0 − s), 0 < s < s0 } . Indeed, using the relations for pn , qn , rn , we find 1  p0 s (z, t) ≤ 2 τ −ξ + 0

1 ≤ 2

  u 0 s  (ξ ) (ξ, τ ) + k0 s  (ξ ) (τ − ξ ) (z,t)

k0 s  (ξ ) (γ )u 0 s  (ξ ) (ξ, τ − γ )dγ dτ dξ

  (z,t)

Rc0 + R(1 + Rt) dτ dξ. (s  (ξ ) − s)2

Here c0 is a positive constant, such that  ϕs ≤ c0

(2.21)

ϕs  , s  > s > 0. (s  − s)2

It is easy to check c0 = 4m. Taking the function s  (ξ ) from (2.20), for n = 0 we have

(2.22)

74

2 The Solvability of Multidimensional Inverse …

 p0 s (z, t)  z ≤ (z − ξ ) 0

z × 0

  4Rc0 + R(1 + Rt)dξ ≤ R 4c0 + s02 (1 + RT ) 0 2 (ν (ξ ) − s)

  (z − ξ )dξ z ≤ a0 R 4c0 + s02 (1 + RT ) 0 , (z, t, s) ∈ F0 . 0 2 (ν (ξ ) − s) ν (z) − s

Proceeding analogously, we obtain z/2 q0 s (z) ≤ 2

0

4Rc0 2 + R z dξ (ν 0 (ξ ) − s)2

≤ R 4c0 + s02 RT



1 r0 s (z, t) ≤ 2

z (ν 0 (z) z  −z

−

s)2

 ≤ s0 R 4c0 + s02 RT

z (ν 0 (z)

− s)3

,

4Rc0 + R(1 + 3Rt) dξ (ν 0 (ξ ) − s)2

  ≤ R 4c0 + s02 (1 + 3RT )

z , (z, t, s) ∈ F0 . (ν 0 (z) − s)2

These estimates imply validity of inequality (2.21) for n = 0. Moreover, for (z, t, s) ∈ F1 we find u 1 − u 0 s (z, t) =  p0 s (z, t) ≤

λ0 a 1 λ0 z ≤ = a 0 λ0 , ν 0 (z) − s 1 − a1 /a0

k1 − k0 s (z, t) = q0 s (z) ≤

λ0 z 4a0 λ0 ≤ , (ν 0 (z) − s)3 (s0 − s)2

v1 − v0 s (z, t) = r0 s (z, t) ≤

λ0 z 0 (ν (z) −

s)2

≤

2a0 λ0 . s0 − s

Choosing a0 so as to have 4a0 λ0 ≤ R, we conclude that inequalities (2.22) are satisfied for n = 0. By way of induction, we show that inequalities (2.21) and (2.22) are also valid for the other values of n if a0 is chosen suitably. Assume that inequalities (2.21) and (2.22) hold for n = 0, 1, 2, . . . , i. Then, for (z, t, s) ∈ Fi+1 we have

2.1 Definition of a Multidimensional Memory …

 pi+1 s (z, t) ≤

1 2

75

   pi s  (ξ ) (ξ, τ ) + qi s  (ξ ) (τ − ξ ) (z,t)

τ −ξ qi s  (ξ ) (γ )u i+1 s  (ξ ) (ξ, τ − γ )

+ 0

+ ki s  (ξ ) (γ ) pi s  (ξ ) (ξ, τ − γ ) dγ dτ dξ   c0 λi ξ λi ξ 1 + ≤ 2 (s  (ξ ) − s)2 (ν i (ξ ) − s) (ν i (ξ ) − s)3 (z,t)

R(1 + s02 ) 4Rλi tξ 2λi tξ dτ dξ + i + i (ν (ξ ) − s)3 (ν (ξ ) − s) (s0 − s)2 z

 (z − ξ )ξ dξ 2 ≤ λi 4c0 + 1 + 6Rt + 2Rs0 t (ν i+1 (ξ ) − s)3 ≤

λi a02

  2c0 + RT (3 + s02 )

0

z ν i+1 (z)

−s

.

Here in the intermediate calculations, the function s  is defined by equality (2.20) with n = i and the inequalities u i s (z, t) ≤ 2R, ki s (z) ≤ R

1 + s02 (s0 − s)2

are used, the latter valid by the induction hypothesis, together with the obvious inequalities ai ≤ a0 and ν i+1 (z) ≤ ν i (z). Similar arguments for qi+1 and ri+1 lead to inequalities qi+1 s (z) z/2

 c0 λi ξ λi R(1 + s0 )ξ + z (s  (ξ ) − s)2 (ν i (ξ ) − s)2 (ν i (ξ ) − s)3 0  λi R(1 + s02 )ξ 4Rλi tξ + dξ + i (ν (ξ ) − s)3 (s0 − s) (ν i (ξ ) − s)2 (s0 − s) z/2   ξ dξ 2 ≤ 2λi 4c0 + RT (2 + s0 + s0 ) i+1 (ν (ξ ) − s)4 1 ≤ 2

0

  ≤ λi a0 4c0 + RT (2 + s0 + s02 )

z (ν i+1 (z)

− s)3

.

76

2 The Solvability of Multidimensional Inverse …

z 

c0 λi ξ (s  (ξ ) − s)2 (ν i (ξ ) − s) −z  3λi R(1 + s02 )tξ λi (1 + 6Rt)ξ dξ + i + (ν i (ξ ) − s)3 (ν (ξ ) − s)(s0 − s)2 z  1  ξ dξ 2 ≤ λi 4c0 + 1 + 3RT (3 + s0 ) i+1 2 (ν (ξ ) − s)3

1 ri+1 s (z, t) ≤ 2

−z

  ≤ λi a0 4c0 + 1 + 3RT (3 + s02 )

z (ν i+1 (z)

− s)2

, (z, t, s) ∈ Fi+1 .

The obtained estimates yield   λi+1 ≤ λi ρ, λi+1 < ∞, ρ = a0 4c0 + 1 + 3RT (3 + s0 + s02 ) max(a0 , 1). Moreover, for (z, t, s) ∈ Fi+2 , we have u i+2 − u 0 s (z, t) ≤

i+1 

 pn s (z, t)

n=0

≤

i+1  n=0

≤

i+1 

 λn ai+2 λn z ≤ n ν (z) − s 1 − ai+2 /an n=0 i+1

λn an (n + 1)2 ≤ λ0 a0

n=0

ki+2 − k0 s (z) ≤

i+1 

ρ n (n + 1)2 ,

n=0 i+1 

qn s (z)

n=0

≤

i+1  n=0

≤

i+1  λn z λn ai+2 1 ≤ n 3 2 (ν (z) − s) (s0 − s) n=0 (1 − ai+2 /an )3

i+1 λ0 a 0  n ρ (n + 1)6 , (s0 − s)2 n=0

2.1 Definition of a Multidimensional Memory …

vi+2 − v0 s (z, t) ≤

i+1 

77

rn s (z, t)

n=0

≤

i+1  n=0

≤

i+1 λn z λn ai+2 1  ≤ n 2 (ν (z) − s) s0 − s n=0 (1 − ai+2 /an )2

i+1 λ0 a 0  n ρ (n + 1)4 . s0 − s n=0

  Now we choose a0 ∈ 0, 2sT0 so as to obtain ρ < 1, λ0 a0

∞ 

ρ n (n + 1)6 ≤ R.

n=0

Then, u i+2 − u 0 s (z, t) ≤ R, ki+2 − k0 s (z) ≤ vi+2 − v0 s (z, t) ≤

R , (s0 − s)2

R , (z, t, s) ∈ Fi+2 . s0 − s

Since the choice of a0 which is independent of the number of the approximation, all the successive approximations (u n , kn , vn ) belong to C(z,t) (F; As ) , F =

∞ 

Fn ,

n=0

and satisfy the inequalities u n − u 0 s (z, t) ≤ R, kn − k0 s (z) ≤ vn − v0 s (z, t) ≤

R , (s0 − s)2

R , (z, t, s) ∈ F. s0 − s

For s ∈ (0, s0 ) the series ∞ ∞ ∞    (u n − u n−1 ), (kn − kn−1 ), (vn − vn−1 ) n=0

n=0

converge uniformly in the norm of the space

n=0

78

2 The Solvability of Multidimensional Inverse …

C(z,t) (PsT ; As ) , PsT = G T



{(z, t)| 0 ≤ z ≤ a(s0 − s)} ,

therefore u n → u, vn → v, kn → k, and the limit functions u, k, v are elements of C(z,t) (PsT ; As ) and satisfy Eqs. (2.11), (2.14), and (2.15). ˆ be any Now we prove that this solution is unique. Let (u, v, k) and (u, ˆ v, ˆ k) two solutions satisfying the inequalities u − u 0 s (z, t) ≤ R, k − k0 s (z) ≤ v − v0 s (z, t) ≤

R , (s0 − s)2

R , (z, t, s) ∈ F. s0 − s

Denote p˜ = u − u, ˆ r˜ = r − rˆ , q˜ = q − q, ˆ   ν(z) − s (ν(z) − s)2 , sup ˜r s (z, t) ,  p ˜ s (z, t) z z (z,t,s)∈F (z,t,s)∈F   (ν(z) − s)3 sup q < ∞, ˜ s (z) z (z,t,s)∈F 

λ := max

sup

  2 −1 where ν(z) = s0 − z/a, a = a0 ∞ . Then, for the functions n=0 1 + 1/(n + 1) p, ˜ r˜ , q˜ can be obtained the relations 1 p(x, ˜ z, t) = 2 τ −ξ +

   p(x, ˜ ξ, τ ) − q(x, ˜ τ − ξ) (z,t)

   q(x, ˜ γ )u(x, ˆ ξ, τ − γ ) + k(x, γ ) p(x, ˜ ξ, τ − γ ) dγ dτ dξ,

0

1 r˜ (x, z, t) = 2

−z  p(x, ˜ z − |ξ |, ξ + t) − q(x, ˜ ξ + |ξ | + t − z) z ξ +|ξ |+t−z

q(x, ˜ γ )u(x, ˆ z − |ξ |, t + ξ − γ )

+ 0

 + k(x, γ ) p(x, ˜ z − |ξ |, t + ξ − γ ) dγ sgn ξ dξ,

2.1 Definition of a Multidimensional Memory …

79

z/2 ˜r (x, ξ, z − ξ ) q(x, ˜ z) = 2 0 z−2ξ 

+

   q(x, ˜ γ )ˆr (x, ξ, z − ξ − γ ) + k(x, γ )˜r (x, ξ, z − ξ − γ ) dγ dξ.

0

Applying to these equations the estimates given above, we find the inequality in the form   λ ≤ λρ  , ρ  := a 4c0 + 1 + 3RT (3 + s0 + s02 ) max(a, 1) < ρ < 1. Consequently, λ = 0. Therefore, u = u, ˆ r = rˆ , q = q. ˆ Theorem 2.1 is proved.



Remark 2.1 The above-constructed functions u and k represent a weak (= general ized) solution to problem (2.8)–(2.10) in the domain Rm × PsT and the functions u  and k, where u  (x, z, t) = −δ(t − z) + u(x, z, t)θ (t − z), a generalized solution to problem (2.1)–(2.3). Consider the set ϒ consisting of all pairs of functions ( f, h) that are the elements

of Ct [0, T ]; As0 , s0 > 0, for which the conditions of Theorem 2.1 are valid with fixed R, T , and s0 . The following stability theorem holds: ¯ ∈ ϒ. Then, the corresponding solutions Theorem 2.2 Let ( f, h) ∈ ϒ and ( f¯, h) ¯ v) (u, k, v) and (u, ¯ k, ¯ to the system of Eqs. (2.11), (2.14), and (2.15) satisfy the following estimates: ¯ s (z) ≤ u − u ¯ s (z, t) ≤ c, k − k v − v ¯ s (z, t) ≤

c , (s0 − s)2

c , (z, t) ∈ PsT , 0 < s < s0 , s0 − s

where   T ¯ ¯ ,  = max max  f − f s0 (t), max h − hs0 (t) max 1, 2 max  f t − f¯t s0 (t), max h t − h¯t s0 (t), max  f tt − f¯tt s0 (t) , t ∈ [0, T ], and the constant c depends only on R, T , and s0 .

(2.23)

80

2 The Solvability of Multidimensional Inverse …

˜ v − v¯ = v, Proof For the differences u − u¯ = u, ˜ k − k¯ = k, ˜ f − f¯ = f˜, and h − ˜ from relations (2.11), (2.14), and (2.15) we derive the equalities h¯ = h, 1 u(x, ˜ z, t) = u˜ 0 (x, z, t) + 2

  ˜ τ − ξ) u(x, ˜ ξ, τ ) − k(x, (z,t)

τ −ξ

 ˜ γ )u(x, ξ, τ − γ ) + k(x, ¯ γ )u(x, k(x, ˜ ξ, τ − γ ) dγ dτ dξ,

+ 0

˜ k(x, z) = k˜0 (x, z) + 2

z/2 v(x, ˜ ξ, z − ξ ) 0

z−2ξ  

+

 ˜ γ )v(x, ξ, z − ξ − γ ) + k(x, ¯ γ )v(x, k(x, ˜ ξ, z − ξ − γ ) dγ dξ, (2.24)

0

1 v(x, ˜ z, t) = v˜0 (x, z, t) + 2

−z ˜ ξ + |ξ | + t − z) u(x, ˜ z − |ξ |, ξ + t) − k(x, z

ξ +|ξ |+t−z

˜ γ )u(x, z − |ξ |, t + ξ − γ ) k(x,

+ 0

 ¯ γ )u(x, + k(x, ˜ z − |ξ |, t + ξ − γ ) dγ sgn ξ dξ, ¯ in which u˜ is calculated by the formula (2.12) with f , h replaced by f¯, h,    ; v(x, ˜ z, t) = u˜ 0t (x, z, t). k˜0 (x, z) = 8 f˜tt (x, t) + h˜ t (x, t)  t=2z

It is obvious that u˜ 0 s0 (z, t) ≤ 2, k˜0 s0 (z) ≤ 16, v˜0 s0 (z, t) ≤ 2, (z, t) ∈ PsT . Theorem 2.1 yields the following estimates: ¯ s (z) ≤ us (z, t) ≤ 2R, k

R(1 + s02 ) R(1 + s0 ) . , vs (z, t) ≤ 2 (s0 − s) s0 − s

(2.25)

2.2 Definition of the Memory Kernel Standing at Some …

81

˜ v˜ the method of Applying to the system of Eq. (2.24) which is linear in u, ˜ k, successive approximation that was used in the proof of Theorem 2.1, we find that, for the same choice of a0 , solutions to system (2.24) satisfy the inequalities u˜ − u˜ 0 s (z, t) ≤ c1 , k˜ − k˜0 s (z) ≤ v˜ − v˜0 s (z, t) ≤

c1  , (s0 − s)2

c1  , (z, t) ∈ PsT , 0 < s < s0 , s0 − s

in which c1 depends only on R, T , and s0 . Hence, taking (2.25) into account, we obtain inequalities (2.23). Denote by Hs0 the set of pairs of functions (u, k) such that u, u t , u z and k belong to As0 and are continuous for (z, t) ∈ G T , T > 0. Theorem 2.3 There exists at most one solution to inverse problem (2.8)–(2.10) in the set Hs0 . Proof The proof is carried out by way of contradiction. Suppose that there are two solutions (u 1 , k1 ) ∈ Hs0 and (u 2 , k2 ) ∈ Hs0 to the inverse problem (2.8)–(2.10). Fix any s ∈ (0, s0 ). Since (u i , ki ) ∈ Hs0 , i = 1, 2, by virtue of Theorem 2.1, there exists such a > 0 that the following relations are valid: u 1 (x, z, t) = u 2 (x, z, t), k1 (x, z) = k2 (x, z), x ∈ Rn , (z, t) ∈ G T , 0 ≤ z ≤ a.

(2.26)

Denote by a ∗ the least upper bound of all possible numbers a ∈ (0, T /2) for which equalities (2.26) hold. Since (u 1 , k1 ) = (u 2 , k2 ), we have a ∗ ∈ (0, T /2). Now consider the problem similar to (2.8)–(2.10), in which the plane z = a ∗ plays the role of the support of the Cauchy data. Since on this plane the traces of the functions u 1 , u 2 and their derivatives u 1z , u 2z coincide, then, in view of Theorem 2.1 there exists a number a ∗∗ ∈ (a ∗ , T /2) such that (u 1 , k1 ) = (u 2 , k2 ), x ∈ Rn , (z, t) ∈ G T , a ∗ ≤ z ≤ a ∗∗ , which fact contradicts the definition of a ∗ . The contradiction proves the validity of Theorem 2.3.

2.2 Definition of the Memory Kernel Standing at Some Derivatives of a Direct Problem Solution In this section, we indicate the statements of several inverse problems, the study of which can be carried out according to the method of Sect. 2.1. Unlike the problem that was considered in the previous section, where the memory kernel is multiplied by the solution of the direct problem, the equations are investigated here, in which

82

2 The Solvability of Multidimensional Inverse …

the memory kernel is multiplied by the second derivative in time of direct problem solution and the first-order differential operator with analytical coefficients. At the beginning, we consider an initial–boundary problem for the wave equation with memory t u tt − u zz − u = k(x, τ )u tt (x, z, t − τ )dτ, 0

(x, t) ∈ Rn+1 , z ∈ R+ ,

(2.27)

u|t 0, i.e., u|z=0 = F(x, t), x ∈ Rn , t < T.

(2.29)

It is assumed that unknown function k(x, t) at any fixed x belongs to class C 2 [0, T ]. In Eq. (2.27) by integrating by parts we “relocate” time derivative and by introducing a new function v(x, z, t) according to the formula u = ρ(x, t)v(x, z, t), ρ(x, t) := exp[k0 (x)t/2], k0 (x) := k(x, 0), from equalities (2.27) and (2.28), we have t vtt − vzz = v + t (∇k0 , ∇v) + H (x, t)v +

h(x, t − τ )v(x, z, τ )dτ, 0

(x, t) ∈ Rn+1 , z > 0,

(2.30)

v|t z. For the regular part of function v(x, z, t) in the domain t > z > 0, x ∈ Rn , the inverse problem (2.30), (2.31), and (2.29) is equivalent to the problem vtt − vzz = v + t (∇k0 , ∇v) + H (x, t)v t−z + h(x, t − τ )v(x, z, τ )dτ,

(2.34)

0

v|z=0 = f (x, t), vz |z=0 = G(x, t),

(2.35)

v|t=z+0 = β(x, z).

(2.36)

In Eqs. (2.34)–(2.36) we replace the variables z, t by z 1 , t1 by the formulas z 1 = t + z, t1 = t − z.  Then, v(x, z, t) = v x, (z 1 − t1 )/2, (z 1 + t1 )/2 := v1 (x, z 1 , t1 ) . The problem (2.34)–(2.36) in new variables is rewritten as the problem of finding the functions v, k from the equations

84

2 The Solvability of Multidimensional Inverse …

∂ 2 v1 (x, z 1 , t1 ) ∂t1 ∂z 1  1 = − v1 + (t1 + z 1 ) (∇k0 , ∇v1 ) /2 + H (x, (t1 + z 1 )/2)v1 + h(x, t1 ) 4 t1 + h(x, τ )v1 (x, z 1 − τ, t1 − τ )dτ , 0

(x, z 1 , t1 ) ∈ {(x, z 1 , t1 ) | x ∈ Rn , 0 ≤ t1 ≤ z 1 } := D, v1 |t1 =z1 = f (x, z 1 ),

(2.37)

∂ 1 1 v1 |t1 =z1 = f t (x, t) |t=z1 + G(x, z 1 ), ∂z 1 2 2 z 1 > 0, x ∈ Rn ,

(2.38)

v1 |t1 =0 = β[x, (1/2)z 1 ], z 1 > 0, x ∈ Rn .

(2.39)

We recall that h is related to k according to the second formula of (2.32). We introduce the function w(x, z 1 , t1 ) =

∂ v1 (x, z 1 , t1 ), t1 < z 1 . ∂z 1

(2.40)

Demanding the continuity of functions v(x, z, t), w(x, z, t) when z 1 = t1 = 0, x ∈ Rn , from (2.38) and (2.39) it is not difficult to express k0 (x), k0t (x) by the known functions: k0 (x) = 2 f (x, 0), k0t (x) = 2 f t (x, t)|t=0 − f 2 (x, 0) + 2g(x, 0). Further we will assume that in the equalities for H (x, z), G(x, t), β(x, z) instead of functions k0 (x), k0t (x) their expressions by means of the latter equations are used. For simplicity we will omit index 1 in z 1 , t1 , v1 . We denote by C zi (As , G) a class of functions which are continuously differentiable i-times in z and continuous in t in the domain G. For fixed (z, t) the norm of the function ω(x, z, t) in As0 , as earlier, we denote by ωs0 (z, t). The norm of the function ω in C (As , G) is defined by the equality ωC(As ,G) = sup ωs0 (z, t). (z,t)∈G

Theorem 2.4 Let f (x, 0), f t (x, t)|t=0 , f xi (x, 0), i = 1, . . . , n,  f (x, 0), g(x, 0) belong to As0 , so > 0, and f (x, t), f t (x, t), f tt (x, t), g(x, t), gt (x, t) belong to C(As0 , [0, T ]), and

2.2 Definition of the Memory Kernel Standing at Some …

85

 max  f (x, t)s0 (t), w0 s0 (z), h 0 (x, z)s0 (z), k0 (x)s0 = R, for (z, t) ∈ G T := {(z, t)|0 ≤ t ≤ z ≤ T }. Then, for any χ > 0 we can find the number a = a(s0 , T, R, n), as0 < T so that for any s ∈ (0, s0 ) there exists a unique solution of the problem (2.37)–(2.39)



 v(x, z, t) ∈ C z1 As0 , Ds , k(x, t) ∈ Ct2 As0 , [0, a(s0 − s)] , Ds is the domain on the plane z, t : Ds := {(z, t)|0 ≤ t ≤ z < a(s0 − s)}, and solution satisfies the following inequalities: v − v0 s (z, t) ≤ χ , k − k 0 s (z) ≤

2χ , (s0 − s)

(2.41)

where h 0 (x, z) := 2 f (x, 0) exp[− f (x, 0)z]g(x, z) − 2 exp[− f (x, 0)z]gt (x, t)|t=z + (1/2) f (x, 0) − H (x, z) f (x, z) −  f (x, z) − 2z (∇ f (x, 0), ∇ f (x, z)) − 2 f tt (x, t)|t=z + (z/2)

n 

f x2i (x, 0),

i=1

v0 := f (x, z), k := k0 (x) + zk0t , (z, t) ∈ Ds . 0

Proof At the beginning, the problem (2.37)–(2.39) is reduced to the closed system of the Volterra-type integro-differential equations in the domain (x, z, t) ∈ D. The equation for v is rewritten with the help of Eq. (2.40), while the equation for k is rewritten with the help of the second equation of (2.32): z v(x, z, t) = v0 (x, z) +

w(x, ξ, t)dξ,

(2.42)

t

z k(x, z) = k (x) +

(z − ξ ) exp [ f (x, 0)ξ ] h(x, ξ )dξ.

0

(2.43)

0

For fixed x ∈ Rn by integrating equality (2.37) on the plane (τ, ξ ) along the line ξ = z from the point (t, z) to the point (z, z) and using the second condition in (2.38), we can get the equation for w(x, z, t):

86

2 The Solvability of Multidimensional Inverse …

1 w(x, z, t) = w0 (x, z) + 4

z [v(x, z, τ ) t

+ (z + τ )∇ f (x, 0)∇v(x, z, τ ) + H [x, (z + τ )/2]v(x, z, τ ) τ + h(x, τ ) + h(x, η)v(x, z − η, τ − η)dη]dτ, (2.44) 0

where w0 (x, z) = (1/2) f t (x, t)|t=z + (1/2)G(x, z). Using Eq. (2.39) differentiated by z, when t = 0 from (2.44), we find z  v(x, z, τ ) + (z + τ ) (∇ f (x, 0), ∇v(x, z, τ )) 0

+ H [x, (z + τ )/2]v(x, z, τ ) + h(x, τ ) τ + h(x, η)v(x, z − η, τ − η)dη dτ = 2 f t (x, 0) 0

+ 2g(x, 0) + (z/2) f (x, 0) + (z 2 /4)

n 

f x2i (x, 0) − 2 f t (x, z)

i=1

− 2 exp[− f (x, 0)z]g(x, z). The equation for h(x, t) can be easily found by differentiation of the latter equation on z and by using the first condition from (2.38): z  w(x, z, τ ) + (z + τ ) (∇ f (x, 0), ∇w(x, z, τ ))

h(x, t) = h 0 (x, t) − 0

+ (∇ f (x, 0), ∇v(x, z, τ )) + Hz [x, (z + τ )/2]v(x, z, τ ) + H [x, (z + τ )/2]w(x, z, τ ) τ (2.45) + h(x, τ ) f (x, z − τ ) + h(x, η)w(x, z − η, τ − η)dη dτ, 0

where the function h 0 (x, t) is determined in Theorem 2.4. Equations (2.42)–(2.45) represent a closed system of integro-differential equations for the functions v, k, w, h in the domain DT = G T × Rn . For convenience we introduce the function vector φ(x, z, t) = (φ1 , φ2 , φ3 , φ4 ) := (v, k, w, h). We rewrite the system (2.42)–(2.45) as an operator equation

2.2 Definition of the Memory Kernel Standing at Some …

87

φ = Mφ,

(2.46)

where M = (M1 , . . . , M4 ) is defined by the right sides of Eqs. (2.42)–(2.45), and φ 0 = M(0), φ 0 = (φ10 , φ20 , φ30 , φ40 ) := (v0 , k 0 , w0 , h 0 ). We define the iterations for Eq. (2.46): φ i+1 = Mφ i , i = 0, 1, 2, . . . , φ i = (φ1i , . . . , φ4i )

(2.47)

and φ i+1 − φ i := ψ i , i = 0, 1, 2, . . . , ψ i = (ψ1i , . . . , ψ4i ). Let the sequence of numbers a0 , a1 , . . . , ai , . . . be determined by the expressions ai+1 = ai /(1 + (i + 1)−2 ), i = 0, 1, 2, . . .. Here a0 is a fixed positive number. The number a0 < T /s0 will be chosen later. With the numerical sequence aσ we link the sequence of enclosed fields Fi = {(z, t, s)|0 < s < s0 , 0 ≤ t ≤ z < ai (s0 − s)}. The following lemma is valid. Lemma 2.1 If the conditions of Theorem 2.4 for any fixed χ > 0 and any i = 0, 1, 2, ... are fulfilled, then there exist a0 ∈ (0, T /s0 ), a0 = a0 (R, s0 , χ , n) and λi = λi (R, s0 , χ , n) > 0, such that for each s ∈ (0, s0 )  

i ψ1 , ψ3i ∈ C(As , Dsi ), ψ2i , ψ4i ∈ C (As , [0, ai (s0 − s)]) , Dsi := {(z, t)|0 ≤ t ≤ z < ai (s0 − s)}, and the following inequalities are valid: ψ1i s (z, t) ≤

λi z λi ai z , ψ σj s (z, t) ≤ , j = 2, 3, ai (s0 − s) − z [ai (s0 − s) − z]2 ψ4i s (z) ≤

λi ai2 z , (z, t, s) ∈ Fi ; [ai (s0 − s) − z]3

φ1i+1 − φ10 s (z, t) ≤ χ , φ i+1 − φ 0j s (z, t) ≤ j φ4i+1 − φ40 s (z) ≤

(2.48)

2χ , j = 2, 3, s0 − s

4χ , (z, t, s) ∈ Fi+1 . (s0 − s)2

(2.49)

As in Sect. 2.1, we use the Nirenberg method and its modification to prove the lemma. Using estimate (2.4) it is not difficult to check that inequalities (2.48), (2.49) are satisfied when i = 0, besides λ0 is proportional to a0 . We determine the validity of these inequalities for any i using the induction method. We assume that if the statement of the lemma is valid for i ≤ σ and prove that, then it is valid for i =

88

2 The Solvability of Multidimensional Inverse …

σ + 1, as well. Using the inductive assumption, we find that ψ σ +1 ∈ C(As , Ds(σ +1) ). Besides, from (2.46), we find ψ1σ +1 s (z, t)

z ≤

ψ3σ s (ξ, t)dξ

t

z ≤ t

λσ aσ ξ dξ λσ z λσ a σ z ≤ a0 , ≤ [aσ (s0 − s) − ξ ]2 [aσ (s0 − s) − z] aσ +1 (s0 − s) − z

ψ2σ +1 s (z)

z ≤

ψ4σ s (ξ )dξ ≤ a0

0

ψ3σ +1 s (z, t)

1 ≤ 4

z

λσ aσ +1 z , [aσ +1 (s0 − s) − z]2

{ψ1σ s (z, ξ ) + 2T R

n  j=1

t

+ Rψ1σ s (z, ξ ) + ψ4σ s (ξ ) +

ξ



∂ σ ψ s (z, ξ ) ∂x j 1

[ψ4σ s (τ )φ1σ +1 |s (z − τ, ξ − τ )

0

+ φ4σ s (τ )ψ1σ s (z − τ, ξ − τ )]dτ }dξ. Using inequality (2.4) to estimate ψ1σ s , 

∂ σ ψ s , j = 1, 2, . . . , n, ∂x j 1

and assuming that s  = s  (ξ ) = (s + s0 − ξ/an )/2, we get 1 ψ3σ +1 s (z, t) ≤ 4

z  t

2λσ z 16naσ2 · [aσ (s0 − s) − ξ ]2 aσ (s0 − s) − z

2λσ z 2Rλσ z 8T Rnaσ · + + aσ (s0 − s) − ξ aσ (s0 − s) − z aσ (s0 − s) − z ⎫

⎪ ξ  2 ⎬ 2 2 4χ + Rs0 λσ aσ τ (χ + R) 2λσ z λσ a σ ξ + + · dτ dξ + ⎪ [aσ (s0 − s) − ξ ]3 (aσ (s0 − s) − τ )3 (s0 − s)2 aσ (s0 − s) − z ⎭ 0

aσ +1 λσ z , ≤ c1 a0 [aσ +1 (s0 − s) − z]2

2.2 Definition of the Memory Kernel Standing at Some …

89

where c1 = c1 (s0 , n, a0 , χ , R, T ). In the latter inequality it is used that aσ2 ≤ aσ +1 a0 are valid for any σ ≥ 1. Moreover, aσ3 ≤ aσ2 +1 a0 for σ ≥ 1. These inequalities can be easily checked by the method of mathematical induction. The latter inequalities are used to estimate the function ψ4σ +1 . Similarly, for ψ4σ +1 we obtain ψ4σ +1 s (z) ≤ c2 a0

aσ2 +1 λσ z , [aσ +1 (s0 − s) − z]3

where c2 = c2 (s0 , n, a0 , χ , R, T ). It is obvious that ci , i = 1, 2 in the latter estimates are monotonically nondecreasing functions of the s0 , a0 parameters. From the estimations made above, it follows that (2.48) is valid when i = σ + 1, if we assume that λσ +1 = a0 cλσ , where c = max(1, s0 , c1 , c2 ). Now we show that inequalities (2.49) are also satisfied when i = σ + 1, if the number a0 is chosen properly. For (x, z, t) ∈ Fσ +2 φ1σ +2 ≤ λ0

−

φ10 s (z, t)

σ +1  j=0

 (a0 c) j

≤

σ +1 

j ψ1 s (z, t)

≤

j=0

j=0

j=0

σ +1 

ψγj s (z, t) ≤

j=0

φ4σ +2 − φ40 s (z) ≤

λjz a j (s0 − s) − z

σ +1  − 1 ≤ λ0 (a0 c) j ( j + 1)2 ,

aj a j+1

φγσ +2 − φγ0 s (z, t) ≤

σ +1 

σ +1 

j

ψ4 s (z) ≤

j=0

σ +1 2λ0  (a0 c) j ( j + 1)4 , γ = 2, 3, s0 − s j=0

σ +1 4λ0  (a0 c) j ( j + 1)6 . (s0 − s)2 j=0

That is why inequalities (2.49) are satisfied when i = σ + 1, if the number a0 is chosen so that ∞  a0 c < 1, λ0 (a0 c) j ( j + 1)6 ≤ χ . (2.50) j=0

It is clear that a0 can be always chosen to be so small that inequalities (2.50) are satisfied. Thus, the validity of the lemma is proven. Further, we assume that the number a0 is chosen from the conditions (2.50). In order to complete the proof of Theorem 2.4, we should note that under the chosen value of a0 the φ i sequence uniformly converges in the norm of space C(As , Ds ), a = lim ai , the limiting function belongs to C(As , Ds ) and provides the

90

2 The Solvability of Multidimensional Inverse …

solution of the operator equation (2.46). The limit transition in the first two inequalities (2.49) when (x, z, t) ∈ F = {(z, t, s)|0 < s < s0 , 0 ≤ t ≤ z < a(s0 − s)} leads to the values coinciding with (2.41). The uniqueness of the constructed solution is established by using the described above technics. We indicate a few settings of inverse problems of memory determination which could be studied by the above methods. Problem 1 Find the functions u(x, z, t), k(x, t) that satisfy the following equalities: t u tt − u zz − u =

k(x, τ )u t (x, z, t − τ )dτ, 0

(x, t) ∈ Rn+1 , z > 0,

(2.51)

u|t t > 0, x ∈ Rn from the equations ∂ 2 v(x, z, t) ∂t∂z ⎤ ⎡ t 1 = − ⎣v + k0 (x)v − g(x)h(x, t) − h(x, τ )v(x, z − τ, t − τ )dτ ⎦ , 4 0

(x, z, t) ∈ {(x, z, t) | x ∈ Rn , 0 ≤ t ≤ z} := D, v |t=z = f (x, z),

∂ 1 v |t=z = f t (x, t)|t=z , z > 0, x ∈ Rn , ∂z 2

v |t=0 =

1 k0 (x)z, z > 0, x ∈ Rn , 2

(2.55) (2.56)

(2.57)

2.2 Definition of the Memory Kernel Standing at Some …

91

in which k0 := k(x, 0) = 1/g(x)[ f t (x, t)|t=0 + g(x)], / h(x, t) := kt (x, t), t > 0. With respect to problem (2.5)–(2.7) the analogue of Theorem 2.4 is valid. Theorem 2.5 Let 

 g(x), 1/g(x), (∂/∂z) f (x, z)|z=0 , (∂ 3 /∂ 2 xi ∂z) f (x, z)|z=0 ∈ As0 ,

s0 > 0, i = 0, 1, . . . , n, 



 f (x, z), (∂/∂z) f (x, z), (∂ 2 /∂ 2 z) f (x, z) ∈ C As0 , [0, T ] ,

when some fixed s0 > 0, T > 0, whereas  max g(x)s0 ,  f (x, z)s0 ,  f z (x, z)s0 , 1/g(x)s0 , g(x)s0 ,  f zz (x, z)s0 = R, for t ∈ [0, T ] and the consistency condition is satisfied: f z (x, z)|z=0 + g(x) = g(x) f z (x, z)|z=0 . Then for any χ > 0 the number a ∈ (0, T /s0 ) can be found such that for any s ∈ (0, s0 ) there exists the unique solution of problem (2.55)–(2.57)



 v(x, z, t) ∈ C z1 As0 , Ds , k(x, t) ∈ Ct1 As0 , [0, a(s0 − s)] , where the domain Ds is determined in Theorem 2.4, and for the solution inequalities (2.41) are valid with the functions v0 = f (x, z), k0 (x) =

1 [ f t (x, t)|t=0 + g(x)]. g(x)

Problem 2 Find the functions u(x, z, t), k(x, t) that satisfy the following equalities: t u tt − u zz − Lu =

k(x, τ )L 0 u(x, z, t − τ )dτ, 0

(x, t) ∈ Rn+1 , z > 0,

(2.58)

u|t 0, and d := max ai j s0 , b0 := max bi s0 , c0 := cs0 . 1≤i, j≤n

1≤i, j≤n

For the operators L 0 , L from (2.19) it follows: L 0 νs  ≤

c1 c2 νs , Lνs  ≤ νs , s  ∈ (0, s),  s−s (s − s  )2 c1 := nb0 + s0 c0 , c2 := 4n 2 d + c1 s0 .

The solution of problem (2) and (2.59) is represented as u(x, z, t) = α(x, z)δ(t − z) + u(x, ˜ z, t)θ (t − z)

(2.61)

and we use the method of separation of singularities. Denoting u(x, ˜ z, z + 0) =: β(x, z), we put (2.61) in (2) and (2.59) and find α(x, z) ≡ α(x) = g(x), β(x, z) = (z/2)Lg(x), x ∈ Rn , z > 0.

(2.62)

It is not difficult, using equalities (2.61) and (2.62), to replace the inverse problem (2)–(2.60) by initial-characteristic problem with Cauchy data on z = 0. It should be noted that when t > z > 0 the equality u = u˜ takes place, then the functions u(x, z, t), k(x, z) satisfy the equations t−z k(x, τ )L 0 u(x, z, t − τ )dτ, u tt − u zz − Lu = L 0 g(x)k(x, t − z) + 0

(x, z, t) ∈ G := {(x, z, t)|x ∈ Rn , t > z > 0},

(2.63)

u|z=0 = f (x, t), u z |z=0 = 0, x ∈ Rn , t > 0,

(2.64)

u|z=0 =

1 z Lg(x), x ∈ Rn , t > 0. 2

(2.65)

2.2 Definition of the Memory Kernel Standing at Some …

93

Further, the problem (2.63)–(2.64) is reduced to the following system of integrodifferential equations for u, u t , k: 1 u(x, z, t) = u 0 (x, z, t) + 2

 (z,t)

[Lu(x, ξ, τ ) − L 0 g(x)k(x, τ − ξ )

τ −ξ +

k(x, γ )L 0 u(x, ξ, τ − γ )dγ ]dτ dξ,

(2.66)

0

u t (x, z, t) = u 0t (x, z, t) +

1 2

z [Lu(x, z − |ξ |, t + ξ ) − L 0 g(x)k(x, t −z t−z+|ξ  |+ξ

k(x, γ )L 0 u(x, z − |ξ |, t + ξ − γ )dγ ]sgnξ dξ,

− z + |ξ | + ξ ) + 0

(2.67)

2 k(x, z) = k0 (x, z) + L 0 g(x)

z/2 1 [Lu t (x, ξ, z − ξ ) + ξ L 0 g(x)k(x, z − 2ξ ) 2 0

z−2ξ 

+

k(x, γ )L 0 u t (x, ξ, z − ξ − γ )dγ ]dξ, 0

where u 0 (x, z, t) = k0 (x, z) =

1 [ f (x, t + z) + f (x, t − z)] , 2

  1 (1/4)z L 2 g(x) − 2 f tt (x, t)|t=z . L 0 g(x)

For the system (2.66)–(2.68) the following theorem is valid. Theorem 2.6 Assume that the consistency conditions f (x, 0) = 0, f t (x, t)|t=0 = (1/2)Lg(x) are satisfied, besides (ai j , bi , c, g)(x) ∈ As0 , 1 ≤ i, j ≤ n,

 [ f (x, t), f t (x, t), f tt (x, t)] ∈ C As0 , [0, T ] ,

(2.68)

94

2 The Solvability of Multidimensional Inverse …

max[ f s0 (t),  f t s0 (t),  f tt s0 (t), L 0 gs0 , 1/L 0 gs0 , L 2 gs0 ] = R, for t ∈ [0, T ], R > 0. Then, a ∈ (0, T /2), as0 < (T /2) can be found such that for any s ∈ (0, s0 ) in the domain G T ∩ {(x, z, t)|x ∈ Rn , 0 ≤ z ≤ a(s0 − s)} there exists the unique solution of the system (2.66)–(2.68), for which (u(x, z, t), u t (x, z, t)) ∈ C(As0 , PsT ) k(x, z) ∈ C(As0 , [0, a(s0 − s)]), PsT := G T ∩ {(z, t)|0 ≤ z ≤ a(s0 − s)}, moreover, u − u 0 s (z, t) ≤ R0 , k − k0 s (t) ≤ u t − u 0t s (z, t) ≤

R0 , (s0 − s)2

R0 , (s0 − s)

(z, t) ∈ PsT , R0 = R0 (R, T, s0 , n) is a constant. Theorems 2.5 and 2.6 are proved similarly to Theorem 2.4.

2.3 Definition of Memory Kernel in a Two-Dimensional System Maxwell Equations Consider the Maxwell system of integro-differential equations: ∇ × H = (∂/∂t)D(x, t) + σ E + j, ∇ × E = −(∂/∂t)B(x, t), (x, t) ∈ R4 , where

(2.69)

t D(x, t) = εE +

ϕ(x, t − τ )E(x, τ )dτ, 0

t B(x, t) = μH +

ψ(x, t − τ )H (x, τ )dτ, 0

E = (E 1 , E 2 , E 3 ), H = (H1 , H2 , H3 ), D = (D1 , D2 , D3 ), B = (B1 , B1 , B3 ), x = (x1 , x2 , x3 ),

(2.70)

2.3 Definition of Memory Kernel in a Two-Dimensional System Maxwell Equations

95

ϕ(x, t), ψ(x, t) are the scalar functions characterizing memory of the medium, the density of the external electric current has the form ⎛ ⎞ 0 0 0 j = j g(x1 )δ(x3 )δ(t), j = ⎝ 1 ⎠ . (2.71) 0 Add to the equations (2.69)–(2.71) the following zero initial conditions: E|t 0, x ∈ R,

(2.80)

u|z=0 = f (x, t), u z |z=0 = 0, t > 0, x ∈ R,

(2.81)

u|t=z+0 = 0, x ∈ R.

(2.82)

In Eqs. (2.80)–(2.82), we make the change of variables z, t to z 1 , t1 by the formulas: z 1 = t + z, t1 = t − z. Then, u(x, z, t) = u(x, (z 1 − t1 )/2, (z 1 + t1 )/2) := u 1 (x, z 1 , t1 ). Relations (2.80)– (2.82) in the new variables are rewritten in the form ∂ 2 u 1 (x, z 1 , t1 ) ∂t1 ∂z 1 ⎤ ⎡ t1 1⎣ + u 1x x − k(x, t1 ) + k(x, τ )u 1 (x, z 1 − τ, t1 − τ )dτ ⎦ = 0, 4 0

0 < t1 < z 1 , x ∈ R, u 1 |t1 =z1 = f (x, z 1 ),

∂ 1 u 1 |t1 =z1 = f z1 (x, z 1 ), z 1 > 0, x ∈ R, ∂z 1 2 u 1 |t1 =0 = 0, x ∈ R.

(2.83) (2.84) (2.85)

Introduce a function ϑ(x, z 1 , t1 ) =

∂ u 1 (x, z 1 , t1 ), t1 < z 1 . ∂z 1

(2.86)

2.4 Differential Properties of the Solution of the Memory …

99

Lemma 2.2 If the conditions f (x, 0) = f z (x, 0) = 0 are fulfilled, then, the inverse problem (2.83)–(2.85) in G T = {(x, z, t) | x ∈ R, 0 ≤ t ≤ z ≤ T } is equivalent to the following system of equations (for simplicity, further, we will omit index 1 in functions and variables) z ϑ(x, η, t)dη = u 0 (x, t),

u(x, z, t) − t

1 ϑ(x, z, t) − 4

z  u x x (x, z, ξ ) + k(x, ξ ) t

ξ −

k(x, τ )u(x, z − τ, ξ − τ )dτ dξ = υ 0 (x, z),

(2.87)

0

z  k(x, z) + ϑx x (x, z, ξ ) − k(x, ξ ) f (x, z − ξ ) 0

ξ −

k(x, τ )ϑ(x, z − τ, ξ − τ )dτ dξ = k 0 (x, z),

0

where 1 ∂ f (x, z), 2 ∂z ∂2 ∂2 ∂i ∂i k 0 (x, z) = −2 2 f (x, z) − 2 f (x, z), f (x, z) = f (x, t) |t=z , ∂z ∂x ∂z i ∂t i i = 0, 1, 2. (2.88) u 0 (x, t) = f (x, t), ϑ 0 (x, z) =

Proof The first equation (2.87) is easily obtained by integrating the equality (2.86) over z from the point (z, t) to the point (z, z) on the plane of the variables (η, ξ ) and using the first condition from (2.84). To obtain the second equation (2.87), it is necessary to integrate Eq. (2.83) with respect to t from the point (z, t) to the point (z, z) and use the second condition from (2.84). Further, pass to limit when t → 0 in the second equation of (2.87). To find the value of ϑ(x, z, 0) we use the differentiated equality (2.85) with respect to z. Then, one gets

100

2 The Solvability of Multidimensional Inverse …

z

ξ [u x x (x, z, ξ ) + k(x, ξ ) −

0

k(x, τ )u(x, z − τ, ξ − τ )dτ ]dξ = −4υ 0 (x, z). 0

Differentiating this equality with respect to z and solving relatively k(x, z) we get the third equation of (2.87). It is easy to show that inverse transformations also take place. Indeed, integrating the third equation (2.87) over z into limits from t to z, changing the order integration, where it is necessary, and taking into account the conditions f (x, 0) = f z (x, 0) = 0, we derive the second equation (2.87). Differentiating with respect to the corresponding variables the first and second equations of the system (2.87), we easily arrive at equalities (2.83)–(2.85). Lemma 2.2 is proved. Definition 2.1 We say that the function h(x) belongs to As , s > 0, if it is representable by Fourier series h(x) =

∞ 

h n exp(−inx)

n=−∞

and its s-norm hs =

&∞ n=−∞

| h n | exp(sn) is finite.

Definition 2.2 If some function u(x, z, t) belongs to As , s > 0, for fixed (z, t) ∈ G and is continuous in domain G, as an element of As , then, we say that it belongs to C(As , G), and by the symbol  u s (z, t) denote its s-norm for fixed (z, t) ∈ G. Theorem 2.7 Let

i i  (∂ /∂t ) f, (∂ 2 /∂ x 2 ) f ∈ C(As0 , [0, T ]), i = 0, 1, 2 for some fixed s0 > 0, T > 0, moreover ' ' ' 2 ' ' 2 ' ( '∂ f ' '∂ f ' '∂ f ' ' ' ' ' ' ≤ R, max  f s0 , ' ' ∂z ' , 2 ' ∂z 2 ' + ' ∂ x 2 ' s0 s0 s0 

R > 0, t ∈ [0, T ] and f (x, +0) = (∂/∂t) f (x, +0) = 0. Then, for an arbitrary χ there exists a number such a ∈ (0, T /s0 ) that for any s ∈ (0, s0 ) and (z, t) ∈ G = {(z, t)|0 ≤ t ≤ z < a(s0 − s)} the solution of the system (2.87) exists, is uniquely and (u, ϑ) ∈ C(As0 , G), k ∈ C(As0 , [0, a(s0 − s)]), besides, for solution the following estimates are fulfilled: u − u 0 s (z, t) ≤ χ , ϑ − ϑ 0 s (z, t) ≤

2χ , s0 − s

2.4 Differential Properties of the Solution of the Memory …

k − k 0 s (z) ≤

101

4χ , (z, t) ∈ G. (s0 − s)2

(2.89)

Proof For the system of Eq. (2.87) we consider successive approximations u σ (x, z, t), ϑ σ (x, z, t), k σ (x, z), defining them for σ = 0, 1, 2, . . . by the formulas u

σ +1

z (x, z, t) −

ϑ σ (x, η, t)dη = u 0 (x, t),

t

ϑ

σ +1

1 (x, z, t) − 4

z  u σx x (x, z, ξ ) t

ξ

σ

+ k (x, ξ ) −

σ



σ

k (x, τ )u (x, z − τ, ξ − τ )dτ dξ = ϑ 0 (x, z),

(2.90)

0

k

σ +1

z  ϑxσx (x, z, ξ ) (x, z) + 0

ξ

σ

− k (x, ξ ) f (x, z − ξ ) −

k σ (x, τ )ϑ σ (x, z − τ, ξ − τ )dτ dξ = k 0 (x, z).

0

Here u 0 (x, t), ϑ 0 (x, z), k 0 (x, z) are defined by the formulas (2.88). In addition, we introduce the differences between two successive approximations: u σ +1 − u σ = u˜ σ , ϑ σ +1 − ϑ σ = ϑ˜ σ , k σ +1 − k σ = k˜ σ , σ = 0, 1, 2, . . .

(2.91)

The functions u˜ σ , ϑ˜ σ , k˜ σ , defined by formula (2.91), as follows from (2.90), satisfy the system of equalities σ

z

u˜ (x, z, t) = t

ϑ˜ σ −1 (x, η, t)dη,

102

2 The Solvability of Multidimensional Inverse …

ϑ˜ σ (x, z, t) =

1 4

z  u˜ σx x−1 (x, z, ξ ) + k˜ σ −1 (x, ξ ) t

ξ  −

 k˜ σ −1 (x, τ )u σ (x, z − τ, ξ − τ ) + k σ −1 (x, τ )u˜ σ −1 (x, z − τ, ξ − τ ) dτ dξ,

0

z 

˜σ

ϑ˜ xσx−1 (x, z, ξ ) − k˜ σ −1 (x, ξ ) f (x, z − ξ )

k (x, z) = − 0

ξ  −

 ˜k σ −1 (x, τ )ϑ σ (x, z − τ, ξ − τ ) + k σ −1 (x, τ )ϑ˜ σ −1 (x, z − τ, ξ − τ ) dτ dξ,

0

σ = 0, 1, 2, . . .

(2.92)

For σ = 0, in these equations it is necessary to formally set u˜ −1 = u 0 , ϑ˜ −1 = ϑ 0 , k˜ −1 = k 0 , k −1 = 0. Let the sequence of numbers a0 , a1 , . . . , aσ , . . . be defined by relations aσ +1 = aσ /(1 + (σ + 1)−2 ), σ = 0, 1, 2, . . .. Here a0 is some fixed positive number generating the sequence aσ . The number a0 < T /s0 will be chosen later. We associate with numerical sequence aσ the following embedded domains   Fσ = (z, t, s)|0 < s < s0 , 0 ≤ t ≤ z < aσ (s0 − s) . The following lemma holds: Lemma 2.3 If the conditions of Theorem 2.7 are fulfilled, for any fixed χ > 0 and for any σ = 0, 1, 2, . . . there exist such numbers a0 ∈ (0, T /s0 ), a0 = (R, s0 , χ ) and λk = λk (R, s0 , χ ) > 0 that for s ∈ (0, s0 ) 

 u˜ σ , ϑ˜ σ ∈ C (As , G σ ) , k˜ σ ∈ C (As , [0, aσ (s0 − s)]) , G σ := {(z, t)|0 ≤ t ≤ z < aσ (s0 − s)}

and the following inequalities are valid: u˜ σ s (x, t) ≤

λσ z λσ a σ z , ϑ˜ σ s (x, t) ≤ , aσ (s0 − s) − z [aσ (s0 − s) − z]2

k˜ σ s (x) ≤

λσ aσ2 z , (z, t, s) ∈ Fσ ; [aσ (s0 − s) − z]3

(2.93)

2.4 Differential Properties of the Solution of the Memory …

u σ +1 − u 0 s (x, t) ≤ χ , ϑ σ +1 − ϑ 0 s (x, t) ≤ k σ +1 − k 0 s (z) ≤

103

2χ , s0 − s

4χ , (z, t, s) ∈ Fσ +1 . (s0 − s)2

(2.94)

Proof We use the method of mathematical induction. Check, first of all, that the statementof Lemma 2.3 is true at σ = 0. Under the conditions of Theorem 2.7,

 we have u˜ 0 , v˜ 0 , k˜ 0 ∈ C As , G σ for each s ∈ (0, s0 ), if a0 s0 ≤ T /2. Using equalities (2.92), we find that for (z, t, s) ∈ F0 the following estimates are valid: z u˜ s (z, t) ≤

z v s (η)dη ≤

o

t

1 v˜ s (z, t) ≤ 4 o

Rη ≤

o

t

z a0 s0 R, a0 (s0 − s) − z

z  u ox x s (ξ ) + k o s (ξ ) t

ξ +



1 k s (τ )u s (ξ − τ )dτ dξ ≤ 4 o

t

0

≤

1 4

z 

o

z  t

4u 0 s0 (ξ ) 2 + R + R ξ dξ e2 (s0 − s)2

 4R 1 4e2 Rz 2 2 + R + R ξ dξ ≤ + (R + R z)z e2 (s0 − s)2 4 (s0 − s)2

 a0 R 2 a0 z 4e + s02 + a0 s03 R . ≤ 2 4 a0 (s0 − s) − z Here, in intermediate inequalities, the estimate of s-norm of the second derivative of an analytic function through s0 -norm of the function itself is used. This has the form: 4hs0 , s < s0 . h x x s ≤ 2 e (s0 − s)2 ) In fact, according to Definition 2.1, we have

104

2 The Solvability of Multidimensional Inverse …

h x x s =

∞ 

|h n |n 2 exp(|n|s)

−∞

  ∞ |h n | exp(|n|s) ≤ sup n 2 exp (−|n|(s0 − s)) n

−∞

  hs0 4hs0 ≤ . sup t 2 exp(−t) = 2 2 (s0 − s) t>0 e (s0 − s)2 ) Similarly to the previous one, the estimate for k˜ 0 s is obtained: k˜ 0 s (z) ≤

z  vx0x s (ξ ) + k 0 s (ξ ) f s (z − ξ ) 0



ξ

z 

k s (τ )v s (z − τ )dτ dξ ≤

+

0

0

0

0

a02 z

≤ a0 (s0 − s) − z

4v 0 s0 (ξ ) 2 dξ + (1 + ξ )R e2 (s0 − s)2

2  2 3 3 a0 s0 R 4e + s0 R + a0 s0 R .

Therefore, estimates (2.93) are true for σ = 0, if λ0 = a0 R [4 + s0 max(1, R) + a0 s0 3R] max(1, s0 ).

(2.95)

At the same time for (z, t, s) ∈ F1 we have u 1 − u 0 s (z, t) = u˜ 0 s (z, t) ≤

λ0 a 1 λ0 z ≤ = λ0 , a0 (s0 − s) a0 − a1

v1 − v0 s (z, t) = v˜0 s (z, t) ≤ ≤

λ0 a 0 z [a0 (s0 − s) − z]2

2λ0 λ0 a 0 a 1 = , 2 (a0 − a1 ) (s0 − s) s0 − s

λ0 a02 z [a0 (s0 − s) − z]3 λ0 a02 a1 4λ0 ≤ = . 3 2 (a0 − a1 ) (s0 − s) (s0 − s)2

k1 − k0 s (z, t) = k 0 s (z) ≤

Therefore, inequalities (2.91) will be satisfied for σ = 0, if a0 is chosen from the condition λ0 < χ .

2.4 Differential Properties of the Solution of the Memory …

105

Thus, the first step of the induction method is justified. Suppose that the statement of the lemma holds for σ ≤ n, and prove that then it is also true for σ = n + 1. Using the inductive hypothesis, we find that (u˜ n+1 , ϑ˜ n+1 , k˜ n+1 ) ∈ C(As , G n+1 ). In addition, from (2.92) we obtain z u˜

n+1

s (z, t) ≤

˜ s (η, t)dη ≤ ϑ

t

z t

λn a n η dη [an (s0 − s) − η]2

z λn a n z ≤ λn a 0 , ≤ an (s0 − s) − z an+1 (s0 − s) − z z 1 4e−2 n+1 ϑ˜ s (z, t) ≤ {  u˜ n s (z, ξ ) + k˜ n s (ξ ) 4 (s − s)2 t

ξ +

[k˜ n s (τ )u n+1 s (z − τ, ξ − τ ) + k n s (τ )u˜ n s (z − τ, ξ − τ )]dτ }dξ.

0

Putting s  = s  (ξ ) = (s + s0 − ξ/an )/2 in this inequality, we get ϑ˜ n+1 s (z, t) z  16e−2 an2 λn an2 ξ 1 2λn z + ≤ 4 [an (s0 − s) − ξ ]2 an (s0 − s) − z [an (s0 − s) − ξ ]3 t ⎫ ⎬ ξ  λn an2 τ (χ + R) 2λn z 4χ + Rs02 dτ dξ + + ⎭ (an (s0 − s) − τ )3 (s0 − s)2 an (s0 − s) − z 0

≤

an+1 λn z a0 (8e−2 + 1 + 2χ + (1 + s02 )R). [an+1 (s0 − s) − z]2

Similar calculations for k˜ n+1 lead to the inequality k˜ n+1 s (z) ≤

2 an+1 λn z a0 (32e−2 + Rs0 (1 + 2s0 ) + χ (1 + 8s0 )). (an (s0 − s) − z)3

From the above estimates it follows that inequalities (2.93) are fulfilled for σ = n + 1, if one puts λn+1 = λn ρ,

106

2 The Solvability of Multidimensional Inverse …

ρ = a0 max[8e−2 + 1 + 2χ + (1 + s02 )R, 32e−2 + Rs0 (1 + 2s02 ) + χ (1 + 8s0 )]. (2.96) Now show that inequalities (2.94) also hold for σ = n + 1, if the number a0 is chosen appropriately. For (z, t, s) ∈ Fn+2 the following inequalities are valid: u

n+1 

n+1 

λσ z a (s − s) − z σ =0 σ =0 σ 0  n+1 n+1 n+1    aσ ρ σ aσ +1 ≤ λ0 ρσ − 1 ≤ λ0 ρ σ (σ + 1)2 , ≤ λ0 a − a a σ σ +1 σ +1 σ =0 σ =0 σ =0

n+2

− u s (z, t) ≤ 0

ϑ n+2 − ϑ 0 s (z, t) ≤ ≤

σ

u˜ s (z, t) ≤

n+1 2λ0  σ ρ (σ + 1)4 , k n+2 − k 0 s (z) s0 − s σ =0

n+1 4λ0  σ ρ (σ + 1)6 . (s0 − s)2 σ =0

Therefore, inequalities (2.94) are satisfied for σ = n + 1, if the number a0 is chosen so that ∞  ρ < 1, λ0 ρ σ (σ + 1)6 ≤ χ , (2.97) σ =0

where numbers λ0 and ρ are defined by the formulas (2.95) and (2.96), respectively. It is clear that one can always choose the parameter a0 so small that inequalities (2.97) hold. Thus, the validity of Lemma 2.3 is established. Further, we assume that the number a0 is chosen from the condition (2.97). Continuing the proof of Theorem 2.7, we note that for the chosen number a0 the sequence (u σ , ϑ σ , k σ ) converges uniformly in the norm of the space C(As , G), a = lim aσ and determines the unique solution of the system of Eq. (2.87), and at the same time the solution of the inverse problem. Limit transition in inequalities (2.94) for (z, t, s) ∈ F = {(z, t, s)|0 < s < s0 , 0 ≤ t ≤ z < a(s0 − s)} leads to estimates that coincide with (2.91). The uniqueness of the found solution is established using the above described technique for evaluating by the standard technique (see Sect. 2.1). Theorem 2.8 Suppose that for t ∈ [0, T ]  

∂i f ∂2 f ∂3 f , , ∂t i ∂ x 2 ∂z∂ x 2

 ∈ C(As0 , [0, T ]), i = 0, 1, 2;

' ' ' 2 ' ' 2 ' '∂ f ' '∂ f ' '∂ f ' ' ' ' , ' ' max  f s0 , ' ' , 2 ' 2 ' + ' ∂z s0 ∂z s0 ' ∂ x 2 's0

' 3 ' ( ' ∂ f ' ' ' ≤ R, ' ∂z∂ x 2 ' s0

2.4 Differential Properties of the Solution of the Memory …

f (x, +0) =

107

∂ f (x, +0) =0 ∂t

and γ ∈ (0, s0 ) is an arbitrary fixed number. Then, for any χ > 0 there exists a positive number C = C(R, s0 , T, χ ) such that for any s ∈ (0, s0 − γ ) the solution defined by Theorem 2.7 is continuously differentiable with respect to (z, t) in domain G γ ⊆ G, G γ = G γ (R, s0 , T, χ , s) = {(z, t)|0 ≤ t ≤ z < a(s0 − γ − s)} and the following estimates hold: ' ' ' ∂u ' c ' us (z, t) ≤ c, ϑs (z, t) = ' ' ∂z ' (z, t) ≤ γ , s ' ' ' ' ' ∂ϑ ' ' ∂u ' c ' ' ' ' max ' ' (z, t), ks (z), ' ' (z, t) ≤ 2 , ∂t s ∂t s γ ' ' ' ∂ϑ ' ' max ' ' ∂z ' (z, t),

' ' ' ∂u x x ' ' (z, t) ≤ c , ' ' ∂z ' γ3 s

' ' ' ∂u x x ' ' (z, t), ' max ' ∂t '

' ' ' ∂ϑx x ' ' (z, t) ≤ c , ' ' ∂t ' γ4 s

s

s

' ' ' ∂k ' ' max ' ' ∂z ' (z), s

(2.98)

' ' ' ∂ϑx x ' ' (z, t) ≤ c , (z, t) ∈ G γ . ' ' ∂z ' γ5 s

Proof Let (u, ϑ, k) be the solution of the system of Eq. (2.87) defined by Theorem 2.7. Below we will propose that the numbers s0 , a, χ , R and the domain G have the same meaning as in Theorem 2.7. First, we find the expressions for the first derivatives of functions ϑ and k with respect to variable z. For this, we differentiate the second and third equations of the system (2.87) with respect to z: ϑz (x, z, t) = (1/2) f zz (x, z) + (1/4) f x x (x, z) + (1/4)k(x, z) z  ϑx x (x, z, ξ ) − k(x, ξ ) f (x, z − ξ ) + (1/4) t

ξ

k(x, τ )ϑ(x, z − τ, ξ − τ )dτ dξ,

− 0

(2.99)

108

2 The Solvability of Multidimensional Inverse …

k z (x, z) = −2 f zz (x, z) − f x x (x, z) z  − (1/2) f zx x − (ϑz )x x (x, z, ξ ) − (3/2)k(x, ξ ) f z (x, z − ξ ) 0



ξ

k(x, τ )ϑz (x, z − τ, ξ − τ )dτ dξ.

−

(2.100)

0

According to (2.89), the following inequalities hold: us (z, t) ≤ R + χ := c1 , ϑs (z, t) ≤ ks (z) ≤

Rs0 + 2χ c2 := , s0 − s s0 − s

Rs02 + 4χ c3 := . 2 (s0 − s) (s0 − s)2

(2.101)

Using these relations, we estimate the functions ϑz and k z , defined by equalities (2.99) and (2.100):

ϑz s (z, t) ≤

s03 R + s0 c3 + (4c2 + s0 c3 + T s0 c2 c3 ) T c4 := , 3 (s0 − s) (s0 − s)3

 2Rs05 + 16c4 + 2s02 Rc3 + c3 c4 T T c5 k z s (z) ≤ := . 5 (s0 − s) (s0 − s)5

(2.102)

(2.103)

Now let s ∈ (0, s0 − γ ) for some γ ∈ (0, s0 ) − (z, t) ∈ G γ . Then, from (2.101)– (2.102) the following estimates follow u z s = ϑs (z, t) ≤ c2 γ −1 , ks (z) ≤ c3 γ −2 , ϑz s (z, t) ≤ c4 γ −3 , k z s (z) ≤ c5 γ −5 . Since u t satisfies the following relation: z u t (x, z, t) = (1/2) f t (x, z) +

ϑt (x, η, z)dη, t

then

u t s (z, t) ≤ (Rs0 + c4 T )γ −2 := c6 γ −2 .

2.5 Conclusions

109

To estimate s-norm of function u x x , we use the inequality u x x s (z, t) ≤ 4e−2 (s  − s)−2 us  (z, t), putting in it s  = s + γ /2. It is clear that s  < s0 − z/a, and then u x x s (z, t) ≤ 16e−2 (R + χ )γ −2 := c7 γ −2 . The estimate for ϑt follows from (2.86) and previous inequalities ϑt s (z, t) ≤

1 (c7 + c3 (1 + c1 T ))γ −2 := c8 γ −2 . 4

Similar inequalities, with the same choice s  , we use to evaluate the first derivatives of functions u x x , ϑx x with respect to the variables t, z: ' ' ' ∂u x x ' −2  −2 −2 −4 −4 ' ' ' ∂t ' (z, t) ≤ 4e (s − s) u t s  (z, t) ≤ 16e c6 γ := c9 γ , s ' ' ' ∂u x x ' −2  −2 −2 −3 −3 ' ' ' ∂z ' (z, t) ≤ 4e (s − s) u z s  (z, t) ≤ 16e c2 γ := c10 γ , s ' ' ' ∂ϑx x ' −2  −2 −4 −4 ' ' ' ∂t ' (z, t) ≤ 4e (s − s)ϑt s  (z, t) ≤ 16e c6 γ := c11 γ , s ' ' ' ∂u x x ' −2  −2 −5 −5 ' ' ' ∂z ' (z, t) ≤ 4e (s − s)ϑz (z, t) ≤ 16e c4 γ := c12 γ . s From the above obtained inequalities it follows the estimates (2.98) at a suitable choice of constant c. Theorem 2.8 is proved. The estimates obtained above for the derivatives of solution of the system of Volterra-type integro-differential equations (2.87) can be successfully used in the numerical solving of the inverse problem.

2.5 Conclusions This chapter studied the questions of solvability of multidimensional kernel determination inverse problems for the second-order hyperbolic integro-differential equations by the method of the Banach’s scale of analytic functions spaces, developed in the works of L. V. Ovsyannikov and L. Nirenberg. There were shown that problems of this chapter are uniquely solvable in the class of functions having finite smoothness in the time variable t and analytical with respect to the spatial variable x, stability estimates are obtained for the solutions, and uniqueness theorems are

110

2 The Solvability of Multidimensional Inverse …

formulated. Problems are investigated in the cases of equations, and when in integral terms the memory is multiplied by the second time derivative and the first-order differential operator in variable x with real analytical coefficients of the direct problem solution. Here, it also considered the inverse problem for a two-dimensional system of Maxwell equations with memory. It is shown that this problem, reformulated in several other terms, can be reduced to the already investigated problem.

References 1. Ovsyannikov, L.V. 1965. Singular operator in the scale of the Banach spaces. Dokl. AN SSSR 163 (4): 819–822 (Russian). 2. Ovsyannikov, L.V. 1971. Nonlinear Cauchy problem in the scales of the Banach spaces. Dokl. AN SSSR 200 (4): 789–792 (Russian). 3. Nirenberg, L. 1974. Topics in nonlinear functional analysis. New York: Courant Institute Math. Sci., New York Univ. 4. Romanov, V.G. 2005. Stability in inverse problems. Moscow: Nauchniy Mir (Russian).

Chapter 3

Global Solvability of Memory Reconstruction Problems

There are the existence theorems for sufficiently small domains of definition of unknown functions in the inverse problems, that is, solvability for such problems is local. An example of the absence of a global solution of the inverse problem is given in [1]. Even with smooth data, the solution of the inverse problem may not exist on any preassigned interval. This phenomenon is connected with the nonlinearity of the problem. However, in problems of determining the kernel of an integral term in second-order hyperbolic equations, where the nonlinearity is convolutional, it is possible to obtain global existence theorems in the space of continuous functions with exponential weight. In this direction, in addition to the works of the authors, we also note [2–7]. In the Sect. 3.1 the problem of finding a memory kernel from a one-dimensional wave equation for the half-line is investigated. In the Sect. 3.2 the results obtained in the previous section were generalized to the case of the n-dimensional wave equation in a half-space. In the Sect. 3.3 the unknown kernel has the form of a diagonal matrix and is determined from the three-dimensional wave equation in R3 . In the Sect. 3.4 the problem of determining two one-dimensional functions that are included in a initial– boundary problem for the integro-differential equation, main part of which coincides with n + 1− dimensional wave operator, is considered. This is one of the versions for setting the inverse problem of determining the kernel of the integral operator in a situation where the shape of the signal from a point source of disturbances causing the medium to oscillate is unknown and to be determined together with the properties of the medium. In the Sect. 3.5 under study is the multidimensional inverse problem of determining the convolutional kernel of the integral term in an integro-differential wave equation. We apply the methods of scales of Banach spaces of real analytic functions of spatial variable and weight norms in the class of continuous functions in time. In this chapter the theorems of global existence, uniqueness, and the stability estimates are obtained. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_3

111

112

3 Global Solvability of Memory Reconstruction Problems

3.1 The Problem of Determining the Memory Kernel from the Integro-Differential Equation of String Vibrations Consider the initial–boundary value problem t u tt − u x x −

k(τ )u(x, t − τ )dτ = 0, (x, t) ∈ R2+ ,

(3.1)

0

 u

t 0 . For given functions k(t), f (t), the problem of finding a function u(x, t) that satisfies (in a generalized sense) the equalities (3.1), (3.2) is called a direct problem. It follows from the theory of hyperbolic equations, the function u(x, t) as a solution to the direct problem has the property u = 0, t < x, x > 0 and in the neighborhood of the characteristic line t = x has following structure: u(x, t) = −δ(t − x) + θ(t − x)v(x, t).

(3.3)

We set the inverse problem: determine the function k(t), included in the integral of Eq. (3.1) according to the given the regular part of the solution of the direct problem at any time t ≥ 0 at the point x = 0: v(0, t) = F(t).

(3.4)

The norm in C[0, T ] and C n [0, T ], n = 1, 2 is defined in the usual ways: g(t)C[0, T ] = max |g(t)| , t∈[0,T ] n 

g(t)C n [0, T ] =

i=0

  i  d g(t)  .  max t∈[0,T ]  dt 

The main results of this section are the following theorems: Theorem 3.1 Let f (t) ∈ C 1 [0, T ], F(t) ∈ C 2 [0, T ] and the matching condition f (0) = −F  (0) is fulfilled. Then, the inverse problem (3.1)–(3.4) has a unique solution k(t) ∈ C[0, T ], T > 0.

3.1 The Problem of Determining the Memory Kernel …

113

Denote by A(k0 ) the set of functions k(t) ∈ C[0, T ], satisfying for some T > 0 the condition kC[0,T ] ≤ k0 with a constant k0 > 0. Theorem 3.2 Let k (i) ∈ A(k0 ), i = 1, 2 be the solutions of the inverse problem (3.1)– (3.4) with data f (i) (t), F (i) (t), i = 1, 2, respectively. Then, there exists the positive constant C, depending on T , k0 ,  f (1) C 1 [0,T ] , F (1) C 2 [0,T ] that the following estimate   k (1) − k (2) C[0,T ] ≤ C  f (1) (t) − f (2) (t)C 1 [0,T ] + F (1) − F (2) C 2 [0,T ] (3.5) is valid. At the beginning, we prove Theorem 3.1. Substituting the function (3.3) into Eq. (3.1), we find that the function v(x, t) in the domain t > x > 0 satisfies the equations t−x vtt − vx x + k(t − x) − k(τ )v(x, t − τ )dτ = 0, (3.6) 0

v|t=x+0 = 0,

(3.7)

vx |x=0 = f (t).

(3.8)

Since the research method allows to simultaneously find functions v(x, t), k(t), it is convenient to interpret the direct and inverse problems as the problem of determining functions v(x, t), k(t) from the equations (3.6)–(3.8), (3.4). Lemma 3.1 Under fulfillment of the conditions of Theorem 3.1, the

problem (3.4), (3.6)–(3.8) for (x, t) ∈ DT , DT = (x, t)|0 ≤ x ≤ t ≤ ≤ T − x is equivalent to the problem of finding the vector of functions v(x, t), vt (x, t), k(t) from the following system of equations: t vt (x, τ )dτ ,

v(x, t) = x

(3.9)

114

3 Global Solvability of Memory Reconstruction Problems

1  F (t − x) 2 1 x − f (t − x) − k(t − x) 2 2 x t−x 1 + k(τ )v(ξ, t − x + ξ − τ )dτ dξ 2

vt (x, t) =

0

1 + 2

0

⎡

(x+t)/2 

⎣−k(t + x − 2ξ) +

t+x−2ξ 

⎤

k(τ )v(ξ, t + x − ξ − τ )dτ ⎦ dξ, (3.10)

x

0

t/2 t−2ξ  k(t) = 2F (t) + 2 f (t) − k(τ )vt (ξ, t − ξ − τ )dτ dξ. 



0

(3.11)

0

In order to prove the lemma 3.1, it is sufficient in (3.6), after substituting the wave operator vtt − vx x with the product     ∂ ∂ ∂ ∂ − (vt + vx ) = + (vt − vx ) ∂t ∂x ∂t ∂x to integrate equality (3.6) along the corresponding characteristics of first-order differential operators for (x, t) ∈ DT . Integration along characteristics of the operator ∂/∂t − ∂/∂x performs from point (x, t) to point ((x + t)/2, (x + t)/2) on the plane of variables (ξ, θ). Using the differentiation with respect to x equality (3.7) to calculate the expression in the form (vt + vx )((x + t)/2, (x + t)/2) = 0, we obtain (vt + vx )(x, t) ⎤ ⎡ (x+t)/2 t+x−2ξ   ⎣−k(t + x − 2ξ) + = k(τ )v(ξ, t + x − ξ − τ )dτ ⎦ dξ. x

(3.12)

0

The integration along the characteristic of the operator ∂/∂t + ∂/∂x is performed from the point (0, t − x) to the point (x, t). Using the condition (3.8) and differentiation with respect to t equality (3.4), we find

3.1 The Problem of Determining the Memory Kernel …

115

(vt − vx )(x, t) x t−x



= F (t − x) − f (t − x) − k(t − x)x +

k(τ )v(ξ, t − x + ξ − τ )dτ dξ. 0

0

(3.13) From equations (3.12) and (3.13) it follows (3.10). In Eq. (3.10), setting x = 0 and using the condition (3.4) we find F  (t) + f (t) ⎤ ⎡ t−2ξ t/2  = ⎣−k(t − 2ξ) + k(τ )v(ξ, t − ξ − τ )dτ ⎦ dξ. 0

(3.14)

0

Differentiating this equality, after simple calculations we arrive at the Eq. (3.11). If the conditions of agreement in Theorem 3.1 are fulfilled, then it is easy to show that the inverse transformations also take place. Indeed, replacing the right and left parts (3.11) t by t − 2η, multiply both sides by dη and integrate over η in the limits from 0 to t. In the repeated integrals of the resulting equality, changing the order of integration and taking into account the relation (3.7), we obtain Eq. (3.14), which, in turn, is equivalent to equality (3.9) with x = 0. Equivalence of equality (3.9) to (3.12), (3.13) and the derivation of relations (3.6)–(3.8) from the last equations are obvious. The obvious equality (3.9) is given for the closure of the system of equations (3.10), (3.11). Thus, Lemma 3.1 is proved. Continue the proof of Theorem 3.1. We write the system of the equations (3.9)– (3.11) in the form of an operator equation ϕ = Aϕ,

(3.15)

where ϕ = [ϕ1 (x, t), ϕ2 (x, t), ϕ3 (t)]∗ = [v(x, t), vt (x, t) + (x/2)k(t − x), k(t)]∗ is a vector function with components ϕi (i = 1, 2, 3), ∗ is as before the sign of transposition, the operator A is defined on the set of functions ϕ ∈ C[DT ] and in accordance with equalities (3.9)–(3.11) has the form

116

3 Global Solvability of Memory Reconstruction Problems

A = (A1 , A2 , A3 ) : t   x ϕ2 (x, τ ) − ϕ3 (τ − x) dτ , A1 ϕ = 2 x

1 1 1 A2 ϕ = F  (t − x) − f (t − x) + 2 2 2

x t−x ϕ3 (τ )ϕ1 (ξ, t − x + ξ − τ )dτ dξ 0

+

0

(x+t)/2 

 −ϕ3 (t + x − 2ξ)

1 2

x t+x−2ξ 

 ϕ3 (τ )ϕ3 (ξ, t + x − ξ − τ )dτ dξ,

+ 0 

A3 ϕ = 2F (t) + 2 f  (t)   t/2 t−2ξ  ξ − ϕ3 (τ ) ϕ2 (ξ, t − ξ − τ ) − ϕ3 (t − 2ξ − τ ) dτ dξ. 2 0

(3.16)

0

Denote by Cρ the Banach space of continuous functions generated by the family of weight norms: ϕρ = max



     −ρt −ρt sup ϕi (x, t)e  , i = 1, 2, sup ϕ3 (t)e  , ρ ≥ 0.

(x,t)∈DT

t∈[0,T ]

It is obviously, when ρ = 0 this space coincides with the space of continuous functions with the usual norm. This norm will be denoted below as ψ. Due to inequality (3.17) e−ρT ϕ ≤ ϕρ ≤ ϕ the norms ϕρ and ϕ are equivalent for any fixed T ∈ (0, ∞). The number ρ will be chosen later. Let Q ρ (ϕ0 , R) be a ball of radius R with center at the point ϕ0 of some weight space Cρ (ρ ≥ 0), where ϕ0 (x, t)

   ∗  1  F (t − x) − f (t − x) , 2 F (t) + f  (t) = (ϕ01 , ϕ02 , ϕ03 )∗ := 0, . 2

It is easy to see that for ϕ ∈ Q ρ (ϕ0 , R) the estimate ϕρ ≤ ϕ0 ρ + R ≤ ϕ0  + R := R0 (R0 is a known number) is valid. Let ϕ(x, t) ∈ Q ρ (ϕ0 , R). We prove that with a suitable choice of ρ > 0 the operator A maps the ball Q ρ (ϕ0 , R) to the same ball, i.e., Aϕ ∈ Q ρ (ϕ0 , R). Using equal-

3.1 The Problem of Determining the Memory Kernel …

ities (3.16) and making up the norm of differences, for (x, t) ∈ DT we have   ||A1 ϕ − ϕ01 ||ρ = sup (A1 ϕ − ϕ01 )e−ρt  (x,t)∈DT

t

  = sup  ϕ2 (x, τ )e−ρτ e−ρ(t−τ ) (x,t)∈DT

x

  x − ϕ3 (τ − x)e−ρ(τ −x) e−ρ(t−τ +x) dτ  2     1 T 1 1 T ≤ ϕ2 ρ + ϕ3 ρ ≤ R0 1 + := α1 . 4 ρ 4 ρ ρ Similarly, we find ||A2 ϕ − ϕ02 ||ρ =  1 = sup  2 (x,t)∈DT ×e

−ρ(x−ξ)

  1 sup (A2 ϕ − ϕ02 )e−ρt)  2 (x,t)∈DT

x t−x ϕ3 (τ )e−ρτ ϕ1 (ξ, t − x + ξ − τ )e−ρ(t−x−τ +ξ) 0

0 (t+x)/2 



dτ dξ +

−ϕ3 (t + x − 2ξ)e−ρ(t+x−2ξ) e−ρ(2ξ−x)

x t+x−2ξ 

ϕ3 (τ )e−ρτ ϕ1 (ξ, t + x − ξ − τ )

+ ×e

0 −ρ(t+x−ξ−τ ) −ρ(ξ−x)

e

  dτ dξ 

  1 T 1 1 3T ϕ3 ρ ϕ1 ρ + ϕ3 ρ + ϕ3 ρ ϕ1 ρ 2 2 4 4 ρ   R0 T 1 3R0 T 1 1 + + := α2 , ≤ R0 2 4 4 ρ ρ   ||A3 ϕ − ϕ03 ||ρ = sup (A3 ϕ − ϕ03 )e−ρt  ≤

t∈[0,T ]

 = sup − t∈[0,T ]

t/2 t−2ξ 

[ϕ3 (τ )e−ρτ ϕ2 (ξ, t − ξ − τ )e−ρ(t−τ −ξ) e−ρξ

0

0

 ξ − ϕ3 (t − 2ξ − τ )e−ρ(t−2ξ−τ ) ϕ3 (τ )e−ρτ e−2ρξ ]dτ dξ  2     T 1 1 T2 T 1 ≤ ϕ3 ρ ϕ2 ρ + ϕ3 2ρ ≤ R02 T 1 + := α3 . 1 8 ρ 8 ρ ρ

117

118

3 Global Solvability of Memory Reconstruction Problems

Choosing ρ ≥ α0 =: (1/R) max(α1 , α2 , α3 ), we obtain that A transfers the ball Q ρ (ϕ0 , R) to the same ball Q ρ (ϕ0 , R). Now let ϕ1 , ϕ2 be any two elements of Q ρ (ϕ0 , R). Then, using auxiliary inequalities of the form  1 1  ϕ ϕ − ϕ2 ϕ2  e−ρt) i j i j         ≤ ϕi1  ϕ1j − ϕ2j  e−ρt) + ϕ2j  ϕi1 − ϕi2  e−ρt) ≤ 2R0 ϕ1 − ϕ2 ρ , (x, t) ∈ DT , we have     (Aϕ1 − Aϕ2 )1  = sup (Aϕ1 − Aϕ2 )1 e−ρt  ρ (x,t)∈DT

t

  x = sup  (ϕ12 − ϕ22 )(x, τ )e−ρτ e−ρ(t−τ ) − (ϕ13 2 (x,t)∈DT x   2 − ϕ3 )(τ − x)e−ρ(τ −x) e−ρ(t−τ +x) dτ       T  ϕ1 − ϕ2  1 =: β1 ϕ1 − ϕ2  1 . ≤ 1+ ρ ρ ρ 4 ρ Similarly, we find     (Aϕ1 − Aϕ2 )2  = sup (Aϕ1 − Aϕ2 )2 e−ρt  ρ (x,t)∈DT

=

x t−x   1  sup ϕ3 (τ )e−ρτ (ϕ11 − ϕ21 )(ξ, t − x + ξ − τ )e−ρ(t−x−τ +ξ)

1 2 (x,t)∈DT

+

ϕ21 (ξ, t (x+t)/2 



+

0

0

 − x + ξ − τ )e−ρ(t−x−τ +ξ) (ϕ13 − ϕ23 )(τ )e−ρτ e−ρ(x−ξ) dτ dξ (ϕ23 − ϕ13 )(t + x − 2ξ)e−ρ(t+x−2ξ) e−ρ(2ξ−x)

x t+x−2ξ 

+ + ×

(ϕ13 (τ )e−ρτ (ϕ11 − ϕ21 )(ξ, t + x − ξ − τ )e−ρ(t+x−ξ−τ )

0 2 ϕ1 (ξ, t + x − ξ − τ )e−ρ(t+x−ξ−τ )   (ϕ11 − ϕ21 )(τ )e−ρτ )e−ρ(ξ−x) dτ dξ 

     1 1 3R0 T  ϕ1 − ϕ2  1 =: β2 ϕ1 − ϕ2  1 , ≤ + R0 T + ρ ρ ρ 2 4 2 ρ     (Aϕ1 − Aϕ2 )3  = sup (Aϕ1 − Aϕ2 )3 e−ρt  ρ t∈[0,T ]

3.1 The Problem of Determining the Memory Kernel …

 = sup − t∈[0,T ]

119

t/2 t−2ξ   1 (ϕ3 (τ )e−ρτ (ϕ12 − ψ22 )(ξ, t − ξ − τ )e−ρ(t−τ −ξ) 0

0

+ ϕ22 (ξ, t − ξ − τ )e−ρ(t−τ −ξ) (ϕ13 − ϕ23 )(τ )e−ρτ )e−ρξ ξ − (ϕ13 (t − 2ξ − τ )e−ρ(t−2ξ−τ ) (ϕ13 − ϕ23 )(τ )e−ρτ ) 2   + ϕ23 e−ρτ (ϕ13 − ϕ23 )(t − 2ξ − τ )e−ρ(t−2ξ−τ ) )e−2ρξ dτ dξ       T  ϕ1 − ϕ2  1 =: β3 ϕ1 − ϕ2  1 . ≤ R0 T 2 + ρ ρ ρ 4 ρ Let β0 := max(β1 , β2 , β3 ). As follows from the above estimates, if the number ρ is chosen from the condition ρ > max(α0 , β0 ), then operator A is contraction one on Q ρ (ϕ0 , R). Consequently, according to the Banach principle, Eq. (3.15) has, and moreover, the only solution in Q ρ (ϕ0 , R) for any fixed T > 0. The Theorem 3.1 is proved. Remark 3.1 The pair of functions v(x, t), k(t), constructed above, is the generalized solution of the problem (3.6)–(3.8), (3.4) in the domain DT , the pair of functions u(x, t), k(t), where u(x, t) = −δ(t − x) + θ(t − x)v(x, t) is the generalized solution to the problem (3.1)–(3.4). Example 3.1 Let f (t) = 0, F(t) = −t 2 . One can easily show that in this case the solution to the inverse problem (3.6)–(3.8), (3.4) uniquely exists:      √ 1 1 √3 3 k(t) = − e 3/2·t − e−1/2 3/2·t cos 6 (3/2)5 · t − sin 6 (3/2)5 · t , 2 2 (3.18) v(x, t) = x 2 − t 2 , the pair of functions u(x, t), k(t), where u(x, t) = −δ(t − x) + θ(t − x)v(x, t) and k(t) is determined by (3.18), is the generalized solution to the problem (3.1)–(3.4). In the case under consideration, Theorem 3.1 holds. Now we prove the conditional stability theorem 3.2. First of all, we note that substituting the expression for vt (x, t) by formula (3.10) into (3.9) for the known function k(t) ∈ A(k0 ) we obtain an integral equation of the second kind of Volterratype with respect to v(x, t): 1 v(x, t) = G(x, t) + 2

 t  x τ −x k(α)v(ξ, τ − x + ξ − α)dαdξ 0

0

0

 k(α)v(ξ, τ + x − ξ − α)dαdξ dτ ,

(x+τ  )/2 τ +x−2ξ 

+ x

0

120

3 Global Solvability of Memory Reconstruction Problems

where 1 G(x, t) = 2

t−x 

1 F (τ ) − f (τ ) − xk(τ ) − 2 

0

τ

 k(α)dα dτ .

0

We represent v(x, t) in the form v(x, t) = v0 (x, t) +

∞ 

vn (x, t),

n=1

in which vn (x, t) are given by formulas: v(x, t) = G(x, t),  t x τ −x 1  k(α)vn−1 (ξ, τ − x + ξ − α)dαdξ vn (x, t) = 2 0

0

0

(x+τ  )/2 τ +x−2ξ 

 k(α)vn−1 (ξ, τ + x − ξ − α)dαdξ dτ , n = 1, 2, . . . .

+ x

0

It is clear, that vn are continuous functions in DT together with the first derivatives with respect to x, t for all n ≥ 0. Doing the same as at the obtaining the estimate (1.17) from Sect. 1.1 we find that in the domain DT the function v(x, t) is modulo bounded with some constant c0 depending on k0 , T,  f C[0,T ] , FC 1 [0,T ] , i.e., |v(x, t)| ≤ c0 , (x, t) ∈ DT . Suppose that the functions f (i) , F (i) (t), k (i) , i = 1, 2 are such that they are defined in Theorem 3.1. The regular part of the solution of the direct problem (3.1)–(3.3) for k = k (i) , f = f (i) , we denote by v (i) (x, t). And also introduce the differences v˜ = v (1) − v (2) , v˜t = vt(1) − vt(2) , k˜ = k (1) − k (2) , f˜ = f (1) − f (2) , F˜ = F (1) − F (2) . Then from relations (3.9)–(3.11) we obtain

3.1 The Problem of Determining the Memory Kernel …

121

t v(x, ˜ t) =

v˜t (x, τ )dτ , x

1 ˜ 1 x˜ F (t − x) − f˜(t − x) − k(t − x) 2 2 2 x t−x  1 ˜ )v (1) (ξ, t − x + ξ − τ ) k(τ + 2

v˜t (x, t) =

0

0

 1 + k (τ )v(ξ, ˜ t − x + ξ − τ ) dτ dξ + 2 (2)

(x+t)/2 

˜ + x − 2ξ) (3.19) {−k(t

x t+x−2ξ 



+

 ˜ )v (1) (ξ, t + x − ξ − τ ) + k (2) (τ )v(ξ, k(τ ˜ t + x − ξ − τ ) dτ }dξ, (3.20)

0

˜ = 2 F˜  (t) + 2 f˜ (t) k(t) t/2 t−2ξ    ˜ )vt(1) (ξ, t − ξ − τ ) + k (2 )(τ )v˜t (ξ, t − ξ − τ ) dτ dξ. (3.21) k(τ − 0

0

The domain DT admits an equivalent description in the form  DT = (ξ, τ )| |τ − t + x| ≤ ξ ≤ (t + x)/2 − |(t + x)/2 − τ |,  (t − x)/2 ≤ τ ≤ t . Let  φ(t) = max

max

0≤x≤T /2−|t−T /2|

|v(x, ˜ t)|,

max

0≤x≤T /2−|t−T /2|

 ˜ |v˜t (x, t)|, |k(t)| ,

t ∈ [0, T ]. Conducting simple estimates in equations (3.19)–(3.21), we find that for (x, t) ∈ DT the integral inequality φ(t)

  ≤ c1  f (1) (t) − f (2) (t)C 1 [0,T ] + F (1) − F (2) C 2 [0,T ] + c2

t φ(τ )dτ 0

122

3 Global Solvability of Memory Reconstruction Problems

is valid, where c1 and c2 are positive constants depending on k0 , T,  f (1) C 1 [0,T ] , F (1) C 2 [0,T ] . From this, using the Gronwall inequality, we obtain estimate (3.5) with some constant C. Here we note that similar results can be obtained for the problem of determining the function k(t) ∈ C 2 [0, T ] from the equation t u tt − u x x −

2 k(τ )u tt (x, t − τ )dτ = 0, (x, t) ∈ R+ ,

(3.22)

0

with initial–boundary conditions (3.2) for a given information (3.4) with respect to the regular part of the solution to the direct problem (3.22), (3.2). In fact, in Eq. (3.22) integrating by parts and introducing new unknown function u(x, ˜ t) by formula  u(x, t) = exp

k0 t 2

 u(x, ˜ t), k0 := k(0)

from (3.22), (3.2) we get t u˜ tt − u˜ x x − h 0 u˜ −

2 h(τ )u(x, ˜ t − τ )dτ = 0, (x, t) ∈ R+ ,

(3.23)

0

u| ˜ t 0. Using the constructed function h(t) the unknown function k(t) is found from the formula t  k(t) = k0 + k (0)t + (t − τ ) exp[k(0)τ /2]h(τ )dτ . 0

Denote by B(k00 ) the set of functions k(t) ∈ C 2 [0, T ], satisfying for some T > 0 the condition kC 2 [0,T ] ≤ k00 with constant k00 > 0. The conditional stability theorem for solution of the inverse problem (3.25)– (3.27), (3.4) is also valid. Theorem 3.4 Let k (i) ∈ B(k00 ), i = 1, 2 be solutions to the inverse problem (3.25)– (3.27), (3.4) with data f (i) (t), F (i) (t), i = 1, 2, respectively. Then, there is a positive constant C, depending on T , k00 ,  f (1) C 1 [0,T ] , F (1) C 2 [0,T ] such that the following estimate   k (1) − k (2) C 2 [0,T ] ≤ C  f (1) (t) − f (2) (t)C 1 [0,T ] + F (1) − F (2) C 2 [0,T ] is true. Remark 3.2 The Neumann boundary condition in (3.2) can be replaced by condition u x |x=0 = δ(t) + η(t)θ(t). Then, considering the function w = u t as a solution to the direct problem and assuming that η(t) satisfies conditions η(0) = 0, η  (t) = f (t), we arrive at above the investigated problem.

124

3 Global Solvability of Memory Reconstruction Problems

3.2 Memory Identification Problem from Integro-Differential Wave Equation Consider n-dimensional wave equation with memory t u tt − u x x − u −

k(τ )u(x, y, t − τ )dτ = 0, (y, t) ∈ Rn+1 , x > 0,

(3.28)

0

in which  is Laplace operator with respect to variables y = (y1 , . . . , yn ), with the following initial and boundary conditions   u t 0,

2 0

  = δ  (t) + f (t)θ(t), ≡ 0, u ˜ x t x > 0 satisfy relations t−x vtt − vx x + k(t − x) + |λ| v − |λ| 2

k(τ )v(x, λ, t − τ )dτ = 0,

2 0

(3.32)

3.3 Inverse Problem for an Integro-Differential Equation of Electrodynamics

  v t=x+0 = 0, vx x=0 = f (t).

125

(3.33)

We pose the inverse problem: to determine k(t), t > 0 from the given trace of the solution of problem (3.32), (3.33) on the line x = 0, for fixed λ = λ0 ∈ Rn and t < T , T > 0, i.e., (3.34) v(0, λ0 , t) = F(t), λ0 ∈ Rn , t < T. Note that the inverse problem (3.32)–(3.34) for λ = 0 is identical to the problem explored in the previous section. We give here typical theorems of unique global solvability and stability for the inverse problem (3.32)–(3.34). Their proofs can easily be carried out according to the scheme described in the previous section and are not given here. Theorem 3.5 Let f (t) ∈ C 1 [0, T ], F(t) ∈ C 2 [0, T ] and the matching conditions F(λ0 , 0) = 0, f (0) =

|λ0 |2 − F  (0) 2

are valid. Then, the inverse problem (3.32)–(3.34) has a unique solution k(t) ∈ C[0, T ], T > 0. We denote, as in Sect. 3.1, by A(k0 ) the set of functions k(t) ∈ C[0, T ], satisfying the condition kC[0,T ] ≤ k0 for some T > 0 with constant k0 > 0. Theorem 3.6 Let k (i) ∈ A(k0 ), i = 1, 2 be solutions to the inverse problem (3.32)– (3.34) with data f (i) (t), F (i) (t), i = 1, 2, respectively. Then, there exists a positive constant C, depending on T , k0 , λ0 ,  f (1) C 1 [0,T ] , F (1) C 2 [0,T ] such that the estimate   k (1) − k (2) C[0,T ] ≤ C  f (1) (t) − f (2) (t)C 1 [0,T ] + F (1) − F (2) C 2 [0,T ] is valid.

3.3 Inverse Problem for an Integro-Differential Equation of Electrodynamics Let the vectors E and H form a solution of the Cauchy problem for the system of Maxwell equations with zero initial data: ∇ × H = (∂/∂t)D(x, t) + σ E + j, ∇ × E = −(∂/∂t)B(x, t), (x, t) ∈ R4 , (E, H )|t 0. The proof of the theorem is completed.

134

3 Global Solvability of Memory Reconstruction Problems

3.4 Global Solvability of the Problem of Determining Two Unknowns in the Inverse Problem for the Integro-Differential Wave Equation Consider the initial–boundary problem t u tt − u yy − u −

k(τ )u(x, y, t − τ )dτ = 0, (x, t) ∈ Rn+1 , y > 0,

(3.50)

0

u|t 0. Therefore, to the relations (3.50), (3.51) the Fourier transform in the variable x is applicable. Denote through u(λ, ˜ y, t), λ = (λ1 , . . . , λn ) the Fourier image of the function u(x, y, t) : 1 u(λ, ˜ y, t) = √ (2π)n

 u(x, y, t)ei(λ,x) d x, (λ, x) = λ1 x1 + . . . + λn xn . Rn

Then, the problem (3.50), (3.51) is transformed to the form t u˜ tt − u˜ yy + |λ| u˜ + |λ| 2

k(τ )u(λ, ˜ y, t − τ )dτ = 0, (λ, t) ∈ Rn+1 , y > 0,

2 0

u| ˜ t 0. The function u(λ, ˜ y, t) as a solution to problem (3.52), (3.53) has the following structure in a neighborhood of the characteristic line t = y: u(λ, ˜ y, t) = −δ(t − y) + θ(t − y)v(λ, y, t),

(3.54)

where v(λ, y, t) is a regular function. Now we pose the inverse problem: determine the functions k(t), f (t), included in equations (3.52), (3.53), respectively, from information on the regular part of the Fourier transform of the direct problem solution at y = 0 for two different values λ of the transformation parameter: v(λi , 0, t) = F(λi , t), t > 0, i = 1, 2.

(3.55)

Theorem 3.8 Let F(λi , t) ∈ C 1 [0, T ], i = 1, 2, F0 (λ1 , λ2 , t) := Ft (λ1 , t) − Ft (λ2 , t) ∈ C 1 [0, T ], and the matching conditions F(λi , 0) = 0, i = 1, 2, λ1 = λ2 are fulfilled. Then, inverse problem (3.52)–(3.55) has the unique solution so that ( f (t), k(t)) ∈ C[0, T ], T > 0. Proof Substituting the function (3.54) into equations (3.52), (3.53) and using the singularity separation method, we find that the function v(λ, y, t) for a fixed λ in the domain t > y > 0 satisfies the equations vtt − v yy

t−y + |λ| v − |λ| k(t − y) + |λ| k(τ )v(λ, y, t − τ )dτ = 0, 2

2

2

(3.56)

0

v|t=y+0 =

|λ|2 y, 2

v y | y=0 = f (t).

(3.57) (3.58)

The following statement establishes the equivalence of the inverse problem (3.52)– (3.58) to some closed system of integral equations of Volterra-type.   Lemma 3.2 Under the conditions of the theorem 3.8, the problem (3.55)–(3.58) for λ = λi = (λi1 , . . . , λin ), i = 1, 2 and (y, t) ∈ DT , DT = ((y, t)|0 ≤ y ≤ t ≤ T − y) is equivalent to the problem of finding functions v(λi , y, t), vt (λi , y, t), k(t), f (t), i = 1, 2 from the following system of equations:

136

3 Global Solvability of Memory Reconstruction Problems

|λi |2 y+ v(λ , y, t) = 2

t vt (λi , y, τ )dτ , i = 1, 2,

i

(3.59)

y

vt (λi , y, t) = Ft (λi , t − y) + |λi |2 + 2

|λi |2 yk(t − y) 2

(y+t)/2 



k(t + y − 2ξ) − v(λi , ξ, t + y − ξ)

y

t+y−2ξ 

 k(τ )v(λi , ξ, t + y − ξ − τ )dτ dξ−

− 0

|λi |2 − 2

t−y)/2 

 k(t − y − 2ξ) − v(λi , ξ, t − y − ξ)

0 t−y−2ξ 

 k(τ )v(λi , ξ, t − y − ξ − τ )dτ dξ

− 0

|λi |2 − 2

y

⎤ t−y  ⎥ ⎢ i k(τ )v(λi , ξ, t − y + ξ − τ )dτ ⎦ dξ, i = 1, 2, ⎣v(λ , ξ, t − y + ξ) + ⎡

0

0

(3.60)

k(t) =

|λ1 |2

2 F0t (λ1 , λ2 , t) − |λ2 |2

|λ1 |2 + |λ2 |2 |λ1 |2 + |λ2 |2 t+ + 4 4

t k(τ )(t − τ )dτ 0

+

|λ1 |2

2 − |λ2 |2

t/2 

|λ1 |2 vt (λ1 , ξ, t − ξ) − |λ2 |2 vt (λ2 , ξ, t − ξ)

0

t−2ξ 

   k(τ ) |λ1 |2 vt (λ1 , ξ, t − ξ − τ ) − |λ2 |2 vt (λ2 , ξ, t − ξ − τ ) dτ dξ,

+ 0

(3.61)

3.4 Global Solvability of the Problem of Determining …

|λ|2 + |λ1 |2 f (t) = −Ft (λ , t) + 2 1

137

t/2  k(t − 2ξ) − v(λ1 , ξ, t − ξ) 0

t−2ξ 

 k(τ )v(λ1 , ξ, t − ξ − τ )dτ dξ.

−

(3.62)

0

Proof In view of the factorization form of the wave operator  vtt − v yy =

∂ ∂ − ∂t ∂y



 (vt + v y ) =

∂ ∂ + ∂t ∂y

 (vt − v y ),

(3.63)

integrating (3.56) along the corresponding characteristics of first-order differential operators for (y, t) ∈ DT from (3.55)–(3.57) we obtain |λ|2 + |λ|2 (vt + v y )(λ, y, t) = 2

(y+t)/2 

 k(t + y − 2ξ) − v(λ, ξ, t + y − ξ)

y t+y−2ξ 

 k(τ )v(λ, ξ, t + y − ξ − τ )dτ dξ,

−

(3.64)

0

(vt − v y )(λ, y, t) = 2Ft (λ, t) −

|λ|2 + |λ|2 k(t − y)y 2

(t−y)/2 

− |λ|

[k(t − y − 2ξ) − v(λ, ξ, t − y − ξ)

2 0

t−y−2ξ 

−

k(τ )v(λ, ξ, t − y − ξ − τ )dτ ]dξ 0

y − |λ|2 0

⎤ t−y ⎣v(λ, ξ, t − y + ξ) + k(τ )v(λ, ξ, t − y + ξ − τ )dτ ⎦ dξ. (3.65) ⎡

0

138

3 Global Solvability of Memory Reconstruction Problems

In obtaining equality (3.65), we also used the relations vt (λ, 0, t) = Ft (λ, t), |λ|2 v y (λ, 0, t) = − Ft (λ, t) 2 ⎤ ⎡ t/2 t−2ξ   + |λ|2 ⎣ k(t − 2ξ) − v(λ, ξ, t − ξ) − v(λ, ξ, t − ξ − τ )dτ ⎦ dξ, (3.66) 0

0

which follow from (3.55), (3.64). Considering the equation (3.66) for λ = λ1 , λ = λ2 , we compose the difference v y (λ1 , 0, t) − v y (λ2 , 0, t). Taking into account (3.58), we obtain |λ1 |2 − |λ2 |2 + Ft (λ2 , t) − Ft (λ1 , t) 2 t t/2 |λ1 |2 − |λ2 |2 + k(τ )dτ − [|λ1 |2 vt (λ1 , ξ, t − ξ) − |λ2 |2 vt (λ2 , ξ, t − ξ) 2 0

0

t−2ξ 

k(τ )(|λ1 |2 vt (λ1 , ξ, t − ξ − τ ) − |λ2 |2 vt (λ2 , ξ, t − ξ − τ ))dτ ]dξ = 0.

+ 0

(3.67) Differentiating the last relation with respect to t, after simple calculations, we arrive at (3.61). Note that the equations (3.59)–(3.61) do not contain the unknown function f (t). After finding the functions v, k the function f (t), according to equality (3.66), is calculated by the formula (3.62). In the formula (3.62) λ = λ1 or λ = λ2 , the calculation result should not depend on the choice of the parameter λ. For definiteness, in this formula we set λ = λ1 . The obvious equality (3.59) is given for the closure systems of the equations (3.60), (3.61).   It is easy to see that inverse transforms also take place. Indeed, in the first integral of the right-hand side of (3.61), we change variable τ to τ  by the formula τ  = (t − τ )/2. Then, on the right-hand and left-hand sides of this equation, replacing t to t − 2ξ, we multiply both sides by dξ and integrate over ξ from 0 to t/2. In the repeated integrals of the resulting equality, we change the order of integration. Using the conditions of the theorem 3.8 after simple calculations, we get (3.67). Equating in this equation functions that depend only on λ1 and λ2 , we arrive at (3.66) for λ = λi , i = 1, 2. Equation (3.66) is obtained after setting y = 0 in the equality for v y (λ, y, t), following from (3.64), (3.65). From these equalities we obtain Eq. (3.60). It is clear that after applying the differential operators (∂/∂t − ∂/∂ y) , (∂/∂t + ∂/∂ y) to relations (3.64), (3.65) we arrive at the equations (3.55)–(3.57).

3.5 About Global Solvability of a Multidimensional Inverse …

139

Further proof of Theorem 3.8 can be carried out quite analogously to the proof of Theorem 3.1 from Sect. 3.1. Remark 3.3 The Neumann boundary condition (3.51) can be replaced by the condition u y | y=0 = δ(t)δ(x) + ω(t)θ(t)δ(x). Then, considering the function w = u t as a solution to the direct problem and assuming that the equalities ω(t) are true for ω(0) = 0, ω  (t) = f (t), we arrive at the problem studied in this section.

3.5 About Global Solvability of a Multidimensional Inverse Problem for an Equation with Memory In this section we obtain the global unique solvability of a multidimensional convolutional kernel determination problem in the class of continuous functions in the time variable and analytic in a part of the space variables. Consider the problem of determining a pair of functions u(x, z, t) and k(x, t) satisfying the equations t u tt − u zz − u =

k(x, τ )u(x, z, t − τ ) dτ , (x, t) ∈ Rm+1 , z > 0,

(3.68)

0

  u 

t 0, of functions ϕ(x), x ∈ Rm , analytic in a neighborhood of the origin such that ∞  s |α| α |D ϕ(x)| < ∞. |x| 0, x ∈ Rm ,  u t=z+0 ≡ 0, x ∈ Rm .

(3.72) (3.73) (3.74)

Thus, (3.68)–(3.70) is reduced to the problem of determining the functions u(x, z, t) and k(x, t) from (3.72)–(3.74). Let C(As0 ; [0, T ]) be the space of continuous with respect to t in [0, T ] with values in As0 functions. Lemma 3.3 Let T > 0 and s0 > 0 be some fixed numbers, condition (2.7) hold, and 

   h(x, t), f t (x, t), h t (x, t), f tt (x, t) ∈ C As0 ; [0, T ] .

Then, the inverse problem (3.72)–(3.74) for (x, z, t) ∈ DT is equivalent to the problem of finding u(x, z, t), u t (x, z, t) and k(x, t) from the system of nonlinear integral Volterra-type equations of the second kind: t u t (x, z, τ ) dτ ,

u(x, z, t) = z

(3.75)

3.5 About Global Solvability of a Multidimensional Inverse …

141

z 1 1 f t (x, t − z) − h(x, t − z) − k(x, t − z) 2 2 2  z  t−z 1 u(x, ξ, t − z + ξ) + + k(x, τ )u(x, ξ, t − z + ξ − τ ) dτ dξ 2

u t (x, z, t) =

0

+

0

(t+z)/2  

1 2

u(x, ξ, t + z − ξ) − k(x, t + z − 2ξ) z

 k(x, τ )u(x, ξ, t + z − ξ − τ ) dτ dξ,

t+z−2ξ 

+

(3.76)

0

k(x, t) = 2 f tt (x, t) + 2h t (x, t) t/2

t−2ξ 

u t (x, ξ, t − 2ξ) +

− 0



k(x, τ )u t (x, ξ, t − ξ − τ ) dτ dξ. (3.77) 0

Proof Taking (3.63) into account, integrate (3.72) along the corresponding characteristics of the differential operators of the first order for (z, t) ∈ G T . Integrate along the characteristic of the operator ∂/∂t − ∂/∂z from(z, t) to ((z + t)/2, (z + t)/2) on the plane of variables (ξ, τ ). Using (3.74) and differentiating with respect to z, we obtain (z+t)/2  

u(x, ξ, t + z − ξ) − k(x, t + z − 2ξ)

(u t + u z )(x, z, t) = z

 k(x, τ )u(x, ξ, t + z − ξ − τ ) dτ dξ.

t+z−2ξ 

+

(3.78)

0

Integrate now (3.72) along the characteristic of ∂/∂t + ∂/∂z from (0, t − z) to (z, t). Using (3.73) and differentiating with respect to t, we get (u t − u z )(x, z, t) = f t (x, t − z) − h(x, t − z) − k(x, t − z)z  z  t−z u(x, ξ, t − z + ξ) + k(x, τ )u(x, ξ, t − z + ξ − τ ) dτ dξ. (3.79) + 0

0

142

3 Global Solvability of Memory Reconstruction Problems

Equalities (3.78) and (3.79) easily yield (3.76). Putting z = 0 in (3.78) and using (3.73), find f t (x, t) + h(x, t) ⎤ ⎡ t−2ξ t/2  = ⎣u(x, ξ, t − ξ) − k(x, t − 2ξ) + k(x, τ )u(x, ξ, t − ξ − τ ) dτ ⎦ dξ. 0

0

(3.80)  

Differentiating (3.80), we arrive at (3.77) by simple computations.

Under the agreement condition (2.7) it is easy to show that the inverse transformations are also valid. Indeed, replacing t with t − 2τ on the right- and left-hand sides of (3.77), multiply both sides by dτ and integrate with respect to τ from 0 to t/2. Changing the order of integration in the iterated integrals of the obtained equality and using (3.74), we obtain (3.80) which, in turn, is equivalent to (3.75) for z = 0. Equivalence of (3.75) to (3.78), (3.79) and derivation of (3.72)–(3.74) from the latter equations are obvious. Equality (3.75) is given for the closure of the system of (3.76) and (3.77). Thus, the lemma is proven. System (3.75)–(3.77) is a closed system of integro-differential equations for u, u t , and k. Note that the operator  in u and u t enters this system only under the integral sign. Introduce v(x, z, t) := u t + 2z k(x, t − z) and rewrite (3.75)–(3.77) as u(x, z, t) =

t 

 z v(x, z, τ ) − k(x, τ − z) dτ , 2

(3.81)

z

v(x, z, t) = v0 (x, z, t)  z  t−z 1 + u(x, ξ, t − z + ξ) + k(x, τ )u(x, ξ, t − z + ξ − τ ) dτ dξ 2 0

+

0

(t+z)/2  

1 2

u(x, ξ, t + z − ξ) − k(x, t + z − 2ξ) z

t+z−2ξ 

+



k(x, τ )u(x, ξ, t + z − ξ − τ ) dτ dξ, 0

(3.82)

3.5 About Global Solvability of a Multidimensional Inverse …

143

t/2 ξ v(x, ξ, t − 2ξ) − k(x, t − 2ξ) k(x, t) = k0 (x, t) − 2 0 t−2ξ 





ξ k(x, τ ) v(x, ξ, t − ξ − τ ) − k(x, t − ξ − τ ) 2

+

 dτ dξ, (3.83)

0

where 1 1 f t (x, t − z) − h(x, t − z), 2 2 k0 (x, t) = 2 f tt (x, t) + 2h t (x, t). v0 (x, z, t) =

(3.84)

We consider system (3.81)–(3.83) in a domain narrower than DT , namely in   T − z for some fixed ε > 0. DTε = G εT × Rm , G εT = (z, t)| 0 ≤ z ≤ t ≤ 1+ε   Denote by Cσ As0 ; G εT the space of continuous in (z, t) in G εT with values in As0 functions which is generated by the family of weight norms σt

gCσ ( As ; G εT ) = sup gs (z, t)e− T −t−z , σ ≥ 0, 0 < s < s0 . (z,t)∈G εT

(3.85)

It is obvious that (3.85) coincides with (3.71) for σ = 0. The number σ will be chosen later. Theorem 3.9 Let T > 0, s0 > 0, and ε > 0 be fixed numbers and the conditions of the Lemma 3.3 be valid. Moreover, ! 1 1 max hs0 (t),  f t s0 (t),  f tt s0 (t), h t s0 (t) = R, t∈[0,T ] 4 4 while R > 0 is a known number. Then, for a and s such that a ∈ (0, T /s0 ) , ε and" s∈  (0, s0 ) there is a 1unique solutionto (3.81)–(3.83) in the domain sT = ε (x, z, t)| 0 ≤ t < 1+ε a(s0 − s) − z for which DT   ε , (u(x, z, t), v(x, z, t)) ∈ C As ; PsT    1 a(s0 − s) . k(x, t) ∈ C As0 ; 0, 1+ε

144

3 Global Solvability of Memory Reconstruction Problems

Moreover, ˜ v − v0 C A ; u − u 0 C ( As ; PsTε ) ≤ R, ( s k − k0 C ( As ; PsTε ) ≤

ε PsT )

≤

R˜ , (s0 − s)2

R˜ , s0 − s

 " 1 ε PsT (z, t)| 0 ≤ z ≤ t < 1+ε = G εT a(s0 − s) − z , where u 0 = 0, k0 and v0 are determined in (3.84), and R˜ = Reσ/ε , σ > 0. Proof Under assumptions of theorem we have   (v0 , k0 ) ∈ C As0 ; G T , k0 Cσ (As ; PsT ) ≤ k0 C(As ; G T ) ≤ R, v0 Cσ (As ; PsT ) ≤ v0 C( As ; G T ) ≤ R,

0 < s < s0 . Let an be the terms of the monotone decreasing sequence an+1 =

an , n = 0, 1, 2, . . . . 1 + 1/(n + 1)2

Put a = lim an = a0 n→∞

∞ #  −1 1 + 1/(n + 1)2 , n=0

where a0 ∈ (0, T /s0 ). Construct a process of successive approximations to system (3.81)–(3.83) by the scheme u n+1 (x, z, t) =

t 

 z vn (x, z, τ ) − kn (x, τ − z) dτ , 2

z

vn+1 (x, z, t) = v0 (x, z, t)  z  t−z 1 u n (x, ξ, t − z + ξ) + + kn (x, τ )u n (x, ξ, t − z + ξ − τ ) dτ dξ 2 0

+

0

(t+z)/2  

1 2

u n (x, ξ, t + z − ξ) − kn (x, t + z − 2ξ) z

t+z−2ξ 

+



kn (x, τ )u n (x, ξ, t + z − ξ − τ ) dτ dξ, 0

3.5 About Global Solvability of a Multidimensional Inverse …

t/2 kn+1 (x, t) = k0 (x, t) −

145

ξ vn (x, ξ, t − 2ξ) − kn (x, t − 2ξ) 2

0



t−2ξ 



ξ kn (x, τ ) vn (x, ξ, t − ξ − τ ) − kn (x, t − ξ − τ ) 2

+

 dτ dξ.

0

Put sn (z, t) =

s + ν n (z, t) n t+z , ν (z, t) = s0 − . 2 an

(3.86)

Introduce pn = u n+1 − u n , rn = vn+1 − vn , and qn = kn+1 − kn , n = 0, 1, 2, . . . .   Then pn , qn , and rn satisfy the relations p0 (x, z, t) =

t 

 z v0 (x, z, τ ) − k0 (x, τ − z) dτ , 2

z

(t+z)/2 

1 r0 (x, z, t) = − 2

k0 (x, t + z − 2ξ) dξ, z

t/2

ξ v0 (x, ξ, t − 2ξ) − k0 (x, t − 2ξ) 2

q0 (x, t) = − 0

   ξ k0 (x, τ ) v0 (x, ξ, t − ξ − τ ) − k0 (x, t − ξ − τ ) dτ dξ, 2

t−2ξ 

+ 0

pn+1 (x, z, t) =

t   z rn (x, z, τ ) − qn (x, τ − z) dτ , 2 z

1 rn+1 (x, z, t) = 2

z   pn (x, ξ, t − z + ξ) 0

+

 t−z   kn (x, τ ) pn (x, ξ, t − z + ξ − τ ) + u n+1 (x, ξ, t − z + ξ − τ )qn (x, τ ) dτ dξ 0

1 + 2

(t+z)/2  

 pn (x, ξ, t + z − ξ) − qn (x, t + z − 2ξ) z

   kn (x, τ ) pn (x, ξ, t + z − ξ − τ ) + qn (x, τ )u n+1 (x, ξ, t + z − ξ − τ ) dτ dξ,

t+z−2ξ 

+ 0

146

3 Global Solvability of Memory Reconstruction Problems

t/2

ξ rn (x, ξ, t − 2ξ) − qn (x, t − 2ξ) 2

qn+1 (x, t) = − 0 t−2ξ 

  ξ qn (x, τ ) vn+1 (x, ξ, t − ξ − τ ) − kn+1 (x, t − ξ − τ ) 2

+ 0

   ξ + kn (x, τ ) rn (x, ξ, t − ξ − τ ) − qn (x, t − ξ − τ ) dτ dξ. 2

Prove that we can choose σ > 0 so that for each n = 0, 1, 2, . . . the following estimates are valid:  λn (σ) = max

sup

(z,t,s)∈Fn

   − σt − σt  pn s (z, t)e bn −t−z , sup rn s (z, t) × (ν n (z, t) − s)e bn −t−z , (z,t,s)∈Fn  ! − σt n qn s (t)(ν (z, t) − s)2 e bn −t−z < ∞, sup

(z,t,s)∈Fn

(3.87)

σt

σt

u n+1 − u 0 s (z, t)e− bn −t−z ≤ R, vn+1 − v0 s (z, t)e− bn −t−z ≤ σt R kn+1 − k0 s (t)e− bn −t−z ≤ (s0 −s) 2 , (z, t, s) ∈ Fn+1 ,

R , s0 −s

where bn : = an (s0 − s), Fn = (z, t, s)| (z, t) ∈

G εT ,

! bn − z, 0 < s < s0 . 0 0, n = 0, 1, 2, . . . .  sn − s We can verify that c0 = 4m.

148

3 Global Solvability of Memory Reconstruction Problems

Taking sn (z, t) from (3.86) (sn (z, t) > s) for n = 0 we obtain q0 s (t)e t/2 ≤ 0

≤

−b

σt 0 −t−z

−

2σξ

Rc0 e b0 −t−z  (s0 (ξ, t − 2ξ) − s)2

    T T − σξ 1+ + T R 2 e b0 −t−z 1 + dξ 4 4

R(1 + T /4)(4c0 + s02 T R) a0 s0 , (z, t, s) ∈ F0 . (ν 0 (z, t) − s)2 σ

Consequently, λ0 (σ) ≤

    T 1 1 1+ a0 s0 max 1, s0 , 4c0 + s 2 T R R =: μ0 R. σ 4 σ

Thus, λ0 (σ) is finite, i.e., (3.87) is valid for n = 0. Moreover, for (z, t, s) ∈ F1 the following estimates hold: u 1 − u 0 s (z, t)e

−b

σt 0 −t−z

=  p0 s (z, t)e

−b

σt 0 −t−z

≤ λ0 (σ), λ0 (σ) 2λ0 (σ) − σt − σt v1 − v0 s (z, t)e b0 −t−z = r0 s (z, t)e b0 −t−z ≤ 0 ≤ , ν (z, t) − s s0 − s λ0 (σ) 4λ0 (σ) − σt − σt ≤ . k1 − k0 s (t)e b0 −t−z = q0 s (t)e b0 −t−z ≤ 0 (ν (z, t) − s)2 (s0 − s)2 For σ > 0 such that σ > 4μ0 , inequalities (3.88) hold for n = 0. Using induction, show that (3.87) and (3.88) hold for other n as well if σ is chosen appropriately. Let (3.87) and (3.88) be valid for n = 0, 1, 2, . . . , i. Then, for (z, t, s) ∈ Fi+1 we have  pi+1 s (z, t)e

σt i+1 −t−z

−b

t  ri s (z, τ )e

≤

στ i+1 −τ −z

−b

e

σ(bi+1 −z)(t−τ ) i+1 −τ −z)(bi+1 −t−z)

− (b

z

 σ(bi+1 −z)(t−τ +z) z − σ(τ −z) − + qi s (τ − z)e bi+1 −τ e (bi+1 −τ )(bi+1 −t−z) dτ 2 t t−z ) λi (σ) λi (σ) t −η − b σ(t−τ − σ(t−η) −t−z ≤ e i+1 dτ + e bi+1 −t−z dη i+1 i+1 2 ν (z, τ ) − s 2 (ν (z, η) − s) z

0

λi (σ)a0 (1 + a0 /2) λi (σ)(1 + a0 /2) bi+1 − t − z ≤ . ≤ i+1 ν (z, t) − s σ σ

3.5 About Global Solvability of a Multidimensional Inverse …

149

Here we used the obvious inequalities ai ≤ a0 , ν i+1 (z, t) ≤ ν i (z, t), and t − η < − (z + η) < bi+1 − (z + η). Similar arguments for ri+1 and qi+1 yield

bi+1 1+ε

ri+1 s (z, t)e

σt i+1 −t−z

−b

≤

1 2

z  0

t−z + 0

1 + 2

(si (ξ, t

σ(z−ξ)(bi+1 +t−z) c0 λi (σ) − e (bi+1 −t+z−2ξ)(bi+1 −t−z) 2 − z + ξ) − s)

R(1 + s02 )λi (σ) λi (σ) + 2R i (s0 − s)2 (ν (ξ, τ ) − s)2

(t+z)/2  

z

−

σ(ξ−z)

−

 e

σ(z−ξ)(bi+1 −t+z+2τ ) i+1 −t+z−2ξ+τ )(bi+1 −t−z)

− (b

 dτ dξ

σ(2ξ−z)

c0 λi (σ)e bi+1 −t−z λi (σ)e bi+1 −t−z + i  2 (si (t + z − ξ) − s) (ν (ξ, τ ) − s)2

t+z−2ξ  

  R(1 + s02 )λi (σ) λi (σ) − b σ(ξ−z) −t−z i+1 e + + 2R i dτ dξ (s0 − s)2 (ν (ξ, τ ) − s)2 0   λi (σ)(bi+1 − t − z) 2 8c0 + 1 + 2RT (3 + s0 ) ≤ 2σ(ν i+1 (z, t) − s)2 ≤

λi (σ) a0 [4c0 + 1/2 + RT (3 + s02 )] , (z, t, s) ∈ Fi+1 . σ ν i+1 (z, t) − s

For the intermediate computations, sn is defined by (3.86) for n = i. We also used the relations 1 1 ≤ i i = 0, 1, 2, . . . , s0 − s ν (z, t) − s 1 + s0 u i σs,i+1 < u i σs,i−1 ≤ 2R, vi σs,i+1 < vi σs,i−1 ≤ R , s0 − s 1 + s02 ki σs,i+1 < ki σs,i−1 ≤ R , (z, t, s) ∈ Fi+1 , (s0 − s)2

 · σs,i+1 <  · σs,i ,

which are true by the inductive assumption. Similar arguments for qi+1 lead to the following inequality (the change of variable t − 2ξ = η was made beforehand): qi+1 s (z)e t  × 0

η + 0

σt i+1 −t−z

−b

−

≤

σ(t−η)

λi (σ) × 2 $

c0 e bi+1 −t−z (si ( t−η , η) − s)2 2

1 t−η i+1 (ν ( 2 , τ ) − s)2

1 t −η + t−η i+1 i+1 ν ( 2 , η) − s 4(ν ( t−η , η) − s)2 2 

R(1 + s0 ) t − η R(1 + s02 ) + s0 − s 4 (s0 − s)2



%

150

3 Global Solvability of Memory Reconstruction Problems

R(1 + s02 ) + (s0 − s)2 ×e

− 2(b (t−η)σ −t−z)



i+1

$

%!

t −η

1

+ ν i+1 ( t−η , t+η−2τ )−s 4(ν i+1 ( t−η , t+η−2τ ) − s)2 2 2 2 2  λi (σ)(bi+1 − t − z) dτ dξ ≤ σ



×

4c0 (4 + a0 ) 4(1 + s0 ) + a0 (1 + s02 ) (4 + a0 )(1 + s02 ) + RT + RT i+1 3 i+1 3 4(ν (z, t) − s) 4(ν (z, t) − s) 4(ν i+1 (z, t) − s)3

≤

λi (σ)a0 c0 (4 + a0 ) + RT [1 + s0 + (2 + 3a0 )(1 + s02 )/4] , (z, t, s) ∈ Fi+1 . σ (ν i+1 (z, t) − s)2

From the above estimates, it follows that λi+1 (σ) ≤ λi (σ)ρ, λi+1 < ∞, a0 max 1 + a0 /2; 1/2 + 4c0 + RT (3 + s02 ); ρ= σ

c0 (4 + a0 ) + RT [1 + s0 + (2 + 3a0 )(1 + s02 )/4] .

At the same time, for (z, t, s) ∈ Fi+2 we obtain u i+2 − u 0 s (z, t)e

σt i+1 −t−z

−b

≤

i+1 

σt

 pn s e− bn −t−z ≤

n=0

vi+2 − v0 s (z, t)e

σt i+1 −t−z

−b

≤

i+1 

λn (σ) ≤ λ0 (σ)

n=0 σt

rn s e− bn −t−z ≤

n=0

≤

i+1 

i+1  n=0

λn (σ) n ν (z, t) −

i+1 i+1 1  λn (σ)an λ0 (σ)  n ≤ ρ [(n + 1)2 + 1], s0 − s n=0 an − an+1 s0 − s n=0

ki+2 − k0 s (t)e

σt i+1 −t−z

−b

≤

i+1 

σt

qn s e− bn −t−z

n=0

≤

i+1  n=0

≤

i+1  λn (σ) λn (σ)an2 1 ≤ (ν n (z, t) − s)2 (s0 − s)2 n=0 (an − an+1 )2

i+1 λ0 (σ)  n ρ [(n + 1)2 + 1]2 , (s0 − s)2 n=0

Choose σ > 0 such that ρ < 1, λ0 (σ)

∞  n=0

ρn [(n + 1)2 + 1]2 ≤ R.

i+1  n=0

s

ρn ,

3.5 About Global Solvability of a Multidimensional Inverse …

151

Then u i+2 − u 0 σs,i+1 ≤ R, vi+2 − v0 σs,i+1 ≤ ki+2 − k0 σs,i+1 ≤

R , s0 − s

R , (z, t, s) ∈ Fi+2 . (s0 − s)2

Note that the norms  · s and  · σs,i are equivalent for i = 0, 1, 2, . . . . Indeed, σ

e− ε  · s (z, t) ≤  · σs,i (z, t) ≤  · s (z, t), (z, t) ∈ Fi+1 . Consequently, (3.88) can be rewritten as σ

u n+1 − u 0 s ≤ Re ε , vn+1 − v0 s ≤ kn+1 − k0 s ≤

σ

Re ε , (s0 −s)2

σ

Re ε , s0 −s

(z, t, s) ∈ Fn+1 .

(3.89)

Since the choice of σ > 0 is independent of the number of approximations, the successive approximations (u n , kn , vn ) belong to C (As ; F) , F =

∞ &

Fn ,

n=0

and ˜ vn − v0 s (z, t) ≤ u n − u 0 s (z, t) ≤ R, kn − k0 s (t) ≤

R˜ , (z, t, s) ∈ F. (s0 − s)2

R˜ , s0 − s

For s ∈ (0, s0 ) the series ∞  n=0

(u n − u n−1 ),

∞ ∞   (vn − vn−1 ), (kn − kn−1 ) n=0

n=0

  ε converge uniformly in the norm of C As ; PsT , therefore un → u, vn → v, and ε and satisfy kn → k, and the limit functions u, k, and v are elements of C As ; PsT (3.81)–(3.83) . ˆ be two arbitrary Prove that the found solution is unique. Let (u, v, k) and (u, ˆ v, ˆ k) solutions satisfying ˜ v − v0 s (y, t) ≤ u − u 0 s (y, t) ≤ R,

R˜ R˜ , k − k0 s (t) ≤ , (z, t, s) ∈ F. s0 − s (s0 − s)2

152

3 Global Solvability of Memory Reconstruction Problems

Put p˜ = u − u, ˆ r˜ = v − v, ˆ and q˜ = k − kˆ and let λ(σ) = max

sup

(z,t,s)∈F

  σt  p ˜ s (z, t)e− b−t−z ,

sup

(z,t,s)∈F

  σt ˜r s (z, t)(ν(z, t) − s)e− b−t−z ,

 ! σt 2 − b−t−z ˜ s (t)(ν(z, t) − s) e < ∞, sup q

(z,t,s)∈F

where b := a(s0 − s), ν(z, t) = s0 −

t+z , a

and a = a0

∞  −1 ' . 1 + 1/(n + 1)2 n=0

Then, p(x, ˜ z, t) =

t   z r˜ (x, z, τ ) − q(x, ˜ τ − z) dτ , 2 z

1 r˜ (x, z, t) = 2

z   p(x, ˜ ξ, t − z + ξ) 0 t−z

   k(x, τ ) p(x, ˜ ξ, t − z + ξ − τ ) + u(x, ˆ ξ, t − z + ξ − τ )q(x, ˜ τ ) dτ dξ

+ 0

1 + 2

(t+z)/2  

 p(x, ˜ ξ, t + z − ξ) − q(x, ˜ t + z − 2ξ) z

   k(x, τ ) p(x, ˜ ξ, t + z − ξ − τ ) + q(x, ˜ τ )u(x, ˆ ξ, t + z − ξ − τ ) dτ dξ,

t+z−2ξ 

+ 0

t/2 q(x, ˜ t) = −

˜r (x, ξ, t − 2ξ) −

ξ q(x, ˜ t − 2ξ) 2

0 t−2ξ 

  ξˆ q(x, ˜ τ ) v(x, ˆ ξ, t − ξ − τ ) − k(x, t − ξ − τ) 2

+ 0

   ξ + k(x, τ ) r˜ (x, ξ, t − ξ − τ ) − q(x, ˜ t − ξ − τ ) dτ dξ. 2

Applying to these equations the estimates presented above, we find λ ≤ λρ , ρ :=

a max 1 + a/2; 1/2 + 4c0 + 2RT + RT (1 + s02 ); σ

c0 (4 + a) + RT [1 + s0 + (2 + 3a)(1 + s02 )/4] < ρ < 1.

ˆ The theorem is proven. Consequently, λ = 0; therefore, u = u, ˆ v = v, ˆ and k = k.

3.5 About Global Solvability of a Multidimensional Inverse …

153

 set ϒ of all pairs of functions ( f, h) representing the elements of Consider the C As0 ; [0, T ] , s0 > 0, for which the conditions of Theorem 3.9 are valid with fixed R, T , s0 , and ε. Then, we have the stability theorem: ¯ ∈ ϒ. Then, for the corresponding soluTheorem 3.10 Let ( f, h) ∈ ϒ, and ( f¯, h) ¯ v) tions (u, k, v) and (u, ¯ k, ¯ to (3.81)–(3.83), we have ¯ s≤ u − u ¯ s ≤ cM, k − k

cM , (s0 − s)2

cM ε , (z, t) ∈ PsT , 0 < s < s0 , s0 − s

v − v ¯ s≤

(3.90)

where  ¯ s0 (t), max  f t − f¯t s0 (t), M = max max h − h  max h t − h¯t s0 (t), max  f tt − f¯tt s0 (t) , t ∈ [0, T ], and the constant c depends on R, T , s0 , σ, and ε. ˜ v − v¯ = v, Proof For the differences u − u¯ = u, ˜ k − k¯ = k, ˜ f − f¯ = f˜, and h − ¯h = h˜ the equalities follow from (3.81)–(3.83) u(x, ˜ z, t) =

t 

 z˜ v(x, ˜ z, τ ) − k(x, τ − z) dτ , 2

(3.91)

z

z  u(x, ˜ ξ, t − z + ξ)

1 v(x, ˜ z, t) = 2

0

   ¯k(x, τ )u(x, ˜ ˜ ξ, t − z + ξ − τ ) + u(x, ξ, t − z + ξ − τ )k(x, τ ) dτ dξ

t−z + 0

(t+z)/2  

1 + 2

˜ t + z − 2ξ) u(x, ˜ ξ, t + z − ξ) − k(x,

z

   ¯ τ )u(x, ˜ τ )u(x, ξ, t + z − ξ − τ ) dτ dξ, k(x, ˜ ξ, t + z − ξ − τ ) + k(x,

t+z−2ξ 

+ 0

(3.92)

154

3 Global Solvability of Memory Reconstruction Problems

˜ t) = − k(x,

t/2

ξ ˜ v(x, ˜ ξ, t − 2ξ) − k(x, t − 2ξ) 2

0 t−2ξ 

  ˜ τ ) v(x, ξ, t − ξ − τ ) − ξ k(x, t − ξ − τ ) k(x, 2 0    ξ˜ ¯ + k(x, τ ) v(x, ˜ ξ, t − ξ − τ ) − k(x, t − ξ − τ ) dτ dξ. 2

+

(3.93)

where  1 ˜ ˜ f t (x, t − z) − h(x, v˜0 (x, z, t) = t − z) , 2  k˜0 (x, t) = 2 f˜tt (x, t) + h˜ t (x, t) . It is obvious that ε . v˜0 s0 (z, t) ≤ M, k˜0 s0 (z) ≤ 4M, (z, t) ∈ PsT

(3.94)

Theorem 3.9 yields the estimates: 2 ˜ ˜ ˜ vs ≤ R(1 + s0 ) , ks ≤ R(1 + s0 ) . us ≤ 2 R, s0 − s (s0 − s)2

  Applying the method of successive approximations used for the proof of Theorem ˜ 3.9 to the system of equations (3.91)–(3.93), which is linear with respect to u, ˜ k, and v˜ we find that, for the same choice of σ, the following inequalities are valid for solution to (3.91)–(3.93): u˜ − u˜ 0 ≤ c1 M, v˜ − v˜0  ≤ k˜ − k˜0  ≤

c1 M , s0 − s

c1 M ε , (z, t) ∈ PsT , 0 < s < s0 , (s0 − s)2

where c1 depends on R, T , s0 , σ, and ε. Hence, due to (3.93), it follows (3.89). Theorem 3.10 is proven.

References

155

3.6 Conclusions In this chapter, the global solvability of memory determination problems is investigated. The local nature of existence theorems proved in the previous chapters is due to the fact that the Volterra integral equations obtained in the process of solving the inverse problem are nonlinear. Turns out, since in these equations the nonlinearity is convolutional, it is possible to solve the resulting Volterra equations in space of continuous functions with an exponential weight. This allowed to prove the global solvability of such problems. Here, the problems of determining memory function from initial–boundary value problems for the integro-differential wave equation and the system of equations of electrodynamics are considered. At the end of this chapter, it is studied the multidimensional inverse problem of determining the convolutional kernel of the integral term in an integro-differential wave equation. It is proven the global unique solvability of this problem in the class of continuous functions in the time variable and analytic in a part of the space variables.

References 1. Romanov, V.G. 2005. Stability in inverse problems, M.: Nauchniy Mir (Russian). 2. Bukhgeim, A.L. 1993. Inverse problems of memory reconstruction. Journal of Inverse and Ill-Posed Problems 1 (3): 193–206. 3. Bukhgeim, A.L., and N.I. Kalinina. 1997. Global convergence of the Newton method in inverse problems of memory reconstruction. Siberian Mathematical Journal 38 (5): 881–895. 4. Bukhgeim, A.L., N.I. Kalinina, and V.B. Kardakov. 2000. Two methods for the inverse problem of memory reconstruction. Siberian Mathematical Journal 41 (4): 634–642. 5. Janno, J., and L. von Wolfersdorf. 1997. Inverse problems for identification of memory kernels in viscoelasticity. Mathematical Methods in the Applied Sciences 20 (4): 291–314. 6. Janno, J., and L. von Wolfersdorf. 1998. Inverse problems for identification of memory kernels in thermo—and poro—viscoelasticity. Mathematical Methods in the Applied Sciences 21 (16): 1495–1517. 7. Janno, J., and L. von Wolfersdorf. 2001. An inverse problem for identification of a time- and space-dependent memory kernel in viscoelasticity. Inverse problems 17 (1): 13–24.

Chapter 4

Stability in Inverse Problems for Determining of Two Unknowns

This chapter discusses the problems of simultaneously identifying two unknowns. In the Sect. 4.1, wave propagation velocity and the memory of the layered medium will be determined. In the Sect. 4.2, the regular part of the instantaneous source physically simulating a point explosion at the boundary of the domain and the memory of the medium are unknowns. For their determination, two observations are used for how the boundary of the domain fluctuates. Main results are stability estimates and uniqueness theorems for the problems under consideration [1, 2].

4.1 The Problem of Determining the Speed of Sound and Memory of Medium In this paragraph, we simultaneously determine the wave propagation velocity and the memory of the layered medium from the two-dimensional integro-differential wave equation with variable coefficient.

4.1.1 Statement of the Problem and Preliminary Transformations Consider for (x, y, t) ∈ R2+ × R the initial–boundary value problem for the equation

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_4

157

158

4 Stability in Inverse Problems for Determining of Two Unknowns

    u tt − a 2 u x x − a 2 u y y −

t k(τ )u x x (x, y, t − τ )dτ = 0, t > 0

(4.1)

0

with conditions   u t 0 . In these equations, the coefficient a = a(y) is a positive function of the class C 2 (R+ ) , R+ = {y ∈ R|y > 0} , and k(t), f (t) are continuous functions for t ∈ R. Given the functions a(y), k(t), f (t) the problem of finding function u(x, y, t), satisfying (in a generalized sense) equalities (4.1), (4.2), we call the direct problem. Suppose that the solution to this problem is given on the boundary of the domain: R2+ × R (4.3) u| y=0 = F(x, t), (x, t) ∈ R2+ . The inverse problem is to determine two functions c(t), k(t) for a given function F(x, t). Denote by u(λ, ˜ y, t) the Fourier transform of the function u(x, y, t) with respect to the variable x:  1 u(x, y, t)eiλx dx. u(λ, ˜ y, t) = √ 2π R

Given the functions a(y), k(t), f (t) the problem (4.1), (4.2) is correctly set and it has a unique solution u(x, y, t) possessing a compact support for any finite t. Eqs. (4.1), (4.2) with respect to the function u(λ, ˜ y, t) can be written as   u˜ tt − a 2 u˜ y y + λ2 a 2 u˜ + λ2

t k(τ )u(λ, ˜ y, t − τ )dτ = 0, (λ, y, t) ∈ R2+ × R, 0

 u˜ 

 ≡ 0, u˜ y  y=0 = δ  (t) + f (t)θ (t). t 0 satisfies equations t−z vtt − vzz − q(λ, z)v(λ, z, t) − λ k(t − z) + λ k(τ )v(λ, z, t − τ )dτ = 0, 2

2

0

 1 v t=z+0 = −H − 2

(4.13)

z q(λ, ξ )dξ, z ∈ R+

(4.14)

0

 (vz + H v) z=0 = f (t), t ∈ R+ .

(4.15)

Note that for the solvability of the inverse problem, as follows from the represen¯ tation (4.12), the function F(λ, t) must have the following structure: ¯ F(λ, t) = −δ(t) + F0 (λ, t)θ (t), (λ, t) ∈ R2+ ,

(4.16)

where the function F0 (λ, t) with respect to the argument t satisfies some smoothness conditions, which will be discussed below. In this regard, the additional information (4.10) for the function v looks like v|z=0 = F0 (λ, t), t ∈ R+ .

(4.17)

Now we narrow down the data of the problem, assuming that the function F0 (λ, t)  t)) is known only for two values of λ1 , λ2 such that λ21 = λ22 . (hence the function F(λ, Thus, the inverse problem (4.1)–(4.3) has been reduced to the problem of determining the functions c(z), k(t) from the relations (4.13)–(4.15) if the solution of the direct problem is known for λ = λi , i = 1, 2 and is given by the equalities (4.17). It turns out that with these data the functions c(z), k(t) are uniquely determined. After finding c(z), the function h(z), defined the correspondence between the variables y and z, as follows from (4.6), is found by the formula z h(z) =

c(ξ )dξ, 0





and a(y) = c h −1 (y) . Due to the fact that Eq. (4.13) describes a wave process propagating with a velocity equal to unity, v(λ, 0, t) for t ∈ [0, T ] and for a fixed λ depends from function c(z) and its derivatives (through function q(λ, z)) only on the interval [0, T /2], and from k(t) on [0, T ]. Therefore, it is natural to expect that the

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

161

inverse is also true: the function c(z) on each interval [0, T /2] and the function k(t) on [0, T ] are determined by F0 (λ, t) only on interval [0, T ]. It turns out that such a local character of the dependence of c(z), k(t) on F0 (λ, t) really takes place, and it, of course, is reflected in the results that in what follows, we intend to prove.

4.1.2 Properties of the Direct Problem Solution We study the properties of the direct problem solution (4.13)–(4.15). Lemma 4.1 Let c(z) ∈ C 2 [0, T /2], k(t) ∈ C[0, T ], f (t) ∈ C[0, T ] for some fixed T > 0. Then, for each fixed value of the parameter λ the solution of problem (4.13)– (4.15) for    (z, t) ∈ DT , DT = (z, t)0 ≤ z ≤ t ≤ T − z belongs to the functional class C 1 (DT ) and for it the following estimate is true:   vC 1 (D(T ) ≤ C c(z)C 2 [0,T /2] + k(t)C[0,T ] +  f (t)C[0,T ] ,

(4.18)

where C depends only on T , λ, c(z)C 2 [0,T /2] and k(t)C[0,T ] . Also function ψ(λ1 , λ2 , t) = vt (λ1 , 0, t) − vt (λ2 , 0, t) for any fixed λ j , j = 1, 2 is a function of the class C 1 [0, T ]. Proof Using the formula (3.63) for vtt − vzz from relations (4.13)–(4.15) for (z, t) ∈ DT by integration along the corresponding characteristics of the first-order differential operators, we obtain the equalities

z+t 1 (vt + vz )(λ, z, t) = − q λ, 2 2 (z+t)/2  q(λ, ξ )v(λ, ξ, t + z − ξ ) + λ2 k(t + z − 2ξ ) + z t+z−2ξ 

−λ

2 0

k(τ )v(λ, ξ, t + z − ξ − τ )dτ dξ,

(4.19)

162

4 Stability in Inverse Problems for Determining of Two Unknowns

v(λ, 0, t) = −H e

Ht

1 − 2

t e

H (t−τ )

t  τ dτ − e H (t−τ ) f (τ )dτ q λ, 2

0

t +

0

τ/2 H (t−τ ) q(λ, ξ )v(λ, ξ, τ − ξ ) + λ2 k(τ − 2ξ ) e

0

(4.20)

0 τ−2ξ

− λ2

k(α)v(λ, ξ, τ − ξ − α)dα dξ dτ,

0

(vt − vz )(λ, z, t)



1 t−z = −2 f (t − z) + 2H v(λ, 0, t − z) − q λ, 2 2 (t−z)/2  2 q(λ, ξ )v(λ, ξ, t − z − ξ ) + λ2 k(t − z − 2ξ ) + λ k(t − z)z + 0 t−z−2ξ 

− λ2

k(τ )v(λ, ξ, t − z − ξ − τ )dτ dξ

0

z −

q(λ, ξ )v(λ, ξ, t − z + ξ ) + λ2

0

t−z

k(τ )v(λ, ξ, t − z + ξ − τ )dτ dξ. 0

(4.21) From (4.19), (4.21), we find the equations for vt , vz , v

t+z 1 vt (λ, z, t) = − f (t − z) + H v(λ, 0, t − z) − q λ, 4 2

t −z λ2 1 + k(t − z)z − q λ, 4 2 2 +

(t−z)/2 

q(λ, ξ )v(λ, ξ, t − z − ξ ) + λ2 k(t − z − 2ξ )

1 2

0 t−z−2ξ 

− λ2

k(τ )v(λ, ξ, t − z − ξ − τ )dτ dξ

0

1 + 2

z 0

q(λ, ξ )v(λ, ξ, t − z + ξ ) − λ2

t−z

k(τ )v(λ, ξ, t − z + ξ − τ )dτ dξ 0

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

163

(z+t)/2 

q(λ, ξ )v(λ, ξ, t + z − ξ ) + λ2 k(t + z − 2ξ )

1 + 2

z t+z−2ξ 

− λ2

k(τ )v(λ, ξ, t + z − ξ − τ )dτ dξ,

(4.22)

0

vz (λ, z, t) = f (t − z) − H v(λ, 0, t − z)



1 t+z 1 t −z λ2 − q λ, + q λ, − k(t − z)z 4 2 4 2 2 (t−z)/2  1 − q(λ, ξ )v(λ, ξ, t − z − ξ ) + λ2 k(t − z − 2ξ ) 2 0 t−z−2ξ 

− λ2

k(τ )v(λ, ξ, t − z − ξ − τ )dτ dξ

0

−

1 2

z

q(λ, ξ )v(λ, ξ, t − z + ξ )

(4.23)

0 t−z

k(τ )v(λ, ξ, t − z + ξ − τ )dτ dξ

− λ2 0

+

1 2

(z+t)/2 

q(λ, ξ )v(λ, ξ, t + z − ξ ) + λ2 k(t + z − 2ξ )

z t+z−2ξ 

− λ2

k(τ )v(λ, ξ, t + z − ξ − τ )dτ dξ,

0

1 v(λ, z, t) = −H − 2

z

t−z q(λ, ξ )dξ −

0

0

t−z f (τ )dτ + H v(λ, 0, τ )dτ 0



t t−z τ −z λ2 z τ +z 1 + q λ, dτ + q λ, − k(τ )dτ 4 2 2 2 z

+

1 2

0

 t (τ−z)/2

q(λ, ξ )v(λ, ξ, τ − z − ξ ) + λ2 k(τ − z − 2ξ )

z

0

164

4 Stability in Inverse Problems for Determining of Two Unknowns τ −2ξ  −z

−λ

k(α)v(λ, ξ, τ − ξ − α − z)dα dξ dτ

2 0

+

1 2

 t z z

q(λ, ξ )v(λ, ξ, τ − z + ξ )

0

τ −z

2 −λ k(α)v(λ, ξ, τ + ξ − α − z)dα dξ dτ 0

1 + 2

t

(τ +z)/2

q(λ, ξ )v(λ, ξ, τ + z − ξ ) + λ2 k(τ + z − 2ξ )

z

0 τ −2ξ  +z

k(α)v(λ, ξ, τ − ξ − α + z)dα dξ dτ.

− λ2

(4.24)

0

Eq. (4.24) is the integral equation of Volterra type in the domain DT and it has the unique continuous solution. It follows from equalities (4.22), (4.23) that this solution is continuously differentiable in DT . Substituting the expression for v(λ, 0, t), defined by formula (4.20) into Eqs. (4.22)–(4.24) and constructing the method of successive approximations for these equations by the usual scheme, which has factorial convergence by argument t, it is easy to establish the validity of the estimate (4.18) in the domain DT . Using equality (4.22), we compose the function ψ(λ1 , λ2 , t) ψ(λ1 , λ2 , t) = H [v(λ1 , 0, t) − v(λ2 , 0, t)] +

 1 2 λ1 − λ22 q1 2

t 2

t/2 + [q(λ1 , ξ )v(λ1 , ξ, t − ξ ) − q(λ2 , ξ )v(λ2 , ξ, t − ξ ) + (λ21 − λ22 )k(t − 2ξ ) 0 t−2ξ 

k(τ )[λ21 v(λ1 , ξ, t − ξ − τ ) − λ22 v(λ2 , ξ, t − ξ − τ )]dτ ]dξ.

− 0

(4.25) The right-hand side of this equality belongs to the class C 1 [0, T ]. Therefore ψ(λ1 , λ2 , t) ∈ C 1 [0, T ], for any fixed λ j , j = 1, 2. As consequence of the Lemma 4.1, we note that under the conditions of lemma , the function F0 (λ, t), entering in equality (4.13), for each fixed λ belongs to the class C 1 [0, T ], and the function F00 (t) ≡ F0 (λ1 , t) − F0 (λ2 , t) belongs to the class C 2 [0, T ].

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

165

4.1.3 Investigation of the Inverse Problem Lemma 4.2 Suppose that the function F(λ, t) has the structure (4.16) and F0 (λ, t) ∈ C 1 [0, T ] for any fixed λ, and F0 (λ1 , t) − F0 (λ2 , t) ≡ F00 (t) ∈ C 2 [0, T ]. In addition, let the function F0t (λ, 0) be increase in λ, λ ∈ R, f (t) ∈ C[0, T ]. Then the inverse problem (4.15)–(4.15), (4.17) in the domain DT is equivalent to the problem of finding the functions v, vt , c(z), c (z), k(t) from the following closed system of the following integral equations: 1 v(λi , z, t) = −H − 2

z

t−z q(λi , ξ )dξ −

0

t−z f (τ )dτ + H v(λi , 0, τ )dτ

0

0



t t−z λi2 z 1 τ +z τ −z − q λi , + q λi , dτ + k(τ )dτ 4 2 2 2 z

+

1 2

0

t

(τ −z)/2

q(λi , ξ )v(λi , ξ, τ − z − ξ ) + λi2 k(τ − z − 2ξ )

z

0 τ −2ξ  −z

k(α)v(λi , ξ, τ − ξ − α − z)dα dξ dτ

− λi2 0

+

1 2

 t z z

q(λi , ξ )v(λi , ξ, τ − z + ξ )

0

τ −z

2 − λi k(α)v(λi , ξ, τ + ξ − α − z)dα dξ dτ 0

1 + 2

t

(τ +z)/2

q(λi , ξ )v(λ, ξ, τ + z − ξ ) + λi2 k(τ + z − 2ξ )

z

0 τ −2ξ  +z

k(α)v(λi , ξ, τ − ξ − α + z)dα dξ dτ, i = 1, 2,

− λi2 0

(4.26)

166

4 Stability in Inverse Problems for Determining of Two Unknowns

vt (λi , z, t)





1 1 t +z t−z − q λi , = − f (t − z) + H v(λi , 0, t − z) − q λi , 4 2 4 2 (t−z)/2  λ2 1 q(λi , ξ )v(λi , ξ, t − z − ξ ) + λi2 k(t − z − 2ξ ) + i k(t − z)z + 2 2 0 t−z−2ξ 

− λi2

k(τ )v(λi , ξ, t − z − ξ − τ )dτ dξ

0

+

1 2

z

q(λi , ξ )v(λi , ξ, t − z + ξ )

0 t−z

k(τ )v(λi , ξ, t − z + ξ − τ )dτ dξ

− λi2 0

+

1 2

(z+t)/2 

q(λi , ξ )v(λi , ξ, t + z − ξ ) + λi2 k(t + z − 2ξ )−

z t+z−2ξ 

− λi2

k(τ )v(λi , ξ, t + z − ξ − τ )dτ dξ, i = 1, 2,

0

z c(z) = c(0) +

  T , c (ξ )dξ, z ∈ 0, 2

(4.27) (4.28)

0

1 c (z) = c (0) + 4

    z   2 c (ξ ) T − 2q0 (ξ )c(ξ ) dξ, z ∈ 0, , c(ξ ) 2

(4.29)

0

q0 (z) = −2F0t (λ1 , 2z) − 2 f (2z) + 2H F0 (λ1 , 2z) + λ21 q1 (z) z q(λ1 , ξ )v(λ1 , ξ, 2z − ξ ) + λ21 k(2z − 2ξ ) +2 (4.30)

0

  T , k(τ )v(λ1 , ξ, 2z − ξ − τ )dτ dξ, z ∈ 0, 2

2(z−ξ  )

−

λ21 0



4.1 The Problem of Determining the Speed of Sound and Memory of Medium

167

 2 2H  (t) k(t) = 2 F00 (t) − 2 F00 2 λ1 − λ 2 λ1 − λ22







1 t t t t t t t 1 − 2 v λ − q λ v λ q λ , , , , − q1 , , 1 1 2 2 2 2 2 2 2 2 2 2 λ1 − λ22

t/2  q(λ1 , ξ )vt (λ1 , ξ, t − ξ ) − q(λ2 , ξ )vt (λ2 , ξ, t − ξ ) − 2 2 λ1 − λ 2 0   − k(t − 2ξ ) λ1 v(λ1 , ξ, ξ ) − λ2 v(λ2 , ξ, ξ ) 2

t−2ξ 

−



 k(τ ) λ21 vt (λ1 , ξ, t − ξ − τ ) − λ22 vt (λ2 , ξ, t − ξ − τ ) dτ dξ, t ∈ [0, T ],

0

(4.31) 

where c(0) =

 2F00 (0)  , c (0) = 2F0 (λ, 0)c(0). 2 λ1 − λ22

(4.32)

Proof At the beginning we establish the validity of equality (4.32). Indeed, setting in Eq. (4.20) t = 0 and using conditions (4.17), we find F0 (λi , 0) = −H, i = 1, 2.

(4.33)

In particular, from this we conclude that F0 |t=0 does not depend on λ. Further, from Eq. (4.25) for t = 0, we obtain  (0) = F00

 1 2 λ1 − λ22 c2 (0). 2

 (0) = F0t (λ1 , 0) − F0t (λ2 , 0), Hence, by virtue of the positivity of the number F00 we arrive at the first equality (4.32). Taking into account the first equality (4.11), from (4.33) we obtain the second equality (4.32). Since F0 |t=0 does not depend on λ, then the second formula from (4.32) uniquely determines c (0). Continuing the proof of the Lemma 4.2, we note that Eqs. (4.26) and (4.27) coincide with the equalities (4.22) and (4.24), respectively, for λ = λi , i = 1, 2. Equation (4.22) is derived from the relations (4.13)–(4.15). Equation (4.24) is obtained from the Eq. (4.22) by integration over t from the point (z, z) to the point (z, t) on the plane of variables (ξ, τ ). In turn, equation (4.24) satisfies equalities (4.13)–(4.16). Further, in Eq. (4.22) we set z = 0 and use condition (4.17) for λ = λ1 . Then, after simple transformations, we obtain equality (4.30). In this equation, for definiteness, we put λ = λ1 . In fact, the result of the calculation should not depend from the choice of the parameter λ. To obtain equality (4.31), we use relation (4.25), which is derived  (t), we from Eq. (4.22) using the conditions (4.17). Noting that ψ(λ1 , λ2 , t) ≡ F00 differentiate equality (4.25) with respect to the variable t. Then we have

168

4 Stability in Inverse Problems for Determining of Two Unknowns





t 1 1 t t t   + q λ1 , v λ1 , , F00 (t) = H F00 + (λ21 − λ22 )q1 4 2 2 2 2 2 t/2



   λ2 − λ22 t t t v λ2 , , + 1 k(t) + q(λ1 , ξ )vt (λ1 , ξ, t − ξ ) −q λ2 , 2 2 2 2 0   − q(λ2 , ξ )vt (λ2 , ξ, t − ξ ) − k(t − 2ξ ) λ21 v(λ1 , ξ, ξ ) − λ22 v(λ2 , ξ, ξ ) t−2ξ 

−



 k(τ ) λ21 vt (λ1 , ξ, t − ξ − τ ) − λ22 vt (λ2 , ξ, t − ξ − τ ) dτ dξ.

0

Solving this equality with respect to k(t), we arrive at the Eq. (4.31). We verify that inverse transformations take place as well. We show this by the example of the Eq. (4.31). In this equation, replacing t by t − τ, we multiply both sides by dτ and integrate over τ from 0 to t/2. In the repeated integrals of the resulting equality, we change the order of integration. Using condition (4.17) in the opposite direction, after simple calculations, we arrive at equation (4.25), from which follow equalities (4.22) for λ = λi , i = 1, 2; z = 0 and differentiated with respect to t equality (4.17). Proof of equivalence (4.30) and (4.30) for λ = λ1 is carried out similarly. The equivalence of equalities (4.28), (4.29) with the last equalities (4.11) is obvious. The system of integral Eqs. (4.26)–(4.31) is closed in the domain DT and defines the unique continuous functions v, vt , c(z), c (z), k(t) for sufficiently small T . Without dwelling on the theorem of local unique solvability of the problem, we pass to the results characterizing stability estimates and unique solvability for an arbitrary T > 0. Denote by A(c0 , k0 ) the set of pairs of functions {c(t), k(t)}, satisfying for some T > 0 the following conditions: 0 < c00 ≤ c(z), c(t)C 2 [0,T /2] ≤ c0 , k(t)C[0,T ] ≤ k0 . Besides,  f (t)C[0,T ] ≤ f 0 , where f 0 > 0 is a known number Theorem 4.1 Let (c(1) , k (1) ) ∈ A(c0 , k0 ), (c(2) , k (2) ) ∈ A(c0 , k0 ) be solutions of inverse problem (4.13)–(4.15), (4.17) with data 

   F0(1) (λ j , t), f (1) (t) , F0(2) (λ j , t), f (2) (t) ,

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

169

j = 1, 2, respectively. Then, there is a positive constant C, depending on λ1 , λ2 , T, c0 , c00 , k0 , f 0 , such that the estimate k (1) − k (2) C[0,T ] + c(1) − c(2) C 2 [0,T /2] + h (1) − h (2) C 3 [0,T /2] ≤ Cμ

(4.34)

is fulfilled, where μ=

2 

(1) (2) F0(1) (λ j , t) − F0(2) (λ j , t)C 1 [0,T ] + F00 − F00 C 2 [0,T ]

j=1

+  f (1) − f (2) C[0,T ] . Theorem 4.1 obviously implies the following uniqueness theorem for any T > 0. Theorem 4.2 The functions c(i) ∈ C 2 [0, T /2], k (i) ∈ C[0, T ] and F0(i) (λ j , t), f (i) (t), i = 1, 2, j = 1, 2 have the same meaning as in Theorem 4.1. If at the same time F0(1) (λ j , t) = F0(2) (λ j , t), j = 1, 2, f (1) (t) = f (2) (t) for t ∈ [0, T ], then, c(1) (z) = c(2) (z), z ∈ [0, T /2], k (1) (t) = k (2) (t), t ∈ [0, T ]. Proof Suppose that the functions f (i) , F (i) (λ j , t), c(i) , k (i) , i = 1, 2, j = 1, 2 are as defined in Theorem 4.1. The solution of problem (4.13)–(4.15) for f = f (i) , c = c(i) , k = k (i) , λ = λ j denote by v (i j) (z, t), i = 1, 2, j = 1, 2. Introduce also notations ˜ = k (1) − k (2) , f˜(t) = f (1) − f (2) , c(z) ˜ = c(1) − c(2) , k(t) (1) (2) ˜ 00 = F00 0 (λ j , t) = F0(1) (λ j , t) − F0(2) (λ j , t), h(z) = h (1) − h (2) , F − F00 , F

v˜ ( j) (z, t) = v (1 j) (z, t) − v (2 j) (z, t), j = 1, 2. Let q0(i) (z), q1(i) (z), q (i j) (z) = q0(i) (z) − λ2j q1(i) (z), q˜ ( j) (z) = q (1 j) − q (2 j) be auxiliary functions and H (i) = −(c(i) ) (0)/(2c(i) (0)) be number corresponding to the functions c(i) (z). Let us write out the corresponding integral relations for the newly introduced functions. From equalities (4.26) and (4.27) it follows that

170

4 Stability in Inverse Problems for Determining of Two Unknowns

 v˜ ( j) (z, t) = − H t−z t−z t−z (1) (2)  j , τ )dτ  F (λ j , τ )dτ + H − F(λ f˜(τ )dτ + H 0

−

0

1 2

z

0



t τ +z τ −z 1 ( j) q˜ ( j) + q˜ ( j) dτ q˜ (ξ )dξ − 4 2 2 z

0

t−z t λ2j z 1 ˜ + k(τ )dτ + 2 2 z

0

+q

(2 j)

(τ −z)/2

( j) q˜ (ξ )v (1 j) (ξ, τ − z − ξ )

0

˜ − 2ξ − z) (ξ )v˜ (ξ, τ − z − ξ ) + λ2j k(τ ( j)

τ −2ξ  −z

−

 (1 j) ˜ k(α)v (ξ, τ − ξ − α − z)

λ2j 0



+ k (α)v˜ ( j) (ξ, τ − ξ − α − z) dα dξ dτ  t z ( j) 1 q˜ (ξ )v (1 j) (ξ, τ − z + ξ ) + q (2 j) (ξ )v˜ ( j) (ξ, τ − z + ξ ) + 2 (2)

z

0

τ −z  2 (1 j) ˜ k(α)v − λj (ξ, τ + ξ − α − z) 0



+ k (2) (α)v˜ ( j) (ξ, τ + ξ − α − z) dα dξ dτ 1 + 2

t

(τ +z)/2

( j) q˜ (ξ )v (1 j) (ξ, τ + z − ξ )

z

+q

0

(2 j)

˜ + z − 2ξ ) (ξ )v˜ ( j) (ξ, τ + z − ξ ) + λ2j k(τ

τ −2ξ  +z

−

 (1 j) ˜ k(α)v (ξ, τ − ξ − α + z)

λ2j 0



+ k (α)v˜ ( j) (ξ, τ − ξ − α + z) dα dξ dτ, (2)

(4.35)

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

171

j

v˜t (z, t)  F (1) (λ j , t − z) + H (2) F(λ  j , t − z) = − f˜(t − z) + H

2 t+z t − z  λj ˜ 1 + q˜ ( j) + k(t − z)z − q˜ ( j) 4 2 2 2 (t−z)/2  ( j) 1 q˜ (ξ )v (1 j) (ξ, t − z − ξ ) + 2 +q

0 (2 j)

˜ − z − 2ξ ) (ξ )v˜ ( j) (ξ, t − z − ξ ) + λ2j k(t

t−z−2ξ 

−

 (1 j) ˜ k(α)v (ξ, t − z − ξ − α)

λ2j 0



+ k (2) (α)v˜ ( j) (ξ, t − z − ξ − α) dα dξ z 1 ( j) + q˜ (ξ )v (1 j) (ξ, t − z + ξ ) + q (2 j) (ξ )v˜ ( j) (ξ, t − z + ξ ) 2 0

−

λ2j

t−z 

 (1 j) ˜ (ξ, t + ξ − α − z) + k (2) (α)v˜ ( j) (ξ, t − z + ξ − α) dα dξ k(α)v 0

+

1 2

(t+z)/2 

[q˜ ( j) (ξ )v (1 j) (ξ, t + z − ξ )

z

˜ + z − 2ξ ) + q (2 j) (ξ )v˜ ( j) (ξ, t + z − ξ ) + λ2j k(t t+z−2ξ 

−

(1 j) ˜ (k(α)v (ξ, t + z − ξ − α) + k (2) (α)v˜ ( j) (ξ, t + z − ξ − α))dα]dξ,

λ2j 0

j = 1, 2. (4.36) Note that in these equalities = H

   (2)  (0) − c˜ (0) c(2) (0) c(0) ˜ c . 2c(1) c(2)

(4.37)

Using equalities (4.28)–(4.31) we find z c(z) ˜ = c(0) ˜ + 0

  T , c˜ (ξ )dξ, z ∈ 0, 2

(4.38)

172

4 Stability in Inverse Problems for Determining of Two Unknowns

(1) 

(2)  2 (c ) (ξ ) + (c(2) ) (ξ ) (c ) (ξ ) c˜ (ξ ) − c(ξ ˜ ) (1) c(1) (ξ ) c (ξ )c(2) (ξ ) 0    T ˜ ) dξ, z ∈ 0, − 2q˜0 (ξ )c(1) (ξ ) − 2q0(2) c(ξ , 2 z

1 c˜ (z) = c˜ (0) + 4 







(4.39)

0t (λ1 , 2z) − 2 f˜(2z) q˜0 (z) = −2 F  F0(1) (λ1 , 2z) + 2H (2) F 0 (λ1 , 2z) + λ21 q˜1 (z) + 2H z (1) ˜ +2 q˜ (ξ )v (11) (ξ, 2z − ξ ) + q (12) (ξ )v˜ (1) (ξ, 2z − ξ ) + λ21 k(2z − 2ξ ) 0 2(z−ξ  )

−

 

˜ )v (11) (ξ, 2z − ξ − τ ) + k (2) (τ )v˜ (1) (ξ, 2z − ξ − τ ) dτ dξ, k(τ

λ21 0

  T z ∈ 0, , 2 (4.40) (2)  2 ˜ =  (t) − 2 H F (1) (t) − 2H 00 (t) − 1 q˜ t F F k(t) 00 2 1 2 λ21 − λ22 λ21 − λ22 00 λ21 − λ22 (1)  t  (1) 1 H − q˜ − 2 2 2 2 λ −λ 1

1

2

+ q (12)

t  1 H− 2 2

t/2

t/2  q (11) (ξ )dξ 0

q˜ (1) (ξ )dξ



0

− q˜ (2)

t/2 t/2  

t  (1) 1 t  1 (21) (22) H − H− q (ξ )dξ + q q˜ (2) (ξ )dξ 2 2 2 2 0

2 − 2 λ1 − λ22

0

t/2

 (1) (11) q˜ (ξ )vt (ξ, t − ξ )

0 (1) (12) (12) +q (ξ )v˜t (ξ, t − ξ ) − q˜ (2) (ξ )vt (ξ, t − ξ )   (2) ˜ − 2ξ ) λ1 v (11) (ξ, ξ ) − λ2 v (12) (ξ, ξ ) + q (22) (ξ )v˜t (ξ, t − ξ ) − k(t   − k (2) (t − 2ξ ) λ1 v˜ (1) (ξ, ξ ) − λ2 v˜ (2) (ξ, ξ ) t−2ξ 

−

  ˜ ) λ2 vt(11) (ξ, t − ξ − τ ) − λ2 vt(12) (ξ, t − ξ − τ ) k(τ 1 2

0    (1) (2) + k (2) (τ ) λ21 v˜t (ξ, t − ξ − τ ) − λ22 v˜t (ξ, t − ξ − τ ) dτ dξ, t ∈ [0, T ].

(4.41)

4.1 The Problem of Determining the Speed of Sound and Memory of Medium

173

We estimate the functions of the system (4.35)–(4.41) in domain DT through value μ, defined in Theorem 4.1. The domain DT allows equivalent description      T  T   DT = (z, t) 0 ≤ z ≤ t ≤ −  − t  , 0 ≤ t ≤ T . 2 2 

Let  ρ(t) = max

max

  ( j) v˜ (z, t) ,

max

0≤z≤T /2−|T /2−t|

0≤z≤T /2−|T /2−t|

 (  c˜ z) ,

max

0≤z≤T /2−|T /2−t|

max

0≤z≤T /2−|T /2−t|

   c˜ (z) ,

   ( j)  v˜t (z, t) ,

max

0≤z≤T /2−|T /2−t|

  ˜  k(t) ,

   ( j)  q˜0 (z) ,

t ∈ [0, T ], j = 1, 2. According to Lemma 4.1, the functions v i j are differentiable in the domain DT and satisfy the estimate  ij v 

C 1 [DT ]

≤ M1 , i, j = 1, 2,

(4.42)

with some constant M1 , depending only on λ1 , λ2 , T , c0 , c00 , k0 , f 0 . Since the (i) (t) are traces of the functions v (i1) (z, t) − v (i2) (z, t) for z = 0, i = functions F00 1, 2, for each of which, as follows from (4.42), (4.25), an estimate similar to (4.42) (i) (t) = F0(i) (λ1 , t) − F0(i) (λ2 , t) we have the inequality holds, then, for F00 (i) F00 (t)C 2 [0,T ] ≤ M2 , i = 1, 2,

in which the constant M2 depends on the same parameters as M1 . From (4.42), (4.17) it follows that the functions F0(i) (λ j , t) must be bounded by the constant M1 : F0(i) (λ j , t)C 1 [0,T ] ≤ M1 , i, j = 1, 2  , The numbers c(i) (0), (c )(i) (0), i=1, 2, defined by formulae (4.32) via λ1 , λ2 , F00 for similar reasons should be bounded by the constant M3 , which depends from the , defined by same parameters as M j , j = 1, 2. Therefore, for the numbers H (i) , H (i)  (i) the numbers c (0), (c ) (0), i = 1, 2, and equality (4.37), the estimates

   (i)   H  ≤ M4 ,  H  ≤ M5 μ are valid with some constants.

174

4 Stability in Inverse Problems for Determining of Two Unknowns

M4 , M5 , depend on the set of numbers λ1 , λ2 , T, c0 , c00 , k0 , f 0 . Recall that μ is defined in Theorem 4.1. Next, the functions q0(i) (z), q1(i) (z), q (i j) (z), defined by c(i) (z) and λ j , i, j = 1, 2, for similar reasons satisfy the inequalities    (i)  q0 (z)

C 2 [0,T /2]

    ≤ M6 , q1(i) (z)

C 2 [0,T /2]

≤ M7 ,

 (i j)  q (z) 2 ≤ M8 , i, j = 1, 2, C [0,T /2] where M6 , M7 , M8 depend on the same parameters as the previous constants. Given the above, we proceed to estimate in the domain DT functions v˜ ( j) , satisfying the integral Eq. (4.35). Note that these equations, like all others, contain terms containing only known and terms with unknown functions. On the right-hand side of Eq. (4.35), the first four terms depend only on known functions and numbers. Therefore, these terms in totality are estimated by the value K 1 μ, with constant K 1 , depending on Mi , i = 1, 8. As it is easily seen the remaining part is estimated in the domain DT by an integral of the form t ρ(τ )dτ,

L1 0

in which L 1 depends only on Mi , i = 1, 8, which in turn depend on the parameters λ1 , λ2 , T, c0 , c00 , k0 , f 0 . Thus, |v˜ ( j) (z, t)| ≤ K 1 μ + L 1

t ρ(τ )dτ, (z, t) ∈ DT , j = 1, 2.

(4.43)

0

From the Eqs. (4.38), (4.39) we can see that the functions c(z), ˜ c˜ (z) are estimated in a similar way t (4.44) |c(z)| ˜ ≤ K 2 μ + L 2 ρ(τ )dτ, |c˜ (z)| ≤ K 3 μ + L 3

t

0

  T . ρ(τ )dτ, z ∈ 0, 2

(4.45)

0

The constants K 2 , K 3 , L 2 , L 3 depend on the same parameters as K 1 , L 1 . We use the inequalities (4.44), (4.45) to estimate the functions ˜ (1)  (1) ˜ (z) + c(2) (z)), q1( ) (z) = 2(c(z)(c ˜ ) (z) + c(2) (z)c˜( ) (z)), q˜1 (z) = c(z)(c

4.2 The Problem of Determining the Memory of Medium …

175

occurring in nonintegral terms in the right-hand sides of Eqs. (4.40), (4.41). Similar reasoning leads to inequalities t |q˜0 (z)| ≤ K 4 μ + L 4

ρ(τ )dτ,

(4.46)

ρ(τ )dτ.

(4.47)

0

˜ |k(t)| ≤ K5μ + L 5

t 0

We use inequalities (4.44)–(4.47) in the estimates of the functions q˜ j , k˜ outside the ( j) integral terms of Eq. (4.36) in order to obtain an estimate for the functions v˜t (z, t) in the form t ( j) |v˜t (z, t)| ≤ K 6 μ + L 6 ρ(τ )dτ, j = 1, 2. (4.48) 0

In the last inequalities, the constants K 4 , K 5 , L 4 , L 5 depend by means of the numbers Mi , i = 1, 8 on the parameters λ1 , λ2 , T, c0 , c00 , k0 , f 0 . From the relations (4.43)–(4.48) it follows that ρ(t) satisfies integral inequality t ρ(τ )dτ,

ρ(t) ≤ K μ + L 0

with the new constants K , L, depending only on λ1 , λ2 , T, c0 , c00 , k0 , f 0 . Hence, using the Gronwall inequality, we obtain the estimate (4.34).

4.2 The Problem of Determining the Memory of Medium and Regular Part of Impulsive Source In this paragraph we suppose the function a(y) as known and consider the initial– boundary value problem (4.1)–(4.3) in which k(t), f (t) are unknown functions.

4.2.1 Statement of the Problem and Preliminary Constructions For simplicity, we set a(y) = 1, y ∈ R+ and rewrite Eqs. (4.49)–(4.51) in the terms of the Fourier transform

176

4 Stability in Inverse Problems for Determining of Two Unknowns

t u˜ tt − u˜ yy + λ u˜ + λ 2

k(τ )u(λ, ˜ y, t − τ )dτ = 0, (λ, y, t) ∈ R2+ × R, (4.49)

2 0

 u˜ 

t y > 0 the regular part of solution to the problem (4.49)– (4.51) satisfies the equations vtt − v yy

t−y + λ v(λ, y, t) − λ k(t − y) + λ k(τ )v(λ, y, t − τ )dτ = 0, (4.52) 2

2

2

0

 λ2 y, v t=y+0 = 2  v y  y=0 = f (t).

(4.53) (4.54)

 t), having representation (4.16). Let F0 (λ, t) be a regular part of function F(λ, Narrowing down the problem data we assume that the function F0 (λ, t) (therefore,  t)) is known only for two values λ = λi , i = 1, 2. Thus, the inverse problem F(λ, (4.1)–(4.3) reduces to the problem of determining the functions k(t), f (t) from relations (4.52)–(4.54), provided that the solution of the direct problem is known for λ = λi , i = 1, 2, and satisfies the following condition:  v  y=0 = F0 (λi , t), t > 0, i = 1, 2.

(4.55)

It turns out that these data uniquely determine the functions k(t), f (t).

4.2.2 Properties of the Direct Problem Solution. Statement and Proof of the Main Results Let us study the properties of the solution of the direct problem (4.52)–(4.54). As well in this case it is true the analog of Lemma 4.1.

4.2 The Problem of Determining the Memory of Medium …

177

Lemma 4.3 Suppose that k(t) ∈ C[0, T ] and f (t) ∈ C[0, T ] for some  T > 0. Then, , D = (y, t)|0 ≤ y ≤ for each fixed value of the parameter λ and for (y, t) ∈ D T T  t ≤ T − y the solution of problem (4.52)–(4.54) belongs to the functional class C 1 (DT ) and the following estimate for the solution holds:   vC 1 (D(T ) ≤ C 1 + k(t)C[0,T ] +  f C[0,T ] ,

(4.56)

where C depends only on T , λ and k(t)C[0,T ] . In addition, for any fixed λ j , j = 1, 2, the function ψ(λ1 , λ2 , t) = vt (λ1 , 0, t) − vt (λ2 , 0, t) is a function of class C 1 [0, T ]. Proof Proceeding in a similar way, as in the previous paragraph, for (y, t) ∈ DT from relations (4.52)–(4.54) one gets λ2 + λ2 (vt + v y )(λ, y, t) = 2

(y+t)/2 

k(t + y − 2ξ ) − v(λ, ξ, t + y − ξ )

y

(4.57)

t+y−2ξ 

−



k(τ )v(λ, ξ, t + y − ξ − τ )dτ dξ, 0

λ2 v(λ, 0, t) = t − 2

t

 t τ/2 k(τ − 2ξ ) − v(λ, ξ, τ − ξ ) f (τ )dτ + λ 2

0

0

0

(4.58)

τ−2ξ

−



k(α)v(λ, ξ, τ − ξ − α)dα dξ dτ, 0

λ2 + λ2 k(t − y)y − 2 f (t − y) + λ2 (vt − v y )(λ, y, t) = 2

(t−y)/2 



k(t − y − 2ξ )

0 t−y−2ξ 

− v(λ, ξ, t − y − ξ ) −

k(τ )v(λ, ξ, t − y − ξ − τ )dτ dξ

0

y − λ2 0



t−y 

v(λ, ξ, t − y + ξ ) + k(τ )v(λ, ξ, t − y + ξ − τ )dτ dξ. 0

(4.59)

178

4 Stability in Inverse Problems for Determining of Two Unknowns

From (4.57), (4.59), we obtain λ2 λ2 λ2 + k(t − y)y − f (t − y) + vt (λ, y, t) = 2 2 2

(t−y)/2 

k(t − y − 2ξ )

0 t−y−2ξ 

− v(λ, ξ, t − y − ξ ) −

k(τ )v(λ, ξ, t − y − ξ − τ )dτ dξ

0

λ2 + 2

y

v(λ, ξ, t − y + ξ ) +

0

+

λ2 2

t−y

k(τ )v(λ, ξ, t − y + ξ − τ )dτ dξ 0

(y+t)/2 

k(t + y − 2ξ ) − v(λ, ξ, t + y − ξ )

y t+y−2ξ 

−

k(τ )v(λ, ξ, t + y − ξ − τ )dτ dξ,

0

v y (λ, y, t) = −

λ2 λ2 k(t − y)y + f (t − y) − 2 2

(4.60)

(t−y)/2 



k(t − y − 2ξ )

0 t−y−2ξ 

− v(λ, ξ, t − y − ξ ) −

k(τ )v(λ, ξ, t − y − ξ − τ )dτ dξ

0

λ2 − 2

y



t−y

v(λ, ξ, t − y + ξ ) + k(τ )v(λ, ξ, t − y + ξ − τ )dτ dξ

0

+

λ2 2

0

(y+t)/2 

k(t + y − 2ξ ) − v(λ, ξ, t + y − ξ )

y t+y−2ξ 

−

k(τ )v(λ, ξ, t + y − ξ − τ )dτ dξ,

0

(4.61)

4.2 The Problem of Determining the Memory of Medium …

t

λ2 y v(λ, y, t) = 2

λ2 k(τ − y)dτ + 2

y

1 − 2

t

t

179

(τ +y)/2

k(τ + y − 2ξ )dξ dτ y

y

λ2 f (τ − y)dτ + v(λ, 0, t − y) + 2

y

 t y y

v(λ, ξ, τ − y + ξ )

0

τ −y

k(α)v(λ, ξ, τ − y + ξ − α)dα dξ dτ

+ 0

t

λ2 − 2

(τ +y)/2

+v(λ, ξ, τ + y − ξ )

y

y

τ +y−2ξ 

+

k(α)v(λ, ξ, τ + y − ξ − α)dα dξ dτ.

0

(4.62) In the domain DT , the equation (4.62) is an integral equation of Volterra type and determines a unique continuous solution. It follows from relations (4.60), (4.61) that this solution is continuously differentiable in DT . Substituting the expression for v(λ, 0, t), given by formula (4.58), into equation (4.62) and constructing, for Eqs. (4.60)–(4.62), the usual scheme of the successive approximation method, which has factorial convergence with respect to the argument t, it is easy to prove that the estimate (4.56) holds in the domain DT . Using relation (4.60), we construct the function ψ(λ1 , λ2 , t) as follows λ21 − λ22 2 t/2 + [(λ21 − λ22 )k(t − 2ξ ) − λ21 v(λ1 , ξ, t − ξ ) + λ22 v(λ2 , ξ, t − ξ )

ψ(λ1 , λ2 , t) =

0 t−2ξ 

−

k(τ )[λ21 v(λ1 , ξ, t − ξ − τ ) − λ22 v(λ2 , ξ, t − ξ − τ )]dτ ]dξ. 0

(4.63) The right-hand side of (4.63) belongs to the class C 1 [0, T ]. Therefore, ψ(λ1 , λ2 , t) ∈ C 1 [0, T ] for any λ j , j = 1, 2.

180

4 Stability in Inverse Problems for Determining of Two Unknowns

As consequence of the Lemma, it is easily noted that under the assumptions of Lemma 4.3 the function F0 (λi , t) appearing in relation (4.55), belongs, for each fixed λ, to the class C 1 [0, T ], and the function F00 (t) ≡ F0 (λ1 , t) − F0 (λ2 , t) to the class C 2 [0, T ]. In addition, formulas (4.60), (4.62) imply that the following relations hold: F0 (λi , 0) = 0, F0t (λi , 0) =

λi2 − f (0), i = 1, 2. 2

In what follows, we assume that the function F0 (λ, t) satisfies these necessary conditions. The following lemma expresses the equivalence of the inverse problem to some system of nonlinear integral equations. Lemma 4.4 Suppose that the assumptions of Lemma 4.3, the necessary conditions on the function F0 and the relation λ21 = λ22 hold. Then, the inverse problem (4.52)–(4.55) in the domain DT is equivalent to the problem of finding the functions v, vt , k, f from the following system of integral equations: λ2j

v(λ j , y, t) = F0 (λ j , t − y) +

2

y+



λ2j 2

k(τ − ξ )

D(y,t)

τ −ξ − v(λ j , ξ, τ ) −



(4.64)

k(α)v(λ j , ξ, τ − α)dα dτ dξ, j = 1, 2, 0

vt (λ j , y, t) = F0t (λ j , t − y) +

λ2j

(t+y)/2 



2

k(t − |y − ξ | − ξ )

0 t−|y−ξ  |−ξ

− v(λ j , ξ, t − |y − ξ |) − 1 − k 2



k(α)v(λ j , ξ, t − |y − ξ | − α)dα 0

   

t − y  t − y  t − y  t − y  − ξ − v λ j , ξ, + ξ − + ξ − 2 2  2 2  

t−y  t−y  2 +ξ − 2 −ξ

 

t − y  t − y  − − α dα k(α)v λ j , ξ, + ξ −  2 2 0

 t−y dξ, × 1 − sign ξ − 2 



(4.65) for j=1,2 and

4.2 The Problem of Determining the Memory of Medium …

t 4  λ1 − λ42 2 2 λ21 + λ22 t + k(ξ )(t − ξ ) k(t) = 2 F (t) + 00tt 2 4 λ1 − λ22 λ21 − λ22 0

t −ξ t +ξ t −ξ t +ξ 2 2 + λ1 vt λ1 , , − λ2 vt λ2 , , 2 2 2 2 ξ

 t −ξ t +ξ + k(τ ) λ21 vt λ1 , , −τ 2 2 0

 t −ξ t +ξ , − τ dτ dξ. − λ22 vt λ2 , 2 2 λ2 λ2 f (t) = −F0t (λ, t) + + 2 2

t



−

(4.66)

t −ξ t +ξ , k(ξ ) − v λ, 2 2

0

ξ

181



t −ξ t +ξ , − τ dτ dξ, k(τ ) − v λ, 2 2

(4.67)

0

where     t−y  t − y  t+y   < τ < t − |ξ − y| , 0 < ξ < D(y, t) = (ξ, τ ) + ξ − . 2 2  2 The proof of this lemma differs from the proof of the similar lemma from the previous section. Therefore, here we will give it. Proof First we note that the system of the integral Eqs. (4.64)–(4.66) is closed in the domain DT and determines the unique continuous functions v, vt , k. After determination of these functions, the unknown function f (t) can be calculated by the formula (4.67). In formula (4.67), λ = λ1 or λ = λ2 and the result of calculation must not depend on the choice of the parameter λ. Now we consider the auxiliary problem of determining the function ϕ(y, t), satisfying relations: ϕtt − ϕ yy = γ (y, t), 0 < y < t < T − y, ϕ(0, t) = ϕ0 (t), 0 ≤ y ≤ T, ϕ(y, y) = ϕ1 (y), 0 ≤ y ≤ T, in which γ , ϕ0 , ϕ1 are continuous functions and ϕ0 (0) = ϕ1 (0). It is easily verified that the generalized solution of this problem is given by the formula

182

4 Stability in Inverse Problems for Determining of Two Unknowns

ϕ(y, t) = ϕ0 (t − y) + ϕ1

t+y 2



− ϕ1

t−y 2

+



1 2

γ (ξ, τ )dτ dξ, (4.68) D(y,t)

which can be verified directly. By formula (4.68) the solution of problem (4.52), (4.53),(4.55) satisfies the integral Eq. (4.64) in the domain DT . To obtain the Eq. (4.65) we write the integral over the domain D(y, t) as the repeated integral 

(t+y)/2 

t−|y−ξ  |

γ (ξ, τ )dτ dξ =

γ (ξ, τ )dτ dξ 0

D(y,t)

(t−y)/2+|ξ −(t−y)/2|

and, differentiating relation (4.64) with respect to the variable t, we obtain Eq. (4.65). Differentiating relation (4.64) with respect to the variable y, we find   1 1 v y (λ j , y, t) = −F0t λ j , t − y + λ2j + λ2j 2 2   − v λ j , ξ, t − |y − ξ | t−|y−ξ  |−ξ

− 0

1 + k 2



(t+y)/2 



k(t − |y − ξ | − ξ )

0

 

k(α)v λ j , ξ, t − |y − ξ | − α dα sign(ξ − y)

   

t − y  t − y  t − y  t − y  − ξ − v λ ξ − + ξ − + , ξ, j  2 2  2 2  

t−y  t−y  2 +ξ − 2 −ξ

 



t − y  t − y  − α dα + ξ − k(α)v λ j , ξ,  2 2 0

 t−y × 1 − sign ξ − dξ, 2 

−

(4.69) for j = 1, 2. In this equation, setting y = 0 and using relation (4.54) we obtain equality (4.67). Considering relation (4.54) for λ = λ j , j = 1, 2, and forming the difference, we get (4.70) v y (λ1 , 0, t) − v y (λ2 , 0, t) = 0. Using relation (4.70) for y = 0, from Eq. (4.69) we obtain

4.2 The Problem of Determining the Memory of Medium …

183

F0t (λ1 , t) − F0t (λ2 , t) λ2 − λ22 + = 1 2

t/2  2 (λ1 − λ22 )k(t − 2ξ ) − λ21 v(λ1 , ξ, t − ξ ) + λ22 v(λ2 , ξ, t − ξ ) 0

t−2ξ 

−



 k(τ ) λ21 v (λ1 , ξ, t − ξ − τ ) − λ22 v (λ2 , ξ, t − ξ − τ ) dτ dτ dξ.

0

(4.71) Differentiating this equality with respect to the variable t, and carrying out simple calculations, we obtain Eq. (4.66). Note that the unknown function f (t) does not appear in relations (4.64)–(4.66). It is readily verified that the inverse transformations also hold. Indeed, integrating Eq. (4.65) over t between the limits from y to t and changing, where it is necessary, the order of integration, we obtain Eq. (4.64) from which, in turn, we derive relations (4.52), (4.53), (4.55) for λ = λ j . Using the formula ξ  = (t − ξ )/2, we perform the change of variable ξ to ξ  in the integrals on the right-hand side of (4.66). Further, on both sides of this equality, we replace t by t − 2t, multiply both sides by dt, and integrate over t between the limits from 0 to t/2. Then changing the order of integration in the repeated integrals of the resulting equality, after simple calculations, we obtain Eq. (4.71). The equivalence of relations (4.67) and (4.54) is proved in a similar way. The Lemma 4.4 is proved. Omitting the theorem on the local unique solvability of our problem, now we pass to formulate the results concerning stability estimates and unique solvability for an arbitrary T > 0. Denote by B(k0 ) the set of pairs of functions (k(t), f (t)) satisfying the following condition for some T > 0 :   max kC[0,T ] ,  f C[0,T ] ≤ k0 , k0 > 0.     Theorem 4.3 Suppose that k (1) , f (1) ∈ B(k0 ), k (2) , f (2) ∈ B(k0 ) are solutions of the inverse problem (4.52)–(4.55) with given F0(1) (λ j , t), F0(2) (λ j , t), j = 1, 2, (i) ≡ F0(i) (λ1 , t) − F0(i) (λ2 , t) and λ21 = λ22 . Then, there exists a posirespectively, F00 tive constant C depending on λ1 , λ2 , T, k0 , such that the following estimate holds: k (1) − k (2) C[0,T ] +  f (1) − f (2) C[0,T ] ⎛ ⎞ 2  (4.72) (1) (2) 00 00 ≤C⎝ F0(1) (λ j , t) − F0(2) (λ j , t)C 1 [0,T ] +  F −F C 2 [0,T ] ⎠ . j=1

Theorem 4.3 readily implies the following uniqueness theorem for any T > 0.

184

4 Stability in Inverse Problems for Determining of Two Unknowns

    Theorem 4.4 Suppose that the functions k (i) , f (i) ∈ C[0, T ] and F0(i) λ j , t , i = 1, 2, j = 1, 2, have the same meaning as in Theorem 4.3. If F0(1) (λ j , t) = F0(2) (λ j , t), j = 1, 2 for t ∈ [0, T ], then k (1) = k (2) , f (1) = f (2) for t ∈ [0, T ]. The proof of Theorem 4.3 can be carried out quite similarly to the proof of the corresponding theorem from the previous section. Therefore, we will omit it.

4.3 Conclusions This chapter investigated two inverse problems of simultaneously determining two unknown functions. In both problems, it is shown that both unknown functions of one variable are uniquely determined by specifying the Fourier transform of the direct problem on the boundary of the half-space with respect to the variable x for two different transformation parameters. The conditional stability estimates and uniqueness theorems of the solution to the problems were obtained.

References 1. Yakhno, V.G. 1990. Inverse Problems for Differential Equations of Elasticity. Novosibirsk: Nauka (Russian). 2. Karchevsky, A.L., and V.G. Yakhno. 1999. One-dimensional inverse problems for systems of elasticity with a source of explosive type. Journal of Investigation Ill - Posed Problems 7 (4): 347–364.

Chapter 5

Two-Dimensional Special Kernel Determination Problems

This chapter is devoted to the inverse problems of recovering a memory kernel in hyperbolic integro-differential equations, having a finite Fourier series with respect to one of the spatial variables. In the first section it is supposed that the unknown kernel is a trigonometric polynomial with respect to the spatial variables with continuous coefficients with respect to the time variable. In the second section it is determined the two-dimensional spatial part of the three-dimensional kernel and wherein it is supposed that the unknown function is a trigonometric polynomial with respect to the spatial variable y with continuous coefficients with respect to the variable x. The direct problem for a hyperbolic integro-differential equation in both cases is the initial–boundary value problem for the half-space x > 0 with the zero initial Cauchy data and a special Neumann’s data at x = 0. Local solvability theorems and stability estimates for the solution to the inverse problem are obtained [1] and [2].

5.1 A Problem of Identification of a Two-Dimensional Kernel In this section it is assumed that the time convolution kernel depends on one of the spatial variables, and it has a form of trigonometric polynomial with respect to this variable with continuous coefficients in t-variable.

5.1.1 Setting of the Problem We consider the integro-differential equation for (x, y, t) ∈ R3+ © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_5

185

186

5 Two-Dimensional Special Kernel Determination Problems

∂ 2u − u = ∂t 2

t k(y, τ )u(x, y, t − τ )dτ

(5.1)

0

with the initial and boundary conditions   u

t 0 . Boundary condition in (5.2) is generalized function and, that is why the solution to the problem (5.1) and (5.2) is understood as a linear functional over the space of infinitely differentiable functions with compact support (test functions),   i.e., u(x, y, t) ∈ D  R3+ . Therefore, equalities in equations (1.1 ), (1.2) are understood as the equality of values of the left and right sides for test functions. We note that the kernel in equation (5.1) does not depend on the spatial variable which is perpendicular to the axis where the boundary condition in (5.2) is given. Moreover, we assume that this kernel can be represented in the form of a finite Fourier series N  k(y, t) = ks (t)eisy (5.3) 2

2

s=−N

with a fixed integer N ≥ 0. Denote by  (N , T, K ) the set of functions k(y, t) for which the coefficients ks , |s| ≤ N , are continuous functions on the interval [0, T ] and satisfy the conditions |ks (t)| ≤ K , t ∈ [0, T ], −N ≤ s ≤ N .

(5.4)

For k(y, t) ∈  (N , T, K ) the solution of problem (5.1) and (5.2) is a 2π −periodic function of y and can be represented by the Fourier series u(x, y, t) =

∞ 

u s (x, t)eisy ,

(5.5)

s=−∞

whereas it follows from relations (5.1)–(5.3) the coefficients satisfy the following equations

5.1 A Problem of Identification of a Two-Dimensional Kernel

t  N ∂2 ∂2 2 u − + m (x, t) = ks (τ )u m−s (x, t − τ )dτ, m ∂t 2 ∂x2 s=−N 0   (x, t) ∈ R2+ = (x, t) ∈ R2 | x > 0 ∂u m |x=0 = δ0m δ  (t), m = 0, ±1, ±2, . . . . u m |t 0} , m = 0, ±1, ±2, . . . , ±N . (5.7) Definition Functions ks (t) ∈ C [0, ∞) , s = 0, ±1, ±2, . . . , ±N are called a solution to inverse problem (5.6) and (5.7), if the the corresponding solution to direct problem (5.7) u m (x, t) ∈ D  R2+ (from the class of generalized functions, i.e., distributions) satisfies equalities (1.7) for f m (t) ∈ D  (R+ ) , m = 0, ±1, ±2, . . . , ±N . Remark It turns out that the solution to direct problem (1.6) at m = 0 consists of a singular and a regular generalized function. With that at m = ±1, ±2, . . . the solution to this problem are only regular generalized functions. The regular parts of the solutions are smooth in some domain (see Lemma 2.1).

5.1.2 Investigation of the Direct Problem We begin studying the inverse problem with consideration of some properties of solution to direct problems (5.6). Concerning the solution to direct problems (5.6) the following lemma is valid. Lemma 5.1 Let T be an arbitrary positive number, ks (y, t) ∈  (N , T, K ) and D(T ) = ((x, t)| 0 ≤ x ≤ T − t) . Then the solution to problem (5.6) exists and can be represented in D(T ) in the form u m (x, t) = −δ0m δ(t − x) + vm (x, t)θ (t − x), m = 0, ±1, ±2, . . . ,

(5.8)

here θ (t) is the Heaviside step-function: θ (t) = 1 for t ≥ 0; θ (t) = 0 for t < 0 and vm (x, t) are continuously differentiable functions in the domain D  (T ) = ((x, t)| 0 ≤ x ≤ t ≤ T − x) . Moreover, this solution is unique and there exist

188

5 Two-Dimensional Special Kernel Determination Problems

positive constants C1 = C1 (N , T, K ) ≥ 1 and C2 = C2 (N , T, K ), continuously depending on K , T such that KT2 , −N ≤ m ≤ N , 2  n+1 KT2 (2N + 1)n t n |vm (x, t)| ≤ C1 , 2n+1 · n!

|vm (x, t)| ≤ C1

if

(5.9)

(x, t) ∈ D  (T ), N n < |m| ≤ (n + 1)N , n = 1, 2, . . . ,

and    ∂vm (x, t)  K T (1 + m 2 T 2 )  ≤ [1 + (2N + 1)C2 ],   ∂t 2    ∂vm (x, t)   ≤ (1 + m 2 T 2 ) − N ≤ m ≤ N ,   ∂t (K T )n+1 (2N + 1)n t n max {C1 , (2N + 1)C2 } , 2n+1 · n! if

(5.10)

(x, t) ∈ D  (T ), N n < |m| ≤ (n + 1)N , n = 1, 2, . . . .

Proof It follows from the hyperbolic equation theory that the solution to the problem (5.1) and (5.2) vanishes for all (x, y, t) satisfying the condition x > t > 0 because the initial data are zero and the boundary source is located on the axis x = 0,t = 0. Hence all u m (x, t) = 0 for x > t > 0. To separate the singular part of the solution to direct problem, we represent it in the form u m (x, t) = αm δ(t − x) + vm (x, t)θ (t − x), m = 0, ±1, ±2, . . . , where αm are unknown constants, vm (x, t) are unknown regular functions. Substituting this equality into equations (5.6) and equating the coefficients for the same singularities, we find αm = −δ0m . In view of representation (5.8), we rewrite equations (5.6) with respect to vm (x, t). For this, we use the property of the Dirac function, from which it follows equality N  s=−N

t δ0(m−s)

ks (τ )δ(t − x − τ )dτ = θ (N − |m|)km (t − x), t > x > 0. 0

5.1 A Problem of Identification of a Two-Dimensional Kernel

189

Taking this fact into consideration, we conclude that equations (5.6) with respect to vm (x, t) can be rewritten in the form

∂2 ∂2 2 − 2 + m vm (x, t) ∂t 2 ∂x

t−x N  ks (τ )vm−s (x, t − τ )dτ, = −θ (N − |m|)km (t − x) + s=−N 0

t > x > 0, m = 0, ±1, ±2, . . . , ∂vm |x=0 = 0, m = 0, ±1, ±2, . . . . vm |t≤x ≡ 0, ∂x

(5.11)

For the sake of convenience, we continue all functions vm (x, t), for x < 0 as even functions: vm (−x, t) = vm (x, t). Then problem (5.11) is equivalent to the following integral equation vm (x, t) = −

θ (N − |m|) 2





J0 m (t − τ )2 − (x − ξ )2 km (τ − |ξ |)dτ dξ

♦(x,t)

+

1 2





τ−|ξ |  N J0 m (t − τ )2 − (x − ξ )2 ks (α)vm−s (ξ, τ − α)dαdτ dξ,

♦(x,t)

0

s=−N

(x, t) ∈ D  (T ), m = 0, ±1, ±2, . . . ,

(5.12)

where J0 (ζ ) is the Bessel function and

x +t x −t ≤ξ ≤ . ♦(x, t) = (ξ, τ ) | |ξ | ≤ τ ≤ t − |x − ξ |, 2 2 Recall that the Bessel function Jν (ζ ) for a fixed integer ν ≥ 0 is defined by the formula 2 j+ν ∞  ζ (−1) j Jν (ζ ) = . j!( j + ν)! 2 j=0 From this formula follows the estimate    Jν (ζ )  1    ζ ν  ≤ 2ν ν! , |ζ | ≤ 2.

(5.13)

190

5 Two-Dimensional Special Kernel Determination Problems

Hereafter we will also use another estimate for Jν [3]: |Jν (ζ )| ≤ 1 for all ζ ∈ R. Consider for equations (5.12) the method of successive approximations. Define ∞  vmj (x, t), (5.14) vm (x, t) = j=0

where vm0 (x, t)

θ (N − |m|) =− 2



   J0 m (t − τ )2 − (x − ξ )2 km (τ − |ξ |)dτ dξ,

♦(x,t)

vmj (x, t) 1 = 2



τ −|ξ | N      j−1 2 2 J0 m (t − τ ) − (x − ξ ) ks (α)vm−s (ξ, τ − α)dαdτ dξ,

♦(x,t)

0

s=−N



(x, t) ∈ D (T ), j = 1, 2, . . . , m = 0, ±1, ±2, . . . .

(5.15)

Obviously all functions vm (x, t) are continuous in D  (T ). Moreover, estimating these functions, we get the following estimates j

  0 v (x, t) ≤ θ (N − |m|) m 2      KT2   , × J0 m (t − τ )2 − (x − ξ )2  |km (τ − |ξ |)| dτ dξ ≤ θ (N − |m|) 2 ♦(x,t)

2  1  v (x, t) ≤ K T m 2

t  N 0

  0 K 2 T 4 (2N + 1)t max vm−s (ξ, τ ) dτ ≤ θ (2N − |m|) , 22 |ξ |≤ T2 s=−N

(x, t) ∈ D  (T ). Continuing these estimates, we easily obtain that 2  j  v (x, t) ≤ θ (( j + 1)N − |m|) K T m 2  (x, t) ∈ D (T ), j = 0, 1, 2, . . . .



K T 2 (2N + 1)t 2

j

1 , j!

5.1 A Problem of Identification of a Two-Dimensional Kernel

191

Since t ≤ T series (5.14) uniformly converges in (x, t) ∈ D  (T ) for all m. Hence, its sum is a continuous function in D  (T ). Moreover, the following estimates hold ∞   KT2    j v0 (x, t) ≤ 2 j=0

j ∞  K T 2 (2N + 1))t 1 KT2 ≤ C1 , 2 j! 2 j=0

j ∞ ∞ 2     j K T 2 (2N + 1)t 1 v (x, t) ≤ K T |vm (x, t)| ≤ m 2 j=n 2 j! j=n   n+1 (2N + 1)n t n KT2 C1 , (x, t) ∈ D  (T ), ≤ 2n+1 n! N n < |m| ≤ (n + 1)N , n = 0, 1, 2, . . . ,

|v0 (x, t)| ≤

(5.16)

  where C1 = exp K T 2 (2N + 1)/2 . Now differentiating equations (5.12) with respect to t and x, we easily check that functions vm (x, t) are continuously differentiable in D  (T ). We check it for the derivatives with respect to t only. The expressions for these derivatives will be useful in the analysis of the inverse problem. From (5.12), we find x+t

θ (N − |m|) ∂vm (x, t) =− ∂t 2 t−|x−ξ  |

+

2  km (t − |ξ | − |x − ξ |) x−t 2

 Hm (t − τ, x − ξ )km (τ − |ξ |)dτ dξ

|ξ | x+t

1 + 2

2

t−|ξ|−|x−ξ | s=N 

x−t 2

1 + 2

0

ks (τ )vm−s (ξ, t − |x − ξ | − τ )dτ dξ

s=−N

 Hm (t − τ, x − ξ ) ♦(x,t)

τ−|ξ | s=N  0



ks (α)vm−s (ξ, τ − α)dαdτ dξ,

s=−N

(x, t) ∈ D (T ), m = 0, ±1, ±2, . . . ,

(5.17)

192

5 Two-Dimensional Special Kernel Determination Problems

where Hm (t, x) =

 ∂  2 J1 (ζ ) J0 m t − x 2 = −m 2 t |ζ =m √t 2 −x 2 . ∂t ζ

It follows from the definition of the Bessel function that J1 (ζ )/ζ is a continuous function for all ζ ∈ [0, ∞) and lim

ζ →0

1 J1 (ζ ) = . ζ 2

(5.18)

From relation (5.12) we see that the derivatives ∂vm (x, t)/∂t are, indeed, continuous in D  (T ) for all m. Since from (5.13) it follows that |J1 (ζ )/ζ | ≤ 1/2 for all ζ ∈ R, we have |Hm (t, x)| ≤ m 2 T /2. Therefore we can estimate ∂vm (x, t)/∂t as follows    ∂vm (x, t)      ∂t ≤

⎡

K T (1 + m T ) ⎣ θ (N − |m|) + 2 2

2

t  N 0

s=−N

⎤ max

ξ ∈ (x,t,τ )

|vm−s (ξ, τ )| dτ ⎦ ,

(x, t) ∈ D  (T ), m = 0, ±1, ±2, . . . , where (x, t, τ ) = {ξ : (ξ, τ ) ∈ ♦(x, t)} . Using (5.9), we get the following estimates    ∂vm (x, t)  K T (1 + m 2 T 2 ) ≤    ∂t 2 t   |vs (ξ, τ )| , × 1 + (2N + 1) max max ξ ∈ (x,t,τ ),|s|≤N

0

 K T (1 + m 2 T 2 ) |vs (ξ, τ )| dτ ≤ max ξ ∈ (x,t,τ ),N 0, |m| ≤ N .

(5.46)

Using these, from the differentiated with respect t equation (5.44) after some transformations, we get  (t) = gm t

2 =−

t−ξ    pm (ξ ) k0 (t − 2ξ ) + pm (ξ ) Hm (t − τ, ξ )k0 (τ − ξ )dτ dξ ξ

0 t 2

N 

+

ps (ξ )

s=−N

0

+

t−ξ 

k0 (t − ξ − τ )vm−s (ξ, τ )dτ dξ ξ τ−ξ

t−ξ 

N 

ps (ξ )

s=−N

Hm (t − τ, ξ ) ξ

k0 (α)vm−s (ξ, τ − α)dαdτ dξ, |m| ≤ N .

0

By differentiating this equality and denoting for pm (x) :

t 2

= x we obtain the integral equation

pm (x) = =

−2gm (2x)

x −2

 m2ξ pm (ξ ) k0 (2x − 2ξ ) − k0 (2x − 2ξ ) 2

0 2x−ξ 

+

 Hm (2x − τ, ξ )k0 (τ − ξ )dτ dξ

ξ

−

N  s=−N

2x−ξ    ps (ξ ) vm−s (ξ, 2x − ξ ) + k0 (2x − ξ − τ )vm−s (ξ, τ )dτ dξ ξ

208

5 Two-Dimensional Special Kernel Determination Problems 2x−2ξ  N m2ξ  + ps (ξ ) k0 (τ )vm−s (ξ, 2x − ξ − τ )dτ dξ 2 s=−N 0

2x−ξ 

−

Hm (2x

ξ

N 

− τ, ξ )

τ −ξ ps (ξ )

s=−N

k0 (α)vm−s (ξ, τ − α)dαdτ dξ, |m| ≤ N , 0

(5.47) where Hm (t, x) =

  J1 (ζ ) J2 (ζ ) ∂ Hm (t, x) = −m 2 − m2t 2 2 . √ ∂t ζ ζ ζ = t 2 −x 2

It should be noted that integral equations (5.47) determine vm (x, t) as a function of pm (x), m = −N , . . . , N . Hence, introducing the operator U = (U−N , . . . , U N ) by the right sides of (5.47), we can rewrite relations (5.47) as the operator equations pm (x) = Um ( p−N (x), . . . , p N (x)) , m = 0, ±1, ±2, . . . , ±N , t ∈ [0, T /2] . (5.48) Let pm0 (x) = −2gm (2x), m = 0, ±1, ±2, . . . , ±N . Denote by 0 (N , X, P0 ) the set of functions pm (x), −N ≤ m ≤ N , satisfying the conditions    pm (x) − p 0 (x) ≤ P0 , P0 = m C[0,X ]

  max  pm0 (x)C[0,X ] ,

−N ≤m≤N

m = 0, ±1, ±2, . . . , ±N . If functions pm (x) ∈ 0 (N , X, P0 ) for −N ≤ m ≤ N then, obviously, pm (x) ∈  (N , X, 2P0 ) . For operator equations (5.48) the following theorem holds. Theorem 5.2 Let presentation (5.45) be valid. Moreover, let data (5.46) satisfy the following conditions: gm (t) ∈ C 2 [0, T ], gm (0) = gm (0) = 0, m = 0, ±1, ±2, . . . , ±N . and G=

  max gm (t)C[0,T ] .

−N ≤m≤N

Then, there exists a number T0 ∈ (0, T ) such that operator equations (5.48) have a unique solution, belonging to the set 0 (N , T0 /2, 2G) .

References

209

The proof of the theorem can be fulfilled quite identically to the corresponding theorem from the previous section. Therefore, we omit it here.

5.3 Conclusions This chapter dealt with the inverse problems of recovering a memory kernel in hyperbolic integro-differential equations, having a special dependence with respect to one of the spatial variables. In the first section it is supposed that the unknown kernel is a trigonometric polynomial in the spatial variables with continuous coefficients with respect to the time variable. In the second section it is determined the twodimensional spatial part of the three-dimensional kernel, and wherein it is supposed that the unknown function is a trigonometric polynomial with respect to one of the spatial variables with continuous coefficients with respect to another. Local solvability theorems and stability estimates for the solution to the inverse problem are obtained.

References 1. Durdiev, D.K., and U.D. Durdiev. 2016. The problem of kernel determination from viscoelasticity system integro-differential equations for homogeneous anisotropic media. Nanosystems: Physics, Chemistry, Mathematics 7 (3): 405–409. 2. Durdiev, U.D., and Z.D. Totieva. 2019. A problem of determining a special spatial part of 3D memory kernel in an integro-differential hyperbolic equation. Mathematical methods in Applied Sciences 42 (18): 7440–7451. 3. Romanov, V.G. 2016. A problem of recovering a special two-dimention potential in a hyperbolic equation. Eurasian Journal of Mathematical and Computer Applications 7 (6): 573–588.

Chapter 6

Some Inverse Problems of Viscoelasticity System with the Known Kernel

In this chapter, we consider various inverse problems of viscoelasticity system with the known kernel. The inverse problems for the system of elasticity equations have been considered by many authors. A detailed review of the sources, available in Russian and foreign literature before 1990, is given in the monograph [1], which presents the results of unique solvability, stability estimates, and numerical methods for constructing solutions for many formulations of inverse problems of isotropic and anisotropic elasticity. These studies were continued in the article [2]. It deals with direct and inverse coefficient problems of anisotropic electroelastic medium of cubic structure. Namely, the elastic modulus and piezoelectric modulus are determined from some information about the solution of direct problems. Note also works [3–9], in which the Lame parameters are defined for the equations of isotropic elasticity. The modules of elasticity and piezo module (polarized ceramic) of anisotropic electro-elasticity are also defined there. One can mention the work [10], related to taking into account the memory effect. Under study is an inverse problem of determining a spatially varying source term in a thermoelastic medium with a memory effect. Based on Carleman’s estimate, authors establish a Hölder stability for the inverse source problem only by making a displacement measurement on a given subdomain for sufficiently large times, provided the source is to be known near the boundary. The uniqueness of such an inverse problem is yielded as a direct result. In the first section of this chapter, we consider a multidimensional inverse problem related to the determination of the function ρ(x2 , x3 ), which is the coefficient of the system of integro-differential equations of viscoelasticity (the system of differential Lame equations with memory) in the linear approximation. Some information regarding the solution of the direct linearized problem is given.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_6

211

212

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

The linearization method in inverse problems for hyperbolic equations has been used by some authors. One of the first works was the work of Lavrentiev and Romanov [11], and later, the inverse Lamb problem was solved in the linear approximation [12]. Kabanikhin [13] applied the linearization method to solve two-dimensional inverse problems for the wave equation. In the monograph, Yakhno [1] the linearization principle has found its application in the study of multidimensional inverse problems of determining density for a system of dynamic equations of elasticity. This monograph describes the arguments that allow us to introduce the so-called linearized statements of direct and inverse problems. First, the one-dimensional inverse problem of determining the density for a system of equations of isotropic elasticity in half-space was studied by Alekseev [14, 15]. Various statements of coefficient inverse problems of elasticity theory are presented in the works of Blagoveshchensky [16–18], Romanov and Volkova [19]. The results of these studies were reflected in the works of Yakhno and Melnikova [20], in particular, the existence, uniqueness, and stability theorems of multidimensional problems of determining the density function for systems of equations of isotropic and anisotropic elasticity. The results obtained are the basis of the theory of inverse problems with pulsed perturbation sources at the boundary of media. In many articles published over the past ten years, the works [21, 22] can be noted as close to the material of this chapter. In [21] for an integro-differential equation (two-dimensional viscoelasticity problem), the problem of determining the density, elastic modulus, and the spatial part of the kernel is studied, and the unknown functions differ from the given constants only inside the unit circle D = {x ∈ R2 | |x| < 1}. As additional information, one considered a set of solutions to the direct problem, which corresponds to the pulse sources localized on straight lines. Solution traces for a finite time interval are specified on the boundary of the domain D. The problem of determining the Lame coefficients and two kernels of the viscoelasticity system in the three-dimensional case for a given density of the medium is solved in [22]. The second section deals with the problem of determining the thermal-expansion coefficient α(z), z ∈ [0, Z ], occurring in the system of integro-differential termoviscoelasticity equations. The medium density and the Lame parameters are assumed to be functions of one variable. The integrand h(t), t ∈ [0; T ] is known. The inverse problem is replaced by the equivalent integral equation for unknown functions. The theorem of unique solvability is proved, and the stability estimate of solving the inverse problem is obtained. In the third section, we investigate the problem of determining the moduli of elasticity c11 (x3 ), c12 (x3 ), c44 (x3 ), x3 > 0, occurring in the system of integrodifferential viscoelasticity equations for homogenous anisotropic medium. The density of medium is constant. The matrix kernel k(t) = diag(k1 , k2 , k3 )(t), t ∈ [0, T ] is known. Additional information is the Fourier transform of the first and third component of the displacements vector for x3 = 0. The results are the theorems of the existence of a unique solution to the inverse problems and the theorems of stability. Note that the problems discussed in this chapter are primarily interesting for practical purposes (in numerical implementation and analyses of direct problem solution with various kernels).

6.1 The Multidimensional Linearized Problem of Determining…

213

6.1 The Multidimensional Linearized Problem of Determining the Density Function for a Viscoelasticity System 6.1.1 Statement of Problems and Main Results The purpose of the research presented in this section is to solve the inverse problem of determining the density of the medium ρ(x2 , x3 ), x2 ∈ R, x3 > 0 for the viscoelasticity system (three-dimensional case) from some information about the solution of the direct problem. The Dirac delta function, defined at the boundary of the considered spatial domain, is used as the source of wave propagation. This study is based on the results of monographs [1, 23]. The main results are theorems on necessary and sufficient conditions for the existence of a unique solution to the inverse problem, the stability theorem in a special class of functions, and the convergence theorem. The direct problem is pre-investigated to the extent necessary for the study of the inverse problem. The fundamental solution for the hyperbolic operator with memory is studied. Let’s consider a system of integro-differential equations in the domain  = {(x1 , x2 , t) ∈ R3 , x3 > 0}  ∂ Ti0j ∂ 2 u i0 = , i = 1, 2, 3, ∂t 2 ∂x j j=1 3

ρ0 (x3 )

 u i0 t 0 is fixed.

(6.6)

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

214

We also assume that functions ρ = ρ(x3 ), μ = μ(x3 ), λ = λ(x3 ) are known and belong to the set :   2

: = ρ(x3 ), λ(x3 ), μ(x3 ) ∈ C ([0, ∞))ρ(x3 ) > 0, μ(x3 ) > 0, λ(x3 ) > 0;

ρ  (+0) = μ (+0) = λ (+0) = 0 .

The system of Eqs. (6.1)–(6.5) arises in the theory of viscoelastic media with variable density ρ and the Lame coefficients λ, μ. In applications the integral term responsible for the prehistory of the process enters the system of equations by means of the formula (6.4). Direct problem for an isotropic viscoelasticity system (the case of vertically inho  mogeneous media): determine the vector function u 0 (x, t) = u 01 (x, t), u 02 (x, t), u 03 (x, t) , satisfying the Eqs. (6.1)–(6.3) in the domain  for given k(t) with (6.6) and ρ0 (x3 ), μ0 (x3 ), λ0 (x3 ) ∈ . Let ρ1 (x2 , x3 ), k(t) are known functions, k(t) satisfies conditions (6.6), u 0 (x, t) is the solution (6.1)–(6.3). Direct linearized problem system: determine the  for an isotropic viscoelasticity  vector function u 1 (x, t) = u 11 (x, t), u 12 (x, t), u 13 (x, t) , satisfying in the domain  the following equations  ∂ Ti1j ∂ 2 u i1 ∂ 2 u i0 = − ρ (x , x ) , i = 1, 2, 3, 1 2 3 ∂t 2 ∂x j ∂t 2 j=1 3

ρ0 (x3 )

(6.7)

 u i1 t 0 the inverse problem is correct; (2) if the data are such that the solution of the original (correct) problem exists, then, at s → ∞ a sequence of solutions the problem of the family with data Hs (ν, t) tends to the solution of the original (ill-posed) problem.

6.1.2 Direct Problem of Isotropic Viscoelasticity and Properties of Its Solution To prove Lemma 6.1, we first prove one auxiliary statement. Lemma 6.3 Let T > 0 be fixed, V 1 (x, t) = (V11 (x, t), V21 (x, t), V31 (x, t)) is defined   1  in domain x ∈ R3 , t ∈ [0, T ], belongs to C (0, T ); W21 (R3 ) , ∂∂tV (x, t) ∈ C (0, T );  W20 (R3 ) and satisfy the equalities:  ∂ Ti∗j ∂ 2 Vi1 = , i = 1, 2, 3, ∂t 2 ∂x j j=1 3

ρ0 (x3 )

(6.16)

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

218

 Vi1 t 0, k1 (t) ∈ C 2 [0, T ]. Then, for sufficiently small T , there exists a unique solution of the inverse problem 1*, which belongs to C[−T /2, T /2]. ˜ 0 , T ) = {q(y) : q C[−T /2,T /2] ≤ q0 . We define the set Q(q Theorem 6.13 (On the stability of the inverse problem 1*). Let q (1) (y), q (2) (y) ∈ ˜ 0 , T ) two solutions of the inverse problem 1* with data {g1(1) (t), k1(1) (t)}, Q(q {g1(2) (t), k1(2) (t)}, respectively. Then there is a positive number C ∗ = C ∗ (q0 , a(0), a  (0), M1 , ρ, T ), M1 = max{ g1(1) (t) 2 (T ), k1(1) (t) 2 (T ), g1(2) (t) 2 (T ), (2) k1 (t) 2 (T )}, which is a valid stability estimate

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation

253

  q (1) − q (2) C[−T /2,T /2] ≤ C ∗ g1(1) − g1(2) 2 (T ) + k1(1) − k1(2) 2 ((T ) . (6.160) Now we prove Theorem 6.12. Proof Differentiate by t the equality (6.158) and (6.159): y+t

∂w1∗ 1 (y, t) = ∂t 2

2  q(ξ ) w1∗ (ξ, t − |y − ξ |) − F(t − |y − ξ | − |ξ |) y−t 2

t−|y−ξ  |−|ξ |

 r1 (η) w1∗ (ξ, t − |y − ξ | − η)

−

(6.161)

0

− F(t − |y − ξ | − |ξ | − η) dη dξ, y+t

∂w2 1 (y, t) = ∂t 2

2  a(0) q(ξ ) + w2 (ξ, t − |y − ξ |) 2 y−t 2

 a(0)  + w2 (ξ, t − |y − ξ | − η) dη dξ, r1 (η) 2

t−|y−ξ  |−|ξ |

−

(6.162)

0 y+t

1 ∂ 2 w1∗ (y, t) = 2 ∂t 2

2  ∂w ∗ 1 q(ξ ) (ξ, t − |y − ξ |) − f (t − |y − ξ | − |ξ |) ∂t

y−t 2

t−|y−ξ  |−|ξ |

r1 (η) ×

−

 ∂w ∗

0

1

∂t

(ξ, t − |y − ξ | − η)

(6.163)

− f (t − |y − ξ | − η − |ξ |) dη dξ,  

 ∂ 2 w2 y+t y−t a(0) q +q (y, t) = ∂t 2 8 2 2 y+t

2 ∂w2 a(0)  1 q(ξ ) (ξ, t − |y − ξ |) − r (t − |y − |ξ |) + (6.164) 2 ∂t 2 1 y−t 2

t−|y−ξ  |−|ξ |

∂w2 (ξ, t r1 (η)

− 0

∂t

− |y − ξ | − η)dη dξ,

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

254

Note that from the presentation (6.147) it follows that w˜  (0, t) = (w1∗ ) (0, t) + w2  (0, t) +

a  (0) L[k1 , g1 (t)]. 2a(0)

On the other hand, from (6.145) we get w(0, ˜ t) = L[k1 , g1 (t)] − g1 (t)

t +

k1 (t − τ )g1 (τ )dτ = (w1∗ ) (0, t) + w2  (0, t) +

a(0) , 2

therefore,

a  (0) L[k1 , g1 (t)], 2a(0)

0

(w1∗ ) (0, t)



+ w2 (0, t) =

g1 (t)

−



 +

t

⎡0

(6.165) k1 (t

− τ )g1 (τ )dτ

a (0) ⎣  g1 (t) + 2a(0)

t

⎤ k1 (t − τ )g1 (τ )dτ ⎦ .

(6.166)

0

From (6.165) for t = 0 we find the value a  (0) = 4g1 (0), because from (6.145) we get a(0) = 2g1 (0). Thus, the values a(0), a  (0) are known. From (6.163), (6.164) for y = 0 and (6.166) we obtain an integral equation for q(y): ⎤ ⎡ 2y 2a  (0) ⎣  q(y) = 2 g1 (2|y|) + sign(y) k1 (2|y| − τ )g1 (τ )dτ ⎦ a (0) 0 ⎤ ⎡ 2y 4 ⎣  − g1 (2|y|) + sign(y) k1 (2|y| − τ )g1 (τ )dτ ⎦ a(0) 0

y  ∂w ∗ 4 1 − q(ξ ) sign(y) (ξ, 2|y| − ξ ) − f (2|y| − 2ξ ) a(0) ∂t 0

2|y|−2ξ 

r1 (η)

−

 ∂w ∗ 1

∂t

0

4 sign(y) − a(0)

(ξ, 2|y| − ξ − η) − f (2|y| − 2ξ − η) dη dξ

y ∂w2 a(0)  q(ξ ) (ξ, 2|y| − ξ ) − r (2|y| − 2ξ ) ∂t 2 1 0

2|y|−2ξ 

r1 (η)

− 0

∂w2 (ξ, 2|y| − ξ − η)dη dξ. ∂t (6.167)

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation

255

Let ♦(y, t) = {(ξ, τ ) : |ξ | ≤ τ ≤ t − |y − ξ |} and consider the equalities (6.158), (6.159), (6.161), (6.161), (6.167) in the domain ♦(0, T ). These equations are a closed system of nonlinear integral of the second kind with respect to func  ∗equations  ∂w2  ∂w1 ∗ tions w1 (y, t), w2 (y, t), ∂t (y, t), ∂t (y, t), q(y). Express (6.158), (6.159), (6.161), (6.161), (6.167) as the operator equation ϕ = Aϕ, where

(6.168)

  ∂w2 ∂w1∗ (y, t), (y, t), q(y) ϕ := w1∗ (y, t), w2 (y, t), ∂t ∂t

is a vector function with components ϕi (i = 1, 2, 3, 4, 5), while the operator A defined on the set of functions ϕ ∈ C(♦(0, T )) in accordance with (6.158), (6.159), (6.161), (6.161), (6.167) is of the form A = (A1 , A2 , A3 , A4 , A5 ): y+t

1 A1 ϕ = ϕ01 + 2

t−|y−ξ | 

2

 ϕ5 (ξ ) ϕ1 (ξ, τ ) − F(τ − |ξ |)

|ξ |

y−t 2

τ −

r1 (τ − η)ϕ1 (ξ, η)dη dτ dξ,

|ξ | y+t

1 A2 ϕ = ϕ02 + 2

2

ϕ5 (ξ ) y−t 2

τ −

t−|y−ξ | 

r1 (τ

|ξ |

 a(0) 2

+ ϕ2 (ξ, τ )

− η)ϕ2 (ξ, τ )dη dτ dξ,

|ξ | y+t

2  1 ϕ5 (ξ ) ϕ1 (ξ, t − |y − ξ |) − F(t − |y − ξ | − |ξ |) A3 ϕ = ϕ03 + 2 y−t 2

r1 (η)ϕ1 (ξ, t − |y − ξ | − η)dη dξ,

t−|y−ξ  |−|ξ |

− 0

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

256

y+t

1 A4 ϕ = ϕ04 + 2

2  a(0) ϕ5 (ξ ) + ϕ2 (ξ, t − |y − ξ |) 2 y−t 2

r1 (η)ϕ2 (ξ, t − |y − ξ | − η)dη dξ,

t−|y−ξ  |−|ξ |

− 0

4 A5 ϕ = ϕ05 − sign(y) a(0)

y  ϕ5 (ξ ) ϕ2 (ξ, 2|y| − ξ ) − f (2|y| − 2ξ ) 0

4 sign(y)dξ r1 (η)ϕ2 (ξ, 2|y| − ξ − η)dη − a(0)

2|y|−2ξ 

− 0

y ϕ1 (ξ )ϕ4 (ξ, 2|y| − ξ ) × 0 2|y|−2ξ 

r1 (η)ϕ4 (ξ, 2|y|

−

− ξ − η)dη dξ,

0

where

y+t

ϕ01

1 = 2

2 y−t 2

t−|y−ξ  |τ

|ξ |

r1 (τ − η)F(η − |ξ |)dηdτ dξ ;

|ξ | y+t

ϕ02 =

a(0) 4

2

r1 (t − |y − ξ | − |ξ |)dξ ; y−t 2

y+t

ϕ03

1 = 2

2  y−t 2

t−|y−ξ |−|ξ | 0

r1 (η)F(t − |y − ξ | − |ξ | − η)dηdξ ;

y+t

ϕ04

a(0) =− 4

2 y−t 2

r1 (t − |y − ξ | − |ξ |)dξ ;

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation

ϕ05

257

⎤ ⎡ 2y 2a  (0) ⎣  = 2 g1 (2|y|) + sign(y) k1 (2|y| − τ )g1 (τ )dτ ⎦ a (0) 0 ⎤ ⎡ 2y 4 ⎣  − g1 (2|y|) + sign(y) k1 (2|y| − τ )g1 (τ )dτ ⎦ a(0) 0

−

4 sign(y) a(0)

y  0

y + 2sign(y)

2|y|−2ξ 0

r1 (η) f (2|y| − 2ξ − η)dηdξ

r1 (2|y| − 2τ )dτ.

0

Let ϕ0 = [ϕ01 , ϕ02 , ϕ03 , ϕ04 , ϕ05 ], ϕ (T ) = max

max

(y,t)∈♦(0,T )

|ϕi |, i = 1, 2, 3, 4;

max

y∈[−T /2,T /2]

|ϕ5 | .

The system (6.168) is a closed system of Volterra integral equations of the second kind with continuous free terms and kernels with respect to unknown functions. As a small parameter, the system (6.168) contains an integration interval that does not exceed the number T . Therefore, for small T the Banach principle is applicable to it, which ensures the existence of a unique system’s solution. Define the set of functions Q = {ϕ| ϕ − ϕ0 ≤ ϕ0 } in the space C(♦(0, T )). There is an estimate for ϕ ∈ Q: ϕ ≤ 2 ϕ0 . We show that for sufficiently small T , the operator A maps the set Q to itself. Evaluating the integrals included in the formulas of the system (6.168), we find:  A1 ϕ − ϕ01 ≤ T 2 ϕ0 2 ϕ0 + F + T r  ,  a(0) + 2 ϕ0 + T r1 , A2 ϕ − ϕ02 ≤ T 2 ϕ0 2  A3 ϕ − ϕ03 ≤ T ϕ0 2 ϕ0 + F + T r1 ,  a(0) + 2 ϕ0 + T r1 , A4 ϕ − ϕ04 ≤ T ϕ0 2 A5 ϕ − ϕ05 ≤

 4 T ϕ0 2 ϕ0 + f + T r1 . a(0)

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

258

Hence Aϕ − ϕ0 (T ) ≤ α ϕ0 , where α = μ ·



a(0) 2

  4T . + 2 ϕ0 + T r1 + f max{1, T } , μ= max T, T 2 , a(0)

Obviously, there is a number T ∗ such that α < 1 and for all T ∈ (0, T ∗ ) the operator A maps the set Q to itself. Take two arbitrary elements ϕ 1 and ϕ 2 of Q(ϕ0 , ϕ0 ). Using the auxiliary inequalities           1 1 ϕ ϕ − ϕ 2 ϕ 2  ≤ ϕ 1  ϕ 1 − ϕ 2  + ϕ 2  ϕ 1 − ϕ 2  ≤ 4 ϕ0 ϕ 1 − ϕ 2  , i

j

i

j

i

we arrive

j

j

j

i

i

  1    Aϕ − Aϕ 2  ≤ α˜ ϕ 1 − ϕ 2  ,

where α˜ =

 a(0) 1 μ· + 4 ϕ0 + T r1 + f max{1, T } . 2 2

Note that α < 1 implies α˜ < 1. Therefore, we can conclude that the operator A is a contraction mapping on Q(ϕ0 , ϕ0 ). Then according to the Banach principle the Eq. (6.168) has a unique solution in Q(ϕ0 , ϕ0 ) for all T ∈ (0, T ∗ ). The system of Eq. (6.168) is solved by the method of successive approximations. Hence, we can find a unique vector function ϕ in domain ♦(0, T ) for all T ∈ (0, T ∗ ) and thus determine the function q(y) ∈ C[−T /2, T /2]. Theorem 6.12 is proved. Now we prove Theorem 6.13. Proof It is known that the domain ♦(0, T ) has an equivalent description

T T (y, t)| 0 ≤ t ≤ T, 0 ≤ |y| ≤ − |t − | . 2 2 ( j)

Let ϕ ( j) be vector functions that are solutions of (6.168) with the data {k1 (t), respectively, i.e., ϕ ( j) = Aϕ ( j) for j = 1, 2. Let  |ϕi(1) (ξ, τ ) − ϕi(2) (ξ, τ )|, i = 1, 4; ϕ(t) ˜ = max max t t

( j) g1 (t)},

0≤|ξ |≤ 2 −|τ − 2 |

max |ϕ5(1) (ξ ) − ϕ5(2) (ξ )|}

ξ ∈[− 2t , 2t ]

Note that the estimate r1 (t) 2 (T ) ≤ k(t) 2 (T ) exp (T k(t) 2 (T ))

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation ( j)

259

( j)

follows from (6.137). The known functions k1 , g1 appear in the free terms of integral equations (6.168) in the functions r1 ( j) (t), F ( j) (t), f ( j) (t), j = 1, 2. Passing in these expressions to the differences k1(1) − k1(2) , g1(1) − g1(2) as in [1], we find the estimate 

ϕ(t) ˜ ≤ C1 g

(1)

(2)

− g C 2 [0,T ] + k

(1)

t



(2)

− k C 2 [0,T ] + C2

ϕ(τ ˜ )dτ, 0

where C1 = C1 (a(0), a  (0), M1 , ρ, T ), C2 = C2 (q0 , a(0), a  (0), M1 , ρ, T ). Using Gronwall’s inequality, we find   ϕ(t) ˜ ≤ C1 exp (C2 T ) g1(1) − g1(2) 2 (T ) + k1(1) − k1(2) 2 (T ) .

(6.169)

From (6.169) it follows the estimate (6.160). Theorem 6.13 is proved. The stability Theorem 6.13 implies the validity of the uniqueness theorem for any fixed T . Theorem 6.14 Suppose that the conditions of the Theorem 6.13 hold. If the solution of the inverse problem* exists and belongs to C[−T /2, T /2], then it is unique. Further, we consider q(y) to be a known function. Let us prove Theorems 6.6–6.9. d We define K (y) = dy ln a(y) as the solution of the Cauchy problem for the Riccati equation: 1 a  (0) . (6.170) K  (y) + K 2 (y) = 2q(y), K (0) = 2 a(0) The existence solution to the Cauchy problem (6.170) under the con   of a unique   (0)  dition T  aa(0)  + T q0 ≤ 1 for all T ∈ (0, min (T ∗ , T1 ) (here T1 is a positive root of the corresponding square equation with respect to T) follows from the contractionmapping principle for the integral equation: a  (0) +2 K (y) = a(0)

y 0

1 q(ξ )dξ + 2

y K 2 (ξ )dξ. 0

We find c44 (ψ1−1 (y)) by the known function a(y): ⎞ ⎛ y  a(y) = a(0) exp ⎝ K (ξ )dξ ⎠ ⇒ c44 (ψ1−1 (y)) = 0

1 . ρa 2 (y)

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

260

Now we determine ψ1−1 (y). Differentiating with respect to y the equality −1  ψ 1 (y)

y= 0



we find 1=

ρ c44 (ψ1−1 (y))

c44 (ξ ) dξ ρ

1 dψ1−1 (y) dψ1−1 (y) ⇒ = . dy ρa(y) dy

y Finally, ψ1−1 (y) = ρ1 0 a(ξ1 ) dξ . The latter formula defines the function ψ1−1 (y) for all y ∈ (0, T /2) by known function a(y) ∈ C 2 [0, T /2] with positive values. Theorem 6.6 is proved. By integrating from 0 to y the Eq. (6.170) for K (1) (y) and K (2) (y), then subtracting one equation from the other, we can get the inequality: |K

(1)

(y) − K

(2)

1 (y)| ≤ 2

y

|K (1) (ξ ) − K (2) (ξ )||K (1) (ξ ) + K (2) (ξ )|dξ

0

y +2

(6.171) |q (1) (ξ ) − q (2) (ξ )|dξ.

0

Using Gronwall’s inequality to (6.171), we find K (1) − K (2) C[−T /2,T /2] ≤ C3 (h 0 , h 1 , ρ, T ) q (1) − q (2) C[−T /2,T /2]   ≤ C3 C ∗ g1(1) − g1(2) 2 (T ) + k1(1) − k1(2) 2 (T ) . (6.172) From the definition of functions K (1) (y) and K (2) (y) it follows that ⎛ ⎜ a ( j) (y) = a(0) exp ⎝

⎞

( j)

ψ 1 (x 3 )

⎟ K ( j) (ξ )dξ ⎠ ,

j = 1, 2,

0

hence

⎛ ( j)

c44 (x3 ) =

1 a 2 (0)ρ

⎜ exp ⎝−2

( j)

ψ 1 (x 3 )

⎞

⎟ K ( j) (ξ )dξ ⎠ ,

0

j = 1, 2.

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation

261

Making up the difference 1

(1) (2) c44 (x3 ) − c44 (x3 ) =

a 2 (0)ρ

⎛ ⎞ (1) ψ 1 (x 3 )  ⎜ ⎟ exp ⎝−2 K (1) (ξ )dξ ⎠

⎛ ⎜ − exp ⎝−2

0

⎞  ⎟ (2) K (ξ )dξ ⎠

ψ1(2) (x3 )

 0

and referring to identical calculations [2], we obtain the required stability estimate (6.130). Thus, we proved Theorem 6.7. Let  1/4   U30 ψ2−1 (z), t c11 (+0)   , p(z) := v(z, ˆ t) := . p(z) c11 ψ2−1 (z) Then, the inverse problems (6.128)–(6.131) in terms of the new functions and variable z take the form  ∂ 2 vˆ ∂ 2 vˆ = L k3 , 2 + q(z) ˜ vˆ , z > 0, t ∈ R, ∂t 2 ∂z

(6.173)

v| ˆ t 0,

(6.176)

where q(z) ˜ =

p (z) p(z)

−2



p (z) p(z)

2

, b(z) = √

1 . ρ·c11 (ψ2−1 (z))

Using in Eqs. (6.173)–(6.176) the same substitutions as in the case of the inverse problem 1, we obtain the problem similar to the problem (6.138)–(6.140): ∂ 2 wˆ ∂ 2 wˆ = + q(z) ˜ wˆ − ∂t 2 ∂z 2

t

r3 (t − τ )w(z, ˆ τ )dτ, z > 0, t ∈ R,

0

w| ˆ t 0,

(6.180)

where ˆ t)], v(z, ˆ t) = L[r3 , w(z, ˆ t)], w(z, ˆ t) = L[k3 , v(z, t r3 (t) = −k3 (t) −

k3 (t − τ )r3 (τ )dτ. 0

Note that the inverse problem 2 (6.177)–(6.180) is similar to the inverse problem (6.138)–(6.141). Therefore, Theorems 6.8 and 6.9 are proved like Theorems 6.6 and 6.7. ∂U 0 For further reasoning, we will need a structure of functions U30 , ∂ x33 . Since the direct problem (6.177)–(6.179) is similar to the problem (6.138)–(6.140) for x3 ∈ R, t ∈ R we find U30 (x3 , t) = θ (t − |ψ2 (x3 )|)U˜ (x3 , t), where

  b(0) + W˜ (ψ2 (x3 ), t) U˜ (x3 , t) = p(ψ2 (x3 ))L r3 , 2

(6.181)

is even function to the variable x3 , U˜ (x3 , t) ∈ C 1 ((T )), (T ) = {(x3 , t)| |ψ2 (x3 )| ≤ t ≤ T } (here function W˜ (ψ2 (x3 ), t) is an analogue of the regular part of function w(y, ˜ t) in inverse problem 1). From (6.181), given the structure of function W˜ (ψ2 (x3 ), t), we obtain ∂U30 = D1 (x3 )δ(t − |ψ2 (x3 )|) + θ (t − |ψ2 (x3 )|)D2 (x3 , t), ∂ x3 D1 (x3 ) = −

ρ ∂ U˜ (x3 , t) . (6.182) U˜ (x3 , |ψ2 (x3 )|)sign(x3 ), D2 (x3 , t) = c11 (x3 ) ∂ x3

Let us study inverse problem 3. Let  1/4   U11 ψ1−1 (y), t c44 (+0)   v1 (y, t) := , s(y) := . s(y) c44 ψ1−1 (y) Then (6.124)–(6.127) in terms of the new functions and variable y takes the form for y > 0, t ∈ R:  ∂ 2v ∂ 2 v1 1 = L k1 , + q(y)v1 2 2 ∂t ∂y

 i ∂U30 ∂  0 c˜12 (y) , + + (c44 U3 )  x3 =ψ1−1 (y) ρs(y) ∂ x3 ∂ x3

(6.183)

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation

v1 |t 0,

(6.186)

where  f 1 (t) = −i

263

y=+0

c˜12 (y) = c12 (ψ1−1 (y)), c44 (+0) 1 a  (0) L[k1 , g3 (t)] − L[k1 , g2 (t)]. ρ 2 a(0)

Let L[k1 , v1 (y, t)] = W (y, t) ⇒ v1 (y, t) = L[r1 , W (y, t)]. Then t ∂2 W ∂2W = + q(y)W − r1 (t − τ )W (y, τ )dτ ∂t 2 ∂ y2 0   i ∂U30 ∂  0 + L k1 , c˜12 (y) + (c44 U3 )  , x3 =ψ1−1 (y) ρs(y) ∂ x3 ∂ x3

(6.187)

W |t 0.

(6.190)

We seek a solution to the direct problem (6.187)–(6.189) in the form: W (y, t) = W1 (y, t) + W2 (y, t), where W1 (y, t) is the solution of the problem ∂ 2 W1 ∂ 2 W1 = + q(y)W1 − 2 ∂t ∂ y2

t

r1 (t − τ )W1 (y, τ )dτ,

(6.191)

0

W1 |t 0. So, − Dˆ 1 (y1∗ (0, t)) > 0 due to (6.181), (6.182) and the fact that W˜ (ψ2 (x3 ), |ψ2 (x3 )|) ≡ 0. Then (6.207) can be rewritten as c˜12 (y1∗ (0, t)) =

∗ y 1 (0,t)

˜ c˜12 (ξ ) K˜ (ξ, t)dξ + F(t),

(6.209)

0

F  (0, t) − L[k1 , g2 (t)] + W1 (0, t) ˜ , F(t) = . Dˆ 1 (y1∗ (0, t))y1∗  (0, t) Dˆ 1 (y1∗ (0, t))y1∗  (0, t) (6.210) Equation (6.209) is the second-kind Volterra integral equation. Here K˜ (ξ, t) ∈ C((T )), (T ) = {(ξ, t) : 0 ≤ ξ ≤ t ≤ T /2}. This obviously follows from the ˜ continuity in this domain of the right side of (6.208). We show that F(t) ∈ C[0, T ]. Let us differentiate (6.205) with respect to the variable t for y = 0, taking into account the parity of the integrand functions: K˜ (ξ, t) = −

K (ξ, t)

6.3 One-Dimensional Inverse Coefficient Problem of Anisotropic Viscoelasticity Equation



269

∗ y 1 (0,t)

 i dc44 ˜ 2c44 (x3 )D1 (x3 )G˜ t (0, t, ξ, |d(ξ )|) + U (x3 , t − |ξ |) ρs(ξ ) dx3 0   + c44 (x3 ) D2 (x3 , t − |ξ |) + D1 (x3 )k1 (t − |ξ | − |d(ξ )|)

F (0, t) =

t−|ξ  |

+

 dc

k1 (t − |ξ | − s)

|d(ξ )|

44

dx3

∗ y 1 (0,t) t−|ξ  |

G˜ t (0, t, ξ, τ )

+2 0

|d(ξ )|

  U˜ (x3 , s) + c44 (x3 )D2 (x3 , s) ds

dξ x3 =ψ1−1 (ξ )

 dc44 ˜ i U (x3 , τ ) ρs(ξ ) dx3

  + c44 (x3 ) D2 (x3 , τ ) + D1 (x3 )k1 (τ − |d(ξ )|) τ k1 (τ − s)

+ |d(ξ )|

 dc

44

dx3

  U˜ (x3 , s) + c44 (x3 )D2 (x3 , s) ds

dτ dξ. x3 =ψ1−1 (ξ )

(6.211) The continuity of the right side of (6.211) implies the continuity of the left side in the domain of integration. Further, W1 (0, t) ∈ C[0, T ] by virtue of the analogy with (6.154) and the continuity of the right side of (6.161) for y = 0. Thus, Eq. (6.209) is a second-kind Volterra integral equation for the function c12 (y1∗ (0, t)) with a continuous kernel and a continuous free term. As is known, in this case, there is a unique continuous solution. After determining c˜12 (y1∗ (0, t)) ∈ C[0, T /2] we find c12 (x3 ) = c˜12 (ψ1 (x3 )) for x3 ∈ [0, ψ1−1 (T /2)]. Since the functions c11 , c12 , c44 belongs to the class , for the solvability of the inverse problem, (to determine the elastic moduli) it is necessary to require the inequality specified in [2] and following from (6.207) for t = 0: c˜12 (+0) Dˆ 1 (0))y1∗  (0, +0) = −g2 (+0) g1 (+0) i = −g2 (+0) ⇒ −c˜12 (+0) 2ρc11 (+0) g1 (+0) + g3 (+0) g1 (+0) ⇒ max [−g2 (+0), 2g2 (+0)] < 2ρ[g1 (+0) + g3 (+0)] by virtue of c11 > max [−c12 , 2c12 ]. Theorem 6.10 is proved. Let us prove Theorem 6.11. Proof For simplicity, we denote y˜ = y1∗ (0, t). Now we consider two integral equations of (6.209) type with its sets of given (1) (1) (2) (2) (x3 ), c44 (x3 ), k1(1) (t), g2(1) (t), g3(1) (t) and c44 (x3 ), c11 (x3 ), k1(2) (t), functions c11 (2) (2) g2 (t), g3 (t), respectively:

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

270

( j) c˜12 ( y˜ )

 y˜ =

( j)

c12 (ξ ) K˜ j (ξ, t)dξ + F˜ ( j) (t),

j = 1, 2.

0

We have 1 c˜12 ( y˜ )

−

2 c˜12 ( y˜ )

=

 y˜  

 (1) (2) c12 (ξ ) − c12 (ξ ) K˜ 1 (ξ, t)

0

(6.212)

  (2) + K (1) (ξ, t) − K (2) (ξ, t) c12 (ξ ) dξ + F˜ (1) (t) − F˜ (2) (t). Taking into account (w1∗ (0, t)) ≡ W1 (0, t) + f 1 (t), F(t) ≡ (6.158) and (6.161) the estimate follows: (1)

t 0

f 1 (s)ds, from

(2)

W  1 (0, t) − W  1 (0, t) (T ) ≤ C4 (h 0 , h 1 , M3 , ρ, T )  (1) (2) · c44 − c44 2 (X ) + g2(1) − g2(2) (T )  + g3(1) − g3(2) (T ) + k1(1) − k1(2) 2 (T ) , (6.213) from equality (6.211) we obtain (1) (2) F  1 (0, t) − F  1 (0, t) (T ) ≤ C5 (h 0 , h 1 , M3 , ρ, T, max G˜ t ) (y,t)∈B1 (0,T )  (1) (2) (1) (2) − c11 2 (X ) × c44 − c44 1 (X ) + c11  + g3(1) − g3(2) 2 (T ) + k1(1) − k1(2) (T ) (6.214) by virtue of the following estimates, taking into account k3 (t) ≡ 0: (1) (2) − c11 (X ), D1(1) ( y˜ ) − D1(2) ( y˜ ) (T /2) ≤ C6 (h 0 , h 1 , ρ) · c11  (1) (1) (2) U˜ (x3 , t) − U˜ (x3 , t) C 1 ((T )) ≤ C7 (h 0 , h 1 , ρ, T, M2 ) g3 − g3(2) 2 (T )  (1) (2) + c11 − c11 2 (X ) .

Consider F˜ (1) (t) − F˜ (2) (t):  (1) F  (1) (0, t) − L[k1(1) , g  (1) 2 (t)] + W 1 (0, t) F˜ (1) (t) − F˜ (2) (t) = 1 Dˆ 1(1) ( y˜ ) y˜ (1)

−

F1(2) (0, t) − L[k1(2) , g2(2) (t)] + W1(2) (0, t) . Dˆ 1(2) ( y˜ ) y˜ (2)

References

271

Passing in these expressions to the differences as in [1], taking into account (6.213), (6.214) and estimates   (1) (2) (1) (2) Dˆ 1(1) ( y˜ ) − Dˆ 1(2) ( y˜ ) (T /2) ≤ C8 (h 0 , h 1 , ρ) c44 − c44 (X ) + c11 − c11 (X ) ,   (1) (2) (1) (2) y˜ (1) − y˜ (2) (T2 ) ≤ C9 (h 0 , h 1 ) c44 − c44 (X ) + c11 − c11 (X ) , using Gronwall’s lemma in (6.212) , we arrive at the estimate (6.132). Theorem 6.11 is proved.

6.4 Conclusions In this chapter, inverse coefficient problems for a system of viscoelasticity equations are studied. The known kernels of integral operators (modeling the memory phenomenon) are assumed to be both scalar and diagonal matrix type. Sufficient conditions for unique solvability of inverse problems in some classes of functions are formulated; stability estimates for the unknown coefficients are obtained. For example, the paragraph shows that inverse coefficient problems are solvable only if the third (vertical) memory component is equal to zero. These problems are of interest for analyzing the effect of the memory function under imposed conditions on the properties of a viscoelastic medium. To this end, it is possible to numerically implement systems of integral equations, that is, to make software packages and conduct a series of computational experiments.

References 1. Yakhno, V.G. 1990. Inverse problems for differential equations of elasticity. Novosibirsk: Nauka (Russian). 2. Yakhno, V.G., and I.Z. Merazhov. 2000. Some direct problems and one-dimensional inverse electroelasticity problem for “slow” waves. Siberian Advances in Mathematics 10 (1): 87–150. 3. Bogachev, I.V., A.O. Vatulyan, and O.V. Yavruyan. 2012. Identification of properties of inhomogeneous electroelastic medium. Applied Mathematics and Mechanics 76 (5): 860–866. 4. Nakamura, G., and G. Uhlmann. 1993. Identification of Lame parameters by boundary observations. American Journal of Mathematics 115: 1161–1189. 5. Nakamura, G., and G. Uhlmann. 1994. Global uniqueness for an inverse boundary value problems arising in elasticity. Inventiones Mathematicae 118: 457–474. 6. Uhlmann, G. 1992. Inverse boundary value problems and applications. Asterisque 207: 153– 211. 7. Yakhno, V.G. 1998. Multidimensional inverse problems in ray formulation for hyperbolic equations. Journal of Inverse and Ill-Posed Problems 6 (4): 373–386. 8. Yakhno, V.G. 1989. A multidimensional inverse dynamic problem of isotropic elasticity. Doklady Mathematics 34 (1): 35–36.

272

6 Some Inverse Problems of Viscoelasticity System with the Known . . .

9. Yakhno, V.G. 1989. A two-dimensional inverse problem for a system of dynamical Lame equations. Doklady Mathematics 34 (7): 616–617. 10. Wu, B., and J. Liu. 2012. Determination of an unknown source for a thermoelastic system with a memory effect. Inverse Problems 28: 095012. 11. Lavrent’ev, M.M., and V.G. Romanov. 1966. On three linearized problems for hyperbolic equations. Doklady Akademii Nauk SSSR 171 (6): 1279–1281. 12. Romanov, V.G. 1983. The inverse lamb problem in the linear approximation. In: The collection “Numerical methods in seismological studies”, 51–78. Novosibirsk: Nauka (Russian). 13. Kabanikhin, S.I. 1988. Projection—Difference methods for determining the coefficients of hyperbolic equations. Novosibirsk: Nauka (Russian). 14. Alekseev, A.S. 1962. Some inverse problems of the theory of wave propagation. Izvestiya Akademii Nauk, Seriya Geograficheskaya 11: 1514–1531 (Russian). 15. Alekseev, A.S. 1967. Inverse dynamic problem of seismics. In: Some methods and applications for interpritation of geophysic data, 9–84. M.: Nauka (Russian). 16. Blagoveshchenskii, A.S. 1971. The local method of solution of the nonstationary inverse problem for an inhomogeneous string. Proceedings of the Steklov Institute of Mathematics 115: 30–41. 17. Blagoveshchenskii, A.S. 1971. The quasi-two-dimensional inverse problem for the wave equation. Proceedings of the Steklov Institute of Mathematics 115: 63–76. 18. Blagoveshchenskii, A.S. 2001. Inverse problems of wave processes. The Netherlands: VSP. 19. Romanov, V.G., and E.A. Volkova. 1982. Inverse dynamic problems of anisotropic elastic medium. Docl. USSR Academy of Sciences 267 (4): 780–783 (Russian). 20. Melnikova, T.V., and V.G. Yakhno. 1985. One-dimensional inverse dynamic problem of isotropic elasticity for a spherically symmetric model of the earth, 44. Novosibirsk (Preprint/as USSR, Sib.ed.) (Russian). 21. Romanov, V.G. 2012. A two-dimensional inverse problem for the viscoelasticity equation. Siberian Mathematical Journal 53 (6): 1128–1138. 22. Romanov, V.G. 2014. On the determination of the coefficients in the viscoelasticity equations. Siberian Mathematical Journal 55 (3): 503–510. 23. Romanov, V.G. 1987. Inverse problems of mathematical physics. Utrecht: VNU Science Press. 24. Mizohata, S. 1979. The theory of partial differential equations. Cambridge University Press. 25. Courant, R., and D. Hilbert. 1962. Methods of mathematical physics, II, 830. New York-London: Interscience Publ. 26. Romanov, V.G. 2011. The problem of determinig the kernel of electrodinamics equations for dispersion media. Doklady Mathematics 84 (2): 613–616.

Chapter 7

Kernel Identification Problems in a Viscoelasticity System

In this chapter, we study one and two-dimensional problems of determining the memory kernel from a system of viscoelasticity and thermoviscoelasticity equations. The first two sections are devoted to inverse problems of determining the one and two-dimensional kernels of the system of viscoelastic equations. Wherein, under the assumption that the coefficients of the system depend on one spatial variable and the applied concentrated force is directed along one of the coordinate axes, the system is reduced to second-order equation. Based on the methods of Chaps. 2 and 3, the global and local solvability theorems are proved and estimates of the stability of the solution are established. In the third section the problem of determining the one-dimensional matrix kernel of the system of viscoelasticity is studied. This problem is reduced to the sequential determination of the diagonal elements of the matrix kernel from second-order hyperbolic equations. In the fourth section, the problem of recovering the memory kernel of thermoviscoelasticity equation is investigated. The research of this problem is also based on the methods used in Chap. 1.

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation For (x1 , x2 , t) ∈ R3 , x3 > 0 consider the following system of integro-differential equations: 3  ∂ Ti j ∂ 2ui ρ 2 = , i = 1, 2, 3, (7.1) ∂t ∂x j j=1

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_7

273

274

7 Kernel Identification Problems in a Viscoelasticity System

at the initial and boundary conditions: u i |t 0, included in the integral of equation (7.1) by (6.4), if additional information about the solution of problem (7.1)–(7.3) (7.4) u 1 (x3 , t)|x3 =+0 = g(t), t > 0 is given. Let

x3 y = ψ(x3 ) := 0

 dξ , ν(x3 ) := ν(ξ )

μ(x3 ) . ρ(x3 )

The main results of this section are the following theorems of global unique solvability and stability. Theorem 7.1 Let g(t) =

a 1 δ(t) + θ (t)g0 (t), a = [μ(+0)ρ(+0)]− 2 2

(7.5)

and g0 ∈ C 2[0, T ] , θ (t) is the Heaviside function. In addition, we also assume (ρ, μ) ∈ C 3 0, ψ −1 (T /2 . Then, there exists a unique solution of the inverse problem (7.1)–(7.4) k(t) ∈ C 2 [0, T ] for any fixed T > 0. Let K (h 0 ) is the set of the functions k(t) ∈ C 2 [0, T ], satisfying for t ∈ [0, T ] the inequality k(t)C 2 [0,T ] ≤ h 0 with a fixed positive constant h 0 . This constant is defined in (7.37). Theorem 7.2 Let k 1 (t) ∈ K (h 0 ), k 2 (t) ∈ K (h 0 ) are two solutions of the inverse problem (7.1)–(7.4) with data sets 

 ρ 1 (ψ −1 (y)), μ1 (ψ −1 (y)), g01 (t) ,

 2 −1  ρ (ψ (y)), μ2 (ψ −1 (y)), g02 (t) , respectively. Then, there is a positive number C = C(h 0 , h 00 , T ),

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation

h 00 = max ρ 1 (y)C 3 [0, ψ −1 (T /2)] , μ1 (y) C 3 [0,

psi −1 (T /2)] ,

275

g01 (t)C 2 [0, T ] ,

ρ 2 (y)C 3 [0,ψ −1 (T /2)] , μ2 (y) C 3 [0,ψ −1 (T /2)] , g02 (t)C 2 [0, T ] such that the following stability estimate is valid:  k (t) − k (t)C 2 [0,T ] ≤ C ρ 1 − ρ 2 C 3 [0,ψ −1 (T /2)] 1 2 1 2 +μ − μ C 3 [0,ψ −1 (T /2)] + g0 − g0 C 2 [0,T ] . 1

2

(7.6)

These theorems will be proved below. We rewrite (7.1)–(7.3) with respect to the nonzero component of the displacement vector u 1 (x3 , t):

  ∂ 2u1 ∂ ∂u 1 μ(x3 ) , (7.7) ρ(x3 ) 2 = L k, ∂t ∂ x3 ∂ x3 u 1 |t 0 takes the form  ∂ 2v ∂ 2v = L k, + q(y)v , ∂t 2 ∂ y2

(7.10)

v|t 0,   2  (y) (y) − 2 ss(y) . where q(y) = ss(y)   t Let v(y, t) + 0 k(t − τ )v(y, τ )dτ exp (−k(0)t/2) = w(y, t), then

(7.12) (7.13)

276

7 Kernel Identification Problems in a Viscoelasticity System

t v(y, t) = exp (k(0)t/2) w(y, t) +

r (t − τ ) exp (k(0)τ/2) w(y, τ )dτ, 0

where

t k(t − τ )r (τ )dτ.

r (t) = −k(t) −

(7.14)

0

In terms of the new functions w(y, t) and r (t) (7.10)– (7.13) take the form ∂ 2w ∂ 2w = + H (y)w − ∂t 2 ∂ y2

t h(t − τ )w(y, τ )dτ,

(7.15)

0

w|t 0 and near the characteristic line t = y has the structure w(y, t) =

a δ(t − y) + θ (t − y)w(y, ˜ t), 2

(7.19)

where w(y, ˜ t) is a regular function. Substituting the function (7.19) into equations (7.15)– (7.18) and using the method of singularity extraction, we find that the function w(y, ˜ t) in the domain t > y > 0 satisfies the equations (w = w˜ for t > y):

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation

∂ 2w ∂ 2w a = + H (y)w − h(t − y) − 2 2 ∂t ∂y 2

w|t=y+0

ar (0) a =− + 4 4

277

t h(t − τ )w(y, τ )dτ,

(7.20)

y

y H (ξ )dξ := β(y),

(7.21)

0

 ∂w  = 0, ∂ y  y=+0 t w| y=+0 = g˜ 0 (t) +

k0 (t − τ )g˜ 0 (τ )dτ +

(7.22)

a k0 (t). 2

(7.23)

0

  (y, t) are continuous functions, we Provided that for y = t = 0 w(y, t), ∂w ∂y can express r (0) and r  (0) from (7.21)–(7.23) in terms of the known numbers as r (0) =

4 4 4 g0 (0), r  (0) = −q(0) − 2 g02 (0) + g˜ 0 (0). a a a

(7.24)

To derive these equalities, we used the relations 



t

k (t) = −r (t) − r (0)k(t) −

r  (t − τ )k(τ )dτ , k  (0) = −r  (0) + r 2 (0).

0

Further suppose that in H (y), the values of r (0) and r  (0) are replaced with (7.24). Lemma 7.1 Under the conditions of Theorem 7.1 the problem (7.20)–(7.23) for to the problem of deter(y, t) ∈ DT , DT = ((y, t)|0 ≤ y ≤  t ≤ T − y) is equivalent   (y, t), h(t), k (t), k (t), k mining the functions w(y, t), ∂w 0 0 0 (t) from the following ∂t system of integral equations: t w(y, t) = β(y) + y

∂w (y, τ )dτ, ∂τ

(7.25)

278

7 Kernel Identification Problems in a Viscoelasticity System

1 ∂w a (y, t) = (g˜ 0 (t − y) − r (0)g˜ 0 (t − y)) − H ∂t 2 8

y+t 2



t−y  a  1 a k0 (t − y − τ )g˜ 0 (τ )dτ − h(t − y)y + k0 (t − y) + 4 4 2 0 ⎡ ⎤ t−y y  1 ⎢ ⎥ + h(τ )w(ξ, t − y + ξ − τ )dτ ⎦ dξ ⎣ H (ξ )w(ξ, t − y + ξ ) − 2 0

+

1 2

0

t+y 2 

H (ξ )w(ξ, t + y − ξ ) − y

t+y−2ξ 

−

a h(t + y − 2ξ ) 2

h(τ )w(ξ, t + y − ξ − τ )dτ dξ,

0

h(t) =

1  H 2 −

(7.26)

 t 2 

4 g˜  (t) − r (0)g˜ 0 (t) + a 0

− 2k0 (t) −

4 + a

t 2

0

⎡

4 a

t





  r 2 (0) t t 1 − r  (0) g˜ 0 (t) − H β 2 2 2 2

k0 (t − τ )g˜ 0 (τ )dτ −

0

2 a

t h(τ )β

t −τ 2

0

∂w ⎢ (ξ, t − ξ ) − ⎣ H (ξ ) ∂t

t−2ξ  0

 dτ

(7.27)

⎤

∂w ⎥ (ξ, t − ξ − τ )dτ ⎦ dξ, h(τ ) ∂t

 t r 2 (0)  − r (0) t + (t − τ )k0 (τ )dτ, k0 (t) = −r (0) + 2

(7.28)

0

k0 (t) =

r 2 (0) − r  (0) + 2

t

k0 (τ )dτ,

(7.29)

0

k0 (t)

 t r 2 (0)  − r (0) k0 (t) − h(t − τ )k0 (τ )dτ. = −h(t) + 4 0

(7.30)

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation

279

Proof To prove the Lemma 7.1, we use the equalities (3.63). Taking this into account, we integrate (7.20) along the corresponding characteristics of the differential operators of the first order for (y, t) ∈ DT . First, we integrate along the characteristic of operator ∂t∂ − ∂∂y from point (y, t) to the point ((y + t)/2, (y + t)/2) in the plane   of the variables (ξ, τ ). Using the equality ∂t∂ + ∂∂y w((y + t)/2, (y + t)/2) = (a/4)H ((y + t)/2), which follows from (7.21) after differentiation with respect to y, we obtain

∂ ∂ + ∂t ∂y



a w(y, t) = H 4

a − h(t + y − 2ξ ) − 2

y+t 2



(y+t)/2  

+

H (ξ )w(ξ, t + y − ξ ) y

t+y−2ξ 

h(τ )w(ξ, t + y − ξ − τ )dτ dξ.

(7.31)

0

Now we integrate along the characteristic of operator ∂t∂ + ∂∂y from point (0, t − y) to the point (y, t). Using the equalities (7.22) and (7.23), we find

 ∂ ∂ − w(y, t) ∂t ∂y = g˜ 0 (t − y) − r (0)g˜ 0 (t − y) t−y a  a − h(t − y)y + k0 (t − y) + k0 (t − y − τ )g˜ 0 (τ )dτ 2 2 0 ⎡ ⎤ y t−y ⎣ H (ξ )w(ξ, t − y + ξ ) − h(τ )w(ξ, t − y + ξ − τ )dτ ⎦ dξ. 0

(7.32)

0

From (7.31) and (7.32) it is easy to obtain equation (7.26). In (7.31), setting y = 0 and using conditions (7.22) and (7.23), we find

 t t a  a  − r (0)g˜ 0 (t) + k0 (t) + k0 (t − τ )g˜ 0 (τ )dτ = H 2 4 2 0 ⎤ ⎡ t−2ξ t/2  a h(τ )w(ξ, t − ξ − τ )dτ ⎦ dξ. + ⎣ H (ξ )w(ξ, t − ξ ) − h(t − 2ξ ) − 2

g˜ 0 (t)

0

0

Differentiating this equation, after simple computations, we arrive at equation (7.27).

280

7 Kernel Identification Problems in a Viscoelasticity System

Equations (7.28)–(7.30) are used to close the system of integral equations (7.25)– (7.27). Under the conditions of Theorem 7.1, the equivalence of the system of integral equations (7.25)–(7.30) and the inverse problem (7.20)–(7.23) is established in the usual way.  Let us prove Theorem 7.1. The proof of the theorem is quite analogous to the proof of the Theorem 3.1 from Sect. 3.1. Express (7.25)–(7.30) as the operator equation ϕ = Aϕ, where

(7.33)

 ϕ = ϕ1 (y, t), ϕ2 (y, t), ϕ3 (t), ϕ4 (t), ϕ5 (t), ϕ6 (t)  a a ∂w (y, t) + h(t − y)y − k0 (t − y), h(t) = w(y, t), ∂t 4 4 − 2k0 (t), k0 (t), k0 (t), k0 (t) + h(t) − r00 k0 (t)

is the vector function with components ϕi (i = 1, 2, . . . , 6) and r00 := r 4(0) − r  (0), while the operator A defined on the set of functions ϕ ∈ C[DT ] in accordance with (7.25)–(7.30), is the form 2

A = (A1 , A2 , A3 , A4 , A5 , A6 ) : t  ϕ2 (y, τ ) A1 ϕ = ϕ01 + y

a a y (ϕt−y ϕ − (τ − y) + 2ϕ (τ − y) + 2r ϕ (τ − y)) + (τ − y) dτ, 3 6 00 4 5 12 4  y  A2 ϕ = ϕ02 +

1 2

ϕ5 (t − y − τ )g˜ 0 (τ )dτ + 0

1 2

H (ξ )ϕ1 (ξ, t − y + ξ ) 0

t−y  1 ((ϕ3 (τ ) + 2ϕ6 (τ ) + 2r00 ϕ4 (τ ))ϕ1 (ξ, t − y + ξ − τ )dτ dξ − 3 0 t+y

1 + 2

2 

(7.34) H (ξ )ϕ1 (ξ, t + y − ξ )

y

a − (ϕ3 (t + y − 2ξ ) + 2ϕ6 (t + y − 2ξ ) + 2r00 ϕ4 (t + y − 2ξ )) 6 t+y−2ξ  1 (ϕ3 (τ ) + 2ϕ6 (τ ) + 2r00 ϕ4 (τ ))ϕ1 (ξ, t + y − ξ − τ )dτ dξ, − 3 0

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation

4 A3 ϕ = ϕ03 + 3a

281

t (ϕ6 (t − τ ) − ϕ3 (t − τ ) + r00 ϕ4 (t − τ )) g˜ 0 (τ )dτ 0

+

2 3a

t (ϕ3 (τ ) + 2ϕ6 (τ ) + 2r00 ϕ4 (τ )) β

t −τ 2

 dτ

0

4 a

t/2

 aξ ϕ3 (t − 2ξ ) + 2ϕ6 (t − 2ξ ) − 2r00 ϕ4 (t − 2ξ ) H (ξ ) ϕ2 (ξ, t − ξ ) − 12

0 t−2ξ   1 a ϕ5 (t − 2ξ ) − (ϕ3 (τ ) + 2ϕ6 (τ ) + 2r00 ϕ4 (τ )) 4 3 0

aξ (ϕ3 (t − 2ξ − τ ) + 2ϕ6 (t − 2ξ − τ ) × ϕ2 (ξ, t − ξ − τ ) − 12  a + 2r00 ϕ4 (t − 2ξ − τ )) + ϕ5 (t − 2ξ − τ ) dτ dξ, 4

+

1 A4 ϕ = ϕ04 + 3

t (t − τ ) (ϕ6 (τ ) − ϕ3 (τ ) + r00 ϕ4 (τ )) dτ, 0

1 A5 ϕ = ϕ05 + 3

t (ϕ6 (τ ) − ϕ3 (τ ) + r00 ϕ4 (τ )) dτ, 0

1 A6 ϕ = ϕ06 − 3

t (ϕ3 (t − τ ) + 2ϕ6 (t − τ ) + 2r00 ϕ4 (t − τ )) ϕ4 (τ )dτ, 0

where ϕ0 (y, t) = (ϕ01 , ϕ02 , ϕ03 , ϕ04 , ϕ05 , ϕ06 )

 

  a y+t 1 1  t g˜ 0 (t − y) − r (0)g˜ 0 (t − y) − H , H := β(y), 2 8 2 2 2

  t t 1 4   g˜ 0 (t) − r (0)g˜ 0 (t) + r00 g˜ 0 (t) + H β , −r (0) + r00 t, r00 , 0 . − a 2 2 2

(7.35) Denote by Cσ the Banach space of continuous functions with the family of weighted norms

282

7 Kernel Identification Problems in a Viscoelasticity System

ϕσ = max

!     −σ t  −σ t    , sup ϕ4,5,6 (t)e sup ϕ1,2 (y, t)e , σ ≥ 0. t∈[0,T ]

(y,t)∈DT

It is obvious that for σ = 0 this is the space of continuous functions with the usual norm. We denote this norm by ϕ. The norms ϕσ and ϕ are equivalent for every fixed T ∈ (0, ∞) due to (3.17) |ϕσ . We choose the number σ below. Take the ball Q σ (ϕ0 , ϕ0 ) =: {ϕ|ϕ − ϕ0 σ ≤ ϕ0 } of radius ϕ0  centered at the point ϕ0 of some weighted space Cσ with σ ≥ 0, where the function ϕ0 is defined using (7.35) and   ϕ0  = max ϕ0 j  . 1≤ j≤6

Observe that ϕ ∈ Q σ (ϕ0 , ϕ0 ) satisfies the estimate ϕσ ≤ ϕ0 σ + ϕ0  ≤ 2ϕ0 . Let ϕ(y, t) ∈ Q σ (ϕ0 , ϕ0 ) and verify that for a suitable choice of σ > 0 the operator A maps balls into balls, i.e., Aϕ ∈ Q σ (ϕ0 , ϕ0 ). Indeed, calculating the norm of the differences ||A1 ϕ − ϕ01 ||σ like a proof of the Theorem 3.1, in view of (7.34), for (y, t) ∈ DT we have 1 ||A1 ϕ − ϕ01 ||σ ≤ 2ϕ0 μ1 (a, T, |r00 |) , σ 1 ||A2 ϕ − ϕ02 ||σ ≤ 2ϕ0 μ2 (a, T, |r00 |, G 0 , H0 , ϕ0 ) , σ 1 ||A3 ϕ − ϕ03 ||σ ≤ 2ϕ0 μ3 (a, T, |r00 |, G 0 , H0 , ϕ0 , β0 ) , σ 1 ||A4 ϕ − ϕ04 ||σ ≤ 2ϕ0 μ4 (T, |r00 |) , σ 1 ||A5 ϕ − ϕ05 ||σ ≤ 2ϕ0 μ5 (|r00 |) , σ 1 ||A6 ϕ − ϕ06 ||σ ≤ 2ϕ0 μ6 (T, |r00 |, |r (0)|, ϕ0 ) , σ where H0 := max |H (y)|, y∈[0,T /2]

β0 := max |β(y)|, y∈[0,T /2]

G 0 := max |g˜ 0 (t)| . We t∈[0,T ]

obtained the last inequality using the fourth equation of (7.34). Choosing " # σ ≥ σ0 := 2 max μ j , 1≤ j≤6

we infer that A maps the ball Q σ (ϕ0 , ϕ0 ) into the ball Q σ (ϕ0 , ϕ0 ).

7.1 The Problem of Determining the One-Dimensional Kernel of Viscoelasticity Equation

283

Take two arbitrary elements ϕ 1 and ϕ 2 . Using the auxiliary inequalities         1 1 ϕ ϕ − ϕ 2 ϕ 2  e−σ t ≤ ϕ 1  ϕ 1 − ϕ 2  e−σ t + ϕ 2  ϕ 1 − ϕ 2  e−σ t i j i j i j j j i i $ $ ≤ 4ϕ0  $ϕ 1 − ϕ 2 $σ , (y, t) ∈ DT , we find

$ $ $ $ $ $ $(Aϕ 1 − Aϕ 2 )$ ≤ $ϕ 1 − ϕ 2 $ ≤ σ00 $ϕ 1 − ϕ 2 $ , σ σ σ σ

where

" σ00 = max μ1 (a, T, |r00 |), μ2 (a, T, |r00 |, G 0 , H0 , 2ϕ0 ), μ3 (a, T, |r00 |, G 0 , H0 , 2ϕ0 , β0 ), μ4 (T, |r00 |), # μ5 (|r00 |), μ6 (T, |r00 |, |r (0)|, 2ϕ0 )

The resulting estimates imply that if σ satisfies σ > σ ∗ := max {σ0 , σ00 }, then, A is a contraction mapping on Q σ (ϕ0 , ϕ0 ). Then, according to the Banach principle, (7.33) has a unique solution in Q σ (ϕ0 , ϕ0 ) for each fixed T > 0. Since k0 (t) := exp(r (0)t/2)k(t), we find k(t) = exp[−r (0)t/2]k0 (t).

(7.36)

The proof of Theorem 7.1 is complete. Now let us prove the Theorem 7.2. Since the Theorem 7.1 holds, the solution (7.33) belongs to the set Q σ (ϕ0 , ϕ0 ) and ϕi σ ≤ 2ϕ0 , i = 1, 2, · · · , 6. Thus, max |k(t)| ≤ 2ϕ0  exp (|r (0)|T ) := h 0 .

t∈[0,T ]

(7.37)



j Take the vector ϕ j , for j = 1, 2 with the data ρ j (ψ −1 (y)), μ j (ψ −1 (y)), g0 (t) % %   respectively; i.e., ϕ j = Aϕ j . The known functions ρ j ψ −1 (y) , μ j ψ −1 (y) , appear in the free terms of these integral equations in the compositions H j (y), q j (y), s j (y), j = 1, 2. Passing in these expressions to the differences ρ 1 − ρ 2 , μ1 − μ2 , as in [1], by the arguments in the proof of Theorem 7.1 for σ ≥ σ ∗ we arrive at the estimate ∗ $ $ $ $ 1 $ϕ − ϕ 2 $ ≤ C0 γ + σ $ϕ 1 − ϕ 2 $ , (7.38) σ σ σ where γ := ρ 1 − ρ 2 C 3 [0,ψ −1 (T /2] + μ1 − μ2 C 3 [0,ψ −1 (T /2] + g01 − g02 C 2 [0,T ] and the constant C0 depends on the same parameter as C in Theorem 7.2. Now (3.17) and (7.38) imply the estimate

284

7 Kernel Identification Problems in a Viscoelasticity System

$ 1 $ $k − k 2 $ ≤ C1 γ , 0

0

with C1 = σ C0 /(σ − σ ∗ ).    Considering (7.36) for k 1 , k01 and k 2 , k02 and using (7.38), we obtain (7.6).

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation Consider the system (7.1)–(7.3) with t Ti j (x, t) = σi j [u](x, t) +

k(x2 , t − τ )σi j [u](x, τ )dτ,

(7.39)

0

σi j are stresses (6.5). We assume that ρ = ρ(x3 ), μ = μ(x3 ), λ = λ(x3 ) are known and belong to the set . Under this assumptions from equalities (7.1)–(7.3),(7.39) it follows that u 1 (x, t) ≡ u 1 (x3 , t) ≡ 0, u 2 ≡ u 3 ≡ 0 and the function λ(x3 ) will not enter the remaining equations. The inverse problem: determine the kernel k(x2 , t), t > 0, included in the integral of equation (7.1) by (7.39), if additional information about the solution of problem (7.1)–(7.3), (7.39) is known u 1 (x2 , x3 , t)|x3 =+0 = g(x2 , t), t > 0.

(7.40)

We rewrite (7.1)–(7.3) with (7.39) with respect to the nonzero component of the displacement vector u 1 (x2 , x3 , t): ∂ 2u1 ∂ ρ(x3 ) 2 = ∂t ∂ x3



   ∂u 1 ∂u 1 ∂ μ(x3 )L k, + μ(x3 ) L k, , ∂ x3 ∂ x2 ∂ x2

(7.41)

u 1 |t 0, (x, t) ∈ R2 , = L k, 2 + q(y)v + ν ∂t 2 ∂y ∂x ∂x

s  (y) s(y)

(7.44)

v(x, y, t)|t 0, x ∈ R,

(7.47)



s  (y) s(y)

2

, a = [μ(+0)ρ(+0)]− 2 .   In the equation (7.44) the function ν φ −1 (y) is denoted by ν. Consider the operator equation for the function v(x, y, t)

where q(y) :=

−2

1

L [k, v] = exp (k(x, 0)t/2) w(x, y, t). It is easy to see that v is expressed by the formula  % v(x, y, t) = L r, exp (k(x, 0)t/2) w , where

(7.48)

t k(x, t − τ )r (x, τ )dτ.

r (x, t) = −k(x, t) −

(7.49)

0

With respect to the functions w(x, y, t), v(x, y, t) and r (x, t) for y > 0, (x, t) ∈ R2 , the equations (7.44)–(7.47) take the form

286

7 Kernel Identification Problems in a Viscoelasticity System

∂ 2w ∂ 2w = + H (x, y)w − ∂t 2 ∂ y2

t h(x, t − τ )w(x, y, τ )dτ 0

+ν 2 exp (r (x, 0)t/2)





(7.50)

∂v ∂ L k, , ∂x ∂x

w(x, y, t)|t y > 0. Let β(x, y) := w(x, y, y). For a fixed x ∈ R we obtain a differential equation for β(x, y) ∂ a aν 2 β(x, y) = H (x, y) + ∂y 4 4 with initial condition



y ∂ 2k (x, 0) + 2 ∂x 2

∂k ∂x

2

y2 (x, 0) 4



a β(x, 0) = − r (x, 0). 4

we find aν 2 a β(x, y) = − r (x, 0) + 4 4



y2 ∂ 2k + (x, 0) 4 ∂x2

  y ∂k 2 y3 a (x, 0) H (x, ξ )dξ. + ∂x 12 4 0

(7.56) Hence, the functions w(x, y, t), k(x, t) in domain t > y > 0 satisfy the equations ∂ 2w a ∂ 2w = + H (x, y)w − h(x, t − y) 2 2 ∂t ∂y 2 t + k00 (x, y, t) − h(x, t − τ )w(x, y, τ )dτ y

+ ν 2 exp (r (x, 0)t/2)

(7.57)

 ∂ ∂ v˜ L 0 k, r00 + , ∂x ∂x

w(x, y, t)|t=y = β(x, y),

(7.58)

 ∂w  = 0, ∂ y  y=+0

(7.59)

%  a k0 (x, t) + L k0 , g˜ 0 , 2

(7.60)

w| y=+0 =

288

7 Kernel Identification Problems in a Viscoelasticity System

where k00 (x, y, t) :=   

2 ∂ 2k ∂k ∂k y y2 ν2a ∂k (x, 0) (x, t − y) , (x, 0) + (x, 0) k(x, t − y) + 2 2 ∂x 4 ∂x ∂x ∂x2

(7.61) 

∂ r (x, t − y) exp (k(x, 0)y/2) . r00 (x, y, t) := ∂x Provided that for y = t = 0 w(x, y, t), we can express r (x, 0) and values of functions as r (x, 0) =

∂r (x, 0) ∂t



∂w ∂y



(7.62)

(x, y, t) are continuous functions,

from (7.58)–(7.60) in terms of the available

4 ∂r 4 ∂ g˜ 0 4 g0 (x, 0), (x, 0) = −q(0) − 2 g02 (x, 0) + (x, 0). a ∂t a a ∂t

(7.63)

To derive this, we use the relations ∂k ∂r (x, t) = − (x, t) − r (x, 0)k(x, t) − ∂t ∂t

t 0

∂r (x, t − τ )k(x, τ )dτ, ∂t

∂k ∂r (x, 0) = − (x, 0) + r 2 (x, 0). ∂t ∂t Assume henceforth that in the relations for H (x, y) we replace r (x, 0) and ∂r (x, 0) ∂t with their values (7.58). (x, 0) are known functions, their values are Further, we assume that r (x, 0), ∂r ∂t calculated by formulas (7.63). It follows that the functions k(x, 0), ∂∂kx (x, 0), (x, 0), ∂2k (x, 0), ∂k (x, 0), and hence k00 (x, y, t), r00 (x, y, t) at t = y are also known. ∂x2 ∂t We construct a system of integro-differential equations for functions w, v, ˜ h, k0 . Using Dalamber’s formula we find from (7.57), (7.59), (7.60) w(x, y, t) = w0 (x, y, t) 1 + 2

y t+y−ξ  "

H (x, ξ )w(x, ξ, τ ) −

a h(x, τ − ξ ) + k00 (x, ξ, τ ) 2

0 t−y+ξ τ−ξ

h(x, γ )w(x, ξ, τ − γ )dγ

− 0

+ ν 2 exp (r (x, 0)τ/2) x ∈ R, 0 < y < t,

 # ∂ ∂ v˜ L 0 k, r00 + (x, ξ, τ ) dτ dξ, ∂x ∂x

(7.64)

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

where

289

1"a (k0 (x, t + y) + k0 (x, t − y)) + g˜ 0 (x, t + y) 2 2 t+y  + k0 (x, τ )g˜ 0 (x, t + y − τ )dτ

w0 (x, y, t) =

0 t−y  # + g˜ 0 (x, t − y) + k0 (x, τ )g˜ 0 (x, t − y − τ )dτ . 0

Passing to the limit in the formula (7.64) at t → y + 0 taking into account the conditions (7.59), (7.60) we find 2β(x, y) − w0 (x, y, y + 0) =

y 2y−ξ  " 0

H (x, ξ )w(x, ξ, τ ) −

ξ

a h(x, τ − ξ ) 2

τ−ξ

+ k00 (x, ξ, τ ) −

h(x, γ )w(x, ξ, τ − γ )dγ 0



∂ ∂ v˜ L 0 k, r00 + + ν 2 exp (r (x, 0)τ/2) ∂x



∂x

# (x, ξ, τ ) dτ dξ.

Differentiating by y the last equality, we obtain: ∂β 1 dw0 (x, y) − (x, y, y + 0) ∂y 2 dy y " a H (x, ξ )w(x, ξ, 2y − ξ ) − h(x, 2(y − ξ )) = 2 0

(7.65)

2(y−ξ  )

+ k00 (x, ξ, 2y − ξ ) −

h(x, γ )w(x, ξ, 2y − ξ − γ )dγ 0



∂ ∂ v˜ L 0 k, r00 + + ν 2 exp (r (x, 0)(2y − ξ )/2) ∂x

∂x



# (x, ξ, 2y − ξ ) dξ.

The equation for h is obtained by differentiating equality (7.65) by y after replacing the variable in the second integral 2(y − ξ ) by ξ  :

290

7 Kernel Identification Problems in a Viscoelasticity System

2 h(x, y) = h 0 (x, y) + a

y k0 (x, ξ ) 0

+

1 a

y/2

"

0

H (x, ξ )

∂ 2 g˜ 0 (x, y − ξ )dξ ∂t 2

∂w ∂k00 (x, ξ, y − ξ ) + (x, ξ, y − ξ ) ∂t ∂t

1 − h(x, y − 2ξ )β(x, ξ ) 2 y−2ξ  (7.66) ∂w (x, ξ, y − ξ − τ )dτ + ν 2 exp (r (x, 0)(y − ξ )/2) h(x, τ ) − ∂t 0   ∂r ∂  ∂ v˜  ∂ 00 L 0 k(x, y − ξ ), + (x, ξ, y − ξ ) × ∂x ∂t ∂ x ∂t  ∂ v˜ r (x, 0) ∂ L 0 k(x, y − ξ ), (r00 + )(x, ξ, y − ξ ) + 2 ∂x ∂x  # 1 ∂ ∂ v˜ + k(x, y − 2ξ )(r00 + )(x, ξ, ξ ) dξ, 2 ∂x ∂x where   ∂ 2 g˜ 0 ∂ 2 k0 2 ∂k0 ∂ g˜ 0 (x, y) + k0 (x, y) (x, 0) + h 0 (x, y) = (x, y) + (x, y) g0 (x, 0) a ∂t ∂t ∂x2 ∂t 2 −

2 2 ∂2 2 β(x, y/2) + H (x, y/2)β(x, y/2) + k00 (x, y/2, y/2) a ∂ y2 a a

  ∂r00 ∂ 2β 2 2 (x, y/2, y/2) + 2 (x, y/2) . + ν exp(r (x, 0)y/4) a ∂x ∂x To calculate the function ∂w ∂t , use the equality (7.64). Applying an equivalent description of the integration domain in the form of {(ξ, τ )|t − y ≤ τ ≤ t + y, 0 ≤ ξ ≤ y − |t − τ |}, we differentiate the equation by t. Replacing the integration variable τ − t = ξ  , we find

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

291

∂w ∂w0 (x, y, t) = (x, y, t) ∂t ∂t y " a 1 sgn(ξ ) H (x, y − |ξ |)w(x, y − |ξ |, ξ + τ ) − h(x, ξ + t − y + |ξ |) + 2 2 −y

t−y+ξ  +|ξ |

+ k00 (x, y − |ξ |, t + |ξ |) −

h(x, γ )w(x, y − |ξ |, t + ξ − γ )dγ 0

 # 

ξ +t ∂ ∂ v˜ 2 L 0 k(x, ξ + t), (r00 + )(x, y − |ξ |, ξ + t) dξ, + ν exp r (x, 0) 2 ∂x ∂x

(7.67)

where 1 " a ∂k0 ∂k0 ∂w0 (x, y, t) = ( (x, y, t)(x, t + y) + (x, y, t)(x, t − y)) ∂t 2 2 ∂t ∂t + g0 (x, 0)[k0 (x, t + y) + k0 (x, t − y)] ∂ g˜ 0 ∂ g˜ 0 (x, y, t)(x, t + y) + (x, t − y) ∂t ∂t t+y t−y   # ∂ g˜ 0 ∂ g˜ 0 k0 (x, τ ) k0 (x, τ ) (x, t + y − τ )dτ + (x, t − y − τ )dτ . + ∂t ∂t

+

0

0

To make the system closed (7.64), (7.66), (7.67) we use the following obvious equality: r0 (x, t) = r (x, 0) + t +

∂r (x, 0)t ∂t

(y − ξ )(h(x, ξ ) + r (x, 0) 0

∂r ∂r0 (x, t) = (x, 0) + ∂t ∂t

t  0

k0 (x, t) = −r (x, 0) +

r 2 (x, 0) ∂r0 (x, ξ ) − r0 (x, ξ ))dξ, ∂t 4

(7.68)

 r 2 (x, 0) ∂r0 (x, ξ ) − r0 (x, ξ ) dξ, (7.69) h(x, ξ ) + r (x, 0) ∂t 4

 r 2 (x, 0) 2

 ∂r − (x, 0) t + ∂t

t (t − ξ ) 0

∂ 2 k0 (x, ξ )dξ, ∂t 2

 2 r 2 (x, 0) ∂r ∂k0 ∂ k0 (x, t) = − (x, 0) + (x, ξ )dξ, ∂t 2 ∂t ∂t 2

(7.70)

t

0

(7.71)

292

7 Kernel Identification Problems in a Viscoelasticity System

  r 2 (x, 0) ∂r ∂ 2 k0 − (x, 0) k0 (x, t) (x, t) = − h(x, t) + 2 4 ∂t ∂t t − h(x, t − ξ )k0 (x, ξ )dξ,

(7.72)

0

following from the above introduced notation for k, r, k0 , r0 and equation (7.55), we find v(x, ˜ y, t) = exp (k(x, 0)t/2) L 0 [r0 , w] ,

(7.73)

after differentiating by t: ∂ v˜ (x, y, t) = exp (k(x, 0)t/2) ∂t 

k(x, 0) × L 0 [r0 , w] + L 0 [r0 , wt ] + r0 (x, t − y)β(x, y) . 2

(7.74)

Let c0 (x) := r (x,0) − ∂r 4 ∂t (x, 0). In further material, the system (7.57)–(7.60) will be considered in the domain 2

DT = G T × R, G T = {(y, t)| 0 ≤ y ≤ t ≤ T − y} , T > 0. Now we introduce a vector function ϕ with components ϕ1 (x, y, t) = w(x, y, t) −

1"a (k0 (x, t + y) + k0 (x, t − y)) 2 2

t+y 

+

k0 (x, τ )g˜ 0 (x, t + y − τ )dτ 0 t−y 

# k0 (x, τ )g˜ 0 (x, t − y − τ )dτ ,

+ 0

˜ y, t), ϕ2 (x, y, t) = v(x,  1 " a  ∂k0 ∂k0 ∂w (x, y, t) − (x, t + y) + (x, t − y) ∂t 2 2 ∂t ∂t + g0 (x, 0)[k0 (x, t + y) + k0 (x, t − y)]

ϕ3 (x, y, t) =

t+y  ∂ g˜ 0 (x, t + y − τ )dτ + k0 (x, τ ) ∂t 0

t−y  # ∂ g˜ 0 (x, t − y − τ )dτ , + k0 (x, τ ) ∂t 0

ϕ4 (x, y, t) =

∂ v˜ (x, y, t), ∂t

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

ϕ5 (x, y) = 2h(x, y) − c0 (x)k0 (x, y) −

293

 2 ∂k0 ∂ g˜ 0 g0 (x, 0) (x, y) + k0 (x, y) (x, 0) , a ∂t ∂t

∂r0 (x, y), ∂t ∂k0 (x, y) ϕ8 (x, y) = k0 (x, y), ϕ9 (x, y) = ∂t

ϕ6 (x, y) = r0 (x, y), ϕ7 (x, y) =

and the vector function ϕ 0 with components ϕ10 (x, y, t) = ϕ30 (x, y, t) = ϕ50 (x, y) =

 1 g˜ 0 (x, t + y) + g˜ 0 (x, t − y) , 2

 ∂ g˜ 0 1  ∂ g˜ 0 (x, t + y) + (x, t − y) , 2 ∂t ∂t

2  ∂ 2 g˜ 0 ∂2 (x, y) − 2 β(x, y/2) + H (x, y/2)β(x, y/2) + k00 (x, y/2, y/2) 2 a ∂t ∂y +ν 2 exp(r (x, 0)y/4)

 ∂r

00

∂x

ϕ60 (x, y) = r (x, 0) +

(x, y/2, y/2) +

∂r (x, 0)y, ϕ70 (x) = rt (x, 0), ∂t

ϕ80 (x, y) = −r (x, 0) + ϕ90 (x) =

 ∂ 2β (x, y/2) , ∂x2

 r 2 (x, 0) 2

−

 ∂r (x, 0) y, ∂t

r 2 (x, 0) ∂r − (x, 0). 2 ∂t

Excluding from the equation (7.66) the unknown function k0tt (x, y) (this function is included in the h 0 ) using (7.72), we rewrite the system of equations (7.64), (7.66)–(7.74) in terms of vector functions ϕ and ϕ 0 ˜ ϕ = ϕ 0 + Aϕ,

(7.75)

˜ which is constructed similarly to the operator A (Sect. 7.1). Note with nonlinear operator A, that

294

7 Kernel Identification Problems in a Viscoelasticity System

1"a (ϕ8 (x, t + y) + ϕ8 (x, t − y)) 2 2 t+y t−y   # + ϕ8 (x, τ )g˜ 0 (x, t + y − τ )dτ + ϕ8 (x, τ )g˜ 0 (x, t − y − τ )dτ ,

w(x, y, t) = ϕ1 (x, y, t) +

0

0

1"a wt (x, y, t) = ϕ3 (x, y, t) + (ϕ9 (x, t + y) + ϕ9 (x, t − y)) 2 2 + g0 (x, 0)[ϕ8 (x, t + y) + ϕ8 (x, t − y)] t+y t−y   # ∂ g˜ 0 ∂ g˜ 0 (x, t + y − τ )dτ + (x, t − y − τ )dτ , + ϕ8 (x, τ ) ϕ8 (x, τ ) ∂t ∂t 0

0

1 h(x, y) = [ϕ5 (x, y) + c0 (x)ϕ8 (x, y)] 2  1 ∂ g˜ 0 g0 (x, 0)ϕ9 (x, y) + ϕ8 (x, y) (x, 0) . + a ∂t

(7.76)

As in Sect. 2.1 let us consider the Banach function space As (r ), s > 0, analytical in the  vicinity of the origin point. If s ∈ (0, s) and the following inequality holds $ α $ $ ∂ $ $ $ ϕ(x) $ ∂xα $

ϕs (r ) ≤ αα   α s−s s 

(7.77)

for all α. The main result of this work is a theorem of local unique solvability in the class of analytical functions with the variable x.  

Theorem 7.3 Let (ρ, μ, λ) ∈ , g˜ 0 (x, 0), ∂∂tg˜0 (x, 0) ∈ As0 , 

   ∂ 2 g˜ 0 ∂ g˜ 0 (x, t), (x, t) ∈ Ct [0, T ]; As0 , g˜ 0 (x, t), 2 ∂t ∂t ! $ ∂ g˜ $ $ ∂ 2 g˜ $ $ 0$ $ 0$ max g˜ 0 s0 (t), $ $ (t), $ 2 $ (t) ≤ R, s0 ∂t s0 ∂t

# " max βs0 (y), H s0 (y), max(1, ν02 ) exp(±r (x, 0)t/2)s0 (t) ≤ R, " # max ϕi0 s0 (t) ≤ R, i = 5, 6, 7, 8, 9, ν02 = max ν 2 (φ −1 (y)), t ∈ [0, T ], y ∈ [0, T /2], R > 0. 0≤y≤ T2

Then there is b ∈ (0, T /2) , that for any s ∈ (0, s0 ) in the domain & sT = DT {(x, y, t)| 0 ≤ y ≤ b(s0 − s)} there is a unique solution of the system of equations (7.75), for which   ϕ1 (x, y, t), ϕ3 (x, y, t) ∈ C(y,t) PsT ; As0 ,

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

295

  ϕi (x, t) ∈ Ct [0, b(s0 − s)]; As0 , i = 5, 6, 7, 8, 9, PsT = G T with

'

{(y, t)| 0 ≤ y ≤ b(s0 − s)} ,

R ϕi − ϕi0 s ≤ R, i = 1, 6, 8, ϕi − ϕi0 s ≤ , i = 3, 7, 9, s0 − s ϕ5 − ϕ50 s ≤

R , (y, t) ∈ PsT . (s0 − s)2

Under the conditions of the Theorem 7.3 ϕi0 ∈ C(y,t) (G T ; As0 ), ϕi0 s (y, t) ≤ R, i = 1, 3, (y, t) ∈ G T , 0 < s < s0 . Let bn be members of a monotonically decreasing sequence (2.17) and the number b is defined by equation (2.18). The number b0 ∈ (0, T /2) will be chosen appropriately. We construct a process of successive approximations with n = 0, 1, 2, ... by the scheme ( n. ϕ n+1 = ϕ 0 + Aϕ We define the function s  (y) by the formula s  (y) =

s + γ n (y) y . , γ n (y) = s0 − 2 bn

Let ψ n = ϕ n+1 − ϕ n . Then the following relations are true: ( 0, ψ 0 = Aϕ ψ1n+1 (x, y, t) 1 = 2

y t+y−ξ  H (x, ξ )wˆ n (x, ξ, τ ) −

a ˆn n (x, ξ, τ ) h (x, τ − ξ ) + kˆ00 2

0 t−y+ξ τ−ξ

−



 hˆ n (x, γ )wn+1 (x, ξ, τ − γ ) − h n (x, γ )wˆ n (x, ξ, τ − γ ) dγ

0

∂ " L 0 [ϕ8n+1 (x, τ ) exp (−r (x, 0)(τ )/2) , + ν 2 exp (r (x, 0)τ/2) ∂x

n n + ∂ψ2 (x, ξ, τ )] rˆ00 ∂x τ−ξ

− 0



#

∂ϕ2n n n (x, ξ, τ − γ )dγ dτ dξ, ψ8 (x, γ ) exp(−r (x, 0)γ /2) r00 + ∂x

(7.78)

296

7 Kernel Identification Problems in a Viscoelasticity System

ψ3n+1 (x, y, t) 1 = 2 −

y

" sgn(ξ ) H (x, y − |ξ |)wˆ n (x, y − |ξ |, ξ + τ )

−y

a ˆn h (x, ξ + t − y + |ξ |) 2

n (x, y − |ξ |, t + |ξ |) − + kˆ00

t−y+ξ  +|ξ |



hˆ n (x, γ )wn+1 (x, y − |ξ |, t + ξ − γ )

0

 − h n (x, γ )wˆ n (x, y − |ξ |, t + ξ − γ ) dγ + ν 2 exp (r (x, 0)(ξ + t)/2)

n ∂ " n + ∂ψ2 (x, y − |ξ |, ξ + t)] L 0 [ϕ8n+1 (x, ξ + t) exp (−r (x, 0)(ξ + t)/2) , rˆ00 × ∂x ∂x t−y+ξ  +|ξ |

n #

n + ∂ϕ2 (x, y − ξ, ξ + t − γ )dγ dξ, ψ8n (x, γ ) exp(−r (x, 0)γ /2) r00 ∂x

−

0 n+1 ψ5 (x, y) y 2 n ψ8 (x, ξ )g˜ 0tt (x, y − ξ ) − hˆ n (x, y − ξ ))ϕ8n+1 (x, ξ ) =

a

0

 y/2 1 n (x, ξ, y − ξ ) + h n (x, y − ξ ))ψ8n (x, ξ ) dξ + H (x, ξ )wˆ tn (x, ξ, y − ξ ) + kˆ00t a 0

1 − hˆ n (x, y − 2ξ )β(x, ξ ) − 2

y−2ξ 

hˆ n (x, τ )wtn+1 (x, ξ, y − ξ − τ )

0

 − h n (x, τ )wˆ tn (x, ξ, y − ξ − τ ) dτ + ν 2 exp (r (x, 0)(y − ξ )/2) 

∂ψ4n ∂ " n+1 n L 0 [ϕ8 (x, y − ξ ) exp (−r (x, 0)(y − ξ )/2) , rˆ00t + (x, ξ, y − ξ )] × ∂x ∂x

y−2ξ 

− 0

+

n # n + ∂ϕ4 (x, ξ, y − ξ − τ )dτ ψ8n (x, τ ) exp (−r (x, 0)τ/2) r00t ∂x

n r (x, 0) ∂ " n + ∂ψ2 (x, ξ, y − ξ )] L 0 [ϕ8n+1 (x, τ ) exp (−r (x, 0)(y − ξ )/2) , rˆ00 2 ∂x ∂x y−2ξ 

− 0



# ∂ϕ2n n n (x, ξ, y − ξ − τ )dγ ψ8 (x, γ ) exp(−r (x, 0)τ/2) r00 + ∂x

n 1 ∂  n + ∂ψ2 (x, ξ, ξ ) exp (−r (x, 0)(y − 2ξ )/2) ϕ8n+1 (x, y − 2ξ ) rˆ00 + 2 ∂x ∂x  

n n + ∂ϕ2 (x, ξ, ξ ) − ψ8n (x, y − 2ξ ) r00 dξ, ∂x

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

ψ6n+1 (x, y) =



y (y − ξ )

r 2 (x, 0) n hˆ n (x, ξ ) + r (x, 0)ψ7n (x, ξ ) − ψ6 (x, ξ ) 4

297

 dξ,

0

ψ7n+1 (x, y) =

y 

r 2 (x, 0) n hˆ n (x, ξ ) + r (x, 0)ψ7n (x, ξ ) − ψ6 (x, ξ ) 4

 dξ,

0

ψ8n+1 (x, y) =

y (y − ξ ) 0



× c0 (x)ψ8n (x, ξ ) − L[ϕ8n , hˆ n ] −

ξ

 ψ8n (x, ξ − τ )h n+1 (x, τ )dτ dξ,

0

ψ9n+1 (x, y) =

y

(c0 (x)ψ8n (x, ξ ) − L[ϕ8n , hˆ n ] −

0

ξ ψ8n (x, ξ − τ )h n+1 (x, τ )dτ )dξ, 0

n (x, y, t), rˆ n (x, y, t), w ˆ n (x, y, t), wˆ tn (x, y, t), hˆ n (x, y) have the form according where kˆ00 00 to (7.61), (7.62), (7.76) in which the corresponding components of the vector function ψ are supplied. We show that one can choose b0 ∈ (0, T /2) so that for ∀n = 0, 1, 2, . . . inequalities are satisfied  γ n (y) − s , ψ1n s (y, t) sup λn = max y (y,t,s)∈Fn  (γ n (y) − s)2 sup , ψ3n s (y, t) y (y,t,s)∈Fn (7.79)   (γ n (y) − s)3 γ n (y) − s n n sup , sup , ψ5 s (y) ψ6,8 s (y) y y (y,t,s)∈Fn (y,t,s)∈Fn  n 2

n  (y) (γ (y) − s) sup ψ7,9 < ∞, s y (y,t,s)∈Fn

ϕ1n+1 − ϕ10 s (y, t) ≤ R, ϕin+1 − ϕi0 s (y) ≤ R, i = 6, 8, R R , ϕin+1 − ϕi0 s (y) ≤ , i = 7, 9, s0 − s s0 − s R , (y, t, s) ∈ Fn+1 , ϕ5n+1 − ϕ50 s (y) ≤ (s0 − s)2

ϕ3n+1 − ϕ30 s (y, t) ≤

where Fn = {(y, t, s)|(y, t) ∈ G T , 0 ≤ y ≤ bn (s0 − s), 0 < s < s0 }.

(7.80)

298

7 Kernel Identification Problems in a Viscoelasticity System

Indeed, using the relations for ψin at n = 0 we obtain: y ψ10 s (y) ≤

a 0  + T h 0  w 0  (y − ξ ) Rw0 s + h 0 s + k00 s s s 2

0

 

∂ϕ20 ∂ 0 0 +R L 0 ϕ8 (x, τ ) exp(−r (x, 0)τ/2), r00 + s (ξ, τ ) dξ, ∂x ∂x 



  a 1 4 w0 s ≤ R 1 + + 2T R , h 0 s ≤ R , + R 1+ 2 2 a

where

ϕ20 s (y, t) ≤ R(1 + RT )w0 s ≤ R(1 + RT )R 0  (y, t) ≤ a R 2 k00 s 2

 4 1 + R 1+ , 2 a

   ∂ϕ80 T + s , s ∈ (0, s0 ), (y, t) ∈ G T . R 1+ 2 ∂x ∂ϕ 0

∂ϕ 0

0  , 2 , ∂ L  : Let us use inequality (2.35) to estimate  ∂ x8 s , r00 s ∂x s ∂x 0 s 0 ∂ϕ R R2 0  (y, t) ≤ , r00 ,  8 s (y) ≤  s ∂x s (y) − s s  (y) − s   0 ∂ 0 + ∂ϕ2  (y, t)  L 0 ϕ80 exp(−r (x, 0)(t ± y ∓ ξ )/2), r00 s ∂x ∂x    ϕ20 s  ϕ20 s  1 0 2 0 r00 s  +  + 2T R r00 s  +  ≤  s (y) − s s (y) − s s (y) − s

≤

R 2 + ϕ20 s  (s  (y) − s)2

(1 + 2T R 2 ).

Taking the function s  (ξ ) from (7.78) for n = 0, taking into account the above inequalities we have: y ψ10 s (y, t) ≤ μ0 (a, T, R, s0 ) 0

y−ξ y dξ ≤ b0 μ0 (a, T, R, s0 ) 0 . (γ 0 (y) − s)2 γ (y) − s

Similarly, 1 ψ30 s (y, t) ≤ 2

y −y

μ1 (a, T, R, s0 ) y dξ ≤ μ1 (a, T, R, s0 ) 0 , 0 2 (γ (y) − s) (γ (y) − s)2

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

a ψ50 s (y) ≤ 2

y R

299

 2 0 R + h s dξ a

0

1 + a

y/2 Rwt0 s +

1 0  + T h 0  w 0  Rh 0 s + k00t s s t s 2

0

  " ∂ ∂ϕ40 0 0 + R  L 0 ϕ8 (x, y − ξ ) exp(−r (x, 0)(y − ξ )/2), r00t + s ∂x ∂x 0 ∂ 0 + ∂ϕ2 ] L 0 [ϕ80 (x, y − ξ ) exp(−r (x, 0)(y − ξ )/2), r00 s ∂x ∂x    0 #

1 ∂ 0 + ∂ϕ2 +  ϕ80 (x, y − ξ ) exp(−r (x, 0)(y − ξ )/2) r00 s dξ, 2 ∂x ∂x

+ R

where   a wt0 s ≤ R 1 + + R(1 + T ) , ϕ40 s ≤ R(1 + T R)(Rw0 s + wt0 s ) + R 3 , 2   0 0  (y, t) ≤ Rk 0  + a R 2 R(1 + T R) + R 2 +  ∂ϕ9  (y) , k00t s s 00t s 2 ∂x 0  (y, t) ≤ r00t s

∂ϕ 0 R 2 (1 + R) R ,  9 s (y) ≤  , s ∈ (0, s0 ), (y, t) ∈ G T .  s (y) − s ∂x s (y) − s

It follows from the above estimates that ψ50 s (y) ≤ μ2 (a, T, R, s0 )

y (γ 0 (y) − s)2

≤ s0 μ2 (a, T, R, s0 )

y (γ 0 (y) − s)3

.

Doing the same, we have y ψ60 s (y) ≤

(y − ξ )(h 0 s + 2R 2 )dξ 0



y 4 , ≤ s0 RT 1/2 + 2R + R 1 + a γ 0 (y) − s

y ψ70 s (y) ≤ 0



y 4 (h 0 s + 2R 2 )dξ ≤ s02 R 1/2 + 2R + R 1 + , a (γ 0 (y) − s)2

300

7 Kernel Identification Problems in a Viscoelasticity System

y ψ80 s (y) ≤

(y − ξ )(R 2 + h 0 s (1 + T R))dξ 0

≤ s0 RT y ψ90 s (y) ≤



4 R + 1/2 + R 1 + a



 (1 + T R)

y , γ 0 (y) − s



4 (1 + T R))dξ (R 2 + R 1/2 + R 1 + a

0

≤ s02 R(R + (1/2 + R(1 +

4 y . ))(1 + T R)) 0 a γ (y) − s

These estimates imply the fulfillment of the inequalities (7.79) with n = 0. In addition, for (y, t, s) ∈ F1 we find ϕ11 − ϕ10 s (y, t) = ψ10 s (y, t) ≤

λ0 y λ0 b1 = λ0 b0 , ≤ 0 1 − b1 /b0 γ (y) − s

ϕ31 − ϕ30 s (y, t) = ψ30 s (y, t) ≤ ϕ51 − ϕ50 s (y) = ψ50 s (y, t) ≤ ϕ61 − ϕ60 s (y) = ψ60 s (y, t) ≤

λ0 y 4λ0 b0 ≤ , 0 3 (γ (y) − s) (s0 − s)3

λ0 y λ0 b1 = λ0 b0 , ≤ 1 − b1 /b0 γ 0 (y) − s

ϕ71 − ϕ70 s (y) = ψ70 s (y, t) ≤ ϕ81 − ϕ80 s (y) = ψ80 s (y, t) ≤

λ0 y 2λ0 b0 , ≤ s0 − s (γ 0 (y) − s)2

λ0 y 2λ0 b0 , ≤ 0 2 s0 − s (γ (y) − s)

λ0 y λ0 b1 = λ0 b0 , ≤ 1 − b1 /b0 γ 0 (y) − s

ϕ91 − ϕ90 s (y) = ψ90 s (y, t) ≤

λ0 y 2λ0 b0 ≤ . 0 2 s (γ (y) − s) 0−s

If we choose b0 so that 4λ0 b0 ≤ R, the inequalities (7.80) will be satisfied for n = 0. We show by induction that inequalities (7.79), (7.80) also occur for other n if b0 is chosen appropriately. Let inequalities (7.79), (7.80) be true for n = 0, 1, 2, . . . , j. Then, for (y, t, s) ∈ F j+1

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

j+1 ψ1 s (y, t) ≤



y

(y − ξ ) Rwˆ j s +

a ˆj j h s + kˆ00 s 2

0

+ T (hˆ j s w j+1 s + h j s wˆ j s ) j " ∂ ∂ψ2 j+1 j ]s + + R  L 0 [ϕ8 exp(−r (x, 0)t/2), rˆ00 + ∂x ∂x   j  #

∂ϕ ∂ j j T ψ8 exp(−r (x, 0)t/2) r00 + 2 s dξ. ∂x ∂x

It’s easy to see that λj y 1 j j , wˆ j s (y, t) ≤ ψ1 s + (a + T R)ψ8 s ≤ (1 + (a + T R)/2) j 2 γ (y) − s hˆ j s (y) ≤

1 2 j j j j ψ s + Rψ8 s + R(ψ9 s + ψ8 s ) 2 5 a

≤ (1/2 + R(1 + 2/a)s0 + 2Rs0 )

a j kˆ00 s (y, t) ≤ R 2 2 ≤

λj y (γ j (y) − s)3

,



 j ψ8 s j (1 + T R/2)ψ8 s +  s (y) − s

λj y a 2 R (3 + T R/2) j , 2 (γ (y) − s)2

j

ˆr00 s (y, t) ≤ R j

λj y ,  (s (y) − s)(γ j (y) − s) j

ψ2 s (y, t) ≤ R(wˆ j s (1 + 2RT ) + T ψ6 s w j s ) ≤ λj y , R((1 + 2RT )(1 + a/2 + T R/2) + T R(2 + a + 2T R)) j γ (y) − s  j ∂ψ2 ∂ j+1 0  L 0 ϕ8 exp(−r (x, 0)t/2), rˆ00 + s (y, t) ∂x ∂x   j ψ2 s λj y 1 j (1 + T R) ˆr00 s +  ≤  , ≤ μ3 (a, T, R) j s (y) − s s (y) − s (γ (y) − s)3 where

j

w j s (y, t) ≤ ϕ1 s +

a j j ϕ s + T Rϕ8 s ≤ R(2 + a + 2T R). 2 8

301

302

7 Kernel Identification Problems in a Viscoelasticity System 

Here and in further intermediate calculations the function s is determined by the equality (7.78) at n = j and the following inequalities are used

j

j

ϕi s ≤ 2R, i = 1, 6, 8, ϕ5 s ≤ R

1 + s02 (s0 − s)2

j

, ϕi s ≤ R

1 + s0 , i = 3, 7, 9, s0 − s

fair under the inductive assumption, as well as obvious inequalities b j ≤ b0 , γ j+1 (y) j ≤ γ j (y), ϕ2 s (y, t) ≤ R(1 + 2RT )w j s ≤ R 2 (1 + 2RT )(2 + a + 2T R), j

ϕ4 s ≤ R(Rw j s (1 + 2T R) + wt s (1 + 2RT ) + 2R 2 ) ≤ j

wt s ≤ j+1

Finally, for ψ1

μ4 (a, T, R, s0 ) , s0 − s

R(1 + s0 )(1 + a R/2) + 2R 2 s0 (1 + T ) . s0 − s

we get

j+1 |ψ1 s (y, t) ≤

y (y − ξ )μ5 (a, T, R, s0 ) 0

λjξ (γ j (ξ ) − s)3

dξ

y . ≤ λ j b02 μ5 (a, T, R, s0 ) j+1 γ (y) − s Similar arguments for ψi

j+1

, i = 3, 5, 6, 7, 8, 9, give us the following inequalities

1 j+1 ψ3 s (y, t) ≤ 2

y μ6 (a, T, R, s0 ) −y

≤ λ j b0 μ6 (a, T, R, s0 )

j+1 ψ5 s (y, t) ≤

μ7 (a, T, R, s0 )

λjξ (γ j (ξ ) − s)4

≤ λ j b0 μ7 (a, T, R, s0 )

j+1

y , (γ j+1 (y) − s)2

y 0

ψ6

λjξ dξ j (γ (ξ ) − s)3

dξ

y , (γ j+1 (y) − s)3

y s (y) ≤

(y − ξ ) 0

×

λ j ξ(1/2 + R(1 + 2/a)s0 + 2Rs0 )

(γ j (y) − s)3 y ≤ λ j μ8 (b0 , a, R, s0 ) j+1 , (γ (y) − s)

+

λjξ R

λjξ R + j (γ j (ξ ) − s)2 γ (ξ ) − s

 dξ

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

j+1

ψ7

s (y)

y ≤

λ j ξ(1/2 + R(1 + 2/a)s0 + 2Rs0 ) (γ j (y) − s)3

0

≤ λ j b0−1 μ4 8(b0 , s0 , R)

j+1

ψ8

j+1

ψ9

303

+

λjξ R

λjξ R + j j 2 (γ (ξ ) − s) γ (ξ ) − s

 dξ

y , (γ j+1 (y) − s)2

y , s (y) ≤ λ j μ9 (b0 , s0 , R, T ) j+1 γ (y) − s

s (y) ≤ λ j b0−1 μ9 (b0 , s0 , R, T )

y (γ j+1 (y) − s)2

, (y, t, s) ∈ Fi+1 .

From the estimates obtained, it follows that λi+1 ≤ λi ρ, λi+1 ≤ ∞, ρ = max(μ5 , μ6 , μ7 , max(1, b0−1 )μ8 , max(1, b0−1 )μ9 )). However, for (y, t, s) ∈ Fi+2 we have j+2

ϕi

− ϕi0 s ≤

j+1 

ψin s ≤

n=0

≤

j+1  n=0

j+1  n=0

λn y γ n (y) − s

j+1 j+1   λn b j+2 ≤ λn bn (n + 1)2 ≤ λ0 b0 ρ n (n + 1)2 , i = 1, 6, 8, 1 − bi+2 /bn n=0

j+2

ϕ5

n=0

− ϕ50 s (y) ≤

j+1 

ψ5n s (y)

n=0

≤

j+1  n=0

j+2

ϕi

j+1 j+1  λn b j+2 λn y 1 λ0 b0  n ≤ ≤ ρ (n + 1)6 , (γ n (y) − s)3 (s0 − s)2 (1 − bi+2 /bn )3 (s0 − s)2 n=0

− ϕi0 s ≤

j+1 

ψin s ≤

n=0

≤

1 s0 − s

j+1  n=0

j+1  n=0

n=0

λn y (γ n (y) − s)2

j+1 λ0 b0  n ≤ ρ (n + 1)4 , i = 3, 7, 9. s0 − s (1 − bi+2 /bn )2

λn b j+2

Choose now b0 ∈ (0, T /(2s0 )) such that

n=0

304

7 Kernel Identification Problems in a Viscoelasticity System

ρ ≤ 1, λ0 b0

∞ 

ρ n (n + 1)6 ≤ R.

n=0

Then

j+2

ϕ1

j+2

0  (y) ≤ R, − ϕ10 s (y, t) ≤ R, ϕ6,8 − ϕ6,8 s j+2

ϕ5 j+2

ϕ3

− ϕ30 s (y, t) ≤

− ϕ50 s (y) ≤

R , (s0 − s)2

R R j+2 0  (y) ≤ , ϕ7,9 − ϕ7,9 , (y, t, s) ∈ F j+2 . s s0 − s s0 − s

Since the choice of b0 does not depend on the number of approximations, successive approximation ϕin , i = 1, 3, 5, 6, 7, 8, 9 belong C(y,t) (F; As ) , F =

∞ '

Fn

n=0

and for them we have the inequality n − ϕ 0  (y) ≤ R, ϕ1n − ϕ10 s (y, t) ≤ R, ϕ6,8 6,8 s

ϕ5n − ϕ50 s (y) ≤

R , (s0 − s)2

R R n − ϕ 0  (y) ≤ , ϕ7,9 , (y, t, s) ∈ F. ϕ3n − ϕ30 s (y, t) ≤ 7,9 s s0 − s s0 − s If s ∈ (0, s0 ), the series

∞ 

(ϕin − ϕin−1 )

n=0

converge uniformly in norm of space C(y,t) (PsT ; As ) , PsT = G T

'

{(y, t)| 0 ≤ y ≤ b(s0 − s)} ,

so ϕin → ϕi . The limit function ϕi are elements of C(y,t) (PsT ; As ) and satisfy the equations (7.75). Now we prove that the existing solution is unique. Let ϕi and ϕ¯i be any two solutions satisfying inequalities ϕ1 − ϕ10 s (y, t) ≤ R, ϕi − ϕi0 s (y) ≤ R, i = 6, 8, ϕ5 − ϕ50 s (y) ≤ ϕ3 − ϕ30 s (y, t) ≤

R , (s0 − s)2

R R , ϕi − ϕi0 s (y) ≤ , i = 7, 9, (y, t, s) ∈ F. s0 − s s0 − s

7.2 The Problem of Determining the Multidimensional Kernel of Viscoelasticity Equation

305

Let ψ˜i = ϕi − ϕ¯i and λ := max

 sup

(y,t,s)∈F

ψ˜ 1 s (y, t)

 sup

(y,t,s)∈F

ψ˜ 6,8 s (y)

 γ (y) − s (γ (y) − s)2 ˜ , sup , ψ3 s (y, t) y y (y,t,s)∈F

 γ (y) − s (γ (y) − s)2 , sup , ψ˜ 7,9 s (y) y y (y,t,s)∈F 

sup

(y,t,s)∈F

ψ˜ 5 s (y)



(γ (y) − s)3 , < ∞, y

∞  )

1 + 1/(n + 1)2

where γ (y) = s0 − y/b, b = b0

−1

. Then for functions ψ˜i we can get

n=0

the relations ψ˜ 1 (x, y, t) =

1 2

y t+y−ξ  H (x, ξ )w(x, ˜ ξ, τ ) −

a˜ h(x, τ − ξ ) + k˜00 (x, ξ, τ ) 2

0 t−y+ξ

−

τ −ξ  

 ˜ h(x, γ )w(x, ¯ ξ, τ − γ ) − h(x, γ )w(x, ˜ ξ, τ − γ ) dγ

0 2

+ ν exp (r (x, 0)τ/2)     ∂ " ∂ ψ˜ 2 L 0 ϕ¯ 8 (x, τ ) exp (−r (x, 0)(τ )/2) , r˜00 + × (x, ξ, τ ) ∂x ∂x τ −ξ 

− 0



#

∂ϕ2 (x, ξ, τ − γ )dγ dτ dξ, ψ˜ 8 (x, γ ) exp(−r (x, 0)γ /2) r00 + ∂x

ψ˜ 3 (x, y, t) =

1 2

y

" sgn(ξ ) H (x, y − |ξ |)w˜ n (x, y − |ξ |, ξ + τ )

−y

a n (x, y − |ξ |, t + |ξ |) − (h˜ n (x, ξ + t − y + |ξ |) + k˜00 2 t−y+ξ  +|ξ |



−

 ˜ h(x, γ )w(x, ¯ y − |ξ |, t + ξ − γ ) − h(x, γ )w˜ n (x, y − |ξ |, t + ξ − γ ) dγ

0

+ν 2 exp (r (x, 0)(ξ + t)/2)

×

    ˜ ∂ " n + ∂ ψ2 (x, y − |ξ |, ξ + t) L 0 ϕ¯8 (x, ξ + t) exp (−r (x, 0)(ξ + t)/2) , r˜00 ∂x ∂x

306

7 Kernel Identification Problems in a Viscoelasticity System t−y+ξ  +|ξ |



#

∂ϕ2 (x, y − ξ, ξ + t − γ )dγ dξ, ψ˜ 8 (x, γ ) exp(−r (x, 0)γ /2) r00 + ∂x

− 0

ψ˜ 5 (x, y) =

y 

2 ˜ ψ˜ 8 (x, ξ )g˜ 0tt (x, y − ξ ) − h(x, y − ξ ))ϕ¯ 8 (x, ξ ) a

0

y/2  1 H (x, ξ )w˜ t (x, ξ, y − ξ ) + k˜00t (x, ξ, y − ξ ) + h(x, y − ξ )ψ˜ 8 (x, ξ ) dξ + a 1˜ − h(x, y − 2ξ )β(x, ξ ) − 2

0 y−2ξ 



˜ h(x, τ )w¯ t (x, ξ, y − ξ − τ )

0

 − h(x, τ )w˜ t (x, ξ, y − ξ − τ ) dτ + ν 2 exp (r (x, 0)(y − ξ )/2)    "  ∂ ∂ ψ˜ 4 n L 0 ϕ¯8 (x, y − ξ ) exp (−r (x, 0)(y − ξ )/2) , r˜00t + × (x, ξ, y − ξ ) ∂x ∂x

 # ∂ϕ4 ψ˜ 8 (x, τ ) exp (−r (x, 0)τ/2) r00t + (x, ξ, y − ξ − τ )dτ ∂x 0     r (x, 0) ∂ " ∂ ψ˜ 2 L 0 ϕ¯8 (x, τ ) exp (−r (x, 0)(y − ξ )/2) , r˜00 + + (x, ξ, y − ξ ) 2 ∂x ∂x y−2ξ 

−



# ∂ϕ2 (x, ξ, y − ξ − τ )dγ ψ˜ 8 (x, γ ) exp(−r (x, 0)τ/2) r00 + ∂x 0  

 1 ∂ ∂ ψ˜ 2 exp (−r (x, 0)(y − 2ξ )/2) ϕ¯8 (x, y − 2ξ ) r˜00 + + (x, ξ, ξ ) 2 ∂x ∂x  

∂ϕ2 (x, ξ, ξ ) dξ, − ψ˜ 8 (x, y − 2ξ ) r00 + ∂x y−2ξ 

−

ψ˜ 6 (x, y) =

y

˜ (y − ξ )(h(x, ξ ) + r (x, 0)ψ˜ 7 (x, ξ ) −

r 2 (x, 0) ψ˜ 6 (x, ξ ))dξ, 4

0

ψ˜ 7 (x, y) =

y

˜ (h(x, ξ ) + r (x, 0)ψ˜ 7 (x, ξ ) −

r 2 (x, 0) ψ˜ 6 (x, ξ ))dξ, 4

0

ψ˜ 8 (x, y) =

y 0

˜ − (y − ξ )(c0 (x)ψ˜ 8 (x, ξ ) − L[ϕ8 , h]

ξ 0

¯ τ ))dξ, ψ˜ 8 (x, ξ − τ )h(x,

7.3 The Problem of Determining the One-Dimensional Matrix Kernel…

ψ˜ 9 (x, y) =

y

˜ − (c0 (ψ˜ 8 (x, ξ ) − L[ϕ8 , h]

0

ξ

307

¯ ψ˜ 8 (x, ξ − τ )h(x, τ ))dξ.

0

Applying to them the above estimates, we find the inequality λ ≤ λρ  , ρ  = max(μ5 , μ6 , μ7 , max(1, b−1 )μ8 , max(1, b−1 )μ9 )) < ρ < 1, here μ5,6,7 = μ5,6,7 (b, a, T, R, s0 ), μ8,9 = μ8,9 (b, T, R, s0 ). Hence, λ = 0. Therefore ϕi = ϕ¯i . The Theorem 7.3 is proved.

7.3 The Problem of Determining the One-Dimensional Matrix Kernel of the System of Viscoelasticity Applied geophysics takes an interest in one-dimensional inverse problem of viscoelasticity with a source which can be described as ∇δ(x − x0 )δ(t) (a source of explosive type). This source is interpretated as an instantaneous center of compression. We notice that there are explosions which have some direction. A distinctive feature of this paper is the use, first, of the explosive-type source as the right side of the considered system of viscoelasticity, and secondly, the after-effect kernel in the form of a diagonal matrix. Among the papers using this type of source, one can note [2], in which one-dimensional inverse coefficient elasticity problems are investigated, existence, uniqueness, and stability theorems for the solution of problems are proved. For x = (x1 , x2 , x3 ), (x1 , x2 ) ∈ R2 , x3 > 0, t ∈ R, let us consider the following system of integro-differential equations of dynamical viscoelasticity: ρ(x3 )

3  ∂ Ti j ∂ 2ui = + Fi (x, t), i = 1, 2, 3; 2 ∂x j ∂t

(7.81)

j=1

under the following initial and boundary conditions: u i |t 0, t ∈ R takes the form   ∂ 2v j + a˜ j δ(y)δ(t), j = 1, 2, = L k , + q(y)v j j ∂t 2 ∂ y2

∂ 2v j

(7.97)

 ∂ 2 v3 ∂ 2 v3 = L k , + q(z)v ˜ 2 3 ∂t 2 ∂z 2   1 ∂ v˜1 ∂ , μ(x3 ) + + (λ(x3 )(v˜1 + v˜2 )) x3 =ψ2−1 (z) ∂ x3 ∂ x3 ρ(ψ2−1 (z)) p(z) v j |t 0, where a˜ j =

(7.98)

(7.100) (7.101)

aj , j = 1, 2; ρ(+0)μ(+0)

  2   2 s  (y) p  (z) s (y) p (z) q(y) = , q(z) ˜ = . −2 −2 s(y) s(y) p(z) p(z) In (7.97), (7.98) the following equations were used, based on the properties of the Delta function: 1 1 δ(x3 ) = δ(x3 ), ρ(x3 ) ρ(+0) 

δ(y) ψ1 (x3 )  δ(x3 ) = δ(ψ1−1 (y)) = = δ(y) = δ(y)  −1 dx3 x=+0 dψ1 (y) | y=+0 dy     Let L k j , v j (·, t) exp −k j (0)t/2 = w j (·, t), j = 1, 2, 3.



ρ(+0) . μ(+0)

7.3 The Problem of Determining the One-Dimensional Matrix Kernel…

311

Then as is easy to see,   v j (·, t) = exp k j (0)t/2 w j (·, t) +

t

  r j (t − τ ) exp k j (0)τ/2 w j (·, τ )dτ,

0

where r j (t) is determined by (7.14) for each j. In terms of the new functions w j (·, t) and r j (t), Eqs. (7.97)–(7.101) for y > 0, z > 0, t ∈ R take the form ∂ 2w j

=

∂t 2

∂ 2w j ∂ y2

t + H j (y)w j −

h j (t − τ )w j (y, τ )dτ + a˜ j δ(y)δ(t), j = 1, 2, (7.102) 0

∂ 2 w3 ∂ 2 w3 ((z)w3 − = +H 2 ∂t ∂z 2

t h 2 (t − τ )w3 (z, τ )dτ + 0

 ∂ v˜1 ∂ , L k02 , μ(x3 ) + (λ(x3 )(v˜1 + v˜2 )) x3 =ψ2−1 (z) ∂ x3 ∂ x3 ρ(ψ2−1 (z)) p(z) 

exp (r2 (0)t/2) 

(7.103)

w j |t 0 there exists a unique inverse problem solution {k1 (t), k2 (t)} ∈ C 2 [0, T ]. Theorem 7.5 Suppose that k 1 (t), k 2 (t) ∈ K (h 0 ) are solutions to inverse problem with the data " # ρ 1 (ψ −1 (y)), μ1 (ψ −1 (y)), g 1 (t) , " # ρ 2 (ψ −1 (y)), μ2 (ψ −1 (y)), g 2 (t) ,

7.3 The Problem of Determining the One-Dimensional Matrix Kernel…

315

respectively. Then there is a positive number C = C(h 0 , h 00 , T ), where h 00 = max ρ 1 (y)C 3 0,ψ −1 (T /2)% , μ1 (y)C 3 0,ψ −1 (T /2)% , g 1 (t)C 3 [0,T ] ,

2 2 2   % % ρ (y)C 3 0,ψ −1 (T /2) , μ (y)C 3 0,ψ −1 (T /2) , g (t)C 3 [0,T ] , such that we have the stability estimate 

k 1 (t) − k 2 (t)C 2 [0,T ] ≤ C ρ 1 − ρ 2 C 3 0,ψ −1 (T /2)% + μ1 − μ2 C 3 0,ψ −1 (T /2)% + g 1 − g 2 C 3 [0,T ] .

(7.119)

The proofs of these theorems are similar to the proofs of Theorems 7.1, 7.2. Now let’s consider the direct problem of determining w3 (y, t). Let    ˜ t) = L k02 , exp (r2 (0)t/2) μ ∂ v˜1 + ∂ (λ(v˜1 + v˜2 )) L(z, −1 x3 =ψ2 (z) ∂ x3 ρ(ψ2−1 (z)) p(z) ∂ x3 L 2 (z, t), := ( L 1 (z, t) + ( where

(7.120)

 exp (r2 (0)t/2)  ∂ v˜1 ∂ ( μ + + v ˜ )) , v ˜ L 1 (z, t) = (λ( 1 2 x3 =ψ2−1 (z) ∂ x3 ρ(ψ2−1 (z)) p(z) ∂ x3 ( L 2 (z, t) t k02 (t − τ )

=

 exp (r2 (0)τ/2)  ∂ v˜1 ∂ μ + + v ˜ )) dτ. v ˜ (λ( 1 2 x3 =ψ2−1 (z) ∂ x3 ρ(ψ −1 (z)) p(z) ∂ x3 2

0

 Here  a˜ j θ (t − |y|) v˜ j (x3 , t) = s(ψ1 (x3 )) · exp(k j (0)τ/2) 2 y+t

2 t−|y−ξ  |

1 + 2

t

F j (ξ, τ )dτ dξ + y−t 2

×



 a˜

|ξ | y+τ

1 θ (τ − |y|) + 2 2 j

r j (t − τ ) exp(k j (0)τ/2)

(7.121)

0 | 2 τ −|y−ξ  y−τ 2

|ξ |

 F j (ξ, η)dηdξ dτ

y=ψ1 (x3 )

,

316

7 Kernel Identification Problems in a Viscoelasticity System

where τ−|ξ |

F j (ξ, τ ) = H j (ξ )w j (ξ, τ ) −

h j (s)w j (ξ, τ − s)ds, j = 1, 2. 0

Notice, that v˜ j (x3 , t) has a structure   (j (y, t) v˜ j (x3 , t) = θ (t − |y|)V

y=ψ1 (x3 )

for j = 1, 2, since for t → |y| + 0 the measure of the set (y, t) tends to zero. It follows (j (y, t) ∈ C((T )). from Theorem7.4 that V Then we have  ( (y, t)  ∂V ∂ v˜ j ρ(x3 )  (j (y, |y|) + θ(t − |y|) j sign(y)δ(t − |y|)V =− y=ψ1 (x3 ) ∂ x3 μ(x3 ) ∂y .  ∂ v˜ j ρ(x3 )  = s (ψ1 (x3 )) ∂ x3 μ(x3 )

Indeed,

y+t

 2  a˜ 1 j × exp(k j (0)t/2) θ(t − |y|) + 2 2 y−t 2

t−|y−ξ  |

F j (ξ, τ )dτ dξ



|ξ |

t r j (t − τ ) exp(k j (0)τ/2)

+ 0

×

y+τ

 a˜

1 j θ(τ − |y|) + 2 2

2 y−τ 2

τ −|y−ξ  |

 F j (ξ, η)dηdξ dτ y=ψ1 (x 3 )

|ξ |

   a˜ ρ(x3 ) j + s(ψ1 (x3 )) exp(k j (0)t/2) − sign(y)δ(t − |y|) 2 μ(x3 )  1 + 2 a˜ j + 2

y+t

ρ(x3 ) μ(x3 )



2

F j (ξ, t − |y − ξ |)sign(ξ − y)dξ y−t 2

ρ(x3 ) 1 sign(y)r j (t − |y|) exp(k j (0)|y|/2) − μ(x3 ) 2

ρ(x3 ) μ(x3 )

2 r j (t − τ ) exp(k j (0)τ/2)

0



y+τ

t ×



F j (ξ, t − |y − ξ |)sign(ξ − y)dξ dτ y−τ 2

. y=ψ1 (x 3 )

(7.122)

7.3 The Problem of Determining the One-Dimensional Matrix Kernel…

317

((z), ( We extend the functions w3 (z, t), H L(z, t) as even functions for z < 0. Then the problem of determining the w3 (z, t) (7.103)–(7.105) is equivalent to the problem of determining the even function w3 (z, t) (respect to z) satisfying the equalities (7.103), (7.104). It follows from the theory of hyperbolic equations that the solution of the problem (7.103), (7.104) is equivalent to the relation: 1 w3 (z, t) = 2

 

((ξ )w3 (ξ, τ ) − H

R2

τ

 h 2 (τ − s)w3 (s, τ )ds θ(t − τ − |z − ξ |)dτ dξ

0

+

1 2



( L(ξ, τ )θ(t − τ − |z − ξ |)dτ dξ.

(7.123)

R2

Since ( L 1 (ξ, τ ) contains the term with Delta function, we consider separately the double integrating of this term in equation (7.123): ∗

 θ (t − τ − |z − ξ |)δ(τ − |d(ξ )|)Q(ξ, τ )dξ dτ =

z i(z,t) n 

Q(ξ, |d(ξ )|)dξ dτ,

i=1z (z,t) ∗i

R2

where

(7.124)

d(ξ ) = ψ1 (ψ2−1 (ξ )), Q(ξ, τ ) = − 1

exp (r2 (0)τ/2) ρ(ψ2−1 (ξ ))μ(ψ2−1 (ξ )) p(ξ )

s(d(ξ ))sign(d(ξ ))

a   1 exp (k1 (0)τ/2) μ(ψ2−1 (ξ )) + λ(ψ2−1 (ξ )) × 2  a2 + exp (k2 (0)τ/2) λ(ψ2−1 (ξ )) , 2 z ∗i (z, t), z i∗ (z, t) are solutions of the equation t − |z − ξ | = |d(ξ )| with respect to ξ and z ∗i (z, t) ≤ z i∗ (z, t). As we can see, the right side (7.123) is continuous. The remaining terms on the right side of the equations (7.121), (7.122) will also be continuous by double integrating. It is obvious that the question of continuity L 2 (z, t) has already been solved, since this component part contains triple integrating. The only question is what will be the domain of continuity w3 (z, t). Let’s introduce the following symbols: (z, t) = {(ξ, τ ) : 0 ≤ τ ≤ t − |z − ξ |},  = {(ξ, τ ) : |d(ξ )| ≤ τ },  = {(z, t) : 0 < t ≤ ψ(z)}.

318

7 Kernel Identification Problems in a Viscoelasticity System

The function ψ(z) is defined by the values z and d(z) exactly as in work [3] (we omit the definition due to its bulkiness and refer the reader to this work). The following formulas follow from the definition of functions ψ(z), d(z) and sets (z, t), ,  : ' (z, t)  = ∅, (z, t) ∈ , (z, t)

'

=

n 2

Di (z, t), (z, t) ∈ R 2 \ ,

i=1

Di (z, t) = {(ξ, τ ) : z ∗i (z, t) ≤ ξ ≤ z i∗ (z, t), |d(ξ )| ≤ τ ≤ t − |z − ξ |}. Integrating terms with θ (t − |d(z)|) in L 1 (ξ, τ ), in particular, we obtain: 

 θ (t − τ − |z − ξ |)θ(t − |d(ξ )|)P(ξ, τ )dξ dτ = θ (t − ψ(z))

(z,t)

R2

= θ (t − ψ(z))

n  

&

P(ξ, τ )dξ dτ 

P(ξ, τ )dξ dτ

i=1D (z,t) i ∗

= θ (t − ψ(z))

z i(z,t) t−|z−ξ  | n 

P(ξ, τ )dξ dτ,

i=1z (z,t) |d(ξ )| ∗i

P(ξ, τ ) =

×

exp (r2 (0)τ/2)

ρ(ψ2−1 (ξ )) p(ξ )

s(d(ξ ))

  dλ  a˜ 1 a˜ 2 exp (k1 (0)τ/2) + exp (k2 (0)τ/2) . 2 2 dx3 x =ψ −1 (ξ ) 3 2

Other terms c θ (t − |d(z)|) will be presented in the same way. ∂ v˜

Thus, taking into account the structure v˜ j (x3 , t) and ∂ x j , in the domain t > ψ(z) we find 3 w3 (z, t) =



1 2

(z,t)

1 + 2



&



(z,t)

((ξ )w3 (ξ, τ ) − H



&

τ

 h 2 (τ − s)w3 (s, τ )ds dτ dξ

0

( L(ξ, τ )dτ dξ.

(7.125)



The equation (7.125) is a Volterra integral equation& of the second kind with a continuous kernel and a continuous free term in the domain (z, t) , t ∈ [0, T ] for any positive number t. From the theory of integral equations it follows that (7.125) has a unique solution in the

7.4 The Problem of Determining the One-Dimensional Kernel…

319

& domain (z, t) . This solution can be found by the method of successive approximations. Each iteration leads to a continuous function. The limit function is a continuous solution of the equation (7.125).

7.4 The Problem of Determining the One-Dimensional Kernel Thermoviscoelasticity Equation For x = (x1 , x2 , x3 ), (x1 , x2 ) ∈ R2 , x3 > 0, t ∈ R let’s consider the following system of integro-differential equations of dynamical thermoviscoelasticity: ρ

3  ∂ Ti j ∂ 2ui = , i = 1, 2, 3, ∂x j ∂t 2 j=1

∂H = kH, ∂t

(7.126)

(7.127)

under the following initial and boundary conditions:

u i |t 0, θ1 (t) = tθ(t), θ(t) = 1 (1 , T of the Dirac delta function; T for t ≥ 0, θ (t) = 0, for t < 0, Ti j is the stress tensor (6.67), (1.87). Here ρ = ρ(x3 ), μ = μ(x3 ), λ = λ(x3 ) ∈ 1 . The inverse problem: find the kernel h(t), t > 0, appearing in (7.126), by using formula (6.67) if the following additional information about the solution of problem (7.126)–(7.131) is known: (7.132) u 3 (x3 , t)|x3 =+0 = g(t), t > 0, g(t) is given function. For the function H (x3 , t) (see Sect. 1.3) equalities (7.126), (7.128), (7.130) are equivalent to the relations ∂ 2u3 ρ(x3 ) 2 ∂t

320

7 Kernel Identification Problems in a Viscoelasticity System

  ∂ ∂u 3 (λ(x3 ) + 2μ(x3 )) = L h, − (3λ(x3 ) + 2μ(x3 ))R(H (x3 , t)) , ∂ x3 ∂ x3

(7.133)

u 3 |t 0, 2 ∂y ∂t ∂y

(7.136)

v|t 0,

(7.139)

where q(y) :=

  2 s (y) s  (y) −2 , λ˜ (y) := λ(ψ −1 (y)), s(y) s(y)

((y, t) := H (ψ −1 (y), t), μ(y) ˜ := μ(ψ −1 (y)), H t   ((y, τ ))dτ . ((y, t)) + h(t − τ )R( H Q[h](y, t) := (3λ˜ (y) + 2μ(y)) ˜ R( H 0

7.4 The Problem of Determining the One-Dimensional Kernel…

321

   Let v(y, t) + 0t h(t − τ )v(y, τ )dτ exp (−h(0)t/2) = w(y, t). Then as is easy to see t r (t − τ ) exp (h(0)τ/2) w(y, τ )dτ,

v(y, t) = exp (h(0)t/2) w(y, t) + 0

where r (t) is determined by 6.76. In terms of the functions w(y, t) and r (t), equations (7.136)–(7.139) take the form 3 ∂ 2w ∂ 2w ∂ = + B(y)w − a(y)a(0) Q 0 [h](y, t) − 2 2 ∂y ∂t ∂y

t

˜ − τ )w(y, τ )dτ, h(t

(7.140)

0

w|t 0 and, in a neighborhood of the characteristic line t = y, has the following structure: w(y, t) = −a(0)δ(t − y) + θ (t − y)w(y, ˜ t), where w(y, ˜ t) is a regular function. The function w(y, ˜ t) satisfies the equations (w = w˜ for t > y):

(7.144)

322

7 Kernel Identification Problems in a Viscoelasticity System

3 ∂ 2w ∂ 2w ˜ − y) − a(y)a(0) ∂ Q 0 [h](y, t) = + B(y)w + a(0)h(t 2 2 ∂y ∂t ∂y t ˜ − τ )w(y, τ )dτ, − h(t

(7.145)

y

w|t=y+0 =

a(0)r (0) a(0) − 2 2

y B(ξ )dξ := β(y),

(7.146)

0

 ∂w  = a(0)Q 0 [h](0, t), ∂ y  y=+0

(7.147)

t w| y=+0 = g˜ 0 (t) +

h 0 (t − τ )g˜ 0 (τ )dτ − a(0)h 0 (t).

(7.148)

0

  Provided that, for y = t = 0 the functions w(y, t) and ∂w ∂ y (y, t) are continuous, using

equations (7.146)–(7.148), we can (as in Sect. 7.1) easily express r (0), r  (0) in terms of the known values: r (0) = −

1 2  2 g0 (0), r  (0) = −q(0) − 2 g02 (0) − g˜ (0). a(0) a(0) 0 a (0)

(7.149)

Lemma 7.3 Let the function g(t) be expressed as 1

g(t) = −a(0)δ(t) + θ (t)g0 (t), a(0) = {[λ(+0) + 2μ(+0)]ρ(+0)}− 2 and let g0 (t) ∈ C 2 [0, T ] , θ (t) be the Heaviside function. In addition, let   ((T /2, T /2)]. (ρ, λ, μ) ∈ C 3 0, ψ −1 (T /2 , α(z) ∈ C[0, H

is equivaThen, for (y, t) ∈ DT , DT := ((y, t)|0 ≤ y ≤ t ≤ T − y) problem   (2.14)–(2.17) ˜ (y, t), h(t), h 0 (t), lent to the problem of determining the vector functions w(y, t), ∂w ∂t 



h 0 (t), h 0 (t) from the following system of integral equations: t w(y, t) = β(y) + y

∂w (y, τ )dτ, ∂τ

(7.150)

7.4 The Problem of Determining the One-Dimensional Kernel…

323

∂w 1 (y, t) = (g˜ 0 (t − y) − r (0)g˜ 0 (t − y)) ∂t 2 

a(0) ˜ a(0) y+t a(0)  + − B h(t − y)y − h (t − y) 4 2 2 2 0 1 t−y 

 a( y+t 1 y+t y+t 2 )a(0) + − h 0 (t − y − τ )g˜ 0 (τ )dτ Q 0 [h] , 2 2 2 2 +

1 2

0



y 

a(0)  a (ξ )Q 0 [h](ξ, t − y + ξ ) 4a(ξ )

B(ξ )w(ξ, t − y + ξ ) + 0

t−y  ˜ )w(ξ, t − y + ξ − τ )dτ dξ h(τ − 0



t+y

1 + 2

2 

a(0)  a (ξ )Q 0 [h](ξ, t + y − ξ ) 4a(ξ )

B(ξ )w(ξ, t + y − ξ ) + y

˜ + y − 2ξ ) − + a(0)h(t

t+y−2ξ 

˜ )w(ξ, t + y − ξ − τ )dτ dξ, h(τ

0

(7.151)

 ˜ = 1 B  t − 2h  (t) h(t) 0 2 2   

  r 2 (0) t t 2 1 − r  (0) g˜ 0 (t) − B β + g˜ 0 (t) − r (0)g˜ 0 (t) + a(0) 2 2 2 2 

t a( 2t ) d 2 t t Q 0 [h] , + +4 2 h 0 (t − τ )g˜ 0 (τ )dτ 2 2 a(0) a (0) dt 0

+

1 a(0)

t

˜ )β h(τ

t −τ 2

 dτ

0 t

2 − a(0)

2  0

t−2ξ 

− 0

∂w (ξ, t − ξ ) − B(ξ ) ∂t



a(0)  ∂ a (ξ ) Q 0 [h](ξ, t − ξ ) 4a(ξ ) ∂t

˜ ) ∂w (ξ, t − ξ − τ )dτ dξ, h(τ ∂t 

h 0 (t) = −r (0) +

r 2 (0) 2

 − r  (0) t +

t 0

(7.152) (t − τ )h 0 (τ )dτ,

(7.153)

324

7 Kernel Identification Problems in a Viscoelasticity System

t r 2 (0)   − r (0) + h 0 (τ )dτ, h 0 (t) = 2 0  ˜ + h 0 (t) = −h(t)

(7.154)

 t r 2 (0) ˜ − τ )h 0 (τ )dτ. − r  (0) h 0 (t) − h(t 4

(7.155)

0

Proof of the Lemma 7.3 is similar to the proof of the Lemma 7.1. The main results of this section are the following theorems on global unique solvability and stability of the solution of the inverse problem. (1 , k, γ be fixed (0 , T Theorem 7.6 Let the assumptions of the lemma 7.3 hold, and let T, T (0 . Then there exists a unique solution of the inverse problem (1 > T positive numbers, where T (7.126)–(7.131) h(t) ∈ C 2 [0, T ] for any fixed T > 0. Let K (h ∗0 ) be the set of functions h(t) ∈ C 2 [0, T ], satisfying the inequality h(t)C 2 [0,T ] ≤ h ∗0 , where t ∈ [0, T ] and h ∗0 is a fixed positive constant. Theorem 7.7 Let h 1 (t), h 2 (t) ∈ K (h ∗0 ) be solutions of the inverse problem (7.126)–(7.131) with the data " # ρ 1 (ψ −1 (y)), λ1 (ψ −1 (y)), μ1 (ψ −1 (y)), α 1 (z), g01 (t) , "

# ρ 2 (ψ −1 (y)), λ2 (ψ −1 (y)), μ2 (ψ −1 (y)), α 2 (z), g02 (t) ,

(0 , T (1 , T ) and respectively. Then, there exists a positive number C = C(h ∗0 , h ∗00 , γ , T h ∗00 = max ρ j (y)C 3 0,ψ −1 (T /2)% , λ j (y)C 3 0,ψ −1 (T /2)% ,

j μ j (y)C 3 0,ψ −1 (T /2)% , α j (z)C[0, H ((T /2,T /2)] , g0 (t)C 2 [0,T ] , j = 1, 2 ,

such that the following stability estimate holds:  h 1 (t) − h 2 (t)C 2 [0,T ] ≤ C ρ 1 − ρ 2 C 3 0,ψ −1 (T /2)% + λ1 − λ2 C 3 0,ψ −1 (T /2)% 1 2 1 2  % % + α 1 − α 2 C 0, H ((T /2,T /2) + μ − μ C 3 0,ψ −1 (T /2) + g0 − g0 C 2 [0,T ] .

(7.156) Proofs of the Theorem 7.6, 7.7 are similar to the proofs of the Theorems 7.1, 7.2.

7.5 Conclusions

325

7.5 Conclusions In this chapter, the methods discussed in Chaps. 1, 2 and 3 were developed. It is shown that the resolvent of the memory kernel makes it possible to reduce the viscoelasticity equation to the studied model problems of determining kernel. For one-dimensional problems, global solvability is investigated and stability estimates in the appropriate classes of functions are obtained. In addition, methods have been developed for finding matrix diagonal kernel in the case of a source of explosive type, which is relevant in geophysics. Sufficient conditions for the global unique solvability of the problem of determining the diagonal memory kernel are formulated. For scalar kernels depending on two or more variables, the method of scales of Banach spaces, discussed in Chap. 2, is used. The theorem of local unique solvability of the finding of a two-dimensional kernel for the viscoelasticity equations is formulated.

References 1. Yakhno, V.G. 1990. Inverse problems for differetial equations of elasticity. Novosibirsk: Nauka (Russian). 2. Karchevsky, A.L., and V.G. Yakhno. 1999. One-dimensional inverse problems for systems of elasticity with a source of explosive type. Journal of Investigation Ill - Posed Problems 7 (4): 347–364. 3. Yakhno, V.G., and I.Z. Merazhov. 2000. Some direct problems and one-dimensional inverse electroelasticity problem for “slow” waves. Siberian Advance Mathematical 10 (1): 87–150.

Chapter 8

Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

In this chapter, the Dirichlet-to-Neumann maps method is used to solve multidimensional inverse problems. In the first section, we study the problem of determining the multidimensional coefficient at the lowest derivative of a second-order hyperbolic equation for a half space. The next two sections investigate the problem of finding the multidimensional kernel of an integral in an integro-differential equation. Considered in these sections, problems resemble that of [1, 2] in their forms; however, in the proof it applies the Fourier transform in the second section and Laplace transform in the third section in time to reduce the problems to a family of stationary problems and employ the technique developed for elliptic equations.

8.1 The Dirichlet-to-Neumann Maps Method for a Half Space Consider the problem of determining the coefficient q(x, y) of the equation: u tt = u x x +  y u − q(x, y)u, y ∈ Rn , x ∈ R+ , t ∈ [0, T ],

(8.1)

in the case when for the solution of the initial–boundary value problem with the data u(x, y, 0) = 0, u t (x, y, 0) = 0, (x, y) ∈ Rn+1 = {(x, y) : y ∈ Rn , x > 0}, + u x (0, y, t) = g(y, t),

y ∈ Rn , t ∈ [0, T ],

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_8

(8.2) (8.3)

327

328

8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

the additional information u(0, y, t) = f g (y, t),

y ∈ Rn , t ∈ [0, T ].

(8.4)

is given. The uniqueness of the inverse problem solution in the class L ∞ (Rn+1 + ) can be proved only in the overdetermined statement. Rakesh [3] extended the SylvesterUllman method [4] to the inverse problem (8.1)–(8.4). Following [4] let us briefly describe the scheme of the study. We define the Dirichlet-to-Neumann map q : L 2 (Rn × [0, T ]) → H 1 (Rn × [0, T ]), which each function g(y, t) ∈ L 2 (Rn × [0, T ]) maps to the function f g (y, t) ∈ H 1 (Rn × [0, T ]) which is the trace of the solution of the direct problem (8.1)– (8.3) on the plane x = 0. The definitions of the classes of functions L 2 (G) and H 1 (G) = W21 (G), where G is a connected open domain in Rn the reader can find in the book [5]. We denote by B(r ) ⊂ R n+1 an open ball of radius r with the center at zero: ⎧ ⎫   n ⎨ ⎬   B(r ) = (x, y) : x 2 + yi2 < r . ⎩ ⎭ i=1

Theorem 8.1 Suppose that the functions q1 , q2 belong to the space L ∞ (R+ × Rn ) and are constant outside the closed ball B(r ). Then, if (π + 1)r < T, and from the condition q1 (g) = q2 (g), g ∈ L 2 (Rn × [0, T ]), it follows q1 ≡ q2 . Proof Suppose that outside the ball B(r ) the functions q1 and q2 are equal to the same constant k (the general case will follow from Lemma 8.3). Let g(y, t) be a sufficiently smooth function that has a compact support for every t ∈ [0, T ]. We denote p = q2 − q1 , w = u 1 − u 2 , where u j is the solution of the direct problem (8.1)–(8.3) with boundary data g(y, t) and a fixed coefficient q(x, y) = q j (x, y), j = 1, 2. Note that since g(y, t) has a compact support for a fixed t ∈ [0, T ], the solutions u j will also have a compact support for a fixed t ∈ [0, T ]. From the condition q1 (g) = q2 (g), it follows that wtt = wx x +  y w − q1 w + pu 2 , x > 0, y ∈ Rn , t ∈ [0, T ],

(8.5)

w(x, y, 0) = 0, wt (x, y, 0) = 0, x > 0, y ∈ Rn ,

(8.6)

8.1 The Dirichlet-to-Neumann Maps Method for a Half Space

w(0, y, t) = 0, wx (0, y, t) = 0, y ∈ Rn , r ∈ [0, T ].

329

(8.7)

Note that p(x, y) = 0 outside the ball B(r ). Let ε = (T − (π + 1)r )/6, T∗ = T − π(r + ε), the function v(x, y, t) satisfies the following conditions vtt = vx x +  y v − q1 (x, y)v, (x, y ∈ H = {(x, y) ∈ B(r ) : x > 0}, v(x, y, T∗ ) = 0, vt (x, y, T∗ ) = 0, (x, y) ∈ H

(8.8) (8.9)

and is quite smooth. Then, using (8.5) and integrating in parts, we get T∗

T∗ pu 2 v dxdydt =

0

(wtt − wx x −  y w + q1 w)vdxdydt 0

H

H

T∗ =

(vtt − vx x −  y v + q1 v)wdxdydt 0

+

H

 T∗ t=T∗ ∂v ∂w  w −v dSdt + (wt v − wvt ) dxdy, t=0 ∂ν ∂ν 0 ∂H

H

∂ where ∂ν is the normal derivative on the surface of the hemisphere H. It can be shown [3] that

w(x, y, t) = 0, Hence,

∂w (x, y, t) = 0 if (x, y) ∈ ∂H, t ∈ [0, T∗ ]. ∂ν

T∗ p(x, y)u 2 (x, y, t)v(x, y, t)dxdydt = 0

(8.10)

0 ∂H

for all g ∈ C ∞ (Rn × [0, T ]), for all corresponding solutions u 2 of the problem (8.1)– (8.3) and all smooth v satisfying (8.8)–(8.9). It follows from (8.10) that p(x, y) = 0. We will highlight the sets of functions u 2 and v that concentrate along some lines lying in space (x, y). The construction of such functions is based on the following Lemma [3]. Lemma 8.1 Let z ∈ Rn+1 be an arbitrary unit vector, σ > 0, the function α(x, y) ∈ C ∞ (Rn+1 ) is finite, and

330

8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

supp{α} ⊂ N = {(x, y) ∈ B(2 ) : x < − }

(8.11)

Then, for any locally bounded functions q(x, y), we can construct functions v(x, y, t) and u 2 (x, y, t) of the forms v(x, y, t) = α ((x, y) + t z) e−iσ ( (x,y),z +t) + β1 (x, y, t), u 2 (x, y, t) = α ((x, y) + t z) e+iσ ( (x,y),z +t) + β2 (x, y, t),

(8.12)

such that u 2 is the solution of (8.1)–(8.3) with the appropriate choice of g(y, t), v satisfies (8.8)–(8.9), and there are estimates β1  L 2 (H ×[0,T∗ ]) ≤

c c , β2  L 2 (R+n+1 ×[0,T ]) ≤ , σ σ

in which the constant c depends only on T , α, and q. Using Lemma 8.1, we define the functions v and u 2 as (8.12) and substitute them in (8.10): T∗ 0= p(x, y)u 2 (x, y, t)v(x, y, t)dxdydt 0 ∂H

T∗ =

p(x, y)α 2 ((x, y) + t z)dxdydt + γ (z, σ ). 0 ∂H

It can be shown [3] that |γ (z, σ )| ≤

c , σ

where the constant c is independent on σ . Therefore, T∗ 0 = lim

σ →∞

p(x, y)u 2 (x, y, t)v(x, y, t)dxdydt 0 ∂H

T∗ p(x, y)α 2 ((x, y) + t z)dxdydt

= 0 ∂H

 T∗ = R n+1

 p((x, y) − t z)dt α 2 (x, y)dxdy.

0

Here we used the fact that p(x, y) = 0 outside H . In view of the completeness of the class of functions α 2 (x, y) with the condition (8.11) in L 2 (N ), it follows that

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

331

T∗ p(a − t z)dt = 0 0

for all a ∈ N , z ∈ Rn+1 , z = 1. Due to the finiteness of p(x, y), we get ∞ p(a − t z)dt = 0 0

for all a ∈ N , z ∈ Rn+1 , z = 1. It remains to apply Lemma 9.1.2 [6] and Lemma 9.1.3 [3] to complete the proof of the Theorem 8.1. Lemma 8.2 Let p(x, y) be a function bounded on Rn+1 with a compact support whose convex hull is separated from some infinite subset A ⊂ R n+1 . Then if for all a ∈ A and z ∈ Rn+1 , z = 1 the equation ∞ p(a + sz)dt = 0, 0

is fair, then p(x, y) = 0. Lemma 8.3 . Suppose that k j ( j = 1, 2) are the real numbers and for j = 1, 2 the functions u j (x, y, t) satisfy the relations j

u tt = u xj x +  y u j − k j u j , y ∈ Rn , x ≥ 0, t ∈ [0, T ], j

u j (x, y, 0) = 0, u t (x, y, 0) = 0, y ∈ Rn , x ≥ 0, ∂u j (0, y, t) = g(y, t), y ∈ Rn , t ∈ [0, T ]. ∂x If u 1 (0, 0, t) = u 2 (0, 0, t) for t ∈ [0, T ] for all g ∈ C 2 (Rn × [0, T ]), such that supp g ⊂ {y : y < 2T } × [0, T ], then k1 = k2 .

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator In this section, we consider the problem of reconstructing the spatially inhomogeneous memory k(x, t) in the integro-differential wave equation over a given

332

8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Dirichlet–Neumann operator. An estimate of the conditional stability of the solution to the problem is obtained. The presentation of this section follows [7].

8.2.1 The Problem Statement and Results Let be a bounded domain in Rn , n ≥ 3, with a smooth boundary ∂ . Given the domain × R considers the following integro-differential equation with memory: t u tt (x, t) − u(x, t) −

u(x, τ )k(x, t − τ )dτ = 0.

(8.13)

−∞

Here k(x, t) is a sufficiently smooth memory function in × (0, ∞). We assume k(x, t) extended by zero to t < 0. Suppose that for a certain class of functions g(x, t) given on the boundary ∂ × R, there exists a sufficiently smooth solution u(x, t) of the Eq. (8.13) with the boundary condition   u(x, t)

∂ ×R

= g(x, y).

(8.14)

On the indicated class of functions we define the nonstationary Dirichlet-toNeumann operator H that assigns to g(x, t) the normal derivative ∂ν u of a solution to (8.13), (8.14) on the boundary ∂ × R. We consider the inverse problem consisting in reconstructing an unknown memory k(x, t) given the Dirichlet-to-Neumann operator H . We prove conditional stability of a solution to the problem; i.e., we show that for some class of functions k(x, t) we may guarantee closeness of two different memories, provided that the corresponding Dirichlet-to-Neumann operators are close. We suppose that the function k(x, t) belongs to the well-posedness set ¯ × [0, ∞)) | kC 2 ( ×[0,∞)) ≤ K }, W (K ) := {k(x, t) ∈ C 2 ( ¯ where K some positive number. Theorem 8.2 Let k j (x, t), j = 1, 2, be two (different) functions of the class W (K ). There is γ0 > 0 depending on and K and such that, for every γ ≥ γ0 , the corresponding Dirichlet-to-Neumann operators H j are well defined as operators from Hγ1 (∂ × R) into Hγ0 (∂ × R) (see the definition of the spaces and the corresponding norms in the next section) and the estimate k1 − k2 γ ,0 ≤ w(H1 − H2 ), holds with w(ε) ∼ C ,K ,γ ,n (C / ln ε−1 )1/(n+2) as ε → 0.

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

333

Here and in the sequel, the letters C with indices stand for constants whose indices indicate the quantities which the constants depend on.

8.2.2 The Direct Problem Assume that γ > 0. Denote by Hγs ( × R) the space of functions u(x, t) such that e−γ t u(x, t) ∈ Hγs ( × R) and denote the corresponding norm by  · γ ,s . Define Hγs (∂ × R) similarly, with the corresponding norm denoted by · |γ ,s . Solvability of the direct problem is established in the following theorem Theorem 8.3 Assume that K > 0. There are positive constants γ0 and C depending ¯ × (0, ∞)), kC ≤ on and K and such that, for arbitrary γ ≥ γ0 , k(x, t) ∈ C( K , and g(x, t) ∈ Hγ1 (∂ × R), there is a unique solution u(x, t) ∈ Hγ1 ( × R) to (8.13) with the boundary condition (8.14); moreover, the solution satisfies the estimate C (8.15) γ u2γ ,1 + ∂ν u 2γ ,0 ≤ g 2γ ,1 . γ Thus, the Dirichlet-to-Neumann operator acts correctly from Hγ1 (∂ × R) into × R). The bulk of the proof is contained in the following assertion which is a consequence of a more general result on solvability of boundary value problems for hyperbolic equations ([8], Theorems 11.1 and 13.5): Hγ0 (∂

Theorem 8.4 There are γ0 ≥ 0 and C depending on and such that, for arbitrary γ ≥ γ0 , g(x, t) ∈ Hγ1 (∂ × R), and f (x, t) ∈ Hγ0 (∂ × R), there is a unique solution u(x, t) ∈ Hγ1 ( × R) to the equation u tt − u = f (x, t)

(8.16)

given condition (8.14); moreover, the following estimate is valid:

γ u2γ ,1

+

∂ν u 2γ ,0

≤C

1  f γ ,0 + g 2γ ,1 γ

 (8.17) .

Proof Proof of Theorem 8.3. We seek a solution to (8.13) and (8.14) in the form u = u 0 + v, where u 0 is a solution to the problem u 0tt − u 0 = 0, u 0 |∂ ×R = g(x, t), and v satisfies the equation vtt − v = k ∗ (v + u 0 )

(8.18)

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

and the homogeneous condition v|∂ ×R = 0.

(8.19)

Here ∗ stands for convolution in time. According to Theorem 8.4, for γ ≥ γ0 there is a well-defined bounded linear operator E : Hγ0 ( × R) → Hγ1 ( × R) which assigns to a function f ∈ Hγ0 (∂ × R) a solution u to Eq. (8.16) with the homogeneous boundary conditions. Moreover, the norm E depends only on the domain

and γ0 . To solve the problem (8.18), (8.19), it suffices to solve the equation v = E(k ∗ v) + E(k ∗ u 0 )

(8.20)

in the space Hγ1 ( × R). Simple calculations using Young’s inequality show that k ∗ v2γ ,0 ≤ (e−γ t k) ∗ (e−γ t v)20,0 =



|(e−γ t k) ∗ (e−γ t v)|2 dtdx

R

≤

⎞2 ⎛ ∞ ⎞ ⎛∞ ⎝ |e−γ t k|dt ⎠ ⎝ |e−γ t v|2 dt ⎠ dx ≤ kC2 1 v2γ ,0 . γ2 0

0

Increasing γ0 of Theorem 8.4, if necessary, so as to have EK /γ0 ≤ q for some q, 0 < q < 1, we obtain E(k ∗ v)γ ,1 ≤ qvγ ,0 , γ ≥ γ0 , which guarantees unique solvability of Eq. (8.20). It is obvious that a solution to (8.20) is a unique solution to (8.18). Prove (8.16). Using (8.17), from (8.18) we derive γ v2γ ,1 + ∂ν v 2γ ,0 ≤

 CkC2   C k ∗ v2γ ,0 + k ∗ u 0 |2γ ,0 ≤ v2γ ,0 + u 0 |2γ ,0 . 3 γ γ

Increasing γ0 once again so as to have γ0 − C K /γ03 ≥ 1 and applying the estimate (8.18) for u 0 : u 0 2γ ,1 ≤ C g 2γ ,1 for γ ≥ γ0 we obtain γ u2γ ,1 + ∂ν u 2γ ,0 ≤

C g 2γ ,1 γ

with some new constant C. Theorem 8.3 is proven.

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

335

8.2.3 Auxiliary Assertions In Theorem 8.2 we claim existence of some γ > 0. First we take γ0 to be the quantity of Theorem 8.3. During the proof (of auxiliary assertions as well as the theorem) we possibly increase it. The Fourier-Laplace transform. Given a function u(x, t) such that ue−γ t ∈ L 2 ( × R) (or ue−γ t ∈ L 2 (∂ × Rn )), introduce the Fourier-Laplace transform in time as follows: 1 (Fu)(x, θ ) = u(x, ˆ θ ) := √ e−itθ u(x, t)dt, 2π where θ = σ − iγ , σ ∈ R. Then the Parseval identity holds: ˆ σ − iγ ) L 2 (∂ ×R) , u(x, t) γ ,0 = u(x,

(8.21)

where uˆ is considered as a function of x and σ . Applying the Fourier-Laplace transform to Eq. (8.13) and the boundary condition (8.14), we obtain ˆ θ ) + θ 2 )u(x, ˆ θ ) = 0, (8.22) u(x.θ ˆ ) + (k(x, ˆ θ ). u(x, ˆ θ )|∂ ×R = g(x,

(8.23)

Thus, u(x, ˆ θ ) is a solution to the family of the stationary problems (8.22), (8.23) depending on the parameter θ = σ − iγ , σ ∈ R. It follows from unique solvability of the problem (8.13), (8.14) that the Dirichlet problem (8.22), (8.23) has a unique solution for almost every θ = σ − iγ , σ ∈ R. Therefore, the inverse Fourier-Laplace transform of a solution to the family of the Dirichlet problems (8.22), (8.23) (with a suitable g(x, ˆ θ ) is a solution to the nonstationary problem (8.13), (8.14). ¯ j = 1, 2. Consider the followThe basic identity. Assume that q j (x) ∈ C 2 ( ), ing equation in : v j (x) + q j v j (x) = 0.

(8.24)

Suppose that the Dirichlet problem for (8.24) in has at most one solution. For each q j (x), define the so-called (stationary) Dirichlet-to-Neumann operator  j as follows:  j g := ∂ν v j |∂ , where v j is a solution to (8.24) given the condition v|∂ = g. The operator  j is a bounded operator from H 1 (∂ ) into H 0 (∂ ). This fact is a consequence of the general results on solvability of elliptic equations in the scale H s ( ) (see, for instance, [9], Chap. 2, Theorem 7.4) and the trace theorem (see, for instance, [9], Chap. 1, Theorem 9.4).

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Lemma 8.4 The following identity holds for sufficiently smooth solutions v j to (8.24), j = 1, 2:

(q2 − q1 )v1 v2 dx =

v1 (1 − 2 )v2 dS.

(8.25)

∂

Here and in the sequel, the notation like 1 v1 means that we apply the operator 1 to the trace v1 |∂ . In the case of real potentials q j (x), (14) was proven in [10]. The case of general elliptic operators was considered in [11]. Proof Multiply (8.24) written for v1 by v2 and integrate the result over : ( · (v1 v2 ) − v1 · v2 + q1 v1 v2 ) dx = 0.

Interchanging v1 and v2 , we similarly obtain ( · (v2 v1 ) − v2 · v1 + q2 v2 v1 ) dx = 0.

Subtracting these equalities from one another, applying the Gauss-Ostrogradskii formula, and recalling that

(q2 − q1 ) v1 v2 dx =

((1 v1 )v2 − (2 v2 )v1 ) dS. ∂

To complete the proof, we have to “transpose” the operator 2 on the right-hand ¯ 2 be the Dirichlet-to-Neumann operator corresponding to side from v2 to v1 . Let  the potential q¯2 . The bar over q2 stands for complex conjugation. We prove that ¯ 2 are mutually adjoint with respect to the inner product of the operators 2 and  L 2 (∂ ). Suppose that v(x) is a solution to the equation v∗ + q¯2 v∗ = 0, and v∗ (x) is a solution to the equation v∗ + q¯2 v∗ = 0. Multiplying the equations in v and v∗ by v¯∗ and v, ¯ respectively, we obtain ¯ − v∗ · v¯ + q¯2 v∗ v¯ = 0  · (v v¯∗ ) − v · v¯∗ + q2 v v¯∗ = 0;  · (v∗ v) Take the complex conjugation of the first of the equalities, subtract them from one another, and integrate the result over on using the Gauss-Ostrogradskii formula: ∂

∂v v∗ dS = ∂ν

∂

∂v∗ vdS. ¯ ∂ν

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

337

Hence, by the definition of  j

(2 v)v∗ dS =

∂

i.e.,

(2 v∗ )vdS, ¯ ∂

¯ 2 v∗ , v), (v∗ , 2 v) = (

where the parentheses stand for the inner product of L 2 (∂ ). For arbitrary data f, g ∈ H 1 (∂ ) we can find solutions v and v∗ to the corresponding equations such that v|∂ = g and v∗ |∂ = f . Thus, ¯ 2 f, g), ( f, 2 g) = ( ¯ 2 are mutually adjoint with respect to the inner product of L 2 (∂ ). 2 i.e., 2 and  ¯ 2 are mutually adjoint with respect to the inner product of L 2 (∂ ). and  Since v¯2 satisfies the equation v¯2 + q¯2 v¯2 = 0, we have 2 v2 =

∂ v¯2 ∂v2 ¯ 2 v¯2 . = = ∂ν ∂ν

¯ 2 are mutually adjoint, we write down Using the fact that the operators 2 and 

v1 2 v2 dS = ∂

v1 2 v2 dS =

∂

¯ 2 v¯2 dS v1 

∂

¯ 2 v¯2 ) = (2 v1 , v¯2 ) = = (v1 , 

(2 v1 )v2 dS ∂

which proves (8.25). Special solutions. Consider the equation ( + θ 2 )v(x) + q(x)v(x) = 0.

(8.26)

To prove Theorem 8.1, we need some generalization of one theorem by Sylvester and Uhlmann [4] (see also [12]). ¯ for some integer s ≥ 0. Then there are Theorem 8.5 Suppose that q(x) ∈ C s ( ) n 2 constants C√ 1 and C 2 such that, for every ζ ∈ C satisfying the conditions ζ · ζ + θ = 0, |ζ | ≥ 2, and |ζ | ≥ C1 qC s ( ) ¯ , equation (8.26) has a solution of the form v(x, ζ ) = eζ ·x (1 + w(x, ζ )),

(8.27)

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

where w(x, ζ ) satisfies the estimate w(·, ζ ) H s ( ) ≤

C2 qC s ( ) ¯ . |ζ |

(8.28)

The constants C1 and C2 depend only on and s. In the statement of Theorem 8.5, the dot stands for the “inner product without complex conjugation”; i.e., x · y = x1 y1 + . . . + xn yn . Theorem 8.5 was proven in [12] for s = 0. Our proof essentially repeats that of [12] with slight modifications. Proof Inserting (8.27) into (8.26) and recalling that ζ · ζ + θ 2 = 0, we obtain the following equation in w: w + 2ζ · w = −q(1 + w).

(8.29)

Consider the operator P :=  + 2ζ · . According to Theorem 10.3.7 of [13], there is a bounded linear operator E : L 2 ( ) → L 2 ( ) such that PE f = f ∀ f ∈ L 2 ( ); moreover, for each constant coefficient differential operator Q(D) weaker than P(D) the operator Q(D)E is bounded in L 2 ( ). However, we need the operator Q(D)E to be bounded in H s ( ). To show this, we repeat the proof of Theorem 10.3.7 [13] with minor changes. Introduce some “local” notations needed only in this proof (cf. [13]). Given a linear differential operator P(D), D = −i∂, with constant coefficients, define the ˜ ), ξ ∈ Rn , by the relation function P(ξ P˜ 2 (ξ ) =



|∂ α P(ξ )|2 .

|α|≥0

Suppose that k(ξ ), ξ ∈ Rn , is a tempered weight function (see [13], Definition 10.1.1) for a precise definition; we do not present the definition, only pointing out that the functions 1 and P˜ that will serve as k(ξ ) satisfy this definition). Define the space B p,k to be the set of tempered distributions u such that the Fourier transform uˆ := (2π )−n/2



e−iξ x u(x)dx

Rn

is a function and u p,k

⎛ ⎞1/ p = ⎝ |k(ξ )u(ξ ˆ ) p |dξ ⎠ < ∞ Rn

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

339

ˆ )u(ξ (for p = ∞ we put u∞,k := ess sup |k(ξ ˆ )|). A fundamental solution E of loc of the operator P is called regular if E belongs to the least local extension B∞, P˜ the space B∞, P˜ ; i.e., E belongs to the space containing B∞, P˜ and possessing the properties loc loc , ϕ ∈ C0∞ (Rn ) ⇒ uϕ ∈ B∞, ; u ∈ B∞, P˜ P˜ loc loc ⇒ u ∈ B∞, . ∀u ∈ D  ∀ϕ ∈ C0∞ (Rn ) uϕ ∈ B∞, P˜ P˜

According to Theorem 10.2.1 of [13], every differential operator with constant coefficients possesses a regular fundamental solution. Fix s and take f ∈ H s ( ). By the well-known results of analysis (see, for instance, [5], Chap. 3, Sect. 3.4, Theorem 1) there is a compactly-supported extension of f to an ε-neighborhood ε of with the class preserved; i.e., there is f 0 ∈ H s (Rn ) such that f = f 0 on , f 0 ≡ 0 beyond ε , and  f 0  H s (Rn ) ≤ C ,s  f  H s (Rn ) with some constant C ,s independent of f . Define E f as the restriction of E 0 ∗ f 0 to , where E 0 is a regular fundamental solution of P, and ∗ stands for convolution. Then P E f = f (see [13], Chap. 4, Sect. 4.4). In line with [13], put F0 = E 0 ψ, where ψ ∈ C0∞ (Rn ) and ψ = 1 in a neighborhood of − = {x − y| x, y ∈ }. Then F0 ∈ B∞, P˜ and F0 ∗ f 0 = E 0 ∗ f 0 in by Theorem 4.2.4 of [13]. Consequently (see [13], Theorem 2, p. 40), ∂ α |ζ |E f  L 2 ( ) ≤ |ζ |∂ α (F0 ∗ f 0 )2,1 ≤ |ζ |F0 ∞,1 ∂ α f 0  L 2 (Rn ) ≤ C ,s F0 ∞, P˜ sup ξ

|ζ |  f  H s ( ) , ˜ ) P(ξ

where C ,s depends only on and s. It was proven in [12] that F0 ∞, P˜ is bounded by a constant depending only on , n, and the order of P. We have thus obtained an operator E : H s ( ) → H s ( ) such that PE f = f and E f  H s ( ) ≤

|ζ | C ,n,s sup  f  H s ( ) . ˜ ) |ζ | ξ P(ξ

Below we show that sup ξ

|ζ | ≤ 1; ˜ P(ξ )

(8.30)

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

therefore, our estimate would take the form E f  H s ( ) ≤

C ,n,s  f  H s ( ) . |ζ |

(8.31)

Return to (8.29). Each solution to the equation w = −E(q(1 + w))

(8.32)

is a solution to (8.29). Since qw H s ( ) ≤ Cs qC s w H s ( ) from (8.31) we obtain Eqw H s ( ) ≤ 1/2w H s ( ) for |ζ | ≥ 2C ,n,s Cs qC s . This implies solvability of (8.32) together with the needed estimate. To complete the proof, it remains to establish (8.30), which amounts to the inequality P˜ 2 (ξ ) ≥ |ζ |2 , ξ ∈ Rn , Suppose that ζ = λ + iμ, with λ, μ ∈ Rn . Calculate the derivatives ∂ξα P(ξ ): P(ξ ) = −ξ 2 + 2iζ · ξ = −ξ 2 − 2μ · ξ + 2iλ · ξ ; ∂ξ j P(ξ ) = −2ξ j + 2iζ j = −2ξ j − 2μ j + 2iλ j ;  −2 for i = j; ∂ξ j ∂ξi P(ξ ) = 0 otherwise. Hence, P˜ 2 (ξ ) = (ξ 2 + 2μ · ξ )2 + 4(λ · ξ )2 + 4

n 

((ξ j + μ j )2 + λ2j ) + 12

j=1

≥ (ξ + 2μ · ξ + μ − μ ) + 4(ξ + μ)2 + 4λ2 2

2

2 2

= ((ξ + μ)2 − μ2 )2 + 4(ξ + μ)2 + 4λ2 . Since min((x − μ2 )2 + 4x) = 4(μ2 − 1) ≥ μ2 x≥0

for μ2 ≥ 2, under the conditions of the theorem we have P˜ 2 (ξ ) ≥ μ2 + 4λ2 ≥ μ2 + λ2 . Theorem 8.5 is proven.

8.2 Stability of Memory Reconstruction from the Dirichlet-to-Neumann Operator

341

Corollary 8.1 Under the conditions of the preceding theorem v(·, ζ ) H s ( ) ≤ C ,s eC |ζ | . Proof The estimate follows from the chain of the inequalities v(·, ζ ) H s ( ) ≤ Cs eζ x C s 1 + w H s ≤ Cs |ζ |2 eC |ζ | (1 H s + w H s )

 C2 2 C |ζ | s s 1 H + qC ≤ Cs |ζ | e |ζ |

 C2 ≤ C ,s eC |ζ | . ≤ Cs |ζ |2 eC |ζ | 1 H s + C1

8.2.4 Proof of the Main Result To proof Theorem 8.2 which is the main result of this section, we suppose that k j (x, t) ∈ W (K ), j = 1, 2, are two (different) memories. Consider Eq. (8.22) with k = k j . Denote a solution to (8.22) by uˆ j (x, t) and denote the corresponding stationary Dirichlet-to-Neumann operators by θj . By Lemma 8.4, we have the identity

(kˆ2 (x, θ ) − kˆ1 (x, θ ))uˆ 1 (x, θ )uˆ 2 (x, θ )dx =



uˆ 1 (x, θ )(θ1 − θ2 )uˆ 2 (x, θ )dS.

∂

By Theorem 8.5, Eq. (8.22) with k = k j possesses a solution uˆ j (x, ζ j , θ ) = eζ j ·x (1 + w j (x, ζ j , θ )), where ζ j ∈ Cn , ζ j · ζ j + θ 2 = 0, |ζ j | ≥ C1 k j (·, θ )C 2 ( ) , |Imζ j | ≥ satisfies the estimate w j (·, ζ j , θ ) H s ( ) ≤

C2 k j (·, θ )C 2 ( ) . |ζ j |

√ 2 and w j

(8.33)

Inserting the special solutions uˆ j into the identity, we obtain

(kˆ2 (x, θ) − kˆ1 (x, θ))e(ζ1 +ζ2 )·x dx



= ∂

uˆ 1 (x, ζ1 , θ)(θ1 − θ2 )uˆ 2 (x, ζ2 , θ)dS +



(kˆ2 (x, θ) − kˆ1 (x, θ))e(ζ1 +ζ2 )·x (8.34)

× (w1 (x, ζ1 , θ)w2 (x, ζ2 , θ) + w1 (x, ζ1 , θ) + w2 (x, ζ2 , θ))dx.

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Now, turn to choosing ζ j . Take an arbitrary ξ ∈ Rn . Choose ζ j so as to have ζ1 + ζ2 = −iξ, ζ j · ζ j + θ 2 = 0, √ |ζ j | ≥ C1 kˆ j (·, θ )C 2 ( ) , |Imζ j | ≥ 2.

(8.35)

We seek ζ j in the form ζ j = −iξ/2 + (−1) j (iμ + λ), where μ, λ ∈ Rn and ξ · μ = ξ · λ = 0. The first condition of (8.35) is obviously satisfied. The second leads to the equation ξ2 − μ2 + λ2 + (−1) j (ξ · μ) + 2i(λ · μ) − (−1) j i(ξ · λ) 4 = −θ 2 = −(σ − iγ )2 = −(σ 2 − γ 2 − 2iσ γ )

ζj · ζj = −

which amounts to the system ξ2 + μ2 − λ2 = (σ 2 − γ 2 ), λ · μ = σ γ . 4

(8.36)

As a solution, take μ and λ in Rn such that μ2 = σ 2 + r 2 and λ2 = ξ4 + γ 2 + √ r 2 , r ≥ 2. This is possible, since for such choice of μ2 and λ2 the first equation becomes an identity while the second can be fulfilled by choosing the angle between μ and λ as |μ| ≥ |σ | and |λ| ≥ γ . Consider the last but one condition in (8.35). We have 2

ξ2 ξ2 ξ2 + μ2 + λ2 = + 2μ2 + γ 2 − σ 2 = + σ 2 + 2r 2 + γ 2 4 2 2 ξ2 + |θ |2 = 2r 2 . = 2

|ζ j |2 =

(8.37)

To fulfill the last but one condition in (8.35), take the constant γ0 in the statement ˆ θ )C 2 ( ) for every k ∈ W (K ); this is obviously of Theorem 8.2 so that γ0 ≥ C1 k(·, possible. Finally, observe that the √ last condition in (8.35) is satisfied by virtue of √ ξ · μ = 0 and |μ| ≥ 2 for r ≥ 2. Introduce the notation ˜ θ ) := (2π )−n/2 (kˆ2 (x, θ ) − kˆ1 (x, θ ))e−iξ x dx k(ξ,

and pass from identity (8.34) to the inequality ˜ θ )| ≤ C eC |ζ1 | (θ1 − θ2 )uˆ 2 (x, ζ2 , θ ) L 2 (∂ ) + |k(ξ,

C ,K ,γ . |ζ j |

(8.38)

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343

Here we have used the Corollary to Theorem 8.5 and the trace theorem, uˆ 1  L 2 (∂ ) ≤ C uˆ 1  H 1 ( ) ≤ C eC |ζ1 | , together with the estimate

(kˆ2 (x, θ ) − kˆ1 (x, θ ))(w1 w2 + w1 + w2 )dx



≤ C ,γ K

C2 K C2 K 2 C2 K + + 2 |ζ1 | |ζ2 | |ζ1 ||ζ2 |

 ≤ C ,K

1 , |ζ j |

which ensues from (8.33). (Here we suppose that |ζ j | > 1 which can be fulfilled by choosing γ0 ≥ 1.) Introduce the function g(x, ˆ ζ2 , θ ), x ∈ ∂ , as follows: g(x, ˆ ζ2 , θ ) :=

u(x, ˆ ζ2 , θ ) , C |ζ |

2 e (1 + σ 2 )

where the constant C is taken from the Corollary to Theorem 8.5. From the same Corollary and the trace theorem we obtain g(·, ˆ ζ2 , θ ) H 1 (∂ ) ≤

C 1 + σ2

for σ ∈ R and every r in the definition of ζ j . (Observe that ζ2 depends on θ and r , and so gˆ depends actually on x, σ, and r.) Consequently,

g(·, ˆ ζ2 , θ )

H 1 (∂ )

(1 + σ )dσ ≤ 2

R

R

C dσ = C < ∞, 1 + σ2

which implies that the inverse Fourier-Laplace transform 1 g(x, t) = √ 2π

∞

ˆ h(x, ζ2 , σ − iγ )eit (σ −iγ ) dσ

∞

belongs to Hγl (∂ × R) and its norm g γ ,1 does not exceed some constant C depending only on the domain . (The vector ζ2 depends on θ = σ − iγ and r ; therefore, the function g(x, t) depends also on r as on a parameter. However, the norm g γ ,1 is bounded uniformly in r ; therefore, we do not indicate the dependence of g on r .)

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Integrating the square of (8.38) over {(σ, ξ ) ∈ Rn+1 | σ 2 + ξ 2 ≤ ρ 2 } with ρ > 0, we obtain √ ˜ θ )|2 dξ dσ ≤ C ,K eC ρ 2 +γ 2 +2r 2 (1 + ρ 2 )2 |k(ξ, σ 2 +ξ 2 ≤ρ 2



C ,K ρ n+1 . r2

(θ1 − θ2 )g(x, ˆ ζ2 , θ )2L 2 (∂ ) dσ dξ +

× σ 2 +ξ 2 ≤ρ 2

(8.39)

Continue (8.39) by using the equality F(H1 − H2 )g = (θ1 − θ2 )gˆ :

˜ θ )|2 dξ dσ ≤ C ,K ,γ eC |k(ξ,

√

ρ 2 +2r 2

(1 + ρ 2 )2 ρ n

σ 2 +ξ 2 ≤ρ 2

×

(F(H1 − H2 )g)(x, θ )2L 2 (∂ ) dσ +

σ ∈R

C ,K ρ n+1 . r2

Applying the Parseval identity (8.21) and the estimate g γ ,1 ≤ C , we now find that

˜ θ )|2 dξ dσ ≤ C ,K ,γ ,n eC |k(ξ,

√

ρ 2 +2r 2

σ 2 +ξ 2 ≤ρ 2

H1 − H2 2 +

C ,K ρ n+1 . (8.40) r2

Since k(x, t) ∈ C 2 ( × (0, ∞)) and kC 2 ≤ K , the function e−γ t k(x, t) belongs to H l/2 ( × (0, ∞)). It follows from the results of ([9], p. 71) that the function e−γ t k(x, t) extended by zero beyond × (0, ∞) belongs to the space H 1/2 (R4 ); moreover, e−γ t k H 1/2 (R4 ) ≤ C K . Using the equivalent norm in the space H 1/2 (R4 ), we infer that ˜ θ )|2 (1 + |ξ |2 + |σ |2 )1/2 dξ dσ ≤ C K |k(ξ, R

which implies that

˜ θ )|2 dξ dσ ≤ 0|k(ξ,

σ 2 +ξ 2 ≤ρ 2

C K C K . ≤ (1 + ρ 2 )1/2 ρ

(8.41)

Combining (8.40) and (8.41), we obtain Rn+1

√  ρ n+1 1 2 C ρ 2 +2r 2 2 ˜ . |k(ξ, θ )| dξ dσ ≤ C ,K ,γ ,n e H1 − H2  + 2 + r ρ

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Recovering a Time- and Space-Dependent Kernel in a Hyperbolic …

345

Putting r = ρ (n+2)/2 , estimate the expression in the parentheses by the quantity eC

√

ρ 2 +2ρ (n+2)

H1 − H2 2 +

which behaves as eC ρ

n/2+1

H1 − H2 2 +

2 ρ

2 ρ

at large ρ. Squaring both sides of our estimate, we obtain

 1 C ρ n/2+1 2 ˜ k L 2 (Rn+1 ) ≤ C ,K ,γ ,n e H1 − H2  + 1/2 ρ at large ρ; whence, by Corollary 2.1 of [14], k1 − k2 γ ,0 ≤ ω(H1 − H2 ), where ω(ε) ∼ C ,K ,γ ,n (C / log ε−1 )1/(n+2) as ε → 0. Theorem 8.2 is proven completely.

8.3 Recovering a Time- and Space-Dependent Kernel in a Hyperbolic Integro-Differential Equation from a Restricted Dirichlet-to-Neymann Operator In this section, it is proved that a space- and time-dependent kernel occurring in a hyperbolic integro-differential equation in three space dimensions can be uniquely reconstructed from the restriction of the Dirichlet-to-Neumann operator of the equation into a set of Dirichlet data of the form of products of a fixed time-dependent coefficient times arbitrary space-dependent functions. This problem was considered by Janno [15]. Note that determination of a time- and space-dependent heat flux relaxation function from a parabolic integro-differential equation by means of this technique was investigated in [16].

8.3.1 Problem Formulation and Results First we introduce some notation. Given a Banach space X, σ ∈ R, k ∈ {0, 1, 2, ...} and p ∈ [1, ∞) we define the following sets of abstract functions with values in X : Wσk, p ((0, ∞); X ) = {z : (0, ∞) → X : e−σ z(·) ∈ W k, p ((0, ∞); X )}

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Cσk ([0, ∞); X ) = {z : [0, ∞) → X : e−σ z(·) ∈ C k ([0, ∞); X )}. Here W k, p ((0, ∞); X ) is the abstract Sobolev space, and C k ([0, ∞); X ) consists of abstract functions bounded and continuous in [0, ∞) together with their derivap 0, p tives up to the order k. In case k = 0 we set L σ ((0, ∞); X ) := Wσ ((0, ∞); X ) 0 and Cσ ([0, ∞); X ) := Cσ ([0, ∞); X ). Moreover, in case X = C, we use the simk, p k, p plified notation Wσ (0, ∞) := Wσ ((0, ∞); C) and Cσk [0, ∞) := Cσk ([0, ∞); C). Finally, given k ∈ {0, 1, 2, ...} and μ > 0, we denote by C k,μ ( ) the space of functions which are Hölder-continuous of degree λ in together with their derivatives up to the order k. Let be a three-dimensional domain with a Lipschitz-boundary . In case is filled by the isotropic nonhomogeneous viscoelastic material, the constitutive relations and the system of equations of motion of the material point of contain timeand space-dependent relaxation functions a(t, x) and b(t, x), which correspond to the shear and bulk moduli, respectively (see [17], p. 122–127). We suppose that these functions are unknown. Unfortunately, inverse problems to determine both a(t, x) and b(t, x) in the viscoelasticity system are very complex. Therefore we essentially simplify the situation taking into consideration a partially theoretical scalar model of such kind. If the displacement field is solenoidal, then the system of equations of motion is reduced to three independent equations of divergence type, which can be summarized as t ρ(x)∂t2 v(t, x)

=

div[a(t − τ, x)∇∂τ v(τ, x)]dτ, x ∈ , t ∈ R

(8.42)

−∞

where v is an arbitrary linear combination of coordinates of the displacement and ρ is the density. We will pose and study an inverse problem that consists in determining the kernel a(t, x) in (8.42) for t ≥ 0 and x ∈ . Let us assume v(t, x) = 0 for t ≤ 0. Then the Eq. (8.42) for twice differentiable v(t, ·) is equivalent to the integrated equation t ρ(x)∂t v(t, x) =

(t − τ )div[a(t − τ, x)∇∂τ v(τ, x)]dτ, x ∈ , t ∈ R. (8.43) 0

We interpret the latter equation as a generalized form of (8.42) and complement it with the initial and Dirichlet boundary conditions v(0, x) = 0, x ∈ , v(t, x) = ψ(t, x), x ∈ , t > 0.

(8.44)

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Recovering a Time- and Space-Dependent Kernel in a Hyperbolic …

347

Let us associate with v the secondary variable t s(t, x) =

a(t − τ, x)∇∂τ v(τ, x)dt −∞

and denote by ν(x) the outer normal of the boundary of at the point x ∈ . Given ρ, a, and ψ, the solution v of the initial–boundary value problem (8.43), (8.44) determines the normal component of s at the boundary: t h(t, x) := −ν(x) · s(t, x) =

a(t − τ, x)∂ν ∂t v(τ, x)dτ, x ∈ , t > 0. (8.45) 0

The operator D, which depends on ρ and a and maps the Dirichlet data ψ via the solution v to the Neumann data h, is called the Dirichlet-to-Neumann operator of the problem (8.43), (8.44). We begin by eliminating the derivative in the Eq. (8.43) by the change of variable t u(t, x) = ∂t v(t, x) ⇔ v(t, x) =

u(τ, x)dτ.

(8.46)

0

Defining ϕ(t, x) := ∂t ψ(t, x), the problem (8.43), (8.44) is transformed to t ρ(x)u(t, x) =

(t − τ )div[a(t − τ, x)∇u(τ, x)]dτ,

(8.47)

0

u(t, x) := ϕ(t, x), x ∈ , t > 0.

(8.48)

For the obtained problem have the following theorem which will be be proved in Sect. 8.3.2. Theorem 8.6 Let

ρ ∈ L ∞ ( ), ρ(x) ≥ ρ0 > 0, x ∈

(8.49)

and ((0, ∞); W 1,∞ ( )) ∩ Wσ2,1 ((0, ∞); L ∞ ( )), a(0, x) ≥ a0 > 0, x ∈ , a ∈ Wσ1,1 0 0 (8.50) with some σ0 ∈ R. Moreover, let 3

((0, ∞); H 2 ()), ϕ(0, x) = ∂t ϕ(0, x) = . . . = ∂t4 ϕ(0, x) ≡ 0 (8.51) ϕ ∈ Wσ5,1 1

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

with some σ1 ∈ R. Then there exists σa , which depends on a and satisfies the inequalities σa > 0, σa ≥ σ0 , such that the problem (8.47), (8.48) has a solution u in the space Cσu ([0, ∞); H 2 ()), where σu = max{σ1 , σa }. Moreover, u(0, x) ≡ 0. The solution is unique in the space L 1σu ((0, ∞); H 2 ( )). Due to (8.45) and (8.46) the Neumann data h has the form t a(t − τ, x)∂ν u(τ, x)dτ, x ∈ , t > 0.

h(t, x) =

(8.52)

0

Theorem 8.6 with the trace theorem implies the following result. Corollary 8.2 Under the assumptions of Theorem 8.6, h ∈ Cσ1u ([0, ∞); L 2 ()).  be the operator which assigns to a function ϕ the function h via the solution Let D  transforms u of the problem (8.47), (8.48). Observing Corollary 8.2 we see that D 1 2 the set of functions ϕ satisfying (8.51) into the space Cσu ([0, ∞); L ()), provided  t ψ. Therefore, ρ and a meet the conditions (8.49) and (8.50). Moreover, Dψ = D∂ the operator D transforms the space 

3

ϕ ∈ Wσ6,1 ((0, ∞); H 2 ()), ψ(0, x) = ∂t ψ(0, x) = . . . = ∂t5 ψ(0, x) ≡ 0 1



into the space Cσ1u ([0, ∞); L 2 ()), provided ρ and a meet the conditions (8.49) and (8.50). Let us choose a function f such that 

(0, ∞), f = 0, I m f = 0, f (0) = f (0) = . . . = f I V (0) = 0, (8.53) f ∈ Wσ5,1 1 where σ1 ∈ R. For given f , ρ, and a satisfying (8.53), (8.49) and (8.50), respectively, 3 let us define the operator λa : H 2 () → Cσ1u ([0, ∞); L 2 ()) by the following relation ⎛ t ⎞  f (x)g(x)) = D ⎝ f (τ )dτ g(x)⎠ . λa g := D( 0

The main result is the following theorem, which asserts that a is uniquely recovered by λa , in other words, by the restriction of the Dirichlet-to-Neumann operator t D in the set of Dirichlet data of the form ψ(t, x) = 0 f (τ )dτ g(x) with fixed f . Theorem 8.7 Let  be some neighborhood of . Assume that (8.49) and (8.53) are valid and ρ ∈ C 0,μ ( ) with some μ > 0. Furthermore, let a1 and a2 be two functions satisfying (8.50) and the relations a j ∈ L 1σ0 ((0, ∞); W 2,∞ ( ) ∩ C 1,μ ( )), Im a j = 0, j = 1, 2.

(8.54)

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Recovering a Time- and Space-Dependent Kernel in a Hyperbolic …

349

Then the equality λa1 = λa2 implies a1 = a2 . The proof of this theorem will be given in Sect. 8.3.4. It uses a preliminary result concerning the boundary identifiability proved in Sect. 8.3.3.

8.3.2 Auxiliary Results Let X be a complex Banach space and z ∈ L 1σ ((0, ∞); X ) with some σ ∈ R. Then the Laplace transform of z, i.e., ∞ Z ( p) = Lt→ p (z) =

e− pt z(t)dt

(8.55)

0

exists in the half-plane Re p > σ and is holomorphic there (see [17]). Lemma 8.5 Let z ∈ L 1σ ((0, ∞); X ) with some σ ∈ R. Then Z ( p) X ≤ C(z, Re p) for Re p > σ where

∞ C(z, s) =

e−(s−σ )t e−σ t z(t) X dt → 0 as s → ∞.

(8.56)

(8.57)

0

If, in addition,z ∈ Wσk,1 ((0, ∞); X ) with some k ∈ {1, 2, . . .} then  p k Z ( p) − p k−1 z(0) − p k−2 z  (0) − . . . − z (k−1) (0)x ≤ C(z (k) , Re p) for Re p > σ,

(8.58)

which in case z(0) = z  (0) = . . . = z (k−1) (0) = 0 implies | p|k Z ( p) ≤ C(z (k) , Re p) for Re p > σ.

(8.59)

The proof of the lemma above can be found in [16]. Lemma 8.6 Let Z : C → X be holomorphic for Re p > σ with some σ ∈ R and | p|2 Z ( p) X ≤ c0 for Re p > σ.

(8.60)

with a constant c0 . Then there exists a function z ∈ Cσ ([0, ∞); X ) such that Z ( p) = Lt→ p (z). Moreover, z(0) = 0. The assertion of the lemma above follows from (see ([17], Proposition 0.2).

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Lemma 8.7 Let a satisfy (8.50) with some σ0 ∈ R and A( p, x) = Lt→ p (a(·, x)). Then (8.61) | p|A( p, ·)W 1,∞ ( ) ≤ c1 for Re p > σ0 with a constant c1 and p A( p, ·) → a(0, ·) as Re p → ∞ in W 1,∞ ( ) uniformly with respect to Im p. (8.62) Moreover, there exists σa , satisfying the inequalities σa > 0, σa ≥ σ0 , such that | p||A( p, x)| ≥ k > 0 for Re p > σa , x ∈ ,

(8.63)

Re p a(0, x) − | p a(0, x) − p 2 A( p, x)| ≥ k > 0 for Re p > σa , x ∈ . (8.64) Proof Using the estimate (8.58) for a we obtain  p A( p, ·) − a(0, ·)W 1,∞ ( ) ≤ C (∂t a, Re p) for Re p > σ0 ,

(8.65)

   p 2 A( p, ·) − pa(0, ·) − ∂t a(0, ·) L ∞ ( ) ≤ C ∂t2 a, Re p for Re p > σ0 . (8.66) In view of (8.57) and the assumption a(0, x) = a0 > 0 the relation (8.65) implies (8.61)-(8.63) and (8.66) yields (8.64). Lemma 8.8 Let ϕ satisfy (8.51) with some σ1 ∈ R and ( p, x) = Lt→ p (ϕ(·, x)). Then (8.67) | p|5 ( p, ·) H 23 () ≤ c2 for Re p > σ1 with a constant c2 . The assertion of this lemma follows from (8.59) and (8.57).

8.3.3 Direct Problem The direct problem (8.47), (8.48) is formally equivalent to the following elliptic boundary value problem derived by means of the Laplace transform: (L A U ) ( p, x) ≡ −div(A( p, x)∇U ( p, x)) + pρ(x)U ( p, x) = 0, x ∈ , (8.68) U ( p, x) = ( p, x), x ∈ .

(8.69)

Here p ∈ C, A( p, x) = Lt→ p (a(·, x)), ( p, x) = Lt→ p (ϕ(·, x)) and U ( p, x) = Lt→ p (u(·, x)).

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Recovering a Time- and Space-Dependent Kernel in a Hyperbolic …

351

Proposition 8.1 Let ρ satisfy (8.49) and a satisfy (8.50) with some σ0 ∈ R. Moreover, let ( p, x) and F( p, x) be given functions such that 3

( p, ·) ∈ H 2 (), F( p, ·) ∈ L 2 ( ) for Re p > σ1

(8.70)

with some σ1 ∈ R. Then the problem (L A U ) ( p, x) = F( p, x), x ∈ , U ( p, x) = ( p, x), x ∈ 

(8.71)

has a unique solution U ( p, ·) ∈ H 2 ( ) for every Re p > σu with σu = max{σa , σ1 } and σa from Lemma 8.7. This solution satisfies the estimate   U ( p, ·) H 2 ( ) ≤ c3 | p|2 F( p, ·) L 2 ( ) + | p|( p, ·) H 23 () , Re p > σu , (8.72) where the coefficient c3 depends on ρ, a, , but is independent of p,  and F. 3

Proof Due to the assumption ( p, ·) ∈ H 2 () for Re  p > σ1 , there exists a function ( p, x) such that  ( p, ·) ∈ H 2 ( ) and  ( p, x)  = ( p, x) for Re p > σ1 . x∈ , the problem (8.71) reduces to the following problem with Denoting W = U −  homogeneous boundary condition for W :    ( p, x), x ∈ , W ( p, x) = 0, x ∈ . (8.73) (L A W ) ( p, x) = F( p, x) − L A  The sesquilinear form, associated with the operator L A in the space H01 ( ), reads l p (W1 , W2 ) =

A( p, x)∇W1 (x)∇W 2 (x)dx + p

ρ(x)W1 (x)W 2 (x)dx.

(8.74)

 2 By (8.49) and the assertion (8.61) of Lemma 8.7, l p is bounded in H01 ( ) if Re p > σ0 . We are going to prove the coercitivity of l p . For any W ∈ H01 ( ) we have     l p (W, W )  ≥  1 a(0, x)|∇W (x)|2 dx + p ρ(x)|W (x)|2 dx − p

(8.75)   1 2 2   − pa(0, x) − p A( p, x) |∇W (x)| dx. | p|2

In order to estimate the first term on the right-hand side of (8.75) from below we use the relation |c1 p −1 + c2 p| ≥ Re p(c1 | p|−2 + c2 ), which, as easily can be c1 , c2 = 0 and Re p > 0. Applying this relation with c1 = verified, holds for any 2 a(0, x)|∇W (x)| dx and c2 = ρ(x)|W (x)|2 dx in (8.75) we derive

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

  l p (W, W )  ≥ 1 | p|2

[Re pa(0, x) − | pa(0, x) − p 2 A( p, x)|]|∇W (x)|2 dx



+ Re p

ρ(x)|W (x)|2 dx

for Re p > 0. Applying the estimate (8.64) of Lemma 8.7 and the assumed inequality ρ(x) ≥ ρ0 > 0 we obtain   l p (W, W )  ≥ c4

 1 2 2 ∇W  L 2 ( ) + W  L 2 ( ) , Re p > σa | p|2

(8.76)

 2 with a coefficient c4 depending on a and ρ. Thus, l p is coercive in H01 ( ) for ( p, ·) ∈ H 2 ( ) for Re p > Re p > σa . By (8.49), (8.70),  and the 2relation −1  (8.61)  ( p, ·) ∈ L ( ) ⊂ H ( ) holds for Re p > σu . σ1 the inclusion F( p, ·) − L A  Consequently, due to the Lax-Milgram lemma the problem (8.73) has a unique solution W ( p, ·) ∈ H01 ( ) for Re p > σu . This yields the existence and uniqueness of the solution U ( p, ·) of the problem (8.71) in the space H 1 ( ) for any Re p > σu . To prove the assertion U ( p, ·) ∈ H 2 ( ) for Re p > σu we rewrite the problem (8.71) in the form U ( p, x) = FU ( p, x), x ∈ , U ( p, x) = ( p, x), x ∈ 

(8.77)

for Re p > σu , where  is the Laplacian and FU ( p, x) = A( p, x)−1 [ pρ(x)U ( p, x) − ∇ A( p, x) · ∇U ( p, x) − F( p, x)].

(8.78)

In view of the assumptions of Proposition 8.1 and the assertions of Lemma 8.7 we see that FU ( p, ·) ∈ L 2 ( ) for Re p > σu . Taking this relation and the assumption ( p, ·) ∈ H 2/3 () into account, the well-known theory of smoothness of the solution of the Dirichlet problem for the Poisson equation (see, e.g., [18], Theorem 27.2) yields U ( p, ·) ∈ H 2 ( ) for Re p > σu and the estimate U ( p, ·) H 2 ( ) ≤ c5 (FU ( p, ·) L 2 ( ) + ( p, ·) H 3/2 () ) , Re p > σu ,

(8.79)

where c5 is a constant depending on. It remains to derive (8.72). We substitute W ( p, ·) for W1 and W2 in the formula (8.74), where W is the solution of (8.73), and apply the divergence theorem for the  first integral in the right-hand side of this formula to get l p (W, W ) = [F( p, x) − )( p, x)]W ( p, x)dx. Thereupon we estimate this expression from above by (L A  means of the Cauchy-Schwartz inequality, use the definition of L A and combine the obtained result with the estimate from below (8.76). This leads to the relation   ( p, ·) H 2 ( ) , W ( p, ·) H k ( ) ≤ c6 | p|k F( p, ·) L 2 ( ) + | p|

(8.80)

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353

for Re p > σu , where k ∈ {0; 1} and c6 depend on a, ρ. Note that we can choose ( p, ·) ∈ H 2 ( ), satisfying the relation  ( p, x)|x∈ = ( p, x), so that the inequal ( p, ·) H 2 ( ) ≤ c7 ( p, ·) H 3/2 () holds for Re p > σu with a constant c7 , ity  which depends on but is independent of .Using the latter inequality, the estimate  (8.80) and the obvious relation 1 < σu−1 | p| for Re p > σu in the formula U = W +  we obtain U ( p, ·) H k ( ) ≤ c8 | p|k (F( p, ·) L 2 ( ) + | p|( p, ·) H 3/2 () ),

(8.81)

for Re p > σu , where k ∈ {0; 1} and c8 depend on a, ρ, . Applying the assumptions imposed on ρ, F, the assertions (8.61), (8.63) of Lemma 8.7 and the estimate (8.81) in (8.78) we derive FU ( p, ·) L 2 ( ) ≤ c9 | p|2 (F( p, ·) L 2 ( ) + | p|( p, ·) H 3/2 () ),

(8.82)

for Re p > σu , where c9 depends on a, ρ, . Finally, (8.79) with (8.82) implies (8.72). Proposition 8.2 Let the assumptions of Proposition 8.1 hold for ρ, a, and . Also let A( p, ·) and ( p, ·) be holomorphic in Re p > σu with values in W 1,∞ ( ) and H 3/2 (), respectively. Then the solution U ( p, ·) of (8.68), (8.69) is holomorphic in Re p > σu with values in H 2 ( ). Proof Let p and q be arbitrary numbers such that Re p > σu , Re q > σu . Define Aq ( p, x) := A(q, x) − A( p, x), q ( p, x) := (q, x) − ( p, x), Uq ( p, x) : = U (q, x) − U ( p, x). From (8.68) and (8.69) we obtain the following problem for Uq :  (L A Uq )( p, x) = (div) Aq ( p, x)∇[Uq ( p, x) + U ( p, x)] − (q − p)ρ(x)[Uq ( p, x) + U ( p, x)], Uq ( p, x)|x∈ = q ( p, x).

(8.83)

Applying estimate (8.72) of Proposition 8.1 to this problem we obtain Uq ( p, ·) H 2 ( ) ≤ c3 | p|2

! " Aq ( p, ·)W 1,∞ ( ) + |q − p|ρ L ∞ ( )

" ! × Uq ( p, ·) H 2 ( ) + U ( p, ·) H 2 ( ) + | p|q ( p, ·) H 3/2 () . This estimate due the relations Aq ( p, ·)W 1,∞ ( ) → 0 and q ( p, ·) H 3/2 () → 0 as q → p, following from the assumptions of the proposition, yields Uq ( p, ·) H 2 ( ) → 0 as q → p.

(8.84)

Further, let A ( p, x) and  ( p, x) be the derivatives of A( p, x) and ( p, x) with # be the solution of the problem respect to p, respectively, and let U

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

#( p, x)|x∈ =  ( p, x). #)( p, x) = div(A ( p, x)∇U ( p, x)), U (L A U #( p, x) = U (q,x)−U ( p,x) − U #( p, x) #q ( p, x) := Uq ( p,x) − U Denoting U q− p q− p subtracting (8.85) from (8.83) divided by q − p we obtain the problem

and

 $ Aq ( p, x) Aq ( p, x)  − A ( p, x) ∇U ( p, x) + ∇Uq ( p, x) q−p q−p − ρ(x)Uq ( p, x), q ( p, x) = −  ( p, x). q−p

#)( p, x) = div (L A U

#q ( p, x)|x∈ U

(8.85)



Using the estimate (8.72) for this problem we have #q ( p, ·) H 2 ( ) U % %  % Aq ( p, ·) %  % − A ≤ c3 | p|2 U ( p, ·) H 2 ( ) % ( p, ·) % q−p % 1,∞ W ( ) &% ' % % Aq ( p, ·) % % + Uq ( p, ·) H 2 ( ) × % % q − p % 1,∞ + ρ L ∞ ( ) W ( ) % % $ % q ( p, ·) %  % . + | p| % % q − p −  ( p, ·)% 3/2 H ()

(8.86)

Due to the assumptions of the proposition we have % % % Aq ( p, ·) %  % % % q − p − A ( p, ·)% 1,∞ → 0 W ( ) % % % q ( p,·) % and % q− −  ( p, ·)% p

H 3/2 ()

→ 0 as q → p.

#q ( p, ·) H 2 ( ) → 0 Using these relations as well as (8.84) in (8.86) we obtain U U (q,·)−U ( p,·) #( p, ·) as q → p in H 2 ( ). This yields →U as q → p, or equivalently, q− p the differentiability, hence holomorphy of U ( p, ·) at p. Proof of Theorem 8.6 In virtue of Proposition 8.1 the problem (8.68), (8.69) has a unique solution U ( p, .) ∈ H 2 ( ) for any Re p > σu . Applying the estimate (8.72) to this solution and observing Lemma 8.8 we obtain | p|2 U ( p, ·) H 2 ( ) ≤ c2 c3 for Re p > σu . Further, since A and  are the Laplace transforms of abstract functions a and ϕ 3 with values in W 1,∞ ( ) and H 2 (), respectively, A( p, ·) ∈ W 1,∞ ( ) and ( p, ·) ∈ 3 H 2 () are holomorphic in the half-plane Re p > σu . Proposition 8.2 yields the holomorphy of U ( p, ·) ∈ H 2 ( ) for Re p > σu . Summing up, the assumptions of Lemma 8.6 are valid for U . Consequently, there exists a function u ∈ Cσu ([0, ∞); H 2 ( ))

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355

such that L t→ p (u(·, x)) = U ( p, x) and u(0, x) ≡ 0. Applying the inverse transform to (8.68), (8.69) we see that u satisfies the problem (8.47), (8.48). This proves the existence assertion of Theorem 8.6. To prove the uniqueness assertion let us suppose that (8.47), (8.48) has two solutions u j ∈ L 1σu ((0, ∞); H 2 ( )), j = 1, 2. Denote U j ( p, x) = L t→ p (u j (·, x)), j = 1, 2. Then U j ( p, ·) ∈ H 2 ( ), j = 1, 2 solve (8.68), (8.69) for Re p > σu . Proposition 8.1 implies U 1 ( p, ·) = U 2 ( p, ·) for Re p > σu . Finally, by the uniqueness of the inverse transform the relation u 1 (t, ·) = u 2 (t, ·) for almost any t ∈ (0, ∞) follows.

8.3.4 Uniqueness on the Boundary In this section we will prove that the assumption λa1 = λa2 implies the equalities a1 = a2 and ∂ν a1 = ∂ν a2 on the boundary . We begin by introducing some additional notation. Let u a,g denotes the solution of (8.47), (8.48) corresponding to the kernel a and the boundary condition ϕ(t, x) = f (t)g(x). Define Ua,g ( p, x) = L t→ (u a,g (·, x)) and F( p) = L t→ p ( f ). Then Ua,g solves (8.68), (8.69) with the boundary condition ( p, x) = F( p)g(x). Further, let a stands for the operator that assigns to every function g ∈ H 3/2 () the Laplace transform of λa g, namely ⎛ (a g)( p, x) = L t→ p ((λa g)(·, x)) = L t→ p ⎝

·

⎞ a(· − τ, x)∂ν u a,g (τ, x)dx ⎠

0

(8.87)

= A( p, x)∂ν Ua,g ( p, x), x ∈ , Re p > σu .

Finally, for any pair of functions a1 and a2 satisfying (2.9) we define A j ( p, x) = L t→ p (a j (., x)), j = 1, 2 and σ12 := max{σ1 , σa1 , σa2 }. Let us prove some lemmas. First one is an analog of the Alessandrini’s equality for the inverse conductivity problem [10]. Lemma 8.9 Let (8.49), (8.53) hold, a1 , a2 satisfy (8.50) and g1 , g2 ∈ H 3/2 (). Then [A1 ( p, x) − A2 ( p, x)]∇Ua1 ,g1 ( p, x) · ∇Ua2 ,g2 ( p, x)dx



= F( p)

(8.88) [(a1 − a2 )g1 ]( p, x)g2 (x)dx , Re p > σ12 ,



where dx is the Lebesgue surface measure of . Proof. Let a, b be arbitrary functions satisfying (8.50), A = L t→ p (a), B=L t→ p (b) and g, γ be arbitrary functions in H 3/2 (). Multiplying the Eq. (8.68) for Ua,g by

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

Ub,γ , using the divergence formula and the definition of a we obtain the equality E(A, g; B, γ ) = 0, where E(A, g; B, γ ) =

[A( p, x)∇Ua,g ( p, x) · ∇Ub,γ ( p, x)

+ pρ(x)Ua,g ( p, x)Ub,γ ( p, x)]dx − F( p) (a g)( p, x)γ (x)dx 

This equality implies the relation E(A1 , g1 ; A2 , g2 ) − E(A2 , g2 ; A1 , g1 ) − E(A2 , g1 ; A2 , g2 ) + E(A2 , g2 ; A2 , g1 ), which is identical to (8.88). Lemma 8.10 Let z ∈  and B be a neighborhood of z. Let α and β be given functions such that α ∈ W 1,∞ ( ) ∩ C 1,μ (B), β ∈ L ∞ ( ) ∩ C 0,μ (B) with some μ > 0 and α(x) ≥ α0 > 0, β(x) ≥ ρ0 > 0 for x ∈ ∪ B. Then, for any y ∈ B \ , there exists a function ω y : ∪ B \ {y} → R satisfying the following conditions: 1. w y solves the equation − div(α(x)∇w y (x)) + β(x)w y (x) = 0, x ∈ ∪ B \ {y};

(8.89)

2. w y (x)|x∈ belongs to H 2 ( ). Moreover, for any x ∈ B \ {y} the relation w y (x) =

α(y) + ω y (x) 4π |x − y|

(8.90)

is valid, where |ω y (x)| ≤ c10 |x − y|δ−1 , |∂xi ω y (x)| ≤ c10 |x − y|δ−2 , i = 1, 2, 3

(8.91)

with a coefficient c10 and an exponent δ ∈ (0, 1) independent of x ∈ B \ {y} and y ∈ B \ . The proof of this lemma is given in [16]. Now we can prove the main result of the section. Theorem 8.8 Let the assumptions of Theorem 8.7 be satisfied for ρ, f , and a1 , a2 . Then λa1 = λa2 implies a1 (t, z) = a2 (t, z) and ∂ν a1 (t, z) = ∂ν a2 (t, z) for z ∈  and a.e. t ∈ (0, ∞).

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357

Proof The proof uses in an adapted form the method of singular solutions due to Alessandrini [19]. First, we note that the assumption λa1 = λa2 implies a1 = a2 . This, by Lemma 8.9 yields [A1 − A2 ]( p, x)∇Ua1 ,g1 ( p, x) · ∇Ua2 ,g2 ( p, x)dx = 0 for Re p > σ12 .

(8.92)

Let us choose some p0 ∈ R, p0 > σ12 , such that F( p0 ) = 0. In view of the assumptions I m a j = 0, a j (0, x) ≤ a0 > 0 and the assertions (8.62), (8.63) of Lemma 8.7, the functions A j satisfy the inequality A j ( p0 , x) ≥ pk0 > 0 for x ∈ . Moreover, due to (8.54) we have A j ( p0 , .) ∈ W 2,∞ ( ) ∩ C 1,μ (ε ), j = 1, 2. Let z be an arbitrary point on . Observing the properties of A j ( p0 , x) and the assumptions imposed on ρ, we can choose a neighborhood Bz ⊂ ε of z such that the assumptions of Lemma 8.10 are fulfilled for B = Bz , β = p0 ρ and α = A j ( p0 , .) with j ∈ {1, 2}. Let w j,y with j ∈ {1, 2} be a function which fulfills the assertions of Lemma 8.10 for α = α j = A j ( p0 , .), β = p0 ρ. Thus, − div(A j ( p0 , x)∇w j,y (x)) + p0 ρ(x)w j,y (x) = 0, x ∈ ∪ Bz \ {y}, j = 1, 2, (8.93)

Observing the assertions of Lemma 8.10 and the inequality α j (y) ≥ can verify that ∇w1,y (x) · ∇w2,y (x) ∼

α1 (y)α2 (y) as y → x |x − y|4

k p0

> 0, one

(8.94)

Let us set g j (x) := w j,y (x)|x∈ for j = 1, 2. Then the function Ua j ,g j ( p, x) solves (8.68), (8.69) with A = A j and ( p, x) = F( p)g j (x) = F( p)w j,y (x)|x ∈ . Since (8.68) in case p = p0 coincides with (8.93), we have Ua j ,g j ( p0 , x) = F( p0 )w j,y (x) for x ∈ . Using this relation in (8.92) and taking the inequality F( p0 ) = 0 into account we obtain I y := [A1 − A2 ]( p0 , x)∇w1,y (x) · ∇w2,y (x)dx = 0 (8.95)

On the other hand, by A j ( p0 , .) ∈ C 1,μ (Bz ) we have (A1 − A2 )( p0 , x) = (A1 − A2 )( p0 , z) + ∇(A1 − A2 )(z) · (x − z) + O(|x − z|1+μ )

for x ∈ Bz . Thus, I y = I y0 + I y1 + I y2 , where I y0 = (A1 − A2 )( p0 , z)

∪Bz

∇w1,y (x) · ∇w2,y (x)dx.

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

I y1 = ∇(A1 − A2 )( p0 , z)

(x − z)∇w1,y (x) · ∇w2,y (x)dx.

∪Bz

I y2 =

O(|x − z|1+μ )∇w1,y (x) · ∇w2,y (x)dx

∪Bz

+

[A1 − A2 ]( p0 , x)∇w1,y (x) · ∇w2,y (x)dx.

/Bz

Observing that the singularity of w j,y (x) is located at x = y, where w j,y (x) satisfies the relation (8.94), and the distance between y and \ Bz is bounded away from 0 as y → z, we see that the term I y2 is bounded as y → z. Moreover, supposing (A1 − A2 )( p0 , z) = 0, we have I y ∼ I y0 and |I y0 | → ∞ as y → z. But this contradicts (8.95). Thus, (A1 − A2 )( p0 , z) = 0. Further, supposing ∇(A1 − A2 )( p0 , z) = 0, we have I y ∼ I y1 and |I y1 | → ∞ as y → z. Again, this contradicts (8.95). Consequently, ∇(A1 − A2 )( p0 , z) = 0. Since z ∈  was chosen arbitrarily, we have proven the equalities A1 ( p, ·) = A2 ( p, ·), ∂ν A1 ( p, ·) = ∂ν A2 ( p, ·)

(8.96)

in C 1,λ () for any p > σ12 such that F( p) = 0. The set p : p > σ12 , F( p) = 0 has an accumulation point because of the assumption f = 0. Therefore, by means of the analytic continuation we can extend the equalities (8.96) in C 1,λ () to all Rep > σ12 . Finally, by uniqueness of the inverse transform we obtain the assertion of the theorem. In this section, we prove Theorem 8.7. For this end we need the following lemmas. Lemma 8.11 Let (8.49), (8.53) hold and a satisfy (8.50), (8.54). Moreover, let g ∈ H 3/2 (). Define (8.97) Va,g ( p, x) := A1/2 ( p, x)Ur,g ( p, x), where A1/2 ( p, x) is the principal value of the square root of A( p, x), namely

 ArgA( p, x) ArgA( p, x) + i sin . A1/2 ( p, x) = |A( p, x)|1/2 cos 2 2 Then Va,g ( p, ·) belongs to H 2 ( ) and solves the problem − Va,g ( p, x) + K ( p, x)Va,g ( p, x) = 0, x ∈ ,

(8.98)

Va,g ( p, x) = F( p)A1/2 ( p, x)g(x), x ∈ 

(8.99)

for Re p > σu , where K ( p, ·) belongs to L ∞ ( ) and is given by

8.3

Recovering a Time- and Space-Dependent Kernel in a Hyperbolic …

! " K ( p, x) = A1/2 ( p, x) + pρ(x)A−1/2 ( p, x) A−1/2 ( p, x)

359

(8.100)

for Re p > σa . Proof In view of the assumption (8.54) we obtain A( p, ·) ∈ W 2,∞ ( ) for Re p > σ0 . This together with the assertion (8.63) of Lemma 8.7 yields   κ 1/2 > 0, x ∈ A1/2 ( p, ·) ∈ W 2,∞ ( ),  A1/2 ( p, x) ≥ | p|1/2

(8.101)

for Re p > σa . Applying these relations and (8.49) in (8.100) we deduce K ( p, ·) ∈ L ∞ ( ) for Re p > σa . Further, due to Theorem 8.6 Ua,g ( p, ·) ∈ H 2 ( ) for Re p > σu . This in view of (8.97) and (8.101) yields Va,g ( p, ·) ∈ H 2 ( ) for Re p > σu . Finally, observing that Ua,g ( p, x) solves the problem (8.68), (8.69) with ( p, x) = F( p)g(x), the change of the unknown by (8.97) in this problem results in (8.98) and (8.99) with K of the form (8.100). Lemma 8.12 Let the assumptions of Theorem 8.7 be valid for f , ρ, a1 , and a2 . Assume that g1 , g2 ∈ H 3/2 () and K j , j = 1, 2, are defined by (8.100) with A replaced by A j . Let A1 ( p, x) = A2 ( p, x), ∂ν A1 ( p, x) = ∂ν A2 ( p, x) for any x ∈  and Re p > σ12 . Then [K 1 ( p, x) − K 2 ( p, x)] Va1 ,g1 ( p, x)Va2 ,g2 ( p, x)dx



!

= F( p)

 " a1 − a2 g1 ( p, x)g2 (x)dx , Re p > σ12 .

(8.102)



Proof Let a, b be arbitrary functions satisfying (8.50), A = Lt→ p (a), B = Lt→ p (b), 3 K be given by (8.100) and g, γ be arbitrary functions in H 2 (). Multiplying the Eq. (8.98) for Va,g by Vb,γ and using the divergence formula we obtain the equality # E(A, g; B, γ ), where # E(A, g; B, γ ) =





−

! " ∇Va,g ( p, x)∇Vb,γ ( p, x) + K ( p, x)Va,g ( p, x)Vb,γ ( p, x) dx ∂ν Va,g Vb,γ dx .



This equality implies the relation # 2 , g2 ; A1 , g1 ) − E(A # 2 , g1 ; A2 , g2 ) + E(A # 2 , g2 ; A2 , g1 ) = 0, # 1 , g1 ; A2 , g2 ) − E(A E(A

which is identical to

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8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

[K 1 − K 2 ] Va,g Vb,γ dx



=

!

" ∂ν Va1 ,g1 Va2 ,g2 − ∂ν Va2 ,g2 Va1 ,g1 + ∂ν Va2 ,g2 Va2 ,g1 − ∂ν Va2 ,g1 Va2 ,g2 dx



(8.103)

By (8.99) and the proved equalities A( p, x) = A1 ( p, x) = A2 ( p, x), ∂ν A1 ( p, x) = ∂ν A2 ( p, x) for x ∈  we have Va2 ,g1 ( p, x) = Va1 ,g1 ( p, x) = F( p)A1/2 ( p, x)g1 (x), Va2 ,g2 ( p, x) = Va1 ,g2 ( p, x) = F( p)A1/2 ( p, x)g2 (x), ∂ν Va1 ,g1 ( p, x) − ∂ν Va2 ,g1 ( p, x) ! " ! " = ∂ν A1 ( p, x)1/2 Ua1 ,g1 ( p, x) − ∂ν A2 ( p, x)1/2 Ua2 ,g1 ( p, x) ! " +A1/2 ( p, x) A1 ( p, x)∂ν Ua1 ,g1 ( p, x) − A2 ( p, x)∂ν Ua2 ,g1 ( p, x)  " ! = A1/2 ( p, x) a1 − a2 g1 ( p, x) for x ∈ . Using these relations in (8.103) we derive (8.102). Lemma 8.13 Let z, ζ, η ∈ R3 and ω > 0 satisfy the relations |z|2 + ω2 . 2

(8.104)

  z + ωη + ζ, ξ2 = i − ωη − ζ, 2 2

(8.105)

z · η = z · ζ = η · ζ = 0, |η| = 1, |ζ |2 = Define ξ1 = i

z

Furthermore, let α j ∈ L ∞ ( ), j = 1, 2. Then there exists M ≥ 0 depending on α1 , α2 such that the equations − wˆ j (x) + α j (x)wˆ j (x) = 0, j = 1, 2

(8.106)

have solutions ωˆ j (x), j = 1, 2, which belong to H2 ( ) and have the form ! " wˆ j (x) = ex·ξ j 1 + ψ j (x) ,

(8.107)

where ψ j , j = 1, 2 satisfy the estimates   % % %ψ j % 2 ≤  1  α j  L ∞ ( ) for ξ j  > M, j = 1, 2. L ( ) ξ j 

(8.108)

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361

The proof of the above lemma is given in [16]. Proof of Theorem 8.7. To prove this theorem, we will apply in an adapted form, the method due to Sylvester and Uhlmann [4]. Assume that λa1 = λa2 . Let us fix some p0 ∈ R, p0 > σ12 , such that F( p0 ) = 0. Further, let us choose z ∈ R3 and ω > 0 and let ζ, η ∈ R3 satisfy the relations (8.104). Define α j (x) := K j ( p0 , x) for be the functions satisfying the assertions of Lemma 8.13 with these α j . By virtue of j = 1, 2, where K j is given by (8.100) with A replaced by A j . Finally, let wˆ j , j = 1, 2, be the functions satisfying the assertions of Lemma 8.13 with these α j . By virtue Theorem 8.8 we have A1 ( p, x) = A2 ( p, x) =: A( p, x) for x ∈ , Re p > σ12 . Let us set g j (x) :=

w ˆ j (x) for x ∈ , j = 1, 2. F( p0 )A1/2 ( p0 , x)

(8.109)

The functions g j , j = 1, 2 belong to H 3/2 () because of the relations ωˆ j ∈ H ( ), F( p0 ) = 0 and (8.101). The problem (8.98), (8.99) for Va j ,g j reads 2

− Va j ,g j ( p, x) + K j ( p, x)Va j ,g j ( p, x) = 0, x ∈ , Va j ,g j ( p, x) = F( p)A1/2 ( p, x)g j (x), x ∈ ,

(8.110) (8.111)

where Re p > σ12 . Due to the relation α j = K j ( p0 , ·) the Eq. (8.110) for Va j ,g j coincides with (8.106) for ωˆ j in case p = p0 . Further, in view of (8.109) and (8.111) we have Va j ,g j ( p0 , x) = wˆ j (x) for x ∈  and j = 1, 2. Recall that wˆ j solves (8.106). Thus, by the uniqueness of the solution of the boundary value problem (8.110), (8.111), we obtain the equality Va j ,g j ( p0 , x) = wˆ j (x) for x ∈ and j = 1, 2. By the assumptions of Theorem 8.7, the supposed relation λa1 = λa2 and Theorem 8.8 the assumptions of Lemma 8.12 are satisfied. Moreover, λa1 = λa2 implies a1 = a2 . Hence, the equality (8.102) holds with 0 in the right-hand side. In case p = p0 it reads [K 1 ( p0 , x) − K 2 ( p0 , x)] wˆ 1 (x)wˆ 2 (x)dx = 0.

This, in view of (8.107) and (8.105), implies [K 1 ( p0 , x) − K 2 ( p0 , x)] ei x·z [1 + ψ1 (x)] [1 + ψ2 (x)] dx = 0.

(8.112)

  From (8.104) and (8.105) we see that that ξ j  → ∞ as ω → ∞ for fixed z. Hence, by and (8.108) ψ j  L 2 ( ) → 0 as ω → ∞ for fixed z. Thus, passing to the limit ω → ∞ in (8.112) we deduce

362

8 Dirichlet-to-Neumann Maps Method in Kernel Determining Problems

[K 1 ( p0 , x) − K 2 ( p0 , x)] ei x·z dx = 0.

Since this equality holds for arbitrary z ∈ R3 , we obtain K 1 ( p0 , x) = K 2 ( p0 , x) for x ∈ . 1/2 Due to this equality, (8.100), and Theorem 8.8, the difference A1 ( p0 , ·) − 1/2 A2 ( p0 , ·) solves the problem ( ) ( ) 1/2 1/2 1/2 1/2 −  A1 ( p0 , x) − A2 ( p0 , x) + α( ˜ p0 , x) A1 ( p0 , x) − A2 ( p0 , x) = 0,

(8.113) 1/2

1/2

A1 ( p0 , x) − A2 ( p0 , x) = 0, x ∈ ,

(8.114)

where α( ˜ p0 , x) = & ×

1 1/2 A2 ( p0 , x)

p0 ρ(x) 1/2

A1 ( p0 , x)

+

1 1/2

A2 ( p0 , x)

A2 ( p0 , x) |∇ A2 ( p0 , x)|2 − p0 ρ(x) + 2 4 A2 ( p0 , x)

' .

Let us study the asymptotic behavior of α˜ as p0 → ∞. Observe that (8.54) and the assertions (8.56), (8.57) of Lemma 8.5 imply A2 ( p0 , ·) L ∞ ( ) → 0 as p0 → ∞. Moreover, by the assertions (8.61) and (8.63) of Lemma 8.7 |∇ A2 ( p0 , x)|2 1 | p0 ∇ A2 ( p0 , x)|2 c1 → 0 as p0 → ∞. = ≤ |A2 ( p0 , x)| p0 p0 |A2 ( p0 , x)| κ p0 Consequently, in view of (8.49) α( ˜ p0 , x) ∼

p0 ρ(x) 1/2

A2 ( p0 , x)

&

1 1/2

A1 ( p0 , x)

+

1 1/2

A2 ( p0 , x)

' as p0 → ∞

uniformly in x ∈ . The assumptions Imaj = 0, j = 1, 2 and a j (0, x) ≥ a0 > 0 ˜ p0 , x) ≥ 0 for p0 > σ˜ imply that A j ( p0 , x) > 0 for sufficiently large p0 . Hence, α( with some sufficiently large σ˜ . This implies that the solution of the problem (8.113), (8.114) is unique if p0 > σ˜ . Consequently, A1 ( p0 , x) = A2 ( p0 , x) for x ∈ and p0 > σ˜ such that F( p0 ) = 0. The set { p | p > σ˜ , F( p) = 0} has an accumulation point in view of the assumption f = 0. Hence, by means of analytic continuation we can extend the equality A1 ( p, ·) = A2 ( p, ·) for all Re p > σ12 . Finally, by the uniqueness of the inverse transform we derive a1 = a2 .

References

363

8.4 Conclusions In this chapter we have considered some inverse problems for hyperbolic equation using the Dirichlet–Neumann operator which maps Dirichlet data into Neumann data. The uniqueness of solutions is proved, and estimates of the conditional stability of multidimensional inverse problems are obtained. This method makes it possible to study the problems of recovering the coefficients at the lowest derivative and kernels of convolution-type integral operators that model the spatial inhomogeneities of a medium.

References 1. Nachman A. 1988. Reconstructions from boundary measurements. Annals of Mathematic (2), 128(3): 531–576 . 2. Rakesh, and Symes, W.W. 1988. Uniqueness for an inverse problem for the wave equation. Communication of Partial Differential Equations 13(1): 87–96. 3. Rakesh. 1993. An inverse problem for the wave equation in the half plane. Inverse Problems 9(3): 433–441. 4. Sylvester, J., and Uhlmann, G. 1987. A global uniqueness theorem for an inverse boundary value problem. Annals of Mathematic (2), 1(251): 153–169. 5. Mikhailov, V.P. 1976. Partial Differential Equations. Moscow: Nauka [in Russian]. 6. Hamaker, C., Smith, K.T., Solmin, D.C, Wagner, S.C. 1980. The divergent beam x-ray transform. Rocky Mountain Journal of Mathematics, Numerical Mathematic 253–283. 7. Bukhgeim, A.L., G.V. Dyatlov, and V.M. Isakov. 1997. Stability of memory reconstruction from the Dirichlet-Neumann operator. Siberian Mathematic Journal 38 (4): 636–646. 8. Volevich, L.R., and S.G. Gindikin. 1980. The method of energy estimates in the mixed problem. Russian Mathematics Surveys 35 (5): 57–137. 9. Lions, J.L., and E. Magenes. 1972. Non-Homogeneous Boundary Value Problems and Applications, vol. 1. Berlin: Springer-Verlag, Berlin Heidelberg. 10. Alessandrini, G. 1988. Stable determination of conductivity by boundary measurements. Application of Analysis 1–3 (27): 153–172. 11. Isakov, V. 1995. Inverse Problems in PDE. Berlin: Springer-Verlag. 12. Isakov, V. 1991. Completeness of products of solutions and some inverse problems for PDE. Journal Differential Equations 92 (2): 305–316. 13. Hermander, L. 1986. The Analysis of Linear Partial Differential Operators [Russian translation]. Moscow: Mir. 14. Bukhgeim, A.L. 1983. Volterra Equations and Inverse Problems. Novosibirsk: Nauka [in Russian]. 15. Janno, J. 2004. Recovering a time- and space-dependent kernel in a hyperbolic integrodifferential equation from a restricted Dirichlet-to-Neymann operator. Electronic Journal of Differential Equations 2004 (67): 1–16. 16. Janno, J. 2004. Determination of a time- and space-dependent heat flux relaxation function by means of a restricted Dirichlet-to-Neumann operator. Mathematical Methods of Application Science 27 (11): 1241–1260. 17. Prüss, J. 1993. Evolutionary Integral Equations and Applications. Basel: Birkh-auser. 18. Treves, F. 1975. Basic Linear Partial Differential Equations. New York: Academic Press. 19. Alessandrini, G. 1990. Singular solutions of elliptic equations and the determination of conductivity by boundary measurements. Journal of Differential Equation 84: 252–273.

Chapter 9

Open Problems

In this chapter we will list some outstanding research problems. Problem 1 Many applied problems (e.g., in viscoelasticity, electrodynamics in dispersed media) lead to the consideration of equations of the forms ˜ = u tt − Lu

t

˜ k(τ ) Lu(x, t − τ )dτ, x ∈ R3 , t > 0

(9.1)

r (τ )u τ τ (x, t − τ )dτ, x ∈ R3 , t > 0,

(9.2)

0

or ˜ = u tt − Lu

t 0

in which L˜ is an uniformly elliptic operator defined by (0.18) (i.e., ai j satisfy the condition (0.15)), k(t) is a relaxation kernel. In Eq. (9.2) r (t) is the resolvent of the kernel k(t). Equations (9.1) and (9.2) are equivalent. In fact, considering (9.1) as a Volterra ˜ and solving it, we get (9.2). integral equation of the second kind with respect to Lu In this case, r (t) and k(t) are related by the formula t k(t − τ )r (τ )dτ.

r (t) = −k(t) − 0

The study of an inverse problem of determining the relaxation kernel in the formulation of Sect. 1.2 for a more general elliptic operator L˜ than the operator L (see (1.43)) is an open problem. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 D. K. Durdiev and Z. D. Totieva, Kernel Determination Problems in Hyperbolic Integro-Differential Equations, Infosys Science Foundation Series in Mathematical Sciences, https://doi.org/10.1007/978-981-99-2260-4_9

365

366

9 Open Problems

Problem 2 In the paper [1] the questions of uniqueness of finding the multidimensional kernel of the integral term of the equation t u tt − u =

k(x, τ )u(x, t − τ )dτ, x ∈  ⊂ R3 , t > 0

(9.3)

−∞

for a bounded domain  with a smooth boundary from a given Dirichlet–Neumann operator were studied (by Dirichlet-to-Neumann maps method). It would be interesting to study the possibility of using this method for Eq. (9.3) ˜ L instead of  and to find sufficient conditions for the coeffiwith the operators L, cients of these operators, under which the kernel k(x, t) is uniquely determined. Problem 3 An interesting problem arises for the inverse problem studied in Sect. 1.5 for the case when f (x, z) ≡ 0, x ∈ R, 0 < z < H, gi (t, x) = δ(t, x), (t, x) ∈ R2 , i = 1, 2 with initial–boundary conditions (1.149), where δ(t, x) is the Dirac function, concentrated at the point t = 0, x = 0. In this case, the problem of finding the kernels of the system of two-dimensional Maxwell integro-differential equations (1.148) with conditions (1.149), (1.150) is an open problem and the study of which is important in applications. Problem 4 In all problems of Chap. 2, it is assumed that the unknown kernel does not depend on one of the spatial variables, on the orthogonal to which the Neumann boundary condition is specified. This condition is a delta δ-function and its derivatives. One should also study such problems for distributed incoming data for different spatial domains. It is also interesting to consider the problem of Sect. 2.3 in the case of the three-dimensional Maxwell integro-differential system (2.69). Problem 5 An analysis of dynamic equations describing processes with memory shows that to the right sides of equations and systems of hyperbolic equations are added operators of the convolution type of some function depending on the time variable and the elliptic part of the corresponding hyperbolic operators on the left side of equations. In applications, it is important to study the inverse problem of Sect. 3.4 with the equation t u tt − u −

k(τ )u(x, y, t − τ )dτ = 0, (x, t) ∈ Rn+1 , y > 0, 0

in which  is n-dimensional Laplacian in the variable x = (y, x1 , . . . , xn ), instead of (3.49).

9 Open Problems

367

Problem 6 In Sect. 4.1, it is investigated the problem of determining the wave propagation velocity and the memory of the layered medium from equalities (4.1)–(4.3). The problem of this section is also important from the point of view of applications and is open if we consider instead of (4.1) the equation     u tt − a 2 u x x − a 2 u y y t =

k(τ )

     a 2 (y)u x x − a 2 (y)u y y (x, y, t − τ )dτ = 0,

0

(x, y, t) ∈ R2+ × R

(9.4)

with conditions (4.2) and (4.3). Problem 7 Section 4.2 discusses the problem of determining the regular part of the instantaneous source physically simulating a point explosion at the boundary of the domain and the memory of the medium, i.e., functions f (t), k(t). 7.1. The problem of this section is also important from the point of the view of applications and is open if we consider instead of (4.1) Eq. (9.4) with initial–boundary and overdetermination conditions (4.2) and (4.3), respectively. 7.2. It is also important to study the existence of a solution to the inverse problem of Sect. 4.2 in the global sense by the method of Chap. 3. Problem 8 Chapter 5 is devoted to the inverse problems of recovering two multidimensional memory kernels in hyperbolic integro-differential equations, having a finite Fourier series with respect to one of the spatial variables. Such problems with equations of the types ∂ 2u − u = ∂t 2 and ∂ 2u − u = ∂t 2

t k(y, τ )u(x, y, t − τ )dτ 0

t k(x, y, τ )u(x, y, t − τ )dτ 0

instead of (5.1) and (5.37), respectively, are open. Problem 9 Prove global unique solvability of the problem of determining the twodimensional kernel of the thermoviscoelasticity equations system (7.126)–(7.131) for x = (x1 , x2 , x3 ), (x1 , x2 ) ∈ R2 , x3 > 0 t ∈ R, where t Ti j (x, t) = σi j [u] +

h(x2 , t − τ )σi j [u](x, τ ) dτ, 0

(9.5)

368

9 Open Problems

The inverse problem: find the kernel h(x2 , t), t > 0, appearing in (7.126) with (9.5) if the following additional information about the solution of problem (7.126)–(7.131) is known: u 3 (x2 , x3 , t)|x3 =+0 = g(x2 , t), t > 0, g(x2 , t) is given function. One can obtain conclusions from the results of Sects. 3.5 and 7.4. Problem 10 Prove unique solvability of the three-dimensional linearized problem of determining the kernel in viscoelasticity equation for x = (x1 , x2 , x3 ), (x1 , x2 ) ∈ R2 , x3 > 0, t ∈ R  ∂ 2u ∂ 2u + ρ 2 = ∂t ∂ x 2j j=1 3

t k(x1 , x2 , t − τ )

3  ∂ 2u j=1

0

∂ x 2j

(x, τ )dτ,

(9.6)

under the following initial and boundary conditions: u |t 0, appearing in (9.6) if the following additional information about the solution of problem (9.6)–(9.8) is known: u(x, t)|x3 =+0 = g(x1 , x2 , t), t > 0, g(x1 , x2 , t) is given function. If we suppose that k(x1 , x2 , t) = k0 (x1 , x2 , t) + εk1 (x1 , x2 , t), where k0 (x1 , x2 , t) ≡ k0 (x2 , t) is known and ε is a small parameter, then one can solve this problem from results of Sect. 6.1.

Reference 1. Bukhgeim, A.L., G.V. Dyatlov, and V.M. Isakov. 1997. Stability of memory reconstruction from the Dirichlet–Neumann operator. Siberian Mathematical Journal 38 (4): 636–646.