On the inverse conductivity problem

Applied Mathematics and Computation 94 (1998) 91±96

On the inverse conductivity problem B unyamin Yildiz *, Ali Sever Atat urk Universitesi, Fen-Edebiyat Fak ultesi, Matematik B ol um u, 25240 Erzurum, Turkey

Abstract It is shown here that for the boundary value problem div…aru† ˆ d…x † in Rn ; u…x† ! 0 as jxj ! 1 in order to identify the coecient a, one needs the additional data u…x; x † ˆ g…x; x †; where x 2 C1 ; x 2 C2 ; C1 ; C2 are two open nonempty subsurfaces of @X (X is a domain with analytic boundary which contains a bounded set V ) and a 2 H 2;p p P n=2, and a satis®es conditions of Lemma 2. Here we also prove the uniqueness of a entering to the problem the additional data. Ó 1998 Elsevier Science Inc. All rights reserved. Keywords: Inverse problem; Potential theory; Conductivity equation

1. Introduction In a number of practical applications for the following homogeneous boundary value problem div…aru† ˆ d…x † in Rn ; u…x† ! 0 as jxj ! 1;

…1†

instead of the so-called Dirichlet to Neumann operator Ka : uj@X ! a…@=@v†uj@X (v is unit normal vector) one uses a slightly di€erent set of data which are generated by exterior sources. Problem (1) lays a mathematical foundation to electrical impedance tomography, which is a new and promising method of prospecting the interior of the human body, by surface electromagnetic measurements on surface one prescribes current sources (like electrodes) and measure voltage (or vice versa) for ®xed or all positions of the sources. The same

*

Corresponding author. E-mail: [email protected].

0096-3003/98/$19.00 Ó 1998 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 9 7 ) 1 0 0 3 3 - 9

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B. Yildiz, A. Sever / Appl. Math. Comput. 94 (1998) 91±96

mathematical model works in a variety of applications in geophysics, mines and rocks detection, water search, etc. [1,2]. The uniqueness of a entering the problem (1) has been proved in Ref. [3] for n ˆ 2 by using the Phillips device. In this paper, we shall prove the uniqueness of a for the problem (1) and show that one needs an additional data which is generated by exterior sources to identity a in this problem, and prepare the ground for numerical implementation of this problem. The latter will be given in a forthcoming paper. 2. Preliminaries In this section we formulate the inverse conductivity problem and state some of the theorems that will be used later. If X is a region in Rn ; u represents a voltage (i.e. ru is the electric ®eld), and j denotes the corresponding current density; then Ohm's law takes the form aru ˆ j;

…2†

where a denotes conductivity, the reciprocal of the resistivity. If we assume further that the current only enters or leaves the region through its boundary, then the current ®eld f must have divergence equal to zero. Combining this observation with Eq. (2) yields the conductivity equation div…aru† ˆ 0 in X;

…3†

for a unique determination of u one can prescribe at the boundary the Dirichlet data u ˆ g on @X: …4† Here we assume that a is a scalar function, 0 < e < a, which is measurable and bounded. In this case we have a unique solution u 2 H …1=2† …X† to direct problem (3) and (4) provided g 2 H …1=2† …@X†. Often we can assume that a is constant near @X and X is a bounded domain in Rn with the boundary class of C 2 , then if g 2 C 2 @X, the solution u 2 C 1 near @X, so the following classical data a

@u ˆh @v

on C;

where C is a part of @X, is well de®ned. Then, the inverse conductivity problem may be stated as follows: ®nd a, given the data h for one g (one boundary measurement) or for all g (many boundary measurements). Now, we state some of the theorems that will be used later. Theorem 1. Let n P 3 and let the conductivity coecient a 2 H 2;p …X†; p P n=2. Then a is uniquely identi®ed by its Dirichlet to Neumann map [4].

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93

Theorem 2. Let aj 2 H 2;p …V † where V is a neighborhood of @X …j ˆ 1; 2†: Then the aj produce the same Dirichlet to Neumann maps if and only if a1 ˆ a2 ; ra1 ˆ ra2 on @X and Z Z a1 ru1 ru2 ˆ a2 ru1 ru2 ; X

X

for all solutions uj 2 H 1;2 …X† to the equations div…aj ruj † ˆ 0 in X [5]. Recently Nachman [8] completely solved the uniqueness question for the plane inverse conductivity problem by using a method of scattering theory. Theorem 3. Let a1 ; a2 satisfy one of the three conditions: (i) they are picewise analytic in X and their analiticity domains X…k; aj † have picewise analytic boundaries, (ii) they are constant on X…k; aj †, (iii) aj ˆ a0 ‡ v…Dj †a# j where a0 is # 2 given and a# j are unknown C …X†-functions, aj 6ˆ 0 on @Dj and Dj is an unknown open subset of X with the Lipschitz boundary and with connected X n Dj : Let KC1 ˆ KC2 , then in the all three cases we have a1 ˆ a2 [5]. Lemma 1. Under the conditions of Theorem 3 we have Z Z a#1 rv1 :rv2 ˆ a#2 rv1 :ru2 D1

X

for all solutions v1 ; v2 to the equations div…aj ruj † ˆ 0 near D0 [5]. Theorem 4 (Analytic Continuation) [6]. A function f …x† with range in C m and domain in C n is analytic in the open set U if f is continuous and has continuous ®rst derivatives with respect to the independent variables. Here the assumption of continuity and its derivate is unnecessary by Hartog's theorem.

3. Statement and proof of result Let us consider the problem (1) with the additional data u…x; x † ˆ g…x; x †;

x 2 C1 and x 2 C2 :

…5†

We ®rst, show that to identify a we need the additional condition given by (i) and second, prove the uniqueness of a entering in this boundary value problem. By choosing a ˆ ÿ1 outside a bounded set V and X is a domain with analytic boundary containing V . We have the Green's function G…x; x † of the operator

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B. Yildiz, A. Sever / Appl. Math. Comput. 94 (1998) 91±96

L: div…r:† as a solution to problem (1) which is analytic when x; x 2 R3 n V ; x 6ˆ x . Using Eq. (5) g…x; x † is analytic on C1  C2 : Indeed g…x; x † is analytic for all x 2 @X …x 6ˆ x † since @X is analytic for x 2 C2 and g…x; x † is analytic on C1 . This implies that g is analytic on the whole boundary of X by using analytic continuation. Therefore we have Dirichlet data on @X. On the other hand if we solve the exterior problem ÿ Du ˆ 0 in R3 n X; uj@X ˆ g…x; x † …u ! 1 as jxj ! 1†: we get the solution G…x; x † for all x 2 R3 n X; and x 2 C2 but since we also know that G is symmetric then we have G…x ; x† for all x on the boundary of X again by analytic continuation. Now, by solving the exterior problem ÿ Du ˆ d…x †; uj@X ˆ G…x ; x†;

on @X …u ! 1 as jxj ! 1†;

we have G…x ; x† for all x 2 R3 n X which means we know the Neuman data. Consequently, we know the Dirichlet to Neuman data. We can now proceed to show that the coecient a entering the problem (1) is unique. Suppose that there exists a1 ; a2 such that u1 …x; x † and u2 …x; x † are the solution to div…a1 ru1 † ˆ d…x †

in R3 ;

u1 …x† ! 0 as jxj ! ‡1 and

div…a2 ru2 † ˆ d…x †

in R3 ;

u2 …x† ! 0 as jxj ! ‡1; respectively. Using the completeness of the product of harmonic solutions [7] we have Z …a1 ÿ a2 †ru1 …:; x †ru2 …:; x † ˆ 0; V

for all solutions of the global problem (1). (Here the Dirichlet-to-Neumann map g inside @X:) Now we need to show that Z …a1 ÿ a2 †rv1 …x†rv2 …x† dx ˆ 0; V

for all solutions vj of the equation div…aj rvj † ˆ 0 near V …j ˆ 1; 2†: To this end, let us subtract the following equations div…a1 ru1 † ˆ d…x †; div…a2 ru2 † ˆ d…x †:

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95

We have, div…a2 ru† ˆ div……a1 ÿ a2 †ru1 †; where u ˆ u1 ÿ u2 : Using the de®nition of weak solution to the above equation, we have, for u 2 C1, Z Z a2 ruru ˆ …a1 ÿ a2 †ru1 ru: V

V

Using as a test function v2 …x† where v2 …x† satis®es the equation div…a1 rv2 † ˆ 0 near V , we have Z Z a2 ru…x; x †rv2 …x† dx ˆ …a1 ÿ a2 †ru1 …x; x †rv2 …x† dx V

…6†

V

since u ˆ 0 near @V . On the other hand since div…a2 rv2 † ˆ 0

near V ;

using the de®nition of a weak solution and u as a test function we have Z a2 rv2 …x†ru…x; x † dx ˆ 0: V

So, from equality (6), we get Z …a1 ÿ a2 †ru1 …x; x †rv2 …x† dx ˆ 0; V

for all solutions u1 …x; x †; v2 …x† with x outside X. Now by using the well-known Runge approximation property and an argument similar to Lemma 1, we can extend these relations onto all solutions v1 …x† and v2 …x† such that div…aj rvj † ˆ 0

near V …j ˆ 1; 2†:

Therefore, Z …a1 ÿ a2 †rv1 …x†rv2 …x† ˆ 0;

…7†

…8†

V

for all solutions vj to Eq. (7). Now we have a1 ˆ a2 ; ra1 ˆ ra2 ; and the ortogonoality relation (8). Therefore, by using the Theorem 2, we have K1 ˆ K2 and so using Theorem 1 we have a1 ˆ a2 for the problem (1).

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B. Yildiz, A. Sever / Appl. Math. Comput. 94 (1998) 91±96

References [1] G.V. Keller, F.C. Frischknect, Electrical Methods in Geophysical Prospecting, Permagon Press, Oxford, 1977. [2] A. Wexler, C.J. Mandel, An impedance computed tomography algorithm and system for ground water and hazardous waste imaging, Presented at the Second Annual Canadian± American Conference on Hydrogeology, Ban€, June 1985. [3] A. Sever, V. Isakov, Numerical implementation of an integral equation method for the inverse boundary value problem, J. Inverse Problems (to appear). [4] J. Sylvester, G. Uhlmann, A global uniqueness theorem for an inverse boundary problem, Ann. Math. 125 (1987) 153±169. [5] V. Isakov, Manuscript of the book in preparation. [6] F. John, Partial Di€erential Equations, 4th ed., 1982. [7] V. Isakov, Inverse Source problems, A.M.S., Math. Surveys Monographs, vol. 34, 1990. [8] A.I. Nachmann, Global uniqueness for a two dimensional inverse boundary value problem, Ann. Math. (to appear).