Anisotropic inverse conductivity and scattering problems

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1 Anisotropic inverse conductivity and scattering probles Kiwoon Kwon Dongwoo Sheen To appear in Inverse Probles, 00 Abstract Uniqueness in inverse conductivity and scattering probles is considered. In case the ediu consists of two discontinuous constant anisotropic conductive parts, the easureents of potential and induced currents on the boundary of surrounding body are enough to guarantee uniqueness to deterine conductivity and region of ebedded unknown aterial under a very weak condition. The analogous uniqueness result is also obtained for an inverse scattering proble in the case that the ediu is coposed of two anisotropic and hoogeneous aterials. Key words: anisotropic ediu, inverse proble, conductivity, scattering 1 Introduction An inverse conductivity proble, soeties called an electrical ipedance toography proble, is to deterine the interior conductivity profile or the location, shape, and size of an obstacle by easuring a pair or any pairs of current and voltage on the boundary or its part. To describe physically ore rigorously, let u denote the electric potential (voltage) in a bounded ediu Ω with a Lipschitz boundary Ω in R with the conductivity of the ediu γ = (γ jk ) j,k=1,. Then the potential f on Ω and u satisfies L γ u := (γ u) = 0 in Ω, (1.1a) u = f on Ω. (1.1b) Then current flux on Ω is given by ν γ u, where ν denotes the unit outward noral to Ω. A atheatical interpretation of an inverse conductivity proble proposed by Calderon [6] is to deterine γ fro the inforation on the Dirichlet-to-Neuann ap Λ γ : H 1/ (Ω) H 1/ (Ω), It is well-known that the knowledge of Λ γ suffices to deterine continuous, isotropic conductivity uniquely in R n for n = [5, 30], and for n 3 [1, 9, 37, 38]. For the case of discontinuous (e.g., piecewise constant) isotropic conductivity, the question of unique deterination has been considered in [4, 1, 0, 1, ]. However, ost of such results cannot be extended directly to the corresponding proble for anisotropic edia, although any interesting aterials are in fact anisotropic. For instance, see [5, 6, 7, 8, 36, 39] and the bibliographies therein. For the continuous conductivity case, anisotropic inverse probles have been studied in [, 3, 7, 30, 36, 37] School of Mechanical and Aerospace Engineering, Seoul National University, Seoul , Korea; E-ail: kkwon@nasc.snu.ac.kr Departent of Matheatics, Seoul National University, Seoul , Korea; E-ail: sheen@snu.ac.kr 1

2 K. Kwon and D. Sheen For discontinuous anisotropic edia, uniqueness has been known if the difference of discontinuities (or refractive indices) is positive-definite [17], and in the two diensional case this condition is weakened so that the deterinant of obstacle conductivity is different fro that of background aterial if the conductivity atrix is sall in a certain sense [16]. In this article, the sallness condition in [16] is reoved in case the ediu consists of two regions of hoogeneous conductive aterials with the deterinants of conductivity of the obstacle and surrounding ediu being different. One of the popular tools in the analysis of uniqueness of inverse probles is the use of singular solutions. In what follows, we liit to anisotropic cases only. Denoting by Φ j,x, j =, 3, the fundaental solutions to the differential operator L γj, with x being the singular point, Isakov [17] observed that Φ j,x satisfies the equation (.4) in place of u j s. He then proceeded to induce a contradiction by showing that the left hand side in (.4) tends to infinity as x converges to a certain point, say, the origin O. In two diension under certain conditions, Ikehata [16] used coplex functions of O( 1 x ) growth, instead. In our paper, real functions of O(log x ) growth are used, which have the sae growth as fundaental solutions. Our analysis can be also applied to inverse scattering probles with discontinuous anisotropic hoogeneous edia. which is to deterine aterial properties (including refractive indices) or the shape and size of an obstacle by easuring scattered fields at a large distance, or equivalently easuring far-field patterns. Let u and u s denote the scattered and total waves, respectively, generated by the incident wave e ikx d of wave nuber k in the direction d S 1. Then the following equations hold: L γ,k,q u := (γ u) + k q u = 0 in R, (1.a) u(x) = e ikx d + u s (x), (1.b) ( ) li r 1 u s r r ikus = 0, (1.c) where γ is the conductivity and q is the refractive index in the ediu. Here, q 1 is assued to be an L (R ) function with copact support. The far-field pattern u then satisfies ( u s (x) = x 1 e ik x u (ˆx) + O( 1 ) x ), with ˆx = x x. An inverse scattering proble is to find γ and q where the far field patterns u ( ; d) are given for all incident direction d S 1 with a fixed frequency k [7, 8]. The theory of inverse scattering probles for isotropic edia has been well-developed atheatically and nuerically (see [7, 8, 9, 19, 3, 4, 3, 35] and the references therein). Corresponding theory for anisotropic edia needs ore attention [10, 11, 14, 33, 34]. For discontinuous anisotropic edia, the far-field pattern deterines the ediu structure uniquely provided that the jup of discontinuity is positive-definite [18]. This condition can be weakened in two diension such that the deterinants of the background and obstacle are different as far as the anisotropy near the boundary of the obstacle is known [31]. In the present paper, uniqueness is obtained without knowing the anisotropy at the boundary for discontinuous anisotropic hoogeneous ediu in R. Fundaental solutions of general elliptic partial differential and Lippann-Schwinger-type equations are used in the analysis of inverse scattering probles [17, 31], respectively. We again ake use of real-valued functions of O(log x ) which have the sae type of singularity as fundaental solutions. The relation between

3 Anisotropic inverse conductivity and scattering probles 3 the far-field operator and Dirichlet-to-Neuann ap enables to apply to inverse scattering proble the analysis carried out for our inverse conductivity proble [17, 19]. We analyze in detail the inverse conductivity proble in discontinuous anisotropic edia in the section to follow. Then in the subsequent section, we will apply the analysis to the inverse scattering proble of discontinuous anisotropic coefficient. Anisotropic Inverse Conductivity Proble.1 Uniqueness Result This section develops a uniqueness result related with the inverse proble of (1.1). Particularly, we restrict our attention to the case of the discontinuous anisotropic conductivity of the for γ = A + (B A)χ D, where A is a known syetric positive-definite atrix, B a syetric positive-definite atrix to be deterined, and D a Lipschitz doain, also to be deterined, copactly contained in Ω R such that Ω\D is connected. Furtherore, assue that A and B are constant atrices. We will then prove the following theore. Theore.1. Let γ j = A+(B j A)χ Dj for j = 1,, where D 1 and D are Lipschitz doains copactly ebedded in Ω R such that Ω \ D j, j = 1, are connected, and B 1 and B are constant syetric positive-definite atrices. If det(b j ) det(a), then Λ γ1 = Λ γ iplies D 1 = D and B 1 = B.. The Proof of Theore.1: D 1 = D We begin by siplifying the types of γ j, j = 1,. Since A is syetric positive-definite, there exist an invertible atrix C such that A = C t C. Identify C with the linear transforation x x = C x, and denoting Ω = C 1 (Ω), D = C 1 (D), f( x) = f(c x), and γ( x) = C t AC 1 + (C t BC 1 C t AC 1 )χ D(C x). Then Proble (1.1) can be reduced to x (I + ( B I)χ D) x ũ = 0 in Ω, ũ = f on Ω, where B = C t BC 1 is a syetric positive-definite atrix. Therefore, without loss of generality, the atrix A ay be assued to be the identity atrix I in the original equations (1.1). The tildes will be dropped fro now on. Furtherore, consider an orthogonal atrix P such that B 1 = P t ΣP, where Σ is a diagonal atrix whose diagonal eleents consist of eigenvalues of B 1. Again, by the identification of P 1 with the induced linear transforation x x = P 1 x, Equations (.3) can be reduced into a sipler setting in which B 1 is a diagonal atrix with positive diagonal entries while B is a syetric, positive-definite atrix.

4 4 K. Kwon and D. Sheen Suing up, we henceforth assue that ( λ1 0 A = I, B 1 = 0 λ ), λ 1, λ > 0, and B is a positive-definite atrix in Equations (.3). Also, without loss of generality, we ay assue λ 1 λ if λ 1 λ > 1 (.3) λ 1 λ if λ 1 λ < 1. ( ) 0 1 Indeed, this can be done, if necessary, by using the transforation x x = Q x, where Q = ; 1 0 Q t B 1 Q 1 then switches the order of diagonal entries of B 1. It is thus enough to prove Theore.1 for the case where γ j = I + (B j I)χ Dj, j = 1, with B 1 being a diagonal atrix satisfying (.3). Next, assue that D 1 D, say, D 1 is not contained in D. Following [17], choose a point p D 1 \ D, which also lies in the intersection of the convex hull of D 1 D and the closure of the connected coponent of Ω \ (D 1 D ) that is in contact with Ω. Consider a disk B r (p) with center p of radius r > 0 to be specified in a oent. Set D 3 = D 1 B r (p). Also, let D 4 be the union of D 1 D and all the coponents of Ω \ (D 1 D ) that do not touch Ω. Finally, let D 5 = Ω \ D 4, the connected coponent of Ω\(D 1 D ) whose boundary contains Ω. Choose the radius r such that B r (p) Ω, B r (p) D = and D 4 B r (p) is the iage of a Lipschitz function ψ : [0, 1] R. Without loss of generality, we ay assue p to be the origin O. For the notational siplicity write B 3 = B 1, and define γ 3 by γ 3 = I + (B 3 I)χ D3 (= I + (B 1 I)χ D3 ). Recall the following lea due to Isakov [17]. Lea.. For j =, 3, let u j satisfy L γj u j = 0 in a neighborhood of D 4 for j =, 3. Then (B 1 I) u u 3 dy = D 1 (B I) u u 3 dy. D (.4) Notice that the restrictions of γ j to B r (O) are the constant atrices. Indeed, they are the identity atrix and the atrix B 1 for j = and j = 3, respectively, which we will denote by γ and γ 3. Then the following lea is a suitable odification of a result in [15]. Lea.3. Suppose that a function Ũ j L 1 loc (R ) H 1 loc (R \ {O}), for j =, 3, satisfies (γ j Ũ j ) = 0 in R \ O. (.5) Then there exists a faily of functions {ũ j,ξ }, ξ B r/ (O) \ D 4, such that sup ũ j,ξ ( ) Ũ j ( ξ) H1 (D 4) <, ξ B r/ (O)\D 4 ũ j,ξ Ω = 0, (γ j ũ j,ξ) = 0 in D 4,ξ = {y d(y, D 4 ) < d(ξ, D 4 )/}. (.6a) (.6b) (.6c)

5 Anisotropic inverse conductivity and scattering probles 5 Proof. Let ξ B r/ (O) \ D 4 be arbitrary. First, a ollification arguent will provide a sequence of C ( Ω)- functions {φ j,ξ } 1 such that For each, let w j,ξ and and z j,ξ φ j,ξ ( ) Ũ j ( ξ) in H 1 loc(ω\{ξ}) as. H 1 (Ω) be the unique solutions of the following probles: (γ j w j,ξ ) = (γ j φ j,ξ ) in Ω, (.7a) (γ j z j,ξ ) = 0 in Ω, z j,ξ By elliptic regularity and the trace theore, we get z j,ξ w j,ξ = 0 on Ω, (.7b) = φ j,ξ on Ω. z j,ξ H 1 (Ω) C φ j,ξ φ j,ξ H 1/ ( Ω) C φ j,ξ φ j,ξ H 1 (Ω\B r(o)), (.8) fro which we have z j,ξ H 1 (Ω) such that Setting v j,ξ = w j,ξ φ j,ξ z j,ξ, we see that z j,ξ z j,ξ as in H 1 (Ω), sup z j,ξ H1 (Ω) <. ξ B r/ (O)\D 4 (γ j v j,ξ ) = ((γ j γ j ) (φ j,ξ + z j,ξ ) ) in Ω, (.9a) Since γ j = γ j in B r (O), the estiate (.8) leads to v j,ξ v j,ξ = 0 on Ω. (.9b) v j,ξ H 1 (Ω) C{ φ j,ξ φ j,ξ H 1 (Ω\B r(o)) + z j,ξ z j,ξ H 1 (Ω)\B r(o)} C φ j,ξ φ j,ξ H 1 (Ω\B r(o)). Fro this it follows that, for all j, ξ, there is a v j,ξ H 1 (Ω) such that v j,ξ v j,ξ in H 1 (Ω) as, sup v j,ξ H 1 (Ω) <. ξ B r/ (O)\D 4 For each j, let ũ j,ξ = Ũ j ( ξ) + v j,ξ + z j,ξ. It is then iediate to see that {ũ j,ξ } satisfies the properties of (.6). This copletes the proof. Since O lies in the convex hull of D 1 D, there is a straight line passing through O such that D 1 B r (O) is contained in one side of this straight line. We ay assue this straight line to be the y 1 axis and the y axis to be a vertical line to this line. Assue further that the y axis passes through D 1 B r (O), if necessary, after

6 6 K. Kwon and D. Sheen rotating the y 1 axis. Choose the positive directions of the y axis toward D 1 B r (O). Since D 1 is Lipschitz, there are two cones K i and K e with angles 0 θ i < π at O satisfying [13] B r (O) K i D 1, B r (O) K e D 1 = {O}. More explicitly, let us denote such cones by K i = {(y 1, y ) R π θi < arg(y 1 + iy ) < π + θi }, with K e being the reflection of K i to the y 1 -axis. Here and in what follows, the local branch of the arguent function arg is chosen as the negative y axis. For j =, 3, let ũ j = 1 log(y 1 + τ j y ), λ where τ = 1, τ 3 = 1 λ. Then one can verify that the function ũ j L 1 loc (R ) Hloc 1 (R \{O}) satisfies (.5). Then let {ũ j,ξ } ξ Br/ (O)\D 4, j =, 3, be given such that (.6) is fulfilled with ũ j in place of Ũ j, respectively, as in Lea.3. Recalling fro (.6c) that D 4,ξ is a neighborhood of D 4, we see that the orthogonality relation (.4) applies to ũ j,ξ, j =, 3, with ξ K e B r/ (O), which yields (B 1 I) ũ,ξ ũ 3,ξ dy (.10) D 1 B r(o) = (B 1 I) ũ,ξ ũ 3,ξ dy D 1 = (B I) ũ,ξ ũ 3,ξ dy D D 1\B r(o) D 1\B r(o) (B 1 I) ũ,ξ ũ 3,ξ dy (B 1 I) ũ,ξ ũ 3,ξ dy. Notice that the last two integrals are bounded regardless of ξ K e B r/ (O) by (.6a). Consequently, the integral (.10) is bounded. In the eanwhile, we are going to show in the following that the integral (.10) is unbounded as ξ O, which leads to a contradiction. Decopose (.10) into the su of I 1, I, I 3 and I 4, where I 1 = (B 1 I) ũ (y ξ) ũ 3 (y ξ) dy, I = I 3 = I 4 = D 1 B r(o) D 1 B r(o) D 1 B r(o) D 1 B r(o) (B 1 I) (ũ,ξ (y) ũ (y ξ) ) ũ 3 (y ξ) dy, (B 1 I) ũ (y ξ) ( ũ 3,ξ (y) ũ 3 (y ξ) ) dy, (B 1 I) ( ũ,ξ (y) ũ (y ξ) ) ( ũ 3,ξ (y) ũ 3 (y ξ) ) dy. Due to (.6a), an upper bound of I 4 is easily obtained, say, I 4 < C 0. Next, again by (.6a) and Young s inequality, there exist C 1 > 0 and C > 0 such that I + I 3 C 1 ɛ D 1 B r(o) for all ξ K i B r/ (O) and ɛ > 0. Using the following inequality, dy y ξ + C ɛ I 1 + I + I 3 + I 4 I 1 I I 3 I 4 dy I 1 C 1 ɛ y ξ C ɛ C 0, D 1 B r(o) (.11a)

7 Anisotropic inverse conductivity and scattering probles 7 we will show that (.11a) is unbounded as ξ O. For s < r/, choose ξ s K e B r/ (O) such that ξ s = (0, s). Depending on s, denote by (ρ s, θ s ) the polar coordinate syste with origin at ξ s such that y ξ s = (ρ s cos θ s, ρ s sin θ s ), where ρ s (y) = y ξ s and θ s (y) = arg(y ξ s ). Let M s = ax y S y ξ s, where S = D 1 B r (O) consists of a set of two points for sufficiently sall r. Notice that M s converges to r as s converges to 0. Take the two points P 1, P K i such that arg(p 1 ) = π θi, arg(p ) = π + θi and P 1 = P = s. Then we have arg(p 1 ξ s ) = π θi, arg(p ξ s ) = π + θi and P 1 ξ s = P ξ s = s 1 + cos θ i. Let us decopose D 1 B r (O) into the three regions: E s = {y D 1 B r (O) s 1 + cos θ i ρ s (y) M s }, F s = {y D 1 B r (O) s ρ s (y) < s 1 + cos θ i }, G s = {y D 1 B r (O) ρ s (y) > M s }. (.1) By (.11) and (.1), I 1 + I + I 3 + I 4 is bounded below as follows: I 1 + I + I 3 + I 4 (B 1 I) ũ (y ξ) ũ 3 (y ξ) dy C 1ɛ E s E s (B 1 I) ũ (y ξ) ũ 3 (y ξ) dy C 1ɛ F s (B 1 I) ũ (y ξ) ũ 3 (y ξ) dy C 1ɛ G s dy y ξ F s G s dy y ξ dy y ξ C ɛ C 0. (.13) We will show that the integrals over F s and G s are uniforly bounded, while that over E s is unbounded. Since G s G r for all s r 1, the integral over G s is bounded uniforly in s as follows: (B 1 I) ũ (y ξ) ũ 3 (y ξ)dy + C dy 1ɛ G s G s y ξ ax( λ 1 1, λ 1 ) ũ (y ξ) ũ 3 (y ξ) dy + C1 ɛ G r G r dy y ξ (C 3 + C 1 ɛ) G r r (C 3 + C 1 ɛ)π. (.14) Turn to estiate the integrals over E s and F s. (B 1 I) ũ (y ξ) ũ 3 (y ξ)dy C dy 1ɛ E s E s y ξ Ms 1 β(s) [ ] λ 1 λ = s 1+cos θ i ρ s α(s) λ cos θ s + λ 1 sin 1 dθ s dρ s θ s Ms 1 β(s) C 1 ɛ dθ s dρ s, (.15) s 1+cos θ i ρ s α(s)

8 8 K. Kwon and D. Sheen and (B 1 I) ũ (y ξ) ũ 3 (y ξ)dy + C dy 1ɛ F s F s y ξ s 1+cos θ i 1 β(s) [ ] λ 1 λ = λ cos θ s + λ 1 sin 1 dθ s dρ s θ s s + C 1 ɛ ρ s α(s) s 1+cos θ i 1 β(s) s ρ s α(s) dθ s dρ s, (.16) where α(s) = in y D 1 B r (O) arg(y ξ s ), β(s) = ax y D 1 B r (O) arg(y ξ s ). y ξ s = ρ s y ξ s = ρ s Notice that β [ ] λ 1 λ λ cos θ + λ 1 sin θ 1 dθ α [ (β α) 1 + ] [ λ 1 λ π 1 + in(λ 1, λ ) More explicitly, a change of variables iplies, for 0 α < π < β π, β α (λ 1 1)λ cos θ + λ 1 (λ 1) sin θ λ cos θ + λ 1 sin dθ θ = λ 1 λ (π arctan( λ1 λ tan(π β)) arctan( λ1 λ tan α) ] λ 1 λ := C 4. (.17) in(λ 1, λ ) ) (β α), where the range of arctan is contained in ( π, π ). Define the continuous functions δ : [0, π ] [ π, π] R, and δ : [0, r] R by ( λ λ1 λ π arctan( 1 λ tan(π β)) arctan( ( δ(α, β) = λ λ1 λ π arctan( 1 λ tan(π β)) π ) λ1 λ (π π arctan( λ 1 λ tan α) ) λ 1 λ tan α) (β α), 0 α < π ) < β π, ( ) β π, α = π, π < β π, ( π α), 0 α < π, β = π, and δ(s) = δ(α(s), β(s)). Using the constant C 4 in (.17), an upper bound of the integral (.16) is obtained: s 1+cos θ i 1 s 1+cos θ i δ(ρ s )dρ s s ρ s + C 1 1ɛ dρ s s ρ s ( ) (C 4 + C 1 ɛ) log 1 + cos θ i. (.18) Finally, let us consider the integral (.15). If λ 1 λ > 1, by (.3), ( ) ( ) λ1 λ1 arctan tan(π β) < π β and arctan tan α < α, λ λ

9 Anisotropic inverse conductivity and scattering probles 9 resulting in δ(s) (λ 1 λ 1)(β(s) α(s)) > 0. On the other hand if λ 1 λ < 1, again by (.3), ( ) ( ) λ1 λ1 arctan tan(π β) > π β and arctan tan α > α, λ λ which leads to δ(s) (λ 1 λ 1)(β(s) α(s)) < 0. Therefore, as s varies, δ(s) does not change its sign for fixed λ 1, λ. In E s, it is obvious that 0 α(s) π θi and π + θi β(s) π, a lower bound of δ(s) being obtained: δ(s) λ 1 λ 1 θ i > 0. The choice of ɛ = λ1λ 1 C 1 gives a lower bound of the integral (.15) over E s : (B 1 I) ũ (y ξ) ũ 3 (y ξ)dy C 1ɛ E s Ms s 1+cos θ i Ms λ 1λ 1 1 [ λ 1 λ 1 C 1 ɛ] dρ s ρ s s 1+cos θ i Suing up (.13), (.14), (.18) and (.19), we get I 1 + I + I 3 + I 4 λ 1λ 1 Ms s 1+cos θ i ( C 3 + λ 1λ 1 ) π E s dy y ξ 1 ρ s dρ s, (.19) ( 1 ρ dρ C 4 + λ ) 1λ 1 log( (1 + cos θ i )) C λ 1 λ 1 C 0. (.0) All ters in the right side of (.0) are constant except the first one that tends to λ1λ 1 1 ρ dρ to blow up. Hence I 1 + I + I 3 + I 4 becoes unbounded as ξ s approaches to O, which is a contradiction originated fro the assuption that D 1 D. This proves that D 1 = D. r 0.3 The Proof of Theore.1: B 1 = B It reains to prove B 1 = B. Let D = D 1 = D and take ˆf η (ξ) = η ξ, for a fixed vector η R. Then by the uniqueness of the Dirichlet proble, η ξ is the unique solution of Next take a haronic function v in Ω \ D such that (B 1 ˆv) = 0 in D, ˆv = ˆf η on D. v η := ˆv η = ˆf η on D, v η ν := ν B 1 ˆv η on D.

10 10 K. Kwon and D. Sheen Let f η be the trace on Ω of v η H 1 (Ω \ D) and u η j be the solution of (1.1) replacing f by f η for j = 1,, respectively. By uniqueness, u η 1 (ξ) = vη (ξ)χ Ω\D (ξ) + η ξχ D (ξ). Suppose that the Dirichlet-to-Neuann aps Λ γ are identical corresponding to γ = I + (B 1 I)χ D and γ = I + (B I)χ D. Then u η 1 = uη in Ω \ D by unique continuation. Thus the transission condition iplies that u η satisfies (B u η ) = 0 in D, u η = ˆf η on D, and hence u η (ξ) = η ξ in D. Moreover, ν (B 1 B ) η = uη 1 ν + uη ν + = 0 on D for all η R. Consequently, get B 1 = B. Therefore we prove Theore.1. 3 Anisotropic Inverse Scattering Proble In this section we apply the analysis carried out for the inverse conductivity proble (1.1) to the inverse scattering proble (1.). As the arguent of the proof is identical to that in Section, it will be briefly sketched. 3.1 Uniqueness Result Analogously to the conductivity case, we consider the case γ = A + (B A)χ D and q = 1 + (ρ 1)χ D, where A is a known syetric positive-definite atrix, B a positive-definite atrix, and D a Lipschitz doain copactly ebedded in R such that R \ D is connected. Moreover, we assue that the atrices A and B are constant. The uniqueness of our inverse scattering proble is stated in the following theore. Theore 3.1. Let γ j = A + (B j A)χ Dj, q j = 1 + (ρ j 1)χ Dj and u j be the solution of (1.) for γ = γ j and q = q j for j = 1,. Assue det(b j ) det(a), j = 1,. Then u 1 ( ; d) = u ( ; d) for all direction d S 1 (3.1) iplies D 1 = D, B 1 = B and ρ 1 = ρ. 3. Proof of Theore 3.1: a sketch First of all, inverse conductivity and scattering probles are closely related with the following approxiation property Leale:consca, which can be found in [18].

11 Anisotropic inverse conductivity and scattering probles 11 Lea 3.. Let Ω be an open set where R \Ω is connected, D Ω be a Lipschitz doain such that Ω\D is connected and v be a solution to the equation (γ v) + k qv = 0 in Ω. Then v is the H 1 (D)-liit of linear cobination of solutions u( : d), d S 1 satisfying (1.). With the aid of this lea we can apply the analysis carried out for our inverse conductivity proble in the previous section to analyze our inverse scattering proble. We then follow the arguents in Section to prove Theore 3.1. The sae changes of variables convert A, B 1, B into I, a diagonal atrix, and a syetric positive-definite atrix, respectively. Let D 3, D 4, D 5 and p be identical to those appeared in the proof of Theore.1. Analogously to Lea., the following lea holds: (See also [18]) Lea 3.3. If (3.1) holds, then ( (B1 I) u u 3 + k q 1 u 3 u ) ( dx = (B I) u u 3 + k q u 3 u ) dx, D 1 D for all u j, j =, 3, such that in soe neighborhood of D 4. Siilarly to Lea.3, we also have (γ j u j ) + k q j u j = 0 j =, 3, Lea 3.4. Suppose that u j, j =, 3 satisfy Then there exist {u j,x } x Br/ such that u j L 1 loc(ω) H 1 loc(ω\{a}) and (γ j u j ) = 0 in D 4. sup u j,x ( ) u j ( x) H 1 (D 4) < ; u j,x Ω = 0 x B(r/) and γ j u j,x + k q j u j,x = 0 in D 4,x = {y d(y, D 4 ) < d(x, D 4 )/}. Proof. The proof is identical to that of Lea.3 except (.7) and (.9). Take φ j,x, z j,x in the sae anner as in Lea.3. Let w j,x H 1 (Ω) be such that and v j,x = w j,x φ j,x (γ j w j,x ) + k qw j,x = (γ j u j,x ) in Ω, z j,x. Then v j,x satisfies w j,x = 0 on Ω, (γ j v j,x ) + k qv j,x = ((B j B j χ Dj ) (φ j,x + z j,x )) in Ω, The rest of the proof follows by the sae arguent as in Lea.3. v j,x = 0 on Ω.

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