UNIQUENESS ON RECOVERY OF PIECEWISE CONSTANT CONDUCTIVITY AND INNER CORE WITH ONE MEASUREMENT

. We consider the recovery of piecewise constant conductivity and an unknown inner core in inverse conductivity problem. We ﬁrst show the unique recovery of the conductivity in a one layer structure without inner core by one measurement on any surface enclosing the unknown medium. Then we recover the unknown inner core in a one layer structure. We then show that in a two layer structure, the conductivity can be uniquely recovered by using one measurement.


(Communicated by Hongyu Liu)
Abstract. We consider the recovery of piecewise constant conductivity and an unknown inner core in inverse conductivity problem. We first show the unique recovery of the conductivity in a one layer structure without inner core by one measurement on any surface enclosing the unknown medium. Then we recover the unknown inner core in a one layer structure. We then show that in a two layer structure, the conductivity can be uniquely recovered by using one measurement.
1. Introduction. where B is the inclusion embedded in R d with a C 1,α (0 < α < 1) smooth boundary, χ(B) (resp. χ(R d \ B)) is the characteristic function of B (resp. R d \ B). The parameter σ stands for the conductivity of the inclusion, which is supposed to be different from the background conductivity 1. We also call σ the contrast. The function h is a harmonic function in R d representing the background electrical potential, and u represents the perturbed electrical potential. The inverse conductivity problem can be defined as finding the inclusion B (and its conductivity σ) given h and boundary measurement u| ∂Ω . There are lots of existing works on recovery of the inclusion B by using finite measurements or infinitely many measurements. The global uniqueness results with finite many measurements are only obtained when the inclusion is restricted to convex polyhedrons and balls in three-dimensional space and polygons and disks in the plane (see [8,13,16,17,23] and the references therein). For uniqueness of finding the inclusion with infinitely many measurements, there is a well-established theory. We refer to survey papers and books [15,24,25,26,27]. We also refer to [3,1,9,10,14,20,22] for uniqueness results in recovering of obstacles in acoustic scattering problems.
In previous works on recovery the inclusion by one measurement, they all assume that the contrast σ is a given constant in advance. The main focus of those works is on how to recover the shape of the inclusion. As far as now, only few special types of inclusion, like disk and ball, polyhedral and polygon, has been proved to be reconstructed by using one measurement. In this paper, instead of considering the recovery of the shape, we consider the recovery of the piecewise constant conductivity together with the inner core. We mainly use the Green function theory, transmission condition and unique continuation theorem to prove the uniqueness by using one measurement. We first prove that the conductivity in a one layer structure without inner core can be recovered by using one measurement. Then we prove that the unknown inner core in a one layer structure can also be recovered by using one measurement. Finally, we show that in a two layer structure, the conductivity can be uniquely recovered by using one measurement. The method in fact can be extended to the recovery of piecewise constant material parameters and obstacle in acoustic scattering problem and electromagnetic scattering problem. We mention that our method can also be used to recover piecewise constant conductivities in multi-layered structures, which will be our future work. On one hand, this method may be used to construct the generalized polarization tensors vanishing structures, which were widely studied in [5,6,7], for enhancement of approximating cloaking effects. On the other hand, the method can also be used for devising nano-shell structures in the application related to plasmon resonances [11].
1.2. Integral representation. Let Γ(x) be the fundamental solution to the Laplacian In what follows, we denote by S B : H −1/2 (∂B) → H 1 (R d \ ∂B) the single layer potential operator given by where (K B ) * is the adjoint operator of K B . Here and throughout this paper, the subscripts ± indicate the limits from the outside and inside of a given inclusion B, respectively. We mention that for Lipschitz domain B, the eigenvalues of K B lies in (−1/2, 1/2], where for C 1,α (0 < α < 1) domain B the corresponding eigenfunction associated with eigenvalue 1/2 is 1, i.e., K B [1] = 1/2. We refer to [4,12,18] for relevant discussion. In what follows, we suppose that all bounded domains which are mentioned in the sequel are C 1,α smooth domains. We mention that if B is a ball then K * B is a self-adjoint operator. On the other hand, if B is not a ball then K * B is generally not a self-adjoint operator, but S B K * B is self-adjoint and satisfies the Calderón's identity In the following let L 2 0 (∂B) be the set of functions which are in L 2 (∂B) and with zero average on ∂B. We recall that −S B is a positive definite operator on L 2 (∂B) for d = 3 and on L 2 0 (∂B) for d = 2 (see, e.g. [4] for details). Furthermore, there exists a self-adjoint operator A B : The proof can be seen from the proof of Theorem 1 in [19] (see also (34) in [2]). In the sequel, we shall only consider the uniqueness results in three dimensional case. However, we want to point out that, most of the results also work in the two dimensional case, and some results only need to have some mild modifications in the two dimensional case.
2.1. One-layer medium. Suppose that σ = k in B, where k is a constant, and B is a simply connected inclusion which has C 1,α (0 < α < 1) smooth boundary. Note that (1.1) is equivalent to It can be found (see, e.g., [4]) that the solution to (2.1) can be represented by where φ has the form where λ is defined by .
We shall present some unique recovery results. Firstly we show the simplest example which has no core inside the inclusion B.
Next, suppose U ⊂⊂ B is a inner core inside the inclusion B, which has vanishing boundary conditions. In this case, the conductivity problem can be formulated as stands for Dirichlet or Neumann boundary conditions. Under this circumstance, we also have the uniqueness result on recovering both the core U and the contrast k.
Theorem 2.2. Let u j be the solution to (2.5) with conductivity k and inner core U j , j = 1, 2, respectively. Let U c 12 be the unbounded connected component of stands for the Neumann condition, then suppose that 2.2. Two-layer medium. In this subsection, we suppose that σ satisfies where B and D are simply connected inclusions which have C 1+α , 0 < α < 1, smooth boundary, denoted by ∂B and ∂D, respectively. Suppose D ⊂⊂ B. Then (1.1) is equivalent to It can be seen that the solution to (2.7) can be represented by (see [4]) By using the uniqueness of harmonic function in D with the same boundary condition on ∂D, the solution to (2.7) can then be represented by where φ 1 , φ 2 and φ 3 satisfy The transmission condition further shows that σ (1) ν · ∇u| − = ν · ∇u| + , on ∂B.
We mention that the assumption of matrix M defined in (2.14) is indeed restriction of measurement u on ∂Ω, or essentially the restriction on the background potential h. To ensure the uniqueness of recovering the two-layer piecewise constant conductivity from one measurement, some restriction on h is quite necessary. In fact, if h is chosen specially, then there is possibility that two different piecewise constant conductivities can generate the same boundary data. One important application of such phenomenon is for enhancement of near cloaking effect (see [5] for details).
3. Proof of the main results. In this section, we shall prove the main theorems stated in the last section.
Proof of Theorem 2.1. Suppose k 1 = k 2 . Firstly, from unique continuation of harmonic function, it is easy to see that u 1 = u 2 in R 3 \ B. Thus there holds In what follows, we define g := ∂u1 ∂ν + and f := ∂h ∂ν on ∂B. By using transmission condition on ∂B, there holds By using (2.2) and the trace formula (1.6) one can find that where j = 1, 2. Then by using (3.2) and (3.3) there holds By straightforward computation one has and thus by assumption k 1 = k 2 again one has g = f . By definition of g and f , one immediately has (3.6) ( Recall that λI + K * B is invertible on L 2 (∂B) for any |λ| > 1/2 and λ = 1/2, thus from (3.6) one has ∂h ∂ν = f = 0. Due to the fact that h is harmonic function in R 3 one has h = c and so u 1 = u 2 = 0, which is a contradiction. Thus k 1 = k 2 .
The proof is complete.
Proof of Theorem 2.2. Without loss of generality we suppose that U * := U 12 \ U 1 and U * is nonempty. We mention that the method is similar to the obstacle recovery result in [21]. We first consider the Neumann condition case, i.e., B[u] = ∂u ∂ν in (2.5). Since k 1 = k 2 in B \ U 12 and u 1 = u 2 on ∂B, by unique continuation theorem one can find that u 1 = u 2 and ∂u1 ∂ν = ∂u2 ∂ν on ∂U 12 . Together with the conditions that ∂u1 ∂ν = 0 on ∂U * ∩ ∂U 1 and ∂u2 ∂ν = 0 on ∂U * ∩ ∂U 2 , one can see that u 1 satisfies From (3.7) one immediately has that u 1 = C in U * . By using unique continuation theorem one thus has u 1 = C in R 3 \ U 12 , which is a contradiction. Thus U 1 = U 2 . For Dirichlet boundary condition, one can follow exactly the same strategy one can prove that u 1 is harmonic in U * and has vanishing boundary condition on ∂U * , by using min-max value principle, where one can avoid the smoothness restriction on U 12 , one can thus show that u 1 = 0 in R 3 \ U 12 , which makes a contradiction and thus U 1 = U 2 .
The proof is complete.
Proof of Theorem 2.3. Since u 1 = u 2 on ∂Ω, by using unique continuation one can easily find that u 1 = u 2 in R 3 \ B. Define f := u 1 | ∂B = u 2 | ∂B and g := ν · ∇u 1 | + = ν · ∇u 2 | + on ∂B. Then u 1 can be represented by (2.8). Let H = L 2 (∂B) × L 2 (∂D) and the Neumann-Poincaré-type operator K * : H → H by It is shown in [2] that the L 2 -adjoint of K * , K, is given by Then from the first two equations in (2.9), (2.10) and (2.11) one can get (3.10) 1g ) T , whereg := ν · ∇u 1 | − on ∂D and h := (−ν · ∇h| ∂B , ν · ∇h| ∂D ) T . I is the identity operator on H. The superscript T means transpose of a vector. Define then there holds the Calderón's identity SK * = KS (see [2]). By applying S on both sides of (3.10) and using the Calderón's identity, one thus has (3.12) Similarly, by applying S D on both sides of the third equation in (2.9) one has By definition of f and (2.10), one additionally has (3.14) By using Green's formula and transmission condition there holds for x ∈ ∂D that Similarly, for x ∈ ∂B, one has Thus (3.12) and (3.13) can be rewritten by Next, suppose ϕ ∈ L 2 0 (∂D) and define ξ 1 ∈ L 2 (∂D) by Let w 1 be a harmonic function in B which satisfies w 1 | ∂D = ξ 1 . Then by straightforward calculations, one has (3.20) where ζ 1 and l 1 are given by , on ∂D, (3.21) By taking inner product of both sides of (3.18) with (w| ∂B , ϕ) and using (3.20), and some straightforward calculations one then has By using Definition 2.1 one can further decompose w 1 and l 1 by 1 w (1) + w (2) , and respectively. Thus (3.22) can be rewritten by Similarly, by considering u 2 in the same way one has where (3.26) By using the assumption that M , defined in (2.14) and (2.15), is invertible, one thus has t 1 = t 2 = t 3 = 0 and so σ 1 = σ 2 .
The proof is complete.

4.
Remarks and future works. In this paper, we consider the inverse medium problem of recovering the unknown piecewise constant conductivity and inner core. We have presented the uniqueness for such recovery. We want to mention that the uniqueness does not just work for two-layer medium case, but also work for piecewise constant case. This will be our forthcoming work. We also stress that the uniqueness can also be, in some intuitive way, used to numerically recover the contrast and the inner core. In fact, from (3.3) we have If B is a disk or a ball, then it is quite easy to recover k by using spectral expansion of K * B . Otherwise, supposeh ∈ H 1/2 (∂B) such that ∂B ∂h ∂νh = 0, and ∂B ∂u ∂ν +h = 0.
Then by multiplying both sides of (4.1) withh and integrating on ∂B, there holds ∂B ∂u ∂ν +h , and k can be recovered. For recovery of the inner core, the numerical method can be applied by using the same scheme as in [21].