LOCATING ANISOTROPIES IN ELECTRICAL IMPEDANCE TOMOGRAPHY

ERKKI SOMERSALO

Abstract. In this article, we consider the problem of finding the support of an inhomogenous possibly anisotropic inclusion in a background of constant electric conductivity from the electrical impedance tomography data at the boundary of a bounded body. The article discusses the linear sampling method applied to this problem. A practical algorithm for solving this problem is suggested.

1. Introduction

In this article, the following electrical impedance tomography problem (EIT) is considered: On the surface of a body with unknown impedance distribution, one applies a set of prescribed electric currents and measures the corresponding voltages on the surface. From this information, one seeks to estimate the internal structure of the body. Potential application areas of the EIT range from medical imaging and monitoring to industrial process monitoring and nondestructive material testing. Often, the materials encountered in applications are anisotropic, i.e., the electromagnetic properties of the medium depend on the direction. It is well known that in general, the anisotropic EIT problem allows no unique solution. Hence, in the presence of anisotropies one has either to use a priori information complementary to that obtained from the measurements or one has to confine to more modest goals than recovering the full information of the material parameters. In this article, the latter approach is taken. More precisely, the following inverse problem is studied: Assume that in an isotropic body with otherwise known electric properties, there is an unknown possibly anisotropic inclusion. Given the EIT data on the surface of the body, estimate the support of the unknown inclusion. In recent years, a number of articles have been published, where the goal is to determine the shape of an inclusion based on either far-field or near field measurements with various probing modalities, see e.g. [1], [2], [4], [3], [7], [9], [10], [11], [6] and the references in these articles. The starting point of the present work is the linear sampling method originally introduced in the article [4]. The ideas here come close to those presented in [1] and [2].

This article was written during the author’s visit at Mathematical Sciences Research Institute (MSRI), Berkeley in August and September, 2001. Research at MSRI is supported in part by NSF grant DMS-9701755. The author wishes to thank Dr. Fioralba Cakoni, prof. David Colton and prof. Rainer Kress for useful discussions as well as MSRI for the hospitality and for bringing us together.

2. The inverse problem

Let B ⊂ ℝ^{n}, n = 2, 3, denote a bounded simply connected domain with a C^{2} smooth connected boundary. When low-frequecy time-harmonic electromagnetic field is induced in the body, within the quasi-static approximation of Maxwell’s equations the electric field is written in terms of the voltage potential u that satisfies

∇ · γ∇u = 0 in B.

Here, γ = γ(x) ∈ ℂ^{n×n} is the admittance distribution that in terms of the conductivity σ(x) ∈ ℝ^{n×n}, permittivity ε(x) ∈ ℝ^{n×n} and frequency ω > 0 is given as

γ = σ − iωε.

Assume that D ⊂ B is an open set that has a smooth boundary and consists of one or several simply connected components. Furthermore, assume that ∂D ∩ ∂B = ∅. We assume that the permittivity of the material outside D is so low that it can be neglected, i.e., γ ≈ σ in B \\overline{D}. Furthermore, we assume that the material is isotropic outisde D. To simplify the discussion, we assume that in fact, σ = 1 outside D. It is not hard to see that the results in this article can be generalized in a straightforward manner to cover the case where σ is a strictly positive scalar function outside D. These assumptions lead to the following model: By denoting by χ_{D} the characteristic function of the set D, the admittance γ is of the form

γ(x) = 1 + h(x)χ_{D}(x).

Furthermore, we assume that the perturbation h ∈ C^{1}(B, ℂ^{n×n}) is symmetric and satisfies in addition that for some positive constants α, β > 0 and for all ζ ∈ ℂ^{n},

(1)  Re(z\overline{ζ} · γ(x)ζ) ≥ α|ζ|^{2}, for all x ∈ B, for some z ∈ ℂ
(2)  Im(\overline{ζ} · h(x)ζ) ≤ −β|ζ|^{2}, x in an open set O ⊂ D

To describe the measurement, we fix the following notation. By H_{0}^{s}(∂B) (with the notation H_{0}^{0}(∂B) = L_{0}^{2}(∂B)) we denote the class of Sobolev functions of smoothness index s over the boundary with the restriction

τf = ⟨f, 1⟩ = 0,

the brackets denoting the natural pairing between H^{s}(∂B) and H^{−s}(∂B). since τ : H^{s}(∂D) → ℂ is continuous, H_{0}^{s}(∂B) is a closed subspace.

Assume that one applies an electric current f ∈ H_{0}^{−1/2}(∂B) on the surface of the body B. Then the voltage potential u satisfies

(3)  ∇ · γ∇u = 0
∂u|_{∂B} = f,

where we have denoted the normal derivative of u at the boundary as ∂u|_{∂B}. One can see, by a standard argument using the Lax-Milgram lemma and Fredholm theory that the problem (3) has a unique solution u ∈ H_{0}^{1}(B), where we have denoted

H_{0}^{1}(B) = {u ∈ H^{1}(B) | u|_{∂B} ∈ H_{0}^{1/2}(∂B)} .

For later reference, let us denote by T the continuous solution operator

T : H_{0}^{−1/2}(∂B) → H_{0}^{1}(B), f ↦ u.

Further, we define the Neumann-to-Dirichlet map Λ as

Λ : H_{0}^{−1/2}(∂B) → H_{0}^{1/2}(∂B), f ↦ u|_{∂B},

where u is the unique solution of the problem (3). We shall also consider the forward problem when no inclusion is present, i.e., the boundary value problem

(4)  Δv = 0
∂v|_{∂B} = f,

and the corresponding solution operator and Neumann-to-Dirichlet maps,

T_{0} : H_{0}^{−1/2}(∂B) → H_{0}^{1}(B), f ↦ v,
Λ_{0} : H_{0}^{−1/2}(∂B) → H_{0}^{1/2}(∂B), f ↦ v|_{∂B}.

The inverse problem studied in this article can be formulated as follows:

Problem 2.1. Given the Neumann-to-Dirichlet map Λ, determine the support D of the perturbation.

In the following section, we discuss the linear sampling method that gives a practical way of estimating the support.

3. The linear sampling method

The linear sampling method discussed in this article is based on the use of certain singular solutions. Therefore, let us define the singular solution B ∋ x ↦ Φ(x, y, α̂), where y ∈ B is a parameter and α̂ ∈ ℝ^{n} is a unit vector, as a solution of the following homogenous Neumann problem,

(5)  ΔΦ = α̂ · ∇δ(x − y)
∂Φ|_{∂B} = 0.

Physically, the singular solution corresponds to the electromagnetic potential created by a dipole source at y pointing in the direction α̂. In terms of the operator T_{0}, we may write the singular solution as

Φ(x, y, α̂) = α̂ · Ψ⃗(x − y) − T_{0}(α⃗ · ∂Ψ⃗(x − y)) − c_{y},

where

Ψ⃗(x − y) = −\frac{1}{2^{n−1}π} \frac{x − y}{|x − y|^{n}} , n = 2, 3,

and c_{y} is a constant equal to the integral of x ↦ α · Ψ⃗(x − y) over the boundary ∂B. We shall use the notation Φ_{y}(x) = Φ(x, y, α̂), suppressing the dependence on the direction as this plays little role in the discussion to ensue. The starting point of the linear sampling method lies in the following observation that we formulate as a lemma for later reference. In the following, we shall use the notation ϕ_{y}(x) = Φ_{y}(x)|_{∂B}.

Lemma 3.1. Assume that y ∈ B \ \overline{D}. Then ϕ_{y} ∉ Ran(Λ − Λ_{0}).

Proof. Assume on the contrary that ϕ_{y} ∈ Ran(Λ − Λ_{0}), and let ϕ_{y} = (Λ − Λ_{0})ψ_{y}, ψ_{y} ∈ H_{0}^{−1/2}(∂B). Let u_{y} = T ψ_{y} and v_{y} = T_{0}ψ_{y} denote the solutions of the problems (3) and (4) with the boundary data ψ_{y}, respectively. Set w_{y} = u_{y} − v_{y}. Now we observe that

∂w_{y} = 0, w_{y}|_{∂B} = u_{y} − v_{y} = (Λ − Λ_{0})ψ_{y} = ϕ_{y},

i.e., the Cauchy data of w_{y} and Φ_{y} coincide on ∂B. By Holmgren’s Uniqueness Theorem, we have w_{y} = Φ_{y} in B \ (\overline{D} ∪ \{y\}). The claim of the lemma follows now, since w_{y} has no singularity at y while Φ_{y} is singular. □ Unfortunately, the converse is not in general true: When y ∈ D, there is no guarantee that ϕ_{y} ∈ Ran(Λ − Λ_{0}). As in the case of inverse scattering problems, we have to confine to an approximate solution of the equation (Λ − Λ_{0})ψ_{y} = ϕ_{y}. To this end, we need to introduce some notations and definitions, and prove a number of auxiliary results. To get a handle of the following definition, assume for a while that ϕ_{y} = (Λ − Λ_{0})ψ_{y} for some ψ_{y} ∈ H_{0}^{−1/2}(∂B) and y ∈ D. By denoting again u_{y} = T ψ_{y} and v_{y} = T_{0}ψ_{y}, we observe as in the proof of the previous lemma that in B \ \overline{D}, u_{y} − v_{y} = Φ_{y}, and in D,

∇ · γ∇u_{y} = Δv_{y} = 0.

At the boundary ∂D, the solutions must satisfy

u_{y}|_{∂D}^{−} − v_{y}|_{∂D}^{−} = Φ_{y}|_{∂D},
∂_{γ}u_{y}|_{∂D}^{−} − ∂v|_{∂D}^{−} = ∂Φ_{y}|_{∂D}.

Above, the notation ∂_{γ}u_{y}|_{∂D}^{−} = n·γ∇u_{y}|_{∂D}^{−} for the conormal derivative was used, and the subscript ”-” indicates that the traces are from inside of ∂D. Hence, in order to investigate the equation (Λ − Λ_{0})ψ_{y} = ϕ_{y}, it is natural to study the following interior transmission problem of impedance tomography.

Problem 3.2. The interior transmission problem (ITP) of electrical impedance tomography with boundary data (f, g) ∈ H^{1/2}(∂D) × H^{−1/2}(∂D) is to find functions (u, w) ∈ (H^{1}(D) × H^{1}(D))/ℂ satisfying the equations

∇ · γ∇u = Δv = 0 in D,

with

u|_{∂D} − v|_{∂D} = f,
∂_{γ}u|_{∂D} − ∂|_{∂D}v = g.

Observe that above, the solution of the interior transmission problem can be unique only up to an additive constant. The space (H^{1}(D) × H^{1}(D))/ℂ consists of equivalence classes of the relation

(u, v) ∼ (u′, v′) if and only if (u, v) = (u′ − c, v′ − c), c ∈ ℂ.

It turns out that under the assumptions made about the admittance γ, the ITP has a unique solution. We formulate this as a lemma.

Lemma 3.3. Assume that the admissivity γ satisfies the conditions (1) and (2). Then the interior transmission problem 3.2 has a unique solution.

Proof: Tho show that there are at most one solution, assume that (u, v) satisfy the homogenous interior transmission problem, i.e., (f, g) = (0, 0). By integrating by parts we obtain

0 = ∫_{∂D}(\overline{v}∂v − v\overline{∂v})dS = ∫_{∂D}(\overline{u}∂_{γ}u − u\overline{∂_{γ}u})dS

= 2i ∫_{D} Im(\overline{∇u} · h∇u)dx.

By the assumption (2), we obtain that u is constant in an open set of D, and by the unique continuation property it is a constant in the whole of D. Thus v has the same Cauchy data with a constant solution on ∂D and so also v is constant in D, which completes the proof of uniqueness. To prove the existence, we refer to the article [2], where the interior transmission problem for the scattering case was studied. It is not hard to see, that the same argument goes through here as well. We leave the details out. □ Although the ITP is closely related to the existence of the solution of the equation (Λ − Λ_{0})ψ_{y} = ϕ_{y}, there is no equivalence, since we cannot in general extend the solutions (u, v) of the ITP from D to the whole domain B. Therefore, we consider only such solutions that have the extension property. This is the motivation for the following considerations. Let Ω ⊂ ℝ^{n} denote a ball that contains \overline{B} in its interior. Further, let G(x, y) denote Green’s function of the Laplacian,

G(x, y) = \{ \begin{matrix}\frac{1}{2π} log\frac{1}{|x − y|}, & n = 2, \\ \frac{1}{4π} \frac{1}{|x − y|}, & n = 3. \end{matrix}

We define the potential operator

S_{D} : L^{2}(∂Ω) → C^{∞}(D), ω ↦ ∫_{∂Ω} G(x, z)ω(z)dS(z), x ∈ D.

Further, we define a particular class of harmonic functions in D as

ℍ(D) = {u | u = S_{D}ω, ω ∈ L^{2}(∂Ω)}.

We also define the operators

K : L^{2}(∂Ω) → H^{1/2}(∂D), ω ↦ ∫_{∂Ω} G(x, z)ω(z)dS(z), x ∈ ∂D,

and, finally

L : L^{2}(∂Ω) → H_{0}^{−1/2}(∂B), ω ↦ ∂ ∫_{∂Ω} G(x, z)ω(z)dS(z), x ∈ ∂B.

Observe that by Green’s formula, the integral of Lω over ∂B automatically vanishes. The class ℍ(D) has the following approximation property of harmonic functions in D.

Theorem 3.4. For each ε > 0 and v ∈ H^{1}(D) satisfying Δv = 0 in the weak sense there is v^{ε} ∈ ℍ(D) such that ∥v − v^{ε}∥_{H^{1}(D)} < ε.

The proof of this theorem is based on the following density result.

Lemma 3.5. The operator K has dense range in both L^{2}(∂D) as in H^{1}(∂D).

Proof: The proof is quite similar to the corresponding one in the article [8]. To prove the denseness in L^{2}(∂D), assume that η ∈ L^{2}(∂D) is such that for all ω ∈ L^{2}(∂Ω),

(Kω, η)_{L^{2}(∂D)} = (ω, K^{∗}η)_{L^{2}(∂Ω)} = 0,

implying that

K^{∗}η(x) = ∫_{∂D} G(y, x)η(y)dS(y) = ∫_{∂D} G(x, y)η(y)dS(y) = 0

for x ∈ ∂Ω. Then, the function w defined as

w(x) = ∫_{∂D} G(x, y)η(y)dS(y), x ∈ ℝ^{n},

is harmonic both in ℝ^{n} \ \overline{D} and D. It has vanishing Dirichlet boundary data on ∂Ω and w → 0 as |x| → ∞, so w = 0 in ℝ^{n} \\overline{Ω}, and by the unique continuation principle, w = 0 in ℝ^{n} \ \overline{D}. By the continuity of the single layer potential, this implies also that w|_{∂D}^{−} = 0 and hence w = 0 in D. The conclusion η = 0 follows from the well known jump relation η = ∂w|_{∂D}^{−} − ∂w|_{∂D}^{+} of the normal derivatives.

To prove the denseness in H^{1}(∂D), we equip it with the inner product

(η, μ)_{H^{1}(∂D)} = ∫_{∂D} (\overline{η(x)}μ(x) + \overline{Gradη(x)} · Gradμ(x))dS(x),

where Grad is the surface gradient on ∂D. By denoting by K^{†} the adjoint of K as a mapping from L^{2}(∂Ω) to H^{1}(∂D), assume that we have

(Kω, η)_{H^{1}(∂D)} = (ω, K^{†}η)_{L^{2}(∂Ω)} = 0

for all ω ∈ L^{2}(∂Ω) or explicitly,

K^{†}η(x) = ∫_{∂D} (G(y, x)η(y) + Grad_{y}G(y, x) · Gradη(y))dS(y) = 0,

when x ∈ ∂Ω. To rewrite the second term in a more convenient form, assume, for a while that η is smooth. Then, by Gauss’ surface divergence theorem and by using G(x, y) = G(y, x), we obtain for x ∉ ∂D,

∫_{∂D} Grad_{y}G(y, x) · Gradη(y)dS(y) = − ∫_{∂D} G(y, x) · DivGradη(y)dS(y)

= ∫_{∂D} Grad_{y}G(x, y) · Gradη(y)dS(y),

and by extension, this holds for all η ∈ H^{1}(∂D). Furthermore, since ∇_{x}G(x, y) = −∇_{y}G(x, y), we obtain

∫_{∂D} Grad_{y}G(x, y) · Gradη(y)dS(y) = ∫_{∂D} ∇_{y}G(x, y) · Gradη(y)dS(y)

= −∇ ∫_{∂D} G(x, y) · Gradη(y)dS(y).

Hence, we define

w(x) = ∫_{∂D} G(x, y)η(y)dS(y) − ∇ · ∫_{∂D} G(x, y)Gradη(y)dS(y), x ∈ ℝ^{n} \ ∂D,

which is harmonic both inside and outside of D, and by the above considerations, w|_{∂Ω} = K^{†}η = 0. As above, we conclude that w = 0 in ℝ^{n} \ \overline{D}. On ∂D, the jump relations for vector potentials (see [5], Theorem 6.12),

w|_{∂D}^{±}(x) = ∫_{∂D} (G(x, y)η(y) − ∇_{x}G(x, y) · Gradη(y)))dS(y).

This expression is the adjoint of the single layer operator

S : L^{2}(∂D) → H^{1}(∂D), η ↦ ∫_{∂D} G(x, y)η(y)dS(y),

see e.g. [5], (pp. 43–44) or [8]. Hence, for all ψ ∈ L^{2}(∂D), we have

(Sψ, η)_{H^{1}(∂D)} = (ψ, w|_{∂D})_{L^{2}(∂D)} = 0.

By the uniqueness of the interior Dirichlet problem for the Laplacian, we now deduce that S is injective, so choosing ψ = S^{−1}η we obtain that η = 0. □ With the aid of the above lemma, we prove Theorem 3.4.

Proof of Theorem 3.4: Clearly, since H^{1}(∂D) is dense in H^{1/2}(∂D), the range of K is also dense in H^{1/2}(∂D). Let v ∈ H^{1}(D) be harmonic and ε > 0 be given. We choose first ω ∈ L^{2}(∂Ω) such that

∥Kω − v|_{∂D}∥_{H^{1/2}(∂D)} < δ

for some δ = δ(ε) > 0 to be defined later. Let

v^{ε} = S_{D}ω ∈ ℍ(D).

Then, by the continuity of the Dirichlet problem with respect to the boundary data, we have

∥v − v^{ε}∥_{H^{1}(D)} ≤ C∥Kω − v|_{∂D}∥_{H^{1/2}(∂D)} < Cδ,

and the claim follows by choosing δ = ε/C. □ It is clear that if we define the space ℍ(B) as functions of the form u = S_{B}ω using obvious notations, the harmonic functions in B can be approximated by ℍ(B)– functions. This observation gives us the following density result.

Lemma 3.6. The range of the operator L is dense in H_{0}^{−1/2}(∂B).

Proof: Let ε > 0 and ψ ∈ H_{0}^{−1/2}(∂B) be given. Define a harmonic function v in B as v = T_{0}ψ. By the observation above, we can find v^{ε} = S_{B}ω ∈ ℍ(B) with

∥v − v^{ε}∥_{H^{1}(B)} < δ,

where δ = δ(ε) is fixed later. But by the trace theorem,

∥∂v − ∂v^{ε}∥_{H^{−1/2}(∂B)} = ∥ψ − Lω∥_{H^{−1/2}(∂B)} ≤ C∥v − v^{ε}∥_{H^{1}(B)} < Cδ,

so by choosing δ = ε/C the desired approximation follows. □ The counterpart of Lemma 3.1 that we want to prove for y ∈ D is the following.

Lemma 3.7. Assume that y ∈ D. Then the interior transmission problem with the boundary data (f, g) = (Φ_{y}|_{∂D}, ∂Φ_{y}|_{∂D}) has a unique solution (u, v) with v ∈ ℍ(D) if and only if ϕ_{y} = (Λ − Λ_{0})Lω for some ω ∈ L^{2}(∂Ω).

Proof: Assume that (u, v) is the unique solution of the ITP with v = S_{D}ω. We can extend v to the whole B by setting v = S_{B}ω, and extend u to whole B by defining u = v + Φ_{y} in B \ \overline{D}. From the ITP boundary conditions, it follows now that u thus defined satisfies the equation ∇ · γ∇u = 0 in the weak sense in B, and at the boundary ∂B,

∂u|_{∂B} = ∂(v + Φ_{y})|_{∂B} = ∂v|_{∂B} = Lω.

Hence, we have

ψ_{y} = (u − v)|_{∂B} = (Λ − Λ_{0})Lω.

To prove the converse, let ψ_{y} = (Λ−Λ_{0})Lω for some ω ∈ L^{2}(∂Ω). We define u = T Lω and v = T_{0}Lω. As in the proof of Lemma 3.1, we see that u − v = Φ_{y} in B \ \overline{D}, and hence (u, v) satisfy the ITP with the boundary data (f, g) = (Φ_{y}|_{∂D}, ∂Φ_{y}|_{∂D}). On the other hand, the function

v_{0} = v − S_{B}ω

is harmonic in B and at the boundary,

∂v_{0} = ∂v − Lω = 0,

implying that v_{0}=constant. By the definition of T_{0}, we also see that the integral of v and thus v_{0} over the boundary vanishes, so v_{0} = 0 and the claim follows. □ The above lemma does not help us much since in general, the unique solution of the ITP is not such that v ∈ ℍ(D). However, we can always find an approximate solution, as the following theorem states.

Theorem 3.8. Assume that y ∈ D. Then for any ε > 0 the equation (Λ − Λ_{0})Lω = ϕ_{y} has an approximate solution in ω_{y}^{ε} ∈ L^{2}(∂Ω), i.e., ω_{y}^{ε} is the satisfies the estimate

∥(Λ − Λ_{0})Lω_{y}^{ε} − ϕ_{y}∥_{H^{1/2}(∂B)} < ε. (6)

Furthermore, when y approaches the boundary ∂D, ∥ω_{y}^{ε}∥_{L^{2}(∂Ω)} → ∞.

Proof: Let (u_{y}, v_{y}) be the solution of the interior transmission problem with the transmission data (f, g) = (Φ_{y}|_{∂D}, ∂Φ_{y}|_{∂D}). First, let v_{y}^{ε} ∈ ℍ(D) = S_{D}ω be an approximation of v_{y} such that

∥v_{y} − v_{y}^{ε}∥_{H^{1}(D)} < δ,

where δ = δ(ε) is fixed later. We extend v_{y}^{ε} to the whole of B as v = S_{B}ω. Having v_{y}^{ε} in B, we define u_{y}^{ε} in B as

u_{y}^{ε} = χ_{D}u_{y} + (1 − χ_{D})(Φ_{y} + v_{y}^{ε}).

We observe that on ∂B, we have (∂u_{y}^{ε} − ∂v_{y}^{ε})|_{∂B} = ∂Φ_{y}|_{∂B} = 0. Let us denote

∂u_{y}^{ε}|_{∂B} = ∂v_{y}^{ε}|_{∂B} = Lω_{y}^{ε}.

Further, let us denote w_{y}^{ε} = T Lω_{y}^{ε}. We show that when δ is small, w_{y}^{ε} and u_{y}^{ε} are close to each other. To this end, let us define

r_{y}^{ε} = w_{y}^{ε} − u_{y}^{ε}.

This residual satisfies the equations

Δr_{y}^{ε} = 0 in B \ \overline{D},
∇ · γ∇r_{y}^{ε} = 0 in D,

the boundary condition

∂r_{y}^{ε}|_{∂B} = 0,

as well as the transmission conditions

r_{y}^{ε}|_{∂D}^{+} − r_{y}^{ε}|_{∂D}^{−} = (v_{y} − v_{y}^{ε})|_{∂D},
∂r_{y}^{ε}|_{∂D}^{+} − ∂_{γ}r_{y}^{ε}|_{∂D}^{+} = (∂v_{y} − ∂v_{y}^{ε})|_{∂D}^{.}

By using Green’s formula and the trace theorem, it is not hard to see that the function r^{ε} satisfies the estimate

∥r_{y}^{ε}∥_{H^{1}(B)} ≤ C(∥v_{y} − v_{y}^{ε}∥_{H^{1/2}(∂D)} + ∥∂v_{y} − ∂v_{y}^{ε}∥_{H^{−1/2}(∂D)}
≤ C∥v_{y} − v_{y}^{ε}∥_{H^{1}(D)} ≤ Cδ,

and in particular,

∥r_{y}^{ε}∥_{H^{1/2}(∂B)} ≤ C∥r_{y}^{ε}∥_{H^{1}(B)} ≤ Cδ.

Now we have the estimate

∥Φ_{y} − (Λ − Λ_{0})Lω_{y}^{ε}∥_{H^{1/2}(∂B)} ≤ ∥Φ_{y} − (w_{y}^{ε} − v_{y}^{ε})∥_{H^{1/2}(∂B)}
≤ ∥r_{y}^{ε}∥_{H^{1/2}(∂B)} ≤ Cδ,

so by choosing δ = ε/C the claim follows. To prove the second claim of the theorem, assume that y ∈ D, and let (v_{y}, w_{y}) ∈ (H^{1}(D) × H^{1}(D))/ℂ be the solution of the interior transmission problem with the transmission data (f, g) = (Φ_{y}|_{∂D}, ∂Φ_{y}|_{∂D}). We show first that as y approaches the boundary ∂D, then ∥v_{y}∥_{H^{1}(D)} → ∞. to show this, assume first the contrary, sup_{y∈D} ∥v_{y}∥_{H^{1}(D)} < ∞. In particular, it follows that ∥v_{y}∥_{H^{1/2}(∂D)} ≤ C and ∥∂v_{y}∥_{H^{−1/2}(∂D)} ≤ C. We define in B the function W_{y} as

W_{y} = χ_{D}w_{y} + (1 − χ_{D})Φ_{y},

satisfying the equations

∇ · γ∇W_{y} = 0 in D,
ΔW_{y} = 0 in B \ \overline{D}

with the transmission data

W_{y}|_{∂D}^{+} − W_{y}|_{∂D}^{−} = −v|_{∂D}^{−},
∂W_{y}|_{∂D}^{+} − ∂_{γ}W_{y}|_{∂D}^{−} = −∂v|_{∂D}^{−},

and the boundary condition

∂W_{y}|_{∂B} = 0.

An application of Green’s formula leads now to the conlusion that

∥W_{y}∥_{H^{1}(B)} ≤ C(∥v_{y}∥_{H^{1/2}(∂D)} + ∥∂v_{y}∥_{H^{−1/2}(∂D)}) ≤ C

for all y ∈ D. In particular, we see that

sup_{y∈D} ∥Φ_{y}∥_{H^{1}(B\D)} < ∞,

which is a contradiction. Thus, we must have ∥v_{y}∥_{H^{1}(D)} → ∞ as claimed. Let v_{y}^{ε} ∈ ℍ(D) be an approximation of v_{y} in H^{1}(D). It follows now that also ∥v_{y}^{ε}∥_{H^{1}(D)} → ∞. By the construction, ∥Lω_{y}^{ε}∥_{H^{−1/2}(∂B)} → ∞ whis is possible only if ∥ω_{y}^{ε}∥_{L^{2}(∂Ω)} → ∞ as y appraches the boundary ∂D. The proof is complete. □

Finally, let us briefly discuss the case when we try to find the an approximate solution when y ∉ D. As it is customary in the linear sampling approach, we consider the Tikhonov regularized approximation of the solution to the equation (Λ−Λ_{0})ψ_{y} = ϕ_{y}. To this end, we need the following result.

Lemma 3.9. Under the assumptions about γ made in Section 2, the operator Λ−Λ_{0} : H_{0}^{−1/2}(∂B) → H^{1/2}(∂B) is injective and has a dense range.

Proof: To show the injectivity, assume that (Λ − Λ_{0})ψ = 0. Set, as usual, u = T ψ and v = T_{0}ψ, yielding that w = u−v is harmonic in B \\overline{D} and has vanishing Cauchy data on ∂B. Hence, u = v in B \ \overline{D}. It follows then that the pair (u|_{D}, v|_{D}) is a solution of the interior transmission problem 3.2 with vanishing boundary data, and so Lemma 3.3 implies that u = v = 0 in D and consequently in the whole of B. Hence, we deduce that also ψ = 0. To prove the density, assume the contrary. Then there is an element 0 ≠ η ∈ H_{0}^{1/2}(∂B)^{∗} with

⟨(Λ − Λ_{0})ψ, η⟩ = 0

for all ψ ∈ H_{0}^{−1/2}(∂B). Since ⟨(Λ − Λ_{0})ψ, c⟩ = 0 for all constants c, we may assume that η ∈ H_{0}^{−1/2}(∂B). Further, since Λ−Λ_{0} is symmetric, we deduce that (Λ−Λ_{0})η = 0, and the injectivity implies η = 0. This contradction proves the claim. □ The above lemma guarantees that we may apply the standard theory of minimumnorm solutions. In particular (see e.g. [5]), for every δ > 0 there is a unique ψ_{y}^{δ} ∈ H^{−1/2}(∂B) that minimizes the functional

F_{α}(ψ) = ∥(Λ − Λ_{0})ψ − ϕ_{y}∥_{H^{1/2}(∂B)}^{2} + α∥ψ∥_{H^{−1/2}(∂B)}^{2}, (7)

with the Morozov discrepancy constraint

∥(Λ − Λ_{0})ψ − ϕ_{y}∥_{H^{1/2}(∂B)} ≤ δ, (8)

used to fix the parameter α = α(δ). By Lemma 3.1, we observe that as δ → 0+, we must have ∥ψ∥_{H^{−1/2}(∂B)} → ∞. In terms of the regularization parameter, we have α → 0 as δ → 0. What is more, by Lemma 3.6, for every ε > 0 we can always find an ω_{y}^{δ,ε} ∈ L^{2}(∂Ω) such that

∥(Λ − Λ_{0})(ψ_{y}^{δ} − Lω_{y}^{δ,ε})∥_{H^{1/2}(∂B)} < ε.

We can summarize these results in the following theorem that is the counterpart of Theorem 3.8 when y ∉ D.

Theorem 3.10. Assume that y ∈ B \ \overline{D} Then for every δ > 0 and ε > 0 there is an ω_{y}^{δ,ε} ∈ L^{2}(∂Ω) such that

∥(Λ − Λ_{0})Lω_{y}^{δ,ε} − ϕ_{y}∥_{H^{1/2}(∂B)} < δ + ε,

for which ∥ω_{y}^{δ,ε}∥_{L^{2}(∂Ω)} → ∞ as δ → 0+.

By comparing Theorems 3.8 and 3.10 it is not obvious how the linear sampling algorithm should be implemented. In the articles [3] and [7] (see also the review article [6] for further references) the linear sampling method in inverse scattering has been studied numerically. Based on those works, one can suggest the following procedure. Given a ’noise level’ δ > 0, one seeks to minimize the functional (7) under the constraint (8), with the parameter y varying in a given grid inside B. The norm of the solution ψ or alternatively, the size of the regularization parameter α = α(δ) is used then as a cut-off indicator.

References

[1] Brühl, M. and Hanke, M.: Numerical implementation of two non-iterative methods locating inclusions in impedance tomography. Inverse Problems 16 (2000) 1029–1042.

[2] Cakoni, F., Colton, D. and Haddar, H.: The linear sampling method for anisotropic media. Preprint

[3] Colton, D., Piana, M. and Potthast, R.: A simple method using Morozov’s discrepancy principle for solving inverse scattering problems. Inverse Problems 13 (1997) 1477–1493.

[4] Colton, D. and Kirsch, A.: A simple method for solving inverse scattering problems in the resonance region. Inverse Problems 12 (1996) 383–393.

[5] Colton, D. and Kress, R.: Inverse Acoustic and Electromagnetic Scattering Theory. 2nd ed., Springer-Verlag, Berlin Heidelberg New York 1998.

[6] Colton, D., Coyle, J. and monk, P.: Recent development in inverse acoustic scattering theory. SIAM Review 42 (2000) 369–414.

[7] Colton, D., Giebermann, K. and Monk, P.: A regularized sampling method for solving threedimensional inverse scattering problems. SIAM J. Comput. 21 (2000) 2316–2330.

[8] Colton, D. and Kress, R.: On the Denseness of Herglotz Wave Functions and Electromagnetic Herglotz Pairs in sobolev Spaces. Preprint, 2001.

[9] Ikehata, M.: Enclosing a polygonal cavity in a two-dimensional bounded domain from Cauchy data. Inverse Problems 15 (1999), 1231–1241.

[10] Ikehata, M. and Siltanen, S.: Numerical method for finding convex hull of an inclusion in conductivity from boundary measurements. Inverse Problems 16 (2000) 1043–1052.

[11] Kirsch, A.: Factorization of the far field operator for the inhomogenous medium case and an application in inverse scattering theory. Inverse Problems 15 (1999) 413–429.

Erkki Somersalo, Helsinki University of Technology, Institute of Mathematics, P.O. Box 1100, FIN–02015 HUT, Finland

E-mail address: erkki.somersalo@hut.fi