Inverse scattering by an inhomogeneous penetrable obstacle in a piecewise homogeneous medium
###### Abstract

This paper is concerned with the inverse problem of scattering of time-harmonic acoustic waves by an inhomogeneous penetrable obstacle in a piecewise homogeneous medium. The well-posedness of the direct problem is first established by using the integral equation method. We then proceed to establish two tools that play an important role for the inverse problem: one is a mixed reciprocity relation and the other is a priori estimates of the solution on some part of the interfaces between the layered media. For the inverse problem, we prove in this paper that both the penetrable interfaces and the possible inside inhomogeneity can be uniquely determined from a knowledge of the far field pattern for incident plane waves.

Keywords: Uniqueness, piecewise homogeneous medium, penetrable obstacle, unique continuation principle, Holmgren’s uniqueness theorem, inverse scattering.

## 1 Introduction

In this paper, we consider the scattering of time-harmonic acoustic plane waves by an inhomogeneous, penetrable obstacle in a piecewise homogeneous surrounding medium. In many practical applications, the background medium might not be homogeneous and then may be modeled as a layered medium. We might think of a problem in medical imaging where we have a damaged tissue (which can be modeled as a penetrable obstacle) inside the human body. It is clear that the human body is not a homogeneous structure and may be modeled as a piecewise homogeneous medium. Therefore, one possible model would then be a penetrable obstacle buried in a piecewise homogeneous medium.

For simplicity, and without loss of generality, in this paper we restrict ourself to the case where the obstacle is buried in a two-layered piecewise homogeneous medium, as shown in Figure 1. Precisely, let be an open bounded region with a boundary such that the background is divided by means of a closed surface into two connected domains and . Here, is a unbounded homogeneous medium, is a bounded homogeneous medium and is a penetrable obstacle. Let denote the complement of , that is, . Choose a large ball centered at the origin such that and let .

The problem of scattering of time-harmonic acoustic waves in a two-layered background medium in can be modeled by

 Δu+k20u=0 inΩ0, (1.1) Δv+k21v=0 inΩ1, (1.2) Δw+k22nw=0 inΩ2, (1.3) u−v=0,∂u∂ν−λ0∂v∂ν=0 onS0, (1.4) v−w=0,∂v∂ν−λ1∂w∂ν=0 onS1, (1.5) limr→∞r(∂us∂r−ik0us)=0, r=|x|, (1.6)

where is the unit outward normal to the interface and , is the refractive index of an inhomogeneous medium with and . Here, the total field is given as the sum of the unknown scattered wave which is required to satisfy the Sommerfeld radiation condition (1.6) and the incident plane wave , is the positive wave number given by in terms of the frequency and the sound speed in the corresponding medium . The distinct wave numbers correspond to the fact that the medium consists of several physically different materials. On the interfaces and , the so-called ”transmission conditions” (1.4) and (1.5) with two positive constants and are imposed, respectively, which represent the continuity of the medium and equilibrium of the forces acting on them.

The direct problem is to look for a set of functions satisfying (1.1)-(1.6). We will establish the well-posedness of the direct problem, employing the integral equation method in the next section. We will also establish a mixed reciprocity relation and prove a priori estimates of the solution on some part of the interfaces (). These results will play an important role in the proof of the uniqueness results in the inverse problem.

It is well known that has the following asymptotic representation

 us(x,d)=eik0|x||x|{u∞(ˆx,d)+O(1|x|)}as |x|→∞ (1.7)

uniformly for all directions , where the function defined on the unit sphere is known as the far field pattern with and denoting, respectively, the observation direction and the incident direction.

The inverse problem we consider in this paper is, given the wave numbers (), the positive constants and the far field pattern for all incident plane waves with incident direction , to determine the interfaces () and the refractive index . Precisely, we will study the uniqueness issue and prove that the interfaces () and the inhomogeneity can be uniquely identified from a knowledge of the far field pattern. It should be remarked that the uniqueness issue for inverse problems is of theoretical interest and is required in order to proceed to efficient numerical methods of solutions.

For the unique determination of an inhomogeneity with compact support in a homogeneous background medium, we refer the reader to Hähner [5], Nachman [13], Novikov [14] and Ramm [15, 16]. We also refer the reader to the monographs of Colton and Kress [4] and Kirsch [7] and the habilitationsschrift of Hähner [6] for a comprehensive discussion. Kirsch and Päivärinta [9] gave the first uniqueness result for determining the interior inhomogeneity in the case of a known inhomogeneous background medium.

However, to the authors’ knowledge, there are few uniqueness results for the inverse obstacle scattering in a piecewise homogeneous medium. In the case when the obstacle is impenetrable, based on a generalization of the mixed reciprocity relation, Liu, Zhang and Hu [12] proved that if the interface is known then the obstacle and its physical property can be uniquely determined by the far field pattern. Later, motivated by Kirsch and Kress [8] and Kirsch and Päivärinta [9] for the uniqueness proof of determining a penetrable boundary, Liu and Zhang [11] extended this result to the case when the interface is unknown and proved that the interface can also be uniquely recovered. Note that all the results in [12, 11] are also available for the case of a multilayered medium and can be proved similarly. Thus, Liu and Zhang [11] have in fact proved a uniqueness result for the case when the obstacle is penetrable with a homogeneous interior. This result has also been proved by Athanasiadis, Ramm and Stratis [1] and Yan [19] using a different method.

In this paper, we consider the case where the obstacle is penetrable with an inhomogeneous interior and use the ideas in [8, 9, 11] to prove the unique determination of the interfaces () in Section 3. In Section 4, we will show that the refractive index is also uniquely determined. To do this, we first establish a completeness result, that is, the normal derivatives , corresponding to incident plane waves with all directions , are complete in . Based on this result, we then establish an orthogonality relation that enables us to prove the unique determination of the inhomogeneity with help of the ideas from the fundamental work of Sylvester and Uhlmann [17].

## 2 The direct scattering problem

In this section we will establish the well-posedness of the direct problem via the integral equation method. We also prove a mixed reciprocity relation and a priori estimates on some part of the interfaces of the solution with the help of its explicit representation in a combination of layer and volume potentials. We assume hereafter that , , , , are given positive numbers and that is not a Neumann eigenvalue of in . In this paper, we shall use to denote generic constants whose values may change in different inequalities but always bounded away from infinity.

###### Remark 2.1.

The assumption on holds provided the refractive index satisfies one of the following two conditions:

(1) on some non-empty open subset of ;

(2) is compactly supported in .

Both cases can be proved by the unique continuation principle.

As incident fields , plane wave and point source , (cf. (2.9) below) are of special interest. Denote by the scattered field for an incident plane wave with incident direction and by the corresponding far field pattern. The scattered field for an incident point source with source point is denoted by and the corresponding far field pattern by .

The direct problem is to look for a set of functions , and satisfying the following boundary value problem

 Δu+k20u=0 in Ω0, (2.1) Δv+k21v=0 in Ω1, (2.2) Δw+k22nw=0 in Ω2, (2.3) u−v=f,∂u∂ν−λ0∂v∂ν=g onS0, (2.4) v−w=p,∂v∂ν−λ1∂w∂ν=q onS1, (2.5) limr→∞r(∂u∂r−ik0u)=0 r=|x| (2.6)

where , , and are given functions from Hölder spaces with exponent .

###### Remark 2.2.

For the scattering problem, if the incident field is the plane wave or the point source with then , and if the incident field is the point source with then .

###### Theorem 2.3.

The boundary value problem admits at most one solution.

###### Proof.

Clearly, it is enough to show that in , in and in for the corresponding homogeneous problem, that is, on and on . Applying Green’s first theorem over , we obtain that

 ∫∂BRu∂¯¯¯u∂νds=∫ΩR(uΔ¯¯¯u+|∇u|2)dx+∫S0u∂¯¯¯u∂νds.

Using Green’s first theorem over and and taking into account the transmission conditions (2.4) and (2.5), we have

 ∫∂BRu∂¯¯¯u∂νds = ∫ΩR(uΔ¯¯¯u+|∇u|2)dx+λ0∫Ω1(vΔ¯¯¯v+|∇v|2)dx (2.8) +λ0λ1∫Ω2(wΔ¯¯¯¯w+|∇w|2)dx.

Making use of equations (2.1)-(2.3) and taking the imaginary part of (2.8) we obtain, on noting that are positive numbers and , that

 I∫∂BRu∂¯¯¯u∂νds=−k22λ0λ1I∫Ω2¯¯¯n|w|2dx≥0.

Thus, by Rellich’s Lemma [4], it follows that in . By the unique continuation principle, we have in . Holmgren’s uniqueness theorem [10] and the homogeneous transmission boundary conditions (2.4) imply that in . Finally, the transmission boundary conditions (2.5) and the assumption on give that in , which completes the proof of the theorem. ∎

For the proof of existence of solutions we need the fundamental solution of the Helmholtz equation with wave number given by

 Φj(x,y)=eikj|x−y|4π|x−y|,x,y∈R3,x≠y. (2.9)

For and , define the single- and double-layer potentials and , respectively, by

 (˜Si,jϕ)(x):=∫SiΦj(x,y)ϕ(y)ds(y), x∈R3∖Si, (˜Ki,jϕ)(x):=∫Si∂Φj(x,y)∂ν(y)ϕ(y)ds(y), x∈R3∖Si

and the normal derivative operators and by

 (˜K′i,jϕ)(x):=∂∂ν(x)∫SiΦj(x,y)ϕ(y)ds(y), x∈R3∖Si, (˜Ti,jϕ)(x):=∂∂ν(x)∫Si∂Φj(x,y)∂ν(y)ϕ(y)ds(y), x∈R3∖Si.

The restrictions on of these operators will be denoted by , , and (), respectively. To prove the existence of solutions, we also need the volume potential

 (Vϕ)(x):=k22∫Ω2Φ2(x,y)[n(y)−1]ϕ(y)dy, x∈R3

and its normal derivative operator denoted by . For mapping properties of these operators in the classical spaces of continuous and Hölder continuous functions we refer the reader to the monographs of Colton and Kress [3, 4].

###### Theorem 2.4.

The boundary value problem has a unique solution. Further, the solution satisfies the estimate

 ∥u∥1,α,¯¯¯ΩR+∥v∥1,α,¯¯¯Ω1+∥w∥1,α,¯¯¯Ω2≤C(∥f∥1,α,S0+∥g∥0,α,S0+∥p∥1,α,S1+∥q∥0,α,S1) (2.10)

for some positive constant .

###### Proof.

Following [3] and [4] we seek the unique solution in the form

 u = λ0˜K0,0ψ0+˜S0,0ϕ0inΩ0, (2.11) v = ˜K0,1ψ0+˜S0,1ϕ0+λ1˜K1,1ψ1+˜S1,1ϕ1inΩ1, (2.12) w = (2.13)

with four densities , , , . Then from the jump relations we see that the potentials and given by solve the boundary value problem provided the densities satisfy the system of integral equations:

 ψ0+λ(λ0K0,0−K0,1)ψ0+λ(S0,0−S0,1)ϕ0−λλ1˜K1,1ψ1−λ˜S1,1ϕ1=λfonS0, (2.14) ϕ0+λλ0(T0,1−T0,0)ψ0+λ(λ0K′0,1−K′0,0)ϕ0 +λλ0λ1˜T1,1ψ1+λλ0˜K′1,1ϕ1=−λgonS0, (2.15) ψ1+μ˜K0,1ψ0+μ˜S0,1ϕ0 +μ(λ1K1,1−K1,2)ψ1+μ(S1,1−S1,2)ϕ1−μVw=μponS1, (2.16) ϕ1−μ˜T0,1ψ0−μ˜K′0,1ϕ0 +μλ1(T1,2−T1,1)ψ1+μ(λ1K′1,2−K′1,1)ϕ1+μλ1V′w=−μqonS1, (2.17)

where and .

Define the product space

 X:=C1,α(¯¯¯¯¯¯¯ΩR)×C1,α(¯¯¯¯¯¯Ω1)×C1,α(¯¯¯¯¯¯Ω2)×C1,α(S0)×C0,α(S0)×C1,α(S1)×C0,α(S1)

and introduce the operator given in the matrix form:

 ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝000λ0˜K0,0˜S0,000000˜K0,1˜S0,1λ1˜K1,1˜S1,100−V00˜K1,2˜S1,2000λ(λ0K0,0−K0,1)λ(S0,0−S0,1)−λλ1˜K1,1−λ˜S1,1000λλ0(T0,1−T0,0)λ(λ0K′0,1−K′0,0)λλ0λ1˜T1,1λλ0˜K′1,100−μVμ˜K0,1μ˜S0,1μ(λ1K1,1−K1,2)μ(S1,1−S1,2)00μλ1V′−μ˜T0,1−μ˜K′0,1μλ1(T1,2−T1,1)μ(λ1K′1,2−K′1,1)⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

which is compact since all its entries are compact. Then the system can be rewritten in the abbreviated form

 (I+A)U=R, (2.18)

where , and is the identity operator. Thus, the Riesz-Fredholm theory is applicable to establish the existence of solutions to the system (2.18) if we can prove the uniqueness of solutions to the system. To this end, let be a solution of the homogeneous system corresponding to (2.18) (that is, the system (2.18) with ). Then it is enough to show that .

We first prove that on . From the system (2.18) or (2.11)-(2.17) with (since ) it is known that defined in (2.11)-(2.13) satisfy the problem with . Thus, by the uniqueness Theorem 2.3, in , in and in .

Now define

 ˜v := ˜K0,1ψ0+˜S0,1ϕ0+λ1˜K1,1ψ1+˜S1,1ϕ1inΩ0, ˜u := −˜K0,0ψ0−1λ0˜S0,0ϕ0inΩ.

Then by the jump relations for single- and double-layer potentials we have

 ˜v−v=ψ0, (2.19) ∂˜v∂ν−∂v∂ν=−ϕ0, ∂u∂ν+λ0∂˜u∂ν=−ϕ0onS0. (2.20)

Thus, and solve the homogeneous transmission problem

 Δ˜v+k20˜v=0inΩ0,Δ˜u+k21˜u=0inΩ

with the transmission conditions

 ˜v−˜u=0,∂˜v∂ν=λ0∂˜u∂ν% onS0.

It is clear that satisfies the radiation condition (2.6). By [3, Lemma 3.40], in and in . Consequently, from the boundary conditions (2.19) and (2.20) we conclude that on .

In a completely similar manner, we can also prove that on . Thus, we have established the injectivity of the operator and, by the Riesz-Fredholm theory, exists and is bounded in . The estimate (2.10) follows from (2.18). The proof is thus complete. ∎

We now establish the following mixed reciprocity relation which is needed for the inverse problem.

###### Lemma 2.5.

( Mixed reciprocity relation.) For the scattering of plane waves with and point-sources from the obstacle we have

 Φ∞(ˆx,z)=⎧⎪ ⎪⎨⎪ ⎪⎩14πus(z,−ˆx),z∈Ω0,ˆx∈S,λ04πv(z,−ˆx),z∈Ω1,ˆx∈S.
###### Proof.

By Green’s second theorem and the radiation condition (2.6) we get

 14π∫S0(us(y,z)∂us(y,−ˆx)∂ν(y)−us(y,−ˆx)∂us(y,z)∂ν(y))ds(y)=0 (2.21)

for and By [4, Theorem 2.5], we have the representation

 Φ∞(ˆx,z)=14π∫S0(us+(y;z)∂e−ik0ˆx⋅y∂ν(y)−e−ik0ˆx⋅y∂us+(y;z)∂ν(y))ds(y) (2.22)

for and Adding to gives

 Φ∞(ˆx,z)=14π∫S0(us(y,z)∂u(y,−ˆx)∂ν(y)−u(y,−ˆx)∂us(y,z)∂ν(y))ds(y)

for and

We first consider the case . Use the boundary condition on and Green’s second theorem to deduce that for ,

 Φ∞(ˆx,z) = 14π∫S0(us(y,z)∂u(y,−ˆx)∂ν(y)−u(y,−ˆx)∂us(y,z)∂ν(y))ds(y) (2.23) = λ014π∫S1((v(y,z)−Φ0(y,z))∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂(v(y,z)−Φ0(y,z))∂ν(y))ds(y) (2.26) +λ014π∫Ω1((k21−k20)Φ0(y,z)v(y,−ˆx))dy +(1−λ0)14π∫S0(v(y,−ˆx)∂Φ0(y,z)∂ν(y))ds(y).

Now Green’s second theorem gives that for and

 14π∫S0(ui(y,−ˆx)∂Φ0(z,y)∂ν(y)−Φ0(z,y)∂ui(y,−ˆx)∂ν(y))ds(y)=0. (2.27)

From Green’s formula (see Theorem 2.4 in [4]) it follows that

 14πus(z,−ˆx)=14π∫S0(us(y,−ˆx)∂Φ0(z,y)∂ν(y)−Φ0(z,y)∂us(y,−ˆx)∂ν(y))ds(y) (2.28)

for and Adding (2.27) to the equation (2.28) we deduce, with the help of the transmission condition on and Green’s second theorem, that for and

 14πus(z,−ˆx) = 14π∫S0(u(y,−ˆx)∂Φ0(z,y)∂ν(y)−Φ0(z,y)∂u(y,−ˆx)∂ν(y))ds(y) (2.29) = λ014π∫S1(v(y,−ˆx)∂Φ0(z,y)∂ν(y)−Φ0(z,y)∂v(y,−ˆx)∂ν(y))ds(y) (2.32) +λ014π∫Ω1((k21−k20)Φ0(y,z)v(y,−ˆx))dy +(1−λ0)14π∫S0(v(y,−ˆx)∂Φ0(y,z)∂ν(y))ds(y)

By (2.23) and (2.29) together with the transmission condition on and the equation (2.3) we deduce that

 Φ∞(ˆx,z)−14πus(z,−ˆx) = λ014π∫S1(v(y,z)∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂v(y,z)∂ν(y))ds(y) = λ0λ114π∫S1(w(y,z)∂w(y,−ˆx)∂ν(y)−w(y,−ˆx)∂w(y,z)∂ν(y))ds(y) = λ0λ114π∫Ω2(w(y,z)Δw(y,−ˆx)−w(y,−ˆx)Δw(y,z))ds(y)=0.

We now consider the case . Using the transmission condition on and , the equations and and Green’s second theorem, we obtain that

 Φ∞(ˆx,z) = 14π∫S0(us(y,z)∂u(y,−ˆx)∂ν(y)−u(y,−ˆx)∂us(y,z)∂ν(y))ds(y) = λ04π∫S0((v(y,z)+Φ1(y,z))∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂(v(y,z)+Φ1(y,z))∂ν(y))ds(y) = λ04π∫S0((v(y,z)+Φ1(y,z))∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂(v(y,z)+Φ1(y,z))∂ν(y))ds(y) = λ04π∫S0(Φ1(y,z)∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂Φ1(y,z)∂ν(y))ds(y) +λ04π∫S1(v(y,z)∂v(y,−ˆx)∂ν(y)−v(y,−ˆx)∂v(y,z)∂ν(y))ds(y) = λ04π∫S0(Φ1(y,