A modification of the factorization method for scatterers with different physical properties

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1 A modification of the factorization method for scatterers with different physical properties Takashi FURUYA arxiv: v2 [math.ap] 25 Oct 2018 Abstract We study an inverse acoustic scattering problem by the Factorization Method when the unknown scatterer consists of two objects with different physical properties. Especially, we consider the following two cases: One is the case when each object has the different boundary condition, and the other one is when different penetrability. Our idea here is to modify the far field operator depending on the cases to avoid unnecessary a priori assumptions. 1 Introduction Sampling methods are proposed for reconstruction of shape and location in inverse acoustic scattering problems. In the last twenty years, sampling methods such as the Linear Sampling method of Colton and Kress [4], the Singular Sources Method of Potthast [16], the Factorization Method of Kirsch [5], have been introduced and intensively studied. As an advantage of these sampling methods, the numerical implementation are so simple and fast. However, as disadvantage of sampling methods except the Factorization Method, only sufficient conditions are given for the identification of unknown scatterers. To overcome this drawback, that is, to provide necessary and sufficient conditions, the Factorization Method was introduced and developed by a lot of researchers. However, for rigorous justification of the original Factorization Method, we have to assume that the wave number of the incident wave is not an eigenvalue of the Laplacian on an obstacle with respect to the boundary condition of the scattering problem. Kirsch and Liu [9] eliminated this problem for the case of a single obstacle by assuming that a small ball is in the interior of the unknown obstacle. They modified the original far field operator by adding the far field operator corresponding to a small ball so that the Factorization Method can be applied to it. On the other hands, in the case of a scatterer consisting of two objects with different physical properties, this problem has been still open. For recent works discussing this case, we refer to [1, 2, 6, 11, 17]. In this paper, we study the Factorization Method for a scatterer consisting of two objects with different physical properties. Especially, we consider 1

2 the following two cases: One is the case when each object has the different boundary condition, and the other one is when different penetrability. For recent works discussing such a scatterer, we refer to [8, 10, 13]. We remark that these works have to assume that the wave number of the incident wave is not an eigenvalue of the Laplacian on impenetrable obstacles included in a scatterer. Our aim of this paper is to eliminate this restriction by developing the idea of [9]. We begin with the formulations of the scattering problems. Let k > 0 be the wave number and for θ S 2 be incident direction. Here, S 2 = {x R 3 : x = 1} denotes the unit spherer in R 3. We set u i (x := e ikθ x, x R 3, (1.1 where i in the left hand side stands for incident plane wave. Let Ω R 3 be a boundedopen set and let its exterior R 3 \Ω beconnected. We assume that Ω consists of two bounded domains, i.e., Ω = Ω 1 Ω 2 such that Ω 1 Ω 2 =. We consider the following two cases. The first case. Ω 1 is an impenetrable obstacle with Dirichlet boundary condition, and Ω 2 with Neumann boundary condition. Find u s H 1 loc (R3 \Ω such that u s +k 2 u s = 0 in R 3 \Ω, (1.2 u s = u i on Ω 1, (1.3 u s = ui on Ω 2, (1.4 ν Ω2 ν Ω2 ( u s lim r r r ikus = 0, (1.5 where r = x, and (1.5 is the Sommerfeld radiation condition. Here, H 1 loc (R3 \ Ω = {u : R 3 \ Ω C : u B H 1 (B for all open balls B} denotes the local Sobolov space of one order. ν Ω2 (x denotes the unit normal vector at x Ω 2. We refer to Theorem 7.15 in [15] for the well posedness of the problem (1.2 (1.5, and refer to [8] and [13] for the factorization method in this case. The second case. Ω 1 is a penetrable medium modeled by a contrast function q L (Ω 1 (that is, Ω 1 = suppq, and Ω 2 is an impenetrable obstacle with Dirichlet boundary condition. Find u s H 1 loc (R3 \Ω 2 such that u s +k 2 (1+qu s = k 2 qu i in R 3 \Ω 2, (1.6 2

3 u s = u i on Ω 2, (1.7 ( u s lim r r r ikus = 0. (1.8 Note that weextendq byzero outsideω 1. Thewell posednessoftheproblem (1.6 (1.8 and its factorization method was shown in [10]. In both cases, it is well known that the scattered wave u s has the following asymptotic behavior: ( u s (x,θ = eik x 1 4π x u (ˆx,θ+O x 2, x, ˆx := x x. (1.9 The function u is called the far field pattern of u s. With the far field pattern u, we define the far field operator F : L 2 (S 2 L 2 (S 2 by Fg(ˆx := u (ˆx,θg(θds(θ, ˆx S 2. (1.10 S 2 We write the far field operator of the problem (1.2 (1.5 as F = FΩ Mix 1,Ω 2, and(1.6 (1.8as F = FΩ Mix 1 q,ω 2, respectively. Theinversescattering problem we consider is to reconstruct Ω from the far field pattern u (ˆx,θ for all ˆx,θ S 2. In other words, given the far field operator F, reconstruct Ω. Our contribution in this paper is, in both cases, to give the characterization of Ω 1 without a priori assumptions for the wave number k > 0. But we have to know the topological properties of Ω. More precisely, an inner domain B 1 of Ω 1 (based on [9], and an outer domain B 2 of Ω 2 ([8], have to be a priori known. Furthermore, we take an additional domain B 3 in the interior of B 2. By adding artificial far field operators corresponding to B 1, B 2, and B 3, we modify the original far field operator F. In the first case, we give the following characterization: Assumption 1.1. Let bounded domain B 1 and B 2 be a prior known. Assume that B 1 Ω 1, Ω 2 B 2, Ω 1 B 2 =. B 1 B 3 Ω 2 Ω 1 Neumann Dirichlet B 2 Figure 1: 3

4 Theorem 1.2. For ˆx S 2, z R 3, define φ z (ˆx := e ikz ˆx. (1.11 Let Assumption 1.1 hold. Take a positive number λ 0 > 0, and a bounded domain B 3 with B 3 B 2. (See Figure 1. Then, for z R 3 \B 2 z Ω 1 (φ z,ϕ n L 2 (S 2 2 <, (1.12 λ n n=1 where (λ n,ϕ n is a complete eigensystem of F # given by F # := ReF + ImF, (1.13 where F := FΩ Mix 1,Ω 2 + FB Dir 2 + F Imp B 1 B 3,iλ 0. Here, FB Dir 2 and F Imp B 1 B 3,iλ 0 are the far field operators for the pure Dirichlet boundary condition on B 2, and for the pure impedance boundary condition on B 1 B 3 with an impedance function iλ 0, respectively. Latter, we explain artificial far field operators FB Dir 2 and F Imp B 1 B 3,iλ 0 in Section 2, and prove Theorem 1.2 in Section 3. In the second case, we give the following characterization: Assumption 1.3. Let a bounded domain B 2 be a priori known. Assume the following assumptions: (i q L (Ω 1 with Imq 0 in Ω 1. (ii q is locally bounded below in Ω 1, i.e., for every compact subset M Ω 1, there exists c > 0 (depend on M such that q c in M. (iii Ω 2 B 2, Ω 1 B 2 =. (iv There exists t (π/2,3π/2 and C > 0 such that Re(e it q C q a.e. in Ω 1. Theorem 1.4. Let Assumption 1.3 hold. Take a positive number λ 0 > 0, and a bounded domain B 3 with B 3 B 2. (See Figure 2. Then, for z R 3 \B 2 (φ z,ϕ n L z Ω 1 2 (S 2 2 <, (1.14 λ n n=1 4

5 Ω 1 Ω 2 B 2 B 3 Obstacle Medium Figure 2: where (λ n,ϕ n is a complete eigensystem of F # given by F # := Re ( e it F + ImF, (1.15 where F := FΩ Mix 1 q,ω 2 + FB Dir 2 ( F Imp B 3,iλ 0. Here, the function φ z is given by We prove Theorem 1.4 in Section. We can also give the characterization by replacing (iv in Assumption 1.3 with (iv There exists t [0,π/2 (3π/2,2π] and C > 0 such that Re(e it q C q a.e. in Ω 1. For details, see Assumption 4.5 and Theorem 4.6. Let us compare our works (Theorems 1.2 and 1.4 with previous works from the mathematical point of view of a priori assumptions. For Theorem 1.2werefertoTheorem2.5of[13], andfortheorems1.4werefertotheorem 3.9 (b of [10]. These previous works also gave the characterization of Ω 1 by assuming the existence of outer domain B 2 of Ω 2 and that the wave number k 2 is not an eigenvalue on an obstacle, while, in our work we can choose arbitrary wave number k > 0 by introducing extra artificial domains such as B 1, B 2, and B 3, which are not so difficult topological assumptions. This paper is organized as follows. In Section 2, we recall a factorization of the far field operator and its properties. In Section 3 and Section 4, we prove Theorems 1.2 and 1.4, respectively. 2 A factorization for the far field operator In Section 2, we briefly recall a factorization for the far field operators and its properties. First, we consider a factorization of the far field operator for the pure boundary condition. Let B be a bounded open set and let R 3 \B be connected. Later, we will use the result of this section by regarding B as 5

6 auxiliary domains, like B 1, B 2, and B 3 in Theorems 1.2 and 1.4. We define G Dir B : H1/2 ( B L 2 (S 2 by G Dir B f := v, (2.1 where v is the far field pattern of a radiating solution v (that is, v satisfies the Sommerfeld radiation condition such that v +k 2 v = 0 in R 3 \B, (2.2 v = f on B. (2.3 Let λ 0 > 0. We also define G Imp B,iλ 0 : H 1/2 ( B L 2 (S 2 in the same way as G Dir B by replacing (2.3 with v ν B +iλ 0 v = f on B. (2.4 We define the boundary integral operators S B : H 1/2 ( B H 1/2 ( B and N B : H 1/2 ( B H 1/2 ( B by S B ϕ(x := ϕ(yφ(x, yds(y, x B, (2.5 N B ψ(x := B ν B (x B ψ(y Φ(x,y ds(y, x B, (2.6 ν B (y where Φ(x,y := eik x y 4π x y. We also define S B,i and N B,i by the boundary integral operators (2.5 and (2.6, respectively, corresponding to the wave number k = i. It is well known that S B,i is self-adjoint and positive coercive, and N B,i is self-adjoint and negative coercive. For details of the boundary integral operators, we refer to [7] and [15]. The following properties of far field operators FB Dir and F Imp B,iλ 0 are given by previous works in [7] and [9]: Lemma 2.1 (Lemma 1.14 in [7], Theorem 2.1 and Lemma 2.2 in [9]. (a The far field operators FB Dir and F Imp B,iλ 0 have a factorization of the form F Dir B = G Dir B S B GDir B, F Imp B,iλ 0 = G Imp B,iλ 0 T Imp B,iλ 0 G Imp B,iλ 0. (2.7 (b The operators S B : H 1/2 ( B H 1/2 ( B and T Imp B,iλ 0 : H 1/2 ( B H 1/2 ( B is of the form S B = S B,i +K, where K and K are some compact operators. 6 T Imp B,iλ 0 = N B,i +K, (2.8

7 (c Im ϕ,s B ϕ 0 for all ϕ H 1/2 ( B. Furthermore, if we assume that k 2 is not a Dirichlet eigenvalue of in B, then Im ϕ,s B ϕ < 0 for all ϕ H 1/2 ( B with ϕ 0. (d Im T Imp B,iλ 0 ϕ,ϕ > 0 for all ϕ H 1/2 ( B with ϕ 0. Secondly, we consider the far field operator FΩ Mix 1,Ω 2 for the problem (1.2 (1.5. Recall that Ω = Ω 1 Ω 2, and Ω 1 is an impenetrable obstacle with Dirichlet boundary condition, and Ω 2 with Neumann boundary condition. We define G Mix : H 1/2 ( Ω 1 H 1/2 ( Ω 2 L 2 (S 2 by ( G Mix f := v, (2.9 g where v is the far field pattern of a radiating solution v such that v +k 2 v = 0 in R 3 \Ω, (2.10 v = f on Ω 1, v ν Ω2 = g on Ω 2. (2.11 The following properties of F Mix are given by previous works in [7]: Lemma 2.2 (Theorem 3.4 in [7]. (a The far field operator F Mix has a factorization of the form F Mix = G Mix T Mix G Mix. (2.12 (b The middle operator TΩ Mix 1,Ω 2 : H 1/2 ( Ω 1 H 1/2 ( Ω 2 H 1/2 ( Ω 1 H 1/2 ( Ω 2 is of the form T Mix = ( SΩ1,i 0 where K is some compact operator. 0 N Ω2,i +K, (2.13 (c Im T Mix ϕ,ϕ 0 for all ϕ H 1/2 ( Ω 1 H 1/2 ( Ω 2. Thirdly, we consider the far field operator FΩ Mix 1 q,ω 2 for the problem (1.6 (1.8. Here, Ω 1 is a penetrable medium modeled by a contrast function q L (Ω 1, and Ω 2 is an impenetrable obstacle with Dirichlet boundary condition. We define G Mix : L 2 (Ω 1 H 1/2 ( Ω 2 L 2 (S 2 by ( G Mix f := v, (2.14 g 7

8 where v is the far field pattern of a radiating solution v such that v +k 2 (1+qv = k 2 q q f in R 3 \Ω 2, (2.15 v = g on Ω 2. (2.16 The following properties of F Mix are given by previous works in [10]: Lemma 2.3 (Theorem 3.2 and Theorem 3.3 in [10]. (a The far field operator F Mix has a factorization of the form F Mix = G Mix M Mix G Mix. (2.17 (b The middle operator MΩ Mix 1 q,ω 2 : L 2 (Ω 1 H 1/2 ( Ω 2 L 2 (Ω 1 H 1/2 ( Ω 2 is of the form M Mix = where K is some compact operator. ( q k 2 q 0 0 S Ω2,i +K, (2.18 (c Im ϕ,mω Mix 1 q,ω 2 ϕ 0 for all ϕ L 2 (Ω 1 H 1/2 ( Ω 2. ( (d If MΩ Mix ϕ1 1 q,ω 2 ϕ = 0, ϕ = L 2 (Ω 1 H 1/2 ( Ω 2, then ϕ 1 = 0. ϕ 2 Finally, we give the following functional analytic theorem behind the factorization method. The proof is completely analogous to previous works, e.g., Theorem 2.15 in [7], Theorem 2.1 in [12], and Theorem 2.1 in [13]. Theorem 2.4. Let X U X be a Gelfand triple with a Hilbert space U and a reflexive Banach space X such that the imbedding is dense. Furthermore, let Y be a second Hilbert space and let F : Y Y, G : X Y, T : X X be linear bounded operators such that We make the following assumptions: (1 G is compact with dense range in Y. F = GTG. (2.19 (2 There exists t [0,2π] such that Re(e it T has the form Re(e it T = C+ K with some compact operator K and some self-adjoint and positive coercive operator C, i.e., there exists c > 0 such that ϕ,cϕ c ϕ 2 for all ϕ X. (2.20 8

9 (3 Im ϕ,tϕ 0 or Im ϕ,tϕ 0 for all ϕ X. Furthermore, we assume that one of the following assumptions: (4 T is injective. (5 Im ϕ,tϕ > 0 or Im ϕ,tϕ < 0 for all ϕ Ran(G with ϕ 0. Then, the operator F # := Re(e it F + ImF is positive, and the ranges of G : X Y and F 1/2 # : Y Y coincide with each other. Remark that, in this paper, the real part and the imaginary part of an operator A are self-adjoint operators given by Re(A = A+A 2 and Im(A = A A. (2.21 2i 3 The first case In section 3, we prove Theorem 1.2. Let Assumption 1.1 hold. We define R 1 : H 1/2 ( Ω 1 H 1/2 ( Ω 2 H 1/2 ( Ω 1 H 1/2 ( B 2 by R 1 ( f1 g 1 := where v 1 is a radiating solution such that ( f1 v 1 B2, (3.1 v 1 +k 2 v 1 = 0 in R 3 \Ω, (3.2 v 1 = f 1 on Ω 1, Then, from the definition of R 1, we obtain v 1 ν Ω2 = g 1 on Ω 2. (3.3 G Mix = G Dir Ω 1,B 2 R 1, (3.4 where G Dir Ω 1,B 2 : H 1/2 ( Ω 1 H 1/2 ( B 2 L 2 (S 2 is also defined for the pure Dirichlet boundary condition on Ω 1 and B 2 in the same way as G Mix. (See (2.9. Next, we define R 2 : H 1/2 ( B 2 H 1/2 ( Ω 1 H 1/2 ( B 2 by R 2 f 2 := ( v2 Ω1 f 2 where v 2 is a radiating solution such that 9, (3.5

10 Then, from the definition of R 2, we obtain v 2 +k 2 v 2 = 0 in R 3 \B 2, (3.6 v 2 = f 2 on B 2, (3.7 G Dir B 2 = G Dir Ω 1,B 2 R 2. (3.8 Here, take a positive number λ 0 > 0, and a bounded domain B 3 with B 3 B 2. We define R 3 : H 1/2 ( B 1 B 3 H 1/2 ( Ω 1 H 1/2 ( B 2 by ( v3 Ω1 R 3 f 3 := v B2, (3.9 3 where v 3 is a radiating solution such that v 3 +k 2 v 3 = 0 in R 3 \B 1 B 3, (3.10 v 3 ν B1 B 3 +iλ 0 v 3 = f 3 on B 1 B 3. (3.11 Then, from the definition of R 3, we obtain G Imp B 1 B 3,iλ 0 = G Dir Ω 1,B 2 R 3. (3.12 By (3.4, (3.8, (3.12, and the factorization of the far field operator in Section 2, we have FΩ Mix 1,Ω 2 +FB Dir 2 +F Imp B 1 B 3,iλ 0 = G Dir Ω 1,B 2 TG Dir Ω 1,B 2, (3.13 [ ] where T := R 1 TΩ Mix 1,Ω 2 R1 R 2SB 2 R2 R 3T Imp B 1 B 3,iλ 0 R3. The following properties of G Dir Ω 1,B 2 are given by the same argument in Theorem 1.12 and Lemma 1.13 in [7]: Lemma 3.1. (a The operator G Dir Ω 1,B 2 : H 1/2 ( Ω 1 H 1/2 ( B 2 L 2 (S 2 is compact with dense range in L 2 (S 2. (b For z R 3 \B 2 z Ω 1 φ z Ran(G Dir Ω 1,B 2, (3.14 where the function φ z is given by (1.11. To prove Theorem 1.2, we apply Theorem 2.4 to this case. First of all, we show the following lemma: 10

11 ( I 0 Lemma 3.2. (a R 1 0 0, R 2 P 2, R 3 are compact. Here, P 2 : H 1/2 ( B 2 H 1/2 ( Ω 1 H 1/2 ( B 2 is defined by P 2 h := ( 0 h. (3.15 (b R3 is injective. ( I 0 Proof. (a The mappings R 1 : H 0 0 1/2 ( Ω 1 H 1/2 ( Ω 2 H 1 ( Ω 1 H 1 ( B 2, R 2 P 2 : H 1/2 ( B 2 H 1 ( Ω 1 H 1 ( B 2, and R 3 : H 1/2 ( B 1 B 3 H 1 ( Ω 1 H 1 ( B 2 are bounded since they are ( ( ( f1 0 given by v2 Ω1 ( v3 Ω1 g 1 v B2, f 2, and f v B2, 3 respectively. By Rellich theorem, they are compact. ( φ (bletφ H 1/2 ( Ω 1 andψ H 1/2 ( B 2. AssumethatR3 = ψ 0. Using the same argument as done in Theorem 2.5 in [14], one knows the existence of a radiating solution w such that w +k 2 w = 0 in R 3 \Ω 1 B 2, (3.16 w +k 2 w = 0 in Ω 1 \B 1, in B 2 \B 3, (3.17 w + w = 0, w + w = 0, w + ν Ω1 w ν Ω1 = φ on Ω 1, (3.18 w + ν B2 w ν B2 = ψ on B 2, (3.19 w ν B1 +iλ 0 w = 0 on B 1, w ν B3 +iλ 0 w = 0 on B 3, (3.20 wherethe subscripts+and denote the trace from theexterior and interior, respectively. (See Figure 3. 11

12 B 1 B 3 Ω 1 B 2 Figure 3: By using the boundary conditions (3.11, (3.18, (3.19, (3.20, and Green s theorem, we have ( 0 = f 3,R3 φ ( v Ω1 ( 3 = φ ψ v B2, 3 ψ = v 3 φds+ v 3 ψds Ω 1 B 2 ( w+ = v 3 Ω 1 B 2 ν w v 3 ds ν Ω 1 B 2 ν (w + w ds [ ] [ ] v3 = Ω 1 B 2 ν w w v3 v 3 ds ν Ω 1 B 2 ν w w + + v 3 ds ν [ ] [ ] v3 w v3 w = w v 3 ds+ w v 3 ds B 1 ν B1 ν B1 B 3 ν B3 ν B3 = f 3 wds, (3.21 B 1 B 3 which proves that w = 0 in B 1 B 3. Holmgren s uniqueness theorem (See e.g., Theorem 2.3 in [4] implies that w vanishes in Ω 1 \ B 1 and B 2 \ B 3. Equations (3.18 and (3.19 yield w + = 0 on Ω 1 B 2 which implies that w vanishes alsooutsideof Ω 1 andb 2 by theuniquenessof theexterior Dirichlet problem. Therefore, equations (3.18 and (3.19 yield φ = 0 and ψ = 0. By Lemma 3.2, the middle operator T of (3.13 has the following properties: Lemma 3.3. (a Re ( e iπ T has the form Re ( e iπ T = C+K with some selfadjoint and positive coercive operator C and some compact operator K. (b Im ϕ,tϕ < 0 for all ϕ H 1/2 ( Ω 1 H 1/2 ( B 2 with ϕ 0. 12

13 Proof. (a By Lemma 2.1 (b, Lemma 2.2 (b, and Lemma 3.2 (a, Re ( e iπ T ( = Re R 1 TΩ Mix 1,Ω 2 R1 +R 2 SB 2 R2 +R 3 T Imp B 1 B 3,iλ 0 R3 ( SΩ1,i 0 = R 1 R1 +R 0 N 2 S B2,iR2 +K Ω2,i ( ( ( I 0 SΩ1,i 0 I 0 = +P N Ω2,i S B2,iP2 +K ( SΩ1,i 0 = +K, ( S B2,i where K and K are some compact operators. Since the boundary integral operators S Ω1,i and S B2,i are self-adjoint and positive coercive, (a holds. (b By Lemma2.1 (c (d, Lemma2.2 (c, andlemma3.2(b, especially, by the strictly positivity of the operator ImT Imp B 1 B 3,iλ 0, and the injectivity of R 3, for all ϕ H 1/2 ( Ω 1 H 1/2 ( B 2 with ϕ 0, we have Im ϕ,tϕ = Im T Mix R 1ϕ,R 1ϕ +Im R 2ϕ,S B2 R 2ϕ Im T Imp B 1 B 3,iλ 0 R 3ϕ,R 3ϕ < 0. (3.23 Therefore, by Lemma 3.3, we can apply Theorem 2.4 to this case. From Lemma 3.1 (b, and applying Theorem 2.4, we obtain Theorem 1.2. Remark 3.4. Unknown obstacle Ω 2 may consist of finitely many connected components whose closures are mutually disjoint. Furthermore, the boundary condition on Ω 2 can notbeonly Neumannbutalso Dirichlet, impedance, and not only impenetrable obstacles but also penetrable mediums, and their mixed situations by the same argument in Theorem 1.2. In all cases, we can choose arbitrary wave numbers k > 0. Remark 3.5. If we assume that k 2 is not a Dirichlet eigenvalue of in artificial domains B 1, B 2, then we do not need to take an additional domain B 3. In such a case, we only use FB Dir 1 B 2 as artificial far field operators since FB Dir 1 B 2 has a role to keep the strictly positivity of the imaginary part of the middle operator of F. (See Lemma 2.1 (c. That is, we can give the following characterization by the same argument in Theorem 1.2: Theorem 3.6. In addition to Assumption 1.1, we assume that k 2 is not a Dirichlet eigenvalue of in B 1, B 2. Take a positive number λ 0 > 0. 13

14 Then, for z R 3 \B 2 z Ω 1 (φ z,ϕ n L 2 (S 2 2 <, (3.24 λ n n=1 where (λ n,ϕ n is a complete eigensystem of F # given by F # := ReF + ImF, (3.25 where F := F Mix +F Dir B 1 B 2. Here, the function φ z is given by (1.11. Remark 3.7. We can also give the characterization of the Neumann part Ω 2 if we assume Ω 1 B 1, B 2 Ω 2, B 1 Ω 2 = by the same argument in Theorem 1.2 (See Figure 4. B 1 B 2 Ω 1 Dirichlet Ω 2 Neumann Figure 4: 4 The second case In Section 4, we prove Theorem 1.4. Let Assumption 1.3 hold. We define G Mix Ω 1 0,B 2 : L 2 (Ω 1 H 1/2 ( B 2 L 2 (S 2 by ( G Mix f Ω 1 0,B 2 := v, (4.1 g where v is the far field pattern of a radiating solution v such that v +k 2 v = k 2 q q f in R 3 \B 2, (4.2 v = g on B 2. (4.3 Note that we extend q by zero outside Ω 1. Next, we define R 1 : L 2 (Ω 1 H 1/2 ( Ω 2 L 2 (Ω 1 H 1/2 ( B 2 by ( ( f1 f R 1 := 1 + q v 1 g 1 v B2, (

15 where v 1 is a radiating solution such that v 1 +k 2 (1+qv 1 = k 2 q q f 1 in R 3 \Ω 2, (4.5 Then, from the definition of R 1, we obtain v 1 = g 1 on Ω 2. (4.6 G Mix = G Mix Ω 1 0,B 2 R 1. (4.7 We define R 2 : H 1/2 ( B 2 L 2 (Ω 1 H 1/2 ( B 2 by ( 0 R 2 f 2 :=. (4.8 Then, from the definition of R 2, we obtain f 2 G Dir B 2 = G Mix Ω 1 0,B 2 R 2. (4.9 Here, take a positive number λ 0 > 0, and a bounded domain B 3 with B 3 B 2. We define R 3 : H 1/2 ( B 3 H 1/2 ( B 2 by where v 3 is a radiating solution such that R 3 f 3 := v 3 B2, (4.10 v 3 +k 2 v 3 = 0 in R 3 \B 3, (4.11 v 3 ν B3 +iλ 0 v 3 = f 3 on B 3. (4.12 Then, from the definition of R 3, and (4.9, we obtain G Imp B 3,iλ 0 = G Dir B 2 R 3 = G Mix Ω 1 0,B 2 R 2 R 3. (4.13 By (4.7, (4.9, (4.13, and the factorization of the far field operator in Section 2, we have where M := F Mix +F Dir B 2 +F Imp B 3,iλ 0 = G Mix Ω 1 0,B 2 MG Mix Ω 1 0,B 2, (4.14 [ R 1 M Mix R 1 R 2S B 2 R 2 R 2R 3 T Imp B 3,iλ 0 R 3 R 2 The following properties are given by the same argument in Theorem 3.2 (c in [10]: 15 ].

16 Lemma 4.1. (a The operator G Mix Ω 1 0,B 2 : L 2 (Ω 1 H 1/2 ( B 2 L 2 (S 2 is compact with dense range in L 2 (S 2. (b For z R 3 \B 2 z Ω 1 φ z Ran(G Mix Ω 1 0,B 2, (4.15 where the function φ z is given by (1.11. To prove Theorem 1.4, we apply Theorem 2.4 to this case with F = FΩ Mix 1 q,ω 2 +FB Dir 2 +F Imp B 3,iλ 0. First, we show the following lemma: ( I 0 Lemma 4.2. (a R 1, R are compact. (b R 1 is injective. (c R3 is injective. ( I 0 Proof. (a Themappings R 1 : L (Ω 1 H 1/2 ( Ω 2 H 1 (Ω 1 H 1 ( B 2, and R 3 : H 1/2 ( B 3 H 1 ( B 2 are bounded since they are ( ( f1 q v1 given by g 1 v B2, and f 3 v B2 3, respectively. By Rellich 1 theorem, they are compact. (b Assume that ( ( f1 f R 1 = 1 + q v 1 g 1 v B2 = 0. ( Equation (4.5 yields that v 1 +k 2 v 1 = 0 in R 3 \B 2, (4.17 v 1 = 0 on B 2. (4.18 By the uniqueness of the exterior Dirichlet problem, v 1 vanishes outside of B 2. Therefore, f 1 = 0. Furthermore, the analyticity of v 1 yields that v 1 also vanishes in B 2 \Ω 2, which implies that g 1 = 0. (c The injectivity of R3 follows from the same argument as done in the proof of Lemma 3.2 in [9]. By Lemma 4.2, the middle operator M of (4.14 has the following properties: 16

17 Lemma 4.3. (a Re ( e it M has the form Re ( e it M = C +K with some self-adjoint and positive coercive operator C, and some compact operator K. (b Im ϕ,m ϕ 0 for all ϕ L 2 (Ω 1 H 1/2 ( B 2. (c M is injective. Proof. (a By Lemma 2.1 (b, Lemma 2.3 (b, and Lemma 4.2 (a, Re ( e it M ( = Re e it R 1 MΩ Mix 1 q,ω 2 R1 eit R 2 S B2 R2 eit R 2 R 3 T Imp B 3,iλ 0 R3 R 2 ( Re( eit q = R 1 k 2 q 0 R1 R 2 (cos ts B2,iR2 +K 0 (cos ts Ω2,i ( ( I 0 Re( eit q ( = k 2 q 0 I (cos ts Ω2,i 0 0 R 2 (cos ts B2,iR 2 +K = ( Re( eit q k 2 q 0 0 ( cos ts B2,i +K, (4.19 where K and K are some compact operators. The first term of the right hand side in (4.19 is self-adjoint and positive coercive since ( cos t > 0 when t (π/2,3π/2, and Assumption 1.3 (iv yields ϕ,re ( e it q k 2 q ϕ = ϕ 2Re(e it q Ω 1 k 2 dx q ϕ 2 C q Ω 1 k 2 q dx = C k 2 ϕ 2 L 2 (Ω 1. (4.20 (bbylemma2.1(c, Lemma2.3(c(d, forallϕ L 2 (Ω 1 H 1/2 ( B 2 Im ϕ,m ϕ = Im R 1ϕ,M Mix R 1ϕ Im R 2ϕ,S B2 R 2ϕ +Im T Imp B 3,iλ 0 R 3R 2ϕ,R 3R 2ϕ 0. (

18 ( φ (c Let φ L 2 (Ω 1 and ψ H 1/2 ( B 2. Assume that M = 0. ψ Inequality (4.21 yields that ( ( Im T Imp B 3,iλ 0 R3R 2 φ,r ψ 3R2 φ = 0, (4.22 ψ ( φ which implies that R3 R 2 = 0 from Lemma 2.1 (d. By Lemma 4.2 ψ (c, and the definition of R 2, we have ψ = 0. Therefore, ( ( M φ = R 0 1 MΩ Mix 1 q,ω 2 R1 φ = 0. ( From Lemma 4.2 (b and Lemma 2.3 (d, we obtain ( ( R1 φ 0 =. ( Finally, we will show φ = 0. Let f 1 L 2 (Ω 1. Take radiating solutions v 1 and w such that v 1 +k 2 (1+qv 1 = k 2 q q f 1 in R 3 \Ω 2, (4.25 By (4.24, 0 = = ( f 1 0 By (4.25 and (4.27, v 1 = 0 on Ω 2, (4.26 w +k 2 (1+qw = q φ in R 3 \Ω 2, (4.27 Ω 1 f 1 φdx+ Ω 1 v 1 q φdx = (,R1 φ 0 w = 0 on Ω 2. (4.28 ( f 1 + q v 1 = v B2 1 ( φ, 0 Ω 1 v 1 q φdx. (4.29 ( v 1 w +k 2 (1+qw dx Ω 1 ( v 1 +k 2 (1+qv 1 +k 2 q f 1 wdx q Ω 1 = k 2 q f 1 wdx Ω 1 q + ( wv 1 w( v 1 dx. (4.30 Ω 1 18

19 By using Green s theorem, (4.26, and (4.28, ( wv 1 w( v 1 dx = ( wv 1 w( v 1 dx Ω 1 R 3 \Ω 2 [ w = v 1 w v ] ds Ω 2 ν Ω2 ν Ω2 = 0. (4.31 By (4.29 (4.31, φ = k 2 q q w in Ω 1. (4.32 From (4.32, (4.27, and (4.28, we obtain w+k 2 w = 0 in R 3 \Ω 2, (4.33 w = 0 on Ω 2, (4.34 which proves that w vanishes in R 3 \Ω 2 by the uniqueness of the exterior Dirichlet problem. Therefore, equation (4.32 yields that φ = 0. Therefore, by Lemma 4.3, we can apply Theorem 2.4 to this case with F = FΩ Mix 1 q,ω 2 +FB Dir 2 +F Imp B 3,iλ 0. From Lemma4.1 (b, andapplyingtheorem 2.4, we obtain Theorem 1.4. Remark 4.4. We can also consider various situations on Ω 2 like Remark 3.4, and replace the assumption of taking B 3 with that k 2 is not a Dirichlet eigenvalue of in an artificial domain B 2 like Remark 3.5. We can also give the characterization by replacing (iv in Assumption 1.3 with (iv There exists t [0,π/2 (3π/2,2π] and C > 0 such that Re(e it q C q a.e. in Ω 1. by the same argument in Theorem 1.4: Assumption 4.5. Let a bounded domain B 2 be a priori known. Assume the following assumptions: (i q L (Ω 1 with Imq 0 in Ω 1. (ii q is locally bounded below in Ω 1, i.e., for every compact subset M Ω 1, there exists c > 0 (depend on M such that q c in M. 19

20 (iii Ω 2 B 2, Ω 1 B 2 =. (iv There exists t [0,π/2 (3π/2,2π] and C > 0 such that Re(e it q C q a.e. in Ω 1. Theorem 4.6. Let Assumption 4.5 hold. Take a positive number λ 0 > 0. Then, for z R 3 \B 2 z Ω 1 (φ z,ϕ n L 2 (S 2 2 <, (4.35 λ n n=1 where (λ n,ϕ n is a complete eigensystem of F # given by F # := Re ( e it F + ImF, (4.36 where F := F Mix +F Imp B 2,iλ 0. Here, the function φ z is given by (1.11. Conclusion In this paper, we give the characterization of the unknown domain Ω 1 in a scatterer consisting of two objects with different physical properties without the assumption of the wave number k > 0. To realize it, we modify the original far field operator F by adding artificial far field operators corresponding to an inner domain B 1, an outer domain B 2, and an additional domain B 3. This idea is mainly based on [9], which treats only a scattering by an obstacle with the pure Dirichlet or Neumann boundary condition. In Section 4 of [9], numerical examples are given to compare modification method (which use the artificial far field operator corresponding to an inner domain with previous method numerically, where we find that the modification method provides numerically a better reconstruction than previous one. Therefore, we expect that even in a scatterer consisting of two objects with different physical properties, our modification method (which use several artificial far field operators would also provide a better reconstruction than previous ones such as [10, 13]. Acknowledgments First, the author would like to express his deep gratitude to Professor Mitsuru Sugimoto, who always supports him in his study. Secondly, he thanks to Professor Masaru Ikehata and Professor Sei Nagayasu, who read this paper carefully and gave him many helpful comments. The author also thanks 20

21 to Professor Andreas Kirsch, and Professor Xiaodong Liu, and the referees who gave him valuable comments which helped to improve this paper. References [1] K. A. Anagnostopoulos and A. Charalambopoulos and A. Kleefeld, The factorization method for the acoustic transmission problem, Inverse Problems 29, (2013, [2] O. Bondarenko and A. Kirsch and X. Liu, The factorization method for inverse acoustic scattering in a layered medium, Inverse Problems 29, (2013, [3] D. Colton and A. Kirsch, A simple method for solving inverse scattering problems in the resonance region, Inverse Problems 12, (1996, [4] D. Colton and R. Kress, Inverse acoustic and electromagnetic scattering theory, Third edition. Applied Mathematical Sciences, 93 Springer, New York, (2013. [5] A. Kirsch, Characterization of the shape of a scattering obstacle using the spectral data of the far field operator, Inverse Problems 14, (1998, [6] A. Kirsch and A. Kleefeld, The factorization method for a conductive boundary condition, J. Integral Equations Appl. 24, (2012, [7] A. Kirsch and N. Grinberg, The factorization method for inverse problems, Oxford University Press, (2008. [8] A. Kirsch and N. Grinberg, The factorization method for obstacles with a priori separated sound-soft and sound-hard parts, Math. Comput. Simulation 66 (2004, [9] A. Kirsch and X. Liu, A modification of the factorization method for the classical acoustic inverse scattering problems, Inverse Problems 30, (2014, [10] A. Kirsch and X. Liu, Direct and inverse acoustic scattering by a mixedtype scatterer, Inverse Problems 29 (2013, [11] A. Kirsch and X. Liu, The factorization method for inverse acoustic scattering by a penetrable anisotropic obstacle, Math. Methods Appl. Sci. 37, (2014,

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