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The boundary integral method for the Helmholtz equation with cracks inside a bounded domain
Acta Mathematica Scientia 2015,35B(3):539–551 http://actams.wipm.ac.cn
THE BOUNDARY INTEGRAL METHOD FOR THE HELMHOLTZ EQUATION WITH CRACKS INSIDE A BOUNDED DOMAIN∗
H)
Jun GUO (
School of Mathematics and Statistics, South-Central University for Nationalities, Wuhan 430074, China E-mail : [email protected]
ww)
†
Lili FAN (
School of Mathematics and Computer Science, Wuhan Polytechnic University, Wuhan 430023, China E-mail : [email protected]
îI)
Guozheng YAN (
School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China E-mail : yan [email protected] Abstract We consider a kind of scattering problem by a crack Γ that is buried in a bounded domain D, and we put a point source inside the domain D. This leads to a mixed boundary value problem to the Helmholtz equation in the domain D with a crack Γ. Both sides of the crack Γ are given Dirichlet-impedance boundary conditions, and different boundary condition (Dirichlet, Neumann or Impedance boundary condition) is set on the boundary of D. Applying potential theory, the problem can be reformulated as a system of boundary integral equations. We establish the existence and uniqueness of the solution to the system by using the Fredholm theory. Key words
Boundary integral equation; Helmholtz equation; Fredholm theory
2010 MR Subject Classification
1
35D05; 35P25
Introduction
The direct and inverse scattering problem for cracks was initiated by Kress [9] in 1995. In that article, Kress considered the scattering problem for a perfectly conducting crack and used the integral equation method to solve both the direct and inverse problems for a sound-soft crack. The case of a sound-hard crack was considered by M¨onch [14] in 1997. In 2000, Kirsch and Ritter [10] used the factorization method to reconstruct the shape of the crack from the ∗ Received
Jun 17, 2013; revised Mar 2, 2014. The second author is supported by the grant from the National Natural Science Foundation of China (11301405). The third author is supported by the grants from the National Natural Science Foundation of China (11171127 and 10871080). † Corresponding author.
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knowledge of the far-field pattern as a continued work of Kress. Also in 2000, Ammari [1] considered the electromagnetic scattering problem with cracks. In 2003, Cakoni and Colton [4] discussed the direct and inverse scattering for a open crack coated on one side by a material and determined the shape of the open arc by the linear sampling method. In 2008, Ming-Kuo [11] studied the direct and inverse scattering problem for a crack with impedance boundary condition. The combination of a crack and a bounded impenetrable and penetrable obstacles were discussed by Guozheng Yan and Mao Yao in [20] and [19], respectively. In 2012, Guozheng Yan and Jianguo Ye [18] considered a scattering problem by a crack that is buried in a piecewise homogeneous medium. However, there are a few results about the interior scattering by a crack that is buried in a bounded domain. This means that the waves propagate in a bounded area constituted by a certain substance with a crack inside. In some industrial applications of non-destructive testing, it is important to test the structural integrity of cavities using acoustic or electromagnetic waves emitted and measured by sources and receivers, respectively, placed inside the cavity (see [15– 17]). In this article, we consider a kind of interior scattering problem. Briefly speaking, we consider the scattering problem by a crack Γ that is buried in a bounded domain D, and we put a point source inside the domain D. We take Dirichlet-impedance boundary conditions on two sides of the crack, and set different boundary condition (Dirichlet, Neumann or Impedance boundary condition) on the boundary of D. This leads to a mixed boundary value problem for the Helmholtz equation for a crack buried inside the domain D. The goal of this article is to establish the existence and uniqueness of the solution to this scattering problem using the boundary integral method [8, 18–20] in some suitable function spaces. Note that this is a classical method for solving elliptic equations. In practical applications, it also facilitates numerical implementation. For a detailed description of this method, please refer to the monographs [3, 6, 12]. We can reformulate the problem as a boundary integral system, and show that the corresponding boundary integral operators have the Fredholm property in some suitable Sobolev space. Furthermore, we can establish the existence and uniqueness of the solution to the boundary integral system. The plan of this article is as follows. In the next section, we introduce the scattering problem, establish uniqueness of the solution, and reformulate the problem as a boundary integral system using the potential theory. In Section 3, in some suitable Sobolev spaces, we prove the main results, that is, the existence and uniqueness of the solution to the corresponding boundary integral system.
2
Boundary Integral System
Let D be a bounded and closed subset of R2 (or R3 ) with a C 2 boundary ∂D. For simplicity, we discuss the problem just in R2 space. An arc (or crack) Γ is located inside the domain D. For further consideration, we extend the arc Γ to an arbitrary piecewise smooth, simply connected and closed curve ∂Ω enclosing a bounded domain Ω which is completely contained in the domain D such that the normal vector ν on Γ coincides with the normal vector on ∂Ω (except a finite numbers of points), and directed into the exterior of Ω which we again denote by ν. The scattering of a point source by a crack Γ in the domain D leads a mixed boundary
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value problem for the Helmholtz equation: ∆U + k 2 U = 0, U− = 0,
541
¯ ∪ {z}), in D\(Γ on Γ,
∂U+ + ikλU+ = 0, ∂ν BU = 0,
(2.1) on Γ, on ∂D.
The total wave U = Φ(·, z)+us is the superposition of the given point source Φ(·, z), z ∈ D, and us is the scattered wave. ‘B’ denotes Dirichlet, Neumann, or impedance boundary condition. We assume that the wave number k and impedance coefficient λ are positive, that is, k > 0 ∂u− + ∂u and λ > 0. u+ (u− ) and ∂u ∂ν ( ∂ν ) denote the limit of u and ∂ν on the boundary ∂Ω from the exterior (interior). Remark 2.1 For convenience, in the following discussion, (·)± means the limit approaching the boundary to the corresponding domain from outside and inside, respectively. In order to formulate the above scattering problem more precisely, we need to properly define the trace space on Γ. The classic reference for such trace space is given by Lions and Magenes [13]. Here, we use the notation of Mclean [12]. As we known, L2 (∂Ω), H 1/2 (∂Ω), and H −1/2 (∂Ω) denote the usual Sobolev spaces, then we define the following spaces [12]: L2 (Γ) = {u|Γ : u ∈ L2 (∂Ω)},
H 1/2 (Γ) = {u|Γ : u ∈ H 1/2 (∂Ω)},
˜ 1/2 (Γ) = {u ∈ H 1/2 (∂Ω) : suppu ⊆ Γ}, ¯ H ′ ˜ 1/2 (Γ) , the dual space of H ˜ 1/2 (Γ), H −1/2 (Γ) = H ′ ˜ −1/2 (Γ) = H 1/2 (Γ) , the dual space of H 1/2 (Γ), H
and we have the chain
˜ 1/2 (Γ) ⊂ H 1/2 (Γ) ⊂ L2 (Γ) ⊂ H ˜ −1/2 (Γ) ⊂ H −1/2 (Γ). H For the scattered field us , the problem (2.1) is a special case of the following problem (2.2): Given f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), h ∈ H 1/2 (∂D) (or h1 ∈ H −1/2 (∂D) or h2 ∈ H −1/2 (∂D)) ¯ such that find u ∈ H 1 (D\Γ) 2 ¯ in D\Γ, ∆u + k u = 0, u− = f, on Γ, (2.2) ∂u+ + ikλu+ = g, on Γ, ∂ν Bu = 0, on ∂D, where the boundary condition Bu = 0 on ∂D is one of the following three situations: u − h = 0, ∂u − h1 = 0, ∂ν
on ∂D, on ∂D,
∂u + ikλ2 u − h2 = 0 ∂ν with λ2 < 0.
on ∂D,
(2a) (2b) (2c)
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Remark 2.2 The problem (2.1) is a special case of the problem (2.2) with f = −Φ(·, z), ∂Φ(·,z) g = − ∂Φ(·,z) − ikλΦ(·, z), h = −Φ(·, z), h1 = − ∂Φ(·,z) − ikλ2 Φ(·, z). ∂ν ∂ν , h2 = − ∂ν Firstly we show that the uniqueness of the solution to the problem (2.2). Theorem 2.3 The problem (2.2) has at most one solution. Proof let u be a solution to the homogeneous problem (2.2), that is, u satisfies the ¯ problem (2.2) with f = g = h (or h1 , or h2 ) = 0. Obviously, this solution u ∈ H 1 (D\Γ) ¯ satisfies the following transmission boundary condition on the complementary part ∂Ω\Γ of ∂Ω: u+ = u− , (2.3) ∂u+ = ∂u− . ∂ν ∂ν ¯ and Ω yields An application of the Green formula for u and u ¯ in D\Ω Z Z Z ∂u ¯ ∂u ∂u ¯+ ∂u+ ds − u+ ds, (u△¯ u−u ¯△u)dx = u − u¯ −u ¯+ ∂ν ∂ν ∂ν ∂ν ¯ D\Ω ∂Ω ∂D and Z
(u△¯ u−u ¯△u)dx =
Ω
Z
∂u ¯− ∂u− u− − u¯− ds. ∂ν ∂ν ∂Ω
By using the boundary conditions in (2.2) and (2.3), the sum of the above two equations is Z −2ikλ |u+ |2 ds = 0 Γ
for (2.2) and (2a), or −2ikλ
Z
|u+ |2 ds = 0
Γ
for (2.2) and (2b), or 2ikλ2
Z
2
|u| ds − 2ikλ
∂D
Z
|u+ |2 ds = 0
Γ
for (2.2) and (2c). For any situation we can conclude u+ = 0 on Γ. From the boundary + condition in (2.2), we have ∂u ∂ν = 0 on Γ. The Holmgren’s uniqueness theorem [7] means that ¯ The transmission boundary condition of (2.3) implies u− = 0 and ∂u− = 0 on u = 0 in D\Ω. ∂ν ¯ then we again obtain u = 0 in Ω by the Holmgren’s uniqueness theorem. So, we show ∂Ω\Γ, ¯ and this completes the proof of the theorem. that u = 0 in D\Γ We are now ready to prove the existence of a solution to the problem (2.2) by using integral equation approaching. By Green representation formula, we have Z Z ∂u ∂Φ(x, y) ∂u ∂Φ(x, y) u(x) = Φ(x, y) − u dsy − Φ(x, y) − u dsy (2.4) ∂ν ∂ν ∂D ∂ν ∂Ω ∂ν
¯ and for x ∈ D\Ω,
u(x) =
Z
∂Ω
∂u ∂Φ(x, y) Φ(x, y) − u dsy ∂ν ∂ν
for x ∈ Ω, where Φ(x, y) =
i (1) H (k|x − y|) 4 0
(2.5)
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(1)
is the fundamental solution to the Helmholtz equation in R2 and H0 is a Hankel function of the first kind of order zero. By making use of the known jump relations of single-and doublelayer potentials across the boundary ∂Ω (see [3]) and approaching the boundary ∂Ω from inside Ω, that is, x ∈ Ω and x → ∂Ω, we obtain u− (x) = SΩΩ
∂u− − KΩΩ u− , ∂ν
∂u− (x) ′ ∂u− = KΩΩ − TΩΩ u− . ∂ν ∂ν ¯ we first consider approaching the boundary ∂Ω, then we have In the domain D\Ω, u+ (x) = SDΩ
∂u+ ∂u − KDΩ u − SΩΩ + KΩΩ u+ , ∂ν ∂ν
∂u+ (x) ∂u ′ ′ ∂u+ = KDΩ − TDΩ u − KΩΩ + TΩΩ u+ . ∂ν ∂ν ∂ν ¯ we consider approaching the boundary ∂D, and obtain Again in the domain D\Ω, u− (x) = SDD
∂u ∂u+ − KDD u − SΩD + KΩD u+ , ∂ν ∂ν
∂u− ∂u ∂u+ ′ ′ = KDD − TDD u − KΩD + TΩD u+ , ∂ν ∂ν ∂ν ′ where SΩΩ , KΩΩ , KΩΩ , and TΩΩ are boundary integral operators defined by Z SΩΩ ϕ(x) = 2 ϕ(y)Φ(x, y)dsy , x ∈ ∂Ω; Z∂Ω Φ(x, y) dsy , x ∈ ∂Ω; KΩΩ ϕ(x) = 2 ϕ(y) ∂νy ∂Ω Z ∂Φ(x, y) ′ KΩΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; ∂νx ∂Ω Z ∂ ∂Φ(x, y) TΩΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω. ∂νx ∂Ω ∂νy
(2.6) (2.7)
(2.8) (2.9)
(2.10) (2.11)
′ ′ The operators SDD , KDD , KDD , and TDD have the similar meaning as SΩΩ , KΩΩ , KΩΩ , and ′ TΩΩ , respectively. The operators SDΩ , KDΩ , KDΩ , and TDΩ are given by Z SDΩ ϕ(x) = 2 ϕ(y)Φ(x, y)dsy , x ∈ ∂Ω; ∂D Z Φ(x, y) KDΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; ∂νy ∂D Z ∂Φ(x, y) ′ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; KDΩ ∂νx ∂D Z ∂ ∂Φ(x, y) TDΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω. ∂νx ∂D ∂νy
However, the operator SΩD is the operator applied a function with supp⊆ ∂Ω and evaluated ′ on ∂D, with analogous definitions for operators such as KΩD , KΩD , TΩD , and the following ′ upcoming operators such as SΓD ,KΓD , KΓD , TΓD and so on. Restrict formula (2.6) on the arc Γ and use formula (2.8), we have ∂u− ∂u+ ∂u 2f = SΩΓ − − KΩΓ (u− − u+ ) + (u− − u+ )|Γ + SDΓ − KDΓ u. (2.12) ∂ν ∂ν ∂ν
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Using (2.8) and (2.9), we restrict ∂u ∂ν + ikλu on the arc Γ and obtain ∂u+ g= + ikλu+ |Γ ∂ν ∂u+ ∂u+ ∂u− ∂u− ′ = KΩΓ − − TΩΓ (u− − u+ ) + ikλSΩΓ − ∂ν ∂ν ∂ν ∂ν ∂u ∂u − − ′ + TΩΓ u− − ikλSΩΓ − ikλKΩΓ (u− − u+ ) − KΩΓ ∂ν ∂ν ∂u ∂u ′ + ikλKΩΓ u− + KDΓ − TDΓ u + ikλSDΓ − ikλKDΓ u. ∂ν ∂ν Combining (2.6) and (2.7), we have ∂u− ′ ∂u− −ikλ SΩΓ − KΩΓ u− − KΩΓ + TΩΓ u− ∂ν ∂ν ∂u− = − ikλu− − |Γ ∂ν ∂u+ ∂u+ ∂u− − |Γ − ikλu+ + |Γ . = −ikλ(u− − u+ )|Γ − ∂ν ∂ν ∂ν Using (2.14), we can rewrite (2.13) as ∂u− ∂u+ ∂u− ∂u+ ′ 2g = KΩΓ − − TΩΓ (u− − u+ ) + ikλSΩΓ − ∂ν ∂ν ∂ν ∂ν ∂u− ∂u+ −ikλKΩΓ (u− − u+ ) − ikλ(u− − u+ )|Γ − − |Γ ∂ν ∂ν ∂u ′ ∂u +KDΓ − TDΓ u + ikλSDΓ − ikλKDΓ u. ∂ν ∂ν From (2.12) and the boundary conditions in (2.2), we can reformulate (2.15) as ∂u+ ∂u− ′ 2g = KΩΓ − − TΩΓ (u− − u+ ) − 2ikλ(u− − u+ )|Γ ∂ν ∂ν ∂u− ∂u+ ′ ∂u − − |Γ + KDΓ − TDΓ u + 2f. ∂ν ∂ν ∂ν
(2.13)
(2.14)
(2.15)
(2.16)
Using the Green formula, we have Z
∂Ω
(u− (y)
−u(x),
∂Φ(x, y) ∂u− (y) − Φ(x, y))dsy = 0, ∂ν ∂ν
in
Ω,
in
¯ D\Ω.
(2.17)
From (2.10), upon using (2.17), we obtain
∂u ∂u− ∂u+ − KDD u + SΩD − ∂ν ∂ν ∂ν ∂u− −KΩD (u− − u+ ) − SΩD + KΩD u− ∂ν ∂u− ∂u+ ∂u − − KΩD (u− − u+ ) + SDD − KDD u. = SΩD ∂ν ∂ν ∂ν
u(x)|∂D = SDD
Similarly, we can rewrite (2.11) as ∂u ∂u− ∂u+ ∂u ′ ′ |∂D = KΩD − − TΩD (u− − u+ ) + KDD − TDD u. ∂ν ∂ν ∂ν ∂ν
(2.18)
(2.19)
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Now, for the problem (2.2) and (2a), we define ∂u+ ∂u ∂u− |∂D = a, − |Γ = b, ∂ν ∂ν ∂ν
545
(u− − u+ )|Γ = c,
and extend b and c by zero to the whole ∂Ω in the following 0, on ∂Ω\Γ, 0, on ∂Ω\Γ, ˜b = c˜ = b, on Γ, c, on Γ, ˜ −1/2 (Γ), c ∈ H ˜ 1/2 (Γ) (see [4]). then b ∈ H Letting r = h + KDD h,
p = 2f + KDΓ h,
q = 2g − 2f + TDΓ h,
and
SDD
SΓD
−KΓD
A= SDΓ
SΓΓ
I − KΓΓ
′ KΓΓ −I
−(TΓΓ + 2ikλI)
′ KDΓ
,
and combining (2.12), (2.16), and (2.18), we have a boundary integral system r(x) a A b = p(x) . q(x) c For the problem (2.2) and (2b), we define ∂u− ∂u+ u|∂D = a1 , − |Γ = b1 , ∂ν ∂ν
(2.20)
(u− − u+ )|Γ = c1 ,
˜ −1/2 (Γ), and ˜b1 and c˜1 are zero extension of b1 and c1 to the whole ∂Ω, respectively, then b1 ∈ H 1/2 ˜ c1 ∈ H (Γ). Denote ′ r1 = h1 − KDD h1 ,
p1 = 2f − SDΓ h1 ,
′ q1 = 2g − 2f − KDΓ h1 ,
and
−TDD
′ KΓD
A1 = SΓΓ −KDΓ ′ −TDΓ KΓΓ −I
−TΓD I − KΓΓ −(TΓΓ + 2ikλI)
,
then combining (2.12), (2.16), and (2.19), we have a boundary integral system a1 r1 (x) A1 b1 = p1 (x) . c1 q1 (x) Similarly, for the problem (2.2) and (2c), we define ∂u− ∂u+ u|∂D = a2 , − |Γ = b2 , (u− − u+ )|Γ = c2 , ∂ν ∂ν
(2.21)
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˜ −1/2 (Γ), and ˜b2 and c˜2 are zero extension of b2 and c2 to the whole ∂Ω, respectively, then b2 ∈ H ˜ 1/2 (Γ). c2 ∈ H ′ ′ Let r2 = h2 − KDD h2 − ikλ2 SDD h2 , p2 = 2f − SDΓ h2 , q2 = 2g − 2f − KDΓ h2 , and ′ ′ −TDD − ikλ2 (KDD + KDD ) + k 2 λ22 SDD KΓD + ikλ2 SΓD −TΓD − ikλ2 KΓD , A2 = −KDΓ − ikλ2 SDΓ SΓΓ I − KΓΓ ′ ′ −TDΓ − ikλ2 KDΓ KΓΓ −I −(TΓΓ + 2ikλI) then combining (2.12), (2.16), and (2.19), we have a boundary integral system a2 r2 (x) = p2 (x) . A2 b 2 c2 q2 (x)
(2.22)
To the operators appeared in A, A1 , and A2 , we list their mapping properties in the following (see [5]): ˜ −1/2 (Γ) → H 1/2 (Γ), SΓΓ : H ′ ˜ −1/2 (Γ) → H −1/2 (Γ), KΓΓ :H
˜ 1/2 (Γ) → H 1/2 (Γ), KΓΓ : H ˜ 1/2 (Γ) → H −1/2 (Γ); TΓΓ : H
SDD : H −1/2 (∂D) → H 1/2 (∂D),
KDD : H 1/2 (∂D) → H 1/2 (∂D),
′ KDD : H −1/2 (∂D) → H −1/2 (∂D), ˜ −1/2 (Γ) → H 1/2 (∂D), SΓD : H
TDD : H 1/2 (∂D) → H −1/2 (∂D); ˜ 1/2 (Γ) → H 1/2 (∂D), KΓD : H
′ ˜ −1/2 (Γ) → H −1/2 (∂D), KΓD :H
˜ 1/2 (Γ) → H −1/2 (∂D); TΓD : H
SDΓ : H −1/2 (∂D) → H 1/2 (Γ),
KDΓ : H 1/2 (∂D) → H 1/2 (Γ),
′ KDΓ : H −1/2 (∂D) → H −1/2 (Γ),
TDΓ : H 1/2 (∂D) → H −1/2 (Γ).
If we define ˜ −1/2 (Γ) × H ˜ 1/2 (Γ) H = H −1/2 (∂D) × H and the dual space H ∗ = H 1/2 (∂D) × H 1/2 (Γ) × H −1/2 (Γ), then the operator A maps H continuously into H ∗ . If we define ˜ −1/2 (Γ) × H ˜ 1/2 (Γ) W = H 1/2 (∂D) × H and the dual space W ∗ = H −1/2 (∂D) × H 1/2 (Γ) × H −1/2 (Γ), then the operators A1 and A2 map W continuously into W ∗ . Remark 2.4 Once the unknown Cauchy data are determined from (2.20) (or (2.21), or (2.22)), then the representation formula (2.4) and (2.5) determine the corresponding weak solution to the problem (2.2).
3
Existence and Uniqueness of the Solution
If the integral system (2.20), (2.21), and (2.22) has a unique solution, our problem (2.2) will has a unique solution (see [2]). In this section, we will show that the matrix operators A, A1 ,
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and A2 are Fredholm with index zero, and they are injective operators, that is, kernA = {0}, kernA1 = {0}, and kernA2 = {0}. Then, using Theorem 2.3, the solvability of the problem (2.20), (2.21), and (2.22) is true. Refer to [20], it is easy to obtain the solvability of the boundary integral system (2.20), so, we omit its proof here. In the following, we will focus on the solvability of the problem (2.21). Theorem 3.1 The boundary integral system (2.21) has a unique solution. Proof There are two main steps to prove the theorem: Step 1 is to show that the operator A1 is Fredholm with index zero; Step 2 is to prove that KernA1 = {0}. Step 1: In fact, the operators −TDD , −TΩΩ , and SΩΩ are positive and bounded up to a compact perturbation [12], and we denote by LD , LΩ , and LS the compact operators LD : H 1/2 (∂D) → H −1/2 (∂D), LΩ : H 1/2 (∂Ω) → H −1/2 (∂Ω), LS : H −1/2 (∂Ω) → H 1/2 (∂Ω), such that ¯ ≥ C||ψ||2 1/2 Re(h−(TDD + LD )ψ, ψi) H (∂D) ,
for ψ ∈ H 1/2 (∂D),
(3.1)
¯ ≥ C||ψ||2 1/2 Re(h−(TΩΩ + LΩ )ψ, ψi) H (∂Ω) ,
for ψ ∈ H 1/2 (∂Ω),
(3.2)
¯ ≥ C||ψ||2 −1/2 Re(h(SΩΩ + LS )ψ, ψi) H (∂Ω) ,
for ψ ∈ H −1/2 (∂Ω),
(3.3)
where h , i denote the duality between H −1/2 (∂D) and H 1/2 (∂D), or H −1/2 (∂Ω) and H 1/2 (∂Ω). ′ Let K0 and K0′ be the operators defined as KΩΩ and KΩΩ , respectively, with kernel Φ(x, y) 1 ′ replaced by Φ0 (x, y) = − 2π ln |x − y|. Then, Kc = KΩΩ − K0 and Kc′ = KΩΩ − K0′ are compact because they have continuous kernels [8]. It is easy to show that K0 and K0′ are adjoint because their kernels are real, that is, (K0 ψ, φ) = (ψ, K0′ φ)
for ψ ∈ H 1/2 (∂Ω) and φ ∈ H −1/2 (∂Ω).
We define
−TDD
′ KΩD
−TΩD
A′1 = −KDΩ −TDΩ
SΩΩ
I − KΩΩ
′ KΩΩ −I
−(TΩΩ + 2ikλI)
,
then A′1 : H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω) → H −1/2 (∂D) × H 1/2 (∂Ω) × H −1/2 (∂Ω). Note that ˜b1 ∈ H −1/2 (∂Ω), c˜1 ∈ H 1/2 (∂Ω), so, we can decompose A′1 into two parts, that is, a1 a1 a1 ′ ˜ ˜ A1 b1 = A10 b1 + A1c ˜b1 (3.4) c˜1 c˜1 c˜1
with
a1 −(TDD + LD )a1 ˜ A10 (SΩΩ + LS )˜b1 + (I − K0 )˜ c1 b1 = (K0′ − I)˜b1 − (TΩΩ + LΩ )˜ c˜1 c1 − 2ikλ˜ c1
(3.5)
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and
′ ˜ a1 LD a1 + KΩD b1 − TΩD c˜1 ˜ ˜ A1c b1 = −KDΩ a1 − LS b1 − Kc c˜1 , c˜1 −TDΩ a1 + Kc′ ˜b1 + LΩ c˜1
(3.6)
where A1c : H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω) → H −1/2 (∂D) × H 1/2 (∂Ω) × H −1/2 (∂Ω) is compact. Furthermore, we have (A10 ξ~′ , ξ~′ ) = (−(TDD + LD )a1 , a1 ) + ((SΩΩ + LS )˜b1 , ˜b1 ) + ((I − K0 )˜ c1 , ˜b1 ) ′ +((K − I)˜b1 , c˜1 ) + (−(TΩΩ + LΩ )˜ c1 , c˜1 ) − (2ikλ˜ c1 , c˜1 ).
(3.7)
0
Here, (u, v) denotes the scalar product on L2 (∂D) or L2 (∂Ω) defined by and ξ~′ = (a1 , ˜b1 , c˜1 )T .
R
∂D
u¯ v ds or
R
∂Ω
u¯ v ds,
From (3.1), (3.2), and (3.3), we obtain Re[(−(TDD + LD )a1 , a1 ) + ((SΩΩ + LS )˜b1 , ˜b1 ) + (−(TΩΩ + LΩ )˜ c1 , c˜1 )] ≥ c||a1 ||2 1/2 + c||˜b1 ||2 −1/2 + c||˜ c1 ||2 1/2 . H
(∂D)
H
(∂Ω)
H
(∂Ω)
(3.8)
Furthermore, as K0 and K0′ are adjoint, we have Re[((I − K0 )˜ c1 , ˜b1 ) + ((K0′ − I)˜b1 , c˜1 )] = Re[(˜ c1 , (I − K ′ )˜b1 ) + ((K ′ − I)˜b1 , c˜1 )] 0
= Re[(˜ c1 , (I −
0
K0′ )˜b1 )
c1 , (K0′ − I)˜b1 ] = 0. + (˜
(3.9)
It is obviously Re[−2ikλ(˜ c1 , c˜1 )] = 0.
(3.10)
Combining (3.8), (3.9), and (3.10), we obtain the fact that the sesquilinear form defined by A10 is coercive, that is, ¯′ Re(hA10 ξ~′ , ~ξi) ≥ C||ξ~′ ||2 ,
for
ξ~′ ∈ H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω).
˜ −1/2 (Γ), and c˜1 is the extension by zero of Recalling that ˜b1 is the extension by zero of b1 ∈ H ˜ 1/2 (Γ), so we can rewrite the above inequality as c1 ∈ H ~ ξi) ~¯ ≥ C||ξ|| ~ 2, Re(hA10,Γ ξ,
˜ −1/2 (Γ) × H ˜ 1/2 (Γ), for ξ~ ∈ H 1/2 (∂D) × H
where A10,Γ is the restriction to Γ of A10 defined for ξ~ = (a1 , b1 , c1 )T ∈ W . The corresponding restriction A1c,Γ : W → W ∗ of A1c clearly remains compact. Whence the operator A1 is Fredholm with index zero. Step 2: In this part, we show that KernA1 = {0}. For simplicity, we still use ξ~ = (a1 , b1 , c1 )T to denote a solution of the homogeneous equation A1 ξ~ = ~0, which means that ′ −TDD a1 + KΓD b1 − TΓD c1 = 0, (3.11) −KDΓ a1 + SΓΓ b1 + (I − KΓΓ )c1 = 0, ′ −TDΓ a1 + (KΓΓ − I)b1 − (TΓΓ + 2ikλI)c1 = 0,
then we need to prove that ξ~ = ~0.
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Remark 3.2 For convenience, here we used the same notation ξ~ = (a1 , b1 , c1 )T , but with different meaning. For this purpose, we define potential v(x) = −KD a1 + SΩ˜b1 − KΩ c˜1 ,
¯ x ∈ R2 \(∂D ∪ Γ),
(3.12)
where KD ϕ(x) =
Z
ϕ(y)
∂D
SΩ ϕ(x) =
∂Φ(x, y) dsy , ∂ν(y)
Z
ϕ(y)Φ(x, y)dsy ,
Z
ϕ(y)
∂Ω
KΩ ϕ(x) =
∂Ω
∂Φ(x, y) dsy . ∂ν(y)
¯ By the jump relationships of This potential v(x) satisfies Helmholtz equation in D\Γ. ¯ letting x → ∂D, single-and double-layer potentials on the boundary ∂D, in the domainD\Ω, we have ∂v− (x) ′ = −TDD a1 + KΓD b1 − TΓD c1 , ∂ν
(3.13)
then the first equation in (3.11) implies that ∂v− (x) |∂D = 0. ∂ν
(3.14)
Letting x → Γ, we have 2v+ (x) = −KDΓ a1 + SΓΓD b1 − KΓΓ c1 − c1 , 2
∂v+ (x) ′ = −TDΓ a1 + KΓΓ b1 − TΓΓ c1 − b1 . ∂ν
So, 2
∂v+ + ikλv+ ∂ν
= ikλ(−KDΓ a1 + SΓΓD b1 − KΓΓ c1 + c1 ) ′ −TDΓ a1 + KΓΓ b1 − b1 − TΓΓ c1 − 2ikλc1 ,
then combining the second and third equation in (3.11), we have ∂v+ + ikλv+ |Γ = 0. ∂ν
(3.15)
In the domain Ω, letting x → Γ, we obtain 2v− (x) = −KDΓ a1 + SΓΓ b1 − KΓΓ c1 + c1 , then the second equation in (3.11) implies that v− (x)|Γ = 0.
(3.16)
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From (3.14), (3.15), and (3.16), the potential v(x) satisfies the following problem ¯ ∆v + k 2 v = 0, in D\Γ, on Γ, v− = 0, ∂v+ + ikλv+ = 0, on Γ, ∂ν ∂v− (x) = 0, on ∂D. ∂ν
(3.17)
¯ It is not difficult to check that the From Theorem 2.3, we know that v(x) = 0 in D\Γ. potential v(x) also satisfies the following exterior scattering problem ¯ ∆v + k 2 v = 0, in R2 \D, (3.18) ∂v+ = 0, in ∂D, ∂ν and v(x) satisfies the Sommerfield radiation condition √ ∂v lim r − ikv = 0 r→∞ ∂r
uniformly in xˆ = x/|x| with r = |x|. ¯ The jump relationships of single From Theorem 3.13 in [3], we have v(x) = 0 in R2 \D. and double layer potential imply that a1 = (v− − v+ )|∂D = 0, ∂v− ∂v+ b1 = − |Γ = 0, ∂ν ∂ν and c1 = (v− − v+ )|Γ = 0. So, we conclude that ξ~ = (a1 , b1 , c1 )T = (0, 0, 0)T . Therefore, we complete the proof of Theorem 3.1.
Corollary 3.3 Assume that f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), and h1 ∈ H −1/2 (∂D), then ¯ to the problem (2.2) and (2b) such that there exists a unique solution u ∈ H 1 (D\Γ) kukH 1 (D\Γ) ¯ ≤ c(kf kH 1/2 (Γ) + kgkH −1/2 (Γ) + kh1 kH −1/2 (∂D) ). Proof From the process for proving Theorem 3.1, the operator A1 is bijective, so, the inverse operator A−1 1 is bounded, and the operators appeared in the right-hand side of (2.21) are bounded as well. The Green representation formula (2.4) and (2.5) now yield the above estimate for the weak solution u. For problem (2.22), because the matrix operator A2 can be treated similarly as A1 , the proving process is just a slight modification of Theorem 3.1, so we omit the special proof and give the result: Theorem 3.4 The boundary integral system (2.22) has a unique solution.
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Corollary 3.5 Assume that f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), and h2 ∈ H −1/2 (∂D), then ¯ to the problem (2.2) and (2c), such that there exists a unique solution u ∈ H 1 (D\Γ) kukH 1 (D\Γ) ¯ ≤ c(kf kH 1/2 (Γ) + kgkH −1/2 (Γ) + kh2 kH −1/2 (∂D) ). Remark 3.6 In this article, we just consider the direct scattering of a point source by a crack buried in a bounded domain D. The inverse problem, that is, the unique determination of the location and shape of the crack from the known Cauchy data on the boundary ∂D, will be our future work. References [1] Ammari H, Bao G, Wood A W. An integral equation method for the electromagnetic scattering from cavities. Math Meth Appl Sci, 2000, 23: 1057–1072 [2] Colton D L, Kress R. Inverse Acoustic and Electromagnetic Scattering Theory. Berlin: Springer-Verlag, 1998 [3] Colton D L, Kress R. Integral Equation Methods in Scattering Theory. Wiley, New York: Springer-Verlag, 1983 [4] Cakoni F, Colton D L. The linear sampling method for cracks. Inverse Problem, 2003, 19: 279–295 [5] Cakoni F, Colton D L. Qualitative Method in Inverse Scattering Theory. Berlin: Springer, 2006 [6] Hsiao G, Wendland W L. On integral equation method for the plane mixed boundary value problem for the Laplacian. Math Methods Appl Sci, 1979: 265–321 [7] Kress R. Acoustic Scattering: Special Theoretical Tools. London: Acamedia, 2001 [8] Kress R. Linear Integral Equations (second ed.). Berlin: Springer-verlag, 1999 [9] Kress R. Frechet differentiability of the far field operator for scattering from a crack. J Inverse Ill-posed Probl, 1995, 3: 305–313 [10] Kirsch A, Ritter R. The linear sampling method for inverse scattering from an open arc. Inverse Problems, 2000, 16: 85–109 [11] Lee K M. Inverse scattering problem for an impedance crack. Wave Motion, 2008, 45: 254–263 [12] Mclean W. Strongly Elliptic Systems and Boundary Integral Equations. Cambridge: Cambridge University Press, 2000 [13] Lions J, Magenes E. Non-homogeneous Boundary Value Problems and Applications. New York: Springer, 1972 [14] Monch L. On the inverse acoustic scattering problem by an open arc: On the sound-hard case. Inverse Problems, 1997, 13: 1379–1392 [15] Peter J, Potthast R. Testing the integrity of some cavity-the Cauchy problem and the range test. Applied Numer Math, 2008, 58: 899–914 [16] Qin H H, Colton D L. The inverse scattering problem for cavities. Applied Numer Math, 2012, 62: 699–708 [17] Qin H H, Colton D L. The inverse scattering problem for cavities with impedance boundary condition. Adv Comput Math, 2012, 36: 157–174 [18] Yan G Z, Ye J G. Boundary integral methods for scattering problems with cracks buried in a piecewise homogeneous medium. Math Meth Appl Sci, 2012, 35: 84–96 [19] Yan G Z, Yao M. Mathematical basis of scattering problems from penetrable obstacles and cracks. J Math Phys, 2010, 51: 123520 [20] Yan G Z, Yao M. The Method of boundary integral equations for a mixed scattering problem. J Differential Equations, 2009, 246: 4618–4631
THE BOUNDARY INTEGRAL METHOD FOR THE HELMHOLTZ EQUATION WITH CRACKS INSIDE A BOUNDED DOMAIN∗
H)
Jun GUO (
School of Mathematics and Statistics, South-Central University for Nationalities, Wuhan 430074, China E-mail : [email protected]
ww)
†
Lili FAN (
School of Mathematics and Computer Science, Wuhan Polytechnic University, Wuhan 430023, China E-mail : [email protected]
îI)
Guozheng YAN (
School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China E-mail : yan [email protected] Abstract We consider a kind of scattering problem by a crack Γ that is buried in a bounded domain D, and we put a point source inside the domain D. This leads to a mixed boundary value problem to the Helmholtz equation in the domain D with a crack Γ. Both sides of the crack Γ are given Dirichlet-impedance boundary conditions, and different boundary condition (Dirichlet, Neumann or Impedance boundary condition) is set on the boundary of D. Applying potential theory, the problem can be reformulated as a system of boundary integral equations. We establish the existence and uniqueness of the solution to the system by using the Fredholm theory. Key words
Boundary integral equation; Helmholtz equation; Fredholm theory
2010 MR Subject Classification
1
35D05; 35P25
Introduction
The direct and inverse scattering problem for cracks was initiated by Kress [9] in 1995. In that article, Kress considered the scattering problem for a perfectly conducting crack and used the integral equation method to solve both the direct and inverse problems for a sound-soft crack. The case of a sound-hard crack was considered by M¨onch [14] in 1997. In 2000, Kirsch and Ritter [10] used the factorization method to reconstruct the shape of the crack from the ∗ Received
Jun 17, 2013; revised Mar 2, 2014. The second author is supported by the grant from the National Natural Science Foundation of China (11301405). The third author is supported by the grants from the National Natural Science Foundation of China (11171127 and 10871080). † Corresponding author.
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knowledge of the far-field pattern as a continued work of Kress. Also in 2000, Ammari [1] considered the electromagnetic scattering problem with cracks. In 2003, Cakoni and Colton [4] discussed the direct and inverse scattering for a open crack coated on one side by a material and determined the shape of the open arc by the linear sampling method. In 2008, Ming-Kuo [11] studied the direct and inverse scattering problem for a crack with impedance boundary condition. The combination of a crack and a bounded impenetrable and penetrable obstacles were discussed by Guozheng Yan and Mao Yao in [20] and [19], respectively. In 2012, Guozheng Yan and Jianguo Ye [18] considered a scattering problem by a crack that is buried in a piecewise homogeneous medium. However, there are a few results about the interior scattering by a crack that is buried in a bounded domain. This means that the waves propagate in a bounded area constituted by a certain substance with a crack inside. In some industrial applications of non-destructive testing, it is important to test the structural integrity of cavities using acoustic or electromagnetic waves emitted and measured by sources and receivers, respectively, placed inside the cavity (see [15– 17]). In this article, we consider a kind of interior scattering problem. Briefly speaking, we consider the scattering problem by a crack Γ that is buried in a bounded domain D, and we put a point source inside the domain D. We take Dirichlet-impedance boundary conditions on two sides of the crack, and set different boundary condition (Dirichlet, Neumann or Impedance boundary condition) on the boundary of D. This leads to a mixed boundary value problem for the Helmholtz equation for a crack buried inside the domain D. The goal of this article is to establish the existence and uniqueness of the solution to this scattering problem using the boundary integral method [8, 18–20] in some suitable function spaces. Note that this is a classical method for solving elliptic equations. In practical applications, it also facilitates numerical implementation. For a detailed description of this method, please refer to the monographs [3, 6, 12]. We can reformulate the problem as a boundary integral system, and show that the corresponding boundary integral operators have the Fredholm property in some suitable Sobolev space. Furthermore, we can establish the existence and uniqueness of the solution to the boundary integral system. The plan of this article is as follows. In the next section, we introduce the scattering problem, establish uniqueness of the solution, and reformulate the problem as a boundary integral system using the potential theory. In Section 3, in some suitable Sobolev spaces, we prove the main results, that is, the existence and uniqueness of the solution to the corresponding boundary integral system.
2
Boundary Integral System
Let D be a bounded and closed subset of R2 (or R3 ) with a C 2 boundary ∂D. For simplicity, we discuss the problem just in R2 space. An arc (or crack) Γ is located inside the domain D. For further consideration, we extend the arc Γ to an arbitrary piecewise smooth, simply connected and closed curve ∂Ω enclosing a bounded domain Ω which is completely contained in the domain D such that the normal vector ν on Γ coincides with the normal vector on ∂Ω (except a finite numbers of points), and directed into the exterior of Ω which we again denote by ν. The scattering of a point source by a crack Γ in the domain D leads a mixed boundary
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value problem for the Helmholtz equation: ∆U + k 2 U = 0, U− = 0,
541
¯ ∪ {z}), in D\(Γ on Γ,
∂U+ + ikλU+ = 0, ∂ν BU = 0,
(2.1) on Γ, on ∂D.
The total wave U = Φ(·, z)+us is the superposition of the given point source Φ(·, z), z ∈ D, and us is the scattered wave. ‘B’ denotes Dirichlet, Neumann, or impedance boundary condition. We assume that the wave number k and impedance coefficient λ are positive, that is, k > 0 ∂u− + ∂u and λ > 0. u+ (u− ) and ∂u ∂ν ( ∂ν ) denote the limit of u and ∂ν on the boundary ∂Ω from the exterior (interior). Remark 2.1 For convenience, in the following discussion, (·)± means the limit approaching the boundary to the corresponding domain from outside and inside, respectively. In order to formulate the above scattering problem more precisely, we need to properly define the trace space on Γ. The classic reference for such trace space is given by Lions and Magenes [13]. Here, we use the notation of Mclean [12]. As we known, L2 (∂Ω), H 1/2 (∂Ω), and H −1/2 (∂Ω) denote the usual Sobolev spaces, then we define the following spaces [12]: L2 (Γ) = {u|Γ : u ∈ L2 (∂Ω)},
H 1/2 (Γ) = {u|Γ : u ∈ H 1/2 (∂Ω)},
˜ 1/2 (Γ) = {u ∈ H 1/2 (∂Ω) : suppu ⊆ Γ}, ¯ H ′ ˜ 1/2 (Γ) , the dual space of H ˜ 1/2 (Γ), H −1/2 (Γ) = H ′ ˜ −1/2 (Γ) = H 1/2 (Γ) , the dual space of H 1/2 (Γ), H
and we have the chain
˜ 1/2 (Γ) ⊂ H 1/2 (Γ) ⊂ L2 (Γ) ⊂ H ˜ −1/2 (Γ) ⊂ H −1/2 (Γ). H For the scattered field us , the problem (2.1) is a special case of the following problem (2.2): Given f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), h ∈ H 1/2 (∂D) (or h1 ∈ H −1/2 (∂D) or h2 ∈ H −1/2 (∂D)) ¯ such that find u ∈ H 1 (D\Γ) 2 ¯ in D\Γ, ∆u + k u = 0, u− = f, on Γ, (2.2) ∂u+ + ikλu+ = g, on Γ, ∂ν Bu = 0, on ∂D, where the boundary condition Bu = 0 on ∂D is one of the following three situations: u − h = 0, ∂u − h1 = 0, ∂ν
on ∂D, on ∂D,
∂u + ikλ2 u − h2 = 0 ∂ν with λ2 < 0.
on ∂D,
(2a) (2b) (2c)
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Remark 2.2 The problem (2.1) is a special case of the problem (2.2) with f = −Φ(·, z), ∂Φ(·,z) g = − ∂Φ(·,z) − ikλΦ(·, z), h = −Φ(·, z), h1 = − ∂Φ(·,z) − ikλ2 Φ(·, z). ∂ν ∂ν , h2 = − ∂ν Firstly we show that the uniqueness of the solution to the problem (2.2). Theorem 2.3 The problem (2.2) has at most one solution. Proof let u be a solution to the homogeneous problem (2.2), that is, u satisfies the ¯ problem (2.2) with f = g = h (or h1 , or h2 ) = 0. Obviously, this solution u ∈ H 1 (D\Γ) ¯ satisfies the following transmission boundary condition on the complementary part ∂Ω\Γ of ∂Ω: u+ = u− , (2.3) ∂u+ = ∂u− . ∂ν ∂ν ¯ and Ω yields An application of the Green formula for u and u ¯ in D\Ω Z Z Z ∂u ¯ ∂u ∂u ¯+ ∂u+ ds − u+ ds, (u△¯ u−u ¯△u)dx = u − u¯ −u ¯+ ∂ν ∂ν ∂ν ∂ν ¯ D\Ω ∂Ω ∂D and Z
(u△¯ u−u ¯△u)dx =
Ω
Z
∂u ¯− ∂u− u− − u¯− ds. ∂ν ∂ν ∂Ω
By using the boundary conditions in (2.2) and (2.3), the sum of the above two equations is Z −2ikλ |u+ |2 ds = 0 Γ
for (2.2) and (2a), or −2ikλ
Z
|u+ |2 ds = 0
Γ
for (2.2) and (2b), or 2ikλ2
Z
2
|u| ds − 2ikλ
∂D
Z
|u+ |2 ds = 0
Γ
for (2.2) and (2c). For any situation we can conclude u+ = 0 on Γ. From the boundary + condition in (2.2), we have ∂u ∂ν = 0 on Γ. The Holmgren’s uniqueness theorem [7] means that ¯ The transmission boundary condition of (2.3) implies u− = 0 and ∂u− = 0 on u = 0 in D\Ω. ∂ν ¯ then we again obtain u = 0 in Ω by the Holmgren’s uniqueness theorem. So, we show ∂Ω\Γ, ¯ and this completes the proof of the theorem. that u = 0 in D\Γ We are now ready to prove the existence of a solution to the problem (2.2) by using integral equation approaching. By Green representation formula, we have Z Z ∂u ∂Φ(x, y) ∂u ∂Φ(x, y) u(x) = Φ(x, y) − u dsy − Φ(x, y) − u dsy (2.4) ∂ν ∂ν ∂D ∂ν ∂Ω ∂ν
¯ and for x ∈ D\Ω,
u(x) =
Z
∂Ω
∂u ∂Φ(x, y) Φ(x, y) − u dsy ∂ν ∂ν
for x ∈ Ω, where Φ(x, y) =
i (1) H (k|x − y|) 4 0
(2.5)
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(1)
is the fundamental solution to the Helmholtz equation in R2 and H0 is a Hankel function of the first kind of order zero. By making use of the known jump relations of single-and doublelayer potentials across the boundary ∂Ω (see [3]) and approaching the boundary ∂Ω from inside Ω, that is, x ∈ Ω and x → ∂Ω, we obtain u− (x) = SΩΩ
∂u− − KΩΩ u− , ∂ν
∂u− (x) ′ ∂u− = KΩΩ − TΩΩ u− . ∂ν ∂ν ¯ we first consider approaching the boundary ∂Ω, then we have In the domain D\Ω, u+ (x) = SDΩ
∂u+ ∂u − KDΩ u − SΩΩ + KΩΩ u+ , ∂ν ∂ν
∂u+ (x) ∂u ′ ′ ∂u+ = KDΩ − TDΩ u − KΩΩ + TΩΩ u+ . ∂ν ∂ν ∂ν ¯ we consider approaching the boundary ∂D, and obtain Again in the domain D\Ω, u− (x) = SDD
∂u ∂u+ − KDD u − SΩD + KΩD u+ , ∂ν ∂ν
∂u− ∂u ∂u+ ′ ′ = KDD − TDD u − KΩD + TΩD u+ , ∂ν ∂ν ∂ν ′ where SΩΩ , KΩΩ , KΩΩ , and TΩΩ are boundary integral operators defined by Z SΩΩ ϕ(x) = 2 ϕ(y)Φ(x, y)dsy , x ∈ ∂Ω; Z∂Ω Φ(x, y) dsy , x ∈ ∂Ω; KΩΩ ϕ(x) = 2 ϕ(y) ∂νy ∂Ω Z ∂Φ(x, y) ′ KΩΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; ∂νx ∂Ω Z ∂ ∂Φ(x, y) TΩΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω. ∂νx ∂Ω ∂νy
(2.6) (2.7)
(2.8) (2.9)
(2.10) (2.11)
′ ′ The operators SDD , KDD , KDD , and TDD have the similar meaning as SΩΩ , KΩΩ , KΩΩ , and ′ TΩΩ , respectively. The operators SDΩ , KDΩ , KDΩ , and TDΩ are given by Z SDΩ ϕ(x) = 2 ϕ(y)Φ(x, y)dsy , x ∈ ∂Ω; ∂D Z Φ(x, y) KDΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; ∂νy ∂D Z ∂Φ(x, y) ′ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω; KDΩ ∂νx ∂D Z ∂ ∂Φ(x, y) TDΩ ϕ(x) = 2 ϕ(y) dsy , x ∈ ∂Ω. ∂νx ∂D ∂νy
However, the operator SΩD is the operator applied a function with supp⊆ ∂Ω and evaluated ′ on ∂D, with analogous definitions for operators such as KΩD , KΩD , TΩD , and the following ′ upcoming operators such as SΓD ,KΓD , KΓD , TΓD and so on. Restrict formula (2.6) on the arc Γ and use formula (2.8), we have ∂u− ∂u+ ∂u 2f = SΩΓ − − KΩΓ (u− − u+ ) + (u− − u+ )|Γ + SDΓ − KDΓ u. (2.12) ∂ν ∂ν ∂ν
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Using (2.8) and (2.9), we restrict ∂u ∂ν + ikλu on the arc Γ and obtain ∂u+ g= + ikλu+ |Γ ∂ν ∂u+ ∂u+ ∂u− ∂u− ′ = KΩΓ − − TΩΓ (u− − u+ ) + ikλSΩΓ − ∂ν ∂ν ∂ν ∂ν ∂u ∂u − − ′ + TΩΓ u− − ikλSΩΓ − ikλKΩΓ (u− − u+ ) − KΩΓ ∂ν ∂ν ∂u ∂u ′ + ikλKΩΓ u− + KDΓ − TDΓ u + ikλSDΓ − ikλKDΓ u. ∂ν ∂ν Combining (2.6) and (2.7), we have ∂u− ′ ∂u− −ikλ SΩΓ − KΩΓ u− − KΩΓ + TΩΓ u− ∂ν ∂ν ∂u− = − ikλu− − |Γ ∂ν ∂u+ ∂u+ ∂u− − |Γ − ikλu+ + |Γ . = −ikλ(u− − u+ )|Γ − ∂ν ∂ν ∂ν Using (2.14), we can rewrite (2.13) as ∂u− ∂u+ ∂u− ∂u+ ′ 2g = KΩΓ − − TΩΓ (u− − u+ ) + ikλSΩΓ − ∂ν ∂ν ∂ν ∂ν ∂u− ∂u+ −ikλKΩΓ (u− − u+ ) − ikλ(u− − u+ )|Γ − − |Γ ∂ν ∂ν ∂u ′ ∂u +KDΓ − TDΓ u + ikλSDΓ − ikλKDΓ u. ∂ν ∂ν From (2.12) and the boundary conditions in (2.2), we can reformulate (2.15) as ∂u+ ∂u− ′ 2g = KΩΓ − − TΩΓ (u− − u+ ) − 2ikλ(u− − u+ )|Γ ∂ν ∂ν ∂u− ∂u+ ′ ∂u − − |Γ + KDΓ − TDΓ u + 2f. ∂ν ∂ν ∂ν
(2.13)
(2.14)
(2.15)
(2.16)
Using the Green formula, we have Z
∂Ω
(u− (y)
−u(x),
∂Φ(x, y) ∂u− (y) − Φ(x, y))dsy = 0, ∂ν ∂ν
in
Ω,
in
¯ D\Ω.
(2.17)
From (2.10), upon using (2.17), we obtain
∂u ∂u− ∂u+ − KDD u + SΩD − ∂ν ∂ν ∂ν ∂u− −KΩD (u− − u+ ) − SΩD + KΩD u− ∂ν ∂u− ∂u+ ∂u − − KΩD (u− − u+ ) + SDD − KDD u. = SΩD ∂ν ∂ν ∂ν
u(x)|∂D = SDD
Similarly, we can rewrite (2.11) as ∂u ∂u− ∂u+ ∂u ′ ′ |∂D = KΩD − − TΩD (u− − u+ ) + KDD − TDD u. ∂ν ∂ν ∂ν ∂ν
(2.18)
(2.19)
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Now, for the problem (2.2) and (2a), we define ∂u+ ∂u ∂u− |∂D = a, − |Γ = b, ∂ν ∂ν ∂ν
545
(u− − u+ )|Γ = c,
and extend b and c by zero to the whole ∂Ω in the following 0, on ∂Ω\Γ, 0, on ∂Ω\Γ, ˜b = c˜ = b, on Γ, c, on Γ, ˜ −1/2 (Γ), c ∈ H ˜ 1/2 (Γ) (see [4]). then b ∈ H Letting r = h + KDD h,
p = 2f + KDΓ h,
q = 2g − 2f + TDΓ h,
and
SDD
SΓD
−KΓD
A= SDΓ
SΓΓ
I − KΓΓ
′ KΓΓ −I
−(TΓΓ + 2ikλI)
′ KDΓ
,
and combining (2.12), (2.16), and (2.18), we have a boundary integral system r(x) a A b = p(x) . q(x) c For the problem (2.2) and (2b), we define ∂u− ∂u+ u|∂D = a1 , − |Γ = b1 , ∂ν ∂ν
(2.20)
(u− − u+ )|Γ = c1 ,
˜ −1/2 (Γ), and ˜b1 and c˜1 are zero extension of b1 and c1 to the whole ∂Ω, respectively, then b1 ∈ H 1/2 ˜ c1 ∈ H (Γ). Denote ′ r1 = h1 − KDD h1 ,
p1 = 2f − SDΓ h1 ,
′ q1 = 2g − 2f − KDΓ h1 ,
and
−TDD
′ KΓD
A1 = SΓΓ −KDΓ ′ −TDΓ KΓΓ −I
−TΓD I − KΓΓ −(TΓΓ + 2ikλI)
,
then combining (2.12), (2.16), and (2.19), we have a boundary integral system a1 r1 (x) A1 b1 = p1 (x) . c1 q1 (x) Similarly, for the problem (2.2) and (2c), we define ∂u− ∂u+ u|∂D = a2 , − |Γ = b2 , (u− − u+ )|Γ = c2 , ∂ν ∂ν
(2.21)
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˜ −1/2 (Γ), and ˜b2 and c˜2 are zero extension of b2 and c2 to the whole ∂Ω, respectively, then b2 ∈ H ˜ 1/2 (Γ). c2 ∈ H ′ ′ Let r2 = h2 − KDD h2 − ikλ2 SDD h2 , p2 = 2f − SDΓ h2 , q2 = 2g − 2f − KDΓ h2 , and ′ ′ −TDD − ikλ2 (KDD + KDD ) + k 2 λ22 SDD KΓD + ikλ2 SΓD −TΓD − ikλ2 KΓD , A2 = −KDΓ − ikλ2 SDΓ SΓΓ I − KΓΓ ′ ′ −TDΓ − ikλ2 KDΓ KΓΓ −I −(TΓΓ + 2ikλI) then combining (2.12), (2.16), and (2.19), we have a boundary integral system a2 r2 (x) = p2 (x) . A2 b 2 c2 q2 (x)
(2.22)
To the operators appeared in A, A1 , and A2 , we list their mapping properties in the following (see [5]): ˜ −1/2 (Γ) → H 1/2 (Γ), SΓΓ : H ′ ˜ −1/2 (Γ) → H −1/2 (Γ), KΓΓ :H
˜ 1/2 (Γ) → H 1/2 (Γ), KΓΓ : H ˜ 1/2 (Γ) → H −1/2 (Γ); TΓΓ : H
SDD : H −1/2 (∂D) → H 1/2 (∂D),
KDD : H 1/2 (∂D) → H 1/2 (∂D),
′ KDD : H −1/2 (∂D) → H −1/2 (∂D), ˜ −1/2 (Γ) → H 1/2 (∂D), SΓD : H
TDD : H 1/2 (∂D) → H −1/2 (∂D); ˜ 1/2 (Γ) → H 1/2 (∂D), KΓD : H
′ ˜ −1/2 (Γ) → H −1/2 (∂D), KΓD :H
˜ 1/2 (Γ) → H −1/2 (∂D); TΓD : H
SDΓ : H −1/2 (∂D) → H 1/2 (Γ),
KDΓ : H 1/2 (∂D) → H 1/2 (Γ),
′ KDΓ : H −1/2 (∂D) → H −1/2 (Γ),
TDΓ : H 1/2 (∂D) → H −1/2 (Γ).
If we define ˜ −1/2 (Γ) × H ˜ 1/2 (Γ) H = H −1/2 (∂D) × H and the dual space H ∗ = H 1/2 (∂D) × H 1/2 (Γ) × H −1/2 (Γ), then the operator A maps H continuously into H ∗ . If we define ˜ −1/2 (Γ) × H ˜ 1/2 (Γ) W = H 1/2 (∂D) × H and the dual space W ∗ = H −1/2 (∂D) × H 1/2 (Γ) × H −1/2 (Γ), then the operators A1 and A2 map W continuously into W ∗ . Remark 2.4 Once the unknown Cauchy data are determined from (2.20) (or (2.21), or (2.22)), then the representation formula (2.4) and (2.5) determine the corresponding weak solution to the problem (2.2).
3
Existence and Uniqueness of the Solution
If the integral system (2.20), (2.21), and (2.22) has a unique solution, our problem (2.2) will has a unique solution (see [2]). In this section, we will show that the matrix operators A, A1 ,
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and A2 are Fredholm with index zero, and they are injective operators, that is, kernA = {0}, kernA1 = {0}, and kernA2 = {0}. Then, using Theorem 2.3, the solvability of the problem (2.20), (2.21), and (2.22) is true. Refer to [20], it is easy to obtain the solvability of the boundary integral system (2.20), so, we omit its proof here. In the following, we will focus on the solvability of the problem (2.21). Theorem 3.1 The boundary integral system (2.21) has a unique solution. Proof There are two main steps to prove the theorem: Step 1 is to show that the operator A1 is Fredholm with index zero; Step 2 is to prove that KernA1 = {0}. Step 1: In fact, the operators −TDD , −TΩΩ , and SΩΩ are positive and bounded up to a compact perturbation [12], and we denote by LD , LΩ , and LS the compact operators LD : H 1/2 (∂D) → H −1/2 (∂D), LΩ : H 1/2 (∂Ω) → H −1/2 (∂Ω), LS : H −1/2 (∂Ω) → H 1/2 (∂Ω), such that ¯ ≥ C||ψ||2 1/2 Re(h−(TDD + LD )ψ, ψi) H (∂D) ,
for ψ ∈ H 1/2 (∂D),
(3.1)
¯ ≥ C||ψ||2 1/2 Re(h−(TΩΩ + LΩ )ψ, ψi) H (∂Ω) ,
for ψ ∈ H 1/2 (∂Ω),
(3.2)
¯ ≥ C||ψ||2 −1/2 Re(h(SΩΩ + LS )ψ, ψi) H (∂Ω) ,
for ψ ∈ H −1/2 (∂Ω),
(3.3)
where h , i denote the duality between H −1/2 (∂D) and H 1/2 (∂D), or H −1/2 (∂Ω) and H 1/2 (∂Ω). ′ Let K0 and K0′ be the operators defined as KΩΩ and KΩΩ , respectively, with kernel Φ(x, y) 1 ′ replaced by Φ0 (x, y) = − 2π ln |x − y|. Then, Kc = KΩΩ − K0 and Kc′ = KΩΩ − K0′ are compact because they have continuous kernels [8]. It is easy to show that K0 and K0′ are adjoint because their kernels are real, that is, (K0 ψ, φ) = (ψ, K0′ φ)
for ψ ∈ H 1/2 (∂Ω) and φ ∈ H −1/2 (∂Ω).
We define
−TDD
′ KΩD
−TΩD
A′1 = −KDΩ −TDΩ
SΩΩ
I − KΩΩ
′ KΩΩ −I
−(TΩΩ + 2ikλI)
,
then A′1 : H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω) → H −1/2 (∂D) × H 1/2 (∂Ω) × H −1/2 (∂Ω). Note that ˜b1 ∈ H −1/2 (∂Ω), c˜1 ∈ H 1/2 (∂Ω), so, we can decompose A′1 into two parts, that is, a1 a1 a1 ′ ˜ ˜ A1 b1 = A10 b1 + A1c ˜b1 (3.4) c˜1 c˜1 c˜1
with
a1 −(TDD + LD )a1 ˜ A10 (SΩΩ + LS )˜b1 + (I − K0 )˜ c1 b1 = (K0′ − I)˜b1 − (TΩΩ + LΩ )˜ c˜1 c1 − 2ikλ˜ c1
(3.5)
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and
′ ˜ a1 LD a1 + KΩD b1 − TΩD c˜1 ˜ ˜ A1c b1 = −KDΩ a1 − LS b1 − Kc c˜1 , c˜1 −TDΩ a1 + Kc′ ˜b1 + LΩ c˜1
(3.6)
where A1c : H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω) → H −1/2 (∂D) × H 1/2 (∂Ω) × H −1/2 (∂Ω) is compact. Furthermore, we have (A10 ξ~′ , ξ~′ ) = (−(TDD + LD )a1 , a1 ) + ((SΩΩ + LS )˜b1 , ˜b1 ) + ((I − K0 )˜ c1 , ˜b1 ) ′ +((K − I)˜b1 , c˜1 ) + (−(TΩΩ + LΩ )˜ c1 , c˜1 ) − (2ikλ˜ c1 , c˜1 ).
(3.7)
0
Here, (u, v) denotes the scalar product on L2 (∂D) or L2 (∂Ω) defined by and ξ~′ = (a1 , ˜b1 , c˜1 )T .
R
∂D
u¯ v ds or
R
∂Ω
u¯ v ds,
From (3.1), (3.2), and (3.3), we obtain Re[(−(TDD + LD )a1 , a1 ) + ((SΩΩ + LS )˜b1 , ˜b1 ) + (−(TΩΩ + LΩ )˜ c1 , c˜1 )] ≥ c||a1 ||2 1/2 + c||˜b1 ||2 −1/2 + c||˜ c1 ||2 1/2 . H
(∂D)
H
(∂Ω)
H
(∂Ω)
(3.8)
Furthermore, as K0 and K0′ are adjoint, we have Re[((I − K0 )˜ c1 , ˜b1 ) + ((K0′ − I)˜b1 , c˜1 )] = Re[(˜ c1 , (I − K ′ )˜b1 ) + ((K ′ − I)˜b1 , c˜1 )] 0
= Re[(˜ c1 , (I −
0
K0′ )˜b1 )
c1 , (K0′ − I)˜b1 ] = 0. + (˜
(3.9)
It is obviously Re[−2ikλ(˜ c1 , c˜1 )] = 0.
(3.10)
Combining (3.8), (3.9), and (3.10), we obtain the fact that the sesquilinear form defined by A10 is coercive, that is, ¯′ Re(hA10 ξ~′ , ~ξi) ≥ C||ξ~′ ||2 ,
for
ξ~′ ∈ H 1/2 (∂D) × H −1/2 (∂Ω) × H 1/2 (∂Ω).
˜ −1/2 (Γ), and c˜1 is the extension by zero of Recalling that ˜b1 is the extension by zero of b1 ∈ H ˜ 1/2 (Γ), so we can rewrite the above inequality as c1 ∈ H ~ ξi) ~¯ ≥ C||ξ|| ~ 2, Re(hA10,Γ ξ,
˜ −1/2 (Γ) × H ˜ 1/2 (Γ), for ξ~ ∈ H 1/2 (∂D) × H
where A10,Γ is the restriction to Γ of A10 defined for ξ~ = (a1 , b1 , c1 )T ∈ W . The corresponding restriction A1c,Γ : W → W ∗ of A1c clearly remains compact. Whence the operator A1 is Fredholm with index zero. Step 2: In this part, we show that KernA1 = {0}. For simplicity, we still use ξ~ = (a1 , b1 , c1 )T to denote a solution of the homogeneous equation A1 ξ~ = ~0, which means that ′ −TDD a1 + KΓD b1 − TΓD c1 = 0, (3.11) −KDΓ a1 + SΓΓ b1 + (I − KΓΓ )c1 = 0, ′ −TDΓ a1 + (KΓΓ − I)b1 − (TΓΓ + 2ikλI)c1 = 0,
then we need to prove that ξ~ = ~0.
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Remark 3.2 For convenience, here we used the same notation ξ~ = (a1 , b1 , c1 )T , but with different meaning. For this purpose, we define potential v(x) = −KD a1 + SΩ˜b1 − KΩ c˜1 ,
¯ x ∈ R2 \(∂D ∪ Γ),
(3.12)
where KD ϕ(x) =
Z
ϕ(y)
∂D
SΩ ϕ(x) =
∂Φ(x, y) dsy , ∂ν(y)
Z
ϕ(y)Φ(x, y)dsy ,
Z
ϕ(y)
∂Ω
KΩ ϕ(x) =
∂Ω
∂Φ(x, y) dsy . ∂ν(y)
¯ By the jump relationships of This potential v(x) satisfies Helmholtz equation in D\Γ. ¯ letting x → ∂D, single-and double-layer potentials on the boundary ∂D, in the domainD\Ω, we have ∂v− (x) ′ = −TDD a1 + KΓD b1 − TΓD c1 , ∂ν
(3.13)
then the first equation in (3.11) implies that ∂v− (x) |∂D = 0. ∂ν
(3.14)
Letting x → Γ, we have 2v+ (x) = −KDΓ a1 + SΓΓD b1 − KΓΓ c1 − c1 , 2
∂v+ (x) ′ = −TDΓ a1 + KΓΓ b1 − TΓΓ c1 − b1 . ∂ν
So, 2
∂v+ + ikλv+ ∂ν
= ikλ(−KDΓ a1 + SΓΓD b1 − KΓΓ c1 + c1 ) ′ −TDΓ a1 + KΓΓ b1 − b1 − TΓΓ c1 − 2ikλc1 ,
then combining the second and third equation in (3.11), we have ∂v+ + ikλv+ |Γ = 0. ∂ν
(3.15)
In the domain Ω, letting x → Γ, we obtain 2v− (x) = −KDΓ a1 + SΓΓ b1 − KΓΓ c1 + c1 , then the second equation in (3.11) implies that v− (x)|Γ = 0.
(3.16)
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From (3.14), (3.15), and (3.16), the potential v(x) satisfies the following problem ¯ ∆v + k 2 v = 0, in D\Γ, on Γ, v− = 0, ∂v+ + ikλv+ = 0, on Γ, ∂ν ∂v− (x) = 0, on ∂D. ∂ν
(3.17)
¯ It is not difficult to check that the From Theorem 2.3, we know that v(x) = 0 in D\Γ. potential v(x) also satisfies the following exterior scattering problem ¯ ∆v + k 2 v = 0, in R2 \D, (3.18) ∂v+ = 0, in ∂D, ∂ν and v(x) satisfies the Sommerfield radiation condition √ ∂v lim r − ikv = 0 r→∞ ∂r
uniformly in xˆ = x/|x| with r = |x|. ¯ The jump relationships of single From Theorem 3.13 in [3], we have v(x) = 0 in R2 \D. and double layer potential imply that a1 = (v− − v+ )|∂D = 0, ∂v− ∂v+ b1 = − |Γ = 0, ∂ν ∂ν and c1 = (v− − v+ )|Γ = 0. So, we conclude that ξ~ = (a1 , b1 , c1 )T = (0, 0, 0)T . Therefore, we complete the proof of Theorem 3.1.
Corollary 3.3 Assume that f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), and h1 ∈ H −1/2 (∂D), then ¯ to the problem (2.2) and (2b) such that there exists a unique solution u ∈ H 1 (D\Γ) kukH 1 (D\Γ) ¯ ≤ c(kf kH 1/2 (Γ) + kgkH −1/2 (Γ) + kh1 kH −1/2 (∂D) ). Proof From the process for proving Theorem 3.1, the operator A1 is bijective, so, the inverse operator A−1 1 is bounded, and the operators appeared in the right-hand side of (2.21) are bounded as well. The Green representation formula (2.4) and (2.5) now yield the above estimate for the weak solution u. For problem (2.22), because the matrix operator A2 can be treated similarly as A1 , the proving process is just a slight modification of Theorem 3.1, so we omit the special proof and give the result: Theorem 3.4 The boundary integral system (2.22) has a unique solution.
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Corollary 3.5 Assume that f ∈ H 1/2 (Γ), g ∈ H −1/2 (Γ), and h2 ∈ H −1/2 (∂D), then ¯ to the problem (2.2) and (2c), such that there exists a unique solution u ∈ H 1 (D\Γ) kukH 1 (D\Γ) ¯ ≤ c(kf kH 1/2 (Γ) + kgkH −1/2 (Γ) + kh2 kH −1/2 (∂D) ). Remark 3.6 In this article, we just consider the direct scattering of a point source by a crack buried in a bounded domain D. The inverse problem, that is, the unique determination of the location and shape of the crack from the known Cauchy data on the boundary ∂D, will be our future work. References [1] Ammari H, Bao G, Wood A W. An integral equation method for the electromagnetic scattering from cavities. Math Meth Appl Sci, 2000, 23: 1057–1072 [2] Colton D L, Kress R. Inverse Acoustic and Electromagnetic Scattering Theory. Berlin: Springer-Verlag, 1998 [3] Colton D L, Kress R. Integral Equation Methods in Scattering Theory. Wiley, New York: Springer-Verlag, 1983 [4] Cakoni F, Colton D L. The linear sampling method for cracks. Inverse Problem, 2003, 19: 279–295 [5] Cakoni F, Colton D L. Qualitative Method in Inverse Scattering Theory. Berlin: Springer, 2006 [6] Hsiao G, Wendland W L. On integral equation method for the plane mixed boundary value problem for the Laplacian. Math Methods Appl Sci, 1979: 265–321 [7] Kress R. Acoustic Scattering: Special Theoretical Tools. London: Acamedia, 2001 [8] Kress R. Linear Integral Equations (second ed.). Berlin: Springer-verlag, 1999 [9] Kress R. Frechet differentiability of the far field operator for scattering from a crack. J Inverse Ill-posed Probl, 1995, 3: 305–313 [10] Kirsch A, Ritter R. The linear sampling method for inverse scattering from an open arc. Inverse Problems, 2000, 16: 85–109 [11] Lee K M. Inverse scattering problem for an impedance crack. Wave Motion, 2008, 45: 254–263 [12] Mclean W. Strongly Elliptic Systems and Boundary Integral Equations. Cambridge: Cambridge University Press, 2000 [13] Lions J, Magenes E. Non-homogeneous Boundary Value Problems and Applications. New York: Springer, 1972 [14] Monch L. On the inverse acoustic scattering problem by an open arc: On the sound-hard case. Inverse Problems, 1997, 13: 1379–1392 [15] Peter J, Potthast R. Testing the integrity of some cavity-the Cauchy problem and the range test. Applied Numer Math, 2008, 58: 899–914 [16] Qin H H, Colton D L. The inverse scattering problem for cavities. Applied Numer Math, 2012, 62: 699–708 [17] Qin H H, Colton D L. The inverse scattering problem for cavities with impedance boundary condition. Adv Comput Math, 2012, 36: 157–174 [18] Yan G Z, Ye J G. Boundary integral methods for scattering problems with cracks buried in a piecewise homogeneous medium. Math Meth Appl Sci, 2012, 35: 84–96 [19] Yan G Z, Yao M. Mathematical basis of scattering problems from penetrable obstacles and cracks. J Math Phys, 2010, 51: 123520 [20] Yan G Z, Yao M. The Method of boundary integral equations for a mixed scattering problem. J Differential Equations, 2009, 246: 4618–4631