Mathematical basis of scattering problems from penetrable obstacles and cracks

Mathematical basis of scattering problems from penetrable obstacles andcracksGuozheng Yan and Mao Yao Citation: J. Math. Phys. 51, 123520 (2010); doi: 10.1063/1.3525831 View online: http://dx.doi.org/10.1063/1.3525831 View Table of Contents: http://jmp.aip.org/resource/1/JMAPAQ/v51/i12 Published by the AIP Publishing LLC. Additional information on J. Math. Phys.Journal Homepage: http://jmp.aip.org/ Journal Information: http://jmp.aip.org/about/about_the_journal Top downloads: http://jmp.aip.org/features/most_downloaded Information for Authors: http://jmp.aip.org/authors

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JOURNAL OF MATHEMATICAL PHYSICS 51, 123520 (2010)

Mathematical basis of scattering problems from penetrableobstacles and cracks

Guozheng Yan1,a) and Mao Yao2

1Faculty of Mathematics and Statistics, Central China Normal University, Wuhan 430079,People’s Republic China2Faculty of Science, Wuhan University of Science and Enginering, Wuhan 430074, People’sRepublic China

(Received 26 March 2010; accepted 18 November 2010; published online 29 December 2010)

We consider a kind of scattering problem which models the scattering of an electro-magnetic time-harmonic plane wave by an infinite cylinder having an open arc and apenetrable obstacle in R2 as cross section. To this end, we solve a scattering problemfor the Helmholtz equation in R2 where the scattering object is a combination of acrack � and a penetrable obstacle D, and we have Dirichlet-Impedance type bound-ary condition on � and transmission boundary condition on ∂ D. Applying potentialtheory, the problem can be reformulated as a boundary integral system. We establishthe existence and uniqueness of a weak solution to the system by using the modifiedFredholm theory. C© 2010 American Institute of Physics. [doi:10.1063/1.3525831]

I. INTRODUCTION

In this paper we consider the scattering of an electromagnetic time-harmonic plane wave by aninfinite cylinder having a mixture of an open arc � and a penetrable obstacle D (∂ D ∈ C2) in R2 ascross section. We assume that the crack is coated on one side by a material with surface impedanceμ. This corresponds to the situation when the boundary or more generally a portion of the boundaryis coated with an unknown material in order to avoid detection. Assuming that the electric field ispolarized in the T M mode (see, Refs. 5 and 6), this leads to a mixed boundary value problem forthe Helmholtz equation defined in the exterior of an open arc and a penetrable obstacle in R2.

The direct and inverse scattering problems for cracks were initiated by Kress in 1995.8 In thepaper, Kress considered the direct and inverse scattering problem for a perfectly conducting crackand used the integral equation method to solve both the direct and inverse problems for a sound-softcrack. The scattering problem in the unbounded domain is thus converted into a boundary integralequation. In 1997, Monch10 extended this approach to a Neumann crack. Later in 2000, Kress’swork was continued by Kirsch and Ritter in Ref. 7 who used the factorization method to reconstructthe shape of the crack from the knowledge of the far-field pattern, and in the same year these resultswere generalized to the scattering problem with cracks for Maxwell equations in Ref. 1 by Ammariet al. In 2003, using integral equation method Cakoni and Colton in Ref. 3 considered the directscattering problems for cracks (possibly) coated on one side by a material with surface impedanceλ and recovering the open arc by the linear sampling method. Extending to the impedance problem,Ming-Kuo9 discussed the direct and inverse scattering problem for an impedance crack in 2008.However, studying an inverse problem (see Refs. 2–4, 6, and 7) always requires a solid knowledgeof the corresponding direct problem. Therefore, in this paper we just consider the direct scatteringproblem for a mixture of an open arc � and a penetrable obstacle D (∂ D ∈ C2) in R2.

The aim of this study is to establish the existence and uniqueness of a solution to this directscattering problem by using the method of boundary integral equations (see, Refs. 5, 6, and 12 and

a)Electronic mail: [email protected].

0022-2488/2010/51(12)/123520/17/$30.00 C©2010 American Institute of Physics51, 123520-1


http://dx.doi.org/10.1063/1.3525831

http://dx.doi.org/10.1063/1.3525831

http://dx.doi.org/10.1063/1.3525831

123520-2 G. Yan and M. Yao J. Math. Phys. 51, 123520 (2010)

......................................................................

..........

DΩui

u

Γ

FIG. 1. Scattering from penetrable obstracles and cracks.

the reference therein). The difficult thing is to show the boundary integral operator A (in Sec. II)is a Fredholm operator with index zero since the boundary is a combination of an open arc and apenetrable closed curve and we have a complicated mixed boundary conditions.

This paper is organized as follows. In Sec. II, we will introduce the direct scattering problem,establish uniqueness to the problem and reformulate the problem as a boundary integral system byusing single- and double-layer potentials. In Sec. III, we will show the main results on the existenceand uniqueness of the solution to the boundary integral system by using the potential theory andFredholm theory.

II. BOUNDARY INTEGRAL EQUATIONS

Consider the scattering of time-harmonic electromagnetic plane waves from an infinite cylinderhaving an open arc � (smooth) and a penetrable obstacle D (∂ D ∈ C2) in R2 as cross section. Forfurther considerations we extend the arc � to an arbitrary smooth, simply connected, closed curve∂� (∂� ∈ C2) enclosing a bounded domain � such that the normal vector ν on � coincides withthe normal vector on ∂�. Notice that ν is the unit normal vector defined almost everywhere on ∂�

and ∂ D (except a finite number of points) and directed into the exterior of � and D. We assume thatthe domain D is completely contained in �, i.e., D ⊂ � and ∂ D ∩ ∂� = ∅. Throughout this paper,we shall assume that ∂ D ∈ C2 and ∂� ∈ C2.

The arc � can be parameterized as

� = {z(s) : s ∈ [s0, s1]},where z : [s0, s1] → R2 is an injective piecewise C1 function. We denote the outside of � withrespect to the chosen orientation by �+ and inside by �− (see, Fig. 1).

Dirichlet-Impedance type boundary condition on � and transmission boundary conditions on∂ D lead to the following problem:

Given g ∈ H 1/2(�), h ∈ H−1/2(�), f1 ∈ H 1/2(∂ D), and f2 ∈ H−1/2(∂ D), find u ∈ H 1(D) ∪H 1

loc(R2\(D ∪ �)), such that⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

�u + k20u = 0 in D

�u + k2u = 0 in R2\(D ∪ �)u− = g on �

∂u+∂ν

+ ikμu+ = h on �

λu+ − λ0u− = f1 on ∂ D∂u+∂ν

− ∂u−∂ν

= f2 on ∂ D

(1)

and the scattered field u is required to satisfy the Sommerfeld radiation condition,

limr→∞

√r (

∂u

∂r− iku) = 0, (2)

uniformly in x = x/|x | with r = |x | (see, Refs. 5 and 6).We suppose that the wavenumbers k0, k, coefficients μ, λ, and λ0 are positive. Notice that

u±(x) = limh→0+

u(x ± hν) for x ∈ � (or x ∈ ∂ D) and ∂u±∂ν

= limh→0+

ν · ∇u(x ± hν) for x ∈ � (or

x ∈ ∂ D ). Here H 1loc(R2\(D ∪ �)), H 1(D), H 1/2(�), H−1/2(�), H 1/2(∂ D), and H−1/2(∂ D) are the

usual Sobolev spaces and trace spaces.


123520-3 Mathematical basis of scattering problems J. Math. Phys. 51, 123520 (2010)

We define the following spaces (see, Ref. 3):

H 1/2(�) = {u|� : u ∈ H 1/2(∂�)}, (3)

H 1/2(�) = {u ∈ H 1/2(�) : suppu ⊆ �}, (4)

H−1/2(�) = (H 1/2(�)

)′, the dual space of H 1/2(�), (5)

H−1/2(�) = (H 1/2(�)

)′, the dual space of H 1/2(�), (6)

and we have the chain

H 1/2(�) ⊂ H 1/2(�) ⊂ L2(�) ⊂ H−1/2(�) ⊂ H−1/2(�).

First we establish uniqueness for the direct scattering problem.

Theorem 2.1: The problem (1) has at most one solution.

Proof: Denote by BR a sufficiently large ball with radius R containing � and by ∂ BR itsboundary. Let u be a solution to the homogeneous problem (1), i.e., u satisfies the problem (1) withg = h = f1 = f2 = 0. We want to show that u = 0.

It is easy to check that this solution u ∈ H 1(D) ∪ H 1(BR\(D ∪ �)) satisfies the followingtransmission conditions on the complementary part ∂�\� of ∂�:{

u+ = u−∂u+∂ν

= ∂u−∂ν

, (7)

where “±” denote the limit approaching ∂� from outside and inside �, respectively.Applying Green’s formula for u and u in �\D and BR\�, we have∫

�\D(u u + ∇u · ∇u)dx

=∫

∂�\�u−

∂ u−∂ν

ds −∫

∂ Du+

∂ u+∂ν

ds (8)

(where on ∂ D we change the direction of ν from inside D to the exterior of D) and∫BR\�

(u u + ∇u · ∇u)dx

=∫

�

u+∂ u+∂ν

ds +∫

∂�\�u+

∂ u+∂ν

ds +∫

∂ BR

u∂ u

∂νds.

Using boundary conditions and the above transmission boundary condition (7), we have∫∂ BR

u∂ u

∂νds = (

∫�\D

+∫

BR\�)(−k2|u|2 + |∇u|2)dx

+∫

�

ikλ|u+|2ds −∫

∂ Du+

∂ u+∂ν

ds, (9)

where ν is directed into the exterior of the corresponding domains.Furthermore, f1 = 0 and f2 = 0 imply that{

λu+ − λ0u− = 0∂u+∂ν

− ∂u−∂ν

= 0. (10)

So,

−∫

∂ Du+

∂ u+∂ν

ds

= −λ0

λ

∫∂ D

u−∂ u−∂ν

ds



= −λ0

λ

∫D

(−k20 |u|2 + |∇u|2)dx

= λ0k20

λ

∫D

|u|2dx − λ0

λ

∫D

|∇u|2dx .

Since k0, k, λ, and λ0 are positive, from (9) we have

I m(k∫

∂ BR

u∂ u

∂νds) ≥ 0. (11)

Hence, from Theorem 2.12 in Ref. 5, a unique continuation argument we obtain that u = 0 inR2\(D ∪ �). In addition, Applying Green’s theorem in the domain D, we also have u = 0 forx ∈ D. So we complete the proof of this lemma. �We are now ready to prove the existence of a solution for the above scattering problem by using anintegral equation approaching. By Green’s representation formula, we have

u(x) =

⎧⎪⎪⎨⎪⎪⎩

∫∂ D[ ∂u

∂ν�0(x, y)) − u ∂�0(x,y)

∂ν]dsy, f or x ∈ D∫

∂�[ ∂u∂ν

�(x, y) − u ∂�(x,y)∂ν

]dsy − ∫∂ D[ ∂u

∂ν�(x, y) − u ∂�(x,y)

∂ν]dsy,

f or x ∈ � \ D∫∂�

[u ∂�(x,y)∂ν

− ∂u∂ν

�(x, y)]dsy, f or x ∈ R2 \ �

,

where

�0(x, y) = i

4H (1)

0 (k0|x − y|) (12)

and

�(x, y) = i

4H (1)

0 (k|x − y|) (13)

are the fundamental solutions to the Helmholtz equation in R2 with respect to the wavenumber k0

and k, respectively, and H (1)0 is a Hankel function of the first kind of order zero.

Remark: In the above formula we choose the normal ν directed into the exterior of the domainD or �.

Pay attention to the boundary ∂ D first.By making use of the known jump relationships of the single- and double-layer potentials across

the boundary ∂ D (see, Refs. 2, 5, and 6) and approaching the boundary ∂ D from inside D we obtain

u−(x) = S0DD

∂u−∂ν

− K 0DDu− (14)

and

∂u−(x)

∂ν= K 0′

DD

∂u−∂ν

− T 0DDu−, (15)

where “u−” and “ ∂u−∂ν

” denote the limit approaching ∂ D from inside D, and S0DD , K 0

DD , K 0′DD , and

T 0DD are the boundary integral operators,

S0DD : H−1/2(∂ D) → H 1/2(∂ D), K 0

DD : H 1/2(∂ D) → H 1/2(∂ D),

K 0′DD : H−1/2(∂ D) → H−1/2(∂ D), T 0

DD : H 1/2(∂ D) → H−1/2(∂ D),

defined by

S0DDϕ(x) = 2

∫∂ D

ϕ(y)�0(x, y)dsy, K 0DDϕ(x) = 2

∫∂�

ϕ(y)∂�0(x, y)

∂νydsy,

K 0′DDϕ(x) = 2

∫∂ D

ϕ(y)∂�0(x, y)

∂νxdsy, T 0

DDϕ(x) = 2∂

∂νx

∫∂ D

ϕ(y)∂�0(x, y)

∂νydsy .



Remark: For simplicity, in the following discussion we use (·)± to denote the limit approaching∂ D (or ∂�) from outside and inside D (or �), respectively.

Similarly, in the domain � \ D let x approach the boundary ∂ D from inside � \ D, we obtain

u+(x) = K DDu+ − SDD∂u+∂ν

+ 2∫

∂�

[∂u−∂ν

�(x, y) − u−∂�(x, y)

∂ν]dsy (16)

and

∂u+(x)

∂ν= TDDu+ − K

′DD

∂u+∂ν

+ 2∂

∂ν(x)

∫∂�

[∂u−∂ν

�(x, y) − u−∂�(x, y)

∂ν]dsy, (17)

where SDD , K DD , K′DD , and TDD are the boundary integral operators,

SDD : H−1/2(∂ D) → H 1/2(∂ D), K DD : H 1/2(∂ D) → H 1/2(∂ D),

K′DD : H−1/2(∂ D) → H−1/2(∂ D), TDD : H 1/2(∂ D) → H−1/2(∂ D),

defined by

SDDϕ(x) = 2∫

∂ Dϕ(y)�(x, y)dsy, K DDϕ(x) = 2

∫∂ D

ϕ(y)∂�(x, y)

∂νydsy,

K′DDϕ(x) = 2

∫∂ D

ϕ(y)∂�(x, y)

∂νxdsy, TDDϕ(x) = 2

∂

∂νx

∫∂ D

ϕ(y)∂�(x, y)

∂νydsy .

The last term in (16) can be reformulated as

2∫

∂�

[∂u−(y)

∂ν�(x, y) − u−(y)

∂�(x, y)

∂ν]dsy

= 2∫

∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

+2∫

∂�

[∂u+(y)

∂ν�(x, y) − u+(y)

∂�(x, y)

∂ν]dsy . (18)

Since x ∈ ∂ D and y ∈ ∂� in (18), we have the following result (see, Refs. 5 and 6).

Lemma 2.1: By using Green’s formula and the Sommerfeld radiation condition (2), we obtain∫∂�

[∂u+(y)

∂ν�(x, y) − u+(y)

∂�(x, y)

∂ν]dsy = 0. (19)

Proof: Denote by BR a sufficiently large ball with radius R containing � and use Green’s formulainside BR\�. Furthermore notice that x ∈ ∂ D, y ∈ ∂� and the Sommerfeld radiation condition (2),we can prove this lemma. �

So (16) can be written as

u+(x) = 2∫

∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

+K DDu+ − SDD∂u+∂ν

. (20)

Similarly, (17) can be written as

∂u+(x)

∂ν



= 2∂

∂ν(x)

∫∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

+ TDDu+ − K′DD

∂u+∂ν

. (21)

From the boundary conditions in (1), on boundary ∂ D, we have

λu+ − λ0u− = f1, or λu+ = f1 + λ0u− (22)

and

∂u+∂ν

− ∂u−∂ν

= f2, or∂u+∂ν

= f2 + ∂u−∂ν

. (23)

By using (14), (15), (20), and (21), we can rewrite (22) as

f1(x) = λu+ − λ0u−

= λ(K DDu+ − SDD∂u+∂ν

) − λ0(S0DD

∂u−∂ν

− K 0DDu−)

+ 2λ

∫∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

= K DDλu+ − λSDD∂u+∂ν

+ λ0 K 0DDu− − λ0S0

DD

∂u−∂ν

+ 2λ

∫∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

= K DD( f1 + λ0u−) − λSDD( f2 + ∂u−∂ν

) + λ0 K 0DDu− − λ0S0

DD

∂u−∂ν

+ 2λ

∫∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

= λ0 K DDu− + λ0 K 0DDu− − λSDD

∂u−∂ν

− λ0S0DD

∂u−∂ν

+ 2λ

∫∂�

[(∂u−(y)

∂ν− ∂u+(y)

∂ν)�(x, y) − (u−(y) − u+(y))

∂�(x, y)

∂ν]dsy

+ K DD f1 − λSDD f2. (24)

We use [u] = u− − u+ and [ ∂u∂ν

] = ∂u−∂ν

− ∂u+∂ν

to denote the jump of u and ∂u∂ν

across the crack �,respectively. Then we have the following lemma.

Lemma 2.2: If u is a solution of (1), then [u] ∈ H 1/2(�) and [ ∂u∂ν

] ∈ H−1/2(�).

Proof: The proof of this lemma can be found in Ref. 3. �Define,

∂u−∂ν

|∂ D = a, u−|∂ D = b (25)

and

[∂u

∂ν]|� = (

∂u−∂ν

− ∂u+∂ν

)|� = c, [u]|� = (u− − u+)|� = d. (26)

Then zero extend c and d to the whole ∂� in the following:

c ={

0, on ∂�\�c, on �

, c ={

0, on ∂�\�d, on �

.



Now, we can rewrite (24) as

λ

λ0SDDa + S0

DDa − K DDb − K 0DDb − λ

λ0S�Dc + λ

λ0K�Dd = r0(x), (27)

where the operator S�D and K�D are the operators applied to a function with supp ⊆ � and evaluatedon ∂ D. We have mapping properties (see, Refs. 2 and 3),

S�D : H−1/2(�) → H 1/2(∂ D), K�D : H 1/2(�) → H 1/2(∂ D),

and

r0 = − 1

λ0f1(x) + 1

λ0K DD f1 − λ

λ0SDD f2. (28)

For convenience in the following discussion in Sec. III, we introduce operators S0 and K0 as

S0ϕ(x) = 2∫

∂ Dϕ(y)(x, y)dsy, K0ϕ(x) = 2

∫∂ D

ϕ(y)∂(x, y)

∂νydsy,

where

(x, y) = 1

2πln

1

|x − y| (29)

is the fundamental solution of the Laplace equation in R2.Now, we can rewrite (27) as

λ + λ0

λ0S0a + λ

λ0(SDD − S0)a + (S0

DD − S0)a

−2K0b − (K DD − K0)b − (K 0DD − K0)b

− λ

λ0S�Dc + λ

λ0K�Dd = r0. (30)

Remark: Refering to Refs. 5, 6, and 12, we know that the operators SDD − S0, S0DD − S0,

K DD − K0, and K 0DD − K0 are compact in corresponding Sobolev spaces (for details, we can see

the discussion in Sec. III).

Similarly, from the boundary conditions in (1) on ∂ D, i.e.,

∂u+∂ν

− ∂u−∂ν

= f2,

we have

K′DDa + K 0′

DDa − λ0

λTDDb − T 0

DDb − K′�Dc + T�Dd = r1, (31)

where

r1(x) = −1

λTDD f1 + K

′DD f2 − f2

and the operators K′�D and T�D have mapping properties (see Refs. 2 and 3),

K′�D : H−1/2(�) → H−1/2(∂ D), T�D : H 1/2(�) → H−1/2(∂ D).

Like (30) we introduce boundary integral operators K′0 and T0 as

K′0 : H−1/2(∂ D) → H−1/2(∂ D), T0 : H 1/2(∂ D) → H−1/2(∂ D),

defined by

K′0ϕ(x) = 2

∫∂ D

ϕ(y)∂(x, y)

∂νxdsy, T0ϕ(x) = 2

∂

∂νx

∫∂ D

ϕ(y)∂(x, y)

∂νydsy .



Then we can rewrite (31) as

(K′DD − K

′0)a + (K 0′

DD − K′0)a + 2K

′0a

−λ + λ0

λT0b − λ0

λ(TDD − T0)b − (T 0

DD − T0)b

−K′�Dc + T�Dd = r1. (32)

Remark: Similarly, the operators K′DD − K

′0, K 0′

DD − K′0, TDD − T0, and T 0

DD − T0 are compactin corresponding Sobolev spaces.

Now, pay attention to the boundary ∂� and let x approach the boundary from both sides,respectively.

First, making use of the known jump relationships of the single- and double-layer potentialsacross the boundary ∂� (see, Refs. 5 and 6) and approaching the boundary ∂� from inside �\D,we obtain (for x ∈ ∂�)

u−(x) = S��

∂u−∂ν

− K��u− + 2∫

∂ D[∂u(y)

∂ν�(x, y) − u(y)

∂�(x, y)

∂ν]dsy (33)

and

∂u−(x)

∂ν= K

′��

∂u−∂ν

− T��u− + 2∂

∂ν(x)

∫∂ D

[∂u(y)

∂ν�(x, y) − u(y)

∂�(x, y)

∂ν]dsy, (34)

where S��, K��, K′��, and T�� are the boundary integral operators,

S�� : H−1/2(∂�) → H 1/2(∂�), K�� : H 1/2(∂�) → H 1/2(∂�),

K′�� : H−1/2(∂�) → H−1/2(∂�), T�� : H 1/2(∂�) → H−1/2(∂�),

defined by

S��ϕ(x) = 2∫

∂�

ϕ(y)�(x, y)dsy, K��ϕ(x) = 2∫

∂�

ϕ(y)�(x, y)

∂νydsy,

K′��ϕ(x) = 2

∫∂�

ϕ(y)∂�(x, y)

∂νxdsy, T��ϕ(x) = 2

∂

∂νx

∫∂�

ϕ(y)∂�(x, y)

∂νydsy .

Then, let x approach the boundary ∂� from inside R2\�, we obtain (for x ∈ ∂�)

u+(x) = −S��

∂u+∂ν

+ K��u+ (35)

and

∂u+(x)

∂ν= −K

′��

∂u+∂ν

+ T��u+. (36)

From (33)–(36), restricting u to boundary �−, we have

2g(x) = S��(∂u−∂ν

− ∂u+∂ν

)|� − K��(u− − u+)|� + (u− − u+)|�

− 2∫

∂ D

∂u

∂ν�ds|� + 2

∫∂ D

∂�

∂νuds|�, x ∈ �−. (37)

By calculation,

2∫

∂ D

∂u

∂ν�ds|� = 2

∫∂ D

∂u+∂ν

�ds|�

= 2∫

∂ D( f2 + ∂u−

∂ν)�ds|�



= 2∫

∂ Df2�ds|� + 2

∫∂ D

∂u−∂ν

�ds|�

= 2∫

∂ Df2�ds|� + SD�a (38)

and

2∫

∂ D

∂�

∂νuds|� = 2

λ

∫∂ D

∂�

∂νf1ds|� + λ0

λK D�b, (39)

where the operators SD� and K D� have similar meaning as before.From (38) and (39), we can rewrite (37) as

− SD�a + λ0

λK D�b + S��c − K��d + d = r2(x), (40)

where

r2(x) = 2g(x) + 2∫

∂ Df2�ds|� − 2

λ

∫∂ D

∂�

∂νf1ds|� (41)

and S�� is the operator applied to a function with supp ⊆ � and evaluated on �, with analogousdefinition for SD� , K D� , and K�� . We have mapping properties (see, Refs. 2 and 3),

SD� : H−1/2(∂ D) → H 1/2(�), S�� : H−1/2(�) → H 1/2(�),

K D� : H 1/2(∂ D) → H 1/2(�), K�� : H 1/2(�) → H 1/2(�).

On the outside of �, restricting u on �+, and using (33) and (34) we have

2(∂u+∂ν

+ ikμu+)

= K′��(

∂u−∂ν

− ∂u+∂ν

) − T��(u− − u+) + ikμS��(∂u−∂ν

− ∂u+∂ν

)

−ikμK��(u− − u+) − ikμ(u− − u+) − (∂u−∂ν

− ∂u+∂ν

)

−2ikμ

∫∂ D

∂u

∂ν�ds + 2ikμ

∫∂ D

u∂�

∂νds

−2∂

∂ν(x)

∫∂ D

∂u

∂ν�ds + 2

∂

∂ν(x)

∫∂ D

u∂�

∂νds. (42)

Similar calculation as before, we can rewrite (42) as

− K′D�a + λ0

λTD�b + K

′��c − c − T��d − 2ikμd = r3(x), (43)

where

r3(x) = −2h(x) + 2ikμg(x) + 2∂

∂ν(x)

∫∂ D

f2�ds|� − 2

λ

∂

∂ν(x)

∫∂ D

f1∂�

∂νds|�.

If we define

A =

⎛⎜⎜⎝

λλ0

SDD + S0DD −K DD − K 0

DD − λλ0

S�Dλλ0

K�D

K′DD + K 0′

DD − λ0λ

TDD − T 0DD −K

′�D T�D

−SD�λ0λ

K D� S�� −K�� + I−K

′D�

λ0λ

TD� K′�� − I −T�� − 2ikμI

⎞⎟⎟⎠ (44)



and

�p =

⎛⎜⎜⎝

r0(x)r1(x)r2(x)r3(x)

⎞⎟⎟⎠ , (45)

then combine (27), (31), (40), and (43), we have a boundary integral system,

A

⎛⎜⎜⎝

abcd

⎞⎟⎟⎠ = �p. (46)

Remark: If the above system has a unique solution, our problem (1) will have a unique solution(see, Refs. 5 and 6).

III. EXISTENCE AND UNIQUENESS

Based on the idea of the paper,2 we show the existence and uniqueness of a solution to theintegral system (46).

Theorem 3.1: The operator A is Fredholm with index zero.

Proof: Pay attention to first row and second row in A which are from (27) and (31). Now,replacing (27) and (31) with (30) and (32), respectively, we find that all the operators in the two rowsexcept S0, T0, K0, and K

′0 are compact. As we know, S0 and −T0 are positive and bounded below up

to compact perturbations (see, Ref. 10), that is, there exist compact operators,

Dc : H−1/2(∂ D) → H 1/2(∂ D) (47)

and

Dt : H 1/2(∂ D) → H−1/2(∂ D), (48)

such that

Re(〈(S0 + Dc)ψ, ψ〉) ≥ C ||ψ ||2H−1/2(∂ D), for ψ ∈ H−1/2(∂ D), (49)

Re(〈−(T0 + Dt )ϕ, ϕ〉) ≥ C ||ϕ||2H 1/2(∂ D), for ϕ ∈ H 1/2(∂ D), (50)

where 〈, 〉 denote the duality between H−1/2(∂ D) and H 1/2(∂ D).For convenience, in the following discussion we define S∗

0 = S0 + Dc and T ∗0 = −(T0 + Dt ).

Now, pay attention to the third and fourth rows in A which are from (40) and (43). We considerthe two operators S�� and T�� which are related to the operators S�� and T��. Similarly, S�� and−T�� are positive and bounded below up to compact perturbations (see, Ref. 10)), that is, there existcompact operators,

�c : H−1/2(∂�) → H 1/2(∂�), (51)

�t : H 1/2(∂�) → H−1/2(∂�), (52)

such that

Re(〈(S�� + �c)ψ, ψ〉) ≥ C ||ψ ||2H−1/2(∂�), for ψ ∈ H−1/2(∂�), (53)

Re(〈−(T�� + �t )ϕ, ϕ〉) ≥ C ||ϕ||2H 1/2(∂�), for ϕ ∈ H 1/2(∂�). (54)

Define S∗1 = S�� + �c and T ∗

1 = −(T�� + �t ), then S∗1 and T ∗

1 are bounded below up and positive.Take a ∈ H−1/2(∂ D) and b ∈ H 1/2(∂ D). Let c ∈ H−1/2(∂�) and d ∈ H 1/2(∂�) be the extension

by zero to ∂� of c ∈ H−1/2(�) and d ∈ H 1/2(�), respectively.



Denote ξ = (a, b, c, d)T ,

H = H−1/2(∂ D) × H 1/2(∂ D) × H−1/2(�) × H 1/2(�) (55)

and the dual space,

H∗ = H 1/2(∂ D) × H−1/2(∂ D) × H 1/2(�) × H−1/2(�). (56)

Pay more attention to (30), (32), and the definitions of S∗0 , S∗

1 , T ∗0 , and T ∗

1 , then we can rewrite Aξ

as the following:

Aξ = A

⎛⎜⎜⎝

abcd

⎞⎟⎟⎠ = A0

⎛⎜⎜⎝

abcd

⎞⎟⎟⎠ + AC

⎛⎜⎜⎝

abcd

⎞⎟⎟⎠ = A0ξ + ACξ, (57)

with

A0ξ =

⎛⎜⎜⎝

λ+λ0λ0

S∗0 a − 2K0b

2K′0a + λ+λ0

λT ∗

0 bS∗

1 c − K��d + dK

′��c − c + T ∗

1 d − 2ikμd

⎞⎟⎟⎠ , (58)

ACξ =

⎛⎜⎜⎝

q0

q1

q2

q3

⎞⎟⎟⎠ , (59)

and

q0 = −λ + λ0

λ0Dca + λ

λ0(SDD − S0)a + (S0

DD − S0)a

−(K DD − K0)b − (K 0DD − K0)b − λ

λ0S�Dc + λ

λ0K�Dd,

q1 = (K′DD − K

′0)a + (K 0′

DD − K′0)a + λ0 + λ

λDt b

−λ0

λ(TDD − T0)b − (T 0

DD − T0)b − K′�Dc + T�Dd,

q2 = −SD�a + λ0

λK D�b − �cc,

q3 = −K′D�a + λ0

λTD�b + �t d.

It is easy to check that AC : H → H∗ is compact and A0 : H → H∗ defines a sesquilinear form, i.e.,

〈A0ξ, ξ 〉H,H∗

= λ + λ0

λ0(S∗

0 a, a) − 2(K0b, a) + 2(K′0a, b)

+λ + λ0

λ(T ∗

0 b, b) + (S∗1 c, c) − (K��d, c)|� + (d, c)

(K′��c, d)|� − (c, d) + (T ∗

1 d, d) − 2ikμ(d, d). (60)

Here (u, v) denotes the scalar product on L2(∂ D) or L2(∂�) defined by∫∂ D uvds or

∫∂�

uvds, and(u, v)|� is the scalar product on L2(�).



Notice that λ and λ0 are positive and use the properties of the operators S∗0 , S∗

1 , T ∗0 , and T ∗

1 , wehave

Re[λ + λ0

λ0(S∗

0 a, a)] ≥ c||a||2H−1/2(∂ D), (61)

Re(S∗1 c, c) ≥ c||c||2H−1/2(�), (62)

Re[λ + λ0

λ(T ∗

0 b, b)] ≥ c||b||2H−1/2(∂ D), (63)

andRe(T ∗

1 d, d) ≥ c||d||2H 1/2(�). (64)

Similarly,

Re[−2(K0b, a) + 2(K′0a, b)]

= Re[−2(K0b, a) + 2(a, K0b)]

= Re[−2(K0b, a) + 2(K0b, a)] = 0, (65)

Re[−(K��d, c)|� + (K′��c, d)|�]

= Re[−(K��d, c)|� + (c, K��d)|�]

= Re[−(K��d, c)|� + (K��d, c)|�] = 0, (66)

and

Re[(d, c) − (c, d) − 2ikμ(d, d)] = Re[(d, c) − (d, c) − 2ikμ(d, d)] = 0, (67)

where μ is positive.So the operator A0 is coercive, i.e.,

Re(〈(A − AC )ξ, ξ〉H,H∗ ) ≥ C ||ξ ||2H , for ξ ∈ H, (68)

whence the operator A is Fredholm with index zero.

Theorem 3.2: The operator A is injective or Kern A = {0}.

Proof: To this end let ψ = (a, b, c, d)T ∈ H be a solution of the homogeneous equation Aψ = �0,we want to prove that ψ ≡ �0.

However, Aψ = �0 means that⎧⎪⎪⎨⎪⎪⎩

− λλ0

SDDa − S0DDa + K DDb + K 0

DDb + λλ0

S�Dc − λλ0

K�Dd = 0−K

′DDa − K 0′

DDa + λ0λ

TDDb + T 0DDb + K

′�Dc − T�Dd = 0

−SD�a + λ0λ

K D�b + S��c − K��d + d = 0−K

′D�a + λ0

λTD�b + K

′��c − c − T��d − 2ikμd = 0

. (69)

Define the potentials v1 and v2 as

v1(x) = − 1

λ0SDa + 1

λK Db + 1

λ0S�c − 1

λ0K�d, (70)

v2(x) = 1

λ0S0

Da − 1

λ0K 0

Db, (71)

where c and d have the same meaning as before and

SDϕ(x) =∫

∂ Dϕ(y)�(x, y)dsy, S�ϕ(x) =

∫∂�

ϕ(y)�(x, y)dsy,



K Dϕ(x) =∫

∂ Dϕ(y)

∂�(x, y)

∂ν(y)dsy, K�ϕ(x) =

∫∂�

ϕ(y)∂�(x, y)

∂ν(y)dsy,

S0Dϕ(x) =

∫∂ D

ϕ(y)�0(x, y)dsy, K 0Dϕ(x) =

∫∂ D

ϕ(y)∂�0(x, y)

∂ν(y)dsy .

The potential v1(x) satisfies Helmholtz equation with wavenumber k, and v2(x) satisfies Helmholtzequation with wavenumber k0.

Pay attention to boundary ∂ D, and let x approach the boundary from both sides. When x ∈R2\D, we consider x → ∂ D to v1(x) and use the jump relationships (see, Refs. 5 and 6), then

v1(x) = 1

2{− 1

λ0SDDa + 1

λK DDb + 1

λb + 1

λ0S�Dc − 1

λ0K�Dd}, x ∈ ∂ D. (72)

Similarly, when x ∈ D, we consider x → ∂ D to v2(x), then

v2(x) = 1

2{ 1

λ0S0

DDa − 1

λ0K 0

DDb + 1

λ0b}, x ∈ ∂ D. (73)

From first row of (69), on boundary ∂ D we have

λv1(x) − λ0v2(x)

= 1

2{− λ

λ0SDDa − S0

DDa + K DDb + K 0DDb + λ

λ0S�Dc − λ

λ0K�Dd}

= 0. (74)

When x ∈ R2\D, we consider x → ∂ D to ∂v1(x)∂ν

, then

∂v1(x)

∂ν= 1

2{− 1

λ0K DDa + 1

λ0a + 1

λTDDb + 1

λ0K

′�Dc − 1

λ0T�Dd}, x ∈ ∂ D.

Similarly, when x ∈ D, we consider x → ∂ D to ∂v2(x)∂ν

, then

∂v2(x)

∂ν= 1

2{ 1

λ0K 0′

DDa + 1

λ0a − 1

λ0T 0

DDb}, x ∈ ∂ D. (75)

From second row of (69), on boundary ∂ D we have

∂v1(x)

∂ν− ∂v2(x)

∂ν

= 1

2{−K

′DDa − K 0′

DDa + λ0

λTDDb + T 0

DDb + K′�Dc − T�Dd}

= 0. (76)

Now, pay attention to boundary ∂�, especially to �, we consider x → �− to v1(x), and use thirdrow of (69), then

v1(x)

= 1

2{− 1

λ0SD�a + 1

λK D�b − 1

λ0S��c − 1

λ0K��d + λ0

d}

= 0, x ∈ �−. (77)

Similarly, we consider x → �+ to v1(x) and ∂v1(x)∂ν

, then

v1(x) = 1

2{− 1

λ0SD�a + 1

λK D�b + 1

λ0S��c − 1

λ0K��d − 1

λ0d}, (78)

∂v1(x)

∂ν= 1

2{− 1

λ0K

′D�a + 1

λTD�b + 1

λ0K

′��c − 1

λ0c − 1

λ0T��d}. (79)



Combine (78) and (79), and first using the fourth row of (69), then using the third row of (69), wehave

[∂v1(x)

∂ν+ ikμv1(x)]|�+

= 1

2{[− 1

λ0K

′D�a + 1

λTD�b + 1

λ0K

′��c − 1

λ0c − 1

λ0T��d]

+ ikμ[− 1

λ0SD�a + 1

λK D�b + 1

λ0S��c − 1

λ0K��d − 1

λ0d]}

= 1

2{2ikμd

λ0+ ikμ[− 1

λ0SD�a + 1

λK D�b + 1

λ0S��c − 1

λ0K��d − 1

λ0d]}

= 1

2{ikμ[− 1

λ0SD�a + 1

λK D�b + 1

λ0S��c − 1

λ0K��d + 1

λ0d]}

= 0. (80)

From (74), (76), (77), and (80), the potentials v1(x) and v2(x) satisfy the following boundary valueproblem:

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

�v2 + k20v2 = 0 in D

�v1 + k2v1 = 0 in R2\(D ∪ �)v1 = 0 on �−

∂v1∂ν

+ ikμv1 = 0 on �+

λv1 − λ0v2 = 0 on ∂ D∂v1∂ν

− ∂v2∂ν

= 0 on ∂ D

, (81)

and v1(x) satisfies the Sommerfeld radiation condition,

limr→∞

√r (

∂v1

∂r− ikv) = 0, (82)

uniformly in x = x/|x | with r = |x |.The uniqueness result Theorem 2.1 in Sec. II implies that

v1(x) = 0, x ∈ R2\(D ∪ �), (83)

and

v2(x) = 0, x ∈ D. (84)

By jump relationships, we have

c = (∂v−

1

∂ν− ∂v+

1

∂ν)|� = 0 (85)

and

d = (v−1 − v+

1 )|� = 0. (86)

In the following, we use the idea in (pp. 99–102 of Ref. 6) to prove that a = b = 0.Consider the following transmission problem:

⎧⎪⎪⎨⎪⎪⎩

�w + k2w = 0 in D�w + k2

0w = 0 in R2\Dλ0w+ − λw− = α on ∂ D

∂w+∂ν

− ∂w−∂ν

= β on ∂ D

, (87)

and w satisfies the Sommerfeld radiation condition.



Lemma 3.1: The problem (87) has a unique solution.

Proof: We can use the same method in Theorem 2.1 in Sec. II to prove this lemma.

Remark: Integral equation method can be used to get the existence of a solution for (87) (fordetails, see appendix at the end of the paper).

Similar to (70) and (71), we use the same a and b in (69) as density functions to define thepotentials w1 and w2, i.e.,

w1(x) = 1

λ0S0

Da − 1

λ0K 0

Db, (88)

w2(x) = − 1

λ0SDa + 1

λK Db, (89)

where the operators SD , S0D , K D , and K 0

D have the same meaning as before.We can prove that w1 and w2 satisfy the following homogeneous transmission problem (see the

appendix): ⎧⎪⎪⎨⎪⎪⎩

�w2 + k2w2 = 0 in D�w1 + k2

0w1 = 0 in R2\Dλ0w1 − λw2 = 0 on ∂ D

∂w1∂ν

− ∂w2∂ν

= 0 on ∂ D

, (90)

and w1 satisfies the Sommerfeld radiation condition.Lemma 3.1 implies that

w1(x) = 0, x ∈ R2\D, (91)

and

w2(x) = 0, x ∈ D. (92)

Using v1(x) and w2(x), we have

a = λ0(∂v+

1

∂ν− ∂w−

2

∂ν)|∂ D = 0 (93)

and

b = λ(v+1 − w−

2 )|∂ D = 0. (94)

(We also can use v2(x) and w1(x) to get a = 0 = b.)So,

ψ = (a, b, c, d)T = (0, 0, 0, 0)T . (95)

In conclusion, we complete the proof of the theorem.Combine Theorems 3.1 and 3.2, we have the following theorem.

Theorem 3.3: The boundary integral system (46) has a unique solution.

ACKNOWLEDGMENTS

We are very grateful to the anonymous referees for their very useful comments and suggestions.This research is supported in part by grants from Ministry of Education of the People’s Republic ofChina, Grant No. 107081. This research is supported by NSFC Grant No. 10871080 and Laboratoryof Nonlinear Analysis of CCNU.



APPENDIX: CHECK (90)

Consider the following transmission problem:⎧⎪⎪⎨⎪⎪⎩

�w + k2w = 0 in D�w + k2

0w = 0 in R2\Dλ0w+ − λw− = α on ∂ D

∂w+∂ν

− ∂w−∂ν

= β on ∂ D

, (A1)

and w satisfies the Sommerfeld radiation condition.From (88), when x ∈ R2\D and let x → ∂ D, we have

w1(x) = 1

2{ 1

λ0S0

DDa − 1

λ0K 0

DDb − 1

λ0b}, f or x ∈ ∂ D. (A2)

From (89), when x ∈ D and let x → ∂ D,

w2(x) = 1

2{− 1

λ0SDDa + 1

λK DDb − 1

λb}, f or x ∈ ∂ D. (A3)

Pay attention to the first row in (69) and notice that c = d = 0, we obtain

λ0w1(x) − λw2(x)

= 1

2{S0

DDa − K 0DDb + λ

λ0SDDa − K DDb}

= 0, f or x ∈ ∂ D. (A4)

Similarly, from (88), when x ∈ R2\D and let x → ∂ D, we have

∂w1(x)

∂ν= 1

2{ 1

λ0K 0′

DDa − 1

λ0a − 1

λ0T 0

DDb} (A5)

and when x ∈ D and let x → ∂ D for w2(x), then

∂w2(x)

∂ν= 1

2{− 1

λ0K

′DDa − 1

λ0a + 1

λTDDb}. (A6)

Pay attention to the second row in (69) and notice that c = d = 0, we have

∂w1(x)

∂ν− ∂w2(x)

∂ν

= 1

2{ 1

λ0K 0′

DDa − 1

λ0T 0

DDb + 1

λ0K

′DDa − 1

λTDDb}

= 1

2λ0{K 0′

DDa − T 0DDb + K

′DDa − λ0

λTDDb}

= 0. (A7)

So, w1(x) and w2(x) satisfy (90).

Remark: It is possible to extend our results to the case that the boundary (∂ D or ∂�) is Lipschitz.

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Mathematical basis of scattering problems from penetrable obstacles and cracks

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