# The Mathematical Basis of the Inverse Scattering Problem for Cracks from Near-Field Data

Authors: Yao Mao; Yongguang Chen; Jun Guo
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

We consider the acoustic scattering problem from a crack which has Dirichlet boundary condition on one side and impedance boundary condition on the other side. The inverse scattering problem in this paper tries to determine the shape of the crack and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. We firstly establish a near-field operator and focus on the operator’s mathematical analysis. Secondly, we obtain a uniqueness theorem for the shape and surface impedance. Finally, by using the operator’s properties and modified linear sampling method, we reconstruct the shape and surface impedance.


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## Body

## 1. Introduction

In this paper, we consider the scattering of an electromagnetic time-harmonic plane wave by an infinite cylinder having an open crack as cross section inR2. We assume that the cylinder is coated on one side by a material with surface impedance λ. This corresponds to the situation when the boundary or more generally a portion of the boundary is coated with an unknown material in order to avoid detection. Assuming that the electric field is polarized in the TM mode (see [1–3]), this leads to a mixed boundary value problem for the Helmholtz equation defined in the exterior of an open arc in R2.Briefly speaking, letΓ⊂R2 be an oriented piecewise smooth nonintersecting crack without cups; that is, Γ=ρ(s):s∈[s0,s1], where ρ:[s0,s1]→R2 is an injective piecewise C1 function and the crack Γ is contained in a closed curve Λ. Then the mixed boundary value problem for the Helmholtz equation in R2 can be formulated as follows: (1)Δu+k2u=0,inR2∖Γ-,u+=0,onΓ,z∈Λ,∂u-∂ν+iλu-=0,onΓ,z∈Λ,where k>0 is the wave number and λ>0 is the surface impedance. u is the total wave of the scattered wave us and the incident wave ui=Φ(x,z); that is, u=ui+us, and Φ(x,z) is the fundamental solution to the Helmholtz equation defined by (2)Φx,z=i4H01kx-z,with H0(1) being a Hankel function of the first kind of order zero. The scattered field us is required to satisfy the Sommerfeld radiation condition (3)limr→∞⁡r∂us∂ν-ikus=0uniformly in x^=x/x with r=x.Remark 1.
u
±
=
l
i
m
h
→
0
+
u
(
x
±
h
ν
) for x∈Γ, and ∂u±/∂ν=limh→0+ν·∇u(x±hν) for x∈Γ (for the details, see Section 2). In the following discussion, (·)± means the limit approaching the boundary from outside and inside the domain.The inverse scattering problem in this paper is trying to determine the shape of the arc (or crack) and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve, the results are as follows.Inverse Problem (Ip). In this paper, the inverse problem we are concerned about is to determine the crack Γ and the surface impedance λ from these measurements us(x,z) for x,z∈Λ.In 1995, Kress considered the inverse scattering problem for cracks with sound-soft boundary condition in [4]. The case of a sound-hard crack was considered by Monch in 1997 in [5]. Both of the authors used Newton’s method to reconstruct the shape of the crack from a knowledge of the far-field pattern. In 2003, Cakoni and Colton considered an inverse scattering problem by cracks, and they reconstructed the cracks by using the linear sampling method in [1]. In 2005, Colton and Haddar discussed similar inverse scattering problem by cracks, and they recovered the cracks by using the reciprocity gap functional method in [6]. Zeev and Cakoni considered the inverse scattering problem for a crack embedded in a known inhomogeneous background and recovered the crack (with a point source as incident field) in 2009 in [3]. More related research works can be found in [2, 3, 7] and the references therein.This paper is arranged as follows. In the next section, we formulate the scattering problem mathematically and prove that the associated near-field operator is injective with dense range under appropriate assumptions. In Section3, we show that the crack Γ and the surface impedance λ are uniquely determined from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. We modify the linear sampling method and reconstruct the shape of the crack (or the surface impedance coefficient) in Section 4.


## 2. The Formulation of the Problem

We suppose that the crackΓ can be extended to an arbitrary piecewise smooth, simply connected, and closed curve ∂Ω enclosing a bounded domain Ω such that the normal vector ν on Γ coincides with outward normal vector on ∂Ω which we again denote by ν. Λ is a closed curve; we denote by D the domain surrounded by Λ. We suppose that Ω is completely contained in D, and we assume the normal vector ν on ∂ν and ∂D is mapped to the exteriors of the domain Ω and the domain D, respectively.In order to formulate our scattering problem more precisely, we need to properly define the trace space on∂Ω and ∂D. Let U be a bounded domain and let Σ be an open subset of the boundary ∂U. If L2(∂U), H1/2(∂U), and H-1/2(∂U) denote the usual Sobolev spaces, we define the following spaces [8]: (4)L2Σ=uΣ:u∈L2∂U,H1/2Σ=uΣ:u∈H1/2∂U,H~1/2Σ=u∈H1/2∂U:supp⁡u⊆Σ-,H-1/2Σ=H~1/2Σ′,the  dual  space  ofH~1/2Σ,H~-1/2Σ=H1/2Σ′,the  dual  space  ofH1/2Σ,and we have the chain (5)H~1/2Σ⊂H1/2Σ⊂L2Σ⊂H~-1/2Σ⊂H-1/2Σ.Then problem (1) can be rewritten as (6)Δus+k2us=0,inR2∖Γ-,u+s=-Φ·,z,onΓ,z∈Λ,∂u-s∂ν+iλu-s=-∂Φ·,z∂ν-iλΦ·,z,onΓ,z∈Λ,and us is required to satisfy Sommerfeld radiation condition (3).Remark 2.
By using similar method in [1], we can obtain the existence and uniqueness of solution to the direct problem (6). Here we use the point source as incident wave, while the plane wave was used as the incident in [1].We define the near-field operatorF:L2(Λ)→L2(Λ) by (7)Fgx=∫Λusx,zgzdsz,g∈L2Λ,x∈Λ,where the function us(x,z) is the solution of problem (6). According to Green’s representation formula, we have (8)usx,z=∫Γu+sy,z∂Φx,y∂νy-∂u+sy,z∂νyΦx,ydsy+∫∂Ω∖Γu+sy,z∂Φx,y∂νy-∂u+sy,z∂νyΦx,ydsy,x∈R2∖Ω,(9)0=∫Γu-sy,z∂Φx,y∂νy-∂u-sy,z∂νyΦx,ydsy+∫∂Ω∖Γu-sy,z∂Φx,y∂νy-∂u-sy,z∂νyΦx,ydsy,x∈R2∖Ω.On the boundary ∂Ω∖Γ¯, we have (10)u+sx,z=u-sx,z,∂u+sx,z∂νx=∂u-sx,z∂νx.Then, by substituting (9) into (8), we have (11)usx,z=∫Γu+sy,z∂Φx,y∂νy-∂u+sy,z∂νyΦx,ydsy-∫Γu-sy,z∂Φx,y∂νy-∂u-sy,z∂νyΦx,ydsy,x∈R2∖Ω.By changing the order of x and z, we have (12)usz,x=∫Γu+sy,x∂Φz,y∂νy-∂u+sy,x∂νyΦz,ydsy-∫Γu-sy,x∂Φz,y∂νy-∂u-sy,x∂νyΦz,ydsy,z∈R2∖Ω.Then we have the following result.Lemma 3.
For the problem (6), one has us(x,z)=us(z,x) for x,z∈Λ.Proof.
Applying Green’s second theorem and (3), we obtain (13)0=∫Γu+sy,z∂u+sy,x∂νy-u+sy,x∂u+sy,z∂νydsy-∫Γu-sy,z∂u-sy,x∂νy-u-sy,x∂u-sy,z∂νydsy,(14)0=∫ΓΦy,z∂Φy,x∂νy-Φy,x∂Φy,z∂νydsy+∫∂Ω∖ΓΦy,z∂Φy,x∂νy-Φy,x∂Φy,z∂νydsy.Substituting (13) into (11) and (14) into (12), respectively, we have (15)usx,z-usz,x=∫Γw+sy,z∂w+sy,x∂νy-w+sy,x∂w+sy,z∂νydsy-∫Γw-sy,z∂w-sy,x∂νy-w-sy,x∂w-sy,z∂νydsy,where ws=us+Φ. By using the boundary conditions in (6), we have (16)w+sΓ=0,∂w-s∂ν+iλw-sΓ=0.This implies that us(x,z)-us(z,x)=0. So, we complete the proof of this lemma.Theorem 4.
The near-field operatorF defined by (7) is injective and has dense range.Proof.
Fromus(x,z)=us(z,x), the L2 adjoint of F is given by (17)F∗hx=∫Λusz,x¯hzdsz=∫Λusx,zhz¯dsz¯,x∈Λ,h∈L2Λ.Then we have (F∗h)(x)=(Fg)(x)¯, where g(z)=h(z)¯. Thus, operator F is injective if and only if F∗ is injective. Since N(F∗)⊥=F(L2(Λ))¯ in a Hilbert space, our proof will be finished by only showing that operator F is injective.
LetFg=0; we need to show that g=0. Define (18)vx=∫Λusx,zgzdsz,x∈R2∖Γ-.It is easy to verify that v satisfies the exterior problem (19)Δv+k2v=0,inR2∖D-,v=0,onΛ,limr→∞⁡r∂v∂r-ikv=0.This exterior Dirichlet problem has only zero solution (see [9]). Then the unique continuation principle now yields v=0 in R2∖Γ-. Therefore, (20)v+Γ=0,∂v-∂ν+iλv-Γ=0.
Now define(21)wx=∫ΛΦx,zgzdsz,x∈R2∖Γ-.By the boundary conditions (6) and (20) together with the jump relationship of the single-layer potential on the boundary Λ, we conclude that w satisfies the following problem: (22)Δw+k2w=0,inR2∖D-,Δw+k2w=0,inD∖Γ-,w+-w-=0,onΛ,∂w+∂ν-∂w-∂ν=-g,onΛ,w+=0,onΓ,∂w-∂ν+iλw-=0,onΓ,limr→∞⁡r∂w∂r-ikw=0.On the boundary ∂Ω, due to the jump relationship of the single-layer potential, we have w+=w- and ∂w+/∂ν=∂w-/∂ν. So, we get w-Γ=0 and (∂w-/∂ν)Γ=0 by the boundary condition of (22) on Γ. Then Holmgren’s principle and uniqueness continuation principle imply that w=0 on D. From this, we obtain that w+Λ=w-Λ=0.
Therefore, in the domainR2∖D-, we have the following problem for w: (23)Δw+k2w=0,inR2∖D-,w+=0,onΛ,limr→∞⁡r∂w∂r-ikw=0.The problem has only zero solution which implies that (∂w+/∂ν)Λ=0; thus -g=(∂w+/∂ν)Λ-(∂w-/∂ν)Λ=0. Hence, F is injective.


## 3. Uniqueness for the Inverse Problem

Based on the idea of [10, 11], we firstly conclude that Γ is uniquely determined from usΛ without knowing λ a priori. Secondly, we show that the surface impedance λ can be uniquely determined by usΛ (see [12]).Theorem 5.
Assume thatΓ1 and Γ2 are two cracks with corresponding impedance λ1 and impedance λ2 such that for a fixed wave number the corresponding scattering fields u1s(·,z) and u2s(·,z) coincide on Λ for all point sources z∈Λ. Then Γ1=Γ2 and λ1=λ2.Proof.
We consider problem (6) with z∈Λ replaced by z∈R2∖Γ-i. By using the same method in Lemma 3 in Section 2, we have (24)uisx,z=uisz,x∀x,z∈R2∖Γ-i,where i=1,2.
In the domainG=R2∖(Γ-1∪Γ-2), let ws=u1s-u2s. Then ws satisfies problem (19) replacing the corresponding domain D with G; that is, (25)Δws+k2ws=0,inG,ws=0,onΛ,limr→∞⁡r∂ws∂r-ikws=0,where we used the condition u1s(x,z)=u2s(x,z) on Λ for all point sources z∈Λ.
The only zero solution of this problem and the unique continuation principle imply thatws=0 in G; that is, u1s(x,z)=u2s(x,z) for all x∈G and z∈Λ.
By reciprocity (24) we have that u1s(z,x)=u2s(z,x) for all z∈Λ and x∈G. Then again arguing as above we have that u1s(z,x)=u2s(z,x) for all z, x∈G.
Now we assume thatΓ1≠Γ2. Without loss of generality, there exists x∗∈Γ1 but x∗∉Γ2. Choose h>0 such that the sequence xn=x∗+h/nν(x∗), n=1,2,…, is contained in G, where the unit normal vector to the boundary ∂Ω1 is directed into the exterior of Ω1. Considering un,js(j=1,2) as the solution of problem (6) with z∈Λ replaced by xn∈G corresponding to λ=λj and Γ=Γj, then un,1s(x)=un,2s(x) for x∈G. For simplicity, we use uns(x) to denote the solution in G; that is, (26)unsx=un,1sx=un,2sx,x∈G.
Consider the crackΓ2; then uns(x)=un,2s(x), and, on the two sides of Γ2, we know that un+s(·)=-Φ(·,xn) and ∂un-s(·)/∂ν+iλ2un-s(·)=-∂Φ(·,xn)/∂ν+iλ2Φ(·,xn) on Γ2 are uniformly bounded.
The well-posedness of the solution to the corresponding problem implies that the limitunsx∗Hloc1(R2∖Γ-2) is bounded as n→∞, so by the trace theorem uns(x∗)H1/2(Br(x∗)∩Γ1) is uniformly bounded with respect to n, where Br(x∗) is a small neighborhood centered at x∗ not intersecting Γ2.
On the other hand, consider the solution of (6) with respect to the crack Γ1; in this case uns(x)=un,1s(x). From the boundary condition un+s(·)=-Φ(·,xn) on the crack Γ1, we have that unsH1/2(Br(x∗)∩Γ1)→∞ as n→∞ since Φ(·,xn)H1/2(Br(x∗)∩Γ1)→∞ as n→∞. This is a contradiction. Therefore Γ1=Γ2.
Now letΓ=Γ1=Γ2 and assume that λ1(x)≠λ2(x) for x∈Γ. Then, from relation (24) and the unique continuation principle, we know that ws=0 in R2∖Γ-. Then w±s=0 and ∂w±s/∂ν=0 on Γ; that is, u1±s=u2±s and ∂u1±s/∂ν=∂u2±s/∂ν on Γ. Let w(·,z)=u1s+Φ(·,z); from the boundary condition (6), we have (27)∂w-∂ν+iλjw-=0,j=1,2,onΓ,and thus (28)λ1-λ2w-·,z=0onΓ.Hence w-(·,z)=0 on Γ since λ1(x)≠λ2(x). Notice that w+(·,z)=u1+s+Φ(·,z)=0 on Γ; then we have (29)Δw+k2w=0,inR2∖Γ-∪z,w±=0,onΓ,limr→∞⁡r∂w∂r-ikw=0.This problem has only zero solution; that is, w=0 in R2∖(Γ-∪z).
Now chooseh>0 sufficiently small such that w(xn,z)=0; that is, u1s(xn)=-Φ(xn,z), where xn=z+h/nν(z), n=1,2,…, and ν is the unit outward normal to Λ. Let n→∞; then the limit of u1s(xn)Hloc1(R2∖Γ-) is bounded because that problem (6) is well posed, but Φ(xn,z)Hloc1(R2∖Γ-) is unbounded which leads to a contraction. So, we complete the proof of the theorem.


## 4. The Linear Sampling Method

The inverse scattering problem in this paper is trying to determine the shape of the crack and the surface impedance coefficient from the near-field measurements of the scattered waves, while the source point is placed on a closed curve. In this part, we provide the mathematical basis to reconstruct the crackΓ from the knowledge of us(x,z) for x,z∈Λ by using the linear sampling method; that is, we want to determine Γ from a knowledge of us(x,z) for x,z∈Λ, where Λ is a circle centered at the origin; that is, Λ={x∈R2,x=rΛ>0}. Based on the ideas of [2, 3], we introduce the near-field equation (30)∫Λusx,zgzdsz=ΦLx,g∈L2Λ,x∈Λ;that is, (31)Fgzx=ΦLx,g∈L2Λ,x∈Λ,where (32)ΦLx=∫LΦx,yαLydsy+∫L∂Φx,y∂νyβLydsyand L is smooth nonintersecting arc and αL(y)∈H~-1/2(L) and βL(y)∈H~1/2(L).We want to characterize the crackΓ by using the behavior of an approximate solution g of the near-field equation (30).Now consider the following problem:(33)Δu+k2u=0,inR2∖Γ-,u+=p,onΓ,∂u-∂ν+iλu-=q,onΓ,limr→∞⁡r∂w∂r-ikw=0,for p∈H1/2(Γ) and q∈H-1/2(Γ). From [1], we know that this problem has a unique solution u∈Hloc1(R2∖Γ-) such that uHloc1(R2∖Γ-)≤c(pH1/2(Γ)+qH-1/2(Γ)), where c>0 is a constant and does not depend on p and q.To understand the near-field equation better, we define an operatorB:H1/2(Γ)×H-1/2(Γ)→L2(Λ) which maps the boundary data (p,q) to the solution u on Λ. We have the following conclusions about this operator B.Theorem 6.
OperatorB is injective and compact and has dense range in L2(Λ).Proof.
LetB(p,q)=0; that is, u=0 on Λ; we want to show that p=0 and q=0. From u=0 on Λ, we know that u satisfies problem (19) which has only zero solution on R2∖D-. By the unique continuation principle, we have that u=0 in R2∖Γ- and thus u±Γ=0 and (∂u±/∂ν)Γ=0. So, from the boundary conditions in (33), we can get p=0 and q=0, which implies that operator B is injective.
Define(34)vnx=∑-nncmJmkreimθ,where (r,θ) are polar coordinates of x, r=x, and cm∈l2. Clearly vn satisfies (33) with p=vnΓ and q=(∂vn/∂ν+iλvn)Γ. Since vnΛ=∑-nn‍cmJm(krΛ)eimθ, the completeness of the trigonometric sequence in L2[0,2π] shows that operator B has dense range.
We now show that operatorB is compact. Choose a disk M={x∈R2,x≤r<rΛ} such that Ω⊂M⊂D. Using Green’s representation formula for u, we can decompose operator B as B=B1B2, where B2:H1/2(Γ)×H-1/2(Γ)→H1/2(∂M)×H-1/2(∂M) is defined by B2(p,q)=(u∂M,(∂u/∂ν)∂M) and B1:H1/2(∂M)×H-1/2(∂M)→L2(Λ) is defined by (35)B1p1,q1=∫∂Mp1y∂Φx,y∂νy-q1yΦx,ydsy,x∈Λ.The regularity of the solution to problem (33) implies that operator B2 is bounded. So, operator B is compact since operator B1 is compact.Theorem 7.
Forx∈Λ, the integral expression ΦL(x) is in the range of B if and only if L⊂Γ.Proof.
IfL⊂Γ, then ΦL(x) is the solution of problem (33) with (36)p=ΦLxΓ+12β~LxΓ,q=∂ΦLx∂νxΓ+12α~LxΓ+iλΦLx-12β~LxΓ,where α~L and β~L are zero extension of αL and βL to the whole boundary ∂Ω. So, B(p,q)=ΦL(x)Λ, which implies that ΦL(x) is in the range of B.
IfL⊄Γ and ΦL(x) is in the range of B, then there exists a solution u with u+=p and ∂u-/∂ν+iλu-=q for x∈Γ, such that uΛ=ΦL(x)Λ. From [1], we know that this solution has the form (37)u=∫ΓΦx,yαydsy+∫Γ∂Φx,y∂νyβydsy,where α(y)∈H~-1/2(Γ) and β(y)∈H~1/2(Γ). Since uΛ=ΦL(x)Λ, the unique continuation principle implies that u=ΦL(x) in R2∖(Γ-∪L-). Now let x0∈L, x0∉Γ, and let Bϵ(x0) be a small ball with center at x0 such that Bϵ(x0)∩Γ=∅. Hence, u is analytic in Bϵ(x0), while ΦL(x) has a singularity at x0 which is a contradiction. This completes the proof of this theorem.To further understand the near-field operatorF, we define function wg by(38)wgx≔∫ΛΦx,ygydsy,x∈R2∖Λand define an operator S:L2(Λ)→H1/2(Γ)×H-1/2(Γ) given by (39)Sg=wgxΓ,∂wgx∂ν+iλwgxΓ.Then by superposition we have the following relation: (40)Fg=-BSg.Theorem 8.
OperatorS is bounded and injective and has dense range in H1/2(Γ)×H-1/2(Γ).Proof.
From the definition of operatorS, we know that S is bounded. To prove that S is injective, we let Sg=0 and want to prove that g=0.
It is easy to check thatwg defined in (38) satisfies problem (22).
By using the same arguments as that in proving thatF is injective in Theorem 6, we have g=0; that is, operator S is injective.
Next, we will show thatS has dense range in H1/2(Γ)×H-1/2(Γ). Let χ=(p,q)∈H~-1/2(Γ)×H~1/2(Γ) such that Sg,χ=0; here 〈·,·〉 denotes the duality pairing between H~-1/2(Γ)×H~1/2(Γ) and H1/2(Γ)×H-1/2(Γ). This means that(41)∫Γ∫ΛΦx,zgzdszpx¯dsx+∫Γ∫Λ∂Φx,z∂νx+iλΦx,zgzdszqx¯dsx=∫Λ∫ΓΦx,zpx¯dsxgzdsz+∫Λ∫Γ∂Φx,z∂νx+iλΦx,zqx¯dsxgzdsz=0,for all g∈L2(Λ). Then we have (42)∫ΓΦx,zpx¯dsx+∫Γ∂Φx,z∂νx+iλΦx,zqx¯dsx=0,z∈Λ.If we define (43)vz=∫ΓΦx,zpx¯dsx+∫Γ∂Φx,z∂νx+iλΦx,zqx¯dsx,z∈R2∖Γ-,then v(z) satisfies problem (19). The same analysis as before shows that v(z)=0 for z∈R2∖Γ-. Therefore, by the jump relationships of single potential and double potential across the crack Γ, we get (44)q¯=v+-v-Γ=0,and then (45)p¯=∂v+∂ν-∂v-∂νΓ=0.So, we have shown that χ=(0,0), which implies that operator S has dense range in H1/2(Γ)×H-1/2(Γ).We are now in the position to give the main result of this paper.Theorem 9.
Assume thatΓ is an oriented nonintersecting piecewise smooth arc without cups. Then, if F is the near-field operator corresponding to the scattering problem (6), then one has the following results:(1)
IfL⊂Γ, then for every ϵ>0 there exists a solution gϵL∈L2(Λ) satisfying (46)FgϵL+ΦLL2Λ<ϵ.(2)
IfL⊄Γ, then for every ϵ>0 and δ>0 there exists a function gϵ,δL∈L2(Λ) such that (47)Fgϵ,δL+ΦLL2Λ<ϵ+δ,limδ→0⁡gϵ,δLL2Λ=∞.Proof.
IfL⊂Γ, by using Theorem 7, there exists χ=(p,q)∈H1/2(Γ)×H-1/2(Γ) such that Bχ=ΦL(x) for x∈Λ. From Theorem 8, for every ϵ0>0, there exists a function gϵ0L∈L2(Λ) such that (48)Sgϵ0L-χH1/2Γ×H-1/2Γ<ϵ0.By using Theorem 6, operator B is bounded and we have (49)BSgϵ0L-BχL2Λ<c1ϵ0,where c1 is a constant; that is, (50)FgϵL+ΦLL2Λ<ϵ,where ϵ=c1ϵ0.
Next, we assume thatL⊄Γ. In this case, by Theorem 7,  ΦL(x) for x∈Λ is not in the range of B. But from Theorem 6 we know that operator B has dense range in L2(Λ). Hence, for every δ>0, we can construct a unique Tikhonov regularized solution χρ,L∈H1/2(Γ)×H-1/2(Γ) of Bχ=ΦL, such that (51)Bχρ,L-ΦLL2Λ<δ,where ρ is the regularization parameter (chosen by a regular regularization strategy, e.g., the Morozov discrepancy principle). Then we have χρ,LH1/2(Γ)×H-1/2(Γ)→∞ as ρ→0. By Theorem 8,  S has dense range, so for ϵ>0 sufficiently small there exists gϵ,ρL such that (52)Sgϵ,ρL-χρ,LH1/2Γ×H-1/2Γ<ϵc1.Combining (51) and (52), we obtain that for every ϵ>0 and δ>0 there exists gϵ,ρL∈L2(Λ) such that (53)Fgϵ,ρL+ΦLL2Λ=BSgϵ,ρL-ΦLL2Λ<BSgϵ,ρL-Bχρ,LL2Λ+Bχρ,L-ΦLL2Λ<ϵ+δ.Since limδ→0ρ(δ)=0, we have that limδ→0χρ,LH1/2(Γ)×H-1/2(Γ)→∞. From (52), we have that limδ→0Sgϵ,δLH1/2(Γ)×H-1/2(Γ)→∞. By the definition of operator S given by (39), we have that limδ→0gϵ,δLL2(Λ)→∞. Then we complete the proof of this theorem.Remark 10.
In numerical analysis, we can choose some suitable smooth arcs as a set such asL and then consider near-field equation (54)Fgx=ΦLx,L∈L.If L⊂Γ, we can find a bounded solution to the near-field equation (30) with discrepancy ϵ, whereas if L⊄Γ, then there exists solution of the near-field equation (with discrepancy ϵ+δ) with arbitrary large norm in the limit as δ→0. Then the arc can be characterized by the behavior of this solution. But how to determine the surface impedance is a problem we need to study further.Remark 11.
Applying reciprocity gap functional method to reconstruct a crack, we need to know the near-field Cauchy datau and ∂u/∂ν of the total field (see [6]). Qin and Colton used a method that may be called a modified linear sampling method to recover a cavity by using the near-field data (see [2, 7]). We combine these two methods to recover a crack (which has empty inner product) by using the measurement of near-field data u. In the process of recovering the crack, the near-field equation that we introduced is different.



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*Source: 863086-2015-06-04.xml*