Reconstruction of an unknown cavity with Robin boundary condition inside a heat conductor
aa r X i v : . [ m a t h - ph ] M a y Reconstruction of an unknown cavity with Robin boundarycondition inside a heat conductor
Gen Nakamura , Haibing Wang ∗ Department of Mathematics, Inha University, Incheon 402-751, Korea Department of Mathematics, Southeast University, Nanjing 210096, P.R. ChinaSeptember 20, 2018
Abstract
Active thermography is a non-destructive testing technique to detect the internal structure of a heatconductor, which is widely applied in industrial engineering. In this paper, we consider the problem ofidentifying an unknown cavity with Robin boundary condition inside a heat conductor from boundarymeasurements. To set up the inverse problem mathematically, we first state the corresponding forwardproblem and show its well-posedness in an anisotropic Sobolev space by the integral equation method.Then, taking the Neumann-to-Dirichlet map as mathematically idealized measured data for the activethermography, we present a linear sampling method for reconstructing the unknown Robin-type cavityand give its mathematical justification by using the layer potential argument. In addition, we analyze theindicator function used in this method and show its pointwise asymptotic behavior by investigating thereflected solution of the fundamental solution. From our asymptotic analysis, we can establish a pointwisereconstruction scheme for the boundary of the cavity, and can also know the distance to the unknowncavity as we probe it from its inside.
Keywords.
Inverse boundary value problem; Heat equation; Robin boundary condition; Reconstruc-tion scheme; Asymptotic behavior.
MSC(2000):
Active thermography is a widely used non-destructive testing technique in industrial engineering, which aimsto detect the internal information of a heat conductor [7, 21, 36, 37]. The principle of active thermography isas follows. If there is an anomaly inside the conductor, it affects the propagation of the heat flow inside theconductor and as a result it also directly affects the temporal behavior of the surface temperature distribution.By measuring the surface temperature distribution and by solving some inverse problem, the informationon the anomaly can be calculated. The information we want to know are the size, location and shape ofanomalies and their physical properties such as heat conductivities. The measurement of active thermographyis a non-contact, very fast and large area measurement which is conducted by injecting a heat flux to theconductor by a flash lamp or heater and measuring the corresponding distribution of temperature on thesurface of the conductor by an infrared light camera. ∗ Corresponding author, E-mail: [email protected]
1n this paper, we want to recover an unknown cavity with Robin boundary condition inside a heatconductor via active thermography. To begin with, we give the mathematical formulation as follows. LetΩ ⊂ R n ( n = 2 ,
3) be a heat conductor and D a cavity with Robin boundary condition embedded in Ω. Weassume that the boundaries ∂ Ω and ∂D of Ω and D , respectively, are of class C . Injecting a heat flux f on ∂ Ω over some time interval (0 , T ), the corresponding temperature distribution u ( x, t ) in Ω \ D × (0 , T ) canbe modeled by the following initial-boundary value problem: ( ∂ t − ∆) u = 0 in (Ω \ D ) × (0 , T ) ,∂ ν u − λu = 0 on ∂D × (0 , T ) ,∂ ν u = f on ∂ Ω × (0 , T ) ,u = 0 at t = 0 , (1.1)where λ = λ ( x ) ∈ C ( ∂D ) is the real-valued impedance and ν on ∂D (or ∂ Ω) is the unit normal vector directedinto the exterior of D (or Ω). We will show that the initial-boundary value problem (1.1) is well-posed in asuitable Sobolev space. Then we idealize a set of many pairs of the heat flux f and the corresponding surfacetemperature distribution u | ∂D × (0 , T ) which is the measured data for active thermography as the Neumann-to-Dirichlet map Λ D given by Λ D : f u | ∂D × (0 , T ) . Thus, our inverse problem for thermography is toreconstruct D from Λ D .When D is a usual cavity with Neumann boundary condition or an inclusion, the corresponding inverseproblem has been extensively studied. In [3, 14–16], the uniqueness and stability estimate are established.In [4, 8, 9], Newton-type iteration algorithms based on domain derivatives are studied. As for non-iterativereconstruction schemes, we can consult the papers [12, 13, 22, 23, 25, 26, 32, 34, 41] and the referencestherein, where the reconstruction schemes called the dynamical probe method and the enclosure methodare extensively studied. A reconstruction scheme for unknown cavities via Feynma-Kac type formula is alsoproposed in [28]. Recently, the authors established a linear sampling-type method for the heat equation toidentify unknown cavities with Neumann boundary condition [20]. This method was extended to the inclusioncase in [35].In this paper, we are concerned with the reconstruction of an unknown cavity where a Robin boundarycondition is prescribed. Some results on uniqueness and stability for this inverse problem can be found in[2, 24]. Assuming that D is given, the determination of the Robin coefficient λ from measured data wasalso considered; see [19, 27] and the references therein. In this work, assuming that both the geometricalinformation of the cavity D and the Robin coefficient λ on its boundary are unknown, we establish a linearsampling method to identify the unknown Robin-type cavity D from the boundary measurements Λ D . Aswe know, the linear sampling method was originally proposed for inverse scattering problems in [10], andwas further investigated from both the theoretical and numerical aspects; for example, see [1, 5, 6, 29–31, 33, 38, 39]. Roughly speaking, this method in the heat equation case is based on the characterization of theapproximate solvability of the so-called Neumann-to-Dirichlet map gap equation (Λ D − Λ ∅ )Ψ = Γ ( x, t ; y, s ),where Λ ∅ is the Neumann-to-Dirichlet map when there are not any cavities inside Ω, and Γ ( x, t ; y, s ) is theGreen function for the heat operator ∂ t − ∆ in Ω × (0 , T ) with homogeneous Neumann boundary condition onits boundary. By giving this characterization, we define a mathematical testing machine called an indicatorfunction to reconstruct the boundary of D . When we probe it from the inside of D , we don’t need to let thediscrepancy of the Neumann-to-Dirichlet map gap equation tend to zero.The new ingredients of this paper consists of two parts. First, we prove the well-posedness of the forwardproblem (1.1) in an anisotropic Sobolev space by the integral equation method. This is necessary for ourpurpose to investigate the inverse problem using layer potential argument, although the well-posedness canalso be justified in the usual function space W (0 , T ) by the argument in [40]. Second, compared with ourprevious works in [20, 35] for the Neumann cavity and inclusion cases, we provide a further investigation ofthe linear sampling method for the heat equation and show the asymptotic behavior of the indicator functionby carefully analyzing that of the reflected solution of the fundamental solution. Since the reflected solution isthe compensating term of the Green function for the related initial-boundary value problem, we actually give2 pointwise short time asymptotic behavior of the Green function near ∂D . As a consequence, by observingthe asymptotic behavior of the indicator function, we can have the followings:(i) pointwise reconstruction of the shape and location of an unknown cavity;(ii) information about the distance to the boundary of the cavity.The more precise meaning of the above item (ii) is that we can know the distance to ∂D when we probe itfrom inside D by using the indicator function.The rest of the paper is organized as follows. In Section 2, we show the well-posedness of the forwardproblem in an anisotropic Sobolev space by the integral equation method. Then, in Section 3, we presenta mathematical justification of the linear sampling method, while the asymptotic behavior of the indicatorfunction is provided in Section 4. Finally, in Section 5, we give some concluding remarks. In this section, we show the unique solvability of the forward problem by the integral equation method. Thenwe explicitly define the Neumann-to-Dirichlet map Λ D and state our inverse problem. Since our analysis forthe inverse problem is based on the layer potential argument, it is necessary to establish the well-posednessof the forward problem in an anisotropic Sobolev space, instead of the usual function space W (0 , T ).Let us start by introducing the anisotropic Sobolev spaces. For p, q ≥ H p,q ( R n × R ) := L ( R ; H p ( R n )) ∩ H q ( R ; L ( R n )) . For p, q ≤ H p,q by duality H p,q ( R n × R ) := ( H − p, − q ( R n × R )) ′ . Throughout this paper,we denote X × (0 , T ) and ∂X × (0 , T ) by X T and ( ∂X ) T , respectively, where X is a bounded domain in R n and ∂X denotes its boundary. By H p,q ( X T ) we denote the space of restrictions of elements of H p,q ( R n × R )to X T . The space H p,q (( ∂X ) T ) is defined analogously. We also introduce the following function spaces:˜ H , ( X T ) := n u ∈ H , ( X × ( −∞ , T )) (cid:12)(cid:12) u ( x, t ) = 0 for t < o ,H , ( X T ; ∂ t − ∆) := n u ∈ H , ( X T ) (cid:12)(cid:12) ( ∂ t − ∆) u ∈ L ( X T ) o . Then our forward problem is formulated as follows.
Forward problem:
Given f ∈ H − , − (( ∂ Ω) T ) and λ ∈ C ( ∂D ), find a unique solution u ∈ ˜ H , ((Ω \ D ) T ) to the problem (1.1).Denote by Γ( x, t ; y, s ) := π ( t − s )) n/ exp (cid:18) − | x − y | t − s ) (cid:19) , t > s, , t ≤ s the fundamental solution of the heat operator ∂ t − ∆. For convenience, we sometimes write it as Γ ( y, s ) ( x, t ).Define the following heat layer potentials:( V ij ϕ )( x, t ) := Z t Z S i Γ( x, t ; y, s ) ϕ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ S j × (0 , T ) , ( K ij ϕ )( x, t ) := Z t Z S i ∂ Γ( x, t ; y, s ) ∂ν ( y ) ϕ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ S j × (0 , T ) , ( N ij ϕ )( x, t ) := Z t Z S i ∂ Γ( x, t ; y, s ) ∂ν ( x ) ϕ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ S j × (0 , T ) , ( W ij ϕ )( x, t ) := − ∂∂ν ( x ) Z t Z S i ∂ Γ( x, t ; y, s ) ∂ν ( y ) ϕ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ S j × (0 , T ) .
3n this paper we take i, j = 1 , S = ∂D and S = ∂ Ω.We now show the unique solvability of (1.1) in ˜ H , ((Ω \ D ) T ) by the integral equation method. Theorem 2.1
Suppose that λ ∈ C ( ∂D ) and ≤ λ ≤ λ ≤ λ , where λ and λ are two given constants.Then, there exists a unique solution u ∈ ˜ H , ((Ω \ D ) T ) to the initial-boundary value problem (1.1) for any f ∈ H − , − (( ∂ Ω) T ) . Proof.
To prove the uniqueness of solutions in ˜ H , ((Ω \ D ) T ), we first establish the uniqueness in thefunction space W (0 , T ) defined by W (0 , T ) := n w : w ∈ L ((0 , T ); H (Ω \ D )) , ∂ t w ∈ L (cid:16) (0 , T ); (cid:0) H (Ω \ D ) (cid:1) ′ (cid:17)o . Consider the bilinear form associated with (1.1) b ( u, v, t ) := Z Ω \ D ∇ u · ∇ v dx + Z ∂D λuv dσ ( x )for u, v ∈ H (Ω \ D ) and t ∈ [0 , T ]. It is easy to verify that | b ( u, v, t ) | ≤ c k u k H (Ω \ D ) k v k H (Ω \ D ) , b ( u, u, t ) ≥ k u k H (Ω \ D ) − k u k L (Ω \ D ) , where c is a positive constant. Then the uniqueness of solutions to (1.1) in W (0 , T ) can be justified in astandard way; see, for example, [40, § u ∈ ˜ H , ((Ω \ D ) T ) to ( ∂ t − ∆) u = 0 with zeroinitial condition is in W (0 , T ), we also have the uniqueness in ˜ H , ((Ω \ D ) T ).To prove the existence of the solution in ˜ H , ((Ω \ D ) T ), we make use of the integral equation method[11]. By Green’s representation theorem for (1.1), we have u ( x, t ) = Z t Z ∂ (Ω \ D ) Γ( x, t ; y, s ) ∂u ( y, s ) ∂ν ( y ) dσ ( y ) ds − Z t Z ∂ (Ω \ D ) ∂ Γ( x, t ; y, s ) ∂ν ( y ) u ( y, s ) dσ ( y ) ds = − Z t Z ∂D Γ( x, t ; y, s ) ∂u ( y, s ) ∂ν ( y ) dσ ( y ) ds + Z t Z ∂D ∂ Γ( x, t ; y, s ) ∂ν ( y ) u ( y, s ) dσ ( y ) ds + Z t Z ∂ Ω Γ( x, t ; y, s ) ∂u ( y, s ) ∂ν ( y ) dσ ( y ) ds − Z t Z ∂ Ω ∂ Γ( x, t ; y, s ) ∂ν ( y ) u ( y, s ) dσ ( y ) ds. (2.1)Set ( u , u , u ) T = (cid:0) ∂ ν u | ( ∂D ) T , u | ( ∂D ) T , u | ( ∂ Ω) T (cid:1) T . Then, using the jump relations of layer potentials, wededuce the following system of boundary integral equations: V
11 12 I − K K − I + N W + λI − W V − K
12 12 I + K u u u = V fN fV f . (2.2)Define the space H := H − , − (( ∂D ) T ) × H , (( ∂D ) T ) × H , (( ∂ Ω) T ) with the norm k ( u , u , u ) T k H := (cid:18) k u k H − , − (( ∂D ) T ) + k u k H , (( ∂D ) T ) + k u k H , (( ∂ Ω) T ) (cid:19) / for ( u , u , u ) T ∈ H . We show the unique solvability of the system (2.2) in H . To accomplish this, wedefine the operator A := (cid:18) V
11 12 I − K − I + N W + λI (cid:19) (2.3)4nd the space H := H − , − (( ∂D ) T ) × H , (( ∂D ) T ) equipped with the norm k ( u , u ) T k H := (cid:18) k u k H − , − (( ∂D ) T ) + k u k H , (( ∂D ) T ) (cid:19) / for ( u , u ) T ∈ H . Let H ′ := H , (( ∂D ) T ) × H − , − (( ∂D ) T ) be the dual space of H with the duality between H and H ′ defined by (cid:28)(cid:18) ϕ ϕ (cid:19) , (cid:18) ψ ψ (cid:19)(cid:29) = h ϕ , ψ i + h ψ , ϕ i for ( ϕ , ϕ ) T ∈ H ′ , ( ψ , ψ ) T ∈ H . In the following, we first prove the invertibility of A : H → H ′ . Define a bilinear form a ( · , · ) on H × H associated with the operator A by a (cid:0) ( u , u ) T , ( v , v ) T (cid:1) := (cid:28) A (cid:18) u u (cid:19) , (cid:18) v v (cid:19)(cid:29) = h V u , v i + h ( 12 I − K ) u , v i + h v , ( − I + N ) u i + h v , W u i + h v , λu i for ( u , u ) T , ( v , v ) T ∈ H . There exists a positive constant C such that | a (cid:0) ( u , u ) T , ( v , v ) T (cid:1) | ≤ C k ( u , u ) T k H k ( v , v ) T k H . (2.4)Note that a (cid:0) ( u , u ) T , ( u , u ) T (cid:1) = h V u , u i + h− K u , u i + h u , + N u i + h u , W u i + h u , λu i = (cid:28)(cid:18) V − K N W (cid:19) (cid:18) u u (cid:19) , (cid:18) u u (cid:19)(cid:29) + h u , λu i , where the operator A := (cid:18) V − K N W (cid:19) is positive on H (see [11, Theorem 3.11]). Then we have the estimate a (cid:0) ( u , u ) T , ( u , u ) T (cid:1) ≥ C k ( u , u ) T k H , (2.5)where C is a positive constant. From (2.4) and (2.5) we conclude that A : H → H ′ is an isomorphism.In addition, it follows from Corollary 3.14 in [11] that12 I + K : H , (( ∂ Ω) T ) → H , (( ∂ Ω) T )is also an isomorphism. Since Γ( x, t ; y, s ) is smooth enough for x = y, ≤ s ≤ t ≤ T , the following operators K : H , (( ∂ Ω) T ) → H , (( ∂D ) T ) , W : H , (( ∂ Ω) T ) → H − , − (( ∂D ) T ) ,V : H − , − (( ∂D ) T ) → H , (( ∂ Ω) T ) , K : H , (( ∂D ) T ) → H , (( ∂ Ω) T )are compact. Therefore, the system (2.2) is Fredholm with index zero. To show the unique solvability of(2.2), it suffices to show its uniqueness.Let ( ϕ , ϕ , ϕ ) T ∈ H be the solution to the homogeneous form of (2.2). That is, V
11 12 I − K K − I + N W + λI − W V − K
12 12 I + K ϕ ϕ ϕ = . (2.6)5efine w ( x, t ) := − Z t Z ∂D Γ( x, t ; y, s ) ϕ ( y, s ) dσ ( y ) ds + Z t Z ∂D ∂ Γ( x, t ; y, s ) ∂ν ( y ) ϕ ( y, s ) dσ ( y ) ds − Z t Z ∂ Ω ∂ Γ( x, t ; y, s ) ∂ν ( y ) ϕ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ ( R n \ ( ∂D ∪ ∂ Ω)) T . (2.7)Note that w ( x, t ) satisfies ∂ t w − ∆ w = 0 in ( R n \ Ω) T (2.8)and w = 0 at t = 0 . (2.9)Letting x tend to the boundary ∂ Ω, we derive from the jump relations of layer potentials that w + = − V ϕ + K ϕ − ( 12 I + K ) ϕ . Throughout this paper we use ‘+’ and ‘ − ’ to denote the limits taken from the exterior and interior of adomain, respectively. It implies from the third equation of (2.6) that w + = 0 on ( ∂ Ω) T . (2.10)Then, using the same argument as in [20, Appendix A], we can prove that w = 0 in ( R n \ Ω) T . It followsthat ∂ ν w + = 0 on ( ∂ Ω) T , and hence ∂ ν w − = 0 on ( ∂ Ω) T due to the jump relations of layer potentials.On the other hand, the function w defined by (2.7) also satisfies the heat equation in (Ω \ D ) T . Usingthe jump relations of layer potentials again, we derive that w = − V ϕ + K ϕ + 12 ϕ − K ϕ on ( ∂D ) T ,∂ ν w = 12 ϕ − N ϕ − W ϕ + W ϕ on ( ∂D ) T . In terms of the first two equations of (2.6), we have w = ϕ and ∂ ν w = λϕ on ( ∂D ) T . In conclusion, thefunction w defined by (2.7) satisfies ( ∂ t − ∆) w = 0 in (Ω \ D ) T ,∂ ν w − λw = 0 on ( ∂D ) T ,∂ ν w = 0 on ( ∂ Ω) T ,w = 0 at t = 0 . (2.11)By the uniqueness of solutions to (1.1), we have w = 0 in (Ω \ D ) T , and hence w − = 0 on ( ∂ Ω) T . (2.12)Combining (2.10) and (2.12), we obtain ϕ = w − − w + = 0 on ( ∂ Ω) T . As a consequence, we have (cid:18) V
11 12 I − K − I + N W + λI (cid:19) (cid:18) ϕ ϕ (cid:19) = (cid:18) (cid:19) . Now by the invertibility of A , we conclude that ϕ = ϕ = 0 on ( ∂D ) T . Due to the Fredholm theory, theunique solvability of the system (2.2) is justified. Moreover, according to the mapping properties of heatlayer potentials, the function u expressed by (2.1) is the desired solution in ˜ H , ((Ω \ D ) T ). The proof iscomplete. ✷ Based on our Theorem 2.1 and Lemma 2.4 in [11], we can define the Neumann-to-Dirichlet map Λ D byΛ D : H − , − (( ∂ Ω) T ) → H , (( ∂ Ω) T ) , f u f | ( ∂ Ω) T . (2.13)6f there is not any cavity inside Ω, i.e., D = ∅ , the unique solvability of the forward problem in ˜ H , (Ω T )can be found in [11]. In this case, we denote the Neumann-to-Dirichlet map by Λ ∅ . Taking the Neumann-to-Dirichlet map Λ D as the measured data, our inverse problem for (1.1) is formulated as follows: Inverse Problem:
Reconstruct D from Λ D . In this section, we present a linear sampling-type method for the inverse problem formulated above and giveits mathematical justification by using the layer potential argument. Let Γ y, s ) ( x, t ) := Γ ( x, t ; y, s ) be theGreen function of the heat operator in Ω T with Neumann boundary condition on its boundary ( ∂ Ω) T . Thenthe linear sampling method for the heat equation is based on the characterization of the approximate solutionto the Neumann-to-Dirichlet map gap equation(Λ D − Λ ∅ ) g = Γ y, s ) ( x, t ) , ( x, t ) ∈ ( ∂ Ω) T , (3.1)where s ∈ (0 , T ) is a fixed time and y ∈ Ω is the sampling point. By this characterization, we can define anindicator function as a mathematical testing machine to reconstruct the location and shape of D .To begin with, let us define the operators S, H, A and F as follows. • Define S : H − , − (( ∂ Ω) T ) → ˜ H , ( D T ) and H : H − , − (( ∂ Ω) T ) → H − , − (( ∂D ) T ) by S : f u f | D T , (3.2)and H : f ( ∂ ν u f − λu f ) | ( ∂D ) T , (3.3)respectively, where u f is the solution to ( ∂ t − ∆) u f = 0 in Ω T ,∂ ν u f = f on ( ∂ Ω) T ,u f = 0 at t = 0 . (3.4) • Define A : H − , − (( ∂D ) T ) → H , (( ∂ Ω) T ) by A : g z g | ( ∂ Ω) T , (3.5)where z g is the solution to ( ∂ t − ∆) z g = 0 in (Ω \ D ) T ,∂ ν z g − λz g = g on ( ∂D ) T ,∂ ν z g = 0 on ( ∂ Ω) T ,z g = 0 at t = 0 . (3.6) • Define F : H − , − (( ∂ Ω) T ) → H , (( ∂ Ω) T ) by F := Λ D − Λ ∅ .To characterize the solution to (3.1), we first investigate the operator F and prove the following lemmas. Lemma 3.1
The operator F can be factorized as F = − AH . Proof.
For f ∈ H − , − (( ∂ Ω) T ), let u and v be such that ( ∂ t − ∆) u = 0 in (Ω \ D ) T ,∂ ν u − λu = 0 on ( ∂D ) T ,∂ ν u = f on ( ∂ Ω) T ,u = 0 at t = 0 , ( ∂ t − ∆) v = 0 in Ω T ,∂ ν v = f on ( ∂ Ω) T ,v = 0 at t = 0 , respectively. Define w as the solution to ( ∂ t − ∆) w = 0 in (Ω \ D ) T ,∂ ν w − λw = − ( ∂ ν v − λv ) on ( ∂D ) T ,∂ ν w = 0 on ( ∂ Ω) T ,w = 0 at t = 0 . Then it holds that ( ∂ t − ∆)( u − v − w ) = 0 in (Ω \ D ) T ,∂ ν ( u − v − w ) − λ ( u − v − w ) = 0 on ( ∂D ) T ,∂ ν ( u − v − w ) = 0 on ( ∂ Ω) T ,u − v − w = 0 at t = 0 . The uniqueness result in Theorem 2.1 says that u − v = w in ˜ H , ((Ω \ D ) T ). Hence, we have A ( − Hf ) = w | ( ∂ Ω) T = ( u − v ) | ( ∂ Ω) T = (Λ D − Λ ∅ ) f = F f, which completes the proof. ✷ Lemma 3.2
The operator H : H − , − (( ∂ Ω) T ) → H − , − (( ∂D ) T ) is continuous and has a dense range. Proof.
We prove the result using the layer potential argument. For f ∈ H − , − (( ∂ Ω) T ), we consider thefollowing initial-boundary value problem: ( ∂ t − ∆) u = 0 in Ω T ,∂ ν u = f on ( ∂ Ω) T ,u = 0 at t = 0 . (3.7)Express its solution by a single-layer heat potential u ( x, t ) = V ψ := Z t Z ∂ Ω Γ( x, t ; y, s ) ψ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ Ω T (3.8)with an unknown density ψ ∈ H − , − (( ∂ Ω) T ). Using the jump relations of layer potentials, the problem(3.7) is reformulated as the boundary integral equation( 12 I + N ) ψ = f on ( ∂ Ω) T , (3.9)where the operator I + N : H − , − (( ∂ Ω) T ) → H − , − (( ∂ Ω) T ) is an isomorphism (see [11, Corollary3.14]).Define ˜ V ψ := ( ∂ ν ( V ψ ) − λ ( V ψ )) | ( ∂D ) T . Then the operator H can be represented as H = ˜ V ( 12 I + N ) − . Thus, the continuity of H is evident. To prove the denseness property of H , it suffices to show that theoperator ˜ V : H − , − (( ∂ Ω) T ) → H − , − (( ∂D ) T ) has a dense range.Indeed, by direct calculations, we have˜ V ψ ( x, t ) = Z t Z ∂ Ω M ( x, t ; y, s ) ψ ( y, s ) dσ ( y ) ds, ( x, t ) ∈ ( ∂D ) T (3.10)8ith M ( x, t ; y, s ) := ∂ ν ( x ) Γ( x, t ; y, s ) − λ ( x ) Γ( x, t ; y, s ). Denote by ˜ V ∗ : H , (( ∂D ) T ) → H , (( ∂ Ω) T )the transpose of ˜ V in the sense that h ˜ V ψ, η i = h ψ, ˜ V ∗ η i for any ψ ∈ H − , − (( ∂ Ω) T ) and η ∈ H , (( ∂D ) T ).It follows that ˜ V ∗ η ( y, s ) = Z Ts Z ∂D M ( x, t ; y, s ) η ( x, t ) dσ ( x ) dt, ( y, s ) ∈ ( ∂ Ω) T . (3.11)Then, to show that the range of ˜ V is dense, we are led to prove the injectivity of ˜ V ∗ , that is, η = 0 in H , (( ∂D ) T ) if we have ˜ V ∗ η = 0 on ( ∂ Ω) T . To this end, we define w ( y, s ) = Z Ts Z ∂D M ( x, t ; y, s ) η ( x, t ) dσ ( x ) dt, ( y, s ) ∈ ( R n \ ∂D ) T . (3.12)Noticing that w satisfies ∂ s w + ∆ w = 0 in ( R n \ Ω) T ,w = 0 on ( ∂ Ω) T ,w = 0 at s = T, we have w = 0 in ( R n \ Ω) T by the same argument as in [20, Appendix A]. Using the unique continuationprinciple, we further have w = 0 in ( R n \ D ) T , (3.13)and therefore ∂w + ∂ν = 0 , w + = 0 on ( ∂D ) T . (3.14)Note that w ( y, s ) = Z Ts Z ∂D M ( x, t ; y, s ) η ( x, t ) dσ ( x ) dt = Z Ts Z ∂D (cid:0) ∂ ν ( x ) Γ( x, t ; y, s ) − λ ( x )Γ( x, t ; y, s ) (cid:1) η ( x, t ) dσ ( x ) dt = Z T − s Z ∂D (cid:0) ∂ ν ( x ) Γ( y, T − s ; x, τ ) − λ ( x )Γ( y, T − s ; x, τ ) (cid:1) ˜ η ( x, τ ) dσ ( x ) dτ, where ˜ η ( x, τ ) = η ( x, T − τ ). Using the jump relations of layer potentials, we have ∂w + ∂ν − ∂w − ∂ν = λη, w + − w − = η on ( ∂D ) T . (3.15)It implies from (3.14) and (3.15) that ∂w − ∂ν − λw − = 0 on ( ∂D ) T . (3.16)Observe that w also meets (cid:26) ∂ s w + ∆ w = 0 in D T ,w = 0 at s = T. Then, by the uniqueness of solutions to the backward problem, we obtain that w = 0 in D T . (3.17)Thus, it can be concluded from (3.13), (3.15) and (3.17) that η = 0. This completes the proof. ✷ Lemma 3.3
The operator A : H − , − (( ∂D ) T ) → H , (( ∂ Ω) T ) is injective, compact and has a dense range. roof. The injectivity can be easily seen from the unique continuation principle for the heat operator ∂ t − ∆.We now show the denseness. Let g j ∈ H − , − (( ∂D ) T ) ( j ∈ N ) be such that the linear hull of { g j } is densein H − , − (( ∂D ) T ). Then it is enough to prove f = 0 if Z ( ∂ Ω) T ϕ j f dσ ( x ) dt = 0for f ∈ H − , − (( ∂ Ω) T ) and all ϕ j := Ag j = z g j | ( ∂ Ω) T . Here we recall that z g j is the solution to (3.6) with g = g j .Let v ∈ ˜ H , ((Ω \ D ) T ) be the solution to the following backward problem ( ∂ t + ∆) v = 0 in (Ω \ D ) T ,∂ ν v − λv = 0 on ( ∂D ) T ,∂ ν v = f on ( ∂ Ω) T ,v = 0 at t = T, and set z j = z g j . Then, we have0 = Z (Ω \ D ) T ( v ∆ z j − z j ∆ v ) dxdt = Z ( ∂ Ω) T ( ∂ ν z j v − ∂ ν v z j ) dσ ( x ) dt − Z ( ∂D ) T ( ∂ ν z j v − ∂ ν v z j ) dσ ( x ) dt = − Z ( ∂D ) T g j vdσ ( x ) dt. So v = 0 on ( ∂D ) T . By the boundary condition of v , we further have ∂ ν v | ( ∂D ) T = 0. Therefore, v = 0 in(Ω \ D ) T , and then f = ∂ ν v | ( ∂ Ω) T = 0.Finally, let us prove the compactness of A . Indeed, there exists a unique density ϕ ∈ H − , − (( ∂D ) T )such that the solution z g to (3.6) with g ∈ H − , − (( ∂D ) T ) is given by z g ( x, t ) = Z t Z ∂D Γ ( x, t ; y, s ) ϕ ( y, s ) dσ ( y ) ds. Notice that Γ ( x, t ; y, s ) is smooth enough for x ∈ ∂ Ω , y ∈ ∂D and 0 ≤ s ≤ t ≤ T . It can be concluded that Ag = z g | ( ∂ Ω) T ∈ C ∞ ( ∂ Ω × [0 , T ]), and hence A is compact. This completes the proof. ✷ We are now in a position to state our main results, which motivate the linear sampling method forreconstructing D . Theorem 3.4
Let s ∈ (0 , T ) be fixed. For y ∈ D , there exists a function g y ∈ H − , − (( ∂ Ω) T ) satisfying k F g y − Γ y, s ) k H , (( ∂ Ω) T ) < ε (3.18) such that lim y → ∂D k g y k H − , − (( ∂ Ω) T ) = ∞ (3.19) and lim y → ∂D k Sg y k ˜ H , ( D T ) = ∞ . (3.20) Proof.
According to Lemma 3.2, for any ε > g y ∈ H − , − (( ∂ Ω) T ) such that (cid:13)(cid:13)(cid:13) Hg y + (cid:16) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:17)(cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) < ε k A k , (3.21)10here k A k is the norm of the operator A defined by (3.5). Since Γ y, s ) satisfies the heat equation in (Ω \ D ) T and ∂ ν Γ y, s ) | ( ∂ Ω) T = 0, it holds that A (cid:16) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:17) = Γ y, s ) | ( ∂ Ω) T . By Lemma 3.1, we have (cid:13)(cid:13)(cid:13)
F g y − Γ y, s ) (cid:13)(cid:13)(cid:13) H , (( ∂ Ω) T ) = (cid:13)(cid:13)(cid:13) − AHg y − A (cid:16) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:17)(cid:13)(cid:13)(cid:13) H , (( ∂ Ω) T ) ≤ k A k (cid:13)(cid:13)(cid:13) Hg y + (cid:16) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:17)(cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) < ε. Hence, due to the boundedness of H , we have k g y k H − , − (( ∂ Ω) T ) ≥ C k Hg y k H − , − (( ∂D ) T ) ≥ C (cid:18)(cid:13)(cid:13)(cid:13) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) − (cid:13)(cid:13)(cid:13) Hg y + (cid:16) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:17)(cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) (cid:19) ≥ C (cid:13)(cid:13)(cid:13) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) − Cε k A k , (3.22)where C is a positive constant.We next prove (cid:13)(cid:13)(cid:13) ∂ ν Γ y, s ) − λ Γ y, s ) (cid:13)(cid:13)(cid:13) H − , − (( ∂D ) T ) → ∞ as y → ∂D. (3.23)Since Γ ( x, t ; y, s ) − Γ( x, t ; y, s ) ∈ C ∞ (Ω T ), that is, Γ has the same singularity as Γ at ( x, t ) = ( y, s ), weonly need to prove that (cid:13)(cid:13) ∂ ν Γ ( y, s ) − λ Γ ( y, s ) (cid:13)(cid:13) H − , − (( ∂D ) T ) → ∞ as y → ∂D. (3.24)We consider here the case for n = 3. Under the assumption that ∂D is of class C , for any point x ∈ ∂D there exists a C -function Φ such that D ∩ B ( x , r ) = { x ∈ B ( x , r ) : x > Φ( x , x ) } , where B ( x , r ) is the ball with radius r > x . We choose a new orthonormal basis { e j } , j =1 , , , centered at x with e = − ν , where ν is the unit outward normal vector to the boundary at x . Thevectors e and e lie in the tangent plane to ∂D at x . Let x be the local coordinate defined by the basis { e j } . We introduce the local transformation of coordinate η = F ( x ) as follows: η ′ = x ′ , η = x − Φ( x ′ ) with x ′ = ( x , x ) , η ′ = ( η , η ) . We note that Φ(0) = ∇ x ′ Φ(0) = 0. For x ∈ ∂D and y ∈ D , let η := F ( x ) , ξ := F ( y ) with η = ( η ′ , ξ = (0 , ξ ). Then it holds that | x − y | = | F − ( η ) − F − ( ξ ) | ≤ c | η − ξ | = c ( ξ + | η ′ | ) . We now pick up the dominant part of (3.24), that is, k ∂ η Γ ( ξ,s ) − λ Γ ( ξ,s ) k H − , − (( D ) T ) , (3.25)and ignore all the other terms which are bounded as y → ∂D , where D := [ − l, l ] × [ − l, l ] with 0 < l ≪ s = 0 and introduce an auxiliary function ϕ ( η ′ , t ) = cχ ( η ′ , t ) t − α e − c | η ′| t < α < /
2, where χ ( η ′ , t ) is a smooth cut-off function such that it vanishes near t = T and η j = ± l ( j = 1 ,
2) but equals one in a neighborhood of the origin ( η ′ , t ) = (0 , ϕ ∈ H , (( D ) T ) ⊂ H , (( D ) T ) and k ϕ k H , (( D ) T ) ≤ c .Indeed, by direct calculations, we have ϕ t = cχ ( η ′ , t ) e − c | η ′| t (cid:18) − αt − α − + c | η ′ | t t − α − (cid:19) + cχ t ( η ′ , t ) t − α e − c | η ′| t ,ϕ η j = − cc χ ( η ′ , t ) t − α − e − c | η ′| t η j + cχ η j ( η ′ , t ) t − α e − c | η ′| t . So, for 0 < t ≪ ϕ t and ϕ η j can be estimated by | ϕ t | ≤ c t − α − e − ˜ c | η ′| t , | ϕ η j | ≤ c t − α − / e − ˜ c | η ′| t . Note that Z R (cid:18) e − ˜ c | η ′| t (cid:19) dη ′ = Z R e − ˜ c | η ′| t dη ′ = O ( t ) . Then we have Z T Z D | ϕ η j | dxdt ≤ c Z T ( t − α − / ) t dt = c Z T t − α dt < ∞ . (3.26)In addition, we deduce that (cid:18)Z D [ ∂ / t ϕ ] dx (cid:19) / ≤ (Z R (cid:18) / Z t ( t − τ ) − / ∂ τ ϕ ( x, τ ) dτ (cid:19) dx ) / ≤ / Z t ( t − τ ) − / (cid:18)Z R | ∂ τ ϕ ( x, τ ) | dx (cid:19) / dτ ≤ c Z t ( t − τ ) − / τ − α − / dτ ≤ c t − α +1 / , (3.27)which guarantees the integrability of ∂ / t ϕ in L under our condition 0 < α < /
2. Thus, we conclude that ϕ ∈ H , (( D ) T ) ⊂ H , (( D ) T ) and k ϕ k H , (( D ) T ) ≤ c .Next, we compute the norm of ∂ η Γ ( ξ,s ) − λ Γ ( ξ,s ) using the duality. In fact, we have (cid:13)(cid:13) ∂ η Γ ( ξ,s ) − λ Γ ( ξ,s ) (cid:13)(cid:13) H − , − (( ∂D ) T ) ≥ c π / Z T Z D (cid:16) c t ξ − λ (cid:17) t − / e − c ξ | η ′| t (cid:18) cχt − α e − c | η ′| t (cid:19) dη ′ dt ≥ c π / Z T (cid:16) c t ξ − λ (cid:17) t − / − α e − c ξ t dt = c π / Z ∞ ξ T (cid:16) c τ ξ − α − λ ξ − α +13 (cid:17) τ α − e − c τ dτ → ∞ as ξ → y → ∂D ) . Now we can conclude from (3.22) and (3.23) that the blow-up properties (3.19) and (3.20) hold. Theproof is complete. ✷ To further investigate the behavior of the density g y as y approaches to ∂D from the exterior of D , weneed the following lemma. Lemma 3.5
For any fixed s ∈ (0 , T ) , the Green function Γ y, s ) ( x, t ) is not in the range of A if y ∈ Ω \ D . roof. Suppose that Γ y, s ) ( x, t ) is in the range of A . Then there exists a function g ∈ H − , − (( ∂D ) T )such that w g | ( ∂ Ω) T = Γ y, s ) | ( ∂ Ω) T , (3.28)where w g ∈ ˜ H , ((Ω \ D ) T ) is the solution to ( ∂ t − ∆) w g = 0 in (Ω \ D ) T ,∂ ν w g − λw g = g on ( ∂D ) T ,∂ ν w g = 0 on ( ∂ Ω) T ,w g = 0 at t = 0 . From (3.28) and ∂ ν w g | ( ∂ Ω) T = ∂ ν Γ y, s ) | ( ∂ Ω) T = 0, it follows that w g = Γ y, s ) in (cid:0) Ω \ ( D ∪ { y } ) (cid:1) T and hence k Γ y, s ) k ˜ H , ((Ω \ D ) T ) = k w g k ˜ H , ((Ω \ D ) T ) < ∞ . Here y ∈ Ω \ D can be either y ∈ ∂D or y ∈ Ω \ D , and foreach case it holds k Γ y, s ) k ˜ H , ((Ω \ D ) T ) = ∞ . This gives a contradiction and completes the proof. ✷ In contrast to Theorem 3.4 for y ∈ D , we now establish the following blowup property of the density g y for y D . Theorem 3.6
Fix s ∈ (0 , T ) and let y ∈ Ω \ D . Then, for every ε > and δ > , there exists a function g yε, δ ∈ H − , − (( ∂ Ω) T ) satisfying k F g yε, δ − Γ y, s ) k H , (( ∂ Ω) T ) < ε + δ (3.29) such that lim δ → k g yε, δ k H − , − (( ∂ Ω) T ) = ∞ (3.30) and lim δ → k Sg yε, δ k ˜ H , ( D T ) = ∞ . (3.31) Proof.
By Lemma 3.3, for arbitrary δ > f yδ ∈ H − , − (( ∂D ) T ) such that k Af yδ − Γ y, s ) k H , (( ∂ Ω) T ) < δ. Since Γ y, s ) is not in the range of A due to Lemma 3.5, we have k f yδ k H − , − (( ∂D ) T ) → ∞ as δ → . (3.32)Recall that the range of H is dense in H − , − (( ∂D ) T ). We can find a function g yε, δ ∈ H − , − (( ∂ Ω) T ) suchthat k Hg yε, δ + f yδ k H − , − (( ∂D ) T ) < ε/ ( k A k + 1) , (3.33)and hence k Af yδ + AHg yε, δ k H , (( ∂ Ω) T ) < ε. So, it can be derived that k F g yε, δ − Γ y, s ) k H , (( ∂ Ω) T ) = k − AHg yε, δ − Γ y, s ) k H , (( ∂ Ω) T ) ≤ k AHg yε, δ + Af yδ k H , (( ∂ Ω) T ) + k Af yδ − Γ y, s ) k H , (( ∂ Ω) T ) ≤ ε + δ. Moreover, we obtain from (3.32) and (3.33) that k Hg yε, δ k H − , − (( ∂D ) T ) → ∞ as δ → , k g yε, δ k H − , − (( ∂ Ω) T ) → ∞ , k Sg yε, δ k ˜ H , ( D T ) → ∞ as δ → . This completes the proof. ✷ By Theorems 3.4 and 3.6, Sg y can be taken as an indicator function for reconstructing the boundary ofthe cavity D , where g y is the approximate solution to (3.1) for the sampling point y ∈ Ω. In this section, we show a short time asymptotic behavior of the indicator function Sg y as y approaches to theboundary ∂D from the interior of D . To do this, we analyze the asymptotic behavior of the reflected solutionof the fundamental solution. As we know, the reflected solution is the compensating term of the Greenfunction for the related initial-boundary value problem. This means that we actually provide a pointwiseshort time asymptotic behavior of the Green function near ∂D . From our asymptotic analysis, we canestablish a pointwise reconstruction scheme for the location and shape of the cavity, and can also know thedistance to the unknown cavity when we probe it from the interior of D . Without loss of generality, we onlyconsider the three dimensional case.Let ¯Γ ( y, s ) ( x, t ) be the reflected solution of Γ y, s ) ( x, t ). That is, ( ∂ t − ∆)¯Γ ( y, s ) ( x, t ) = 0 in D T , ( ∂ ν − λ )¯Γ ( y, s ) ( x, t ) = − ( ∂ ν − λ )Γ y, s ) ( x, t ) on ( ∂D ) T , ¯Γ ( y, s ) ( x, t ) = 0 for x ∈ D, t ≤ s. Then, by the well-posedness of the above initial-boundary value problem, we obtain from (3.21) that (cid:13)(cid:13) Sg y − ¯Γ ( y, s ) (cid:13)(cid:13) ˜ H , ( D T ) < C ε, (4.1)where C is a positive constant. Denote by R ( y, s ) ( x, t ) the reflected solution of Γ ( y,s ) ( x, t ), i.e., ( ∂ t − ∆) R ( y, s ) ( x, t ) = 0 in D T , ( ∂ ν − λ ) R ( y, s ) ( x, t ) = − ( ∂ ν − λ )Γ ( y, s ) ( x, t ) on ( ∂D ) T ,R ( y, s ) ( x, t ) = 0 for x ∈ D, t ≤ s. (4.2)Since Γ y, s ) ( x, t ) − Γ ( y, s ) ( x, t ) for ( y, s ) ∈ D T is smooth in Ω T , there exists a positive constant C such that (cid:13)(cid:13) ¯Γ ( y, s ) ( x, t ) − R ( y, s ) ( x, t ) (cid:13)(cid:13) ˜ H , ( D T ) < C . (4.3)Combining (4.1) and (4.3) yields (cid:13)(cid:13) ( Sg y ) ( x, t ) − R ( y, s ) ( x, t ) (cid:13)(cid:13) ˜ H , ( D T ) < C ε + C . (4.4)This implies that Sg y and R ( y, s ) have the same blow-up behavior as y approaches to ∂D . Hence, to get theasymptotic behavior of Sg y , we only need to show the corresponding result of R ( y, s ) ( x, t ).To begin with, we locally flatten the boundary ∂D . For any point z ∈ ∂D , there is a diffeomorphismΦ : R → R which transforms z to the origin such thatΦ( D ) ⊂ R − := (cid:8) ξ = ( ξ , ξ , ξ ) ∈ R : ξ < (cid:9) . Without loss of generality, we assume ∈ ∂D with ν ( ) = e and locally flatten the boundary ∂D around . Let ξ = Φ( x ) and η = Φ( y ). Denote by J ( x ) the Jacobian matrix of Φ. We have J ( x ) = ∇ Φ( x ) , J ( ) = I, ν ( x ) = J T ν ( ξ ) | J T ν ( ξ ) | . R ( ξ, t ; η, s ) := R ( x, t ; y, s ) and ˜Γ( ξ, t ; η, s ) := Γ( x, t ; y, s ). We deduce from (4.2) that ˜ R ( ξ, t ; η, s )near satisfies ( ∂ t − ∇ ξ · M ( ξ ) ∇ ξ ) ˜ R ( ξ, t ; η, s ) = 0 , t ∈ (0 , T ) , ξ < ,e · ( M ( ξ ) ∇ ξ ˜ R )( ξ, t ; η, s ) − λ (Φ − ( ξ )) | J T ν ( ξ ) | ˜ R ( ξ, t ; η, s )= − (cid:16) e · ( M ( ξ ) ∇ ξ ˜Γ)( ξ, t ; η, s ) − λ (Φ − ( ξ )) | J T ν ( ξ ) | ˜Γ( ξ, t ; η, s ) (cid:17) , t ∈ (0 , T ) , ξ = 0 , ˜ R ( ξ, t ; η, s ) = 0 , t ≤ s, ξ < , (4.5)where M ( ξ ) = ( JJ T )(Φ − ( ξ )). Let W + ( ξ, t ; η, s ) be such that ( ∂ t − ∆ ξ ) W + ( ξ, t ; η, s ) = 0 , t ∈ (0 , T ) , ξ < , ( ∂ ξ − λ ( )) W + ( ξ, t ; η, s ) = − ( ∂ ξ − λ ( )) ˜Γ( ξ, t ; η, s ) , t ∈ (0 , T ) , ξ = 0 ,W + ( ξ, t ; η, s ) = 0 , t ≤ s, ξ < . (4.6)In the following, we first estimate the difference ˜ R ( ξ, t ; η, s ) − W + ( ξ, t ; η, s ), and then analyze theasymptotic behavior of W + ( ξ, t ; η, s ). As a consequence, the asymptotic behavior of ˜ R ( ξ, t ; η, s ) is obtained. Lemma 4.1
Let η = ( η , η , η ) T , where η = − ǫ/ and | η | ≤ cǫ for small ǫ > and some constant c > .Then there exists a positive constant C such that (cid:12)(cid:12)(cid:12) ˜ R ( η, s + ǫ ; η, s ) − W + ( η, s + ǫ ; η, s ) (cid:12)(cid:12)(cid:12) ≤ Cǫ − as ǫ → . (4.7) Proof.
Define P ( ξ, t ; η, s ) := ˜ R ( ξ, t ; η, s ) − W + ( ξ, t ; η, s ) . We derive from (4.5) and (4.6) that near thefunction P ( ξ, t ; η, s ) satisfies ( ∂ t − ∆ ξ ) P ( ξ, t ; η, s ) = ∇ · ( M − I ) ∇ ˜ R, t ∈ (0 , T ) , ξ < , ( ∂ ξ − λ ( )) P ( ξ, t ; η, s ) = (cid:16) ∂ ξ ( ˜ R + ˜Γ) − e · [ M ( ξ ) ∇ ξ ( ˜ R + ˜Γ)] (cid:17) − (cid:0) λ ( ) − λ (Φ − ( ξ )) | J T ν ( ξ ) | (cid:1) ( ˜ R + ˜Γ) , t ∈ (0 , T ) , ξ = 0 ,P ( ξ, t ; η, s ) = 0 , t ≤ s, ξ < . (4.8)Denote by ˆΓ ( η, s ) ( ξ, t ) := ˆΓ( ξ, t ; η, s ) the solution to the following problem: − ( ∂ t + ∆ ξ )ˆΓ( ξ, t ; η, s ) = δ ( t − s ) δ ( ξ − η ) , t ∈ (0 , T ) , ξ < , ( ∂ ξ − λ ( )) ˆΓ( ξ, t ; η, s ) = 0 , t ∈ (0 , T ) , ξ = 0 , ˆΓ( ξ, t ; η, s ) = 0 , t ≥ s, ξ < . (4.9)Let Q be a small bounded domain in the lower half-plane such that ∈ ∂Q ∩ ∂ R − ⊂ ∂ (Φ( D )) and Q totallylies in Φ( D ). We decompose the boundary of Q into two disjoint parts such that ∂Q = ∂Q ∪ ∂Q with ∂Q = ∂Q ∩ ∂ R − . Since we only consider the singularity locally at , we have P ( η, s ) ( ξ, t ) = − Z ts dτ Z ∂Q (cid:16) P ( η, s ) ( z, τ ) ∂ ν ( z ) ˆΓ ( ξ, t ) ( z, τ ) − ∂ ν ( z ) P ( η, s ) ( z, τ )ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dσ ( z )+ Z ts dτ Z ∂Q (cid:16) ν ( z ) · ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ˆΓ ( ξ, t ) ( z, τ ) (cid:17) ds ( z ) − Z ts dτ Z Q (cid:16) ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ∇ ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz = − Z ts dτ Z ∂Q (cid:16) P ( η, s ) ( z, τ ) ∂ z ˆΓ ( ξ, t ) ( z, τ ) − ∂ z P ( η, s ) ( z, τ )ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz ′ + Z ts dτ Z ∂Q (cid:16) e · ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ˆΓ ( ξ, t ) ( z, τ ) (cid:17) ds ( z ) − Z ts dτ Z Q (cid:16) ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ∇ ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz + O (1) . (4.10)15o estimate (4.10), we need the following estimates [17, 18]: (cid:12)(cid:12)(cid:12) ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( τ − s ) − / exp (cid:16) − | z − ˆ η | c ( τ − s ) (cid:17) , (cid:12)(cid:12)(cid:12) ∇ z ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( τ − s ) − exp (cid:16) − | z − ˆ η | c ( τ − s ) (cid:17) , (cid:12)(cid:12)(cid:12) ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( τ − s ) − / exp (cid:16) − | z − η | c ( τ − s ) (cid:17) , (cid:12)(cid:12)(cid:12) ∇ z ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( τ − s ) − exp (cid:16) − | z − η | c ( τ − s ) (cid:17) , (cid:12)(cid:12)(cid:12) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( t − τ ) − / exp (cid:16) − | z − ξ | c ( t − τ ) (cid:17) , (cid:12)(cid:12)(cid:12) ∇ z ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ c ( t − τ ) − exp (cid:16) − | z − ξ | c ( t − τ ) (cid:17) , where ˆ η = ( η , η , − η ) T and c j ( j = 1 , · · · ,
4) are positive constants.On the plane z = 0, it holds that (cid:12)(cid:12)(cid:12) P ( η, s ) ( z, τ ) ∂ z ˆΓ ( ξ, t ) ( z, τ ) − ∂ z P ( η, s ) ( z, τ )ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:16) ∂ z ( ˜ R + ˜Γ) − e · [ M ( z ) ∇ z ( ˜ R + ˜Γ)] (cid:17) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:0) λ ( ) − λ (Φ − ( z )) | J T ν ( z ) | (cid:1) ( ˜ R + ˜Γ) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:16) e · [( I − M ( z )) ∇ z ( ˜ R + ˜Γ)] (cid:17) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:2) λ ( )(1 − | J T ν ( z ) | ) + ( λ ( ) − λ (Φ − ( z ))) | J T ν ( z ) | (cid:3) ( ˜ R + ˜Γ) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) ≤ C | z ′ | (cid:16)(cid:12)(cid:12)(cid:12) ∇ ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ∇ ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12)(cid:17) (cid:12)(cid:12)(cid:12) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) . (4.11)Set z ′ := ( z , z ), η ′ := ( η , η ) and ξ ′ := ( ξ , ξ ). Using Lemma 3 in [17, Chapter 1], we derive that Z ts dτ Z ∂Q | z ′ | (cid:16)(cid:12)(cid:12)(cid:12) ∇ ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ∇ ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12)(cid:17) (cid:12)(cid:12)(cid:12) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) dz ′ ≤ C Z ts dτ Z ∂Q ( | z ′ − η ′ | + | η ′ | ) (cid:16)(cid:12)(cid:12)(cid:12) ∇ ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ∇ ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12)(cid:17) (cid:12)(cid:12)(cid:12) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) dz ′ ≤ C Z ts dτ Z ∂Q | z ′ − η ′ | ( τ − s ) − exp (cid:18) − | z ′ − η ′ | + η c ( τ − s ) (cid:19) ( t − τ ) − / exp (cid:18) − | z ′ − ξ ′ | + ξ c ( t − τ ) (cid:19) dz ′ + C Z ts dτ Z ∂Q | η ′ | ( τ − s ) − exp (cid:18) − | z ′ − η ′ | + η c ( τ − s ) (cid:19) ( t − τ ) − / exp (cid:18) − | z ′ − ξ ′ | + ξ c ( t − τ ) (cid:19) dz ′ ≤ C η − ξ − Z ts dτ Z ∂Q ( τ − s ) − / exp (cid:18) − | z ′ − η ′ | c ( τ − s ) (cid:19) ( t − τ ) − / exp (cid:18) − | z ′ − ξ ′ | c ( t − τ ) (cid:19) dz ′ + C | η ′ | η − ξ − Z ts dτ Z ∂Q ( τ − s ) − exp (cid:18) − | z ′ − η ′ | c ( τ − s ) (cid:19) ( t − τ ) − / exp (cid:18) − | z ′ − ξ ′ | c ( t − τ ) (cid:19) dz ′ ≤ C η − ξ − ( t − s ) exp (cid:18) − | ξ ′ − η ′ | c ( t − s ) (cid:19) + C | η ′ | η − ξ − ( t − s ) / exp (cid:18) − | ξ ′ − η ′ | c ( t − s ) (cid:19) . (4.12)Noticing that Z ts dτ Z ∂Q | z ′ | (cid:16)(cid:12)(cid:12)(cid:12) ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ˜Γ ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12)(cid:17) (cid:12)(cid:12)(cid:12) ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) dz ′ . has a less singularity than (4.12), we omit the estimate of this integral here.In addition, the integral Z ts dτ Z ∂Q (cid:16) e · ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ˆΓ ( ξ, t ) ( z, τ ) (cid:17) ds ( z )16an be estimated in the same way as in (4.12). As for the volume integral in (4.10), we have that (cid:12)(cid:12)(cid:12)(cid:12)Z ts dτ Z Q (cid:16) ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ∇ ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ C Z ts dτ Z Q | z | (cid:12)(cid:12)(cid:12) ∇ ˜ R ( η, s ) ( z, τ ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ∇ ˆΓ ( ξ, t ) ( z, τ ) (cid:12)(cid:12)(cid:12) dz ≤ C ( t − s ) − exp (cid:18) − | ξ − ˆ η | c ( t − s ) (cid:19) + C | η | ( t − s ) − / exp (cid:18) − | ξ − ˆ η | c ( t − s ) (cid:19) ≤ C ( t − s ) − exp (cid:18) − | ξ + η | c ( t − s ) (cid:19) + C | η | ( t − s ) − / exp (cid:18) − | ξ + η | c ( t − s ) (cid:19) . (4.13)Take ξ = η with | η | ≤ cǫ , ξ + η = − ǫ, t − s = ǫ . Then we finally get the following estimates for ǫ → (cid:12)(cid:12)(cid:12)(cid:12)Z ts dτ Z ∂Q (cid:16) P ( η, s ) ( z, τ ) ∂ ˆΓ ( ξ, t ) ( z, τ ) − ∂ P ( η, s ) ( z, τ )ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz ′ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ǫ − , (cid:12)(cid:12)(cid:12)(cid:12)Z ts dτ Z Q (cid:16) ( M ( z ) − I ) ∇ ˜ R ( η, s ) ( z, τ ) · ∇ ˆΓ ( ξ, t ) ( z, τ ) (cid:17) dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ǫ − , which lead to the estimate (4.7) by (4.10). The proof of this lemma is complete. ✷ Next we derive the expression of W + ( ξ, t ; η, s ), and then show its asymptotic behavior. Lemma 4.2
The solution W + ( ξ, t ; η, s ) to (4.6) can be expressed by W + ( ξ, t ; η, s ) = 116 π i Z σ + i ∞ σ − i ∞ e tτ dτ Z R e iξ ′ · ζ ′ Θ + λ Θ(Θ − λ ) e − ( τs + iζ ′ · η ′ ) e Θ( ξ + η ) dζ ′ , (4.14) where ξ ′ := ( ξ , ξ ) , η ′ := ( η , η ) , λ := λ ( ) and Θ := p τ + | ζ ′ | with Re Θ ≥ . Proof.
Define H ( ξ, t ; η, s ) := W + ( ξ, t ; η, s ) + ˜Γ( ξ, t ; η, s ). We know from (4.6) that H ( ξ, t ; η, s ) satisfies ( ∂ t − ∆ ξ ) H ( ξ, t ; η, s ) = δ ( t − s ) δ ( ξ − η ) , t ∈ (0 , T ) , ξ < , ( ∂ ξ − λ ( )) H ( ξ, t ; η, s ) = 0 , t ∈ (0 , T ) , ξ = 0 ,H ( ξ, t ; η, s ) = 0 , t ≤ s, ξ < . (4.15)Denote by ˆ H the Laplace transform of H with respect to t . Then we have ( ( τ − ∆ ξ ) ˆ H ( ξ, t ; η, s ) = e − τs δ ( ξ − η ) , ξ < , ( ∂ ξ − λ ) ˆ H ( ξ, t ; η, s ) = 0 , ξ = 0 . (4.16)We now look for the solution to (4.16) in the form ofˆ H ( ξ, t ; η, s ) = ˆ H ± ( ξ, t ; η, s ) , ± ( ξ − η ) > , where ˆ H ± ( ξ, t ; η, s ) satisfy ( τ − ∆ ξ ) ˆ H ± ( ξ, t ; η, s ) = 0 , ± ( ξ − η ) > , ˆ H + − ˆ H − = 0 , ξ = η ,∂ ξ ( ˆ H + − ˆ H − ) = − e − τs δ ( ξ ′ − η ′ ) , ξ = η , ( ∂ ξ − λ ) ˆ H + ( ξ, t ; η, s ) = 0 , ξ = 0 . (4.17)17enote by ϕ ± the Fourier transform of ˆ H ± with respect to ξ ′ = ( ξ , ξ ) and let ζ ′ = ( ζ , ζ ) be the Fouriervariable corresponding to ξ ′ . Then we have ( τ + | ζ ′ | − ∂ ξ ) ϕ ± = 0 , ± ( ξ − η ) > ,ϕ + − ϕ − = 0 , ξ = η ,∂ ξ ( ϕ + − ϕ − ) = − e − τs e − iζ ′ · η ′ , ξ = η , ( ∂ ξ − λ ) ϕ + = 0 , ξ = 0 . (4.18)Since the operator ∂ ξ − Θ has the fundamental solutions e ± Θ ξ , we seek the solution to (4.18) in the formof ϕ + = c +1 e Θ ξ + c +2 e − Θ ξ , ϕ − = ce Θ ξ . The transmission conditions at ξ = η give (cid:26) ϕ + − ϕ − = c +1 e Θ η + c +2 e − Θ η − ce Θ η , − e − ( τs + iζ ′ · η ′ ) = Θ c +1 e Θ η − Θ c +2 e − Θ η − Θ ce Θ η . (4.19)The boundary condition at ξ = 0 leads to0 = (Θ c +1 − Θ c +2 ) − λ ( c +1 + c +2 ) = (Θ − λ ) c +1 − (Θ + λ ) c +2 , which implies c +2 = Θ − λ Θ + λ c +1 . (4.20)From (4.19) and (4.20), we obtain c +1 = 12 Θ − e Θ η Θ + λ Θ − λ e − ( τs + iζ ′ · η ′ ) , c +2 = 12 Θ − e Θ η e − ( τs + iζ ′ · η ′ ) . Thus, ϕ + is expressed by ϕ + = 12Θ e − ( τs + iζ ′ · η ′ ) (cid:26) e Θ( ξ + η ) Θ + λ Θ − λ + e Θ( η − ξ ) (cid:27) . (4.21)We note that in the brace of (4.21) the second term comes from the fundamental solution, while the first termcorresponds to the reflected solution W + . So (4.14) is obtained by taking the inverse Fourier and Laplacetransforms of (4.21). The proof is complete. ✷ Finally, we show the pointwise asymptotic behavior of W + ( ξ, t ; η, s ). Lemma 4.3
Let η = ( η , η , η ) T , where η = − ǫ/ for small ǫ > . Then we have W + ( η, s + ǫ ; η, s ) = ǫ − e / π / + O ( ǫ − ) ∼ ǫ − e / π / → ∞ as ǫ → . (4.22) Proof.
To analyze the asymptotic behavior of W + , let us first calculate I = I ( ζ ′ , t, s ) := Z σ + i ∞ σ − i ∞ e ( t − s ) τ Θ + λ Θ(Θ − λ ) e Θ( ξ + η ) dτ, ξ , η < σ − i ∞ to σ + i ∞ into a closed contour so that we can apply the residuetheorem. The contour used here is described in Figure 4.1.The contributions from the arcs ABC and
F GH are negligibly small as R → ∞ by Jordan’s lemma. LetΓ ρ be the circle with radius ρ centered at ( −| ζ ′ | , τ + | ζ ′ | = ρe iθ , we can show that Z Γ ρ e ( t − s ) τ Θ + λ Θ(Θ − λ ) e Θ( ξ + η ) dτ = Z − ππ e ( t − s )( −| ζ ′ | + ρe iθ ) √ ρe iθ/ + λ √ ρe iθ/ ( √ ρe iθ/ − λ ) e Θ( ξ + η ) iρe iθ dθ → ρ → . (4.24)18 e τ ABDE G HC Im τ F −| ζ ′ | Γ ρ Figure 4.1: In the contour marked with arrow, R is the radius of the half circle and Γ ρ is the circle withradius ρ centered at ( −| ζ ′ | , R be the radius of the half circle in the contour. By taking r = −| ζ ′ | − τ , we have Z −−→ CD e ( t − s ) τ Θ + λ Θ(Θ − λ ) e Θ( ξ + η ) dτ = Z R −| ζ ′ | e − ( t − s )( r + | ζ ′ | ) i √ r + λ i √ r ( i √ r − λ ) e i √ r ( ξ + η ) ( − dr )= ie − ( t − s ) | ζ ′ | Z R −| ζ ′ | e − ( t − s ) r − r + 2 iλ √ r + λ √ r ( r + λ ) e i √ r ( ξ + η ) dr → ie − ( t − s ) | ζ ′ | Z ∞ e − ( t − s ) r − r + 2 iλ √ r + λ √ r ( r + λ ) e i √ r ( ξ + η ) dr as R → ∞ . (4.25)Similar to the derivation of (4.25), we also have Z −−→ EF e ( t − s ) τ Θ + λ Θ(Θ − λ ) e Θ( ξ + η ) dτ → ie − ( t − s ) | ζ ′ | Z ∞ e − ( t − s ) r − r − iλ √ r + λ √ r ( r + λ ) e − i √ r ( ξ + η ) dr as R → ∞ . (4.26)Notice that ( λ − | ζ ′ | ,
0) is contained in the interior of the contour. By the residue theorem, we get I = 8 πλ ie ( t − s )( λ −| ζ ′ | ) e λ ( ξ + η ) − ie − ( t − s ) | ζ ′ | Re (cid:18)Z ∞ e − ( t − s ) r − r + 2 iλ √ r + λ √ r ( r + λ ) e i √ r ( ξ + η ) dr (cid:19) . (4.27)Define L = L ( t, s, ξ , η ; λ ) := Re (cid:18)Z ∞ e − ( t − s ) r − r + 2 iλ √ r + λ √ r ( r + λ ) e i √ r ( ξ + η ) dr (cid:19) . Then W + can be represented by W + ( ξ, t ; η, s ) = 116 π i Z R e i ( ξ ′ − η ′ ) · ζ ′ Idζ ′ = (cid:18) λ π e λ ( t − s )+ λ ( ξ + η ) − π L (cid:19) Z R e − ( t − s ) | ζ ′ | e i ( ξ ′ − η ′ ) · ζ ′ dζ ′ . (4.28)19y direct calculations, we obtain that Z R e − ( t − s ) ζ e i ( ξ − η ) ζ dζ = ( √ t − s ) − Z R e − ( ζ ′ ) e i ( √ t − s ) − ( ξ − η ) ζ ′ dζ ′ = ( √ t − s ) − e − ( t − s ) − ( ξ − η ) Z R e − (˜ ζ ) d ˜ ζ = √ π ( √ t − s ) − e − ( t − s ) − ( ξ − η ) . Similarly, we have Z R e − ( t − s ) ζ e i ( ξ − η ) ζ dζ = √ π ( √ t − s ) − e − ( t − s ) − ( ξ − η ) . Therefore, it follows from (4.28) that W + ( ξ, t ; η, s ) = (cid:18) λ π ( t − s ) e λ ( t − s )+ λ ( ξ + η ) − π ( t − s ) L (cid:19) e − ( t − s ) − | ξ ′ − η ′ | . (4.29)In the sequel, we compute L . By setting p = √ r , we have L = Re Z ∞ e − ( t − s ) p − p + 2 iλ p + λ p ( p + λ ) e ip ( ξ + η ) pdp = Re Z + ∞−∞ − p + λ p + λ e − ( t − s ) p + ip ( ξ + η ) dp − λ Im Z + ∞−∞ pp + λ e − ( t − s ) p + ip ( ξ + η ) dp. (4.30)To compute the integrals in (4.30), we introduce the contour described in Figure 4.2 so that the residuetheorem can be applied. Re σ Im σ −i λ Figure 4.2: The contour marked with arrow contains the point (0 , − λ ) in its interior.Note that − ( t − s ) p + ip ( ξ + η ) = − (cid:18) √ t − s p − i √ t − s ( ξ + η ) (cid:19) − ( ξ + η ) t − s ) , and define ˜ p := √ t − s p − i √ t − s ( ξ + η ) .
20t holds that − p = − i ( t − s ) − ( ξ + η ) (cid:26) ( t − s ) − / ˜ p + i t − s ) − ( ξ + η ) (cid:27) − ( ξ + η ) t − s ) − ˜ p ( t − s ) − = − i ( t − s ) − / ( ξ + η )˜ p + 14 ( ξ + η ) ( t − s ) − − ˜ p ( t − s ) − = A + ib ˜ p, where A := 14 ( ξ + η ) ( t − s ) − − ˜ p ( t − s ) − , b := − ( t − s ) − / ( ξ + η ) . Using the residue theorem and noticing that the integrals on Re p = ± R go to zero as R → ∞ , we obtainfrom (4.30) that L = 4 πλ e ( t − s ) λ e λ ( ξ + η ) + ( t − s ) − / e − ( ξ η t − s ) Z + ∞−∞ e − ˜ p − A + λ − b ˜ p ( − A + λ ) + b ˜ p d ˜ p − λ e − ( ξ η t − s ) Z + ∞−∞ e − ˜ p ( A − λ ) b + 2 b ˜ p ( t − s ) − ( − A + λ ) + b ˜ p d ˜ p. (4.31)Now take ξ = η with ξ = η = − ǫ/ t − s = ǫ . Then b = ǫ − and A = ǫ − ( − ˜ p ). By directcalculations, we have − A + λ − b ˜ p ( − A + λ ) + b ˜ p = − ǫ λ ˜ p + 2 ǫ λ − λ / p + (1 / ǫ λ )˜ p + ( ǫ λ − / , ( A − λ ) b + 2 b ˜ p ( t − s ) − ( − A + λ ) + b ˜ p = ˜ p − λ ǫ + 1 / p + (1 / ǫ λ )˜ p + ( ǫ λ − / . Hence, it can be easily seen that Z + ∞−∞ e − ˜ p − A + λ − b ˜ p ( − A + λ ) + b ˜ p d ˜ p = −√ π + O ( ǫ ) , Z + ∞−∞ e − ˜ p ( A − λ ) b + 2 b ˜ p ( t − s ) − ( − A + λ ) + b ˜ p d ˜ p = O (1)for ǫ →
0. Consequently, we have L = 4 πλ − e − / √ πǫ − + O (1) as ǫ → . Thus, in terms of (4.29), we finally obtain W + ( η, s + ǫ ; η, s ) = ǫ − e / π / + O ( ǫ − ) ∼ ǫ − e / π / → ∞ as ǫ → . This completes the proof. ✷ Based on the above lemmas, we conclude that W + is the dominant part of ˜ R as ǫ →
0. More explicitly,the pointwise asymptotic behavior of ˜ R ( ξ, t ; η, s ) can be stated as follows. Theorem 4.4
Let η = ( η , η , η ) T , where η = − ǫ/ for small ǫ > . Then we have ˜ R ( η, s + ǫ ; η, s ) = ǫ − e / π / + O ( ǫ − ) ∼ ǫ − e / π / → ∞ as ǫ → . (4.32) Remark 4.5
For any fixed discrepancy in (3.18) , combining (4.4) with (4.32) , we finally have the followingasymptotic behavior ( Sg y )( y, s + ǫ ) = ǫ − e / π / + O ( ǫ − ) as ǫ → . (4.33)21 Concluding remarks
This paper investigated the inverse problem of identifying an unknown Robin-type cavity inside a heat con-ductor from boundary measurements. The so-called linear sampling method was established to reconstructthe shape and location of the cavity. Based on our well-posedness analysis of the corresponding forward prob-lem, we gave rigorously a theoretical justification of this reconstruction scheme by using the layer potentialargument. Further, we proved a short time asymptotic behavior of the reflected solution of the fundamentalsolution, and hence the asymptotic behavior of the indicator function used in this method was obtained.Since the reflected solution is the compensating term of the Green function for the related initial-boundaryvalue problem, we actually provided a short time asymptotic behavior of the Green function. The asymptoticbehavior naturally yields a pointwise reconstruction scheme for the boundary of the cavity. We would alsolike to emphasize that from the asymptotic behavior we can know the distance to the unknown cavity as weprobe it from its inside. However, to establish a pointwise reconstruction formula for the Robin coefficient λ ,we need to carefully examine the lower order term where the information about the Robin coefficient shouldbe involved. This is one of our future works. Also, we intend to utilize this asymptotic behavior to generatea good numerical performance of this reconstruction scheme. Acknowledgement:
The first author is partially supported by National Research Foundation of Korea(No. 49771-01). The second author is supported by National Natural Science Foundation of China (Nos.11301075, 91330109) and Natural Science Foundation of Jiangsu Province of China (No. BK20130594).
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