Simultaneous recovery of a locally rough interface and its buried obstacles and homogeneous medium
SSimultaneous recovery of a locally rough interfaceand its buried obstacles and homogeneous medium
Jiaqing Yang ∗ Jianliang Li † Bo Zhang ‡ Abstract
Consider the inverse scattering of time–harmonic point sources by an infinite rough in-terface with buried obstacles in the lower half-space. The interface is assumed to be a localperturbation of a plane, which allows us to reduce the model problem to an equivalent inte-gral equation formulation defined in a bounded domain. The well–posedness is thus obtainedby employing the classical Fredholm theory. For the inverse problem, a global uniquenesstheorem is established which shows that the locally rough interface, the wave number inthe lower half-space and the buried obstacle can be uniquely determined by near-field datameasured only above the interface.
Keywords: inverse acoustic scattering, uniqueness, rough interfaces, buried obstacles.
This paper concerns the inverse scattering of time–harmonic acoustic point sources by an infinitelocally rough interface with an obstacle embedded in the lower half-space. This type of problemis motivated by a wide variety of diverse scientific areas such as radar techniques, underwaterexploration and non-destructive testing. In these areas one tries to reconstruct the locally roughinterface, the wave–number in the lower half-space, and the shape and the location, even theboundary condition of the embedded obstacle from the measurements of the scattered field insome certain domain.As seen in Figure 1, we denote the scattering interface by a curve Γ := { ( x , x ) ∈ R : x = f ( x ) } where f is assumed to be a Lipschitz continuous function with compact support,which means that the scattering interface Γ is a local perturbation of the planar interface Γ := { ( x , x ) ∈ R : x = 0 } . The scattering interface separates the whole space R into twohalf-spaces: the upper half-space Ω := { ( x , x ) ∈ R : x > f ( x ) } and the lower half-spaceΩ := { ( x , x ) ∈ R : x < f ( x ) } . And in the lower half-space Ω , we assume that there has anembedded object. In this paper, we investigate two cases, one is the case where the embeddedobject is impenetrable, denoted by D which is described by ∂D := { ( x ( t ) , x ( t )) : t ∈ [0 , π ) } ∗ School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, China( [email protected] ) † School of Mathematics and Statistics, Changsha University of Science and Technology, Changsha 410114,P.R. China ( [email protected] ) ‡ LSEC, NCMIS and Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing100190, China and School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049,China ( [email protected] ) a r X i v : . [ m a t h . A P ] F e b igure 1: The physical configuration of the scattering problem.with ∂D ∈ C ,α for some H¨older exponent 0 < α ≤
1; the other is the case where the embeddedobject is penetrable, known as refractive index, denoted by n ( x ) which is always positive andin our case n ( x ) = 1 for x ∈ R \ D for some D ⊂ Ω . Then the scattering of a time–harmonicacoustic point source by the complex scatterers (Γ , D ) can be modelled by ∆ u + κ u = 0 in R \ (cid:0) D ∪ { z } (cid:1) , B u = 0 on ∂D, lim r →∞ √ r (cid:18) ∂u s ∂r − i κu s (cid:19) = 0 for r = | x | , (1.1)and the scattering of a time–harmonic acoustic point source by (Γ , n ( x )) can be formulated by ∆ u + κ n ( x ) u = 0 in R \ { z } , lim r →∞ √ r (cid:18) ∂u s ∂r − i κu s (cid:19) = 0 for r = | x | , (1.2)where u = u ( x, z ) denotes the total field in R \ (cid:0) D ∪ { z } ) which consists of the incident pointsource u i ( x, z ) for z ∈ Ω and the scattered field u s ( x, z ) in Ω , and consists of the transmittedfield u s ( x, z ) in Ω . Here, κ = κ ( x ) denotes the wave–number satisfying κ = κ in Ω and κ = κ in Ω \ D where κ and κ are two positive constants. And B stands for the boundary conditionimposed on the boundary ∂D of the buried object D satisfying that B u := u when D is a sound-soft obstacle, B u := ∂ ν u with the outward normal ν directing into Ω \ D when D is a sound-hardobstacle, and B u := ∂ ν u + i λu with the impedance coefficient λ be a real-valued, continuous andnonnegative function when D is a mixed-type obstacle. In this paper, we consider two kinds ofincident wave, one is point source Φ κ ( x, z ), the other is a class of hyper-singular point source ∂ x l Φ κ ( x, z ) for l = 1 ,
2. Here, Φ κ ( x, z ) := i4 H (1)0 ( κ | x − z | ) is the fundamental solution of theHelmholtz equation satisfying ∆Φ κ ( x, z ) + κ Φ κ ( x, z ) = − δ ( x − z ) in R , where H (1)0 denotesthe Hankel function of the first kind of order zero, and δ is the Kronecker delta function. Itshould be pointed out that all these two kinds of incident wave belong to L p loc ( R ) for 1 < p < κ ( · , z ) ∈ H ,p loc ( R ) for 1 < p <
2. 2or the forward problem (1.1) or (1.2), given the unbounded rough interface Γ, the boundedburied obstacle D or n ( x ), and the incident wave, the direct scattering problem is to show thatthere admits a unique solution which depends continuously on the incidence in the sense ofsome suitable norm. For scattering by rough surfaces with no buried obstacles, there exists twomain approaches to establish the well–posedness, one is the variational method in [6, 24]; theother is the integral equation method in [19, 22, 26] which is based on the generalized Fredholmtheory developed in [11, 12]. However, the generalized Fredholm theory cannot be immediatelytransferred to the forward scattering problem because of the existence of the buried obstacle.For the forward scattering problem (1.1) or (1.2), by considering the difference of the solutionsassociated to (1.1) or (1.2) with and without the buried object corresponding to the same time–harmonic incident point source, we can first convert the problem into an equivalent Lippmann–Schwinger type integral equation defined in a bounded domain which avoids the unboundednessof the interface. Then with the help of the classical Fredholm theory we are able to establishthe well–posedness. It is worth pointing out that, compared with the variational method andthe integral equation method based on the generalized Fredholm theory, employing the noveltechnique proposed in this paper can obtain the L p (1 < p <
2) solution which treats the casewhere the incidence belongs to L p space, including a more extensive situation.The purpose of the inverse scattering problem is to determine the locally rough interfaceΓ, the wave–number κ in the lower half-space, and the embedded obstacle D with the natureof the boundary condition B or the embedded refractive index n ( x ) from a knowledge of thenear-field data measured above the interface. Due to the penetrability of the interface and theinterconnection of an unbounded rough interface and a bounded obstacle, the above inverseproblem is quite challenging and there is little available results. For the special case where theinfinite rough surface is perfectly conducting and there has no buried obstacles, the authors in [10]have established the unique determination of the rough surface from a knowledge of the scattereddata associated with incoming plane waves when the medium is lossy. For the electromagneticscattering of a point source by a perfectly electrically conducting obstacle which is embeddedin a two-layered lossy medium separated by an unbounded rough surface, a global uniquenessresult is established in [21] which shows that the obstacle and the unbounded rough surfacecan be uniquely determined by the scattered field. However, as far as we know, there is nouniqueness results for the general case where the medium is not lossy and the buried obstacle issound-soft, sound-hard, mixed-type or a penetrable refractive index. Motivated by [23,25] wherethe authors consider the inverse scattering of an inhomogeneous cavity and a bounded scatterer,respectively, we first prove that the locally rough interface, the wave–number in the lower half-space, and the embedded obstacle with its boundary condition or the refractive index can beuniquely determined by the near-field data measured only on a line segment above the interface.In the proof of this global uniqueness result, we employ two key ingredients. One is the well–posedness of an interior transmission problem in a small domain associated with the Helmholtzequation. Here, the smallness of the domain guarantees that the lowest transmission eigenvalueis large so that a given wave–number can not be the eigenvalue of the interior transmissionproblem. The other ingredient is a priori estimate of the solution to the interior transmissionproblem in the sense of L p with p ∈ (1 , In this section, we introduce some function spaces and the existence of the background Greenfunction which describes the scattering of a point source or a hyper–singular point source by theplanar interface Γ with no buried obstacles. For a bounded domain Ω ⊂ R with Lipschitz boundary ∂ Ω, let W m,p (Ω) be the usual Sobolevspace for index m ∈ N and p ∈ [1 , ∞ ). It is a Banach space equipped with the W m,p -norm || u || m,p := (cid:88) | α |≤ m || ∂ α u || pL p (Ω) /p . For p = 2, the Sobolev space W m, (Ω) is denoted by H m (Ω), which is a Hilbert space under theinner product ( u, v ) m := (cid:80) | α |≤ m ( ∂ α u, ∂ α v ) L (Ω) . For m = 0, W ,p (Ω) is reduced to the usual L p (Ω) space which consists of all L p –integrable functions on Ω. Moreover, we define H (Ω) := { u ∈ D (cid:48) (Ω) | u ∈ H (Ω) , ∆ u ∈ L (Ω) } which is also a Hilbert space with respect to the inner product( u, v ) H (Ω) = ( u, v ) L (Ω) + ( ∇ u, ∇ v ) L (Ω) + (∆ u, ∆ v ) L (Ω) for u, v ∈ H (Ω) . In this subsection, we introduce the existence of the background Green function of Helmholtzequation in a two–layered background medium [16]. Suppose the incident wave u i ( x, z ) is inducedby a point source Φ κ ( x, z ) or a hyper-singular point source ∂ x l Φ κ ( x, z ) for κ = κ or κ and l = 1 ,
2. Then the scattering of u i ( x, z ) by the planar Γ can be modelled by ∆ G + κ G = 0 in R \ { z } lim r →∞ √ r (cid:18) ∂G s0 ∂r − i κ G s0 (cid:19) = 0 for r = | x | (2.1)4here κ is the wave–number satisfying κ := κ in R := { ( x , x ) ∈ R : x > } and κ := κ in R − := { ( x , x ) ∈ R : x < } , and G s0 ( x, z ) denotes the scattered field Ψ ( x, z ) or Ψ ( x, z )when the observation point x and source point z belong to the same half–space, denotes thetransmitted filed Ψ ( x, z ) or Ψ ( x, z ) when the observation point x and source point z belongto the different half-space. More precisely, we have G ( x, z ) = Ψ ( x, z ) + u i ( x, z ) for x ∈ R , z ∈ R ,G ( x, z ) = Ψ ( x, z ) for x ∈ R − , z ∈ R ,G ( x, z ) = Ψ ( x, z ) for x ∈ R , z ∈ R − ,G ( x, z ) = Ψ ( x, z ) + u i ( x, z ) for x ∈ R − , z ∈ R − . Now we are in position to give the formulation of Ψ j ( x, z ), j = 1 , .., u i ( x, z ) = Φ κ ( x, z ) for z ∈ R and u i ( x, z ) = Φ κ ( x, z ) for z ∈ R − , by means ofthe Fourier transform, we haveΨ ( x, z ) = i4 π (cid:90) + ∞−∞ β β − β β + β e i β ( x + z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = i2 π (cid:90) + ∞−∞ β + β e i( β z − β x ) e i ξ ( x − z ) dξ Ψ ( x, z ) = i2 π (cid:90) + ∞−∞ β + β e i( β x − β z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = i4 π (cid:90) + ∞−∞ β β − β β + β e − i β ( x + z ) e i ξ ( x − z ) dξ, where β , β are defined by β j = (cid:113) κ j − ξ for | κ j | > | ξ | , i (cid:113) ξ − κ j for | κ j | < | ξ | , for j = 1 ,
2. When the incident wave is given by the first class of hyper-singular point sources,which means u i ( x, z ) = ∂ x Φ κ ( x, z ) for z ∈ R and u i ( x, z ) = ∂ x Φ κ ( x, z ) for z ∈ R − , we haveΨ ( x, z ) = − π (cid:90) + ∞−∞ β β − β β + β ξe i β ( x + z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = − π (cid:90) + ∞−∞ β + β ξe i( β z − β x ) e i ξ ( x − z ) dξ Ψ ( x, z ) = − π (cid:90) + ∞−∞ β + β ξe i( β x − β z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = − π (cid:90) + ∞−∞ β β − β β + β ξe − i β ( x + z ) e i ξ ( x − z ) dξ. For the last case where the incident wave is given by the second class of hyper-singular pointsource, which means u i ( x, z ) = ∂ x Φ κ ( x, z ) for z ∈ R and u i ( x, z ) = ∂ x Φ κ ( x, z ) for z ∈ R − ,5e obtain Ψ ( x, z ) = 14 π (cid:90) + ∞−∞ β − β β + β e i β ( x + z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = 12 π (cid:90) + ∞−∞ β β + β e i( β z − β x ) e i ξ ( x − z ) dξ Ψ ( x, z ) = − π (cid:90) + ∞−∞ β β + β e i( β x − β z ) e i ξ ( x − z ) dξ Ψ ( x, z ) = − π (cid:90) + ∞−∞ β − β β + β e − i β ( x + z ) e i ξ ( x − z ) dξ. Using the dominated convergence theorem to Ψ j ( x, z ) and their derivatives shows that Ψ j ( x, z ) ∈ C ∞ ( R × R \ { z } × { z } ) for j = 1 , ..., L p space for < p < This section is devoted to the well-posedness of the direct scattering problem (1.1) and (1.2).The uniqueness can be obtained by a direct application of the proposition 2.1 in [8]. For theexistence, we first consider a special case where D = ∅ or n ( x ) ≡
1, which is equivalent to aLippmann–Schwinger integral equation defined in a bounded domain by introducing a specialinterface. Based on this result, we then investigate the general case where D (cid:54) = ∅ or n ( x ) (cid:54)≡ L p (1 < p <
2) estimate of the solution to the problem, which will play a central rolein the proof of the global uniqueness result for the inverse problem in the next section.
Theorem 3.1.
The direct scattering problem (1.1) or (1.2) has at most one solution.
Proof.
Let u i = 0, from the proposition 2.1 in [8], it is sufficient to provelim r →∞ (cid:90) ∂B r (cid:32)(cid:12)(cid:12)(cid:12)(cid:12) ∂u∂ν (cid:12)(cid:12)(cid:12)(cid:12) + | u | (cid:33) ds = 0 , (3.1)where B r := { x ∈ R : | x | < r } . The equation (3.1) is a direct consequence of combination ofthe Sommerfeld radiation condition, the Green’s first theorem with the boundary condition on ∂D or the assumption n ( x ) > D = ∅ or n ( x ) ≡ In this subsection, we study the scattering of point sources by the locally rough interface Γ butwith no buried obstacle. Being different from the variational method in [6, 24] and the integralequation method in [19, 22, 26], we employ a novel technique to prove the existence. Simplyspeaking, by means of introducing a special interface, we can deduce an equivalent Lippmann–Schwinger integral equation defined in a bounded domain, which indeed has a unique solution in L p space by applying the classical Fredholm theory. By the novel technique, we can transformthe scattering from unbounded, rough interface to a Lippmann–Schwinger integral equationdefined in a bounded domain, which is the innovation of the technique.6s mentioned above, this subsection is devoted to investigate the scattering of point sourcesby the locally rough interface Γ without any embedded obstacle, which can be modelled by ∆ G Γ + κ G Γ = 0 in R \ { z } lim r →∞ √ r (cid:18) ∂G sΓ ∂r − i κ Γ G sΓ (cid:19) = 0 for r = | x | (3.2)where G Γ = G Γ ( · , z ) is the total field which satisfies G Γ ( · , z ) = u i ( · , z ) + G sΓ ( · , z ) in Ω and G Γ ( · , z ) = G sΓ ( · , z ) in Ω for the case z ∈ Ω . Here, κ Γ denotes the wave–number satisfies κ Γ = κ in Ω and κ Γ = κ in Ω .The uniqueness of (3.2) is a direct consequence of Theorem 3.1. To show the existence of(3.2), we introduce a special interface Γ R given byΓ R := { ( x , x ) ∈ R : x = 0 for | x | ≥ R and x = − (cid:113) R − x for | x | < R } , which is also a locally rough interface with the local perturbation described by the lower semi–circle. Now we consider the scattering of an incident point source u i by the locally rough surfaceΓ R , which reads ∆ G R + κ R G R = 0 in R \ { z } lim r →∞ √ r (cid:18) ∂G s R ∂r − i κ R G s R (cid:19) = 0 for r = | x | (3.3)where G R = G R ( · , z ) stands for the total field which consists of the incident point source u i ( · , z )for z ∈ Ω ,R and the scattered field G s R ( · , z ) in Ω ,R , and consists of the transmitted field G s R ( · , z )in Ω ,R . And κ R is the wave–number satisfying κ R = κ in Ω ,R and κ R = κ in Ω ,R . Here,Ω ,R and Ω ,R denote the upper and lower half-spaces separated by the locally rough surfaceΓ R .To obtain the existence of (3.2), it requires to prove the uniquely solvable of (3.3). To thisend, we recall that G ( · , z ) describes the scattering of u i ( · , z ) by the planar interface Γ , whichmeans that the filed G ( · , z ) contains the information of the plane Γ , note that the field G R ( · , z )contains the information of the surface Γ R , if we consider the difference G ( · , z ) := G R ( · , z ) − G ( · , z ) which, roughly speaking, only contains the information of the local perturbation B which is the domain bounded below by Γ R and bounded above by Γ , namely, B := { ( x , x ) ∈ R : − (cid:113) R − x < x < | x | < R } . A direct derivation using (2.1) and (3.3) implies that the difference G ( · , z ) satisfies the followingproblem ∆ G + κ G = ϕ in R , lim r →∞ √ r (cid:18) ∂G∂r − i κ G (cid:19) = 0 for r = | x | (3.4)where the function ϕ is defined as ϕ := ηG R ( · , z ) in B with η := κ − κ and ϕ := 0 in R \ B . With the well–posedness of (2.1), we can reduce the well–posedness of (3.3) to theuniquely solvable of (3.4). For the problem (3.4), we have the following result which shows thatit can be equivalently formulated to a Lippmann–Schwinger equation defined in the domain B .7 heorem 3.2. Let u i ( x, z ) ∈ L p loc ( R ) for < p < and fixed z ∈ R , if G s R ∈ W ,p loc ( R ) is thescattered field associated to the problem (3.3), then G R | B := ( G s R + u i ) | B is a solution to thefollowing Lippmann–Schwinger equation G R ( x, z ) + η (cid:90) B G ( x, y ) G R ( y, z ) dy = G ( x, z ) x ∈ B , (3.5) where G ( x, y ) denotes the solution of (2.1) associated with the incidence u i ( x, y ) = Φ κ ( x, y ) for y ∈ B . Conversely, if G R | B ∈ L p ( B ) is a solution to the Lippmann–Schwinger equation (3.5),then G s R := G R − u i can be extended to a solution to the problem (3.3) such that G s R ∈ W ,p loc ( R ) . Proof.
Let x ∈ B be an arbitrary point, we choose a sufficient small ε > B ε ( x ) with x as the centre and ε as the radius contains in the domain B . If G s R is the scatteredfield associated to the problem (3.3), applying the second Green’s theorem to G and G in thedomain B \ B ε ( x ) yields (cid:90) B \ B ε ( x ) (∆ G ( y, z ) G ( y, x ) − ∆ G ( y, x ) G ( y, z )) dy = (cid:40)(cid:90) Γ \ Γ R − (cid:90) Γ R \ Γ − (cid:90) ∂B ε ( x ) (cid:41) (cid:18) ∂G ( y, z ) ∂ν ( y ) G ( y, x ) − ∂ G ( y, x ) ∂ν ( y ) G ( y, z ) (cid:19) ds ( y ) , (3.6)where ν ( y ) denotes the upward unit normal vector when y ∈ Γ or y ∈ Γ R , and denotes theoutward unit normal vector when y ∈ ∂B ε ( x ). Employing (2.1) and (3.4) implies that the lefthand side of (3.6) will trend to η (cid:90) B G ( y, x ) G R ( y, z ) dy (3.7)as ε →
0. For ease of notations, we denote the right hand side of (3.6) by I , I , and I ,respectively. Substituting G ( y, x ) = Φ κ ( y, x )+Ψ ( y, x ) into the term I gives that I = I + I with definitions I := (cid:90) ∂B ε ( x ) (cid:18) ∂G ( y, z ) ∂ν ( y ) Φ κ ( y, x ) − ∂ Φ κ ( y, x ) ∂ν ( y ) G ( y, z ) (cid:19) ds ( y ) ,I := (cid:90) ∂B ε ( x ) (cid:18) ∂G ( y, z ) ∂ν ( y ) Ψ ( y, x ) − ∂ Ψ ( y, x ) ∂ν ( y ) G ( y, z ) (cid:19) ds ( y ) . A direct manipulation, using the mean value theorem, shows thatlim ε → I = G ( x, z ) . (3.8)A straightforward derivation using the equations that G and Ψ satisfy gives that I = η (cid:90) B ε ( x ) Ψ ( y, x ) G R ( y, z ) dy → ε → , (3.9)where we use the smooth of G R ( y, z ) and Ψ ( y, x ) for x ∈ B , y ∈ B ε ( x ) and z ∈ R . Hence,( ?? ) and ( ?? ) yield lim ε → I = G ( x, z ) . (3.10)8or I and I , let B R (cid:48) be the circle of radius R (cid:48) > R and center at the origin, and let B ± R (cid:48) denotethe upper and lower semi-circle, respectively. Application of the second Green’s theorem for G and G in the domain B + R (cid:48) and B − R (cid:48) \ B leads to I − I = (cid:90) ∂B R (cid:48) (cid:20)(cid:18) ∂G ( y, z ) ∂ν ( y ) − i κ G ( y, z ) (cid:19) G ( y, x ) − (cid:18) ∂ G ( y, x ) ∂ν ( y ) − i κ G ( y, x ) (cid:19) G ( y, z ) (cid:21) ds ( y )where ν ( y ) stands for the outward unit normal vector when y ∈ ∂B R (cid:48) . To prove (3.5), we firstshow that (cid:90) ∂B R (cid:48) | G ( y, z ) | ds ( y ) = O (1) , (cid:90) ∂B R (cid:48) | G ( y, x ) | ds ( y ) = O (1) , R (cid:48) → ∞ . (3.11)It follows from the Sommerfeld radiation condition that (cid:90) ∂B + R (cid:48) (cid:34)(cid:12)(cid:12)(cid:12)(cid:12) ∂G ( y, z ) ∂ν ( y ) (cid:12)(cid:12)(cid:12)(cid:12) + κ | G ( y, z ) | + 2 κ Im (cid:32) ∂G ( y, z ) ∂ν ( y ) G ( y, z ) (cid:33)(cid:35) ds ( y ) → , (3.12) (cid:90) ∂B − R (cid:48) (cid:34) κ κ (cid:12)(cid:12)(cid:12)(cid:12) ∂G ( y, z ) ∂ν ( y ) (cid:12)(cid:12)(cid:12)(cid:12) + κ κ | G ( y, z ) | + 2 κ Im (cid:32) ∂G ( y, z ) ∂ν ( y ) G ( y, z ) (cid:33)(cid:35) ds ( y ) → , (3.13)as R (cid:48) → ∞ . Applying the Green’s first theorem for G and G in B + R (cid:48) and B − R (cid:48) \ B implies (cid:90) ∂B R (cid:48) ∂G ( y, z ) ∂ν ( y ) G ( y, z ) ds ( y ) = (cid:40)(cid:90) Γ \ Γ R − (cid:90) Γ R \ Γ (cid:41) ∂G ( y, z ) ∂ν ( y ) G ( y, z ) ds ( y )+ (cid:90) B R (cid:48) \ B (cid:0) |∇ G ( y, z ) | − κ | G ( y, z ) | (cid:1) dy (3.14)We now insert the imaginary part of (3.14) into (3.12)-(3.13) and find that (cid:90) ∂B + R (cid:48) (cid:32)(cid:12)(cid:12)(cid:12)(cid:12) ∂G ( y, z ) ∂ν ( y ) (cid:12)(cid:12)(cid:12)(cid:12) + κ | G ( y, z ) | (cid:33) ds ( y ) + (cid:90) ∂B − R (cid:48) (cid:32) κ κ (cid:12)(cid:12)(cid:12)(cid:12) ∂G ( y, z ) ∂ν ( y ) (cid:12)(cid:12)(cid:12)(cid:12) + κ κ | G ( y, z ) | (cid:33) ds ( y ) → κ Im (cid:40)(cid:90) Γ R \ Γ − (cid:90) Γ \ Γ R (cid:41) ∂G ( y, z ) ∂ν ( y ) G ( y, z ) ds ( y ) , R (cid:48) → ∞ , which implies the first part in (3.11) holds. Similarly, we can prove that the second part in (3.11)holds. Thus, it follows from the Cauchy-Schwarz inequality using the Sommerfeld radiationcondition and (3.11) that I − I = 0. Hence, using (3.6), (3.7), (3.10) and the definition of G ( x, z ) := G R ( x, z ) − G ( x, z ) shows that G R | B is a solution G R ( x, z ) + η (cid:90) B G ( y, x ) G R ( y, z ) dy = G ( x, z ) x ∈ B . It is easily verified from the formulation of G that the reciprocity relation G ( y, x ) = G ( x, y )holds for x, y ∈ B , x (cid:54) = y , thus, we conclude that G R | B is a solution to (3.5).9onversely, let G R ( · , z ) ∈ L p ( B ) be a solution of (3.5) and define G R by G R ( x, z ) := G ( x, z ) − η (cid:90) B G ( x, y ) G R ( y, z ) dy x ∈ R (3.15)which gives that the scattered field G s R := G R − u i in Ω ,R and G s R := G R in Ω ,R belongs to W ,p loc ( R ) from the smoothness of G ( x, z ) for x (cid:54) = z and [14]. In the domain R , it is easily tosee ∆ G R + κ G R = 0 in R \{ z } ; in the domain B , we have ∆ G R + κ G R = ηG R = ( κ − κ ) G R which gives ∆ G R + κ G R = 0 in B ; and in the domain Ω ,R , we conclude ∆ G R + κ G R = 0;thus, we have G R satisfies the equation ∆ G R + κ R G R = 0 in R \ { z } . Since G satisfies theSommerfeld radiation condition, we conclude that G s R also satisfies the Sommerfeld radiationcondition. So G R given by (3.15) is a solution to the problem (3.2). The proof is finished.With the equivalence theorem 3.2, we can transform the well–posedness of the problem (3.3)into solving the Lippmann–Schwinger type equation (3.5) in L p ( B ). To this end, we define theintegral operator T : L p ( B ) → L p ( B ) by T ϕ ( x ) := (cid:90) B G ( x, y ) ϕ ( y ) dy, thus, the equation (3.5) can be rewritten in the following operator form( I + ηT ) G R = G in L p ( B ) . (3.16)where I : L p ( B ) → L p ( B ) is the identity operator. Now we are able to obtain the followingexistence result for the Lippmann–Schwinger type equation (3.16). Theorem 3.3.
For < p < , there exists a unique solution G R ∈ L p ( B ) to (3.16) such that (cid:107) G R (cid:107) L p ( B ) ≤ C (cid:107) G (cid:107) L p ( B ) . (3.17) Proof.
By the fact that the operator T : L p ( B ) → W ,p ( B ) is bounded and the Sobolevcompact embedding theorem, we conclude that T : L p ( B ) → L p ( B ) is compact. Hence, I + ηT : L p ( B ) → L p ( B ) is a Fredholm operator. By the Riesz-Fredholm theory, the existenceof a solution to (3.16) can be established from the uniqueness of (3.16). Let ( I + ηT ) ϕ = 0,then ϕ ( x ) = − η (cid:90) B G ( x, y ) ϕ ( y ) dy for x ∈ B , (3.18)which implies that ∆ ϕ + κ ϕ = 0 in B . Furthermore, we can extend ϕ to R \ B by theright hand side of (3.18). Thus, ϕ satisfies the Helmholtz equation ∆ ϕ + κ R ϕ = 0 in R andSommerfeld radiation condition. Then we conclude ϕ = 0 from the uniqueness of the problem(3.3) which is a direct consequence of theorem 3.1. Thus, the operator I + ηT : L p ( B ) → L p ( B )is injective, so it is bijective from the Riesz-Fredholm theory and has a bounded inverse whichimplies that (3.17) holds. The proof is completed.With the unique solvability of the problem (3.3), we are able to show the existence of asolution to (3.2). Recalling that the locally rough surface Γ is described by a Lipschitz continuousfunction denoted by f which has compact support, we choose a sufficient large R such that10upp f ⊂ [ − R, R ] and the local perturbation of Γ lies totally above the local perturbation of Γ R .For some fixed z ∈ Ω , we consider the scattering of the point source u i ( · , z ) by Γ R , which isrepresented by the problem (3.3). The displacement G R ( · , z ) contains the information of thesurface Γ R , and the field G Γ ( · , z ) contains the information of the surface Γ. If we consider thedifference v ( · , z ) := G Γ ( · , z ) − G R ( · , z ) which, roughly speaking, only contains the informationof the bounded domain B which is bounded below by Γ R and bounded above Γ, which reads B := { ( x , x ) ∈ R : − (cid:113) R − x < x < f ( x ) for | x | < R } . It follows from (3.2) and (3.3) that the difference v satisfies ∆ v + κ R v = ϕ in R lim r →∞ √ r (cid:18) ∂v∂r − i κ R v (cid:19) = 0 for r = | x | (3.19)where the function ϕ = − ηG Γ ( · , z ) in B and ϕ = 0 in R \ B . Since the problem (3.3) isuniquely solvable, we can reduce the well–posedness of (3.2) to the well–posedness of (3.19).In fact, by the similar arguments, it is not difficult to find that the problem (3.19) can beequivalently formulated as a Lippmann–Schwinger equation defined in the domain B . Theorem 3.4.
Let u i ( x, z ) ∈ L p loc ( R ) for < p < and fixed z ∈ Ω , if G sΓ ∈ W ,p loc ( R ) is thescattered field associated to the problem (3.2), then G Γ | B := ( G sΓ + u i ) | B is a solution to thefollowing Lippmann–Schwinger equation G Γ ( x, z ) − η (cid:90) B G R ( x, y ) G Γ ( y, z ) dy = G R ( x, z ) x ∈ B , (3.20) where G R ( x, y ) denotes the solution to the problem (3.3) associated with incidence u i ( x, y ) =Φ κ ( x, y ) . Conversely, if G Γ | B ∈ L p ( B ) is a solution to the Lippmann–Schwinger equation(3.20), then G sΓ := G Γ − u i can be extended to a solution to the problem (3.2) such that G sΓ ∈ W ,p loc ( R ) . For the solvability of (3.20), we have the following result.
Theorem 3.5.
For < p < , there exists a unique solution G Γ ∈ L p ( B ) to (3.20) such that (cid:107) G Γ (cid:107) L p ( B ) ≤ C (cid:107) G R (cid:107) L p ( B ) . (3.21)The proof of Theorem 3.4 and Theorem 3.5 are analogous to Theorem 3.2 and Theorem 3.3,we omit it here. D (cid:54) = ∅ or n ( x ) (cid:54)≡ Based on the well-posedness of (3.2) for the special case D = ∅ or n ( x ) ≡
1, we are able todeal with the general case D (cid:54) = ∅ or n ( x ) (cid:54)≡
1. For some fixed source location z ∈ Ω , the field G Γ ( · , z ) includes the information of the surface Γ, and the field u ( · , z ) includes the informationof the surface Γ and the embedded obstacle D or the refractive index n ( x ). Similarly, if weconsider the difference w ( · , z ) := u ( · , z ) − G Γ ( · , z ) which, roughly speaking, only includes the11nformation of D or n ( x ). For the impenetrable case, using (1.1) and (3.2), it is easy to see thatthe difference w solves ∆ w + κ w = 0 in R \ D B w = −B G Γ on ∂D lim r →∞ √ r (cid:18) ∂w∂r − i κw (cid:19) = 0 for r = | x | . (3.22)For the penetrable case, it follows from (1.2) and (3.2) that the difference w satisfies ∆ w + κ nw = κ (1 − n ) G Γ ( · , z ) in R lim r →∞ √ r (cid:18) ∂w∂r − i κw (cid:19) = 0 for r = | x | . (3.23)Thus, to establish the well–posedness of (1.1) or (1.2), it suffices to prove that the problem(3.22) or (3.23) is uniquely solvable. Theorem 3.6.
The problem (1.1) admits a unique solution.Proof.
As mentioned above, it is enough to show (3.22) is uniquely solvable. The uniqueness of(3.22) follows immediately from theorem 3.1. Now we restrict ourselves to the existence. Fora sound-soft buried obstacle D , one tries to seek a solution in the form of a combined acousticdouble and single-layer potential w ( x ) = (cid:90) ∂D (cid:18) ∂ G Γ ( x, y ) ∂ν ( y ) − i G Γ ( x, y ) (cid:19) ψ ( y ) ds ( y ) for x ∈ R \ D (3.24)with some unknown density ψ ∈ C ( ∂D ). Here G Γ ( x, y ) denotes the background Green functionwhich describes the scattering of a point source Φ κ ( x, y ) located at y ∈ Ω by the interface Γwith no buried object. The unique solvability of G Γ ( x, y ) is a direct result of the well–posednessof the problem (3.2).Then from the boundary condition imposed on the buried obstacle D , we see that thepotential (3.24) solve the problem (3.22) provided the density ψ is a solution of a second-kindintegral equation as follows ( I + K − i S ) ψ = − G Γ on ∂D with the single- and double-layer operators S and K given by( Sψ )( x ) := 2 (cid:90) ∂D G Γ ( x, y ) ψ ( y ) ds ( y ) x ∈ ∂D ( Kψ )( x ) := 2 (cid:90) ∂D ∂ G Γ ( x, y ) ∂ν ( y ) ψ ( y ) ds ( y ) x ∈ ∂D. From the fact ∂D ∈ C ,α with some H¨older exponent 0 < α ≤ S, K : C ( ∂D ) → C ,α ( ∂D ) are bounded. Combined with the compact embeddingtheorem from C ,α ( ∂D ) to C ( ∂D ), we conclude that I + K − i S : C ( ∂D ) → C ( ∂D ) is a Fredholmoperator. By a similar argument with theorem 3.11 in [9], the existence of a density ψ can be12stablished with the aid of the Riesz-Fredholm theory for compact operators. Furthermore, wehave the following estimate (cid:107) ψ (cid:107) C ( ∂D ) ≤ C (cid:107) G Γ (cid:107) C ( ∂D ) . (3.25)The ideas used in this existence proof can be immediately extended to other boundary conditionsby making the obvious modifications in (3.24). The proof of the existence is completed. Theorem 3.7.
The problem (1.2) admits a unique solution.Proof.
The uniqueness follows from theorem 3.1. For the existence, we recall w ( x, z ) := u ( x, z ) − G Γ ( x, z ) satisfies the problem (3.23). Using Theorem 8.3 in [9] implies that (3.23) is equivalentto the following Lippmann-Schwinger equation w ( x, z ) = − κ (cid:90) R G Γ ( x, y ) m ( y )( w ( y, z ) + G Γ ( y, z )) dy (3.26)where m := 1 − n and G Γ ( x, y ) denotes the solution of (3.2) associated with incidence u i ( x, y ) =Φ κ ( x, y ). Replacing w ( x, z ) by u ( x, z ) − G Γ ( x, z ) in (3.26) gives that u ( x, z ) + κ (cid:90) R G Γ ( x, y ) m ( y ) u ( y, z ) dy = G Γ ( x, z ) . (3.27)By the similar arguments with theorem 3.3, we obtain that (3.27) is uniquely solvable. This section considers the uniqueness of the inverse problem, which consists of two subsections,one is devoted to the case where D is an impenetrable obstacle; the other is devoted to the casewhere D is a penetrable inhomogeneous medium. The purpose of this subsection is to prove the global uniqueness result in determining thelocally rough surface, the wave number in the lower half-space, the buried obstacle and itsboundary condition from the measurement of the scattered field on a line segment denoted byΓ b,a := { x ∈ R : x = b, | x | ≤ a } where b > sup x ∈ R f ( x ). Theorem 4.1. (Global uniqueness) Given κ > , let u s1 ( x, y ) and u s2 ( x, y ) be the scatteredfields of the problem (1.1) associated with the incident point source u i ( x, y ) = Φ κ ( x, y ) and withrespect to the complex scatterers (Γ , κ , , D , B ) and (Γ , κ , , D , B ) , respectively. Assumethat u s1 ( x, y ) = u s2 ( x, y ) for all x, y ∈ Γ b,a , then Γ = Γ , κ , = κ , , D = D and B = B .Proof. We divide the proof in four steps.
Step 1 : In this step, we demonstrate a generalized mixed reciprocity which will be used instep 2. For z , z ∈ Ω ∪ (Ω \ D ) with z (cid:54) = z , let u ( · , z ) and u ( · , z ) be the solutions to (1.1)associated with two different incident sources u i ( · , z ) and u i ( · , z ), respectively. And we definetwo functions E ε ( z ) and F ε ( z ) by E ε ( z ) := (cid:90) ∂B ε ( z ) (cid:20) ∂u i ( y, z ) ∂ν ( y ) u ( y, z ) − ∂u ( y, z ) ∂ν ( y ) u i ( y, z ) (cid:21) ds ( y )13 ε ( z ) := (cid:90) ∂B ε ( z ) (cid:20) ∂u i ( y, z ) ∂ν ( y ) u ( y, z ) − ∂u ( y, z ) ∂ν ( y ) u i ( y, z ) (cid:21) ds ( y ) , where B ε ( z ) and B ε ( z ) are two circles centered at z and z , respectively, with radius ε > B ε ( z ) ∩ B ε ( z ) = ∅ . It follows from the same procedure as theorem 2.3in [23] that lim ε → E ε ( z ) = lim ε → F ε ( z ) for z , z ∈ Ω ∪ (Ω \ D ) . (4.1)Throughout, we consider two classes of incident wave Φ κ ( x, z ) and ∂ x l Φ κ ( x, z ) for l = 1 , u i ( x, z ) :=Φ κ ( x, z ) and (cid:101) u i ( x, z, l ) := ∂ x l Φ κ ( x, z ). Moreover, to distinguish corresponding total waves andscattered waves, we will always use u ( x, z ), u s ( x, z ) and (cid:101) u ( x, z, l ), (cid:101) u s ( x, z, l ) to denote the totalfields and the scattered fields associated with (1.1) corresponding to the incident wave u i ( x, z )and (cid:101) u i ( x, z, l ), respectively.Assume that u i ( · , z ) = Φ κ ( · , z ) and (cid:101) u i ( · , z , l ) = ∂ x l Φ κ ( · , z ), then it follows from (4.1)that (cid:101) u ( z , z , l ) = lim ε → E ε ( z ) = lim ε → F ε ( z ) , which implies (cid:101) u ( z , z , l ) = lim ε → (cid:90) ∂B ε ( z ) (cid:20) ∂ (cid:101) u i ( y, z , l ) ∂ν ( y ) u ( y, z ) − ∂u ( y, z ) ∂ν ( y ) (cid:101) u i ( y, z , l ) (cid:21) ds ( y ) . (4.2)The equation (4.2) plays a key role in the proof of the uniqueness of Γ. Step 2 : In this part, we show that the locally rough interface Γ is uniquely determined bythe scattered field u s ( x, y ) for all x, y ∈ Γ b,a .Assume on the contrary that Γ (cid:54) = Γ which implies that, without loss of generality, thereexists z ∗ ∈ Γ and z ∗ / ∈ Γ . Furthermore, we can choose ε > δ > z n := z ∗ + δ n ν ( z ∗ ) for n = 1 , , ... (4.3)is contained in B ε ( z ∗ ) for all n and B ε ( z ∗ ) ∩ Ω (Γ ) = ∅ .Let (cid:101) u ( · , z n , l ), (cid:101) u ( · , z n , l ) denote the solutions to (1.1) with (Γ , κ , D, B ) replaced by(Γ , k , , D , B ), (Γ , k , , D , B ), respectively. Then we aim to show that (cid:101) u ( y, z n , l ) = (cid:101) u ( y, z n , l ) for y ∈ Ω (Γ ) ∩ Ω (Γ ) \ { z n } (4.4)holds for l = 1 , n ∈ N . In view of (4.2), we have (cid:101) u m ( y, z n , l ) = lim ε → (cid:90) ∂B ε ( z n ) (cid:20) ∂ (cid:101) u i ( x, z n , l ) ∂ν ( x ) u m ( x, y ) − ∂u m ( x, y ) ∂ν ( x ) (cid:101) u i ( x, z n , l ) (cid:21) ds ( x ) . (4.5)for m = 1 ,
2. For y ∈ Γ b,a , it is apparent from u s1 ( · , y ) = u s2 ( · , y ) on Γ b,a and the analyticityof the scattered field that u s1 ( · , y ) = u s2 ( · , y ) on Γ b := { ( x , x ) ∈ R : x = b } , which implies u s1 ( · , y ) = u s2 ( · , y ) on Ω (Γ ) ∩ Ω (Γ ) due to the uniqueness of the Dirichlet problem and the14nalytic continuation principle. Hence, u ( · , y ) = u ( · , y ) on Ω (Γ ) ∩ Ω (Γ ) \{ y } . Thus, for any z n ∈ Ω (Γ ) ∩ Ω (Γ ), we can choose sufficient small ε > ∂B ε ( z n ) ⊂ Ω (Γ ) ∩ Ω (Γ ),so we have u ( · , y ) = u ( · , y ) on ∂B ε ( z n ) which yields (4.4) holds for y ∈ Γ b,a from (4.5). Theanalyticity, the uniqueness of the Dirichlet problem and the analytic continuation principle willgive that (4.4) holds for y ∈ Ω (Γ ) ∩ Ω (Γ ) \ { z n } . Define v ( · , z n , l ) := (cid:101) u ( · , z n , l ) | D z ∗ , v ( · , z n , l ) := (cid:101) u ( · , z n , l ) | D z ∗ where D z ∗ := B ε ( z ∗ ) ∩ Ω (Γ ) ∩ Ω (Γ ) which can be chosen as a Lipschitz domain with a sufficiently small ε > (cid:107) v ( · , z n , l ) (cid:107) L ( D z ∗ ) + (cid:107) v ( · , z n , l ) (cid:107) L ( D z ∗ ) ≤ C (4.6)holds uniformly for l = 1 , n ∈ N , and some positive constant C independent of n ∈ N .According to the definitions, it is clear that v and v solve the following interior transmissionproblem ∆ v + κ , v = 0 in D z ∗ , ∆ v + κ v = 0 in D z ∗ ,v − v = f l on ∂D z ∗ ,∂ ν ( v − v ) = f l on ∂D z ∗ . (4.7)with f l ∈ H ( ∂D z ∗ ) and f l ∈ H − ( ∂D z ∗ ). Furthermore, it follows from (4.4) that f l = 0 and f l = 0 on B ε ( z ∗ ) ∩ Γ .Actually, we can choose ε sufficiently small to ensure the uniqueness of the interior trans-mission problem (4.7) since the smallest real eigenvalue of (4.7) trends to + ∞ as ε → f l and f l satisfy the condition : for f l ∈ H ( ∂D z ∗ ) and f l ∈ H − ( ∂D z ∗ ), there exists afunction h ∈ H ( D z ∗ ) satisfying h = f l and ∂ ν h = f l on ∂D z ∗ . In this case, the following L estimate holds (cid:107) v (cid:107) L ( D z ∗ ) + (cid:107) v (cid:107) L ( D z ∗ ) ≤ C ( (cid:107) f l (cid:107) H ( ∂D z ∗ ) + (cid:107) f l (cid:107) H − ( ∂D z ∗ ) ) (4.8)where (cid:107) f l (cid:107) H ( ∂D z ∗ ) and (cid:107) f l (cid:107) H − ( ∂D z ∗ ) are defined by (cid:107) f l (cid:107) H ( ∂D z ∗ ) + (cid:107) f l (cid:107) H − ( ∂D z ∗ ) ≤ inf h satisfies condition (cid:0) (cid:107) h (cid:107) H ( D z ∗ ) + (cid:107) ∆ h (cid:107) L ( D z ∗ ) (cid:1) . (4.9)The interior transmission problem (4.7) motivates us to construct a function h := (1 − χ )( (cid:101) u ( · , z n , l ) − (cid:101) u ( · , z n , l )) in D z ∗ , (4.10)where χ ∈ C ( R ) is a cut-off function which satisfies χ = 1 in B ε ( z ∗ ) for some ε < ε and χ = 0 in R \ B ε ( z ∗ ). Obviously, the construction satisfies h = f l and ∂ ν h = f l on ∂D z ∗ . By (4.8), (4.9) and (4.10), together with the fact that (cid:101) u m ( · , z n , l ) ( m = 1 ,
2) satisfies theHelmholtz equation in D with D := D z ∗ \ B ε ( z ∗ ), we conclude (cid:107) v (cid:107) L ( D z ∗ ) + (cid:107) v (cid:107) L ( D z ∗ ) ≤ C ( ε , κ , κ , ) (cid:0) (cid:107) (cid:101) u ( · , z n , l ) (cid:107) H ( D ) + (cid:107) (cid:101) u ( · , z n , l ) (cid:107) H ( D ) (cid:1) . (4.11)15wing to the bounded embedding of W ,p into H for 1 < p <
2, we have that (cid:107) (cid:101) u m ( x, z n , l ) (cid:107) H ( D ) ≤ C (cid:107) (cid:101) u m ( x, z n , l ) (cid:107) W ,p ( D ) (4.12)holds for m, l = 1 ,
2. And utilizing the triangle inequality leads to (cid:107) (cid:101) u m ( x, z n , l ) (cid:107) W ,p ( D ) ≤ (cid:107) (cid:101) u m ( x, z n , l ) − (cid:101) G Γ m ( x, z n , l ) (cid:107) W ,p ( D ) + (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) W ,p ( D ) , (4.13)where (cid:101) G Γ m ( · , z n , l ) denotes the solution to (3.2) with Γ replaced by Γ m ( m = 1 ,
2) and theincident wave is given by (cid:101) u i ( x, z n , l ) = ∂ x l Φ κ ( x, z n ). Noting that the difference (cid:101) u m ( x, z n , l ) − (cid:101) G Γ m ( x, z n , l ) solves the problem (3.22), then the estimate (3.25) and the fact that a positivedistance from D to ∂D and ∂D imply (cid:107) (cid:101) u m ( x, z n , l ) − (cid:101) G Γ m ( x, z n , l ) (cid:107) W ,p ( D ) ≤ C (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) C ( ∂D m ) ≤ (cid:101) C, (4.14)where (cid:101) C is independent of n ∈ N .For the second term in (4.13), employing of (3.20) yields that (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) W ,p ( D ) ≤ C (cid:16) (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) L p ( B ,m ) + (cid:107) (cid:101) G R ( x, z n , l ) (cid:107) W ,p ( D ) (cid:17) (4.15)where B ,m := Ω (Γ R ) ∩ Ω (Γ m ) denotes the region between Γ R and Γ m . Now we aim to showthat (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) L p ( B ,m ) is uniformly bounded. To this end, noticing that for the incidentwave given by (cid:101) u i ( x, z n , l ) = ∂ x l Φ κ ( x, z n ) = 1 | x − z n | + O (1)which shows that (cid:107) (cid:101) u i ( x, z n , l ) (cid:107) L p ( B ,m ) is uniformly bounded for 1 < p < l = 1 , n ∈ N .Therefore, (cid:107) (cid:101) G Γ m ( x, z n , l ) (cid:107) L p ( B ,m ) ≤ (cid:107) (cid:101) u i ( x, z n , l ) (cid:107) L p ( B ,m ) + (cid:107) (cid:101) G sΓ m ( x, z n , l ) (cid:107) L p ( B ,m ) ≤ C (4.16)In view of a positive distance from z n to Γ R , we conclude that there exists a constant C which is independent of n ∈ N such that (cid:107) (cid:101) G R ( x, z n , l ) (cid:107) W ,p ( D ) ≤ C (4.17)holds uniformly for any n ∈ N .Noticing that (4.11)-(4.17), we conclude (4.6) holds and it implies that (cid:107) (cid:88) l =1 , ν l ( z ∗ ) v ( · , z n , l ) (cid:107) L ( D z ∗ ) + (cid:107) (cid:88) l =1 , ν l ( z ∗ ) v ( · , z n , l ) (cid:107) L ( D z ∗ ) ≤ C (4.18)where ν ( z ∗ ) := ( ν ( z ∗ ) , ν ( z ∗ )) denotes the upward unit normal at z ∗ . By linearity, it is obviousthat (cid:80) l =1 , ν l ( z ∗ ) v ( · , z n , l ) is the restriction of the solution to (1.1) with Γ replaced by Γ on thedomain D z ∗ and the incident wave given by (cid:98) u i ( x, z n ) := (cid:88) l =1 , ν l ( z ∗ ) (cid:101) u i ( x, z n , l ) . (cid:107) (cid:88) l =1 , ν l ( z ∗ ) v ( · , z n , l ) (cid:107) L ( D z ∗ ) ≥ (cid:107) (cid:98) u i ( · , z n ) (cid:107) L ( D z ∗ ) − (cid:107) (cid:88) l =1 , ν l ( z ∗ ) v ( · , z n , l ) − (cid:98) u i ( · , z n ) (cid:107) L ( D z ∗ ) (4.19)A direct calculation leads to (cid:107) (cid:98) u i ( · , z n ) (cid:107) L ( D z ∗ ) = O ( n ) for sufficiently large n . And it is obviousthat the second term of (4.19) is uniformly bounded for any n . Hence (cid:107) (cid:88) l =1 , ν l ( z ∗ ) v ( · , z n , l ) (cid:107) L ( D z ∗ ) → + ∞ which contradicts with (4.18). So we conclude Γ = Γ . Step 3 : In this part, we prove that the wave number in the lower half-space is uniquelydetermined by the scattered field u s ( x, y ) for all x, y ∈ Γ b,a .Let Γ := Γ = Γ and assume κ , (cid:54) = κ , . We choose z ∗ ∈ Γ and define z n as (4.3).Recalling that (cid:101) u ( · , z n , l ), (cid:101) u ( · , z n , l ) denote the solutions of the scattering problems (1.1) with(Γ , κ , D, B ) replaced by (Γ , κ , , D , B ), (Γ , κ , , D , B ), respectively, with the point sources (cid:101) u i ( x, z n , l ) = ∂ x l Φ κ ( x, z n ). By the same analysis as in step 1, we conclude that (cid:101) u ( x, z n , l ) = (cid:101) u ( x, z n , l ) in Ω . (4.20)For z ∗ ∈ Γ, we choose a Lipschitz domain D (cid:48) z ∗ := B ε ( z ∗ ) ∩ Ω and define (cid:101) v ( x, z n , l ) := (cid:101) u ( x, z n , l ) | D (cid:48) z ∗ and (cid:101) v ( x, z n , l ) := (cid:101) u ( x, z n , l ) | D (cid:48) z ∗ . Then (cid:101) v and (cid:101) v satisfy the following interiortransmission problem ∆ (cid:101) v + κ , (cid:101) v = 0 in D (cid:48) z ∗ , ∆ (cid:101) v + κ , (cid:101) v = 0 in D (cid:48) z ∗ , (cid:101) v − (cid:101) v = g l on ∂D (cid:48) z ∗ ,∂ ν ( (cid:101) v − (cid:101) v ) = g l on ∂D (cid:48) z ∗ . with g l ∈ H ( ∂D z ∗ ) and g l ∈ H − ( ∂D z ∗ ). From (4.20), we deduce that g l = 0 and g l = 0 on B ε ( z ∗ ) ∩ Γ. The remaining part of this proof is completely analogous to the step 1, so we omitit here.
Step 4 : In this step, we show that the embedded obstacle and its boundary condition areuniquely determined by the scattered field u s ( x, y ) for all x, y ∈ Γ b,a .Let κ := κ , = κ , and assume that D (cid:54) = D . Then, without loss of generality, thereexists z ∗ ∈ ∂D but z ∗ / ∈ ∂D . Thus, we define z n := z ∗ + 1 n ν ( z ∗ ) for n ∈ N , where ν ( z ∗ ) is the outward unit normal vector at z ∗ .Now we show that u s1 ( y, x ) = u s2 ( y, x ) for x, y ∈ Ω \ D ∪ D and y (cid:54) = x. (4.21)For each fixed y ∈ Γ b,a , we have u s1 ( x, y ) = u s2 ( x, y ) for x ∈ Γ b,a , then it follows from theanalyticity of the scattered field on Γ b that u s1 ( x, y ) = u s2 ( x, y ) for x ∈ Γ b , which implies17 s1 ( x, y ) = u s2 ( x, y ) for x ∈ U b := { ( x , x ) ∈ R : x > b } because of the uniqueness of theDirichlet problem. And using the analytic continuation principle gives that u s1 ( x, y ) = u s2 ( x, y )for x ∈ Ω which implies that u ( x, y ) = u ( x, y ) for x ∈ Ω and y ∈ Γ b,a . Hence, we obtain u ( x, y ) | + = u ( x, y ) | + and ∂u ( x, y ) ∂ν ( x ) (cid:12)(cid:12)(cid:12) + = ∂u ( x, y ) ∂ν ( x ) (cid:12)(cid:12)(cid:12) + on Γ , where ·| + indicates the limit of the functions which are approached from Ω . Since u ( x, y ), u ( x, y ) and their normal derivatives are continuous across the interface Γ, we arrive at u ( x, y ) | − = u ( x, y ) | − and ∂u ( x, y ) ∂ν ( x ) (cid:12)(cid:12)(cid:12) − = ∂u ( x, y ) ∂ν ( x ) (cid:12)(cid:12)(cid:12) − on Γ , where ·| − indicates the limit of the functions which are approached from Ω \ D ∪ D . Thus, itfollows from the Holmgren’s theorem (Theorem 2.3 in [9]) that u ( x, y ) = u ( x, y ) for x ∈ Ω \ D ∪ D and y ∈ Γ b,a . (4.22)Employing (4.1) yields that u ( y, x ) = u ( y, x ) for y ∈ Γ b,a and x ∈ Ω \ D ∪ D . (4.23)By a similar argument, we have u ( y, x ) = u ( y, x ) for y, x ∈ Ω \ D ∪ D and y (cid:54) = x, (4.24)which implies that (4.21) holds. Hence u s1 ( z ∗ , z n ) = u s2 ( z ∗ , z n ) . (4.25)On one hand, since u s2 ( z ∗ , · ) is continuously differentiable in a neighborhood of z ∗ / ∈ D due tothe reciprocity (4.1), we have lim n →∞ B u s1 ( z ∗ , z n ) = B u s2 ( z ∗ , z ∗ ) . (4.26)On the other hand, using the boundary condition on ∂D , we find thatlim n →∞ B u s1 ( z ∗ , z n ) = − lim n →∞ B Φ κ ( z ∗ , z n ) = ∞ , which contradicts (4.26), and there D = D .Finally, we show that B = B . We set D := D = D and assume that B (cid:54) = B . Since wecan see the Neumann boundary condition as a special case of the impedance boundary conditionwhere the impedance coefficient vanishes, we just need to consider two cases: For the first case, B is Dirichlet boundary condition and B is impedance boundary condition; For the secondcase, both of B and B are impedance boundary conditions but the impedance coefficients aredifferent, which means λ (cid:54) = λ . By (4.24), we set u ( y, x ) = u ( y, x ) = u ( y, x ) for y, x ∈ Ω \ D and y (cid:54) = x . For the first case, we have B u = u = 0 on ∂D and B u = ∂u∂ν + i κ λu = 0 on ∂D which indicates that ∂u∂ν = u = 0 on ∂D . Thus, by the Holmgren’s theorem, we obtain u ( y, x ) = 0 for y, x ∈ Ω \ D and y (cid:54) = x . So u s ( y, x ) = − u i ( y, x ) = − Φ κ ( y, x ) for y, x ∈ Ω \ D y (cid:54) = x , which contradicts the fact that u s ( y, x ) is smooth at y = x but Φ κ ( y, x ) is singularat y = x .For the second case, since ∂u∂ν + i κ λ u = ∂u∂ν + i κ λ u = 0hence, we have ( λ − λ ) u = 0 on ∂D , which shows that u = 0 on ∂D from λ (cid:54) = λ . Thus, wehave ∂u∂ν = 0 on ∂D . By Holmgren’s theorem, we get u ( y, x ) = 0 for y, x ∈ Ω \ D and y (cid:54) = x .Thus, we obtain the same contradiction as the first case. Therefore, B = B . The proof iscompleted. In this subsection, we investigate the case where D is a penetrable inhomogeneous medium. Aglobal uniqueness is obtained which shows that the locally rough interface, the wave numberin the lower half-space, and the inhomogeneous medium can be uniquely determined by thescattered field measured on Γ b,a . Theorem 4.2. (Global uniqueness) Given κ > , let u s1 ( x, y ) and u s2 ( x, y ) be the scatteredfields of the problem (1.2) associated with the incidence u i ( x, y ) = Φ κ ( x, y ) and the complexscatterers (Γ , κ , , n ) and (Γ , κ , , n ) , respectively. If u s1 ( x, y ) = u s2 ( x, y ) for all x, y ∈ Γ b,a ,then Γ = Γ , κ , = κ , , n = n .Proof. It follows from the same arguments as Theorem 4.1 that Γ = Γ and κ , = κ , , and wedenote Γ := Γ = Γ , κ := κ , = κ , for simiplicity. The remaining part is devoted to show n = n . Let D be an open disks containing the supports of 1 − n and 1 − n and κ is neithera Dirichlet eigenvalue of ∆ u + κ n u = 0 in D , nor a Dirichlet eigenvalue of ∆ u + κ n u = 0in D . Since u s1 ( x, y ) = u s2 ( x, y ) for all x, y ∈ Γ b,a , from the analyticity of the scattered field,uniqueness of the Dirichlet problem and the analytic continuation principle, we conclude that u ( x, y ) = u ( x, y ) in R \ D for all y ∈ Γ b,a . We define w ( x, y ) := u ( x, y ) − u ( x, y ), which satisfies w = ∂w∂ν = 0 on ∂D (4.27)and ∆ w + κ n w = κ ( n − n ) u in D. (4.28)Let ˆ u := u ( · , ˆ y ) for ˆ y ∈ Γ b,a , multiplying ˆ u on both sides of (4.28) yields κ ( n − n )ˆ u u = (∆ w + κ n w )ˆ u = ∆ w ˆ u − ∆ˆ u w in D. Employing the Green’s theorem and (4.27) leads to (cid:90) D ( n ( x ) − n ( x )) u ( x, ˆ y ) u ( x, y ) dx = 0 for all y, ˆ y ∈ Γ b,a . (4.29)19o show n = n , it requires to prove that the set of total fields { u ( · , y ) | D : y ∈ Γ b,a } satisfyingthe problem (1.2) is dense in the set S := { v ∈ H ( D ) : ∆ v + κ nv = 0 in D } (4.30)with respect to the L ( D ) norm. Thus, using this dense result yields (cid:90) D ( n − n ) v v dx = 0 (4.31)for all solutions v , v ∈ H ( D ) of ∆ v + κ n v = 0, ∆ v + κ n v = 0 in D . From (4.31) andstandard discussion in [1], we conclude n = n .Now, it remains to prove the dense result. To this end, we consider the problem (cid:40) ∆ u + κ nu = 0 in Du = g on ∂D (4.32)where g ∈ H ( ∂D ). The well-posedness of (4.32) is easily obtained by the assumption on D . Hence, to show { u ( · , y ) | D : y ∈ Γ b,a } is dense in the set S , it is sufficient to show that { u ( · , y ) | ∂D : y ∈ Γ b,a } is dense in H ( ∂D ). For some ϕ ∈ H − ( ∂D ), we assume that (cid:90) ∂D u ( x, y ) ϕ ( x ) ds ( x ) = 0 for all y ∈ Γ b,a (4.33)then we aim to show that the condition (4.33) implies ϕ = 0. From (4.33), we have (cid:90) Γ b,a ψ ( y ) (cid:90) ∂D u ( x, y ) ϕ ( x ) ds ( x ) ds ( y ) = 0 for any ψ ∈ L (Γ b,a ) (4.34)which implies (cid:90) ∂D ϕ ( x ) (cid:90) Γ b,a u ( x, y ) ψ ( y ) ds ( y ) ds ( x ) = 0 for any ψ ∈ L (Γ b,a ) . (4.35)To prove ϕ = 0, it is enough to prove that the operator L : L (Γ b,a ) → H ( ∂D ) given by( Lψ )( x ) := (cid:90) Γ b,a u ( x, y ) ψ ( y ) ds ( y ) (4.36)has dense range in H ( ∂D ). It suffices to show that the adjoint operator L ∗ of L is injectivein the extension of L inner product. Utilizing (4.36), we can obtain L ∗ : H − ( ∂D ) → L (Γ b,a )given by ( L ∗ φ )( y ) = (cid:90) ∂D u ( x, y ) φ ( x ) ds ( x ) = (cid:90) ∂D u ( y, x ) φ ( x ) ds ( x ) for y ∈ Γ b,a (4.37)where we use the reciprocity u ( x, y ) = u ( y, x ) for x ∈ ∂D and y ∈ Γ b,a . Let ( L ∗ φ )( y ) = 0 for y ∈ Γ b,a , thus, we have q = 0 on Γ b,a where q ( y ) := (cid:90) ∂D u ( y, x ) φ ( x ) ds ( x ) for y ∈ R \ ∂D. (4.38)20y means of the analyticity of q , uniqueness of the Dirichlet problem and the analytic continua-tion principle, we conclude that q = 0 in R \ D . Moreover, since u ( · , x ) satisfies ∆ u + κ nu = 0in D , we have (cid:40) ∆ q + κ nq = 0 in Dq = 0 on ∂D (4.39)Noting that κ is not a Dirichlet eigenvalue of D , hence q = 0 in D . Since u ( · , x ) = Φ κ ( · , x ) + u s ( · , x ) where u s is analytic, we obtain the jump relation ∂q∂ν (cid:12)(cid:12)(cid:12) − − ∂q∂ν (cid:12)(cid:12)(cid:12) + = φ (4.40)where ·| + and ·| − stands for the limits from the exterior and interior of D . Hence, it followsfrom (4.40) that φ = 0, the proof is completed. In this work, we studied the direct and inverse scattering of an incident point source from alocally rough interface with some obstacles buried in the lower half-space. By introducing aspecial rough interface, the direct problem was reduced to an equivalent Lippmann–Schwingertype integral equation defined in a bounded domain which was shown to have a unique solution.For the inverse problem, we first prove that the locally rough interface, the wave number in thelower half-space and the buried obstacle with its boundary condition are uniquely determinedby the near-field measured only above the interface.The main contribution of this paper is to give a novel method to first obtain (as far as weknow) the well–posedness of the direct problem, the global uniqueness results of the inverseproblem. Future studies are devoted to reconstruct the interface, together with the buriedobstacle. Furthermore, we are currently trying to extend the method to the case of elastic wave.
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