Study Of Some Scattering And Inverse Scattering Problems With Near Field Measurements In A Half-Plane

Scattering and inverse scattering theory of sound wave and electromagneticwave plays an important role in mathematical physics. Usually, direct scattering problem is to find scattered field or far field of the scattered field with given incident wave and scatterer; However, inverse scattering problem is to find boundary of the scatterer or physical parameter with given scattered field or its far field. There are integral equation method, finite element method and PML method for solving direct scattering problems, and iterative techniques,analytic continuation methods and sampling method for inverse scatteringproblems. The theory and numerical methods of scattering problems and inverse problems are still arresting in the research of applied mathematics and computational mathematics. In this paper, some scattering and inverse scattering problems with near field measurements in a half-plane are discussed. What we have done is as follows:â… . Reconstruction of the shape of object with near field measurements in a half-planeOur model problem arises in the study of near field optical microscope. We assume that there is a sample on a flat impenetrable substrate, and the substrate is relatively thick so that only one face needs to be considered, thus defining an interface between two half-space. We take a monofrequency line source which is parallel to x₃-axis as our incident field. Moreover, we assumethat the sample and substrate are invariant along with direction x₃, and consider our problem in a plane which is perpendicular to the x₃-axis. In this case, we can reduce the problem to a two dimensional model. To describe the shape of the sample D, let T be a smooth curve in the upper half-plane R₊²:={(x₁,x₂) :x₃≥0} (?) R² such that only its two end points a = (a₁,0) and b = (b₁,0) with a₁ > b₁ lie on the boundary (?)R₊². Then the boundary (?)D of the sample can be written as (?)D = T∪ba. Denote T := {(x₁,0) : -∞< x₁ < b₁} and T₊ := {(x₁,0) : a₁ < x₁ < +∞} and let D_e be the unbounded domain above (?)De := T_∪T∪T₊. For x = (x₁,x₂)∈R² we denote by x^ρ= (x₁,-x₂) its reflection about the x₁-axis. Let T₁ be a smooth curve in unbounded domain D_e with two end points c = (c₁, 0) and d = (d₁, 0), such that c₁ > a₁ and b₁ > d₁. Settings of the model problem are shown in Figure 1.1. Inverse medium scattering problemWe assume that the sample is penetrable and its index of refraction n(x) = n₀ > 1 for x∈D and n(x) = 1 for x∈D_e. Givenωⁱ(x,y)=Φ(x,y), a point source at y, y∈T₁, as the incident field, whereΦ(x, y) := iH₀¹ (k|x- y|)/4 for x≠y is the fundamental solution to the Helmholtz equation in two dimensions, and H₀¹ is the Hankel function of order zero and of the first kind. Then the direct scattering problem is to findω(x,y) =ωⁱ(x,y) +ω^ρ(x,y) +ω^s(x,y) satisfyingwhereω^ρ(x, y) = -Φ(x, y^ρ) is the solution of the scattering problem for the half-plane, i.e., the absence of D.On the other hand, the inverse scattering problem can be stated as: using the near field measurement datato reconstruct the unknown boundary T of domain D.In order to transform the model problem, we denote the mirror image of T and T₁ about (?)R₊², i.e. about x₁-axis, by T^ρ:= {x^ρ: x∈T} and T₁^ρ:= {x^ρ: x∈T₁} respectively. Then T∪T^ρforms the boundary of a bounded domain D, and T₁∪T₁^ρforms the boundary of a bounded domain M. For convenience, we denote T∪T^ρby (?)D and denote T₁∪T₁^ρby (?)M. Specially we assume that (?)D∈C² and (?)M∈C².Let the index of refraction m(x) = n₀ > 1 for x∈D and m(x) = 1 for x∈R²\D. Let us consider the following problem: find a solutionω(x,y) =ωⁱ(x, y) +ω^ρ(x,y) +ω^s(x, y)∈C²(R²) satisfying By symmetric continuation, we proveTheorem 1. Problem (1)-(3) and problem (4)-(5) are equivalent.Thus, we reformulate the model problem defined in a half-plane as one defined in a whole plane. Before reconstruction, we introduce mixed reciprocityrelation to reduce the near field problem with line source incident to the corresponding far-field problem with plane wave incident field.Theorem 2. Let u^s(x, d) be the solution to the plane wave incident problemwhere uⁱ(x,d) = exp(ikx·d), u^ρ(x,d) = -exp(ikx·d^ρ),Ω:= {x∈R² : |x| = 1}. letω^s(x,z) be the solution to the line source incident problem (4)-(5) , where z∈R²,z≠x. Then we have the relationwhereγ= exp(iÏ€/4)/(?).Define the far-field operator F : L²(Ω)→L²(Ω) bywhere u^∞(x, d) is the far-field pattern of the scattered field of problem (6)-(8).We also define the near field operator E : L²((?)M)→L²((?)M) by whereω^s(y, x) is the scattered field of the problem (4)-(5).The following lemma in [91] will be useful later:Lemma 1. Let u^s be the solution of the following problemand lt u(x,d) = e^ikx·d+ u^s(x, d) be the total field. Assume v is the solution of the following boundary value problemThen we havewhereγ= exp(iÏ€/4)/(?), and u_D denotes the total field of the problem (9)-(11).Next we will give the main step towards our final results to reconstruct the support of the object D by establishing a connection between the far-field operator F and the near field operator E.Theorem 3. The far-field operator F : L²(Ω)→L²(Ω) can be factorized as where G_(?)M : L²((?)M)→L²(Ω) is the far-field solution operator of the exterior boundary value problem (12)-(14) with replacing (?)D by (?)M, defined by (15), E : L²((?)M)→L²((?)M) is the near field operator and J : L²(Ω)→L²((?)M) is an operator defined bywhere u_M is the total field of the plane wave scattering problem (9)-(11).In numerical experiments, since the domain M is known, we give an approximationof the near field operator E by integral equation method and construct the operators G_(?)M and J by solving the Helmholtz equation with plane wave incident in the exterior of M. Prom Theorem 3, we give the approximationof the far-field operator F. By using of factorization method in [40], we immediately obtain a characterization of the domain D. Two examples are presented, one is the reconstruction for ellipse-shape with k = 3 and the other one is peanut-shape with k = 7. It is shown that the results for reconstruction become better and better when the amount of nodes increases.2. Inverse obstacle scattering problemAssume the sample is impenetrable, then our direct scattering problem is to findω(x, y) =ωⁱ(x, y) +ω^ρ(x, y) +ω^s(x, y) satisfyingAgain, we reformulate the model problem defined in a half-plane as one defined in a whole plane, and then we reconstruct the boundary of obstacle by using of mixed reciprocity relation and factorization method. Similar results as Theorem 1, Theorem 2 and Theorem 3 are obtained. Here, we do not describe them in detail. Analogously, two examples are presented, one is ellipse-shape with k = 5 and the other one is peanut-shape with k = 7. It is shown that the results for reconstruction in inverse obstacle scattering problem are better than inverse medium scattering problem.â…¡. Construction of Green function for circular domain of background medium and applicationsIn order to express a solution of direct scattering problem exactly, the Green function is considered to be constructed. We assume that some medium with circular shape is embedded in homogeneous background medium, and our problem is to find the Green function of this inhomogeneous background medium problem, i.e. find Green function g(r,θ|ρ,θ') governed by the planar Helmholtz equation in terms of polar coordinateswhere -∞< r,ρ<∞, 0≤θ,θ'≤2Ï€, with the boundary conditionandwhere k_b denotes the wave number bywhere B_a := {x∈R² : |x| < a},ωis frequency,μis magnetic permeability, and the piecewise constant electric permittivity is defined byWe consider the model problem in the sense of generalized function. Notingthatwe assume that the solution g has the formSubstituting the two equations above into (16) and simplifying, we find thatThe homogeneous solution of it is:and ifρ> a,Where J_n is Bessel function of first kind with order n, and H_n¹ denotes Hankel function of first kind with order n. We now use the requirements that the solution is continuous at r =ρ, boundary condition (17) andto calculate the coefficients in (19) and (20), and we have where r< = min(r,ρ), r> = max(r,ρ). We also obtain a reciprocity relation about this Green function.Theorem 4. The Green function for circular domain of background medium satisfies the following reciprocity relationandIn our numerical experiments, we make a contrast between the Green function solved by numerical method and the previous Green function to show the feasibility of it, and Theorem 4 is also validated. Moreover, we apply the Green function to the linear sampling method in [25], and simplify the process of solving inverse scattering problem.â…¢. MUSIC algorithm for locating inclusions in inhomogeneous background mediumBased on the assumption of penetrable sample, we assume that the sample contains some small inclusions. The geometry of each of them may take the form of C_j+αB_j,j = 1,…, l, where B_j is a bounded, smooth domain containing the origin. The point C_j∈R² denotes the location of the inclusions, satisfyingThe value ofαis the common order of magnitude of the diameters of theinclusions. I_α=∪_j=1^l(C_j +αB_j) denotes the collection of inclusions.Then the direct scattering problem is to findω(x, z) =Φⁱ(x, z)+Φ^ρ(x, z)+ w_Φ^s(x,z) satisfying whereΦ_ρ(x,z) =-Φ(x,z^ρ), wave numberThe inverse scattering problem can be stated as: using the near field measurementdatato locate C_j,j = 1,…,l.By using of symmetric continuation and the uniqueness for the solution, the problem (21)-(23) is equivalent to findω(x,z) =gⁱ(x,z) + g^ρ(x,z) +ω_g^s(x, z), satisfyingwhere gⁱ(x, z) = g(x, z), g^ρ(x, z) = -g(x, z^ρ), wave numberthe index of refraction We show the solution of problem (24)-(25) by making use of the Lippmann-Schwinger equation,Noting the symmetry of D, Theorem 5 and the asymptotic equations in [55],j = 1,…2l, where C_l+i = C_i^p, B_l+i = B_i^p,i = 1,…,l,Ï„_j = k₂²(ε₀/ε₂ -)α²|B_j|,Ï„_l+i =Ï„_i, we getwhereÏ„_j =Ï„_jk₂²/2k₁², thenÏ„_l+i =Ï„_i, i = 1,…, l.Let us introduce the multi-static response matrix when x,z∈(?)M definedbywhere S₀ = [S₁,…,S_l]∈C^n×l, S_j = (g(z₁,C_j) - g(z₁^ρ, C_j),…,g(z_n,C_j) -g(z_n^ρ,C_j))^?, F₀ = diag(Ï„₁,…,Ï„_l)∈R^l×l,For any point C∈D we define the vector f_C∈Cⁿ byWe note that f_C1,…,f_Cl are the columns of the matrix S₀. According to [61], the following proposition holds. Proposition 5. There exists n₀∈N such that for any n≤n₀, the vector f_c, defined by (27) satisfyingAnalogously to [55], let the singular value decomposition of the matrix W^s be defined by W^s = V∑V^?, V∈C^n×n is unitary and∑= diag{σ₁,σ₂,…,σ_n}, whereσ₁≥σ₂≥…≥σ_l > 0,σ_i= 0,i = l + 1,…,n. The first m columns of V, V_s := {v₁,v₂,…,v_l}, provide an orthonormal basis for the space of W^s, and the rest of the matrix V, V_N:={v_l+1, v_l+2,…,v_n}, provide a basis for the left null space of W^s. Then the best rank approximation for W^s is V_sV_s^? W^s with error E = W^s - V_sV_s(?)W^s.From Proposition 5, we define P = I - V_sV_s^? : C→N(W^s), we can form an image of C_j, j = 1,…,l, by plotting the quantityat each point C. The resulting plot will have large peaks at the positions C_j,j = 1,…, l.In our experiments, we reconstruct the locations with two inclusions and three inclusions respectively, and give the numerical examples when there is a small perturbation in multi-static response matrix.Due to the complexity of the scattering problems and the inverse scatteringproblems, there are still many problems to solve. How to improve on the method for computing is our aim for further work.