44 CHAPTER 3. CAVITY SCATTERING

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1 44 CHAPTER 3. CAVITY SCATTERING For the TE polarization case, the incident wave has the electric field parallel to the x 3 -axis, which is also the infinite axis of the aperture. Since both the incident field and medium are uniform along the x 3 -axis, i.e., no variation of any kind with respect to x 3, the scattered electric field, and thus the total electric field, are also parallel to the x 3 -axis, i.e., E = [0, 0, u]. It is therefore convenient to formulate the problem in terms of the electric field since it has only one component. It deduces from (3.5) and (3.6) that the total electric field satisfies (3.8) (3.9) u + κ 2 u = 0 above g S, u = 0 on g S. For the case of TM polarization, the magnetic field has only a x 3 -component, i.e., H = [0, 0, u], and therefore it is convenient to formulate the problem in terms of the magnetic field. It follows from (3.5) and (3.7) that the total magnetic field satisfies (3.0) (3.) (κ 2 u ) + u = 0 above g S, n u = 0 on g S. 3.3 TE polarization Let an incoming plane wave u inc = exp(iαx iβx 2 ) be incident on the perfect electrically conducting surface g S from above, where α = κ 0 sin θ, β = κ 0 cos θ, θ ( π/2, π/2) is the angle of incidence with respect to the positive x 2 -axis, and κ 0 = ω µ 0 ε 0 is the wavenumber of the free space. Denote the reference field u ref as the solution of the homogeneous equation in the upper half space: u ref + κ 2 0u ref = 0 in R 2 + together with the boundary condition u ref = 0 on {x 2 = 0}. It can be shown that the reference field consists of the incident field u inc and the reflected field u r : u ref = u inc + u r, where u r = exp(iαx + iβx 2 ). v: The total field u is composed of the reference field u ref and the scattered field u = u ref + v.

2 3.3. TE POLARIZATION 45 It can be verified that the scattered field satisfies v + κ 2 v = (κ 2 κ 2 0)u ref above g S, v = u ref on g S. Particularly, it can be verified that the scattered field satisfies (3.2) v + κ 2 0v = 0 in R 2 +. In addition, the scattered field is required to satisfy the radiation condition ( ) v lim ρ ρ ρ iκ 0v = 0, ρ = x. By taking the Fourier transform of (3.2) with respect to x, we have 2 x 2ˆv(ξ, x 2 ) + (κ 2 0 ξ 2 )ˆv(ξ, x 2 ) = 0 for x 2 > 0, whose solution, which satisfies the radiation condition, is easily obtained: where ˆv(ξ, x 2 ) = ˆv(ξ, 0)e iβ 0(ξ)x 2, { κ 2 β 0 (ξ) = 0 ξ 2 for κ 0 > ξ, i ξ 2 κ 2 0 for κ 0 < ξ. Taking the inverse Fourier transform of the solution yields which gives v(x, x 2 ) = n v x2 =0 = e iβ 0(ξ)x2ˆv(ξ, 0)e iξx dξ, iβ 0ˆv(ξ, 0)e iξx Hence we define a Dirichlet-to-Neumann or boundary operator. For w H /2 (R), define (3.3) T (w) = iβ 0 (ξ)e iξx Therefore we have the following boundary condition (3.4) n u = T (u) + g on,

3 46 CHAPTER 3. CAVITY SCATTERING where Define g = 2iβe iαx. H /2 () = {w H /2 () : supp w } or equivalently H /2 () = {w H /2 () : w H /2 (R), w = 0 on R \ and w = w }. In other words, w is called an extension of w to H /2 (R). We denote by H /2 () the dual space of H /2 () and by H /2 () the dual space of H /2 (). We next present two important properties of the boundary operator T. Lemma The boundary operator T : H /2 () H /2 () is continuous. Lemma For w H /2 (), it holds the following identities Im T (w) w = Re β 0 (ξ) ˆ w dξ, Re T (w) w = Im β 0 (ξ) ˆ w as By combining (3.8), (3.9), and (3.4), the scattering problem can be formulated u + κ 2 u = 0 in Ω, u = 0 on S, n u = T (ũ) + g on. The scattering problem has an equivalent weak formulation: Find u H 0(Ω) = {w H (Ω), w = 0 on S, w H /2 ()} such that (3.5) a(u, v) = g, v for all v H 0(Ω), where the bilinear form is defined by a(u, v) = u wdx Ω Ω κ 2 u wdx and the linear functional g, v = ( 2iβe iαx )v(x )dx. T (ũ)vdx,

4 3.3. TE POLARIZATION 47 Lemma (Fredholm alternative). Let V be a Hilbert space. Let W be a Hilbert space which contains W. Let a(u, v) be a continuous bilinear form on V V which satisfies Rea(u, u) c u 2 V c 2 u 2 W for all u V. Consider the variational problem a(u, v) = (g, v) for all v V, g V. Suppose that the injection of V into W is compact. Then the variational problem satisfies the Fredholm alternative, i.e., () either it admits a unique solution in V ; (2) it has a finite dimensional kernel, and a unique solution up to any element in this kernel, when the duality product of the right-hand side g vanishes on every element in this kernel. Theorem The scattering problem (3.5) attains a unique solution in H 0(Ω). Proof. The proof consists of two parts: uniqueness and existence. We first prove the uniqueness. In order to establish uniqueness of solution to the scattering problem, it suffices to show that u = 0 in Ω if g = 0 (no source term). From Ima(u, u) = 0, we get Im T (ũ)u = 0. It follows from Lemma that Re β 0 (ξ) ũ 2 dξ = 0, which implies that ũ = 0 for ξ κ0. Since ũ has a compact support on the x axis, ũ is analytical with respect to ξ. Therefore ũ = 0 for all ξ, and hence ũ = 0 on {x 0 = 0}. The transparent boundary condition on yields further that n u = 0 on. By the Holmgren uniqueness theorem, u = 0 in {x 2 > 0}. A unique continuation result concludes that u = 0 in Ω. Next is to consider the existence. An application of Lemma yields that Re T (ũ)u 0.

5 48 CHAPTER 3. CAVITY SCATTERING Thus there exist two positive constants c and c 2 such that Rea(u, u) c u 2 L 2 (Ω) c 2 u 2 L 2 (Ω). There the Fredholm alternative holds. The existence then follows from the uniqueness of the solution. 3.4 TM polarization Similarly, we can derive the variational formulation in TM polarization. However, in this case, the function n u rather than the function u is compactly supported. Consequently, the transparent boundary condition becomes slightly more complicated. The scattered field v satisfies the homogeneous Helmholtz equation in {x 2 > 0} as well as the boundary condition n v = 0 on g. By taking the Fourier transform of the model equation in {x 2 > 0} and observing that the function n v is compactly supported on the x axis, we obtain after some similar computation that v(x, 0) = 2πi β 0 (ξ) x 2ˆv(ξ, 0)e iξx Define for f H /2 (R), T TM (f) = 2πi β 0 (ξ) ˆf(ξ)e iξx We arrive at the transparent boundary condition v = T TM ( n v). Lemma The following identity holds ) n vt TM ( n v = i γ β 0 (ξ) n v 2, where n v H /2 (R) is an extension of n v with n v = 0 on g. Introduce a functional space H 0(Ω) = {w H (Ω), n w H ( ) /2 (), w = T TM n w on },