| In this paper, we study the scattering and inverse scattering problems for some types cavity-analysis and numerical computation, electromagnetic scattering and inverse scattering problems is an important field in mathematical physics.The direct scattering problem is to determine the scattering field from a cavity of the incident field and the differential equation governing the wave motion. The corresponding references are [24] [29] [44] [49] [54] [55] [56] [57] [83] [87] [89] etc.. The inverse scattering problem is to reconstruct the differential equation or the domain of definition from a knowledge of the behavior of the scattering solution. For the references, we refer to [6] [30] [31] [35] [66] [82] [84]etc.Recently, electromagnetic scattering and inverse scattering problems in special media have been payed attention. Chiral media have attracted much attention due to many associated interesting phenomena in optical and electromagnetic activities and their potential applications in various fields. Ammari. Nedelec. Gerlach. Am-mari,G.Bao. Athanasiadis and D.Y.Zhang, F.M.Ma, e.t.[1] [2] [3] [4] [5] [62] [93] [94][25] [26] [27] [28] have studied some concerned problems.In this paper, the theoretical analysis and numerical methods of electromag-netic scattering and inverse scattering problems for chiral media are discussed, and in this work we give a fast numerical method for scattering problem a simple rect-angular open cavity, and we carry out our method for some numerical example. We also give a numerical method for scattering problems with multiple rectangular complex open cavities. And we propose an idea to solve inverse scattering problems in which we reconstruct the shape of rectangular open cavity.What we have done is as follows:1.Problem Formulation Of TM Polarization1.Case 1.(see Fig.4).Problem 1.(construct approximation of u1 and us).GivenΩ0={(x,y)|y>0}.The total field u0 inΩ0 takes the following form u0(x,y)=ui(x,y)+ur(x,y)+us(x,y). (1) Here ui(x,y)is the incident wave ui(x,y)=ei(αx-βy), (2) ur(x,y)is the reflection wave ur(x,y)=-ei(αx+βy), (3) and us(x,y)is the scattered field which satisfies(as the total field u0)the Helmholtz equation△us(x,y)+k02us(x,y)=0,inΩ0 (4) along with the Sommerfeld radiation condition (?)((?)u0/(?)r-ik0u0)0,r=(?), uniformly in all directions,where k0 is the wave number,α=k0 sin(θ)andβ= k0 cos(θ)with the incident angleθ.The boundarv condition on the line v=0 is where u1 is the total field inΩ1.InΩ0,by definition we have that△us+k02us=0. (7) Let u(ξ,y)=Fxus(x,y),(8) where Fx is Fourier Transform on x.By using FT,we have that: This is a OPE on y for fixedξ,so that we obtain that u(ξ,y)=C1(ξ)ei((?))+C2(ξ)e-(?)y,-∞<ξ<+∞. (10) From the radiation condition we know C2(ξ)=0 so that and which y=0 so u(ξ,0)=C1(ξ)then u(ξ,0)=Fxus(x,0),-∞0}, Ω1={(x,y),0L}, Now,the bounded cavity is assumed to be rectangular cavity (Ω1)(Fig.4.)Ω1={(x,y)|00}.InΩ0 The total field u takes the following form u0(x,y)=ui(x,y)+ur(x,y)+us(x,y). (49) Here ui(x,y)is the incident wave ui(x,y)=ei(αx-βy), (50) ur(x,y)is the reflection wave ur(x,y)=ei(αx+βy), (51) and us(x,y)is the scattered field.Which satisfies(as the total field u0) the Helmholtz equation By Fourier transformation on x us(ξ,y)=Fxus(x,y), (53) we have that forγ(ξ)=(?) we can find us(ξ,y)=C1(ξ)eiγ0y+C2(ξ)e-iγ0y, (55) Using radiation condition,we obtained that u0(ξ,y)=C1(ξ)eγ0(ξ)y,(56) on interface conditionΓ0.we have that u0=u1 and Therefore, Furthermore, So that Defined then This mean So that inΩ0: From the fig,4.we knowΓ0={(x,0),00},Ω1={(x,y),0L}.Similar to TM case,we have that where for find Cn we must study u0 and u1 in interface conditionΓ0 onΓ0 we have for y=0,we get that and we know that forφ(x)=(?)us/(?)y(x,0).so that:therefore. here, In addition, From u1(x,0)=us(x,0)+ui(x,0)+ur(x,0),and put g(x)=ui(x,0)+ur(x,0)we can obtain that From(29),we can see that Due to we have that for calculate dm,.where and. Hence. 3.Numerical Method of Inverse ProblemsProblem 4.(solve inverse scattering problems).GivenΠFs-1(eiγ0(ξ)hFxφ(x))=d0(x),-b≤x≤b(74) ΠFs-1(iγ0(ξ)eiγ0(ξ)hFxφ((x))=d1(x),-b≤x≤b. (75) We can getφ(x)=us(x,0),-∞ |
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