The direct scattering problem of obliquely incident electromagnetic waves by a penetrable homogeneous cylinder

# The direct scattering problem of obliquely incident electromagnetic waves by a penetrable homogeneous cylinder

Drossos Gintides dgindi@math.ntua.gr Department of Mathematics, National Technical University of Athens, Greece. Leonidas Mindrinos leonidas.mindrinos@univie.ac.at Computational Science Center, University of Vienna, Austria.
###### Abstract

In this paper we consider the direct scattering problem of obliquely incident time-harmonic electromagnetic plane waves by an infinitely long dielectric cylinder. We assume that the cylinder and the outer medium are homogeneous and isotropic. From the symmetry of the problem, Maxwell’s equations are reduced to a system of two 2D Helmholtz equations in the cylinder and two Helmholtz equations in the exterior domain coupled on the boundary. We prove uniqueness and existence of this differential system by formulating an equivalent system of integral equations using the direct method. We transform this system into a Fredholm type system of boundary integral equations taking advantage of Maue’s formula for hypersingular operators. Applying a collocation method we derive an efficient numerical scheme and provide accurate numerical results using as test cases transmission problems corresponding to analytic fields derived from fundamental solutions.

Keywords direct electromagnetic scattering, oblique incidence, integral equation method, hypersingular operator

## 1 Introduction

An interesting area of electromagnetism for its applications and the arising theoretical problems is the scattering process from obliquely incident time-harmonic plane waves by an infinitely long cylinder. The basic waves in the propagation domain satisfy Maxwell’s equations [1, 3, 16, 18] and due to the symmetry of the problem it is equivalent to find two scalar fields satisfying a pair of two-dimensional Helmholtz equations with different wavenumbers. The complication appears in the boundary conditions. Even for the case of a perfect conductor, in the boundary conditions appear tangential derivatives which make the analysis more difficult. There are many studies providing analytical or numerical solutions [2, 13, 19, 20, 21, 22, 24, 25]. The proposed methods are based on specific geometries or well known numerical schemes without examining the well-posedness of the corresponding boundary value problem.

Recently, Wang and Nakamura [23] used a more elegant theoretical analysis to prove well-posedness of the problem based on the integral equation approach. They proved theoretical and numerical results for the case of homogeneous impedance cylinder using integral equations. For the theoretical analysis they used properties of the Cauchy singular integrals and proved that the derived system is of Fredholm type with index zero. For the numerical results they applied a specific decomposition of the kernels and formulations using Hilbert’s and Symm’s integral operators. Considering trigonometric interpolation, they introduced an efficient numerical scheme.

The case for general dielectric cylinders is not considered yet, however the same authors, in a later work [17], investigated a more complicated model having also a non-homogeneous part, in the sense that the permittivity and the permeability of the exterior medium are non-constants and smooth in a bounded domain surrounding the cylinder. The main theoretical analysis providing uniqueness and existence in non-homogeneous materials is much harder. For the well-posedness they used the Lax-Phillips method [7].

In this work, we examine the case of infinite dielectric cylinder illuminated by a transverse magnetic polarized electromagnetic plane wave, known as oblique incidence. More precisely, in the second Section starting from Maxwell’s equations we describe initially the derivation of the mathematical model for the scattering process from obliquely incident time-harmonic plane waves for the case of infinite inhomogeneous cylinder. We assume that transmission conditions hold on the boundary. The boundary conditions involve normal and tangential derivatives of the fields.

In Section 3, we formulate the direct problem in differential form. We derive the Helmholtz equations and the exact form of the boundary conditions in the case of homogeneous cylinder. We prove that the problem is uniquely solvable using Green’s formulas and Rellich’s lemma. Considering the direct method, initially applied in transmission problems in [5, 8, 9], we formulate the problem into an equivalent system of integral equations. We show that this system is of Fredholm type in an appropriate Sobolev space setting. Due to uniqueness of the boundary value problem existence follows from Fredholm alternative. The system consists of compact, singular and hypersingular operators. We consider Maue’s formula [14], as in the case of the normal derivative of the double layer potential, to reduce the hypersingularity of the tangential derivative of the double layer potential.

In the last Section we investigate numerically the problem by a collocation method based on Kress’s method for two dimensional integral equation with strongly singular operators [10]. We transform the system of integral equations to a linear system by parametrizing the operators and considering well-known quadrature rules. We derive accurate numerical results for the four fields, interior and exterior, and we compute numerically the far-field patterns of the two exterior fields computed for a specific boundary value problem. Namely, we consider boundary data corresponding to analytic fields derived from point sources, where the interior and exterior fields have singularities outside of their domain of consideration.

## 2 Formulation of the direct scattering problem for an inhomogeneous cylinder

We consider the scattering problem of an electromagnetic wave by a penetrable cylinder in . Let then we model the cylinder as where is a bounded domain in with smooth boundary The cylinder is oriented parallel to the -axis and is its horizontal cross section. We assume constant permittivity and permeability for the exterior domain The interior domain is characterized by the electric constants and for all

We define for the magnetic field and electric field and equivalently the interior fields and for Then, these fields satisfy the Maxwell’s equations

 ∇×Eext+μ0∂Hext∂t =0, ∇×Hext−ϵ0∂Eext∂t =0, x∈Ωext, (1) ∇×Eint+μ∂Hint∂t =0, ∇×Hint−ϵ∂Eint∂t =0, x∈Ωint.

On the boundary we consider transmission conditions

 ^n×Eint=^n×Eext,^n×Hint=^n×Hext,x∈Γ,

where is the outward normal vector, directed into .

In order to take advantage of the symmetry of the specific medium, we probe the cylinder with an incident transverse magnetic (TM) polarized electromagnetic plane wave, the so-called oblique incidence in the literature. An arbitrary time-harmonic incident electromagnetic plane wave has the form:

 Einc(x,t;^d,^p) =1k20√ϵ0∇×∇×(^peik0x⋅^d)e−iωt, Hinc(x,t;^d,^p) =1ik0√μ0∇×(^peik0x⋅^d)e−iωt,

where is the frequency, is the wave number in the exterior domain, is the polarization vector and the vector describing the incident direction, satisfying

In the following, due to the linearity of the problem we suppress the time-dependence and we consider the fields only as functions of the space variable . In order to describe the incident fields for the specific TM polarization, we define by the incident angle with respect to the negative axis and by the polar angle of (in spherical coordinates), then, and assuming that Hence, we obtain

 Einc(x;^d,^p) =1√ϵ0^d×^p×^deik0x⋅^d=1√ϵ0^peik0x⋅^d, Hinc(x;^d,^p) =1√μ0^d×^peik0x⋅^d=1√μ0(sinϕ,−cosϕ,0)eik0x⋅^d.

Taking into account the cylindrical symmetry of the medium and the -independence of the electric coefficients we express the incident fields as separable functions of and Thus, we define and and it follows that the incident fields can be decomposed to

 Einc(x;^d,^p)=einc(x,y)e−iβz,Hinc(x;^d,^p)=hinc(x,y)e−iβz, (2)

where

 einc(x,y) =1√ϵ0^peiκ0(xcosϕ+ysinϕ), hinc(x,y) =1√μ0(sinϕ,−cosϕ,0)eiκ0(xcosϕ+ysinϕ).

Now, we are in position to transform equations (1) into a system of equations only for the -component of the electric and magnetic fields. Firstly, we see that for the specific illumination of the form (2), using separation of variables, also the scattered fields take the form:

 Esc(x;^d,^p)=esc(x,y)e−iβz,Hsc(x;^d,^p)=hsc(x,y)e−iβz,x∈Ωext,

where and Then, the exterior fields are given by

 Eext(x;^d,^p) =(esc(x,y)+einc(x,y))e−iβz=eext(x,y)e−iβz, x∈Ωext, Hext(x;^d,^p) =(hsc(x,y)+hinc(x,y))e−iβz=eext(x,y)e−iβz, x∈Ωext.

Equivalently, the interior fields are represented by

 Eint(x;^d,^p)=eint(x,y)e−iβz,Hint(x;^d,^p)=hint(x,y)e−iβz,x∈Ωint,

where and

For any field of the form

 E(x;^d,^p)=e(x,y)e−iβz,H(x;^d,^p)=h(x,y)e−iβz,x∈\mathbbmR3,

we consider the Maxwell’s equations in for arbitrary and (remark here the space dependence of Then, following [17] we obtain the relations

 e1(x,y)=−1k2(iβ∂e3∂x(x,y)−iμω∂h3∂y(x,y)), (3) e2(x,y)=−1k2(iβ∂e3∂y(x,y)+iμω∂h3∂x(x,y)), h1(x,y)=−1k2(iβ∂h3∂x(x,y)+iϵω∂e3∂y(x,y)), h2(x,y)=−1k2(iβ∂h3∂y(x,y)−iϵω∂e3∂x(x,y)).

Substituting (3) in (1), we have that the pair satisfies the equations

 k2ϵω∇⋅(ϵωk2∇e3)+k2ϵωJ∇(βk2)⋅∇h3+k2e3 =0, k2μω∇⋅(μωk2∇h3)−k2μωJ∇(βk2)⋅∇e3+k2h3 =0,

where

 J=(−01−10).

The interior and the exterior domains are characterized by different wavenumbers, given by

 k2(x)={k2int(x):=μ(x,y)ϵ(x,y)ω2−β2,x∈Ωint,k2ext(x):=μ0ϵ0ω2−β2=κ20,x∈Ωext.

In this section, for completeness in the formulation of the direct problem we keep the space dependence of Later, we consider only the case of constant parameters. Here, we have to assume that in order to have Thus, the fields and satisfy

 Δeext3+κ20eext3=0,Δhext3+κ20hext3=0,x∈Ωext, (4)

and the interior fields

 k2int(x)ϵ(x)∇⋅(ϵ(x)k2int(x)∇eint3)+k2int(x)ϵ(x)ωJ∇(βk2int(x))⋅∇hint3+k2int(x)eint3=0,x ∈Ωint, (5) k2int(x)μ(x)∇⋅(μ(x)k2int(x)∇hint3)−k2int(x)μ(x)ωJ∇(βk2int(x))⋅∇eint3+k2int(x)hint3=0,x ∈Ωint.

Now, we are going to derive the exact form of the boundary conditions. We introduce the notations: and where denote the unit vectors in Let be a local coordinate system, where is the outward normal vector and the outward tangent vector on Then, from (3) we obtain

 ^τ⋅et =−1k2(iμω^n⋅∇th3+iβ^τ⋅∇te3), (6) ^τ⋅ht =−1k2(−iϵω^n⋅∇te3+iβ^τ⋅∇th3),

using that

We observe, setting zero to the -component of in that

 ^n×E=−e3^τ+(n1e2−n2e1)^z,^n×H=−h3^τ+(n1h2−n2h1)^z.

Then from (3) and (6), we derive

 ^n×Eext=−eext3^τ+^τ⋅eextt^z,^n×Hext=−hext3^τ+^τ⋅hextt^z,

for the exterior fields, where and

 ^n×Eint=−eint3^τ+^τ⋅eintt^z,^n×Hint=−hint3^τ+^τ⋅hintt^z.

for the interior fields, where .

Here, we observe that the tangential forms of the fields can be written in terms of and two linear independent vectors. Thus, the boundary condition

 ^n×Eint=^n×Eext,x∈Γ,

is equivalent to the system

 eint3=eext3,^τ⋅eintt=^τ⋅eextt,x∈Γ,

and equivalently for the magnetic fields

 hint3=hext3,^τ⋅hintt=^τ⋅hextt,x∈Γ.

We define

 ∂∂n=^n⋅∇t,∂∂τ=^τ⋅∇t

and we rewrite the above boundary conditions as

 eint3 =eext3, x∈Γ, (7) μ(x)k2int(x)ω∂hint3∂n+βk2int(x)∂eint3∂τ =μ0κ20ω∂hext3∂n+βκ20∂eext3∂τ, x∈Γ,

and

 hint3 =hext3, x∈Γ, (8) ϵ(x)k2int(x)ω∂eint3∂n−βk2int(x)∂hint3∂τ =ϵ0κ20ω∂eext3∂n−βκ20∂hext3∂τ, x∈Γ.

To ensure that the scattered fields are outgoing, the components must satisfy in addition the radiation conditions in

 limr→∞√r(∂esc3∂r−iκ0esc3)=0,limr→∞√r(∂hsc3∂r−iκ0hsc3)=0, (9)

where uniformly over all directions.

Thus, the direct transmission problem for oblique incident wave, is to find the fields and which satisfy equations (4) and (5), the transmission conditions (7) and (8) and the radiation conditions (9).

We remark here that since we consider TM polarized wave, see equation (2), the incident fields for are simplified to

 einc3(x,y)=1√ϵ0sinθeiκ0(xcosϕ+ysinϕ),hinc3(x,y)=0. (10)

## 3 The direct problem for a homogeneous cylinder using the integral equation method

From now on, In this section we consider the simplified version where and are constant in the interior domain. To simplify the following analysis, we set and

 u0(x) =esc3(x), v0(x) =hsc3(x), x∈Ω0, u1(x) =eint3(x), v1(x) =hint3(x), x∈Ω1.

In the following, counts for the exterior () and interior domain (), respectively. Then, the direct scattering problem, presented in the previous section, is modified to

 Δuj+κ2juj =0, Δvj+κ2jvj =0, x∈Ωj, (11)

for where with boundary conditions

 u1 =u0+einc3, x∈Γ, (12a) ~μ1ω∂v1∂n+β1∂u1∂τ =~μ0ω∂v0∂n+β0∂u0∂τ+β0∂einc3∂τ, x∈Γ, (12b) v1 =v0, x∈Γ, (12c) ~ϵ1ω∂u1∂n−β1∂v1∂τ =~ϵ0ω∂u0∂n+~ϵ0ω∂einc3∂n−β0∂v0∂τ, x∈Γ, (12d)

 limr→∞√r(∂u0∂r−iκ0u0)=0,limr→∞√r(∂v0∂r−iκ0v0)=0. (13)
###### Theorem

If is not an interior Dirichlet eigenvalue, then the problem (11) - (13) has at most one solution.

###### Proof:

It is sufficient to show that if solve the homogeneous problem (11) - (13), that is for then, in and in Let be a disk with radius boundary centered at the origin and containing We set see Figure 1.

The boundary conditions of the homogeneous problem read

 u1 =u0, x∈Γ, (14) ~μ1∂v1∂n−~μ0∂v0∂n =−β1ω∂u1∂τ+β0ω∂u0∂τ, x∈Γ, v1 =v0, x∈Γ, ~ϵ1∂u1∂n−~ϵ0∂u0∂n =β1ω∂v1∂τ−β0ω∂v0∂τ, x∈Γ.

We apply Green’s first identity in and considering (11) we obtain

 ~ϵ1∫Γu1∂¯¯¯u1∂nds =~ϵ1∫Ω1(|∇u1|2+u1Δ¯¯¯u1)dx (15) =~ϵ1∫Ω1(|∇u1|2−κ21|u1|2)dx, ~μ1∫Γv1∂¯¯¯v1∂nds =~μ1∫Ω1(|∇v1|2+v1Δ¯¯¯v1)dx =~μ1∫Ω1(|∇v1|2−κ21|v1|2)dx.

Similarly, Green’s first identity in together with equations (14) and (15) gives

 ~ϵ0∫Γru0∂¯u0∂nds =~ϵ0∫Ωr(|∇u0|2+u0Δ¯¯¯u0)dx+~ϵ0∫Γu0∂¯¯¯u0∂nds =~ϵ0∫Ωr(|∇u0|2−κ20|u0|2)dx +∫Γu0(~ϵ1∂¯¯¯u1∂n−β1ω∂¯¯¯v1∂τ+β0ω∂¯¯¯v0∂τ)ds =~ϵ0∫Ωr(|∇u0|2−κ20|u0|2)dx +~ϵ1∫Ω1(|∇u1|2−κ21|u1|2)dx−β1ω∫Γu1∂¯¯¯v1∂τds+β0ω∫Γu0∂¯¯¯v0∂τds

and

 ~μ0∫Γrv0∂¯¯¯v0∂nds =~μ0∫Ωr(|∇v0|2+v0Δ¯¯¯v0)dx+~μ0∫Γv0∂¯¯¯v0∂nds =~μ0∫Ωr(|∇v0|2−κ20|v0|2)dx +∫Γv0(~μ1∂¯¯¯v1∂n−β0ω∂¯¯¯u0∂τ+β1ω∂¯¯¯u1∂τ)ds =~μ0∫Ωr(|∇v0|2−κ20|v0|2)dx +~μ1∫Ω0(|∇v1|2−κ21|v1|2)dx−β0ω∫Γv0∂¯¯¯u0∂τds+β1ω∫Γv1∂¯¯¯u1∂τds.

We add the above two equations and noting that

 −∫Γu1∂¯¯¯v1∂τds=¯∫Γv1∂¯¯¯u1∂τds,∫Γu0∂¯¯¯v0∂τds=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∫Γv0∂¯¯¯u0∂τds,

we obtain

 Im(~ϵ0∫Γru0∂¯¯¯u0∂nds+~μ0∫Γrv0∂¯¯¯v0∂nds)=0,

or equivalently, using the radiation conditions (see equation 2.12 [23])

 limr→∞∫Γr(ϵ0|u0|2+~ϵ0∣∣∣∂u0∂n∣∣∣2+μ0|v0|2+~μ0∣∣∣∂v0∂n∣∣∣2)ds=0.

Thus

 limr→∞∫Γr|u0|2ds=limr→∞∫Γr|v0|2ds=0,

and by Rellich’s lemma it fol