Scattering of Electromagnetic Waves by Thin Dielectric Planar Structures

SIAM J. MATH. ANAL. c© 2006 Society for Industrial and Applied MathematicsVol. 38, No. 4, pp. 1329–1342

SCATTERING OF ELECTROMAGNETIC WAVES BY THINDIELECTRIC PLANAR STRUCTURES∗

HABIB AMMARI† , HYEONBAE KANG‡ , AND FADIL SANTOSA§

Abstract. The problem of scattering of electromagnetic waves by a thin dielectric planarstructure is considered. If the squared index of refraction of the scatterer scales as 1/h, where his the thickness of the structure, we show that an approximate solution to the scattering problemcan be obtained by a perturbation method. The approximate solution consists of two terms, thezeroth order term and the first order corrector, both of which can be found by solving 2-D integralequations for 3-D problems. We provide error estimates for the approximation. Therefore, themethod described in this work can be viewed as a computational approach which can potentiallygreatly simplify scattering calculations for problems involving thin scatterers.

Key words. electromagnetic scattering, Maxwell’s equations, approximate solution, asymp-totics, error estimates

AMS subject classifications. 35B40, 65R20, 78A45

DOI. 10.1137/040618382

1. Introduction. To understand how light behaves in a thin-film structure witha high-refractive-index contrast, we need to solve the time-harmonic Maxwell’s equa-tion. The structure is usually surrounded by air. Thus, the domain in which Max-well’s equation must be solved will be all of R

3. The thin-film structure is modeledby prescribing the index of refraction to a subdomain of R

3.The standard approach for performing the required simulation of wave propaga-

tion in such a structure is the finite-difference time-domain (FDTD) method [2], withabsorbing boundary conditions. While the computation proceeds in a straightforwardmanner, it can be very time consuming.

In this paper, we consider a scattering problem in which an incident wave strikesa thin-film structure. Our goal is to derive an asymptotic expansion for the solution ofMaxwell’s equation. The approach is to exploit the fact that the scattering structureis thin, and the index of refraction is large. In this work, we assume that the squaredindex of refraction is O(1/h), where h is the scatterer thickness. Under a regularityassumption on the index of refraction, we show that an approximate solution, con-sisting of a leading term and first order (in h) corrector, can be obtained by solving a2-D integral equation for a 3-D problem. Thus, the expansion described here can bethought of as a computational approach.

This work is an extension of [5] which considered the scattering problem in thecontext of the scalar Helmholtz equation. In that work, a perturbation approach

∗Received by the editors November 5, 2004; accepted for publication (in revised form) July 10,2006; published electronically December 18, 2006. This research was partially supported by CNRS-KOSEF grant 14889 and F01-2003-000-00016-0 and Korea Science and Engineering Foundation grantR02-2003-000-10012-0.

http://www.siam.org/journals/sima/38-4/61838.html†Centre de Mathematiques Appliquees, Ecole Polytechnique, 91128 Palaiseau Cedex, France

([email protected]).‡School of Mathematical Sciences, Seoul National University, Seoul 151-747, Korea (hkang@math.

snu.ac.kr).§School of Mathematics, University of Minnesota, 127 Vincent Hall, 206 Church Street, SE, Min-

neapolis, MN 55455 ([email protected]). The research of this author was partially supportedby the National Science Foundation.

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1330 HABIB AMMARI, HYEONBAE KANG, AND FADIL SANTOSA

based on the thickness h is developed to reduce the complexity of the computation byone dimension. The present paper addresses the case of Maxwell’s equations, whichis the correct model for the propagation phenomena. We start with the Lippmann–Schwinger formulation of the electromagnetic scattering problem which involves a 3-D(volume) integral equation. Based on delicate estimates on the kernels involved in thisintegral formulation, we carefully derive the asymptotic expansion of the electric fieldin h and provide rigorous error estimates. Using our asymptotic expansion in h, wethen simplify the scattering calculations to solving a 2-D (surface) integral equation.

The paper is organized as follows. We give a description of the problem we wishto solve in the next section. The perturbation approach and its rigorous justificationare presented in section 3. Our justification uses a nontrivial bound on the C1,α-normof the electric field proved in Appendix A.

The results in this paper are expected to lead to a very effective computationalalgorithm for solving the electromagnetic scattering problem for thin-film devices inthe parameter regime, where the approximation is accurate.

2. Problem statement. Let E and H denote the electric and magnetic fields.The governing Maxwell’s equations are given by{

∇× E = ikH,

∇×H = −ikn(x, z)E,

where k > 0 is the normalized frequency. The total field consists of the incident fieldand the scattered field. We write E = Ei + Es, where Ei is a given incident wave.The scattered electric field Es satisfies the Sommerfeld radiation condition. We callthe function n(x, z) the squared index of refraction.

Here x = (x1, x2) and z is the third direction. The squared index of refractionn(x, z) is defined by

n(x, z) :=

⎧⎪⎪⎨⎪⎪⎩1 for |z| > h

2,

n0(x)

hfor |z| < h

2.

Therefore, the thin scattering structure has been incorporated into the definition ofn(x, z). Let Ω be a bounded domain in R

2. We denote Ωh := Ω × (−h/2, h/2).We assume that n0 is a C2 function, and that (1 − n0/h) is supported in Ωd

for some d > 0, where Ωd := {x ∈ Ω : dist(x, ∂Ω) > d}, �n0(x) ≥ C > 0, and�n0(x) ≥ 0.

The problem we wish to solve is to determine the total field E given the incidentwave Ei and a scatterer n(x, z). We will seek an approximate solution for this problemby exploiting the fact that h will be small. The approach we take is to write the fieldE as a perturbation series in h.

3. Derivation of the asymptotic expansion. For x, x′ ∈ R2 and z, z′ ∈ R,

let the fundamental solution Φ be defined by

Φ(x, z, x′, z′) :=1

4π

eik|(x,z)−(x′,z′)|

|(x, z) − (x′, z′)| .

According to [3], we can rewrite the problem as the following Lippmann–Schwingerequation:

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THIN DIELECTRIC STRUCTURES 1331

E(x, z) = Ei(x, z) − k2

∫Ω

∫ h/2

−h/2

Φ(x, z, x′, z′)

(1 − n0(x

′)

h

)E(x′, z′)dx′dz′

−∇x,z

∫Ω

∫ h/2

−h/2

Φ(x, z, x′, z′)∇n0(x

′)

n0(x′)· E(x′, z′)dx′dz′.

(3.1)

Consequently, it suffices to construct an approximation of the electric field E insideΩh and insert it into the right-hand side of (3.1) to obtain an expansion of E for allpoints outside the thin structure Ωh.

After scaling z = hζ and z′ = hζ ′, (3.1) can be written as

E(x, hζ) = Ei(x, hζ) − k2

∫Ω

∫ 1/2

−1/2

Φ(x, hζ, x′, hζ ′)(h− n0(x′))E(x′, hζ ′)dx′dζ ′

−(

h∇x

∇ζ

)∫Ω

∫ 1/2

−1/2

Φ(x, hζ, x′, hζ ′)∇n0(x

′)

n0(x′)· E(x′, hζ ′)dx′dζ ′.

(3.2)

For a scalar function f defined in Ωh, define the integral operators K1 and K2 by

K1(f)(x, ζ) :=

∫Ω

∫ 1/2

−1/2

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′)dx′dζ ′,

K2(f)(x, ζ) :=

(h∇x

∇ζ

)∫Ω

∫ 1/2

−1/2

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′)dx′dζ ′.

Then the integral representation (3.2) can be rewritten as

E(x, hζ) = Ei(x, hζ) + k2K1(n0E)(x, ζ) − hk2K1(E)(x, ζ) −K2(n−10 ∇n0 · E)(x, ζ).

(3.3)

We now investigate the behavior of the integral operators K1 and K2 as h → 0.We begin by proving the following lemma.

Lemma 3.1. Let 0 < α < 1. Suppose that f is C1,α in Ωh; then

K1(n0f)(x, ζ) =

∫Ω

Φ(x, 0, x′, 0)n0(x′)f(x′, 0)dx′ − h

2

(ζ2 +

1

4

)n0(x)f(x, 0)(3.4)

+ O

(h1+α‖f‖C1,α(Ωh)

).

Suppose that g ∈ L∞(Ωh); then

K1(g)(x, ζ) =

∫Ω

Φ(x, 0, x′, 0)g(x′, 0)dx′ + O(h‖g‖L∞(Ωh)).(3.5)

Proof. Observe that

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′) − Φ(x, 0, x′, 0)f(x′, 0)

= Φ(x, hζ, x′, hζ ′)

(f(x′, 0) + hζ ′

∂f

∂z(x′, 0) + O(h1+α)

)− Φ(x, 0, x′, 0)f(x′, 0)

=[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]f(x′, 0) + Φ(x, hζ, x′, hζ ′)

(hζ ′

∂f

∂z(x′, 0) + O(h1+α)

).

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Let S := Ω × (−1/2, 1/2). We then have∫S

[Φ(x, hζ, x′, hζ ′)n0(x

′)f(x′, hζ ′) − Φ(x, 0, x′, 0)n0(x′)f(x′, 0)

]dx′dζ ′

=

∫S

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]n0(x

′)f(x′, 0)dx′dζ ′

+

∫S

Φ(x, hζ, x′, hζ ′)n0(x′)hζ ′

∂f

∂z(x′, 0)dx′dζ ′ + O(h1+α)

∫S

Φ(x, hζ, x′, hζ ′)n0(x′)dx′dζ ′

:= I1 + I2 + I3.

Since ∣∣∣∣∫S

Φ(x, hζ, x′, hζ ′)n0(x′)dx′dζ ′

∣∣∣∣ ≤ C

regardless of h, one can see that I3 = O(h1+α).To estimate I2, we first observe that∫ 1/2

−1/2

ζ ′∂f

∂z(x′, 0)dζ ′ = 0,

and hence

I2 = h

∫S

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]n0(x

′)ζ ′∂f

∂z(x′, 0)dx′dζ ′.

Therefore, we get

|I2| ≤ ‖f‖C1(Ωh)h

∫S

∣∣∣∣Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

∣∣∣∣dx′dζ ′.

For a given (x, ζ) ∈ S and h > 0, let Sh := {(x′, ζ ′) | |x − x′| > 2h|ζ − ζ ′|}. If(x′, ζ ′) ∈ Sh, then

(|x− x′|2 + h2|ζ − ζ ′|2)1/2 = |x− x′|(

1 + O

(h2|ζ − ζ ′|2|x− x′|2

)),

and hence

Φ(x, hζ, x′, hζ ′) =1

4π

eik(|x−x′|2+h2|ζ−ζ′|2)1/2

(|x− x′|2 + h2|ζ − ζ ′|2)1/2

=1

4πeik|x−x′|

(1 + O

(h2|ζ − ζ ′|2|x− x′|

))(1

|x− x′| + O

(h2|ζ − ζ ′|2|x− x′|3

)).

Thus we get

∣∣∣∣Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

∣∣∣∣ ≤ Ch2|ζ − ζ ′|2|x− x′|2 + C

h2|ζ − ζ ′|2|x− x′|3 + C

h4|ζ − ζ ′|4|x− x′|4 .

(3.6)

One can easily show that∫ 1/2

−1/2

∫|x−x′|>2h|ζ−ζ′|

h2|ζ − ζ ′|2|x− x′|2 dx′dz′ ≤ Ch2,

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THIN DIELECTRIC STRUCTURES 1333∫ 1/2

−1/2

∫|x−x′|>2h|ζ−ζ′|

h2|ζ − ζ ′|2|x− x′|3 dx′dz′ ≤ Ch,

and ∫ 1/2

−1/2

∫|x−x′|>2h|ζ−ζ′|

h4|ζ − ζ ′|4|x− x′|4 dx′dz′ ≤ Ch2.

Thus we get ∫Sh

∣∣∣∣Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

∣∣∣∣dx′dζ ′ ≤ Ch.

On the other hand, one can also easily see that∫S\Sh

∣∣∣∣Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

∣∣∣∣dx′dζ ′

≤∫ 1/2

−1/2

∫|x−x′|≤2h|ζ−ζ′|

1

|x− x′| + h|ζ − ζ ′| +1

|x− x′|dx′dζ ′ ≤ Ch,

and hence we obtain

|I2| ≤ Ch2.

We now estimate I1. Since n0(x′)− h is compactly supported in Ω, we can write

I1 =

∫ 1/2

−1/2

∫BR(x)

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

](n0(x

′) − h)f(x′, 0)dx′dζ ′

+ h

∫S

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]f(x′, 0)dx′dζ ′,

where BR(x) is a disk of radius R centered at x containing Ω. One can prove thatthe second term in the right-hand side of above identity is O(h2) as before, while thefirst term equals∫BR(x)

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

][(n0(x

′) − h)f(x′, 0) − (n0(x) − h)f(x, 0)]dx′dζ ′

+ (n0(x) − h)f(x, 0)

∫BR(x)

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]dx′dζ ′.

Since |(n(x′)− h)f(x′, 0)− (n(x)− h)f(x, 0)| ≤ C|x− x′|, one can show that the firstterm of the above is O(h2) in a similar way to the estimate of I2. On the other hand,it is proved in [5] that∫

BR(x)

[Φ(x, hζ, x′, hζ ′) − Φ(x, 0, x′, 0)

]dx′dζ ′ = −h

2

(ζ2 +

1

4

)+ O(h2) as h → 0.

Thus we have

I1 = −h

2

(ζ2 +

1

4

)n0(x)f(x, 0) + O(h2).

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Combining all these estimates we obtain (3.4).Estimate (3.5) can be proved in a similar but simpler way, so we omit the

proof.We now derive the leading-order term in the asymptotic expansion of the integral

operator K2.Lemma 3.2. Let 0 < α < 1. Define K1

2 (f)(x) by

K12 (f)(x) :=

∫Ω

∇xΦ(x, 0, x′, 0)[f(x′, 0) − f(x, 0)]dx′ + f(x, 0)

∫Ω

∇xΦ(x, 0, x′, 0)dx′,

(3.7)

and K2(f)(x, ζ) by

K2(f)(x, ζ) :=

(K1

2 (f)(x)

( ζ2 − 1)f(x, 0)

).(3.8)

Then, for any C1,α-function f such that f(·, z) is supported in Ωd for each z,

K2(f)(x, ζ) = hK2(f)(x, ζ) + O(h1+α),(3.9)

where O(h1+α) is bounded by Ch1+α‖f‖C1,α(Ωh).Proof. Note that

∇x

∫S

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′)dx′dζ ′

=

∫S

∇xΦ(x, hζ, x′, hζ ′)

[f(x′, hζ ′) − f(x, hζ ′)

]dx′dζ ′

+

∫S

∇xΦ(x, hζ, x′, hζ ′)f(x, hζ ′)dx′dζ ′.

Thus,

∇x

∫S

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′)dx′dζ ′ − K12 (f(·, 0))(x)

=

∫S

[∇xΦ(x, hζ, x′, hζ ′) −∇xΦ(x, 0, x′, 0)

][f(x′, hζ ′) − f(x, hζ ′)

]dx′dζ ′

+

∫S

∇xΦ(x, 0, x′, 0)

[f(x′, hζ ′) − f(x, hζ ′) − f(x′, 0) + f(x, 0)

]dx′dζ ′

+

∫S

[∇xΦ(x, hζ, x′, hζ ′) −∇xΦ(x, 0, x′, 0)

]f(x, hζ ′)dx′dζ ′

+

∫S

∇xΦ(x, 0, x′, 0)

[f(x, hζ ′) − f(x, 0)

]dx′dζ ′

:= I1 + I2 + I3 + I4.

Since


=1

4πeik(|x−x′|2+h2|ζ−ζ′|2)1/2

[ik

x− x′

|x− x′|2 + h2|ζ − ζ ′|2 − x− x′

(|x− x′|2 + h2|ζ − ζ ′|2)3/2

],

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we get for |x− x′| ≥ 2h|ζ − ζ ′|


=ik

4πeik|x−x′|

[1 + O

(h2|ζ − ζ ′|2|x− x′|

)][ik

x− x′

|x− x′|2 − x− x′

|x− x′|3 + O

(h2|ζ − ζ ′|2|x− x′|4

)]= ∇xΦ(x, 0, x′, 0) + O

(h2|ζ − ζ ′|2|x− x′|4

).

It then follows that∫Sh

[∇xΦ(x, hζ, x′, hζ ′) −∇xΦ(x, 0, x′, 0)

][f(x′, hζ ′) − f(x, hζ ′)

]dx′dζ ′

≤ C‖f‖C1(Ωh)

∫ 1/2

−1/2

∫|x−x′|≥2h|ζ−ζ′|

h2|ζ − ζ ′|2|x− x′|4 |x− x′|dx′dζ ′

≤ Ch.

On the other hand, we get∫S\Sh

[∇xΦ(x, hζ, x′, hζ ′) −∇xΦ(x, 0, x′, 0)

][f(x′, hζ ′) − f(x, hζ ′)

]dx′dζ ′

≤ C‖f‖C1(Ωh)

∫ 1/2

−1/2

∫|x−x′|≥2h|ζ−ζ′|

1

|x− x′|2 |x− x′|dx′dζ ′

≤ Ch.

Therefore, we obtain

|I1| ≤ Ch.

Observe that∣∣∣∣f(x′, hζ ′) − f(x, hζ ′) − f(x′, 0) + f(x, 0)

∣∣∣∣ =

∣∣∣∣∫ hζ′

0

[∂

∂zf(x′, z) − ∂

∂zf(x, z)

]∣∣∣∣≤ C‖f‖C1,α(Ωh)h|x− x′|α.

It then follows that

|I2| ≤ C‖f‖C1,α(Ωh)h

∫ 1/2

−1/2

∫Ω

1

|x− x′|2 |x− x′|αdx′dζ ′ ≤ C‖f‖C1,α(Ωh)h.

If x ∈ supp(f), then Bd(x) ⊂ Ω. Moreover,∫Bd(x)

∇xΦ(x, hζ, x′, hζ ′)dx′ =

∫Bd(x)

∇xΦ(x, 0, x′, 0)dx′ = 0 for all ζ and ζ ′.

Therefore we get

|I3| =

∫ 1/2

−1/2

∫Ω\Bd(x)

[∇xΦ(x, hζ, x′, hζ ′) −∇xΦ(x, 0, x′, 0)

]f(x, hζ ′)dx′dζ ′

≤ C‖f‖L∞(Ωh)h2

∫ 1/2

−1/2

|ζ − ζ ′|2dζ ′ ≤ C‖f‖L∞(Ωh)h2.

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Finally, we arrive at

|I4| = C‖f‖C1(Ωh)h

∫Ω\Bd(x)

1

|x− x′|2 dx′ ≤ C‖f‖C1(Ωh)h.

So far we proved that

∇x

∫S

Φ(x, hζ, x′, hζ ′)f(x′, hζ ′)dx′dζ ′ = K12 (f(·, 0))(x) + O(hα‖f‖C1,α(Ωh)) as h → 0.

We now investigate the behavior of

∂

∂z

∫S


as h → 0.Elementary computations show that

∂

∂zΦ(x, hζ, x′, hζ ′)

=1

4πeik(|x−x′|2+h2|ζ−ζ′|2)1/2

[ik

h2(ζ − ζ ′)

|x− x′|2 + h2|ζ − ζ ′|2 − h2(ζ − ζ ′)

(|x− x′|2 + h2|ζ − ζ ′|2)3/2

]= − 1

4π

h2(ζ − ζ ′)

(|x− x′|2 + h2|ζ − ζ ′|2)3/2 + O

(h2|ζ − ζ ′|

|x− x′|2 + h2|ζ − ζ ′|2

).

Since − 14π

t(|x−x′|2+t2)3/2 is the Poisson kernel for the half space and f(·, hζ ′) has a

compact support in Ω, we get

− 1

4π

∫Ω

h2(ζ − ζ ′)

(|x− x′|2 + h2|ζ − ζ ′|2)3/2 f(x′, hζ ′)dx′ = −h

2f(x, 0)

ζ − ζ ′

|ζ − ζ ′| + O(h2),

where the O(h2)-term is bounded by h2‖f‖C1(Ωh). On the other hand,∫Ω

h2|ζ − ζ ′||x− x′|2 + h2|ζ − ζ ′|2 |f(x′, hζ ′)|dx′ ≤ C‖f‖L∞(Ωh)h

2|ζ − ζ ′| |ln(h|ζ − ζ ′|)|.

It then follows that

∂

∂z

∫S


= −h

2f(x, 0)

∫ 1/2

−1/2

ζ − ζ ′

|ζ − ζ ′|dζ′ + O(h2|lnh|)

= h

(ζ

2− 1

)f(x, 0) + O(h2|lnh|).

This completes the proof.Remark 3.3. It is worth noticing that the above lemma applies if f is C1,α in

the z variable and only C1 in the x variable.The following lemma is also of use to us.Lemma 3.4. The operator

T : f ∈ C0(Ω) �→ f(x) − k2

∫Ω

Φ(x, 0, x′, 0)n0(x′)f(x′) dx′ ∈ C0(Ω)

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is invertible.Proof. As proved in [5], the operator

f ∈ L2(Ω) �→∫

Ω

Φ(x, 0, x′, 0)n0(x′)f(x′) dx′ ∈ L2(Ω)

is compact. Since n0(x) is compactly supported in Ω, we can actually prove that

f ∈ C0(Ω) �→∫

Ω

Φ(x, 0, x′, 0)n0(x′)f(x′) dx′ ∈ C0(Ω)

is compact. Thus, by the Fredholm alternative, it suffices to show that T is injectiveon C0(Ω). Let f ∈ C0(Ω) be a solution to

f(x) − k2

∫Ω

Φ(x, 0, x′, 0)n0(x′)f(x′) dx′ = 0.

Define the function u(x, z) in R3 \ Ω by

u(x, z) =

∫Ω

Φ(x, z, x′, 0)n0(x′)f(x′) dx′.

It follows that ⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

(Δ + k2)u(x) = 0 in R3 \ Ω,

u|+ − u|− = 0 on Ω,

∂u

∂z

∣∣∣∣+

(x) − ∂u

∂z

∣∣∣∣−

(x) = n0(x)f(x) on Ω,

u satisfies the Sommerfeld radiation condition.

Let BR be a sphere of center the origin and of radius R which is such that Ω ⊂ BR.Integrating by parts we obtain that∫

BR\Ω|∇u|2 − k2

∫BR

|u|2 −∫∂BR

∂u

∂νu−

∫Ω

[∂u

∂z

]u = 0,

where [∂u/∂z] = ∂u/∂z|+ − ∂u/∂z|− denotes the jump of ∂u/∂z across Ω. Sinceu = (1/k2)f and [∂u/∂z] = n0f on Ω we get∫

Ω

[∂u

∂z

]u =

1

k2

∫Ω

n0(x)|f(x)|2 dx.

Thus,

�∫∂BR

∂u

∂νu = −� 1

k2

∫Ω

n0(x)|f(x)|2 dx ≤ 0,

since �n0(x) ≥ 0. By the Rellich lemma (see [3]), the above inequality implies thatu ≡ 0 in R

3 \ Ω, and hence n0f = [∂u/∂z] = 0 which gives f = 0 on Ω since�n0(x) ≥ C > 0.

We finally establish the following bound on the C1,α-norm of E.

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Proposition 3.5. Let E be the solution of the Lippmann–Schwinger equation(3.1). Then for any 0 < α < 1, there exists a constant C independent of h such that

‖E‖C1,α(Ωh) ≤ C|log h|2.(3.10)

We will give a proof of Proposition 3.5 in Appendix A.At this point we have all the necessary ingredients to state and prove our main

result in this paper. Let

K1(f)(x) =

∫Ω

Φ(x, 0, x′, 0)f(x′) dx′, x ∈ Ω,(3.11)

and K2(f) be as in (3.8). Equation (3.3), together with Lemmas 3.1 and 3.2, yieldsthe pointwise approximation

E(x, hζ) = Ei(x, hζ) + k2K1(n0E0)(x, 0)

(3.12)

− h

[k2K1(E0)(x) +

k2

2

(ζ2 +

1

4

)n0(x)E0(x) + K2(n

−10 ∇n0 · E0)(x, ζ)

]+ O(h1+α),

where E0(x) = E(x, 0) and O(h1+α) ≤ Ch1+α||E||C1,α(Ωh) for some positive constant

C. Using Lemma 3.4, we define E(0) and E(1) as the unique solutions to the integralequations posed on Ω:

T (E(0)) := E(0) − k2K1(n0E(0))(x) = Ei(x, 0)

and

T (E(1)) = −[k2K1(E

(0))(x) +k2

2

(ζ2 +

1

4

)n0(x)E(0)(x) + K2(n

−10 ∇n0 · E(0))(x, ζ)

]+ ζ∂zE

i(x, 0).

Remark 3.6. Note that E(0) is independent of ζ while E(1)(x, ζ) is a secondorder polynomial in the stretched variable ζ.

Let

E(x, hζ) = Ei(x, hζ) + k2K1(n0(E(0) + hE(1))(x, 0)

− h

[k2K1(E

(0))(x) +k2

2

(ζ2 +

1

4

)n0(x)E(0)(x) + K2(n

−10 ∇n0 · E(0))(x, ζ)

],

(3.13)

for x ∈ Ω and ζ ∈ (−1/2, 1/2). The following error estimate is our main result in thispaper.

Theorem 3.7. Let E be defined by (3.13) in Ω× (−1/2, 1/2). For any 0 < ν < 1we have

||E(x, hζ) − E(x, hζ)||C0(Ω×[−1/2,1/2]) ≤ Ch1+ν ,

for some positive constant C independent of h and E.

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Proof. Let 0 < ν < 1. From (3.12) we immediately obtain that

T

(E0 − (E(0) + hE(1))

)= O(h1+α),

where α = (1 + ν)/2, which shows that

||E0 − (E(0) + hE(1))||C0(Ω) ≤ Ch1+α||E||C1,α(Ωh).

Inserting this approximation of E0 into the right-hand side of (3.12) and using thefact from Lemma 3.4 that T has a bounded inverse together with the definition(3.13) of E, we arrive at the desired estimate since, by Proposition 3.5, the quan-

tity h1−α

2 ||E||C1,α(Ωh) is uniformly bounded in h.To conclude the paper we make the following remarks.Remark 3.8. To obtain an approximate solution to the scattering problem, we

find E(0) and E(1) by solving integral equations in Ω ⊂⊂ R2. The electric field within

the thin structure is easily obtained using (3.13). Should we wish to evaluate thefield outside the scatterer, we can use the Lippman–Schwinger equation. Thus, theanalysis described in this work can be viewed as a computational method. The methodis efficient since only 2-D integral equations need to be solved to obtain the desiredapproximate solution. We note that in [5] numerical calculations were carried outto assess the accuracy of the method for the scalar problem in two dimensions. Seealso [6].

Remark 3.9. If n0 is constant, then the leading-order term E(0) satisfies theintegral equation

E(0)(x, 0) − k2n0

∫Ω

Φ(x, 0, x′, 0)E(0)(x′, 0) dx′ = Ei(x, 0), x ∈ Ω,

and therefore, the tangential component of E(0) on Ω × {0} is continuous while ez ×(∇ × E(0)) on Ω × {0} has a jump given by −k2n0ez × (ez × E(0)), where ez is thebasis vector in the z-direction. These jump conditions are exactly those derived byBouchitte in [1].

Remark 3.10. We should emphasize the fact that the asymptotic expansion de-rived in this paper is not valid in the case of piecewise constant media such as holesin the slab. The regularity assumption on n0 is essential to first write the Lippmann–Schwinger equation on the electric field E and then to establish the asymptotic ex-pansion (3.12) using Lemmas 3.1 and 3.2. In fact, the estimate of the remainderin (3.12) may not be valid if the regularity assumption on n0 does not hold. Thisimportant case in practice requires further delicate analysis and will be the subject ofa forthcoming work.

Remark 3.11. We believe that the approach presented in this paper can be easilygeneralized to deal with a dielectric layer having a curved mean surface.

Appendix A. Proof of Proposition 3.5. Here we prove Proposition 3.5.Define

Tf(x, z) :=

∫Ω

∫ h/2

−h/2

Φ(x, z, x′, z′)f(x′, z′)dx′dz′, (x, z) ∈ Ωh.

We show the following regularity properties of the operator T .Lemma A.1. Let m(x) be a C1-function compactly supported in Ω. For any

α ∈ (0, 1), there exists a constant C independent of h and f such that

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(i) ‖T (mf)‖C2,α(Ωh) ≤ C‖m‖C1(Ω)‖f‖Cα(Ωh),(ii) ‖T (mf)‖C1(Ωh) ≤ Ch|log h| ‖m‖L∞(Ω)‖f‖L∞(Ωh),(iii) ‖T (mf)‖C2(Ωh) ≤ Ch|log h| ‖m‖C1(Ω)‖f‖C1(Ωh).Before proving Lemma A.1, let us first show that Proposition 3.5 follows from

Lemma A.1. Equation (3.1) can be written as

E = Ei − k2T

((1 − n0

h

)E

)−∇x,zT

(∇n0

n0· E

).(A.1)

Since the solution E to (A.1) is uniformly bounded in L∞(Ωh), it follows from LemmaA.1(ii) that ‖E‖C1(Ωh) ≤ C|log h|, and hence from (A.1)(i), (iii) we obtain that

‖E‖C1,α(Ωh) ≤ C|log h|2 for some constant C independent of h.Proof of Lemma A.1. We first note that

|DsΦ(x, z, x′, z′)| ≤ C

|(x, z) − (x′, z′)|1+s,(A.2)

for s = 0, 1, . . . , where Ds denotes any sth order derivative.Let (x, z) = y, (x′, z′) = y′, and Dij denote the second order partial derivative

with respect to yi and yj . Then it is known that for i, j = 1, 2, 3,

DijT (mf)(y) =

∫Ωh

DijΦ(y − y′)m(x′)(f(y′) − f(y))dy

− f(y)

∫∂Ωh

DiΦ(y − y′)m(x′)νj(y′)dσ(y′)

:= I1(y) − f(y)I2(y),(A.3)

where ν = (ν1, ν2, ν3) is the outward unit normal to ∂Ωh. See Lemma 4.2 in [4, p. 55].By a standard Holder estimate (see, for example, the proof of Lemma 4.4 in [4, p. 57]),we can show that

|I1(y) − I1(y)| ≤ C‖m‖L∞(Ω)‖f‖Cα(Ωh)|y − y|α, y, y ∈ Ωh.

Observe that ∂Ωh consists of two parts: Ω × {h/2,−h/2} and ∂Ω × [−h/2, h/2].Since m has a compact support in Ω, we have

I2(y) =

∫Ω×{h/2,−h/2}

DiΦ(y − y′)m(x′)νj(y′)dx′.

Moreover, νj = 0 on ∂Ω × {h/2,−h/2} if j = 3, and then we have only to considerthe quantity I2 when j = 3. If j = 3, then

I2(y) =

∫Ω

DiΦ(x, z, x′, h/2)m(x′)dx′ −∫

Ω

DiΦ(x, z, x′,−h/2)m(x′)dx′.

Suppose that i = 1 or 2. Then, since m has a compact support in Ω andDiΦ(x, z, x′, z′) = −D′

iΦ(x, z, x′, z′) where D′ denotes the derivative with respectto y′ variables, we have

I2(y) =

∫Ω

Φ(x, z, x′, h/2)Dim(x′)dx′ −∫

Ω

Φ(x, z, x′,−h/2)Dim(x′)dx′.(A.4)

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Thus, we have

|I2(y) − I2(y)| ≤ C‖m‖C1(Ω)|y − y|α, y, y ∈ Ωh,

for any 0 < α < 1.Suppose now that i = 3. In this case

I2(y) = −∫∂Ω

∂

∂z′Φ

(x, z, x′,

h

2

)m(x′)dx′ +

∫∂Ω

∂

∂z′Φ

(x, z, x′,−h

2

)m(x′)dx′.

Using once again the fact that m has a compact support and Ωh = Ω × (−h/2, h/2),we can write I2(y) as

I2(y) = −∫∂Ωh

∇y′Φ(y − y′) · ν(y)m(x′)dσ(y′).

Since (Δ + k2)Φ(y − y′) = δ(y − y′), we obtain from the divergence theorem that

I2(y) = −m(x) + k2

∫Ωh

Φ(y − y′)m(x′)dy′ −∫

Ωh

∇y′Φ(y − y′) · ∇m(x′)dy′.(A.5)

Thus it follows from (A.2) and standard arguments that

|I2(y) − I2(y)| ≤ C‖m‖C1(Ω)|y − y|α,

and hence

|f(y)I2(y) − f(y)I2(y)| ≤ C‖m‖C1(Ω)‖f‖Cα(Ωh)|y − y|α.

This completes the proof of (i).To prove (ii), we first compute∫

Ωh

|DjΦ(x, z, x′, z′)|dx′dz′ ≤ C

∫ h/2

−h/2

∫Ω

1

|x− x′|2 + |z − z′|2 dx′dz′

≤ C

∫ h/2

−h/2

(1 + |log |z − z′||)dz′

≤ Ch|log h|,

for h small enough. Therefore, we get

|DjT (mf)(x, z)| ≤ Ch|log h|‖mf‖L∞(Ωh).

To prove (iii) we use (A.3), (A.4), and (A.5). Then the same estimates lead usto (iii). For example, from (A.4) we get

|I2(y)| ≤ C

∫ h/2

−h/2

∫Ω

∣∣∣∣ ∂

∂z′Φ(x, z, x′, z′)

∣∣∣∣ |∇m(x′)|dx′dz′

≤ Ch|log h|‖∇m‖L∞(Ω).

This completes the proof.

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REFERENCES

[1] G. Bouchitte, Analyse limite de la diffraction d’ondes electromagnetiques par une structuremince, C. R. Acad. Sci. Paris Ser. II Mec. Phys. Chim. Sci. Univers Sci. Terre, 311 (1990),pp. 51–56.

[2] G. Cohen, Higher-Order Numerical Methods for Transient Wave Equations, Springer-Verlag,Berlin, 2002.

[3] D. Colton and R. Kress, Inverse Acoustic and Electromagnetic Scattering Theory, Appl. Math.Sci. 93, Springer-Verlag, New York, 1992.

[4] D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, 2nded., Springer-Verlag, Berlin, 1983.

[5] S. Moskow, F. Santosa, and J. Zhang, An approximate method for scattering by thin struc-tures, SIAM J. Appl. Math., 66 (2005), pp. 187–205.

[6] J. H. Richmond, Scattering by thin dielectric strips, IEEE Trans. Antennas and Propagation,33 (1985), pp. 64–68.

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Scattering of Electromagnetic Waves by Thin Dielectric Planar Structures

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Transcript of Scattering of Electromagnetic Waves by Thin Dielectric Planar Structures