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Analysis of the non-reflecting boundary condition for the time-harmonic electromagnetic wave propagation in waveguides Seungil Kim Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University, Seoul 130-701, Republic of Korea Abstract In this paper, we study the non-reflecting boundary condition for the time- harmonic Maxwell’s equations in homogeneous waveguides with an inhomoge- neous inclusion. We analyze a series representation of solutions to the Maxwell’s equations satisfying the radiating condition at infinity, from which we develop the so-called electric-to-magnetic operator for the non-reflecting boundary con- dition. Infinite waveguides are truncated to a finite domain with a fictitious boundary on which the non-reflecting boundary condition based on the electric- to-magnetic operator is imposed. As the main goal, the well-posedness of the reduced problem will be proved. This study is important to develop numerical techniques of accurate absorbing boundary conditions for electromagnetic wave propagation in waveguides. Key words: Maxwell’s equations, non-reflecting boundary condition, electric-to-magnetic operator, waveguide 1. Introduction In this paper, we will study the non-reflecting boundary condition for the electric fields E satisfying the second-order Maxwell’s equations in a waveguide × × E ω 2 ǫE = J in Ω , ν × E = 0 on Ω , (1.1) where Ω is a semi-infinite waveguide in R 3 with Lipschitz boundary such that Ω ∩{(x,y,z ) R 3 : z> b} × (b, ) and Ω ∩{z< b} is bounded with b> 0 (see Figure 1). Here ν represents the outward unit normal vector on the boundary of Ω , and Θ is a simply connected Email address: [email protected] (Seungil Kim) Preprint submitted to Elsevier April 4, 2017

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Page 1: Analysis of the non-reflecting boundary condition for the time …sikim.khu.ac.kr/paper/rev_maxwell6.pdf · 2017-04-04 · Analysis of the non-reflecting boundary condition for

Analysis of the non-reflecting boundary condition for

the time-harmonic electromagnetic wave propagation in

waveguides

Seungil Kim

Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee

University, Seoul 130-701, Republic of Korea

Abstract

In this paper, we study the non-reflecting boundary condition for the time-harmonic Maxwell’s equations in homogeneous waveguides with an inhomoge-neous inclusion. We analyze a series representation of solutions to the Maxwell’sequations satisfying the radiating condition at infinity, from which we developthe so-called electric-to-magnetic operator for the non-reflecting boundary con-dition. Infinite waveguides are truncated to a finite domain with a fictitiousboundary on which the non-reflecting boundary condition based on the electric-to-magnetic operator is imposed. As the main goal, the well-posedness of thereduced problem will be proved. This study is important to develop numericaltechniques of accurate absorbing boundary conditions for electromagnetic wavepropagation in waveguides.

Key words: Maxwell’s equations, non-reflecting boundary condition,electric-to-magnetic operator, waveguide

1. Introduction

In this paper, we will study the non-reflecting boundary condition for theelectric fields E satisfying the second-order Maxwell’s equations in a waveguide

∇×∇× E − ω2ǫE = iωJ in Ω∞,

ν × E = 0 on ∂Ω∞,(1.1)

where Ω∞ is a semi-infinite waveguide in R3 with Lipschitz boundary such that

Ω∞ ∩ (x, y, z) ∈ R3 : z > −b = Θ× (−b,∞)

and Ω∞∩z < −b is bounded with b > 0 (see Figure 1). Here ν represents theoutward unit normal vector on the boundary of Ω∞, and Θ is a simply connected

Email address: [email protected] (Seungil Kim)

Preprint submitted to Elsevier April 4, 2017

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Figure 1: Semi-infinite waveguide Ω∞.

and smooth (or piecewise smooth with no reentrance corners) bounded domainin R

2. We assume that the cross-section denoted by Θ of the semi-infinitewaveguide Ω∞ perpendicular to the z-axis is uniform for z > −b. Electric fieldsfor this model problem are confined within the perfectly conducting boundary∂Ω∞ and propagate along the z-axis. Also, ω stands for a positive angularfrequency and ǫ ∈ L∞(Ω∞) is the electric permittivity satisfying ℑ(ǫ) ≥ 0 andℜ(ǫ) > ǫ0 for a positive constant ǫ0 > 0 and ǫ−1 has a compact support, sayingǫ = 1 for z > −b. In addition, J represents the current source density satisfying∇ · J = 0 and vanishing for z > −b.

The numerical study for electromagnetic wave propagation in the unboundeddomain Ω∞ requires domain truncation to a finite computational domain e.g,Ω = Ω∞ ∩ z < 0 with a non-reflecting boundary condition on the fictitiousboundary ΓE = Θ×0, which can guarantee that the reduced problem has thesame radiation solution without reflection. The goal of the paper is to studythe non-reflecting radiation boundary condition on ΓE satisfied by radiatingelectric fields in the semi-infinite waveguide and analyze the well-posedness ofthe truncated problem supplemented with the non-reflecting radiation boundarycondition. This result will play a crucial role for developing absorbing bound-ary conditions such as perfectly matched layers (PMLs) and complete radiationboundary conditions (CRBCs) and studying their convergence to the exact ra-diation condition in a subsequent work.

In case of no inhomogeneity inclusion in waveguides, the Maxwell’s equa-tions can be reduced to the scalar Helmholtz equation. It coincides with theacoustic wave propagation problem, for which the Dirichlet-to-Neumann oper-ator can give an approach for defining the non-reflecting boundary condition.This simplified model has been extensively investigated as well as ones with ap-proximate absorbing boundary conditions including truncated DtN [5, 18, 26]PML [4] and CRBCs [21, 22, 23]. When ǫ is not constant, however, compo-nents of electromagnetic fields can not be separated in general and the full-waveanalysis based on the vector equations such as (1.1) is required. To the bestof the author’s knowledge, radiation conditions for vector electromagnetic fieldsin waveguides have not been studied, although non-reflecting radiation bound-ary conditions for exterior scattering problems have been studied in [2] with the

2

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Dirichlet-to-Neumann operator and in [24, 27] based on the electric-to-magneticoperator.

On the other hand, it is worth noting that there are studies about electro-magnetic wave propagation in waveguides with more general ǫ, that can be afunction of x and y but does not have a variation along the axis of waveguides.In such a case, propagation constants for a given wavenumber are involved ina non-selfadjoint eigenvalue problem, which is a more difficult problem thanthe homogeneous one. Thus, the research for the general model is focused oncomputation of propagation constants of the Maxwell’s equations for a givenwavenumber as done in [13, 28, 29]. Some studies [6, 20] are interested inpossible wavenumbers for a given propagation constant. Also, a study on elec-tromagnetic wave propagation in periodic media can be found in [1].

This paper consists of two parts. One is devoted to defining the electric-to-magnetic operator suitable for electromagnetic fields in waveguides by using aseries representation of radiating electromagnetic fields. For this, it is requiredto understand the tangential traces and tangential component traces of functionsbelonging to H(curl,Ω) studied in [3, 11, 10]. In particular, characterization ofν×u and u on ΓE , when ν×u for u ∈ H(curl,Ω) vanishes on ∂Θ×(−b, 0), willbe presented (see e.g., [8, 9]). The other is for establishing the well-posedness ofthe truncated problem supplemented with the non-reflecting radiation boundarycondition based on the electric-to-magnetic operator on ΓE . The main ingre-dient for this is compact perturbation arguments assuming the model problemadmits at most one solution. It is well-known that waveguides, in general, mayhave eigenvalues associated with exponentially decaying eigenfunctions (see e.g.,[14, 16]). Therefore, in order to establish well-posedness of the model problem,throughout the paper we assume that ω2 is not an eigenvalue of the problem.

The paper is organized as follows. In Section 2, we study a series representa-tion of radiating electromagnetic fields by using TE/TM mode decomposition.Section 3 is devoted to defining the electric-to-magnetic operator, that serves asa base for the non-reflecting radiation boundary condition for electromagneticfields. Section 4 studies the Helmholtz decomposition, and the well-posednessof the truncated problem is established in Section 5.

2. Series representation of radiating electromagnetic fields

In order to discuss the radiation condition on ΓE for electric fields, we willrely on a series representation of solutions and so we first consider the first-orderMaxwell’s equations involving both of electric fields E and magnetic fields H

∇× E = iωH,

∇×H = −iωǫE,

∇ ·H = 0,

∇ · ǫE = 0

(2.1)

in Θ × (−b,∞) with the perfectly conducting boundary conditions ν × E = 0and ν · H = 0 on ∂Θ × (−b,∞). The outward unit normal vector ν on the

3

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boundary can be written as ν = (νtT , 0)

t with νT ∈ R2. Here we note that

ǫ = 1 in Θ × (−b,∞) by the assumption. The radiating solutions toward thepositive z-axis are superpositions of modes of the form

E(x, y, z) = E(x, y)eiβz and H(x, y, z) = H(x, y)eiβz ,

where ℜ(β) ≥ 0 and ℑ(β) ≥ 0.

2.1. Preliminaries and notations

It is useful to introduce the transverse components ET ,HT and longitudinalcomponents Ez, Hz

E(x, y) := (Ex, Ey, Ez)t := (Et

T , Ez)t and H(x, y) := (Hx, Hy, Hz) := (Ht

T , Hz)t.

Also, for vector-valued functions φ = (φx, φy, φz)t = (φt

T , φz)t, we introduce

the following notations for divergence and curl operators

φ⊥T = (φy,−φx)

t = RφT ,

∇T · φT = ∂xφx + ∂yφy ,

∇⊥T φz = (∂yφz ,−∂xφz)

t = R∇Tφz,

∇⊥T · φT = ∂xφy − ∂yφx = ∇T · RφT ,

where R =

[0 1−1 0

]is the rotation by −90. Then we have the identity

∇× φ =

[∂yφz−∂xφz

]−

[∂zφy−∂zφx

]

∂xφy − ∂yφx

=

[∇

⊥T φz − ∂zφ

⊥T

∇⊥T · φT

],

ez ×

[φT

φz

]=

[−φ⊥

T

0

],

ez × (∇ × φ) =

[∇Tφz − ∂zφT

0

],

where ez = (0, 0, 1)t. The transverse components and longitudinal componentsof the propagating mode satisfy

∇⊥TEz − iβE⊥

T = iωHT , (2.2)

∇⊥T · ET = iωHz, (2.3)

∇⊥THz − iβH⊥

T = −iωǫET , (2.4)

∇⊥T ·HT = −iωǫEz, (2.5)

∇T ·HT + iβHz = 0, (2.6)

∇T · ǫET + iβǫEz = 0. (2.7)

The boundary conditions for the transverse and longitudinal components canbe written as

Ez = 0 and ν⊥T · ET = 0 on ∂Θ× (−b,∞)

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and∂Hz

∂ν= 0 and νT ·HT = 0 on ∂Θ× (−b,∞).

Therefore, it is shown that the longitudinal components Ez and Hz are Dirichletand Neumann eigenfunctions, respectively

∆TEz + (ω2 − β2)Ez = 0 in Θ,

Ez = 0 on ∂Θ,(2.8)

∆THz + (ω2 − β2)Hz = 0 in Θ,

∂Hz

∂νT= 0 on ∂Θ.

(2.9)

Here ∆T = ∂2x + ∂2y is the transverse Laplace operator. Also, a simple compu-tation using (2.2)-(2.7) shows that if β 6= ±ω, then the transverse componentsare determined by the longitudinal components

ET =i

ω2 − β2(β∇TEz + ω∇⊥

THz), (2.10)

HT =i

ω2 − β2(−ω∇⊥

TEz + β∇THz). (2.11)

By linearity of the problem, we assume that one of Ez or Hz vanishes. WhenEz = 0 and Hz is a Neumann eigenfunction, the electric fields are perpendicularto the propagating direction and so it is called a transverse electric mode (TEmode). On the other hand, in case that Hz = 0 and Ez is a Dirichlet eigenfunc-tion, the magnetic fields are perpendicular to the propagating direction and soit is called a transverse magnetic mode (TM mode).

Here we note that there is no propagating mode whose propagation constant±β equals the angular frequency ω. Indeed, if so, we have that Ez = 0 from(2.8) and Hz is constant from (2.9). By the Faraday’s law of the integral form

∂Θ

E · ds = −

Θ

iωH · dA

on any cross-section of Ω∞ for z > −b and ν ×E = 0 on ∂Θ, we see that

0 = −area(Θ)iωHz,

and consequently Hz also vanishes. This mode such that Ez = Hz = 0 is calleda transverse electromagnetic mode (TEM mode). Now, due to (2.3) and (2.7)it holds that

∇⊥T · ET = 0 and ∇T · ET = 0,

and the first equality implies that there is a potential function φ such that∇Tφ = ET since Θ is simply connected. Combining it with the second equalityand the boundary condition ν⊥

T ·ET = 0 on ∂Θ, we conclude that ET = 0 in Θas ∆Tφ = 0 in Θ and φ is constant on ∂Θ. Finally, (2.2) shows that HT = 0 inΘ.

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As discussed above, homogeneous waveguides with simply connected cross-section can support TE and TM modes but not TEM modes. It is found in [19]that it is allowed for a finite number of TEMmodes to propagate in homogeneouswaveguides with non-simply connected cross-section such as coaxial waveguidesmade of more than two separated conductors. However the above analysisbased on transverse mode decomposition can not be available to the case ofnon-constant ǫ, in which all modes have a longitudinal component in E and H.

2.2. Series representation of radiating electric and magnetic fields

In order to find TE-polarized solutions, let Yn be normalized Neumann eigen-functions associated with Neumann eigenvalues λ2n in Θ,

∆TYn + λ2nYn = 0 in Θ,

∂Yn∂νT

= 0 on ∂Θ.

It is well-known that eigenvalues λ2n are all non-negative real numbers and theyare arranged in increasing order toward infinity, 0 = λ20 < λ21 ≤ λ22 ≤ . . ., andYn

∞n=0 is an orthonormal basis for L2(Θ). Then the longitudinal component

Hz of the magnetic fields H can be written as the Fourier series

Hz(x, y, z) =

∞∑

n=1

AneiβnzYn(x, y) (2.12)

with βn =√ω2 − λ2n. As mentioned in the previous subsection, we note that

the constant eigenfunction Y0 associated with the zero eigenvalue does not makea contribution to the series of Hz. Here there exists an integer Nλ such thatλn ≤ ω for n ≤ Nλ, which corresponds to a propagating mode, and λn > ωfor n > Nλ, which pertains to an exponentially decaying evanescent mode sincesuch βn is i

√λ2n − ω2 := iβn.

By the formulas (2.10) and (2.11), it can be shown that the transverse com-ponents ET and HT are given by

ET =∞∑

n=1

Aniω

λ2neiβnz∇

⊥T Yn and HT =

∞∑

n=1

Aniβnλ2n

eiβnz∇TYn (2.13)

and hence combining (2.12) and (2.13) gives the series solution of the TE po-larization

E =

∞∑

n=1

Aneiβnz

[iωλ2n∇

⊥T Yn0

]and H =

∞∑

n=1

Aneiβnz

[iβn

λ2n∇TYnYn

].

Similarly, for TM modes we introduce normalized Dirichlet eigenfunctionsVn associated with Dirichlet eigenvalues µ2

n,

∆Vn + µ2nVn = 0 in Θ,

Vn = 0 on ∂Θ.

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In this case, µ2n are all positive and they are arranged in increasing order, 0 <

µ21 ≤ µ2

2 ≤ . . ., and Vn∞n=1 forms an orthonormal basis for L2(Θ). Thus, the

longitudinal component Ez of E can be written in terms of these basis functions

Ez(x, y, z) =∞∑

n=1

BneiγnzVn(x, y)

with γn =√ω2 − µ2

n. Also, we can find an integer Nµ separating the propagat-ing regime and evanescent regime such that µn ≤ ω for 1 ≤ n ≤ Nµ and µn > ωfor n > Nµ. For n > Nµ, γn = iγn for some γn > 0. As done before, we can seethat TM polarized electric and magnetic fields for the Maxwell’s equations aregiven by

E =

∞∑

n=1

Bneiγnz

[ iγn

µ2n∇TVnVn

]and H =

∞∑

n=1

Bneiγnz

[−iωµ2n∇

⊥T Vn

0

].

By combining two polarized solutions, it is obtained that general radiatingsolutions of the Maxwell’s equations in Θ× (−b,∞) are of the forms

E =

∞∑

n=1

Aneiβnz

[iωλ2n∇

⊥T Yn0

]+Bne

iγnz

[ iγn

µ2n∇TVnVn

], (2.14)

H =

∞∑

n=1

Aneiβnz

[iβn

λ2n∇TYnYn

]+Bne

iγnz

[−iωµ2n∇

⊥T Vn

0

]. (2.15)

The next subsection is devoted to justifying that the series (2.14) and (2.15)of transverse components ET and HT are expanded in terms of complete or-thonormal bases as functions in

L2N (Θ) := z ∈ (L2(Θ))2 | ν⊥

T · z = 0 on ∂Θ,

L2T (Θ) := z ∈ (L2(Θ))2 | νT · z = 0 on ∂Θ,

respectively.

2.3. Orthonormal basis for L2N (Θ)

Let us define

Y⊥n :=

1

λn∇

⊥T Yn and Vn :=

1

µn∇TVn

for n ≥ 1. We claim that Y⊥n ,Vn

∞n=1 is an orthonormal basis for L2

N (Θ). Bystraightforward computations based on the fact that Yn

∞n=0 and Vn

∞n=1 are

the orthonormal bases consisting of eigenfunctions in L2(Θ), it is easy to see

〈Y⊥n ,Y

⊥m〉Θ = δn,m, 〈Vn,Vm〉Θ = δn,m, 〈Vn,Y

⊥m〉Θ = 0,

where 〈·, ·〉Θ is the L2-inner product on Θ and δn,m is the Kronecker delta.

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To prove the completeness of the set Y⊥n ,Vn

∞n=1, assume that z ∈ L2

N (Θ).Then there exists ψ1 ∈ H1(Θ)/R such that

〈∇⊥T ψ1,∇

⊥T ξ〉Θ = 〈z,∇⊥

T ξ〉Θ for all ξ ∈ H1(Θ)/R

and ψ1 can be written as a series in terms of Neumann eigenfunctions Yn

ψ1 =

∞∑

n=0

AnYn and ∇⊥T ψ1 =

∞∑

n=1

An∇⊥T Yn.

Similarly, we can find a unique solution ψ2 ∈ H10 (Θ) satisfying

〈∇Tψ2,∇T ξ〉Θ = 〈z,∇T ξ〉Θ for all ξ ∈ H10 (Θ)

and express ψ2 as a series in terms of Dirichlet eigenfunctions Vn

ψ2 =

∞∑

n=1

BnVn and ∇Tψ2 =

∞∑

n=1

Bn∇TVn.

Since ∇⊥T · (z − ∇

⊥T ψ1 −∇Tψ2) = 0 in Θ, there exists φ ∈ H1(Θ)/R such

that∇Tφ = z−∇

⊥T ψ1 −∇Tψ2.

Here we note that

ν⊥T ·∇Tφ = ν⊥

T · z− νT ·∇Tψ1 − ν⊥T ·∇Tψ2 = 0 on ∂Θ

and hence φ is constant on ∂Θ, from which we can choose φ such that φ = 0 on∂Θ.

In addition, since ∇T · (z−∇⊥T ψ1−∇Tψ2) = 0 in Θ, it holds that ∆Tφ = 0

in Θ and φ = 0 on ∂Θ, which leads to z−∇⊥T ψ1−∇Tψ2 = 0. As a consequence

it follows that

z =

∞∑

n=1

An∇⊥T Yn +Bn∇TVn.

Therefore Y⊥n ,Vn

∞n=1 is an orthonormal basis for L2

N (Θ).More generally, denoting Hs(Θ) = (Hs(Θ))2 for s ≥ 0, we can define

HsN (Θ) = u ∈ L2

N (Θ) : u ∈ Hs(Θ). In particular, since Θ is smooth(piecewise smooth with no reentrance corners), the H1-norm for functions inH1

N (Θ) is equivalent to the norm

(‖u‖2L2(Θ) + ‖∇T · u‖2L2(Θ) + ‖∇⊥T · u‖L2(Θ))

1/2

(see e.g., [17]) and hence H1N (Θ) is equipped with the norm

‖u‖2H1(Θ) :=

∞∑

n=1

(λ2n + 1)|An|2 + (µ2

n + 1)|Bn|2

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for u =∑∞

n=1AnY⊥n + BnVn. By the interpolation technique (see e.g., [7, 12,

25]), it can be shown that

‖u‖2Hs(Θ) :=

∞∑

n=1

(λ2n + 1)s|An|2 + (µ2

n + 1)s|Bn|2

for 0 ≤ s ≤ 1. The dual space of HsN (Θ) with the pivot space L2

N (Θ) is denotedby H−s

N (Θ) and it is characterized by functions u =∑∞

n=1AnY⊥n + BnVn

satisfying

‖u‖2H−s(Θ) :=

∞∑

n=1

(λ2n + 1)−s|An|2 + (µ2

n + 1)−s|Bn|2 <∞

for 0 ≤ s ≤ 1.

2.4. Orthonormal basis for L2T (Θ)

For an orthonormal basis for L2T (Θ), we define

Yn :=1

λn∇TYn and V⊥

n :=1

µn∇

⊥T Vn

for n ≥ 1 and we will show that Yn,V⊥n

∞n=1 forms an orthonormal basis for

L2T (Θ). First, the following orthonormality

〈Yn,Ym〉Θ = δn,m, 〈V⊥n ,V

⊥m〉Θ = δn,m, 〈V⊥

n ,Ym〉Θ = 0

can be easily verified by the definition of Yn and Vn.For the completeness of the set Yn,V

⊥n

∞n=1 we assume that z ∈ L2

T (Θ).Now, there exists ψ1 ∈ H1(Θ)/R such that

〈∇Tψ1,∇T ξ〉Θ = 〈z,∇T ξ〉Θ for all ξ ∈ H1(Θ)/R

and the solution can be expressed as

ψ1 =

∞∑

n=0

AnYn and ∇Tψ1 =

∞∑

n=1

An∇TYn.

Analogously, there exists ψ2 ∈ H10 (Θ) such that

〈∇⊥T ψ2,∇

⊥T ξ〉Θ = 〈z,∇⊥

T ξ〉Θ for all ξ ∈ H10 (Θ)

and it is obtained that

ψ2 =

∞∑

n=1

BnVn and ∇⊥T ψ2 =

∞∑

n=1

Bn∇⊥T Vn.

Since ∇⊥T · (z − ∇Tψ1 −∇

⊥T ψ2) = 0 in Θ, there exists φ ∈ H1(Θ)/R such

that∇Tφ = z−∇Tψ1 −∇

⊥T ψ2.

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Here we note that

νT ·∇Tφ = νT · z− νT ·∇Tψ1 − νT ·∇⊥T ψ2 = 0 on ∂Θ.

In addition, since ∇T · (z −∇Tψ1 −∇⊥T ψ2) = 0 in Θ, we have ∆Tφ = 0 in Θ

and νT ·∇Tφ = 0 on ∂Θ, which yields that φ is constant and hence

z−∇Tψ1 −∇⊥T ψ2 = 0.

As a consequence it follows that

z =∞∑

n=1

An∇TYn +Bn∇⊥T Vn

and it completes the proof that Yn,V⊥n

∞n=1 is an orthonormal basis for L2

T (Θ).As done before, we can also define Hs

T (Θ) = u ∈ L2T (Θ) : u ∈ Hs(Θ) for

0 ≤ s ≤ 1 with the norm

‖u‖2Hs(Θ) :=

∞∑

n=1

(λ2n + 1)s|An|2 + (µ2

n + 1)s|Bn|2

for u =∑∞

n=1AnYn + BnV⊥n and its dual space with the pivot space L2

T (Θ)

denoted by H−sT (Θ) as the set of functions u =

∑∞n=1AnYn +BnV

⊥n such that

‖u‖2H−s(Θ) :=∞∑

n=1

(λ2n + 1)−s|An|2 + (µ2

n + 1)−s|Bn|2 <∞.

3. Electric-to-Magnetic operator

Recall that Ω is the bounded domain obtained from the semi-infinite waveg-uide Ω∞ by truncating at z = 0. We denote the boundary of Ω on z = 0 by ΓE

and the complement of ΓE in ∂Ω by ΓT , that is,

ΓE = Θ× 0 ⊂ ∂Ω and ΓT = ∂Ω \ ΓE ⊂ ∂Ω.

We begin by defining Sobolev spaces to be used throughout the remaining paper.Let

H(curl,Ω) = u ∈ (L2(Ω))3 : ∇× u ∈ (L2(Ω))3,

H0,ΓT (curl,Ω) = u ∈ H(curl,Ω) : ν × u = 0 on ΓT .

For the Lipschitz domain Ω, the tangential trace space is rigorously studied in[10] and it is known to be given by

H−1/2(Div, ∂Ω) := u ∈ (H−1/2(∂Ω))3 : ν · u = 0, there exists η ∈ H−1/2(∂Ω)

such that 〈u, γ(∇φ)〉1/2,∂Ω = 〈η, φ〉1/2,∂Ω for all φ ∈ H2(Ω),

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where γ : (H1(Ω))3 → (H1/2(∂Ω))3 is the trace operator and 〈·, ·〉1/2,∂Ω is the

duality pairing between H−1/2(∂Ω) and H1/2(∂Ω). It can be also found in [10]

that the tangential trace operator γτ : H(curl,Ω) → H−1/2(Div, ∂Ω) definedby γτ (u) = ν × u is surjective.

Further, we define trace spaces H−1/2T (Div,ΓE) and H

−1/2N (Curl,ΓE) of

H0,ΓT (curl,Ω) on the part ΓE (identified with Θ) of the boundary, as follows:noting that H−1/2(ΓE) is the set of functions φ =

∑∞n=0 AnYn satisfying

‖φ‖2H−1/2(ΓE) :=

∞∑

n=0

(λ2n + 1)−1/2|An|2 <∞,

we define

H−1/2T (Div,ΓE) := φ ∈ H

−1/2T (ΓE) : ∇T · φ ∈ H−1/2(ΓE)

identified with a subspace in H−1/2(ΓE)× 0 ⊂ (H−1/2(ΓE))3 with the norm

‖φ‖2H

−1/2T (Div,ΓE)

:= ‖φ‖2H−1/2(ΓE) + ‖∇T · φ‖2H−1/2(ΓE)

=∞∑

n=1

(λ2n + 1)1/2|An|2 + (µ2

n + 1)−1/2|Bn|2

for φ =∑∞

n=1AnYn +BnV⊥n .

Similarly, we define

H−1/2N (Curl,ΓE) = φ ∈ H

−1/2N (ΓE) : ∇⊥

T · φ ∈ H−1/2(ΓE)

identified with a subspace in H−1/2(ΓE)× 0 ⊂ (H−1/2(ΓE))3 with the norm

‖φ‖2H

−1/2N (Curl,ΓE)

:= ‖φ‖2H−1/2(ΓE) + ‖∇⊥

T · φ‖2H−1/2(ΓE)

=∞∑

n=1

(λ2n + 1)1/2|An|2 + (µ2

n + 1)−1/2|Bn|2

for φ =∑∞

n=1AnY⊥n +BnVn.

It is shown in [8, 9] that the restrictions of tangential traces and tangential

component traces of functions in H0,ΓT (curl,Ω) to ΓE are in H−1/2T (Div,ΓE)

and H−1/2N (Curl,ΓE), respectively. Furthermore, the restricted tangential trace

operator γτ,ΓE and tangential component trace operator πτ,ΓE ,

γτ,ΓE : H0,ΓT (curl,Ω) → H−1/2T (Div,ΓE) defined by γτ,ΓE (u) = ν × u,

πτ,ΓE : H0,ΓT (curl,Ω) → H−1/2N (Curl,ΓE) defined by πτ,ΓE (u) = uT = (ν × u)× ν,

are linear, continuous and surjective.

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As the last ingredient Sobolev space for studying the radiation condition for

electric fields, we require the dual space H−1/2N (Div,ΓE) of H

−1/2N (Curl,ΓE)

with the pivot space L2N (ΓE), which is defined by

H−1/2N (Div,ΓE) := φ ∈ H

−1/2N (ΓE) : ∇T · φ ∈ (H

1/200 (ΓE))

with the norm

‖φ‖2H

−1/2N (Div,ΓE)

:= ‖φ‖2H−1/2(ΓE) + ‖∇T · φ‖2

(H1/200

(ΓE))′

=

∞∑

n=1

(λ2n + 1)−1/2|An|2 + (µ2

n + 1)1/2|Bn|2

for φ =∑∞

n=1AnY⊥n + BnVn. Here H

1/200 (ΓE) is a subspace of H1/2(ΓE)

of functions vanishing on ∂ΓE in a special sense (see [25]). Also, (H1/200 (ΓE))

represents the dual space ofH1/200 (ΓE) and consists of functions φ =

∑∞n=1AnVn

satisfying

‖φ‖2(H

1/200

(ΓE))′:=

∞∑

n=1

(µ2n + 1)−1/2|An|

2 <∞.

Now, we shall define the electric-to-magnetic operator

Gω : H−1/2T (Div,ΓE) → H

−1/2N (Div,ΓE)

byGω(g) = f,

where f is the tangential trace of the magnetic fields corresponding to the electricfields solving the problem

∇×∇× E − ω2ǫE = 0 in Θ× (0,∞),

ν × E = 0 on ∂Θ× (0,∞),

ν × E = g on ΓE = Θ× 0

(3.1)

with the radiation condition at z = ∞, i.e, f = ν ×H = 1iων × (∇× E) on ΓE .

In other words, for a given tangential trace g = γτ,ΓE(E) on ΓE of radiatingelectric fields

g = ν × E =

∞∑

n=1

Aniω

λ2n∇TYn −Bn

iγnµ2n

∇⊥T Vn

=∞∑

n=1

Aniω

λnYn −Bn

iγnµn

V⊥n

we define

f = ν ×H =∞∑

n=1

−Aniβnλ2n

∇⊥T Yn −Bn

µ2n

∇TVn

=

∞∑

n=1

−Aniβnλn

Y⊥n −Bn

µnVn.

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Equivalently, for ζ ∈ H−1/2T (Div,ΓE) such that ζ =

∑∞n=1 AnYn + BnV

⊥n we

have

Gω(ζ) =

∞∑

n=1

−βnωAnY

⊥n +

ω

γnBnVn. (3.2)

Lemma 3.1. The electric-to-magnetic operator Gω is a continuous operator

from H−1/2T (Div,ΓE) to H

−1/2N (Div,ΓE).

Proof. Noting that βn, γn 6= ω for all n ≥ 1, we see that

|β2n| = |λ2n − ω2| ≤ C|λ2n + 1|,

|γ2n| = |µ2n − ω2| ≥ C|µ2

n + 1|

for some positive constant C. Therefore, for ζ =∑∞

n=1 AnYn + BnV⊥n in

H−1/2T (Div,ΓE)

‖Gω(ζ)‖2

H−1/2N (Div,ΓE)

=

∞∑

n=1

(λ2n + 1)−1/2 |βn|2

ω2|An|

2 + (µ2n + 1)1/2

ω2

|γn|2|Bn|

2

≤ C

∞∑

n=1

(λ2n + 1)1/2|An|2 + (µ2

n + 1)−1/2|Bn|2

= C‖ζ‖2H

−1/2T (Div,ΓE)

,

which shows that Gω is continuous from H−1/2T (Div,ΓE) to H

−1/2N (Div,ΓE).

Now, the problem (1.1) can be interpreted as the Maxwell’s equations inthe truncated domain Ω equipped with the radiation condition on the fictitiousboundary ΓE based on the electric-to-magnetic operator,

∇×∇× E − ω2ǫE = iωJ in Ω,

ν × E = 0 on ΓT ,

ν × (∇× E) = iωGω(ν × E) on ΓE .

For J in (L2(Ω))3 the corresponding variational problem is one that seeks forE ∈ H0,ΓT (curl,Ω) satisfying

A(E ,φ) = (iωJ,φ)Ω for all φ ∈ H0,ΓT (curl,Ω), (3.3)

where

A(E ,φ) = (∇× E,∇× φ)Ω − ω2(ǫE,φ)Ω + iω〈Gω(ν × E),φT 〉ΓE

for E and φ in H0,ΓT (curl,Ω). Also, (·, ·)Ω is the L2-inner product on Ω and

〈·, ·〉ΓE represents the duality pairing betweenH−1/2N (Div,ΓE) andH

−1/2N (Curl,ΓE)

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with L2N (ΓE) as the pivot space. The well-posedness of the variational problem

(3.3) will be investigated in the subsequent sections as the main result.

At last, we introduce the divergence-free subspace of H−1/2T (Div,ΓE),

H−1/2T (Div0,ΓE) = ζ ∈ H

−1/2T (Div,ΓE) : ∇T · ζ = 0

for Helmholtz decomposition of functions in H0,ΓT (curl,Ω).

Lemma 3.2. The imaginary part of the operator Gω is non-negative in H−1/2T (Div0,ΓE)

in the sense of

ℑ(〈Gω(ζ), ζ⊥〉ΓE ) ≥ 0

for ζ ∈ H−1/2T (Div0,ΓE).

Proof. Suppose that ζ ∈ H−1/2T (Div0,ΓE) has the expression ζ =

∑∞n=1 BnV

⊥n .

We here note that ζ⊥ =∑∞

n=1(−Bn)Vn belongs to H−1/2N (Curl,ΓE). Since

γn = iγn with γn =√λ2n − ω2 > 0 for n > Nµ, by using (3.2) we have that

ℑ(〈Gω(ζ), ζ⊥〉ΓE ) =

∞∑

n=Nµ+1

ω

γn|Bn|

2 ≥ 0,

which completes the proof.

4. Helmholtz decomposition

Recall that we assume ω2 is not an eigenvalue of the Maxwell’s equations,which guarantees that the problem has at most one solution. For the Helmholtzdecomposition of functions in H0,ΓT (curl,Ω), let us define the Sobolev spaces

S = φ ∈ H1(Ω) : φ = 0 on ΓT ,

X0 = E ∈ H0,ΓT (curl,Ω) : ω2(ǫE,∇φ)Ω − iω〈Gω(ν × E),∇Tφ〉ΓE = 0 for all φ ∈ S

= E ∈ H0,ΓT (curl,Ω) : ∇ · ǫE = 0 in Ω and ν · E = −i

ω∇T ·Gω(ν × E) on ΓE.

We begin with the Rellich lemma for the uniqueness of solutions.

Lemma 4.1. Suppose that E is a radiating solution of the Maxwell’s equations

in the semi-infinite straight waveguide Θ× (0,∞) satisfying the condition ∇T ·E⊥T = 0 on Θ× 0. Let H = 1/(iω)∇× E. If

ℜ (〈ν × E,H〉Γr ) ≤ 0 and ℑ (〈ν × E,H〉Γr ) ≤ 0 (4.1)

with Γr = Θ× r for any r ≥ 0, then E = 0 in Θ× (0,∞).

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Proof. Noting that the radiating electric and magnetic fields are given by (2.14)and (2.15), respectively, the divergence-free condition of the tangential trace ofE on the boundary Θ×0 implies that the radiating solutions E and H are ofthe form

E =

∞∑

n=1

Bneiγnz

[ iγn

µnVn

Vn

]and H =

∞∑

n=1

Bneiγnz

[−iωµn

V⊥n

0

]

for z ≥ 0. Thus, we have

〈ν × E,H〉Γr =

∞∑

n=1

|Bneiγnr|2

ωγnµ2n

for r ≥ 0, which shows that

ℜ (〈ν × E,H〉Γr ) =

Nµ∑

n=1

|Bn|2ωγnµ2n

≥ 0,

ℑ (〈ν × E,H〉Γr ) =

∞∑

n=Nµ+1

|Bn|2e−2γnr

ωγnµ2n

≥ 0.

Finally, due to (4.1) we can conclude that Bn = 0 for all n ≥ 1 and hence E = 0in Θ× (0,∞).

Lemma 4.2. The only function in both ∇S and X0 is zero, i.e., ∇S∩X0 = 0.

Proof. Suppose that ∇p ∈ ∇S ∩X0 for p ∈ S. Then by the definition of X0, psolves the problem

−ω2(ǫ∇p,∇ξ)Ω + iω〈Gω(ν ×∇p),∇T ξ〉ΓE = 0 for all ξ ∈ S.

For ξ = p, we have

iω〈Gω(ν ×∇p),∇T p〉ΓE = ω2(ǫ∇p,∇p)Ω. (4.2)

Since Gω(ν ×∇p) represents the tangential trace ν×H on ΓE of the radiatingmagnetic fields H of the problem (3.1) on the semi-infinite waveguide Θ×(0,∞)when the radiating electric solution E is given by ∇p on ΓE , it follows that

〈ν ×H,E〉ΓE = 〈Gω(ν ×∇p),∇T p〉ΓE = −iω(ǫ∇p,∇p)Ω.

Since〈ν × E,H〉ΓE = −〈ν ×H,E〉ΓE = −iω(ǫ∇p,∇p)Ω,

both real and imaginary parts of 〈ν × E,H〉ΓE are non-positive and it thenfollows from Lemma 4.1 that E = 0 in Θ × (0,∞) and in particular ∇p = 0on ΓE . By (4.2), it is obtained that (ǫ∇p,∇p)Ω = 0 and finally ∇p = 0 in Ω,which completes the proof.

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Theorem 4.3. For any u ∈ H0,ΓT (curl,Ω), there exist unique w ∈ X0 and

p ∈ S such that u = w+∇p, i.e., H0,ΓT (curl,Ω) = X0 ⊕∇S. Furthermore, w

and p satisfy the estimates

‖w‖H(curl,Ω) ≤ C‖u‖H(curl,Ω) and ‖∇p‖H(curl,Ω) ≤ C‖u‖H(curl,Ω).

Proof. For u ∈ H0,ΓT (curl,Ω), we first verify the existence of p ∈ S satisfying

A(∇p,∇ξ) = A(u,∇ξ) for all ξ ∈ S, (4.3)

and‖∇p‖H(curl,Ω) ≤ C‖u‖H(curl,Ω).

Once we have it, it immediately follows from (4.3) that w = u −∇p is in X0,which justifies the decomposition of u = w+∇p ∈ X0 +∇S. In addition, thetriangle inequality shows the continuity of the projection,

‖w‖H(curl,Ω) ≤ ‖u‖H(curl,Ω) + ‖∇p‖H(curl,Ω) ≤ C‖u‖H(curl,Ω).

Finally, the uniqueness of w ∈ X0 and p ∈ S is a consequence of Lemma 4.2.For establishing that there exists a solution p in S satisfying (4.3), we notice

that solving the problem (4.3) is equivalent to finding a solution p ∈ S to theproblem

a1(p, ξ) = A(u,∇ξ) for all ξ ∈ S, (4.4)

where

a1(p, ξ) = −ω2(ǫ∇p,∇ξ)Ω + iω〈Gω(ν ×∇p),∇T ξ〉ΓE .

Due to the continuity ofGω, the tangential trace operator γτ,ΓE fromH0,ΓT (curl,Ω)

to H−1/2T (Div,ΓE) and the tangential component trace operator πτ,ΓE from

H0,ΓT (curl,Ω) to H−1/2N (Curl,ΓE), we see that a1(·, ·) is continuous in S × S

and A(u, ·) is a bounded anti-linear functional in S.

Since ν ×∇p = −∇⊥T p is in H

−1/2T (Div0,ΓE), by Lemma 3.2 we have

|ℜ(a1(p, p))| = ω2(ℜ(ǫ)∇p,∇p)Ω + ωℑ(〈Gω(ν ×∇p),∇T p〉ΓE ) ≥ C‖∇p‖2L2(Ω),

which leads to the coercivity of a1(·, ·) in S × S. Therefore, the Lax-Milgramtheorem shows that there exists a solution p ∈ S to the problem (4.4) satisfying

‖∇p‖H(curl,Ω) ≤ ‖p‖H1(Ω) ≤ C sup06=ξ∈S

|a1(p,∇ξ)|

‖ξ‖H1(Ω)≤ C‖u‖H(curl,Ω),

which completes the proof.

We close this section with the compactness of the embedding of X0 in(L2(Ω))3, whose proof is based on the idea in [27].

Lemma 4.4. The space X0 is compactly embedded in (L2(Ω))3.

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Proof. As seen in the above, it is also true that H0,ΓT (curl,Ω) can be decom-posed into

H0,ΓT (curl,Ω) = X10 ⊕∇S,

where X10 is the set of divergence-free components but with ǫ ≡ 1. Here X1

0 iscompactly embedded in (L2(Ω))3. In fact, let wj

∞j=1 be a bounded sequence

of functions in X10. For each j = 1, 2, . . ., we define an extension wj of wj to

Ω∞ by finding the radiating solution to the problem in the outside of Ω in Ω∞

∇×∇× vj − ω2vj = 0 in Θ× (0,∞),

ν × vj = 0 on ∂Θ× (0,∞),

ν × vj = ν ×wj on Θ× 0,

that is, wj = wj in Ω and vj in Ω∞ \ Ω. Since the tangential component ofwj are continuous across ΓE = Θ × 0, wj is in Hloc(curl,Ω∞). In addition,by invoking the definition of X1

0 and Gω and using the Maxwell’s equations, wecan show that

ν ·wj = −i

ω∇T ·Gω(ν ×wj) = −

1

ω2∇T · (ν × (∇× vj))

=1

ω2ν · (∇× (∇ × vj)) = ν · vj ,

from which it follows that wj belongs to Hloc(div,Ω∞).Now, let χ(z) be a cutoff function such that χ(z) = 1 for z < 0 and χ(z) = 0

for z > 1, and Ω1 = Ω∞ ∩ z < 1. Since

‖wj‖H(curl,Ω1\Ω) ≤ C‖ν ×wj‖H−1/2T (Div,ΓE)

≤ C‖wj‖H(curl,Ω)

and wj is divergence-free, χwj∞j=1 is bounded in H0(curl,Ω1) ∩ H(div,Ω1)

with respect to the norm

(‖wj‖2L2(Ω1)

+ ‖∇×wj‖2L2(Ω1)

+ ‖∇ ·wj‖2L2(Ω1)

)1/2,

which is compactly embedded in (L2(Ω1))3. Here

H0(curl,Ω1) = w ∈ (L2(Ω1))3 : ∇×w ∈ (L2(Ω1))

3,ν ×w = 0 on ∂Ω1

H(div,Ω1) = w ∈ (L2(Ω1))3 : ∇ ·w ∈ L2(Ω1).

Therefore, there exists a subsequence χwjk converging to w in (L2(Ω1))3 and

so we conclude that wjk converges to w|Ω in (L2(Ω))3.To prove the lemma, let uj be a bounded sequence of X0. By the Helmholtz

decomposition with ǫ = 1, it can be written as uj = u1j +∇p1j with u1

j ∈ X10

and p1j ∈ S such that ‖u1j‖H(curl,Ω) ≤ C‖uj‖H(curl,Ω). Since X1

0 is compactly

embedded in (L2(Ω))3, there exists a subsequence, still denoted by u1j

∞j=1,

which is Cauchy in (L2(Ω))3. By using the fact that uj ∈ X0,

(ǫ(uj − uk),uj − uk)Ω = (ǫ(uj − uk), (uj −∇p1j)− (uk −∇p1k))Ω

= (ǫ(uj − uk),u1j − u1

k)Ω,

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which shows that

ǫ0‖uj − uk‖2L2(Ω) ≤ ‖ǫ‖L∞(Ω∞)‖uj − uk‖L2(Ω)‖u

1j − u1

k‖L2(Ω)

and hence we have ‖uj −uk‖L2(Ω) ≤ C‖u1j −u1

k‖L2(Ω). Since u1j

∞j=1 is Cauchy

in (L2(Ω))3, so is uj∞j=1, which completes the proof.

5. Well-posedness

In this section, we establish that the variational problem (3.3) is well-posedin H0,ΓT (curl,Ω). The proof follows the same general line used for scatteringproblems in the exterior domain in [27].

Based on the Helmholtz decomposition in Theorem 4.3, we will find eachcomponent of the solution E = w +∇p for w ∈ X0 and p ∈ S. Test functionsφ ∈ H0,ΓT (curl,Ω) also can be decomposed as φ = ψ +∇q for some ψ ∈ X0

and q ∈ S by the Helmholtz decomposition and hence the problem (3.3) isequivalent to finding w ∈ X0 and p ∈ S satisfying

A(w+∇p,ψ +∇q) = (iωJ,ψ +∇q)Ω for all ψ ∈ X0 and q ∈ S.

5.1. Curl-free component ∇p in ∇S

To find the curl-free component ∇p of E, we take test functions φ = ∇q forq ∈ S (i.e., ψ = 0) and use the fact A(w,∇q) = 0 for q ∈ S due to the definitionof the divergence-free component w ∈ X0 of E, which reveals that p ∈ S is, infact, the unique solution to the problem

A(∇p,∇q) = (iωJ,∇q)Ω for all q ∈ S (5.1)

and satisfies‖∇p‖H(curl,Ω) ≤ C‖J‖L2(Ω) (5.2)

as we examined the problem (4.3) in the proof of Theorem 4.3.

5.2. Divergence-free component w in X0

Next, we will show that there exists the divergence-free component w of Ethat solves the problem

A(w,ψ) = (iωJ,ψ)Ω −A(∇p,ψ) for all ψ ∈ X0 (5.3)

and depends continuously on the data. We do this by showing that the sesquilin-ear form A(·, ·) on X0×X0 satisfies the inf-sup conditions based on the standardcompact perturbation arguments in the following lemma.

Lemma 5.1. There exists a positive constant C such that

‖w‖H(curl,Ω) ≤ C sup06=ψ∈X0

|A(w,ψ)|

‖ψ‖H(curl,Ω)(5.4)

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and

‖w‖H(curl,Ω) ≤ C sup06=ψ∈X0

|A(ψ,w)|

‖ψ‖H(curl,Ω)(5.5)

for all w ∈ X0.

For proving the above lemma, we need to investigate Gω in more detail bybreaking it into two parts as follows,

Gω = GYω +GV

ω ,

whereGY

ω : H−1/2T (Div,ΓE) → H

−1/2N (Div,ΓE)

is defined by

GYω (ζ) =

∞∑

n=1

−βnωAnY

⊥n

andGV

ω : H−1/2T (Div,ΓE) → H

−1/2N (Div,ΓE)

is defined by

GVω (ζ) =

∞∑

n=1

ω

γnBnVn

for ζ =∑∞

n=1 AnYn + BnV⊥n ∈ H

−1/2T (Div,ΓE).

Lemma 5.2. The operator GVω γτ,ΓE : X0 → H

−1/2N (Div,ΓE) is compact.

Proof. For u ∈ X0, let ζ = (ν × u)T = γτ,ΓE (u) =∑∞

n=1(AnYn + BnV⊥n ) on

ΓE . Since there exists a positive constant C such that µ2n +1 ≤ Cµ2

n for n ≥ 1,we have

‖GVω (ζ)‖

2

H−1/2N (Div,ΓE)

=∞∑

n=1

(µ2n + 1)1/2

∣∣∣∣ω

γnBn

∣∣∣∣2

≤ C

∞∑

n=1

(µ2n + 1)−1/2µ2

n

∣∣∣∣ω

γnBn

∣∣∣∣2

= C‖∇T ·GVω (ζ)‖

2

(H1/200

(ΓE))′.

Thus, by the definition of X0 it is obtained that

‖GVω (ζ)‖

2

H−1/2N (Div,ΓE)

≤ C‖∇T ·GVω (ζ)‖

2

(H1/200

(ΓE))′

= C‖∇T ·Gω(ζ)‖2

(H1/200

(ΓE))′≤ C‖ν · u‖2H−1/2(ΓE)

≤ C(‖∇ · (ǫu)‖2L2(Ω) + ‖ǫu‖2L2(Ω)) ≤ C‖u‖2L2(Ω).

(5.6)

Now, for a bounded sequence un in X0, there exists a subsequence unj suchthat unj converges in (L2(Ω))3 by Lemma 4.4. Then GV

ω (ν×unj ) is Cauchy by

(5.6) and hence it converges in H−1/2N (Div,ΓE), which completes the proof.

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For analyzing the other operator GYω , we introduce the operator associated

with ω = iGY

i : H−1/2T (Div,ΓE) → H

−1/2N (Div,ΓE)

defined by

GYi (ζ) :=

∞∑

n=1

−βi,niAnY

⊥n

for ζ =∑∞

n=1 AnYn+BnV⊥n ∈ H

−1/2T (Div,ΓE), where βi,n =

√i2 − λ2n = βi,ni

with βi,n =√1 + λ2n > 0.

Lemma 5.3. The operator GYi is non-positive in H

−1/2T (Div,ΓE) in the sense

of

〈GYi (ζ), ζ

⊥〉ΓE ≤ 0

for ζ ∈ H−1/2T (Div,ΓE).

Proof. For ζ ∈ H−1/2T (Div,ΓE) with ζ =

∑∞n=1 AnYn + BnV

⊥n , it follows that

〈GYi (ζ), ζ

⊥〉ΓE =

∞∑

n=1

〈−βi,nAnY⊥n , AnY

⊥n − BnVn〉ΓE

= −∞∑

n=1

βi,n|An|2 ≤ 0,

which completes the proof.

Lemma 5.4. The operator

GYω −

i

ωGY

i : H−1/2T (Div,ΓE) → H

−1/2N (Div,ΓE)

is compact.

Proof. We first show that (GYω−

iωG

Yi )(ζ) is inH1

N (ΓE) for ζ ∈ H−1/2T (Div,ΓE).

Let ζ ∈ H−1/2T (Div,ΓE) be given by ζ =

∑∞n=1(AnYn + BnV

⊥n ). Noting that

|βn − βi,n| = O(λ−1n

), we have

‖(GYω −

i

ωGY

i )(ζ)‖2H1

N (ΓE) =

∞∑

n=1

(λ2n + 1)

∣∣∣∣βn − βi,n

ωAn

∣∣∣∣2

≤ C

∞∑

n=1

(λ2n + 1)1/2|An|2 ≤ C‖ζ‖2

H−1/2T (Div,ΓE)

.

Now, the compactness of the given operator will be established once we show

that H1N (ΓE) is compactly embedded in H

−1/2N (Div,ΓE). To verify this claim,

assume that un is a bounded sequence in H1N (ΓE). Since H

1N (ΓE) is compactly

embedded in L2N (ΓE), there is a subsequence unj of un converging to u in

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L2N (ΓE). Also, since the inclusion form H1/2(ΓE) to L

2(ΓE) is compact, ∇T ·unj ∈ H1/2(ΓE) has a subsequence (still denoted by unj for a simple notation)in L2(ΓE) converging to v in L2(ΓE), and clearly v = ∇T ·u, which implies thatunj converges to u in H0

N (Div,ΓE). Finally, since H0N (Div,ΓE) is continuously

embedded in H−1/2N (Div,ΓE), unj converges to u in H

−1/2N (Div,ΓE).

Now, we are in a position to analyze the well-posedness of the problem (5.3)in X0.

Proof of Lemma 5.1. We first note that (3.2) shows that 〈Gω(ν ×w),ψT 〉ΓE =〈Gω(ν × ψ), wT 〉ΓE and hence it holds that

A(w,ψ) = A(ψ, w).

Therefore, it is clear that (5.4) implies (5.5).In order to establish (5.4), we define two sesquilinear forms in X0 ×X0,

a2(w,ψ) = (∇×w,∇×ψ)Ω + ω2(ǫw,ψ)Ω − 〈GYi (ν ×w),ψT 〉ΓE

b2(w,ψ) = −2ω2(ǫw,ψ)Ω + iω〈(GYω −

i

ωGY

i )(ν ×w),ψT 〉ΓE + iω〈GVω (ν ×w),ψT 〉ΓE ,

so that we haveA(w,ψ) = a2(w,ψ) + b2(w,ψ).

Here we note that by Lemma 5.3

ℜ(a2(w,w)) = ‖∇×w‖2L2(Ω) + ω2(ℜ(ǫ)w,w)Ω ≥ C‖w‖2H(curl,Ω)

for w ∈ X0, which leads to the coercivity of a2(·, ·) in X0. On the other hand,we claim that b2 : X0 → X0

′ defined by b2(w) = b2(w, ·) is compact, where X0′

is the set of bounded anti-linear functionals from X0 with the norm

‖l‖X0′ = sup

06=ψ∈X0

|l(ψ)|

‖ψ‖H(curl,Ω)for l ∈ X0

′.

Indeed, by the compactness results of the operators in Lemma 4.4, Lemma 5.4and Lemma 5.2, for a bounded sequence wn in X0, there exists a subsequencewnj such that wnj , (G

Yω − i

ωGYi )(ν ×wnj ) and G

Vω (ν ×wnj ) converge in their

corresponding spaces, (L2(Ω))3 or H−1/2N (ΓE). Setting ζnj ,nk

= wnj −wnk, by

linearity we see that

‖b2(ζnj ,nk)‖X0

′ ≤ C

(‖ζnj ,nk

‖L2(Ω) + ‖(GYω −

i

ωGY

i )(ν × ζnj ,nk)‖

H−1/2N (Div,ΓE)

+ ‖GVω (ν × ζnj ,nk

)‖H

−1/2N (Div,ΓE)

),

which implies that b2(wnj ) is Cauchy and so it converges.

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Now, for w ∈ X0 we have

‖w‖H(curl,Ω) ≤ C sup06=ψ∈X0

|a2(w,ψ)|

‖ψ‖H(curl,Ω)= C sup

06=ψ∈X0

|A(w,ψ)− b2(w,ψ)|

‖ψ‖H(curl,Ω)

≤ C

(sup

06=ψ∈X0

|A(w,ψ)|

‖ψ‖H(curl,Ω)+ ‖b2(w)‖X0

).

(5.7)Once we prove the uniqueness of solutions to the problem (5.3), the compactperturbation arguments (see e.g., [15]) applied to (5.7) leads to the inf-supcondition (5.4).

Thus, we are only left with showing that the problem (5.3) admits at mostone solution in order to establish the well-posedness of the problem (5.3). As-suming A(w,ψ) = 0 for all ψ ∈ X0, by the definition of X0, it also holds that

A(w,ψ +∇q) = 0 for all ψ ∈ X0 and q ∈ S, (5.8)

that is, A(w,φ) = 0 for all φ ∈ H0,ΓT (curl,Ω). Choosing w as a test functionand taking the imaginary parts of both sides shows that

−ω2(ℑ(ǫ)w,w)Ω + ωℜ(〈Gω(ν ×w),wT 〉ΓE ) = 0.

On the other hand, if the transverse component wT of the extension of w toΘ× (0,∞) as a radiating solution to (1.1) is written as

wT =

∞∑

n=1

AneiµnzY⊥

n +BneiγnzVn for z ≥ 0,

then

〈Gω(ν ×w),wT 〉ΓE =∞∑

n=1

−βnω

|An|2 +

−ω

γn|Bn|

2

and so

ℜ(〈Gω(ν ×w),wT 〉ΓE =

Nλ∑

n=1

−βnω

|An|2 +

Nµ∑

n=1

−ω

γn|Bn|

2 ≤ 0,

which implies that An = 0 for 1 ≤ n ≤ Nλ and Bn = 0 for 1 ≤ n ≤ Nµ sinceℑ(ǫ) ≥ 0. Thus the extended solution w to the problem (5.8) consists of onlyevanescent modes and hence it decays exponentially, from which it then followsthat the solution w is an eigenfunction to (1.1). However, by the assumptionthat ω2 is not an eigenvalue of the problem (1.1), w must be a trivial solution,which completes the proof of the inf-sup condition (5.4) for A(·, ·).

5.3. Main result

We present the main result in the following theorem.

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Theorem 5.5. Assume that ω2 is not an eigenvalue of the problem (3.3). Thenthere exists a unique solution E in H0,ΓT (curl,Ω) to the problem (3.3) satisfying

‖E‖H(curl,Ω) ≤ C‖J‖L2(Ω).

Proof. The proof immediately follows from the results of the preceding subsec-tions. Let p and w be the unique solutions to the problem (5.1) and (5.3),respectively, and define E = w +∇p in H0,ΓT (curl,Ω). By the definition of pand w, it is straightforward to show that

A(E ,ψ +∇q) = (iωJ,ψ +∇q)Ω for all ψ ∈ X0 and q ∈ S

and in addition (5.2) and Lemma 5.1 gives

‖E‖H(curl,Ω) ≤ ‖w‖H(curl,Ω) + ‖∇p‖H(curl,Ω) ≤ C‖J‖L2(Ω),

which completes the proof.

Remark 5.6. The analysis can be carried over to the case in which the right-

hand-side J is replaced by anti-linear functionals F ∈ (H0,ΓT (curl,Ω))′, which

leads to the inf-sup condition

‖u‖H(curl,Ω) ≤ C sup06=φ∈H0,ΓT

(curl,Ω)

|F(φ)|

‖φ‖H(curl,Ω)

= C sup06=φ∈H0,ΓT

(curl,Ω)

|A(u,φ)|

‖φ‖H(curl,Ω).

6. Acknowledgment

The author wishes to express the sincere gratitude to the referees for thevaluable comments and suggestions. This research of the author was supportedby Basic Science Research Program through the National Research Foundationof Korea (NRF) funded by the Ministry of Education, Science and Technology(2015R1D1A1A01057350).

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