International Scholarly Research Notices

International Scholarly Research Notices / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 983809 | 25 pages |

On Energy Conditions for Electromagnetic Diffraction by Apertures

Academic Editor: G. Ghirardi
Received24 Sep 2011
Accepted02 Nov 2011
Published06 Feb 2012


The diffraction of light is considered for a plane screen with an open bounded aperture. The corresponding solution behind the screen is given explicitly in terms of the Fourier transforms of the tangential components of the electric boundary field in the aperture. All components of the electric as well as the magnetic field vector are considered. We introduce solutions with global finite energy behind the screen and describe them in terms of two boundary potential functions. This new approach leads to a decoupling of the vectorial boundary equations in the aperture in the case of global finite energy. For the physically admissible solutions, that is, the solutions with local finite energy, we derive a characterisation in terms of the electric boundary fields.

1. Introduction

This paper deals with the classical diffraction problem for electromagnetic waves passing a bounded aperture in an ideally conducting plane screen. We treat the problem within the exact theory, that is, we consider the corresponding solutions of the time harmonic Maxwell equations that fulfil the correct boundary conditions on the screen.

The problem of diffraction of electromagnetic waves by an infinite slit has been treated by the Fourier method in the papers [1, 2]. In [1] especially representations of the solutions that fulfil a certain energy condition have been given in terms of distributional electric boundary fields satisfying special regularity properties. In [2] mapping properties of the corresponding boundary operators between Sobolev spaces have been studied. These Sobolev spaces have been chosen such that the corresponding diffraction solutions satisfy the correct physical energy condition.

While the slit problem treated in [1, 2] can be decoupled into two scalar problems by considering two kinds of polarisations of the electromagnetic field, in the case of a bounded aperture such a decoupling is not possible in general. However, for the latter case we derive a new kind of decoupling of the vectorial system which can be performed if and only if the condition of global finite energy in part (b) of Definition 2.5 is fulfilled, see Theorem 3.2.

Finally we study the condition of local finite energy which covers all physically admissible solutions. Here we give a characterisation of solutions with local finite energy in terms of a regularity property of the electric boundary fields, see Theorem 4.1. This is done in a self-contained way by using the Paley-Wiener theorem for distributions defined on the bounded aperture as well as a special contour integration method in the Fourier domain.

2. Electromagnetic Diffraction by an Aperture in a Plane Screen

We start with an informal physical description of the electromagnetic diffraction problem and fix some notations which will be used in the sequel. Then we will develop a more general mathematical frame with boundary distributions in Sobolev spaces in order to obtain diffraction solutions satisfying physical energy conditions.

Monochromatic light waves with a fixed wavenumber 𝑘>0 satisfy the first-order system of Maxwell-Helmholtz equations𝑖𝑘𝐸𝑥+×𝐵𝑥=0,𝑖𝑘𝐵𝑥+×𝐸𝑥=0.(2.1) In the whole paper we consider a real wavenumber 𝑘>0, although the results can be generalised to the case of a complex wavenumber 𝑘0 with Re𝑘0 and Im𝑘0. We assume that the electromagnetic field with components 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3,𝑒1𝑥,𝑥3𝑒2𝑥,𝑥3𝑒3𝑥,𝑥3=𝐸𝑥,𝑥3,𝑏1𝑥,𝑥3𝑏2𝑥,𝑥3𝑏3𝑥,𝑥3=𝐵𝑥,𝑥3,(2.2) consists of functions defined in the upper half-space𝑥==𝑥,𝑥33𝑥2,𝑥3.>0(2.3)

The diffraction problem is considered for an open bounded apertureΩ(𝑥,0)𝑥2(2.4) in the screen plane 𝑥3=0. In the sequel we will suppress the notation of the third component 0 for the points in the screen plane, and interpret Ω as well as the screen Ω𝑐=2Ω as subsets of 2. For describing the whole electromagnetic field in terms of its boundary values, for the moment we assume that these are functions, given for 𝑥2 by𝑒1,0𝑒(𝑥)2,0(𝑒𝑥)3,0(𝑥)=lim𝑥30𝐸𝑥,𝑥3,𝑏1,0𝑏(𝑥)2,0(𝑏𝑥)3,0(𝑥)=lim𝑥30𝐵𝑥,𝑥3.(2.5)

The screen Ω𝑐 is assumed to be an ideal conducting wall. This implies the physical boundary conditions𝑒1,0(𝑥)=𝑒2,0(𝑥)=0𝑥Ω𝑐.(2.6) In the general case the boundary fields (2.5) have to be replaced by appropriate distributions, and the limit 𝑥30 will be performed in the Fourier domain instead of the half-space .

For this purpose we write again 𝑥=(𝑥,𝑥3), with 𝑥=(𝑥1,𝑥2)2 and fixed 𝑥3>0, and assume that each field component 𝑢(,𝑥3) represents a tempered distribution in 𝒮(2) with Fourier transform𝜉̂𝑢1,𝜉2,𝑥3=12𝜋2𝑢𝑥1,𝑥2,𝑥3𝑒𝑖(𝜉1𝑥1+𝜉2𝑥2)𝑑𝑥1𝑑𝑥2,𝜉1,𝜉2.(2.7) Then we obtain from the first-order Maxwell-Helmholtz equations the following Fourier transformed Maxwell-Helmholtz system: for all 𝑥30 and fixed 𝜉1,𝜉2 we have𝑖𝑘̂𝑒1̂𝑒2̂𝑒3+𝑖𝜉2̂𝑏3𝑑𝑑𝑥3̂𝑏2𝑑𝑑𝑥3̂𝑏1𝑖𝜉1̂𝑏3𝑖𝜉1̂𝑏2𝑖𝜉2̂𝑏1=000,̂𝑏𝑖𝑘1̂𝑏2̂𝑏3+𝑖𝜉2̂𝑒3𝑑𝑑𝑥3̂𝑒2𝑑𝑑𝑥3̂𝑒1𝑖𝜉1̂𝑒3𝑖𝜉1̂𝑒2𝑖𝜉2̂𝑒1=000.(2.8) Here we have replaced the partial derivative with respect to 𝑥3 by the ordinary derivative 𝑑/𝑑𝑥3.

We define1𝐶=𝑘𝜕2𝜕𝑥1𝜕𝑥2𝑘2+𝜕2𝜕𝑥21𝑘2+𝜕2𝜕𝑥22𝜕2𝜕𝑥1𝜕𝑥2.(2.9) With the two-dimensional Laplace operator Δ we have𝐶2𝑘=2+Δ1001.(2.10) For the action of 𝐶 in the Fourier domain we obtain multiplication by the matrix1𝐶(𝜉)=𝑘𝜉1𝜉2𝑘2𝜉21𝑘2𝜉22𝜉1𝜉2𝜉,𝜉=1,𝜉22.(2.11) We replace (2.8) with the ordinary differential equations𝑑𝑑𝑥3̂𝑒1̂𝑒2𝐶̂𝑏=𝑖1̂𝑏2,𝑑𝑑𝑥3̂𝑏1̂𝑏2𝐶=𝑖̂𝑒1̂𝑒2,(2.12) and the two algebraic conditionŝ𝑒3=1𝑘𝜉2̂𝑏1𝜉1̂𝑏2,̂𝑏3=1𝑘𝜉1̂𝑒2𝜉2̂𝑒1.(2.13) For all fixed 𝜉2 we supplement the system of differential equations (2.12) by the initial conditionŝ𝑒𝑗,0(𝜉)=̂𝑒𝑗̂𝑏(𝜉,0),𝑗,0̂𝑏(𝜉)=𝑗(𝜉,0),𝑗=1,2,(2.14) and put 𝑚(𝜉)=𝑘2||𝜉||2,||𝜉||𝑖𝑘,||𝜉||2𝑘2,||𝜉||>𝑘.(2.15) Then the general solution of the homogeneous linear system (2.12) iŝ𝑒1𝜉,𝑥3̂𝑒2𝜉,𝑥3̂𝑏1𝜉,𝑥3̂𝑏2𝜉,𝑥3=cos𝑚(𝜉)𝑥3𝐸𝑖sin𝑚(𝜉)𝑥3𝑖𝑚(𝜉)𝐶(𝜉)sin𝑚(𝜉)𝑥3𝐶𝑚𝑚(𝜉)(𝜉)cos(𝜉)𝑥3𝐸̂𝑒1,0(𝜉)̂𝑒2,0(̂𝑏𝜉)1,0̂𝑏(𝜉)2,0(𝜉)(2.16) with the 2×2 unit matrix 𝐸.

But the terms cos(𝑚(𝜉)𝑥3) and sin(𝑚(𝜉)𝑥3) are exponentially increasing for fixed 𝑥3>0 and |𝜉|. For avoiding that the Fourier transformed fields are also exponentially increasing we have to require the following algebraic conditions for |𝜉|>𝑘:̂𝑏𝑚1,0̂𝑏2,0=𝐶̂𝑒1,0̂𝑒2,0.(2.17) By using (2.17), for 𝑥30 and |𝜉|>𝑘 we can replace (2.13) witĥ𝑒3,01=𝑚𝜉1̂𝑒1,0+𝜉2̂𝑒2,0,̂𝑏3,01=𝑘𝜉2̂𝑒1,0𝜉1̂𝑒2,0.(2.18) From the general solution and (2.17), (2.18) we obtain the following decay conditions for 𝑥3>0:̂𝑒𝑗𝜉,𝑥3=𝑒𝑖𝑥3𝑚(𝜉)̂𝑒𝑗,0̂𝑏(𝜉),𝑗𝜉,𝑥3=𝑒𝑖𝑥3𝑚(𝜉)̂𝑏𝑗,0(𝜉),𝑗=1,2,3.(2.19) With 𝑥=(𝑥1,𝑥2)2, 𝑥=(𝑥,𝑥3)3 and 𝑥3>0 there holds the important and well-known Sommerfeld-Weyl integral representation𝐹𝑘𝑥,𝑥3𝑒=𝑖𝑘|𝑥|||𝑥||=𝑖2𝜋2𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑚(𝜉)𝑑𝜉.(2.20) The left-hand side in (2.20) is the singular solution of the three-dimensional Helmholtz equation (Δ+𝑘2)𝐹𝑘=4𝜋𝛿. For this reason it is natural to require the algebraic conditions (2.17), (2.18) also in the case |𝜉|<𝑘, such that (2.19) is generally valid for 𝑥3>0 and 𝜉2, |𝜉|𝑘.

Distributions 𝑢𝒟(2) with compact support in the screen plane are tempered, and it follows from the Paley-Wiener theorem that ̂𝑢 is a smooth function which has polynomial growth on 2. This is used in the following theorem, which results if we regard (2.17), (2.18), and (2.19) and apply the Fourier inversion formula for 𝑥3>0 to each component ̂𝑒𝑗(,𝑥3) and ̂𝑏𝑗(,𝑥3).

Theorem 2.1. Let there be given 𝑒1,0,𝑒2,0𝒮(2) with support in the bounded region Ω. Then the following functions 𝑒𝑗,𝑏𝑗 constitute a 𝐶-solution of the Maxwell-Helmholtz system (2.1) in the upper half-space , 𝑗=1,2,3; 𝑒1𝑥,𝑥3=12𝜋2̂𝑒1,0(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑒𝑑𝜉,2𝑥,𝑥3=12𝜋2̂𝑒2,0(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑏𝑑𝜉,1𝑥,𝑥31=2𝜋𝑘2𝜉1𝜉2̂𝑒1,0𝑘(𝜉)+2𝜉21̂𝑒2,0(𝜉)𝑒𝑚(𝜉)𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑏𝑑𝜉,2𝑥,𝑥3=12𝜋𝑘2𝑘2𝜉22̂𝑒1,0(𝜉)+𝜉1𝜉2̂𝑒2,0(𝜉)𝑒𝑚(𝜉)𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑒𝑑𝜉,3𝑥,𝑥31=2𝜋2𝜉1̂𝑒1,0(𝜉)+𝜉2̂𝑒2,0(𝜉)𝑚𝑒(𝜉)𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑏𝑑𝜉,3𝑥,𝑥31=2𝜋𝑘2𝜉2̂𝑒1,0(𝜉)𝜉1̂𝑒2,0𝑒(𝜉)𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉.(2.21)

Proof. The calculation of the partial derivatives of 𝑒𝑗 and 𝑏𝑗 can be interchanged with integration. This can be used to check the Maxwell-Helmholtz equations independent from the previous representations of the Fourier-transforms ̂𝑒𝑗 and ̂𝑏𝑗 in terms of ̂𝑒1,0 and ̂𝑒2,0.

Definition 2.2. The electromagnetic field in the half-space 𝑥3>0 behind the screen is completely determined by the electric boundary components 𝑒1,0, 𝑒2,0. We call 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, the half-space solution determined by the boundary distributions 𝑒1,0,𝑒2,0𝒮(2) with compact support in Ω.

Remark 2.3. Assume that 𝑒1,0,𝑒2,0 are smooth functions with compact support in Ω and that 𝐸, 𝐵 is the corresponding electromagnetic field with the components given in Theorem 2.1. Then we obtain from Fourier’s inversion formula that for all 𝑥2lim𝑥30𝑒𝑗𝑥,𝑥3=𝑒𝑗,0(𝑥),𝑗=1,2.(2.22)

Remark 2.4. Assume again that 𝑒1,0,𝑒2,0 are smooth functions with compact support in Ω. Then we obtain from Theorem 2.1 and the Sommerfeld-Weyl integral (2.20) for all 𝑥2 and all 𝑥3>0 that 𝑏1𝑥,𝑥3𝑖=𝜕2𝜋𝑘2𝜕𝑥1𝜕𝑥2Ω𝑒1,0(𝑦)𝐹𝑘𝑥𝑦,𝑥3+𝑖𝑑𝑦𝑘2𝜋𝑘2+𝜕2𝜕𝑥21Ω𝑒2,0(𝑦)𝐹𝑘𝑥𝑦,𝑥3𝑏𝑑𝑦,2𝑥,𝑥3𝑖=𝑘2𝜋𝑘2+𝜕2𝜕𝑥22Ω𝑒1,0(𝑦)𝐹𝑘𝑥𝑦,𝑥3+𝑖𝑑𝑦𝜕2𝜋𝑘2𝜕𝑥1𝜕𝑥2Ω𝑒2,0(𝑦)𝐹𝑘𝑥𝑦,𝑥3𝑑𝑦.(2.23) Equations (2.23) involve the boundary conditions (2.6) for the electric field components on the ideal conducting plane screen Ω𝑐. In order to obtain a coupled system of boundary integro-differential equations we pass for 𝑥2 to the limit 𝑥30 and obtain 𝑏1,0𝑖(𝑥)=𝜕2𝜋𝑘2𝜕𝑥1𝜕𝑥2Ω𝑒1,0𝑒(𝑦)𝑖𝑘|𝑥𝑦|||||+𝑖𝑥𝑦𝑑𝑦𝑘2𝜋𝑘2+𝜕2𝜕𝑥21Ω𝑒2,0𝑒(𝑦)𝑖𝑘|𝑥𝑦|||||𝑏𝑥𝑦𝑑𝑦,2,0𝑖(𝑥)=𝑘2𝜋𝑘2+𝜕2𝜕𝑥22Ω𝑒1,0𝑒(𝑦)𝑖𝑘|𝑥𝑦|||||+𝑖𝑥𝑦𝑑𝑦𝜕2𝜋𝑘2𝜕𝑥1𝜕𝑥2Ω𝑒2,0(𝑒𝑦)𝑖𝑘|𝑥𝑦|||||𝑥𝑦𝑑𝑦.(2.24) In general the electric boundary fields 𝑒1,0 and 𝑒2,0 are unknown distributions with compact support in Ω, whereas 𝑏1,0 and 𝑏2,0 are given distributions in the aperture Ω. In order to select physical admissible solutions of the diffraction problem we need some conditions for its electromagnetic energy content, especially in local volume elements 𝐺. Recall that =2×+.

Definition 2.5. Let 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, be the half-space solution determined by the boundary distributions 𝑒1,0,𝑒2,0𝒮(2) with compact support in Ω.(a)The solution is called physical admissible if and only if it satisfies the local energy condition123𝑗=1𝐺||𝑒𝑗𝑥||2+||𝑏𝑗𝑥||2𝑑𝑥<(2.25)for every bounded domain 𝐺.(b)The solution satisfies the stronger global energy condition if and only if 123𝑗=1||𝑒𝑗𝑥||2+||𝑏𝑗𝑥||2𝑑𝑥<(2.26)for some >0, and therewith for all >0, with the layer={(𝑥1,𝑥2,𝑥3)|𝑥3<}.
In the following two sections we determine the solutions with global as well as those with local finite energy in terms of an appropriate functional analytical setting for the electric boundary fields 𝑒1,0 and 𝑒2,0.

3. The Global Energy Condition

Throughout the rest of this paper let the open aperture Ω=𝜅𝑗=1Ω𝑗2 be a finite union of nonempty bounded Lipschitz domains Ω𝑗 in the screen plane, such that the compact sets Ω𝑗 are pairwise disjoint.

By 𝐻𝑠(2), 𝑠, we denote the Sobolev space of tempered distributions , for which the Fourier transform is locally integrable with𝐻𝑠(2)=2||||||(𝜉)2||𝜉||1+2𝑠𝑑𝜉1/2<(3.1) (cf. [3], Chapter  8.8). 𝐻𝑠(2) is the norm on the Banach space 𝐻𝑠(2).

For 𝑠 the Sobolev space 𝐻𝑠(Ω) is given by𝐻𝑠(Ω)=𝐻𝑠2suppΩ.(3.2) Here supp denotes the support of the distribution in the compact set Ω. The space 𝐻𝑠(Ω) is equipped with the norm of 𝐻𝑠(2), which makes it into a Banach space (cf. [4, Chapter  3]). The Lipschitz property of 𝜕Ω guarantees that 𝐻𝑠(Ω) is the closure of 𝒟(Ω) in 𝐻𝑠(2). For a more general result see [4, Theorem  3.29].

Theorem 3.1. Let 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, be the half-space solution determined by the boundary distributions 𝑒1,0,𝑒2,0𝒮(2) with compact support in Ω. Then the diffraction solution has global finite energy if and only if𝑒1,0,𝑒2,0,𝑒2,0𝑒1,0𝐻1/2(Ω)(3.3) and ̂𝑒1,0(𝜉)𝜉1+̂𝑒2,0(𝜉)𝜉2=0 for all 𝜉=(𝜉1,𝜉2)2 with |𝜉|=𝑘.

Proof. In order to perform the energy evaluation with Parseval’s theorem we define with the Fourier transformed electromagnetic boundary fields ̂𝑒𝑗,0,̂𝑏𝑗,0, 𝑗=1,2,3, the quantity 𝑊01(𝜉)=23𝑗=1||̂𝑒𝑗,0||(𝜉)2+||̂𝑏𝑗,0||(𝜉)2.(3.4) It follows from a lengthy calculation with |𝑧|2=𝑧𝑧 for the values 𝑧 of the Fourier transformed boundary fields as well as from the algebraic relations (2.17), (2.18), which are valid for |𝜉|𝑘, that 𝑊0||(𝜉)=̂𝑒1,0||(𝜉)2+||̂𝑒2,0||(𝜉)2+||̂𝑒3,0||(𝜉)2,||𝜉||||<𝑘,̂𝑒3,0||(𝜉)2+||̂𝑏3,0||(𝜉)2,||𝜉||>𝑘.(3.5)
With 𝑉1={𝜉2||𝜉|<𝑘}, 𝑉2={𝜉2||𝜉|>𝑘} we conclude from Parseval’s theorem, (2.19) and the definition (2.15) of 𝑚(𝜉) for >01()=23𝑗=102||𝑒𝑗𝑥,𝑥3||2+||𝑏𝑗𝑥,𝑥3||2𝑑𝑥𝑑𝑥3=𝑉1𝑊0(𝜉)𝑑𝜉+𝑉21𝑒2|𝜉|2𝑘22||𝜉||2𝑘2𝑊0(𝜉)𝑑𝜉.(3.6) Using for all 𝜉2 with |𝜉|𝑘 the estimate 12||̂𝑒1,0||(𝜉)2+||̂𝑒2,0||(𝜉)2+||̂𝑒3,0||(𝜉)2+||̂𝑏3,0||(𝜉)2𝑊0||(𝜉)̂𝑒1,0||(𝜉)2+||̂𝑒2,0||(𝜉)2+||̂𝑒3,0||(𝜉)2+||̂𝑏3,0||(𝜉)2,(3.7) and observing that we have uniformly in 𝑟>𝑘 for appropriate constants 𝛼,𝛽>0𝛼1+𝑟21𝑒2𝑟2𝑘22𝑟2𝑘2𝛽1+𝑟2,(3.8) we conclude from (3.6) that () is finite if and only if 𝑒1,0,𝑒2,0,𝑒3,0,𝑏3,0=𝑖𝑘𝑒2,0𝑒1,0𝐻1/22.(3.9) Note that with 𝑒1,0 and 𝑒2,0 also (𝑒2,0,𝑒1,0)𝑇 has support in Ω.
Assume first that ()< and that ̂𝑒1,0(𝜉)𝜉1+̂𝑒2,0(𝜉)𝜉20 for a certain 𝜉=(𝜉1,𝜉2)2 with |𝜉|=𝑘. Since 𝑒1,0,𝑒2,0 have compact support, we conclude from the Paley-Wiener theorem that ̂𝑒1,0,̂𝑒2,02 are smooth and especially continuous. Thus |̂𝑒1,0(𝜉)𝜉1+̂𝑒2,0(𝜉)𝜉2|2𝛿>0 in a bounded domain 𝜉1,𝜉2=𝑟(cos𝜑,sin𝜑)with𝜑1<𝜑<𝜑2,𝑘<𝑟<𝑘+𝜀,(3.10) and hence |̂𝑒3,0|2 because of (2.18) is not integrable. Due to (3.1) this violates the necessary condition 𝑒3,0𝐻1/2(2) in (3.9).
For showing the other direction of the equivalence stated in the theorem we assume that 𝑒1,0,𝑒2,0,𝑒2,0𝑒1,0𝐻1/2(Ω),(3.11)̂𝑒1,0𝜉1,𝜉2𝜉1+̂𝑒2,0𝜉1,𝜉2𝜉2=0,(3.12) for all 𝜉1,𝜉2 with 𝜉21+𝜉22=𝑘2. In order to prove (3.9) it remains to show that 𝑒3,0𝐻1/2(2). Since 𝑒1,0, 𝑒2,0 have compact support in Ω, we obtain from the Paley-Wiener theorem that the Fourier transforms ̂𝑒1,0, ̂𝑒2,0 can be continuated to entire functions. We also denote these entire functions by ̂𝑒1,0,̂𝑒2,0. We define the entire function 𝑓 by 𝑓𝑧1,𝑧2=̂𝑒1,0𝑧1,𝑧2𝑧1+̂𝑒2,0𝑧1,𝑧2𝑧2.(3.13) Using (3.12) and (2.18), it follows for example from Theorem A.1 in the appendix that |̂𝑒3,0|2 is a locally integrable function.
From the first equation in (2.18) and the Cauchy-Schwarz inequality we obtain the estimate ||̂𝑒3,0||(𝜉)2||𝜉||2||𝜉||2𝑘2||̂𝑒1,0||(𝜉)2+||̂𝑒2,0||(𝜉)2,||𝜉||>𝑘.(3.14) Now 𝑒3,0𝐻1/2(2) follows from the assumption 𝑒1,0,𝑒2,0𝐻1/2(Ω), because |𝑒3,0|2 is locally integrable and |𝜉|2/(|𝜉|2𝑘2) is bounded for sufficiently large |𝜉|.

Next we present a characterisation of the global finite energy solutions in terms of two boundary potential functions. In the following we use the vectorial differential operator (2.9) satisfying (2.10).

Theorem 3.2. Boundary potential functions(a)Given are two functions 𝑢1,𝑢2𝐻1/2(Ω), in the following called boundary potential functions, satisfying the regularity condition 𝑢2𝑢1𝐻1/2(Ω).(3.15)Define 𝑒1,0𝑒2,0𝑢=𝐶1𝑢2.(3.16)Then the corresponding electromagnetic field 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, determined by 𝑒1,0 and 𝑒2,0 according to Theorem 2.1, has global finite energy. (b)Let 𝑒1,0, 𝑒2,0 be given such that the half-space solution in Theorem 2.1 has global finite energy. If the set 2Ω is connected, then the boundary fields 𝑒1,0 and 𝑒2,0 can be represented by two boundary potential functions 𝑢1,𝑢2𝐻1/2(Ω) satisfying (3.15) and (3.16) in part (a).(c)From the assumptions of part (a), or alternatively part (b), one obtains in the distributional sense 𝑏1,0𝑖(𝑥)=𝑘2𝜋2+ΔΩ𝑢1𝑒(𝑦)𝑖𝑘|𝑥𝑦|||||𝑏𝑥𝑦𝑑𝑦,2,0𝑖(𝑥)=𝑘2𝜋2+ΔΩ𝑢2𝑒(𝑦)𝑖𝑘|𝑥𝑦|||||𝑥𝑦𝑑𝑦(3.17)on the whole screen plane 2 with 𝑏1,0,𝑏2,0,𝑏2,0𝑏1,0𝐻1/22.(3.18)

Proof. Part (a) follows in the Fourier domain by representing ̂𝑒1,0, ̂𝑒2,0, ̂𝑒3,0 and ̂𝑏3,0 in terms of ̂𝑢1 and ̂𝑢2 as ̂𝑒1,0=𝑘̂𝑢2𝜉1𝑘𝜉2̂𝑢1𝜉1̂𝑢2,̂𝑒2,0=+𝑘̂𝑢1𝜉2𝑘𝜉2̂𝑢1𝜉1̂𝑢2,̂𝑒3,0𝑚=𝑘𝜉2̂𝑢1𝜉1̂𝑢2,̂𝑏3,0=𝜉1̂𝑢1+𝜉2̂𝑢2.(3.19) These equations and the assumptions imply that 𝑒1,0,𝑒2,0,𝑏3,0𝐻1/2(2), and regarding |𝑚(𝜉)|𝛼1+|𝜉|2 uniformly in 𝜉2 for a certain constant 𝛼>0 also 𝑒3,0𝐻1/2(2). Since 𝑒1,0, 𝑒2,0 have compact support in Ω like 𝑢1, 𝑢2, the proof of part (a) follows from the fact that condition (3.9) is equivalent to ()<.
For proving part (b) we assume that the diffraction solution corresponding to 𝑒1,0,𝑒2,0𝐻1/2(Ω) has global finite energy. We conclude from Theorem 3.1 and Theorem A.1 in the appendix that we obtain entire functions 𝑣1,𝑣22 by 𝑣1𝜉1,𝜉21=+𝑘̂𝑒2,0𝜉1,𝜉2𝜉2𝑘̂𝑒1,0(𝜉)𝜉1+̂𝑒2,0(𝜉)𝜉2𝜉21+𝜉22𝑘2,𝑣2𝜉1,𝜉21=𝑘̂𝑒1,0𝜉1,𝜉2+𝜉1𝑘̂𝑒1,0(𝜉)𝜉1+̂𝑒2,0(𝜉)𝜉2𝜉21+𝜉22𝑘2.(3.20) From (3.20) we get for 𝑗=1,2 that 2||𝑣𝑗(||𝜉)2||𝜉||1+21/2𝑑𝜉<.(3.21) Thus for 𝑗=1,2 the inverse Fourier transform 𝑢𝑗 of 𝑣𝑗 lies in 𝐻1/2(2).
Now we show that 𝑢1 and 𝑢2 have their support in Ω and hence, by reason of (3.21), 𝑢1,𝑢2𝐻1/2(Ω). To this aim we choose some 𝑅>0 such that Ω𝐵𝑅=𝑥2.|𝑥|<𝑅(3.22) Since 𝑒1,0,𝑒2,0 are supported in Ω, we have supp𝑒1,0,supp𝑒2,0𝐵𝑅. Therefore we obtain from the Paley-Wiener theorem and (3.20) that 𝑢1 and 𝑢2 are also supported in 𝐵𝑅, because 𝑣1, 𝑣2, as ̂𝑒1,0, ̂𝑒2,0, are of exponential type not larger than 𝑅.
Equations (3.20) can be resolved with respect to ̂𝑒1,0, ̂𝑒2,0. This gives (3.16). Together with (2.10) it follows that in the complement of Ω there holds 𝑘2𝑢+Δ1𝑢2=𝐶2𝑢1𝑢2𝑒=𝐶1,0𝑒2,0=00.(3.23) Now we make use of the fact that the complement of Ω is connected. Since supp𝑢𝑗𝐵𝑅 for 𝑗=1,2, Holmgren’s unique continuation principle, applied to the two scalar Helmholtz equations in (3.23), implies that 𝑢1, 𝑢2 have their support in Ω.
Finally, since 𝑣𝑗=̂𝑢𝑗 for 𝑗=1,2, from (3.20) we obtain for𝜉2𝜉1̂𝑢2+𝜉2̂𝑢1𝜉=𝑘1̂𝑒1,0+𝜉2̂𝑒2,0𝜉21+𝜉22𝑘2.(3.24) The asymptotic behaviour of this expression for |𝜉| shows the validity of the regularity condition (3.15).
The convolution integrals in part (c) can be rewritten in the Fourier domain by using the Sommerfeld-Weyl integral representation (2.20) in the limit 𝑥30. The resulting relations in the Fourier domain follow from the representations of the components 𝑏1 and 𝑏2 in Theorem 2.1.
The validity of (3.18) is a consequence of the algebraic relations between ̂𝑢1, ̂𝑢2, ̂𝑏1,0, and ̂𝑏2,0.

Remark 3.3. Consider the Sobolev spaces 𝐻𝑠(Ω) given for 𝑠 by 𝐻𝑠(Ω)=𝐹|Ω𝐹𝐻𝑠2,(3.25) with the restriction 𝐹|Ω of the tempered distribution 𝐹𝒮(2) to the subspace 𝒮(Ω) of the Schwartz space 𝒮(2), where 𝒮(Ω) is the closure of the set 𝒟(Ω) in 𝒮(2) with respect to the topology of 𝒮(2) [5], §1 in Section 5.
𝐻𝑠(Ω) is a Banach space with respect to the norm 𝐻𝑠(Ω) given by𝑓𝐻𝑠(Ω)=inf𝐹𝐻𝑠(2)𝐹𝐻𝑠2.isacontinuationof𝑓(3.26) The magnetic boundary fields in part (c) of Theorem 3.2 corresponding to global finite energy solutions may also be reinterpreted as distributions restricted to Ω. In this case we obtain 𝑏1,0,𝑏2,0,𝑏2,0𝑏1,0𝐻1/2(Ω).(3.27) In general the conditions (3.27) are weaker than the conditions (3.18) in Theorem 3.2, and we assume that they are fulfilled for diffraction solutions with local finite energy.
The conditions (3.27), where 𝑏1,0 and 𝑏2,0 are considered only in the aperture Ω, reflects the physical fact that 𝑏1,0 and 𝑏2,0 are only prescribed in Ω. Namely, 𝑏1,0 and 𝑏2,0 are the tangential magnetic components of the incoming electromagnetic wave in the aperture Ω.

4. The Local Energy Condition

In this section we derive the following characterisation for the diffraction solutions of local finite energy.

Theorem 4.1. Let be 𝑒1,0,𝑒2,0𝒮(2) and supp𝑒1,0,supp𝑒2,0Ω. Let 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, be defined as in Theorem 2.1. Then the diffraction solution 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, has local finite energy if and only if 𝑒1,0,𝑒2,0,𝑒2,0𝑒1,0𝐻1/2(Ω).(4.1)

For proving Theorem 4.1 firstly we formulate the subsequent Lemma 4.2. Then, using this lemma, we give the proof of the theorem. Afterwards we prove the lemma.

In the sequel we will make use of the following notations. For 𝑟>0 we define the open ball 𝐵𝑟 by𝐵𝑟=𝑥2.|𝑥|<𝑟(4.2) For 𝑟>0 and >0 the open cylinder 𝑍𝑟, is defined by𝑍𝑟,𝑥==𝑥,𝑥33𝑥𝐵𝑟,0<𝑥3.<(4.3)

Lemma 4.2. Let be 𝑒1,0,𝑒2,0𝒮(2) and supp𝑒1,0,supp𝑒2,0Ω. Let 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, be the diffraction solution given in Theorem 2.1. Let be 𝑅>0 with Ω𝐵𝑅 and let be 𝑅>𝑅 and 𝐻>0. Then one has the following equivalences and implications.
(a)𝑒𝑗𝐿2(𝑍𝑅,𝐻)𝑒𝑗𝐿2(2×(0,𝐻))for𝑗{1,2}. (b)𝑏3𝐿2(𝑍𝑅,𝐻)𝑏3𝐿2(2×(0,𝐻)). (c)𝑒𝑗𝐿2(2×(0,𝐻))𝑒𝑗,0𝐻1/2(Ω)for𝑗{1,2}. (d)𝑏3𝐿2(2×(0,𝐻))𝑏3,0𝐻1/2(Ω). (e)𝑒1,0,𝑒2,0𝐻1/2(Ω)𝑒3𝐿2(𝑍𝑅,𝐻). (f)𝑒1,0,𝑒2,0,𝑒2,0𝑒1,0𝐻1/2(Ω)𝑏1,𝑏2𝐿2(𝑍𝑅,𝐻).

Proof of Theorem 4.1. Let there be given 𝑒1,0,𝑒2,0𝒮(2) with support in Ωand let 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, be defined as in Theorem 2.1.
Firstly, we assume that the diffraction solution 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, has local finite energy and show the validity of (4.1). Since the diffraction solution has local finite energy especially we have 𝑒1,𝑒2,𝑏3𝐿2(𝑍𝑅,𝐻). From the parts (a) and (b) of Lemma 4.2 we find that 𝑒1,𝑒2,𝑏3𝐿2(2×(0,𝐻)). From the parts (c) and (d) of the lemma we thus obtain 𝑒1,0,𝑒2,0,𝑏3,0𝐻1/2(Ω). Now the validity of (4.1) follows from𝑏3,0=𝑖𝑘𝑒2,0𝑒1,0(4.4) (cf. (3.9)).
For proving the other direction, we assume that relation (4.1) is valid. Because of (4.4) from the parts (c) and (d) of Lemma 4.2 it follows that 𝑒1,𝑒2,𝑏3𝐿2(2×(0,𝐻)). Regarding the other three field components, from the parts (e) and (f) of the lemma we obtain 𝑒3,𝑏1,𝑏2𝐿2(𝑍𝑅,𝐻). Since 𝑅>𝑅 and 𝐻>0 can be chosen arbitrarily large, the diffraction solution 𝑒𝑗,𝑏𝑗, 𝑗=1,2,3, has local finite energy.

We have yet to prove Lemma 4.2.

Proof of Lemma 4.2. Let the assumptions of Lemma 4.2 be fulfilled.
Proof of Part (a)
Let be 𝑗{1,2}. Obviously we have only to show the validity of the implication𝑒𝑗𝐿2𝑍𝑅,𝐻𝑒𝑗𝐿22.×(0,𝐻)(4.5)
We set 𝜀=(𝑅𝑅)/2 and 𝑅=𝑅+𝜀. For 𝜑[0,2𝜋) we consider the rotation matrix 𝐴𝜑=cos𝜑sin𝜑sin𝜑cos𝜑(4.6) and define the function 𝑒𝑗(𝜑)by 𝑒𝑗(𝜑)(𝑥,𝑥3)=𝑒𝑗(𝐴𝜑𝑥,𝑥3)for 𝑥2 and 𝑥3>0. We show that 𝑒𝑗(𝜑)𝐿2𝑅[,××(0,𝐻)for𝜑0,2𝜋),(4.7) regardless of whether 𝑒𝑗𝐿2(𝑍𝑅,𝐻) or not.
From (4.7) we have 𝑒𝑗𝐿2𝐴𝜑𝑅[,××(0,𝐻)for𝜑0,2𝜋),(4.8) where 𝐴𝜑((𝑅,)×) is the image of the set (𝑅,)×under the linear mapping 𝐴𝜑, that is, the set (𝑅,)×rotated by the angle 𝜑.
Since 𝑅<𝑅, by considering a sufficient number of angles 𝜑, from (4.8) we find that there is a polygon 𝑃𝐵𝑅 with 𝑒𝑗𝐿2((2𝑃)×(0,𝐻)). Thus 𝑒𝑗𝐿22×(0,𝐻)𝑍𝑅,𝐻,(4.9) and therefore (4.5) holds true.
It remains to prove (4.7). To this end we use a contour integration technique which one of the authors had already used earlier in the treatment of diffraction problems [6, 7].
From the representation of 𝑒𝑗 given in Theorem 2.1 we conclude that 𝑒𝑗(𝜑)𝑥1,𝑥2,𝑥3=12𝜋𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2,𝑥1,𝑥2,𝑥3,(4.10) where the function 𝛾𝜑 is given by 𝛾𝜑(𝜉)=̂𝑒𝑗,0𝐴𝜑𝜉.(4.11) By assumption, it holds that supp𝑒𝑗,0Ω𝐵𝑅. Thus, from the Paley-Wiener theorem and the fact that 𝐴𝜑 is an orthogonal matrix it follows that ||𝛾𝜑||||𝜉||(𝜉)𝑐1+2𝑁/2𝑒𝑅|Im𝜉|,𝜉2,(4.12) for some constants 𝑐>0 and 𝑁0.
We split the inner integral in (4.10) into one integral over the interval (,0) and one over (0,). Firstly we consider the integral over (,0).
For 𝑟>0 we define the curve 𝜂𝑟 by 𝜂𝑟=𝑟𝑒𝑖𝛼𝜋2.𝛼𝜋(4.13) Let Γ𝑟 be the closed contour consisting of the parts {𝑖𝑡0𝑡<𝑟}, 𝜂𝑟 and (𝑟,0). By Cauchy’s theorem we have Γ𝑟𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0,(4.14) here for nonreal values of 𝜉1 the function 𝑚 is defined by analytic continuation of the function 𝜉1𝑚(𝜉1,𝜉2), where 𝜉1<0 and |𝜉|𝑘, and 𝑥,𝜉=𝑥1𝜉1+𝑥2𝜉2.
Because of (4.12), for 𝑥1>𝑅 it holds that lim𝑟𝜂𝑟𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0.(4.15) From the last two equations we obtain 0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0𝑖𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1,𝑥1>𝑅.(4.16)
Now we treat the integral over (0,). Let be the principal branch of the square root function (branch cut (,0)). For 𝑧<0 we set 𝑧=𝑖|𝑧|,thatis, is thought to be defined on the negative real half-axis by continuation from the lower complex half-plane. In what follows, the function 𝑚 is defined by 𝑚(𝜉)=𝑘2𝜉21𝜉22. Regarding the integration variable 𝜉2 we distinguish the two cases |𝜉2|<𝑘 and |𝜉2|>𝑘.
Firstly we assume that |𝜉2|<𝑘. We have 0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=𝑘2𝜉220𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1+𝑘2𝜉22𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1+2𝑖𝑘2𝜉220𝛾𝜑𝑥(𝜉)sin3𝑚𝑒(𝜉)𝑖𝑥,𝜉𝑑𝜉1.(4.17) By contour integration analogous as in the derivation of (4.16), applied to the next-to-last integral, eventually we get 0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0𝑖𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1+2𝑖𝑘2𝜉220𝛾𝜑𝑥(𝜉)sin3𝑚𝑒(𝜉)𝑖𝑥,𝜉𝑑𝜉1||𝜉for2||<𝑘,𝑥1>𝑅.(4.18)
In the case |𝜉2|>𝑘 it holds that 𝑚(𝜉)=𝑚(𝜉)for 𝜉1>0, and contour integration yields 0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1=0𝑖𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1||𝜉for2||>𝑘,𝑥1>𝑅.(4.19)
In the following, for 𝜉1or 𝜉1{𝑖𝑡|0<𝑡<} we set 𝑚(𝜉)=𝑘2𝜉21𝜉22, where the square root is chosen in the way that Re𝑚(𝜉)0and Im𝑚(𝜉)0. This definition is in accordance with (2.15). In this notation, from (4.18) and (4.19) we obtain 0𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2=𝑘𝑘0𝑖𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2+2𝑖𝑘𝑘𝑘2𝜉220𝛾𝜑(𝑥𝜉)sin3𝑒𝑚(𝜉)𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2+[]𝑘,𝑘𝑖𝜉22𝑘20𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2+[]𝑘,𝑘𝑖𝑖𝜉22𝑘2𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2for𝑥1>𝑅.(4.20) Together with (4.16) we thus find 𝛾𝜑(𝜉)𝑒𝑖𝑥3𝑚(𝜉)+𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2=2𝑖𝑘𝑘𝑘2𝜉220𝛾𝜑𝑥(𝜉)sin3𝑒𝑚(𝜉)𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉22𝑖𝑘𝑘0𝑖𝛾𝜑𝑥(𝜉)sin3𝑒𝑚(𝜉)𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉22𝑖[]𝑘,𝑘𝑖𝑖𝜉22𝑘2𝛾𝜑𝑥(𝜉)sin3𝑒𝑚(𝜉)𝑖𝑥,𝜉𝑑𝜉1𝑑𝜉2for𝑥1>𝑅.(4.21)
Now we show that each of the three addends in the right-hand side of (4.21), considered as a function of 𝑥=(𝑥,𝑥3), lies in 𝐿2((𝑅,)××(0,𝐻)). Then, because of the representation (4.10), the stated relation (4.7) is proved.
We begin with the first addend. Let 𝐴={𝜉2||𝜉|𝑘,𝜉10}, let 𝜒𝐴 be the characteristic function of the set 𝐴 and let ̌ denote the inverse Fourier transform. There is a constant 𝑐>0, which does not depend on 𝑥3(0,𝐻), such that 𝜒𝐴𝛾𝜑𝑥sin3𝑚()̌𝐿2(2)=𝜒𝐴𝛾𝜑𝑥sin3𝑚()𝐿2(2)𝑐.(4.22) From this it follows that the first addend in the right-hand side of (4.21) is quadratically integrable over the set 2×(0,𝐻) and thus especially quadratically integrable over (𝑅,)××(0,𝐻).
Now we come to the second addend. We define 𝑅 by 𝑅=(𝑅+𝑅)/2=𝑅+(𝜀/2). From (4.12) we find that there is a constant 𝑐1>0 with ||𝛾𝜑𝑖𝑡,𝜉2||𝑐1𝑒𝑅𝑡for𝜉2[]𝑘,𝑘,𝑡0.(4.23) Since for 𝜉2[𝑘,𝑘] and 𝑡0 the quantity 𝑚(𝑖𝑡,𝜉2) is a real number and therefore |sin(𝑥3𝑚(𝑖𝑡,𝜉2))|1 for 𝑥3(0,𝐻), we conclude that ||||0𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1||||𝑑𝑡𝑐10𝑒(𝑅𝑥1)𝑡𝑐𝑑𝑡=1𝑥1𝑅for𝜉2[]𝑘,𝑘,𝑥3(0,𝐻),𝑥1>𝑅.(4.24) Thus we have 𝑘𝑘||||0𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1||||𝑑𝑡2𝑑𝜉22𝑘𝑐21𝑥1𝑅2for𝑥3(0,𝐻),𝑥1>𝑅.(4.25) Because the inverse Fourier transform (here with regard to the variable 𝜉2) is isometric on 𝐿2, we see that ||||𝑘𝑘0𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1𝑑𝑡𝑒𝑖𝜉2𝑥2𝑑𝜉2||||2𝑑𝑥24𝜋𝑘𝑐21𝑥1𝑅2for𝑥3(0,𝐻),𝑥1>𝑅.(4.26) It follows that 𝐻0𝑅||||𝑘𝑘0𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1𝑑𝑡𝑒𝑖𝜉2𝑥2𝑑𝜉2||||2𝑑𝑥2𝑑𝑥1𝑑𝑥3<.(4.27) The substitution 𝑡=𝑖𝜉1 in the innermost integral now shows that the second addend in the right-hand side of (4.21) indeed lies in 𝐿2((𝑅,)××(0,𝐻)).
Regarding the third addend, from (4.12) we find |||||𝜉22𝑘2𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1|||||𝑑𝑡𝑐𝜉22𝑘21+𝑡2+𝜉22𝑁/2𝑒(𝑅𝑥1)𝑡𝑑𝑡for𝜉2[]𝑘,𝑘,𝑥3(0,𝐻),𝑥1>𝑅.(4.28) Now let be 𝑅=𝑅+(𝜀/2) as above. By reason of 1+𝑡2+𝜉22𝑁/21+𝑡2𝑁/21+𝜉22𝑁/2,(4.29) we conclude that there are constants 𝑐2,𝑐3>0such that |||||𝜉22𝑘2𝛾𝜑𝑖𝑡,𝜉2𝑥sin3𝑚𝑖𝑡,𝜉2𝑒𝑡𝑥1|||||𝑑𝑡𝑐21+𝜉22𝑁/2𝜉22𝑘2𝑒(𝑅𝑥1)𝑡=𝑐𝑑𝑡2𝑥1𝑅1+𝜉22𝑁/2𝑒(𝑅𝑥1)𝜉22𝑘2𝑐3𝑥1𝑅𝑒(𝜀/4)𝜉22𝑘2for𝜉2[]𝑘,𝑘,𝑥3(0,𝐻),𝑥1>𝑅.(4.30) This estimate corresponds to formula (4.24), used in the case of the second addend. Continuing as in this latter case, it is seen that the third addend in the right-hand side of (4.21) lies in 𝐿2((𝑅,)××(0,𝐻)) too.
Proof of Part (b)
The preceding proof of part (a) is based on the fact that for 𝑗{1,2} it holds that ̂𝑒𝑗𝜉,𝑥3=𝑒𝑖𝑥3𝑚(𝜉)̂𝑒𝑗,0(𝜉)for𝜉2,𝑥3>0,(4.31)supp𝑒𝑗,0Ω.(4.32) The relation (4.31), which is equivalent to the representation of 𝑒𝑗 given in Theorem 2.1, has led to (4.10).
Since the third magnetic component 𝑏3 fulfills conditions which are analogous to (4.31) and (4.32), the proof of part (b) of the lemma is completely analogous to the proof of part (a). The condition for 𝑏3 that is analogous to (4.31) is given in (2.19); note that (2.19) also holds for |𝜉|<𝑘. The condition supp𝑏3,0Ωis a direct consequence of 𝑏3,0=𝑖𝑘𝑒2,0𝑒1,0,(4.33) following from the second equation in (