 Next Article in Journal
Effective Congruences for Mock Theta Functions
Previous Article in Journal
On the Distribution of the spt-Crank
Article

# Scattering of Electromagnetic Waves by Many Nano-Wires

Department of Mathematics, Kansas State University, Manhattan, KS 66506-2602, USA
Mathematics 2013, 1(3), 89-99; https://doi.org/10.3390/math1030089
Received: 10 May 2013 / Revised: 11 July 2013 / Accepted: 11 July 2013 / Published: 18 July 2013

## Abstract

Electromagnetic wave scattering by many parallel to the zaxis, thin, impedance, parallel, infinite cylinders is studied asymptotically as a 0. Let Dm be the cross-section of the mth cylinder, a be its radius and ${\text{x ^}}_{\text{m}}{\text{= (x}}_{\text{m1}}{\text{, x}}_{\text{m2}}\text{)}$ be its center, 1 m M , M = M (a). It is assumed that the points, ${\stackrel{^}{x}}_{m}$ , are distributed, so that $N\left(\Delta \right)=\frac{1}{2\pi a}\underset{\Delta }{\int }N\left(\stackrel{^}{x}\right)d\stackrel{^}{x}\left[1+o\left(1\right)\right]$ where N (∆) is the number of points, ${\stackrel{^}{x}}_{m}$ , in an arbitrary open subset, ∆, of the plane, xoy. The function, $N\left(\stackrel{^}{x}\right)\text{}\ge 0$, is a continuous function, which an experimentalist can choose. An equation for the self-consistent (effective) field is derived as a 0. A formula is derived for the refraction coefficient in the medium in which many thin impedance cylinders are distributed. These cylinders may model nano-wires embedded in the medium. One can produce a desired refraction coefficient of the new medium by choosing a suitable boundary impedance of the thin cylinders and their distribution law.

## 1. Introduction

There is a large literature on electromagnetic (EM) wave scattering by an array of parallel cylinders (see, e.g., , where there are many references given, and ). Electromagnetic wave scattering by many parallel to the $z -$axis, thin, circular, of radius a, infinite cylinders, on the boundary of which an impedance boundary condition holds, is studied in this paper asymptotically as $a → 0$. The cylinders are thin in the sense, $k a ≪ 1$, where k is the wave number in the exterior of the cylinders.
The novel points in this paper include:
(1) The asymptotically exact, as $a → 0$, solution of the EM wave scattering problem by one impedance cylinder is given. The solution to the EM wave scattering problem by many thin impedance cylinders is given. The limiting behavior of this solution is found in the limit $a → 0$ when the number, $M = M ( a )$, of the cylinders tends to infinity at a suitable rate. The equation for the limiting (as $a → 0$) effective (self-consistent) field in the medium obtained by embedding into it many thin impedance cylinders is derived.
(2) This theory is a basis for a method for changing the refraction coefficient in a medium by embedding into this medium many thin impedance cylinders. The thin cylinders model nano-wires embedded in the medium. The basic physical result of this paper is the formula, (43), which shows how the embedded thin cylinders change the refraction coefficient, $n 2 ( x )$. This formula serves for designing a desired refraction coefficient.
The results of the paper are summarized in three theorems.
The author considered earlier scalar wave scattering by small particles of an arbitrary shape ([3,4,5,6,7,8,9]) and EM wave scattering by many thin perfectly conducting cylinders (). The introduction of the impedance boundary condition is of principal importance, because it gives much more flexibility in creating a desired refraction coefficient; see formula (43).
Let $D m$, $1 ≤ m ≤ M$, be a set of non-intersecting domains on a plane, P, which is the $x o y$ plane. Let $x ^ m ∈ D m$, $x ^ m = ( x m 1 , x m 2 )$, be a point inside $D m$ and $C m$ be the cylinder with the cross-section $D m$, and the axis, parallel to the z-axis, passing through $x ^ m$. We assume that $x ^ m$ is the center of the disc $D m$, if $D m$ is a disc of radius a.
Let us assume that on the boundary of the cylinders, an impedance boundary condition holds; see Equation (5) below. Let $a = 0 . 5$diam$D m$. Our basic assumptions are
$k a ≪ 1$
where k is the wave number in the region exterior to the union of the cylinders and
$N ( Δ ) = 1 2 π a ∫ Δ N ( x ^ ) d x ^ [ 1 + o ( 1 ) ] , a → 0$
where $N ( Δ ) = ∑ x ^ m ∈ Δ 1$ is the number of the cylinders in an arbitrary open subset of the plane P, $N ( x ^ ) ≥ 0$ is a continuous function, which can be chosen as we wish, and $2 π a$ is the arc length of a circle of radius, a. The points, $x ^ m$, are distributed in an arbitrarily large, but fixed, bounded domain on the plane P. We denote by Ω the union of domains $D m$, by $Ω ′$, its complement in P, and by $D ′$ the complement of D in P. The complement in $R 3$ of the union C of the cylinders $C m$, we denote by $C ′$.
The EM wave scattering problem consists of finding the solution to Maxwell’s equations:
$∇ × E = i ω μ H$
$∇ × H = - i ω ϵ E$
in $C ′$, such that:
$E t = ζ [ n , H ] on ∂ C$
where $∂ C$ is the union of the surfaces of the cylinders $C m$, $E t$ is the tangential component of E on the boundary of C, n is the unit normal to $∂ C$ directed out of the cylinders, μ and ϵ are constants in $C ′$, ω is the frequency, $k 2 = ω 2 ϵ μ$ and k is the wave number. The ζ in Equation (5) is the boundary impedance. The boundary impedance satisfies the physical restriction Re$ζ ≥ 0$ (see ).
The usual impedance boundary condition is $E t = ζ [ n , H t ]$, because we use the unit normal pointing out of C, but $[ n , H ] = [ n , H t ]$ if $H 3 = 0$, as we assume in this paper.
Denote by $n 0 2 = ϵ μ$, so $k 2 = ω 2 n 0 2$. The solution to Equations (3)–(5) must have the following form:
$E ( x ) = E 0 ( x ) + v ( x ) , x = ( x 1 , x 2 , x 3 ) = ( x , y , z ) = ( x ^ , z )$
where $E 0 ( x )$ is the incident field and v is the scattered field satisfying the radiation condition:
$r ∂ v ∂ r - i k v = o ( 1 ) , r = ( x 1 2 + x 2 2 ) 1 / 2$
We assume that
$E 0 ( x ) = k - 1 e i κ y + i k 3 z ( - k 3 e 2 + κ e 3 ) , κ 2 + k 3 2 = k 2$
${ e j }$, $j = 1 , 2 , 3$ are the unit vectors along the Cartesian coordinate axes, $x , y , z$. We consider EM waves with $H 3 : = H z = 0$, that is, E-waves, or TH waves,
$E = ∑ j = 1 3 E j e j , H = H 1 e 1 + H 2 e 2 = ∇ × E i ω μ$
. One can prove (see Appendix) that the components of E can be expressed by the formulas:
$E j = i k 3 κ 2 U x j e i k 3 z , j = 1 , 2 , E 3 = U e i k 3 z , U = κ k u$
where $u x j : = ∂ u ∂ x j$, $u = u ( x , y )$ solves the problem:
$( Δ 2 + κ 2 ) u = 0 in Ω ′$
$( u n + i ξ u ) | ∂ Ω = 0 , u n : = ∇ u · n , ξ : = ω μ κ 2 ζ k 2$
$u = e i κ y + w$
and w satisfies the radiation condition, (7).
If $k 3 = 0$, then $E 0 = e 3 e i k y$,
$E 1 = E 2 = 0 , E 3 = u ; H 1 = u y i ω μ , H 2 = - u x i ω μ , H 3 = 0 ; ξ = ω μ ζ ; κ 2 = k 2$
This is the case of the EM wave incident perpendicular to the axis of the cylinders with the E-field parallel to this axis
If $k 3 ≠ 0$, then one takes $κ = 0$, $k 3 = k$ and $E 0 = e i k z e 2$. This is the case of the EM wave incident along the axis of the cylinders with the E-field parallel to the y-axis.
If $k 3 ≠ 0$, then $E t$ in the condition (5) does not lead to Equation (12). If $E t$ is approximately replaced by $E 3$ in condition (5), then Equation (12) holds.
Our main results come from the analysis of problem, (11)–(13).
The unique solution to Equations (3)–(8) is given by the formulas:
$E 1 = i k 3 κ 2 U x e i k 3 z , E 2 = i k 3 κ 2 U y e i k 3 z , E 3 = U e i k 3 z$
$H 1 = k 2 i ω μ κ 2 U y e i k 3 z , H 2 = - k 2 i ω μ κ 2 U x e i k 3 z , H 3 = 0$
where $U x : = ∂ U ∂ x$, $U y = ∂ U ∂ y$ and $u = u ( x ^ ) = u ( x , y )$ solves the scalar two-dimensional problem, (11)–(13), if $k 3$ = 0. These formulas are derived in the Appendix.
Problem (11)–(13) has a unique solution (see, for example, ), provided that Re$ζ ≥ 0$ or, equivalently, that Im$ξ ≥ 0$. This corresponds to the assumption that the material inside the cylinders is passive, that is, the energy absorption is non-negative. Our goal is to derive an asymptotic formula for this solution as $a → 0$. Our results include formulas for the solution to the scattering problem, the derivation of the equation for the effective field in the medium obtained by embedding many thin perfectly conducting cylinders and a formula for the refraction coefficient in this limiting medium. This formula shows that by choosing a suitable distribution of the cylinders, one can change the refraction coefficient, for example, one can make it smaller than the original one.
The paper is organized as follows.
In Section 2 we derive an asymptotic formula for the solution to Equations (11)–(13) when $M = 1$, that is, for scattering by one cylinder. The result is formulated in Theorem 1.
In Section 3 we derive a linear algebraic system for finding some numbers that define the solution to problem (11)–(13) with $M > 1$. This gives a numerical method for solving the EM wave scattering problem by many thin impedance cylinders.
Furthermore, in Section 3 we derive an integral equation for the effective (self-consistent) field in the medium with $M ( a ) → ∞$ cylinders as $a → 0$. The result is formulated in Theorem 2.
At the end of Section 3 this result is applied to the problem of changing the refraction coefficient of a given material by embedding many thin perfectly conducting cylinders into it. An analytic formula for the refraction coefficient is derived. The result is formulated in Theorem 3.
In Section 4 conclusions are formulated.
In Appendix formulas (14) and (15) are derived.

## 2. EM Wave Scattering by One Thin Perfectly Conducting Cylinder

Consider problem (11)–(13) with $Ω = D 1$, $Ω ′$ being the complement to $D 1$ in $R 2$. Our theory and the results formulated in Theorems 1, 2 and 3 are valid for the cylinders with an arbitrary cross-section. We assume that the point $x ^ 1$ is a point inside $D 1$.
Let us look for a solution of the form:
$u ( x ^ ) = e i κ y + ∫ S 1 g ( x ^ , t ) σ ( t ) d t , g ( x ^ , t ) : = g ( κ | x ^ - t | ) = i 4 H 0 ( 1 ) ( κ | x ^ - t | )$
where $S 1$ is the boundary of $D 1$, $H 0 ( 1 )$ is the Hankel function of the order of one, with index, zero, and σ is to be found from the boundary condition, (12). Let $r = | x ^ - t |$. It is known that as $r → 0$, one has:
$g ( κ r ) = α ( κ ) + 1 2 π ln 1 r + o ( 1 ) , α : = α ( κ ) : = i 4 + 1 2 π ln 2 κ$
and:
$g ( κ r ) = i 4 H 0 ( 1 ) ( k r ) = i 4 2 π k r e i ( k r - π 4 ) 1 + o ( 1 ) , r → ∞$
Thus:
$u = u 0 + g ( x ^ , 0 ) Q + o ( 1 r ) , | x ^ | = r → ∞ ; Q : = ∫ S 1 σ ( t ) d t$
and $u 0$ here is the same plane wave as in Equation (16).
One sees from formula (19) that it is sufficient to find one number, Q, in order to solve the scattering problem, (11)–(13), for one thin cylinder.
This is the central idea of our work. Below, we derive an explicit analytical formula for the number, Q. This formula (formula (24)) is asymptotically exact.
Let us explain how to derive formula (19). The asymptotics of the integral term in Equation (16) is:
$g ( x ^ , 0 ) ∫ S 1 e - i κ β · t σ ( t ) d t , β : = x ^ / | x ^ |$
where $0 ∈ D 1$ is the origin. Since $κ a ≪ 1$, it follows that:
$e - i κ β · t = 1 + O ( a ) a → 0$
The remainder, $o ( 1 r )$, in Equation (19) comes from the radiation condition for the function, g. When one studies the scattering problem by a single thin cylinder, one can take an arbitrary point inside $D 1$ as an origin.
In the usual approach to the scattering problem, one has to find an unknown function, $σ ( t )$, rather than one number, Q, in order to solve the scattering problem, (11)–(13). The function, (16), satisfies Equations (11) and (13) for any σ, and if σ is such that function (16) satisfies boundary condition (12), then u solves problem, (11)–(13). We assume σ to be sufficiently smooth (Hölder-continuous is sufficient).
The solution to problem (11)–(13) is known to be unique (see, e.g., ). The exact boundary condition (12) yields:
$- u 0 n ( s ) - i ξ u 0 = i ξ α Q + i ξ ∫ S 1 g 0 ( s , t ) σ ( t ) d t + ( A σ - σ ) / 2$
where $A σ : = ∫ S 1 ∂ g 0 ( s , t ) ∂ n s d t ,$ and the formula for the limiting value on $S 1$ of the exterior normal derivative of the simple layer potential, $∫ S 1 g 0 ( x , t ) σ ( t ) d t$, was used and
$u 0 ( s ) : = e i κ s 2 , s ∈ S 1 ; g 0 ( s , t ) : = 1 2 π ln 1 r s t , r s t : = | s - t |$
If $k a ≪ 1$ and $k 2 = κ 2 + k 3 2$, then
$u 0 ( s ) = 1 + O ( κ a ) , u 0 n = i κ n 2 + O ( κ a )$
Equation (20) is uniquely solvable for σ if a is sufficiently small (see [11,12]).
We are interested in finding the asymptotic formula for Q as $a → 0$, because $u ( x ^ )$ in Equation (16) can be well approximated in the region, $| x ^ | ≫ a$, by the formula,
$u ( x ^ ) = u 0 ( x ^ ) + g ( x ^ , 0 ) Q + o ( 1 ) , a → 0$
. To find the asymptotic of Q as $a → 0$, let us integrate Equation (20) over $S 1$; keep the main terms of the asymptotic as $a → 0$, take into account that
$∫ S 1 d t n 2 ( t ) = 0 , ∫ S 1 g 0 ( s , t ) d s = O ( a | log a | ) a → 0$
use formulas (22) and obtain
$Q = i ξ u 0 ( x ^ 1 ) | S 1 | ( 1 + o ( 1 ) ) , a → 0$
where $x ^ 1$ is a point inside $D 1$, $| S 1 |$ is the length of $S 1$ and $r s t = | s - t |$. If $S 1$ is the circle of radius, a, then $| S 1 | = 2 π a$.
If $S 1$ is an arbitrary curve, we assume that the length $| S 1 | = c a$, where a is the same as in Equation (1), and $c > 0$ is a constant. The reader can check the estimate, $∫ S 1 g 0 ( s , t ) d s = O ( a | log a | )$, as $a → 0$, where $s , t ∈ S 1$. From formulas (24) and (19) the asymptotic solution to the scattering problem (11)–(13) in the case of one circular cylinder of radius a, as $a → 0$, is
$u ( x ^ ) ∼ u 0 ( x ^ ) + i 2 π a ξ u 0 ( x ^ 1 ) g ( x ^ , x ^ 1 ) , a → 0 , | x ^ - x ^ 1 | > a$
Here, the sign, ∼, stands for the asymptotic equivalence as $a → 0$, the point; $x ^$ is an arbitrary point on the plane that cannot be at distances less than a from the point $x ^ 1 ∈ D 1$.
Let us formulate the result.
Theorem 1. Electromagnetic wave, scattered by a single cylinder, is calculated by formulas (14) and (15) in which $u = u ( x ^ ) : = u ( x 1 , x 2 )$ is given by formula (25).
If the cylinder is not circular, then formula (25) takes the form:
$u ( x ^ ) ∼ u 0 ( x ^ ) + i | S 1 | ξ u 0 ( x ^ 1 ) g ( x ^ , x ^ 1 ) , a → 0 , | x ^ - x ^ 1 | > a$
where $| S 1 |$ is the length of $S 1$.

## 3. Wave Scattering by Many Thin Cylinders

Problem (11)–(13) should be solved when Ω is a union of many small domains, $D m$, $Ω = ∪ m = 1 M D m$. We assume that $D m$ is a circle of radius a centered at the point, $x ^ m$.
Let us look for u of the form
$u ( x ^ ) = u 0 ( x ^ ) + ∑ m = 1 M ∫ S m g ( x ^ , t ) σ m ( t ) d t$
We assume that the points, $x ^ m$, are distributed in a bounded domain D on the plane $P = x o y$ by formula (2). The field $u 0 ( x ^ )$ is the same as in Section 2, $u 0 ( x ^ ) = e i κ y$, and Green’s function, g, is the same as in formulas (16)–(18). It follows from Equation (2) that $M = M ( a ) = O ( ln 1 a ) .$ We define the effective field, acting on the $D j$ by the formula
$u e = u e ( j ) = u ( x ^ ) - ∫ S j g ( x ^ , t ) σ j ( t ) d t , | x ^ - x ^ j | > a$
which can also be written as
$u e ( x ^ ) = u 0 ( x ^ ) + ∑ m = 1 , m ≠ j M ∫ S m g ( x ^ , t ) σ m ( t ) d t$
We assume that the distance $d = d ( a )$ between neighboring cylinders is much greater than a:
$d ≫ a , lim a → 0 a d ( a ) = 0$
Let us rewrite Equation (27) as
$u = u 0 + ∑ m = 1 M g ( x ^ , x ^ m ) Q m + ∑ m = 1 M ∫ S m [ g ( x ^ , t ) - g ( x ^ , x ^ m ) ] σ m ( t ) d t$
where
$Q m : = ∫ S m σ m ( t ) d t$
As $a → 0$, the second sum in Equation (30) (let us denote it $Σ 2$) is negligible compared with the first sum in Equation (30), denoted $Σ 1$:
$| Σ 2 | ≪ | Σ 1 | , a → 0$
The idea of the proof of this is similar to the one given in  for a quite different problem of scalar wave scattering in $R 3$. Let us sketch this proof.
Let us check that
$| g ( x ^ , x ^ m ) Q m | ≫ | ∫ S m [ g ( x ^ , t ) - g ( x ^ , x m ) ] σ m ( t ) d t | , a → 0$
If $k | x ^ - x ^ m | ≫ 1$, and $k > 0$ is fixed, then
$| g ( x ^ , x ^ m ) | = O ( 1 | x ^ - x ^ m | 1 / 2 ) , | g ( x ^ , t ) - g ( x ^ , x m ) | = O ( a | x ^ - x ^ m | 1 / 2 )$
and $Q m ≠ 0$; so, estimate Equation (33) holds.
If
$| x ^ - x ^ m | ∼ d ≫ a$
then
$| g ( x ^ , x ^ m ) | = O ( 1 ln 1 a ) , | g ( x ^ , t ) - g ( x ^ , x m ) | = O ( a d )$
as follows from the asymptotic of $H 0 1 ( r ) = O ( ln 1 r )$ as $r → 0$ and from the formulas $d H 0 1 ( r ) d r = - H 1 1 ( r ) = O ( 1 r )$ as $r → 0$. Thus, Equation (33) holds for $| x ^ - x ^ m | ≫ d ≫ a$.
Consequently, the scattering problem is reduced to finding the numbers $Q m$, $1 ≤ m ≤ M$.
Let us estimate $Q m$ asymptotically as $a → 0$. To do this, we use the exact boundary condition on $S m$ and an argument similar to the one given in the case of wave scattering by one cylinder. The role of the incident field, $u 0$, is played now by the effective field, $u e$. The result is a formula, similar to Equation (24):
$Q j = i 2 π a ξ j u e ( x ^ j ) , a → 0$
The formula, similar to Equation (25), is
$u ( x ^ ) ∼ u 0 ( x ^ ) + i 2 π a ∑ m = 1 M g ( x ^ , x ^ m ) ξ m u e ( x ^ m ) , a → 0$
The numbers, $u e ( x ^ m )$, $1 ≤ m ≤ M$, in Equation (35) are not known. Setting $x ^ = x ^ j$ in Equation (35), neglecting the $o ( 1 )$ term and using the definition (28) of the effective field, one gets a linear algebraic system for finding numbers $u e ( x ^ m )$:
$u e ( x ^ j ) = u 0 ( x ^ j ) + i 2 π a ∑ m ≠ j g ( x ^ j , x ^ m ) ξ m u e ( x ^ m ) , 1 ≤ j ≤ M$
This system can be solved numerically if the number M is not very large, say $M ≤ O ( 10 3 )$.
If M is very large, $M = M ( a ) → ∞ , a → 0$, then we derive a linear integral equation for the limiting effective field in the medium obtained by embedding many cylinders.
Passing to the limit $a → 0$ in system Equation (36) is done by the method used in  in the problem of wave scattering by many small bodies. Consider a partition of the domain D into a union of $P$ small squares $Δ p$ of size $b = b ( a )$, $b ≫ d ≫ a$. For example, one may choose $b = O ( a 1 / 4 )$, $d = O ( a 1 / 2 )$, so that there are many discs $D m$ in the square $Δ p$. We assume that squares $Δ p$ and $Δ q$ do not have common interior points if $p ≠ q$. Let $y ^ p$ be the center of $Δ p$. One can also choose as $y ^ p$ any point $x ^ m$ in a domain $D m ⊂ Δ p$. Since $u e$ is a continuous function, one may approximate $u e ( x ^ m )$ by $u e ( y ^ p )$, provided that $x ^ m ⊂ Δ p$. The error of this approximation is $o ( 1 )$ as $a → 0$. Let $ξ ( x ^ )$ be a continuous function in D, such that $ξ ( x ^ m ) = ξ m$. Let us rewrite the sum in Equation (36) as follows:
$2 π a ∑ m ≠ j g ( x ^ j , x ^ m ) ξ m u e ( x ^ m ) = ∑ x j ∉ Δ p p = 1 P g ( x ^ j , y ^ p ) ξ ( y ^ p ) u e ( y ^ p ) 2 π a ∑ x m ∈ Δ p 1$
and use formula (2) in the form
$2 π a ∑ x m ∈ Δ p 1 = N ( y ^ p ) | Δ p | [ 1 + o ( 1 ) ] , a → 0$
Here, $| Δ p |$ is the area of the square $Δ p$.
From Equations (37) and (38), one obtains:
$2 π a ∑ m ≠ j g ( x ^ j , x ^ m ) ξ m u e ( x ^ m ) = ∑ x ^ j ∉ Δ p p = 1 P g ( x ^ j , y ^ p ) N ( y ^ p ) ξ ( y ^ p ) u e ( y ^ p ) | Δ p | [ 1 + o ( 1 ) ]$
The sum in the right-hand side of formula (39) is the Riemannian sum for the integral,
$lim a → 0 ∑ p = 1 P g ( x ^ j , y ^ p ) N ( y ^ p ) ξ ( y ^ p ) u e ( y ^ p ) | Δ p | = ∫ D g ( x ^ , y ^ ) N ( y ^ ) ξ ( y ^ ) u ( y ^ ) d y$
where $u ( x ^ ) = lim a → 0 u e ( x ^ ) .$ Therefore, system (36) in the limit, $a → 0$, yields the integral equation for the limiting effective field,
$u ( x ^ ) = u 0 ( x ^ ) + i ξ ∫ D g ( x ^ , y ^ ) N ( y ^ ) ξ ( y ^ ) u ( y ^ ) d y ^$
Note that the function, $ξ ( y ^ )$, can be chosen by the experimenter as he wishes. This is an advantage of having impedance cylinders with the impedance that can be chosen as one wishes.
Let us formulate this result.
Theorem 2. The effective field in the limiting medium satisfies Equation (41).
One obtains system (36) if one solves Equation (41) by a collocation method. Convergence of this method to the unique solution of Equation (41) is proved in . Existence and uniqueness of the solution to Equation (41) are proved as in , where a three-dimensional analog of this equation was studied.
One has $( Δ 2 + κ 2 ) g ( x ^ , y ^ ) = - δ ( x ^ - y ^ )$. Using this relation and applying the operator $Δ 2 + κ 2$ to Equation (41) yields the following differential equation for $u ( x ^ )$:
$Δ 2 u ( x ^ ) + κ 2 u ( x ^ ) + i ξ ( x ^ ) N ( x ^ ) u ( x ^ ) = 0 x ^ ∈ R 2$
This is a Schrödinger-type equation, and $u ( x ^ )$ is its scattering solution corresponding to the incident wave, $u 0 = e i κ y$.
Let us assume that $N ( x ^ ) = N$ and $ξ ( x ^ )$ are constants. One concludes from Equation (42) that the limiting medium, obtained by embedding many perfectly conducting circular cylinders, has new parameter, $κ N 2 : = κ 2 + i ξ N$. This means that $k 2 = κ 2 + k 3 2$ is replaced by $k ˜ 2 : = k 2 + i ξ N$. The quantity, $k 3 2$, is not changed. One has $k ˜ 2 = ω 2 n 2$, $k 2 = ω 2 n 0 2$. Consequently, $n 2 / n 0 2 = ( k 2 + i ξ N ) / k 2$. Therefore, the new refraction coefficient, $n 2$, is
$n 2 = n 0 2 ( 1 + i ξ N k - 2 ) , ξ = ω μ κ 2 ζ k 2$
Let us formulate this result.
Theorem 3. If $N ( x ^ ) = N$ and $ξ ( x ^ ) = ξ$ are constants in D, then the refraction coefficient in the limiting medium is given by formula (43).
Since the number, $N > 0$, and the impedance, ζ, are at our disposal, Equation (43) shows that, choosing suitable N, one can create a medium with a desired refraction coefficient. It is of interest that if $ξ ( x ^ )$ is not a constant, one can create a refraction coefficient depending on $x ^$ in a desired way and having a desired absorption as a function of the position, $x ^$.
In practice, one does not go to the limit, $a → 0$, but chooses a sufficiently small a. As a result, one obtains a medium with a refraction coefficient $n a 2$, which differs from Equation (43) a little, $lim a → 0 n a 2 = n 2 .$

## 4. Conclusions

Asymptotic, as $a → 0$, solution is given for the EM wave scattering problem by many perfectly conducting parallel cylinders of radius, a. The equation for the effective field in the limiting medium obtained when $a → 0$ and the distribution of the embedded cylinders is given by formula (2). The presented theory gives formula (43) for the refraction coefficient in the limiting medium. This formula shows how the distribution of the cylinders influences the refraction coefficient.

## References

1. Martin, P. Multiple Scattering; Cambridge University Press: Cambridge, UK, 2006. [Google Scholar]
2. Mei, C.; Vernescu, B. Homogenization Methods for Multiscale Mechanics; World Science: New Jersey, NJ, USA, 2010. [Google Scholar]
3. Ramm, A.G. Distribution of particles which produces a “smart” material. J. Stat. Phys. 2007, 127, 915–934. [Google Scholar] [CrossRef]
4. Ramm, A.G. Electromagnetic wave scattering by small bodies. Phys. Lett. A 2008, 372, 4298–4306. [Google Scholar] [CrossRef]
5. Ramm, A.G. Wave scattering by many small particles embedded in a medium. Phys. Lett. A 2008, 372, 3064–3070. [Google Scholar] [CrossRef]
6. Ramm, A.G. A collocation method for solving integral equations. Int. J. Comput. Sci. Math. 2009, 3, 222–228. [Google Scholar] [CrossRef]
7. Ramm, A.G. A method for creating materials with a desired refraction coefficient. Int. J. Mod. Phys. B 2010, 24, 5261–5268. [Google Scholar] [CrossRef]
8. Ramm, A.G. Materials with a desired refraction coefficient can be created by embedding small particles into the given material. Int. J. Struct. Chang. Solids 2010, 2, 17–23. [Google Scholar]
9. Ramm, A.G. Wave scattering by many small bodies and creating materials with a desired refraction coefficient. Afrika Mat. 2011, 22, 33–55. [Google Scholar] [CrossRef]
10. Ramm, A.G. Scattering of electromagnetic waves by many thin cylinders. Results Phys. 2011, 1, 13–16. [Google Scholar] [CrossRef]
11. Ramm, A.G. Scattering by Obstacles; D.Reidel: Dordrecht, The Netherlands, 1986. [Google Scholar]
12. Ramm, A.G. Wave Scattering by Small Bodies of Arbitrary Shapes; World Science Publishers: Singapore, Singapore, 2005. [Google Scholar]
13. Ramm, A.G. Many-body wave scattering by small bodies and applications. J. Math. Phys. 2007, 48, 103511. [Google Scholar] [CrossRef]
14. Landau, L.; Lifshitz, E. Electrodynamics of Continuous Media; Pergamon Press: London, UK, 1984. [Google Scholar]

## Appendix

Let us derive formulas, (14) and (15). Look for the solution to Equations (3) and (4) of the form:
$E 1 = e i k 3 z E ˜ 1 ( x , y ) , E 2 = e i k 3 z E ˜ 2 ( x , y ) , E 3 = e i k 3 z u ( x , y )$
$H 1 = e i k 3 z H ˜ 1 ( x , y ) , H 2 = e i k 3 z H ˜ 2 ( x , y ) , H 3 = 0$
where $k 3 = c o n s t$. Equation (3) yields
$u y - i k 3 E ˜ 2 = i ω μ H ˜ 1 , - u x + i k 3 E ˜ 1 = i ω μ H ˜ 2 , E ˜ 2 , x = E ˜ 1 , y$
where, for example, $E ˜ j , x : = ∂ E ˜ j ∂ x$. Equation (4) yields
$i k 3 H ˜ 2 = i ω ϵ E ˜ 1 , i k 3 H ˜ 1 = - i ω ϵ E ˜ 2 , H ˜ 2 , x - H ˜ 1 , y = - i ω ϵ u$
Excluding $H ˜ j$, $j = 1 , 2$, from Equation (46) and using Equation (47), one gets
$E ˜ 1 = i k 3 κ 2 u x , E ˜ 2 = i k 3 κ 2 u y , E ˜ 3 = u$
$H ˜ 1 = k 2 u y i ω μ κ 2 , H ˜ 2 = - k 2 u x i ω μ κ 2 , H ˜ 3 = 0$
Since $E j = E ˜ j e i k 3 z$ and $H j = H ˜ j e i k 3 z$, formulas (14) and (15) follow immediately from Equations (48) and (49).