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In this paper, we investigate the problem of electromagnetic (EM) wave scattering by one and many small perfectly conducting bodies and present a numerical method for solving it. For the case of one body, the problem is solved for a body of arbitrary shape, using the corresponding boundary integral equation. For the case of many bodies, the problem is solved asymptotically under the physical assumptions
a <<
d << λ, where
a is the characteristic size of the bodies,

Many real-world electromagnetic (EM) problems like EM wave scattering, EM radiation, etc. [

In [

N ( Δ ) = 1 a 3 ∫ Δ N ( x ) d x [ 1 + o ( 1 ) ] , a → 0, (1)

in which Δ is an arbitrary open subset of the domain Ω that contains all the small bodies, N ( Δ ) is the number of the small bodies in Δ , and N ( x ) is the distribution function of the bodies

N ( x ) ≥ 0, N ( x ) ∈ C ( Ω ) . (2)

In Sections 2 and 3, the theory of EM wave scattering by one and many small perfectly conducting bodies is presented. The numerical methods for solving these problems are also described in details. Furthermore, error analysis for the numerical methods of solving the EM scattering problem are also provided. In Section 4 these methods are tested and numerical results are discussed.

Let D be a bounded perfectly conducting body, a = 1 2 diam D , S be its C^{2}-smooth

boundary, and D ′ : = ℝ 3 \ D . Let ϵ and μ be the dielectric permittivity and magnetic permeability constants of the medium in D ′ . Let E and H denote the electric and magnetic fields, respectively, E 0 be the incident field and v E be the scattered field. The problem of electromagnetic wave scattering by one perfectly conducting body can be stated as follows

∇ × E = i ω μ H , in D ′ : = ℝ 3 \ D , (3)

∇ × H = − i ω ϵ E , in D ′ , (4)

[ N , [ E , N ] ] = 0 , on S : = ∂ D , (5)

E = E 0 + v E , (6)

E 0 = E e i k α ⋅ x , E ⋅ α = 0 , α ∈ S 2 , (7)

∂ v E ∂ r − i k v E = o ( 1 r ) , r : = | x | → ∞ , (8)

where ω > 0 is the frequency, k = 2 π / λ = ω ϵ μ is the wave number, k a ≪ 1 , λ is the wave length, E is a constant vector, and α is a unit vector that indicates the direction of the incident wave E 0 . This incident wave satisfies the relation ∇ ⋅ E 0 = 0 . The scattered field v E satisfies the radiation Condition (8). Here, N is the unit normal vector to the surface S, pointing out of D. By [ ⋅ , ⋅ ] the vector product is denoted and α ⋅ x is the scalar product of two vectors.

The solution to problem (3)-(8) can be found in the form

E ( x ) = E 0 ( x ) + ∇ × ∫ S g ( x , t ) J ( t ) d t , g ( x , t ) : = e i k | x − t | 4 π | x − t | , (9)

see [

J 2 + A J : = J ( s ) 2 + ∫ S [ N s , [ ∇ s g ( s , t ) , J ( t ) ] ] d t = − [ N s , E 0 ] , (10)

or, equivalently

( I + 2 A ) J = F , (11)

where F : = − 2 [ N s , E 0 ] . Equation (11) is of Fredholm type since A is compact, see [

Once we have J, E can be computed by formula (9) and H can be found by the formula

H = ∇ × E i ω μ . (12)

If D is sufficiently small, then Equation (11) is uniquely solvable in C ( S ) and its solution J is tangential to S, see [

E ( x ) = E 0 ( x ) + [ ∇ g ( x , x 1 ) , Q ] + ∇ × ∫ S [ g ( x , t ) − g ( x , x 1 ) ] J ( t ) d t , (13)

where x 1 ∈ D , an arbitrary point inside the small body D, and

Q : = ∫ S J ( t ) d t . (14)

Since

| ∇ g ( x , x 1 ) | = O ( k d + 1 d 2 ) , d = | x − x 1 | , (15)

| g ( x , t ) − g ( x , x 1 ) | = O ( ( k d + 1 d 2 ) a ) , a = | t − x 1 | , (16)

and

| ∇ [ g ( x , t ) − g ( x , x 1 ) ] | = O ( a k 2 d + a k d 2 + a d 3 ) , (17)

the second term in (13) is much greater than the last term

| [ ∇ g ( x , x 1 ) , Q ] | ≫ | ∇ × ∫ S [ g ( x , t ) − g ( x , x 1 ) ] J ( t ) d t | , a → 0. (18)

Then, the asymptotic formula for E when a tends to zero is

E ( x ) = E 0 ( x ) + [ ∇ x g ( x , x 1 ) , Q ] , a → 0, (19)

where | x − x 1 | ≫ a , x 1 ∈ D . Thus, when D is sufficiently small, instead of finding J, we can just find one pseudovector Q.

The analytical formula for Q is derived as follows, see [

∫ S J ( s ) 2 d s + ∫ S d s ∫ S d t [ N s , [ ∇ s g ( s , t ) , J ( t ) ] ] = − ∫ S [ N s , E 0 ] d s . (20)

This is equivalent to

Q 2 + ∫ S d t ∫ S d s ∇ s g ( s , t ) N s ⋅ J ( t ) − ∫ S d t J ( t ) ∫ S d s ∂ g ( s , t ) ∂ N s = − ∫ D ∇ × E 0 d x . (21)

When a → 0 , this equation becomes

Q 2 + e p ∫ S d t ∫ S d s ∂ g ( s , t ) ∂ s p N q ( s ) J q ( t ) + 1 2 ∫ S d t J ( t ) = − | D | ∇ × E 0 , 1 ≤ p , q ≤ 3, (22)

where in the second term, summations over the repeated indices are understood, e p , 1 ≤ p ≤ 3 , are the orthogonal unit vectors in ℝ 3 , | D | is the volume of D, | D | = c D a 3 , and in the third term we use this estimate

∫ S d s ∂ g ( s , t ) ∂ N s ≃ ∫ S d s ∂ g 0 ( s , t ) ∂ N s = − 1 2 , g 0 ( s , t ) : = 1 4 π | s − t | . (23)

Let

Γ p q ( t ) : = ∫ S d s ∂ g ( s , t ) ∂ s p N q ( s ) , (24)

then Equation (22) can be rewritten as follows

Q 2 + e p ∫ S d t Γ p q ( t ) J q ( t ) + Q 2 = − | D | ∇ × E 0 , (25)

or

Q + Γ Q = − | D | ∇ × E 0 , (26)

where Γ is a 3 × 3 constant matrix and it is defined by

Γ Q = e p ∫ S d t Γ p q ( t ) J q ( t ) , (27)

in which summations are understood over the repeated indices. Thus, Q can be written as

Q = − | D | ( I + Γ ) − 1 ∇ × E 0 , a → 0 , (28)

where I : = I 3 , the 3 × 3 identity matrix. This formula is asymptotically exact as a → 0 .

In this section, we consider the EM wave scattering problem by a small perfectly conducting spherical body. Instead of solving the problem (3)-(8) directly, we will solve its corresponding boundary integral Equation (10) for the unknown vector J

J ( s ) 2 + ∫ S [ N s , [ ∇ s g ( s , t ) , J ( t ) ] ] d t = − [ N s , E 0 ] . (29)

Then the solution E to the EM wave scattering problem by one perfectly conducting body can be computed by either the exact formula (9) or the asymptotic formula (19).

Scattering by a sphere has been discussed in many papers, for example [

Suppose S is a smooth surface of a spherical body. Let S be partitioned into P non-intersecting subdomains S i j ,1 ≤ i ≤ m θ ,1 ≤ j ≤ m ϕ , using spherical coordinates, where m θ is the number of intervals of θ between 0 and 2π and m ϕ defines the number of intervals of ϕ between 0 and π. Then P = m θ m ϕ + 2 , which includes the two poles of the sphere. m θ is defined in this way:

m θ = m ϕ + | ϕ − π 2 | 6 m ϕ . This means the closer it is to the poles of the sphere, the

more intervals for θ are used. Then the point ( θ i , ϕ j ) in S i j is chosen as follows

θ i = i 2 π m θ , 1 ≤ i ≤ m θ , (30)

ϕ j = j π m ϕ + 1 , 1 ≤ j ≤ m ϕ . (31)

Note that there are many different ways to distribute collocation points. However, the one that we describe here will guarantee convergence to the solution to (29) with fewer collocation points used from our experiment. Furthermore, one should be careful when choosing the distribution of

collocation points on a sphere. If one chooses ϕ j = j π m ϕ , 1 ≤ j ≤ m ϕ , then when

j = m ϕ , ϕ j = π and thus there is only one point for this ϕ regardless of the value of θ as shown in (32). The position of a point in each S i j can be computed by

( x , y , z ) i j = a ( c o s θ i s i n ϕ j , s i n θ i s i n ϕ j , c o s ϕ j ) , (32)

and the outward-pointing unit normal vector N to S at this point is

N i j = N ( θ i , ϕ j ) = ( cos θ i sin ϕ j , sin θ i sin ϕ j , cos ϕ j ) . (33)

For a star-shaped body with a different shape, only the normal vector N needs to be recomputed. Rewrite the integral Equation (29) as

J ( s ) 2 + ∫ S ∇ s g ( s , t ) N s ⋅ J ( t ) d t − ∫ S ∂ g ( s , t ) ∂ N s J ( t ) d t = − [ N s , E 0 ] . (34)

This integral equation can be discretized as follows

J ( i ) + 2 ∑ j ≠ i P [ ∇ s g ( i , j ) N s ( i ) ⋅ J ( j ) − J ( j ) ∇ s g ( i , j ) ⋅ N s ( i ) ] Δ j = F ( i ) , 1 ≤ i ≤ P ,

(35)

in which by i the point ( x i , y i , z i ) is denoted, F ( i ) : = − 2 [ N s , E 0 ] ( i ) , and Δ j is the surface area of the subdomain j. This is a linear system with unknowns J ( i ) : = ( X i , Y i , Z i ) , 1 ≤ i ≤ P . This linear system can be rewritten as follows

X i + ∑ j ≠ i P a i j X j + b i j Y j + c i j Z j = F x ( i ) , (36)

Y i + ∑ j ≠ i P a ′ i j X j + b ′ i j Y j + c ′ i j Z j = F y ( i ) , (37)

Z i + ∑ j ≠ i P a ″ i j X j + b ″ i j Y j + c ″ i j Z j = F z ( i ) , (38)

where by the subscripts x , y , z the corresponding coordinates are denoted, e.g. F ( i ) = ( F x , F y , F z ) ( i ) , and

a i j : = 2 [ ∇ g ( i , j ) x N x ( i ) − ∇ g ( i , j ) ⋅ N ( i ) ] Δ j , (39)

b i j : = 2 ∇ g ( i , j ) x N y ( i ) Δ j , (40)

c i j : = 2 ∇ g ( i , j ) x N z ( i ) Δ j , (41)

for i ≠ j ; when i = j : a i i = 1 , b i i = 0 , and c i i = 0 ,

a ′ i j : = 2 ∇ g ( i , j ) y N x ( i ) Δ j , (42)

b ′ i j : = 2 [ ∇ g ( i , j ) y N y ( i ) − ∇ g ( i , j ) ⋅ N ( i ) ] Δ j , (43)

c ′ i j : = 2 ∇ g ( i , j ) y N z ( i ) Δ j , (44)

for i ≠ j ; when i = j : a ′ i i = 0 , b ′ i i = 1 , and c ′ i i = 0 ,

a ″ i j : = 2 ∇ g ( i , j ) z N x ( i ) Δ j , (45)

b ″ i j : = 2 ∇ g ( i , j ) z N y ( i ) Δ j , (46)

c ″ i j : = 2 [ ∇ g ( i , j ) z N z ( i ) − ∇ g ( i , j ) ⋅ N ( i ) ] Δ j , (47)

for i ≠ j ; when i = j : a ″ i i = 0 , b ″ i i = 0 , and c ″ i i = 1 .

Solving system (36)-(38) yields X i , Y i , Z i , and J ( i ) can be computed by J ( i ) = ( X i , Y i , Z i ) , 1 ≤ i ≤ P .

Recall the boundary integral Equation (10)

J ( s ) 2 + ∫ S [ N s , [ ∇ s g ( s , t ) , J ( t ) ] ] d t = − [ N s , E 0 ] . (48)

Integrate both sides of this equation over S and get

Q + Γ Q = − | D | ∇ × E 0 , (49)

see Section 2. Once J is found from solving (48), Q can be computed by Q = ∫ S J ( t ) d t . Then one can validate the values of J and Q by checking the following things

・ Is J tangential to S as shown in Section 2? One needs to check J ( s ) ⋅ N s .

・ Is Q = ∫ S J ( t ) d t correct? The relative error of Q can be computed as follows

Error = | Q + Γ Q − R H S | | R H S | , (50)

where R H S : = − | D | ∇ × E 0 . This will give the error of the numerical method for the case of one body.

Furthermore, one can also compare the value of the asymptotic Q a in formula (28) with the exact Q e defined in (14) by

Error = | Q e − Q a | | Q e | , (51)

and check the difference between the asymptotic E a in (19) and the exact E e defined in (9) by computing this relative error

Error = | E e − E a | | E e | . (52)

In this section, we present a general method for solving the EM wave scattering problem by one perfectly conducting body, whose surface is parametrized by f ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) .

・ Step 1: One needs to partition the surface of the body into P non-intersecting subdomains. In each subdomain, choose a collocation point. The position of the collocation points can be computed using f ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) , see for example (30)-(32).

・ Step 2: Find the unit normal vector N of the surface from the function f.

・ Step 3: Solve the linear system (36)-(38) for X i , Y i , and Z i ,1 ≤ i ≤ P . Then vector J in the boundary integral Equation (10) is computed by J ( i ) : = ( X i , Y i , Z i ) at the point i on the surface.

・ Step 4: Compute the electric field E using (9).

Consider a bounded domain Ω containing M small bodies D m , 1 ≤ m ≤ M , and S m are their corresponding smooth boundaries. Let D : = ∪ m = 1 M D m ⊂ Ω and D ′ be the complement of D in ℝ 3 . We assume that S = ∪ m = 1 M S m is C^{2}-smooth. ϵ is the dielectric permittivity constant and μ is the magnetic permeability constant of the medium. Let E and H denote the electric and magnetic fields, respectively. E 0 is the incident field and v is the scattered field. The problem of electromagnetic wave scattering by many small perfectly conducting bodies involves solving the following system

∇ × E = i ω μ H , in D ′ : = ℝ 3 \ D , D : = ∪ m = 1 M D m , (53)

∇ × H = − i ω ϵ E , in D ′ , (54)

[ N , [ E , N ] ] = 0 , on S , (55)

E = E 0 + v , (56)

E 0 = E e i k α ⋅ x , E ⋅ α = 0 , α ∈ S 2 . (57)

where v satisfies the radiation Condition (8), ω > 0 is the frequency, k = 2 π / λ

is the wave number, k a ≪ 1 , a : = 1 2 m a x m diam D m , and α is a unit vector that

indicates the direction of the incident wave E 0 . Furthermore,

ϵ = ϵ 0 , μ = μ 0 in Ω ′ : = ℝ 3 \ Ω . (58)

Assume that the distribution of small bodies D m , 1 ≤ m ≤ M , in Ω satisfies the following formula

N ( Δ ) = 1 a 3 ∫ Δ N ( x ) d x [ 1 + o ( 1 ) ] , a → 0, (59)

where N ( Δ ) is the number of small bodies in Δ , Δ is an arbitrary open subset of Ω , and N ( x ) is the distribution function

N ( x ) ≥ 0, N ( x ) ∈ C ( Ω ) . (60)

Note that E solves this equation

∇ × ∇ × E = k 2 E , k 2 = ω 2 ϵ μ , (61)

if μ = const. Once we have E, then H can be found from this relation

H = ∇ × E i ω μ . (62)

From (62) and (61), one can get (54). Thus, we need to find only E which satisfies the boundary condition (55). It was proved in [

E ( x ) = E 0 ( x ) + ∑ m = 1 M ∇ × ∫ S m g ( x , t ) J m ( t ) d t , (63)

where J m are unknown continuous functions that can be found from the boundary condition. Let

Q m : = ∫ S m J m ( t ) d t . (64)

When a → 0 , the asymptotic solution for the electric field is given by

E ( x ) = E 0 ( x ) + ∑ m = 1 M [ ∇ g ( x , x m ) , Q m ] , a → 0. (65)

Therefore, instead of finding J m ( t ) , ∀ t ∈ S ,1 ≤ m ≤ M , to get the solution E, one can just find Q m . This allows one to solve the EM scattering problem with a very large number of small bodies which is impossible to do before. The analytic formula for Q m can be derived by using formula (28) and replacing E 0 in this formula by the effective field E e ( x m ) acting on the m-th body

Q m = − | D m | ( I + Γ ) − 1 ∇ × E e ( x m ) , 1 ≤ m ≤ M , x m ∈ D m , (66)

where the effective field acting on the m-th body is defined as

E e ( x m ) = E 0 ( x m ) + ∑ j ≠ m M [ ∇ g ( x m , x j ) , Q j ] , 1 ≤ m ≤ M . (67)

When a → 0 , the effective field E e ( x ) is asymptotically equal to the field E ( x ) in (65) as proved in [

Let E e m : = E e ( x m ) , where x m is a point in D m . From (66), and (67), one gets

E e m = E 0 m − ∑ j ≠ m M [ ∇ g ( x m , x j ) , ( I + Γ ) − 1 ∇ × E e j ] | D j | , 1 ≤ m ≤ M . (68)

For finding the solution to EM wave scattering in the case of many small perfectly conducting bodies, we need to find E e m in (68). Apply the operator ( I + Γ ) − 1 ∇ × to both sides of (68) and let A m : = ( I + Γ ) − 1 ∇ × E e m . Then

A m = A 0 m − ( I + Γ ) − 1 ∑ j ≠ m M | D j | ( ∇ x × [ ∇ g ( x , x j ) , A j ] ) | x = x m , 1 ≤ m ≤ M , (69)

Solving this system yields A m , for 1 ≤ m ≤ M . Then E can be computed by

E ( x ) = E 0 ( x ) + ∑ m = 1 M [ ∇ g ( x , x m ) , Q m ] , (70)

where

Q m = − | D m | A m , 1 ≤ m ≤ M . (71)

Equation (69) can be rewritten as follows

A m = A 0 m − ∑ j ≠ m M τ [ k 2 g ( x m , x j ) A j + ( A j ⋅ ∇ x ) ∇ g ( x , x j ) | x = x m ] | D j | , (72)

where 1 ≤ m ≤ M , τ : = ( I + Γ ) − 1 , and A m are vectors in ℝ 3 .

Let A i : = ( X i , Y i , Z i ) then one can rewrite the system (72) as

X i + ∑ j ≠ i M a i j X j + b i j Y j + c i j Z j = F x ( i ) , (73)

Y i + ∑ j ≠ i M a ′ i j X j + b ′ i j Y j + c ′ i j Z j = F y ( i ) , (74)

Z i + ∑ j ≠ i M a ″ i j X j + b ″ i j Y j + c ″ i j Z j = F z ( i ) , (75)

in which by the subscripts x , y , z the corresponding coordinates are denoted, e.g. F ( i ) = ( F x , F y , F z ) ( i ) , where F ( i ) : = A 0 i and

a i j : = [ k 2 g ( i , j ) + ∂ x ∇ g ( i , j ) x ] | D j | τ ( 1,1 ) , (76)

b i j : = ∂ y ∇ g ( i , j ) x | D j | τ ( 1,1 ) , (77)

c i j : = ∂ z ∇ g ( i , j ) x | D j | τ ( 1,1 ) , (78)

for i ≠ j , here τ ( 1,1 ) is the entry (1,1) of matrix τ in (72); when i = j : a i i = 1 , b i i = 0 , and c i i = 0 ;

a ′ i j : = ∂ x ∇ g ( i , j ) y | D j | τ ( 2 , 2 ) , (79)

b ′ i j : = [ k 2 g ( i , j ) + ∂ y ∇ g ( i , j ) y ] | D j | τ ( 2 , 2 ) , (80)

c ′ i j : = ∂ z ∇ g ( i , j ) y | D j | τ ( 2 , 2 ) , (81)

for i ≠ j ; when i = j : a ′ i i = 0 , b ′ i i = 1 , and c ′ i i = 0 ;

a ″ i j : = ∂ x ∇ g ( i , j ) z | D j | τ ( 3 , 3 ) , (82)

b ″ i j : = ∂ y ∇ g ( i , j ) z | D j | τ ( 3 , 3 ) , (83)

c ″ i j : = [ k 2 g ( i , j ) + ∂ z ∇ g ( i , j ) z ] | D j | τ ( 3 , 3 ) , (84)

for i ≠ j ; when i = j : a ′ i i = 0 , b ′ i i = 0 , and c ′ i i = 1 .

The error of the solution to the EM wave scattering problem by many small perfectly conducting bodies can be estimated as follows. From the solution E of the electromagnetic scattering problem by many small bodies given in (63)

E ( x ) = E 0 ( x ) + ∑ m = 1 M ∇ × ∫ S m g ( x , t ) J m ( t ) d t , (85)

we can rewrite it as

E ( x ) = E 0 ( x ) + ∑ m = 1 M [ ∇ g ( x , x m ) , Q m ] + ∑ m = 1 M ∇ × ∫ S m [ g ( x , t ) − g ( x , x m ) ] J m ( t ) d t . (86)

Comparing this with the asymptotic formula for E when a → 0 given in (65)

E ( x ) = E 0 ( x ) + ∑ m = 1 M [ ∇ g ( x , x m ) , Q m ] , (87)

we have the error of this asymptotic formula is

Error = | ∑ m = 1 M ∇ × ∫ S m [ g ( x , t ) − g ( x , x m ) ] J m ( t ) d t | (88)

~ 1 4π ( a k 2 d + a k d 2 + a d 3 ) ∑ m = 1 M | Q m | , (89)

where d = min m | x − x m | and

Q m = − | D m | ( I + Γ ) − 1 ∇ × E e ( x m ) , 1 ≤ m ≤ M , x m ∈ D m , a → 0 , (90)

because

| ∇ [ g ( x , t ) − g ( x , x m ) ] | = O ( a k 2 d + a k d 2 + a d 3 ) , a = m a x m | t − x m | . (91)

To illustrate the idea of the numerical method, we use the following physical parameters to solve the EM wave scattering problem by one small perfectly conducting sphere, i.e. solving the linear system (36)-(38)

・ Speed of wave, c = ( 3.0 E + 10 ) cm / sec .

・ Frequency, ω = ( 5.0 E + 14 ) Hz .

・ Wave number, k = ( 1.05 E + 05 ) cm − 1 .

・ Wave length, λ = ( 6.00 E − 05 ) cm .

・ Direction of incident plane wave, α = ( 0 , 1 , 0 ) .

・ Magnetic permeability, μ = 1 .

・ Vector E = ( 1,0,0 ) .

・ Incident field vector, E 0 : E 0 ( x ) = E e i k α ⋅ x .

・ The body is a sphere of radius a, centered at the origin.

We use GMRES iterative method, see [

Γ p q ( t ) : = ∫ S ∂ g ( s , t ) ∂ s p N q ( s ) d s , 1 ≤ p , q ≤ 3, (92)

where

N = ( c o s θ s i n ϕ , s i n θ s i n ϕ , c o s ϕ ) (93)

and

∂ g ( s , t ) ∂ s p ≃ ∂ g 0 ( s , t ) ∂ s p = − s p − t p 4 π | s − t | 3 , g 0 ( s , t ) : = 1 4 π | s − t | . (94)

We choose a coordinate system centered at the center of the sphere such that t = ( 0 , 0 , a ) and s = a N . Then

Γ p q ( t ) : = − a 2 4 π ∫ 0 2π d θ ∫ 0 π d ϕ sin ϕ ( s p − t p ) N q a 3 8 sin 3 ϕ 2 , 1 ≤ p , q ≤ 3. (95)

When

p = q = 1 : Γ 11 ( t ) = − 1 / 3 , (96)

p = q = 2 : Γ 22 ( t ) = − 1 / 3 , (97)

p = q = 3 : Γ 33 ( t ) = 1 / 6 , (98)

p ≠ q : Γ p q ( t ) = 0. (99)

Therefore, matrix Γ is

Γ ≃ [ − 1 / 3 0 0 0 − 1 / 3 0 0 0 1 / 6 ] (100)

For example,

P = 766 , a = 1.0 E − 09 | |||
---|---|---|---|

1.0E−21 * | |||

Q e | 0.0000 + 0.0000i | 0.0000 + 0.0000i | 0.0000 + 0.3925i |

Q a | 0.0000 + 0.0000i | 0.0000 + 0.0000i | 0.0000 + 0.3760i |

| x − x 1 | | E e ( x ) | ||
---|---|---|---|

1.73E−08 | 1.0000 + 0.0010i | 0.0001 + 0.0000i | 0.0004 + 0.0000i |

1.73E−07 | 0.9999 + 0.0105i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

1.73E−06 | 0.9945 + 0.1045i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

| x − x 1 | | E a ( x ) | ||
---|---|---|---|

1.73E−08 | 1.0000 + 0.0010i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

1.73E−07 | 0.9999 + 0.0105i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

1.73E−06 | 0.9945 + 0.1045i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

In this case, we also verify the following things:

1) Is J tangential to S?

In fact, this vector J is tangential to the surface S of the body, J ⋅ N s = O ( 10 − 14 ) .

2) How accurate is the asymptotic formula (28) for Q?

We check the accuracy of the asymptotic formula for Q in (28) by comparing it with the exact formula (14), see Section 2.2, and the relative error is 4.21 E − 02 . The more collocation points used, the little this relative error is.

3) How accurate is the asymptotic formula (19) for E?

The accuracy of the asymptotic formula for E in (19) can be checked by comparing it with the exact formula (9) at several points x outside of the body, | x − x 1 | ≫ a where x 1 is the center of the body, see the error analysis in Section 2.2. The relative errors are given in

Furthermore, the numerical results also depend on the number of collocation points used. The more collocation points used, the more accurate the results is.

In this section, we consider the EM wave scattering problem by a small perfectly conducting ellipsoid body. The method for solving the problem in this setting is the same as that of Section 2.1 except that one needs to recompute the unit normal vector N.

To get the solution of this problem, one can follow the steps in Section 2.1. In particular, one needs to solve the linear system (36)-(38).

To illustrate the idea, we use the same physical parameters as described in Section 4.1, except that the body now is an ellipsoid. Let S be its smooth surface. The way we partition S into many subdomains S i j is the same as the way we partition a spherical body as described in Section 2.1. Then, the position of the collocation point in each subdomain S i j is defined by

( x , y , z ) i j = ( a c o s θ i s i n ϕ j , b s i n θ i s i n ϕ j , c c o s ϕ j ) , (101)

| x − x 1 | | E e vs E a |
---|---|

1.73E−08 | 4.67E−04 |

1.73E−07 | 4.67E−07 |

1.73E−06 | 4.70E−10 |

P = 1386 , | x − x 1 | = 1.73 E − 05 | ||||
---|---|---|---|---|

a | 1.00E−07 | 1.00E−08 | 1.00E−09 | 1.00E−10 |

E e vs E a | 1.08E−06 | 1.08E−09 | 1.08E−12 | 1.12E−15 |

Q e vs Q a | 1.96E−02 | 1.96E−02 | 1.96E−02 | 1.89E−02 |

where a, b and c are the lengths of the semi-principal axes of the ellipsoid. The outward-pointing normal vector n to S at this point is

n i j = n ( θ i , ϕ j ) = 2 ( cos θ i sin ϕ j a , sin θ i sin ϕ j b , cos ϕ j c ) , (102)

and the corresponding unit normal vector N is

N i j = N ( θ i , ϕ j ) = n i j / | n i j | . (103)

For example,

As for the case of one body, we need to verify the following things:

a) Is J tangential to S?

In fact, this vector J is tangential to the surface S of the body, J ⋅ N s = O ( 10 − 13 ) .

b) Are Q and J correct? We check the relative error described in Section 2.2,

Error = | Q + Γ Q − R H S | | R H S | = 14 % . The more collocation points used, the smaller

this error is, for example, with P = 1762 collocation points, this error is only 3.6%.

c) How accurate is the asymptotic formula (19) for E?

The accuracy of the asymptotic formula for E in (19) can be checked by comparing it with the exact formula (9) at several points x outside of the body, | x − x 1 | ≫ m a x ( a , b , c ) where x 1 is the center of the body. The relative errors

| x − x 1 | | E e ( x ) | ||
---|---|---|---|

1.01E−07 | 0.9998 + 0.0010i | −0.0000 + 0.0000i | −0.0000 − 0.0000i |

1.01E−06 | 0.9999 + 0.0105i | −0.0000 + 0.0000i | −0.0000 − 0.0000i |

1.01E−05 | 0.9945 + 0.1045i | −0.0000 + 0.0000i | −0.0000 − 0.0000i |

| x − x 1 | | E a ( x ) | ||
---|---|---|---|

1.01E−07 | 1.0000 + 0.0010i | −0.0000 + 0.0000i | 0.0000 − 0.0000i |

1.01E−06 | 0.9999 + 0.0105i | −0.0000 + 0.0000i | 0.0000 − 0.0000i |

1.01E−05 | 0.9945 + 0.1045i | −0.0000 + 0.0000i | 0.0000 − 0.0000i |

are given in

In this section, we consider the EM wave scattering problem by a small perfectly conducting cubic body. Again, the method for solving the problem in this setting is the same as that of Section 2.1 except that one needs to recompute the unit normal vector N.

One can follow the steps outlined in Section 2.1 to solve this problem. That means, one needs to solve the linear system (36)-(38).

For illustration purpose, we use the same physical parameters as described in Section 4.1, except that the body now is a cube. Suppose the cube is placed in the first octant where the origin is one of its vertices, one can use the standard unit vectors in ℝ 3 as the unit normal vectors to the surfaces of the cube.

For example,

As before, for the case of one body, we need to verify the following things:

a) Is J tangential to S?

| x − x 1 | | E e vs E a |
---|---|

1.01E−07 | 1.73E−04 |

1.01E−06 | 1.73E−07 |

1.01E−05 | 1.73E−10 |

P = 1052 , | x − x 1 | = 1.73 E − 07 | ||||
---|---|---|---|---|

a | 1.00E−07 | 1.00E−08 | 1.00E−09 | 1.00E−10 |

b | 1.00E−08 | 1.00E−09 | 1.00E−10 | 1.00E−11 |

c | 1.00E−08 | 1.00E−09 | 1.00E−10 | 1.00E−11 |

E e vs E a | 2.65E−02 | 2.76E−05 | 2.76E−08 | 2.76E−11 |

| x − x 1 | | E e ( x ) | ||
---|---|---|---|

1.73E−04 | −0.5000 − 0.8660i | −0.0000 + 0.0000i | −0.0000 + 0.0000i |

1.73E−05 | 0.5000 + 0.8660i | 0.0000 + 0.0000i | 0.0000 + 0.0000i |

1.73E−06 | 0.9945 + 0.1045i | 0.0006 + 0.0000i | 0.0006 + 0.0000i |

| x − x 1 | | E a ( x ) | ||
---|---|---|---|

1.73E−04 | −0.5000 − 0.8660i | 0.0000 − 0.0000i | 0.0000 + 0.0000i |

1.73E−05 | 0.5000 + 0.8660i | −0.0000 + 0.0000i | −0.0000 − 0.0000i |

1.73E−06 | 0.9945 + 0.1045i | −0.0000 + 0.0000i | −0.0000 − 0.0000i |

In fact, this vector J is tangential to the surface S of the body, J ⋅ N s = O ( 10 − 13 ) .

b) How accurate is the asymptotic formula (28) for Q?

We check the relative error described in Section 2.2,

Error = | Q + Γ Q − R H S | | R H S | = 1.13 % .

c) How accurate is the asymptotic formula (19) for E?

The accuracy of the asymptotic formula for E in (19) can be checked by comparing it with the exact formula (9) at several points x outside of the body, | x − x 1 | ≫ a where x 1 is the center of the body. The relative errors are given in

| x − x 1 | | E e vs E a |
---|---|

1.73E−03 | 1.19E−08 |

1.73E−04 | 1.19E−07 |

1.73E−05 | 1.52E−06 |

1.73E−06 | 8.64E−04 |

M = 600 , | x − x 1 | = 1.73 E − 06 | |||
---|---|---|---|

a | 1.00E−07 | 1.00E−08 | 1.00E−09 |

E e vs E a | 8.64E−04 | 6.49E−07 | 6.32E−10 |

a. From this table, we can see that the smaller the side of the cube is, compared to the distance from the point of interest to the center of the body, the more accurate the asymptotic formula of E is.

To illustrate the idea, consider a domain Ω as a unit cube placed in the first octant such that the origin is one of its vertices. This domain Ω contains M small bodies. Suppose these small bodies are particles. We use GMRES iterative method, see [

・ Speed of wave, c = ( 3.0 E + 10 ) cm / sec .

・ Frequency, ω = ( 5.0 E + 14 ) Hz .

・ Wave number, k = ( 1.05 E + 05 ) cm − 1 .

・ Wave length, λ = ( 6.00 E − 05 ) cm .

・ Direction of incident plane wave, α = ( 0,1,0 ) .

・ Magnetic permeability, μ = 1 .

・ Volume of the domain Ω that contains all the particles, | Ω | = 1 cm^{3}.

・ The distance between two neighboring particles, d = ( 1.00 E − 07 ) cm .

・ Vector E = ( 1,0,0 ) .

・ Vector A 0 : A 0 m : = ( I + Γ ) − 1 ∇ × E 0 ( x ) | x = x m = ( I + Γ ) − 1 ∇ × E e i k α ⋅ x | x = x m .

Note that the distance d satisfies the assumption d ≪ λ . The radius a of the particles is chosen variously so that it satisfies the assumption k a ≪ 1 . For illustration purpose, the problem of EM wave scattering by many small perfectly conducting bodies is solved with M = 27 and 1000 particles.

For example,

M = 27 , d = 1.0 E − 07 , a = 1.0 E − 09 | ||
---|---|---|

1.00E+00+1.01E−14i | 5.69E−17−1.01E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.19E−14i | 0.00E+00+0.00E+00i | 0.00E+00+0.00E+00i |

1.00E+00+1.01E−14i | −5.69E−17+1.01E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.05E−02i | 1.24E−16−1.19E−14i | −1.36E−29−5.20E−36i |

1.00E+00+1.05E−02i | 0.00E+00+0.00E+00i | −4.80E−30−5.20E−36i |

1.00E+00+1.05E−02i | −1.24E−16+1.19E−14i | −1.22E−30−5.20E−36i |

1.00E+00+2.09E−02i | 1.54E−16−1.01E−14i | −3.40E−30−1.04E−35i |

1.00E+00+2.09E−02i | 0.00E+00+0.00E+00i | −2.43E−30−1.04E−35i |

1.00E+00+2.09E−02i | −1.54E−16+1.01E−14i | −1.20E−30−1.04E−35i |

1.00E+00+1.19E−14i | 6.63E−17−1.19E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.40E−14i | 4.80E−30+0.00E+00i | 0.00E+00+0.00E+00i |

1.00E+00+1.19E−14i | −6.63E−17+1.19E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.05E−02i | 1.47E−16−1.40E−14i | −4.80E−30−5.20E−36i |

1.00E+00+1.05E−02i | 2.61E−30+0.00E+00i | −2.61E−30−5.20E−36i |

1.00E+00+1.05E−02i | −1.47E−16+1.40E−14i | −9.24E−31−5.20E−36i |

1.00E+00+2.09E−02i | 1.82E−16−1.19E−14i | −2.43E−30−1.04E−35i |

1.00E+00+2.09E−02i | 9.24E−31+0.00E+00i | −1.85E−30−1.04E−35i |

1.00E+00+2.09E−02i | −1.82E−16+1.19E−14i | −1.01E−30−1.04E−35i |

1.00E+00+1.01E−14i | 5.69E−17−1.01E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.19E−14i | 2.43E−30+0.00E+00i | 0.00E+00+0.00E+00i |

1.00E+00+1.01E−14i | −5.69E−17+1.01E−14i | 0.00E+00+0.00E+00i |

1.00E+00+1.05E−02i | 1.24E−16−1.19E−14i | −1.22E−30−5.20E−36i |

1.00E+00+1.05E−02i | 1.85E−30+0.00E+00i | −9.24E−31−5.20E−36i |

1.00E+00+1.05E−02i | −1.24E−16+1.19E−14i | −5.03E−31−5.20E−36i |

1.00E+00+2.09E−02i | 1.54E−16−1.01E−14i | −1.20E−30−1.04E−35i |

1.00E+00+2.09E−02i | 9.98E−31+0.00E+00i | −1.01E−30−1.04E−35i |

1.00E+00+2.09E−02i | −1.54E−16+1.01E−14i | −6.54E−31−1.04E−35i |

the point i in the cube. The norm of this asymptotic solution E is 5.20 E + 00 and the error of the solution is 8.16 E − 10 . This error is computed using (89).

The slope of the curve in

M = 27, d = 1.0E−07 | ||||
---|---|---|---|---|

a | 1.00E−08 | 1.00E−09 | 1.00E−10 | 1.00E−11 |

a/d | 1.00E−01 | 1.00E−02 | 1.00E−03 | 1.00E−04 |

Norm of E | 5.20E+00 | 5.20E+00 | 5.20E+00 | 5.20E+00 |

Error of E | 8.16E−06 | 8.16E−10 | 8.16E−14 | 8.16E−18 |

M = 1000 , d = 1.0 E − 07 | ||||
---|---|---|---|---|

a | 1.00E−08 | 1.00E−09 | 1.00E−10 | 1.00E−11 |

a/d | 1.00E−01 | 1.00E−02 | 1.00E−03 | 1.00E−04 |

Norm of E | 3.16E+01 | 3.16E+01 | 3.16E+01 | 3.16E+01 |

Error of E | 3.02E−04 | 3.02E−08 | 3.02E−12 | 3.02E−16 |

M = 1000 particles, when the distance between neighboring particles is d = ( 1.0 E − 07 ) cm , and with different radius a. From these table and figure, one can see that the relative error of the asymptotic solution in this case is also very small, less than 3.02 E − 04 , when the ratio a / d < 1.0 E − 01 . In this case, the error of the asymptotic E is greater than that of the previous case when M = 27 . However, this time, the error is also decreasing quickly and linearly when the ratio a/d decreases from 1.0 E − 01 to 1.0 E − 04 . Therefore, the asymptotic formula (65) for the solution E is applicable when a ≪ d .

In this paper, we present a numerical method for solving the EM wave scattering by one and many small perfectly conducting bodies. One of the advantages of this method is that it is relatively easy to implement. Furthermore, one can get an asymptotically exact solution to the problem when the characteristic size of the bodies tends to zero. To illustrate the applicability and efficiency of the method, we use it to solve the EM wave scattering problem by one and many small perfectly conducting bodies. Numerical results of these experiments are presented and error analysis of the asymptotic solutions for the case of one and many bodies are also discussed. For the case of one small body, one can always find the exact solution using the described method. For the case of many small bodies, the accuracy of our method is high if a ≪ d ≪ λ .

The problem of EM wave scattering is much harder to treat, compared to scalar wave scattering [

Tran, N.T. (2017) Numerical Method for Solving Electromagnetic Wave Scattering by One and Many Small Perfectly Conducting Bodies. American Journal of Computational Mathematics, 7, 413-434. https://doi.org/10.4236/ajcm.2017.74030