Page 472 - Electromagnetics

By the Fourier integral theorem we have
2 2 ˜

(∇ + k )G(r|r ; ω) =−δ(r − r ). (A.46)
This is known as the Helmholtz equation. Here
1
2
k = ω − j2ω (A.47)
v
is called the wavenumber.
˜
To solve the Helmholtz equation we write G in terms of a 3-dimensional inverse Fourier
transform. Substitution of
1 ∞
˜ ˜ r jk·r 3
G(r|r ; ω) = G (k|r ; ω)e d k,
(2π) 3
−∞
1 ∞
jk·(r−r ) 3
δ(r − r ) = e d k,
(2π) 3
−∞
into (A.46) gives
1 ∞ 2 ˜ r jk·r 2 ˜ r jk·r jk·(r−r ) 3

∇ G (k|r ; ω)e + k G (k|r ; ω)e + e d k = 0.
(2π) 3
−∞
Here
k = ˆ xk x + ˆ yk y + ˆ zk z
2
2
2
2
2
with |k| = k + k + k = K . Carrying out the derivatives and invoking the Fourier
x y z
integral theorem we have
2 ˜ r
2
(K − k )G (k|r ; ω) = e − jk·r .

˜
Solving for G and substituting it into the inverse transform relation we have
1 ∞ e jk·(r−r )
˜ 3
G(r|r ; ω) = d k. (A.48)
(2π) 3 −∞ (K − k)(K + k)
To compute the inverse transform integral in (A.48) we write the 3-D transform variable
in spherical coordinates:
3
2

k · (r − r ) = KR cos θ, d k = K sin θ dK dθ dφ,
where R =|r − r | and θ is the angle between k and r − r . Hence (A.48) becomes

2
1 ∞ K dK 2π π
˜ jK R cos θ
G(r|r ; ω) = dφ e sin θ dθ
(2π) 3 0 (K − k)(K + k) 0 0
2 ∞ K sin(KR)
= dK,
2
(2π) R 0 (K − k)(K + k)
or, equivalently,
1 ∞ e jK R
˜
G(r|r ; ω) = KdK −
2 jR(2π) 2 −∞ (K − k)(K + k)
1 ∞ e − jkR
− KdK.
2 jR(2π) 2 −∞ (K − k)(K + k)
© 2001 by CRC Press LLC