Page 467 - Electromagnetics

To calculate a(x, y, t) and b(x, y, t), we must use the boundary conditions (A.17) and
(A.18). To apply (A.17), we put z = 0 into (A.28) to give

a(x, y, t) + b(x, y, t) = f (x, y, t). (A.29)
Using (A.18) is a bit more complicated since we must compute ∂ψ/∂z, and z is a pa-
rameter in the limits of the integral describing ψ. To compute the derivative we apply
Leibnitz’ rule for differentiation:
d θ(α) dθ dφ θ(α) ∂ f
f (x,α) dx = f (θ(α), α) − f (φ(α), α) + dx. (A.30)
dα φ(α) dα dα φ(α) ∂α
Using this on the integral term in (A.28) we have

z z−ζ z z−ζ
∂ c t+ c c ∂ t+ c
S(x, y,ζ,τ) dτ dζ = S(x, y,ζ,τ) dτ dζ,
∂z 2 0 z−ζ 2 0 ∂z z−ζ
t−
c t− c
which is zero at z = 0.Thus
1 1

= g(x, y, t) =− a (x, y, t) + b (x, y, t)
∂ψ

∂z z=0 c c

where a = ∂a/∂t and b = ∂b/∂t. Integration gives
t
− a(x, y, t) + b(x, y, t) = c g(x, y,τ) dτ. (A.31)
−∞
Equations (A.29) and (A.31) represent two algebraic equations in the two unknown
functions a and b. The solutions are
t
2a(x, y, t) = f (x, y, t) − c g(x, y,τ) dτ,
−∞
t

2b(x, y, t) = f (x, y, t) + c g(x, y,τ) dτ.
−∞
Finally, substitution of these into (A.28) gives us the solution to the inhomogeneous wave
equation
z
c t+ z−ζ 1 z z
c
ψ(x, y, z, t) = S(x, y,ζ,τ) dτ dζ + f x, y, t − + f x, y, t + +
2 0 z−ζ 2 c c
t−
c
z
c t+ c

+ g(x, y,τ) dτ. (A.32)
2 z
t−
c
This is known as the D’Alembert solution. The terms f (x, y, t ∓ z/c) contribute to ψ
as waves propagating away from the plane z = 0 in the ±z-directions, respectively. The
integral over the forcing term S is seen to accumulate values of S over a time interval
determined by z − ζ.
The boundary conditions could have been applied while still in the temporal frequency
domain (but not the spatial frequency domain, since the spatial position z is lost). But to
do this, we would need the boundary conditions to be in the temporal frequency domain.
This is easily accomplished by transforming them to give

˜ = f (x, y,ω),
˜
ψ(x, y, z,ω)
z=0
∂
˜ = ˜ g(x, y,ω).
∂z ψ(x, y, z,ω) z=0
© 2001 by CRC Press LLC