Page 505 - Electromagnetics

1 ν
x
J ν (x) → ,
ν! 2
ν

(ν − 1)! 2
N ν (x) →− .
π x
Because J ν (x) and N ν (x) oscillate for large argument, they can represent standing waves
along the radial direction. However, N ν (x) is unbounded for small x and is inappropriate
for regions containing the origin. The Hankel functions become complex exponentials for
large argument, hence represent traveling waves. Finally, K ν (x) is unbounded for small x
and cannot be used for regions containing the origin, while I ν (x) increases exponentially
for large x and cannot be used for unbounded regions.

Examples. Consider the boundary value problem for Laplace’s equation
2
∇ V (ρ, φ) = 0 (A.126)
in the region

0 ≤ ρ ≤∞, 0 ≤ φ ≤ φ 0 , −∞ < z < ∞,
where the boundary conditions are

V (ρ, 0) = 0, V (ρ, φ 0 ) = V 0 .

Since there is no z-dependence we let k z = 0 in (A.120) and choose a z = 0. Then
2
2
2
k = k − k = 0 since k = 0. There are two possible solutions, depending on whether
z
c
k φ is zero. First let us try k φ = 0. Using (A.123) and (A.125) we have
V (ρ, φ) = [A φ sin(k φ φ) + B φ cos(k φ φ)][a ρ ρ −k φ + b ρ ρ ]. (A.127)
k φ
Assuming k φ > 0 we must have b ρ = 0 to keep the solution finite. The condition
V (ρ, 0) = 0 requires B φ = 0.Thus
V (ρ, φ) = A φ sin(k φ φ)ρ −k φ .

Our final boundary condition requires
V (ρ, φ 0 ) = V 0 = A φ sin(k φ φ 0 )ρ −k φ .

Because this cannot hold for all ρ, we must resort to k φ = 0 and

V (ρ, φ) = (a φ φ + b φ )(a ρ ln ρ + b ρ ). (A.128)
Proper behavior as ρ →∞ dictates that a ρ = 0. V (ρ, 0) = 0 requires b φ = 0.Thus
V (ρ, φ) = V (φ) = b φ φ. The constant b φ is found from the remaining boundary condition:
V (φ 0 ) = V 0 = b φ φ 0 so that b φ = V 0 /φ 0 . The final solution is

V (φ) = V 0 φ/φ 0 .

It is worthwhile to specialize this to φ 0 = π/2 and compare with the solution to the same
problem found earlier using rectangular coordinates. With a = 0 in (A.116) we have

2 ∞ −k y x sin k y y
V (x, y) = e dk y .
π 0 k y

© 2001 by CRC Press LLC