Page 509 - Electromagnetics

converges, and the constants are
2 a ρ
C m = 2 f (ρ)J ν p νm ρ dρ
2
a J (p νm ) 0 a
ν+1
by (A.136). Here p νm is the mth root of J ν (x). An alternative form of the series uses

p , the roots of J (x), and is given by

νm
ν
∞

ρ

f (ρ) = D m J ν p νm , 0 ≤ ρ ≤ a,ν > −1.
m=1 a
In this case the expansion coefficients are found using the orthogonality relationship
p p a ν
a 2 2
νm νn 2
J ν ρ J ν ρ ρ dρ = δ mn 1 − 2 J (p ),
νm
ν
0 a a 2 p νm
and are
2 a p νm
D m = f (ρ)J ν ρ ρ dρ.
2
2
a 2 1 − ν 2 J (p ) 0 a

ν
νm
p νm
Solutions in spherical coordinates. If into Helmholtz’s equation
1 ∂ 2 ∂ψ(r,θ,φ) 1 ∂ ∂ψ(r,θ,φ)
r + sin θ +
2
2
r ∂r ∂r r sin θ ∂θ ∂θ
2
1 ∂ ψ(r,θ,φ) 2
+ + k ψ(r,θ,φ) = 0
2
2
r sin θ ∂φ 2
2
2
we put ψ(r,θ,φ) = R(r)%(θ)#(φ) and multiply through by r sin θ/ψ(r,θ,φ), we obtain
2
2
sin θ d 2 dR(r) sin θ d d%(θ) 2 2 2 1 d #(φ)
r + sin θ + k r sin θ =− .
R(r) dr dr %(θ) dθ dθ #(φ) dφ 2
Since the right side depends only on φ while the left side depends only on r and θ, both
2
sides must equal some constant µ :
2
sin θ d 2 dR(r) sin θ d d%(θ) 2 2 2 2
r + sin θ + k r sin θ = µ , (A.137)
R(r) dr dr %(θ) dθ dθ
2
d #(φ) 2
+ µ #(φ) = 0. (A.138)
dφ 2
We have thus separated off the φ-dependence. Harmonic ordinary differential equation
(A.138) has solutions

A φ sin µφ + B φ cos µφ, µ = 0,
#(φ) =
a φ φ + b φ , µ = 0.
(We could have used complex exponentials to describe #(φ), or some combination of
exponentials and trigonometric functions, but it is conventional to use only trigonometric
2
functions.) Rearranging (A.137) and dividing through by sin θ we have
1 d 2 dR(r) 2 2 1 d d%(θ) µ 2
r + k r =− sin θ + .
2
R(r) dr dr sin θ%(θ) dθ dθ sin θ
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