Class 5 in Electrodynamics

Class 5 in Electrodynamics
Based on course by Yuri Lyubarsky and Edited By Avry Shirakov
Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel
This exercise pool is intended for an undergraduate course in “Electrodynamics 1”.
Boundary Value problem with cylindrical symmetry
In cylindrical coordinates the Laplace eq. solution is of the form Φ(ρ, φ, z) = R(ρ)Q(φ)Z(z) (separation of variables
method), while the Laplace eq. in cylindrical coordinates is:
∂ 2 φ 1 ∂φ
1 ∂2φ ∂2φ
+
+ 2 =0
+
∂ρ2
ρ ∂ρ
ρ2 ∂φ2
∂z
this leads to following ordinary differential eq.
∂2Z
− κ2 Z = 0 ;( Z(z) = e±κz with solutions of the form cosh(κz), sinh(κz))
∂z 2
∂2Q
+ κ2 Q = 0 ;( Q(φ) = e±iνφ with solutions of the form cos(νφ), sin(νφ))
∂z 2
∂ 2 R 1 ∂R
ν2
2
+
+ κ − 2 R=0
∂ρ2
ρ ∂ρ
ρ
Assuming κ real and positive and defining x = κρ, we rewrite the eq. above
∂2R
1 ∂R
ν2
+
+ 1− 2 R=0
∂x2
x ∂x
x
This is A Bessel eq., and its solutions are Bessel functions of the ν order. If we assume a solution of a power series of
the form
∞
X
α
R(x) = x
aj xj
j=0
1
a2j−2 for j = 1, 2, 3... (all odd powers have vanishing coefficients). The
4j(j + α)
recursion formula can be iterated to obtain
we find that α = ±ν and a2j = −
a2j =
(−1)j Γ(α + 1)
a0
22j j!Γ(j + α + 1)
Γ is a Gamma function and for integer n is given by Γ(n) = (n − 1)!. Usually a0 is taken as a0 = [2α Γ(α + 1)]
result the solution looks like,
∞
x 2j
x ν X
(−1)j
Jν (x) =
2 j=0 j!Γ(j + ν + 1) 2
J−ν (x) =
∞
x −ν X
2
j=0
−1
as a
x 2j
(−1)j
j!Γ(j − ν + 1) 2
For an integer ν these solutions are linearly dependant, and moreover there is a relation connecting between them
J−ν (x) = (−1)ν Jν (x)
So we need another solution which is linearly independent, this solution is called Neumann function;
Nν (x) =
Jν (x) cos(νπ) − J−ν (x)
sin(νπ)
The set [Jν (x), Nν (x)] is linearly independent, while Jν (x) converges for all finite values of x and Nν (x) not(it involves
ln(x)).
2
FIG. 1: Bessel Jν (x) functions [Wolfram]