s lies in the x-y plane, an angle φ up from the x axis. The relevant equations are:
x = s cos φ
y = s sin φ z = z |
s = √ (x^{2} + y^{2})
φ = arctan(y/x) z = z |
Laplace's equation for the potential in the absence of sources is:
Ñ^{2}Φ = 0
Ñ^{2} in cylindrical coordinates is:
1 | ∂ | ∂ | 1 | ∂^{2} | ∂^{2} | |||||
Ñ^{2} | = | s | + | + | ||||||
s | ∂s | ∂s | s^{2} | ∂φ^{2} | ∂z^{2} |
We'll presume that our solution is separable, that is Φ( s, φ, z ) = S(s)Q(φ)Z(z). We can check the separability assumption later. Laplace's equation is a second order differential equation, so if we find two unique solutions we've done it. Generally speaking, if the boundary conditions are separable, there's a good chance the solution is separable. If the boundary conditions are not separable, most likely we're hosed. Now,
Ñ^{2}Φ(s,φ,z) = Ñ^{2}S(s)Q(φ)Z(z).
Q is independent of s and z, so it can be pulled through the s and z partial derivatives. Similarly S is independent of φ and z, so it can be pulled through the φ and z partial derivatives. Z is independent of s and φ, so it can be pulled through the s and φ partial derivatives.
ZQ | ∂ | ∂ | ZS | ∂^{2} | ∂^{2}Z | ||||||
Ñ^{2}SQZ | = | s | S + | Q + SQ | = 0 | ||||||
s | ∂s | ∂s | s^{2} | ∂φ^{2} | ∂z^{2} |
Now multiply the entire equation by s^{2} / SQZ:
s | ∂ | ∂ | 1 | ∂^{2} | ∂^{2}Z | ||||||
Ñ^{2}SQZ | = | s | S + | Q + s^{2}Z | = 0 | ||||||
S | ∂s | ∂s | Q | ∂φ^{2} | ∂z^{2} |
Now there's a portion which depends only on φ. We'll set that portion equal to a separation variable, -η^{2}.
∂^{2}Q | |
= -η^{2}Q | |
∂φ^{2} |
s | ∂ | ∂ | ∂^{2}Z | |||
s | S + s^{2}Z | = η^{2} | ||||
S | ∂s | ∂s | ∂z^{2} |
We can solve the Q equation immediately - this is just sines and cosines.
We'll choose to represent the solutions as Q = A_{η}cos
ηφ
+ B_{η} sin ηφ.
The remaining equation still needs a bit of work. We'll divide through
by s^{2}, resulting in:
1 | ∂ | ∂ | η^{2} | ∂^{2}Z | ||||
s | S - | + Z | = 0 | |||||
sS | ∂s | ∂s | s^{2} | ∂z^{2} |
Now, the z portion can be set to another seperation variable, k^{2}. This leaves us with two equations:
∂^{2}Z | |
= k^{2}Z | |
∂z^{2} |
1 | ∂ | ∂ | η^{2} | ||||
s | S - | S + k^{2}S | = 0 | ||||
s | ∂s | ∂s | s^{2} |
We can solve the Z portion immediately - this is Z = a_{k}e^{kz} + b_{k}e^{-kz}. Now we divide the s equation through by k^{2}, then change variable s -> ks, which eliminates k from the equation entirely.
1 | ∂ | ∂ | η^{2} | ||||
s | S - | S + S | = 0 | ||||
s | ∂s | ∂s | s^{2} |
This is Bessel's equation. The solutions are Bessel functions, Neumann functions, and Hankel functions, and we've officially entered Graduate Student Hell.