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 = √ (x2 + y2)
φ = arctan(y/x)
z = z
Laplace's equation for the potential in the absence of sources is:
Ñ2Φ = 0
Ñ2 in cylindrical coordinates is:
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,
Ñ2Φ(s,φ,z) = Ñ2S(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.
|Ñ2SQZ||=||s||S +||Q + SQ||= 0|
Now multiply the entire equation by s2 / SQZ:
|Ñ2SQZ||=||s||S +||Q + s2Z||= 0|
Now there's a portion which depends only on φ. We'll set that portion equal to a separation variable, -η2.
|s||S + s2Z||= η2|
We can solve the Q equation immediately - this is just sines and cosines.
We'll choose to represent the solutions as Q = Aηcos
+ Bη sin ηφ.
The remaining equation still needs a bit of work. We'll divide through
by s2, resulting in:
|s||S -||+ Z||= 0|
Now, the z portion can be set to another seperation variable, k2. This leaves us with two equations:
|s||S -||S + k2S||= 0|
We can solve the Z portion immediately - this is Z = akekz + bke-kz. Now we divide the s equation through by k2, then change variable s -> ks, which eliminates k from the equation entirely.
|s||S -||S + S||= 0|
This is Bessel's equation. The solutions are Bessel functions, Neumann functions, and Hankel functions, and we've officially entered Graduate Student Hell.