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Three Dimensional Boundary Value Problems in Cylindrical Coordinates

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

Ѳ in cylindrical coordinates is:

		1	∂		∂		1	∂2		∂2
Ñ2	=			s		+			+
		s	∂s		∂s		s2	∂φ2		∂z2

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,

ѲΦ(s,φ,z) = ѲS(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		∂2Z
Ñ2SQZ	=			s		S +			Q + SQ		= 0
		s	∂s		∂s		s2	∂φ2		∂z2

Now multiply the entire equation by s² / SQZ:

		s	∂		∂		1	∂2		∂2Z
Ñ2SQZ	=			s		S +			Q + s2Z		= 0
		S	∂s		∂s		Q	∂φ2		∂z2

Now there's a portion which depends only on φ. We'll set that portion equal to a separation variable, -η².

∂2Q
	= -η2Q
∂φ2

s	∂		∂		∂2Z
		s		S + s2Z		= η2
S	∂s		∂s		∂z2

We can solve the Q equation immediately - this is just sines and cosines. We'll choose to represent the solutions as Q = A_ηcos ηφ + B_η sin ηφ. The remaining equation still needs a bit of work. We'll divide through by s²,  resulting in:
 
 

1	∂		∂		η2		∂2Z
		s		S -		+ Z		= 0
sS	∂s		∂s		s2		∂z2

Now, the z portion can be set to another seperation variable, k². This leaves us with two equations:

∂2Z
	= k2Z
∂z2

1	∂		∂		η2
		s		S -		S + k2S	= 0
s	∂s		∂s		s2

We can solve the Z portion immediately - this is Z = a_ke^kz + b_ke^-kz. Now we divide the s equation through by k², then change variable s -> ks, which eliminates k from the equation entirely.

1	∂		∂		η2
		s		S -		S + S	= 0
s	∂s		∂s		s2

This is Bessel's equation. The solutions are Bessel functions, Neumann functions, and Hankel functions, and we've officially entered Graduate Student Hell.

A Worked Problem