Template:Separation of variables in cylindrical coordinates

We now separate variables, noting that since the problem has circular symmetry we can write the potential as

[math]\displaystyle{ \phi(r,\theta,z)=\zeta(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta} }[/math]

We now solve for the function [math]\displaystyle{ \rho_{n}(r) }[/math]. Using Laplace's equation in polar coordinates we obtain

[math]\displaystyle{ \frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r} \frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left( \frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0 }[/math]

where [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ k_{m} }[/math] or [math]\displaystyle{ \kappa_{m}, }[/math] depending on whether [math]\displaystyle{ r }[/math] is greater or less than [math]\displaystyle{ a }[/math]. We can convert this equation to the standard form by substituting [math]\displaystyle{ y=\mu r }[/math] (provided that [math]\displaystyle{ \mu\neq 0 }[/math]to obtain

[math]\displaystyle{ y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n} }{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0 }[/math]

The solution of this equation is a linear combination of the modified Bessel functions of order [math]\displaystyle{ n }[/math], [math]\displaystyle{ I_{n}(y) }[/math] and [math]\displaystyle{ K_{n}(y) }[/math] (Abramowitz and Stegun 1964).