Laplace's equation in two dimensions_clyindrical coordinates

[haidara]

Homework Statement

an infinitely long thin conducting cylindrical shell(radius R) of surface charge density

$$\sigma$$=$$\sigma_{1}$$sin(2$$\Phi$$)+$$\sigma_{2}$$cos($$\Phi$$).
what are the four boundary conditions for this problem?
using the four boundary conditions and the identification of the coefficients of sin(n$$\Phi$$)and cos(n$$\Phi$$)find the expression of the potential inside and outside the cylindrical shell.

Homework Equations

the general solution of laplace's equation in this case can be written:
V(r,$$\Phi$$)=$$\sum^{n=1}_{\infty}$$[$$A_{n}$$cos(n$$\Phi$$)+
$$B_{n}$$sin(n$$\Phi$$))r$$^{n}$$+(C$$_{}n$$cos(n$$\Phi$$)+D$$_{}n$$sin(n$$\Phi$$)r$$^{-n}$$]+A$$_{}0$$ln(r)+C$$_{}0$$.
take C$$_{0}$$=0 inside the cylinder

The Attempt at a Solution

 

[NateD]

for r=RV(R,\Phi)=\sum^{n=1}_{\infty}[A_{n}cos(n\Phi)+ B_{n}sin(n\Phi))R^{n}+(C_{}ncos(n\Phi)+D_{}nsin(n\Phi)R^{-n}]+A_{}0ln(R)the four boundary conditions are: 1. V(R,\Phi)=0 (continuity of potential)2. \frac{\partial V}{\partial r}(R,\Phi)=\frac{1}{2\pi}\int_{0}^{2\pi} \sigma(r,\Phi)d\Phi (Gauss's law)3. \frac{\partial V}{\partial \Phi}(R,\Phi)=0 (no current through the cylinders surface)4. \frac{\partial^2 V}{\partial r^2}(R,\Phi)=0(Laplace's equation)using these boundary conditions, the coefficients of sin(n\Phi) and cos(n\Phi) can be identified and the potential inside and outside the cylindrical shell can be found.inside the cylinderV(r,\Phi)=\sum^{n=1}_{\infty}[A_{n}cos(n\Phi)+B_{n}sin(n\Phi))r^{n}+(C_{}ncos(n\Phi)+D_{}nsin(n\Phi)r^{-n}]+A_{}0ln(r)outside the cylinderV(r,\Phi)=\sum^{n=1}_{\infty}[A_{n}cos(n\Phi)+B_{n}sin(n\Phi))r^{-n}+(C_{}ncos(n\Phi)+D_{}nsin(n\Phi)r^{n}]+A_{}0ln(r)