A Semi-infinite right circular cylinder problem A semi-infinite right circular cylinder whose axis lie along the $z$ axis, has its base on the $x$-$y$ plane.The base is maintained at a constant potential $V_0$  and the side of the cylinder is maintained at zero potential. Write down the Laplace equation to determine the potential inside the cylinder. Separate the PDE into a set of ODEs and write down the boundary conditions.
Please help....
 A: Clearly, there is no angular dependence, so the potential will be a function of $r$ and $z$.  We may separate variables by letting the potential $V(r,z)= R(r)Z(z)$  The equation for the potential is then
$$\frac{1}{rR} \frac{dR}{dr} + \frac{1}{R} \frac{d^2 R}{d r^2} + \frac{1}{Z} \frac{d^2 Z}{d z^2} = 0$$
We deal with the $R$ equation first:
$$\frac{1}{rR} \frac{dR}{dr} + \frac{1}{R} \frac{d^2 R}{d r^2} = -\lambda^2$$
where $\lambda$ is some separation constant.  Then 
$$r R'' + R' + \lambda^2 r R = 0$$
where, for $z>0$, $R(a) = 0$.  The solution to this equation is a Bessel function of the first kind, of order zero: $R(r) = A \, J_0(\lambda r)$.  By requiring that the potential be zero at $r=a$ leads to $\lambda = j_{0,n}/a$, where $j_{0,n}$ is the $n$th zero of the Bessel function.  
Because we have a semi-infinite cylinder with a boundary condition at $z=0$, then the solution for $Z(z)$ must take the form
$$Z(z) = B_n \, e^{- j_{0,n} z/a}$$
The general solution is
$$V(r,z) = \sum_{n=1}^{\infty} B_n \, J_0\left ( j_{0,n} \frac{r}{a} \right ) \, e^{- j_{0,n} z/a}$$
Because $V(r,0) = V_0$, and the fact that the $J_0\left ( j_{0,n} \frac{r}{a} \right )$ form an orthogonal basis set over $L^2[0,a]$, we may determine the $B_n$ from
$$B_n = \frac{\displaystyle V_0 \int_0^a dr \, r \, J_0\left ( j_{0,n} \frac{r}{a} \right )}{\displaystyle\int_0^a dr \, r \, J_0\left ( j_{0,n} \frac{r}{a} \right )^2} = \frac{2 V_0}{j_{0,n} J_1(j_{0,n})} $$
