Laplace equation outside finite cylinder I am trying to find a solution to Laplace's equation outside a finite cylinder of radius $a$ and height $h$ with the boundary condition that $u=\frac{c}{\rho}$. where $c$ is a constant and $\rho$ is the radial coordinate. 
So we are trying to solve $\nabla^2 u=0$ everywhere. 
Question 1: Is it safe to apply separation of variables to solve this? I'm not entirely sure that would be the case due to the non-trivial geometry and BCs
Question 2: Assuming we can carry out separation of variables we write $u=R(\rho)Z(z)$ and we realize that there will be no $\phi$ component because of the symmetry. Hence we get 
$$\frac{d^2Z}{dz^2}-k^2Z=0$$
$$\frac{d^2R}{d\rho^2}+\frac{1}{\rho}\frac{dR}{d\rho}+\left(k^2-\frac{m^2}{\rho^2}\right)R=0$$
Does anyone have any insight as to how this could be solved?
 A: It's a linear homogeneous PDE, so separation of variables should work here. 
Using the Ansatz $u(\rho,z)=R(\rho)(Z(z)$, cylindrical coordinates for $\nabla^2u$ and ignoring the obsolete term in $\phi$ we get:
$$\frac{u_{\rho}}{\rho}+u_{\rho \rho}+u_{zz}=0$$
So that:
$$\frac{R'}{\rho R}+\frac{R''}{R}+\frac{Z''}{Z}=0$$
Separation:
$$\frac{R'}{\rho R}+\frac{R''}{R}=-\frac{Z''}{Z}=-k^2$$
where $k$ is a Real number.
The ODEs:
$$Z''-k^2Z=0$$
$$\rho R''+R'+k^2 \rho R=0$$ 
So I'm not sure where you get your second ODE and $m$ from? Two variables ($z$ and $\rho$) require only one separating constant, not two.
The second ODE is a Sturm Liouville equation and it solves to:
$$R(\rho)=C_1J_0(k \rho)+C_2Y_0(k \rho)$$
A: Because your cylinder has finite height, you cannot get very far with separation of variables.  You can do it of course (see another answer) but there's no reason to believe your potential field is everywhere cylindrical, and as you suspect you're gonna have trouble with the boundary conditions, especially on the "caps" of your cylinder.   
A: @ZeroTheHero: Of course the solution has cylindrical symmetry -- provided the boundary condition at the cylinder is symmetric. For example, consider a charged conducting cylinder. The potential is constant on  the cylinder (hence symmetric) and $u =Q/r$ a long distances, with $Q$ the charge on the cylinder.
@Gert has it right, at least for any problem on which the boundary condition is independent of the azimuthal angle. This is not required: the symmetry of the solution is irrelevant (in spite of @ZeroTheHero); rather than symmetry the choice of cylindrical coordinates is so that the boundaries correspond to linear conditions on one variable, eg, the side of the cylinder corresponds to $\rho=a$. If the boundary condition does depend on \phi then you use $J_m$ and $Y_m$ summed over all m, not just $m=0$. This is standard textbook material.
But you may need to break the problem into various regions to apply boundary conditions. I f the cylinder is at  $\rho=a$ between $z=0$ and $z=h$ then I count five regions:

*

*$z\ge h$, $\rho\le a$

*$z\ge h$, $\rho\ge a$

*$z\le 0$, $\rho\le a$

*$z\le 0$, $\rho\ge a$

*$0\le z\le h$, $\rho\ge a$
and in addition to the boundary condition at the cylinder there are boundary conditions (of continuity and smoothness of electric field) at the boundary between any of these regions.
