Solving Laplace equation in Cylindrical coordinates with azimuthal symmetry?

Question

I am having some trouble solving what the potential field looks like for a finite cylinder along the $z$ direction.

Physical Scenario

Consider a cylindrical electrode with radius $R_0$ that extends along $z$ from $z = 0$ to $ z= z_0$. This electrode is at a fixed potential $\phi_0$. The electrode is an empty shell.

What have I tried

I want to solve Laplace's equation: $$\nabla ^2\phi(r,z) = 0$$

with boundary conditions: $$\phi(R_0, z) = \phi_0, z \in [0,z_0]$$

and $$\lim_{z \to \pm\infty} \phi(r,z) = 0.$$

Given the azimuthal symmetry in my configuration then Laplace equation is: $$\frac{\partial ^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{\partial ^2 \phi}{\partial z^2} = 0$$

And after separation of variables the problem reduces to: $$\frac{d^2Z}{dz^2} - k^2Z = 0$$ $$\frac{d^2R}{dr^2} + \frac{1}{r}\frac{dR}{dr} + k^2R = 0$$ where $k$ is just a constant.

The solutions would then be of the form: $$Z(z) = A\exp(kz)+B\exp(-kz)$$ $$R(r) = J_0(kr)$$

where $J_0$ is a Bessel function of first kind.

What is my Problem?

As $\phi$ vanishes when $z$ tends to infinity and minus infinity, wouldn't these conditions set both $A$ and $B$ in $Z(z)$ to zero?

I am also very confused as to how the value of $k$ is dealt with. Is the solution $Z \times R$ summed over values of $k$? How do the set of boundary conditions come into place, would they restrict $k$?

My end goal is to take advantage of the in-built C++ Bessel functions to get the solution for the potential.

EDIT

The potential $\phi_0$ is a constant on the surface of the electrode. My end goal is to simulate the electrostatic potential caused in a Penning-Malmberg Trap. As of this question I am only interested in a single electrode.

Answer 1

You are partly correct. If $\phi(z,r)$ vanishes at $z\pm \infty$ then $A=B=0$ and $\phi$ is identically zero. Furthermore, this is in conflict with the value of $\phi(z,R_0)=\phi_0$ unless $\phi_0=0$. The problem is that you have chosen the wrong set of solutions to Laplace's equation.

All solutions of Laplace oscillate in one direction and decay expoentially in the other. You want ones that oscillate along the $z$ direction (so you can match the $\phi(z,R_0) = \phi_0$ when $0<z<z_0$ boundary conditions) and decay or grow along the radial direction. In other words you need solutions like $e^{ikz}I_0(kr)$ or $e^{ikz} K_0(kr)$ where $I$ and $K$ are the modified Bessel functions. You will need to use $I_0(kr)$ for $r<R_0$ and $K_0(kr)$ for $r>R_0$.

The problem is that although the physical situation is clear, the BC's are not. You have a nice boundadry condition at $r=R_0$ for $z\in[0,z_0]$ there is no simple BC for $z\notin[0,z_0]$ Even if there were it would still be a hard problem. For example think of a case when the electrode is strip of width $w$ lying parallel in the $x,y$ plane and parallel to the x axis. On the strip you have the BC $\phi(x,y,0)=\phi_0$ for $0<y<w$ and by symmetry you have vanishing electric field off the strip so $\partial_z \phi(x,y,z)|_{z=0}=0$ for $y\notin [0,w]$. This is a mixed boundary problem (Dirichlet on the strip and Neumann offf the strip) and is hard. You, however, do not even know the analogue of the Neumann BC off your cylinder.