[NOTES/EM-04002] Poisson Equation in Cylindrical coordinates

Poisson equation in cylindrical coordinates
The Laplacian in cylindrical coordinates, \(\rho, \phi,z\) is given  by
\begin{equation}
 \nabla^2 = \PP{\rho} +  \frac{1}{\rho} \pp{\rho} + \frac{1}{\rho^2}\PP{\phi}  
+ \PP{z} \label{EQ01}
\end{equation}
So the Laplace equation, \(\nabla^2 \Phi=0\) , in the cylindrical coordinates takes the form
\begin{equation}
 \Big\{ \PP{\rho} +  \frac{1}{\rho} \pp{\rho} + \frac{1}{\rho^2}\PP{\phi}  
+ \PP{z}\Big\} \Phi =0. \label{EQ02}
\end{equation}
A Special Case : Problems with cylindrical symmetry
In this section Also we restrict our attention to those problems which can be solved by separation of variables in cylindrical coordinates. We consider the boundary value problems with translational symmetry along \(z\)- axis. This means that the potential \(\Phi\) is assumed to be independent of \(z\), \(\Phi\equiv\Phi(\rho,\phi)\). The problems with nontrivial \(z\) dependence require use of Bessel functions and will be taken up separately. Thus  we begin with substituting \(\Phi(\rho,\phi) = P(\rho) Q(\phi)\) in Laplace equation and divide by \(\Phi\). This gives
\begin{equation}\label{EQ03}
 \frac{1}{P(\rho)} \PP[P(\rho)]{\rho} +  \frac{1}{\rho}\frac{1}{P(\rho)}
\pp[P(\rho)]{\rho} + \frac{1}{\rho^2}\frac{1}{Q(\phi)}\PP[Q(\phi)]{\phi} =0.
\end{equation}
Multiply by \(\rho^2\), we get
\begin{equation}\label{EQ04}
 \frac{\rho^2}{P(\rho)} \PP[P(\rho)]{\rho} +  \rho\frac{1}{P(\rho)}
\pp[P(\rho)]{\rho} + \frac{1}{Q(\phi)}\PP[Q(\phi)]{\phi} =0.
\end{equation}
The first two terms in the above equation depend on \(\rho\) only and the last term depends on \(\phi\) alone, hence each  of these must separately a constant giving
\begin{eqnarray}\label{EQ05}
 \frac{\rho^2}{P(\rho)} \PP[P(\rho)]{\rho} +  \rho\frac{1}{P(\rho)}
\pp[P(\rho)]{\rho} = K \\
\frac{1}{Q(\phi)}\PP[Q(\phi)]{\phi}= -K.\label{EQ06}
\end{eqnarray}
where \(K\) is a separation constant. The above equations can be rearranged as
\begin{eqnarray}\label{EQ07}
 {\rho^2} \PP[P(\rho)]{\rho} +  \rho\pp[P(\rho)]{\rho} -K P(\rho) = 0 \\
\PP[Q(\phi)]{\phi} + K Q(\phi)=0 \label{EQ08}
\end{eqnarray}
Case $\underline{K\ne 0}$
The solution of \EqRef{EQ07}, \eqref{EQ08} for \(K\ne 0\) takes the form
\begin{eqnarray}
 P(\rho) = A_k \rho^{\surd K} + B_k \rho^{-\surd K}, \qquad Q(\phi) = C_k
\cos(\sqrt{K}\phi) + D_k \sin(\sqrt{K} \phi) \label{EQ09}\\
\Phi_K =  [A_k \rho^{\surd K} + B_k \rho^{-\surd K}]\times[C_k
\cos(\sqrt{K}\phi) + D_k \sin(\sqrt{K} \phi)] \label{EQ10}
\end{eqnarray}
Case $\underline{K=0}$
In this case the solutions of the \(\rho\) and \(\phi\) equations are \begin{equation} P_0(\rho) = A_0 + B_0 \log \rho, \qquad     Q(\phi) = C_0 + D_0 \phi.          \end{equation}
Therefore
\begin{equation}\label{EQ12} \Phi_0(\rho,\phi) = (A_0 + B_0 \log \rho)\times( C_0 + D_0\phi). \end{equation}
A general solution will be a linear combination of solutions \eqref{EQ10} and \eqref{EQ12}.
Single-valuedness of potential
Note that the values \(\rho, \phi\) and \(\rho, \phi+2
\pi\) correspond correspond  to the same point in space. Hence the potential
should be equal for \(\phi\) and \(\phi+2\pi \). This means that
\(Q(\phi)=Q(\phi+2\pi)\). Noting the expression for \(Q(\phi)\) in
\eqref{EQ09}, we get
\begin{equation}
       \cos(\sqrt{K}(\phi+2\pi)) + D_k \sin(\sqrt{K} (\phi+2\pi)) =
\cos(\sqrt{K}\phi) + D_k \sin(\sqrt{K} \phi)
\end{equation}
and this equation has to be satisfied for all \(\phi\). Since sine and cosine functions are linearly independent, we must have \(\sqrt{K} (2\pi) = 2n\pi \), or \(\sqrt{K}=n, n=\pm1,\pm2,...\).
Final Form of general solution
We can now write down the most general solution of the Laplace equation in cylindrical coordinates (assuming \(z\)- translational invariance) as
\begin{equation}\label{EQ14}
 \Phi(\rho,\phi) = \Phi_0 + \sum_{n=1}^\infty \Phi_n(\rho,\phi),
\end{equation}
where \(\Phi_0\) is same as before and \(\Phi_n\) is given by
\begin{eqnarray}\label{EQ15}
\Phi_0(\rho,\phi) &=& (A_0 + B_0 \log \rho)\times( C_0 + D_0\phi),\\
 \Phi_n(\rho,\phi)&=&(A_n \rho^{n} +B_n\rho^{-n}) \times (C_n\cos(n\phi)
+ D_n\sin(n\phi)), \quad n=\pm1,\pm2,....\label{EQ16}
\end{eqnarray}
We are now ready to solve boundary value problems for cylindrical geometry and with translational invariance along the \(z\)- axis.
Three simple cases
We will discuss three simple boundary value problems involving regions bounded by infinitely long cylinders and can be solved using the solution in \eqref{EQ15}-\eqref{EQ16}.
Demanding the potential be finite everywhere in the region of interest, it is easy to see that following constraints.

1. If we have to solve the problem inside an infinitely long cylinder, the constants \(B_0\) and \(B_n,n=1,2,...\) must be set equal to zero. Otherwise the potential becomes infinite at the origin \(\rho=0\).
2. If the solution is required outside an infinitely long cylinder, then the constants \(B_0\) and \(A_n, n=1,2,...\) must all be set equal to zero. Otherwise the potential will become infinite as \(\rho \to \infty\).
3. Suppose the volume of interest where Laplace equation is to be solved is bounded by two coaxial, infinitely long, cylinders. In this case there are no further restrictions on the constants \(A_0,B_0, A_n,B_n\) and we must start with form in \EqRef{EQ14} and determine the constants from the boundary conditions given in the problem.