Legendre series solutions for the Laplace equation: can Neumann boundary conditions be applied?

Question

In Jackson E&M, it is shown in equation (3.33) that the Laplace equation with azimuthal symmetry can be expanded in Legendre series in spherical coordinates,

$$ \Phi(r,\theta)=\sum_{l=0}^\infty\ [A_l r^l + B_l r^{-(l+1)}]\,P_l\,(\cos\theta). $$

The expansion coefficients $A_l,B_l$ can be determined from boundary conditions. I see how to do this for Dirichlet problem, but is there an example of finding the coefficients in a simple Neumann problem?

EDIT: I guess my original question appears kind of trivial. Precisely speaking, I am wondering about for what Neumann boundary condtions does the solution admit the above series expansion. Is it true that for all Neumann boundary conditions satisfying the well-known solvability condition, the solution can be written into the series expansion?

Answer 1

When solving the Laplace equation in a ball (say unit radius) with Neumann boundary conditions and azimutal symmetry, you are given: $$ \partial_r\phi(r=1) = \sigma(\theta) $$ When expanding in Legendre polynomials, you get: $$ \phi = \sum_{l=0}^\infty \phi_lr^lP_l(\cos\theta) $$ and you need to match $\phi_l$ with the boundary condition: $$ \sum_{l=0}^\infty l\phi_lP_l(\cos\theta) = \sigma(\theta) $$ Expanding $f$ in Legendre polynomials: $$ \sigma(\theta) = \sum_{l=0}^\infty \sigma_lP_l(\cos\theta) $$ you can calculate the $A$ by identifying: $$ \phi_l = \frac{\sigma_l}{l} $$ This works for all coefficients except for the one $l=0$. This forces $\sigma_l=0$ and $\phi_0$ cannot be found so the problem is under and over determined.

Looking back at the original problem, this is to be expected. You are solving the linear problem: $$ \begin{align} \Delta \phi &= 0 \\ \partial_r\phi(r=1) &= \sigma(\theta) \end{align} $$

Immediately, you can check that all constant potentials are solution without looking into Legendre polynomials. Physically, you are specifying the surface charge of the sphere and you want to define the potential in the ball, so as usual, the potential is defined up to an additive constant. What is well defined is the electric field $E=-\nabla \phi$ and you can check that this is the case here.

Now the consistency condition $\sigma_0=0$ is also to be expected. The boundary as written say that the electric field outside the sphere is zero. Since the external field is zero, by Gauss' law, this means that the total charge of the sphere is null, i.e. $\sigma_0=0$.

This second problem can be amended if physically, you are interested in looking at the field generated by a spherical shell with zero potential at infinity. You therefore have a potential $\phi$ inside and $\psi$ outside. Your problem is therefore: $$ \begin{align} \Delta \phi &= 0 & \Delta \psi &= 0 \\ \phi(r=1) &= \psi(r=1) & \partial_r\phi(r=1) &= \sigma +\partial_r\psi(r=1) \end{align} $$

Using the expansion which automatically solves the bulk Laplace equations: $$ \begin{align} \phi &= \sum_{l=0}^\infty \phi_lr^lP_l(\cos\theta) & \psi &= \sum_{l=0}^\infty \psi_lr^{-l-1}P_l(\cos\theta) \end{align} $$ the boundary conditions become: $$ \begin{align} \phi_l &= \psi_l \\ l\phi_l +(l+1)\psi_l &= \sigma_l \end{align} $$ so now $\sigma_0$ is not necessarily zero. Note that the first problem problem is also solved thanks to the fixing of potential to be null at infinity.

Hope this helps.