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?

  • $\begingroup$ Just differentiate and match? Sorry but I am not understanding the confusion. $\endgroup$ Jun 13, 2023 at 16:07
  • $\begingroup$ How do you know you can match? Consider the interior problem for a sphere. Then $B_l=0$. Differentiation kills $P_0$, and you don't have a complete set of polynomials to match the Neumann boundary condition. For Dirichlet there is no such problem at least in this specific case because you have the complete set of polynomials to match. $\endgroup$
    – 111
    Jun 13, 2023 at 16:27
  • $\begingroup$ The disappearance of $A_0$ is so standard and well-known that the textbooks cover how to deal with it. There is a derivation of what it should be in the case of Dirichlet, and IIRC in Neumann, it is a free parameter. $\endgroup$ Jun 13, 2023 at 16:30
  • $\begingroup$ Can you give a reference? I don't think I have read about it in Jackson. I don't see how one can match any Neumann boundary condition without constraint with an incomplete set of polynomials. $\endgroup$
    – 111
    Jun 13, 2023 at 16:36

1 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.

  • $\begingroup$ Thanks! I understand how to solve your special example when charge density is specified. But I am still interested in the general problem in the EDIT: when only Neumann boundary condition is specified and when it can be solved by Legendre series expansion. Note that specifying the Neumann boundary condition is not the same as specifying the surface charge density in general. $\endgroup$
    – 111
    Jun 18, 2023 at 14:52
  • $\begingroup$ If your domain is a sphere (otherwise spherical harmonics are not appropriate) your “general” problem, ie Laplace equation with Neumann boundary conditions is exhaustively treated in the first half of the question. $\endgroup$
    – LPZ
    Jun 18, 2023 at 19:35

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