I'm trying to solve Laplace's equation in an (axisymmetric) external spherical domain. The controlling equation is:

$$\nabla^2 f = 0$$

$f$ must dissappear at infinity, and at the surface of the sphere ($r=1$) we have:

$$\frac{\partial f}{\partial r} = \sigma h(f+1)$$

Here $\sigma$ is a positive parameter, and $h$ is an angular "step" function:

$$h=\begin{cases} 1 & \mu \leq \tau \\ 0 & \mu \gt \tau \end{cases}$$

Here $\mu$ is simply a transformed angular coordinate ($\mu = \cos{\theta}$) and $\tau$ is a parameter. The most important case is $\tau = 0$, but it can take any value from $-1$ to $1$.

The solution to Laplace's equation in spherical coordiantes for this situation is:

$$ f(r,\mu) = \sum_0^\infty A_nr^{-(n+1)}P_n({\mu})$$

Here $P_n$ are the Legendre Polynomials and the $A_n$ are the coefficients that define the solution.

Inserting this into the boundary condition at the surface ($r=1$) we get:

$$ \sum_0^\infty -(n+1)A_nP_n(\mu)= \sigma h\left(1 + \sum_0^\infty A_nP_n(\mu)\right) $$

The next step here is to multiply by a generic Legendre Polynomial (here $P_m$), and to integrate from $-1$ to $1$, to take advantage of orthogonality. Unfortunately, the factor $h$ ruins orthogonality on the RHS, so we end up with:

$$ \frac{-2(m+1)}{2m+1}A_m = \sigma \left[ \int_{-1}^\tau P_m \, d\mu +\sum_{n=0}^\infty A_n\left( \int_{-1}^\tau P_m(\mu) P_n(\mu) \, d\mu \right)\right] $$

So we end up with an infinite linear system of equations for the $A_n$ (note than in the above the $A_n$ are the same as the $A_m$, just different counters).

For now what I've been doing is truncating the series (which basically forces all $A_n$ above a certain $n$ to be $0$) and solving the now finite system computationally. However, choosing a maximum $n$ has been somewhat arbitrary, and for non-small values of $\sigma$ this procedure seems to converge incorrectly.

Is there a smart to way to solve this system? It's be best to have procedures that work for all $\sigma$ and $\tau$, but I'd be satisfied with a procedure that handles $\tau = 0$ and\or limiting cases of $\sigma$ ($\sigma \to \infty$ or $\sigma \to 0$).

  • $\begingroup$ I have two questions. First why do you write $\nabla f = 0$ for the Laplace equation ? Isn't it $\nabla^2 f = 0$ or $\Delta f =0 $ ? Second : What exactly is the definition of the word "axisymmetric" that you use ? And related to this question, what are the definitions of the two independent angles in your spherical coordinates system ? $\endgroup$ – JJacquelin May 2 '15 at 9:59
  • $\begingroup$ @JJacquelin Ah thanks that was a typo. Axisymmetric is another way to say cylindrically symmetric. My $f$ is independent from the azimuthal angle. The other angle, $\theta$ starts at the positive z axis,and goes from $0$ to $\pi$. For simplicity I have a coordinate transform of the form $\mu =\cos{\theta}$ $\endgroup$ – Sciencertobe May 2 '15 at 12:44
  • $\begingroup$ May i ask you why you search a solution only with Legendre polynomials (i.e. of integer indexes) ? Why not the solution on the form of Legendre functions $P_\nu(\cos(\theta))$ or $Q_\nu(\cos(\theta))$ with real $\nu$ ? For example $f=-1+C\frac{P_{\sigma-1}(\cos(\theta))}{r^{\sigma}}$ is convenient to met the Robin boundary condition. $\endgroup$ – JJacquelin May 2 '15 at 17:08
  • $\begingroup$ It doesn't actually need to be Legendre Polynomials, I just used them as a base since they are the most commonly given solutions. I'll look into Legendre functions and see if they simplify the math. Thanks! Note, however, that your proposed solution doesn't actually satisfy my boundary conditions since it goes to $-1$ at infinity, and fails to cancel out the step function $h$ in the Robin BC. $\endgroup$ – Sciencertobe May 2 '15 at 19:23
  • $\begingroup$ I agree. The given example of function never pretended to solve the whole problem. Just an idea to extend the possibilities for solving it. I didn't knew if some hidden requirements were behind the use of the Legendre polynomials. Now, I know that it is not the case. $\endgroup$ – JJacquelin May 2 '15 at 20:39

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