As I understand it, the Laplacian operator is linear, and thus $\nabla^2(f +g) =\nabla^2f + \nabla^2g.$ I was wondering if this might be exploitable to solve Poisson-like equations in spherical / elliptical geometry, and if not where this breaks down. I'll illustrate with an example - with some symmetry, we know the solution to Laplace's equation ($\nabla^2 p_{e} = 0$) has solutions of the form
$p_{e}(r,\theta) = (Ar^L + Br^{-1(L+1)})P_L(\cos\theta).$
As discussed recently here, this is homogeneous and Green's function methods would usually be employed to find the non-homogeneous form. But instead, can one combine this with a known solution of a specific problem to find a unique solution to an inhomogeneous version? Under spherical symmetry, the non-homogeneous solution to $\nabla^2 P_S = A$ is straightforward to find, and if we know $p_s = p_o$ at $r_o$, then the solution is simply
$p_{s}(r) = p_o + \frac{A}{6}(r^2 - r_o^2)$.
I want to solve an equation similar to this, but on an ellipsoid so that if $r_o$ is the semi-major axis and $e$ the eccentricity, then the boundary condition is $p_{f}(r_{o}\sqrt{1 - e^2 + e^2\cos^2 \theta},\theta) = p_o$. Can I then write
$\nabla^2p_{F} = \nabla^2p_{e}(r,\theta) + \nabla^2p_s(r) = A$
and just find the constants? If so, then by geometry of problem $B_{L} = 0$, and I'm looking for a solution of the form
$p_{f} = \Sigma A_Lr^L P_L(\cos\theta) + \frac{Ar^2}{6} + D$.
From boundary conditions, we know
$\Sigma A_L(r_o \sqrt{1 - e^2 + e^2 \cos^2 \theta}) P_L(\cos\theta) = p_o -\frac{Ar_o^2}{6}\left(1 - e^2 + e^2 \cos^2 \theta \right) - D$
and after re-writing $\cos \theta = x$, we have $f(x) = (p_o - D) - \frac{Ar_o^2}{6}\left(1 - e^2 + e^2x^2 \right)$, and exploiting Legendre polynomials we determine the co-efficients to eventually yield a new identity
$p_{f}(r,\theta) = p_o + \frac{A}{6} \left(r^2 \left(1 - \frac{2e^2P_{2}(\cos\theta)}{3(1 - e^2 \sin^2 \theta)}\right) - r_o^2 \left(1 - \frac{2e^2}{3} \right) \right)$
Superficially at least this looks reasonable - when $e=0$, it collapses back to the $p_{s}$ form, as expected. It also gives reasonable gradients and satisfies that boundary condition. However, I am convinced it can't be correct, because $\nabla^2 p_{f} \neq A $. Can someone explain why this approach doesn't work, and whether it is a wasted effort to proceed? I am relatively new to Legendre polynomial approaches and may have mixed them up, but I suspect the problem here is a more fundamental one and I'd be grateful if anyone could point out the flaw in my approach!
I think I can answer part of this question in so much as I believe I know why this particular example doesn't work. Let's say we have
$p_{e}(r,\theta) = (Ar^L)P_L(\cos\theta)$ then it's relatively straightforward to show $\nabla^2p_e = 0$.
However, this is not the case for a function with the form $p_n(r,\theta) = (Ar^L)(1 - e^2\sin^2\theta)P_L(\cos\theta)$ where instead $\nabla^2p_n = f(\theta)$, which is the problem encountered above.