Talk:Uniqueness theorem for Poisson's equation

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All right, so I rewrote the uniqueness theorem (as known in physics) here as the previous page did not contain any formulas but had a somewhat confusing text-only proof. If anyone wants to clean up and make things prettier, please do go ahead! Also, I changed the page type from math to physics, as this is what this page discusses anyways. For people who want just the math definition, there already exists uniqueness page which is just that. IlyaV (talk) 19:21, 3 April 2010 (UTC)[reply]

Boundary at infinity

I'm reverting the edits about boundary at infinities.

I think boundaries at infinities are already covered in the 4 boundary conditions aforementioned. If $\phi$ or $\mathbf {\nabla } \phi$ don't converge at infinity rapidly enough then the uniqueness theorem doesn't hold. There is nothing in the solution of the Laplace equation that necessitates that in

$\varphi =\sum \limits _{n,m}{\left({{a}_{n,m}}{{r}^{n}}+{\frac {{b}_{n,m}}{{r}^{n+1}}}\right)}P_{n}^{m}\left(\cos \theta \right)\cos m\phi +\sum \limits _{n,m}{\left({{c}_{n,m}}{{r}^{n}}+{\frac {{d}_{n,m}}{{r}^{n+1}}}\right)}P_{n}^{m}\left(\cos \theta \right)\sin m\phi$

${a}_{n,m}$ or ${c}_{n,m}$ should be 0. Simply, if they are zero, then the uniqueness theorem holds, if they aren't, then it doesn't hold. Which is to say, $\phi =r$ is a valid solution to the Laplace equation with non-zero (actually infinite potential) boundaries at infinity. I suspect one can also construct (unphysical) cases where $\phi$ and it's gradient both go to non-zero constants asymptotically AND satisfy Laplace equation (again, uniqueness theorem doesn't hold then).

I think the only thing that can be added about "boundaries at infinity" is that since energy of the fields is proportional to $\mathbf {\nabla } \phi ^{2}$ , if $\mathbf {\nabla } \phi$ doesn't go rapidly enough to 0 as $r\to \infty$ this would imply infinite energy which is unphysical. Again, this is not a "new" boundary condition, it's simply a statement that for infinitely large systems with non-infinite energies in the field, the electric field at infinity must be 0 (Neumann boundary condition).

All the best,

IlyaV (talk) 00:23, 25 April 2010 (UTC)[reply]