Talk:Uniqueness theorem for Poisson's equation

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)

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, so the difference in energies of the two solutions would be infinite which is unphysical (the two systems would have to have very different Hamiltonians, indeed, difference in Hamiltonians of two solutions is sufficient for the uniqueness theorem to not hold; of course it's also very unphysical). Again, this is not a "new" boundary condition, it's simply a statement that for infinitely large systems, the electric field at infinity must be 0 (Neumann boundary condition).

All the best

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

Reply: In real world,the boundary conditions are not necessary because all real charges are allocated within bounded domain and the potential hence decreases rapidly enough.But here we are talking about mathematical model,like an infinitely large grounded conductor plate,which is NOT finite.Another example is an infinitely long charged conductor pipe,which has infinite energy and its field doesn't even go to zero.It's unphysical,but it is still a good model under certain circumstances.If you want everything to be purely PHYSICAL,then there should not be the concept of "plane" and "particle" alike.

Netheril96 (talk)

I think you misunderstood my argument. What you are saying is that there has to be ${a}_{n,m}=0$ or ${c}_{n,m}=0$ but there is absolutely no reason for this to be true unless boundaries (at infinity) force that. But the boundaries can only be one of the 4 boundaries already mentioned. If boundaries at infinity are not established, the Uniqueness Theorem doesn't hold. To avoid over-reverting, I won't revert until our argument is settled, but please do reply. If I don't get a reply within a week I'll revert back (no hard feelings). IlyaV (talk) 17:53, 29 April 2010 (UTC)

Reply:

OK,I did misunderstand you.The boundary conditions at infinity ARE included in the 4 mentioned,but you didn't understand what I meant,either.What I emphasize is that it requires a different demonstration at infinity.If the solution of Laplace's equation could be something that is proportional to ${\frac {1}{\sqrt {r}}}$,then the uniqueness theorem would still hold for bounded domain,but for a infinitely large domain

$\oint _{S}{\varphi {\frac {\partial \varphi }{\partial n}}}dS\propto {\frac {1}{\sqrt {r}}}\cdot {\frac {1}{\sqrt {{r}^{3}}}}\cdot {{r}^{2}}=1$

This wouldn't converge to zero when r approaches infinity.Then uniqueness theorem would not hold for infinite domain with boundary conditions set.Therefore it is necessary to prove the infinite one separately.

Netheril96

There is another issue.The uniqueness theorem in fact doesn't not require the boundary conditions at all the surfaces set.It requires only the outermost boundary condtions.For inner interfaces,the two surface integrals from two sides automatically cancel out,that is to say

$\oint _{{S}_{1}}{{{\varepsilon }_{1}}{{\varphi }_{1}}}\nabla {{\varphi }_{1}}\cdot d\mathbf {S} +\oint _{{S}_{1}}{{{\varepsilon }_{2}}{{\varphi }_{2}}}\nabla {{\varphi }_{2}}\cdot d\mathbf {S} =\oint _{{S}_{1}}{\left({{\varepsilon }_{1}}{{\varphi }_{1}}{\frac {\partial {{\varphi }_{1}}}{\partial n}}-{{\varepsilon }_{2}}{{\varphi }_{2}}{\frac {\partial {{\varphi }_{2}}}{\partial n}}\right)}dS$

From the interface condition

${\begin{aligned}&{{\varphi }_{1}}={{\varphi }_{2}}\\&{{\varepsilon }_{1}}{\frac {\partial {{\varphi }_{1}}}{\partial n}}={{\varepsilon }_{2}}{\frac {\partial {{\varphi }_{2}}}{\partial n}}\\\end{aligned}}$

Thus the integrals cancel out.You need only to specify the boundary condition at the outermost surface.

(Here the two $\varphi$ means the difference of two possible solution at either side of an inner interface S₁)

Netheril96