Existence of Solutions to PDEs - How do I know I've got them all?

280 Views Asked by At

I'm taking a very computational course in partial differential equations. Because of this emphasis, I'm feeling very underwhelmed by the course, and have a lot of questions that really aren't answerable in the current state of affairs. My professor basically tells me to take an advanced course in real analysis for a rigorous treatment, but that's a long way away for me (at least two years), I was hoping that someone could answer this question here.

In the course, we have only looked at three equations and some minor generalizations on them - the heat, wave, and potential equations. I understand these to be characteristic of larger classes of PDEs, but I don't know anything at all about them except sometimes I can separate variables. For each of them, the method has been identical. Separate variables, solve two (or sometimes even three) ODEs. Then by superposition, sum them up. Determine coefficients with Fourier sums or integrals. Wash, rinse, repeat.

The question is this: How do I know that that is all the solutions? There is no existence/uniqueness theorem for PDEs. How can I know that without some more advanced technique for solving PDEs that I couldn't find others? Does it follow from existence/uniqueness of ODEs? What about for those larger classes of PDEs?

1

There are 1 best solutions below

0
On

Uniqueness for Laplace's equation on a domain $D$ (with, say, piecewise-differentiable boundary $\partial D$) with, say, Dirichlet boundary conditions $u=f$ on $\partial D$ is easy: suppose there are two such solutions, $u$ and $v$. Take $$ -\nabla^2 (u-v) = 0, $$ multiply by $u-v$ and integrate over $D$, and use the divergence theorem: $$ 0 = -\int_D (u-v)\nabla^2 (u-v) \, dV = -\int_{\partial D} (u-v)[\nabla(u-v)] \cdot n \, dS + \int_D [\nabla(u-v)]^2 \, dV. $$ The surface integral is $0$ because on $\partial D$, $u-v = f-f=0 $. The quantity inside the remaining volume integral is nonnegative. If $u,v$ are continuous and $u \neq v$ somewhere, then the integral has to be positive by the usual argument, which is a contradiction.

Exactly the same proof can be made to work for Neumann boundary conditions, and Poisson's equation (note $u-v$ satisfies Laplace's equation if $u,v$ satisfy Poisson's).

(The above works in any number of dimensions provided that the normal vector exists almost everywhere and the solution is taken to have continuous first and second derivatives.)

For the other two equations you mentioned, the important theorem studied in any first rigorous course in PDEs is the Cauchy-Kovalevskaya theorem; once you understand that, you understand basically everything you need for local existence. Global existence is much harder: this is what most of the rest of the course could look at...

Now, important remark on uniqueness: if the solution is unique, it really doesn't matter how we find it. Provided it satisfies the PDE and the boundary conditions, we can find it however we like, with whatever dirty Fourier series tricks we like. (Caveat: provided the infinite series converge well enough.)