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

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?

• You'll need a full course on functional analysis to get to a detailed treatment of those questions. – Raskolnikov Apr 27 '15 at 23:03
• I have to echo your professor, unfortunately. The details involved in answering this question, even in the case of linear PDE, are very technical. In fact, the details involved in asking this question, even in the case of linear PDE, are also very technical. (I mean, what even is a solution to a PDE, really? You've already seen cases where a PDE has non-differentiable solutions, and you may even have seen solutions that are not functions…) – Eric Stucky Apr 27 '15 at 23:10
• All of our solutions have been infinite sums of things which were differentiable, and so obviously functions. When I said computational, I meant really really computational. – Alfred Yerger Apr 27 '15 at 23:12
• For easy PDE, we can prove unicity of the solution in regard to the boundaries conditions. Hence, if your method find a solution for each boundary condition, and have proved it is unique, you then have all the possible solutions... – Tryss Apr 27 '15 at 23:15

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).