So here’s an attempt to summarize how to prove that a solution exists for linear, second order partial differential equations. This is some of my favorite mathematics, though at first I was really turned off by all this talk about Sobolev spaces and functional analysis, instead of just solving the darn thing. So I encourage those who have never seen any PDE theory yet to dig in (though maybe some ODE familiarity would be helpful). Unfortunately, I could not come up with any helpful figures/pictures for today (suggestions welcome!), so you get a potpourri of shots.
First, let me define a second order linear differential operator L, but let me do so by describing what it does to a function (since a second order linear operator is just something that eats functions and gives back functions, in a linear manner):
One famous example of such an operator is the Laplacian, denoted , for which is 1 when j = k, and 0 otherwise, and s and c are zero. Put more simply, the Laplacian is the sum of the pure second derivatives. Now we provide first an outline, and then a few details of a proof of existence of solutions.
How to solve Lu = f:
1. Associate to the operator L a bilinear form B. That is, a function B that eats two functions and gives a number which is linear in each coordinate.
2. Remember from functional analysis that, under certain conditions, there exists a unique u so that for any v, where is a linear functional (eats functions, gives a number).
3. Define .
4. Pat yourself on the back for defining B in a clever way so that the u produced actually solves Lu = f in a weak sense.
Again, with slightly more detail:
1. We’ll define
where everything is a function of x, but that really clutters up the equation. A reason you might have thought to do this yourself is that it takes a little pressure off of u. That is, we only need to require that u has one derivative, and that derivative only has to be defined under an integral sign. If u happened to have two derivatives, we could integrate the first term by parts, and then we would have . In this beautiful, integrable, differentiable world, if we could show that for all functions v, then it would be reasonable to conclude that Lu = f.
2. The Lax-Milgram theorem says, roughly, that if H is a Hilbert space, then for a bilinear form and a (continuous) linear functional , there is a unique
so that for each . There are some other non-trivial hypotheses on B, which may be translated into hypotheses on L, but let’s revel in success now.
3.-4. These steps were actually described pretty well above.
So we need functional analysis to use this great Lax-Milgram theorem. We need Sobolev spaces because our favorite function spaces from calculus- , the m-times differentiable functions- are not Hilbert spaces. Certain Sobolev spaces are, though, and they provide precisely the integrability conditions we need to get precise conditions to guarantee solutions for large classes of PDE.