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.

An equally relevant comment: Milgram was at Syracuse from 1947 to at least 1951 (I could not ascertain when he left, it was sometime in the 1950s). The theorem was published in 1954, so it’s possible that he got the idea here.

I’ve made the same claim about Tibor Rado, who was a visiting lecturer at Harvard and Rice in 1929, before going to The Ohio State University in 1930, and publishing his solution to Plateau’s problem. His entire biography is actually very interesting. It involves escaping from prison camps and befriending Russian Eskimos: http://www-history.mcs.st-andrews.ac.uk/Biographies/Rado.html

Yes, I actually read the biography of Rado couple of years ago when using some results from his book “Length and area”. The book reminds me of Federer’s treatise; I find both of them delightfully unreadable.

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