20 October 2008

Solvable PDEs have measure zero

PDEs are very hard to solve.

I heard it claimed that it's very lucky, if you're someone who does mathematical finance, that the Black-Scholes partial differential equation, when used to price a call option, has a closed-form solution. (Well, if you include the error function in "closed form".)

The payoff of the thing being priced gives the initial condition for the Black-Scholes PDE and for a simple perturbation of that, there is probably not such a simple solution. And options markets probably would have developed very differently if there hadn't been an exact solution, because solving such an equation numerically, while possible, is a lot slower.

A friend of mine who knows more about PDEs than I do said that, basically, the set of exactly solvable PDEs has measure zero. Of course this isn't a theorem, but the point is that people who study PDEs don't expect there to be exact solutions.

3 comments:

thecooper said...

One might point out that knowing particular closed-form solutions to PDE doesn't necessarily tell you much about the PDE itself.

If you have a uniqueness theorem, then it does--because all solutions are your closed-form one. But in the presence of nonuniquess, knowing the behaviour of one solution doesn't tell you much about the PDE itself.

That's why the prototypical PDE theorems are a priori estimates--they're theorems about the equation, not theorems about its solutions.

CarlBrannen said...

More generally, mathematicians form a set of measure zero.

Veky said...

> And options markets probably would have developed very differently if there hadn't been an exact solution, because solving such an equation numerically, while possible, is a lot slower.

I find it very difficult to believe. It's not the case that we have some particularly nice (and fast to compute) functions, that we use to construct closed-form solutions, it's rather a historic coincidence what we consider elementary.

For example, erf is not on the same footing as sin or log, but it's merely a matter of tradition. We "invented" erf to compute an important integral, and now we know enough about it to compute it very fast. If there was economic advantage in solving that PDE quickly, we would soon invent and research another "special function" or even a family of them (remember Bessel functions?;).