# The PDE $u_t=u_{xx}-u_{yy}$: The simplest linear second-order PDE that isn't elliptic, parabolic, or hyperbolic.

I know that there have been several questions on here and stackexchange about linear PDE's which don't fall into the standard classification, but I had a more focused question which I haven't seen answered. The PDE $u_t=u_{xx}-u_{yy}$ (or, equivalently, $u_t=u_{xy}$), is the simplest linear second-order PDE that is not elliptic, parabolic, or hyperbolic. Its time-invariant solutions are solutions to the one-dimensional wave equation.

My question is,

How does this PDE compare to the standard parabolic and hyperbolic PDE's in the following categories:

• Fundamental solution: is there an integral solution which is a convolution with a fundamental solution?

• Smoothness: are solutions infinitely smooth or does the smoothness depend on boundary conditions?

• Propagation speed: infinite or finite?

I only recently read Evan's PDE book, and the only question I really took a crack at was the first. I looked for a solution involving exponentials in t, but I couldn't find one. I am interested in this question purely from a classification standpoint. Thanks!

Edit: As the comments below indicate, there can be no general solution with arbitrary initial conditions analogous to that for parabolic or hyperbolic equations. Also, solutions need not be infinitely smooth. Since this answer was pieced together from the comments, I'm making this community wiki, in case anyone would like to add to it later.

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I don't know if this particular equation has been studied before, but there was a lot of work on linear constant coefficient PDE's, notably by Leon Ehrenpreis, back in the 60's and maybe 70's. You might want to dig through those to see what was done back then. Also, the most obvious way to try to analyze your questions is via separation of variables. – Deane Yang Dec 8 '12 at 16:34
If one specialises to solutions independent of x, one obtains the backwards heat equations $u_t = -u_{yy}$, which has basically no good solvability properties forward in time. So I doubt that there is any meaningful way to solve the initial value problem. – Terry Tao Dec 8 '12 at 17:54
Solutions need not be infinitely smooth. A nonsmooth solution $u(x,y)$ to the equation $u_{xx}-u_{yy}=0$ (and these certainly exist) will be a solution to your equation. – Robert Bryant Dec 8 '12 at 19:46
As the comments by Terry Tao and Robert Bryant indicate, the understanding of a broad class of solutions to a PDE usually requires both identifying what boundary or initial value conditions to impose and in what function space to look for solutions. Standard PDE's come from physical models or variational problems that tell you what these should be. With a PDE that is not based on any such model or variational problem, it is much less clear what to do. – Deane Yang Dec 8 '12 at 20:44

Exponential solutions are easy to find. Just plug $\exp(ax+by+ct)$, and you will see that every $(a,b,c)$ that satisfies $c=a^2-b^2$ gives you a solution. This polynomial is called the symbol of the differential operator. For the general theory of linear PDE with constant coefficients, the best source is the first two volumes of Hormander, Analysis of Linear Differential Operators.