# Proof that solutions of the wave equation with smooth initial data have to be smoot

Is it possible to show that the $n$-dimensional initial value problem of the wave equation \begin{align*} u_{tt}(x,t)-\Delta u(x,t)&=0\qquad \mbox{in } \mathbb{R}^n \times (0,\infty)\\ u(x,0)&=g(x)\qquad \mbox{on } \mathbb{R}^n\\ u_t(x,0)&=h(x)\qquad \mbox{on } \mathbb{R}^n\\ \end{align*} with $g,h\in C^\infty$ does lead to $u\in C^\infty$ solutions without explicitly solving it or using the fact that the system ist hyperbolic.

In other words is there a theorem like: For every linear PDE with constant coefficients $C^\infty$ initial data does lead to $C^\infty$ solutions.

## 1 Answer

The Cauchy-Kowalevski theorem gives local existence of analytic solutions of analytic PDEs: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem Of course, a linear constant coefficient PDE is analytic, but this theorem is far more general and allows nonlinear PDEs as well.

In general, if you allow the coefficients to be non-constant smooth functions (not necessarily analytic), then it is possible to construct PDEs of this type for which there are no smooth solutions. There are many papers along these lines, one is "An example of a smooth linear PDE without solution" by Hans Lewy.

• From what I have found the theorem is always stated in the general form with analytic coefficients and analytic initial data. Is it possible to reduce the initial data to $C^\infty$ if i have constant coefficients and achieve $C^\infty$ solutions? – lecovee Jan 28 '17 at 11:50
• I strongly suspect this is true, but I don't know of any proofs that do not use some structure of the PDE (parabolicity or ellitpicity, etc.). With linear constant coefficient PDEs, every derivative of the solution satisfies the same PDE, so if you can show (weak) solutions are continuous then you are done. This is probably how a proof would work. – Jeff Jan 29 '17 at 19:48