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Nov 6, 2013 at 10:08 comment added Willie Wong Since you mentioned constant coefficients, see also this article. In Petrowsky's notation your $x$ is his $x_0$. You can fix $x_1$ to be $t$ and $n = 1$. His result concerns explicitly analyticity (and lack thereof) in the $x_0$ variable of solutions.
Nov 6, 2013 at 9:45 comment added Willie Wong Also, since you mentioned Petrowsky, maybe this article is relevant.
Nov 6, 2013 at 9:34 comment added Willie Wong In general the odd derivatives are only dispersive and the even ones are dissipative, assuming $a_k \in \mathbb{R}$. You can see this by taking the Fourier transform formally. As long as you have some dissipation you expect regularisation. But if you don't have dissipation, you have all sorts of bad examples. In addition to what @DelioM. wrote, consider the homogeneous Schrodinger equation in 1 dimension, where $n = 2$ and $a_1 = 0$ and $a_2 = i$. You can solve it both forward and backward in time from some $C^2$ but not $C^\omega$ initial data.
Nov 6, 2013 at 8:49 comment added Delio Mugnolo Are you assuming the initial data to be analytic? Otherwise, I believe that your conjecture is false - simply take $n=3$ and $a_0=a_1=a_2=0$, $a_3=1$.
Nov 5, 2013 at 16:43 history edited Amritanshu Prasad CC BY-SA 3.0
corrected spelling in header
Nov 5, 2013 at 15:57 history asked Andrew CC BY-SA 3.0