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Let $n>1$ and $u$ be a solution of a linear PDE with constant coefficients $$ u_t-\sum_{k=0}^n a_k \partial_x^k u=0,\quad a_k\in \mathbb C,\quad a_n\ne0, $$ in some neighborhood of a point $(x_0,t_0)$ on a plane. Is it true that the function $u(x,t_0)$ is analytic in some neighborhood of $x_0$?

In some special cases, such as the heat equation it follows from results of Petrovskii, namely that solutions of homogeneous parabolic equations with analytic coefficients are analytic with respect to space variables. But in the general case I'm unable to find a reference.

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    $\begingroup$ Are you assuming the initial data to be analytic? Otherwise, I believe that your conjecture is false - simply take $n=3$ and $a_0=a_1=a_2=0$, $a_3=1$. $\endgroup$
    – Delio Mugnolo
    Commented Nov 6, 2013 at 8:49
  • $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Nov 6, 2013 at 9:34
  • $\begingroup$ Also, since you mentioned Petrowsky, maybe this article is relevant. $\endgroup$
    – Willie Wong
    Commented Nov 6, 2013 at 9:45
  • $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Nov 6, 2013 at 10:08

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