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Let's consider a (partial) differential equation with analytic coefficients. The initial conditions may be non-analytic*.

Is it possible a solution $f$, which is non-analytic at any point of the domain?

*Even if the initial conditions are analytic, it is possible that the solution is not [1].

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    $\begingroup$ You certainly need to assume something about the PDE for this question to have a chance of making sense. For example, your PDE could be $\partial_t u =0$, with initial conditions $u(t=0) = f$.... On the other hand, for the heat equation, for example, a solution immediately becomes analytic, even for very rough initial data.. $\endgroup$
    – Otis Chodosh
    Commented Aug 14, 2012 at 17:28
  • $\begingroup$ @Otis: I just wanted to know if it is possible to have pde with analytic coefficients and solutions like that in the question. If the answer depends on the particular pde and the initial conditions, I think that the answer should contain the relevant assumptions, not the question. It is more likely that these conditions are known by the person who knows the answer, not by the person who asks the question. $\endgroup$
    – Cristi Stoica
    Commented Aug 15, 2012 at 12:21
  • $\begingroup$ You should probably add some motivation as to what sort of PDE's you are interested in, otherwise I doubt there exists a "good" answser. I took a brief glance at the paper you linked to, and if I understand correctly, the paper gives an example of a system of linear PDE's with analytic coefficients and initial data, but non-analytic solutions. This should be contrasted with the Cauchy-Kowalewski theorem, which says that for a linear PDE (for a single function) analytic data and coefficeints imply analyticity of solutions and ( in light of @Bazin's answer) suprisingly existence. $\endgroup$
    – Otis Chodosh
    Commented Aug 15, 2012 at 23:50
  • $\begingroup$ In particular, you don't specify if you're interested in systems, single PDE's, linear PDE's, nonlinear PDE's, etc. It seems that the only place that a nice theory might have a hope of existing is in the setting of linear PDEs for one function. Can you give some more precise details, or motivation for the question? $\endgroup$
    – Otis Chodosh
    Commented Aug 15, 2012 at 23:52
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    $\begingroup$ @Otis: Thanks for the comments. I am interested in the most general settings, systems or singles, linear or nonlinear, elliptic, parabolic, hyperbolic etc. My hope was that if somebody has something to say about the question, in a general or in a particular setting, to be free to say, because I am interested in the general question. So my hope was to receive answers of the form "in general, ...", or, "in the particular case of ... equations, ...". I see that Rafe accuses me of being rude, his interpretation of what I said is wrong, but if I made you feel that I was rude with you I apologize. $\endgroup$
    – Cristi Stoica
    Commented Aug 16, 2012 at 6:57

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Points off for the rude response to Otis' sincere attempt to help you!

Deanne is correct, but you can see this simply even with the operator $\partial_{x_1}$ in $\mathbb R^n$. If $u(0, x_2, \ldots, x_n)$ is nowhere analytic, then the solution of $\partial u/ \partial_{x_1} = 0$ is constant in $x_1$ but never analytic in the orthogonal directions.

On the other hand, there is a class of operators known as analytic hypoelliptic (which includes both the Laplacian for any analytic metric and the associated heat operator) for which such a statement is true: $L u = f$ and $f$ analytic implies $u$ analytic.

However as the paper you reference indicates, all sorts of wild things can happen in general; that paper gives an example of a constant coefficient system and analytic ``initial data'' where the solution is not analytic. Actually, that is a rather nonstandard sort of initial value problem where Cauchy data for one component is posed on one hypersurface and Cauchy data for the second component is posed on a transverse hypersurface.

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  • $\begingroup$ Rafe, thanks for a much better answer than mine, especially the example of $\partial_1$. $\endgroup$
    – Deane Yang
    Commented Aug 16, 2012 at 2:02
  • $\begingroup$ Thanks for the answer. On the other hand, while I agree that Otis's attempt was sincere, I disagree that my reply was rude at all, and I even don't see what may have appeared to you impolite in what I wrote. I tried to explain better what I need, and what I need differs from particularizing the problem and risking, by adding conditions, to miss what I am interested in, and making the problem too localized. $\endgroup$
    – Cristi Stoica
    Commented Aug 16, 2012 at 6:48
  • $\begingroup$ My apologies then. $\endgroup$
    – Rafe Mazzeo
    Commented Aug 16, 2012 at 15:58
  • $\begingroup$ Rafe, can you detect analytic hypoellipticity via the symbol? (Am I mistaken in remembering that you can detect hypoellipticity via the symbol?) Cool answer, thanks! $\endgroup$
    – Otis Chodosh
    Commented Aug 16, 2012 at 19:19
  • $\begingroup$ @Otis: For constant coefficient operators, Petrowsky proved that analytic hypoellipticity is equivalent to ellipticity of the symbol. He also proved that elliptic operators with analytic coefficients are analytic hypoelliptic. $\endgroup$
    – timur
    Commented Apr 20, 2013 at 19:17
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Take the Lewy operator $\mathcal L$ in $\mathbb R^3_{x,y,t}$ $$ \mathcal L=\partial_x+i\partial_y+i(x+iy)\partial_t. $$ There exists a set $S$ of second category in $C^\infty(\mathbb R^3)$ such that for all $f\in S$, the equation $\mathcal L u=f$ has no distribution solution. So even with polynomial (even affine here) coefficients, a complex vector field could have terrible pathologies. A good explanation of this phenomenon was given by Lars Hörmander and developed further by several authors. The symbol of the Hans Lewy operator is $a+ib$ and for each point $X\in \mathbb R^3$, you can find $\Xi\in \mathbb R^3$ such that at $(X,\Xi)$, $$ a=b=0\quad\text{and the Poisson bracket {$a,b$} is positive}. $$

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    $\begingroup$ Thank you for this example. If I understand well, this is an equation with no solution at all, not a solution which is non-analytic at any point. $\endgroup$
    – Cristi Stoica
    Commented Aug 15, 2012 at 12:23
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This can't happen if the PDE is either elliptic or parabolic. If the PDE is hyperbolic and you start with initial data that is nowhere analytic, then it seems plausible that the resulting solution is also nowhere analytic. But I don't have a rigorous proof of this. The idea would be to assume that there is a point where the solution is analytic in a neighborhood and try to derive a contradiction by propagating the analyticity backwards back to the given intial data.

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