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It seems that Cauchy–Kovalevskaya is not commonly used in books on PDE theory. I am thinking about applying it somewhere interesting.

It is known that if $L$ is a uniformly elliptic operator, with analytic coefficients, and $f:U\to \mathbb R$ is analytic, then if the Dirichlet problem (with boundary of $U$ reasonably regular) $$ \begin{cases} Lu=f, \\ u|_{\partial U} = g \end{cases} $$ has a unique solution $u,$ then the solution is analytic.

I am wondering - can we prove this using Cauchy–Kovalevskaya?

I have done some attempts, but I could not find a way to overcome the obvious issue that the boundary condition given is insufficient for Cauchy–Kovalevskaya. Also, even if Cauchy–Kovalevskaya gives an analytic solution locally, how is that going to be proven to equal to the true solution to the Dirichlet problem?

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    $\begingroup$ Why would you use a theorem for a PDE initial value problem on a PDE boundary value problem? $\endgroup$
    – Phil Harmsworth
    Commented Oct 14, 2023 at 8:42
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    $\begingroup$ @PhilHarmsworth Why not? There is no clear boundary between initial value and boundary value problems, really. $\endgroup$
    – Ma Joad
    Commented Oct 14, 2023 at 10:01
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    $\begingroup$ The difference between specifying only the value of a function and specifying both the value and its derivative along all or part of the boundary is quite significant. Specifying data on the boundary of a compact domain versus on an open hypersurface is also quite significant. The fundamental difference is that the boundary value problem is a global one but, in the analytic category, the initial value problem is local. $\endgroup$
    – Deane Yang
    Commented Oct 16, 2023 at 16:02
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    $\begingroup$ Notice that the Cauchy-Kovalevskaya theorem does not preclude the possibility of a non-analytic solution to the initial value problem. There is also no obvious way to use ellipticity to strengthen it. The theorem is rarely used because it is by modern standards a rather weak theorem. $\endgroup$
    – Deane Yang
    Commented Oct 16, 2023 at 16:11
  • $\begingroup$ Gaetano Fichera used to say that boundary value problems are globally defined problems, while initial value ones are local problems. In such a way he explained the profound difference between the hypotheses required by his solution of the boundary value problem for holomorphic functions of several variables and the result of Hans Levy on the Cauchy problem for the same function class, which additionally requires pseudoconvexity of the (open) hypersurface where the initial data is specified. Thus, for general PDEs or systems of PDEs, BVPs and the Cauchy problem are deeply different. $\endgroup$
    – Daniele Tampieri
    Commented Feb 3 at 15:07

1 Answer 1

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Standard regularity theory guarantees that $u$ is smooth. If you want to use Cauchy-Kovaleskaya you would need to know that both $g$ and $\frac{\partial u}{\partial \nu}$ are real analytic: these are the Cauchy data for a second order pde. Assuming that somehow you can prove that $\frac{\partial u}{\partial \nu}$ is real analytic, then you can use Cauchy-Kovaleskaya coupled with Holmgren's uniqueness result to conclude only that $u$ is real analytic near the boundary.

Concerning the analyticity of solutions see one of my older posts.

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    $\begingroup$ Are you giving the right link? I could not find things about proving analyticity. It is about physics, not pure maths. $\endgroup$
    – Ma Joad
    Commented Oct 13, 2023 at 15:41
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    $\begingroup$ Petrowsky proved this for differential operators with constant coefficients and for these operators he proved that ellipticity is a necesseray and sufficient condition. For elliptic operators with variable operators, regularity in Gevrey class (includes the nalytic situation) you can find a proof in the 3rd volume of Lions-Magenes "Non-homogenous boundary value problems..." $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 14, 2023 at 9:59
  • $\begingroup$ Thank you. I will have a look. $\endgroup$
    – Ma Joad
    Commented Oct 15, 2023 at 9:41
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    $\begingroup$ My guess is that any proof that the normal derivative is analytic first proves the solution is analytic. $\endgroup$
    – Deane Yang
    Commented Oct 16, 2023 at 16:07
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    $\begingroup$ @DeaneYang My guess the same $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 16, 2023 at 17:29

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