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Suppose $u$ is a harmonic function of a domain $\Omega\subset \mathbb{R}^n$ and $u$ is continuous up to the boundary. If $\partial\Omega$ has an open smooth portion, can $u$ be extended to a harmonic function outside this smooth portion?

I have a very vague claim that if this portion is analytic, then we can extend $u$ by schwarz reflection principle. But I don't know anything about the smooth case. Can anyone give me a hint?

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    $\begingroup$ The Schwartz reflection principle would only work if the harmonic conjugate of $u$ were constant along that particular arc on $\partial\Omega$ (or equivalently, if the normal derivative of $u$ vanishes there). $\endgroup$
    – John Pardon
    Commented Jul 6, 2012 at 21:33
  • $\begingroup$ I am not sure your are right. As I said, I don't have a rigorous proof of the analytic case. My point is we can use adapted Schwartz reflection principle to extend u. The harmonic conjugate of u were not necessary to be constant. Suppose $\Omega$ is simply connected domain, then u is real part of holomorphic function $f$. We can always extend $f$ respect to an analytic curve. So we can extend $u$ $\endgroup$
    – Slm2004
    Commented Jul 6, 2012 at 23:57
  • $\begingroup$ Schwartz refers to the mathematician Laurent Schwartz; the Schwartz space of functions is named after him. Schwarz refers to Hermann Schwarz; the Schwarz reflection principle is named after him. $\endgroup$
    – Jiaze Zhang
    Commented Dec 12, 2021 at 14:47

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Certainly not: think of $\Omega$ the unit disk, and $u$ the harmonic extension to $\Omega$ of any continuous, nowhere differentiable function on $\partial \Omega$.

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  • $\begingroup$ I understood the question as: is there a harmonic extension of $u$ on a strictly larger domain $\Omega'$. Or maybe you want a continuous extension to $\Omega'$ which is harmonic on $\Omega'\setminus\partial \Omega$? $\endgroup$
    – Pietro Majer
    Commented Jul 6, 2012 at 17:06
  • $\begingroup$ Yes. If the boundary behavior of u at this smooth portion is not nice enough, of course we can not extend it. What if I add assumption that $\frac{\partial u}{\partial v}$ exists on this portion? $\endgroup$
    – Slm2004
    Commented Jul 6, 2012 at 23:49

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