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Let $\Omega$ be a smooth bounded domain of $\mathbb{R}^N$ where $N\geq 2$. Consider the Laplace equation with a Neumann boundary condition
$$ -\Delta u = 0 \quad\mbox{in } \Omega, \qquad \frac{\partial u}{\partial \mathbf{n}} = \varphi(x) \quad\mbox{on } \partial \Omega, $$ where $\varphi$ is a sufficiently smooth function defined on $\partial\Omega$ for which the solutions are in $C^{2}(\overline{\Omega})$. Assume the existence of a positive solution $u_0 > 0$ in $\overline{\Omega}$, meaning that the solution $u_0$ is strictly positive in $\overline{\Omega}$.

My question is that: can $u_0$ be extended to some domain $\Omega^\prime \supset \overline{\Omega}$ in the manner that $u_0 > 0$ in $\Omega^\prime$, and $-\Delta u_0 = 0$ in $\Omega^\prime$ ?

If yes, then I think the Harnack inequality provides us with the assertion that $\sup_{\Omega}u \leq c \inf_{\Omega}u$ for any positive solution $u$ of this problem, where $c>0$ is a constant which does not depend on $\varphi$. Is it true ?

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  • $\begingroup$ If $\Omega$ has a real-analytic boundary, that would require $\varphi$ to be at least real-analytic on the boundary, right? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Mar 16, 2021 at 16:04
  • $\begingroup$ Thank you for your comment. I think you mean a reflection argument. But, I thought $-\Delta u=0$ in $\Omega$ and $u\in C^2(\overline{\Omega})$ imply that $-\Delta u = 0$ in $\Omega^\prime$, where $\Omega^\prime \supset \overline{\Omega}$, since $\overline{\Omega}$ is a closed set. But, this is not true, you mean. Thanks. $\endgroup$
    – kichr
    Commented Mar 16, 2021 at 23:13
  • $\begingroup$ I did not refer to reflection. If $u_0$ extends to a larger domain, it is real-analytic on a neighbourhood of the boundary, and so $\varphi$, its normal derivative along the boundary, is real-analytic, too. (But it is indeed instructive to take a look at the simpler case when $\Omega$ is the half-space.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Mar 17, 2021 at 10:51
  • $\begingroup$ Thank you for your further comment. I understand what you meant. You meant that if $u_0$ is extended to a larger domain $\Omega^\prime$ as a harmonic function in $\Omega^\prime$, then $u_0$ should be real analytic in $\Omega^\prime$, so that so is $\frac{\partial u_0}{\partial \mathbf{n}}$, since $\partial \Omega \subset \Omega^\prime$. Consequently, it is necessary that $\varphi$ is real analytic. It's nice. Your comments are very helpful for me. Thanks again ! $\endgroup$
    – kichr
    Commented Mar 17, 2021 at 12:25

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