Timeline for Extending harmonic functions defined in the closure of a bounded smooth domain to some larger domain
Current License: CC BY-SA 4.0
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| Mar 17, 2021 at 12:25 | comment | added | kichr | 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 ! | |
| Mar 17, 2021 at 10:51 | comment | added | Mateusz Kwaśnicki | 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.) | |
| Mar 16, 2021 at 23:13 | comment | added | kichr | 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. | |
| Mar 16, 2021 at 16:04 | comment | added | Mateusz Kwaśnicki | If $\Omega$ has a real-analytic boundary, that would require $\varphi$ to be at least real-analytic on the boundary, right? | |
| Mar 16, 2021 at 8:52 | history | edited | gmvh |
Added top-level tag
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| Mar 16, 2021 at 8:27 | review | First posts | |||
| Mar 16, 2021 at 8:51 | |||||
| Mar 16, 2021 at 8:19 | history | asked | kichr | CC BY-SA 4.0 |