Timeline for Elliptic partial differential equations with Robin boundary condition and domain of fractional power of Robin Laplacian operator
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| S Sep 27, 2022 at 13:20 | history | suggested | Hannes | CC BY-SA 4.0 |
Fixed several typos (there were $v$s in the definition of the norms of $u$) and some layout
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| Sep 27, 2022 at 12:43 | comment | added | Hannes | @username That's what I meant, agreed. | |
| Sep 27, 2022 at 12:42 | review | Suggested edits | |||
| S Sep 27, 2022 at 13:20 | |||||
| Sep 26, 2022 at 17:12 | comment | added | username | @Hannes Let us call $N(u) = \sqrt{ \| \nabla u \|^2_{L^2(\Omega)} + \| u \|^2_{L^2(\partial\Omega)} }$ and $\| u \| = \sqrt{ \nabla u \|^2_{L^2(\Omega)} + \| u \|^2_{L^2(Omega) }}$. To show $N(u)\leq C \| u \|$ yo{u need at least a continuous embedding of $H^1(\Omega)$ in $L^2(\partial \Omega)$. It happens to be compact. In the other direction, yes, you use Rellich-Kondrachov (for $\| u \|_{L^2(\Omega)}=1$). | |
| Sep 26, 2022 at 14:42 | comment | added | monotone operator | @Glorfindel Thank you for your correction!!! | |
| Sep 26, 2022 at 13:18 | comment | added | Hannes | @username you only need that $H^1(\Omega) \hookrightarrow L^2(\Omega)$ is compact, or am I mistaken? //// For the second question of OP, this is the Kato square root property which is always true for operators induced by symmetric forms as in the present case, see Kato's "Perturbation Theory", Chapter 6, §2.6. | |
| Sep 26, 2022 at 7:28 | comment | added | username | Whether the norms are equivalent doesn't depend on Robin bdy conditions. It is clear that they are, you just need to know that the embedding $H^1(\Omega) \rightharpoonup L^2(\partial\Omega)$ is compact. One direction is immediate : bounding bdy $L^2$ with $H^1$. In the other direction, supppose bdy $L^2$ +grad goes to zero while $H^1=1$. Take a weakly convergent subsequence. then the $L^2$ bdy norm converges strongly to zero, so the limit is constant (zero gradient in a connected domain) and with zero trace, so it is null, and $L^2$ norm $=1$, a contradiction. The usual proof.. | |
| Sep 26, 2022 at 7:10 | history | edited | Glorfindel | CC BY-SA 4.0 |
added 67 characters in body
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| S Sep 26, 2022 at 4:08 | review | First questions | |||
| Sep 26, 2022 at 7:10 | |||||
| S Sep 26, 2022 at 4:08 | history | asked | monotone operator | CC BY-SA 4.0 |