Timeline for A basic stability question
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 22, 2018 at 15:23 | history | edited | A random mathematician | CC BY-SA 4.0 |
Improved the title, previous title did not accurately describe the question
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| Aug 16, 2018 at 21:48 | vote | accept | A random mathematician | ||
| Aug 16, 2018 at 4:20 | vote | accept | A random mathematician | ||
| Aug 16, 2018 at 20:27 | |||||
| Aug 15, 2018 at 18:14 | answer | added | Willie Wong | timeline score: 2 | |
| Aug 15, 2018 at 17:11 | comment | added | Willie Wong | The answer to which would still be "no" because you can replace $u, v$ by $\lambda u, \lambda v$. So you still need to fix a scale. | |
| Aug 15, 2018 at 17:02 | comment | added | Willie Wong | If you are so happy to move goal posts, you may as well add that $\Omega$ is bounded (to rule out exterior of a ball), and that $|V_i| = 1$ (in fact, a fortiori if $V_i$ is the direction of the gradient of $\nabla u_i$, and $u_i$ is $C^1$, then $V_i \in C^0$). Fixing the size of $V_i$ gets rid of the scaling issue also. In fact, you might as well re-phrase your question as: let $u, v\in C^1(\Omega)$ be two functions that agree on $\partial\Omega$ with non-vanishing gradient. Can $\|u-v\|_{H^1}$ be bounded by $\| \nabla u / |\nabla u| - \nabla v / |\nabla v|\|_{L^2}$? | |
| Aug 15, 2018 at 16:43 | history | edited | A random mathematician | CC BY-SA 4.0 |
added 24 characters in body
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| Aug 15, 2018 at 15:51 | comment | added | A random mathematician | I added the assumption $|V_i|>0$. | |
| Aug 15, 2018 at 15:48 | history | edited | A random mathematician | CC BY-SA 4.0 |
added 15 characters in body
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| Aug 15, 2018 at 15:22 | history | edited | A random mathematician | CC BY-SA 4.0 |
added 26 characters in body
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| Aug 15, 2018 at 13:44 | comment | added | Willie Wong | Please be more thoughtful about tags going forward. Your PDE is obviously not linear (absolute value function is not linear). Your PDE is not elliptic (a fortiori since my answer below; but I don't see any reason for you to believe that it is elliptic to start with). There's no hard dependencies on function spaces, and I don't see any special function entering the discussion. | |
| Aug 15, 2018 at 13:35 | history | edited | Willie Wong |
edited tags
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| Aug 15, 2018 at 13:35 | answer | added | Willie Wong | timeline score: 2 | |
| Aug 14, 2018 at 23:48 | history | asked | A random mathematician | CC BY-SA 4.0 |