Timeline for Showing a normal-derivative operator is a (sort of) contraction (related to Crandall-Liggett and PDEs)
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 2, 2014 at 13:14 | comment | added | TheBook | Yes, $y$ is normal. Sorry, yes, as you wrote, just take $|f|$ in the definition to be some given constant. As for $u^m$, that should be $u_t^m$ (and/or $u_s^m$). I missed out the subscript to avoid repetition. | |
| May 29, 2014 at 7:46 | comment | added | username | Just to make sure I understand your notations: $y$ is the normal to the boundary? What is $f$ in the definition of $D(A)$ : as it does not appear in the problem, do you mean $|v|_{L^\infty(\partial\Omega)}\leq M$, is it that $g=f$? In your attempt, line 1, what is $u^m$? | |
| May 28, 2014 at 10:09 | history | edited | TheBook | CC BY-SA 3.0 |
edited title
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| May 7, 2014 at 9:26 | history | edited | TheBook | CC BY-SA 3.0 |
added 311 characters in body
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| May 6, 2014 at 14:10 | history | asked | TheBook | CC BY-SA 3.0 |