Timeline for Proof of Friedrichs inequality in a domain with simple geometry
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2013 at 17:14 | history | edited | Maciej Skorski | CC BY-SA 3.0 |
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| Nov 15, 2013 at 17:14 | vote | accept | Maciej Skorski | ||
| May 17, 2012 at 23:26 | comment | added | Michael Renardy | By the way, the name is Friedrichs, not Friedrich. | |
| Apr 19, 2012 at 20:17 | answer | added | Bazin | timeline score: 2 | |
| Mar 12, 2011 at 3:45 | comment | added | J.C. Ottem | Granted this question has an easy answer, but is this really worth the votes to close? | |
| Mar 8, 2011 at 15:25 | comment | added | Willie Wong | Ack, the outer exponent in the far right of the displayed equation should be $1/2$, not $2$. | |
| Mar 8, 2011 at 15:24 | comment | added | Willie Wong | This question is perhaps not suitable for this website. Anyway, a quick sketch for your specific case: using a density argument it suffices to prove for $C^\infty\cap H^1$. For smooth functions, use the fundamental theorem of calculus to get $$ | u(r,\theta) | \leq |u(1,\theta)| + \int_1^2 |\partial_r u(s,\theta)| ds \leq |u(1,\theta)| + \left(\int_1^2 | DU(s,\theta) |^2 ds\right)^2 $$ Now square, integrate over $r,\theta$, and you are done. | |
| Mar 8, 2011 at 15:13 | history | edited | Willie Wong |
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| Mar 8, 2011 at 13:51 | history | asked | Maciej Skorski | CC BY-SA 2.5 |