Timeline for Caccioppoli inequality in $ \mathbb{R}^2 $
Current License: CC BY-SA 4.0
9 events
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| Dec 19, 2021 at 17:07 | comment | added | username | You are mixing two different things : 1/ Caccioppoli estimates and 2/ $L^1$ right hand side. To be clear: you can derive Caccioppoli estimates in dimension 2 with $q>1$. But what you are asking is not related to Caccioppoli estimates particularly you are asking about the case when $d=2$ and $q=1$. Let us start from the beginning : how do you define a solution when $q=1$? | |
| Dec 18, 2021 at 16:58 | history | edited | Luis Yanka Annalisc | CC BY-SA 4.0 |
edited title
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| Dec 18, 2021 at 1:53 | comment | added | Luis Yanka Annalisc | @username I have already known how to do the case when $ q>1 $. I do not understand what you say well. I know that by using dulity arguments and the $ W^{1,p} $ where $ p>2 $, we can get the estimates for $ p<2 $, but I do not know how to use the Gerhing Lemma. Can you give me more precise explanation？By the way, Is the generalization right? | |
| Dec 18, 2021 at 1:44 | history | edited | Luis Yanka Annalisc | CC BY-SA 4.0 |
added 25 characters in body
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| Dec 17, 2021 at 18:38 | comment | added | username | So, as I said: if $d=2$ and $F$ is in $L^q(\Omega)$, $q>1$, then yes the Caccioppoli inequality holds. In the limit case $q=1$, The answer is more complicated. You first have to define your solution as a duality solution, or a renormalized solution, as Lax--Milgram doesn't apply. You obtain a solution in $W^{1,s}$, with $s<2$. Then, if $A$ is bounded above, you can gain just a little by a Gerhing Lemma type argument, and obtain an actual $H^1$ solution. | |
| Dec 17, 2021 at 10:51 | comment | added | Luis Yanka Annalisc | @username Now the problem is edited. When $ d=2 $, $ q=2\times 2/(2+2)=1 $, I want to ask if we can get the inequality for it. | |
| Dec 17, 2021 at 10:46 | history | edited | Luis Yanka Annalisc | CC BY-SA 4.0 |
edited body
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| Dec 17, 2021 at 10:32 | comment | added | username | There is a bit of muddling between $G$, $F$ and $f$ and $p$ and $q$. The question is what $p$ you want : if you want $p>1$, it works verbatim since you don't have to use the strongest embedding availble: just use the one given for $p/(p-1)$. If you want $p=1$, you are in trouble. | |
| Dec 17, 2021 at 8:28 | history | asked | Luis Yanka Annalisc | CC BY-SA 4.0 |