Let $ \Omega\subset\mathbb{R}^d $ be a Lipschitz domain and $ A=(a_{ij}(y)):\Omega\to\mathbb{R}^{d\times d} $ be a matrix valued function with uniformly elliptic conditions i.e. $ \lambda|\xi|^2\leq a_{ij}(y)\xi_i\xi_j\leq\lambda^{-1} |\xi|^2 $ with $ \lambda>0 $. Consider the equation
$$ -\operatorname{div}(A\nabla u)=F+\operatorname{div}(G) $$
where $ F\in L^q(\Omega) $ and $ G\in L^p(\Omega) $ with $ p,q $ to be determined. For $ \psi\in C_0^{\infty}(\Omega) $, we can choose test function $ \psi^2 u $ and get
$$ \int_{\Omega}|\nabla u|^2|\psi|^2dx\leq C\left\{\int_{\Omega}|u|^2|\nabla\psi|^2dx+\int_{\Omega}|G|^2|\psi|^2dx+\int_{\Omega}|F||u||\psi|^2dx\right\}. $$
If $ B\subset 2B\subset\Omega $ with $ B(x,r) $, we can choose $ \psi $ be the cut off function $ \phi\in C_{0}^{\infty}(2B) $ with $ \phi\equiv 1 $ in $ B $ and $ |\nabla\phi|\leq\frac{C}{r} $. When $ d\geq 3 $, we can get that for $ q=2d/(d+2) $,
\begin{eqnarray}
\int_{\Omega}\phi^2|F||u|dx&\leq&\left(\int_{\Omega}(\phi|u|)^{2d/(d-2)}dx\right)^{(d-2)/(2d)}\left(\int_{\Omega}(\phi|F|)^qdx\right)^{1/q}\\
&\leq&\left(\int_{\Omega}|\nabla(\phi u)|^2dx\right)^{1/2}\left(\int_{\Omega}(\phi|F|)^qdx\right)^{1/q}\\
&\leq&\frac{1}{4}\int_{\Omega}|\nabla u|^2\phi^2 dx+\frac{1}{4}\int_{\Omega}|\nabla\phi|^2|u|^2dx+C\left(\int_{\Omega}|\phi F|^qdx\right)^{2/q}.
\end{eqnarray}
Then we have
$$ \left(\int_{B}|\nabla u|^2dx\right)^{1/2}\leq\frac{C}{r}\left(\int_{2B}|u|^2dx\right)^{1/2}+C\left(\int_{2B}|G|^2dx\right)^{1/2}+r\left(\int_{2B}|F|^qdx\right)^{1/q},$$
where $ C $ is a constant independent of $ u $.
I want to ask can we generalize the Caccioppoli inequality to the dimension $ d=2 $? As the Sobolev index $ 2d/(d-2) $ may fail, I can not go on. Can you give me some hints or references?
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asked Dec 17, 2021 at 8:28
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$\begingroup$ 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. $\endgroup$– usernameCommented Dec 17, 2021 at 10:32
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$\begingroup$ @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. $\endgroup$– Luis Yanka AnnaliscCommented Dec 17, 2021 at 10:51
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$\begingroup$ 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. $\endgroup$– usernameCommented Dec 17, 2021 at 18:38
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$\begingroup$ @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? $\endgroup$– Luis Yanka AnnaliscCommented Dec 18, 2021 at 1:53
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$\begingroup$ 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 $q=1$? $\endgroup$– usernameCommented Dec 19, 2021 at 17:07
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