Consider smooth positive solutions $u_m$ of
$$-\Delta u_m(x) = u_m(x)^p \quad \mbox{ in } \Omega$$ with $u_m=0$ on $ \partial \Omega$. My interest is in obtaining some sort of global integral estimates independent of $m$. Here $p>1$ and you can assume its close to $1$. The reason I am asking about integral estimates is that I do not want to apply any blow up arguments. The real equation has some other terms with parameters that go to infinity; but they drop out on various integral type estimates.
thanks
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$\begingroup$ Do you assume that $u>0$ inside $\Omega$? Also why don't you just integrate against $u_m$ and apply Sobolev inequality (should work if $p$ is only slightly above $1$) $\endgroup$– fedjaCommented Jun 18, 2018 at 21:31
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$\begingroup$ yes i am assuming $u>0$. Maybe i missing something but I thought this doesn't give me anything. So i was under the impression the best initial estimate i can get is something like $ \int_\Omega u_m(x)^p \delta(x) dx \le C $ ($C$ independent of $m$) where $ \delta$ is the distance to the boundary. $\endgroup$– Math604Commented Jun 18, 2018 at 22:19
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$\begingroup$ then in the case of the above equation one can use some weighted estimates of Souplet, Quittner to start a bootstrap (it is presicely this part is causing problems with the extra advection terms that i didn't add) $\endgroup$– Math604Commented Jun 18, 2018 at 22:22
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$\begingroup$ Yes, that is what you obtain from integrating against the first eigenfunction of the Laplacian. However Sobolev tells you that you can estimate some $L^q$-norm with fixed $q>2$ by some close to $1$ power of $L^{p+1}$-norm and, thereby, by Holder by some power of $L^{p/3}$-norm (if $p$ is close enough to $1$), but the latter is controlled by your expression for domains with decent boundary. Am I talking nonsense? (this happens sometimes :lol:) $\endgroup$– fedjaCommented Jun 19, 2018 at 1:55
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$\begingroup$ i pretty much talk nonsense continuously these days... If you add a little detail to your above comments I can attempt to see if i understand (sorry) $\endgroup$– Math604Commented Jun 19, 2018 at 3:44
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OK, let me elaborate. We assume that $\Omega$ is bounded with smooth boundary (this can be relaxed a bit, but we still need something for the naive argument below to work). Let $v$ be the first eigenfunction of the Laplacian normalized by $\int_\Omega v=1$. Then, integrating against $v$ and transferring the Laplacian to $v$, we get $$ \lambda \int_\Omega uv=\int_\Omega u^pv\ge \left[\int_\Omega uv\right]^p $$ whence we can control $\int_\Omega uv$ and, thereby $\int_\Omega u^pv$. Since $v$ is comparable to the distance to the boundary, we can also use Holder to control $$ \int_\Omega u^{p/3}\le \left[\int_\Omega u^pv\right]^{1/3}\left[\int_\Omega v^{-1/2}\right]^{2/3}\,. $$ Now integrate against $u$ itself. We'll get $$ \int_\Omega|\nabla u|^2\le\int_\Omega u^{p+1}=\|u\|_{p+1}^{p+1}\,. $$ The left hand side dominates $\|u\|_q^2$ with $\frac 1q=\frac 12-\frac 1n$ (Let's say $n>2$, otherwise you can just take any finite $q$ you want).
Thus $$ \|u\|_q\le C\|u\|_{p+1}^{\frac{p+1}2}=C\|u\|_{p+1}^{1+\delta(p)} $$ with $\delta(p)\to 0$ as $p\to 1$.
Let $N=\|u\|_{p+1}$. Since $\|u\|_s$ is a logarithmically convex function of $1/s$ (Holder again), we conclude that $$ \|u\|_{p/3}\ge cN^{1-\Delta} $$ with $\Delta=\frac{\frac 3p-\frac 1{p+1}}{\frac 1{p+1}-\frac 1q}\delta(p)\to 0$ as $p\to 1$. Once $\Delta<1$, we can control $\|u\|_q$ by $\|u\|_{p/3}$ and we are done. (I hope that by "integral estimates" you meant some $L^q$ control with respect to the volume measure; otherwise you have to tell me what exactly you wanted).
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$\begingroup$ thanks you very much for the detailed answer... let me attempt to digest it and see if i understand. $\endgroup$– Math604Commented Jun 19, 2018 at 16:17
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$\begingroup$ I think i understand the proof. Thank you very much. $\endgroup$– Math604Commented Jun 19, 2018 at 18:33
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$\begingroup$ @Math604 You are welcome. Does it work in your setting or you need something even more robust? In the latter case, just post the actual setup. $\endgroup$– fedjaCommented Jun 19, 2018 at 21:20
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$\begingroup$ so this was exactly the type of thing i was looking for... now whether it works in my situation might be a problem. I think unless i can get a $C^{0,1}$ estimate on the first eigenfunction (indepenent of the parameter) this might be a problem. But i can make some other assumptions (maybe) to get around this. So the actually equation i was working with is $ -\Delta u_m(x) + \lambda_m a(x) \cdot \nabla u_m(x) = u_m(x)^p$ where $a$ is smooth and divergence free but where $ \lambda_m \rightarrow \infty$. $\endgroup$– Math604Commented Jun 19, 2018 at 22:47
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$\begingroup$ @Math604 OK, let me think a bit then. If I find some way around this, I'll let you know :-) $\endgroup$– fedjaCommented Jun 19, 2018 at 23:52