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Let $u_n\in W_{0}^{1,p}(\Omega)$ be a positive weak solution of the equation: $$ -\Delta_p u=\frac{f_n(x)}{(u+\frac{1}{n})^\delta}\text{ in }\Omega. $$

Let $p=N$ and $f\in L^m(\Omega)$ for some $m>1$. Then how $u_n$ is uniformly bounded in $W_{0}^{1,p}(\Omega) $. This result is proved in Lemma 4.5 in the following article: https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf

Can you help me with it. It is written as a simple fact.

Here $\Omega$ is a bounded smooth domain in $\mathbb{R}^N$, $f$ is a nonnegative but not identically zero function in $\Omega$.

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Okay, as you written the result is proved in the article for $1 < p < N$. They say that the case $p=N$ can be proved similarly. You are now asking about this case.

Remember that $\Omega \subset \mathbb{R}^N$ and $\Delta_N u = \text{div}(|\nabla u|^{N-2} \nabla u).$ The Sobolev inequality tells us that $W^{1,N}_0$ is embedded in $L^q$ for all $q\geq 1$.

Then

$$\begin{aligned}\|\nabla u_n\|_{L^N}^N = \int |\nabla u_n|^{N-2} \nabla u \cdot \nabla u= -\int \Delta_N u_n \cdot u_n &=\int \frac{f_n u_n}{(u_n+\frac{1}{n})^\gamma} \\ &\leq \int f_n u_n^{1-\gamma} \\ &\leq \|f\|_{L^m} \|u_n^{1-\gamma}\|_{L^{m'}} \\ &=\|f\|_{L^m} \|u_n\|_{L^{(1-\gamma)m'}}^{1-\gamma} \\ &\leq C \|f\|_{L^m} \|\nabla u_n\|_{L^N}^{1-\gamma} \end{aligned}$$

Hence

$$\|\nabla u_n\|_{L^N} \leq (C \|f\|_{L^m})^{\frac{1}{N+\gamma-1}}.$$

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  • $\begingroup$ Ok. I agree but one confusion is that suppose we have given $m>1$ and $\gamma\in(0,1)$. Then to us ethe above embedding theorem one need that $(1-\gamma)m'\geq 1$. Why this is true? $\endgroup$ – Mathlover Jan 15 at 19:09
  • $\begingroup$ Please help me with this argument. I have understood all the arguments except this. Thank you very much. $\endgroup$ – Mathlover Jan 15 at 19:13
  • $\begingroup$ I have got. If this is less than 1 then using Holder inequality one can get the uniform bound. Thanks. $\endgroup$ – Mathlover Jan 18 at 7:05

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