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



boundary integral estimates for elliptic pde
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).


