I am interested in nonnegative solutions of 
$-div( e^{-\gamma(x)} \nabla u(x)) = e^{-\gamma(x)} u(x)^p$ in $\Omega$ with $ u=0$ on $ \partial \Omega$. 


Or instead the equation $ -\Delta u + \nabla \gamma(x) \cdot \nabla u = u^p$. I would like to know if there are known Pohozaev type results regarding the non-existence of non-zero non-negative solutions. 


I have tried a Pohozaev argument but I need to impose a smallness condition on $ \nabla \gamma$ and I am not sure this should be required. I would prefer some structural assumptions on $ \gamma$ besides a smallness assumption. 


thanks for your remarks. 



Pohozaev result for equations with weights
Here is one for any $p>0$. Doing the usual change of unknown $v(x)=u(x)\exp(-\gamma(x)/2)$, you obtain
$$
-\Delta v + \left(\exp\left(-\frac{\gamma}{2}\right) \Delta \left(\exp\left(\frac{\gamma}{2}\right)\right)\right) v = v^p \exp\left(\frac{(p-1)\gamma}{2}\right) 
$$ 
Note that 
$$
\exp\left(-\frac{\gamma}{2}\right) \Delta \left(\exp\left(\frac{\gamma}{2}\right)\right)= \frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla\gamma\right|^2
$$
Let $\lambda(\gamma)$ be the first dirichlet eigenvalue of 
$$
-\Delta \psi + \left(\frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla\gamma\right|^2\right) \psi = \lambda(\gamma)\psi 
$$
It always exists if $\gamma \in C^2(\bar\Omega)$, but it could be positive or negative. However the corresponding eigenvector $\psi\in H^1\_0(\Omega)$ can be taken postive. 


Integrate by parts against $\psi$ to obtain
$$
\int\_\Omega \left(\lambda(\gamma)v - \exp\left(\frac{(p-1)\gamma}{2}\right)v^{p}\right) \psi =0. 
$$
If $\lambda(\gamma)<0$ the map
$$
z\to \lambda(\gamma)z -\exp\left(\frac{(p-1)\gamma}{2}\right) z^{p}
$$
is negative over $\mathbb{R}^{+}$ , and therefore a non-trivial $v$ leads to a contradiction, for any $p$. So one structural condition is 



> 
> if $\lambda(\gamma)\leq 0$, then there is no nontrivial solution, for any $p$. 
> 
> 
> 


An explicit corollary is :



> 
> Let $\lambda\_{+}(\Omega)$ be the first dirichlet eigenvalue of the Laplacian on $\Omega$. If
>  $$
> \frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla \gamma\right|^2 \leq -\lambda\_{+}(\Omega) 
> $$
>  there is no non-trivial positive solutions.
> 
> 
> 


In the special case $p=1$, there could be non trivial solutions, of course, depending on $\Omega$. If $\lambda\_{+}(\Omega)>1$, which happens for example when $\Omega\subset(0,\sqrt{d}\pi)^{d}$, choose $\gamma=2\ln(\psi)$ where $\psi$ is the solution of 
$$-\Delta \psi = \lambda\_{+}(\Omega^\prime) \psi \mbox{ on } \Omega^\prime,\quad \psi=0 \mbox{ on } \partial\Omega^\prime, \int\_\Omega \psi=1
$$ for some $\Omega^\prime$ containing $\Omega$. The problem then become $$ -\Delta v = (\lambda\_{+}(\Omega^\prime) +1)v \mbox{ in } \Omega.$$
 As $\Omega^\prime$ grows, $(\lambda\_{+}(\Omega^\prime) +1)$ will reach $\lambda\_{+}(\Omega)$ at some point, leading to a non-trivial solution.
In general, if  

$$
\frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla \gamma\right|^2 = \lambda\_{+}(\Omega)-1,
$$
there is a non trivial solution for $p=1$.


