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.

  • $\begingroup$ I forgot to say that I just multiplied the equation by $x \nabla u(x)$ and then integrated as in the case when $ \gamma=0$. So maybe my question is does anyone know the correct function to multiply the equation by and then proceed... thanks $\endgroup$ – Craig Jan 9 '14 at 21:02

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 $p=1$.

  • $\begingroup$ Thanks Athanagor. I didn't realize you answered my question. I will look into your answer and I think this maybe exactly what I am looking for. I will get back to you after I read details. $\endgroup$ – Craig Feb 10 '14 at 4:51

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