1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $p=1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.