Let $\Omega \subset \mathbb{R}^N$, $N\geq 5$. Now assume the biharmonic problem with singular term as follow \begin{cases} ‎\Delta^2u=‎‎\lambda ‎‎\dfrac{u}{|x|^4}‎‎+u^{‎p}‎ & \mathrm{in}‎\hspace{.2cm}‎ \Omega \\ u=-\Delta u =0 & \mathrm{‎on}‎\hspace{.2cm}‎ \partial \Omega ‎ \end{cases}‎

Where $\lambda=\dfrac{N^2(N-4)^2}{16}$ is the optimal constant of Rellich inequality and $p=\dfrac{N+4}{N-4}$ is the critical sobolev exponent. I know that there exist at least one positive solution but I am not able to show it by classical methods . (obviously $u \equiv 0$ is a trival solution).

I know that by using improved Hardy-Rellich inequality with remider term, we get

$$\|u\|^2= \int_{\Omega} \Big((\Delta u)^2 - \lambda \dfrac{u^2}{|x|^4}\Big) \,\mathrm{d}x $$

as an equivalent norm on the space $W_0^{2,2}(\Omega)$.

Now I define functional $J: W_0^{2,2}(\Omega) \to \mathbb{R} $ as follow

$$J(u)=\dfrac{1}{2} \int_{\Omega}(\Delta u)^2 - \dfrac{\lambda}{2} \int_{\Omega} \dfrac{u^2}{|x|^4} \,\mathrm{d}x - \dfrac{1}{p+1} \int_{\Omega} u^{p+1} \,\mathrm{d}x $$ I know critical points of this functional are weak solutions of problem.

By definition of the above norm I can write functional as follow.

$$J(u)=\dfrac{1}{2} \|u\|^2 - \dfrac{1}{p+1} \int_{\Omega} u^{p+1} \,\mathrm{d}x $$

If $1\leq p < \dfrac{N+4}{N-4} $ then By the help of semicontinuty of norm and compactness of imbedding $W_0^{2,2} (\Omega) \hookrightarrow L^{p}(\Omega)$ I can minimize this functional on the unite sphere of $L^{p+1}(\Omega)$ by classical methods of calculus of variation.

But $p=\dfrac{N+4}{N-4}$ is critical sobolev imbedding and the above argument break down.

Can some one help me to show the existence of at least one positive solution.

So thanks for your help.

  • $\begingroup$ Are you sure that norm is equivalent to the $W_0^{2,2}$ one? I think this may be wrong in two ways? Also this looks like it may be close to some sort of Brezis-Nirenberg type result (is $ \lambda$ was small). Can you type out the equivalent second order result? $\endgroup$ – Math604 Aug 26 '15 at 19:11

Your Answer

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

Browse other questions tagged or ask your own question.