2
$\begingroup$

I am reading an article and there, author claim that $$‎L(.)=\Delta ‎^2(.)-‎\frac{‎‎\lambda‎‎}{|x|^4}(.): W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \to W_0^{-2,2}(\Omega) ‎$$ is coercive if ‎‎$ ‎0\leq ‎‎\lambda<\Lambda_N=(‎\frac{N^2(N-4)^2}{16})‎$ , because of the following inequality:

For all ‎‎‎$ u‎ ‎\in‎ W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) $ و ‎‎$‎N>4‎$‎‎ ‎ $$ ‎\Lambda_N ‎\int_{\Omega}‎\frac{u^2}{|x|^4}\mathrm{d}x ‎\leq ‎\int_{\Omega} ‎|\Delta u|^2 ‎\mathrm{d}x‎ $$ where $‎\Lambda_N=(‎\frac{N^2(N-4)^2}{16})‎$ is optimal constant.

My question is this that how the inequality with the optimal constant conclude that operator is coercive for ‎‎$ ‎0\leq ‎‎\lambda<\Lambda_N=(‎\frac{N^2(N-4)^2}{16})‎$.

my try:

the norm on the hilbet space $\mathbb{H}=W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ is $$\langle u,v\rangle_{‎\mathbb{‎‎H}}=\int_{\Omega} \Delta u \Delta v dx$$

I must show that $$\langle Lu,u \rangle_{\mathbb H} \geq c ||u||_{\mathbb H}^2 $$ for a positive constant $c$.

with simple calculations and using above inequality, I have showed that $$\langle Lu,u \rangle_{L^2} \geq c ||\Delta u||_{L^2}=c||u||_{\mathbb H}^2 $$ for a positive $ c $

but I must show that $$\langle Lu,u \rangle_{\mathbb H} \geq c||u||_{\mathbb H}^2 $$ .

$\endgroup$
1
$\begingroup$

It appears that you are using the notation $\langle \cdot, \cdot \rangle_{\mathbb{H}}$ for the inner product on you Hilbert space $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $\langle \cdot,\cdot\rangle_{L^2}$ for the dual pairing with the space $W^{-2,2}_0(\Omega)$, which is the co-domain of the operator $L$. The term $\langle Lu,u\rangle_{\mathbb{H}}$ then does not make sense unless $u$ has more differentiability and even then will not be coercive.

The appropriate bi-linear form for coercivity, i.e. for existence of solutions via the Lax-Milgram theorem, is $\langle Lu,u\rangle_{L^2}$ which you have already shown is coercive.

$\endgroup$

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.