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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 $$\langle Lu,u \rangle_{\mathbb H} \geq c||u||_{\mathbb H}^2 $$ .

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1 Answer 1

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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.

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