Show that there is an operator on $H^{2}$ and it's compact.

Question

Let $H^{2}=W_{0}^{2,2}(\Omega)$.

Define $(u,v)=\int_{\Omega} (\triangle u\triangle v+2v\triangle u)\mathrm{d}S$ as an inner product on $H^{2}$.

Define $a(u;v)=\int_{\Omega} (\nabla u\cdot \nabla v-2uv)\mathrm{d}S$ as a bounded linear functional on $H^{2}$.

Show there is an operator $A:H^{2}\rightarrow H^{2}$ such that $a(u;v)=(Au,v) \forall v\in H^{2}$.

Show that $A$ is compact.

How should one approach this problem?

If there is an operator $A$ such that $a(u;v)=(Au,v)$, then $\nabla u\nabla v-2uv=\triangle (Au)\triangle v+2v\triangle(Au)$. So the question is how to find such an $A$?

Answer 1

Due to your boundary conditions, you have $$a(u;v) = \int_\Omega -\Delta u \, v - 2 \, u \, v \, \mathrm{d}S.$$ We define $B\,u := -\Delta u - 2 \, u$. Hence, $B : H^2 \to L^2$ is continuous. Since the embedding of $L^2$ into $(H^2)^*$ is compact, $B : H^2 \to (H^2)^*$ is compact. Now your $A$ is just the composition of $B$ with the inverse of the Riesz map $J : H^2 \to (H^2)^*$. Thus, it is compact.