Inner product? Yes or no?

Question

I define an "inner product" on $H_0^2(U)$ where $U \subset R^n$ is bounded open set:

$$\langle u,v\rangle = \int_U \Delta u \Delta v dx.$$

I need this when trying to find a weak solution for my PDE problem: $$\Delta^2 u= f \textrm{ on } U, \quad u=\frac{\partial u}{\partial n}=0 \textrm{ on } \partial U.$$

I have problems with condition $\langle u,u\rangle=0$ if and only if $u=0$.

If we take into account $u=\frac{\partial u}{\partial n}=0 $ on $\partial U$, i suppose that this condition holds on space $V:=\{ u \in H^2(U), u=\frac{\partial u}{\partial n}=0 $ on $\partial U \}$ but i am not sure.

Also in my opinion $V \neq H_0^2(U) = \{ u \in H^2(U), u=0 $ on $\partial U \}$ and this is trubling me. Any contribution appreciated.

Answer 1

Yes it is an inner product when $\Omega$ is conneted. Note that the condition $\langle u,u\rangle=0$ implies that $u$ is weakly harmonic. Using Weyl's lemma of these lecture notes, $u$ can be represented by a $C^\infty(\Omega)$ function. Then conclude as in this thread.

Answer 2

I suggest you to see link Or link Containing discussions about weak solution of your problem.