I am trying to prove the existence of a weak solution of the problem:
$$
-\Delta^2 u = f \in L^2(U)\\ \\
u|_{\partial U}=\Delta u|_{\partial U} = 0
$$
on the bounded open set $U\subset\mathbb{R}^n$ which has smooth boundary. The weak formulation follows from multiplying by the test function $v$ and integration by parts:
$$
\int_U fv\,dx= \int_U \, (\Delta^2 u) v\, dx= \int_U \Delta u \Delta v \, dx + \int_{\partial U} (v \frac{\partial \Delta u}{\partial n} – \frac{\partial v}{\partial n}\Delta u)dS.
$$
The second boundary vanishes for $\Delta u|_{\partial U} = 0$ and to make the first boundary term vanish we require $v$ to be in the Sobolev space $H^1_0(U)\cap H^2(U)$, so that $v=0$ on the boundary.
Therefore $u\in H^1_0(U)\cap H^2(U)$ is a weak solution if:
$$
\int_U \Delta u \Delta v \, dx = \int_U fv\,dx\,\,\,\, \forall v\in H^1_0(U)\cap H^2(U).
$$
Now I want to use the Lax-Milgram theorem on the bilinear form $B[u,v]=\int_U \Delta u \Delta v \, dx$. My problem is: what norm on $H^1_0(U)\cap H^2(U)$ should I use? At first I thought I could use either one of the norms of $H^1_0(U)$ or $H^2(U)$, since clearly $H^1_0(U)\cap H^2(U)$ is a closed subspace of both spaces. However I realized that this argument must be wrong, otherwise I might as well use the norm of the subspace $H^{8}(U)$ and conclude there exists a weak solution in that space.
Or should I use Lax Milgram for both $H^1_0(U)$ and $H^2(U)$ and conclude that the weak solution is in their intersection?
My second problem: when proving the coercivity of $B[u,v]$, I think I need an inequality like $\int_U |\Delta u|^2 dx \geq C\int_U u^2 dx$ for some constant $C$. I know this holds for $u\in H^2_0(U)$ but I don't see why this should hold in $H^1_0(U)\cap H^2(U)$?
Best Answer
Let $H=H_0^1\cap H^2$. A good norm to work with is $$\|u\|_H=\|\Delta u\|_2$$
As you can see here, this norm is equivalently to the usual one. Also, the inequality you are looking for is true, in fact, you have that $$\int |\Delta u|^2\geq\lambda_1^2\int|u|^2,\ \forall\ u\in H$$
where $\lambda_1$ is the firt eigenvalue associated with $(-\Delta, H_0^1)$. To prove this inequality, note that
\begin{eqnarray} \int |\nabla u|^2 &=& -\int u\Delta u \nonumber \\ &\leq& \|u\|_2\|\Delta u\|_2 \nonumber \end{eqnarray}
From the last inequality, you can conclude by using Poincare inequality. Now you can easily apply Lax-Milgram.