Let $U\subset\mathbb{R}^{n}$ be an open subset (with sufficiently smooth boundary) and consider the boundary value problem $$\begin{cases}\Delta^{2}u=f \ \text{on $U$} \\ u=\Delta u=0 \ \text{on $\partial U$} \end{cases}.$$ To find a unique weak solution, I want to apply the Riesz representation theorem or Lax-Milgram theorem to an appropriate Sobolev space. My problem is finding this appropriate Sobolev space, that is, I'm trying to find the weak formulation of the above problem.
I multiplied the above equation with a sufficiently smooth test function $v$ and found that \begin{align*} \int_{U}\Delta^{2}u \cdot v&=\int_{\partial U}\frac{\partial\Delta u}{\partial n}\cdot v-\int_{\partial U}\Delta u\cdot\frac{\partial v}{\partial n}+\int_{U}\Delta u\cdot\Delta v\\ &=\int_{\partial U}\frac{\partial\Delta u}{\partial n}\cdot v+\int_{U}\Delta u\cdot\Delta v. \end{align*} I have seen that $H^{2}(U)\cap H_{0}^{1}(U)$ is the Sobolev space we are looking for, but how do I deduce that? Is there a general approach?
Of course we need $v\in H^{2}(U)$ in order to apply $\Delta$ to it. Furthermore, it looks like we are forcing the integral on the boundary to be zero by requiring $v=0$ on $\partial U$, i.e. $v\in H_{0}^{1}(U)$. Also, in order to apply the above mentioned theorems, one needs $u$ and $v$ to lie in the same Sobolev space. But if $u\in H^{2}(U)\cap H_{0}^{1}(U)$ is a weak solution, then we do not have $\Delta u=0$ on $\partial U$.