I am doing Exercise 6.3 from Evans' PDE. I know this problem is already addressed in Unique weak solution to the biharmonic equation. But I wish to check my proof works.
Let $U \subset \mathbb{R}^n$ be an open, bounded subset with smooth boundary. Consider the boundary-value problem for the biharmonic equation $$ \begin{cases} \Delta^2 u = f \ \ \ \ \ \ \ \text{in $U$}; \\ u = \frac{\partial u}{\partial \nu} = 0 \ \ \ \text{at $\partial U$}. \end{cases} $$ We call a function $u \in H^2_0(U)$ is a weak solution of this problem if $\int_U \Delta u \Delta v = \int_U fv$ for all $v \in H^2_0(U)$. Given $f \in L^2$, prove that there exists a unique weak solution to the above equation.
My idea: Fix $f \in L^2$. We could first solve for a weak solution $w \in H^1_0(U)$ such that $\Delta w = f$ weakly by using Lax-Milgram. In particular, for $v \in C_c^\infty(U)$, we have $(f,v) = \int Dw \cdot Dv$. But by invoking regularity theorm, we know that $w \in H^2(U)$. Therefore, do once more integration by parts, we have $(f,v) = \int Dw \cdot Dv = -\int w \Delta v$. Next we could solve for $u \in H^1_0(U)$ such that $\Delta u = w$. Likewise $u \in H^2(U)$, for any $v' \in C_c^\infty(U)$, we have $\int w v' = \int Du \cdot Dv' = -\int (\Delta u) v'$. Letting $v' := \Delta v$ gives that $(f,v) = \int \Delta u \Delta v$. The uniqueness is given by Lax-Milgram. The remaining is to approximate $H^2_0$ function by $C_c^\infty$ functions.
But I quite doubt it, and I didn't use the condition that $\frac{\partial u}{\partial \nu} = 0$ at the boundary as well. Could anyone take a look at my idea and tell me what's going wrong?
Edit: I think the boundary conditions are used to define the weak solutions. But I am still not sure if my approach works or not.
