= Solution
The difference $w$ of two <weak solutions> lies in the <clamped second-order Sobolev space>. Testing with $w$ yields $\int_U(\Delta w)^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, <integration by parts> gives
$$
\|\nabla w\|_2^2=-\int_Uw\Delta w=0.
$$
The <Poincare inequality> for zero boundary values now implies $\|w\|_2=0$. \b[The <clamped biharmonic problem> has at most one <weak solution>.] Unlike the <Neumann Poisson problem>, no additive constant is allowed by these boundary traces.
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