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I'm trying to show that if $u \in C^4(\Omega)\cap C^1(\overline{\Omega})$ satisfies the weak formulation of the biharmonic problem $$ \int_\Omega \Delta u\Delta \varphi = \int_\Omega f \varphi \ \ \ \ \ \forall \varphi \in H_0^2(\Omega) $$ then $u$ is a classical solution of $$ \Delta^2 u = f \ \text{ in } \Omega \\ u = \frac{\partial u}{\partial \eta} = 0 \ \text{ in } \partial \Omega $$

Here I'm assuming the domain is well behaved (open, bounded and with smooth boundary) and $f \in L^2$.

My doubt is that i get a bit lost still in what aproaches i can use, following this answer i was thinking of doing something like

$$ 0 = \int \Delta u \Delta \varphi - f \varphi \\ =\int_\partial \nabla \varphi \Delta u \cdot\eta \ - \int_\partial \varphi \nabla(\Delta u) \cdot \eta \ + \int (\Delta^2 u- f)\varphi $$

and from there i could conclude that $\Delta^2 u = f$ but I'd have to handle two integrals for the boundary conditions and i can't conclude from there, am I missing something?

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Actually forgot a vital piece of information, $u \in H_0^2$, so therefore the boundary conditions arise from the space where the weak solution lies.

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