Consider the following problem: $$-\Delta \phi + \Delta^2 \phi = 4\pi u^2, \ \Omega$$ $$ \Delta \phi = 0, \partial \Omega$$ $$\phi = 0, \partial\Omega$$ How can I prove that the space of weak solutions of this problem is $H^2\cap H^{1}_0(\Omega)$?
It's clear for me that $\phi \in H^1_0$, but why $\Delta \phi = 0$ leads to the space $H^2\cap H^{1}_0(\Omega)$?
The 'correct' choice of the solution space has to fulfill two basic requirements:
It is clear that the first point is satisfied by $V = H^2(\Omega) \cap H_0^1(\Omega)$.
Now, let $\phi$ be strong / classical solution with regularity $\phi \in C^4(\bar\Omega)$ and it satisfies the equations in a pointwise way. Then, it is clear that $\phi \in V$. For an arbitrary $v \in V$, we have $$ \int_\Omega -\Delta \phi \cdot v \, \mathrm{d}x = \int_\Omega \nabla\phi \cdot \nabla v \, \mathrm{d}x - \int_{\partial\Omega} \frac{\partial\phi}{\partial n} \cdot \underbrace{v}_{=0} \, \mathrm{d}s $$ and $$ \int_\Omega \Delta^2 \phi \cdot v \, \mathrm{d}x = \int_\Omega \Delta \phi \cdot \Delta v \, \mathrm{d}x + \int_{\partial\Omega} \frac{\partial\Delta\phi}{\partial n} \cdot \underbrace{v}_{=0} - \frac{\partial v}{\partial n} \cdot \underbrace{\Delta \phi}_{=0} \, \mathrm{d}s. $$ By adding these identities you can plug in the right-hand side of your PDE and arrive at the weak formulation. (Note that this process is very similar to the incorporation of a (natural) Neumann boundary condition for Poisson's equation)
Along the same lines one can show that a weak solution with enough regularity is also a strong solution.