Solution in $H^2 \cap H^{1}_{0}$ 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 $H^2\cap H^{1}_0(\Omega)$?
 A: The 'correct' choice of the solution space has to fulfill two basic requirements:

*

*It must be complete (and typically, reflexive) space such that you can get good existence results.

*Weak and strong solutions should (up to regularity issues) coincide.

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.
