I am interested in the solvability of 
$$ \Delta^2 u + u = f(x) \mbox{ in } \Omega $$ with $ \partial\_\nu u = \Delta u=0$ on $ \partial \Omega$ where $ f(x)$ is some smooth bounded function on $ \Omega$ (a bounded smooth domain in $ R^N$). I have tried the variational approach but cannot obtain the correct boundary conditions. I assume one must be able to use a Fredholm alternative approach to solve (but I am unable). 



Elliptic pde with bilaplacian; boundary conditions.
You will not get a direct variational structure (because of the boundary conditions) but there is a mixed approach that will work on your case: Set $-\Delta u=v$ and obtain the following system:
$$
\begin{equation}
\left\{
\begin{array}{rl}
-\Delta v+u=f & \text{in }\Omega , \\
v=0 & \text{on }\partial \Omega%
\end{array}%
\right.\ \text{ and }\ \left\{
\begin{array}{rl}
-\Delta u=v & \text{in }\Omega , \\
\partial\_n u=0 & \text{on }\partial \Omega .%
\end{array}
\right.
\end{equation}
$$
Now you can define a bilinear form $B:\big(W^{1,2}\times W^{1,2}\_0\big)^2\rightarrow \mathbb R$ by
$$
B\big((u,v),(\phi,\psi)\big):=\int\_\Omega \big(\nabla u\cdot\nabla \phi+\nabla v\cdot\nabla \psi-v\,\phi-(f-u)\;\psi\big)dx
$$
which will provide you with a solution $(u,v)\in W^{1,2}\times W^{1,2}\_0$. Since the domain has a smooth boundary you can then use standard regularity results (see for example the book of Evans on PDE...) and argument that $u$ is a strong solution to the original problem.


If you want to use Fredholm-type arguments you need to prove that the spectrum of the bilaplace with these boundary conditions lies on the positive axis. You then get that the resolvent $R(-1,\Delta ^2)=-1-\Delta^2$ is invertible in $L^2$.


All these assuming you are searching for solutions in the Hilbert space setting....


