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).
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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....
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$\begingroup$ Thank you very much tks for the nice answer. I am not sure I exactly follow what you are suggesting. Can I apply the Lax-Milgram theorem directly to $B$? Note that $$B((u,v),(u,v)) = \int_\Omega | \nabla u|^2 + | \nabla v|^2 - fv dx$$ and so it looks like $B$ isn't co-ercive on $ W^{1,2} \times W_0^{1,2}$. I then though maybe I should work on $ \dot{W}^{1,2} \times W_0^{1,2}$ (the dot means the zero average subspace). But then i think we solve the sytem up to some average (which reduces to the desired pde if the average is zero)... or maybe I am completely missing the point ? thanks $\endgroup$– Math604Commented Jun 30, 2015 at 8:34
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$\begingroup$ @Math604 Yes, this is what I had in mind, it should work similarly to the Neumann problem for the Laplacian (I did not do all the details here so let me know if you encounter any difficulties...). $\endgroup$ Commented Jun 30, 2015 at 8:58
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$\begingroup$ @Math604 By the way, where does this problem come from? $\endgroup$ Commented Jun 30, 2015 at 9:39
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$\begingroup$ The problem is not really coming from anywhere besides the fact I wanted to solve a linear fourth order problem which was not the usual Dirichlet or Navier boundary condition. In particular I didn't want to impose $ u=0$ on $ \partial \Omega$. Regarding the approach on the zero average subspace; I am still getting some terms in the pde which involve a zero average ( i will attempt to edit my answer some details). In any case I will accept your answer. Thanks again. $\endgroup$– Math604Commented Jun 30, 2015 at 18:56
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$\begingroup$ Working on the zero average subspace I end up solving the pde $-\Delta v + u =f$ in $ \Omega$ with $ v=0$ on $ \partial \Omega$ and $ -\Delta u = v-(v)_\Omega$ in $\Omega$ with $\partial_\nu u=0$ on $\partial \Omega$. (here $ (v)_\Omega$ is the average of $v$ over $\Omega$) $\endgroup$– Math604Commented Jun 30, 2015 at 19:06