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Jun 30, 2015 at 19:06 comment added Math604 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 $\Omega$)
Jun 30, 2015 at 18:56 comment added Math604 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.
Jun 30, 2015 at 18:56 vote accept Math604
Jun 30, 2015 at 9:39 comment added Thanasis Stylianou @Math604 By the way, where does this problem come from?
Jun 30, 2015 at 8:58 comment added Thanasis Stylianou @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...).
Jun 30, 2015 at 8:34 comment added Math604 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
Jun 30, 2015 at 8:02 history answered Thanasis Stylianou CC BY-SA 3.0