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I found the next exercise in a book. I'm stuck at proving the coercivity of the bilinear functional in the variational formulation of the problem.

Suppose that $\Omega$ is a regular $\mathcal{C}^1$ bounded open set. Prove the existence and uniqueness for the following problem with Fourier limit conditions $$ \begin{cases} -\Delta u&=f & \text{in }\Omega \\ \frac{\partial u}{\partial n}+u&=g & \text{in } \partial \Omega \end{cases}$$ where $f \in L^2(\Omega)$ and $g$ is the trace on $\partial \Omega$ of a function in $H^1(\Omega)$.

Can you please give me a hint on how to prove the coercivity? Thank you.

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You have to show that $a(v,v)=\int_{\Omega}|\nabla v|^2+\int_{\partial\Omega}|v|^2dx\geq \alpha ||v||_{W^1(\Omega)}$, right? In order to bound by below the integral on $\partial\Omega$ we may have to work with the definition of the integral on the boundary (with partitions of unity etc...). –  Davide Giraudo Jan 29 '12 at 17:28
    
@DavideGiraudo: That's right, that's what I am trying to prove. I will try the approach you mention, although I'm not very familiar with the partitions of unity. –  Beni Bogosel Jan 29 '12 at 17:46
    
I'm not sure it will work, but it's the only thing I see. The problem is much simpler when we have $-\Delta u+cu=f$ in $\Omega$ where $c$ is a bounded positive function with $c(x)\geq c_0>0$ for almost every $x$. Maybe we have to approach the solution by this way, for example we have an unique $u_k\in H^1(\Omega)$ such that $-\Delta u_k+\frac 1k= f$ in $\Omega$ and $\frac{\partial u_k}{\partial n}+u_k=g$ on $\partial \Omega$. –  Davide Giraudo Jan 29 '12 at 17:49
    
Maybe in the last approach (where I missed a $u_k$ after $\frac 1k$) we can show that the sequence $\{u_k\}$ is bounded in $H^1(\Omega)$. Then we extract a weakly convergent subsequence. This will give the existence. For uniqueness, if $u_1$ and $u_2$ are two solution, their difference $w=u_1-u_2$ is solution of $\int_{\Omega}\nabla w\cdot \nabla wdx-\int_{\partial\Omega}wvd\sigma$ and taking $v=w$ we get $w=0$. –  Davide Giraudo Jan 29 '12 at 19:08

1 Answer 1

up vote 4 down vote accepted

The argument is basically the same as a proof of the Poincare inequality in $H_0^1$: Assume that $a$ (as in Davide's comment) is not coercive, then we can take $u_n\in H^1$ with $\| u_n\|_{H^1}=1$ such that $$\| \nabla u_n\|_{L^2(\Omega)}^2 + \| u_n \|_{L^2(\partial\Omega)}^2 \leq n^{-1} \| u_n\|_{H^1}^2=n^{-1}$$

First notice that the inequality gives that $\nabla u_n, u_n|_{\partial\Omega} \to 0$ strongly in the corresponding $L^2$ spaces. By basic results on Sobolev spaces, if we can prove that there is $u\in L^2(\Omega)$ with $u_n \to u$ strongly in this last space, we could conclude two things: $u\in H^1$ and $u$ is constant on each connected component of $\Omega$. So $u$ is locally constant, by smoothness of the domain each connected component must have a nontrivial piece of the boundary, so if $u\neq 0$ in a component we would have a non-zero trace, which is absurd since the trace is a bounded linear operator and $u_n|_{\partial\Omega} \to 0$. We conclude that $u=0$ on $\Omega$. This is of course a contradiction since $u_n\to u$ in $H^1$ and $\|u_n\|_{H^1}=1$.

Now all we have to do is show such an $u$. By reflexivity there is $u\in H^1$ with (really a subsequence) $u_n\rightharpoonup u$ in $H^1$, and by Rellich's theorem $u_n \to u$ in $L^2$.

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$\nabla u=0$, ok it's clear now (and should have been earlier) but how do you deduce that $u$ is constant (since $\Omega$ is not supposed to be connected)? –  Davide Giraudo Jan 29 '12 at 21:04
    
@Davide I've rewritten the answer, is everything clear now? –  Jose27 Jan 29 '12 at 21:35
    
Now it's clear. Thanks! It's a nice answer. –  Davide Giraudo Jan 29 '12 at 21:39

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