Let $\beta:\mathbb{R}\to\mathbb{R}$ be a Lipschitz function such that $0<a\leq \beta^\prime\leq b$, for some constants $a,b$. Give the weak formulation of the problem \begin{equation} \left\{\begin{array}{cl}-\Delta u=f&in\ \Omega\\\partial_\nu u+\beta(u)=0&on\ \partial\Omega,\end{array}\right. \end{equation} where $\Omega\subset\mathbb{R}^n$ is a bounded domain and $f\in L^2(\Omega)$. Show that there exists a weak solution that it is unique.

My guess is that you have to use Lax-Milgram in order to prove existence and uniqueness of solutions, but I don't know how to deal with the nonlinear term. The weak formulation of the problem reads \begin{equation} a(u,v)=\varphi(v)\ \forall v\in H^1(\Omega), \end{equation} where \begin{equation} \begin{array}{l} a(u,v)=\int_\Omega\nabla u\nabla v+\int_{\partial\Omega}\beta(u)v,\\ \varphi(v)=\int_\Omega fv. \end{array} \end{equation} $\varphi$ is clearly continuous, but how can I prove continuity and coercivity of the bilinear form $a(\cdot,\cdot)$?

  • $\begingroup$ It's not a bilinear form anymore, so you're going to have to do something else. (I don't know what it is, however.) $\endgroup$ – Ian May 20 '15 at 15:57
  • $\begingroup$ True, it is not bilinear anymore... Thus you cannot apply Lax-Milgram in this case. Then...? $\endgroup$ – Marc May 20 '15 at 16:00
  • $\begingroup$ Not verbatim, no. However, I think you can show that $v \mapsto a(u,v)$ is a bounded linear functional on $H^1$ if $u \in H^1$. So you can apply Riesz to that (much like in the proof of Lax-Milgram) and maybe work from there. $\endgroup$ – Ian May 20 '15 at 16:02

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