We are given the equation \begin{cases} -\Delta u=0&x\in \Omega\\ \partial_\nu u+\beta(u)=0&x\in\partial\Omega \end{cases} where $\Omega$ is bounded bounded smooth boundary and $0<a\leq \beta'(z)\leq b<\infty$ where $\beta$: $R\to R$.

The question ask me to find a unique weak solution $u\in H^1(\Omega)$. To do so, I first try to write weak formulation of this problem. Take arbitrary $v\in H^1(\Omega)$ and integration by parts we have $$ \int_\Omega \nabla u\nabla v\,dx+\int_{\partial \Omega}\beta(u)v\,dH^{N-1}=\int_\Omega fv\,dx $$

Update: based on @Behaviours suggest, I made the following progress.

Define $B(x)$ to be the anti-derivative of $\beta(x)$ above and I obtain that $$ \frac{1}{2}ax^2+k_1x+k_2\leq B(x)\leq \frac{1}{2}bx^2+k_3x+k_4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ for some constant $k_1\ldots k_4\in R$

Hence, I conclude that $B(x)$ is BOUNDED BELOW.

Now define our energy function to be $$ E[u]:=\frac{1}{2}\int_\Omega |\nabla u|^2dx+\int_{\partial \Omega}B(u)dH^{N-1}-\int_\Omega fu\,dx $$ So, if we have a minimizer in $H^1(\Omega)$, we would be done.

Now the problem is how to prove the existence of minimizer. This should be a standard direct method in CoV.

The key is obtain $E[u]$ is bounded below and also we have the coercivity. Usually, we could use the first term to control the last term in $E[u]$ by viewing of Poincare inequality. However, in this question we have no Poincare inequality so we can not control last term with the first term. I then assume that $f=0$ because the homogenizies problem is easy to solve. Now I have new energy function

$$ E[u]:=\frac{1}{2}\int_\Omega |\nabla u|^2dx+\int_{\partial \Omega}B(u)dH^{N-1} $$

This energy functional is bounded below and hence I could obtain a minimizing sequence $(u_n)\subset H^1(\Omega)$ such that $$ E[u_n]\to \alpha $$ where $\alpha$ is the min value.

Next I try to prove that $$\|u_n\|_{H^1(\Omega)}\leq C<\infty\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$ for some constant $C>0$, and I got stuck here...

From $(1)$ and take in the trace estimation, we have $$\int_{\partial \Omega}B(u)dH^{N-1}\leq C_1\|u\|_{H^1(\Omega)}^2+C_2\|u\|_{H^1(\Omega)}+C_3 $$ and hence we have $$ E[u_n]\geq (\frac{1}{2}-C_1)\|\nabla u\|_{L^2}^2-C_2\|u\|_{L^2}^2 -C_2\|u\|_{H^1} $$ which can not conclude $(2)$ as I want...

So how can I conclude $(2)$?


The variational approach is the way to go. You know that the Laplacian corresponds to minimizing the Dirichlet energy $\int_\Omega |\nabla u|^2 $; the question is how to fit $\int_{\partial \Omega}\beta(u)v\,dH^{N-1}$ into the variational problem. Let's try a generic boundary energy term: $$\int_{\partial \Omega}B(u) \,dH^{N-1}$$ with $\Phi$ to be determined. Substitute $u$ with $u+v$ and keep only the first order term with $v$, i.e., the first variation: $$\int_{\partial \Omega}B'(u)v \,dH^{N-1}$$ This looks good. You just need $B'(u)=\beta(u)$. The condition that $\beta$ is increasing gives you convexity of $B$. Furthermore, the two-sided bound on $\beta'$ says that $B$ is strongly convex with quadratic growth at infinity, i.e., the nicest kind of convex functions. It remains to study the functional $$ E(u)=\frac12\int_\Omega |\nabla u|^2+\int_{\partial \Omega}B(u) \,dH^{N-1}$$ which, by the standard trace theorem, is well-defined on $H^1(\Omega)$ (there is a bounded trace operator $H^1(\Omega)\to L^2(\partial \Omega)$).

Minimizing sequence is bounded

Take a sequence $(u_n)$ such that $E(u_n)\to\inf_{H^1} E$. Let $\mu_n$ be the mean of $u_n$ on $\Omega$. Since the sequence $(\nabla u_n)$ is bounded in $L^2(\Omega)$, by the Poincaré lemma the sequence $(u_n-\mu_n)$ is bounded in $H^1(\Omega)$. It remains to show that $(\mu_n)$ is a bounded sequence.

Let $T:H^1(\Omega)\to L^2(\partial\Omega)$ be the trace operator. Since it is a bounded operator, the sequence of functions $T(u-\mu_n)=T(u_n)-\mu_n$ is bounded in $L^2(\partial\Omega)$. On the other hand, $T(u_n)$ are themselves bounded in $L^2(\partial\Omega)$, since their norm is controlled by $E(u_n)$. Hence $(\mu_n)$ is bounded.

  • $\begingroup$ I updated my post. Please help. Thank you! $\endgroup$ – spatially Dec 24 '14 at 1:40
  • $\begingroup$ hmm... I still can not get the part of coercivity... Can you help me about that? Thank you! $\endgroup$ – spatially Dec 25 '14 at 18:02
  • $\begingroup$ See the addition to the answer. $\endgroup$ – user147263 Dec 26 '14 at 0:49
  • $\begingroup$ Thanks! It indeed makes sense now $\endgroup$ – spatially Dec 26 '14 at 16:56

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