Proof of the coercivity for Robin boundary problem

3 minute read

This is an exercise for finite element method most likely. Our question is:

Is Robin (read as “rho-bon”) boundary problem well-posed?

Say our equation is:

\[-\Delta u = f \quad \text{in } \Omega\]

with Robin boundary condition for all the boundary $\partial \Omega$:

\[\alpha u +\frac{\partial u}{\partial n} = g\]

where $\alpha$ is a positive constant ($\alpha$ has to be positive a.e. on the boundary), and $\partial u/\partial n = \nabla u\cdot n$ is the normal derivative on boundary. Multiply the equation by a test function $v\in H^1$,

\[-\int_{\Omega} \Delta u\,v = \int_{\Omega} fv\]

and perform integration by parts:

\[\int_{\Omega} \nabla u\cdot \nabla v - \int_{\partial \Omega}(\nabla u\cdot n )v= \int_{\Omega} fv\]

plug in the Robin boundary data:

\[\int_{\Omega} \nabla u\cdot \nabla v - \int_{\partial \Omega}(g-\alpha u)v= \int_{\Omega} fv\]


\[\int_{\Omega} \nabla u\cdot \nabla v + \int_{\partial \Omega}\alpha u v= \int_{\Omega} fv + \int_{\partial \Omega}gv\]

The left hand side $a(u,v) = \displaystyle \int_{\Omega} \nabla u\cdot \nabla v + \int_{\partial \Omega}\alpha u v$, right hand side is the $L(v)$. The proof of continuities of the right hand side and the first term in $a(u,v)$ is natural, just following the proof for Dirichlet and Neumann boundary problems. For second term on the left we would like to use the trace inequality:

\[\int_{\partial \Omega}\alpha u v \leq \alpha \|u\|_{L^2(\partial \Omega)} \|v\|_{L^2(\partial \Omega)} \leq C \alpha \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}\]


\[a(u,v) \leq C \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}.\]

The coercivity of $a(u,v)$ is a little bit more difficult to prove. We can use the “compactness argument” for Gelfand’s triple to prove it. The trick is using proof by contradiction, by the compactness argument, “compact $\Leftrightarrow$ sequential compact”:

Suppose $a(u,v)$ is not coercive, then () there exists a sequence ${v_n}\subset H^1$, such that: \(\|v_n\|_{H^1(\Omega)} = 1, \text{ and } a(v_n,v_n) \to 0.\tag{1}\)

Now since ${v_n}$ is a bounded sequence in $H^1$ (consider the distance being induced by the $H^1$-norm), then there exists a weakly convergent subsequence $v_{n_j}\to v$ in $H^1$ and $v_{n_j}\to v$ in $L^2$-norm (thanks user33869 for pointing out my previous error). Weak convergence is:

\[\int_{\Omega}(v-v_{n_j})w + \int_{\Omega}\nabla(v-v_{n_j})\cdot \nabla w \to 0, \; \text{ for all } w\in H^1(\Omega).\]

This implies that:

\[\int_{\Omega} |\nabla v|^2 = \lim_{j\to \infty}\int_{\Omega} \nabla v_{n_j}\cdot \nabla v \leq \lim_{j\to \infty}\left( \int_{\Omega} |\nabla v_{n_j}|^2 \right)^{1/2} \left(\int_{\Omega} |\nabla v |^2\right)^{1/2}\]

This is

\[|v|_{H^1(\Omega)} \leq \lim_{j\to \infty} |v_{n_j}|_{H^1(\Omega)}\tag{2}\]

Using another assumption now

\[a(v_n,v_n) = \int_{\Omega} |\nabla v_n|^2 + \int_{\partial \Omega}\alpha v_n^2 \to 0 \tag{3}\]

This implies that

\[\int_{\Omega} \vert \nabla v_n \vert^2 \to 0.\]

By (2), letting $n_j \to \infty$ yields $\|\nabla v\|^2 = 0$. Now we have


Now using $v_{n_j}\to v$ in $L^2$-norm, together with (3),

\[\|v\|_{L^2(\Omega)} = \lim\limits_{j\to \infty} \|v_{n_j}\|_{L^2(\Omega)}=1\]

The $L^2$-norm of gradient of $v$ being zero implies $v$ is a constant, The $L^2$-norm of $v$ being 1 implies $v=c$ where the constant $c\neq 0$. However, again by (3):

\[\int_{\partial \Omega}\alpha v^2 = \lim_{n\to \infty}\int_{\partial \Omega}\alpha v_{n_j}^2 = 0\]

together with the positivity assumption on $\alpha$, this says $v=0$ on $\partial \Omega$. Constradition.

Therefore there does not exist such sequence satisfying (1). The negation of the statement (1) is: For all sequence in $H^1(\Omega)$, if $|v_n |_{H^1(\Omega)}$ is bounded, $a(v_n,v_n)$ is always bounded away from 0. This can be translated to any function $v$ in $H^1$, for we could always find a bounded sequence that goes to $v$ by the completeness of a Hilbert space. Thus the coercivity is proved.