MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Mistake in variational formulation of Neumann problem?

Consider the problem

$$\left\{\begin{array}{ll} -\Delta u+c\cdot \nabla u=f, & \Omega \\ \partial_{\nu} u=0, & \partial \Omega \end{array} \right.$$

with $\Omega$ smooth, $c\in C^1(\bar{\Omega})$ and $f\in L^2(\Omega)$. Assume further that $div c=0$ and that $c\cdot \nu\geq c_0>0$.

The variational formlation of this problem consists of finding $u\in H^1(\Omega)$ such that $$\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx=\int_{\Omega} fv dx, \qquad \forall v\in H^1(\Omega)$$

Let $V=H^1(\Omega)$ and consider the bilinear form $$B(u,v)=\int_{\Omega} \nabla u\cdot \nabla v+(c\cdot \nabla u)v dx$$

I am wondering whether there is something wrong with the following argument showing existence and uniqueness of a solution to the previous problem using the Lax-Milgram theorem. By direct computation, from the above assumptions we easily conclude that

$$\int_{\Omega} (c\cdot \nabla u)u dx=\frac{1}{2}\int_{\Omega} c\cdot \nabla (u^2) dx=\int_{\partial \Omega} u^2c\cdot \nu d\sigma$$

Recall also that in $H^1(\Omega)$, the norm $$||u||_{1,\partial}=\int_{\partial \Omega} u^2\,d\sigma+\int_{\Omega} |\nabla u|^2\,dx$$ is equivalent to the standard norm $||u||_{1,2}$

Thus, if $c\cdot \nu\geq c_0>0$ then $$B(u,u)=\int_{\Omega} |\nabla u|^2+(c\cdot \nabla u)u \,dx \geq \int_{\Omega} |\nabla u|^2+c_0\int_{\partial \Omega} u^2\,d\sigma\geq C||u||_{1,2}$$

so that $B$ is $V$-coercive. Checking that both B and $Lv=\int_{\Omega} fv\,dx$ are continuous, one concludes that the problem has a unique solution $u\in H^1(\Omega)$.

-
Overall, the argument is OK. Note only that $\|\cdot\|_{1,\partial}$ is equivalent to the $H^1$-norm only if you assume in addition that $\Omega$ is bounded. – Denis Serre May 14 2012 at 20:25
If $\Omega$ is bounded, you cannot have div(c)=0 and $c\cdot\nu>0$! – Michael Renardy May 14 2012 at 20:53
I think it would give you a better perspective if you first prove that the operator is Fredholm with index zero without assuming anything on $c$, and then look at conditions on $c$ that would kill the kernel. – timur May 14 2012 at 21:56