How to solve this variational problem related to the Couette Flow? Let $\Omega\subset \mathbb R^3$ be a solid of revolution around the $z$-axis that does not meet the $z$-axis. I don't understand how a solution is proven to exist for the following constrained variational problem:
$$  J(u) = \frac{\int_{\Omega} r^{-4} u_\theta u_r}{\int_\Omega |\nabla u|^2 }\to \min  $$
under the constraint that $u\neq 0 $ belongs to the set of divergence-free axisymmetric $H^1$ functions with zero trace on $\partial\Omega$:
$$ X := \{ u \in H^1_{0,\sigma}(\Omega) : u \text{ is axisymmetric} \} $$
A vector field is axisymmetric if it has the form
$$ u(r,z,\theta) = u_r(r,z)e_r + u_\theta(r,z) e_\theta + u_z(r,z) e_z$$
that is, the components in cyllindrical coordinates don't depend on $\theta$.
My source (Tsai Tai-Peng's "lectures on the Navier-Stokes equations", page 31) says that

The constraint is the divergence free condition $\operatorname{div} u=0$, or (2.35) which says that 
  $(\partial_r + r^{-1})u_r + \partial_z u_z = 0.$ Since our domain is away from the $z$-axis, $J[u]$ is bounded from below. Also $\min J<0$ by choosing $u_\theta= -u_r$.

I'm OK with this. (Actually I havent checked if the divergence free condition can be satisfied if $u_ r= -u_\theta $) but the next sentence is simply

Hence, there is a minimizer, $u$.

My Question
Why does it follow, and is it so obvious? I don't think $J$ is convex or coercive...what are my next options to look at?
 A: Here is a proposal, how this could be achieved. I have no idea whether these arguments work in the cylindrical coordinates setting.
First, take a minimizing sequence $(u_n)$. Since $\inf J<0$, let me assume $J(u_n)<0$  and $u_n\ne0$ for all $n$. Then we can normalize the sequence to have $\|\nabla u_n\|_{L^2}=1$. Hence the sequence is bounded in $H^1$. 
Then (after extracting a subsequence) if necessary, we have $u_n \rightharpoonup u$ 
in $H^1$ and (due to compact embeddings) $u_n \to u$ in $L^2$. This
should be enough to pass to the limit in numerator, 
$$
\int_\Omega r^{-4} u_{n,\theta}u_{n,r} \to \int_\Omega r^{-4} u_{\theta}u_{r} \le0.
$$
In addition, $X$ should be a closed subspace, so $u\in X$.
And we should be able to pass to the lim-inf in the functional:
since the numerators along the sequence are negative we have
$$
\liminf J(u_n) =  \left(\lim \int_\Omega r^{-4} u_{n,\theta}u_{n,r}\right)\cdot \limsup (\|\nabla u_n\|_{L^2}^{-2})
= \left(\int_\Omega r^{-4} u_{\theta}u_{r} \right)\cdot ( \liminf \|\nabla u_n\|_{L^2})^{-2}
\ge  \left(\int_\Omega r^{-4} u_{\theta}u_{r} \right)( \|\nabla u\|_{L^2})^{-2} = J(u).
$$
(This inequality shows that $t \mapsto - f(t)^{-2}$ is lower-semicontinuous for lower semicontinuous and non-negative $f$.)
