# 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 comment for the downvote on an old question? In fact, I’ll take a package deal, a comment for the downvotes on many of my old questions would be nice Commented Nov 9, 2021 at 2:57

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 $$f$$.)
• Due to the negativity of the numerator, this inf-problem has the flavor of problem: $\sup_{u\in H^1_0} \|u\|_{L^2}^2 \|\nabla u\|_{L^2}^{-2}$, which results in an eigenvalue problem.