Superfluid-Mott insulator transition in Bose-Hubbard model in terms of vortex condensation

Question

I have heard that there is some effective field theoretic type understanding of the superfluid-Mott insulator transition in Bose-Hubbard model. It says if the system is in a superfluid phase where the $U(1)$ symmetry is broken, then by proliferating vortices we can destroy the phase coherence and obtain a Mott insulator. 

I understand that there is no phase coherence in a Mott insulator and having many vortices will mess up the phase coherence, but I do not understand why the mechanism to destroy phase coherence has to be proliferating vortices. To be more precise, suppose we can describe the superfluid phase by the following Lagrangian:
\begin{equation}
\mathcal{L}=|\partial_\mu\phi|^2+\frac{r}{2}|\phi|^2-\frac{u}{4!}|\phi|^4
\end{equation}
where $r$ and $u$ are positive so that the $U(1)$ symmetry is broken. Now if you asked me how to go from this $U(1)$ symmetry broken phase to a $U(1)$ symmetric phase, I would say we need to decrease $r$ until $r$ starts to be negative. However, when $r$ is decreased, I do not expect there will be more and more vortices in the ground state since here the energy of a single vortex always diverges logarithmically with the system size. Then how is decreasing $r$ related to vortex condensation?

By the way, if that vortex condensation picture is somehow correct, should one see vortices in experiments with Bosons in an optical lattice?