Integral identity for critical points of the Ginzburg-Landau functional

Question

I am reading a paper of Comte and Mironescu [CM96], where they discuss critical points $v = v_{\epsilon}: G \to \mathbf{C}$ of the (non-magnetic) Ginzburg–Landau functional $E_\epsilon(v) = \frac{1}{2} \int_G \lvert \nabla v \rvert^2 + \frac{1}{4 \epsilon^2} \int_G (1 - \lvert v \rvert^2)^2$, defined on a starshaped domain $G \subset \mathbf{R}^2$.

How is the identity \eqref{1} proved? The authors refer to a paper of Bethuel, Brezis and Helein [BBH93] (and specifically the proof of Proposition 1), but I can't find the relevant argument: no identity of the sort seems to be used there.

On page 203, in the course of a derivation of an estimate for such a critical point $v = \lvert v \rvert \mathrm{e}^{\mathrm{i} \varphi}$, they use an integral identity stating that \begin{equation} \label{1}\tag{1} \int_{\partial B_{C\epsilon}(x_j^\epsilon)} \lvert v \rvert^2 \frac{\partial \varphi}{\partial \nu} = 0. \end{equation} The integral is over the boundary of a disk $B_{C\epsilon}(x_j^\epsilon)$ of radius $C\epsilon$ around a 'bad point' $x_j^\epsilon$. The points $x_1^\epsilon,\dots,x_J^\epsilon$ are called 'bad' because the discs around them contain the zero set of $v$, where $\varphi$ is not defined.

In contrast, $\lvert v \rvert \geq 1/2$ in their complement $G \setminus \cup_j B_{C\epsilon}(x_j^\epsilon)$. Note however that even in this 'good' region, $\varphi$ is only defined modulo $2\pi$, as $G \setminus \cup_j B_{C \epsilon}(x_j^\epsilon)$ is not simply connected.

[BBH93] F. Bethuel, H. Brezis, F. Helein. Asymptotics for minimizers of a Ginzburg–Landau functional, Calculus of Variations and PDE 1 (1993), pp. 123-148.

[CM96] M. Comte and P. Mironescu. Remarks on nonminimizing solutions of Ginzburg-Landau type equation, Asymptotic Analysis 13 (1996) pp. 199-215.

Answer 1

If I'm not miscalculating, the variational equation for $\varphi$ gives $\delta E_\epsilon/\delta\varphi = \partial^k (\rho^2 \partial_k \varphi) = 0$. Writing your integral identity as a flux through $\partial B$, and using the divergence theorem (Stokes' lemma more generally), you'll find $$ \int_{\partial B} d\sigma^k (\rho^2 \partial_k \varphi) = \int_B d^2x \partial^k (\rho^2 \partial_k \varphi) = 0 , $$ where the $B$-volume integrand vanishes as it coincides with the variational equation for $\varphi$. I haven't checked, but there might also be a Noether's theorem interpretation for the conserved current $\partial^k f_k = 0$ with $f_k = \rho^2 \partial_k \varphi$, perhaps corresponding to the symmetry $\varphi \mapsto \varphi + \text{const}$.