# Generation of vortices in the Ginzburg–Landau heat flow

Michal Kowalczyk, Xavier Lamy

► To cite this version:

Michal Kowalczyk, Xavier Lamy. Generation of vortices in the Ginzburg–Landau heat flow. Annales de l'Institut Henri Poincaré (C), Analyse non linéaire, 2023, 41 (6), pp.1509-1526. 10.4171/aihpc/96 . hal-04188293

HAL Id: hal-04188293

https://hal.science/hal-04188293v1

Submitted on 7 Jan 2025

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Distributed under a Creative Commons Attribution 4.0 International License
GENERATION OF VORTICES IN THE
GINZBURG-LANDAU HEAT FLOW

MICHAL KOWALCZYK AND XAVIER LAMY

ABSTRACT. We consider the Ginzburg-Landau heat flow on the two-
dimensional flat torus, starting from an initial data with a finite num-
ber of nondegenerate zeros – but possibly very high initial energy. We
show that the initial zeros are conserved and the flow rapidly enters a
logarithmic energy regime, from which the evolution of vortices can be
described by the works of Bethuel, Orlandi and Smets.

# 1. INTRODUCTION

In the flat two-dimensional torus $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$ we consider $u(t,x)$, a solution of the Ginzburg-Landau heat flow

$$
\begin{equation} \tag{1}
\begin{aligned}
\partial_t u - \varepsilon^2 \Delta u &= (1 - |u|^2)u & t \ge 0, x \in \mathbb{T}^2, \\
u(0, x) &= u_0(x),
\end{aligned}
\end{equation}
$$

with $u_0 \in C^1(\mathbb{T}^2)$. The initial condition $u_0$ may have a finite number of zeros. More precisely, we assume that there exists $\alpha_0 > 0$ such that

$$
(2) \quad |u_0(x)| + |\det \nabla u_0(x)| \ge \alpha_0.
$$

This implies in particular that the zeros of $u_0$ are nondegenerate and the topological degree of the vector field $u_0$ at each zero is 1 or $-1$.

We will denote the energy associated with (1) by

$$
E_{\varepsilon}(u) = \int_{\mathbb{T}^2} |\nabla u|^2 + \frac{1}{4\varepsilon^2} (1 - |u|^2)^2.
$$

Note that (1) is the $L^2$ gradient flow of $E_\epsilon$ up to a factor $\epsilon^2$, hence $E_\epsilon$
is decreasing along the flow. The Ginzburg-Landau heat flow has been
extensively studied [BCPS95, JS98, Lin96, SS04, Ser07b, Ser07a, BOS05,
BOS07a, BOS07b], in the case of initial data $u_0 = u_{0\epsilon}$ satisfying a loga-
rithmic energy bound $E_\epsilon(u_{0\epsilon}) \le M \ln(1/\epsilon)$. This bound enables to identify
vortices, the zeros of $u_{0\epsilon}$, and to describe their evolution. More precisely,

<footnote>M. K. was partially funded by Chilean research grants FONDECYT 1210405 and ANID projects ACE210010 and FB210005 and National Science Centre, Poland (Grant No. 2020/37/B/ST1/02742). Part of this work was done during his stay in the Institut de Mathématiques de Toulouse.</footnote>

<footnote>X.L. was partially funded by ANR project ANR-22-CE40-0006-01. Part of this work was conducted during X.L.’s stay at the CMM, with the financial support of CNRS and the CMM.</footnote>
in [JS98, Lin96, SS04], well-prepared initial data are considered, with a finite number of vortices of degree $\pm 1$ and correspondingly quantized energy. These works establish via different methods that, in the accelerated time-scale $s = (\varepsilon^2/\ln(1/\varepsilon))t$, vortices move according to the gradient flow of a renormalized energy analyzed in [BBH94], for as long as no collisions happen. This limitation is removed in the works [Ser07a, BOS07a, BOS07b], where splittings and collisions of vortices are described rigorously. Specifically, [Ser07a] describes the global-in-time motion of vortices, taking collisions into account, in bounded domains with Dirichlet or Neumann boundary conditions. Initial well-preparedness is also relaxed: initial vortices are of degree $\pm 1$, but the energy quantization assumption is less stringent; moreover, splitting of higher degree vortices into vortices of degree $\pm 1$ is described under specific assumptions. In [BOS07a, BOS07b], the domain is the whole plane and a global motion law allowing for splittings and collisions is obtained, for initial data satisfying the logarithmic bound $E_\varepsilon(u_{0\varepsilon}) \le M\ln(1/\varepsilon)$. In case of $N_\varepsilon \gg 1$ initial vortices, evolution of the vortex density is described by a mean-field equation first obtained rigorously in [Ser17].

Here we are interested in initial data that may have much higher energy, and wish to describe the emergence of vortices. This is mentioned as an open problem in [BOS08, Problem 5]. Our methods are strongly inspired by similar results on the emergence of sharp transitions in the Allen-Cahn heat flow [Che04].

Our first main result concerns the evolution of the zeros of $u$.

**Theorem 1.1.** There exists $C_0 > 0$, depending on $u_0$, such that, for all $\varepsilon > 0$ sufficiently small (depending on $u_0$), if $\mathcal{Z}(t)$ denotes the set of zeros of $u(t)$, we have

$$ \#\mathcal{Z}(t) = \#\mathcal{Z}(0), \quad \text{for } 0 \le t \le T_\varepsilon := \ln \frac{1}{\varepsilon} - \frac{1}{2} \ln \ln \frac{1}{\varepsilon} - C_0. $$

In other words, no new zeros of the vector field $u(t)$ are generated up to $t = T_\varepsilon$. Additionally, if $z_j(t)$ is the evolution of the j-th zero $z_j^0$ of $u_0$, then $|z_j(t) - z_j^0| \lesssim \varepsilon\sqrt{\ln(1/\varepsilon)}$, and the topological degree of $u(t)$ at $z_j(t)$ is preserved. Finally, at $t = T_\varepsilon$ we have

$$ (3) \quad |u(T_\varepsilon, x)| \ge \frac{1}{2} \quad \text{for } \text{dist}(x, \mathcal{Z}(0)) \ge \varepsilon \sqrt{\ln \frac{1}{\varepsilon}}. $$

Above and throughout the paper the symbol $A \lesssim B$ for two nonnegative quantities $A, B$ means that there exists a constant $C > 0$, depending only on $u_0$, such that $A \le CB$.

An immediate corollary of Theorem 1.1 is that, if $u_0$ does not vanish, then $u(t)$ does not vanish for $0 \le t \le T_\varepsilon$.

**Corollary 1.2.** If $\mathcal{Z}(0) = \emptyset$ then $\mathcal{Z}(t) = \emptyset$ for $t \in [0, T_\varepsilon]$.

This means that up to time $T_\varepsilon$ the Ginzburg-Landau heat flow does not undergo a Berezinsky-Kosterlitz-Thouless phase transition. If one allows
the initial condition $u_0$ to depend on $\epsilon$, one may however observe creation of zeros, as in [RS95, Proposition 4.1].

Our second main result is a logarithmic energy bound at the time $t = T_\epsilon$ given by Theorem 1.1.

**Theorem 1.3.** For all sufficiently small $\epsilon > 0$ (depending on $u_0$), we have

$$E_{\epsilon}(u(t)) \lesssim \ln \frac{1}{\epsilon}, \quad \forall t \geq T_{\epsilon}.$$

Theorem 1.3 shows that the evolution enters an energy regime where the analysis of [BOS05, BOS07a, BOS07b] can be applied. The present context is actually slightly different, because we work on the torus $\mathbb{T}^2$ instead of $\mathbb{R}^2$, but the results of [BOS05, BOS07a, BOS07b] should apply to $\mathbb{T}^2$, with appropriate modifications. Reciprocally, the results of the present paper could be adapted to $\mathbb{R}^2$, with appropriate conditions at infinity, at the price of minor technical complications.

In particular, the work [BOS07b] describes the evolution of the vortices of $u$ as functions of the accelerated time-variable

$$s = \frac{\epsilon^2}{\ln \frac{1}{\epsilon}} t.$$

The vortices $a_k(s)$ evolve according to the gradient flow of a renormalized energy $W(a)$, combined with a finite number of collision or branching times. Note that in the torus $\mathbb{T}^2$, the renormalized energy $W(a)$ would be slightly different than the one considered in [BOS07b]. The initial conditions for the vortices $a_k(s)$ as $s \to 0^+$ are identified via the jacobian $Ju = \det(\nabla u)$ at the initial time [BOS08, Proposition 2]. We therefore complement Theorem 1.3 with our third main result, which characterizes the jacobian at time $t = T_\epsilon$.

**Theorem 1.4.** We have, as $\epsilon \to 0$,

$$Ju(T_{\epsilon}) = \det(\nabla u(T_{\epsilon})) \to \sum_{j=1}^{N} \hat{d}_{j} \delta_{z_{j}^{0}},$$

in the sense of distributions, where $z_1^0, \dots, z_N^0$ are the zeros of $u_0$, and $\hat{d}_j \in \{\pm 1\}$ its topological degree at $z_j^0$.

Now we can be more specific about the initial conditions for the later evolution of the vortices $a_k(s)$, as described in [BOS08, Proposition 2]. Letting $d_k$ denote the topological degree of $u(s)$ at $a_k(s)$ for small $s > 0$, the initial conditions $a_k^0 = \lim_{s \to 0^+} a_k(s)$ must satisfy

$$\sum_{k=1}^{L} d_k \delta_{a_k^0} = \sum_{j=1}^{N} \hat{d}_j \delta_{z_j^0}.$$

This implies in particular that $\{a_k^0\} = \{z_j^0\}$. But the points $a_k^0$ may not be disjoint: this description does not prevent a priori a single initial zero $z_j^0$ to spontaneously split into several vortices $\{a_k\}$, because at $s = 0$ the energy
is not yet quantized (in the sense of [BOS07b, Theorem 1.5]). In fact initial splitting into two vortices can easily be ruled out, but it is not clear whether splitting into three or more vortices can occur.

However, note that in the setting of Corollary 1.2, if there are no initial zeros, we can directly conclude that no later vortices appear. A complete proof of this fact would require adapting [BOS08] to our torus-based setting.

The main idea of this paper is that on the time scale considered, the effect of diffusion in the Ginzburg-Landau equation is dominated by the nonlinear effect. This means that the modulus of any initial data instantaneously (on the fast time scale $s = \varepsilon^2 t/(\ln 1/\varepsilon)$) approaches 1 except possibly on small regions where the initial data is close to 0. The methods are elementary and provide explicit pointwise estimates on $u(t, x)$, which directly imply the stated results. To control diffusive effects, the key tool is Lemma 2.2, which is a type of Gronwall inequality (new to our best knowledge). The organization of the paper follows that of the presentation of the results which are proven in the same order in the consecutive sections.

## 2. ZEROS OF $u$: PROOF OF THEOREM 1.1

Denote by $\Phi: \mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^2$ the flow of the ODE $y' = (1 - |y|^2)y$, that is,

$$ \partial_t \Phi = (1 - |\Phi|^2)\Phi, \quad \Phi(0, X) = X, $$

given explicitly by

$$ (4) \qquad \Phi(t, X) = \frac{e^t X}{\sqrt{1 + |X|^2(e^{2t} - 1)}}. $$

We want to estimate how far $u$ is from

$$ v(t, x) = \Phi(t, g(t, x)) $$

for some well-chosen map $g$ with $g(0, x) = u_0(x)$. To this end we define $w = e^{-t}(u - v)$, so that

$$ u = v + e^t w. $$

Using the equations satisfied by $u$ and $\Phi$ we obtain

$$ (5) \qquad \partial_t w - \varepsilon^2 \Delta w = -2(v \cdot w)v - |v|^2 w - e^{-t} \mathcal{N}(v, e^t w) - e^{-t} \mathcal{R}, $$

$$ \mathcal{N}(v, X) = |X|^2 v + 2(v \cdot X)X + |X|^2 X, $$

$$ \mathcal{R} = \partial_t v - \varepsilon^2 \Delta v - (1 - |v|^2)v $$

$$ = D\Phi(t, g)(\partial_t g - \varepsilon^2 \Delta g) - \varepsilon^2 D^2\Phi(t, g)\nabla g \cdot \nabla g $$

In view of (5), it is natural to choose, as in [Che04], $g(t) = e^{\varepsilon^2 t \Delta} u_0$, that is, $g$ solves

$$ (6) \qquad \partial_t g - \varepsilon^2 \Delta g = 0, \quad g(0, x) = u_0(x), $$

and therefore

$$ (7) \qquad \mathcal{R} = -\varepsilon^2 D^2 \Phi(t, g) \nabla g \cdot \nabla g. $$
The rest of the article is devoted to obtaining good pointwise estimates on $e^t w = u - v$.

**Lemma 2.1.** If $w$ solves

$$ \partial_t w - \varepsilon^2 \Delta w = -2(v \cdot w)v - |v|^2 w + F, \quad t > 0, x \in \Omega, $$

with $w(0, x) = 0$, then

$$ \|w(t)\|_{L^\infty} \le \int_0^t \|F(s)\|_{L^\infty} ds. $$

*Proof of Lemma 2.1.* Multiplying the equation by $w/|w|$ we obtain

$$ \begin{aligned} \partial_t |w| &= \varepsilon^2 \frac{w}{|w|} \cdot \Delta w - |v|^2 |w| - 2 \frac{(v \cdot w)^2}{|w|} + F \cdot \frac{w}{|w|} \\ &\le \varepsilon^2 \frac{w}{|w|} \cdot \Delta w + |F| \\ &\le \varepsilon^2 \Delta |w| + |F|, \end{aligned} $$

so by comparison principle we have $|w| \le \rho$ where $\rho$ solves $\partial_t \rho - \varepsilon^2 \Delta \rho = |F|$ and $\rho(0, x) = 0$, that is, $\rho(t) = \int_0^t e^{\varepsilon^2(t-s)\Delta} |F(s)| ds$, where $e^{t\Delta}$ denotes the heat semigroup on the torus $\mathbb{T}^2$. Since the $L^\infty$-norm is nonincreasing under the action of that semigroup, we deduce the announced bound. $\square$

We apply Lemma 2.1 to our map $w$ and $F = -e^{-t} \mathcal{N}(v, e^t w) - e^{-t} \mathcal{R}$. We have $|g| \le |u_0| \lesssim 1$, so $|v| \lesssim 1$ and $|\mathcal{N}(v, X)| \lesssim |X|^2$. Thus we obtain

$$ (8) \quad \|e^t w(t)\|_{L^\infty} \lesssim \int_0^t e^{(t-s)} \|e^s w(s)\|_{L^\infty}^2 ds + \int_0^t e^{t-s} \|\mathcal{R}(s)\|_{L^\infty} ds $$

Recall

$$ \mathcal{R} = -\varepsilon^2 D^2 \Phi(t, g) \nabla g \cdot \nabla g, $$

and $|\nabla g| \le |\nabla u_0| \lesssim 1$, hence

$$ \|\mathcal{R}(t)\|_{\infty} \lesssim \varepsilon^2 \sup_{|X| \lesssim 1} |D^2 \Phi(t, X)|. $$

Direct calculation gives

$$ \begin{aligned} |D^2 \Phi(t, X)| &\lesssim \frac{e^t |X|(e^{2t} - 1)}{(1 + |X|^2(e^{2t} - 1))^{3/2}} \\ &= e^t (e^{2t} - 1)^{1/2} \frac{(|X|^2(e^{2t} - 1))^{1/2}}{(1 + |X|^2(e^{2t} - 1))^{3/2}} \\ &\lesssim e^t (e^{2t} - 1)^{1/2}, \end{aligned} $$

so

$$ \begin{aligned} \int_0^t e^{t-s} \|\mathcal{R}(s)\|_{L^\infty} ds &\lesssim \varepsilon^2 e^t \int_0^t (e^{2s} - 1)^{1/2} ds \\ &\lesssim \varepsilon^2 e^t (e^{2t} - 1)^{1/2}, \end{aligned} $$
where we have used

$$
\begin{align*}
\int_0^t (e^{2s} - 1)^{1/2} ds &= \int_0^{(e^{2t}-1)^{1/2}} \frac{x^2}{1+x^2} dx \\
&= (e^{2t}-1)^{1/2} - \arctan((e^{2t}-1)^{1/2}) \\
&\le (e^{2t}-1)^{1/2}.
\end{align*}
$$

Plugging this into (8) we deduce

$$
(9) \quad \|e^t w(t)\|_{L^\infty} \lesssim \int_0^t e^{(t-s)} \|e^s w(s)\|_{L^\infty}^2 ds + \epsilon^2 e^t (e^{2t}-1)^{1/2}.
$$

**Lemma 2.2.** Assume $f, h$ are continuous positive functions on $(0, \infty)$ satisfying

$$
\begin{align*}
\limsup_{t \searrow 0} \frac{f(t)}{h(t)} &\le 1, \\
\text{and } f(t) &\le c \int_0^t e^{t-s} f(s)^2 ds + h(t) \quad \forall t > 0,
\end{align*}
$$

for some constant $c > 0$. If $T > 0$ is such that

$$
\sup_{0<t<T} \int_0^t e^{t-s} \frac{h(s)}{h(t)} h(s) ds \le \frac{1}{8c},
$$

then

$$
f(t) \le 2h(t) \quad \forall t \in (0, T).
$$

If in addition $h$ is nondecreasing it suffices to check that

$$
\int_0^T e^{T-s} h(s) ds \le \frac{1}{8c}.
$$

Proof of Lemma 2.2. For all $t > 0$ we have

$$
\begin{align*}
\frac{f(t)}{h(t)} &\le c \int_0^t e^{t-s} \left( \frac{f(s)}{h(s)} \right)^2 \frac{h(s)}{h(t)} h(s) ds + 1 \\
&\le c \int_0^t e^{t-s} \frac{h(s)}{h(t)} h(s) ds F(t)^2 + 1,
\end{align*}
$$

where

$$
F(t) = \sup_{0<s<t} \frac{f(s)}{h(s)}.
$$

Since $F$ is nondecreasing and using that the integral in the right-hand side is bounded by $1/8c$ for $0 < t < T$ we deduce

$$
F(t) \le \frac{1}{8} F(t)^2 + 1 \quad \forall t \in (0, T),
$$
so $F$ takes values into

$$ \left\{ x \in \mathbb{R} : \frac{x^2}{8} - x + 1 \ge 0 \right\} = (-\infty, 4 - 2\sqrt{2}] \cup [4 + 2\sqrt{2}, +\infty). $$

Since $F$ is continuous on $(0, T)$ and $F(0^+) \le 1 < 4 - 2\sqrt{2}$ we deduce that $F(t) \le 4 - 2\sqrt{2} \le 2$ for all $t \in (0, T)$. $\square$

We apply Lemma 2.2 to

$$ f(t) = \|e^t w(t)\|_{L^\infty}, \quad h(t) = A\varepsilon^2 e^t (e^{2t} - 1)^{1/2}, $$

where $A \ge 1$ is the constant hidden in the sign $\lesssim$ in (9). By Lemma 2.1 the map $w$ satisfies $\|e^t w(t)\|_{L^\infty} \lesssim t$ so $\limsup_{0+}(f/h) = 0$, and thanks to (9) we deduce that

$$ \begin{array}{ll} (10) & \|e^t w(t)\|_{L^\infty} \lesssim \varepsilon^2 e^t (e^{2t} - 1)^{1/2} \\ & \text{for } 0 \le t \le T = \ln \frac{1}{\varepsilon} - \ln(16A^2), \end{array} $$

since $h$ is nondecreasing and for this value of $T$ we have

$$ 8A \int_0^T e^{T-s} h(s) \, ds \le 8A^2 \varepsilon^2 e^{2T} \le \frac{1}{2}. $$

Estimate (10) tells us that $u$ is close to $v$. Note in particular that (10) is valid up to $t = \ln \frac{1}{\varepsilon} - \frac{1}{2} \ln \frac{1}{\varepsilon}$ if $\varepsilon$ is small enough. From (10) we also deduce a bound on $\nabla w$, using again the equation (5).

**Lemma 2.3.** If $w$ solves $\partial_t w - \varepsilon^2 \Delta w = F$ with $w(0) = 0$ in the torus $\mathbb{T}^2$, then we have

$$ \|\nabla w(t)\|_{L^\infty} \lesssim \frac{1}{\varepsilon} \int_0^t \frac{\|F(s)\|_{L^\infty}}{\sqrt{t-s}} \, ds. $$

*Proof of Lemma 2.3.* We can consider $w$ and $F$ as periodic maps defined on $\mathbb{R}^2$, then $w$ is given by the Duhamel formula

$$ w(t) = \int_{0}^{t} H_{\varepsilon\sqrt{t-s}} * F(s) \, ds, $$

where the convolution is on $\mathbb{R}^2$ and $H_\delta(x) = \delta^{-2}H(x/\delta)$, $H(x) = (4\pi)^{-1}e^{-|x|^2/4}$. Therefore we have

$$ \|\nabla w(t)\|_{L^\infty} \lesssim \int_0^t \|\nabla H_{\varepsilon\sqrt{t-s}}\|_{L^1} \|F(s)\|_{L^\infty} \, ds, $$

and the estimate follows from

$$ \|\nabla H_{\varepsilon\sqrt{t-s}}\|_{L^1} \lesssim \frac{1}{\varepsilon\sqrt{t-s}} \|\nabla H\|_{L^1}. $$

$\square$
Applying Lemma 2.3 to the equation (5) satisfied by $w$ and using (10) to estimate the right-hand side we obtain

$$ (11) \quad \|\nabla w\|_{L^{\infty}} \lesssim \varepsilon\sqrt{t}(e^{2t} - 1)^{1/2}, $$

for all $t \le T = \ln \frac{1}{\varepsilon} - \ln(16A^2)$.

All assertions of Theorem 1 will follow from the bounds (10)-(11) on $e^t w = u - v$ and the explicit expression of $v = \Phi(t, g)$. First we need to gather some information on $g = e^{\varepsilon^2 t \Delta} u_0$. To that end we use the nondegeneracy assumption (2). It implies that $u_0$ has a finite number of zeroes, all of degree $\pm 1$. We denote

$$ \{u_0 = 0\} = \{z_1^0, \dots, z_N^0\}. $$

Since $g(t, x) = \tilde{g}(\varepsilon^2 t, x)$ where $\tilde{g}(\tilde{t}) = e^{\tilde{t}\Delta} u_0$ is $C^1$ in $[0, \infty) \times \mathbb{T}^2$, we deduce that there exists $t_0, \beta_0, r_0 > 0$ such that, for all $t \le t_0/\varepsilon^2$,

$$ |g(t)| + |\det(\nabla g(t))| \ge \frac{\alpha_0}{2}, $$

$$ |g(t, x)| \ge \beta_0 \quad \text{for } \text{dist}(x, \{z_j^0\}) \ge r_0, $$

$$ g(t) \text{ is invertible and } |\nabla g(t)^{-1}| \lesssim 1 \text{ on } B(z_j^0, r_0). $$

In each disk $B(z_j^0, r_0)$, the map $g(t)$ has exactly one zero $\hat{z}_j(t)$, so

$$ \{g(t) = 0\} = \{\hat{z}_1(t), \dots, \hat{z}_N(t)\}, $$

and we have

$$ (12) \quad \operatorname{dist}(\cdot, \{\hat{z}_j(t)\}) \lesssim |g(t)| \lesssim \operatorname{dist}(\cdot, \{\hat{z}_j(t)\}) $$

Thanks to the implicit function theorem, the maps $t \mapsto \hat{z}_j(t)$ are $C^1$, and

$$ \frac{d}{dt} \hat{z}_j(t) = -\nabla g(t, \hat{z}_j)^{-1} \partial_t g(t, \hat{z}_j), $$

hence

$$ \left| \frac{d}{dt} \hat{z}_j \right| \lesssim \| \partial_t g \|_{\infty} \lesssim \varepsilon^2 \| \Delta g \|_{L^{\infty}}. $$

Viewing $g$ and $u_0$ as periodic maps defined on $\mathbb{R}^2$, $g$ is given by the formula

$$ g(t) = H_{\varepsilon\sqrt{t}} * u_0, $$

where the convolution is on $\mathbb{R}^2$ and $H_\delta(x) = \delta^{-2}H(x/\delta)$, $H(x) = (4\pi)^{-1}e^{-|x|^2/4}$, so

$$ \|\Delta g\|_{L^\infty} \le \|\nabla H_{\varepsilon\sqrt{t}}\|_{L^1} \|\nabla u_0\|_{L^\infty} \lesssim \frac{1}{\varepsilon\sqrt{t}}, $$

and we infer

$$ (13) \quad \left| \frac{d}{dt} \hat{z}_j \right| \lesssim \frac{\varepsilon}{\sqrt{t}}, \quad |z_j(t) - z_j^0| \lesssim \varepsilon \sqrt{t}. $$

Next we combine these properties of $g(t)$ with the explicit expression $v = \Phi(t, g)$ and the bounds (10)-(11) on $e^t w = u - v$ to obtain the desired properties on $u$. We denote by $C > 0$ a generic constant depending on $u_0$ and which
may change from line to line. We start by bounding the modulus $|u|$ from below: using (4) and (10) we obtain, for $0 \le t \le \ln(1/\epsilon) - C$,

$$ egin{aligned} |u| &\ge |v| - e^t|w| \ge \frac{e^t|g|}{\sqrt{1 + |g|^2(e^{2t} - 1)}} - C\epsilon^2 e^{2t} \\ &\ge \frac{1}{2}\min(e^t|g|, 1) - C\epsilon^2 e^{2t}. \end{aligned} $$

The last quantity is positive whenever $e^t|g| \ge 1$ and $e^{2t} < 1/(2C\epsilon^2)$, or $e^t|g| \le 1$ and $|g|^2 > 2C\epsilon^2$. Hence we deduce that

$$ |u| > 0 \quad \text{in} \quad \{|g| \ge C\epsilon\} \quad \text{for} \quad 0 \le t \le \ln \frac{1}{\epsilon} - C. $$

In the case without initial zeros, this proves in particular Corollary 1.2. Moreover, combining this with (12) we have $|u(t)| > 0$ outside the disks $B(\hat{z}_j(t), C\epsilon)$. By homotopy invariance of the topological degree, $u(t)$ must have at least one zero $z_j(t) \in B(\hat{z}_j(t), C\epsilon)$. Next we verify that this zero is unique.

Recall that $g(t)$ is invertible on $B(z_0, r)$, and maps $B(z_j(t), C\epsilon)$ into $B(0, K\epsilon)$, for some constant $K$ depending on $u_0$, thanks to (12). The flow map $\Phi(t) = \Phi(t, \cdot)$ is invertible from $B(0, K\epsilon)$ onto $B(0, R)$ given by $R = \Phi(t, K\epsilon)$, with inverse $\Phi(t)^{-1} = \Phi(-t)$. Therefore $v(t) = \Phi(t) \circ g(t)$ is invertible on $B(z_j(t), C\epsilon)$, and

$$ \sup_{v(B(\hat{z}_j(t), C\epsilon))} |\nabla v(t)^{-1}| \lesssim \sup_{|X| \le \Phi(t, K\epsilon)} |\nabla \Phi(-t, X)|. $$

For $t \le \ln \frac{1}{\epsilon} - C$ we have

$$ \Phi(t, K\epsilon) = \frac{e^t K\epsilon}{\sqrt{1 + K^2\epsilon^2(e^{2t} - 1)}} \le e^t K\epsilon \le \frac{1}{2}, $$

provided $C$ is large enough, and, for $|X| \le 1/2$,

$$ \nabla \Phi(-t, X) = \frac{e^{-t}}{\sqrt{1 - |X|^2(1 - e^{-2t})}} \left( I + \frac{1 - e^{-2t}}{1 - |X|^2(1 - e^{-2t})} X \otimes X \right), $$

so we infer

$$ \sup_{v(B(\hat{z}_j(t), C\epsilon))} |\nabla v(t)^{-1}| \lesssim e^{-t}. $$

We use this to show that $u(t)$ is invertible on $B(\hat{z}_j(t), C\epsilon)$. Since the equation $y = u(x) = v(x) + e^t w(x)$ is equivalent to $x = v^{-1}(y - e^t w(x))$, it suffices to check that the map $F: x \mapsto v^{-1}(y - e^t w(x))$ is a contraction on $B(\hat{z}_j(t), C\epsilon)$, for $|y| < \delta$. Here $\delta > 0$ is a small constant such that $v^{-1}$ is well defined on $B(0, 2\delta)$. Thanks to (10) we have $|e^t w| \le \delta$ provided $C$ is large enough, and

$$ \sup_{B(\hat{z}_j(t), C\epsilon)} |\nabla F| \lesssim e^{-t} \|e^t \nabla w\|_{\infty} \lesssim \|\nabla w\|_{\infty}. $$
Since $\|\nabla w\|_{\infty} \lesssim \varepsilon\sqrt{t}e^t$ thanks to (11), we deduce that $F$ is a contraction for

$$0 \le t \le T_{\varepsilon} = \ln \frac{1}{\varepsilon} - \frac{1}{2} \ln \ln \frac{1}{\varepsilon} - C_0,$$

if the constant $C_0$ is large enough, depending on $u_0$. By the above discussion this shows that $u(t)$ is invertible on $B(\hat{z}_j(t), C\varepsilon)$, and

$$\{u(t) = 0\} = \{z_1(t), \dots, z_N(t)\},$$

for some $z_j(t) \in B(\hat{z}_j(t), C\varepsilon)$. This proves Theorem 1.1, except for its last assertion (3). To verify (3), we note that (10) ensures $|u - v| = |e^t w| \le 1/4$ for $t = T_{\varepsilon}$, so it suffices to check that $|v(T_{\varepsilon}, x)| \ge 3/4$ for $\text{dist}(x, \{z_j^0\}) \gtrsim \varepsilon\sqrt{\ln(1/\varepsilon)}$. We have

$$|v(T_{\varepsilon})| = \frac{e^{T_{\varepsilon}}|g|}{\sqrt{1 + |g|^2(e^{2T_{\varepsilon}} - 1)}} = \frac{1}{\sqrt{1 + (1 - |g|^2)e^{-2T_{\varepsilon}}|g|^{-2}}} \\ = \frac{1}{\sqrt{1 + (1 - |g|^2)e^{C_0} \left( \frac{\varepsilon\sqrt{\ln(1/\varepsilon)}}{g} \right)^2}}.$$

If $|g| \ge M\varepsilon\sqrt{\ln(1/\varepsilon)}$ for some large enough $M > 0$, we deduce $|v(T_{\varepsilon})| \ge 3/4$. Thanks to (12) this implies that $|v(T_{\varepsilon}, x)| \ge 3/4$ for $\text{dist}(x, \{z_j^0\}) \gtrsim \varepsilon\sqrt{\ln(1/\varepsilon)}$ and concludes the proof of Theorem 1.1.

### 3. ENERGY OF $u$: PROOF OF THEOREM 1.3

First, we seek to obtain more precise estimates for $w$ away from the bad disks $B(z_j^0, C\varepsilon\sqrt{\ln(1/\varepsilon)})$. To this end we localize the equation by setting

$$\tilde{w} = \chi^2 w,$$

for some appropriate smooth cut-off function $0 \le \chi(x) \le 1$, to be chosen later. From the equation (5) satisfied by $w$ we deduce

$$(14) \quad \partial_t \tilde{w} - \varepsilon^2 \Delta \tilde{w} = -2(v \cdot \tilde{w})v - |v|^2 \tilde{w} - e^{-t} \chi^2 \mathcal{N}(v, e^t w) - e^{-t} \chi^2 \mathcal{R} \\ - \varepsilon^2 (\Delta \chi^2) w - 2\varepsilon^2 \nabla \chi^2 \cdot \nabla w.$$

Applying Lemma 2.1 to the equation (14) satisfied by $\tilde{w}$, and using (10)-(11) to estimate the two last terms, we deduce

$$\|e^t \tilde{w}\|_{L^\infty} \lesssim \int_0^t e^{t-s} \|e^s \tilde{w}\|_{L^\infty}^2 ds + \int_0^t e^{t-s} \|\chi^2 \mathcal{R}(s)\|_{L^\infty} ds \\ + (\varepsilon\sqrt{t} \|\nabla \chi\|_{L^\infty} + \varepsilon^2 \|\nabla^2 \chi\|_{L^\infty}) \varepsilon^2 e^t (e^{2t} - 1)^{1/2},$$

for all $t \le \ln(1/\varepsilon) - C$. Applying Lemma 2.2 we therefore have

$$(15) \quad \|e^t \chi^2 w\|_{L^\infty} \lesssim \int_0^t e^{t-s} \|\chi^2 \mathcal{R}(s)\|_{L^\infty} ds \\ + (\varepsilon\sqrt{t} \|\nabla \chi\|_{L^\infty} + \varepsilon^2 \|\nabla^2 \chi\|_{L^\infty}) \varepsilon^2 e^t (e^{2t} - 1)^{1/2},$$
provided $t \le \ln(1/\epsilon) - C$ and $\epsilon\sqrt{t}\|\nabla\chi\|_{L^\infty} + \epsilon^2\|\nabla^2\chi\|_{L^\infty} \le 1$. Using the properties (12) of $g$, the fact that $|\hat{z}_j(t) - z_j^0| \lesssim \epsilon\sqrt{\ln(1/\epsilon)}$ for $t \le \ln(1/\epsilon)$ thanks to (13), and letting

$$D(x) = \mathrm{dist}(x, \{z_j^0\}),$$

we have

$$D \lesssim |g| \lesssim D \quad \text{in} \quad \left\{ D \ge M\epsilon\sqrt{\ln\frac{1}{\epsilon}} \right\},$$

for $t \le \ln(1/\epsilon)$. Here $M > 0$ is a large constant that depends only on $u_0$. Since $|\nabla g| \lesssim 1$, recalling the explicit formulas (7) and (4) we deduce

$$ \begin{aligned} |\mathcal{R}| & \lesssim \epsilon^2 \frac{e^t|g|(e^{2t}-1)}{(1+|g|^2(e^{2t}-1))^{3/2}} \\ & \lesssim \epsilon^2 e^t (e^{2t}-1) \frac{D}{(1+C^{-2}D^2(e^{2t}-1))^{3/2}}, \end{aligned} $$

in $\{D \ge M\epsilon\sqrt{\ln(1/\epsilon)}\}$ for $t \le \ln(1/\epsilon) - C$. Therefore, choosing cut-off functions $\chi$ satisfying

$$\mathbf{1}_{2\lambda \le D \le 3\lambda} \le \chi \le \mathbf{1}_{\lambda \le D \le 4\lambda}, \quad |\nabla\chi| \lesssim \frac{1}{\lambda}, \quad |\nabla^2\chi| \lesssim \frac{1}{\lambda^2},$$

for some $\lambda \gtrsim \epsilon\sqrt{\ln(1/\epsilon)}$, from (15) we infer

$$ (16) \quad |e^t w| \lesssim \epsilon^2 e^t D \int_0^t \frac{e^{2s}-1}{(1+C^{-2}D^2(e^{2s}-1))^{3/2}} ds \\ \qquad + \frac{\epsilon}{D} \sqrt{1+t} \epsilon^2 e^t (e^{2t}-1)^{1/2} $$

in $\{D \ge M\epsilon\sqrt{\ln(1/\epsilon)}\}$ for $t \le \ln(1/\epsilon) - C$ and $\epsilon$ small enough. For any $\alpha \in (0, 1/2)$ we have

$$ \begin{aligned} & \int_0^t \frac{e^{2s}-1}{(1+\alpha^2(e^{2s}-1))^{3/2}} ds \\ & = \frac{1}{\alpha^2} \int_1^{(1+\alpha^2(e^{2t}-1))^{1/2}} \frac{(x^2-1)}{x^2(x^2-1+\alpha^2)} dx \le \frac{1}{\alpha^2} \int_1^\infty \frac{dx}{x^2}, \end{aligned} $$

thanks to the change of variable $x = (1 + \alpha^2(e^{2s} - 1))^{1/2}$. Hence from (16) we deduce

$$(17) \quad \frac{1}{\epsilon}|e^t w| \lesssim \epsilon e^t \frac{\sqrt{1+t}}{D}$$

in $\{D \ge M\epsilon\sqrt{\ln(1/\epsilon)}\}$ for $t \le \ln(1/\epsilon) - C$ and $\epsilon$ small enough. Using Lemma 2.3 and (17) to estimate the right-hand side of (14), we also obtain gradient bounds

$$(18) \quad e^t |\nabla w| \lesssim \epsilon e^t \sqrt{1+t} \frac{\sqrt{1+t}}{D}$$
in {$D \ge M\varepsilon\sqrt{\ln(1/\varepsilon)}$} for $t \le \ln(1/\varepsilon) - C$ and $\varepsilon$ small enough.

Next we refine these estimates by including the effect of the second term $-|v|^2\tilde{w}$ in the right-hand side of (14). We choose as above a cut-off function $\chi$ supported in $\{\lambda \le D \le 4\lambda\}$, with $|\nabla\chi| \lesssim \lambda^{-1}$ and $|\nabla^2\chi| \lesssim \lambda^{-2}$, for some $\lambda \ge M\varepsilon\sqrt{\ln\frac{1}{\varepsilon}}$. Combining (10)-(11) and (17)-(18) to bound the two last terms in (14), we have

$$ (19) \quad \partial_t \tilde{w} - \varepsilon^2 \Delta \tilde{w} = -2(v \cdot \tilde{w})v - |v|^2 \tilde{w} + \tilde{F}, \\ |\tilde{F}| \lesssim e^{-t} \|e^t \tilde{w}\|_{L^\infty}^2 + e^{-t} \|\chi^2 \mathcal{R}\|_{L^\infty} \\ \qquad + \frac{\varepsilon^3}{\lambda} \sqrt{1+t} \min \left( \frac{\sqrt{1+t}}{\lambda}, (e^{2t}-1)^{\frac{1}{2}} \right). $$

Arguing as in the proof of Lemma 2.1 but retaining the second term in the right-hand side of (19), we have

$$ \partial_t |\tilde{w}| + |v|^2 |\tilde{w}| - \varepsilon^2 \Delta |w| \le |\tilde{F}|. $$

In the support of $\chi$ we have

$$ \begin{aligned} |v|^2 &= \frac{e^{2t}|g|^2}{1 + |g|^2(e^{2t} - 1)} = 1 - \frac{1 - |g|^2}{1 + |g|^2(e^{2t} - 1)} \\ &\ge \max \left( 1 - \frac{e^{-2t}}{C^2\lambda^2}, 0 \right) \end{aligned} $$

hence

$$ \partial_t |\tilde{w}| + \max \left( 1 - \frac{e^{-2t}}{C^2 \lambda^2}, 0 \right) |\tilde{w}| - \varepsilon^2 \Delta |w| \le |\tilde{F}|. $$

We rewrite this as

$$ \partial_t e^{h(t)} |\tilde{w}| - \varepsilon^2 \Delta e^{h(t)} |\tilde{w}| \le e^{h(t)} |\tilde{F}|, $$

where

$$ \begin{aligned} h(t) &= \int_0^t \max \left( 1 - \frac{e^{-2t}}{C^2 \lambda^2}, 0 \right) ds \\ &= \begin{cases} 0 & \text{for } 0 < t < t_\lambda, \\ t - t_\lambda - \frac{1}{2C^2 \lambda^2} (e^{-2t} - e^{-2t_\lambda}) & \text{for } t > t_\lambda, \end{cases} \end{aligned} $$
where $t_λ = \ln(1/(Cλ))$ is such that $1 - e^{-2t_λ}/(C^2λ^2) = 0$. Arguing again as in Lemma 2.1 we deduce

$$
\begin{align*}
\|\tilde{w}(t)\|_{L^{\infty}} &\le \int_0^t e^{h(s)-h(t)} \|\tilde{F}(s)\|_{L^{\infty}} ds \\
&= \int_0^{t_\lambda} \|\tilde{F}(s)\|_{L^{\infty}} ds \\
&\quad + \int_{t_\lambda}^t e^{s-t} e^{\frac{1}{2C^2\lambda^2}(e^{-2t}-e^{-2s})} \|\tilde{F}(s)\|_{L^{\infty}} ds \\
&\le \int_0^{t_\lambda} \|\tilde{F}(s)\|_{L^{\infty}} ds + C \int_{t_\lambda}^t e^{s-t} \|\tilde{F}(s)\|_{L^{\infty}} ds
\end{align*}
$$

hence, from the bound on $\tilde{F}$ in (19), and estimating the term $\chi^2\mathcal{R}$ exactly as before (because the worst term in (19) is the last one anyway),

$$
\begin{align*}
\|e^t \tilde{w}(t)\|_{L^\infty} &\lesssim \int_0^t e^{t-s} \|e^s \tilde{w}(s)\|_{L^\infty}^2 ds + \int_0^t e^{t-s} \|\chi^2 \mathcal{R}(s)\|_{L^\infty} ds \\
&\quad + \frac{\varepsilon^3}{\lambda} e^t \int_0^{t_\lambda} \sqrt{1+s} (e^{2s}-1)^{1/2} ds + \frac{\varepsilon^3}{\lambda^2} \int_{t_\lambda}^t (1+s) e^s ds \\
&\lesssim \int_0^t e^{t-s} \|e^s \tilde{w}(s)\|_{L^\infty}^2 ds + \frac{\varepsilon^2}{\lambda} e^t + \frac{\varepsilon^3}{\lambda^2} e^t (1+t).
\end{align*}
$$

Applying Lemma 2.2 we obtain

$$
(20) \quad \frac{1}{\varepsilon} |e^t w| \lesssim \varepsilon e^t \frac{1 + (\varepsilon/D)(1+t)}{D},
$$

and with the help of Lemma 2.3 the gradient bound

$$
(21) \quad |e^t \nabla w| \lesssim \varepsilon e^t \sqrt{t} \frac{1 + (\varepsilon/D)(1+t)}{D}.
$$

These bounds are valid in $\{D \ge M\varepsilon\sqrt{\ln(1/\varepsilon)}\}$ for $t \le \ln(1/\varepsilon) - C$ and $\varepsilon$ small enough. Using

$$
\begin{align*}
\int_{\{D \ge M\varepsilon \sqrt{\ln \frac{1}{\varepsilon}}\}} \frac{1 + (\varepsilon^2/D^2)(1+t)^2}{D^2} dx \\
&\lesssim \int_{\varepsilon \sqrt{\ln \frac{1}{\varepsilon}}}^{1} \frac{dr}{r} + \varepsilon^2 (1+t)^2 \int_{\varepsilon \sqrt{\ln \frac{1}{\varepsilon}}}^{1} \frac{dr}{r^3} \lesssim \ln \frac{1}{\varepsilon},
\end{align*}
$$

and $u-v=e^t w$, we deduce the energy bounds

$$
\begin{align*}
\int_{\{D \ge M\varepsilon \sqrt{\ln \frac{1}{\varepsilon}}\}} \frac{|u-v|^2}{\varepsilon^2} dx &\lesssim \varepsilon^2 e^{2t} \ln \frac{1}{\varepsilon}, \\
\int_{\{D \ge M\varepsilon \sqrt{\ln \frac{1}{\varepsilon}}\}} |\nabla u - \nabla v|^2 dx &\lesssim \varepsilon^2 e^{2t} \ln \frac{1}{\varepsilon},
\end{align*}
$$
and, using (10)-(11) to estimate the contributions from $\{D \lesssim \varepsilon\sqrt{\ln(1/\varepsilon)}\}$,

$$ (22) \quad \int_{\Omega} \left( |\nabla u - \nabla v|^2 + \frac{|u - v|^2}{\varepsilon^2} \right) dx \lesssim \varepsilon^2 e^{2t} t \ln \frac{1}{\varepsilon}. $$

Note that this upper bound is $\lesssim \ln(1/\varepsilon)$ at $t = T_\varepsilon \le \ln(1/\varepsilon\sqrt{\ln(1/\varepsilon)})$. Next, we derive energy bounds for $v$. We have

$$ \begin{aligned} 1 - |v|^2 &= 1 - \frac{e^{2t}|g|^2}{1 + |g|^2(e^{2t} - 1)} = \frac{1 - |g|^2}{1 + |g|^2(e^{2t} - 1)} \\ &\le \frac{1}{1 + |g|^2(e^{2t} - 1)}, \end{aligned} $$

and since $|g|$ is of the same order as $\text{dist}(\cdot, \{\hat{z}_j(t)\})$ thanks to (12) we deduce

$$ \begin{aligned} \frac{1}{\varepsilon^2} \int_{\Omega} (1 - |v|^2)^2 dx &\le \frac{1}{\varepsilon^2} \int_{\Omega} \frac{1}{(1 + |g|^2(e^{2t} - 1))^2} dx \\ &\lesssim \frac{1}{\varepsilon^2} \int_0^1 \frac{1}{(1 + C^{-2}r^2(e^{2t} - 1))^2} r dr \\ &\lesssim \frac{1}{\varepsilon^2 e^{2t} - 1} \end{aligned} $$

We also have

$$ |\nabla v| = |D_X \Phi(t, g) \nabla g| \lesssim \frac{e^t}{(1 + |g|^2(e^{2t} - 1))^{1/2}}, $$

hence

$$ \begin{aligned} \int_{\Omega} |\nabla v|^2 &\lesssim e^{2t} \int_0^1 \frac{1}{1 + C^{-2}r^2(e^{2t} - 1)} r dr \\ &\lesssim \frac{e^{2t}}{e^{2t} - 1} \ln(1 + C^{-2}(e^{2t} - 1)) \end{aligned} $$

Gathering the above and recalling $T_\varepsilon = \ln(1/\varepsilon) - \frac{1}{2}\ln\ln(1/\varepsilon) - C_0$, we obtain

$$ \int_{\Omega} \left( |\nabla v|^2 + \frac{1}{\varepsilon^2}(1 - |v|^2)^2 \right) dx \lesssim \ln \frac{1}{\varepsilon} \quad \text{at } t = T_{\varepsilon}. $$

Combining this with the bounds (22) concludes the proof of Theorem 1.3.

### 4. JACOBIAN OF $u$: PROOF OF THEOREM 1.4

Define $u_\varepsilon(x) = u(T_\varepsilon, x)$, where $T_\varepsilon = \ln(1/\varepsilon) - (1/2)\ln\ln(1/\varepsilon) - C_0$ for a large enough constant $C_0$. We consider the jacobian

$$ J u_{\varepsilon} = \det(\nabla u_{\varepsilon}), $$

and show, as $\varepsilon \to 0$, the convergence

$$ (23) \quad J u_{\varepsilon} \to \pi \sum_{j=1}^{N} \hat{d}_{j} \delta_{z_{j}^{0}}, $$
in the sense of distributions, where $\hat{d}_j \in \{\pm 1\}$ is the topological degree of $u_0$ at $z_j^0$.

Note that one can check, by direct calculation, that $Jv(T_\varepsilon)$ converges to this sum of Dirac masses. But the bounds we have obtained on $e^t w = u - v$ are not enough to directly infer (23). Instead we invoke the compactness result of [JS02, Theorem 3.1]: thanks to the energy bound

$$E_\varepsilon(\mathbf{u}_\varepsilon) \lesssim \ln \frac{1}{\varepsilon},$$

there exists a sequence $\varepsilon_n \to 0$, integers $\tilde{d}_k \in \mathbb{Z} \setminus \{0\}$ and distinct points $a_k \in \mathbb{T}^2$ such that

$$J u_{\varepsilon_n} \to \pi \sum_{k=1}^{M} \tilde{d}_k \delta_{a_k}.$$

We show next that we must have $M = N$, $\{a_k\} = \{z_j^0\}$, without loss of generality $a_j = z_j^0$ for $j = 1, \dots, N$, and $\tilde{d}_j = \hat{d}_j$. Therefore the limit is unique and this proves (23).

First we prove that $\{a_k\} \subset \{z_j^0\}$. This is a consequence of the bounds obtained above on the map $u$, and the fact that the limit of $J u_\varepsilon$ provides a lower bound for $E_\varepsilon(\mathbf{u}_\varepsilon)/\ln(1/\varepsilon)$ [JS02, Theorem 4.1]. By that lower bound, for any $\delta > 0$ we must have

$$E_{\varepsilon_n}(\mathbf{u}_{\varepsilon_n}; B(a_k, \delta)) \geq \pi |d_k| \ln \frac{1}{\varepsilon_n} + o\left(\ln \frac{1}{\varepsilon_n}\right),$$

as $n \to \infty$. Note that $|d_k| \ge 1$. Therefore, to show that $\{a_k\} \subset \{z_j^0\}$ it suffices to obtain an upper bound of the form

$$E_{\varepsilon}(\mathbf{u}_{\varepsilon}; B(a, \delta)) \leq \frac{\pi}{2} \ln \frac{1}{\varepsilon} \quad \text{for } \varepsilon \ll 1,$$

for any $a \notin \{z_j^0\}$ and some $\delta > 0$. Recall that we have $u = v + e^t w$, and the pointwise bounds from § 3,

$$\begin{aligned} |\nabla v|^2 + \frac{1}{\varepsilon^2} (1 - |v|^2)^2 &\lesssim \frac{1}{1 + D^2 e^{2t}} \left( e^{2t} + \frac{1}{\varepsilon^2} \frac{1}{1 + D^2 e^{2t}} \right) \\ |\nabla e^t w|^2 + \frac{1}{\varepsilon^2} |e^t w|^2 &\lesssim \varepsilon^2 e^{2t} \frac{t^2}{D^2}, \end{aligned}$$

in $\{D \ge C\varepsilon \ln^{1/2}(1/\varepsilon)\}$ and for $1 \le t \le \ln(1/\varepsilon) - C_0$. For $t = T_\varepsilon = \ln(1/\varepsilon) - (1/2) \ln \ln(1/\varepsilon) - C_0$ we deduce

$$|\nabla u|^2 + \frac{1}{\varepsilon^2} (1 - |u|^2)^2 \lesssim \frac{1 + (\varepsilon/D)^2 e^{4C_0} \ln^2(1/\varepsilon) + e^{-2C_0} \ln(1/\varepsilon)}{D^2},$$

in $\{D \ge C\varepsilon \ln^{1/2}(1/\varepsilon)\}$. Hence at time $t = T_\varepsilon$, for any $\delta \ge \varepsilon \ln(1/\varepsilon)$ and for $\text{dist}(a, \{z_j^0\}) \ge 4\delta$, we have

$$E_{\varepsilon}(\mathbf{u}_{\varepsilon}; B(a, \delta)) \lesssim 1 + \left(\frac{\varepsilon}{\delta}\right)^2 e^{4C_0} \ln^2 \frac{1}{\varepsilon} + e^{-2C_0} \ln \frac{1}{\varepsilon} \le \frac{\pi}{2} \ln \frac{1}{\varepsilon},$$
for $\varepsilon \ll 1$, provided $C_0$ is chosen large enough. By the above discussion, this
implies that $\{a_k\} \subset \{z_j^0\}$.

Therefore we may write

$$
(24) \quad J u_{\varepsilon_n} \rightarrow \pi \sum_{j=1}^{N} \tilde{d}_j \delta_{z_j^0},
$$

where $\tilde{d}_j \in \mathbb{Z}$. Here we allow the possibility that $\tilde{d}_j = 0$ because we have not proven yet that $\{z_j^0\} \subset \{a_k\}$. To prove (23), it suffices to show that $\tilde{d}_j = \hat{d}_j$. To that end, note that for all small $\varepsilon > 0$ we have

$$
\frac{1}{\pi} \int_{B(z_j^0, r)} \det(\nabla u_{\varepsilon}) = \deg(u_{\varepsilon}, \partial B(z_j^0, r)) = \hat{d}_j,
$$

for any small $r > 0$ and $j = 1, \dots, N$. This is because $t \mapsto u(t)$ is smooth
and $u(t)$ doesn't vanish on $\partial B(z_j^0, r)$ for small $\varepsilon > 0$ and all $t \in [0, T_\varepsilon]$, so
the degree of $u_\varepsilon = u(T_\varepsilon)$ is equal to the degree of $u_0 = u(0)$, which is $\hat{d}_j$
by definition. Therefore, testing (24) with a test function $\varphi \approx \mathbf{1}_{B(z_j^0, r)}$, we
obtain $\tilde{d}_j \approx \hat{d}_j$, hence $\tilde{d}_j = \hat{d}_j$ because these are integers. This concludes
the proof of (23).

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DEPARTAMENTO DE INGENIERÍA MATEMÁTICA AND CENTRO DE MODELAMIENTO MATEMÁTICO (UMI 2807 CNRS), UNIVERSIDAD DE CHILE, CASILLA 170 CORREO 3, SANTIAGO, CHILE AND INSTITUTE OF MATHEMATICS, POLISH ACADEMY OF SCIENCES, UL. ŚNIADECKICH 8, 00-656, WARSAW, POLAND

Email address: kowalczy@dim.uchile.cl

INSTITUT DE MATHÉMATIQUES DE TOULOUSE; UMR 5219, UNIVERSITÉ DE TOULOUSE; CNRS, UPS IMT, F-31062 TOULOUSE CEDEX 9, FRANCE.

Email address: Xavier.Lamy@math.univ-toulouse.fr