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\title{On the convergence of solutions for the Ginzburg--Landau equation and system}
\alttitle{Résultats de convergence pour l’équation de Ginzburg--Landau et ses généralisations aux systèmes}

\author{\firstname{Rejeb} \lastname{Hadiji}\IsCorresp}
\address{Univ. Paris Est Creteil, Univ. Gustave Eiffel, CNRS, LAMA UMR8050,
F-94010 Creteil, France}
\email{rejeb.hadiji@u-pec.fr}

\author{\firstname{Jongmin} \lastname{Han}}
\address{Department of Mathematics, Kyung Hee University, Seoul, 130-701, Korea}
\email{jmhan@khu.ac.kr}
\thanks{Jongmin Han was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (RS-2024-00357675)}
\CDRGrant[National Research Foundation of Korea]{RS-2024-00357675}

\subjclass{35B40, 35J60, 35Q60}

\keywords{\kwd{Two component Ginzburg--Landau equations} \kwd{non-symmetric potential} \kwd{asymptotic behavior of solutions}}
\altkeywords{\kwd{\'Equations de Ginzburg--Landau à deux composantes} \kwd{potentiel non symétrique} \kwd{comportement asymptotique des solutions}}

\editornote{Article submitted by invitation}
\alteditornote{Article soumis sur invitation}

\begin{abstract}
Let $(u_\varepsilon)$ be a family of solutions of the Ginzburg--Landau equation with boundary condition $u_\varepsilon = g$ on $\partial \Omega$ and of degree~$0$. Let $u_0$ denote the harmonic map satisfying $u_0 = g$ on $\partial \Omega$. We show that, if there exists a constant $C_1 > 0$ such that for $\varepsilon$ sufficiently small we have $\frac{1}{2} \int_\Omega \lvert{\nabla u_\varepsilon}\rvert^2 \, \mathrm{d}x \leq C_1 \leq
\frac{1}{2} \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x,$ then $C_1 = \frac{1}{2} \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x$ and $u_\varepsilon \to u_0$ in $H^1(\Omega)$. We also prove that if there is a constant $C_2$ such that for $\varepsilon$ small enough we have $\frac12 \int_\Omega \lvert{\nabla u_\varepsilon}\rvert^2 \, \mathrm{d}x \geq C_2 > \frac12 \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x,$ then $\lvert{u_{\varepsilon}}\rvert$ does not converge uniformly to $1$ on $\overline{\Omega} $. We obtain analogous results for both symmetric and non-symmetric two-component Ginzburg--Landau systems.
\end{abstract}

\begin{altabstract}
Soit $(u_\varepsilon)$ une famille de solutions de l’équation de Ginzburg--Landau
avec la condition au bord $u_\varepsilon = g$ sur $\partial \Omega$ et le degré de $g$ est $0$.
On note $u_0$ l’application harmonique vérifiant $u_0 = g$ sur $\partial \Omega$.
Nous montrons que, s’il existe une constante $C_1 > 0$ telle que, pour $\varepsilon$
suffisamment petit,
$
\frac{1}{2} \int_\Omega \lvert{\nabla u_\varepsilon}\rvert^2 \, \mathrm{d}x \leq C_1 \leq
\frac{1}{2} \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x,
$
alors
$
C_1 = \frac{1}{2} \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x
$
et
$
u_\varepsilon \to u_0$
dans $H^1(\Omega)$.
Nous prouvons également que s’il existe une constante $C_2$ telle que, pour
$\varepsilon$ suffisamment petit,
$
\frac{1}{2} \int_\Omega \lvert{\nabla u_\varepsilon}\rvert^2 \, \mathrm{d}x \geq C_2 >
\frac{1}{2} \int_\Omega \lvert{\nabla u_0}\rvert^2 \, \mathrm{d}x,
$
alors $\lvert{u_{\varepsilon}}\rvert$ ne converge pas uniformément vers~$1$ sur~$\overline{\Omega}$.
Nous obtenons des résultats analogues pour des systèmes de Ginzburg--Landau à deux
composantes, symétriques et non symétriques. 
\end{altabstract}

\begin{document}
%\input{CR-pagedemetas}
%\end{document}
\maketitle

\section{Introduction}

Let $\Om \subset \rtwo$ be a smooth bounded domain. Let
\[
g \colon \pa \Om \to S^1=\braces[\big]{z \in \cpx \st \abs{z}=1}
\]
be a smooth map that has a nonnegative integer-valued degree $\deg (g,\pa\Om)=d$. Let us define
\[
H^1_g (\Om) = \braces[\big]{u \in H^1(\Om; \cpx) \st u=g \ \text{on $\pa \Om$}}.
\]
For $\ve>0$, we consider the Ginzburg--Landau energy functional
\begin{equation}\label{eq:ftnal Eb}
G_{\ve} (u) = \frac{1}{2} \int_\Om \abs{\nabla u}^2 \dif x + \frac{1}{4\ve^2} \int_\Om \parens[\big]{1-\abs{u}^2}^2 \dif x.
\end{equation}
The Euler--Lagrange equations for $G_{\ve}$ are the Ginzburg--Landau equations
\begin{equation}\label{eq:GL}
\begin{cases}
	-\Delta u = \frac{1}{\ve^2} u \parens[\big]{1-\abs{u}^2} & \text{in $\Omega$},
\\	u = g & \text{on $\pa\Omega$}.
\end{cases}
\end{equation}
In~\cite{BBH93,BBH94}, Bethuel, Brezis and H\'{e}lein studied the convergence of minimizers. In particular, if $\deg (g, \pa \Om)=0$, they proved the following.

\begin{theo}[\cite{BBH93}]\label{thma:BBH} %%% Theorem A.
Let $u_{\ve}$ be a minimizer of $G_{\ve}$ over $H^1_g (\Om)$ where $\Om$ is a star-shaped domain. If $d=0$, then $u_{\ve} \to u_{0}$ in $C^k_{\loc} (\Om)$ for any nonnegative integer $k$ as $\ve \to 0$ such that $u_0$ is a unique solution of
\begin{equation}\label{eq:u0 ftnal}
u_0 = \argmin_{u \in H^1_g (\Om;S^1)} J_g(u)
\quad \text{where $J_g(u) = \frac{1}{2}\int_\Om \abs{\nabla u}^2\dif x$}.
\end{equation}
The function $u_0$ satisfies
\begin{equation}\label{eq:u0 GL}
\begin{cases}
	-\Delta u = u \abs{\nabla u}^2 & \text{on $\Om$},
\\	u = g & \text{on $\pa\Omega$},
\\	\abs{u} = 1 & \text{on $\Om$}.
\end{cases}
\end{equation}
\end{theo}

See also~\cite{BBH94} for the nonzero-degree case, \cite{HaSh06} for a potential having a zero of infinite order, and~\cite{BMR94} for the quantization effect on the whole plane. According to~\cite[Remark~A.1]{BBH94}, the conclusion of Theorem~\ref{thma:BBH} can still hold even when $u_\ve$ is not a minimizer. Indeed, we have the following.

\begin{theo}[{\cite[p.~144]{BBH94}}]\label{thmb:BBH-Rmk} %%% Theorem B.
Assume $\deg (g,\pa \Om)=0$ and let $u_{\ve}$ be a solution of~\eqref{eq:GL}. If
\begin{equation}\label{eq:u to u0 in H1}
u_\ve \to u_0
\quad \text{in $H^1(\Om)$},
\end{equation}
then the conclusion of Theorem~\ref{thma:BBH} is valid.
\end{theo}

Theorem~\ref{thmb:BBH-Rmk} tells us that the strong convergence~\eqref{eq:u to u0 in H1} is a key ingredient in the proof of Theorem~\ref{thma:BBH}.

Let $(u_\varepsilon)$ be a sequence of solutions to~\eqref{eq:GL}. In this work, we establish that
\[
\frac{1}{2} \int_\Omega \abs{\nabla u_\varepsilon}^2 \dif x
\]
admits the critical lower bound
\[
\frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x,
\]
beyond which the sequence $(u_\varepsilon)$ cannot be lifted to a smooth function, see the proof of Theorem~\ref{thm:main1}.

We provide another sufficient condition for Theorem~\ref{thma:BBH} by identifying an equivalent formulation of~\eqref{eq:u to u0 in H1}. We also introduce a two-component generalization of~\eqref{eq:ftnal Eb} and~\eqref{eq:GL}, from which we derive analogous results.

Two facts used in the proof of Theorem~\ref{thma:BBH} will also play a central role in this paper.

First, if $u_\ve$ is a solution of~\eqref{eq:GL}, then
\begin{equation}\label{u-veleq-1}
\abs{u_\ve} \leq 1
\quad \text{on $\overline{\Om}$}.
\end{equation}
We can prove the inequality~\eqref{u-veleq-1} by applying the maximum principle to the following identity:
\begin{equation}\label{eq:1-u^2 eqn}
- \Delta \parens[\big]{1- \abs{u_\ve}^2} = - \frac{2}{\ve^2} \abs{u_\ve}^2 \parens[\big]{1-\abs{u_\ve}^2} + 2 \abs{\nabla u_\ve}^2
\quad \text{on $\Om$}.
\end{equation}
See~\cite[Proposition~2]{BBH93}.

Second, if the domain $\Om$ is star-shaped, then for any solution $u_\varepsilon$ of~\eqref{eq:GL}, the potential
\[
\frac{1}{\varepsilon^{2}} \int_\Omega \parens[\big]{1 - \abs{u_\varepsilon}^{2}}^{2} \dif x
\]
is bounded. See~\cite[Theorem~III.2]{BBH94} and~\cite{Str94}. Moreover, it is proved in~\cite{Lin97} (see also~\cite{DelP-F98}) that the potential is also bounded provided that
\begin{equation}\label{E-bded}
G_\varepsilon(u_\varepsilon) \leq k \ln\frac{1}{\varepsilon}
\end{equation}
for some constant $k>0$.

In what follows, we suppose that~\eqref{E-bded} is valid or $\Om$ is star-shaped. We have then
\begin{equation}\label{potential-bded}
\frac{1}{\ve^2} \int_\Om \big(1- \abs{u_\ve}^2 \big)^2\dif x \le \ga_0.
\end{equation}
Here, $\ga_0$ depends only on $\Om$ and $g$. The first result of this paper is the following theorem.

\begin{theo}\label{thm:main1}
Suppose that
\begin{equation}\label{eq:deg zero}
\deg (g,\partial \Omega)=0.
\end{equation}
Let $u_\ve$ be a solution of~\eqref{eq:GL}.
\begin{enumerate}\romanenumi
\item \label{thm:main1_i} If there exists a constant $C_1$ such that, for $\ve$ small enough, we have
\begin{equation}\label{G < C_1}
\frac{1}{2} \int_\Omega \abs{\nabla u_\ve}^2 \dif x \leq C_1 \leq
\frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x,
\end{equation}
then
\begin{equation}\label{eq:C1-1}
C_1 = \frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x
\end{equation}
and
\[
u_\ve \to u_0
\quad \text{in $H^1(\Omega)$}.
\]
Thus, Theorem~\ref{thma:BBH} holds true by Theorem~\ref{thmb:BBH-Rmk}.
\item \label{thm:main1_ii} If there exists a constant $C_2$ such that, for $\ve$ small enough, we have
\begin{equation}\label{G bdd by C1-C2}
\frac{1}{2} \int_\Omega \abs{\nabla u_\ve}^2 \dif x \geq C_2 > \frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x,
\end{equation}
then
\begin{equation}\label{no-uniform-convergence-2}
\text{$\abs{u_{\ve}}$ does not converge uniformly to $1$ on $\overline{\Omega}$}.
\end{equation}
\end{enumerate}
\end{theo}

By using Theorem~\ref{thm:main1}, we prove the next theorem where we find a condition that is equivalent to~\eqref{eq:u to u0 in H1}.

\begin{theo}\label{thm:main2}
Let us assume~\eqref{eq:deg zero} and let $u_\ve$ be a solution for~\eqref{eq:GL}. Then,
\begin{equation}\label{eq:potential to zero}
\lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1- \abs{u_\ve}^2 \big)^2 \dif x=0
\end{equation}
if and only if
\begin{equation}\label{eq:lim nabla u}
u_\ve \to u_0
\quad \text{in $H^1(\Om)$}.
\end{equation}
\end{theo}

As a two-component generalization of~\eqref{eq:ftnal Eb}, let us consider
\begin{equation}\label{eq:ftnal G2V}
F_\ve (u, v) = \frac12 \int_\Om \big(\abs{\nabla u}^2+\abs{\nabla v}^2\big) \dif x + \frac{1}{4\ve^2}\int_\Om V\parens[\big]{\abs{u}^2,\abs{v}^2} \dif x
\end{equation}
for $(u_\ve,v) \in H^1_{g_1} (\Om) \times H^1_{g_2}(\Om)$. Here, $g_1, g_2 \colon \pa \Om \to S^1$ are smooth maps such that
\begin{equation}\label{eq:deg of g sys}
d_i=\deg (g_i,\pa\Om)
\end{equation}
is a nonnegative integer for each $i=1,2$. We assume that $\Om$ is star-shaped. The potential function~$V$ is given two cases:
\begin{align*}
	\text{symmetric case:} \quad & V_s\parens[\big]{\abs{u}^2,\abs{v}^2}=\parens[\big]{2-\abs{u}^2-\abs{v}^2}^2,
\\	\text{non-symmetric case:} \quad & V_n\parens[\big]{\abs{u}^2,\abs{v}^2}=\parens[\big]{2-\abs{u}^2-\abs{v}^2}^2 + \parens[\big]{1-\abs{u}^2}^2.
\end{align*}
In each case, $F_\ve$ has a minimizer $(u_\ve,v_\ve)$ over $H^1_{g_1} (\Om) \times H^1_{g_2}(\Om)$. The potential appears in the semi-local gauge field theories~\cite{Hi92,VaAch91}.

The Euler--Lagrange equations are given as follows: for $V=V_s$
\begin{equation}\label{eq:2-GL sym}
\begin{dcases}
	-\Delta u = \frac{1}{\ve^2} u \big(2-\abs{u}^2-\abs{v}^2\big) & \text{in $\Omega$},
\\	-\Delta v = \frac{1}{\ve^2}v \big(2-\abs{u}^2-\abs{v}^2\big) & \text{in $\Omega$},
\\	u =g_1, \quad v =g_2 & \text{on $\pa\Omega$},
\end{dcases}
\end{equation}
and for $V=V_n$
\begin{equation}\label{eq:2-GL nonsym}
\begin{dcases}
	-\Delta u = \frac{1}{\ve^2} u \big(2-\abs{u}^2-\abs{v}^2\big) + \frac{1}{\ve^2} u \parens[\big]{1- \abs{u}^2} & \text{in $\Omega$},
\\	-\Delta v = \frac{1}{\ve^2}v \big(2-\abs{u}^2-\abs{v}^2\big) & \text{in $\Omega$},
\\	u = g_1, \quad v = g_2 & \text{on $\pa\Omega$}.
\end{dcases}
\end{equation}
Now, we want to extend Theorem~\ref{thm:main1} for solutions of~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym}. Since~\eqref{u-veleq-1} and~\eqref{potential-bded} play important roles in the proof Theorem~\ref{thm:main1}, a natural question arises: can we have inequalities for solutions of~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym} analogous to~\eqref{u-veleq-1} and~\eqref{potential-bded}? The answer is not easy. In fact, although the systems~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym} appear to be simple extensions of~\eqref{eq:GL}, the nature of their solutions is quite different, as we shall see.

First, one may expect that if $(u_\ve,v_\ve)$ is a solution of~\eqref{eq:2-GL nonsym}, then
\begin{equation}\label{eq:uv less that 1}
\abs{u_\ve}\le 1
\quad \text{and}
\quad \abs{v_\ve}\le 1
\quad \text{on $\overline{\Om}$}.
\end{equation}

We recall that~\eqref{u-veleq-1} was obtained using the maximum principle applied to the equation~\eqref{eq:1-u^2 eqn}. However, since~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym} are systems of equations, it is not possible to derive such an estimate by simply applying the maximum principle. Instead, weaker versions of~\eqref{eq:uv less that 1} were established in~\cite{HHS23ANONA,HHS-nonsym-deg-0}.

\begin{lemm}[{\cite[Lemma~2.2]{HHS23ANONA}, \cite[Lemma~2.1]{HHS-nonsym-deg-0}}]\label{lem:L-infty variant}
\mbox{}
\begin{enumerate}\romanenumi
\item \label{lem:L-infty variant_i} If $(u_\ve,v_\ve)$ is a solution pair of~\eqref{eq:2-GL sym}, then we have
\begin{equation}\label{eq:L-infty var sym}
\abs{u_\ve}^2 + \abs{v_\ve}^2 \leq 2
\quad \text{on $\overline\Om$}.
\end{equation}
\item \label{lem:L-infty variant_ii} If $(u_\ve,v_\ve)$ is a solution pair of~\eqref{eq:2-GL nonsym}, then we have
\begin{equation}\label{eq:L-infty var}
\abs{u_\ve}^2 \leq \frac{3}{2}
\quad \text{and}
\quad \abs{v_\ve}^2 \leq 2
\quad \text{on $\overline\Om$}.
\end{equation}
Moreover, either $\abs{u_\ve}\le 1$ or $\abs{v_\ve} \le 1$ on $\overline\Om$.
\end{enumerate}
\end{lemm}

The first statement~\eqref{lem:L-infty variant_i} gives no information on the individual upper bounds of $\abs{u_\ve}$ and $\abs{v_\ve}$ although theirs sums are bounded by $2$. The second statement provides no information on the bounds of $\abs{u_\ve}^2 + \abs{v_\ve}^2$ and the upper bounds of $\abs{u_\ve}$ and $\abs{v_\ve}$ are rather rough compared to~\eqref{eq:uv less that 1}. Since the pointwise estimate $\abs{u_\ve}\le 1$ for solutions of~\eqref{eq:GL} are crucial in various analysis of solutions, it is very interesting to prove~\eqref{eq:uv less that 1} or to make analysis of solutions of~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym} without appealing the property of~\eqref{eq:uv less that 1}.

Second difference among solutions of~\eqref{eq:GL}, \eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym} is the Pohozaev identity. Analogous to~\eqref{potential-bded}, we can prove that if $\Om$ is star-shaped, then
\begin{align}
	&	\text{$(u_\ve,v_\ve)$: solution of~\eqref{eq:2-GL sym}}
		\myimplies
		\frac{1}{\ve^2}\int_\Om \parens[\big]{2-\abs{u_\ve}^2-\abs{v_\ve}^2}^2 \dif x \le \ga_1, \label{eq:Pohozaev sys sym}
\\	&	\text{$(u_\ve,v_\ve)$: solution of~\eqref{eq:2-GL nonsym}}
		\myimplies
		\frac{1}{\ve^2}\int_\Om \parens[\big]{2-\abs{u_\ve}^2-\abs{v_\ve}^2}^2 \dif x
		+ \frac{1}{\ve^2}\int_\Om \parens[\big]{1-\abs{u_\ve}^2}^2 \dif x \le \ga_2, \label{eq:Pohozaev sys nonsym}
\end{align}
for some constants $\ga_1$ and $\ga_2$. Since we do not know the signs of $1-\abs{u_\ve}^2$ and $1-\abs{v_\ve}^2$, \eqref{eq:Pohozaev sys sym} does not imply
\begin{equation}\label{eq:pot u and v}
\frac{1}{\ve^2}\int_\Om \parens[\big]{1-\abs{u_\ve}^2}^2 \dif x < \infty
\qquad \text{and} \qquad
\frac{1}{\ve^2}\int_\Om \parens[\big]{1-\abs{v_\ve}^2}^2 \dif x < \infty.
\end{equation}
Indeed, these quantities can diverge for some solutions of~\eqref{eq:2-GL sym} although they satisfy~\eqref{eq:Pohozaev sys sym}. See Theorem~\ref{theorem_C} below. On the other hand, solutions of~\eqref{eq:2-GL nonsym} always satisfy not only~\eqref{eq:Pohozaev sys nonsym} but also~\eqref{eq:pot u and v}. See the proof of Theorem~\ref{thm:main4}\eqref{thm:main4_ii} below.

To state the main results on the solutions of~\eqref{eq:2-GL sym} and~\eqref{eq:2-GL nonsym}, we assume that $d_1=d_2=0$ in~\eqref{eq:deg of g sys} and $\Om$ is star-shaped. We set
\begin{align*}
	\calY(g_1,g_2)	& \coloneqq H^1_{g_1} (\Om;S^1) \times H^1_{g_2} (\Om;S^2),
\\	\calX(g_1, g_2)	& \coloneqq \braces[\Big]{(u_,v) \in H^1_{g_1}(\Om; \mathbb C) \times H^1_{g_2} (\Om; \mathbb C) \st \abs{u}^2 + \abs{v}^2= 2 \ \text{a.e.\ on $\Om$}},
\end{align*}
and
\[
I_{(g_1,g_2)} (u, v) =\frac12 \int_\Om \big(\abs{\nabla u}^2+\abs{\nabla v}^2 \big)\dif x = J_{g_1}(u)+J_{g_2} (v).
\]
Let us consider the following minimization problems:
\begin{align}
	\alpha(g_1,g_2) & \coloneqq \inf \braces[\big]{I_{(g_1,g_2)} (u,v) \st (u,v) \in \calY (g_1,g_2)}, \label{eq:min prob alpha}
\\	\beta(g_1,g_2) & \coloneqq \inf \braces[\big]{I_{(g_1,g_2)} (u,v) \st (u,v) \in \calX (g_1,g_2)}. \label{eq:min prob beta}
\end{align}
The problem~\eqref{eq:min prob alpha} has a unique solution $(u_0,v_0)$ on $\calY(g_1,g_2)$ that satisfies
\[
\begin{cases}
	-\Delta u_0 = u_0 \abs{\nabla u_0}^2 & \text{on $\Om$},
\\	u_0 = g_1 & \text{on $\pa\Omega$},
\\	\abs{u_0} = 1 & \text{on $\Om$},
\end{cases}
\qquad\qquad\qquad
\begin{cases}
	-\Delta v_0 = v_0 \abs{\nabla v_0}^2 & \text{on $\Om$},
\\	v_0 = g_2 & \text{on $\pa\Omega$},
\\	\abs{v_0} = 1 & \text{on $\Om$}.
\end{cases}
\]
If $(u_*,v_*)$ is a solution of~\eqref{eq:min prob beta}, then $(u_*,v_*)$ satisfies
\[
\begin{dcases}
	-\Delta u_* = \frac12 u_* (\abs{\nabla u_*}^2+ \abs{\nabla v_*}^2) & \text{on $\Om$}, \qquad u_* =g_1 \ \text{on $\pa\Omega$},
\\	-\Delta v_* = \frac12 v_* (\abs{\nabla u_*}^2+ \abs{\nabla v_*}^2) & \text{on $\Om$}, \qquad v_* =g_2 \ \text{on $\pa\Omega$},
\\	2 = \abs{u_*}^2+\abs{v_*}^2 \ \text{a.e.} & \text{on $\Om$}.
\end{dcases}
\]
Since $\calY(g_1,g_2) \subset \calX(g_1,g_2)$, it is obvious that
\begin{equation}\label{eq:alpha >= beta}
\alpha (g_1,g_2) \ge \beta (g_1,g_2).
\end{equation}
The next theorem tells us that~\eqref{eq:alpha >= beta} has a close relation with some properties of solutions of~\eqref{eq:2-GL sym}.

\begin{theo}[{\cite[Theorem~1.3(iii)]{HHS23ANONA}}]\label{theorem_C} %%% Theorem C.
Suppose that
\begin{equation}\label{eq:deg zero g1 g2}
\deg (g_1,\pa \Om)= \deg (g_2,\pa \Om)=0.
\end{equation}
Let $(u_\ve,v_\ve)$ be a minimizer of~\eqref{eq:ftnal G2V} with $V=V_s$. If $\al(g_1,g_2)>\beta(g_1,g_2)$, then
\[
\lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1- \abs{u_\ve}^2 \big)^2 \dif x = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1- \abs{v_\ve}^2 \big)^2 \dif x =\infty.
\]
\end{theo}

Now, we extend Theorem~\ref{thm:main1} for solutions of~\eqref{eq:2-GL nonsym} as follows.

\begin{theo}\label{thm:main3}
Let $\Om$ be star-shaped. Suppose that $d_1=d_2=0$ such that~\eqref{eq:deg zero g1 g2} holds. Let $(u_\ve,v_\ve)$ be a solution for~\eqref{eq:2-GL nonsym} and $(u_0,v_0)$ be a unique minimizer of $I_{(g_1,g_2)}$ on $\calY(g_1,g_2)$.
\begin{enumerate}\romanenumi
\item \label{thm:main3_i} If there is a constant $C_3$ such that we have for $\ve$ small enough
\begin{equation}\label{G < C_1-sys-nonsym}
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_\ve}^2 + \abs{\nabla v_\ve}^2 \big)\dif x \leq C_3 \leq
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_0}^2 + \abs{\nabla v_0}^2 \big)\dif x,
\end{equation}
then
\begin{equation}\label{eq:C1-1-sys-nonsym}
C_3 = \int_\Omega \big(\abs{\nabla u_0}^2 + \abs{\nabla v_0}^2 \big)\dif x
\end{equation}
and
\[
(u_\ve,v_\ve) \to (u_0,v_0)
\quad \text{in $H^1(\Om)\times H^1(\Om)$}.
\]
\item \label{thm:main3_ii} If there is a constant $C_4$ such that for $\ve$ small enough we have
\begin{equation}\label{G bdd by C1-C2-sys-nonsym}
\frac12 \int_\Om \big(\abs{\nabla u_\ve}^2 + \abs{\nabla v_\ve}^2 \big)\dif x \geq C_4 > \frac12 \int_\Om \big(\abs{\nabla u_0}^2 + \abs{\nabla v_0}^2 \big) \dif x,
\end{equation}
then
\begin{equation}\label{eq:potential zero nonsym}
\lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \bracks[\Big]{\big(2- \abs{u_\ve}^2 - \abs{v_\ve}^2 \big)^2 + \big(1- \abs{u_\ve}^2 \big)^2} \dif x > 0.
\end{equation}
\end{enumerate}
\end{theo}

Next, we deal with solutions of~\eqref{eq:2-GL sym}. In view of Theorem~\ref{theorem_C}, we obtain the following theorem.

\begin{theo}\label{thm:main4}
Let $\Om$ be star-shaped. Suppose that $d_1=d_2=0$ such that~\eqref{eq:deg zero g1 g2} holds. Let $(u_\ve,v_\ve)$ be a solution for~\eqref{eq:2-GL sym} and $(u_*,v_*)$ be a minimizer of $I_{(g_1,g_2)}$ on $\calX(g_1,g_2)$.
\begin{enumerate}\romanenumi
\item \label{thm:main4_i} If there is a constant $C_5$ such that we have for $\ve$ small enough
\begin{equation}\label{G < C_1-sys-sym}
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_\ve}^2 + \abs{\nabla v_\ve}^2 \big)\dif x \leq C_5 \leq
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_*}^2 + \abs{\nabla v_*}^2 \big)\dif x,
\end{equation}
then
\begin{equation}\label{eq:C1-1-sys-sym}
C_5 = \int_\Omega \big(\abs{\nabla u_*}^2 + \abs{\nabla v_*}^2 \big)\dif x
\end{equation}
and there exists $(\tilu,\tilv) \in \calX(g_1,g_2)$ such that
\[
(u_\ve,v_\ve) \to (\tilu,\tilv)
\quad \text{in $H^1(\Om)\times H^1(\Om)$}.
\]
If $\al(g_1,g_2)=\beta(g_1,g_2)$, then $(\tilu,\tilv)=(u_0,v_0)$.
\item \label{thm:main4_ii} Assume that
\begin{align}
	& \al(g_1,g_2) = \beta(g_1,g_2), \label{eq:al=beta}
\\	& \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2\big)^2\dif x \le \ga_3, \label{eq:gamma3}
\\	& \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{v_\ve}^2\big)^2\dif x \le \ga_4. \label{eq:gamma4}
\end{align}
If there is a constant $C_6$ such that for $\ve$ small enough we have
\begin{equation}\label{G bdd by C1-C2-sys-sym}
\frac12 \int_\Om \big(\abs{\nabla u_\ve}^2 + \abs{\nabla v_\ve}^2 \big)\dif x
\geq C_6 > \frac12 \int_\Om \big(\abs{\nabla u_0}^2 + \abs{\nabla v_0}^2 \big) \dif x + \sqrt{\ga_1\ga_3} + \sqrt{\ga_1\ga_4},
\end{equation}
then
\[
\text{either $\abs{u_\ve}$ or $\abs{v_\ve}$ does not converges uniformly to 1 on $\overline\Om$.}
\]
\end{enumerate}
\end{theo}

We will prove Theorems~\ref{thm:main1} and~\ref{thm:main2} in Section~\ref{sec:thm pf main1 first}. The proofs of Theorems~\ref{thm:main3} and~\ref{thm:main4} are given in Sections~\ref{sec:thm pf main3} and~\ref{sec:thm pf main4}, respectively.

\section{Proofs of Theorems~\ref{thm:main1} and~\ref{thm:main2}}\label{sec:thm pf main1 first}

Throughout this section, we assume~\eqref{eq:deg zero} and prove Theorems~\ref{thm:main1} and~\ref{thm:main2}. Then, we can write
\[
g = e^{i \vp_0}
\quad \text{where $\vp_0 \colon \pa \Om \rightarrow \rone$}.
\]
Moreover, the function $u_0$ is lifted by a harmonic function $\varphi$ such that
\[
\begin{dcases}
	\Delta \vp = 0 \quad \text{in $\Om$} & \quad \text{and} \qquad \vp=\vp_0 \quad \text{on $\pa \Om$},
\\	u_0 = e^{i \varphi} & \quad \text{and} \qquad \int_\Omega \abs{\nabla u_0}^2 \dif x =\int_\Omega \abs{\nabla \vp}^2 \dif x.
\end{dcases}
\]

\begin{proof}[Proof of Theorem~\ref{thm:main1}(\ref{thm:main1_i})]
Suppose that~\eqref{G < C_1}. Since $\norm{u_\ve}_\infty \leq 1$, up to a subsequence, we have $u_\ve \rightharpoonup \tilde{u}$ in $H_g^1(\Omega)$ for some $\tilu \in H_g^1(\Omega)$. By~\eqref{potential-bded}, $\abs{\tilu} =1$ a.e.\ on $\Om$ and consequently ${\tilde{u}} \in H^1_g(\Om; S^1)$. Since $u_0$ is a minimizer of $J_g$, we are led to
\begin{equation}\label{eq:lim-inf}
\frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x
\leq \frac{1}{2} \int_\Omega \abs{\nabla \tilde{u}}^2 \dif x
\leq \liminf_{\ve \to 0} \frac{1}{2}\int_\Omega \abs{\nabla u_\ve}^2 \dif x
\leq C_1 \le \frac{1}{2}\int_\Om \abs{\nabla u_0}^2 \dif x.
\end{equation}
Thus, \eqref{eq:C1-1} is true. Since $u_\ve \to u_0$ weakly in $H^1(\Om)$, we deduce that
\begin{equation}\label{eq:strong conv u to u0}
\int_\Om \abs{\nabla u_\ve - \nabla u_0}^2 \dif x
= \int_\Om \abs{\nabla u_\ve}^2 \dif x + \int_\Om \abs{\nabla u_0}^2 \dif x -2 \int_\Om \nabla u_\ve \cdot \nabla u_0\dif x \to 0.
\end{equation}
Hence, $u_\ve \to u_0$ in $H_g^1(\Om)$. Thus, Theorem~\ref{thma:BBH} holds true by Theorem~\ref{thmb:BBH-Rmk}.
\end{proof}

In the above proof, we prove the following corollary.

\begin{coro}\label{cor:liminf}
If $u_\ve$ is any solution for~\eqref{eq:GL}, then
\[
\liminf_{\ve \to 0} \int_\Om \abs{\nabla u_\ve}^2 \dif x \ge \int_\Om \abs{\nabla u_0}^2 \dif x.
\]
\end{coro}

\begin{proof}
If we assume the contrary, up to a subsequence, we may assume that
\[
\frac{1}{2} \int_\Omega \abs{\nabla u_\ve}^2 \dif x \leq C_1 < \frac{1}{2} \int_\Omega \abs{\nabla u_0}^2 \dif x
\]
for some $C_1$. Then, we get a contradiction by arguing as in~\eqref{eq:lim-inf}.
\end{proof}

To prove Theorem~\ref{thm:main1}\eqref{thm:main1_ii}, we need two lemmas.

\begin{lemm}\label{lem:unif conv}
Let $u_\ve$ be a solution for~\eqref{eq:GL}. If $\abs{u_\ve} \to 1$ uniformly on $\overline\Om$, then
\begin{equation}\label{eq:nabla u L2}
\int_\Om \abs{\nabla u_\ve}^2 \dif x \le 2 \int_\Om \abs{\nabla u_0}^2 \dif x.
\end{equation}
\end{lemm}

\begin{proof}
Since $\abs{u_\varepsilon} \to 1$ uniformly on $\overline\Omega$, we may assume that
\begin{equation}\label{eq:u_bigger_half}
\text{$\abs{u_\varepsilon} \geq \frac{1}{2}$ on $\Omega$ for $\varepsilon > 0$ small enough.}
\end{equation}
Then, $u_\varepsilon / \abs{u_\varepsilon}$ can be lifted by a smooth function $\zeta_\varepsilon$ such that
\[
\frac{u_\varepsilon}{\abs{u_\varepsilon}} = e^{i \zeta_\varepsilon}
\quad \text{on $\Omega$}.
\]
Hence, we can write
\[
u_\varepsilon = \rho_\varepsilon e^{i \zeta_\varepsilon}
\quad \text{with $\rho_\varepsilon = \abs{u_\varepsilon}$}.
\]
Then, $\zeta_\ve = \vp_0$ on $\pa \Om$ and
\begin{equation}\label{eq:nabla form}
\abs{\nabla u_\varepsilon}^2 = \abs{\nabla \rho_\varepsilon}^2 + \rho_\varepsilon^2 \abs{\nabla \zeta_\varepsilon}^2
\end{equation}
and the equation~\eqref{eq:GL} is transformed into a system of $\rho_\ve$ and $\zeta_\ve$:
\begin{align}
	\divergence \parens{\rho_\varepsilon^2 \nabla \zeta_\varepsilon} & = 0 && \text{in $\Omega$}, \label{eq-rho-zeta-1}
\\	\hspace{3.5cm} -\Delta \rho_\ve + \rho_\ve\abs{\nabla \zeta_\varepsilon}^2 & = \frac{1}{\ve^2}\rho_\ve(1- \rho_\ve^2) \quad && \text{in $\Om$}. \hspace{3.5cm} \label{eq-rho-zeta-2}
\end{align}
Multiplying~\eqref{eq-rho-zeta-1} by $\rho_\ve-1$, we obtain
\begin{equation}\label{eq-rho-2-1}
\int_\Omega \abs{\nabla \rho_\varepsilon}^2 \dif x +
\int_\Omega\rho_\ve^2 \abs{\nabla \zeta_\varepsilon}^2 \dif x - \int_\Omega \rho_\ve \abs{\nabla \zeta_\varepsilon}^2 \dif x
= \frac{1}{\ve^2}\int_\Omega \rho_\ve(\rho_\varepsilon - 1)(1- \rho_\ve^2) \dif x \leq 0.
\end{equation}
Hence, it comes from~\eqref{eq:u_bigger_half}, \eqref{eq:nabla form} and~\eqref{eq-rho-2-1} that
\[
\int_\Om \abs{\nabla u_\varepsilon}^2 \dif x \le \int_\Omega \rho_\ve \abs{\nabla \zeta_\varepsilon}^2 \dif x \le 2 \int_\Omega \rho_\ve^2 \abs{\nabla \zeta_\varepsilon}^2 \dif x.
\]
On the other hand, multiplying~\eqref{eq-rho-zeta-1} by $\zeta_\ve -\vp$, we have
\[
\int_\Om \rho_\ve^2 \abs{\nabla \zeta_\ve}^2 \dif x = \int_\Om \rho_\ve^2 \nabla \zeta_\ve \cdot \nabla \vp \dif x \le \parens*{\int_\Om \rho_\ve^2 \abs{\nabla \zeta_\ve}^2 \dif x}^{\frac12} \parens*{\int_\Om \rho_\ve^2 \abs{\nabla \vp}^2 \dif x}^{\frac12}.
\]
In this integration, we used the fact $u_0=u_\varepsilon = g$ on $\partial \Omega$, i.e., $\vp=\zeta_\ve=\vp_0$ on $\pa \Om$. Hence, we conclude that
\[
\int_\Om \abs{\nabla u_\varepsilon}^2 \dif x \le 2 \int_\Om \rho_\ve^2 \abs{\nabla \vp}^2 \dif x \le 2 \int_\Om \abs{\nabla \vp}^2 \dif x.
\qedhere
\]
\end{proof}

In fact, we can obtain a better, sharp estimate.

\begin{lemm}\label{lem:limsup}
Let $u_\ve$ be a solution for~\eqref{eq:GL}. If\/ $\abs{u_\ve} \to 1$ uniformly on $\overline\Om$, then
\begin{equation}\label{eq:limsup}
\limsup_{\ve \to 0} \int_\Om \abs{\nabla u_\ve}^2 \dif x \le \int_\Om \abs{\nabla u_0}^2 \dif x.
\end{equation}
\end{lemm}

\begin{proof}
Let us assume the contrary. Then, there exists a constant $C_2>0$ and a subsequence, still denoted by $u_\ve$, such that
\begin{equation}\label{eq:larger than B1}
\frac12 \int_\Omega \abs{\nabla \vp}^2 \dif x = \frac12 \int_\Om \abs{\nabla u_0}^2 \dif x< C_2 \le \frac12 \int_\Om \abs{\nabla u_\ve}^2 \dif x.
\end{equation}
Since $\abs{u_\varepsilon} \to 1$ uniformly on $\overline\Omega$, we may keep the notations in the proof of Lemma~\ref{lem:unif conv}. Given $\delta \in \parens[\big]{0,\frac14}$, if $\ve$ is small enough, then
\begin{equation}\label{eq:delta chosen}
\frac{1}{2} < \rho_\ve <\rho_\ve^2 +\delta
\quad \text{i.e.,}
\quad \frac{1+\sqrt{1-4\delta}}{2} <\rho_\ve <1.
\end{equation}
Let
\[
\psi_\varepsilon = \zeta_\varepsilon - \varphi.
\]
Then, by~\eqref{eq:nabla form} and~\eqref{eq:larger than B1},
\begin{equation}\label{eq:nabla u formula}
C_2 \le \frac{1}{2} \int_\Omega \abs{\nabla u_\varepsilon}^2 \dif x
= \frac{1}{2} \int_\Omega \abs{\nabla \rho_\varepsilon}^2 \dif x +\frac{1}{2} \int_\Omega \rho_\varepsilon^2 \abs[\big]{\nabla (\vp +\psi_\ve)}^2 \dif x.
\end{equation}
We rewrite~\eqref{eq-rho-zeta-1} and~\eqref{eq-rho-zeta-2} as
\begin{align}
	\divergence \parens*{\rho_\varepsilon^2 \nabla(\varphi + \psi_\varepsilon)} & = 0 && \text{in $\Omega$}, \label{eq-rho-psi-1}
\\	\hspace{3cm} -\Delta \rho_\ve + \rho_\ve\abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 & = \frac{1}{\ve^2}\rho_\ve(1- \rho_\ve^2) \quad && \text{in $\Om$}. \hspace{3cm} \label{eq-rho-psi-2}
\end{align}
Multiplying~\eqref{eq-rho-psi-2} by $\rho_\varepsilon - 1$ and integrating it over $\Omega$, and using the boundary condition $\rho_\varepsilon =1$ on $\partial\Omega$, we obtain
\begin{multline}\label{eq-rho-psi-2-1}
\frac12\int_\Omega \abs{\nabla \rho_\varepsilon}^2 \dif x +
\frac12 \int_\Omega\rho_\ve^2 \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x - \frac12 \int_\Omega \rho_\ve \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x
\\	\begin{aligned}
		& = \frac{1}{2\ve^2}\int_\Omega \rho_\ve(\rho_\varepsilon - 1)(1- \rho_\ve^2) \dif x
	\\	& \leq 0.
	\end{aligned}
\end{multline}
Then, from~\eqref{eq:delta chosen}, \eqref{eq:nabla u formula} and~\eqref{eq-rho-psi-2-1}, it follows that
\begin{equation}\label{eq-C2-final}
\begin{split}
C_2
	& \le \frac12
\int_\Omega\rho_\ve \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x
\\	& \leq \frac12 \int_\Omega (\rho_\varepsilon^2+\delta) \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x
\\	& \le \frac12 \int_\Omega \rho_\varepsilon^2 \big(\abs{\nabla \varphi}^2 +2 \nabla \varphi \cdot \psi_\varepsilon +\abs{\nabla \psi_\varepsilon}^2 \big)\dif x + \frac12 \delta \norm{\nabla \zeta_\ve}_2^2
\\	& \le \frac12 \int_\Omega \abs{\nabla \varphi}^2 +\frac12 \int_\Omega \rho_\varepsilon^2 \big(2 \nabla \varphi \cdot \psi_\varepsilon +\abs{\nabla \psi_\varepsilon}^2 \big)\dif x + \frac12 \delta \norm{\nabla \zeta_\ve}_2^2.
\end{split}
\end{equation}
Multiplying~\eqref{eq-rho-psi-1} by $\psi_\varepsilon$, integrating it over $\Omega$, and using the boundary condition $\psi_\varepsilon = 0$ on $\partial\Omega$, we obtain
\begin{equation}\label{eq-rho-psi-1-1}
\int_\Omega \rho_\varepsilon^2 \abs{\nabla \psi_\varepsilon}^2 \dif x +\int_\Omega \rho_\varepsilon^2 \nabla \varphi \cdot \nabla \psi_\varepsilon \dif x =0.
\end{equation}
Furthermore, by~\eqref{eq:nabla u L2} and~\eqref{eq:delta chosen},
\begin{equation}\label{eq:nabla zeta}
\norm{\nabla \zeta_\ve}_2^2 \le 4 \int_\Om \rho_\ve^2 \abs{\nabla \zeta_\ve}^2 \dif x \le 8 \int_\Om \abs{\nabla u_0}^2 \dif x.
\end{equation}
Hence, by~\eqref{eq-C2-final}, \eqref{eq-rho-psi-1-1} and~\eqref{eq:nabla zeta}, we are led to
\[
0< C_2 - \frac12 \int_\Omega \abs{\nabla \varphi}^2 \le - \int_\Omega \rho_\varepsilon^2 \abs{\nabla \psi_\varepsilon}^2 \dif x+ 4 \delta \norm{\nabla u_0}_2^2 \le 4 \delta \norm{\nabla u_0}_2^2.
\]
Letting $\delta \to 0$, we arrive at a contradiction.
\end{proof}

\begin{lemm}\label{lem:pot to zero implies unif conv}
Let $u_\ve$ be a solution for~\eqref{eq:GL} that satisfies~\eqref{eq:potential to zero}. Then, $\abs{u_\ve} \to 1$ uniformly on $\overline\Om $.
\end{lemm}
\begin{proof}
See~\cite[Step A.1, B.2]{BBH93}.
\end{proof}

\begin{lemm}\label{lem:H1 conv implies unif conv}
Let $u_\ve$ be a solution for~\eqref{eq:GL}. If $u\to u_0$ in $H^1(\Om)$, then $\abs{u_\ve} \to 1$ uniformly on $\overline\Om $.
\end{lemm}

\begin{proof}
By multiplying~\eqref{eq:1-u^2 eqn} by $1-\abs{u_\ve}^2$, we obtain
\begin{equation}\label{eq:1-u^2 int}
2 \int_\Om \abs{\nabla u_\ve}^2 \parens[\big]{1-\abs{u_\ve}^2}\dif x
= \frac{2}{\ve^2} \int_\Om \abs{u_\ve}^2 \big(1-\abs{u_\ve}^2 \big)^2 \dif x + \int_\Om \abs[\big]{\nabla \parens[\big]{1-\abs{u_\ve}^2}}^2 \dif x.
\end{equation}
Given $\delta \in \parens[\big]{0,\frac14}$, let
\[
\Om_\ve^\delta = \braces[\big]{x \in \Om \st 1-\abs{u_\ve}^2 >\delta}.
\]
By~\eqref{potential-bded},
\[
\ga_0 \ge \frac{1}{\ve^2} \int_{\Om_\ve^\delta} \big(1- \abs{u_\ve}^2\big)^2 \dif x \ge \frac{(1-\delta)^2}{\ve^2} \abs{\Om_\ve^\delta}.
\]
Hence, for all $\delta \in \parens[\big]{0,\frac14}$, $\abs{\Om_\ve^\delta} \to 0$ as $\ve\to 0$. Since $u\to u_0$ in $H^1(\Om)$, it follows that for each fixed $\delta \in \parens[\big]{0,\frac14}$,
\[
\int_{\Om_\ve^\delta} \abs{\nabla u_\ve}^2 \dif x \le 2 \int_{\Om_\ve^\delta} \abs{\nabla u_\ve - \nabla u_0}^2 \dif x + 2 \int_{\Om_\ve^\delta} \abs{\nabla u_0}^2 \dif x \to 0
\]
as $\ve \to 0$. Since $u\to u_0$ in $H^1(\Om)$, we have $\norm{\nabla u_\ve}_2^2 \le C$ for some $C$. Now, we see that as $\ve \to 0$,
\[
\int_\Om \abs{\nabla u_\ve}^2 \parens[\big]{1-\abs{u_\ve}^2}\dif x \le \delta \int_{\Om \backslash \Om_\ve^\delta} \abs{\nabla u_\ve}^2 \dif x + \int_{\Om_\ve^\delta} \abs{\nabla u_\ve}^2 \dif x \le C \delta +o(1).
\]
So, we deduce from~\eqref{eq:1-u^2 int} that for all $\delta \in \parens[\big]{0,\frac14}$,
\[
\limsup_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \abs{u_\ve}^2 \big(1-\abs{u_\ve}^2 \big)^2 \dif x +\limsup_{\ve \to 0} \int_\Om \abs[\big]{\nabla \parens[\big]{1-\abs{u_\ve}^2}}^2 \dif x \le C \delta.
\]
Letting $\delta \to 0$, we obtain that
\begin{equation}\label{eq:(1-u^2)^2 vs (1-u^2)^3}
\begin{split}
0
	& = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \abs{u_\ve}^2 \big(1-\abs{u_\ve}^2 \big)^2 \dif x
\\	& = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2 \big)^2 \dif x - \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2 \big)^3 \dif x.
\end{split}
\end{equation}
and
\begin{equation}\label{eq:nabla 1-u^2}
\lim_{\ve \to 0} \int_\Om \abs[\big]{\nabla \parens[\big]{1-\abs{u_\ve}^2}}^2 \dif x =0.
\end{equation}
By using~\eqref{potential-bded}, \eqref{eq:nabla 1-u^2} and the Gagliardo--Nirenberg inequality
\[
\norm{u}_3^3 \le C \, \norm{u}_2^2 \, \norm{\nabla u}_2
\quad \text{for $u \in H^1_0(\Om)$},
\]
we are led to
\[
\begin{split}
\frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2 \big)^3 \dif x
	& \le \frac{C}{\ve^2} \parens*{\int_\Om \big(1-\abs{u_\ve}^2 \big)^2 \dif x} \parens*{\int_\Om \abs[\big]{\nabla \parens[\big]{1-\abs{u_\ve}^2}}^2 \dif x}^{\frac12}
\\	& \le C \ga_0 \parens*{\int_\Om \abs[\big]{\nabla \parens[\big]{1-\abs{u_\ve}^2}}^2 \dif x}^{\frac12} \to 0.
\end{split}
\]
In the sequel, we conclude from~\eqref{eq:(1-u^2)^2 vs (1-u^2)^3} that
\begin{equation}\label{eq:pot to zero in pf}
\lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2 \big)^2 \dif x = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \big(1-\abs{u_\ve}^2 \big)^3 \dif x = 0,
\end{equation}
which implies by Lemma~\ref{lem:pot to zero implies unif conv} that $\abs{u_\ve} \to 1$ uniformly on $\overline\Om$. This finishes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main1}(\ref{thm:main1_ii})]
Let us assume the contrary. Then, $\abs{u_\ve} \to 1$ uniformly on $\overline\Om $. Hence, \eqref{eq:limsup} holds by Lemma~\ref{lem:limsup} which contradicts~\eqref{G bdd by C1-C2}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main2}]
Suppose that~\eqref{eq:potential to zero} holds. Then, $\abs{u_\ve} \to 1$ uniformly on $\Om$ by Lemma~\ref{lem:pot to zero implies unif conv}. Moreover, by Corollary~\ref{cor:liminf} and Lemma~\ref{lem:limsup}, we have
\[
\lim_{\ve \to \infty} \int_\Om \abs{\nabla u_\ve}^2 \dif x = \int_\Om \abs{\nabla u_0}^2 \dif x.
\]
Since $u_\ve \to u_0$ weakly in $H^1(\Om)$, we deduce from~\eqref{eq:strong conv u to u0} that $u_\ve \to u_0$ in $H_g^1(\Om)$.

Conversely, suppose that~\eqref{eq:lim nabla u} is true. Since $\abs{u_\ve} \to 1$ uniformly on $\overline\Om$ by Lemma~\ref{lem:H1 conv implies unif conv}, we may assume that $\abs{u_\ve}^2 \ge 1/2$ and use notations in Lemmas~\ref{lem:unif conv} and~\ref{lem:limsup}. Multiplying~\eqref{eq-rho-psi-2} by $\rho_\ve -1$, we obtain
\[
\begin{split}
\int_\Omega \abs{\nabla \rho_\varepsilon}^2 \dif x +\frac{1}{\ve^2} \int_\Omega \rho_\ve(1- \rho_\varepsilon)(1- \rho_\ve^2) \dif x
	& = \int_\Omega (\rho_\ve - \rho_\ve^2) \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x 
\\	& \le \norm{1- \rho_\ve}_\infty \int_\Omega \abs[\big]{\nabla (\varphi + \psi_\varepsilon)}^2 \dif x \to 0.
\end{split}
\]
Here, we used the fact that $u_\ve \to u_0$ in $H^1(\Om)$ such that $\norm[\big]{\nabla (\varphi + \psi_\varepsilon)}_2$ is bounded as $\ve \to 0$. As a consequence,
\[
\begin{split}
0
	& = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Omega \rho_\ve(1- \rho_\varepsilon)(1- \rho_\ve^2) \dif x
\\	& = \lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Omega \frac{\rho_\ve}{1+\rho_\ve} (1- \rho_\ve^2)^2\dif x
\\	& = \lim_{\ve \to 0} \frac{1}{2\ve^2} \int_\Om (1-\rho_\ve^2)^2 \dif x
\\	& = 0
\end{split}
\]
and the proof is complete.
\end{proof}

\section{Proof of Theorem~\ref{thm:main3}}\label{sec:thm pf main3}

Throughout this section, we assume~\eqref{eq:deg zero g1 g2} and prove Theorem~\ref{thm:main3}. We also assume that $\Om$ is star-shaped. We can write
\[
g_1 = e^{i \vp_0}
\quad \text{and}
\quad g_2= e^{i \psi_0}
\quad \text{where $\vp_0, \psi_0 \colon \pa \Om \rightarrow \rone$}.
\]
The functions $u_0$ and $v_0$ are lifted by harmonic functions $\varphi$ and $\psi$ respectively such that
\begin{equation}\label{eq:phi sys}
\begin{dcases}
	\Delta \vp = 0 \quad \text{in $\Om$} & \quad \text{and} \qquad \vp=\vp_0 \quad \text{on $\pa \Om$},
\\	u_0 = e^{i \varphi} & \quad \text{and} \qquad \int_\Omega \abs{\nabla u_0}^2 \dif x =\int_\Omega \abs{\nabla \vp}^2 \dif x,
\end{dcases}
\end{equation}
and
\begin{equation}\label{eq:psi sys}
\begin{dcases}
	\Delta \psi = 0 \quad \text{in $\Om$} & \quad \text{and} \qquad \psi=\psi_0 \quad \text{on $\pa \Om$},
\\	v_0 = e^{i \psi} & \quad \text{and} \qquad \int_\Omega \abs{\nabla v_0}^2 \dif x =\int_\Omega \abs{\nabla \psi}^2 \dif x.
\end{dcases}
\end{equation}

\begin{proof}[Proof of Theorem~\ref{thm:main3}(\ref{thm:main3_i})]
Suppose that~\eqref{G < C_1-sys-nonsym} is valid. Since $\norm{u_\ve}_\infty + \norm{v_\ve}_\infty \leq 3$ by Lemma~\ref{lem:L-infty variant}\eqref{lem:L-infty variant_ii}, up to a subsequence, we have $(u_\ve,v_\ve) \rightharpoonup (\tilde{u},\tilv)$ in $H^1 (\Omega) \times H^1(\Omega)$ for some $(\tilu,\tilv) \in H^1_{g_1}(\Omega) \times H^1_{g_2}(\Omega)$. By~\eqref{eq:Pohozaev sys nonsym}, $\abs{\tilu} =1$ and $\abs{\tilv} =1$ a.e.\ on $\Om$ and consequently ${\tilde{u}} \in H^1_{g_1}(\Om; S^1)$ and ${\tilde{v}} \in H^1_{g_2} (\Om; S^1)$. Since $(u_0,v_0)$ is a unique minimizer of $I_{(g_1,g_2)}$ on $\calY(g_1,g_2)$, we are led to
\[
\begin{split}
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_0}^2+\abs{\nabla v_0}^2 \big) \dif x
	& \leq \frac{1}{2} \int_\Omega \big(\abs{\nabla \tilu}^2+\abs{\nabla \tilv}^2 \big)\dif x
\\	& \leq \liminf_{\ve \to 0} \frac{1}{2}\int_\Omega \big(\abs{\nabla u_\ve}^2+\abs{\nabla v_\ve}^2 \big) \dif x
\\	& \leq C_3
\\	& \le \frac{1}{2} \int_\Omega \big(\abs{\nabla u_0}^2 + \abs{\nabla v_0}^2 \big)\dif x.
\end{split}
\]
Thus, \eqref{eq:C1-1-sys-nonsym} is true. Moreover, $u_\ve \to u_0$ in $H_{g_1}^1(\Om)$ and $v_\ve \to v_0$ in $H_{g_2}^1(\Om)$ as in the proof of Theorem~\ref{thm:main1}\eqref{thm:main1_i}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main3}(\ref{thm:main3_ii})]
Let us assume the contrary so that
\begin{equation}\label{eq:potential zero nonsym pf}
\lim_{\ve \to 0} \frac{1}{\ve^2} \int_\Om \bracks[\Big]{\big(2- \abs{u_\ve}^2 - \abs{v_\ve}^2 \big)^2 + \big(1- \abs{u_\ve}^2 \big)^2} \dif x = 0.
\end{equation}
If~\eqref{eq:potential zero nonsym pf} is valid, then it follows from~\cite[Lemma~2.5]{HHS-nonsym-deg-0} that $\abs{u_\ve}\to 1$ and $\abs{v_\ve}\to 1$ uniformly on $\overline\Om$. So, we may assume that $\abs{u_\ve}^2 \ge 1/2$ and $\abs{v_\ve}^2 \ge 1/2$ on $\Om$. We can write
\begin{equation}\label{eq:zeta xi}
u_\varepsilon = \rho_\varepsilon e^{i \zeta_\varepsilon}
\qquad \text{and} \qquad
v_\varepsilon = \sigma_\varepsilon e^{i \xi_\varepsilon},
\end{equation}
where $\rho_\ve = \abs{u_\ve}$ and $\sigma_\ve = \abs{v_\ve}$. Set
\begin{equation}\label{eq:eta chi}
\eta_\varepsilon = \zeta_\varepsilon -\varphi
\qquad \text{and} \qquad
\chi_\varepsilon = \xi_\varepsilon - \psi.
\end{equation}
Then, \eqref{eq:2-GL nonsym} is written as
\begin{align}
	\divergence \big(\rho_\varepsilon^2 \nabla(\varphi+\eta_\varepsilon)\big) &= 0, \label{eq:rho-nonsys}
\\	-\Delta \rho_\varepsilon + \rho_\varepsilon \abs{\nabla \varphi + \nabla \eta_\varepsilon}^2 &= \frac{1}{\varepsilon^2}\rho_\varepsilon \parens{2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2} + \frac{1}{\varepsilon^2}\rho_\varepsilon \parens{1 - \rho_\varepsilon^2}, \label{eq:zeta-nonsys}
\\	\divergence \big(\sigma_\varepsilon^2 \nabla(\psi+\chi_\varepsilon)\big) &= 0, \label{eq:sigma-nonsys}
\\	-\Delta \sigma_\varepsilon + \sigma_\varepsilon \abs{\nabla \psi+\chi_\varepsilon}^2 &= \frac{1}{\varepsilon^2}\sigma_\varepsilon \parens{2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2}. \label{eq:xi-nonsys}
\end{align}
By multiplying~\eqref{eq:zeta-nonsys} by $\rho_\ve-1$ and~\eqref{eq:xi-nonsys} by $\sigma_\ve-1$, we obtain from~\eqref{G bdd by C1-C2-sys-nonsym}
\[
\begin{split}
C_4
	& \le \frac12 \int_\Om \big(\abs{\nabla u_\ve}^2 + \abs{\nabla v_\ve}^2 \big) \dif x
\\	& = \frac{1}{2}\int_\Omega
\parens[\big]{\abs{\nabla \rho_\varepsilon}^2 + \abs{\nabla \sigma_\varepsilon}^2 + \rho_\varepsilon^2 \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 + \sigma_\varepsilon^2 \abs{\nabla \psi+\nabla \chi_\varepsilon}^2}
\\	& = \frac12 \int_\Omega \parens[\big]{\rho_\varepsilon \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 + \sigma_\varepsilon \abs{\nabla \psi+\nabla \chi_\varepsilon}^2} \dif x + D_1+D_2+D_3,
\end{split}
\]
where
\begin{equation}\label{eq:D123}
\begin{dcases}
	D_1 = \frac{1}{\varepsilon^2} \int \rho_\varepsilon (\rho_\varepsilon - 1)(2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2),
\\	D_2 = \frac{1}{\varepsilon^2} \int \sigma_\varepsilon (\sigma_\varepsilon - 1)(2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2),
\\	D_3 = \frac{1}{\varepsilon^2} \int \rho_\varepsilon (\rho_\varepsilon - 1)(1 - \rho_\varepsilon^2).
\end{dcases}
\end{equation}
Then, $D_j\to 0$ for each $j=1,2,3$ as $\ve \to0$. Indeed, by H\"{o}lder's inequality and the condition~\eqref{eq:potential zero nonsym pf}, we can show that $D_1 \to 0$ and $D_3 \to 0$ as $\ve \to 0$. Moreover, as $\ve \to 0$, we have
\[
\begin{split}
o(1)
	& = \frac{1}{\varepsilon^2} \int_\Om (2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2)^2 \dif x
\\	& = \frac{1}{\varepsilon^2}\int_\Om (1 - \rho_\varepsilon^2)^2 \dif x + \frac{2}{\varepsilon^2}\int_\Om (1 - \rho_\varepsilon^2)(1 - \sigma_\varepsilon^2) \dif x + \frac{1}{\varepsilon^2}\int_\Om (1 - \sigma_\varepsilon^2)^2 \dif x
\\	& = o(1) + \frac{2}{\varepsilon^2}\int_\Om (1 - \rho_\varepsilon^2)(1 - \sigma_\varepsilon^2) \dif x + \frac{1}{\varepsilon^2}\int_\Om (1 - \sigma_\varepsilon^2)^2 \dif x.
\end{split}
\]
Hence, by H\"{o}lder's inequality, we obtain
\[
\frac{1}{\varepsilon^2}\int_\Om (1 - \sigma_\varepsilon^2)^2 \dif x \le o(1)+ 2 \bracks*{\frac{1}{\varepsilon^2}\int_\Om (1 - \rho_\varepsilon^2)^2 \dif x}^{\frac12}
\bracks*{\frac{1}{\varepsilon^2}\int_\Om (1 - \sigma_\varepsilon^2)^2 \dif x}^{\frac12}.
\]
Thus, $\norm{1-\sigma_\ve^2}_2\to 0$ and then H\"{o}lder's inequality implies that $D_2 \to 0$.

We have shown that as $\ve \to 0$,
\begin{equation}\label{eq:A1 A2 nonsym}
\begin{split}
C_4
	& \le o(1) + \frac{1}{2}\int_\Omega \rho_\varepsilon \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 \dif x + \frac{1}{2}\int_\Omega
\sigma_\varepsilon \abs{\nabla \psi+\nabla \chi_\varepsilon}^2 \dif x
\\	& \eqqcolon o(1)+A_1+A_2.
\end{split}
\end{equation}
Let $\delta \in \parens[\big]{0,\frac14}$ be given and we may assume~\eqref{eq:delta chosen}. So, we have
\[
\begin{split}
A_1
	& \le \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 \dif x + \frac{\delta}{2}\int_\Omega \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 \dif x
\\	& = \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2 \dif x + \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \big(2 \nabla \vp \cdot \nabla \eta_\varepsilon + \abs{\nabla \eta_\varepsilon}^2 \big) \dif x + \frac{\delta}{2}\int_\Omega \abs{\nabla \vp +\nabla \eta_\ve}^2 \dif x.
\end{split}
\]
By multiplying~\eqref{eq:rho-nonsys} by $\psi_\ve$, we obtain
\begin{equation}\label{eq:vp psi sys nonsym}
\int_\Om \rho_\ve^2 \abs{\nabla \eta_\ve}^2 \dif x+ \int_\Om \rho_\ve^2 \nabla \vp \cdot \nabla \eta_\ve \dif x=0.
\end{equation}
So,
\begin{equation}\label{eq:A1 ineq 3}
A_1 \le \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2 \dif x - \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \abs{\nabla \eta_\ve}^2\dif x + \frac{\delta}{2}\int_\Omega \abs{\nabla \vp +\nabla \eta_\varepsilon}^2 \dif x.
\end{equation}
On the other hand, \eqref{eq:vp psi sys nonsym} implies that
\[
\begin{split}
\int_\Om \rho_\ve^2 \abs{\nabla \vp +\nabla \eta_\ve}^2 \dif x
	& = \int_\Om \rho_\ve^2 (\nabla \vp +\nabla \eta_\ve) \cdot \nabla \vp \dif x
\\	& \le \parens*{\int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp+\nabla \eta_\ve}^2\dif x}^{\frac12} \parens*{\int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2\dif x}^{\frac12}.
\end{split}
\]
Since $\rho_\ve^2\ge 1/2$, this inequality implies that
\[
\frac12 \int_\Om \abs{\nabla \vp +\nabla \eta_\ve}^2 \dif x \le \int_\Om \rho_\ve^2 \abs{\nabla \vp +\nabla \eta_\ve}^2 \dif x \le \int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2\dif x.
\]
Hence, we can rewrite~\eqref{eq:A1 ineq 3} as
\[
A_1 \le \frac{1}{2}\int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2 \dif x + \delta \int_\Omega \rho_\varepsilon^2 \abs{\nabla \vp}^2 \dif x.
\]
By a similar argument, we also obtain
\[
A_2 \le \frac{1}{2}\int_\Omega \sigma_\varepsilon^2 \abs{\nabla \psi}^2 \dif x + \delta \int_\Omega \sigma_\varepsilon^2 \abs{\nabla \psi}^2 \dif x.
\]
In the sequel, we deduce from~\eqref{eq:A1 A2 nonsym} that
\[
C_4 \le o(1) + \frac{1}{2}\int_\Omega \big(\rho_\varepsilon^2 \abs{\nabla \vp}^2+\sigma_\ve^2 \abs{\nabla \psi}^2 \big) \dif x + \delta \int_\Omega \big(\rho_\ve^2 \abs{\nabla \vp}^2 + \sigma_\ve^2 \abs{\nabla \psi}^2 \big) \dif x.
\]
Letting $\ve \to 0$, we are led to
\[
C_4 \le \frac{1}{2}\int_\Omega \big(\abs{\nabla \vp}^2+ \abs{\nabla \psi}^2 \big) \dif x + \frac{\delta}{2}\int_\Omega \big(\abs{\nabla \vp}^2 + \abs{\nabla \psi}^2 \big) \dif x.
\]
Finally, by taking the limit $\delta \to 0$, we get a contradiction from the assumption~\eqref{G bdd by C1-C2-sys-nonsym}.
\end{proof}

\section{Proof of Theorem~\ref{thm:main4}}\label{sec:thm pf main4}

This section is devoted to the proof of Theorem~\ref{thm:main4}. Throughout this section, we assume that~\eqref{eq:deg zero g1 g2} holds and $\Om$ is star-shaped.

\begin{proof}[Proof of Theorem~\ref{thm:main4}(\ref{thm:main4_i})]
Suppose that~\eqref{G < C_1-sys-nonsym} is valid. Since $\norm{u_\ve}_\infty + \norm{v_\ve}_\infty \leq 2$ by Lemma~\ref{lem:L-infty variant}\eqref{lem:L-infty variant_i}, up to a subsequence, we have $(u_\ve,v_\ve) \rightharpoonup (\tilde{u},\tilv)$ in $H^1 (\Omega) \times H^1(\Omega)$ for some $(\tilu,\tilv) \in H^1_{g_1}(\Omega) \times H^1_{g_2}(\Omega)$. By~\eqref{eq:Pohozaev sys sym}, $\abs{\tilu}^2+\abs{\tilv}^2 =2$ a.e.\ on $\Om$ and thus $(\tilu,\tilv) \in \calX(g_1,g_2)$. Since $(u_*,v_*)$ is a minimizer of $I_{(g_1,g_2)}$ on $\calX(g_1,g_2)$, we are led to
\[
\begin{split}
\frac{1}{2} \int_\Omega \big(\abs{\nabla u_*}^2+\abs{\nabla v_*}^2 \big) \dif x
	& \leq \frac{1}{2} \int_\Omega \big(\abs{\nabla \tilu}^2+\abs{\nabla \tilv}^2 \big)\dif x
\\	& \leq \liminf_{\ve \to 0} \frac{1}{2}\int_\Omega \big(\abs{\nabla u_\ve}^2+\abs{\nabla v_\ve}^2 \big) \dif x
\\	& \leq C_5
\\	& \le \frac{1}{2} \int_\Omega \big(\abs{\nabla u_*}^2 + \abs{\nabla v_*}^2 \big)\dif x.
\end{split}
\]
Thus, \eqref{eq:C1-1-sys-sym} is obtained. As in the proof of Theorem~\ref{thm:main1}\eqref{thm:main1_i}, it also holds that $u_\ve \to \tilu$ in $H_{g_1}^1(\Om)$ and $v_\ve \to \tilv$ in $H_{g_2}^1(\Om)$. Furthermore, if $\al(g_1,g_2)=\beta(g_1,g_2)$, then it is easy to see that $(u_*,v_*)=(\tilu,\tilv)=(u_0,v_0)$. This completes the proof.
\end{proof}

\begin{rema}\label{rmk:tilu vs u*}
We do not know the uniqueness of solution to the problem~\eqref{eq:min prob beta}. If this problem has a unique solution, then we obtain $(u_*,v_*)=(\tilu,\tilv)$ in the proof of Theorem~\ref{thm:main4}\eqref{thm:main4_i}.
\end{rema}

\begin{proof}[Proof of Theorem~\ref{thm:main4}(\ref{thm:main4_ii})]
Let us assume the contrary so that $\abs{u_\ve} \to 1$ and $\abs{v_\ve}\to 1$ uniformly on $\overline\Om$. Then, $\abs{u_*}=1$ and $\abs{v_*}=1$. Since $\al(g_1,g_2)=\beta (g_1,g_2)$ by~\eqref{eq:al=beta}, it follows that $(u_*,v_*)=(u_0,v_0)$. So, we can use the notations~\eqref{eq:phi sys} and~\eqref{eq:psi sys}. Moreover, we may assume that $\abs{u_\ve}^2 \ge 1/2$ and $\abs{v_\ve}^2 \ge 1/2$ on $\Om$, and take the notations~\eqref{eq:zeta xi} and~\eqref{eq:eta chi}. We can rewrite~\eqref{eq:2-GL sym} as
\begin{align}
	\divergence \big(\rho_\varepsilon^2 \nabla(\varphi+\eta_\varepsilon)\big) &= 0, \label{eq:rho-sys}
\\	-\Delta \rho_\varepsilon + \rho_\varepsilon \abs{\nabla \varphi + \nabla \eta_\varepsilon}^2 &= \frac{1}{\varepsilon^2}\rho_\varepsilon \parens{2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2}, \label{eq:zeta-sys}
\\	\divergence \big(\sigma_\varepsilon^2 \nabla(\psi+\chi_\varepsilon)\big) &= 0, \label{eq:sigma-sys}
\\	-\Delta \sigma_\varepsilon + \sigma_\varepsilon \abs{\nabla \psi+\chi_\varepsilon}^2 &= \frac{1}{\varepsilon^2}\sigma_\varepsilon \parens{2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2}. \label{eq:xi-sys}
\end{align}
By proceeding as in the proof of Theorem~\ref{thm:main3}\eqref{thm:main3_ii}, we obtain
\[
C_6 \le \frac12 \int_\Omega \parens[\big]{\rho_\varepsilon \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 + \sigma_\varepsilon \abs{\nabla \psi+\nabla \chi_\varepsilon}^2} \dif x + D_1+D_2,
\]
where $D_1$ and $D_2$ are defined by~\eqref{eq:D123}. By~\eqref{eq:Pohozaev sys sym}, \eqref{eq:gamma3}, \eqref{eq:gamma4} and H\"{o}lder's inequality, we obtain
\begin{align*}
	D_1 & = \frac{1}{\varepsilon^2} \int_\Om \frac{\rho_\ve}{\rho_\ve+1} (\rho_\varepsilon^2 - 1)(2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2) \le \sqrt{\ga_1\ga_3},
\\	D_2 & = \frac{1}{\varepsilon^2} \int_\Om \frac{\sigma_\ve}{\sigma_\ve+1} (\sigma_\varepsilon^2 - 1)(2 - \rho_\varepsilon^2 - \sigma_\varepsilon^2) \le \sqrt{\ga_1\ga_4}.
\end{align*}
So,
\[
C_6 \le \frac{1}{2}\int_\Omega \rho_\varepsilon \abs{\nabla \varphi+\nabla \eta_\varepsilon}^2 \dif x + \frac{1}{2}\int_\Omega
\sigma_\varepsilon \abs{\nabla \psi+\nabla \chi_\varepsilon}^2 \dif x + \sqrt{\ga_1\ga_3} + \sqrt{\ga_1\ga_4}.
\]
Furthermore, by arguing as in the proof of Theorem~\ref{thm:main3}\eqref{thm:main3_ii}, we are led to
\[
C_6 \le \frac{1}{2}\int_\Omega \big(\abs{\nabla \vp}^2+ \abs{\nabla \psi}^2 \big) \dif x + \sqrt{\ga_1\ga_3} + \sqrt{\ga_1\ga_4},
\]
which contradicts the assumption~\eqref{G bdd by C1-C2-sys-sym}.
\end{proof}

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations.

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