#  Commuting maps with the Mean Transform under Jordan product

F. Chabbabi  Department of Mathematics, FS, Abdelmalek Essaadi University,
Tetouan, Morocco<EMAIL_ADDRESS>
###### Abstract.

In this article, we give a complete characterization of the bijective maps
which commute with the mean transform under Jordan product. The main result is
the following : Let $H,K$ be two complex Hilbert spaces and $\Phi:\
B(H)\to\mathcal{B}(K)$ be a bijective map, then

$\mathcal{M}(\Phi(A)\circ\Phi(B))=\Phi(\mathcal{M}(A\circ B))\;\;\text{for
all}\;\;A,B\in\mathcal{B}(H)$

if and only if there exists a unitary or anti-unitary operator $U:H\to K$ such
that,

$\Phi(T)=UTU^{*}\;\text{for all}\;T\in\mathcal{B}(H).$

###### Key words and phrases:

normal, quasi-normal operators, polar decomposition, mean transform, Jordan
product

###### 2010 Mathematics Subject Classification:

47A05, 47A10, 47B49, 46L40

##  1\. Introduction

Let $H$ and $K$ be two complex Hilbert spaces and let $\mathcal{B}(H,K)$ be
the Banach space of all bounded linear operators from $H$ into $K$. In the
case $K=H$, $\mathcal{B}(H,H)$ is simply denoted by $\mathcal{B}(H)$ and it is
a Banach algebra.

For an arbitrary operator $T\in\mathcal{B}(H,K)$, we denote by
$\mathcal{R}(T)$, $\mathcal{N}(T)$ and $T^{*}$ the range, the null subspace
and the adjoint operator of $T$, respectively. For $T\in\mathcal{B}(H)$, the
spectrum of $T$ is denoted by $\sigma(T)$.

An operator $T\in\mathcal{B}(H,K)$ is a partial isometry when $T^{*}T$ is an
orthogonal projection (or, equivalently $TT^{*}T=T$). In particular $T$ is an
isometry if $T^{*}T=I$, and $T$ is unitary if it is a surjective isometry. An
operator $T\in\mathcal{B}(H)$ is said to be normal if $T^{*}T=TT^{*}$, and
quasi-normal if $TT^{*}T=T^{*}TT$. As usual, we denote the module of
$T\in\mathcal{B}(H)$ by $|T|=(T^{*}T)^{1/2}$, and $T=V|T|$ is the unique polar
decomposition of $T$, where $V$ is a partial isometry satisfying
$\mathcal{N}(V)=\mathcal{N}(T)$. In general $V$ and $|T|$ does not commute and
it is the case if $T$ is quasi-normal.

From the polar decomposition, the Aluthge transform is defined in ([1, 14]),
by

$\Delta(T)=|T|^{\frac{1}{2}}V|T|^{\frac{1}{2}}.$

The Aluthge transform has been well studied by many authors, it is a good tool
for studying sum class of operators (see [3, 4, 9, 6, 17]). In the same way,
the mean transform of the operator $T$ was introduced in [8, 13], by

$\mathcal{M}(T):=\frac{1}{2}(V|T|+|T|V).$

The mean transform has also well been studied in many articles, it is the
arithmetic mean of $T=V|T|$ and it’s Duggal transform $\tilde{T}=|T|V$, it has
nice properties, for example we can cite [5, 13, 10]. The fixed point of mean
transform are the quasi-normal operator as the Aluthge transform.

The main result of the present paper is the following theorem which gives a
nice characterization of the commuting maps with the mane transform under the
Jordan product.

######  Theorem 1.1.

Let $H$ and $K$ be two complex Hilbert space, with $\dim H\geq 3$. Let
$\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map. Then $\Phi$
satisfies the condition

$\mathcal{M}(\Phi(A)\circ\Phi(B))=\Phi(\mathcal{M}(A\circ B))\;\;\text{for
all}\;\;A,B\in\mathcal{B}(H),$

if and only if there exists a unitary or anti-unitary operator $U:H\to K$,
such that

$\Phi(T)=UTU^{*}\;\;\;\;\text{ for every}\;T\in\mathcal{B}(H).$

######  Remark 1.1.

Observe that, even if the hypothesis on the map $\Phi$ is purely algebraic,
the conclusion gives automatically the continuity of the map. Also, the
linearity of $\Phi$ is not assumed, we get it automatically.

##  2\. Auxiliary results of the mean transform

For $x,y\in H$ we denote by $x\otimes y$ the rank one operator (or $0$)
defined by

$(x\otimes y)u=<u,y>x\>\mbox{ for }\>u\in H.$

Every rank one operator has the previous form, and $x\otimes y$ is an
orthogonal projection (i.e. $T^{2}=T=T^{*})$ if and only if $x=y$ and
$\|x\|=1$.

The following results can be found in [5].

######  Proposition 2.1.

[5] Let $x,y\in H$ be two non-zero vectors. Let $T=x\otimes y$, then

$\mathcal{M}({T})=\mathcal{M}({x\otimes
y})=\frac{1}{2}(x+\dfrac{<x,y>}{\|y\|^{2}}y)\otimes y.$

######  Proposition 2.2.

[5] Let $T\in B(H)$. Then

$\mathcal{M}(T)=0\;\;\iff T=0.$

Next lemma gives a characterization of the nilpotent operator of order two.

######  Lemma 2.1.

[5] Let $T\in\mathcal{B}(H)$. Then

$\mathcal{M}(T)=\frac{T}{2}\>\text{ if and only if}\>T^{2}=0.$

######  Theorem 2.1.

Let $T\in\mathcal{B}(H)$. Then

$\mathcal{M}(T^{2})=T\>\text{ if and only if $T$ is an orthogonal
projection}.$

###### Proof.

It is clear that, if $T$ is a projection, then $\mathcal{M}(T^{2})=T=T^{*}$.
So we need to show the direct meaning.

Let $T^{2}=V_{2}|T^{2}|$ be the polar decomposition of $T^{2}$ and suppose
that $\mathcal{M}(T^{2})=T$. We will divided the proof on a few steps.

Step 1 :  The operator $|T^{2}|V_{2}|T^{2}|$ is positive. From the assumption,
we have

(2.1) $\dfrac{T^{2}+|T^{2}|V_{2}}{2}=T\;\;\text{ and
}\;\;\dfrac{(T^{2})^{*}+V_{2}^{*}|T^{2}|}{2}=T^{*}.$

Thus

$\dfrac{|T^{2}|+V_{2}^{*}|T^{2}|V_{2}}{2}=V_{2}^{*}T\geq 0,$

is fact that the operators $|T^{2}|$ and $V_{2}^{*}|T^{2}|V_{2}$ are positive.
Therefore

$T^{*}|T^{2}|=T^{*}V_{2}^{*}T^{2}=T^{*}(V_{2}^{*}T)T\geq 0.$

In particular $T^{*}|T^{2}|$ is self adjoint. Hence $T^{*}|T^{2}|=|T^{2}|T$.
Multiplying on the left by $T^{*}$, we get that

$|T^{2}|V_{2}|T^{2}|=|T^{2}|TT=T^{*}|T^{2}|T\geq 0.$

Step 2 :  $2I-T^{*}$ is injective.

Let $x\in H$ such that $(2I-T^{*})x=0$. Then $T^{*}x=2x$ and thus
$(T^{2})^{*}x=4x$. By the assumption we have $|T^{2}|V_{2}=2T-T^{2}$. We take
the adjoint, we get that

$V_{2}^{*}|T^{2}|=2T^{*}-(T^{2})^{*}.$

It follows that $V_{2}^{*}|T^{2}|x=2T^{*}x-(T^{2})^{*}x=0$. Thus
$|T^{2}|V_{2}^{*}|T^{2}|x=0$. Since

$|T^{2}|V_{2}^{*}|T^{2}|=|T^{2}|V_{2}|T^{2}|\geq 0\;\;\text{ ( see Step 1)},$

then

$|T^{2}|V_{2}|T^{2}|x=0.$

Multiplying the preceding equality on the left by $V_{2}$, we get that

$T^{4}x=V_{2}|T^{2}|V_{2}|T^{2}|x=0.$

It follows that

$16\|x\|^{2}=<(T^{4})^{*}x,x>=<x,T^{4}x>=0.$

By consequence $x=0$ and then $2I-T^{*}$ is injective.

Step 3 :  The inclusion
$\mathcal{N}(T^{2})\subseteq\mathcal{N}(T^{*})\subseteq\mathcal{N}((T^{2})^{*})$
holds.

Let $x\in\mathcal{N}(T^{2})$, we have $T^{2}x=|T^{2}|x=0$, hence by equation
(2.1)

$V_{2}^{*}|T^{2}|x=(2I-T^{*})T^{*}x=0,$

and since $2I-T^{*}$ is injective then $T^{*}x=0$, thus
$x\in\mathcal{N}(T^{*})\subseteq\mathcal{N}((T^{2})^{*})$.

Step 4 :  $T$ is a projection.

Now we are in position to prove that $T$ is a projection. First we show that
$V_{2}\geq 0$. Let $x\in H$ and consider the following decomposition
$H=\mathcal{N}(|T^{2}|)\oplus\overline{\mathcal{R}(|T^{2}|)}$. Put
$x=x_{1}+x_{2}$ with
$x_{1}\in\mathcal{N}(|T^{2}|),\;\;x_{2}\in\overline{\mathcal{R}(|T^{2}|)}$.
Then we have

$\displaystyle\langle V_{2}x,x\rangle$ $\displaystyle=$ $\displaystyle\langle
V_{2}x_{1}+V_{2}x_{2},x_{1}+x_{2}\rangle$ $\displaystyle=$
$\displaystyle\langle V_{2}x_{2},x_{2}\rangle+\langle V_{2}x_{2},x_{1}\rangle$
$\displaystyle=$ $\displaystyle<V_{2}x_{2},x_{2}>+<x_{2},V_{2}^{*}x_{1}>$
$\displaystyle=$ $\displaystyle<V_{2}x_{2},x_{2}>\;\;(\text{ since
}\;\;\mathcal{N}(V_{2})\subseteq\mathcal{N}(V_{2}^{*})).$

Since $|T^{2}|V_{2}|T^{2}|\geq 0$ ( see Step 1),
$<V_{2}x,x>=<V_{2}x_{2},x_{2}>\geq 0$. Thus $V_{2}$ a positive partial
isometry, so $V_{2}$ is an orthogonal projection and
$T^{2}=V_{2}|T^{2}|=V_{2}^{*}V_{2}|T^{2}|=|T^{2}|$ is positive. Therefore
$\mathcal{M}(T^{2})=T^{2}=T=T^{*}$. This completes the proof. ∎

######  Corollary 2.1.

Let $T\in\mathcal{B}(H)$. Then the following equivalence holds:

$T\;\text{ is a projection }\;\iff(\mathcal{M}(T))^{2}=T.$

###### Proof.

We need to show the second direction. Suppose that $(\mathcal{M}(T))^{2}=T$.
Then $\mathcal{M}((\mathcal{M}(T))^{2})=\mathcal{M}(T)$. By Theorem 2.1
replacing $T$ by $\mathcal{M}(T)$, we get that $\mathcal{M}(T)$ is a
projection. Hence $\mathcal{M}(T)=(\mathcal{M}(T))^{2}=T$ is also a
projection. ∎

######  Lemma 2.2.

[5] Let $T\in\mathcal{B}(H)$ and $T=V|T|$ be the polar decomposition of $T$.
The following equivalence holds:

$\mathcal{M}(T)\;\;\text{ is self-adjoint }\;\iff V\;\;\text{ is self-
adjoint}.$

######  Lemma 2.3.

Let $T\in\mathcal{B}(H)$. Then $\mathcal{M}(T)$ is a projection if and only if
$T$ is also a projection. In this case $\mathcal{M}(T)=T$.

###### Proof.

It is clear that we need to show the first direction. Let $T=V|T|$ be the
polar decomposition of $T$ and suppose that $\mathcal{M}(T)$ is a projection.
In particular is a self-adjoint. By the preceding lemma, $V$ is a self adjoint
partial isometry. Hence $V^{3}=V$ and

$\overline{\mathcal{R}(T)}=\overline{\mathcal{R}(V)}=\overline{\mathcal{R}(V^{*})}=\overline{\mathcal{R}(T^{*})}=\overline{\mathcal{R}(|T|)}.$

Hence $V^{2}$ is the projection on
$\overline{\mathcal{R}(V)}=\overline{\mathcal{R}(V^{2})}$.

On the other hand, we have also
$\mathcal{N}(\mathcal{M}(T))=\mathcal{N}(V)=\mathcal{N}(V^{2})$. Since
$\mathcal{M}(T)$ and $V^{2}$ are both projections with the same kernel, then
$V^{2}=\mathcal{M}(T)$ and thus

$V=V^{3}=V\mathcal{M}(T)=V(\dfrac{V|T|+|T|V}{2})=\dfrac{|T|+V|T|V}{2}.$

From this equality, we get $V=\dfrac{|T|+V|T|V}{2}\geq 0$ as sum of two
positive operators. It follows that $V$ is positive partial isometry. In
particular $V$ is also projection and then $V=V^{2}$. Now from the polar
decomposition of $T$, we have $T=V|T|=V^{2}|T|=|T|$ is positive. This implies
that $\mathcal{M}(T)=T$ is also a projection. ∎

######  Lemma 2.4.

Let $T\in\mathcal{B}(H)$. Then

$\mathcal{M}(T\circ P)=P\;\;\text{for all rank one
projection}\;P,\quad\mbox{if and only if},\quad T=I.$

###### Proof.

Clearly, if $T=I$ then for every rank one projection $P$, we have
$\mathcal{M}(T\circ P)=P$. So we need to show the direct implication. Indeed,
let $x\in H$ be a unit vector, and $P=x\otimes x$. Let $S=T\circ P$, it is
clear that $S$ is an operator of at most of rank two. So $\mathcal{M}(S)$ is
at most of rank four. Using the functional trace $Tr$ well defined on the set
of finite rank operator, then we get $Tr(S)=Tr(\mathcal{M}(S))=Tr(P)=1$, thus
$\langle Tx,x\rangle=Tr(T\circ P)=Tr(S)=1$. Hence we conclude that the numeral
range $\langle Tx,x\rangle=\|x\|^{2}$ for all $x\in H$, and thus $T=I$. ∎

##  3\.  The proof of main theorem

The proof of Theorem 1.1 will be divided in many intermediary lemmas and it
will be given at the end of this section. Along of this section, we assume
that $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map, which
satisfies the condition :

(3.1) $\mathcal{M}(\Phi(A)\circ\Phi(B))=\Phi(\mathcal{M}(A\circ
B))\;\;\text{for all}\;\;A,B\in\mathcal{B}(H).$

As an immediate consequence of (3.1) and Lemma 2.1 and Theorem 2.4, we derive
the following result :

######  Lemma 3.1.

The following two statements are hold :

  1. (i)

$\Phi(0)=0$,

  2. (ii)

$\Phi(I)=I$.

###### Proof.

(i). Since $\Phi$ is onto, there is $A\in\mathcal{B}(H)$ such that
$\Phi(A)=0$. By (3.1) we have

$0=\mathcal{M}((\Phi(0)\circ\Phi(A))=\Phi(0).$

Therefore, $\Phi(0)=0$.
(ii) For simplicity, let us denote $T=\Phi(I)$. We take $A=B=I$ in (3.1), we
get that

(3.2) $\mathcal{M}(T^{2})=\Phi(I)=T.$

Using Theorem 2.1, we get that $T^{2}=T=T^{*}$.

To complete the proof, it must to show $T$ is injective. Pick a $y\in K$ such
that $Ty=0$, we have also $T^{*}y=0$. Since $\Phi$ is onto, there exists
$B\in\mathcal{B}(H)$ such that $\Phi(B)=y\otimes y$. By (3.1), we get

$0=\mathcal{M}(\frac{1}{2}(Ty\otimes y+y\otimes
T^{*}y)=\mathcal{M}(\Phi(I)\Phi(B))=\Phi(\mathcal{M}(B)).$

Since $\Phi$ is bijective and $\Phi(0)=0$, we have $\mathcal{M}(B)=0$. By
Proposition 2.2, we get that $B=0$. Therefore $y\otimes y=\Phi(B)=0$ and
$y=0$. Hence $T$ is an injective projection, then $\Phi(I)=T=I$. ∎

As a consequence of the previous lemma, we get the following result.

######  Lemma 3.2.

Let $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ be a bijective map satisfying
(3.1). Then

  1. (i)

$\mathcal{M}(\Phi(B))=\Phi(\mathcal{M}(B))$, for all $B\in\mathcal{B}(H)$.
In particular, $\Phi$ preserves the set of quasi-normal operators in both
directions.

  2. (ii)

$\Phi(A^{2})=(\Phi(A))^{2}$ for all $A$ quasi-normal.

  3. (iii)

$\Phi$ preserves the set of orthogonal projections in both directions.

  4. (iv)

$\Phi$ preserves the orthogonality between the projections:

$P\perp Q\Leftrightarrow\Phi(P)\perp\Phi(Q).$

  5. (v)

$\Phi$ preserves the order relation between projections in both directions:

$Q\leq P\Leftrightarrow\Phi(Q)\leq\Phi(P).$

  6. (vi)

$\Phi(P+Q)=\Phi(P)+\Phi(Q)$ for all orthogonal projections $P,Q$ such that
$P\perp Q$.

  7. (vii)

$\Phi$ preserves the set of rank one projections in both directions.

###### Proof.

(i). Taking $B=I$ in (3.1), then we get
$\mathcal{M}(\Phi(B))=\Phi(\mathcal{M}(B))$. Since the quasi-normal operators
are exactly the fixed point of the mean transform, then (i) holds.

(ii). Let $A$ be a quasi-normal operator. Since $\Phi$ preserves the set of
quasi-normal operators, then $\Phi(A),\Phi(A^{2})$ and $(\Phi(A))^{2}$ are
also quasi-normal. By (3.1) with $B=A$, we get
$\mathcal{M}((\Phi(A))^{2})=\Phi(\mathcal{M}(A^{2}))$. Hence
$(\Phi(A))^{2}=\Phi(A^{2})$, it is also the fact that the quasi-normal
operators are the fixed point of $\mathcal{M}$.

(iii). It is an immediate consequence of (ii) and the fact that a idempotent
quasi-normal is orthogonal projection.

Throughout the remaining of the proof $P$ and $Q$ are orthogonal projections.

(iv). Assume that $P,Q$ are orthogonal (i.e. $PQ=0$ this equivalent also to
$P\circ Q=0$). Since $\Phi$ preserves the set of orthogonal projections, then
$\Phi(P),\Phi(Q)$ are orthogonal projection. By condition (3.1), we get

$\mathcal{M}(\Phi(P)\circ\Phi(Q))=\Phi(\mathcal{M}(P\circ Q))=\Phi(0)=0.$

Thus $\Phi(Q)\circ\Phi(P)=0$. The converse holds since the inverse $\Phi^{-1}$
satisfies the same condition as $\Phi$.

(v). Suppose that $Q\leq P$ or equivalently $Q\circ P=P\circ Q=Q$. Then by
(3.1), we get

(3.3) $\Phi(Q)\circ\Phi(P)=\mathcal{M}(\Phi(P)\circ\Phi(Q))=\Phi(Q).$

Since $\Phi^{-1}$ has the same assumption as $\Phi$, then we deduce that
$\Phi$ preserves the order relation between the orthogonal projections in both
directions.

(vi). We have $P,Q\leq P+Q$. Then $\Phi(P),\Phi(Q)\leq\Phi(P+Q)$. So
$\Phi(P)+\Phi(Q)\leq\Phi(P+Q)$. Since $\Phi$ and $\Phi^{-1}$ both satisfy the
same conditions, it follows that $\Phi(P)+\Phi(Q)=\Phi(P+Q)$.

(vii). Let $P=x\otimes x$ be a rank one projection. Then $\Phi(P)$ is a non
zero projection. Let $y\in K$ be a unit vector such that $y\otimes
y\leq\Phi(P)$. Thus $\Phi^{-1}(y\otimes y)\leq P$. Since $P$ is a minimal
projection and $\Phi^{-1}(y\otimes y)$ is a non zero projection, then
$\Phi^{-1}(y\otimes y)=P$. Therefore $\Phi(P)=y\otimes y$ is a rank one
projection. This complete the proof. ∎

######  Lemma 3.3.

There exists a bijective multiplicative function $h:\mathbb{C}\to\mathbb{C}$
such that : for every quasi-normal $A\in\mathcal{B}(H)$ and
$\alpha\in\mathbb{C}$, we have $\Phi(\alpha A)=h(\alpha)\Phi(A)$.

In particular, $h(0)=0$, $h(1)=1$ and $h(-1)=-1$.

###### Proof.

Let $x\in H$ be a unit vector and $\alpha\in\mathbb{C}$. Let $T=\Phi(\alpha
x\otimes x)$. By the assumption and condition (3.1), we have

$\mathcal{M}(\Phi(\alpha x\otimes x)\circ x\otimes x))=\Phi(\alpha x\otimes
x).$

Therefore

(3.4) $\mathcal{M}(Tx\otimes x+x\otimes T^{*}x)=2T.$

Taking the norm and using the triangular inequality, we get

$2\|T\|=\|\mathcal{M}(Tx\otimes x+x\otimes T^{*}x)\|\leq\|Tx\otimes x+x\otimes
T^{*}x\|\leq\|Tx\|+\|T^{*}x\|\leq 2\|T\|.$

Which implies the following,

$2\|T\|=\|Tx\otimes x+x\otimes T^{*}x\|,\;\text{ and
}\;\;\|T\|=\|Tx\|=\|T^{*}x\|.$

Let denote $S=Tx\otimes x+x\otimes T^{*}x$, form the preceding equality we get

$4\|T\|^{2}=\|S\|^{2}=\|S^{*}S\|=r(S^{*}S)\leq Tr(S^{*}S).$

On the other hand, we calculate $S^{*}S$ then

$S^{*}S=\big{(}\|Tx\|^{2}x\otimes x+<x,Tx>x\otimes T^{*}x+<Tx,x>T^{*}x\otimes
x+T^{*}x\otimes T^{*}x\big{)}.$

It follows that

$\displaystyle 4\|T\|^{2}$ $\displaystyle\leq$ $\displaystyle Tr(S^{*}S)$
$\displaystyle=$ $\displaystyle Tr\big{(}\|Tx\|^{2}x\otimes x+<x,Tx>x\otimes
T^{*}x+<Tx,x>T^{*}x\otimes x+T^{*}x\otimes T^{*}x\big{)}$ $\displaystyle=$
$\displaystyle\|Tx\|^{2}+2|<Tx,x>|^{2}+\|T^{*}x\|^{2}$ $\displaystyle=$
$\displaystyle 2\|T\|^{2}+2|<Tx,x>|^{2}\leq 4\|T\|^{2}.$

Therefore

$\|T\|^{2}\leq|<Tx,x>|^{2},\;\;\text{ and
}\;\;\|T\|=|<Tx,x>|=\|Tx\|=\|T^{*}x\|.$

By consequence,

$\|Tx-<Tx,x>x\|^{2}=\|Tx\|^{2}-|<Tx,x>|^{2}=0,$

and thus $Tx=<Tx,x>x$. With the same way, $T^{*}x=<T^{*}x,x>x$. By equation
(3.4), it follows that

$\Phi(\alpha x\otimes x)=T=<Tx,x>x\otimes x=h_{x}(\alpha)x\otimes x,$

where $h_{x}(\alpha)=<Tx,x>$ for all $\alpha\in\mathbb{R}$. In particular
$h_{x}(0)=0$ and $h_{x}(1)=1$.

Now, by condition (3.1), we have for all unit vector $x\in H$,

$\mathcal{M}(\Phi(\alpha I)\circ x\otimes x)=\Phi(\alpha x\otimes
x)=h_{x}(\alpha)x\otimes x.$

By Lemma 2.4, $\Phi(\alpha I)x\circ x=h_{x}(\alpha)x\circ x$, this implies
that $\Phi(\alpha I)x$ and $x$ are colinear for all unit vector $x$ of $H$. By
classical arguments, $\Phi(\alpha I)$ is a scalar multiple of the identity, so
there exists a function $h:\mathbb{C}\to\mathbb{C}$ such that $\Phi(\alpha
I)=h(\alpha)I$ for all $\alpha\in\mathbb{C}$. By the assumption on $\Phi$, the
function $h$ is bijective and multiplicative. Moreover $\Phi(\alpha
A)=h(\alpha)\Phi(A)$ for every quasi-normal operator $A$. ∎

######  Lemma 3.4.

Let $P,Q$ two orthogonal projections and $\alpha,\beta\in\mathbb{C}$, then we
have

$\Phi(\alpha P+\beta Q)=h(\alpha)\Phi(P)+h(\beta)\Phi(Q).$

###### Proof.

If $\alpha=0$ or $\beta=0$, then the result follows from preceding lemma. So
suppose that $\alpha\neq 0$ and $\beta\neq 0$. By Lemma 3.2 $\Phi(P)$ and
$\Phi(Q)$ are also two orthogonal projections, using condition (3.1), we get
the following:

$\displaystyle\mathcal{M}(\Phi(\alpha P+\beta Q)\circ\Phi(P))$
$\displaystyle=$ $\displaystyle\Phi(\mathcal{M}({\alpha}P+\beta Q)\circ P))$
$\displaystyle=$ $\displaystyle\Phi(\alpha P)=h(\alpha)\Phi(P).$

Therefore

$\mathcal{M}(\Phi(\alpha P+\beta Q)\circ\Phi(P))=h(\alpha)\Phi(P).$

By Lemma 2.3,

(3.5) $\Phi(\alpha P+\beta Q)\circ\Phi(P)=\Phi(\alpha P+\beta
Q)\Phi(P)=\Phi(P)\Phi(\alpha P+\beta Q)=h(\alpha)\Phi(P).$

Similarly,

(3.6) $\Phi(\alpha P+\beta Q)\circ\Phi(Q)=\Phi(\alpha P+\beta
Q)\Phi(Q)=\Phi(Q)\Phi(\alpha P+\beta Q)=h(\beta)\Phi(Q).$

Again the condition (3.1) implies that

$\displaystyle\Phi(\alpha P+\beta Q)$ $\displaystyle=$
$\displaystyle\Phi(\mathcal{M}((\alpha P+\beta Q)\circ(P+Q)))$
$\displaystyle=$ $\displaystyle\mathcal{M}(\Phi(\alpha P+\beta
Q)\circ\Phi(P+Q))$ $\displaystyle=$ $\displaystyle\mathcal{M}(\Phi(\alpha
P+\beta Q)\circ(\Phi(P)+\Phi(Q))$ $\displaystyle=$
$\displaystyle\mathcal{M}(\Phi(\alpha P+\beta Q)\circ\Phi(P)+\Phi(\alpha
P+\beta Q)\circ\Phi(Q))$ $\displaystyle=$
$\displaystyle\mathcal{M}(h(\alpha)\Phi(P)+h(\beta)\Phi(Q))$ $\displaystyle=$
$\displaystyle h(\alpha)\Phi(P)+h(\beta)\Phi(Q).$

∎

######  Lemma 3.5.

The function $h:\mathbb{C}\to\mathbb{C}$ is a automorphism of complex fields
$\mathbb{C}$.

###### Proof.

Let $x,y$ be two unit and orthogonal vectors, and let $A=x\otimes y+y\otimes
x$. First, note that $A$ is self adjoint operator of rank two, and we have

$A^{2}=x\otimes x+y\otimes y,$

which is a non-trivial orthogonal projection. Since the dimensional of $H$ is
greater than 3, therefore the spectrum of $A$ is $\sigma(A)=\\{-1,0,1\\}$.
Hence we can find $f_{1},f_{2}$ two unit and orthogonal vectors from $H$
($\|f_{1}\|=\|f_{2}\|=1$ and $<f_{1},f_{2}>=0$) such that

$A=f_{1}\otimes f_{1}-f_{2}\otimes f_{2}.$

In particular

$\Phi(A)=\Phi(f_{1}\otimes f_{1}-f_{2}\otimes f_{2})=\Phi(f_{1}\otimes
f_{1})-\Phi(f_{2}\otimes f_{2}),$

this implies that $\Phi(A)$ is self-adjoint as sum of two rank one projection.

Now, let us consider the rank one projections $P=x\otimes x,\;Q=y\otimes y$.
For $\alpha,\beta\in\mathbb{C}$, we denote by $B=\alpha P+\beta Q$. Clearly

$A\circ P=\frac{1}{2}A\;\;\text{ and }\;\;A\circ Q=\frac{1}{2}A,$

by condition (3.1) we get the following equalities :

$h(\frac{1}{2})\Phi(A)=\Phi(\frac{1}{2}A)=\mathcal{M}(\Phi(A)\circ\Phi(P))=\Phi(A)\circ\Phi(P).$

And

$h(\frac{1}{2})\Phi(A)=\Phi(\frac{1}{2}A)=\mathcal{M}(\Phi(A)\circ\Phi(Q))=\Phi(A)\circ\Phi(Q).$

Since $A\circ B=\dfrac{\alpha+\beta}{2}A$ then,

$\displaystyle h(\dfrac{\alpha+\beta}{2})\Phi(A)$ $\displaystyle=$
$\displaystyle\Phi(\mathcal{M}(A\circ B))$ $\displaystyle=$
$\displaystyle\mathcal{M}(\Phi(A)\circ\Phi(B)))$ $\displaystyle=$
$\displaystyle\mathcal{M}(\Phi(A)\circ\Phi(\alpha P+\beta Q)))$
$\displaystyle=$
$\displaystyle\mathcal{M}(\Phi(A)\circ(h(\alpha)\Phi(P)+h(\beta)\Phi(Q)))$
$\displaystyle=$
$\displaystyle\mathcal{M}(h(\alpha)\Phi(A)\circ\Phi(P)+h(\beta)\Phi(A)\circ\Phi(Q))$
$\displaystyle=$
$\displaystyle\mathcal{M}(h(\alpha)h(\frac{1}{2})\Phi(A)+h(\beta)h(\frac{1}{2})\Phi(A))$
$\displaystyle=$
$\displaystyle\mathcal{M}(\big{(}{h(\frac{\alpha}{2})+h(\frac{\beta}{2})}\big{)}\Phi(A))$
$\displaystyle=$
$\displaystyle(h(\frac{\alpha}{2})+h(\frac{\beta}{2}))\Phi(A).$

Consequently

$h(\frac{\alpha}{2})+h(\frac{\beta}{2})=h(\frac{\alpha+\beta}{2}),$

and thus

$h(\alpha^{\prime}+\beta^{\prime})=h(\alpha^{\prime})+h(\beta^{\prime}),\;\;\text{
for all}\;\alpha^{\prime},\beta^{\prime}\in\mathbb{C}.$

∎

The following result has been chowed by Uhlhorn, it gives a nice
characterization of a bijective map $\Psi:P_{1}(H)\to P_{1}(H)$ which
preserves the orthogonality. This result can be reformulate as :

######  Theorem 3.1 (Uhlhorn’s Theorem).

Let $\Phi:P_{1}(H)\to P_{1}(K)$ be a bijective map, with
$dim\leavevmode\nobreak\ H\geq 3$. Assume that $\Phi$ satisfies the following
property

$PQ=0\iff\Phi(P)\Phi(Q)=0\;\;(P,Q\in P_{1}(H)).$

Then there exists a unitary or anti-unitary operator $U:H\to K$, such that
$\Phi$ is of the form :

(3.7) $\Phi(P)=UPU^{*}\;\;\;\text{ for all }\;\;P\in P_{1}(H).$

######  Lemma 3.6.

The function $h$ is the identity or the complex conjugate. Moreover for every
unit vector $y\in K$ and $x\in H$ such that $\Phi(x\otimes x)=y\otimes y$, and
for every self-adjoint operator $A\in\mathcal{B}(H)$, we have

(3.8) $\langle\Phi(A)y,y\rangle=\langle Ax,x\rangle.$

###### Proof.

Let $A\in\mathcal{B}(H)$ be a self-adjoint operator, for an arbitrary unit
vectors $x\in H$ and $y\in K$, such that $\Phi(x\circ x)=y\otimes y$, put
$P=x\otimes x$, then $A\circ P$ is also self-adjoint of rank less than 2. And
So there exists $\alpha,\beta\in\mathbb{C}$ and two orthogonal projections
$P_{1}$ and $P_{2}$ such that $A\circ P=\alpha P_{1}+\beta P_{2}$. In
particular we have

$Tr(A\circ P)=\alpha+\beta=\langle Ax,x\rangle.$

By the Lemma 3.4, we have

$\Phi(A\circ P)=\Phi(\alpha P_{1}+\beta
P_{2})=h(\alpha)\Phi(P_{1})+h(\beta)\Phi(P_{2}),$

so

$\Phi(A\circ P)=h(\alpha)+h(\beta)=h(\alpha+\beta)=h(\langle Ax,x\rangle).$

On the other hand,

$Tr(\Phi(A)\circ\Phi(P)=\langle\Phi(A)y,y\rangle$

and by the assumption we have

$\langle\Phi(A)y,y\rangle=Tr(\mathcal{M}(\Phi(A)\circ\Phi(P)))=Tr(\mathcal{M}(\Phi(A\circ
P)))=Tr(\Phi(A\circ P))=h(\langle Ax,x\rangle).$

From the preceding argument, it follows that for any self-adjoint operator
$A\in\mathcal{B}(H)$ we have

(3.9) $h(W(A))=W(\Phi(A)).$

Now, let $a,b\in\mathbb{R}$ such that $a<b$, and let $A\in\mathcal{B}(H)$ be a
self-adjoint operator such that the numerical range $W(A)=\\{\langle
Au,u\rangle\;\;:\;\|u\|=1\\}=[a,b]$, for example we take $A=aP+bQ$ where $P,Q$
are two orthogonal projections. By the preceding (3.9)

$h([a,b])=h(W(A))=W(\Phi(A)),$

is bounded, according to the numerical range of an operator
$T\in\mathcal{B}(K)$ is always bounded. Which implies that the automorphism
$h:\mathbb{C}\to\mathbb{C}$ is bounded on all segment $[a,b]\subset\mathbb{R}$
and also bounded in all rectangle set $\mathcal{R}=[a,b]+i[c,d]$. By
Proposition 1.1 in [11] $h$ is either the identity
$h(z)=Id_{\mathbb{C}}(z)=z$, for all $z\in\mathbb{C}$ or the conjugate complex
$h(z)=\bar{Id_{\mathbb{C}}}(z)=\bar{z}$ for all $z\in\mathbb{C}$.

In particular, we conclude that for every self-adjoint operator $A\in B(H)$
and for every unit vector $y\in K$ and $x\in H$ such that $\Phi(x\otimes
x)=y\otimes y$ we have

$\langle\Phi(A)y,y\rangle=\langle Ax,x\rangle\in\mathbb{R}.$

∎

######  Corollary 3.1.

The maps $\Phi:\mathcal{B}(H)\to\mathcal{B}(K)$ preserves the set self adjoint
operator. Moreover, the exist an unitary or anti-unitary operator $U:H\to K$
such that

$\Phi(A)=UAU^{*}\;\;\;\text{for all}\;\;A\in\mathcal{B}_{s}(H).$

###### Proof.

The map $\Phi$ satisfies the conditions of Uhlhorn’s, hence $\Phi$ takes the
form (3.7) on the set $P_{1}(H)$ of rank one operators.

Let $A\in\mathcal{B}(H)$ be a self-adjoint operator, and let $x\in H$ be an
arbitrary unit vectors from $H$, we have such that $\Phi(x\otimes x)=Ux\otimes
Ux$, and thus by equation (3.8)

(3.10) $\displaystyle\langle U^{*}\Phi(A)Ux,x\rangle$ $\displaystyle=$
$\displaystyle\langle\Phi(A)Ux,Ux\rangle$ (3.11) $\displaystyle=$
$\displaystyle\langle Ax,x\rangle$

Therefore $U^{*}\Phi(A)U=A$ and this completes the proof. ∎

In the rest of the manuscript, we replace the map $\Phi$ by the map $T\to
U^{*}\Phi(T)U$ which satisfies the same conditions as $\Phi$ and noted it too
by $\Phi$. Then we can suppose that $\Phi(A)=A$ for every self-adjoint
operator $A\in\mathcal{B}(H)$. To finish the proof, it must to show that
$\Phi$ is the identity on $\mathcal{B}(H)$.

Let $A\in\mathcal{B}(H)$ and let $x\in H$ be an arbitrary unit vector from
$H$. Put

$T=2A\circ x\otimes x=Ax\otimes x+x\otimes A^{*}x,$

thus $T^{*}=x\otimes Ax+A^{*}x\otimes x$, in particular $T$ and $T^{*}$ are of
rank two, moreover there image

$\mathcal{R}(T)\subseteq span\\{x,Ax\\},\;\;\text{and
}\;\;\mathcal{R}(T^{*})\subseteq span\\{x,A^{*}x\\}.$

First we have

$M(\Phi(T))=2M(\Phi(A\circ x\otimes x))=2M(\Phi(A)\circ\Phi(x\otimes
x))=2M(\Phi(A)\circ x\otimes x).$

Hence

(3.12) $Tr(\Phi(T))=2\langle\Phi(A)x,x\rangle$

On the other hand, we have

$T\circ(x\otimes x-\frac{I}{2})=\langle Ax,x\rangle x\otimes x.$

By the assumption on $\Phi$ and $h$, we get that

$M(\Phi(T)\circ(x\otimes x-\frac{I}{2}))=h(\langle Ax,x\rangle)x\otimes x.$

By Lemma 2.3,

(3.13) $\Phi(T)\circ(x\otimes x-\frac{I}{2})=h(\langle Ax,x\rangle)x\otimes
x.$

Therefore, from the equation (3.13),

$\displaystyle\Phi(T)$ $\displaystyle=$ $\displaystyle 2\Phi(T)\circ(x\otimes
x)-2h(\langle Ax,x\rangle)x\otimes x$ $\displaystyle=$
$\displaystyle\Phi(T)x\otimes x+x\otimes\Phi(T)^{*}x-2h(\langle
Ax,x\rangle)x\otimes x.$

Multiplying the last equation on the right by $x\otimes x$, then we get

$\Phi(T)x\otimes x=\Phi(T)x\otimes x+\langle\Phi(T)x,x\rangle x\otimes
x-2h(\langle Ax,x\rangle)x\otimes x.$

This implies the following equation

(3.14) $\langle\Phi(T)x,x\rangle=2h(\langle Ax,x\rangle).$

Now, from equation (3.13) we get

$\displaystyle h(<Ax,x>)$ $\displaystyle=$ $\displaystyle
Tr(\Phi(T)\circ(x\otimes x-\frac{I}{2})$ $\displaystyle=$ $\displaystyle
Tr(\Phi(T)(x\otimes x-\frac{I}{2}))$ $\displaystyle=$
$\displaystyle\langle\Phi(T)x,x\rangle-\dfrac{Tr(\Phi(T))}{2}.$

Then, from equations (3.12) we get

(3.15) $Tr(\Phi(T))=2\langle\Phi(T)x,x\rangle-2h(\langle
Ax,x\rangle)=2h(\langle Ax,x\rangle).$

From equation (3.12) and (3.15), we conclude that

(3.16) $\langle\Phi(A)x,x\rangle=h(\langle Ax,x\rangle).$

The last equality holds for every unit vector $x\in H$ and every
$A\in\mathcal{B}(H)$. Let us distinct the following two cases :

Case 1 :  If the function $h=Id_{\mathbb{C}}$, then from the equation (3.16)
we have,

(3.17) $\langle\Phi(A)x,x\rangle=\langle Ax,x\rangle,\text{ for all unit
vector $x$ and $A\in\mathcal{B}(H)$}.$

This show that the map $\Phi$ is the identity.

Case 2 :  Suppose that the function $h=\overline{Id}_{\mathbb{C}}$ is the
complex conjugate. Hence from equation (3.16), we get

(3.18) $\langle\Phi(A)x,x\rangle=\langle A^{*}x,x\rangle,\text{ for all unit
vector $x$ and $A\in\mathcal{B}(H)$}.$

Which implies that $\Phi(A)=A^{*}$ for all $A\in\mathcal{B}(H)$.

Now, let us consider $A=x\otimes x^{\prime}$ with $x,x^{\prime}$ are unit,
independent and non-orthogonal vectors in $H$. Then $A^{*}=x^{\prime}\otimes
x$. By Proposition 2.1, we have

$\mathcal{M}(\Phi(A))=\mathcal{M}(A^{*})=\frac{1}{2}(x^{\prime}+<x^{\prime},x>x)\otimes
x),$

and

$\Phi(\mathcal{M}(A))=(\mathcal{M}(A))^{*}=\frac{1}{2}(x^{\prime}\otimes(x+<x^{\prime},x>x^{\prime})),$

which contradicts with the fact that $\Phi$ commute with the mean transform.
Hence, we conclude that the function $h$ must be the identity, and $\Phi$ must
be linear and of the form

$\Phi(T)=UTU^{*},\;\;\text{ for all }T\in\mathcal{B}(H),$

where $U:H\to K$ is a unitary or anti-unitary operator.

Acknowledgments.

I wish to thank Professor Mostafa Mbekhta for the interesting discussions as
well as his useful suggestions for the improvement of this paper.

## References

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# On a system of multi-component Ginzburg-Landau vortices

Rejeb Hadiji and Jongmin Han and Juhee Sohn Univ Paris Est Creteil, CNRS,
LAMA, F-94010 Creteil, France.
Univ Gustave Eiffel, LAMA, F-77447 Marne-la-Vallée, France.
email<EMAIL_ADDRESS>of Mathematics, Kyung Hee University,
Seoul, 02447, Korea.
email<EMAIL_ADDRESS>of General Education, Kookmin University, Seoul
02707, Korea.
email<EMAIL_ADDRESS>
###### Abstract

We study the asymptotic behavior of solutions for $n$-component Ginzburg-
Landau equations as ${\varepsilon}\to 0$. We prove that the minimizers
converges locally in any $C^{k}$-norm to a solution of a system of generalized
harmonic map equations.

MSC2000 : 35B40, 35J60, 35Q60.

Keywords : $n$-component Ginzburg-Landau equations, semi-local gauge field
model, asymptotic behavior of solutions

##  1 Introduction

The classical Ginzburg-Landau model describes the macroscopic theory for
phenomena on superconductivity at low temperature. It is derived from the
Helmholtz free energy that consists of a complex order parameter $u$ and the
magnetic potential $A$ [4]. In the physical literature, $|u|^{2}$ represents
the density of the superconducting electron pairs in the superconducting
material. The state $|u|^{2}=0$ implies that material remains in normal
conducting state. Whereas the material is superconducting if $|u|^{2}=1$.

The Ginzburg-Landau theory also gives a good insight for understanding various
topological defects arising from cosmology. The superconducting state is
achieved at the vacuum level of the potential and it can be regarded as a
broken symmetry during a phase transition. For instance, this idea is realized
by the (special) relativistic extension of the planar Ginzburg-Landau theory,
called the Abelian-Higgs model [8]. This model describes a charged scalar
field $\psi$ interacting with a ${\mathrm{U}(1)}$ gauge field $A$, and allows
vortex solutions which are charged magnetically but electrically neutral.

The Abelian-Higgs model is considered in the (2+1)-dimensional Minkowski space
$\mathbb{R}^{1,2}$ with the metric $\text{diag}(-1,1,1)$. The metric is used
to raise or lower indices. The Lagrangian density for the model is defined as

${\mathcal{L}}_{0}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}D_{\mu}u\overline{D^{\mu}u}-\frac{1}{8{\varepsilon}^{2}}(|u|^{2}-1)^{2}.$
(1.1)

Here, $\mu,\nu=0,1,2$, ${\varepsilon}>0$ is the Higgs coupling constant,
$u:\mathbb{R}^{1,2}\to\mathbb{C}$ is the Higgs field and
$A_{\mu}:\mathbb{R}^{1,2}\to\mathbb{R}$ is ${\mathrm{U}(1)}$ gauge fields. In
addition, $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ represents
the electromagnetic field, $D_{\mu}u=\partial_{\mu}u+iA_{\mu}u$ is the
covariant derivative. The Lagrangian ${\mathcal{L}}_{0}$ is invariant under
the local ${\mathrm{U}(1)}$ gauge transform

$u\mapsto e^{i\alpha}u,\quad A_{\mu}\mapsto A_{\mu}-\partial_{\mu}\alpha$

for any smooth function $\alpha:\mathbb{R}^{1,2}\to\mathbb{R}$. The Euler-
Lagrange equations are

$\left\\{\begin{aligned}
D_{\mu}D^{\mu}u&~{}=\frac{1}{{\varepsilon}^{2}}(|u|^{2}-1)u\\\
\partial^{\nu}F_{\mu\nu}&~{}=\frac{i}{2{\varepsilon}^{2}}\big{[}u\overline{D_{\mu}u}-\overline{u}D_{\mu}u\big{]},\end{aligned}\right.$
(1.2)

which are the (special) relativistic Ginzburg-Landau equations.

As a direct generalization of (1.1), Vachaspati and Achucarro proposed a new
model in [12] that contains $n$ scalar fields:

$\Psi=\big{(}u_{1},\cdots
u_{n}\big{)},{\quad\mbox{where}\quad}u_{k}:\mathbb{R}^{1,2}\to\mathbb{C}{\quad\mbox{for}\quad}k=1,\cdots,n.$

The Lagrangian ${\mathcal{L}}_{0}$ is modified as

${\mathcal{L}}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}D_{\mu}\Psi\overline{D^{\mu}\Psi}-\frac{1}{8{\varepsilon}^{2}}\big{(}|\Psi|^{2}-n\big{)}^{2}.$
(1.3)

It is easy to show that ${\mathcal{L}}$ is invariant under the local
${\mathrm{U}(1)}$ gauge transformation

$\Psi\mapsto e^{i\alpha}\Psi,\quad A_{\mu}\mapsto
A_{\mu}-\partial_{\mu}\alpha$

for any smooth function $\alpha:\mathbb{R}^{1,2}\to\mathbb{R}$. It is also
invariant under the global ${\mathrm{SU}(n)}$ gauge transformation

$\Psi\mapsto e^{i\alpha_{k}\tau^{k}}\Psi,$

where $\alpha_{k}\in\mathbb{R}$ for $k=1,\cdots,n^{2}-1$ and
$\\{\tau^{k}\\}_{k=1}^{n^{2}-1}$ are the generators of the Lie algebra
$\mathfrak{su}(n)$. The Euler-Lagrange equations are obtained as

$\left\\{\begin{aligned}
D_{\mu}D^{\mu}u_{k}&~{}=\frac{1}{{\varepsilon}^{2}}\Big{(}\sum_{j=1}^{n}|u_{j}|^{2}-n\Big{)}u_{k},\\\
\partial^{\nu}F_{\mu\nu}&~{}=\frac{i}{2{\varepsilon}^{2}}\sum_{j=1}^{n}\big{(}u_{j}\overline{D_{\mu}u_{j}}-\overline{u}_{j}D_{\mu}u_{j}\big{)},\end{aligned}\right.$
(1.4)

which we will refer to as $n$-component Ginzburg-Landau equations.

The model (1.3) is useful in the study of some issues in cosmology, for
instance, the formation of cosmic strings that have both local and global
natures [12]. So, the string solutions for (1.3) are called semilocal. The
semilocal string solutions for (1.3) are similar to those for (1.1) that
reflect the local gauge transformation. But they have additional features that
have some resemblance to global defect. For further physical implication of
the model (1.3), one may refer to [7, 12].

In this article, we are interested in static solutions for (1.2) and (1.4). In
particular, we assume that the electromagnetic fields vanish, that is,
$A_{\mu}=0$ for $\mu=0,1,2$. Then, the main topic is to study the asymptotic
behavior of solutions for (1.2) and (1.4) as ${\varepsilon}\to 0$. For the
case (1.2), there have been lots of researches on this topic after the seminal
work of Bethuel-Brezis-Helein [1, 2]. More specifically, let
${\Omega}\subset{\mathbb{R}^{2}}$ be a smooth bounded simply connected domain.
Given a smooth map

$g:\partial{\Omega}\to S^{1}=\\{z\in\mathbb{C}:|z|=1\\}\quad\text{with}\quad
d=\deg(g,\partial{\Omega})$

the Ginzburg-Landau equations (1.2) with $A_{\mu}=0$ reduce to

$\left\\{\begin{aligned} -\Delta
u&=\frac{1}{{\varepsilon}^{2}}u(1-|u|^{2}){\quad\mbox{in}\quad}\Omega,\\\
u&=g{\quad\mbox{on}\quad}\partial\Omega.\end{aligned}\right.$ (1.5)

The associated energy functional is

$E_{{\varepsilon},{\Omega}}^{b}(u)=\frac{1}{2}\int_{\Omega}|\nabla
u|^{2}dx+\frac{1}{4{\varepsilon}^{2}}\int_{\Omega}(1-|u|^{2})^{2}dx.$ (1.6)

Then, $E_{{\varepsilon},{\Omega}}^{b}$ has a minimizer
$u_{{\varepsilon},g}^{b}$ in

$H^{1}_{g}({\Omega};\mathbb{C})=\big{\\{}u\in
H^{1}({\Omega};\mathbb{C}):u=g\mbox{ on }\partial{\Omega}\big{\\}},$

that is

$E_{{\varepsilon},{\Omega}}^{b}(u_{{\varepsilon},g}^{b})=\inf\\{E_{{\varepsilon},{\Omega}}^{b}(u):u\in
H^{1}_{g}({\Omega};\mathbb{C})\\}.$ (1.7)

We often write $H^{1}_{g}$ instead of $H^{1}_{g}({\Omega};\mathbb{C})$ if
there is no confusion.

The asymptotic behavior of solutions of (1.5) as ${\varepsilon}\to 0$ has
attracted lots of interest for three decades. The behavior of global
minimizers for $u_{{\varepsilon},g}^{b}$ was studied in detail by Bethuel,
Brezis and Hélein [1, 2]. If $d=0$, $u_{{\varepsilon},g}^{b}$ converges to a
harmonic map that minimizes

$\int_{\Omega}|\nabla u|^{2}dx$

over the space

$H^{1}_{g}({\Omega};S^{1})=\\{u\in W^{1,2}({\Omega},S^{1}):u=g\mbox{ on
}\partial{\Omega}\\}.$

This problem has a solution $u_{0}^{b}$ satisfying

$\left\\{\begin{aligned} -\Delta u_{0}^{b}&=u_{0}^{b}|\nabla
u_{0}^{b}|^{2}\quad\text{on}\quad{\Omega},\\\
u_{0}^{b}&=g{\quad\mbox{on}\quad}\partial\Omega,\\\ |u_{0}^{b}|&=1\quad\text{
on }{\Omega}.\end{aligned}\right.$ (1.8)

When $d\neq 0$, the analysis is more delicate because
$H^{1}_{g}({\Omega};S^{1})=\emptyset$. There exists a set
$\\{a_{1},\cdots,a_{d}\\}\subset{\Omega}$ such that up to a subsequence,
$u_{{\varepsilon},g}^{b}$ converges to the map $u_{*}$ that satisfies the
harmonic map equation [2, 6]

$\left\\{\begin{aligned} -\Delta u_{*}&=u_{*}|\nabla
u_{*}|^{2}\quad\text{on}\quad{\Omega}\backslash\\{a_{1},\cdots,a_{d}\\},\\\
u_{*}&=g{\quad\mbox{on}\quad}\partial\Omega,\\\ |u_{*}|&=1\quad\text{ on
}~{}{\Omega}\backslash\\{a_{1},\cdots,a_{d}\\}.\end{aligned}\right.$ (1.9)

Moreover, we have

$u_{*}(x)=\frac{x-a_{1}}{|x-a_{1}|}\cdots\frac{x-a_{d}}{|x-a_{d}|}e^{i\psi(x)},$
(1.10)

where $\psi$ is harmonic on ${\Omega}$ and the singularities
$a_{1},\cdots,a_{d}$ of $u_{*}$ minimizes the associated renormalized energy.
We refer to [10] for the study of the Ginzburg-Landau model with and without a
magnetic fields that describe superconductivity.

Now, we turn to the $n$-component Ginzburg-Landau equations (1.4). If we
assume $A_{\mu}=0$, then (1.4) is reduced to

$\left\\{\begin{aligned} -\Delta
u_{i}&=\frac{1}{{\varepsilon}^{2}}u_{i}\Big{(}n-\sum_{j=1}^{n}|u_{j}|^{2}\Big{)}{\quad\mbox{in}\quad}\Omega,\\\
u_{i}&=g_{i}{\quad\mbox{on}\quad}\partial\Omega,\end{aligned}\right.$ (1.11)

for each $i=1,\cdots,n$. One may regard (1.11) as a direct extension of (1.5)
to $n$-component equations. Here, $g_{1},\cdots,g_{n}:\partial{\Omega}\to
S^{1}$ are smooth maps such that

$\deg(g_{j}):=\deg(g_{j},\partial{\Omega})=d_{j}\in\mathbb{N}\cup\\{0\\}{\quad\mbox{for}\quad}j=1,\cdots,n.$
(1.12)

The system (1.11) is the Euler-Lagrange equations of the functional

$E_{{\varepsilon},{\Omega}}(u_{1},\cdots,u_{n})=\frac{1}{2}\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j}|^{2}dx+\frac{1}{4{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j}|^{2}\Big{)}^{2}dx$
(1.13)

for a pair of maps $(u_{1},\cdots,u_{n})\in H^{1}_{g_{1}}\times\cdots\times
H^{1}_{g_{n}}$. For simplicity, we write $E_{\varepsilon}^{b}$ and
$E_{\varepsilon}$ instead of $E_{{\varepsilon},{\Omega}}^{b}$ and
$E_{{\varepsilon},{\Omega}}$ if there is no confusion on the domain
${\Omega}$. It is easy to check that

$\inf\big{\\{}E_{\varepsilon}(u_{1},\cdots,u_{n}):(u_{1},\cdots,u_{n})\in
H^{1}_{g_{1}}\times\cdots\times H^{1}_{g_{n}}\big{\\}}$ (1.14)

is achieved by some $(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\in
H^{1}_{g_{1}}\times\cdots\times H^{1}_{g_{n}}$. We also denote by
$u_{j,{\varepsilon}}^{b}$ a minimizer for $E_{\varepsilon}^{b}$ on
$H^{1}_{g_{j}}$.

The purpose of this paper is to study the asymptotic behavior of minimizers
$(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ as ${\varepsilon}\to 0$. In
this study, there is a remarkable difference between (1.5) and (1.11). First,
we consider the possible limit equation which could be a $n$-component
generalization of (1.9) in some sense. For (1.5), if $d\neq 0$, then the limit
function $u_{*}$ in (1.10) has $d$ singularities. However, even for the case
$(d_{1},\cdots,d_{n})\neq(0,\cdots,0)$, the limit functions for (1.11) turns
out to have no singularities. This difference is related to the nonexistence
of singular harmonic map in $H^{1}$. In fact, we have
$H^{1}_{g}({\Omega};S^{1})=\emptyset$ for the single equation (1.5). For the
system (1.11), it is natural to define a function space analogously:

$\displaystyle{\mathcal{X}}(g_{1},\cdots,g_{n};{\Omega})$ $\displaystyle=$
$\displaystyle\Big{\\{}(u_{1},\cdots,u_{n})\in
H^{1}_{g_{1}}({\Omega};\mathbb{C})\times\cdots\times
H^{1}_{g_{n}}({\Omega};\mathbb{C})~{}:~{}\sum_{j=1}^{n}|u_{j}|^{2}=n~{}\text{
a.e. on }{\Omega}\Big{\\}}.$

We write ${\mathcal{X}}(g_{1},\cdots,g_{n})$ if there is no confusion on
domains.

The asymptotic behavior of minimizers for $E_{\varepsilon}$ is closely related
to the maps in ${\mathcal{X}}(g_{1},\cdots,g_{n})$. Since it is expected that
$|u_{1,{\varepsilon}}|^{2}+\cdots+|u_{n,{\varepsilon}}|^{2}\to n$ as
${\varepsilon}\to 0$, the limit functions, say, $(u_{1}^{*}\cdots,u_{n}^{*})$
will satisfy $|u_{1}^{*}|^{2}+\cdots+|u_{n}^{*}|^{2}=n$ a.e. on ${\Omega}$.
Thus, it is important to know whether ${\mathcal{X}}(g_{1},\cdots,g_{n})$ is
nonempty or not. We can see that if
$\Psi=(u_{1},\cdots,u_{n})\in{\mathcal{X}}(g_{1},\cdots,g_{n})$, then

$\tilde{\Psi}=\Big{(}\frac{u_{1}}{\sqrt{n}},\cdots,\frac{u_{n}}{\sqrt{n}}\Big{)}~{}\in~{}S^{2n-1}.$

Since the homotopy group $\pi_{1}(S^{2n+1})$ is trivial, there is no
topological obstruction and we expect that the limit problem will not have
singularities. More precisely, the next theorem tells us that
${\mathcal{X}}(g_{1},\cdots,g_{n})$ is nonempty if $n\geq 2$.

######  Theorem 1.1.

If $d_{j}\geq 0$ for all $1\leq j\leq n$, then
${\mathcal{X}}(g_{1},\cdots,g_{n})\neq\emptyset$. Furthermore, if

$I(u_{1},\cdots,u_{n})=\sum_{j=1}^{n}\int_{\Omega}|\nabla u_{j}|^{2}dx,$
(1.15)

the minimization problem

$\displaystyle\beta(g_{1},\cdots,g_{n})$
$\displaystyle=\inf\big{\\{}I(u_{1},\cdots,u_{n}):(u_{1},\cdots,u_{n})\in{\mathcal{X}}(g_{1},\cdots,g_{n})\big{\\}},$
(1.16)

is achieved in ${\mathcal{X}}(g_{1},\cdots,g_{n})$.

###### Proof.

We choose smooth functions $v_{j}\in H_{g_{j}}^{1}({\Omega};\mathbb{C})$ such
that $\inf_{\Omega}(|v_{1}|^{2}+\cdots+|v_{n}^{2}|)>0$ on ${\Omega}$. Set

$w_{j}=\frac{\sqrt{n}v_{j}}{\big{(}\sum_{k=1}^{n}|v_{k}|^{2}\big{)}^{1/2}}\quad\mbox{for
each}\quad j=1,\cdots,n.$

Then, $(w_{1},\cdots,w_{n})\in{\mathcal{X}}(g_{1},\cdots,g_{n})\neq\emptyset$.
Moreover,

$\int_{\Omega}|\nabla w_{j}|^{2}~{}dx\leq C\int_{\Omega}\Big{(}\frac{|\nabla
v_{j}|^{2}}{\sum_{k}|v_{k}|^{2}}+\frac{|v_{j}|^{2}(\sum_{k}\nabla|v_{k}|^{2})^{2}}{(\sum_{k}|v_{k}|^{2})^{3}}\Big{)}dx\leq
C.$

Thus, $\beta(g_{1},\cdots,g_{n})<\infty$ and it is achieved in
${\mathcal{X}}(g_{1},\cdots,g_{n})$ by a standard variational argument. ∎

The associated Euler-Lagrange equations for (1.15) are

$\left\\{\begin{aligned} -\Delta u_{j}&=\frac{1}{n}u_{j}\sum_{k=1}^{n}|\nabla
u_{k}|^{2}{\quad\mbox{in}\quad}{\Omega},\quad
u_{j}=g_{j}{\quad\mbox{on}\quad}\partial{\Omega},\\\
&\sum_{j=1}^{n}|u_{j}|^{2}=n{\quad\mbox{in}\quad}{\Omega}.\end{aligned}\right.$
(1.17)

Indeed, if $(u_{1},\cdots,u_{n})$ is a critical point of $I$, then we are led
to

$0=i^{\prime}(0)=-\sum_{j=1}^{n}\int_{\Omega}\Big{[}\Delta
u_{j}+\frac{1}{n}u_{j}\sum_{k=1}^{n}|\nabla u_{k}|^{2}\Big{]}\psi_{j}dx$

by using the fact $\nabla(\sum_{j=1}^{n}|u_{j}|^{2})=0$. Here, we set
$i(t)=I\big{(}u_{1}(t),\cdots,u_{n}(t)\big{)}$ for
$(\psi_{1},\cdots,\psi_{n})\in H^{1}_{0}({\Omega})\times\cdots\times
H^{1}_{0}({\Omega})$ and

$u_{j}(t)=\frac{n(u_{j}+t\psi_{j})}{\sqrt{n\sum_{k=1}^{n}|u_{k}+t\psi_{k}|^{2}}}~{}\in~{}{\mathcal{X}}(g_{1},\cdots,g_{n}).$

We will refer to (1.17) as $n$-component harmonic map equations. The condition
$|u_{*}|=1$ in (1.9) makes its zeros singular. Whereas the condition
$|u_{1}|^{2}+\cdots+|u_{n}|^{2}=n$ in (1.17) do not force their zeros singular
since it does not mean $|u_{1}|=\cdots=|u_{n}|=1$. We also note from (1.17)
that $u_{1},\cdots,u_{n}$ share the same singularities if any.

Now, we are ready to state the first main result of this paper.

######  Theorem 1.2.

Let $(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ be a minimizer for
(1.14). Then, as ${\varepsilon}\to 0$, up to a subsequence,

$\displaystyle(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\to(u_{1}^{*},\cdots,u_{n}^{*}){\quad\mbox{in}\quad}C^{1,\alpha}(\bar{\Omega})\times\cdots\times
C^{1,\alpha}(\bar{\Omega}).$ (1.18)

Here, $(u_{1}^{*},\cdots,u_{n}^{*})$ is a minimizer of $I$ defined by (1.15)
and satisfies a system (1.17) of harmonic map equations. Moreover, for any
positive integer $k$,

$\displaystyle(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\to(u_{1}^{*},\cdots,u_{n}^{*}){\quad\mbox{in}\quad}C^{k}_{loc}({\Omega})\times\cdots\times
C^{k}_{loc}({\Omega}),$ (1.19)
$\displaystyle\frac{1}{{\varepsilon}^{2}}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}\to\frac{1}{n}\sum_{j=1}^{n}|\nabla
u_{j}^{*}|^{2}{\quad\mbox{in}\quad}C^{k}_{loc}({\Omega})\times\cdots\times
C^{k}_{loc}({\Omega}).$ (1.20)

We will show that although the degrees of the minimizers
$(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ are not $0$, our energy
$E_{\varepsilon}(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ is still
bounded as ${\varepsilon}\to 0$. This gives a different situation for the
Ginzburg-Landau energy $E_{\varepsilon}^{b}$ in (1.6). It is known that
$E_{\varepsilon}^{b}(u_{{\varepsilon},g}^{b})$ grows logarithmically in the
limit ${\varepsilon}\to 0$ if $\deg(g,\partial{\Omega})\neq 0$. See [9] for
optimal lower bounds for the energy $E_{\varepsilon}^{b}$ and many
applications.

The convergence (1.19) and (1.20) imply that $u_{1}^{*},\cdots,u_{n}^{*}$ are
smooth. The condition $|u_{1}^{*}|^{2}+\cdots+|u_{n}^{*}|^{2}=n$ makes each
$u_{j}^{*}$ smooth at its zeros. It is an interesting question whether
$|u_{j}^{*}|=1$ on $\Omega$ for some $j$. This is not true if $d_{j}>0$. As
the following theorem says, we are not sure that this can happen even for the
case $d_{j}=0$.

######  Theorem 1.3.

Let $(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ be a minimizing
sequence for (1.13).

  * (i)

If $d_{j}>0$ for some $1\leq j\leq n$, then we have

$\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{j,{\varepsilon}}|^{2}\big{)}^{2}dx\to\infty.$

  * (ii)

Suppose that $d_{j}=0$ for all $j=1,\cdots,n$. If
${\alpha}(g_{1},\cdots,g_{n})>\beta(g_{1},\cdots,g_{n})$, then there is some
$k\in\\{1,\cdots,n\\}$ such that

$\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{k,{\varepsilon}}|^{2}\big{)}^{2}dx\to\infty.$

Here, ${\alpha}(g_{1},\cdots,g_{n})$ is defined by Remark 4.2.

  * (iii)

For $n=2$, if either one of $d_{1}$ and $d_{2}$ is positive or
${\alpha}(g_{1},g_{2})>\beta(g_{1},g_{2})$, then

$\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{j,{\varepsilon}}|^{2}\big{)}^{2}dx\to\infty,\quad\mbox{for
each $j=1,2$.}$

This paper is organized as follows. In Section 2 and 3, we prove Theorem 1.2.
The basic idea is based on [1]. We deal with interior and boundary estimates
respectively in Section 2 and 3. We will focus on how the argument in [1] for
single equation (1.5) can be generalized to the system (1.11) nontrivially. In
Section 4, we prove Theorem 1.3. We consider another minimization problem for
(1.15) on a smaller space and see how it is related to the question that
$|u_{j}^{*}|=1$ for some $j$. We also study some additional properties of
solutions for (1.11) and (1.17).

We list some notations and facts that are used hereafter. The vector $\nu$
stands for the outward unit normal vector field on a given domain. We write
$B_{r}(x)$ or $B(x,r)$ to denote the ball of radius $r$ centered at a point
$x$. We often use the following fact: there exists
${\gamma}_{0}={\gamma}_{0}({\Omega})>0$ such that

$meas({\Omega}\cap B_{r}(x))\geq{\gamma}_{0}r^{2}\quad\quad\forall
x\in{\Omega},~{}\forall r\leq 1.$ (1.21)

##  2 Interior Estimates

In this section, we will prove Theorem 1.2. In what follows, we often use the
following lemma.

######  Lemma 2.1 ([1, Lemma A.1]).

Let $-\Delta u=f$ in an open set $\Omega\subset{\mathbb{R}^{2}}$. Given any
$K\Subset\Omega$, we have

$\|\nabla u\|_{L^{\infty}(K)}^{2}\leq
C_{K}\|u\|_{L^{\infty}(\Omega)}\Big{(}\|u\|_{L^{\infty}(\Omega)}+\|f\|_{L^{\infty}(\Omega)}\Big{)},$

where $C$ depends only on $\Omega$ and $K$. In addition, if $u=0$ on
$\partial{\Omega}$, then

$\|\nabla u\|_{L^{\infty}(K)}^{2}\leq
C\|u\|_{L^{\infty}(\Omega)}\|f\|_{L^{\infty}(\Omega)},$

where $C$ depends only on $\Omega$.

The next lemma provides $L^{\infty}$-estimates for $u_{j,{\varepsilon}}$ and
their gradients for $j=1,\cdots,n$.

######  Lemma 2.2.

Let $(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ be any solution of the
system (1.11). Then, for $0<{\varepsilon}<1$, we have

$\displaystyle\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}<n\quad\hbox{on}\quad{\Omega},$
(2.1) $\displaystyle\|\nabla
u_{j,{\varepsilon}}\|_{L^{\infty}({\Omega})}\leq\frac{C_{0}}{{\varepsilon}}{\quad\mbox{for}\quad}j=1,\cdots,n.$
(2.2)

Here, $C_{0}$ is a constant that depends only on $g_{j}$ and ${\Omega}$.

###### Proof.

Let $f=n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}$. Then,

$\Delta
f\leq\frac{2}{{\varepsilon}^{2}}\Big{(}\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}f-2\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\leq\frac{2}{{\varepsilon}^{2}}\Big{(}\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}f{\quad\mbox{in}\quad}{\Omega}.$

Since $f=0$ on $\partial{\Omega}$, we have $f>0$ in ${\Omega}$ by the strong
maximum principle.

To show (2.2), let us decompose
$u_{j,{\varepsilon}}=u_{j,{\varepsilon}}^{1}+u_{j,{\varepsilon}}^{2}$ where

$\left\\{\begin{aligned} -\Delta
u_{j,{\varepsilon}}^{1}&=\frac{1}{{\varepsilon}^{2}}u_{j,{\varepsilon}}\big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\big{)}{\quad\mbox{in}\quad}\Omega,\quad
u_{j,{\varepsilon}}^{1}=0{\quad\mbox{on}\quad}\partial{\Omega},\\\ -\Delta
u_{j,{\varepsilon}}^{2}&=0{\quad\mbox{in}\quad}\Omega,\quad
u_{j,{\varepsilon}}^{2}=g_{j}{\quad\mbox{on}\quad}\partial\Omega.\end{aligned}\right.$

Then, $\|\nabla u_{j,{\varepsilon}}^{1}\|_{\infty}\leq C/{\varepsilon}$ by
Lemma 2.1 and $\|\nabla u_{j,{\varepsilon}}^{2}\|_{\infty}\leq C$ by elliptic
estimates for $j=1,\cdots,n$. ∎

######  Proposition 2.3.

Let $(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})$ be a minimizer for the
problem (1.14). Then,

$(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\to(u_{1}^{*},\cdots,u_{n}^{*}){\quad\mbox{in}\quad}H^{1}_{g_{1}}\times\cdots\times
H^{1}_{g_{n}}{\quad\mbox{as}\quad}{\varepsilon}\to 0,$

where $(u_{1}^{*},\cdots,u_{n}^{*})$ is a minimizer for (1.16). Moreover,

$\lim_{{\varepsilon}\rightarrow
0}\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}^{2}dx=0.$
(2.3)

###### Proof.

Let $({\widetilde{u}}_{1},\cdots,{\widetilde{u}}_{n})\in
H^{1}_{g_{1}}\times\cdots\times H^{1}_{g_{n}}$ be any minimizer for (1.16).
Since $\sum_{j=1}^{n}|{\widetilde{u}}_{j}|^{2}=n$ a.e. on ${\Omega}$ and
$E_{\varepsilon}(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\leq
E_{\varepsilon}({\widetilde{u}}_{1},\cdots,{\widetilde{u}}_{n})$, we have

$\displaystyle\frac{1}{2}\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}dx+\frac{1}{4{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}^{2}dx$
(2.4) $\displaystyle\leq{}$
$\displaystyle\frac{1}{2}\int_{\Omega}\sum_{j=1}^{n}|\nabla{\widetilde{u}}_{j}|^{2}dx.$

Hence,

$\\{(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\\}$ is uniformly bounded
in $H^{1}\times\cdots\times H^{1}$ (2.5)

and there exists an $n$ tuple $(u_{1}^{*},\cdots,u_{n}^{*})\in
H^{1}_{g_{1}}\times\cdots\times H^{1}_{g_{n}}$ such that up to a subsequence,

$(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\to(u_{1}^{*},\cdots,u_{n}^{*})$

weakly in $H^{1}\times\cdots\times H^{1}$ and strongly in
$L^{p}\times\cdots\times L^{p}$ for all $p\geq 1$. Moreover,
$\sum_{j=1}^{n}|u_{j}^{*}|^{2}=n$ a.e. on $\Omega$ by (2.4) and thus
$(u_{1}^{*},\cdots,u_{n}^{*})\in{\mathcal{X}}(g_{1},\cdots,g_{n})$. Since
$E_{\varepsilon}(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\leq
E_{\varepsilon}(u_{1}^{*},\cdots,u_{n}^{*})$, we obtain

$\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}dx\leq\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j}^{*}|^{2}dx.$

This implies that

$\displaystyle\int_{\Omega}\sum_{j=1}^{n}|\nabla u_{j,{\varepsilon}}-\nabla
u_{j}^{*}|^{2}dx$ $\displaystyle\leq~{}$ $\displaystyle
2\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j}^{*}|^{2}dx-2\int_{\Omega}\sum_{j=1}^{n}\nabla
u_{j,{\varepsilon}}\cdot\nabla u_{j}^{*}dx~{}\to~{}0.$

So,
$(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\to(u_{1}^{*},\cdots,u_{n}^{*})$
strongly in $H^{1}\times\cdots\times H^{1}$. Taking $\liminf_{{\varepsilon}\to
0}$ on (2.4), we are led to

$\int_{\Omega}\sum_{j=1}^{n}|\nabla
u_{j}^{*}|^{2}dx\leq\int_{\Omega}\sum_{j=1}^{n}|\nabla{\widetilde{u}}_{j}|^{2}dx,$

and we conclude that $(u_{1}^{*},\cdots,u_{n}^{*})$ is also a solution for the
minimization problem (1.16). Moreover, (2.3) is a direct consequence of (2.4).
∎

######  Lemma 2.4.

$\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\to n$ uniformly on
$\overline{\Omega}$.

###### Proof.

Fix $x_{0}\in{\Omega}$. By (2.2), we have that for
$|x-x_{0}|<{\varepsilon}\rho_{\varepsilon}$,

$|u_{j,{\varepsilon}}(x)|^{2}\leq\big{(}|u_{j,{\varepsilon}}(x_{0})|+C_{0}\rho_{\varepsilon}\big{)}^{2}{\quad\mbox{for}\quad}j=1,\cdots,n.$

Since $|u_{j,{\varepsilon}}|\leq\sqrt{n}$ for all $j=1,\cdots,n$, we obtain

$\displaystyle n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}(x)|^{2}$
$\displaystyle\geq
n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}(x_{0})|^{2}-2n\sqrt{n}C_{0}\rho_{\varepsilon}-nC_{0}^{2}\rho_{\varepsilon}^{2}$
$\displaystyle\geq\frac{1}{2}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}(x_{0})|^{2}\Big{)},$

where the last inequality holds if and only if

$a_{\varepsilon}(x_{0}):=\frac{1}{2}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}(x_{0})|^{2}\Big{)}\geq
2n\sqrt{n}C_{0}\rho_{\varepsilon}+nC_{0}^{2}\rho_{\varepsilon}^{2},$

in other words,

$0<\rho_{\varepsilon}\leq\frac{-n\sqrt{n}+\sqrt{n^{3}+na_{\varepsilon}(x_{0})}}{nC_{0}}=:b_{\varepsilon}(x_{0}).$

By taking $\rho_{\varepsilon}=b_{\varepsilon}$, we deduce from (1.21) and
(2.3) that

$\displaystyle o(1)$
$\displaystyle=\frac{1}{{\varepsilon}^{2}}\int_{B_{{\varepsilon}\rho_{\varepsilon}}(x_{0})\cap{\Omega}}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}^{2}dx\geq{\gamma}_{0}a_{\varepsilon}^{2}(x_{0})b_{\varepsilon}^{2}(x_{0}).$

Hence, $a_{\varepsilon}(x_{0})\to 0$ uniformly. ∎

######  Lemma 2.5.

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{H^{2}_{loc}\times\cdots\times
H^{2}_{loc}}\leq C$.

###### Proof.

We note that

$\Delta\Big{(}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}=~{}2\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}+2\sum_{j=1}^{n}\nabla
u_{j,{\varepsilon}}\cdot\nabla(\Delta u_{j,{\varepsilon}}).$

Hence, we have

$\displaystyle\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}=~{}$
$\displaystyle\frac{1}{2}\sum_{j=1}^{n}\Delta\Big{(}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}+\frac{1}{{\varepsilon}^{2}}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{(}n-\sum_{i=1}^{n}|u_{i,{\varepsilon}}|^{2}\Big{)}$
$\displaystyle\qquad-\frac{2}{{\varepsilon}^{2}}\Big{(}\sum_{j=1}^{n}u_{j,{\varepsilon}}\cdot\nabla
u_{j,{\varepsilon}}\Big{)}^{2}.$

Since $\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\to n$ uniformly on
$\overline{\Omega}$, at least one of $u_{j,{\varepsilon}}$ satisfies that
$|u_{j,{\varepsilon}}|\geq 1/2$ on ${\Omega}$ for all small ${\varepsilon}$.
If $|u_{k,{\varepsilon}}|\geq 1/2$ for some $k\in\\{1,\cdots,n\\}$, then

$\frac{1}{{\varepsilon}^{2}}\Big{(}n-\sum_{i=1}^{n}|u_{i,{\varepsilon}}|^{2}\Big{)}=\frac{|\Delta
u_{k,{\varepsilon}}|}{|u_{k,{\varepsilon}}|}\leq 2|\Delta
u_{k,{\varepsilon}}|.$

Hence, we can write

$\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}\leq~{}\frac{1}{2}\sum_{j=1}^{n}\Delta\Big{(}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}+2\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}|\Delta u_{k,{\varepsilon}}|,$

which implies that

$\frac{1}{2}\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}\leq\frac{1}{2}\sum_{j=1}^{n}\Delta\Big{(}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}+C\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{4}.$ (2.6)

Fix $x_{0}$ and $r={\mathrm{dist}}(x_{0},\partial{\Omega})/4$. Let $\zeta$ be
a smooth function such that $\zeta=1$ on $B_{r}(x_{0})$,
${\mathrm{supp}}~{}\zeta\subset B_{2r}(x_{0})$ and $0\leq\zeta\leq 1$.
Multiplying (2.6) by $\zeta^{2}$ and using the Sobolev embedding
$W^{1,1}({\Omega})\hookrightarrow L^{2}({\Omega})$, we are led to

$\displaystyle\frac{1}{2}\int_{B_{2r}(x_{0})}\zeta^{2}\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}dx$
$\displaystyle\leq~{}$ $\displaystyle
C\int_{B_{2r}(x_{0})}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}dx+C\int_{B_{2r}(x_{0})}\sum_{j=1}^{n}\big{(}\zeta|\nabla
u_{j,{\varepsilon}}|^{2}\big{)}^{2}dx$ $\displaystyle\leq~{}$ $\displaystyle
C\int_{B_{2r}(x_{0})}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}dx+C\int_{B_{2r}(x_{0})}\zeta\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|\,|D^{2}u_{j,{\varepsilon}}|dx.$

In the sequel, by (2.5)

$\frac{1}{2}\int_{B_{2r}(x_{0})}\zeta^{2}\sum_{j=1}^{n}|D^{2}u_{j,{\varepsilon}}|^{2}dx\leq
C\int_{{\Omega}}\sum_{j=1}^{n}|\nabla u_{j,{\varepsilon}}|^{2}dx\leq C.$

Now, the standard covering argument shows that each $u_{j,{\varepsilon}}$ is
uniformly bounded in $H^{2}_{loc}({\Omega})$. ∎

Let

$f_{\varepsilon}=\frac{1}{{\varepsilon}^{2}}\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}.$
(2.7)

Then, we can rewrite (1.11) as

$\left\\{\begin{aligned} -\Delta
u_{j,{\varepsilon}}&=f_{\varepsilon}u_{j,{\varepsilon}}{\quad\mbox{in}\quad}{\Omega},\\\
u_{j,{\varepsilon}}&=g_{j}{\quad\mbox{on}\quad}\partial{\Omega}.\end{aligned}\right.$
(2.8)

A simple calculation gives

$\left\\{\begin{aligned} -{\varepsilon}^{2}\Delta
f_{\varepsilon}+2\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}f_{\varepsilon}&=2\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}{\quad\mbox{in}\quad}{\Omega},\\\
f_{\varepsilon}&=0{\quad\mbox{on}\quad}\partial{\Omega}.\end{aligned}\right.$
(2.9)

Given a compact set $K\subset{\Omega}$, we define

${\mathcal{A}}_{K}=\Big{\\{}\tilde{K}\subset{\Omega}~{}|~{}K\Subset\tilde{K}\Subset{\Omega}\Big{\\}}.$

For $\tilde{K}\in{\mathcal{A}}_{K}$, we define a set
${\mathcal{F}}(K,\tilde{K})$ of smooth functions by

${\mathcal{F}}(K,\tilde{K})=\Big{\\{}{\varphi}\in
C^{\infty}({\Omega})~{}|~{}0\leq{\varphi}\leq 1,{\varphi}\equiv 1~{}\mbox{on
$K$ and ${\mathrm{supp}}~{}{\varphi}\subset\tilde{K}$}\Big{\\}}.$

In what follows, $C_{K}$, or simply $C$, denotes a generic constant depending
on a compact set $K$ but independent of ${\varepsilon}$. For simplicity, we
also often write $\|(u_{1},\cdots,u_{n})\|_{X\times\cdots\times X}$ as
$\|(u_{1},\cdots,u_{n})\|_{X}$ for a function space $X$.

######  Lemma 2.6.

For any $p\geq 1$, we have
$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{2,p}_{loc}\times\cdots\times
W^{2,p}_{loc}}\leq C$ and $\|f_{\varepsilon}\|_{C^{0}_{loc}}\leq C$.

###### Proof.

By Lemma 2.4, we may assume that $\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\geq
1$ on ${\Omega}$. Given $K\Subset{\Omega}$, we denote $K=K_{0}$ and choose
compact sets $K_{j}\in{\mathcal{A}}_{K_{j-1}}$ for $j=1,2,3$ and
${\varphi}\in{\mathcal{F}}(K_{3},K_{2})$. Set
$F_{\varepsilon}={\varphi}f_{\varepsilon}$. Then,

$\displaystyle-{\varepsilon}^{2}\Delta F_{\varepsilon}$
$\displaystyle=-{\varepsilon}^{2}f_{\varepsilon}\Delta{\varphi}-2{\varepsilon}^{2}\nabla{\varphi}\cdot\nabla
f_{\varepsilon}-{\varepsilon}^{2}{\varphi}\Delta f_{\varepsilon}$
$\displaystyle=-\Big{(}n-\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}\Delta{\varphi}+4\sum_{j=1}^{n}u_{j,{\varepsilon}}(\nabla{\varphi}\cdot\nabla
u_{j,{\varepsilon}})$
$\displaystyle\qquad-2F_{\varepsilon}\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}+2{\varphi}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}$
$\displaystyle\leq-2F_{\varepsilon}+C_{K}\Big{(}1+\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}.$

Hence,

$-{\varepsilon}^{2}\Delta F_{\varepsilon}+2F_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}{\quad\mbox{on}\quad}K_{3}.$ (2.10)

For $p>2$, multiplying (2.10) by $F_{\varepsilon}^{p-1}$, we are led to

$\displaystyle{\varepsilon}^{2}(p-1)\int_{K_{3}}|\nabla
F_{\varepsilon}|^{2}F_{\varepsilon}^{p-2}dx+2\int_{K_{3}}F_{\varepsilon}^{p}dx$
$\displaystyle\leq~{}$ $\displaystyle
C_{K}\Big{(}\int_{K_{3}}\Big{(}1+\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}^{p}dx\Big{)}^{\frac{1}{p}}\Big{(}\int_{K_{3}}F_{\varepsilon}^{p}dx\Big{)}^{\frac{p-1}{p}}$
$\displaystyle\leq~{}$ $\displaystyle
C_{K}\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{H^{2}(K_{3})}\|F_{\varepsilon}\|_{L^{p}(K_{3})}^{p-1}~{}\leq~{}C_{K}\|F_{\varepsilon}\|_{L^{p}(K_{3})}^{p-1}.$

Here, we used Lemma 2.5 in the last inequality. Then, we deduce from Young’s
inequality that $\|F_{\varepsilon}\|_{L^{p}({K_{3}})}\leq C_{K}$. This implies
by (2.8) that $\|\Delta u_{j,{\varepsilon}}\|_{L^{p}(K_{2})}\leq C_{K}$ for
any $p>2$ and $j=1,\cdots,n$. So,

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{2,p}(K_{1})}\leq
C_{K}$

by the interior regularity. Moreover,
$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{C^{1}(K_{1})}\leq
C_{K}$ by the Sobolev embedding.

Now choose ${\widetilde{\phi}}\in{\mathcal{F}}(K,K_{1})$ and set
${\widetilde{F}}_{\varepsilon}={\widetilde{\phi}}f_{\varepsilon}$. Then, as
above, we have

$-{\varepsilon}^{2}\Delta{\widetilde{F}}_{\varepsilon}+2{\widetilde{F}}_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}{\quad\mbox{on}\quad}K_{1}.$

By applying the maximum principle to this inequality, we are led to

$\max_{K_{1}}{\widetilde{F}}_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}\|\nabla
u_{j,{\varepsilon}}\|_{L^{\infty}(K_{1})}^{2}\Big{)}\leq C_{K}.$

In particular, we conclude that $\|f_{\varepsilon}\|_{L^{\infty}(K)}\leq
C_{K}$. ∎

######  Lemma 2.7.

Given a nonnegative integer $k$ and a real number $p\geq 1$, we have

$\displaystyle\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{k+2,p}_{loc}\times\cdots\times
W^{k+2,p}_{loc}}\leq
C{\quad\mbox{and}\quad}\|f_{\varepsilon}\|_{C^{k}_{loc}}\leq C.$ (2.11)

###### Proof.

We use an induction on $k$. The case $k=0$ follows from Lemma 2.6. Suppose
that (2.11) is true for $k$. By the Sobolev embedding,
$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{C^{k+1}_{loc}}\leq C$.
By the induction hypothesis, we note from (2.9) that

$-{\varepsilon}^{2}\Delta\partial^{k}f_{\varepsilon}=\partial^{k}\Big{(}-2\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}f_{\varepsilon}+2\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}=O(1)$

on every compact set. Then, by applying Lemma 2.1, we obtain that

$\|\partial^{k+1}f_{\varepsilon}\|_{L^{\infty}_{loc}}\leq
C{\varepsilon}^{-1}.$ (2.12)

By (2.8) and (2.12), on every compact set,

$-\Delta\partial^{k+1}u_{j,{\varepsilon}}=\partial^{k+1}(f_{\varepsilon}u_{j,{\varepsilon}})=O({\varepsilon}^{-1}).$

Hence, by Lemma 2.1 again,

$\|\partial^{k+2}u_{j,{\varepsilon}}\|_{L^{\infty}_{loc}}\leq
C{\varepsilon}^{-1/2}.$ (2.13)

By (2.12) and (2.13), we get

$-{\varepsilon}^{2}\Delta\partial^{k+1}f_{\varepsilon}=\partial^{k+1}\Big{(}-2\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}f_{\varepsilon}+2\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}=O({\varepsilon}^{-1})$

on every compact set. Applying Lemma 2.1 to this equation, we are led to

$\|\partial^{k+2}f_{\varepsilon}\|_{L^{\infty}_{loc}}\leq
C{\varepsilon}^{-2}.$ (2.14)

Now, given $K\Subset{\Omega}$, we denote $K=K_{0}$ and choose compact sets
$K_{j}\in{\mathcal{A}}_{K_{j-1}}$ for $j=1,2,3$ and
${\varphi}\in{\mathcal{F}}(K_{2},K_{3})$. Set
$F_{\varepsilon}={\varphi}\partial^{k+1}f_{\varepsilon}$. It follows from
(2.12) and (2.14) that

$\displaystyle-{\varepsilon}^{2}\Delta F_{\varepsilon}$
$\displaystyle=-{\varepsilon}^{2}\partial^{k+1}f_{\varepsilon}\Delta{\varphi}-2{\varepsilon}^{2}\nabla{\varphi}\cdot\nabla\partial^{k+1}f_{\varepsilon}-{\varepsilon}^{2}{\varphi}\Delta\partial^{k+1}f_{\varepsilon}$
$\displaystyle=O(1)-{\varepsilon}^{2}{\varphi}\Delta\partial^{k+1}f_{\varepsilon}.$

On the other hand, by (2.9)

$\displaystyle-{\varepsilon}^{2}{\varphi}\Delta\partial^{k+1}f_{\varepsilon}$
$\displaystyle=-2\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}F_{\varepsilon}-2{\varphi}\sum_{i+l=k+1,~{}l<k+1}\partial^{i}\Big{(}\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}\Big{)}\partial^{l}f_{\varepsilon}$
$\displaystyle\qquad+2{\varphi}\partial^{k+1}\Big{(}\sum_{j=1}^{n}|\nabla
u_{j,{\varepsilon}}|^{2}\Big{)}$
$\displaystyle\leq-2F_{\varepsilon}+C_{K}\Big{(}1+\sum_{j=1}^{n}|D^{k+2}u_{j,{\varepsilon}}|\Big{)}.$

Hence,

$-{\varepsilon}^{2}\Delta F_{\varepsilon}+2F_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}|D^{k+2}u_{j,{\varepsilon}}|\Big{)}{\quad\mbox{on}\quad}K_{3}.$
(2.15)

For $p>2$, multiplying (2.15) by $F_{\varepsilon}^{p-1}$, we are led to

$\displaystyle{\varepsilon}^{2}(p-1)\int_{K_{3}}|\nabla
F_{\varepsilon}|^{2}F_{\varepsilon}^{p-2}dx+2\int_{K_{3}}F_{\varepsilon}^{p}dx$
$\displaystyle\leq~{}$ $\displaystyle
C_{K}\Big{(}\int_{K_{3}}\Big{(}1+\sum_{j=1}^{n}|D^{k+2}u_{j,{\varepsilon}}|\Big{)}^{p}dx\Big{)}^{\frac{1}{p}}\Big{(}\int_{K_{3}}F_{\varepsilon}^{p}dx\Big{)}^{\frac{p-1}{p}}$
$\displaystyle\leq~{}$ $\displaystyle
C_{K}\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{k+2,p}(K_{3})}\|F_{\varepsilon}\|_{L^{p}(K_{3})}^{p-1}~{}\leq~{}C_{K}\|F_{\varepsilon}\|_{L^{p}(K_{3})}^{p-1},$

which implies that $\|F_{\varepsilon}\|_{L^{p}({K_{3}})}\leq C_{K}$. As a
consequence, by (2.8)

$\|\Delta\partial^{k+1}u_{j,{\varepsilon}}\|_{L^{p}(K_{2})}\leq
C_{K}{\quad\mbox{for}\quad}j=1,\cdots,n.$

So,

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{k+3,p}(K_{1})}\leq
C_{K}.$ (2.16)

In particular,
$\|\partial^{k+2}(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{C^{0}(K_{1})}\leq
C_{K}$.

Now choose ${\widetilde{\phi}}\in{\mathcal{F}}(K,K_{1})$ and set
${\widetilde{F}}_{\varepsilon}={\widetilde{\phi}}\partial^{k+1}f_{\varepsilon}$.
As above, we obtain

$-{\varepsilon}^{2}\Delta{\widetilde{F}}_{\varepsilon}+2{\widetilde{F}}_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}|D^{k+2}u_{j,{\varepsilon}}|\Big{)}{\quad\mbox{on}\quad}K_{1}.$

Then, we deduce from the maximum principle that

$\max_{K_{1}}{\widetilde{F}}_{\varepsilon}\leq
C_{K}\Big{(}1+\sum_{j=1}^{n}\|D^{k+2}u_{j,{\varepsilon}}\|_{L^{\infty}(K_{1})}\Big{)}\leq
C_{K}.$

In the sequel, we conclude that

$\|\partial^{k+1}f_{\varepsilon}\|_{L^{\infty}(K)}\leq C_{K}.$ (2.17)

Therefore, (2.11) is true for $k+1$ by (2.16) and (2.17). This completes the
proof. ∎

##  3 Boundary Estimates and Proof of Theorem 1.2

######  Lemma 3.1.

For any solution of problem (1.11) there is a constant $C$ such that for all
${\varepsilon}>0$,

$\int_{\partial{\Omega}}\sum_{j=1}^{n}\Big{|}\frac{\partial
u_{j,{\varepsilon}}}{\partial\nu}\Big{|}^{2}d\sigma\leq C.$ (3.1)

Here, $\nu$ is the outward unit normal vector field on $\partial{\Omega}$.

###### Proof.

For simplicity, we drop the subscript ${\varepsilon}$. Let us multiply (1.11)
by $V\cdot\nabla u_{i}$, where $V=(V_{1},V_{2})$ is a smooth vector field on
${\Omega}$ such that $V=\nu$ on $\partial{\Omega}$. By integrating them by
part, we obtain

$\text{(LHS)}=\int_{\Omega}\Delta u_{i}(V\cdot\nabla
u_{i})~{}dx=\int_{\partial{\Omega}}\Big{|}\frac{\partial
u_{i}}{\partial\nu}\Big{|}^{2}d\sigma-\frac{1}{2}\int_{\partial{\Omega}}|\nabla
u_{i}|^{2}d\sigma+O(1)$

since $\nabla u_{i}$ is uniformly bounded in $L^{2}({\Omega})$. On the other
hand,

$\text{(RHS)}=\frac{1}{2{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j}|^{2}\Big{)}\big{(}V\cdot\nabla|u_{i}|^{2}\big{)}dx.$

Adding these for each $i=1,\cdots,n$, we have

$\displaystyle\sum_{i=1}^{n}\Big{(}\int_{\partial{\Omega}}\Big{|}\frac{\partial
u_{i}}{\partial\nu}\Big{|}^{2}d\sigma-\frac{1}{2}\int_{\partial{\Omega}}|\nabla
u_{i}|^{2}d\sigma\Big{)}$ $\displaystyle=$
$\displaystyle~{}\frac{1}{2{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j}|^{2}\Big{)}\sum_{i=1}^{n}\big{(}V\cdot\nabla|u_{i}|^{2}\big{)}dx+O(1)$
$\displaystyle=$
$\displaystyle~{}\frac{1}{4{\varepsilon}^{2}}\int_{\Omega}\Big{(}n-\sum_{j=1}^{n}|u_{j}|^{2}\Big{)}^{2}(\nabla\cdot
V)dx+O(1)=O(1),$

where the last inequality comes from (2.4). Combining these identities, we
conclude that

$\sum_{i=1}^{n}\int_{\partial{\Omega}}\Big{|}\frac{\partial
u_{i}}{\partial\nu}\Big{|}^{2}d\sigma=\sum_{i=1}^{n}\int_{\partial{\Omega}}\Big{|}\frac{\partial
g_{i}}{\partial\tau}\Big{|}^{2}d\sigma+O(1)=O(1).$

Here, $\tau$ is the tangential vector field on $\partial{\Omega}$. ∎

In this section, the estimates of solutions for problem (1.11) would be proved
up to the boundary.

######  Lemma 3.2.

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{H^{2}({\Omega})\times\cdots\times
H^{2}({\Omega})}\leq C$.

###### Proof.

In view of Lemma 2.5, it suffices to prove uniform $H^{2}$-boundedness near
each $x_{0}\in\partial{\Omega}$. We may assume $x_{0}=0$ and change local
coordinates $(x_{1},x_{2})\to(y_{1},y_{2})=(x_{1},x_{2}-h(x_{1}))$ where $h$
represent $\partial{\Omega}$ locally and $h^{\prime}(0)=0$. We set
${\widetilde{u}}_{i,{\varepsilon}}(y_{1},y_{2})=u_{i,{\varepsilon}}(x_{1},x_{2})$
and $\tilde{g}_{i}(y_{1},y_{2})=g_{i}(x_{1},x_{2})$ on
$U=\big{\\{}(y_{1},y_{2}):y_{2}>0\big{\\}}\cap B_{r}(0)$ for some $r>0$, and
rewrite (1.11) as

$\left\\{\begin{aligned}
-L{\widetilde{u}}_{i,{\varepsilon}}&=\frac{1}{{\varepsilon}^{2}}{\widetilde{u}}_{i,{\varepsilon}}\Big{(}n-\sum_{j=1}^{n}|{\widetilde{u}}_{j,{\varepsilon}}|^{2}\Big{)}{\quad\mbox{on}\quad}U,\\\
{\widetilde{u}}_{i,{\varepsilon}}&=\tilde{g}_{i,{\varepsilon}}{\quad\mbox{on}\quad}\\{y_{2}=0\\}\cap\partial
U.\end{aligned}\right.$

Here,

$\displaystyle L=\sum_{k,l=1}^{2}\frac{\partial}{\partial
y_{l}}\Big{(}a_{kl}\frac{\partial}{\partial y_{k}}\Big{)}\mbox{ with
$a_{11}=1$, $a_{12}=a_{21}=-h^{\prime}$ and $a_{22}=1+(h^{\prime})^{2}$}$

is a strongly elliptic operator if $r$ is small enough. For simplicity, we
write ${\widetilde{u}}_{j,{\varepsilon}}$ as $u_{j}$ in what follows.

Let $A=\frac{1}{2}\sum_{j=1}^{n}|\nabla u_{j}|^{2}$. By a direct calculation,

$\displaystyle L[A]$
$\displaystyle=\sum_{j=1}^{n}\sum_{k,l,r=1}^{2}\Big{(}a_{kl}(u_{j})_{y_{k}y_{r}}(u_{j})_{y_{l}y_{r}}+(u_{j})_{y_{r}}\cdot
L\big{[}(u_{j})_{y_{r}}\big{]}\Big{)}$
$\displaystyle\geq\alpha\sum_{j=1}^{n}|D^{2}u_{j}|^{2}+\sum_{j=1}^{n}\sum_{r=1}^{2}(u_{j})_{y_{r}}\cdot
L\big{[}(u_{j})_{y_{r}}\big{]},$

where ${\alpha}$ is the ellipticity constant of $L$. We note that

$\displaystyle\sum_{j=1}^{n}\sum_{r=1}^{2}(u_{j})_{y_{r}}\cdot
L\big{[}(u_{j})_{y_{r}}\big{]}$ $\displaystyle=~{}$
$\displaystyle\sum_{j=1}^{n}\sum_{k,l,r=1}^{2}(u_{j})_{y_{r}}\cdot\Big{\\{}\big{(}L[u_{j}]\big{)}_{y_{r}}-\big{[}(a_{kl})_{y_{r}}(u_{j})_{y_{k}}\big{]}_{y_{l}}\Big{\\}}$
$\displaystyle=~{}$
$\displaystyle-\frac{1}{{\varepsilon}^{2}}\sum_{j=1}^{n}|\nabla
u_{j}|^{2}\Big{(}n-\sum_{i=1}^{n}|u_{i}|^{2}\Big{)}+\frac{2}{{\varepsilon}^{2}}\sum_{r=1}^{2}\Big{(}\sum_{j=1}^{n}u_{j}\cdot(u_{j})_{y_{r}}\Big{)}\Big{(}\sum_{i=1}^{n}u_{i}\cdot(u_{i})_{y_{r}}\Big{)}$
$\displaystyle-\sum_{j=1}^{n}\sum_{k,l,r=1}^{2}(u_{j})_{y_{r}}\big{[}(a_{kl})_{y_{l}y_{r}}(u_{j})_{y_{k}}+(a_{kl})_{y_{r}}(u_{j})_{y_{k}y_{l}}\big{]}$
$\displaystyle\geq~{}$
$\displaystyle-\frac{1}{{\varepsilon}^{2}}\sum_{j=1}^{n}|\nabla
u_{j}|^{2}\Big{(}n-\sum_{i=1}^{n}|u_{i}|^{2}\Big{)}-C\sum_{j=1}^{n}(|\nabla
u_{j}|^{2}+|D^{2}u_{j}|^{2}).$

By Lemma 2.4, given $y\in U$, if ${\varepsilon}$ is small, then we can find
$k=k(y)$ such that $|u_{k}(y)|\geq 1/2$. Then,

$|L[u_{k}](y)|=\frac{1}{{\varepsilon}^{2}}|u_{k}(y)|\Big{(}n-\sum_{j=1}^{n}|u_{j}(y)|^{2}\Big{)}\geq\frac{1}{2{\varepsilon}^{2}}\Big{(}n-\sum_{j=1}^{n}|u_{j}(y)|^{2}\Big{)}.$

Hence,

$\displaystyle\frac{1}{{\varepsilon}^{2}}\Big{(}n-\sum_{j=1}^{n}|u_{j}(y)|^{2}\Big{)}$
$\displaystyle\leq
2|L[u_{k}](y)|=2\Big{|}\Big{(}a_{kl}(y)(u_{k})_{y_{k}}(y)\Big{)}_{y_{l}}\Big{|}$
$\displaystyle\leq C\sum_{j=1}^{n}(|\nabla u_{j}(y)|+|D^{2}u_{j}(y)|).$

As a consequence, by Young’s inequality, we are led to

$-L[A]+\frac{\alpha}{2}\sum_{j=1}^{n}|D^{2}u_{j}|^{2}\leq
C\Big{(}1+\sum_{j=1}^{n}|\nabla u_{j}|^{4}\Big{)}.$

Let us choose $\zeta\in{\mathcal{F}}(B_{r/2}(0),B_{r}(0))$. Then,

$\frac{\alpha}{2}\sum_{j=1}^{n}\int_{U}\zeta^{2}|D^{2}u_{j}|^{2}dx\leq
C+C\sum_{j=1}^{n}\int_{U}\zeta^{2}|\nabla
u_{j}|^{4}dx+\int_{U}\zeta^{2}L[A]dx.$ (3.2)

By integrating by parts, we obtain

$\displaystyle\int_{U}\zeta^{2}L[A]dx$
$\displaystyle=\int_{U}\zeta^{2}\big{[}a_{kl}A_{y_{l}}\big{]}_{y_{k}}dx$
$\displaystyle=\int_{U}AL[\zeta^{2}]dx+2\int_{\\{y_{2}=0\\}}a_{12}A(\zeta^{2})_{y_{1}}+\int_{\\{y_{2}=0\\}}a_{22}A(\zeta^{2})_{y_{2}}$
$\displaystyle\quad+\int_{\\{y_{2}=0\\}}\zeta^{2}A(a_{12})_{y_{1}}-\int_{\\{y_{2}=0\\}}\zeta^{2}a_{22}A_{y_{2}}.$

The first four terms are uniformly bounded by Lemma 3.1. Furthermore,

$\displaystyle\int_{\\{y_{2}=0\\}}\zeta^{2}a_{22}A_{y_{2}}$
$\displaystyle=~{}$
$\displaystyle\sum_{j=1}^{n}\Big{(}\int_{\\{y_{2}=0\\}}\zeta^{2}a_{22}(u_{j})_{y_{1}}\cdot(u_{j})_{y_{1}y_{2}}+\zeta^{2}a_{22}(u_{j})_{y_{2}}\cdot(u_{j})_{y_{2}y_{2}}\Big{)}$
$\displaystyle=~{}$ $\displaystyle\sum_{j=1}^{n}\big{(}(I)+(II)\big{)}.$

By Lemma 3.1,

$(I)=-\int_{\\{y_{2}=0\\}}\Big{(}(\zeta^{2}a_{22})_{y_{1}}(u_{j})_{y_{1}}\cdot(u_{j})_{y_{2}}+\zeta^{2}a_{22}(g_{j})_{y_{1}y_{1}}\cdot(u_{j})_{y_{2}}\Big{)}=O(1).$

Since $Lu_{j}=0$ on $\\{y_{2}=0\\}$ and $a_{11}=1$, we have

$\displaystyle(II)$
$\displaystyle=-\int_{\\{y_{2}=0\\}}\zeta^{2}\Big{(}(u_{j})_{y_{2}}\cdot(g_{j})_{y_{1}y_{1}}+(a_{12})_{y_{1}}|(u_{j})_{y_{2}}|^{2}$
$\displaystyle\qquad+~{}(a_{21})_{y_{2}}(u_{j})_{y_{2}}\cdot(g_{j})_{y_{1}}+(a_{22})_{y_{2}}|(u_{j})_{y_{2}}|^{2}\Big{)}-\int_{\\{y_{2}=0\\}}\zeta^{2}a_{12}(|(u_{j})_{y_{2}}|^{2})_{y_{1}}$
$\displaystyle=O(1)+\int_{\\{y_{2}=0\\}}(\zeta^{2}a_{12})_{y_{1}}|(u_{j})_{y_{2}}|^{2}~{}=~{}O(1).$

Here, we also used Lemma 3.1. In the sequel, we can rewrite (3.2) as

$\frac{\alpha}{2}\sum_{j=1}^{n}\int_{U}\zeta^{2}|D^{2}u_{j}|^{2}dx\leq
C+C\sum_{j=1}^{n}\int_{U}\zeta^{2}|\nabla u_{j}|^{4}dx.$

Now, using this inequality and employing the same argument of the proof of
Lemma 2.5, we can show

$\sum_{j=1}^{n}\int_{B_{r/2}(x_{0})}|D^{2}u_{j}|^{2}dx\leq C.$

This completes the proof. ∎

######  Proposition 3.3.

For any $p\geq 1$, we have

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{2,p}({\Omega})\times\cdots\times
W^{2,p}({\Omega})}\leq C_{p}$

Moreover, if $f_{\varepsilon}$ is defined by (2.7), then
$\|f_{\varepsilon}\|_{C^{0}(\Omega)}\leq C$.

###### Proof.

By Lemma 3.2 and the Sobolev embedding,

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{W^{1,p}({\Omega})\times\cdots\times
W^{1,p}({\Omega})}\leq C\quad\forall p\geq 1.$

We keep the notation (2.7). By Lemma 2.4 and (2.9), we obtain

$-2{\varepsilon}^{2}\Delta f_{\varepsilon}+f_{\varepsilon}\leq
4\sum_{j=1}^{n}|\nabla u_{j,{\varepsilon}}|^{2}\quad\mbox{on ${\Omega}$}.$
(3.3)

For $q>1$, since $f_{\varepsilon}=0$ on $\partial{\Omega}$, multiplying this
equation by $f_{\varepsilon}^{q-1}$, we have

$\int_{\Omega}f_{\varepsilon}^{q}dx\leq 4\sum_{j=1}^{n}\int_{\Omega}|\nabla
u_{j,{\varepsilon}}|^{2}f_{\varepsilon}^{q-1}dx\leq 4\sum_{j=1}^{n}\|\nabla
u_{j,{\varepsilon}}\|^{2}_{2q}\|f_{\varepsilon}\|^{q-1}_{q},$

which implies that $\|f_{\varepsilon}\|_{q}\leq C$ for some $C=C(n,q)>0$.
Hence, each $\Delta u_{j,{\varepsilon}}$ is uniformly bounded in
$L^{q}({\Omega})$ for any $q\geq 1$. By applying the elliptic regularity to
(2.8), we obtain that

$\|(u_{1,{\varepsilon}},\cdots,u_{n,{\varepsilon}})\|_{C^{1,\alpha}({\Omega})\times\cdots\times
C^{1,\alpha}({\Omega})}\leq C\quad\forall{\alpha}\in(0,1).$

By the maximum principle, we obtain from (3.3) that
$\|f_{\varepsilon}\|_{C^{0}(\Omega)}\leq C$. In particular, the right hand
side of (2.8) is uniformly bounded in $C^{0}({\Omega})$. Hence, each
$u_{j,{\varepsilon}}$ is also uniformly bounded in $W^{2,p}({\Omega})$ for any
$p\geq 1$. ∎

Proof of Theorem 1.2. The proof of (1.19) follows readily from Proposition 2.3
and Proposition 3.3. To show (1.20), let $f_{\varepsilon}$ be defined by (2.7)
and set

$h_{\varepsilon}=\sum_{j=1}^{n}|u_{j,{\varepsilon}}|^{2}{\quad\mbox{and}\quad}f_{*}=\frac{1}{n}\sum_{j=1}^{n}|\nabla
u_{j}^{*}|^{2}.$

Then, $-\Delta u_{j}^{*}=u_{j}^{*}f_{*}$ and
$\|h_{\varepsilon}-n\|_{C^{k}_{loc}({\Omega})}={\varepsilon}^{2}\|f_{\varepsilon}\|_{C^{k}_{loc}({\Omega})}\to
0$. We note that

$-\sum_{j=1}^{n}(u_{j,{\varepsilon}}\Delta u_{j,{\varepsilon}}-u_{j}^{*}\Delta
u_{j}^{*})=n(f_{\varepsilon}-f_{*})+f_{\varepsilon}(h_{\varepsilon}-n).$

Hence, as ${\varepsilon}\to 0$,

$\displaystyle\|f_{\varepsilon}-f_{*}\|_{C^{k}_{loc}({\Omega})}$
$\displaystyle\leq~{}$
$\displaystyle\frac{1}{n}\Big{(}\|f_{\varepsilon}(h_{\varepsilon}-n)\|_{C^{k}_{loc}({\Omega})}+\sum_{j=1}^{n}\|u_{j,{\varepsilon}}\Delta
u_{j,{\varepsilon}}-u_{j}^{*}\Delta
u_{j}^{*}\|_{C^{k}_{loc}({\Omega})}\Big{)}\to 0.$

This completes the proof. ∎

##  4 Further Properties of Solutions

In this section, we study some properties of solutions of the $n$-component
Ginzburg-Landau equations (1.11) and the generalized harmonic map equations
(1.17). First, the next proposition tells us that if $g_{j}$ is a rotation of
$g_{1}$ for each $j$ and ${\varepsilon}$ is not so small, then $u_{j}$ is a
rotation of $u_{1}$ for any solution pair $(u_{1},\cdots,u_{n})$ of (1.11).
Consequently, each of $u_{j}$ and $u_{k}$ is a rotation of the other.

######  Proposition 4.1.

Let $(u_{1},\cdots,u_{n})$ be a solution of (1.11). Assume that for each $j$,
there exists ${\gamma}_{j}\in\mathbb{R}$ such that
$g_{j}=e^{i\gamma_{j}}g_{1}$. Then, $u_{j}=e^{i\gamma_{j}}u_{1}$ for all
${\varepsilon}>\sqrt{n/{\lambda}_{1}}$. Here, ${\lambda}_{1}$ is the first
eigenvalue of $-\Delta$ on ${\Omega}$ with the Dirichlet boundary condition.

###### Proof.

Set ${\widetilde{u}}_{j}=e^{-i{\gamma}_{j}}u_{j}$. Then, ${\widetilde{u}}_{j}$
satisfies

$\displaystyle-\Delta{\widetilde{u}}_{j}$
$\displaystyle=\frac{1}{{\varepsilon}^{2}}{\widetilde{u}}_{j}\Big{(}n-\sum_{k=1}^{n}|{\widetilde{u}}_{k}|^{2}\Big{)}{\quad\mbox{in}\quad}\Omega,$
$\displaystyle{\widetilde{u}}_{j}$
$\displaystyle=g_{1}{\quad\mbox{on}\quad}\partial\Omega.$

If $w_{j}={\widetilde{u}}_{j}-u_{1}$, then

$\displaystyle-\Delta
w_{j}+\frac{1}{{\varepsilon}^{2}}\sum_{k=1}^{n}|u_{k}|^{2}w_{j}$
$\displaystyle=\frac{n}{{\varepsilon}^{2}}w_{j},{\quad\mbox{in}\quad}\Omega,$
$\displaystyle w_{j}$ $\displaystyle=0{\quad\mbox{on}\quad}\partial\Omega.$

Hence, it follows that

$\int_{\Omega}|\nabla
w_{j}|^{2}+\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\sum_{k=1}^{n}|u_{k}|^{2}|w_{j}|^{2}=\frac{n}{{\varepsilon}^{2}}\int_{\Omega}|w_{j}|^{2}\leq\frac{n}{{\lambda}_{1}{\varepsilon}^{2}}\int_{\Omega}|\nabla
w_{j}|^{2}.$

If ${\varepsilon}>\sqrt{n/{\lambda}_{1}}$, then $w_{j}$ is a constant. Since
$w_{j}=0$ on $\partial{\Omega}$, we have $w_{j}=0$. ∎

It is not clear whether the conclusion of Proposition 4.1 is valid for
arbitrary ${\varepsilon}$.

######  Remark 4.2.

Suppose that $d_{j}=0$ for all $1\leq j\leq n$. Since
$H^{1}_{g_{j}}({\Omega};S^{1})\neq\emptyset$, the minimization problem

$\alpha(g_{1},\cdots,g_{n})=\inf\Big{\\{}I(u_{1},\cdots,u_{n}):u_{j}\in
H^{1}_{g_{j}}({\Omega};S^{1})\Big{\\}}$

is achieved by a unique pair $(u_{1}^{0},\cdots,u_{n}^{0})$ satisfying

$\left\\{\begin{aligned} -\Delta u_{j}^{0}&=u_{j}^{0}|\nabla
u_{j}^{0}|^{2}&&\mbox{on }{\Omega},\\\ |u_{j}^{0}|&=1&&\mbox{on }{\Omega},\\\
u_{j}^{0}&=g_{j}&&\mbox{on }\partial{\Omega}.\end{aligned}\right.$ (4.1)

Since $d_{j}=0$ for each $j=1,\cdots,n$, we have that
$u_{j}^{0}=e^{i{\varphi}_{j}}$ for a harmonic function ${\varphi}_{j}$ in
${\Omega}$. In general, we have
$\alpha(g_{1},\cdots,g_{n})\geq\beta(g_{1},\cdots,g_{n})$ since
$(u_{1}^{0},\cdots,u_{n}^{0})\in{\mathcal{X}}(g_{1},\cdots,g_{n})$. An
interesting question is whether
$\alpha(g_{1},\cdots,g_{n})=\beta(g_{1},\cdots,g_{n})$ or not. The next
proposition gives us a necessary condition that they are equal.

######  Proposition 4.3.

Assume that $d_{j}=0$ for all $j\in\\{1,\cdots,n\\}$. If
$\alpha(g_{1},\cdots,g_{n})=\beta(g_{1},\cdots,g_{n})$ and it is achieved by
$(u_{1},\cdots,u_{n})$, then $|\nabla u_{j}|=|\nabla u_{k}|$ for all
$j,k\in\\{1,\cdots,n\\}$.

###### Proof.

By assumption, $u_{1},\cdots,u_{n}\in
H_{g_{1}}^{1}(\Omega;S^{1})\times\cdots\times H_{g_{n}}^{1}(\Omega;S^{1})$ and

$\beta=I(u_{1},\cdots,u_{n}).$

Since $|u_{j}|=1$, we can write $u_{j}=e^{i{\varphi}_{j}}$. By plugging this
in (1.17), we are led to

$-i\Delta{\varphi}_{j}=\Big{(}\frac{1}{n}-1\Big{)}|\nabla{\varphi}_{j}|^{2}+\frac{1}{n}\sum_{k\neq
j}^{n}|\nabla{\varphi}_{k}|^{2}{\quad\mbox{in}\quad}{\Omega}.$

Thus, for all $j=1\cdots,n$

$\Delta{\varphi}_{j}=0{\quad\mbox{and}\quad}|\nabla{\varphi}_{j}|^{2}=\frac{1}{n-1}\sum_{k\neq
j}^{n}|\nabla{\varphi}_{k}|^{2}.$

In particular,

$|\nabla{\varphi}_{j}|^{2}-|\nabla{\varphi}_{1}|^{2}=\frac{1}{n-1}(|\nabla{\varphi}_{1}|^{2}-|\nabla{\varphi}_{j}|^{2}),$

which implies that $|\nabla{\varphi}_{j}|^{2}=|\nabla{\varphi}_{1}|^{2}$, or
equivalently $|\nabla u_{j}|^{2}=|\nabla u_{1}|^{2}$ for all $1\leq j\leq n$.
∎

######  Remark 4.4.

Let us consider the case $n=2$ for simplicity with $d_{1}=d_{2}=0$. We can
write $u_{j}=e^{i{\varphi}_{j}}$ and $g_{j}=e^{i\tilde{{\varphi}}_{j}}$ for
$j=1,2$. According to the proof of Proposition 4.3, if
${\alpha}(g_{1},g_{2})=\beta(g_{1},g_{2})$, then it is necessary that

$\left\\{\begin{aligned}
&\Delta{\varphi}_{1}=0=\Delta{\varphi}_{2}{\quad\mbox{on}\quad}{\Omega},\\\
&|\nabla{\varphi}_{1}|=|\nabla{\varphi}_{2}|{\quad\mbox{on}\quad}{\Omega},\\\
&{\varphi}_{1}=\tilde{\varphi}_{1},~{}{\varphi}_{2}=\tilde{\varphi}_{2}{\quad\mbox{on}\quad}\partial{\Omega}.\end{aligned}\right.$
(4.2)

So, if there are no solutions for (4.2), we may conclude that
${\alpha}(g_{1},g_{2})>\beta(g_{1},g_{2})$. For instance, if
${\widetilde{\varphi}}_{1}$ is a constant and ${\widetilde{\varphi}}_{2}$ is
not a constant function, then there are no solution of (4.2). Another example
is the case that
${\widetilde{\varphi}}_{1}={\lambda}{\widetilde{\varphi}}_{2}$ with
$|{\lambda}|\neq 1$. Indeed, if
${\widetilde{\varphi}}_{1}={\lambda}{\widetilde{\varphi}}_{2}$, then

$-\Delta\Big{(}\frac{{\varphi}_{2}}{{\lambda}}\Big{)}=0~{}~{}\mbox{on}~{}~{}{\Omega}{\quad\mbox{and}\quad}\frac{{\varphi}_{2}}{{\lambda}}={\widetilde{\varphi}}_{1}~{}~{}\mbox{on}~{}~{}\partial{\Omega}.$

Since harmonic maps with the same boundary values are equal, we have
${\varphi}_{1}={\varphi}_{2}/{\lambda}$. Thus,
$|\nabla{\varphi}_{1}|=|\nabla{\varphi}_{2}|$ only when $|{\lambda}|=1$.

As a final subject of this section, we prove Theorem 1.3.

Proof of Theorem 1.3:

(i) Suppose to the contrary that there is $k\in\\{1,\cdots,n\\}$ such that
$d_{k}>0$ and

$\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{k,{\varepsilon}}|^{2}\big{)}^{2}dx\leq
C.$

By Theorem 1.2, we have $u_{k,{\varepsilon}}\to u_{k}^{*}\in H^{1}({\Omega})$
with $|u_{k}^{*}|=1$. This implies that $u_{k}^{*}\in
H^{1}_{g_{k}}({\Omega};S^{1})$, which is impossible since
$H^{1}_{g_{k}}({\Omega};S^{1})=\emptyset$ if $d_{k}>0$.

(ii) Suppose that $d_{1}=\cdots=d_{n}=0$. If

$\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{j,{\varepsilon}}|^{2}\big{)}^{2}dx\leq
C\quad\text{for all}\quad j=1,\cdots,n,$

then $u_{j,{\varepsilon}}\to u_{j}^{*}$ in $H^{1}$ and $|u_{j}^{*}|=1$ for
each $j=1,\cdots,n$. So, we have
$\alpha(g_{1},\cdots,g_{n})=\beta(g_{1},\cdots,g_{n})$ that contradicts the
assumption.

(iii) If we set

$X_{\varepsilon}^{2}=\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{1,{\varepsilon}}|^{2}\big{)}^{2}dx\quad\mbox{and}\quad
Y_{\varepsilon}^{2}=\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\Big{(}1-|u_{2,{\varepsilon}}|^{2}\Big{)}^{2}dx,$

we are led by (2.4) and the Cauchy-Schwartz inequality that

$\displaystyle C$
$\displaystyle\geq\frac{1}{{\varepsilon}^{2}}\int_{\Omega}\big{(}2-|u_{1,{\varepsilon}}|^{2}-|u_{2,{\varepsilon}}|^{2}\big{)}^{2}dx$
$\displaystyle=X_{\varepsilon}^{2}+Y_{\varepsilon}^{2}+\frac{2}{{\varepsilon}^{2}}\int_{\Omega}\big{(}1-|u_{1,{\varepsilon}}|^{2}\big{)}\big{(}1-|u_{2,{\varepsilon}}|^{2}\big{)}dx$
$\displaystyle\geq\big{(}X_{\varepsilon}-Y_{\varepsilon}\big{)}^{2}.$

So, either both $X_{\varepsilon}$ and $Y_{\varepsilon}$ are bounded, or they
are unbounded. By (i) and (ii), at least one of $X_{\varepsilon}$ and
$Y_{\varepsilon}$ is unbounded and we get the desired conclusion. ∎

### Acknowledgements.

Jongmin Han was supported by Basic Science Research Program through the
National Research Foundation of Korea(NRF) funded by the Ministry of Education
(2018R1D1A1B07042681). Juhee Sohn was supported by the National Research
Foundation of Korea(NRF) grant funded by the Korea government(MSIT)
(2021R1G1A1003396).

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