# The qualitative behavior at a vortex point for the Chern-Simon-Higgs
equation

Jiayu Li  School of Mathematical Sciences, University of Science and
Technology of China, Hefei, 230026, P. R. China<EMAIL_ADDRESS>and  Lei
Liu  School of Mathematics and Statistics & Hubei Key Laboratory of
Mathematical Sciences, Central China Normal University, Wuhan 430079, P. R.
China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the qualitative behavior at a vortex blow-up point for
Chern-Simon-Higgs equation. Roughly speaking, we will establish an energy
identity at a each such point, i.e. the local mass is the sum of the bubbles.
Moreover, we prove that either there is only one bubble which is a singular
bubble or there are more than two bubbles which contains no singular bubble.
Meanwhile, we prove that the energies of these bubbles must satisfy a
quadratic polynomial which can be used to prove the simple blow-up property
when the multiplicity is small. As is well known, for many Liouville type
system, Pohozaev type identity is a quadratic polynomial corresponding to
energies which can be used directly to compute the local mass at a blow-up
point. The difficulty here is that, besides the energy’s integration, there is
a additional term in the Pohozaev type identity of Chern-Simon-Higgs equation.
We need some more detailed and delicated analysis to deal with it.

###### Key words and phrases:

Chern-Simon-Higgs equation, Blow up, Energy identity, Vortex point

###### 2010 Mathematics Subject Classification:

The work was carried out when Lei Liu was visiting the China-France
Mathematics Center, University of Science and Technology of China. He would
like to thank the center for the hospitality and the good working conditions.

##  1\. introduction

The present model deals with a simplified form of the so-called anyon model, a
classical field theory defined on $(2+1)$ dimensional Minkowski apace where
the Lagrangian couples Maxwell and Chern-Simons terms coming from a gauge
field with a scalar field. Generally, the Euler-Lagrange equations of this
model are complicated. For a simplified version without the Maxwell term,
Hong-Kim-Pac [13] and Jackiw-Weinberg [14] proposed a special choice of
$6^{th}$ order potential where there may exist time-independent vortex
solutions satisfying a first order Bogomolny type self-dual equations.

Mathematically, Chern-Simons-Higgs 6-th model can be described as follows. Let
$M$ be a compact Riemann surface with volume $1$ and $L$ be a line bubble over
$M$. For a unitary connection $D_{A}=d+A$ on $L$ with curvature $F_{A}$, and a
section $\phi$ of $L$, Hong-Kim-Pac [13] and Jackiw-Weinberg [14] introduced
Chern-Simons-Higgs 6-th functional as follows:

$\displaystyle CS(A,\phi)$
$\displaystyle=\int_{M}\left(|D_{A}\phi|^{2}+\frac{\epsilon^{2}}{|\phi|^{2}}|F_{A}|^{2}+\frac{1}{4\epsilon^{2}}|\phi|^{2}(1-|\phi|^{2})^{2}\right)dV$
$\displaystyle=\int_{M}\left(|(D_{1}+iD_{2})\phi|^{2}+\left(\frac{\epsilon}{|\phi|}F-\frac{1}{2\epsilon}|\phi|(1-|\phi|^{2})\right)^{2}\right)dV+2\pi\
deg\ L.$

The absolute minimizers of above functional satisfy the Bogomolny type self-
dual equations

(1.1) $\displaystyle\begin{cases}(D_{1}+iD_{2})\phi&=0,\\\
\frac{1}{2}|\phi|^{2}(1-|\phi|^{2})&=\epsilon^{2}F.\end{cases}$

Putting

$v:=\log|\phi|^{2},$

we reduce the above system to the following single scalar equation that

(1.2) $\Delta
v=\frac{4}{\epsilon^{2}}e^{v}(e^{v}-1)+4\pi\sum_{j=1}^{J}N_{j}\delta_{p_{j}},\
\ in\ \ M,$

where $p_{j}\in M$, $N_{j}$ is a positive integer, $\delta_{p_{j}}$ is the
Dirac measure at $p_{j}$, $j=1,...,J$. For more details of deriving this
equation, we refer to [13, 14, 10, 22, 23].

In this paper, we want to study the compactness problem for a sequence of
solutions of (1.2). Let $v_{n}$ satisfies

(1.3) $\Delta
v_{n}(x)=\lambda_{n}e^{v_{n}(x)}(e^{v_{n}(x)}-1)+4\pi\sum_{j=1}^{J}N_{j}\delta_{p_{j}},$

where $N_{1},...,N_{J}$ are positive integers, $\sum_{j=1}^{J}N_{j}=\bar{N}$
and $\lim_{n\to\infty}\lambda_{n}\to+\infty$.

Let $u_{0}$ be the corresponding Green function, i.e.

$\displaystyle\begin{cases}\Delta
u_{0}&=-4\pi\bar{N}+4\pi\sum_{j=1}^{J}N_{j}\delta_{p_{j}},\\\
\int_{M}u_{0}dM&=0.\end{cases}$

Then $u_{n}=v_{n}-u_{0}$ satisfies

(1.4) $-\Delta
u_{n}=\lambda_{n}e^{u_{0}}e^{u_{n}}(1-e^{u_{0}}e^{u_{n}})-4\pi\bar{N}.$

Denote

$\bar{u}_{n}=\int_{M}u_{n}dM,\ \ w_{n}(x):=u_{n}(x)-\bar{u}_{n}.$

In [16, 7], they proved the following theorem that

######  Theorem 1.1 ([16, 7]).

Let $M$ be a compact Riemann surface with volume $1$. Then passing to a
subsequence, the following alternatives hold.

  * (1)

If $\lim_{n\to\infty}\lambda_{n}e^{\bar{u}_{n}}=+\infty$, we have $u_{n}\to-
u_{0}$ in the sense of $W^{1,p}(M),\ (1<p<2)$ as $n\to\infty$.

  * (2)

If $\lim_{n\to\infty}\lambda_{n}e^{\bar{u}_{n}}=c>0$, we have $w_{n}\to w_{0}$
weakly in $W^{1,p}(M),\ (1<p<2)$ as $n\to\infty$, where $w_{0}$ satisfies

$-\Delta
w_{0}=ce^{u_{0}}e^{w_{0}}-4\pi\bar{N}+4\pi\sum_{i=1}^{I}\alpha_{i}\delta_{p_{k_{i}}},$

with $\int_{M}w_{0}dM=0$, $1\leq\alpha_{i}\leq N_{k_{i}}$, $N_{k_{i}}$ is the
multiplicity of the vortex $p_{k_{i}}$.

  * (3)

If $\lim_{n\to\infty}\lambda_{n}e^{\bar{u}_{n}}=c=0$, we have
$w_{n}\to\tilde{w}$ in $W^{1,p}(M),\ (1<p<2)$ as $n\to\infty$, where
$\tilde{w}$ is a positive Green function satisfying

$-\Delta\tilde{w}=-4\pi\bar{N}+4\pi\sum_{i=1}^{I}\beta_{i}\delta_{q_{i}},$

with $\int_{M}\tilde{w}dM=0$, $1\leq\beta_{i}\leq\bar{N}$ and
$\sum_{i=1}^{I}\beta_{i}=\bar{N}$.

Recently, Lin-Yan [19, 20] studied doubly periodic solutions $u_{n}$ for (1.4)
in $\Omega$ with the boundary condition

(1.5) $u_{n}\mbox{ is doubly periodic on }\partial\Omega.$

Similar to classical theory of Liouville equations, they [19, 20] proved that
if the blow-up point is not a vortex point, then there is only one bubble,
i.e. it must be simple. Furthermore, some interesting bubbling solutions were
also constructed in their papers.

In this paper, we continue to study the blow-up picture of (1.4) by focusing
on following two issues:

  * (1)

If the blow-up point is not a vortex point, the boundary condition (1.5) can
be changed to a general one, e.g. $osc_{\partial\Omega}u_{n}\leq C$, which is
more nature from the blow-up theory [4, 7, 16]?

  * (2)

If the blow-up point is a vortex point, can we get a energy identity, i.e. the
local mass is the sum of bubbles? Furthermore, if the simple blow-up property
still hold at a vortex blow-up point?

Integrating (1.4), we get

(1.6)
$\int_{M}\lambda_{n}e^{u_{0}}e^{u_{n}}(1-e^{u_{0}}e^{u_{n}})dM=4\pi\bar{N}.$

By a standard maximal argument in [5], it is easy to see that
$v_{n}=u_{0}+u_{n}\leq 0$. Furthermore, if
$\lim\lambda_{n}e^{\bar{u}_{n}}=c<\infty$, by Proposition 4.1 of [7], we get
that

$v_{n}(x)\leq-C$

for some positive constant $C>0$, which immediately implies that

(1.7)
$\int_{M}\lambda_{n}e^{u_{0}}e^{u_{n}}dM\leq\frac{4\pi\bar{N}}{1-e^{-C}}.$

Under the assumptions of Theorem 1.1 and (1.6)-(1.7), at each blow-up point
$q$ (i.e. there exists $x_{n}\to q$ such that $w_{n}(x_{n})\to+\infty$), after
a suitable rescaling argument, Lan-Li [16] proved that there is a bubble
$\varphi(x)$ splitting off, which satisfies the equation

(1.8)
$\displaystyle\begin{cases}-\Delta\varphi(x)=e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right),\ \ x\in\mathbb{R}^{2},\\\ \delta e^{\varphi(x)}\leq 1,\
\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right)dx+\int_{\mathbb{R}^{2}}e^{\varphi(x)}dx\leq
C,\end{cases}$

where $\delta\geq 0$ is a constant, or satisfy the equation

(1.9)
$\displaystyle\begin{cases}-\Delta\varphi(x)=|x-x_{0}|^{2N_{j}}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N_{j}}e^{\varphi(x)}\right),\
\ x\in\mathbb{R}^{2},\\\ \delta|x-x_{0}|^{2N_{j}}e^{\varphi(x)}\leq 1,\\\
\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N_{j}}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N_{j}}e^{\varphi(x)}\right)dx+\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N_{j}}e^{\varphi(x)}dx\leq
C;\end{cases}$

where $\delta\geq 0$ is a constant (maybe different from the previous one),
$x_{0}\in\mathbb{R}^{2}$.

In this paper, near a blow-up point $q$, if there exist $x_{n}\to q$ and
$\mu_{n}\to 0$ such that

(1.10) $\varphi_{n}(x):=w_{n}(x_{n}+\mu_{n}x)-w_{n}(x_{n})\to\varphi(x),\ \
in\ \ C^{2}_{loc}(\mathbb{R}^{2}),$

where $\varphi(x)$ satisfies (1.9) or (1.8), then we say $\varphi(x)$ or
$(x_{n},\mu_{n})$ is a bubble. Specially, we say $\varphi(x)$ or
$(x_{n},\mu_{n})$ a singular bubble if $\varphi(x)$ satisfies (1.9).

Denote

$\alpha_{n}(s_{n};x_{n}):=\frac{1}{2\pi}\int_{B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}e^{u_{0}(x)}e^{w_{n}}[1-e^{\bar{u}_{n}}e^{u_{0}(x)}e^{w_{n}}]dx.$

Our first main result is

######  Theorem 1.2.

Under the assumptions and notations of Theorem 1.1, assuming
$c=\lim_{n\to\infty}\lambda_{n}e^{\bar{u}_{n}}<+\infty$, we have:

  * (1)

if $c>0$, then $I=0$, that is $w_{n}\to w_{0}$ converges strongly in
$C^{2}(M)$.

  * (2)

if $c=0$ and $w_{n}$ (or $u_{n}$) blows up at a point $q_{i}$ which is not a
vortex point, then there is only one bubble $\varphi(x)$ satisfying (1.8),
i.e. the simple blow-up property holds. Moreover, we have following energy
identity that

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;q_{i})=b,$

where

$b=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right)dx\geq 4.$

  * (3)

if $c=0$ and $w_{n}$ (or $u_{n}$) blows up at a vortex point $p_{i}$, then the
following two alternatives hold.

    * (3-1)

If there is a singular bubble $\varphi(x)$ near the blow-up point $p_{i}$,
then $\varphi(x)$ is the only bubble, i.e. the simple blow-up property holds.
Moreover, we have following energy identity

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;p_{i})=b,$

where

$b=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N_{i}}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N_{i}}e^{\varphi(x)}\right)dx\geq
4(1+N_{i})$

and $N_{i}$ is the multiplicity of vortex point $p_{i}$.

    * (3-2)

If there is a non-singular bubble $\varphi^{1}(x)$ near the blow-up point
$p_{i}$, then there must exist more than two bubbles
$\\{\varphi^{k}\\}_{k=1}^{L}$. Moreover $\\{\varphi^{k}\\}_{k=1}^{L}$ contains
no singular bubble and satisfies the following energy identity

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;p_{i})=\sum_{k=1}^{L}b_{k},$

where

$b_{k}=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{\varphi^{k}(x)}\left(1-\delta
e^{\varphi^{k}(x)}\right)dx\geq 4.$

Furthermore, energies of these bubbles satisfy the following relation that

(1.11)
$\sum_{k=1}^{L}b_{k}^{2}=\left(\sum_{k=1}^{L}b_{k}\right)^{2}-4N_{i}\sum_{k=1}^{L}b_{k}.$

To prove Theorem 1.2, we first need to extend some lemmas in the blow-up
theory of Liouville equations to Chern-Simon-Higgs equation, including a
selection of bubbling areas, fast decay lemma and two kinds of Pohozaev type
identities. This kind of selection of bubbling areas was firstly used in Li-
Shafrir [18] for Liouville equations where they showed at the centers
$\Sigma_{n}$ of bubbling disks, the singularity is $-2log\
dist(x,\Sigma_{n})$, i.e. $w_{n}(x)+2\log\ dist(x,\Sigma_{n})$ is uniformly
bounded from above. However, this process can not be extended easily to Chern-
Simon-Higgs equation, since the latter is more complicated due to the
different singularities at centers of bubbling disks. We need to introduce a
new test function and some classification discussion, see Proposition 2.4. To
prove fast decay lemma, we need to estimate the oscillation of

$w_{n}(x)+2\log dist(x,\Sigma_{n})+2N\log|x|$

in a annulus containing zero point, where one can see that the term $\log|x|$
will take trouble to us. For Pohozaev type identity, it is well known that for
Liouville equation, this identity is a quadratic polynomial with respect to
the energy which implies immediately the value of local mass. For Chern-Simon-
Higgs equation, on one hand, there are two kinds of identities, local type
together with global type. On the other hand, besides the energy term, there
is a additional integration

$\int_{B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}e^{u_{0}(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}e^{u_{0}(x)}e^{w_{n}}]dx$

in both two kinds of Pohozaev type identities, see Lemma 2.10. This is a main
difficulty in the blow-up analysis of Chern-Simon-Higgs equation. To overcome
this obstacle, we will use some ideas in the blow-up theory of harmonic maps
[12] by proving that there is no energy loss in neck domains.

Secondly, since the main proof is a induction argument on the bubbling disks,
we need to explore the positional structures of these bubbling areas. This
will be solved by using a argument of local and global type Pohozaev
identities, where the relation (1.11) follows, see Section 2. We want to
remark that in case $(3-1)$ of Theorem 1.2, by using this type argument, we
can prove the simple blow-up phenomenon. But generally, we can not exclude
case $(3-2)$ of Theorem 1.2.

In the blow-up theory of Liouville type system, there is an important question
that if the simple blow-up property holds near a blow-up point, i.e. there is
only one bubble at each blow-up point? This kind result was firstly prove by
Li [17] for Liouville equation. Later, Jost-Lin-Wang [15] proved it also holds
for Toda system. For Liouville equation with quantized singular sources, see
[1, 2, 3, 8, 9, 24, 25].

For Chern-Simon-Higgs equation, by Lin-Yan [19] and above Theorem 1.2, we know
the simple blow-up property holds near a non-vortex blow-up point. At a vortex
blow-up point, Theorem 1.2 tells us that either $(3-1)$ or $(3-2)$ happens and
generally we can not exclude the second case. But, when the sum of
multiplicities $\bar{N}=\sum_{j=1}^{J}N_{j}$ is small, we can use identity
(1.11) in Theorem 1.2 to show that there is one bubble near a vortex blow-up
point which must be a singular bubble. The following theorem may be useful
when one want to construct non-simple bubbling solutions.

######  Theorem 1.3.

Denote the blow-up set by $S:=\\{q_{1},...,q_{I}\\}$. Assume $S$ is not empty.
Then we have

  * (1)

When $\bar{N}\leq 3$, $S$ contains no vortex point.

  * (2)

When $\bar{N}=5$, if $q_{1}$ is a vortex point, then $S=\\{q_{1}\\}$ and its
multiplicity $N_{1}=1$. In this case, the simple blow-up property holds.

  * (3)

When $\bar{N}=6$, if $S=\\{q_{1}\\}$ where $q_{1}$ is a vortex point with
multiplicity $N_{1}=1$, then the simple blow-up property holds.

  * (4)

When $\bar{N}=7$, if $S=\\{q_{1}\\}$ where $q_{1}$ is a vortex point, then
$N_{1}=1$ or $N_{1}=2$ and the simple blow-up property holds.

At the end of this part, we want to ask the following:

Question: For Chern-Simon-Higgs equation, if the simple blow-up property holds
at a vortex blow-up point?

The rest of the paper is organized as follows. In Section 2, we will prove
some basic lemmas which will be used in our proof, such as a selection of
bubbling areas, oscillation lemma, fast decay lemma and two kinds of Pohozaev
type identities. The proof of Theorem 1.2 and Theorem 1.3 will be given in
Section 3.

##  2\. Some basic lemmas

In this section, we will first divide Theorem 1.2 into two local versions,
i.e. Theorem 2.1 and Theorem 2.2. Secondly, we will establish some lemmas
which will be used in our proof, including a selection of bubbling areas,
oscillation lemma, fast decay lemma and two kinds of Pohozaev type identities.
Lastly, we will prove a local energy identity.

For simplicity of notations, we assume $0$ is a vortex point with multiplicity
$N$. Writing Green function

$u_{0}(x)=2N\log|x|+\zeta(x)$

where $\zeta(x)$ is a smooth function near point $0$, then we consider the
following system

(2.1) $\displaystyle-\Delta u_{n}$
$\displaystyle=\lambda_{n}|x|^{2N}e^{\zeta(x)}e^{u_{n}}(1-|x|^{2N}e^{\zeta(x)}e^{u_{n}}),\
\ in\ \ B_{1}(0),$

where $\lambda_{n}$ is a sequence of numbers which converges to $+\infty$, and

$\int_{B_{1}(0)}\lambda_{n}|x|^{2N}e^{\zeta(x)}e^{u_{n}}(1-|x|^{2N}e^{\zeta(x)}e^{u_{n}})dx+\int_{B_{1}(0)}\lambda_{n}|x|^{2N}e^{\zeta(x)}e^{u_{n}}dx\leq\Lambda$

and

$|x|^{2N}e^{\zeta(x)}e^{u_{n}}\leq 1,\ \ osc_{\partial B}u_{n}\leq C.$

Denote

$\bar{u}_{n}:=\frac{1}{\pi}\int_{B_{1}(0)}u_{n}(x)dx,\ \
w_{n}(x):=u_{n}(x)-\bar{u}_{n},$

then we have

(2.2) $\displaystyle-\Delta w_{n}$
$\displaystyle=\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}(1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}),\
\ in\ \ B_{1}(0),$

and

(2.3)
$\int_{B_{1}(0)}\lambda_{n}|x|^{2N}e^{\zeta(x)}e^{\bar{u}_{n}+w_{n}(x)}(1-|x|^{2N}e^{\zeta(x)}e^{\bar{u}_{n}+w_{n}(x)})dx+\int_{B_{1}(0)}\lambda_{n}|x|^{2N}e^{\zeta(x)}e^{\bar{u}_{n}+w_{n}(x)}dx\leq\Lambda$

and

(2.4) $|x|^{2N}e^{\zeta(x)}e^{\bar{u}_{n}+w_{n}(x)}\leq 1,\ \ osc_{\partial
B}w_{n}\leq C.$

Assume $0$ is the only blow-up point in $B_{1}(0)$ for sequence $w_{n}(x)$,
i.e.

(2.5) $\max_{K\subset\subset\overline{B_{1}(0)}\setminus\\{0\\}}|w_{n}(x)|\leq
C,\ \ \max_{\overline{B_{1}(0)}}|w_{n}(x)|\to+\infty.$

We will prove the following two theorems of a local version of Theorem 1.2.

######  Theorem 2.1.

Let $w_{n}$ be a sequence of solutions of (2.2) satisfying (2.3), (2.4) and
(2.5). Assume

$c=\lim_{n\to\infty}\lambda_{n}e^{\bar{u}_{n}}<\infty.$

If $0$ is a vortex point, i.e. $N>0$, then we have $c=0$ and the following two
alternatives hold:

  * (1)

if there is a singular bubble $\varphi(x)$ near the blow-up point $0$, then
$\varphi(x)$ is the only bubble, i.e. the simple blow-up property holds.
Moreover, we have the following energy identity

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=b,$

where

$b=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right)dx\geq
4(1+N).$

  * (2)

if there is a non-singular bubble $\varphi^{1}(x)$ near the blow-up point $0$,
then there must exist more than two bubbles $\\{\varphi^{i}\\}_{i=1}^{m}$ near
the blow-up point $0$. Moreover $\\{\varphi^{i}\\}_{i=1}^{m}$ contains no
singular bubble and satisfies the following energy identity

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=\sum_{i=1}^{m}b_{i},$

where

$b_{i}=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{\varphi^{i}(x)}\left(1-\delta_{i}e^{\varphi^{i}(x)}\right)dx\geq
4,\ \ i=1,...,m.$

Furthermore, energies $b_{1},...,b_{m}$ satisfy the following relation that

$\sum_{i=1}^{m}b_{i}^{2}=\left(\sum_{i=1}^{m}b_{i}\right)^{2}-4N\sum_{i=1}^{m}b_{i}.$

When $0$ is not a vortex point, i.e. $N=0$, we have

######  Theorem 2.2.

Let $w_{n}$ be a sequence of solutions of (2.2) satisfying (2.3), (2.4) and
(2.5). If $0$ is not a vortex point, i.e. $N=0$, then $c=0$ and there is only
one bubble $\varphi(x)$ near the blow-up point $0$, satisfying (1.8) and
following energy identity that

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=b,$

where

$b=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right)dx\geq 4.$

Next, we establish some lemmas which will be used in our later proof. We start
by proving the following lemma for a lower bound of energy concentration.

######  Lemma 2.3.

Let $w_{n}$ be a sequence of solutions of (2.2) satisfying (2.3), (2.4) and
(2.5). Then we have $c=0$ and

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)\geq 4(1+N).$

###### Proof.

By Theorem 1.1, we know $w_{n}\to w_{0}$ weakly in $W^{1,p}(B_{1}(0))$ and
strongly in $C^{2}_{loc}(B_{1}(0)\setminus\\{0\\})$, where $w_{0}$ satisfies

$-\Delta w_{0}=ce^{u_{0}}e^{w_{0}}+4\pi\beta\delta_{0},\ \ in\ \ B_{1}(0)$

and $\beta\geq 1$. Moreover,

$\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}(1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}})\to
ce^{u_{0}}e^{w_{0}}+4\pi\beta\delta_{0}$

in the sense of measure which implies that

$\alpha_{n}(\rho;0)=2\beta+o(1),$

where $\lim_{\rho\to 0}\lim_{n\to\infty}o(1)=0$.

For any $\rho\in(0,\frac{1}{2})$, multiplying (2.2) by $x\nabla w_{n}$ and
integrating by parts, we get

$\displaystyle\int_{\partial B_{\rho}(0)}r\left(\frac{1}{2}|\nabla
w_{n}|^{2}-|\frac{\partial w_{n}}{\partial r}|^{2}\right)d\theta=$
$\displaystyle\int_{\partial
B_{\rho}(0)}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]d\theta$
$\displaystyle-\int_{B_{\rho}(0)}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
$\displaystyle-\int_{B_{\rho}(0)}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
(2.6)
$\displaystyle-\int_{B_{\rho}(0)}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}(x\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx.$

Since we can write

$w_{0}(x)=-2\beta\log|x|+\phi(x)$

where $\phi(x)$ is a smooth function, then

$\int_{\partial B_{\rho}(0)}r\left(\frac{1}{2}|\nabla
w_{n}|^{2}-|\frac{\partial w_{n}}{\partial
r}|^{2}\right)d\theta=-4\pi\beta^{2}+o(1)=-\pi\alpha_{n}^{2}(\rho;0)+o(1).$

Noting that $|x\cdot\nabla\zeta|\leq C\rho$ yields

$\lim_{\rho\to
0}\lim_{n\to\infty}\int_{B_{\rho}(0)}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}(x\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx=0,$

we have

$\displaystyle-\pi\alpha_{n}^{2}(0;\rho)$
$\displaystyle\leq-\int_{B_{\rho}(0)}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
$\displaystyle\quad-\int_{B_{\rho}(0)}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx+o(1)$
$\displaystyle\leq-2(1+N)\int_{B_{\rho}(0)}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx=-4\pi(1+N)\alpha_{n}(\rho;0)+o(1)$

which implies that

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)\geq 4(1+N).$

Now, we claim that $c=0$. In fact, if not, by Theorem 1.1, we have $\beta\leq
N$. Then there holds

$4(1+N)\leq\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=2\beta\leq 2N.$

This is a contradiction. Thus, $c=0$. ∎

Next, we prove a proposition for a selection of bubbling areas. Different from
Liouville equations [18] where the singularity near the center $p$ of each
bubbling disk is $-2\log|x-p|$, Chern-Simon-Higgs equations will involve two
kinds of singularities which is important in our later proof.

######  Proposition 2.4.

Let $w_{n}$ be a sequence of solutions of (2.2) satisfying (2.3), (2.4) and
(2.5). Then there exist a sequence of finite points
$\Sigma_{n}:=\\{x^{1}_{n},x^{2}_{n},...,x^{m}_{n}\\}$ and a sequence of
positive numbers $\beta^{1}_{n},\beta^{2}_{n},...,\beta^{m}_{n}$ such that

  * (1)

$x^{i}_{n}\to 0$ and $\beta_{n}^{i}\to 0$ as $n\to\infty$,
$\beta^{i}_{n}\leq\frac{1}{2}dist(x^{i}_{n},\Sigma_{n}\setminus\\{x^{i}_{n}\\})$,
$i=1,...,m$;

  * (2)

$w_{n}(x^{i}_{n})=\max_{B_{\beta_{n}^{i}}(x^{i}_{n})}w_{n}(x)\to+\infty$ as
$n\to\infty$, $i=1,...,m$;

  * (3)

Take

$\displaystyle\mu_{n}^{i}:=\begin{cases}\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}e^{\zeta(x_{n}^{i})}\right)^{-\frac{1}{2(1+N)}},\
\ &if\ \
|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\leq
C;\\\
\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}|x^{i}_{n}|^{2N}e^{\zeta(x_{n}^{i})}\right)^{-\frac{1}{2}},\
\ &if\ \
|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\to\infty.\end{cases}$

Then $\frac{\mu_{n}^{i}}{\beta^{i}_{n}}\to 0$ as $n\to\infty$, $i=1,...,m$.

  * (4)

In each $B_{\beta_{n}^{i}}(x^{i}_{n})$, we define the scaled function

$\varphi^{i}_{n}(x):=w_{n}(x^{i}_{n}+\mu_{n}^{i}x)-w_{n}(x_{n}^{i}).$

Then we have the following two cases:

    * (4-1)

if
$|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\leq
C$, then $\varphi^{i}_{n}(x)\to\varphi^{i}(x)$ in
$C^{2}_{loc}(\mathbb{R}^{2})$ as $n\to\infty$, where $\varphi^{i}(x)$
satisfies (1.9).

    * (4-2)

if
$|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\to\infty$,
then $\varphi^{i}_{n}(x)\to\varphi^{i}(x)$ in $C^{2}_{loc}(\mathbb{R}^{2})$ as
$n\to\infty$, where $\varphi^{i}(x)$ satisfies (1.8).

  * (5)

There is at most one point $x^{i}_{n}\in\Sigma_{n}$ such that

$|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\leq
C.$

Furthermore, we have the following:

    * (5-1)

if there is one point $x^{i}_{n}\in\Sigma_{n}$ such that
$|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\leq
C,$ then we have

$0\in B_{\beta_{n}^{i}}(x_{n}^{i})\ \ and\ \
\lim\frac{|x_{n}^{j}|}{|x_{n}^{i}|}=\infty,\ \forall j\neq i.$

Moreover, for any positive constant $c_{0}>0$, there holds

$\displaystyle w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C(c_{0}),\ \
\forall|x-x_{n}^{i}|\geq c_{0}\mu_{n}^{i}.$

    * (5-2)

if for any $x^{i}_{n}\in\Sigma_{n}$, there holds
$|x^{i}_{n}|\left(\lambda_{n}e^{\bar{u}_{n}+w_{n}(x^{i}_{n})}\right)^{\frac{1}{2(1+N)}}\to\infty,$
then

$\displaystyle w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\ \ \forall\
x\in B_{1}(0).$

###### Proof.

Step 1: Construction of the first blow-up point $x_{n}^{1}$ and $\mu_{n}^{1}$.

Choose $x_{n}\in B_{\frac{1}{2}}(0)$ such that $x_{n}\to 0$ and

$w_{n}(x_{n})=\max_{\bar{B}}w_{n}(x)\to+\infty.$

Denote $\gamma_{n}:=w_{n}(x_{n})$ and

$\varphi_{n}(x)=w_{n}(x_{n}+\mu_{n}x)-\gamma_{n},$

where $\mu_{n}$ is a positive number going to zero which will be choosed
later.

A direct computation yields

$\displaystyle-\Delta\varphi_{n}(x)$
$\displaystyle=\mu_{n}^{2}\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|x_{n}+\mu_{n}x|^{2N}e^{\zeta(x_{n}+\mu_{n}x)}e^{\varphi_{n}(x)}(1-e^{\bar{u}_{n}+\gamma_{n}}|x_{n}+\mu_{n}x|^{2N}e^{\zeta(x_{n}+\mu_{n}x)}e^{\varphi_{n}(x)})$
$\displaystyle=\mu_{n}^{2(1+N)}\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|\frac{x_{n}}{\mu_{n}}+x|^{2N}e^{\zeta(x_{n}+\mu_{n}x)}e^{\varphi_{n}(x)}\left(1-\mu_{n}^{2N}e^{\bar{u}_{n}+\gamma_{n}}|\frac{x_{n}}{\mu_{n}}+x|^{2N}e^{\zeta(x_{n}+\mu_{n}x)}e^{\varphi_{n}(x)}\right),\
\ in\ \ B_{\frac{1}{2}\mu_{n}^{-1}}(0).$

Case 1: $\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|x_{n}|^{2(1+N)}\leq C$.

In this case, we first claim:

$\lambda_{n}^{-N}e^{\bar{u}_{n}+\gamma_{n}}\leq C.$

In fact, if not, passing to a subsequence, there holds
$\lim_{n\to\infty}\lambda_{n}^{-N}e^{\bar{u}_{n}+\gamma_{n}}=\infty$. Taking

$\mu_{n}=\left(\lambda_{n}e^{2(\bar{u}_{n}+\gamma_{n})}e^{2\zeta(x_{n})}\right)^{-\frac{1}{4N+2}},$

then we have

$\displaystyle\frac{|x_{n}|}{\mu_{n}}$
$\displaystyle=\left(|x_{n}|^{4N+2}\lambda_{n}e^{2(\bar{u}_{n}+\gamma_{n})}e^{2\zeta(x_{n})}\right)^{\frac{1}{4N+2}}$
$\displaystyle=\left(|x_{n}|^{2N+2}\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|x_{n}|^{2N}e^{\bar{u}_{n}+\gamma_{n}}e^{2\zeta(x_{n})}\right)^{\frac{1}{4N+2}}\leq
C\left(|x_{n}|^{2N}e^{\bar{u}_{n}+\gamma_{n}}e^{2\zeta(x_{n})}\right)^{\frac{1}{4N+2}}\leq
C$

where the last inequality follows from the fact that
$v_{n}(x_{n})=\bar{u}_{n}+\gamma_{n}+u_{0}(x_{n})\leq 0$, and

$\mu_{n}^{2(1+N)}\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}=e^{-\frac{2(N+1)}{2N+1}\zeta(x_{n})}\left(\lambda_{n}^{N}e^{-\bar{u}_{n}-\gamma_{n}}\right)^{\frac{1}{2N+1}}\to
0,$

which implies $\mu_{n}\to 0$, because
$\lambda^{-N}_{n}e^{\bar{u}_{n}+\gamma_{n}}\to\infty$.

Passing to a subsequence, we assume
$\lim_{n\to\infty}\frac{x_{n}}{\mu_{n}}=-x_{0}$. Since $\varphi_{n}(x)\leq 0,\
\varphi_{n}(0)=0$, by standard elliptic estimates, we know
$\varphi_{n}(x)\to\varphi(x),\ in\ C^{2}_{loc}(\mathbb{R}^{2})$ where
$\varphi(x)$ satisfies

$-\Delta\varphi(x)=-|x-x_{0}|^{4N}e^{2\varphi(x)},\ \ in\ \ \mathbb{R}^{2},$

where $\varphi(x)\leq 0$ and $\varphi(0)=0$. This is a contradiction to
maximal principle. We proved the claim.

Now, we choose
$\mu_{n}^{2(1+N)}\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}e^{\zeta(x_{n})}=1$,
i.e.

$\mu_{n}:=\left(\frac{1}{\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}e^{\zeta(x_{n})}}\right)^{\frac{1}{2(1+N)}}\to
0.$

It is easy to see that

$\frac{|x_{n}|}{\mu_{n}}\leq C$

and

$\mu_{n}^{2N}e^{\bar{u}_{n}+\gamma_{n}}e^{\zeta(x_{n})}\leq
C\left(\frac{e^{\bar{u}_{n}+\gamma_{n}}}{\lambda_{n}^{N}}\right)^{\frac{1}{1+N}}\leq
C.$

W.L.O.G, we assume

$\frac{x_{n}}{\mu_{n}}\to-x_{0},\ \
\mu_{n}^{2N}e^{\bar{u}_{n}+\gamma_{n}}e^{\psi(x_{n})}\to\delta.$

By a standard elliptic estimate, we know that

$\varphi_{n}(x)\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where

$\displaystyle-\Delta\varphi(x)=|x-x_{0}|^{2N}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right),\
\ x\in\mathbb{R}^{2},$

with

$\delta|x-x_{0}|^{2N}e^{\varphi(x)}\leq 1$

and

$\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right)dx+\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N}e^{\varphi(x)}dx\leq
C.$

By classification result in Lan-Li [16], we know

$\int_{\mathbb{R}^{2}}|x-x_{0}|^{2N}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right)dx=2\pi
b,$

and

$\varphi(x)=-b\log|x|+C+O(|x|^{-\varepsilon}),\ \ |x|\geq 2,$

where $\varepsilon\in(0,1)$, $b\geq 4(1+N)$ are two constants.

Taking $R_{n}\to\infty$ such that

$\lim\|\varphi_{n}(x)-\varphi(x)\|_{L^{\infty}(B_{2R_{n}}(0))}=0,$

which implies that

$w_{n}(x)+2(1+N)\log|x-x_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\ \
|x-x_{n}|\leq 2R_{n}\mu_{n}.$

Then, for any $c_{0}\mu_{n}\leq|x-x_{n}|\leq 2R_{n}\mu_{n}$ where $c_{0}>0$ is
a positive constant, since

$|x|\leq|x-x_{n}|+|x_{n}|\leq(1+\frac{C}{c_{0}})|x-x_{n}|,$

we get

$w_{n}(x)+2\log|x-x_{n}|+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq
C(c_{0}).$

Choosing $x_{n}^{1}=x_{n}$ and $\beta_{n}^{1}=R_{n}\mu_{n}$, we get the first
blow-up point.

Case 2: $\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|x_{n}|^{2(1+N)}\to\infty$.

In this case, we choose

$\mu_{n}:=\left(e^{-\gamma_{n}-\bar{u}_{n}-\log\lambda_{n}}|x_{n}|^{-2N}e^{-\zeta(x_{n})}\right)^{\frac{1}{2}},$

i.e.
$\gamma_{n}=-2\log\mu_{n}-2N\log|x_{n}|-\bar{u}_{n}-\log\lambda_{n}-\zeta(x_{n})$.

It is easy to see that

$\frac{|x_{n}|}{\mu_{n}}=\left(\lambda_{n}e^{\bar{u}_{n}+\gamma_{n}}|x_{n}|^{2(1+N)}e^{\zeta(x_{n})}\right)^{\frac{1}{2}}\to\infty.$

By a standard elliptic estimate, we know that

$\varphi_{n}(x)\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where

$\displaystyle-\Delta\varphi(x)=e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right),\ \ x\in\mathbb{R}^{2},$

with

$\delta e^{\varphi(x)}\leq 1$

and

$\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right)dx+\int_{\mathbb{R}^{2}}e^{\varphi(x)}dx\leq
C.$

By classification result in Lan-Li [16], we know

$\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right)dx=2\pi b,$

and

$\varphi(x)=-b\log|x|+C+O(|x|^{-\varepsilon}),\ \ |x|\geq 2,$

where $\varepsilon\in(0,1)$, $b\geq 4$ are two constants.

Take $R_{n}\to\infty$, such that

$R_{n}\mu_{n}\to 0,\ \ \frac{|x_{n}|}{R_{n}\mu_{n}}\to\infty,$

and

$\lim\|\varphi_{n}(x)-\varphi(x)\|_{L^{\infty}(B_{2R_{n}}(0))}=0,$

which implies

$w_{n}(x)+2\log|x-x_{n}|+2N\log|x_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\
\ |x-x_{n}|\leq 2R_{n}\mu_{n}.$

This immediately implies

$w_{n}(x)+2\log|x-x_{n}|+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\ \
|x-x_{n}|\leq 2R_{n}\mu_{n},$

because

$|x|\leq|x-x_{n}|+|x_{n}|\leq 2|x_{n}|.$

Choosing $x_{n}^{1}=x_{n}$ and $\beta_{n}^{1}=R_{n}\mu_{n}$, we get the first
blow-up point.

Step 2: Construction of second blow-up point $x_{n}^{2}$ and $\mu_{n}^{2}$.

If

$w_{n}(x)+2\log|x-x_{n}^{1}|+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\
\ \forall\ |x-x_{n}^{1}|\geq\beta_{n}^{1},$

then we have done. Otherwise, we have

$\sup_{\overline{B_{1}(0)\setminus
B_{\beta_{n}^{1}}(x_{n}^{1})}}w_{n}(x)+2\log|x-x_{n}^{1}|+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}=+\infty.$

Suppose $y_{n}$ is a maximal point such that

$w_{n}(y_{n})+2\log|y_{n}-x_{n}^{1}|+2N\log|y_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}=+\infty,\
|y_{n}-x_{n}^{1}|\geq 2R_{n}\mu_{n}.$

Denote $t_{n}:=\frac{1}{2}|y_{n}-x_{n}^{1}|$ and

$\psi_{n}(x):=w_{n}(x)+2\log[t_{n}-|x-y_{n}|]+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}},\
\ x\in B_{t_{n}}(y_{n}).$

Supposing $p_{n}\in B_{t_{n}}(y_{n})$ such that
$\psi_{n}(p_{n})=\sup_{\overline{B_{t_{n}}(y_{n})}}\psi_{n}(x),$ then

$\psi_{n}(p_{n})\geq\psi_{n}(y_{n})=w_{n}(y_{n})+2\log\frac{1}{2}|y_{n}-x_{n}^{1}|+2N\log|y_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\to+\infty.$

Denoting $s_{n}:=\frac{1}{2}[t_{n}-|p_{n}-y_{n}|]$, then we have

(2.7) $\psi_{n}(p_{n})=w_{n}(p_{n})+2\log
2s_{n}+2N\log|p_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\to+\infty$

and for any $x\in B_{s_{n}}(p_{n})$, there holds

$\displaystyle\psi_{n}(x)$
$\displaystyle=w_{n}(x)+2\log[t_{n}-|x-y_{n}|]+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq\psi_{n}(p_{n})=w_{n}(p_{n})+2\log
2s_{n}+2N\log|p_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}.$

Since

$t_{n}-|x-y_{n}|\geq s_{n},\ \ \forall x\in B_{s_{n}}(p_{n}),$

then we have

(2.8) $w_{n}(x)\leq w_{n}(p_{n})+2N\log\frac{|p_{n}|}{|x|}+2\log 2,\ \ \forall
x\in B_{s_{n}}(p_{n}).$

Denote $\tilde{\gamma}_{n}:=w_{n}(p_{n})$ and

$\varphi_{n}(x)=w_{n}(p_{n}+\tilde{\mu}_{n}x)-\tilde{\gamma}_{n},$

where $\tilde{\mu}_{n}$ is a positive number going to zero and satisfying
$\frac{s_{n}}{\tilde{\mu}_{n}}\to+\infty$, which will be choosed later.

A direct computation yields

$\displaystyle-\Delta\varphi_{n}(x)=\tilde{\mu}_{n}^{2(1+N)}\lambda_{n}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}|\frac{p_{n}}{\tilde{\mu}_{n}}+x|^{2N}e^{\zeta(p_{n}+\tilde{\mu}_{n}x)}e^{\varphi_{n}(x)}\left(1-\tilde{\mu}_{n}^{2N}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}|\frac{p_{n}}{\tilde{\mu}_{n}}+x|^{2N}e^{\zeta(p_{n}+\tilde{\mu}_{n}x)}e^{\varphi_{n}(x)}\right),\
\ in\ \ B_{\frac{s_{n}}{\tilde{\mu}_{n}}}(0).$

We have to consider the following cases.

Case 1-1: If Case 1 happens, we claim first:

(2.9)
$|p_{n}|e^{\frac{\tilde{\gamma}_{n}+\bar{u}_{n}+\log\lambda_{n}}{2(1+N)}}\to+\infty.$

In fact, since $\frac{|x_{n}^{1}|}{\mu_{n}}\leq C$, we have

$\displaystyle|p_{n}|\geq|p_{n}-x^{1}_{n}|-|x_{n}^{1}|\geq
R_{n}\mu_{n}-|x_{n}^{1}|\geq\frac{1}{2}R_{n}\mu_{n}$

and

$\displaystyle
2|s_{n}|=t_{n}-|p_{n}-y_{n}|=\frac{1}{2}|y_{n}-x^{1}_{n}|-|p_{n}-y_{n}|\leq\frac{1}{2}|p_{n}-x^{1}_{n}|\leq\frac{1}{2}|p_{n}|(1+\frac{|x^{1}_{n}|}{|p_{n}|})\leq|p_{n}|.$

Combing this with the fact (2.7) that

$w_{n}(p_{n})+2\log
s_{n}+2N\log|p_{n}|+\bar{u}_{n}+\log\lambda_{n}\to+\infty,$

we proved (2.9).

We claim secondly:

(2.10) $w_{n}(x)\leq w_{n}(p_{n})+2(1+N)\log 2,\ \ \forall\ x\in
B_{s_{n}}(p_{n}).$

In fact, since $|p_{n}|\geq 2s_{n}$ and for any $x\in B_{s_{n}}(p_{n})$, there
holds

$\frac{|p_{n}|}{|x|}\leq\frac{|p_{n}|}{|p_{n}|-|x-p_{n}|}\leq\frac{|p_{n}|}{|p_{n}|-s_{n}}\leq
2,$

then (2.10) follows immediately from (2.8).

Now, we take

$\tilde{\mu}_{n}:=\left(\frac{e^{-\tilde{\gamma}_{n}-\bar{u}_{n}-\log\lambda_{n}}e^{-\zeta(p_{n})}}{|p_{n}|^{2N}}\right)^{\frac{1}{2}},$

i.e.

$\tilde{\gamma}_{n}=-2\log\tilde{\mu}_{n}-2N\log|p_{n}|-\bar{u}_{n}-\log\lambda_{n}-\zeta(p_{n}).$

Then (2.7) yields $\tilde{\mu}_{n}\to 0$ and

$\displaystyle\frac{s_{n}}{\tilde{\mu}_{n}}=e^{\frac{1}{2}[\tilde{\gamma}_{n}+2\log
s_{n}+2N\log|p_{n}|+\bar{u}_{n}+\log\lambda_{n}+\zeta(p_{n})]}\to+\infty,$

which also implies that $\frac{|p_{n}|}{\tilde{\mu}_{n}}\to+\infty$.

By a standard elliptic analysis, we know that

$\varphi_{n}(x)\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where

$\displaystyle-\Delta\varphi(x)=e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right),\ \ x\in\mathbb{R}^{2},$

with

$\delta e^{\varphi(x)}\leq 1$

and

$\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta|x-x_{0}|^{2N}e^{\varphi(x)}\right)dx+\int_{\mathbb{R}^{2}}e^{\varphi(x)}dx\leq
C.$

By classification result in Lan-Li [16], we know

$\int_{\mathbb{R}^{2}}e^{\varphi(x)}\left(1-\delta
e^{\varphi(x)}\right)dx=2\pi b,$

and

$\varphi(x)=-b\log|x|+C+O(|x|^{-\varepsilon}),\ \ |x|\geq 2,$

where $\varepsilon\in(0,1)$, $b\geq 4$ are two constants.

Assume

$\varphi(\bar{x})=\max_{\mathbb{R}^{2}}\varphi(x).$

Take $R_{n}\to+\infty$, such that $R_{n}\tilde{\mu}_{n}=o(1)s_{n}$ and

$\|\varphi_{n}(x)-\varphi(x)\|_{L^{\infty}(B_{4R_{n}})}\to 0.$

Let $q_{n}\in B_{R_{n}}(0)$ such that

$\varphi_{n}(\bar{x}+q_{n})=\max_{\overline{B_{R_{n}}(0)}}\varphi_{n}(x).$

Now, setting

$x^{2}_{k}:=p_{n}+\tilde{\mu}_{n}(\bar{x}+q_{n}),\ \
\beta_{n}^{2}:=R_{n}\tilde{\mu}_{n},$

one can see that it satisfies all the conditions of proposition.

If Case 2 happens, then we have to consider the following two cases:

Case 2-1: $\lambda_{n}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}|p_{n}|^{2(1+N)}\leq
C$.

Similar to case 1, we first have

$\lambda_{n}^{-N}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}\leq C.$

In this case, taking

$\tilde{\mu}_{n}:=\left(\frac{1}{\lambda_{n}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}e^{\zeta(p_{n})}}\right)^{\frac{1}{2(1+N)}}\to
0,$

it is easy to see that

$\frac{|p_{n}|}{\tilde{\mu}_{n}}\leq C,\ \
\tilde{\mu}_{n}^{2N}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}\leq
C\left(\frac{e^{\bar{u}_{n}+\tilde{\gamma}_{n}}}{\lambda_{n}^{N}}\right)^{\frac{1}{1+N}}\leq
C.$

Since

$w_{n}(p_{n})+2(1+N)\log|p_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,$

then (2.7) yields

$\frac{s_{n}}{|p_{n}|}\to\infty$

and

$\frac{s_{n}}{\tilde{\mu}_{n}}=e^{\frac{1}{2(1+N)}\left(w_{n}(p_{n})+2(1+N)\log
s_{n}+\bar{u}_{n}+\log\lambda_{n}+\zeta(p_{n})\right)}\to\infty.$

W.L.O.G, we assume

$\frac{p_{n}}{\tilde{\mu}_{n}}\to-x_{0},\ \
\tilde{\mu}_{n}^{2N}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}e^{\zeta(p_{n})}\to\delta.$

From (2.8), we see that

$|\Delta\varphi_{n}(x)|\leq
C|\frac{p_{n}}{\tilde{\mu}_{n}}+x|^{2N}e^{\varphi_{n}(x)}\leq
C|\frac{p_{n}}{\tilde{\mu}_{n}}+x|^{2N}e^{2N\log\frac{p_{n}}{p_{n}+\tilde{\mu}_{n}x}}\leq
C\frac{|p_{n}|^{2N}}{\tilde{\mu}_{n}^{2N}}\leq C.$

Then a standard elliptic estimate yields $\varphi_{n}(x)\to\varphi(x)$ in the
sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi(x)$ satisfies (1.9). By
classification result in Lan-Li [16], we have

(2.11) $\varphi(x)=-b\log|x|+C+O(|x|^{-\varepsilon}),\ \ |x|\geq 2,$

where $\varepsilon\in(0,1)$, $b\geq 4(N+1)$ are two constants.

Taking $R_{n}\to\infty$ such that $R_{n}\tilde{\mu}_{n}=o(1)s_{n}$ and

$\lim\|\varphi_{n}(x)-\varphi(x)\|_{L^{\infty}(B_{4R_{n}}(0))}=0,$

which implies that

$w_{n}(x)+2(1+N)\log|x-p_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\ \
|x-p_{n}|\leq 4R_{n}\tilde{\mu}_{n}.$

Then, for any $c_{0}\tilde{\mu}_{n}\leq|x-p_{n}|\leq 4R_{n}\tilde{\mu}_{n}$
where $c_{0}>0$ is a positive constant, since

$|x|\leq|x-p_{n}|+|p_{n}|\leq|x-p_{n}|+C\tilde{\mu}_{n}\leq(1+\frac{1}{Cc_{0}})|x-p_{n}|,$

we get

$w_{n}(x)+2\log|x-p_{n}|+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C.$

Similarly to Case 1-1, we can take $x^{2}_{n}$ and $\beta^{2}_{n}$.

Case 2-2:
$\lambda_{n}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}|p_{n}|^{2(1+N)}\to\infty$.

In this case, we choose

$\tilde{\mu}_{n}:=\left(e^{-\tilde{\gamma}_{n}-\bar{u}_{n}-\log\lambda_{n}}|p_{n}|^{-2N}e^{-\zeta(p_{n})}\right)^{\frac{1}{2}},$

i.e.
$\tilde{\gamma}_{n}=-2\log\tilde{\mu}_{n}-2N\log|p_{n}|-\bar{u}_{n}-\log\lambda_{n}-\zeta(p_{n})$.

It is easy to see that

$\frac{|p_{n}|}{\tilde{\mu}_{n}}=\left(\lambda_{n}e^{\bar{u}_{n}+\tilde{\gamma}_{n}}|p_{n}|^{2(1+N)}e^{\zeta(p_{n})}\right)^{\frac{1}{2}}\to\infty.$

W.L.O.G, we may assume $s_{n}\leq\frac{1}{2}|p_{n}|$. Otherwise, we take
$\tilde{s}_{n}=\min\\{s_{n},\frac{1}{2}|p_{n}|\\}$ and consider in
$B_{\tilde{s}_{n}}(p_{n})$. Then we have

$\frac{s_{n}}{\tilde{\mu}_{n}}=e^{\frac{1}{2}\left(w_{n}(p_{n})+2\log
s_{n}+2N\log|p_{n}|+\bar{u}_{n}+\log\lambda_{n}+\zeta(p_{n})\right)}\to\infty,$

where the last limit follows from (2.7).

Since for any $x\in B_{s_{n}}(p_{n})$, there holds
$|x|\geq\frac{1}{2}|p_{n}|$. By (2.8), we have

$w_{n}(x)\leq w_{n}(p_{n})+C,\ \ \forall\ x\in B_{s_{n}}(p_{n}).$

By a standard elliptic estimate, we know that

$\varphi_{n}(x)\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi(x)$ satisfies
(1.8). By classification result in Lan-Li [16], we know

$\varphi(x)=-b\log|x|+C+O(|x|^{-\varepsilon}),\ \ |x|\geq 2,$

where $\varepsilon\in(0,1)$, $b\geq 4$ are two constants. Similarly to Case
1-1, we can take $x^{2}_{n}$ and $\beta^{2}_{n}$.

Step 3: By above two steps, we have defined the selection process.
Continuously, we consider the function

$w_{n}(x)+2\log
dist(x,\\{x_{n}^{1},x_{n}^{2}\\})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq
C,\ \ \forall\ |x-x_{n}^{i}|\geq\beta_{n}^{i},\ i=1,2.$

If it is uniformly bounded, then we stop and conclude the proposition for
$m=2$. Otherwise, using the same argument, we get $x_{n}^{3}$ and
$\beta_{n}^{3}$. Since each bubble area contributes a positive energy, this
process must stop after finite steps. We proved the proposition.

∎

######  Definition 2.5.

For $x_{n}^{i}\in\Sigma_{n}$, we call $(x_{n}^{i},\mu_{n}^{i})$ a bubble.
Specially, we call it a singular bubble if the first case of conclusion $(4)$
in Proposition 2.4 happens.

By using the singularities we got in Proposition 2.4 and a blow-up argument,
we prove following the oscillation estimate.

######  Lemma 2.6.

Let $w_{n}$ be a sequence of solutions of (2.2) satisfying (2.3), (2.4) and
(2.5). Denote

$osc_{\Omega}u:=\sup_{x,y\in\Omega}(u(x)-u(y))$

and $d_{n}(x):=dist(x,\Sigma_{n})$. Then we have:

$osc_{\frac{1}{2}d_{n}(x)}w_{n}\leq C,\ \ \forall\ x\in B\setminus\Sigma_{n},$

where $C$ is a constant independent of $x,\ n$.

###### Proof.

Since $0$ is the only blow-up point in $B_{1}(0)$ and $osc_{\partial
B}w_{n}\leq C$, by a standard estimate, we know

$osc_{B\setminus B_{\delta}}w_{n}\leq C(\delta),$

for any $\delta\in(0,1)$. Hence we just need to show the lemma in
$B_{\frac{1}{20}}$. Let $\phi_{n}(x)$ be the solution of

$\displaystyle\begin{cases}-\Delta\phi_{n}(x)=0,\ \ &in\ \ B_{1}(0),\\\
\phi_{n}(x)=w_{n}(x),\ \ &on\ \ \partial B_{1}(0).\end{cases}$

It is clear that

$osc_{B}\phi_{n}(x)\leq osc_{\partial B}w_{n}(x)\leq C.$

Let $\eta_{n}(x):=w_{n}(x)-\phi_{n}(x)$ and let

$G(x,y)=-\frac{1}{2\pi}\log|x-y|+H(x,y)$

be Green’s function on $B_{1}$ with respect to the Dirichlet boundary, where
$H(x,y)$ is a smooth harmonic function. Then

$\eta_{n}(x)=\int_{B}G(x,y)(-\Delta w_{n}(y))dy.$

Let $x_{0}\in B_{\frac{1}{20}}$ and $r_{n}:=dist(x_{0},\Sigma_{n})$. For any
$x_{1},x_{2}\in B_{\frac{1}{2}r_{n}}(x_{0})$, then

$\displaystyle\eta_{n}(x_{1})-\eta_{n}(x_{2})$
$\displaystyle=\int_{B}\left(G(x_{1},y)-G(x_{2},y)\right)(-\Delta w_{n}(y))dy$
$\displaystyle=\frac{1}{2\pi}\int_{B}\log\frac{|x_{1}-y|}{|x_{2}-y|}\Delta
w_{n}(y)dy+O(1).$

We divide the integral into two parts, i.e.
$B_{1}=B_{\frac{3}{4}r_{n}}(x_{0})\cup(B_{1}\setminus
B_{\frac{3}{4}r_{n}}(x_{0}))$. Noting that

$\left|\log\frac{|x_{1}-y|}{|x_{2}-y|}\right|\leq C,\ \ y\in B_{1}\setminus
B_{\frac{3}{4}r_{n}}(x_{0}),$

we have

$\left|\int_{B_{1}\setminus
B_{\frac{3}{4}r_{n}}(x_{0})}\log\frac{|x_{1}-y|}{|x_{2}-y|}\Delta
w_{n}(y)dy\right|\leq C\int_{B_{1}}|\Delta w_{n}(y)|dy\leq C.$

A direct computation yields

$\displaystyle\left|\int_{B_{1}\setminus
B_{\frac{3}{4}r_{n}}(x_{0})}\log\frac{|x_{1}-y|}{|x_{2}-y|}\Delta
w_{n}(y)dy\right|$
$\displaystyle=\left|\int_{B_{\frac{3}{4}r_{n}}(x_{0})}\log\frac{|x_{1}-y|}{|x_{2}-y|}\lambda_{n}e^{\bar{u}_{n}}|y|^{2N}e^{\zeta(y)}e^{w_{n}(y)}(1-e^{\bar{u}_{n}}|y|^{2N}e^{\zeta(y)}e^{w_{n}(y)})dy\right|$
$\displaystyle\leq
C\int_{B_{\frac{3}{4}r_{n}}(x_{0})}\left|\log\frac{|x_{1}-y|}{|x_{2}-y|}\right|\lambda_{n}e^{\bar{u}_{n}}|y|^{2N}e^{w_{n}(y)}dy$
$\displaystyle=C\int_{B_{\frac{3}{4}}(0)}\left|\log\frac{|x_{1}-x_{0}-r_{n}y|}{|x_{2}-x_{0}-r_{n}y|}\right|\lambda_{n}e^{\bar{u}_{n}}|x_{0}+r_{n}y|^{2N}e^{w_{n}(x_{0}+r_{n}y)}r_{n}^{2}dy$
$\displaystyle=C\int_{B_{\frac{3}{4}}(0)}\left|\log\frac{|\frac{x_{1}-x_{0}}{r_{n}}-y|}{|\frac{x_{2}-x_{0}}{r_{n}}-y|}\right|\lambda_{n}e^{\bar{u}_{n}}|x_{0}+r_{n}y|^{2N}e^{w_{n}(x_{0}+r_{n}y)}r_{n}^{2}dy.$

Now, we have to consider the following three cases:

Case 1: if there is no singular bubble or there is a singular bubble
$(x_{n}^{i},\mu_{n}^{i})$ such that
$r_{n}=dist(x_{0},\Sigma_{n})=|x_{0}-x_{n}^{i}|\geq\mu_{n}^{i}$.

In this case, we claim:

(2.12) $w_{n}(x_{0}+r_{n}y)+2\log
r_{n}+2N\log|x_{0}+r_{n}y|+\log\lambda_{n}e^{\bar{u}_{n}}\leq C,\ \ \forall\
y\in B_{\frac{3}{4}}(0).$

In fact, if there is no singular bubble, since

(2.13) $dist(x_{0}+r_{n}y,\Sigma_{n})\geq
dist(x_{0},\Sigma_{n})-dist(x_{0},x_{0}+r_{n}y)\geq\frac{1}{4}r_{n},\ \
\forall\ y\in B_{\frac{3}{4}}(0),$

then (2.12) follows immediately from Proposition 2.4.

If there is a singular bubble $(x_{n}^{i},\mu_{n}^{i})$, by Proposition 2.4,
(2.13) and the fact that

(2.14) $dist(x_{0}+r_{n}y,x^{i}_{n})\geq
dist(x_{0}+r_{n}y,\Sigma_{n})\geq\frac{1}{4}r_{n}=\frac{1}{4}dist(x_{0},\Sigma_{n})\geq\frac{1}{4}\mu_{n}^{i},\
\ \forall\ y\in B_{\frac{3}{4}}(0),$

we get (2.12).

Therefore, we get

$\left|\int_{B_{1}\setminus
B_{\frac{3}{4}r_{n}}(x_{0})}\log\frac{|x_{1}-y|}{|x_{2}-y|}\Delta
w_{n}(y)dy\right|\leq\int_{B_{\frac{3}{4}}(0)}\left|\log\frac{|\frac{x_{1}-x_{0}}{r_{n}}-y|}{|\frac{x_{2}-x_{0}}{r_{n}}-y|}\right|dy\leq
C,$

which implies

$|\eta_{n}(x_{1})-\eta_{n}(x_{2})|\leq C.$

Then the conclusion of the lemma follows.

Case 2: if there is a singular bubble $(x_{n}^{i},\mu_{n}^{i})$ such that
$r_{n}=dist(x_{0},\Sigma_{n})=|x_{0}-x_{n}^{i}|\leq\mu_{n}^{i}$.

In this case, there must be that

$r_{n}=dist(x_{0},\Sigma_{n})=|x_{0}-x_{n}^{i}|.$

By Proposition 2.4, we know

$\varphi_{n}(x)=w_{n}(x_{n}^{i}+\mu_{n}^{i}x)-w_{n}(x_{n}^{i})\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi(x)$ satisfies

$|\varphi(x)|\leq C+b\log(1+|x|).$

It is easy to see that, when $|x_{0}-x_{n}^{i}|\leq\mu_{n}^{i}$, there holds

$osc_{\frac{1}{2}d_{n}(x_{0})}w_{n}(x)=osc_{B_{\frac{|x_{0}-x_{n}^{i}|}{2}}(x_{0})}w_{n}(x)=osc_{B_{\frac{|x_{0}-x_{n}^{i}|}{2\mu_{n}^{i}}}(\frac{x_{0}-x_{n}^{i}}{\mu_{n}^{i}})}\varphi_{n}(x)\leq
osc_{B_{2}(0)}\varphi_{n}(x)\leq C,$

which implies immediately the conclusion of the lemma.

Case 3: if there is a singular bubble $(x_{n}^{i},\mu_{n}^{i})$ and
$r_{n}=dist(x_{0},\Sigma_{n})=|x_{0}-x_{n}^{j}|$ where $j\neq i$.

Then we have the two following cases.

$(i)$ $\frac{r_{n}}{\mu_{n}^{i}}\to+\infty$. In this case, it is easy to see
that (2.14) still holds and then the conclusion of the lemma follows as in
Case 1.

$(i)$ $\frac{r_{n}}{\mu_{n}^{i}}\leq C$. In this case, by proposition 2.4, we
know $\frac{|x_{n}^{i}-x_{n}^{j}|}{\mu_{n}^{i}}\to+\infty$, which implies

$dist(x_{0}+r_{n}y,x_{n}^{i})\geq|x_{0}-x_{n}^{i}|-\frac{3}{4}r_{n}\geq|x_{n}^{i}-x_{n}^{j}|-|x_{0}-x_{n}^{j}|-\frac{3}{4}r_{n}=|x_{n}^{i}-x_{n}^{j}|-\frac{7}{4}r_{n}\geq\mu_{n}^{i}.$

Then the conclusion of the lemma follows as in Case 1.

We proved this lemma. ∎

Before stating the next lemma, we first extend definitions of fast decay and
slow decay of [21] for Liouville type system to Chern-Simon-Higgs equation.

######  Definition 2.7.

We say $w_{n}$ has fast decay on $\partial B_{r_{n}}(x_{n})$ (resp.
$B_{r_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$) if

$w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-N_{n},\
\forall x\in\partial B_{r_{n}}(x_{n})\ \ (resp.\ \forall x\in
B_{r_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})),$

for some $N_{n}\to\infty$ as $n\to\infty$.

Next, we derive a fast decay lemma. For classical Liouville equation or Toda
system without singular sources (i.e. $N=0$), it is a direct corollary of
oscillation Lemma 2.6, see [21]. Here, in our case, we need to deal with the
singular term $2N\log|x|$.

######  Lemma 2.8.

Let $\Sigma^{1}_{n}\subset\Sigma_{n}$ be a subset of $\Sigma_{n}$ with

$\Sigma^{1}_{n}\subset B_{r_{n}}(x_{n})\subset B_{1}(0).$

Suppose

$dist(\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n}))=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})).$

Then for any $s_{n}\geq 2r_{n}$ with
$s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial B_{r_{n}}(x_{n}))$,
if $u_{n}$ has fast decay on $\partial B_{s_{n}}(x_{n})$, then for any
$\beta_{n}\to\infty$ with
$\beta_{n}s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n}))$, there exists $\alpha_{n}\to\infty$ with
$\alpha_{n}=o(1)\beta_{n}$ such that $w_{n}$ has fast decay in
$B_{\alpha_{n}s_{n}}\setminus B_{s_{n}}$, i.e.

$w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-N_{n},\
\forall s_{n}\leq|x-x_{n}|\leq\alpha_{n}s_{n},$

for some $N_{n}\to\infty$, and

$\int_{B_{\alpha_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}(x)}dx=o(1).$

###### Proof.

Since $w_{n}(x)$ has fast decay on $\partial B_{s_{n}}(x_{n})$, we have

$w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-N_{n},\
\forall|x-x_{n}|=s_{n},$

for some $N_{n}\to+\infty$.

Next, we claim: for any fixed $\Lambda>0$, there holds

(2.15) $w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-
N_{n}+C(\Lambda),\ \forall s_{n}\leq|x-x_{n}|\leq\Lambda s_{n}.$

In fact, by Lemma 2.6, for any $\Lambda>1$, we get

(2.16) $osc_{B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})}w_{n}(x)\leq
C(\Lambda).$

Noting that for any $x\in B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$,
there hold

(2.17) $\displaystyle dist(x,\Sigma_{n})$
$\displaystyle\leq|x-x_{n}|+dist(x_{n},\Sigma_{n})\leq(\Lambda+\frac{1}{2})s_{n},$
(2.18) $\displaystyle dist(x,\Sigma_{n})$ $\displaystyle\geq
s_{n}-r_{n}\geq\frac{1}{2}s_{n},$

which implies

(2.19) $osc_{B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})}\log
dist(x,\Sigma_{n})\leq C(\Lambda).$

Now, we consider the following two cases:

Case 1: $|x_{n}|\geq 2s_{n}$.

In this case, for any $x\in B_{\Lambda s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})$, there holds

$\displaystyle|x|$ $\displaystyle\leq|x-x_{n}|+|x_{n}|\leq\Lambda
s_{n}+|x_{n}|\leq(1+\frac{\Lambda}{2})|x_{n}|.$

Taking any point $p_{n}\in\partial B_{s_{n}}(x_{n})$, since
$|p_{n}|\geq|x_{n}|-|p_{n}-x_{n}|=|x_{n}|-s_{n}\geq\frac{1}{2}|x_{n}|$, thus,
we obtain

$\displaystyle w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq w_{n}(p_{n})+2\log
dist(p_{n},\Sigma_{n})+2N\log|p_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}+2N\log\frac{|x|}{|p_{n}|}+C(\Lambda)$
$\displaystyle\leq-N_{k}+C(\Lambda),\ \forall s_{n}\leq|x-x_{n}|\leq\Lambda
s_{n},$

which implies (2.15) immediately.

Case 2: $|x_{n}|\leq 2s_{n}$.

In this case, we first choose a point $y_{n}\in\partial B_{s_{n}}(x_{n})$ such
that $|y_{n}|\geq s_{n}$. Then we have

$w_{n}(y_{n})+2\log
dist(y_{n},\Sigma_{n})+2N\log|y_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-
N_{k}.$

Noting that for any $x\in B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$,
if $|x|\leq\frac{1}{2}s_{n}$, then by (2.16)-(2.18), we have

$\displaystyle w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq w_{n}(y_{n})+2\log
dist(y_{n},\Sigma_{n})+2N\log|y_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq-N_{k}+C(\Lambda).$

For $x\in B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$, if
$|x|\geq\frac{1}{2}s_{n}$, since

$\frac{1}{2}s_{n}\leq|x|\leq|x-x_{n}|+|x_{n}|\leq(\Lambda+2)s_{n},$

we get

$\displaystyle osc_{x\in B_{\Lambda s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n}),\
|x|\geq\frac{1}{2}s_{n}}\log|x|\leq C(\Lambda).$

Combing this with (2.16)-(2.18), we have

$\displaystyle w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq w_{n}(y_{n})+2\log
dist(y_{n},\Sigma_{n})+2N\log|y_{n}|+\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq-N_{k}+C(\Lambda).$

Thus, we proved (2.15).

By (2.15), it is easy to see that, for any $\beta_{n}\to\infty$, there exists
$\alpha^{1}_{n}\to\infty,\ \alpha^{1}_{n}=o(1)\beta_{n}$ such that

(2.20) $w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-\frac{1}{2}N_{n},\
\forall s_{n}\leq|x-x_{n}|\leq\alpha^{1}_{n}s_{n}.$

Now, we can choose $\alpha_{n}\to\infty,\ \ \alpha_{n}=o(1)\alpha^{1}_{n}$ and
$e^{-\frac{1}{2}N_{n}}\log\alpha_{n}=o(1)$. Then we have

$\displaystyle\int_{B_{\alpha_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}(x)}dx$
$\displaystyle\leq
Ce^{-\frac{1}{2}N_{n}}\int_{B_{\alpha_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}e^{-2\log dist(x,\Sigma_{n})}dx$ $\displaystyle\leq
Ce^{-\frac{1}{2}N_{n}+2\log 2}\int_{B_{\alpha_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}e^{-2\log|x-x_{n}|}dx$ $\displaystyle\leq
Ce^{-\frac{1}{2}N_{n}+2\log 2}\log\alpha_{n}=o(1),$

where the second inequality follows from the fact that

$dist(x,\Sigma_{n}^{1})\geq|x-x_{n}|-dist(x_{n},\Sigma^{1}_{n})\geq|x-x_{n}|-r_{n}\geq|x-x_{n}|-\frac{1}{2}s_{n}\geq\frac{1}{2}|x-x_{n}|.$

We prove the lemma.

∎

######  Lemma 2.9.

Let $\Sigma^{1}_{n}\subset\Sigma_{n}$ be a subset of $\Sigma_{n}$ with

$\Sigma^{1}_{n}\subset B_{r_{n}}(x_{n})\subset B_{1}(0).$

Suppose

$dist(\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n}))=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})).$

For any $s_{n}\geq 2r_{n}$ and $R_{n}\to\infty$ satisfying

$R_{n}s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})),$

if $w_{n}$ has fast decay in $B_{R_{n}s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$
and

(2.21) $\int_{B_{R_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}dx=o(1),$

then we have

$\nabla
w_{n}(x)=-\frac{x-x_{n}}{|x-x_{n}|^{2}}\alpha_{n}(s_{n};x_{n})+\frac{o(1)}{|x-x_{n}|},\
\ x\in\partial B_{\sqrt{R_{n}}s_{n}}(x_{n}).$

###### Proof.

The proof of this lemma is more or less standard. Here, for reader’s
convenience, we give a detailed proof.

Let $\phi_{n}(x)$ be the solution of

$\displaystyle\begin{cases}-\Delta\phi_{n}(x)=0,\ \ &in\ \ B_{1}(0),\\\
\phi_{n}(x)=w_{n}(x),\ \ &on\ \ \partial B_{1}(0).\end{cases}$

A standard elliptic theory yields that $\|w_{n}(x)\|_{C^{1}(B_{1}\setminus
B_{\delta})}\leq C(\delta)$ for any $\delta>0$. Then it is clear that

$\|\phi_{n}(x)\|_{C^{1}(B_{1}(0))}\leq\|w_{n}(x)\|_{C^{1}(\partial B)}\leq C.$

Let $\eta_{n}(x):=w_{n}(x)-\phi_{n}(x)$, then

$\eta_{n}(x)=\int_{B}G(x,y)(-\Delta w_{n}(y))dy,$

where $G(x,y)$ is the Green function as in Lemma 2.6. Thus,

$\nabla\eta_{n}(x)=-\frac{1}{2\pi}\int_{B}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy.$

Next, we divide the integral domain $B_{1}(0)$ as follows:

$\displaystyle B_{1}(0)=$ $\displaystyle B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})\cup B_{R_{n}s_{n}}(x_{n})\setminus
B_{\frac{3}{2}\sqrt{R_{n}}s_{n}}(x_{n})\cup
B_{\frac{3}{2}\sqrt{R_{n}}s_{n}}(x_{n})\setminus
B_{\frac{1}{2}\sqrt{R_{n}}s_{n}}(x_{n})$ $\displaystyle\cup
B_{\frac{1}{2}\sqrt{R_{n}}s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})\cup
B_{s_{n}}(x_{n}).$

For $x\in\partial B_{\sqrt{R_{n}}s_{n}}(x_{n})$ and $y\in B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})$, since

$\frac{x-y}{|x-y|^{2}}=\frac{o(1)}{|x-x_{n}|},$

then we immediately have

$-\frac{1}{2\pi}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy=\frac{o(1)}{|x-x_{n}|}.$

For $y\in B_{R_{n}s_{n}}(x_{n})\setminus
B_{\frac{3}{2}\sqrt{R_{n}}s_{n}}(x_{n})$, since

$|\frac{x-y}{|x-y|^{2}}|\leq\frac{C}{|x-x_{n}|},$

by (2.21), we have

$\left|\frac{1}{2\pi}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy\right|\leq\frac{C}{|x-x_{n}|}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}(-\Delta w_{n}(y))dy=\frac{o(1)}{|x-x_{n}|}.$

For $y\in B_{\frac{3}{2}\sqrt{R_{n}}s_{n}}(x_{n})\setminus
B_{\frac{1}{2}\sqrt{R_{n}}s_{n}}(x_{n})$, noting that

$B_{\frac{3}{2}\sqrt{R_{n}}s_{n}}(x_{n})\setminus
B_{\frac{1}{2}\sqrt{R_{n}}s_{n}}(x_{n})\subset\bigcup_{j=1}^{J}B_{\frac{|x-x_{n}|}{2}}(z_{j}),$

where $z_{j}\in\partial B_{\sqrt{R_{n}}s_{n}}(x_{n})$, $J$ is a universal
constant independent of $n$, since $w_{n}$ has fast decay in
$B_{R_{n}s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$, we have

$|\Delta w_{n}(x)|\leq
C\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}\leq\frac{o(1)}{|x-x_{n}|^{2}}.$

Then,

$\displaystyle\left|\frac{1}{2\pi}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy\right|\leq\frac{o(1)}{|x-x_{n}|^{2}}\sum_{j=1}^{J}\int_{B_{\frac{|x-x_{n}|}{2}}(z_{j})}\frac{1}{|x-y|}dy=\frac{o(1)}{|x-x_{n}|}.$

For $y\in B_{\frac{1}{2}\sqrt{R_{n}}s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$,
since

$|\frac{x-y}{|x-y|^{2}}|\leq\frac{C}{|x-x_{n}|},$

by (2.21), we have

$\left|\frac{1}{2\pi}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy\right|\leq\frac{C}{|x-x_{n}|}\int_{B_{1}(0)\setminus
B_{R_{n}s_{n}}(x_{n})}(-\Delta w_{n}(y))dy=\frac{o(1)}{|x-x_{n}|}.$

For $y\in B_{s_{n}}(x_{n})$, since

$\frac{x-y}{|x-y|^{2}}=\frac{x-x_{n}}{|x-x_{n}|^{2}}+\frac{o(1)}{|x-x_{n}|},$

we have

$-\frac{1}{2\pi}\int_{B_{s_{n}}(x_{n})}\frac{x-y}{|x-y|^{2}}(-\Delta
w_{n}(y))dy=-\frac{x-x_{n}}{|x-x_{n}|^{2}}\alpha(s_{n};x_{n})+\frac{o(1)}{|x-x_{n}|}.$

This immediately implies the conclusion of lemma. ∎

Now, at this position, we derive two kinds of Pohozaev identities (local type
and global type) for Chern-Simon-Higgs equation which play an important role
in our proof.

######  Lemma 2.10.

Let $\Sigma^{1}_{n}\subset\Sigma_{n}$ be a subset of $\Sigma_{n}$ with

$\Sigma^{1}_{n}\subset B_{r_{n}}(x_{n})\subset B_{1}(0).$

Suppose

$dist(\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n}))=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})).$

Then for any $s_{n}\geq 2r_{n}$ with
$s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial B_{r_{n}}(x_{n}))$,
if $u_{n}$ has fast decay on $\partial B_{s_{n}}(x_{n})$, we have the
following Pohozaev identities:

  * (1)

local type: if $\lim\frac{|x_{n}|}{s_{n}}=\infty$, then

$\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi\alpha^{2}_{n}(s_{n};x_{n})+o(1).$

  * (2)

global type: if $\frac{|x_{n}|}{s_{n}}\leq C$, then

$\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi\alpha^{2}_{n}(s_{n};x_{n})-4\pi
N\alpha_{n}(s_{n};x_{n})+o(1).$

###### Proof.

We divide the proof into two steps according two cases in the lemma.

Step 1: If $\lim\frac{|x_{n}|}{s_{n}}=\infty$.

By Lemma 2.8, we know there exists

$R_{n}\to\infty,\ \ R_{n}s_{n}=o(1)|x_{n}|,\ \
R_{n}s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})),$

such that $w_{n}$ has fast decay in $B_{R_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})$, i.e.

$w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-N_{n},\
\forall s_{n}\leq|x-x_{n}|\leq R_{n}s_{n},$

for some $N_{n}\to\infty$ and

(2.22) $\int_{B_{R_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}(x)}dx=o(1).$

Multiplying (2.2) by $(x-x_{n})\nabla w_{n}$ and integrating by parts, we get
the following Pohozaev type identity that

$\displaystyle\int_{\partial B_{s}(x_{n})}r\left(\frac{1}{2}|\nabla
w_{n}|^{2}-|\frac{\partial w_{n}}{\partial r}|^{2}\right)d\theta=$
$\displaystyle\int_{\partial
B_{s}(x_{n})}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]d\theta$
$\displaystyle-\int_{B_{s}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
$\displaystyle-\int_{B_{s}(x_{n})}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N-2}(x\cdot(x-x_{n}))e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
(2.23)
$\displaystyle-\int_{B_{s}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}((x-x_{n})\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$

for any $s_{n}\leq s\leq R_{n}s_{n}$, where $(r,\theta)$ is the polar
coordinate system centered at point $x_{n}$.

Now, we take $s=\sqrt{R_{n}}s_{n}$ in (2). Since $w_{n}$ has fast decay on
$\partial B_{\sqrt{R_{n}}s_{n}}(x_{n})$, we have

$\displaystyle\int_{\partial
B_{s}(x_{n})}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]\leq
C\int_{\partial B_{s}(x_{n})}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}\leq
Ce^{-N_{n}}=o(1).$

Noting that for any $x\in B_{\sqrt{R_{n}}s_{n}}(x_{n})$, there holds

$|x-x_{n}|\leq\sqrt{R_{n}}s_{n}\leq o(1)|x_{n}|\leq o(1)(|x|+|x-x_{n}|),$

which implies

$|x-x_{n}|=o(1)|x|.$

Then

$\displaystyle\left|\int_{B_{s}(x_{n})}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N-2}(x\cdot(x-x_{n}))e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx\right|$
$\displaystyle\leq
o(1)\int_{B_{s}(x_{n})}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}dx=o(1).$

Since $|\nabla\zeta|\leq C$, then

$\left|\int_{B_{s}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}((x-x_{n})\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx\right|\leq
Cs.$

Combining these together, by (2), we arrived at

$\displaystyle\int_{\partial
B_{\sqrt{R_{n}}s_{n}}(x_{n})}r\left(\frac{1}{2}|\nabla
w_{n}|^{2}-|\frac{\partial w_{n}}{\partial r}|^{2}\right)d\theta$
$\displaystyle=-\int_{B_{\sqrt{R_{n}}s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx+o(1)$
(2.24)
$\displaystyle=-\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx+o(1),$

where the last equality follows from (2.22).

By lemma 2.9, we know

$\nabla
w_{n}(x)=-\frac{x-x_{n}}{|x-x_{n}|^{2}}\alpha_{n}(s_{n};x_{n})+\frac{o(1)}{|x|},\
\ x\in\partial B_{\sqrt{R_{n}}s_{n}}.$

Then we conclude

(2.25)
$\displaystyle\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx=\pi\alpha^{2}_{n}(s_{n};x_{n})+o(1),$

which yields the first conclusion of lemma.

Step 2: If $\frac{|x_{n}|}{s_{n}}\leq C_{1}$.

By Lemma 2.8, we know there exists

$R_{n}\to\infty,\ \
R_{n}s_{n}=o(1)dist(\Sigma_{n}\setminus\Sigma^{1}_{n},\partial
B_{r_{n}}(x_{n})),$

such that $w_{n}$ has fast decay in $B_{2R_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})$, i.e.

$w_{n}(x)+2\log
dist(x,\Sigma_{n})+2N\log|x|+\log\lambda_{n}e^{\bar{u}_{n}}\leq-N_{n},\
\forall s_{n}\leq|x-x_{n}|\leq 2R_{n}s_{n},$

for some $N_{n}\to\infty$ and

(2.26) $\int_{B_{2R_{n}s_{n}}(x_{n})\setminus
B_{s_{n}}(x_{n})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}(x)}dx=o(1).$

Multiplying (2.2) by $x\nabla w_{n}$ and integrating by parts, we get the
following Pohozaev type identity that

$\displaystyle\int_{\partial B_{s}(0)}r\left(\frac{1}{2}|\nabla
w_{n}|^{2}-|\frac{\partial w_{n}}{\partial r}|^{2}\right)d\theta=$
$\displaystyle\int_{\partial
B_{s}(0)}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]d\theta$
$\displaystyle-\int_{B_{s}(0)}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
$\displaystyle-\int_{B_{s}(0)}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
(2.27)
$\displaystyle-\int_{B_{s}(0)}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}(x\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$

for any $0\leq s\leq 1$, where $(r,\theta)$ is the polar coordinate system
centered at point $0$.

Take $s=\sqrt{R_{n}}s_{n}$ in (2). Noting that

$B_{s_{n}}(x_{n})\subset B_{\sqrt{R_{n}}s_{n}}(0)\subset
B_{2R_{n}s_{n}}(x_{n})$

and

$dist(x,\Sigma_{n})=dist(x,\Sigma^{1}_{n})\geq\sqrt{R_{n}}s_{n}-|x_{n}|-r_{n}\geq\frac{1}{2}\sqrt{R_{n}}s_{n},\
\ \forall\ x\in\partial B_{s}(0),$

when $n$ is big, since $w_{n}$ has fast decay in
$B_{2R_{n}s_{n}}(x_{n})\setminus B_{s_{n}}(x_{n})$, we have

$\displaystyle\int_{\partial
B_{s}(0)}r\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]d\theta\leq
Ce^{-N_{n}}\frac{s^{2}}{\min_{\partial B_{s}(0)}dist^{2}(x,\Sigma_{n})}=o(1).$

By (2.26), we have

$\int_{B_{\sqrt{R_{n}}s_{n}}(0)}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx=\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx+o(1)$

and

$\int_{B_{\sqrt{R_{n}}s_{n}}(0)}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx=\int_{B_{s_{n}}(x_{n})}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx+o(1).$

Since $|\nabla\zeta|\leq C$, then

$\left|\int_{B_{s}(0)}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}(x\cdot\nabla\zeta)e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx\right|\leq
C\sqrt{R_{n}}s_{n}=o(1).$

Then we arrive at

$\displaystyle\int_{\partial
B_{\sqrt{R_{n}}s_{n}}(0)}r\left(\frac{1}{2}|\nabla w_{n}|^{2}-|\frac{\partial
w_{n}}{\partial r}|^{2}\right)d\theta$
$\displaystyle=-\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx$
(2.28)
$\displaystyle\quad-\int_{B_{s_{n}}(x_{n})}2N\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}[1-e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}]dx.$

By lemma 2.9, we know

$\nabla
w_{n}(x)=-\frac{x-x_{n}}{|x-x_{n}|^{2}}\alpha_{n}(s_{n};x_{n})+\frac{o(1)}{|x-x_{n}|}=-\frac{x}{|x|^{2}}\alpha_{n}(s_{n};x_{n})+\frac{o(1)}{|x|},\
\ x\in\partial B_{\sqrt{R_{n}}s_{n}}(0).$

Then we conclude

(2.29)
$\displaystyle\int_{B_{s_{n}}(x_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}]dx=\pi\alpha^{2}_{n}(s_{n};x_{n})-4\pi
N\alpha_{n}(s_{n};x_{n})+o(1).$

We finished the proof of this lemma.

∎

Let $x^{i}_{n}\in\Sigma_{n}$. By Proposition 2.4, we know

$\varphi^{i}_{n}(x)=w_{n}(x^{i}_{n}+\mu_{n}^{i}x)-w_{n}(x^{i}_{n})\to\varphi^{i}(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi^{i}(x)$
satisfies (1.8) or (1.9). Denote

$\int_{\mathbb{R}^{2}}|x-x_{0}^{i}|^{2N}e^{\varphi^{i}(x)}\left(1-\delta_{i}|x-x_{0}^{i}|^{2N}e^{\varphi(x)}\right)dx=2\pi
b_{i}$

or

$\int_{\mathbb{R}^{2}}e^{\varphi^{i}(x)}\left(1-\delta_{i}e^{\varphi^{i}(x)}\right)dx=2\pi
b_{i}$

corresponding to the fact that $(x_{n}^{i},\mu_{n}^{i})$ is a singular bubble
or not.

At the end of this section, we prove following local energy identity.

######  Proposition 2.11.

Denote

$\tau^{i}_{n}:=\frac{1}{2}dist(x^{i}_{n},\Sigma_{n}\setminus\\{x^{i}_{n}\\}).$

Then the following two alternatives hold:

  * (1)

if $(x^{i}_{n},\mu_{n}^{i})$ is a singular bubble, then $w_{n}(x)$ has fast
decay on $\partial B_{\tau^{i}_{n}}(x^{i}_{n})$ and

$\alpha_{n}(\tau_{n}^{i};x^{i}_{n})=b_{i}+o(1)$

and

$\int_{B_{\tau^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{i}^{2}-4\pi Nb_{i}+o(1).$

  * (2)

if $(x^{i}_{n},\mu_{n}^{i})$ is not a singular bubble, denoting
$s_{n}^{i}:=\min\\{\tau_{n}^{i},\frac{3}{4}|x_{n}^{i}|\\}$, then $w_{n}(x)$
has fast decay on $\partial B_{s^{i}_{n}}(x^{i}_{n})$ and

$\alpha_{n}(s_{n}^{i};x^{i}_{n})=b_{i}+o(1)$

and

$\int_{B_{s^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{i}^{2}+o(1).$

###### Proof.

We divide the proof into two cases.

Case 1: If $(x^{i}_{n},\mu_{n}^{i})$ is a singular bubble.

In this case, by Proposition 2.4, we know $\frac{|x^{i}_{n}|}{\mu^{i}_{n}}\leq
C$ and

$\varphi_{n}(x):=w_{n}(x_{n}^{i}+\mu_{n}^{i}x)-w_{n}(x_{n}^{i})\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi(x)$ satisfies
(1.9).

Now, we claim:

(2.30) $\alpha_{n}(\tau^{i}_{n};x_{n}^{i})=b_{i}+o(1).$

In fact, by the proof of Proposition 2.4, we can choose $R_{n}\to\infty$ such
that $R_{n}\mu_{n}^{i}=o(1)\tau^{i}_{n}$, $w_{n}(x)$ has fast decay on
$\partial B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})$ and

(2.31) $\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})=b_{i}+o(1),$

where $b_{i}\geq 4(1+N)$.

Denote

(2.32) $\bar{w}_{n}(r):=\frac{1}{2\pi r}\int_{\partial
B_{r}(x^{i}_{n})}w_{n}(x)d\sigma,$

where $(r,\theta)$ is the polar coordinates system centered at point
$x^{i}_{n}$. Then we have

$\displaystyle\frac{d}{dr}\bar{w}_{n}=\frac{1}{2\pi
r}\int_{B_{r}(x_{n}^{i})}\Delta w_{n}(x)dx=-\frac{\alpha_{n}(r;x^{i}_{n})}{r}$
$\displaystyle\leq-\frac{b_{i}+o(1)}{r}$ (2.33)
$\displaystyle\leq-\frac{4+4N+o(1)}{r},\ \ \forall\ \mu_{n}^{i}R_{n}\leq
r\leq\tau^{i}_{n},$

which implies

$\bar{w}_{n}(r)+2(1+N)\log r,\ \ \forall\ \mu_{n}^{i}R_{n}\leq
r\leq\tau^{i}_{n}$

and

$\bar{w}_{n}(r)+3+2N\log r,\ \ \forall\ \mu_{n}^{i}R_{n}\leq
r\leq\tau^{i}_{n}$

are decreasing functions with respect to $r$.

This together with the facts that $w_{n}(x)$ has fast decay on $\partial
B_{\mu_{n}^{i}R_{n}}(x^{i}_{n})$, $\frac{|x^{i}_{n}|}{\mu^{i}_{n}}\leq C$ and

$\frac{1}{2}|x-x_{n}^{i}|\leq|x|\leq 2|x-x_{n}^{i}|,\ \ \forall\
\mu_{n}^{i}R_{n}\leq|x-x_{n}^{i}|\leq\tau^{i}_{n},$

taking arbitrary point $p_{n}\in\partial B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})$, for
any $\mu_{n}^{i}R_{n}\leq|x-x_{n}^{i}|\leq\tau^{i}_{n}$, by Lemma 2.6, we
immediately get that

$\displaystyle w_{n}(x)+2\log r+2N\log|x|+2\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq\bar{w}_{n}(r)+2\log r+2N\log
r+2\log\lambda_{n}e^{\bar{u}_{n}}+C,\ \ (\mbox{here }r=|x-x_{n}^{i}|)$
$\displaystyle\leq\bar{w}_{n}(\mu_{n}^{i}R_{n})+2\log(\mu_{n}^{i}R_{n})+2N\log(\mu_{n}^{i}R_{n})+2\log\lambda_{n}e^{\bar{u}_{n}}+C$
$\displaystyle\leq w_{n}(p_{n})+2\log
dist(p_{n},\Sigma_{n})+2N\log|p_{n}|+2\log\lambda_{n}e^{\bar{u}_{n}}+C\leq-
N_{n}+C,$

which implies that $w_{n}(x)$ has fast decay in
$B_{\tau_{n}^{i}}(x^{i}_{n})\setminus B_{\mu_{n}^{i}R_{n}}(x^{i}_{n})$.

Moreover, we have

$\displaystyle\alpha_{n}(\tau^{i}_{n};x_{n}^{i})-\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})$
$\displaystyle\leq\int_{B_{\tau^{i}_{n}}(x_{n}^{i})\setminus
B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}dx$
$\displaystyle\leq C\int_{B_{\tau^{i}_{n}}(x_{n}^{i})\setminus
B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\bar{w}_{n}}dx$
$\displaystyle\leq C\int_{B_{\tau^{i}_{n}}(x_{n}^{i})\setminus
B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x-x_{n}^{i}|^{2N}e^{\bar{w}_{n}}dx$
$\displaystyle\leq
Ce^{\bar{w}_{n}(\mu_{n}^{i}R_{n})+(3+2N)\log(\mu_{n}^{i}R_{n})}\int_{B_{\tau^{i}_{n}}(x^{i}_{n})\setminus
B_{\mu_{n}^{i}R_{n}}(x^{i}_{n})}\lambda_{n}e^{\bar{u}_{n}}|x-x_{n}^{i}|^{-3}dx$
$\displaystyle\leq
Ce^{\bar{w}_{n}(\mu_{n}^{1}R_{n})+2(1+N)\log(\mu_{n}^{1}R_{n})}$
$\displaystyle\leq Ce^{w_{n}(p_{n})+2\log
dist(p_{n},\Sigma_{n})+2N\log|p_{n}|}=o(1),$

where the second and last inequality follows from Lemma 2.6.

Thus, we get

$\alpha_{n}(\tau^{i}_{n};x_{n}^{i})=\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})+o(1)=b_{i}+o(1).$

Moreover, using global type Pohozaev identity of Lemma 2.10 in
$B_{R_{n}\mu^{i}_{n}}(x^{i}_{n})$, we get

$\displaystyle\int_{B_{\tau^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx$
$\displaystyle=\int_{B_{R_{n}\mu^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{i}^{2}-4\pi Nb_{i}+o(1).$

Case 2: If $(x^{i}_{n},\mu_{n}^{i})$ is not a singular bubble.

In this case, we know $\frac{|x_{n}^{i}|}{\mu_{n}^{i}}\to\infty$. By
Proposition 2.4, we know

$\varphi_{n}(x):=w_{n}(x_{n}^{i}+\mu_{n}^{i}x)-w_{n}(x_{n}^{i})\to\varphi(x)$

in the sense of $C^{2}_{loc}(\mathbb{R}^{2})$, where $\varphi(x)$ satisfies
(1.8). We choose $R_{n}\to\infty$ such that
$R_{n}\mu_{n}^{i}=o(1)\tau_{n}^{i}$ and $R_{n}\mu_{n}^{i}=o(1)|x_{n}^{i}|$,
$w_{n}(x)$ has fast decay on $\partial B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})$ and

(2.34) $\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})=b_{i}+o(1),$

where $b_{i}\geq 4$.

Denote $s_{n}^{i}:=\min\\{\tau_{n}^{i},\frac{3}{4}|x_{n}^{i}|\\}$, by (2), we
have

$\displaystyle\frac{d}{dr}\bar{w}_{n}=\frac{1}{2\pi
r}\int_{B_{r}(x_{n}^{i})}\Delta w_{n}(x)dx=-\frac{\alpha_{n}(r;x^{i}_{n})}{r}$
$\displaystyle\leq-\frac{b^{i}+o(1)}{r}$ (2.35)
$\displaystyle\leq-\frac{4+o(1)}{r},\ \ \forall\ \mu_{n}^{i}R_{n}\leq r\leq
s^{i}_{n},$

which implies

$\bar{w}_{n}(r)+3\log r,\ \ \forall\ \mu_{n}^{i}R_{n}\leq r\leq s^{i}_{n}$

is a decreasing function with respect to $r$. Noting that for any
$\mu_{n}^{i}R_{n}\leq|x-x_{n}^{i}|\leq s^{i}_{n}$, there holds

$\frac{1}{4}|x_{n}^{i}|\leq|x|\leq 2|x_{n}^{i}|.$

Taking any point $p_{n}\in\partial B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})$, we have

$\bar{w}_{n}(r)+3\log
r+2N\log|x|\leq\bar{w}_{n}(\mu_{n}^{i}R_{n})+3\log(\mu_{n}^{i}R_{n})+2N\log|p_{n}|+C,\
\forall\ \mu_{n}^{i}R_{n}\leq r\leq s^{i}_{n},$

which implies that for any $\mu_{n}^{i}R_{n}\leq|x-x_{n}^{i}|\leq s^{i}_{n}$,
there holds

$\displaystyle w_{n}(x)+2\log r+2N\log|x|+2\log\lambda_{n}e^{\bar{u}_{n}}$
$\displaystyle\leq\bar{w}_{n}(r)+2\log
r+2N\log|x|+2\log\lambda_{n}e^{\bar{u}_{n}}+C$
$\displaystyle\leq\bar{w}_{n}(\mu_{n}^{i}R_{n})+2\log(\mu_{n}^{i}R_{n})+2N\log|p_{n}|+\log\frac{\mu_{n}^{i}R_{n}}{r}+2N\log\frac{|x|}{|p_{n}|}+2\log\lambda_{n}e^{\bar{u}_{n}}+C$
$\displaystyle\leq\bar{w}_{n}(\mu_{n}^{i}R_{n})+2\log(\mu_{n}^{i}R_{n})+2N\log|p_{n}|+2\log\lambda_{n}e^{\bar{u}_{n}}+C\leq-
N_{n}+C.$

Moreover, we have

$\displaystyle\alpha_{n}(s^{i}_{n};x_{n}^{i})-\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})$
$\displaystyle\leq C\int_{B_{s^{i}_{n}}(x_{n}^{i})\setminus
B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\bar{w}_{n}}dx$
$\displaystyle\leq
C|x_{n}^{i}|^{2N}\lambda_{n}e^{\bar{u}_{n}}\int_{B_{s^{i}_{n}}(x_{n}^{i})\setminus
B_{\mu_{n}^{i}R_{n}}(x_{n}^{i})}e^{\bar{w}_{n}}dx$ $\displaystyle\leq
C|x_{n}^{i}|^{2N}\lambda_{n}e^{\bar{u}_{n}}e^{\bar{w}_{n}(\mu_{n}^{i}R_{n})+3\log(\mu_{n}^{i}R_{n})}\int_{B_{s^{i}_{n}}(x^{i}_{n})\setminus
B_{\mu_{n}^{i}R_{n}}(x^{i}_{n})}|x-x_{n}^{i}|^{-3}dx$ $\displaystyle\leq
C|x_{n}^{i}|^{2N}\lambda_{n}e^{\bar{u}_{n}}e^{\bar{w}_{n}(\mu_{n}^{i}R_{n})+2\log(\mu_{n}^{i}R_{n})}$
$\displaystyle=C\left(\frac{|x_{n}^{i}|}{|p_{n}|}\right)^{2N}\lambda_{n}e^{\bar{u}_{n}}e^{\bar{w}_{n}(\mu_{n}^{i}R_{n})+2\log(\mu_{n}^{i}R_{n})+2N\log|p_{n}|}=o(1).$

Thus, we get

$\alpha_{n}(s^{i}_{n};x_{n}^{i})=\alpha_{n}(\mu_{n}^{i}R_{n};x_{n}^{i})+o(1)=b_{i}+o(1).$

Moreover, using local type Pohozaev identity of Lemma 2.10 in
$B_{R_{n}\mu^{i}_{n}}(x^{i}_{n})$, we get

$\displaystyle\int_{B_{s^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx$
$\displaystyle=\int_{B_{R_{n}\mu^{i}_{n}}(x^{i}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{i}^{2}+o(1).$

We proved this proposition.

∎

##  3\. Proof of theorems

In this section, we will prove Theorem 2.1, Theorem 2.2 and Theorem 1.2 based
on a induction argument on the numbers of bubbles. Here, we need some
information of positional structure of bubbles which will be got by Pohozaev
type identities. As a corollary of Theorem 1.2, the proof of Theorem 1.3 will
also be given in this section.

Before proving our main theorems and in order to describe the induction
process better, we first give some definitions for positional structure of
bubbling disks.

######  Definition 3.1.

We call $B_{s_{n}}(x_{n})$ a bubbling disk if $w_{n}(x)$ has fast decay on
$\partial B_{s_{n}}(x_{n})$ and

$\alpha_{n}(s_{n};x_{n})=\sum_{i=k_{1}}^{k_{2}}b_{i}+o(1),$

where $b_{i}$ is the energy of the bubble $\varphi^{i}(x)$ in Proposition 2.4.

######  Definition 3.2.

Let $B_{s_{n}^{i}}(y^{i}_{n}),\ i=1,...,m$ be bubbling disks which are
pairwise disjoint. Let $\\{i_{1},i_{2},...,i_{I}\\}\subset\\{1,2,...,m\\},\
I\geq 2,$ be a subset and $B_{r_{n}}(x_{n})$ be a ball such that

$\cup_{i=i_{1}}^{i_{I}}B_{s_{n}^{i}}(y^{i}_{n})\subset B_{r_{n}}(x_{n}),\ \
r_{n}=o(1)dist(\partial
B_{r_{n}}(x_{n}),\\{y_{n}^{1},...,y_{n}^{m}\\}\setminus\\{y_{n}^{i_{1}},...,y_{n}^{i_{I}}\\}).$

We call $B_{s_{n}^{i_{j}}}(y^{i_{j}}_{n}),\ 1\leq j\leq I$, a locally smallest
bubbling disk in $B_{r_{n}}(x_{n})$ if there exists a positive constant $C>1$
such that

$\limsup_{n}\frac{s_{n}^{i_{k}}}{s_{n}^{i_{j}}}\geq C^{-1},\ k=1,...,I.$

######  Definition 3.3.

Let $B_{s_{n}^{i}}(y^{i}_{n}),\ i=1,...,m$ be bubbling disks which are
pairwise disjoint. Denote $S_{n}:=\\{y_{n}^{1},...,y_{n}^{m}\\}$. For a point
$x_{n}$, we call $B_{s_{n}^{j}}(y^{j}_{n})$ a nearest bubbling disk
corresponding to $x_{n}$, if

$dist(x_{n},S_{n}\setminus\\{x_{n}\\})=|x_{n}-y_{n}^{j}|.$

For a bubbling disk $B_{s^{i}_{n}}(y^{i}_{n})$, we call
$B_{s_{n}^{j}}(y^{j}_{n})$ is a almost nearest bubbling disk corresponding to
$B_{s^{i}_{n}}(y^{i}_{n})$ if

$1\leq\frac{|y^{j}_{n}-y^{i}_{n}|}{dist(y_{n}^{i},S_{n}\setminus\\{y_{n}^{i}\\})}\leq
C.$

######  Remark 3.4.

By above definitions, we immediately have the following facts that

  * (1)

For $x_{n}^{i}\in\Sigma_{n}$, by Proposition 2.11, we know
$B_{s_{n}^{i}}(x_{n}^{i})$ is a bubbling disk, where
$s_{n}^{i}:=\min\\{\tau_{n}^{i},\frac{3}{4}|x_{n}^{i}|\\}$ and
$\tau_{n}^{i}:=\frac{1}{2}dist(x_{n}^{i},\Sigma_{n}\setminus\\{x_{n}^{i}\\})$.

  * (2)

Let $B_{\tau_{n}^{i}}(y^{i}_{n}),\ i=1,...,m$ be bubbling disks where
$\tau_{n}^{i}:=\frac{1}{2}dist(y^{i}_{n},S_{n}\setminus\\{y_{n}^{i}\\})$ and
$S_{n}:=\\{y_{n}^{1},...,y_{n}^{m}\\}$. For $B_{\tau_{n}^{i}}(y^{i}_{n})$,
there exist some almost nearest bubbling disks (at least one which is the
nearest one), denoted by
$\\{B_{\tau_{n}^{i_{j}}}(y^{i_{j}}_{n})\\}_{j=1}^{J},\
i_{1},...,i_{j}\in\\{1,...,i-1,i+1,...,m\\}$. Then there exists a positive
constant $C$ such that

$\bigcup_{j=1}^{J}B_{\tau_{n}^{i_{j}}}(y^{i_{j}}_{n})\subset
B_{C\tau_{n}^{i}}(y^{i}_{n}),\ \ \tau_{n}^{i}=o(1)dist(\partial
B_{C\tau_{n}^{i}}(y^{i}_{n}),S_{n}\setminus\\{y_{n}^{i},y_{n}^{i_{1}},...,y_{n}^{i_{J}}\\}).$

  * (3)

If $B_{\tau_{n}^{i}}(y^{i}_{n})$ is a locally smallest bubbling disk in
$B_{C\tau_{n}^{i}}(y^{i}_{n})$ and $B_{\tau_{n}^{j}}(y^{j}_{n})\
(j\in\\{i_{1},...,i_{J}\\})$ is a almost nearest bubbling disk corresponding
to $B_{\tau^{i}_{n}}(y^{i}_{n})$, then we have

$C^{-1}\tau_{n}^{i}\leq\tau_{n}^{j}=\frac{1}{2}dist(y^{j}_{n},S_{n}\setminus\\{y_{n}^{j}\\})\leq\frac{1}{2}|y_{n}^{j}-y_{n}^{i}|\leq
C\tau_{n}^{i}.$

Next, we prove two propositions to simplify the structure of bubbling disks.

######  Proposition 3.5.

Let $x_{n}^{1}\in\Sigma_{n}$ and $(x_{n}^{1},\mu_{n}^{1})$ be not a singular
bubble. Then either $\Sigma_{n}=\\{x_{n}^{1}\\}$ or
$\lim\frac{|x_{n}^{1}|}{\tau_{n}^{1}}\leq C$.

###### Proof.

Suppose $\Sigma_{n}$ contains more than two points, we will show
$\lim\frac{|x_{n}^{1}|}{\tau_{n}^{1}}\leq C$. We prove by a contradiction
argument. If not, then passing to a subsequence, there holds
$\lim\frac{|x_{n}^{1}|}{\tau_{n}^{1}}=\infty$. Then by Proposition 2.11, we
know that $B_{\tau_{n}^{1}}(x^{1}_{n})$ is a bubbling disk, i.e.
$s_{n}^{1}=\tau_{n}^{1}$.

Since, there are more than two bubbles, for $B_{\tau_{n}^{1}}(x_{n}^{1})$,
there must be a nearest bubbling disk denoted by $B_{\tau_{n}^{2}}(x_{n}^{2})$
and maybe some almost nearest bubbling disks denoted by
$B_{\tau_{n}^{i}}(x_{n}^{i}),\ i=3,...,I$. It is easy to see that

$\frac{1}{2}|x_{n}^{2}-x_{n}^{1}|=\tau_{n}^{1}=o(1)|x_{n}^{1}|,\ \
|x_{n}^{i}-x_{n}^{1}|\leq C|x_{n}^{2}-x_{n}^{1}|\leq
C\tau_{n}^{1}=o(1)|x_{n}^{1}|,\ i=2,...,I,$

and

$\cup_{i=1}^{I}B_{\tau_{n}^{i}}(x_{n}^{i})\subset
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$

for some positive constant $\Lambda>0$. Moreover, since

$\tau_{n}^{i}\leq 2|x_{n}^{i}-x_{n}^{1}|=o(1)|x_{n}^{1}|=o(1)|x_{n}^{i}|,\ \
i=2,...,I,$

we know $\\{(x_{n}^{i},\mu_{n}^{i})\\}_{i=1}^{I}$ are not singular bubbles and

$\tau_{n}^{1}=o(1)dist(\partial
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1}),\Sigma_{n}\setminus\\{x_{n}^{1},...,x_{n}^{I}\\})$

according to the definition of almost nearest bubbling disk.

Since $2\tau_{n}^{i}\leq|x_{n}^{i}-x_{n}^{1}|\leq C\tau_{n}^{1},\ \
i=2,...,I$, we shall consider following two cases:

Case 1: if $\lim\frac{\tau_{n}^{i}}{\tau_{n}^{1}}\geq C^{-1},\ i=2,...,I$, for
some positive constant $C>0$.

In this case, we have $C^{-1}\leq\frac{\tau_{n}^{1}}{\tau_{n}^{i}}\leq C,\
i=2,...,I$. Now, we claim: $w_{n}(x)$ has fast decay on $\partial
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$ and

(3.1)
$\alpha_{n}(\Lambda_{n}\tau_{n}^{1};x_{n}^{1})=\sum_{i=1}^{I}\alpha_{n}(\tau_{n}^{i};x_{n}^{i})+o(1)=\sum_{i=1}^{I}b_{i}+o(1).$

For (3.1), we just need to show the first equality because the second one
follows from Proposition 2.11.

In fact, denoting
$\Omega_{n}:=B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})\setminus\left(\cup_{i=1}^{I}B_{\tau_{n}^{i}}(x_{n}^{i})\right)$,
by Lemma 2.6, we firstly have

$osc_{\Omega_{n}}w_{n}(x)\leq C.$

Secondly, for any $x\in\Omega_{n}$, it is easy to see that

$\frac{1}{2}|x_{n}^{1}|\leq|x_{n}^{1}|-|x-x_{n}^{1}|\leq|x|\leq|x-x_{n}^{1}|+|x_{n}^{1}|\leq
2|x_{n}^{1}|$

and

$C^{-1}\tau_{n}^{1}\leq\tau_{n}^{i}\leq|x-x_{n}^{i}|\leq|x-x_{n}^{1}|+|x_{n}^{1}-x_{n}^{i}|\leq
C\tau_{n}^{1},\ i=1,...,I,$

which implies

$osc_{\Omega_{n}}\log
dist(x,\\{x^{1}_{n},...,x_{n}^{I}\\})+osc_{\Omega_{n}}\log|x|\leq C.$

Since $w_{n}(x)$ has fast decay on $\partial B_{\tau_{n}^{i}}(x_{n}^{i})$, we
get that $w_{n}(x)$ has fast decay in $\Omega_{n}$ and

$\displaystyle\int_{\Omega_{n}}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}dx$
$\displaystyle\leq
o(1)\int_{\Omega_{n}}\frac{1}{|dist(x,\\{x_{n}^{1},x_{n}^{2},...,x_{n}^{I}\\})|^{2}}dx$
(3.2) $\displaystyle\leq
o(1)\sum_{i=1}^{I}\int_{\Omega_{n}}\frac{1}{|x-x_{n}^{i}|^{2}}dx=o(1),$

which implies (3.1) immediately.

Now we estimate the integration

$2\int_{B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx.$

On one hand, by using (3) and local type Pohozaev identity of Lemma 2.10 in
$B_{\tau_{n}^{i}}(x_{n}^{i}),\ i=1,...,I$, we obtain

$\displaystyle
2\int_{B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
$\displaystyle=2\sum_{i=1}^{I}\int_{B_{\tau_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\pi\sum_{i=1}^{I}\alpha_{n}^{2}(x_{n}^{i};\tau_{n}^{i})+o(1)=\pi\sum_{i=1}^{I}b_{i}^{2}+o(1).$

On the other hand, using local type Pohozaev identity of Lemma 2.10 in
$B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$, we have

$2\int_{B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}})dx=\pi\alpha_{n}^{2}(x_{n}^{1};\Lambda\tau_{n}^{1})+o(1)=\pi\left(\sum_{i=1}^{I}b_{i}\right)^{2}+o(1).$

Then we get

$\sum_{i=1}^{I}b_{i}^{2}=\left(\sum_{i=1}^{I}b_{i}\right)^{2},$

which is a contradiction because we have at least two bubbles, i.e. $I\geq 2$.

Case 2: There exists $i\in\\{2,...,I\\}$ such that
$\lim\frac{\tau_{n}^{i}}{\tau_{n}^{1}}=0$.

In this case, there exists a locally smallest bubbling disk (denoted by
$B_{\tau_{n}^{2}}(x_{n}^{2})$) in $B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$, such
that

$\frac{\tau_{n}^{1}}{\tau_{n}^{2}}\to\infty,\ \
\frac{\tau_{n}^{j}}{\tau_{n}^{2}}\geq C^{-1}>0,\ j\in\\{2,...,I\\}.$

Then we know there must be more than three bubbles, i.e. $I\geq 3$. Otherwise,
$\tau_{n}^{2}=\tau_{n}^{1}$ which is a contradiction.

Furthermore, for $B_{\tau_{n}^{2}}(x_{n}^{2})$, there must exist a nearest
bubble (denoted by $B_{\tau_{n}^{3}}(x_{n}^{3})$) and maybe some almost
nearest bubbles denoted by
$B_{\tau_{n}^{4}}(x_{n}^{4}),...,B_{\tau_{n}^{I}}(x_{n}^{I})$. By Remark 3.4,
we get

$|x_{n}^{3}-x_{n}^{2}|=2\tau_{n}^{2},\ \
1\leq\frac{|x_{n}^{j}-x_{n}^{2}|}{|x_{n}^{3}-x_{n}^{2}|}\leq C,\ \
C^{-1}\tau_{n}^{2}\leq\tau_{n}^{j}\leq\frac{1}{2}|x_{n}^{j}-x_{n}^{2}|\leq
C\tau_{n}^{2},\ j=3,...,I$

and there exists a constant $\Lambda_{1}>100$ such that

$\displaystyle\cup_{j=2}^{I}B_{\tau_{n}^{j}}(x_{n}^{j})\subset
B_{\Lambda_{1}\tau_{n}^{2}}(x_{n}^{2}),\ \ \tau_{n}^{2}=o(1)dist(\partial
B_{\Lambda_{1}\tau_{n}^{2}}(x_{n}^{2}),\Sigma_{n}\setminus\Sigma_{n}^{1}),$

where $\Sigma_{n}^{1}:=\\{x_{n}^{2},...,x_{n}^{I}\\}$.

Similarly to $case\ 1$, we can prove that $w_{n}(x)$ has fast decay on
$\partial B_{\Lambda_{1}\tau_{n}^{2}}(x_{n}^{2})$ and

$\alpha_{n}(\Lambda_{1}\tau_{n}^{2};x_{n}^{2})=\sum_{j=2}^{I}b_{j}+o(1).$

Moreover, by using local Pohozaev identity in each bubbling disk
$B_{\tau_{n}^{j}}(x_{n}^{j}),\ j=2,...,I$ and
$B_{\Lambda_{1}\tau_{n}^{2}}(x_{n}^{2})$, we obtain

$\sum_{j=2}^{I}b_{j}^{2}=\left(\sum_{j=2}^{I}b_{j}\right)^{2}+o(1),$

which is also a contradiction because $\Sigma_{n}^{1}$ contains at least two
points, i.e. $I\geq 3$.

We proved proposition. ∎

######  Proposition 3.6.

Let $x_{n}^{1}\in\Sigma_{n}$ and $(x_{n}^{1},\mu_{n}^{1})$ be a singular
bubble. Then $\Sigma_{n}=\\{x_{n}^{1}\\}$, i.e. there is only one bubble.

###### Proof.

We prove by a contradiction argument. If not,
$\Sigma_{n}=\\{x_{n}^{1},...,x_{n}^{m}\\}$ have more than two points. In this
case, by Proposition 2.4, we know $|x_{n}^{1}|=o(1)|x_{n}^{i}|,\ i=2,...,m$.
Then

$\tau_{n}^{i}\leq\frac{1}{2}|x_{n}^{i}-x_{n}^{1}|\leq\frac{3}{4}|x_{n}^{i}|,\
i=2,...,m.$

By Proposition 2.11, we know $B_{\tau_{n}^{i}}(x_{n}^{i}),\ i=1,...,m$ are all
bubbling disks, i.e. $s_{n}^{i}=\tau_{n}^{i}$.

For bubbling disk $B_{\tau_{n}^{1}}(x^{1}_{n})$, we can find a nearest bubble
(denoted by $(x_{n}^{2},\tau_{n}^{2})$) and maybe almost nearest bubbles
(denoted by $(x_{n}^{i},\tau_{n}^{i}),\ i=3,...,I$). By definition, we know

$1\leq\frac{|x^{i}_{n}-x^{1}_{n}|}{|x^{2}_{n}-x^{1}_{n}|}\leq C,\ \
\tau_{n}^{i}\leq\frac{1}{2}|x_{n}^{i}-x_{n}^{1}|\leq C\tau_{n}^{1},\
i=2,...,I$

and

$\cup_{i=1}^{I}B_{\tau_{n}^{i}}(x_{n}^{i})\subset
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$

for some positive constant $\Lambda>0$. Moreover, according to the definition
of almost nearest bubbling disk, we know

$\tau_{n}^{1}=o(1)dist(\partial
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1}),\Sigma_{n}\setminus\\{x_{n}^{1},...,x_{n}^{I}\\}).$

Since $\lim\frac{\tau_{n}^{i}}{\tau_{n}^{1}}\leq C,\ i=2,...,I$, we shall
consider the following cases:

Case I: if $\lim\frac{\tau_{n}^{i}}{\tau_{n}^{1}}\geq C^{-1},\ i=2,...,I$, for
some positive constant $C$.

In this case, we have $C^{-1}\leq\frac{\tau_{n}^{i}}{\tau_{n}^{1}}\leq C,\
i=2,...,I$. Now, we claim: $w_{n}(x)$ has fast decay on $\partial
B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$ and

(3.3)
$\alpha_{n}(\Lambda_{n}\tau_{n}^{1};x_{n}^{1})=\sum_{i=1}^{I}\alpha_{n}(\tau_{n}^{i};x_{n}^{i})+o(1)=\sum_{i=1}^{I}b_{i}+o(1).$

In fact, denoting
$\Omega_{n}:=B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})\setminus\left(\cup_{i=1}^{I}B_{\tau_{n}^{i}}(x_{n}^{i})\right)$,
by Lemma 2.6, we firstly have

$osc_{\Omega_{n}}w_{n}(x)\leq C.$

Secondly, for any $x\in\Omega_{n}$, it is easy to see that

$\frac{1}{2}\tau_{n}^{1}\leq|x-x_{n}^{1}|-|x_{n}^{1}|\leq|x|\leq|x-x_{n}^{1}|+|x_{n}^{1}|\leq(1+\Lambda)\tau_{n}^{1}$

and

$C^{-1}\tau_{n}^{1}\leq\tau_{n}^{i}\leq|x-x_{n}^{i}|\leq|x-x_{n}^{1}|+|x_{n}^{1}-x_{n}^{i}|\leq
C\tau_{n}^{1},\ i=1,...,I,$

which implies

$osc_{\Omega_{n}}\log
dist(x,\\{x^{1}_{n},...,x_{n}^{I}\\})+osc_{\Omega_{n}}\log|x|\leq C.$

Since $w_{n}(x)$ has fast decay on $\partial B_{\tau_{n}^{i}}(x_{n}^{i})$, we
get that $w_{n}(x)$ has fast decay in $\Omega_{n}$ and

$\displaystyle\int_{\Omega_{n}}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}dx$
$\displaystyle\leq
o(1)\int_{\Omega_{n}}\frac{1}{|dist(x,\\{x_{n}^{1},x_{n}^{2},...,x_{n}^{I}\\})|^{2}}dx$
(3.4) $\displaystyle\leq
o(1)\sum_{i=1}^{I}\int_{\Omega_{n}}\frac{1}{|x-x_{n}^{i}|^{2}}dx=o(1),$

which implies (3.3) immediately.

Similarly to Proposition 3.5, on one hand, by using (3) and Proposition 2.11,
we obtain

$\displaystyle
2\int_{B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}})dx$
$\displaystyle=2\sum_{i=1}^{I}\int_{B_{\Lambda\tau_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}})dx=\pi\sum_{i=1}^{I}b_{i}^{2}-4\pi
Nb_{1}+o(1).$

On the other hand, using global type Pohozaev identity of Lemma 2.10 in
$B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})$, we have

(3.5)
$2\int_{B_{\Lambda\tau_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{w_{n}})dx=\pi\left(\sum_{i=1}^{I}b_{i}\right)^{2}-4\pi
N\sum_{i=1}^{I}b_{i}+o(1).$

Then we get

$\sum_{i=1}^{I}b_{i}^{2}-4Nb_{1}=\left(\sum_{i=1}^{I}b_{i}\right)^{2}-4N\sum_{i=1}^{I}b_{i},$

which implies

$\sum_{2\leq i,j\leq I,i\neq j}b_{i}b_{j}=(2N-b_{1})\sum_{i=2}^{I}b_{i}.$

This is a contradiction since $b_{1}\geq 4N+4$ and $2N-b_{1}<0$.

Case II: if there exists $i\in\\{2,...,I\\}$ such that
$\lim\frac{\tau_{n}^{i}}{\tau_{n}^{1}}=0$.

By a almost same argument as in case 2 of Proposition 3.5, we will also get a
contradiction.

We proved the proposition.

∎

Now, at the end of this section, we prove our main theorems.

###### Proof of Theorem 2.1..

Firstly, by Lemma 2.3, we get $c=0$. Next, We divide the proof into the
following two steps.

Step 1: If $\Sigma_{n}$ contains one point $\\{x_{n}^{1}\\}$ and
$(x_{n}^{1},\mu_{n}^{1})$ is a singular bubble, then conclusion $(2-1)$
follows.

By Proposition 3.6, we know $\Sigma_{n}=\\{x_{n}^{1}\\}$. Since
$(x_{n}^{1},\mu_{n}^{1})$ is the only singular bubble and
$\tau_{n}^{1}\geq\frac{1}{2}$, by Proposition 2.11, we have

$\alpha_{n}(\frac{1}{2};x^{1}_{n})=b_{1}+o(1),$

which implies

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=b_{1}.$

Thus, conclusion $(2-1)$ follows immediately.

Step 2: If $\Sigma_{n}$ contains a point $x_{n}^{i}$ such that
$(x_{n}^{i},\mu_{n}^{i})$ is not a singular bubble, then conclusion $(2-2)$
follows.

For simplicity of notation, we assume $i=1$. In this case, by proposition 3.5,
we know either $\Sigma_{n}=\\{x_{n}^{1}\\}$ or
$\lim\frac{|x_{n}^{1}|}{\tau_{n}^{1}}\leq C$.

We first claim that $\Sigma_{n}$ contains more than two points. In fact, if
not, since $\Sigma_{n}$ contains only one point $\\{x_{n}^{1}\\}$, then we
have $\tau_{n}^{1}\geq\frac{1}{2}$ which implies
$s_{n}^{1}=\frac{3}{4}|x_{n}^{1}|$.

On one hand, since $(x_{n}^{1},\mu_{n}^{1})$ is not a singular bubble, by
Proposition 2.11, we have $w_{n}(x)$ has fast decay on $\partial
B_{s_{n}^{1}}(x_{n}^{1})$ and

$\alpha_{n}(s_{n}^{1};x^{1}_{n})=b_{1}+o(1)$

and

$\int_{B_{s^{1}_{n}}(x^{1}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{1}^{2}+o(1),$

where $b_{1}\geq 4$.

On the other hand, since $\lim\frac{|x_{n}^{1}|}{s_{n}^{1}}\leq C$, by global
type Pohozaev identity of Lemma 2.10 in $B_{s^{1}_{n}}(x^{1}_{n})$, we get

$\int_{B_{s^{1}_{n}}(x^{1}_{n})}2\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}[1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta(x)}e^{w_{n}}]dx=\pi
b_{1}^{2}-4\pi Nb_{1}+o(1).$

This yields

$\pi b_{1}^{2}-4\pi Nb_{1}=\pi b_{1}^{2}+o(1),$

which is a contradiction. Thus, $\Sigma_{n}$ must contain more than two
points.

By Proposition 3.5 and 3.6, we know that $\Sigma_{n}$ contains no singular
bubble and

$\lim\frac{|x_{n}^{i}|}{\tau_{n}^{i}}\leq C,\ \ i=1,...,m.$

By Proposition 2.11, we know that $\\{B_{s_{n}^{i}}(x_{n}^{i})\\}_{i=1}^{m}$
are pairwise disjoint bubbling disks where
$s_{n}^{i}:=\min\\{\tau_{n}^{i},\frac{3}{4}|x_{n}^{i}|\\}$. Then we shall
consider the following cases.

Case 1: If there exists $i\in\\{1,...,m\\}$ such that
$\lim\frac{|x_{n}^{i}|}{\tau_{n}^{i}}=0$.

For simplicity of notation, we assume $i=1$. In this case, we see that
$s_{n}^{1}=\frac{3}{4}|x_{n}^{1}|$. On one hand, by Proposition 2.11, we have

(3.6)
$2\int_{B_{s_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\pi
b_{1}^{2}+o(1).$

On the other hand, since $|x_{n}^{1}|=o(1)\tau_{n}^{1}$, then
$s_{n}^{1}=o(1)dist(\partial
B_{s_{n}^{1}}(x_{n}^{1}),\Sigma_{n}\setminus\\{x_{n}^{1}\\})$. Using global
type Pohozaev identity of Lemma 2.10 in $B_{s_{n}^{1}}(x_{n}^{1})$, we have

(3.7)
$2\int_{B_{s_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\pi
b_{1}^{2}-4\pi Nb_{1}+o(1).$

Thus, we get

$b_{1}^{2}=b_{1}^{2}-4Nb_{1}+o(1),$

which is a contradiction.

Case 2: $\lim\frac{|x_{n}^{i}|}{\tau_{n}^{i}}\geq C^{-1},\ \ i=1,...,m$ for
some constant $C>0$.

In this case, we first take a smallest bubbling disk denoted by
$B_{s_{n}^{1}}(x_{n}^{1})$. Next, we claim: $w_{n}(x)$ has fast decay on
$\partial B_{\frac{1}{2}}(x_{n}^{1})$ and

(3.8)
$\alpha_{n}(\frac{1}{2};x_{n}^{1})=\sum_{i=1}^{m}\alpha_{n}(s_{n}^{i};x_{n}^{i})+o(1)=\sum_{i=1}^{m}b_{i}+o(1)$

and

$\displaystyle
2\int_{B_{\frac{1}{2}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
(3.9)
$\displaystyle=\sum_{i=1}^{m}2\int_{B_{s_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx.$

We prove by a induction argument on the numbers of bubbling disks
$B_{s_{n}^{i}}(x_{n}^{i}),\ i=1,...,m$.

Step 2-1: We prove for two bubbling disks, i.e. $m=2$.

In this case, we have two bubbling disks $B_{s_{n}^{1}}(x_{n}^{1})$ and
$B_{s_{n}^{2}}(x_{n}^{2})$, where $B_{s_{n}^{1}}(x_{n}^{1})$ is a smallest
bubbling disk satisfying

$C^{-1}\tau_{n}^{1}\leq s_{n}^{1}\leq\tau_{n}^{1},\ \
C^{-1}s_{n}^{1}\leq|x_{n}^{1}|\leq Cs_{n}^{1}.$

It is easy to see that

$\tau_{n}^{1}=\tau_{n}^{2},\ \ C^{-1}s_{n}^{1}\leq s_{n}^{2}\leq Cs_{n}^{1},\
\ C^{-1}s_{n}^{1}\leq|x_{n}^{2}|\leq Cs_{n}^{1}.$

By the proof of Proposition 3.5 or Proposition 3.6, we can show that there
exists a positive constant $\Lambda>100$ such that
$B_{s_{n}^{1}}(x_{n}^{1})\cup B_{s_{n}^{2}}(x_{n}^{2})\subset B_{\Lambda
s_{n}^{1}}(x_{n}^{1})$, $w_{n}(x)$ has fast decay in $B_{\Lambda
s_{n}^{1}}(x_{n}^{1})\setminus\cup_{i=1}^{2}B_{s_{n}^{i}}(x_{n}^{i})$ and

$\alpha_{n}(\Lambda
s_{n}^{1};x_{n}^{1})=\sum_{i=1}^{2}\alpha_{n}(s_{n}^{i};x_{n}^{i})+o(1)=\sum_{i=1}^{2}b_{i}+o(1)$

and

$\displaystyle 2\int_{B_{\Lambda
s_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
$\displaystyle=\sum_{i=1}^{2}2\int_{B_{s_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx.$

Moreover, using a local and global type Pohozaev identity, we have the
following relation

$b_{1}^{2}+b_{2}^{2}=(b_{1}+b_{2})^{2}-4N(b_{1}+b_{2}),$

which implies that

$b_{1}+b_{2}\geq 4(1+N).$

By (2), we have

$\displaystyle\frac{d}{dr}\bar{w}_{n}=\frac{1}{2\pi
r}\int_{B_{r}(x_{n}^{1})}\Delta w_{n}(x)dx$
$\displaystyle=-\frac{\alpha_{n}(r;x^{1}_{n})}{r}$
$\displaystyle\leq-\frac{4(1+N)+o(1)}{r},\ \ \forall\ \Lambda s_{n}^{1}\leq
r=|x-x_{n}^{1}|\leq\frac{1}{2}.$

Similarly to the proof of Proposition 2.11, we can prove that $w_{n}(x)$ has
fast decay in $B_{\frac{1}{2}}(x_{n}^{1})\setminus B_{\Lambda
s_{n}^{1}}(x_{n}^{1})$ and

$\int_{B_{\frac{1}{2}}(x_{n}^{1})\setminus\cup_{i=1}^{2}B_{s_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}dx=o(1)$

which immediately implies (3.8) and (3). We proved for $m=2$.

Step 2-2: Assuming when the number of bubbling disks is less than $m-1$, we
can show that $w_{n}(x)$ has fast decay on $\partial
B_{\frac{1}{2}}(x_{n}^{1})$ and (3.8), (3) hold. Next, we prove it still holds
for $m$ bubbling disks.

In this case, there holds $C^{-1}\tau_{n}^{1}\leq|x_{n}^{1}|\leq
C\tau_{n}^{1}$ which implies $C^{-1}\tau_{n}^{1}\leq s_{n}^{1}\leq
C\tau_{n}^{1}$. Recall that $B_{s_{n}^{1}}(x_{n}^{1})$ is a smallest bubbling
disk. Since $\Sigma_{n}$ has more than two points, for
$B_{s_{n}^{1}}(x_{n}^{1})$, there must be a nearest bubbling disk denoted by
$B_{s_{n}^{2}}(x_{n}^{2})$ and maybe some almost nearest bubbling disks
denoted by $B_{s_{n}^{i}}(x_{n}^{i}),\ i=3,...,I$. Noting that

$C^{-1}s_{n}^{1}\leq s_{n}^{i}\leq\tau_{n}^{i}\leq|x_{n}^{i}-x_{n}^{1}|\leq
C|x_{n}^{2}-x_{n}^{1}|\leq C\tau_{n}^{1}\leq Cs_{n}^{1},\ i=2,...,I,$

then there exists a positive constant $\Lambda>0$ such that

$\cup_{i=1}^{I}B_{s_{n}^{i}}(x_{n}^{i})\subset B_{\Lambda
s_{n}^{1}}(x_{n}^{1}),\ \ and\ \ 0\in B_{\Lambda s_{n}^{1}}(x_{n}^{1}).$

Noting that $s_{n}^{1}\leq s_{n}^{i}\leq Cs_{n}^{1},\ i=2,...,I$, similar to
the case 1 of Proposition 3.5 or case I of Proposition 3.6, we can show that

$\alpha_{n}(\Lambda s_{n}^{1};x_{n}^{1})=\sum_{i=1}^{I}b_{i}+o(1)$

and

$\displaystyle 2\int_{B_{\Lambda
s_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
$\displaystyle=\sum_{i=1}^{I}2\int_{B_{s_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\sum_{i=1}^{I}b_{i}^{2}+o(1).$

Writing $B_{\Lambda s_{n}^{1}}(x_{n}^{1})$ as a new bubbling disk, together
with the left bubbling disks $B_{s_{n}^{i}}(x_{n}^{i}),\ i=I+1,...,m$ and the
induction assumption, we get $w_{n}(x)$ has fast decay on $\partial
B_{\frac{1}{2}}(x_{n}^{1})$ and

$\alpha_{n}(\frac{1}{2};x_{n}^{1})=\alpha_{n}(\Lambda
s_{n}^{1};x_{n}^{1})+\sum_{i=I}^{m}\alpha_{n}(\Lambda
s_{n}^{i};x_{n}^{i})+o(1)=\sum_{i=1}^{m}b_{i}+o(1)$

and

$\displaystyle
2\int_{B_{\frac{1}{2}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
$\displaystyle=2\int_{B_{\Lambda
s_{n}^{1}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx$
$\displaystyle\quad+2\sum_{i=I+1}^{m}\int_{B_{s_{n}^{i}}(x_{n}^{i})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx.$

We finished the proof of induction, i.e. $w_{n}(x)$ has fast decay on
$\partial B_{\frac{1}{2}}(x_{n}^{1})$ and (3.8), (3) hold.

We continue to show the conclusion of $(2-2)$. In fact, we first easy to see
that (3.8) implies

$\lim_{\rho\to
0}\lim_{n\to\infty}\alpha_{n}(\rho;0)=\sum_{i=1}^{m}b_{i}+o(1).$

Secondly, on one hand, by (3) and Proposition 2.11, we have

$\displaystyle
2\int_{B_{\frac{1}{2}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\pi\sum_{i=1}^{m}b_{i}^{2}+o(1).$

On the other hand, since $w_{n}(x)$ has fast decay on $\partial
B_{\frac{1}{2}}(x_{n}^{1})$, by global type Pohozaev identity, we obtain

$\displaystyle
2\int_{B_{\frac{1}{2}}(x_{n}^{1})}\lambda_{n}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}}(1-\frac{1}{2}e^{\bar{u}_{n}}|x|^{2N}e^{\zeta}e^{w_{n}})dx=\pi(\sum_{i=1}^{m}b_{i})^{2}-4\pi
N\sum_{i=1}^{m}b_{i}+o(1).$

Thus, we have the following relation

$\sum_{i=1}^{m}b_{i}^{2}=\left(\sum_{i=1}^{m}b_{i}\right)^{2}-4N\sum_{i=1}^{m}b_{i}.$

We finished the proof of Theorem 2.1.

∎

###### Proof of main Theorem 2.2..

Since blow-up point $0$ is not a vortex point, i.e. $N=0$, by Theorem 1.1, we
know that $c=0$. By Lemma 2.10, we see that there is only one kind of Pohozaev
type identity in this case (i.e. local and global type Pohozaev identity are
the same). If there are more than two bubbles, denoting their energies by
$b_{i},\ i=1,...,m$, then by Theorem 2.1, we get

$\sum_{i=1}^{m}b_{i}^{2}=(\sum_{i=1}^{m}b_{i})^{2},$

which is a contradiction for $m\geq 2$ and $b_{i}\geq 4,\ i=1,...,m$. Thus
there must be one bubble. We proved Theorem 2.2.

∎

###### Proof of main Theorem 1.2..

Conclusions of Theorem 1.2 is a direct consequence of Theorem 2.1 and Theorem
2.2. ∎

###### Proof of main Theorem 1.3..

If the vortex point $p_{i}$ is a blow-up point, then Theorem 1.2 tells us that
the local mass

$\lim_{\rho\to 0}\lim_{n\to\infty}\alpha_{n}(\rho;p_{i})\geq 4(1+N_{i}).$

Since the total energy is $2\bar{N}$ (divided by $2\pi$), thus we have

$2(1+N_{i})\leq\bar{N},$

which implies $\bar{N}\geq 4$.

For $\bar{N}=5$, if $q_{1}$ is a vortex point, from inequality

$2(1+N_{1})\leq\bar{N},$

we have $N_{1}=1$ and $S=\\{q_{1}\\}$. We claim the simple blow-up property
holds, i.e. $(3-2)$ of Theorem 1.2 will not happen. Otherwise, in this case,
there are just two bubbles with energies $b_{i}\geq 4,\ i=1,2$ satisfying
$b_{1}+b_{2}=2\bar{N}=10$ and
$b_{1}^{2}+b_{2}^{2}=(b_{1}+b_{2})^{2}-4(b_{1}+b_{2})$. Solving directly, we
get

$b_{i}=5\pm\sqrt{5},\ i=1,2,$

which is a contradiction that $5-\sqrt{5}<4$. Thus, the simple blow-up
property holds.

For $\bar{N}=6$, if $S=\\{q_{1}\\}$ and $q_{1}$ is a vortex point with
multiplicity $N_{1}=1$, since the total energy is $2\bar{N}=12$, if there are
more than two bubbles, then we have the following two cases. $(i)$ if there
are three bubbles, then there must be $b_{1}=b_{2}=b_{3}=4$. This is a
contradiction to equality
$b_{1}^{2}+b_{2}^{2}+b_{3}^{2}=(b_{1}+b_{2}+b_{3})^{2}-4(b_{1}+b_{2}+b_{3})$.
$(ii)$ if there are two bubbles, then we have $b_{1}+b_{2}=2\bar{N}=12$ and
$b_{1}^{2}+b_{2}^{2}=(b_{1}+b_{2})^{2}-4(b_{1}+b_{2})=96$ which implies

$b_{i}=6\pm 2\sqrt{3},\ i=1,2.$

This is also a contradiction since $6-2\sqrt{3}<4$.

For $\bar{N}=7$, if $q_{1}$ is a vortex point, from inequality

$2(1+N_{1})\leq\bar{N},$

we have following the two cases. $(1)\ N_{1}=1.$ In this case, since the total
energy is $2\bar{N}=14$, if there are more than two bubbles, then we have
following two cases. $(1-i)$ If there are three bubbles whose energies denoted
by $b_{1},\ b_{2},\ b_{3}$. Then we have $b_{1}+b_{2}+b_{3}=14$ and
$b_{1}^{2}+b_{2}^{2}+b_{3}^{2}=(b_{1}+b_{2}+b_{3})^{2}-4(b_{1}+b_{2}+b_{3})=140$.
One can check by computer that there is no solution $(b_{1},b_{2},b_{3})$ such
that $b_{i}\geq 4,\ i=1,2,3$, which is a contradiction. $(1-ii)$ If there are
two bubbles, then we have $b_{1}+b_{2}=2\bar{N}=14$ and
$b_{1}^{2}+b_{2}^{2}=(b_{1}+b_{2})^{2}-4(b_{1}+b_{2})=140$ which implies

$b_{i}=7\pm\sqrt{21},\ i=1,2.$

This is also a contradiction since $7-\sqrt{21}<4$.

$(2)\ N_{1}=2$. Similarly to first case, we have the following two cases.
$(2-i)$ If there are three bubbles whose energies denoted by $b_{1},\ b_{2},\
b_{3}$. Then we have $b_{1}+b_{2}+b_{3}=14$ and
$b_{1}^{2}+b_{2}^{2}+b_{3}^{2}=(b_{1}+b_{2}+b_{3})^{2}-8(b_{1}+b_{2}+b_{3})=84$.
One can check by computer that there is no solution $(b_{1},b_{2},b_{3})$ such
that $b_{i}\geq 4,\ i=1,2,3$, which is a contradiction. $(2-ii)$ If there are
two bubbles, then we have $b_{1}+b_{2}=2\bar{N}=14$ and
$b_{1}^{2}+b_{2}^{2}=(b_{1}+b_{2})^{2}-8(b_{1}+b_{2})=84$ which is also a
contradiction since $196=(b_{1}+b_{2})^{2}\leq 2(b_{1}^{2}+b_{2}^{2})\leq
168$.

We finished all the proof. ∎

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