# The blow-up analysis on $\mathbf{B}_{2}^{(1)}$ affine Toda system: Local
mass and affine Weyl group

Leilei Cui  ,  Jun-cheng Wei  ,  Wen Yang  and  Lei Zhang  Leilei Cui
School of Mathematics and Statistics
Central China Normal University
Wuhan, 430079, China<EMAIL_ADDRESS>Jun-cheng Wei
Department of Mathematics, The University of British Columbia
Vancouver, BC Canada V6T 1Z2<EMAIL_ADDRESS>Wen Yang
Wuhan Institute of Physics and Mathematics
Innovation Academy for Precision Measurement Science and Technology
Chinese Academy of Sciences
Wuhan 430071, P. R. China<EMAIL_ADDRESS>Lei Zhang
Department of Mathematics
University of Florida
1400 Stadium Rd
Gainesville FL 32611<EMAIL_ADDRESS>
###### Abstract.

It has been established that the local mass of blow-up solutions to Toda
systems associated with the simple Lie algebras
$\mathbf{A}_{n},~{}\mathbf{B}_{n},~{}\mathbf{C}_{n}$ and $\mathbf{G}_{2}$ can
be represented by a finite Weyl group. In particular, at each blow-up point,
after a sequence of bubbling steps (via scaling) is performed, the
transformation of the local mass at each step corresponds to the action of an
element in the Weyl group. In this article, we present the results in the same
spirit for the affine $\mathbf{B}_{2}^{(1)}$ Toda system with singularities.
Compared with the Toda system with simple Lie algebras, the computation of
local masses is more challenging due to the infinite number of elements of the
affine Weyl group of type $\mathbf{B}_{2}^{(1)}$. In order to give an explicit
expression for the local mass formula we introduce two free integers and write
down all the possibilities into 8 types. This shows a striking difference to
previous results on Toda systems with simple Lie algebras. The main result of
this article seems to provide the first major advance in understanding the
relation between the blow-up analysis of affine Toda system and the affine
Weyl group of the associated Lie algebras.

Key Words: Affine Toda system, Pohozaev identity, Liouville equation, sinh-
Gordon equation, Affine Weyl group.

AMS subject classification: 35B44, 35J61, 35R01

###### Contents

  1. 1 Introduction and main results
  2. 2 Bubbling analysis and Pohozaev identity
  3. 3 The structure of $\Gamma(\mu_{1},\mu_{2},\mu_{3})$ and affine Weyl group of $\mathbf{B}_{2}^{(1)}$
  4. 4 Some technical lemmas
  5. 5 Local masses on blow-up areas away from the origin
    1. 5.1 Local masses on the bubbling disk centered at $x^{k}_{j}\neq 0$
    2. 5.2 Local mass in a group that does not contain $0$
  6. 6 Proof of Theorem 1.1 and Theorem 1.3
  7. 7 Blow-up analysis for the sinh-Gordon equation
  8. 8 Appendix: classification results

##  1\. Introduction and main results

The main focus of this article is to derive the relation between the
quantization result from the blow up analysis and the associated algebraic
structure for the following system

(1.1) $\begin{cases}\Delta
u_{i}+\sum\limits_{j=1}^{3}a_{ij}e^{u_{j}}=4\pi\gamma_{i}\delta_{0}\quad&\mbox{in}\quad
B_{1}(0)\subseteq\mathbb{R}^{2},\qquad i=1,2,3,\\\
u_{1}+u_{2}+2u_{3}=0\quad&\mathrm{in}\quad B_{1}(0),\end{cases}$

where $\gamma_{i}>-1$ for $i=1,2,3$, $\delta_{0}$ stands for the Dirac measure
at the point $0$ and

(1.2) $A:=(a_{ij})_{3\times 3}=\left(\begin{matrix}1&0&-1\\\ 0&1&-1\\\
-\frac{1}{2}&-\frac{1}{2}&1\end{matrix}\right).$

Up to a constant one sees that the coefficient matrix $A$ is exactly the
Cartan matrix of affine Lie algebra $\mathbf{B}_{2}^{(1)}$ and we call (1.1)
the affine $\mathbf{B}_{2}^{(1)}$ Toda system. Let

$u_{1}=-\omega+\eta,\quad u_{2}=-\omega-\eta,\quad u_{3}=\omega,$

then (1.1) is equivalent to

(1.3)
$\begin{cases}\Delta\omega+e^{\omega}-\frac{1}{2}e^{-\omega+\eta}-\frac{1}{2}e^{-\omega-\eta}=4\pi\gamma_{3}\delta_{0},\\\
\\\
\Delta\eta+\frac{1}{2}e^{-\omega+\eta}-\frac{1}{2}e^{-\omega-\eta}=2\pi(\gamma_{1}-\gamma_{2})\delta_{0}.\end{cases}$

When $\gamma_{1}=\gamma_{2}=0$, system (1.3) is related to minimal surface
into $\mathbb{S}^{4}$ without superminimal points [18] and also appeared in
the work of integrable system by Fordy-Gibbons in [19]. If $\eta=0$ then
system (1.3) is reduced to the simplest affine Toda system, i.e., the well-
known sinh-Gordon equation

(1.4) $\Delta u+e^{u}-e^{-u}=0,$

which was originally introduced by Edmond Bour [7] in the study of surfaces of
constant negative curvature as the Gauss-Codazzi equation for surfaces of
curvature $-1$ in $3$-sphere, see [30, 51, 48] for its further applications in
many other mathematical and physical problems. It is also interesting to
remark that system (1.3) shared some similar structures with the rank two
$TT^{*}$ Toda system (1.5) which appeared in the work of Cecotti-Vafa (see
[9]) on topological-anti-topological fusion,

(1.5) $\left\\{\begin{aligned} \Delta
w_{0}-e^{w_{0}}+e^{\frac{1}{2}(w_{1}-w_{0})}&=0,\\\ \Delta
w_{1}+e^{-w_{1}}-e^{\frac{1}{2}(w_{1}-w_{0})}&=0.\end{aligned}\right.$

Indeed, if we write

$v_{1}=w_{0},\quad v_{2}=-w_{1},\quad v_{3}=\frac{1}{2}(w_{1}-w_{0}),$

then $(v_{1},v_{2},v_{3})$ satisfies

$\Delta\left(\begin{matrix}v_{1}\\\ v_{2}\\\
v_{3}\end{matrix}\right)-A\left(\begin{matrix}v_{1}\\\ v_{2}\\\
v_{3}\end{matrix}\right)=0,\quad v_{1}+v_{2}+2v_{3}=0,$

which differs from (1.1) only in the sign of $A$. For more backgrounds and
recent developments on $TT^{*}$ Toda system we refer the readers to [20, 21,
22, 23] and references therein.

From the analytic point of view, one of the fundamental issue is to establish
the compactness result of the solution space of system (1.1). It is a very
challenging problem due to the critical nonlinearity in dimension two and lack
of compactness in general situation. One of the important techniques attacking
this issue is the so called blow up analysis method, which dates back at least
to the famous work of Sacks-Uhlenbeck [50] on the investigation of the blow up
phenomena for two-dimensional harmonic maps. Since then, there are a lot of
works concerning the blow up analysis for harmonic maps [17, 15], minimal
surfaces [52], the Yamabe equation [1, 10], the Liouville equation and systems
[3, 8, 11, 12, 13, 14, 34, 35, 46] and Toda system of simple Lie algebras [27,
28, 29, 38, 36, 39, 40, 41].

Among the aforementioned equations, Liouville equation and Toda system of
simple Lie algebras have the closest relation to the system we considered in
the current article. In the pioneering work of Brezis-Merle [8], Li-Shafrir
[35], Li [34], Bartolucci-Tarantello [4, 3], Lin-Kuo [33], Wei-Zhang [54, 56,
55], the blow up phenomena for the single Liouvlle equation with singularity
has been fully understood

$\Delta u+e^{u}=4\pi\alpha\delta_{0},\quad\alpha>-1.$

After that, the following Toda system has been widely studied (See [5, 6, 16,
28, 29, 27, 31, 32, 36, 38, 39, 40, 37, 41, 45, 47])

(1.6) $\Delta
u_{i}+\sum\limits_{j=1}^{n}a_{ij}e^{u_{j}}=4\pi\alpha_{i}\delta_{0},\quad\alpha_{i}>-1,\quad
i=1,\cdots,n,$

where $A=(a_{ij})_{n\times n}$ is the Cartan matrix of a simple Lie algebra of
rank $n.$ It is known that all the simple Lie algebras are
$\mathbf{A}_{n},~{}\mathbf{B}_{n},~{}\mathbf{C}_{n},~{}\mathbf{D}_{n},~{}\mathbf{E}_{6},$
$\mathbf{E}_{7}$, $\mathbf{E}_{8}$, $\mathbf{F}_{4}$ and $\mathbf{G}_{2}$. In
[29, 27], the authors initiated the study on the blow up analysis for the
$\mathbf{A}_{2}$ Toda system. The first step of the blow up analysis is to
classify the possible values of the local mass at a blow up point $p$, which
is defined as

$\sigma_{i}(p):=\frac{1}{2\pi}\lim\limits_{r\to
0}\lim\limits_{k\to\infty}\int_{B_{r}(p)}e^{u_{i}^{k}(x)}\mathrm{d}x,\quad
i=1,2,$

for a sequence of blow up solutions $\\{(u_{1}^{k},u_{2}^{k})\\}$. Under some
mild assumptions Jost-Lin-Wang [27] proved that the local mass value
$(\sigma_{1}(p),\sigma_{2}(p))$ belongs to the following set

$\left\\{(0,2),(2,0),(0,4),(4,0),(4,4)\right\\}.$

Since then there has appeared several works [36, 38, 39, 41] concerning the
compactness issue for rank two Toda system with simple Lie algebras
$\mathbf{A}_{2}$, $\mathbf{B}_{2}(\mathbf{C}_{2})$ and $\mathbf{G}_{2}$. While
for Toda system with rank $n\geq 3$, Lin-Yang-Zhong [40] considered the cases
for $\mathbf{A}_{n}$, $\mathbf{B}_{n}$, $\mathbf{C}_{n}$ and $\mathbf{G}_{2}$.
Especially, they have shown that the formula of the local mass has a deep
connection with the Weyl group of the corresponding Lie algebra. In more
precise terms, we consider the $\mathbf{A}_{n}$ type Toda system, we set

$\beta_{i}=1+\alpha_{i},\quad i=1,\cdots,n.$

Then the local mass for the blow up solutions to (1.6) at the singular point
$0$ can be represented by

(1.7)
$\sigma_{i}(0)=2\sum_{j=0}^{i-1}\left(\sum_{\ell=1}^{f(j)}\beta_{\ell}-\sum_{\ell=1}^{j}\beta_{\ell}\right)+2N_{i},\quad
i=1,\cdots,n,$

where $N_{i}\in\mathbb{Z}$ for any $i=1,\cdots,n$ and $f$ is a permutation map
from $\\{0,\cdots,n\\}$ to itself, which is exactly isomorphic to the
associated Weyl group of the $\mathbf{A}_{n}$ type Lie algebra. The essential
point of obtaining the above formula (1.7) is that for the Lie algebras of
types $\mathbf{A}_{n}$, $\mathbf{B}_{n}$, $\mathbf{C}_{n}$ and
$\mathbf{G}_{2}$ the solution is related to an complex ODE whose coefficients
are the $W$-invariants of the Toda system. The crucial fact of the
corresponding ODE is that the local monodromy matrix is unitary. It is
interesting to mention that the representation of the underlying Lie algebra
plays a vital role in deriving the ODE. For the other types of Lie algebras,
even though we are only able to get a pseudo differential operator instead of
an ODE (see [2]), one can derive the local mass value for $\mathbf{D}_{4}$ and
$\mathbf{F}_{4}$ Toda systems. The reason is that any sub-system belongs to
one of $\mathbf{A}_{n}$, $\mathbf{B}_{n}$ or $\mathbf{C}_{n}$ type Toda
systems with lower rank, see [31].

In the current article we shall compute the local mass value of the blow up
solutions to (1.1) and try to connect it with the associated algebraic
structure of the affine Lie algebra $\mathbf{B}_{2}^{(1)}$. It is known that
(see [44] for instance) the corresponding algebraic structure of
$\mathbf{B}_{2}^{(1)}$ is the following affine Weyl group: letting $G$ be a
group generated by the following generators

(1.8)
$G=\langle\mathfrak{R}_{1},~{}\mathfrak{R}_{2},~{}\mathfrak{R}_{3}\rangle$

with $\mathfrak{R}_{i}$, $i=1,2,3$ satisfying

(1.9)
$\displaystyle\mathfrak{R}_{1}\circ\mathfrak{R}_{2}=\mathfrak{R}_{2}\circ\mathfrak{R}_{1},$
$\displaystyle|\mathfrak{R}_{1}\circ\mathfrak{R}_{2}|=2,$
$\displaystyle(\mathfrak{R}_{1}\circ\mathfrak{R}_{3})^{2}=(\mathfrak{R}_{3}\circ\mathfrak{R}_{1})^{2},$
$\displaystyle|\mathfrak{R}_{1}\circ\mathfrak{R}_{3}|=4,$
$\displaystyle(\mathfrak{R}_{2}\circ\mathfrak{R}_{3})^{2}=(\mathfrak{R}_{3}\circ\mathfrak{R}_{2})^{2},$
$\displaystyle|\mathfrak{R}_{2}\circ\mathfrak{R}_{3}|=4,$
$\displaystyle\mathfrak{R}_{1}^{2}=\mathfrak{R}_{2}^{2}=\mathfrak{R}_{3}^{2}=\mathfrak{I},$
$\displaystyle|\mathfrak{R}_{i}|=2,~{}i=1,2,3,$

where $\mathfrak{I}$ and $\circ$ denotes the identity and the operation of $G$
respectively, $|\mathfrak{R}|$ represents the order of the element
$\mathfrak{R}$, i.e., $\mathfrak{R}^{|\mathfrak{R}|}=\mathfrak{I}$. To
simplify the notation, we omit the operation notation $"\circ"$ in this
article and denote

${\mathfrak{R}_{12}=\mathfrak{R}_{1}\mathfrak{R}_{2},~{}\mathfrak{R}_{13}=\left(\mathfrak{R}_{1}\mathfrak{R}_{3}\right)^{2},~{}\mathfrak{R}_{23}=\left(\mathfrak{R}_{2}\mathfrak{R}_{3}\right)^{2}.}$

To state the main result, we formulate the problem as follows.

Let ${\bf u}^{k}=(u_{1}^{k},u_{2}^{k},u_{3}^{k})$ be a sequence of blow up
solutions to (1.1) satisfying the following conditions

(1.10) $\begin{cases}(i):&0~{}\mbox{is the only blow up point of }{\bf
u}^{k}~{}\mbox{in}~{}B_{1}(0),~{}\mbox{i.e.},\\\
&\max\limits_{i}\sup\limits_{x\in
B_{1}(0)}{u_{i}^{k}(x)+2\gamma_{i}\log|x|}\to+\infty~{}\mbox{and}~{}\max\limits_{i}\sup\limits_{K\subseteq
B_{1}(0)\setminus\\{0\\}}u_{i}^{k}\leq C(K),\\\
(ii):&|u_{i}^{k}(x)-u_{i}^{k}(y)|\leq C,~{}\forall x,y~{}\mbox{on}~{}\partial
B_{1}(0),~{}i=1,2,3,\\\ (iii):&\int_{B_{1}(0)}e^{u_{i}^{k}}\mathrm{d}x\leq
C,~{}i=1,2,3.\end{cases}$

For this sequence of blow up solutions we define the local mass by

(1.11) $\sigma_{i}=\frac{1}{2\pi}\lim_{r\to
0}\lim_{k\to+\infty}\int_{B_{r}(0)}e^{u_{i}^{k}}\mathrm{d}x,\quad i=1,2,3.$

We shall see in next section that
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ always satisfies the Pohozaev
identity

(1.12)
$(\sigma_{1}-\sigma_{3})^{2}+(\sigma_{2}-\sigma_{3})^{2}=4(\mu_{1}\sigma_{1}+\mu_{2}\sigma_{2}+2\mu_{3}\sigma_{3}),\quad\text{where}~{}\mu_{i}=1+\gamma_{i},~{}i=1,2,3.$

For a given $\bm{\mu}=(\mu_{1},\mu_{2},\mu_{3})$ we introduce the set
$\Gamma(\bm{\mu})=\Gamma(\mu_{1},\mu_{2},\mu_{3})$ for the local mass quantity
$\bm{\sigma}$ via the following way:

  1. (i)

${\bf 0}=(0,0,0)\in\Gamma(\bm{\mu})$.

  2. (ii)

If $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$, then
$\mathfrak{R}\bm{\sigma}\in\Gamma(\bm{\mu})$ for any $\mathfrak{R}\in G$
satisfying (1.8) and (1.9), where each generator $\mathfrak{R}_{i},~{}i=1,2,3$
sends $\bm{\sigma}$ to $\mathfrak{R}_{i}\bm{\sigma}$ with

$(\mathfrak{R}_{i}\bm{\sigma})_{j}=\begin{cases}4\mu_{i}-2\sum\limits_{j=1}^{3}a_{ij}\sigma_{j}+\sigma_{i},\quad&\mbox{if}~{}j=i,\\\
\sigma_{j},\quad&\mbox{if}~{}j\neq i.\end{cases}$

Here, $(a_{ij})_{3\times 3}$ is given in (1.2).

It is easy to see that for any
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$,
$\sigma_{i}$ is a degree one polynomial of $\mu_{1},~{}\mu_{2}$ and $\mu_{3}$.
Now we are able to state the first result of this article

######  Theorem 1.1.

Let $\mathbf{u}^{k}=(u^{k}_{1},u^{k}_{2},u^{k}_{3})$ be a sequence of
solutions of system (1.1) satisfying (1.10) and the local mass
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ be defined by (1.11). Then
there exists
$\hat{\bm{\sigma}}=(\hat{\sigma}_{1},\hat{\sigma}_{2},\hat{\sigma}_{3})\in\Gamma(\bm{\mu})$
such that

${\sigma}_{i}=\hat{\sigma}_{i}+4m_{i},~{}m_{i}\in\mathbb{Z},~{}i=1,2,3.$

In section 3 we can give an equivalent definition of $\Gamma(\bm{\mu})$ in
some concrete way. We have already pointed out $\bm{\sigma}$ satisfies (1.12),
based on this fact we shall prove that (see Proposition 3.4)

(1.13)
$\Gamma(\bm{\mu})=\Gamma_{N}(\bm{\mu}):=\left\\{\bm{\sigma}~{}\big{|}\big{.}~{}\bm{\sigma}~{}\mbox{satisfies}~{}\eqref{1.12}~{}\mbox{and}~{}\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j},~{}n_{ij}\in\mathbb{N}\cup\\{0\\},~{}i=1,2,3\right\\}.$

In fact, we can give explicit formulas for the coefficients
$n_{ij},~{}i,j=1,2,3$, see Theorem 3.7. As a consequence, we can restate the
Theorem 1.1 as the following corollary

######  Corollary 1.2.

Under the setting of Theorem 1.1 we can find
$\hat{\bm{\sigma}}=(\hat{\sigma}_{1},\hat{\sigma}_{2},\hat{\sigma}_{3})\in\Gamma_{{N}}(\bm{\mu})$
such that

${\sigma}_{i}=\hat{\sigma}_{i}+4m_{i},~{}m_{i}\in\mathbb{Z},~{}i=1,2,3.$

######  Remark 1.

By the fourth equation of (1.1), $\gamma_{1},~{}\gamma_{2},~{}\gamma_{3}$ are
forced to verify that $\gamma_{1}+\gamma_{2}+2\gamma_{3}=0$. Actually, the
conclusion of Theorem 1.1 still holds if we replace the fourth equation of
(1.1) by $u_{1}^{k}+u_{2}^{k}+2u_{3}^{k}=C_{k}$ with $C_{k}$ being uniformly
bounded from above for any $k$.

######  Remark 2.

The computation on the local mass for (1.1) without singularity is given by
Liu-Wang [43], they have shown that each $\sigma_{i}$ is multiple of $4$, see
Remark 7 for the explicit expressions. When $u_{1}=u_{2}$, system (1.1) is
reduced to the sinh-Gordon equation (1.4) with singular source (see system
(7.2)). In Section 7 we shall present the results on the computation of its
local mass, which extends the corresponding quantitative results of the sinh-
Gordon equation (1.4) in [30, 24, 25] to the case with singularity.

######  Remark 3.

In [40] we have already seen that the Weyl group of the corresponding Lie
algebra plays an important role in determining the local mass of the blow-up
solutions. From the result of Theorem 1.1 and Theorem 3.7, we find that
similar things also happen for the affine Toda system. However, there is a
major difference between these two types of systems. In the former one, the
Weyl group is just the permutation map from $\\{0,1,\cdots,n\\}$ to itself and
the number of elements is finite; while for the later one, from the work [24]
and [43] we see that there are infinite choices for the values of local mass
due to the fact that the corresponding affine Weyl group has infinite number
of elements. Even though it is not difficult to guess that there are two free
integers in the expression of the local mass for $\mathbf{B}_{2}^{(1)}$ Toda
system (there is only one free integer for the sinh-Gordon equation and
Tzitzéica equation, see [30, 24, 25] and [26, 53] for these two equations
respectively), it is not easy to write all of them in a clean form. Instead,
we have found that there are 8 types of expression formulas according to the
remainder integers after modulo the number $4$, see Theorem 3.7 for details.

Theorem 1.1 has several important applications for the affine Toda system
defined on compact Riemann surface. For example, let $(M,g)$ be a compact
Riemann surface, $\Delta_{g}$ be the Beltrami operator on $M$, we consider a
sequence of solutions $\mathbf{u}^{k}=(u^{k}_{1},u^{k}_{2},u^{k}_{3})$ to the
following system

(1.14)
$\begin{cases}\Delta_{g}u^{k}_{i}+\sum\limits_{j=1}^{3}a_{ij}\rho^{k}_{j}\left(\frac{h^{k}_{j}e^{u^{k}_{j}}}{\int_{M}h^{k}_{j}e^{u^{k}_{j}}\mathrm{d}V_{g}}-\frac{1}{|M|}\right)=4\pi{\sum\limits_{p\in
S}\alpha_{p,i}\left(\delta_{p}-\frac{1}{|M|}\right)},\quad i=1,2,3,\\\
u^{k}_{1}+u^{k}_{2}+2u^{k}_{3}=0\quad\text{on}\quad M,\end{cases}$

where $h^{k}_{1},h^{k}_{2},h^{k}_{3}$ are positive and smooth functions on $M$
with their $C^{3}(M)$ norm being uniformly bounded in $M$, $S$ is a finite set
of $M$, $\alpha_{p,i}>-1$ is the strength of the Dirac mass $\delta_{p}$, and
$\bm{\rho}^{k}=(\rho^{k}_{1},\rho^{k}_{2},\rho^{k}_{3})$ is a sequence of
constant vectors with nonnegative components satisfying
$\lim\limits_{k\to+\infty}\bm{\rho}^{k}=(\rho_{1},\rho_{2},\rho_{3})$. As the
Toda system with simple Lie algebra and Liouville equation, (1.14) can be
regarded as a natural extension of the local version (1.1) to the case on
Riemannian manifold. Equation (1.14) together with (1.1) are closely related
to the nonlinear Klein-Gordon equations ([19]) and the minimal tori in
$\mathbb{S}^{4}$ ([18]). We notice that (1.14) remains the same if $u_{i}$ is
replaced by $u_{i}+c_{i}$ for any constant $c_{i}$. Thus it is reasonable to
assume that each component of $\mathbf{u}^{k}$ is in

$\mathring{H}^{1}(M)=\left\\{f\in
H^{1}(M)~{}\big{|}~{}\int_{M}f\mathrm{d}V_{g}=0\right\\}.$

Simultaneously, from the equation $u_{1}^{k}+u_{2}^{k}+2u_{3}^{k}=0$ one can
easily verify that

(1.15) $\alpha_{p,1}+\alpha_{p,2}+2\alpha_{p,3}=0,\quad\forall p\in S.$

By introducing the following Green function on Riemann surface $M$

$\Delta
G(x,p)+\left(\delta_{p}-\frac{1}{|M|}\right)=0,\quad\int_{M}G(x,p)\mathrm{d}V_{g}=0,$

we decompose

$u_{i}^{k}(x)=\tilde{u}_{i}^{k}(x)-4\pi\sum\limits_{p\in
S}\alpha_{p,i}G(x,p),\quad i=1,2,3.$

Then (1.14) can be rewritten as

(1.16)
$\begin{cases}\Delta_{g}\tilde{u}_{i}^{k}+\sum\limits_{j=1}^{3}a_{ij}\rho^{k}_{j}\left(\frac{\tilde{h}^{k}_{j}e^{\tilde{u}^{k}_{j}}}{\int_{M}\tilde{h}^{k}_{j}e^{\tilde{u}^{k}_{j}}\mathrm{d}V_{g}}-\frac{1}{|M|}\right)=0,\quad
i=1,2,3,\\\
\tilde{u}^{k}_{1}+\tilde{u}^{k}_{2}+2\tilde{u}^{k}_{3}=0\quad\text{on}\quad
M,\end{cases}$

where

$\tilde{h}_{i}^{k}=h_{i}^{k}\exp\left(-4\pi\sum\limits_{p\in
S}\alpha_{p,i}G(x,p)\right),\quad i=1,2,3.$

Let $\mu_{p,i}=1+\alpha_{p,i}$ for $i=1,2,3$ and denote

$\Gamma_{i}=\left\\{2\pi\sum\limits_{p\in R}\sigma_{p,i}+8\pi
m_{i}~{}\big{|}~{}(\sigma_{p,1},\sigma_{p,2},\sigma_{p,3})\in\Gamma(\mu_{p,1},\mu_{p,2},\mu_{p,3}),~{}R\subseteq
S,~{}m_{i}\in\mathbb{Z}\right\\}.$

The second result in this article is the following a priori estimate for the
system (1.16).

######  Theorem 1.3.

Suppose that $\rho_{i}\notin\Gamma_{i}$ for $i=1,2,3$. Then there exists a
constant $C>0$ depending on $\rho_{i}$ such that for any solution
$(\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{3})$ of system (1.16) in
$\mathring{H}^{1}(M)$,

$|\tilde{u}_{i}(x)|\leq C,~{}\forall x\in M,\quad i=1,2,3.$

To prove Theorem 1.3, we shall determine the local mass at each blow-up point
of solutions $\mathbf{\tilde{u}}^{k}$. Here the blow up point $p$ is defined
by

$\exists\mbox{~{}a sequence }p^{k}\to p~{}\mbox{such that
}\max_{i}\tilde{u}_{i}^{k}(p^{k})\to+\infty.$

The set $B$ of all blow up points is called the blow up set. For each point
$p\in B$ we define the local mass by

$\sigma_{i}(p)=\frac{1}{2\pi}\lim\limits_{r\to
0}\lim\limits_{k\to+\infty}\frac{\rho^{k}_{i}\int_{B(p,r)}\tilde{h}^{k}_{i}e^{\tilde{u}^{k}_{i}}\mathrm{d}V_{g}}{\int_{M}\tilde{h}^{k}_{i}e^{\tilde{u}^{k}_{i}}\mathrm{d}V_{g}},\quad
i=1,2,3.$

By Theorem 1.1 we are able to compute $\sigma_{i}(p)$. Together with the
assumption $\rho_{i}\notin\Gamma_{i}$ we shall see that $B=\emptyset$ and it
leads to the compactness of (1.16).

The organization of this article is as follows. In Section 2, we establish a
selection process of a sequence of blow up solutions, an oscillation estimate
outside the blow-up set, and the local Pohozaev identities corresponding to
the system (1.1). In Section 3, we study the Pohozaev identity of blow up
solutions and classify all the possible values of $\Gamma(\bm{\mu})$. In
Section 4, we present several important lemmas for later use. In Section 5 and
Section 6, we discuss the local mass on each bubbling disk centered at $0$ and
not at $0$ respectively, then we prove Theorem 1.1 by grouping these bubbling
areas together. Furthermore, we sketch the proof of Theorem 1.3. In Section 7,
we carry out the blow-up analysis for sinh-Gordon equation with singular
source. In Section 8, we collect some classification results which are used in
our proof.

In this article, we always use the notation $r_{k}=o(1)s_{k}$ to express
$r_{k}/s_{k}\to 0$ as $k\to\infty$, and the notation $r_{k}=O(1)s_{k}$ to
represent $C^{-1}\leq r_{k}/s_{k}\leq C$ as $k\to\infty$ for some constant
$C>0$. For the sake of brevity, we use boldface type for sequence and vectors,
such as $\mathbf{s}=\\{s_{k}\\}$, $\mathbf{r}=\\{r_{k}\\}$,
$\bm{\tau}=\\{\tau_{k}\\}$ and
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$. We will not distinguish
sequence and subsequence in this article, i.e., we shall still denote
$\mathbf{u}^{k}$ the subsequence of itself if necessary.

##  2\. Bubbling analysis and Pohozaev identity

In this section, we analyze the bubbling areas by a standard selection
procedure and establish some type of Pohozaev identity for local mass.

######  Proposition 2.1.

Let $\mathbf{u}^{k}$ be a sequence of solutions of system (1.1) satisfying
(1.10). Then there exists a sequence of finite points
$\Sigma_{k}:=\\{0,x^{k}_{1},\cdots,x^{k}_{m}\\}$ (if $0$ is not a singular
point, then $0$ can be deleted from $\Sigma_{k}$) and a sequence of positive
numbers $l^{k}_{1},\cdots,l^{k}_{m}$ such that

  1. (1)

$x^{k}_{j}\to 0$ and $l^{k}_{j}\to 0$ as $k\to+\infty$,
$l^{k}_{j}\leq\frac{1}{2}\mathrm{dist}(x^{k}_{j},\Sigma_{k}\setminus\\{x^{k}_{j}\\})$,
$j=1,\cdots,m$. Furthermore, $B(x^{k}_{i},l_{i}^{k})\cap
B(x^{k}_{j},l_{j}^{k})=\emptyset$ for $1\leq i,j\leq m,~{}i\neq j$.

  2. (2)

$\max\limits_{i=1,2,3}u^{k}_{i}(x^{k}_{j})=\max\limits_{i=1,2,3}\max\limits_{B(x^{k}_{j},l^{k}_{j})}u^{k}_{i}(x)\to+\infty$
as $k\to+\infty$, $j=1,\cdots,m$. Denote

$\varepsilon_{j}^{k}:=e^{-\frac{1}{2}\max\limits_{i=1,2,3}u^{k}_{i}(x^{k}_{j})},\quad
j=1,\cdots,m.$

Then $R^{k}_{j}:=\frac{l^{k}_{j}}{\varepsilon_{j}^{k}}\to+\infty$ as
$k\to+\infty$, $j=1,\cdots,m$.

  3. (3)

In each $B(x^{k}_{i},l_{i}^{k})$, set

$v^{k}_{i}(y):=u^{k}_{i}(x^{k}_{j}+\varepsilon_{j}^{k}y)+2\log\varepsilon_{j}^{k},\quad
i=1,2,3.$

Then one of the following alternatives holds:

    1. (a)

$v^{k}_{1}$ and $v^{k}_{3}$ (or $v^{k}_{2}$ and $v^{k}_{3}$) converge to a
solution of the following Toda system in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$,

$\begin{cases}-\Delta v=e^{v}-e^{w},\\\ -\Delta
w=-\frac{1}{2}e^{v}+e^{w},\end{cases}~{}~{}\text{in}\quad\mathbb{R}^{2},$

while $v^{k}_{2}$ (or $v^{k}_{1}$) converges to $-\infty$ in
$L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$;

    2. (b)

$v^{k}_{1}$ and $v^{k}_{2}$ converge to solutions of Liouville equation in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$, while $v^{k}_{3}$ converges to
$-\infty$ in $L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$;

    3. (c)

One component of $\mathbf{v}^{k}$ converges to a solution of Liouville
equation in $C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$, while the left ones
converge to $-\infty$ in $L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$.

  4. (4)

There exists a constant $C$ independent of $k$ such that the following
Harnack-type inequality holds:

(2.1)
$\max\limits_{i=1,2,3}\left\\{u^{k}_{i}(x)+2\log\mathrm{dist}(x,\Sigma_{k})\right\\}\leq
C,\quad\forall x\in B_{1}(0).$

###### Proof.

We construct $\Sigma_{k}$ by induction. If
$\gamma_{i}=0~{}\text{for}~{}i=1,2,3$, i.e., the system (1.1) has no
singularity at $0$, we start with $\Sigma_{k}=\emptyset$ and the proof is in
line with that of [43, Lemma 2.1]. Otherwise, we start with
$\Sigma_{k}=\\{0\\}$ and the proof is similar. But it should be mentioned that
all components of ${\mathbf{v}}^{k}$ cannot be bounded at the same time by the
fourth equation of (1.1). ∎

We have the following oscillation estimate generated by the Harnack-type
inequality (2.1).

######  Proposition 2.2.

Let $\Sigma_{k}=\\{0,x^{k}_{1},\cdots,x^{k}_{m}\\}$ be the blow-up set (if $0$
is not a singular point, then $0$ can be deleted from $\Sigma_{k}$). Suppose
that $\mathbf{u}^{k}$ is a sequence of solutions to system (1.1) satisfying
(1.10). Then for any $x_{0}\in B_{1}(0)\setminus\Sigma_{k}$, there exists a
constant $C_{0}$ independent of $x_{0}$ and $k$ such that

$|u^{k}_{i}(x_{1})-u^{k}_{i}(x_{2})|\leq C_{0},~{}~{}\forall x_{1},x_{2}\in
B(x_{0},\mathrm{dist}(x_{0},\Sigma_{k})/2),\quad i=1,2,3.$

###### Proof.

One can prove this result by following the same arguments of [38, Lemma 2.1]
and [43, Lemma 2.2], here we omit the details. ∎

Next we introduce the definitions of fast decay and slow decay.

######  Definition 2.3.

  1. (i)

We say $u^{k}_{i}$ has fast decay on $\partial B(x_{k},r_{k})$ if

$u^{k}_{i}(x)+2\log|x-x_{k}|\leq-N_{k},\quad\forall x\in\partial
B(x_{k},r_{k}),$

for some $N_{k}\to+\infty$ as $k\to+\infty$.

  2. (ii)

We say $u^{k}_{i}$ has slow decay on $\partial B(x_{k},r_{k})$ if

$u^{k}_{i}(x)+2\log|x-x_{k}|\geq-C,\quad\forall x\in\partial B(x_{k},r_{k}),$

for some constant $C>0$ which is independent of $k$.

At the end of this section, we can establish a Pohozaev identity of local
masses for the system (1.1). In Proposition 2.4 and Remark 4, we denote

$\sigma(r,x_{0};u):=\frac{1}{2\pi}\int_{B_{r}(x_{0})}e^{u(x)}\mathrm{d}x,~{}~{}\sigma_{i}^{k}(r,x_{0}):=\sigma(r,x_{0};u^{k}_{i}),~{}~{}\sigma_{i}^{k}(r):=\sigma(r,0;u^{k}_{i}),\quad
i=1,2,3.$

Let $x^{k}_{l}\in\Sigma_{k}$ and
$\tau^{k}_{l}=\frac{1}{2}\mathrm{dist}\left(x^{k}_{l},\Sigma_{k}\setminus\\{x^{k}_{l}\\}\right)$,
then we can derive from system (1.1) and Proposition 2.2 that

(2.2) $u^{k}_{i}(x)=\overline{u}^{k}_{x^{k}_{l},i}(r)+O(1),\quad x\in
B(x^{k}_{l},\tau^{k}_{l}),$

where $r=|x-x^{k}_{l}|$ and

$\overline{u}^{k}_{x^{k}_{l},i}(r)=\frac{1}{2\pi r}\int_{\partial
B(x^{k}_{l},r)}u^{k}_{i}~{}\mathrm{d}S,$

and $O(1)$ is independent of $r$ and $k$.

######  Proposition 2.4.

Let $\Sigma_{k}^{\prime}\subseteq\Sigma_{k}$ be a subset of $\Sigma_{k}$ with
$0\in\Sigma_{k}^{\prime}\subseteq B(x_{k},r_{k})\subseteq B_{1}(0)$, and there
holds

$\mathrm{dist}(\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k}))=o(1)\mathrm{dist}(\Sigma_{k}\setminus\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k})).$

Suppose $u^{k}_{1},u^{k}_{2},u^{k}_{3}$ have fast decay on $\partial
B(x_{k},r_{k})$. Then we have the following Pohozaev identity

(2.3)
$\displaystyle\left(\sigma^{k}_{1}(r_{k},x_{k})-\sigma^{k}_{3}(r_{k},x_{k})-2\gamma_{1}\right)^{2}+\left(\sigma^{k}_{2}(r_{k},x_{k})-\sigma^{k}_{3}(r_{k},x_{k})-2\gamma_{2}\right)^{2}$
$\displaystyle=4\left(\sigma^{k}_{1}(r_{k},x_{k})+\sigma^{k}_{2}(r_{k},x_{k})+2\sigma^{k}_{3}(r_{k},x_{k})\right)+4\left(\gamma_{1}^{2}+\gamma_{2}^{2}\right)+o(1).$

Furthermore, let $\mu_{i}=1+\gamma_{i}$ for $i=1,2,3$, then
${\bm{\sigma}}=(\sigma_{1},\sigma_{2},\sigma_{3})$ satisfies the Pohozaev
identity

(2.4)
$(\sigma_{1}-\sigma_{3})^{2}+(\sigma_{2}-\sigma_{3})^{2}=4(\mu_{1}\sigma_{1}+\mu_{2}\sigma_{2}+2\mu_{3}\sigma_{3}),$

where $\mu_{1}+\mu_{2}+2\mu_{3}=4$ and
$\sigma_{i}=\lim\limits_{k\to+\infty}\sigma^{k}_{i}(r_{k},x_{k})$ for
$i=1,2,3$.

###### Proof.

Without loss of generality, we assume $x_{k}=0$. For fixed $\theta\in(0,1)$,
let $\Omega=B_{\theta}(0)\setminus B_{\varepsilon}(0)$, where $\varepsilon>0$
is small enough. Then applying the standard Pohozaev identity for the first
two equations of (1.1) on $\Omega$, we have

$\displaystyle\frac{1}{2}\int_{\partial\Omega}|\nabla
u^{k}_{i}|^{2}(x\cdot\nu)\mathrm{d}S-\int_{\partial\Omega}(x\cdot\nabla
u^{k}_{i})(\nu\cdot\nabla u^{k}_{i})\mathrm{d}S$
$\displaystyle=-2\int_{\Omega}e^{u^{k}_{i}}\mathrm{d}x+\int_{\partial\Omega}e^{u^{k}_{i}}(x\cdot\nu)\mathrm{d}S-\int_{\Omega}e^{u^{k}_{3}}(x\cdot\nabla
u^{k}_{i})\mathrm{d}x,\quad i=1,2,$

where $\nu$ denotes the unit outer normal vector. Combining the above
equations and the fourth equation of (1.1) we derive that

(2.5) $\displaystyle\frac{1}{2}\int_{\partial\Omega}\left(|\nabla
u^{k}_{1}|^{2}+|\nabla
u^{k}_{2}|^{2}\right)(x\cdot\nu)\mathrm{d}S-\int_{\partial\Omega}\left((x\cdot\nabla
u^{k}_{1})(\nu\cdot\nabla u^{k}_{1})+(x\cdot\nabla u^{k}_{2})(\nu\cdot\nabla
u^{k}_{2})\right)\mathrm{d}S$
$\displaystyle=-2\int_{\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}})\mathrm{d}x+\int_{\partial\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}})(x\cdot\nu)\mathrm{d}S-\int_{\Omega}e^{u^{k}_{3}}\left(x\cdot(\nabla
u^{k}_{1}+\nabla u^{k}_{2})\right)\mathrm{d}x$
$\displaystyle=-2\int_{\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}})\mathrm{d}x+\int_{\partial\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}})(x\cdot\nu)\mathrm{d}S+2\int_{\Omega}e^{u^{k}_{3}}(x\cdot\nabla
u^{k}_{3})\mathrm{d}x$
$\displaystyle=-2\int_{\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}x+\int_{\partial\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})(x\cdot\nu)\mathrm{d}S.$

We calculate the L.H.S. and R.H.S. of (2.5) respectively. By (iii) of (1.10),
we obtain

R.H.S.
$\displaystyle=-2\int_{\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}x+\int_{\partial\Omega}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})(x\cdot\nu)\mathrm{d}S$
$\displaystyle=-2\int_{B_{\theta}(0)\setminus
B_{\varepsilon}(0)}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}x+\theta\int_{\partial
B_{\theta}(0)}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}S$
$\displaystyle\quad-\varepsilon\int_{\partial
B_{\varepsilon}(0)}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}S+o(1)$
$\displaystyle=-4\pi(\sigma^{k}_{1}(\theta)+\sigma^{k}_{2}(\theta)+2\sigma^{k}_{3}(\theta))+\theta\int_{\partial
B_{\theta}(0)}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}S+o(1).$

On the other hand, we have

L.H.S. $\displaystyle=\frac{1}{2}\int_{\partial\Omega}\left(|\nabla
u^{k}_{1}|^{2}+|\nabla
u^{k}_{2}|^{2}\right)(x\cdot\nu)\mathrm{d}S-\int_{\partial\Omega}\left((x\cdot\nabla
u^{k}_{1})(\nu\cdot\nabla u^{k}_{1})+(x\cdot\nabla u^{k}_{2})(\nu\cdot\nabla
u^{k}_{2})\right)\mathrm{d}S$ $\displaystyle=\frac{\theta}{2}\int_{\partial
B_{\theta}(0)}\left(|\nabla u^{k}_{1}|^{2}+|\nabla
u^{k}_{2}|^{2}\right)\mathrm{d}S-\frac{\varepsilon}{2}\int_{\partial
B_{\varepsilon}(0)}(|\nabla u^{k}_{1}|^{2}+|\nabla u^{k}_{2}|^{2})\mathrm{d}S$
$\displaystyle\quad-\theta\int_{\partial
B_{\theta}(0)}\left(\left(\nu\cdot\nabla{u^{k}_{1}}\right)^{2}+\left(\nu\cdot\nabla{u^{k}_{2}}\right)^{2}\right)\mathrm{d}S+\varepsilon\int_{\partial
B_{\varepsilon}(0)}\left(\left(\nu\cdot\nabla{u^{k}_{1}}\right)^{2}+\left(\nu\cdot\nabla{u^{k}_{2}}\right)^{2}\right)\mathrm{d}S.$

Applying the same argument of [42, Lemma 4.1], we deduce the following
estimate

(2.6) $\nabla u^{k}_{i}(x)=-2\gamma_{i}x/|x|^{2}+o(1)~{}~{}\text{near the
origin},$

which implies that

$\displaystyle-\frac{\varepsilon}{2}\int_{\partial
B_{\varepsilon}(0)}\left(|\nabla u^{k}_{1}|^{2}+|\nabla
u^{k}_{2}|^{2}\right)\mathrm{d}S+\varepsilon\int_{\partial
B_{\varepsilon}(0)}\left(\left(\nu\cdot\nabla{u^{k}_{1}}\right)^{2}+\left(\nu\cdot\nabla{u^{k}_{2}}\right)^{2}\right)\mathrm{d}S=4\pi(\gamma_{1}^{2}+\gamma_{2}^{2})+o(1).$

Hence we can rewrite (2.5) as

(2.7) $\displaystyle\frac{\theta}{2}\int_{\partial B_{\theta}(0)}(|\nabla
u^{k}_{1}|^{2}+|\nabla u^{k}_{2}|^{2})\mathrm{d}S-\theta\int_{\partial
B_{\theta}(0)}\left(\left(\nu\cdot\nabla{u^{k}_{1}}\right)^{2}+\left(\nu\cdot\nabla{u^{k}_{2}}\right)^{2}\right)\mathrm{d}S+4\pi(\gamma_{1}^{2}+\gamma_{2}^{2})$
$\displaystyle=-4\pi\left(\sigma^{k}_{1}(\theta)+\sigma^{k}_{2}(\theta)+2\sigma^{k}_{3}(\theta)\right)+\theta\int_{\partial
B_{\theta}(0)}(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}})\mathrm{d}S+o(1).$

Since $u^{k}_{1},u^{k}_{2},u^{k}_{3}$ have fast decay on $\partial
B(0,{r_{k}})$, i.e.,

$u^{k}_{i}(x)+2\log|x|\leq-N_{k},~{}|x|=r_{k},\quad i=1,2,3$

for some $N_{k}\to+\infty$, then by [43, Lemma 2.4], there exists
$R^{k}_{1}\to+\infty$ such that

$R^{k}_{1}\mathrm{dist}(\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k}))=o(1)\mathrm{dist}(\Sigma_{k}\setminus\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k}))$

and

$u^{k}_{i}(x)+2\log|x|\leq-N_{k},\quad r_{k}\leq|x|\leq R^{k}_{1}r_{k},$

for some $N_{k}\to+\infty$. So we can choose $R^{k}\to+\infty$ with $R^{k}\leq
R^{k}_{1}$ such that $u^{k}_{1},u^{k}_{2},u^{k}_{3}$ have fast decay on
$\partial B(0,R^{k}r_{k})$ and

$\int_{B(0,R^{k}r_{k})\setminus B(0,r_{k})}e^{u^{k}_{i}(x)}\mathrm{d}x\leq
Ce^{-N_{k}}\log R_{k}=o(1),\quad i=1,2,3,$

which implies that

$\sigma^{k}_{i}(R^{k}r_{k})=\sigma^{k}_{i}(r_{k})+o(1),\quad i=1,2,3.$

Letting $\theta=\sqrt{R^{k}}r_{k}$, then we have

$-4\pi(\sigma^{k}_{1}(\theta)+\sigma^{k}_{2}(\theta)+2\sigma^{k}_{3}(\theta))=-4\pi(\sigma^{k}_{1}(r_{k})+\sigma^{k}_{2}(r_{k})+2\sigma^{k}_{3}(r_{k}))+o(1)$

and

$\theta\int_{\partial
B_{\theta}(0)}\left(e^{u^{k}_{1}}+e^{u^{k}_{2}}+2e^{u^{k}_{3}}\right)\mathrm{d}S=o(1),$

since $u^{k}_{1},u^{k}_{2},u^{k}_{3}$ have fast decay on $\partial
B(0,\sqrt{R^{k}}r_{k})$. Therefore, (2.7) can be rewritten as

(2.8) $\displaystyle-\frac{\sqrt{R^{k}}r_{k}}{2}\int_{\partial
B_{\sqrt{R^{k}}r_{k}}(0)}\left(|\nabla u^{k}_{1}|^{2}+|\nabla
u^{k}_{2}|^{2}\right)\mathrm{d}S+\sqrt{R^{k}}r_{k}\int_{\partial
B_{\sqrt{R^{k}}r_{k}}(0)}\left(\left(\nu\cdot\nabla{u^{k}_{1}}\right)^{2}+\left(\nu\cdot\nabla{u^{k}_{2}}\right)^{2}\right)\mathrm{d}S$
$\displaystyle=4\pi\left(\sigma^{k}_{1}(r_{k})+\sigma^{k}_{2}(r_{k})+2\sigma^{k}_{3}(r_{k})\right)+4\pi\left(\gamma_{1}^{2}+\gamma_{2}^{2}\right)+o(1).$

Similar to the estimate (2.6) (by [42, Lemma 4.1] and a scaling argument), we
obtain

$\nabla
u^{k}_{1}(x)=-\frac{x}{|x|^{2}}(\sigma^{k}_{1}(r_{k})-\sigma^{k}_{3}(r_{k})-2\gamma_{1})+\frac{o(1)}{|x|},~{}x\in\partial
B_{\sqrt{R^{k}}r_{k}}(0),$

and

$\nabla
u^{k}_{2}(x)=-\frac{x}{|x|^{2}}(\sigma^{k}_{2}(r_{k})-\sigma^{k}_{3}(r_{k})-2\gamma_{2})+\frac{o(1)}{|x|},~{}x\in\partial
B_{\sqrt{R^{k}}r_{k}}(0).$

Therefore, we conclude from (2.8) that

$\left(\sigma^{k}_{1}(r_{k})-\sigma^{k}_{3}(r_{k})-2\gamma_{1}\right)^{2}+\left(\sigma^{k}_{2}(r_{k})-\sigma^{k}_{3}(r_{k})-2\gamma_{2}\right)^{2}=4\left(\sigma^{k}_{1}(r_{k})+\sigma^{k}_{2}(r_{k})+2\sigma^{k}_{3}(r_{k})\right)+4\left(\gamma_{1}^{2}+\gamma_{2}^{2}\right)+o(1).$

Finally, taking $\mu_{i}=1+\gamma_{i}$ for $i=1,2,3$ and letting
$k\to+\infty$, we can rewrite (2.3) as (2.4). This completes the proof of
Proposition 2.4. ∎

######  Remark 4.

Let $\Sigma_{k}^{\prime}\subseteq\Sigma_{k}$ be a subset of $\Sigma_{k}$ with
$0\notin\Sigma_{k}^{\prime}\subseteq B(x_{k},r_{k})\subseteq B_{1}(0)$, and
there holds

$\mathrm{dist}(\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k}))=o(1)\mathrm{dist}(\Sigma_{k}\setminus\Sigma_{k}^{\prime},\partial
B(x_{k},r_{k})).$

Suppose $u^{k}_{1},u^{k}_{2},u^{k}_{3}$ have fast decay on $\partial
B(x_{k},r_{k})$. Then we have the following Pohozaev identity

$\displaystyle\left(\sigma^{k}_{1}(r_{k},x_{k})-\sigma^{k}_{3}(r_{k},x_{k})\right)^{2}+\left(\sigma^{k}_{2}(r_{k},x_{k})-\sigma^{k}_{3}(r_{k},x_{k})\right)^{2}=4\left(\sigma^{k}_{1}(r_{k},x_{k})+\sigma^{k}_{2}(r_{k},x_{k})+2\sigma^{k}_{3}(r_{k},x_{k})\right)+o(1).$

In other words,

$(\sigma_{1}-\sigma_{3})^{2}+(\sigma_{2}-\sigma_{3})^{2}=4(\sigma_{1}+\sigma_{2}+2\sigma_{3}).$

##  3\. The structure of $\Gamma(\mu_{1},\mu_{2},\mu_{3})$ and affine Weyl
group of $\mathbf{B}_{2}^{(1)}$

In this section, we shall compute all the possible local mass values for the
blow up solutions to (1.1) by studying the set $\Gamma(\bm{\mu})$, which is
introduced in the introduction (Section 1). First of all, we prove that each
element of $\Gamma(\bm{\mu})$ verifies the Pohozaev identity (2.4).

######  Proposition 3.1.

For each element $\bm{\sigma}\in\Gamma(\bm{\mu})$, $\bm{\sigma}$ satisfies the
Pohozaev identity (2.4).

###### Proof.

One can get it easily by Vieta’s theorem and Pohozaev identity (2.4). ∎

As indicated by the setting of $G$, there might have several ways defining the
same elements in $\Gamma(\bm{\mu})$. In order to represent each element in the
cheapest way we introduce the following concept

######  Definition 3.2.

We say
$\widetilde{\mathfrak{R}}=\mathfrak{R}_{i_{n}}\mathfrak{R}_{i_{n-1}}\cdots\mathfrak{R}_{i_{1}}$
($i_{a}\in\\{1,2,3\\},~{}1\leq a\leq n$) is a simplest chain with length $n$
if $\widetilde{\mathfrak{R}}$ can not be reduced to a sub-chain
$\mathfrak{R}_{j_{m}}\mathfrak{R}_{j_{m-1}}\cdots\mathfrak{R}_{j_{1}}$ for any
$m<n$, $j_{a}\in\\{1,2,3\\},~{}1\leq a\leq m$.

######  Remark 5.

For example,
$\mathfrak{R}_{3}\mathfrak{R}_{1}\mathfrak{R}_{2}\mathfrak{R}_{1}$ is not a
simplest chain, while
$\mathfrak{R}_{3}\mathfrak{R}_{1}\mathfrak{R}_{2}\mathfrak{R}_{3}$ is a
simplest chain with length $4$.

######  Proposition 3.3.

For any $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$,
$\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j}$ for some
$n_{ij}\in\mathbb{N}\cup\\{0\\}$.

###### Proof.

By induction, one can easily verify that
$\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j}$ is a polynomial of $\mu_{j}$ with
$n_{ij}\in\mathbb{N}\cup\\{0\\}$. In fact, direct computation shows that the
first $3$ levels of $\Gamma(\bm{\mu})$ are given as

$\cancel{(0,0,0)}$ $(4\mu_{1},0,0)$ $(4\mu_{1},4\mu_{2},0)\cdots$
$(4\mu_{1},0,4\mu_{1}+4\mu_{3})\cdots$ $(4\mu_{1},4\mu_{2},0)\cdots$ $\bm{0}$
$(0,4\mu_{2},0)$
$\cancel{(0,0,0)}$$(0,4\mu_{2},4\mu_{2}+4\mu_{3})\cdots$$(4\mu_{1}+8\mu_{3},0,4\mu_{3})\cdots$$(0,0,4\mu_{3})$
$(0,4\mu_{2}+8\mu_{3},4\mu_{3})\cdots$$\cancel{(0,0,0)}$$1$st level $2$nd
level $3$rd level $\cdots$
$\mathfrak{R}_{1}$$\mathfrak{R}_{2}$$\mathfrak{R}_{3}$$\mathfrak{R}_{1}$$\mathfrak{R}_{2}$$\mathfrak{R}_{3}$$\mathfrak{R}_{1}$$\mathfrak{R}_{2}$$\mathfrak{R}_{3}$$\mathfrak{R}_{1}$$\mathfrak{R}_{2}$$\mathfrak{R}_{3}$

For any $\bm{\sigma}\in\Gamma(\bm{\mu})$, there exists a simplest chain
$\widetilde{\mathfrak{R}}=\mathfrak{R}_{i_{n}}\mathfrak{R}_{i_{n-1}}\cdots\mathfrak{R}_{i_{1}}$
with length $n$ such that $\bm{\sigma}=\widetilde{\mathfrak{R}}(\bm{0})$. In
this case, we say that $\bm{\sigma}$ is in the $n$-th level. Now the
conclusion is true for those $\bm{\sigma}$ in the $\mathcal{L}$-th level,
$1\leq\mathcal{L}\leq 3$. By induction, we suppose that the conclusion holds
for all $\bm{\sigma}^{(L)}$ in the $L$-th level, then we shall prove it also
holds in the $(L+1)$-th level. Let
$\bm{\sigma}^{(L)}=(\sigma_{1}^{(L)},\sigma_{2}^{(L)},\sigma_{3}^{(L)})$,
where $\sigma_{i}^{(L)}=4\sum_{j=1}^{3}n_{ij}^{(L)}\mu_{j}$,
$n_{ij}^{(L)}\in\mathbb{N}\cup\\{0\\}$. Then

$\left(\mathfrak{R}_{1}\bm{\sigma}^{(L)}\right)_{1}=\left(4-4n_{11}^{(L)}+8n_{31}^{(L)}\right)\mu_{1}+\left(-4n_{12}^{(L)}+8n_{32}^{(L)}\right)\mu_{2}+\left(-4n_{13}^{(L)}+8n_{33}^{(L)}\right)\mu_{3},$
$\left(\mathfrak{R}_{2}\bm{\sigma}^{(L)}\right)_{2}=\left(-4n_{21}^{(L)}+8n_{31}^{(L)}\right)\mu_{1}+\left(4-4n_{22}^{(L)}+8n_{32}^{(L)}\right)\mu_{2}+\left(-4n_{23}^{(L)}+8n_{33}^{(L)}\right)\mu_{3},$

and

$\left(\mathfrak{R}_{3}\bm{\sigma}^{(L)}\right)_{3}=\left(4n_{11}^{(L)}+4n_{21}^{(L)}-4n_{31}^{(L)}\right)\mu_{1}+\left(4n_{12}^{(L)}+4n_{22}^{(L)}-4n_{32}^{(L)}\right)\mu_{2}+\left(4+4n_{13}^{(L)}+4n_{23}^{(L)}-4n_{33}^{(L)}\right)\mu_{3}.$

Obviously, the coefficients of $\mu_{i}$ in the above components are multiples
of $4$. Thus we shall prove that these coefficients are nonnegative. We divide
the proof into two steps as follows.

Step 1. We claim

(3.1) $4-4n_{11}^{(L)}+8n_{31}^{(L)}\geq
0,\quad-4n_{12}^{(L)}+8n_{32}^{(L)}\geq
0,\quad-4n_{13}^{(L)}+8n_{33}^{(L)}\geq 0,$

and

(3.2) $-4n_{21}^{(L)}+8n_{31}^{(L)}\geq 0,\quad
4-4n_{22}^{(L)}+8n_{32}^{(L)}\geq 0,\quad-4n_{23}^{(L)}+8n_{33}^{(L)}\geq 0.$

Since the proof for (3.1) and (3.2) are almost the same, we shall only give
details for the first one. Using (2.4) and substituting the expression of
$\bm{\sigma}^{(L)}$ (or $\mathfrak{R}_{1}\bm{\sigma}^{(L)}$) into (2.4), we
can obtain the following three equations by comparing the coefficients of
$\mu_{j}^{2},~{}j=1,2,3$,

(3.3)
$\displaystyle\left(n_{11}^{(L)}-n_{31}^{(L)}\right)^{2}+\left(n_{21}^{(L)}-n_{31}^{(L)}\right)^{2}$
$\displaystyle=n_{11}^{(L)},$ (3.4)
$\displaystyle\left(n_{12}^{(L)}-n_{32}^{(L)}\right)^{2}+\left(n_{22}^{(L)}-n_{32}^{(L)}\right)^{2}$
$\displaystyle=n_{22}^{(L)},$ (3.5)
$\displaystyle\left(n_{13}^{(L)}-n_{33}^{(L)}\right)^{2}+\left(n_{23}^{(L)}-n_{33}^{(L)}\right)^{2}$
$\displaystyle=2n_{33}^{(L)}.$

By (3.3), we can regard $n_{11}^{(L)}$ as a solution of the following
quadratic equation

(3.6)
$x^{2}-\left(2n_{31}^{(L)}+1\right)x+\left(n_{31}^{(L)}\right)^{2}+\left(n_{21}^{(L)}-n_{31}^{(L)}\right)^{2}=0.$

Since
$\left(n_{31}^{(L)}\right)^{2}+\left(n_{21}^{(L)}-n_{31}^{(L)}\right)^{2}\geq
0$, then the other solution of (3.6) satisfies

$2n_{31}^{(L)}+1-n_{11}^{(L)}\geq
0~{}~{}\Longleftrightarrow~{}~{}4-4n_{11}^{(L)}+8n_{31}^{(L)}\geq 0.$

In a similar manner, by (3.4) we can view $n_{12}^{(L)}$ as a solution of the
following quadratic equation

(3.7)
$x^{2}-2n_{32}^{(L)}x+\left(n_{32}^{(L)}\right)^{2}+\left(n_{22}^{(L)}-n_{32}^{(L)}\right)^{2}-n_{22}^{(L)}=0.$

It is not difficult to check that

$\left(n_{32}^{(L)}\right)^{2}+\left(n_{22}^{(L)}-n_{32}^{(L)}\right)^{2}-n_{22}^{(L)}\geq
0.$

Hence, the other solution of (3.7) satisfies

$2n_{32}^{(L)}-n_{12}^{(L)}\geq
0~{}~{}\Longleftrightarrow~{}~{}-4n_{12}^{(L)}+8n_{32}^{(L)}\geq 0.$

As $n_{11}^{(L)}$ and $n_{12}^{(L)}$, by (3.5) we can treat $n_{13}^{(L)}$ as
a solution of the following quadratic equation

(3.8)
$x^{2}-2n_{33}^{(L)}x+\left(n_{33}^{(L)}\right)^{2}+\left(n_{23}^{(L)}-n_{33}^{(L)}\right)^{2}-2n_{33}^{(L)}=0.$

Notice that if

(3.9)
$\left(n_{33}^{(L)}\right)^{2}+\left(n_{23}^{(L)}-n_{33}^{(L)}\right)^{2}-2n_{33}^{(L)}\geq
0,$

then the other solution of (3.8) satisfies

$2n_{33}^{(L)}-n_{13}^{(L)}\geq
0~{}~{}\Longleftrightarrow~{}~{}-4n_{13}^{(L)}+8n_{33}^{(L)}\geq 0.$

Therefore, (3.1) is proved. So it remains to verify (3.9). For the cases of
$n_{33}^{(L)}\geq 2$ and $n_{33}^{(L)}=0$, it is trivial. While if
$n_{33}^{(L)}=1$, we must have $\left(n_{23}^{(L)}-1\right)^{2}\geq 1$.
Indeed, comparing the coefficients of $\mu_{2}\mu_{3}$ in the Pohozaev
identity, we find $n_{23}^{(L)}$ is even. Hence (3.1) is proved.

Step 2. We claim that

(3.10) $4n_{11}^{(L)}+4n_{21}^{(L)}-4n_{31}^{(L)}\geq
0,~{}4n_{12}^{(L)}+4n_{22}^{(L)}-4n_{32}^{(L)}\geq
0,~{}4+4n_{13}^{(L)}+4n_{23}^{(L)}-4n_{33}^{(L)}\geq 0.$

As Step 1, we can apply (2.4) to get that $n_{3i}^{(L)},~{}i=1,2,3$ are the
solution to the following three equations respectively

$\left\\{\begin{array}[]{ll}2x^{2}-2\left(n_{11}^{(L)}+n_{21}^{(L)}\right)x+\left(n_{11}^{(L)}\right)^{2}+\left(n_{21}^{(L)}\right)^{2}-n_{11}^{(L)}=0,\\\
2x^{2}-2\left(n_{12}^{(L)}+n_{22}^{(L)}\right)x+\left(n_{12}^{(L)}\right)^{2}+\left(n_{22}^{(L)}\right)^{2}-n_{22}^{(L)}=0,\\\
2x^{2}-2\left(n_{13}^{(L)}+n_{23}^{(L)}+1\right)x+\left(n_{13}^{(L)}\right)^{2}+\left(n_{23}^{(L)}\right)^{2}=0.\end{array}\right.$

Notice that

$\left(n_{11}^{(L)}\right)^{2}+\left(n_{21}^{(L)}\right)^{2}-n_{11}^{(L)}\geq
0,~{}\left(n_{12}^{(L)}\right)^{2}+\left(n_{22}^{(L)}\right)^{2}-n_{22}^{(L)}\geq
0,~{}\left(n_{13}^{(L)}\right)^{2}+\left(n_{23}^{(L)}\right)^{2}\geq 0,$

then by the same argument of Step 1, we conclude that

$n_{11}^{(L)}+n_{21}^{(L)}-n_{31}^{(L)}\geq
0,~{}n_{12}^{(L)}+n_{22}^{(L)}-n_{32}^{(L)}\geq
0,~{}1+n_{13}^{(L)}+n_{23}^{(L)}-n_{33}^{(L)}\geq 0.$

Hence (3.10) is proved and we finish the proof. ∎

######  Proposition 3.4.

Let $\Gamma_{N}(\bm{\mu})$ be defined as

(3.11)
$\Gamma_{N}(\bm{\mu})=\left\\{\bm{\sigma}~{}\big{|}~{}\bm{\sigma}~{}\text{satisfies
the Pohozaev identity~{} \eqref{CYL-Ch-2-Eq-Pohozaev-
identity}},~{}\sigma_{i}=4\sum\limits_{j=1}^{3}n_{ij}\mu_{j},~{}n_{ij}\in\mathbb{N}\cup\\{0\\},~{}i=1,2,3\right\\}.$

Then

$\Gamma_{N}(\bm{\mu})=\Gamma(\bm{\mu}).$

###### Proof.

It is easy to see that $\Gamma(\bm{\mu})\subseteq\Gamma_{N}(\bm{\mu})$ by
Proposition 3.1 and Proposition 3.3. Thus it remains to show
$\Gamma_{N}(\bm{\mu})\subseteq\Gamma(\bm{\mu})$. For any
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma_{N}(\bm{\mu})$, by
Proposition 3.1, $\mathfrak{R}_{i}\bm{\sigma}$ satisfies the Pohozaev identity
(2.4). So we shall prove

(3.12)
$\text{if}~{}\bm{\sigma}\in\Gamma_{N}(\bm{\mu}),~{}\text{then}~{}\mathfrak{R}_{i}\bm{\sigma}\in\Gamma_{N}(\bm{\mu}),\quad
i=1,2,3.$

Let $\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$,
${\sigma}_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j},~{}i=1,2,3$, where
$n_{ij}\in\mathbb{N}\cup\\{0\\}$. Then by the same arguments of Proposition
3.3, we obtain

$\left\\{\begin{aligned} 4-4n_{11}+8n_{31}\geq 0,\\\ -4n_{12}+8n_{32}\geq
0,\\\ -4n_{13}+8n_{33}\geq 0,\end{aligned}\right.\quad\left\\{\begin{aligned}
-4n_{21}+8n_{31}\geq 0,\\\ 4-4n_{22}+8n_{32}\geq 0,\\\ -4n_{23}+8n_{33}\geq
0,\end{aligned}\right.~{}\text{and}~{}\left\\{\begin{aligned}
4n_{11}+4n_{21}-4n_{31}\geq 0,\\\ 4n_{12}+4n_{22}-4n_{32}\geq 0,\\\
4+4n_{13}+4n_{23}-4n_{33}\geq 0.\end{aligned}\right.$

This completes the proof of (3.12).

Next we define a partial order $\preceq$ in $\Gamma_{N}(\bm{\mu})$, we say

$\bm{\sigma}_{1}\preceq\bm{\sigma}_{2}~{}\text{provided}~{}(\bm{\sigma}_{1})_{i}\leq(\bm{\sigma}_{2})_{i},\quad
i=1,2,3.$

Then there must hold

$\text{either}~{}~{}~{}\bm{\sigma}\preceq\mathfrak{R}_{i}\bm{\sigma}~{}~{}~{}\text{or}~{}~{}~{}\mathfrak{R}_{i}\bm{\sigma}\preceq\bm{\sigma},\quad
i=1,2,3.$

For any $\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$, let

$\Gamma_{\bm{\sigma}}:=\left\\{\mathfrak{R}_{i_{1}}\mathfrak{R}_{i_{2}}\cdots\mathfrak{R}_{i_{n}}\bm{\sigma}\mid
n\in\mathbb{N}\cup\\{0\\},~{}i_{k}\in\\{1,2,3\\},~{}1\leq k\leq n\right\\}.$

Thus for any $\bm{\sigma}_{1},\bm{\sigma}_{2}\in\Gamma_{N}(\bm{\mu})$, we have
either

(3.13)
$\Gamma_{\bm{\sigma}_{1}}=\Gamma_{\bm{\sigma}_{2}}~{}~{}~{}\text{or}~{}~{}~{}\Gamma_{\bm{\sigma}_{1}}\cap\Gamma_{\bm{\sigma}_{2}}=\emptyset.$

Now we can prove that $\bm{0}=(0,0,0)\in\Gamma_{\bm{\sigma}}$ for any
$\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$. An element
$\hat{\bm{\sigma}}\in\Gamma_{\bm{\sigma}}$ is minimal if
$\widetilde{\bm{\sigma}}\preceq\hat{\bm{\sigma}}$ for some
$\widetilde{\bm{\sigma}}\in\Gamma_{\bm{\sigma}}$, then
$\widetilde{\bm{\sigma}}=\hat{\bm{\sigma}}$. By well-ordering Principle, we
conclude that $\Gamma_{\bm{\sigma}}$ has a local minimal element
$\bm{\sigma}_{0}=(\sigma_{1,0},\sigma_{2,0},\sigma_{3,0})$, i.e.,

$\bm{\sigma}_{0}\preceq\mathfrak{R}_{i}\bm{\sigma}_{0},\quad i=1,2,3.$

Therefore, we have

$4\mu_{i}-2\sum\limits_{j=1}^{3}a_{ij}\sigma_{j,0}\geq 0,\quad i=1,2,3.$

On the other hand, since $\bm{\sigma}_{0}$ satisfies the Pohozaev identity
(2.4), we get

$0\leq\sigma_{1,0}\left(2\sum\limits_{j=1}^{3}a_{1j}\sigma_{j,0}-4\mu_{1}\right)+\sigma_{2,0}\left(2\sum\limits_{j=1}^{3}a_{2j}\sigma_{j,0}-4\mu_{2}\right)+2\sigma_{3,0}\left(2\sum\limits_{j=1}^{3}a_{3j}\sigma_{j,0}-4\mu_{3}\right)\leq
0,$

which implies that $\sigma_{i,0}=0,~{}i=1,2,3$. Otherwise, we must have

$\displaystyle
2\left(\sigma_{1,0}-\sigma_{3,0}\right)=4\mu_{1},~{}2\left(\sigma_{2,0}-\sigma_{3,0}\right)=4\mu_{2},~{}2\left(-\frac{1}{2}\sigma_{1,0}-\frac{1}{2}\sigma_{2,0}+\sigma_{3,0}\right)=4\mu_{3},$

then $2\mu_{1}+2\mu_{2}+4\mu_{3}=0$, a contradiction. Hence
$\bm{0}\in\Gamma_{\bm{\sigma}}$. By (3.13), we obtain
$\Gamma_{\bm{\sigma}}=\Gamma_{\bm{0}}$ for any
$\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$. This amounts to say that

$\Gamma_{N}(\bm{\mu})=\Gamma(\bm{\mu}).$

The proof of Proposition 3.4 is complete. ∎

In the following of this section we are focused on giving a precise expression
for any element in $\Gamma(\bm{\mu})$.

######  Lemma 3.5.

Let $\gamma_{i}=0$, $i=1,2,3$. For any
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(1,1,1)$ with
$\sigma_{i}=4n_{i}$ and $n_{i}\in\mathbb{N}\cup\\{0\\}$, $i=1,2,3$, we set

$n_{1}-n_{3}\equiv m_{1}^{\prime}~{}(\mathrm{mod}~{}4),\quad n_{2}-n_{3}\equiv
m_{2}^{\prime}~{}(\mathrm{mod}~{}4)\quad\mbox{for
some}~{}m_{1}^{\prime},m_{2}^{\prime}\in\\{0,1,2,3\\}.$

Then $(m_{1}^{\prime},m_{2}^{\prime})$ admits all the following $\mathbf{8}$
types:

$\Big{\\{}(0,0),~{}(0,1),~{}(1,0),~{}(1,1),~{}(2,2),~{}(2,3),~{}(3,2),~{}(3,3)\Big{\\}}.$

In this way, we call that $\bm{\sigma}$ is of type
$(m_{1}^{\prime},m_{2}^{\prime})$.

###### Proof.

Applying the Pohozaev identity (2.4) with $\mu_{i}=1$, $i=1,2,3$, we obtain

(3.14) $(n_{1}-n_{3})^{2}+(n_{2}-n_{3})^{2}=n_{1}+n_{2}+2n_{3}.$

Notice that

(3.15)
$n_{1}-n_{3}=4k_{1}+m_{1}^{\prime},~{}~{}n_{2}-n_{3}=4k_{2}+m_{2}^{\prime}$

for some $k_{1},k_{2}\in\mathbb{Z}$ and
$m_{1}^{\prime},m_{2}^{\prime}=0,1,2,3$. Substituting (3.15) into (3.14), we
conclude that

${m_{1}^{\prime}}^{2}+{m_{2}^{\prime}}^{2}\equiv
m_{1}^{\prime}+m_{2}^{\prime}~{}~{}(\mathrm{mod}~{}4),$

which is solvable if and only if

$(m_{1}^{\prime},m_{2}^{\prime})\in\Big{\\{}(0,0),~{}(0,1),~{}(1,0),~{}(1,1),~{}(2,2),~{}(2,3),~{}(3,2),~{}(3,3)\Big{\\}}.$

We notice that the following $8$ elements satisfy the Pohozaev identity (2.4)
with $\mu_{i}=1$, $i=1,2,3$, and

$\displaystyle(0,0,0)~{}\text{is of type}~{}(0,0),\quad(4,0,0)~{}\text{is of
type}~{}(1,0),$ $\displaystyle(0,4,0)~{}\text{is of
type}~{}(0,1),\quad(4,4,0)~{}\text{is of type}~{}(1,1),$
$\displaystyle(4,4,12)~{}\text{is of type}~{}(2,2),\quad(4,0,8)~{}\text{is of
type}~{}(3,2),$ $\displaystyle(0,4,8)~{}\text{is of
type}~{}(2,3),\quad(0,0,4)~{}\text{is of type}~{}(3,3).$

Therefore, all the $8$ types could happen. ∎

######  Remark 6.

(1) For any $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(1,1,1)$
which is of type $(m_{1}^{\prime},m_{2}^{\prime})$, one can easily check that
$\mathfrak{R}_{i}\bm{\sigma}$ is of type $(n_{1}^{\prime},n_{2}^{\prime})$,
where

$\displaystyle\begin{cases}n_{1}^{\prime}\equiv-
m_{1}^{\prime}+1~{}(\mathrm{mod}~{}4),\quad
n_{2}^{\prime}=m_{2}^{\prime},\qquad&\mathrm{if}~{}i=1,\\\
n_{1}^{\prime}=m_{1}^{\prime},\quad n_{2}^{\prime}\equiv-
m_{2}^{\prime}+1~{}(\mathrm{mod}~{}4),\qquad&\mathrm{if}~{}i=2,\\\
n_{1}^{\prime}\equiv-m_{2}^{\prime}-1~{}(\mathrm{mod}~{}4),\quad
n_{2}^{\prime}\equiv-
m_{1}^{\prime}-1~{}(\mathrm{mod}~{}4),\qquad&\mathrm{if}~{}i=3.\end{cases}$

Precisely, we conclude that

$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(0,0),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(1,0),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(0,1),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(3,3);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(0,1),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(1,1),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(0,0),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(2,3);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(1,0),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(0,0),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(1,1),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(3,2);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(1,1),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(0,1),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(1,0),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(2,2);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(2,2),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(3,2),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(2,3),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(1,1);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(2,3),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(3,3),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(2,2),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(0,1);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(3,2),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(2,2),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(3,3),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(1,0);$
$\text{if}~{}\bm{\sigma}~{}\text{is of
type}~{}(3,3),~{}\mathfrak{R}_{1}\bm{\sigma}~{}\text{is of
type}~{}(2,3),~{}\mathfrak{R}_{2}\bm{\sigma}~{}\text{is of
type}~{}(3,2),~{}\mathfrak{R}_{3}\bm{\sigma}~{}\text{is of type}~{}(0,0).$

(2) For any general $\bm{\sigma}\in\Gamma(\bm{\mu})$,
$\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j},~{}n_{ij}\in\mathbb{N}\cup\\{0\\}$.
Let $N({\sigma}_{i})=4\sum_{j=1}^{3}n_{ij}$ and denote

$\frac{1}{4}\left(N({\sigma}_{1})-N({\sigma}_{3})\right)\equiv
m_{1}^{\prime}~{}(\mathrm{mod}~{}4),\quad\frac{1}{4}\left(N({\sigma}_{2})-N({\sigma}_{3})\right)\equiv
m_{2}^{\prime}~{}(\mathrm{mod}~{}4)\quad\mbox{for
some}~{}m_{1}^{\prime},m_{2}^{\prime}\in\\{0,1,2,3\\}.$

Then $(m_{1}^{\prime},m_{2}^{\prime})$ also admits all the following
$\mathbf{8}$ types:

$\Big{\\{}(0,0),~{}(0,1),~{}(1,0),~{}(1,1),~{}(2,2),~{}(2,3),~{}(3,2),~{}(3,3)\Big{\\}}.$

We will prove this result in Lemma 3.6. In this way, we say that $\bm{\sigma}$
is of type $(m_{1}^{\prime},m_{2}^{\prime})$ as well.

######  Lemma 3.6.

For any $\bm{\sigma}\in\Gamma(\bm{\mu})$, suppose that $\bm{\sigma}$ is of
type $(m_{1}^{\prime},m_{2}^{\prime})$ for some
$m_{1}^{\prime},m_{2}^{\prime}=0,1,2,3$. Then
$(m_{1}^{\prime},m_{2}^{\prime})$ admits all the following $\mathbf{8}$ types:

(3.16)
$\Big{\\{}(0,0),~{}(0,1),~{}(1,0),~{}(1,1),~{}(2,2),~{}(2,3),~{}(3,2),~{}(3,3)\Big{\\}}.$

###### Proof.

We prove this result by induction. Recall the binary tree diagram given in the
proof of Proposition 3.3 and we can directly check that

$\displaystyle(0,0,0)~{}\text{is of type}~{}(0,0),$
$\displaystyle(4\mu_{1},0,0)~{}\text{is of type}~{}(1,0),$
$\displaystyle(0,4\mu_{2},0)~{}\text{is of type}~{}(0,1),$
$\displaystyle(0,0,4\mu_{3})~{}\text{is of type}~{}(3,3),$
$\displaystyle(4\mu_{1},4\mu_{2},0)~{}\text{is of type}~{}(1,1),$
$\displaystyle(4\mu_{1},0,4\mu_{1}+4\mu_{3})~{}\text{is of type}~{}(3,2),$
$\displaystyle(0,4\mu_{2},4\mu_{2}+4\mu_{3})~{}\text{is of type}~{}(2,3),$
$\displaystyle(4\mu_{1}+8\mu_{3},0,4\mu_{3})~{}\text{is of type}~{}(2,3),$
$\displaystyle(0,4\mu_{2}+8\mu_{3},4\mu_{3})~{}\text{is of type}~{}(3,2),$
$\displaystyle(4\mu_{1},4\mu_{2},4\mu_{1}+4\mu_{2}+4\mu_{3})~{}\text{is of
type}~{}(2,2).$

The first $9$ elements are in the $1$-st, $2$-nd and $3$-rd levels, the last
one is in the $4$-th level. Obviously, these elements are of type $(i,j)$
described as in (3.16) and all $8$ types could happen. We suppose that each
element in the $N$-th level is of type $(i,j)$, which is one of (3.16). Then
we shall prove each element in the $(N+1)$-th level is of type
$(i^{\prime},j^{\prime})$, where $(i^{\prime},j^{\prime})$ admits a form
belonging to (3.16).

Suppose $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$ is
an element in the $(N+1)$-level, where
$\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j},~{}n_{ij}\in\mathbb{N}\cup\\{0\\}$.
Without loss of generality, we assume that $\bm{\sigma}$ is of type $(0,0)$.
Then one can easily check that $\mathfrak{R}_{i}\bm{\sigma}$ is of type
$(n_{1}^{\prime},n_{2}^{\prime})$, where for $i=1,2,3$, respectively,

$\begin{cases}n_{1}^{\prime}\equiv-m_{1}^{\prime}+1~{}(\mathrm{mod}~{}4),\\\
n_{2}^{\prime}=m_{2}^{\prime},\end{cases}\qquad\begin{cases}n_{1}^{\prime}=m_{1}^{\prime},\\\
n_{2}^{\prime}\equiv-m_{2}^{\prime}+1~{}(\mathrm{mod}~{}4),\\\
\end{cases}\qquad\begin{cases}n_{1}^{\prime}\equiv-
m_{2}^{\prime}-1~{}(\mathrm{mod}~{}4),\\\ n_{2}^{\prime}\equiv-
m_{1}^{\prime}-1~{}(\mathrm{mod}~{}4).\\\ \end{cases}$

Then we get the same conclusion holds by Remark 6-(1). Thus, we finish the
proof. ∎

######  Theorem 3.7.

For each element $\bm{\sigma}\in\Gamma(\bm{\mu})$, we can express
$\bm{\sigma}$ as $\bm{\sigma}^{t}=\mathbb{F}\bm{\mu}^{t}$, where
$\mathbb{F}=(f_{ij})_{3\times 3}$ is a matrix, $f_{ij}$ are multiples of $4$
by Proposition 3.4, and $t$ denotes the transpose operator, i.e.,

$\begin{aligned} \left(\begin{array}[]{ccc}\sigma_{1}\\\ \sigma_{2}\\\
\sigma_{3}\end{array}\right)\end{aligned}=\mathbb{F}\bm{\mu}^{t}=\begin{aligned}
\left(\begin{array}[]{ccc}f_{11}&f_{12}&f_{13}\\\ f_{21}&f_{22}&f_{23}\\\
f_{31}&f_{32}&f_{33}\end{array}\right)\end{aligned}\begin{aligned}
\left(\begin{array}[]{ccc}\mu_{1}\\\ \mu_{2}\\\
\mu_{3}\end{array}\right).\end{aligned}$

Let

$\Gamma_{\mathbb{F}}(\bm{\mu})=\left\\{\bm{\sigma}~{}|~{}\bm{\sigma}^{t}=\mathbb{F}\bm{\mu}^{t},~{}\text{where}~{}\mathbb{F}~{}\text{admits}~{}\mathbb{F}^{(\ell)},~{}\ell=1,\cdots,8\right\\}.$

Then $\Gamma_{\mathbb{F}}(\bm{\mu})=\Gamma_{N}(\bm{\mu})$. Here
$\mathbb{F}^{(\ell)}$ are defined as the following:

Type $(0,0)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(0,0)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(1)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{4}m_{1}^{2}+m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}&\frac{1}{2}m_{1}^{2}+2m_{1}+\frac{1}{2}m_{2}^{2}\\\
\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}+m_{2}&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}+2m_{2}\\\
\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-m_{2}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}\end{array}\right).\end{aligned}$

Type $(0,1)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(0,1)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(2)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}+\frac{1}{4}&\frac{1}{4}m_{1}^{2}+m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{4}&\frac{1}{2}m_{1}^{2}+2m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}+\frac{1}{2}\\\
\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}+\frac{3}{2}m_{2}+\frac{9}{4}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}+m_{2}-\frac{3}{2}\\\
\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}+\frac{1}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{4}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}-m_{2}+\frac{1}{2}\end{array}\right).\end{aligned}$

Type $(1,0)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(1,0)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(3)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{3}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{9}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}-\frac{3}{4}&\frac{1}{2}m_{1}^{2}+m_{1}+\frac{1}{2}m_{2}^{2}-\frac{3}{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+m_{2}-\frac{3}{4}&\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{4}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+2m_{2}+\frac{1}{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{3}{4}&\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}+\frac{1}{4}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+\frac{1}{2}\end{array}\right).\end{aligned}$

Type $(1,1)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(1,1)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(4)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{3}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}+\frac{5}{2}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{2}&\frac{1}{2}m_{1}^{2}+m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}-1\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{2}&\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{3}{2}m_{2}+\frac{5}{2}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+m_{2}-1\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}-\frac{1}{2}&\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{1}{2}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}+1\end{array}\right).\end{aligned}$

Type $(2,2)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(2,2)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(5)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}+2&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-2&\frac{1}{2}m_{1}^{2}+2m_{1}+\frac{1}{2}m_{2}^{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-2&\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}+m_{2}+2&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}+2m_{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-m_{2}&\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}\end{array}\right).\end{aligned}$

Type $(2,3)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(2,3)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(6)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}+\frac{5}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}-\frac{7}{4}&\frac{1}{2}m_{1}^{2}+2m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}+\frac{1}{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}+\frac{3}{2}m_{2}+\frac{1}{4}&\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}+\frac{5}{4}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}+m_{2}-\frac{3}{2}\\\
\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{3}{4}&\frac{1}{4}m_{1}^{2}-m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}+\frac{1}{4}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}-m_{2}+\frac{1}{2}\end{array}\right).\end{aligned}$

Type $(3,2)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(3,2)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(7)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}+\frac{5}{4}&\frac{1}{4}m_{1}^{2}+\frac{3}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{4}&\frac{1}{2}m_{1}^{2}+m_{1}+\frac{1}{2}m_{2}^{2}-\frac{3}{2}\\\
\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{7}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+m_{2}+\frac{5}{4}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+2m_{2}+\frac{1}{2}\\\
\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-m_{2}+\frac{1}{4}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{3}{4}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+\frac{1}{2}\end{array}\right).\end{aligned}$

Type $(3,3)$. For any $m_{1},m_{2}\in\mathbb{Z}$ with
$(m_{1},m_{2})\equiv(3,3)~{}(\mathrm{mod}~{}4)$, we have

$\mathbb{F}^{(8)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}+\frac{1}{2}&\frac{1}{4}m_{1}^{2}+\frac{3}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}+\frac{1}{2}&\frac{1}{2}m_{1}^{2}+m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}-1\\\
\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{3}{2}m_{2}+\frac{1}{2}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}+\frac{1}{2}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}+m_{2}-1\\\
\frac{1}{4}m_{1}^{2}-\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}+\frac{1}{2}m_{2}-\frac{1}{2}&\frac{1}{4}m_{1}^{2}+\frac{1}{2}m_{1}+\frac{1}{4}m_{2}^{2}-\frac{1}{2}m_{2}-\frac{1}{2}&\frac{1}{2}m_{1}^{2}-m_{1}+\frac{1}{2}m_{2}^{2}-m_{2}+1\end{array}\right).\end{aligned}$

###### Proof.

Let

$\Gamma_{\mathbb{F}}(\bm{\mu})=\left\\{\bm{\sigma}~{}|~{}\bm{\sigma}^{t}=\mathbb{F}\bm{\mu}^{t}~{}\text{where}~{}\mathbb{F}~{}\text{admits}~{}\mathbb{F}^{(\ell)},~{}\ell=1,\cdots,8\right\\}.$

Our aim is to show $\Gamma_{\mathbb{F}}(\bm{\mu})=\Gamma_{N}(\bm{\mu})$.

On one hand, for any $\bm{\sigma}\in\Gamma_{\mathbb{F}}(\bm{\mu})$ with
$\bm{\sigma}^{t}=\mathbb{F}^{(\ell)}\bm{\mu}^{t}$ for some
$\ell\in\\{1,\cdots,8\\}$, one can directly check that $\bm{\sigma}$ satisfies
the Pohozaev identity (2.4) and $f_{ij}^{(\ell)}$ are multiples of $4$. Hence
$\Gamma_{\mathbb{F}}(\bm{\mu})\subseteq\Gamma_{N}(\bm{\mu})$ by the definition
of $\Gamma_{N}(\bm{\mu})$ in (3.11).

On the other hand, we can prove
$\Gamma_{N}(\bm{\mu})\subseteq\Gamma_{\mathbb{F}}(\bm{\mu})$ by induction. For
any $\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$,
$\sigma_{i}=4\sum_{j=1}^{3}n_{ij}\mu_{j}$, $n_{ij}\in\mathbb{N}\cup\\{0\\}$.
Let $N({\sigma}_{i})=4\sum_{j=1}^{3}n_{ij}$ and denote

(3.17)
$\frac{1}{4}\left(N({\sigma}_{1})-N({\sigma}_{3})\right)=m_{1},\quad\frac{1}{4}\left(N({\sigma}_{2})-N({\sigma}_{3})\right)=m_{2}.$

Suppose $\bm{\sigma}$ is of type $(m_{1}^{\prime},m_{2}^{\prime})$, where
$(m_{1},m_{2})\equiv(m_{1}^{\prime},m_{2}^{\prime})$ (mod $4$). We recall the
binary tree diagram given in the proof of Proposition 3.3 and directly check
that

$\displaystyle\bm{0}=(0,0,0)~{}\text{is of type}~{}(0,0),$
$\displaystyle\quad\bm{0}^{t}=\mathbb{F}^{(1)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(0,0);$
$\displaystyle\bm{\sigma}=(4\mu_{1},0,0)~{}\text{is of type}~{}(1,0),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(3)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(1,0);$
$\displaystyle\bm{\sigma}=(0,4\mu_{2},0)~{}\text{is of type}~{}(0,1),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(2)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(0,1);$
$\displaystyle\bm{\sigma}=(0,0,4\mu_{3})~{}\text{is of type}~{}(3,3),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(8)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(-1,-1);$
$\displaystyle\bm{\sigma}=(4\mu_{1},4\mu_{2},0)~{}\text{is of type}~{}(1,1),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(4)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(1,1);$
$\displaystyle\bm{\sigma}=(4\mu_{1},0,4\mu_{1}+4\mu_{3})~{}\text{is of
type}~{}(3,2),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(7)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(-1,-2);$
$\displaystyle\bm{\sigma}=(0,4\mu_{2},4\mu_{2}+4\mu_{3})~{}\text{is of
type}~{}(2,3),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(6)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(-2,-1);$
$\displaystyle\bm{\sigma}=(4\mu_{1}+8\mu_{3},0,4\mu_{3})~{}\text{is of
type}~{}(2,3),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(6)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(2,-1);$
$\displaystyle\bm{\sigma}=(0,4\mu_{2}+8\mu_{3},4\mu_{3})~{}\text{is of
type}~{}(3,2),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(7)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(-1,2);$
$\displaystyle\bm{\sigma}=(4\mu_{1},4\mu_{2},4\mu_{1}+4\mu_{2}+4\mu_{3})~{}\text{is
of type}~{}(2,2),$
$\displaystyle\quad\bm{\sigma}^{t}=\mathbb{F}^{(5)}\bm{\mu}^{t}~{}\text{for}~{}(m_{1},m_{2})=(-2,-2).$

This implies that $\bm{\sigma}\in\Gamma_{\mathbb{F}}(\bm{\mu})$ for those
$\bm{\sigma}$ in the $\mathcal{L}$-th level with
$\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$, $1\leq\mathcal{L}\leq 3$. In particular,
all possibilities $\bm{\sigma}^{t}=\mathbb{F}^{(\ell)}\bm{\mu}^{t}$ could
happen, $\ell=1,\cdots,8$. Suppose that
$\bm{\sigma}\in\Gamma_{\mathbb{F}}(\bm{\mu})$ for those $\bm{\sigma}$ in the
$\mathcal{L}$-th level with $\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$,
$1\leq\mathcal{L}\leq M$. We need to verify
$\mathfrak{R}_{i}\bm{\sigma}\in\Gamma_{\mathbb{F}}(\bm{\mu})$ for any
$\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$ in $M$-th level and any $i=1,2,3$.

Now, suppose that $\bm{\sigma}$ is an element in the $M$-th level of
$\Gamma_{N}(\bm{\mu})$, by assumption we have
$\bm{\sigma}\in\Gamma_{\mathbb{F}}(\bm{\mu})$ with
$\bm{\sigma}^{t}=\mathbb{F}^{(\ell)}\bm{\mu}^{t}$ for some
$\ell\in\\{1,\cdots,8\\}$. Denote

$\sigma_{i}=\sum\limits_{j=1}^{3}f^{(\ell)}_{ij}(m_{1},m_{2})\mu_{j},~{}\text{where}~{}f^{(\ell)}_{ij}(m_{1},m_{2})~{}\text{is
the element of}~{}\mathbb{F}^{(\ell)}.$

Assume that $\bm{\sigma}$ is of type $(m_{1}^{\prime},m_{2}^{\prime})$, where
$(m_{1},m_{2})\equiv(m_{1}^{\prime},m_{2}^{\prime})$ ($\mathrm{mod}$ $4$), and
$\mathfrak{R}_{i}\bm{\sigma}$ is of type $(n_{1}^{\prime},n_{2}^{\prime})$,
where $(n_{1},n_{2})\equiv(n_{1}^{\prime},n_{2}^{\prime})$ ($\mathrm{mod}$
$4$). Here $(m_{1},m_{2})$ and $(n_{1},n_{2})$ are defined as in (3.17)
related to $\bm{\sigma}$ and $\mathfrak{R}_{i}\bm{\sigma}$, respectively. The
corresponding matrices $\mathbb{F}$ of type $(m_{1}^{\prime},m_{2}^{\prime})$
and $(n_{1}^{\prime},n_{2}^{\prime})$ are $\mathbb{F}^{(\ell)}$ and
$\mathbb{F}^{(\ell^{\prime})}$ respectively. One can directly check that

(3.18) $\left(\mathfrak{R}_{i}\bm{\sigma}\right)^{t}=\left\\{\begin{aligned}
&\mathbb{F}^{(\ell_{1})}(1-m_{1},m_{2})\bm{\mu}^{t},\quad&\text{if}~{}i=1,\\\
&\mathbb{F}^{(\ell_{2})}(m_{1},1-m_{2})\bm{\mu}^{t},\quad&\text{if}~{}i=2,\\\
&\mathbb{F}^{(\ell_{3})}(-m_{2}-1,-m_{1}-1)\bm{\mu}^{t},\quad&\text{if}~{}i=3,\end{aligned}\right.$

for some indices $\ell_{i}\in\\{1,\cdots,8\\},~{}i=1,2,3.$

In the following, we just provide the details for the case $l=1$. By
straightforward computation we have

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(3)}(1-m_{1},m_{2})\bm{\mu}^{t},\quad~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(2)}(m_{1},1-m_{2})\bm{\mu}^{t},\quad~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(8)}(-m_{2}-1,-m_{1}-1)\bm{\mu}^{t}.$

Denote $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$, where

$\sigma_{i}=\sum\limits_{j=1}^{3}f^{(1)}_{ij}(m_{1},m_{2})\mu_{j},~{}\text{where}~{}f^{(1)}_{ij}(m_{1},m_{2})~{}\text{is
the element of}~{}\mathbb{F}^{(1)}.$

Then $\mathfrak{R}_{1}\bm{\sigma}$, $\mathfrak{R}_{2}\bm{\sigma}$ and
$\mathfrak{R}_{3}\bm{\sigma}$ is of type $(1,0)$, $(0,1)$ and $(3,3)$
respectively by Remark 6, and hence ${\ell}_{1}=3$, ${\ell}_{2}=2$ and
${\ell}_{3}=8$. On one hand, direct computation shows
$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{A}\bm{\mu}^{t}$ with

$\mathbb{A}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}\left(m_{1}-4\right)^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{4}\left(m_{1}-2\right)^{2}+\frac{1}{4}\left(m_{2}-2\right)^{2}-2&\frac{1}{2}\left(m_{1}-2\right)^{2}+\frac{1}{2}m_{2}^{2}-2\\\
\frac{1}{4}\left(m_{1}-2\right)^{2}+\frac{1}{4}\left(m_{2}+2\right)^{2}-2&\frac{1}{4}m_{1}^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{2}m_{1}^{2}+\frac{1}{2}\left(m_{2}+2\right)^{2}-2\\\
\frac{1}{4}\left(m_{1}-2\right)^{2}+\frac{1}{4}m_{1}^{2}-1&\frac{1}{4}m_{1}^{2}+\frac{1}{4}\left(m_{2}-2\right)^{2}-1&\frac{1}{2}m_{1}^{2}+\frac{1}{2}m_{2}^{2}\end{array}\right).\end{aligned}$

On the other hand, we can rewrite $\mathbb{F}^{(3)}$ as

$\mathbb{F}^{(3)}=\begin{aligned}
\left(\begin{array}[]{lll}\frac{1}{4}\left(m_{1}+3\right)^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{4}\left(m_{1}+1\right)^{2}+\frac{1}{4}\left(m_{2}-2\right)^{2}-2&\frac{1}{2}\left(m_{1}+1\right)^{2}+\frac{1}{2}m_{2}^{2}-2\\\
\frac{1}{4}\left(m_{1}+1\right)^{2}+\frac{1}{4}\left(m_{2}+2\right)^{2}-2&\frac{1}{4}\left(m_{1}-1\right)^{2}+\frac{1}{4}m_{2}^{2}&\frac{1}{2}\left(m_{1}-1\right)^{2}+\frac{1}{2}\left(m_{2}+2\right)^{2}-2\\\
\frac{1}{4}\left(m_{1}+1\right)^{2}+\frac{1}{4}m_{2}^{2}-1&\frac{1}{4}\left(m_{1}-1\right)^{2}+\frac{1}{4}\left(m_{2}-2\right)^{2}-1&\frac{1}{2}\left(m_{1}-1\right)^{2}+\frac{1}{2}m_{2}^{2}\end{array}\right).\end{aligned}$

Comparing the above two matrices, it is easy to observe that

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(3)}(1-m_{1},m_{2})\bm{\mu}^{t}.$

Similarly, we obtain

$\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(2)}(m_{1},1-m_{2})\bm{\mu}^{t}~{}\text{and}~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(8)}(-m_{2}-1,-m_{1}-1)\bm{\mu}^{t}.$

Then we finish the discussion for the case of $\ell=1$. For the other cases,
by direct computation one can get:

  1. (1).

If $\ell=1$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(3)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(2)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(8)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  2. (2).

If $\ell=2$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(4)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(1)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(6)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  3. (3).

If $\ell=3$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(1)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(4)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(7)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  4. (4).

If $\ell=4$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(2)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(3)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(5)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  5. (5).

If $\ell=5$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(7)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(6)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(4)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  6. (6).

If $\ell=6$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(8)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(5)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(2)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  7. (7).

If $\ell=7$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(5)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(8)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(3)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t};$

  8. (8).

If $\ell=8$,

$\left(\mathfrak{R}_{1}\bm{\sigma}\right)^{t}=\mathbb{F}^{(6)}(1-m_{1},m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{2}\bm{\sigma}\right)^{t}=\mathbb{F}^{(7)}(m_{1},1-m_{2})\bm{\mu}^{t},~{}\left(\mathfrak{R}_{3}\bm{\sigma}\right)^{t}=\mathbb{F}^{(1)}(-1-m_{2},-1-m_{1})\bm{\mu}^{t}.$

Therefore (3.18) is proved. This implies that any element
$\bm{\sigma}\in\Gamma_{N}(\bm{\mu})$ in the $(M+1)$-level also belongs to
$\Gamma_{\mathbb{F}}(\bm{\mu})$. Thus,
$\Gamma_{N}(\bm{\mu})\subseteq\Gamma_{\mathbb{F}}(\bm{\mu})$ and the proof of
Theorem 3.7 is complete. ∎

######  Remark 7.

For the case of $\gamma_{1}=\gamma_{2}=0$, one can easily verify that for any
$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$,

$\left\\{\begin{aligned} \sigma_{1}=m_{1}(m_{1}+3)+m_{2}(m_{2}-1),\\\
\sigma_{2}=m_{1}(m_{1}-1)+m_{2}(m_{2}+3),\\\
\sigma_{3}=m_{1}(m_{1}-1)+m_{2}(m_{2}-1),\end{aligned}\right.~{}\text{where}~{}(m_{1},m_{2})\in\mathbb{Z}^{2}\quad\text{and}\quad\left\\{\begin{aligned}
\text{either}~{}&m_{1},m_{2}\equiv 0,1~{}(\mathrm{mod}~{}4),\\\
\text{or}~{}&m_{1},m_{2}\equiv 2,3~{}(\mathrm{mod}~{}4).\end{aligned}\right.$

This expression coincides with that of [43, Theorem 1.1].

##  4\. Some technical lemmas

In this section, we present some important results for preparation. Consider
the following system

(4.1) $\Delta
u^{k}_{i}(x)+\sum\limits_{j=1}^{3}a_{ij}e^{u^{k}_{j}(x)}=4\pi\gamma_{i}\delta_{0}\quad\text{in}\quad
B_{1}(0),\quad i=1,2,3,$

where the coefficient matrix $A:=(a_{ij})_{3\times 3}$ is defined as in (1.2).
By a little abuse of notations, in the sequel we denote for any sequence
$(\mathbf{x},\mathbf{r})=\\{(x^{k},r_{k})\\}$,

$\sigma^{k}_{i}(B(x^{k},r_{k})):=\frac{1}{2\pi}\int_{B(x^{k},r_{k})}e^{u^{k}_{i}(x)}\mathrm{d}x,\quad
i=1,2,3.$

In Lemma 4.1, we assume that

  1. (i)

The Harnack-type inequality holds

(4.2) $u^{k}_{i}(x)+2\log|x|\leq
C\quad\text{for}\quad\frac{1}{2}l_{k}\leq|x|\leq 2s_{k},\quad i=1,2,3.$

  2. (ii)

All components of $\mathbf{u}^{k}$ have fast decay on $\partial B(0,l_{k})$
and

$\lim\limits_{r\to
0}\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(0,rl_{k}))=\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(0,l_{k})).$

For simplicity, we denote
$\sigma_{i}:=\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(0,l_{k}))$ in Lemma
4.1.

######  Lemma 4.1.

Let $\mu_{i}=1+\gamma_{i}$. Assume that (i) and (ii) hold.

  1. (a)

If $u^{k}_{i}$ has slow decay on $\partial B(0,s_{k})$ for some
$i\in\\{1,2,3\\}$, then

$2\mu_{i}-\sum\limits_{j=1}^{3}a_{ij}\sigma_{j}>0.$

  2. (b)

At least one component of $\mathbf{u}^{k}=(u^{k}_{1},u^{k}_{2},u^{k}_{3})$ has
fast decay on $\partial B(0,s_{k})$.

###### Proof.

(a) Performing the following scaling for the system (4.1)

$v^{k}_{i}(y)=u^{k}_{i}(s_{k}y)+2\log{s_{k}}\quad\text{for}\quad y\in
B_{2}(0),\quad i=1,2,3,$

gives

$\displaystyle\Delta v^{k}_{i}(y)+\sum\limits_{j=1}^{3}a_{ij}e^{v^{k}_{j}(y)}$
$\displaystyle=4\pi\gamma_{i}\delta_{0}\quad\text{in}\quad B_{2}(0),\quad
i=1,2,3.$

If there is at least one component of $\mathbf{u}^{k}$ has slow decay on
$\partial B(0,s_{k})$, we denote $J$ the set of consecutive indices such that
$u^{k}_{i}$ $(i\in J)$ has slow decay on $\partial B(0,s_{k})$, then
$v^{k}_{i}$ $(i\in J)$ converges to the solution $v_{i}$ $(i\in J)$ of a Toda
system (or Liouville equation). Precisely, one of the following alternatives
holds:

Case (1). If $J=\\{i\\}$ for $i\in\\{1,2,3\\}$. Then $v^{k}_{j}(y)\to-\infty$
in $L^{\infty}_{\mathrm{loc}}(B_{2}(0)\setminus\\{0\\})$ for
$j\in\\{1,2,3\\}\setminus J$, and $v^{k}_{i}(y)\to v(y)$ in
$C^{2}_{\mathrm{loc}}(B_{2}(0)\setminus\\{0\\})$, where $v(y)$ satisfies

$-\Delta v=e^{v}\quad\text{in}\quad B_{2}(0)\setminus\\{0\\}.$

Case (2). If $J=\\{1,2\\}$. Then $v^{k}_{3}(y)\to-\infty$ in
$L^{\infty}_{\mathrm{loc}}(B_{2}(0)\setminus\\{0\\})$, and
$(v^{k}_{1},v^{k}_{2})\to(v_{1},v_{2})$ in
$C_{\mathrm{loc}}^{2}(B_{2}(0)\setminus\\{0\\})$, where $(v_{1},v_{2})$
satisfies

$\displaystyle\left\\{\begin{array}[]{rr}\Delta v_{1}+e^{v_{1}}=0\\\ \Delta
v_{2}+e^{v_{2}}=0\end{array}\right.\quad\text{in}\quad
B_{2}(0)\setminus\\{0\\}.$

Case (3). If $J=\\{i,3\\}$ for $i=1$ or $2$. Then $v^{k}_{j}(y)\to-\infty$ in
$L^{\infty}_{\mathrm{loc}}(B_{2}(0)\setminus\\{0\\})$ for
$j\in\\{1,2,3\\}\setminus J$, and $(v^{k}_{i},v^{k}_{3})\to(v_{i},v_{3})$ in
$C_{\mathrm{loc}}^{2}(B_{2}(0)\setminus\\{0\\})$, where $(v_{i},v_{3})$
satisfies

$\displaystyle\left\\{\begin{array}[]{rr}\Delta v_{i}+e^{v_{i}}-e^{v_{3}}=0\\\
\Delta
v_{3}-\frac{1}{2}e^{v_{i}}+e^{v_{3}}=0\end{array}\right.\quad\text{in}\quad
B_{2}(0)\setminus\\{0\\}.$

Case (4). If $J=\\{1,2,3\\}$. Then
$(v^{k}_{1},v^{k}_{2},v^{k}_{3})\to(v_{1},v_{2},v_{3})$ in
$C_{\mathrm{loc}}^{2}(B_{2}(0)\setminus\\{0\\})$, where $(v_{1},v_{2},v_{3})$
satisfies

(4.3) $\left\\{\begin{array}[]{ll}-\Delta v_{1}=e^{v_{1}}-e^{v_{3}}\\\ -\Delta
v_{2}=e^{v_{2}}-e^{v_{3}}\\\ -\Delta
v_{3}=-\frac{1}{2}e^{v_{1}}-\frac{1}{2}e^{v_{2}}+e^{v_{3}}\end{array}\right.\quad\text{in}\quad
B_{2}(0)\setminus\\{0\\}.$

We shall prove the point (a) by concerning the Case (4), the proofs for the
other cases are similar and simpler. The strength of the Dirac measure at $0$
for (4.3) can be expressed by

$\displaystyle\lim\limits_{r\to 0}\int_{\partial B(0,r)}\frac{\partial
v_{1}(y)}{\partial\nu}\mathrm{d}S$ $\displaystyle=\lim\limits_{r\to
0}\lim\limits_{k\to\infty}\left(4\pi\gamma_{1}-\int_{B(0,r)}e^{v^{k}_{1}(y)}\mathrm{d}y+\int_{B(0,r)}e^{v^{k}_{3}(y)}\mathrm{d}y\right)$
$\displaystyle=4\pi\gamma_{1}-2\pi(\sigma_{1}-\sigma_{3})=:4\pi\beta_{1},$
$\displaystyle\lim\limits_{r\to 0}\int_{\partial B(0,r)}\frac{\partial
v_{2}(y)}{\partial\nu}\mathrm{d}S$ $\displaystyle=\lim\limits_{r\to
0}\lim\limits_{k\to\infty}\left(4\pi\gamma_{2}-\int_{B(0,r)}e^{v^{k}_{2}(y)}\mathrm{d}y+\int_{B(0,r)}e^{v^{k}_{3}(y)}\mathrm{d}y\right)$
$\displaystyle=4\pi\gamma_{2}-2\pi(\sigma_{2}-\sigma_{3})=:4\pi\beta_{2},$

and

$\displaystyle\lim\limits_{r\to 0}\int_{\partial B(0,r)}\frac{\partial
v_{3}(y)}{\partial\nu}\mathrm{d}S$ $\displaystyle=\lim\limits_{r\to
0}\lim\limits_{k\to\infty}\left(4\pi\gamma_{3}+\frac{1}{2}\int_{B(0,r)}e^{v^{k}_{1}(y)}\mathrm{d}y+\frac{1}{2}\int_{B(0,r)}e^{v^{k}_{2}(y)}\mathrm{d}y-\int_{B(0,r)}e^{v^{k}_{3}(y)}\mathrm{d}y\right)$
$\displaystyle=4\pi\gamma_{3}-2\pi\left(-\frac{1}{2}\sigma_{1}-\frac{1}{2}\sigma_{2}+\sigma_{3}\right)=:4\pi\beta_{3}.$

This implies that there is an extra term written as $4\pi\beta_{i}\delta_{0}$
on the R.H.S. of (4.3) for each $i=1,2,3$. It is known that if
$\beta_{i}\leq-1$ for some $i$, then (4.3) has no solutions. Hence

$\beta_{1}:=\gamma_{1}-\frac{1}{2}(\sigma_{1}-\sigma_{3})>-1,~{}\beta_{2}:=\gamma_{2}-\frac{1}{2}(\sigma_{2}-\sigma_{3})>-1,~{}\beta_{3}:=\gamma_{3}-\frac{1}{2}\left(-\frac{1}{2}\sigma_{1}-\frac{1}{2}\sigma_{2}+\sigma_{3}\right)>-1,$

which implies that

(4.4) $2\mu_{i}-\sum\limits_{j=1}^{3}a_{ij}\sigma_{j}>0,\quad i=1,2,3.$

(b) Since $\mathbf{u}^{k}$ has fast decay on $\partial B(0,l_{k})$,
$\bm{\sigma}$ satisfies the Pohozaev identity (2.4), i.e.,

$(\sigma_{1}-\sigma_{3})^{2}+(\sigma_{2}-\sigma_{3})^{2}=4(\mu_{1}\sigma_{1}+\mu_{2}\sigma_{2}+2\mu_{3}\sigma_{3}),$

which is equivalent to

$\displaystyle\sigma_{1}\left(\sum\limits_{j=1}^{3}a_{1j}\sigma_{j}-2\mu_{1}\right)+\sigma_{2}\left(\sum\limits_{j=1}^{3}a_{2j}\sigma_{j}-2\mu_{2}\right)+2\sigma_{3}\left(\sum\limits_{j=1}^{3}a_{3j}\sigma_{j}-2\mu_{3}\right)=2\mu_{1}\sigma_{1}+2\mu_{2}\sigma_{2}+4\mu_{3}\sigma_{3}\geq
0.$

As a consequence, there exists $i_{0}\in\\{1,2,3\\}$ such that
$\sum_{j=1}^{3}a_{i_{0}j}\sigma_{j}-2\mu_{i_{0}}\geq 0$. However, if all
components of $\mathbf{u}^{k}$ have slow decay on $\partial B(0,s_{k})$, then
by the (a)-th conclusion we obtain (4.4). Contradiction arises. Hence, we
conclude that at least one component of $\mathbf{u}^{k}$ has fast decay on
$\partial B(0,s_{k})$. ∎

######  Lemma 4.2.

Suppose that the Harnack-type inequality (4.2) holds for all components of
$\mathbf{u}^{k}$ over $r\in[\frac{l_{k}}{2},2s_{k}]$. If all components of
$\mathbf{u}^{k}$ have fast decay on all $r\in[l_{k},s_{k}]$, then

$\sigma^{k}_{i}(B(0,s_{k}))=\sigma^{k}_{i}(B(0,l_{k}))+o(1),\quad i=1,2,3.$

###### Proof.

The conclusion is trivial if $s_{k}/l_{k}\leq C$. In the following we assume
that $s_{k}/l_{k}\to+\infty$. We shall prove the lemma by contradiction.
Suppose there exists $l\in\\{1,2,3\\}$ such that

(4.5) $\sigma^{k}_{l}(B(0,s_{k}))>\sigma^{k}_{l}(B(0,l_{k}))+\delta_{1}$

for some $\delta_{1}>0$. We define

$\hat{\sigma}_{i}=\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(0,l_{k})),\quad
i=1,2,3,$

and

$\displaystyle
I_{1~{}(\mathrm{resp.}2,3)}=\left\\{i\in\\{1,2,3\\}~{}\Big{|}~{}2\mu_{i}-\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}<~{}(\mathrm{resp.}>,=)~{}0\right\\}.$

Then we claim

(4.6) $\text{\eqref{CYL-Ch-4-Eq-4} is impossible for}~{}i\in I_{1}\cup
I_{2}\cup I_{3}.$

In fact, we have $I_{1}\cup I_{2}$ is not empty. Otherwise, we have

$\hat{\sigma}_{1}-\hat{\sigma}_{3}=2\mu_{1},\quad\hat{\sigma}_{2}-\hat{\sigma}_{3}=2\mu_{2},\quad-\hat{\sigma}_{1}-\hat{\sigma}_{2}+2\hat{\sigma}_{3}=4\mu_{3}.$

Adding the above three equations we get that $\mu_{1}+\mu_{2}+2\mu_{3}=0$ and
this is impossible due to that $\mu_{i}>0$ for each $i$. Thus $I_{1}\cup
I_{2}$ is not empty. We set

(4.7) $\delta_{2}=\frac{1}{10000}\min\left\\{\min\limits_{i\in I_{1}\cup
I_{2}}\Big{|}2\mu_{i}-\sum\limits_{j=1}^{3}a_{ij}\hat{\sigma}_{j}\Big{|},\delta_{1},1\right\\},$

and choose $\tilde{l}_{k}\in(l_{k},s_{k})$ such that

(4.8)
$\max\limits_{i=1,2,3}\left\\{\sigma^{k}_{i}(\tilde{l}_{k})-\sigma^{k}_{i}(l_{k})\right\\}=\delta_{2}.$

Obviously, $\delta_{2}>0$. Using Harnack-type inequality (4.2) we have
$u^{k}_{i}(x)=\overline{u}^{k}_{i}(|x|)+O(1)$ for $\frac{1}{2}l_{k}\leq|x|\leq
2s_{k}$, see (2.2). By direct computations, we deduce from (4.1) that

(4.9)
$\frac{\mathrm{d}}{\mathrm{d}r}\left(\overline{u}^{k}_{i}(r)+2\log{r}\right)=\frac{2\mu_{i}-\sum_{j=1}^{3}a_{ij}\sigma^{k}_{j}(r)}{r},\quad
l_{k}\leq r\leq s_{k},\quad i=1,2,3,$

where $\sigma^{k}_{i}(r)=\sigma^{k}_{i}(B(0,r)),~{}i=1,2,3$.

Concerning the indices of $I_{1}$, we could get from (4.7) that

$\frac{\mathrm{d}}{\mathrm{d}r}\left(\overline{u}^{k}_{i}(r)+2\log{r}\right)\leq-\frac{\delta_{2}}{r},\quad\text{for}~{}r\in[l_{k},\tilde{l}_{k}],\quad
i\in{I}_{1}.$

By integrating the above equation from $l_{k}$ up to
$r\in[l_{k},\tilde{l}_{k}]$, we have

$\overline{u}^{k}_{i}(r)+2\log{r}\leq\overline{u}^{k}_{i}(l_{k})+2\log{l_{k}}+\delta_{2}\log{\frac{l_{k}}{r}},\quad
i\in{I}_{1},$

which implies that for $|x|=r$,

(4.10) $e^{u^{k}_{i}(x)}\leq O(1)e^{\overline{u}^{k}_{i}(r)}\leq
Ce^{-N_{k}}l_{k}^{\delta_{2}}r^{-2-\delta_{2}},\quad i\in{I}_{1},$

where we have used $\overline{u}^{k}_{i}(l_{k})+2\log{l_{k}}\leq-N_{k}$ by the
assumption of fast-decay. By integrating (4.10) over
$l_{k}\leq|x|\leq\tilde{l}_{k}$, we conclude that

$\int_{l_{k}\leq|x|\leq\tilde{l}_{k}}e^{u^{k}_{i}(x)}\mathrm{d}x\leq 2C\pi
e^{-N_{k}}l_{k}^{\delta_{2}}\int_{l_{k}}^{\tilde{l}_{k}}r^{-1-\delta_{2}}\mathrm{d}r\leq
2C\pi\frac{e^{-N_{k}}}{\delta_{2}}\to 0,\quad i\in{I}_{1},$

as $k\to+\infty$. Hence

(4.11) $\sigma^{k}_{i}(\tilde{l}_{k})=\sigma^{k}_{i}(l_{k})+o(1),\quad
i\in{I}_{1}.$

Similarly, for those $i\in{I}_{2}$, we deduce from (4.7) that

$\frac{\mathrm{d}}{\mathrm{d}r}\left(\overline{u}^{k}_{i}(r)+2\log{r}\right)\geq\frac{\delta_{2}}{r},\quad\text{for}~{}r\in[l_{k},\tilde{l}_{k}],\quad
i\in{I}_{2}.$

By integrating the above equation from $r\in[l_{k},\tilde{l}_{k}]$ up to
$\tilde{l}_{k}$, we get that

$\overline{u}^{k}_{i}(r)+2\log{r}\leq\overline{u}^{k}_{i}(\tilde{l}_{k})+2\log{\tilde{l}_{k}}+\delta_{2}\log{\frac{r}{\tilde{l}_{k}}},\quad\text{for}~{}r\in[l_{k},\tilde{l}_{k}],\quad
i\in{I}_{2}.$

This implies that for $|x|=r\in[l_{k},\tilde{l}_{k}]$, there holds

$e^{u^{k}_{i}(x)}\leq O(1)e^{\overline{u}^{k}_{i}(r)}\leq
Ce^{-N_{k}}{\tilde{l}_{k}}^{-\delta_{2}}r^{-2+\delta_{2}},\quad i\in{I}_{2},$

where we have used
$\overline{u}^{k}_{i}(\tilde{l}_{k})+2\log{\tilde{l}_{k}}\leq-N_{k}$ by the
assumption of fast-decay. Thus

$\int_{l_{k}\leq|x|\leq\tilde{l}_{k}}e^{u^{k}_{i}(x)}\mathrm{d}x\leq 2\pi
Ce^{-N_{k}}{\tilde{l}_{k}}^{-\delta_{2}}\int_{l_{k}}^{\tilde{l}_{k}}r^{-1+\delta_{2}}\mathrm{d}r\leq
2\pi C\frac{e^{-N_{k}}}{\delta_{2}}\to 0,\quad i\in{I}_{2},$

as $k\to+\infty$, which implies that

(4.12) $\sigma^{k}_{i}(\tilde{l}_{k})=\sigma^{k}_{i}(l_{k})+o(1),\quad
i\in{I}_{2}.$

Hence, combining (4.11) with (4.12) we have

$\sigma^{k}_{i}(\tilde{l}_{k})=\sigma^{k}_{i}(l_{k})+o(1),\quad i\in I_{1}\cup
I_{2}.$

It remains to consider the indices $i\in I_{3}$. We set
$\varepsilon_{i}=\lim\limits_{k\to+\infty}\sigma_{i}^{k}(\tilde{l}_{k})-\hat{\sigma}_{i}$.
For $i\in I_{1}\cup I_{2}$, $\varepsilon_{i}=0$. For $i\in I_{3}$, we have
$2\mu_{i}-\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}=0$. Substituting
$\hat{\sigma}_{i}+\varepsilon_{i},~{}i=1,2,3$ into the Pohozaev identity (2.4)
we derive that

$\varepsilon_{1}^{2}+\varepsilon_{2}^{2}+2\varepsilon_{3}^{2}-2\varepsilon_{1}\varepsilon_{3}-2\varepsilon_{2}\varepsilon_{3}=0,$

and it leads to

$\varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=0.$

Contradiction arises. Thus (4.8) does not hold for $i\in I_{3}$ and it proves
(4.6). ∎

To state the last lemma in this section, we introduce the following
definition. Let the Harnack-type inequality (4.2) hold for all components of
$\mathbf{u}^{k}$ over all $r\in[\frac{1}{2}l_{k},2\tau_{k}]$. For a sequence
$s_{k}\leq\tau_{k}$, we define

(4.13)
$\hat{\sigma}_{i}(B(\mathbf{x},\mathbf{s}))=\begin{cases}\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(x^{k},s_{k})),~{}&\text{if}~{}u^{k}_{i}~{}\text{has
fast decay on}~{}\partial B(x^{k},s_{k}),\\\ \lim\limits_{r\to
0}\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(x^{k},rs_{k})),~{}&\text{if}~{}u^{k}_{i}~{}\text{has
slow decay on}~{}\partial B(x^{k},s_{k}),\end{cases}$

where $(\mathbf{x},\mathbf{s})$ stands for the sequence of the pair
$\\{(x^{k},s_{k})\\}$. When $x^{k}=0$, we simply denote
$\hat{\sigma}_{i}(B(\mathbf{x},\mathbf{s}))$ by
$\hat{\sigma}_{i}(\mathbf{s})$.

######  Lemma 4.3.

Let $\hat{\sigma}_{i}(\mathbf{s})$ be defined as in (4.13). There is a
sequence $r_{k}\in[l_{k},\tau_{k}]$ satisfying the following conditions

  1. (1)

$\mathbf{u}^{k}$ has fast decay on $\partial B(0,r_{k})$,

  2. (2)

$\exists~{}i\in\\{1,2,3\\}$, such that
$\hat{\sigma}_{i}(\mathbf{r})\neq\hat{\sigma}_{i}(\bm{\tau})$ ($\mathbf{r}$
and $\bm{\tau}$ stand for the sequence $\\{r_{k}\\}$ and $\\{\tau_{k}\\}$).

Then there exists $s_{k}\in(r_{k},\tau_{k})$ such that

  1. (i)

$s_{k}/r_{k}\to+\infty$, there is at least one component $u^{k}_{i}$ of
$\mathbf{u}^{k}$ such that $u^{k}_{i}$ has slow decay on $\partial
B(0,{s_{k}})$.

  2. (ii)

$\hat{\sigma}_{i}(\mathbf{s})=\hat{\sigma}_{i}(\mathbf{r}),~{}i=1,2,3$
($\mathbf{s}$ stands for the sequence $\\{s_{k}\\}$).

###### Proof.

Since $\mathbf{u}^{k}$ has fast decay on $\partial B(0,{r_{k}})$, by
Proposition 2.4,
$(\hat{\sigma}_{1}(\mathbf{r}),\hat{\sigma}_{2}(\mathbf{r}),\hat{\sigma}_{3}(\mathbf{r}))$
satisfies the Pohozaev identity (2.4). Set
$\delta=\max\limits_{i=1,2,3}\\{\hat{\sigma}_{i}(\bm{\tau})-\hat{\sigma}_{i}(\mathbf{r})\\}$,
then $\delta>0$ by the assumption (2). We decompose $\\{1,2,3\\}:=I_{1}\cup
I_{2}\cup I_{3}$, where

$\displaystyle
I_{1~{}(\mathrm{resp.}2,3)}=\left\\{i\in\\{1,2,3\\}~{}\Big{|}~{}2\mu_{i}-\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\mathbf{r})<~{}(\mathrm{resp.}>,=)~{}0\right\\}.$

Set

(4.14) $\delta_{0}=\frac{1}{10000}\min\left\\{\min\limits_{i\in I_{1}\cup
I_{2}}\Big{|}2\mu_{i}-\sum\limits_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\mathbf{r})\Big{|},\delta,1\right\\}.$

We choose $\ell_{k}\in[r_{k},\tau_{k}]$ such that

(4.15)
$\max\limits_{i=1,2,3}\\{\sigma^{k}_{i}(\ell_{k})-\sigma^{k}_{i}(r_{k})\\}=\delta_{0}.$

Lemma 4.2 and (4.15) implies that $\mathbf{u}^{k}$ can not have fast decay on
$[r_{k},\ell_{k}]$. So there is a sequence $r_{k}\ll s_{k}\ll\ell_{k}$ such
that some component of $\mathbf{u}^{k}$ has slow decay on $\partial
B(0,s_{k})$. So it remains to show

(4.16)
$\hat{\sigma}_{i}(\mathbf{s})=\hat{\sigma}_{i}(\mathbf{r})\quad\text{for}\quad
i=1,2,3.$

We prove (4.16) by contradiction. Suppose it is not true, then

(4.17)
$\varepsilon_{0}=\max\limits_{i=1,2,3}\\{\hat{\sigma}_{i}(\mathbf{s})-\hat{\sigma}_{i}({\mathbf{r}})\\}>0.$

By Lemma 4.2 we conclude that there exists a sequence $r_{k}\ll\hat{s}_{k}\ll
s_{k}$ such that

  1. (i)

some components of $\mathbf{u}^{k}$ have slow decay on $\partial
B(0,\hat{s}_{k})$,

  2. (ii)

$\max\limits_{i=1,2,3}\\{\hat{\sigma}_{i}(\hat{\mathbf{s}})-\hat{\sigma}_{i}({\mathbf{r}})\\}\in[\frac{1}{2}\delta_{0},\delta_{0}]$,
where $\hat{\mathbf{s}}$ stands for the sequence $\\{\hat{s}_{k}\\}$.

We scale $u^{k}_{i}$ by

$v^{k}_{i}(y)=u^{k}_{i}(\hat{s}_{k}y)+2\log{\hat{s}_{k}}.$

Using (i), we have some components of $\mathbf{v}^{k}$ converge and there is a
sequence $R_{k}\to+\infty$ such that

  1. (i)

$R_{k}\hat{s}_{k}\ll s_{k}$, $\mathbf{u}^{k}$ has fast decay on $\partial
B(0,R_{k}\hat{s}_{k})$,

  2. (ii)

Set
$\sigma_{i}=\lim\limits_{k\to+\infty}\sigma^{k}_{i}(B(0,R_{k}\hat{s}_{k}))$,
then $\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ satisfies the Pohozaev
identity (2.4).

Let $\varepsilon_{i}=\sigma_{i}-\hat{\sigma}_{i}(\mathbf{r})$ and then
$\max\limits_{i=1,2,3}\varepsilon_{i}\geq\frac{1}{2}\delta_{0}$. We claim that

(4.18) $\varepsilon_{i}=0\quad\text{for}\quad i\in I_{1}\cup I_{2}.$

First, by the same arguments of Lemma 4.2 we have $I_{1}\cup
I_{2}\neq\emptyset$. Next, we shall prove that $u_{i}^{k}$ has fast decay on
$\partial B(0,\hat{s}_{k})$ for $i\in I_{1}\cup I_{2}$. For $i\in I_{1}$,
using Lemma 4.1-(a) we can verify it. While for $i\in I_{2}$, we shall prove
it by contradiction. Suppose it is not true then we could get some $i_{0}\in
I_{2}\cap J$, where $J$ is the maximal set with consecutive indices such that
$v^{k}_{i}~{}(i\in J)$ converges to a solution of a Toda system (or Liouville
equation)

$\Delta v_{i}+\sum\limits_{j\in
J}a_{ij}e^{v_{j}}=4\pi\left(\gamma_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\hat{\mathbf{s}})\right)\delta_{0}~{}~{}\text{in}~{}\mathbb{R}^{2},\quad
i\in J,$

where
$\gamma_{i}-\frac{1}{2}\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\hat{\mathbf{s}})>-1,~{}i\in
J$. Applying the classification result of [49], we conclude that

$\displaystyle\sigma^{k}_{i_{0}}(R_{k}\hat{s}_{k})$
$\displaystyle=\hat{\sigma}_{i_{0}}(\hat{\mathbf{s}})+\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{i_{0}}}+o(1)$
$\displaystyle\geq\hat{\sigma}_{i_{0}}(\hat{\mathbf{s}})+\mu_{i_{0}}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{i_{0}j}\hat{\sigma}_{j}(\hat{\mathbf{s}})+o(1)$
$\displaystyle\geq\hat{\sigma}_{i_{0}}(\hat{\mathbf{s}})+2\delta_{0}+o(1),$

which contradicts to (4.14) and (4.15). Hence, each $u^{k}_{i}~{}(i\in I_{2})$
has fast decay on $\partial B(0,\hat{s}_{k})$, as required. After that, we can
apply the same arguments (4.7)–(4.12) in the proof of Lemma 4.2 to obtain

$\hat{\sigma}_{i}(\mathbf{R}\hat{\mathbf{s}})=\hat{\sigma}_{i}(\hat{\mathbf{s}})=\hat{\sigma}_{i}(\mathbf{r}),~{}i\in
I_{1}\cup I_{2}.$

Thus (4.18) is proved.

Since both $\hat{\bm{\sigma}}(\mathbf{r})$ and $\bm{\sigma}$ satisfy the
Pohozaev identity (2.4), substituting
$\hat{\sigma}_{i}(\mathbf{r})+\varepsilon_{i}$, $i=1,2,3$ into the Pohozaev
identity (2.4) we derive that

$\varepsilon_{1}^{2}+\varepsilon_{2}^{2}+2\varepsilon_{3}^{2}-2\varepsilon_{1}\varepsilon_{3}-2\varepsilon_{2}\varepsilon_{3}=0,$

which implies that

$\varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=0,$

a contradiction to (4.17). This finishes the proof of Lemma 4.3. ∎

##  5\. Local masses on blow-up areas away from the origin

###  5.1. Local masses on the bubbling disk centered at $x^{k}_{j}\neq 0$

In this subsection, we study the local behavior of $\mathbf{u}^{k}$ in the
ball $B(x^{k}_{t},\tau^{k}_{t})$ where $x^{k}_{t}\neq 0$. Precisely, let
$x^{k}_{t}\in\Sigma_{k}\setminus\\{0\\}$ and set

$\tau^{k}_{t}=\frac{1}{2}\mathrm{dist}(x^{k}_{t},\Sigma_{k}\setminus\\{x^{k}_{t}\\}).$

By Proposition 2.1, $l^{k}_{t}\ll\tau^{k}_{t}$. Notice that the Harnack-type
inequality (2.1) holds for
$B(x^{k}_{t},\tau^{k}_{t})\setminus\\{x^{k}_{t}\\}$, i.e.,

$u^{k}_{i}(x)+2\log{|x-x^{k}_{t}|}\leq C,~{}\text{for}~{}x\in
B(x^{k}_{t},\tau^{k}_{t}),\quad i=1,2,3.$

The local mass of the $i$-th component is given by

$\sigma^{k}_{i,t}(r)=\frac{1}{2\pi}\int_{B(x^{k}_{t},r)}e^{u^{k}_{i}(x)}\mathrm{d}x,\quad
i=1,2,3.$

Since $x^{k}_{t}\neq 0$ and $0\notin B(x^{k}_{t},\tau^{k}_{t})$, the system
(1.1) is reduced to

(5.1) $\left\\{\begin{aligned} \Delta
u^{k}_{i}(x)+\sum\limits_{j=1}^{3}a_{ij}e^{u^{k}_{j}(x)}&=0,\quad i=1,2,3,\\\
u^{k}_{1}+u^{k}_{2}+2u^{k}_{3}&=0,\end{aligned}\right.$

in $B(x^{k}_{t},\tau^{k}_{t})$, where the coefficient matrix
$A=(a_{ij})_{3\times 3}$ is defined as in (1.2).

For convenience, we fix $t$ and simplify the notations by dropping the index
$t$ throughout this subsection. Recall that all the components of
$\mathbf{u}^{k}$ have fast decay on $\partial B(x^{k},{l^{k}})$. In (5.1),
$\gamma_{i}=0$, we have $\mu_{i}=1$. In next proposition we shall assume
$\bm{\mu}=(\mu_{1},\mu_{2},\mu_{3})=(1,1,1)$. For a sequence
$s_{k}\leq\tau^{k}$, we recall that
$\hat{\bm{\sigma}}(\mathbf{s})=(\hat{\sigma}_{1}(\mathbf{s}),\hat{\sigma}_{2}(\mathbf{s}),\hat{\sigma}_{3}(\mathbf{s}))$
is defined as in (4.13).

######  Proposition 5.1.

Let $\mathbf{u}^{k}=(u^{k}_{1},u^{k}_{2},u^{k}_{3})$ be a sequence of
solutions of (5.1) and $\hat{\bm{\sigma}}(\bm{\tau})$ be defined as in (4.13),
then there holds

  1. (1)

At least one component of $\mathbf{u}^{k}$ has fast decay on $\partial
B(x^{k},\tau^{k})$.

  2. (2)

$\hat{\bm{\sigma}}(\bm{\tau})=(\hat{\sigma}_{1}(\bm{\tau}),\hat{\sigma}_{2}(\bm{\tau}),\hat{\sigma}_{3}(\bm{\tau}))\in\Gamma(1,1,1)$,
where $\bm{\tau}$ stands for the sequence $\\{\tau^{k}\\}$.

###### Proof.

(1) has been proved in Lemma 4.1. To prove (2), we divide the argument into
several steps.

Step 1. We prove that
$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))\in\Gamma(1,1,1)$,
where $\bm{l}$ stands for the sequence $\\{l^{k}\\}$. Set

$\varepsilon^{k}=e^{-\frac{1}{2}\max\limits_{i=1,2,3}u^{k}_{i}(x^{k})},$

and let

$v^{k}_{i}(y)=u^{k}_{i}(x^{k}+\varepsilon^{k}y)+2\log{\varepsilon^{k}},\quad
i=1,2,3.$

Suppose that there exists $J\subsetneqq\\{1,2,3\\}$ such that $u^{k}_{i}$ has
slow decay on $\partial B(x^{k},\tau^{k})$ for $i\in J$, while the left ones
have fast decay on $\partial B(x^{k},\tau^{k})$, i.e.,
$v^{k}_{i}(y)\to-\infty$ in $L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$ for
$i\in\\{1,2,3\\}\setminus J$ and $v^{k}_{i}(y)$ converges to $v_{i}(y)$ in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$ $\text{for}~{}i\in J$, where $v_{i}(y)$
satisfies

$\Delta v_{i}(y)+\sum\limits_{j\in
J}a_{ij}e^{v_{j}(y)}=0\quad\text{in}\quad\mathbb{R}^{2},\quad i\in J.$

Furthermore, there exists a sequence $R_{k}$ with $R_{k}\to+\infty$ as
$k\to+\infty$ such that $l^{k}:=R_{k}\varepsilon^{k}\leq\tau^{k}$ and
satisfies

  1. (1).

$\int_{B(0,{R_{k}})}e^{v_{i}}\mathrm{d}y=\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}y+o(1)$,
$i\in J$,

  2. (2).

$v^{k}_{i}(y)+2\log|y|\leq-N_{k}$ for $|y|=R_{k}$, for some $N_{k}\to+\infty$,
$i=1,2,3$,

  3. (3).

$\mathbf{u}^{k}$ has fast decay on $\partial B(x^{k},l^{k})$.

Then one of the following alternatives holds:

  1. (a).

If $J=\emptyset$, then
$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))=(0,0,0)$.

  2. (b).

If $J=\\{i\\}$ for some $i\in\\{1,2,3\\}$, then
$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))=(4,0,0)$
or $(0,4,0)$ or $(0,0,4)$.

  3. (c).

If $J=\\{1,2\\}$, then
$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))=(4,4,0)$.

  4. (d).

If $J=\\{1,3\\}$ or $\\{2,3\\}$, then by the classification result Theorem
8.1-(a), we get
$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))=(16,0,12)$
or $(0,16,12)$.

Therefore, in any case, we always have

$(\hat{\sigma}_{1}(\bm{l}),\hat{\sigma}_{2}(\bm{l}),\hat{\sigma}_{3}(\bm{l}))\in\Gamma(1,1,1).$

Step 2. If $\hat{\sigma}_{i}(\bm{l})=\hat{\sigma}_{i}(\bm{\tau})$ for
$i=1,2,3$, then Proposition 5.1 is proved. Otherwise, there exists
$j\in\\{1,2,3\\}$ such that
$\hat{\sigma}_{j}(\bm{l})\neq\hat{\sigma}_{j}(\bm{\tau})$, then we can apply
Lemma 4.3 to find a sequence $s_{k}$ such that $l^{k}\ll s_{k}\ll\tau^{k}$,
some components of $\mathbf{u}^{k}$ have slow decay on $\partial
B(x^{k},{s_{k}})$ and

$\hat{\sigma}_{i}(\bm{l})=\hat{\sigma}_{i}(\mathbf{s})\quad\text{for}\quad
i=1,2,3,$

where $\mathbf{s}$ stands for the sequence $\\{s_{k}\\}$. Let

$v^{k}_{i}(y)=u^{k}_{i}(x^{k}+s_{k}y)+2\log{s_{k}}.$

Suppose that there exists $J\subsetneqq\\{1,2,3\\}$ such that $u^{k}_{i}$ has
slow decay on $\partial B(x^{k},s_{k})$ for $i\in J$, while the left ones have
fast decay on $\partial B(x^{k},s_{k})$, i.e., $v^{k}_{i}(y)\to-\infty$ in
$L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$ for $i\in\\{1,2,3\\}\setminus J$
and $v^{k}_{i}(y)$ converges to $v_{i}(y)$ in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$ $\text{for}~{}i\in J$, where $v_{i}(y)$
satisfies

$\Delta v_{i}(y)+\sum\limits_{j\in
J}a_{ij}e^{v_{j}(y)}=4\pi\gamma_{i}^{*}\delta_{0}\quad\text{in}\quad\mathbb{R}^{2},\quad
i\in J,$

and
$\gamma_{i}^{*}=-\frac{1}{2}\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\bm{l})>-1$.
Furthermore, there exists a sequence $N_{k}^{*}$ with $N_{k}^{*}\to+\infty$ as
$k\to+\infty$ such that $N_{k}^{*}s_{k}\leq\tau^{k}$ and satisfies

  1. (1).

$\int_{B(0,{N_{k}^{*}})}e^{v_{i}}\mathrm{d}y=\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}y+o(1)$,
$i\in J$,

  2. (2).

$v^{k}_{i}(y)+2\log|y|\leq-N_{k}$ for $|y|=N^{*}_{k}$, for some
$N_{k}\to+\infty$, $i=1,2,3$,

  3. (3).

$\mathbf{u}^{k}$ has fast decay on $\partial B(x^{k},N^{*}_{k}s_{k})$.

Then one of the following alternatives holds:

Case (a). If $J=\\{i_{0}\\}$ for some $i_{0}\in\\{1,2,3\\}$. Notice that
$\hat{\sigma}_{i}(\mathbf{N^{*}s})=\hat{\sigma}_{i}(\bm{l})$ for
$i\in\\{1,2,3\\}\setminus J$, where $\mathbf{N^{*}s}$ stands for the sequence
$\\{N_{k}^{*}s_{k}\\}$, and both $\hat{\bm{{\sigma}}}(\mathbf{N^{*}s})$ and
$\hat{\bm{\sigma}}(\bm{l})$ satisfy the Pohozaev identity (2.4). Then
$\hat{\sigma}_{{i_{0}}}(\mathbf{N^{*}s})$ and $\hat{\sigma}_{i_{0}}(\bm{l})$
are two roots of a quadratic polynomial in $\sigma_{i_{0}}$. From which we can
directly solve
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{{i_{0}}}(\hat{\bm{{\sigma}}}(\bm{l}))$.
We can also deduce this result in another way by applying classification
result of singular Liouville equation [49]:

$\displaystyle\hat{\sigma}_{{i_{0}}}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{{i_{0}}}}\mathrm{d}x+\hat{\sigma}_{i_{0}}(\bm{l})=4\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{{i_{0}}j}\hat{\sigma}_{j}(\bm{l})\right)+\hat{\sigma}_{i_{0}}(\bm{l})=\left(\mathfrak{R}_{{i_{0}}}(\hat{\bm{\sigma}}(\bm{l}))\right)_{{i_{0}}}.$

Then
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{{i_{0}}}(\hat{\bm{{\sigma}}}(\bm{l}))\in\Gamma(1,1,1)$
if $\hat{\bm{\sigma}}(\bm{l})\in\Gamma(1,1,1)$.

Case (b). If $J=\\{1,2\\}$. From the view of the classification result of
[49], we directly deduce that

$\displaystyle\hat{\sigma}_{m}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{m}}\mathrm{d}x+\hat{\sigma}_{m}(\bm{l})=4\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{mj}\hat{\sigma}_{j}(\bm{l})\right)+\hat{\sigma}_{m}(\bm{l})=\left(\mathfrak{R}_{12}(\hat{\bm{\sigma}}(\bm{l}))\right)_{m},~{}\text{for}~{}m=1,2.$

We also notice that
$\hat{\sigma}_{3}(\mathbf{N^{*}s})=\hat{\sigma}_{3}(\bm{l})$, hence
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{12}(\hat{\bm{\sigma}}(\bm{l}))\in\Gamma(1,1,1)$
provided $\hat{\bm{\sigma}}(\bm{l})\in\Gamma(1,1,1)$.

Case (c). If $J=\\{m,3\\}$ for $m=1$ or $2$. From the classification result
Theorem 8.1-(a), we directly deduce that

$\displaystyle\hat{\sigma}_{m}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{m}}\mathrm{d}x+\hat{\sigma}_{m}(\bm{l})=8\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{mj}\hat{\sigma}_{j}(\bm{l})\right)+8\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{3j}\hat{\sigma}_{j}(\bm{l})\right)+\hat{\sigma}_{m}(\bm{l})$
$\displaystyle=\left(\mathfrak{R}_{m3}\left(\hat{\bm{\sigma}}(\bm{l})\right)\right)_{m},~{}\text{for}~{}m=1~{}\text{or}~{}2,$

and

$\displaystyle\hat{\sigma}_{3}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{3}}\mathrm{d}x+\hat{\sigma}_{3}(\bm{l})=4\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{mj}\hat{\sigma}_{j}(\bm{l})\right)+8\left(1-\frac{1}{2}\sum\limits_{j=1}^{3}a_{3j}\hat{\sigma}_{j}(\bm{l})\right)+\hat{\sigma}_{3}(\bm{l})$
$\displaystyle=\left(\mathfrak{R}_{m3}\left(\hat{\bm{\sigma}}(\bm{l})\right)\right)_{3},~{}\text{for}~{}m=1~{}\text{or}~{}2.$

We also notice that
$\hat{\sigma}_{j}(\mathbf{N^{*}s})=\hat{\sigma}_{j}(\bm{l})$ for
$j\in\\{1,2,3\\}\setminus J$, then
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{m3}(\hat{\bm{\sigma}}(\bm{l}))\in\Gamma(1,1,1)$
since $\hat{\bm{\sigma}}(\bm{l})\in\Gamma(1,1,1)$.

Step 3. Let $s_{k,1}=N_{k}^{*}s_{k}$. If

$\hat{\sigma}_{i}(\mathbf{s}_{1})=\hat{\sigma}_{i}(\bm{\tau})\quad\text{for}\quad
i=1,2,3,$

then Proposition 5.1 is proved. Otherwise, we could repeat the argument of
Step 2 to find $s_{k,1}\ll s_{k,j}\ll s_{k,j+1}$ such that
$\hat{\bm{\sigma}}(\mathbf{s}_{j+1})\in\Gamma(1,1,1)$, where
$\mathbf{s}_{j+1}$ stands for the sequence $\\{s_{k,j+1}\\}$. Since the energy
is finite by (iii) of (1.10) and the total gain for the local masses at each
step has a lower bound

$\sum\limits_{i=1}^{3}\left(\hat{\sigma}_{i}(\mathbf{s}_{j+1})-\hat{\sigma}_{i}(\mathbf{s}_{j})\right)\geq
4>0,$

then after finitely many steps we have
$\hat{\sigma}_{i}(\mathbf{s}_{j})=\hat{\sigma}_{i}(\bm{\tau})~{}\text{for}~{}i=1,2,3$.
∎

From the proof of Proposition 5.1, we can further deduce a more general result
which will be used to prove Theorem 1.1 in Section 6. Suppose that there is a
subset $J\subsetneqq\\{1,2,3\\}$ such that each $u^{k}_{i}~{}(i\in J)$ has
slow decay on $\partial B(x^{k},s_{k})$. Let

$v^{k}_{i}(y)=u^{k}_{i}(x^{k}+s_{k}y)+2\log{s_{k}},\quad i=1,2,3.$

Then $v^{k}_{i}(y)\to-\infty$ for $i\in\\{1,2,3\\}\setminus J$ in
$L^{\infty}_{\mathrm{loc}}(\mathbb{R}^{2})$ and $v^{k}_{i}(y)$ converges to
$v_{i}(y)$ in $C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$ for $i\in J$, where
$v_{i}(y)$ satisfies

(5.2) $\Delta v_{i}(y)+\sum\limits_{j\in
J}a_{ij}e^{v_{j}(y)}=4\pi\gamma_{i}^{*}\delta_{0}+4\pi\sum\limits_{l=1}^{N}m_{il}\delta_{q_{l}}\quad\text{in}\quad\mathbb{R}^{2},\quad
i\in J,$

where $0\neq q_{l}\in\mathbb{R}^{2}$, $m_{il}\in\mathbb{N}$ and
$\gamma_{i}^{*}=\gamma_{i}-\frac{1}{2}\sum_{j=1}^{3}a_{ij}\sigma_{j}>-1$. Set

$\sigma^{*}_{j}=\left\\{\begin{array}[]{ll}\sigma_{j},{}&\text{if}~{}j\in\\{1,2,3\\}\setminus
J,\\\
\sigma_{j}+\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{j}}dx,{}&\text{if}~{}j\in
J.\end{array}\right.$

Then we have the following result

######  Proposition 5.2.

If there is $\hat{\bm{\sigma}}\in\Gamma(\bm{\mu})$ such that
$\sigma_{i}=\hat{\sigma}_{i}+4n_{i},~{}n_{i}\in\mathbb{Z}$. Then
$\sigma^{*}_{i}=\hat{\sigma}^{*}_{i}+4n^{*}_{i}$ with
$\hat{\bm{\sigma}}^{*}\in\Gamma(\bm{\mu})$ and $n^{*}_{i}\in\mathbb{Z}$.

###### Proof.

We shall consider the following three cases.

Case (a). If $J=\\{l\\}$ for some $l\in\\{1,2,3\\}$. Then by the
classification result of [49], we have

(5.3) $\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x$
$\displaystyle=4(1+\gamma^{*}_{l})+4N_{l}=4\mu^{*}_{l}+4N_{l},$

where

$\mu^{*}_{l}=1+\gamma^{*}_{l}=1+\gamma_{l}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{lj}{\sigma}_{j}=\mu_{l}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{lj}{\sigma}_{j},$

and $N_{l}\in\mathbb{Z}$. Since $\sigma_{i}=\hat{\sigma}_{i}+4n_{i}$ for some
$n_{i}\in\mathbb{Z}$, we get

$\mu^{*}_{i}=\mu_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}{\sigma}_{j}=\mu_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}\hat{\sigma}_{j}+\overline{n}_{i},\quad\overline{n}_{i}\in\mathbb{Z},\quad
i=1,2,3.$

Then (5.3) can be rewritten as

$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x=4\mu^{*}_{l}+4N_{l}=4\mu_{l}-\sum_{j=1}^{3}a_{lj}\hat{\sigma}_{j}+4\widetilde{N}_{l},\quad\widetilde{N}_{l}\in\mathbb{Z}.$

Thus we have

$\displaystyle\sigma^{*}_{l}=\sigma_{l}+\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x=\hat{\sigma}_{l}+4n_{l}+4\mu_{l}-2\sum_{j=1}^{3}a_{lj}\hat{\sigma}_{j}+4\widetilde{N}_{l}=:\hat{\sigma}_{l}^{*}+4n^{*}_{l},\quad
n_{l}^{*}\in\mathbb{Z}.$

While for $j\neq l$, we set $\hat{\sigma}_{j}=\hat{\sigma}_{j}^{*}$, then

$\displaystyle\sigma^{*}_{j}=\sigma_{j}=\hat{\sigma}_{j}+4n_{j}=:\hat{\sigma}_{j}^{*}+4n_{j}.$

Therefore, we get

$\bm{\sigma}^{*}=\hat{\bm{\sigma}}^{*}+4\mathbb{Z}\quad\text{with}\quad\hat{\bm{\sigma}}^{*}=\mathfrak{R}_{l}(\hat{\bm{\sigma}})\in\Gamma(\bm{\mu}).$

Case (b). If $J=\\{1,2\\}$. Then by the classification result of [49], we have

(5.4) $\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}x$
$\displaystyle=4(1+\gamma^{*}_{i})+4N_{i}=4\mu^{*}_{i}+4N_{i},\quad
N_{i}\in\mathbb{Z},\quad i=1,2,$

where

$\mu^{*}_{i}=1+\gamma^{*}_{i}=1+\gamma_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}{\sigma}_{j}=\mu_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}{\sigma}_{j},\quad
i=1,2.$

Since $\sigma_{i}=\hat{\sigma}_{i}+4n_{i}$ for some $n_{i}\in\mathbb{Z}$,
$i=1,2,3$, we get that

$\mu^{*}_{i}=\mu_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}{\sigma}_{j}=\mu_{i}-\frac{1}{2}\sum\limits_{j=1}^{3}a_{ij}\hat{\sigma}_{j}+\overline{n}_{i},\quad\overline{n}_{i}\in\mathbb{Z},\quad
i=1,2,3.$

Then (5.4) can be rewritten as

$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}x=4\mu^{*}_{i}+4N_{i}=4\left(\mu_{i}-\frac{1}{2}\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}\right)+4\hat{N}_{i},\quad\hat{N}_{i}\in\mathbb{Z},\quad
i=1,2.$

Hence, for $i=1,2$ we get that

$\displaystyle\sigma^{*}_{i}=\sigma_{i}+\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}x=\hat{\sigma}_{i}+4\mu_{i}-2\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}++4n_{i}+4\widetilde{N}_{i}=:\hat{\sigma}_{i}^{*}+4n^{*}_{i},$

with
$\hat{\sigma}_{i}^{*}=4\mu_{i}-2\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}+\hat{\sigma}_{i}=\left(\mathfrak{R}_{12}(\hat{\bm{\sigma}})\right)_{i}$.
While for the third component, we set $\hat{\sigma}_{3}^{*}=\hat{\sigma}_{3}$.
Then we conclude that

$\bm{\sigma}^{*}=\hat{\bm{\sigma}}^{*}+4\mathbb{Z}\quad\text{with}\quad\hat{\bm{\sigma}}^{*}=\mathfrak{R}_{12}(\hat{\bm{\sigma}})\in\Gamma(\bm{\mu}).$

Case (c). If $J=\\{i,3\\}$ for $i=1$ or $2$. Then by the classification result
Theorem 8.1-(c), we have

$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}x=\bar{\sigma}_{i}+4N_{i},\quad
i=1~{}\text{or}~{}2,$

and

$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{3}}\mathrm{d}x$
$\displaystyle=\bar{\sigma}_{3}+4N_{3},$

where

(5.5) $\displaystyle(\bar{\sigma}_{i},\bar{\sigma}_{3})\in\Big{\\{}$
$\displaystyle(0,0),~{}(4\mu_{i}^{*},0),~{}(0,4\mu_{3}^{*}),~{}(4\mu_{i}^{*},4(\mu_{i}^{*}+\mu_{3}^{*})),~{}(4\mu_{i}^{*}+8\mu_{3}^{*},4\mu_{3}^{*}),$
$\displaystyle(4\mu_{i}^{*}+8\mu_{3}^{*},4\mu_{i}^{*}+8\mu_{3}^{*}),~{}(8\mu_{i}^{*}+8\mu_{3}^{*},4\mu_{i}^{*}+4\mu_{3}^{*}),~{}(8\mu_{i}^{*}+8\mu_{3}^{*},4\mu_{i}^{*}+8\mu_{3}^{*})\Big{\\}},$

and

$\mu_{i}^{*}=\gamma_{i}^{*}+1,~{}i=1~{}(\mbox{or}~{}2),3,\quad
N_{i}\in\mathbb{Z}.$

Then we get that

(5.6)
$\sigma_{i}^{*}=\hat{\sigma}_{i}+\bar{\sigma}_{i}+4\bar{N}_{i},~{}i=1~{}\mbox{or}~{}2,\quad\mbox{and}\quad\sigma_{3}^{*}=\hat{\sigma}_{3}+\bar{\sigma}_{3}+4\bar{N}_{3},$

where $\bar{N}_{i}\in\mathbb{Z},~{}i=1~{}(\text{or}~{}2),3$. We set

$\hat{\sigma}_{j}^{*}=\hat{\sigma}_{j}+\bar{\sigma}_{j},~{}j=i,3,\quad\mathrm{and}\quad\hat{\sigma}_{j}^{*}=\hat{\sigma}_{j},~{}j\neq
i,3.$

By (5.5) and (5.6), from direct computation we get that

$\hat{\bm{\sigma}}^{*}\in\Big{\\{}\hat{\bm{\sigma}},~{}\mathfrak{R}_{i}\hat{\bm{\sigma}},~{}\mathfrak{R}_{3}\hat{\bm{\sigma}},~{}\mathfrak{R}_{i}\mathfrak{R}_{3}\hat{\bm{\sigma}},~{}\mathfrak{R}_{3}\mathfrak{R}_{i}\hat{\bm{\sigma}},~{}\mathfrak{R}_{i}\mathfrak{R}_{3}\mathfrak{R}_{i}\hat{\bm{\sigma}},~{}\mathfrak{R}_{3}\mathfrak{R}_{i}\mathfrak{R}_{3}\hat{\bm{\sigma}},~{}\mathfrak{R}_{i3}\hat{\bm{\sigma}}\Big{\\}}.$

Since $\hat{\bm{\sigma}}\in\Gamma(\mu)$ we get that
$\hat{\bm{\sigma}}^{*}\in\Gamma(\mu)$ as well by the definition of
$\Gamma(\mu)$. Thus, we have

${\bm{\sigma}}^{*}={\hat{\bm{\sigma}}}^{*}+4\mathbb{Z}\quad\mbox{with}\quad{\hat{\bm{\sigma}}}^{*}\in\Gamma(\mu).$

This finishes the whole proof. ∎

###  5.2. Local mass in a group that does not contain $0$

In this subsection, we shall collect together some of the bubbling disks
$B(x_{i}^{k},l_{i}^{k})$ into a larger one. We present the procedure as
follows. For any $x_{i}^{k}\neq 0\in\Sigma_{k}$, we define the subset
$S^{k}_{1}$ of $\Sigma_{k}$ by the following way. Let $x_{2}^{k}\neq
0\in\Sigma_{k}$ be the point of $\Sigma_{k}$ such that

$\mathrm{dist}(x_{1}^{k},x_{2}^{k})=\mathrm{dist}\left(x_{1}^{k},\Sigma_{k}\setminus\\{0,x_{1}^{k}\\}\right),$

where $\mathrm{dist}(\cdot,\cdot)$ is the distance function. We set

$S^{k}_{1}=\begin{cases}\\{x_{1}^{k}\\},\quad&\mbox{if}\quad\frac{\mathrm{dist}(x_{1}^{k},x_{2}^{k})}{|x_{1}^{k}|}\quad\mbox{has
a positive lower bound},\\\
\left\\{x_{j}^{k}~{}\mid~{}\mathrm{dist}(x_{1}^{k},x_{j}^{k})\leq
C\mathrm{dist}(x_{1}^{k},x_{2}^{k})\right\\},\quad&\mbox{if}\quad\frac{\mathrm{dist}(x_{1}^{k},x_{2}^{k})}{|x_{1}^{k}|}~{}\to
0.\end{cases}$

Then we have

$\forall x,y\in S_{1}^{k},\quad\mbox{the
ratio}~{}\frac{\mathrm{dist}(x,y)}{|y|}\to 0~{}\mbox{as}~{}k\to+\infty.$

Obviously, one can see that $\Sigma_{k}$ can be decomposed into a disjoint
union of $S_{j}^{k}:$

$\Sigma_{k}=\\{0\\}\cup S_{1}^{k}\cup\cdots\cup S_{m_{0}}^{k},\quad m_{0}\leq
m\quad\mbox{and}\quad m_{0}\in\mathbb{N}.$

Next, we shall use $S_{1}^{k}$ as an example to illustrate the combination
process for the points in $S_{1}^{k}$. We may assume

$S_{1}^{k}=\\{x_{1,1}^{k},\cdots,x_{1,m_{1}}^{k}\\}\subseteq\Sigma_{k}.$

Let $\tau_{S^{k}_{1}}^{k}$ and $\tau_{l}^{k}$ be defined as follows:

$\tau_{S^{k}_{1}}^{k}=\frac{1}{2}\mbox{dist}(x_{1,1}^{k},\Sigma_{k}\setminus
S_{1}^{k})\quad\mbox{and}\quad\tau_{l}^{k}=\frac{1}{2}\mbox{dist}(x_{1,l}^{k},S_{1}^{k}\setminus\\{x_{1,l}^{k}\\})\quad\text{for}\quad
l=1,\cdots,m_{1}.$

Then it is not difficult to see that

$\bigcup_{l=1}^{m_{1}}B(x_{1,l}^{k},\tau_{l}^{k})\subseteq
B(x_{1}^{k},\tau_{S^{k}_{1}}^{k})\quad\mbox{and}\quad\tau_{l}^{k}/\tau_{S^{k}_{1}}^{k}\to
0\quad\mbox{for}\quad l=1,\cdots,m_{1}.$

By Proposition 5.1, the local mass
$\hat{\sigma}_{i}(B(\mathbf{x}_{1,l},\bm{\tau}_{l}))=4m_{l,i},~{}i=1,2,3,~{}l=1,\cdots,m_{1}$
satisfies

$(4m_{l,1},4m_{l,2},4m_{l,3})\in\Gamma(1,1,1)\quad\mbox{for}\quad
l=1,\cdots,m_{1},$

where $(\mathbf{x}_{1,l},\bm{\tau}_{l})$ stands for the sequence of pairs
$\\{(x^{k}_{1,l},\tau^{k}_{l})\\}$, $m_{l,i}\in\mathbb{N}\cup\\{0\\},~{}1\leq
l\leq m_{1}$. In the following proposition we shall compute the possible
values on the blow up mass in $B(x_{1,1}^{k},\tau_{S^{k}_{1}}^{k})$

######  Proposition 5.3.

There holds:

  1. (i)

At least one component of $\mathbf{u}^{k}$ has fast decay on $\partial
B(x^{k}_{1,1},\tau^{k}_{1})$.

  2. (ii)

Let $\sigma_{i}=\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\bm{\tau}_{S^{k}_{1}}))$,
then $\sigma_{i}\in 4\mathbb{N}\cup\\{0\\}$.

###### Proof.

By Proposition 5.1, it is sufficient to consider the following two cases.

Case (1). $\mathbf{u}^{k}$ has fast decay on $\partial
B(x^{k}_{1,1},\tau^{k}_{1})$. Let

$l^{k}(S^{k}_{1})=2\max_{1\leq l\leq
m_{1}}\mbox{dist}(x_{1,1}^{k},x_{1,l}^{k}).$

Then $\mathbf{u}^{k}$ has fast decay on $\partial B(x^{k}_{1,l},\tau_{l}^{k})$
for any $l=1,\cdots,m_{1}.$ Therefore

(5.7) $\displaystyle\sigma^{k}_{i}(B(x^{k}_{1,1},l^{k}(S^{k}_{1})))$
$\displaystyle=\frac{1}{2\pi}\int_{B(x^{k}_{1,1},l^{k}(S^{k}_{1}))}e^{u^{k}_{i}(x)}\mathrm{d}x$
$\displaystyle=\frac{1}{2\pi}\int_{\bigcup_{l=1}^{m_{1}}B(x^{k}_{1,l},\tau^{k}_{l})}e^{u^{k}_{i}(x)}\mathrm{d}x+\frac{1}{2\pi}\int_{B(x^{k}_{1,1},l^{k}(S^{k}_{1}))\setminus\bigcup_{j=1}^{m_{1}}B(x^{k}_{1,j},\tau^{k}_{j})}e^{u^{k}_{i}(x)}\mathrm{d}x$
$\displaystyle=\frac{1}{2\pi}\int_{\bigcup_{l=1}^{m_{1}}B(x^{k}_{1,l},\tau^{k}_{l})}e^{u^{k}_{i}(x)}\mathrm{d}x+o(1)$
$\displaystyle=\sum\limits_{l=1}^{m_{1}}4m_{l,i}+o(1),$

where $m_{l,i}\in\mathbb{N}\cup\\{0\\},~{}i=1,2,3,~{}1\leq l\leq m_{1}$. Here
we have used the fast-decay of $\mathbf{u}^{k}$ outside of
$B(x^{k}_{1,l},\tau^{k}_{l})$, and then the second integral of (5.7) is
$o(1)$.

If it holds that

$\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{l}(S^{k}_{1})))=\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{\bm{\tau}}_{S^{k}_{1}})),\quad\forall
i=1,2,3,$

where $(\mathbf{x}_{1,1},\mathbf{l}(S^{k}_{1}))$ and
$(\mathbf{x}_{1,1},\mathbf{\bm{\tau}}_{S^{k}_{1}})$ stand for the sequences of
pairs $\\{(x^{k}_{1,1},l^{k}(S^{k}_{1}))\\}$ and
$\\{(x^{k}_{1,1},\tau^{k}_{S^{k}_{1}})\\}$ respectively, then Proposition 5.3
is proved. Otherwise,

$\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{\bm{\tau}}_{S^{k}_{1}}))>\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{l}(S^{k}_{1})))\quad\mbox{for
some}~{}i\in\\{1,2,3\\}.$

In this situation we can apply the same arguments of Step 2 and Step 3 in the
proof of Proposition 5.1 to obtain that

$\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{\bm{\tau}}_{S^{k}_{1}}))=\hat{\sigma}_{i}(B(\mathbf{x}_{1,1},\mathbf{s}_{\ell}))\in
4\mathbb{N}\cup\\{0\\},~{}i=1,2,3,$

for some $\mathbf{s}_{\ell}$, where $\mathbf{s}_{\ell}$ stands for the
sequence $\\{s^{k}_{\ell}\\}$ with $l^{k}(S^{k}_{1})\ll
s^{k}_{\ell}\ll\tau^{k}_{S_{1}}$.

Case (2). Some components of $\mathbf{u}^{k}$ have slow decay on $\partial
B(x^{k}_{1,1},\tau^{k}_{1})$. Since $l^{k}(S^{k}_{1})\thicksim\tau^{k}_{l}$
for $1\leq l\leq m_{1}$, some components of $\mathbf{u}^{k}$ have slow decay
on $\partial B(x^{k}_{1,1},l^{k}(S^{k}_{1}))$. Then one of the following
alternatives (similar to that of Step 2 in the proof of Proposition 5.1)
holds:

  1. (2-1).

$u^{k}_{1}$ and $u^{k}_{2}$ have slow decay on $\partial
B(x^{k}_{1,1},l^{k}(S^{k}_{1}))$.

  2. (2-2).

$u^{k}_{1}$ and $u^{k}_{3}$ (or $u^{k}_{2}$ and $u^{k}_{3}$) have slow decay
on $\partial B(x^{k}_{1,1},l^{k}(S^{k}_{1}))$.

  3. (2-3).

One component of $\mathbf{u}^{k}$ has slow decay on $\partial
B(x^{k}_{1,1},l^{k}(S^{k}_{1}))$.

We set $J$ as the maximal set such that each $u^{k}_{i}$ ($i\in J$) has slow
decay on $\partial B(x^{k}_{1,1},l^{k}(S^{k}_{1}))$. Let

$v^{k}_{i}(y)=u^{k}_{i}(x^{k}_{1,1}+l^{k}(S^{k}_{1})y)+2\log{l^{k}(S^{k}_{1})},\quad
i=1,2,3,$

then $v^{k}_{i}(y)\to-\infty$ in $L^{\infty}_{\mathrm{loc}}(\mathbb{R}^{2})$
for $i\in\\{1,2,3\\}\setminus J$ and $v^{k}_{i}(y)$ converges to $v_{i}(y)$ in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$ for $i\in J$, where $v_{i}(y)$
satisfies

$\Delta v_{i}(y)+\sum\limits_{j\in
J}a_{ij}e^{v_{j}(y)}=4\pi\sum\limits_{l=1}^{m_{1}}n_{l,i}\delta_{q_{l}}\quad\text{in}\quad\mathbb{R}^{2},\quad
i\in J,$

where $n_{l,i}=-2\sum_{j=1}^{3}a_{ij}m_{l,j}\in\mathbb{N}\cup\\{0\\}$.
Applying the classification result Theorem 8.1-(b), there exists a sequence
$R_{k}$ with $R_{k}\to+\infty$ as $k\to+\infty$ such that
$R_{k}l^{k}(S^{k}_{1})\leq\tau^{k}_{S^{k}_{1}}$ and

  1. (a).

$\int_{B(0,{R_{k}})}e^{v_{i}}\mathrm{d}y=\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}y+o(1)=4\widetilde{m}_{i}+o(1),~{}i\in
J$, where $\widetilde{m}_{i}\in\mathbb{N}$,

  2. (b).

$v^{k}_{i}(y)+2\log|y|\leq-N_{k}$ for $|y|=R_{k}$, for some
$N_{k}\to+\infty,~{}i=1,2,3$,

  3. (c).

$\mathbf{u}^{k}$ has fast decay on $\partial
B(x^{k}_{1,1},R_{k}l^{k}{(S^{k}_{1})})$.

Hence the local mass

$\sigma^{k}_{i}(B(x^{k}_{1,1},R_{k}l^{k}(S^{k}_{1})))=4\left(\sum\limits_{l=1}^{m_{1}}m_{l,i}+\widetilde{m}_{i}\right)+o(1),\quad
i\in J,$

and

$\sigma^{k}_{i}(B(x^{k}_{1,1},R_{k}l^{k}(S^{k}_{1})))=4\sum\limits_{l=1}^{m_{1}}m_{l,i}+o(1),\quad
i\in\\{1,2,3\\}\setminus J.$

Notice that the total gain of local masses at each step is at least $4$.
Following almost the same argument as Step 3 of the proof to Proposition 5.1,
we could get the conclusion 5.3 in finitely many steps. ∎

##  6\. Proof of Theorem 1.1 and Theorem 1.3

In last section, we have decomposed

$\Sigma_{k}=\\{0\\}\cup S_{1}^{k}\cup\cdots\cup S_{m_{0}}^{k},$

and completed the combination process for the points in each group $S_{l}$
with $1\leq l\leq m_{0}$. In this section, we shall regard $S_{l}^{k}$ as
points and do the further collection. Precisely, we shall collect $\\{0\\}$
and several closer sets $S_{l}^{k}$ at first. Suppose $x_{l}^{k}\in
S_{l}^{k}$. Without loss of generality, we may assume the sets
$S_{1}^{k},\cdots,S_{l_{0}}^{k}$ are the ones with

$C^{-1}|x_{1}^{k}|\leq|x_{l}^{k}|\leq C|x_{1}^{k}|,~{}\mbox{for}~{}1\leq l\leq
l_{0},$

and

$|x_{l}^{k}|\gg|x_{1}^{k}|\quad\mbox{for}\quad l>l_{0}.$

Setting $\tau_{k}=\frac{1}{2}$ if $m_{0}=0$ and
$\tau_{k}=\frac{1}{2}|x^{k}_{1}|$ if $m_{0}>0$. Let
$\sigma_{i}=\hat{\sigma}_{i}(\bm{\tau})$, where $\hat{\sigma}_{i}(\bm{\tau})$
is defined as in (4.13), then $\bm{\sigma}\in\Gamma(\bm{\mu})$. To start the
second step of combination procedure, we compute the possible blow up mass in
the bubbling disk centered at origin.

######  Lemma 6.1.

$\bm{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\in\Gamma(\bm{\mu})$.

###### Proof.

We select $r_{k}\ll\tau_{k}$ such that
$\max\limits_{i=1,2,3}\sigma^{k}_{i}(B(0,r_{k}))\leq 1/k$ and $\mathbf{u}^{k}$
has fast decay on $\partial B(0,r_{k})$. We can apply Lemma 4.3 to find a
sequence $s_{k}$ such that at least one component of $\mathbf{u}^{k}$ has slow
decay on $\partial B(0,s_{k})$ and $\hat{\bm{\sigma}}(\mathbf{s})=0$. If
$\tau_{k}/s_{k}\leq C$ or $\bm{\sigma}=\hat{\bm{\sigma}}(\mathbf{s})$, then
Lemma 6.1 is proved, i.e., $\bm{\sigma}=\bm{0}\in\Gamma(\bm{\mu})$. If
$\tau_{k}/s_{k}\to+\infty$, we can apply the same arguments of Step 2 and Step
3 in the proof of Proposition 5.1. In fact, let

$v^{k}_{i}(y)=u^{k}_{i}({s}_{k}y)+2\log{{s}_{k}}.$

Suppose that there exists $J\subsetneqq\\{1,2,3\\}$ such that $u^{k}_{i}$ has
slow decay on $\partial B(0,s_{k})$ for $i\in J$, while the left ones have
fast decay on $\partial B(0,s_{k})$, i.e., $v^{k}_{i}(y)\to-\infty$ in
$L_{\mathrm{loc}}^{\infty}(\mathbb{R}^{2})$ for $i\in\\{1,2,3\\}\setminus J$
and $v^{k}_{i}(y)$ converges to $v_{i}(y)$ in
$C_{\mathrm{loc}}^{2}(\mathbb{R}^{2})$ $\text{for}~{}i\in J$, where $v_{i}(y)$
satisfies

$\Delta v_{i}(y)+\sum\limits_{j\in
J}a_{ij}e^{v_{j}(y)}=4\pi\gamma_{i}^{*}\delta_{0}\quad\text{in}\quad\mathbb{R}^{2},\quad
i\in J,$

and
$\gamma_{i}^{*}=\gamma_{i}-\frac{1}{2}\sum_{j=1}^{3}a_{ij}\hat{\sigma}_{j}(\mathbf{s})=\gamma_{i}>-1$.
Furthermore, there exists a sequence $N_{k}^{*}$ with $N_{k}^{*}\to+\infty$ as
$k\to+\infty$ such that $N_{k}^{*}s_{k}\leq\tau_{k}$ and satisfies

  1. (1).

$\int_{B(0,{N_{k}^{*}})}e^{v_{i}}\mathrm{d}y=\int_{\mathbb{R}^{2}}e^{v_{i}}\mathrm{d}y+o(1)$,
$i\in J$,

  2. (2).

$v^{k}_{i}(y)+2\log|y|\leq-N_{k}$ for $|y|=N^{*}_{k}$, for some
$N_{k}\to+\infty$, $i=1,2,3$,

  3. (3).

$\mathbf{u}^{k}$ has fast decay on $\partial B(x^{k},N^{*}_{k}s_{k})$.

Then one of the following alternatives holds:

Case (a). If $J$ only contains one element $l$ for some $l\in\\{1,2,3\\}.$
Applying the classification result of [49], we have

$\displaystyle\hat{\sigma}_{l}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x+\hat{\sigma}_{l}(\mathbf{s})=4\left(1+\gamma_{l}\right)+0=4\mu_{l}=\left(\mathfrak{R}_{l}(\bm{0})\right)_{l}.$

Notice that $\hat{\sigma}_{i}(\mathbf{N^{*}s})=\hat{\sigma}_{i}(\mathbf{s})=0$
for $i\in\\{1,2,3\\}\setminus\\{l\\}$, hence
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{l}(\bm{0})\in\Gamma(\bm{\mu})$.

Case (b). If $J=\\{1,2\\}$. By the classification result of [49] again, we
have

$\displaystyle\hat{\sigma}_{l}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x+\hat{\sigma}_{l}(\mathbf{s})=4\mu_{l}+0=\left(\mathfrak{R}_{12}(\bm{0})\right)_{l},\quad
l=1,2.$

While $\hat{\sigma}_{3}(\mathbf{N^{*}s})=\hat{\sigma}_{3}(\mathbf{s})=0$,
therefore
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{12}(\bm{0})\in\Gamma(\bm{\mu})$.

Case (c). If $J=\\{l,3\\}$ for $l=1$ or $2$. Applying the classification
result Theorem 8.1-(a), we have

$\displaystyle\hat{\sigma}_{l}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{l}}\mathrm{d}x+\hat{\sigma}_{l}(\mathbf{s})=8\mu_{l}+8\mu_{3}+0=\left(\mathfrak{R}_{l3}(\bm{0})\right)_{l},\quad
l=1~{}\text{or}~{}2,$

and

$\displaystyle\hat{\sigma}_{3}(\mathbf{N^{*}s})$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{v_{3}}\mathrm{d}x+\hat{\sigma}_{3}(\mathbf{s})=4\mu_{l}+8\mu_{3}+0=\left(\mathfrak{R}_{l3}(\bm{0})\right)_{3},\quad
l=1~{}\text{or}~{}2.$

For the left component $j\in\\{1,2,3\\}\setminus\\{l,3\\}$,
$\hat{\sigma}_{j}(\mathbf{N^{*}s})=\hat{\sigma}_{j}(\mathbf{s})=0$, then
$\hat{\bm{\sigma}}(\mathbf{N^{*}s})=\mathfrak{R}_{l3}(\bm{0})\in\Gamma(\bm{\mu})$,
$l=1$ or $2$.

Next, let $s_{k,1}=N_{k}^{*}s_{k}$. If

$\hat{\sigma}_{i}(\mathbf{s}_{1})=\hat{\sigma}_{i}(\bm{\tau})\quad\text{for}\quad
i=1,2,3,$

then Lemma 6.1 is proved. Otherwise, we could repeat the above argument to
find $s_{k,1}\ll s_{k,j}\ll s_{k,j+1}$ such that
$\hat{\bm{\sigma}}(\mathbf{s_{j+1}})\in\Gamma(\bm{\mu})$. Since the energy is
finite and the total gain for the local masses at each step has a lower bound

$\sum\limits_{i=1}^{3}\left(\hat{\sigma}_{i}(\mathbf{s}_{j+1})-\hat{\sigma}_{i}(\mathbf{s}_{j})\right)\geq\min\limits_{i=1,2,3}4\mu_{i}>0,$

the process will stop after finitely many $j$ steps (by a little abuse of
notations) and we have

$\hat{\sigma}_{i}(\mathbf{s}_{j})=\hat{\sigma}_{i}(\bm{\tau})\quad\text{for}\quad
i=1,2,3.$

Hence

$(\sigma_{1},\sigma_{2},\sigma_{3})=(\hat{\sigma}_{1}(\bm{\tau}),\hat{\sigma}_{2}(\bm{\tau}),\hat{\sigma}_{3}(\bm{\tau}))\in\Gamma(\bm{\mu}),$

The statement of at least one component of $\mathbf{u}^{k}$ has fast decay on
$\partial B(0,\tau_{k})$ follows by Lemma 4.1-(b). ∎

Now we are able to provide the proof of Theorem 1.1

###### Proof of Theorem 1.1..

If $m_{0}=0$, we prove it immediately by Lemma 6.1. So we assume that
$m_{0}\neq 0$. We select $S_{1}^{k},\cdots,S_{l_{0}}^{k}$ as the beginning of
this section and group together

$S=\\{0\\}\cup S_{1}^{k}\cup\cdots\cup S_{l_{0}}^{k}.$

In other words, we view $\\{0\\}$ and $S_{i}^{k}$ ($1\leq i\leq l_{0}$) as
points. Let

${\hat{\tau}^{k}(S)=\frac{1}{2}\|S^{k}_{l_{0}+1}\|\gg l^{k}_{l_{0}}(S),}$

where

${\|S^{k}_{l_{0}+1}\|=\min\limits_{x^{k}\in\Sigma_{k}\setminus
S}\mathrm{dist}(0,x^{k})}\quad\text{and}\quad
l^{k}_{l_{0}}(S):=2\max\limits_{1\leq l\leq l_{0}}\max\limits_{x^{k}\in
S_{l}^{k}}\mathrm{dist}(0,x^{k}).$

Similar to the proof of Proposition 5.3, we shall encounter the following two
cases.

Case (1). $\mathbf{u}^{k}$ has fast decay on $\partial B(0,l^{k}_{l_{0}}(S))$.
Similar to the argument of (5.7) in Proposition 5.3, we deduce that the local
mass

$\displaystyle\hat{\sigma}_{i}(\mathbf{l}_{l_{0}}(S))=\hat{\sigma}_{i}({\bm{\tau}})+\sum\limits_{l=1}^{l_{0}}\hat{\sigma}_{i}\left(B\left(\mathbf{x}_{l},\bm{\tau}_{\mathbf{s}_{l}}\right)\right)=\hat{\sigma}_{i}({\bm{\tau}})+4m_{i},\quad
i=1,2,3,$

where $m_{i}\in\mathbb{N}\cup\\{0\\}$, $\mathbf{l}_{l_{0}}(S)$ and
$\left(\mathbf{x}_{\ell},\bm{\tau}_{S_{\ell}}\right)$ stand for the sequence
of $\\{l^{k}_{l_{0}}(S)\\}$ and the pair $\\{(x^{k}_{l},\tau^{k}_{S_{l}})\\}$,
and $\hat{\bm{\sigma}}({\bm{\tau}})\in\Gamma(\bm{\mu})$ by Lemma 6.1. If

(6.1)
$\hat{\sigma}_{i}(\hat{\bm{\tau}}(S))=\hat{\sigma}_{i}(\mathbf{l}_{l_{0}}(S)),\quad
i=1,2,3,$

then we complete the process of grouping $\\{0\\}$ and
$S_{1}^{k},\cdots,S_{l_{0}}^{k}$, and the proof of Theorem 1.1 is complete by
grouping till $S^{k}_{m_{0}}$. If (6.1) does not hold, by Lemma 4.3 we can
find a sequence $s_{k}$ with $l^{k}_{l_{0}}(S)\ll s_{k}\ll\hat{\tau}^{k}(S)$
such that some components $u^{k}_{i}~{}(i\in J)$ have slow decay on $\partial
B(0,s_{k})$ and

$\hat{\sigma}_{i}(\mathbf{l}_{l_{0}}(S))=\hat{\sigma}_{i}(\mathbf{s}),\quad
i=1,2,3,$

where $\mathbf{s}$ stands for the sequence $\\{s_{k}\\}$. Let

$v^{k}_{i}(y)=u^{k}_{i}(s_{k}y)+2\log{s_{k}}.$

Then we can apply the same argument of Step 2 and Step 3 in the proof of
Proposition 5.1 and the result of Proposition 5.2 to get that

$\hat{\sigma}_{i}(\hat{\bm{\tau}}(S))=\hat{\sigma}_{i}(\mathbf{s}_{\ell})\in\Gamma(\bm{\mu})+4\mathbb{Z},\quad
i=1,2,3,$

for some $\mathbf{s}_{\ell}$, where $\mathbf{s}_{\ell}$ stands for the
sequence $\\{s^{k}_{\ell}\\}$ with $l_{l_{0}}^{k}(S)\ll
s^{k}_{\ell}\ll\hat{\tau}^{k}(S)$. Indeed, we have $v^{k}_{i}(y)\to-\infty$ in
$L^{\infty}_{\mathrm{loc}}(\mathbb{R}^{2})$ for $i\in\\{1,2,3\\}\setminus J$
and $v^{k}_{i}(y)$ converges to $v_{i}(y)$ for $i\in J$ in
$C^{2}_{\mathrm{loc}}(\mathbb{R}^{2})$, where $v_{i}(y)$ satisfies (5.2). All
the possibilities have been discussed in Proposition 5.2. Then there is a
sequence $N_{k}^{*}\to+\infty$ as $k\to+\infty$ such that $\mathbf{u}^{k}$ has