††Lina Wu is partially supported by the China Scholarship Council
(No.201806210165).††Lei Zhang is partially supported by a Simons Foundation
Collaboration Grant

# Classification and a priori estimates for the singular prescribing
$Q$-curvature equation on 4-manifold

Mohameden Ahmedou  ,  Lina Wu  and  Lei Zhang  Mohameden Ahmedou
Department of Mathematics
Giessen University
Arndtstrasse 2, 35392, Giessen, Germany<EMAIL_ADDRESS>Lina Wu
Department of Mathematical Sciences
Tsinghua University
No.1 Qinghuayuan, Haidian District, Beijing China, 100084
<EMAIL_ADDRESS>Lei Zhang
Department of Mathematics
University of Florida
1400 Stadium Rd
Gainesville FL 32611<EMAIL_ADDRESS>
###### Abstract.

On $(M,g)$ a compact riemannian $4-$manifold we consider the prescribed
$Q-$curvature equation defined on $M$ with finite singular sources. We first
prove a classification theorem for singular Liouville equations defined on
$\mathbb{R}^{4}$ and perform a concentration compactness analysis. Then we
derive a quantization result for bubbling solutions and establish a priori
estimate under the assumption that some conformal invariant does take some
quantized values. Furthermore we prove a spherical Harnack inequality around
singular sources provided their strength is not an integer. Such an inequality
implies that in this case singular sources are _isolated simple blow up
points_.

###### Key words and phrases:

Fourth order elliptic equations, Blow-up analysis, A priori estimates

Key Words: Blow up analysis, Conical metrics, Moving plane method,
$Q$-curvature, Spherical Harnack inequality.

AMS subject classification: 53C21, 35C60, 58J60.

###### Contents

  1. 1 Introduction
  2. 2 Entire solutions
    1. 2.1 Asymptotic behavior of entire solution
    2. 2.2 Classification of entire solutions for the case $-3/4<\gamma<0$
  3. 3 preliminaries
  4. 4 Blow-up analysis
  5. 5 A Priori Estimate
  6. 6 A spherical Harnack inequality

##  1\. Introduction

The question of existence of conformal metrics of constant or more generally
prescribed curvature on riemannian manifolds is a recurrent problem in
differential geometry and geometric analysis. Indeed a positive or a negative
answer to such a question has far reaching consequences on the geometry and
topology of the underlying manifold. The Poincaré uniformization’s theorem on
closed surfaces, the Yamabe problem on riemannian manifolds of dimension
$n\geq 3$ and the Nirenberg’s problem on standard spheres $\mathbb{S}^{n}$,
just to name a few, are well known and well studied mathematical problems.
A similar question, which goes back to Picard [37, 38] deals with the
existence of conformal metrics of constant or prescribed curvature on surfaces
with conical singularities. After the pioneering work of Picard at the
beginning of the last century, such a problem has been systematically
investigated by Berger [8], McOwen[41, 42] and Troyanov[46, 45]. More recently
Bartolucci-deMarchis-Malchiodi[5] used a Morse theoretical approach to prove
further existence and multiplicity results.
In this paper, a first part of a series of papers, we address the problem of
existence of conformal conical metrics of constant, or more generally
prescribed $Q-$curvature on four dimensional riemannian manifolds. In the
following we will explain in some details the geometric context of such a
problem:
Given $(M,g)$ a compact four-dimensional Riemannian manifold, the
$Q$-curvature and the Paneitz operator are defined respectively, by

(1.1)
$Q_{g}=-\frac{1}{12}\big{(}\Delta_{g}R_{g}-R_{g}^{2}+3|\rm{Ric}_{g}|^{2}\big{)},$
(1.2) $P_{g}\varphi=\Delta_{g}^{2}\varphi+{\rm
div}_{g}\Big{(}\big{(}\frac{2}{3}R_{g}g-2{\rm Ric}_{g}\big{)}\nabla u\Big{)},$

where ${\rm Ric}_{g}$ is the Ricci tensor and $R_{g}$ is the scalar curvature
of $(M,g)$.
Similar to second order equations a natural question is the following
uniformization statement: given a four-dimensional Riemannian manifold
$(M,g)$, is there a metric $\tilde{g}=e^{2u}g$ in the conformal class of $g$
with constant $Q$-curvature?

Under the conformal change of metric above, the Paneitz operator is
conformally covariant:

(1.3) $P_{\tilde{g}}\varphi=e^{-4u}P_{g}\varphi,$

and the $Q$ curvature of $\tilde{g}$ is given by

(1.4) $P_{g}u+2Q_{g}=2Q_{\tilde{g}}e^{4u}.$

From (1.3) and (1.4), the question above is equivalent to the existence of
solution to this fourth order equation:

(1.5) $P_{g}u+2Q_{g}=2\bar{Q}e^{4u},$

where $\bar{Q}$ is a real constant.
Integrating with respect to the volume element ${\rm d}V_{g}$, we can see that

$\kappa_{P}=\int_{M}Q_{g}{\rm d}V_{g}$

is a constant in the conformal class of $g$, here we also point out the Gauss-
Bonnet-Chern formula that links the local curvature to the global topology of
$M$ is:

$\int_{M}\Big{(}Q_{g}+\frac{1}{8}|W_{g}|^{2}\Big{)}{\rm
d}V_{g}=4\pi^{2}\chi_{M},$

where $W_{g}$ denotes the Weyl’s tensor of $(M,g)$ and $\chi_{M}$ is the Euler
characteristic of $M$. From this equality and the aforementioned conformal
covariance property it is not hard to imagine that $P_{g}$ and $Q_{g}$ are
related to a number of studies such as Moser-Trudinger type inequatilities,
log-determinant formulas and the compactification of locally flat manifolds,
see [7, 9, 10, 16, 17, 19]. In many of these studies the kernel of $P_{g}$ is
usually assumed to consist only of constants:

$None$ ${\rm Ker}\,(P_{g})=\\{constants\\}.$

In this paper consider the following prescribed $Q$-curvature equation
involving singular sources

(1.6)
$P_{g}u+2Q_{g}=2he^{4u}-8\pi^{2}\sum_{j=1}^{N}\gamma_{j}\Big{(}\delta_{q_{j}}-\frac{1}{{\rm
vol}_{g}(M)}\Big{)},$

where $h$ is a smooth positive function, $N\in\mathbb{N}$ is a positive
integer, $q_{1},\cdots,q_{N}$ are distinct points on $M$ and where Dirac
measures $\delta_{q_{j}}$ are located, $\gamma_{j}>-1$ are constants.
Solutions to (1.6) have the following geometric interpretation: Setting
$\tilde{g}:=e^{2u}g$ we obtain a metric conformal to $g$ on
$M\setminus\\{q_{1},\cdots,q_{N}\\}$ which has a conical singularity at each
$q_{i}$. One says that $\tilde{g}$ is represented by the divisor
$D:=\sum_{i=1}^{N}\gamma_{i}q_{i}$. See Fang-Ma [23]. Furthermore due to
Gauss-Bonnet-Chern formula for conic four manifolds, see [17], [16],[14], we
have that

$\tilde{\kappa}_{P}\,:=\int_{M}Q_{\tilde{g}}{\rm
d}V_{\tilde{g}}\,=\,\int_{M}Q_{g}{\rm
d}V_{g}\,+\,8\pi^{2}\sum_{i=1}^{N}\gamma_{i}$

is a conformal invariant.
Considerable progress has been made for the regular case of (1.6), that is
$N=0$ in (1.6). Under the assumption that the Kernel of the Paneitz operator
contains only constants, Chang-Yang [19] proved existence for
$\kappa_{P}<8\pi^{2}$, Djadli-Malchiodi [21] settled the case that
$\kappa_{P}\neq 8\pi^{2}n$ for any $n\in\mathbb{N}$. Li-Li-Liu [30] gave a
necessary condition for existence in the case $\kappa_{P}=8\pi^{2}$, Ahmedou-
Ndiaye[2] developed a Morse theory the case of $\kappa_{P}=8\pi^{2}n$ and
Ndiaye [36] combined the celebrated topological argument of Bahri-Coron [3]
with the _critical point theory at infinity_ in [2] to derive some existence
results. We point out that an essential estimate related to the proof in [21]
is a priori estimate when $\kappa_{P}$ is away from $8\pi^{2}\mathbb{N}$
proved by Malchiodi in [33]. Later in [22], Druet and Robert extended such an
a priori estimate to the following more general equation in the same class:

(1.7) $P_{g}u+2b=2he^{4u},$

where $b$ is a smooth function. If $b=Q_{g}$ is the $Q$-curvature of the
conformal metric $e^{2u}g$. More specially, assuming $h_{k}\to h_{0}$,
$h_{k}\geq c_{0}>0$ and $b_{k}\to b_{0}$, then any sequence of solutions
$\\{u_{k}\\}$ of (1.7) with $h=h_{k}$ and $b=b_{k}$ is uniformly bounded under
the condition $\int_{M}b_{0}{\rm d}V_{g}\neq 8\pi^{2}n$, see also Malchiodi
[33].

However, bubbling can occur when $\int_{M}b_{0}{\rm d}V_{g}=8\pi^{2}n$ for
some positive integer $n$. The understanding of this bubbling phnomenon is
vital for the existence problem. The study of the blow-up profile and other
blow-up phenomena for the Paneitz operator and other elliptic equations has
attracted much interest recently and the reference is too numerous to be
mentioned, we just list a few closely related to our article in our humble
opinion: [1, 11, 18, 20, 25, 26, 27, 30, 34, 40, 35, 39, 48, 49, 51].
Particularly, in [51], the third named author and Weinstein have obtained
sharp estimates on the difference near the blow-up points between a bubbling
sequence of solutions to (1.7) with $h=h_{k}$ and $b=b_{k}$ and standard
bubbles, and obtained the vanishing rate under the assumption that $(M,g)$ may
not be locally conformally flat.

When taking the singularities into the account as in (1.6), we consider the
following more general singular equation:

(1.8)
$P_{g}u+2b=2he^{4u}-8\pi^{2}\sum_{j=1}^{N}\gamma_{j}\Big{(}\delta_{q_{j}}-\frac{1}{vol_{g}(M)}\Big{)},$

where $h$ is a positive smooth function on $M$ and $b\in C^{1}(M)$. Before
stating our first main result, we define a critical set $\Gamma$ as follows:

$\Gamma=\Big{\\{}16\pi^{2}n+16\pi^{2}\sum_{j\in J}(1+\gamma_{j}):\
\,n\in\mathbb{N}\cup\\{0\\}\ \,{\rm and}\
\,J\subset\\{1,\cdots,N\\}\Big{\\}}.$

In order to obtain the a priori estimates and existence results, we mainly
study the blow-up phenomena for (1.8). Let us consider the following
equations:

(1.9)
$P_{g}u_{k}+2b_{k}=2h_{k}e^{4u_{k}}-8\pi^{2}\sum_{j=1}^{N}\gamma_{j}\Big{(}\delta_{q_{j}}-\frac{1}{{\rm
vol}_{g}(M)}\Big{)}\quad{\rm in}\ \,M,$

with normalized total integration:

(1.10) $\int_{M}e^{4u_{k}}{\rm d}V_{g}=1.$

Let $\\{u_{k}\\}$ be a sequence of solutions of (1.9). We say $p$ is a blowup
point of $u_{k}$ if there exists a sequence $p_{k}\to p$ such that

$u_{k}(p_{k})+8\pi^{2}\sum_{j=1}^{N}\gamma_{j}G(p_{k},q_{j})\to\infty.$

$u_{k}$ is called a sequence of blowup solutions if it has a blowup point.
Here $G(x,p)$ is the Green’s function of $P_{g}$ defined in (3.3). For blowup
solutions we assume that coefficient functions are regular enough to have
limits:

(1.11) $\parallel b_{k}-b_{0}\parallel_{C^{1}(M)}\to 0,\quad\parallel
h_{k}-h_{0}\parallel_{C^{1}(M)}\to 0,\quad 0<c_{0}<h_{0}<1/c_{0}.$

Without loss of generality, we assume the integration of $h_{k}e^{u_{k}}$ is
$1$:

Our first main result asserts that a priori estimate holds for $u_{k}$, as
long as $2\int_{M}b_{k}$ does not tend to the following critical set:

$\Gamma=\Big{\\{}16\pi^{2}n+16\pi^{2}\sum_{j\in J}(1+\gamma_{j}):\
\,n\in\mathbb{N}\cap\\{0\\}\ \,{\rm and}\
\,J\subset\\{1,\cdots,N\\}\Big{\\}}.$

######  Theorem 1.1 (A Priori Estimate).

Suppose (P) holds, $b$ and $h$ satisfy (1.11). If $\\{u_{k}\\}$ is a sequence
of solutions of (1.9) under restriction (1.10) and $\int_{M}2b_{0}{\rm
d}V_{g}\in\mathbb{R}^{+}\setminus\Gamma$,

$\Big{|}u_{k}(x)+8\pi^{2}\sum_{j=1}^{N}\gamma_{j}G(x,q_{j})\Big{|}\leq
C,\quad\forall x\in M$

for some $C>0$ independent of $k$.

In particular, the a priori estimate holds for the singular prescribing
$Q$-curvature equations. Indeed Theorem 1.1 is an extension of previous
results of Malchiodi [33], Druet-Robert [22] and Fardoun-Regbaoui [24] for the
regular prescribed $Q$-curvature equation. We point out that the argument in
the regular case uses in a crucial way the explicit form of the bubble, while
our argument uses only the asymptotic behavior of the bubble and is based on a
Pohozaev identity for equations under conformal normal coordinates (see [51]).
One indispensable part of the blowup analysis for the Q curvature equation is
the classification of global solutions on $\mathbb{R}^{4}$. For this purpose
we consider the limiting equation used to describe the profile of bubbling
solutions:

(1.12) $\Delta^{2}u(x)=6|x|^{4\gamma}e^{4u(x)},\quad{\rm in}\
\;\mathbb{R}^{4},\qquad|x|^{4\gamma}e^{4u(x)}\in L^{1}(\mathbb{R}^{4}).$

Clearly if $u$ is a solution of (1.12), so is $u_{\lambda}$ defined by

(1.13) $u_{\lambda}(x)=u(\lambda x)+(1+\gamma)\log\lambda$

for any given $\lambda>0$. Our next main result is

######  Theorem 1.2.

Suppose that $u$ is a solution of (1.12) with $\gamma>-1$ and
$|u(x)|=o(|x|^{2})$ at infinity. Then

  * (i)

$\int_{\mathbb{R}^{4}}6|y|^{4\gamma}e^{4u(y)}{\rm d}y=16\pi^{2}(1+\gamma)$.

  * (ii)

(1.14)
$\begin{split}u(x)=&\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y+C_{0}\\\ =&-2(1+\gamma)\log|x|+c+O(\frac{1}{|x|}),\quad|x|>1\end{split}$

for some $C_{0},c\in\mathbb{R}$,

(1.15) $\left\\{\begin{array}[]{ll}-\Delta
u(x)=\frac{4(1+\gamma)}{|x|^{2}}+O(\frac{1}{|x|^{2+\tau}}),\\\
-\frac{\partial}{\partial x_{i}}\Delta
u(x)=-8(1+\gamma)\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{3+\tau}}),\\\
-\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}\Delta
u(x)=O(\frac{1}{|x|^{4}}),\end{array}\right.$

where $\tau=1$ if $\gamma>-3/4$, and $\tau\in(0,1)$ if $-1<\gamma\leq-3/4$.

  * (iii)

Furthermore, if $-3/4<\gamma<0$, $u$ is radially symmetric about the origin
and is unique up to scaling in (1.13).

Theorem 1.2 is a quantization result that also determines the asymptotic
behavior of $u$ at infinity under a sub-quadratic growth condition. If the
$o(|x|^{2})$ assumption is removed, we have

######  Theorem 1.3.

Let $u$ be a solution of (1.12) with $\gamma>-1$. Then after an orthogonal
transformation, $u(x)$ can be represented by

(1.16)
$\begin{split}u(x)=&\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\sum_{j=1}^{4}a_{j}(x_{j}-x_{j}^{0})^{2}+c_{0}\\\
=&-\sum_{j=1}^{4}a_{j}(x_{j}-x_{j}^{0})^{2}-2(1+\gamma)\log|x|+c_{0}+O(|x|^{-\tau})\end{split}$

for some $\tau>0$ and large $|x|$. The function $\Delta u$ satisfies

(1.17) $\Delta
u(x)=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-2\sum_{j=1}^{4}a_{j},$

where $a_{j}\geq 0$, $c_{0}$ are constants and
$x^{0}=(x_{1}^{0},\cdots,x_{4}^{0})\in\mathbb{R}^{4}$.

Moreover, if $-3/4<\gamma<0$, $u$ is symmetric with respect to the hyperplane
$\\{x:x_{i}=0\\}$ when $a_{i}x_{i}^{0}=0$. In particular, under the assumption
$-3/4<\gamma<0$, if $a_{i}x_{i}^{0}=0$ for all $i=1,\cdots,4$, $u$ is radially
symmetric with respect to the origin.

When we were about to finish writing this article we heard that Jevnikar-Sire-
Yang [28] are working on a similar project independently and their results are
to be posted soon.

Here we briefly outline the strategy of the proofs in our paper. For the proof
of the classification result for globally defined singular equation, we follow
the argument of Lin [31] but we need to take care of all the complications
caused by the singular source. In particular we are able to prove the complete
classification for $\gamma\in(-\frac{3}{4},0]$ and a quantization result for
all $\gamma>-1$. For blowup solutions we first use a small-energy lemma (Lemma
4.1) to prove that there are at most finite blowup points. Then we take
advantage of a Pohozaev identity established by Weinstein-Zhang [51] to
describe a precise asymptotic behavior of blowup solutions around a blowup
point. Then the total integration as well as precise asymptotic behavior of
solutions can be further determined. With this information the critical set
$\Gamma$ can be identified and if the total integral of $b$ is not equal to a
number corresponding to the critical set we obtain a priori estimate. Here it
is worth to remark that Theorem 6.2 asserts that if the strength of the
singular source is not an integer, the blowup solutions satisfy a _spherical
Harnack inequality_ around $q$, this is a new phenomenon and we shall explore
in a forthcoming work. The idea of the proof of such a spherical Harnack
inequaliy is as follows: If the spherical harnack inequality is violated,
there should be finite small bubbling circles around the singular source all
tending to the singular source. Around each tiny bubbling disk there is a
Pohozaev identity, and a ”big” circle that includes all these tiny disks also
has a Pohozaev identity. The comparison of these Pohozaev identities implies
that the strength of the singular source has to be an integer.

The organization of the remainder of this paper is as follows. In section 2 we
analyze the globally defined solutions and proved the quantization and the
classification results stated in Theorem 1.2. Then in Section 3, we list some
useful facts about the conformal normal coordinates and Pohozaev identity and
in section 4, we perform a blow-up analysis near singular points. Section 5 is
devoted to a concentration-compactness theorem and a priori estimate for the
singular prescribing $Q$-curvature equation on 4-manifolds while in section 6
we prove our spherical Harnack inequality. Finally we provide is the appendix
an useful estimate of the difference between the geodesic distance and the
Euclidean one for nearby points on the manifold.

##  2\. Entire solutions of fourth order singular Liouville type equations in
$\mathbb{R}^{4}$

In this section, we will follow the argument of Lin [31] to analyze solutions
of (1.12) and prove Theorem 1.2 and Theorem 1.3.

###  2.1. Asymptotic behavior of entire solution

Our argument is progressive in nature and we shall obtain a rough estimate of
$u$ at infinity. For this purpose we set

(2.1)
$v(x):=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y,$

which is obviously a solution of

(2.2) $\Delta^{2}v(x)=-6|x|^{4\gamma}e^{4u(x)},\quad{\rm in}\
\;\mathbb{R}^{4}.$

The asymptotic behavior of $u$ has a large to do with that of $v$, so in the
first lemma we derive a rough upper bound of $v$. For convenience we set

(2.3) $\alpha=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}|y|^{4\gamma}e^{4u}dy.$

######  Lemma 2.1.

Suppose $u$ is a solution of (1.12) and let $\alpha$ be given as in (2.3).
Then

(2.4) $v(x)\leq\alpha\log|x|+C$

for some constant $C$.

###### Proof.

Since the goal is to describe asymptotic behavior it is natural to assume
$|x|>4$. For such $x$, we decompose $\mathbb{R}^{4}=A_{1}\cup A_{2}$, where

$A_{1}=\Big{\\{}y:|y-x|\leq\frac{|x|}{2}\Big{\\}},\quad
A_{2}=\Big{\\{}y:|y-x|\geq\frac{|x|}{2}\Big{\\}}.$

For $y\in A_{1}$, $\log\frac{|x-y|}{|y|}\leq 0$ because
$|y|\geq|x|-|x-y|\geq\frac{|x|}{2}\geq|x-y|$, Thus

$\int_{A_{1}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\leq 0$

and

$v(x)\leq\frac{3}{4\pi^{2}}\int_{A_{2}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\leq 0.$

To evaluate the integral over $A_{2}$, we first make two trivial observations:

$|x-y|\leq|x|+|y|\leq|x||y|,\quad\mbox{if}\quad|y|\geq 2,$
$\log|x-y|\leq\log|x|+C,\quad\mbox{if }\quad|y|\leq 2$

where $|x|>4$ is used. Consequently

$\begin{split}v(x)&\leq\frac{3}{4\pi^{2}}\int_{A_{2}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\\\ &\leq\frac{3}{4\pi^{2}}\Big{\\{}\log|x|\int_{A_{2}\cap\\{|y|\geq
2\\}}|y|^{4\gamma}e^{4u}{\rm d}y+\int_{A_{2}\cap\\{|y|\leq
2\\}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u}{\rm
d}y\Big{\\}}\\\
&\leq\frac{3}{4\pi^{2}}\Big{\\{}\log|x|\int_{A_{2}}|y|^{4\gamma}e^{4u}{\rm
d}y+C\int_{A_{2}\cap\\{|y|\leq 2\\}}|y|^{4\gamma}e^{4u}{\rm d}y\\\
&\qquad\quad-\int_{A_{2}\cap\\{|y|\leq
2\\}}\big{(}\log|y|\big{)}|y|^{4\gamma}e^{4u}{\rm d}y\Big{\\}}\\\
&\leq\frac{3}{4\pi^{2}}\log|x|\int_{\mathbb{R}^{4}}|y|^{4\gamma}e^{4u}{\rm
d}y+C.\end{split}$

Lemma 2.1 is established. ∎

Before proving a lower bound of $v(x)$ we derive an expression of $\Delta
u(x)$ in integral form.

######  Lemma 2.2.

Suppose $u$ is a solution of (1.12). Then there exists a constant $C_{1}\geq
0$ such that

(2.5) $\Delta
u(x)=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-C_{1}.$

###### Proof.

Let $w(x)=u(x)+v(x)$. Then from the equations of $u$ and $v$ in (1.12) and
(2.2), we have $\Delta^{2}w=0$ in $\mathbb{R}^{4}$. Hence, $\Delta w$ is a
harmonic function in $\mathbb{R}^{4}$. By the mean value property of harmonic
functions, we have, for any $x_{0}\in\mathbb{R}^{4}$ and $r>0$,

$\begin{split}\Delta w(x_{0})=\frac{2}{\pi^{2}r^{4}}\int_{B(x_{0},r)}\Delta
w(y){\rm d}y=\frac{2}{\pi^{2}r^{4}}\int_{\partial B(x_{0},r)}\Delta w(y){\rm
d}\sigma,\end{split}$

where $\frac{\pi^{2}}{2}$ is the volume of the unit ball. That is

(2.6) $\frac{r}{4}\Delta w(x_{0})=\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{|y-x_{0}|=r}\frac{\partial w}{\partial r}(y){\rm
d}\sigma,$

where $\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{|y-x_{0}|=r}f(y){\rm
d}\sigma=\frac{1}{2\pi^{2}r^{3}}\int_{|y-x_{0}|=r}f(y){\rm d}\sigma$ denotes
the integral average of $f$ over $\partial B(x_{0},r)$. Then integrating the
identity above along $r$, we get

$\frac{r^{2}}{8}\Delta w(x_{0})=\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{|y-x_{0}|=r}w{\rm d}\sigma-w(x_{0}).$

Therefore, the Jensen inequality implies

$\begin{split}\exp\Big{(}\frac{r^{2}}{2}\Delta w(x_{0})\Big{)}&\leq
e^{-4w(x_{0})}\exp\Big{(}4\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{|y-x_{0}|=r}w{\rm d}\sigma\Big{)}\\\ &\leq
e^{-4w(x_{0})}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{|y-x_{0}|=r}e^{4w}{\rm d}\sigma\end{split}$

From Lemma 2.1, we have $w(x)=u(x)+v(x)\leq u(x)+\alpha\log|x|+C$, and as a
consequence

$\begin{split}&\int_{0}^{\infty}r^{3-4\alpha+4\gamma}\exp\big{(}\frac{\Delta
w(x_{0})}{2}r^{2}\big{)}{\rm
d}r\leq\int_{\mathbb{R}^{4}}|x|^{-4\alpha+4\gamma}e^{-4w(x_{0})}e^{4w(x)}{\rm
d}x\\\
\leq&C\int_{\mathbb{R}^{4}}|x|^{-4\alpha+4\gamma}e^{4u(x)}|x|^{4\alpha}{\rm
d}x=C\int_{\mathbb{R}^{4}}|x|^{4\alpha}e^{4u}{\rm d}x<+\infty,\end{split}$

which means

$r^{3-4\alpha+4\gamma}\exp\Big{(}\frac{\Delta w(x_{0})}{2}r^{2}\Big{)}\in
L^{1}\big{(}[1,+\infty)\big{)}.$

From here we see $\Delta w(x_{0})\leq 0$ for all $x_{0}\in\mathbb{R}^{4}$.
Liouville’s Theorem implies that there exists some constant $C_{1}\geq 0$ such
that $\Delta w(x)\equiv-C_{1}$ in $\mathbb{R}^{4}$. Lemma 2.2 is established
based on this and

$\Delta
v(x)=\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(x)}{\rm
d}y.$

∎

With the help of the representation for $\Delta u$, we can estimate $v(x)$
from below for $|x|$ large. We will use following result in Lemma 2.3 of [31].

Let $h(x)$ be the solution of

$\left\\{\begin{array}[]{lcl}\Delta^{2}h(x)=f(x),&&{\rm in}\ \;\Omega,\\\
\Delta h(x)=h(x)=0,&&{\rm on}\ \;\partial\Omega,\end{array}\right.$

where $\Omega$ is a bounded domain of $\mathbb{R}^{4}$.

######  Lemma 1.

[31]

Suppose $f\in L^{1}(\bar{\Omega})$.Then for any $\delta\in(0,32\pi^{2})$,
there exists a constant $C_{\delta}>0$ such that

$\int_{\Omega}\exp\Big{(}\frac{\delta|h|}{\parallel
f\parallel_{L^{1}}}\Big{)}{\rm d}x\leq C_{\delta}(diam\;\Omega)^{4},$

where $diam\;\Omega$ denotes the diameter of $\Omega$.

######  Lemma 2.3.

Let $u$ be a solution of (1.12) and $v$ be in (2.1). Then for given
$\varepsilon>0$, there exists a constant $R=R(\varepsilon)$ only depending on
$\varepsilon$ such that

(2.7) $v(x)\geq(\alpha-\varepsilon)\log|x|,\quad|x|>R(\epsilon).$

###### Proof.

We first prove a claim slightly weaker than (2.7): for any $\varepsilon>0$,
there exists $R=R(\varepsilon)>0$ such that

(2.8)
$v(x)\geq(\alpha-\frac{\varepsilon}{2})\log|x|+\frac{3}{4\pi^{2}}\int_{B(x,1)}(\log|x-y|)|y|^{4\gamma}e^{4u(y)}{\rm
d}y.$

To prove (2.8) we consider $\mathbb{R}^{4}$ as a disjoint union of three sets:
$\mathbb{R}^{4}=A_{1}\cup A_{2}\cup A_{3}$, where

$\displaystyle A_{1}$ $\displaystyle=\\{y:|y|<R_{0}\\},$ $\displaystyle A_{2}$
$\displaystyle=\\{y:|x-y|\leq|x|/2,|y|\geq R_{0}\\},$ $\displaystyle A_{3}$
$\displaystyle=\\{y:|x-y|\geq|x|/2,|y|\geq R_{0}\\}.$

Then we choose $R_{0}=R_{0}(\varepsilon)$ sufficiently large so that

$\begin{split}&\frac{3}{4\pi^{2}}\int_{A_{1}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\alpha\log|x|\\\
=&\frac{3}{4\pi^{2}}\log|x|\int_{A_{1}}\frac{\log|x-y|-\log|x|-\log|y|}{\log|x|}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\frac{\varepsilon}{8}\log|x|\\\
\geq&-\frac{\varepsilon}{4}\log|x|\end{split}$

for large $|x|$. Thus we have

(2.9)
$\frac{3}{4\pi^{2}}\int_{A_{1}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\geq(\alpha-\frac{\varepsilon}{4})\log|x|.$

For $x\in A_{2}$ and $|x|$ large, we have
$\frac{|x|}{2}\leq|x|\leq\frac{3}{2}|x|$. Then

(2.10)
$\begin{split}&\int_{A_{2}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\\\ =&\int_{A_{2}}(\log|x-y|)|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\int_{A_{2}}(\log|y|)|y|^{4\gamma}e^{4u(y)}{\rm d}y\\\
\geq&\int_{B(x,1)}(\log|x-y|)|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\log(2|x|)\int_{A_{2}}|y|^{4\gamma}e^{4u(y)}{\rm d}y.\end{split}$

For $y\in A_{3}$, we use two trivial inequalities:
$|x-y|\geq\frac{|x|}{2}\geq\frac{|y|}{4}$ if $|y|\leq 2|x|$ and
$|x-y|\geq|y|-|x|\geq\frac{|y|}{2}$ if $|y|\geq 2|x|$. Clearly in both cases,
we have

$\frac{|x-y|}{|y|}\geq\frac{1}{4},\quad y\in A_{3}.$

Therefore,

(2.11)
$\frac{3}{4\pi^{2}}\int_{A_{3}}\log\big{(}\frac{|x-y|}{|y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y\geq\log\frac{1}{4}\int_{A_{3}}|y|^{4\gamma}e^{4u(y)}{\rm d}y.$

From (2.9), (2.10), (2.11) and $|y|^{4\gamma}e^{4u(y)}\in
L^{1}(\mathbb{R}^{4})$, we obtain (2.8).

Next, we show that

(2.12) $\int_{B(x,1)}(\log|x-y|)|y|^{4\gamma}e^{4u(y)}{\rm d}y\geq-C$

for some positive constant $C$. For this purpose we set

(2.13) $\tilde{u}(x)=u(x)-\gamma\log|x|$

Then $\tilde{u}$ satisfies

(2.14)
$\left\\{\begin{array}[]{lcl}\Delta^{2}\tilde{u}(x)=6e^{4\tilde{u}(x)}-8\pi^{2}\gamma\delta_{0},&&{\rm
in}\ \;\mathbb{R}^{4},\\\ e^{4\tilde{u}}\in
L^{1}(\mathbb{R}^{4}).&&\end{array}\right.$

Let $0<\varepsilon_{0}<\pi^{2}$ and $R_{0}=R_{0}(\varepsilon_{0})$ be
sufficiently large such that

(2.15) $6\int_{B(x,4)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y=6\int_{B(x,4)}e^{4\tilde{u}(y)}{\rm d}y\leq\varepsilon_{0},\quad{\rm for}\
\,|x|\geq R_{0},$

then we let $h$ be the solution of

$\left\\{\begin{array}[]{lcl}\Delta^{2}h(x)=6e^{4\tilde{u}(y)},&&{\rm in}\
\;B(x,4),\\\ h(x)=\Delta h(x)=0,&&{\rm on}\ \;\partial
B(x,4).\end{array}\right.$

From Lemma 1, we can see that for $\varepsilon_{0}>0$ small,

(2.16) $\int_{B(x,4)}e^{24|h|}{\rm d}y\leq c_{1}$

for some constant $c_{1}$ independent of $x$.

Next we set $q(y)=\tilde{u}(y)-h(y)$ for $y\in B(x,4)$, which clearly
satisfies

$\left\\{\begin{array}[]{lcl}\Delta^{2}q(y)=0,&&{\rm in}\ \;B(x,4),\\\
q(y)=\tilde{u}(y),\quad\Delta q(y)=\Delta\tilde{u}(y),&&{\rm on}\ \;\partial
B(x,4).\end{array}\right.$

Let $\tilde{q}(y)=-\Delta q(y)$. Then by Lemma 2.2, we see that for $|x|$
large enough and $y\in\partial B(x,4)$

$\begin{split}\tilde{q}(y)=&-\Delta\tilde{u}(y)=-\Delta
u(y)+\gamma\Delta(\log|y|)=-\Delta u(y)-\frac{2\gamma}{|y|^{2}}.\end{split}$

By setting

$\hat{q}(y)=\tilde{q}(y)+\frac{2\gamma}{|y|^{2}}=-\Delta
u(y)=\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|z-y|^{2}}|z|^{4\gamma}e^{4u(z)}{\rm
d}z+C_{1},$

we obviously have $\hat{q}(y)>0$ on $\partial B(x,4)$, and hence
$\tilde{q}(y)>-2\gamma/|y|^{2}$ on $\partial B(x,4)$. Observing that
$1/|y|^{2}$ is the fundamental solution of $\Delta$. In other words,
$\hat{q}(y)$ is harmonic in $B(x,4)$ with positive boundary value on $\partial
B(x,4)$. The maximum principle implies $\hat{q}>0$ in $B(x,4)$. Thus, by the
Harnack inequality and mean value property of harmonic functions, we have

(2.17) $\begin{split}\tilde{q}(y)&=\hat{q}(y)-\frac{2\gamma}{|y|^{2}}\\\ &\leq
c_{2}\hat{q}(x)-\frac{2\gamma}{|y|^{2}}=-\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta u{\rm
d}\sigma-\frac{2\gamma}{|y|^{2}}=-\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta\tilde{u}{\rm d}\sigma,\quad
y\in\overline{B(x,2)},\end{split}$

with a constant $c_{2}$.

Integrating (2.14) along $r$, we have

$\begin{split}&\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta\tilde{u}{\rm
d}\sigma-\Delta\tilde{u}(x)=\int_{0}^{r}\frac{3}{\pi^{2}s^{3}}\int_{0}^{s}\int_{\partial
B(x,t)}e^{4\tilde{u}}{\rm d}\sigma{\rm d}t{\rm d}s\\\
=&\int_{0}^{r}\frac{3}{\pi^{2}s^{3}}\int_{0}^{s}t^{3}\int_{\partial
B(x,1)}e^{4\tilde{u}}{\rm d}\sigma{\rm d}t{\rm d}s\\\
=&\int_{0}^{r}t^{3}\int_{\partial
B(x,1)}e^{4\tilde{u}}\Big{(}\frac{3}{2\pi^{2}t^{2}}-\frac{3}{2\pi^{2}r^{2}}\Big{)}{\rm
d}\sigma{\rm d}t.\end{split}$

That is

(2.18) $\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta\tilde{u}{\rm
d}\sigma-\Delta\tilde{u}(x)=\frac{3}{2\pi^{2}}\int_{B(x,r)}\Big{(}\frac{1}{|x-y|^{2}}-\frac{1}{r^{2}}\Big{)}e^{4\tilde{u}}{\rm
d}y.$

Next by Lemma 2.2 and (2.18), we can see

$\begin{split}&-\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta\tilde{u}{\rm
d}\sigma=-\Delta\tilde{u}(x)-\frac{3}{2\pi^{2}}\int_{B(x,r)}\frac{1}{|x-y|^{2}}e^{4\tilde{u}}{\rm
d}y+\frac{3}{2\pi^{2}r^{2}}\int_{B(x,r)}e^{4\tilde{u}}{\rm d}y\\\ =&-\Delta
u(x)-\frac{3}{2\pi^{2}}\int_{B(x,r)}\frac{1}{|x-y|^{2}}e^{4\tilde{u}}{\rm
d}y+\frac{3}{2\pi^{2}r^{2}}\int_{B(x,r)}e^{4\tilde{u}}{\rm
d}y-\frac{2\gamma}{|x|^{2}}\\\ =&\frac{3}{2\pi^{2}r^{2}}\int_{|x-y|\geq
r}\frac{1}{|x-y|^{2}}e^{4\tilde{u}}{\rm
d}y+\frac{3}{2\pi^{2}r^{2}}\int_{B(x,r)}e^{4\tilde{u}}{\rm
d}y-\frac{2\gamma}{|x|^{2}}+C_{1}.\end{split}$

In particular, for $r=4$ and $|x|$ large,

(2.19) $-\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(x,4)}\Delta\tilde{u}{\rm d}\sigma\leq
c_{3}.$

Hence, from (2.17), we get

(2.20) $\tilde{q}(y)\leq c_{4},\,\quad y\in\overline{B(x,2)},$

and immediately

$|\tilde{q}(y)|\leq c_{5},\,\quad y\in\overline{B(x,2)}.$

Since $q$ satisfies

$\left\\{\begin{array}[]{lcl}\Delta q(y)=-\tilde{q}(y),&&{\rm in}\
\;B(x,4),\\\ q(y)=\tilde{u}(y),&&{\rm on}\ \;\partial
B(x,4),\end{array}\right.$

by estimates for linear elliptic equations, we have for any $p>1$ and
$\sigma>2$,

(2.21) $\sup_{B(x,1)}q\leq c\big{(}\parallel
q^{+}\parallel_{L^{p}(B(x,2))}+\parallel\tilde{q}\parallel_{L^{\sigma}(B(x,2))}\big{)},$

where $c=c(p,\sigma)$.

On the other hand, we observe that $q^{+}(y)\leq\tilde{u}^{+}(y)+|h(y)|$ for
$y\in B(x,4)$. Then by (2.16), we get

$\begin{split}\int_{B(x,2)}(q^{+})^{p}\leq c_{6}\int_{B(x,2)}e^{2q^{+}}\leq
c_{5}\Big{(}\int_{B(x,2)}e^{4\tilde{u}^{+}}\Big{)}^{\frac{1}{2}}\Big{(}\int_{B(x,2)}e^{4|h|}\Big{)}^{\frac{1}{2}}\leq
c_{7}\Big{(}\int_{B(x,2)}e^{4\tilde{u}^{+}}\Big{)}^{\frac{1}{2}}\end{split}$

Since $e^{4\tilde{u}^{+}}\leq 1+e^{4\tilde{u}}$, we have $\parallel
q^{+}\parallel_{L^{p}(B(x,2))}\leq c_{7}$, which together with (2.20) and
(2.21) implies

(2.22) $\sup_{B(x,1)}q\leq c_{8}.$

In view of $\tilde{u}=h+q$, we now obtain

(2.23) $\tilde{u}(y)\leq h(y)+q(y)\leq c_{8}+|h(y)|,\,\quad
y\in\overline{B(x,2)}.$

Therefore,

(2.24) $\int_{B(x,1)}e^{24\tilde{u}}\leq c_{9}\int_{B(x,1)}e^{24|h|}{\rm
d}y\leq c_{10},$

Then

$\Big{|}\int_{B(x,1)}(\log|x-y|)e^{4\tilde{u}}{\rm
d}y\Big{|}\leq\Big{(}\int_{B(x,1)}(\log|x-y|)^{2}{\rm
d}y\Big{)}^{\frac{1}{2}}\Big{(}\int_{B(x,1)}e^{8\tilde{u}(y)}{\rm
d}y\Big{)}^{\frac{1}{2}}\leq c_{11},$

which means

(2.25) $\Big{|}\int_{B(x,1)}(\log|x-y|)|y|^{4\gamma}e^{4u(y)}{\rm
d}y\Big{|}\leq c_{11},$

where $c_{11}$ is a constant independent of $x$ ($|x|$ large). As a
consequence, (2.8) and (2.25) lead to

$v(x)\geq(\alpha-\frac{\varepsilon}{2})\log|x|-c_{11}\geq(\alpha-\varepsilon)\log|x|$

for $|x|$ large, which is (2.7).

To estimate $\Delta v$, we observe that

$\Delta
v(x)=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}e^{4\tilde{u}(y)}{\rm
d}y.$

We decompose $\mathbb{R}^{4}=B_{1}\cup B_{2}\cup B_{3}$ for $|x|$ large, where

$\displaystyle B_{1}$ $\displaystyle=\\{y:|x-y|\geq|x|/2\\},$ $\displaystyle
B_{2}$ $\displaystyle=\\{y:1\leq|x-y|\leq|x|/2,|y|\geq R_{0}\\},$
$\displaystyle B_{3}$ $\displaystyle=\\{y:|x-y|\leq 1\\}.$

Then

$\int_{B_{1}}\frac{1}{|x-y|^{2}}e^{4\tilde{u}(y)}{\rm
d}y\leq\frac{C}{|x|^{2}}\to 0,\quad{\rm as}\ \,|x|\to\infty,$
$\int_{B_{2}}\frac{1}{|x-y|^{2}}e^{4\tilde{u}(y)}{\rm
d}y\leq\int_{B_{2}}e^{4\tilde{u}(y)}{\rm d}y\to 0,\quad{\rm as}\
\,|x|\to\infty,$

and

$\begin{split}&\int_{B_{3}}\frac{1}{|x-y|^{2}}e^{4\tilde{u}(y)}{\rm d}y\\\
\leq&\Big{(}\int_{|x-y|\leq 1}\big{(}\frac{1}{|x-y|}\big{)}^{\frac{8}{5}}{\rm
d}y\Big{)}^{\frac{5}{8}}\Big{(}\int_{|x-y|\leq 1}e^{4\tilde{u}}{\rm
d}y\Big{)}^{\frac{1}{4}}(\int_{|x-y|\leq 1}e^{24\tilde{u}}{\rm
d}y\Big{)}^{\frac{1}{8}}\\\ \leq&c\Big{(}\int_{|x-y|\leq 1}e^{4\tilde{u}}{\rm
d}y\Big{)}^{\frac{1}{4}}\to 0,\quad{\rm as}\ \,|x|\to\infty,\end{split}$

where we have used (2.24). Lemma 2.3 is established.

∎

With the estimates of $v(x)$ near infinity and the expression of $\Delta u$,
we can show the expression of $u$ in integral form under the condition
$|u(x)|=o(|x|^{2})$ at $\infty$:

######  Lemma 2.4.

Suppose $|u(x)|=o(|x|^{2})$ at $\infty$. Then there exists some constant
$C_{0}$ such that

(2.26)
$u(x)=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y+C_{0}.$

Furthermore, for any given $\varepsilon>0$,

(2.27) $-\alpha\log|x|-C\leq u(x)\leq(-\alpha+\varepsilon)\log|x|,\quad|x|\geq
R(\varepsilon),$

where $R(\varepsilon)$ comes from Lemma 2.3.

###### Proof.

We start from the integral expression of $\Delta u$ in Lemma 2.2:

$\Delta
u(x)=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-C_{1},\quad C_{1}\geq 0,$

and we first prove $C_{1}=0$ by contradiction. If $C_{1}>0$ we have

$\Delta u(x)\leq-C_{1}<0,\quad|x|\geq R_{0},$

where $R_{0}$ is large. Let

(2.28) $h(y)=u(y)+\varepsilon|y|^{2}+A\big{(}|y|^{-2}-R_{0}^{-2}\big{)}.$

Under the assumption of $|u(y)|=o(|y|^{2})$ at $\infty$, we have
$\lim\limits_{|y|\to+\infty}h(y)=+\infty$ for any fixed $\varepsilon>0$ and
$A>0$. So we choose $\varepsilon>0$ small to make

(2.29) $\Delta h(y)=\Delta u(y)+8\varepsilon<-\frac{C_{1}}{2}<0,\quad|y|\geq
R_{0},$

and $A$ sufficiently large such that $\inf\limits_{|y|\geq R_{0}}h(y)$ is
achieved by some $y_{0}\in\mathbb{R}^{4}$ and $|y_{0}|>R_{0}$. Clearly we have
obtained a contradiction to the maximum principle. Hence, $C_{1}=0$ and $u+v$
is harmonic in $\mathbb{R}^{4}$.

From Lemma 2.1 and Lemma 2.3, we know for $|x|$ large enough

$(\alpha-\varepsilon)\log|x|\leq v(x)\leq\alpha\log|x|+C,$

which together with the assumption $|u(x)|=o(|x|^{2})$ at $\infty$ indicates

(2.30) $|u(x)+v(x)|=o(|x|^{2})\quad{\rm at}\ \,\infty.$

Since $u+v$ is harmonic, by the gradient estimates for harmonic functions, we
have

$u(x)+v(x)=\sum_{j=1}^{4}a_{j}x_{j}+a_{0}$

with some constants $a_{j}\in\mathbb{R},\,j=0,\cdots,4$. Therefore, for $|x|$
large enough, we get

$e^{4u(x)}=e^{a_{0}}e^{-4v(x)}e^{\sum_{j=1}^{4}a_{j}x_{j}}\geq
C|x|^{-4\alpha}e^{\sum_{j=1}^{4}a_{j}x_{j}}.$

Since $|y|^{4\gamma}e^{4u(x)}\in L^{1}(\mathbb{R}^{4})$, we have $a_{j}=0$ for
$1\leq j\leq 4$. Therefore,

$u(x)=-v(x)+a_{0}=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y+C_{0},$

and then

$-\alpha\log|x|-C\leq u(x)\leq(-\alpha+\varepsilon)\log|x|,$

for $|x|$ large. Lemma 2.4 is established. ∎

Next we need a Pohozaev identity for $u$ of

(2.31) $\Delta u=Q(x)e^{4u}\ \,{\rm in}\ \,\mathbb{R}^{4}.$

######  Lemma 2.5.

Suppose $u$ is an entire smooth solution of (2.31). Then for any bounded
domain, we have

(2.32) $\begin{split}&\int_{\Omega}\big{(}Q+\frac{1}{4}<x,\nabla
Q>e^{4u}\big{)}{\rm d}x\\\
=&\frac{1}{4}\int_{\partial\Omega}<x,\nu>Q(x)e^{4u}{\rm
d}\sigma+\int_{\partial\Omega}\Big{\\{}\frac{1}{2}|\Delta
u|^{2}<x,\nu>-2\frac{\partial u}{\partial\nu}\Delta u\\\ &\quad-<x,\nabla
u>\frac{\partial\Delta u}{\partial\nu}-<x,\nabla\Delta u>\frac{\partial
u}{\partial\nu}+<x,\nu><\nabla u,\nabla\Delta u>\Big{\\}}{\rm
d}\sigma.\end{split}$

In particular, taking $\Omega=B_{R}$, we have

(2.33) $\begin{split}&\int_{B_{R}}Q(x)e^{4u}{\rm
d}x+\frac{1}{4}\int_{B_{R}}<x,\nabla Q>e^{4u}{\rm d}x\\\
=&\frac{1}{4}\int_{\partial B_{R}}|x|Qe^{4u}{\rm
d}\sigma+\frac{1}{2}\int_{\partial B_{R}}|x||\Delta u|^{2}{\rm
d}\sigma-2\int_{\partial B_{R}}\frac{\partial u}{\partial r}\Delta u{\rm
d}\sigma-\int_{\partial B_{R}}|x|\frac{\partial u}{\partial
r}\frac{\partial\Delta u}{\partial r}{\rm d}\sigma.\end{split}$

###### Proof.

Multiplying (2.31) by $x\cdot\nabla u$, we have

(2.34) $\int_{\Omega}\nabla^{2}(x\cdot\nabla
u)=\int_{\Omega}Q(x)e^{4u}(x\cdot\nabla u).$

After integrating by parts and direct computation, we get

$\begin{split}({\rm RHS})\;{\rm
of}\;(\ref{IBP})=\frac{1}{4}\int_{\partial\Omega}<x,\nu>Q(x)e^{4u}{\rm
d}\sigma-\int_{\Omega}\big{(}Q+\frac{1}{4}<x,\nabla Q>e^{4u}\big{)}{\rm
d}x,\end{split}$

and

$\begin{split}({\rm LHS})\;{\rm
of}\;(\ref{IBP})=&-\frac{1}{2}\int_{\partial\Omega}|\Delta
u|^{2}<x,\nu>+2\int_{\partial\Omega}\frac{\partial u}{\partial\nu}\Delta
u+\int_{\partial\Omega}<x,\nabla u>\frac{\partial\Delta u}{\partial\nu}\\\
&+\int_{\partial\Omega}<x,\nabla\Delta u>\frac{\partial
u}{\partial\nu}-\int_{\partial\Omega}<x,\nu><\nabla u,\nabla\Delta
u>.\end{split}$

Thus we establish (2.32). Taking $\Omega=B_{R}$, we immediately obtain (2.33)
from (2.32).

∎

From the Pohozaev identity we shall determine the exact value of $\alpha$.

######  Lemma 2.6.

Let $u$ be a solution of (1.12). Assume $|u(x)|=o(|x|^{2})$ at $\infty$, then
$\alpha=2(1+\gamma)$.

###### Proof.

Taking $Q(x)=6|x|^{4\gamma}$ in (2.33), we have

(2.35) $\begin{split}&6(1+\gamma)\int_{B_{R}}|x|^{4\gamma}e^{4u}{\rm d}x\\\
=&\frac{1}{4}\int_{\partial B_{R}}6r^{4\gamma+1}e^{4u}{\rm
d}\sigma+\frac{1}{2}\int_{\partial B_{R}}r|\Delta u|^{2}-2\int_{\partial
B_{R}}\frac{\partial u}{\partial r}\Delta u-\int_{\partial
B_{R}}r\frac{\partial u}{\partial r}\frac{\partial\Delta u}{\partial
r}.\end{split}$

In view of Lemma 2.4, we have obtained

$u(x)=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y+C_{0},$

and $e^{u(y)}\geq|y|^{-4\alpha}$ for $|y|$ large enough. Then
$|y|^{4\gamma}e^{4u(y)}\geq|y|^{4(\gamma-\alpha)}$. Hence
$|y|^{4\gamma}e^{4u(y)}\in L^{1}(\mathbb{R}^{4})$ implies $\alpha>1+\gamma$.
On the other hand, by the representation of $u$ and direct calculations, there
hold

$\frac{\partial u}{\partial
r}(x)=-\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\frac{x\cdot(x-y)}{|x||x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y,$ $\Delta
u(x)=-\frac{3}{2\pi^{2}}\int_{\mathbb{R}^{4}}\frac{1}{|x-y|^{2}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y,$

and

$\frac{\partial}{\partial r}\Delta
u(x)=\frac{3}{\pi^{2}}\int_{\mathbb{R}^{4}}\frac{x\cdot(x-y)}{|x||x-y|^{4}}|y|^{4\gamma}e^{4u(y)}{\rm
d}y,$

Recall the definication of $u$, Lemma 2.4 and $\alpha>1+\gamma$, then we have

(2.36) $\begin{split}&\lim_{r\to+\infty}\frac{\partial u}{\partial r}=0,\\\
&\lim_{r\to+\infty}r\frac{\partial u}{\partial r}=-\alpha,\\\
&\lim_{r\to+\infty}r^{2}\Delta u=-2\alpha,\\\
&\lim_{r\to+\infty}r^{3}\frac{\partial}{\partial r}\Delta
u=4\alpha,\end{split}$

where $r=|x|$. Therefore, applying the Pohozaev identity (2.35) and (2.36), we
have

$8\pi^{2}(1+\gamma)\alpha=4\pi^{2}\alpha^{2},$

which leads to $\alpha=2(1+\gamma)$.

∎

Now we can determine the asymptotic behavior of $u$ at infinity using the
exact value of $\alpha$.

######  Lemma 2.7.

Let $u$ be a solution of (1.12) and suppose $|u(x)|=o(|x|^{2})$ at $\infty$.
Then there exist constants $c$ and $a_{j}$ $(j=1,\cdots,4)$ such that for
$|x|$ large enough,

(2.37) $u(x)=-2(1+\gamma)\log|x|+c+O(\frac{1}{|x|}),$

the derivatives of $u$ at infinity are determined in two cases: if
$\gamma>-3/4$,

(2.38) $\left\\{\begin{array}[]{ll}-\Delta
u(x)=\frac{4(1+\gamma)}{|x|^{2}}+\sum_{j=1}^{4}\frac{a_{j}x_{j}}{|x|^{4}}+O(\frac{1}{|x|^{4}}),\\\
-\frac{\partial}{\partial x_{i}}\Delta
u(x)=-8(1+\gamma)\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{4}}),\\\
-\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}\Delta
u(x)=O(\frac{1}{|x|^{4}}).\end{array}\right.$

if $-1<\gamma\leq-3/4$, there exists a constant $\tau\in(0,1)$ such that

(2.39) $\left\\{\begin{array}[]{ll}-\Delta
u(x)=\frac{4(1+\gamma)}{|x|^{2}}+O(\frac{1}{|x|^{2+\tau}}),\\\
-\frac{\partial}{\partial x_{i}}\Delta
u(x)=-8(1+\gamma)\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{3+\tau}}),\\\
-\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}\Delta
u(x)=O(\frac{1}{|x|^{4}}).\end{array}\right.$

###### Proof.

Let $w(x)=u(\frac{x}{|x|^{2}})-2(1+\gamma)\log|x|$, then from the equation of
$u$ and the assumption, we see $w$ satisfies

(2.40)
$\left\\{\begin{array}[]{ll}\Delta^{2}w(x)=6|x|^{4\gamma}e^{4w(x)},\quad{\rm
in}\ \,\mathbb{R}^{4}\setminus\\{0\\},\\\
|w(x)|=o(\log\frac{1}{|x|}),\quad|\Delta w(x)|=o(\frac{1}{|x|^{2}}),\quad{\rm
as}\ \,|x|\to 0.\end{array}\right.$

Let $h(x)$ be a solution of

(2.41)
$\left\\{\begin{array}[]{lcl}\Delta^{2}h(x)=6|x|^{4\gamma}e^{4w(x)},&&{\rm
in}\ \,B_{1}\setminus\\{0\\},\\\ h(x)=w(x),\quad\Delta h(x)=\Delta w(x),&&{\rm
on}\ \,\partial B_{1}.\end{array}\right.$

and $q(x)=w(x)-h(x)$. Then $q(x)$ satisfies

(2.42) $\left\\{\begin{array}[]{lcl}\Delta^{2}q(x)=0,&&{\rm in}\ \,B_{1},\\\
q(x)=\Delta q=0,&&{\rm on}\ \,\partial B_{1},\\\
|q(x)|=o(\log\frac{1}{|x|}),\quad|\Delta q(x)|=o(\frac{1}{|x|^{2}}),&&{\rm
as}\ \,|x|\to 0.\end{array}\right.$

First for $\Delta q$, since its growth near the singular source is weaker than
fundamental solutions, the singularity is removable, thus $\Delta q=0$ in
$B_{1}$. By exactly the same reason we further conclude that $q\equiv 0$ in
$B_{1}$. That means $w(x)=h(x)\in C^{0,\tau}(\bar{B}_{1})$. It suffices to
consider the regularity of $h$ in $B_{1}$.

Note that

$\begin{split}|x|^{4\gamma}e^{4w(x)}&=|x|^{4\gamma}e^{4u(\frac{x}{|x|^{2}})-8(1+\gamma)\log|x|}\\\
&\sim|x|^{4\gamma}|x|^{-8(1+\gamma)}|x|^{4\alpha}\sim|x|^{4\gamma},\quad{\rm
near}\ \,0,\end{split}$

where we used Lemma 2.4 and Lemma 2.6.

Case 1: $\gamma>-\frac{3}{4}$. In this case, there exists some $p>\frac{4}{3}$
such that $|x|^{4\gamma}e^{4w(x)}\in L^{p}(\bar{B}_{1})$. By the standard
elliptic estimates and general Sobolev inequality, we have $h\in
C^{1,\tau}(B_{1})$ for some $0<\tau<1$, which is valid for $w$ as well. Then
it is not difficult to obtain (2.37), and (2.38) follows immediately.

Case 2: $-1<\gamma\leq-\frac{3}{4}$, we can still find some $p>1$ (indeed
$1<p\leq\frac{4}{3}$) such that $|x|^{4\gamma}e^{4w(x)}\in
L^{p}(\bar{B}_{1})$. Then by the regularity theorems of linear elliptic
equations, $h(x)\in C^{0,\tau}(\bar{B}_{1})$ for some $0<\tau<1$. In this
case, we get

(2.43) $u(x)=-2(1+\gamma)\log|x|+c+O(\frac{1}{|x|^{\tau}}),$

and

(2.44) $\left\\{\begin{array}[]{ll}-\Delta
u(x)=\frac{4(1+\gamma)}{|x|^{2}}+O(\frac{1}{|x|^{2+\tau}}),\\\
-\frac{\partial}{\partial x_{i}}\Delta
u(x)=-8(1+\gamma)\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{3+\tau}}),\\\
-\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}\Delta
u(x)=O(\frac{1}{|x|^{4}}).\end{array}\right.$

∎

###  2.2. Classification of entire solutions for the case $-3/4<\gamma<0$

In this subsection, we will show the solution of (2.14) or (1.12) has radial
symmetry and uniqueness property up to scaling if $-3/4<\gamma<0$. Similar to
[31], we will use the method of moving planes.

Suppose that $u$ is a smooth entire solution of (1.12) with
$|u(x)|=o(|x|^{2})$ at $\infty$. Recall $-\Delta u>0$ in $\mathbb{R}^{4}$ and
(2.37)$\sim$(2.38), so we will apply the method of moving planes to $-\Delta
u$. Let $v(x)=-\Delta u(x)$. Then by Lemma 2.7, $v(x)$ has the following
harmonic asymptotic expansion at $\infty$:

(2.45)
$\left\\{\begin{array}[]{ll}v(x)=\frac{4(1+\gamma)}{|x|^{2}}+O(\frac{1}{|x|^{3}}),\\\
v_{x_{i}}(x)=-8(1+\gamma)\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{4}}),\\\
v_{x_{i}x_{j}}(x)=O(\frac{1}{|x|^{4}}).\end{array}\right.$

First, we state some conventional notations for moving planes. For any
$\lambda\in\mathbb{R}$, let
$T_{\lambda}=\\{x\in\mathbb{R}^{4}:x_{1}=\lambda\\}$,
$\Sigma_{\lambda}=\\{x:x_{1}>\lambda\\}$ and $x^{\lambda}=(2\lambda-
x_{1},x_{2},x_{3},x_{4})$ be the reflection point of $x$ with respect to
$T_{\lambda}$. In addition, when applying the method of moving plane, we need
two lemmas in [15], also can be seen in [31].

######  Lemma 2.

[15]

Let $v$ be a positive function defined in a neighborhood at $\infty$
satisfying the asymptotic expansion (2.45). Then there exists
$\bar{\lambda}_{0}<0$ and $R>0$ such that there holds

$v(x)>v(x^{\lambda})$

for $\lambda\leq\bar{\lambda}_{0}$, $x\in\Sigma_{\lambda}$ and $|x|\geq R$.

######  Lemma 3.

[15]

Suppose $v$ satisfies the assumption of Lemma 2, and $v(x)>v(x^{\lambda}_{0})$
for $x\in\Sigma_{\lambda_{0}}$. Assume $v(x)-v(x^{\lambda_{0}})$ is
superharmonic in $\Sigma_{\lambda_{0}}$. Then there exist $\varepsilon>0$ and
$S>0$ such that the followings hold.

  * (i)

$v_{x_{1}}>0$ in $\\{x:|x_{1}-\lambda_{0}|<\varepsilon,|x|>S\\}$.

  * (ii)

$v(x)>v(x^{\lambda})$ in $\\{x:x_{1}\geq\lambda_{0}|,|x|>S\\}$.

for all$\lambda\leq\bar{\lambda}_{0}$, $x\in\Sigma_{\lambda}$ and $|x|\geq R$
for all $x\in\Sigma_{\lambda}$, $\lambda\leq\lambda_{1}$ with
$|\lambda_{1}-\lambda_{0}|<c\varepsilon$, where $c_{0}$ is a small positive
number depending on $\lambda_{0}$ and $v$ only.

###### Proof of Theorem 1.2.

Lemma 2.6 and Lemma 2.7 complete the proof of (i) and (ii) in Theorem (1.2),
respectively. Next we aim to prove the radial symmetry of solutions by the
method of moving planes.

Step 1: We start moving planes along $x_{1}$-direction. For any $\lambda$, we
consider $w_{\lambda}(x)=u(x)-u(x^{\lambda})$ in $\Sigma_{\lambda}$. Then
$w_{\lambda}(x)$ satisfies

(2.46)
$\left\\{\begin{array}[]{lcl}\Delta^{2}w_{\lambda}(x)=b_{\lambda}(x)w_{\lambda}(x),&&{\rm
in}\ \;\Sigma_{\lambda},\\\ w_{\lambda}(x)=\Delta w_{\lambda}(x)=0,&&{\rm on}\
\;\partial T_{\lambda},\end{array}\right.$

where

$b_{\lambda}(x)=6\frac{|x|^{4\gamma}e^{4u(x)}-|x^{\lambda}|^{4\gamma}e^{4u(x^{\lambda})}}{u(x)-u(x^{\lambda})}.$

By (2.45) and Lemma 2.3 in [15], we have $\Delta
w_{\lambda}(x)=v(x^{\lambda})-v(x)<0$ for $x\in\Sigma_{\lambda}$,
$\lambda<\bar{\lambda}_{0}$ and $|x|\geq R$. Since $v(x)>0$ in
$\mathbb{R}^{4}$ and $v(x)\to 0$ as $|x|\to\infty$, there exists
$\bar{\lambda}_{1}<\bar{\lambda}_{0}$ such that

$v(x^{\lambda})<v(x),\quad|x|\leq R,\ \,\lambda<\bar{\lambda}_{1}.$

Therefore, we have

$\Delta w_{\lambda}(x)<0,\quad x\in\Sigma_{\lambda},\
\,\lambda<\bar{\lambda}_{1}.$

By (2.37), $\lim\limits_{|x|\to+\infty}w_{\lambda}(x)=0$. Applying the maximum
principle, we have

$w_{\lambda}(x)>0,\quad x\in\Sigma_{\lambda},\ \,\lambda<\bar{\lambda}_{1}.$

Let

(2.47) $\lambda_{0}:=\sup\big{\\{}\lambda:v(x^{\mu})<v(x)\quad{\rm for\
\,all}\ \,x\in\Sigma_{\lambda}\ \,{\rm and}\ \,\mu\leq\lambda\big{\\}}$

Since $v(x)\to 0$ as $|x|\to\infty$, we can see that $\lambda_{0}<+\infty$.
Next we claim

$u(x)\equiv u(x^{\lambda_{0}}),\quad{\rm for\ \,all\ \,}x\in\Sigma_{\lambda}.$

Suppose $w_{\lambda_{0}}\not\equiv 0$ in $\Sigma_{\lambda}$. By continuity,
$\Delta w_{\lambda_{0}}(x)\leq 0$. Since $w_{\lambda_{0}}\to 0$ as
$|x|\to+\infty$ and $w_{\lambda_{0}}\big{|}_{T_{\lambda_{0}}}=0$. By the
strong maximum principle. we have

(2.48) $w_{\lambda_{0}}(x)>0,\quad x\in\Sigma_{\lambda_{0}}.$

From (1.12), we have

$\Delta^{2}w_{\lambda_{0}}(x)=6\big{(}|x|^{4\gamma}e^{4u(x)}-|x^{\lambda_{0}}|^{4\gamma}e^{4u(x^{\lambda_{0}})}\big{)}.$

(2.48) implies $u(x)>u(x^{\lambda_{0}})$ in $\Sigma_{\lambda_{0}}$. On the
other hand, $|x|<|x^{\lambda_{0}}|$ and $\gamma<0$ imply
$|x|^{4\gamma}>|x^{\lambda_{0}}|^{4\gamma}$. Hence

(2.49) $\Delta^{2}w_{\lambda_{0}}(x)>0,\quad x\in\Sigma_{\lambda_{0}},$

which means $\Delta w_{\lambda_{0}}$ is a subharmonic function. Applying the
strong maximum principle again, we have

(2.50) $\Delta w_{\lambda_{0}}(x)<0,\quad x\in\Sigma_{\lambda_{0}}.$

By the definition of $\lambda_{0}$, there exists
$\lambda_{n}\nearrow\lambda_{0}$ and $x_{n}\in\Sigma_{\lambda_{n}}$ such that

$w_{\lambda_{n}}(x_{n})=\sup_{\Sigma_{\lambda_{n}}}w_{\lambda_{n}}>0.$

By Lemma 2.4 in [15], $\\{x_{n}\\}$ is bounded. Without loss of generality, we
assume $x_{0}=\lim\limits_{n\to+\infty}x_{n}$.

If $x_{0}\in\Sigma_{\lambda_{0}}$, by continuity we have $\Delta
w_{\lambda_{0}}(x_{0})=0$, a contradiction to (2.50).

If $x_{0}\in T_{\lambda_{0}}$, then $\nabla\big{(}\Delta
w_{\lambda_{0}}(x_{0})\big{)}=0$. However, by (2.49) and (2.50), $\Delta
w_{\lambda_{0}}$ is a negative subharmonic function in $\Sigma_{\lambda_{0}}$.
Thus by the Hopf boundary lemma, we have $\frac{\partial}{\partial\nu}\Delta
w_{\lambda_{0}}(x_{0})>0$, which yields a contradiction.

At this point, the claim is proved. Then we can change the $x_{1}$-direction
to another one and finally obtain the radial symmetry.

Step 2: Now we prove the uniqueness of the solution of (1.12) modulus scaling
in (1.13).

Let $w_{1}$ and $w_{2}$ be two radial solutions of (1.12). Then $w_{i}$
satisfies $w_{i}^{{}^{\prime}}(0)=w_{i}^{{}^{\prime}}(0)=0$. Without loss of
generality, we may assume $w_{1}(0)=w_{2}(0)$, on account of the rescale
invariance. By the uniqueness of ODE, we only need to prove
$w_{1}^{{}^{\prime\prime}}(0)=w_{1}^{{}^{\prime\prime}}(0)$.

If $w_{1}^{{}^{\prime\prime}}(0)<w_{2}^{{}^{\prime\prime}}(0)$, then
$w_{1}(r)<w_{2}(r)$ for small $r>0$. We will prove $w_{1}(0)<w_{2}(0)$ for all
$r>0$. Suppose there exists $r_{0}>0$ such that $w_{1}(r_{0})=w_{2}(r_{0})$
and $w_{1}(r)<w_{2}(r)$ for $0<r<r_{0}$. Then by (1.12), we have

$\frac{\partial}{\partial
r}\Delta(w_{1}(r)-w_{2}(r))=6r^{4\gamma}\big{(}e^{4w_{1}(r)}-e^{4w_{2}(r)}\big{)}<0,\quad
1<r\leq r_{0},$

which together with the assumption implies

$\Delta(w_{1}-w_{2})<0,\quad{\rm in}\ \,B(0,r_{0}).$

Since $w_{1}(r_{0})-w_{2}(r_{0})=0$, from the maximum principle, we have
$w_{1}(r)-w_{2}(r)>0$ for $0<r<r_{0}$, which contradicts with
$w_{1}(0)=w_{2}(0)$. Thus, $w_{1}(r)<w_{2}(r)$ for all $r>0$. Hence
$\frac{\partial}{\partial r}\Delta(w_{1}(r)-w_{2}(r))<0$ for all $r>0$, which
means $\Delta(w_{1}(r)-w_{2}(r))$ is decreasing in $r$. Thus
$w_{1}(r)-w_{2}(r)\leq-cr^{2}$ as $r\to+\infty$ for some constant $c>0$, which
yields a contradiction to the assumption $w_{i}(r)=o(r^{2})$ at $\infty$.

Similarly, it is impossible for
$w_{1}^{{}^{\prime\prime}}(0)>w_{2}^{{}^{\prime\prime}}(0)$. Thus, the radial
solution of (1.12) is unique under the rescale $u_{\lambda}(x)=u(\lambda
x)+(1+\gamma)\log\lambda$ for some $\lambda>0$, and it is valid for (2.14)
after scaling.

∎

Next, we consider the case without the assumption $|u(x)|=o(|x|^{2})$ at
$\infty$.

The following lemma is similar to Lemma 3.3 in [31]. But we do not require $u$
is harmonic, and in fact $\Delta u\equiv const.$ is enough. Furthermore, when
we apply this lemma in the proof of Theorem 1.3, it is uncertain that $w$ is
harmonic.

######  Lemma 2.8.

Suppose that $\Delta u=a$ in $\mathbb{R}^{n}$ for a constant $a\in\mathbb{R}$
such that $\exp(u-c|x|^{2})\in L^{1}(\mathbb{R}^{n})$ for some $c>0$. Then $u$
is a polynomial of order at most 2.

###### Proof.

We only need to prove for any unit vector $\xi\in\mathbb{R}^{n}$, there holds

$u_{\xi\xi}(x)\equiv const.\quad{\rm in}\ \,\mathbb{R}^{n}.$

By Liouville’s Theorem, it suffices to prove $u_{\xi\xi}(x)$ is bounded from
above by a constant independent of $x$. Without loss of generality, we may
consider $x=0$ and $\xi=e_{1}$. For simplicity, we denote $B(0,r)$ and
$\partial B(0,r)$ to $B_{r}$ and $\partial B_{r}$ respectively.

Since $\Delta u=a$, then $u_{x_{1}x_{1}}$ is harnomic. Hence for any $r>0$, we
have

(2.51)
$u_{x_{1}x_{1}}(0)=\frac{1}{\sigma_{n}r^{n}}\int_{B_{r}}u_{x_{1}x_{1}}(x){\rm
d}x=\frac{1}{\sigma_{n}r^{n}}\int_{\partial
B_{r}}u_{x_{1}}\frac{x_{1}}{|x|}{\rm d}\sigma,$

where $\sigma_{n}$ is the volume of the unit ball in $\mathbb{R}^{n}$.
Integrating the identity above from $0$ to $r$, we have

(2.52)
$\begin{split}\frac{\sigma_{n}}{n+1}r^{n+1}u_{x_{1}x_{1}}(0)=&\int_{B_{r}}u_{x_{1}}\frac{x_{1}}{|x|}{\rm
d}\sigma=-\int_{B_{r}}u\frac{\partial}{\partial
x_{1}}\big{(}\frac{x_{1}}{|x|}\big{)}{\rm d}x+\int_{\partial
B_{r}}u\frac{x_{1}^{2}}{|x|^{2}}{\rm d}\sigma\\\
=&-\int_{B_{r}}\frac{u}{|x|}{\rm
d}x+\int_{B_{r}}u\frac{x_{1}^{2}}{|x|^{3}}{\rm d}x+\int_{\partial
B_{r}}u\frac{x_{1}^{2}}{|x|^{2}}{\rm d}\sigma.\end{split}$

Note that

$\int_{B_{r}}\frac{u}{|x|}{\rm
d}x=n\sigma_{n}\int_{0}^{r}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{s}}u{\rm d}\sigma\Big{)}s^{n-2}{\rm
d}s.$

Since $\Delta u=a$, then

$\begin{split}a\sigma_{n}s^{n}=&\int_{B_{s}}\Delta u(y){\rm d}y=\int_{\partial
B_{s}}\frac{\partial u}{\partial\nu}{\rm
d}\sigma=s^{n-1}\int_{|\omega|=1}\frac{\partial u}{\partial s}(s\omega){\rm
d}\sigma_{\omega}\\\ =&s^{n-1}\frac{\partial}{\partial
s}\int_{|\omega|=1}u(s\omega){\rm d}\sigma_{\omega}.\end{split}$

That is

$\frac{\partial}{\partial s}\int_{|\omega|=1}u(s\omega){\rm
d}\sigma_{\omega}=a\sigma_{n}s.$

Integrating the identity above from $0$ to $s$, we obtain

$\int_{|\omega|=1}u(s\omega){\rm d}\sigma_{\omega}=\int_{|\omega|=1}u(0){\rm
d}\sigma_{\omega}+\frac{a}{2}\sigma_{n}s^{2},$

which means

$\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{s}}u{\rm
d}\sigma=\frac{1}{n\sigma_{n}}\int_{|\omega|=1}u(s\omega){\rm
d}\sigma_{\omega}=u(0)+\frac{a}{2n}s^{2}.$

Then

(2.53) $\int_{B_{r}}\frac{u}{|x|}{\rm
d}x=n\sigma_{n}\int_{0}^{r}\big{(}u(0)+\frac{a}{2n}s^{2}\big{)}s^{n-2}{\rm
d}s=\frac{n}{n-1}\sigma_{n}r^{n-1}u(0)+\frac{a}{2(n+1)}r^{n+1}.$

On the other hand, by a direct computation, we have

(2.54) $\int_{B_{r}}u\frac{x_{1}^{2}}{|x|^{3}}{\rm
d}x=\frac{\sigma_{n}}{n-1}r^{n-1}+\int_{\partial
B_{r}}u\frac{x_{1}^{2}}{|x|^{2}}{\rm d}\sigma$

and

(2.55) $\int_{\partial B_{r}}u\frac{x_{1}^{2}}{|x|^{2}}{\rm
d}\sigma=\sigma_{n}r^{n-1}$

From (2.52)$\sim$(2.55), we have

$\frac{r^{2}}{n+1}u_{x_{1}x_{1}}(0)=-\frac{n}{n-1}u(0)-\frac{a}{2(n+1)\sigma_{n}}r^{2}+\frac{1}{n-1}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{B_{r}}u{\rm
d}\mu_{1}+\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{r}}{\rm d}\mu_{2},$

where ${\rm d}\mu_{1}=\frac{x_{1}^{2}}{|x|^{3}}{\rm d}x$ and ${\rm
d}\mu_{2}=\frac{x_{1}^{2}}{|x|^{2}}{\rm d}x$ on $\partial B_{r}$. By Jensen’s
inequality, we have

(2.56)
$\begin{split}&\exp\Big{(}\frac{r^{2}}{2(n+1)}u_{x_{1}x_{1}}(0)\Big{)}\\\
\leq&\exp\Big{(}-\frac{n}{2(n-1)}u(0)\Big{)}\exp\Big{(}-\frac{a}{4(n+1)\sigma_{n}}r^{2}\Big{)}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{B_{r}}e^{\frac{u}{2(n-1)}}{\rm
d}\mu_{1}\Big{)}\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{r}}e^{\frac{u}{2}}{\rm
d}\mu_{2}\Big{)}.\end{split}$

For any $c_{1}>-\frac{a}{4(n+1)\sigma_{n}}$, denoting
$\tilde{c}_{1}=c_{1}+\frac{a}{4(n+1)}$, we have

(2.57)
$\begin{split}&\int_{1}^{\infty}\exp\Big{[}\big{(}\frac{r^{2}}{2(n+1)}u_{x_{1}x_{1}}(0)-c_{1}\big{)}r^{2}\Big{]}{\rm
d}r\leq\exp\Big{(}-\frac{n}{2(n-1)}u(0)\Big{)}\\\
&\qquad\Big{\\{}\int_{1}^{\infty}\big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{B_{r}}e^{\frac{u}{n-1}}{\rm
d}\mu_{1}\big{)}e^{-\tilde{c}_{1}r^{2}}{\rm
d}r\Big{\\}}^{\frac{1}{2}}\Big{\\{}\int_{1}^{\infty}\big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B_{r}}e^{u}{\rm
d}\mu_{2}\big{)}e^{-\tilde{c}_{1}r^{2}}{\rm
d}r\Big{\\}}^{\frac{1}{2}}.\end{split}$

By the assumption, we can choose $c_{1}$ large enough such that the right hand
of (2.57) is finite. Therefore, we have

(2.58) $u_{x_{1}x_{1}}(0)\leq 2(n+1)c_{1}.$

(2.58) is valid for all $x\in\mathbb{R}^{n}$. By Liouville’s Theorem, we
obtain $u_{x_{1}x_{1}}(x)\equiv const.$, which indicates $u$ is a polynomial
of order at most 2.

∎

###### Proof of Theorem 1.3.

Suppose that $u$ is a solution of (1.12). Let $v$ be (2.1) and
$w(x)=u(x)+v(x)$. By Lemma 2.2, we have $\Delta w(x)\equiv-C_{1}\leq 0$ in
$\mathbb{R}^{4}$. Then Lemma 2.8 implies there exist constants $c_{0}$ and
$a_{ij}$ $(i,j=1,\cdots,4)$ such that $a_{ij}=a_{ji}$ and

$w(x)=\sum_{i,j,k=1}^{4}(a_{ij}x_{i}x_{j}+b_{k}x_{k})+c_{0}.$

After an orthogonal transformation, we may assume

$u(x)=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\sum_{j=1}^{4}(a_{j}x_{j}^{2}+b_{j}x_{j})+c_{0}.$

Since $|x|^{4\gamma}e^{4u}\in L^{1}(\mathbb{R}^{4})$, we have $a_{j}\geq 0$
for all $j=1,\cdots,4$, and $b_{j}=0$ if $a_{j}=0$. Hence, we can rewrite
$u(x)$ as follows:

$u(x)=\frac{3}{4\pi^{2}}\int_{\mathbb{R}^{4}}\log\big{(}\frac{|y|}{|x-y|}\big{)}|y|^{4\gamma}e^{4u(y)}{\rm
d}y-\sum_{j=1}^{4}a_{j}(x_{j}-x_{j}^{0})^{2}+c_{0}$

and (1.17) holds from the previous argument.

Next, we show the radial symmetry. Let
$\hat{u}(x)=u(x)+\sum_{j=1}^{4}a_{j}(x_{j}-x_{j}^{0})^{2}$. Then

(2.59)
$\Delta^{2}\hat{u}(x)=6|x|^{4\gamma}e^{-4\sum_{j=1}^{4}a_{j}(x_{j}-x_{j}^{0})^{2}}e^{4\hat{u}(x)},\quad{\rm
in}\ \,\mathbb{R}^{4}.$

As in Lemma 2.7, we set $\hat{w}(x)=\hat{u}(\frac{x}{|x|^{2}})-\alpha\log|x|$,
then $|\hat{w}(x)|=o(\log\frac{1}{|x|})$ near 0 from Lemma 2.1 and Lemma 2.3,
and $\hat{w}$ satisfies

(2.60)
$\Delta^{2}\hat{w}(x)=6|x|^{4\gamma}e^{-4\sum_{j=1}^{4}a_{j}(\frac{x_{j}}{|x|^{2}}-x^{0})^{2}}e^{4\hat{w}(x)},\quad{\rm
in}\ \,\mathbb{R}^{4}\setminus\\{0\\}.$

Note that $a_{j}\geq 0$, then we can follow the argument in the proof of Lemma
2.7 to obtain for $|x|$ large,

(2.61) $\hat{u}(x)=-\alpha\log|x|+c_{0}+O(\frac{1}{|x|}),$

and

(2.62)
$\left\\{\begin{array}[]{ll}-\Delta\hat{u}(x)=\frac{2\alpha}{|x|^{2}}+O(\frac{1}{|x|^{3}}),\\\
-\frac{\partial}{\partial
x_{i}}\Delta\hat{u}(x)=-4\alpha\frac{x_{i}}{|x|^{4}}+O(\frac{1}{|x|^{4}}),\\\
-\frac{\partial^{2}}{\partial x_{i}\partial
x_{j}}\Delta\hat{u}(x)=O(\frac{1}{|x|^{4}}).\end{array}\right.$

At this point, we establish (1.16). Finally, as in the proof of Theorem 1.2,
we can apply the method of moving planes to show that $\hat{u}(x)$ is
symmetric with respect to the hyperplane $\\{x:x_{i}=0\\}$ if
$a_{i}x_{i}^{0}=0$. In particular, if $a_{i}x_{i}^{0}=0$ for all
$i=1,\cdots,4=0$, then $u$ is radially symmetric with respect to the origin.

∎

##  3\. Preliminaries for blowup analysis

Let $G:M\times M\setminus{\rm diag}$ denote the Green’s function for the
Paneitz operator

(3.1) $f(x)-\bar{f}=\int_{M}G(x,y)P_{g}f(y){\rm
d}V_{g}(y),\quad\int_{M}G(x,y){\rm d}V_{g}(y)=0,$

where $\bar{f}=\frac{1}{{\rm vol}_{g}(M)}\int_{M}f{\rm d}V_{g}$ is the average
of $f$ over $M$. Then the weak form of (3.1) is

(3.2) $P_{g,y}G(x,y)=\delta_{x}-\frac{1}{{\rm vol}_{g}(M)}.$

Set $R$ be the regular part of the Green function. Then by the Appendix A in
[51], for $y$ in a neighborhood of $x$,

(3.3) $G(x,y)=-\frac{1}{8\pi^{2}}\log d_{g}(x,y)\chi+R(x,y).$

where $\chi$ is a cut-off function to avoid cut locus. Using $G$ we can
decompose $u_{k}$ as the sum of its regular part and singular part

(3.4) $u_{k}(x)=\tilde{u}_{k}(x)-8\pi^{2}\sum_{j=1}^{N}\gamma_{j}G(x,q_{j}).$

Then $\tilde{u}_{k}$ satisfies

(3.5) $P_{g}\tilde{u}_{k}+2b_{k}=2H_{k}e^{4\tilde{u}_{k}}\quad{\rm in}\ \,M,$

where

(3.6) $H_{k}(x)=h_{k}(x)\prod_{j=1}^{N}e^{-32\pi^{2}\gamma_{j}G(x,q_{j})}.$

Clearly, (1.9) and (1.11) imply that

(3.7) $\int_{M}H_{k}e^{4\tilde{u}_{k}}{\rm d}V_{g}\leq C$

We will work with $\tilde{u}_{k}$ in the later blow-up analysis.

Similar with [51], since the metric $g$ may not be locally conformally flat,
we will apply the conformal normal coordinates, whose existence has been
proved in [29]. More specially, for $q\in M$, there exists a normal coordinate
around $q$ such that $g$ can be deformed to $\mathfrak{g}$ which satisfies
$\det\,(\mathfrak{g})=1$. We use $R_{ijkl}$ to denote the curvature tensor
under $\mathfrak{g}$.

We will apply the expansions of the metric $\mathfrak{g}$ and its derivatives
in the conformal normal coordinates (seen in the Appendix B of [51]), which
are

(3.8)
$\begin{split}&\mathfrak{g}_{ab}(x)=\delta_{ab}+\frac{1}{3}R_{aijb}x^{i}x^{j}+O(r^{3}),\\\
&\mathfrak{g}^{ab}(x)=\delta_{ab}-\frac{1}{3}R_{aijb}x^{i}x^{j}+O(r^{3}),\\\
&\partial_{c}\mathfrak{g}^{ab}(x)=-\frac{1}{3}(R_{acib}+R_{aicb})x^{i}+O(r^{2}),\\\
&\partial_{ab}\mathfrak{g}^{ab}(x)=\frac{1}{3}R_{ia,a}x^{i}+O(r^{2}),\end{split}$

In addition, the following Pohozaev identity from the Appendix D in [51] will
play an important role when the blow-up analysis is carried out in the
conformal normal coordinates.

######  Lemma 4.

[51]

For equation $P_{g}u+2b=2he^{4u}$ in $M$ and $\Omega=B(0,r)$, there holds

(3.9)
$\begin{split}&\int_{\Omega}\Big{(}2he^{4u}+\frac{1}{2}x^{i}\partial_{i}he^{4u}\Big{)}\\\
=&\int_{\partial\Omega}\Big{(}\frac{1}{2}x^{i}\nu_{i}he^{4u}-x^{k}\nu_{j}g^{ij}\partial_{i}(\Delta_{g}u)\partial_{k}u+\nu_{j}g^{ij}\Delta_{g}u\partial_{i}u+x^{k}\nu_{j}g^{ij}\Delta_{g}u\partial_{ik}u-\frac{1}{2}x^{i}\nu_{i}(\Delta_{g}u)^{2}\Big{)}\\\
&+\int_{\Omega}\Big{(}\Delta_{g}u\partial_{i}g^{ij}\partial_{j}u+x^{k}\Delta_{g}u\partial_{ik}g^{ij}\partial_{j}u+x^{k}\Delta_{g}u\partial_{k}g^{ij}\partial_{ij}u-2bx^{i}\partial_{i}u\Big{)}\\\
&+2\int_{\partial\Omega}\Big{(}R_{ij,l}(q)x^{l}x^{k}\nu_{i}\partial_{j}u\partial_{k}u+O(r^{3})|\nabla
u|^{2}\Big{)}\\\
&-\int_{\Omega}\Big{(}2R_{ij,l}(q)\big{(}x^{l}\partial_{j}u\partial_{i}u+x^{k}x^{l}\partial_{j}u\partial_{ik}u\big{)}+0(r^{2})|\nabla
u|^{2}+O(r^{4})|\nabla^{2}u|\Big{)}\end{split}$

Note that we use $B(p,r)$ to denote a ball centered at $p$ with radius $r$.
Sometimes if the center is the origin, we use $B_{r}$ instead of $B(0,r)$.

##  4\. Blow-up analysis near the singularity

In this section, we focus on the blow-up analysis near $q_{j}$, and to
simplify the notation, we will omit the superscript $j$. Similar to the
argument in [51], we will work in the conformal normal coordinates near $q$
from [29]. To be specific, we can find some function $w$ defined on $M$, such
that in a small neighborhood $B(q,\delta)$ of $q$, $\delta>0$, we have

(4.1) $\det\,(\hat{g})=1$

in the normal coordinates of the conformal metric $\hat{g}=e^{2w}g$. For
convenience we just use $g$ instead of $\hat{g}$. Note that in a neighborhood
of $q$,

(4.2) $w(x)=O(d_{g}(x,q)^{2}).$

where $d_{g}(x,q)$ stands for the distance between $x$ and $q$ under metric
$g$.

Using the conformal covariance property of $P_{g}$ the function
$\mathfrak{u}_{k}=\tilde{u}_{k}-w$ satisfies

(4.3) $P_{g}\mathfrak{u}_{k}+2\mathfrak{b}_{k}=2H_{k}e^{4\mathfrak{u}_{k}},$

and

(4.4) $\int_{M}H_{k}e^{4\mathfrak{u}_{k}}{\rm d}V_{g}\leq C,$

where $2\mathfrak{b}_{k}=P_{g}w+2b_{k}$ and $\tilde{H}_{k}=H_{k}e^{4w}$. For
simplity, we still denote $\tilde{H}_{k}$ by $H_{k}$.

We still use $G$ to denote the Green’s function for $P_{g}$. Then we have the
following Green’s representation formula

(4.5)
$\mathfrak{u}_{k}(x)=\bar{\mathfrak{u}}_{k}+2\int_{M}G(x,y)H_{k}(y)e^{4\mathfrak{u}_{k}(y)}{\rm
d}V_{g}(y)-2\int_{M}G(x,y)\mathfrak{b}_{k}(y){\rm d}V_{g}(y),$

where $\bar{\mathfrak{u}}_{k}$ is the average of $\mathfrak{u}_{k}$ over
$(M,g)$. Using the expression of $G$ in (3.3, we have

(4.6)
$\mathfrak{u}_{k}(x)=\bar{\mathfrak{u}}_{k}+2\int_{M}\Big{(}-\frac{1}{8\pi^{2}}\log
d_{g}(x,y)\chi\Big{)}H_{k}(y)e^{4\mathfrak{u}_{k}(y)}{\rm
d}V_{g}(y)+\mathfrak{\phi}_{k}(x),$

where

(4.7)
$\mathfrak{\phi}_{k}(x)=2\int_{M}R(x,y)H_{k}(y)e^{4\mathfrak{u}_{k}(y)}{\rm
d}V_{g}(y)-2\int_{M}G(x,y)\mathfrak{b}_{k}(y){\rm d}V_{g}(y).$

Note that $\det\,(g)=1$ in $B(q,\delta)$, we have ${\rm d}V_{g}(y)={\rm d}y$
in $B(q,\delta)$. Taking the difference of (4.6) evaluated at $x$ and $q$, we
get

(4.8) $\begin{split}&\mathfrak{u}_{k}(x)-\mathfrak{u}_{k}(q)\\\
=&\frac{1}{4\pi^{2}}\int_{M}\big{(}\chi(r_{q})\log|y-q|-\chi(r_{x})\log
d_{g}(x,y)\big{)}H_{k}(y)e^{4\mathfrak{u}_{k}(y)}{\rm
d}V_{g}(y)+\mathfrak{\phi}_{k}(x)-\mathfrak{\phi}_{k}(q),\end{split}$

where $r_{q}=|y-q|$ and $r_{x}=d_{g}(x,y)$. Here since the coordinates are
normal, we have $d_{g}(y,q)=|y-q|$.

Thanks to the cut-off function $\chi$, we can replace the integral over $M$ by
an integral over $B(q,2\delta)$:

(4.9) $\begin{split}&\mathfrak{u}_{k}(x)-\mathfrak{u}_{k}(q)\\\
=&\frac{1}{4\pi^{2}}\int_{B(q,4\delta)}\big{(}\chi(r_{q})\log|y-q|-\chi(r_{x})\log
d_{g}(x,y)\big{)}H_{k}(y)e^{4\mathfrak{u}_{k}(y)}{\rm d}V_{g}(y)\\\
&+\mathfrak{\phi}_{k}(x)-\mathfrak{\phi}_{k}(q),\quad x\in
B(q,2\delta).\end{split}$

We next give an upper bound of the mass near $q$ when $\mathfrak{u}_{k}$
cannot blow up at $q$. Before stating such a small energy Lemma we point out
that the function $H_{k}$ can be written in the neighborhood of a singular
source $q$ as

(4.10) $H_{k}(x)=\mathfrak{h}_{k}(x)d_{g}(x,q)^{4\gamma},$

with $\mathfrak{h}_{k}(q)\neq 0$. Using this notation, our result states as
follows:

######  Lemma 4.1.

Let $q$ be a singular source with index $\gamma$. If

(4.11)
$\int_{B(q,2\delta)}2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}{\rm
d}V_{g}<\min\\{8\pi^{2},8\pi^{2}(1+\gamma)\\},$

$\mathfrak{h}_{k}$ is defined in (4.10). Then $\mathfrak{u}_{k}\leq C$ in
$B(q,\delta)$.

###### Proof.

Note that

$P_{g}\mathfrak{u}_{k}=2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}-2\mathfrak{b}_{k}.$

We write $\mathfrak{u}_{k}=u_{1k}+u_{2k}$ where $u_{1k}$ is the solution of

(4.12)
$\left\\{\begin{array}[]{lcl}\Delta^{2}u_{1k}=2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}&&{\rm
in}\ \,B(q,2\delta)\\\ u_{1k}(x)=\Delta u_{1k}(x)=0&&{\rm on}\ \,\partial
B(q,2\delta).\end{array}\right.$

By Lemma 1 from [31] we have

(4.13)
$\int_{B(q,2\delta)}\exp\Big{\\{}\frac{\tilde{\delta}|\mathfrak{u}_{1k}|}{\parallel
2H_{k}e^{4\mathfrak{u}_{k}}\parallel_{L^{1}(B(q,2\delta),g)}}\Big{\\}}{\rm
d}V_{g}\leq C,$

with any $\tilde{\delta}\in(0,32\pi^{2})$ and some constant
$C=C(\tilde{\delta},\delta)$. On one hand in $B(q,2\delta)$,

(4.14)
$u_{1k}(x)=\int_{B(q,\delta)}G_{\delta}(x,y)2\mathfrak{h}_{k}d_{g}(\eta,q)^{4\gamma}e^{4\mathfrak{u}_{k}}{\rm
d}\eta,\quad x\in B(q,2\delta)$

where $G_{\delta}(x,y)$ is the Green’s function of $\Delta^{2}$ on
$B(q,2\delta)$:

$G_{\delta}(x,y)=-\frac{1}{8\pi^{2}}\log|x-y|+R_{\delta}(x,y),$

with

$G_{\delta}(x,y)=\Delta_{y}G_{\delta}(x,y)=0,\quad\mbox{for }x\in
B(q,2\delta),y\in\partial B(q,2\delta).$

In particular for $x\in B(q,\frac{3}{2}\delta)$,

$u_{1k}(x)=-\frac{1}{8\pi^{2}}\int_{B(q,2\delta)}\log|x-q|2\mathfrak{h}_{k}|\eta-q|^{4\gamma}e^{4\mathfrak{u}_{k}}{\rm
d}\eta+O(1),\quad x\in B(q,\frac{3}{2}\delta).$

On the other hand the Green’s representation formula of $\mathfrak{u}_{k}$
gives

$\mathfrak{u}_{k}(x)=u_{1k}(x)+u_{2k}(x)=\overline{\mathfrak{u}_{k}}+\int_{M}G(x,\eta)2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}}{\rm
d}\eta.$

Since the leading term of $G$ and $G_{\delta}$ are both
$-\frac{1}{8\pi^{2}}\log d_{g}(x,q)$, we have

$u_{2k}(x)=\overline{\mathfrak{u}_{k}}+O(1).$

From $\int_{M}e^{4\mathfrak{u}_{k}}{\rm d}V_{g}\leq C$ and Jensen’s inequality

$e^{4\bar{\mathfrak{u}}_{k}}\leq\int_{M}e^{4\mathfrak{u}_{k}}{\rm d}V_{g}\leq
C.$

Therefore, $\mathfrak{u}_{2k}\leq C$ in $B(q,\frac{3}{2}\delta)$. Now we focus
on $u_{1k}$.

If $\gamma>0$, (4.11) is

$\int_{B(q,2\delta)}2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}{\rm
d}V_{g}<8\pi^{2}.$

Since $u_{2k}$ is bounded from above in $B(q,\frac{3}{2}\delta)$, we see from
(4.13) that there exists some $p>1$ to make $e^{4u_{1k}}\in
L^{p}(B(q,2\delta))$:

(4.15) $\parallel
2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}\parallel_{L^{p}(B(q,\frac{3}{2}\delta))}\leq
C.$

The estimate (4.15) leads to a $L^{\infty}$ of $u_{1k}$ in $B(q,\delta)$ based
on two reasons. First the integration of (4.14) gives a $L^{1}$ bound of
$u_{1k}$ in $B(q,\frac{3}{2}\delta)$:

(4.16) $\parallel u_{1k}\parallel_{L^{1}(B(q,\frac{3}{2}\delta))}\leq C.$

Second, the standard interior regularity results in [13] (Theorem 1 in Section
3 of [13]) gives

$\parallel u_{1k}\parallel_{W^{4,p}B(q,\delta)}\leq\parallel
2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}\parallel_{L^{p}(B(q,\frac{3}{2}\delta))}+\parallel
u_{1k}\parallel_{L^{1}(B(q,\frac{3}{2}\delta))}\leq C.$

Thus we have obtained the $L^{\infty}$ bound of $u_{1k}$ in $B(q,\delta)$ by
standard Sobolev embedding theorem.

If $-1<\gamma<0$, (4.11) is

$\int_{B(q,2\delta)}2\mathfrak{h}_{k}d_{g}(x,q)^{-4|\gamma|}e^{4\mathfrak{u}_{k}(x)}{\rm
d}V_{g}<8\pi^{2}(1-|\gamma|).$

Since $\frac{\tilde{\delta}}{8\pi^{2}(1-|\gamma|)}<\frac{4}{1-|\gamma|}$, this
strict inequality makes it possible to choose $\tilde{\delta}\in(0,32\pi^{2})$
and $p>1$ with

$p<\frac{1}{|\gamma|}\quad p<\frac{\tilde{\delta}}{32\pi^{2}(1-|\gamma|)}.$

Then H${\rm\ddot{o}}$lder inequality tells us that (4.15) is also true in this
case and the $L^{\infty}$ bound of $u_{1k}$ over $B(q,\delta)$ follows
immediately. The combination of the $L^{\infty}$ bound of $u_{1k}$ and the
upper bound of $u_{2k}$ implies that $\mathfrak{u}_{k}\leq C$ in
$B(q,\delta)$.

∎

######  Remark 4.2.

If $q$ is a regular point the same proof obviously leads to a similar
conclusion: Suppose $B(q,2\delta)$ has no singular source and
$\int_{B(q,2\delta)}2\mathfrak{h}_{k}e^{4\mathfrak{u}_{k}}<8\pi^{2}$, there is
an upper bound of $\mathfrak{u}_{k}$ in $B(q,\delta)$.

Another immediate consequence of Lemma 4.1 is that blowup sequence only
converges to point measures. The following theorem takes one step further to
assert that the point measure is quantized and the bubbling solutions tend to
$-\infty$ away from blowup points.

######  Theorem 4.3 (Concentration and Quantization).

Let $\\{\mathfrak{u}_{k}\\}$ be a sequence of solution to (4.3) with (4.4).
Assume that $q$ is the only blow-up point of $\mathfrak{u}_{k}$ in
$B(q,\delta)$, then as $k\to+\infty$, along a subsequence, there hold

(4.17) $\mathfrak{u}_{k}\to-\infty,\quad uniformly\ \,on\ \,any\ \,compact\
\,set\ \,of\ \,B(q,\delta)\setminus\\{q\\},$ (4.18)
$2\mathfrak{h}_{k}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}}\to\beta\delta_{q},\quad
weakly\ \,in\ \,the\ \,sense\ \,of\ \,measure\ \,on\ \,M,$

with $\beta=16\pi^{2}(1+\gamma)$ and $\mathfrak{h}_{k}$ is defined in (4.10).

In particular, along a subsequence,

(4.19) $2\int_{B(q,\delta)}H_{k}(x)e^{4\mathfrak{u}_{k}(x)}{\rm d}V_{g}\to
16\pi^{2}(1+\gamma),\quad as\ \,k\to+\infty.$

###### Proof.

Suppose $q$ is the only blowup pint in $B(q,2\delta)$, then we observe that
for any given $K\subset\subset B(q,\delta)$

(4.20) $|\mathfrak{u}_{k}(x)-\mathfrak{u}_{k}(y)|\leq C(K)$

because the $|x-q|$ is comparable to $|y-q|$ and the total integration of
$\mathfrak{h}_{k}e^{4\mathfrak{u}_{k}}$ is bounded. It is easy to obtain
(4.20) from the Green’s representation formula. Then it is also immediate to
observe that

(4.21) $|\nabla_{g}^{j}\mathfrak{u}_{k}(x)|_{g}\leq C(K),\quad{\rm in}\
\,K,\quad j=1,2,3$

because when $x\in K$,

$|\nabla^{j}G(x,y)|\leq C|x-y|^{-j},\quad x,y\in B(q,2\delta),$

and $\mathfrak{u}_{k}$ has a upper bound. These two facts imply (4.21 by
trivial estimates. Then the equation for $\mathfrak{u}_{k}$ further provides
the estimate for the fourth order derivatives of $\mathfrak{u}_{k}$:

$|\nabla^{\alpha}\mathfrak{u}(x)|\leq C(k),\quad x\in K,\quad|\alpha|=4.$

Now we prove (4.17) by contradiction. Suppose that there exists a point
$x_{0}\in B(q,\delta)\setminus\\{q\\}$ such that
$\\{\mathfrak{u}_{k}(x_{0})\\}_{k\in\mathbb{N}}$ is also bounded from below.
By (4.20) we see that $\mathfrak{u}_{k}$ is bounded in $L^{\infty}$ norm in
any compact subset of $B(q,\delta)\setminus\\{q\\}$. This fact and the
gradients estimates of $\mathfrak{u}_{k}$ guarantee that along a subsequence

$\mathfrak{u}_{k}\to\mathfrak{u}_{0}\quad{\rm in}\
\,C_{loc}^{3,\sigma}(B(q,2\delta)\setminus\\{q\\}),$

with some constant $\sigma\in(0,1)$ and the limit function $\mathfrak{u}_{0}$
solves

$P_{g}\mathfrak{u}_{0}(x)+2\mathfrak{b}_{0}(x)=2\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}(x)}\quad{\rm
in}\ \,B(q,2\delta)\setminus\\{q\\}.$

Around $q$ we use $\beta(r)$ and its limit to describe the concentration of
energy:

$\begin{split}&\beta_{k}(r)=\int_{B(q,r)}2\mathfrak{h}_{k}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}{\rm
d}V_{g}\\\ &\beta(r)=\lim_{k\to+\infty}\beta_{k}(r),\quad\beta=\lim_{r\to
0}\beta(r).\end{split}$

From Lemma 4.1 and Remark 4.2 we see that if $q$ is singular point
$\beta\geq\min\\{8\pi^{2}(1+\gamma),8\pi^{2}\\}$, if $q$ is a regular point
$\beta\geq 8\pi^{2}$. Fixing any $r>0$ small, we integrate the equation of
$\mathfrak{u}_{k}$ in $B(q,r)$ to obtain

$\int_{B(q,r)}\big{(}P_{g}\mathfrak{u}_{k}+2\mathfrak{b}_{k}\big{)}{\rm
d}V_{g}=\int_{B(q,r)}H_{k}e^{4\mathfrak{u}_{k}}{\rm d}V_{g}=\beta_{k}(r),$

which implies

$\lim_{r\to
0}\int_{B(q,r)}\big{(}P_{g}\mathfrak{u}_{0}+2\mathfrak{b}_{0}\big{)}{\rm
d}V_{g}=\beta.$

Therefore, $\mathfrak{u}_{0}$ satisfies, in the distribution sense,

$P_{g}\mathfrak{u}_{0}(x)+2\hat{b}_{0}(x)=2\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}(x)}+\beta\delta_{q}\quad{\rm
in}\ \,B(q,2\delta)$

Using the Green’s representation formula for $\mathfrak{u}_{k}$, we have

(4.22) $\mathfrak{u}_{0}(x)=-\frac{\beta}{8\pi^{2}}\log d_{g}(x,q)+v(x)+w(x)$

where the first term comes from the convolution of $-\frac{1}{8\pi^{2}}\log
d_{g}(x,y)\chi$ with $\beta\delta_{q}$, the second term $v$ comes from the
convolution of $-\frac{1}{8\pi^{2}}\log d_{g}(x,y)\chi$ with
$2\mathfrak{h}_{0}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}}$:

(4.23) $v(x)=-\frac{1}{4\pi^{2}}\int_{B(q,\delta)}\log
d_{g}(x,y)\mathfrak{h}_{0}(y)d_{g}(y,q)^{4\gamma}e^{4\mathfrak{u}_{0}(y)}{\rm
d}V_{g}(y),$

and $w$ is the collection of insignificant other terms:

(4.24) $w\in C^{4}(M).$

For $v$ we use (4.23) to denote $v(x)$ in $B(q,\delta)$ and we extend $v$ to
make $v\equiv 0$ on $M\setminus B(q,2\delta)$. Based on the definition of $v$
we now show that $v\in L^{\infty}(B(q,\delta))$. In fact, from (4.23) we have
this lower bound of $v$ in $B(q,\delta)$:

(4.25) $v(x)\geq\frac{1}{4\pi^{2}}\log\frac{1}{\delta}\parallel
V\parallel_{L^{1}(M)}\geq C\quad{\rm in}\ \,B(q,\delta)$

where

$V(x)=\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}(x)}.$

The lower bounds of $\mathfrak{h}_{0}$ and $v$ lead to a lower bound for
$V(x)$:

$V(x)=\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}(x)}\geq
Cd_{g}(x,q)^{4\gamma-\frac{\beta}{2\pi^{2}}}e^{4v(x)+w(x)}\geq\frac{c}{d_{g}(x,q)^{s}}$

with $s=\frac{\beta}{2\pi^{2}}-4\gamma$ and suitable $c>0$. Since
$\|V\|_{L^{1}}<\infty$ we see immediately that $s<4$, which is

(4.26) $\beta<8\pi^{2}(1+\gamma).$

Thus there is no way for $\mathfrak{u}_{k}$ to be bounded from below away from
singular source unless $\gamma>0$. We have proved (4.17) for $\gamma\leq 0$.
For $\gamma>0$ we have an upper bound for $V(x)$:

(4.27) $V(x)\leq\frac{c}{d_{g}(x,q)^{s}}e^{4v(x)}\quad{\rm in}\
\,B(q,\delta),\quad\mbox{if}\quad\gamma>0.$

To proceed with the proof of $v\in L^{\infty}$ we observe from (4.23) and
direct computation that

$\left\\{\begin{array}[]{lcl}\Delta^{2}v(x)=V(x)+\eta(x)&&{\rm in}\
\,B(q,2\delta)\\\ v(x)=\Delta v(x)=0&&{\rm on}\ \,\partial
B(q,2\delta),\end{array}\right.$

where $\eta$ is smooth in $B(q,2\delta)$. Note that the boundary condition of
$v$ on $\partial B(q,2\delta)$ is based on the smooth extension of $v$
mentioned before. Now we employ a standard argument of Brezis-Merle [12] to
obtain $e^{\kappa|v|}\in L^{1}(B(q,2\delta),g)$ for any constant $\kappa>0$.
Indeed, let $0<\varepsilon<1/\kappa$ and $V=V_{1}+V_{2}$ with $\parallel
V_{1}\parallel_{L^{1}(B(q,2\delta))}<\epsilon$ and $V_{2}\in
L^{\infty}(B(q,2\delta))$. Correspondingly we write $v=v_{1}+v_{2}$, where
$v_{1}$ solves

$\left\\{\begin{array}[]{lcl}\Delta^{2}v_{1}(x)=V_{1}(x)&&{\rm in}\
\,B(q,2\delta)\\\ v_{1}(x)=\Delta v_{1}(x)=0&&{\rm on}\ \,\partial
B(q,2\delta),\end{array}\right.$

and $v_{2}$ solves

$\left\\{\begin{array}[]{lcl}\Delta^{2}v_{2}(x)=V_{2}(x)+\eta(x)&&{\rm in}\
\,B(q,2\delta)\\\ v_{2}(x)=\Delta v_{2}(x)=0&&{\rm on}\ \,\partial
B(q,2\delta),\end{array}\right.$

Choosing $\tilde{\delta}=32\pi^{2}-1$ in (4.13), we find
$\int_{B(q,2\delta)}e^{\kappa|v_{1}|}\leq\int_{B(q,2\delta)}e^{\frac{|v_{1}|}{\parallel
V_{1}\parallel_{L^{1}(B(q,2\delta))}}}\leq C$ based on the smallness of
$\epsilon$. Then by standard elliptic regularity theory, we have $v_{2}\in
L^{\infty}(B(q,2\delta))$. Consequently, $e^{\kappa|v|}\in
L^{1}(B(q,2\delta),g)$. Since $\kappa$ is large it is possible to use
H${\rm\ddot{o}}$lder inequality to obtain $V\in
L^{p^{\prime}}(B(q,2\delta),\hat{g})$ for some $p\in(1,\frac{4}{s})$ if $s\geq
0$ and $p\in(1,+\infty)$ if $s\leq 0$. Thus we have proved $v\leq C$ in
$B(q,2\delta)$.

From the $L^{\infty}$ bound of $v$ we can use two positive constants $c_{1}$
and $c_{2}$ to bound $V$ from above and below

(4.28) $\frac{c_{1}}{d_{g}(x,q)^{s}}\leq
V(x)=\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}(x)}\leq\frac{c_{2}}{d_{g}(x,q)^{s}},\quad\mbox{if}\quad\gamma>0.$

Next, we aim to derive a contradiction by taking advantage of the Pohozaev
identity (3.9) in Section 3. Set
$h(x)=\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}$ and $\Omega=B(q,r)$ in (3.9),
then direct computation and (A.1) give rise to

$\int_{\Omega}x^{i}\partial_{i}\big{(}\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}\big{)}e^{4\mathfrak{u}_{0}}=\int_{B(q,r)}\big{(}x^{i}\partial_{i}\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}}+x^{i}\mathfrak{h}_{0}\partial_{i}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}}\big{)}$

and

$\begin{split}\partial_{i}d_{g}(x,q)^{4\gamma}=&4\gamma
d_{g}(x,q)^{4\gamma-1}\partial_{i}d_{g}(x,q)=4\gamma
d_{g}(x,q)^{4\gamma-1}\big{(}\partial_{i}|x-q|+C\big{)}\\\ =&4\gamma
d_{g}(x,q)^{4\gamma-2}\big{(}(x-q)^{i}+C\big{)}\end{split}$

Immediately, together with (A.1) we get

$\begin{split}&\int_{\Omega}\frac{1}{2}(x-q)^{i}\partial_{i}\big{(}\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}\big{)}e^{4\mathfrak{u}_{0}}\\\
=&\int_{B(q,r)}2\gamma\mathfrak{h}_{0}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}}+\int_{B(q,r)}\frac{1}{2}\partial_{\nu}\mathfrak{h}_{0}d_{g}(x,q)^{4\gamma+1}e^{4\mathfrak{u}_{0}}\end{split}$

Therefore, for $r\to 0$

(4.29) ${\rm(LHS)\ \,of\ \,}(\ref{PI-
mfd})=2(1+\gamma)\int_{B(q,r)}\mathfrak{h}_{0}d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{0}}+O(r)=(1+\gamma)\beta+o_{r}(1),$

where $\lim_{r\to 0}o_{r}(1)=0$. Denote the four integrals on the right hand
side of (3.9) by $I_{1}$, $I_{2}$, $I_{3}$ and $I_{4}$, respectively. Thanks
to the expansions of $g$, which are in (3.8), we obtain that

$|I_{2}|\leq
C\int_{B(q,r)}\Big{(}r|\nabla^{2}\mathfrak{u}_{0}||\nabla\mathfrak{u}_{0}|+|x-q||\nabla^{2}\mathfrak{u}_{0}||\nabla\mathfrak{u}_{0}|+r|x-q||\nabla^{2}\mathfrak{u}_{0}|+|x-q||\nabla\mathfrak{u}_{0}|\Big{)},$
$|I_{3}|\leq C\int_{\partial
B(q,r)}\Big{(}|x-q|^{2}|\nabla\mathfrak{u}_{0}|^{2}+O(r^{3})|\nabla\mathfrak{u}_{0}|^{2}\Big{)},$
$|I_{4}|\leq
C\int_{B(q,r)}\Big{(}|x-q||\nabla\mathfrak{u}_{0}|^{2}+|x-q|^{2}|\nabla^{2}\mathfrak{u}_{0}||\nabla\mathfrak{u}_{0}|+O(r^{2})|\nabla\mathfrak{u}_{0}|^{2}+O(r^{4})|\nabla^{2}\mathfrak{u}_{0}|\Big{)},$

Next, we shall estimate $|\nabla^{j}\mathfrak{u}_{0}|$ in $B(q,r)$ for
$j=1,2,3$. Recalling (4.22)$\sim$(4.24), it is important to consider the
$\nabla^{j}v$ in $B(q,r)$ for $j=1,2,3$.By means of the Green’s representation
formula, we observe that

$\begin{split}|\nabla^{j}v(x)|\leq
C\int_{B(q,2\delta)}\frac{1}{d_{g}(x,y)^{j}}V(y)+O(1)\end{split}$

In order to estimate the integral in the inequality above, we decompose
$B(q,2\delta)$ into two parts

$\Omega_{1}=B(q,2\delta)\cap\Big{\\{}d_{g}(x,y)\leq\frac{d_{g}(x,q)}{2}\Big{\\}},\quad\Omega_{2}=B(q,2\delta)\setminus\Omega_{1}.$

In this estimate we use $V(y)=O(1)d_{g}(y,q)^{-s}$ in (4.28). Hence

(4.30) $\int_{\Omega_{1}}\frac{1}{d_{g}(x,y)^{j}}V(y){\rm
d}V_{g}(y)\leq\frac{C}{d_{g}(x,q)^{s}}\int_{B(x,\frac{d_{g}(x,q)}{2})}\frac{1}{d_{g}(x,y)^{j}}{\rm
d}V_{g}(y)\leq Cd_{g}(x,q)^{4-s-j}.$

Using (4.28) again, we obtain that

$\begin{split}\tilde{I}:=\int_{\Omega_{2}}\frac{1}{d_{g}(x,y)^{j}}V(y){\rm
d}V_{g}(y)\leq
C\int_{\Omega_{2}}\frac{1}{d_{g}(x,y)^{j}}\frac{1}{d_{g}(y,q)^{s}}{\rm
d}V_{g}(y).\end{split}$

Fixing some $t$ as follows

(4.31) $t\in\left\\{\begin{array}[]{lcl}(0,\frac{4}{s}),&&s>0,\\\
(1,+\infty),&&s\leq 0,\end{array}\right.$

we have $-ts>-4$. It follows from the H$\ddot{\rm o}$lder inequality that

$\begin{split}\tilde{I}\leq
C\Big{(}\int_{\Omega_{2}}\frac{1}{d_{g}(x,y)^{jt^{*}}}{\rm
d}V_{g}(y)\Big{)}^{\frac{1}{t^{*}}}\Big{(}\int_{\Omega_{1}}\frac{1}{d_{g}(y,q)^{st}}{\rm
d}V_{g}(y)\Big{)}^{\frac{1}{t}}\leq
C\Big{(}\int^{\tilde{c}}_{\frac{d_{g}(x,q)}{2}}\frac{1}{\rho^{jt^{*}-3}}{\rm
d}\rho\Big{)}^{\frac{1}{t^{*}}}\end{split}$

where $t^{*}=\frac{t}{t-1}$ denotes the conjugate of $t$ and $\tilde{c}$ is
some positive constant. Then direct computation and the fact $-ts>-4$ imply
that

(4.32) $\tilde{I}=\int_{\Omega_{2}}\frac{1}{d_{g}(x,y)^{j}}V(y){\rm
d}V_{g}(y)\leq\left\\{\begin{array}[]{lcl}C|\log
d_{g}(x,q)|^{\frac{1}{t^{*}}},&&{\rm if}\ \,jt^{*}=4,\\\
Cd_{g}(x,q)^{\frac{4}{t^{*}}-j}+C,&&{\rm if}\ \,jt^{*}\neq
4.\end{array}\right.$

In view of (4.31), we get that

$t^{*}\in\left\\{\begin{array}[]{lcl}(\frac{4}{4-s},+\infty),&&s>0,\\\
(1,+\infty),&&s\leq 0.\end{array}\right.$

Hence, there holds $\frac{4}{t^{*}}-j<4-s-j$. Consequently, from (4.30) and
(4.32) there exists some $\tau>0$ such that for any $r\in(0,\delta)$

(4.33) $|\nabla^{j}v(x)|\leq Cd_{g}(x,q)^{\tau-j}+C,\quad j=1,2,3,\quad x\in
B(q,r).$

In fact, we may choose $\tau\in(0,1)$ if $jt^{*}=4$, and otherwise
$\tau=\frac{4}{t^{*}}$. At this point, we obtain that

(4.34) $|\nabla^{j}\mathfrak{u}_{0}(x)|\leq Cd_{g}(x,q)^{-j},\quad
j=1,2,3\quad x\in B(q,r).$

Thus by virtue of (A.1) and (4.33), we may adjust $\tau>0$ such that on
$\partial B(q,r)$

$\begin{split}&\frac{\partial\mathfrak{u}_{0}}{\partial
r}=-\frac{\beta}{8\pi^{2}}\frac{1}{|x-q|}+O(r^{\tau})\frac{1}{|x-q|}+O(1),\\\
&\Delta\mathfrak{u}_{0}=-\frac{\beta}{4\pi^{2}}\frac{1}{|x-q|^{2}}+O(r^{\tau})\frac{1}{|x-q|^{2}}+O(1),\\\
&\frac{\partial\Delta\mathfrak{u}_{0}}{\partial
r}=\frac{\beta}{2\pi^{2}}\frac{1}{|x-q|^{3}}+O(r^{\tau})\frac{1}{|x-q|^{3}}+O(1).\end{split}$

Therefore, the estimates of $I_{2}$, $I_{3}$ and $I_{4}$ can be improved:

$\begin{split}|I_{2}|&\leq C\int_{B(q,r)}\big{(}rd^{-3}+d^{-2}+1\big{)}{\rm
d}V_{g}\leq Cr^{2},\\\ |I_{3}|&\leq C\int_{\partial
B(q,r)}r^{2}d_{g}(x,q)^{-2}{\rm d}V_{g}\leq Cr^{3},\\\ |I_{4}|&\leq
C\int_{B(q,r)}d_{g}(x,q)^{-1}{\rm d}V_{g}\leq Cr^{3}.\\\ \end{split}$

Finally for $I_{1}$ and we use the expansions of $g^{ij}$ to obtain

$\begin{split}I_{1}=&\int_{\partial
B(q,r)}\big{(}-r\nu_{i}\partial_{i}(\Delta_{g}\mathfrak{u}_{0})\partial_{\nu}\mathfrak{u}_{0}+\Delta_{g}\mathfrak{u}_{0}\partial_{\nu}\mathfrak{u}_{0}+(x-q)^{k}\nu_{i}\Delta_{g}\mathfrak{u}_{0}\partial_{ik}\mathfrak{u}_{0}-\frac{1}{2}r(\Delta_{g}\mathfrak{u}_{0})^{2}\big{)}\\\
&+O(r)\\\ =&\int_{\partial
B(q,r)}\Big{(}-r\nu_{i}\partial_{i}(\Delta\mathfrak{u}_{0})\partial_{\nu}<x-q,\mathfrak{u}_{0}>_{flat}+\Delta\mathfrak{u}_{0}\partial_{\nu}\mathfrak{u}_{0}-\frac{1}{2}r(\Delta\mathfrak{u}_{0})^{2}\Big{)}+O(r),\end{split}$

where we have used $\det\,(g)=1$ in $B(q,\delta)$ and
$\Delta_{g}u=\partial_{i}(g^{ij}\partial_{j}u)$. Consequently,

(4.35) ${\rm(RHS)\ \,of\ \,}(\ref{PI-
mfd})=I_{1}+o_{r}(1)=\frac{\beta^{2}}{16\pi^{2}}+o_{r}(1).$

Combining (4.29) and (4.35), we derive that $\beta=16\pi^{2}(1+\gamma)$, which
yields a contradiction to (4.26) in the case $\gamma>0$. Therefore
$\mathfrak{u}_{k}\to-\infty$ uniformly on any compact subset of
$B(q,2\delta)\setminus\\{q\\}$,
$\mathfrak{h}_{k}(x)d_{g}(x,q)e^{4\mathfrak{u}_{k}}(x)\to 0$ uniformly on any
compact subset of $B(q,2\delta)\setminus\\{q\\}$ and

$\mathfrak{h}_{k}(x)d_{g}(x,q)e^{4\mathfrak{u}_{k}}(x)\to\beta\delta_{q}\
\,{\rm\ \,weakly\ \,in\ \,the\ \,sense\ \,of\ \,measure\ \,on\ \,}M.$

In the end, we show the quantization $\beta$ is exactly $16\pi^{2}(1+\gamma)$.
To see this, set $c_{k}=\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int\mathfrak{u}_{k}{\rm d}\sigma$ and
$\check{u}_{k}(x)=\mathfrak{u}_{k}(x)-c_{k}$. Then, we have $c_{k}\to-\infty$
and $\check{u}_{k}\to\check{u}$ in $C^{4}(B(q,\delta))$ as $k\to+\infty$.
Moreover, there exists a smooth function $\check{v}$ such that
$\check{u}(x)=\frac{\beta}{8\pi^{2}}\log d_{g}(x,q)+\check{v}$. Taking
advantage of the Pohozaev identity as before, we obtain
$\beta=16\pi^{2}(1+\gamma)$ and

$2\mathfrak{h}_{k}(x)d_{g}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}\to
16\pi^{2}(1+\gamma)\delta_{q}.$

∎

Based on Theorem 4.3 and its proof, we immediately obtain the following
corollaries.

######  Corollary 4.4.

Suppose that $\mathfrak{u}_{k}$ satisfies the assumptions in Theorem 4.3, then
along a subsequence, there holds

$\mathfrak{u}_{k}-c_{k}\to-2(1+\gamma)\log d_{g}(x,q)+\hat{v},\quad in\
\,C^{4}_{loc}(B(q,2\delta)\setminus\\{q\\}),$

where $c_{k}=\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.83337pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.90005pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.75003pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.25003pt}}\\!\int_{\partial B(q,\delta)}\mathfrak{u}_{k}{\rm
d}\sigma_{g}\to-\infty$ and $\hat{v}$ is a smooth function in $B(q,2\delta)$.

######  Corollary 4.5.

Suppose that $\mathfrak{u}_{k}$ satisfy

$P_{g}\mathfrak{u}_{k}(x)+2\mathfrak{b}_{k}=2H_{k}e^{4\mathfrak{u}_{k}}\quad
in\ \,M$

with $\int_{B(q_{j},2\delta)}2H_{k}e^{4\mathfrak{u}_{k}}{\rm
d}V_{g}\to\rho_{j}<16\pi^{2}(1+\gamma_{j})$ for some $j\in\\{1,\cdots,N\\}$.
Then $\\{\mathfrak{u}_{k}\\}$ is uniformly bounded from above on any subset of
$B(q_{j},2\delta)$. In particular, $\\{\mathfrak{u}_{k}\\}$ can not blow up in
$B(q_{j},2\delta)$.

##  5\. Concentration-Compactness Result and A Priori Estimate

In this section, we aim to establish the concentration-compactness principle
and a priori estimate basing on the result in the section 4. Indeed, we will
derive the following concentration-compactness type result for the regular
part of $\\{u_{k}\\}$.

Set

$\rho_{k}=\int_{M}2H_{k}e^{4\mathfrak{u}_{k}}{\rm d}V_{\mathfrak{u}_{k}}.$

######  Theorem 5.1 (Concentration-Compactness).

Let $\\{\tilde{u}_{k}\\}$ be a sequence of solution to (3.5) and (3.7) with
$\rho_{k}\to\rho$. Then there exists a subsequence, still denoted
$\\{\tilde{u}_{k}\\}$, for which one of the following alternative holds:

  * (i)

$\sup_{\varSigma}|\tilde{u}_{k}|\leq C_{\varSigma}$, for any
$\varSigma\subset\subset M$.

  * (ii)

$\sup_{\varSigma}\tilde{u}_{k}\to-\infty$, for any $\varSigma\subset\subset
M$.

  * (iii)

There exist a finite set $S=\\{p^{1},\cdots,p^{m}\\}\subset M$ with
$m\in\mathbb{N}$, and sequences of points
$\\{x_{k}^{1}\\}_{k\in\mathbb{N}},\cdots,\\{x_{k}^{m}\\}_{k\in\mathbb{N}}\subset
M$, such that for all $i=1,\cdots,m$

$x_{k}^{i}\to p^{i},\quad\sup_{\varSigma}\tilde{u}_{k}\to-\infty\ for\ \,any\
\,\varSigma\subset M\setminus S$

and

$2H_{k}e^{4\tilde{u}_{k}}\to\sum_{i=1}^{m}\beta_{i}\delta_{p^{i}}\ weakly\
\,in\ \,the\ \,sense\ \,of\ \,measures\ \,in\ \,M.$

Futhermore, $\beta_{i}\in 16\pi^{2}\mathbb{N}$ if
$p^{i}\notin\\{q_{1},\cdots,q_{N}\\}$, and $\beta_{i}=16\pi^{2}(1+\gamma)$ if
$p^{i}=q_{j}$ for some $j\in\\{1,\cdots,N\\}$.

###### Proof of Theorem 5.1.

We define $S$ to be the set of blow-up points of $\mathfrak{u}_{k}$ in $M$,
that is,

$S=\\{x\in M:\ \,\exists x_{k}\in M,\ \,{\rm s.t.}\ \,x_{k}\to x\ \,{\rm and}\
\,\tilde{u}_{k}(x_{k})\to+\infty\ \,{\rm as}\ \,k\to+\infty\\}.$

We distinguish two cases.

Case 1: $S\neq\emptyset$.

For $p\in S$, Lemma 4.1 say that the mass of $\\{\tilde{u}_{k}\\}$ near $p$ is
no less than $8\pi^{2}(1+\gamma)$. Then finite integral assumption
$\int_{M}H_{k}e^{4\tilde{u}_{k}}{\rm d}V_{g}\leq C$ implies ${\rm
card}\,(S)\leq C$. We may denote $S=\\{p_{1},\cdots,p_{m}\\}$ with some
$n\in\mathbb{N}$. Therefore, there exists $r_{0}\in(0,1)$ such that for any
$p^{i}\in S$, $p^{i}$ is the only blow-up point of $\tilde{u}_{k}$ in
$B(p^{i},r_{0})$. Therefore, from the results in [22] and Theorem 4.3, we
obtain the alternative (iii).

Case 2: $S=\emptyset$.

In this case, we have $\sup_{M}\tilde{u}_{k}\leq C$, which implies
$H_{k}e^{4\tilde{u}_{k}}$ is uniformly bounded in $M$. Taking into account of
the Green’s representation formula,

$\tilde{u}_{k}(x)-\bar{\tilde{u}}_{k}=\int_{M}G(x,y)H_{k}(y)e^{4\tilde{u}_{k}}{\rm
d}V_{g}(y)=O(1).$

Hence, after taking a subsequence, the alternative (i) occurs if
$\limsup_{k\to+\infty}\int_{M}\tilde{u}_{k}{\rm d}V_{g}>-\infty$, the
alternative (ii) holds if $\limsup_{k\to+\infty}\int_{M}\tilde{u}_{k}{\rm
d}V_{g}\to-\infty$,

∎

Immediately, we derive the following two corollaries from Theorem 5.1.

######  Corollary 5.2.

Suppose that $\tilde{u}_{k}$ satisfies the assumption in Theorem 5.1 and
alternative (iii) occurs, then $\rho\in\Gamma$.

######  Corollary 5.3.

Suppose that $\tilde{u}_{k}$ satisfies the assumption in Theorem 5.1. Then for
every $\rho\in\mathbb{R}^{+}\setminus\Gamma$, there exists a constant $C$ only
depending on $\rho$, such that $\tilde{u}_{k}\leq C$. In particular, if we
additionally assume $\int_{M}\tilde{u}_{k}{\rm d}V_{g}=0$ or
$\int_{M}H_{k}e^{4\tilde{u}_{k}}{\rm d}V_{g}\geq c$ for some constant $c>0$,
then

$\parallel\tilde{u}_{k}\parallel_{L^{\infty}(M)}\leq C.$

The following result explains that $\Gamma$ is some critical set to (1.8) or
(1.9).

######  Proposition 5.4 (Critical Set).

Suppose Ker $(P_{g})=\\{constants\\}$ and that $\\{u_{k}\\}$ is a sequence of
solutions to (1.9)$\sim$(1.10) with the coefficients satisfying (1.11). If the
blow-up phenomena occur, then $\int_{M}2b{\rm d}V_{g}\in\Gamma$.

###### Proof of Propertion 5.4 and Theorem 1.1.

Since $\int_{M}H_{k}e^{4\mathfrak{u}_{k}}{\rm
d}V_{g}=\int_{M}h_{k}e^{4u_{k}}{\rm d}V_{g}=\int_{M}b_{k}{\rm d}V_{g}$
Proposition 5.4 and Theorem 1.1 obviously follow from the two corollaries
above.

∎

##  6\. A spherical Harnack inequality

As a corollary of the Concentration-Compactness theorem, we derive a spherical
Harnack inequality near a singular source $q_{j}$ when its strength
$\gamma_{i}$ is not an integer. Recall that if $q_{j}$ is a blow-up point of
$\mathfrak{u}_{k}$, then there exist $\delta>0$ and a sequence
$\\{x_{k}^{j}\\}_{k\in\mathbb{N}}$ such that
$\mathfrak{u}_{k}(x_{k})=\max_{B(q_{j},\delta)}\mathfrak{u}_{k}\to+\infty$ and
$S\cap B(q_{j},\delta)=\\{q_{j}\\}$. Thus, we have

$\varepsilon_{k}^{j}=\exp\Big{(}-\frac{\mathfrak{u}_{k}(x_{k}^{j})}{1+\gamma_{j}}\Big{)}\to
0,\quad{\rm as}\;k\to+\infty.$

######  Corollary 6.1.

Suppose that $\tilde{u}_{k}$ satisfies the assumption in Theorem 5.1 and
$q_{j}$ is a blow-up point of $\tilde{u}_{k}$. If
$\gamma_{j}\notin\mathbb{N}$, then
$d_{{g}}(x_{k}^{j},q_{j})=O(\varepsilon_{k}^{j})$, that is

$y_{k}^{j}:=\frac{x_{k}^{j}-q_{j}}{\varepsilon_{k}^{j}}\to
y^{0}\in\mathbb{R}^{4}$

and $\tilde{u}_{k}$ has only one bubble at $q_{j}$.

###### Proof.

Otherwise, we assume that

$\frac{x_{k}^{j}-q_{j}}{\varepsilon_{k}^{j}}\to+\infty\quad{\rm as}\
\,k\to+\infty.$

For simplicity, we omit the superscript $j$ and work on the $\mathfrak{u}_{k}$
under the conformal normal coordinate. Let $r_{k}=d_{g}(x_{k},q)$ and define
the map $\varphi_{k}:B(0,\delta r_{k}^{-1})\to B(q,\delta)$ by

$\varphi_{k}:y\mapsto r_{k}y+q,$

where on the right hand side we are using the conformal normal coordinates on
$B(q,\delta)$. We use the notation $\breve{f}=\varphi_{k*}f=f\circ\varphi_{k}$
to denote the pull-back of a function $f$ defined on $B(q,\delta)$, and we let
$\breve{g}_{k}=r_{k}^{-2}\varphi_{k*}g$ be the blow-up metric, i.e., a
rescaling of the pull-back metric.

We define

$v_{k}(y)=\breve{\mathfrak{u}}_{k}+(1+\gamma)\log r_{k},\quad y\in
B(0,\delta/r_{k}).$

Then $v_{k}$ satisfies

$\left\\{\begin{array}[]{lcl}P_{g_{k}}v_{k}+2\breve{\mathfrak{b}}_{k}=2\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}},\quad{\rm
in}\ \,B(0,\delta/r_{k}),\\\
\int_{B(0,\delta/r_{k})}\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}}{\rm
d}V_{g_{k}}\leq C,\end{array}\right.$

where $\breve{\mathfrak{b}}_{k}(y)=\mathfrak{b}_{k}(r_{k}y+q)$,
$\breve{\mathfrak{h}}_{k}(y)=\mathfrak{h}_{k}(r_{k}y+q)$.

Denote $e_{k}=\frac{x_{k}-q}{r_{k}}$, then $e_{k}\to e_{0}\in
S^{1}\subset\mathbb{R}^{4}$ and $e_{0}$ is a blow-up point of $\tilde{v}_{k}$.
Applying Theorem 5.1, there exists a finite blow-up set
$\tilde{S}=\\{y^{1},\cdots,y^{l}\\}$ such that $v_{k}\to-\infty$ uniformly on
any compact subset of $\mathbb{R}^{4}\setminus\tilde{S}$ and

$2\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}}\to\sum_{i=1}^{l}\alpha_{i}\delta_{y^{i}},\quad{\rm
weakly\ \,in\ \,the\ \,sense\ \,of\ \,measure},$

where

$\alpha_{i}=\lim_{k\to+\infty}\int_{B(y^{i},r_{0})}2\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}(y)}{\rm
d}V_{g_{k}}\quad{\rm and}\quad B(y^{i},r_{0})\cap\tilde{S}=\\{y^{i}\\}.$

First, we claim that $0$ is not a blow-up point of $v_{k}$. Suppose
$0\in\tilde{S}$, then by means of Theorem 5.1 we have

$\lim_{k\to+\infty}\int_{B(0,r_{0})}\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}(y)}{\rm
d}V_{g_{k}}=16\pi^{2}(1+\gamma).$

On the other hand, since $e_{0}\in S^{1}$ is also a blow-up point of $v_{k}$
in $B(0,R)$ for $R$ large. Then

$\begin{split}16\pi^{2}(1+\gamma)=&\lim_{k\to+\infty}\int_{B(q,\delta)}2\hat{h}_{k}(x)d_{g_{k}}(x,q)^{4\gamma}e^{4\mathfrak{u}_{k}(x)}{\rm
d}V_{g_{k}}\\\
=&\lim_{k\to+\infty}\int_{B(0,\delta/r_{k})}2\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}(y)}{\rm
d}V_{g_{k}}\\\ \geq&\lim_{k\to+\infty}\int_{B(0,r_{0})\cup
B(e_{0},r_{0})}2\breve{\mathfrak{h}}_{k}(y)d_{g_{k}}(y,0)^{4\gamma}e^{4v_{k}(y)}{\rm
d}V_{g_{k}}\\\ =&16\pi^{2}(1+\gamma)+16\pi^{2},\end{split}$

which yields a contradiction. Thus, $0\notin\tilde{S}$.

As a consequence, we have

$16\pi^{2}(1+\gamma)=16\pi^{2}l$

which is impossible, since $\gamma\notin\mathbb{N}$ while $l$ is an integer.
In other words, if $\gamma$ is not an integer, then
$d_{g}(x_{k},q)=O(\varepsilon_{k})$ and $\hat{u}_{k}$ has only one bubble at
$q$.

∎

######  Theorem 6.2 (A Spherical Harnack Inequality
($\gamma\notin\mathbb{N}$)).

Suppose that $\tilde{u}_{k}$ satisfies the assumption in Theorem 5.1 and
$q_{j}$ is a blow-up point of $\tilde{u}_{k}$. If
$\gamma_{j}\notin\mathbb{N}$, then near $q_{j}$ there holds the following
spherical Harnack inequality:

(6.1) $\max_{x\in
B(q_{j},\delta)}\\{\tilde{u}_{k}(x)+(1+\gamma_{j})\log|x-q|\\}\leq C$

with some constant $C$.

###### Proof.

Suppose that (6.1) fails, then there exists a sequence $\\{x_{k}\\}\subset
B(q_{j},\delta)$ such that $\max_{x\in
B(q_{j},\delta)}\\{\tilde{u}_{k}(x)+(1+\gamma_{j})\log|x-q|\\}\to+\infty$. By
means of the selection process and quantization property in [33, 22], we
obtain that $\int_{B(q,\delta)}H_{k}(x)e^{4\tilde{u}_{k}(x)}{\rm d}V_{g}\to
16\pi^{2}n$ for some positive integer $n$. However, from Theorem 5.1, we know
that

(6.2) $\int_{B(q,\delta)}H_{k}(x)e^{4\tilde{u}_{k}(x)}{\rm d}V_{g}\to
16\pi^{2}(1+\gamma),\quad as\ \,k\to+\infty,$

which is impossible.

∎

## Appendix: Comparsion between $d_{g}(x,q)$ and $|x-q|$

In this appendix, we will establish the comparison between the distance
$d_{g}(x,q)$ and its derivatives and their Euclidean counterparts as in the
Appendix B in [51]. We will follow the argument in [51] and give the detail
for completeness. We claim that for $j=0,1,2,3$, there holds

(A.1) $\nabla^{j}\big{(}\log|x-y|-\log d_{g}(x,y)\big{)}=O(r^{2-j}),\quad x\in
B(q,2r)\setminus B(q,r/2).$

We recall that $g$ is the conformal normal metric centered at $q$, and we
identity $x,y\in T_{q}M$ with $\exp_{q}x$ and $\exp_{q}y$ respectively, where
$\exp_{q}$ is the exponential map at $q$ with respect to the metric $g$. Thus,

$d_{g}(x,y)=d(\exp_{q}x,\exp_{q}y),\quad x,y\in B(q,\delta).$

Also, we denote $\nabla_{g}$ by $\nabla$ for convenience. First, let us note
that the following simple estimates on $d$ hold:

(A.2) $\big{|}\nabla^{j}(\log d_{g}(x,y))\big{|}\leq C|x-y|^{-j},\quad
j=1,2,3,4.$

Set

$f(x)=\log|x-y|-\log d_{g}(x,y),\quad x\in B(q,2r)\setminus B(q,r/2).$

We aim to show that

$\big{|}\nabla^{j}f(x)\big{|}\leq C|x-q|^{2-j},\quad j=0,1,2,3,\quad x\in
B(q,2r)\setminus B(q,r/2).$

Let $R$, $R_{ij}$ and $R_{ijkl}$ respectively denote the scalar, Ricci and
Riemann curvature of $g$. From the definitions of $g$ and $R^{i}_{jkl}$, we
obtain that

$\nabla^{j}R^{i}_{jkl}(x)=O(1),\quad j=1,2.$

In conformal normal coordinates, there holds $R(q)=R_{ij}(q)=|\nabla R(q)|=0$.
As a consequence, we have futher

$R(x)=O(r^{2}),\quad R_{ij}(x)=O(r).$

We shall derive an estimate on $\Delta_{g}^{2}f(x)$. By the definition of $g$
and (A.4) in [51], that is

$P_{g,y}\Big{(}-\frac{1}{8\pi^{2}}\chi(r)\log
d_{g}(x,y)\Big{)}=\delta_{x}+E(x,y),\quad{\rm with\ \,}E\ \,{\rm bounded},$

we have that

(A.3) $P_{g}\log d_{g}(x,q)=O(r^{4}),\quad x\in B(q,2r)\setminus B(q,r/2).$

In view of the rough estimates (A.2), we can estimate the term:

$\big{(}P_{g}-\Delta_{g}^{2}\big{)}\log
d_{g}(x,q)=\partial_{m}\Big{(}g^{mi}\big{(}\frac{2}{3}R(x)g_{ij}-2R_{ij}(x)\big{)}g^{lj}\partial_{j}\big{(}\log
d_{g}(x,q)\big{)}\Big{)}=O(r^{-1}).$

Therefore, we can get

$\Delta_{g}^{2}\big{(}\log d_{g}(x,q)\big{)}=O(r^{-1}),\quad x\in
B(q,2r)\setminus B(q,r/2).$

Finally, we consider the term $\Delta_{g}^{2}\big{(}\log|x-q|\big{)}$. Since
$\Delta^{2}\big{(}\log|x-q|\big{)}=0$, it suffices to estimate
$\Delta_{g}^{2}-\Delta^{2}$. For any function $u$, the direct computation
leads to

(A.4)
$\begin{split}\Delta_{g}^{2}u=&g^{ab}g^{ij}\partial_{ijab}u+2\partial_{ija}u\big{(}\partial_{b}g^{ab}g^{ij}+g^{ab}\partial_{b}g^{ij}\big{)}\\\
&+\partial_{ij}u\big{(}\partial_{a}g^{ab}\partial_{b}g^{ij}+2g^{ai}\partial_{ab}g^{bj}+g^{ab}\partial_{ab}g^{ij}+\partial_{a}g^{ia}\partial_{b}g^{bj}\big{)}\\\
&+\partial_{j}u\big{(}\partial_{a}g^{ab}\partial_{ib}g^{ij}+g^{ab}\partial_{iab}g^{ij}\big{)},\end{split}$

where we have used $\det\,(g)=1$. Using the expansion of $g^{ab}$:

$g^{ab}(x)=\delta_{ab}-\frac{1}{3}R_{mabl}(q)x^{a}x^{b}+O(|x-q|^{3}),$

and replacing $u$ by $\log|x-q|$ in (A.4) above, we obtain that

$\Delta_{g}^{2}\big{(}\log|x-q|\big{)}=O(r^{-2}).$

Consequently,

(A.5) $\Delta_{g}^{2}f(x)=O(r^{-2}),\quad x\in B(q,2r)\setminus B(q,r/2).$

An estimate on the $L^{\infty}$-norm of $f(x)$ can easily be seen as follows:

(A.6)
$d_{g}(x,q)=\int_{0}^{x(t)}\sqrt{x_{i}^{\prime}(t)g_{ij}(t)x_{j}^{\prime}(t)}{\rm
d}t=|x-q|\big{(}1+O(r^{2})\big{)},$

which implies

(A.7) $f(x)=O(r^{2}),\quad x\in B(q,2r)\setminus B(q,r/2).$

Applying the elliptic theory to (A.5) and (A.7), we obtain the claim (A.1) and

(A.8) $d_{g}(x,q)=|x-q|(1+O(r^{2})).$

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