Quantization of the Blow-Up Value for the Liouville Equation with Exponential Neumann Boundary Condition

Abstract

In this paper, we analyze the asymptotic behavior of solution sequences of the Liouville-type equation with Neumann boundary condition. In particular, we will obtain a sharp mass quantization result for the solution sequences at a blow-up point.

1 Introduction

Let $\Omega$ be a bounded domain in ${\mathbb R}^2$. The so-called Liouville equation is

$$\begin{aligned} -\Delta u =V(x)e^{2u} \quad \text { in }\Omega , \end{aligned}$$

(1.1)

which was first studied by Liouville in 1853 in [14]. In 1991, Brezis and Merle [1] initiated the study of the blow-up analysis for the Liouville equation. Under the finite energy condition, they first showed that any solution belongs to $L^\infty$, and further, they analyzed the convergence of a sequence of solutions and obtained a concentration–compactness type result. Their results initiate many works on the asymptotic behavior of blow-up solutions, for there are many applications in geometric and physical problems, for example, in the problem of prescribing Gaussian curvature [3, 4, 7], in the theory of the mean field equation [5, 8, 9], and in the Chern–Simons theory [10, 17,18,19,20]. See also the reference therein.

In particular, in the celebrated paper by Li and Shafrir [15], they initiated to evaluate the blow-up value at the blow-up point. They showed at the each blow-up point the blow-up value is quantized, i.e., there is no contribution of mass outside the m disjoint balls (whose radii is going to zero) which contain a contribution of $4\pi m$ mass for some positive integer m. Concerning the mean field equation, this kind of mass quantization leads to the crucial compactness property of solutions. Then, the existence issues can be attacked by variational methods; see [8, 9].

The aim of the present paper is to generalize the blow-up analysis for (1.1) to a Liouville-type equation with Neumann boundary condition. In other words, we consider the following Neumann boundary problem:

$$\begin{aligned} \left\{ \begin{array}{ll} -\,\Delta u =V(x)e^{2u} \quad &{}\text { in }\;\Omega , \\ \displaystyle \frac{\partial u}{\partial n}= h(x)e^{u} \quad &{}\text { on }\;L. \\ \end{array} \right. \end{aligned}$$

(1.2)

Here L is a proper subset of $\partial \Omega$, and V(x) and h(x) are nonnegative functions. This problem plays a very important role in the study of the construction of prescribed Gaussian curvature surfaces with prescribed geodesic curvature on their boundary.

Guo and Liu [11] have analyzed the asymptotic behavior of solutions in the case $V(x)\equiv 0$. Their problem is

$$\begin{aligned} \left\{ \begin{array}{ll} -\,\Delta u =0 \quad &{}\text { in }\;\Omega , \\ \displaystyle \frac{\partial u}{\partial n} +\beta = h(x)e^{u} \quad &{}\text { on }\;\partial \Omega . \\ \end{array} \right. \end{aligned}$$

They obtained a Brezis–Merle type concentration–compactness phenomena and a Li-Shafrir type energy quantization result.

In this paper, we pursue this line of investigation on more general class of system (1.2). Now we introduce some notions firstly. Define

$$\begin{aligned} B^{+}_{R}(x_{0})= & {} \left\{ x\in {\mathbb R}^2 | x-x_{0}=(s,t), s^2+t^2<R^2, t>0 \right\} , \\ L_{R}(x_{0})= & {} \left\{ x\in {\mathbb R}^2 | x-x_{0}=(s,0), |s|<R \right\} =\partial B^{+}_{R}(x_{0}) \cap \partial {\mathbb R}^2_{+}, \\ S^{+}_{R}(x_{0})= & {} \left\{ x\in {\mathbb R}^2 | x-x_{0}=(s,t), s^2+t^2=R^2, t>0 \right\} =\partial B^{+}_{R}(x_{0}) \cap {\mathbb R}^2_{+}. \end{aligned}$$

In addition, we use the notions $B^{+}_{R}$, $L_{R}$ and $S^{+}_{R}$ for $B^{+}_{R}(0)$, $L_{R}(0)$ and $S^{+}_{R}(0)$, respectively. For simplicity, we consider the following Neumann boundary value problem of Liouville equation in $B^+_R$:

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta {u} =V(x)e^{2u} \quad &{}\text { in }\;B^{+}_{R}, \\ \displaystyle \frac{\partial u }{\partial n}= h(x)e^{u} \quad &{}\text { on }\;L_{R}. \\ \end{array} \right. \end{aligned}$$

(1.3)

From [2], we have the following Brezis–Merle type concentration–compactness theorem.

Theorem 1.1

Assume that $\{u_{n}\}$ is a sequence of solution for the following problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta {u_n} =V_n(x)e^{2u_n} \quad &{}\text { in }\;B^{+}_{R}, \\ \displaystyle \frac{\partial u_n }{\partial n}= h_n(x)e^{u_n} \quad &{}\text { on }\;L_{R}, \\ \end{array} \right. \end{aligned}$$

(1.4)

with the energy conditions

$$\begin{aligned} \int _{B^{+}_{R}}e^{2u_{n}}\mathrm{d}x\le C;\quad \quad \int _{L_{R}}e^{u_{n}}\mathrm{d}s \le C. \end{aligned}$$

(1.5)

Here $V_n(x)$ and $h_n(x)$ satisfy

$$\begin{aligned} 0<a<V_{n}(x)<C, \forall x\in B_R^+;\quad \quad 0<b<h_{n}(x)<C, \forall x\in L_R, \end{aligned}$$

(1.6)

for positive numbers a, b and C.

Define the blow-up set as

$$\begin{aligned} S=\left\{ x\in B^{+}_{R}\cup L_{R}, \text { there is a sequence } y_{n}\rightarrow x, \text { such that }u_{n}(y_{n})\rightarrow +\,\infty \right\} . \end{aligned}$$

Then, there exists a subsequence, denoted still by $\{u_{n}\}$, satisfying one of the following alternatives:

• (i) $\{u_{n}\}$ is bounded in $L_{loc}^{\infty } (B^{+}_{R}\cup L_{R})$,

• (ii) $\{u_{n}\} \rightarrow -\,\infty$ uniformly on compacts of $B^{+}_{R}\cup L_{R}$,

• (iii) there exists a finite blow-up set $S=\{p_{1},p_{2},\ldots ,p_{m}\}\subset B^{+}_{R}\cup L_{R}$. Moreover, $u_{n}(x) \rightarrow -\,\infty$ uniformly on compact subsets of $(B^{+}_{R}\cup L_{R}) \backslash S$, and

  $$\begin{aligned} \int _{B^{+}_{R}}V_{n}e^{2u_{n}}\phi \mathrm{d}x + \int _{L_{R}}h_{n}e^{u_{n}}\phi \mathrm{d}s \rightarrow \sum _{i=1}^m \alpha _{i}\phi (p_{i}), \end{aligned}$$

  for every $\phi \in C_o^{\infty }(B^{+}_{R}\cup L_{R})$ with $\alpha _{i}\ge \pi$.

From Theorem 1.1, the blow-up set S is nonempty if $u_{n}$ is blow-up. We can define the blow-up value at each blow-up point. For $p\in S\cap B_R^+$, we define the blow-up value at point p as:

$$\begin{aligned} m(p)=\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{B_r(p)\cap B_R^+}V_ne^{2u_n}\mathrm{d}x; \end{aligned}$$

For $p\in S\cap L_R$, we define the blow-up value at point p as:

$$\begin{aligned} m(p)=\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\left( \int _{B_r(p)\cap B_R^+}V_ne^{2u_n}\mathrm{d}x+\int _{B_r(p)\cap L_R}h_ne^{u_n}\mathrm{d}s\right) . \end{aligned}$$

Li and Shafrir [15] have shown that $m(p)=4m\pi$ for any $p\in S\cap B_R^+$. In this paper, we want further to show that $m(p)=2m\pi$ for any $p\in S\cap L_R$. Our main theorem is the following Li-Shafrir type energy quantization theorem.

Theorem 1.2

For $R>0$, let $\{V_{n}\}$ and $\{h_{n}\}$ be two sequences of functions satisfying

$$\begin{aligned} 0<a\le V_{n}\rightarrow V\in C^0(\overline{B}_{R}^+),\quad 0<b\le h_{n}\rightarrow h\in C^0(L_{R}). \end{aligned}$$

(1.7)

Let $\{u_{n}\}$ be a sequence of solutions of (1.4), (1.5) with the following properties:

$$\begin{aligned}&u_{n}(x_{n})=\max \limits _{\bar{B}_{R}^+} u_{n}\rightarrow +\,\infty , \end{aligned}$$

(1.8)

$$\begin{aligned}&\max \limits _{\bar{B}_{R}^+\backslash B^{+}_{r}(0)} u_{n}\rightarrow -\,\infty , \quad \text { for } 0<r<R. \end{aligned}$$

(1.9)

Then,

$$\begin{aligned} \alpha :=\lim \limits _{n\rightarrow \infty } \left( \int _{B^{+}_{R}} V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}e^{u_{n}}\right) =2\pi m \end{aligned}$$

(1.10)

for some positive integer m.

The proof of Theorem 1.2 follows closely the idea of Li and Shafrir in [15] where they proved the quantization of the blow-up value for the Liouville equation. The approach in [15] is based on a classification result of bubbling equation $-\,\triangle u=e^{2u}$ in ${\mathbb R}^2$ with $\int _{{\mathbb R}^2}e^{2u}<\infty$ and a “$\sup +\inf$” type inequality $u(0)+C_1\inf _{B_1}u\le C_2$ for equation $-\,\triangle u=Ve^{2u}$ in $B_1$. For our problem, we need the corresponding results. On the one hand, besides of the above bubbling equation, there exists the other kind of bubbling equation, i.e.,

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u= V(0)e^{2u} \quad &{}\text { in } {\mathbb R}^2\cap \{t>-\Lambda \} , \\ \displaystyle \frac{\partial u }{\partial n}= h(0)e^{u} \quad &{}\text { on } {\mathbb R}^2\cap \{t=-\Lambda \},\\ \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{{\mathbb R}^2\cap \{t>-\Lambda \}} V(0)e^{2u}\le C,\quad \int _{{\mathbb R}^2\cap \{t=-\Lambda \}}h(0)e^{u}\le C. \end{aligned}$$

We will use the classification result proved in [16] to handle our problem. On the other hand, we need to prove a “$\sup +\inf$” type inequality under Neumann boundary value condition.

The paper is organized as follows. In Introduction, we state the problem and the main theorems. In Sect. 2, we prove a “$\sup +\inf$” type inequality and other auxiliary results. In Sect. 3, we complete the proof of Theorem 1.2.

2 A $\sup +\inf$ Inequality Under Neumann Boundary Value Condition

In this section, we establish a “$\sup +\inf$” inequality under Neumann boundary value condition. We start to show some auxiliary lemmas.

Lemma 2.1

Under the hypotheses of Theorem 1.2, we have $\alpha \ge 2\pi$.

Proof

Let $x_{n}=(s_{n},t_{n})$, $\delta _{n}=e^{-u_{n}(x_{n})}$, then $\delta _{n}\rightarrow 0$ and $x_n \rightarrow 0$. By letting $\tilde{u}_{n}(x)=u_{n}(\delta _{n}x+x_{n})+\log \delta _{n}$, we see that

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}_{n} = V_{n}(\delta _{n}x+x_{n})e^{2\tilde{u}_{n}} \quad &{}\text { in } B_{\frac{R}{2\delta _{n}}}\cap \{t>-\frac{t_{n}}{\delta _{n}}\}, \\ \displaystyle \frac{\partial \tilde{u}_{n} }{\partial n}= h_{n}(\delta _{n}x+x_{n})e^{\tilde{u}_{n}} \quad &{}\text { on }B_{\frac{R}{2\delta _{n}}}\cap \{t=-\frac{t_{n}}{\delta _{n}}\},\\ \tilde{u}_{n}(x) \le \tilde{u}_{n}(0)=0 \quad &{}\text { in } B_{\frac{R}{2\delta _{n}}}\cap \{t\ge -\frac{t_{n}}{\delta _{n}}\}, \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{B_{\frac{R}{2\delta _{n}}}\cap \{t>-\frac{t_{n}}{\delta _{n}}\}} V_{n}(\delta _{n}x+x_{n})e^{2\tilde{u}_{n}}\le C, \qquad \int _{B_{\frac{R}{2\delta _{n}}}\cap \{t=-\frac{t_{n}}{\delta _{n}}\}}h_{n}(\delta _{n}x+x_{n})e^{\tilde{u}_{n}}\le C. \end{aligned}$$

Now we distinguish two cases.

Case (1) $\frac{t_{n}}{\delta _{n}}\rightarrow \Lambda <+\,\infty$.

In this case, by Theorem 1.1, $\{\tilde{u}_{n}\}$ admits a subsequence converging to $\tilde{u}$ in $C_{loc}^{1,\alpha }({\mathbb R}^2\cap \{t\ge -\Lambda \})$, which satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}= V(0)e^{2\tilde{u}} \quad &{}\text { in } {\mathbb R}^2\cap \{t>-\Lambda \} , \\ \displaystyle \frac{\partial \tilde{u} }{\partial n}= h(0)e^{\tilde{u}} \quad &{}\text { on } {\mathbb R}^2\cap \{t=-\Lambda \},\\ \tilde{u}(x)\le \tilde{u}(0)=0 \quad &{}\text { in } {\mathbb R}^2\cap \{t\ge -\Lambda \}, \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{{\mathbb R}^2\cap \{t>-\Lambda \}} V(0)e^{2\tilde{u}}\le C,\quad \int _{{\mathbb R}^2\cap \{t=-\Lambda \}}h(0)e^{\tilde{u}}\le C. \end{aligned}$$

It follows from the classification results in [16] that

$$\begin{aligned} \tilde{u}(s,t)=\log \frac{2\lambda }{\sqrt{V(0)}(\lambda ^2+(s-s_{0})^2+(t+\Lambda +\frac{h(0)\lambda }{\sqrt{V(0)}})^2)}, \quad \lambda >0, s_{0}\in {\mathbb R}, \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb R}^2\cap \{t>-\Lambda \}} V(0)e^{2\tilde{u}}\mathrm{d}x+\int _{{\mathbb R}^2\cap \{t=-\Lambda \}}h(0)e^{\tilde{u}}\mathrm{d}s=2\pi . \end{aligned}$$

Therefore,

$$\begin{aligned} \alpha\ge & {} \lim \limits _{n\rightarrow \infty } \left( \int _{B^{+}_{R\delta _n}(x_n)} V_{n}e^{2u_{n}}+\int _{L_{R\delta _n}(x_n)}h_{n}e^{u_{n}}\right) \\= & {} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{R}\cap \{t>-\frac{t_{n}}{\delta _{n}}\}} V_{n}\left( \delta _{n}x+x_{n}\right) e^{2\tilde{u}_{n}}+\int _{B_{R}\cap \{t=-\frac{t_{n}}{\delta _{n}}\}}h_{n}\left( \delta _{n}x+x_{n}\right) e^{\tilde{u}_{n}}\right) \\= & {} \int _{B_R\cap \{t>-\Lambda \}} V(0)e^{2\tilde{u}}+\int _{B_R\cap \{t=-\Lambda \}}h(0)e^{\tilde{u}}=2\pi +o_R(1). \end{aligned}$$

Let $R\rightarrow \infty$, we get that $\alpha \ge 2\pi$.

Case (2) $\frac{t_{n}}{\delta _{n}}\rightarrow +\,\infty$.

In this case, also by Theorem 1.1, $\{\tilde{u}_{n}\}$ admits a subsequence converging to $\tilde{u}$ in $C_{loc}^{1,\alpha }({\mathbb R}^2)$, which satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}= V(0)e^{2\tilde{u}} \quad &{}\text { in } {\mathbb R}^2, \\ \tilde{u}(x) \le \tilde{u}(0)=0 \quad &{}\text { in } {\mathbb R}^2, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{{\mathbb R}^2} V(0)e^{2\tilde{u}}\le C. \end{aligned}$$

It follows from [6] that

$$\begin{aligned} \tilde{u}(s,t)=\log \frac{\lambda }{1+\frac{V(0)}{4}\lambda ^2((s-s_{0})^2+(t-t_{0})^2)}, \quad \lambda >0, s_{0}, t_{0}\in {\mathbb R}, \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb R}^2} V(0)e^{2\tilde{u}}\mathrm{d}x=4\pi . \end{aligned}$$

Therefore,

$$\begin{aligned} \alpha\ge & {} \lim \limits _{n\rightarrow \infty } \left( \int _{B^{+}_{R\delta _n}(x_n)} V_{n}e^{2u_{n}}+\int _{L_{R\delta _n}(x_n)}h_{n}e^{u_{n}}\right) \\= & {} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{R}\cap \{t>-\frac{t_{n}}{\delta _{n}}\}} V_{n}\left( \delta _{n}x+x_{n}\right) e^{2\tilde{u}_{n}}+\int _{B_{R}\cap \{t=-\frac{t_{n}}{\delta _{n}}\}}h_{n}\left( \delta _{n}x+x_{n}\right) e^{\tilde{u}_{n}}\right) \\= & {} \int _{B_R} V(0)e^{2\tilde{u}}=4\pi +o_R(1). \end{aligned}$$

Let $R\rightarrow \infty$, we get that $\alpha \ge 4\pi$. $\square$

Corollary 2.2

Let $u_{n}$ be a sequence of solutions of (1.4) with $u_{n}(x_{n})=\max \nolimits _{\bar{B}_{R}^+} u_{n}, \forall n$. Assume $\lim \nolimits _{n\rightarrow \infty }(\int _{B^{+}_{R}} V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}(x)e^{u_{n}})=\alpha <2\pi$, then we have $u_{n}(x_{n})\le C$, where the constant C depends only on $\alpha$ and R.

The “$\sup +\inf$” inequality and the Harnark inequality for

$$\begin{aligned} -\Delta u= Ve^{u} \end{aligned}$$

(2.1)

are shown in [15], which is as following lemmas:

Lemma 2.3

[15] Let $V\in L^{\infty }(B_{R})$ satisfy $a\le V(x)\le b$, $\forall x\in B_{R}$, where a, b are positive constants. Suppose that u is a solution to (2.1) in $B_{R}$. Then

$$\begin{aligned} u(0)+C_{1}\inf \limits _{B_{R}^+} u+2(C_{1}+1)\log R \le C_{2}, \end{aligned}$$

where $C_1\ge 1$ and $C_2$ are constants depending only on a and b.

Lemma 2.4

[15] For $R>0,0<R_{0}\le R/4$, we set $\tilde{\Omega }=\{ x\in {\mathbb R}^2| R_{0}<|x|<R \}$. Let u be a solution to (2.1) in $\tilde{\Omega }$ with $\Vert V \Vert _{L^{\infty }(\tilde{\Omega })}\le C_{1}$ and $u(x)+\log |x|\le C_{2}, \forall x\in \tilde{\Omega }$ for some positive constants $C_{1}$ and $C_{2}$. Then, there exists constant $\beta \in (0,1)$ and $C_{3}$ depending only on $C_{1}$ and $C_{2}$ such that

$$\begin{aligned} \sup \limits _{\partial B_{r}} u\le C_{3}+\beta \inf \limits _{\partial B_{r}} u+2(\beta -1)\log r, \quad \forall 2R_{0}\le r\le R/2. \end{aligned}$$

Now we establish the “$\sup +\inf$” inequality and the Harnark inequality for Neumann boundary value problem.

Lemma 2.5

Let $\{V_{n}\}$ and $\{h_{n}\}$ be two sequences of functions satisfying (1.7). Let $\{u_{n}\}$ be a sequence of solutions of (1.4), (1.5) satisfying (1.8)–(1.9). Then, for each $C_{1}>1$, there exists $C_{2}$ such that

$$\begin{aligned} u_{n}(x_n)+\log r+C_{1}\left( \inf \limits _{B_{r}^+(x_n)} u_{n}+\log r \right) \le C_{2} \end{aligned}$$

(2.2)

for $0<r\le R_0\le R$ and for sufficiently large n provided $B^+_{R_0}(x_n)\subset B^+_R$.

Proof

Let $x_n=(s_n, t_n)$, then $x_{n}\rightarrow 0$. Clearly $u_n$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u_{n} =V_{n}e^{2u_{n}} \quad &{}\text { in }B^{+}_{R_0}(x_n), \\ \displaystyle \frac{\partial u_{n} }{\partial n}= h_{n}e^{u_{n}} \quad &{}\text { on }L_{R_0}(x_n). \\ \end{array} \right. \end{aligned}$$

(2.3)

We first use a slightly modified version of the arguments provided in [15]. For $\forall 0<r\le R_0$ and $\forall n\in N$, we define

$$\begin{aligned} \psi (r)=u_{n}(x_n)+\log r+C_{1}\left( \frac{1}{|S_r^+(x_n)|}\int _{S_{r}^+(x_n)}u_{n}+\log r \right) . \end{aligned}$$

We have

$$\begin{aligned} \psi '(r)= & {} \frac{1+C_{1}}{r}+\frac{C_{1}}{|S_r^+(x_n)|}\int _{S_{r}^+(x_n)}\frac{\partial u_{n}}{\partial n}\\= & {} \frac{1+C_{1}}{r}-\frac{C_{1}}{|S_r^+(x_n)|}\left( \int _{B_{r}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_r(x_n)}\frac{\partial u_{n}}{\partial n}\right) \\= & {} \frac{1+C_{1}}{r}-\frac{C_{1}}{|S_r^+(x_n)|}\left( \int _{B_{r}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_r(x_n)}h_{n}e^{u_{n}}\right) . \end{aligned}$$

So that $\psi '(r)=0$ if $\int _{B_{r}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_r(x_n)}h_{n}e^{u_{n}}=\frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}$. Note that $\frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}$ is independent of r. At this point, either $\int _{B_{R_0}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_{R_0}(x_n)}h_{n}e^{u_{n}}\le \frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}$, and then, we take $r_{n}=R_0$, or $\int _{B_{R_0}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_{R_0}(x_n)}h_{n}e^{u_{n}}> \frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}$ and we may choose $r_{n}\in (0,R_0)$, such that $\int _{B_{r_n}^+(x_n)}V_{n}e^{2u_{n}}+\int _{L_{r_n}(x_n)}h_{n}e^{u_{n}}=\frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}$. In any case, $\forall n\in N$, we have $\psi (r)\le \psi (r_{n}),0<r\le R_0$.

For $\forall x\in B_{1}\cap \{t\ge -\frac{t_n}{r_n}\}$, we define $\omega _{n}(x)=u_{n}(x_n+r_{n}x)+\log r_{n}$. We see that $\omega _n(x)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \omega _{n} = V_{n}(r_{n}x+x_n)e^{2\omega _{n}} \quad &{}\text { in } B_{1}\cap \{t> -\frac{t_n}{r_n}\}, \\ \displaystyle \frac{\partial \omega _{n} }{\partial n}= h_{n}(r_{n}x+x_n)e^{\omega _{n}} \quad &{}\text { on }B_{1}\cap \{t= -\frac{t_n}{r_n}\}.\\ \end{array} \right. \end{aligned}$$

Now we argue by contradiction and assume that $\omega _{n}(0)=u_n(x_n)+\log r_n\rightarrow +\infty$. Set $\rho _{n}=e^{-\omega _{n}(0)}$, then $\rho _{n}\rightarrow 0$. Consider the sequence of functions $\widetilde{\omega }_{n}(x)=\omega _{n}(\rho _{n}x)+\log \rho _{n}$. Then $\widetilde{\omega }_{n}$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\,\Delta \widetilde{\omega }_{n} = V_{n}\left( r_{n}\rho _{n}x+x_{n}\right) e^{2\widetilde{\omega }_{n}} \quad &{}\text { in } B_{\frac{1}{\rho _{n}}}\cap \{t> -\frac{t_n}{r_n\rho _{n}}\}, \\ \displaystyle \frac{\partial \widetilde{\omega }_{n} }{\partial n}= h_{n}\left( r_{n}\rho _{n}x+x_{n}\right) e^{\widetilde{\omega }_{n}} \quad &{}\text { on }B_{\frac{1}{\rho _{n}}}\cap \{t= -\frac{t_n}{r_n\rho _{n}}\},\\ \widetilde{\omega }_{n}(x) \le \widetilde{\omega }_{n}(0)=0 \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{B_{\frac{1}{\rho _{n}}}\cap \{t>-\frac{t_{n}}{r_{n}\rho _{n}}\}} V_{n}\left( r_{n}\rho _{n}x+x_{n}\right) e^{2\tilde{\omega }_{n}}\!\le \! C, \quad \int _{B_{\frac{1}{\rho _{n}}}\cap \{t=-\frac{t_{n}}{r_{n}\rho _{n}}\}} h_{n}(r_{n}\rho _{n}x+x_{n})e^{\tilde{\omega }_{n}}\!\le \! C. \end{aligned}$$

Then, as in Lemma 2.1, we have to analyze the following two situations:

Case (1):

: $\frac{t_{n}}{r_{n}\rho _{n}}\rightarrow \Lambda <+\,\infty$.

Case (2):

: $\frac{t_{n}}{r_{n}\rho _{n}}\rightarrow +\,\infty$.

Arguing as in Lemma 2.1, we can drive either

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{\frac{1}{\rho _{n}}}\cap \{t> -\frac{t_n}{r_n\rho _{n}}\}} V_{n}(r_{n}\rho _{n}x+x_{n})e^{2\tilde{\omega }_{n}}+ \int _{B_{\frac{1}{\rho _{n}}}\cap \{t= -\frac{t_n}{r_n\rho _{n}}\}} V_{n}(r_{n}\rho _{n}x+x_{n})e^{\tilde{\omega }_{n}}\right) =2\pi , \end{aligned}$$

or

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{\frac{1}{\rho _{n}}}\cap \{t> -\frac{t_n}{r_n\rho _{n}}\}} V_{n}(r_{n}\rho _{n}x+x_{n}\right) e^{2\tilde{\omega }_{n}}=4\pi . \end{aligned}$$

But

$$\begin{aligned}&\int _{B_{\frac{1}{\rho _{n}}}\cap \{t> -\frac{t_n}{r_n\rho _{n}}\}} V_{n}(r_{n}\rho _{n}x+x_{n})e^{2\tilde{\omega }_{n}}+ \int _{B_{\frac{1}{\rho _{n}}}\cap \{t= -\frac{t_n}{r_n\rho _{n}}\}} V_{n}(r_{n}\rho _{n}x+x_{n})e^{\tilde{\omega }_{n}}\\&\quad = \int _{B_{1}\cap \{t> -\frac{t_n}{r_n}\}}V_{n}(r_{n}x+x_{n})e^{2\omega _{n}}+\int _{B_{1}\cap \{t= -\frac{t_n}{r_n}\}}h_{n}(r_{n}x+x_{n})e^{\omega _{n}}\\&\quad = \int _{B_{r_{n}}^+(x_{n})}V_{n}e^{2u_{n}}+\int _{L_{r_{n}}(x_{n})}h_{n}e^{u_{n}}\\&\quad =\frac{(1+C_{1})|S_r^+(x_n)|}{rC_{1}}< 2\pi ,\\ \end{aligned}$$

for n sufficiently large, which is the desired contradiction. So there is a constant C such that $\omega _{n}(0)=u_n(x_n)+\log r_n\le C$. Consequently, we have

$$\begin{aligned} \psi (r)=u_{n}(x_n)+\log r+C_{1}\left( \frac{1}{|S_r^+(x_n)|}\int _{S_{r}^+(x_n)}u_{n}+\log r \right) \le C(1+C_{1})=C_{2}. \end{aligned}$$

Notice that $u_{n}$ is superharmonic and $\frac{\partial u_{n}}{\partial n}\ge 0$ from (2.3), we have $\inf \limits _{B_{r}^+(x_n)} u_{n}=\inf \limits _{S_{r}^+(x_n)} u_{n}\le \frac{1}{|S_{r}^+(x_n)|}\int _{S_{r}^+(x_n)}u_{n}$. Then, we derive the desired inequality. $\square$

Lemma 2.6

For $R>0,0<R_{0}\le R/4$, we define $T=\{x\in {\mathbb R}^2_{+}|R_{0}<|x-x_0|<R\}$. Assume that $||V||_{L^\infty (T)}\le C_1$ and $||h||_{L^\infty (\partial T\cap \partial {\mathbb R}^2_{+})}\le C_1$. Let u be a solution of

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u = Ve^{2u} \qquad &{}\text { in } T, \\ \displaystyle \frac{\partial u }{\partial n}= he^{u} \quad &{}\text { on }\partial T\cap \partial {\mathbb R}^2_{+},\\ \end{array} \right. \end{aligned}$$

with $u(x)+\log |x-x_0|\le C_{2}, \forall x\in \overline{T}$. Then, there exists constant $\beta \in (0,1)$ and $C_{3}$ such that

$$\begin{aligned} \sup \limits _{S_{r}^+(x_0)} u\le C_{3}+\beta \inf \limits _{S_{r}^+(x_0)} u+(\beta -1)\log r, \quad \forall 2R_{0}\le r\le R/2. \end{aligned}$$

Here $\beta$ and $C_3$ are dependent only on $C_1$ and $C_2$.

Proof

Without loss of generality, we assume that $x_0=0$. For $2R_{0}\le r\le R/2$, by letting $\tilde{u}(x)=u(rx)+\log r$, then $\tilde{u}(x)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u} = V(rx)e^{2\tilde{u}} \quad &{}\text { in } B_{2}^+\backslash B_{\frac{1}{2}}^+, \\ \displaystyle \frac{\partial \tilde{u} }{\partial n}= h(rx)e^{\tilde{u}} \quad &{}\text { on }L_{2}\backslash L_{\frac{1}{2}}.\\ \end{array} \right. \end{aligned}$$

For $\frac{1}{2}\le |x|\le 2$, by the given assumptions we have $\tilde{u}(x)=u(rx)+\log (r|x|)-\log |x|\le C_{2}+\log 2$. It follows that $|V(rx)|e^{2\tilde{u}(x)}\le C$ on $B_{2}^+\backslash B_{\frac{1}{2}}^+$ and $|h(rx)|e^{\tilde{u}(x)}\le C$ on $L_{2}\backslash L_{\frac{1}{2}}$. Define $\omega (x)=\frac{1}{\pi }\int _{B_{2}^+\backslash B_{\frac{1}{2}}^+}\log \frac{4}{|x-y|}V(ry)e^{2\tilde{u}(y)}+ \frac{1}{\pi }\int _{L_{2}\backslash L_{\frac{1}{2}}}\log \frac{4}{|x-y|}h(ry)e^{\tilde{u}(y)}$. Then, $\omega (x)$ is bounded in $B_{2}^+\backslash B_{\frac{1}{2}}^+$ and satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \omega = V(rx)e^{2\tilde{u}} \quad &{}\text { in } B_{2}^+\backslash B_{\frac{1}{2}}^+, \\ \displaystyle \frac{\partial \omega }{\partial n}= h(rx)e^{\tilde{u}} \quad &{}\text { on }L_{2}\backslash L_{\frac{1}{2}}.\\ \end{array} \right. \end{aligned}$$

Let $g=\omega -\tilde{u}$. Then, we have

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta g = 0 \qquad &{}\text { in } B_{2}^+\backslash B_{\frac{1}{2}}^+, \\ \displaystyle \frac{\partial g}{\partial n}= 0 \quad &{}\text { on }L_{2}\backslash L_{\frac{1}{2}}.\\ \end{array} \right. \end{aligned}$$

We conclude that g is bounded below. Then, by the Harnack inequality we get

$$\begin{aligned} \sup \limits _{S_{1}^+} (g+C)\le \beta ^{-1}\inf \limits _{S_{1}^+} (g+C), \end{aligned}$$

for some constants C and $\beta \in (0,1)$. Then, returning to the original u we obtain the desired estimates. $\square$

3 Proof of Theorem 1.2

In this section, we prove the main theorem.

Proof of Theorem 1.2

We divide the proof into two steps.

Step 1. In this step, we want to show: After passing to a subsequence, there exist m sequences of points $\{x_{n}^{(j)}=(s_{n}^{(j)},t_{n}^{(j)})\}_{j=0}^{m-1}$ in $\bar{B}_R^+$ and m sequences of positive numbers $\{k_{n}^{(j)}\}_{j=0}^{m-1}$ with $\lim \nolimits _{n\rightarrow \infty } x_{n}^{(j)}=0$ and $\lim \nolimits _{n\rightarrow \infty } k_{n}^{(j)}=\infty (0\le j\le m-1)$ such that

• (a) For any $0\le j\le m-1$, $u_{n}(x_{n}^{(j)})=\max \limits _{x\in \bar{B}_{k_{n}^{(j)}\delta _{n}^{(j)}}^+(x_{n}^{(j)})} u_{n}(x)\rightarrow \infty$;

• (b) For any $0\le j\le m-1$, $\frac{|x_{n}^{(i)}-x_{n}^{(j)}|}{k_{n}^{(j)}\delta _{n}^{(j)}}\rightarrow \infty , \forall i\ne j$, where $\delta _{n}^{(j)}=e^{-u_{n}(x_{n}^{(j)})}$;

• (c) For any $0\le j\le m-1$,

  $$\begin{aligned} \beta _j:= & {} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(j)}\delta _{n}^{(j)}}^+(x_{n}^{(j)})}V_{n}e^{2u_{n}} +\int _{L_{k_{n}^{(j)}\delta _{n}^{(j)}}(x_{n}^{(j)})}h_{n}e^{u_{n}}\right) \\= & {} \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(j)}\delta _{n}^{(j)}}^+(x_{n}^{(j)})}V_{n}e^{2u_{n}} +\int _{L_{2k_{n}^{(j)}\delta _{n}^{(j)}}(x_{n}^{(j)})}h_{n}e^{u_{n}}\right) . \end{aligned}$$

  Further, when $\frac{t_{n}^{(j)}}{\delta _{n}^{(j)}}\rightarrow \Lambda < \infty$, $\beta _{j}=2\pi$;   And when $\frac{t_{n}^{(j)}}{\delta _{n}^{(j)}}\rightarrow \infty$, $\beta _{j}=4\pi$.

• (d) $\max \limits _{x\in \bar{B}_{R}^+} \{u_{n}(x)+\log \min \limits _{0\le j\le m-1}|x-x_{n}^{(j)}| \}\le C, \forall n.$

Proof

Let $x_{n}^{(0)}=x_{n}=(s^{(0)}_{n},t^{(0)}_{n})$, $\delta _{n}^{(0)}=e^{-u_{n}(x_{n}^{(0)})}$. By letting $\tilde{u}_{n}^{(0)}(x)=u_{n}(\delta _{n}^{(0)}x+x_{n}^{(0)})+\log \delta _{n}^{(0)}$, we see that

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}_{n}^{(0)} = V_{n}(\delta _{n}^{(0)}x+x_{n}^{(0)})e^{2\tilde{u}_{n}^{(0)}} \quad &{}\text { in } B_{\frac{R}{\frac{\delta _{n}}{2}}}\cap \left\{ t>-\frac{t^{(0)}_{n}}{\delta ^{(0)}_{n}}\right\} , \\ \displaystyle \frac{\partial \tilde{u}_{n} ^{(0)} }{\partial n}= h_{n}(\delta _{n}^{(0)}x+x_{n}^{(0)})e^{\tilde{u}_{n}^{(0)}} \quad &{}\text { on }B_{\frac{R}{\frac{\delta _{n}}{2}}}\cap \left\{ t=-\frac{t^{(0)}_{n}}{\delta ^{(0)}_{n}}\right\} ,\\ \tilde{u}_{n}^{(0)}(x) \le \tilde{u}_{n}^{(0)}(0)=0 \quad &{}\text { in } B_{\frac{R}{\frac{\delta _{n}}{2}}}\cap \left\{ t\ge -\frac{t^{(0)}_{n}}{\delta ^{(0)}_{n}}\right\} , \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{B_{\frac{R}{\frac{\delta _{n}}{2}}}\cap \left\{ t>-\frac{t^{(0)}_{n}}{\delta ^{(0)}_{n}}\right\} } V_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{2\tilde{u}_{n}^{(0)}}\le C,\quad \int _{B_{\frac{R}{\frac{\delta _{n}}{2}}}\cap \left\{ t=-\frac{t^{(0)}_{n}}{\delta ^{(0)}_{n}}\right\} }h_{n}\left( \delta _{n}x+x_{n}\right) e^{\tilde{u}_{n}^{(0)}}\le C. \end{aligned}$$

As in Lemma 2.1, we distinguish two cases.

Case (i) $\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\rightarrow \Lambda < \infty$. In this case, $\{\tilde{u}_{n}^{(0)}\}$ admits a subsequence converging to $\tilde{u}^{(0)}$ in $C_{loc}^{1,\alpha }({\mathbb R}^2\cap \{t\ge -\Lambda \})$. Then, we may select $k_{n}^{(0)}\rightarrow \infty$, such that

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(0)}\delta _{n}^{(0)}}^+(x_{n}^{(0)})}V_{n}e^{2u_{n}} +\int _{L_{k_{n}^{(0)}\delta _{n}^{(0)}}(x_{n}^{(0)})}h_{n}e^{u_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(0)}}\cap \{t>-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}V_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{2\tilde{u}_{n}^{(0)}} +\int _{B_{k_{n}^{(0)}}\cap \{t=-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}h_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{\tilde{u}_{n}^{(0)}}\right) \\&\qquad \rightarrow 2\pi . \end{aligned}$$

and

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(0)}\delta _{n}^{(0)}}^+(x_{n}^{(0)})}V_{n}e^{2u_{n}} +\int _{L_{2k_{n}^{(0)}\delta _{n}^{(0)}}(x_{n}^{(0)})}h_{n}e^{u_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(0)}}\cap \{t>-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}V_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{2\tilde{u}_{n}^{(0)}} +\int _{B_{2k_{n}^{(0)}}\cap \{t=-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}h_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{\tilde{u}_{n}^{(0)}}\right) \\&\quad \rightarrow 2\pi . \end{aligned}$$

Case (ii) $\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\rightarrow \infty$. Similarly, $\{\tilde{u}_{n}^{(0)}\}$ admits a subsequence converging to $\tilde{u}^{(0)}$ in $C_{loc}^{1,\alpha }({\mathbb R}^2)$. And also we may select $k_{n}^{(0)}\rightarrow \infty$, such that

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(0)}\delta _{n}^{(0)}}^+(x_{n}^{(0)})}V_{n}e^{2u_{n}} +\int _{L_{k_{n}^{(0)}\delta _{n}^{(0)}}(x_{n}^{(0)})}h_{n}e^{u_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(0)}}\cap \{t>-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}V_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{2\tilde{u}_{n}^{(0)}} +\int _{B_{k_{n}^{(0)}}\cap \{t=-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}h_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{\tilde{u}_{n}^{(0)}}\right) \\&\qquad \rightarrow 4\pi \end{aligned}$$

and

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(0)}\delta _{n}^{(0)}}^+(x_{n}^{(0)})}V_{n}e^{2u_{n}} +\int _{L_{2k_{n}^{(0)}\delta _{n}^{(0)}}(x_{n}^{(0)})}h_{n}e^{u_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(0)}}\cap \{t>-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}V_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{2\tilde{u}_{n}^{(0)}} +\int _{B_{2k_{n}^{(0)}}\cap \{t=-\frac{t_{n}^{(0)}}{\delta _{n}^{(0)}}\}}h_{n}\left( \delta _{n}^{(0)}x+x_{n}^{(0)}\right) e^{\tilde{u}_{n}^{(0)}}\right) \\&\qquad \rightarrow 4\pi . \end{aligned}$$

Next we suppose that we have selected l sequences $\{x_{n}^{(j)}\}_{j=0}^{l-1}$, $\{k_{n}^{(j)}\}_{j=0}^{l-1}(l\ge 1)$ satisfying a), b) and c) for $m=l$. At this point, either $M_{n}=\max \nolimits _{x\in \bar{B}_{R}^+} \{u_{n}(x)+\log \min \nolimits _{0\le j\le l-1}|x-x_{n}^{(j)}| \}\le C, \forall n\in N$, and then, we stop and define $m=l$, or $M_{n}\rightarrow \infty$. We define $\bar{x}_{n}^{(l)}$ as a point where $M_{n}$ is attained. So we have $u_{n}(\bar{x}_{n}^{(l)})\rightarrow \infty$. Letting $\bar{\delta }_{n}^{(l)}=e^{-u_{n}(\bar{x}_{n}^{(l)})}$, $M_{n}\rightarrow \infty$ reads as $\min \nolimits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|/\bar{\delta }_{n}^{(l)}\rightarrow \infty$. First we see that for all $|x|\le \frac{1}{2}\min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|/\bar{\delta }_{n}^{(l)}$, we have

$$\begin{aligned} \min \limits _{0\le j\le l-1}|\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}x-x_{n}^{(j)}|\ge & {} \min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|-\bar{\delta }_{n}^{(l)}|x|\\\ge & {} \frac{1}{2}\min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|. \end{aligned}$$

Define $\tilde{u}_{n}(x)=u_{n}(\bar{\delta }_{n}^{(l)}x+\bar{x}_{n}^{(l)})+\log \bar{\delta }_{n}^{(l)}$. Then, $\tilde{u}_{n}$ satisfies

$$\begin{aligned} \left\{ \begin{array}{llll} -\Delta \tilde{u}_{n}&{}=&{}V_{n}(\bar{\delta }_{n}x+\bar{x}_{n}^{(l)})e^{2\tilde{u}_{n}}, ~~&{} |x|\le \frac{1}{2}\min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|/\bar{\delta }_{n}^{(l)}, ~~\bar{\delta }_{n}x+\bar{x}_{n}^{(l)}\in B_{R}^+,\\ \frac{\partial \tilde{u}_{n}}{\partial n}&{}=&{}h_{n}(\bar{\delta }_{n}x+\bar{x}_{n}^{(l)})e^{\tilde{u}_{n}},~~&{} |x|\le \frac{1}{2}\min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|/\bar{\delta }_{n}^{(l)},~~\bar{\delta }_{n}x+\bar{x}_{n}^{(l)}\in L_{R},\\ \tilde{u}_{n}(0)&{}=&{}0,\\ \tilde{u}_{n}(x)&{}\le &{} 2\log 2, \quad \quad \quad \quad \quad \quad &{}|x|\le \frac{1}{2}\min \limits _{0\le j\le l-1} |\bar{x}_{n}^{(l)}-x_{n}^{(j)}|/\bar{\delta }_{n}^{(l)},~~\bar{\delta }_{n}x+\bar{x}_{n}^{(l)}\in \bar{B}_{R}^+. \end{array} \right. \end{aligned}$$

Let $\bar{x}_{n}^{(l)}=(s^{(l)}_{n},t^{(l)}_{n})$, we distinguish two cases.

Case (1) $\frac{t^{(l)}_{n}}{\bar{\delta }_{n}^{(l)}}\rightarrow t_{0}<+\,\infty$. As before, we conclude that $\tilde{u}_{n}$ converges in $C_{loc}^{1,\alpha }({\mathbb R}^2\cap \{t\ge -t_{0}\})$ to a function $\tilde{u}$ satisfying

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}=V(0)e^{2\tilde{u}}, \quad &{}\text { in } {\mathbb R}^2\cap \{t>-t_{0}\}, \\ \frac{\partial \tilde{u}}{\partial n}=h(0)e^{\tilde{u}}, \quad &{}\text { on } {\mathbb R}^2\cap \{t=-t_{0}\},\\ \tilde{u}(x) \le 2\log 2, \qquad &{}\text { on } {\mathbb R}^2\cap \{t\ge -t_{0}\},\\ \tilde{u}(0)=0, \end{array} \right. \end{aligned}$$

with the energy conditions

$$\begin{aligned} \int _{{\mathbb R}^2\cap \{t>-t_{0}\}}V(0)e^{2\tilde{u}}\le C, \quad \int _{{\mathbb R}^2\cap \{t=-t_{0}\}}h(0)e^{\tilde{u}}\le C. \end{aligned}$$

It follows from the classification results in [16] that

$$\begin{aligned} \tilde{u}(s,t)=\log \frac{2\lambda }{\sqrt{V(0)}(\lambda ^2+(s-s_{0})^2+(t+t_{0}+\frac{h(0)\lambda }{\sqrt{V(0)}})^2)}, \quad t\ge -\,t_{0}. \end{aligned}$$

Since $\tilde{u}(0)=0$ and $\tilde{u}(x)\le 2\log 2$, we have

$$\begin{aligned} \frac{\sqrt{V(0)}}{2(V(0)+h^2(0))}\le \lambda \le \frac{2}{\sqrt{V(0)}}, \quad 0\le |s_{0}|, t_{0}\le \frac{2}{\sqrt{V(0)}}. \end{aligned}$$

We see that $\tilde{u}$ attains its maximum at $\bar{x}=(s_{0},-t_{0})$. We choose $\gamma$ satisfying $|\bar{x}|\le \frac{1}{2\gamma }$. Then for given $k>\frac{2}{\gamma }$ and any $\frac{1}{\gamma } \le |x|\le k$, we have $\tilde{u}(x)<\tilde{u}(\bar{x})$. This implies that, when n is sufficiently large, $u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}x)< u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}\bar{x})$ for any $\frac{1}{\gamma } \le |x|\le k$. On the other hand, we may find $y_{n}^{(l)}$ such that $y_{n}^{(l)}\le \frac{1}{\gamma }$ and $u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}x)\le u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}y_{n}^{(l)}), |x|\le k$. Set $x_{n}^{(l)}=\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}y_{n}^{(l)}$. Thus, for all $|x-x_{n}^{(l)}|\le \frac{1}{2}k \bar{\delta }_{n}^{(l)}$, it follows that

$$\begin{aligned} |x-\bar{x}_{n}^{(l)}|/ \bar{\delta }_{n}^{(l)}\le |x-x_{n}^{(l)}|/ \bar{\delta }_{n}^{(l)} +|x_{n}^{(l)}-\bar{x}_{n}^{(l)}|/ \bar{\delta }_{n}^{(l)}\le \frac{1}{2}k+\frac{1}{\gamma }\le k, \end{aligned}$$

where we have used $k>\frac{2}{\gamma }$. Hence, we have $u_{n}(x)=u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}\frac{x-\bar{x}_{n}^{(l)}}{\bar{\delta }_{n}^{(l)}}) \le u_{n}(\bar{x}_{n}^{(l)}+\bar{\delta }_{n}^{(l)}y_{n}^{(l)})=u_{n}(x_{n}^{(l)})$. Now we set $\delta _{n}^{(l)}=e^{-u_{n}(x_{n}^{(l)})}$, we can also obtain:

$$\begin{aligned} \delta _{n}^{(l)}\le \bar{\delta }_{n}^{(l)} \le 2\delta _{n}^{(l)}. \end{aligned}$$

(3.1)

By (3.1) we can choose $k_{n}^{(l)}\rightarrow +\,\infty$ such that

$$\begin{aligned} u_{n}(x_{n}^{(l)})=\max \limits _{x\in \bar{B}_{k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})} u_{n}(x), \end{aligned}$$

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})}V_{n}e^{2u_{n}} +\int _{L_{k_{n}^{(l)}\delta _{n}^{(l)}}(x_{n}^{(l)})}h_{n}e^{u_{n}}\right) \nonumber \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})}V_{n}e^{2u_{n}} +\int _{L_{2k_{n}^{(l)}\delta _{n}^{(l)}}(x_{n}^{(l)})}h_{n}e^{u_{n}}\right) =2\pi . \end{aligned}$$

(3.2)

In addition, since $\delta _{n}^{(l)}\le \bar{\delta }_{n}^{(l)}$, we have

$$\begin{aligned} \frac{|x_{n}^{(l)}-x_{n}^{(i)}|}{\delta _{n}^{(l)}}\ge \frac{|\bar{x}_{n}^{(l)}-x_{n}^{(i)}|}{\bar{\delta }_{n}^{(l)}}-\frac{|x_{n}^{(l)}-\bar{x}_{n}^{(l)}|}{\bar{\delta }_{n}^{(l)}}\rightarrow +\,\infty , 0\le i\le l-1. \end{aligned}$$

(3.3)

We are left to prove that

$$\begin{aligned} \frac{|x_{n}^{(l)}-x_{n}^{(i)}|}{\delta _{n}^{(i)}}\rightarrow \infty , 0\le i\le l-1. \end{aligned}$$

We argue by contradiction and assume $\frac{x_{n}^{(l)}-x_{n}^{(i)}}{\delta _{n}^{(i)}}\rightarrow \tilde{x}$. Then, we have

$$\begin{aligned} \log \delta _{n}^{(i)}-\log \delta _{n}^{(l)}= & {} u_{n}(x_{n}^{(l)})+\log \delta _{n}^{(i)}\\= & {} u_{n}(x_{n}^{(i)}+\delta _{n}^{(i)}\frac{x_{n}^{(l)}-x_{n}^{(i)}}{\delta _{n}^{(i)}})+\log \delta _{n}^{(i)}\\\rightarrow & {} \tilde{u}(\tilde{x}), \end{aligned}$$

which clearly contradicts to (3.3). Thus, we have proved (b) for $m=l$ and $\frac{|x_{n}^{(i)}-x_{n}^{(j)}|}{\delta _{n}^{(i)}}\rightarrow \infty , \forall i\ne j, 0\le i,j\le l$. It is clear that $B_{k_{n}^{(j)}\delta _{n}^{(j)}}^+(x_{n}^{(j)}), 0\le j\le l$ do not intersect. By further reducing $\{k_{n}^{(j)}\}$, we can assume that c) holds.

Case (2) $\frac{t^{(l)}_{n}}{\bar{\delta }_{n}^{(l)}}\rightarrow +\,\infty$. As before $\tilde{u}_{n}$ converges in $C_{loc}^{1,\alpha }({\mathbb R}^2)$ to a function $\tilde{u}$ satisfying

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \tilde{u}=V(0)e^{2\tilde{u}} \quad &{}\text { in } {\mathbb R}^2, \\ \tilde{u}(x) \le 2\log 2 \quad &{}\text { on } {\mathbb R}^2,\\ \tilde{u}(0)=0, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{{\mathbb R}^2}V(0)e^{2\tilde{u}}\le C. \end{aligned}$$

The proof of Case (2) is similar to the Case 1. We note that the difference with case (1) is the following

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})}V_{n}e^{2u_{n}} +\int _{L_{k_{n}^{(l)}\delta _{n}^{(l)}}(x_{n}^{(l)})}h_{n}e^{u_{n}}\right) \nonumber \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})}V_{n}e^{2u_{n}} +\int _{L_{2k_{n}^{(l)}\delta _{n}^{(l)}}(x_{n}^{(l)})}h_{n}e^{u_{n}}\right) =4\pi . \end{aligned}$$

(3.4)

We omit the proof. So under the two cases, we can obtain b) and c).

We continue in this manner until d) holds. We must stop after a finite step since each time we find a mass of $2\pi$ or $4\pi$ near $x_{n}^{(j)}$. $\square$

Step 2 In this step, we show that the mass contribution outside the chosen neighborhoods of the m centers $x_{n}^{(0)}$, $\ldots$, $x_{n}^{(m-1)}$ tends to zero. Namely,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left( \int _{B^+_{R}{\setminus } \cup _{l=0}^{m-1} B_{k_{n}^{(l)}\delta _{n}^{(l)}}^+(x_{n}^{(l)})}V_{n}e^{2u_{n}}+\int _{L_{R}{\setminus } \cup _{l=0}^{m-1} L_{k_{n}^{(l)}\delta _{n}^{(l)}}(x_{n}^{(l)})}V_{n}e^{2u_{n}}h_{n}e^{u_{n}}\right) =0. \end{aligned}$$

To prove this result, we deal with a slightly more general situation that ours.

Lemma 3.1

For $R>0$, Let $\{V_{n}\}$ and $\{h_{n}\}$ be two sequences of functions satisfying (1.7). Let $\{u_{n}\}$ be a sequence of solutions of (1.4) and (1.5) satisfying (1.8)–(1.9). Assume that $\{x_{n}^{(j)}\}_{j=0}^{m-1}$ are $m (m\ge 1)$ sequences of points, $\{r_{n}^{(j)}\}_{j=0}^{m-1}$ are m sequences of positive numbers which satisfy

$$\begin{aligned}&u_{n}(x_{n}^{(j)})=\max \limits _{x\in \bar{B}_{r_{n}^{(j)}}^+(x_{n}^{(j)})} u_{n}(x)\rightarrow \infty , x_{n}^{(j)}=(s_{n}^{(j)},t_{n}^{(j)}), \quad \forall 0\le j\le m-1, \qquad \quad \end{aligned}$$

(3.5)

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty } \frac{r_{n}^{(j)}}{\delta _{n}^{(j)}}=\infty , \quad \forall 0\le j\le m-1, \end{aligned}$$

(3.6)

where $\delta _{n}^{(j)}=e^{-u_{n}(x_{n}^{(i)})}$,

$$\begin{aligned}&\frac{|x_{n}^{(i)}-x_{n}^{(j)}|}{r_{n}^{(j)}}\rightarrow \infty , \quad \forall i\ne j, 0\le i, j\le m-1, \end{aligned}$$

(3.7)

$$\begin{aligned}&\max \limits _{x\in \bar{B}_{R}^+\backslash \cup _{j=0}^{m-1}B_{r_{n}^{(j)}}^+(x_{n}^{(j)}) } \{u_{n}(x)+\log \min \limits _{0\le j\le m-1}|x-x_{n}^{(j)}| \}\le C, \forall n, \end{aligned}$$

(3.8)

and

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{r_{n}^{(j)}}^+(x_{n}^{(j)})}V_{n}e^{2u_{n}} +\int _{L_{r_{n}^{(j)}}(x_{n}^{(j)})}h_{n}e^{u_{n}}\right) \nonumber \\&\quad =\lim \limits _{n\rightarrow \infty }\left( \int _{B_{2r_{n}^{(j)}}^+(x_{n}^{(j)})}V_{n}e^{2u_{n}} +\int _{L_{2r_{n}^{(j)}}(x_{n}^{(j)})}h_{n}e^{u_{n}}\right) =\beta _{j}, \end{aligned}$$

(3.9)

where $\beta _{j}=2\pi$ when $\frac{t_{n}^{(j)}}{\delta _{n}^{(j)}}\rightarrow \Lambda < \infty$, and $\beta _{j}=4\pi$ when $\frac{t_{n}^{(j)}}{\delta _{n}^{(j)}}\rightarrow \infty$ for all $0\le j\le m-1$. Then,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left( \int _{B^+_{R}}V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}e^{u_{n}}\right) =\sum _{j=1}^m \beta _{j}. \end{aligned}$$

Proof

We will follow the approach of [15] to prove the lemma by induction on m. First we prove the lemma for $m=1$. We also distinguish two cases.

Case (1) $\frac{t_{n}}{\delta ^{(0)}_{n}}\rightarrow \Lambda < \infty$.

In this case, we can assume that $\lim \limits _{n\rightarrow \infty }r_{n}^{(0)}=0$, since otherwise the lemma obviously holds due to (1.9). We also assume that $B^+_{\frac{R}{2}}(x_n)\subset B^+_R$.

By using Lemma 2.6, we obtain that

$$\begin{aligned} \sup \limits _{S_{r}^+(x_n)} u_{n}\le C+\beta \inf \limits _{S_{r}^+(x_n)} u_{n}+(\beta -1)\log r, \quad \forall 2r_{n}^{(0)}\le r\le \frac{R}{2}. \end{aligned}$$

By using Lemma 2.5, we obtain that

$$\begin{aligned} \inf \limits _{S_{r}^+(x_n)} u_{n}\le C-\frac{1}{C_{1}}u_{n}(x_n)-(1+\frac{1}{C_{1}})\log r,\quad \forall 0<r<R. \end{aligned}$$

It follows that

$$\begin{aligned} \sup \limits _{S_{r}^+(x_n)} u_{n}\le C-\frac{\beta }{C_{1}}u_{n}(x_n)-(1+\frac{\beta }{C_{1}})\log r, \quad \forall 2r_{n}^{(0)}\le r\le \frac{R}{2}, \end{aligned}$$

namely

$$\begin{aligned} e^{u_{n}(x)}\le C(\delta _{n}^{(0)})^{\beta /C_{1}}|x-x_n|^{-(1+\frac{\beta }{C_{1}})}, \forall x\in B_{\frac{R}{2}}^+(x_n)\backslash B_{2r_{n}^{(0)}}^+(x_n). \end{aligned}$$

Therefore, we have

$$\begin{aligned}&\int _{B_{R/2}^+(x_n)\backslash B^+_{2r_{n}^{(0)}}(x_n)}V_{n}e^{2u_{n}}+\int _{L_{R/2}(x_n)\backslash L_{2r_{n}^{(0)}}(x_n)}h_{n}e^{u_{n}}\\&\quad \le C(\delta _{n}^{(0)})^{2\beta /C_{1}}\int _{2r_{n}^{(0)}}^{\infty }r^{-2(1+\frac{\beta }{C_{1}})}r\mathrm{d}r +C(\delta _{n}^{(0)})^{\beta /C_{1}}\int _{2r_{n}^{(0)}}^{\infty }s^{-(1+\frac{\beta }{C_{1}})}\mathrm{d}s\\&\quad = C(\frac{\delta _{n}^{(0)}}{2r_{n}^{(0)}})^{2\beta / C_{1}}+C(\frac{\delta _{n}^{(0)}}{2r_{n}^{(0)}})^{\beta / C_{1}} \rightarrow 0. \end{aligned}$$

By (1.9), (3.9) and above formula, we obtain that $\lim \limits _{n\rightarrow \infty }\left( \int _{B^+_{R}}V_{n}e^{2u_{n}}\!+\!\int _{L_{R}}h_{n}e^{u_{n}}\right) \!=\! \beta _{0}=2\pi$.

Case (2) $\frac{t_{n}}{\delta ^{(0)}_{n}}\rightarrow +\infty$. Note that for n sufficiently large, $\overline{B}_{r_{n}^{(j)}}^+(x_{n}^{(j)})$ is contained in the interior of $B^{+}_{R}$. The proof is very similar with case (1). We can use Lemmas 2.3, 2.4, 2.5 and 2.6 to obtain $\lim \nolimits _{n\rightarrow \infty }\left( \int _{B^+_{R}}V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}e^{u_{n}}\right) = \beta _{0}=4\pi$.

Next we proceed the proof by induction. Suppose that lemma holds for $1,2,\ldots ,m-1(m\ge 2)$, we prove that it holds for m. Without loss of generality, we assume that

$$\begin{aligned} d_{n}=|x_{n}^{(0)}-x_{n}^{(1)}|=\min \{|x_{n}^{(i)}-x_{n}^{(j)}|,i\ne j, 0\le i,j \le m-1 \} \end{aligned}$$

and $x_{n}^{(0)}=0$. There exist two cases.

Case 1 For some constant A, we have

$$\begin{aligned} |x_{n}^{(i)}-x_{n}^{(j)}|\le Ad_{n}, \quad 0\le i,j \le m-1. \end{aligned}$$

In this case, we will establish

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{4Ad_{n}}^+(x_n^{(0)})}V_{n}e^{2u_{n}}+\int _{L_{4Ad_{n}}(x_n^{(0)})}h_{n}e^{u_{n}}\right) \nonumber \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2Ad_{n}}^+(x_n^{(0)})}V_{n}e^{2u_{n}}+\int _{L_{2Ad_{n}}(x_n^{(0)})}h_{n}e^{u_{n}}\right) =\sum _{j=0}^{m-1} \beta _{j}. \end{aligned}$$

(3.10)

Once (3.10) is established, we introduce $x_{n}'^{(0)}=x_n^{(0)}$, $r_{n}'^{(0)}=2Ad_{n}$ and $\beta '_{0}=\sum _{j=0}^{m-1} \beta _{j}$. And then we can apply Lemma 3.1 for $m=1$ to obtain

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left( \int _{B^+_{R}}V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}e^{u_{n}}\right) = \beta '_{0}. \end{aligned}$$

We adopt the method applied in [15]. For $x\in \bar{B}_{R/d_{n}}^+$, define $\tilde{u}_{n}(x)=u_{n}(d_{n}x)+\log d_{n}$. Denote

$$\begin{aligned} \tilde{V}_{n}(x)= & {} V_{n}(d_{n}x), \quad x\in \bar{B}_{R/d_{n}}^+, \\ \tilde{h}_{n}(x)= & {} h_{n}(d_{n}x), \quad x\in \bar{B}_{R/d_{n}}^+, \\ \tilde{x}_{n}^{(j)}= & {} x_{n}^{(j)}/d_{_{n}}, \quad 0\le j \le m-1, \\ \tilde{\delta }_{n}^{(j)}= & {} e^{-\tilde{u}_{n}(\tilde{x}_{n}^{(j)})}=\delta _{n}^{(j)}/d_{n}, \quad 0\le j \le m-1, \\ \tilde{r}_{n}^{(j)}= & {} r_{n}^{(j)}/d_{n}, \quad 0\le j \le m-1. \end{aligned}$$

It follows that

$$\begin{aligned}&\tilde{u}_{n}(\tilde{x}_{n}^{(j)})=\max \limits _{x\in \bar{B}_{\tilde{r}_{n}^{(j)}}^+(\tilde{x}_{n}^{(j)})} \tilde{u}_{n}(x)\rightarrow \infty , \quad 0\le j\le m-1, \\&\lim \limits _{n\rightarrow \infty } \frac{\tilde{r}_{n}^{(j)}}{\tilde{\delta }_{n}^{(j)}}=\infty , \quad 0\le j \le m-1, \\&\tilde{r}_{n}^{(j)}\rightarrow 0, \quad 0\le j \le m-1, \\&\max \limits _{x\in \bar{B}_{R/d_{n}}^+\backslash \cup _{j=0}^{m-1}B_{\tilde{r}_{n}^{(j)}}^+(\tilde{x}_{n}^{(j)}) } \{\tilde{u}_{n}(x)+\log \min \limits _{0\le j\le m-1}|x-\tilde{x}_{n}^{(j)}| \}\le C, \forall n, \\&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{\tilde{r}_{n}^{(j)}}^+(\tilde{x}_{n}^{(j)})}\tilde{V}_{n}e^{2\tilde{u}_{n}} +\int _{L_{\tilde{r}_{n}^{(j)}}(\tilde{x}_{n}^{(j)})}\tilde{h}_{n}e^{\tilde{u}_{n}}\right) \\&=\lim \limits _{n\rightarrow \infty }\left( \int _{B_{2\tilde{r}_{n}^{(j)}}^+(\tilde{x}_{n}^{(j)})}\tilde{V}_{n}e^{2\tilde{u}_{n}} +\int _{L_{2\tilde{r}_{n}^{(j)}}(\tilde{x}_{n}^{(j)})}\tilde{h}_{n}e^{\tilde{u}_{n}}\right) =\beta _{j}, \quad 0\le j\le m-1. \end{aligned}$$

We assume that $\tilde{x}_n^{(j)}\rightarrow \tilde{x}^{(j)}$ for $0\le j\le m-1$. Set $S=\{\tilde{x}^{(j)}, \quad 0\le j\le m-1\}$. Note that

$$\begin{aligned} 1\le |\tilde{x}^{(i)}-\tilde{x}^{(j)}|\le A. \end{aligned}$$

Hence we know that the set of blow-up points of $\tilde{u}_n$ in $\bar{B}^+_{4A}(x_n^{(0)})$ is S. Then, it follows from Theorem 1.1 that $u_{n}\rightarrow -\infty$ uniformly on any compact sets of $\bar{B}^+_{4A}(x_n^{(0)})\backslash S$. Now we apply the case $m=1$ of Lemma 3.1 to conclude $\int _{B_{\frac{1}{2}}^+(\tilde{x}^{(j)})}\tilde{V}_{n}e^{2\tilde{u}_{n}}+\int _{L_{\frac{1}{2}}(\tilde{x}^{(j)})}\tilde{h}_{n}e^{\tilde{u}_{n}}\rightarrow \beta _{j}$. Consequently,

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{4A}^+(x_n^{(0)})}\tilde{V}_{n}e^{2\tilde{u}_{n}}+\int _{L_{4A}(x_n^{(0)})}\tilde{h}_{n}e^{\tilde{u}_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2A}^+(x_n^{(0)})}\tilde{V}_{n}e^{2\tilde{u}_{n}}+\int _{L_{2A}(x_n^{(0)})}\tilde{h}_{n}e^{\tilde{u}_{n}}\right) =\sum _{j=0}^{m-1} \beta _{j}. \end{aligned}$$

A simple change of variables leads to (3.10). Then, we derive the desired conclusion.

Case 2 A proper subset J of $\{0,1,2,\ldots ,m-1\}$ containing $\{0,1\}$ and a constant A satisfy

$$\begin{aligned} |x_{n}^{(j)}-x_{n}^{(0)}|\le Ad_{n}, \forall j\in J; \lim \limits _{n\rightarrow \infty }|x_{n}^{(j)}-x_{n}^{(0)}|/d_{n}=\infty , \quad \forall j \notin J. \end{aligned}$$

Without loss of generality, we assume that $J=\{0,1,2,...,k-1\}$. In this case, we consider $\tilde{u}_{n}(x)=u_{n}(d_{n}x)+\log d_{n}$ in $\bar{B}_{4A}^+$. Arguing as in case 1, we obtain:

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\left( \int _{B_{4Ad_{n}}^+(x_n^{(0)})}V_{n}e^{2u_{n}}+\int _{L_{4Ad_{n}}(x_n^{(0)})}h_{n}e^{u_{n}}\right) \\&\quad = \lim \limits _{n\rightarrow \infty }\left( \int _{B_{2Ad_{n}}^+(x_n^{(0)})}V_{n}e^{2u_{n}}+\int _{L_{2Ad_{n}}(x_n^{(0)})}h_{n}e^{u_{n}}\right) =\sum _{j=0}^{k-1} \beta _{j}:=\beta _0'. \end{aligned}$$

We set $r_{n}'^{(0)}=Ad_{n}$ and $x_{n}'^{(0)}=x_n^{(0)}$. If the $m-k+1$ sequences $x_{n}'^{(0)}$, $\{x_{n}^{(j)}\}_{j=k}^{m-1}$ with the radius $r_{n}'^{(0)}$, $\{r_{n}^{(j)}\}_{j=k}^{m-1}$ and the mass $\beta '_{0}$, $\{\beta _{j}\}_{j=k}^{m-1}$ satisfy (3.5)–(3.9), we may apply the case $m-k+1$ of Lemma 3.1. Now we need to verify (3.5)–(3.9). We only need to show (3.7) since others are obvious. Note that

$$\begin{aligned} \frac{|x_{n}'^{(0)}-x_{n}^{(j)}|}{r_{n}'^{(0)}}=\frac{|x_{n}^{(j)}-x_n^{(0)})|}{Ad_{n}}\rightarrow \infty , \quad \forall j \notin J. \end{aligned}$$

Therefore, we obtain:

$$\begin{aligned} \int _{B^+_{R}}V_{n}e^{2u_{n}}+\int _{L_{R}}h_{n}e^{u_{n}}\rightarrow \beta '_{0}+\sum _{j=k}^{m-1} \beta _{j}=\sum _{j=0}^{m-1} \beta _{j}. \end{aligned}$$

$\square$