数学物理学报, 2026, 46(2): 552-583

$G_2$ 型 Toda 系统的完全爆破解的精确估计——献给陈化教授 70 寿辰

敖微微,, 赖珊珊,*

武汉大学数学与统计学院 武汉 430072

Sharp Estimates for Fully Bubbling Solutions of $G_2$ Toda System

Ao Weiwei,, Lai Shanshan,*

School of Mathematics and Statistics, Wuhan University, Wuhan 430072

通讯作者: *赖珊珊, Email:sslai_math@whu.edu.cn

收稿日期: 2025-12-16   修回日期: 2026-01-4  

基金资助: 国家重点研发计划(2022YFA1006800)
国家自然科学基金创新研究群体项目(12221001)
国家自然科学基金重点项目(12131017)
国家自然科学基金面上项目(12471111)

Received: 2025-12-16   Revised: 2026-01-4  

Fund supported: National Key Research and Development Program of Science and Technology(2022YFA1006800)
National Natural Science Foundation of China Group Project(12221001)
National Natural Science Foundation of China Key Project(12131017)
National Natural Science Foundation China General Project(12471111)

作者简介 About authors

敖微微,Email:wwao@whu.edu.cn

摘要

该文旨在得到紧致 Riemann 曲面上的 $G_2$ 型 Toda 系统的完全爆破解的精确估计, 从而充分理解完全爆破解的渐近行为. 作者利用全局解的非退化性, 证明了: 1) 所有完全爆破解均可用一组具有精确误差的整体解序列逼近; 2) 特定函数的梯度在爆破点处必须以足够快的速度趋于零, 从而确定了爆破点的位置; 3) 存在对应的 $\partial_z^2$ 条件.

关键词: $G_2$ 型 Toda 系统; 完全爆破解; 渐近行为估计

Abstract

This paper aims to sharp estimates of fully bubbling solutions of the Toda system with Cartan matrix $G_2$ in a compact Riemann surface, thereby providing a comprehensive understanding of the asymptotic behavior of such solutions. By using the non-degeneracy results of entire solutions, it proves that: 1) All fully bubbling solutions are approximated by a sequence of global solutions with precise error estimates; 2) the gradient of certain functions must approach zero with sufficient rate at the blow-up points, which uniquely determines their locations; 3) a corresponding $\partial_z^2$ condition exists.

Keywords: $G_2$ Toda system; fully bubbling solutions; asymptotic behavior

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本文引用格式

敖微微, 赖珊珊. $G_2$ 型 Toda 系统的完全爆破解的精确估计——献给陈化教授 70 寿辰[J]. 数学物理学报, 2026, 46(2): 552-583

Ao Weiwei, Lai Shanshan. Sharp Estimates for Fully Bubbling Solutions of $G_2$ Toda System[J]. Acta Mathematica Scientia, 2026, 46(2): 552-583

1 绪论

1.1 背景介绍

设 $(M,g)$ 为紧致 Riemann 曲面. 我们考虑如下具有多个奇点源的 Toda 系统

$\begin{equation} \Delta_{g} v_{i}+\sum_{j=1}^{n} k_{i j} {\rm e}^{v_{j}} =4 \pi \sum_{j=1}^{n} \gamma_{i j} \delta_{q_{j}}, i=1, \cdots, n, \end{equation}$

其中 $\Delta_{g}$ 表示 Laplace-Beltrami 算子, $\left(k_{i j}\right)_{i, j=1, \cdots, n}$ 是秩为 $n$ 的简单 Lie 代数的 Cartan 矩阵, $\gamma_{ij}>-1$, $q_{j}$ 是 $M$ 中互不相同的点, $\delta_{q_{j}}$ 是 $q_{j}$ 处的 Dirac 测度.

Toda 系统与几何和物理学中的多个研究领域密切相关, 由此衍生出许多相关问题.

在几何学中, Toda 系统的解与射影空间中的全纯曲线有紧密联系. Doliwa在文献 [22] 中描述了特殊全纯曲线的几何与四种非例外简单 Lie 代数 ($A_n, B_n, C_n$ 和 $D_n$) 对应的开 Toda 系统之间的关系. 具体而言, 射影空间中全纯曲线的 Plücker 公式构成了与相应 Lie 代数的紧实形式相关的开 Toda 系统. 更多细节可参阅文献 [11,25,26,46] 及其中的参考文献.

在物理学中, Toda 系统在规范场论中扮演着重要角色. 在 Chern-Simons 方面, 关于相对论性 Abelian Chern-Simons-Higgs 模型 (对应 $n=1$ 时的 (1.1)系统) 可参考文献 [12,39,41,48]; 对于 $n \geq 2$ 的情形, 可参阅文献 [3,4,40,49]. 关于 non-Abelian 模型的研究, 参见文献 [23,24,54,55] 等.

设 $|M|$ 表示 $M$ 的面积, 为简便起见, 设面积为 $|M|=1$. 引入 Green 函数

$ \left\{\begin{array}{l} \Delta G(x, p)=-\delta_{p}+1, \\ \int_{M} G(x, p){\rm d}x=0, \end{array}\right. $

并采用标准变换 $ u_{i}=v_{i}+4 \pi \sum_{j=1}^{m} \gamma_{j i} G\left(x, q_{j}\right), i=1, 2, $ 可消除 (1.1) 式右侧的奇异性, 从而将 (1.1)式改写为

$\begin{equation} \Delta_g u_i + \sum_{j=1}^{n} k_{i j}\rho_{j} \left( \frac{h_j {\rm e}^{u_j}}{\int_M h_j {\rm e}^{u_j}} -1 \right)=0, \quad i=1, \cdots, n, \end{equation}$

其中 $\rho_i>0$ 为常数, $h_i$ 由下式给出

$ h_{i}(x)=\exp \left(-\sum_{j=1}^{m} 4 \pi \gamma_{j i} G\left(x, q_{j}\right)\right), \quad i=1, \cdots, n, $

是 $M$ 上的非负连续函数.

众所周知, 简单 Lie 代数包括 $A_n$、$B_n$、$C_n$、$D_n$、$G_2$、$F_4$、$E_6$、$E_7$ 和 $E_8$. 当 $K=A_n$ 时, 方程 (1.2) 称为 $\mathrm{SU}(n+1)$ Toda 系统. 在文献 [34] 中, Lin-Wei-Ye 完全分类了具有单个奇源的 $\mathrm{SU}(n+1)$ Toda 系统的解, 并得到了其非退化性结果.

若 (1.2) 式仅包含一个方程, 则其退化为平均场方程

$\begin{equation} \Delta_g u + \rho \left( \frac{h {\rm e}^{u}}{\int_M h {\rm e}^{u}} -1 \right)=0. \end{equation}$

过去 30 年间, 该方程得到了广泛研究.

爆破现象是这类模型的一个基本问题. 对于平均场方程而言, 在爆破点附近, 解可能呈现两种类型的爆破行为. 一种称为简单爆破, 即爆破的局部形态由整体爆破解所控制. 相关的研究课参考文献 [7,13,16,31,32] 等. 另一种是非简单爆破, 这种情况仅出现在 $\gamma \in \mathbb{N}$ 时,最近关于非简单爆破的研究也有很大的进展, 可参考参考文献 [6,18{21,21,29,50{52] 等. 对于爆破解的研究在进一步研究方程解的存在性, 唯一性等方面是非常重要, 可参考文献 [8-10,53] 等.

在标量平均场方程的情形中, 局部 Pohozaev 恒等式是研究爆破分析的有力工具, 因为 Pohozaev 恒等式的数量与自由参数的个数相等. 但当 $n \geq 2$ 时, $\mathrm{SU}(n+1)$ Toda 系统转化为椭圆方程组, 其爆破行为变得更为复杂. 主要困难在于可能出现部分爆破现象. 要理解部分爆破, 首先必须研究完全爆破的情形. 然而, 许多对单个方程非常有效的基本工具 (如极值原理、Pohozaev 恒等式) 却无法直接应用或推广. 此外, $\mathrm{SU}(n+1)$ Toda 系统的解缺乏对称结构, 且涉及过多参数. 在研究爆破解时, 孤立爆破点附近爆破解的渐近行为极为复杂, 且这种复杂性随着方程数量的增加而显著增强. 例如, 对于 $\mathrm{SU}(3)$ 系统, 局部 Pohozaev 恒等式仅提供 3 个方程, 但要描述 $\mathbb{R}^2$ 中所有整体解却需要 8 个参数.

[27,28] 中, 作者开始了对无奇异源的 $A_2$ 型 Toda 系统爆破分析的研究. 爆破分析的第一步是对爆破点 $p$ 处的局部质量的可能取值进行分类. 局部质量定义为

$ \sigma_i(p):=\lim_{r \rightarrow 0} \lim_{k \rightarrow \infty} \frac{\int_{B_r\left(p\right)} \rho_{i k} h_{i k} {\rm e}^{u_{i k}}}{\int_M h_{i k} {\rm e}^{u_{i k}}}, $

其中 $\left\{\left(u_1^k, u_2^k\right)\right\}$ 是 $A_2$ Toda 系统的一列爆破解, $B_r\left(p\right)$ 是以 $p$ 为中心、半径为 $r$ 的球. 在一些温和的假设下, Jost-Lin-Wang 在文献 [28] 中得到了局部质量的分类. 文献 [27] 证明了任何完全爆破解序列在任意爆破点处均为简单爆破. 该证明是 Pohozaev 恒等式的一种非常有效的推广. 但该方法无法推广到含奇异源的情形. Lin-Wei-Zhao 在文献 [38] 中深入探讨了完全爆破解的行为, 进一步 Lin-Wei-Zhang 在文献 [37] 中证明了对于一般的 $n$, 含奇异源的 $\mathrm{SU}(n+1)$ Toda 系统的完全爆破解同样满足上述简单爆破性质.

完全爆破解的精确估计在构造系统的爆破解及研究解的存在性、唯一性等方面具有重要作用.

对于 $n=1$ 的情形 (即平均场方程), 基于[13,15] 中关于多重爆破解精确估计的分析工作, Chen-Lin 分别在[14,16] 中证明了无奇异源及含奇异源平均场方程的度计算公式.

对于 $n \geq 2$ 的情形, 研究则更为复杂, 因为系统爆破的情况更为复杂. 在文献 [38] 中, Lin-Wei-Zhao 对 $\mathrm{SU}(3)$ Toda 系统 (即 $A_2$ Toda 系统) 的完全爆破解获得了更精确的估计. 他们并未采用在标量平均场方程中非常有效但无法处理 Toda 系统解空间中更多自由参数的局部 Pohozaev 恒等式, 而是利用了 $\mathrm{SU}(3)$ Toda 系统整体解的非退化性结果来推导其结论. Lin-Wei-Zhang 在文献 [36] 中将类似定理推广到一般的 $\mathrm{SU}(n+1)$ Toda 系统. 他们证明了关于 $\mathrm{SU}(n+1)$ Toda 系统完全爆破解的三个主要精确估计结果: 所有完全爆破解均可用一组具有精确误差的整体解序列逼近; 特定函数的梯度在爆破点处必须以足够快的速度趋于零; 存在一个 $\partial_z^2$ 条件, 该条件为 $\mathrm{SU}(n+1)$ Toda 系统所独有. 所有这些估计对于理解爆破解相互作用以及未来构造爆破解至关重要.相应的 $B_2$ 型 Toda 系统完全爆破解的精细估计在文献 [1] 中研究. 而对于 $G_2-$Toda 系统的完全爆破的精细估计目前仍然没有研究. 本文将继续这方面的研究.

对于 $B_n$、$C_n$、$G_n$ 以及其他例外 Lie 群 $F_2$、$E_6$、$E_7$ 和 $E_8$ 的 Cartan 矩阵, 近年来已有多项相关研究. 对于秩为 2 的 Toda 系统, 涉及简单 Lie 代数 ${A}_2$、${B}_2\left({C}_2\right)$ 和 ${G}_2$ 的局部质量分类和先验估计等, 可参阅文献 [1,17,30,33,35,42{44] 等.

现在我们考虑 $G_2$ 型 Toda 系统

$\begin{equation} \Delta_{g} v_i +\sum_{j=1}^{2} g_{i j} {\rm e}^{v_{j}}=4 \pi \sum_{j=1}^{2} \gamma_{i j} \delta_{q_{j}}, \end{equation}$

其中 $i=1, 2$, 而矩阵 $\left(g_{i j}\right)$ 由下式给出

$ G_{2}=\begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}. $

沿袭文献 [38] 的思想, 我们将利用文献 [2] 中关于 Cartan 矩阵 $G_2$ 的分类结果和非退化性结论, 得到紧致 Riemann 曲面上的 $G_2$ 型 Toda 系统的完全爆破解的精确估计, 从而充分理解完全爆破解的渐近行为.

1.2 主要结果

我们考虑以下无奇异源的 $G_2$ 型 Toda 系统的完全解

$\begin{equation} \left\{\begin{array}{l} \Delta_{g} u_{1 k}+2 \rho_{1 k}\left(\frac{h_{1 k} {\rm e}^{u_{1 k}}}{\int_{M} h_{1 k} {\rm e}^{u_{1 k}}}-1\right)-\rho_{2 k}\left(\frac{h_{2 k} {\rm e}^{u_{2 k}}}{\int_{M} h_{2 k} {\rm e}^{u_{2 k}}}-1\right)=0, \\[3mm] \Delta_{g} u_{2 k}-3 \rho_{1 k}\left(\frac{h_{1 k} {\rm e}^{u_{1 k}}}{\int_{M} h_{1 k} {\rm e}^{u_{1 k}}}-1\right)+2 \rho_{2 k}\left(\frac{h_{2 k} {\rm e}^{u_{2 k}}}{\int_{M} h_{2 k} {\rm e}^{u_{2 k}}}-1\right)=0, \end{array}\right. \end{equation}$

其中 $\rho_i>0$ 为常数, $h_i$ 是 $M$ 上的非负连续函数.

容易看出, 如果 $u = (u_1, u_2)$ 是 (1.5) 式的一个解, 那么对于任意常数 $c_1$ 和 $c_2$, $u + c = (u_1 + c_1, u_2 + c_2)$ 仍然是一个解. 因此, 不失一般性, 我们可以假设每个分量 $u_i$ 属于 $\stackrel{\circ}{H}(M)$, 其定义为

$ \stackrel{\circ}{H}(M)=\left\{u_{i} \in H(M) \mid \int_{M} u_{i}=0\right\}. $

设 $p_{1}, \cdots, p_{m}$ 是 $u_{k}$ 的爆破点, 并选取足够小的 $r_{0}$, 使得当 $i \neq j$ 时, $B\left(p_{i}, r_{0}\right) \cap B\left(p_{j}, r_{0}\right) = \emptyset$. 记 $p_{k, j}$ 为 $u_{k}$ 在 $B_{p_{j}}\left(r_{0}\right)$ 中的局部最大值点, 其中 $j=1, \cdots, m$. 在整篇论文中, 我们假设

$ (\mathbf{H}) \quad u_{k} \text{ 在每个爆破点 } p_{j} \ (j=1, \cdots, m) \text{ 处完全爆破. } $

关于 "完全爆破" 的准确定义, 参见第 2 节中的定义 2.1. 令 $\varepsilon_{k, j}$ 和 $\varepsilon_{k}$ 分别由公式 (3.1) 和 (3.2) 定义.

我们的主要结果如下

定理 1.1 设 $u_{k}=\left(u_{1 k}, u_{2 k}\right) \in \stackrel{\circ}{H}(M)$ 是方程组 (1.5) 的一列爆破解序列, 且满足假设 $\mathrm{(H)}$. 则成立

(i) 收敛速率

$\begin{aligned} & \rho_{1 k}-24 m \pi=\sum_{j=1}^{m} C_{1 k, j}\left(\Delta \ln h_{1 k}\left(p_{k, j}\right)-2 K\left(p_{j}\right)+8 m \pi\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \\ & \rho_{2 k}-40 m \pi=\sum_{j=1}^{m} C_{2 k, j}\left(\Delta \ln h_{2 k}\left(p_{k, j}\right)-2 K\left(p_{j}\right)+8 m \pi\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \end{aligned} $

其中 $C_{i k, j}$ 是满足 $0<C_{1}<C_{i k, j}<C_{2}<\infty$ 的常数, 而 $K$ 表示 $M$ 的 Gauss 曲率;

(ii) 爆破点 $p_{j}$ 的位置

$\begin{aligned} & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{1 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right), \\ & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{2 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right), \end{aligned}$

其中 $H(x, p)$ 是 $G(x, p)$ 的正则部分;

(iii) $\partial_{z}^{2}$ 条件

$\begin{aligned} & \frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta \ln h_{1 k}-2 K\left(p_{j}\right)+8 m \pi\right) T_{1 k, 1}^{j} \\ & +\frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right)-2 K\left(p_{j}\right)+8 m \pi\right) T_{2 k, 1}^{j}=O\left(\varepsilon_{k}^{\beta}\right), \\ & 3 \pi \partial_{12}\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta \ln h_{1 k}-2 K\left(p_{j}\right)+8 m \pi\right) T_{1 k, 2}^{j} \\ & +\frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right)-2 K\left(p_{j}\right)+8 m \pi\right) T_{2 k, 2}^{j}=O\left(\varepsilon_{k}^{\beta}\right). \end{aligned}$

其中 $T_{i k, l}^{j}$ 是由命题 7.1 中的公式 (7.1) 所定义的常数.

我们还考虑在边界区域 $\Omega$ 中研究方程 (1.5).

$\begin{equation} \left\{\begin{array}{l} \Delta u_{1 k}+2 \rho_{1 k} \frac{h_{1 k} {\rm e}^{u_{1 k}}}{\int_{\Omega} h_{1 k} {\rm e}^{u_{1 k}}}-\rho_{2 k} \frac{h_{2 k} {\rm e}^{u_{2 k}}}{\int_{\Omega} h_{2 k} {\rm e}^{u_{2 k}}}=0, \\ \Delta u_{2 k} -3 \rho_{1 k} \frac{h_{1 k} {\rm e}^{u_{1 k}}}{\int_{\Omega} h_{1 k} {\rm e}^{u_{1 k}}} + 2 \rho_{2 k} \frac{h_{2 k} {\rm e}^{2 k}}{\int_{\Omega} h_{2 k} {\rm e}^{u_{2 k}}}=0, \end{array} \quad \text { 在 } \Omega\text { 中}.\right. \end{equation}$

对于 Dirichlet 问题, 可以证明 $u_k$ 不可能在边界 $\partial \Omega$ 上爆破 (参见[45,47]). 因此, 我们可以得到类似的结果.

定理 1.2 假设 $h_{i k}$ 在 $C^{2}(\bar{\Omega})$ 中收敛到某个正函数 $h_{i}$, 且 $u_{k}=(u_{1 k}, u_{2 k})$ 是方程 (1.6) 的一列爆破解序列, 满足齐次 Dirichlet 边界条件. 设 $S=\left\{p_{1}, \cdots, p_{m}\right\}$ 为其爆破点集, 且满足假设 $\mathrm{(H)}$. 则成立

(i) 收敛速率

$\begin{aligned} & \rho_{1 k}-24 m \pi=\sum_{j=1}^{m} C_{1 k, j} \Delta \ln h_{1 k}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \\ & \rho_{2 k}-40 m \pi=\sum_{j=1}^{m} C_{2 k, j} \Delta \ln h_{2 k}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \end{aligned}$

其中 $C_{i k, j}$ 是满足 $0<C_{1}<C_{i k, j}<C_{2}<\infty$ 的常数;

(ii) 爆破点 $p_{j}$ 的位置

$\begin{aligned} & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{1 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right), \\ & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{2 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right), \end{aligned}$

其中 $H(x, p)$ 是 $G(x, p)$ 的正则部分;

(iii) $\partial_{z}^{2}$ 条件

$\begin{aligned} &\frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right) +\frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right) T_{2 k, 1}^{j}+\Delta \ln h_{1 k} T_{1 k, 1}^{j}\right)=O\left(\varepsilon_{k}^{\beta}\right), \\ &3 \pi \partial_{12}\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right)+ \frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right) T_{2 k, 2}^{j}+\Delta \ln h_{1 k} T_{1 k, 2}^{j}\right)=O\left(\varepsilon_{k}^{\beta}\right), \end{aligned}$

其中 $T_{i k, l}^{j}$ 是由命题 7.1 中的公式 (7.1) 所定义的常数.

1.3 论文结构

为表述上的方便, 我们将先证明欧氏空间中的结果, 即定理 1.2.

在第 2 节中, 我们列举了 $\mathbb{R}^{2}$ 上 $G_{2}$ 型 Toda 系统整体解的一些重要性质. 这部分借鉴了 Lin-Wei-Zhang[37] 的思路, 用于分析整体解的行为.

我们将利用文献 [2] 中对整体解的分类和非退化性结果, 并在第 3 节到第 7 节研究方程 (1.6) 的一列爆破解. 在第 3 节中, 我们推导出爆破解的两个重要估计: 一个在爆破点附近, 另一个远离爆破点. 在第 4 节中, 我们利用参数化的整体解来逼近气泡解, 并获得更精确的内部估计. 第 5 至 7 节包含了爆破速率和爆破位置的计算. 这里我们遵循 Lin-Wei-Zhao[38] 的思路, 利用 $G_{2}$ 型 Toda 系统的核函数对方程组进行局部检验.

在第 8 节中完成对定理 1.2 和定理 1.1 的证明.

2 $\mathbb{R}^2$ 中整体解的性质与完全爆破解

本节我们研究方程 (1.6) 的完全爆破解在爆破点附近的局部形态. 由于这是一个局部问题, 为简便起见, 我们考虑

$\begin{equation} \left\{\begin{array}{l} \Delta u_{1 k}+2 h_{1 k} {\rm e}^{u_{1 k}}-h_{2 k} {\rm e}^{u_{2 k}}=0, \\ \Delta u_{2 k}+2 h_{2 k} {\rm e}^{u_{2 k}}-3 h_{1 k} {\rm e}^{u_{1 k}}=0, \end{array} \quad \text { 在 } B_{1}\text { 中}.\right. \end{equation}$

对于 $u_{k}=\left(u_{1 k}, u_{2 k}\right)$ 和 $h_{k}=\left(h_{1 k}, h_{2 k}\right)$, 我们假设以下条件 ($\mathrm{H}$) 成立

(i) $ C^{-1} \leq h_{i k} \leq C, \left\|h_{i k}\right\|_{C^{2}\left(B_{1}\right)} \leq C, h_{i k}(0)=1, i=1, 2,$ $C>0$ 是与 $k$ 无关的常数;

(ii) $\int_{B_{1}} h_{i k} {\rm e}^{u_{i k}} \leq C$;

(iii) $\left|u_{i k}(x)-u_{i k}(y)\right| \leq C, \text { 对任意 }\ x, y \in \partial B_{1}$ 成立;

(iv) $\max\limits_{K \subset B_{1} /\{0\}} u_{i k} \leq C(K)$, 0 是唯一的爆破点.

为了研究爆破的局部形态, 我们令

$\begin{equation} -2 \ln \varepsilon_{k}=\max _{x \in B_{1}, i=1, 2} u_{i k}(x), \end{equation}$

以及

$\begin{equation} v_{i k}(y)=u_{i k}\left(\varepsilon_{k} y\right)+2 \ln \varepsilon_{k}, i=1, 2, \end{equation}$

则 $v_{k}=\left(v_{1 k}, v_{2 k}\right)$ 满足

$\begin{equation} \left\{\begin{array}{l} \Delta v_{1 k}+2 h_{1 k}\left(\varepsilon_{k} y\right) {\rm e}^{v_{1 k}}-h_{2 k}\left(\varepsilon_{k} y\right) {\rm e}^{v_{2 k}}=0, \\ \Delta v_{2 k}-3 h_{1 k}\left(\varepsilon_{k} y\right) {\rm e}^{v_{1 k}}+2 h_{2 k}\left(\varepsilon_{k} y\right) {\rm e}^{v_{1 k}}=0. \end{array}\right. \end{equation}$

我们的主要假设是 $v_{k}=\left(v_{1 k}, v_{2 k}\right)$ 在 $\mathbb{R}^{2}$ 的所有紧子集上一致收敛到一个 $G_{2}$ 型 Toda 系统.

定义 2.1 如果 $v_{k}$ 在 $C_{\mathrm{loc}}^{2, \alpha}$ 意义下收敛到 $v=\left(v_{1}, v_{2}\right)$, 且 $v$ 是 $\mathbb{R}^{2}$ 中如下 $G_{2}$ 型 Toda 系统的解, 则我们称 (2.1) 式的 $u_{k}$ 为一个完全爆破序列

$\begin{equation} \left\{\begin{array}{l} \Delta v_{1}+2 {\rm e}^{v_{1}}-{\rm e}^{v_{2}}=0, \\ \Delta v_{2}+2 {\rm e}^{v_{2}}-3 {\rm e}^{v_{1}}=0, \\ \int_{\mathbb{R}^{2}} {\rm e}^{v_{i}}<\infty. \end{array}\right. \end{equation}$

本节的主要目的是证明, 一个完全爆破序列 $u_{k}$ 可以用一列整体解 $U_{k}=\left(U_{1 k}, U_{2 k}\right)$ 来精确逼近, 其中 $U_{k}$ 满足以下系统

$\begin{equation} \left\{\begin{array}{l} \Delta U_{1 k}+2 {\rm e}^{U_{1 k}}-{\rm e}^{U_{2 k}}=0, \\ \Delta U_{2 k}+2 {\rm e}^{U_{2 k}}-3 {\rm e}^{U_{1 k}}=0, \end{array} \quad \text { 在 }\ \mathbb{R}^{2} \text{ 上}.\right. \end{equation}$

定理 2.1 设条件 $\mathrm{(H)}$ 成立, 且 $u_{k}$ 是定义 2.1 中描述的完全爆破序列. 则存在一列方程 (2.6) 的整体解 $U_{k}=(U_{1 k}, U_{2 k})$, 使得

$\begin{equation} \left|u_{i k}\left(\varepsilon_{k} y\right)-U_{i k}\left(\varepsilon_{k} y\right)\right| \leq C(\sigma) \varepsilon_{k}^{\sigma}(1+|y|)^{\sigma}, \quad |y| \leq \frac{1}{\varepsilon_{k}}, \end{equation}$

其中 $\sigma \in(0, 1)$, $C(\sigma)>0$ 是与 $k$ 无关的常数. 此外, 存在与 $k$ 无关的常数 $C>0$, 使得对于$i=1, 2$, $ |y| \leq \varepsilon_{k}^{-1}$, 有

$\begin{equation} \left|U_{i k}\left(\varepsilon_{k} y\right)+4 \ln (1+|y|)\right| \leq C. \end{equation}$

推论 2.1 设 $u_{k}$ 和 $\varepsilon_{k}$ 同定理 2.1 中所述, $v_{k}$ 由公式 (2.3) 定义. 则对于 $i=1, 2$, $|y| \leq \varepsilon_{k}^{-1}$, 有

$\begin{equation} \left|v_{i k}(y)+4 \ln (1+|y|)\right| \leq C. \end{equation}$

对于 $A_{2}$ 型 Toda 系统, 类似的定理已在文献 [38] 中由 Lin-Wei-Zhao 证明. 他们采用的方法是固定初始 Cauchy 数据 $v_{1 k}(0)$、$v_{2 k}(0)$、$\partial_{y} v_{1 k}(0)$、$\partial_{y} v_{2 k}(0)$ 和 $\partial_{y y} v_{1 k}(0)$, 然后求解 8 个代数方程.

Lin-Wei-Zhang 在文献 [37] 中提出了另一种方法, 证明了上述定理对一般的 $S U(n+1)$ 型 Toda 系统成立. 他们在 $\mathbb{R}^{2}$ 中的某些特定点上对 $v_{1 k}$ 和 $U_{1 k}$ 进行匹配. 现在我们遵循 Lin-Wei-Zhang 的思想, 通过在 14 个特定选定的点上匹配 $v_{1 k}$ 和 $U_{1 k}$ 来证明定理 2.1.

首先, 我们收集 $G_{2}$ 型 Toda 系统 (2.5) 的整体解 $\left(v_{1}, v_{2}\right)$ 的几个有用性质. 一个重要观察是, $G_{2}$ 型 Toda 系统可以嵌入到 $A_{6}$ 型 Toda 系统中. 换句话说, 在以下群作用下, $G_{2}$ 型 Toda 系统对应于 $A_{6}$ 型 Toda 系统的解: $u_{3}=u_{1}+\log 2, \ u_{4}=u_{1}+\log 2, \ u_{5}=u_{2}, \ u_{6}=u_{1}$. 更多细节可参见文献 [2].

设 $\left(v_{1}, v_{2}\right)$ 满足方程 (2.5). 我们定义

$\begin{equation} \binom{\tilde{u}_{1}}{\tilde{u}_{2}}=G_{2}^{-1}\binom{v_{1}}{v_{2}}. \end{equation}$

则系统 (2.5) 可转化为

$\begin{equation} \left\{\begin{array}{l} \Delta \tilde{u}_{1}+{\rm e}^{2 \tilde{u}_{1}-\tilde{u}_{2}}=0, \\ \Delta \tilde{u}_{2}+{\rm e}^{2 \tilde{u}_{2}-3 \tilde{u}_{1}}=0, \\ \int_{\mathbb{R}^{2}} {\rm e}^{2 \tilde{u}_{1}-\tilde{u}_{2}}<+\infty, \int_{\mathbb{R}^{2}} {\rm e}^{2 \tilde{u}_{2}-3 \tilde{u}_{1}}<+\infty. \end{array}\right. \end{equation}$

文献 [2] 中的分类结果断言

$\begin{equation}\label{2.12} {\rm e}^{-\tilde{u}_{1}}= 2 \left(\lambda_{0}+\sum_{i=1}^{6} \lambda_{i}\left|P_{i}(z)\right|^{2}\right), \end{equation}$

其中

$ \begin{equation*} P_{i}(z)=z^{i}+\sum_{j=0}^{i-1} c_{i j} z^{j}, \end{equation*}$

其中 $c_{i j}$ 是复数, $\lambda_{i} > 0$, 且解依赖于 14 个参数: $\lambda_{4}$、$\lambda_{5}$、$c_{54}$、$c_{52}$、$c_{53}$、$c_{43}$、$c_{61}$ 和 $c_{62}$.

此外, 我们有以下非退化性结果

推论 2.2 (非退化性) 方程组 (2.11)的线性化算子所对应的解集恰好是 14 维的. 更确切地说, 如果 $\phi=\binom{\phi_{1}}{\phi_{2}}$ 满足对某个 $0 \leq \alpha<1$ 有 $|\phi(z)| \leq C(1+|z|)^{\alpha}$, 且

$ \left\{\begin{array}{c} \Delta \phi_{1}+{\rm e}^{2 \tilde{u}_{1}-\tilde{u}_{2}}\left(2 \phi_{1}-\phi_{2}\right)=0, \\ \Delta \phi_{2}+{\rm e}^{2 \tilde{u}_{2}-3 \tilde{u}_{1}}\left(2 \phi_{2}-3 \phi_{1}\right)=0, \end{array}\right.$

则 $\phi$ 属于以下线性空间 $\mathcal{K}$: 由

$\begin{array}{r} \left\{\tilde{u}_{\lambda_{4}}, \tilde{u}_{\lambda_{5}}, \tilde{u}_{c_{54, 1}}, \tilde{u}_{c_{54, 2}}, \tilde{u}_{c_{52, 1}}, \tilde{u}_{c_{52, 2}}, \tilde{u}_{c_{53, 1}}, \tilde{u}_{c_{53, 2}}, \right. \left.\tilde{u}_{c_{43, 1}}, \tilde{u}_{c_{43, 2}}, \tilde{u}_{c_{61, 1}}, \tilde{u}_{c_{61, 2}}, \tilde{u}_{c_{62, 1}}, \tilde{u}_{c_{62, 2}}\right\} \end{array}$

张成. 这些核函数的精确表达式可在文献 [2,推论 2.3] 中找到.

定理 2.1 的证明 回顾 $(v_{1 k}, v_{2 k})$ 满足方程 (2.4), 且 $v_{k}$ 在 $C_{\mathrm{loc}}^{1, \alpha}\left(\mathbb{R}^{2}\right)$ 中收敛到 $v=\left(v_{1}, v_{2}\right)$, 其中 $v$ 满足方程 (2.5). 根据文献 [2 中的分类定理, 存在参数集 $\Lambda = \{\lambda_{4}, \lambda_{5}, c_{54}, c_{52}, c_{53}, c_{43}, c_{61},$ $ c_{62}\}$, 使得 $\tilde{u}$ 由公式 (2.12) 定义. 为强调对 $\Lambda$ 的依赖性, 我们分别记 $v_{i} = v_{i}(z, \Lambda)$ 和 $\tilde{u}_{i} = \tilde{u}_{i}(z, \Lambda)$.

$\binom{\tilde{u}_{1}(z, \Lambda)}{\tilde{u}_{2}(z, \Lambda)}=G_{2}^{-1}\binom{v_{1}(z, \Lambda)}{v_{2}(z, \Lambda)}, $

$\binom{W_{1 k}}{W_{2 k}}=G_{2}^{-1}\binom{v_{1 k}}{v_{2 k}}.$

证明定理 2.1 的主要思路是寻找一列整体解 $v\left(z, \Lambda_{k}\right)$, 使得 $v_{1 k}$ 在 14 个点上与 $v_{1}\left(z, \Lambda_{k}\right)$ 足够接近. 由于 $\Lambda_{k}$ 有 14 个分量, 我们将证明通过精心选择 14 个点, 可以得到 $\Lambda_{k} \to \Lambda$.

$\begin{aligned} \Theta(q)= & \left(\frac{\partial \tilde{u}_{1}}{\partial \lambda_{4}}, \frac{\partial \tilde{u}_{1}}{\partial \lambda_{5}}, \frac{\partial \tilde{u}_{1}}{\partial c_{54, 1}}, \frac{\partial \tilde{u}_{1}}{\partial c_{54, 2}}, \frac{\partial \tilde{u}_{1}}{\partial c_{51, 1}}, \frac{\partial \tilde{u}_{1}}{\partial c_{51, 2}}, \frac{\partial \tilde{u}_{1}}{\partial c_{53, 1}}, \right. \\ & \left. \frac{\partial \tilde{u}_{1}}{\partial c_{53, 2}}, \frac{\partial \tilde{u}_{1}}{\partial c_{43, 1}}, \frac{\partial \tilde{u}_{1}}{\partial c_{43, 2}}, \frac{\partial \tilde{u}_{1}}{\partial c_{61, 1}}, \frac{\partial \tilde{u}_{1}}{\partial c_{61, 2}}, \frac{\partial \tilde{u}_{1}}{\partial c_{62, 1}}, \frac{\partial \tilde{u}_{1}}{\partial c_{62, 2}} \right)^{\prime}. \end{aligned}$

其中 $()^{\prime}$ 表示转置.

下面的矩阵起到关键作用: 对于 $q_{1}, q_{2}, \cdots, q_{14} \in \mathbb{C}$,

$ M = \left( \Theta\left(q_{1}\right), \cdots, \Theta\left(q_{14}\right) \right). $

通过精心选择 $q_{1}, \cdots, q_{14}$, 我们可以使 $M$ 可逆. $q_{1}, \cdots, q_{14}$ 的选择方法在文献 [第 4 节] 中进行了描述.

由于 $v_{k} \to v$ 在 $C^{1, \alpha}$ 中收敛, 我们已经知道在 $\mathbb{R}^{2}$ 的任意固定紧子集上, $W_{i k} \to \tilde{u}_{i}(z, \Lambda)$ 在 $C^{1, \alpha}$ 范数意义下成立. 现在, 我们断言: 如果对 $l=1, \cdots, 14$ 有 $\tilde{u}_{1}\left(q_{l}, \Lambda_{k}\right) = W_{1}^{k}\left(q_{l}\right)$, 那么成立

$\begin{equation} \begin{aligned} &\Lambda_{k} \rightarrow \Lambda, \\ \text { 即 }\quad \lambda_{i}^{k} &\rightarrow \lambda_{i}, \ c_{i j}^{k} \rightarrow c_{i j}. \end{aligned} \end{equation}$

事实上, 由 $q_{l}$ 的定义可知 $M$ 是可逆的, 且已知 $W_{1}^{k}\left(q_{l}\right) = \tilde{u}_{1}\left(q_{l}, \Lambda\right) + o(1)$, 因此可推出 (2.13) 式成立.

定义误差函数

$ \phi_{i}^{k} = W_{i k} - \tilde{u}_{i}(\cdot, \Lambda_{k}). $

由 $W_{i k}$ 和 $\tilde{u}(\cdot, \Lambda_{k})$ 所满足的方程, $\phi_{i}^{k}$ 满足

$ \Delta \phi_{i}^{k}(y) + {\rm e}^{\xi_{i}^{k}(y)} \left( \sum_{j=1}^{2} g_{i j} \phi_{j}^{k} \right) = O\left(\varepsilon_{k} |y|\right) {\rm e}^{\sum_{j=1}^{2} g_{i j} W_{j k}}, $

其中

$ {\rm e}^{\xi_{i}^{k}} = \int_{0}^{1} {\rm e}^{\sum_{j=1}^{2} g_{i j} \left( t W_{j k} + (1-t) \tilde{u}_{j}(\cdot, \Lambda_{k}) \right)} \,{\rm d}t, \quad \left( g_{i j} \right) = \begin{pmatrix} 2 & -1 \\ -2 & 3 \end{pmatrix}. $

由于我们知道 $G_{2}$ 型 Toda 系统可嵌入到 $A_{6}$ 型 Toda 系统中,那么根据文献 [35,定理 4.1], ${\rm e}^{\xi_{i}^{k}}$ 在 $\mathbb{R}^{2}$ 的所有紧子集上一致收敛于 ${\rm e}^{v_{i}(\cdot, \Lambda)}$, 并且

$ {\rm e}^{\xi_{i}^{k}(y)} = O(1+|y|)^{-4+o(1)}, \quad |y| \leq \frac{1}{\varepsilon_{k}}. $

因此我们有

$ \Delta \phi_{i}^{k}(y) + {\rm e}^{\xi_{i}^{k}(y)} \left( \sum_{j=1}^{2} g_{i j} \phi_{j}^{k} \right) = \frac{O\left(\varepsilon_{k}\right)}{(1+|y|)^{3+o(1)}}, \quad |y| \leq \frac{1}{\varepsilon_{k}}. $

为方便起见, 我们使用以下函数来消除 $\phi^{k}$ 在 $\partial B_{\frac{1}{\varepsilon_{k}}}$ 上的振荡

$ \begin{cases} \Delta \psi_{i}^{k}=0, \quad&\text { 在 } B_{\frac{1}{\varepsilon_{k}}}\text { 中}, \\ \psi_{i}^{k}=\phi_{i}^{k}-\frac{\varepsilon_{k}}{2 \pi} \int_{\partial B_{\frac{1}{\varepsilon_{k}}}} \phi_{i}^{k}, \quad&\text { 在 } \partial B_{\frac{1}{\varepsilon_{k}}}\text { 上}. \end{cases} $

显然, 我们有

$\begin{equation} \left|\psi_{i}^{k}(y)\right| \leq C \varepsilon_{k}|y|. \end{equation}$

令 $\tilde{\phi}_{i}^{k} = \phi_{i}^{k} - \psi_{i}^{k}$, 则有

$\begin{equation*} \Delta \tilde{\phi}_{i}^{k}(y)+{\rm e}^{\xi_{i}^{k}(y)}\left(\sum_{j=1}^{2} g_{i j} \tilde{\phi}_{j}^{k}\right)=\frac{O\left(\varepsilon_{k}\right)}{(1+|y|)^{3+o(1)}}, \quad |y| \leq \frac{1}{\varepsilon_{k}}, \end{equation*}$

以及

$\tilde{\phi}_{1}^{k}\left(q_{l}\right)=O\left(\varepsilon_{k}\right), \text{ 对所有 } l=1, \cdots, 14.$

$\begin{equation*} H_{k}=\max _{i=1, 2} \max _{|y| \leq \frac{1}{\varepsilon_{k}}} \frac{\left|\tilde{\phi}_{i}^{k}(y)\right|}{\varepsilon_{k}^{\sigma}(1+|y|)^{\sigma}}. \end{equation*}$

对于固定的 $\sigma \in(0, 1)$, 我们断言 $H_{k}$ 是有界的.

我们通过反证法证明. 假设 $H_{k} \to \infty$, 并设 $y_{k}$ 是最大值点. 令

$\begin{equation*} \hat{\phi}_{i}^{k}(y)=\frac{\tilde{\phi}_{i}^{k}(y)}{H_{k} \varepsilon_{k}^{\sigma}\left(1+\left|y_{k}\right|\right)^{\sigma}}, \end{equation*}$

则有

$\begin{equation} \left|\hat{\phi}_{i}^{k}\right| \leq \frac{(1+|y|)^{\sigma}}{\left(1+\left|y_{k}\right|\right)^{\sigma}}. \end{equation}$

因此 $\hat{\phi}^{k}$ 满足的方程为

$\begin{equation*} \Delta \hat{\phi}_{i}^{k}+{\rm e}^{\xi_{i}^{k}}\left(\sum_{j=1}^{2} g_{i j} \hat{\phi}_{j}^{k}\right)=\frac{O\left(\varepsilon_{k}^{1-\sigma}\right)(1+|y|)^{-3+o(1)}}{H_{k}\left(1+\left|y_{k}\right|\right)^{\sigma}}. \end{equation*}$

情况 1 如果 $y_{k} \to y^{*}$, 由公式 (2.15) 可知 $\hat{\phi}_{i}^{k}$ 在 $\mathbb{R}^{2}$ 的任意固定紧子集上一致有界, 则 $(\hat{\phi}_{1}^{k}, \hat{\phi}_{2}^{k})$ 收敛到 $(\phi_{1}, \phi_{2})$, 且满足

$\begin{equation}\label{2.16} \left\{\begin{array}{l} \Delta \phi_{i}+{\rm e}^{v_{i}(\dot, \Lambda)}\left(\sum\limits_{j=1}^{2} g_{i j} \phi_{j}\right)=0, \\ \left|\phi_{i}\right| \leq C(1+|y|)^{\sigma}, \\ \phi_{1}\left(q_{l}\right)=0, l=1, \cdots, 14. \end{array}\right. \end{equation}$

根据非退化性结果, 我们知道

$\phi_{1}(y)=\sum_{i=1}^{14} x_{i} \Theta_{i}(y).$

如果我们记

$ X = \left( x_{1}, \cdots, x_{14} \right), $

那么由公式 (2.16) 中的最后一个方程可得

$ X M = 0. $

由于 $M$ 是可逆的, 我们得到 $X = 0$, 从而 $\phi_{1} = 0$ 且 $\phi_{2} = 0$. 这与对某个 $i$ 有 $\left| \phi_{i}(y^{*}) \right| = 1$ 矛盾.

情况 2 $y_{k} \rightarrow \infty$.

$\begin{equation} \begin{aligned} & \hat{\phi}_{i}^{k}\left(y_{k}\right)-\hat{\phi}_{i}^{k}(0) \\ =&\int_{B_{\frac{1}{\varepsilon_{k}}}}\left(G_{k}\left(y_{k}, \eta\right)-G_{k}(0, \eta)\right)\left[{\rm e}^{\xi_{i}^{k}(\eta)}\left(\sum_{j=1}^{2} g_{i j} \tilde{\phi}_{j}^{k}(\eta)\right)+O\left(\frac{(1+|\eta|)^{-3+o(1)}}{H_{k}\left(1+\left|y_{k}\right|\right)^{\sigma}}\right)\right] {\rm d} \eta, \end{aligned} \end{equation}$

其中 $G_{k}$ 是 $B_{\frac{1}{\varepsilon_{k}}}$ 上满足 Dirichlet 边界条件的 Green 函数. 现在我们定义

$\begin{aligned} & \Sigma_{1}=\left\{\eta \in B_{\frac{1}{\varepsilon_{k}}}, |\eta|<\frac{|y|}{2}\right\}, \Sigma_{2}=\left\{\eta \in B_{\frac{1}{\varepsilon_{k}}}, |y-\eta|<\frac{|y|}{2}\right\}, \\ & \Sigma_{3}=B_{\frac{1}{\varepsilon_{k}}} /\left(\Sigma_{1} \cup \Sigma_{2}\right). \end{aligned} $

那么对于 $|y|>2$, 我们有

$ \left|G_{k}(y, \eta)-G_{k}(0, \eta)\right| \leq\left\{\begin{array}{l} C(\ln |y|+|\ln | \eta| |), \eta \in \Sigma_{1}, \\ C(\ln |y|+|\ln | y-\eta| |), \eta \in \Sigma_{2}, \\ \frac{C|y|}{|\eta|}, \eta \in \Sigma_{3}. \end{array}\right. $

由上述估计容易验证, 方程 (2.17) 的右边是 $o(1)$, 而其左边是 $1+o(1)$, 这导出矛盾. 因此 $H_{k}$ 有界, 再结合公式 (2.14), 我们即证明了定理 2.1.

3 关于爆破解的初步估计

本节我们将对 $G_{2}$ 型 Toda 系统进行爆破解分析. 我们将推导爆破解的两个估计: 一个在爆破点附近, 另一个远离爆破点. 为简化表述, 我们将先证明定理 1.2.从现在起到第 8 节, 我们假设 $(u_{1 k}, u_{2 k})$ 是方程 (1.6) 的一个爆破解序列, 并满足条件 $(\mathrm{H})$, 且令 $p_{k j}$ 为 $u_{1 k}$ 在 $p_{j}$ 附近的局部最大值点.

定义

$\begin{equation} -2 \ln \varepsilon_{k, j} =u_{1 k}\left(p_{k j}\right)-\ln \int_{\Omega} h_{1 k} {\rm e}^{u_{1 k}}+\ln \left(\rho_{1 k} h_{1 k}\left(p_{k j}\right)\right), \end{equation}$
$\begin{equation} \varepsilon_{k} =\max _{1 \leq j \leq m} \varepsilon_{k, j}. \end{equation}$

同时设

$\begin{array}{cl} {\rm e}^{\alpha_{1 k}}=\int_{\Omega} h_{1 k} {\rm e}^{u_{1 k}}, & \quad {\rm e}^{\alpha_{2 k}}=\int_{\Omega} h_{2 k} {\rm e}^{u_{2 k}}; \\ \tilde{u}_{1 k}=u_{1 k}-\alpha_{1 k}, & \quad \tilde{u}_{2 k}=u_{2 k}-\alpha_{2 k}, \end{array} $

则 $(\tilde{u}_{1 k}, \tilde{u}_{2 k})$ 满足

$\left\{\begin{array}{l} \Delta \tilde{u}_{1 k}+2 \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}}-\rho_{2 k} h_{2 k} {\rm e}^{\tilde{u}_{2 k}}=0, \\ \Delta \tilde{u}_{2 k}+2 \rho_{2 k} h_{2 k} {\rm e}^{\tilde{u}_{2 k}}-3 \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}}=0. \end{array} \quad \text { 在 } \Omega \text{ 上 }.\right. $

根据第 2 节中的简单爆破结果, 即定理 2.1 和推论 2.1, 我们得到以下引理.

引理3.1 (爆破点附近的估计) 在定理 2.1 的假设下, 存在与 $k$ 无关的小常数 $\delta>0$, 使得

$\begin{equation*} \tilde{u}_{i k}+2 \ln \varepsilon_{k, j}-\left[v_{i}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}, \Lambda^{j}\right)-\ln \left(\rho_{i k} h_{i k}\left(p_{k, j}\right)\right)\right]=O(1) \text { 在 } B_{\delta}\left(p_{j}\right)\text { 中}, \end{equation*}$

其中 $v\left(\cdot; \Lambda^{j}\right)$ 是 $G_{2}$ 型 Toda 系统的整体解.

$ \begin{aligned} \bar{u}_{1 k}&=\tilde{u}_{1 k}(x)+\ln \left(\rho_{1 k} h_{1 k}\left(p_{k j}\right)\right), \\ \bar{u}_{2 k}&=\tilde{u}_{2 k}(x)+\ln \left(\rho_{2 k} h_{2 k}\left(p_{k j}\right)\right), \end{aligned} $

则 $\left(\bar{u}_{1 k}, \bar{u}_{2 k}\right)$ 满足

$ \left\{\begin{array}{l} \Delta \bar{u}_{1 k}+\frac{2 h_{1 k}(x)}{h_{1 k}\left(p_{k j}\right)} {\rm e}^{\bar{u}_{1 k}}-\frac{h_{2 k}(x)}{h_{2 k}\left(p_{k j}\right)} {\rm e}^{\bar{u}_{2 k}}=0, \\ \Delta \bar{u}_{2 k}+\frac{2 h_{2 k}(x)}{h_{2 k}\left(p_{k j}\right)} {\rm e}^{\bar{u}_{2 k}}-\frac{3 h_{1 k}(x)}{h_{1 k}\left(p_{k j}\right)} {\rm e}^{\bar{u}_{1 k}}=0, \end{array} \quad \text { 在 } B_{\delta}\left(p_{j}\right) \text{ 中 }.\right. $

那么 $\bar{u}_{1 k}, \bar{u}_{2 k}$ 满足第 2 节中的条件 ($\mathrm{H}$). 因此我们推断, 存在 $\delta>0$ 和 $C>0$, 使得

$\begin{equation*} \left|\bar{u}_{i k}(x)+2 \ln \varepsilon_{k, j}-v_{i}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}, \Lambda^{j}\right)\right| \leq C, \text { 在 }B_{\delta}\left(p_{j}\right) \text { 中 }. \end{equation*}$

因此我们得到了期望的结果.

注 3.1 这里我们将 $v\left(\cdot, \Lambda^{j}\right)$ 视为第 2 节中的 $v(\cdot, \Lambda)$. 因此存在 $v\left(\cdot, \Lambda_{k}^{j}\right)$, 即定理 2.1 中的 $U_{i k}$.

注 3.2 通过考虑 $\bar{u}_{i k}\left(\varepsilon_{k, j} y + p_{k, j}\right) + 2 \ln \varepsilon_{k, j}$, 我们得到

$ \begin{equation*} \varepsilon_{k, j}\left|\nabla \tilde{u}_{i k}\left(p_{k, j}\right)\right| \leq C, \quad \varepsilon_{k, j}^{2}\left|\nabla^{2} \tilde{u}_{i k}\left(p_{k, j}\right)\right| \leq C. \end{equation*}$

由引理 3.1, 我们得到以下推论

推论 3.1 对于 $i=1, 2$ 和 $j=1, \cdots, m$, 成立

$\begin{equation}\label{estimate alpha} \alpha_{i k}+2 \ln \varepsilon_{k, j}=O(1), \end{equation}$
$\begin{equation}\label{estimate eps} C^{-1} \varepsilon_{k, \ell} \leq \varepsilon_{k, j} \leq C \varepsilon_{k, \ell}, \ell \neq j. \end{equation}$

注意到在 $\partial B_{\delta}(p_{j})$ 上 $v_{i}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}, \Lambda^{j}\right) \sim 4 \ln \varepsilon_{k, j}$, 我们得到公式 (3.3). 公式 (3.4) 直接由公式 (3.3) 推出.

对于固定的 $\delta>0$, 我们定义局部质量 $\rho_{i k, j}$ 为

$ \rho_{i k, j} = \rho_{i k} \int_{B_{\delta}(p_{j})} h_{i k} {\rm e}^{\tilde{u}_{i k}}, \quad \text{对于 } i=1, 2. $

由引理 3.1, 我们有

$\begin{equation*} \rho_{i k, j}=\rho_{i k} \int_{B_{\delta}\left(p_{k j}\right)} h_{i k} {\rm e}^{\tilde{u}_{i k}}+O\left(\varepsilon_{k, j}^{2}\right). \end{equation*}$

注意到 $\rho_{i k} = \rho_{i k} \int_{\Omega} h_{i k} {\rm e}^{\tilde{u}_{i k}}$, 容易看出

$ \rho_{i k} = \sum_{j=1}^{m} \rho_{i k, j} + O\left(\varepsilon_{k}^{2}\right). $

现定义

$ \begin{aligned} & w_{1 k}(x)=2 u_{1 k}(x)+u_{2 k}(x)- \sum_{j=1}^{m} \rho_{1 k, j} G\left(x, p_{k, j}\right), \\ & w_{2 k}(x)=3 u_{1 k}(x)+2 u_{2 k}(x)- \sum_{j=1}^{m} \rho_{2 k, j} G\left(x, p_{k, j}\right). \end{aligned} $

引理 3.2 (远离爆破点的估计) 对于 $i=1, 2$, 我们有

$ \left| w_{i k} \right| + \left| \nabla w_{i k} \right| = O\left( \varepsilon_{k} \right), \quad \text{对于 } x \in \Omega \setminus \bigcup_{j=1}^{m} B_{\delta}(p_{j}). $

容易看出, $\left(2 u_{1 k} + u_{2 k}, 3 u_{1 k} + 2 u_{2 k}\right)$ 满足

$ \left\{ \begin{array}{l} -\Delta \left( 2 u_{1 k} + u_{2 k} \right) = \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}}, \\ -\Delta \left( 3 u_{2 k} + 2 u_{1 k} \right) = \rho_{2 k} h_{2 k} {\rm e}^{\tilde{u}_{2 k}}, \end{array} \right. \quad \text{ 在 } \Omega \text{ 中}. $

那么由 Green 表示公式可得此引理. 事实上, 对于 $\ell=0, 1$ 和 $x \in \Omega \setminus \bigcup_{j=1}^{m} B_{\delta}(p_{j})$, 有

$ \begin{aligned} \partial^{\ell}\left(2 u_{1 k}+u_{2 k}\right) = & \int_{\Omega} \partial^{\ell} G(x, z) \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(z)}{\rm d}z \\ = & \sum_{j=1}^{m} \int_{B_{\frac{\delta}{2}}\left(p_{k, j}\right)} \partial^{\ell} G(x, z) \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(z)}{\rm d}z+O\left(\varepsilon_{k}^{2}\right) \\ = & \sum_{j=1}^{m} \int_{B_{\frac{\delta}{2}}\left(p_{k, j}\right)} \partial^{\ell}\left[G(x, z)-G\left(x, p_{k, j}\right)\right] \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(z)}{\rm d}z \\ &+ \sum_{j=1}^{m} \int_{B_{\frac{\delta}{2}}\left(p_{k, j}\right)} \partial^{\ell} G\left(x, p_{k, j}\right) \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(z)}{\rm d}z+O\left(\varepsilon_{k}^{2}\right) \\ = & \sum_{j=1}^{m} \int_{B_{\frac{\delta}{2}}\left(p_{k, j}\right)} O\left(\left|z-p_{k, j}\right|\right) \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}}{\rm d}z+ \sum_{j=1}^{m} \rho_{1 k, j} \partial^{\ell} G\left(x, p_{k, j}\right)+O\left(\varepsilon_{k}\right)\\ = & \sum_{j=1}^{m} \rho_{1 k, j} \partial^{\ell} G\left(x, p_{k, j}\right)+O\left(\varepsilon_{k}\right). \end{aligned} $

其余证明类似, 故省略.

4 爆破解的精确逼近

在本节中, 我们利用整体解 $\left(v_{1}(\cdot, \Lambda_{k}^{j}), v_{2}(\cdot, \Lambda_{k}^{j})\right)$, 对 $2\tilde{u}_{1k} + \tilde{u}_{2k}$ 和 $3\tilde{u}_{1k} + 2\tilde{u}_{2k}$ 在球 $B_{\delta}(p_{kj})$ 中的爆破行为给出精确描述.

为简便起见, 我们设

$ \begin{aligned} & \tilde{G}_{1 k, j}(x)=\rho_{1 k, j} H\left(x, p_{k, j}\right)+\sum_{\ell \neq j} \rho_{1 k, \ell} G\left(x, p_{k, \ell}\right), \\ & \tilde{G}_{2 k, j}(x)=\rho_{2 k, j} H\left(x, p_{k, j}\right)+\sum_{\ell \neq j} \rho_{2 k, \ell} G\left(x, p_{k, \ell}\right), \end{aligned} $

其中 $H(x, y)$ 是 Green 函数 $G(x, y)$ 的正则部分.

$ \begin{aligned} V_{1 k, j} & =v_{1}\left(x, \Lambda_{k}^{j}\right)-\ln \rho_{1 k} h_{1 k}\left(p_{k, j}\right), \\ V_{2 k, j} & =v_{2}\left(x, \Lambda_{k}^{j}\right)-\ln \rho_{2 k} h_{2 k}\left(p_{k, j}\right), \end{aligned} $

以及

$\begin{equation*} U_{1 k, j}=V_{1 k, j}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}\right), U_{2 k, j}=V_{2 k, j}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}\right). \end{equation*}$

对 $x \in B_{\delta}\left(p_{k, j}\right)$, 定义

$ \begin{aligned} & \eta_{1 k, j}(x)=2 \tilde{u}_{1 k}+\tilde{u}_{2 k}+6 \ln \varepsilon_{k, j}-2 U_{1 k, j}-U_{2 k, j}- \tilde{G}_{1 k, j}+ \tilde{G}_{1 k, j}\left(p_{k, j}\right), \\ & \eta_{2 k, j}(x)=3\tilde{u}_{2 k}+2 \tilde{u}_{1 k}+10 \ln \varepsilon_{k, j}-3 U_{2 k, j}-2 U_{1 k, j}- \tilde{G}_{2 k, j}+ \tilde{G}_{2 k, j}\left(p_{k, j}\right). \end{aligned} $

在 $B_{\delta}\left(p_{k, j}\right) \setminus B_{\frac{\delta}{2}}\left(p_{k, j}\right)$ 中, 由 Taylor 展开容易看出

$ \begin{aligned} & 2 U_{1 k, j}+U_{2 k, j}-6 \ln \varepsilon_{k, j} \\ =&-12 \ln \left|x-p_{k, j}\right|+6 \ln \varepsilon_{k, j}+\ln \frac{1}{2\lambda_{6 k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right)}+O\left(\varepsilon_{k, j}\right), \end{aligned} $
$ \begin{aligned} & 3 U_{2 k, j}+2 U_{1 k, j}-10 \ln \varepsilon_{k, j} \\ =&-20 \ln \left|x-p_{k, j}\right|+10 \ln \varepsilon_{k, j}+\ln \frac{\lambda_{6 k}}{128 \lambda_{5 k}^{4} \left( \rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^3\left(\rho_{2 k} h_{2 k}\left(p_{k, j}\right)\right)^{2}}+O\left(\varepsilon_{k, j}\right). \end{aligned} $

因此在 $B_{\delta}\left(p_{k, j}\right) \setminus B_{\frac{\delta}{2}}\left(p_{k, j}\right)$ 中, 有

$\begin{equation} \begin{aligned} \eta_{1 k, j}(x) = & 2 \tilde{u}_{1 k}+\tilde{u}_{2 k}+12 \ln \left|x-p_{k, j}\right|\\ &-6 \ln \varepsilon_{k, j}-\ln \frac{1}{2\lambda_{6 k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right)} \\ &- \tilde{G}_{1 k, j}(x)+ \tilde{G}_{1 k, j}\left(p_{k, j}\right)+O\left(\varepsilon_{k, j}\right) \\ = & w_{1 k}+ \rho_{1 k, j} G\left(x, p_{k, j}\right)-2 \alpha_{1 k}-\alpha_{2 k}- \rho_{1 k, j} H\left(x, p_{k, j}\right)+\tilde{G}_{1 k, j}\left(p_{k, j}\right) \\ &+ 12 \ln \left|x-p_{k, j}\right|-6 \ln \varepsilon_{k, j}-\ln \frac{1}{2\lambda_{6 k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right)}+O\left(\varepsilon_{k, j}\right) \\ = & \frac{1}{2\pi}\left(24 \pi-\rho_{1 k, j}\right) \ln \left|x-p_{k, j}\right|+A_{1 k, j}+O\left(\varepsilon_{k, j}\right). \end{aligned} \end{equation}$

其中 $A_{1 k, j}$ 是一个常数, 由下式给出

$ A_{1 k, j} = -2 \alpha_{1 k} - \alpha_{2 k} + \tilde{G}_{1 k, j}\left(p_{k, j}\right) - 6 \ln \varepsilon_{k, j} - \ln \frac{1}{2\lambda_{6 k} \left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right)}. $

容易验证 $A_{1 k, j} = O(1)$. 此外, 引理 3.2 表明公式 (4.1) 对 $\nabla \eta_{1 k, j}$ 也成立.

类似地, 在 $B_{\delta}\left(p_{k, j}\right) \setminus B_{\frac{\delta}{2}}\left(p_{k, j}\right)$ 中, 有

$\begin{equation} \eta_{2 k, j}=\frac{1}{2\pi}\left(40 \pi-\rho_{2 k, j}\right) \ln \left|x-p_{k, j}\right|+A_{2 k, j}+O\left(\varepsilon_{k, j}\right), \end{equation}$

其中

$ \begin{aligned} A_{2 k, j}= & -3 \alpha_{2 k}-2 \alpha_{1 k}-10 \ln \varepsilon_{k, j}+ \tilde{G}_{2 k, j}\left(p_{k, j}\right) \\ & -\ln \frac{\lambda_{6 k}}{128 \lambda_{5 k}^{4} \left( \rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{3}\left(\rho_{2 k} h_{2 k}\left(p_{k, j}\right)\right)^{2}}. \end{aligned} $

并且阶的估计为 $O(1)$. 公式 (4.2) 对 $\nabla \eta_{2 k, j}$ 同样成立. 为了估计 $\eta_{i k, j}$ 在整个 $B_{\delta}\left(p_{k, j}\right)$ 上的行为, 我们定义, 对于 $|y| \leq \frac{\delta}{\varepsilon_{k, j}}$,

$ \tilde{\eta}_{1 k, j} = \eta_{1 k, j}\left(\varepsilon_{k, j} y + p_{k, j}\right), \quad \tilde{\eta}_{2 k, j} = \eta_{2 k, j}\left(\varepsilon_{k, j} y + p_{k, j}\right). $

由 $\tilde{\eta}_{1 k, j}$ 和 $\tilde{\eta}_{2 k, j}$ 的定义, 容易看出它们满足

$ \left\{ \begin{array}{l} -\Delta \tilde{\eta}_{1 k, j} = \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} D_{1 k, j}(y), \\ -\Delta \tilde{\eta}_{2 k, j} = \rho_{2 k} h_{2 k}\left(p_{k, j}\right) {\rm e}^{V_{2 k, j}} D_{2 k, j}(y), \quad \text{在 } B_{\frac{\delta}{\varepsilon_{k, j}}} \text{ 中}, \\ \tilde{\eta}_{1 k, j} = O(1), \quad \tilde{\eta}_{2 k, j} = O(1), \quad \text{在 } \partial B_{\frac{\delta}{\varepsilon_{k, j}}} \text{ 上}, \end{array} \right. $

其中

$ \begin{aligned} D_{1 k, j} & =\exp \left[\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)+Q_{1 k, j}\left(\varepsilon_{k, j} y+p_{k, j}\right)-Q_{1 k, j}\left(p_{k, j}\right)\right]-1, \\ D_{2 k, j} & =\exp \left[\left(2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}\right)+Q_{2 k, j}\left(\varepsilon_{k, j} y+p_{k, j}\right)-Q_{2 k, j}\left(p_{k, j}\right)\right]-1, \end{aligned} $

以及

$\begin{equation} Q_{1 k, j} =2 \tilde{G}_{1 k, j}-\tilde{G}_{2 k, j}+\ln h_{1 k}, \end{equation}$
$\begin{equation} Q_{2 k, j} =2 \tilde{G}_{2 k, j}-3 \tilde{G}_{1 k, j}+\ln h_{2 k}. \end{equation}$

由于 $Q_{i k, j}\left(\varepsilon_{k, j} y+p_{k, j}\right) - Q_{i k, j}\left(p_{k, j}\right) = O\left(\varepsilon_{k, j} |y|\right)$, 我们在 $B_{\frac{\delta}{\varepsilon_{k, j}}}$ 中有

$ \left\{\begin{array}{l} -\Delta \tilde{\eta}_{1 k, j}= \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)-1}}{2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}}\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)+O\left(\frac{\varepsilon_{k, j}}{1+|y|^{3}}\right), \\ -\Delta \tilde{\eta}_{2 k, j}= \rho_{2 k} h_{2 k}\left(p_{k, j}\right) {\rm e}^{V_{2 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}\right)-1}}{2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}}\left(2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}\right)+O\left(\frac{\varepsilon_{k, j}}{1+|y|^{3}}\right), \\ \tilde{\eta}_{1 k, j}=O(1), \tilde{\eta}_{2 k, j}=O(1), \quad \text { 在 } \partial B_{\frac{\delta}{\varepsilon_{k, j}}}\text{ 上}. \end{array}\right. $

我们首先得到 $\tilde{\eta}_{i k, j}$ 的以下关键估计

引理 4.1 对任意 $\tau \in(0, 1)$, 在 $B_{\frac{\delta}{\varepsilon_{k, j}}}$ 中成立

$ \left| \tilde{\eta}_{i k, j} \right| \leq C_{\tau} \varepsilon_{k, j} (1+|y|)^{\tau}, $

其中 $i=1, 2$.

证明. 证明与定理 2.1 的证明类似. 设

$ \tilde{H}_{k} = \max_{i=1, 2} \max_{|y| \leq \frac{\delta}{\varepsilon_{k, j}}} \frac{\left| \tilde{\eta}_{i k, j}(y) \right|}{\varepsilon_{k} (1+|y|)^{\tau}}. $

对固定的 $\tau \in(0, 1)$, 我们断言 $\tilde{H}_{k}$ 是有界的. 采用反证法证明.

假设 $\tilde{H}_{k} \to \infty$, 并设 $y_{k}$ 是最大值点. 令

$\begin{equation*} \hat{\eta}_{i k, j}(y)=\frac{\tilde{\eta}_{i k, j}(y)}{\tilde{H}_{k} \varepsilon_{k}\left(1+\left|y_{k}\right|\right)^{\tau}}, \end{equation*}$

则有

$\begin{equation} \left|\hat{\eta}_{i k, j}\right| \leq \frac{(1+|y|)^{\tau}}{\left(1+\left|y_{k}\right|\right)^{\tau}}. \end{equation}$

于是 $\hat{\eta}_{i k, j}$ 满足

$ \left\{\begin{array}{l} -\Delta \hat{\eta}_{1 k, j}= \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)-1}}{2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}}\left(2 \hat{\eta}_{1 k, j}-\hat{\eta}_{2 k, j}\right) \\ +O\left(\frac{\varepsilon_{k, j}}{\tilde{H}_{k} \varepsilon_{k}(1+|y|)^{\tau}\left(1+\left|y_{k}\right|\right)^{3}}\right), \\ -\Delta \hat{\eta}_{2 k, j}= \rho_{2 k} h_{2 k}\left(p_{k, j}\right) {\rm e}^{V_{2 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}\right)-1}}{2 \tilde{\eta}_{2 k, j}-3 \tilde{\eta}_{1 k, j}}\left(2 \hat{\eta}_{2 k, j}-3 \hat{\eta}_{1 k, j}\right) \\ +O\left(\frac{\varepsilon_{k, j}}{\tilde{H}_{k} \varepsilon_{k}(1+|y|)^{\tau}\left(1+\left|y_{k}\right|\right)^{3}}\right), \\ \hat{\eta}_{1 k, j}=O\left(\frac{1}{\tilde{H}_{k} \varepsilon_{k}\left(1+\left|y_{k}\right|\right)^{\tau}}\right), \hat{\eta}_{2 k, j}=O\left(\frac{1}{\tilde{H}_{k} \varepsilon_{k}\left(1+\left|y_{k}\right|\right)^{\tau}}\right), \quad \text { 在 } \partial B_{\frac{\delta}{\varepsilon_{k, j}}} \text{ 上 }. \end{array}\right. $

由公式 (4.5) 可知, $\hat{\eta}_{i k, j}$ 在 $\mathbb{R}^{2}$ 的任意固定紧子集上一致有界.

情况 1 如果 $y_{k} \to y^{*}$, 则 $\left(\hat{\eta}_{1 k, j}, \hat{\eta}_{2 k, j}\right)$ 收敛到 $\left(\eta_{1, j}, \eta_{2, j}\right)$, 且满足

$ \left\{\begin{array}{l} \Delta \eta_{i, j}+{\rm e}^{v_{i}\left(; \Lambda^{j}\right)}\left(\sum_{j=1}^{2} g_{i j} \phi_{j}\right)=0, \\ \left|\eta_{i, j}\right| \leq C(1+|y|)^{\tau}, \\ \eta_{1, j}\left(q_{l, j}\right)=0, l=1, \cdots, 14. \end{array}\right. $

类似于定理 2.1 证明中的论证, 我们得到 $\eta_{1, j}=0$ 且 $\eta_{2, j}=0$. 这与对某个 $i$ 有 $\left|\eta_{i, j}\left(y^{*}\right)\right|=1$ 矛盾.

情况 2 $y_{k} \to \infty$. 不失一般性, 假设 $\left| \hat{\eta}_{1 k, j}\left(y_{k}\right) \right| = 1$, 由 Green 表示公式可得

$ \begin{aligned} \hat{\eta}_{1 k, j}\left(y_{k}\right)= & \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} G_{k}\left(y_{k}, z\right)\left\{ \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)-1}}{2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}}\left(2 \hat{\eta}_{1 k, j}-\hat{\eta}_{2 k, j}\right)\right. \\ & \left.+O\left(\frac{\varepsilon_{k, j}}{\tilde{H}_{k} \varepsilon_{k}\left(1+\left|y_{k}\right|\right)^{\tau}(1+|z|)^{3}}\right)\right\}{\rm d}z-\int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} \frac{\partial G_{k}}{\partial \nu}\left(y_{k}, z\right) \hat{\eta}_{1 k, j}(z){\rm d}s, \end{aligned} $

其中 $G_{k}$ 是 $B_{\frac{\delta}{\varepsilon_{k, j}}}$ 上关于 Dirichlet 边界条件的 Green 函数. 由于

$\hat{\eta}_{1 k, j}(0) = \frac{\tilde{\eta}_{1 k, j}(0)}{\tilde{H}_{k} \varepsilon_{k} \left(1 + |y_{k}| \right)^{\tau}} = O\left( \frac{1}{\tilde{H}_{k} \left(1 + |y_{k}| \right)^{\tau}} \right) = o(1),$

我们有

$\begin{equation} \begin{aligned} & \hat{\eta}_{1 k, j}\left(y_{k}\right)-\hat{\eta}_{1 k, j}(0) \\ =& \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}(0)}\left(G_{k}\left(y_{k}, z\right)-G_{k}(0, z)\right)\left\{ \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \frac{{\rm e}^{\left(2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}\right)-1}}{2 \tilde{\eta}_{1 k, j}-\tilde{\eta}_{2 k, j}}\left(2 \hat{\eta}_{1 k, j}-\hat{\eta}_{2 k, j}\right)\right. \\ & \left.+O\left(\frac{\varepsilon_{k, j}}{\tilde{H}_{k} \varepsilon_{k}\left(1+\left|y_{k}\right|\right)^{\tau}(1+|z|)^{3}}\right)\right\}{\rm d}z-\int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)}\left(\frac{\partial G_{k}}{\partial \nu}\left(y_{k}, z\right)-\frac{\partial G_{k}}{\partial \nu}(0, z)\right) \hat{\eta}_{1 k, j}(z){\rm d}s. \end{aligned} \end{equation}$

对于边界积分, 我们观察到, 由公式 (4.1) 可得

$ \hat{\eta}_{1 k, j}(z) = B_{1 k, j} + O\left( \frac{1}{\tilde{H}_{k} \left(1 + |y_{k}| \right)^{\tau}} \right) = B_{1 k, j} + o(1), $

其中 $B_{1 k, j}$ 是常数. 由于

$ \int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} \frac{\partial G_{k}}{\partial \nu}(y, z) \, {\rm d}s = -1, $

我们有

$ \int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} \left( \frac{\partial G_{k}}{\partial \nu} \left( y_{k}, z \right) - \frac{\partial G_{k}}{\partial \nu}(0, z) \right) B_{1 k, j} \, {\rm d}s = 0, $

因此边界积分可估计为

$ \begin{aligned} & \int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} \left( \frac{\partial G_{k}}{\partial \nu} \left( y_{k}, z \right) - \frac{\partial G_{k}}{\partial \nu}(0, z) \right) \hat{\eta}_{1 k, j} \,{\rm d}s \\ & = o(1) \left| \int_{\partial B_{\frac{\delta}{\varepsilon_{k, j}}}(0)} \left( \frac{\partial G_{k}}{\partial \nu} \left( y_{k}, z \right) - \frac{\partial G_{k}}{\partial \nu}(0, z) \right) \, {\rm d}s \right| \\ & = o(1). \end{aligned} $

类似于定理 (2.1) 证明中的估计, 我们知道公式 (4.6) 右边的第一项是 $o(1)$, 而其左边是 $1+o(1)$, 这导致矛盾.

引理 4.2 对于 $0<\tau<1$, 我们有

$ A_{i k, j}=O\left(\varepsilon_{k}\right). $

由 Green 公式,

$ \begin{aligned} & 2 u_{1 k}\left(p_{k, j}\right)+u_{2 k}\left(p_{k j}\right) \\ & = \sum_{\ell=1}^{m} \int_{B_{\delta}\left(p_{k, \ell}\right)} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(x)} G\left(x, p_{k, j}\right){\rm d}x+O\left(\varepsilon_{k}^{2}\right) \\ & = \sum_{\ell=1}^{m} \int_{B_{\delta}\left(p_{k, \ell}\right)} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(x)}\left(\frac{1}{2 \pi} \ln \frac{1}{\left|x-p_{k, j}\right|}+H\left(x, p_{k, j}\right)\right){\rm d}x+O\left(\varepsilon_{k}^{2}\right) \\ & =\frac{1}{2\pi} \int_{B_{\delta}\left(p_{k, \ell}\right)} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(x)} \ln \frac{1}{\left|x-p_{k, j}\right|}+2 \tilde{G}_{1 k, j}\left(p_{k, j}\right)+O\left(\varepsilon_{k}\right). \end{aligned} $

注意

$\begin{equation*} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}}=\rho_{1 k} h_{1 k} \varepsilon_{k, j}^{-2} {\rm e}^{V_{1 k, j}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}\right)}\left[1+D_{1 k, j}\left(\frac{x-p_{k, j}}{\varepsilon_{k, j}}\right)\right], \end{equation*}$

因此, 回顾 $\tilde{\eta}_{i k, j}$ 的定义, 并利用 $0<\tau<1$ 这一事实, 我们得到

$\begin{align*} & \frac{1}{2\pi} \int_{B_{\delta}\left(p_{k, \ell}\right)} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}(x)} \ln \frac{1}{\left|x-p_{k, j}\right|}{\rm d}x \\ =&-\frac{1}{2\pi} \ln \varepsilon_{k, j} \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}}\left(1+D_{1 k, j}\right){\rm d}y \\ & -\frac{1}{2\pi} \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}}\left(1+D_{1 k, j}\right) \ln |y| {\rm d} y \\ =&-12 \ln \varepsilon_{k, j}+2 V_{1 k, j}(0)+V_{2 k, j}(0)-\ln \frac{1}{2\lambda_{6k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right) } +O\left(\varepsilon_{k, j}\right). \end{align*}$

事实上, 直接计算表明

$ \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}}=24 \pi+O\left(\varepsilon_{k, j}^{2}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} D_{1 k, j} \\ = & \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}}\left[O\left(\left|\tilde{\eta}_{1 k, j}\right|+\left|\tilde{\eta}_{2 k, j}\right|\right)+O\left(\varepsilon_{k, j}|y|\right)\right] \\ = & O\left(\varepsilon_{k, j}\right), \end{aligned} $

以及

$\begin{align*} &-\frac{1}{2\pi} \int_{B \frac{\delta}{\varepsilon_{k, j}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \ln |y| \\ = &\frac{1}{2 \pi} \int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \Delta\left(2 V_{1 k, j}+V_{2 k, j}\right) \ln |y| \\ = & 2 V_{1 k, j}(0)+V_{2 k, j}(0) \\ &+ \frac{1}{2 \pi} \int_{\partial B \frac{\delta}{\varepsilon_{k, j}}}\left[\ln |y| \frac{\partial\left(2 V_{1 k, j}+V_{2 k, j}\right)}{\partial \nu}-\left(2 V_{1 k, j}+V_{2 k, j}\right) \frac{\partial \ln |y|}{\partial \nu}\right] \\ = & 2 V_{1 k, j}(0)+V_{2 k, j}(0) \\ &+ \frac{1}{2 \pi} \int_{\partial B \frac{\delta}{\varepsilon_{k, j}}}\left[-\ln |y| \frac{12}{|y|}+\left(12 \ln |y|-\ln \frac{1}{2\lambda_{6k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right) }\right) \frac{1}{|y|}+O\left(\varepsilon_{k, j}\right)\right. \\ = & -\ln \frac{1}{2\lambda_{6k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right) }+2 V_{1 k, j}(0)+V_{2 k, j}(0)+O\left(\varepsilon_{k, j}\right). \end{align*}$

另一方面, 根据我们的选择, 有

$ \begin{aligned} 2 u_{1 k}\left(p_{k, j}\right)+u_{2 k}\left(p_{k, j}\right) & =2 \tilde{u}_{1 k, j}\left(p_{k, j}\right)+\tilde{u}_{2 k, j}\left(p_{k, j}\right)+2 \alpha_{1 k}+\alpha_{2 k} \\ & =2 \alpha_{1 k}+\alpha_{2 k}+2 \tilde{u}_{1 k, j}\left(p_{k, j}\right)+\tilde{u}_{2 k, j}\left(p_{k, j}\right)-6 \ln \varepsilon_{k, j} \\ & =2 \alpha_{1 k}+\alpha_{2 k}+2 V_{1 k, j}(0)+V_{2 k, j}(0)-6 \ln \varepsilon_{k, j}, \end{aligned} $

于是,

$ \begin{aligned} A_{1 k, j} & =-2 \alpha_{1 k}-\alpha_{2 k}+ \tilde{G}_{1 k, j}\left(p_{k, j}\right)-6 \ln \varepsilon_{k, j}-\ln \frac{1}{2\lambda_{6k}\left(\rho_{1 k} h_{1 k}\left(p_{k, j}\right)\right)^{2} \rho_{2 k} h_{2 k}\left(p_{k, j}\right) } \\ & =O\left(\varepsilon_{k}\right). \end{aligned} $

类似地, 我们可以得到 $A_{2 k, j}$ 的估计.

注 4.1 利用引理 4.2, 我们得到, 在 $B_{\frac{\delta}{\varepsilon_{k, j}}} \setminus B_{\frac{\delta}{2 \varepsilon_{k, j}}}$ 中, 有

$ \begin{aligned} & \eta_{1 k, j}=\frac{1}{2\pi}\left(24 \pi-\rho_{1 k, j}\right) \ln \left|x-p_{k, j}\right|+O\left(\varepsilon_{k}\right), \\ & \eta_{2 k, j}=\frac{1}{2\pi}\left(40 \pi-\rho_{2 k, j}\right) \ln \left|x-p_{k, j}\right|+O\left(\varepsilon_{k}\right). \end{aligned} $

5 $\nabla Q_{1 k, j}$ 与 $\nabla Q_{2 k, j}$ 的估计

在本节中, 我们估计由公式 (4.3) 和 (4.4) 定义的 $Q_{1 k, j}$ 和 $Q_{2 k, j}$ 的梯度. 由于问题是在局部考虑的, 为简化记号, 在不引起混淆的情况下, 我们省略下标 $j$, 并记 $\rho_{1k, j}$ 和 $\rho_{2k, j}$ 为 $\rho_{1k}^{0}$ 和 $\rho_{2k}^{0}$.

命题 5.1 对于 $j=1, \cdots, m$, 我们有

$ \begin{aligned} \nabla Q_{1 k} & =O\left(\varepsilon_{k}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right)+O\left(\varepsilon_{k}\right), \\ \nabla Q_{2 k} & =O\left(\varepsilon_{k}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right)+O\left(\varepsilon_{k}\right). \end{aligned} $

我们设

$ \begin{array}{ll} \psi_{1 k, 1}^{c_{43}}=\partial_{c_{43, 1}^{k}} V_{1 k}(y), & \psi_{43, 1}^{c_{43}}=\partial_{c_{43, 1}^{k}} V_{2 k}(y), \\ \psi_{1 k, 2}^{c_{43}}=\partial_{c_{43, 2}^{k}} V_{1 k}(y), & \psi_{43, 2}^{c_{43}}=\partial_{c_{43, 2}^{k}} V_{2 k}(y), \\ \psi_{1 k, 1}^{c_{54}}=\partial_{c_{54, 1}^{k}} V_{1 k}(y), & \psi_{43, 1}^{c_{54}}=\partial_{c_{54, 1}^{k}} V_{2 k}(y), \\ \psi_{1 k, 2}^{c_{54}}=\partial_{c_{54, 2}^{k}} V_{1 k}(y), & \psi_{43, 2}^{c_{54}}=\partial_{c_{54, 2}^{k}} V_{2 k}(y). \end{array} $

由文献 [2,推论 2.3], 容易验证在 $\partial B_{\frac{\delta}{\varepsilon_{k}}}$ 上, 有

$ \begin{aligned} & \psi_{1 k, 1}^{c_{43}}=\frac{3}{r}\cos\theta+O\left(\frac{1}{r^{2}}\right), \quad \psi_{2 k, 1}^{c_{43}}=-\frac{9}{2}\frac{\cos \theta}{r}+O\left(\frac{1}{r^{2}}\right), \\ & \psi_{1 k, 2}^{c_{43}}=\frac{3}{r}\sin\theta+O\left(\frac{1}{r^{2}}\right), \quad \psi_{2 k, 2}^{c_{43}}=-\frac{9}{2} \frac{\sin \theta}{r}+O\left(\frac{1}{r^{2}}\right), \\ & \psi_{1 k, 1}^{c_{54}}=-\frac{2}{r} \cos \theta+O\left(\frac{1}{r^{2}}\right), \quad \psi_{2 k, 1}^{c_{54}}=\frac{4}{r} \cos \theta+O\left(\frac{1}{r^{2}}\right), \\ & \psi_{1 k, 2}^{c_{54}}=-\frac{2}{r} \sin \theta+O\left(\frac{1}{r^{2}}\right), \quad \psi_{2 k, 2}^{c_{54}}=\frac{4}{r} \sin \theta+O\left(\frac{1}{r^{2}}\right), \end{aligned} $
$ \begin{aligned} & \frac{\partial \psi_{1 k, 1}^{c_{43}}}{\partial \nu}=-\frac{3}{r^{2}} \cos \theta+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 1}^{c_{43}}}{\partial \nu}=\frac{9}{2}\frac{\cos \theta}{r^2}+O\left(\frac{1}{r^{3}}\right), \\ & \frac{\partial \psi_{1 k, 2}^{c_{43}}}{\partial \nu}=-\frac{3}{r^{2}} \sin \theta+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 2}^{c_{43}}}{\partial \nu}=\frac{9}{2}\frac{\sin \theta}{r^2}+O\left(\frac{1}{r^{3}}\right), \\ & \frac{\partial \psi_{1 k, 1}^{c_{54}}}{\partial \nu}=\frac{2}{r^{2}} \cos \theta+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 1}^{c_{54}}}{\partial \nu}=-\frac{4}{r^{2}} \cos \theta+O\left(\frac{1}{r^{3}}\right), \\ & \frac{\partial \psi_{1 k, 2}^{c_{54}}}{\partial \nu}=\frac{2}{r^{2}} \sin \theta+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 2}^{c_{54}}}{\partial \nu}=-\frac{4}{r^{2}} \sin \theta+O\left(\frac{1}{r^{3}}\right). \end{aligned} $

通过分部积分, 我们得到以下结果.

$\begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k, 1}^{c_{43}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k, 1}^{c_{43}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \psi_{1 k, 1}^{c_{43}}\right) \tilde{\eta}_{1 k}+\left(-\Delta \psi_{2 k, 1}^{c_{43}}\right) \tilde{\eta}_{2 k}+\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}}\left(\tilde{\eta}_{1 k} \frac{\partial \psi_{1 k, 1}^{c_{43}}}{\partial \nu}-\psi_{1 k, 1}^{c_{43}} \frac{\partial \tilde{\eta}_{1 k}}{\partial \nu}\right) \\ & +\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \left(\tilde{\eta}_{2 k} \frac{\partial \psi_{2 k, 1}^{c_{43}}}{\partial \nu}-\psi_{2 k, 1}^{c_{43}} \frac{\partial \tilde{\eta}_{2 k}}{\partial \nu}\right). \end{aligned}$

根据对 $\eta_{1 k}$ 和 $\eta_{2 k}$ 的估计, 我们得到, 对于 $i=1, 2$, 有

$\begin{equation} \begin{aligned} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \tilde{\eta}_{i k} \frac{\partial \psi_{i k, 1}^{c_{43}}}{\partial \nu}-\psi_{i k, 1}^{c_{43}} \frac{\partial \tilde{\eta}_{i k}}{\partial \nu} =O\left(\varepsilon_{k}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right) +O\left(\varepsilon_{k}^{2}\right). \end{aligned} \end{equation}$

另一方面, 注意到由 $D_{i k, j}$ 的定义, 我们有

$ \begin{aligned} & D_{1 k}(y)=\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right)+\nabla Q_{1 k}\left(p_{k}\right) \varepsilon_{k} y+O\left(\tilde{\eta}_{1 k}^{2}+\tilde{\eta}_{2 k}^{2}\right)+O\left(\varepsilon_{k}^{2}|y|^{2}\right), \\ & D_{2 k}(y)=\left(2 \tilde{\eta}_{2 k}-3 \tilde{\eta}_{1 k}\right)+\nabla Q_{2 k}\left(p_{k}\right) \varepsilon_{k} y+O\left(\tilde{\eta}_{1 k}^{2}+\tilde{\eta}_{2 k}^{2}\right)+O\left(\varepsilon_{k}^{2}|y|^{2}\right). \end{aligned} $

于是可以得到

$\begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k, 1}^{c_{43}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k, 1}^{c_{43}} \\ =& \int_{B_{\frac{\delta}{\varepsilon}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k}(y) \psi_{1 k, 1}^{c_{43}}+\rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k}(y) \psi_{2 k, 1}^{c_{43}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) \psi_{1 k, 1}^{c_{43}}+\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(2\tilde{\eta}_{2 k}-3\tilde{\eta}_{1 k}\right) \psi_{1 k, 1}^{c_{43}} \\ & + \nabla Q_{1 k}\left(p_{k}\right) \varepsilon_{k} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y \psi_{1 k, 1}^{c_{43}} +\nabla Q_{2 k}\left(p_{k}\right) \varepsilon_{k} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y \psi_{2 k, 1}^{c_{43}} \\ & +O\left(\varepsilon_{k}^{2}\right). \end{aligned}$

由 $\tilde{\eta}_{1 k}$ 和 $\tilde{\eta}_{2 k}$ 的方程, 以及公式 (5.1)、(5.2) 和 (5.3), 可得

$ \begin{aligned} & \nabla Q_{1 k}\left(p_{k}\right) \varepsilon_{k} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y \psi_{1 k, 1}^{c_{43}}+\nabla Q_{2 k}\left(p_{k}\right) \varepsilon_{k} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y \psi_{2 k, 1}^{c_{43}} \\ &=O\left(\varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right). \end{aligned} $

类似地, 上述过程也可以应用于 $\binom{\psi_{1 k, 2}^{c_{43}}}{\psi_{2 k, 2}^{c_{2}}}$、$\binom{\psi_{1 k, 1}^{c_{54}}}{\psi_{2 k, 1}^{c_{54}}}$ 和 $\binom{\psi_{1 k, 2}^{c_{54}}}{\psi_{2 k, 2}^{c_{54}}}$. 我们需要证明相应的矩阵是非退化的. 由于

$\begin{equation*} \Delta\left(2 \psi_{1 k, 1}^{c_{43}}+\psi_{2 k, 1}^{c_{43}}\right)+ \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 1}^{c_{43}}=0, \end{equation*}$

则有

$\begin{align*} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 1}^{c_{43}} y_{1}&=-\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \Delta\left(2 \psi_{1 k, 1}^{c_{43}}+\psi_{2 k, 1}^{c_{43}}\right) y_{1} \\ & =\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial y_{1}}{\partial \nu}\left(2 \psi_{1 k, 1}^{c_{43}}+\psi_{2 k, 1}^{c_{43}}\right)-\frac{\partial\left(2 \psi_{1 k, 1}^{c_{43}}+\psi_{2 k, 1}^{c_{43}}\right)}{\partial \nu} y_{1} \\ & = \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial y_1}{\partial \nu} \frac{3}{2}\frac{\cos\theta}{r} - \left( -\frac{3}{2}\frac{\cos\theta}{r^2} \right) y_1 + O\left(\varepsilon_{k}\right)\\ & =3\pi + O\left(\varepsilon_{k}\right). \end{align*}$

类似地,

$ \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y_{2} \psi_{1 k, 1}^{c_{43}}=O\left(\varepsilon_{k}\right), \quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k, 1}^{c_{43}} y_{1}=O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y_{2} \psi_{2 k, 1}^{c_{43}}=O\left(\varepsilon_{k}\right). \end{aligned} $

类似地, 我们可以证明

$ \begin{aligned} &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,2}^{c_{43}}y_{1}=O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,2}^{c_{43}}y_{2}=3\pi+O(\varepsilon_{k}),\\ &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,2}^{c_{43}}y_{1}=O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,2}^{c_{43}}y_{2}=O(\varepsilon_{k}). \end{aligned} $
$ \begin{aligned} &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,1}^{c_{54}}y_{1}=O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,1}^{c_{54}}y_{2}=O(\varepsilon_{k}),\\ &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,1}^{c_{54}}y_{1}=4\pi+O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,1}^{c_{54}}y_{2}=O(\varepsilon_{k}). \end{aligned} $
$ \begin{aligned} &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,2}^{c_{54}}y_{1}=O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{1k}h_{1k}(p_{k}){\rm e}^{V_{1k}}\psi_{1k,2}^{c_{54}}y_{2}=O(\varepsilon_{k}),\\ &\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,2}^{c_{54}}y_{1}=O(\varepsilon_{k}),\quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\rho_{2k}h_{2k}(p_{k}){\rm e}^{V_{2k}}\psi_{2k,2}^{c_{54}}y_{2}=4\pi+O(\varepsilon_{k}). \end{aligned} $

上述计算表明了矩阵的非退化性.

6 $\rho_{1 k, j} - 24\pi$ 与 $\rho_{2 k, j} - 40\pi$ 的估计

引理 6.1 成立以下估计

$\begin{equation}\label{6.1} \rho_{1 k, j}-24 \pi =-\int_{\partial B_{\delta}\left(p_{k, j}\right)} \frac{\partial \eta_{1 k, j}}{\partial \nu}+O\left(\varepsilon_{k}^{2}\right), \end{equation}$
$\begin{equation} \rho_{2 k, j}-40 \pi =-\int_{\partial B_{\delta}\left(p_{k, j}\right)} \frac{\partial \eta_{2 k, j}}{\partial \nu}+O\left(\varepsilon_{k}^{2}\right). \end{equation}$

由局部质量 $\rho_{i k, j}$ 的定义, 即有

$\begin{align*} \rho_{1 k, j} & =\int_{B_{\delta}\left(p_{k, j}\right)} \rho_{1 k} h_{1 k} {\rm e}^{\tilde{u}_{1 k}} \\ & =-\int_{B_{\delta}\left(p_{k, j}\right)} \Delta\left(2 \tilde{u}_{1 k}+\tilde{u}_{2 k}\right) \\ & =-\int_{B_{\delta}\left(p_{k, j}\right)} \Delta \eta_{1 k, j}-\int_{B_{\delta}\left(p_{k, j}\right)} \Delta\left(2 U_{1 k, j}+U_{2 k, j}\right) \\ & =-\int_{\partial B_{\delta}\left(p_{k, j}\right)} \frac{\partial \eta_{1 k, j}}{\partial \nu}+\int_{B_{\frac{\delta}{\varepsilon_{k, j}}}} \rho_{1 k} h_{1 k}\left(p_{k, j}\right) {\rm e}^{V_{1 k, j}} \\ & =24 \pi-\int_{\partial B_{\delta}\left(p_{k, j}\right)} \frac{\partial \eta_{1 k, j}}{\partial \nu}+O\left(\varepsilon_{k, j}^{2}\right). \end{align*}$

公式 (6.2) 可类似证明.

命题 6.1

$ \begin{aligned} & \rho_{1 k, j}-24 \pi=C_{1 k, j} \Delta_{x} Q_{1 k, j}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \\ & \rho_{2 k, j}-40 \pi=C_{1 k, j} \Delta_{x} Q_{2 k, j}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \end{aligned} $

其中 $0 < \frac{1}{C} \leq C_{i k, j} \leq C$, $C$ 为某常数.

证明. 为简化记号, 如前一节所述, 我们省略下标 $j$. 设

$ \begin{array}{ll} \psi_{1 k}^{\lambda_{4}}=\partial_{\lambda_{4}^{k}} V_{1 k}, & \psi_{1 k}^{\lambda_{5}}=\partial_{\lambda_{5}^{k}} V_{1 k}, \\ \psi_{2 k}^{\lambda_{4}}=\partial_{\lambda_{4}^{k}} V_{2 k}, & \psi_{2 k}^{\lambda_{5}}=\partial_{\lambda_{5}^{k}} V_{2 k}. \end{array} $

Taylor 展开给出

$\begin{align*} & \psi_{1 k}^{\lambda_{4}}=\frac{1}{\lambda_{4 k}}+O\left(\frac{1}{r^{2}}\right), &\frac{\partial \psi_{1 k}^{\lambda_{4}}}{\partial \nu}=O\left(\frac{1}{r^{3}}\right), \\ & \psi_{2 k}^{\lambda_{4}}=-\frac{1}{\lambda_{4 k}}+O\left(\frac{1}{r^{2}}\right), &\frac{\partial \psi_{2 k}^{\lambda_{4}}}{\partial \nu}=O\left(\frac{1}{r^{3}}\right), \\ & \psi_{1 k}^{\lambda_{5}}=O\left(\frac{1}{r^{2}}\right), &\frac{\partial \psi_{1 k}^{\lambda_{5}}}{\partial \nu}=O\left(\frac{1}{r^{3}}\right), \\ & \psi_{2 k}^{\lambda_{5}}=\frac{1}{\lambda_{5 k}}+O\left(\frac{1}{r^{2}}\right), &\frac{\partial \psi_{2 k}^{\lambda_{5}}}{\partial \nu}=O\left(\frac{1}{r^{3}}\right). \end{align*}$

分别用 $\psi_{1 k}^{\lambda_{5}}$ 和 $\psi_{2 k}^{\lambda_{5}}$ 检验 $\left( -\Delta \tilde{\eta}_{1 k} \right)$ 和 $\left( -\Delta \tilde{\eta}_{2 k} \right)$, 由 $\psi_{1 k}^{\lambda_{5}}$ 和 $\psi_{2 k}^{\lambda_{5}}$ 的方程, 我们有

$\begin{equation}\label{6.3} \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k}^{\lambda_{5}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k}^{\lambda_{5}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \psi_{1 k}^{\lambda_{5}}\right) \tilde{\eta}_{1 k}+\left(-\Delta \psi_{2 k}^{\lambda_{5}}\right) \tilde{\eta}_{2 k} \\ & +\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}}\left(\frac{\partial \psi_{1 k}^{\lambda_{5}}}{\partial \nu} \tilde{\eta}_{1 k}-\frac{\partial \tilde{\eta}_{1 k}}{\partial \nu} \psi_{1 k}^{\lambda_{5}}\right)+\left(\frac{\partial \psi_{2 k}^{\lambda_{5}}}{\partial \nu} \tilde{\eta}_{2 k}-\frac{\partial \tilde{\eta}_{2 k}}{\partial \nu} \psi_{2 k}^{\lambda_{5}}\right). \end{aligned} \end{equation}$

很容易即可验证

$\begin{equation}\label{6.4} \begin{aligned} & \int_{\partial B \frac{\delta}{\varepsilon_{k}}}\left(\frac{\partial \psi_{1 k}^{\lambda_{5}}}{\partial \nu} \tilde{\eta}_{1 k}-\frac{\partial \tilde{\eta}_{1 k}}{\partial \nu} \psi_{1 k}^{\lambda_{5}}\right) \\ =&O\left(\varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right), \end{aligned} \end{equation}$

以及

$\begin{equation}\label{6.5} \begin{aligned} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}}\left(\frac{\partial \psi_{2 k}^{\lambda_{5}}}{\partial \nu} \tilde{\eta}_{1 k}-\frac{\partial \tilde{\eta}_{1 k}}{\partial \nu} \psi_{2 k}^{\lambda_{5}}\right) =& - \frac{1}{\lambda_{5 k}}\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\tilde{\eta}_{2 k}}{\partial\nu}\\ &+ O\left(\varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right). \end{aligned} \end{equation}$

结合公式 (6.3)、(6.4) 和 (6.5), 我们得到

$\begin{equation} \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k}^{\lambda_{5}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k}^{\lambda_{5}} \\ =& \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \psi_{1 k}^{\lambda_{5}}\right) \tilde{\eta}_{1 k}+\left(-\Delta \psi_{2 k}^{\lambda_{5}}\right) \tilde{\eta}_{2 k} - \frac{1}{\lambda_{5 k}}\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\tilde{\eta}_{2 k}}{\partial\nu}\\ & +O\left(\varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right). \end{aligned} \end{equation}$

另一方面, 利用命题 5.1, 我们得到

$\begin{equation*} \begin{aligned} D_{1 k, j} & =\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) +\frac{1}{2}\nabla^{2} Q_{1 k, j}\left(p_{k}\right) \varepsilon_{k}^{2} y^{2}+O\left(\tilde{\eta}_{1 k}^{2}+\tilde{\eta}_{2 k}^{2}\right) \\ & +O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)|y|\left(\left|\rho_{1 k}^0-24 \pi\right| +\frac{1}{2}\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}|y|+\varepsilon_{k}^{2+\beta}|y|^{2+\beta}\right) \\ & =\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right)+\frac{1}{2}\nabla^{2} Q_{1 k, j}\left(p_{k, j}\right) \varepsilon_{k}^{2} y^{2}+O\left(\varepsilon_{k}^{2}\right)(1+|y|)^{2 \tau} \\ & +O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)(1+|y|)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}|y|+\varepsilon_{k}^{2+\beta}|y|^{2+\beta}\right), \end{aligned} \end{equation*}$
$\begin{equation*} \begin{aligned} D_{2 k, j} & =\left(2 \tilde{\eta}_{2 k}-3 \tilde{\eta}_{1 k}\right) +\frac{1}{2}\nabla^{2} Q_{2 k, j}\left(p_{k}\right) \varepsilon_{k}^{2} y^{2}+O\left(\tilde{\eta}_{1 k}^{2}+\tilde{\eta}_{2 k}^{2}\right) \\ & +O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)|y|\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}|y|+\varepsilon_{k}^{2+\beta}|y|^{2+\beta}\right) \\ & =\left(2 \tilde{\eta}_{2 k}-3 \tilde{\eta}_{1 k}\right) +\frac{1}{2} \nabla^{2} Q_{2 k, j}\left(p_{k, j}\right) \varepsilon_{k}^{2} y^{2}+O\left(\varepsilon_{k}^{2}\right)(1+|y|)^{2 \tau} \\ & +O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)(1+|y|)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}|y|+\varepsilon_{k}^{2+\beta}|y|^{2+\beta}\right). \end{aligned} \end{equation*}$

由 $\tilde{\eta}_{1 k}$ 和 $\tilde{\eta}_{2 k}$ 的方程, 成立以下关系

$ \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k}^{\lambda_{5}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k}^{\lambda_{5}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k} \psi_{1 k}^{\lambda_{5}}+\rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k} \psi_{2 k}^{\lambda_{5}}. \end{aligned} $

由于当 $r \rightarrow \infty$ 时 $\psi_{1 k}^{\lambda_{5}}=O\left(\frac{1}{r^{2}}\right)$, 所以有

$ \begin{aligned} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k} \psi_{1 k}^{\lambda_{5}}=&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(\tilde{\eta}_{1 k}-2\tilde{\eta}_{1 k}\right) \psi_{1 k}^{\lambda_{5}} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^0-40 \pi\right|\right), \end{aligned} $

$\begin{equation} \begin{aligned} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k} \psi_{2 k}^{\lambda_{5}} =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{k}}\left(2 \tilde{\eta}_{2 k}-3\tilde{\eta}_{2 k}\right) \psi_{2 k}^{\lambda_{5}} \\ & +\frac{\varepsilon_{k}^{2}}{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}} \nabla^{2} Q_{2 k, j}\left(p_{k}\right) y^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^0-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right). \end{aligned} \end{equation}$

接下来, 我们断言

$\begin{equation} \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}} \nabla^{2} Q_{2 k, j}\left(p_{k}\right) y^{2} \\ =&\frac{1}{2} \Delta_{x} Q_{2 k, j}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}}|y|^{2}+O(1). \end{aligned} \end{equation}$

事实上, 直接计算表明

$\begin{equation} \begin{aligned} &\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}}\left(y_{1}^{2}-y_{2}^{2}\right) \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}}-\Delta\left(3\psi_{2 k}^{\lambda_{5}} + 2 \psi_{2 k}^{\lambda_{5}}\right)\left(y_{1}^{2}-y_{2}^{2}\right) \\ =&\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\left(y_{1}^{2}-y_{2}^{2}\right)}{\partial \nu}\left(3\psi_{2 k}^{\lambda_{5}}+2 \psi_{2 k}^{\lambda_{5}}\right)-\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\left(3\psi_{2 k}^{\lambda_{5}}+2 \psi_{2 k}^{\lambda_{5}}\right)}{\partial \nu}\left(y_{1}^{2}-y_{2}^{2}\right) \\ =&O(1), \end{aligned} \end{equation}$

$\begin{equation} \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}} y_{1} y_{2} \\ =&-\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \Delta\left(3\psi_{2 k}^{\lambda_{5}}+2 \psi_{2 k}^{\lambda_{5}}\right) y_{1} y_{2} \\ =&\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\left(y_{1}y_{2}\right)}{\partial \nu}\left(3\psi_{2 k}^{\lambda_{5}}+2 \psi_{2 k}^{\lambda_{5}}\right)-\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\left(3\psi_{2 k}^{\lambda_{5}}+2 \psi_{2 k}^{\lambda_{5}}\right)}{\partial \nu}y_{1}y_{2} \\ =&O(1). \end{aligned} \end{equation}$

公式 (6.8) 由公式 (6.9) 和公式 (6.10) 推出.

将公式 (6.8) 代入公式 (6.7),我们得到

$ \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k} \psi_{2 k}^{\lambda_{5}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(2 \tilde{\eta}_{2 k}-3 \tilde{\eta}_{2 k}\right)+\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k, j}\left(p_{j}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}}|y|^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{2 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right). \end{aligned} $

由以上等式, 我们得到

$\begin{equation} \begin{aligned} -\frac{1}{\lambda_{5 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial \tilde{\eta}_{2 k}}{\partial \nu} & =\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{5}}|y|^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{2 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right). \end{aligned} \end{equation}$

紧接着, 我们通过用 $\psi_{1 k}^{\lambda_{4}}$ 和 $\psi_{2 k}^{\lambda_{4}}$ 替换 $\psi_{1 k}^{\lambda_{5}}$ 和 $\psi_{2 k}^{\lambda_{5}}$ 来执行类似过程, 得到

$\begin{align*} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k}^{\lambda_{4}}+\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k}^{\lambda_{4}} \\ =& \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \left( -\Delta \psi_{1 k}^{\lambda_{4}} \right)\tilde{\eta}_{1 k} + \left( -\Delta \psi_{2 k}^{\lambda_{4}} \right)\tilde{\eta}_{2 k}\\ &+ \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \left( \frac{\partial \psi_{1 k}^{\lambda_{4}} }{\partial \nu} \tilde{\eta}_{1 k} - \frac{\partial \tilde{\eta}_{1 k} }{\partial \nu}\psi_{2 k}^{\lambda_{4}} \right) + \left( \frac{\partial \psi_{2 k}^{\lambda_{4}} }{\partial \nu} \tilde{\eta}_{2 k} - \frac{\partial \tilde{\eta}_{2 k} }{\partial \nu}\psi_{2 k}^{\lambda_{4}} \right)\\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \psi_{1 k}^{\lambda_{4}}\right) \tilde{\eta}_{1 k}+\left(-\Delta \psi_{2 k}^{\lambda_{4}}\right) \tilde{\eta}_{2 k} -\frac{1}{\lambda_{4 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial \tilde{\eta}_{1 k}}{\partial \nu}+\frac{1}{\lambda_{4 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial \tilde{\eta}_{2 k}}{\partial \nu} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right), \end{align*}$

$\begin{aligned} & \quad \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k} \psi_{1 k}^{\lambda_{4}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k}^{\lambda_{4}}\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) +\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{1 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k}^{\lambda_{4}}|y|^{2} \\ & \quad+O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right), \end{aligned}$
$\begin{aligned} & \int_{B \frac{\delta}{\varepsilon_{k}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k} \psi_{2 k}^{\lambda_{4}} \\ =& \int_{B \frac{\delta}{\varepsilon_{k}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{4}}\left(2\tilde{\eta}_{2 k}-3\tilde{\eta}_{1 k}\right) +\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{4}}|y|^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right). \end{aligned}$

于是我们得到

$\begin{equation}\label{6.12} \begin{aligned} -\frac{1}{\lambda_{4 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial \tilde{\eta}_{1 k}}{\partial \nu}+\frac{1}{\lambda_{4 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}}\frac{\partial \tilde{\eta}_{2 k}}{\partial \nu} =&\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{1 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k}^{\lambda_{4}}|y|^{2} \\ & +\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{4}}|y|^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right). \end{aligned} \end{equation}$

结合公式 (6.11) 和公式 (6.12), 我们得到

$\begin{aligned} -\frac{1}{\lambda_{4 k}} \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial \tilde{\eta}_{2 k}}{\partial \nu} =& \frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{1 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}|y|^{2} \\ &+\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{1}}|y|^{2} \left(\psi_{2 k}^{\lambda_{4}}+\frac{\lambda_{5 k}}{\lambda_{4 k}} \psi_{2 k}^{\lambda_{5}}\right) \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2}\ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right) \\ & =\frac{\varepsilon_{k}^{2}}{4} \Delta_{x} Q_{2 k}\left(p_{k}\right) \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k}^{\lambda_{1}}|y|^{2} \\ & +O\left(\varepsilon_{k}^{2}\right)+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right), \end{aligned}$

其中最后一个等式使用了 $\left(\psi_{2 k}^{\lambda_{4}}+\frac{\lambda_{5 k}}{\lambda_{4 k}} \psi_{2 k}^{\lambda_{5}}\right) = O\left( \frac{1}{|y|^2} \right)$, 当 $|y| \to \infty$ 时.

由于 ${\rm e}^{V_{1 k}}$ 和 ${\rm e}^{V_{2 k}} \sim |y|^{-4}$, 引理 6.1 与公式 (6.12) 和公式 (6.13) 意味着

$ \begin{aligned} & \rho_{1 k}^{0}-24 \pi=C_{1 k} \Delta_{x} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right) \\ & \rho_{2 k}^{0}-40 \pi=C_{2 k} \Delta_{x} Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|+O\left(\varepsilon_{k}^{2} \ln \varepsilon_{k}\right)\left(\left|\rho_{1 k}^{0}-24 \pi\right|+\left|\rho_{2 k}^{0}-40 \pi\right|\right)+O\left(\varepsilon_{k}^{2}\right) \end{aligned} $

其中 $0<C_{i k}<\infty$.

因此我们得到

$ \begin{aligned} & \rho_{1 k}^{0}-24 \pi=C_{1 k} \Delta_{x} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|+O\left(\varepsilon_{k}^{2}\right), \\ & \rho_{2 k}^{0}-40 \pi=C_{2 k} \Delta_{x} Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2}\left|\ln \varepsilon_{k}\right|+O\left(\varepsilon_{k}^{2}\right). \end{aligned} $

证明完毕.

7 $\nabla^{2} Q_{1 k, j}$ 与 $\nabla^{2} Q_{2 k, j}$ 的估计

命题 7.1 成立以下估计

$ \begin{aligned} & \frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(Q_{2 k}\left(p_{k}\right)-Q_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta Q_{1 k}\left(p_{k}\right) T_{1 k, 1} + \Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 1}\right)=O\left(\varepsilon_{k}^{\beta}\right), \\ & 3 \pi\left(\partial_{12} Q_{2 k}\left(p_{k}\right)-\partial_{12} Q_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta Q_{1 k}\left(p_{k}\right) T_{1 k, 2}+\Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 2}\right)=O\left(\varepsilon_{k}^{\beta}\right), \end{aligned} $

其中

$\begin{equation} \begin{aligned} & T_{1 k, 1}=\int_{\mathbb{R}^{2}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 1}^{c_{31}}|y|^{2}, \\ & T_{1 k, 2}=\int_{\mathbb{R}^{2}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 2}^{c_{31}}|y|^{2}, \\ & T_{2 k, 1}=\frac{1}{3}\int_{\mathbb{R}^{2}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k, 1}^{c_{31}}|y|^{2}, \\ & T_{2 k, 2}=\frac{1}{3}\int_{\mathbb{R}^{2}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k, 2}^{c_{31}}|y|^{2}. \end{aligned} \end{equation}$

我们设

$ \begin{array}{ll} \psi_{1 k, 1}^{c_{31}}=\partial_{c_{31, 1}^{k}} V_{1 k}(y), & \psi_{2 k, 1}^{c_{31}}=\partial_{c_{31, 1}^{k}} V_{2 k}(y), \\ \psi_{1 k, 2}^{c_{31}}=\partial_{c_{31, 2}^{k}} V_{1 k}(y), & \psi_{2 k, 2}^{c_{31}}=\partial_{c_{31, 2}^{k}} V_{2 k}(y), \end{array} $

由 Taylor 展开得到

$ \begin{aligned} & \psi_{1 k, 1}^{c_{53}}=-\frac{15}{2} \frac{\cos (2 \theta)}{r^{2}}+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{1 k, 1}^{c_{53}}}{\partial \nu}=\frac{15}{r^{3}} \cos (2 \theta)+O\left(\frac{1}{r^{4}}\right), \\ & \psi_{2 k, 1}^{c_{53}}=\frac{27}{2} \frac{\cos (2 \theta)}{r^{2}}+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 1}^{c_{53}}}{\partial \nu}=-\frac{27}{r^{3}} \cos (2 \theta)+O\left(\frac{1}{r^{4}}\right), \\ & \psi_{1 k, 2}^{c_{53}}=-\frac{15}{2} \frac{\sin (2 \theta)}{r^{2}}+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{1 k, 2}^{c_{53}}}{\partial \nu}=\frac{15}{r^{3}} \sin (2 \theta)+O\left(\frac{1}{r^{4}}\right), \\ & \psi_{2 k, 2}^{c_{53}}=\frac{27}{2} \frac{\sin (2 \theta)}{r^{2}}+O\left(\frac{1}{r^{3}}\right), \quad \frac{\partial \psi_{2 k, 2}^{c_{53}}}{\partial \nu}=-\frac{27}{r^{3}} \sin (2 \theta)+O\left(\frac{1}{r^{4}}\right). \end{aligned} $

利用前面各节的估计, 我们得到

$ \begin{aligned} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \tilde{\eta}_{1 k}\right) \psi_{1 k, 1}^{c_{53}}+ \frac{1}{3}\left(-\Delta \tilde{\eta}_{2 k}\right) \psi_{2 k, 1}^{c_{53}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}}\left(-\Delta \psi_{1 k, 1}^{c_{53}}\right) \tilde{\eta}_{1 k}+\frac{1}{3}\left(-\Delta \psi_{2 k, 1}^{c_{53}}\right) \tilde{\eta}_{2 k}+O\left(\varepsilon_{k}^{3}\right). \end{aligned} $

由于 $h_{1 k}$ 和 $h_{2 k} \in C^{2, \beta}(\bar{\Omega})$, 根据 $D_{1 k}$ 和 $D_{2 k}$ 的定义, 我们有

$ \begin{aligned} & D_{1 k}=\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right)+\nabla Q_{1 k}\left(p_{k}\right) \varepsilon_{k} y+\frac{1}{2}\nabla^{2} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2} y^{2}+O\left(\varepsilon_{k}^{2+\beta}\right)(1+|y|)^{2+\beta}, \\ & D_{2 k}=\left(2 \tilde{\eta}_{2 k}-3 \tilde{\eta}_{1 k}\right)+\nabla Q_{2 k}\left(p_{k}\right) \varepsilon_{k} y+\frac{1}{2}\nabla^{2} Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2} y^{2}+O\left(\varepsilon_{k}^{2+\beta}\right)(1+|y|)^{2+\beta}. \end{aligned} $

于是,

$ \begin{align*} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k}(y) \psi_{1 k, 1}^{c_{53}} \\ =& \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) \psi_{1 k, 1}^{c_{53}} + \varepsilon_{k}\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\nabla Q_{1 k}\left(p_{k}\right) y \psi_{1 k, 1}^{c_{53}} \\ & +\frac{1}{2} \nabla^{2} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y^{2} \psi_{1 k, 1}^{c_{53}}+O\left(\varepsilon_{k}^{2+\beta}\right)\\ =& \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) \psi_{1 k, 1}^{c_{53}} \\ & +\frac{1}{2} \nabla^{2} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y^{2} \psi_{1 k, 1}^{c_{53}}+O\left(\varepsilon_{k}^{2+\beta}\right). \end{align*}$

在最后一个等式中, 我们利用了 $|\nabla Q_{i k}\left(p_{k}\right)| = O(\varepsilon_{k})$ 以及

$ \begin{aligned} \left|\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 1}^{c_{53}} y \right|& =\left|-\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \Delta\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right) y \right|\\ & =\left|\int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial y}{\partial \nu}\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right)-\frac{\partial\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right)}{\partial \nu} y\right| \\ & =O\left(\varepsilon_{k}\right). \end{aligned} $

类似地,

$ \begin{aligned} & \int_{B \frac{\delta}{\varepsilon_{k}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k}(y) \psi_{2 k, 1}^{c_{53}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(2\tilde{\eta}_{2 k}-3\tilde{\eta}_{1 k}\right) \psi_{2 k, 1}^{c_{53}} \\ & +\frac{1}{2} \nabla^{2} Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y^{2} \psi_{2 k, 1}^{c_{53}}+O\left(\varepsilon_{k}^{2+\beta}\right). \end{aligned} $

显然,

$ \begin{aligned} \nabla^{2} Q_{1 k}\left(p_{k}\right) y^{2} & =\frac{1}{2}\left(\partial_{11}-\partial_{22}\right) Q_{1 k}\left(p_{k}\right)\left(y_{1}^{2}-y_{2}^{2}\right)+\frac{1}{2} \Delta Q_{1 k}\left(p_{k}\right)|y|^{2}+2 \partial_{12} Q_{1 k}\left(p_{k}\right) y_{1} y_{2}, \\ \nabla^{2} Q_{2 k}\left(p_{k}\right) y^{2} & =\frac{1}{2}\left(\partial_{11}-\partial_{22}\right) Q_{2 k}\left(p_{k}\right)\left(y_{1}^{2}-y_{2}^{2}\right)+\frac{1}{2} \Delta Q_{2 k}\left(p_{k}\right)|y|^{2}+2 \partial_{12} Q_{2 k}\left(p_{k}\right) y_{1} y_{2}. \end{aligned} $

可以进一步验证

$ \begin{align*} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(y_{1}^{2}-y_{2}^{2}\right) \psi_{1 k, 1}^{c_{53}} \\ = & -\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \Delta\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right)\left(y_{1}^{2}-y_{2}^{2}\right) \\ = & \int_{\partial B_{\frac{\delta}{\varepsilon_{k}}}} \frac{\partial\left(y_{1}^{2}-y_{2}^{2}\right)}{\partial \nu}\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right)-\frac{\partial\left(2 \psi_{1 k, 1}^{c_{53}}+\psi_{2 k, 1}^{c_{53}}\right)}{\partial \nu}\left(y_{1}^{2}-y_{2}^{2}\right) \\ = & -6 \pi+O\left(\varepsilon_{k}\right), \end{align*}$

类似地, 可以得到

$ \begin{align*} & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y_{1} y_{2} \psi_{1 k, 1}^{c_{53}} =O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(y_{1}^{2}-y_{2}^{2}\right) \psi_{2 k, 1}^{c_{53}} =18 \pi+O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y_{1} y_{2} \psi_{2 k, 1}^{c_{53}} =O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(y_{1}^{2}-y_{2}^{2}\right) \psi_{1 k, 2}^{c_{53}}=O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y_{1} y_{2} \psi_{1 k, 2}^{c_{53}} =-3 \pi+O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(y_{1}^{2}-y_{2}^{2}\right) \psi_{2 k, 2}^{c_{53}}=O\left(\varepsilon_{k}\right), \\ & \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y_{1} y_{2} \psi_{2 k, 2}^{c_{53}} =9 \pi+O\left(\varepsilon_{k}\right). \end{align*}$

此外, 我们用 $T_{i k, j}$ 表示以下常数

$ \begin{aligned} & T_{1 k, 1}=\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 1}^{c_{31}}|y|^{2}, \\ & T_{1 k, 2}=\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} \psi_{1 k, 2}^{c_{31}}|y|^{2}, \\ & T_{2 k, 1}=\frac{1}{3}\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k, 1}^{c_{31}}|y|^{2}, \\ & T_{2 k, 2}=\frac{1}{3}\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} \psi_{2 k, 2}^{c_{31}}|y|^{2}. \end{aligned} $

因此上述结果意味着

$ \begin{aligned} & \frac{1}{2} \nabla^{2} Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} y^{2} \psi_{1 k, 1}^{c_{31}} \\ =&-\frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right) Q_{1 k}\left(p_{k}\right) \varepsilon_{k}^{2}+\frac{1}{4} \Delta Q_{1 k}\left(p_{k}\right) T_{1 k, 1} \varepsilon_{k}^{2}+O\left(\varepsilon_{k}^{3}\right), \\ & \frac{1}{2} \nabla^{2} Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2} \int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} y^{2} \psi_{2 k, 1}^{c_{31}} \\ =&\frac{3}{2}\pi\left(\partial_{11}-\partial_{22}\right) Q_{2 k}\left(p_{k}\right) \varepsilon_{k}^{2}+\frac{1}{4} \Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 1} \varepsilon_{k}^{2}+O\left(\varepsilon_{k}^{3}\right). \end{aligned} $

从而我们得出结论

$ \begin{aligned} & \int_{B \frac{\delta}{\varepsilon_{k}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}} D_{1 k} \psi_{1 k, 1}^{c_{31}}+\frac{1}{3}\rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}} D_{2 k} \psi_{2 k, 1}^{c_{31}} \\ =&\int_{B_{\frac{\delta}{\varepsilon_{k}}}} \rho_{1 k} h_{1 k}\left(p_{k}\right) {\rm e}^{V_{1 k}}\left(2 \tilde{\eta}_{1 k}-\tilde{\eta}_{2 k}\right) \psi_{1 k, 1}^{c_{31}}+\frac{1}{3}\rho_{2 k} h_{2 k}\left(p_{k}\right) {\rm e}^{V_{2 k}}\left(\tilde{\eta}_{2 k}-\tilde{\eta}_{1 k}\right) \psi_{2 k, 1}^{c_{31}} \\ & +\frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(Q_{2 k}\left(p_{k}\right)-Q_{1 k}\left(p_{k}\right)\right) \varepsilon_{k}^{2}+\frac{1}{4}\left(\Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 1}+\Delta Q_{1 k} T_{1 k, 1}\right) \varepsilon_{k}^{2}+O\left(\varepsilon_{k}^{2+\beta}\right). \end{aligned} $

因此, 我们有

$\frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(Q_{2 k}\left(p_{k}\right)-Q_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta Q_{1 k}\left(p_{k}\right) T_{1 k, 1} + \Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 1}\right)=O\left(\varepsilon_{k}^{\beta}\right), $

类似地, 我们有

$ 3 \pi\left(\partial_{12} Q_{2 k}\left(p_{k}\right)-\partial_{12} Q_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta Q_{1 k}\left(p_{k}\right) T_{1 k, 2}+\Delta Q_{2 k}\left(p_{k}\right) T_{2 k, 2}\right)=O\left(\varepsilon_{k}^{\beta}\right), $

证明完成.

8 定理 1.2 与定理 1.1 的证明

8.1 定理 1.2 的证明

将命题 6.1 应用于命题 5.1 可得

$ \nabla Q_{1 k, j}\left(p_{k, j}\right) = O\left(\varepsilon_{k}\right), \quad \nabla Q_{2 k, j}\left(p_{k, j}\right) = O\left(\varepsilon_{k}\right). $

由 $Q_{1 k, j}$ 和 $Q_{2 k, j}$ 的定义, 我们有

$ \begin{aligned} Q_{1 k, j} & = 2 \tilde{G}_{1 k, j} - \tilde{G}_{2 k, j} + \ln h_{1 k}, \\ Q_{2 k, j} & = 2 \tilde{G}_{2 k, j} - 3 \tilde{G}_{1 k, j} + \ln h_{2 k}. \end{aligned} $

因此我们得到

$ \begin{aligned} & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{1 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right), \\ & 8 \pi \nabla_{x} H\left(p_{k, j}, p_{k, j}\right)+8 \pi \sum_{j \neq \ell} \nabla_{x} G\left(p_{k, j}, p_{k, \ell}\right)+\nabla \ln h_{2 k}\left(p_{k, j}\right)=O\left(\varepsilon_{k}\right). \end{aligned} $

这就证明了定理 1.2 的结论 (ii).

类似地, 由命题 7.1 可得

$ \begin{array}{ll} \frac{3}{2} \pi\left(\partial_{11}-\partial_{22}\right)\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right) T_{2 k, 1}^{j}+\Delta \ln h_{1 k} T_{1 k, 1}^{j}\right)=O\left(\varepsilon_{k}^{\beta}\right), \\ 3 \pi \partial_{12}\left(\ln h_{2 k}\left(p_{k}\right)-\ln h_{1 k}\left(p_{k}\right)\right)+\frac{1}{4}\left(\Delta \ln h_{2 k}\left(p_{k}\right) T_{2 k, 2}^{j}+\Delta \ln h_{1 k} T_{1 k, 2}^{j}\right)=O\left(\varepsilon_{k}^{\beta}\right). \end{array} $

这就证明了定理 1.2 的结论 (iii).

最后还需要计算 $\rho_{1 k} - 24 m \pi$ 和 $\rho_{2 k} - 40 m \pi$. 回顾之前有

$ \rho_{1 k}=\sum_{j=1}^{m} \rho_{1 k, j}+O\left(\varepsilon_{k}^{2}\right), $

注意到 $\Delta_{x} Q_{i k, j}\left(p_{k, j}\right) = \Delta \ln h_{i k}\left(p_{k, j}\right)$, 并利用命题 6.1, 我们有

$ \begin{aligned} & \rho_{1 k}-24 m \pi=\sum_{j=1}^{m} C_{1 k, j} \Delta \ln h_{1 k}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right), \\ & \rho_{2 k}-40 m \pi=\sum_{j=1}^{m} C_{2 k, j} \Delta \ln h_{2 k}\left(p_{k, j}\right) \varepsilon_{k, j}^{2}\left|\ln \varepsilon_{k, j}\right|+O\left(\varepsilon_{k}^{2}\right). \end{aligned} $

至此完成了定理 1.2 的全部证明.

8.2 定理 1.1 的证明

现在我们考虑 Riemann 曲面上的情形. 回顾 Riemann 曲面上的 $G_{2}$ 型 Toda 系统

$ \left\{\begin{array}{l} \Delta_g u_{1 k}+2 \rho_{1 k}\left(\frac{h_{1 k} {\rm e}^{u_{1 k}}}{\int_{M} h_{1 k} {\rm e}^{u_{1 k}}}-1\right)-\rho_{2 k}\left(\frac{h_{2 k} {\rm e}^{u_{2 k}}}{\int_{M} h_{2 k} {\rm e}^{u_{2 k}}}-1\right)=0, \\ \Delta_g u_{2 k}-3 \rho_{1 k}\left(\frac{h_{1 k} {\rm e}^{u} 1 k}{\int_{M} h_{1 k} {\rm e}^{u_{1 k}}}-1\right)+2 \rho_{2 k}\left(\frac{h_{2 k} {\rm e}^{2 k}}{\int_{M} h_{2 k} {\rm e}^{u_{2 k}}}-1\right)=0. \end{array}\right. $

设 $\tilde{u}_{i k}$, $\varepsilon_{k, j}$ 及其他记号如前定义. 令

$ \begin{aligned} & w_{1 k}=2 u_{1 k}+u_{2 k}-\sum_{j=1}^{m} \rho_{1 k, j} G\left(x, p_{k, j}\right)-2 \bar{u}_{1 k}-\bar{u}_{2 k}, \\ & w_{2 k}=3 u_{1 k}+2 u_{2 k}-\sum_{j=1}^{m} \rho_{2 k, j} G\left(x, p_{k, j}\right)-3 \bar{u}_{1 k}-2 \bar{u}_{2 k}. \end{aligned} $

其中 $\bar{u}_{i k}$ 是 $u_{i k}$ 在 $M$ 上的平均值, $G(x, p)$ 是 $M$ 上以 $p$ 为奇点的 Laplace 算子 $\Delta$ 的 Green 函数. 类似地, 我们得到以下估计

引理 8.1 对于 $i=1, 2$, 有

$ |w_{i k}| + |\nabla w_{i k}| = O(\varepsilon_k), \quad \text{对于 } x \in M \backslash \bigcup_{j=1}^{m} B_{\delta}(p_j). $

由于前面各节的分析是局部的, 我们引入一些记号以便进行局部计算. 为了方便, 引入局部坐标 $x$ 使得 $p_{k, j}$ 的坐标为 0, 且度量 $g_{i j} = {\rm e}^{\phi} \delta_{i j}$ 满足 $\phi(0) = \nabla \phi(0) = 0$. 此时, $\tilde{u}_{i k}$ 将满足以下方程

$ \left\{\begin{array}{l} \Delta \tilde{u}_{1 k}+2 \rho_{1 k} {\rm e}^{\phi}\left(h_{1 k} {\rm e}^{\tilde{u}_{1 k}}-1\right)-\rho_{2 k} {\rm e}^{\phi}\left(h_{2 k} {\rm e}^{\tilde{u}_{2 k}}-1\right)=0, \\ \Delta \tilde{u}_{2 k}-3 \rho_{1 k} {\rm e}^{\phi}\left(h_{1 k} {\rm e}^{\tilde{u}_{1 k}}-1\right)+2 \rho_{2 k} {\rm e}^{\phi}\left(h_{2 k} {\rm e}^{\tilde{u}_{2 k}}-1\right)=0, \end{array} \quad \text { 对于 }|x| \leq \delta_{0}.\right. $

其中 $\Delta$ 表示 $\mathbb{R}^{2}$ 中的 Laplace 算子. 此外, 我们设

$ \begin{equation*} \hat{u}_{1 k}=\tilde{u}_{1 k}-\left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}(x), \hat{u}_{2 k}=\tilde{u}_{2 k}-\left(2 \rho_{2 k}-3 \rho_{1 k}\right) f_{k}(x), \end{equation*}$

其中 $f_{k}$ 定义为

$ \left\{\begin{array}{l} \Delta f_{k}={\rm e}^{\phi},\quad \text { 对于 }|x| \leq \delta_{0}, \\ f_{k}(0)=\nabla f_{k}(0)=0. \end{array}\right. $

显然 $(\hat{u}_{1 k}, \hat{u}_{2 k})$ 满足以下方程组

$ \begin{equation} \left\{\begin{array}{l} \Delta \hat{u}_{1 k}+2 \rho_{1 k} \hat{h}_{1 k} {\rm e}^{\hat{u}_{1 k}}-\rho_{2 k} \hat{h}_{2 k} {\rm e}^{\hat{u}_{2 k}}=0, \\ \Delta \hat{u}_{2 k}-3 \rho_{1 k} \hat{h}_{1 k} {\rm e}^{\hat{u}_{1 k}}+2 \rho_{2 k} \hat{h}_{2 k} {\rm e}^{\hat{u}_{2 k}}=0, \end{array} \quad \text { 对于 }|x| \leq \delta_{0}, \right. \end{equation}$

其中

$ \begin{equation*} \hat{h}_{1 k}={\rm e}^{\phi} h_{1 k} {\rm e}^{\left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}}, \hat{h}_{2 k}={\rm e}^{\phi} h_{2 k} {\rm e}^{\left(2 \rho_{2 k}-3\rho_{1 k}\right) f_{k}}. \end{equation*}$

因此, 第 4 节到第 8 节的类似过程可应用于 (8.1) 式. 现在注意到

$ \begin{aligned} Q_{1 k, j} & =2 \tilde{G}_{1 k, j}-\tilde{G}_{2 k, j}+\ln \hat{h}_{1 k} \\ & =2 \tilde{G}_{1 k, j}-\tilde{G}_{2 k, j}+\ln h_{1 k}+\phi+\left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}, \\ Q_{2 k, j} & =2 \tilde{G}_{2 k, j}-3 \tilde{G}_{1 k, j}+\ln \hat{h}_{2 k} \\ & =2 \tilde{G}_{2 k, j}-3 \tilde{G}_{1 k, j}+\ln h_{2 k}+\phi+\left(2 \rho_{2 k}-3 \rho_{1 k}\right) f_{k}. \end{aligned} $

利用以下事实

$ \begin{aligned} & \nabla \phi\left(p_{k, j}\right)=\nabla f_{k}\left(p_{k, j}\right)=0, \quad \Delta \phi\left(p_{k, j}\right)=-2 K\left(p_{k, j}\right), \\ & \left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}\left(p_{k, j}\right)=8 m \pi+2\left(\rho_{1 k}-24 m \pi\right)-\left(\rho_{2 k}-40 m \pi\right), \\ & \left(2 \rho_{2 k}-3 \rho_{1 k}\right) f_{k}\left(p_{k, j}\right)=8 m \pi+2\left(\rho_{2 k}-40 m \pi\right)-3\left(\rho_{1 k}-24 m \pi\right), \\ & \left(\partial_{11}-\partial_{22}\right)\left[\left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}-\left(2 \rho_{2 k}-3 \rho_{1 k}\right) f_{k}\right]=O\left(\left|\rho_{1 k}-24 m \pi\right|+\left|\rho_{2 k}-40 m \pi\right|\right), \\ & \partial_{12}\left[\left(2 \rho_{1 k}-\rho_{2 k}\right) f_{k}-\left(2 \rho_{2 k}-3 \rho_{1 k}\right) f_{k}\right]=O\left(\left|\rho_{1 k}-24 m \pi\right|+\left|\rho_{2 k}-40 m \pi\right|\right), \end{aligned} $

即得到定理 1.1.

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