数学物理学报, 2025, 45(5): 1565-1576

奇异摄动积分边值问题中的空间对照结构

武利猛,1,2, 李素红,1, 倪明康,3,*, 陆海波,4

1河北科技师范学院数学与信息科技学院 河北秦皇岛 066004

2河北省海洋动力过程与资源环境重点实验室 河北秦皇岛 066004

3华东师范大学数学科学学院 上海 200241

4上海应用技术大学经管学院 上海 201418

Contrast Structure of Singularly Perturbed Integral Boundary Value Problem

Wu Limeng,1,2, Li Suhong,1, Ni Mingkang,3,*, Lu Haibo,4

1School of Mathematics and Information Technology, Hebei Normal University of Science and Technology, Hebei Qinhuangdao 066004

2Hebei Key Laboratory of Ocean Dynamics, Resources and Environments, Hebei Qinhuangdao 066004

3School of Mathematical Sciences, East China Normal University, Shanghai 200241

4School of Economics and Management, Shanghai Institute of Technology, Shanghai 201418

通讯作者: * 倪明康,E-mail:xiaovikdo@163.com

收稿日期: 2022-08-25   修回日期: 2024-01-10  

基金资助: 国家自然科学基金(11871217)
国家自然科学基金(11901152)
国家自然科学基金(72371097)

Received: 2022-08-25   Revised: 2024-01-10  

Fund supported: NSFC(11871217)
NSFC(11901152)
NSFC(72371097)

作者简介 About authors

武利猛,E-mail:neamou123@163.com;

李素红,E-mail:lisuhong103@126.com;

陆海波,E-mail:hblu1985@163.com

摘要

该文研究了一类带有积分边界条件的奇异摄动边值问题的空间对照结构, 利用奇异摄动几何理论和微积分技巧, 证明了等价边值问题阶梯状空间对照结构解的存在性. 基于解的结构, 利用边界层函数法构造了一致有效的形式渐近解. 通过一个例子, 验证了主要结果.

关键词： 奇异摄动; 渐近解; 空间对照结构

Abstract

In this paper, we consider the contrast structure solution for singularly perturbed boundary value problem with integral boundary condition. By using geometric singular perturbation theory, as well as analysis technique, we first establish the existence result of step-like contrast structure solution for the corresponding equivalent boundary value problem. Furthermore, by virtue of the structure of the solution, we construct the uniformly valid formal asymptotic solution by the boundary function method. Finally, an example is given to show the main result.

Keywords： singular perturbation; asymptotic solution; contrast structure

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本文引用格式

武利猛, 李素红, 倪明康, 陆海波. 奇异摄动积分边值问题中的空间对照结构[J]. 数学物理学报, 2025, 45(5): 1565-1576

Wu Limeng, Li Suhong, Ni Mingkang, Lu Haibo. Contrast Structure of Singularly Perturbed Integral Boundary Value Problem[J]. Acta Mathematica Scientia, 2025, 45(5): 1565-1576

1 引言

奇异摄动边值问题在神经生物学、量子力学、流体动力学、海洋学、应用数学等领域有着重要的应用 [1-9]. 在一些热传导、半导体 [10]和生物医学 [11]等问题的研究过程中, 学者们发现轨线的初始时刻和终端时刻的值不是固定的, 其中一种情形为可移动的积分边界 [12]. Cakir M 和 Amiraliyev G M 讨论了带有积分边界条件的奇异摄动边值问题, 利用有限差分的方法, 构造了具有边界层的数值解 [13]. 文献 [14] 中, Kumar D 和 Kumari P 研究了奇异摄动边值问题

$\begin{equation*} \begin{cases} \displaystyle -\varepsilon u''(x)+p(x)u'(x)+q(x)u(x)+r(x)u(x-1)=\psi(x), \quad 0<x<2,\\ u(x)=\phi(x),\quad -1\leqslant x\leqslant 0,\\ \displaystyle u(2)=k+\varepsilon\int_{0}^{2}s(x)u(x)\mathrm{d}x=k+\varepsilon \omega=E, \end{cases} \end{equation*}$

其中 $\varepsilon \in (0,1]$, $E$ 是一常数. 基于 B-样条函数, 给出了问题一致有效的数值解.

文献 [15] 研究了一类带有积分边界条件的二阶奇异摄动边值问题, 利用边界层函数法构造了一致有效的渐近解. 文献 [16] 研究了奇异摄动边值问题

$\begin{equation*} \begin{cases} \displaystyle \varepsilon^2 y''=f(t,y), \quad 0<t<1,\quad 0<\varepsilon\ll 1,\\ \displaystyle y(0)=\int_{0}^{1}h_1(y(s,\varepsilon)){\rm d}s, \quad y(1)=\int_{0}^{1}h_2(s,\varepsilon){\rm d}s. \end{cases} \end{equation*}$

利用微分不等式理论, 证明了空间对照结构解的存在性, 积分边界条件的引入, 会使问题的研究比普通边值问题变得复杂. 文献 [17] 研究了高阶奇异摄动 Dirichlet 问题

$\begin{equation*} \begin{cases} \displaystyle \varepsilon y_1'=f_1(y_1,y_2,\cdots,y_n,t),\\ \displaystyle \varepsilon y_2'=f_2(y_1,y_2,\cdots,y_n,t),\\ \vdots\\ \displaystyle \varepsilon y_n'=f_n(y_1,y_2,\cdots,y_n,t),\quad 0\leqslant t\leqslant 1,\\ \displaystyle Ay(0)=Ay^0, \quad By(1)=By^1.\\ \end{cases} \end{equation*}$

利用首次积分证明了空间对照结构解的存在性, 基于解的结构和边界层函数法, 构造了一致有效的形式渐近解.

空间对照结构解的研究一直是很多学者关注的热点问题[18-23]. 关于带有积分边界条件和慢变量的高维奇异摄动边值问题的空间对照结构研究至今未见报道, 已有的方法[15-17]很难解决积分边界条件和慢变量引入带来的困难. 本文将利用微积分技巧和几何奇异摄动理论[9,19,21,24,25]研究带有积分边界条件和慢变量的高维奇异摄动边值问题, 证明空间对照结构解的存在性, 同时, 构造一致有效的形式渐近解.

2 奇异摄动边值问题

考虑奇异摄动积分边值问题

$\begin{equation}\label{2.1} \begin{cases} \varepsilon\dfrac{\mathrm{d}y}{\mathrm{d}t}=F(x,y,t),\quad t\in[0,1],\\[3mm] \dfrac{\mathrm{d}x}{\mathrm{d}t}=G(x,y,t),\\[3mm] x(0)=x^0,\quad Ay(0)=A\int^{1}_{0}h(y(s,\varepsilon))\mathrm{d}s\,,\quad By(1)=B\int^{1}_{0}h(y(s,\varepsilon))\mathrm{d}s, \end{cases} \end{equation}$

其中 $\varepsilon>0$ 是一小参数, $A=\begin{pmatrix}E_l & O\\O & O \end{pmatrix}$ 和 $B=\begin{pmatrix}O & O\\ O & E_{n-l}\end{pmatrix}$ 是 $n \times n$ 矩阵, $E_l$ 是 $l \times l$ 单位矩阵, $E_{n-l}$ 是 $(n-l)\times (n-l)$ 单位矩阵, $O$ 是相应的零矩阵, $x \in \mathbb{R}^m$ 和 $y\in\mathbb{R}^n$ 是慢-快变量, $l,\ m$ 和 $\ n$ 是给定的正常数.

令 $y=\begin{pmatrix}y_1 \\y_2\end{pmatrix}$, $h(y)=\begin{pmatrix}h_1(y)\\h_2(y)\end{pmatrix}$, 其中 $y_1\in \mathbb{R}^l, y_2\in \mathbb{R}^{n-l}, h_1(y)\in \mathbb{R}^l, h_2(y)\in \mathbb{R}^{n-l}$. 将问题 (2.1) 化为如下等价的奇异摄动边值问题

$\begin{equation}\label{2.2} \begin{cases} \varepsilon \dfrac{\mathrm{d}y}{\mathrm{d}t}=F(x,y,t),\\[3mm] \dfrac{\mathrm{d}x}{\mathrm{d}t}=G(x,y,t), \\[3mm] \dfrac{\mathrm{d}k_{1}}{\mathrm{d}t}=h_{1}(y), \\[3mm] \dfrac{\mathrm{d}k_{2}}{\mathrm{d}t}=h_{2}(y),\quad x(0)=x^{0},\\ y_1(0)+k_{1}(0)=0\,,\quad k_{1}(1)=0,\quad k_{2}(0)=0, \quad y_2(1)-k_{2}(1)=0. \end{cases} \end{equation}$

全文作如下假设

${[\bf{H}_1]}$ 函数 $F(x,y,t),\ G(x,y,t)$ 和 $h_i(y)$ 在区域 $D=\{(x, y, t)|\| x \| \leqslant C, \ \| y \|\leqslant A, 0\leqslant t\leqslant 1\}$ 上充分光滑, 其中 $C$ 和 $A$ 是正常数, $i=1,2$.

${[\bf{H}_2]}$ 方程 $F(\bar{x},\bar{y},t)=0$ 存在两个孤立根 $\bar{y}=\varphi_{1}(\bar{x},t)$ 和 $\bar{y}=\varphi_{2}(\bar{x},t)$, 同时左初值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{x}^{(-)}}{\mathrm{d}t}=G(\bar{x}^{(-)},\varphi_{1}(\bar{x}^{(-)},t),t),\quad \bar{x}^{(-)}(0)=x^{0},\\ \end{equation*}$

存在唯一解 $\alpha_{1}(t)$ 和右边值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{x}^{(+)}}{\mathrm{d}t}=G(\bar{x}^{(+)},\varphi_{2}(\bar{x}^{(+)},t),t),\quad \bar{x}^{(+)}(t_0)=\bar{x}^{(-)}(t_0),\\ \end{equation*}$

存在唯一解 $\alpha_{2}(t)$, 其中 $\alpha_{1}(t)$ 和 $\alpha_{2}(t)$ 在 $t_{0}\in (0,1)$ 横截相交, $t_{0}$ 是一确定常数.

${[\bf{H}_3]}$ 矩阵 $F_y(\alpha_i(t),\varphi_i(\alpha_i(t),t),t)=0,\ i=1,2$ 有特征根 $\lambda_i(t),i=1,2,\cdots,n$, 满足

$\mathrm{Re}\lambda_i(t)<0,\ i=1,2,\cdots,l,\quad \mathrm{Re}\lambda_i(t)>0,\ i=l+1,\cdots,n.$

3 空间对照结构解的存在性

本节将利用文献 [19,21] 的主要结果证明 (2.1) 式空间对照结构解的存在性. 考虑 (2.2) 式的连接问题

$\begin{equation}\label{3.1} \begin{cases} \varepsilon \dfrac{\mathrm{d}y}{\mathrm{d}\xi}=F(x,y,t), \dfrac{\mathrm{d}x}{\mathrm{d}\xi}=G(x,y,t),\\[3mm] \dfrac{\mathrm{d}k_{1}}{\mathrm{d}\xi}=h_{1}(y), \ \dfrac{\mathrm{d}k_{2}}{\mathrm{d}\xi}=h_{2}(y),\ \dfrac{\mathrm{d}t}{\mathrm{d}\xi}=1,\\ \end{cases} \end{equation}$

(3.1) 式的边值条件可改写为

$B_{\varepsilon}^{L}=\{(y,x,k_1,k_2,t)\mid x=x^0,y_1+k_{1}=0,\ k_{2}=0, \ t=0\},$
$B_{\varepsilon}^{R}=\{(y,x,k_1,k_2,t)\mid k_{1}=0,\ y_2-k_{2}=0,\ t=1\}.$

令 $\tau=\xi/\varepsilon$, 连接问题 (3.1) 可写为

$\begin{equation}\label{3.2} \begin{cases} \dfrac{\mathrm{d}y}{\mathrm{d}\tau}=F(x,y,t),\ \dfrac{\mathrm{d}x}{\mathrm{d}\tau}=\varepsilon G(x,y,t), \\[3mm] \dfrac{\mathrm{d}k_{1}}{\mathrm{d}\tau}=\varepsilon h_{1}(y), \dfrac{\mathrm{d}k_{2}}{\mathrm{d}\tau}=\varepsilon h_{2}(y),\ \dfrac{\mathrm{d}t}{\mathrm{d}\tau}=\varepsilon.\\ \end{cases} \end{equation}$

(3.1) 和 (3.2) 式中的 $\varepsilon=0$, 可得极限慢系统

$\begin{equation}\label{3.3} \begin{cases} 0=F(x,y,t), \dfrac{\mathrm{d}x}{\mathrm{d}\xi}=G(x,y,t), \\[3mm] \dfrac{\mathrm{d}k_{1}}{\mathrm{d}\xi}=h_{1}(y), \ \dfrac{\mathrm{d}k_{2}}{\mathrm{d}\xi}=h_{2}(y),\ \dfrac{\mathrm{d}t}{\mathrm{d}\xi}=1\\ \end{cases} \end{equation}$

和极限快系统

$\begin{equation}\label{3.4} \begin{cases} \dfrac{\mathrm{d}y}{\mathrm{d}\tau}=F(x,y,t),\ \dfrac{\mathrm{d}x}{\mathrm{d}\tau}=0, \\[3mm] \dfrac{\mathrm{d}k_{1}}{\mathrm{d}\tau}=0, \ \dfrac{\mathrm{d}k_{2}}{\mathrm{d}\tau}=0,\ \dfrac{\mathrm{d}t}{\mathrm{d}\tau}=0.\\ \end{cases} \end{equation}$

利用条件 [H$_{2}$] 和 [H$_{3}$], 可得临界流形

$S_1=\{y=\varphi_{1}(x,t)\}, S_2=\{y=\varphi_{2}(x,t)\}.$

$\dim(S_{i})=n+m+1$, 系统(3.4)的线性化系统在临界流形上有 $n+m+1$ 个零根, $n$ 个非零根, 因此 $S_{i}$ 是法向双曲, $i=1,2$. 为保证问题 (2.1) 空间对照结构解的存在性, 接下来的条件是必要的.

{$[\bf{H}_4]$} 稳定流形 $W^{s}(S_1)$ 和 $B_{0}^{L}$ 横截相交, 不稳定流形 $W^{u}(S_1)$ 和稳定流形 $W^{s}(S_2)$ 在 $t_0\in (0,1)$ 处横截相交, 不稳定流形 $W^{u}(S_2)$ 和 $B_{0}^{R}$ 横截相交, 其中 $W^{s}(S_i)=\bigcup\limits_{p\in S_i}W^{s}(p)$, $W^{u}(S_i)=\bigcup\limits_{p\in S_i}W^{u}(p)$, $ i=1,2$.

利用条件 {$[\mathrm{H}_4]$}, 可知存在连接平衡点 $M_1(\varphi_{1}(\alpha_{1}(t),t)$ 和 $M_2(\varphi_{2}(\alpha_{2}(t),t)$ 的异宿轨道.令

$N_0=B_0^L\cap W^s(S_1),\ N_1=B_0^R\cap W^u(S_2),\ N_{0}\rightarrow \omega(N_{0})= \chi^{1}$
$N_1\rightarrow \alpha(N_1)= \chi^2,\ U^i= \chi^i\cdot(T_i-\delta, T_i+\delta), i=1,2,$

其中 $\omega(N_{0})$ 为 $N_{0}$ 的 $\omega$ 极限集, $\alpha(N_1)$ 为 $N_{1}$ 的 $\alpha$ 极限集, $T_i$ 是奇异解在慢流上运动的时间.

问题 (3.1) 的奇异解是指初始点在 $B_0^L$ 和终端点在 $B_0^R$ 一系列退化系统的解. 令 $\bar{p}_0$ 为奇异解在边界流形 $B_0^L$ 上的初始点, $p_3$ 为奇异解在边界流形 $B_0^R$ 上的终端点, 点 $p_i$ 和 $\bar{p}_{i}$ 为奇异解在流形 $S_{i}$ 上的初始点和终端点, 其中 $S_{i}$ 是奇异解经过的第 $i$ 个慢流形, $i$ = 1, 2.

${\bf定理3.1}$ 假设条件 $[\mathrm{H}_{1}]$-$[\mathrm{H}_{4}]$ 成立, 那么对于充分小的 $\varepsilon>0$, 奇异摄动积分边值问题 (2.1) 存在阶梯状空间对照结构解 $x(t,\varepsilon)$ 和 $y(t,\varepsilon)$. 进一步, 满足

$\begin{equation*}\lim\limits_{\varepsilon\rightarrow0}x(t,\varepsilon)= \begin{cases} \alpha_1(t),& 0\leqslant t\leqslant t_0,\\ \alpha_2(t),& t_0\leqslant t\leqslant 1, \end{cases}\quad \lim\limits_{\varepsilon\rightarrow0}y(t,\varepsilon)= \begin{cases} \varphi_{1}(\alpha_1(t),t),& 0<t<t_0,\\ \varphi_2(\alpha_2(t),t),& t_0<t<1. \end{cases} \end{equation*} $

${\bf证}$ 连接问题 (3.1) 的讨论空间为 $\mathbb{R}^{2n+m+1}$, 计算可得 $\dim B_{\varepsilon}^{L}=n,\ \dim B_{\varepsilon}^{R}=n+m,\ \dim S_1=\dim S_2=n+m+1$. $N_0=B_{0}^{L}\cap W^{s}(S_1)$, $N_{0}\rightarrow \omega(N_{0})= \chi^{1}, p_{1}\in \chi^{1}$, $\omega(\bar{p}_{0})=p_{1}$, $U^{1}=\chi^{1}\cdot(T_{1}-\delta, T_{1}+\delta)$. 利用横截相交条件, 计算可知 $\dim W^s(S_1)=n+m+l+1$, $\dim N_0=l$. 显然, $\dim \chi^{1}=l,\ \dim U^{1}=l+1$. 解由点 $p_{1}$ 到点 $\bar{p}_{1}$ 的时间是有限的, 且 $\dim W^u(U^1)=n+1$.

$N_1=B_{0}^{R}\cap W^{u}(S_2)$, $p_{3}\in N_{1}$, 映射 $N_{1}\rightarrow \alpha(N_{1})= \chi^{2},$$ \bar{p}_{2}\in \chi^{2}$, $\alpha(p_{3})=\bar{p}_{2}$, $U^{2}=\chi^{2}\cdot(T_{2}-\delta, T_{2}+\delta)$. 利用横截相交条件, 计算可得 $\dim W^u(S_2)=2n+m-l+1$, $\dim N_1=n+m-l$. 显然, $\dim \chi^{2}=n+m-l,\ \dim U^{2}=n+m-l+1$. 解由点 $\bar{p}_{2}$ 到 $p_{2}$ 的时间是有限的, 且 $\dim W^s(U^2)=n+m+1$.

令 $\sigma=\dim(W^s(U^2)\cap W^u(U^1))$, 不稳定流形 $W^u(U^1)$ 和稳定流形 $W^s(U^2)$ 是横截相交的, 且

$\dim(W^s(U^2)+\dim(W^u(U^1))-2n-m-1=1.$

从而, 存在连接慢流形 $S_1$ 和 $S_2$ 的异宿轨道. 验证可得交换引理[19]的条件全部满足, 奇异摄动积分边值问题 (2.1) 存在阶梯状空间对照结构解.

4 形式渐近解

利用空间对照结构解的结构和边界层函数法 [18,23], 奇异摄动边值问题 (2.2) 的形式渐近解为 $W=(y,x,k_{1},k_{2})^{\mathrm{T}}$

$\begin{equation}\label{4.1} W^{(-)}(t,\varepsilon)=\sum\limits_{n=0}^\infty\varepsilon^n(\bar{W}_{n}^{(-)}(t)+L_nW(\tau_0)+Q_n^{(-)}W(\tau_1)),\quad 0\leqslant t\leqslant t^*, \end{equation}$
$\begin{equation}\label{4.2} W^{(+)}(t,\varepsilon)=\sum\limits_{n=0}^\infty\varepsilon^n(\bar{W}_{n}^{(+)}(t)+Q_{n}^{(+)}W(\tau_1)+R_nW(\tau_2)),\quad t^*\leqslant t\leqslant 1, \end{equation}$

其中 $\tau_{0}=t\varepsilon^{-1},$$ \tau_{1}=(t-t^{*})\varepsilon^{-1},$$ \tau_{2}=(t-1)\varepsilon^{-1}$, $\bar{W}_{n}^{(\mp)}(t)$ 是正则项系数, $L_nW(\tau_0)$ 是在 $t=0$ 处的左边界层项系数, $R_nW(\tau_2)$ 是在 $t=1$ 处的右边界层项系数,

$Q_n^{(\mp)}W(\tau_1)$ 是在 $t^*$ 处的内部层项系数.

假设内部转移点 $t^*(\varepsilon)\in (0,1)$ 的渐近展开为

$t^*=t_0+\varepsilon t_1+\cdots+\varepsilon^k t_n+\cdots.$

将 (4.1) 和 (4.2) 式代入到边值问题 (2.2), 利用边界层函数法, 按 $t$, $\tau_0$, $\tau_1$ 和 $\tau_2$ 尺度分离, 比较 $\varepsilon$ 的同次幂, 可得确定 $\bar{y}_n^{(\mp)}(t),\ $$\bar{x}_n^{(\mp)}(t), \bar{k}_{1n}^{(\mp)}(t)$$\bar{k}_{2n}^{(\mp)}(t),$$L_ny(\tau_0),$$L_nx(\tau_0),\ L_nk_{1}(\tau_0),\ L_nk_{2}(\tau_0)$, $Q_n^{(\mp)}y(\tau_1),$$Q_n^{(\mp)}x(\tau_1)$, $Q_n^{(\mp)}k_{1}(\tau_1),Q_n^{(\mp)}k_{2}(\tau_1)$, $R_ny(\tau_2),\ R_nx(\tau_2)$, $R_nk_{1}(\tau_2), R_nk_{2}(\tau_2)$, $n\geqslant 0$ 的方程和条件.

先给出确定零次正则项 $\bar{y}_0^{(\mp)}(t),$$\bar{x}_0^{(\mp)}(t), \bar{k}_{10}^{(\mp)}(t)$ 和 $\bar{k}_{20}^{(\mp)}(t)$ 的方程和条件

$\begin{equation}\label{4.3} \begin{cases} F(\bar{x}_0^{(\mp)}(t),\bar{y}_0^{(\mp)}(t),t)=0,\\[3mm] \dfrac{\mathrm{d}\bar{x}_{0}^{(\mp)}(t)}{\mathrm{d}t}=G(\bar{x}_0^{(\mp)}(t),\bar{y}_0^{(\mp)}(t),t),\ \dfrac{\mathrm{d}\bar{k}_{10}^{(\mp)}(t)}{\mathrm{d}t}=h_{1}(\bar{y}_0^{(\mp)}(t)), \\[3mm] \dfrac{\mathrm{d}\bar{k}_{20}^{(\mp)}(t)}{\mathrm{d}t}=h_{2}(\bar{y}_0^{(\mp)}(t)),\ \bar{x}_0^{(-)}(0)=x^0. \end{cases} \end{equation}$

利用条件 [$\mathrm{H}_2$],可得

$\bar{y}_0^{(-)}(t)=\varphi_1(\alpha_1(t),t),\ \bar{y}_0^{(+)}(t)=\varphi_2(\alpha_2(t),t),\ \bar{x}_0^{(-)}(t)=\alpha_1(t),\ \bar{x}_0^{(+)}(t)=\alpha_2(t).$

需要指出的是, $\bar{k}_{10}^{(\mp)}(t)$ 和 $\bar{k}_{20}^{(\mp)}(t)$ 中包含待定的参数, 将结合条件 $\bar{k}_{10}^{(+)}(1)+R_0k_{1}(0)=0,$$\bar{k}_{20}^{(-)}(0)+L_0k_{2}(0)=0,$$ \bar{k}_{10}^{(-)}(t_0)=\bar{k}_{10}^{(+)}(t_0),\bar{k}_{20}^{(-)}(t_0)=\bar{k}_{20}^{(+)}(t_0)$ 来确定.

接下来, 给出确定零次左边界层项 $L_0y(\tau_0)$, $L_0x(\tau_0),\ L_0k_{1}(\tau_0),\ L_0k_{2}(\tau_0)$ 的方程和条件

$\begin{equation}\label{4.4} \begin{cases} \dfrac{\mathrm{d}L_0y}{\mathrm{d}\tau_0}=F(\alpha_1(0)+L_0x,\varphi_1(\alpha_1(0),0)+L_0y,0),\\[3mm] \dfrac{\mathrm{d}L_0x}{\mathrm{d}\tau_0}=0,\ \ \dfrac{\mathrm{d}L_0k_{1}}{\mathrm{d}\tau_0}=0, \dfrac{\mathrm{d}L_0k_{2}}{\mathrm{d}\tau_0}=0,\ \bar{x}_0^{(-)}(0)+L_0x(0)=x^0,\\[3mm] \bar{y}_{10}^{(-)}(0)+L_{0}y_1(0)+\bar{k}_{10}^{(-)}(0)+L_0k_{1}(0)=0, \ \bar{k}_{20}^{(-)}(0)+L_0k_2(0)=0,\ \\ L_0y(+\infty)=0,\ \ L_0x(+\infty)=0,\ L_0k_{1}(+\infty)=0,L_0k_{2}(+\infty)=0. \end{cases} \end{equation}$

由边界层函数法, 可知边界层项是指数衰减的, 从而 $L_0x=0,\ L_0k_{1}=0,\ L_0k_{2}=0$, 同样地, 可得 $R_0k_{1}=0$. 利用上述已知项, 以及函数在点 $t^*$ 处的连续性, 可以确定正则项 $\bar{k}_{10}^{(\mp)}(t),\ \bar{k}_{20}^{(\mp)}(t)$.

为保证高阶渐近解的存在性, 给出如下一些条件, 需要指出这些条件是平凡的.

[$\bf{H}_{5}$] 右初值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{k}_{10}^{(+)}}{\mathrm{d}t}=h_1(\varphi_2(\alpha_2(t)),t), \quad \bar{k}_{10}^{(+)}(1)=0 \end{equation*}$

有解 $\mu_1(t)$, 左初值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{k}_{10}^{(-)}}{\mathrm{d}t}=h_1(\varphi_{1}(\alpha_1(t)),t), \quad \bar{k}_{10}^{(-)}(t_0)=\bar{k}_{10}^{(+)}(t_0) \end{equation*}$

有解 $\mu_2(t)$, 其中 $\mu_1(t)$ 和 $\mu_2(t)$ 在点 $t_0$ 处横截相交. 同时, 左初值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{k}_{20}^{(-)}}{\mathrm{d}t}=h_2(\varphi_1(\alpha_1(t)),t), \quad \bar{k}_{20}^{(-)}(0)=0 \end{equation*}$

有解 $\gamma_1(t)$, 右初值问题

$\begin{equation*} \dfrac{\mathrm{d}\bar{k}_{20}^{(+)}}{\mathrm{d}t}=h_2(\varphi_2(\alpha_2(t)),t), \quad \bar{k}_{20}^{(+)}(t_0)=\bar{k}_{20}^{(-)}(t_0) \end{equation*}$

有解 $\gamma_2(t)$, 其中 $\gamma_1(t)$ 和 $\gamma_2(t)$ 在点 $t_0$ 处横截相交, $t_0\in (0,1)$.

接下来, 给出确定零次内部转移层项 $Q_0^{(\mp)}y(\tau_1),\ Q_0^{(\mp)}x(\tau_1)$, $Q_0^{(\mp)}k_{1}(\tau_1)$ 和 $Q_0^{(\mp)}k_{2}(\tau_1)$ 的方程和条件

$\begin{equation}\label{4.5} \begin{cases} \dfrac{\mathrm{d}Q_0^{(\mp)}y}{\mathrm{d}\tau_1}=F(\alpha_{1,2}(t_0)+Q_0^{(\mp)}x,\varphi_{1,2}(\alpha_{1,2}(t_0),t_0)+Q_0^{(\mp)}y,t_0),\\[3mm] \dfrac{\mathrm{d}Q_{0}^{(\mp)}x}{\mathrm{d}\tau_1}=0,\ \dfrac{\mathrm{d}Q_0^{(\mp)}k_{1}}{\mathrm{d}\tau_1}=0, \dfrac{\mathrm{d}Q_0^{(\mp)}k_{2}}{\mathrm{d}\tau_1}=0,\\[3mm] Q_{0}^{(\mp)}y(0)+\varphi_{1,2}(\alpha_{1,2}(t_{0}),t_0)=\beta(t_{0}), Q_{0}^{(\mp)}y(\mp \infty)=0,\ Q_{0}^{(\mp)}x(\mp \infty)=0,\\ Q_{0}^{(\mp)}k_1(\mp \infty)=0,Q_{0}^{(\mp)}k_2(\mp \infty)=0, \end{cases} \end{equation}$

其中 $\beta(t)=\dfrac{1}{2}(\varphi_1(\alpha_1(t),t)+\varphi_2(\alpha_2(t)),t)$.

计算可得, 零次右边界层项 $R_0y(\tau_2), R_0x(\tau_2)$, $R_0k_{1}(\tau_2)$ 和 $R_0k_{2}(\tau_2)$ 满足

$\begin{equation}\label{4.6} \begin{cases} \dfrac{\mathrm{d}R_{0}y}{\mathrm{d}\tau_{2}}=F(\alpha_2(1)+R_0x,\varphi_2(\alpha_2(1),1)+R_0y,1),\ \dfrac{\mathrm{d}R_{0}x}{\mathrm{d}\tau_{2}}=0,\\[3mm] \dfrac{\mathrm{d}R_0k_1}{\mathrm{d}\tau_2}=0,\ \dfrac{\mathrm{d}R_0k_2}{\mathrm{d}\tau_2}=0,\ \bar{y}_{20}^{(+)}(1)+R_{0}y_2(0)-\bar{k}_{20}^{(+)}(1)-R_0k_{2}(0)=0, \\[3mm] \bar{k}_{10}^{(+)}(1)+R_0k_1(0)=0,\ R_0y(-\infty)=0,\\ R_0x(-\infty)=0,\ R_0k_{1}(-\infty)=0,\ R_0k_{2}(-\infty)=0. \end{cases} \end{equation}$

类似于零次左边界层项的讨论, 可得 $R_0x=0,\ R_0k_1=0$, $R_0k_2=0$. 利用条件 [H$_{4}$], 可知 (4.4)-(4.6) 式的解存在. 至此, 已经确定了全部零次项的渐近解. 接下来, 给出确定高阶渐近项的方程和条件.

高阶正则项 $\bar{y}_{n}^{(\mp)}(t),\ \bar{x}_{n}^{(\mp)}(t), \ \bar{k}_{1n}^{(\mp)}(t)$ 和 $\bar{k}_{2n}^{(\mp)}(t), \ n\geqslant 1$ 满足的方程和条件为

$\begin{equation}\label{4.7} \begin{cases} \dfrac{\mathrm{d}\bar{y}_{n-1}^{(\mp)}}{\mathrm{d}t}=\tilde{F}_{x}^{(\mp)}\bar{x}_{n}^{(\mp)}(t)+\tilde{F}_{y}^{(\mp)}\bar{y}_{n}^{(\mp)}(t)+F_{1n}^{(\mp)}(t),\\[3mm] \dfrac{\mathrm{d}\bar{x}_{n}^{(\mp)}}{\mathrm{d}t}=\tilde{G}_{x}^{(\mp)}\bar{x}_{n}^{(\mp)}(t)+\tilde{G}_{y}\bar{y}_{n}^{(\mp)}(t)+F_{2n}^{(\mp)}(t),\\[3mm] \dfrac{\mathrm{d}\bar{k}_{1n}^{(\mp)}}{\mathrm{d}t}=h_{1y}(\varphi_{1,2}(\alpha_{1,2}(t)))\bar{y}_{n}^{(\mp)}(t)+F_{3n}^{(\mp)}(t),\\[3mm] \dfrac{\mathrm{d}\bar{k}_{2n}^{(\mp)}}{\mathrm{d}t}=h_{2y}(\varphi_{1,2}(\alpha_{1,2}(t)))\bar{y}_{n}^{(\mp)}(t)+F_{4n}^{(\mp)}(t),\ \bar{x}_n^{(-)}(0)+L_nx(0)=0,\\[3mm] \bar{y}_{1n}^{(-)}(0)+L_ny_1(0)+k_{1n}^{(-)}(0)+L_nk_1(0)=0,\ \bar{k}_{1n}^{(+)}(1)+R_nk_1(0)=0,\\[3mm] \bar{k}_{2n}^{(-)}(0)+L_nk_2(0)=0,\ \bar{y}_{2n}^{(+)}(1)+R_ny_2(0)-\bar{k}_{2n}^{(+)}(1)-R_nk_2(0)=0, \end{cases} \end{equation}$

其中

$\tilde{F}_{x}^{(\mp)}=F_{x}(\alpha_{1,2}(t),\varphi_{1,2}(\alpha_{1,2}(t),t),t),\ $
$\tilde{F}_{y}^{(\mp)}=F_{y}(\alpha_{1,2}(t),\varphi_{1,2}(\alpha_{1,2}(t),t),t),$
$\tilde{G}_{x}^{(\mp)}=G_{x}(\alpha_{1,2}(t),\varphi_{1,2}(\alpha_{1,2}(t),t),t),\ $
$\tilde{G}_{y}^{(\mp)}=G_{y}(\alpha_{1,2}(t),\varphi_{1,2}(\alpha_{1,2}(t),t),t),$

$F_{in}^{(\mp)}(t),\ i=1,2,3,4$ 是依赖于一些已确定项的已知函数. 需要指出的是 (4.7) 式主要依赖于 $\bar{x}_{n}^{(\mp)}$ 的可解性, $\bar{x}_{n}^{(\mp)}$ 将在高阶边界层项的求解过程中确定.

高阶左边界层项 $L_{n}y(\tau_0)$, $L_{n}x(\tau_0)$, $L_{n}k_{1}(\tau_0)$ 和 $L_{n}k_{2}(\tau_0)$, $n\geqslant 1$ 满足的方程和条件为

$\begin{equation}\label{4.8} \begin{cases} \dfrac{\mathrm{d}L_ny}{\mathrm{d}\tau_0}=\tilde{F}_{x}^{(L)}(\tau_0)L_nx+\tilde{F}_{y}^{(L)}(\tau_0)L_ny+F_{1n}^{(L)}(\tau_0),\\[3mm] \dfrac{\mathrm{d}L_nx}{\mathrm{d}\tau_0}=\tilde{G}_{x}^{(L)}(\tau_0)L_{n-1}x+\tilde{G}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{2n}^{(L)}(\tau_0), \\[3mm] \dfrac{\mathrm{d}L_nk_1}{\mathrm{d}\tau_0}=\tilde{h_1}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{3n}^{(L)}(\tau_0), \\[3mm] \dfrac{\mathrm{d}L_nk_2}{\mathrm{d}\tau_0}=\tilde{h_2}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{4n}^{(L)}(\tau_0),\\ L_ny_1(0)=-\bar{y}_{1n}^{(-)}(0)-\bar{k}_{1n}^{(-)}(0)-L_{n}k_{1}(0), \ L_nx(0)=-\bar{x}_n^{(-)}(0), \\ L_nk_{2}(0)=-\bar{k}_{2n}^{(-)}(0),\ L_ny(+\infty)=0, \ L_nx(+\infty)=0,\\ L_nk_{1}(+\infty)=0,\ L_nk_{2}(+\infty)=0, \end{cases} \end{equation}$

其中 $\tilde{F}_{y}^{(L)}(\tau_0),\tilde{F}_{x}^{(L)}(\tau_0),\tilde{G}_{y}^{(L)}(\tau_0),\ \tilde{G}_{x}^{(L)}(\tau_0)$ 在 $(\alpha_{1}(0)+L_0x,\varphi_1(\alpha_1(0),0)+L_0y,0)$ 处取值. $F_{in}^{(L)}(\tau_0),$$i=1,2,3,4$ 是依赖于一些已确定项的已知函数. 令

$G_{1}(\tau_0)=\tilde{G}_{x}^{(L)}(\tau_0)L_{n-1}x+\tilde{G}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{2n}^{(L)}(\tau_0),$

$G_{1}(\tau_0)$ 是依赖于一些已确定项的已知函数. 方程 $\dfrac{{\rm d}L_nx}{{\rm d}\tau_0}=G_{1}(\tau_0)$ 从 $+\infty$ 到 $\tau_0$ 积分, 计算可得

$L_{n}x(\tau_0)=\int_{+\infty}^{\tau_0}G_{1}(\tau_0)\mathrm{d}\tau_0.$

令 $\tau_0=0$, 结合 (4.8) 式, 有

$\bar{x}_{n}^{(-)}(0)=-\int_{+\infty}^{0}G_{1}(\tau_0)\mathrm{d}\tau_0,$

利用 (4.7) 式, 可确定高阶正则项 $\bar{W}_{n}^{(-)}(t)$. 需要指出的是, 将会利用高阶内部层项和右边界层项来确定 $\bar{k}_{1n}^{(-)}(t)$. 类似于 $L_nx(\tau_0)$ 的求解过程, 可以求解 $L_nk_{1}(\tau_0)$ 和 $L_nk_{2}(\tau_0)$. 关于 $L_ny(\tau_0)$ 的可解性依赖于高阶内部层项.

高阶内部转移层项 $Q_{n}^{(\mp)}y(\tau_1), $$Q_{n}^{(\mp)}x(\tau_1), $$Q_{n}^{(\mp)}k_{1}(\tau_1)$ 和 $Q_{n}^{(\mp)}k_{2}(\tau_1), n\geqslant 1$ 满足的方程和条件为

$\begin{matrix}\label{4.9} \begin{cases} \dfrac{{\rm d}Q_{n}^{(\mp)}y}{{\rm d}\tau_1}=\tilde{F}_{x}^{(Q){(\mp)}}(\tau_1)Q_{n}^{(\mp)}x(\tau_{1})+\tilde{F}_{y}^{(Q){(\mp)}}(\tau_1)Q_{n}^{(\mp)}y(\tau_{1})+ \tilde{F}_{t}^{(Q){(\mp)}}(\tau_1)t_{n}\\ ~~~~~~~~~~~~~\, +F_{1n}^{(Q){(\mp)}}(\tau_1),\\[1mm] \dfrac{{\rm d}Q_{n}^{(\mp)}x}{{\rm d}\tau_1}=\tilde{G}_{x}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}x(\tau_{1})+\tilde{G}_{y}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}y(\tau_{1})+ F_{2n}^{(Q){(\mp)}}(\tau_1),\\[3mm] \dfrac{{\rm d}Q_n^{(\mp)}k_{1}}{{\rm d}\tau_1}=\tilde{h}_{1y}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}y(\tau_{1})+ F_{3n}^{(Q){(\mp)}}(\tau_1),\\[3mm] \dfrac{{\rm d}Q_n^{(\mp)}k_{2}}{{\rm d}\tau_1}=\tilde{h}_{2y}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}y(\tau_{1})+ F_{4n}^{(Q){(\mp)}}(\tau_1),\\[3mm] Q_{n}^{(\mp)}y(0)=\sigma_{n}\big(\rho(t_1,\cdots t_{n})\big),\ Q_{n}^{(\mp)}y(\mp\infty)=0,\\ Q_{n}^{(\mp)}x(\mp\infty)=0,Q_{n}^{(\mp)}k_{1}(\mp\infty)=0,Q_{n}^{(\mp)}k_{2}(\mp\infty)=0, \end{cases} \end{matrix}$

其中 $\tilde{F}_{y}^{(Q){(\mp)}}(\tau_1),\ \tilde{F}_{x}^{(Q){(\mp)}}(\tau_1),\ \tilde{G}_{y}^{(Q){(\mp)}}(\tau_1),\tilde{G}_{x}^{(Q){(\mp)}}(\tau_1)$ 和 $\tilde{F}_{t}^{(Q){(\mp)}}(\tau_1)$ 在

$(\alpha_{1,2}(t_0)+Q_0^{(\mp)}x(\tau_1),\varphi_{1,2}(\alpha_{1,2}(t_0),t_0)+Q_0^{(\mp)}y(\tau_1),t_0), $

处取值. $\sigma_{n}(\rho),\ F_{in}^{Q{(\mp)}}(\tau_1),\ i=1,2,3,4$ 是一些依赖于已确定项的已知函数. 令

$G_{2}(\tau_1)=\tilde{G}_{x}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}x(\tau_{1})+\tilde{G}_{y}^{(Q){(\mp)}}(\tau_1)Q_{n-1}^{(\mp)}y(\tau_{1})+ F_{2n}^{(Q){(\mp)}}(\tau_1),$

$G_{2}(\tau_1)$ 是依赖于一些已确定项的已知函数. 方程 $\dfrac{{\rm d}Q_n^{(\mp)}x}{{\rm d}\tau_1}=G_{2}(\tau_1)$ 从 $\mp\infty$ 到 $\tau_1$ 积分, 计算可得

$Q_{n}^{(\mp)}x(\tau_1)=\int_{\mp\infty}^{\tau_1}G_{2}(\tau_1)\mathrm{d}\tau_1,$

类似地, 可以确定 $Q_{n}^{(\mp)}k_1$ 和 $Q_{n}^{(\mp)}k_2$.

进一步, 给出确定 $t_{n}$ 的方程和条件, (4.9) 式的联合系统为

$\begin{equation}\label{4.10} \dfrac{\mathrm{d}Q_{n}y}{\mathrm{d}\tau_1}=\tilde{F}_{y}^{(Q)}(\tau_1)Q_{n}y(\tau_{1}) +\tilde{F}_{x}^{(Q)}(\tau_1)Q_{n}x(\tau_{1})+\tilde{F}_{t}^{(Q)}(\tau_1)t_{n}+F_{1n}^{(Q)}(\tau_1),\\ \end{equation}$

其中 $\tilde{F}_{y}^{(Q)}(\tau_1),\ \tilde{F}_{x}^{(Q)}(\tau_1)$ 和 $\tilde{F}_{t}^{(Q)}(\tau_1)$ 在

$(q_{11}(\tau_1),q_{12}(\tau_1),t_0), $

处取值, 满足

$q_{11}(\tau_1) \rightarrow \alpha_1(t_0), \tau_1 \rightarrow -\infty,q_{11}(\tau_1) \rightarrow \alpha_2(t_0), \tau_1 \rightarrow +\infty,$
$q_{12}(\tau_1) \rightarrow \varphi_1(\alpha_{1}(t_0),t_0), \tau_1 \rightarrow -\infty,q_{12}(\tau_1) \rightarrow \varphi_2(\alpha_{2}(t_0),t_0), \tau_1\rightarrow +\infty.$

借助于文献 [22,引理 3.7] 的主要结果和指数二分法, 可知 $F(Q_{n}y)=\frac{\mathrm{d}Q_{n}y}{\mathrm{d}\tau_1}-F_{y}(\tau_1)Q_{n}y$ 是 Fredholm 算子, Fredholm 指标 $\dim \mathrm{Ker} F-\dim \mathrm{Ker} F^{*}=0$, 其中 $F_{y}(\tau_1)$ 在 $(q_{11}(\tau_1),q_{12}(\tau_1),t_0)$ 处取值. 计算可得存在唯一的 $\psi_{1}(\tau_1)\in \mathrm{Ker} F^{*}$, 方程 (4.10) 有解等价于

$t_{n}\int_{-\infty}^{+\infty}\psi_{1}^{*}(\tau_1)F_{t}(q_{11}(\tau_1),q_{12}(\tau_1),t_0)\mathrm{d}\tau_1$
$\begin{equation*} \quad \quad \quad = -\int_{-\infty}^{+\infty}\psi_{1}^{*}(\tau_1)(\tilde{F}_{x}^{(Q){(\mp)}}(\tau_1)Q_{n}^{(\mp)}x(\tau_{1})+\tilde{F}_{1n}^{(Q)}(\tau_1))\mathrm{d}\tau_1. \end{equation*}$

结合条件 $[\mathrm{H}_4]$ 和 Melnikov 函数, 可得

$\int_{-\infty}^{\infty}\psi_{1}^*F_{t}(q_{11}(\tau_1),q_{12}(\tau_1),t_0) \mathrm{d}\tau_1 \neq 0.$

至此已经确定了 $t_{n}$, 进一步, 可确定内部层项 $Q_{n}^{(\mp)}W(\tau_1)$.

高阶右边界层项 $R_{n}y(\tau_2)$, $R_{n}x(\tau_2)$, $R_{n}k_{1}(\tau_2)$ 和 $R_{n}k_{2}(\tau_2),$$n\geqslant 1$ 满足的方程和条件为

$\begin{equation}\label{4.11} \begin{cases} \dfrac{\mathrm{d}R_ny}{\mathrm{d}\tau_2}=\tilde{F}_{x}^{(R)}(\tau_2)R_nx+\tilde{F}_{y}^{(R)}(\tau_2)R_ny+F_{1n}^{(R)}(\tau_2),\\[3mm] \dfrac{\mathrm{d}R_nx}{\mathrm{d}\tau_2}=\tilde{G}_{x}^{(R)}(\tau_2)R_{n-1}x+\tilde{G}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{2n}^{(R)}(\tau_2), \\[3mm] \dfrac{\mathrm{d}R_nk_1}{\mathrm{d}\tau_2}=\tilde{h_1}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{3n}^{(R)}(\tau_2), \\[3mm] \dfrac{\mathrm{d}R_nk_2}{\mathrm{d}\tau_2}=\tilde{h_2}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{4n}^{(R)}(\tau_2),\\ R_ny_2(0)=-\bar{y}_n^{(+)}(1)+\bar{k}_{2n}^{(+)}(1)+R_{n}k_{2}(0), \ R_nk_{1}(0)=-\bar{k}_{1n}^{(+)}(1),\\ R_ny(-\infty)=0, \ R_nz(-\infty)=0,\ R_nx(-\infty)=0,\\ R_nk_{1}(-\infty)=0,\ R_nk_{2}(-\infty)=0, \end{cases} \end{equation}$

其中 $\tilde{F}_{y}^{(R)}(\tau_2),\tilde{F}_{x}^{(R)}(\tau_2),\tilde{G}_{y}^{(R)}(\tau_2),\ \tilde{G}_{x}^{(R)}(\tau_2)$ 在 $(\alpha_{2}(1)+R_0x,\varphi_2(\alpha_2(1),1)+R_0y,1)$ 处取值, $F_{in}^{(R)}(\tau_2),\ i=1,2,3,4$ 是一些依赖于已确定项的函数. 令

$G_{3}(\tau_2)=\tilde{h_1}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{3n}^{(R)}(\tau_2),$

其中 $G_{3}(\tau_2)$ 是依赖于一些已确定项的函数. 方程 $\frac{{\rm d}R_nk_{1}}{{\rm d}\tau_2}=G_{3}(\tau_2)$ 从 $-\infty$ 到 $\tau_2$ 积分, 可得

$R_{n}k_{1}(\tau_2)=\int_{-\infty}^{\tau_2}G_{3}(\tau_2)\mathrm{d}\tau_2.$

令 $\tau_2=0$, 利用 (4.11) 式, 有

$\bar{k}_{1n}^{(+)}(1)=-\int_{-\infty}^{0}G_{3}(\tau_2)\mathrm{d}\tau_2,$

利用变量 $y,x,\ k_{1},\ k_{2}$ 的连续性, 可以确定 $\bar{y}_n^{(\mp)}(t)$, $\bar{x}_n^{(\mp)}(t)$, $\bar{k}_{1n}^{(\mp)}$ 和 $\bar{k}_{2n}^{(\mp)}(t)$.

系统 (4.8) 和 (4.11) 中的变量 $L_nx(\tau_0),L_nk_1(\tau_0),L_nk_2(\tau_0)$, $R_nx(\tau_2),$$R_nk_1(\tau_2)$ 和 $R_nk_2(\tau_2)$ 已得到确定, 因此, 只需考虑 (4.8) 和 (4.11) 式中的第一个变量即可. $L'_0y$ 和 $R'_0y$ 是线性齐次方程组 (4.8) 和 (4.11) 的解, 利用文献 [23] 的相关结果, 可知系统 (4.8) 和 (4.11) 关于第一个变量的解是存在的.

${\bf定理 4.1}$ 如果满足条件 $[\mathrm{H}_1]$-$[\mathrm{H}_5]$, 对于充分小的 $\varepsilon>0$, 奇异摄动边值问题 (2.1) 存在阶梯状空间对照结构 $x(t,\varepsilon)$ 和 $y(t,\varepsilon)$, 且满足

$\begin{equation*}x(t,\varepsilon)= \begin{cases} \sum\limits_{k=0}^{n}\varepsilon^{k}\big(\bar x_k^{(-)}(t)+L_kx(\tau_0)+Q_{k}^{(-)}x(\tau_1)\big)+O(\varepsilon^{n+1}),\quad 0\leqslant t \leqslant T_{n}+O(\varepsilon^{n+1}),\\[4mm] \sum\limits_{k=0}^{n}\varepsilon^{k}\big(\bar x_k^{(+)}(t)+Q_{k}^{(+)}x(\tau_1)+R_kx(\tau_2)\big)+O(\varepsilon^{n+1}),\quad T_{n}+O(\varepsilon^{n+1})\leqslant t\leqslant 1, \end{cases} \end{equation*}$
$\begin{equation*}y(t,\varepsilon)= \begin{cases} \sum\limits_{k=0}^{n}\varepsilon^{k}\big(\bar y_k^{(-)}(t)+L_ky(\tau_0)+Q_{k}^{(-)}y(\tau_1)\big)+O(\varepsilon^{n+1}),\quad 0\leqslant t \leqslant T_{n}+O(\varepsilon^{n+1}),\\[4mm] \sum\limits_{k=0}^{n}\varepsilon^{k}\big(\bar y_k^{(+)}(t)+Q_{k}^{(+)}y(\tau_1)+R_ky(\tau_2)\big)+O(\varepsilon^{n+1}),\quad T_{n}+O(\varepsilon^{n+1})\leqslant t\leqslant 1. \end{cases} \end{equation*}$

其中 $T_{n}=t_{0}+\varepsilon t_{1}+\cdots+\varepsilon^n t_{n}.$

5 例子

考虑奇异摄动积分边值问题

$\begin{equation}\label{5.1} \begin{cases} \varepsilon\dfrac{\mathrm{d}y_1}{\mathrm{d}t}=2y_2+\varepsilon t, \\[3mm] \varepsilon\dfrac{\mathrm{d}y_2}{\mathrm{d}t}=(y_1^2-16)(y_1-b(t)),\\[3mm] \dfrac{\mathrm{d}x}{\mathrm{d}t}=x,\\ x(0,\varepsilon)=2,\quad y_1(0,\varepsilon)=2\int_{0}^{1}y_1(s,\varepsilon)\mathrm{d}s\,,\quad y_2(1,\varepsilon)=3\int_{0}^{1}y_2(s,\varepsilon)\mathrm{d}s. \end{cases} \end{equation}$

为简化计算过程, 令 $b(0)=b(1)=0$, $b(\dfrac{1}{2})=0$ 和 $b'(\dfrac{1}{2})\neq 0$. 问题 (5.1) 可改写为

$\begin{equation}\label{5.2} \begin{cases} \varepsilon\dfrac{\mathrm{d}y_1}{\mathrm{d}t}=2y_2+\varepsilon t, \\[3mm] \varepsilon\dfrac{\mathrm{d}y_2}{\mathrm{d}t}=(y_1^2-16)(y_1-b(t)),\\[3mm] \dfrac{\mathrm{d}x}{\mathrm{d}t}=x,\quad \dfrac{\mathrm{d}k_{1}}{\mathrm{d}t}=2y_1,\quad \dfrac{\mathrm{d}k_{2}}{\mathrm{d}t}=3y_2,\\ x(0,\varepsilon)=2,\quad y_1(0)=-k_1(0),\quad\ k_1(1)=0, \quad k_2(0)=0, \quad y_2(1)=k_2(1). \end{cases} \end{equation}$

利用边界层函数法, 可得零次正则项 $\bar{y}_{10}^{(-)}(t)=-4,\ \bar{y}_{10}^{(+)}(t)=4,\ \bar{y}_{20}^{(-\mp)}(t)=0,\ \bar{k}_{10}^{(-)}(t)=-8t,\ \bar{k}_{10}^{(+)}(t)=8t-8,\ \bar{k}_{20}^{(\mp)}(t)=0,\ \bar{x}_{0}^{(\mp)}(t)=2 {\rm e}^{t}.\ $ 内部转移点主项 $t_0$ 满足方程 $I(t_{0})=\int_{-4}^{4}(y_1^2-16)(y_1-b(t_0))\mathrm{d}y=0$, 从而, 可得 $t_{0}=1/2$.

确定 $Q^{(\mp)}_{0}y_1$ 的方程和条件为

$\dfrac{\mathrm{d}Q^{(\mp)}_{0}y_1}{\mathrm{d}\tau_1}=16-(\mp 4+Q^{(\mp)}_{0}y_1)^2, \quad Q^{(\mp)}_{0}y_1(0)=\pm 4, \quad Q^{(\mp)}_0y_1(\mp \infty)=0,$

计算可得

$Q^{(-)}_0y_1=\dfrac{8}{1+{\rm e}^{-8\tau_1}},\quad Q^{(+)}_0y_1=\dfrac{-8{\rm e}^{-8\tau_1}}{1+{\rm e}^{-8\tau_1}}.$

类似地, 计算可得

$Q^{(-)}_0y_2=\dfrac{32 {\rm e}^{8\tau_1}}{(1+{\rm e}^{8\tau_1})^2},\quad Q^{(+)}_0y_2=\dfrac{32{\rm e}^{-8\tau_1}}{(1+{\rm e}^{-8\tau_1})^2}.$
$Q^{(-)}_0x=Q^{(-)}_0k_{1}=Q^{(-)}_0k_{2}=0,\ Q^{(+)}_0x=Q^{(+)}_0k_{1}=Q^{(+)}_0k_{2}=0,$
$L_0y_1=\dfrac{8}{1+{\rm e}^{8\tau_0}},\quad L_0y_2=-\dfrac{32{\rm e}^{-8\tau_0}}{(1+{\rm e}^{-8\tau_0})^2}, \quad L_{0}x=L_0k_{1}=L_0k_{2}=0,$
$R_0y_1=R_0y_2=R_{0}x=R_0k_{1}=R_0k_{2}=0.$

问题 (5.1) 的形式渐近解为

$\begin{equation*}x(t,\varepsilon)= \begin{cases} 2{\rm e}^t+O(\varepsilon), \quad 0\leqslant t \leqslant \dfrac{1}{2},\\ 2{\rm e}^t+O(\varepsilon), \quad \dfrac{1}{2} \leqslant t \leqslant 1, \end{cases} \end{equation*} $
$\begin{equation*}y_1(t,\varepsilon)= \begin{cases} -4+\dfrac{8}{1+{\rm e}^{8\tau_0}}+\dfrac{8}{1+{\rm e}^{-8\tau_1}}+O(\varepsilon), 0\leqslant t \leqslant \dfrac{1}{2}, \\[3mm] 4+\dfrac{-8{\rm e}^{-8\tau_1}}{1+{\rm e}^{-8\tau_1}}+O(\varepsilon), \dfrac{1}{2} \leqslant t \leqslant 1, \end{cases} \end{equation*} $
$\begin{equation*}y_2(t,\varepsilon)= \begin{cases} -\dfrac{32{\rm e}^{-8\tau_0}}{(1+{\rm e}^{-8\tau_0})^2}+\dfrac{32 {\rm e}^{8\tau_1}}{(1+{\rm e}^{8\tau_1})^2}+O(\varepsilon), 0\leqslant t \leqslant \dfrac{1}{2}, \\[3mm] \dfrac{32{\rm e}^{-8\tau_1}}{(1+{\rm e}^{-8\tau_1})^2}+O(\varepsilon), \dfrac{1}{2} \leqslant t \leqslant 1. \end{cases} \end{equation*}$

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