奇异摄动理论和方法源于对天体力学的研究,是处理非线性问题的重要工具,在许多方面都有重要应用,如流体动力学、最优控制和空气动力学^[1-6].随着奇异摄动理论的不断发展,出现了很多新方向和新方法.在一些热传导、半导体^[7]和生物医学^[8]等问题的研究过程中,学者们发现初始时刻和终端时刻的值不是固定的,出现了可移动边界,其中的一种情形为积分边界^[9]. Cakir和Amiraliyev^[10]考虑了带有积分边界条件的奇异摄动边值问题

$ \begin{eqnarray*} \left\{\begin{array}{ll} \varepsilon^2 y''+\varepsilon a(t)y'-b(t)y = f(t), \quad 0<t<l, \ 0<\varepsilon\ll 1, \\ y(0) = y^0, \quad y(l) = y^{l}+\int_{l_0}^{l_1}g(s)y(s){\rm d}s, \quad 0\leq l_0<l_1\leq l, \\ \end{array}\right. \end{eqnarray*} $

利用有限差分的方法,构造了具有边界层的数值解.文献[11]中,谢峰等研究了一类含有积分边界条件的二阶奇异摄动边值问题,利用微分不等式的理论,证明了边值问题空间对照结构解的存在性,并构造了形式渐近解.文献[12]中,作者考虑了二阶非线性奇异摄动边值问题

$ \begin{eqnarray*} \left\{\begin{array}{ll} \varepsilon^2 y'' = f(t, y, \varepsilon y'), \quad 0<t<1, \ 0<\varepsilon\ll 1, \\ y(0) = y^0, \quad y(1) = y^{1}+\int_{0}^{1}g(t)y(t){\rm d}t, \\ \end{array}\right. \end{eqnarray*} $

针对上述积分边界问题,借助于边界层函数法和微分不等式技巧,构造了零阶渐近解,并证明了解的存在性.因为积分边界的引入使得所讨论问题变得复杂,文献[11-12]中,作者仅考虑了含有快变量的系统,没有考虑慢变量.奇异摄动系统也称为快-慢系统,因此对于含有积分边界条件的奇异摄动快-慢系统的研究是十分有意义的.

通过研究发现,关于奇异摄动空间对照结构的研究成果已非常丰富,可参见文献[13-14]及其参考文献,关于带有积分边界条件的奇异摄动问题中的空间对照结构至今未见报道.难点在于已有参考文献中的方法已不再适用,需要运用几何方法进行研究.关于奇异摄动几何方法的研究可参看文献[15-21],学者们利用奇异摄动几何理论讨论了解的存在性、唯一性、孤立波解、同异宿轨的存在性等等.

本文将运用几何理论^[13]研究一类带有积分边界条件的奇异摄动边值问题,不但证明了空间对照结构解的存在性,而且构造了一致有效的形式渐近解.

考虑带有积分边界条件的奇异摄动边值问题

(2.1) $ \begin{equation} \left\{\begin{array}{ll} { } \mu^2\frac{{\rm d}^2y}{{\rm d}t^2} = f(x, y, t), \quad t\in[0, 1], \\ [3mm] { } \frac{{\rm d}x}{{\rm d}t} = g(x, y, t), \\ { } x(0) = x^0, \quad y(0) = y^{0}+\int^{1}_{0}h_{1}(y(s, \mu)){\rm d}s\, , \quad y(1) = y^{1}+\int^{1}_{0}h_{2}(y(s, \mu)){\rm d}s, \end{array}\right. \end{equation} $

其中$ \mu>0 $是一小参数, $ x, \ y\in\mathbb{R} $分别是慢变量和快变量.

由于积分边界条件的引入,使得问题(2.1)的讨论要比常规固定变界的情形要复杂的多,为此将方程(2.1)转化为如下等价的奇异摄动边值问题

(2.2) $ \begin{equation} \left\{\begin{array}{ll} { } \mu \frac{{\rm d}y}{{\rm d}t} = z, { } \mu \frac{{\rm d}z}{{\rm d}t} = f(x, y, t), \quad\\ { } \frac{{\rm d}x}{{\rm d}t} = g(x, y, t), { } \frac{{\rm d}k_{1}}{{\rm d}t} = h_{1}(y), \quad\\ { } \frac{{\rm d}k_{2}}{{\rm d}t} = h_{2}(y), \quad x(0) = x^{0}, \\ y(0) = y^{0}-k_{1}(0)\, , \quad k_{1}(1) = 0, \quad k_{2}(0) = 0, \quad y(1) = y^{1}+k_{2}(1). \end{array}\right. \end{equation} $

对所提问题(2.1)作如下假设

(H$ _1) $  函数$ f(x, y, t), \ g(x, y, t) $和$ h_{i}(y) $在区域$ D = \{(x, y, t)|\mid x \mid \leq A, \ \mid y\mid\leq A, 0\leq t\leq 1\} $上充分光滑,其中$ A $是一正常数, $ i = 1, 2 $.

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

$ \frac{{\rm d}\bar{x}^{(-)}}{{\rm d}t} = g(\bar{x}^{(-)}, \alpha_{1}(\bar{x}^{(-)}, t), t), \quad \bar{x}^{(-)}(0) = x^{0} $

和右初值问题

$ \frac{{\rm d}\bar{x}^{(+)}}{{\rm d}t} = g(\bar{x}^{(+)}, \alpha_{2}(\bar{x}^{(+)}, t), t), \quad \bar{x}^{(+)}(t_0) = \bar{x}^{(-)}(t_0) $

分别存在唯一解$ \varphi_{1}(t) $和$ \varphi_{2}(t) $,其中$ \varphi_{1}(t) $和$ \varphi_{2}(t) $在$ t_{0}\in (0, 1) $横截相交,同时满足$ f_{y}(\varphi_{i}(t), \alpha_{i}(\varphi_{i}(t)), t)>0, \ t\in[0, 1], \ i = 1, 2 $.

本节将利用文献[13, 15]的主要结果证明问题(2.1)空间对照结构解的存在性.首先,研究(2.2)式的连接问题

(3.1) $ \begin{equation} \left\{\begin{array}{ll} { } \mu \frac{{\rm d}y}{{\rm d}\xi} = z, \quad \mu \frac{{\rm d}z}{{\rm d}\xi} = f(x, y, t), \\ { } \frac{{\rm d}x}{{\rm d}\xi} = g(x, y, t), {\quad} \frac{{\rm d}k_{1}}{{\rm d}\xi} = h_{1}(y), \quad \frac{{\rm d}k_{2}}{{\rm d}\xi} = h_{2}(y), \quad \frac{{\rm d}t}{{\rm d}\xi} = 1, \end{array}\right. \end{equation} $

边值条件可改写为

$ B_{\mu}^{L} = \{(y, z, x, k_1, k_2, t)\mid x(0) = x^0, y(0) = y^{0}-k_{1}, \ k_{2}(0) = 0, \ t = 0\}, \quad $

$ B_{\mu}^{R} = \{(y, z, x, k_1, k_2, t)\mid k_{1}(1) = 0, \ y(1) = y^{1}+k_{2}(1), \ t = 1\}. $

令$ \tau = \xi/\mu $,连接问题(3.1)可改写为

(3.2) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}y}{{\rm d}\tau} = z, \quad \frac{{\rm d}z}{{\rm d}\tau} = f(x, y, t), \\ { } \frac{{\rm d}x}{{\rm d}\tau} = \mu g(x, y, t), \quad \frac{{\rm d}k_{1}}{{\rm d}\tau} = \mu h_{1}(y), \quad \frac{{\rm d}k_{2}}{{\rm d}\tau} = \mu h_{2}(y), \quad \frac{{\rm d}t}{{\rm d}\tau} = \mu . \end{array}\right. \end{equation} $

在(3.1)和(3.2)式中分别令$ \mu = 0 $,可得极限慢系统

(3.3) $ \begin{equation} \left\{\begin{array}{ll} 0 = z, \quad 0 = f(x, y, t), \\ { } \frac{{\rm d}x}{{\rm d}\xi} = g(x, y, t), \quad \frac{{\rm d}k_{1}}{{\rm d}\xi} = h_{1}(y), \quad \frac{{\rm d}k_{2}}{{\rm d}\xi} = h_{2}(y), \quad \frac{{\rm d}t}{{\rm d}\xi} = 1 \end{array}\right. \end{equation} $

和极限快系统

(3.4) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}y}{{\rm d}\tau} = z, \quad \frac{{\rm d}z}{{\rm d}\tau} = f(x, y, t), \\ { }\frac{{\rm d}x}{{\rm d}\tau} = 0, \quad \frac{{\rm d}k_{1}}{{\rm d}\tau} = 0, \quad \frac{{\rm d}k_{2}}{{\rm d}\tau} = 0, \quad \frac{{\rm d}t}{{\rm d}\tau} = 0. \end{array}\right. \end{equation} $

由假设(H$ _{2} $),可知临界流形为

$ S_1 = \{z = 0, y = \alpha_{1}(\varphi_{1}(t))\}, \; \; \; \; S_2 = \{z = 0, y = \alpha_{2}(\varphi_{2}(t))\}, $

$ \dim(S_{i}) = 4 $,系统(3.4)对应的线性系统在临界流形上有4个零根,一个正根和一个负根,从而可知$ S_{i} $是法向双曲的, $ i = 1, 2 $.因此,存在连接$ M_1(\alpha_{1}(\varphi_{1}(t)), 0) $和$ M_2(\alpha_{2}(\varphi_{2}(t)), 0) $的异宿轨道.

(H$ _{3} $)  稳定流形$ 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) $横截相交,不稳定流形$ 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 $.

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.

系统(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.如图 1所示, $ \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. $

定理 3.1   如果满足条件$ ({\rm H}_{1}) $–$ ({\rm H}_{3}) $,则对于充分小的$ \mu>0 $,奇异摄动边值问题(2.1)存在阶梯状空间对照结构解$ x(t, \mu) $和$ y(t, \mu) $,即

$ \begin{eqnarray*} \lim\limits_{\mu\rightarrow0}x(t, \mu) = \left\{\begin{array}{ll} \varphi_1(t), & 0\leq t\leq t_0, \\ \varphi_2(t), & t_0\leq t\leq 1, \end{array}\right.\quad \lim\limits_{\mu\rightarrow0}y(t, \mu) = \left\{\begin{array}{ll} \alpha_{1}(\varphi_1(t)), & 0<t<t_0, \\ \alpha_2(\varphi_2(t)), & t_0<t<1. \end{array}\right. \end{eqnarray*} $

证   连接问题(3.1)的讨论空间维数为$ {\rm R}^6 $, $ \dim B_{\mu}^{L} = 2, \ \dim B_{\mu}^{R} = 3, \ \dim S_1 = \dim S_2 = 4 $. $ N_0 = B_{0}^{L}\cap W^{s}(S_1) $,映射$ 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) $.由横截相交假设可知, $ \dim W^s(S_1) = 5 $, $ \dim N_0 = 1 $.因此,稳定流形$ W^{s}(S_1) $与$ B_{0}^{L} $横截相交于一个一维流形.显然, $ \dim \chi^{1} = 1, \ \dim U^{1} = 2 $.解由点$ p_{1} $到达$ \bar{p}_{1} $的时间是有限的,同时有$ \dim W^u(U^1) = 3 $.

$ 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) $.由假设可知, $ \dim W^u(S_2) = 5 $, $ \dim N_1 = 2 $.因此, $ S_2 $的不稳定流形与$ B_{0}^{R} $横截相交于一个二维流形.显然, $ \dim \chi^{2} = 2, \ \dim U^{2} = 3 $.解由点$ \bar{p}_{2} $到达$ p_{2} $的时间是有限的,同时$ \dim W^s(U^2) = 4 $.

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) $横截相交,且

$ \dim(W^s(U^2)+\dim(W^u(U^1))-6 = 1. $

因此,存在连接慢流形$ S_1 $和慢流形$ S_2 $的异宿轨道,进一步交换引理^[13]的全部条件都满足,奇异摄动边值问题(2.1)存在阶梯状空间对照结构解.

根据解的结构,本节将利用边界层函数法^[1],构造奇异摄动边值问题(2.2)的渐近解.假设渐近级数为$ \omega = (y, z, x, k_{1}, k_{2})^{{\rm T}} $

(4.1) $ \begin{equation} \omega^{(-)}(t, \mu) = \sum\limits_{k = 0}^\infty\mu^k(\bar{\omega}_{k}^{(-)}(t)+L_k\omega(\tau_0)+Q_k^{(-)}\omega(\tau_1)), \quad 0\leq t\leq t^*, \end{equation} $

(4.2) $ \begin{equation} \omega^{(+)}(t, \mu) = \sum\limits_{k = 0}^\infty\mu^k(\bar{\omega}_{k}^{(+)}(t)+Q_{k}^{(+)}\omega(\tau_1)+R_k\omega(\tau_2)), \quad t^*\leq t\leq 1, \end{equation} $

其中$ \tau_{0} = t\mu^{-1}, \quad \tau_{1} = (t-t^{*})\mu^{-1}, \quad \tau_{2} = (t-1)\mu^{-1} $, $ \bar{\omega}_{k}^{(\mp)}(t) $是正则项的系数, $ L_k\omega(\tau_0) $是左边界层项的系数, $ R_k\omega(\tau_2) $是右边界层项的系数, $ Q_k^{(\mp)}\omega(\tau_1) $内部转移层项的系数.

为确定内部转移点\ $ t^*(\mu)\in [0, 1] $,假设其渐近级数为

$ t^* = t_0+\mu t_1+\cdots+\mu^k t_k+\cdots. $

把形式渐近解(4.1)和(4.2)代入边值问题(2.2),按快慢尺度$ t $, $ \tau_0 $, $ \tau_1 $和$ \tau_2 $分离,比较$ \mu $的同次幂,可得确定$ \bar{y}_k^{(\mp)}(t), \ \bar{z}_k^{(\mp)}(t), $$ \bar{x}_k^{(\mp)}(t), \bar{k}_{1k}^{(\mp)}(t) $$ \bar{k}_{2k}^{(\mp)}(t), $$ L_ky(\tau_0), \ L_kz(\tau_0) $, $ L_kx(\tau_0), L_kk_{1}(\tau_0), $$ L_kk_{2}(\tau_0) $, $ Q_k^{(\mp)}y(\tau_1), \ Q_k^{(\mp)}z(\tau_1), $$ Q_k^{(\mp)}x(\tau_1) $, $ Q_k^{(\mp)}k_{1}(\tau_1), Q_k^{(\mp)}k_{2}(\tau_1) $, $ R_ky(\tau_2), R_kz(\tau_2), R_kx(\tau_2) $, $ R_kk_{1}(\tau_2), R_kk_{2}(\tau_2) $, $ k\geq0 $的方程和条件.

先给出确定零次正则项$ \bar{y}_0^{(\mp)}(t), \ \bar{z}_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

(4.3) $ \begin{equation} \left\{\begin{array}{ll} \bar{z}_{0}^{(\mp)}(t) = 0, \ f(\bar{x}_0^{(\mp)}(t), \bar{y}_0^{(\mp)}(t), t) = 0, \\ { } \frac{{\rm d}\bar{x}_{0}^{(\mp)}(t)}{{\rm d}t} = g(\bar{x}_0^{(\mp)}(t), \bar{y}_0^{(\mp)}(t), t), \quad \frac{{\rm d}\bar{k}_{10}^{(\mp)}(t)}{{\rm d}t} = h_{1}(\bar{y}_0^{(\mp)}(t)), \\ { } \frac{{\rm d}\bar{k}_{20}^{(\mp)}(t)}{{\rm d}t} = h_{2}(\bar{y}_0^{(\mp)}(t)), \end{array}\right. \end{equation} $

由假设($ {\rm H}_2 $),可知

$ \begin{eqnarray*} \bar{z}_0^{(\mp)} = 0, \ \ \bar{y}_0^{(\mp)} = \left\{\begin{array}{ll} { } \alpha_{1}(\varphi_1(t))\, 0\leq t<t_0, \\ { } \alpha_{2}(\varphi_2(t))\ , t_0<t\leq 1, \end{array}\right. \bar{x}_0^{(\mp)} = \left\{\begin{array}{ll} \varphi_1(t)\, 0\leq t\leq t_0, \\ \varphi_2(t)\, t_0\leq t\leq 1, \end{array}\right. \end{eqnarray*} $

关于$ \bar{k}_{10}^{(\mp)}(t), \ \bar{k}_{20}^{(\mp)}(t) $,将结合条件$ \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,需要指出解中包含边界层中的未知项.

确定零次左边界层项$ L_0y(\tau_0), \ L_0z(\tau_0) $, $ L_0x(\tau_0), \ L_0k_{1}(\tau_0), \ L_0k_{2}(\tau_0) $的方程和条件为

(4.4) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}L_{0}y}{{\rm d}\tau_{0}} = L_0z, \ \frac{{\rm d}L_0z}{{\rm d}\tau_0} = f(\varphi_1(0)+L_0x, \alpha_1(\varphi(0))+L_0y, 0), \\ { } \frac{{\rm d}L_0x}{{\rm d}\tau_0} = 0, \ \ \frac{{\rm d}L_0k_{1}}{{\rm d}\tau_0} = 0, \frac{{\rm d}L_0k_{2}}{{\rm d}\tau_0} = 0, \\ { } \bar{y}_{0}^{(-)}(0)+L_{0}y(0) = -\bar{k}_{10}^{(-)}(0)+L_0k_{1}(0)+y^{0}, \ \bar{x}_0^{(-)}(0)+L_0x(0) = x^0, \\ { } L_0y(+\infty) = 0, \ L_0z(+\infty) = 0, \ L_0x(+\infty) = 0, \ L_0k_{1}(+\infty) = 0, L_0k_{2}(+\infty) = 0, \end{array}\right. \end{equation} $

由边界层函数法的性质可知,边界层项都是指数衰减的,因此$ L_0x = 0, \ L_0k_{1} = 0, \ L_0k_{2} = 0 $.同理, $ R_0k_{1} = 0 $,基于此,正则项$ \bar{k}_{10}^{(\mp)}(t), \ \bar{k}_{20}^{(\mp)}(t) $可同时确定.

类似地,确定零次内部转移层项$ Q_0^{(\mp)}y(\tau_1), \ Q_0^{(\mp)}z(\tau_1), $$ Q_0^{(\mp)}x(\tau_1) $, $ Q_0^{(\mp)}k_{1}(\tau_1) $和$ Q_0^{(\mp)}k_{2}(\tau_1) $的方程和条件为

(4.5) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}Q_{0}^{(\mp)}y}{{\rm d}\tau_1} = Q_0^{(\mp)}z, \ \frac{{\rm d}Q_0^{(\mp)}z}{{\rm d}\tau_1} = f(\varphi_{1, 2}(t_0)+Q_0^{(\mp)}x, \alpha_{1, 2}(\varphi_{1, 2}(t_0))+Q_0^{(\mp)}y, t_0), \\ { } \frac{{\rm d}Q_{0}^{(\mp)}x}{{\rm d}\tau_1} = 0, \ \frac{{\rm d}Q_0^{(\mp)}k_{1}}{{\rm d}\tau_1} = 0, \frac{{\rm d}Q_0^{(\mp)}k_{2}}{{\rm d}\tau_1} = 0, \\ Q_{0}^{(\mp)}y(0)+\alpha_{1, 2}(\varphi_{1, 2}(t_{0})) = \beta(t_{0}), Q_{0}^{(\mp)}y(\mp \infty) = 0, Q_{0}^{(\mp)}z(\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{array}\right. \end{equation} $

其中$ \beta(t_{0}) = \frac{1}{2}(\alpha_1(\varphi_1(t))+\alpha_2(\varphi_2(t))) $.作变量代换

$ \tilde{y}^{(\mp)} = \alpha_{1, 2}(\varphi_{1, 2}(t_0))+Q_0^{(\mp)}y, \; \; \tilde{z}^{(\mp)} = Q_0^{(\mp)}z, \; \; \tilde{x}^{(\mp)} = \varphi_{1, 2}(t_0)+Q_0^{(\mp)}x, $

$ \tilde{k}_1^{(\mp)} = Q_0^{(\mp)}k_1, \; \; \tilde{k}_2^{(\mp)} = Q_0^{(\mp)}k_2, $

可得

(4.6) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}\tilde{y}^{(\mp)}}{{\rm d}\tau_1} = \tilde{z}^{(\mp)}, \ \frac{{\rm d}\tilde{z}^{(\mp)}}{{\rm d}\tau_1} = f(\tilde{x}^{(\mp)}, \tilde{y}^{(\mp)}, t_0), \\ { } \frac{{\rm d}\tilde{x}^{(\mp)}}{{\rm d}\tau_1} = 0, \ \frac{{\rm d}\tilde{k}_1^{(\mp)}}{{\rm d}\tau_1} = 0, \frac{{\rm d}\tilde{k}_2^{(\mp)}}{{\rm d}\tau_1} = 0, \\ \tilde{y}^{(\mp)}(0) = \beta(t_{0}), \tilde{y}^{(\mp)}(\mp \infty) = \alpha_{1, 2}(\varphi_{1, 2}(t_{0})), \\ \tilde{z}^{(\mp)}(\mp \infty) = 0, \tilde{x}^{(\mp)}(\mp \infty) = \varphi_{1, 2}(t_0), \ \tilde{k}_1^{(\mp)}(\mp \infty) = 0, \tilde{k}_2^{(\mp)}(\mp \infty) = 0, \end{array}\right. \end{equation} $

利用文献[1]的主要结果,可知确定内部转移层点$ t^* $的主项$ t_0 $的方程为

$ I(t_0) = \int_{\alpha_{1}(\varphi_1(t_0))}^{\alpha_{2}(\varphi_2(t_0))}f(\varphi_{1, 2}(t_0), y, t_0){\rm d}y = 0. $

为保证解的存在性,接下来的条件需要满足,需要指出这些条件都是平凡的.

($ {\rm H}_{4} $)  $ I'(t_0)\neq 0 $,左初值问题

$ \frac{{\rm d}\bar{k}_{10}^{(-)}}{{\rm d}t} = h_1(\alpha_{1}(\varphi_1(t))), \quad \bar{k}_{10}^{(-)}(t_0) = \bar{k}_{10}^{(+)}(t_0) $

有解$ \beta_1(t) $,右初值问题

$ \frac{{\rm d}\bar{k}_{10}^{(+)}}{{\rm d}t} = h_1(\alpha_2(\varphi_2(t))), \quad \bar{k}_{10}^{(+)}(1) = 0 $

有解$ \beta_2(t) $,其中$ \beta_1(t) $和$ \beta_2(t) $在$ t_0 $处横截相交.\同时,左初值问题

$ \frac{{\rm d}\bar{k}_{20}^{(-)}}{{\rm d}t} = h_2(\alpha_1(\varphi_1(t))), \quad \bar{k}_{20}^{(-)}(0) = 0 $

有解$ \gamma_1(t) $,同时右初值问题

$ \frac{{\rm d}\bar{k}_{20}^{(+)}}{{\rm d}t} = h_2(\alpha_2(\varphi_2(t))), \quad \bar{k}_{20}^{(+)}(t_0) = \bar{k}_{20}^{(-)}(t_0) $

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

确定零次右边界层项$ R_0y(\tau_2), R_0z(\tau_2), R_0x(\tau_2) $, $ R_0k_{1}(\tau_2) $和$ R_0k_{2}(\tau_2) $的方程和条件为

(4.7) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}R_{0}y}{{\rm d}\tau_{2}} = R_0z, \ \frac{{\rm d}R_0z}{{\rm d}\tau_2} = f(\varphi_2(1)+R_0x, \alpha_2(\varphi_2(1))+R_0y, 1), \\ { } \frac{{\rm d}R_{0}x}{{\rm d}\tau_{2}} = 0, \ \frac{{\rm d}R_0k_1}{{\rm d}\tau_2} = 0, \frac{{\rm d}R_0k_2}{{\rm d}\tau_2} = 0, \\ \alpha_2(\varphi_2(1))+R_{0}y(0) = \bar{k}_{20}^{(+)}(1)+R_0k_{2}(0)+y^{1}, R_0y(-\infty) = 0, \ R_0z(-\infty) = 0, \\ R_0x(-\infty) = 0, \ R_0k_{1}(-\infty) = 0, R_0k_{2}(-\infty) = 0, \end{array}\right. \end{equation} $

类似于零次左边界层项的讨论,可知$ R_0x = 0, \ R_0k_1 = 0 $, $ R_0k_2 = 0 $.结合条件(H$ _{3} $),可知问题(4.4)–(4.7)的解存在.至此,已确定了形式渐近解的全部零次主项,接下来,给出确定渐近解高阶项的方程和条件.

确定高次正则项$ \bar{y}_{n}^{(\mp)}(t), \ \bar{z}_{n}^{(\mp)}(t), \ \bar{x}_{n}^{(\mp)}(t), \ \bar{k}_{1n}^{(\mp)}(t), $和$ \bar{k}_{2n}^{(\mp)}(t), \ n\geq 1 $的方程和条件为

(4.8) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}\bar{y}_{n-1}^{(\mp)}}{{\rm d}t} = \bar{z}_{n}^{(\mp)}(t), \\ { } \frac{{\rm d}\bar{z}_{n-1}^{(\mp)}}{{\rm d}t} = \tilde{f}_{x}\bar{x}_{n}^{(\mp)}(t)+\tilde{f}_{y}\bar{y}_{n}^{(\mp)}(t)+F_{1n}^{(\mp)}(t), \\ { } \frac{{\rm d}\bar{x}_{n}^{(\mp)}}{{\rm d}t} = \tilde{g}_{x}\bar{x}_{n}^{(\mp)}(t)+\tilde{g}_{y}\bar{y}_{n}^{(\mp)}(t)+F_{2n}^{(\mp)}(t), \\ { } \frac{{\rm d}\bar{k}_{1n}^{(\mp)}}{{\rm d}t} = h_{1y}(\alpha_{1, 2}(\varphi_{1, 2}(t)))\bar{y}_{n}^{(\mp)}(t)+F_{3n}^{(\mp)}(t), \\ { }\frac{{\rm d}\bar{k}_{2n}^{(\mp)}}{{\rm d}t} = h_{2y}(\alpha_{1, 2}(\varphi_{1, 2}(t)))\bar{y}_{n}^{(\mp)}(t)+F_{4n}^{(\mp)}(t), \end{array}\right. \end{equation} $

其中

$ \tilde{f}_{x} = f_{x}(\varphi_{1, 2}(t), \alpha_{1, 2}(\varphi_{1, 2}(t)), t), \ \tilde{f}_{y} = f_{y}(\varphi_{1, 2}(t), \alpha_{1, 2}(\varphi_{1, 2}(t)), t), $

$ \tilde{g}_{x} = g_{x}(\varphi_{1, 2}(t), \alpha_{1, 2}(\varphi_{1, 2}(t)), t), \ \tilde{g}_{y} = g_{y}(\varphi_{1, 2}(t), \alpha_{1, 2}(\varphi_{1, 2}(t)), t), $

$ F_{in}^{(\mp)}(t), \ i = 1, 2, 3, 4 $是依赖于一些已确定项的已知函数.需要指出问题(4.8)的解依赖于$ \bar{x}_{n}^{(\mp)} $的可解性,将在高次边界层项确定时给出.

确定高次左边界层项$ L_{n}y(\tau_0) $$ L_{n}z(\tau_0), $$ L_{n}x(\tau_0) $, $ L_{n}k_{1}(\tau_0) $和$ L_{n}k_{2}(\tau_0), n\geq 1 $的方程和条件为

(4.9) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}L_ny}{{\rm d}\tau_0} = L_{n}z, \\ { } \frac{{\rm d}L_nz}{{\rm d}\tau_0} = \tilde{f}_{x}^{(L)}(\tau_0)L_nx+\tilde{f}_{y}^{(L)}(\tau_0)L_ny+F_{1n}^{(L)}(\tau_0), \\ { } \frac{{\rm d}L_nx}{{\rm d}\tau_0} = \tilde{g}_{x}^{(L)}(\tau_0)L_{n-1}x+\tilde{f}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{2n}^{(L)}(\tau_0), \\ { } \frac{{\rm d}L_nk_1}{{\rm d}\tau_0} = \tilde{h_1}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{3n}^{(L)}(\tau_0), \\ { } \frac{{\rm d}L_nk_2}{{\rm d}\tau_0} = \tilde{h_2}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{4n}^{(L)}(\tau_0), \\ \bar{y}_n(0)+L_ny(0) = -\bar{k}_{1n}(0)-L_{n}k_{1}(0), \ L_nk_{2}(0) = -\bar{k}_{2n}^{(-)}(0), \\ L_nx(0) = -\bar{x}_n^{(-)}(0), \ L_ny(+\infty) = 0, \ L_nz(+\infty) = 0, \ L_nx(+\infty) = 0, \\ L_nk_{1}(+\infty) = 0, \ L_nk_{2}(+\infty) = 0, \end{array}\right. \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) $在$ (\varphi_{1}(0)+L_0x, \alpha_1(\varphi_1(0))+L_0y, 0) $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{f}_{y}^{(L)}(\tau_0)L_{n-1}y+F_{2n}^{(L)}(\tau_0) $,对方程$ \frac{{\rm d}L_nx}{{\rm d}\tau_0} = G_{1}(\tau_0) $由$ +\infty $到$ \tau_0 $积分,可知

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

令$ \tau_0 = 0 $,利用(4.9)式,可得

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

其中$ G_{1}(\tau_0) $是依赖于一些已确定项的已知函数.利用方程(4.8),可确定高次项$ \bar{\omega}_{n}^{(-)}(t) $.需要指出$ \bar{k}_{1n}^{(-)}(t) $中包含任意常数参数,需要在内部层和右边界层项的计算中确定.类似于$ L_nx(\tau_0) $的计算过程, $ L_nk_{1}(\tau_0) $和$ L_nk_{2}(\tau_0) $同样可以确定,关于变量$ L_ny(\tau_0) $和$ L_nz(\tau_0) $解的存在性接下来给出说明.

考虑确定内部层高次项$ Q_{n}^{(\mp)}y(\tau_1), $$ Q_{n}^{(\mp)}z(\tau_1), $$ Q_{n}^{(\mp)}x(\tau_1), $$ Q_{n}^{(\mp)}k_{1}(\tau_1) $和$ Q_{n}^{(\mp)}k_{2}(\tau_1), n\geq 1 $的方程和条件

(4.10) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}Q_{n}^{(\mp)}y}{{\rm d}\tau_1} = Q_{n}^{(\mp)}z(\tau_1), \\ { } \frac{{\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}\\ {\quad} +F_{1n}^{(Q){(\mp)}}(\tau_1), \\ { } \frac{{\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), \\ { } \frac{{\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), \\ { } \frac{{\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), \\ { } 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)}z(\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{array}\right. \end{equation} $

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

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

取值, $ \sigma_{n}(\rho), \ F_{in}^{Q{(\mp)}}(\tau_1), \ i = 1, 2, 3, 4 $是依赖于一些已确定项的已知函数.令$ 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) $,对方程$ \frac{{\rm d}Q_n^{(\mp)}x}{{\rm d}\tau_1} = G_{2}(\tau_1) $由$ \mp\infty $到$ \tau_1 $积分,可知

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

其中$ G_{2}(\tau_1) $是依赖于一些已确定项的已知函数.类似地,可确定$ Q_{n}^{(\mp)}k_1 $和$ Q_{n}^{(\mp)}k_2 $.讨论确定$ t_{n} $的方程和条件,考虑系统(4.10)的联合系统

(4.11) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}Q_{n}y}{{\rm d}\tau_1} = Q_{n}z(\tau_1), \\ { } \frac{{\rm d}Q_nz}{{\rm 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{array}\right. \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) $取值,同时满足

$ q_{11}(\tau_1)\rightarrow\alpha_1(\varphi_{1}(t_0)), \tau_1\rightarrow -\infty, q_{11}(\tau_1)\rightarrow\alpha_2(\varphi_{2}(t_0)), \tau_1\rightarrow +\infty, $

$ q_{12}(\tau_1)\rightarrow0, \tau_1\rightarrow -\infty, q_{12}(\tau_1)\rightarrow0, \tau_1\rightarrow +\infty, $

令$ Q_{n}\lambda = \big(Q_{n}^{{\rm T}}y, \ Q_{n}^{{\rm T}}z\big)^{{\rm T}} $, $ \tilde{f}_n^{Q}(\tau_1) = \big(0, \ F_{1n}^{Q{\rm T}}(\tau_1)\big)^{{\rm T}} $,

$ I(y, z, x, t) = \left(\begin{array}{c} z\\ f(x, y, t)\end{array}\right). $

利用指数二分法,算子$ F(Q_{n}\lambda) = \frac{{\rm d}Q_{n}\lambda}{{\rm d}\tau_1}-I_{\lambda}(\tau_1)Q_{n}\lambda $是Fredholm型算子, Fredholm指标$ \dim {\rm Ker} F-\dim {\rm Ker} F^{*} = 0 $,其中$ I_{\lambda}(\tau_1) $在$ (q_{11}(\tau_1), q_{12}(\tau_1), t_0) $取值.利用文献[16]的引理3.7,可知存在唯一的函数$ \psi_{1}(\tau_1)\in {\rm Ker} F^{*} $,方程(4.11)有解等价于

$ t_{n}\int_{-\infty}^{+\infty}\psi_{1}^{*}(\tau_1)B{\rm d}\tau_1 = -\int_{-\infty}^{+\infty}\psi_{1}^{*}(\tau_1)\tilde{f}_n^{Q}(\tau_1){\rm d}\tau_1. $

借助于假设$ {\rm H}_3 $和melnikov函数,计算可得$ \int_{-\infty}^{\infty}\psi_{1}^*B {\rm d}\tau_1 \neq 0 $,其中

$ B = \left(\begin{array}{cc} 0\\ f_{t}(q_{11}(\tau_1), q_{12}(\tau_1), t_0)\end{array}\right). $

到此为止,已确定了$ t_{n} $,利用假设$ {\rm H}_3 $,可确定内部转移层项$ Q_{n}^{(\mp)}\omega(\tau_1) $.利用变量$ y, z, x, \ k_{1}, \ k_{2} $的表达式和连续性,可确定正则项$ \bar{y}_n^{(+)}(t), \bar{z}_n^{(+)}(t) $, $ \bar{x}_n^{(+)}(t) $, $ \bar{k}_{1n}^{(+)} $和$ \bar{k}_{2n}^{(+)}(t) $.

确定右边界层高次项$ R_{n}y(\tau_2) $, $ R_{n}z(\tau_2), $$ R_{n}x(\tau_2) $, $ R_{n}k_{1}(\tau_2) $和$ R_{n}k_{2}(\tau_2), n\geq 1 $的方程和条件为

(4.12) $ \begin{equation} \left\{\begin{array}{ll} { } \frac{{\rm d}R_ny}{{\rm d}\tau_2} = R_{n}z, \\ { } \frac{{\rm d}R_nz}{{\rm d}\tau_2} = \tilde{f}_{x}^{(R)}(\tau_2)R_nx+\tilde{f}_{y}^{(R)}(\tau_2)R_ny+F_{1n}^{(R)}(\tau_2), \\ { } \frac{{\rm d}R_nx}{{\rm d}\tau_2} = \tilde{g}_{x}^{(R)}(\tau_2)R_{n-1}x+\tilde{f}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{2n}^{(R)}(\tau_0), \\ { } \frac{{\rm d}R_nk_1}{{\rm d}\tau_2} = \tilde{h_1}_{y}^{(R)}(\tau_2)R_{n-1}y+F_{3n}^{(R)}(\tau_2), \\{ } \frac{{\rm d}R_nk_2}{{\rm d}\tau_2} = \tilde{h_2}_{x}^{(R)}(\tau_2)R_{n-1}y+F_{4n}^{(R)}(\tau_2), \\ R_ny(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{array}\right. \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) $在$ (\varphi_{2}(1)+R_0x, \alpha_2(\varphi_2(1))+R_0y, 1) $取值, $ F_{in}^{(R)}(\tau_2), $$ i = 1, 2, 3, 4 $依赖于一些已确定项的已知函数.令$ G_{3}(\tau_2) = \tilde{h_1}_{y}^{(R)}(\tau_2)L_{n-1}y+F_{3n}^{(R)}(\tau_2) $,对方程$ \frac{{\rm d}R_nk_{1}}{{\rm d}\tau_2} = G_{3}(\tau_2) $由$ -\infty $到$ \tau_2 $积分,有

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

令$ \tau_2 = 0 $,利用方程(4.12),可得

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

其中$ G_{3}(\tau_2) $是依赖于一些已确定项的已知函数.借助于变量$ \bar{k}_{1n}(t) $的表达式和连续性,则可确定$ \bar{\omega}_{n}^{(+)}(t) $.

需要指出方程(4.9)和(4.12)是线性非齐次微分方程,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) $已由方程确定.因此,只需考虑方程(4.9)和(4.12)前两个变量就可以. $ (L'_0y, L'_0z) $和$ (R'_0y, R'_0z) $是方程(4.9)和(4.12)对应的线性齐次方程的解,利用文献[1]的结果,可知方程(4.9)和(4.12)前两个变量的解存在.

定理 4.1   如果满足条件$ ({\rm H}_1) $–$ ({\rm H}_4) $,那么对充分小的$ \mu>0 $,奇异摄动边值问题(2.1)存在阶梯状空间对照结构解$ x(t, \mu) $和$ y(t, \mu) $且满足

$ x(t, \mu) = \left\{\begin{array}{ll} { } \sum\limits_{k = 0}^{n}\mu^{k}\big(\bar x_k^{(-)}(t)+L_kx(\tau_0)+Q_{k}^{(-)}x(\tau_1)\big)+O(\mu^{n+1}), & 0\leq t \leq T_{n}+O(\mu^{n+1}), \\ { } \sum\limits_{k = 0}^{n}\mu^{k}\big(\bar x_k^{(+)}(t)+Q_{k}^{(+)}x(\tau_1)+R_kx(\tau_2)\big)+O(\mu^{n+1}), & T_{n}+O(\mu^{n+1})\leq t\leq 1, \end{array}\right. $

$ y(t, \mu) = \left\{\begin{array}{ll} { } \sum\limits_{k = 0}^{n}\mu^{k}\big(\bar y_k^{(-)}(t)+L_ky(\tau_0)+Q_{k}^{(-)}y(\tau_1)\big)+O(\mu^{n+1}), & 0\leq t \leq T_{n}+O(\mu^{n+1}), \\ { } \sum\limits_{k = 0}^{n}\mu^{k}\big(\bar y_k^{(+)}(t)+Q_{k}^{(+)}y(\tau_1)+R_ky(\tau_2)\big)+O(\mu^{n+1}), & T_{n}+O(\mu^{n+1})\leq t\leq 1, \end{array}\right. $

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

考虑奇异摄动边值问题

(5.1) $ \begin{equation} \left\{\begin{array}{ll} { } \mu^{2}\frac{{\rm d}^{2}y}{{\rm d}t^{2}} = (y^2-4)(y-a(t)), \\ { } \frac{{\rm d}x}{{\rm d}t} = t, \\ { } x(0, \mu) = 1, \quad y(0, \mu) = \int_{0}^{1}y(s, \mu){\rm d}s\, , \quad y(1, \mu) = \int_{0}^{1}2y(s, \mu){\rm d}s. \end{array}\right. \end{equation} $

为了简化计算,不妨假设$ a(0) = a(1) = 0 $, $ a(\frac{1}{2}) = 0 $和$ a'(\frac{1}{2})\neq 0 $.原问题可改写为

(5.2) $ \begin{equation} \left\{\begin{array}{ll} { } \mu\frac{{\rm d}y}{{\rm d}t} = z, \quad \mu\frac{{\rm d}z}{{\rm d}t} = (y^2-4)(y-a(t)), \\ { } \frac{{\rm d}x}{{\rm d}t} = t, \quad \frac{{\rm d}k_{1}}{{\rm d}t} = y, \quad \frac{{\rm d}k_{2}}{{\rm d}t} = 2y, \\ y(0) = -k_1(0), \quad x(0) = 1, \quad k_1(1) = 0, \quad k_2(0) = 0, \quad y(1) = k_2(1). \end{array}\right. \end{equation} $

确定正则零次项的方程为条件为

$ \bar{y}_0^{(\mp)}(t) = \left\{\begin{array}{ll} -2, &0\leq t< t_{0}, \\ 2, & t_{0}< t \leq 1, \end{array}\right.\ \bar{z}_0^{(\mp)}(t) = \left\{\begin{array}{ll} 0, & 0\leq t \leq t_{0}, \\ 0, &t_{0} \leq t \leq 1, \end{array}\right. \ \bar{x}_0^{(\mp)}(t) = \left\{\begin{array}{ll} { }\frac{t^2}{2}+1, & 0\leq t\leq t_{0}, \\ { } \frac{t^2}{2}+1, &t_{0}\leq t \leq 1, \end{array}\right. $

$ \bar{k}_{10}^{(\mp)}(t) = \left\{\begin{array}{ll} -2t, \quad \quad 0\leq t< t_{0}, \\ 2t-2, \quad t_{0}< t \leq 1, \end{array}\right.\ \bar{k}_{20}^{(\mp)}(t) = \left\{\begin{array}{ll} -4t, \quad \quad 0\leq t \leq t_{0}, \\ 4t-4, \quad t_{0} \leq t \leq 1, \end{array}\right. $

转移点$ t^* $的主项$ t_0 $满足方程$ I(t_{0}) = \int_{-2}^{2}(y^2-4)(y-a(t_0)){\rm d}y = 0 $,有$ t_{0} = 1/2 $.

确定内部转移层零次项$ Q^{(\mp)}_{0}y $的方程和条件为

$ \frac{{\rm d}Q^{(\mp)}_{0}y}{{\rm d}\tau_1} = -\frac{\sqrt{2}}{2}\big((\mp 2+Q^{(\mp)}_{0}y)^2-4\big), \quad Q^{(\mp)}_{0}y(0) = \pm2, \quad Q^{(\mp)}_0y(\mp \infty) = 0, $

可得

$ Q^{(-)}_0y = \frac{4}{1+e^{-2\sqrt{2}\tau_1}}, \quad Q^{(+)}_0y = \frac{-4e^{-2\sqrt{2}\tau_1}}{1+e^{-2\sqrt{2}\tau_1}}. $

同理,可知

$ Q^{(-)}_0x = Q^{(-)}_0k_{1} = Q^{(-)}_0k_{2} = 0, \ Q^{(+)}_0x = Q^{(+)}_0k_{1} = Q^{(+)}_0k_{2} = 0, $

$ L_0y = \frac{4}{1+e^{2\sqrt{2}\tau_0}}, \quad L_{0}x = L_0k_{1} = L_0k_{2} = 0, $

$ R_0y = \frac{-4e^{2\sqrt{2}\tau_2}}{1+e^{2\sqrt{2}\tau_2}}, \quad R_{0}x = R_0k_{1} = R_0k_{2} = 0. $

从而可得问题(5.1)的形式渐近解为

$ x(t, \mu) = \left\{\begin{array}{ll} { }\frac{t^2}{2}+O(\mu), \quad 0\leq t \leq \frac{1}{2}, \\ { }\frac{t^2}{2}+O(\mu), \quad \frac{1}{2} \leq t \leq 1, \end{array}\right. $

$ y(t, \mu) = \left\{\begin{array}{ll} { } -2+\frac{4}{1+e^{2\sqrt{2}\tau_0}}+\frac{4}{1+e^{-2\sqrt{2}\tau_1}}+O(\mu), \quad \ 0\leq t \leq \frac{1}{2}, \\ { } 2+\frac{-4e^{-2\sqrt{2}\tau_1}}{1+e^{-2\sqrt{2}\tau_1}}+\frac{-4e^{2\sqrt{2}\tau_2}}{1+e^{2\sqrt{2}\tau_2}}+O(\mu), \quad \quad \frac{1}{2} \leq t \leq 1. \end{array}\right. $