奇摄动问题是国际数学界研究的热点问题, 奇摄动方程组渐近解与小迟滞问题渐近解都是热点问题之一, 如O'malley^[1]研究的方程组的初值问题、方程组的边值问题以及具有一个小参数的微分-差分方程的解, 陈育森研究的双参数非线性积分微分方程组奇摄动边值问题^[2], 唐荣荣研究的一类非线性方程组的奇摄动问题^[3], 吴钦宽研究的伴有边界摄动非线性积分微分方程系统的奇摄动^[4], 欧阳成研究的一类非线性方程组的奇摄动初值问题^[5]和有一个参数的小延迟的微分-差分方程渐近解^[6], 莫嘉琪研究的一个参数的奇摄动时滞反应扩散方程渐近解^[7].本文把方程组和迟滞问题同时考虑, 研究含有小迟滞的奇摄动方程组初值问题的渐近解.

问题

$ \begin{eqnarray} \frac{dx}{dt}=f(x, x'(t-\varepsilon), y'(t-\varepsilon)), \end{eqnarray} $

(1.1)

$ \varepsilon\frac{dy}{dt}=g(x, x'(t-\varepsilon), y'(t-\varepsilon)), $

(1.2)

$ \begin{eqnarray} x(t)=\phi(t), -\varepsilon\leq t\leq0, \end{eqnarray} $

(1.3)

$ \begin{eqnarray} y(t) =\psi(t), -\varepsilon\leq t\leq0, \end{eqnarray} $

(1.4)

其中 $0 < \varepsilon\ll 1$.

为了叙述方便, 现作如下假设:

[H₁] (1.1)-(1.4) 的退化问题

$ \begin{eqnarray*} && X'_{0}(t)=f( X_{0}(t), X_{0}'(t), Y_{0}'(t ) ), t \geq 0, \\ && g ( X_{0}(t), X_{0}'(t), Y_{0}'(t )) =0, t \geq 0, \\ && X_{0}(0) =\phi(0), Y_{0}(0)=\psi(0)\end{eqnarray*} $

在某个区间 $0\leq t\leq T$上存在唯一的连续可微解 $X_{0}(t), Y_{0}(t)$;

[H₂] $f, g $在所考虑的区域内关于其变元无限可微;

[H₃]存在正常数 $k>0, $有

$ \mid f_{2}'(X_{0}(0), \phi'(0), \psi'(0))\mid < e^{-k} < 1, \mid f_{3}'(X_{0}(0), \phi'(0), \psi'(0))\mid < e^{-k} < 1; $

当 $\mid y\mid < \mid X_{0}'(0)\mid+\mid X_{0}'(0)-\phi'(0)\mid$, 有

$ \mid f_{2}'(X_{0}(0), y, \psi'(0))\mid < e^{-k} < 1; $

当 $\mid z\mid < \mid Y_{0}'(0)\mid+\mid Y_{0}'(0)-\psi'(0)\mid$, 有

$ \mid f_{3}'(X_{0}(0), \phi'(0), z)\mid < e^{-k} < 1; $

[H₄] $f_{3}'(X_{0}(t), X_{0}'(t), $ $Y_{0}'(t ))\neq0, $ $ 1-f_{2}'(X_{0}(t), X_{0}'(t), $

$ Y_{0}'(t ))+ \frac{f_{3}'(X_{0}(t), X_{0}'(t), Y_{0}'(t ))g_{2}'(X_{0}(t), X_{0}'(t), Y_{0}'(t ))}{g_{3}'(X_{0}(t), X_{0}'(t), Y_{0}'(t ))} \neq0. $

设问题(1.1)-(1.4) 的解为

$ \begin{eqnarray}x(t)&=&X(t)+\varepsilon U(\tau)=\sum\limits_{i=0}^{\infty} X_{i}(t)\varepsilon^{i}+\varepsilon \sum\limits_{i=0}^{\infty}u_{i}(\tau)\varepsilon^{i};\nonumber\\ y(t)&=& Y(t)+\varepsilon V(\tau)=\sum\limits_{i=0}^{\infty} Y_{i}(t)\varepsilon^{i}+\varepsilon \sum\limits_{i=0}^{\infty}v_{i}(\tau)\varepsilon^{i}, \end{eqnarray} $

(2.1)

其中 $\tau=\frac{t}{\varepsilon}$, 问题(1.1)-(1.4) 在远离 $t=0$且 $t>0$处外部解为

$ \begin{equation}X(t)=\sum\limits_{i=0}^{\infty} X_{i}(t)\mu^{i}, Y(t)=\sum\limits_{i=0}^{\infty} Y_{i}(t)\varepsilon^{i}, \end{equation} $

(2.2)

在 $t=0$处 $x(t), y(t)$边界层校正项分别为

$ U(\tau)=\sum\limits_{i=0}^{\infty}u_{i}(\tau) \varepsilon^{i}, V(\tau)=\sum\limits_{i=0}^{\infty}v_{i}(\tau)\varepsilon^{i}. $

把(2.2) 式代入(1.1), (1.2) 式得

$ \begin{eqnarray}&& \sum\limits_{i=0}^{\infty} X'_{i}(t)\varepsilon^{i}=f(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t), \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon), \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) ), t \geq 0, \end{eqnarray} $

(2.3)

$ \begin{eqnarray} && \varepsilon\sum\limits_{i=0}^{\infty} Y'_{i}(t)\varepsilon^{i}=g(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t), \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon), \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) ), t \geq 0, \end{eqnarray} $

(2.4)

所以

$ \begin{eqnarray} && X'_{i}(t)=f_{1}'( X_{0}(t), X_{0}'(t), Y_{0}'(t ))X_{i}(t) +f_{2}'( X_{0}(t), X_{0}'(t), Y_{0}'(t ))X'_{i}(t)\nonumber\\ && +f_{3}'( X_{0}(t), X_{0}'(t), Y_{0}'(t )) Y'_{i}(t)+A_{i}(t), t \geq 0, \end{eqnarray} $

(2.5)

$ \begin{eqnarray} && g_{1}'( X_{0}(t), X_{0}(t), Y_{0}'(t ))X_{i}(t) +g_{2}'( X_{0}(t), X_{0}'(t), Y_{0}'(t )) X'_{i}(t)\nonumber \\ && +g_{3}'( X_{0}(t), X_{0}'(t), Y_{0}'(t )) Y'_{i}(t)+B_{i}(t)=0, t \geq 0, \end{eqnarray} $

(2.6)

这里 $i=1, 2, 3, \cdots, $且 $A_{i}(t) , B_{i}(t)$是逐次确定的光滑函数.把(2.1) 式代入(1.3), (1.4) 式, 得

$ \begin{equation}X_{i}(0)=u_{i-1}(0), Y_{i}(0)=v_{i-1}(0), i=1, 2, 3, \cdots.\end{equation} $

(2.7)

下面求 $u_{i}(0), v_{i}(0)(i=0, 1, 2, \cdots).$把(2.1) 式代入(1.1), (1.2) 式得

$ \begin{eqnarray*}&& \sum\limits_{i=0}^{\infty}\dot{u}_{i}(\tau)\varepsilon^{i}=f(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t) +\varepsilon\sum\limits_{i=0}^{\infty}u_{i}(\tau)\varepsilon^{i}, \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon) +\sum\limits_{i=0}^{\infty}\dot{u}_{i}(\tau-1)\varepsilon^{i}, \\ && \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) + \sum\limits_{i=0}^{\infty}\varepsilon^{i}\dot{v}_{i}(\tau-1) ) -f(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t) , \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon), \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) ), \\ && t \geq 0, \\ && \sum\limits_{i=0}^{\infty}\dot{v}_{i}(\tau)\varepsilon^{i}=g(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t) +\varepsilon\sum\limits_{i=0}^{\infty}u_{i}(\tau)\varepsilon^{i}, \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon) +\sum\limits_{i=0}^{\infty}\dot{u}_{i}(\tau-1)\varepsilon^{i}, \\ && \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) + \sum\limits_{i=0}^{\infty}\varepsilon^{i}\dot{v}_{i}(\tau-1) ) -g(\sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}(t) , \sum\limits_{i=0}^{\infty}\varepsilon^{i} X_{i}'(t-\varepsilon), \sum\limits_{i=0}^{\infty}\varepsilon^{i} Y_{i}'(t-\varepsilon) ), \\ && t \geq 0.\end{eqnarray*} $

当 $0 < \tau \leq1$时,

$ \begin{eqnarray*}&&\dot{u}_{0}(\tau) = f(X_{0}(0) , \phi'(0), \psi'(0))-f(X_{0}(0), X_{0}'(0), Y'_{0}(0)), \\ && \dot{v}_{0}(\tau) = g(X_{0}(0) , \phi'(0), \psi'(0))-g(X_{0}(0), X_{0}'(0), Y'_{0}(0)); \end{eqnarray*} $

当 $\tau\geq1$时,

$ \dot{u}_{0}(\tau) = f(X_{0}(0) , X'(0)+\dot{u}_{0}(\tau-1), Y'(0)+\dot{v}_{0}(\tau-1))-f(X_{0}(0), X_{0}'(0), Y'_{0}(0)), \\ \dot{v}_{0}(\tau) = g(X_{0}(0) , X'(0)+\dot{u}_{0}(\tau-1), Y'(0)+\dot{v}_{0}(\tau-1))-g(X_{0}(0), X_{0}'(0), Y'_{0}(0)). $

所以 $\dot{u}_{0}(\tau), \dot{v}_{0}(\tau)$为分段常数.对于非负整数 $p, p\leq\tau\leq p+1.$记

$ \dot{u}_{0}(\tau)=M^{0}_{p}, \dot{v}_{0}(\tau)=N^{0}_{p}, $

所以

$ {u}_{0}(\tau)={u}_{0}(0)+\int_{0}^{\tau}\dot{u}_{0}( s)ds={u}_{0}(0)+\sum\limits_{j=0}^{p-1}M^{0}_{j}+(\tau-p)M^{0}_{p}, \\ {v}_{0}(\tau)={v}_{0}(0)+\int_{0}^{\tau}\dot{v}_{0}( s)ds={v}_{0}(0)+\sum\limits_{j=0}^{p-1}N^{0}_{j}+(\tau-p)N^{0}_{p}. $

因为当 $\tau\rightarrow\infty$时, ${u}_{0}(\tau)\rightarrow 0, {v}_{0}(\tau)\rightarrow 0$, 所以

$ u_{0}(0)=-\sum\limits_{j=0}^{\infty}M^{0}_{j}, v_{0}(0)=-\sum\limits_{j=0}^{\infty}N^{0}_{j}. $

同时

$ \dot{u}_{i}(\tau )= M^{i}_{p}(\tau), \dot{v}_{i}(\tau )= N^{i}_{p}(\tau)\ \ \ (p\leq\tau\leq p+1, \ \ i=1, 2, \cdots), $

这里 $M^{i}_{p}(\tau), N^{i}_{p}(\tau)$是依次确定的函数, 并且当 $\tau$取整数值时, $\dot{u}_{i}(\tau), \dot{v}_{i}(\tau)$一般是间断的.所以

$ u_{i}(\tau)=u_{i}(0)+\sum\limits_{l=0}^{p-1}(\int_{l}^{l+1} M^{i}_{l}(s)ds)+\int_{p}^{\tau} M^{i}_{p}(s)ds\ \ \ (p\leq\tau\leq p+1 ), \\v_{i}(\tau)=v_{i}(0)+\sum\limits_{l=0}^{p-1}(\int_{l}^{l+1} N^{i}_{l}(s)ds)+\int_{p}^{\tau} N^{i}_{p}(s)ds\ \ \ (p\leq\tau\leq p+1 ). $

当 $\tau\rightarrow\infty$时, $u_{i}(\tau)\rightarrow0, v_{i}(\tau)\rightarrow0, $所以

$ u_{i}(0)=-\sum\limits_{l=0}^{\infty}\int_{l}^{l+1} M^{i}_{l}(s)ds, \ \ v_{i}(0)=-\sum\limits_{l=0}^{\infty}\int_{l}^{l+1} N^{i}_{l}(s)ds. $

又因为[H₃], $\dot{u}_{i}(\tau), \dot{v}_{i}(\tau)(i=0, 1, 2\cdots)$都是指数性小项, 且

$ \lim\limits_{\tau\rightarrow\infty} u_{i}(\tau)=0, \lim\limits_{\tau\rightarrow\infty} v_{i}(\tau)=0, $

所以 $u_{i}(\tau), v_{i}(\tau)(i=0, 1, 2\cdots)$都是指数性小项, 且 $v_{i}(0)(i=0, 1, 2\cdots)$是有限值^[1].这样就可以利用(2.5)-(2.7) 式与[H₄]依次得 $X_{i}(t), Y_{i}(t)(i=1, 2, \cdots)$.因此原方程的内部解、外部解都可依次解出, 从而得到原方程的渐近解.

定理  在[H₁]-[H₄]下, 在 $0\leq t\leq T$上, 问题(1.1)-(1.4) 有唯一解

$ x(t)=X(t)+\varepsilon U(\tau)=\sum\limits_{i=0}^{\infty} X_{i}(t)\varepsilon^{i}+\varepsilon \sum\limits_{i=0}^{\infty}u_{i}(\tau)\varepsilon^{i}, \\y(t)=Y(t)+\varepsilon V(\tau)=\sum\limits_{i=0}^{\infty} Y_{i}(t)\varepsilon^{i}+\varepsilon \sum\limits_{i=0}^{\infty}v_{i}(\tau)\varepsilon^{i}, $

其中 $\tau=\frac{t}{\varepsilon}$.