0 引言

关于时滞微分方程的边值问题已受到广泛而深入的讨论^[1-5], 但主要都集中在连续情形下的奇异摄动时滞问题^[6-10], 对一些含不连续系数或不连续源项的奇摄动时滞问题研究相对较少.近几十年来, 在物理、化学、生物以及环境学等许多领域的数学模型都可归结为带有不连续系数或不连续源项的奇摄动时滞问题, 并备受研究者重视, 但研究主要集中在数值方法方面^[11-12].2015年, Subburayan^[12]研究了一类具有不连续对流系数的二阶线性奇摄动时滞问题，并结合文献[13]给出了分段线性插值Shishkin网格的数值方法.

本文主要研究下述含不连续源项的时滞微分方程的奇摄动边值问题

$ {\rm{ \mathsf{ ε} }}x'' + f\left( {t,x\left( t \right)} \right)x'\left( t \right) = g\left( {t,x\left( t \right),x\left( {t - \tau } \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right), $

(1)

$ x\left( t \right) = \varphi \left( t \right),t \in \left[ { - \tau ,0} \right];x\left( 1 \right) = A. $

(2)

其中

$ \begin{array}{l} f\left( {t,x\left( t \right)} \right) = \left\{ \begin{array}{l} {f_1}\left( {t,x\left( t \right)} \right),t \in \left( {0,\tau } \right),\\ {f_2}\left( {t,x\left( t \right)} \right),t \in \left( {\tau ,1} \right), \end{array} \right.{f_1}\left( {\tau ,x\left( \tau \right)} \right) \ne {f_2}\left( {\tau ,x\left( \tau \right)} \right),\\ g\left( {t,x\left( t \right),x\left( {t - \tau } \right)} \right) = \left\{ \begin{array}{l} {g_1}\left( {t,x\left( t \right),\varphi \left( {t - \tau } \right)} \right),t \in \left( {0,\tau } \right),\\ {g_2}\left( {t,x\left( t \right),x\left( {t - \tau } \right)} \right),t \in \left( {\tau ,1} \right). \end{array} \right. \end{array} $

1 上下解引理

考虑如下非线性边值问题

$ x''\left( t \right) = f\left( {t,x\left( t \right),x\left( {t - \tau } \right),x'\left( t \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right), $

(3)

$ x\left( t \right) = \varphi \left( t \right),t \in \left[ { - \tau ,0} \right];x\left( 1 \right) = A. $

(4)

其中

$ f\left( {t,x\left( t \right),x\left( {t - \tau } \right),x'\left( t \right)} \right) = \left\{ \begin{array}{l} {f_1}\left( {t,x\left( t \right),\varphi \left( {t - \tau } \right),x'\left( t \right)} \right),t \in \left( {0,\tau } \right),\\ {f_2}\left( {t,x\left( t \right),x\left( {t - \tau } \right),x'\left( t \right)} \right),t \in \left( {\tau ,1} \right), \end{array} \right.{f_1}\left( \tau \right) \ne {f_2}\left( \tau \right). $

引理1  假设(H₁)f(t, x, y, z) 于[0, τ)∪(τ, 1]×R³上连续, 且对∀r > 0, 存在[0, +∞) 上的正值连续函数h(s) 满足

$ \int_0^\infty {\frac{s}{{1 + h\left( s \right)}}{\rm{d}}s} = \infty , $

且当t∈[0, 1], |x|≤r, |y|≤r, |z| > ∞时有

$ \left| {f\left( {t,x,y,z} \right)} \right| \le h\left( {\left| z \right|} \right). $

(H₂)f(t, x, y, z) 关于y单调不增.

(H₃) 存在函数α(t), β(t)∈C[-τ, 1]∩C²[0, τ)∪(τ, 1], 满足

$ \begin{array}{*{20}{c}} {a\left( t \right) \le \beta \left( t \right),t \in \left[ { - \tau ,1} \right];\alpha '\left( {{\tau ^ - }} \right) \le \alpha '\left( {{\tau ^ + }} \right),\beta '\left( {{\tau ^ - }} \right) \ge \beta '\left( {{\tau ^ + }} \right),}\\ {\alpha ''\left( t \right) \ge f\left( {t,\alpha \left( t \right),x\left( {t - \tau } \right),\alpha '\left( t \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right),}\\ {\beta ''\left( t \right) \le f\left( {t,\beta \left( t \right),x\left( {t - \tau } \right),\beta '\left( t \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right)} \end{array} $

则∀φ(t)∈C[-τ, 0], 当α(t)≤φ(t)≤β(t), t∈[-τ, 0]，α(1)≤A≤β(1) 时, 边值问题(3)~(4) 有解x(t)∈C¹[-τ, 1]∩C²([0, τ)∪(τ, 1]), 满足

$ \alpha \left( t \right) \le x\left( t \right) \le \beta \left( t \right),t \in \left[ {0,1} \right]. $

(5)

证明  取r=max{$\mathop {\max }\limits_{t \in \left[ { - \tau ,1} \right]} $|α(t)|, $\mathop {\max }\limits_{t \in \left[ { - \tau ,1} \right]} $|β(t)|}, 根据(H₁), 存在相应的正值连续函数h(s) 和正数M₀使得$\int_{2r}^{{M_0}} {\frac{s}{{1 + h(s)}}{\rm{d}}s > 2r} $.再令M₁=max{M₀, $\mathop {\max }\limits_{t \in \left[ {0,1} \right]} $|α(t)|, $\mathop {\max }\limits_{t \in \left[ {0,1} \right]} $|β(t)|}, 参考文献[14], 可定义函数f(t, x, y, z) 为

$ \bar f\left( {t,x,y,z} \right) = \left\{ \begin{array}{l} f\left( {t,x,{y_{\beta \left( \tau \right)}},z} \right),\;\;\;\;y > \beta \left( t \right),\\ f\left( {t,x,y,z} \right),\;\;\;\;\;\;\;\;\alpha \left( t \right) \le y \ge \beta \left( t \right),\\ f\left( {t,x,{y_{\alpha \left( \tau \right)}},z} \right),\;\;\;\;y < \alpha \left( t \right). \end{array} \right. $

其中

$ {y_{\beta \left( \tau \right)}} = \beta \left( {t - \tau } \right),{y_{\alpha \left( \tau \right)}} = \alpha \left( {t - \tau } \right). $

又令

$ \begin{array}{*{20}{c}} {F\left( {t,x,y,z} \right) = \left\{ \begin{array}{l} \bar f\left( {t,{x_\beta },y,z} \right) + \frac{{x - \beta \left( t \right)}}{{1 + x - \beta \left( t \right)}},\;\;\;\;\;\;x > \beta \left( t \right),\\ \bar f\left( {t,x,y,z} \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\alpha \left( t \right) \le y \ge \beta \left( t \right),\\ \bar f\left( {t,{x_\alpha },y,z} \right) - \frac{{\alpha \left( t \right) - x}}{{1 + \alpha \left( t \right) - x}},\;\;\;\;\;\;x < \alpha \left( t \right). \end{array} \right.}\\ {\tilde f\left( {t,x,y,z} \right) = \left\{ \begin{array}{l} F\left( {t,x,y,{z_M}} \right),\;\;\;\;\;\;z > M,\\ F\left( {t,x,y,z} \right),\;\;\;\;\;\;\;\;\left| z \right| \le M,\\ F\left( {t,x,y,{z_{ - M}}} \right),\;\;\;\;z < - M. \end{array} \right.} \end{array} $

其中x_β=β(t), x_α=α(t); z_M=M, z_-M=-M.则${\tilde f}$(t, x, y, z) 在[0, τ)∪(τ, 1]×R³上连续有界.当t∈[0, 1], |x|≤r, |y|≤r, |z|≤M时, |f(t, x, y, z)|≤h(|z|), 并且在集合E={(t, x, y, z)|0≤t≤1, α(t)≤x≤β(t), α(t)≤y≤β(t), |z|≤M|}上有f(t, x, y, z)=${\tilde f}$(t, x, y, z) 成立.参考文献[15]可知, 边值问题

$ \left\{ \begin{array}{l} x''\left( t \right) = \tilde f\left( {t,x\left( t \right),x\left( {t - \tau } \right),x'\left( t \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right),\\ x\left( t \right) = \varphi \left( t \right),t \in \left[ { - \tau ,0} \right],\\ x\left( 1 \right) = A \end{array} \right. $

有解.

下面先证明解x(t) 满足不等式(5).以下只证明x(t)≤β(t)(因为α(t)≤x(t) 可类似处理).用反证法, 设存在一点t₀∈[0, 1]使得x(t₀) > β(t₀), 则由于x(0)=φ(0)≤β(0), x(1)=A≤β(1), 函数x(t)-β(t) 必在(0, 1) 内的某点ξ处取正的极大值, 从而x(ξ) > β(ξ), x′(ξ)=β′(ξ), x″(ξ)≤β″(ξ).但依据(H₂) 和(H₃), 当ξ∈(0, τ) 时，有

$ \begin{array}{l} x''\left( \xi \right) - \beta ''\left( \xi \right) \ge {{\tilde f}_1}\left( {\xi ,x\left( \xi \right),\varphi \left( {\xi - \tau } \right),x'\left( \xi \right)} \right) - {f_1}\left( {\xi ,\beta \left( \xi \right),\varphi \left( {\xi - \tau } \right),\beta '\left( \xi \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\tilde f}_1}\left( {\xi ,\beta \left( \xi \right),\varphi \left( {\xi - \tau } \right),\beta '\left( \xi \right)} \right) + \frac{{x\left( \xi \right) - \beta \left( \xi \right)}}{{1 + x\left( \xi \right) - \beta \left( \xi \right)}} - {f_1}\left( {\xi ,\beta \left( \xi \right),\varphi \left( {\xi - \tau } \right),\beta '\left( \xi \right)} \right) > 0, \end{array} $

这与x″(ξ)-β″(ξ)≤0矛盾.

同理当ξ∈(τ, 1) 时也推出矛盾.

当ξ=τ时, x(t)-β(t) 在τ处取得正的极大值, 从而有

$ \begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\tilde f}_1}\left( {\xi ,\beta \left( \xi \right),\varphi \left( {\xi - \tau } \right),\beta '\left( \xi \right)} \right) + \frac{{x\left( \xi \right) - \beta \left( \xi \right)}}{{1 + x\left( \xi \right) - \beta \left( \xi \right)}} - {f_1}\left( {\xi ,\beta \left( \xi \right),\varphi \left( {\xi - \tau } \right),\beta '\left( \xi \right)} \right) > 0,\\ \begin{array}{*{20}{c}} {x\left( {{\tau ^ \pm }} \right) \ge \beta \left( {{\tau ^ \pm }} \right),x'\left( {{\tau ^ \pm }} \right) = \beta '\left( {{\tau ^ \pm }} \right),x''\left( {{\tau ^ \pm }} \right) \le \beta ''\left( {{\tau ^ \pm }} \right),}\\ \begin{array}{l} x''\left( {{\tau ^ - }} \right) - \beta ''\left( {{\tau ^ - }} \right) \ge {{\tilde f}_1}\left( {{\tau ^ - },x\left( {{\tau ^ - }} \right),\varphi \left( 0 \right),x'\left( {{\tau ^ - }} \right)} \right) - {f_1}\left( {{\tau ^ - },\beta \left( {{\tau ^ - }} \right),\beta \left( 0 \right),\beta '\left( {{\tau ^ - }} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\bar f}_1}\left( {{\tau ^ - },\beta \left( {{\tau ^ - }} \right),\varphi \left( 0 \right),\beta '\left( {{\tau ^ - }} \right)} \right) + \frac{{x\left( {{\tau ^ - }} \right) - \beta \left( {{\tau ^ - }} \right)}}{{1 + x\left( {{\tau ^ - }} \right) - \beta \left( {{\tau ^ - }} \right)}} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{f_1}\left( {{\tau ^ - },\beta \left( {{\tau ^ - }} \right),\beta \left( 0 \right),\beta '\left( {{\tau ^ - }} \right)} \right) > 0, \end{array} \end{array} \end{array} $

类似可得x″(τ⁺)-β″(τ⁺) > 0, 这与x″(τ^±)≤β″(τ^±) 矛盾.

再证|x′(t)|≤M(0≤t≤1), 若此不等式不成立, 即存在一点t₁∈[0, 1], 使得|x′(t₁)| > M.由中值定理, 必存在ξ∈(0, 1) 使得x′(ξ)=x(1)-x(0), 故|x′(ξ)|≤2r.又从x′(t) 的连续性知, 存在τ₁, τ₂∈[0, 1], 使得

$ \left| {x'\left( {{\tau _1}} \right)} \right| = 2r,\left| {x'\left( {{\tau _2}} \right)} \right| = M, $

且当τ₁ < t < τ₂(或τ₂ < t < τ₂) 时, 有

$ \left| {\int_{2r}^M {\frac{s}{{1 + h\left( s \right)}}{\rm{d}}s} } \right| = \left| {\int_{{\tau _1}}^{{\tau _2}} {\frac{{x''\left( t \right)x'\left( t \right)}}{{1 + h\left( {\left| {x'\left( t \right)} \right|} \right)}}} } \right| \le 2r, $

这与M的取法矛盾, 故|x′(t)|≤M(0≤t≤1).即有|x(t)|≤M(0≤t≤1), 从而x(t) 就是原问题的解.

2 渐近解的构造

由于原问题在t=τ处间断, 所以会产生内部层, 因此可以把问题(1)~(2) 看成是以下两个问题的光滑连接.

左问题P_L:

$ \varepsilon x''\left( t \right) + {f_1}\left( {t,x\left( t \right)} \right)x'\left( t \right) = {g_1}\left( {t,x\left( t \right),\varphi \left( {t - \tau } \right)} \right),t \in \left( {0,\tau } \right), $

(6)

$ x\left( 0 \right) = \varphi \left( 0 \right),{x_L}\left( \tau \right) = \gamma \left( \varepsilon \right). $

(7)

右问题P_R:

$ \varepsilon x''\left( t \right) + {f_2}\left( {t,x\left( t \right)} \right)x'\left( t \right) = {g_2}\left( {t,x\left( t \right),x\left( {t - \tau } \right)} \right),t \in \left( {\tau ,1} \right), $

(8)

$ {x_R}\left( \tau \right) = \gamma \left( \varepsilon \right),x\left( 1 \right) = A. $

(9)

其中γ(ε) 是与ε有关的待定参数.

现作如下假设:

(ⅰ) 函数f₁(t, x), g₁(t, x, y) 在[0, τ]×R³上连续, f₂(t, x), g₂(t, x, y) 在[τ, 1]×R³上连续, 且g_i(i=1, 2) 关于y单调不增;

(ⅱ) 初值问题f₁(t, x(t))x′(t)=g₁(t, x(t), φ(t-τ)), x(0)=φ(0) 有解φ₁(t)∈[0, τ];

(ⅲ) 初值问题f₂(t, x(t))x′(t)=g₂(t, x(t), x(t-τ)), x(1)=A有解φ₂(t)∈C²[0, 1];

(ⅳ) f₁(t, x)≤-σ₁ < 0, (t, x)∈[0, τ]×R; f₂(t, x)≥σ₂>0, (t, x)∈[τ, 1]×R.

首先考虑左问题的渐近解构造, 假设左问题P_L的形式渐近解表达式为

$ {x_L}\left( t \right) = {{\bar x}_L}\left( t \right) + {V_L}\left( \eta \right),\eta = \frac{{t - \tau }}{\varepsilon }, $

(10)

其中

$ {{\bar x}_L}\left( t \right) = {{\bar x}_{0L}}\left( t \right) + \varepsilon {{\bar x}_{1L}}\left( t \right) + {\varepsilon ^2}{{\bar x}_{2L}}\left( t \right) + \cdots , $

(11)

$ {V_L}\left( \eta \right) = {V_{0L}}\left( \eta \right) + \varepsilon {V_{1L}}\left( \eta \right) + {\varepsilon ^2}{V_{2L}}\left( \eta \right) + \cdots , $

(12)

$ \gamma \left( \varepsilon \right) = {\gamma _0} + \varepsilon {\gamma _1} + {\varepsilon ^2}{\gamma _2} + \cdots , $

(13)

将式(10) 代入式(6) 并分离变量，再根据ε同次幂系数相等, 可得到一系列递推等式.为了简单起见, 以下只考虑零阶近似.

由上可得正则项x_0L(t) 满足

$ {f_1}\left( {t,{{\bar x}_{0L}}\left( t \right)} \right){{\bar x'}_{0L}}\left( t \right) = {g_1}\left( {t,{{\bar x}_{0L}}\left( t \right),\varphi \left( {t - \tau } \right)} \right), $

(14)

且由(ⅱ) 可知x_0L(t)=φ₁(t), t∈[0, τ].

同时可得边界层项V_0L(η) 满足如下边值问题

$ \frac{{{{\rm{d}}^2}{V_{0L}}\left( \eta \right)}}{{{\rm{d}}{\eta ^2}}} + {f_1}\left( {\tau ,{\varphi _1}\left( \tau \right) + {V_{0L}}\left( \eta \right)} \right)\frac{{{\rm{d}}{V_{0L}}\left( \eta \right)}}{{{\rm{d}}\eta }} = 0, $

(15)

$ {V_{0L}}\left( 0 \right) = {\gamma _0} - {\varphi _1}\left( \tau \right),{V_{0L}}\left( { - \frac{\tau }{\varepsilon }} \right) = 0. $

(16)

由文献[16]中引理1以及条件(ⅰ), (ⅱ), (ⅳ) 可知, 边值问题(15)~(16) 存在解V_0L(η) 且满足指数估计, 即存在正常数K₁, 使得

$ \left| {{V_{0L}}\left( \eta \right)} \right| \le {K_1}\exp \left( {{\sigma _1}\eta } \right). $

因此, 可得左问题的零阶近似渐近解形式为

$ {x_L}\left( t \right) = {\varphi _1}\left( t \right) + {V_{0L}}\left( \eta \right),t \in \left[ {0,\tau } \right]. $

(17)

同左问题, 可得右问题的零阶近似渐近解形式为

$ {x_R}\left( t \right) = {\varphi _2}\left( t \right) + {V_{0R}}\left( \eta \right),\;\;\;t \in \left[ {\tau ,1} \right]. $

(18)

其中正则项x_0R(t) 满足

$ {f_2}\left( {t,{{\bar x}_{0R}}\left( t \right)} \right){{\bar x'}_{0R}}\left( t \right) = {g_2}\left( {t,{{\bar x}_{0R}}\left( t \right),{{\bar x}_{0R}}\left( {t - \tau } \right)} \right). $

(19)

且由(ⅲ) 知x_0R(t)=φ₂(t), t∈[τ, 1].

边界层项V_0R(η) 满足边值问题

$ \frac{{{\rm{d}}{V_{0R}}\left( \eta \right)}}{{{\rm{d}}{\eta ^2}}} + {f_2}\left( {\tau ,{\varphi _2}\left( \tau \right) + {V_{0R}}\left( \eta \right)} \right)\frac{{{\rm{d}}{V_{0R}}\left( \eta \right)}}{{{\rm{d}}\eta }} = 0, $

(20)

$ {V_{0R}}\left( 0 \right) = {\gamma _0} - {\varphi _2}\left( \tau \right),\;\;\;{V_{0R}}\left( {\frac{{1 - \tau }}{\varepsilon }} \right) = 0. $

(21)

由文献[16]中引理2以及条件(ⅰ), (ⅲ), (ⅳ) 可知, 边值问题(20)~(21) 存在解V_0R(η) 且满足指数估计, 即存在正常数K₂, 使得

$ \left| {{V_{0R}}\left( \eta \right)} \right| \le {K_2}\exp \left( { - {\sigma _2}\eta } \right). $

为了使左问题的解与右问题的解在t=τ处光滑连接, 须有

$ \frac{{{\rm{d}}{x_L}}}{{{\rm{d}}t}}\left( \eta \right) = \frac{{{\rm{d}}{x_R}}}{{{\rm{d}}t}}\left( \eta \right). $

(22)

(17), (18) 代入代(22) 得

$ \begin{array}{*{20}{c}} {\frac{{{\rm{d}}{V_{0L}}}}{{{\rm{d}}\eta }}\left| {_{\eta = 0}} \right. = \frac{{{\rm{d}}{V_{0R}}}}{{{\rm{d}}\eta }}\left| {_{\eta = 0}} \right.,}\\ {\frac{{{\rm{d}}{x_{0L}}}}{{{\rm{d}}t}}\left| {_{t = \tau }} \right. + \frac{{{\rm{d}}{V_{1L}}}}{{{\rm{d}}\eta }}\left| {_{\eta = 0}} \right. = \frac{{{\rm{d}}{x_{0R}}}}{{{\rm{d}}t}}\left| {_{t = \tau }} \right. + \frac{{{\rm{d}}{V_{1R}}}}{{{\rm{d}}\eta }}\left| {_{\eta = 0}} \right..} \end{array} $

(23)

令$\omega = \frac{{{\rm{d}}{V_{0L}}}}{{{\rm{d}}\eta }}$, 根据条件(16), 对式(15) 积分可得

$ \frac{{{\rm{d}}{V_{0L}}}}{{{\rm{d}}\eta }} = \omega = \frac{{{\gamma _0} - {\varphi _1}\left( \tau \right)}}{{\int_{ - \infty }^0 {\exp \left( { - \int_0^\eta {{f_1}\left( {\tau ,{\varphi _1}\left( \tau \right) + {V_{0L}}\left( s \right)} \right){\rm{d}}s} } \right){\rm{d}}\eta } }}\exp \left( { - \int_0^\eta {{f_1}\left( {\tau ,{\varphi _1}\left( \tau \right) + {V_{0L}}\left( s \right)} \right){\rm{d}}s} } \right). $

(24)

同理, 由式(20) 和(21) 可推出

$ \frac{{{\rm{d}}{V_{0R}}}}{{{\rm{d}}\eta }} = \frac{{{\gamma _0} - {\varphi _2}\left( \tau \right)}}{{\int_{ + \infty }^0 {\exp \left( { - \int_0^\eta {{f_2}\left( {\tau ,{\varphi _2}\left( \tau \right) + {V_{0R}}\left( s \right)} \right){\rm{d}}s} } \right){\rm{d}}\eta } }}\exp \left( { - \int_0^\eta {{f_2}\left( {\tau ,{\varphi _2}\left( \tau \right) + {V_{0R}}\left( s \right)} \right){\rm{d}}s} } \right). $

(25)

将式(24), (25) 代入式(23) 得

$ \frac{{{\gamma _0} - {\varphi _1}\left( \tau \right)}}{{{\gamma _0} - {\varphi _2}\left( \tau \right)}} = \frac{{\int_{ - \infty }^0 {\exp \left( { - \int_0^\eta {{f_1}\left( {\tau ,{\varphi _1}\left( \tau \right) + {V_{0L}}\left( s \right)} \right){\rm{d}}s} } \right){\rm{d}}\eta } }}{{\int_{ + \infty }^0 {\exp \left( { - \int_0^\eta {{f_2}\left( {\tau ,{\varphi _2}\left( \tau \right) + {V_{0R}}\left( s \right)} \right){\rm{d}}s} } \right){\rm{d}}\eta } }}. $

(26)

易知式(26) 的右边是负值, 则γ₀可以由式(26) 决定.同理γ_i(i≥1) 也可由此递推得到.

因此, 可得问题(1)~(2) 的形式渐近解.

定理1  假设条件(ⅰ)~(ⅳ) 成立, 则对于充分小的ε > 0, 边值问题(1)~(3) 有一个C¹光滑解x(t) 满足

$ x\left( t \right) = {{\bar x}_0}\left( t \right) + {V_0}\left( {\frac{{t - \tau }}{\varepsilon }} \right) + O\left( \varepsilon \right), $

其中

$ {{\bar x}_0}\left( t \right) = \left\{ \begin{array}{l} {\varphi _1}\left( t \right),\;\;\;\;\;t \in \left[ {0,\tau } \right),\\ {\gamma _0},\;\;\;\;\;\;\;\;\;\;t = \tau ,\\ {\varphi _2}\left( t \right),\;\;\;\;\;t \in \left( {\tau ,1} \right], \end{array} \right.\;\;{V_0}\left( {\frac{{t - \tau }}{\varepsilon }} \right) = \left\{ \begin{array}{l} {V_{0L}}\left( {\frac{{t - \tau }}{\varepsilon }} \right),\;\;\;t \in \left[ {0,\tau } \right),\\ 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t = \tau ,\\ {V_{0R}}\left( {\frac{{t - \tau }}{\varepsilon }} \right),\;\;\;t \in \left( {\tau ,1} \right]. \end{array} \right. $

3 定理的证明

现证明定理1, 首先构造合适的上下解.记

$ \tilde x\left( t \right) = \left\{ \begin{array}{l} {{\tilde x}_L}\left( t \right) = {\varphi _1}\left( t \right) + {V_{0L}}\left( \eta \right) + \varepsilon {{\bar x}_{1L}}\left( t \right) + \varepsilon {V_{1L}}\left( \eta \right) + {\varepsilon ^2}{V_{2L}}\left( \eta \right),t \in \left[ {0,\tau } \right),\\ {{\tilde x}_R}\left( t \right) = {\varphi _2}\left( t \right) + {V_{0R}}\left( \eta \right) + \varepsilon {{\bar x}_{1R}}\left( t \right) + \varepsilon {V_{1R}}\left( \eta \right) + {\varepsilon ^2}{V_{2R}}\left( \eta \right),t \in \left( {\tau ,1} \right]. \end{array} \right. $

根据近似解的构造过程和假设(ⅰ) 知, 存在正常数M使得

$ \left| {\varepsilon x''\left( t \right) + f\left( {t,x\left( t \right)x'\left( t \right) - g\left( {t,x\left( t \right),x\left( {t - \tau } \right)} \right)} \right)} \right| \le M{\varepsilon ^2},t \in \left[ {0,1} \right], $

(27)

$ \begin{array}{l} \left| {\frac{{\partial f}}{{\partial x}}\left( {t,x} \right)\bar x' - \frac{{\partial g}}{{\partial x}}\left( {t,x,\left[ x \right]} \right) - \frac{{\partial g}}{{\partial \left[ x \right]}}\left( {t,x,\left[ x \right]} \right)} \right| \le K{\varepsilon ^2},\\ \left( {t,x,\left[ x \right]} \right) \in \left[ {0,1} \right] \times {\left[ { - \vartheta - K{\varepsilon ^2},\vartheta + K{\varepsilon ^2}} \right]^2}, \end{array} $

(28)

其中x=x(t), [x]=x(t-τ), ϑ=${\rm{max}}\{ |\tilde x\left( t \right)|,|\tilde x\left( {t - \tau } \right)|,0 \le t \le 1\} $.此外,

$ {{\tilde x}_L}\left( \tau \right) = {{\tilde x}_R}\left( \tau \right) = {\gamma _0} + \varepsilon {\gamma _1},{{\tilde x'}_L}\left( \tau \right) = {{\tilde x'}_R}\left( \tau \right). $

(29)

令

$ \alpha \left( t \right) = \left\{ \begin{array}{l} {{\tilde x}_L}\left( t \right) - {\lambda _L}\left( t \right){\varepsilon ^2},\;\;\;\;\;t \in \left[ {0,\tau } \right),\\ {\gamma _0} + \varepsilon {\gamma _1} - \mu {\varepsilon ^2},\;\;\;\;\;\;\;t = \tau ,\\ {{\tilde x}_R}\left( t \right) - {\lambda _R}\left( t \right){\varepsilon ^2},\;\;\;\;\;t \in \left( {\tau ,1} \right], \end{array} \right.\;\;\;\beta \left( t \right) = \left\{ \begin{array}{l} {{\tilde x}_L}\left( t \right) + {\lambda _L}\left( t \right){\varepsilon ^2},\;\;\;\;\;t \in \left[ {0,\tau } \right),\\ {\gamma _0} + \varepsilon {\gamma _1} + \mu {\varepsilon ^2},\;\;\;\;\;\;\;t = \tau ,\\ {{\tilde x}_R}\left( t \right) + {\lambda _R}\left( t \right){\varepsilon ^2},\;\;\;\;\;t \in \left( {\tau ,1} \right], \end{array} \right. $

其中

$ \begin{array}{l} \lambda \left( t \right) = \left\{ \begin{array}{l} {\lambda _L}\left( t \right) = \frac{{M + 1}}{K}\left( {\left( {L + 1} \right)\exp \left( {\frac{{k\left( {t - \tau } \right)}}{{{\sigma _1}}}} \right) - 1} \right),\;\;\;\;t \in \left[ {0,\tau } \right],\\ {\lambda _R}\left( t \right) = \frac{{M + 1}}{K}\left( {\left( {L + 1} \right)\exp \left( {\frac{{k\left( {\tau - t} \right)}}{{{\sigma _2}}}} \right) - 1} \right),\;\;\;\;t \in \left( {\tau ,1} \right]. \end{array} \right.\\ \;\;\;\;\;\;L = \max \left\{ {\exp \left( {\frac{{k\tau }}{{{\sigma _1}}}} \right),\exp \left( {\frac{{k\left( {1 - \tau } \right)}}{{{\sigma _2}}}} \right)} \right\},\mu = \frac{{\left( {M + 1} \right)L}}{K}. \end{array} $

容易验证, λ(t) 是一个正值连续函数且具有如下性质

$ {{\lambda '}_L}\left( t \right) > 0,\;\;\;t \in \left[ {0,\tau } \right];\;\;\;{{\lambda '}_R}\left( t \right) < 0,\;\;\;t \in \left[ {\tau ,1} \right], $

(30)

$ {\sigma _1}{{\lambda '}_L}\left( t \right) - K{\lambda _L}\left( t \right) = M + 1 = - {\sigma _2}{{\lambda '}_R}\left( t \right) - K{\lambda _R}\left( t \right). $

(31)

再依据式(29)~(30), 有

$ \begin{array}{*{20}{c}} {\alpha \left( t \right),\beta \left( t \right) \in C\left( {\left[ {0,1} \right]} \right),\alpha \left( t \right) < \beta \left( t \right),t \in \left[ {0,1} \right],}\\ {\alpha '\left( {{\tau ^ - }} \right) \le \alpha '\left( {{\tau ^ + }} \right),\beta '\left( {{\tau ^ - }} \right) \ge \beta '\left( {{\tau ^ + }} \right),\alpha \left( 0 \right) < \varphi \left( 0 \right) < \beta \left( 0 \right),\alpha \left( 1 \right) < A < \beta \left( 1 \right).} \end{array} $

下面验证不等式εα″(t)+f(t, α(t))α′(t)≥g(t, α(t), α(t-τ)).仅考虑区间(τ, 1) 上, 在(0, τ) 上可类似证明.根据式(27), (28) 以及(30), (31) 可得

$ \begin{array}{l} \varepsilon \alpha ''\left( t \right) + f\left( {t,\alpha \left( t \right)} \right)\alpha '\left( t \right) - g\left( {t,\alpha \left( t \right),\alpha \left( {t - \tau } \right)} \right) = \\ \varepsilon {{\tilde x''}_R}\left( t \right) - {\varepsilon ^3}{{\lambda ''}_R}\left( t \right) + {f_2}\left( {t,{{\tilde x}_R}\left( t \right) - {\varepsilon ^2}{\lambda _R}\left( t \right)} \right)\left( {{{\tilde x'}_R}\left( t \right) - {\varepsilon ^2}{{\lambda '}_R}\left( t \right)} \right) - \\ {g_2}\left( {t,{{\tilde x}_R}\left( t \right) - {\varepsilon ^2}{\lambda _R}\left( t \right),{{\tilde x}_R}\left( {t - \tau } \right) - {\varepsilon ^2}{\lambda _R}\left( {t - \tau } \right)} \right) = \\ \varepsilon {{\tilde x''}_R} + {f_2}\left( {t,{{\tilde x}_R}} \right){{\tilde x'}_R} - {g_2}\left( {t,{{\tilde x}_R},\left[ {{{\tilde x}_R}} \right]} \right) - \left( {\frac{{\partial {f_2}}}{{\partial x}}\left( {t,{{\tilde x}_R} - {\theta _1}{\varepsilon ^2}{\lambda _R}} \right){{\tilde x'}_R}} \right){\lambda _R}{\varepsilon ^2} + \\ \left( {\frac{{\partial {g_2}}}{{\partial x}}\left( {t,{{\tilde x}_R} - {\theta _2}{\varepsilon ^2}{\lambda _R},\left[ {{{\tilde x}_R}} \right] - {\varepsilon ^2}\left[ {{\lambda _R}} \right]} \right)} \right){\lambda _R}{\varepsilon ^2} + \\ \left( {\frac{{\partial {g_2}}}{{\partial \left[ x \right]}}\left( {t,{{\tilde x}_R},\left[ {{{\tilde x}_R}} \right] - {\theta _3}{\varepsilon ^2}\left[ {{\lambda _R}} \right]} \right)} \right){\lambda _R}{\varepsilon ^2} - {f_2}\left( {t,{{\tilde x}_R} - {\varepsilon ^2}{\lambda _R}} \right){{\lambda '}_R}{\varepsilon ^2} - {\varepsilon ^3}{{\lambda ''}_R} \ge \\ - M{\varepsilon ^2} - K{\lambda _R}{\varepsilon ^2} - {\sigma _2}{{\lambda '}_R}{\varepsilon ^2} - {{\lambda '''}_R}{\varepsilon ^3} = \\ \left( {1 - {{\lambda '''}_R}\varepsilon } \right){\varepsilon ^2} > 0. \end{array} $

其中0 < θ₁, θ₂, θ₃ < 1.

类似可得对充分小的ε > 0, 不等式

$ \varepsilon \beta ''\left( t \right) + f\left( {t,\beta \left( t \right)} \right)\beta '\left( t \right) \le g\left( {t,\beta \left( t \right),\beta \left( {t - \tau } \right)} \right),t \in \left( {0,\tau } \right) \cup \left( {\tau ,1} \right) $

成立.

以上证明了α(t), β(t) 分别是问题(1)~(3) 的下解和上解.由上下解引理可知, 问题(1)~(2) 存在解

$ x\left( t \right) \in {C^1}\left[ {0,1} \right] \cap {C^2}\left( {\left( {0,t} \right) \cup \left( {\tau ,1} \right)} \right), $

且∀t∈[0, 1]有α(t)≤x(t)≤β(t), 证毕.

4 应用举例

考虑如下问题

$ \left\{ \begin{array}{l} - \varepsilon u''\left( x \right) + 3u'\left( x \right) - u\left( {x - 1} \right) = 0,x \in \left( {0,1} \right),\\ - \varepsilon u''\left( x \right) + 4u'\left( x \right) - 2u\left( {x - 1} \right) = 0,x \in \left( {1,2} \right),\\ u\left( x \right) = 1,x \in \left[ { - 1,0} \right],u\left( 2 \right) = 2 \end{array} \right. $

(32)

将问题(32) 看成是以下两个问题的光滑连接.

左问题：

$ \left\{ \begin{array}{l} - \varepsilon u''\left( x \right) + 3u'\left( x \right) - 1 = 0,x \in \left( {0,1} \right),\\ u\left( 0 \right) = 1,u\left( 1 \right) = \gamma \left( \varepsilon \right). \end{array} \right. $

(33)

右问题:

$ \left\{ \begin{array}{l} - \varepsilon u''\left( x \right) + 4u'\left( x \right) - 2u\left( {x - 1} \right) = 0,x \in \left( {1,2} \right),\\ u\left( 1 \right) = \gamma \left( \varepsilon \right),u\left( 2 \right) = 2. \end{array} \right. $

(34)

其中γ(ε) 是与ε有关的待定参数.可令

$ \gamma \left( \varepsilon \right) = {\gamma _0} + \varepsilon {\gamma _1} + {\varepsilon ^2}{\gamma _2} + \cdots , $

根据前面的构造方法, 可得左右问题的退化解分别为φ(x)∈[0, 1], ψ(x)∈[0, 2], 且

$ \varphi \left( x \right) = \frac{1}{3}x + 1,\psi \left( x \right) = - \frac{1}{{12}}{x^2} - \frac{1}{2}x + \frac{{10}}{3}. $

再由式(15)~(16) 以及(20)~(21) 可求出左右问题边界层函数的零阶近似V_0L(η), V_0R(η) 分别为

$ {V_{0L}}\left( \eta \right) = \left( {{\gamma _0} - \frac{4}{3}} \right)\exp \left( {3\eta } \right),{V_{0R}}\left( \eta \right) = \left( {{\gamma _0} - \frac{{11}}{4}} \right)\exp \left( { - 4\eta } \right), $

再根据式(26) 可求出γ₀=$\frac{{15}}{7}$.

从而问题(32) 的零阶近似解可表示为

$ u\left( x \right) = \left\{ \begin{array}{l} \frac{1}{3}x + 1 + \frac{{17}}{{21}}\exp \left( {3\eta } \right) + O\left( \varepsilon \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \left[ {0,1} \right),\\ \frac{{15}}{7} + O\left( \varepsilon \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 1,\\ - \frac{1}{{12}}{x^2} - \frac{1}{2}x + \frac{{10}}{3} - \frac{{17}}{{28}}\exp \left( { - 4\eta } \right) + O\left( \varepsilon \right),\;\;\;\;\;\;\;x \in \left( {1,2} \right] \end{array} \right. $