奇异摄动理论和方法是处理带有小参数的非线性系统的重要工具, 它最初起源于对天体力学的研究. 它处理相关问题一般具有以下方法, 如边界层函数法、匹配法、平均化法、多重尺度法等, 针对不同的问题可以选择不同的方法^[1-4]. 众所周知, 奇异摄动初边值问题中的临界情况一直都是非常复杂, 其原因在于零次近似的正则部分和边界层部分的求解都是十分困难的. 因此, 时至今日仍有很多奇异摄动中的临界问题有待解决. 不过目前, 一些临界问题得到解决, 如临界情况下的拟线性不同初边值问题^[5-7]、时滞边值问题^[8]等.

然而, 对于奇异摄动的临界情况下的弱非线性问题却很少, 此时, 如果加上可移动边界中的一种积分边界^[9], 则使得问题变得更有意义且更加复杂. 受此启发, 本文研究一类临界情况下的弱非线性带有积分初边值问题. 在研究中发现, Cakir和Amiraliyev考虑了带有积分边界条件的奇异摄动边值问题^[10]

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

文献[11]中, 作者研究了一类具有积分边值问题的二阶奇异摄动问题, 证明了解的存在性, 并构造了渐近解. 文献[12]中, 作者研究了一类具有积分边界条件的奇异摄动边值问题, 并利用奇异摄动几何理论证明了解的存在唯一性, 构造了渐近解.

本文将运用边界层函数法研究一类带有积分初边值条件的弱非线性临界情况下的奇异摄动问题, 不但构造了一致有效的形式渐近解, 而且证明了解的存在性、唯一性, 并给出一个例子来验证了我们的结果.

考虑以下弱非线性临界情况带有积分初边值问题

(2.1) $ \begin{equation} \mu \frac{{{\rm d}z}}{{{\rm d}t}} = f(\mu z, y, t)z + \mu g(z, y, t), \;\;t \in [a, b], \end{equation} $

(2.2) $ \begin{equation} \mu \frac{{{\rm d}y}}{{{\rm d}t}} = z, \end{equation} $

(2.3) $ \begin{equation} y(a) = {y^0} + \int_a^b {{q_1}} \left( {y\left( {s, \mu } \right)} \right){\rm d}s, \;\;\;\;\;y(b) = {y^1} + \int_a^b {{q_2}} \left( {y\left( {s, \mu } \right)} \right){\rm d}s. \end{equation} $

其中$ 0 < \mu \ll 1 $是一小参数.

由于积分初边值条件的引入, 使得原本就很复杂的临界问题(2.1)–(2.2)变得更加复杂. 因此, 为了降低复杂程度, 将方程组(2.1)–(2.2)转化为如下等价奇异初边值问题

(2.4) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {\mu \frac{{{\rm d}z}}{{{\rm d}t}} = f(\mu z, y, t)z + \mu g(z, y, t), \;}\\ {\mu \frac{{{\rm d}y}}{{{\rm d}t}} = z, }\\ { \frac{{{\rm d}{k_1}}}{{{\rm d}t}} = {q_1}(y), \;\;\;\;\;\; \frac{{{\rm d}{k_2}}}{{{\rm d}t}} = {q_2}(y).} \end{array}} \right. \end{eqnarray} $

(2.5) $ \begin{equation} y(a)={y^0}-{k_1}(a), \;\;\;{k_1}(b)=0, \;\;\;{k_2}(a)=0, \;\;\;y(b)={y^1}+{k_2}(b). \end{equation} $

对以上复杂临界问题(2.1)–(2.2)作如下假设.

[H$ _1 $] 函数$ f(\mu z, y, t), g(z, y, t) $和$ {q_i}(y), i = 1, 2 $在区域$ D $上充分光滑, 其中$ A, B $是一正常数, $ D = \left\{ {(z, y, t)\left| {\left| z \right| \le A, \left| y \right| \le B, a \le t \le b} \right.} \right\} $.

令$ \mu {\rm{ = }}0 $可得复杂临界问题(2.1)–(2.2) 的退化方程

(2.6) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {f\left( {0, \bar y, t} \right)\bar z = 0}, \\ {\bar z = 0}, \end{array}} \right. \end{eqnarray} $

得到退化方程的解

(2.7) $ \begin{equation} \left\{ {\begin{array}{*{20}{l}} {\bar y = \alpha (t)}, \\ {\bar z = 0}. \end{array}} \right. \end{equation} $

复杂临界问题(2.1)–(2.2)方程右端等于零, 求平衡点

(2.8) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {f(\mu z, y, t)z + \mu g(z, y, t) = 0, }\\ {z = 0.} \end{array}} \right. \end{eqnarray} $

得到方程的解

(2.9) $ \begin{equation} \left\{ {\begin{array}{*{20}{l}} {y = {y^*}, }\\ {z = 0.} \end{array}} \right. \end{equation} $

使方程(2.1), (2.2)右端在点$ (0, {y^*}) $泰勒展开, 可得系数矩阵为

$ A = \left( {\begin{array}{*{20}{c}} {f(0, {y^*}, t)} &0\\ 1 &0 \end{array}} \right). $

其中$ A $的特征值为$ \lambda_{1}=0, \lambda_{2}=f(0, {y^*}, t) $, 并且满足如下条件.

[H$ _2 $] 特征值满足$ {\lambda _2} = f(0, {y^*}, t) < 0, $且$ f\left( {0, \alpha (t), t} \right) \ne 0, t \in \left[ {a, b} \right]. $

因此, 方程(2.1)–(2.2)具有临界稳定情形.

使用边界层函数法构造渐近解, 系统(2.4)–(2.5)渐近级数$ \big( {x = {{(z, y, {k_1}, {k_2})}^{\rm T}}} \big) $

(2.10) $ \begin{equation} x(t, \mu ) = \bar x(t, \mu ) + Lx({\tau _0}, \mu ), \;\;{\tau _0} = \frac{{t - a}}{\mu }\; \ge 0. \end{equation} $

正则级数$ \bar x(t, \mu ) $有如下形式

(2.11) $ \begin{equation} \bar x(t, \mu ) = {\bar x_0}(t) + \mu {\bar x_1}(t) + \cdots + {\mu ^k}{\bar x_k}(t) + \cdots . \end{equation} $

边界层级数$ Lx({\tau _0}, \mu ) $有如下形式

(2.12) $ \begin{equation} Lx({\tau _0}, \mu ) = {L_0}x({\tau _0}) + \mu {L_1}x({\tau _0}) + \cdots + {\mu ^k}{L_k}x({\tau _0}) + \cdots . \end{equation} $

利用边界层函数法可得正则部分的零次方程为

(2.13) $ \left\{ \begin{array}{l} f(0, {{\bar y}_0}, t){{\bar z}_0} = 0, \\ {{\bar z}_0} = 0, \\ \frac{{{\rm d}{{\bar k}_{10}}(t)}}{{{\rm d}t}} = {q_1}({{\bar y}_0}), \\ \frac{{{\rm d}{{\bar k}_{20}}(t)}}{{{\rm d}t}} = {q_2}({{\bar y}_0}).\end{array} \right.$

通过(2.13)式可求得

(2.14) $ \begin{equation} {\bar z_0}(t) = 0, \;\;\;\;{\bar y_0}(t) = \alpha (t), \end{equation} $

其中$ \alpha (t) $未知.

求解边界层部分的零次主项的方程为

(2.15) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{L_0}z({\tau _0})}}{{{\rm d}{\tau _0}}} = f(0, \alpha (a) + {L_0}y({\tau _0}), a){L_0}z({\tau _0}), }\\ { \frac{{{\rm d}{L_0}y({\tau _0})}}{{{\rm d}{\tau _0}}} = {L_0}z({\tau _0}), }\\ { \frac{{{\rm d}{L_0}{k_1}({\tau _0})}}{{{\rm d}{\tau _0}}} = 0, }\\ { \frac{{{\rm d}{L_0}{k_2}({\tau _0})}}{{{\rm d}{\tau _0}}} = 0.} \end{array}} \right. \end{eqnarray} $

相对应的零次初边值条件为

(2.16) $ \begin{eqnarray} \begin{array}{l} \alpha (a) + {L_0}y(0) = {y^0} - {{\bar k}_{10}}(a) - {L_0}{k_1}(0), \;\;\;\;\;\;\;\;{{\bar k}_{10}}(b) = 0;\\ {{\bar k}_{20}}(a) + {L_0}{k_2}(0) = 0, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\bar y}_0}(b) = {y^1} + {{\bar k}_{20}}(b);\\ {L_0}x( + \infty ) = 0. \end{array} \end{eqnarray} $

由(2.15), (2.16)式求得: $ {L_0}{k_1}({\tau _0}) = {L_0}{k_2}({\tau _0}) = 0 $, 从而可求得$ {\bar k_{20}}(a) = 0. $

求解正则部分一次主项的方程为

(2.17) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar z}_0}}}{{{\rm d}t}}{\rm{ = }}\bar f {{\bar z}_1} + \bar g, }\\ { \frac{{{\rm d}{{\bar y}_0}}}{{{\rm d}t}} = {{\bar z}_1}, }\\ { \frac{{{\rm d}{{\bar k}_{11}}(t)}}{{{\rm d}t}} = {q_{1y}}(\alpha (t)){{\bar y}_1}, }\\ { \frac{{{\rm d}{{\bar k}_{21}}(t)}}{{{\rm d}t}} = {q_{2y}}(\alpha (t)){{\bar y}_1}.} \end{array}} \right. \end{eqnarray} $

其中符号"$ - $"表示在$ (0, \alpha (t), t) $点取值.

联立(2.13), (2.16), (2.17)式可得$ {\bar k_{10}}(t), {\bar k_{20}}(t) $的表达式

(2.18) $ \begin{equation} {\bar k_{10}}(t) = \int_{\alpha (b)}^{\alpha (t)} { - \frac{{{q_1}(\eta )f(0, \eta , t)}}{{g(0, \eta , t)}}} {\rm d}\eta . \end{equation} $

(2.19) $ \begin{equation} {\bar k_{20}}(t) = \int_{\alpha (a)}^{\alpha (t)} { - \frac{{{q_2}(\sigma )f(0, \sigma , t)}}{{g(0, \sigma , t)}}} {\rm d}\sigma . \end{equation} $

其中$ \alpha (a), \alpha (b) $未知.

考虑(2.16), (2.17)式联立得到的如下边值问题

(2.20) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{d\alpha (t)}}{{{\rm d}t}} = - \frac{{g(0, \alpha (t), t)}}{{f(0, \alpha (t), t)}}, }\\ {\alpha (b) = {y^1} + {{\bar k}_{20}}(b).} \end{array}} \right. \end{eqnarray} $

[H$ _3 $] 假设方程(2.20)有解$ \alpha (t) = {\omega _0}. $

(2.21) $ \begin{equation} \alpha (t) = {\omega _0}(t, {\bar k_{20}}(b)). \end{equation} $

联立(2.18), (2.19), (2.21)式可求得

(2.22) $ \begin{equation} {\bar k_{20}}(b) = \gamma _0^*, \;\;\;\;\alpha (a) = \beta _0^*, \;\;\;\;{\bar k_{10}}(a) = \xi _0^*.\; \end{equation} $

以上$ \gamma _0^*, \beta _0^*, \xi _0^* $为已知常数, 即$ \alpha (t) = {\omega _0}(t, \gamma _0^{\rm{*}}), {\bar z_1}(t) = {\omega '_0}(t, \gamma _0^{\rm{*}}) $已知. 则$ \alpha (b) $也已知, 即可求得$ {\bar k_{10}}(t), {\bar k_{20}}(t). $

通过(2.15)式求得

(2.23) $ \begin{eqnarray} \tilde z({\tau _0}) = \int_{\beta _0^*}^{\tilde y} {f(0, s, a){\rm d}s} \end{eqnarray} $

其中$ \tilde z({\tau _0}) = {L_0}z({\tau _0}), \tilde y = \alpha (a) + {L_0}y({\tau _0}). $

(2.24) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}\tilde y}}{{{\rm d}{\tau _0}}} = \int_{\beta _0^*}^{\tilde y} {f(0, s, a){\rm d}s, } }\\ {\tilde y(0) = \delta _0^*.} \end{array}} \right. \end{eqnarray} $

其中$ \delta _0^*= {y^0} - \xi _0^* $已知常数. 因此, 通过(2.24)式求得$ {L_0}y({\tau _0}) $, 再由$ {L_0}y'({\tau _0}){\rm{ = }}{L_0}z({\tau _0}) $, $ {L_0}z({\tau _0}) $已知.

零次主项$ {\bar y_0}(t), {\bar z_0}(t), {\bar k_{10}}(t), {\bar k_{20}}(t), {L_0}y({\tau _0}), {L_0}z({\tau _0}), {L_0}{k_1}({\tau _0}), {L_0}{k_2}({\tau _0}) $全部求出.

求解边界层级数一次主项方程为

(2.25) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{L_1}z({\tau _0})}}{{{\rm d}{\tau _0}}} = \tilde f{L_1}z({\tau _0}) + {{\tilde f}_y}{L_0}z({\tau _0}){L_1}y({\tau _0}) + {{\tilde f}_y}{L_0}z({\tau _0}){{\bar y}_1(a)} + {h_1}, }\\ { \frac{{{\rm d}{L_1}y({\tau _0})}}{{{\rm d}{\tau _0}}} = {L_1}z({\tau _0}), }\\ { \frac{{{\rm d}{L_1}{k_1}({\tau _0})}}{{{\rm d}{\tau _0}}} = {q_1}(\tilde y) - {q_1}(\alpha (a)), }\\ { \frac{{{\rm d}{L_1}{k_2}({\tau _0})}}{{{\rm d}{\tau _0}}} = {q_2}(\tilde y) - {q_2}(\alpha (a)).} \end{array}} \right. \end{eqnarray} $

其中$ {h_1}({\tau _0}) = \big( {{{\tilde f}_z} + {{\tilde f}_y}\alpha '{\tau _0} + {{\tilde f}_t}{\tau _0}} \big){L_0}z({\tau _0}) - \overline{\overline{ f}}{\bar z_1} + \hat g - \overline{\overline{ g}}. $符号"$ \sim $"表示在$ (0, \alpha (a) + {L_0}y({\tau _0}), a) $点取值, "$ = $"表示在$ (0, \alpha (a), a) $取值, 而则"$ \wedge $"表示在$ ({L_0}z({\tau _0}), \alpha (a) + {L_0}y({\tau _0}), a) $取值.

相对应的一次初边值条件为

(2.26) $ \begin{eqnarray} \begin{array}{l} {{\bar y}_1}(a) + {L_1}y(0) = - {{\bar k}_{11}}(a) - {L_1}{k_1}(0), \;\;\;\;\;\;\;\;{{\bar k}_{11}}(b) = 0;\\ {{\bar k}_{21}}(a) = - {L_2}{k_2}(0), \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\bar y}_1}(b) = {{\bar k}_{21}}(b);\\ {L_1}x( + \infty ) = 0. \end{array} \end{eqnarray} $

通过(2.25), (2.26)式可求得

(2.27) $L_{1} k_{i}\left(\tau_{0}\right)=\int_{+\infty}^{\tau_{0}}\left[q_{i}\left(\alpha(a)+L_{0} y(s)\right)-q_{i}(\alpha(a))\right] \mathrm{d} s . \quad i=1,2. $

因此, $ {\bar k_{21}}(a) = - {L_1}{k_2}(0) $已知.

求解正则部分的二次项的方程为

(2.28) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar z}_1}}}{{{\rm d}t}}\left( t \right) = \bar f {{\bar z}_2} + \left( {{{\bar f}_y}{{\bar z}_1} + {{\bar g}_y}} \right){{\bar y}_1} + {{\bar g}_z} {{\bar z}_1}, \;\;\;}\\ { \frac{{{\rm d}{{\bar y}_1}}}{{{\rm d}t}}\left( t \right) = {{\bar z}_2}, }\\ { \frac{{{\rm d}{{\bar k}_{12}}(t)}}{{{\rm d}t}} = {q_{1y}}(\alpha (t)){{\bar y}_2} + {S_{12}}, }\\ { \frac{{{\rm d}{{\bar k}_{22}}(t)}}{{{\rm d}t}} = {q_{2y}}(\alpha (t)){{\bar y}_2} + {S_{22}}.} \end{array}} \right. \end{eqnarray} $

其中$ {S_{i2}} = {q_{iy}}(\alpha (t))\bar y_1^2, \;i = 1, 2. $

通过(2.28)式变换可得

(2.29) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar y}_1}}}{{{\rm d}t}} + {\theta _1}{{\bar y}_1} = {\phi _1}, }\\ {{{\bar y}_1}(b) = {{\bar k}_{21}}(b).} \end{array}} \right. \end{eqnarray} $

其中$ {\theta _1}(t) = \frac{{\left( {{{\bar f}_y} \cdot {{\bar z}_1} + {{\bar g}_y}} \right)}}{{\bar f}}, {\phi _1}(t) = \frac{{\frac{{{\rm d}{{\bar z}_n}}}{{{\rm d}t}} - {{\bar g}_z}{{\bar z}_1}}}{{\bar f}}. $

(2.30) $ \begin{equation} {\bar y_1}(t) = {e^{\int_b^t { - {\theta _1}(\lambda ){\rm d}\lambda } }}\left( {\int_b^t {{\phi _1}(s){e^{\int_b^s {{\theta _1}(p){\rm d}p} }}{\rm d}s + {{\bar k}_{21}}\left( b \right)} } \right). \end{equation} $

其中$ {\bar k_{21}}(b) $未知, 为了便于计算方便, 可将(2.30)式右端表示成$ {\bar y_1}\left( t \right) = {\omega _1}\left( {t, {{\bar k}_{21}}\left( b \right)} \right). $

由(2.17), (2.26), (2.28)式可得

(2.31) $ \begin{equation} {\bar k_{21}}(t) = \int_{{\omega _1}\left( {a, {{\bar k}_{21}}(b)} \right)}^{{{\bar y}_1}(t)} {\bar f\frac{{{q_{2y}}(\alpha (t))s}}{{{{\bar z'}_1} - {{\bar g}_z} \cdot {{\bar z}_1} - \left( {{{\bar f}_y}{{\bar z}_1} + {{\bar g}_y}} \right)s}}}{\rm d}s + {\bar k_{21}}(a). \end{equation} $

(2.32) $ \begin{equation} {\bar k_{11}}(t) = \frac{{{q_{1y}}\left( {\alpha (t)} \right)}}{{{q_{2y}}\left( {\alpha (t)} \right)}}\left( {{{\bar k}_{21}}(t) - {{\bar k}_{21}}(b)} \right). \end{equation} $

联立(2.31), (2.32)式求得

(2.33) $ \begin{equation} {\bar k_{21}}(b) = \gamma _1^*, \;\;\;\;{\bar y_1}(a) = \beta _1^*, \;\;\;\;{\bar k_{11}}(a) = \xi _1^*.\; \end{equation} $

以上$ \gamma _1^*, \beta _1^*, \xi _1^* $为已知常数. 联立上述方程可求得$ {\bar k_{11}}(t), {\bar k_{21}}(t), {\bar y_1}(t). $从而能求得$ {L_1}y(0) = \delta _1^*. $

为了求得$ {L_1}y({\tau _0}) $, 将(2.25)式前两个方程写成二阶方程

(2.34) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{d^2}{L_1}y({\tau _0})}}{{{\rm d}\tau _0^2}} = \tilde f \frac{{{\rm d}{L_1}y({\tau _0})}}{{{\rm d}{\tau _0}}} + {\varphi _1}{L_1}y({\tau _0}) + {\psi _1}.}\\ {{L_1}y(0) = \delta _1^*, \;\;\;\;\;\;{L_1}y( + \infty ) = 0.} \end{array}} \right. \end{eqnarray} $

$ {\varphi _1} = {\tilde f_y}{L_0}z({\tau _0}), \;\;\;\;\;\;{\psi _1}{\rm{ = }}{\tilde f_y}{L_0}z({\tau _0})\beta _1^* + {h_1}. $

由于方程(2.34)是非自共轭的, 不能直接套用公式. 但是, 可把(2.34)式化成自共轭方程后就能用公式

(2.35) $ \begin{equation} {L_1}y({\tau _0}) = \delta _1^*\frac{{\tilde z({\tau _0})}}{{\tilde z(0)}} + \tilde z({\tau _0}) \int_0^{{\tau _0}} {\frac{{{\rm d}\eta }}{{{{\tilde z}^2}(\eta )p(\eta )}}} \int_\infty ^\eta {\tilde z(\sigma )p(\sigma )} {\psi _1}(\sigma ){\rm d}\sigma . \end{equation} $

这里$ p = {e^{ - \int_0^{{\tau _0}} {{\varphi _1}(s){\rm d}s} }} $是自共轭因子.

一次主项$ {\bar y_1}(t), {\bar z_1}(t), {\bar k_{11}}(t), {\bar k_{21}}(t), {L_1}y({\tau _0}), {L_1}z({\tau _0}), {L_1}{k_1}({\tau _0}), {L_1}{k_2}({\tau _0}) $, 全部求出.

至此, 已确定了形式渐近解的全部零次及一次主项, 接下来, 给出确定渐近解$ n $次项的方程和条件.

用归纳法可以证明渐近级数(2.10)中的以后各项$ {\bar x_n}(t), {L_n}x({\tau _0})(n \ge 2) $都能惟一确定. 不妨认为我们已经求出了前$ n-1 $项, 所以在求第$ n $项的方程式中$ {F_n}(t), {S_{i, n}}(t) $$ (i = 1, 2), $$ {h_n}({\tau _0}), {Q_{l, n - 1}}({\tau _0}) $$ (l = 1, 2) $都是已知函数.

求解$ n $次正则项$ {\bar y_n}, {\bar z_n}, {\bar k_{1n}}, {\bar k_{2n}}, \;\;n \ge 2 $的方程为

(2.36) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar z}_{n - 1}}}}{{{\rm d}t}} = \bar f \cdot {{\bar z}_n} + \left( {{{\bar f}_y} \cdot {{\bar z}_1} + {{\bar g}_y}} \right){{\bar y}_{n - 1}} + {F_{n - 1}}, }\\ { \frac{{{\rm d}{{\bar y}_{n - 1}}}}{{{\rm d}t}} = {{\bar z}_n}, }\\ { \frac{{{\rm d}{{\bar k}_{1n}}}}{{{\rm d}t}} = {q_{1y}}\left( {\alpha (t)} \right){{\bar y}_n} + {S_{1, n}}, }\\ { \frac{{{\rm d}{{\bar k}_{2n}}}}{{{\rm d}t}} = {q_{2y}}\left( {\alpha (t)} \right){{\bar y}_n} + {S_{2, n}}.} \end{array}} \right. \end{eqnarray} $

$ {F_{n - 1}} $是关于$ {\bar z_i}(t), {\bar y_j}(t)\left( {i \le n \!-\! 1, j \le n \!-\! 2} \right) $的复合函数, $ {S_{m, n}}\left( {m = 1, 2} \right) $是关于$ {\bar y_l}\left( {l \le n \!-\! 1} \right) $的复合函数.

对应的$ n $次初边值条件为

(2.37) $ \begin{eqnarray} \begin{array}{l} {{\bar y}_n}(a) + {L_n}y(0) = - {{\bar k}_{1n}}(a) - {L_1}{k_n}(0), \;\;\;\;\;\;\;\;{{\bar k}_{1n}}(b) = 0;\\ {{\bar k}_{2n}}(a) = - {L_n}{k_2}(0), \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\bar y}_n}(b) = {{\bar k}_{2n}}(b);\\ {L_n}x( + \infty ) = 0. \end{array} \end{eqnarray} $

求解$ n $次边界层项$ {L_n}y\left( {{\tau _0}} \right), {L_n}z\left( {{\tau _0}} \right), {L_n}{k_1}\left( {{\tau _0}} \right), {L_n}{k_2}\left( {{\tau _0}} \right), \;\;n \ge 2 $的方程为

(2.38) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{L_n}z({\tau _0})}}{{{\rm d}{\tau _0}}} = \tilde f{L_n}z({\tau _0}) + {{\tilde f}_y}{L_0}z({\tau _0}){L_n}y({\tau _0}) + {{\tilde f}_y}{L_0}z({\tau _0}){{\bar y}_n(a)} + {h_n}, }\\ { \frac{{{\rm d}{L_n}y({\tau _0})}}{{{\rm d}{\tau _0}}} = {L_n}z({\tau _0}), }\\ { \frac{{{\rm d}{L_n}{k_1}({\tau _0})}}{{{\rm d}{\tau _0}}} = {q_{1y}}(\tilde y){L_{n - 1}}y({\tau _0}) + {Q_{1, n - 1}}, }\\ { \frac{{{\rm d}{L_n}{k_2}({\tau _0})}}{{{\rm d}{\tau _0}}} = {q_{2y}}(\tilde y){L_{n - 1}}y({\tau _0}) + {Q_{2, n - 1}}.} \end{array}} \right. \end{eqnarray} $

其中$ {h_n} $是关于$ {L_i}y({\tau _0}), {L_j}z({\tau _0}), {\bar z_k}(a), {\bar y_k}(a), {\tau _0}, \left( {i, j, k \le n - 1} \right) $等所组成的已知复合函数, $ {Q_{1, n - 1}}, {Q_{2, n - 1}} $是关于$ {L_s}y({\tau _0}), {\tau _0}, \left( {s \le n - 2} \right) $已知复合函数.

由(2.37), (2.38)式可求得

(2.39) $ \begin{equation} L_{n} k_{i}\left(\tau_{0}\right)=\int_{+\infty}^{\tau_{0}}\left[q_{1 y}(\tilde{y}) L_{n-1} y(s)+Q_{1, n-1}(s)\right] \mathrm{d} s, \quad i=1,2. \end{equation} $

因此, $ {\bar k_{2n}}(a) = - {L_n}{k_2}(0) $已知.

求解$ n+1 $次正则项的方程为

(2.40) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar z}_n}}}{{{\rm d}t}} = \bar f {{\bar z}_{n + 1}} + \left( {{{\bar f}_y} \cdot {{\bar z}_1} + {{\bar g}_y}} \right){{\bar y}_n} + {F_n}, }\\ { \frac{{{\rm d}{{\bar y}_n}}}{{{\rm d}t}} = {{\bar z}_{n + 1}}, }\\ { \frac{{{\rm d}{{\bar k}_{1, n + 1}}}}{{{\rm d}t}} = {q_{1y}}\left( {\alpha (t)} \right){{\bar y}_{n + 1}} + {S_{1, n + 1}}, }\\ { \frac{{{\rm d}{{\bar k}_{2, n + 1}}}}{{{\rm d}t}} = {q_{2y}}\left( {\alpha (t)} \right){{\bar y}_{n + 1}} + {S_{2, n + 1}}.} \end{array}} \right. \end{eqnarray} $

其中$ {F_n} $是关于$ {\bar z_i}(t), {\bar y_j}(t)\left( {i \le n, j \le n - 1} \right) $的复合函数, $ {S_{m, n + 1}}\left( {m = 1, 2} \right) $是关于$ {\bar y_l}\left( {l \le n} \right) $的复合函数.

此时, 由于归纳法假设, 我们前$ n-1 $项已求出, 故$ {\bar y_{n - 1}}(t) $已知. 由(2.36)式可知$ {\bar y'_{n - 1}}(t) = {\bar z_n}(t) $, 则$ {\bar z_n}(t) $求出.

通过(2.40)式变换可得

(2.41) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar y}_n}}}{{{\rm d}t}} + {\theta _n}{{\bar y}_n} = {\phi _n}, }\\ {{{\bar y}_n}(b) = {{\bar k}_{2n}}(b).} \end{array}} \right. \end{eqnarray} $

其中$ {\theta _n}(t) = \frac{{\left( {{{\bar f}_y} \cdot {{\bar z}_1} + {{\bar g}_y}} \right)}}{{\bar f}}, {\phi _n}(t) = \frac{{\frac{{{\rm d}{{\bar z}_n}}}{{{\rm d}t}} - {F_n}}}{{\bar f}}. $

(2.42) $ \begin{equation} {\bar y_n}(t) = {e^{\int_b^t { - {\theta _n}(\lambda ){\rm d}\lambda } }}\left( {\int_b^t {{\phi _n}(s) {e^{\int_b^s {{\theta _n}(p){\rm d}p} }}{\rm d}s + {{\bar k}_{2n}}\left( b \right)} } \right). \end{equation} $

其中$ {\bar k_{2n}}(b) $未知, 为了计算方便, 令上式$ {\bar y_n}(t) = {\omega _n}\left( {t, {{\bar k}_{2n}}(b)} \right). $

由(2.36), (2.37), (2.40)式可得

(2.43) $ \begin{equation} {\bar k_{2n}}(t) = \int_{{\omega _n}\left( {a, {{\bar k}_{2n}}(b)} \right)}^{{{\bar y}_n}(t)} {\bar f\frac{{{q_{2y}}\left( {\alpha (t)} \right)s + {S_{2, n}}}}{{{{\bar z'}_n} - {F_n} - \left( {{{\bar f}_y}{{\bar z}_1} + {{\bar g}_y}} \right)s}}}{\rm d}s + {\bar k_{2n}}(a). \end{equation} $

(2.44) $ \begin{equation} {\bar k_{1n}}(t) = \frac{{{q_{1y}}\left( {\alpha (t)} \right){{\bar y}_n} + {S_{1, n}}}}{{{q_{2y}}\left( {\alpha (t)} \right){{\bar y}_n} + {S_{2, n}}}}\left( {{{\bar k}_{2n}}(t) - {{\bar k}_{2n}}(b)} \right). \end{equation} $

联立(2.43), (2.44)式得

(2.45) $ \begin{equation} {\bar k_{2n}}(b) = \gamma _n^*, \;\;\;\;{\bar y_n}(a) = \beta _n^*, \;\;\;\;{\bar k_{1n}}(a) = \xi _n^*.\; \end{equation} $

以上$ \gamma _n^*, \beta _n^*, \xi _n^* $为已知常数. 联立上述方程可求得$ {\bar k_{1n}}(t), {\bar k_{2n}}(t), {\bar y_n}(t). $从而能求得$ {L_n}y(0) = \delta _n^*. $

为了求得$ {L_n}y({\tau _0}) $, 将(2.38)式前两个方程写成二阶方程

(2.46) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{d^2}{L_n}y({\tau _0})}}{{{\rm d}\tau _0^2}} = \tilde f \frac{{{\rm d}{L_n}y({\tau _0})}}{{{\rm d}{\tau _0}}} + {\varphi _n}{L_n}y({\tau _0}) + {\psi _n}.}\\ {{L_n}y(0) = \delta _n^*, \;\;\;\;\;\;{L_n}y( + \infty ) = 0.} \end{array}} \right. \end{eqnarray} $

$ {\varphi _n} = {\tilde f_y}{L_1}z({\tau _0}), \;\;\;\;\;\;{\psi _n}{\rm{ = }}{\tilde f_y}{L_{n - 1}}z({\tau _0}) \beta _n^*+ {h_n}. $

由于方程(2.46)是非自共轭的, 不能直接套用公式. 但是, 可把(2.45)式化成自共轭方程后就能用公式

(2.47) $ \begin{eqnarray} {L_n}y({\tau _0}) = \delta _n^*\frac{{\tilde z({\tau _0})}}{{\tilde z(0)}} + \tilde z({\tau _0})\int_0^{{\tau _0}} {\frac{{{\rm d}\eta }}{{{{\tilde z}^2}(\eta )r(\eta )}}} \int_\infty ^\eta {\tilde z(\sigma )r(\sigma )} {\psi _n}(\sigma ){\rm d}\sigma . \end{eqnarray} $

这里$ r = {e^{ - \int_0^{{\tau _0}} {{\varphi _n}(s){\rm d}s} }} $是自共轭因子.

由$ \frac{{{\rm d}{L_n}y({\tau _0})}}{{{\rm d}{\tau _0}}} = {L_n}z({\tau _0}), {L_n}z({\tau _0}) $可求得.

因此, $ n $次近似解已全部求出, 并且可知系统(2.1), (2.2)的边界层厚度为$ O\left( \mu \right) $. 至此, 渐近展开式(2.10)可完全确定, 用$ {X_n}\left( {t, \mu } \right) $表示(2.10)前$ n+1 $项阶部分和

(2.48) $ \begin{equation} {X_n}(t, \mu ) = \sum\limits_{k = 0}^n {{\mu ^k}} \left( {{{\bar x}_k}(t) + {L_k}x({\tau _0})} \right). \end{equation} $

定理3.1  当满足条件[H$ _1 $]–[H$ _3 $]时, 必存在常数$ {\mu _0} > 0 $和$ c > 0 $, 使得当$ \mu \in (0, {\mu _0}] $时, 问题(2.4)–(2.5)的解$ x(t, \mu ) $在$ a \le t \le b $上存在唯一且满足不等式

(3.1) $ \begin{equation} \left\| {x(t, \mu ) - {X_n}(t, \mu )} \right\| \le C{\mu ^{n + 1}}, \;\;\;\;t \in [a, b]. \end{equation} $

证  令$ \zeta = z - {Z_{n + 1}}, \eta = y - {Y_{n + 1}}, \omega = {k_1} - {K_{1, n + 1}}, \xi = {k_2} - {K_{2, n + 1}} $, 其中$ (z\;y\;{k_1}\;{k_2}) $是(2.4)–(2.5)式的解, 而$ ({Z_{n + 1}}\;{Y_{n + 1}}\;{K_{1, n + 1}}\;{K_{2, n + 1}}) $由(3.2)式确定

(3.2) $ \begin{equation} {X_n}(t, \mu ) = \sum\limits_{k = 0}^n {{\mu ^k}} [{\bar x_k}(t) + {L_k}x({\tau _0})]. \end{equation} $

因此, 我们得到下面方程组

(3.3) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {\mu \frac{{{\rm d}\zeta }}{{{\rm d}t}} = {f_z}(0, \alpha (t), t)\zeta + {f_y}(0, \alpha (t), t)\eta + {G_1}(\zeta , \eta , t, \mu ), }\\ {\mu \frac{{{\rm d}\eta }}{{{\rm d}t}} = \zeta + {G_2}(t, \mu ), }\\ { \frac{{{\rm d}\omega }}{{{\rm d}t}} = {q_{1y}}(\alpha (t))\eta + {G_3}(\eta , t, \mu ), }\\ { \frac{{{\rm d}\xi }}{{{\rm d}t}} = {q_{2y}}(\alpha (t))\eta + {G_4}(\eta , t, \mu ), }\\ {\zeta \left( {0, \mu } \right) = 0, \eta \left( {0, \mu } \right) = 0, \omega \left( {0, \mu } \right) = 0, \xi \left( {0, \mu } \right) = 0.} \end{array}} \right. \end{eqnarray} $

而

$ \begin{eqnarray*} &&{G_1}(\zeta , \eta , t, \mu ) = f(\mu (\zeta + {Z_{n + 1}}), \eta + {Y_{n + 1}}, t)(\zeta + {Z_{n + 1}}) + \mu g(\zeta + {Z_{n + 1}}, \eta + {Y_{n + 1}}, t)\\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ - \mu \frac{{{\rm d}{Z_{n + 1}}}}{{{\rm d}t}} - {f_z}(0, \alpha (t), t)\zeta - {f_y}(0, \alpha (t), t)\eta , \\ &&{G_2}(t, \mu ) = {Z_{n + 1}} - \mu \frac{{{\rm d}{Y_{n + 1}}}}{{{\rm d}t}}, \\ &&{G_3}(\eta , t, \mu ) = {q_1}(\eta + {Y_{n + 1}}) - \frac{{{\rm d}{K_{1, n + 1}}}}{{{\rm d}t}} - {q_{1y}}(\alpha (t))\eta , \\ &&{G_4}(\eta , t, \mu ) = {q_2}(\eta + {Y_{n + 1}}) - \frac{{{\rm d}{K_{2, n + 1}}}}{{{\rm d}t}} - {q_{2y}}(\alpha (t))\eta . \end{eqnarray*} $

$ {G_1}(\zeta , \eta , t, \mu ), {G_i}(\eta , t, \mu ) $$ (i = 3, 4) $, 有如下两个重要性质

(1) $ {G_1}(0, 0, t, \mu ) = O({\mu ^{n + 2}}), {G_i}(0, t, \mu ) = O({\mu ^{n + 2}}); $

(2) $ \forall \mu > 0, \exists \delta (\mu ), $使得当$ \left| {{\zeta _1}} \right| < {c_1}\delta , \left| {{\zeta _2}} \right| < {c_2}\delta , \left| {{\eta _1}} \right| < {c_3}\delta , \left| {{\eta _2}} \right| < {c_4}\delta $时有

$ \left| {{G_1}({\zeta _1}, {\eta _1}, t, \mu ) - {G_1}({\zeta _2}, {\eta _2}, t, \mu )} \right| \le \mu \left| {{\zeta _1} - {\zeta _2}} \right| + \mu \left| {{\eta _1} - {\eta _2}} \right|. $

$ \left| {{G_i}({\eta _1}, t, \mu ) - {G_i}({\eta _2}, t, \mu )} \right| \le \mu \left| {{\eta _1} - {\eta _2}} \right|. $

为了更加方便地分析, 对$ {G_1} $进行一些转换

$ \begin{eqnarray*} &&f(\mu (\zeta + {Z_{n + 1}}), \eta + {Y_{n + 1}}, t) = f(\mu {Z_{n + 1}}, {Y_{n + 1}}, t) +\mu {f_z}(\mu {Z_{n + 1}}, {Y_{n + 1}}, t)\zeta \\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \;\;\, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \mu {f_y}(\mu {Z_{n + 1}}, {Y_{n + 1}}, t)\eta + {h_1}(\zeta , \eta , t, \mu ), \\ &&g(\zeta + {Z_{n + 1}}, \eta + {Y_{n + 1}}, t) = f({Z_{n + 1}}, {Y_{n + 1}}, t) + {f_z}({Z_{n + 1}}, {Y_{n + 1}}, t)\zeta + {f_y}({Z_{n + 1}}, {Y_{n + 1}}, t)\eta \\ &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + {h_2}(\zeta , \eta , t, \mu ). \end{eqnarray*} $

$ {h_i}(\zeta , \eta , t, \mu ), $$ i = 1, 2 $是$ \zeta , \eta $压缩系数为$ O(\mu ) $阶的压缩算子, 并且$ {h_i}(0, 0, t, \mu ) = 0. $

$ {G_1}(\zeta , \eta , t, \mu ) = \mu {a_1}(t, \mu )\zeta + \mu {b_1}(t, \mu )\eta + {c_1}(t, \mu )\zeta \eta + {h_1}(\zeta , \eta , t, \mu ){d_1}(t, \mu ) + {Q_1}(\zeta , \eta , t, \mu ), $

其中$ {a_1}, {b_1}, {c_1} $为有界函数, $ {Q_1}(\zeta , \eta , t, \mu ) $是$ \zeta , \eta $压缩系数为$ O(\mu ) $阶的压缩算子, 并且$ {Q_1}(0, 0, $$ t, \mu ) = O({\mu ^{n + 2}}). $

由(3.3)式可得

(3.4) $ \begin{equation} \eta (t, \mu ) = \frac{1}{\mu }\int_0^t {\left( {\zeta (s, \mu ) + {G_2}(s, \mu )} \right){\rm d}s.} \end{equation} $

把它代入(3.3)式, 有

(3.5) $ \begin{equation} \mu \frac{{{\rm d}\zeta }}{{{\rm d}t}} = {f_z}(0, \alpha (t), t)\zeta + {f_y}(0, \alpha (t), t){\mu ^{ - 1}}\int_0^t {\left( {\zeta (s, \mu ) + {G_2}(s, \mu )} \right){\rm d}s} + {G_1}(\zeta , \eta , t, \mu ). \end{equation} $

把(3.5)式写成等价的积分方程

(3.6) $ \begin{eqnarray} \zeta (t, \mu ) &=& \int_0^t {\left( {{f_y}(0, \alpha (s), s){\mu ^{ - 2}}\int_0^s {\left( {\zeta (p, \mu ) + {G_2}(p, \mu )} \right){\rm d}p} } \right)}{\rm d}s \\ &&+ \int_0^t {{\mu ^{ - 1}}\left( {{G_1}(\zeta , \eta , s, \mu ) + {f_z}(0, \alpha (s), s)\zeta (s, \mu )} \right)}{\rm d}s = J\left( {\zeta , \eta , t, \mu } \right). \end{eqnarray} $

不难看出算子$ J $具有与$ {G_1} $相同性质.

由于(3.3)式中的后两个关于$ \omega , \xi $的方程右端只和$ \eta $有关, 并且关于$ t $连续, 而(3.3)式中的前两个方程均不含有$ \omega , \xi $, 故只需证明$ \eta $存在且唯一, $ \omega $和$ \xi $也一定存在且唯一.

最后对(3.6)式采用逐次逼近法, 可证明方程(3.3)解的存在和估计$ \left| {\zeta (t, \mu )} \right| \le C{\mu ^{n + 1}}, $$ \left| {\eta (t, \mu )} \right| \le C{\mu ^{n + 1}}, \left| {\omega (t, \mu )} \right| \le C{\mu ^{n + 1}}, \left| {\xi (t, \mu )} \right| \le C{\mu ^{n + 1}}. $由此不但推出了解$ y, z $的存在性, 并且给出了余项估计式(3.1). 证毕.

讨论奇异摄动初边值问题

(4.1) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {\mu \frac{{{\rm d}z}}{{{\rm d}t}} = \left( {\mu z - 2} \right)z + \mu y, \;\;0 \le t \le 2, }\\ {\mu \frac{{{\rm d}y}}{{{\rm d}t}} = z, }\\ {y(0, \mu ) = 1 + \int_0^2 {y(s, \mu ){\rm d}s, \;y(2, \mu ) = 2 + \int_0^2 { - y(s, \mu ){\rm d}s.\;\;} \;} } \end{array}} \right. \end{eqnarray} $

为了简化计算, 可把(4.1)式改写为

(4.2) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {\mu \frac{{{\rm d}z}}{{{\rm d}t}} = \left( {\mu z - 2} \right)z + \mu y, \;\;\;0 \le t \le 2, }\\ {\mu \frac{{{\rm d}y}}{{{\rm d}t}} = z, }\\ { \frac{{{\rm d}{k_1}}}{{{\rm d}t}} = y, \;\; \frac{{{\rm d}{k_2}}}{{{\rm d}t}} = - y, \;}\\ {y(0, \mu ) = 1 - {k_1}(0), \;\;{k_1}(2) = 0, \;\;{k_2}(0) = 0, \;\;y(2, \mu ) = 2 + {k_2}(2).} \end{array}} \right. \end{eqnarray} $

求解正则零次项、一次项的方程为

(4.3) $ \begin{equation} {\mu ^0}:\left\{ {\begin{array}{*{20}{l}} {0 = - 2{{\bar z}_0}, }\\ {0 = {{\bar z}_0}, }\\ { \frac{{{\rm d}{{\bar k}_{10}}(t)}}{{{\rm d}t}} = {{\bar y}_0}(t), \frac{{{\rm d}{{\bar k}_{20}}(t)}}{{{\rm d}t}} = - {{\bar y}_0}(t).} \end{array}} \right. \end{equation} $

(4.4) $ \begin{equation} {\mu ^1}:\left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{{\bar z}_0}}}{{{\rm d}t}} = \bar z_0^2 - 2{{\bar z}_1} + {{\bar y}_0}, }\\ { \frac{{{\rm d}{{\bar y}_0}}}{{{\rm d}t}} = {{\bar z}_1}, }\\ { \frac{{{\rm d}{{\bar k}_{11}}}}{{{\rm d}t}} = {{\bar y}_1}, \;\;\frac{{{\rm d}{{\bar k}_{21}}}}{{{\rm d}t}} = - {{\bar y}_1}.\;\;} \end{array}} \right. \end{equation} $

求解边界层零次项的方程为

(4.5) $ \begin{eqnarray} {\mu ^0}:\left\{ {\begin{array}{*{20}{l}} {\frac{{{\rm d}{L_0}z({\tau _0})}}{{{\rm d}{\tau _0}}} = - 2{L_0}z({\tau _0}), \;\;\;{\tau _0} = \frac{t}{\mu }, }\\ {\frac{{{\rm d}{L_0}y({\tau _0})}}{{{\rm d}{\tau _0}}} = {L_0}z({\tau _0}), }\\ {\frac{{{\rm d}{L_0}{k_1}({\tau _0})}}{{{\rm d}{\tau _0}}} = 0, \;\;\frac{{{\rm d}{L_0}{k_2}({\tau _0})}}{{{\rm d}{\tau _0}}} = 0.} \end{array}} \right. \end{eqnarray} $

以及零次初边值条件为

(4.6) $ \begin{eqnarray} \begin{array}{l} {{\bar y}_0}(0) + {L_0}y(0) = {1} - {{\bar k}_{10}}(0) - {L_0}{k_1}(0), \;\;\;\;\;\;\;\;{{\bar k}_{10}}(2) = 0;\\ {{\bar k}_{20}}(0) + {L_0}{k_2}(0) = 0, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{{\bar y}_0}(2) = {2} + {{\bar k}_{20}}(2);\\ {L_0}x( + \infty ) = 0. \end{array} \end{eqnarray} $

通过以上方法可解出零次主项为

(4.7) $ \begin{eqnarray} \begin{array}{l} {{\bar y}_0}(t) = \frac{2}{{3 - 2{e^{ - 1}}}}{e^{\frac{1}{2}(t - 2)}}, \;\;{{\bar z}_0}(t) = 0, \;\\ {{\bar k}_{10}}(t) = \frac{4}{{3 - 2{e^{ - 1}}}}\left( {{e^{\frac{1}{2}(t - 2)}} - 1} \right), \;\;{{\bar k}_{20}}(t) = \frac{{ - 4}}{{3 - 2{e^{ - 1}}}}\left( {{e^{\frac{1}{2}(t - 2)}} - 1} \right) + \frac{{4{e^{ - 1}} - 4}}{{3 - 2{e^{ - 1}}}}, \\ \;{L_0}{k_1}({\tau _0}) = {L_0}{k_2}({\tau _0}) = 0, \\ {L_0}y({\tau _0}) = \left( {1 - \frac{{6{e^{ - 1}} - 4}}{{3 - 2{e^{ - 1}}}}} \right){e^{ - 2{\tau _0}}}, \;\;{L_0}z({\tau _0}) = \left( { \frac{{12{e^{ - 1}} + 8}}{{3 - 2{e^{ - 1}}}} - 2} \right){e^{ - 2{\tau _0}}}. \end{array} \end{eqnarray} $

从而求得问题(4.1)形式渐近解为

(4.8) $ \begin{equation} y\left({t, \mu}\right)=\frac{2}{{3-2{e^{-1}}}}{e^{\frac{1}{2}(t-2)}}+\left({1-\frac{{6{e^{-1}}-4}}{{3-2{e^{-1}}}}}\right){e^{-2{\tau_0}}}+O\left(\mu\right), \;\;0\le t\le 2.\\ \end{equation} $

(4.9) $ \begin{equation} z\left( {t, \mu } \right) = \left( {\frac{{12{e^{ - 1}} + 8}}{{3 - 2{e^{ - 1}}}} - 2} \right){e^{ - 2{\tau _0}}} + O\left( \mu \right), \;\;\;\;0 \le t \le 2. \end{equation} $

通过观察下面图 1、图 2可知, 当使用2.1中边界层函数法求方程(4.1)的形式渐近解时, 分别让$ \mu = 0.1, 0.01 $可以发现求得的零次近似解与方程的精确解有很好的逼近. 通过图 2显示, 当$ \mu = 0.01 $时, 零次近似解的图像与方程的精确解的图像已经几乎完全重合. 这一情况不仅说明形式渐近解的零次近似可以非常好的逼近精确解, 而且说明形式渐近解的$ n(n \ge 1) $次近似在$ \mu $减小的同时, 对形式渐近解的整体影响也在减小. 而当小参数从$ \mu = 0.1 $到$ \mu = 0.01 $时, 边界层厚度$ O\left( \mu \right) $也在变薄.

下面分别给出了在$ \mu = 0.1, 0.01 $时问题(4.1)形式渐近解的(4.8)式与问题(4.1)的精确解的图像.

时至今日, 在奇异摄动问题的研究领域中, 临界情况一直都是研究难度很大的一种问题. 受此启发, 本文通过利用边界层函数法, 在临界条件稳定的情况下, 考虑了一类弱非线性的方程, 以及加上复杂而又有意义的积分初边值条件^[9-12], 得到了方程的近似解的表达式, 并给出例子和图像证明了上述方法求出的近似解相对于精确解有非常好的近似.

而在研究的过程中, 发现方程(2.1)中$ f, g $的结构不同, 会有不同的情况出现.

(1) 当方程(2.1)右端$ f $为$ f\left( {z, y, t} \right) $, 边界层部分零次近似解方程为

(5.1) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} { \frac{{{\rm d}{L_0}z\left( {{\tau _0}} \right)}}{{{\rm d}{\tau _0}}} = f\left( {{L_0}z\left( {{\tau _0}} \right), \alpha (a) + {L_0}y\left( {{\tau _0}} \right), a} \right){L_0}z\left( {{\tau _0}} \right), }\\ { \frac{{{\rm d}{L_0}y\left( {{\tau _0}} \right)}}{{{\rm d}{\tau _0}}} = {L_0}z\left( {{\tau _0}} \right).} \end{array}} \right. \end{eqnarray} $

由(5.1)式可得

(5.2) $ \begin{equation} \frac{{{\rm d}{L_0}z\left( {{\tau _0}} \right)}}{{{\rm d}{L_0}y\left( {{\tau _0}} \right)}} = f\left( {{L_0}z\left( {{\tau _0}} \right), \alpha (a) + {L_0}y\left( {{\tau _0}} \right), a} \right) \end{equation} $

观察发现(5.2)式具有很强的非线性, 所以, 无法得到边界层$ {L_0}y\left( {{\tau _0}} \right), {L_0}z\left( {{\tau _0}} \right) $之间的关系式, 后面也就无法进行. 因此, 此类方程具有很大的挑战性, 值得进一步研究.

(2) 当方程(2.1)右端由于$ \mu $在$ g\left( {z, y, t} \right) $前面, 因此, 对于$ g $为$ g\left( {\mu z, y, t} \right) $, 或者是$ g\left( {z, y, t} \right) $, 系统(2.1), (2.2)依然是临界稳定情形, 且不会改变对方程(2.1)形式渐近解的构造和求解方法.

$ (3) $当弱非线性系统方程(2.1)右端$ f, g $为$ f\left( {z, y, t} \right), f\left( {z, \mu y, t} \right), g\left( {z, y, t} \right), g\left( {z, \mu y, t} \right) $以及$ f\left( {\mu z, y, t} \right) $, $ g\left( {\mu z, y, t} \right) $时, 无法满足带有无穷大的边界值条件. 下面给出证明.

(I) 当$ f, g $为$ f\left( {z, y, t} \right), f\left( {z, \mu y, t} \right), g\left( {z, y, t} \right), g\left( {z, \mu y, t} \right) $, 例如系统(5.3)

(5.3) $ \begin{eqnarray} \left\{ {\begin{array}{*{20}{l}} {\mu \frac{{{\rm d}z}}{{{\rm d}t}} = f(z, y, t)z + \mu g(z, y, t).\;\;t \in [0, T], }\\ {\mu \frac{{{\rm d}y}}{{{\rm d}t}}{\rm{ = }}z, }\\ {y(0, \mu ) = {y^0}, \;\;\;\;z(0, \mu ) = \frac{{z_{ - 1}^0}}{\mu }.} \end{array}} \right. \end{eqnarray} $

由于带有无穷大初值, 故对于渐近解的边界层级数$ Ly\left( {{\tau _0}} , \mu\right), Lz\left( {{\tau _0}} , \mu\right) $与(2.12)式是不一样的, 有下面形式

(5.4) $ \begin{eqnarray} \begin{array}{l} Ly({\tau _0}, \mu ) = {L_0}y({\tau _0}) + \mu {L_1}y({\tau _0}) + \cdots + {\mu ^k}{L_k}y({\tau _0}) + \cdots , \;\;{\tau _0} = \frac{t}{\mu }, \\ Lz({\tau _0}, \mu ) = {\mu ^{ - 1}}{L_{ - 1}}z({\tau _0}) + {L_0}z({\tau _0}) + \mu {L_1}z({\tau _0}) + \cdots + {\mu ^k}{L_k}z({\tau _0}) + \cdots . \end{array} \end{eqnarray} $

将(5.4)式代入方程(5.3)可得$ f, g $如下形式

(5.5) $ \begin{eqnarray} \begin{array}{l} f\left( {\bar z(t, \mu ) + Lz({\tau _0}), \bar y(t, \mu ) + Ly({\tau _0}), t} \right), \\ g\left( {\bar z(t, \mu ) + Lz({\tau _0}), \bar y(t, \mu ) + Ly({\tau _0}), t} \right). \end{array} \end{eqnarray} $

通过观察计算发现, $ f $和$ g $无法在小参数$ \mu = 0 $处进行泰勒展开. 因此, 无法进行下去, 故一般不产生无穷初边值.

(II) 当$ f, g $为$ f\left( {\mu z, y, t} \right), g\left( {\mu z, y, t} \right) $, 方程(2.1)带有(5.3)式相同的无穷初边值, 通过计算求得$ {L_{ - 1}}z\left( {{\tau _0}} \right) = 0 $, 故带有无穷初边值条件不成立.