#### 期刊菜单

Singular Perturbation for Boundary Value Problem of a Second-Order Semi-Linear System with Discontinuous Source Terms
DOI: 10.12677/AAM.2019.88158, PDF, HTML, XML, 下载: 1,021  浏览: 1,273

Abstract: In this paper, a singularly perturbed boundary value problem of second-order semi-linear system with discontinuous source terms is discussed. Firstly, we use the boundary function method and sewing method to construct formal asymptotic solution to the original problem. Then, the uniformly validity of the solution obtained is proved by the lower and upper solutions theorem. Finally, an example is given as an illustration.

1. 引言

$\left\{\begin{array}{l}-E{u}^{″}+a\left(x,u\right)=f,\text{\hspace{0.17em}}x\in {\Omega }_{1}\cup {\Omega }_{2},\\ u\left(0\right)=p,\text{\hspace{0.17em}}u\left(1\right)=q.\end{array}$

$\left\{\begin{array}{l}-{\epsilon }^{2}{{y}^{″}}_{1}+{a}_{1}\left(x,{y}_{1},{y}_{2}\right)={f}_{1}\left(x\right),\\ -{\epsilon }^{2}{{y}^{″}}_{2}+{a}_{2}\left(x,{y}_{1},{y}_{2}\right)={f}_{2}\left(x\right),\\ {y}_{1}\left(0\right)={p}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(0\right)={p}_{2},\\ {y}_{1}\left(1\right)={q}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(1\right)={q}_{2}.\end{array}$ (1.1)

$\epsilon {y}^{\prime }=z$，则(1.1)式等价于如下的四维吉洪诺夫系统：

$\left\{\begin{array}{l}\epsilon {{y}^{\prime }}_{1}={z}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\epsilon {{y}^{\prime }}_{2}={z}_{2}\\ \epsilon {{z}^{\prime }}_{1}={a}_{1}\left(x,{y}_{1},{y}_{2}\right)-{f}_{1}\left(x\right),\\ \epsilon {{z}^{\prime }}_{2}={a}_{2}\left(x,{y}_{1},{y}_{2}\right)-{f}_{2}\left(x\right),\\ {y}_{1}\left(0\right)={p}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(0\right)={p}_{2},\\ {y}_{1}\left(1\right)={q}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(1\right)={q}_{2}.\end{array}$ (1.2)

(H1)函数 ${f}_{1}\left(x\right),{f}_{2}\left(x\right)$ 定义如下：

${f}_{i}\left(x\right)=\left\{\begin{array}{l}{f}_{i1}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,c\right),\\ {f}_{i2}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(c,1\right],\end{array}$

(H2)方程组 ${a}_{i}\left(x,{y}_{1},{y}_{2}\right)={f}_{i1}\left(x\right),x\in \left[0,c\right)$${a}_{j}\left(x,{y}_{1},{y}_{2}\right)={f}_{j2}\left(x\right),x\in \left(c,1\right]$，分别存在唯一解 ${\phi }_{i}\left(x\right)\in {C}^{2}\left[0,c\right],{\psi }_{j}\left(x\right)\in {C}^{2}\left[c,1\right]\text{\hspace{0.17em}}\left(i,j=1,2\right)$

(H3)存在 ${\sigma }_{1}>0$${\sigma }_{2}>0$ 使得对任意 $\left(x,{y}_{1},{y}_{2}\right)\in \left[0,1\right]×{R}^{2}$ 成立

$\frac{\partial {a}_{1}}{\partial {y}_{1}}\left(x,{y}_{1},{y}_{2}\right)\ge {\sigma }_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial {a}_{2}}{\partial {y}_{2}}\left(x,{y}_{1},{y}_{2}\right)\ge {\sigma }_{2},$

$\frac{\partial {a}_{1}}{\partial {y}_{2}}\left(x,{y}_{1},{y}_{2}\right)\le \frac{{\sigma }_{1}}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial {a}_{2}}{\partial {y}_{1}}\left(x,{y}_{1},{y}_{2}\right)\le \frac{{\sigma }_{2}}{2}.$

$\left(\begin{array}{cc}0& {E}_{2}\\ A\left(x\right)& 0\end{array}\right)$

$\left\{\begin{array}{l}{{y}^{\prime }}_{1}={z}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{y}^{\prime }}_{2}={z}_{2},\\ {{z}^{\prime }}_{1}={a}_{1}\left(0,{\phi }_{1}+{z}_{1},{\phi }_{2}+{z}_{2}\right),\\ {{z}^{\prime }}_{2}={a}_{2}\left(0,{\phi }_{1}+{z}_{1},{\phi }_{2}+{z}_{2}\right),\end{array}$ (1.3)

$\left\{\begin{array}{l}{{y}^{\prime }}_{1}={z}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{{y}^{\prime }}_{2}={z}_{2},\\ {{z}^{\prime }}_{1}={a}_{1}\left(1,{\psi }_{1}+{z}_{1},{\psi }_{2}+{z}_{2}\right),\\ {{z}^{\prime }}_{2}={a}_{2}\left(1,{\psi }_{1}+{z}_{1},{\psi }_{2}+{z}_{2}\right),\end{array}$ (1.4)

${W}^{s}=\left\{\left(y,z\right)|y=\Phi \left(z\right),y\in {G}^{-}\right\}$ ；记系统(1.4)的不稳定流形为 ${W}^{u}=\left\{\left(y,z\right)|y=\Psi \left(z\right),y\in {G}^{+}\right\}$

$B\left(x\right)=\left(\begin{array}{cc}{B}_{11}\left(x\right)& {B}_{12}\left(x\right)\\ {B}_{21}\left(x\right)& {B}_{22}\left(x\right)\end{array}\right),$

${B}^{-1}\left(x\right)\cdot \left(\begin{array}{cc}0& {E}_{2}\\ A& 0\end{array}\right)\cdot B\left(x\right)=\left(\begin{array}{cc}{C}_{-}\left(x\right)& 0\\ 0& {C}_{+}\left(x\right)\end{array}\right).$

(H4) $\mathrm{det}{B}_{11}\left(0\right)\ne 0,\mathrm{det}{B}_{22}\left(1\right)\ne 0$$\left({p}_{1}-{\phi }_{1}\left(0\right),{p}_{2}-{\phi }_{2}\left(0\right)\right)\in {G}^{-},\left({q}_{1}-{\psi }_{1}\left(1\right),{q}_{2}-{\psi }_{2}\left(1\right)\right)\in {G}^{+}$

2. 构造形式渐近解

$\left\{\begin{array}{l}-{\epsilon }^{2}{{\stackrel{¯}{y}}^{″}}_{1}+{a}_{1}\left(x,{\stackrel{¯}{y}}_{1},{\stackrel{¯}{y}}_{2}\right)={f}_{11}\left(x\right),\\ -{\epsilon }^{2}{{\stackrel{¯}{y}}^{″}}_{2}+{a}_{2}\left(x,{\stackrel{¯}{y}}_{1},{\stackrel{¯}{y}}_{2}\right)={f}_{21}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,c\right),\\ {\stackrel{¯}{y}}_{1}\left(0\right)={p}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{y}}_{2}\left(0\right)={p}_{2},\\ {\stackrel{¯}{y}}_{1}\left(c\right)={\gamma }_{1}\left(\epsilon \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{y}}_{2}\left(c\right)={\gamma }_{2}\left(\epsilon \right).\end{array}$ (2.1)

$\left\{\begin{array}{l}-{\epsilon }^{2}{{\stackrel{˜}{y}}^{″}}_{1}+{a}_{1}\left(x,{\stackrel{˜}{y}}_{1},{\stackrel{˜}{y}}_{2}\right)={f}_{12}\left(x\right),\\ -{\epsilon }^{2}{{\stackrel{˜}{y}}^{″}}_{2}+{a}_{2}\left(x,{\stackrel{˜}{y}}_{1},{\stackrel{˜}{y}}_{2}\right)={f}_{22}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(c,1\right],\\ {\stackrel{˜}{y}}_{1}\left(1\right)={q}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{y}}_{2}\left(1\right)={q}_{2},\\ {\stackrel{˜}{y}}_{1}\left(c\right)={\gamma }_{1}\left(\epsilon \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{y}}_{2}\left(c\right)={\gamma }_{2}\left(\epsilon \right).\end{array}$ (2.2)

$\left\{\begin{array}{l}{\stackrel{¯}{y}}_{1}={\stackrel{¯}{u}}_{1}\left(x\right)+{\stackrel{¯}{v}}_{1}\left(\tau \right)+{\stackrel{¯}{w}}_{1}\left(\eta \right),\\ {\stackrel{¯}{y}}_{2}={\stackrel{¯}{u}}_{2}\left(x\right)+{\stackrel{¯}{v}}_{2}\left(\tau \right)+{\stackrel{¯}{w}}_{2}\left(\eta \right),\end{array}$ (2.3)

${\stackrel{¯}{u}}_{j}\left(x\right)={\stackrel{¯}{u}}_{j0}\left(x\right)+\epsilon {\stackrel{¯}{u}}_{j1}\left(x\right)+{\epsilon }^{2}{\stackrel{¯}{u}}_{j2}\left(x\right)+\cdot \cdot \cdot ,$ (2.3a)

${\stackrel{¯}{v}}_{j}\left(\tau \right)={\stackrel{¯}{v}}_{j0}\left(\tau \right)+\epsilon {\stackrel{¯}{v}}_{j1}\left(\tau \right)+{\epsilon }^{2}{\stackrel{¯}{v}}_{j2}\left(\tau \right)+\cdot \cdot \cdot ,$ (2.3b)

${\stackrel{¯}{w}}_{j}\left(\eta \right)={\stackrel{¯}{w}}_{j0}\left(\eta \right)+\epsilon {\stackrel{¯}{w}}_{j1}\left(\eta \right)+{\epsilon }^{2}{\stackrel{¯}{w}}_{j2}\left(\eta \right)+\cdot \cdot \cdot ,$ (2.3c)

$\left\{\begin{array}{l}-{\epsilon }^{2}{{\stackrel{¯}{u}}^{″}}_{1}+{a}_{1}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)={f}_{11}\left(x\right),\\ -{\epsilon }^{2}{{\stackrel{¯}{u}}^{″}}_{2}+{a}_{2}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)={f}_{21}\left(x\right),\end{array}$ (2.4)

$\left\{\begin{array}{l}-{{\stackrel{¯}{v}}^{″}}_{1}\left(\tau \right)+{a}_{1}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}\right)-{a}_{1}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)=0,\\ -{{\stackrel{¯}{v}}^{″}}_{2}\left(\tau \right)+{a}_{2}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}\right)-{a}_{2}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)=0,\end{array}$ (2.5)

$\left\{\begin{array}{l}-{{\stackrel{¯}{w}}^{″}}_{1}\left(\eta \right)+{a}_{1}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1}+{\stackrel{¯}{w}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}+{\stackrel{¯}{w}}_{1}\right)-{a}_{1}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}\right)=0,\\ -{{\stackrel{¯}{w}}^{″}}_{2}\left(\eta \right)+{a}_{2}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1}+{\stackrel{¯}{w}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}+{\stackrel{¯}{w}}_{1}\right)-{a}_{2}\left(x,{\stackrel{¯}{u}}_{1}+{\stackrel{¯}{v}}_{1},{\stackrel{¯}{u}}_{2}+{\stackrel{¯}{v}}_{2}\right)=0.\end{array}$ (2.6)

${a}_{1}\left(x,{y}_{1},{y}_{2}\right),{a}_{2}\left(x,{y}_{1},{y}_{2}\right)$ 进行 $\epsilon$ 幂级数展开：

$\left\{\begin{array}{l}{a}_{1}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)={a}_{1}\left(x,{\phi }_{1},{\phi }_{2}\right)+\epsilon \left(\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{11}\left(x\right)+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{21}\left(x\right)+{\stackrel{¯}{g}}_{1}\left(x\right)\right)+\cdot \cdot \cdot ,\\ {a}_{2}\left(x,{\stackrel{¯}{u}}_{1},{\stackrel{¯}{u}}_{2}\right)={a}_{2}\left(x,{\phi }_{1},{\phi }_{2}\right)+\epsilon \left(\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{1}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{11}\left(x\right)+\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{2}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{21}\left(x\right)+{\stackrel{¯}{h}}_{1}\left(x\right)\right)+\cdot \cdot \cdot ,\end{array}$ (2.7)

${\epsilon }^{0}:\left\{\begin{array}{l}{a}_{1}\left(x,{\phi }_{1},{\phi }_{2}\right)={f}_{11}\left(x\right),\\ {a}_{2}\left(x,{\phi }_{1},{\phi }_{2}\right)={f}_{21}\left(x\right),\end{array}$ (2.8)

${\epsilon }^{i}:\left\{\begin{array}{l}{{\stackrel{¯}{u}}^{″}}_{1i}=\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{1\left(i+2\right)}\left(x\right)+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{2\left(i+2\right)}\left(x\right)+{\stackrel{¯}{g}}_{i+2}\left(x\right),\\ {{\stackrel{¯}{u}}^{″}}_{2i}=\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{1}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{1\left(i+2\right)}\left(x\right)+\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{2}}\left(x,{\phi }_{1},{\phi }_{2}\right)\cdot {\stackrel{¯}{u}}_{2\left(i+2\right)}\left(x\right)+{\stackrel{¯}{h}}_{i+2}\left(x\right),\end{array}$ (2.9)

${\epsilon }^{0}:\left\{\begin{array}{l}{{\stackrel{¯}{v}}^{″}}_{10}={a}_{1}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)-{a}_{1}\left(0,{\phi }_{1}\left(0\right),{\phi }_{2}\left(0\right)\right),\\ {{\stackrel{¯}{v}}^{″}}_{20}={a}_{2}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)-{a}_{2}\left(0,{\phi }_{1}\left(0\right),{\phi }_{2}\left(0\right)\right),\\ {\stackrel{¯}{v}}_{10}\left(0\right)={p}_{1}-{\phi }_{1}\left(0\right),\\ {\stackrel{¯}{v}}_{20}\left(0\right)={p}_{2}-{\phi }_{2}\left(0\right),\end{array}$ (2.10)

${\epsilon }^{i}:\left\{\begin{array}{l}{{\stackrel{¯}{v}}^{″}}_{1i}=\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)\cdot {\stackrel{¯}{v}}_{1i}\left(\tau \right)+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)\cdot {\stackrel{¯}{v}}_{2i}\left(\tau \right)+{\stackrel{¯}{F}}_{i}\left(\tau \right),\\ {{\stackrel{¯}{v}}^{″}}_{2i}=\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{1}}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)\cdot {\stackrel{¯}{v}}_{1i}\left(\tau \right)+\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{2}}\left(0,{\phi }_{1}\left(0\right)+{\stackrel{¯}{v}}_{10},{\phi }_{2}\left(0\right)+{\stackrel{¯}{v}}_{20}\right)\cdot {\stackrel{¯}{v}}_{2i}\left(\tau \right)+{\stackrel{¯}{G}}_{i}\left(\tau \right),\end{array}$ (2.11)

${\stackrel{¯}{v}}_{10}\left(\tau \right)=O\left({\text{e}}^{-\sqrt{{\delta }_{1}}\tau }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{v}}_{20}\left(\tau \right)=O\left({\text{e}}^{-\sqrt{{\delta }_{2}}\tau }\right).$

${\stackrel{¯}{v}}_{1i}\left(\tau \right)=O\left({\text{e}}^{-\left(\sqrt{{\delta }_{1}}-{\stackrel{¯}{k}}_{1i}\right)\tau }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{v}}_{2i}\left(\tau \right)=O\left({\text{e}}^{-\left(\sqrt{{\delta }_{2}}-{\stackrel{¯}{k}}_{2i}\right)\tau }\right).$

${\epsilon }^{0}:\left\{\begin{array}{l}{{\stackrel{¯}{w}}^{″}}_{10}={a}_{1}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)-{a}_{1}\left(c,{\phi }_{1}\left(c\right),{\phi }_{2}\left(c\right)\right),\\ {{\stackrel{¯}{w}}^{″}}_{20}={a}_{2}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)-{a}_{2}\left(c,{\phi }_{1}\left(c\right),{\phi }_{2}\left(c\right)\right),\\ {\stackrel{¯}{w}}_{10}\left(0\right)={\gamma }_{10}-{\phi }_{1}\left(c\right),\\ {\stackrel{¯}{w}}_{20}\left(0\right)={\gamma }_{20}-{\phi }_{2}\left(c\right),\end{array}$ (2.12)

${\epsilon }^{i}:\left\{\begin{array}{l}{{\stackrel{¯}{w}}^{″}}_{1i}=\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)\cdot {\stackrel{¯}{w}}_{1i}\left(\eta \right)+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)\cdot {\stackrel{¯}{w}}_{2i}\left(\eta \right)+{\stackrel{¯}{H}}_{i}\left(\eta \right),\\ {{\stackrel{¯}{w}}^{″}}_{2i}=\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{1}}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)\cdot {\stackrel{¯}{w}}_{1i}\left(\eta \right)+\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{2}}\left(c,{\phi }_{1}\left(c\right)+{\stackrel{¯}{w}}_{10},{\phi }_{2}\left(c\right)+{\stackrel{¯}{w}}_{20}\right)\cdot {\stackrel{¯}{w}}_{2i}\left(\eta \right)+{\stackrel{¯}{W}}_{i}\left(\eta \right),\end{array}$

${\stackrel{¯}{w}}_{10}\left(\eta \right)=O\left({\text{e}}^{\sqrt{{\delta }_{1}}\eta }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{w}}_{20}\left(\eta \right)=O\left({\text{e}}^{\sqrt{{\delta }_{2}}\eta }\right),$

${\stackrel{¯}{w}}_{1i}\left(\eta \right)=O\left({\text{e}}^{\left(\sqrt{{\delta }_{1}}-{\stackrel{¯}{l}}_{1i}\right)\eta }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{w}}_{2i}\left(\eta \right)=O\left({\text{e}}^{\left(\sqrt{{\delta }_{2}}-{\stackrel{¯}{l}}_{2i}\right)\eta }\right),$

$\left\{\begin{array}{l}{\stackrel{˜}{y}}_{1}={\stackrel{˜}{u}}_{1}\left(x\right)+{\stackrel{˜}{v}}_{1}\left(\eta \right)+{\stackrel{˜}{w}}_{1}\left(\xi \right),\\ {\stackrel{˜}{y}}_{2}={\stackrel{˜}{u}}_{2}\left(x\right)+{\stackrel{˜}{v}}_{2}\left(\eta \right)+{\stackrel{˜}{w}}_{2}\left(\xi \right),\end{array}$ (2.13)

${\stackrel{˜}{v}}_{10}\left(\eta \right)=O\left({\text{e}}^{-\sqrt{{\delta }_{1}}\eta }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{v}}_{20}\left(\eta \right)=O\left({\text{e}}^{-\sqrt{{\delta }_{2}}\eta }\right),$

${\stackrel{˜}{w}}_{10}\left(\xi \right)=O\left({\text{e}}^{\sqrt{{\delta }_{1}}\xi }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{w}}_{20}\left(\xi \right)=O\left({\text{e}}^{\sqrt{{\delta }_{2}}\xi }\right),$

${\stackrel{˜}{v}}_{1i}\left(\eta \right)=O\left({\text{e}}^{-\left(\sqrt{{\delta }_{1}}-{\stackrel{˜}{k}}_{1i}\right)\eta }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{v}}_{2i}\left(\eta \right)=O\left({\text{e}}^{-\left(\sqrt{{\delta }_{2}}-{\stackrel{˜}{k}}_{2i}\right)\eta }\right),$

${\stackrel{˜}{w}}_{1i}\left(\xi \right)=O\left({\text{e}}^{\left(\sqrt{{\delta }_{1}}-{\stackrel{˜}{l}}_{1i}\right)\xi }\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{w}}_{2i}\left(\xi \right)=O\left({\text{e}}^{\left(\sqrt{{\delta }_{2}}-{\stackrel{˜}{l}}_{2i}\right)\xi }\right),$

$\left\{\begin{array}{l}{{\stackrel{¯}{y}}^{\prime }}_{1}\left(c\right)={{\stackrel{˜}{y}}^{\prime }}_{1}\left(c\right),\\ {{\stackrel{¯}{y}}^{\prime }}_{2}\left(c\right)={{\stackrel{˜}{y}}^{\prime }}_{2}\left(c\right).\end{array}$ (2.14)

$\left\{\begin{array}{l}{\frac{\text{d}{\stackrel{¯}{w}}_{10}}{\text{d}\eta }|}_{\eta =0}={\frac{\text{d}{\stackrel{˜}{v}}_{10}}{\text{d}\eta }|}_{\eta =0},\\ {\frac{\text{d}{\stackrel{¯}{w}}_{20}}{\text{d}\eta }|}_{\eta =0}={\frac{\text{d}{\stackrel{˜}{v}}_{20}}{\text{d}\eta }|}_{\eta =0}.\end{array}$ (2.15)

$\frac{\text{d}{\stackrel{¯}{w}}_{10}}{\text{d}\eta }=V$，对(2.12)变形为：

$2V\frac{\text{d}V}{\text{d}\eta }=2\left[{a}_{1}\left(\cdot \right)-{f}_{11}\left(c\right)\right]\cdot V.$

${\int }_{-\infty }^{0}2V\frac{\text{d}V}{\text{d}\eta }\text{d}\eta ={\int }_{-\infty }^{0}2{a}_{1}\left(\cdot \right)\frac{\text{d}{\stackrel{¯}{w}}_{10}}{\text{d}\eta }\text{d}\eta -{\int }_{-\infty }^{0}2{f}_{11}\left(c\right)\frac{\text{d}{\stackrel{¯}{w}}_{10}}{\text{d}\eta }\text{d}\eta ,$

$\begin{array}{c}{V}^{2}\left(0\right)=2{a}_{1}\left(\cdot \right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)-2{f}_{11}\left(c\right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-2{\int }_{-\infty }^{0}\left[\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{¯}{w}}_{10}\left(\eta \right)\cdot \text{d}\eta .\end{array}$

$\begin{array}{c}{{\left(\frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }\right)}^{2}|}_{\eta =0}=2{a}_{1}\left(\cdot \right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)-2{f}_{11}\left(c\right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-2{\int }_{-\infty }^{0}\left[\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{¯}{w}}_{10}\left(\eta \right)\cdot \text{d}\eta .\end{array}$

$\begin{array}{c}{{\left(\frac{\text{d}{\stackrel{˜}{v}}_{10}}{\text{d}\eta }\right)}^{2}|}_{\eta =0}=2{a}_{1}\left(\ast \right)\cdot {\stackrel{˜}{v}}_{10}\left(0\right)-2{f}_{12}\left(c\right)\cdot {\stackrel{˜}{v}}_{10}\left(0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{\int }_{0}^{+\infty }\left[\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{1}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{2}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{˜}{v}}_{10}\left(\eta \right)\cdot \text{d}\eta .\end{array}$

$\begin{array}{l}2{a}_{1}\left(\cdot \right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)-2{f}_{11}\left(c\right)\cdot {\stackrel{¯}{w}}_{10}\left(0\right)-2{\int }_{-\infty }^{0}\left[\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{¯}{w}}_{10}\left(\eta \right)\cdot \text{d}\eta \\ =2{a}_{1}\left(\ast \right)\cdot {\stackrel{˜}{v}}_{10}\left(0\right)-2{f}_{12}\left(c\right)\cdot {\stackrel{˜}{v}}_{10}\left(0\right)+2{\int }_{0}^{+\infty }\left[\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{1}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{2}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{˜}{v}}_{10}\left(\eta \right)\cdot \text{d}\eta .\end{array}$

$\begin{array}{l}{a}_{1}\left(c,{\gamma }_{10},{\gamma }_{20}\right)\cdot \left({\gamma }_{10}-{\phi }_{1}\left(c\right)\right)-{f}_{11}\left(c\right)\cdot \left({\gamma }_{10}-{\phi }_{1}\left(c\right)\right)\\ \text{ }+{a}_{1}\left(c,{\gamma }_{10},{\gamma }_{20}\right)\cdot \left({\psi }_{1}\left(c\right)-{\gamma }_{10}\right)+{f}_{12}\left(c\right)\cdot \left({\psi }_{1}\left(c\right)-{\gamma }_{10}\right)\\ ={\int }_{0}^{+\infty }\left[\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{1}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{˜}{y}}_{2}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{˜}{v}}_{10}\left(\eta \right)\cdot \text{d}\eta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{\int }_{-\infty }^{0}\left[\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{1}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{1}}{\partial {\stackrel{¯}{y}}_{2}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{¯}{w}}_{10}\left(\eta \right)\cdot \text{d}\eta \\ ={P}_{1}\left({\gamma }_{10},{\gamma }_{20}\right).\end{array}$

$\begin{array}{l}{a}_{2}\left(c,{\gamma }_{10},{\gamma }_{20}\right)\cdot \left({\gamma }_{20}-{\phi }_{2}\left(c\right)\right)-{f}_{21}\left(c\right)\cdot \left({\gamma }_{20}-{\phi }_{2}\left(c\right)\right)\\ \text{ }+{a}_{2}\left(c,{\gamma }_{10},{\gamma }_{20}\right)\cdot \left({\psi }_{2}\left(c\right)-{\gamma }_{20}\right)+{f}_{22}\left(c\right)\cdot \left({\psi }_{2}\left(c\right)-{\gamma }_{20}\right)\\ ={\int }_{0}^{+\infty }\left[\frac{\partial {a}_{2}}{\partial {\stackrel{˜}{y}}_{1}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{2}}{\partial {\stackrel{˜}{y}}_{2}}\left(\ast \right)\cdot \frac{\text{d}{\stackrel{˜}{v}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{˜}{v}}_{20}\left(\eta \right)\cdot \text{d}\eta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{\int }_{-\infty }^{0}\left[\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{1}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{10}\left(\eta \right)}{\text{d}\eta }+\frac{\partial {a}_{2}}{\partial {\stackrel{¯}{y}}_{2}}\left(\cdot \right)\cdot \frac{\text{d}{\stackrel{¯}{w}}_{20}\left(\eta \right)}{\text{d}\eta }\right]\cdot {\stackrel{¯}{w}}_{20}\left(\eta \right)\cdot \text{d}\eta \\ ={P}_{2}\left({\gamma }_{10},{\gamma }_{20}\right).\end{array}$

(H5)假设方程关于 ${\gamma }_{10},{\gamma }_{20}$ 的方程组

$\left\{\begin{array}{l}\left({a}_{1}\left(c,{\gamma }_{10},{\gamma }_{20}\right)-{f}_{11}\left(c\right)\right)\cdot \left({\gamma }_{10}-{\phi }_{1}\left(c\right)\right)+\left({a}_{1}\left(c,{\gamma }_{10},{\gamma }_{20}\right)+{f}_{12}\left(c\right)\right)\cdot \left({\psi }_{1}\left(c\right)-{\gamma }_{10}\right)={P}_{1}\left({\gamma }_{10},{\gamma }_{20}\right)\\ \left({a}_{2}\left(c,{\gamma }_{10},{\gamma }_{20}\right)-{f}_{21}\left(c\right)\right)\cdot \left({\gamma }_{20}-{\phi }_{2}\left(c\right)\right)+\left({a}_{2}\left(c,{\gamma }_{10},{\gamma }_{20}\right)+{f}_{22}\left(c\right)\right)\cdot \left({\psi }_{1}\left(c\right)-{\gamma }_{20}\right)={P}_{2}\left({\gamma }_{10},{\gamma }_{20}\right)\end{array}$

${\stackrel{^}{y}}_{i}\left(x\right)=\left\{\begin{array}{l}{\stackrel{¯}{y}}_{i}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,c\right),\\ {\gamma }_{10},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=c,\\ {\stackrel{˜}{y}}_{i}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(c,1\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2.\end{array}$ (2.16)

3. 主要结论及证明

$\begin{array}{l}{y}_{i}\left(x\right)\in C\left[0,1\right]\cap {C}^{2}\left(\left[0,c\right)\cup \left(c,1\right]\right),\\ {y}_{i}\left(x\right)={\stackrel{^}{y}}_{i}\left(x\right)+O\left({\epsilon }^{m+1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\end{array}$ (3.1)

${\alpha }_{i}\left(x\right)=\left\{\begin{array}{l}{\stackrel{¯}{z}}_{i}\left(x\right)-\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,c\right),\\ {\gamma }_{i}-\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }x=c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\\ {\stackrel{˜}{z}}_{i}\left(x\right)-\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(c,1\right],\end{array}$

${\beta }_{i}\left(x\right)=\left\{\begin{array}{l}{\stackrel{¯}{z}}_{i}\left(x\right)+\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,c\right),\\ {\gamma }_{i}+\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }x=c,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\\ {\stackrel{˜}{z}}_{i}\left(x\right)+\gamma {\epsilon }^{m+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(c,1\right],\end{array}$

$\left\{\begin{array}{l}{\stackrel{¯}{z}}_{i}=\underset{j=0}{\overset{m}{\sum }}\left({\stackrel{¯}{u}}_{ij}\left(x\right)+{\stackrel{¯}{v}}_{ij}\left(\tau \right)+{\stackrel{¯}{w}}_{ij}\left(\eta \right)\right)\cdot {\epsilon }^{j},\\ {\stackrel{˜}{z}}_{i}=\underset{j=0}{\overset{m}{\sum }}\left({\stackrel{˜}{u}}_{ij}\left(x\right)+{\stackrel{˜}{v}}_{ij}\left(\eta \right)+{\stackrel{˜}{w}}_{ij}\left(\xi \right)\right)\cdot {\epsilon }^{j},\end{array}i=1,2.$

$\left\{\begin{array}{l}{\alpha }_{1}\left(0\right)\le {p}_{1}\le {\beta }_{1}\left(0\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\alpha }_{2}\left(0\right)\le {p}_{2}\le {\beta }_{2}\left(0\right),\\ {\alpha }_{1}\left(1\right)\le {q}_{1}\le {\beta }_{1}\left(1\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\alpha }_{2}\left(1\right)\le {q}_{2}\le {\beta }_{2}\left(1\right).\end{array}$

$x\in \left[0,c\right)$，由条件(H3)及中值定理得

$\begin{array}{l}{a}_{1}\left(x,{\alpha }_{1}\left(x\right),{\stackrel{¯}{y}}_{2}\right)-{a}_{1}\left(x,{\stackrel{¯}{z}}_{1}\left(x\right),{\stackrel{¯}{z}}_{2}\right)\\ =\frac{\partial {a}_{1}}{\partial {y}_{1}}\cdot \left({\alpha }_{1}-{\stackrel{¯}{z}}_{1}\right)+\frac{\partial {a}_{1}}{\partial {y}_{2}}\cdot \left({\stackrel{¯}{y}}_{2}-{\stackrel{¯}{z}}_{2}\right)\le -{\sigma }_{1}\gamma {\epsilon }^{m+1}+\frac{{\sigma }_{1}}{2}\gamma {\epsilon }^{m+1}=-\frac{{\sigma }_{1}}{2}\gamma {\epsilon }^{m+1}.\end{array}$

$|{\epsilon }^{2}\cdot \frac{{\text{d}}^{2}{\stackrel{¯}{z}}_{1}}{\text{d}{x}^{2}}-\left({a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)-{f}_{11}\left(x\right)\right)|=O\left({\epsilon }^{m+1}\right).$

$|{\epsilon }^{2}\cdot \frac{{\text{d}}^{2}{\stackrel{¯}{z}}_{1}}{\text{d}{x}^{2}}-\left({a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)-{f}_{11}\left(x\right)\right)|\le {\mu }_{1}\cdot {\epsilon }^{m+1}.$

$\begin{array}{l}{\epsilon }^{2}{{\alpha }^{″}}_{1}-{a}_{1}\left(x,{\alpha }_{1},{\stackrel{¯}{y}}_{2}\right)+{f}_{11}\left(x\right)\\ ={\epsilon }^{2}\frac{\text{d}{\stackrel{¯}{z}}_{1}^{2}}{\text{d}{x}^{2}}-{a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)+{f}_{11}\left(x\right)+{a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)-{a}_{1}\left(x,{\alpha }_{1},{\stackrel{¯}{y}}_{2}\right)\\ \ge \frac{{\sigma }_{1}}{2}\gamma {\epsilon }^{m+1}-{\mu }_{1}{\epsilon }^{m+1}=\left(\frac{{\sigma }_{1}}{2}\gamma -{\mu }_{1}\right)\cdot {\epsilon }^{m+1}\ge 0.\end{array}$

$x\in \left[0,c\right)$，存在 ${\epsilon }_{2},{\mu }_{2}>0$$\gamma \ge \frac{2{\mu }_{1}}{{\sigma }_{1}}$，当 $0<\epsilon <{\epsilon }_{2}$ 时，

$\begin{array}{l}{\epsilon }^{2}{{\beta }^{″}}_{1}-{a}_{1}\left(x,{\beta }_{1},{\stackrel{¯}{y}}_{2}\right)+{f}_{11}\left(x\right)\\ ={\epsilon }^{2}\frac{\text{d}{\stackrel{¯}{z}}_{1}^{2}}{\text{d}{x}^{2}}-{a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)+{f}_{11}\left(x\right)+{a}_{1}\left(x,{\stackrel{¯}{z}}_{1},{\stackrel{¯}{z}}_{2}\right)-{a}_{1}\left(x,{\beta }_{1},{\stackrel{¯}{y}}_{2}\right)\\ \le -\frac{{\sigma }_{1}}{2}\gamma {\epsilon }^{m+1}+{\mu }_{2}{\epsilon }^{m+1}\le 0.\end{array}$

${\epsilon }^{2}{{\alpha }^{″}}_{1}-{a}_{1}\left(x,{\alpha }_{1},{\stackrel{˜}{y}}_{2}\right)+{f}_{12}\left(x\right)\ge 0,$

${\epsilon }^{2}{{\beta }^{″}}_{1}-{a}_{1}\left(x,{\beta }_{1},{\stackrel{˜}{y}}_{2}\right)+{f}_{12}\left(x\right)\le 0.$

${\epsilon }^{2}{{\alpha }^{″}}_{2}-{a}_{2}\left(x,{\stackrel{¯}{y}}_{1},{\alpha }_{2}\right)+{f}_{21}\left(x\right)\ge 0,\text{\hspace{0.17em}}{\epsilon }^{2}{{\alpha }^{″}}_{2}-{a}_{2}\left(x,{\stackrel{¯}{y}}_{1},{\beta }_{2}\right)+{f}_{22}\left(x\right)\ge 0$

${\epsilon }^{2}{{\beta }^{″}}_{2}-{a}_{2}\left(x,{\stackrel{¯}{y}}_{1},{\beta }_{2}\right)+{f}_{21}\left(x\right)\le 0,\text{\hspace{0.17em}}{\epsilon }^{2}{{\beta }^{″}}_{2}-{a}_{2}\left(x,{\stackrel{¯}{y}}_{1},{\beta }_{2}\right)+{f}_{22}\left(x\right)\le 0.$

4. 例子

$\left\{\begin{array}{l}-{\epsilon }^{2}{{y}^{″}}_{1}+5{y}_{1}-2{y}_{2}={f}_{1}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right)\cup \left(\frac{1}{2},1\right],\\ -{\epsilon }^{2}{{y}^{″}}_{2}-2{y}_{1}+8{y}_{2}={f}_{2}\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right)\cup \left(\frac{1}{2},1\right],\\ {y}_{1}\left(0\right)=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(0\right)=2,\\ {y}_{1}\left(1\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(1\right)=1,\end{array}$ (4.1)

${f}_{1}\left(x\right)=\left\{\begin{array}{l}x+2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ 4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(\frac{1}{2},1\right],\end{array}$ ${f}_{2}\left(x\right)=\left\{\begin{array}{l}x-2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ 2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }x\in \left(\frac{1}{2},1\right].\end{array}$

$\left\{\begin{array}{l}-{\epsilon }^{2}{{y}^{″}}_{1}+5{y}_{1}-2{y}_{2}=x+2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ -{\epsilon }^{2}{{y}^{″}}_{2}-2{y}_{1}+8{y}_{2}=x-2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ {y}_{1}\left(0\right)=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(0\right)=2,\\ {y}_{1}\left(\frac{1}{2}\right)={\gamma }_{1}\left(\epsilon \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(\frac{1}{2}\right)={\gamma }_{2}\left(\epsilon \right).\end{array}$ (4.2)

$\left\{\begin{array}{l}-{\epsilon }^{2}{{y}^{″}}_{1}+5{y}_{1}-2{y}_{2}=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(\frac{1}{2},1\right],\\ -{\epsilon }^{2}{{y}^{″}}_{2}-2{y}_{1}+8{y}_{2}=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(\frac{1}{2},1\right],\\ {y}_{1}\left(1\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(1\right)=1,\\ {y}_{1}\left(\frac{1}{2}\right)={\gamma }_{1}\left(\epsilon \right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}_{2}\left(\frac{1}{2}\right)={\gamma }_{2}\left(\epsilon \right).\end{array}$ (4.3)

${\stackrel{¯}{V}}_{10}\left(\tau \right)=-\frac{11}{15}{\text{e}}^{-3\tau }+\frac{7}{5}{\text{e}}^{-2\tau },\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{¯}{V}}_{20}\left(\tau \right)=\frac{22}{15}{\text{e}}^{-3\tau }+\frac{7}{10}{\text{e}}^{-2\tau },$

${\stackrel{¯}{W}}_{10}\left(\eta \right)=\left(2{\gamma }_{10}-\frac{7}{20}\right){\text{e}}^{2\eta }+\left({\gamma }_{20}-\frac{11}{90}\right){\text{e}}^{3\eta },$

${\stackrel{¯}{W}}_{20}\left(\eta \right)=\left({\gamma }_{10}-\frac{7}{40}\right){\text{e}}^{2\eta }-\left(2{\gamma }_{20}-\frac{22}{90}\right){\text{e}}^{3\eta }.$

${\stackrel{˜}{V}}_{10}\left(\eta \right)=\left(2{\gamma }_{10}-1\right){\text{e}}^{-2\eta }+{\gamma }_{20}{\text{e}}^{-3\eta },$

${\stackrel{˜}{V}}_{20}\left(\eta \right)=-2{\gamma }_{20}{\text{e}}^{-3\eta }+\left({\gamma }_{10}-\frac{1}{2}\right){\text{e}}^{-2\eta },$

${\stackrel{˜}{W}}_{10}\left(\xi \right)=-\frac{3}{5}{\text{e}}^{-2\xi }-\frac{2}{5}{\text{e}}^{-3\xi },\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\stackrel{˜}{W}}_{20}\left(\xi \right)=-\frac{3}{10}{\text{e}}^{-2\xi }+\frac{4}{5}{\text{e}}^{-3\xi }.$

${\stackrel{^}{y}}_{1}\left(x\right)=\left\{\begin{array}{l}\frac{5x+6}{18}-\frac{11}{15}{\text{e}}^{-\frac{3x}{\epsilon }}+\frac{7}{5}{\text{e}}^{-\frac{2x}{\epsilon }}+\frac{11}{720}{\text{e}}^{\frac{6x-3}{2\epsilon }}+\frac{13}{40}{\text{e}}^{\frac{2x-1}{\epsilon }},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ \frac{27}{80},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{1}{2},\\ 1-\frac{13}{40}{\text{e}}^{\frac{1-2x}{\epsilon }}-\frac{3}{5}{\text{e}}^{\frac{2x-2}{\epsilon }}+\frac{11}{180}{\text{e}}^{\frac{3-6x}{2\epsilon }}-\frac{2}{5}{\text{e}}^{\frac{3x-3}{\epsilon }},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(\frac{1}{2},1\right].\end{array}$

${\stackrel{^}{y}}_{2}\left(x\right)=\left\{\begin{array}{l}\frac{7x-6}{36}+\frac{22}{15}{\text{e}}^{-\frac{3x}{\epsilon }}+\frac{7}{10}{\text{e}}^{-\frac{2x}{\epsilon }}-\frac{11}{360}{\text{e}}^{\frac{6x-3}{2\epsilon }}+\frac{13}{180}{\text{e}}^{\frac{2x-1}{\epsilon }},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left[0,\frac{1}{2}\right),\\ \frac{11}{180},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{1}{2},\\ \frac{1}{2}-\frac{13}{180}{\text{e}}^{\frac{1-2x}{\epsilon }}-\frac{3}{10}{\text{e}}^{\frac{2x-2}{\epsilon }}-\frac{11}{90}{\text{e}}^{\frac{3-6x}{2\epsilon }}+\frac{4}{5}{\text{e}}^{\frac{3x-3}{\epsilon }},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\in \left(\frac{1}{2},1\right].\end{array}$

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