## The Contrast Structure for the Singularly Perturbed Problem with Slow-Fast Layers and Discontinuous Righthand Side

Chen Huaxiong,1, Wang Yanyan1, Ni Mingkang2

 基金资助: 国家自然科学基金.  61703447河南省高等学校重点科研项目.  19A110038河南省高等学校重点科研项目.  18A110039周口师范学院科研基金.  ZKNUC2016012周口师范学院科研基金.  ZKNUC2017020周口师范学院科研基金.  ZKNUB2201806

 Fund supported: the NSFC.  61703447the Key Research Projects in University of Henan Province.  19A110038the Key Research Projects in University of Henan Province.  18A110039the Research Fund Projects of Zhoukou Normal University.  ZKNUC2016012the Research Fund Projects of Zhoukou Normal University.  ZKNUC2017020the Research Fund Projects of Zhoukou Normal University.  ZKNUB2201806

Abstract

This paper discusses the contrast structure solution for the singularly perturbed problem with slow-fast layers and discontinuous righthand side. By applying the boundary function method, the asymptotic solution of this problem is constructed. Then using the sewing connection method, the existence of the solution is shown and the asymptotic solution is proved to be uniformly valid. Finally, an example is given to illustrate the main results.

Keywords： Singular perturbation ; Contrast structure ; Boundary layer function method ; Sewing connection method

Chen Huaxiong, Wang Yanyan, Ni Mingkang. The Contrast Structure for the Singularly Perturbed Problem with Slow-Fast Layers and Discontinuous Righthand Side. Acta Mathematica Scientia[J], 2019, 39(4): 865-874 doi:

## 1 引言

$\begin{eqnarray}\label{60} \left\{\begin{array}{ll} \epsilon x''+\mu ax+bx=0, \, \ 0<t<1, \\ x(0)=1, \, \ x'(0)=0, \end{array}\right. \end{eqnarray}$

$\begin{eqnarray}\label{61}\left\{\begin{array}{ll}\epsilon^{4}y''={\rm sgn}(-t)A(y, t)\epsilon y'+B(y, t), \quad -1\leq t\leq 1, \\y(-1, \epsilon)=a, \, \ y(1, \epsilon)=b, \end{array}\right.\end{eqnarray}$

$\epsilon y'=z$,则方程(1.2)可化为等价的方程组

$\begin{eqnarray}\label{62}\left\{\begin{array}{ll}\epsilon y'=z, \\\epsilon^{3}z'={\rm sgn}(-t)A(y, t)z+B(y, t), \\y(-1, \epsilon)=a, \, \ y(1, \epsilon)=b.\end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{63}\left\{\begin{array}{ll}\epsilon y'^{(-)}=z^{(-)}, \\\epsilon^{3}z'^{(-)}=A(y^{(-)}, t)z^{(-)}+B(y^{(-)}, t), \\y^{(-)}(-1, \epsilon)=a, \, \ y^{(-)}(0, \epsilon)=p(\epsilon).\end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{64}\left\{\begin{array}{ll}\epsilon y'^{(+)}=z^{(+)}, \\\epsilon^{3}z'^{(+)}=-A(y^{(+)}, t)z^{(+)}+B(y^{(+)}, t), \\y^{(+)}(0, \epsilon)=p(\epsilon), \, \ y^{(+)}(1, \epsilon)=b, \end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{65}y'^{(-)}(0, \epsilon)=y'^{(+)}(0, \epsilon).\end{eqnarray}$

## 2 渐近解的构造

$$$\label{68} \bar{x}^{(\mp)}(t, \epsilon)=\bar{x}_{0}^{(\mp)}(t)+\epsilon \bar{x}_{1}^{(\mp)}(t)+\epsilon^{2}\bar{x}_{2}^{(\mp)}(t)+\cdots,$$$

$$$\label{69} Lx(\tau, \epsilon)=L_{0}x(\tau)+\epsilon L_{1}x(\tau)+\epsilon^{2}L_{2}x(\tau)+\cdots,$$$

$$$\label{610} Rx(\tau_{1}, \epsilon)=R_{0}x(\tau_{1})+\epsilon R_{1}x(\tau_{1})+\epsilon^{2}R_{2}x(\tau_{1})+\cdots,$$$

$\begin{eqnarray}\label{616} \left\{\begin{array}{ll} \displaystyle\frac{\rm d}{{\rm d}t}\bar{y}_{k-1}^{(\mp)}=\bar{z}_{k}^{(\mp)}, \\ A^{(\mp)}(\bar{y}_{0}^{(\mp)}, t)\bar{z}_{k}^{(-)}+B_{y}(\bar{y}_{0}^{(\mp)}, t)\bar{y}_{k}^{(\mp)}+f_{k}^{(\mp)}(t)=0, \end{array}\right. \end{eqnarray}$

(2)左边界层级数(2.2)各项系数的确定.

$\begin{eqnarray}\label{618} \left\{\begin{array}{ll} \displaystyle\frac{{\rm d}L_{0}y}{{\rm d}\tau}=L_{0}z, \, \ \bar{A}(\tau)L_{0}z+\bar{B}(\tau)=0, \\ \varphi_{1}(-1)+L_{0}y(0)=a, \, \ L_{0}y(+\infty)=0, \end{array}\right. \end{eqnarray}$

$\begin{eqnarray}\label{619} \left\{\begin{array}{ll} \displaystyle\frac{{\rm d}L_{0}y}{{\rm d}\tau}=-\frac{\bar{B}(\tau)}{\bar{A}(\tau)}\equiv G(L_{0}y), \\ L_{0}y(0)=a-\varphi_{1}(-1), \, \ L_{0}y(+\infty)=0, \end{array}\right. \end{eqnarray}$

$\begin{eqnarray}\label{620} \displaystyle\frac{{\rm d}G(L_{0}y)}{{\rm d} L_{0}y}(0)=-\frac{\bar{B}_{y}\bar{A}-\bar{B}\bar{A}_{y}}{\bar{A}^{2}}\Big|_{L_{0}y=0}<0. \end{eqnarray}$

$\begin{eqnarray}\label{621}\left\{\begin{array}{ll}\displaystyle\frac{{\rm d}L_{k}y}{{\rm d}\tau}=L_{k}z, \\\bar{A}(\tau)L_{k}z+[\bar{A}_{y}(\tau)L_{0}z+\bar{B}_{y}(\tau)]L_{k}y+\bar{h}_{k}(\tau)=0, \\\bar{y}^{(-)}_{k}(-1)+L_{k}y(0)=0, \, \ L_{k}y(+\infty)=0, \end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{622}\left\{\begin{array}{ll}\displaystyle\frac{{\rm d}L_{k}y}{{\rm d}\tau}=-\frac{[\bar{A}_{y}(\tau)L_{0}z+\bar{B}_{y}(\tau)]L_{k}y+\bar{h}_{k}(\tau)}{\bar{A}(\tau)}, \\\bar{y}^{(-)}_{k}(-1)+L_{k}y(0)=0, \, \ L_{k}y(+\infty)=0.\end{array}\right.\end{eqnarray}$

(3)内部层级数(2.4), (2.5)各项系数的确定.

$\begin{eqnarray}\label{638}\left\{\begin{array}{ll}\displaystyle\frac{{\rm d}Q^{(\mp)}_{k}y}{{\rm d}\tau_{0}}=Q^{(\mp)}_{k-2}z, \\\displaystyle\frac{{\rm d}Q^{(\mp)}_{k-2}z}{{\rm d}\tau_{0}}=\tilde{A}^{(\mp)}Q^{(\mp)}_{k-2}z+\tilde{A}^{(\mp)}_{y}Q_{-2}^{(\mp)}zQ_{k}^{(\mp)}y+\tilde{h}^{(\mp)}_{k}(\tau_{0}), \\Q^{(\mp)}_{k}y(0)=p_{k}-\bar{y}_{k}^{(\mp)}(0), \, \ Q^{(\mp)}_{k}y(\mp\infty)=0, \end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{639}Q^{(-)}_{k-2}z(0)=A^{(\mp)}(p_{0}, 0)[p_{k}-\bar{y}_{k}^{(\mp)}(0)]+\int^{0}_{\mp\infty}\tilde{h}^{(\mp)}_{k}(\tau_{0}){\rm d}\tau_{0}.\end{eqnarray}$

$\begin{eqnarray}\label{647}\bar{z}^{(-)}_{k-2}(0)+ Q^{(-)}_{k-2}z(0)=\bar{z}^{(+)}_{k-2}(0)+ Q^{(+)}_{k-2}z(0), \, \ k\geq2.\end{eqnarray}$

$\begin{eqnarray}\label{6300}\left\{\begin{array}{ll}\epsilon^{1-\alpha} \displaystyle\frac{\rm d}{{\rm d}\tau}y^{(-)}=z^{(-)}, \\\epsilon^{3-\alpha}\displaystyle\frac{\rm d}{{\rm d}\tau}z^{(-)}=A(y^{(-)}, t)z^{(-)}+B(y^{(-)}, t).\end{array}\right.\end{eqnarray}$

$\alpha=1$时,左问题(1.4)的附加方程为

$\begin{eqnarray}\label{6301} \left\{\begin{array}{ll} \displaystyle\frac{\rm d}{{\rm d}\tau}y^{(-)}=z^{(-)}, \\ A(y^{(-)}, t)z^{(-)}+B(y^{(-)}, t)=0. \end{array}\right. \end{eqnarray}$

$\alpha=3$时,考虑到$\epsilon$的奇异性,则左问题(1.4)的附加方程为

$\begin{eqnarray}\label{6302} \left\{\begin{array}{ll} \displaystyle\frac{\rm d}{{\rm d}\tau}y^{(-)}=z^{(-)}, \\ \displaystyle\frac{\rm d}{{\rm d}\tau}z^{(-)}=A(y^{(-)}, t)z^{(-)}. \end{array}\right. \end{eqnarray}$

## 4 例子

$\begin{eqnarray}\label{6102}\left\{\begin{array}{ll}\epsilon^{4}y''={\rm sgn}(-t)\epsilon y'+(y-1-t^{2})(y+1+t^{2}), \\y(-1, \epsilon)=-2, \, \ y(1, \epsilon)=2.\end{array}\right.\end{eqnarray}$

$\begin{eqnarray}\label{11618} \left\{\begin{array}{ll} \displaystyle\frac{{\rm d}L_{0}y}{{\rm d}\tau}=L_{0}z, \, \ L_{0}z+L_{0}y(L_{0}y-4)=0, \\ L_{0}y(0)=0, \, \ L_{0}y(+\infty)=0\end{array}\right. \end{eqnarray}$

$\begin{eqnarray}\label{11628} \left\{\begin{array}{ll} \displaystyle\frac{{\rm d}Q^{(-)}_{0}y}{{\rm d}\tau_{0}}=Q^{(-)}_{-2}z, \, \ \displaystyle\frac{{\rm d}Q^{(-)}_{-2}z}{{\rm d}\tau_{0}}=Q^{(-)}_{-2}z, \\ Q^{(-)}_{0}y(0)=p_{0}+1, \, \ Q^{(-)}_{0}y(-\infty)=0. \end{array}\right. \end{eqnarray}$

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