时标上一类二维动力系统的非振荡解的存在性

Existence of Nonoscillatory Solutions of Two-Dimensional Time Scale Systems

摘要: 时标理论的研究是将微分方程理论和差分方程理论有效的结合起来,且其结果比微分方程和差分方程理论应用更为广泛。本文通过Schauder不动点定理，Knaster不动点定理以及双重广义积分来证明时标上一类由一阶动力学方程构成的二维动力系统非振荡解的存在性问题，并通过一些实例来验证所得到的主要结果。

Abstract: The study of time-scale theory combines the theory of differential equations with the theory of difference equations effectively, and the results obtained are more extensively used than those of differential equations and difference equations. In this paper, we employ Schauder’s fixed point theorem, Knaster’s fixed point theorem and double improper integrals to establish the existence of nonoscillatory solutions to the two-dimensional time scale system composed of first-order dynamic equations, and verify the main obtained results through some examples.

1. 引言

在本文中，研究时标上由一阶动力学方程构成的二维动力系统

$\{\begin{array}{l}{x}^{\Delta }\left(t\right)=f\left(t,y\left(t\right)\right)\\ y^{\Delta }\left(t\right)=g\left(t,x\left(t\right)\right)\end{array}$ (1.1)

的非振荡解的存在性，其中 $f\left(t,u\right)$ 和 $g\left(t,u\right)$ 是关于u的非减函数且对 $t\in \mathbb{T}$， $u\ne 0$，有 $uf\left(t,u\right)>0$， $ug\left(t,u\right)>0$。本文假设时标 $\mathbb{T}$ 无上界。文中出现的 $t\ge t_0$ 等价于 $t\in {\left[t_0,\infty \right)}_{\mathbb{T}}:=\left[t_0,\infty \right)\cap \mathbb{T}$。当 $x,y$ 均为非振荡的，即终于正或终于负，则称 $\left(x,y\right)$ 为非振荡解，反之为振荡解。

本文的想法来源于W. T. Li [1] 的二维非线性微分动力系统

$\{\begin{array}{l}{x}^{\prime }\left(t\right)=a\left(t\right)f\left(y\left(t\right)\right)\\ y^{\prime }\left(t\right)=b\left(t\right)g\left(x\left(t\right)\right)\end{array}$ (1.2)

的非振荡解的存在性和Özkan Öztürk [2] 的时标上的三维动力系统

$\{\begin{array}{l}{x}^{\Delta }\left(t\right)=a\left(t\right)f\left(y\left(t\right)\right)\\ y^{\Delta }\left(t\right)=b\left(t\right)g\left(z\left(t\right)\right)\\ z^{\Delta }\left(t\right)=c\left(t\right)h\left(x\left(t\right)\right)\end{array}$ (1.3)

的非振荡解的存在性。

在本文中，回顾了一些与时标相关的文章以及概念，更多细节可以参阅文献 [3] - [8]。

2. 几个引理

本文将动力系统(1.1)的所有非振荡解构成的集合设为S,则(1.1)的所有非振荡解属于下列两种情况之一：

$\begin{array}{l}{S}^{+}:=\left\{\left(x,y\right)\in S:\underset{t\to \infty }{\lim }x\left(t\right)y\left(t\right)>0\right\}\\ S^{-}:=\left\{\left(x,y\right)\in S:\underset{t\to \infty }{\lim }x\left(t\right)y\left(t\right)<0\right\}\end{array}$

引理2.1：若 $\left(x,y\right)$ 是动力系统(1.1)的一个非振荡解，本文只考虑 $S^{+}$ 这种情况，假设x是终于正的即 $\underset{t\to \infty }{\lim }x\left(t\right)>0$ (x是终于负的证明与之类似)。

非振荡解其中x是终于正的，则x,y均为终于正的单调递增函数且该解属于下列四种情况之一：

$\begin{array}{l}{S}_{B,B}^{+}:=\left\{\left(x,y\right)\in S^{+}:\underset{t\to \infty }{\lim }x\left(t\right)=c,\underset{t\to \infty }{\lim }y\left(t\right)=d\right\}\\ S_{B,\infty }^{+}:=\left\{\left(x,y\right)\in S^{+}:\underset{t\to \infty }{\lim }x\left(t\right)=c,\underset{t\to \infty }{\lim }y\left(t\right)=\infty \right\}\end{array}$

$\begin{array}{l}{S}_{\infty ,B}^{+}:=\left\{\left(x,y\right)\in S^{+}:\underset{t\to \infty }{\lim }x\left(t\right)=\infty ,\underset{t\to \infty }{\lim }y\left(t\right)=d\right\}\\ S_{\infty ,\infty }^{+}:=\left\{\left(x,y\right)\in S^{+}:\underset{t\to \infty }{\lim }x\left(t\right)=\infty ,\underset{t\to \infty }{\lim }y\left(t\right)=\infty \right\}\end{array}$

引理2.2 [9] ：若X为巴拿赫空间，Y是X的非空有界凸闭子集，且存在一个紧算子 $F:Y\to Y$，则F在X上有一个不动点，称该不动点为Schauder不动点。

引理2.3 [10] ：若 $\left(Y,\le \right)$ 为完全格，且存在一个保序算子 $F:Y\to Y$, 则F在Y上有一个不动点，称该不动点为Knaster不动点。

3. 主要结论

定理3.1 $S_{B,B}^{+}\ne \varnothing$ 当且仅当存在 $L>0$, $K_1>0$， $K_2>0$ 使得

${\displaystyle {\int }_{t_0}^{\infty }f\left(s,L+{\displaystyle {\int }_{t_0}^{s}g\left(u,K_1\right)\Delta u}\right)\Delta s<\infty }$ (3.1)

和

${\displaystyle {\int }_{t_0}^{\infty }g\left(s,K_2\right)\Delta s<\infty }$ (3.2)

成立。

证明：先证必要性，假设 $S_{B,B}^{+}\ne \varnothing$ 即存在 $\left(x,y\right)\in S_{B,B}^{+}$，则 $x,y$ 均为终于正的并且单调递增，当 $t\to \infty$ 时，有 $x\to c$， $y\to d$，其中 $0<c,d<\infty$。故存在 $t_1\ge t_0$，使得当 $t\ge t_1$ 时，有 $\frac{c}{2}\le x\left(t\right)\le c$ 成立。

由 $g\left(t,u\right)$ 关于u的单调性和对(1.1)的第二个方程由 $t_1$ 到t进行积分，得

$y\left(t\right)=y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}g\left(s,x\left(s\right)\right)\Delta s}\ge y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}g\left(s,\frac{c}{2}\right)\Delta s}$ (3.3)

当 $t\to \infty$ 时，可得

${\displaystyle {\int }_{t_1}^{\infty }g\left(s,\frac{c}{2}\right)\Delta s}\le d-y\left(t_1\right)<\infty$

对(1.1)的第一个方程由 $t_1$ 到t进行积分并利用(3.3)式，可得

$c\ge x\left(t\right)\ge x\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}f\left(s,y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{s}g\left(u,\frac{c}{2}\right)\Delta u}\right)\Delta s}$

当 $t\to \infty$ 时，有

${\displaystyle {\int }_{t_1}^{t}f\left(s,y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{s}g\left(u,\frac{c}{2}\right)\Delta u}\right)\Delta s}\le c-x\left(t_1\right)<\infty$

再证充分性，假设(3.1)式和(3.2)式成立，则存在 $t_1\ge t_0$， $L>0$， $K_1>0$， $K_2>0$ 使得 ${\displaystyle {\int }_{t_1}^{\infty }f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,K_1\right)\Delta u}\right)\Delta s<\frac{K_1}{2}}$ 和 ${\displaystyle {\int }_{t_0}^{\infty }g\left(s,K_1\right)\Delta s<\infty }$ 成立，其中 $K_1=K_2$。

设X是 ${\left[t_1,\infty \right)}_{\mathbb{T}}$ 上全体连续有界实值函数构成的，且以最大模为范数 $\Vert x\Vert =\underset{t\ge t_1}{\sup }\left|x\left(t\right)\right|$，即X为巴拿赫空间。定义X的子集Y如下：

$Y=\left\{x\left(t\right)\in X:\frac{K_1}{2}\le x\left(t\right)\le K_1,t\ge t_1\right\}$

显然，Y是X的一个非空的有界凸闭子集。定义如下一个算子 $F:Y\to X$，

$\left(Fx\right)\left(t\right)=K_1-{\displaystyle {\int }_t^{\infty }f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,x\left(u\right)\right)\Delta u}\right)\Delta s}$

接下来，我们将要证明F满足引理2.2的条件：

1) 先证F为自映射算子，

$K_1\ge \left(Fx\right)\left(t\right)\ge K_1-{\displaystyle {\int }_t^{\infty }f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,K_1\right)\Delta u}\right)\Delta s}\ge \frac{K_1}{2}$

2) 再证F在Y上连续，取Y上的一组函数列 $x_n$，使得 $x_n\to x$， $x\in Y$ 有

$\left|\left(F{x}_n\right)\left(t\right)-\left(Fx\right)\left(t\right)\right|\le {\displaystyle {\int }_t^{\infty }\left|f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,x_n\left(u\right)\right)\Delta u}\right)-f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,x\left(u\right)\right)\Delta u}\right)\right|\Delta s}$

由f，g的连续性和Lebesgue控制收敛定理可得，当 $n\to \infty$ 时， $F{x}_n\to Fx$ 即F在Y上连续。

3) 最后证明FY相对紧致，由于

$0<{\left(Fx\right)}^{\Delta }\left(t\right)\le f\left(t,L+{\displaystyle {\int }_{t_1}^{t}g\left(u,K_1\right)\Delta u}\right)<\infty$

故FY相对紧致。

由引理2.2得，存在 $\bar{x}\in Y$ 使得 $\bar{x}=F\bar{x}$，故有

$\bar{x}\left(t\right)=K_1-{\displaystyle {\int }_t^{\infty }f\left(s,L+{\displaystyle {\int }_{t_1}^{s}g\left(u,\bar{x}\left(u\right)\right)\Delta u}\right)\Delta s}$

和

${\bar{x}}^{\Delta }\left(t\right)=f\left(t,L+{\displaystyle {\int }_{t_1}^{t}g\left(u,\bar{x}\left(u\right)\right)\Delta u}\right)$

令

$\bar{y}\left(t\right)=L+{\displaystyle {\int }_{t_1}^{t}g\left(u,\bar{x}\left(u\right)\right)\Delta u}$

当 $t\to \infty$ 时，有 $\bar{x}\left(t\right)\to K_1$， $\bar{y}\left(t\right)\to \beta$，其中 $0<K_1,\beta <\infty$，故 $\left(\bar{x},\bar{y}\right)$ 是(1.1)的非振荡，因此 $S_{B,B}^{+}\ne \varnothing$，证毕。

定理3.2： $S_{B,\infty }^{+}\ne \varnothing$ 当且仅当存在 $K_1>0$， $K_2>0$ 使得

${\displaystyle {\int }_{t_0}^{\infty }f\left(s,{\displaystyle {\int }_{t_0}^{s}g\left(u,K_1\right)\Delta u}\right)\Delta s<\infty }$ (3.4)

和

${\displaystyle {\int }_{t_0}^{\infty }g\left(s,K_2\right)\Delta s=\infty }$ (3.5)

成立。

证明：先证必要性，假设 $S_{B,\infty }^{+}\ne \varnothing$ 即存在 $\left(x,y\right)\in S_{B,\infty }^{+}$，则 $x,y$ 均为终于正的并且单调递增，当 $t\to \infty$ 时，有 $x\to c$， $y\to \infty$，其中 $0<c<\infty$。故存在 $t_1\ge t_0$，使得当 $t\ge t_1$ 时，有 $\frac{c}{2}\le x\left(t\right)\le c$ 成立。对(1.1)的第一个方程从 $t_1$ 到t进行积分，得

$\frac{c}{2}\le x\left(t\right)=x\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}f\left(s,y\left(s\right)\right)\Delta s}\le c$

当 $t\to \infty$ 时，有

${\displaystyle {\int }_{t_1}^{\infty }f\left(s,y\left(s\right)\right)\Delta s}\le c-x\left(t_1\right)<\infty$ (3.6)

由 $g\left(t,u\right)$ 关于u的单调性和对(1.1)的第二个方程由 $t_1$ 到t进行积分，得

$y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}g\left(s,c\right)\Delta s}\ge y\left(t\right)=y\left(t_1\right)+{\displaystyle {\int }_{t_1}^{t}g\left(s,x\left(s\right)\right)\Delta s}$

当 $t\to \infty$ 时，有

${\displaystyle {\int }_{t_1}^{\infty }g\left(s,c\right)\Delta s}=\infty$ (3.7)

由(3.6)和(3.7)可得

${\displaystyle {\int }_{t_1}^{\infty }f\left(s,{\displaystyle {\int }_{t_1}^{s}g\left(u,\frac{c}{2}\right)\Delta u}\right)\Delta s\le {\displaystyle {\int }_{t_1}^{\infty }f\left(s,y\left(s\right)\right)\Delta s}<\infty }$

再证充分性，若(3.4)和(3.5)成立，则存在 $t_1\ge t_0$， $K_1>0$， $K_2>0$ 使得 ${\displaystyle {\int }_{t_1}^{\infty }f\left(s,{\displaystyle {\int }_{t_1}^{s}g\left(u,K_1\right)\Delta u}\right)\Delta s<K}$ 和 ${\displaystyle {\int }_{t_0}^{\infty }g\left(s,K_2\right)\Delta s=\infty }$ 成立，其中 $\frac{K_1}{2}=K_2=K$。设X是 ${\left[t_1,\infty \right)}_{\mathbb{T}}$ 上全体连续有界实值函数构成的，且以最大模为范数 $\Vert x\Vert =\underset{t\ge t_1}{\sup }\left|x\left(t\right)\right|$ 有序的巴拿赫空间。定义X的子集 $\Omega$ 如下：

$\Omega =\left\{x\left(t\right)\in X:K\le x\left(t\right)\le 2K,t\ge t_1\right\}$

定义如下一个算子 $F:\Omega \to X$，

$\left(Fx\right)\left(t\right)=K+{\displaystyle {\int }_{t_1}^{t}f\left(s,{\displaystyle {\int }_{t_1}^{s}g\left(u,x\left(u\right)\right)\Delta u}\right)\Delta s}$

显然，F是递增的且对任意 $\Omega$ 的子集 $\omega$ 都有 $\inf \omega \in \Omega$， $\sup \omega \in \Omega$，因此 $\left(\Omega ,\le \right)$ 是一个完全格。下证 $F:\Omega \to \Omega$，由于

$K\le \left(Fx\right)\left(t\right)\le K+{\displaystyle {\int }_{t_1}^{t}f\left(s,{\displaystyle {\int }_{t_1}^{s}g\left(u,2K\right)\Delta u}\right)\Delta s}\le 2K$

故F为自映射算子。

由引理2.3得，存在 $\bar{x}\in \Omega$ 使得 $\bar{x}=F\bar{x}$，故有

$\bar{x}\left(t\right)=K+{\displaystyle {\int }_{t_1}^{t}f\left(s,{\displaystyle {\int }_{t_1}^{s}g\left(u,\bar{x}\left(u\right)\right)\Delta u}\right)\Delta s}$

和

${\bar{x}}^{\Delta }\left(t\right)=f\left(t,{\displaystyle {\int }_{t_1}^{t}g\left(u,\bar{x}\left(u\right)\right)\Delta u}\right)$

令

$\bar{y}\left(t\right)={\displaystyle {\int }_{t_1}^{t}g\left(u,\bar{x}\left(u\right)\right)\Delta u}$

当 $t\to \infty$ 时，有 $\bar{x}\left(t\right)\to \alpha$， $\bar{y}\left(t\right)\to \infty$，其中 $0<\alpha <\infty$，故 $\left(\bar{x},\bar{y}\right)$ 是(1.1)的非振荡解，因此 $S_{B,\infty }^{+}\ne \varnothing$，证毕。

定理3.3： $S_{\infty ,B}^{+}\ne \varnothing$ 当且仅当存在 $K_1>0$， $K_2>0$ 使得

${\displaystyle {\int }_{t_0}^{\infty }g\left(s,{\displaystyle {\int }_{t_0}^{s}f\left(u,K_1\right)\Delta u}\right)\Delta s<\infty }$

和

${\displaystyle {\int }_{t_0}^{\infty }f\left(s,K_2\right)\Delta s=\infty }$

成立。

证明：该证明与定理3.2的证明相似，故省略。

4. 例子

例4.1：令 $q>1$ 且 $\mathbb{T}:=\left\{q^n:n\in {\mathbb{N}}_0\right\}$，考虑下列系统：

$\{\begin{array}{l}{x}^{\Delta }\left(t\right)=\frac{q+1}{q^2{t}^3}{e}_{\frac{ty\left(t\right)-t+1}{\left(q-1\right)t^2+\left(2-q\right)t}}\left(t,1\right),\\ y^{\Delta }\left(t\right)=\frac{1}{q\left(2t^2-1\right)}x\left(t\right).\end{array}$ (4.1)

首先证明对于任意的 $K>0$ 都有 ${\displaystyle {\int }_1^{\infty }g\left(t,K\right)\Delta t<\infty }$ 成立，即

$\begin{array}{c}{\displaystyle {\int }_1^{\infty }g\left(t,K\right)\Delta t}={\displaystyle {\int }_1^{\infty }\frac{K}{q\left(2t^2-1\right)}}\Delta t\\ =\frac{K}{q}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{2t^2-1}}\left(q-1\right)t\le \frac{\left(q-1\right)K}{q}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{t}}\\ =\frac{\left(q-1\right)K}{q}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{q^n}}=K<\infty \end{array}$

其次证明 ${\displaystyle {\int }_1^{\infty }f\left(t,\frac{1}{2}+{\displaystyle {\int }_1^{t}g\left(u,\frac{1}{2}\right)\Delta u}\right)\Delta t<\infty }$ 成立，即

$\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{q+1}{q^2{t}^3}f\left(t,\frac{1}{2}+{\displaystyle {\int }_1^t\frac{1}{q\left(2t^2-1\right)}g\left(u,\frac{1}{2}\right)\Delta u}\right)\Delta t}\le {\displaystyle {\int }_1^{\infty }\frac{q+1}{q^2{t}^3}f\left(t,1\right)\Delta t}\\ ={\displaystyle {\int }_1^{\infty }\frac{q+1}{q^2{t}^3}{e}_{\frac{1}{t+\left(q-1\right)\left(t-1\right)t}}\left(t,1\right)\Delta t}\\ ={\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{q+1}{q^2{t}^3}{e}_{\frac{1}{t+\left(q-1\right)\left(t-1\right)t}}\left(t,1\right)\left(q-1\right)t}\\ =\frac{q^2-1}{q^2}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{t^2}}\exp \left\{\frac{1}{\left(q-1\right)s}{\displaystyle {\int }_1^t\ln \left(1+\left(q-1\right)s\right)\frac{\frac{1}{s}}{1+\left(q-1\right)\left(s-1\right)s}\Delta s}\right\}\end{array}$

$\begin{array}{l}<\frac{q^2-1}{q^2}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{t^2}}\exp \left\{\ln q{\displaystyle {\int }_1^t\frac{1}{\left(q-1\right)s}\Delta s}\right\}\\ =\frac{q^2-1}{q^2}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{t^2}}\exp \left\{\ln q{\displaystyle \underset{m=0}{\overset{n}{\sum }}1}\right\}\\ =\frac{q^2-1}{q^2}{\displaystyle \underset{t\in {\left[1,\infty \right)}_{q^{{\mathbb{N}}_0}}}{\sum }\frac{1}{t}}\\ =\frac{q^2-1}{q^2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{q^n}=\frac{q+1}{q}<\infty }\end{array}$

因此，由定理3.1可得 $S_{B,B}^{+}\ne \varnothing$,且 $\left(x\left(t\right),y\left(t\right)\right)=\left(2-\frac{1}{t^2},1-\frac{1}{t}\right)$ 为系统(4.1)的非振荡解。

例4.2：令 $\mathbb{T}:=\left\{n^2:n\in \mathbb{N}\right\}$，考虑下列系统：

$\{\begin{array}{l}{x}^{\Delta }\left(t\right)=\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}y\left(t\right),\\ y^{\Delta }\left(t\right)=e_{\frac{\left(x\left(t\right)-1\right)\left(t^2+1\right)+1}{t^2+1+t^2\left(1+2\sqrt{t}\right)}}\left(t,1\right).\end{array}$ (4.2)

首先证明， ${\displaystyle {\int }_1^{\infty }g\left(t,1\right)\Delta t=\infty }$ 成立，即

$\begin{array}{c}{\displaystyle {\int }_1^{\infty }g\left(t,1\right)\Delta t}={\displaystyle {\int }_1^{\infty }\frac{e_1\left(t,1\right)}{e_{1-\frac{1}{t^2+1}}\left(t,1\right)}\Delta t}\\ ={\displaystyle {\int }_1^{\infty }{e}_{\frac{\frac{1}{t^2+1}}{1+\left(1+2\sqrt{t}\right)\left(1-\frac{1}{t^2+1}\right)}}}\left(t,1\right)\Delta t\\ \ge {\displaystyle {\int }_1^{\infty }\exp \left\{{\displaystyle {\int }_1^t\frac{1}{1+2\sqrt{s}}\ln \left(1\right)\Delta s}\right\}\Delta t}=\infty \end{array}$

其次证明， ${\displaystyle {\int }_1^{\infty }f\left(t,{\displaystyle {\int }_1^{t}g\left(u,1\right)\Delta u}\right)\Delta t<\infty }$，即

$\begin{array}{c}{\displaystyle {\int }_1^{\infty }f\left(t,{\displaystyle {\int }_1^{t}g\left(u,1\right)\Delta u}\right)\Delta t}\\ ={\displaystyle {\int }_1^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}\left({\displaystyle {\int }_1^t\frac{e_1\left(u,1\right)}{e_{1-\frac{1}{u^2+1}}\left(u,1\right)}\Delta u}\right)\Delta t}\\ ={\displaystyle {\int }_1^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}\left({\displaystyle {\int }_1^t{e}_{\frac{1}{u^2+1+u^2\left(1+2\sqrt{u}\right)}}\left(u,1\right)\Delta u}\right)\Delta t}\\ ={\displaystyle {\int }_1^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}\left({\displaystyle {\int }_1^t\exp \left\{{\displaystyle {\int }_1^u\frac{1}{1+2\sqrt{s}}\ln \left(2+2\sqrt{s}\right)}\frac{1}{s^2+1+s^2\left(1+2s\right)}\Delta s\right\}\Delta u}\right)\Delta t}\end{array}$

$\begin{array}{c}\le {\displaystyle {\int }_1^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}\left({\displaystyle {\int }_1^{t}u\Delta u}\right)\Delta t}\\ \le {\displaystyle {\int }_1^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^2}{t\left(t^2+1\right)\left[{\left(\sqrt{t}+1\right)}^4+1\right]}{t}^2\Delta t}\\ \le {\displaystyle {\int }_1^{\infty }\frac{1}{t{\left(\sqrt{t}+1\right)}^2}\frac{t+{\left(\sqrt{t}+1\right)}^2}{{\left(\sqrt{t}+1\right)}^2}\Delta t}\\ \le {\displaystyle {\int }_1^{\infty }\frac{2}{t\sigma \left(t\right)}\Delta t}=2<\infty \end{array}$

因此，由定理3.2可得 $S_{B,\infty }^{+}\ne \varnothing$,且 $\left(x\left(t\right),y\left(t\right)\right)=\left(1-\frac{1}{t^2+1},t\right)$ 为系统(4.1)的非振荡解。