ABSTRACT

By introducing parameter function, combined with the integral method in mathematical analysis and complete square method and Riccati transform, some new interval oscillation criteria for a class of second-order nonlinear differential equations with multiple time-varying delays are obtained. The oscillation criteria are more general. These results are different from most known ones in the sense that they are based on the information only on a sequence of sub-intervals of $I=\left[t_0,\infty \right]$, rather than on the whole half-line. An example is given to illustrate the feasibility of the main results.

Keywords:Interval Oscillation, Parameter Function, Riccati Transform, Multiple Time-Varying Delays, Differential Equations

一类二阶非线性多时滞微分方程的 区间振动准则

黄秋语，孙莉^*，马婷婷，王广瓦

江苏师范大学数学与统计学院，江苏 徐州

收稿日期：2020年2月25日；录用日期：2020年3月9日；发布日期：2020年3月16日

摘 要

引入参数函数，结合数学分析中的积分方法和完全平方法以及Riccati变换，得到一类二阶非线性多时滞微分方程的新的区间振动准则。该振动准则更具有一般性。区别于已知依赖于整个大区间的性质结果，这里得出的振动准则是仅仅依赖于该区间上的子区间列的性质。我们还给出了一个例子说明主要结果的有效性。

关键词 :区间振动，参数方程，Riccati变换，多时滞，微分方程

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

微分方程理论是数学学科的一个重要分支，其中泛函微分方程理论是一个重要研究方向，可以参看文献 [1] [2]。微分方程振动理论可广泛应用于种群动力学、物理科学、通讯技术、神经网络科学和计算机科学等学科方向。微分方程振动理论及其应用受到很多学者的关注，出现大量的研究论文和专著，请参考文献 [3] - [11]。文献 [12] [13] [14] [15] [16] 研究了各种方程的区间振动性。

受以上文献的启发，本文主要讨论如下较为一般的二阶非线性多时滞微分方程的区间振动性质：

${\left[a\left(t\right){\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)}^{\prime }\right]}^{\prime }+q\left(t\right)f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)=0,t\ge t_0$ (1)

为此，我们拟引入参数函数，并结合数学分析中的积分方法、完全平方法以及Riccati变换，得到方程(1)的新的区间振动准则。这里的振动准则比已有文献中的准则更具有一般性。区别于已知依赖于整个大区间的性质结果，这里得出的振动准则仅仅依赖于该区间上的子区间列的性质。

关于方程(1)，我们假设下面的几个条件始终是成立的：

(I₁) $a\in C\left(\left[t_0,\infty \right),{ℝ}^{+}\right)$，其中 $t_0\in ℝ=\left(-\infty ,\infty \right)$ ；

(I₂) $0\le p\left(t\right)\le 1,q\left(t\right)\ge 0$，且 $q\left(t\right)$ 不最终恒为0；

(I₃) $\tau \in C\left(\left[t_0,\infty \right),ℝ\right),{\tau }^{\prime }\left(t\right)\ge 0$ ；

(I₄) $\exists \sigma \in C^1\left(\left[t_0,\infty \right),ℝ\right),\sigma \left(t\right)\le {\sigma }_i\left(t\right)\le t,{\sigma }^{\prime }\left(t\right)>0,i\in I_m=\left\{1,2,\cdots ,m\right\}$ ；

(I₅) $f\in C\left({ℝ}^m,ℝ\right)$，且当 $x_1,x_2,\cdots ,x_m$ 具有相同符号时， $f\left(x_1,x_2,\cdots ,x_m\right)$ 与诸 $x_i$ 同号， $i\in I_m,\exists k_1>0,i_0\in I_m$，使得 $\left|\frac{f\left(x_1,x_2,\cdots ,x_m\right)}{x_{i_0}}\right|\ge k_1>0\left(x_{i_0}\ne 0\right)$ ；

(I₆) $g\left(t\right)>k_2,k_2>0$。

2. 预备知识和引理

定义1 (H函数)：称函数 $H=H\left(t,s\right)$ 属于函数类X，记作 $H\in X$，假如 $H\in C\left(D,{ℝ}^{\text{+}}\right)$ 满足：

$H\left(t,t\right)=0,H\left(t,s\right)>0,t>s$, 其中 $D=\left\{\left(t,s\right):-\infty <s\le t<\infty \right\}$,

且H在D上有导函数 $\frac{\partial H}{\partial t}$ 和 $\frac{\partial H}{\partial s}$ 满足

$\frac{\partial H}{\partial t}\left(t,s\right)=h_1\left(t,s\right)\sqrt{H\left(t,s\right)},\frac{\partial H}{\partial s}\left(t,s\right)=-h_2\left(t,s\right)\sqrt{H\left(t,s\right)},h_1,h_2=L_{loc}\left(D,ℝ\right)$

定义2 (解)：称函数 $x\left(t\right)$ 是方程(1)的解，如果函数 $x\left(t\right):\left[T_0,\infty \right)\to ℝ,T_0\ge t_0$，使得 $x\left(t\right)$ 和 $a\left(t\right){\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)}^{\prime }$ 都是连续并且可微的，而且对 $t\ge T_0$，满足方程(1)。

本文中我们着重讨论研究的是方程(1)的非平凡解 $x\left(t\right)$，若该解满足 $\sup \left\{\left|x\left(t\right)\right|:t\ge T\right\}>0$ 那么对于所有的 $T\ge T_0$ 结论都是成立的。

定义3 (振动)：如果方程(1)的非平凡解有任意大的零点，即无论自变量多大，该解的函数都与坐标轴x轴有交点，那么就称这个非平凡解是振动的；若不然就称它是非振动的。方程(1)称为振动的，如果它的所有的非平凡解都是振动的。

引理1：假设 $x\left(t\right)$ 是方程(1)的最终正解，即存在足够大的 $T_0\ge t_0$，使得 $t\ge T_0$ 时 $x\left(t\right)>0$，那么对于它的任何子区间 $\left[c,b\right)\subset \left[T_0,\infty \right)$ 以及任何非递减的函数 $\rho \in C^1\left(I,{ℝ}^{+}\right)$ 和 $H\in C\left(D,{ℝ}^{+}\right)$ 都有：

$\begin{array}{l}{\displaystyle {\int }_c^b{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-\rho \left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(b,s\right)\text{d}s}\\ \le H\left(b,s\right)\omega \left(c\right)+\frac{1}{4}{\displaystyle {\int }_c^b\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}{\left[h_2\left(b,s\right)-\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(b,s\right)}\right]}^2\text{d}s\end{array}$ (2)

成立，其中 $\omega \left(t\right)=\rho \left(t\right)\frac{a\left(t\right)z^{\prime }\left(t\right)}{z\left(\sigma \left(t\right)\right)}$，$z\left(t\right)=x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)$。

证明：假设 $x\left(t\right)$ 是方程的最终正解，则存在一个很大的 $T_0\ge t_0$，使 $t\ge T_0$ 时， $x\left(t\right)>0$，进而由假设可知

$x\left(\tau \left(t\right)\right)>0,x\left(\sigma \left(t\right)\right)>0,x\left({\sigma }_i\left(t\right)\right)>0$ ( $i\in I_m$ ),

显然在 $\left[c,b\right)\subset \left[T_0,\infty \right)$ 上，以上各不等式依然成立。令

$z\left(t\right)=x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right),$

可以证明为 $a\left(t\right)z^{\prime }\left(t\right)$ 减函数。事实上，由(1)式

${\left[a\left(t\right){\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)}^{\prime }\right]}^{\prime }+q\left(t\right)f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)=0$

可知

${\left[a\left(t\right){\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)}^{\prime }\right]}^{\prime }=-q\left(t\right)f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right).$

根据已有的条件可以知道

$k_1{k}_2>0,f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)>0,q\left(t\right)>0,g\left(x\left(t\right)\right)>0$

因为 ${\left[a\left(t\right)\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)\right]}^{\prime }<0$，所以 $a\left(t\right)z^{\prime }\left(t\right)$ 为减函数。

我们还可以证明

${\left(a\left(t\right)z^{\prime }\left(t\right)\right)}^{\prime }+k_1{k}_{2}q\left(t\right)\left[1-p\left({\sigma }_{i_0}\left(t\right)\right)\right]z\left(\sigma \left(t\right)\right)\le 0.$ (3)

事实上，

$\begin{array}{l}{\left(a\left(t\right)z^{\prime }\left(t\right)\right)}^{\prime }+k_1{k}_{2}q\left(t\right)\left[1-p\left({\sigma }_{i_0}\left(t\right)\right)\right]z\left(\sigma \left(t\right)\right)\\ =-q\left(t\right)f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)\\ +k_1{k}_{2}q\left(t\right)z\left(\sigma \left(t\right)\right)-k_1{k}_{2}q\left(t\right)p\left({\sigma }_{i_0}\left(t\right)\right)z\left(\sigma \left(t\right)\right)\\ =q\left(t\right)(k_1{k}_{2}z\left(\sigma \left(t\right)\right)-f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)\\ -k_1{k}_{2}p\left({\sigma }_{i_0}\left(t\right)\right)z\left(\sigma \left(t\right)\right))\end{array}$ (*)

$\begin{array}{l}=q\left(t\right)[k_1{k}_2\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)-f\left(x\left({\sigma }_1\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)\\ -k_1{k}_{2}p\left({\sigma }_{i_0}\left(t\right)\right)\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)]\\ =q\left(t\right)[k_1{k}_{2}x\left(t\right)+k_{1}p\left(t\right)x\left(\tau \left(t\right)\right)-f\left(x\left({\sigma }_1\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)\\ -k_1{k}_{2}p\left({\sigma }_{i_0}\left(t\right)\right)p\left(t\right)x\left(\tau \left(t\right)\right)]\end{array}$

由条件(I₅)

$f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)\ge k_1{x}_i\left(t\right)\left(f>0,x_i\left(t\right)>0\right)$

$f\left(x\left({\sigma }_1\left(t\right)\right),x\left({\sigma }_2\left(t\right)\right),\cdots ,x\left({\sigma }_m\left(t\right)\right)\right)g\left(x\left(t\right)\right)\ge k_1{k}_2{x}_i\left(t\right)g\left(x\right)\ge k_1{k}_2{x}_i(t)$

所以

$\begin{array}{c}(\ast )\le q\left(t\right)\left[k_1{k}_{2}p\left(t\right)x\left(\tau \left(t\right)\right)-k_1{k}_{2}p\left({\sigma }_i\left(t\right)\right)-k_1{k}_{2}p\left({\sigma }_i\left(t\right)\right)p\left(t\right)x\left(\tau \left(t\right)\right)\right]\\ =q\left(t\right)\left(-k_1{k}_{2}p\left({\sigma }_i\left(t\right)\right)-k_1{k}_{2}p\left(t\right)x\left(\tau \left(t\right)\right)\left(1-p\left({\sigma }_i\left(t\right)\right)\right)\right)\end{array}$

又

$1-p\left({\sigma }_i\left(t\right)\right)>0,q\left(t\right)>0,-k_1{k}_{2}p\left({\sigma }_i\left(t\right)\right)-k_1{k}_{2}p\left(t\right)x\left(\tau \left(t\right)\right)\left(1-p\left({\sigma }_i\left(t\right)\right)\right)<0$

所以 $(\ast )\le 0$，故(3)式得证。

令 $\omega \left(t\right)=\rho \left(t\right)\frac{a\left(t\right)z^{\prime }\left(t\right)}{z\left(\sigma \left(t\right)\right)}$，在子区间 $\left[c,b\right)$ 上，有

$\begin{array}{c}{\omega }^{\prime }\left(t\right)=\frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\omega \left(t\right)+\rho \left(t\right)\frac{{\left(a\left(t\right)z^{\prime }\left(t\right)\right)}^{\prime }}{z\left(\sigma \left(t\right)\right)}-\rho \left(t\right)\frac{a\left(t\right)z^{\prime }\left(t\right)}{z^2\left(\sigma \left(t\right)\right)}{z}^{\prime }\left(\sigma \left(t\right)\right){\sigma }^{\prime }\left(t\right)\\ \le \frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\omega \left(t\right)-k_1{k}_2\rho \left(t\right)q\left(t\right)\left[1-p\left({\sigma }_{i_0}\left(t\right)\right)\right]-\frac{{\sigma }^{\prime }\left(t\right)}{\rho \left(t\right)a\left(\sigma \left(t\right)\right)}{\omega }^2(t)\end{array}$

$k_1{k}_2\rho \left(t\right)q\left(t\right)\left[1-p\left({\sigma }_{i_0}\left(t\right)\right)\right]\le -\omega \text{'}\left(t\right)+\frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\omega \left(t\right)-\frac{{\sigma }^{\prime }\left(t\right)}{\rho \left(t\right)a\left(\sigma \left(t\right)\right)}{\omega }^2\left(t\right),$ (4)

将上式中的t替换成s，两边同时乘以 $H\left(t,s\right)$，并对s从c到 $t\left(t\in \left[c,b\right)\right)$ 积分，得出

$\begin{array}{l}{\displaystyle {\int }_c^t{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(t,s\right)\text{d}s}\\ \le -{\displaystyle {\int }_c^{t}H\left(t,s\right){\omega }^{\prime }\left(s\right)\text{d}s}+{\displaystyle {\int }_c^{t}H\left(t,s\right)\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\omega \left(s\right)\text{d}s}-{\displaystyle {\int }_c^{t}H\left(t,s\right)}\frac{{\sigma }^{\prime }\left(s\right)}{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{\omega }^2\left(s\right)\text{d}s\\ =H\left(t,c\right)\omega \left(c\right)-{\displaystyle {\int }_c^t\left[h_2\left(t,s\right)\sqrt{H\left(t,s\right)}-H\left(t,s\right)\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\right]\omega \left(s\right)\text{d}s}\\ -{\displaystyle {\int }_c^{t}H\left(t,s\right)\frac{{\sigma }^{\prime }\left(s\right)}{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{\omega }^2\left(s\right)\text{d}s}\end{array}$

$\begin{array}{l}=H\left(t,s\right)\omega \left(c\right)-{{\displaystyle {\int }_c^t\left[\sqrt{\frac{H\left(t,s\right){\sigma }^{\prime }\left(t\right)}{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}}\omega \left(s\right)+\frac{1}{2}\sqrt{\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}\left(h_2\left(t,s\right)-\sqrt{H\left(t,s\right)}\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\right)\right]}}^2\text{d}s\\ +\frac{1}{4}{\displaystyle {\int }_c^t\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}{\left[h_2\left(t,s\right)-\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(t,s\right)}\right]}^2\text{d}s\\ \le H\left(t,c\right)\omega \left(c\right)+\frac{1}{4}{\displaystyle {\int }_c^t\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}{\left[h_2\left(t,s\right)-\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(t,s\right)}\right]}^2\text{d}s\end{array}$

对上式中的t，使 $t\to b^{-}$，由此可知(2)式成立，引理得证。

引理2：假设 $x\left(t\right)$ 是方程(1)的最终正解，即存在足够大的 $T_0\ge t_0$，使得 $t\ge T_0$ 时， $x\left(t\right)\ge 0$，$\omega \left(t\right)$ 如引理1所定义，那么对于任何子区间 $\left(a,c\right]\subset \left[T_0,\infty \right)$ 以及任何非递减函数 $\rho \in C^1\left(I,{ℝ}^{+}\right)$ 和 $H\in C\left(D,{ℝ}^{+}\right)$ 有下列不等式成立

$\begin{array}{l}{\displaystyle {\int }_a^c{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,a\right)\text{d}s}\\ \le H\left(c,a\right)\omega \left(c\right)+\frac{1}{4}{\displaystyle {\int }_a^c\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}\left[h_1\left(s,a\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,a\right)}\right]\text{d}s\end{array}$ (5)

证明方法类似引理1的证明，此处略。

3. 主要结果及其证明

借助于如上两个引理，我们可以证明下面的主要定理。

定理1：假设 $H,h_1,h_2$ 如前所定义，若对于任一 $t\ge l\ge t_0$，存在 $H\in C\left(D,{ℝ}^{+}\right)$ 以及 $\rho \in C^1\left(I,{ℝ}^{+}\right)$ 满足

$\underset{t\to \infty }{\lim }\sup {\displaystyle {\int }_l^t\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,l\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(s,l\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,l\right)}\right]}^2\right\}\text{d}s}>0.$ (6)

和

$\underset{t\to \infty }{\lim }\sup {\displaystyle {\int }_l^t\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(t,s\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(t,s\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(t,s\right)}\right]}^2\right\}\text{d}s}>0.$ (7)

则方程(1)是振动的。

证明：我们这里使用反证法证明。若不然，假设 $x\left(t\right)$ 是方程(1)的一个非振动正解，而且 $x\left(t\right)$ 最终为正。不失一般性，我们不妨可以假设，存在充分大的 $T_0\ge t_0$ 使得对所有的 $t\ge T_0$ 都有， $x\left(t\right)>0,x\left(\tau \left(t\right)\right)>0,$ $x\left(\sigma \left(t\right)\right)>0,x\left({\sigma }_i\left(t\right)\right)>0\left(i\in I_m\right)$ ，令 $a=T_0$，在(6)式中，令 $l=a$，则存在 $c>a$ 满足

${\displaystyle {\int }_a^t\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,a\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_1\left(s,a\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,a\right)}\right]}^2\right\}\text{d}s}>0.$ (8)

在(7)式中，令 $l=c$，则存在 $b>c$ 满足

${\displaystyle {\int }_c^b\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(b,s\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(b,s\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(b,s\right)}\right]}^2\right\}\text{d}s}>0.$ (9)

由(8)式移项可知

${\displaystyle {\int }_a^t{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,a\right)\text{d}s}>{\displaystyle {\int }_a^t\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_1\left(s,a\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,a\right)}\right]}^2\text{d}s}.$

由(9)式移项可知

${\displaystyle {\int }_c^b{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(b,s\right)\text{d}s}>{\displaystyle {\int }_c^b\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(b,s\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(b,s\right)}\right]}^2\text{d}s}.$

如上两式左边和右边分别乘以 $\frac{\text{1}}{H\left(c,a\right)}$ 和 $\frac{\text{1}}{H\left(b,c\right)}$ 再相加，可得

$\begin{array}{l}\frac{\text{1}}{H\left(c,a\right)}{\displaystyle {\int }_a^c{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,a\right)\text{d}s}\\ +\frac{\text{1}}{H\left(b,c\right)}{\displaystyle {\int }_a^c{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(b,s\right)\text{d}s}\\ >\frac{\text{1}}{4H\left(c,a\right)}{\displaystyle {\int }_a^c\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}}{\left[h_1\left(s,a\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,a\right)}\right]}^2\text{d}s\\ +\frac{\text{1}}{4H\left(b,c\right)}{\displaystyle {\int }_a^c\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(b,s\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(b,s\right)}\right]}^2}\text{d}s\end{array}$ (10)

而(2)式除以 $H\left(b,c\right)$，(5)式除以 $H\left(c,a\right)$，所得两式相加得到下式

$\begin{array}{l}\frac{\text{1}}{H\left(c,a\right)}{\displaystyle {\int }_a^c{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(s,a\right)\text{d}s}\\ \text{+}\frac{\text{1}}{H\left(b,c\right)}{\displaystyle {\int }_c^b{k}_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)H\left(b,s\right)\text{d}s}\\ <\frac{\text{1}}{4H\left(c,a\right)}{\displaystyle {\int }_a^c\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}{\left[h_1\left(s,a\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(s,a\right)}\right]}^2\text{d}s}\\ +\frac{\text{1}}{4H\left(b,c\right)}{\displaystyle {\int }_a^c\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{{\sigma }^{\prime }\left(s\right)}{\left[h_2\left(b,s\right)+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\sqrt{H\left(b,s\right)}\right]}^2}\text{d}s\end{array}$

显然此式与(10)式矛盾，故假设不成立，从而定理1得证。

令 $H\left(t,s\right)=H\left(t-s\right)$，如前所述，则有 $h_1\left(t-s\right)=h_2\left(t-s\right)$，并记其为 $h\left(t-s\right)$，则得到如下定理：

定理2：若对任一 $t>t_0$，存在 $a,c\in ℝ$ 使 $t\le a<c$ 且存在 $H\in C\left(D,{ℝ}^{+}\right),\rho \in C^1\left(I,{ℝ}^{+}\right)$ 满足

$\begin{array}{l}{\displaystyle {\int }_a^c{k}_1{k}_2\left[\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)-\rho \left(2c-s\right)q\left(2c-s\right)\left(1-p\left({\sigma }_{i_0}\left(2c-s\right)\right)\right)\right]H\left(s-a\right)\text{d}s}\\ >\frac{1}{4}{\displaystyle {\int }_a^c\left[\frac{\rho \left(s\right)}{{\sigma }^{\prime }\left(s\right)}+\frac{\rho \left(2c-s\right)}{{\sigma }^{\prime }\left(2c-s\right)}\right]h^2\left(s-a\right)\text{d}s}\\ +\frac{1}{2}{\displaystyle {\int }_a^c\left[\frac{{\rho }^{\prime }\left(s\right)}{{\sigma }^{\prime }\left(s\right)}+\frac{{\rho }^{\prime }\left(2c-s\right)}{{\sigma }^{\prime }\left(2c-s\right)}\right]h\left(s-a\right)\sqrt{H\left(s-a\right)}\text{d}s}\\ +\frac{1}{4}{\displaystyle {\int }_a^c\left[\frac{\rho \left(s\right)}{{\sigma }^{\prime }\left(s\right)\rho \left(s\right)}+\frac{{{\rho }^{\prime }}^2\left(2c-s\right)}{{\sigma }^{\prime }\left(2c-s\right)\rho \left(2c-s\right)}\right]H\left(s-a\right)\text{d}s}\end{array}$

则方程(1)是振动的。

又令 $H\left(t,s\right)={\left(t-s\right)}^{\lambda }$，$\lambda >1$ 是常数。此时， $h_1\left(t,s\right)=\lambda {\left(t-s\right)}^{\frac{\lambda }{2}-1}$，$h_2\left(t,s\right)=-\lambda {\left(t-s\right)}^{\frac{\lambda }{2}-1}$，则得到如下定理：

定理3：若对任一 $t\ge l\ge t_0$，存在 $\rho \in C^1\left(I,{ℝ}^{+}\right)$ 以及一常数 $\lambda >1$ 满足

$\underset{t\to \infty }{\lim }\sup \frac{1}{t^{\lambda -1}}{\displaystyle {\int }_l^t{\left(s-l\right)}^{\lambda }\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[\frac{\lambda }{s-l}+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\right]}^2\right\}\text{d}s>0}$

$\underset{t\to \infty }{\lim }\sup \frac{1}{t^{\lambda -1}}{\displaystyle {\int }_l^t{\left(s-l\right)}^{\lambda }\left\{k_1{k}_2\rho \left(s\right)q\left(s\right)\left(1-p\left({\sigma }_{i_0}\left(s\right)\right)\right)-\frac{\rho \left(s\right)a\left(\sigma \left(s\right)\right)}{4{\sigma }^{\prime }\left(s\right)}{\left[\frac{\lambda }{t-s}+\frac{{\rho }^{\prime }\left(s\right)}{\rho \left(s\right)}\right]}^2\right\}\text{d}s}>0$

则方程(1)是振动的。

4. 例子

本节给出一个具体的例题来说明我们结论的有效性。

考虑如下二阶中立型时滞微分方程

${\left[\frac{1}{t}{\left(x\left(t\right)+\frac{1}{t+2}x\left(t-2\right)\right)}^{\prime }\right]}^{\prime }+\frac{t}{{\left(t^2-1\right)}^2}x\left(t-1\right)\left(1+x^2\left(t-2\right)\right)\left(1+x\left(t\right)\right)=0$

在此方程中，

$t\ge 2,a\left(t\right)=\frac{1}{t},p\left(t\right)=\frac{1}{t+2},x\left(\tau \left(t\right)\right)=x\left(t-2\right),q\left(t\right)=\frac{t}{{\left(t^2-1\right)}^2}$

$f\left(x\left({\sigma }_1\left(t\right),x\left({\sigma }_2\left(t\right)\right)\right)\right)=x\left(t-1\right)\left(1+x^2\left(t-2\right)\right),g\left(x\left(t\right)\right)=1+x(t)$

容易验证这些函数满足条件(I₁)~(I₆)，取 $\rho \left(t\right)=t^{-2}$ 和 $\lambda =2$，事实上易证明(I₁)~(I₄)以及(I₆)满足条件，下面证明(I₅)也满足：

$f\left(x\left({\sigma }_1\left(t\right),x\left({\sigma }_2\left(t\right)\right)\right)\right)=x\left(t-1\right)\left(1+x^2\left(t-2\right)\right),x\left({\sigma }_1\left(t\right)\right)=x\left(t-1\right)>0,{\left(1+x\left({\sigma }_2\left(t\right)\right)\right)}^2>0$

所以

$f\left(x\left({\sigma }_1\left(t\right),x\left({\sigma }_2\left(t\right)\right)\right)\right)=x\left(t-1\right)\left(1+x^2\left(t-2\right)\right)$

与 $x\left({\sigma }_1\left(t\right)\right)=x\left(t-1\right)>0,1+x^2\left({\sigma }_2\left(t\right)\right)>1>0$ 同号，而且有

$\left|\frac{f\left(x\left({\sigma }_1\left(t\right),x\left({\sigma }_2\left(t\right)\right)\right)\right)}{x\left(t-1\right)}\right|=\left|\frac{x\left(t-1\right)\left(1+x^2\left(t-2\right)\right)}{x\left(t-1\right)}\right|=\left|1+x^2\left(t-2\right)\right|\ge 1=k_1>0,$

$g\left(x\left(t\right)\right)=1+x\left(t\right)>1=k_2>0.$

经计算此方程满足定理3的条件，所以此方程是振动的。

基金项目

本文受国家自然科学基金项目(NSFC11501260)，江苏高校优势学科建设工程项目(PAPD)，江苏高校品牌专业建设工程资助项目(PPZY2015A013)和江苏省大学生创新创业训练计划项目(201810320015Z)资助。

文章引用

黄秋语,孙 莉,马婷婷,王广瓦. 一类二阶非线性多时滞微分方程的区间振动准则
Interval Oscillation Criteria for a Class of Second-Order Nonlinear Differential Equations with Multiple Time-Varying Delays[J]. 应用数学进展, 2020, 09(03): 293-300. https://doi.org/10.12677/AAM.2020.93035

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