## 具拟线性中立项的二阶变时滞动力方程的振动定理

1 梧州学院大数据与软件工程学院 广西梧州 543002

2 梧州学院广西高校行业软件技术重点实验室 广西梧州 543002

3 梧州学院广西高校图像处理与智能信息系统重点实验室 广西梧州 543002

## Oscillation Theorems of Second-Order Variable Delay Dynamic Equations with Quasilinear Neutral Term

Qin Guijiang,1,2, Yang Jiashan,1,3

1 School of Data Science and Software Engineering, Wuzhou University, Guangxi Wuzhou 543002

2 Guangxi Colleges and Universities Key Laboratory of Professional Software Technology, Wuzhou University, Guangxi Wuzhou 543002

3 Guangxi Colleges and Universities Key Laboratory of Image Processing and Intelligent Information System, Wuzhou University, Guangxi Wuzhou 543002

 基金资助: 国家自然科学基金.  51765060广西自然科学基金青年项目.  2018JJB170034广西高校中青年教师基础能力提升项目.  2018KY0543广西高校中青年教师基础能力提升项目.  2020KY17007梧州学院重点科研项目.  2016B008梧州学院重点科研项目.  2020B005

 Fund supported: the NSFC.  51765060the NSF of Guangxi.  2018JJB170034the Basic Ability Improvement Project of Young and Middle-aged Teachers in Guangxi Universities.  2018KY0543the Basic Ability Improvement Project of Young and Middle-aged Teachers in Guangxi Universities.  2020KY17007the Key Project of Wuzhou University.  2016B008the Key Project of Wuzhou University.  2020B005

Abstract

The objective of this paper is to discuss the oscillation of a class of second-order dynamic equations with a quasi-linear neutral term on the time measurement chain. Under the regularity condition, by using the generalized Riccati transformation and the classical inequality, and combining with the time scales theory, some new oscillation theorems for the equations are established. The results obtained extend, improve and enrich the part of the study results established in previous literatures. Finally, examples are given to illustrate the applications of the obtained theorems.

Keywords： Oscillation ; Time scales ; Quasilinear neutral ; Riccati transformation ; Variable delay dynamic equations

Qin Guijiang, Yang Jiashan. Oscillation Theorems of Second-Order Variable Delay Dynamic Equations with Quasilinear Neutral Term. Acta Mathematica Scientia[J], 2021, 41(5): 1492-1503 doi:

## 1 引言

\begin{align} \left\{a(t)\left[(x(t)+p(t)x^{\alpha}(\tau(t)))^{\Delta}\right]^{\beta}\right\}^{\Delta} +q(t)f(x(\delta(t))) = 0, t \in {\Bbb T} \end{align}

(H$_{1}$) 常数$0<\alpha \leqslant 1 $$\beta >0 均为两个正奇数之商. (H _{2} ) 函数 p, q \in C_{rd}({\Bbb T}, {\Bbb R}) , 且 0\leqslant p(t)<1, q(t)>0 . (H _{3} ) 函数 \tau, \delta 均为定义在时间测度链 {\Bbb T}$$ {\Bbb T}$上的滞量函数, 且$\tau(t)\leqslant t$, $\lim\limits_{t\to+\infty}\tau(t) = +\infty; $$\delta(t)\leqslant t , \lim\limits_{t \to +\infty} \delta(t) = +\infty , 并且滞量函数 \delta$$ \Delta$ -可微且严格递增的, $\delta({\Bbb T})\subset {\Bbb T}$是一时间测度链.

(H$_{4}$) 函数$f\in C({\Bbb R}, {\Bbb R})$, $uf(u) >0(u\neq0)$, 并且存在常数$L>0$使得$f(u)/u\ge L(u\neq0)$.

(H$_{5}$) 函数$a \in C_{rd}({\Bbb T}, (0, +\infty))$, 且$\int_{t_{0}}^{+\infty} a^{-1/\beta}(t) \Delta t = +\infty$.

## 2 方程振动的判别定理

$$$\left[a(t)z^{\Delta}(t)^{\beta}\right]^{\Delta}\le-Lq(t)x(\delta(t))<0.$$$

$$$z(\delta(\sigma(t)))\ge z(\delta(t))\ge z(\tau(\delta(t)))\ge z(\tau(\delta(t_{1})))\ge \min\{ 1, z(\tau(\delta(t_{1}))) \} = K\in(0, 1].$$$

$\begin{eqnarray} x(\delta(t))&\ge &z(\delta(t))- p(\delta(t))z^{\alpha}(\tau(\delta(t)))\\ &\ge &z(\delta(t))- p(\delta(t))z^{\alpha-1}(\tau(\delta(t)))z(\delta(t)) \ge \left[ 1-\frac {p(\delta(t))}{K^{1-\alpha}} \right] z(\delta(t)) . \end{eqnarray}$

$$$\left\{ \begin{array}{ll} \left[(z(\delta(t)))^{\lambda} \right]^{\Delta}\ge \lambda z^{\Delta}(\delta(t)) \delta^{\Delta}(t)(z(\delta(t)))^{\lambda-1}, &\lambda>1, \\ \left[(z(\delta(t)))^{\lambda} \right]^{\Delta}\ge \lambda z^{\Delta}(\delta(t)) \delta^{\Delta}(t)(z(\delta(\sigma (t))))^{\lambda-1}, &0<\lambda\le 1. \end{array}\right.$$$

$z^{\Delta}(t) >0$及(2.5)式, 当$t_{1}\in [t_{0}, +\infty]_{{\Bbb T}}$时, 可得

$$$\frac{a(t)(z^{\Delta}(t))^{\beta}}{z(\delta(t))} \ge \frac{a(\sigma(t))(z^{\Delta}(\sigma(t)))^{\beta}}{z(\delta(\sigma(t)))}, z^{\Delta}(\delta(t))\ge\left(\frac{a(\sigma(t))}{a(\delta(t))}\right)^{1/\beta}z^{\Delta}(\sigma(t)).$$$

$$$v(t) = \varphi(t)\frac {a(t)(z^{\Delta}(t))^{\beta} }{z(\delta(t))}, t\in [t_{1}, +\infty]_{{\Bbb T}},$$$

$\begin{eqnarray} v^{\Delta}(t)& = &\frac {\varphi(t)}{z(\delta(t))}\left[ a(t)(z^{\Delta}(t))^{\beta} \right]^{\Delta} +a(\sigma(t))(z^{\Delta}(\sigma(t)))^{\beta}\frac {z(\delta(t))\varphi^{\Delta}(t)-\varphi(t)[z(\delta(t))]^{\Delta}}{z(\delta(t))z(\delta(\sigma(t)))} \\ & \le &-\varphi(t)\frac {Lq(t)x(\delta(t))}{z(\delta(t))}+\varphi^{\Delta}(t)\frac {a(\sigma(t))(z^{\Delta}(\sigma(t)) )^{\beta}}{z(\delta(\sigma(t)))}{}\\ &&-a(\sigma(t))(z^{\Delta}(\sigma(t)))^{\beta}\frac {\varphi(t)z^{\Delta}(\delta(t))\delta^{\Delta}(t)}{z(\delta(t))z(\delta(\sigma(t)))}\\ &\le& -L\varphi(t)q(t)[1-K^{\alpha-1}p(\delta(t))]+\varphi^{\Delta}(t)\frac {a(\sigma(t))(z^{\Delta}(\sigma(t))) ^{\beta}}{z(\delta(\sigma(t)))}\\ &&-\frac{a(\sigma(t))(z^{\Delta}(\sigma(t)))^{\beta+1}\varphi(t)\delta^{\Delta}(t)}{[z(\delta(\sigma(t)))]^{2}}\left( \frac{a(\sigma(t))}{a(\delta(t))} \right)^{1/\beta}. \end{eqnarray}$

$\begin{eqnarray} v^{\Delta}(t)&\le &-L\varphi(t)q(t)\left[ 1-\frac{p(\delta(t))}{K^{1-\alpha}} \right]+ \frac{\varphi^{\Delta}(t)v(\sigma(t))}{\varphi(\sigma(t))}{}\\ &&- \frac {[z(\delta(\sigma(t)))]^{(1-\beta)/\beta}\varphi(t)\delta^{\Delta}(t)}{[\varphi(\sigma(t))]^{(1+\beta)/\beta}[a(\delta(t))]^{1/\beta}} [v(\sigma(t))]^{\frac{\beta+1}\beta} {}\\ & \le& -L\varphi(t)q(t)\left[ 1-\frac{p(\delta(t))}{K^{1-\alpha}} \right]+ \frac{\varphi^{\Delta}(t)v(\sigma(t))}{\varphi(\sigma(t))}{}\\ &&- \frac {K^{(1-\beta)/\beta}\varphi(t)\delta^{\Delta}(t)[v(\sigma(t))]^{(1+\beta)/\beta}} {[\varphi(\sigma(t))]^{(1+\beta)\beta}[a(\delta(t))]^{1/\beta}} . \end{eqnarray}$

$$$v^{\Delta}(t)\le -L\varphi(t)q(t)\left[ 1-\frac{p(\delta(t))}{K^{1-\alpha}} \right]+ \frac{\beta^{\beta}\varphi(t)a(\delta(t))}{(\beta+1)^{(\beta+1)}K^{1-\beta}[\delta^{\Delta}(t)]^{\beta}}\left( \frac{\varphi_{+}^{\Delta}(t)}{\varphi(t)} \right)^{\beta+1},$$$

$\begin{eqnarray} &&\int_{t_{1}}^{t}\varphi(s)\bigg\{Lq(s) \bigg[ 1-\frac{p(\delta(s))}{K^{1-\alpha}} \bigg]+ \frac{\delta^{\delta}\varphi(s)a(\delta(s))}{(\delta+1)^{(\delta+1)} K^{1-\beta}[\delta^{\Delta}(s)]^{\beta}} \bigg( \frac{\varphi_{+}^{\Delta}(s)}{\varphi(s)} \bigg)^{\beta+1} \bigg\}\Delta s {}\\ &\le& -v(t)+v(t_{1})\le v(t_{1}), \end{eqnarray}$

$M = \frac 1{K_{1}}\in (0, 1]$, 则由上式可得$z^{\Delta}(\sigma(t))\le \frac 1{[M a(\sigma(t))]^{1/\beta}}, t\in [t_{1}, +\infty]_{T}$, 所以

$$$\frac 1{(z^{\Delta}(\sigma(t)))^{\beta-1}}\ge[M a(\sigma(t))]^{\frac{\beta-1}\beta}.$$$

$$$\limsup\limits_{t\to+\infty}\int_{t_0}^{t}q(s)\left(1-\frac{p(\delta(s))}{K^{1-\alpha}}\right)\Delta s = +\infty,$$$

设方程(1.1)存在一个非振动解$x(t)$, 同定理1, 不妨设$x(t)$为最终正解, 则存在$t_1\in[t_0, +\infty]_{{\Bbb T}}$, 使得当$t\in[t_1, +\infty]_{{\Bbb T}}$时, 有$x(t)>0, x(\tau(t))>0, x(\delta(t))>0$, 于是, 当$t_1\in[t_0, +\infty]_{{\Bbb T}}$时(2.12)和(2.14)式成立.

$$$\int_{u}^{t}L\varphi(s)q(s)\left(1-\frac{p(\delta(s))}{K^{1-\alpha}}\right)\Delta s \le v(u)+ \int_{u}^{t}\frac{\beta^{\beta}\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}K^{1-\beta} [\delta^{\Delta}(s)]^{\beta}}\left(\frac{\varphi_{+}^{\Delta}(s)}{\varphi(s)} \right)^{\beta+1}\Delta s,$$$

$$$\zeta(u)-\omega\eta(u)\le L^{-1}v(u).$$$

$$$\liminf\limits_{t\to+\infty}\int_{t_1}^{t}\left\{\frac{K^{(1-\beta)/\beta}\varphi(s)\delta^{\Delta}(s)[v(\sigma(s))]^{ (1+\beta)/\beta }}{[\varphi(\sigma(s))]^{(1+\beta)/\beta }[a(\delta(s))]^{1/\beta}}- \frac{\varphi^{\Delta}(s)v(\sigma(s))}{\varphi(\sigma(s))}\right\}\Delta s\le M_1,$$$

$$$\liminf\limits_{t\to+\infty}\int_{t_1}^{t}\frac{\varphi(s)\delta^{\Delta}(s)[v(\sigma(s))]^ {(1+\beta)/\beta }}{[\varphi(\sigma(s))]^ {(1+\beta)/\beta } [a(\delta(s))]^{1/\beta}} \Delta s<+\infty .$$$

$$$\lim\limits_{n\to+\infty}\int_{t_1}^{T_n}\frac{\varphi(s)\delta^{\Delta}(s)[v(\sigma(s))]^ {(1+\beta)/\beta }}{[\varphi(\sigma(s))]^ {(1+\beta)/\beta } [a(\delta(s))]^{1/\beta}} \Delta s = +\infty,$$$

$$$\lim\limits_{n\to+\infty}\int_{t_1}^{T_n}\frac{\varphi^{\Delta}(s)v(\sigma(s))}{\varphi(\sigma(s))}\Delta s = +\infty.$$$

$$$\frac{\int_{t_1}^{T_n}\frac{\varphi^{\Delta}(s)v(\sigma(s))}{\varphi(\sigma(s))}\Delta s}{ \int_{t_1}^{T_n}\frac{K^{(1-\beta)/\beta}\varphi(s)\delta^{\Delta}(s)[v(\sigma(s))]^{(1+\beta)/\beta}}{[\varphi(\sigma(s))]^{ (1+\beta)/\beta }[a(\delta(s))]^{1/\beta}}\Delta s }>1-\varepsilon>0.$$$

$\begin{eqnarray} &&\int_{t_1}^{T_n} \frac{\varphi^{\Delta}(s)v(\sigma(s))}{\varphi(\sigma(s))}\Delta s{}\\ & = &K^{\frac{\beta-1}{\beta+1}}\int_{t_1}^{T_n}\left\{\frac{K^{(1-\beta)/\beta}\varphi(s)\delta^{\Delta}(s)[v(\sigma(s))]^{(1+\beta)/\beta}}{[\varphi(\sigma(s))]^{(1+\beta)/\beta} [a(\delta(s))]^{1/\beta}} \right\}^{\beta/(\beta+1)}\frac{[a(\delta(s))]^{1/(\beta+1)}\varphi^{\Delta}(s)}{[\varphi(s)\delta^{\Delta}(s)]^{\beta/(\beta+1)}}\Delta s{}\\ &\le& K^{\frac{\beta-1}{\beta+1}} \left\{\int_{t_1}^{T_n}\frac{K^{(1-\beta)/\beta}\varphi(s)\delta^{\Delta}(s) [v(\sigma(s))]^{(1+\beta)/\beta}}{[\varphi(\sigma(s))]^{(1+\beta)/\beta}[a(\delta(s))]^{1/\beta}} \Delta s \right\}^{\beta/(\beta+1)} {}\\ &&\times \left\{ \int_{t_1}^{T_n}\frac{a(\delta(s))[\varphi_{+}^{\Delta}(s)]^{\beta+1}}{[\varphi(s)\delta^{\Delta}(s)]^{\beta}}\Delta s \right\}^{1/(\beta+1)}. \end{eqnarray}$

$\beta \ge 1$时, 完全类似于定理2, 可得如下定理.

$$$\limsup\limits_{t\to+\infty}\int_{t_0}^{t}q(s)\left[1-\frac{p(\delta(s))}{K^{1-\alpha}}\right]{\rm d} s = +\infty,$$$

$$$\limsup\limits_{t\to+\infty}\int_{t_0}^{t}A(\delta(s))q(s)\left[1-\frac{p(\delta(s))}{K^{1-\alpha}}\right] {\rm d} s = +\infty,$$$

类似于定理1的证明, 可得当$t\ge t_1 $$z(t)>0, z(t)\ge x(t) , 且 z'(t)>0 . 注意到(2.7)式, 由方程(2.31), 当 s\ge t_1 时, 有 此式两边从 t$$ u$ ($u\ge t$$u, t\in[t_1, +\infty)$) 积分, 得

$u\to+\infty$, 得

## 3 例子分析

$$$[x(t)+\frac16x(\frac{t}{4} )]'' +\frac{\gamma }{t^2}x(\frac{t}{2} ) = 0, t\ge1,$$$

\begin{align} \left\{t^{-\frac57}\left[(x(t)+\frac ctx^{\frac13}(\frac t4))^{\Delta}\right]^{\frac13}\right\}^{\Delta}+\frac{\gamma}{t^2}f(x(\frac t2)) = 0, t\in{\Bbb T}, t\ge1, \end{align}

$$$\left\{t^{\frac25}[(x(t)+\frac 1tx^{\frac13}(\frac t2))^{\Delta}] ^{\frac35}\right\}^{\Delta}+\frac{1}{t^{\frac{11}{10}}} x(\frac t2) = 0, t\in{\Bbb T} = 2^{{\Bbb Z}}, t\ge t_0 = 2,$$$

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