数学物理学报, 2021, 41(5): 1492-1503 doi:

论文

具拟线性中立项的二阶变时滞动力方程的振动定理

覃桂茳,1,2, 杨甲山,1,3

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

通讯作者: 杨甲山, E-mail: syxyyjs@163.com

收稿日期: 2021-01-18  

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

Received: 2021-01-18  

Fund supported: the NSFC.  51765060
the NSF of Guangxi.  2018JJB170034
the Basic Ability Improvement Project of Young and Middle-aged Teachers in Guangxi Universities.  2018KY0543
the Basic Ability Improvement Project of Young and Middle-aged Teachers in Guangxi Universities.  2020KY17007
the Key Project of Wuzhou University.  2016B008
the Key Project of Wuzhou University.  2020B005

作者简介 About authors

覃桂茳,E-mail:57841824@qq.com , E-mail:57841824@qq.com

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

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本文引用格式

覃桂茳, 杨甲山. 具拟线性中立项的二阶变时滞动力方程的振动定理. 数学物理学报[J], 2021, 41(5): 1492-1503 doi:

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

微分方程的理论由于在自然科学和工程技术中具有广泛的应用而得到了快速发展, 研究成果也非常丰富[1-18]. 笔者考虑时间测度链上具有一个拟线性中立项的二阶动力方程

$ \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} $

的振动性. 为了叙述方便, 总假设以下条件(H$ _{1} $)–(H$ _{5} $) 是成立的:

(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 $.

方程(1.1)的解及其振动的定义, 英文可参见文献[1-4], 中文可参见文献[5-12], 我们感兴趣的是方程(1.1)的最终不恒为0的解. 最近, 时间测度链上动力方程的振动性引起了学者们的广泛关注, 各种类型的一阶、二阶及二阶以上的动力方程都有丰富的振动性研究成果[1-12]. 当$ \alpha = 1 $时(或相近的情形, 如$ \alpha = 1 $, $ p(t)\equiv 0 $$ \alpha = 1 $$ \beta = 1 $等), 方程(1.1) 就成为具有线性中立项的动力方程, 这类方程已有大量振动性研究成果[1-5]. 当$ \alpha \ne 1 $时, 方程(1.1) 就是具有非线性中立项的动力方程. 由于中立项是非线性的, 研究较为困难, 因此很多学者采取增加一些限制条件(如增加限制条件: 当$ u \neq 0 $时, $ g(u)/u \le \eta $, 其中常数$ 0<\eta \le 1 $等), 将中立项转化为线性的来处理[6-12]. 本文将不额外增加限制条件, 运用不同于文献[13, 16, 17] 的方法, 寻找方程(1.1)的更为精细便捷的振动条件, 特别是当$ {\Bbb T} = {\Bbb R} $时的特殊情形的结果推广且改进了现有文献中的结论.

2 方程振动的判别定理

引理1[7]   若$ x(t) $$ \Delta $ -可微的且最终为正或最终为负时, 则

$ \begin{equation} (x^{\lambda}(t))^{\Delta} = \lambda x^{\Delta}(t) \int_{0}^{1}\left[hx(\delta(t))+(1-h)x(t)\right]^{\lambda-1}{\rm d}h . \end{equation} $

引理2[7]   设$ A>0 $, $ B>0 $$ \alpha>0 $均为常数, 则当$ x>0 $时, 有

引理3[7] (时间测度链上的Hölder不等式)   设$ a, b\in{\Bbb T} $$ a< b $, $ f, g \in C_{rd}([a, b], {\Bbb R}) $, 则

这了方便, 引入记号:

$ \begin{equation} z(t) = x(t)+p(t)x^{\alpha}(\tau(t)), \phi_{+}(t) = \max\{\phi(t), 0\}\ . \end{equation} $

定理1   如果存在函数$ \varphi\in C^{1}({\Bbb T}, (0, +\infty)) $使得当$ 0<\beta\le1 $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{t_{0}}^{t}\varphi(s)\Big\{Lq(s)\Big[1-\frac{p(\delta(s))}{K^{1-\alpha}}\Big] -\frac{\beta^{\beta}\alpha(\delta(s))}{(\beta+1)^{\beta+1}K^{1-\beta}[\delta^{\Delta}(s)]^{\beta}}\Big(\frac{\varphi_{+}^{\Delta}(s)}{\varphi(s)}\Big)^{\beta+1}\Big\}\Delta s = +\infty, \end{equation} $

$ \beta\ge1 $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{t_{0}}^{t}\varphi(s)\left\{ Lq(s)\left[1-\frac{p(\delta(s))}{K^{1-\alpha}}\right]-\frac{\alpha^{1/\beta}(\delta(s))}{4M^{(\beta-1)/\beta}\delta^{\Delta}(s)}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}\right)^{2}\right\}\Delta s = +\infty, \end{equation} $

其中$ K, M\in (0, 1] $是任意常数(特别地, 当$ \alpha = 1 $$ K = 1 $; 当$ \beta = 1 $$ M = 1 $), 则方程(1.1) 是振动的.

  设方程(1.1)存在一个非振动解$ x(t) $, 不妨设$ x(t) $为最终正解(当$ x(t) $为最终负解时类似可证), 则存在$ t_{1}\in [t_{0}, +\infty]_{{\Bbb T}} $, 使得$ x(t)>0 $, $ x(\tau(t))>0(t\in [t_{1}, +\infty]_{{\Bbb T}}) $. 于是由$ z(t) $的定义知, 当$ t\in [t_{1}, +\infty]_{{\Bbb T}} $时, $ z(t)>0 $$ z(t) \ge x(t) $. 由方程(1.1) 及条件(H$ _{4} $), 可得

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

利用条件(H$ _{5} $), 由(2.5)式则不难推出, $ z^{\Delta}(t)>0(t\in[t_{1}, +\infty]_{T}) $. 注意到$ \delta(t) $是严格递增的, 于是有

$ \begin{equation} 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]. \end{equation} $

注意到$ z(t)\ge x(t) $, $ 0<\alpha\le 1 $及(2.6) 式, 当$ t\in [t_{1}, +\infty]_{T} $时, 由函数$ z(t) $的定义进一步可得

$ \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} $

特别地, 当$ \alpha = 1 $时可取$ K = 1 $, (2.7)式显然是成立的. 此外, 根据引理1, 对实数$ \lambda>0 $[9]

$ \begin{equation} \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. \end{equation} $

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

由此得

$ \begin{equation} \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)). \end{equation} $

考虑到时滞$ \delta(t) $的影响, 我们引入如下广义的Riccati变换:

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

显然有$ v(t)>0(t\in [t_{1}, +\infty]_{{\Bbb T}}) $. 注意到(2.5), (2.7)–(2.9) 式, 由(2.10) 式可得

$ \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} $

根据$ \beta $的取值范围, 我们考虑两种情形: (ⅰ) $ 0<\beta\le 1 $; (ⅱ) $ \beta\ge1 $.

情形(ⅰ)    $ 0<\beta\le 1. $注意到(2.10), (2.6)式, 由(2.11)式可得

$ \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} $

利用引理2中的不等式, 由(2.12)式可进一步推出

$ \begin{equation} 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}, \end{equation} $

于是有

$ \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} $

这与条件(2.3)式矛盾.

情形(ⅱ)   $ \beta\ge 1 $. 由于$ z(t)>0 $, $ z^{\Delta}(t)>0(t\in [t_{1}, +\infty]_{T}) $, 所以当$ t\in [t_{1}, +\infty]_{T} $时, 由(2.5) 式可得

$ 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

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

特别地, 当$ \beta = 1 $时我们可取$ M = 1 $, 此时(2.15)式显然成立. 利用(2.15) 式及完全平方公式(也可用引理2中的不等式), 则由(2.11) 式可得

因此

这与条件(2.4)矛盾. 定理证毕.

注1   从定理1的条件可看出, 当$ 0< \beta <1 $$ \beta >1 $时方程(1.1) 的振动准则是不同的, 但$ \beta = 1 $时条件(2.3) 和(2.4) 是一致的. 特别地, 在定理1中取$ \varphi(t) = 1 $, 则得到非常简洁的动力方程(1.1) 的Leighton型振动定理.

推论1   如果对任意常数$ K\in (0, 1] $(特别地, 当$ \alpha = 1 $$ K = 1 $)

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

则方程(1.1)是振动的.

如果定理1的条件(2.3)或者(2.4)不成立, 则方程(1.1)的振动准则如下:

定理2   设$ 0< \beta \le1 $, 如果有函数$ \varphi\in C^1({\Bbb T}, (0, +\infty)) $$ \zeta, \eta\in C^1_{rd}({\Bbb T}, {\Bbb R}) $, 使得

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{u}^{t}\varphi(s)q(s) \left( 1-\frac{p(\delta(s))}{K^{1-\alpha}} \right) \Delta s\ge\zeta(u) \end{equation} $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{u}^{t}\frac{\varphi(s)a(\delta(s))}{[\delta^{\Delta}(s)]^{\beta}}\left(\frac{\varphi_{+}^{\Delta}(s)}{\varphi(s)} \right)^{\beta+1}\Delta s\le\eta(u) \end{equation} $

对一切$ u\ge t_1\ge t_0(u, t_1\in[t_0, +\infty]_{{\Bbb T}}) $成立, 并且上述两式中的函数$ \zeta, \eta $满足

$ \begin{equation} \liminf\limits_{t\to+\infty}\int_{t_1}^{t}\frac{\varphi(s)\delta^{\Delta}(s)[\zeta(\sigma (s))-\omega\eta(\sigma(s))]^{(\beta+1)/\beta}_{+}}{[\varphi(\sigma(s))]^{(\beta+1)/\beta} [a(\delta(s))]^{1/\beta}} \Delta s = +\infty, \end{equation} $

其中$ t_1\ge t_0 $为某常数, $ \omega = \frac{\beta^{\beta}}{L(\beta+1)^{ \beta+1}K^{1-\beta}} $, 而$ K\in(0, 1) $是任意常数(特别地, 当$ \alpha = 1 $$ K = 1 $), 则方程(1.1)是振动的.

  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.

首先, 由(2.14)式可以看出, 当$ t\ge u\ge t_1\ge t_0(t, u\in[t_0, +\infty]_{{\Bbb T}}) $时有

$ \begin{equation} \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, \end{equation} $

对(2.20) 式两边取上极限, 注意(2.17), (2.18) 两式, 则有

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

此外, 对(2.12)式两边积分, 得

利用(2.17)式, 从上式可以推导出

$ \begin{equation} \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, \end{equation} $

这里$ M_1 = v(t_1)-L\zeta(t_1) $为常数. 于是, 由(2.22)式我们就可得到

$ \begin{equation} \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 . \end{equation} $

若(2.23)式不成立, 则必存在序列$ \left\{T_n \right\}^{+\infty}_{n = 1}: T_n \in[t_1, +\infty)_{T} $, 使得$ \lim\limits_{n\to+\infty}T_n = +\infty $, 且满足

$ \begin{equation} \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, \end{equation} $

这样一来, 根据(2.22)式, 则必有

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

于是, 根据(2.22), (2.24)及(2.25)式, 当$ n $充分大时有

所以, 对任意的$ 0<\varepsilon<1 $, 当$ n $充分大时, 容易得到

$ \begin{equation} \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. \end{equation} $

其次, 利用引理3(时间测度链上的Hölder不等式), 可得

$ \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} $

于是, 分别利用(2.26)式和条件(2.18), 由(2.27)式进一步可得

这与(2.25)式矛盾! 这就意味着(2.23)式是成立的. 因此, 注意到(2.21) 和(2.23) 式, 可得

这与定理的条件(2.19)矛盾! 定理2证毕.

$ \beta \ge 1 $时, 完全类似于定理2, 可得如下定理.

定理3   设$ \beta \ge 1 $, 如果有函数$ \varphi\in C^1({\Bbb T}, (0, +\infty)) $$ \zeta, \eta\in C^1_{rd}({\Bbb T}, {\Bbb R}) $, 使得

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{u}^{t}\varphi(s)q(s)\left(1-\frac{p(\delta(s))}{K^{1-\alpha}}\right)\Delta s\ge\zeta(u) \end{equation} $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{u}^{t}\frac{a^{1/\beta}(\delta(s))}{\delta^{\Delta}(s)}\left(\frac{\varphi^{\Delta}(s)}{\varphi(s)}\right)^2\Delta s\ge\eta(u) \end{equation} $

对一切$ u\ge t_1\ge t_0 $ ($ u, t_1\in[t_0, +\infty]_{{\Bbb T}} $)成立, 并且上述两式中的函数$ \zeta, \eta $满足

$ \begin{equation} \liminf\limits_{t\to+\infty}\int_{t_1}^{t}\frac{\varphi(s)\delta^{\Delta}(s)[\zeta(\sigma(s))-\omega\eta(\sigma(s))]^2}{\varphi^2(\sigma(s))[a(\delta(s))]^{1/\beta}}\Delta s = +\infty, \end{equation} $

其中$ t_1\ge t_0 $为某常数, $ \omega = \frac 1{4LM^{(\beta-1)/\beta}} $, 而$ K, M\in(0, 1] $是任意常数(特别地, 当$ \alpha = 1 $$ K = 1 $; 当$ \beta = 1 $$ M = 1 $), 则方程(1.1) 是振动的.

作为特殊情形, 当$ {\Bbb T} = {\Bbb R} $时, 则由定理1–定理3及推论1可以得到下列具有非线性中立项的微分方程

$ \begin{equation} \left\{a(t)[(x(t)+p(t)x^{\alpha}(\tau(t)))^{ '}]^{\beta}\right\}^{'}+q(t)f(x(\delta(t))) = 0, t\ge t_0>0 \end{equation} $

的振动准则. 例如, 由定理1和推论1分别可得:

推论2   如果存在函数$ \varphi\in C^1({\Bbb R}, (0, +\infty)) $使得当$ 0<\beta\le1 $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{t_0}^{t}\varphi(s)\left\{ Lq(s)\left[1-\frac{p(\delta(s))}{K^{1-\alpha}}\right]- \frac{\beta^{\beta}a(\delta(s))}{( \beta+1)^{\beta+1}K^{1-\beta}[\delta^{'}(s)]^{\beta}}\left(\frac{\varphi_{+}^{'}(s)}{\varphi(s)}\right)^{\beta+1}\right\}{\rm d} s = +\infty, \end{equation} $

$ \beta\ge1 $

$ \begin{equation} \limsup\limits_{t\to+\infty}\int_{t_0}^{t}\varphi(s)\left\{ Lq(s)\left[1-\frac{p(\delta(s))}{K^{1-\alpha}}\right]- \frac{\beta^{\beta}a(\delta(s))}{4M^{(\beta-1)/\beta}\delta^{'}(s)}\left(\frac{\varphi^{'}(s)}{\varphi(s)}\right)^{2}\right\}{\rm d} s = +\infty, \end{equation} $

其中$ K, M\in(0, 1] $是任意常数(特别地, 当$ \alpha = 1 $$ K = 1 $; 当$ \beta = 1 $$ K = 1 $), 则方程(2.31)是振动的.

推论3   如果对任意常数$ K\in(0, 1] $ (特别地, 当$ \alpha = 1 $$ K = 1 $)

$ \begin{equation} \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, \end{equation} $

则方程(2.31)是振动的.

注2   显然, 推论2的条件较文献[13]定理1中的条件简洁得多, 其特殊情形即当$ a(t) = 1 $, $ \beta = 1 $$ f(u) = u $时的结果也较文献[16]中的定理2.1的条件简洁得多; 而推论3就是方程(2.31) 的Leighton型振动定理. 进一步, 若条件(2.34)不成立, 我们还可得到方程(3.31)的另一个Leighton型振动定理.

推论4   设$ \beta>1 $, 如果对任意常数$ K\in(0, 1] $ (特别地, 当$ \alpha = 1 $$ K = 1 $)

$ \begin{equation} \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, \end{equation} $

其中$ A(t) = \int_{t_0}^{t}a^{-1/\beta}(s){\rm d} s $, 则方程(2.31)是振动的.

  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

由上式, 进一步可得

于是

上述不等式两端同乘以$ Lq(s)(1-K^{\alpha-1}p(\delta(s)))z(\delta(s)) $后再积分, 并记

则可导出

从上式可以看出, 这与条件(2.35)矛盾!推论4证毕.

3 例子分析

例1   作为$ {\Bbb T} = {\Bbb R} $的特殊情形, 我们考虑如下二阶微分方程

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

其中, $ \gamma >0 $为常数. 令$ a(t)\equiv1 $, $ p(t) = \frac 16 $, $ q(t) = \frac{\gamma }{t^2} $, $ \tau(t) = \frac{t}{4} $, $ \delta(t) = \frac{t}{2} $, $ f(u) = u $, $ \alpha = \beta = 1 $. 显然条件(H$ _{1} $)–(H$ _{5} $) 均满足. 取$ \varphi(t) = t $, 则当$ \gamma>\frac35 = 0.6 $时, 有

所以, 由推论2知, 当$ \gamma>0.6 $时方程(3.1)是振动的.

注3   我们也可以用文献[14]中的定理3.4来判定方程(3.1)的振动性: 同样取$ \varphi(t) = t $, 因为当$ \gamma>\frac{10}6 = 1.\dot 6 $时, 有

所以, 当$ \gamma>1.\dot 6 $时方程(3.1)是振动的.

此外, 我们还可以用文献[15]中的定理2.1来判定方程(3.1) 的振动性: 还是取$ \varphi(t) = t $, 因为当$ \gamma>\frac56 = 0.8\dot3 $时, 有

所以当$ \gamma>0.8\dot3 $时方程(3.1)是振动的. 这就说明, 我们所得到的方程(1.1) 振动定理是较为“精细”的, 其特殊情形即当$ {\Bbb T} = {\Bbb R} $, $ \alpha = 1 $时的结果改进了文献[14, 15] 中的相关定理.

例2   设$ c\ge0 $, $ \gamma>0 $为常数, 考虑如下二阶动力方程

$ \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} $

这相当于方程(1)中$ a(t) = t^{-\frac57} $, $ p(t) = \frac ct $, $ q(t) = \frac{\gamma}{t^2} $, $ \tau(t) = \frac t4 $, $ \delta(t) = \frac t2 $, $ \alpha = \frac13 $, $ \beta = \frac13 $.$ f(u) = u[1+\ln(2+u^2)] $, 注意到

容易看出条件(H$ _{1} $)–(H$ _{5} $)均满足. 又因为对$ \varphi(t) = t $及任意常数$ 0<K\le1 $, 有$ t\to+\infty $

所以由定理1知方程(3.2)是振动的.

例3   考虑具拟线性中立项的二阶动力方程

$ \begin{equation} \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, \end{equation} $

显然这是二阶非线性-2差分方程, 相当于方程(1.1) 中$ a(t) = t^{\frac25} $, $ \tau(t) = \delta(t) = \frac t2 $, $ p(t) = \frac1t $, $ q(t) = t^{-\frac{11}{10}} $, $ \alpha = \frac13 $, $ \beta = \frac35 $, $ f(u) = u $. 由于

所以条件(H$ _1 $)–(H$ _5 $)均满足. 又因为对$ \varphi(t) = 1 $及任意常数$ 0<K\le1 $, 总存在$ t_1 = \frac4{K^{2/3}}\ge2 $, 当$ s\ge t_1 $

于是当$ t\ge u\ge t_1 $

并且函数$ \zeta, \eta $满足

这就说明, 定理2的条件(2.17)–(2.19)均满足, 所以, 由定理2知, 方程(3.3) 是振动的.

注4   由于方程(3.2)和(3.3)均为具有非线性中立项的动力方程, 因此现有文献如[1-12]中定理都不适用于方程(3.2)和(3.3). 因此本文定理改进且丰富了现有文献中的相关结果.

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