二阶中立型时滞微分方程的振动准则

Oscillation Criterion of Second Order Differential Equations with Neutral Delay

摘要: 考虑二阶中立型时滞微分方程：的振动性，在现有文献基础上，利用广义Riccati变换、函数单调性和经典不等式，对方程做了进一步研究，建立新准则，改进了文献的某些结果。并且给出了例子说明主要结果的先进性。

Abstract: In this paper, we study the oscillation of second order differential equations with neutral delay using generalized Riccati transformation, the classical inequality and functional monotonicity, some new oscillation criterion are obtained, and improved the results of the references. At last, some examples are given to illustrate the advancement of our results.

1. 引言

本文研究的是一类二阶半线性中立型时滞微分方程

${\left(r\left(t\right){\left|{\phi }^{\prime }\left(t\right)\right|}^{\alpha -1}{\phi }^{\prime }\left(t\right)\right)}^{\prime }+q\left(t\right){\left|x\left(\delta \left(t\right)\right)\right|}^{\beta -1}x\left(\delta \left(t\right)\right)=0$， $t>t_1>t_0$ (1)

的振动性，其中 $\phi \left(t\right)=x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)$，并且满足以下条件： $\alpha >0$， $\beta >0$ ；时滞函数 $\delta \in C^1\left(\left[t_0,\infty \right),R\right)$， $\tau \in C^1\left(\left[t_0,\infty \right),R\right)$，任意 $t\ge t_0$，都有 $\tau \left(t\right)\le t$， $\delta \left(t\right)\le t$， ${\delta }^{\prime }\left(t\right)>0$， $\underset{t\to \infty }{\lim }\delta \left(t\right)=\underset{t\to \infty }{\lim }\tau \left(t\right)=\infty$ ； $r\in C^1\left(\left[t_0,\infty \right),\left(0,\infty \right)\right)$，记 $\pi \left(t\right)={\displaystyle {\int }_t^{\infty }{r}^{-\frac{1}{\alpha }}}\left(t\right)\text{d}s$，并且满足 $\pi \left(t_0\right)<\infty$ ； $q\in C\left(\left[t_0,\infty \right),\left(0,\infty \right)\right)$， $p\in C\left(\left[t_0,\infty \right),\left(0,\infty \right)\right)$， $q\left(t\right)>0$， $0\le p\left(t\right)<1$ ； $p\left(t\right)<\frac{\pi \left(t\right)}{\pi \left(\tau \left(t\right)\right)}$。

若方程(1)的解既不最终为正，也不是最终为负，则它为振动的；否则称它为非振动的。若方程(1)所有的解都是振动的，则称方程(1)是振动的。对二阶半线性中立型微分方程振动性建立了一系列准则并给出证明方法 [1] - [11]，但研究所得结果还不够完善，本文对文献 [1] 的振动准则进行研究，进一步改进其结果，得到一个新的振动准则。为了能更好地研究这方面的知识，需要引入以下定理。

引理1.1 [1] 设 $\alpha >0,\beta >0$ 是两个奇数商，如果

${\displaystyle {\int }_{t_0}^{\infty }{\left(\frac{1}{r\left(t\right)}{\displaystyle {\int }_{t_0}^{t}Q\left(s\right){\pi }^{\beta }\left(\sigma \left(s\right)\right)\text{d}s}\right)}^{\frac{1}{\alpha }}\text{d}t=\infty }$ (2)

成立，则方程 ${\left({\left(r\left(t\right){\left(x\left(t\right)+p\left(t\right)x\left(\tau \left(t\right)\right)\right)}^{\prime }\right)}^{\alpha }\right)}^{\prime }+q\left(t\right)x^{\beta }\left(\sigma \left(t\right)\right)=0$ 是振动的。

引理1.2 [11] 设 $\pi \left(t_0\right)<\infty$，若存在函数 $\rho \in C\left(\left[t_0,\infty \right),R\right)$ 满足

$\underset{t\to \infty }{\lim \sup }{\displaystyle {\int }_{t_0}^t\left(\rho \left(s\right)G_1-\frac{{\left({{\rho }^{\prime }}_{+}\left(s\right)\right)}^{\alpha +1}r\left(\sigma \left(t\right)\right)}{{\left(\alpha +1\right)}^{\alpha +1}{\left(\rho \left(s\right){\sigma }^{\prime }\left(s\right)\right)}^{\alpha }}\right)\text{d}s}=\infty$ (3)

且

$\underset{t\to \infty }{\lim \sup }{\displaystyle {\int }_{t_0}^t\left({\pi }^{\alpha }\left(s\right)G_2\left(s\right)-\frac{\eta }{\pi \left(s\right)r^{\frac{1}{\alpha }}\left(s\right)}\right)\text{d}s}=\infty$ (4)

其中

$G_1\left(t\right)=g\left(t\right){\left(1-p\left(\sigma \left(t\right)\right)\right)}^{\alpha }$， $G_2\left(t\right)=g\left(t\right){\left(1-p\left(t\right)\frac{\pi \left(\tau \left(\sigma \left(t\right)\right)\right)}{\pi \left(\sigma \left(t\right)\right)}\right)}^{\alpha }$， (5)

${{\rho }^{\prime }}_{+}\left(t\right)=\max \left\{0,{\rho }^{\prime }\left(t\right)\right\}$， $\eta ={\left(\frac{\alpha }{\alpha +1}\right)}^{\alpha +1}$， $\pi \left(t\right)={\displaystyle {\int }_t^{\infty }{\left(\frac{1}{r\left(s\right)}\right)}^{\frac{1}{\alpha }}\text{d}s}$ (6)

则方程 ${\left(r\left(t\right){\left|{\phi }^{\prime }\left(t\right)\right|}^{\alpha -1}{\phi }^{\prime }\left(t\right)\right)}^{\prime }+g\left(t\right){\left|x\left(\sigma \left(t\right)\right)\right|}^{\alpha -1}x\left(\sigma \left(t\right)\right)=0$ 振动。

受文献 [1] 和 [11] 启发，建立新的准则，主要对条件改进得到相对完善的结果，使得只要文献 [1] 中的 $\alpha >0,\beta >0$ 且 $\alpha ,\beta$ 可取偶数而不仅仅限于 $\alpha >0,\beta >0$ 的两个奇数商。

2. 主要结果

定理2.1：设(1)~(5)都成立，若存在函数 $\rho \in C\left(\left[t_0,\infty \right),R\right)$，满足

$\underset{t\to \infty }{\lim \sup }{\displaystyle {\int }_{t_0}^t\left(\rho \left(t\right)Q_1\left(t\right)-\frac{{\alpha }^{\alpha }}{{\beta }^{\alpha }{\left(\alpha +1\right)}^{\alpha +1}}\frac{{\left({{\rho }^{\prime }}_{+}\left(t\right)\right)}^{\alpha +1}r\left(\delta \left(t\right)\right)}{{\left(\rho \left(t\right){\delta }^{\prime }\left(t\right)\right)}^{\alpha }}\right)\text{d}t}=\infty$ (7)

且

$\underset{t\to \infty }{\lim \sup }{\displaystyle {\int }_{t_1}^t{\left(\left(\frac{1}{r\left(s\right)}\right){\displaystyle {\int }_{t_1}^t{\pi }^{\beta }\left(u\right)Q\left(u\right)\text{d}u}\right)}^{\frac{1}{\alpha }}\text{d}s}=\infty$ (8)

其中

$Q_1\left(t\right)=q\left(t\right){\left(1-p\left(\delta \left(t\right)\right)\right)}^{\beta }$， $Q\left(t\right)=q\left(t\right){\left(1-p\left(\delta \left(t\right)\right)\frac{\pi \left(\tau \left(\delta \left(t\right)\right)\right)}{\pi \left(\delta \left(t\right)\right)}\right)}^{\beta }$， (9)

${{\rho }^{\prime }}_{+}\left(t\right)=\max \left\{0,{\rho }^{\prime }\left(t\right)\right\}$， (10)

则方程(1)是振动的。

证明：设方程有一个非振动解 $x\left(t\right)$，不妨设当 $t\ge t_1$ 时，使得 $x\left(t\right)>0$， $x\left(\sigma \left(t\right)\right)>0$， $x\left(\tau \left(t\right)\right)>0$ 成立。

由方程(1)可得

${\left(r\left(t\right){\left|{\phi }^{\prime }\left(t\right)\right|}^{\alpha -1}{\phi }^{\prime }\left(t\right)\right)}^{\prime }=-q\left(t\right){\left|x\left(\delta \left(t\right)\right)\right|}^{\beta -1}x\left(\delta \left(t\right)\right)\le 0$ (11)

在 $\left[t_1,t\right]$ 上， $\left(r\left(t\right){\left|{\phi }^{\prime }\left(t\right)\right|}^{\alpha -1}{\phi }^{\prime }\left(t\right)\right)$ 是非増函数， ${\phi }^{\prime }\left(t\right)$ 是定号的，那么分 ${\phi }^{\prime }\left(t\right)>0$， ${\phi }^{\prime }\left(t\right)<0$ 两种情况展开讨论：

(I) 假设 ${\phi }^{\prime }\left(t\right)>0$， $t\ge t_1\ge t_0$，

因为 $\tau \left(t\right)\le t$，所以 $\phi \left(t\right)\ge \phi \left(\tau \left(t\right)\right)$，即有

$x\left(t\right)=\phi \left(t\right)-p\left(t\right)x\left(\tau \left(t\right)\right)\ge \phi \left(t\right)-p\left(t\right)\phi \left(\tau \left(t\right)\right)\ge \left(1-p\left(t\right)\right)\phi \left(t\right)$ (12)

将(12)代入(1)得，

${\left(r\left(t\right){\phi }^{\prime }{\left(t\right)}^{\alpha }\right)}^{\prime }\le -Q_1\left(t\right){\left(\phi \left(\delta \left(t\right)\right)\right)}^{\beta }\le 0$ (13)

考虑广义Riccati变换，得到

$\nu \left(t\right)=\frac{\rho \left(t\right)r\left(t\right){\left({\phi }^{\prime }\left(t\right)\right)}^{\alpha }}{{\phi }^{\beta }\left(\delta \left(t\right)\right)}>0$， $t\ge t_1$ (14)

对(14)式中分子，分母同时乘上 $\rho \left(t\right)$，再对t进行求导，结合(13)得

$\begin{array}{c}{\nu }^{\prime }\left(t\right)\le -\rho \left(t\right)q\left(t\right){\left(1-p\left(\delta \left(t\right)\right)\right)}^{\beta }+\frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\nu \left(t\right)-\frac{\beta \rho \left(t\right)r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }\left({\phi }^{\prime }\left(\delta \left(t\right)\right)\right){\delta }^{\prime }\left(t\right)}{{\phi }^{\beta +1}\left(\delta \left(t\right)\right)}\\ \le -\rho \left(t\right)Q_1\left(t\right)+\frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\nu \left(t\right)-\frac{\beta {\delta }^{\prime }\left(t\right)}{{\left(\rho \left(t\right)r\left(\delta \left(t\right)\right)\right)}^{\frac{1}{\alpha }}}{\nu }^{\frac{\alpha +1}{\alpha }}\left(t\right)\end{array}$ (15)

由经典不等式 $Ay-B{y}^{\frac{\alpha +1}{\alpha }}\le \frac{{\alpha }^{\alpha }}{{\left(\alpha +1\right)}^{\alpha +1}}\cdot \frac{A^{\alpha +1}}{B^{\alpha }}$，上式变成

$\begin{array}{c}{\nu }^{\prime }\left(t\right)\le -\rho \left(t\right)Q_1\left(t\right)+\frac{{\alpha }^{\alpha }}{{\left(\alpha +1\right)}^{\alpha +1}}{\left(\frac{{\rho }^{\prime }\left(t\right)}{\rho \left(t\right)}\right)}^{\alpha +1}\frac{\rho \left(t\right)r\left(\delta \left(t\right)\right)}{{\left(\beta {\delta }^{\prime }\left(t\right)\right)}^{\alpha }}\\ \le -\rho \left(t\right)Q_1\left(t\right)+\frac{{\alpha }^{\alpha }}{{\beta }^{\alpha }{\left(\alpha +1\right)}^{\alpha +1}}\frac{{\left({{\rho }^{\prime }}_{+}\left(t\right)\right)}^{\alpha +1}r\left(\delta \left(t\right)\right)}{{\left(\rho \left(t\right){\delta }^{\prime }\left(t\right)\right)}^{\alpha }}\end{array}$ (16)

对(16)式在 $\left[t_1,t\right]$ 上积分，即有

${\displaystyle {\int }_{t_1}^t{\nu }^{\prime }\left(t\right)\text{d}t}=\nu \left(t\right)-\nu \left(t_1\right)\le {\displaystyle {\int }_{t_1}^t\left(-\rho \left(t\right)Q_1\left(t\right)+\frac{{\alpha }^{\alpha }}{{\beta }^{\alpha }{\left(\alpha +1\right)}^{\alpha +1}}\frac{{\left({{\rho }^{\prime }}_{+}\left(t\right)\right)}^{\alpha +1}r\left(\delta \left(t\right)\right)}{{\left(\rho \left(t\right){\delta }^{\prime }\left(t\right)\right)}^{\alpha }}\right)\text{d}t}$ (17)

即

${\displaystyle {\int }_{t_1}^t\left(\rho \left(t\right)Q_1\left(t\right)-\frac{{\alpha }^{\alpha }}{{\beta }^{\alpha }{\left(\alpha +1\right)}^{\alpha +1}}\frac{{\left({{\rho }^{\prime }}_{+}\left(t\right)\right)}^{\alpha +1}r\left(\delta \left(t\right)\right)}{{\left(\rho \left(t\right){\delta }^{\prime }\left(t\right)\right)}^{\alpha }}\right)\text{d}t}\le \nu \left(t_1\right)-\nu \left(t\right)\le \nu \left(t_1\right)$ (18)

显然式(18)与条件(7)矛盾。

(II) ${\phi }^{\prime }\left(t\right)<0$， $t\ge t_1\ge t_0$，由于

$\phi \left(t\right)\ge -{\displaystyle {\int }_t^{\infty }{r}^{-\frac{1}{\alpha }}\left(t\right)r^{\frac{1}{\alpha }}\left(t\right){\phi }^{\prime }\left(t\right)\text{d}t}\ge -\pi \left(t\right)r^{\frac{1}{\alpha }}\left(t\right){\phi }^{\prime }\left(t\right)$ (19)

则

${\left(\frac{\phi \left(t\right)}{\pi \left(t\right)}\right)}^{\prime }\ge 0$

所以

$\frac{\phi \left(t\right)}{\pi \left(t\right)}\ge \frac{\phi \left(\tau \left(t\right)\right)}{\pi \left(\tau \left(t\right)\right)}$ (20)

即有

$x\left(t\right)\ge \phi \left(t\right)-p\left(t\right)\phi \left(\tau \left(t\right)\right)\ge \left(1-p\left(t\right)\frac{\pi \left(\tau \left(t\right)\right)}{\pi \left(t\right)}\right)\phi \left(t\right)$ (21)

由方程(1)和不等式(21)得

$-{\left(r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }\right)}^{\prime }\le -Q\left(t\right){\phi }^{\beta }\left(t\right)$ (22)

考虑到 $r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }$ 的单调性，则有

$r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }\ge r\left(t_1\right){\left(-{\phi }^{\prime }\left(t_1\right)\right)}^{\alpha }=\mu >0$， $t\ge t_1$ (23)

由式(19)可得

$\phi \left(t\right)\ge -\pi \left(t\right)r^{\frac{1}{\alpha }}\left(t\right){\phi }^{\prime }\left(t\right)\ge -\pi \left(t\right)r^{\frac{1}{\alpha }}\left(t_1\right){\phi }^{\prime }\left(t_1\right)=\pi \left(t\right){\mu }^{\frac{1}{\alpha }}$ (24)

又结合式(22)和式(24)可得

$-{\left(r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }\right)}^{\prime }\le -Q\left(t\right){\phi }^{\beta }\left(t\right)\le -{\pi }^{\beta }\left(t\right){\mu }^{\frac{\beta }{\alpha }}Q\left(t\right)$ (25)

对式(25)从 $t_1$ 到t积分

$\begin{array}{c}-r\left(t\right){\left(-{\phi }^{\prime }\left(t\right)\right)}^{\alpha }\le -r\left(t_1\right){\left(-{\phi }^{\prime }\left(t_1\right)\right)}^{\alpha }-{\mu }^{\frac{\beta }{\alpha }}{\displaystyle {\int }_{t_1}^t{\pi }^{\beta }\left(s\right)Q\left(s\right)\text{d}s}\\ \le -{\mu }^{\frac{\beta }{\alpha }}{\displaystyle {\int }_{t_1}^t{\pi }^{\beta }\left(s\right)Q\left(s\right)\text{d}s}\end{array}$ (26)

对上式从 $t_1$ 到t积分得

$\phi \left(t\right)\le \phi \left(t_1\right)-{\mu }^{\frac{\beta }{2\alpha }}{\displaystyle {\int }_{t_1}^t{\left(\frac{1}{r\left(t\right)}{\displaystyle {\int }_{t_1}^t{\pi }^{\beta }\left(u\right)Q\left(u\right)\text{d}u}\right)}^{\frac{1}{\alpha }}\text{d}s}$ (27)

即

${\displaystyle {\int }_{t_1}^t{\left(\left(\frac{1}{r\left(t\right)}\right){\displaystyle {\int }_{t_1}^t{\pi }^{\beta }\left(u\right)Q\left(u\right)\text{d}u}\right)}^{\frac{1}{\alpha }}\text{d}s}\le {\mu }^{-\frac{\beta }{2\alpha }}\phi \left(t_1\right)-{\mu }^{-\frac{\beta }{2\alpha }}\phi \left(t\right)<{\mu }^{-\frac{\beta }{2\alpha }}\phi \left(t_1\right)$ (28)

显然式(28)与条件(8)矛盾。所以假设不成立，即方程(1)是振动的。

3. 二阶中立型时滞微分方程的振动准则的应用

例：考虑方程：

${\left(t^4{\left({\left(x\left(t\right)+\frac{1}{9}\times x\left(\frac{t}{3}\right)\right)}^{\prime }\right)}^2\right)}^{\prime }+kt{x}^4\left(\frac{t}{3}\right)=0$ (29)

的解振动性

证明：取 $r\left(t\right)=t^4$， $\pi \left(t\right)={\displaystyle {\int }_t^{\infty }{\left(\frac{1}{t^4}\right)}^{\frac{1}{2}}\text{d}t}=\frac{1}{t}$， $p\left(t\right)=\frac{1}{9}$， $q\left(t\right)=k{t}^8$， $\delta \left(t\right)=\frac{t}{3}$， $\tau \left(t\right)=\frac{t}{3}$， $k>0$，

显然

$Q_1\left(t\right)=k{t}^8{\left(1-\frac{1}{9}\right)}^4={\left(\frac{8}{9}\right)}^{4}k{t}^8$， $Q\left(t\right)=k{t}^8{\left(1-\frac{1}{9}\times \frac{9}{t}\times \frac{t}{3}\right)}^4=\frac{16}{81}k{t}^8$， (30)

则有

${\displaystyle {\int }_{t_0}^t\left({\left(\frac{8}{9}\right)}^{4}k{t}^8\right)\text{d}t}=\frac{2048}{729}k\left(t^9-t_0^9\right)$ (31)

$\begin{array}{c}{\displaystyle {\int }_{t_0}^t{\left(\left(\frac{1}{t^4}\right){\displaystyle {\int }_{t_0}^t{t}^{-4}\frac{16}{81}k{t}^8\text{d}t}\right)}^{\frac{1}{2}}\text{d}t}={\displaystyle {\int }_{t_0}^t{\left(\left(\frac{1}{t^4}\right)\left(\frac{16}{405}k\left(t^5-t_0^5\right)\right)\right)}^{\frac{1}{2}}\text{d}t}\\ =\frac{8}{405}k\left(t^2-t_0^2\right)+\frac{16}{1215}k{t}_0^5\left(t^{-5}-t_0^{-5}\right)\end{array}$ (32)

显然当 $t\to \infty$ 时，方程(29)条件(7)和(8)成立，故方程(29)是振动的。

注：本例子是建立在文献 [1] 中方程的基础上，主要是针对于 $\alpha ,\beta$ 是偶数情况下，文献 [1] 的定理准则不能判断方程(29)的振动性。本例子检验了 $\alpha =2,\beta =4$ 时，仍可判断方程(29)是振动的，体现了本准则的优越性。

基金项目

教育部产学合作协同育人项目(No. 201901104007)，茂名市科技计划(No. 2022056)。

NOTES

^*通讯作者。