据文献可知,起源于偏微分方程的Emden-Fowler型泛函微分方程在科学理论研究和工程技术应用中发挥着重要作用.带有阻尼项后的二阶Emden-Fowler方程更广泛,已被应用在天体物理、气体动力学、物理化学、高速计算机无损传输、智能机器人设计和神经动力系统理论与工程等高新技术领域中^[1-4].

本文研究一类带阻尼项的二阶非线性中立型广义Emden-Fowler时滞微分方程

(1.1) $\begin{equation}(r(t)\phi_\alpha(z'(t)))'+g(t)\phi_\alpha(z'(t))+f(t, \phi_\beta(x(\sigma(t))))=0, ~~t\geq t_0\geq 0 \end{equation}$

的振动性,其中$z(t)=x(t)+p(t)x(\tau(t));~ \phi_\alpha(u)=|u|^{\alpha-1}u, u\in {\Bbb R} ;~ r\in C^1([t_0, \infty), (0, \infty));$$ g, p, \tau, \sigma \in C([t_0, \infty), {\Bbb R} );$$\tau(t)\leq t, \sigma(t)\leq t, \lim\limits_{t\to \infty}\tau(t)=\lim\limits_{t\to \infty}\sigma(t)=\infty;$$f\in C([t_0, \infty)\times {\Bbb R} , {\Bbb R} );$$ \alpha>0, \beta>0$为常数.

如通常所述,我们称$ x\in C([T_x, \infty ), {\Bbb R} )$是方程(1.1)的解,其中$ T_x=\min \{ \tau(T), \sigma(T)\}, $$T\geq t_0$,是指$ r\phi_\alpha(z')\in C^1([T_x, \infty ), {\Bbb R})$且在$ [T_x, \infty)$上满足方程(1.1).本文仅考虑方程(1.1)的非平凡解,即方程(1.1)在$ [ T_x, \infty)$上的解$ x(t)$满足$ \sup \{ |x(t)|:t\geq T\}>0, $对一切$ T\geq T_x$成立.方程(1.1)的解称为振动的,如果它有任意大的零点,否则,称它为非振动的.若方程(1.1)的一切解均振动,则称方程(1.1)振动.

近十几年来,对于二阶泛函微分方程振动性理论的研究,又产生了一大批新的研究成果,可参见文献[5-35]及其参考文献,但对于带有阻尼项的二阶泛函微分方程振动性的研究结果较少且几乎其阻尼项系数(例如$g(t)\geq0$)均为非负.那么这一限定是否可在不失物理意义的前提下进一步放宽,并可得到方程(1.1)的振动结果呢？为此,我们先对近年来研究的带阻尼和不带阻尼项的经典方程简要分析如下.

Liu等^[7]、曾云辉等^[8]、Li等^[9-10]、Luo等^[11]和吴英柱等^[12]分别研究了二阶中立型广义Emden-Fowler微分方程

(1.2) $(r(t)|z'(t)|^{\alpha-1}z'(t))'+q(t)|x(\sigma(t))|^{\beta-1}x(\sigma(t))=0, ~~t\geq t_0>0$

的振动性和渐近性,其中$z(t)=x(t)+p(t)x(\tau(t)), \alpha\geq\beta>0$或$\beta\geq \alpha=1, $$0\leq p(t)\leq 1, $$q(t)\geq 0, r(t)>0, r'(t)\geq 0, $$\tau(t)\leq t, 0<\sigma(t)\leq t, \sigma'>0, $$ \lim\limits_{t\to \infty}\tau(t)=\lim\limits_{t\to \infty}\sigma(t)=\infty. $

Agarwal等^[13]、Grace等^[14]和Bohner等^[15]先后研究了二阶中立型Emden-Fowler时滞微分方程

(1.3) $(r(t)((x(t)+p(t)x(\tau(t)))')^\alpha)'+q(t)x^\gamma (\sigma(t))=0, ~~t\geq t_0, $

分别对$\gamma\geq \alpha, \gamma<\alpha, \gamma<\alpha=1$和$\gamma=\alpha$等情况给出了多个振动定理,其中$ \alpha, \gamma>0$是两正奇数之比的常数.

Wang等^[16]、Sun等^[17]、罗红英等^[21]和吴英柱^[22]分别将方程(1.3)扩展到方程

(1.4) $(r(t)|z'(t)|^{\alpha-1}z'(t))'+f(t, x(\sigma(t)))=0, ~~t\geq t_0, $

其中$ z(t)=x(t)+p(t)x(\tau(t)), ~r, p\in C([t_0, \infty), {\Bbb R}), $$ 0\leq p(t)\leq1, \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)>0, $$ \lim\limits_{t\to \infty} \tau(t)=\lim\limits_{t\to \infty}\sigma(t)=\infty, $$ f\in C([t_0, \infty)\times {\Bbb R}, {\Bbb R}), uf(t, u)\geq 0, $$f(t, u)/u^\beta\geq q(t)\geq0, u\neq0, $$1<\beta \leq \alpha $为常数.

显然,方程(1.2)-(1.4)均不带阻尼项.对于带阻尼项的二阶Emden-Fowler型方程振动性与渐近性的研究成果也已相继出现,例如Erbe等^[23]和Zhang等^[24]先后研究了时间尺度上二阶阻尼时滞动力方程

(1.5) $(a(t)(x^\Delta(t))^\gamma)^\Delta+r(t)(x^\Delta(t))^\gamma+q(t)x^\gamma(g(t))=0, ~~t\in [t_0, \infty)_{\mathbb{T}}$

的振动性,其中$\gamma>0$是两正奇数之比的常数, $r(t)\geq 0$右稠连续.

Saker等^[25], Rogovchenko和Tuncay ^[26]也分别对二阶阻尼动力方程

(1.6) $(a(t)x^\Delta(t))^\Delta+r(t)x^\Delta(t)+q(t)f(x(t))=0, ~~t\in [t_0, \infty)_{\mathbb{T}}$

给出了不同的振动性定理,其中$ r(t)\geq0$右稠连续.

方程(1.5)先后被张全信等^[27-29],孙一冰等^[30]和杨甲山、李同兴^{[31, 32]}等许多学者拓展为更一般的形式

(1.7) $\begin{equation}(a(t)|z^\Delta(t)|^{\gamma-1}z^\Delta(t))^\Delta+p(t)|z^\Delta(t)|^{\gamma-1}z^\Delta(t) +q(t)|x(\delta(t))|^{\beta-1}x(\delta(t))=0, \end{equation} $

其中$t\in [t_0, \infty)_{\mathbb{T}}, z(t)=x(t)+r(t)x(\tau(t))$, $ \gamma, \beta>0$为常数, $a(t), r(t), p(t), q(t)$都是正值右稠连续函数.

李文娟等^[33]将方程(1.2)的类型拓展到了带阻尼项的中立型时滞微分方程

(1.8) $\begin{equation}(r(t)|z'(t)|^{\alpha-1}z'(t))' +p(t)|z'(t)|^{\alpha-1}z'(t)+q(t)|x(\sigma(t))|^{\beta-1}x(\sigma(t))=0, ~~t\geq t_0>0, \end{equation}$

其中$z(t)=x(t)+g(t)x(\tau(t)), r\in C^1([t_0, \infty), (0, \infty)), p, q\in C([t_0, \infty), [0, \infty)), $$ \alpha>0, \beta>0$为常数,在$ 0\leq g(t)\leq 1, p(t)\geq0, q(t)\geq0, r'(t)>0$等基本假设条件下,获得了多个振动定理,推广了上述有关文献的部分结果.

可见,方程(1.2)-(1.8)所带的阻尼项系数(例如方程(1.5), (1.6)中的$r(t)$和方程(1.7), (1.8)中的$p(t)$)都是非负函数.但是,我们发现,当方程(1.1)中的阻尼项系数$ g(t)\leq0$时还是可以具有物理意义的.事实上,由文献[34]知,当$ r(t)>0, r'(t)\geq 0$时,二阶微分方程

$(r(t)\phi_\alpha(x'(t)))'+g(t)\phi_\alpha(x'(t))+f(t, x(t))=0$

与二阶微分方程

$(\phi_\alpha(x'(t)))'+\frac{r'(t)+g(t)}{r(t)}\phi_\alpha(x'(t))+\frac{f(t, x(t))}{r(t)}=0$

等价.而又当$r'(t)+g(t)\geq 0$时,后一方程是阻尼系数为$ (r'(t)+g(t))/r(t)\geq 0$的有阻尼简谐振动系统$(\alpha=1)$数学模型^[36]的推广,有着实在的物理意义.顺便指出,文献[35]研究了带阻尼的分数阶微分方程

(1.9) $ \begin{equation}[r(t)(D_{0+}^\alpha y)(t)]'+p(t)(D_{0+}^\alpha y)(t)+q(t)f\left(\int_0^t (t-s)^{-\alpha}y(s){\rm d} s\right)=0 \end{equation}$

的振动性,其中$\alpha\in(0, 1), D_{0+}^\alpha y$是$y$的$\alpha$阶导数,在其假设条件$A_1), A_2)$及

$\omega(t)=\exp \left(\int_{t_0}^t \frac{r'(s)+p(s)}{r(s)}{\rm d} s\right) , \quad \int_{t_0}^\infty\frac{1}{\omega(t)}{\rm d} t=\infty$

之下建立了方程(1.9)的振动定理.显然这里有$ (r'(t)+p(t))/r(t)<0$的情况(如其例4.1),但这时的“阻尼”已不再是实际物理意义的阻尼了^[36].

因此,为使方程(1.1)所带的阻尼系数广泛而不失物理意义,本文考虑以下假设条件

${\rm (H_1)}\quad r'(t)+g(t)\geq 0 , 0\leq p(t)\leq 1 $且非$ p(t)\equiv 1$;

${\rm (H_2)}\quad \sigma'(t)\geq0$;

${\rm (H_3)}~~ $存在不恒为零的$q\in C([t_0, \infty), [0, \infty)), $满足$ f(t, u)/u\geq q(t)\geq 0, u\neq0, t\geq t_0$.

本文将引入指数函数变换并借助于Riccati变换、积分平均和不等式技巧研究方程(1.1)的振动性和渐近性,建立新的振动准则,推广、改进和统一文献[7-33]及其引文中关于带或不带阻尼项方程的相应结果.

首先,引入指数函数变换

(1.10) $\begin{equation}\varphi(t)=\exp\left(\int_{t_0}^{t} g(u)/r(u){\rm d} u\right), \end{equation}$

用$\varphi(t)$乘以方程(1.1)的两端,则方程(1.1)变为等价的不显含阻尼项的方程

$(R(t)| z'(t)|^{\alpha-1}z'(t))'+\varphi(t)f(t, |x(\sigma(t))|^{\beta-1}x(\sigma(t)))=0, ~t\ge t_0, ~~~~~~~~~~~~~~~~~~~~(E_0)$

其中$R(t)=r(t)\varphi(t).$

我们通过方程$(E_0)$,在两种情况

(1.11) $\begin{equation}\int_{t_0}^{\infty} (1/R(t))^{1/\alpha}{\rm d} t=\infty, \end{equation}$

(1.12) $\begin{equation}\int_{t_0}^{\infty} (1/R(t))^{1/\alpha}{\rm d} t<\infty\end{equation}$

下,分别讨论方程(1.1)的振动性和渐近性,首先给出以下几个重要引理.

引理1.1  设条件${\rm (H_3)}$和$(1.11)$成立,如果$x(t)$是方程$(1.1)$的最终正解,则$z'(t)>0$.

证  因为$x(t)$是方程(1.1)在$[t_0, \infty)$上的最终正解,则存在$t_1\ge t_0, $使得当$t\ge t_1$时有$x(t)>0, x(\tau(t))>0, x(\sigma(t))>0, $由条件${\rm (H_3)}$和$(E_0), $我们得到

(1.13) $ \begin{equation} z(t)\geq x(t)>0, (\varphi(t)r(t)|z'(t)|^{\alpha-1} z'(t))'\leq 0, t\geq t_1. \end{equation}$

因此$\varphi(t)r(t)|z'(t)|^{\alpha-1} z'(t)$是非增函数且$z'(t)$最终保号,于是$z'(t)$仅有两种可能.我们断言$z'(t)>0, t>t_1$.否则,假设$z'(t)\leq0, t>t_1$.由(1.13)式知,存在常数$K>0$,使得

$-R(t)(-z'(t))^\alpha\le-R(t_1)(-z'(t_1))^\alpha=-K<0, t>t_1, $

$z'(t)\le-K^{1/\alpha}(R(t))^{-1/\alpha}, t>t_1.$

从$t_1$到$t$积分上式,我们得到

$z(t)\le z(t_1)-K^{1/\alpha}\int_{t_1}^{t} (R(s))^{-1/\alpha}{\rm d}s, t>t_1.$

上式中令$t\to\infty, $由条件(1.11)得$z(t)\to-\infty.$此与(1.13)式矛盾,故结论成立.

引理1.2  设条件${\rm (H_1)}$-${\rm (H_3)}$和$(1.11)$成立,如果$x(t)$是方程$(1.1)$的最终正解,则存在$T\geq t_0$和某个常数$m_\lambda \in(0, 1]$(特别,当$\alpha=\beta$时$m_\lambda=1$),使得不等式

(1.14) $\begin{equation} w'(t)+{Q}(t)+\frac{\lambda m_\lambda \sigma'(t)}{R^{\frac{1}{\lambda}}(t)}w^{\frac{\lambda+1}{\lambda}}(t)\le0, ~t\ge T \end{equation} $

成立,其中

$w(t)=\frac{R(t)(z'(t))^\alpha}{z^\beta(\sigma(t))}, t\ge T, $

$\lambda=\min \{\alpha, \beta\}, Q(t)=\varphi(t)q(t)[1-p(\sigma(t))]^\beta, ~R(t)$由方程$(E_0)$定义.

证  因为$x(t)$是方程(1.1)在$[t_0, \infty)$上的最终正解,则存在$T\ge t_0$,使得当$t\ge T$时,有$x(t)>0, $$x(\tau(t))>0, x(\sigma(t))>0$.由引理1.1,不妨设这时也有$z'(t)>0.$又由条件${\rm (H_3)}$知方程(1.1)的等价方程$(E_0)$变为

$(R(t)(z'(t))^\alpha)'+\varphi(t)q(t)x^\beta(\sigma(t))\leq0, ~~t\ge T.$

由于$z(t)=x(t)+p(t)x(\tau(t))$,则

$x(t)=z(t)-p(t)x(\tau(t)), ~~z(\tau(\sigma(t)))\leq z(\sigma(t)), $

$x(\sigma(t))=z(\sigma(t))-p(\sigma(t))x(\tau(\sigma(t)))\geq z(\sigma(t))[1-p(\sigma(t))], ~t\ge T, $

将其代入上式,我们有

$\begin{equation} (R(t)(z'(t))^\alpha)'+Q(t)z^\beta(\sigma(t))\leq0, ~~t\ge T, \end{equation} $

其中$Q(t)=\varphi(t)q(t)[1-p(\sigma(t))]^\beta.$从而,由$w(t)$的定义又得

(1.16) $ \begin{equation} w'(t)\leq -Q(t)-R(t)(z'(t))^\alpha\frac{\beta\sigma'(t)z'(\sigma(t))}{z^{\beta+1}(\sigma(t))}, ~~t\geq T. \end{equation}$

当$\alpha\le\beta$时,由(1.13)式,我们得到$R(t)(z'(t))^\alpha$为减函数,即

$R(t)(z'(t))^\alpha\le R(\sigma(t))(z'(\sigma(t)))^\alpha, $

亦即

$z'(\sigma(t))\ge\left(\frac{R(t)}{R(\sigma(t))}\right)^{\frac{1}{\alpha}}z'(t), t\ge T.$

将上式代入(1.16)式,可得

$\begin{eqnarray} w'(t)&\le& -Q(t)-\frac{\beta\sigma'(t)R(t)(z'(t))^{\alpha+1}}{z^{\beta+1}(\sigma(t))}\left(\frac{R(t)}{R(\sigma(t))}\right)^{\frac{1}{\alpha}}\nonumber\\&\le&-Q(t)-\frac{\beta\sigma'(t)z^{\frac{\beta}{\alpha}-1}(\sigma(t))}{R^{\frac{1}{\alpha}}(\sigma(t))}w^{\frac{\alpha+1}{\alpha}}(t), ~t\geq T.\nonumber\end{eqnarray}$

又知

$z^{\frac{\beta-\alpha}{\alpha}}(\sigma(t))\ge z^{\frac{\beta-\alpha}{\alpha}}(\sigma(T))\geq m_\alpha, ~~t\geq T, $

其中$m_\alpha=\min {\{z^{\frac{\beta-\alpha}{\alpha}}(\sigma(T)), 1\}}$.代入上式并注意到$\sigma(t)\leq t $和条件${\rm (H_1)}$,有$R'(t)=\varphi(t)(g(t)+r'(t))\geq0$,从而$R(\sigma(t))\leq R(t)$,可得

(1.17) $\begin{equation}w'(t)\leq -Q(t)-\frac{\beta m_\alpha \sigma'(t)}{R^{\frac{1}{\alpha}}(t)}w^{\frac{\alpha+1}{\alpha}}(t).\end{equation}$

当$\alpha >\beta $时,又由(1.13)式知

$(R(t)(z'(t))^\alpha)'=R'(t)(z'(t))^\alpha+\alpha R(t)(z'(t))^{\alpha-1}z''(t)\leq 0.$

再由$R'(t)\geq 0$及上式,可得$z''(t)\le0$,从而, $z'(t)$单调减,于是,有

$z'(t)\le z'(\sigma(t)), ~~(z'(t))^{1-\frac{\alpha}{\beta}}\ge (z'(T))^{1-\frac{\alpha}{\beta}}\ge m_\beta, ~t\geq T, $

其中$m_\beta=\min\{(z'(T))^{1-\frac{\alpha}{\beta}}, 1 \}$.再将上式代入(1.16)式,可得

(1.18) $\begin{equation}w'(t)\le-Q(t)-\frac{\beta m_\beta \sigma'(t)}{R^{\frac{1}{\beta}}(t)}w^{\frac{\beta+1}{\beta}}(t).\end{equation}$

综合(1.17)和(1.18)式,得

$w'(t)\le-Q(t)-\frac{\lambda m_\lambda \sigma'(t)}{R^{\frac{1}{\lambda}}(t)}w^{\frac{\lambda+1}{\lambda}}(t), ~~t\geq T, $

其中$m_\lambda=\min{\{m_\alpha, m_\beta \}}, \alpha=\beta$时, $m_\lambda=1, $$\lambda=\min{\{\alpha, \beta\}}.$故(1.14)式成立.

引理1.3  设$A>0, B\ge0, \lambda>0$且均为常数,则当$u>0$时,有

(1.19) $\begin{equation}Bu-Au^{\frac{\lambda+1}{\lambda}}\le \frac{\lambda^\lambda}{(\lambda+1)^{\lambda+1}}\frac{B^{\lambda+1}}{A^\lambda}.\end{equation}$

引理1.4  设$X>0, Y>0, \lambda>0$为任意实数,则有

(1.20) $\begin{equation}X^\lambda+Y^\lambda\geq C_\lambda (X+Y)^\lambda, ~~~~C_\lambda=\left\{\begin{array}{ll}1, ~&0<\lambda\leq 1, \\2^{1-\lambda}, ~&\lambda > 1, \end{array}\right.\end{equation}$

当且仅当$X=Y, \lambda\geq1$时第一式等号成立.

引理1.3与引理1.4的证明从略.

定理2.1  设条件${\rm (H_1)}$-${\rm (H_3)}$和$(1.11)$满足.如果存在函数$\rho(t)\in C^1([t_0, \infty), (0, \infty))$,使得对任意常数$m\in(0, 1]$ (当$\alpha=\beta$时, $m=1)$,有

(2.1) $\begin{equation}\int_{t_0}^{\infty}\left[\rho(t)Q(t)-\frac{R(t)(\rho'(t))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\rho(t)\sigma'(t))^\lambda}\right]{\rm d} t=\infty\end{equation}$

成立,其中$\lambda=\min\{\alpha, \beta\}, ~~Q(t)=\varphi(t)q(t)[1-p(\sigma(t))]^\beta$,则方程$(1.1)$振动.

证  假设$x(t)$是方程(1.1)的非振动解,不失一般性,设$x(t)$为$[t_0, \infty)$上的最终正解($x(t)<0$的情况类似可证),则由引理1.2知,存在某个常数$m_\lambda \in(0, 1]$,使得(1.14)式成立.对其两边同时乘以$\rho(t)$,再对$ t$从$T$到$t$积分,可得

$\begin{eqnarray}\int_{T}^{t}\rho(s)Q(s){\rm d} s &\le&-\int_{T}^{t}\rho(s)w'(s){\rm d} s-\int_{T}^{t}\rho(s)\frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}w^{\frac{\lambda+1}{\lambda}}(s){\rm d} s \nonumber\\&\le&\rho(T)w(T)-\rho(t)w(t)+\int_{T}^{t} \left[\rho'(s)w(s)-\rho(s)\frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}w^{\frac{\lambda+1}{\lambda}}(s)\right]{\rm d} s, t\ge T.\nonumber\end{eqnarray}$

由上式及引理1.3可得

$\begin{eqnarray}\int_{T}^{t}\rho(s)Q(s){\rm d} s &\le&\rho(T)w(T)+\int_{T}^{t}\frac{\lambda^\lambda}{(\lambda+1)^{\lambda+1}}(\rho'(s))^{\lambda+1}\frac{R(s)}{(m_\lambda\sigma'(s)\rho(s))^\lambda}{\rm d} s \nonumber\\&\le&\rho(T)w(T)+\int_{T}^{t} \frac{(\rho'(s))^{\lambda+1}R(s)}{(\lambda+1)^{\lambda+1}(m_\lambda\sigma'(s)\rho(s))^\lambda}{\rm d} s, t\ge T.\nonumber\end{eqnarray}$

可以看出

$\int_{t_0}^{\infty}\left[\rho(t)Q(t)-\frac{R(t)(\rho'(t))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m_\lambda\rho(t)\sigma'(t))^\lambda}\right]{\rm d} t<\infty.$

这与定理条件(2.1)式矛盾,故方程(1.1)振动.

推论2.1  设$(1.11)$式成立,且$ g(t)\geq0, r'(t)>0, \sigma'(t)\geq 0, f(t, u)=q(t)u.$若存在函数$\rho(t)\in C^1([t_0, \infty), (0, \infty))$,使得对任意常数$m\in(0, 1]$ (当$\alpha=\beta$时, $m=1)$,有

$\int_{t_0}^{\infty}\left[\rho(s)Q(s)-\frac{R(s)(\rho'(s))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\rho(s)\sigma'(s))^\lambda}\right]{\rm d}t=\infty$

成立,其中$\lambda=\min\{\alpha, \beta\}, ~~Q(t)=\varphi (t)q(t)[1-p(\sigma(t))]^\beta, $则方程$(1.1)$振动.

注2.1  推论$2.1$即文献[33]中定理$1$,因此推论$2.1$既包含和统一了文献[8](其中对应本文$g(t)=0)$的定理$2.1$和定理$3.1$,又改进了文献[7]仅关于$\alpha\geq\beta>0$时方程的解振动或渐近于零的定理$2.1$.特别地,当$g(t)<0$时本文定理$2.1$的结论对于本文所列文献均为新的.

注2.2  当取$\rho(t)$为正常数时,定理$2.1$的振动条件$(2.1)$成为

$\int_{t_0}^\infty \varphi(t)q(t)[1-p(\sigma(t))]^\beta {\rm d} t=\infty, $

其中$\varphi(t)$由$(1.10)$式确定.因此,定理$2.1$是著名的Leighton振动定理^[37](关于方程$(r(t)x'(t))'+q(t)x(t)=0)$的推广.

例2.1  考虑方程

(2.2) $\begin{eqnarray}&&[t|z'(t)|^{\alpha-1}z'(t)]'-|z'(t)|^{\alpha-1}z'(t)\nonumber\\&&+t^\beta|x(\sigma(t))|^{\beta-1}x(\sigma(t))[1+\ln(1+(|x(\sigma(t))|^{\beta-1}x(\sigma(t)))^4)]=0, ~~t\geq1\end{eqnarray}$

的振动性,其中$z(t)=x(t)+(1-1/t)x(t-1), \sigma(t)=t/2;\alpha>0, \beta>0$为常数.

对应于定理2.1,这里有

$r(t)=t, ~~ p(t)=1-\frac{1}{t}, ~~g(t)=-1, ~~\tau(t)=t-1, ~~\sigma(t)=\frac{t}{2}, ~~q(t)=t^\beta, ~~t_0=1, $

则有

$\varphi(t)=\exp\left(\int_{1}^{t}\frac{-1}{s}{\rm d}s\right)=\frac{1}{t}, ~~R(t)=\varphi(t)r(t)=1, $

$Q(t)=\varphi(t)q(t)[1-p(\sigma(t))]^\beta=t^{-1}t^\beta[1-(1-\frac{2}{t})]^\beta=2^\beta t^{-1} , $

显然满足条件${\rm (H_1)}-{\rm (H_3)}$和(1.11)式.若取$\rho(t)=1, $则有$\rho'(t)=0, $所以,有

$\int_{t_0}^{\infty}\left[\rho(t)Q(t)-\frac{R(t)(\rho'(t))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\rho(t)\sigma'(t))^\lambda}\right]{\rm d} t=\int_{t_0}^{\infty}\frac{2^\beta}{t}{\rm d} t=\infty.$

因此(2.1)式成立,故由定理2.1可得方程(2.2)振动.

定理2.2  设条件${\rm (H_1)}$-${\rm (H_3)}$和$(1.11)$成立.若对任意常数$m\in(0, 1]$(当$\alpha=\beta$时$m=1)$均有

(2.3) $\begin{equation}\liminf\limits_{t\to \infty}\frac{1}{Q_1(t)}\int_{t}^{\infty} [Q_1(s)]^{\frac{\lambda+1}{\lambda}}\frac{\lambda m\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}{\rm d} s>\frac{\lambda}{(\lambda+1)^{\frac{\lambda+1}{\lambda}}}\end{equation}$

成立,其中$\lambda=\min\{\alpha, \beta\}, ~~Q_1(t)=\int_{t}^{\infty} Q(s){\rm d}s, ~~Q(t)=\varphi(t)q(t)[1-p(\sigma(t))]^\beta, $则方程$(1.1)$振动.

证  假设$x(t)$是方程(1.1)的非振动解.不失一般性,设$x(t)$为$[t_0, \infty)$上的最终正解($x(t)<0$的情况类似可证),则由引理1.2知(1.14)成立,即存在某个$m_\lambda \in (0, 1]$$(\alpha=\beta$时$m_\lambda=1)$和$T\geq t_0, $使得

$w'(t)+Q(t)+\frac{\lambda m_\lambda \sigma'(t)}{R^{\frac{1}{\lambda}}(t)}w^{\frac{\lambda+1}{\lambda}}(t)\le0, ~~t\ge T.$

再从$t$到$\infty$积分上式,得

$w(\infty)-w(t)+\int_{t}^{\infty} Q(s){\rm d} s+\int_{t}^{\infty} \frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}w^{\frac{\lambda+1}{\lambda}}(s){\rm d} s\le0, t\ge T.$

从而,有

$w(t)\ge\int_{t}^{\infty} Q(s){\rm d} s+\int_{t}^{\infty} \frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}w^{\frac{\lambda+1}{\lambda}}(s){\rm d} s, ~~t\ge T.$

令$Q_1(t)=\int_{t}^{\infty} Q(s){\rm d} s, $将上式两边同除以$Q_1(t)$,得

(2.4) $\begin{equation}\frac{w(t)}{Q_1(t)}\ge 1+\frac{1}{Q_1(t)}\int_{t}^{\infty} \frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}[Q_1(s)]^{\frac{\lambda+1}{\lambda}}[\frac{w(s)}{Q_1(s)}]^{\frac{\lambda+1}{\lambda}}{\rm d} s, t\ge T.\end{equation}$

再令

$\eta=\inf\limits_{t\geq T}\frac{w(t)}{Q_1(t)}, ~~\delta=\frac{\lambda}{(\lambda+1)^{\frac{\lambda+1}{\lambda}}}, $

由条件(2.3)式,可得

$\liminf\limits_{t\to \infty}\frac{1}{Q_1(t)}\int_{t}^{\infty} [Q_1(s)]^{\frac{\lambda+1}{\lambda}}\frac{\lambda m_\lambda\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}{\rm d} s>\frac{\lambda}{(\lambda+1)^{\frac{\lambda+1}{\lambda}}}=\delta, $

因此,由上式和(2.4)式,得

$\eta>1+\eta^{\frac{\lambda+1}{\lambda}}\delta.$

在引理1.3的不等式(1.19)中,取$A=\delta, B=1, u=\eta, $得

$\eta-\eta^{\frac{\lambda+1}{\lambda}}\delta\le\frac{\lambda^\lambda}{(\lambda+1)^{\lambda+1}}\frac{1}{\delta^\lambda}=1, $

这与上式结论$\eta>1+\eta^{\frac{\lambda+1}{\lambda}}\delta$矛盾.因此,方程(1.1)振动.

推论2.2  设条件${\rm (H_1)}, {\rm (H_3)}$和$(1.11)$成立,又设$f(t, u)=q(t)u.$若对任意常数$m\in(0, 1]$ (当$\alpha=\beta$时$m=1)$均有

$\lim\limits_{t\to \infty}\inf\frac{1}{Q_1(t)}\int_{t}^{\infty} [Q_1(s)]^{\frac{\lambda+1}{\lambda}}\frac{\lambda m\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}{\rm d} s>\frac{\lambda}{(\lambda+1)^{\frac{\lambda+1}{\lambda}}}$

成立,其中$\lambda, Q_1(t)$由$(2.3)$式定义,则方程$(1.1)$振动.

注2.3  显然,文献[7, 33]中的定理$2$和文献[8]中的定理3.2均为本文推论2.2的特例,其中文献[7-8]中对应的阻尼系数$g(t)=0$,文献[33]中对应方程$(1.8)$的阻尼系数$p(t)\geq0$.再者,文献[7-8]是关于$\alpha\geq\beta >0$或$\beta\geq\alpha >0$时方程的解振动或渐近收敛于零的结果,而我们的结果是方程更广泛且一切解为振动的.由下例可见本文所列文献的结果均对其无效.

例2.2  考虑方程

(2.5) $\begin{eqnarray}&&[t^{1+\lambda/2}|z'(t)|^{\alpha-1}z'(t)]'-(1+\frac{\lambda}{2})t^{\lambda/2}|z'(t)|^{\alpha-1}z'(t)\nonumber\\&&+2^\beta [(|x(t/2)|^{\beta-1}x(t/2))^5+|x(t/2)|^{\beta-1}x(t/2)]=0, ~~t\geq 1\end{eqnarray}$

的振动性,其中$z(t)=x(t)+\frac{1}{2}x(t-1), \alpha >0, \beta >0$为常数, $\lambda=\min \{ \alpha, \beta \}$.

易知,这里有$r(t)=t^{1+\lambda/2}, ~~g(t)=-(1+\frac{\lambda}{2})t^{\lambda/2}<0, ~~\tau(t)=t-1, ~~\sigma(t)=\frac{t}{2}, ~~q(t)=2^\beta, $从而$\sigma'(t)=\frac{1}{2}, ~~g(t)+r'(t)=0.$故方程(2.5)满足假设条件${\rm (H_1)}-{\rm (H_3)}.$又易知

$\varphi(t)=\exp\left(\int_{1}^{t}\frac{g(u)}{r(u)}{\rm d} u\right)=t^{-(1+\lambda/2)}, \quad R(t)=\varphi(t)r(t)=1, $

所以, (1.11)式满足.又由于

$Q_1(t)=\int_t^\infty Q(s){\rm d} s=\int_t^\infty \varphi(s)q(s)[1-\frac{1}{2}]^\beta {\rm d} s=\int_t^\infty s^{-(1+\lambda/2)} {\rm d} s=\frac{2}{\lambda}t^{-\lambda/2}, $

所以,对于任意的$m\in (0, 1]$,有

$\begin{eqnarray*}\frac{1}{Q_1(t)}\int_{t}^{\infty} [Q_1(s)]^\frac{\lambda+1}{\lambda}\frac{\lambda m\sigma'(s)}{R^{\frac{1}{\lambda}}(s)}{\rm d} s&=&\frac{\lambda^2 m t^{\frac{\lambda}{2}}}{4}(\frac{2}{\lambda})^{\frac{\lambda+1}{\lambda}}\int_{t}^{\infty}s^{-\frac{\lambda+1}{2}}{\rm d} s\\&=&\left\{ \begin{array}{ll} \infty, &0< \lambda\leq 1, \\[2mm] \frac{\lambda^2m}{2(\lambda-1)}(\frac{2}{\lambda})^{\frac{\lambda+1}{\lambda}}t^{\frac{1}{2}}, &\lambda>1. \end{array}\right.\end{eqnarray*}$

故条件(2.3)式满足.因此,由定理2.2知,方程(2.5)振动.

注2.4  推论2.2,从而定理2.2完全包含了文献[33]的定理2.

下面再讨论当正则条件(1.11)不成立,即非正则条件(1.12)成立时方程(1.1)的振动性.

定理2.3  设条件${\rm (H_1)}$-${\rm (H_3)}$和$(1.12)$满足且有$ \tau'(t)\geq 0, p'(t)\geq 0$.如果存在函数$\rho(t), $$\eta(t)\in C^1([t_0, \infty), (0, \infty)), $$\eta'(t)\geq 0$,使得对任意常数$m\in(0, 1]$ (当$\alpha=\beta$时$m=1)$,有$(2.1)$式和

(2.6) $\begin{equation}\int_{t_0}^{\infty}\left[\frac{1}{\eta (t)R(t)} \int_{t_0}^t \eta(s)\varphi(s)q(s){\rm d} s\right]^{\frac{1}{\alpha}}{\rm d} t=\infty\end{equation}$

成立,其中$\varphi(t), R(t)$分别由$(1.10)$式, $(E_0)$式定义.则方程$(1.1)$的每一解$x(t)$振动或

$\lim\limits_{t\to\infty}x(t)=0.$

证  假设$x(t)$是方程(1.1)的非振动解.类似于引理1.1证明中的(1.13)式知, $z'(t)$最终保号且仅有两种可能.

当为$z'(t)>0$时,注意到条件(2.1)式成立,所以完全类似于定理2.1的证明推出矛盾.故知方程(1.1)在$[t_0, \infty)$上无最终正解.

当为$z'(t)<0$时,因有

$\tau'(t)>0, p'(t)\ge 0, z'(t)=x'(t)+p'(t)x(\tau(t))+p(t)x'(\tau(t))\tau'(t)<0, $

所以必有$ x'(t)\leq 0.$又因为$z(t)>0, z'(t)<0, $故有$\lim\limits_{t\to\infty}z(t)=a\ge0.$我们可断定$a=0.$否则,有$\lim\limits_{t\to\infty}x(t)=\frac{a}{1+c}>0, $其中$ c=\lim _{t\to \infty}p(t).$故存在常数$M>0$,使得最终有$x^\beta(\sigma(t))>M$,从而由方程$ (E_0)$知,存在$ T>t_0, $使得

$(R(t)(-z'(t))^\alpha)'\geq \varphi(t)q(t)x^\beta (\sigma(t))\geq M\varphi(t)q(t), ~~t\geq T.$

定义$V(t)=\eta(t)R(t)(-z'(t))^\alpha$,则显然有$V(t)\ge0, ~t\geq T.$又注意到$\eta'(t)\geq0$,由上式,得

(2.7) $\begin{equation} V'(t)=\eta'(t)R(t)(-z'(t))^\alpha +\eta(t)(R(t)(-z'(t))^\alpha)'\geq M \eta(t)\varphi(t)q(t), ~~t\geq T.\end{equation}$

对(2.7)式两端从$T$到$t$积分,可得

$V(t)\ge V(T)+M\int_{T}^{t} \eta(s)\varphi(s)q(s){\rm d} s\ge M\int_{T}^{t} \eta(s)\varphi(s)q(s){\rm d} s, $

即

$\eta(t)R(t)(-z'(t))^\alpha \ge M\int_{T}^{t} \eta(s)\varphi (s)q(s){\rm d} s, $

从而,有

$-z'(t)\ge M^{\frac{1}{\alpha}}\left[\frac{1}{\eta(t)R(t)}\int_{T}^{t} \eta(s)\varphi (s)q(s){\rm d} s\right]^{\frac{1}{\alpha}}.$

再对上式两端从$T$到$t$积分,得

$z(t)\leq z(T)-M^{\frac{1}{\alpha}}\int_{T}^{t}\left[\frac{1}{\eta(s)R(s)}\int_{T}^{s}\eta(\xi)\varphi (\xi)q(\xi){\rm d} \xi\right]^{\frac{1}{\alpha}}{\rm d} s.$

由条件(2.6)式知上式与$z(t)>0, ~t\geq T$矛盾.故必有$\lim\limits_{t\to\infty}z(t)=\lim\limits_{t\to\infty}x(t)=0.$

在定理2.3中取$ \rho(t)=\eta(t)\equiv1$,立即可得

推论2.3  设条件${\rm (H_1)}$-${\rm (H_3)}$和$(1.12)$成立且有$\tau'(t)\geq0, ~p'(t)\geq0.$如果

(2.8) $\begin{equation} \int_{t_0}^\infty \varphi(t)q(t)\left[1-p(\sigma(t))\right]^\beta {\rm d} t=\infty\end{equation}$

和

(2.9) $\begin{equation} \int_{t_0}^\infty \left[\frac{1}{\varphi (t)r(t)}\int_{t_0}^t\varphi(s)q(s){\rm d} s\right]^{\frac{1}{\alpha}}{\rm d} t=\infty\end{equation}$

均成立,其中$ \varphi(t)$由$(1.10)$式定义,则方程$(1.1)$的每一个解$ x(t)$振动或$\lim\limits_{x\to\infty}x(t)=0.$

注2.5  推论$2.3$是著名的Leighton振动定理^[37](关于方程$(r(t)x'(t))'+q(t)x(t)=0)$在非正则条件下的推广.显然,本文定理$2.3$已完全包含和改进了文献[7]中定理$3$、文献[8]中定理$2.3$和文献[33]中定理$3$.

例2.3  讨论方程

(2.10) $\begin{eqnarray}&&[t^{2\alpha+1}|z'(t)|^{\alpha-1}z'(t)]'-(1+\frac{\alpha}{2})t^{2\alpha}|z'(t)|^{\alpha-1}z'(t)\nonumber\\&&+2^\beta t^{1+\alpha}[(|x(\frac{t}{2})|^{\beta-1}x(\frac{t}{2}))^5+|x(\frac{t}{2})|^{\beta-1}x(\frac{t}{2})]=0, ~~t\geq 1\end{eqnarray}$

的振动性,其中$z(t)=x(t)+\frac{1}{2}x(t-1), \alpha>0, \beta>0$为常数.

这里$r(t)=t^{2\alpha+1}, $$p(t)=\frac{1}{2}, $$g(t)=-(1+\frac{\alpha}{2})t^{2\alpha}, $$f(t, u)=2^\beta t^{\alpha+1}(u^3+u), $$u=|x(\frac{t}{2})|^{\beta-1}x(\frac{t}{2}), $$\sigma(t)=\frac{t}{2}, \tau(t)=t-1.$所以, $q(t)=2^\beta t^{\alpha+1}.$易知$ {\rm (H_1)}$-${\rm (H_3)}$满足.又因为

$\varphi(t)=\exp \left(\int_1^t \frac{g(u)}{r(u)}d u\right)=\frac{1}{t^{1+\frac{\alpha}{2}}}, R(t)=\varphi(t)r(t)=t^{\frac{3\alpha}{2}}, \varphi(t)q(t)\left[1-p(\sigma(t))\right]^\beta= t^{\frac{\alpha}{2}}, $

所以,易知(1.12)式和(2.8)式满足.又由于当$t\geq 2$时, $ t-t^{-\frac{\alpha}{2}}\geq1, $所以

$\begin{eqnarray*}\int_1^\infty\left[\frac{1}{R(t)}\int_1^{t}\varphi(s)q(s){\rm d} s\right]^{\frac{1}{\alpha}}{\rm d} t&=&\int_{1}^{\infty}\left[\frac{2^\beta}{t^{3\alpha/2}}\int_{1}^{t} s^{\frac{\alpha}{2}}{\rm d} s\right]^\frac{1}{\alpha}{\rm d} t\\&=&\left(\frac{2^{\beta+1}}{2+\alpha}\right)^{\frac{1}{\alpha}}\int_{1}^\infty \frac{(t-t^{-\alpha/2})^{\frac{1}{\alpha}}}{t}{\rm d} t\\&\geq &\left(\frac{2^{\beta+1}}{2+\alpha}\right)^{\frac{1}{\alpha}}\int_{2}^\infty \frac{1}{t}{\rm d} t =\infty, \end{eqnarray*}$

所以, (2.9)式也成立.故由推论2.3知,方程(2.10)的每个解$x(t)$振动或$\lim\limits_{t\to\infty}x(t)=0.$

例2.4  讨论方程

(2.11) $\begin{equation}\left(e ^{2t}z'(t)\right)'-z'(t)+\frac{2+e ^2}{2e }(e^{2t}-1)x(t-1)=0, ~~t\geq 0\end{equation}$

的振动性,其中$z=x(t)+\frac{1}{2}x(t-2)$.

这里有$r(t)=e^{2t}, p(t)=\frac{1}{2}, g(t)=-1, q(t)=\frac{2+e^2}{2e}(e^{2t}-1), $$ \tau(t)=t-2, \sigma(t)=t-1, $$ \alpha=\beta=1, \tau'(t)=1, p'(t)=0, $$ r'(t)+g(t)=-1+2e^{2t}>1, $

$\varphi(t)=\exp\left(\int_0^t \frac{g(u)}{r(u)}{\rm d} u\right)=\exp\left(\frac{1}{2}(e^{-2t}-1)\right), $

$R(t)=\varphi(t)r(t)=e^{2t}\exp\left(\frac{1}{2}(e^{-2t}-1)\right)>e^{2t}, $

所以

$\int_0^\infty \frac{1}{R(t)}{\rm d} t< \int_0^\infty e^{-2t}{\rm d} t=\frac{1}{2}, ~~\int_0^\infty \varphi(t)q(t)[1-p(\sigma(t))] {\rm d} t\geq \frac{2+e^2}{4e}\int_0^\infty (e^{2t}-1){\rm d} t=\infty . $

又因为由L'Hopital法则,得

$\lim\limits_{t\to \infty}\frac{\int_0^t \varphi(s)q(s){\rm d} s}{R(t)}=\lim\limits_{t\to \infty}\frac{\varphi(t)q(t)}{R'(t)}=\frac{2+e^2}{4e}\lim\limits_{t\to \infty}\frac{e^{2t}-1}{e^{2t}}=\frac{2+e^2}{4e}>0, $

所以

$ \int_0^\infty \left[\frac{1}{R(t)}\int_0^t \varphi(s)q(s){\rm d} s\right]{\rm d} t=\infty.$

综上所述,推论2.3的条件满足,因此,方程(2.11)的每一解$x(t)$振动或有$ \lim\limits_{t\to \infty}x(t)=0.$

事实上,容易验证$x(t)=e^{-t}$恰是方程(2.11)渐近收敛于零的非振动解.

注2.6  文献[13, 18-20]等对于方程$(1.3)$在非正则条件下,区分$\alpha, \gamma$的不同情况给出了若干个有效的振动定理,但它们均不适用于方程$(2.11)$,因此,方程$(1.1)$即使退化为线性方程时,所列文献中的结果也是对其无效的.