考虑Emden-Fowler型半线性中立型泛函微分方程

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

(1.1)

的振动性问题, 其中$\alpha>0, y(t)=x(t)+p(t)x(\tau(t)), p(t)\in{C([t_0, \infty), {\Bbb R})}$.

如果不作说明, 本文总假设下列条件成立:

(H₁)$-\mu\leq{p(t)}\leq{1}, \mu\in(0, 1)$;

(H₂)$r(t)\in{C^{1}([t_0, \infty), {\Bbb R})}, r(t)>0, r'(t)\geq{0}, R(t)=\int_{t_0}^{t}r^{-\frac{1}{\alpha}}(s){\rm d}s, \pi(t)=\int_{t}^{\infty}r^{-\frac{1}{\alpha}}(s){\rm d}s$;

(H₃)$\tau(t)\in{C^{1}([t_0, \infty), {\Bbb R})}, \tau(t)\leq{t}, \lim\limits_{{t}\rightarrow\infty}\tau(t)=\infty$;

(H₄)$\sigma(t)\in{C^{1}([t_0, \infty), {\Bbb R})}, \sigma(t)>0, \sigma'(t)>0, \sigma(t)\leq{t}, \lim\limits_{{t}\rightarrow\infty}\sigma(t)=\infty$;

(H₅)$f(t, x)\in{C([t_0, \infty)\times{{\Bbb R}}, {\Bbb R})}, $存在函数$q(t)\in{C([t_0, \infty), [0, \infty))}$和常数$\beta>0$使得

$ f(t, x)\cdot{\rm sgn}x\geq{q(t)}|x|^{\beta}, x\neq{0}, t\geq{t_0}. $

函数$x(t)\in{C^{1}([T_x, \infty), {\Bbb R})}, T_x\geq{t_0}$称为方程(1.1)的一个解, 如果它具有性质$r(t)|y'(t)|^{\alpha-1}$.$y'(t)\in{C^1}([T_x, \infty), {\Bbb R})$且在区间$[T_x, \infty)$上满足方程(1.1), 如通常一样, 方程(1.1)的一个解称为振动的.如果它有任意大的零点.否则称为非振动.方程(1.1)称为振动的, 如果它的一切解均为振动.

中立型泛函微分方程出现在高速计算机无损传输线路的网络设计中, 在自动控制理论、神经动力系统理论及电子网络等高新技术领域中有着广泛应用^[20-21].半线性中立型时滞微分方程

$ (r(t)|y'(t)|^{\alpha-1}y'(t))'+q(t)|x(\sigma(t))|^{\alpha-1}x(\sigma(t))=0 $

(1.2)

与其相应的线性方程

$ (r(t)y'(t))'+q(t)x(\sigma{t})=0 $

具有许多相似的性质, 其中$y(t)=x(t)+p(t)x(\tau(t))$.近年来, 方程(1.2)及其相关方程的振动性研究取得了丰硕成果, 可以参看文献[1-10].

众所周知, Emden-Fowler方程

$ x''(t)+q(t)|x(t)|^{\lambda}{\rm sgn}x(t)=0, \lambda>0 $

在核物理、天体物理及气体动力学等许多重要领域有应用.最近文献[14-15]考虑了中立型Emden-Fowler方程

$ (r(t)y'(t))'+q(t)|x(\sigma(t))|^{\lambda-1}x(\sigma(t))=0, \lambda>0, $

(1.3)

其中$y(t)=x(t)+p(t)x(\tau(t))$, 分别给出了当$\int_{t_0}^{\infty}\frac{1}{r(t)}{\rm d}t=\infty$和$\int_{t_0}^{\infty}\frac{1}{r(t)}{\rm d}t<\infty$时方程(1.3)的振动准则, 同类的工作可以参看文献[11-13, 16-19]及其参考文献.但是, 我们看到方程(1.2)当$\alpha=1$时不等于方程(1.3).因此, 这两类方程的振动结果不能通用.然而, 方程(1.1)在$\alpha=\beta$时包含了方程(1.2);在$\alpha=1$时包含了方程(1.3).

本文目的是建立广义中立型Emden-Fowler方程(1.1)的若干新的振动准则.在第二节中考虑$0\leq{p(t)}\leq1$的情况, 当$\int_{t_0}^{\infty}r^{-\frac{1}{\alpha}}(t){\rm d}t=\infty$时, 新的振动准则推广和改进了文献[3, 8, 11, 14, 17, 19]的结果.当$\int_{t_0}^{\infty}r^{-\frac{1}{\alpha}}(t){\rm d}t<\infty$时, 我们的结果推广和改进了文献[15]和[19]的结果.在第三节中考虑$-\mu\leq{p(t)}\leq{0}$的情况, 所得准则推广和改进了韩振来等在文献[4]中建立的最新结果.

在下面出现的函数不等式中, 如不作特别说明, 我们总假设它对一切充分大的t成立.

定理2.1  设

$ \int_{t_0}^{\infty}{r^{-\frac{1}{\alpha}}(t)}{\rm d}t=\infty, $

(2.1)

且存在函数$\rho(t)\in{C^{1}([t_0, \infty), (0, \infty))}$和任意常数K>0, 使得

$ \int_{t_0}^{\infty}[\rho(t)Q(t)-K\rho'(t)\xi^{c}(t)]{\rm d}t=\infty, $

(2.2)

其中

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

(2.3)

$ \xi(t)= \bigg[\int_{T}^{\sigma(t)}{r^{-\frac{1}{\alpha}}(s)}{\rm d}s\bigg]^{-1}, \quad \sigma(t)\geq{T}\geq{t_0}, $

(2.4)

$ c=\min\{\alpha, \beta\}, K=\left\{\begin{array}{ll} 1, \qquad &\alpha=\beta, \\ \mbox{正常数}, \quad&\alpha\neq\beta, \end{array}\right. $

(2.5)

则方程(1.1)振动.

证  设方程(1.1)有非振动解x(t), 不失一般, 我们设x(t)>0, 故由$(H_5)$我们得到

$ (r(t)|y'(t)|^{\alpha-1}y'(t))'\leq-q(t)x^{\beta}(\sigma(t))<0. $

(2.6)

因此, $r(t)|y'(t)|^{\alpha-1}y'(t)$是减函数, 存在两种情况, 即$y'(t)>0$和$y'(t)<0$, 类似于文献[19, 定理2.1]的证明, 由条件(2.1)知$y'(t)<0$是不可能的, 故我们有

$ y'(t)>0, ~~t\geq{T}\geq{t_0}. $

(2.7)

由(2.7)式得

$ x(t)\geq(1-p(t))y(t). $

(2.8)

联合(2.6)-(2.8)式可得

$ (r(t)(y'(t))^{\alpha})'+Q(t)y^{\beta}(\sigma(t))\leq{0}, ~~t\geq{T}, $

(2.9)

其中Q(t)由(2.3)式定义.

注意到$r^{\frac{1}{\alpha}}(t)y'(t)$非增, 我们有

$ y(\sigma(t))>\int_{T}^{\sigma(t)}r^{-\frac{1}{\alpha}}(s) r^{\frac{1} {\alpha}}(s)y'(s){\rm d}s\geq r^{\frac{1}{\alpha}}(t)y'(\sigma(t))\xi^{-1}(t), ~~ t\geq\sigma(t)\geq{T}, $

即

$ \xi^{\alpha}(t)\geq r(t)\bigg(\frac{y'(t)} {y(\sigma(t))}\bigg)^{\alpha}. $

(2.10)

另一方面,

$ y(\sigma(t))=y(T)+\int_{T}^{\sigma(t)}r^{-\frac{1}{\alpha}}(s) r^{\frac{1}{\alpha}}(s)y'(s){\rm d}s\leq{y(T)}+r^{\frac{1}{\alpha}}(T)y'(T)\xi^{-1}(t), $

易知, 存在常数$L_1\in(0, 1)$使得

$ {L_1}y(\sigma(t))\leq{y(\sigma(t))}-y(T), ~~\sigma(t)\geq{T}. $

因此, 记$L=\frac{r^{\frac{1}{\alpha}}(T)y'(T)}{L_1}$, 我们有

$ y(\sigma(t))\leq{L\xi^{-1}(t)}, ~~t\geq{\sigma(t)}\geq{T}. $

(2.11)

定义

$ w(t)=\rho(t)r(t)\frac{(y'(t))^\alpha}{y^{\beta}(\sigma(t))}, ~~t\geq{T}. $

(2.12)

利用(2.9)和(2.10)式, 我们有

$ w'(t)\leq-\rho(t)Q(t)+\rho'(t)\xi^{\alpha}(t)y^{\alpha-\beta}(\sigma(t)). $

(2.13)

注意到(2.7)式, 当$\beta>{\alpha}$时, 我们有

$ y^{\alpha-\beta}(\sigma(t))= \Big(\frac{1}{y(\sigma(t))}\Big)^{\beta-\alpha}\leq{L_1}, ~~L_1>0. $

(2.14)

利用(2.11)式, 当$\alpha>\beta$时, 有

$ y^{\alpha-\beta}(\sigma(t))\leq L^{\alpha-\beta}\xi^{\alpha-\beta}(t). $

(2.15)

联合(2.13)-(2.15)式, 可得

$ w'(t)\leq-\rho(t)Q(t)+K\rho'(t)\xi^{c}(t), $

(2.16)

其中c由(2.5)式定义, 对(2.16)式积分有

$ w(t)\leq{w(T)}-\int_T^t[\rho(s)Q(s)-K\rho'(s)\xi^{c}(s)]{\rm d}s, $

在上式中, 令$t\rightarrow{\infty}$, 注意到(2.2)式, 我们得到$\lim\limits_{t\rightarrow{\infty}}w(t)=-\infty$.这与(2.12)式矛盾.定理2.1证毕.

推论1(Leighton型振动准则)  设$\alpha>0, \beta>0$且

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

则方程(1.1)振动.

证  只须在定理2.1中令$\rho(t)=1$.

推论1是Leighton振动准则对方程(1.1)的推广.

例2.1  考虑中立型时滞方程

$ [|(x(t)+(1-{\rm e}^{-t})x(t-1))'|^{\alpha-1}(x(t)+(1-{\rm e}^{-t}) x(t-1))']'+{\rm e}^{\mu\beta{t}}|x(\mu{t})|^{\beta-1}x(\mu{t})=0, $

(2.17)

这里$r(t)=1, Q(t)=q(t)[1-p(\sigma(t))]^{\beta}=1$, 易知推论1的条件成立, 故方程(2.17)对任意$\alpha>0, \beta>0$都是振动.方程(2.17)即为文献[19]中的例3.1.文献[19]中的定理2.1要求当$\alpha>\beta>1$时, 方程(2.17)才振动.

利用推论1也能判定文献[19]中的方程(3.2)的振动性, 而文献[19]中的定理2.2要求$\alpha>\beta>1$才能确定其振动, 因此, 我们的推论1推广和改进了文献[19]的此两定理.

例2.2  考虑中立型Emden-Fowler方程

$ \Big(\frac{1}{t^{1+\frac{\alpha}{\beta}}}|y'(t)|^{\alpha-1}y'(t)\Big)' +\frac{1}{t^2}|x\Big(\frac{t}{3}\Big)|^{\beta-1}x \Big(\frac{t}{3}\Big)=0, $

(2.18)

其中$\alpha>0, \beta>0, y(t)=x(t)+\frac{1}{2}x(t-1)$, 此时

$ Q(t)=\Big(\frac{1}{2}\Big)^{\beta}\frac{1}{t^2}, ~~\xi^{-1}(t)= \int_{T}^{\frac{t}{3}}s^{\frac{1}{\alpha}+\frac{1}{\beta}}{\rm d}s= \frac{1}{d}\Big[\Big(\frac{t}{3}\Big)^{d}-T^{d}\Big], $

其中$d=1+\frac{1}{\alpha}+\frac{1}{\beta}$, 因此, 存在T_1>T, 使得

$ \xi^{-1}(t)>\frac{1}{2d}\Big(\frac{t}{3}\Big)^{d}, ~~t\geq{T_1}, $

即

$ \xi(t)<\frac{2d\cdot3^d}{t^d}, ~~t\geq{T_1}. $

故有

$ \xi^{\alpha}(t)<\frac{(2d\cdot3^d)^{\alpha}}{t^{1+\alpha+\frac{\alpha}{\beta}}}, ~~\xi^{\beta}(t)<\frac{(2d\cdot3^d)^{\beta}}{t^{1+\beta+\frac{\beta}{\alpha}}}, ~~t\geq{T_1}. $

对于条件(2.2), 取$\rho(t)=t, $则$\int_{t_0}^{\infty}\rho(t)Q(t){\rm d}t=\infty$且

$ \int_{t_0}^{\infty}K\rho'(t)\xi^{\alpha}(t){\rm d}t=\int_{T_1}^{\infty}K\frac{(2d\cdot3^d)^{\alpha}}{t^{1+\alpha+\frac{\alpha}{\beta}}}{\rm d}t+\int_{t_0}^{T_1}K\frac{(2d\cdot3^d)^{\alpha}}{t^{1+\alpha+\frac{\alpha}{\beta}}}{\rm d}t<\infty, $

即

$ \int_{t_0}^{\infty}K\rho'(t)\xi^{\beta}(t){\rm d}t= \bigg\{\int_{t_0}^{T_1}+\int_{T_1}^{\infty}\bigg\}K\frac{(2d\cdot3^d)^{\beta}}{t^{1+\beta+\frac{\beta}{\alpha}}}{\rm d}t<\infty. $

因此, 对任意$\alpha>0, \beta>0$时, 方程(2.18)振动.

定理2.2  设(2.1)式成立, 存在函数$\rho(t)\in{C^{1}([t_0, \infty), (0, \infty))}$和任意常数K>0, 使得

$ \int_{t_0}^{\infty}[\rho(t)Q_{K}(t)-\frac{(\rho'(t))^{c+1}r(\theta(t))}{(c+1)^{c+1}[\rho(t)\sigma'(t)]^c}]{\rm d}t=\infty, $

(2.19)

其中Q(t)由(2.3)式定义且$Q_{K}(t)=K^{c}Q(t)$, c由(2.5)式定义, $\theta(t)=\left\{\begin{array}{ll} \sigma(t), \quad &\beta\geq\alpha, \\ t, \qquad &\alpha>\beta, \end{array}\right.$则方程(1.1)振动.

证  设方程(1.1)有非振动解x(t), 不失一般性, 可设x(t)>0对于x(t) < 0的情况可以类似地证明.如同定理2.1的证明中一样, 我们有(2.7), (2.8)和(2.9)式成立.

定义函数w(t)同(2.12)式, 我们有

$ w'(t)\leq-\rho(t)Q(t)+\frac{\rho'(t}{\rho(t)}w(t)-\rho(t)r(t)(y'(t))^{\alpha}\frac{\beta{\sigma'(t)}y'(\sigma(t))}{y^{\beta+1}(\sigma(t))}, t\geq{T}, $

(2.20)

注意到$r^{\frac{1}{\alpha}}(t)y'(t)$是减函数, 利用不等式$y'(\sigma(t))\geq{y'(t)(\frac{r(t)}{r(\sigma(t))})^{\frac{1}{\alpha}}}, $由(2.20)式, 我们得到

$ w'(t)\leq-\rho(t)Q(t)+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{\beta{\sigma'(t)}} {[\rho(t)r(t)]^{\frac{1}{\alpha}}} \Big(\frac{r(t)}{r(\sigma(t))}\Big)^{\frac{1}{\alpha}}[y(\sigma(t))]^{\frac{\beta}{\alpha}-1}w^{\frac{\alpha+1}{\alpha}}(t). $

(2.21)

因$y'(t)>0, y(t)$为增函数, 故当${\beta}\geq{\alpha}$时, $[y(\sigma(t))]^{\frac{\beta}{\alpha}-1}\geq{k_1}>0$为常数, 则

$ w'(t)\leq-\rho(t)Q(t)+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{\alpha{k_1}{\sigma'(t)}}{[\rho(t)r(\sigma(t))]^{\frac{1}{\alpha}}}w^{\frac{\alpha+1}{\alpha}}(t), \beta\geq\alpha. $

(2.22)

对于${\alpha}>{\beta}$情况, 由$r'(t)\geq0$和$(r(t)(y'(t))^{\alpha})'\leq0.$我们得到$y''(t)\leq0, $故$y'(t)$为减函数, 由(2.20)式可得

$ \begin{array}{*{20}{c}} {w'(t)}&{ \le - \rho(t)Q(t)+\frac{{\rho '(t)}}{{\rho(t)}}w(t)- \frac{{\beta \sigma '(t)}}{{{{[\rho(t)r(t)]}^{\frac{1}{\beta }}}}}{{[\frac{1}{{y'(t)}}]}^{\frac{\alpha }{\beta } - 1}}{w^{1+\frac{1}{\beta }}}(t)}\\ {}&{ \le - \rho(t)Q(t)+\frac{{\rho '(t)}}{{\rho(t)}}w(t)- \frac{{\beta {k_2}\sigma '(t)}}{{{{[\rho(t)r(t)]}^{\frac{1}{\beta }}}}}{w^{1+\frac{1}{\beta }}}(t), } \end{array} $

(2.23)

其中k_2>0为常数.联合(2.22)和(2.23)式, 我们有

$ w'(t)\leq{-\rho(t)Q(t)}+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{cK\sigma'(t)}{[\rho(t)r(\theta(t))]^{\frac{1}{c}}}w^{\frac{c+1}{c}}(t), $

(2.24)

其中$c=\min \{ \alpha, \beta \}, K=\left\{ {\begin{array}{*{20}{l}} {1, \qquad }&{\alpha=\beta, }\\ {正常数\quad }&{\alpha \ne \beta, } \end{array}} \right.\theta(t)=\left\{ {\begin{array}{*{20}{l}} {\sigma(t), \quad }&{\beta \ge \alpha, }\\ {t, \qquad }&{\alpha >\beta.} \end{array}} \right.$

对(2.24)式利用不等式

$ Bu-Au^{\frac{c+1}{c}}\leq\frac{c^{c}}{(c+1)^{c+1}}\frac{B^{c+1}}{A^c}, ~~A>0, B\geq{0}, c>0, u>0, $

则(2.24)式成为

$ w'(t)\leq-\rho(t)Q(t)+\frac{(\rho'(t))^{c+1}r(\theta(t))}{(c+1)^{c+1}(K\rho(t)\sigma'(t))^c}, ~~t\geq{T}. $

(2.25)

对(2.25)式积分, 有

$ \begin{eqnarray*} w(t)&\leq&{w(T)}-\int_{T}^{t} \bigg[\rho(s)Q(s)-\frac{(\rho'(s))^{c+1}r(\theta(s))}{(c+1)^{c+1}(K\rho(s)\sigma'(s))^c}\bigg]{\rm d}s \\ &=&w(T)-\frac{1}{K^c}\int_{T}^{t} \bigg[\rho(s)Q_{K}(s)-\frac{(\rho'(s))^{c+1}r(\theta(s))}{(c+1)^{c+1}(\rho(s)\sigma'(s))^c}\bigg]{\rm d}s. \end{eqnarray*} $

在上式中, 令$t\rightarrow\infty$, 由(2.19)式, 我们有$\lim\limits_{t\rightarrow\infty}w(t)=-\infty, $此与$w(t)>0$矛盾.定理2.2证毕.

注1  前面的推论1也是本定理取$\rho(t)=1$时的特例, 文献[8]考虑的方程是方程(1.1)中$\alpha=\beta$且$p(t)=0$时的特例, 而文献[8]的定理是定理2.2中取$\rho(t)=R^{\alpha}(\sigma(t))$时的特例.

例2.3  考虑方程

$ (|y'(t)|^{\alpha-1}y'(t))'+\frac{1}{t(\ln{t})^{c+1}}|x(\mu{t})|^{\beta-1}x(\mu{t})=0, $

(2.26)

其中$\alpha>0, \beta>0, y(t)=x(t)+\frac{1}{2}x(t-1), \mu\in(0, 1).$此时$Q(t)=\frac{1}{2^{\beta}t(\ln{t})^{c+1}}, $取$\rho(t)=(\ln{t})^c.$则$\rho'(t)=\frac{c}{t}(\ln{t})^{c-1}$, 我们有

$ \int_{t_0}^{\infty}\rho(t)Q_{K}(t){\rm d}t=\int_{t_0}^{\infty}\frac{K^c}{2^{\beta}t\ln{t}}{\rm d}t=\infty, $

$ \int_{t_0}^{\infty}\frac{(\rho'(t))^{c+1}r(\theta(t))}{(c+1)^{c+1} [\rho(t)\sigma'(t)]^c}{\rm d}t= \Big(\frac{c}{c+1}\Big)^{c+1}\frac{1}{\mu^c}\int_{t_0}^{\infty}\frac{1}{t^{1+c}\ln{t}}{\rm d}t<\infty. $

因此条件(2.19)成立.故对任意$\alpha>0, \beta>0, $方程(2.26)振动.但是方程(2.26)不满足条件(2.2).故定理2.1不适用, 即定理2.2改进了定理2.1.

现考虑当条件(2.1)不成立时, 我们给出下面的定理.

定理2.3  设

$ p'(t)\geq{0}, ~~\tau'(t)\geq{0}, ~~\sigma(t)\leq\tau(t), $

(2.27)

若(2.19)式成立, 且

$ \int_{t_0}^{\infty}\bigg[\pi^{\lambda}(t)Q_{m}(t)-\frac{L(r(t))^{\frac{\alpha-\lambda-1}{\alpha}}}{\pi(t)} \bigg]{\rm d}t=\infty, $

(2.28)

其中$\pi(t)$由$(H_2)$定义, $\lambda=\max\{\alpha, \beta\}, L=\frac{\lambda^{2\lambda+1}}{(\lambda+1)^{\lambda+1}}, Q_{m}(t)=m^{\lambda}Q_{1}(t), Q_{1}(t)$由(2.32)式定义.m由(2.37)式定义, 则方程(1.1)振动.

证  设方程(1.1)有非振动解$x(t)$, 不失一般, 可设$x(t)>0, $当$x(t)<0$时可以用类似的方法证明.由方程(1.1)和假设$(H_5)$, 我们有

$ (r(t)|y'(t)|^{\alpha-1}y'(t))'\leq-q(t)x^{\beta}(\sigma(t))\leq{0}. $

故$r(t)|y'(t)|^{\alpha-1}y'(t)$非增.因此, 存在两种情况: $y'(t)>0$和$y'(t)<0.$

因(2.19)式成立, 由定理2.2的证明知$y'(t)>0$不可能.即我们有

$ y'(t)<0, t\geq{T}\geq{t_0}. $

(2.29)

注意到(2.27)式和等式

$ y'(t)=x'(t)+p'(t)x(\tau(t))+p(t)x'(\tau(t))\tau'(t), $

我们得到$x'(t)<0$, 于是

$ y(t)\leq{x(\tau(t))}+p(t)x(\tau(t))=(1+p(t))x(\tau(t)). $

(2.30)

联合(2.27)和(2.30)式, 我们有

$ x(\sigma(t))\geq{x(\tau(t))}\geq\frac{y(t)}{1+p(t)}>(1-p(t))y(t). $

(2.31)

由(2.29)和(2.31)式, 由方程(1.1)可得

$ (r(t)(-y'(t))^{\alpha})'-Q_{1}(t)y^{\beta}(t)>0, $

(2.32)

其中$Q_{1}(t)=q(t)(1-p(t))^{\beta}.$

定义函数

$ u(t)=\frac{r(t)(-y'(t))^{\alpha}}{y^{\beta}(t)}, ~~t\geq{T}, $

(2.33)

则u(t)>0且

$ u'(t)=\frac{(r(t)(-y'(t))^{\alpha})'}{y^{\beta}(t)}+\frac{\beta{r(t)}(-y'(t))^{\alpha+1}}{y^{\beta+1}(t)}. $

(2.34)

注意到(2.29)式, 当$\alpha>\beta$时, 我们有

$ u'(t)>Q_{1}(t)+\frac{\beta}{r^{\frac{1}{\alpha}}(t)}(y(t))^{\frac{\beta-\alpha}{\alpha}}u^{\frac{\alpha+1}{\alpha}}(t) >Q_{1}(t)+\frac{\beta{m_1}}{r^{\frac{1}{\alpha}}(t)}u^{\frac{\alpha+1}{\alpha}}(t), ~~t\geq{T}, $

(2.35)

其中$m_1=[y(T)]^{\frac{\beta-\alpha}{\alpha}}.$

另一方面, 注意到$r'(t)\geq{0}$和$(r(t)(-y'(t))^{\alpha})'>0$, 因此, $y'(t)<0, $当$\beta>\alpha$时, 我们有

$ u'(t)>Q_{1}(t)+\frac{\beta}{r^{\frac{1}{\beta}}(t)}[-y'(t)]^{\frac{\beta-\alpha}{\beta}}u^{\frac{\beta+1}{\beta}}(t) >Q_{1}(t)+\frac{\beta{m_2}}{r^{\frac{1}{\beta}}(t)}u^{\frac{\beta+1}{\beta}}(t), ~~t\geq{T}, $

(2.36)

其中$m_2=[-y'(T)]^{\frac{\beta-\alpha}{\beta}}.$

联合(2.35)和(2.36)式, 我们得到

$ u'(t)>Q_{1}(t)+\frac{m}{r^{\frac{1}{\lambda}}(t)}u^{\frac{\lambda+1}{\lambda}}(t), ~~t\geq{T}, $

(2.37)

其中$Q_{1}(t)$由(2.32)式定义, $\lambda=\max\{\alpha, \beta\}, m=\left\{\begin{array}{ll} \alpha, \qquad &\alpha=\beta, \\ \mbox{正常数}, \quad &\alpha\neq\beta.\end{array}\right.$下面证明:存在常数l>0, 使得

$ 0<u(t)[\pi^{\alpha}(t)+\pi^{\beta}(t)]<l, ~~t\geq{T}. $

(2.38)

事实上, 因$r^{\frac{1}{\alpha}}(t)(-y'(t))$是增函数, 我们有

$ r^{\frac{1}{\alpha}}(s)(-y'(s))\geq{r^{\frac{1}{\alpha}}(t)(-y'(t))}, ~~s\geq{t}\geq{T}. $

对上式从t到$\infty$积分产生

$ \int_{t}^{\infty}y'(s){\rm d}s\leq-r^{\frac{1}{\alpha}}(t)(-y'(t))\int_{t}^{\infty}r^{-\frac{1}{\alpha}}(\tau){\rm d}\tau. $

因此

$ 0<y(t)-r^{\frac{1}{\alpha}}(t)(-y'(t))\pi(t), $

即

$ y(t)>r^{\frac{1}{\alpha}}(t)(-y'(t))\pi(t). $

(2.39)

由(2.39)式, 有

$ y^{\alpha}(t)>r(t)(-y'(t))^{\alpha}\pi^{\alpha}(t). $

利用(2.33)式, 我们得到

$ y^{\alpha-\beta}(t)>u(t)\pi^{\alpha}(t). $

由(2.29)式, 当$\alpha>\beta$时, 我们有

$ 0<u(t)\pi^{\alpha}(t)\leq{l_1}, $

(2.40)

其中$l_1=[y(T)]^{\alpha-\beta}.$由(2.39)式可得

$ y^{\beta}(t)>r^{\frac{\beta}{\alpha}}(t)(-y'(t))^{\beta}\pi^{\beta}(t). $

即有

$ \bigg[\frac{1}{r^{\frac{1}{\alpha}}(t)(-y'(t))}\bigg]^{\beta-\alpha}>\frac{r(t)(-y'(t))^{\alpha}}{y^{\beta}(t)}\pi^{\beta}(t). $

因此, 当$\beta>\alpha$时有

$ 0<u(t)\pi^{\beta}(t)\leq{l_2}, $

(2.41)

其中$l_2=[r^{\frac{1}{\alpha}}(T)(-y'(T))]^{\alpha-\beta}.$

取$l=l_1+l_2, $由(2.40)和(2.41)式, 我们得到(2.38)式.

下面用$\pi^{\lambda}(t)$乘不等式(2.37)式产生

$ \begin{array}{*{20}{c}} {{\pi ^\lambda }(t)u(t)- {\pi ^\lambda }(\tau)u(\tau)+\lambda \int_T^t {{\pi ^{\lambda - 1}}}(s){r^{ - \frac{1}{\alpha }}}(s)u(s){\rm{d}}s}&{}\\ { > \int_T^t {{\pi ^\lambda }}(s){Q_1}(s){\rm{d}}s+m\int_T^t {{\pi ^\lambda }}(s){r^{ - \frac{1}{\lambda }}}(s){u^{\frac{{\lambda+1}}{\lambda }}}(s){\rm{d}}s.}&{} \end{array} $

(2.42)

利用不等式

$ Bu-Au^{\frac{\lambda+1}{\lambda}}\leq\frac{(\lambda)^{\lambda}}{(\lambda+1)^{\lambda+1}}\frac{B^{\lambda+1}}{A^{\lambda}}, ~~A>0, B>0, u>0, \lambda>0, $

(2.42')

我们有

$ \begin{array}{l} \;\;\;\lambda \int_T^t {{\pi ^{\lambda - 1}}}(s){r^{ - \frac{1}{\alpha }}}(s)u(s){\rm{d}}s - m\int_T^t {{\pi ^\lambda }}(s){r^{ - \frac{1}{\lambda }}}(s){u^{\frac{{\lambda+1}}{\lambda }}}(s){\rm{d}}s\\ \le \int_T^t {\frac{{{{(\lambda)}^{2\lambda+1}}}}{{{{(\lambda+1)}^{\lambda+1}}}}} \frac{{{{[r(s)]}^{\frac{{\alpha - \lambda - 1}}{\alpha }}}}}{{{m^\lambda }\pi(s)}}{\rm{d}}s. \end{array} $

(2.43)

注意到(2.38)式, 将(2.43)式代入(2.42)式, 我们得到

$ \int_{T}^{t} \bigg(\pi^{\lambda}(s)Q_{1}(s)-\frac{\lambda^{2\lambda+1}}{(\lambda+1)^{\lambda+1}}\frac{[r(s)]^{\frac{\alpha-\lambda-1}{\alpha}}}{m^{\lambda}\pi(s)}\bigg){\rm d}s\leq{l}-\pi^{\lambda}(\tau)u(\tau) $

或者

$ \int_{T}^{t}\frac{1}{m^{\lambda}} \bigg[\pi^{\lambda}(s)Q_{m}(s)-\frac{L[r(s)]^{\frac{\alpha-\lambda-1}{\alpha}}}{\pi(s)}\bigg]{\rm d}s\leq{l}-\pi^{\lambda}(\tau)u(\tau), $

(2.44)

其中$L=\frac{\lambda^{2\lambda+1}}{(\lambda+1)^{\lambda+1}}, Q_{m}(s)=m^{\lambda}Q_{1}(s).$在上式中, 令$t\rightarrow\infty$我们看到(2.44)式与(2.28)式矛盾, 定理2.3证毕.

注2  最近, 文献[15]仅考虑方程(1.1)当$\alpha=1$时的情况.文献[19]仅考虑当$\alpha\geq\beta$的情况, 而且只得到方程(1.1)的振动或者趋向于零.我们的定理2.3的结果对任意的$\alpha>0, \beta>0$均成立.

例2.4  考虑方程

$ (t^{2\alpha}|y'(t)|^{\alpha-1}y'(t))'+t^{1+\lambda}{\rm e}^{t}|x(t-2)|^{\beta-1}x(t-2)=0, ~~t\geq{2}. $

(2.45)

此时$r(t)=t^{2\alpha}, \pi(t)=\frac{1}{t}, $其中$y(t)=x(t)+\frac{1}{3}x(t-1), $则(2.27)式满足$Q_{1}(t)=q(t)(1-p(t))^{\beta}=(\frac{2}{3})^{\beta}t^{1+\lambda}{\rm e}^{t}.$

$ \begin{eqnarray*} &&\int_{t_0}^{\infty}\bigg[\pi^{\lambda}(t)Q_{m}(t)-\frac{L[r(t) ]^{\frac{\alpha-\lambda-1}{\alpha}}}{\pi(t)}\bigg]{\rm d}t \\ &=&\int_{t_0}^{\infty}\bigg[m^{\lambda}\Big(\frac{2}{3}\Big)^{\beta}t^{1+\lambda}{\rm e}^{t}t^{-\lambda}-Lt^{2(\alpha-\lambda-1)+1}\bigg]{\rm d}t \\ &=&\int_{t_0}^{\infty}t\bigg[m^{\lambda}\Big(\frac{2}{3}\Big)^{\beta}{\rm e}^{t}-Lt^{2(\alpha-\lambda-1)}\bigg]{\rm d}t=\infty. \end{eqnarray*} $

另一方面, 在(2.19)式中取$\rho(t)=1, $显然, (2.19)式成立.因此, 对任意$\alpha>0, \beta>0$成立, 由定理2.3知方程(2.45)振动.

引理3.1  设(2.1)式成立, x(t)是方程(1.1)的最终正解.则存在$t_1\geq{t_0}$使得

$ y(t)>0, y'(t)>0, ~~t\geq{t_1} $

(3.1)

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

证  设x(t)是方程(1.1)的最终正解, 由$H_5$, 我们有

$ (r(t)|y'(t)|^{\alpha-1}y'(t))'\leq-q(t)x^{\beta}(\sigma(t))\leq-q(t)y^{\beta}(\sigma(t))<0. $

(3.2)

因此$r(t)|y'(t)|^{\alpha-1}y'(t)$是减函数且最终定号.即有$y'(t)$最终定号且y(t)最终定号.于是y(t)只有两种可能情况:y(t)>0和y(t) < 0.

(ⅰ)若y(t)>0, 则我们断言y'(t)>0, 事实上, 如果y'(t) < 0.则方程(1.1)成为

$ (r(t)(-y'(t))^{\alpha})'-q(t)x^{\beta}(\sigma(t))\geq{0}, t\geq{T}, $

故有

$ r(t)(-y'(t))^{\alpha}\geq{r(T)(-y'(T))^{\alpha}}\equiv{b}>0. $

即

$ y'(t)\leq-\Big(\frac{b}{r(t)}\Big)^{\frac{1}{\alpha}}, ~~t\geq{T}. $

对上式积分产生

$ 0<y(t)\leq{y(T)}-b^{\frac{1}{\alpha}}\int_{T}^{t}r^{-\frac{1}{\alpha}}(s){\rm d}s, $

令$t\rightarrow\infty$, 由(2.1)式我们得到$\lim\limits_{t\rightarrow\infty}y(t)=-\infty.$此与$y(t)>0$矛盾.因此, $y'(t)>0, $即(3.1)式成立.

(ⅱ)若y(t) < 0.则x(t)有两种可能:x(t)有界或者x(t)无界.

如果x(t)无界, 则存在序列$\{t_n\}, t_n\geq{T}, \lim\limits_{n\rightarrow\infty}t_n=\infty.$

令

$ x(t_n)=\max_{[T, \tau(t_n)]}\{x(s)\}. $

因y(t) < 0, 我们有

$ x(t)=y(t)-p(t)x(\tau(t))<-p(t)x(\tau(t))<x(\tau(t)), ~~t\geq{T}. $

故x(t)是减函数.注意到

$ x(\tau(t_n))=\max_{[T, \tau(t_n)]}\{x(s)\}\leq\max_{[T, t_n]}\{x(s)\}=x(t_n)<x(\tau(t_n)). $

此矛盾说明x(t)有界.因此, 根据有界性, 可得

$ \begin{eqnarray*} 0&\geq&\lim\limits_{t\rightarrow\infty}\sup{y(t)}\geq\lim\limits_{t\rightarrow\infty}\sup{x(t)}+\lim\limits_{t\rightarrow\infty}\inf{p(t)x(\tau(t))} \\ &\geq&\lim\limits_{t\rightarrow\infty}\sup{x(t)}-\mu\lim\limits_{t\rightarrow\infty}\sup{x(\tau(t))} \\ &=&(1-\mu)\lim\limits_{t\rightarrow\infty}\sup{x(t)}\geq{0}. \end{eqnarray*} $

因此, $\lim\limits_{t\rightarrow\infty}x(t)=0.$引理3.1证毕.

引理3.2  设(2.1)式成立, 且

$ \int_{t_0}^{\infty}q(t)\sigma^{\beta}(t){\rm d}t=\infty, $

(3.3)

若x(t)是方程(1.1)最终正解使得(3.1)式成立, 则存在$T\geq{t_0}$使得

$ y(t)\geq{ty'(t)}, ~~t\geq{T}, $

(3.4)

且$\frac{y(t)}{t}$是严格的减函数.

证  引理3.2的证明类似于文献[4, 引理2.2], 我们省略.

定理3.1  设(2.1)和(3.3)式成立, 存在$\rho(t)\in{C^1}([t_0, \infty), (0, \infty))$使得

$ \int_{t_0}^{\infty} \bigg[\rho(t)q_{m}(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{r(t)(\rho'(t))^{c+1}}{(c+1)^{c+1}\rho^{c}(t)}\bigg]{\rm d}t=\infty, $

(3.5)

则方程(1.1)的每一解x(t)振动或$\lim\limits_{t\rightarrow\infty}x(t)=0.$

证  设x(t)是方程(1.1)的非振动解, 不失一般性, 设x(t)最终为正, 由引理3.1知存在$t_1\geq{t_0}$使得(3.1)式成立或$\lim\limits_{t\rightarrow\infty}x(t)=0.$

现设(3.1)成立.定义函数

$ w(t)=\rho(t)\frac{r(t)(y'(t))^{\alpha}}{y^{\beta}(t)}, ~~t\geq{t_1}, $

(3.6)

则$w(t)>0.$注意到(3.2)式, 我们有

$ \begin{eqnarray*} w'(t)&\leq&-\rho(t)q(t) \Big(\frac{y(\sigma(t))}{y(t)}\Big)^{\beta}+\frac{\rho'(t)}{\rho(t)}w(t)-\rho(t)r(t)(y'(t))^{\alpha}{\frac{\beta y'(t)}{y^{\beta+1}(t)}} \\ &=&-\rho(t)q(t)\Big(\frac{y(\sigma(t))}{y(t)}\Big)^{\beta}+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{\beta}{[\rho(t)r(t)]^{\frac{1}{\alpha}}}[y(t)]^{\frac{\beta-\alpha}{\alpha}}w^{\frac{\alpha+1}{\alpha}}(t). \end{eqnarray*} $

故当$\beta\geq\alpha$时, 我们得到

$ w'(t)\leq-\rho(t)q(t)\Big(\frac{y(\sigma(t))}{y(t)}\Big)^{\beta}+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{\alpha{m_1}}{[\rho(t)r(t)]^{\frac{1}{\alpha}}}w^{\frac{\alpha+1}{\alpha}}(t), ~~t\geq{t_1}, $

(3.7)

其中$m_1=[y(t_1)]^{\frac{\beta-\alpha}{\alpha}}.$

注意到$r'(t)\geq{0}$和$(r(t)(y'(t))^{\alpha})'\leq{0}.$故$y''(t)\leq{0}.$

因此, 当$\alpha>\beta$时, 我们有

$ \begin{array}{l} w'\left(t \right)\le - \rho(t)q(t){(\frac{{y(\sigma(t))}}{{y(t)}})^\beta }+\frac{{\rho '(t)}}{{\rho(t)}}w(t)- \frac{\beta }{{{{[\rho(t)r(t)]}^{\frac{1}{\beta }}}}}{[y'(t)]^{\frac{{\beta - \alpha }}{\alpha }}}{w^{\frac{{\beta+1}}{\beta }}}(t)\\ \;\;\;\;\;\;\;\;\le - \rho(t)q(t){(\frac{{y(\sigma(t))}}{{y(t)}})^\beta }+\frac{{\rho '(t)}}{{\rho(t)}}w(t)- \frac{{\beta {m_2}}}{{{{[\rho(t)r(t)]}^{\frac{1}{\beta }}}}}{w^{\frac{{\beta+1}}{\beta }}}(t), t \ge {t_1}, \end{array} $

(3.8)

其中$m_2=[\frac{1}{y'(t_1)}]^{\frac{\alpha-\beta}{\beta}}.$

由引理3.2, 我们得到$\frac{y(\sigma(t))}{y(t)}\geq\frac{\sigma(t)}{t}.$由(3.7)和(3.8)式可得

$ w'(t)\leq-\rho(t)q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}+\frac{\rho'(t)}{\rho(t)}w(t)-\frac{c m}{[\rho(t)r(t)]^{\frac{1}{c}}}w^{\frac{c+1}{c}}(t), ~~t\geq{t_1}, $

(3.9)

其中$c=\min\{\alpha, \beta\}, m=\left\{\begin{array}{ll} 1, \qquad&\alpha=\beta, \\ \mbox{正常数}, \quad &\alpha\neq\beta. \end{array}\right.$

在(3.9)式中利用不等式(2.42)', 我们有

$ w'(t)\leq-\rho(t)q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}+\frac{(\rho'(t))^{c+1}r(t)}{(c+1)^{c+1}(m\rho(t))^c}, ~~t\geq{t_1}, $

(3.10)

对(3.10)式积分得

$ w(t)\leq{w(t_1)}-\frac{1}{m^c}\int_{t_1}^{t} \Big[\rho(s)q_{m}(s)\Big(\frac{\sigma(s)}{s}\Big)^{\beta}-\frac{(\rho'(s))^{c+1}r(s)}{(c+1)^{c+1}\rho^{c}(s)}\Big]{\rm d}s, $

令$t\rightarrow\infty, $由(3.5)式得$\lim\limits_{t\rightarrow\infty}w(t)=-\infty.$此与$w(t)>0$矛盾.定理3.1证毕.

例3.1  考虑方程

$ \Big(\frac{1}{t^{\beta-c}}|y'(t)|^{\alpha-1}y'(t)\Big)'+\frac{\ln{t}}{t^{1+\beta}} \Big|x \Big(\frac{t}{2}\Big)\Big|^{\beta-1}x\Big(\frac{t}{2}\Big)=0, ~~t\geq1, $

(3.11)

其中$y(t)=x(t)-\frac{1}{3}x(t-1).$

$ \int_{t_0}^{\infty}q(t)\sigma^{\beta}(t){\rm d}t=\int_{t_0}^{\infty} \frac{\ln{t}}{t^{1+\beta}}\Big(\frac{t}{2}\Big)^{\beta}{\rm d}t=\frac{1}{2^{\beta}}\ln\ln{t}\mid_{t_0}^{\infty}=\infty, $

其中取$\rho(t)=t^{\beta}$, 则

$ \begin{eqnarray*} &&\int_{t_0}^{\infty} \bigg[\rho(t)q_{m}(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{(\rho'(t))^{c+1}r(t)}{(c+1)^{c+1}\rho^{c}(t)}\bigg]{\rm d}t \\ &=&\int_{t_0}^{\infty}\bigg[m^{c}\frac{\ln{t}}{t^{1+\beta}}t^{\beta} \Big(\frac{1}{2}\Big)^{\beta}-\frac{\beta^{c+1}t^{\beta{c}+\beta-c-1}}{(c+1)^{c+1}t^{\beta-c}t^{\beta{c}}}\bigg]{\rm d}t \\ &=&\int_{t_0}^{\infty}\frac{1}{t}\bigg[\frac{m^{c}}{2^{\beta}}\ln{t}- \frac{(\beta)^{c+1}}{(c+1)^{c+1}}\bigg]{\rm d}t=\infty. \end{eqnarray*} $

由定理3.1知方程(3.11)振动或者趋向于零.

定理3.2  设(2.1)和(3.3)式成立, 存在$\varphi(t)\in{C^1}([t_0, \infty), (0, \infty))$和常数m>0使得

$ \int_{t_0}^{\infty} \bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}- \frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg]\exp\bigg[(c+1)m^{\frac{c}{c+1}} \int_T^t \frac{\varphi(s)}{r^{\frac{1}{c}}(s)}{\rm d}s\bigg]{\rm d}t=\infty, $

(3.12)

则方程(1.1)的解x(t)振动或者$\lim\limits_{t\rightarrow\infty}x(t)=0.$

证  设x(t)是方程(1.1)的非振动解, 不失一般性, 可设x(t)最终为正, 由引理3.1知存在$t_1\geq{t_0}$使得(3.1)式成立或者$\lim\limits_{t\rightarrow\infty}x(t)=0.$现设(3.1)式成立.

定义函数

$ u(t)=\frac{r(t)(y'(t))^{\alpha}}{y^{\beta}(t)}, ~~t\geq{t_1}, $

(3.13)

则$u(t)>0.$

利用(3.9)式中取$\rho(t)=1.$我们得到

$ u'(t)\leq-q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{cm}{r^{\frac{1}{c}}(t)}u^{\frac{c+1}{c}}(t), ~~t\geq{t_1}, $

(3.14)

其中$c=\min\{\alpha, \beta\}, m=\left\{\begin{array}{ll} 1, \qquad &\alpha=\beta, \\ \mbox{正常数}, \quad &\alpha\neq\beta.\end{array}\right.$(3.14)式即为

$ u'(t)\leq- \bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg]-\frac{1}{r^{\frac{1}{c}}(t)}[\varphi^{c+1}(t)+cmu^{\frac{c+1}{c}}(t)]. $

记$u_m(t)=m^{\frac{c}{c+1}}u(t), $则

$ u'(t)\leq- \bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg]-\frac{1}{r^{\frac{1}{c}}(t)}[\varphi^{c+1}(t)+cu_m^{\frac{c+1}{c}}(t)]. $

(3.15)

利用Young不等式

$ ab\leq\frac{1}{p}\cdot a^{p}+\frac{1}{q}\cdot b^{q}, $

其中$a>0, b>0, \frac{1}{p}+\frac{1}{q}=1.$

令$p=c+1, q=\frac{c+1}{c}, c>0.$则(3.15)式成为

$ (c+1)ab\leq{a^{c+1}}+cb^{\frac{c+1}{c}}. $

(3.16)

在(3.16)式中令$a=\varphi(t), b=u_{m}(t)$则

$ \varphi^{c+1}(t)+cu_m^{\frac{c+1}{c}}(t)\geq(c+1)\varphi(t)u_{m}(t). $

因此

$ u'(t)\leq- \bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg] -(c+1)r^{-\frac{1}{c}}(t)\varphi(t)u_{m}(t). $

即

$ u'(t)+(c+1)r^{-\frac{1}{c}}(t)\varphi(t)u_{m}(t)\leq- \bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg]. $

故有

$ \begin{eqnarray*} &&\bigg(\exp\bigg[(c+1)m^{\frac{c}{c+1}}\int_T^{t}r^{-\frac{1}{c}}(s)\varphi(s){\rm d}s\bigg]u(t)\bigg)' \\ &\leq&-\bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}-\frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg] \exp\bigg[(c+1)m^{\frac{c}{c+1}}\int_T^{t}r^{-\frac{1}{c}}(t)\varphi(s){\rm d}s\bigg]. \end{eqnarray*} $

积分得

$ \begin{eqnarray*} 0&<&\exp\bigg[(c+1)m^{\frac{c}{c+1}}\int_T^{t}r^{-\frac{1}{c}}(s)\varphi(s){\rm d}s\bigg]u(t)\\ &\leq&{u(T)} -\int_T^{t}\bigg[q(s)\Big(\frac{\sigma(s)}{s}\Big)^{\beta}-\frac{\varphi^{c+1}(s)}{r^{\frac{1}{c}}(s)}\bigg] \exp\bigg[(c+1)m^{\frac{c}{c+1}}\int_T^{s}r^{-\frac{1}{c}}(\tau)\varphi(\tau){\rm d}\tau\bigg]{\rm d}s. \end{eqnarray*} $

上式与条件(3.12)矛盾.故(3.1)式不成立.则$\lim\limits_{t\rightarrow\infty}x(t)=0.$定理3.2证毕.

例3.2  考虑方程

$ (|y'(t)|^{\alpha-1}y'(t))'+\frac{\ln{t}}{t^{1+c}}|x \Big(\frac{t}{2}\Big)|^{\beta-1}x\Big(\frac{t}{2}\Big)=0, $

(3.17)

其中

$ y(t)=x(t)-\frac{1}{3}x(t-1), ~~\int_{t_0}^{\infty}q(t)\sigma^{\beta}(t){\rm d}t=\infty, $

$ \int_{t_0}^{\infty}\bigg[q(t)\Big(\frac{\sigma(t)}{t}\Big)^{\beta}- \frac{\varphi^{c+1}(t)}{r^{\frac{1}{c}}(t)}\bigg]\exp\bigg[(c+1)m^{\frac{c}{c+1}} \int_T^t\frac{\varphi(s)}{r^{\frac{1}{c}}(s)}{\rm d}s\bigg]{\rm d}t=\infty. $

由定理3.2知方程(3.17)的解振动或者$\lim\limits_{t\rightarrow\infty}x(t)=0.$

注3  本文中定理对任意$\alpha>0, \beta>0$适用, 文献[1-19]中的定理均不能适用于本文的例子.