本文考虑非线性二阶中立型分布时滞微分方程

(1.1) $\left(r(t)|z'(t)|^{\alpha-1}z'(t)\right)'+\int^{d}_{c}f\left(t, x\left(\sigma(t, \xi)\right)\right){\rm d}\xi=0$

(其中$z(t)=x(t)+\int^{b}_{a}p(t, \xi)x\left(\tau(t, \xi)\right){\rm d}\xi, ~t\geq t_{0}>0, ~\alpha>0, ~0\leq a<b, ~0\leq c<d $).在本文中,我们总假设下列条件成立

(H$_{1})~~~\tau(t, \xi)\in C\left([t_{0}, \infty)\times[a, b], (0, \infty)\right), ~\tau(t, \xi)\leq t, ~\xi\in[a, b], $当$t\rightarrow\infty$时$\tau(t, \xi)\rightarrow\infty$;

(H$_{2})~~\sigma(t, \xi)\in C\left([t_{0}, \infty)\times[c, d], (0, \infty)\right), ~\sigma(t, \xi)$是$\xi$的减函数, $\sigma(t, \xi)\leq t, ~\xi\in[c, d], $当$t\rightarrow\infty$时$\sigma(t, \xi)\rightarrow\infty, ~g(t)=\sigma(t, d)$且$g'(t)>0;$

(H$_{3})~~p(t, \xi)\geq0, ~0\leq p(t)=\int^{b}_{a}p(t, \xi){\rm d}\xi<1; $

(H$_{4})~~r(t)>0, ~r'(t)\geq 0, \int^{\infty}_{t_{0}}\left(\frac{1}{r(t)}\right)^{\frac{1}{\alpha}}{\rm d}t=\infty;$

(H$_{5})~~f(t, u)\in C\left([t_{0}, \infty)\times{\Bbb R}, {\Bbb R}\right), ~uf(t, u)>0, ~u\neq0.$存在函数$q(t, \xi)\in C\big([t_{0}, \infty)\times[c, d], $$ [0, \infty)\big)$使得

$|f(t, u)|\geq q(t, \xi)|u^{\beta}|, ~\beta> 0.$

方程(1.1)有如下重要的特例

(ⅰ)在相应的离散情况下,方程(1.1)成为

(1.2) $\left(r(t)|z'(t)|^{\alpha-1}z'(t)\right)'+f\left(t, x\left(\sigma(t)\right)\right)=0, $

其中$z(t)=x(t)+p(t)x\left(\tau(t)\right), ~f(t, x)sgn x\geq q(t)|x|^{\beta} , ~x\neq0, ~\beta>0.$

(ⅱ)方程(1.2)中当$\alpha=\beta$时有特例

(1.3) $\left(r(t)|z'(t)|^{\alpha-1}z'(t)\right)'+q(t)|x\left(\sigma(t)\right)|^{\alpha-1}x\left(\sigma(t)\right)=0, $

(ⅲ)方程(1.2)中当$\alpha=1$时有特例

(1.4) $\left(r(t)\left(x(t)+p(t)x\left(\tau(t)\right)\right)'\right)'+q(t)|x\left(\sigma(t)\right)|^{\beta}sgnx\left(\sigma(t)\right)=0.$

方程(1.3)和(1.4)分别称为半线性中立型微分方程和中立型Emden-Fowler微分方程.我们看到两者是互不包含的,但是,它们都是方程(1.1)和(1.2)的特例.

近年来,随着二阶中立型微分方程在自然科学和工程技术中的应用日益广泛,许多学者对其振动性的充分条件的研究有着浓厚的兴趣.我们推荐文献[1-13]及其引文.其中Candan在文献[1]中建立了方程(1.1)在$\alpha=\beta$的条件下的若干振动准则. Dong在文献[2]中给出了当$\alpha=\beta$时方程(1.2)的某些振动结果.此后,文献[3]和[8]进一步研究了方程(1.2)当$~f(t, x)= q(t)|x|^{\beta}sgn x$时的振动性,分别给出了$\alpha\geq\beta$和$\alpha\leq\beta$时方程(1.2)的振动准则.此外,文献[4-6]给出了半线性微分方程(1.3)振动的若干充分条件.而Erbe等在文献[7]中证明了中立型Emden-Fowler微分方程(1.4)当$\beta>1$时的Philos型振动准则.

本文的目的是研究方程(1.1)在一般情况下的振动准则.利用广义Riccati变换和分析技巧,建立了一个新的对任意$\alpha>0$和$\beta>0$均成立的广义Riccati不等式.亦由此证明了方程(1.1)的四个新的振动定理.所得结果推广改进并统一了文献[1-8]中相应的振动结果.它们既适用于连续分布时滞方程(1.1),也适用于离散时滞方程(1.2).而且,也同样适用于半线性中立型微分方程(1.3)和中立型Emden-Fowler微分方程(1.4).

函数$x(t)$称为方程(1.1)的一个解,如果函数$z(t)$和$r(t)|z'(t)|^{\alpha-1}z'(t)$是连续可微且在区间$[t_{0}, \infty)$上$x(t)$满足方程(1.1).方程(1.1)的一个非平凡解称为振动的,如果它有任意大的零点,否则,称它为非振动的.方程(1.1)的一切解均振动,则称方程(1.1)为振动的.

定理2.1  设

(2.1) $\begin{equation}\int^{\infty}_{t_{0}}Q(t){\rm d}t=\infty, \end{equation}$

其中

(2.2) $\begin{equation}Q(t)=\int^{d}_{c}\left[1-p\left(\sigma(t, \xi)\right)\right]^{\beta}q(t, \xi){\rm d}\xi, \end{equation}$

则方程(1.1)振动.

证  设$x(t)$为方程(1.1)的一个非振动解.不失一般性,不妨设$x(t)$为最终正解.则存在$t_{1}\geq t_{0}, $使得当$t\geq t_{1}$时有$x(t)>0, ~x\left(\tau(t, \xi)\right)>0~(\xi\in[a, b]), ~x\left(\sigma(t, \xi)\right)>0~(\xi\in[c, d])$.如果$x(t)$为最终负解,可用同样的方法来讨论.由条件(H$_{5})$我们得到

(2.3) $\begin{eqnarray}\left(r(t)|z'(t)|^{\alpha-1}z'(t)\right)'&=&-\int^{d}_{c}f\left(t, x\left(\sigma(t, \xi)\right)\right){\rm d}\xi\\&\leq&-\int^{d}_{c}q(t, \xi)x^{\beta}\left(\sigma(t, \xi)\right){\rm d}\xi\leq0. \end{eqnarray}$

因此, $r(t)|z'(t)|^{\alpha-1}z'(t)$是非增函数且$z'(t)$最终保号,于是$z'(t)$仅有两种可能.最终有$z'(t)<0$或$z'(t)>0.$

(ⅰ)假设$z'(t)<0, ~ t\geq t_{1}.$则由(2.3)式我们有

$(r(t)|z'(t)|^{\alpha-1}z'(t))'=(-r(t)(-z'(t))^{\alpha})'\leq0, $

即

$(r(t)(-z'(t))^{\alpha})'\geq0.$

利用函数$r(t)(-z'(t))^{\alpha}$是非减函数,可知存在常数$C>0$使得

(2.4) $r(t)(-z'(t))^{\alpha}\geq r(t_{2})(-z'(t_{2}))^{\alpha}\equiv C>0, ~t\geq t_{2}\geq t_{1}. $

用$r(t)$除(2.4)式两边,从$t_{2}$到$t$积分得

$z(t)\leq z(t_{2})-C^{\frac{1}{\alpha}}\int^{t}_{t_{2}}r^{-\frac{1}{\alpha}}(s){\rm d}s, ~~t\geq t_{2}.$

上式中令$t\rightarrow\infty$,由条件(H$_{4})$得$z(t)\rightarrow-\infty$.此式与$z(t)>0$矛盾,故假设不成立.

(ⅱ)假设$z'(t)>0, ~t\geq t_{1}$.因$z(t)\geq x(t)$且$z(t)$是增的,则有

$\begin{eqnarray*}z(t)&=&x(t)+\int^{b}_{a}p(t, \xi)x\left(\tau(t, \xi)\right){\rm d}\xi\\&\leq &x(t)+\int^{b}_{a}p(t, \xi)z\left(\tau(t, \xi)\right){\rm d}\xi\\&\leq &x(t)+\int^{b}_{a}p(t, \xi)z(t){\rm d}\xi, \end{eqnarray*}$

由上式,我们有

$\left(1-p(t)\right)z(t)\leq x(t), t\geq t_{2}^{\ast}\geq t_{1}$

或者

(2.5) $\left[\left(1-p\left(\sigma(t, \xi)\right)\right)z\left(\sigma(t, \xi)\right)\right]^{\beta}\leq x^{\beta}\left(\sigma(t, \xi)\right), t\geq t_{3}\geq t_{2}^{\ast}, ~\xi\in[c, d]. $

将(2.5)式带入(2.3)式,注意到$\sigma(t, \xi)$是$\xi$的减函数,我们得到

$\left(r(t)\left(z'(t)\right)^{\alpha}\right)'\leq-\int^{d}_{c}q(t, \xi)\left[1-p\left(\sigma(t, \xi)\right)\right]^{\beta} z^{\beta}\left(\sigma(t, d)\right){\rm d}\xi$

或者

(2.6) $\left(r(t)\left(z'(t)\right)^{\alpha}\right)'\leq-Q(t)z^{\beta}\left(g(t)\right).$

另一方面,因$r(t)\left(z'(t)\right)^{\alpha}$是减函数,我们有

$r(t)\left(z'(t)\right)^{\alpha}\leq r\left(g(t)\right)\left(z'(g(t))\right)^{\alpha}$

或者

(2.7) $\left(\frac{r(t)}{r\left(g(t)\right)}\right)^{\frac{1}{\alpha}}\leq\frac{z'\left(g(t)\right)}{z'(t)}.$

定义函数

(2.8) $W(t)=\frac{r(t)\left(z'(t)\right)^{\alpha}}{z^{\beta}\left(g(t)\right)}, ~t\geq t_{3}.$

显然, $W(t)>0.$利用(2.6)-(2.8)式,求导产生

(2.9) $\begin{eqnarray}W'(t)&=&\frac{\left(r(t)\left(z'(t)\right)^{\alpha}\right)'}{z^{\beta}\left(g(t)\right)}-\frac{\beta r(t)\left(z'(t)\right)^{\alpha}z'\left(g(t)\right)g'(t)}{z^{\beta+1}\left(g(t)\right)}\\&\leq&-Q(t)-\frac{\beta g'(t) \left(r^{\frac{1}{\alpha}}(t)z'(t)\right)^{\alpha+1}}{r^{\frac{1}{\alpha}}\left(g(t)\right)z^{\beta+1}\left(g(t)\right)}.\end{eqnarray}$

因为$z(t)$和$z'(t)$均为正.我们有

(2.10) $\begin{eqnarray}W'(t)+Q(t)\leq0.\end{eqnarray}$

对(2.10)式从$t_{3}$到$t$积分,利用(2.2)式可得

$ W(t)\leq W(t_{3})-\int^{t}_{t_{3}}Q(s){\rm d}s\rightarrow\infty, ~~t\rightarrow\infty. $

此与$W(t)>0$矛盾.定理2.1证毕.

注2.1  定理2.1将对二阶线性微分方程

$(r(t)x'(t))'+q(t)x(t)=0$

的Leighton-Wintner振动定理推广到非线性中立型微分方程(1.1),而且使其成为特例.其次,最近Candan在文献[2]中建立的定理2.1仅对方程(1.1)中当$\alpha=\beta$时才成立,而本文中定理2.1则对任意的$\alpha>0, ~\beta>0$都成立.

下面的引理给出方程(1.1)的一个新的Riccati不等式.

引理2.1  设$x(t)$是方程(1.1)的最终正解, $W(t)$由(2.8)式定义.则下面的不等式成立

(2.11) $\begin{eqnarray}W'(t)\leq-Q(t)-\frac{\rho kg'(t)}{r^{\frac{1}{\rho}}\left(\theta(t)\right)}W^{\frac{\rho+1}{\rho}}(t), \end{eqnarray}$

其中$\rho=\min\{\alpha, \beta\}, $且$k= \left \{\begin{array}{ll}1, & \alpha=\beta, \\ \mbox{正常数, }&\alpha\neq\beta, \end{array}\right.$和$\theta(t)= \left \{\begin{array}{ll}t, & \alpha>\beta, \\ g(t) , & \alpha\leq\beta. \end{array}\right.$

证  如同定理2.1的证明,我们得到(2.9)式成立,即

(2.12) $\begin{eqnarray}W'(t)&\leq&-Q(t)-\frac{\beta g'(t) \left(r^{\frac{1}{\alpha}}(t)z'(t)\right)^{\alpha+1}}{r^{\frac{1}{\alpha}}\left(g(t)\right)z^{\beta+1}\left(g(t)\right)} \\&\leq&-Q(t)-\frac{\beta g'(t)}{r^{\frac{1}{\alpha}}\left(g(t)\right)}\left[z\left(g(t)\right)\right]^{\frac{\beta-\alpha}{\alpha}}W^{\frac{\alpha+1}{\alpha}}(t) .\end{eqnarray}$

注意到$z\left(g(t)\right)$是增函数,当$\beta\geq\alpha$时,存在常数$k_{1}>0$和$t_{4}\geq t_{3}$使得当$t\geq t_{4}$时, $\left[z\left(g(t)\right)\right]^{\frac{\beta-\alpha}{\alpha}}\geq k_{1}, $故(2.12)式成为

(2.13) $\begin{eqnarray}W'(t)\leq-Q(t)-\frac{\alpha k_{1} g'(t)}{r^{\frac{1}{\alpha}}(g(t))}W^{\frac{\alpha+1}{\alpha}}(t), ~t\geq t_{4}, ~\beta\geq\alpha .\end{eqnarray}$

易知,当$\beta=\alpha$时, $k_{1}$可为1.

其次,因为$\left(r(t)(z'(t))^{\alpha}\right)'\leq0$和$r'(t)\geq0$固有$z''(t)\leq0$即$z'(t)$是减函数.则当$\alpha>\beta$时$\left[z'(t)\right]^{\frac{\beta-\alpha}{\beta}}$是增函数,于是存在常数$k_{2}>0$和$t_{5}\geq t_{4}$使得$\left[z'(t)\right]^{\frac{\beta-\alpha}{\beta}}\geq k_{2}, ~t\geq t_{5}$因此，不等式(2.9)也可产生

(2.14) $\begin{eqnarray}W'(t)&\leq&-Q(t)-\frac{\beta g'(t)}{r^{\frac{1}{\beta}}(t)}\left[z'(t)\right]^{\frac{\beta-\alpha}{\beta}}W^{\frac{\beta+1}{\beta}}(t) \\&\leq&-Q(t)-\frac{\beta k_{2} g'(t)}{r^{\frac{1}{\beta}}(t)}W^{\frac{\beta+1}{\beta}}(t), ~t\geq t_{5}, ~\alpha>\beta .\end{eqnarray}$

$k=\min\{k_{1}, k_{2}\}, $联合(2.13)和(2.14)式,我们得到不等式(2.11)对一切$\alpha>0, ~\beta>0$成立.引理2.1证毕.

定义函数序列$\left\{Q_{n}\right\}$如下

(2.15) $Q_{1}(t)=\int^{\infty}_{t}Q(s){\rm d}s, $

(2.15') $Q_{n}(t)=\int^{\infty}_{t}Q_{n-1}^{\frac{\rho+1}{\rho}}(s)R(s){\rm d}s+Q_{1}(t), ~t\geq t_{0}, n=2, ~3, ~4, ~\cdots.$

利用归纳法,我们有$Q_{n}(t)\leq Q_{n+1}(t), ~t\geq t_{0}, n=1, ~2, ~3, ~\cdots.$

为简单计算,在不等式(2.11)中,记

(2.16) $R(t)=\frac{\rho kg'(t)}{r^{\frac{1}{\rho}}(\theta(t))}.$

则(2.11)式成为

(2.17) $W'(t)+Q(t)+R(t)W^{\frac{\rho+1}{\rho}}(t)\leq0.$

引理2.2  设$x(t)$是方程(1.1)的最终正解,则

(ⅰ) $Q_{n}(t)\leq W(t), $其中$W(t)$和$Q_{n}(t)$分别由(2.8)、(2.15)和$(2.15')$式定义.

(ⅱ)存在正函数$\overline{Q}(t)$使得$\lim\limits_{ t\rightarrow\infty}Q_{n}(t)=\overline{Q}(t), ~t\geq T\geq t_{0}$且

(2.18) $\overline{Q}(t)=\int^{\infty}_{t}R(s)\left[\overline{Q}(s)\right]^{\frac{\rho+1}{\rho}}{\rm d}s+Q_{1}(t), ~t\geq T .$

证  在引理2.1的证明中,我们得到(2.11)式成立,即(2.17)式成立.对(2.17)积分可得

(2.19) $W(t')-W(t)+\int^{t'}_{t}Q(s){\rm d}s+\int^{t'}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s\leq0.$

易知

(2.20) $W(t')-W(t)+\int^{t'}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s\leq0.$

(2.20)式给出

(2.21) $\int^{\infty}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s<\infty, ~t\geq T.$

否则,若(2.21)式发散,则对固定的$t\geq T$,我们有

$W(t')\leq W(t)-\int^{t'}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s \rightarrow-\infty, ~t'\rightarrow\infty, $

此与$W(t)>0$矛盾.又由(2.17)式知$W'(t)\leq0$,因此, $\lim\limits_{n\rightarrow\infty}W(t)=l\geq0.$注意到(2.21)式和条件(H$_{4}$),我们有$l=0.$因此,由(2.19)式我们得到

(2.22) $W(t)\geq\int^{\infty}_{t}Q(s){\rm d}s+\int^{\infty}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s=Q_{1}(t)+\int^{\infty}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s .$

则$W(t)\geq Q_{1}(t).$

假设$W(t)\geq Q_{n-1}(t)$成立.由上式和不等式(2.22)得

$W(t)\geq Q_{1}(t)+\int^{\infty}_{t}R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s\geq Q_{1}(t)+\int^{\infty}_{t}R(s)Q_{n-1}^{\frac{\rho+1}{\rho}}(s){\rm d}s=Q_{n}(t).$

由归纳法,可以得到$W(t)\geq Q_{n}(t), ~t\geq t_{0}, ~(n=2, ~3, ~4, ~\cdots).$注意到$W(t)$有上界且由引理2.1证明过程得$Q_{n}(t)\leq Q_{n+1}(t), ~t\geq t_{0}, n=1, ~2, ~3, ~\cdots$故函数序列$\left\{Q_{n}\right\}$单调增加有上界. $\left\{Q_{n}\right\}$收敛到$\overline{Q}(t).$在$(2.15')$式中令$n\rightarrow\infty, $利用勒贝格单调收敛定理,我们得到(2.18)式.引理2.2证毕.

定理2.2  设

(2.23) $\liminf\limits_{t\rightarrow\infty}\frac{1}{Q_{1}(t)}\int^{\infty}_{t}\left[Q_{1}(s)\right]^{\frac{\rho+1}{\rho}}R(s){\rm d}s>\frac{\rho}{(\rho+1)^{\frac{\rho+1}{\rho}}} , $

其中$\rho, ~Q_{1}(t)$和$R(t)$分别由(2.11)、(2.15)和(2.16)式定义,则方程(1.1)振动.

证  设$x(t)$是方程(1.1)的非振动解.如同引理2.1和引理2.2的证明,我们得到(2.22)式.由此可得

(2.24) $\frac{W(t)}{Q_{1}(t)} \geq1+\frac{1}{Q_{1}(t)}\int^{\infty}_{t}R(s)\left[Q_{1}(s)\right]^{\frac{\rho+1}{\rho}} \left[\frac{W(s)}{Q_{1}(s)}\right]^{\frac{\rho+1}{\rho}}{\rm d}s , ~~t\geq T.$

令

(2.25) $\inf\limits_{t\geq T}\frac{W(t)}{Q_{1}(t)}=\lambda.$

则可知$\lambda\geq1.$另一方面,由(2.23)式知存在常数$c>0$使得

(2.26) $\liminf\limits_{t\rightarrow\infty}\frac{1}{Q_{1}(t)}\int^{\infty}_{t}\left[Q_{1}(s)\right]^{\frac{\rho+1}{\rho}}R(s){\rm d}s>c> \frac{\rho}{(\rho+1)^{\frac{\rho+1}{\rho}}}.$

联合(2.24)-(2.26)式,我们有

(2.27) $\lambda\geq1+\lambda^{\frac{\rho+1}{\rho}}c.$

利用不等式

(2.28) $Bu-Au^{\frac{\rho+1}{\rho}}\leq\frac{\rho^{\rho}}{(\rho+1)^{\rho+1}}\frac{B^{\rho+1}}{A^{\rho}}, ~A>0, ~B\geq0, , ~\rho>0.$

我们得到

(2.29) $\lambda-c\lambda^{\frac{\rho+1}{\rho}}\leq\frac{\rho^{\rho}}{(\rho+1)^{\rho+1}}\frac{1}{c^{\rho}}.$

注意到(2.26)式的后部分,由不等式(2.29)可得

$\lambda-c\lambda^{\frac{\rho+1}{\rho}}<1, $

即

(2.30) $\lambda<1+c\lambda^{\frac{\rho+1}{\rho}}.$

我们看到(2.27)与(2.30)式矛盾.假设不成立.定理2.2证毕.

注2.2  注意到,方程(1.1)中当$\alpha=\beta$时, (2.16)中$\rho=\alpha, ~k=1, ~\theta(t)=g(t)=\sigma_{1}(t).$因此文献[1]中的定理2.3是本文定理2.2的特例.即定理2.2将文献[1]中的$\alpha=\beta$时的结果推广到对任意$\alpha>0, ~\beta>0$都成立.此外,也推广和改进了文献[2]的结果.

定理2.3  设存在函数$\varphi (t)\in C^{1}([t_{0}, \infty), {\Bbb R}^{+})$使得条件

(2.31) $\int^{\infty}_{t_{0}}\left[\varphi(t)Q(t)-\frac{r(\theta(t))(\varphi'(t))^{\rho+1}}{(\rho+1)^{\rho+1}(k\varphi(t)g'(t))^{\rho}}\right]{\rm d}t=\infty$

成立,其中常数$\rho, ~k$函数$Q(t), ~\theta(t)$和$g(t)$的定义同上文,则方程(1.1)振动.

证  设$x(t)$是方程(1.1)的非振动解.不失一般性,设$x(t)$为最终正解($x(t)<0$的情况类似的分析成立).如同引理2.1的证明,利用定义(2.16)可得(2.17)式成立,即

$W'(t)+Q(t)+R(t)W^{\frac{\rho+1}{\rho}}(t)\leq0, ~~t\geq T\geq t_{0}.$

上式乘以$\varphi(t)$再积分可得

(2.32) $\int^{t}_{T}\varphi(s)Q(s){\rm d}s\leq -\int^{t}_{T}\varphi(s)W'(s){\rm d}s-\int^{t}_{T}\varphi(s)R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s .$

利用分部积分和(2.16)式中$R(t)$的定义,我们有

(2.33) $\int^{t}_{T}\varphi(s)Q(s){\rm d}s\leq\varphi(T)W(T)+ \int^{t}_{T}\left(\varphi'(s)W(s)-\frac{\rho k\varphi(s)g'(s)}{r^{\frac{1}{\rho}}(\theta(s))} W^{\frac{\rho+1}{\rho}}(s)\right){\rm d}s.$

在(2.33)式右端积分中利用不等式(2.28)可得

$\int^{t}_{T}\varphi(s)Q(s){\rm d}s\leq\varphi(T)W(T)+ \int^{t}_{T}\frac{r\left(\theta(s)\right)\left(\varphi'(s)\right)^{\rho+1}}{(\rho+1)^{\rho+1}\left(k\varphi(s) g'(s)\right)^{\rho}}{\rm d}s, $

即

(2.34) $\int^{t}_{T}\left(\varphi(s)Q(s)-\frac{r\left(\theta(s)\right)\left(\varphi'(s)\right)^{\rho+1}}{(\rho+1)^{\rho+1}\left(k\varphi (s) g'(s)\right)^{\rho}}\right){\rm d}s\leq\varphi(T)W(T).$

我们得到(2.34)与(2.31)式矛盾.假设不成立.定理2.3证毕.

注2.3  定理2.3推广和改进了文献[3]中的定理2.1,由于文献[3]的证明中使用了普通的Riccati不等式.本文使用了我们证明的推广了的广义Riccati不等式(2.11).而且文献[3]的振动定理只适用于$\alpha\geq\beta>0$和离散时滞的情况.而本文结果适用于任意$\alpha>0, ~\beta>0$和连续分布时滞方程.此外,本文定理2.3还推广和改进了文献[8]的定理2.1和定理3.1,以及文献[4-6]中相应的振动结果.

下面我们给出方程(1.1)的Philos型振动定理.

定义集合

$D=\left\{(t, s):t\geq s\geq t_{0}\right\};~D_{0}=\left\{(t, s):t> s\geq t_{0}\right\}. $

函数$H\in C^{1}(D, {\Bbb R})$称为属于$\Delta$类,如果它满足

(ⅰ) $H(t, t)=0, ~t\geq t_{0};~H(t, s)>0, ~(t, s)\in~D_{0};$

(ⅱ) $\frac{\partial H(t, s)}{\partial s}\leq0.$存在$\psi\in C^{1}\left([t_{0}, \infty), {\Bbb R}^{+}\right)$和$h\in C(D_{0}, {\Bbb R})$使得

(2.35) $\frac{\partial H(t, s)}{\partial s}+\frac{\psi'(s)}{\psi(s)}H(t, s)=-h(t, s)H^{^{\frac{\rho}{\rho+1}}}(t, s),$

其中$\rho=\min\{\alpha, \beta\}.$

定理2.4  设存在函数$\psi\in C^{1}\left([t_{0}, \infty), {\Bbb R}^{+}), ~h\in C(D_{0}, {\Bbb R}\right)$和$ H\in\Delta, ~T\geq t_{0}$使得

(2.36) $\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T}\left[H(t, s)\psi(s)Q(s)- \frac{\psi(s)r\left(\theta(s)\right)}{(\rho+1)^{\rho+1}\left(kg'(s)\right)^{\rho}}|h(t, s)|^{\rho+1}\right]{\rm d}s=\infty $

成立,其中其中常数$\rho, ~k$函数$Q(t), ~\theta(t)$和$g(t)$的定义如上文,则方程(1.1)振动.

证  设$x(t)$是方程(1.1)的非振动解.不失一般性,设$x(t)$为最终正解($x(t)<0$的情况类似的分析成立).如同引理2.1的证明,利用定义(2.16)可得(2.17)式

$W'(t)+Q(t)+R(t)W^{\frac{\rho+1}{\rho}}(t)\leq0, ~~t\geq T\geq t_{0}.$

对$t>s$,上式乘以$H(t, s)\psi(s)$再从$T$到$t$积分可得

(2.37) $\int^{t}_{T}H(t, s)\psi(s)Q(s){\rm d}s\leq -\int^{t}_{T}H(t, s)\psi(s)W'(s){\rm d}s-\int^{t}_{T}H(t, s)\psi(s)R(s)W^{\frac{\rho+1}{\rho}}(s){\rm d}s. $

对(2.37)式分部积分和利用(2.35)式我们有

(2.38) $\begin{eqnarray}&&\int^{t}_{T}H(t, s)\psi(s)Q(s){\rm d}s\\&\leq &H(t, T)\psi(T)W(T) + \int^{t}_{T}\psi(s)\left[|h(t, s)|H^{\frac{\rho}{\rho+1}}(t, s)W(s)- H(t, s)R(s)W^{\frac{\rho+1}{\rho}}(s)\right]{\rm d}s.\end{eqnarray}$

在(2.38)式右端积分中利用不等式(2.28)和(2.16)式可得

(2.39) $\int^{t}_{T}H(t, s)\psi(s)Q(s){\rm d}s\leq H(t, T)\psi(T)W(T) + \int^{t}_{T}\frac{\psi(s)r\left(\theta(s)\right)|h(t, s)|^{\rho+1}}{(\rho+1)^{\rho+1}\left(kg'(s)\right)^{\rho}}] {\rm d}s .$

由不等式(2.39)可知

(2.40) $\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T}\left[H(t, s)\psi(s)Q(s)- \frac{\psi(s)r\left(\theta(s)\right)|h(t, s)|^{\rho+1}}{(\rho+1)^{\rho+1}\left(kg'(s)\right)^{\rho}}\right]{\rm d}s\leq\psi(T)W(T).$

我们得到(2.40)与(2.36)式矛盾.假设不成立.定理2.4证毕.

推论2.1  若(2.36)式由下列两条件代替

(2.41) $\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T}H(t, s)\psi(s)Q(s){\rm d}s=\infty$

和

(2.42) $\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T} \frac{\psi(s)r\left(\theta(s)\right)}{\left(g'(s)\right)^{\rho}}|h(t, s)|^{\rho+1}{\rm d}s<\infty, $

则定理2.4的结论仍然成立.

注2.4  定理2.4在三方面改进了文献[3]的定理2.2: (ⅰ)不要求$\alpha\geq\beta$; (ⅱ)简化了文献[3]中(2.14)式的计算; (iii)可适用于方程(1.1).它在三方面改进了文献[6]的定理5: (1)不要求$\alpha\leq\beta;$ (2)不要求$p(t)\equiv0$; (3)可适用于连续分布时滞方程(1.1).定理2.4也将文献[7]的定理2.1对方程(1.4)的振动结果推广且改进为方程(1.1)的结果.

本节给出四个例子说明主要结果的应用.

例3.1  考虑下面的二阶中立型泛函微分方程

$\left(|z'(t)|^{\alpha-1}z'(t)\right)'+\int^{5\pi}_{3\pi}\frac{3}{8}|x(t-\frac{\xi}{2})|^{\beta-1}x(t-\frac{\xi}{2}){\rm d}\xi=0, ~~~~~~~~~~~~~~~~~~~~~~~~({\rm E}_{1}) $

其中$z(t)=x(t)+\int^{\frac{5\pi}{2}}_{\frac{3\pi}{2}}\frac{1}{4}x(t-\xi){\rm d}\xi, ~\alpha>0, ~\beta>0.$

比较方程(E$_{1})$和(1.1),我们看到$r(t)=1, ~a=\frac{3\pi}{2}, ~b=\frac{5\pi}{2}, ~c=3\pi, ~d=5\pi, $$p(t, \xi)=\frac{1}{4}, $$f(t, x)=\frac{3}{8} |x(t)|^{\beta-1}x(t).$显然,定理2.1的条件都满足,故方程(E$_{1})$振动.事实上,当$\alpha=\beta=1$时, $x(t)=\cos t$是方程(E$_{1})$的一个解.

例3.2  考虑下面的二阶中立型泛函微分方程

$\left(|z'(t)|^{\alpha-1}z'(t)\right)'+\int^{1}_{0}\frac{\xi}{t^{1+\frac{\rho}{2}}}|x(t-\xi-1)|^{\beta-1}x(t-\xi-1){\rm d}\xi=0, ~~~~~~~~~~~~~~~~~~~~~({\rm E}_{2}) $

其中$z(t)=x(t)+\int^{1}_{0}\frac{1}{2}x(t-\frac{\xi}{2}-1){\rm d}\xi, ~\alpha>0, ~\beta>0, ~\rho=\min\{\alpha, \beta\}.$

比较方程(E$_{2})$和(1.1),我们可知$r(t)=1, ~p(t, \xi)=\frac{1}{2}, ~f(t, u)=q(t, \xi) |u|^{\beta-1}u, ~q(t, \xi)=\frac{\xi}{t^{1+\frac{\rho}{2}}}, ~\sigma(t, \xi)=t-\xi-1, ~g(t)=t-2.$因此, $R(t)=\frac{\rho kg'(t)}{r^{\frac{1}{\rho}}(\theta(t))}=\rho k.~Q(t)=\int^{d}_{c}[1-p(\sigma(t, \xi))]^{\beta}q(t, \xi){\rm d}\xi= \int^{1}_{0}(\frac{1}{2})^{\beta}\frac{\xi}{t^{1+\frac{\rho}{2}}}{\rm d}\xi=\frac{(\frac{1}{2})^{\beta+1}}{t^{1+\frac{\rho}{2}}} , ~Q_{1}(t)=\int^{\infty}_{t}Q(s){\rm d}s=\frac{1}{\rho}(\frac{1}{2})^{\beta}\frac{1}{t^{\frac{\rho}{2}}}.$下面验证(2.23)式成立

$\begin{eqnarray*}\liminf\limits_{t\rightarrow\infty}\frac{1}{Q_{1}(t)}\int^{\infty}_{t}\left[Q_{1}(s)\right]^{\frac{\rho+1}{\rho}}R(s) {\rm d}s&=&\liminf\limits_{t\rightarrow\infty}2^{\beta}\rho t^{\frac{\rho}{2}}\int^{\infty}_{t}\left[\frac{\frac{1}{\rho}(\frac{1}{2})^{\beta}}{s^{\frac{\rho}{2}}}\right]^{\frac{\rho+1}{\rho}}\rho k{\rm d}s\\&=&\rho k\left[\frac{1}{\rho}(\frac{1}{2})^{\beta}\right]^{\frac{1}{\rho}}\liminf\limits_{t\rightarrow\infty}t^{\frac{1}{2}}=\infty.\end{eqnarray*}$

故(2.23)式成立.由定理2.2知,方程(E$_{2})$振动.

例3.3  考虑下面的二阶中立型泛函微分方程

$\left(|z'(t)|^{\alpha-1}z'(t)\right)'+\int^{2}_{1}\frac{\xi}{t(\ln t)^{1+\rho}}|x(t-2\xi-4)|^{\beta-1}x(t-2\xi-4){\rm d}\xi=0, ~~~~~~~~~~~~~~~~~~~~~({\rm E}_{3}) $

其中$z(t)=x(t)+\int^{1}_{0}\frac{2}{3}x(t+\xi-1){\rm d}\xi, ~\alpha>0, ~\beta>0, ~\rho=\min\{\alpha, \beta\}.$

比较方程(E$_{3})$和(1.1),易得$r(t)=1, ~p(t, \xi)=\frac{2}{3}, ~f(t, u)=q(t, \xi) |u|^{\beta-1}u, ~q(t, \xi)=\frac{\xi}{t(\ln t)^{1+\rho}}, ~\sigma(t, \xi)=t-2\xi-4, ~g(t)=t-8.$因此

$Q(t)=\int^{d}_{c}[1-p(\sigma(t, \xi))]^{\beta}q(t, \xi){\rm d}\xi=\int^{2}_{1}(\frac{1}{3})^{\beta}\frac{\xi}{t(\ln t)^{1+\rho}}{\rm d}\xi=\frac{\frac{1}{2}(\frac{1}{3})^{\beta-1}}{t(\ln t)^{1+\rho}}.$

下面验证(2.31)式是否成立,我们取$\varphi(t)=(\ln t)^{\rho}$则$\varphi'(t)=\frac{\rho}{t}(\ln t)^{\rho-1}.$欲证下式成立

$\int^{\infty}_{t_{0}}\left[\varphi(t)Q(t)-\frac{r(\theta(t))(\varphi'(t))^{\rho+1}}{(\rho+1)^{\rho+1}(k\varphi(t)g'(t))^{\rho}}\right]{\rm d}t=\infty, $

只需证明

$\int^{\infty}_{t_{0}}\varphi(t)Q(t){\rm d}t=\infty $

和

$\int^{\infty}_{t_{0}}\frac{r(\theta(t))(\varphi'(t))^{\rho+1}}{(\varphi(t)g'(t))^{\rho}}{\rm d}t<\infty.$

事实上,我们有

$\int^{\infty}_{t_{0}}\varphi(t)Q(t){\rm d}t=\int^{\infty}_{t_{0}}=\frac{\frac{1}{2}(\frac{1}{3})^{\beta-1}}{t\ln t}{\rm d}t=\infty$

和

$\int^{\infty}_{t_{0}}\frac{r(\theta(t))(\varphi'(t))^{\rho+1}}{(\varphi(t)g'(t))^{\rho}}{\rm d}t= \int^{\infty}_{t_{0}}\frac{(\frac{\rho}{\rho+1})^{\rho+1}}{k^{\rho}t^{\rho+1}\ln t}{\rm d}t<\infty.$

故(2.31)式成立.利用定理2.3知,方程(E$_{3})$振动.

例3.4  考虑下面的二阶中立型泛函微分方程

$\left(|z'(t)|^{\alpha-1}z'(t)\right)'+\int^{1}_{0}2t\xi|x(t-\xi-1)|^{\beta-1}x(t-\xi-1){\rm d}\xi=0, ~~~~~~~~~~~~~~~~~~~~~({\rm E}_{4}) $

其中$z(t)=x(t)+\int^{\frac{\pi}{2}}_{0}\frac{1}{2}x(t-\xi-\frac{\pi}{2}){\rm d}\xi, ~\alpha>0, ~\beta>0.$

比较方程(E$_{4})$和(1.1),易得$r(t)=1, ~p(t, \xi)=\frac{1}{2}, ~f(t, u)=q(t, \xi) |u|^{\beta-1}u, $$q(t, \xi)=2t\xi, $$\sigma(t, \xi)=t-\xi-1, ~g(t)=t-2.$因此, $Q(t)=\int^{1}_{0}[1-p(\sigma(t, \xi))]^{\beta}q(t, \xi){\rm d}\xi= (\frac{1}{2})^{\beta}t.$

下面验证(2.41)式和(2.42)式是否成立,我们取$\psi(t)=1, ~H(t, s)=(t-s)^{1+\rho}, $$\rho=\min\{\alpha, \beta\}.$故有

$\begin{eqnarray*}\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T}H(t, s)\psi(s)Q(s){\rm d}s&= &\limsup\limits_{t\rightarrow\infty}\frac{(\frac{1}{2})^{\beta}}{(t-T)^{1+\rho}}\int^{t}_{T}(t-s)^{1+\rho}s{\rm d}s\\&=&\limsup\limits_{t\rightarrow\infty}\frac{1}{2^{\beta}}\left(\frac{T(t-T)}{2+\rho}-\frac{(t-T)^{2+\rho}}{3+\rho}\right)= \infty, \end{eqnarray*}$

利用(2.35)式知$h(t, s)=1+\rho$,我们有

$\limsup\limits_{t\rightarrow\infty}\frac{1}{H(t, T)}\int^{t}_{T} \frac{\psi(s)r(\theta(s))}{(g'(s))^{\rho}}|h(t, s)|^{\rho+1}{\rm d}s= \limsup\limits_{t\rightarrow\infty}\frac{1}{(t-T)^{1+\rho}}\int^{t}_{T}(1+\rho)^{1+\rho}{\rm d}s <\infty.$

由推论2.1知方程(E$_{4})$振动.

注3.1  本文中定理对任意的$\alpha>0, ~\beta>0$都适用.文献[1-13]中的定理均不能适用于本文的例子.