考虑三阶中立型分布时滞微分方程

$\begin{equation}(r(t)(c(t)[x(t)+\int_{a}^{b}p(t, \xi)x[{\tau}(t, \xi)]d\xi]')')'+\int_{a}^{b}q(t, \xi)f(x[g(t, \xi)])d\xi=0, t\geq t_0.\end{equation}$

(1.1)

本文中总假设下列条件成立

(H$_1)$ $\displaystyle r(t), c(t)\in C([t_0, \infty), (0, \infty)), \int_{t_0}^{\infty}\frac{dt}{r(t)}=\int_{t_0}^{\infty}\frac{dt}{c(t)}=\infty.$

(H$_2)$ $\displaystyle p(t, \xi), q(t, \xi)\in C([t_0, \infty)\times[a, b], (0, \infty)), \int_{a}^{b}p(t, \xi)d\xi\leq p<1.$

(H$_3)$ $\tau(t, \xi), g(t, \xi)\in C([t_0, \infty)\times[a, b], (0, \infty)), \tau(t, \xi)\leq t, g(t, \xi)\leq t, \xi \in [a, b], $

$\liminf\limits_{t \rightarrow \infty \xi \in [a, b]} \tau(t, \xi)=\liminf\limits_{t \rightarrow \infty \xi \in [a, b]}g(t, \xi)=\infty, $

$g(t, \xi)$关于$t$和$\xi$均为增函数.

(H$_4)$ $ f(u) \in C(R, R), \frac{f(u)}{u} \geq \delta ＞0, u \neq 0.$

我们仅限于考虑方程(1.1) 的非平凡解, 即存在某一半直线$[t_x, \infty]$使得对任意$T \geq t_x, $满足不等式$\sup{\mid x(t)\mid:t>T}＞0$的解$x(t)$.方程(1.1) 的解称为振动, 如果它有任意大的零点, 否则, 成为非振动.

当今, 时滞微分方程解的振动性, 为人们讨论的热点问题, 有关三阶泛函微分方程解的振动性的研究, 有文献[1-15], 尤其1978年, Schot ^[15]在《美国物理杂志》上, 发表了题为《急动度--加速度的时间变率的论文》, 介绍了位移对时间的三阶导数--急动度的历史背景, 并且考虑了它在凸轮和星形轮等间歇运动机械设计等方面的应用.急动度作为加速度随时间的变化率是力学中的基本概念, 具有重要理论意义.因此, 近年来, 三阶泛函微分方程解的振动性和渐进性研究日益受到重视, 并且取得不少重要成果.例如最新结果可以参看文[1-12].但是, 其中多数结果是三阶时滞微分方程的, 而关于三阶中立型的振动结果很少.最近文[10]研究三阶时滞微分方程

$\begin{equation}(b(t)(a(t)x'(t))')'+q(t)f(x(t-\sigma))=0.\end{equation}$

(1.2)

文中证明了方程(1.2) 每一解振动或者收敛到零的若干充分条件.显然, 当方程(1.1) 中$p(t, \xi)\equiv 0, q(t, \xi)$与$\xi$无关, $g(t, \xi)=t-\sigma, b-a=1$时, 即为方程(1.2).我们也注意到文[13]研究了与方程(1.1) 相应的下列二阶方程的特例

$\begin{equation}(r(t)[x(t)+c(t)x(t-\tau)]')'+\int_{a}^{b}p(t, \xi)x[g(t, \xi)]d\xi=0, \end{equation}$

(1.3)

给出了方程(1.3) 解的振动准则.本文目的是建立三阶中立型方程(1.1) 每一解振动或者收敛到零的充分条件.我们将文[13]关于二阶方程(1.3) 的结果推广到三阶方程.我们也给出例子说明本文定理的应用.

引理1 设$x(t)$是(1.1) 式最终正解, 定义函数

$\begin{equation}z(t)=x(t)+\int_{a}^{b}p(t, \xi)x[\tau(t, \xi)]d\xi, \end{equation}$

(2.1)

则$z(t)$只能有下列两种可能

(Ⅰ) $z(t)＞0, z'(t)＞0, (c(t)z'(t))'＞0, $

(Ⅱ) $z(t)＞0, z'(t)＜0, (c(t)z'(t))'＞0.$

记作$z(t)\in$ Ⅰ或者$z(t)\in$ Ⅱ.

证 设$x(t)$是(1.1) 式最终正解, 故存在$t_1\geq t_0, $当$t > t_1$时, 使得$x[\tau(t, \xi)]＞0, x[g(t, \xi)]＞0, \xi \in [a, b], t>t_1.$由(2.1) 式知方程(1.1) 为

$(r(t)(c(t)z^{'}(t))^{'})^{'}+\int_{a}^{b}q(t, \xi)f(x[g(t, \xi)])d\xi=0, $

由$q(t, \xi)\geq 0, x[g(t, \xi)]＞0$及(H$_{4}), $得到$z(t)＞0, (r(t)(c(t)z'(t))')'\leq 0, t \geq t_1.$因$((r(t)(c(t)z'(t))')'＜0, $故$r(t)(c(t)z'(t))'$单调减小且最终定号, 则$(c(t)z'(t))'$最终定号.我们断言

$\begin{equation}(c(t)z'(t))'＞0, t\geq t_2 \geq t_1.\end{equation}$

(2.2)

否则, 若$(c(t)z'(t))'\leq 0$, 则存在$t_3\geq\texttt{} t_2, $当$t>t_3$时, 使得$r(t)(c(t)z'(t))'\leq r(t_3)(c(t_3)z'(t_3))'\leq 0, t\geq t_3.$用$r(t)$除上式, 并从$t_3$到$t$对其积分产生

$c(t)z'(t)-c(t_3)z'(t_3)\leq r(t_3)(c(t_3)z'(t_3))' \int_{t_3}^{t} \frac{ds}{r(s)}.$

在上式中, 令$t \rightarrow \infty$, 利用(H$_1)$, 我们有$c(t)z'(t)\rightarrow -\infty$.因此存在$t_4\geq t_3, $当$t>t_4$时, 使得

$c(t)z'(t)\leq c(t_4)z'(t_4)＜0, t\geq t_4.$

用$c(t)$除上式, 再从$t_4$到$t$积分, 利用假设(H$_1)$, 我们得到$z(t) \rightarrow -\infty, t \rightarrow \infty.$因此与$z(t)＞0$矛盾.因此, (2.2) 式成立.引理1证毕.

引理2 设$x(t)$是(1.1) 式的最终正解, $z(t)\in$Ⅰ, 则存在$T\geq t_0$使得

$z'[g_2(t)]\geq \frac{R[g_2(t)]r(t)}{c[g_2(t)]}(c(t)z'(t))', t\geq T, $

其中

$\begin{equation}R(t)=\int_{T}^{t}\frac{1}{r(s)}ds, g_2(t)=g(t, a).\end{equation}$

(2.3)

证 设$z(t)\in$Ⅰ, 由引理1, 有

$ z(t)＞0, c(t)z'(t)＞0, r(t)(c(t)z'(t))'＞0, (r(t)(c(t)z'(t))')'\leq 0, t\geq T.$

因此

$c(t)z'(t)=c(T)z'(T)+\int_{T}^{t}\frac{r(s)(c(s)z'(s))'}{r(s)}ds\geq r(t)(c(t)z'(t))'R(t), t\geq T.$

由$(r(t)(c(t)z'(t))')'\leq 0$, 我们得到$z'[g_2(t)]\geq \frac{R[g_2(t)]r(t)}{c[g_2(t)]}(c(t)z'(t))', t\geq T.$引理2证毕.

引理3 设$x(t)$是(1.1) 式的最终正解, $z(t)\in$ Ⅱ.若

$\begin{equation}\int_{t_0}^{\infty}\frac{1}{c(v)}\int_{v}^{\infty}\frac{1}{r(u)}\int_{u}^{\infty}\int_{a}^{b}q(s, \xi)d\xi dsdudv=\infty, \end{equation}$

(2.4)

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

证 设$z(t)\in$ Ⅱ, 则$z(t)＞0, z'(t)＜0, $故有$\lim\limits_{t\rightarrow \infty}z(t)=l\geq 0.$我们断言$l=0$.事实上, 若$l＞0, $则对任意$\varepsilon > 0, $最终有$l＜z(t)＜l+\varepsilon, $取$\varepsilon＜\frac{l(1-p)}{p}, $利用(H$_2)$和(H$_3)$, 有

$ x(t)=z(t)-\int_{a}^{b}p(t, \xi)x[\tau(t, \xi)]d\xi \geq z(t)-\int_{a}^{b}p(t, \xi)z[\tau(t, \xi)]d\xi\\ > l-z[\tau(t, a)]\int_{a}^{b}p(t, \xi)d\xi >l-pz[\tau(t, a)]>l-p(l+\varepsilon)=k(l+\varepsilon)>kz(t)$

最终成立, 其中$k=\frac{l-p(l+\varepsilon)}{l+\varepsilon}＞0$.

利用(H$_4)$和$(1.1)$式, 产生

$ (r(t)(c(t)z'(t))')'\leq -k\delta \int_{a}^{b}q(t, \xi)z[g(t, \xi)]d\xi\nonumber\\ \leq -k\delta z[g(t, b)]\int_{a}^{b}q(t, \xi)d\xi=-k\delta q(t)z[g_1(t)], t\geq T, $

其中利用

$\begin{equation}z'(t)＜0, q(t)=\displaystyle\int_{a}^{b}q(t, \xi)d\xi, g_{1}(t)=g(t, b).\end{equation}$

(2.5)

从$t$到$\infty$对(2.5) 式积分产生

$r(t)(c(t)z'(t))'\geq \int_{t}^{\infty}k\delta q(s)z[g_{1}(s)]ds\geq k\delta l\int_{t}^{\infty}q(s)ds.$

用$r(t)$除上式, 从$t$到$\infty$积分, 有

$-c(t)z'(t)\geq k\delta l\int_{t}^{\infty}\frac{1}{r(u)}\int_{u}^{\infty}q(s)dsdu, t\geq T.$

在上式中双边除$c(t)$, 从$T$到$\infty$积分, 得到

$z(T)\geq k\delta l\int_{T}^{\infty}\frac{1}{c(v)}\int_{v}^{\infty}\frac{1}{r(u)}\int_{u}^{\infty}q(s)dsdudv, t\geq T.$

此与条件(2.4) 式矛盾, 故$l=0, $因$0＜x(t)\leq z(t), $则$\lim\limits_{t \to \infty}x(t)=0.$引理3证毕.

现在, 我们利用Philos型积分平均条件^[14]给出方程(1.1) 新的振动结果.为此, 引进函数类$X.$

令$D_{0}=\{(t, s):t＞s\geq t_{0}\}, D=\{(t, s):t\geq s\geq t_{0}\}.$函数$H \in C(D, R)$称为属于$X, $如果

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

(ⅱ) $H$在$D_{0}$上关于第二变量有连续非正偏导数, 且$-\frac{\partial H(t, s)}{\partial s}=h(t, s)\sqrt{H(t, s)}, (t, s)\in D_{0}.$

定理1 设$(\hbox{H}_1)-(\hbox{H}_4)$和(2.4) 式成立且存在$\rho (t) \in C'([t_{0}, \infty], R^{+})$和$H \in X$, 使得

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}H(t, s)[\delta(1-p)\rho(s)q(s)-\frac{\rho(s)c[g_{2}(s)]}{4R[g_{2}(s)][g'_{2}(s)]}Q^{2}(t, s)]ds=\infty, \end{equation}$

(2.6)

其中$g_{2}(t), q(t)$由(2.3) 和(2.5) 式定义,

$\begin{equation}Q(t, s)=\frac{h(t, s)}{\sqrt{H(t, s)}}-\frac{\rho'(s)}{\rho(s)}, \end{equation}$

(2.7)

则方程(1.1) 的每一解$x(t)$振动或者收敛到零.

证 设$x(t)$是方程(1.1) 的非振动解, 不失一般, 设$x(t)$最终为正, 则有$x(t)＞0, t\geq t_{1}, x[\tau(t, \xi)]＞0, x[g(t, \xi)]＞0, (t, \xi) \in [t_{1}, \infty)\times[a, b].$当$x(t)$最终为负时, 可以类似地处理.令$z(t)$由(2.1) 式定义, 则利用引理1知$z(t)\in$Ⅰ或者$z(t)\in$ Ⅱ.

首先, 设$z(t)\in$Ⅰ, 即$z'(t)＞0, t\geq t_{1}.$由(H$_{3})$知, $\exists t_{2}\geq t_{1}$使得$\tau(t, \xi)>t_{1}, (t, \xi)\in [t_{2}, \infty) \times [a, b], $则当$t>t_{2}$时, $z(\tau(t, \xi))<z(t).$故有

$x(t)=z(t)-\int_{a}^{b}P(t, \xi)x[\tau(t, \xi)]d\xi \geq z(t)-\int_{a}^{b}P(t, \xi)z[\tau(t, \xi)]d\xi\geq (1-p)z(t), t\geq t_{2}.$

故由上式和(H$_4)$, (H$_3)$和方程(1.1) 产生

$\begin{equation}(r(t)(c(t)z'(t))')'+\delta (1-p)q(t)z[g_{2}(t)]\leq 0, t\geq t_{2}, \end{equation}$

(2.8)

其中$g_{2}(t)$和$q(t)$分别由(2.3) 和(2.5) 式定义, 令

$\begin{equation}W(t)=\rho(t)\frac{r(t)(c(t)z'(t))'}{z[g_{2}(t)]}, t\geq t_{2}, \end{equation}$

(2.9)

则由(2.8) 式和引理2, 有

$\begin{equation}W'(t)\leq -\delta (1-p)\rho(t)q(t)+\frac{\rho'(t)}{\rho(t)}W(t)-\frac{R[g_{2}(t)]g'_{2}(t)}{\rho(t)c[g_{2}(t)]} W^{2}(t).\end{equation}$

(2.10)

作Riccati变换$A_{1}(s)=\frac{\rho'(s)}{\rho(s)}, A_{2}(s)=\frac{R[g_{2}(s)]g'_{2}(s)}{\rho(s)c[g_{2}(t)]}.$则从(2.10) 式得到

$\begin{align} & \int_{{{t}_{2}}}^{t}{H}(t, s)\delta (1-p)\rho (s)q(s)ds \\ & \le \int_{{{t}_{2}}}^{t}{H}(t, s)[-{W}'(s)+{{A}_{1}}(s)W(s)-{{A}_{2}}(s){{W}^{2}}(s)]ds \\ & =-H(t, s)W(s)\mid _{{{t}_{2}}}^{t}+\int_{{{t}_{2}}}^{t}{\{\frac{\partial H(t, s)}{\partial s}W(s)+H(}t, s)[{{A}_{1}}(s)W(s)-{{A}_{2}}(s){{W}^{2}}(s)]\}ds \\ & \rm{=}\mathit{H}\rm{(}\mathit{t}\rm{, }{{\mathit{t}}_{\rm{2}}}\rm{)}\mathit{W}\rm{(}{{\mathit{t}}_{\rm{2}}}\rm{)}\mathop{\int }_{{{t}_{2}}}^{t}[\sqrt{H(t, s)}(h(t, s)-\sqrt{H(t, s)}{{A}_{1}}(s))W(s)+H(t, s){{A}_{2}}(s){{W}^{2}}(s)]ds \\ & \rm{=}\mathit{H}\rm{(}\mathit{t}\rm{, }{{\mathit{t}}_{\rm{2}}}\rm{)}\mathit{W}\rm{(}{{\mathit{t}}_{\rm{2}}}\rm{)}-\mathop{\int }_{{{t}_{2}}}^{t}{{[\sqrt{H(t, s){{A}_{2}}(s)}W(s)+\frac{1}{2}\frac{{{Q}_{1}}(t, s)}{\sqrt{{{A}_{2}}(s)}}]}^{2}}ds+\int_{{{t}_{2}}}^{t}{\frac{Q_{1}^{2}(t, s)}{4{{A}_{2}}(s)}}ds, \\ \end{align}$

(2.11)

其中

$\begin{equation}Q_{1}(t, s)=h(t, s)-\sqrt{H(t, s)}A_{1}(s).\end{equation}$

(2.12)

因此$\displaystyle\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[H(t, s)\delta (1-p)\rho(s)q(s)-\frac{Q_{1}^{2}(t, s)}{4A_{2}(s)}]ds\leq W(t_{2})$.

此即

$\begin{equation}\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[H(t, s)\delta (1-p)\rho(s)q(s)-\frac{\rho(s)c[g_{2}(s)]}{4R[g_{2}(t)]g'_{2}(t)}Q^{2}(t, s)]ds \leq W(t_{2}).\end{equation}$

(2.13)

上式与条件(2.6) 矛盾.其次, 若$z(t)\in$ Ⅱ, 注意到条件(2.4) 成立, 故由引理3知$x(t)$收敛到零.定理1证毕.

推论1 设定理1的条件(2.6) 用下列条件代替

$\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}H(t, s)\rho(s)q(s)ds=\infty, $

(2.6)1

$\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}\frac{\rho(s)c[g_{2}(s)]}{R[g_{2}(s)]g'_{2}(s)}Q^{2}(t, s)ds＜ \infty. $

(2.6)2

其它条件不变, 则定理1的结论不变.

当定理1中的条件(2.6) 不易验证时, 我们有下面的定理.我们将使用定理1的符号和证明.

定理2 设除(2.6) 以外定理1的假设均成立, 又设

$\begin{equation}0＜\mathop {\inf }\limits_{s \ge {t_0}} [\mathop {\lim }\limits_{t \to \infty }\inf\frac{H(t, s)}{H(t, t_0)}]\leq \infty \end{equation}$

(2.14)

且

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}\frac{Q^{2}(t, s)}{H(t, t_{0})}ds＜\infty.\end{equation}$

(2.15)

令$\phi(t) \in C([t_0, \infty), R)$使得

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}\phi_{+}^{2}(s)A_{2}(s)ds＜\infty, \end{equation}$

(2.16)

其中$\phi_{+}(t)=\max\{\phi(t), 0\}, $且

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}(\delta (1-p)H(t, s)\rho(s)q(s)-\frac{Q_{1}^{2}(t, s)}{4A_{2}(s)})ds\geq \mathop {\sup }\limits_{t \ge {t_0}} \phi(t), \end{equation}$

(2.17)

则方程(1.1) 的每一个解振动或者收敛到零.

证 设$x(t)$是(1.1) 的最终正解, $z(t)$由(2.1) 式定义, 首先, 设$z(t)\in$ Ⅰ, 则在定理1的证明中, 有(2.11) 式成立, 即

$ \int_{t_{2}}^{t}H(t, s)\delta (1-p)\rho(s)q(s)ds\\ \leq H(t, t_{2})W(t_{2})-\int_{t_{2}}^{t}[\sqrt{H(t, s)A_{2}(s)}W(s)+\frac{1}{2}\frac{Q_{1}(t, s)}{\sqrt{A_{2}(s)}}]^{2}ds+\int_{t_{2}}^{t} \frac{Q_{1}^{2}(t, s)}{4A_{2}(s)}ds.$

因此

$ \mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[\delta (1-p)H(t, s)\rho(s)q(s)-\frac{Q_{1}^{2}(t, s)}{4A_{2}(s)}]ds \\ \leq W(t_{2})-\mathop {\lim }\limits_{t \to \infty }\inf\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[\sqrt{H(t, s)A_{2}(s)}W(s)+\frac{1}{2}\frac{Q_{1}(t, s)}{\sqrt{A_{2}(s)}}]^{2}ds. $

利用(2.17) 式, 上式产生

$W(t_{2})\geq \phi(t_{2})+\mathop {\lim }\limits_{t \to \infty }\inf\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[\sqrt{H(t, s)A_{2}(s)}W(s)+\frac{1}{2}\frac{Q_{1}(t, s)}{\sqrt{A_{2}(s)}}]^{2}ds.$

故有

$\begin{equation}0\leq \mathop {\lim }\limits_{t \to \infty }\inf\frac{1}{H(t, t_{2})}\int_{t_{2}}^{t}[\sqrt{H(t, s)A_{2}(s)}W(s)+\frac{1}{2}\frac{Q_{1}(t, s)}{\sqrt{A_{2}(s)}}]^{2}ds＜ \infty.\end{equation}$

(2.18)

定义函数

$ \alpha(t)=\frac{1}{H(t, t_{1})}\int_{t_{1}}^{t}H(t, s)A_{2}(s)W^{2}(s)ds, \\ \beta(t)=\frac{1}{H(t, t_{1})}\int_{t_{1}}^{t}\sqrt{H(t, s)}Q_{1}(t, s)W(s)ds, $

则由(2.18) 式知

$\mathop {\lim }\limits_{t \to \infty } \inf [\alpha (t) + \beta (t)] ＜ \infty .$

我们断言

$\begin{equation}\int_{t_{1}}^{\infty}A_{2}(t, s)W^{2}(s)ds＜\infty.\end{equation}$

(2.19)

否则, 如果相反, 则有

$\begin{equation}\int_{t_{1}}^{\infty}A_{2}(t, s)W^{2}(s)ds=\infty.\end{equation}$

(2.20)

利用条件(2.14), 存在$\eta＞0$, 使得

$\mathop {\inf }\limits_{s \ge {t_0}} [\mathop {\lim }\limits_{t \to \infty } \inf \frac{{H(t,s)}}{{H(t,{t_0})}}] ＞ \eta .$

(2.21)

令$\mu$是任意正数, 则由(2.20) 式知存在$t_2\geq t_1$, 使得$\int_{t_{1}}^{t}A_{2}(t, s)W^{2}(s)ds＞\frac{\mu}{\eta}, t\geq t_2.$因此对$t\geq t_2$, 有

$\begin{eqnarray} \alpha(t)=\frac{1}{H(t, t_1)}\int_{t_1}^{t}H(t, s)\frac{d}{ds}[\int_{t_1}^{s}A_2(u)W^2(u)du] \nonumber\\ =\frac{1}{H(t, t_1)}\int_{t_1}^{t}-\frac{\partial H(t, s)}{\partial s}[\int_{t_1}^{s}A_2(u)W^2(u)du]ds\nonumber\\ \geq \frac{1}{H(t, t_1)}\int_{t_2}^{t}-\frac{\partial H(t, s)}{\partial s}[\int_{t_1}^{s}A_2(u)W^2(u)du]ds \nonumber\\ \geq \frac{\mu}{\eta}\frac{1}{H(t, t_1)}\int_{t_2}^{t}-\frac{\partial H(t, s)}{\partial s}ds=\frac{\mu}{\eta}\frac{H(t, t_2)}{H(t, t_1)}.\end{eqnarray}$

(2.22)

此时, 由于函数类$X$的函数$H$在$D_{0}$上, 关于第二个变量有连续非正偏导数, 故函数$H$关于第二个变量是减函数.利用(2.21) 式, 存在$t_3\geq t_2$, 使得

$\begin{equation}\frac{H(t, t_2)}{H(t, t_1)}\geq \frac{H(t, t_{2})}{H(t, t_{0})} \geq \mu, t\geq t\geq t_3.\end{equation}$

(2.23)

联合(2.22), (2.23) 式, 我们得到$\alpha(t)\geq \mu, t\geq t_{3}.$因$\mu$是任意的, 有

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\alpha(t)=\infty.\end{equation}$

(2.24)

这一情况的证明的剩余部分是类似于文[11]和[13]中的定理2.因此, 我们省略.

其次, 设$z(t)\in$ Ⅱ, 注意到条件(2.4) 成立, 则由引理3知$x(t)$收敛到零.定理2证毕.

定理3 设定理2的全部条件成立, 除了将条件(2.15) 改为

$\begin{equation}\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}H(t, s)p(s)q(s)ds＜\infty, \end{equation}$

(2.25)

则方程(1.1) 的每一个振动或者收敛到零.

证 定理3的证明类似于定理2的证明.因此, 我们省略.

下面的例子说明我们结果的应用.

例 考虑三阶中立型分布时滞微分方程

$\begin{equation}(\frac{1}{(t+1)^\alpha}[x(t)+\int_{-1}^{0}(\frac{1}{2}+\frac{1}{3}e^{-2t}+\xi)x(t+\frac{\xi}{2})d\xi]')''+\int_{-1}^{0}e^{t+\xi} [2+sinx(t+\xi)]x(t+\xi)d\xi=0, t>1.\end{equation}$

(2.26)

此时

$ r(t)=1, c(t)=\frac{1}{(t+1)^{\alpha}}, p(t, \xi)=\frac{1}{2}+\frac{1}{3}e^{-2t}+\xi, \\ \int_{-1}^{0}p(t, \xi)d\xi\leq \frac{1}{3}＜1, q(t, \xi)=e^{t+2\xi}\geq 0, q(t)=\int_{-1}^{0}q(t, \xi)d\xi=\frac{1}{2}(1-\frac{1}{e^2})e^{t}, \\ \tau(t, \xi)=t+\frac{\xi}{2}\leq t, g(t, \xi)= t+\xi\leq t, \xi \in [-1, 0], f(u)=[2+\sin u]u\geq u, \delta=1, $

则对于任意$t\geq1$, 有

$\int_{1}^{\infty}\frac{1}{r(t)}ds=\int_{1}^{\infty}\frac{1}{c(t)}ds=\infty, R(t)=\int_{1}^{t}\frac{1}{r(t)}ds=t-1, \int_{1}^{\infty}q(s)ds=\infty.$

因此, $(\hbox{H}_1)$--$(\hbox{H}_4)$和(2.4) 式成立.为应用推论1, 剩下只需验证条件$(2.6)_{1}$和$(2.6)_{2}$满足即可.我们取$H(t, s)=(t-s)^2, \rho(t)=1, t\geq s\geq 1, $有

$\mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}H(t, s)\rho(s)q(s)ds\geq \mathop {\lim }\limits_{t \to \infty }\sup \frac{1}{t^2}\int_{1}^{t}(t-s)^2\frac{1}{2}(1-\frac{1}{e^2})e^sds=\infty, \\ \mathop {\lim }\limits_{t \to \infty }\sup\frac{1}{H(t, t_{0})}\int_{t_{0}}^{t}\frac{\rho(s)c[g_2(s)]}{R[g_2(s)]g'_2(s)}Q^2(t, s)ds\geq \mathop {\lim }\limits_{t \to \infty }\sup \frac{1}{t^2}\int_{1}^{t}\frac{4}{(s-2)s^\alpha}ds<\infty. $

故$(2.6)_{1}$和$(2.6)_{2}$满足, 由推论1知方程(2.26) 的每一解振动或者收敛到零.

注1  文[10]中的定理1--3是本文相应结果中, 当$p(t, \xi)\equiv 0, q(t, \xi)\equiv q(t), g(t, \xi)=t-\sigma, [a, b]=[-1, 0]$时的特例.

注2  本文的定理1--3是三阶中立型方程(1.1) 的Philos型振动定理, 它们将文[13]关于二阶三阶中立型方程的振动结果推广到相应的三阶方程.

注3  参考文献中的振动结果均不能适用本文例子.