由于微分方程在自然科学和工程技术、生物医药及金融学等领域都有广泛的应用,因而微分方程的振动性等定性理论的研究引起了国内外学者们的广泛兴趣,见文献[1-12]及其参考文献.笔者考虑如下形式的二阶广义Emden-Fowler型延迟非线性的微分方程

(1.1) $ \begin{equation} [\alpha(t)\phi_1(z'(t))]'+q(t)f(\phi_2(x(\delta(t)))) = 0, \quad t\geq t_0 \end{equation} $

的振荡性, (1.1)式中, $ z(t) = x(t)+p(t)x(\tau(t)), \phi_1(u) = |u|^{\lambda-1}u, \phi_2(u) = |u|^{\beta-1}u $ ($ \lambda, \beta>0 $为实常数).并总假设以下条件(H$ _1) $–(H$ _4) $成立.

(H$ _1)\ a\in C^1([t_0, +\infty), (0, +\infty)) $;

(H$ _2)\ p, q\in C([t_0, +\infty), {\Bbb R} ) $,且$ q(t)>0, p(t)\geq 0 $;

(H$ _3) $滞量函数$ \tau, \delta:[t_0, \infty)\rightarrow(0, +\infty) $,满足$ \tau(t)\leq t, \lim\limits_{t\rightarrow+\infty}\tau(t) = +\infty, \delta(t)\leq t $,和$ \lim\limits_{t\rightarrow+\infty}\delta(t) = +\infty $,并且$ \tau\circ\delta = \delta\circ\tau, \tau'(t)\geq\tau_0 $ (这里$ \tau_0>0 $为常数);

(H$ _4)\ f\in C(\mathbb{(R, R)}) $,当$ u\neq0 $时$ uf(u)>0 $,并且$ \frac{f(u)}{u}\geq L $,这里$ L>0 $是常数.

如果称函数$ x(t)\in C^1([T_x, +\infty), {\Bbb R} )(T_x\geq t_0) $是方程$ (1.1) $的一个解,则这个函数$ x(t) $满足$ \alpha(t)\phi_1(z'(t))\in C^1([T_x, +\infty), {\Bbb R} ) $且在区间$ [T_x, +\infty) $满足方程(1.1).本文只讨论方程(1.1)的非平凡解,方程(1.1)的一个解称为是振荡的,如果它既不最终为正也不最终为负,否则称它是非振荡的;方程(1.1)称为是振荡的,如果它的所有解都是振荡的.

方程(1.1)含有许多典型的微分方程,如下列方程

(1.2) $ \begin{equation} [\alpha(t)|x'(t)|^{\gamma-1}x'(t)]'+q(t)|x(t)|^{\gamma-1}x(\delta(t)) = 0, \end{equation} $

(1.3) $ \begin{equation} \{{\alpha(t)|[x(t)+p(t)x(\tau(t))]'|^{\gamma-1}[x(t)+p(t)x(\tau(t))]'}\}'+q(t)|x(\delta(t))|^{\beta-1}x(\delta(t)) = 0 \end{equation} $

等等,对于这些特殊类型的Emden-Fowler型延迟微分方程的振荡性,已有一些研究成果(参见文献[1-12]及其参考文献).如文献[1-2]研究了一类较为典型的二阶Emden-Fowler型延迟微分方程(1.2)的振荡性,其主要结果之一如下.

定理A^[2]  设方程(1.2)满足条件: $ \alpha'(t)\geq 0, \delta'(t)>0 $,且

$ \int_{t_0}^{+\infty}\alpha^{\frac{-1}{\gamma}}(t){\rm d}t = +\infty. ~~~~~~~~~~~~~({\rm C}_1) $

记$ R(t) = \int _{t_0}^t\alpha^{\frac{-1}{\gamma}}(t){\rm d}t $,如果

$ \begin{eqnarray*} \int_{t_0}^{+\infty}\left[R^\gamma(\delta(t))q(t)-\Big(\frac{\gamma}{\gamma+1}\Big)^{\gamma+1}\frac{\delta'(t)}{R(\delta(t))\alpha^{\frac{1}{\gamma}}(\delta(t))}\right]{\rm d}t = +\infty, \end{eqnarray*} $

则方程(1.2)是振荡的.

这里,我们看到条件"$ \alpha'(t)\geq0 $"似乎有点苛刻.以此为契机,文献[3]研究了更一般的二阶Emden-Fowler型延迟微分方程(1.3)的振荡性,在正则和非正则两种情形下得到了方程(1.3)的一系列振荡准则,改进了文献[2]的结果,其主要结论之一如下.

定理B^[3]  设方程(1.3)满足条件: $ \alpha'(t)\geq0, \delta'(t)>0, 0\leq p(t)<1, \gamma\geq\beta $,且(C$ _1) $成立.如果有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)) $,使得

$ \begin{eqnarray*} \int_{t_0}^{+\infty}\bigg[\varphi(s)q(s)[1-p(\delta(s))]^\beta-\frac{\alpha(\delta(s))(\varphi'(s))^{\beta+1}}{(\beta+1)^{\beta+1}(k\varphi(s)\delta'(s))^\beta}\bigg]{\rm d}s = +\infty, \end{eqnarray*} $

这里$ k>0 $是某常数,则方程(1.3)是振荡的.

对于非正则情形,有如下结论.

定理C^[3]  设方程(1.3)满足条件: $ \alpha'(t)\geq0, \tau'(t)>0, 0\leq p(t)<1, \gamma\geq\beta, p'(t)\geq0 $,且

$ \int_{t_0}^{+\infty}\alpha^{\frac{-1}{\gamma}}(t){\rm d}t<+\infty, ~~~~~~~~~~~~({\rm C}_2) $

并且有$ \int_{t_0}^{+\infty}q(t)[1-p(\delta(t))]^\beta {\rm d}t = +\infty $和$ \int_{t_0}^{+\infty}\Big(\frac{1}{\alpha(t)}\int_{t_0}^tq(s){\rm d}s\Big)^{\frac{1}{\gamma}}{\rm d}t = +\infty $,则方程(1.3)的每一个解$ x(t) $或者振荡或者满足$ \lim\limits_{t\rightarrow+\infty}x(t) = 0 $.

显然,定理B和定理C中有限制条件"$ \alpha'(t)\geq0 $及$ 0\leq p(t)<1 $",且$ \gamma<\beta $时文献[3]没有给出方程(1.3)的振荡准则.而且在非正则的情形,定理C的结论是不确定的,因此应用时就会有很多限制.例如,对下列Emden-Fowler型延迟微分方程

$ \Big\{\frac{1}{t^{\frac{5}{7}}}\Big[\Big(x(t)+(5-\sin t^2)x(\frac{t}{2})\Big)'\Big]^{\frac{1}{3}}\Big\}'+\frac{1}{t^2}f(x(\frac{t}{2})) = 0. ~~~~~~~~~~~~({\rm E}_1) $

由于不满足$ 0\leq p(t)<0 $,因此定理B不能用于上述方程.再例如,对于著名的Euler方程

$ (t^2x'(t))'+mx(t) = 0, t\geq1, ~~~~~~~~~~~~({\rm E}_2) $

定理C也不能确定其是否振荡.其它文献如[4-12]中的定理也不能确定方程(E$ _1) $和(E$ _2) $是否振荡.

本文的目的是利用双黎卡提(Riccati)变换技术及各种分析技巧来研究方程(1.1)的振荡性,在正则和非正则的情形下(即在条件(C$ _1) $和(C$ _2) $下),得到了该方程的一系列新的振荡准则,推广、改进且丰富了现有文献中关于Emden-Fowler型延迟微分方程的振荡性结果.

首先给出如下引理.

引理2.1^[6]  设X, Y为非负实数,则当$ 0<\lambda\leq1 $时$ X^\lambda+Y^\lambda\geq(X+Y)^\lambda $,等号成立当且仅当$ X = Y $.

引理2.2^[7]  设X, Y为非负实数,当$ \lambda>1 $时$ X^\lambda+Y^\lambda\geq2^{1-\lambda}(X+Y)^\lambda $,等号成立当且仅当$ X = Y $.

引理2.3^[8]  设$ A>0, B>0 $和$ \alpha>0 $均为常数,则当$ x>0 $时, $ Ax-Bx^{\frac{\alpha+1}{\alpha}}\leq\frac{\alpha^\alpha A^{\alpha+1}}{(\alpha+1)^{\alpha+1}B^\alpha} $.其次,我们引入如下三个记号(使用时不再说明): $ L_0 = \Bigg\{\begin{array}{ll} L, &0<\beta\leq 1, \\ L2^{1-\beta}, &\beta>1, \end{array}\Bigg., $$ Q(t) = \min\{q(t), q(\tau(t))\}, $$ A(t) = \int_{t_0}^t\alpha^{\frac{-1}{\lambda}}(t){\rm d}t $.

定理2.1  设条件(C$ _1 $)成立且$ 0\leq p(t)\leq p_0<+\infty $(这里$ p_0 $为常数), $ \delta(t)\leq\tau(t) $且$ \delta'(t)>0 $,如果存在函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)) $使得

(2.1) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\Big[L_0Q(s)-\frac{k\alpha^{\omega_1}(\delta(s))}{(\delta'(s))^{\omega_2}}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\omega_2+1}\Big]{\rm d}s = +\infty, \end{equation} $

其中常数$ k = \Bigg\{\begin{array}{ll} \frac{1+\frac{p_0^\beta}{\tau_0}}{(\lambda+1)^{\lambda+1}}, &\lambda = \beta, \\ c, &\lambda\neq\beta, \end{array}\Bigg., $$ c $为正常数.而$ \omega_1 = \min\{1, \frac{\beta}{\lambda}\}, \omega_2 = \min\{\lambda, \beta\} $,则方程(1.1)是振荡的.

证  反证法.设方程(1.1)存在一个非振荡解$ x(t) $,不妨设$ x(t) $为方程(1.1)的最终正解(当$ x(t) $为最终负解时类似可证),则$ \exists t_1\geq t_0 $,使得当$ t\geq t_1 $时,有$ x(t)>0, x(\tau(t))>0, $$ x(\delta(t))>0 $,于是由$ z(t) $的定义知, $ z(t)>0 $且$ z(t)\geq x(t)(t\geq t_1) $.由方程(1.1),得

(2.2) $ \begin{equation} [\alpha(t)\phi_1(z'(t))]' = -q(t)f(\phi_2(x(\delta(t))))\leq-Lq(t)(x(\delta(t)))^\beta<0. \end{equation} $

注意到条件(C$ _1) $,于是由(2.2)式就不难推出$ z'(t)>0(t\geq t_1) $.应用(2.2)式,可得

(2.3) $ \begin{equation} \frac{[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'}{\tau'(t)}+Lq(\tau(t))(x(\delta(\tau(t))))^\beta\leq0, t\geq t_1, \end{equation} $

因此,综合(2.2)和(2.3)两式,当$ t\geq t_1 $时,就有

(2.4) $ \begin{equation} [\alpha(t)\phi_1(z'(t))]'+Lq(t)(x(\delta(t)))^\beta+p_0^\beta Lq(\tau(t))(x(\delta(\tau(t))))^\beta+p_0^\beta\frac{[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'}{\tau'(t)}\leq0. \end{equation} $

当$ 0<\beta\leq1 $时,注意到$ \tau'(t)\geq\tau_0>0, \tau\circ\delta = \delta\circ\tau $及$ z(t)\leq x(t)+p_0x(\tau(t)) $以及引理2.1,则(2.4)式进一步可写成

$ \begin{eqnarray*} [\alpha(t)\phi_1(z'(t))]'+\frac{p_0^\beta}{\tau_0}[\alpha(\tau(t))\phi_1(z' (\tau(t)))]'&\leq& -Lq(t)(x(\delta(t)))^\beta-p_0^\beta Lq(\tau(t))(x(\delta(\tau(t))))^\beta \\ &\leq&-LQ(t)[(x(\delta(t)))^\beta+p_0^\beta(x(\delta(\tau(t))))^\beta]\\ &\leq&-LQ(t)[x(\delta(t))+p_0x(\delta(\tau(t)))]^\beta \\ &\leq&-LQ(t)z^\beta(\delta(t))\leq0. \end{eqnarray*} $

当$ \beta>1 $时,注意到引理2.2,则类似地可得

$ \begin{eqnarray*} [\alpha(t)\phi_1(z'(t))]'+\frac{p_0^\beta}{\tau_0} [\alpha(\tau(t))\phi_1(z'(\tau(t)))]' &\leq&-LQ(t)[(x(\delta(t)))^\beta+p_0^\beta(x(\delta(\tau(t))))^\beta] \\ &\leq&-L2^{1-\beta}Q(t)[x(\delta(t))+p_0x(\delta(\tau(t)))]^\beta\\ &\leq&-L2^{1-\beta}Q(t)z^\beta(\delta(t))\leq0. \end{eqnarray*} $

注意到记号$ L_0 $的定义,则上述两式可写成

(2.5) $ \begin{equation} [\alpha(t)\phi_1(z'(t))]'+\frac{p_0^\beta}{\tau_0}[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'\leq-L_0Q(t)z^\beta(\delta(t))\leq0. \end{equation} $

情形(a)  $ \lambda\leq\beta $.首先,作Riccati变换:

(2.6) $ \begin{equation} w(t) = \varphi(t)\frac{\alpha(t)\phi_1(z'(t))}{\phi_2(z(\delta(t)))} = \varphi(t)\frac{\alpha(t)(z'(t))^\lambda}{z^\beta(\delta(t))}, t\geq t_1, \end{equation} $

则$ w(t)>0(t\geq t_1) $,注意到$ \tau'(t)\geq \tau_0>0 $,由(2.6)式就有

(2.7) $ \begin{eqnarray} w'(t)& = &\varphi'(t)\frac{\alpha(t)(z'(t))^\lambda}{z^\beta(\delta(t))}+\varphi(t)\frac{[\alpha(t)(z'(t))^\lambda]'z^\beta(\delta(t))-\alpha(t)(z'(t))^\lambda\beta z^{\beta-1}(\delta(t))z'(\delta(t))\delta'(t)}{z^{2\beta}(\delta(t))} \\ & = &\frac{\varphi'(t)}{\varphi(t)}w(t)+\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'-\beta\frac{\varphi(t)\alpha(t)\delta'(t)(z'(t))^\lambda z'(\delta(t))}{z^{\beta+1}(\delta(t))}. \end{eqnarray} $

因为函数$ \alpha(t)[z'(t)]^\lambda $单调减少,所以$ \alpha(\delta(t))[z'(\delta(t))]^\lambda\geq\alpha(t)[z'(t)]^\lambda $,亦即$ z'(\delta(t))\geq\Big(\frac{\alpha(t)}{\alpha(\delta(t))}\Big)^{\frac{1}{\lambda}}z'(t) $.此外,由$ z(t)>0, z'(t)>0 $知,存在常数$ \eta>0 $使得当$ t\geq t_1 $时,有$ z(t)\geq\eta $且$ z(\delta(t))\geq\eta $.于是,综合(2.7)式和(2.6)式,并注意到引理2.3中的不等式,得

(2.8) $ \begin{eqnarray} w'(t)&\leq&\frac{\varphi'(t)}{\varphi(t)}w(t)+\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'-\beta\frac{\varphi(t)\alpha(t)\delta'(t)(z'(t))^\lambda }{z^{\beta+1}(\delta(t))}\Big(\frac{\alpha(t)}{\alpha(\delta(t))}\Big)^{\frac{1}{\lambda}}z'(t) \\ & = &\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\varphi'(t)}{\varphi(t)}w(t)-\frac{\beta\delta'(t)z^{\frac{\beta-\lambda}{\lambda}}(t)}{[\varphi(t)\alpha(\delta(t))]^{\frac{1}{\lambda}}}w^{\frac{\lambda+1}{\lambda}}(t) \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\varphi'_+(t)}{\varphi(t)}w(t)-\frac{\beta\delta'(t)\eta^{\frac{\beta-\lambda}{\lambda}}}{[\varphi(t)\alpha(\delta(t))]^{\frac{1}{\lambda}}}w^{\frac{\lambda+1}{\lambda}}(t) \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1}. \end{eqnarray} $

其次,再作Riccati变换:

(2.9) $ \begin{equation} v(t) = \varphi(t)\frac{\alpha(\tau(t))\phi_1(z'(\tau(t)))}{\phi_2(z(\delta(t)))} = \varphi(t)\frac{\alpha(\tau(t))(z'(\tau(t)))^\lambda}{z^\beta(\delta(t))}, t\geq t_1. \end{equation} $

则$ v(t)>0(t\geq t_1) $.类似于上面的推导过程,并注意到$ \alpha(\delta(t))[z'(\delta(t))]^\lambda\geq\alpha(\tau(t))[z'(\tau(t))]^\lambda $和$ z(\delta(t))\geq\eta $,得

(2.10) $ \begin{eqnarray} v'(t)& = &\varphi(t)\frac{[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'z^\beta(\delta(t))-\alpha(\tau(t))(z'(\tau(t)))^\lambda\beta z^{\beta-1}(\delta(t))z'(\delta(t))\delta'(t)}{z^{2\beta}\delta(t)} \\ && +\varphi'(t)\frac{\alpha(\tau(t))(z'(\tau(t)))^\lambda}{z^\beta(\delta(t))} \\ &\leq&\frac{\varphi'(t)}{\varphi(t)}v(t)+\frac{\varphi(t)}{z^\beta(\delta(t))} [\alpha(\tau(t))(z'(\tau(t)))^\lambda]'\\ &&-\beta\frac{\varphi(t)\alpha(\tau(t))(z'(\tau(t)))^\lambda\delta'(t)}{z^{\beta+1}(\delta(t))}\Big(\frac{\alpha(\tau(t))}{\alpha(\delta(t))}\Big)^{\frac{1}{\lambda}}z'(\tau(t)) \\ &\leq & \frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'+\frac{\varphi'_+(t)}{\varphi(t)}v(t)-\frac{\beta\delta'(t)\eta^{\frac{\beta-\lambda}{\lambda}}}{[\varphi(t)\alpha(\delta(t))]^{\frac{1}{\lambda}}}v^{\frac{\lambda+1}{\lambda}}(t) \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'+\frac{\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1}. \end{eqnarray} $

于是,综合(2.8)和(2.10)两式,并注意到(2.5)式,就有

$ \begin{eqnarray*} &&w'(t)+\frac{p_0^\beta}{\tau_0}v'(t)\\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1} \\ &&+\frac{p_0^\beta}{\tau_0}\Big\{\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'+\frac{\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1}\Big\} \\ & = &\frac{\varphi(t)}{z^\beta(\delta(t))}\Big\{[\alpha(t)(z'(t))^\lambda]' +\frac{p_0^\beta}{\tau_0}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'\Big\}\\ &&+\frac{(1+\frac{p_0^\beta}{\tau_0})\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1} \\ &\leq&-L_0\varphi(t)Q(t)+\frac{(1+\frac{p_0^\beta}{\tau_0})\lambda^\lambda\varphi(t)\alpha(\delta(t))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(t))^\lambda}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\lambda+1}. \end{eqnarray*} $

两边从$ t_1 $到$ t(t\geq t_1) $积分,得

$ \begin{eqnarray*} &&\int_{t_1}^t\varphi(s)\Big[L_0Q(s)-\frac{(1+\frac{p_0^\beta}{\tau_0})\lambda^\lambda\alpha(\delta(s))}{(\lambda+1)^{\lambda+1}\beta^\lambda\eta^{\beta-\lambda}(\delta'(s))^\lambda}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\lambda+1}\Big]{\rm d}s \\ &\leq&-\Big\{[w(t)-w(t_1)]+\frac{p_0^\beta}{\tau_0}[v(t)-v(t_1)]\Big\}\leq\Big\{w(t_1)+\frac{p_0^\beta}{\tau_0}v_1(t_1)\Big\}. \end{eqnarray*} $

这与(2.1)式矛盾.

情形(b)  $ \lambda>\beta $.作Riccati变换如(2.6)式,则(2.7)式成立.注意到$ \alpha(t)\phi_1(z'(t)) = \alpha(t)(z'(t))^\lambda $单调减少,所以$ \alpha(t)(z'(t))^\lambda\leq\alpha(\tau(t))(z'(\tau(t)))^\lambda\leq\eta_1, t\geq t_1 $, (常数$ \eta_1>0 $),即

(2.11) $ \begin{equation} \frac{1}{(z'(t))^{\frac{\lambda-\beta}{\beta}}}\leq\frac{\alpha^{\frac{\lambda-\beta}{\beta\lambda}}(t)}{\eta_1^{\frac{\lambda-\beta}{\beta\lambda}}}, \frac{1}{(z'(\tau(t)))^{\frac{\lambda-\beta}{\beta}}}\leq\frac{\alpha^{\frac{\lambda-\beta}{\beta\lambda}}(\tau(t))}{\eta_1^{\frac{\lambda-\beta}{\beta\lambda}}}. \end{equation} $

于是由(2.7)式,并注意到(2.11)式的第1个式子,得

(2.12) $ \begin{eqnarray} w'(t)&\leq&\frac{\varphi'(t)}{\varphi(t)}w(t)+\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'-\beta\frac{\varphi(t)\alpha(t)\delta'(t)(z'(t))^\lambda}{z^{\beta+1}(\delta(t))}\Big(\frac{\alpha(t)}{\alpha(\delta(t))}\Big)^{\frac{1}{\lambda}}z'(t) \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\varphi'_+(t)}{\varphi(t)}w(t)-\frac{\beta\delta'(t)(w(t))^{\frac{\beta+1}{\beta}}}{\varphi^{\frac{1}{\beta}}(t)\alpha^{\frac{1}{\lambda}}(\delta(t))\alpha^{\frac{\lambda-\beta}{\beta\lambda}}(t)(z'(t))^{\frac{\lambda-\beta}{\beta}}} \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\varphi'_+(t)}{\varphi(t)}w(t)-\frac{\beta\delta'(t)(w(t))^{\frac{\beta+1}{\beta}}}{\varphi^{\frac{1}{\beta}}(t)\alpha^{\frac{1}{\lambda}}(\delta(t))\eta_1^{\frac{\lambda-\beta}{\beta\lambda}}} \\ &\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1}. \end{eqnarray} $

再作Riccati变换如(2.9)式,同(2.10)式的推导过程类似,注意到(2.11)式的第2个式子可得

(2.13) $ \begin{equation} v'(t)\leq\frac{\varphi(t)}{z^\beta(t)}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'+\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1}. \end{equation} $

于是,综合(2.12)和(2.13)两式,并分别注意到(2.5)式,可得

$ \begin{eqnarray*} w'(t)+\frac{p_0^\beta}{\tau_0}v'(t)&\leq&\frac{\varphi(t)}{z^\beta(\delta(t))}[\alpha(t)(z'(t))^\lambda]'+\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1} \\ &&+\frac{p_0^\beta}{\tau_0}\Big\{\frac{\varphi(t)}{z^\beta(t)}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'+\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1}\Big\} \\ & = &\frac{\varphi(t)}{z^\beta(t)}\Big\{[\alpha(t)(z'(t))^\lambda]' +\frac{p_0^\beta}{\tau_0}[\alpha(\tau(t))(z'(\tau(t)))^\lambda]'\\ &&+(1+\frac{p_0^\beta}{\tau_0})\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1}\Big\} \\ &\leq&-L_0\varphi(t)Q(t)+(1+\frac{p_0^\beta}{\tau_0})\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\varphi(t)\alpha^{\frac{\beta}{\lambda}}(\delta(t))}{(\beta+1)^{\beta+1}(\delta'(t))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(t)}\Big)^{\beta+1}, \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&\int_{t_1}^t\varphi(s)\Big[L_0Q(s)-(1+\frac{p_0^\beta}{\tau_0})\frac{\eta_1^{\frac{\lambda-\beta}{\lambda}}\alpha^{\frac{\beta}{\lambda}}(\delta(s))}{(\beta+1)^{\beta+1}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(t)}{\varphi(s)}\Big)^{\beta+1}\Big]{\rm d}s \\ &\leq&-\Big\{[w(t)-w(t_1)]+\frac{p_0^\beta}{\tau_0}[v(t)-v(t_1)]\Big\}\leq\Big\{w(t_1)+\frac{p_0^\beta}{\tau_0}v_1(t_1)\Big\}. \end{eqnarray*} $

这与(2.1)式矛盾.定理证毕.

由定理2.1可以得到以下一系列结果.

推论2.2  设条件(C$ _1) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda, \delta(t)\leq\tau(t), $$ \delta'(t)>0 $,如果

(2.14) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\Big[L_0A(\delta(s))Q(s)-\frac{(1+\frac{p_0^\lambda}{\tau_0})\delta'(s)}{(\lambda+1)^{\lambda+1}A^\lambda(\delta(s))\alpha^{\frac{1}{\lambda}}(\delta(s))}\Big]{\rm d}s = +\infty. \end{equation} $

则方程(1.1)是振荡的.

证  在定理2.1中取$ \varphi(t) = A(\delta(t)) $即得.

推论2.3  设条件(C$ _1) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda\leq1, \delta(t)\leq\tau(t), $$ \delta'(t)>0 $,如果

(2.15) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{A^{1-\lambda}(\delta(t))}\int_{t_0}^tA(\delta(s))Q(s){\rm d}s>\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}, (\lambda<1), \end{equation} $

(2.16) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{\ln A(\delta(t))}\int_{t_0}^tA(\delta(s))Q(s){\rm d}s>\frac{(1+\frac{P_0}{\tau_0}}{4L_0}, (\lambda = 1), \end{equation} $

则方程(1.1)是振荡的.

证  当$ \lambda<1 $时,由(2.15)式知:存在常数$ \varepsilon>0 $,对充分大的$ t $,有

$ \frac{1}{A^{1-\lambda}(\delta(t))}\int_{t_0}^t A(\delta(s))Q(s){\rm d}s\geq\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0 (\lambda+1)^{\lambda+1}(1-\lambda)}+\varepsilon, $

即

$ \int_{t_0}^tA(\delta(s))Q(s){\rm d}s-\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}A^{1-\lambda}(\delta(t))\geq\varepsilon A^{1-\lambda}(\delta(t)), $

于是

$ \begin{eqnarray*} &&\int_{t_0}^t\Big[A(\delta(s))Q(s)-\frac{(1+\frac{p_0^\lambda}{\tau_0})\delta'(s)}{L_0(\lambda+1)^{\lambda+1}A^\lambda(\delta(s))\alpha^{\frac{1}{\lambda}}(\delta(s))}\Big]{\rm d}s \\ & = &\int_{t_0}^tA(\delta(s))Q(s){\rm d}s-\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}}\Big[\frac{A^{1-\lambda}(\delta(s))}{1-\lambda}\Big]_{t_0}^t \\ & = &\int_{t_0}^tA(\delta(s))Q(s){\rm d}s-\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}A^{1-\lambda}(\delta(t))+\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}A^{1-\lambda}(\delta(t_0)) \\ &\geq&\varepsilon A^{1-\lambda}(\delta(t))+\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}A^{1-\lambda}(\delta(t_0))\rightarrow+\infty(t\rightarrow+\infty), \end{eqnarray*} $

即(2.14)式成立,于是由推论2.2知,方程(1.1)是振荡的.

当$ \lambda = 1 $时,注意到$ [\ln A(\delta(s))]' = \frac{\delta'(s)}{A(\delta(s))\alpha(\delta(s))} $,则类似可证.证毕.

推论2.4  设条件(C$ _1) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda\leq1, $$ \delta(t)\leq\tau(t), $$ \delta'(t)>0 $,如果

(2.17) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{Q(t)A^{\lambda+1}(\delta(t))\alpha^{\frac{1}{\lambda}}(\delta(t))}{\delta'(t)}>\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}}, (\lambda<1), \end{equation} $

(2.18) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{Q(t)A^2(\delta(t))\alpha(\delta(t))}{\delta'(t)}>\frac{(1+\frac{p_0}{\tau_0})}{4L_0}, (\lambda = 1), \end{equation} $

则方程(1.1)是振荡的.

证  当$ \lambda<1 $时,由(2.15)式知:存在常数$ \varepsilon>0 $,对充分大的$ t $,有

$ \begin{eqnarray*} \frac{Q(t)A^{\lambda+1}(\delta(t))\alpha^{\frac{1}{\lambda}}(\delta(t))}{\delta'(t)}\geq\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}}+\varepsilon, \end{eqnarray*} $

两边同乘以$ \frac{\delta'(t)}{A^\lambda(\delta(t))\alpha^{\frac{1}{\lambda}}(\delta(t))} $,可得

$ \begin{eqnarray*} L_0Q(t)A(\delta(t))-\frac{(1+\frac{p_0^\lambda}{\tau_0})\delta'(t)}{(\lambda+1)^{\lambda+1}A^\lambda(\delta(t))\alpha^{\frac{1}{\lambda}}(\delta(t))}\geq\frac{L_0\varepsilon\delta'(t)}{A^\lambda(\delta(t))\alpha^{\frac{1}{\lambda}}(\delta(t))} = \frac{L_0\varepsilon}{1-\lambda}(A^{1-\lambda}(\delta(t)))', \end{eqnarray*} $

由此式不难看出(2.14)式成立,于是由推论2.2知,方程(1.1)是振荡的.

当$ \lambda = 1 $时,注意到$ [\ln A(\delta(s))]' = \frac{\delta'(s)}{A(\delta(s))\alpha(\delta(s))} $,则类似可证.证毕.

注2.1  由定理2.1看出,在$ \lambda>\beta $和$ \lambda<\beta $的情形,方程(1.1)的振荡准则是有差异的.若$ \beta = 1 $且$ \lambda = 1 $,则定理2.1即为文献[9]中的定理2.1.

定理2.5  设条件(C$ _2) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \delta(t)\leq\tau(t), \delta'(t)>0 $,如果存在函数$ \varphi\in C^1([t_0+\infty), (0, +\infty)), $使得(2.1)式成立,且有

(2.19) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_T^t\Big\{L_0Q(s)\pi(s)\zeta^\lambda(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{(\frac{\lambda}{\lambda+1})^{\lambda+1}}{\zeta(s)\alpha^{\frac{1}{\lambda}}(s)}{\rm d}s\Big\} = +\infty, \end{equation} $

其中常数$ T\geq t_0 $足够大,函数

$ \pi(t) = \left\{\begin{array}{ll} k, &\lambda>\beta, \\ 1, &\lambda = \beta, \\ k\zeta^{\beta-\lambda}(t), \quad &\lambda<\beta, \end{array} \right. $

($ k>0 $为某常数), $ \zeta(t) = \int_t^{+\infty}\alpha^{\frac{-1}{\lambda}}(s){\rm d}s $则方程(1.1)是振荡的.

证  反证法,设方程(1.1)存在一个非振荡解$ x(t) $,不妨设$ x(t) $为方程(1.1)的最终正解(当为$ x(t) $最终负解时类似可证),则$ \exists t_1\geq t_0 $,使得当$ t\geq t_1 $时,有$ x(t)>0, x(\tau(t))>0, $$ x(\delta(t))>0 $,由定理2.1的证明知函数$ \alpha(t)[z'(t)]^\gamma $严格单调减少且最终定号,从而$ z'(t) $最终为正或最终为负.所以我们只需考虑下列两种情形

(a)当$ t\geq t_1 $时$ z'(t)>0 $; (b)当$ t\geq t_1 $时$ z'(t)<0 $.

情形(a)  $ z'(t)>0(t\geq t_1) $.此时同定理2.1的证明,得与(2.1)式矛盾.

情形(b)  $ z'(t)<0(t\geq t_1) $.首先,定义函数$ v(t) $为

(2.20) $ \begin{equation} v(t) = \frac{\alpha(\tau(t))\phi_1(z'(\tau(t)))}{\phi_1(z(t))} = \frac{\alpha(\tau(t))(-z'(\tau(t)))^{\lambda-1}z'(\tau(t))}{z^\lambda(t)}, t\geq t_1, \end{equation} $

则$ v(t)<0(t\geq t_1) $.由于$ \alpha(t)[-z'(t)]^{\lambda-1}z'(t) $是单调减少的,所以有

$ \begin{eqnarray*} \alpha(\tau(t))[-z'(\tau(t))]^{\lambda-1}z'(\tau(t))\geq\alpha(t)[-z'(t)]^{\lambda-1}z'(t), \end{eqnarray*} $

即

$ \begin{eqnarray*} \alpha(\tau(t))[-z'(\tau(t))]^\lambda\leq\alpha(t)[-z'(t)]^\lambda, \end{eqnarray*} $

亦即$ z'(t)\leq\Big(\frac{\alpha(\tau(t))}{\alpha(t)}\Big)^{\frac{1}{\lambda}}z'(\tau(t)) $.注意到$ z'(t)<0 $,于是由(2.20)式可得

(2.21) $ \begin{eqnarray} v'(t) & = &\frac{[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'}{z^\lambda(t)}-\frac{\alpha(\tau(t))(-z'(\tau(t)))^{\lambda-1}z'(\tau(t))\lambda z^{\lambda-1}(t)z'(t)}{z^{2\lambda}(t)} \\ &\leq&\frac{[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'}{z^\lambda(t)}-\frac{\lambda\alpha(\tau(t))(-z'(\tau(t)))^{\lambda-1}z'(\tau(t))}{z^{\lambda+1}(t)}\Big(\frac{\alpha(\tau(t))}{\alpha(t)}\Big)^{\frac{1}{\lambda}}z'(\tau(t)) \\ & = &\frac{[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'}{z^\lambda(t)}-\frac{\lambda(-v(t))^{\frac{\lambda+1}{\lambda}}}{\alpha^{\frac{1}{\lambda}}(t)}. \end{eqnarray} $

再定义函数$ w(t) $为

$ \begin{eqnarray*} w(t) = \frac{\alpha(t)\phi_1(z'(t))}{\phi_1(z(t))} = \frac{\alpha(t)(-z'(t))^{\lambda-1}z'(t)}{z^\lambda(t)}, t\geq t_1, \end{eqnarray*} $

则$ w(t)<0(t\geq t_1) $,按与上面同样类似的方法可得

(2.22) $ \begin{equation} w'(t)\leq\frac{[\alpha(t)\phi_1(z'(t))]'}{z^\lambda(t)}-\frac{\lambda(-w(t))^{\frac{\lambda+1}{\lambda}}}{\alpha^{\frac{1}{\lambda}}(t)}. \end{equation} $

由于(2.5)式仍然成立,且$ z(\delta(t))\geq z(t) $,所以综合(2.21)和(2.22)两式,得

(2.23) $ \begin{eqnarray} w'(t)+\frac{p_0^\beta}{\tau_0}v'(t) &\leq&\frac{1}{z^\lambda(t)}\Big\{[\alpha(t)\phi_1(z'(t))]'+\frac{p_0^\beta}{\tau_0}[\alpha(\tau(t))\phi_1(z'(\tau(t)))]'\Big\} \\ &&-\frac{\lambda}{\alpha^{\frac{1}{\lambda}}(t)}\Big[(-w(t))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(t))^{\frac{\lambda+1}{\lambda}}\Big] \\ &\leq&-\frac{L_0Q(t)z^\beta(\delta(t))}{z^\lambda(t)}-\frac{\lambda}{\alpha^{\frac{1}{\lambda}}(t)}\Big[(-w(t))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(t))^{\frac{\lambda+1}{\lambda}}\Big] \\ &\leq&-L_0Q(t)z^{\beta-\lambda}(t)-\frac{\lambda}{\alpha^{\frac{1}{\lambda}}(t)}\Big[(-w(t))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(t))^{\frac{\lambda+1}{\lambda}}\Big]. \end{eqnarray} $

当$ \lambda>\beta $时,由$ z(t)>0, z'(t)<0(t\geq t_1) $知$ z(t)\leq z(t_1) $,即$ z^{\beta-\lambda}(t)\geq z^{\beta-\lambda}(t_1) = k $.当$ \lambda = \beta $时, $ z^{\beta-\lambda}(t) = 1 $.当$ \lambda<\beta $时,再次利用$ \alpha(t)(-z'(t))^{\lambda-1}z'(t) $的单调减少性,得当$ s\geq t_1 $时,有

$ \begin{eqnarray*} \alpha(s)(-z'(s))^{\lambda-1}z'(s)\leq\alpha(t_1)(-z'(t_1))^{\lambda-1}z'(t_1) = -M. \end{eqnarray*} $

上式中$ M = -\alpha(t_1)(-z'(t_1))^{\lambda-1}z'(t_1)>0 $是常数,于是$ \alpha(s)(-z'(s))^\lambda\geq M $即$ z'(s)\leq-M^{\frac{1}{\lambda}}\alpha^{\frac{-1}{\lambda}}(s) $,进一步就有$ z(u)-z(t)\leq-M^{\frac{1}{\lambda}}\int_t^u\alpha^{\frac{-1}{\lambda}}(s){\rm d}s $,即$ z(t)\geq z(u)+M^{\frac{1}{\lambda}}\int_t^u\alpha^{\frac{-1}{\lambda}}(s){\rm d}s\geq M^{\frac{1}{\lambda}}\int_t^u\alpha^{\frac{-1}{\lambda}}(s){\rm d}s $,令$ u\rightarrow+\infty $,得$ z(t)\geq M^{\frac{1}{\lambda}}\int_t^u\alpha^{\frac{-1}{\lambda}}(s){\rm d}s = M^{\frac{1}{\lambda}}\zeta(t) $,亦即$ z^{\beta-\lambda}(t)\geq k\zeta^{\beta-\lambda}(t) $,这里$ k = M^{\frac{\beta-\lambda}{\lambda}}>0 $是常数.

综合上述3种情形并注意到函数$ \pi(t) $的定义,根据(2.23)式,则有

(2.24) $ \begin{equation} L_0Q(t)\pi(t)\leq-w'(t)-\frac{p_0^\beta}{\tau_0}v'(t)-\frac{\lambda}{\alpha^{\frac{1}{\lambda}}(t)}\Big[(-w(t))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(t))^{\frac{\lambda+1}{\lambda}}\Big]. \end{equation} $

将(2.24)式两边同乘以$ \zeta^\lambda(t) $并从$ t_1 $到$ t(t\geq t_1) $积分,注意到$ \zeta'(t) = -\alpha^{\frac{-1}{\lambda}}(t) $以及引理2.3,得

$ \begin{eqnarray*} &&\int_{t_1}^tL_0Q(s)\pi(s)\zeta^\lambda(s){\rm d}s\\ &\leq&-\int_{t_1}^t\zeta^\lambda(s)w'(s){\rm d}s-\frac{p_0^\beta}{\tau_0}\int_{t_1}^t\zeta^\lambda(s)v'(s){\rm d}s -\int_{t_1}^t\frac{\lambda\zeta^\lambda(s)}{\alpha^{\frac{1}{\lambda}}(s)}\Big[(-w(s))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s \\ & = &-\zeta^\lambda(t)w(t)+\zeta^\lambda(t_1)w(t_1) -\frac{p_0^\beta}{\tau_0}[\zeta^\lambda(t)v(t)-\zeta^\lambda(t_1)v(t_1)] \\ &&+\int_{t_1}^t\Big[\lambda\zeta^{\lambda-1}(s)\alpha^{\frac{-1}{\lambda}}(s)(-w(s))-\frac{\lambda\zeta^\lambda(s)}{\alpha^{\frac{1}{\lambda}}(s)}(-w(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s \\ &&+\frac{p_0^\beta}{\tau_0}\int_{t_1}^t\Big[\lambda\zeta^{\lambda-1}(s)\alpha^{\frac{-1}{\lambda}}(s)(-v(s))-\frac{\lambda\zeta^\lambda(s)}{\alpha^{\frac{1}{\lambda}}(s)}(-v(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s \\ &\leq&-\zeta^\lambda(t)w(t)+\zeta^\lambda(t_1)w(t_1)-\frac{p_0^\beta}{\tau_0} [\zeta^\lambda(t)v(t)-\zeta^\lambda(t_1)v(t_1)] +(1+\frac{p_0^\beta}{\tau_0})\int_{t_2}^t\frac{(\frac{\lambda}{\lambda+1})^{\lambda+1}}{\zeta(s)\alpha^{\frac{1}{\lambda}}(s)}{\rm d}s, \end{eqnarray*} $

即

(2.25) $ \begin{eqnarray} &&\int_{t_1}^t\Big\{L_0Q(s)\pi(s)\zeta^\lambda(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{(\frac{\lambda}{\lambda+1})^{\lambda+1}}{\zeta(s)\alpha^{\frac{1}{\lambda}}(s)}\Big\}{\rm d}s \\ &\leq&\Big[-w(t)\zeta^\lambda(t)-\frac{p_0^\beta}{\tau_0}v(t)\zeta^\lambda(t)\Big]+w(t_1)\zeta^\lambda(t_1)+\frac{p_0^\beta}{\tau_0}v(t_1)\zeta^\lambda(t_1). \end{eqnarray} $

此外,再次利用$ \alpha(t)(-z'(t))^{\lambda-1}z'(t) $的单调减少性,对$ s\geq t\geq t_1 $,有$ \alpha(s)(-z'(s))^{\lambda-1}z'(s)\leq\alpha(t)(-z'(t))^{\lambda-1}z'(t) $,即$ z'(s)\leq\frac{\alpha^{\frac{1}{\lambda}}(t)z'(t)}{\alpha^{\frac{1}{\lambda}}(s)} $,两边对$ s $从$ t $到$ u(u\geq t) $积分,于是得$ z(u)-z(t)\leq\alpha^{\frac{1}{\lambda}}(t)z'(t)\int_t^u\alpha^{\frac{1}{\lambda}}(s){\rm d}s $,从而$ z(t)+\alpha^{\frac{1}{\lambda}}(t)z'(t)\int_t^u\alpha^{\frac{1}{\lambda}}(s){\rm d}s\geq0 $,令$ u\rightarrow+\infty $,则有

$ z(t)+\alpha^{\frac{1}{\lambda}}(t)z'(t)\int_t^{+\infty} \alpha^{\frac{-1}{\lambda}}(s){\rm d}s\geq0, t\geq t_1. $

因此, $ -1\leq\frac{\alpha^{\frac{1}{\lambda}}(t)z'(t)}{z(t)}\zeta(t)\leq0 $,于是由函数$ w(t) $的定义知

(2.26) $ \begin{equation} -1\leq w(t)\zeta^\lambda(t)\leq0, t\geq t_1. \end{equation} $

同理可得

(2.27) $ \begin{equation} -1\leq v(t)\zeta^\lambda(t)\leq0, t\geq t_1. \end{equation} $

结合(2.26)和(2.27)两式,由(2.25)式就得到一个与(2.19)式矛盾的结果.定理证毕.

定理2.6  设条件(C$ _2) $成立,并且有$ 0\leq p(t)\leq p_0\leq+\infty $ ($ p_0 $为常数), $ \delta(t)\leq\tau(t) $, $ \delta'(t)>0 $且(2.1)式成立,如果存在常数$ \omega>0 $使得

(2.28) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_T^th^\omega(t, s)\Big[L_0Q(s)\pi(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{\omega^{\lambda+1}\xi^{\lambda+1}(s)\alpha(s)}{(\lambda+1)^{\lambda+1}h^{\lambda+1}(t, s)}\Big]{\rm d}s>0, \end{equation} $

其中常数$ T\geq t_0 $足够大,函数$ h(t, s) = \int_s^t\xi(\tau){\rm d}\tau $ (这里函数$ \xi(t) $是$ [t_0, +\infty) $上非负的连续函数,且在其任一有限子区间内不恒为0),而函数$ \pi(t) $和$ \zeta(t) $的定义如定理2.5,则方程(1.1)是振荡的.

证  同定理2.5的证明,可得(2.24)式,现将(2.24)式两边同乘以$ h^\omega(t, s) $后,再从$ t_1 $到$ t(t\geq t_1) $积分,并注意到$ [h^\omega(t, s)]'_s = -\omega h^{\omega-1}(t, s)\xi(s) $及引理2.3,则可得

$ \begin{eqnarray*} &&\int_{t_1}^tL_0h^\omega(t, s)Q(s)\pi(s){\rm d}s\\ &\leq&-\int_{t_1}^th^\omega(t, s)w'(s){\rm d}s-\frac{p_0^\beta}{\tau_0}\int_{t_1}^th^\omega(t, s)v'(s){\rm d}s \\ &&-\int_{t_1}^t\frac{\lambda h^\omega(t, s)}{\alpha^{\frac{1}{\lambda}}(s)}\Big[(-w(s))^{\frac{\lambda+1}{\lambda}}+\frac{p_0^\beta}{\tau_0}(-v(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s \\ &\leq & h^\omega(t, t_1)w(t_1)+\int_{t_1}^th^{\omega-1}(t, s)\Big[\omega\xi(s)(-w(s))-\frac{\lambda h(t, s)}{\alpha^{\frac{1}{\lambda}}(s)}(-w(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s \\ &&+\frac{p_0^\beta}{\tau_0}\Big\{h^\omega(t, t_1)v(t_1)+\int_{t_1}^th^{\omega-1}(t, s)\Big[\omega\xi(s)(-v(s))-\frac{\lambda h(t, s)}{\alpha^{\frac{1}{\lambda}}(s)}(-v(s))^{\frac{\lambda+1}{\lambda}}\Big]{\rm d}s\Big\} \\ &\leq & h^\omega(t, t_1)w(t_1)+\frac{p_0^\beta}{\tau_0}h^\omega(t, t_1)v(t_1)+(1+\frac{p_0^\beta}{\tau_0})\int_{t_1}^t\frac{\omega^{\lambda+1}\xi^{\lambda+1}(s)\alpha(s)}{(\lambda+1)^{\lambda+1}h^{\lambda-\omega+1}(t, s)}{\rm d}s. \end{eqnarray*} $

整理得

$ \begin{eqnarray*} &&\int_{t_1}^th^\omega(t, s)\Big[L_0Q(s)\pi(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{\omega^{\lambda+1}\xi^{\lambda+1}(s)\alpha(s)}{(\lambda+1)^{\lambda+1}h^{\lambda+1}(t, s)}\Big]{\rm d}s\\ &\leq & h^\omega(t, t_1)w(t_1)+\frac{p_0^\beta}{\tau_0}h^\omega(t, t_1)v(t_1). \end{eqnarray*} $

这与条件(2.28)矛盾.定理证毕.

推论2.7  设条件(C$ _2) $成立且$ 0\leq p(t)\leq P_0<+\infty $ ($ p_0 $为常数), $ \delta(t)\leq\tau(t), \delta'(t)>0 $且(2.1)式成立,如果存在常数$ \omega>0 $使得

(2.29) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_T^t(t-s)^\omega\Big[L_0Q(s)\pi(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{\omega^{\lambda+1}\alpha(s)}{(\lambda+1)^{\lambda+1}(t-s)^{\lambda+1}}\Big]{\rm d}s>0, \end{equation} $

其中常数$ T\geq t_0 $足够大,函数$ \pi(t) $和$ \zeta(t) $的定义如定理2.5,则方程(1.1)是振荡的.

证  在定理2.6中取$ \xi(t)\equiv1 $即得.

类似地,可得

推论2.8  设条件(C$ _2) $成立且$ 0\leq p(t)\leq P_0<+\infty $($ p_0 $为常数), $ \beta = \lambda, \delta(t)\leq\tau(t) $,且$ \delta'(t)>0 $,如果(2.14)–(2.18)式之一成立,并且(2.19)、(2.28)和(2.29)式之一成立,则方程(1.1)是振荡的.

例2.1  对常数$ \alpha>0 $,考虑二阶时滞微分方程

(2.30) $ \begin{equation} (x(t)+\frac{9}{10}x(\frac{t}{4}))''+\frac{\alpha}{t^2}x(\frac{t}{4}) = 0, t\geq1, \end{equation} $

这里$ \beta = \lambda = 1, \alpha(t)\equiv1, p(t) = \frac{9}{10}, q(t) = \frac{\alpha}{t^2}, \tau(t) = \delta(t) = \frac{t}{4}, f(u) = u $.显然,条件(H$ _1) $–(H$ _4) $及(C$ _1) $均满足.若取$ \varphi(t) = t $,由定理2.1(注意到此时$ L = 1, p_0 = \frac{9}{10}, \tau_0 = \frac{1}{4} $),则

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\Big[L_0Q(s)-\frac{k\alpha^{\omega_1}(\delta(s))}{(\delta'(s))^{\omega_2}}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\omega_2+1}\Big]{\rm d}s \\ & = &\limsup\limits_{t\rightarrow+\infty}\int_1^ts\Big[\frac{\alpha}{s^2}-\Big(1+\frac{\frac{9}{10}}{\frac{1}{4}}\Big)\frac{1}{2^2(\frac{1}{4})}\Big(\frac{1}{s}\Big)^2\Big]{\rm d}s = \Big(\alpha-\frac{23}{5}\Big)\limsup\limits_{t\rightarrow+\infty}\int_1^t\frac{1}{s}{\rm d}s. \end{eqnarray*} $

所以由定理2.1知,当$ \alpha>\frac{23}{5} = 4.6 $时方程(2.30)是振荡的.

注2.2  若用文献[10]的定理2.1来判定,则由于

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\Big \{\varphi(s)q(s)[1-p(\delta(s))]^\gamma-\frac{(\varphi'(s))^{\gamma+1} \alpha(\delta(s))}{k(\gamma+1)^{\gamma+1}[\varphi(s)\delta'(s)]^\gamma} \Big\}{\rm d}s = \big(\alpha-10\big)\limsup\limits_{t\rightarrow+\infty}\int_1^t \frac{1}{s}{\rm d}s. \end{eqnarray*} $

所以当$ \alpha>10 $时方程(2.30)是振荡的.因此,本文定理2.1改进了文献[10]中的定理2.1.

例2.2  考虑二阶泛函微分方程(E$ _1) $,即下列方程

$ \Big\{\frac{1}{t^{\frac{5}{7}}}\Big[\Big(x(t)+(5-\sin t^2)x(\frac{t}{2})\Big)'\Big]^{\frac{1}{3}}\Big\}'+\frac{1}{t^2}f(x(\frac{t}{2})) = 0, t\geq1, ~~~~~~~~~~~~({\rm E}_1) $

这里$ t_0 = 1, \lambda = \frac{1}{3}, \beta = 1, \alpha(t) = \frac{1}{t^{\frac{5}{7}}}, p(t) = 5-\sin t^2, \tau(t) = \frac{t}{2}, \delta(t) = \frac{t}{2}, q(t) = \frac{1}{t^2}. $我们取$ f(u) = u[1+\ln(1+u^4)] $,由于

$ \begin{eqnarray*} 0<p(t) = 5-\sin t^2\leq6 = p_0, \frac{f(u)}{u} = 1+\ln(1+u^4)\geq1 = L, . \end{eqnarray*} $

$ \int_{t_0}^{+\infty}\alpha^{\frac{-1}{\lambda}}(t){\rm d}t = \int_1^{+\infty}t^{\frac{15}{7}}{\rm d}t = +\infty $.显然条件(H$ _1) $–(H$ _4) $及(C$ _1) $全部满足.又因为$ \lambda\leq\beta, $$ Q(t) = \min{\{q(t), q(\tau(t))\}} = \frac{1}{t^2} $,取$ \varphi(t) = t $,则

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\Big[L_0Q(s)-\frac{k\alpha^{\omega_1}(\delta(s))}{(\delta'(s))^{\omega_2}}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\omega_2+1}\Big]{\rm d}s = \limsup\limits_{t\rightarrow+\infty}\int_1^t\Big(\frac{1}{s}-\frac{k}{s^{\frac{22}{21}}}\Big){\rm d}s = +\infty, \end{eqnarray*} $

所以定理2.1的条件均满足,故由定理2.1知,方程(E$ _1) $是振荡的.

注2.3  由于方程(E$ _1) $的中立项系数函数$ p(t)>1 $且$ \lambda\neq\beta $,因此文献[1-6, 9-12]等中的定理都不能用于方程(E$ _1) $.

例2.3  考虑Euler方程(E$ _2) $,即下列方程

$ (t^2x'(t))'+mx(t) = 0, t\geq1, ~~~~~~~~~~~~({\rm E}_2) $

其中$ m>0 $是常数.取$ \varphi(t) = 1 $,注意到$ \pi(t) = 1, \zeta(t) = \frac{1}{t} $,则

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\Big[L_0Q(s)-\frac{k\alpha^{\omega_1}(\delta(s))}{(\delta'(s))^{\omega_2}}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\omega_2+1}\Big]{\rm d}s = \limsup\limits_{t\rightarrow+\infty}\int_1^tm {\rm d}s = +\infty, \end{eqnarray*} $

且当$ m>\frac{1}{4} $时

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\int_T^t\Big\{L_0Q(s)\pi(s) \zeta^\lambda(s)-\Big(1+\frac{p_0^\beta}{\tau_0}\Big)\frac{(\frac{\lambda} {\lambda+1})^{\lambda+1}}{\zeta(s)\alpha^{\frac{1}{\lambda}}(s)}\Big\}{\rm d}s = (m-\frac{1}{4})\limsup\limits_{t\rightarrow+\infty}\int_1^t\frac{1}{s}{\rm d}s = +\infty. \end{eqnarray*} $

所以由定理2.5知当$ m>\frac{1}{4} $时方程(E$ _2) $是振荡的,这就是我们熟知的结果.

注2.4  显然,文献[1-12]等中的定理或者只能得到"方程(E$ _2) $的每一个解$ x(t) $或者振荡或者满足$ \lim\limits_{t\rightarrow+\infty}x(t) = 0 $",或者不能用于方程(E$ _2) $.

在上一节中的定理2.1中,由于有条件$ \delta(t)\leq\tau(t) $,因此当$ \delta(t)\geq\tau(t) $时上一节的结论就不一定成立.但我们可以修改定理2.1证明过程中的Riccati变换分别为

$ w(t) = \varphi(t)\frac{\alpha(t)\phi_1(z'(t))}{\phi_2(z(\tau(t)))} = \varphi(t)\frac{\alpha(t)(z'(t))^\lambda}{z^\beta(\tau(t))}, t\geq t_1 $

和

$ v(t) = \varphi(t)\frac{\alpha(\tau(t))\phi_1(z'(\tau(t)))}{\phi_2(z(\tau(t)))} = \varphi(t)\frac{\alpha(\tau(t))(z'(\tau(t)))^\lambda}{z^\beta(\tau(t))}, t\geq t_1. $

利用与上一节完全类似的分析方法,可得方程(1.1)的如下振荡准则(限于篇幅,不作证明,仅列出结果).

定理3.1  设条件(C$ _1) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \delta(t)\geq\tau(t) $如果存在函数$ \varphi\in C^1([t_0, +\infty), (0+\infty)) $使得

(3.1) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\Big[L_0Q(s)-\frac{k\alpha^{\omega_1}(\tau(s))}{\tau_0^{\omega_2}}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\omega_2+1}\Big]{\rm d}s = +\infty, \end{equation} $

其中常数$ k, \omega_1, \omega_2 $的定义如定理2.1,则方程(1.1)是振荡的.

推论3.2  设条件(C$ _1) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda, \delta(t)\geq\tau(t) $,如果

(3.2) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\Big[L_0A(\tau(s))Q(s)-\frac{(1+\frac{p_0^\lambda}{\tau_0})\tau_0}{(\lambda+1)^{\lambda+1}A^\lambda(\tau(s))\alpha^{\frac{1}{\lambda}}(\tau(s))}\Big]{\rm d}s = +\infty, \end{equation} $

则方程(1.1)是振荡的.

推论3.3  设条件(C$ _1) $成立,且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda\leq1, \delta(t)\geq\tau(t) $,如果

(3.3) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty} \frac{1}{A^{1-\lambda}(\delta(t))}\int_{t_0}^tA(\tau(s))Q(s){\rm d}s>\frac{1+\frac{p_0^\lambda}{\tau_0}}{L_0(\lambda+1)^{\lambda+1}(1-\lambda)}, (\lambda<1 ). \end{equation} $

(3.4) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{\ln A(\delta(t))}\int_{t_0}^tA(\tau(s))Q(s){\rm d}s>\frac{1+\frac{p_0}{\tau_0}}{4L_0}, (\lambda = 1), \end{equation} $

则方程(1.1)是振荡的.

推论3.4  设条件(C$ _1) $成立,且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda\leq1, \delta(t)\geq\tau(t) $,如果

(3.5) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{Q(t)A^{\lambda+1}(\tau(t))\alpha^{\frac{1}{\lambda}}(\tau(t))}{\tau_0}>\frac{(1+\frac{p_0^\lambda}{\tau_0})}{L_0(\lambda+1)^{\lambda+1}}, (\lambda<1), \end{equation} $

(3.6) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{Q(t)A^2(\tau(t))\alpha(\tau(t))}{\tau_0}>\frac{(1+\frac{p_0}{\tau_0})}{4L_0}, (\lambda = 1), \end{equation} $

则方程(1.1)是振荡的.

定理3.5  设条件(C$ _2) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \delta(t)\geq\tau(t) $,如果存在函数$ \varphi\in C^1([t_0, +\infty), (0+\infty)) $使得(3.1)式成立,并且(2.19)、(2.28)和(2.29)式之一成立,则方程(1.1)是振荡的.

推论3.6  设条件(C$ _2) $成立且$ 0\leq p(t)\leq p_0<+\infty $ ($ p_0 $为常数), $ \beta = \lambda, \delta(t)\geq\tau(t) $,如果(3.2)–(3.6)式之一成立,并且(2.19)、(2.28)和(2.29)式之一成立,则方程(1.1)是振荡的.