近年来, 中立型时滞泛函微分方程的振动性等定性理论的研究引起了国内外学者的广泛兴趣和高度关注, 并发表了许多论文和专著, 见文献[1-17]及其参考文献.笔者考虑具有一个非线性中立项的二阶非线性变时滞Emden-Fowler型泛函微分方程

$ \{a(t)|[x(t)+p(t)x^{\alpha}(\tau(t))]'|^{\beta-1}[x(t)+p(t)x^{\alpha}(\tau(t))]'\}' +q(t)f(|x(\delta(t))|^{\gamma-1}x(\delta(t)))=0, t \ge t_{0} $

(1.1)

的振动性, 为了叙述方便, 本文总假设下列条件成立.

(H$_{1}$)  常数$0<\alpha\leqslant1$及$\gamma>0$均为两个正奇数的商, 而$\beta>0$为实常数; 函数$a\in C^{1}([t_{0}, +\infty), (0, +\infty))$, $p, q\in C([t_{0}, +\infty), {\mathbb{R}})$, 且$0\leqslant p(t)<1, q(t)>0$.

(H$_{2}$)  滞量函数$\tau, \delta:[t_{0}, +\infty)\rightarrow(0, +\infty)$, $\tau(t)\leqslant t$且$\lim\limits_{t\rightarrow+\infty}\tau(t)=+\infty$; $\delta(t)\leqslant t, $ $\lim\limits_{t\rightarrow+\infty}\delta(t)=+\infty$且$\delta'(t)>0$.

(H$_{3}$)  函数$f\in C({\mathbb{R}}, {\mathbb{R}})$, 且当$u\neq0$时$uf(u)\neq0$, $f(u)/u \ge L$ (这里$L>0$为常数).

我们称函数$x(t)\in C^{1}([T_{x}, +\infty), {\mathbb{R}})(T_{x} \ge t_{0})$是方程(1.1)的一个解, 如果函数$x(t)$满足

$ a(t)|[x(t)+p(t)x^{\alpha}(\tau(t))]'|^{\beta-1}[x(t)+p(t)x^{\alpha}(\tau(t))]'\in C^{1}([T_{x}, +\infty), {\mathbb{R}}) $

且在半区间$[T_{x}, +\infty)$上满足方程(1.1);方程(1.1)的一个解$x(t)$称为是振动的, 如果它既不最终为正也不最终为负, 否则称它是非振动的; 方程(1.1)称为是振动的, 如果它的所有解都是振动的.本文中我们感兴趣的是方程(1.1)的非平凡解.

由于方程(1.1)是具有一个拟线性中立项的微分方程, 这就给研究带来了极大的困难.而对于具有非线性中立项的, 学者们要么回避研究这类方程, 要么是增加一些条件, 将方程的这个非线性中立项转化为线性的来研究^{[1-3, 5-17]}.仅文献[4]直接研究了具有非线性中立项的一阶微分方程

$ [x(t)-px^{\alpha}(t-\tau)]'+q(t)\prod\limits^{m}_{j=1}|x(t-\sigma_{j})|^{\beta_{j}}{\rm sgn}|x(t-\sigma_{j})|=0 $

的振动性, 得到了其解振动的一些判别准则.近年来, 许多文献研究了方程(1.1)的特殊情形即当$\alpha=1$(相当于中立项是线性的)时的振动准则.如Sun等^[6]在条件

$ \int^{+\infty}_{t_{0}}a^{-1/\beta}(t){\rm d}t=+\infty $

(1.2)

下研究了二阶半线性时滞微分方程

$ [a(t)|x'(t)|^{\beta-1}x'(t)]'+q(t)|x(\delta(t))|^{\beta-1}x(\delta(t))=0 $

(1.3)

的振动性, 得到(1.3)式振动的若干充分条件.最近, 黄记洲等^[7], 曾云辉等^[8]在条件(1.2)或

$ \int^{+\infty}_{t_{0}}a^{-1/\beta}(t){\rm d}t<+\infty $

(1.4)

下研究了二阶Emden-Fowler型微分方程

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

(1.5)

的振动性, 得到了方程(1.5)若干新的振动准则, 推广并改进了文献[6]的一些结果, 但文献[7]有限制条件"$a'(t) \ge 0$", 且当$\beta<\gamma$时没有方程(1.5)的振动准则.本文将利用Riccati变换技术和大量不等式技巧来研究具有非线性中立项的微分方程(1.1)的振动性, 在(1.2)或(1.4)式成立的条件得到了方程(1.1)振动的一些新的准则, 推广并改进了现有文献中的一些结果, 同时使得其它许多已知结果均为我们结果的特例.

首先给出几个引理.其中引理2.1的结果由函数$f(x)=x^{\lambda}(0<\lambda\leqslant1)$的凹凸性容易得到, 故略去其证明; 而引理2.2和引理2.3是众所周知的结果, 也可在有关的文献中找到.

引理2.1  设$X, Y$为非负实数, 则当$0<\lambda\leqslant1$时$X^{\lambda}+Y^{\lambda}\leqslant2^{1-\lambda}(X+Y)^{\lambda}$.

引理2.2  (贝努利不等式)对任意实数$x>-1$, 当$0\leqslant r\leqslant1$时, $(1+x)^{r}\leqslant1+rx$, 当$r\leqslant0$或$r \ge 1$时, $(1+x)^{r} \ge 1+rx$.

引理2.3  (杨氏不等式)设$a>0$及$b>0$均为常数, 则$ab\leqslant\frac{1}{p}a^{p}+\frac{1}{q}b^{q}$, 其中$p>1, q>1$且$\frac{1}{p}+\frac{1}{q}=1$.

本文引入下列记号, 用到时不再进一步说明

$ z(t)=x(t)+p(t)x^{\alpha}(\tau(t)), \;\;\; \varphi_{+}(t)=\max\{\varphi(t), 0\}, \\ A(t)=\int^{+\infty}_{t}a^{-1/\beta}(s){\rm d}s,\;\;\; \Theta(t)=\int^{t}_{t_{0}}a^{-1/\beta}(s){\rm d}s. $

定理 2.1  设条件(1.2)成立, 若存在函数$\varphi\in C^{1}([t_{0}, +\infty), (0, +\infty))$使得当$\beta \ge \gamma$时有

$ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\varphi(s)Q(s)- \frac{\varphi(s)a^{\gamma/\beta}(\delta(s))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(s))^{\gamma}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\gamma+1}\right\}{\rm d}s=+\infty, \end{equation} $

(2.1)

当$\beta<\gamma$时有

$ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\varphi(s)Q(s)- \frac{(\beta/\gamma)^{\beta}\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k^{(\gamma-\beta)}_{1}(\delta'(s))^{\beta}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\beta+1}\right\}{\rm d}s=+\infty, \end{equation} $

(2.2)

其中函数$Q(t)=q(t)\left[1-\left(\alpha2^{1-\alpha}+\frac{(2^{1-\alpha})} {k_{1}}\right)p(\delta(t))\right]^{\gamma}$, $k_{i}>0\ (i=1, 2)$为常数, 则方程(1.1)是振动的.

证  反证法.设方程(1.1)存在一个非振动解$x(t)$, 若$x(t)$为最终正解, 则存在$t_{1} \ge t_{0}$, 使得当$t \ge t_{1}$时, 有$x(t)>0, x(\tau(t))>0, x(\delta(t))>0$, 于是$z(t)>0$且$z(t) \ge x(t)(t \ge t_{1})$.由方程(1.1), 得

$ \begin{equation} [a(t)|z'(t)|^{\beta-1}z'(t)]'\leqslant-Lq(t)x^{\gamma}(\delta(t))<0, \end{equation} $

(2.3)

注意到条件(1.2), 由(2.3)式容易推出$z'(t)>0\ (t \ge t_{1})$.

由$z(t)$的定义及引理2.1和引理2.2, 得

$ \begin{eqnarray} x(t)&=&z(t)-p(t)x^{\alpha}(\tau(t))\\ &=&z(t)-p(t)[1+x^{\alpha}(\tau(t))]+p(t) \\ & \ge&z(t)-2^{1-\alpha}p(t)[1+x(\tau(t))]^{\alpha}+p(t)\\ & \ge &z(t)-2^{1-\alpha}p(t)[1+\alpha x(\tau(t))]+p(t) \\ &=&z(t)-\alpha2^{1-\alpha}p(t)x(\tau(t))+(1-2^{1-\alpha})p(t)\\ & \ge &z(t)-\alpha2^{1-\alpha}p(t)z(\tau(t))+(1-2^{1-\alpha})p(t) \\ & \ge &[1-\alpha2^{1-\alpha}p(t)]z(t)-(2^{1-\alpha}-1)p(t). \end{eqnarray} $

(2.4)

作广义的Riccati变换如下

$ \begin{equation} w(t)=\varphi(t)\frac{a(t)|z'(t)|^{\beta-1}z'(t)}{|z(\delta(t))|^{\gamma-1}z(\delta(t))} =\varphi(t)\frac{a(t)(z'(t))^{\beta}}{z^{\gamma}(\delta(t))}, \;\;\;t \ge t_{1}, \end{equation} $

(2.5)

显然有$w(t)>0(t \ge t_{1})$.注意到(2.3), (2.4)式及$a(t)(z'(t))^{\beta}\leqslant a(\delta(t))(z'(\delta(t)))^{\beta}$, 由(2.5)式, 就有

$ \begin{eqnarray} w'(t)&=&\varphi'(t)\frac{a(t)(z'(t))^{\beta}}{z^{\gamma}(\delta(t))}+ \varphi(t)\frac{[a(t)(z'(t))^{\beta}]'z^{\gamma}(\delta(t))-a(t)(z'(t))^{\beta}\gamma z^{\gamma-1}(\delta(t))z'(\delta(t))\delta'(t)}{z^{2\gamma}(\delta(t))} \\ &\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)-\varphi(t)\frac{Lq(t)x^{\gamma}(\delta(t))}{z^{\gamma}(\delta(t))}- \gamma\frac{\varphi(t)a(t)(z'(t))^{\beta}\delta'(t)}{z^{\gamma+1}(\delta(t))}\frac{a^{1/\beta}(t)z'(t)}{a^{1/\beta}(\delta(t))} \\ &\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)-L\varphi(t)q(t)\left[\frac{(1-\alpha2^{1-\alpha}p(\delta(t)))z(\delta(t)) -(2^{1-\alpha}-1)p(\delta(t))}{z(\delta(t))}\right]^{\gamma} \\ &&-\gamma\frac{\varphi(t)a^{(\beta+1)/\beta}(t)(z'(t))^{\beta+1}\delta'(t)}{z^{\gamma+1}(\delta(t))a^{1/\beta}(\delta(t))}. \end{eqnarray} $

(2.6)

下面根据$\beta$和$\gamma$的取值分两种情形考虑: (ⅰ) $\beta \ge \gamma$, (ⅱ) $\beta<\gamma$.

情形(ⅰ)  $\beta \ge \gamma$.因为$z(t)>0, z'(t)>0(t \ge t_{1})$, 所以

$ z(\delta(t)) \ge z(\delta(t_{1}))=k_{1}, \;\;\;t \ge t_{1}, $

(2.7)

其中$k_{1}>0$为常数.又由$z'(t)>0$及(2.3)式知, 当$t \ge t_{1}$时, $a(t)(z'(t))^{\beta}\leqslant a(t_{1})(z'(t_{1}))^{\beta}=k_{2}$ (其中$k_{2}>0$为常数), 由此得$z'(t)\leqslant\frac{k^{1/\beta}_{2}}{a^{1/\beta}(t)}$, 即

$ (z'(t))^{(\gamma-\beta)/\gamma} \ge \frac{k^{(\gamma-\beta)/\beta\gamma}_{2}}{a^{(\gamma-\beta)/\beta\gamma}(t)}, \;\;\;t \ge t_{1}. $

(2.8)

于是, 由(2.6)式, 并注意到(2.5)、(2.7)及(2.8)式, 可得

$ \begin{eqnarray*} w'(t)&\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)- L\varphi(t)q(t)\left[(1-\alpha2^{1-\alpha}p(\delta(t)))-\frac{(2^{1-\alpha}-1)p(\delta(t))}{k_{1}}\right]^{\gamma} \\ &&-\frac{\gamma(z'(t))^{(\gamma-\beta)/\gamma}\delta'(t)a^{(\gamma-\beta)/\beta\gamma}(t)}{\varphi^{1/\gamma}(t)a^{1/\beta}(\delta(t))} w^{\frac{\gamma+1}{\gamma}}(t) \\ &\leqslant&\frac{\varphi'_{+}(t)}{\varphi(t)}w(t)-L\varphi(t)Q(t)- \frac{\gamma k^{(\gamma-\beta)/\beta\gamma}_{2}\delta'(t)}{\varphi^{1/\gamma}(t)a^{1/\beta}(\delta(t))}w^{\frac{\gamma+1}{\gamma}}(t). \end{eqnarray*} $

在引理2.3中的不等式$ab-\frac{1}{p}a^{p}\leqslant\frac{1}{q}b^{q}$中令$p=\frac{\gamma+1}{\gamma}, q=\gamma+1$, 且

$ a=\frac{k^{(\gamma-\beta)/\beta(\gamma+1)}_{2}[(\gamma+1)\delta'(t)]^{\gamma/(\gamma+1)}} {\varphi^{1/(\gamma+1)}(t)a^{\gamma/\beta(\gamma+1)}(\delta(t))}w(t), \;\;\; b=\frac{\varphi^{1/(\gamma+1)}(t)a^{\gamma/\beta(\gamma+1)}(\delta(t))} {k^{(\gamma-\beta)/\beta(\gamma+1)}_{2}[(\gamma+1)\delta'(t)]^{\gamma/(\gamma+1)}}\frac{\varphi'_{+}(t)}{\varphi(t)}, $

就有

$ \frac{\varphi'_{+}(t)}{\varphi(t)}w(t)- \frac{\gamma k^{(\gamma-\beta)/\beta\gamma}_{2}\delta'(t)}{\varphi^{1/\gamma}(t)a^{1/\beta}(\delta(t))}w^{\frac{\gamma+1}{\gamma}}(t) \leqslant\frac{\varphi(t)a^{\gamma/\beta}(\delta(t))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(t))^{\gamma}} \left(\frac{\varphi'_{+}(t)}{\varphi(t)}\right)^{\gamma+1}. $

于是

$ w'(t)\leqslant-L\varphi(t)Q(t)+\frac{\varphi(t)a^{\gamma/\beta}(\delta(t))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(t))^{\gamma}} \left(\frac{\varphi'_{+}(t)}{\varphi(t)}\right)^{\gamma+1}, $

所以

$ \int^{t}_{t_{1}}\left\{L\varphi(s)Q(s)- \frac{\varphi(s)a^{\gamma/\beta}(\delta(s))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(s))^{\gamma}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\gamma+1}\right\}{\rm d}s\leqslant-w(t)+w(t_{1})\leqslant w(t_{1}), $

这与(2.1)式矛盾.

情形(ⅱ)  $\beta<\gamma$.注意到(2.5)、(2.7)式及引理2.3, 由(2.6)式, 得

$ \begin{eqnarray*} w'(t)&\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)- L\varphi(t)q(t)\left[(1-\alpha2^{1-\alpha}p(\delta(t)))-\frac{(2^{1-\alpha}-1)p(\delta(t))}{k_{1}}\right]^{\gamma} \\ &&-\frac{\gamma z^{(\gamma-\beta)/\beta}(\delta(t))\delta'(t)}{\varphi^{1/\beta}(t)a^{1/\beta}(\delta(t))}w^{\frac{\beta+1}{\beta}}(t) \\ &\leqslant&-L\varphi(t)Q(t)+\frac{(\beta/\gamma)^{\beta}\varphi(t)a(\delta(t))}{(\beta+1)^{\beta+1}k^{(\gamma-\beta)}_{1}(\delta'(t))^{\beta}} \left(\frac{\varphi'_{+}(t)}{\varphi(t)}\right)^{\beta+1}, \end{eqnarray*} $

上式两边从$t_{1}$到$t(t \ge t_{1})$积分, 得

$ \int^{t}_{t_{1}}\left\{L\varphi(s)Q(s)- \frac{(\beta/\gamma)^{\beta}\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k^{(\gamma-\beta)}_{1}(\delta'(s))^{\beta}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\beta+1}\right\}{\rm d}s\leqslant-w(t)+w(t_{1})\leqslant w(t_{1}), $

这与(2.2)式矛盾.

另一方面, 如果$x(t)$是方程(1.1)的最终负解, 则令$y(t)=-x(t)$, 于是方程(1.1)就可转化为

$ \{a(t)|[y(t)+p(t)y^{\alpha}(\tau(t))]'|^{\beta-1}[y(t)+p(t)y^{\alpha}(\tau(t))]'\}' +q(t)f^{*}(|y(\delta(t))|^{\gamma-1}y(\delta(t)))=0, (1.1)^{*} $

其中函数$f^{*}(|y(\delta(t))|^{\gamma-1}y(\delta(t)))=-f(|y(\delta(t))|^{\gamma-1}(-y(\delta(t))))$.很明显, $f^{*}(u)=-f(-u)$具有与$f(u)$完全相同的性质, 并且$y(t)$是方程$(1.1)^{*}$的最终正解, 因此, 用与上面完全相同的方法可推出矛盾.定理证毕.

推论 2.1  设条件(1.2)成立, 如果

$ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{\Theta^{\beta}(\delta(s))q(s)-\frac{[\beta/(\beta+1)]^{\beta+1}\delta'(s)} {a^{1/\beta}(\delta(s))\Theta(\delta(s))}\right\}{\rm d}s=+\infty, \end{equation} $

(2.9)

则方程(1.3)是振动的.

证  在方程(1.1)中令$f(u)=u, p(t)\equiv0, \beta=\gamma$, 并在定理2.1中取$\varphi(t)=\Theta^{\beta}(\delta(t))$即得.

注 2.1  推论2.1就是Sun等在文献[6]得到的关于方程(1.3)振动的判别定理.若在定理2.1中取$\varphi(t)=\Theta(\delta(t))$, 还可得到以下一系列结果.

推论 2.2  设条件(1.2)成立, 如果$\beta=\gamma$且

$ \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\Theta(\delta(s))Q(s)-\frac{\delta'(s)} {(\beta+1)^{\beta+1}a^{1/\beta}(\delta(s))\Theta^{\beta}(\delta(s))}\right\}{\rm d}s=+\infty, $

(2.10)

其中函数$Q(t)$同定理2.1, 则方程(1.1)是振动的.

推论 2.3  设条件(1.2)成立, 且$\beta=\gamma\leqslant1$.如果当$\beta<1$时

$ \liminf\limits_{t\rightarrow+\infty}\frac{1}{\Theta^{1-\beta}(\delta(t))}\int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s >\frac{1}{L(\beta+1)^{\beta+1}(1-\beta)}, $

(2.11)

当$\beta=1$时

$ \liminf\limits_{t\rightarrow+\infty}\frac{1}{\ln\Theta(\delta(t))}\int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s >\frac{1}{4L}, $

(2.12)

其中函数$Q(t)$同定理2.1, 则方程(1.1)是振动的.

证  当$\beta<1$时, 由(2.11)式知, 存在常数$\varepsilon>0$, 对充分大的实数$t$, 有

$ \frac{1}{\Theta^{1-\beta}(\delta(t))}\int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s \ge \frac{1}{L(\beta+1)^{\beta+1}(1-\beta)}+\varepsilon, $

这就意味着

$ \int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s-\frac{\Theta^{1-\beta}(\delta(t))}{L(\beta+1)^{\beta+1}(1-\beta)} \ge \varepsilon\Theta^{1-\beta}(\delta(t)), $

于是, 由上式并注意到条件(1.2), 则有

$ \begin{eqnarray*} &&\int^{t}_{t_{0}}\left\{\Theta(\delta(s))Q(s)-\frac{\delta'(s)}{L(\beta+1)^{\beta+1}a^{1/\beta}(\delta(s))\Theta^{\beta}(\delta(s))}\right\}{\rm d}s \\ &=&\int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s-\frac{1}{L(\beta+1)^{\beta+1}}\left[\frac{\Theta^{1-\beta}(\delta(s))}{1-\beta}\right]^{t}_{t_{0}} \\ &=&\int^{t}_{t_{0}}\Theta(\delta(s))Q(s){\rm d}s-\frac{\Theta^{1-\beta}(\delta(t))}{L(\beta+1)^{\beta+1}(1-\beta)} +\frac{\Theta^{1-\beta}(\delta(t_{0}))}{L(\beta+1)^{\beta+1}(1-\beta)} \\ & \ge &\varepsilon\Theta^{1-\beta}(\delta(t))+\frac{\Theta^{1-\beta}(\delta(t_{0}))}{L(\beta+1)^{\beta+1}(1-\beta)}\rightarrow+\infty\;\;\;(t\rightarrow+\infty), \end{eqnarray*} $

所以, 由推论2.2知方程(1.1)是振动的.

当$\beta=1$时, 由于$\frac{{\rm d}}{{\rm d}s}[\ln\Theta(\delta(s))]=\frac{\delta'(s)}{\Theta(\delta(s))a(\delta(s))}, $余下部分类似可证.证毕.

推论 2.4  设条件(1.2)成立, 且$\beta=\gamma\leqslant1$.如果

$ \liminf\limits_{t\rightarrow+\infty}\frac{Q(t)\Theta^{\beta+1}(\delta(t))a^{1/\beta}(\delta(t))}{\delta'(t)}>\frac{1}{L(\beta+1)^{\beta+1}}, $

(2.13)

其中函数$Q(t)$同定理2.1, 则方程(1.1)是振动的.

证  若$\beta<1$, 则由(2.13)式知, 存在常数$\varepsilon>0$, 对充分大的实数$t$, 有

$ \frac{Q(t)\Theta^{\beta+1}(\delta(t))a^{1/\beta}(\delta(t))}{\delta'(t)} \ge \frac{1}{L(\beta+1)^{\beta+1}}+\varepsilon, $

上式两边同乘以$\frac{\delta'(t)}{\Theta^{\beta}(\delta(t))a^{1/\beta}(\delta(t))}$, 得

$ \begin{eqnarray*} &&L\Theta(\delta(t))Q(t)-\frac{\delta'(t)}{(\beta+1)^{\beta+1}a^{1/\beta}(\delta(t))\Theta^{\beta}(\delta(t))} \\ & \ge &\frac{L\varepsilon\delta'(t)}{\Theta^{\beta}(\delta(t))a^{1/\beta}(\delta(t))} =\frac{L\varepsilon}{1-\beta}\frac{{\rm d}}{{\rm d}t}\left[\Theta^{1-\beta}(\delta(t))\right], \end{eqnarray*} $

由上式容易验证(2.10)成立, 于是由推论2.2知, 方程(1.1)是振动的.

当$\beta=1$时, 由于$\frac{{\rm d}}{{\rm d}t}[\ln\Theta(\delta(t))]=\frac{\delta'(t)}{\Theta(\delta(t))a(\delta(t))}$, 余下部分类似可证.证毕.

定理 2.2  设条件(1.4)成立, $1-\alpha2^{1-\alpha}p(t)\frac{A(\tau(t))}{A(t)}>0$, 且存在函数$\varphi\in C^{1}([t_{0}, +\infty), $ $(0, +\infty))$使得当$\beta \ge \gamma$时(2.1)式成立, 当$\beta<\gamma$时(2.2)式成立.如果

$ \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{LA^{\beta}(s)\Psi(s)\pi(s)-\frac{[\beta/(\beta+1)]^{\beta+1}} {a^{1/\beta}(s)A(s)}\right\}{\rm d}s=+\infty, $

(2.14)

其中$\Psi(t)=q(t)\left[1-\left(\alpha2^{1-\alpha}\frac{A(\tau(\delta(t)))}{A(\delta(t))}+\frac{(2^{1-\alpha}-1)}{m_{1}A(t)}\right)p(\delta(t))\right]^{\gamma}$,

$ \pi(t)=\left\{\begin{array}{ll} m^{\gamma-\beta}_{1}A^{\gamma-\beta}(t), &\beta<\gamma, \\ 1, &\beta=\gamma, \\ m^{\gamma-\beta}_{2}, &\beta>\gamma, \end{array} \right. $

而$m_{1}>0$及$m_{2}>0$为某常数, 则方程(1.1)是振动的.

证  反证法, 设方程(1.1)存在一个非振动解$x(t)$, 不妨设$x(t)$为最终正解(当$x(t)$为最终负解时类似可证), 则$\exists t_{1} \ge t_{0}$, 使得当$t \ge t_{1}$时, 有$x(t)>0, x(\tau(t))>0, x(\delta(t))>0$, 从而$z(t)>0(t \ge t_{1})$.由定理2.1的证明知$a(t)|z'(t)|^{\beta-1}z'(t)$是严格单调减少的且最终定号, 从而$z'(t)$最终为正或最终为负.因此, 只需考虑下列两种情形

(a) 当$t \ge t_{1}$时$z'(t)>0$; (b)当$t \ge t_{1}$时$z'(t)<0$.

情形(a) $z'(t)>0(t \ge t_{1})$.同定理2.1的证明, 可得这种情形下方程(1.1)是振动的.

情形(b) $z'(t)<0(t \ge t_{1})$.定义函数$v(t)$为

$ v(t)=\frac{a(t)|z'(t)|^{\beta-1}z'(t)}{z^{\beta}(t)}=\frac{a(t)(-z'(t))^{\beta-1}z'(t)}{z^{\beta}(t)}, t \ge t_{1}, $

(2.15)

则有$v(t)>0(t \ge t_{1})$.由(2.3)式知, 对$s \ge t \ge t_{1}$, 有$a(s)|z'(s)|^{\beta-1}z'(s)\leqslant a(t)|z'(t)|^{\beta-1}z'(t)$, 即$a(s)(-z'(s))^{\beta} \ge a(t)(-z'(t))^{\beta}$, 亦即$z'(s)\leqslant\frac{a^{1/\beta}(t)z'(t)}{a^{1/\beta}(s)}$, 于是进一步就有

$ z(u)-z(t)\leqslant a^{1/\beta}(t)z'(t)\int^{u}_{t}a^{-1/\beta}(s){\rm d}s, $

令$u\rightarrow+\infty$, 得$z(t)+a^{1/\beta}(t)z'(t)A(t) \ge 0$, 即$-1\leqslant\frac{a^{1/\beta}(t)z'(t)}{z(t)}A(t)<0$, 由此进一步可得

$ -1\leqslant v(t)A^{\beta}(t)<0. $

(2.16)

另一方面, 由

$ \frac{{\rm d}}{{\rm d}t}\frac{z(t)}{A(t)}=\frac{z'(t)A(t)-z(t)[-a^{-1/\beta}(t)]}{A^{2}(t)}=\frac{a^{1/\beta}(t)z'(t)A(t)+z(t)}{a^{1/\beta}(t)A^{2}(t)} \ge 0, $

得$\frac{z(\tau(t))}{A(\tau(t))}\leqslant\frac{z(t)}{A(t)}$, 即

$ z(\tau(t))\leqslant\frac{A(\tau(t))}{A(t)}z(t). $

(2.17)

于是, 由$z(t)$的定义及引理2.1, 引理2.2及(2.17)式, 得

$ \begin{eqnarray} x(t)&=&z(t)-p(t)x^{\alpha}(\tau(t))\\ &=&z(t)-p(t)[1+x^{\alpha}(\tau(t))]+p(t) \\ & \ge&z(t)-2^{1-\alpha}p(t)[1+x(\tau(t))]^{\alpha}+p(t)\\ & \ge&z(t)-2^{1-\alpha}p(t)[1+\alpha x(\tau(t))]+p(t) \\ &=&z(t)-\alpha2^{1-\alpha}p(t)x(\tau(t))+(1-2^{1-\alpha})p(t)\\ & \ge &z(t)-\alpha2^{1-\alpha}p(t)z(\tau(t))+(1-2^{1-\alpha})p(t) \\ & \ge &\left[1-\alpha2^{1-\alpha}p(t)\frac{A(\tau(t))}{A(t)}\right]z(t)-(2^{1-\alpha}-1)p(t). \end{eqnarray} $

(2.18)

由(2.15)式, 并依次注意到(2.3)式和(2.18)式及$z(\delta(t)) \ge z(t)$, 得

$ \begin{eqnarray} v'(t)&=&\frac{[a(t)|z'(t)|^{\beta-1}z'(t)]'}{z^{\beta}(t)} -\frac{a(t)|z'(t)|^{\beta-1}z'(t)\beta z^{\beta-1}(t)z'(t)}{z^{2\beta}(t)} \\ &\leqslant&-Lq(t)\frac{[(1-\alpha2^{1-\alpha}p(\delta(t))\frac{A(\tau(\delta(t)))}{A(\delta(t))})-\frac{(2^{1-\alpha}-1)}{z(t)}p(\delta(t))]^{\gamma}} {z^{\beta-\gamma}(t)} -\frac{\beta(-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}. \end{eqnarray} $

(2.19)

由(2.3)式知, 当$s \ge t_{1}$时, $a(s)|z'(s)|^{\beta-1}z'(s)\leqslant a(t_{1})|z'(t_{1})|^{\beta-1}z'(t_{1})=-m^{\beta}_{1}$ (常数$m_{1}>0$), 即$z'(s)\leqslant-m_{1}a^{-1/\beta}(s)$, 两边积分, 得$z(u)-z(t)\leqslant-m_{1}\int^{u}_{t}a^{-1/\beta}(s){\rm d}s$, 令$u\rightarrow+\infty$, 则有

$ z(t) \ge m_{1}\int^{+\infty}_{t}a^{-1/\beta}(s){\rm d}s=m_{1}A(t). $

当$\beta>\gamma$时, 由$z(t)>0, z'(t)<0(t \ge t_{1})$知, $z(t)\leqslant z(t_{1})=m_{2}$, 即$\frac{1}{z^{\beta-\gamma}(t)} \ge \frac{1}{m^{\beta-\gamma}_{2}}$.

当$\beta<\gamma$时, $\frac{1}{z^{\beta-\gamma}(t)} \ge m^{\gamma-\beta}_{1}A^{\gamma-\beta}(t)$.

当$\beta=\gamma$时, $\frac{1}{z^{\beta-\gamma}(t)}=1$.

于是, 综合上述3种情形, 注意到函数$\pi(t)$及$\Psi(t)$的定义, 由(2.19)式, 可得

$ v'(t)\leqslant-L\Psi(t)\pi(t)-\frac{\beta(-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}. $

(2.20)

将上式中的$t$改成$s$, 再两边同乘以$A^{\beta}(s)$, 并对$s$从$t_{1}$到$t(t \ge t_{1})$积分, 并利用分部积分公式及引理2.3, 可得

$ \begin{eqnarray*} &&\int^{t}_{t_{1}}LA^{\beta}(s)\Psi(s)\pi(s){\rm d}s\\ &\leqslant&-\int^{t}_{t_{1}}A^{\beta}(s)v'(s){\rm d}s -\int^{t}_{t_{1}}A^{\beta}(s)\frac{\beta(-v(s))^{(\beta+1)/\beta}}{a^{1/\beta}(s)}{\rm d}s \\ &=&-\left[A^{\beta}(s)v(s)\right]^{t}_{t_{1}}+\int^{t}_{t_{1}}v(s)\beta A^{\beta-1}(s)A'(s){\rm d}s -\int^{t}_{t_{1}}\frac{\beta A^{\beta}(s)}{a^{1/\beta}(s)}(-v(s))^{(\beta+1)/\beta}{\rm d}s \\ &=&A^{\beta}(t_{1})v(t_{1})-A^{\beta}(t)v(t)+\int^{t}_{t_{1}}\left[\frac{\beta A^{\beta-1}(s)}{a^{1/\beta}(s)}(-v(s)) -\frac{\beta A^{\beta}(s)}{a^{1/\beta}(s)}(-v(s))^{(\beta+1)/\beta}\right]{\rm d}s \\ &\leqslant& A^{\beta}(t_{1})v(t_{1})-A^{\beta}(t)v(t)+ \int^{t}_{t_{1}}\frac{[\beta/(\beta+1)]^{\beta+1}}{a^{1/\beta}(s)A(s)}{\rm d}s, \end{eqnarray*} $

注意到(2.16)式, 于是就有

$ \int^{t}_{t_{1}}\left\{LA^{\beta}(s)\Psi(s)\pi(s)-\frac{[\beta/(\beta+1)]^{\beta+1}}{a^{1/\beta}(s)A(s)}\right\}{\rm d}s\leqslant A^{\beta}(t_{1})v(t_{1})+1, $

这与(2.7)式矛盾.定理证毕.

定理 2.3  设条件(1.4)成立, $1-\alpha2^{1-\alpha}p(t)\frac{A(\tau(t))}{A(t)}>0$, 且存在函数$\varphi\in C^{1}([t_{0}, +\infty), $ $(0, +\infty))$使得当$\beta \ge \gamma$时(2.1)式成立, 当$\beta<\gamma$时(2.2)式成立.如果

$ \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\Psi(s)\pi(s)A^{\beta+1}(s){\rm d}s=+\infty, $

(2.21)

其中函数$\Psi(t)$及$\pi(t)$的定义如定理2.2, 而$m_{1}>0$为某常数, 则方程(1.1)是振动的.

证  同定理2.2的证明, 可得(2.20)式, 将(2.20)式中的$t$改成$s$, 两边同乘以$A^{\beta+1}(s)$, 再对$s$从$t_{1}$到$t(t \ge t_{1})$积分, 并利用分部积分公式, 可得

$ \begin{eqnarray} &&\int^{t}_{t_{1}}L\Psi(s)\pi(s)A^{\beta+1}(s){\rm d}s\\ &\leqslant&-\int^{t}_{t_{1}}A^{\beta+1}(s)v'(s){\rm d}s -\int^{t}_{t_{1}}\frac{\beta A^{\beta+1}(s)}{a^{1/\beta}(s)}(-v(s))^{(\beta+1)/\beta}{\rm d}s \\ &=&-\left[A^{\beta+1}(s)v(s)\right]^{t}_{t_{1}}+\int^{t}_{t_{1}}v(s)(\beta+1)A^{\beta}(s)A'(s){\rm d}s \\ && -\int^{t}_{t_{1}}\frac{\beta A^{\beta+1}(s)}{a^{1/\beta}(s)}(-v(s))^{(\beta+1)/\beta}{\rm d}s \\ &=&A^{\beta+1}(t_{1})v(t_{1})+A^{\beta+1}(t)(-v(t))+(\beta+1)\int^{t}_{t_{1}}\frac{A^{\beta}(s)(-v(s))}{a^{1/\beta}(s)}{\rm d}s \\ && -\beta\int^{t}_{t_{1}}\frac{A^{\beta+1}(s)(-v(s))^{\frac{\beta+1}{\beta}}}{a^{1/\beta}(s)}{\rm d}s. \end{eqnarray} $

(2.22)

利用(2.16)式可得

$ |A^{\beta+1}(t)(-v(t))|=|A^{\beta}(t)v(t)|A(t)\leqslant A(t)<+\infty, $

$ \begin{eqnarray*} \left|\int^{t}_{t_{1}}\frac{A^{\beta}(s)(-v(s))}{a^{1/\beta}(s)}{\rm d}s\right| &\leqslant&\int^{t}_{t_{1}}\frac{|A^{\beta}(s)v(s)|}{a^{1/\beta}(s)}{\rm d}s \leqslant\int^{t}_{t_{1}}\frac{1}{a^{1/\beta}(s)}{\rm d}s\\ &\leqslant&\int^{+\infty}_{t_{1}}\frac{1}{a^{1/\beta}(s)}{\rm d}s<+\infty, \end{eqnarray*} $

$ \begin{eqnarray*} \left|\int^{t}_{t_{1}}\frac{A^{\beta+1}(s)(-v(s))^{\frac{\beta+1}{\beta}}}{a^{1/\beta}(s)}{\rm d}s\right| &\leqslant&\int^{t}_{t_{1}}\frac{|A^{\beta}(s)v(s)|^{\frac{\beta+1}{\beta}}}{a^{1/\beta}(s)}{\rm d}s \leqslant\int^{t}_{t_{1}}\frac{1}{a^{1/\beta}(s)}{\rm d}s\\ &\leqslant&\int^{+\infty}_{t_{1}}\frac{1}{a^{1/\beta}(s)}{\rm d}s<+\infty, \end{eqnarray*} $

这样一来, 由(2.22)式就有

$ \limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{1}}L\Psi(s)\pi(s)A^{\beta+1}(s){\rm d}s=+\infty, $

这与(2.21)式矛盾, 定理证毕.

注 2.2  从定理2.1-2.3的结果可以看出, 当$\beta<\gamma$和$\beta>\gamma$时方程(1.1)的振动准则是不同的, 但当$\beta=\gamma$时(2.1)式和(2.2)式是一样的.若$f(u)=u$且$\alpha=1$ (即中立项是线性的), 则由定理2.1可得文献[7]中的定理2.2, 但我们的结果没有限制条件"$a'(t) \ge 0$", 而且$\beta \ge \gamma$和$\beta\leqslant\gamma$两种情形均有方程(1.1)的振动结果; 若$\alpha=1$且$\gamma=1$, 则由定理2.2即得文献[5]中的定理2.1, 但我们没有限制条件"对$t \ge t_{0}$有$p'(t) \ge 0, \delta(t)\leqslant t-\tau\ (\tau>0$为常数)".因此, 本文所得定理的条件是较宽松的.此外, 在条件(1.4)下且$\alpha=1$ (即中立项是线性的)的情形, 文献[7]和^[8]只能得出方程所有解$x(t)$或者振动或者$\lim\limits_{t\rightarrow+\infty}x(t)=0$, 而不能确定方程的振动性.

例 3.1  考虑如下二阶微分方程

$ (x(t)+\frac{1}{5}x(\frac{t}{5}))''+\frac{q_{0}}{t^{2}}x(t)=0, t \ge 1, $

(3.1)

其中常数$q_{0}>0$.令$a(t)=1, p(t)=1/5, q(t)=q_{0}/t^{2}, \tau(t)=t/5, \delta(t)=t, f(u)=u, \alpha=1, $ $\beta=1, \gamma=1$, 容易验证条件(H$_{1}$)-(H$_{3}$)及(1.2)均满足.现取$\varphi(t)=t$, 注意到$L=1$, 则当$q_{0}>5/16=0.3125$时

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\varphi(s)Q(s)- \frac{\varphi(s)a^{\gamma/\beta}(\delta(s))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(s))^{\gamma}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\gamma+1}\right\}{\rm d}s \\ &=&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\left\{s\frac{q_{0}}{s^{2}}\left[1-\frac{1}{5}\right] -\frac{s}{2^{2}}\left(\frac{1}{s}\right)^{2}\right\}{\rm d}s =\frac{4}{5}\left(q_{0}-\frac{5}{16}\right)\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\frac{1}{s}{\rm d}s=+\infty, \end{eqnarray*} $

所以条件(2.1)成立, 于是由定理2.1知, 当$q_{0}>0.3125$时方程(3.1)是振动的.

注 3.1  若用文献[10]中的定理3.4来判定方程(3.1)的振动性, 则因为当$q_{0}>2.5$时, 有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{\frac{\varphi(s)q(s)}{2^{\gamma-1}}- \frac{(1+p^{\gamma}_{0}/\tau_{0})(\varphi'_{+}(s))^{\gamma+1}a(\delta(s))}{(\gamma+1)^{\gamma+1}[\tau_{0}\varphi(s)]^{\gamma}}\right\}{\rm d}s \\ &=&(q_{0}-2.5)\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\frac{1}{s}{\rm d}s=+\infty, \end{eqnarray*} $

所以由文献[10]中的定理3.4知, 当$q_{0}>2.5$时方程(3.1)是振动的.此外, 我们也可以用文献[11]中的定理3.1来判定方程(3.1)的振动性:对常数$\varepsilon\in(0, 1)$, 由于当$q_{0}>\frac{6+\varepsilon}{20(1-\varepsilon)}$时, 有

$ \begin{eqnarray*} &&\int^{+\infty}_{t_{0}}\left\{\frac{(1-\varepsilon)^{\alpha}\varphi(t)q(t)}{[1+p_{0}(1+\varepsilon)]^{\alpha}} -\frac{(\varphi_{+}'(t))^{\alpha+1}a(\delta(t))}{(\alpha+1)^{\alpha+1}[\varphi(t)\delta'(t)]^{\alpha}}\right\}{\rm d}t \\ &=&\frac{5(1-\varepsilon)}{6+\varepsilon}\left[q_{0}-\frac{6+\varepsilon}{20(1-\varepsilon)}\right]\int^{+\infty}_{1}\frac{1}{t}{\rm d}t=+\infty, \end{eqnarray*} $

所以由文献[11]中的定理3.1知, 当$q_{0}>\frac{6+\varepsilon}{20(1-\varepsilon)}$时方程(3.1)是振动的.本例说明, 本文所获得的振动准则是较"精确的", 其特殊情形即当$\alpha=1$时的结果改进了文献[10]和[11]中的有关结论.

例 3.2  考虑具拟线性中立项的二阶微分方程

$ \Big\{t\Big[(x(t)+\frac{1}{5}\sqrt[3]{x(t/2)})'\Big]^{5/3}\Big\}'+\frac{q_{0}}{t}x^{7/5}(t/3)=0, \;\;\; t \ge 1, $

(3.2)

其中常数$q_{0}>0$.令$\alpha=1/3, \beta=5/3, \gamma=7/5, a(t)=t, p(t)=1/5, q(t)=q_{0}/t, f(u)=u, $ $\tau(t)=t/2, \delta(t)=t/3$, 则条件(H$_{1}$)-(H$_{3}$)及(1.2)式显然均满足.取$\varphi(t)=1$, 注意到$\beta>\gamma$且$L=1$, 用定理2.1的第1种情形, 因为

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\varphi(s)Q(s)- \frac{\varphi(s)a^{\gamma/\beta}(\delta(s))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(s))^{\gamma}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\gamma+1}\right\}{\rm d}s \\ &=&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\left\{\frac{q_{0}}{s}\left[1-\left(\frac{1}{3}2^{2/3} +\frac{(2^{2/3}-1)}{k_{1}}\right)\frac{1}{5}\right]^{7/5}-0\right\}{\rm d}s=+\infty, \end{eqnarray*} $

所以由定理2.1知, 方程(3.2)是振动的.

例 3.3  考虑具非线性中立项的二阶时滞微分方程

$ \Big[t^{2}(x(t)+\frac{1}{t^{2}}x^{3/5}(t/2))'\Big]'+\lambda f(x(t/2))=0, \;\;\;t \ge 1, $

(3.3)

这里常数$\lambda>0$.令$\alpha=3/5, \beta=1, \gamma=1, a(t)=t^{2}, p(t)=1/t^{2}, q(t)=\lambda, \tau(t)=t/2, \delta(t)=t/2$, 并令$f(u)=u[\ln(1+u^{4})]$, 则条件(H$_{1}$)-(H$_{3}$)及(1.4)式显然均满足.取$\varphi(t)=1$, 则由定理2.2(注意到此时$L=1, \beta=\gamma=1$, 且$A(t)=\int^{+\infty}_{t}a^{-1/\beta}(s){\rm d}s=1/t$), 有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{L\varphi(s)Q(s)- \frac{\varphi(s)a^{\gamma/\beta}(\delta(s))}{(\gamma+1)^{\gamma+1}k^{(\gamma-\beta)/\beta}_{2}(\delta'(s))^{\gamma}} \left(\frac{\varphi'_{+}(s)}{\varphi(s)}\right)^{\gamma+1}\right\}{\rm d}s \\ &=&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\left\{\lambda\left[1-\left(\frac{3}{5}2^{2/5} +\frac{(2^{2/5}-1)}{k_{1}}\right)\frac{4}{s^{2}}\right]-0\right\}{\rm d}s=+\infty, \end{eqnarray*} $

且当$\lambda>1/4$时

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{t_{0}}\left\{LA^{\beta}(s)\Psi(s)\pi(s)-\frac{[\beta/(\beta+1)]^{\beta+1}} {a^{1/\beta}(s)A(s)}\right\}{\rm d}s \\ &=&\limsup\limits_{t\rightarrow+\infty}\int^{t}_{1}\left\{\frac{\lambda}{s}\left[1- \left(\frac{3}{5}2^{2/5}\frac{(s/4)^{2}}{(s/2)^{2}}+\frac{(2^{2/5}-1)s}{k}\right)\frac{4}{s^{2}}\right]-\frac{1}{4s}\right\}=+\infty, \end{eqnarray*} $

所以, 定理2.2的条件全部满足, 因此当$\lambda>1/4$时方程(3.3)是振动的.

注 3.2  由于方程(3.2)及(3.3)是具有非线性中立项的微分方程, 并且方程(3.2)中$\beta\neq\gamma$, 所以文献[1-3, 5-16]中的定理均不能用于方程(3.2)及(3.3).