本文研究具有次线性中立项的二阶广义Emden-Fowler型时滞微分方程

(1.1) $ \begin{equation} \left(a(t)|z'(t)|^{\beta-1}z'(t)\right)'+f(t, |x(\sigma(t))|^{\gamma-1}x(\sigma(t))) = 0, \; \; t\geq t_0\geq 0 \end{equation} $

的振动性, 其中$ z(t) = x(t)+g(t, x^\alpha(\tau(t)) $.

本文总假设以下条件成立:

$ {\rm (H_1)}\; \beta>0, \gamma>0 $为常数, $ 0<\alpha\leq1 $为两正奇数之比的常数.

$ {\rm (H_2)}\; $函数$ a\in C^1([t_0, \infty), (0, \infty)), g\in C([t_0, \infty)\times {{\Bbb R}} , {{\Bbb R}} ) $; 当$ v\neq0 $时, 有$ 0\leq g(t, v)/v\leq p_\alpha(t), p_\alpha\in C([t_0, \infty), [0, \infty)), 0\leq p_1(t)<1, \lim\limits_{t\to \infty} p_\alpha(t) = 0(0<\alpha<1); f\in C({{\Bbb R}} , {{\Bbb R}} ), $当$ u\neq0, t\geq t_0 $时, $ f(t, u)/u\geq q(t)\geq0 $且连续函数$ q(t) $不恒等于零.

$ {\rm (H_3)}\; \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)\geq0 $且$ \lim\limits_{t\rightarrow \infty}\tau(t) = \lim\limits_{t\rightarrow \infty}\sigma(t) = \infty. $

通常, 我们称$ x\in C([T_x, \infty ), {{\Bbb R}} ) $是方程(1.1)的解, 其中$ T_x = \min \{ \tau(T), \sigma(T)\}, $$ T\geq t_0, $是指$ a|z'|^{\beta = 1}z'\in C^1([T_x, \infty ), \Bbb R) $且在$ [T_x, \infty) $上满足方程(1.1). 本文仅考虑方程(1.1)的非平凡解$ x(t), $即方程(1.1)在$ [ T_x, \infty) $上的解$ x(t) $满足$ \sup \{ |x(t)|:t\geq T\}>0, $对一切$ T\geq T_x $成立. 方程(1.1)的解称为振动的, 如果它有任意大的零点, 否则, 称它为非振动的. 若方程(1.1)的一切解均振动, 则称方程(1.1)振动.

由于来源于数学物理的Emden-Fowler型微分方程的振动性在理论物理、物理化学、辐传动力学和自动工程技术领域广泛的应用价值而引起的研究兴趣经久不衰^[1-4], 近年来的研究成果仍大量涌现, 参见文献[3-20]及其参考文献. 易知方程(1.1)推广于经典的Emden-Fowler方程

(1.2) $ \begin{equation} x''(t)+\frac{a}{t}x'(t)+bt^{m-1}x^n(t) = 0, \; \; t\geq t_0>0, \end{equation} $

其中$ n\neq0, n\neq1, a, b, m $为常数, 以及著名的Euler方程

(1.3) $ \begin{equation} (t^2x'(t))'+q_0x(t) = 0, \; \; t\geq 1, \end{equation} $

这里, $ q_0>0 $为常数. 方程(1.1)还有以下多个熟知的特例

(1.4) $ \begin{equation} (a(t)((x(t)+p(t)x (\tau(t)))')^\beta)'+q(t)x^\gamma (\sigma(t)) = 0, \; \; t\geq t_0>0, \end{equation} $

其中$ \beta>0, \gamma>0 $均为两正奇数之比的常数, $ \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)>0, 0\leq p(t)<1, $$ q(t)\geq 0. $

(1.5) $ \begin{equation} (a(t)((x(t)+p(t)x ^\alpha(\tau(t)))')^\beta)'+q(t)x^\gamma (\sigma(t)) = 0, \end{equation} $

其中$ 0<\alpha\leq1, \beta>0, \gamma>0 $均为两正奇数之比的常数, $ \tau(t)\leq t, \sigma(t)\leq t, $$ \sigma'(t)>0. $

(1.6) $ \begin{equation} \left(a(t)|z'(t)|^{\beta-1}z'(t)\right)'+q(t)f(|x(\sigma(t))|^{\gamma-1}x(\sigma(t))) = 0, \end{equation} $

这里$ z(t) = x(t)+p(t)x^\alpha(\tau(t)) $; $ 0<\alpha\leq1 $为两正奇数之比的常数; $ \beta>0, \gamma>0, a(t)>0, $$ 0\leq p(t)<1, q(t)>0, f(u)/u\geq L>0, \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)>0 $.

2014年, Agarwal等在文献[3]中从方程(1.2)引出了方程(1.4)并在非正则条件

(1.7) $ \begin{equation} A(t) = \int_{t}^\infty a^{-1/\beta}(s){\rm d}s<\infty \end{equation} $

之下, 区分$ \gamma\geq\beta, $$ \gamma<\beta $和$ \gamma<\beta = 1, $分别建立了方程(1.4)的多个振动定理, 并应用于方程(1.2), 得到了当$ a = 2, m = -1, n = 1/3 $时方程(1.2)振动的结论.

我们在文献[4]中通过对显含阻尼项和非线性中立项的Emden-Fowler方程在正则和非正则条件下建立的振动定理, 对方程(1.2)给出了更系统的结果即其推论2.3如下.

定理1.1^{[4, 推论2.3]}  设$ 0<n\neq1 $为两正奇数之比的常数, $ a\geq0, b>0, m\geq-1 $为常数且$ a+m\geq0. $则当$ a\leq 1 $时方程$ (1.2) $振动; 当$ a>1 $时方程$ (1.2) $的每个解$ x(t) $振动或$ \lim\limits_{t\to \infty}x(t) = 0.\; \; $此外, 当$ b = a-2, \; a>2, n = m+2, m = -\frac{\alpha}{\beta}, \beta>\alpha, $或$ m = \frac{\alpha}{\beta}, $其中$ \alpha, \beta $为正奇数时, 方程$ (1.2) $的每一解振动或渐近于零. 事实上, 这时$ x(t) = t^{-1} $是其渐近于零的非振动解. 实际上, 还可以找到方程$ (1.2) $若干个$ x(t) = t^{-\lambda}, \lambda>0 $型的渐近于零的非振动解, 只需其中常数满足关系"$ \frac{m+1}{n-1} = \lambda, \; \; b = \lambda(a-\lambda-1) $"即可.

2017年Bohner等^[8], 2019年Prabaharan和Dharuman^[22]分别独立地对于$ \beta = \gamma $时建立了类似于方程(1.4)只有一个式子成立时的振动定理如下.

定理1.2^{[8, 定理2.1]}  设$ (1.7) $式成立且$ \beta = \gamma, \sigma'(t)>0, 0\leq p(t)\leq \frac{A(t)}{A(\tau(t)} $. 如果有

(1.8) $ \begin{equation} \int_{}^{\infty}\left(\frac{1}{a(t)}\int^t q(s)\left[1-p(\sigma(t))\frac{A(\tau(\sigma(s)))}{A(\sigma(s))}\right]^\beta A^\beta(\sigma(s)){\rm d}s\right)^{\frac{1}{\beta}}{\rm d}t = \infty, \end{equation} $

则方程$ (1.4) $振动.

定理1.3^{[8, 定理2.2]}  设$ (1.7) $式成立且$ \beta = \gamma, \sigma'(t)>0, 0\leq p(t)\leq \frac{A(t)}{A(\tau(t)} $. 如果对于所有足够大的$ t_1\geq t_0, $有

(1.9) $ \begin{equation} \limsup\limits_{t\to\infty}A^\beta(t)\int_{t_1}^t q(s)\left(1-p(\sigma(t))\frac{A(\tau(\sigma(s)))}{A(\sigma(s))}\right)^\beta {\rm d}s>1, \end{equation} $

则方程$ (1.4) $振动.

但易知, 定理1.2不适用于二阶Euler方程(1.3), 因为这里$ a(t) = t^2, p(t)\equiv0, q(t)\equiv q_0, $$ \beta = \gamma = 1, $$ A(t) = \frac{1}{t}, $

$ \label{eq:a1} \int_{}^{\infty}\left(\frac{1}{a(t)}\int^t q(s)\left[1-p(\sigma(t))\frac{A(\tau(\sigma(s)))}{A(\sigma(s))}\right]^\beta A^\beta(\sigma(s)){\rm d}s\right)^{\frac{1}{\beta}}{\rm d}t = q_0\int^\infty \frac{\ln t}{t^2}{\rm d}t<\infty, $

即(1.8)式不成立.同样地, 定理1.3虽然可用于方程(1.3), 但结果是只有当$ q_0>1 $时方程(1.3)振动, 这显然逊色于"当$ q_0>\frac{1}{4} $时方程(1.3)振动"的著名结果.

显然, 方程(1.4)所含的中立项$ z(t) $是$ x(t) $的线性函数. 关于这一类型方程的振动性研究仍然非常活跃, 并不断有新的成果出现^[3-13]. 人们自然更感兴趣的是中立项$ z(t) $是$ x(t) $的非线性函数甚至将方程(1.4) 的首项拓展为$ z(t) $的更一般非线性函数的情况, 文献[12-13]对此给出了有效尝试. 例如文献[13]对于二阶非线性中立型广义Emden-Fowler时滞微分方程

$ (g(t, z(t), z'(t))'+f(t, |x(\sigma(t))|^{\beta-1}x(\sigma(t))) = 0, \; t\geq t_0, $

其中, $ z(t) = x(t)+p(t)x(\tau(t)), $在满足$ L_1a(t)|v|^\alpha\leq g(t, u, v)\leq L_2a(t)|v|^\alpha, $$ \alpha>0, \beta>0, $$ f(t, u)/u\geq q(t)\geq0 $时, 构造了等价的双Riccati变换, 分别在正则和非正则条件下获得了4个振动定理.

对于方程(1.5)和(1.6), 因为所含中立项$ z(t) = x(t)+p(t)x^\alpha(\tau(t)), 0<\alpha\leq1 $是$ x(t) $的次线性函数, 已有文献[14-22]采用不同的不等式缩放技巧, 得到了不同形式的振动定理, 其中具代表性的是2014年Agarwal等在文献[14]中对于特殊情况$ \beta = \gamma $, 建立的依赖于两个式子的振动定理, 其中之一如下.

定理1.4^{[14, 定理2.2]}  设$ 0<\alpha\leq1, \beta = \gamma = 1 $, $ 1-\alpha 2^{1-\alpha}p(t)A(\tau(t))/A(t)>0 $和$ (1.7) $式成立. 若存在可微函数$ \rho \in C^1([t_0, \infty), (0, \infty)) $, 使得对任意常数$ M>0, K>0 $, 有

(1.10) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t\left(\rho(s) q(s)\left[1-(\alpha2^{1-\alpha}+\frac{2^{1-\alpha}-1}{M})p(\sigma(s))\right]-\frac{a(\sigma(s))(\rho'(s))^{2}}{4\rho(s)\sigma'(s)}\right){\rm d} s = \infty \end{equation} $

和

(1.11) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t\left( q(s)A(s)\left[1-(\alpha2^{1-\alpha}\frac{A(\tau(\sigma(s)))}{A(\sigma(s))}+\frac{2^{1-\alpha}-1}{KA(s)})p(\sigma(s))\right]-\frac{1}{4a(s)A(s)}\right){\rm d} s = \infty, \end{equation} $

则方程$ (1.5) $振动.

值得注意的是, 在文献[14]的前提假设和各定理中虽然均未提及条件"$ \lim\limits_{t\to \infty} p(t) = 0 $" 成立, 但当$ 0<\alpha<1 $时, 这是必需的(严格地说"$ \liminf\limits_{t\to \infty}p(t) = 0 $"是必需的), 否则不能保证其(2.6)式的右端非负. 所以作者在最后的注3.1中指出不得不需要这一必要条件成立! 其后已有多篇文献, 如文献[15-17, 22]对方程(1.5), 文献[18-21]对方程(1.6), 甚至文献[23-24]对于3阶带次线性中立项的Emden-fowler时滞微分方程建立了多个振动定理. 但遗憾的是他们仍采用类似于文献[14]关键不等式的缩放方法, 或默认或忽略了条件"$ \lim\limits_{t\to \infty} p(t) = 0 $", 并在$ \beta\neq\alpha $时, 还将对应于定理1.4中的"任意常数$ M>0 $"或"任意常数$ K>0 $" 误认为"某常数", 导致出现了文献[18-19]中的例2, 文献[21]中的例0.9–0.11, 文献[23]中的例1, 例2和文献[24]中的例3的方程均为"$ 0<p(t) = c\leq1 $"的失效情况. 很明显, 例如文献[23]中例1, 只有当$ M>(1/4)^{5/2} $而不是对于任意的$ M>0 $时核心式子成立, 因此断定方程振动是不充分的. 为了克服这些困难, 本文将采用更加谨慎的不等式缩放技巧, 寻找方程(1.1)更精细简便的振动性充分判据, 从而更系统地解决方程(1.2)和(1.3)的振动性问题.

为此, 需要如下至关重要的引理^[13].

引理1.1  设$ A>0, B\geq 0, \lambda>0 $为常数, 则对任意的$ u\geq 0, $有

(1.12) $ \begin{equation} Bu-Au^{\frac{\lambda+1}{\lambda}}\leq \frac{\lambda^\lambda}{(\lambda+1)^{\lambda+1}}\frac{B^{\lambda+1}}{A^\lambda}. \end{equation} $

定理2.1  设条件$ \rm {(H_1)} $–$ \rm{(H_3)} $及$ (1.7) $式成立. 如果存在函数$ \varphi\in C^1([t_0, \infty), (0, \infty)), $使得对于任意常数$ M, K, K_1, K_2\in (0, 1] $ (特别, 当$ \alpha = 1 $时$ M = 1 $, 当$ \beta = \gamma $时$ K = K_1 = K_2 = 1), $有

(2.1) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t\left[ \varphi(s)q(s)P_1(M, \sigma(s))-\frac{\eta(s)\varphi(s)}{(\lambda+1)^{\lambda+1}(\sigma'(s))^\lambda}\left(\frac{\varphi'_+(s)}{\varphi(s)}\right)^{\lambda+1}\right]{\rm d}s = \infty \end{equation} $

和

(2.2) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t\left[K_2 A^\mu(s)q(s)P_2(K_1, \sigma(s))-\left(\frac{\beta}{\beta+1}\right)^{\beta+1}\frac{1}{a^{1/\beta}(s)A(s)}\right]{\rm d}s = \infty, \end{equation} $

其中$ P_1(M, t) = \left(1-\frac{p_\alpha(t)}{M^{1-\alpha}}\right)^\gamma, $$ \lambda = \min \{\beta, \gamma\}, $$ \eta(t) = \left\{ \begin{array}{ll} \frac{ \lambda^\lambda a(\sigma(t))}{\gamma^\lambda M^{\gamma-\lambda}}, \; \; \lambda = \beta<\gamma, \\ \frac{a^{\lambda/\beta}(\sigma(t))}{K}, \; \lambda = \gamma\leq \beta, \end{array} \right. $$ \varphi'_+(t) = \max\{\varphi'(t), 0\}, $$ \mu = \max\{ \beta, \gamma \}, $$ P_2(K_1, t) = \left( 1-\frac{p_\alpha(t)}{K_1^{1-\alpha}}\frac{A^\alpha(\tau(t))}{A(t)}\right)^\gamma $, 则方程$ (1.1) $振动.

证  假设方程(1.1)存在一个非振动解$ x(t), $不妨设该解是最终正解(当$ x(t) $是最终负解时用类似方法可以证明), 则存在$ t_1\geq t_0, $得当$ t>t_1 $时, 有$ x(t)>0, x(\tau(t))>0, x(\sigma(t))>0. $由$ z(t) $的定义可得$ z(t)>0, $$ t\geq t_1 $且$ z(t)\geq x(t), t\geq t_1. $代入方程(1.1), 得

(2.3) $ \begin{equation} \left[a(t)|z'(t)|^{\beta-1}z'(t) \right]' = -f\left( t, |x(\sigma(t))|^{\gamma-1}x(\sigma(t))\right)\leq 0. \end{equation} $

所以$ a(t)|z'(t)|^{\beta-1}z'(t) $是非增函数且$ z'(t) $最终保号, 以此, 必有$ z'(t)\geq 0, $$ t\geq t_1 $或$ z'(t)\leq 0, $$ t\geq t_1. $下面分别讨论之.

(Ⅰ) 当有$ z'(t)\geq 0, t\geq t_1 $时, 函数$ z(t) $为非减函数, 因此, 存在某个$ M_0 = \min \{ z(\tau(t_1)), 1\} = M\in (0, 1], $使得

(2.4) $ \begin{equation} z(t)\geq z(\tau(t))\geq M_0. \end{equation} $

由上式和$ z(t) $的定义, 可得

(2.5) $ \begin{equation} x(t) = z(t)-g(t, x^\alpha (\tau(t)))\geq z(t)-p_\alpha(t)z^\alpha(\tau(t))\geq \left(1-\frac{p_\alpha(t)}{M_0^{1-\alpha}}\right)z(t), \end{equation} $

特别, 当$ \alpha = 1 $时, 取$ M_0 = 1, $时(2.5)式自然成立.又由(H$ _2 $), 不妨设当$ t\geq t_1 $时(2.5)式右端为正.

作Riccati变换如下

(2.6) $ \begin{equation} \omega(t) = \varphi(t)\frac{a(t)(z'(t))^\beta}{z^\gamma(\sigma(t))}, t\geq t_1. \end{equation} $

于是, 显然有$ \omega(t)>0, t\geq t_1 $. 由于$ a(t)(z'(t))^\beta $非增, 于是有

$ a(t)(z'(t))^\beta\leq a(\sigma(t))(z'(\sigma(t)))^\beta, t\geq t_1, $

从而, 有

(2.7) $ \begin{equation} z'(\sigma(t))\geq \left(\frac{a(t)}{a(\sigma(t))} \right)^{1/\beta}z'(t). \end{equation} $

故由(2.5)–(2.7)式, 可得, 当$ t\geq t_1 $时, 有

(2.8) $ \begin{eqnarray} \omega'(t)& = &\frac{\varphi'(t)}{\varphi(t)}\omega(t)+\varphi(t)\frac{[a(t)(z'(t))^\beta]'z^\gamma(\sigma(t))-a(t)(z'(t))^\beta \gamma z^{\gamma-1}(\sigma(t))z'(\sigma(t))\sigma'(t)}{z^{2\gamma}(\sigma(t))}{}\\ &\leq& \frac{\varphi'(t)}{\varphi(t)}\omega(t)-\varphi(t)\frac{q(t)x^\gamma(\sigma(t))}{z^\gamma(\sigma(t))}-\varphi(t)\gamma\frac{a(t)(z'(t))^\beta \sigma'(t)}{z^{\gamma+1}(\sigma(t))}\frac{a^{1/\beta}(t)z'(t)}{a^{1/\beta}(\sigma(t))} {}\\ &\leq &\frac{\varphi'(t)}{\varphi(t)}\omega(t)-\varphi(t)q(t)P_1(M_0, \sigma(t))-\gamma\frac{\varphi(t)a^{(\beta+1)/\beta}(t)(z'(t))^{\beta+1}\sigma'(t)}{z^{\gamma+1}(\sigma(t))a^{1/\beta}(\sigma(t))}. \end{eqnarray} $

根据$ \beta $和$ \gamma $的取值分两种情况讨论如下.

(ⅰ) 当$ \beta<\gamma $时, 注意到(2.4)和(2.6)式, (2.8)式成为

(2.9) $ \begin{eqnarray} \omega'(t)&\leq& \frac{\varphi'(t)}{\varphi(t)}\omega(t)-\varphi(t)q(t)P_1(M_0, \sigma(t))-\gamma \frac{z^{(\gamma-\beta)/\beta}(\sigma(t))\sigma'(t)}{\varphi^{1/\beta}(t)a^{1/\beta}(\sigma(t))}\omega^{\frac{\beta+1}{\beta}}(t) {}\\ &\leq &\frac{\varphi'_+(t)}{\varphi(t)}\omega(t)-\varphi(t)q(t)P_1(M_0, \sigma(t))-\frac{\gamma M_0^{(\gamma-\beta)/\beta}\sigma'(t)}{\varphi^{1/\beta}(t)a^{1/\beta}(\sigma(t))}\omega^{\frac{\beta+1}{\beta}}(t), t\geq t_1. \end{eqnarray} $

对于(2.9)式右端运用引理1.1中的不等式(1.12), 其中令

$ B = \frac{\varphi'_+(t)}{\varphi(t)}, \; \; A = \frac{\gamma M_0^{(\gamma-\beta)/\beta}\sigma'(t)}{\varphi^{1/\beta}(t)a^{1/\beta}(\sigma(t))}, \; \; u = \omega(t), $

得

$ \frac{\varphi'_+(t)}{\varphi(t)}\omega(t)-\frac{\gamma M_0^{(\gamma-\beta)/\beta}\sigma'(t)}{\varphi^{1/\beta}(t)a^{1/\beta}(\sigma(t))}\omega^{\frac{\beta+1}{\beta}}(t) \leq \frac{\varphi(t)a(\sigma(t))(\beta/\gamma)^\beta}{M_0^{\gamma-\beta}(\beta+1)^{\beta+1}(\sigma'(t))^\beta}\left( \frac{\varphi'_+(t)}{\varphi(t)}\right)^{\beta+1}, t\geq t_1, $

将其代入(2.9)式, 得

$ \omega'(t)\leq -\varphi(t)q(t)P_1(M_0, \sigma(t))+\frac{(\beta/\gamma)^\beta\varphi(t)a(\sigma(t))}{M_0^{\gamma-\beta}(\beta+1)^{\beta+1}(\sigma'(t))^\beta}\left( \frac{\varphi'_+(t)}{\varphi(t)}\right)^{\beta+1}, t\geq t_1, $

上式两边从$ t_1 $到$ t(t\geq t_1) $积分, 得

$ \begin{eqnarray*} &&\int_{t_1}^t \left[ \varphi(s)q(s)P_1(M_0, \sigma(s))-\frac{(\beta/\gamma)^\beta\varphi(s)a(\sigma(s))}{M_0^{\gamma-\beta}(\beta+1)^{\beta+1}(\sigma'(s))^\beta}\left( \frac{\varphi'_+(s)}{\varphi(s)}\right)^{\beta+1} \right]{\rm d}s\\ &\leq & -\omega(t)+\omega(t_1)\leq \omega(t_1), \end{eqnarray*} $

注意到这时$ \beta = \lambda, $知上式与(2.1)式矛盾.

(ⅱ) 当$ \beta\geq \gamma $时, 由$ z'(t)>0, t\geq t_1 $及(2.3)式可知, 存在某个$ k_1 = \max\{1, a(t_1)(z'(t_1))^\beta \}, $使得当$ t\geq t_1 $时, 有

$ a(t)(z'(t))^\beta \leq a(t_1)(z'(t_1))^\beta \leq k_1. $

令$ k_2 = \frac{1}{k_1}\in (0, 1], $则有$ z'(t)\leq \frac{1}{a^{1/\beta}(t)k_2^{1/\beta}}, $从而, 有

(2.10) $ \begin{equation} (z'(t))^{(\gamma-\beta)/\gamma}\geq \frac{1}{a^{(\gamma-\beta)/\gamma\beta}(t)k_2^{(\gamma-\beta)/\gamma\beta}}, t\geq t_1. \end{equation} $

显然, 当$ \beta = \gamma $时, 取$ k_2 = 1 $时(2.10)式仍成立. 注意到(2.6)式, 将(2.10)式代入(2.8)式, 可得

(2.11) $ \begin{eqnarray} \omega'(t)&\leq& \frac{\varphi'(t)}{\varphi(t)}\omega(t)-\varphi(t)q(t)P_1(M_0, \sigma(t))-\gamma \frac{a^{(\gamma-\beta)/\gamma\beta}(t)(z'(t))^{(\gamma-\beta)/\gamma}\sigma'(t)}{\varphi^{1/\gamma}(t)a^{1/\beta}(\sigma(t))}\omega^{\frac{\gamma+1}{\gamma}}(t) {}\\ &\leq &\frac{\varphi'_+(t)}{\varphi(t)}\omega(t)-\varphi(t)q(t)P_1(M_0, \sigma(t))-\frac{\gamma \sigma'(t)}{\varphi^{1/\gamma}(t)a^{1/\beta}(\sigma(t))k_2^{(\gamma-\beta)/\beta\gamma}}\omega^{\frac{\gamma+1}{\gamma}}(t), \; t\geq t_1.{\qquad} \end{eqnarray} $

同(ⅰ)一样, 对(2.11)式右端再运用引理1.1的不等式(1.12), 可得

$ \omega'(t)\leq -\varphi(t)q(t)P_1(M_0, \sigma(t))+\frac{\varphi(t)a^{\gamma/\beta}(\sigma(t))}{K_0(\gamma+1)^{\gamma+1}(\sigma'(t))^\gamma}\left(\frac{\varphi'_+(t)}{\varphi(t)} \right)^{\gamma+1}, $

其中$ K_0 = k_2^{(\beta-\gamma)/\beta}\in(0, 1] $(特别当$ \beta = \gamma $时$ K_0 = 1 $). 对上式两边从$ t_1 $到$ t\geq t_1 $积分, 得

$ \begin{eqnarray*} &&\int_{t_1}^t\left[\varphi(s)q(s)P_1(M_0, \sigma(s))-\frac{\varphi(s)a^{\gamma/\beta}(\sigma(s))}{K_0 (\gamma+1)^{\gamma+1}(\sigma'(s))^\gamma}\left( \frac{\varphi'_+(s)}{\varphi(s)}\right)^{\gamma+1} \right]{\rm d}s \\ &\leq& -\omega(t)+\omega(t_1)\leq \omega(t_1), \end{eqnarray*} $

这时$ \gamma = \lambda, $上式也与(2.1)式矛盾.

(Ⅱ) 当$ z'(t)\leq 0, t\geq t_1 $时, 首先由(2.3)式可知$ a(t)|z'(t)|^{\beta-1}z'(t) $是减函数, 从而存在某个常数$ K_1^\beta = \min\{a(t_1)(-z'(t_1))^\beta, 1\}\in(0, 1], $使得当$ t\geq t_1 $时, 有

$ -a(t)(-z'(t))^\beta\leq -a(t_1)(-z'(t_1))^{\beta} \leq \max\{ -a(t_1)(-z'(t_1))^\beta, -1\} = -K_1^\beta, $

因此有$ z'(t)\leq -\frac{K_1}{a^{1/\beta}(t)}, $两边从$ t $到$ \infty $积分, 可得$ z(t)\geq K_1 A(t). $显然又有$ z(t)\leq z(t_1)\leq k_2 = \max\{ 1, z(t_1)\}, $即存在某个$ k_3 = \frac{1}{k_2}\in(0, 1], $综上所述, 可得

(2.12) $ \begin{equation} K_1 A(t)\leq z(t)\leq \frac{1}{k_3}, \; \; t\geq t_1, \; K_1, k_3\in(0, 1]. \end{equation} $

再引入第二个Riccati变换

(2.13) $ \begin{equation} 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\geq t_1. \end{equation} $

显然, $ v(t)<0, \; t\geq t_1. $又由(2.3)式知, 对于任意的$ s\geq t\geq t_1, $有$ a(s)(-z'(s))^\beta\geq a(t)(-z'(t))^\beta, $从而, 有$ z'(s)\leq \frac{a^{1/\beta}(t)z'(t)}{a^{1/\beta}(s)}, $对$ s $从$ t $到$ \infty $积分，得

$ z(t)+a^{1/\beta}(t)z'(t)A(t)\geq 0, $

即

$ -1\leq \frac{a^{1/\beta}(t)z'(t)}{z(t)}A(t)<0, $

从而, 有

(2.14) $ \begin{equation} -1\leq v(t)A^\beta(t)<0. \end{equation} $

又因为

$ \left( \frac{z(t)}{A(t)}\right)' = \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)}\geq0, $

于是, 函数$ z(t)/A(t) $非减, 所以, 有

(2.15) $ \begin{equation} z(\tau(t))\leq \frac{A(\tau(t))}{A(t)}z(t), \; \; t\geq t_1. \end{equation} $

因为$ 0<\alpha\leq1 $并注意到(2.15)和(2.12)式, 又得到

(2.16) $ \begin{eqnarray} x(t)&\geq & z(t)-p_\alpha(t)z^\alpha(\tau(t)){}\\ &\geq& z(t)-p_\alpha(t)\left( \frac{A(\tau(t))}{A(t)}\right)^\alpha z^\alpha(t){}\\ & = & \left( 1-p_\alpha(t)\left( \frac{A(\tau(t))}{A(t)}\right)^\alpha z^{\alpha-1}(t)\right)z(t){}\\ &\geq& \left(1-p_\alpha(t)\left( \frac{A(\tau(t))}{A(t)}\right)^\alpha K_1^{\alpha-1}A^{\alpha-1}(t)\right)z(t){}\\ & = &\left( 1-\frac{p_\alpha(t)}{K_1^{1-\alpha}}\frac{A^\alpha(\tau(t))}{A(t)}\right)z(t), \; \; t\geq t_1. \end{eqnarray} $

对(2.13)式求导, 并注意到(2.16)式, 得

(2.17) $ \begin{eqnarray} v'(t)& = &\frac{\left( a(t)|z'(t)|^{\beta-1}z'(t)\right)'}{z^\beta(t)}-\frac{a(t)|z'(t)|^{\beta-1}z'(t)\beta z^{\beta-1}(t)z'(t)}{z^{2\beta}(t)}{}\\ &\leq& \frac{-q(t)x^\gamma(\sigma(t))}{z^\beta(t)}-\frac{\beta(-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}{}\\ &\leq &-q(t)P_2(K_1, \sigma(t))\frac{z^\gamma(t)}{z^\beta(t)}-\frac{\beta(-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}{}\\ & = &-q(t)\frac{P_2(K_1, \sigma(t))}{z^{\beta-\gamma}(t)}-\frac{\beta(-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}, \; \; t\geq t_1. \end{eqnarray} $

再注意到当$ \beta<\gamma $和$ \beta\geq \gamma $时, 分别有

$ \frac{1}{z^{\beta-\gamma}(t)} = (z(t))^{\gamma-\beta}\geq K_1^{\gamma-\beta}A^{\gamma-\beta}(t);\; \; \frac{1}{z^{\beta-\gamma}(t)}\geq k_3^{\beta-\gamma}. $

于是, 存在某个

$ K_2 = \left\{ \begin{array}{ll} K_1^{\gamma-\beta}, \; \; &\beta\leq\gamma, \\ k_3^{\beta-\gamma}, \; \; &\beta \geq \gamma. \end{array} \right. $

将$ \mu = \max\{\beta, \gamma\} $和$ K_2 $代入(2.17)式, 得

(2.18) $ \begin{equation} A^\beta(t)v'(t)\leq -K_2q(t)P_2(K_1, \sigma(t))A^\mu(t)-A^\beta(t)\frac{\beta (-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}, \; \; t\geq t_1. \end{equation} $

将(2.18)式中的$ t $改为$ s, $并对$ s $从$ t_1 $到$ t\geq t_1 $积分, 注意到分部积分公式, 得

(2.19) $ \begin{eqnarray} && \int_{t_1}^t K_2 A^\mu(s)q(s)P_2(K_1, \sigma(s)){\rm d}s{}\\ &\leq& -\int_{t_1}^t A^\beta(s)v'(s){\rm d}s-\int_{t_1}^t A^\beta(s)\frac{\beta(-v(s))^{(\beta+1)/\beta}}{a^{1/\beta}(s)}{\rm d}s {}\\ & = & A^\beta(t_1)v(t_1)-A^\beta(t)v(t)+\int_{t_1}^t \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.{\qquad} \end{eqnarray} $

对(2.19)式右端的被积式利用引理1.1中的不等式(1.12), 可得

$ \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}\leq \left(\frac{\beta}{\beta+1}\right)^{\beta+1}\frac{1}{a^{1/\beta}(s)A(s)}, $

将上式代入(2.19)式, 并注意到(2.14)式, 得

$ \begin{eqnarray*} &&\int_{t_1}^t\left[K_2 A^\mu(s)q(s)P_2(K_1, \sigma(s))-\left(\frac{\beta}{\beta+1}\right)^{\beta+1}\frac{1}{a^{1/\beta}(s)A(s)}\right]{\rm d}s {\nonumber}\\ &\leq & (A^\beta(t_1)v(t_1)-A^\beta(t)v(t))\leq (A^\beta(t_1)v(t_1)+1). \end{eqnarray*} $

这与(2.2)式矛盾. 综合以上(Ⅰ)与(Ⅱ)两种矛盾情况可知, 方程(1.1)振动.

注2.1  显然, 当本文的$ \alpha, \beta, \gamma $退化成$ "\alpha = 1, \beta = \gamma = 1" $时, 定理$ 2.1 $与定理$ 1.4 $相同, 所以, 这时定理$ 2.1 $包含和改进了文献[14]的定理$ 2.2 $和诸多关于带线性中立项的方程$ (1.4) $的多个结果, 如文献[10]中的定理$ 2.1 $等. 而当$ 0<\alpha<1, \beta = \gamma = 1 $时, 本文定理$ 2.1 $比定理$ 1.4 $更简便, 因为本文中的$ (1-p_\alpha(t)/M^{1-\alpha}), M\in(0, 1] $和$ 1-\frac{p_\alpha(t)}{K_1^{1-\alpha}}\frac{A^\alpha(\tau(t))}{A(t)}, K_1\in(0, 1] $对应于定理$ 1.4 $的$[1-(\alpha2^{1-\alpha}+\frac{2^{1-\alpha}}{M})p(t)] $和$ 1-(\alpha2^{1-\alpha}\frac{A(\tau(t))}{A(t)}+\frac{2^{1-\alpha}-1}{KA(t)})p(t), \; M>0, K>0 $任意. 所以定理$ 2.1 $推广和改进了文献[14]的定理$ 2.2 $.

例2.1   讨论具有非线性中立项的二阶广义Emden-Fowler方程

(2.20) $ \begin{equation} \left(t^2|z'(t)|^{\beta-1}z'(t)\right)'+q_0x\left(\frac{t}{3}\right) = 0, \; \; t\geq 1 \end{equation} $

的振动性, 其中$ z(t) = x(t)+\frac{c}{t(1+bx^2(t))}x^{\frac{3}{5}}(\frac{t}{2}), $常数$ b\geq 0, c\geq0, q_0>0, \beta $分别为$ 1 $和$ \frac{3}{2} $.

易知, 对应于方程(1.1), 这里有

$ a(t) = t^2, p_\alpha(t) = \frac{c}{t}, \alpha = \frac{3}{5}, \gamma = 1, \beta = 1, \frac{3}{2};\tau(t) = \frac{t}{2}, \sigma(t) = \frac{t}{3}, t_0 = 1. $

则, 显然条件(H$ _1) $–(H$ _3) $及(1.7)式成立且分别有$ A(t) = \frac{1}{t}, \frac{3}{t^2}. $取$ \varphi(t) = 1, $则对于任意取定的常数$ M\in(0, 1], $取$ t_1 = \frac{6c}{M^{2/5}}, $则当$ t>t_1 $时, 有$ 1-\frac{3c}{M^{2/5}t}>\frac{1}{2.} $于是, 又有

$ \limsup\limits_{t\to\infty}\int_{t_1}^tq(s)P_1(M, \sigma(s)){\rm d}s = \limsup\limits_{t\to\infty}\int_{t_1}^tq_0\left[1-\frac{3c}{M^{2/5}s} \right]{\rm d}s\geq \limsup\limits_{t\to\infty}\int_{t_1}^t\frac{q_0}{2}{\rm d}s = \infty, $

即(2.1)式成立. 又当$ \beta = \gamma = 1 $时, $ \mu = 1, K_1 = K_2 = 1, $有

$ A(s)q(s)P_2(1, \sigma(s)) = \frac{q_0}{s}\left( 1-\frac{3c}{s}\frac{(6/s)^{3/5}}{3/s}\right) = \frac{q_0}{s}\left( 1-\frac{c6^{3/5}}{s^{3/5}}\right), $

$ \begin{eqnarray*} &&\limsup\limits_{t\to\infty}\int_1^t\left[A^\mu(s)q(s)P_2(1, \sigma(s))-\left( \frac{\beta}{\beta+1}\right)^{\beta+1} \frac{1}{a^{1/\beta}(s)A(s)}\right]{\rm d}s {\nonumber}\\ & = & \limsup\limits_{t\to\infty}\int_1^t\left[\frac{q_0}{s}\left( 1-\frac{c6^{3/5}}{s^{3/5}}\right)-\frac{1}{4s} \right]{\rm d}s {\nonumber}\\ & = & \limsup\limits_{t\to\infty}\int_1^t \frac{1}{s}\left[ q_0\left( 1-\frac{c6^{3/5}}{s^{3/5}}\right)-\frac{1}{4}\right]{\rm d}s = \infty\; \; (q_0>\frac{1}{4}), \end{eqnarray*} $

即(2.2)式成立, 故由定理2.1知, 当$ \beta = \gamma = 1, q_0>\frac{1}{4} $时, 方程(2.20)振动. 这是关于广义Euler方程著名振动定理的推广.

同理, 当$ \beta = \frac{3}{2}>\gamma = 1 $时, $ \mu = \frac{3}{2}, $对于任意取定的$ K_1, K_2\in(0, 1], $易知

$ \begin{eqnarray*} &&K_2A^\mu(s)q_0P_2(K_1, \sigma(s))-\bigg(\frac{\beta}{\beta+1}\bigg)^{\beta+1}\frac{1}{a^{1/\beta}(s)A(s)}\\ & = &\frac{1}{s}\left[K_2 3^{\frac{2}{3}}q_0s^{\frac{1}{2}} \bigg(1-\frac{6^{1/5}3^{4/15}c}{K_1^{2/5}s^{13/15}}\bigg)-\frac{(3/5)^{5/2}}{3} \right], \end{eqnarray*} $

所以, 当$ q_0>0 $时(2.2)式也成立, 因此, 方程(2.20)也振动.

显然, 本文所列文献对方程(2.20)无效.

定理2.2   设条件(H$ _1) $–(H$ _3) $及$ (1.7) $式成立. 如果对任意常数$ K\in (0, 1] $ (特别当$ \alpha = 1 $时$ K = 1), $有

(2.21) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t A^{\mu+1}(s)q(s)P_2(K, \sigma(s)){\rm d}s = \infty \end{equation} $

或

(2.22) $ \begin{equation} \limsup\limits_{t\to \infty}\int_{t_0}^t \left[\frac{1}{a(u)}\int_{t_0}^uA^{\gamma}(s)q(s)P_2(K, \sigma(s)){\rm d}s\right]^{\frac{1}{\beta}} = \infty \end{equation} $

成立, 其中$ \mu = \max\{ \beta, \gamma \}, P_2(K, t) $由$ (2.2) $式定义, 则方程$ (1.1) $振动.

证  假设方程(1.1)有一个非振动解$ x(t), $不妨设该解是最终正解(当$ x(t) $是最终负解时用类似方法可以证明), 则完全类似于定理2.1的证明, 不妨设存在$ t_1\geq t_0, $使得当$ t\geq t_1 $时, 有$ z(t)>0, z(\tau(t))>0, z(\sigma(t))>0 $且$ z'(t)>0 $或$ z'(t)<0. $下面分别讨论之.

(Ⅰ) 当$ z'(t)>0, t\geq t_1 $时, 类似于(2.5)式知, 存在某个$ K\in(0, 1] $, 使得

(2.23) $ \begin{equation} x(t)\geq \left(1-\frac{p(t)}{K^{1-\alpha}} \right)z(t), \; t\geq t_1, \end{equation} $

将其代入(2.3)式, 得

(2.24) $ \begin{equation} (a(t)(z'(t))^\beta)'\leq -q(t)\left(1-\frac{p(\sigma(t))}{K^{1-\alpha}} \right)^\gamma z^\gamma(\sigma(t)), \; t\geq t_1. \end{equation} $

由于$ A(t)\to 0\; (t\to\infty), $所以, 存在$ t_2\geq t_1, $当$ t\geq t_2 $时, 有$ A(\tau(\sigma(t)))<1 $和$ \frac{A(\tau(\sigma(t)))}{A(\sigma(t))}\geq 1, $所以, 有

$ 1-\frac{p(\sigma(t))}{K^{1-\alpha}}\geq 1-\frac{p(\sigma(t))}{K^{1-\alpha}}\frac{A(\tau(\sigma(t)))}{A(\sigma(t))} \geq 1-\frac{p(\sigma(t))}{K^{1-\alpha}}\frac{A^\alpha(\tau(\sigma(t)))}{A(\sigma(t))}, \; \; t\geq t_2. $

将上式代入(2.24)式, 得

(2.25) $ \begin{equation} (a(t)(z'(t))^\beta)'\leq -q(t)P_2(K, \sigma(t)) z^\gamma(\sigma(t)), \; t\geq t_2. \end{equation} $

又由于$ A(t)\to 0\; (t\to\infty) $和(1.7)式知, 当(2.21)或(2.22)式成立时, 总有

(2.26) $ \begin{equation} \limsup\limits_{t\to\infty}\int_{t_1}^t q(s)P_2(K, \sigma(s)){\rm d}s = \infty. \end{equation} $

再将(2.25)式两端从$ t_2 $到$ t\geq t_2 $积分, 得

$ \begin{eqnarray*} a(t)(z'(t))^\beta &\leq& a(t_2)(z'(t_2))^\beta-\int_{t_2}^tq(s)P_2(K, \sigma(s))z^\gamma(\sigma(s)){\rm d}s{\nonumber}\\ &\leq& a(t_2)(z'(t_2))^\beta-z^\gamma(\sigma(t_2))\int_{t_2}^tq(s)P_2(K, \sigma(s)){\rm d}s, \; \; t\geq t_2. \end{eqnarray*} $

注意到(2.26)式, 在上式两端令$ t\to\infty $取上极限, 得与$ z'(t)>0, t\geq t_1 $的矛盾.

(Ⅱ) 当$ z'(t)<0, t\geq t_1 $时, 注意到$ -a(t)(-z'(t))^\beta $单调减, 则有

(2.27) $ \begin{equation} -z(t)\leq \int_{t}^\infty\frac{-(a(s)(-z'(s))^\beta)^{1/\beta}}{a^{1/\beta}(s)}{\rm d}s\leq -(a(t_1)(-z'(t_1)^\beta)^{1/\beta}A(t) = -LA(t), \end{equation} $

其中$ L = (a(t_1)(-z'(t_1)^\beta)^{1/\beta}>0 $. 将上式和(2.16)式代入方程(1.1), 得

(2.28) $ \begin{eqnarray} (-a(t)(-z'(t))^\beta)' &\leq&-q(t)x^\gamma(\sigma(t)){}\\ &\leq &-q(t)P_2(K_1, \sigma(t))z^\gamma(t) {}\\ &\leq &-q(t)P_2(K_1, \sigma(t))L^\gamma A^\gamma(t). \end{eqnarray} $

(ⅰ) 当(2.22)式成立时, 将上式从$ t_1 $到$ t $积分, 得

$ -a(t)(-z'(t))^\beta \leq -a(t_1)(-z'(t_1))^\beta-L^\gamma \int_{t_1}^t q(s)P_2(K_1, \sigma(s))A^\gamma(s){\rm d}s, $

从而, 有

$ (-z'(t))^\beta \geq L^\gamma \frac{1}{a(t)}\int_{t_1}^t q(s)P_2(K_1, \sigma(s))A^\gamma(s){\rm d}s, $

$ z'(t)\leq -L^{\frac{\gamma}{\beta}}\left[ \frac{1}{a(t)}\int_{t_1}^t q(s)P_2(K_1, \sigma(s))A^\gamma(s){\rm d}s\right]^{\frac{1}{\beta}}, $

$ z(t)\leq z(t_1)-L^{\frac{\gamma}{\beta}}\int_{t_1}^t\left[ \frac{1}{a(u)}\int_{t_1}^u q(s)P_2(K_1, \sigma(s))A^\gamma(s){\rm d}s\right]^{\frac{1}{\beta}}{\rm d}u. $

由(2.22)式知, 上式与$ z(t)>0, t\geq t_1 $矛盾.

(ⅱ) 当(2.21)式成立时, 我们可以证明(2.2)式也成立. 事实上, 当(1.7)式成立时, (2.21)式等价于

(2.29) $ \begin{equation} \limsup\limits_{t\to\infty}\int_{t_0}^t\left(K_2A^{\mu+1}(s)q(s)P_2(K, \sigma(t))-(\frac{\beta}{\beta+1})^{\beta+1}\frac{1}{a^{1/\beta}(s)}\right){\rm d}s = \infty. \end{equation} $

又因为

$ \frac{K_2A^{\mu}(s)q(s)P_2(K, \sigma(t))-(\frac{\beta}{\beta+1})^{\beta+1}\frac{1}{a^{1/\beta}(s)A(s)}}{K_2A^{\mu+1}(s)q(s)P_2(K, \sigma(t))-(\frac{\beta}{\beta+1})^{\beta+1}\frac{1}{a^{1/\beta}(s)}} = \frac{1}{A(s)}\to\infty\; (t\to\infty), $

所以, 由广义积分的比值审敛法知, 当(2.29)式成立时(2.2)式也成立. 因此, 完全类似于定理2.1的证明中"$ z'(t)<0, \; t\geq t_1 $"的情形也得到矛盾.

综和(Ⅰ)和(Ⅱ)两种情况可知, 方程(1.1)振动.

注2.2   定理$ 2.2 $已包含和改进了文献[5]关于方程$ (1.4) $在$ p(t)\equiv0, \beta = \gamma $时的定理$ 2.2 $, 因为文献[5]要求本文的$ (2.1), (2.21) $和$ (2.22) $式同时成立且结论是每个解振动或渐近于$ 0 $. 同时本文定理$ 2.2 $还包含了文献[7]关于方程$ (1.4) $在$ p(t)\equiv0, \beta = \gamma $时的定理$ 2 $及其引文中的相关定理, 包含了文献[8]中关于方程$ (1.4) $当$ \alpha = 1, \beta = \gamma $时的定理$ 2.1 $, 也包含了文献[15]关于方程$ (1.5) $当$ \beta = 1 $时的定理$ 2.4 $和定理$ 2.6 $, 又包含了文献[22]关于方程$ (1.5) $当$ \beta = \gamma, m = 1 $时的定理$ 1 $.

定理2.3   设条件(H$ _1) $–(H$ _3) $及$ (1.7) $式成立且$ \beta = \gamma $. 如果对任意常数$ K\in (0, 1] $ (特别当$ \alpha = 1 $时$ K = 1), $有

(2.30) $ \begin{equation} \limsup\limits_{t\to \infty}\left( A^{\beta}(t)\int_{t_0}^tq(s)P_2(K, \sigma(s)){\rm d}s\right)>1, \end{equation} $

其中$ P_2(K_1, t) $由$ (2.2) $式定义, 则方程$ (1.1) $振动.

证    假设方程(1.1)有一个非振动解$ x(t) $, 不妨设该解是最终正解(当$ x(t) $是最终负解时用类似方法可以证明), 则完全类似于定理2.1的证明, 不妨设存在$ t_1\geq t_0, $使得当$ t\geq t_1 $时, 有$ z(t)>0, z(\tau(t))>0, z(\sigma(t))>0 $且$ z'(t)>0 $或$ z'(t)<0. $下面分别讨论之.

(Ⅰ) 当$ z'(t)>0, t\geq t_1 $时, 类似于定理2.2证明中的(Ⅰ), 有(2.25)式成立, 由(2.30)式知(2.26)式也成立, 于是类似可得与$ z'(t)>0, t\geq t_1 $的矛盾.

(Ⅱ) 当$ z'(t)<0, t\geq t_1 $时, 注意到$ -a(t)(-z'(t))^\beta<0 $单调减, 则有$ a(t)(-z'(t))^\beta>0 $单调增, 所以有

(2.31) $ \begin{equation} z(t)\geq \int_t^\infty \frac{(a(s)(-z'(s))^\beta)^{1/\beta}}{a^{1/\beta}}{\rm d}s\geq -z'(t)a^{\frac{1}{\beta}}(t)A(t). \end{equation} $

将(2.16)式代入方程(1.1)并注意到$ \beta = \gamma, $得

$ (-a(t)(-z'(t))^\beta)'\leq -q(t)x^\beta(\sigma(t))\leq -q(t)P_2(K_1, \sigma(t))z^\beta(t), $

即

$ (a(t)(-z'(t))^\beta)'\geq q(t)P_2(K_1, \sigma(t))z^\beta(t), $

从$ t_1 $到$ t\geq t_1 $积分并注意到(2.31)式和$ z^\beta(t) $单调减, 可得

$ \begin{eqnarray*} a(t)(-z'(t))^\beta &\geq &z^\beta(t)\int_{t_1}^t q(s)P_2(K_1, \sigma(s)){\rm d}s\\ &\geq &a(t)(-z'(t))^\beta A^\beta(t)\int_{t_1}^t q(s)P_2(K_1, \sigma(s)){\rm d}s, \end{eqnarray*} $

亦即

$ A^\beta(t)\int_{t_1}^tq(s)P_2(K_1, \sigma(s)){\rm d}s\leq 1. $

这与(2.30)式矛盾.

综合(Ⅰ)和(Ⅱ)两种情形的证明知, 方程(1.1)振动.

注2.3   显然，本文定理$ 2.3 $已包含了文献[7]中的定理$ 3 $, 文献[8]中的定理$ 2.2 $和文献[22]中的定理$ 2 $.

例2.2  讨论具有非线性中立项的二阶广义Emden-Fowler方程

(2.32) $ \begin{equation} \left(t^2|z'(t)|^{\beta-1}z'(t)\right)'+q_0t^mf(x^\gamma(\frac{t}{2})) = 0, \; \; t\geq 1 \end{equation} $

的振动性, 其中$ z(t) = x(t)+\frac{c}{t(1+bx^2(t))}x^{\frac{1}{3}}(\varepsilon t), \; f(u) = u\left( 1+\kappa \ln(1+u^4)\right), $常数$ b\geq 0, $$ c\geq0, $$ 0<\varepsilon\leq1, q_0>0, m\geq-1, \kappa\geq0;\alpha = \frac{1}{3}, \beta>0;\gamma>0 $为两正奇数之比的常数.

对应于方程(1.1), 这里有

$ a(t) = t^2, p_\alpha(t) = \frac{c}{t}, q(t) = q_0t^m, \tau(t) = \varepsilon t, \sigma(t) = \frac{t}{2};\; A(t) = \frac{1}{t}; $

$ P_1(M, \sigma(t)) = \left( 1-\frac{2c}{M^{2/3}t}\right)^\gamma, \; P_2(K_1, \sigma(t)) = \left( 1-\frac{2^{2/3}c}{(\varepsilon K_1)^{2/3}t^{2/3}}\right)^\gamma. $

所以, 显然条件(H$ _1 $)–(H$ _3) $及(1.7)式满足. 对于任意给定的常数$ M\in(0, 1], $有

$ \limsup\limits_{t\to\infty}\int_1^tq(s)P_1(M, \sigma(s)){\rm d}s = \limsup\limits_{t\to\infty}\int_1^t q_0s^m\left(1-\frac{2c}{M^{2/3}s} \right)^\gamma {\rm d}s = \infty, $

即当取$ \varphi(t) = 1 $时, (2.1)式成立. 下面再考察(2.2), (2.21), (2.22)或(2.30)式. 因为, 对于任意给定的常数$ K_1, K_2\in(0, 1], $有

(2.33) $ \begin{eqnarray} &&K_2A^\mu(t)q(t)P_2(K_1, \sigma(t))-\left( \frac{\beta}{\beta+1}\right)^{\beta+1}\frac{1}{a^{1/\beta}(t)A(t)} {}\\ & = &q_0K_2\left( 1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma t^{m-\mu}-\left( \frac{\beta}{\beta+1}\right)^{\beta+1}t^{1-\frac{2}{\beta}} {}\\ & = &\frac{1}{t}\left[q_0K_2\left(1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma t^{m+1-\mu}-\left(\frac{\beta}{\beta+1}\right)^{\beta+1}t^{2(1-\frac{1}{\beta})}\right]. \end{eqnarray} $

所以, 当

(ⅰ) $ m+1-\mu>2(1-\frac{1}{\beta}) $或(ⅱ) $ \gamma = \beta = 1, m = 0, q_0>\frac{1}{4} $ (这时可取$ K_2 = 1 $)时, 由(2.33)式可知(2.2)式满足, 又(2.1)式已成立, 故由定理2.1知方程(2.32)振动.

又因为, 对于任意给定的$ 0<l<1 $, 存在充分大的$ u_l>1, $使得当$ s\geq u_l $时, 有

$ \left( 1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma\geq l, $

所以, 当$ t\geq u_1 $时, 有

$ A^{\mu+1}(t)q(t)P_2(K_1, \sigma(t)) = q_0 \left(1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma t^{m-\mu-1}\geq lq_0 t^{m-\mu-1}. $

因此, 当

(ⅲ) $ m\geq\mu $时(2.21)式成立, 因此, 由定理2.2知方程(2.32)振动.

再注意到, 当$ t\geq u_1 $时, 有

$ \frac{1}{a(u)}\int_1^u A^\gamma(s)q(s)P_2(K_1, \sigma(s)){\rm d}s = \frac{1}{u^2}\left[\int_1^{u_l}+\int_{u_l}^u \right] q_0\left( 1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma s^{m-\gamma}{\rm d}s, $

而当$ m-\gamma+1>0 $时, 又有

$ \frac{1}{u^2}\int_{u_l}^uq_0\left( 1-c\left(\frac{2}{\varepsilon K_1t}\right)^{\frac{2}{3}}\right)^\gamma s^{m-\gamma}{\rm d}s\geq \frac{lq_0}{u^2}\frac{ u^{m-\gamma+1}}{m-\gamma+1}, $

所以, 当

(ⅳ) $ m\geq \gamma $时, (2.22)式成立, 故由定理2.2知, 方程(2.32)振动.

又对于上述任意的$ 0<l<1, u_l, $有

(2.34) $ \begin{equation} \lim\limits_{t\to\infty}A^\beta(t)\int_1^tq(s)P_2(K_1, \sigma(s)){\rm d}s\geq \lim\limits_{t\to\infty}\frac{lq_0}{t^\beta}\int_{u_l}^t s^m {\rm d}s = \left\{ \begin{array}{ll} 0, \; &m = -1, \\ \infty, &m>-1, \beta<m+1, \\ { } \frac{lq_0}{\beta}, \; &m>-1, \beta = m+1. \end{array} \right. \end{equation} $

因此, 如果$ \gamma = \beta, $则当

(v) $ m>-1, \beta<m+1 $或(vi) $ m>-1, q_0>\beta = m+1\; ( $令$ l\to1) $时，(2.30)式满足. 故由定理2.3知方程(2.32)振动.

注2.4   从例$ 2.2 $可见本文定理2.1–2.3作用各有千秋且不能相互替代, 定理$ 2.1 $更全面精细, 定理$ 2.2 $和定理$ 2.3 $更简捷方便. 本文所列文献的结果对方程$ (2.32) $均无效.

我们再来考察著名的Emden-Fowler方程(1.2)的振动性. 从定理1.1已知, 当$ 0\leq a\leq 1 $时方程(1.2)振动, 当$ a>1 $时方程(1.2)的每一个解$ x(t) $振动或有$ \lim\limits_{t\to\infty}x(t) = 0, $并在某些条件下确有$ x(t) = t^{-\lambda}\; (\lambda>0) $型的渐近于0的解. 但当$ a>1 $时方程(1.2)是否还有确定的振动性吗？

为了回答这个问题, 我们利用文献[4]的方法, 通过指数变换将方程(1.2)化为其等价方程

(2.35) $ \begin{equation} (t^ax'(t))'+bt^{a+m-1}x^n(t) = 0, \; \; t\geq t_0, \end{equation} $

其中$ a\geq0, b>0, m $为常数, $ 0<n\neq1 $为两正奇数之比的商. 为简便起见, 不妨取$ t_0 = 1. $

对应于方程(1.1), 这里

$ \beta = 1, \gamma = n, p_\alpha(t)\equiv0, q(t) = bt^{a+m-1}, \tau(t) = \sigma(t) = t, $

所以

$ P_1(M, t) = P_2(K, t) = 1, A(t) = \int_t^\infty \frac{{\rm d}s}{s^a} = \frac{1}{a-1}t^{1-a}\; \; (a>1). $

当$ a+m\geq0 $时, 有

(2.36) $ \begin{equation} \int_1^\infty q(s){\rm d}s = \int_1^\infty bs^{a+m-1}{\rm d}s = \infty, \end{equation} $

即取$ \varphi(t) = 1 $时, (2.1)式满足, 而对于任意取定的$ K_2\in(0, 1], $有

(2.37) $ \begin{eqnarray} &&\int_1^\infty \left[K_2A^\mu(s)q(s)-\frac{1}{4a(s)A(s)}\right]{\rm d}s{}\\ & = &\int_1^\infty\left[ K_2\left(\frac{s^{1-a}}{a-1} \right)^\mu bs^{a+m-1}-\frac{a-1}{4s}\right]{\rm d}s{}\\ & = & \left\{ \begin{array}{ll} { }\int_1^\infty\frac{1}{s}\left[ \frac{bK_2}{a-1}s^{m+1}-\frac{a-1}{4}\right]{\rm d}s = \infty, \; \; &0<n<1, m+1>0, \\ { }\int_1^\infty\frac{1}{s}\left[ \frac{bK_2}{(a-1)^n}s^{a+m-n(a-1)}-\frac{a-1}{4}\right]{\rm d}s = \infty, \; &n>1, a+m>n(a-1), \end{array} \right. \end{eqnarray} $

这时, 易知(2.2)式满足.

又因为

(2.38) $ \begin{eqnarray} &&\int_1^\infty A^{\mu+1}(s)q(s){\rm d}s{}\\ & = &\int_1^\infty \frac{b}{(a-1)^{\mu+1}}s^{a+m-1-(a-1)(\mu+1)}{\rm d}s{}\\ & = & \left\{ \begin{array}{ll} { } \frac{b}{(a-1)^2}\int_1^\infty s^{m+1-a}{\rm d}s = \infty, \; \; &0<n<1, m+1>a-1, \\ { }\frac{b}{(a-1)^{n+1}}\int_1^\infty s^{m-n(a-1)}{\rm d}s = \infty, \; &n>1, m+1>n(a-1), \end{array} \right. \end{eqnarray} $

所以, 此时(2.21)式满足.

再由于

(2.39) $ \begin{eqnarray} &&\int_1^\infty \left[\frac{1}{a(t)}\int_1^t A^\gamma(s)q(s){\rm d}s\right]{\rm d}t{}\\ & = &\frac{b}{(a-1)^n}\int_1^\infty \left[\frac{1}{t^a}\int _1^t s^{a+m-1-n(a-1)}{\rm d}s \right]{\rm d}t{}\\ & = & \left\{ \begin{array}{ll} { }\frac{b}{(a-1)^n}\int_1^\infty \frac{\ln t}{t^a}{\rm d}t<\infty, \; \; &n(a-1) = a+m, \\ { }\frac{b}{(a-1)^n}\int_1^\infty\frac{t^{a+m-n(a-1)}-1}{t^a(a+m-n(a-1))}{\rm d}t = \infty, \; &m+1\geq n(a-1), \end{array} \right. \end{eqnarray} $

因此, 当(2.39)的第二式成立时(2.22)式满足.

综合以上分析, 知(2.36)和(2.37)式成立时, 由定理2.1, (2.38)或(2.39)式的第二式成立时由定理2.2, 均得方程(2.35), 即方程(1.2)振动. 故有以下结论.

推论2.1  设$ 0<n\neq1 $为两正奇数之比的常数, $ a>1, b>0, m>-1 $为常数. 则当

(ⅰ) $ 0<n<1, m+1>0\; \rm{或}\; n>1, a+m>n(a-1), $

(ⅱ) $ 0<n<1, m+1\geq a-1\; \rm{或}\; n>1, m+1\geq n(a-1), $

(ⅲ) $ m+1\geq n(a-1) $

之一满足时, Emden-Fowler方程$ (1.2) $振动.

注2.5   推论$ 2.1 $完善了定理$ 1.1 $关于Emden-Fowler方程$ (1.2) $当$ a>1 $时的振动性结果, 但不能替代之. 如果取$ n = 1 $, 则方程$ (2.35) $就包含了Euler方程(1.3) (即$ a = 2, $$ n = 1, $$ m = -1, b = q_0), $这时注意到$ (2.36) $和$ (2.37) $式或直接应用定理$ 2.3 $, 立即可得如下两个重要结论. 特别是, 其中后者将著名结果{Euler}方程$ (1.3) $振动的"充分条件$ q_0>\frac{1}{4} $"确定为"充要条件".

推论2.2   设$ n = 1, \; a>1. $则当(ⅰ) $ m = -1, b>(a-1)^2 $或者(ⅱ) $ m>-1 $满足时, 方程$ (2.35) $振动.

推论2.3   Euler方程$ (1.3) $振动的充分必要条件是

$ q_0>\frac{1}{4}. $

证    充分性. 这时, 对应于方程(2.35), 有$ n = 1, a = 2, m = -1, b = q_0. $易知(2.36)和(2.37)式仍成立, 所以, (2.1)和(2.2)式成立. 因此, 方程(1.3)振动.

必要性. 只需证明对任意的$ q_0\in(0, \frac{1}{4}], $方程(1.3)存在非振动解即可. 事实上, 对于任意的$ q_0\in(0, \frac{1}{4}], $有且仅有一个$ \lambda\in(0, \frac{1}{2}], $使得$ q_0 = \lambda(1-\lambda) $且易知$ x(t) = t^{-\lambda} $是方程(1.3)渐近于零的非振动解. 证毕.