本文研究二阶非线性中立型时滞微分方程

(1.1) $ \begin{equation} (g(t, z(t), z'(t)))'+f(t, y(\sigma(t))) = 0, \quad t\geq t_0, \end{equation} $

其中$ z(t) = x(t)+p(t)x(\tau(t)), y(t) = \left|x(t)\right|^{\beta} {\rm sgn}x(t), $$ \tau (t), \sigma (t)\in {C^1([t_0, \infty), (0, \infty))}, $$ p(t)\in{C^1([t_0, \infty), [0, 1])}, $$ f(t, u)\in{C([t_{0}, \infty)\times {\Bbb R} , {\Bbb R} )}, $$ g(t, u, v), \frac{\partial g(t, u, v)}{\partial t}, \frac{\partial g(t, u, v)}{\partial u} \in C([t_{0}, \infty)\times {\Bbb R} \times {\Bbb R} , {\Bbb R} ), $$ \frac{\partial g(t, u, v)}{\partial v} \in{C([t_{0}, \infty)\times {\Bbb R} \times \{{\Bbb R} \backslash\{ 0\}\}, {\Bbb R} )}, \; \beta>0. $

本文将用到下列条件

(A$ _1)\; \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)> 0, $$ \lim\limits_{t\to\infty}\tau(t) = \lim\limits_{t\to\infty}\sigma(t) = \infty. $

(A$ _2)\; \frac{f(t, u)}{u}\geq q(t)\; (u\neq0), $其中$ q(t)\in{C([t_0, \infty), [0, \infty))}. $

(A$ _3)\; vg(t, u, v)\geq 0, v\frac{\partial g(t, u, v)}{\partial t}\geq 0, v\frac{\partial g(t, u, v)}{\partial u}\geq 0, \frac{\partial g(t, u, v)}{\partial v}>0\; (v\neq 0), t\geq t_{0}, $且存在常数$ L_{2}\geq L_{1}>0, $使得

$ L_{1}r(t)|v|^{\alpha}\leq |g(t, u, v)|\leq L_{2}r(t)|v|^{\alpha}, $

其中$ r(t)\in C^{1}([t_{0}, \infty), (0, \infty)), \; r'(t)\geq 0, \; t\geq t_{0}, \alpha >0. $

(A$ _4)\; g(t, u, v) = a(t, u)r(t)|v|^{\alpha-1}v, \; $其中$ a(t, u)\in C^{1}([t_{0}, \infty)\times {\Bbb R} , (0, \infty)), $$ r(t)\in C^{1}([t_{0}, \infty), $$ (0, \infty)), $$ r'(t)\geq 0, \alpha >0, $且存在常数$ L_{2}\geq L_{1}>0, $使得

$ L_{1}\leq a(t, u)\leq L_{2}, \; \; u'(t)(a(t, u(t)))'\geq 0, u(t)\in C^{1}([t_{0}, \infty), {\Bbb R} ). $

函数$ x(t)\in{C^1([T_x, \infty), {\Bbb R} )}, $$ T_x\geq t_0 $称为方程(1.1)的解,若它满足：$ g(t, z(t), z'(t))\in C^1([T_x, \infty), {\Bbb R} ) $且在$ [T_x, \infty) $上满足方程(1.1).本文仅考虑方程(1.1)的非平凡解,即对一切$ T\geq T_x $使得$ \sup\{\left|x(t)\right|: t\geq T\}>0 $的解.称方程(1.1)的解是非振动的,如果它在$ [T_x, \infty) $上最终为正或最终为负,否则称之为振动的.方程(1.1)称为振动的,如果它的每一个解都是振动的.

本文在条件(A$ _{3}) $或(A$ _{4}) $成立时考虑以下两种情况

(1.2) $ \begin{equation} \int_{t_0}^{\infty} r^{-1/{\alpha}}(s) {\rm d}s = \infty \end{equation} $

和

(1.3) $ \begin{equation} \int_{t_0}^{\infty} r^{-1/{\alpha}}(s) {\rm d}s < \infty. \end{equation} $

近年来,二阶半线性泛函微分方程和Emden-Fowler型微分方程的振动性问题受到很大关注,例如可参见文献[1-25]及其引文.

Dzurina和Stavroulakis^[1], Sun和Meng^[2]研究了二阶半线性时滞微分方程

(1.4) $ \begin{equation} (r(t)\left|u'(t)\right|^{\alpha-1}u'(t))'+q(t)\left|u(\sigma(t))\right|^{\alpha-1}u(\sigma(t)) = 0, \quad \alpha>0 \end{equation} $

的振动性,文献[2]证明了如下结果.

定理1.1  设条件(A$ _{1}) $和$ (1.2) $式成立, $ r'(t)\geq 0, \alpha >0, $且

$ \int_{t_0}^{\infty} \bigg[R^{\alpha}(\sigma(t))q(t)-(\frac{\alpha}{\alpha+1})^{\alpha+1}\frac{\sigma'(t)}{R(\sigma(t))r^{1/\alpha}(\sigma(t))}\bigg] {\rm d}t = \infty, $

其中$ R(t) = \int_{t_0}^{t} r^{-1/\alpha}(s){\rm d}s $,则方程$ (1.4) $振动.

对二阶半线性中立型方程

(1.5) $ \begin{equation} (r(t)\left|z'(t)\right|^{\alpha-1}z'(t))'+f(t, x(\sigma(t))) = 0, \quad \alpha>0, \end{equation} $

其中$ z(t) = x(t)+p(t)x(\tau(t)), $$ f(t, u)\geq q(t)\left|u\right|^{\beta-1}u, u\neq 0, $文献[3]建立了当$ \beta = \alpha $时的Hill-Nehari型振动准则,文献[23]给出了更广泛的振动定理.

Erbe等^[4]和Li等^[5]研究了Emden-Fowler中立型方程

(1.6) $ \begin{equation} (r(t)[x(t)+p(t)x(\tau(t))]')'+q(t)\left|x(\sigma(t))\right|^{\beta} {\rm sgn} x(\sigma(t)) = 0. \end{equation} $

文献[4]给出了当$ \beta>1 $且$ (1.2) $式成立时的Philos型振动定理.文献[5]证明了当$ (1.3) $式成立时该方程振动的若干充分条件(相对于方程$ (1.1) $,这时都有$ \alpha = 1 $).

文献[6]研究了广义Emden-Fowler中立型方程

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

其中$ z(t) = x(t)+p(t)x(\tau(t)), $并利用$ \mathrm {Riccati} $方法和积分平均技巧建立了方程$ (1.7) $当$ \alpha\geq\beta>0 $时的若干振动准则,推广和改进了所列文献中的相关结果.

近来,文献[7]在$ \alpha\geq \beta $的条件下给出了方程$ (1.7) $不同于文献[6]的新振动条件,同时也证明了当$ \alpha\leq \beta $时方程$ (1.7) $的若干振动定理.

显然,方程(1.4)–(1.7)是方程$ (1.1) $的特殊情况.它们均源于二阶Emden-Fowler型线性方程,并逐步发展为半线性、超线性、次线性方程或非导数项抽象化的方程.但对导数项抽象化的研究仍未见到有本质性突破的成果,其原因显然在于此项非线性化拓展的难度.最近文献[24-25]在研究导数项的非线性化方面做了有益的尝试,该文分别研究了以下广义中立型Emden-Fowler方程的振动性.

(1.8) $ \begin{equation} (a(t)[x(t)+p(t)f(x(\delta(t)))]')'+q(t)g(\varphi(x(\sigma(t))))-r(t)h(\varphi(x(t-\tau))) = 0, \; \; t\geq t_{0}, \end{equation} $

其中$ \varphi(s) = |s|^{\alpha} {\rm sgn} s, \alpha>0, f\in C({\Bbb R} , {\Bbb R} ), \frac{u}{f(u)}\geq k>0, u\neq 0, $以及

(1.9) $ \begin{equation} (r(t)\left|z'(t)\right|^{\alpha-1}z'(t))'+p(t)\left|z'(t)\right|^{\alpha-1}z'(t)+q(t)\left|x(\sigma(t))\right|^{\beta-1} x(\sigma(t)) = 0, \; t\geq t_{0}, \end{equation} $

其中$ z(t) = x(t)+g(t)x(\tau(t)), r(t)\in C^{1}([t_{0}, \infty), {\Bbb R} ), p(t), q(t)\in C([t_{0}, \infty), [0, \infty)), \alpha>0, \beta>0. $

在保留Emden-Fowler型方程本质属性的前提下,本文将利用双Riccati变换法建立含一般非线性导数项方程的振动准则.所得结果与文献[24-25]的互不包含,同时推广、改进和统一了文献[1-7, 23]中的相关结果.本文所得结果既适合于线性方程$ (f(t, u) = q(t)u, g(t, u, v) = r(t)v^{\alpha}, \alpha = \beta = k = 1) $、半线性方程$ (f(t, u) = q(t)u, g(t, u, v) = r(t)|v|^{\alpha-1}v, \alpha = \beta) $,也适合于超线性方程$ (f(t, u) = q(t)u, g(t, u, v) = r(t)|v|^{\alpha-1}v, \beta>\alpha = 1) $和次线性方程$ (f(t, u) = q(t)u, g(t, u, v) = r(t)|v|^{\alpha-1}v, 0<\beta<\alpha = 1) $.随后,本文对每一个抽象定理给出具体的例子以示所得结果的意义.

定理2.1  设条件(A$ _1) $, (A$ _{2}) $及$ (1.2) $式成立, (A$ _3) $或(A$ _{4}) $成立,且存在$ \rho(t)\in C^1([t_0, \infty), $$ (0, \infty)) $$ (\rho'(t)\geq0) $和某个常数$ M>0 $,使得对任意常数$ m\in (0, M] $,有

(2.1) $ \begin{equation} \int_{t_0}^{\infty} \bigg[\rho(t)Q_1(t)-\frac{r(\theta(t))(\rho'(t))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\rho(t)\sigma'(t))^{\lambda}}\bigg] {\rm d}t = \infty, \end{equation} $

其中

(2.2) $ \begin{equation} Q_1(t) = (1-p(\sigma(t)))^{\beta}q(t), \; \; \theta(t) = \left\{\begin{array}{ll} \sigma(t), \; &\alpha \leq \beta, \\ t, \; \; \; \; \; &\alpha > \beta, \end{array}\right. \quad \lambda = \min\{\alpha, \beta\}, \end{equation} $

则方程$ (1.1) $振动.

证  假设$ x(t) $是方程(1.1)的非振动解.不失一般性,存在$ t_1\geq t_0 $,使得

$ x(t)>0, \quad x(\tau(t))>0, \quad x(\sigma(t))>0, \quad t\geq t_1 $

(对$ x(t)<0 $的情况可作类似分析).于是,有

$ z(t)\geq x(t)>0, \quad (g(t, z(t), z'(t)))'\leq 0, \quad t\geq t_1. $

因此, $ g(t, z(t), z'(t)) $非增,且由条件(A$ _{3}) $或(A$ _{4}) $,可以断言存在$ t_{2}\geq t_{1}, $使得$ z'(t)\geq 0, t\geq t_{2}. $若不然,则存在$ t_{3}\geq t_{2}, $使得$ z'(t)<0, t\geq t_{3}. $于是,存在常数$ k>0 $,使得$ g(t, z(t), z'(t))\leq -k, $$ t\geq t_{3}, $进而有$ -L_{2}r(t)(-z'(t)) ^{\alpha}\leq -k, \; t\geq t_{3}. $即有

$ z'(t)\leq -(\frac{k}{L_{2}r(t)})^{1/\alpha}, \; t\geq t_{3}. $

积分上式,并由$ (1.2) $式,得

$ z(t)\leq z(t_{3})-(\frac{k}{L_{2}})^{1/\alpha} \int_{t_3}^{t}(\frac{1}{r(t)})^{1/\alpha}{\rm d}s\rightarrow -\infty\; \; (t\rightarrow \infty), $

这与$ z(t)>0, t\geq t_{1} $矛盾.因此,存在$ T\geq t_2 $,使得$ x(t)\geq(1-p(t))z(t), \quad t\geq T. $故由方程(1.1)可得

(2.3) $ \begin{equation} (g(t, z(t), z'(t)))'+Q_1(t)z^\beta(\sigma(t))\leq 0, \; \; t\geq T, \end{equation} $

其中$ Q_1(t) $由(2.2)式定义.

此时,定义双Riccati变换函数

(2.4) $ \begin{equation} w(t) = \frac{g(t, z(t), z'(t))}{z^{\beta}(\sigma(t))}, \; \; v(t) = \frac{r(t)(z'(t))^\alpha}{z^\beta(\sigma(t))}, \quad t\geq T. \end{equation} $

则易知, $ w(t)\geq0, v(t)\geq0 $,且有$ L_{2}^{-1}w(t)\leq v(t)\leq L_{1}^{-1}w(t), \; t\geq T, $

$ L_{2}^{-\frac{\alpha+1}{\alpha}}w^{\frac{\alpha+1}{\alpha}}(t)\leq v^{\frac{\alpha+1}{\alpha}}(t)\leq L_{1}^{-\frac{\alpha+1}{\alpha}}w^{\frac{\alpha+1}{\alpha}}(t), \; t\geq T $

和

(2.5) $ \begin{equation} w'(t)\leq -Q_1(t)-\beta L_{1}\sigma '(t)z'(\sigma(t))\frac{r(t)(z'(t))^\alpha}{z^{\beta+1}(\sigma(t))}. \end{equation} $

又由$ (2.3) $式知$ g(t, z(t), z'(t)) $关于$ t\geq t_{1} $非增,从$ \sigma(t)\leq t, t\geq t_{1} $和条件(A$ _{3}) $ (或(A$ _{4}) $)知

$ L_{1}r(t)(z'(t))^{\alpha}\leq g(t, z(t), z'(t))\leq g(\sigma (t), z(\sigma(t)), z'(\sigma(t)))\leq L_{2}r(\sigma(t))(z'(\sigma(t)))^{\alpha}, t\geq T, $

从而又有

(2.6) $ \begin{equation} r^{1/\alpha}(\sigma(t))z'(\sigma(t)\geq (\frac{L_{1}}{L_{2}})^{1/\alpha}r^{1/\alpha}(t)z'(t), \; t\geq T. \end{equation} $

下面分三种情况讨论(2.5)式.

(ⅰ) $ \alpha = \beta. $将$ (2.6) $式代入(2.5)式得到

(2.7) $ \begin{equation} w'(t)\leq -Q_1(t)-\alpha \bigg (\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}\frac{\sigma'(t)}{r^{1/\alpha}(\sigma(t))} w^{\frac{\alpha+1}{\alpha}}(t), \quad t\geq T. \end{equation} $

(ⅱ) $ \alpha<\beta. $因为$ z(\sigma(t)) $是增函数,故存在常数$ M_1>0 $,使得当$ t\geq T $时,有

$ M_{1} = (z(\sigma(T)))^{\frac{\beta-\alpha}{\alpha}}\leq (z(\sigma(t)))^{\frac{\beta-\alpha}{\alpha}}, $

由$ (2.5) $式并注意到$ (2.6) $式,得

(2.8) $ \begin{eqnarray} w'(t)&\leq& -Q_1(t)-\beta L_{1}\bigg(\frac{L_{1}}{L_{2}}\bigg)^{\frac{1}{\alpha}}\frac{\sigma'(t)}{r^{1/\alpha}(\sigma(t))}[z(\sigma(t))]^\frac{\beta-\alpha}{\alpha} \bigg[\frac{r(t)(z'(t))^{\alpha}}{z^{\beta}(\sigma(t))}\bigg]^{\frac{\alpha+1}{\alpha}}\\ &\leq & -Q_1(t)-\bigg(\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}\frac{\alpha\sigma'(t)M_1}{r^{1/\alpha}(\sigma(t))}w^\frac{\alpha+1}{\alpha}(t).\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{eqnarray} $

(ⅲ) $ \alpha>\beta. $若条件(A$ _{3}) $成立,由$ (2.3) $式知$ (g(t, z(t), z'(t)))' = \frac{\partial g}{\partial t}+\frac{\partial g}{\partial u}v+\frac{\partial g}{\partial v}z''(t)\leq 0 $,以及当$ v>0 $时, $ \frac{\partial g}{\partial t}\geq 0, \frac{\partial g}{\partial u}v\geq 0 $和$ \frac{\partial g}{\partial v}>0, $知$ z''(t)\leq 0 $,即$ z'(t) $非增.若条件(A$ _{4}) $成立,又由$ (2.3) $式和$ (g(t, z(t), z'(t)))' = (a(t, z(t)))'r(t)(z'(t))^{\alpha}+a(t, z(t))(r(t)(z'(t))^{\alpha})'\leq 0 $及$ (a(t, z(t)))'\geq 0, $知$ (r(t)(z'(t))^{\alpha})'\leq 0 $.所以, $ r(t)(z'(t))^{\alpha} $及$ r^{1/\alpha}(t)z'(t) $非增,又$ r^{1/\alpha}(t) $非减,所以仍得$ z'(t) $非增.则$ z'(\sigma(t))\geq z'(t) $且存在常数$ M_2>0 $,使得

$ M_{2} = (z'(T))^{(\beta-\alpha)/\beta}\leq (z'(t))^{(\beta-\alpha)/\beta}, \quad t\geq T. $

代入$ (2.5) $式,得

(2.9) $ \begin{eqnarray} w'(t)&\leq& -Q_1(t)-\frac{\beta L_{1}\sigma'(t)}{r^{1/\beta}(t)}[z'(t)]^\frac{\beta-\alpha}{\beta} \bigg(\frac{r(t)(z'(t))^{\alpha}}{z^{\beta}(\sigma(t))}\bigg)^{\frac{\beta+1}{\beta}} \\ &\leq & -Q_1(t)-\bigg(\frac{L_{1}^{\beta}}{L_{2}^{\beta+1}}\bigg)^{\frac{1}{\beta}}\frac{\beta\sigma'(t)M_2}{r^{1/\beta}(t)}w^\frac{\beta+1}{\beta}(t), \; \; t\geq T. \end{eqnarray} $

联合(2.7)–(2.9)式,对任意$ \alpha>0 $和$ \beta>0 $,有

(2.10) $ \begin{equation} w'(t)\leq -Q_1(t)-\frac{\lambda M_{0}\sigma'(t)}{r^{1/\lambda}(\theta(t))}w^{(\lambda+1)/\lambda}(t), \quad t\geq T, \end{equation} $

其中$ \lambda, \theta(t) $和$ Q_1(t) $由(2.2)式定义,

$ M_{0} = \min \bigg\{\bigg (\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}, M_{1}\bigg(\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}, M_{2}\bigg(\frac{L_{1}^{\beta}}{L_{2}^{\beta+1}}\bigg)^{\frac{1}{\beta}}, M\bigg\}, \quad M>0 $

由(2.1)式确定.

对(2.10)式乘以(2.1)式中的$ \rho(t) $,在$ [T, t] $上分部积分,产生

$ \int_{T}^{t} \rho(s)Q_1(s){\rm d}s \leq \rho(T)w(T)+ \int_{T}^{t} \bigg[\rho'(s)w(s)-\frac{\lambda M_{0}\rho(s)\sigma'(s)}{r^{1/\lambda}(\theta(s))}w^{(\lambda+1)/\lambda}(s)\bigg] {\rm d}s. $

注意到上式右端积分中$ \rho'(s)\geq 0, s\geq T $并利用不等式

(2.11) $ \begin{equation} Bu-Au^{(\lambda+1)/\lambda} \leq \frac{\lambda^{\lambda}}{(\lambda+1)^{\lambda+1}}\frac{B^{\lambda+1}}{A^\lambda}, \lambda>0, A>0, B\geq 0, \end{equation} $

容易得到

(2.12) $ \begin{equation} \int_{T}^{t} \bigg[\rho(s)Q_1(s)-\frac{r(\theta(s))(\rho'(s))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(M_{0}\rho(s)\sigma'(s))^\lambda} \bigg] {\rm d}s \leq\rho(T)w(T), \end{equation} $

其中$ M_{0}\in (0, M]. $在(2.12)式中令$ t\to\infty $,立即看到这与(2.1)式成立矛盾.定理2.1证毕.

在定理2.1中取$ \rho(t)\equiv 1 $,则方程(1.1)有如下的Leighton型振动准则.

推论2.1  设$ \alpha>0, \beta>0 $,条件(A$ _{1} $), (A$ _{2}) $和$ (1.2) $式成立,又条件(A$ _{3}) $ (或(A$ _{4})) $与

(2.13) $ \begin{equation} \int_{t_0}^{\infty} [1-p(\sigma(t))]^\beta q(t){\rm d}t = \infty \end{equation} $

成立,则方程$ (1.1) $振动.

显然,关于线性方程

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

的Leighton振动定理^[21]是推论$ 2.1 $的特例.

易见,当$ \alpha = \beta >0, p(t)\equiv 0, f(t, u) = q(t)u, g(t, u, v) = r(t)|v|^{\alpha-1}v, r'(t)\geq 0 $时方程$ (1.1) $退化为半线性方程$ (1.4) $.于是,便得如下推论.

推论2.2  设条件(A$ _{1}) $和$ (1.2) $式成立, $ r'(t)\geq 0, \alpha>0, $且存在$ \rho(t)\in {C^1([t_0, \infty), (0, \infty))} $$ (\rho'(t)\geq0), $使得

$ \int_{t_0}^{\infty} \bigg[\rho(t)q(t)-\frac{r(\sigma(t))(\rho'(t))^{\alpha+1}}{(\alpha+1)^{\alpha+1}(\rho(t)\sigma'(t))^\alpha} \bigg] {\rm d}t = \infty, $

则方程$ (1.4) $振动.

不难看出,定理$ 1.1 $是推论$ 2.2 $取$ \rho(t) = R^\alpha(\sigma(t)), $其中$ R(t) = \int_{t_0}^{t} r^{-1/\alpha}(s){\rm d}s $时的特例.

对方程$ (1.1) $,若取$ \alpha = 1, f(t, u) = q(t)u, g(t, u, v) = r(t)v, $则成为Emden-Fowler中立型微分方程$ (1.6) $,于是又得如下推论.

推论2.3  设(A$ _{1}) $和$ (1.2) $式成立, $ r'(t)\geq 0, $且存在$ \rho(t)\in {C^1([t_0, \infty), (0, \infty))}\; (\rho'(t)\geq0) $和某个常数$ M>0 $,使得对任意常数$ m\in (0, M], $有

$ \int_{t_0}^{\infty} \bigg[\rho(t)Q_{1}(t)-\frac{r(\sigma(t))(\rho'(t))^2}{4m\rho(t)\sigma'(t)} \bigg] {\rm d}t = \infty, \quad \beta>1 $

或

$ \int_{t_0}^{\infty} \bigg[\rho(t)Q_{1}(t)-\frac{r(t)(\rho'(t))^{\beta+1}}{(\beta+1)^{\beta+1}(m\rho(t)\sigma'(t))^\beta} \bigg] {\rm d}t = \infty, \quad 0<\beta<1 $

成立,其中$ Q_{1}(t) $由$ (2.2) $式定义,则方程$ (1.6) $振动.

当$ g(t, u, v) = r(t)|v|^{\alpha-1}v $或$ a(t, u)\equiv 1 $时,方程$ (1.1) $成为方程$ (1.5), $且显然(A$ _{3}) $和(A$ _{4}) $均成立,其中$ r'(t)\geq 0, \alpha>0, \beta>0, L_{1} = L_{2} = 1. $于是有如下推论.

推论2.4  设(A$ _{1} $), (A$ _{2}) $和$ (1.2) $式成立, $ \alpha>0, \beta>0, r'(t)\geq 0 $且存在$ \rho(t)\in $$ C^{1}([t_{0}, \infty), $$ (0, \infty)) $$ (\rho'(t)\geq0) $和某个常数$ M>0 $,使得对任意常数$ m\in (0, M] $有$ (2.1) $和$ (2.2) $式成立,则方程$ (1.5) $振动.

显然,由推论2.4知,本文定理2.1已包含并改进了文献[23]的定理2.2,因为那里要求相当于$ m>0 $为任意常数.

注2.1  综上所述,本文定理$ 2.1 $适应于上述线性方程、半线性方程、超线性方程、次线性方程和非线性方程,推广和改进了文献[6]的定理$ 2.1 $和文献[23]的定理$ 2.2 $,因为文献[6]仅考虑了$ \alpha \geq \beta $的情况,而且不能包含经典的振动结果.同时也统一了文献[7]的定理$ 2.1 $和定理$ 3.1 $,因为它们都是本文定理$ 2.1 $取$ f(t, u) = q(t)u $或$ g(t, u, v) = r(t)|v|^{\alpha-1}v $时的特例.需要指出的是,文献[7]定理$ 2.1 $的充分条件$ (2.2) $式中取“$ K = (z'(t_{1}))^{(\beta-\alpha)/\beta} $”是不妥的,因为$ z(t) $是方程中含未知解$ x(t) $的中立项,所以应用时无法验证该条件$ (2.2) $是否成立.此外,这里不得不指出,现有文献[7, 24-27]中较普遍存在的问题是,关于振动定理充分条件中涉及的常数$ M, K, c $等往往未说明具有一定的“任意性”.因为用反证法证明定理时,推得存在某个常数$ M_0>0 $,使得与已知条件矛盾,则该条件中对应的常数$ M>0 $就必须是任意的,否则,就会出现逻辑错误.

例2.1  考虑方程

(${{\rm{E}}_1}$) $ \bigg[t^{\lambda/4}(\pi+\arctan z(t))\frac{2+|z'(t)|^{\alpha}}{1+|z'(t)|^{\alpha}}\left|z'(t)\right|^{\alpha-1}z'(t) \bigg]'+t^{-(1+\lambda /2)}[ky(\mu t)+ly^3(\mu t)] = 0, $

其中$ z(t) = x(t)+\frac{1}{2}x(t-1), y(t) = {\left|x(t)\right|}^\beta {\rm sgn} x(t), $$ \alpha>0, \; \beta>0, \; \lambda = \min\{\alpha, \beta\}, k>0, $$ l\geq 0, $$ 0<\mu<1 $为常数, $ t\geq t_0\geq 1. $

对应于方程$ (1.1) $和定理$ 2.1 $,此时$ g(t, u, v) = r(t)(\pi+\arctan u)\frac{2+|v|^{\alpha}}{1+|v|^{\alpha}}|v|^{\alpha-1}v, r(t) = t^{\lambda/4}. $于是,取$ L_{1} = \pi/2, L_{2} = 3\pi $,则有

$ \; L_{1}r(t)|v|^{\alpha} \leq |g(t, u, v)|\leq L_{2}r(t)|v|^{\alpha}, \; v\in R, t\geq t_{0}, $

$ v\frac{\partial g}{\partial t} = \frac{\lambda}{4} t^{(\lambda/4)-1}(\pi+\arctan u)\frac{2+|v|^{\alpha}}{1+|v|^{\alpha}}|v|^{\alpha +1}\geq 0, v\frac{\partial g}{\partial u} = \frac{r(t)}{1+u^{2}}\frac{2+|v|^{\alpha}}{1+|v|^{\alpha}}|v|^{\alpha+1}\geq 0, $

$ \frac{\partial g}{\partial v} = t^{\lambda/4}(\pi +\arctan u)\frac{2\alpha|v|^{\alpha}+2\alpha|v|^{2\alpha}+\alpha|v|^{3\alpha}}{|v|(1+|v|^{\alpha})^{2}}>0, v\neq 0. $

即$ g(t, u, v) $满足条件(A$ _{3}) $.又知

$ f(t, u)\geq kt^{-(1+\lambda/2)}u, \; t\geq t_{0}\geq 1 $

满足条件(A$ _{2}) $.再取$ m>0 $为任意常数且由

$ p(t) = \frac{1}{2}, q(t) = \frac{k}{t^{1+\lambda/2}}, \tau(t) = t-1, \sigma(t) = \mu t, \; Q_1 = k(\frac{1}{2})^\beta\frac{1}{t^{1+\lambda/2}}, \rho(t) = t^{\lambda/2}, $

易得

$ \begin{eqnarray*} && \int_{t_0}^{\infty}\bigg [\rho(t)Q_1(t)-\frac{r(\theta(t))(\rho'(t))^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\rho(t)\sigma'(t))^\lambda} \bigg] {\rm d}t\\ & = & \int_{t_0}^{\infty}\bigg [\frac{(1/2)^\beta k}{t}-\frac{((\theta(t))^{\lambda/4}(\lambda/2)^{\lambda+1}t^{(\lambda-2)(\lambda+1)/2}}{(\lambda+1)^{\lambda+1}(\mu m)^\lambda t^{\lambda^{2}/2}}\bigg] {\rm d}t \\ &\geq& \int_{t_0}^{\infty}\bigg[\frac{(1/2)^\beta k}{t}-\frac{(\lambda/2)^{\lambda+1}}{(\lambda+1)^{\lambda+1}(\mu m)^{\lambda}t^{1+\lambda/4}}\bigg]{\rm d}t = \infty. \end{eqnarray*} $

故由定理$ 2.1 $知方程(E$ _1) $振动.

注2.2  易见方程(E$ _{1}) $不满足条件(A$ _{4}), $但去掉因式$ (2+|z'(t)|^{\alpha}) /(1+|z'(t)|^{\alpha}) $后便满足条件(A$ _{4}), $故方程仍振动,而这时不满足条件(A$ _{3}), $因此条件(A$ _{3}) $与条件(A$ _{4}) $互不包含.又易知去掉两个因式$ \pi +\arctan z(t) $和$ (2+|z'(t)|^{\alpha})/(1+|z'(t)|^{\alpha}) $后方程(E$ _{1}) $均满足条件(A$ _{3}) $和(A$ _{4}), $故由定理$ 2.1 $知方程(E$ _{1}) $仍振动,但不难知道文献[1-25]中的振动定理均不能判定在这两种情况下方程(E$ _1) $的振动性.

下面再给出当$ (2.13) $式不成立时方程$ (1.1) $的另一个振动准则.此时,对应于$ (2.2) $式和$ (2.10) $式,下设

(2.14) $ \begin{equation} \overline{Q}_1(t) = \int_{t}^{\infty} Q_1(s){\rm d}s, \quad A_{0}(t) = \frac{\lambda M_{0}\sigma'(t)}{r^{1/\lambda}(\theta(t))}, \quad t\geq T, \end{equation} $

立即可得如下振动准则.

定理2.2  设条件(A$ _1 $), (A$ _{2}) $及$ (1.2) $式成立,且(A$ _3) $或(A$ _{4}) $成立.若存在某个常数$ M>0 $,使得对任意常数$ m\in (0, M] $,有

(2.15) $ \begin{equation} \liminf\limits_{t\to\infty} \frac{1}{\overline{Q}_1(t)}\int_{t}^{\infty} \overline{Q}_1^{{(\lambda+1)}/\lambda}(s)\frac{\sigma'(s)}{r^{1/\lambda}(\theta(s))}{\rm d}s>\frac{1}{m(\lambda+1)^{(\lambda+1)/\lambda}}, \end{equation} $

其中$ \overline{Q}_1(t) $由$ (2.14) $式定义, $ \lambda $由$ (2.2) $式定义,则方程$ (1.1) $振动.

证  假设$ x(t) $是方程(1.1)的非振动解.不妨假设$ x(t) $最终为正,类似于定理2.1的证明,定义函数$ w(t) $同(2.4)式,则存在某个正数$ M_{0}\in (0, M] $使得$ (2.10) $式成立,即

(2.16) $ \begin{equation} w'(t)\leq -Q_1(t)-A_{0}(t)w^{(\lambda+1)/\lambda}(t), \quad t\geq T, \end{equation} $

其中$ A_{0}(t) $由$ (2.14) $式定义.积分上式可得

(2.17) $ \begin{equation} w(t)\geq \overline{Q}_1(t)+ \int_{t}^{\infty} A_{0}(s)w^{(\lambda+1)/\lambda}(s){\rm d}s. \end{equation} $

其中$ \overline{Q}_1(t) $由(2.14)式定义.由(2.16)式知$ w(t) $非增,故有界.因此,上式右端积分收敛,用$ \overline{Q}_1(t) $除(2.17)式产生

(2.18) $ \begin{equation} \frac{w(t)}{\overline{Q}_1(t)}\geq 1+\frac{1}{\overline{Q}_1(t)}\int_{t}^{\infty} \frac{\lambda M_0\sigma '(s)}{r^{1/\lambda}(\theta(s))}\overline{Q}_1^{(\lambda+1)/\lambda}(s) \bigg[\frac{w(s)}{\overline{Q}_1(s)}\bigg]^{(\lambda+1)/\lambda} {\rm d}s, \quad t\geq T. \end{equation} $

另一方面,对于这个$ M_0\in (0, M] $,由(2.15)式知,存在常数$ \xi>0 $,使得

(2.19) $ \begin{equation} \liminf\limits_{t\to\infty} \frac{1}{\overline{Q}_1(t)} \int_{t}^{\infty} [\overline{Q}_1(s)]^{(\lambda+1)/\lambda}\frac{\lambda M_0\sigma '(s)}{r^{1/\lambda}(\theta(s))}{\rm d}s >\lambda M_0\xi >\frac{\lambda }{(\lambda+1)^{(\lambda+1)/\lambda}}. \end{equation} $

令$ \eta = \inf_{t \geq T}\{w(t)/\overline{Q}_1(t)\}, $则$ \eta \geq 1 $.由(2.18)式,有

(2.20) $ \begin{equation} \eta \geq 1+ \lambda M_0\xi\eta^{(\lambda+1)/\lambda}. \end{equation} $

但是,利用不等式(2.11),其中令$ B = 1 $, $ A = \lambda M_0\xi, u = \eta $及(2.19)式,可得

(2.21) $ \begin{equation} \eta- \lambda M_0\xi\eta^{(\lambda+1)/\lambda} \leq \frac{\lambda^\lambda}{(\lambda+1)^{\lambda+1}} \frac{1}{(\lambda M_0\xi)^\lambda} < 1. \end{equation} $

显然, (2.21)式与(2.20)式矛盾.故方程(1.1)无非振动解.定理2.2证毕.

例2.2  考虑带超线性或次线性项的中立型广义Emden-Fowler方程

(${{\rm{E}}_2}$) $ [(\pi+\arctan z^{3}(t))t^{1/4}z'(t)]' +t^{-3/2}\left|x(\mu t)\right|^\beta {\rm sgn} x(\mu t) = 0, \quad t\geq t_0, $

其中$ z(t) = x(t)+\frac{1}{2}x(t/2), \beta>0 $, $ 0<\mu<1 $.

这里有$ \; \; a(t, u) = \pi+\arctan z^{3}, $当$ \beta \geq 1 $ (即超线性)时显然有

$ \frac{\pi}{2}\leq a(t, z)\leq \frac{3\pi}{2}, \; \; (a(t, z(t)))'z'(t) = \frac{3z^{2}(t)(z'(t))^{2}}{1+z^{6}(t)}\geq 0, $

$ \alpha = 1, \; \lambda = 1, \; r(t) = t^{1/4}, \; p(t) = 1/2, \; q(t) = t^{-3/2}, \; \sigma(t) = \mu t, $

$ Q_1(t) = (1/2)^\beta t^{-3/2}, \quad \overline{Q}_1(t) = (1/2)^{\beta-1}t^{-1/2}, \quad r(\theta(t)) = [\theta(t)]^{1/4} = (\mu t)^{1/4}. $

于是条件(A$ _{1} $), (A$ _{2}) $, (A$ _{4}) $和$ (1.2) $式成立,对任意常数$ m>0 $, (2.15)式右边的$ 1/m(\lambda+1)^{(\lambda+1)/\lambda} = 1/4m. $现有$ 1/\overline{Q}_1(t) = 2^{\beta-1}t^{1/2} $,从而$ (2.15) $式的左边为

$ \liminf\limits_{t\to\infty} 2^{\beta-1}t^{1/2} \int_{t}^{\infty} \frac{\mu }{(\mu s)^{1/4}} (\frac{1}{4})^{\beta-1} \frac{1}{s}{\rm d}s = \infty. $

因此$ (2.15) $式成立(当$ 0<\beta<1 $ (即次线性)时, $ \lambda = \beta $,也可得(2.15)式成立).故由定理$ 2.2 $知方程(E$ _2) $振动.

注2.3  文献[3]研究了方程$ (1.1) $中$ g(t, u, v) = r^{\alpha}(t)|v|^{\alpha-1}v, \alpha = \beta $的情况,文献[7]讨论了方程$ (1.1) $中$ g(t, u, v) = r^{\alpha}(t)|v|^{\alpha-1}v, f(t, u) = q(t)u, \beta \geq\alpha $的情况.本文定理$ 2.2 $适用于方程$ (1.1) $中含中立项$ g(t, z(t), z'(t)) $具有更一般非线性性态和任意$ \alpha >0 $和$ \beta>0 $的情况.

下面建立方程$ (1.1) $的Philos^[22]型振动准则.

令$ D = \{(t, s): t\geq s\geq t_0\}, D_0 = \{(t, s): t > s\geq t_0\}, $设$ H(t, s)\in C(D, {\Bbb R} ) $满足：$ H(t, t) = 0, t\geq t_0, H(t, s)>0, (t, s)\in D_0 $.又设$ h(t, s)\in C(D_0, {\Bbb R} ), \rho(t) \in C^1([t_0, \infty), (0, \infty)) $\\ $ (\rho'(t)\geq0), $使得

(2.22) $ \begin{equation} \frac{\partial H(t, s)}{\partial s}+\frac{\rho'(s)}{\rho(s)}H(t, s) = -h(t, s)H^{\lambda/(\lambda+1)}(t, s), \end{equation} $

其中$ \lambda = \min\{\alpha, \beta\}. $若$ H(t, s) $具有上述性质,则称$ H\in X. $

  定理2.3  设(A$ _1 $), (A$ _2) $及$ (1.2) $式成立, (A$ _{3}) $或(A$ _{4}) $成立,且$ H\in X $.若存在某个常数$ M>0 $和$ T\geq t_{0} $,使得对于任意常数$ m\in (0, M] $,有

(2.23) $ \begin{equation} \limsup\limits_{t\to\infty} \frac{1}{H(t, T)} \int_{T}^{t} \bigg[H(t, s)\rho(s)Q_1(s)-\frac{\rho(s)r(\theta(s))\left|h(t, s)\right|^{\lambda+1}}{(\lambda+1)^{\lambda+1}(m\sigma'(s))^\lambda}\bigg] {\rm d}s = \infty, \end{equation} $

其中$ Q_1(t), \theta(t), \lambda $的定义同$ (2.2) $式,则方程$ (1.1) $振动.

证  假若方程(1.1)存在非振动解$ x(t) $.不失一般性,可设$ x(t) $最终为正(对最终为负的情况可作类似的分析).定义函数$ w(t) $同(2.4)式,利用定理2.1证明中的方法,可以得到不等式(2.10).用$ H(t, s)\rho(s) $乘(2.10)式,并从$ T $到$ t $积分,立即可得

(2.24) $ \begin{eqnarray} \int_{T}^{t} H(t, s)\rho(s)Q_1(s){\rm d}s & \leq& -\int_{T}^{t} H(t, s)\rho(s) w'(s){\rm d}s - \int_{T}^{t} H(t, s)\rho(s)A_{0}(s)w^{(\lambda+1)/\lambda}(s){\rm d}s \\ & = &B(t)+\int_{T}^{t}\bigg[\frac{\partial H(t, s)}{\partial s}+\frac{\rho'(s)}{\rho(s)}H(t, s) \bigg]\rho(s)w(s){\rm d}s\\ &&-\int_{T}^{t} H(t, s)\rho(s)A_{0}(s)w^{(\lambda+1)/\lambda}(s){\rm d}s, \end{eqnarray} $

其中$ B(t) = H(t, T)\rho(T)w(T) $, $ A_{0}(t) $由(2.14)式定义.利用(2.22)和(2.24)式,产生

(2.25) $ \begin{eqnarray} & &\int_{T}^{t} H(t, s)\rho(s)Q_1(s){\rm d}s \\ &\leq & B(t)+\int_{T}^{t} [\left|h(t, s)\right|H^{\lambda/(\lambda+1)}(t, s)\rho(s)w(s)- H(t, s)\rho(s)A_{0}(s)w^{(\lambda+1)/\lambda}(s) ]{\rm d}s. \end{eqnarray} $

在(2.25)式右端积分中代入(2.14)式并利用(2.11)式,可以得到

(2.26) $ \begin{equation} \frac{1}{H(t, T)} \int_{T}^{t} \bigg [H(t, s)\rho(s)Q_1(s)-\frac{r(\theta(s))\rho(s)}{(\lambda+1)^{\lambda+1}(M_{0} \sigma'(s))^\lambda} \left|h(t, s)\right|^{\lambda+1} \bigg]{\rm d}s \leq \rho(T)w(T), \end{equation} $

其中$ M_{0}\in (0, M] $.在上式中令$ t \to \infty $取上极限,立即看到与(2.23)式矛盾.定理2.3证毕.

推论2.5  设条件(A$ _1 $), (A$ _2) $及$ (1.2) $式成立,条件(A$ _{3}) $或(A$ _{4}) $成立,且$ H\in X $.若存在某个$ T\geq t_{0} $,使得

(2.27) $ \begin{equation} \limsup\limits_{t\to\infty} \frac{1}{H(t, T)} \int_{T}^{t} H(t, s)\rho(s)Q_1(s){\rm d}s = \infty, \end{equation} $

(2.28) $ \begin{equation} \limsup\limits_{t\to\infty} \frac{1}{H(t, T)} \int_{T}^{t} \frac{r(\theta(s))\rho(s)}{(\sigma'(s))^\lambda} \left|h(t, s)\right|^{\lambda+1} {\rm d}s<\infty, \end{equation} $

则方程$ (1.1) $振动.

注2.4  本文定理$ 2.3 $推广和改进了文献[6]的定理$ 2.2 $,文献[4]的定理$ 2.1 $,文献[13]的定理$ 5 $和文献[22]的定理$ 1 $.

例2.3  考虑超线性中立型方程

(${{\rm{E}}_3}$) $ \bigg[t(1+\frac{1}{1+\sqrt{1+(z'(t))^{2}}})z'(t)\bigg]' +t^\xi\bigg [\xi \frac{(2-\cos t)}{t}+\sin t\bigg]\bigg[y(\frac{t}{2})+2y^3(\frac{t}{2})\bigg] = 0, $

其中$ z(t) = x(t)+\frac{1}{2}x(t-1), y(t) = \left|x(t)\right|^\beta {\rm sgn}x(t), \beta>1, \xi>0 $为常数, $ t\geq 1. $

对应于方程$ (1.1) $和推论$ 2.5 $,这里有

$ \lambda = \alpha = 1, r(t) = t, p(t) = \frac{1}{2}, \sigma(t) = \frac{t}{2}, \tau(t) = t-1, Q_1(t) = (\frac{1}{2})^\beta t^\xi [\xi \frac{2-\cos t}{t}+\sin t], $

$ g(t, u, v) = t(1+\frac{1}{1+\sqrt{1+v^{2}}})v, \; r(t)|v|\leq |g(t, u, v)|\leq \frac{3}{2}r(t)|v|, $

$ \; v\frac{\partial g}{\partial t} = \frac{2+\sqrt{1+v^{2}}}{1+\sqrt{1+v^{2}}}v^{2}\geq 0, \; \frac{\partial g}{\partial u} = 0, \frac{\partial g}{\partial v} = \frac{t(2+v^{2}+\sqrt{1+v^{2}})}{1+v^{2}+\sqrt{1+v^{2}}}>0. $

从而条件(A$ _{1} $)–(A$ _{3}) $及$ (1.2) $式均成立.取$ \rho(t)\equiv 1, H(t, s) = (t-s)^2, $则有$ h(t, s) = 2, $以及

(2.29) $ \begin{eqnarray} \int_{1}^{t} Q_1(s){\rm d}s & = &(\frac{1}{2})^\beta \int_{1}^{t} {\rm d}[s^\xi (2-\cos s)] \\ & = & (\frac{1}{2})^\beta [t^\xi (2-\cos t)-(2-\cos 1)] \geq (\frac{1}{2})^\beta (t^\xi-2). \end{eqnarray} $

由$ (2.29) $式不难得到,当$ t>1 $时有

$ \begin{eqnarray*} &&\frac{1}{(t-1)^2} \int_{1}^{t} (t-s)^2\rho(s)Q_1(s){\rm d}s\\ & = &\frac{1}{(t-1)^2} \int_{1}^{t} 2(t-s) \bigg(\int_{1}^{s} Q_1(u){\rm d}u\bigg){\rm d}s\\ &\geq &\frac{(1/2)^{\beta-1}}{(t-1)^2} \int_{1}^{t} (t-s)(s^\xi-2){\rm d}s\\ & = &\frac{(1/2)^{\beta-1}}{(t-1)^2} \bigg[\frac{t^{\xi+2}}{(\xi+1)(\xi+2)}-t^{2}+\frac{2\xi+1}{\xi+1}t-\frac{\xi+1}{\xi+2}\bigg]\rightarrow \infty \; \; (t \rightarrow \infty), \end{eqnarray*} $

于是知$ (2.27) $式成立.另一方面,又有

$ \frac{1}{(t-1)^2} \int_{1}^{t} \frac{r(\theta(s))\rho(s)}{(\sigma'(s))^\lambda} \left|h(t, s)\right|^{\lambda+1} {\rm d}s = \frac{1}{(t-1)^2} \int_{1}^{t} 4s{\rm d}s = \frac{2(t^{2}-1)}{(t-1)^{2}} \rightarrow 2\; \; (t\rightarrow \infty), $

因此知$ (2.28) $式成立.由推论$ 2.5 $知方程(E$ _3) $振动.

称方程$ (1.1) $为非正则的,若$ (1.3) $式成立.现在考虑非正则条件下方程$ (1.1) $的振动性.

定理2.4  设条件(A$ _1 $), (A$ _{2} $), (A$ _4 $), (1.3)式和$ (2.1) $式成立,又设

(2.30) $ \begin{equation} p'(t)\geq 0, \quad \tau'(t)\geq 0, \quad \sigma(t) \leq \tau(t). \end{equation} $

若存在某个常数$ M>0 $,使得对于任意常数$ m\geq M $,有

(2.31) $ \begin{equation} \int_{t_0}^{\infty} [\underline{R}^\mu(t)Q_2(t)-\frac{m}{\underline{R}(t)r^{1/\alpha}(t)}] {\rm d}t = \infty, \end{equation} $

其中$ \mu = \max\{\alpha, \beta\}, \underline{R}(t) = \int_{t}^{\infty} r^{-1/\alpha}(s){\rm d}s, Q_2(t) = q(t)(1-p(t))^\beta $,则方程$ (1.1) $振动.

证  假若方程(1.1)存在非振动解$ x(t) $.不失一般性,可设$ x(t) $最终为正.于是最终有$ [a(t, z(t))r(t)\left|z'(t)\right|^{\alpha-1}z'(t)]'\leq 0 $.因此, $ z'(t) $最终为正或最终为负.若前者成立,则与条件(2.1)式矛盾.于是可设存在$ t_1 \geq t_0 $,使得$ z'(t)<0, $$ t\geq t_1. $类似于文献[5]定理2.1或文献[6]定理2.5的证明,知$ x'(t) $的符号最终非正,即存在$ t_{2}\geq t_{1} $使得$ x'(t)\leq0\; (t\geq t_{2}). $若不然, $ x'(t) $的符号最终为正,则必存在$ t_{3}\geq t_{1} $,使得当$ t\geq t_{3} $时,有$ x'(t)> 0, x'(\tau(t))> 0, $从而有

$ z'(t) = x'(t)+p'(t)x(\tau (t))+p(t)x'(\tau (t))\tau'(t)\geq 0, t\geq t_{3}, $

这与$ z'(t)<0, \; (t\geq t_{1}) $矛盾.所以有

$ z(t)\leq x(\tau(t))+p(t) x(\tau(t)) = (1+p(t)) x(\tau(t)). $

因为$ x'(t)\leq0 , t\geq t_{2} $,于是有

$ x(\sigma(t))\geq x(\tau(t)) \geq \frac{z(t)}{1+p(t)} \geq (1-p(t))z(t), \; \; t\geq t_{2}. $

将上式代入方程(1.1),得

$ [a(t, z(t))r(t)|z'(t)|^{\alpha-1}z'(t)]'+ q(t)(1-p(t))^\beta z^\beta(t) \leq 0, \; \; t\geq t_{2}. $

注意到$ Q_{2}(t) $的定义,上式又成为

(2.32) $ \begin{equation} [a(t, z(t))r(t)(-z'(t))^\alpha]'- Q_2(t)z^\beta(t)\geq 0, \; \; t\geq t_2. \end{equation} $

再定义双Riccati变换函数

(2.33) $ \begin{equation} \overline{w}(t) = \frac{a(t, z(t))r(t)(-z'(t))^\alpha}{z^\beta(t)}, \; \; \overline{v}(t) = \frac{r(t)(-z'(t))^\alpha}{z^\beta(t)}, \quad t\geq t_2. \end{equation} $

则有$ \overline{w}(t)>0, \overline{v}(t)>0 $,且由条件(A$ _{4}) $及$ (2.32) $式有

$ L_{1}\overline{v}(t)\leq \overline{w}(t)\leq L_{2}\overline{v}(t), \; \; L_{2}^{-1}\overline{w}(t)\leq \overline{v}(t)\leq L_{1}^{-1}\overline{w}(t), \; \; t\geq t_{2}, $

(2.34) $ \begin{equation} \; \overline{w}'(t) \geq Q_2(t)+\frac{\beta L_{1}r(t)(-z'(t))^{\alpha+1}}{z^{\beta+1}(t)}, \quad t\geq t_2. \end{equation} $

下面分三种情况讨论(2.34)式.

(ⅰ) $ \alpha = \beta $.则(2.34)式成为

(2.35) $ \begin{equation} \overline{w}'(t) \geq Q_2(t)+\alpha (L_{1}^{\alpha}/L_{2}^{\alpha+1})^{1/\alpha}r^{-1/\alpha}(t)\overline{w}^{\frac{\alpha+1}{\alpha}}(t), \quad t\geq t_2. \end{equation} $

(ⅱ) $ \alpha > \beta $.因$ z(t) $是减函数,故$ [z(t)]^{(\beta-\alpha)/\alpha}(t\geq t_{2}) $是增函数.记$ \overline{M}_{1} = (z(t_{2}))^{(\beta-\alpha)/\alpha} $,于是由(2.34)式得

(2.36) $ \begin{eqnarray} \overline{w}'(t)&\geq& Q_2(t)+ \beta (\frac{L_{1}^{\alpha}}{L_{2}^{\alpha+1}})^{1/\alpha}r^{\frac{-1}{\alpha}}(t)[z(t)]^{\frac{\beta-\alpha}{\alpha}}\overline{w}^{\frac{\alpha+1}{\alpha}}(t) \\ & \geq& Q_{2}(t)+\beta \overline{M}_{1} (\frac{L_{1}^{\alpha}}{L_{2}^{\alpha+1}})^{1/\alpha}r^{\frac{-1}{\alpha}}(t)\overline{w}^{\frac{\alpha+1}{\alpha}}(t), \; \; t\geq t_2. \end{eqnarray} $

(ⅲ) $ \alpha < \beta $.由(2.33)和(2.34)式知,当$ t\geq t_{2} $时,有

$ \begin{eqnarray*} \overline{w}'(t)&\geq &Q_{2}(t)+\beta L_{1}r(t)(-z'(t))^{\alpha+1}/z^{\beta+1}(t) \\ &\geq &Q_{2}(t)+\beta (\frac{L_{1}^{\beta}}{L_{2}^{\beta+1}})^{1/\beta} r^{-1/\alpha}(t)[r^{1/\alpha}(t)(-z'(t))]^{(\beta-\alpha)/\beta}\overline{w}^{(\beta+1)/\beta}(t). \end{eqnarray*} $

又由(2.32)式知,当$ t\geq t_{2} $时,有

$ [a(t, z(t))r(t)(-z'(t))^{\alpha}]' = [a(t, z(t))]'r(t)(-z'(t))^{\alpha}+a(t, z(t))[r(t)(-z'(t))^{\alpha}]'\geq 0. $

由条件(A$ _{4}) $知$ [a(t, z(t))]'\leq 0, $又因$ r(t)(-z'(t))^{\alpha}\geq 0, a(t, z(t))>0, $故$ [r(t)(-z'(t))^{\alpha}]'\geq 0, $从而$ r^{1/\alpha}(t)(-z'(t)) $是增函数,记$ \overline{M}_{2} = [r^{1/\alpha}(t_{2})(-z'(t_{2})]^{(\beta-\alpha)/\beta}, $则有

(2.37) $ \begin{equation} \overline{w}'(t) \geq Q_2(t)+ \beta \overline{M}_2 (\frac{L_{1}^{\beta}}{L_{2}^{\beta+1}})^{\frac{1}{\beta}}r^{\frac{-1}{\alpha}}(t) \overline{w}^{\frac{\beta+1}{\beta}}(t), \; \; t\geq t_{2}. \end{equation} $

联合(2.35)–(2.37)式,对任意的$ \alpha>0, \beta>0 $,有

(2.38) $ \begin{equation} \overline{w}'(t) \geq Q_2(t)+\overline{M}_{0}r^{-1/\alpha}(t)\overline{w}^{(\mu+1)/\mu}(t), \quad t\geq t_2, \end{equation} $

其中$ \mu = \max \{\alpha, \beta\}, \overline{M}_{0} = \min \{\alpha (\frac{L_{1}^{\alpha}}{L_{2}^{\alpha+1}})^{\frac{1}{\alpha}}, \beta \overline {M}_{1}(\frac{L_{1}^{\alpha}}{L_{2}^{\alpha+1}})^{\frac{1}{\alpha}}, \beta \overline{M}_{2}(\frac{L_{1}^{\beta}}{L_{2}^{\beta+1}})^{\frac{1}{\beta}}, (\frac{\mu^{2\mu+1}}{(\mu+1)^{\mu+1}M})^{\frac{1}{\mu}}\} $, $ M $由(2.31)式确定.

用$ \underline{R}^\mu(s) $乘(2.38)式,从$ t_2 $到$ t $分部积分可得

(2.39) $ \begin{eqnarray} \int_{t_2}^{t} \underline{R}^\mu(s)Q_2(s){\rm d}s &\leq& \int_{t_2}^{t} \underline{R}^\mu(s)\overline{w}'(s){\rm d}s- \overline{M}_{0} \int_{t_2}^{t} \underline{R}^\mu(s)r^{-1/\alpha}(s)\overline{w}^{(\mu+1)/\mu}(s){\rm d}s \\ & = &\underline{R}^\mu(t)\overline{w}(t)- \underline{R}^\mu(t_2)\overline{w}(t_2) \\ &&+ \int_{t_2}^{t} \underline{R}^\mu(s) \bigg[\frac{\mu \overline{w}(s)}{\underline{R}(s)r^{1/\alpha}(s)}-\frac{\overline{M}_{0}\overline{w}^{(\mu+1)/\mu}(s)}{r^{1/\alpha}(s)}\bigg]{\rm d}s. \end{eqnarray} $

利用不等式(2.11)和(2.39)式,则有

(2.40) $ \begin{equation} \int_{t_2}^{t} [\underline{R}^\mu(s)Q_2(s)-\overline{M}\underline{R}^{-1}(s)r^{-1/\alpha}(s)]{\rm d}s \leq \underline{R}^\mu(t)\overline{w}(t)- \underline{R}^\mu(t_2)\overline{w}(t_2), \end{equation} $

其中$ \overline{M} = \mu^{2\mu+1}/((\mu+1)^{\mu+1}\overline{M}_{0}^{\mu})\geq M. $

下面证明(2.40)式右端有界.事实上,在上述(iii)中已证$ r^{1/\alpha}(t)(-z'(t)) $是增函数,从而有

$ r^{1/\alpha}(s)(-z'(s)) \geq r^{1/\alpha}(t)(-z'(t)), \quad s \geq t \geq t_2, $

$ -z'(s) \geq r^{1/\alpha}(t)(-z'(t))r^{-1/\alpha}(s), \quad s \geq t \geq t_2. $

上式对$ s $积分,可得

(2.41) $ \begin{equation} z(t) \geq r^{1/\alpha}(t)(-z'(t))\underline{R}(t)\; \; \mbox {或}\; \; z^\alpha(t) \geq [ r^{1/\alpha}(t)(-z'(t))]^{\alpha} \underline{R}^{\alpha}(t). \end{equation} $

当$ \alpha \geq \beta $时,因$ z'(t)<0 $,故存在常数$ l_1>0 $,使得$ l_1 \geq z^{\alpha-\beta}(t) \geq L_{2}^{-1}\underline{R}^{\alpha}(t)\overline{w}(t)>0, \; t\geq t_2. $因此, (2.40)式右端有界.

另一方面,当$ \beta > \alpha $时,由(2.41)式得到

$ z^{\beta}(t) \geq r^{\beta/\alpha}(t)(-z'(t))^{\beta}\underline{R}^{\beta}(t), \; t\geq t_{2}, $

即

$ 1 \geq \frac{ r^{1-1+\beta/\alpha}(t)(-z'(t))^{\alpha+\beta-\alpha}}{z^{\beta}(t)}\underline{R}^{\beta}(t), \; t\geq t_{2}. $

注意到$ r^{1/\alpha}(t)(-z'(t)) $是增函数,故存在$ l_2>0 $,使得

$ l_2 \geq \bigg[\frac{1}{r^{1/\alpha}(t)(-z'(t))}\bigg]^{\beta-\alpha} \geq L_{2}^{-1}\underline{R}^{\beta}(t) \overline{w}(t)>0, t\geq t_{2}. $

因此, (2.40)式右端同样有界.

故对任意$ \alpha>0, \beta>0 $, (2.40)式右端有界,此与(2.31)式矛盾.定理2.4证毕.

注2.5  文献[2, 5, 6, 23]仅对方程$ (1.1) $的特殊情况给出了相应方程一切解振动或渐近于零的类似充分条件,因此,本文定理$ 2.4 $推广和改进了这些结果.还需要指出的是,文献[23]定理$ 2.3 $的证明中当$ \beta>\alpha, $其$ (2.36) $式成立时用到了$ (-y'(t)) $ (对应于本文的$ (-z'(t))) $为增函数是错误的.事实上“$ (-y'(t))'\geq 0, t\geq T $”不成立.否则,存在$ k>0 $使得$ -y'(t)\geq k>0, t\geq T, $积分之,得$ y(t)-y(T)\leq -k(t-T)\rightarrow -\infty\ (t\rightarrow\infty) $与$ y(t)>0, t\geq T $矛盾.

例2.4  考虑中立型方程

(${{\rm{E}}_4}$) $ [t^{2\alpha}(\pi+\arctan z(t))\left|z'(t)\right|^{\alpha-1}z'(t)]'+ t^{2\alpha+\beta} \left|x(t-2)\right|^{\beta-1} x(t-2) = 0, \quad t \geq t_0 \geq 2, $

其中$ z(t) = x(t)+\frac{1}{2}x(t-1), \alpha>0, \beta>0. $

这里有$ a(t, z(t)) = \pi +\arctan z(t), \ r(t) = t^{2\alpha}, \ p(t) = 1/2, \ q(t) = t^{2\alpha+\beta}; $$ \sigma(t) = t-2<t-1 = \tau(t); $$ z'(t)(a(t, z(t)))' = \frac{(z'(t))^{2}}{1+z^{2}(t)}\geq 0. $故满足条件(A$ _{1} $), (A$ _{2}) $和(A$ _{4}). $又$ Q_1(t) = Q_2(t) = \frac{1}{2^{\beta}}t^{2\alpha+\beta}. $显然$ (1.3) $式和$ (2.30) $式均成立.若取$ \rho(t)\equiv 1 $,则$ (2.1) $式成立.再证$ (2.31) $式也成立.

事实上,此时有$ \underline{R}(t) = 1/t $,所以对任意的常数$ m>0, $有

$ \begin{eqnarray*} \int_{t_0}^{\infty}\bigg [\underline{R}^{\mu}(t)Q_2(t)-\frac{m}{\underline{R}(t)r^{1/\alpha}(t)}\bigg]{\rm d}t & = &\int_{t_0}^{\infty}\bigg [\frac{1}{2^\beta}t^{2\alpha+\beta-\mu}-\frac{m}{t}\bigg]{\rm d}t\\ & = & \left\{\begin{array}{ll} \int_{t_0}^{\infty}\bigg[\frac{1}{2^\beta}t^{\alpha+\beta}-\frac{m}{t}\bigg]{\rm d}t = \infty, \; &\alpha \geq \beta, \\ \int_{t_0}^{\infty}\bigg[ \frac{1}{2^{\beta}}t^{2\alpha}-\frac{m}{t}\bigg]{\rm d}t = \infty, \; \; &\alpha < \beta. \end{array}\right. \end{eqnarray*} $

因此, $ (2.31) $式成立,由定理$ 2.4 $知方程(E$ _4) $振动.

注2.6  不难看出文献[1-25]及其引文中的定理均不能判定方程(E$ _4) $的振动性.