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  浙江大学学报(理学版)  2016, Vol. 43 Issue (3): 257-263  DOI:10.3785/j.issn.1008-9497.2016.03.002
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引用本文 [复制中英文]

杨甲山. 2016. 具可变时滞的二阶非线性中立型泛函微分方程的振动性[J]. 浙江大学学报(理学版), 43(3): 257-263. DOI: 10.3785/j.issn.1008-9497.2016.03.002.
[复制中文]
YANG Jiashan. 2016. Oscillation of certain second-order nonlinear neutral functional differential equations with variable delay[J]. Journal of Zhejiang University(Science Edition), 43(3): 257-263. DOI: 10.3785/j.issn.1008-9497.2016.03.002.
[复制英文]

基金项目

广西壮族自治区教育厅科研项目(2013YB223); 国家青年科研基金资助项目(61503171); 梧州学院2014年校级科研重大项目(2014A003); 硕士学位授予单位立项建设项目(桂学位[2013]4号).

作者简介

杨甲山(1963-), ORCID: http://orcid.org/0000-0002-0340-097X, 男, 教授, 主要从事微分方程的理论与应用研究, E-mail: syxyyjs@163.com

文章历史

收稿日期:2015-08-16
具可变时滞的二阶非线性中立型泛函微分方程的振动性
杨甲山     
梧州学院 信息与电子工程学院, 广西 梧州 543002
摘要: 研究一类非线性的具有可变时滞的二阶中立型泛函微分方程的振动性,利用Riccati变换技术及不等式分析技巧,获得了该方程振动的2个新的判别准则,所举例子说明这些准则是方程振动的"sharp"条件.
关键词: 振动性    变时滞    泛函微分方程    Riccati变换    
Oscillation of certain second-order nonlinear neutral functional differential equations with variable delay
YANG Jiashan     
School of Information and Electronic Engineering, Wuzhou University, Wuzhou 543002, Guangxi Zhuang Autonomous Region, China
Abstract: We study the oscillatory behavior of a class of second-order nonlinear neutral functional differential equations with variable delay. By using the generalized Riccati transformation and the inequality technique, we establish two new oscillation criteria for the oscillation of the equations. The examples are provided to illustrate that our result gives a sharper estimate for the oscillation of the equations.
Key words: oscillation    variable delay    functional differential equation    Riccati transformation    

微分方程在自然科学及工程技术等领域有着非常广泛的应用,如著名的二阶Emden-Fowler型微分方程x″(t)+at-1x′(t)+btm-1xn(t)=0已广泛应用于数学物理、理论物理(特别是核物理)、生物工程、信息技术及工程机械等领域.近年来具变时滞的中立型泛函微分方程的振动性研究引起了国内外学者的广泛兴趣[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24].笔者考虑如下形式的具变时滞的二阶非线性中立型微分方程

$ {\{ a(t)[(x(t) + p(t)x(t)))']^\gamma }\} ' + q(t)f(x(\delta (t))) = 0,\;t \geqslant {t_0} $ (1)

的振动性,其中,函数a,p,qC([t0,+∞),R);常数γ为2个正奇数之商;函数fC(R,R)并且uf(u)>0(u≠0),本文假设下列条件成立:

(H1) aC1([t0,+∞),(0,+∞)),q(t)>0,p(t)≥0.

(H2) τ,δ:[t0,+∞)→(0,+∞)并满足:τ(t)≤t,$\mathop {\lim }\limits_{t \to + \infty } \tau (t) = + \infty $;δ(t)≤t,$\mathop {\lim }\limits_{t \to + \infty } \delta (t) = + \infty $;τ°δ=δ°ττ′(t)≥τ0>0(这里τ0为常数).

(H3) 当u≠0时,f(u)/uL(这里常数L>0).

如果x(t)满足a(t)[(x(t)+p(t)x(τ(t)))′]γC1([Tx,+∞),R),且在区间[Tx,+∞)满足式(1),则称函数x(t)∈C1([Tx,+∞),R)(Txt0)是方程(1)的解,本文只关注方程(1)的非平凡解.如果方程(1)的解x(t)既不最终为正也不最终为负,则称解x(t)是振动的,否则是非振动的;如果方程(1)的所有解都是振动的,则称其是振动的.本文将分别在条件

$ \int_{{t_0}}^{ + \infty } {{a^{ - 1/\gamma }}(t){\text{d}}t} = + \infty $ (2)

$ \int_{{t_0}}^{ + \infty } {{a^{ - 1/\gamma }}(t){\text{d}}t} < + \infty $ (3)

成立的情况下建立方程(1)的振动性判别准则,改善对方程(1)的中立项系数函数的限制条件:0≤p(t)<1,得到这些准则的特殊情形,推广并改进了最近文献中的一系列结果.

引理1A>0,B>0和λ>0均为常数,则

(i)当x>0时,$Ax - B{x^{\frac{{\lambda + 1}}{\lambda }}} \leqslant \frac{{{\lambda ^\lambda }{A^{\lambda + 1}}}}{{{{(\lambda + 1)}^{\lambda + 1}}{B^\lambda }}}$;

(ii)当x<0时,$Ax + B{x^{\frac{{\lambda + 1}}{\lambda }}} \geqslant \frac{{{\lambda ^\lambda }{A^{\lambda + 1}}}}{{{{(\lambda + 1)}^{\lambda + 1}}{B^\lambda }}}$.

可由数学分析法证明之.

1 主要结果及其证明

引入记号

$ \begin{gathered} z(t) = x(t) + p(t)x(\tau (t)),\hfill \\ Q(t) = \min \{ q(t),\;q(\tau (t))\} ,\hfill \\ {\varphi _ + }(t) = \max \{ \varphi (t),\;0\} . \hfill \\ \end{gathered} $

定理1 设式(2)成立且0≤p(t)≤p0<+∞(p0为常数),若存在函数φC1([t0,+∞),(0,+∞))使得

$ \mathop {\lim \sup }\limits_{t \to + \infty } \int_T^t {\left[{\frac{{L\varphi (s)Q(s)\psi (s,\;{t_1})}}{{{{[b\theta (s)]}^{\gamma - 1}}}} - \frac{{\varphi (s)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{{\left( {\frac{{{\varphi _ + }'(s)}}{{\varphi (s)}}} \right)}^{\gamma + 1}} \times \left( {a(s) + \frac{{{p_0}a(\tau (s))}}{{\tau _0^{\gamma + 1}}}} \right)} \right]{\text{d}}s = + \infty } , $ (5)

其中常数Tt0足够大,b>0,函数

$ \begin{gathered} \psi (t,\;{t_0}) = \left( {\int_{{t_1}}^{\delta (t)} {{a^{ - 1/\gamma }}(s){\text{d}}s} } \right){\left( {\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s){\text{d}}s} } \right)^{ - 1}},\hfill \\ \theta (t) = \int_{{t_0}}^t {{a^{ - 1/\gamma }}(s){\text{d}}s} ,\hfill \\ \end{gathered} $

则方程(1)是振动的.

证明 用反证法:设方程(1)有一个最终正解x(t)(当x(t)为最终负解时类似可证),则∃t1t0,当tt1时,有x(t)>0,x(τ(t))>0,x(δ(t))>0.由z(t)的定义,有z(t)>0,z(t)≥x(t)(tt1).由方程(1),

$ [a(t){(z'(t))^\gamma }]' = - q(t)f(x(\delta (t))) \leqslant - Lq(t)x(\delta (t)) < 0, $ (6)

由此式容易证得z′(t)>0(tt1).应用式(6),当tt1时,有

$ \frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'}}{{\tau '(t)}} + Lq(\tau (t))x(\delta (\tau (t))) \leqslant 0, $ (7)

综合式(6)与(7),当tt1时,得

$ \begin{gathered} [a(t){(z'(t))^\gamma }]' + Lq(t)x(\delta (t)) + \hfill \\ {p_0}Lq(\tau (t))x(\delta (\tau (t))) + \hfill \\ {p_0}\frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'}}{{\tau '(t)}} \leqslant 0,\hfill \\ \end{gathered} $

由于τ′(t)≥τ0>0,τ°δ=δ°τz(t)≤x(t)+p0x(τ(t)),根据函数Q(t)的定义,上式可进一步写成

$ \begin{gathered} [a(t){(z'(t))^\gamma }]' + \frac{{{p_0}}}{{{\tau _0}}}[a(\tau (t)){(z'(\tau (t)))^\gamma }]' \leqslant \hfill \\ - LQ(t)[x(\delta (t)) + {p_0}x(\delta (\tau (t)))] \leqslant \hfill \\ - LQ(t)z(\delta (t)) \leqslant 0,\hfill \\ \end{gathered} $ (8)

$ w(t) = \varphi (t)\frac{{a(t){{(z'(t))}^\gamma }}}{{{z^{\gamma (t)}}}},\;t \geqslant {t_1} $ (9)

w(t)>0(tt1),利用式(9)及引理1(i),可得

$ \begin{gathered} w'(t) = \varphi '(t)\frac{{a(t){{(z'(t))}^\gamma }}}{{{z^\gamma }(t)}} + \varphi (t) \times \hfill \\ \frac{{[a(t){{(z'(t))}^\gamma }]'{z^\gamma }(t) - a(t){{(z'(t))}^\gamma }\gamma {z^{\gamma - 1}}(t)z'(t)}}{{{z^{2\gamma }}(t)}} \hfill \\ = \frac{{\varphi '(t)}}{{\varphi (t)}}w(t) + \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(t){(z'(t))^\gamma }]' - \hfill \\ \gamma \frac{{\varphi (t)a(t){{(z'(t))}^{\gamma + 1}}}}{{{z^{\gamma + 1}}(t)}} \leqslant \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(t){(z'(t))^\gamma }]' + \hfill \\ \frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}w(t) - \frac{{\gamma {w^{(\gamma + 1)/\gamma }}(t)}}{{{{[\varphi (t)a(t)]}^{1/\gamma }}}} \leqslant \hfill \\ \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(t){(z'(t))^\gamma }]' + \frac{{\varphi (t)a(t)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}}. \hfill \\ \end{gathered} $ (10)

再令

$ v(t) = \varphi (t)\frac{{a(\tau (t)){{(z'(\tau (t)))}^\gamma }}}{{{z^\gamma }(\tau (t))}},\;t \geqslant {t_1}, $ (11)

v(t)>0(tt1).由于τ′(t)≥τ0>0,z′(t)>0,由引理1(i),类似地可得

$ \begin{gathered} v'(t) = \varphi '(t)\frac{{a(\tau (t)){{(z'(\tau (t)))}^\gamma }}}{{{z^\gamma }(\tau (t))}} + \hfill \\ \varphi (t)\frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'{z^\gamma }(\tau (t))}}{{{z^{2\gamma }}(\tau (t))}} - \hfill \\ \frac{{\varphi (t)a(\tau (t)){{(z'(\tau (t)))}^\gamma }\gamma {z^{\gamma - 1}}(\tau (t))z'(\tau (t))\tau '(t)}}{{{z^{2\gamma }}(\tau (t))}} \hfill \\ \leqslant \frac{{\varphi (t)}}{{{z^\gamma }(\tau (t))}}[a(\tau (t)){(z'(\tau (t)))^\gamma }]' + \frac{{\varphi '(t)}}{{\varphi (t)}}v(t) - \hfill \\ \gamma {\tau _0}\frac{{\varphi (t)a(\tau (t)){{(z'(\tau (t)))}^{\gamma + 1}}}}{{{z^{\gamma + 1}}(\tau (t))}} \leqslant \hfill \\ \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(\tau (t)){(z'(\tau (t)))^\gamma }]' + \frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}v(t) - \hfill \\ \frac{{\gamma {\tau _0}{{(v(t))}^{(\gamma + 1)/\gamma }}}}{{{{(\varphi (t)a(\tau (t)))}^{1/\gamma }}}} \hfill \\ \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(\tau (t)){(z'(\tau (t)))^\gamma }]' + \hfill \\ \frac{{\varphi (t)a(\tau (t))}}{{{{(\gamma + 1)}^{\gamma + 1}}\tau _0^\gamma }}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}}. \hfill \\ \end{gathered} $ (12)

于是,综合式(10),(12),并注意到式(8),可得

$ \begin{gathered} w'(t) + \frac{{{p_0}}}{{{\tau _0}}}v'(t) \leqslant \frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(t){(z'(t))^\gamma }]' + \hfill \\ \frac{{\varphi (t)a(t)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}} + \frac{{{p_0}}}{{{\tau _0}}}\left[{\frac{{\varphi (t)}}{{{z^\gamma }(t)}}[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]' + \frac{{\varphi (t)a(\tau (t))}}{{{{(\gamma + 1)}^{\gamma + 1}}\tau _0^\gamma }}{{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)}^{\gamma + 1}}} \right] \hfill \\ = \frac{{\varphi (t)}}{{{z^\gamma }(t)}}\left\{ {[a(t){{(z'(t))}^\gamma }]' + \frac{{{p_0}}}{{{\tau _0}}}[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'} \right\} + \hfill \\ \frac{{\varphi (t)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}}\left( {a(t) + \frac{{{p_0}a(\tau (t))}}{{\tau _0^{\gamma + 1}}}} \right) \leqslant \hfill \\ - \frac{{\varphi (t)}}{{{z^\gamma }(t)}}LQ(t)z(\delta (t)) + \frac{{\varphi (t)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}}^{\gamma + 1}\left( {a(t) + \frac{{{p_0}a(\tau (t))}}{{\tau _0^{\gamma + 1}}}} \right). \hfill \\ \end{gathered} $ (13)

由式(6)知,a(t)[z′(t)]γ(tt1)是单调减少的,因此有

$ z(t) \geqslant z(t) - z({t_1}) = \int_{{t_1}}^t {\frac{{{a^{1/\gamma }}(s)z'(s)}}{{{a^{1/\gamma }}(s)}}} ds \geqslant {a^{1/\gamma }}(t)z'(t)\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} ds, $

由此可推得

$ \frac{{\text{d}}}{{{\text{dt}}}}\frac{{z(t)}}{{\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s}} = \frac{{z'(t)\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s - z(t)\int_{{t_1}}^t {{a^{ - 1/\gamma }}(t)} }}{{{{\left( {\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s} \right)}^2}}} \leqslant 0, $

于是$\frac{{z(t)}}{{\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s}} \leqslant \frac{{z(\delta (t))}}{{\int_{{t_1}}^{\delta (t)} {{a^{ - 1/\gamma }}(s)} {\text{d}}s}}$,注意到Ψ(t,t1)的定义,即得

$ \frac{{z(\delta (t))}}{{z(t)}} \geqslant \psi (t,\;{t_1}). $ (14)

另外,利用a(t)[z′(t)]γ的单调递减性,当tt1时,有a(t)[z′(t)]γa(t1)[z′(t1)]γ=M0(M0>0为常数),即z′(t)≤M01/γa-1/γ(t),因此

$ \begin{gathered} z(t) \leqslant z({t_1}) + M_0^{1/\gamma }\int_{{t_1}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s \leqslant \hfill \\ z({t_1}) + M_0^{1/\gamma }\int_{{t_0}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s,\hfill \\ \end{gathered} $

所以,存在充分大的t2t1及常数b>0,当tt2时,就有

$ z(t) \leqslant b\int_{{t_0}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s = b\theta (t) $ (15)

将式(14),(15)代入式(13),得

$ \begin{gathered} w'(t) + \frac{{{p_0}}}{{{\tau _0}}}v'(t) \leqslant - \frac{{L\varphi (t)Q(t)\psi (t,\;{t_1})}}{{{{[b\theta (t)]}^{\gamma - 1}}}} + \hfill \\ \frac{{\varphi (t)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{\left( {\frac{{\varphi {'_ + }(t)}}{{\varphi (t)}}} \right)^{\gamma + 1}}\left( {a(t) + \frac{{{p_0}a(\tau (t))}}{{\tau _0^{\gamma + 1}}}} \right),\hfill \\ \end{gathered} $

所以

$ \begin{gathered} \int_{{t_2}}^{{t_1}} {\left[{\frac{{L\varphi (s)Q(s)\psi (s,\;{t_1})}}{{{{[b\theta (s)]}^{\gamma - 1}}}} - \frac{{\varphi (s)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{{\left( {\frac{{\varphi {'_ + }s)}}{{\varphi (s)}}} \right)}^{\gamma + 1}} \times \left( {a(s) + \frac{{{p_0}a(\tau (s))}}{{\tau _0^{\gamma + 1}}}} \right)} \right]} ds \hfill \\ \leqslant - w(t) + w({t_2}) - \frac{{{p_0}}}{{{\tau _0}}}v(t) + \frac{{{p_0}}}{{{\tau _0}}}v({t_2}) \leqslant w({t_2}) + \frac{{{p_0}}}{{{\tau _0}}}v({t_2}),\hfill \\ \end{gathered} $

这与式(5)矛盾.定理1证毕.

定理2 设式(3)成立,且0≤p(t)≤p0<+∞(p0为常数),若存在函数φC1([t0,+∞),(0,+∞))使得式(5)成立,且

$ \mathop {\lim \sup }\limits_{t \to + \infty } \int_T^t {\left[{LQ(s)\eta (s){\zeta ^\gamma }(s) - \left( {1 + \frac{{{p_0}}}{{{\tau _0}}}} \right)\frac{{{\gamma ^{\gamma + 1}}{a^{ - 1/\gamma }}(s)}}{{{{(\gamma + 1)}^{\gamma + 1}}\zeta (s)}}} \right]} {\text{d}}s = + \infty $ (16)

其中常数Tt0足够大,函数

$ \eta (t) = \left\{ {\begin{array}{*{20}{l}} {k,\;\gamma > 1,} \\ {1,\;\gamma = 1,} \\ {k{\zeta ^{1 - \gamma }}(t),\;\gamma < 1,} \end{array}} \right.\;\;k > 0为常数, $

函数$\zeta (t) = \int_t^{ + \infty } {{a^{ - 1/\gamma }}(s)} {\text{d}}s$,则方程(1)是振动的.

证明 用反证法:设方程(1)有一个最终正解x(t)(当x(t)为最终负解时类似可证),则∃t1t0,当tt1时,有x(t)>0,x(τ(t))>0,x(δ(t))>0,根据定理1的证明,a(t)[z′(t)]γ严格单调递减且最终定号,进而z′(t)最终为正或为负.因此只需考虑如下2种情形:

情形i z′(t)>0(tt1).证明同定理1,可得到一个与式(5)矛盾的结果.

情形ii z′(t)<0(tt1).定义函数w(t)如下:

$ w(t) = \frac{{a(t){{(z'(t))}^\gamma }}}{{{z^\gamma }(t)}},\;t \geqslant {t_1}, $ (17)

w(t)<0(tt1),并且

$ \begin{gathered} w'(t) = \frac{{[a(t){{(z'(t))}^\gamma }]'}}{{{z^\gamma }(t)}} - \frac{{a(t){{(z'(t))}^\gamma }\gamma {z^{\gamma - 1}}(t)z'(t)}}{{{z^{2\gamma }}(t)}} \hfill \\ = \frac{{[a(t){{(z'(t))}^\gamma }]'}}{{{z^\gamma }(t)}} - \frac{{\gamma a(t){{(z'(t))}^{\gamma + 1}}}}{{{z^{\gamma + 1}}(t)}} \hfill \\ = \frac{{[a(t){{(z'(t))}^\gamma }]'}}{{{z^\gamma }(t)}} - \frac{{\gamma {{(w(t))}^{(\gamma + 1)/\gamma }}}}{{{a^{1/\gamma }}(t)}}. \hfill \\ \end{gathered} $ (18)

利用a(t)[z′(t)]γ的单调递减性,当stt1时,有a(s)[z′(s)]γa(t)[z′(t)]γ,进而

$ z'(s) \leqslant {a^{1/\gamma }}(t)z'(t){a^{ - 1/\gamma }}(s), $

此式两边从tu(ut)对s积分,可得

$ z(u) - z(t) \leqslant {a^{1/\gamma }}(t)z'(t)\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s, $

所以$z(t) + {a^{1/\gamma }}(t)z'(t)\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s \geqslant 0$,再令u→+∞,则

$ z(t) + {a^{1/\gamma }}(t)z'(t)\int_t^{ + \infty } {{a^{ - 1/\gamma }}(s)} {\text{d}}s \geqslant 0,\;t \geqslant {t_1}. $

所以,$\frac{{{a^{1/\gamma }}(t)z'(t)}}{{z(t)}}\zeta (t) \geqslant - 1$,由式(17),则有

$ - 1 \leqslant {w^{1/\gamma }}(t)\zeta (t) \leqslant 0,\;t \geqslant {t_1}. $ (19)

定义函数v(t)如下:

$ v(t) = \frac{{a(\tau (t)){{(z'(\tau (t)))}^\gamma }}}{{{z^\gamma }(t)}},\;t \geqslant {t_1}, $ (20)

v(t)<0(tt1).再利用a(t)[z′(t)]γ的单调递减性,有

$ a(\tau (t)){[z'(\tau (t))]^\gamma } \geqslant a(t){[z'(t)]^\gamma }, $

进一步,

$ z'(t) \leqslant {\left( {\frac{{a(\tau (t))}}{{a(t)}}} \right)^{1/\gamma }}z'(\tau (t)). $

于是,由式(20),并利用z′(t)<0,可得

$ \begin{gathered} v'(t) = \frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'}}{{{z^\gamma }(t)}} - \hfill \\ \frac{{a(\tau (t)){{(z'(\tau (t)))}^\gamma }\gamma {z^{\gamma - 1}}(t)z'(t)}}{{{z^{2\gamma }}(t)}} \leqslant \hfill \\ \frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'}}{{{z^\gamma }(t)}} - \hfill \\ \frac{{\gamma a(\tau (t)){{(z'(\tau (t)))}^\gamma }}}{{{z^{\gamma + 1}}(t)}}{\left( {\frac{{a(\tau (t))}}{{a(t)}}} \right)^{1/\gamma }}z'(\tau (t)) = \hfill \\ \frac{{[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'}}{{{z^\gamma }(t)}} - \frac{{\gamma {{(v(t))}^{(\gamma + 1)/\gamma }}}}{{{a^{1/\gamma }}(t)}}. \hfill \\ \end{gathered} $ (21)

a(τ(t))[z′(τ(t))]γa(t)[z′(t)]γ,得v(t)≥w(t),利用式(19),同样有

$ - 1 \leqslant {v^{1/\gamma }}(t)\zeta (t) \leqslant 0,\;t \geqslant {t_1}. $ (22)

由式(18)和(21),并应用式(8)(由定理1的证明知,此时式(8)仍然成立)及z(δ(t))≥z(t),可得

$ \begin{gathered} w'(t) + \frac{{{p_0}}}{{{\tau _0}}}v'(t) \leqslant \frac{1}{{{z^\gamma }(t)}}\left\{ {[a(t){{(z'(t))}^\gamma }]' + \frac{{{p_0}}}{{{\tau _0}}}[a(\tau (t)){{(z'(\tau (t)))}^\gamma }]'} \right\} - \hfill \\ \frac{\gamma }{{{a^{1/\gamma }}(t)}}\left[{(w{{(t)}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(t))}^{(\gamma + 1)/\gamma }}} \right] \leqslant - \frac{{LQ(t)z(\delta (t))}}{{{z^\gamma }(t)}} - \hfill \\ \frac{\gamma }{{{a^{1/\gamma }}(t)}}\left[{(w{{(t)}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(t))}^{(\gamma + 1)/\gamma }}} \right] \leqslant - LQ(t){z^{1 - \gamma }}(t) - \hfill \\ \frac{\gamma }{{{a^{1/\gamma }}(t)}}\left[{(w{{(t)}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(t))}^{(\gamma + 1)/\gamma }}} \right]. \hfill \\ \end{gathered} $ (23)

γ>1时,因为z(t)>0,z′(t)<0(tt1),所以z(t)≤z(t1),即z1-γ(t)≥z1-γ(t1)=k

γ=1时,显然有z1-γ(t)=1.

γ<1时,由于a(t)[z′(t)]γ是单调减小的,所以当stt1时,有

$ a(s){[z'(s)]^\gamma } \leqslant a(t){[z'(t)]^\gamma }, $

$ z'(s) \leqslant {\{ a(t){[z'(t)]^\gamma }\} ^{1/\gamma }}{a^{ - 1/\gamma }}(s), $

因此

$ z(u) - z(t) \leqslant {\{ a(t){[z'(t)]^\gamma }\} ^{1/\gamma }}\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s, $

$ \begin{gathered} z(t) \geqslant - {\{ a(t){[z'(t)]^\gamma }\} ^{1/\gamma }}\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s \geqslant \hfill \\ - {\{ a({t_1}){[z'({t_1})]^\gamma }\} ^{1/\gamma }}\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s = \hfill \\ M\int_t^u {{a^{ - 1/\gamma }}(s)} {\text{d}}s,\hfill \\ \end{gathered} $

这里$M = - {\{ a({t_1}){[z'({t_1})]^\gamma }\} ^{1/\gamma }} = - {a^{1 - \gamma }}({t_1})z'({t_1}) > 0$为常数,令u→+∞,则由上式得

$ z(t) \geqslant M\int_t^{ + \infty } {{a^{ - 1/\gamma }}(s)} {\text{d}}s = M\zeta (t), $

即${z^{1 - \gamma }}(t) \geqslant {M^{1 - \gamma }}{\zeta ^{1 - \gamma }}(t)$.

综合上述3种情形及函数η的定义,由式(23),得

$ \begin{gathered} LQ(t)\eta (t) \leqslant - w'(t) - \frac{{{p_0}}}{{{\tau _0}}}v'(t) - \hfill \\ \frac{\gamma }{{{a^{1/\gamma }}(t)}}\left[{(w{{(t)}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(t))}^{(\gamma + 1)/\gamma }}} \right],\hfill \\ \end{gathered} $

上式两边乘以ζγ(t)并从t1t(tt1)积分,采用分部积分法,ζ′(t)=-a-1/γ(t),由式(19)和(22),以及引理1(ii),得

$ \begin{gathered} \int_{{t_1}}^t {LQ(s)\eta (s){\zeta ^\gamma }(s)} {\text{d}}s \leqslant \hfill \\ - \int_{{t_1}}^t {{\zeta ^\gamma }(s)w'(s)} {\text{d}}s - \frac{{{p_0}}}{{{\tau _0}}}\int_{{t_1}}^t {{\zeta ^\gamma }(s)v'(s)} {\text{d}}s - \hfill \\ \int_{{t_1}}^t {\frac{{\gamma {\zeta ^\gamma }(s)}}{{{a^{1/\gamma }}(s)}}\left[{{{(w(s))}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(s))}^{(\gamma + 1)/\gamma }}} \right]} {\text{d}}s = \hfill \\ - [{\zeta ^\gamma }(s)w(s)]_{{t_1}}^t + \int_{{t_1}}^t {\gamma {\zeta ^{\gamma - 1}}(s)\zeta '(s)} w(s){\text{d}}s - \hfill \\ \frac{{{p_0}}}{{{\tau _0}}}\left\{ {[{\zeta ^\gamma }(s)v(s)]_{{t_1}}^t - \int_{{t_1}}^t {\gamma {\zeta ^{\gamma - 1}}(s)\zeta '(s)} v(s){\text{d}}s} \right\} - \hfill \\ \int_{{t_1}}^t {\frac{{\gamma {\zeta ^\gamma }(s)}}{{{a^{1/\gamma }}(s)}}\left[{{{(w(s))}^{(\gamma + 1)/\gamma }} + \frac{{{p_0}}}{{{\tau _0}}}{{(v(s))}^{(\gamma + 1)/\gamma }}} \right]} {\text{d}}s = \hfill \\ - {\zeta ^\gamma }(t)w(t) + {\zeta ^\gamma }({t_1})w({t_1}) - \hfill \\ \frac{{{p_0}}}{{{\tau _0}}}[{\zeta ^\gamma }(t)v(t) + {\zeta ^\gamma }({t_1})v({t_1})] - \hfill \\ \int_{{t_1}}^t {\left[{\gamma {\zeta ^{\gamma - 1}}(s){a^{ - 1/\gamma }}(s)w(s) + \frac{{\gamma {\zeta ^\gamma }(s)}}{{{a^{1/\gamma }}(s)}}{{(w(s))}^{(\gamma + 1)/\gamma }}} \right]} {\text{d}}s - \hfill \\ \frac{{{p_0}}}{{{\tau _0}}}\int_{{t_1}}^t {\left[{\gamma {\zeta ^{\gamma - 1}}(s){a^{ - 1/\gamma }}(s)v(s) + \frac{{\gamma {\zeta ^\gamma }(s)}}{{{a^{1/\gamma }}(s)}}{{(v(s))}^{(\gamma + 1)/\gamma }}} \right]} {\text{d}}s \leqslant \hfill \\ 1 + {\zeta ^\gamma }({t_1})w({t_1}) + \frac{{{p_0}}}{{{\tau _0}}}[1 + {\zeta ^\gamma }({t_1})v({t_1})] + \hfill \\ \int_{{t_1}}^t {\frac{{{\gamma ^{\gamma + 1}}{a^{ - 1/\gamma }}(s)}}{{{{(\gamma + 1)}^{\gamma + 1}}\zeta (s)}}} {\text{ds + }}\frac{{{p_0}}}{{{\tau _0}}}\int_{{t_1}}^t {\frac{{{\gamma ^{\gamma + 1}}{a^{ - 1/\gamma }}(s)}}{{{{(\gamma + 1)}^{\gamma + 1}}\zeta (s)}}} {\text{ds}},\hfill \\ \hfill \\ \hfill \\ \end{gathered} $

所以,

$ \begin{gathered} \int_{{t_1}}^t {\left[{LQ(s)\eta (s){\zeta ^\gamma }(s) - \left( {1 + \frac{{{p_0}}}{{{\tau _0}}}} \right)\frac{{{\gamma ^{\gamma + 1}}{a^{ - 1/\gamma }}(s)}}{{{{(\gamma + 1)}^{\gamma + 1}}\zeta (s)}}} \right]} {\text{d}}s \hfill \\ \leqslant 1 + {\zeta ^\gamma }({t_1})w({t_1}) + \frac{{{p_0}}}{{{\tau _0}}}[1 + {\zeta ^\gamma }({t_1})v({t_1})],\hfill \\ \end{gathered} $

这与式(16)矛盾.定理2证毕.

注1 笔者给出的二阶非线性中立型变时滞泛函微分方程(1)的振动准则分别是在积分$\int_{{t_0}}^{ + \infty } {{{[a(t)]}^{ - 1/\gamma }}} {\text{d}}t$收敛和发散的情况下得到的,其结果改进了对中立项系数函数0≤p(t)<1的限制.值得注意的是,方程(1)的特殊情形即当γ=1时,文献[11]得到其振动的判别准则,但有限制条件“δ′(t)>0且δ(t)≤τ(t)”,而本文却不需要这一条件.此外,从下面几个例子可以看出,即使是中立项系数函数,当0≤p(t)<1时,所得到的振动准则也是较“精细”的,几乎是方程(1)振动的“sharp”条件.因此本文定理推广并改进了现有文献中的一系列结果,具有更为广泛的意义.

2 实例分析

例1 考虑如下二阶微分方程(α>0为常数):

$ \left( {x(t) + \frac{2}{3}x(t - 1)} \right)'' + \frac{\alpha }{{{t^2}}}x(t) = 0,\;t \geqslant 1, $ (24)

这里γ=1,a(t)≡1,p(t)=2/3,q(t)=α/t2τ(t)=t-1,δ(t)=tf(u)=u.容易验证,条件(H1)~(H3)均满足.现取φ(t)=t,注意到L=1,p0=2/3,τ0=1,Ψ(s,t1)=1,则

$ \begin{gathered} \mathop {\lim \sup }\limits_{t \to + \infty } \int_T^t {\left[{\frac{{L\varphi (s)Q(s)\psi (s,\;{t_1})}}{{{{[b\theta (s)]}^{\gamma - 1}}}} - \frac{{\varphi (s)}}{{{{(\gamma + 1)}^{\gamma + 1}}}}{{\left( {\frac{{\varphi {'_ + }(s)}}{{\varphi (s)}}} \right)}^{\gamma + 1}} \times \left( {a(s) + \frac{{{p_0}a(\tau (s))}}{{\tau _0^{\gamma + 1}}}} \right)} \right]} {\text{d}}s \hfill \\ = \mathop {\lim \sup }\limits_{t \to + \infty } \int_1^t {\left[{\frac{{s\alpha }}{{{s^2}}} - \frac{s}{{{2^2}}}{{\left( {\frac{1}{s}} \right)}^2}\left( {1 + \frac{2}{3}} \right)} \right]} {\text{d}}s = \left( {\alpha - \frac{5}{{12}}} \right)\mathop {\lim \sup }\limits_{t \to + \infty } \int_1^t {\frac{1}{s}{\text{d}}s} ,\hfill \\ \end{gathered} $

所以由定理1知,当α>5/12≈0.416 67时,方程(24)是振动的.

注2 现改用文献[12]中的定理2.1来判定:因为

$ \begin{gathered} \mathop {\lim \sup }\limits_{t \to + \infty } \int_{{t_0}}^t {\left\{ {\varphi (s)q(s){{[1 - p(\delta (s))]}^\gamma } - \frac{{{{(\varphi {'_ + }(s))}^{\gamma + 1}}a(\delta (s))}}{{k{{(\gamma + 1)}^{\gamma + 1}}{{[\varphi (s)\delta '(s)]}^\gamma }}}} \right\}{\text{d}}s} \hfill \\ = \left( {\frac{\alpha }{3} - \frac{1}{4}} \right)\mathop {\lim \sup }\limits_{t \to + \infty } \int_1^t {\frac{1}{s}{\text{d}}s} ,\hfill \\ \end{gathered} $

所以当${\frac{\alpha }{3} - \frac{1}{4} > 0}$,即当$\alpha > \frac{3}{4} = 0.75$时,方程(24)是振动的.本文定理1的结果不仅包括而且改进了文献[12]中的定理2.1.

例2 考虑如下二阶非线性变时滞泛函微分方程

$ \begin{gathered} \left\{ {\frac{1}{{{t^{1/3}}}}{{\left[{\left( {x(t) + (5 - \sin {t^2})x\left( {\frac{t}{2}} \right)} \right)'} \right]}^{1/3}}} \right\}' + \hfill \\ \frac{1}{{{t^{1/4}}}}f\left( {x\left( {\frac{t}{2}} \right)} \right) = 0,\;\;t \geqslant 1,\hfill \\ \end{gathered} $ (25)

此处$\begin{gathered} {t_0} = 1,\;\gamma = \frac{1}{3},\;a(t) = \frac{1}{{{t^{1/3}}}},\;p(t) = 5 - \sin {t^2},\hfill \\ \tau (t) = \frac{t}{2},\;\delta (t) = \frac{t}{2},\;q(t) = \frac{1}{{{t^{1/4}}}}. \hfill \\ \end{gathered} $

f(u)=u[1+ln(1+u4)],因为

$ \begin{gathered} 0 < p(t) = 5 - \sin {t^2} \leqslant 6 = {p_0},\hfill \\ \frac{{f(u)}}{u} = 1 + \ln (1 + {u^4}) \geqslant 1 = L,\hfill \\ \end{gathered} $

且$\int_{{t_0}}^{ + \infty } {{a^{ - 1/\gamma }}(t)} {\text{d}}t = \int_1^{ + \infty } t {\text{d}}t = + \infty $,

所以条件(H1)~(H3)全部满足.又因为

$ Q(t) = \min \{ q(t),\;q(\tau (t))\} = {t^{ - 1/4}}, $

且$\theta (t) = \int_{{t_0}}^t {{a^{ - 1/\gamma }}(s)} {\text{d}}s = \int_1^t s {\text{d}}s = ({t^2} - 1)/2$.

为简化计算,取φ(t)=1,T=3,则

$ \begin{gathered} \mathop {\lim \sup }\limits_{t \to + \infty } \int_T^t {\left[{k\varphi (s)Q(s){\theta ^{ - \gamma }}(\tau (s)) - \left( {1 + \frac{{{p_0}}}{{{\tau _0}}}} \right)\frac{{\varphi (s)a(\tau (s))}}{{{{(\gamma + 1)}^{\gamma + 1}}\tau _0^\gamma }}{{\left( {\frac{{\varphi '(s)}}{{\varphi (s)}}} \right)}^{\gamma + 1}}} \right]} {\text{d}}s \hfill \\ = \mathop {\lim \sup }\limits_{t \to + \infty } \int_3^t {\left[{k\frac{1}{{{s^{1/4}}}}{{\left( {\frac{1}{8}({s^2} - 4)} \right)}^{ - 1/3}} - 0} \right]{\text{d}}s} \hfill \\ = \mathop {\lim \sup }\limits_{t \to + \infty } \int_3^t {\frac{{2k}}{{{s^{1/4}}{{({s^2} - 4)}^{1/3}}}}{\text{d}}s = + \infty } ,\hfill \\ \end{gathered} $

所以定理1的条件全部满足,于是由定理1知方程(25)是振动的.

注3 笔者注意到,由于方程(25)的中立项系数函数p(t)>1,因此文献[1, 2, 3, 5, 6, 7, 8, 11, 12, 13, 14, 15, 16, 19, 20, 21, 22]等中的定理对方程(25)均不适用.

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