研究如下的具有非线性中立项的二阶延迟Emden-Fowler型微分方程

(1.1) $ \begin{equation} \{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\geqslant t_0 \end{equation} $

的振动性.考虑下列条件(C$ _1) $–(C$ _4) $

(C$ _1) $实常数$ \alpha, \gamma $均为2个正奇数之商,且$ 0 < \alpha\leqslant1 $,而$ \beta $为任意正的实常数;

(C$ _2) $函数$ a\in C^1([t_0, +\infty), (0, +\infty)), p, q\in C([t_0, +\infty), \mathbb{R}) $并且$ q(t) > 0, 0\leqslant p(t) < 1; $

(C$ _3) $滞量函数$ \tau, \delta:[t_0, +\infty)\rightarrow(0, +\infty) $满足: $ \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 $,并且$ \delta'(t) > 0; $

(C$ _4) $函数$ f\in C(\mathbb{R}, \mathbb{R}) $,满足$ uf(u) > 0(u\neq0) $,且存在常数$ L > 0 $使得$ \frac{f(u)}{u}\geqslant L(u\neq0) $.方程(1.1)的解及其振动的定义,可参见文献[1-2].方程(1.1)包含了许多特殊类型的方程,如下列方程

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

(1.3) $ \begin{equation} \{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, \end{equation} $

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

等等,对于这些典型的Emden-Fowler型微分方程,现有文献得到了诸多振动性成果.如Dzurina等^[3], Sun等^[4]研究了二阶半线性微分方程$ (1.2) $的振动性,但文献[3-4]有较严格的条件"$ a'(t)\geqslant0 $及$ \delta'(t) > 0 $".在此基础上,黄记洲等^[5]研究了较一般的方程$ (1.3) $的振动性,得到了若干新结果,部分地改进并拓展了文献[3-4]的结果,其基本振动结果如下.

定理A^[5]  设$ \beta\geqslant\gamma, a'(t)\geqslant0, \delta'(t) > 0, 0\leqslant p(t) < 1, $且有

(1.5) $ \begin{equation} \int_{t_0}^{+\infty}a^{-\frac{1}{\beta}}(t){\rm d}t = +\infty, \end{equation} $

如果$ \exists\varphi\in C^1([t_0, +\infty), (0, +\infty)) $,使得

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

其中常数$ k > 0 $,则方程(1.3)振动.

显然,定理A也有较为严格的条件"$ a'(t)\geqslant0 $及$ \delta'(t) > 0 $, "且$ \beta < \gamma $时作者没有得到方程(1.3)的振动准则.紧接着, Yang等^[6]研究了更一般的方程$ (1.4) $的振动性,得到了若干新振动准则,其基本结论如下.

定理B^[6]  设条件$ (1.5) $成立, $ \tau'(t)\geqslant\tau_0 > 0 $,且$ 0\leqslant p(t)\leqslant p_0 < +\infty $ ($ p_0 $为常数), $ \delta\circ\tau = \tau\circ\delta, \delta(t)\leqslant\tau(t) $并且$ \delta'(t) > 0, $如果存在函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)) $使得当$ \beta\leqslant\gamma $时

$ \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s) \bigg[L_0Q(s)-\Big(1+\frac{p_0^\gamma}{\tau_0}\Big)\frac{(\frac{\beta}{\gamma})^\beta a(\delta(s))}{(\beta+1)^{\beta+1}\eta^{\gamma-\beta}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}\bigg]{\rm d}s = +\infty, $

当$ \beta > \gamma $时

$ \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\varphi(s)\bigg[L_0Q(s)-\Big(1+\frac{p_0^\gamma}{\tau_0}\Big )\frac{\eta^{\frac{\beta-\gamma}{\beta}}a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}(\delta'(s))^\gamma} \Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s = +\infty, $

其中$ \eta > 0 $为常数,则方程$ (1.4) $振动.

可以看到,定理B中没有了限制条件$ a'(t)\geqslant0, $中立项的系数函数$ p(t) $放宽到了$ 0\leqslant p(t)\leqslant p_0 < +\infty, $且当$ \beta > \gamma $和$ \beta < \gamma $时都有振动准则,遗憾的是对时滞函数却有新的要求"$ \delta\circ\tau = \tau\circ\delta, \delta(t)\leqslant\tau(t) $".

因此,对微分方程

(1.6) $ \begin{equation} \bigg[t^{-2}|(x(t)+\frac{1}{2}x(\sqrt{t}))'|^{\frac{1}{3}}(x(t)+\frac{1}{2}x(\sqrt{t}))' \bigg]'+2t^{-2}|x(\frac{t}{2})|x(\frac{t}{2}) = 0, t\geqslant1 \end{equation} $

而言,由于不满足条件"$ a'(t)\geqslant0, \beta\geqslant\gamma $",也不满足条件"$ \delta\circ\tau = \tau\circ\delta, \delta(t)\leqslant\tau(t) $",所以定理A和定理B都不适用于方程$ (1.6). $

如果方程是非正则的(即条件$ (1.5) $不满足,亦即该积分收敛),黄记洲等^[5]得到如下结果.

定理C^[5]  设$ \beta\geqslant\gamma, a'(t)\geqslant0, $且$ \tau'(t) > 0, 0\leqslant p(t) < 1, p'(t)\geqslant0, $如果

(1.7) $ \begin{equation} \int_{t_0}^{+\infty}a^{\frac{-1}{\beta}}(t){\rm d}t<+\infty, \end{equation} $

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

定理C就是文献[5]中的定理2.4.显然,定理C中的条件"$ p'(t)\geqslant0, \tau'(t) > 0 $"也较为苛刻,而且其结论是不确定的(其它文献,如文献[16]所得到的结论也是如此),因此在应用时就会有种种制约,自然也就不能确定非正则Euler (欧拉)方程

(1.8) $ \begin{equation} (t^2x'(t))'+mx(t) = 0, t\geqslant1 \end{equation} $

是否振动了.

本文将利用Riccati变换技术并结合经典不等式来研究方程(1.1)的振动性,在方程是正则和非正则两种情形下得到了该方程的一系列新型的Philos型振动条件,放宽了对方程的条件限制,改进且拓展了已有的结果.

引理2.1^[7]  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 (伯努利不等式)^[7]  对实数$ x > -1 $,当$ 0\leqslant r\leqslant1 $时, $ (1+x)^r\leqslant1+rx $,当$ r\leqslant0 $或$ r\geqslant1 $时, $ (1+x)^r\geqslant1+rx $.

引理2.3^[7]  设$ A > 0, $$ B > 0 $和$ \alpha > 0 $均为常数,则当$ x > 0 $时, $ Ax-Bx^{\frac{\alpha+1}{\alpha}}\leqslant\frac{\alpha^\alpha A^{\alpha+1}}{(\alpha+1)^{\alpha+1}B^\alpha}. $

引理2.4^[7]  设$ \lambda\geqslant1 $是两个正奇数之商, $ X, Y $为实数且$ XY\geqslant0 $,则

$ \begin{eqnarray*} X^{1+\frac{1}{\lambda}}-(X-Y)^{1+\frac{1}{\lambda}}\leqslant Y^{\frac{1}{\lambda}}\bigg[(1+\frac{1}{\lambda})X-\frac{1}{\lambda}Y\bigg]. \end{eqnarray*} $

考虑集合$ D = \{(t, s)|t\geqslant s\geqslant t_0\} $及$ D_0 = \{(t, s)|t > s\geqslant t_0\}, $如果二元函数$ H(t, s)\in C({\Bbb D}, \mathbb{R}), $且当$ t\geqslant t_0 $时$ H(t, t) = 0; $当$ (t, s)\in D_0 $时$ H(t, s) > 0 $且$ H(t, s) $对第二个变量$ s $有连续非正的偏导数:$ \frac{\partial H(t, s)}{\partial s}\leqslant0, $则记为$ H\in\Omega. $

引入记号:$ z(t) = x(t)+p(t)x^\alpha(\tau(t)), \varphi_+(t) = \max\{\varphi(t), 0\}. $

定理2.1  如果(C$ _1) $–(C$ _4) $及$ (1.5) $式成立,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)) $及$ H\in\Omega $,使得当$ \beta\geqslant\gamma $时

(2.1) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s) \bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma} \Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s = +\infty, \end{equation} $

当$ \beta < \gamma $时

(2.2) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s) \bigg[L\varphi(s)Q(s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}\bigg]{\rm d}s = +\infty, \end{equation} $

其中函数$ Q(t) = q(t)\{1-[\alpha2^{1-\alpha}+k_1^{-1}(2^{1-\alpha}-1)]p(\delta(t))\}^\gamma, $而$ t_1\geqslant t_0, k_i > 0\ (i = 1, 2) $为常数,则方程(1.1)振动.

证  若不然,则方程(1.1)存在一个非振动解$ x(t) $.如果$ x(t) $为最终正解,则存在$ t_1\geqslant t_0, $使得$ x(t) > 0, x(\delta(t)) > 0(t\geqslant t_1) $,由函数$ z(t) $的定义得$ z(t)\geqslant x(t) > 0(t\geqslant t_1) $,由方程(1.1),得

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

于是,由(2.3)式并利用条件(1.5),则不难推出$ z'(t) > 0(t\geqslant t_1) $.由$ z(t) $的定义并分别利用引理2.1和引理2.2,则可得

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

令

(2.5) $ \begin{equation} w(t) = \varphi(t)\frac{a(t)(z'(t))^\beta}{z^\gamma(\delta(t))}, t\geqslant t_1, \end{equation} $

则有$ w(t) > 0(t\geqslant t_1) $.再由$ (2.5) $式,并利用(2.3), (2.4)式及$ a(t)(z'(t))^\beta\leqslant a(\delta(t))(z'(\delta(t)))^\beta $,得

(2.6) $\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^{\frac{1}{\beta}}(t)z'(t)}{a^{\frac{1}{\beta}}(\delta(t))}\\&\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)-L\varphi(t)q(t)\bigg[\frac{(1-\alpha2^{1-\alpha}p(\delta(t)))z(\delta(t))-(2^{1-\alpha}-1)p(\delta(t))}{z(\delta(t))}\bigg]^\gamma\\&&-\gamma\frac{\varphi(t)a^{\frac{\beta+1}{\beta}}(t)(z'(t))^{\beta+1}\delta'(t)}{z^{\gamma+1}(\delta(t))a^{\frac{1}{\beta}}(\delta(t))}.\end{eqnarray} $

情形(a)  $ \beta\geqslant\gamma $.由于当$ t\geqslant t_1 $时$ z(t) > 0, z'(t) > 0 $,所以当$ t\geqslant t_1 $时,有

(2.7) $ \begin{equation} z(\delta(t))\geqslant z(\delta(t_1)) = k_1, \end{equation} $

这里$ k_1 > 0 $为常数.另一方面,由$ z'(t) > 0 $及(2.3)式知,当$ t\geqslant t_1 $时,有$ a(t)(z'(t))^\beta\leqslant a(t_1)(z'(t_1))^\beta = k_2 $ (这里$ k_2 > 0 $为常数),即$ z'(t)\leqslant\frac{k_2^{\frac{1}{\beta}}}{a^{\frac{1}{\beta}}(t)}, $亦即

(2.8) $ \begin{equation} (z'(t))^{\frac{\gamma-\beta}{\gamma}}\geqslant\frac{k_2^{\frac{\gamma-\beta}{\beta\gamma}}}{a^{\frac{\gamma-\beta}{\beta\gamma}}(t)}. \end{equation} $

于是,分别利用(2.5), (2.7)及(2.8)式,由(2.6)式,则当$ t\geqslant t_1 $时,得

(2.9) $ \begin{eqnarray} w'(t)&\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)-L\varphi(t)q(t)\bigg[(1-\alpha2^{1-\alpha}p(\delta(t)))-\frac{(2^{1-\alpha}-1)p(\delta(t))}{k_1}\bigg]^\gamma \\ &&-\frac{\gamma(z'(t))^{\frac{\gamma-\beta}{\gamma}}\delta'(t)a^{\frac{\gamma-\beta}{\beta\gamma}}(t)}{\varphi^{\frac{1}{\gamma}}(t)a^{\frac{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_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(t)}{\varphi^{\frac{1}{\gamma}}(t)a^{\frac{1}{\beta}}(\delta(t))}w^{\frac{\gamma+1}{\gamma}}(t). \end{eqnarray} $

分别注意到$ \frac{\partial H(t, s)}{\partial s}\leqslant0 $及引理2.3,由上式就可导出

(2.10) $ \begin{eqnarray} &&\int_{t_1}^tLH(t, s)\varphi(s)Q(s){\rm d}s\\ &\leqslant&-\int_{t_1}^tH(t, s)w'(s){\rm d}s+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^tH(t, s)\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}{\rm d}s, \end{eqnarray} $

整理上式可得

(2.11) $ \begin{eqnarray} &&\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\\ &\leq &H(t, t_1)w(t_1)\leqslant H(t, t_0)w(t_1), \end{eqnarray} $

即

$ \begin{eqnarray*} \frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\leqslant w(t_1), \end{eqnarray*} $

这与(2.1)式矛盾.

情形(b)  $ \beta < \gamma $.由(2.6)式,并利用(2.5), (2.7)式,则当$ t\geqslant t_1 $时,可得

(2.12) $ \begin{eqnarray} w'(t)&\leqslant&\frac{\varphi'(t)}{\varphi(t)}w(t)-L\varphi(t)q(t)\bigg[(1-\alpha2^{1-\alpha}p(\delta(t))) -\frac{(2^{1-\alpha}-1)p(\delta(t))}{k_1}\bigg]^\gamma\\ &&-\gamma\frac{z^{\frac{\gamma-\beta}{\beta}}(\delta(t)) \delta'(t)}{\varphi^{\frac{1}{\beta}}(t)a^{\frac{1}{\beta}}(\delta(t))}w^{\frac{\beta+1}{\beta}}(t) \\ &\leqslant&-L\varphi(t)Q(t)+\frac{\varphi'_+(t)}{\varphi(t)}w(t)-\frac{k_1^{\frac{\gamma-\beta}{\beta}}\gamma\delta'(t)}{\varphi^{\frac{1}{\beta}}(t)a^{\frac{1}{\beta}}(\delta(t))}w^{\frac{\beta+1}{\beta}}(t). \end{eqnarray} $

分别注意到$ \frac{\partial H(t, s)}{\partial s}\leqslant0 $及引理2.3,由上式就可推出

(2.13) $ \begin{eqnarray} &&\int_{t_1}^tLH(t, s)\varphi(s)Q(s){\rm d}s\\ &\leqslant&-\int_{t_1}^tH(t, s)w'(s){\rm d}s+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^tH(t, s)\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}{\rm d}s, \end{eqnarray} $

因此

(2.14) $ \begin{eqnarray} &&\int_{t_1}^t\bigg[LH(t, s)\varphi(s)Q(s)-\frac{(\frac{\beta}{\gamma})^\beta H(t, s)\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}\bigg]{\rm d}s\\ &\leqslant &H(t, t_1)w(t_1)\leqslant H(t, t_0)w(t_1), \end{eqnarray} $

这就意味着

$ \begin{eqnarray*} \frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{(\frac{\beta}{\gamma})^\beta \varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}\bigg]{\rm d}s\leqslant w(t_1), \end{eqnarray*} $

这与(2.2)式矛盾.

如果$ x(t) $为最终负解,则令$ y(t) = -x(t) $,则方程(1.1)就转化为

$ \{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)^* $

上式中函数$ F(|y(\delta(t))|^{\gamma-1}y(\delta(t))) = -f(|y(\delta(t))|^{\gamma-1}(-y(\delta(t)))), $易知函数$ F(u) = -f(-u) $与$ f(u) $具有完全相同的性质,注意到$ y(t) $是方程$ (1.1)^* $的最终正解,于是,用与上面完全相同的方法可推出矛盾.定理证毕.

定理2.2  如果条件(C$ _1) $–(C$ _4) $及(1.5)式成立,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)) $及$ H\in\Omega $,使得当$ \beta\geqslant\gamma $时

(2.15) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(H(t, s)\delta'(s))^\gamma}\bigg]{\rm d}s = +\infty, \end{equation} $

当$ \beta < \gamma $时

(2.16) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))|h(t, s)|^{\beta+1}}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(H(t, s)\delta'(s))^\beta}\bigg]{\rm d}s = +\infty, \end{equation} $

其中函数$ h(t, s) = \frac{\partial H(t, s)}{\partial s}w(s)+H(t, s)\frac{\varphi'_+(s)}{\varphi(s)}, Q(t) $如定理2.1,常数$ t_1\geqslant t_0, k_i > 0\ (i = 1, 2) $,则方程(1.1)振动.

证  同定理2.1的证明可得(2.9)式和(2.12)式.于是,当$ \beta\geqslant\gamma $时,由(2.9)式得

$ L\varphi(s)Q(s)\leqslant-w'(s)+\frac{\varphi'_+(t)}{\varphi(t)}w(t)-\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s), s\geqslant t_1, $

两边同乘以$ H(t, s) $后积分,并注意到函数$ h(t, s) $的定义及引理2.3,就可推得

$ \begin{eqnarray*} &&\int_{t_1}^tLH(t, s)\varphi(s)Q(s){\rm d}s\\ &\leqslant&-\int_{t_1}^tH(t, s)w'(s){\rm d}s+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)\bigg]{\rm d}s\\ & = &H(t, t_1)w(t_1)+\int_{t_1}^th(t, s)w(s){\rm d}s-\int_{t_1}^t\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}H(t, s)\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^t\bigg[|h(t, s)|w(s)-\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}H(t, s)\delta'(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^t\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(H(t, s)\delta'(s))^\gamma}{\rm d}s, \end{eqnarray*} $

整理得

$ \begin{eqnarray*} &&\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(H(t, s)\delta'(s))^\gamma}\bigg]{\rm d}s\\ &\leqslant &H(t, t_1)w(t_1)\leqslant H(t, t_0)w(t_1), \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} \frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(H(t, s)\delta'(s))^\gamma}\bigg]{\rm d}s\leqslant w(t_1), \end{eqnarray*} $

这与条件(2.15)矛盾.

当$ \beta < \gamma $时,由(2.12)式得

$ \begin{eqnarray*} L\varphi(s)Q(s)\leqslant-w'(s)+\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s), s\geqslant t_1, \end{eqnarray*} $

则类似于上面的情形,可得

$ \begin{eqnarray*} &&\int_{t_1}^tL\varphi(s)Q(s)H(t, s){\rm d}s\\ &\leqslant&-\int_{t_1}^tH(t, s)w'(s){\rm d}s+\int_{t_1}^tH(t, s)\bigg[\frac{\varphi'_+(s)}{\varphi(s)}w(s)-\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s)\bigg]{\rm d}s\\ & = &H(t, t_1)w(t_1)+\int_{t_1}^th(t, s)w(s){\rm d}s-\int_{t_1}^t\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma H(t, s)\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s){\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^t\bigg[|h(t, s)|w(s)-\frac{ k_1^{\frac{\gamma-\beta}{\beta}}\gamma H(t, s)\delta'(s)}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\beta+1}{\beta}}(s)\bigg]{\rm d}s\\ &\leqslant & H(t, t_1)w(t_1)+\int_{t_1}^t\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))|h(t, s)|^{\beta+1}}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(H(t, s)\delta'(s))^\beta}{\rm d}s, \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} &&\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))|h(t, s)|^{\beta+1}}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(H(t, s)\delta'(s))^\beta}\bigg]{\rm d}s\\ &\leqslant& H(t, t_1)w(t_1)\leqslant H(t, t_0)w(t_1), \end{eqnarray*} $

因此

$ \begin{eqnarray*} \frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))|h(t, s)|^{\beta+1}}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(H(t, s)\delta'(s))^\beta}\bigg]{\rm d}s\leqslant w(t_1), \end{eqnarray*} $

这与条件(2.16)矛盾.定理证毕.

如果定理2.1中的条件(2.1)或(2.2)不成立,亦或定理2.2中的条件(2.15)或(2.16)不成立,则方程(1.1)有下列振动准则.

定理2.3  如果条件(C$ _1) $–(C$ _4) $及$ (1.5) $式成立,且$ \beta\geqslant\gamma $,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)), $$ \xi\in L^2([t_0, +\infty), \mathbb{R}) $及$ H\in\Omega $,使得当$ s\geq t_1 $时

(2.17) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}{\rm d}s<+\infty, \end{equation} $

(2.18) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_u^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\geqslant\xi(u), \end{equation} $

并且函数$ \xi $满足

(2.19) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)[\xi(s)]_+^{\frac{\gamma+1}{\gamma}}}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s = +\infty, \end{equation} $

其中函数$ Q(s) $如定理2.1,常数$ t_1\geqslant t_0 $,则方程(1.1)振动.

证  同定理2.1的证明,当$ \beta\geqslant\gamma $时可得(2.10)式和(2.11)式.于是由(2.11)式,当$ t\geqslant u\geqslant t_1\geqslant t_0 $时,有

$ \begin{eqnarray*} \int_u^tLH(t, s)\varphi(s)Q(s){\rm d}s\leqslant H(t, t_0)w(u)+\int_u^t\frac{H(t, s)\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}{\rm d}s, \end{eqnarray*} $

即

$ \begin{eqnarray*} \frac{1}{H(t, t_0)}\int_u^t\bigg[LH(t, s)\varphi(s)Q(s)-\frac{H(t, s)\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\leqslant w(u), \end{eqnarray*} $

注意到(2.18)式,则当$ u\geqslant t_1 $时,有

(2.20) $ \begin{equation} \xi(u)\leqslant w(u),~~~~ \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tLH(t, s)\varphi(s)Q(s){\rm d}s\geqslant\xi(t_1). \end{equation} $

再根据(2.10)式,可得

$ \begin{eqnarray*} &&\frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(t, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)-H(t, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s)\bigg]{\rm d}s\\ &\leqslant & w(t_1)-\frac{1}{H(t, t_0)}\int_{t_1}^tLH(t, s)\varphi(s)Q(s){\rm d}s, \end{eqnarray*} $

注意到(2.20)式,于是由上式就可导出

(2.21) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\bigg[\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(t, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s)-H(t, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s)\bigg]{\rm d}s\leqslant C_1, \end{equation} $

这里$ C_1 = w(t_1)-\xi(t_1) $是常数.这样一来,就可断定

(2.22) $ \begin{equation} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(t, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s<+\infty. \end{equation} $

若不然,则存在序列$ \{T_n\}_{n = 1}^{+\infty}:T_n\in[t_1, +\infty) $,它满足$ \lim\limits_{n\rightarrow+\infty}T_n = +\infty, $并使得

(2.23) $ \begin{equation} \lim\limits_{n\rightarrow+\infty}\frac{1}{H(T_n, t_0)}\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s = +\infty. \end{equation} $

于是,结合(2.21)和(2.23)式就可推得

(2.24) $ \begin{equation} \lim\limits_{n\rightarrow+\infty}\frac{1}{H(T_n, t_0)}\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s = +\infty, \end{equation} $

因此,当正整数$ n $充分大时,就有

$ \begin{eqnarray*} \frac{1}{H(T_n, t_0)}\bigg[\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s-\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s\bigg]<C_1+1. \end{eqnarray*} $

这样一来,对充分大的正整数$ n $及任意的$ \varepsilon\in(0, 1) $,由上式就能导出

(2.25) $ \begin{equation} \frac{\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s}{\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s}>1-\varepsilon>0. \end{equation} $

此外,利用Hölder不等式$ \int_a^bf(x)g(x){\rm d}x\leqslant\Big[\int_a^b(f(x))^p{\rm d}x\Big]^{\frac{1}{p}}\Big[\int_a^b(g(x))^q{\rm d}x\Big]^{\frac{1}{q}} $ (其中$ p > 1 $且$ \frac{1}{p}+\frac{1}{q} = 1 $),我们可推得

(2.26) $ \begin{eqnarray} &&\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s\\ & = &\int_{t_1}^{T_n}\bigg[\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}\bigg]^{\frac{\gamma}{\gamma+1}}w(s)\frac{\bigg[\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))\bigg]^{\frac{\gamma}{\gamma+1}}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}}{\bigg[\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)\bigg]^{\frac{\gamma}{\gamma+1}}}{\rm d}s\\ &\leqslant&\bigg[\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s\bigg]^{\frac{\gamma}{\gamma+1}}\\ &&\times\bigg[\int_{t_1}^{T_n}\frac{[\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))]^\gamma\Big(H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}}{[\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)]^\gamma}{\rm d}s\bigg]^{\frac{1}{\gamma+1}}\\ & = &\frac{k_2^{\frac{\beta-\gamma}{\beta(\gamma+1)}}}{\gamma^{\frac{\gamma}{\gamma+1}}}\bigg[\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s\bigg]^{\frac{\gamma}{\gamma+1}}\\ &&\times \bigg[\int_{t_1}^{T_n}\frac{H(T_n, s)\varphi(s)a^{\frac{1}{\beta}}(\delta(s))\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}}{( \delta'(s))^\gamma}{\rm d}s\bigg]^{\frac{1}{\gamma+1}}. \end{eqnarray} $

由(2.26)式,并分别利用(2.25)和(2.17)式,进一步可推得

$ \begin{eqnarray*} 0&<&\frac{(1-\varepsilon)^\gamma}{H(T_n, t_0)}\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s <\frac{\Big[\int_{t_1}^{T_n}H(T_n, s)\frac{\varphi'_+(s)}{\varphi(s)}w(s){\rm d}s\Big]^{\gamma+1}}{H(T_n, t_0)\Big[\int_{t_1}^{T_n}\frac{\gamma k_2^{\frac{\gamma-\beta}{\beta\gamma}}\delta'(s)H(T_n, s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}w^{\frac{\gamma+1}{\gamma}}(s){\rm d}s\Big]^{\gamma}}\\ &\leqslant&\frac{k_2^{\frac{\beta-\gamma}{\beta}}}{\gamma^\gamma}\frac{1}{H(T_n, t_0)}\int_{t_1}^{T_n}\frac{H(T_n, s)\varphi(s))a^{\frac{1}{\beta}}(\delta(s))}{(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}{\rm d}s<+\infty, \end{eqnarray*} $

这与(2.24)式矛盾,这就证明了(2.22)式成立.注意到(2.20)式和(2.22)式,于是可得

$ \begin{eqnarray*} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)[\xi(s)]_+^{\frac{\gamma+1}{\gamma}}}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s&\leqslant&\liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)w^{\frac{\gamma+1}{\gamma}}(s)}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s\\ &<&+\infty, \end{eqnarray*} $

这与(2.19)式矛盾.定理证毕.

当$ \beta < \gamma $时,利用定理2.1证明所得到的(2.13)式和(2.14)式,并使用与定理2.3相同的方法,可得如下定理.

定理2.4  如果条件(C$ _1) $–(C$ _4) $及$ (1.5) $式成立,且$ \beta < \gamma $,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)), $$ \xi\in L^2([t_0, +\infty), \mathbb{R}) $及$ H\in\Omega $,使得当$ s\geqslant t_1 $时

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\varphi(s)a(\delta(s))}{(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}{\rm d}s<+\infty, \end{eqnarray*} $

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_u^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(\delta'(s))^\beta}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\beta+1}\bigg]{\rm d}s\geqslant\xi(u), \end{eqnarray*} $

并且函数$ \xi $满足

$ \begin{eqnarray*} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)[\xi(s)]_+^{\frac{\beta+1}{\beta}}}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s = +\infty, \end{eqnarray*} $

其中函数$ Q(s) $如定理2.1中定义,常数$ t_1\geqslant t_0 $,则方程(1.1)振动.

结合定理2.2证明过程中所得到的有关不等式,类似地可得如下定理.

定理2.5  如果条件(C$ _1) $–(C$ _4) $及$ (1.5) $式成立,且$ \beta\geqslant\gamma $,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)), $$ \xi\in L^2([t_0, +\infty), \mathbb{R}) $及$ H\in\Omega $,使得当$ s\geqslant t_1 $时

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(H(t, s)\delta'(s))^\gamma}{\rm d}s<+\infty, \end{eqnarray*} $

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_u^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\gamma+1}}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(H(t, s)\delta'(s))^\gamma}\bigg]{\rm d}s\geqslant\xi(u), \end{eqnarray*} $

并且函数$ \xi $满足

$ \begin{eqnarray*} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)[\xi(s)]_+^{\frac{\gamma+1}{\gamma}}}{\varphi^{\frac{1}{\gamma}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s = +\infty, \end{eqnarray*} $

其中函数$ Q(s) $如定理2.1中定义, $ h(t, s) $如定理2.2,常数$ t_1\geqslant t_0 $,则方程(1.1)振动.

定理2.6  如果条件(C$ _1) $–(C$ _4) $及$ (1.5) $式成立,且$ \beta < \gamma $,若有函数$ \varphi\in C^1([t_0, +\infty), (0, +\infty)), $$ \xi\in L^2([t_0, +\infty), \mathbb{R}) $及$ H\in\Omega $,使得当$ s\geqslant t_1 $时

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))|h(t, s)|^{\beta+1}}{(H(t, s)\delta'(s))^\beta}{\rm d}s<+\infty, \end{eqnarray*} $

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_u^t\bigg[L\varphi(s)Q(s)H(t, s)-\frac{(\frac{\beta}{\gamma})^\beta\varphi(s)a(\delta(s))|h(t, s)|^{\beta+1}}{(\beta+1)^{\beta+1}k_1^{(\gamma-\beta)}(H(t, s)\delta'(s))^\beta}\bigg]{\rm d}s\geqslant\xi(u), \end{eqnarray*} $

并且函数$ \xi $满足

$ \begin{eqnarray*} \liminf\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^t\frac{H(t, s)\delta'(s)[\xi(s)]_+^{\frac{\beta+1}{\beta}}}{\varphi^{\frac{1}{\beta}}(s)a^{\frac{1}{\beta}}(\delta(s))}{\rm d}s = +\infty, \end{eqnarray*} $

其中函数$ Q(s) $如定理2.1中定义, $ h(t, s) $如定理2.2,常数$ t_1\geqslant t_0 $,则方程(1.1)振动.

注1  在定理2.1 –定理2.6中,选取适当的不同的函数$ \varphi(t) $和$ H(t, s) $,就可得到含非线性中立项的方程(1.1)的种种不同的振动准则.此外,从定理2.1 –定理2.6可以看出,在$ \beta < \gamma $和$ \beta > \gamma $两种不同的情形下,方程(1.1)的振动准则是不一样的,但当$ \beta = \gamma $时定理2.1中的(2.1)式和(2.2)式,定理2.2中的(2.15)式和(2.16)式、定理2.3和定理2.4以及定理2.5和定理2.6中相应的条件却完全相同.

定理2.7  如果条件(C$ _1) $–(C$ _4) $及$ (1.7) $式成立,且$ \theta(t)-\alpha2^{1-\alpha}p(t)\theta(\tau(t)) > 0 $,并且当$ \beta\geqslant\gamma $时条件(2.1)成立,当$ \beta < \gamma $时条件(2.2)成立.如果存在函数$ K\in\Omega $使得

(2.27) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\int_{t_1}^t\bigg[L\overline{Q}(s)K(t, s)-\frac{a(s)}{(\beta+1)^{\beta+1}K^\beta(t, s)}\Big|\frac{\partial K(t, s)}{\partial s}\Big|^{\beta+1}\bigg]{\rm d}s>0, \end{equation} $

其中常数$ t_1\geqslant t_0 $,函数

$ \overline{Q}(t) = q(t)\bigg[\Big(1-\alpha2^{1-\alpha}p(\delta(t))\frac{\theta(\tau(\delta(t)))} {\theta(\delta(t))}\Big)^\gamma\rho(t)-\frac{(2^{1-\alpha}-1)^\gamma}{k\theta^\beta(t)}p^\gamma(\delta(t))\bigg], $

$ \theta(t) = \int_t^{+\infty}a^{-\frac{1}{\beta}}(s){\rm d}s, $

$ \rho(t) = \left\{\begin{array}{ll} k^{\frac{\gamma-\beta}{\beta}}\theta^{(\gamma-\beta)}(t), & \beta<\gamma, \\ m, &\beta>\gamma, \\ 1, &\beta = \gamma, \end{array} \right.\ \mbox{(这里$k>0, m>0$是某常数), } $

则方程(1.1)振动.

证  若不然,则方程$ (1.1) $存在一个非振动解$ x(t) $,不妨设$ x(t) $为最终正解,则有$ x(t) > 0 $, $ x(\tau(t)) > 0, x(\delta(t)) > 0, t\geqslant t_1\geqslant t_0 $,且$ z(t)\geqslant x(t) > 0\ (t\geqslant t_1), $于是容易由定理2.1证明知: $ a(t)|z'(t)|^{\beta-1}z'(t) $严格单调减少且最终定号,从而$ z'(t) $也最终定号,故只须考虑两种情形

(ⅰ) $ z'(t) > 0\ (t\geqslant t_1) $; (ⅱ) $ z'(t) < 0\ (t\geqslant t_1) $.

情形(ⅰ)   $ z'(t) > 0\ (t\geqslant t_1) $.同定理$ 2.1 $的证明,当$ \beta\geqslant\gamma $时可得到一个与条件$ (2.1) $矛盾的结果,当$ \beta < \gamma $时可得到一个与条件(2.2)矛盾的结果.

情形(ⅱ)   $ z'(t) < 0\ (t\geqslant t_1) $.定义函数$ u(t) $如下

(2.28) $ \begin{equation} u(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\geqslant t_1, \end{equation} $

则$ u(t) < 0\ (t\geqslant t_1) $.由$ a(t)|z'(t)|^{\beta-1}z'(t) = a(t)(-z'(t))^{\beta-1}z'(t) $在$ [t_1, +\infty) $上严格单减,得

$ \begin{eqnarray*} a(s)(-z'(s))^{\beta-1}z'(s)\leqslant a(t)(-z'(t))^{\beta-1}z'(t), s\geqslant t\geqslant t_1, \end{eqnarray*} $

即$ z'(s)\leqslant a^{\frac{1}{\beta}}(t)z'(t)a^{-\frac{1}{\beta}}(s) $,由此得

$ z(l)-z(t)\leqslant a^{\frac{1}{\beta}}(t)z'(t)\int_t^la^{-\frac{1}{\beta}}(s){\rm d}s, $

所以

$ z(t)+a^{\frac{1}{\beta}}(t)z'(t)\int_t^la^{-\frac{1}{\beta}}(s){\rm d}s\geqslant0, $

令$ l\rightarrow+\infty $,并利用函数$ \theta(t) $的定义,得

(2.29) $ \begin{equation} z(t)+a^{\frac{1}{\beta}}(t)z'(t)\theta(t)\geqslant0, \end{equation} $

所以

(2.30) $ \begin{equation} -1\leqslant\frac{a^{\frac{1}{\beta}}(t)z'(t)}{z(t)}\theta(t)<0. \end{equation} $

注意到(2.29)式,有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\bigg[\frac{z(t)}{\theta(t)}\bigg] = \frac{z'(t)\theta(t)-z(t)[-a^{-\frac{1}{\beta}}(t)]}{\theta^2(t)} = \frac{a^{\frac{1}{\beta}}(t)z'(t)\theta(t)+z(t)}{a^{\frac{1}{\beta}}(t)\theta^2(t)}\geqslant0, \end{eqnarray*} $

这就是说$ \frac{z(t)}{\theta(t)} $是单调增加的,从而有$ \frac{z(\tau(t))}{\theta(\tau(t))}\leqslant\frac{z(t)}{\theta(t)} $,即

(2.31) $ \begin{equation} z(\tau(t))\leqslant\frac{\theta(\tau(t))}{\theta(t)}z(t). \end{equation} $

另一方面,利用引理2.1、引理2.2及(2.31)式,可推得

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

此外,由(2.28)式得

(2.33) $ \begin{eqnarray} u'(t)& = &\frac{[a(t)|z'(t)|^{\beta-1}z'(t)]'}{z^\beta(t)}-\frac{a(t)(-z'(t))^{\beta-1}z'(t)\cdot\beta z^{\beta-1}(t)z'(t)}{z^{2\beta}(t)} \\ &\leqslant&-Lq(t)\frac{x^\gamma(\delta(t))}{z^\beta(t)}-\frac{\beta}{a^{\frac{1}{\beta}}(t)}(-u(t))^{\frac{\beta+1}{\beta}}. \end{eqnarray} $

根据(2.3)式,有

$ \begin{eqnarray*} a(s)(-z'(t))^{\beta-1}z'(s)\leqslant a(t_1)(-z'(t_1))^{\beta-1}z'(t_1) = -k, s\geqslant t_1, \end{eqnarray*} $

(这里$ k > 0 $是常数),所以$ z'(s)\leqslant-k^{\frac{1}{\beta}}a^{-\frac{1}{\beta}}(s) $,两边积分得$ z(l)-z(t)\leqslant-k^{\frac{1}{\beta}}\int_t^la^{-\frac{1}{\beta}}(s){\rm d}s, $即

$ \begin{eqnarray*} z(t)\geqslant z(l)+k^{\frac{1}{\beta}}\int_t^la^{-\frac{1}{\beta}}(s){\rm d}s\geqslant k^{\frac{1}{\beta}}\int_t^la^{-\frac{1}{\beta}}(s){\rm d}s, \end{eqnarray*} $

令$ l\rightarrow+\infty $,得$ z(t)\geqslant k^{\frac{1}{\beta}}\theta(t), $即

(2.34) $ \begin{equation} \frac{1}{z^\beta(t)}\leqslant\frac{1}{k\theta^\beta(t)}. \end{equation} $

当$ \beta = \gamma $时, $ z^{\gamma-\beta}(t) = 1. $当$ \beta > \gamma $时,由$ z(t) > 0, z'(t) < 0(t\geqslant t_1) $知, $ z(t)\leqslant z(t_1) = b $,因此$ z^{\gamma-\beta}(t)\geqslant b^{\gamma-\beta}. $当$ \beta < \gamma $时,由(2.34)式知$ z^{\gamma-\beta }(t)\geqslant k^{\frac{\gamma-\beta}{\beta}}\theta^{\gamma-\beta}(t). $利用函数$ \rho(t) $的定义有

(2.35) $ \begin{equation} z^{\gamma-\beta}(t)\geqslant\rho(t), \end{equation} $

再由(2.32), (2.34)和(2.35)式及函数$ \overline{Q}(t) $的定义,得

(2.36) $ \begin{eqnarray} q(t)\frac{x^\gamma(\delta(t))}{z^\beta(t)}&\geqslant & q(t)\frac{\Big(1-\alpha2^{1-\alpha}p(\delta(t))\frac{\theta(\tau(\delta(t)))}{\theta(\delta(t))}\Big)^\gamma z^\gamma(\delta(t))+(1-2^{1-\alpha})^\gamma p^\gamma(\delta(t))}{z^\beta(t)} \\ &\geqslant & q(t)\bigg[\Big(1-\alpha2^{1-\alpha}p(\delta(t))\frac{\theta(\tau(\delta(t)))}{\theta(\delta(t))}\Big)^\gamma z^{\gamma-\beta}(t)-\frac{(2^{1-\alpha}-1)^\gamma}{z^\beta(t)}p^\gamma(\delta(t))\bigg] \\ &\geqslant & \overline{Q}(t). \end{eqnarray} $

将(2.36)式代入(2.33)式,得

$ \begin{eqnarray*} u'(t)\leqslant-L\overline{Q}(t)-\frac{\beta}{a^{\frac{1}{\beta}}(t)}(-u(t))^{\frac{\beta+1}{\beta}}, t\geqslant t_1, \end{eqnarray*} $

两边同乘以$ K(t, s) $后积分,得

$ \begin{eqnarray*} \int_{t_1}^tL\overline{Q}(s)K(t, s){\rm d}s &\leqslant&-\int_{t_1}^tK(t, s)u'(s){\rm d}s-\int_{t_1}^t\frac{\beta K(t, s)}{a^{\frac{1}{\beta}}(t)}(-u(s))^{\frac{\beta+1}{\beta}}{\rm d}s\\ &\leqslant & K(t, t_1)v(t_1)+\int_{t_1}^t\bigg[\Big(-\frac{\partial K(t, s)}{\partial s}\Big)(-u(s))-\frac{\beta K(t, s)}{a^{\frac{1}{\beta}}(t)}(-u(s))^{\frac{\beta+1}{\beta}}\bigg]{\rm d}s, \end{eqnarray*} $

应用引理2.3,由上式进一步可得

$ \begin{eqnarray*} \int_{t_1}^tL\overline{Q}(s)K(t, s){\rm d}s\leqslant K(t, t_1)v(t_1)+\int_{t_1}^t\frac{a(s)}{(\beta+1)^{\beta+1}K^\beta(t, s)}\Big|\frac{\partial K(t, s)}{\partial s}\Big|^{\beta+1}{\rm d}s, \end{eqnarray*} $

整理,得

$ \begin{eqnarray*} \int_{t_1}^t\bigg[L\overline{Q}(s)K(t, s)-\frac{a(s)}{(\beta+1)^{\beta+1}K^\beta(t, s)}\Big|\frac{\partial K(t, s)}{\partial s}\Big|^{\beta+1}\bigg]{\rm d}s\leqslant K(t, t_1)v(t_1), \end{eqnarray*} $

这与条件(2.27)矛盾,证毕.

定理2.8  如果条件(C$ _1) $–(C$ _4) $及$ (1.7) $式成立,且$ \theta(t)-\alpha2^{1-\alpha}p(t)\theta(\tau(t)) > 0 $,并且当$ \beta\geqslant\gamma $时条件(2.1)成立,当$ \beta < \gamma $时条件(2.2)成立.进一步,设$ \beta\geqslant1 $是两个正奇数的商,并且存在函数$ K\in\Omega $及$ \xi, \eta\in C^1([t_0, +\infty), (0, +\infty)) $且$ \eta(t)\geqslant\frac{1}{a(t)\theta^\beta(t)} $,使得

(2.37) $ \begin{equation} \limsup\limits_{t\rightarrow+\infty}\frac{1}{K(t, t_0)}\int_{t_1}^t\bigg[K(t, s)\xi(s)\Phi(s)-\frac{K(t, s)a(s)\xi(s)}{(\beta+1)^{\beta+1}}\Big(\frac{ \xi'(s)}{\xi(s)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(s)} \Big)^{\beta+1}\bigg]{\rm d}s = +\infty, \end{equation} $

其中函数$ \Phi(t) = L\overline{Q}(t)-[\eta(t)a(t)]'+a(t)\eta^{\frac{\beta+1}{\beta}}(t), $而函数$ \theta(t) $及$ \overline{Q}(t) $如定理2.7.则方程(1.1)是振动的.

证  同定理2.7,只要考虑下列两种情形

(ⅰ) $ z'(t) > 0\ (t\geqslant t_1) $; (ⅱ) $ z'(t) < 0\ (t\geqslant t_1) $.

情形(ⅰ)   $ z'(t) > 0\ (t\geqslant t_1) $.同定理$ 2.1 $的证明,当$ \beta\geqslant\gamma $时可得到一个与条件(2.1)矛盾的结果,当$ \beta < \gamma $时可得到一个与条件(2.2)矛盾的结果.

情形(ⅱ)  $ z'(t) < 0\ (t\geqslant t_1) $.由定理2.1的证明知(2.30)式成立,注意到$ \beta\geqslant1 $为两个正奇数之商,于是有

(2.38) $ \begin{equation} \frac{a(t)(z'(t))^\beta}{z^\beta(t)}+\frac{1}{\theta^\beta(t)}\geqslant0. \end{equation} $

定义函数$ u(t) $如下

(2.39) $ \begin{equation} u(t) = \xi(t)\bigg[\frac{a(t)|z'(t)|^{\beta-1}z'(t)}{z^\beta(t)}+\eta(t)a(t)\bigg] = \xi(t)\bigg[\frac{a(t)(z'(t))^\beta}{z^\beta(t)}+\eta(t)a(t)\bigg], t\geqslant t_1, \end{equation} $

则由条件(2.38)及函数$ \eta(t) $的定义知$ u(t)\geqslant0\ (t\geqslant t_1) $.由(2.39)式及(2.36)式可导出

(2.40) $ \begin{eqnarray} u'(t)& = &\xi'(t)\bigg[\frac{a(t)|z'(t)|^{\beta-1}z'(t)}{z^\beta(t)}+\eta(t)a(t)\bigg] \\ &&+\xi(t)\bigg\{\frac{[a(t)(z'(t))^\beta]'}{z^\beta(t)}-\frac{a(t)(z'(t))^\beta\beta z^{\beta-1}(t)z'(t)}{z^{2\beta}(t)}+[\eta(t)a(t)]'\bigg\} \\ &\leqslant&\frac{\xi'(t)}{\xi(t)}u(t)+\xi(t)\bigg\{-Lq(t)\frac{x^\gamma(\delta(t))}{z^\beta(t)}-\frac{a(t)(z'(t))^\beta\beta z^{\beta-1}(t)z'(t)}{z^{2\beta}}+[\eta(t)a(t)]'\bigg\} \\ &\leqslant&\frac{\xi'(t)}{\xi(t)}u(t)-\xi(t)L\overline{Q}(t)-\beta\bigg[\xi(t)\frac{a(t)(z'(t))^{\beta+1}}{z^{\beta+1}(t)}\bigg]+\xi(t)[\eta(t)a(t)]' \\ & = &\frac{\xi'(t)}{\xi(t)}u(t)-L\xi(t)\overline{Q}(t)-\frac{\beta\xi(t)}{a^{\frac{1}{\beta}}(t)}\bigg[\frac{u(t)} {\xi(t)}-\eta(t)a(t)\bigg]^{\frac{\beta+1}{\beta}}+\xi(t)[\eta(t)a(t)]'. \end{eqnarray} $

令$ \lambda = \beta, X = \frac{u(t)}{\xi(t)}, Y = \eta(t)a(t) $,代入引理2.4中的不等式

$ (X-Y)^{\frac{1}{\lambda}+1}\geqslant X^{\frac{1}{\lambda}+1}-\Big(\frac{1}{\lambda}+1\Big)XY^{\frac{1}{\lambda}}+\frac{1}{\lambda}Y^{\frac{1}{\lambda}+1}, $

得

$ \begin{eqnarray*} \bigg[\frac{u(t)}{\xi(t)}-\eta(t)a(t)\bigg]^{\frac{\beta+1}{\beta}}\geqslant \Big(\frac{u(t)}{\xi(t)}\Big)^{\frac{\beta+1}{\beta}}-\frac{\beta+1}{\beta} [\eta(t)a(t)]^{\frac{1}{\beta}}\frac{u(t)}{\xi(t)}+\frac{1}{\beta}[\eta(t)a(t)]^{\frac{\beta+1}{\beta}}, \end{eqnarray*} $

将上式代入(2.40)式,则有

(2.41) $ \begin{eqnarray} u'(t)&\leqslant&-L\xi(t)\overline{Q}(t)+\xi(t)[\eta(t)a(t)]'+\frac{\xi'(t)}{\xi(t)}u(t)\\ &&-\frac{\beta\xi(t)}{a^{\frac{1}{\beta}}(t)}\bigg[\Big(\frac{u(t)}{\xi(t)}\Big)^{\frac{\beta+1}{\beta}}-\frac{\beta+1}{\beta}[\eta(t)a(t)]^{\frac{1}{\beta}}\frac{u(t)}{\xi(t)}+\frac{1}{\beta}[\eta(t)a(t)]^{\frac{\beta+1}{\beta}}\bigg] \\ & = &-\xi(t)\bigg[L\overline{Q}(t)-(\eta(t)a(t))'+a(t)\eta^{\frac{\beta+1}{\beta}}(t)\bigg]+\bigg[\frac{\xi'(t)}{\xi(t)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(t)}\bigg]u(t)-\frac{\beta[u(t)]^{\frac{\beta+1}{\beta}}}{[a(t)\xi(t)]^{\frac{1}{\beta}}} \\ & = &-\xi(t)\Phi(t)+\bigg[\frac{\xi'(t)}{\xi(t)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(t)}u(t)\bigg]-\frac{\beta[u(t)]^{\frac{\beta+1}{\beta}}}{[a(t)\xi(t)]^{\frac{1}{\beta}}}. \end{eqnarray} $

(2.41)式两边同乘以$ K(t, s) $后积分,注意到$ \frac{\partial K(t, s)}{\partial s}\leqslant0 $,并应用引理2.3,可得

$ \begin{eqnarray*} &&\int_{t_1}^tK(t, s)\xi(s)\Phi(s){\rm d}s\\ &\leqslant&-\int_{t_1}^tK(t, s)u'(s){\rm d}s+\int_{t_1}^tK(t, s)\bigg[\Big(\frac{\xi'(t)}{\xi(s)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(s)}\Big)u(s)-\frac{\beta(u(s))^{\frac{\beta+1}{\beta}}}{(a(s)\xi(s))^{\frac{1}{\beta}}}\bigg]{\rm d}s\\ &\leqslant & K(t, t_1)u(t_1)+\int_{t_1}^tK(t, s)\bigg[\Big(\frac{\xi'(t)}{\xi(s)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(s)}\Big)u(s)-\frac{\beta(u(s))^{\frac{\beta+1}{\beta}}}{(a(s)\xi(s))^{\frac{1}{\beta}}}\bigg]{\rm d}s\\ &\leqslant & K(t, t_1)u(t_1)+\int_{t_1}^t\frac{K(t, s)a(t)\xi(t)}{(\beta+1)^{\beta+1}}\Big(\frac{\xi'(t)}{\xi(t)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(t)}\Big)^{\beta+1}{\rm d}s, \end{eqnarray*} $

所以

$ \begin{eqnarray*} \frac{1}{k(t, t_0)}\int_{t_1}^t\bigg[k(t, s)\xi(s)\Phi(s)-\frac{k(t, s)a(s)\xi(s)}{(\beta+1)^{\beta+1}}\Big(\frac{\xi'(s)}{\xi(s)}+\frac{\beta+1}{\eta^{-\frac{1}{\beta}}(s)}\Big)^{\beta+1}\bigg]{\rm d}s\leqslant u(t_1), \end{eqnarray*} $

这与条件(2.37)矛盾,证毕.

注2  显然,本文的定理2.7和定理2.8没有文献[5]中定理2.4中的条件"$ a'(t)\geqslant0, $$ \beta\geqslant\gamma, $$ p'(t)\geqslant0 $且$ \tau'(t) > 0 $",而且本文得到的结果是确定性的结论,应用非常方便.

例3.1  对常数$ m > 0, $考虑如下微分方程

(3.1) $ \begin{equation} \bigg[x(t)+\frac{1}{10}x(\frac{t}{4})\bigg]''+\frac{m}{t^2}x(t) = 0, t\geqslant1, \end{equation} $

令$ a(t)\equiv1, p(t) = \frac{1}{10}, q(t) = mt^{-2}, \tau(t) = \frac{t}{4}, \delta(t) = t, f(u) = u, \alpha = \beta = \gamma = 1, t_0 = 1 $则容易验证条件(C$ _1) $–(C$ _4) $及$ (1.5) $式都是满足的.在定理$ 2.1 $中取$ \varphi(t) = t, H(t, s) = (t-s)^2, t_1 = 2, $注意到$ L = 1 $,则当$ m > \frac{5}{18} = 0.2\dot{7} $时,有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\\ & = &\frac{9}{10}\Big(m-\frac{5}{18}\Big)\limsup\limits_{t\rightarrow+\infty}\frac{1}{(t-1)^2}\int_2^t\frac{(t-s)^2}{s}{\rm d}s = +\infty, \end{eqnarray*} $

即$ (2.1) $式成立,于是根据定理$ 2.1, $当$ m > 0.2\dot{7} $时方程$ (3.1) $是振动的.

注3  对方程$ (3.1) $,我们也可以用文献[8]中的定理$ 3.4 $来判定其振动性:由于

$ \begin{eqnarray*} \limsup\limits_{t\rightarrow+\infty}\int_{t_0}^t\bigg\{\frac{\varphi(s)q(s)}{2^{\gamma-1}}-\frac{(1+\frac{p_0^\gamma}{\tau_0})[\varphi'_+(s)]^{\gamma+1}a(\delta(s))}{(\gamma+1)^{\gamma+1}[\tau_0\varphi(s)]^\gamma}\bigg\}{\rm d}s = (m-1.4)\limsup\limits_{t\rightarrow+\infty}\int_1^t\frac{1}{s}{\rm d}s, \end{eqnarray*} $

因此当$ m > 1.4 $时方程$ (3.1) $是振动的.这就说明它是本文所获得的振动准则的特殊情形,其"精细"程度与文献[8]的结果相比要好得多.

例3.2  考虑方程$ (1.6) $,即下列二阶变时滞微分方程

$ \begin{eqnarray*} \bigg[\frac{1}{t^2}\Big|\Big(x(t)+\frac{1}{2}x(\sqrt{t})\Big)'\Big|^{\frac{1}{3}}\Big(x(t)+\frac{1}{2}x(\sqrt{t}\Big)'\bigg]'+\frac{2}{t^2}|x(\frac{t}{2})|x(\frac{t}{2}) = 0, t\geqslant1, \end{eqnarray*} $

相当于方程$ (1.1) $中$ \alpha = 1, \beta = \frac{4}{3}, \gamma = 2, t_0 = 1, a(t) = \frac{1}{t^2}, $$ p(t) = \frac{1}{2}, q(t) = \frac{2}{t^2}, f(u) = u, $$ \tau(t) = \sqrt{t}, $$ \delta(t) = \frac{t}{2}. $由于$ \int_{t_0}^{+\infty}a^{-\frac{1}{\beta}}(t){\rm d}t = \int_1^{+\infty}t^{\frac{3}{2}}{\rm d}t = +\infty, $因此条件(C$ _1) $–(C$ _4) $及$ (1.5) $式都满足.现在取$ \varphi(t) = t, H(t, s) = (t-s)^2, $注意到$ Q(t) = q(t)\{1-[\alpha2^{1-\alpha}+k_1^{-1}(2^{1-\alpha}-1)]p(\delta(t))\}^\gamma = \frac{1}{2t}, $则有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\\ & = &\limsup\limits_{t\rightarrow+\infty}\frac{1}{(t-1)^2}\int_1^t(t-s)^2\Big(\frac{1}{s}-\frac{2^5}{9\sqrt{k_2}}\frac{1}{s^5}\Big){\rm d}s = +\infty, \end{eqnarray*} $

因此定理$ 2.1 $的条件全部满足,由此知方程$ (1.6) $是振动的.

注4  显然,由于不满足条件"$ a'(t)\geqslant0, \beta\geqslant\gamma $",因此定理A和定理B都不能用于方程(1.6);又由于不满足条件"$ \delta\circ\tau = \tau\circ\delta, \delta(t)\leqslant\tau(t) $",因此定理C也不能用于方程(1.6).

例3.3  对常数$ m > 0 $,考虑二阶微分方程

(3.2) $ \begin{equation} \bigg\{t\bigg[\Big(x(t)+\frac{1}{4}x^{\frac{1}{3}}(\frac{t}{2})\Big)'\bigg]^{\frac{5}{3}}\bigg\}' +\frac{m}{t}|x(\frac{t}{3})|x^{\frac{1}{9}}(\frac{t}{3}) = 0, t\geqslant1, \end{equation} $

这是具非线性中立项的二阶时滞微分方程,令$ \alpha = \frac{1}{3}, \beta = \frac{5}{3}, \gamma = \frac{10}{9}, t_0 = 1, a(t) = t, $$ p(t) = \frac{1}{4}, $$ q(t) = \frac{m}{t}, f(u) = u, \tau(t) = \frac{t}{2}, \delta(t) = \frac{t}{3}, $则条件(C$ _1) $–(C$ _4) $及$ (1.5) $式显然满足.为了计算简单,在定理$ 2.1 $中取$ \varphi(t) = 1, H(t, s) = t-s, t_1 = 2 $,注意$ L = 1, \beta > \gamma $,则

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s)\bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\\ & = &\limsup\limits_{t\rightarrow+\infty}\frac{1}{(t-1)}\int_2^t(t-s)\bigg\{\frac{m}{s}\bigg[1-\Big(\frac{1}{3}2^{\frac{2}{3}}+k_1^{-1}(2^{\frac{2}{3}}-1)\Big)\frac{1}{4}\bigg]^{\frac{10}{9}}-0\bigg\}{\rm d}s = +\infty, \end{eqnarray*} $

因此定理2.1的条件全部满足,由此知方程(3.2)振动.

注5  由于方程(3.2)的中立项是非线性的,并且$ \beta\neq\gamma, $因此文献[1-18]中的定理对方程(3.2)都不适用.

例3.4  考虑方程$ (1.8) $,即下面的Euler微分方程

$ \begin{eqnarray*} (t^2x'(t))'+mx(t) = 0, t\geqslant1, \end{eqnarray*} $

其中常数$ m > 0 $.这相当于方程(1.1)中$ a(t) = t^2, $$ p(t)\equiv0, $$ q(t) = m, $$ \tau(t) = \delta(t) = t, $$ \beta = \gamma = 1, $$ t_0 = 1 $.容易验证条件(C$ _1 $)–(C$ _4) $及(1.7)式成立.取$ \varphi(t) = 1, H(t, s) = (t-s)^2, $$ K(t, s) = t-s, t_1 = 2, $则有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\frac{1}{H(t, t_0)}\int_{t_1}^tH(t, s) \bigg[L\varphi(s)Q(s)-\frac{\varphi(s)a^{\frac{\gamma}{\beta}}(\delta(s))}{(\gamma+1)^{\gamma+1}k_2^{\frac{\gamma-\beta}{\beta}}(\delta'(s))^\gamma}\Big(\frac{\varphi'_+(s)}{\varphi(s)}\Big)^{\gamma+1}\bigg]{\rm d}s\\ & = &\limsup\limits_{t\rightarrow+\infty}\frac{1}{(t-1)^2}\int_2^tm(t-s)^2{\rm d}s = +\infty, \end{eqnarray*} $

且当$ m > \frac{1}{4} $时,有

$ \begin{eqnarray*} &&\limsup\limits_{t\rightarrow+\infty}\int_{t_1}^t\bigg[L\overline{Q}(s)K(t, s)-\frac{a(s)}{(\beta+1)^{\beta+1}K^\beta(t, s)}\Big|\frac{\partial K(t, s)}{\partial s}\Big|^{\beta+1}\bigg]{\rm d}s\\ \\ & = &\limsup\limits_{t\rightarrow+\infty}\int_2^t(t-s)\bigg[m-\frac{1}{4}\frac{s^2}{(t-s)^2}\bigg]{\rm d}s>0, \end{eqnarray*} $

因此,根据定理2.7,当$ m > \frac{1}{4} $时方程(1.8)是振动的.这就是我们熟悉的结果.

显然,当方程(1.1)中的$ \alpha = \beta = \gamma = 1 $,或者$ \alpha = 1 $且$ \beta = \gamma $,或者$ \gamma = 1 $时的特殊情形,本文所得到的这些振动准则改进且推广了一系列已有的结果.