Oscillation of Second Order Nonlinear Differential Equations with Neutral Delay

Zhang Zhiyu ,1, Yu Yuanhong2, Li Shuping3, Qiao Shizhu1

 基金资助: 国家自然科学基金.  11701528国家自然科学基金.  11647034山西省自然科学基金.  2011011002-3

 Fund supported: the NSFC.  11701528the NSFC.  11647034the Natural Science Foundation of Shanxi Province.  2011011002-3 Abstract

In this paper, the oscillatory behavior of solutions to a nonlinear second-order neutral differential equation is to study. Using double Riccati transformation and the technique of inequations, some new sufficient conditions are obtained for the solutions of all oscillations and the results generalize, improve and unify the oscillation theorems of half linear functional differential equations, nonlinear equations and generalized Emden-Fowler type equations in the literature recently. At last, some examples are given to illustrate the effectiveness of our results.

Keywords： Second order nonlinear differential equation ; Half linear differential equation ; Emden-Fowler equation ; Neutral ; Riccati transformation ; Oscillation criterion

Zhang Zhiyu, Yu Yuanhong, Li Shuping, Qiao Shizhu. Oscillation of Second Order Nonlinear Differential Equations with Neutral Delay. Acta Mathematica Scientia[J], 2019, 39(4): 797-811 doi:

1 引言

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

(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

(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)), $$r'(t)\geq 0, \alpha >0, 且存在常数 L_{2}\geq L_{1}>0, 使得 函数 x(t)\in{C^1([T_x, \infty), {\Bbb R} )},$$ 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.

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

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

Dzurina和Stavroulakis, Sun和Meng研究了二阶半线性时滞微分方程

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

Erbe等和Li等研究了Emden-Fowler中立型方程

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

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

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

2 主要结果

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

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

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

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

(ⅰ) $\alpha = \beta. $$(2.6) 式代入(2.5)式得到 \begin{equation} w'(t)\leq -Q_1(t)-\alpha \bigg (\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}\frac{\sigma'(t)}{r^{1/\alpha}(\sigma(t))} w^{\frac{\alpha+1}{\alpha}}(t), \quad t\geq T. \end{equation} (ⅱ) \alpha<\beta. 因为 z(\sigma(t)) 是增函数,故存在常数 M_1>0 ,使得当 t\geq T 时,有 (2.5) 式并注意到 (2.6) 式,得 \begin{eqnarray} w'(t)&\leq& -Q_1(t)-\beta L_{1}\bigg(\frac{L_{1}}{L_{2}}\bigg)^{\frac{1}{\alpha}}\frac{\sigma'(t)}{r^{1/\alpha}(\sigma(t))}[z(\sigma(t))]^\frac{\beta-\alpha}{\alpha} \bigg[\frac{r(t)(z'(t))^{\alpha}}{z^{\beta}(\sigma(t))}\bigg]^{\frac{\alpha+1}{\alpha}}\\ &\leq & -Q_1(t)-\bigg(\frac{L_{1}^{\alpha+1}}{L_{2}^{\alpha+2}}\bigg)^{\frac{1}{\alpha}}\frac{\alpha\sigma'(t)M_1}{r^{1/\alpha}(\sigma(t))}w^\frac{\alpha+1}{\alpha}(t).\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{eqnarray} (ⅲ) \alpha>\beta. 若条件(A _{3}) 成立,由 (2.3) 式知 (g(t, z(t), z'(t)))' = \frac{\partial g}{\partial t}+\frac{\partial g}{\partial u}v+\frac{\partial g}{\partial v}z''(t)\leq 0 ,以及当 v>0 时, \frac{\partial g}{\partial t}\geq 0, \frac{\partial g}{\partial u}v\geq 0$$ \frac{\partial g}{\partial v}>0, $$z''(t)\leq 0 ,即 z'(t) 非增.若条件(A _{4}) 成立,又由 (2.3) 式和 (g(t, z(t), z'(t)))' = (a(t, z(t)))'r(t)(z'(t))^{\alpha}+a(t, z(t))(r(t)(z'(t))^{\alpha})'\leq 0$$ (a(t, z(t)))'\geq 0, $$(r(t)(z'(t))^{\alpha})'\leq 0 .所以, r(t)(z'(t))^{\alpha}$$ 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

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

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

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

$g(t, u, v) = r(t)|v|^{\alpha-1}v $$a(t, u)\equiv 1 时,方程 (1.1) 成为方程 (1.5), 且显然(A _{3}) 和(A _{4}) 均成立,其中 r'(t)\geq 0, \alpha>0, \beta>0, L_{1} = L_{2} = 1. 于是有如下推论. 推论2.4 设(A _{1} ), (A _{2})$$ (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)) $$(\rho'(t)\geq0) 和某个常数 M>0 ,使得对任意常数 m\in (0, M]$$ (2.1) $$(2.2) 式成立,则方程 (1.5) 振动. 显然,由推论2.4知,本文定理2.1已包含并改进了文献的定理2.2,因为那里要求相当于 m>0 为任意常数. 注2.1 综上所述,本文定理 2.1 适应于上述线性方程、半线性方程、超线性方程、次线性方程和非线性方程,推广和改进了文献的定理 2.1 和文献的定理 2.2 ,因为文献仅考虑了 \alpha \geq \beta 的情况,而且不能包含经典的振动结果.同时也统一了文献的定理 2.1 和定理 3.1 ,因为它们都是本文定理 2.1$$ f(t, u) = q(t)u $$g(t, u, v) = r(t)|v|^{\alpha-1}v 时的特例.需要指出的是,文献定理 2.1 的充分条件 (2.2) 式中取“ K = (z'(t_{1}))^{(\beta-\alpha)/\beta} ”是不妥的,因为 z(t) 是方程中含未知解 x(t) 的中立项,所以应用时无法验证该条件 (2.2) 是否成立.此外,这里不得不指出,现有文献[7, 24-27]中较普遍存在的问题是,关于振动定理充分条件中涉及的常数 M, K, c 等往往未说明具有一定的“任意性”.因为用反证法证明定理时,推得存在某个常数 M_0>0 ,使得与已知条件矛盾,则该条件中对应的常数 M>0 就必须是任意的,否则,就会出现逻辑错误. 例2.1 考虑方程 \bigg[t^{\lambda/4}(\pi+\arctan z(t))\frac{2+|z'(t)|^{\alpha}}{1+|z'(t)|^{\alpha}}\left|z'(t)\right|^{\alpha-1}z'(t) \bigg]'+t^{-(1+\lambda /2)}[ky(\mu t)+ly^3(\mu t)] = 0, 其中 z(t) = x(t)+\frac{1}{2}x(t-1), y(t) = {\left|x(t)\right|}^\beta {\rm sgn} x(t),$$ \alpha>0, \; \beta>0, \; \lambda = \min\{\alpha, \beta\}, k>0, $$l\geq 0,$$ 0<\mu<1$为常数, $t\geq t_0\geq 1.$

$g(t, u, v)$满足条件(A$_{3})$.又知

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

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

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

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

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

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

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

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

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

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

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

$(2.29)$式不难得到,当$t>1$时有

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

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

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, $$t\geq t_1. 类似于文献定理2.1或文献定理2.5的证明,知 x'(t) 的符号最终非正,即存在 t_{2}\geq t_{1} 使得 x'(t)\leq0\; (t\geq t_{2}). 若不然, x'(t) 的符号最终为正,则必存在 t_{3}\geq t_{1} ,使得当 t\geq t_{3} 时,有 x'(t)> 0, x'(\tau(t))> 0, 从而有 这与 z'(t)<0, \; (t\geq t_{1}) 矛盾.所以有 因为 x'(t)\leq0 , t\geq t_{2} ,于是有 将上式代入方程(1.1),得 注意到 Q_{2}(t) 的定义,上式又成为 \begin{equation} [a(t, z(t))r(t)(-z'(t))^\alpha]'- Q_2(t)z^\beta(t)\geq 0, \; \; t\geq t_2. \end{equation} 再定义双Riccati变换函数 \begin{equation} \overline{w}(t) = \frac{a(t, z(t))r(t)(-z'(t))^\alpha}{z^\beta(t)}, \; \; \overline{v}(t) = \frac{r(t)(-z'(t))^\alpha}{z^\beta(t)}, \quad t\geq t_2. \end{equation} 则有 \overline{w}(t)>0, \overline{v}(t)>0 ,且由条件(A _{4})$$ (2.32)$式有

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

(ⅰ) $\alpha = \beta$.则(2.34)式成为

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

(ⅱ) $\alpha > \beta$.$z(t)$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

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

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

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

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

$\underline{R}^\mu(s)$乘(2.38)式,从$t_2 $$t 分部积分可得 \begin{eqnarray} \int_{t_2}^{t} \underline{R}^\mu(s)Q_2(s){\rm d}s &\leq& \int_{t_2}^{t} \underline{R}^\mu(s)\overline{w}'(s){\rm d}s- \overline{M}_{0} \int_{t_2}^{t} \underline{R}^\mu(s)r^{-1/\alpha}(s)\overline{w}^{(\mu+1)/\mu}(s){\rm d}s \\ & = &\underline{R}^\mu(t)\overline{w}(t)- \underline{R}^\mu(t_2)\overline{w}(t_2) \\ &&+ \int_{t_2}^{t} \underline{R}^\mu(s) \bigg[\frac{\mu \overline{w}(s)}{\underline{R}(s)r^{1/\alpha}(s)}-\frac{\overline{M}_{0}\overline{w}^{(\mu+1)/\mu}(s)}{r^{1/\alpha}(s)}\bigg]{\rm d}s. \end{eqnarray} 利用不等式(2.11)和(2.39)式,则有 \begin{equation} \int_{t_2}^{t} [\underline{R}^\mu(s)Q_2(s)-\overline{M}\underline{R}^{-1}(s)r^{-1/\alpha}(s)]{\rm d}s \leq \underline{R}^\mu(t)\overline{w}(t)- \underline{R}^\mu(t_2)\overline{w}(t_2), \end{equation} 其中 \overline{M} = \mu^{2\mu+1}/((\mu+1)^{\mu+1}\overline{M}_{0}^{\mu})\geq M. 下面证明(2.40)式右端有界.事实上,在上述(iii)中已证 r^{1/\alpha}(t)(-z'(t)) 是增函数,从而有 上式对 s 积分,可得 \begin{equation} z(t) \geq r^{1/\alpha}(t)(-z'(t))\underline{R}(t)\; \; \mbox {或}\; \; z^\alpha(t) \geq [ r^{1/\alpha}(t)(-z'(t))]^{\alpha} \underline{R}^{\alpha}(t). \end{equation} \alpha \geq \beta 时,因 z'(t)<0 ,故存在常数 l_1>0 ,使得 l_1 \geq z^{\alpha-\beta}(t) \geq L_{2}^{-1}\underline{R}^{\alpha}(t)\overline{w}(t)>0, \; t\geq t_2. 因此, (2.40)式右端有界. 另一方面,当 \beta > \alpha 时,由(2.41)式得到 注意到 r^{1/\alpha}(t)(-z'(t)) 是增函数,故存在 l_2>0 ,使得 因此, (2.40)式右端同样有界. 故对任意 \alpha>0, \beta>0 , (2.40)式右端有界,此与(2.31)式矛盾.定理2.4证毕. 注2.5 文献[2, 5, 6, 23]仅对方程 (1.1) 的特殊情况给出了相应方程一切解振动或渐近于零的类似充分条件,因此,本文定理 2.4 推广和改进了这些结果.还需要指出的是,文献定理 2.3 的证明中当 \beta>\alpha,$$ (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) $$y(t)>0, t\geq T 矛盾. 例2.4 考虑中立型方程 [t^{2\alpha}(\pi+\arctan z(t))\left|z'(t)\right|^{\alpha-1}z'(t)]'+ t^{2\alpha+\beta} \left|x(t-2)\right|^{\beta-1} x(t-2) = 0, \quad t \geq t_0 \geq 2, 其中 z(t) = x(t)+\frac{1}{2}x(t-1), \alpha>0, \beta>0. 这里有 a(t, z(t)) = \pi +\arctan z(t), \ r(t) = t^{2\alpha}, \ p(t) = 1/2, \ q(t) = t^{2\alpha+\beta};$$ \sigma(t) = t-2<t-1 = \tau(t); $$z'(t)(a(t, z(t)))' = \frac{(z'(t))^{2}}{1+z^{2}(t)}\geq 0. 故满足条件(A _{1} ), (A _{2}) 和(A _{4}).$$ 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.

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