## Oscillation Criteria of Second-Order Generalized Emden-Fowler Delay Differential Equations with a Sub-Linear Neutral Term

Zhang Zhiyu,

 基金资助: 国家自然科学基金.  11701528国家自然科学基金.  11747098山西省自然科学基金.  2011011002-3

 Fund supported: the NSFC.  11701528the NSFC.  11747098the NSF of Shanxi Province.  2011011002-3

Abstract

In this paper, we study the oscillation of second order generalized Emden-Fowler delay differential equations with a sub-linear neutral term. Under the irregularity condition, by using Riccati transformation and the inequalities technique, several simple new oscillation criteria of this kind of equations to ensure that every solution oscillates are established. These oscillation criteria generalize and improve the classical research results including those adapted to Euler equations established in previous literatures. Finally, some examples to verify the wide application of these oscillation criteria are given in this paper.

Keywords： Emden-Fowler equation ; Euler equation ; Second order ; Delay differential equation ; Sub-linear neutral term ; Oscillation criterion ; Riccati transformation

Zhang Zhiyu. Oscillation Criteria of Second-Order Generalized Emden-Fowler Delay Differential Equations with a Sub-Linear Neutral Term. Acta Mathematica Scientia[J], 2021, 41(3): 811-826 doi:

## 1 引言

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

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

${\rm (H_2)}\;$函数$a\in C^1([t_0, \infty), (0, \infty)), g\in C([t_0, \infty)\times {{\Bbb R}} , {{\Bbb R}} )$; 当$v\neq0$时, 有$0\leq g(t, v)/v\leq p_\alpha(t), p_\alpha\in C([t_0, \infty), [0, \infty)), 0\leq p_1(t)<1, \lim\limits_{t\to \infty} p_\alpha(t) = 0(0<\alpha<1); f\in C({{\Bbb R}} , {{\Bbb R}} ), $$u\neq0, t\geq t_0 时, f(t, u)/u\geq q(t)\geq0 且连续函数 q(t) 不恒等于零. {\rm (H_3)}\; \tau(t)\leq t, \sigma(t)\leq t, \sigma'(t)\geq0$$ \lim\limits_{t\rightarrow \infty}\tau(t) = \lim\limits_{t\rightarrow \infty}\sigma(t) = \infty.$

$$$(a(t)((x(t)+p(t)x ^\alpha(\tau(t)))')^\beta)'+q(t)x^\gamma (\sigma(t)) = 0,$$$

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

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

假设方程(1.1)存在一个非振动解$x(t),$不妨设该解是最终正解(当$x(t)$是最终负解时用类似方法可以证明), 则存在$t_1\geq t_0,$得当$t>t_1$时, 有$x(t)>0, x(\tau(t))>0, x(\sigma(t))>0. $$z(t) 的定义可得 z(t)>0,$$ t\geq t_1 $$z(t)\geq x(t), t\geq t_1. 代入方程(1.1), 得 $$\left[a(t)|z'(t)|^{\beta-1}z'(t) \right]' = -f\left( t, |x(\sigma(t))|^{\gamma-1}x(\sigma(t))\right)\leq 0.$$ 所以 a(t)|z'(t)|^{\beta-1}z'(t) 是非增函数且 z'(t) 最终保号, 以此, 必有 z'(t)\geq 0,$$ t\geq t_1 $$z'(t)\leq 0,$$ t\geq t_1.$下面分别讨论之.

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

$$$z(t)\geq z(\tau(t))\geq M_0.$$$

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

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

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

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

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

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

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

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

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

$\mu = \max\{\beta, \gamma\} $$K_2 代入(2.17)式, 得 $$A^\beta(t)v'(t)\leq -K_2q(t)P_2(K_1, \sigma(t))A^\mu(t)-A^\beta(t)\frac{\beta (-v(t))^{(\beta+1)/\beta}}{a^{1/\beta}(t)}, \; \; t\geq t_1.$$ 将(2.18)式中的 t 改为 s, 并对 s$$ t_1 $$t\geq t_1 积分, 注意到分部积分公式, 得 \begin{eqnarray} && \int_{t_1}^t K_2 A^\mu(s)q(s)P_2(K_1, \sigma(s)){\rm d}s{}\\ &\leq& -\int_{t_1}^t A^\beta(s)v'(s){\rm d}s-\int_{t_1}^t A^\beta(s)\frac{\beta(-v(s))^{(\beta+1)/\beta}}{a^{1/\beta}(s)}{\rm d}s {}\\ & = & A^\beta(t_1)v(t_1)-A^\beta(t)v(t)+\int_{t_1}^t \left( \frac{\beta A^{\beta-1}(s)}{a^{1/\beta}(s)}(-v(s))-\frac{\beta A^\beta(s)}{a^{1/\beta}(s)}(-v(s))^{(\beta+1)/\beta}\right){\rm d}s.{\qquad} \end{eqnarray} 对(2.19)式右端的被积式利用引理1.1中的不等式(1.12), 可得 将上式代入(2.19)式, 并注意到(2.14)式, 得 这与(2.2)式矛盾. 综合以上(Ⅰ)与(Ⅱ)两种矛盾情况可知, 方程(1.1)振动. 注2.1 显然, 当本文的 \alpha, \beta, \gamma 退化成 "\alpha = 1, \beta = \gamma = 1" 时, 定理 2.1 与定理 1.4 相同, 所以, 这时定理 2.1 包含和改进了文献[14]的定理 2.2 和诸多关于带线性中立项的方程 (1.4) 的多个结果, 如文献[10]中的定理 2.1 等. 而当 0<\alpha<1, \beta = \gamma = 1 时, 本文定理 2.1 比定理 1.4 更简便, 因为本文中的 (1-p_\alpha(t)/M^{1-\alpha}), M\in(0, 1]$$ 1-\frac{p_\alpha(t)}{K_1^{1-\alpha}}\frac{A^\alpha(\tau(t))}{A(t)}, K_1\in(0, 1]$对应于定理$1.4 $$[1-(\alpha2^{1-\alpha}+\frac{2^{1-\alpha}}{M})p(t)]$$ 1-(\alpha2^{1-\alpha}\frac{A(\tau(t))}{A(t)}+\frac{2^{1-\alpha}-1}{KA(t)})p(t), \; M>0, K>0$任意. 所以定理$2.1$推广和改进了文献[14]的定理$2.2$.

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

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

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

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

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

$$$(a(t)(z'(t))^\beta)'\leq -q(t)P_2(K, \sigma(t)) z^\gamma(\sigma(t)), \; t\geq t_2.$$$

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

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

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

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

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

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

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

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

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

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

(v) $m>-1, \beta<m+1$或(vi) $m>-1, q_0>\beta = m+1\; ( $$l\to1) 时，(2.30)式满足. 故由定理2.3知方程(2.32)振动. 注2.4 从例 2.2 可见本文定理2.1–2.3作用各有千秋且不能相互替代, 定理 2.1 更全面精细, 定理 2.2 和定理 2.3 更简捷方便. 本文所列文献的结果对方程 (2.32) 均无效. 我们再来考察著名的Emden-Fowler方程(1.2)的振动性. 从定理1.1已知, 当 0\leq a\leq 1 时方程(1.2)振动, 当 a>1 时方程(1.2)的每一个解 x(t) 振动或有 \lim\limits_{t\to\infty}x(t) = 0, 并在某些条件下确有 x(t) = t^{-\lambda}\; (\lambda>0) 型的渐近于0的解. 但当 a>1 时方程(1.2)是否还有确定的振动性吗？ 为了回答这个问题, 我们利用文献[4]的方法, 通过指数变换将方程(1.2)化为其等价方程 $$(t^ax'(t))'+bt^{a+m-1}x^n(t) = 0, \; \; t\geq t_0,$$ 其中 a\geq0, b>0, m 为常数, 0<n\neq1 为两正奇数之比的商. 为简便起见, 不妨取 t_0 = 1. 对应于方程(1.1), 这里 所以 a+m\geq0 时, 有 $$\int_1^\infty q(s){\rm d}s = \int_1^\infty bs^{a+m-1}{\rm d}s = \infty,$$ 即取 \varphi(t) = 1 时, (2.1)式满足, 而对于任意取定的 K_2\in(0, 1], \begin{eqnarray} &&\int_1^\infty \left[K_2A^\mu(s)q(s)-\frac{1}{4a(s)A(s)}\right]{\rm d}s{}\\ & = &\int_1^\infty\left[ K_2\left(\frac{s^{1-a}}{a-1} \right)^\mu bs^{a+m-1}-\frac{a-1}{4s}\right]{\rm d}s{}\\ & = & \left\{ \begin{array}{ll} { }\int_1^\infty\frac{1}{s}\left[ \frac{bK_2}{a-1}s^{m+1}-\frac{a-1}{4}\right]{\rm d}s = \infty, \; \; &0<n<1, m+1>0, \\ { }\int_1^\infty\frac{1}{s}\left[ \frac{bK_2}{(a-1)^n}s^{a+m-n(a-1)}-\frac{a-1}{4}\right]{\rm d}s = \infty, \; &n>1, a+m>n(a-1), \end{array} \right. \end{eqnarray} 这时, 易知(2.2)式满足. 又因为 \begin{eqnarray} &&\int_1^\infty A^{\mu+1}(s)q(s){\rm d}s{}\\ & = &\int_1^\infty \frac{b}{(a-1)^{\mu+1}}s^{a+m-1-(a-1)(\mu+1)}{\rm d}s{}\\ & = & \left\{ \begin{array}{ll} { } \frac{b}{(a-1)^2}\int_1^\infty s^{m+1-a}{\rm d}s = \infty, \; \; &0<n<1, m+1>a-1, \\ { }\frac{b}{(a-1)^{n+1}}\int_1^\infty s^{m-n(a-1)}{\rm d}s = \infty, \; &n>1, m+1>n(a-1), \end{array} \right. \end{eqnarray} 所以, 此时(2.21)式满足. 再由于 \begin{eqnarray} &&\int_1^\infty \left[\frac{1}{a(t)}\int_1^t A^\gamma(s)q(s){\rm d}s\right]{\rm d}t{}\\ & = &\frac{b}{(a-1)^n}\int_1^\infty \left[\frac{1}{t^a}\int _1^t s^{a+m-1-n(a-1)}{\rm d}s \right]{\rm d}t{}\\ & = & \left\{ \begin{array}{ll} { }\frac{b}{(a-1)^n}\int_1^\infty \frac{\ln t}{t^a}{\rm d}t<\infty, \; \; &n(a-1) = a+m, \\ { }\frac{b}{(a-1)^n}\int_1^\infty\frac{t^{a+m-n(a-1)}-1}{t^a(a+m-n(a-1))}{\rm d}t = \infty, \; &m+1\geq n(a-1), \end{array} \right. \end{eqnarray} 因此, 当(2.39)的第二式成立时(2.22)式满足. 综合以上分析, 知(2.36)和(2.37)式成立时, 由定理2.1, (2.38)或(2.39)式的第二式成立时由定理2.2, 均得方程(2.35), 即方程(1.2)振动. 故有以下结论. 推论2.1 设 0<n\neq1 为两正奇数之比的常数, a>1, b>0, m>-1 为常数. 则当 (ⅰ) 0<n<1, m+1>0\; \rm{或}\; n>1, a+m>n(a-1), (ⅱ) 0<n<1, m+1\geq a-1\; \rm{或}\; n>1, m+1\geq n(a-1), (ⅲ) m+1\geq n(a-1) 之一满足时, Emden-Fowler方程 (1.2) 振动. 注2.5 推论 2.1 完善了定理 1.1 关于Emden-Fowler方程 (1.2)$$ a>1$时的振动性结果, 但不能替代之. 如果取$n = 1$, 则方程$(2.35)$就包含了Euler方程(1.3) (即$a = 2, $$n = 1,$$ m = -1, b = q_0),$这时注意到$(2.36)$$(2.37)$式或直接应用定理$2.3$, 立即可得如下两个重要结论. 特别是, 其中后者将著名结果{Euler}方程$(1.3)$振动的"充分条件$q_0>\frac{1}{4}$"确定为"充要条件".

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

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