Theory and Modern Applications

# Oscillatory behavior of second-order nonlinear neutral differential equations

## Abstract

We shall consider a class of second-order nonlinear neutral differential equations. Some new oscillation criteria are established by using the Riccati transformation technique. One example is given to show the applicability of the main results.

## 1 Introduction

In this paper, we study the oscillation of a class of second-order nonlinear differential equations,

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } + f\bigl(t,x\bigl(\sigma (t)\bigr)\bigr) = 0,\quad t \ge t_{0} > 0,$$
(1)

where $$z(t) = x(t) - p(t)x(\tau (t))$$, $$\alpha > 0$$, and Î± is the ratio of two odd integers. The following assumptions are satisfied:

($$H_{1}$$):

$$r,p \in C([t_{0},\infty ),R), r(t) > 0, 0 \le p(t) \le p_{0} < 1$$.

($$H_{2}$$):

$$\tau \in C([t_{0},\infty ),R), \tau (t) \le t, \lim_{t \to \infty } \tau (t) = \infty$$.

($$H_{3}$$):

$$\sigma \in C^{1}([t_{0},\infty ),R),\sigma (t) \le t, \sigma '(t) > 0, \lim_{t \to \infty } \sigma (t) = \infty$$.

($$H_{4}$$):

$$f \in C(R,R)$$, $$uf(t,u) > 0$$ for all $$u \ne 0$$, and there exists a function $$q(t) \in C([t_{0},\infty ], [0,\infty ))$$ such that $$\vert f(t,u) \vert \ge q(t) \vert u^{\alpha } \vert$$.

Second-order and third-order differential equations are widely used in population dynamics, physics, technology and other fields. Many scholars have studied the oscillation of second-order differential equations [1â€“10]. Similarly, many scholars have studied the oscillation of third-order differential equations [11â€“14]. On this basis, this paper studies the second-order neutral differential Eq. (1), Some new oscillation criteria are established by using the Riccati transformation technique.

## 2 Lemmas

In order to establish the oscillation criterion of Eq.Â (1), we will give three lemmas.

### Lemma 2.1

Assume that

$$\int _{ t_{0}}^{ \infty } r^{ - \frac{1}{\alpha }} (t)\,dt = \infty$$
(2)

and$$x(t)$$is an eventually positive solution of Eq.Â (1). Then$$z(t)$$has the following two possible cases:

1. (i)

$$z(t) > 0$$, $$z'(t) > 0$$, $$( r(t)(z'(t))^{\alpha } )^{\prime } \le 0$$;

2. (ii)

$$z(t) < 0$$, $$z'(t) > 0$$, $$( r(t)(z'(t))^{\alpha } )^{\prime } \le 0$$.

### Proof

Since $$x(t)$$ is an eventually positive solution of (1), there exists a $$t_{1} \ge t_{0}$$ such that $$x(t) > 0$$, for $$t \ge t_{1}$$. From (1), we have

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } \le 0$$

hence $$r(t)(z'(t))^{\alpha }$$ is decreasing function and of one sign, therefore $$z'(t)$$ is also of one sign, that is, there exists a $$t_{2} \ge t_{1}$$ such that, for $$t \ge t_{2}$$, $$z'(t) > 0$$ or $$z'(t) < 0$$.

If $$z'(t) > 0$$, we have (i) or (ii). Now, we prove that $$z'(t) < 0$$ will not happen.

If $$z'(t) < 0$$, we have

$$r(t) \bigl( - z'(t)\bigr)^{\alpha } \ge r(t_{2}) \bigl( - z'(t_{2})\bigr)^{\alpha } = K \ge 0,$$

where $$K = r(t_{2})( - z'(t_{2}))^{\alpha } \ge 0$$, that is,

$$z'(t) \le - k^{\frac{1}{\alpha }} r^{ - \frac{1}{\alpha }} (t).$$

Integrating this inequality from $$t_{2}$$ to t, we have

$$z(t) \le z(t_{2}) - k^{\frac{1}{\alpha }} \int _{ t_{2}}^{ t} r^{ - \frac{1}{\alpha }} (s)\,ds$$

by condition (2), $$\lim_{t \to \infty } z(t) = - \infty$$. We will consider the following two cases.

Case 1. If $$x(t)$$ is unbounded, then there exists a sequence $$\{ t_{m} \}$$, such that $$\lim_{m \to \infty } t_{m} = \infty$$ and $$\lim_{m \to \infty } x(t_{m}) = \infty$$, here $$x(t_{m}) = \max \{ x(s):t_{0} \le s \le t_{m} \}$$. Hence, we have

\begin{aligned} x\bigl(\tau (t_{m})\bigr) &= \max \bigl\{ x(s): t{}_{0} \le s \le \tau (t_{m}) \bigr\} \\ &\le \max \bigl\{ x(s): t{}_{0} \le s \le t_{m} \bigr\} = x(t_{m}). \end{aligned}

We get

$$z(t_{m}) = x(t_{m}) - p(t_{m})x\bigl(\tau (t_{m})\bigr)\ge \bigl[1 - p(t_{m})\bigr]x(t_{m}) > 0.$$

This contradicts $$\lim_{t \to \infty } z(t) = - \infty$$.

Case 2. If $$x(t)$$ is bounded, then $$z(t)$$ is bounded, this contradicts $$\lim_{t \to \infty } z(t) = - \infty$$.

Hence, $$z(t)$$ satisfies one of the cases (i) and (ii).â€ƒâ–¡

### Lemma 2.2

Assume that$$x(t)$$is a positive solution of Eq.Â (1) and$$z(t)$$satisfies case (i) of Lemma2.1, then

$$z(t) \ge R(t)r^{\frac{1}{\alpha }} (t)z'(t),\qquad \biggl( \frac{z(t)}{R(t)} \biggr)^{\prime } \le 0,$$

where$$R(t) = \int _{ T}^{ t} r^{ - \frac{1}{\alpha }} (s)\,ds$$, $$T \ge t_{0}$$.

### Proof

For $$t > T \ge t_{0}$$, we have

$$z(t) = z(T) + \int _{ T}^{ t} \frac{r^{\frac{1}{\alpha }} (s)z'(s)}{r^{\frac{1}{\alpha }} (s)}\,ds\ge r^{\frac{1}{\alpha }} (t)z'(t) \int _{ T}^{ t} r^{ - \frac{1}{\alpha }} (s)\,ds = R(t)r^{\frac{1}{\alpha }} (t)z'(t).$$

Thus, we conclude that

$$\biggl( \frac{z(t)}{R(t)} \biggr)^{\prime } = \frac{z'(t)R(t) - z(t)R'(t)}{R^{2}(t)}\le \frac{z'(t)R(t) - R(t)r^{\frac{1}{\alpha }} (t)z'(t)r^{ - \frac{1}{\alpha }} (t)}{R^{2}(t)} = 0.$$

â€ƒâ–¡

### Lemma 2.3

Assume that$$x(t)$$is an eventually positive solution of (1) and

$$\mathop{\lim \sup}_{t \to \infty } \int _{ \tau ^{ - 1}(\sigma (t))}^{ t} \biggl( \frac{1}{r(s)} \int _{ s}^{ t} q(u)\,du \biggr)^{\frac{1}{\alpha }} \,ds > p_{0}.$$
(3)

Then the impossibility for$$z(t)$$satisfies case (ii) of Lemma2.1.

### Proof

Assume that $$z(t)$$ satisfies case (ii) of Lemma 2.1, we have

$$- z(t) = - x(t) + p(t)x\bigl(\tau (t)\bigr) < p(t)x\bigl(\tau (t)\bigr)\le p_{0}x\bigl(\tau (t)\bigr).$$

That is,

$$x\bigl(\tau (t)\bigr) \ge - \frac{1}{p_{0}}z(t).$$

We deduce that

$$x(t) \ge - \frac{1}{p_{0}}z\bigl(\tau ^{ - 1}(t)\bigr),\qquad x\bigl(\sigma (t)\bigr) \ge - \frac{1}{p_{0}}z\bigl(\tau ^{ - 1}\bigl(\sigma (t) \bigr)\bigr).$$

From (1) and ($$H_{4}$$), we have

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } + q(t) \bigl(x\bigl(\sigma (t)\bigr)\bigr)^{\alpha } \le 0.$$

We get

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } + q(t) \biggl( - \frac{1}{p_{0}}\biggr)^{\alpha } z^{\alpha } \bigl(\tau ^{ - 1}\bigl(\sigma (t)\bigr)\bigr) \le 0.$$

Integrating this inequality from s to t, we conclude that

$$r(t) \bigl(z'(t)\bigr)^{\alpha } - r(s) \bigl(z'(s) \bigr)^{\alpha } - \frac{1}{p_{0}^{\alpha }} \int _{ s}^{ t} q(u)z^{\alpha } \bigl(\tau ^{ - 1}\bigl(\sigma (u)\bigr)\bigr)\,du \le 0.$$

That is,

$$- z'(s) \le \frac{1}{p_{0}} \biggl( \frac{1}{r(s)} \int _{ s}^{ t} q(u)z^{\alpha } \bigl(\tau ^{ - 1}\bigl(\sigma (u)\bigr)\bigr)\,du \biggr)^{\frac{1}{\alpha }}.$$

Integrating this inequality from $$\tau ^{ - 1}(\sigma (t))$$ to t, we get

$$z\bigl(\tau ^{ - 1}\bigl(\sigma (t)\bigr)\bigr) - z(t) \le \frac{1}{p_{0}}z\bigl(\tau ^{ - 1}\bigl(\sigma (t)\bigr)\bigr) \int _{ \tau ^{ - 1}(\sigma (t))}^{ t} \biggl( \frac{1}{r(s)} \int _{ s}^{ t} q(u)\,du \biggr)^{\frac{1}{\alpha }} \,ds.$$

Since $$z(t) < 0$$, we have

$$\int _{ \tau ^{ - 1}(\sigma (t))}^{ t} \biggl( \frac{1}{r(s)} \int _{ s}^{ t} q(u)\,du \biggr)^{\frac{1}{\alpha }} \,ds \le p_{0}.$$

This contradicts (3). Thus the impossibility for $$z(t)$$ satisfies case (ii) of Lemma 2.1.â€ƒâ–¡

## 3 Oscillation results

### Theorem 3.1

Assume that (2) and (3) be satisfied. If there exists a positive function$$\rho \in C^{1}([t_{0},\infty ),(0,\infty ))$$, such that, for all sufficiently large$$T \ge t_{0}$$,

$$\int _{ t_{0}}^{ \infty } \biggl[\rho (t)\bar{Q}(t) - \frac{r(t)(\rho '(t))^{\alpha + 1}}{(\alpha + 1)^{\alpha + 1}\rho ^{\alpha } (t)}\biggr]\,dt = \infty,$$
(4)

where$$\bar{Q}(t) = Q(t)\frac{R^{\alpha } (\sigma (t))}{R^{\alpha } (t)}$$), $$Q(t) = q(t)[1 + \bar{p}(\sigma (t))]^{\alpha }$$, $$\bar{p}(t) = p(t)\frac{R(\tau (t))}{R(t)}$$, then Eq.Â (1) is oscillatory.

### Proof

Assume that $$x(t) > 0$$. From Lemma 2.1, $$z(t)$$ satisfies one of the cases (i) and (ii).

Case (i). Suppose that case (i) holds, from Lemma 2.2, we have

$$\frac{z(t)}{R(t)} \le \frac{z(\tau (t))}{R(\tau (t))}.$$

That is,

$$z\bigl(\tau (t)\bigr) \ge R\bigl(\tau (t)\bigr)\frac{z(t)}{R(t)}.$$

We get

$$z(t) = x(t) - p(t)x\bigl(\tau (t)\bigr)\le x(t) - p(t)z\bigl(\tau (t)\bigr)\le x(t) - p(t)R\bigl(\tau (t)\bigr)\frac{z(t)}{R(t)}.$$

That is,

$$x(t) \ge \biggl[1 + p(t)\frac{R(\tau (t))}{R(t)}\biggr]z(t) = \bigl[1 + \bar{p}(t) \bigr]z(t),$$

where $$\bar{p}(t) = p(t)\frac{R(\tau (t))}{R(t)}$$.

From (1), we conclude that

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } + q(t)x^{\alpha } \bigl(\sigma (t)\bigr) \le 0.$$

Then we have

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } + q(t)\bigl[1 + \bar{p}\bigl(\sigma (t)\bigr)\bigr]^{\alpha } z^{\alpha } \bigl(\sigma (t)\bigr) \le 0.$$

That is,

$$\bigl( r(t) \bigl(z'(t)\bigr)^{\alpha } \bigr)^{\prime } \le - Q(t)z^{\alpha } \bigl(\sigma (t)\bigr),$$
(5)

where $$Q(t) = q(t)[1 + \bar{p}(\sigma (t))]^{\alpha }$$.

We define a function $$w(t)$$ of the generalized Riccati transformation by

$$w(t) = \frac{\rho (t)r(t)(z'(t))^{\alpha }}{z^{\alpha } (t)}.$$

Then $$w(t) > 0$$, from Lemma 2.2, we have $$\frac{z(\sigma (t))}{R(\sigma (t))} \ge \frac{z(t)}{R(t)}$$, that is, $$\frac{z(\sigma (t))}{z(t)} \ge \frac{R(\sigma (t))}{R(t)}$$.

Using the inequality [2]

$$Bu - Au^{\frac{\theta + 1}{\theta }} \le \frac{\theta ^{\theta }}{(\theta + 1)^{\theta + 1}}\frac{B^{\theta + 1}}{A^{\theta }},\quad \theta > 0, A > 0, B \in R,$$

we have

\begin{aligned} w'(t) &= \rho '(t)\frac{r(t)(z'(t))^{\alpha }}{z^{\alpha } (t)} + \rho (t) \frac{(r(t)(z'(t))^{\alpha } )'}{z^{\alpha } (t)} - \rho (t)\frac{\alpha r(t)(z'(t))^{\alpha + 1}}{z^{\alpha + 1}(t)} \\ &\le \frac{\rho '(t)}{\rho (t)}w(t) - \rho (t)Q(t)\frac{z^{\alpha } (\sigma (t))}{z^{\alpha } (t)} - \frac{\alpha }{(\rho (t)r(t))^{{1 / \alpha }}} w^{\frac{\alpha + 1}{\alpha }} (t) \\ &\le \frac{\rho '(t)}{\rho (t)}w(t) - \rho (t)Q(t)\frac{R^{\alpha } (\sigma (t))}{R^{\alpha } (t)} - \frac{\alpha }{(\rho (t)r(t))^{{1 / \alpha }}} w^{\frac{\alpha + 1}{\alpha }} (t) \\ &\le - \rho (t)\bar{Q}(t) + \frac{\rho '(t)}{\rho (t)}w(t) - \frac{\alpha }{(\rho (t)r(t))^{{1 / \alpha }}} w^{\frac{\alpha + 1}{\alpha }} (t) \\ &= - \rho (t)\bar{Q}(t) + \frac{r(t)(\rho '(t))^{\alpha + 1}}{(\alpha + 1)^{\alpha + 1}\rho ^{\alpha } (t)}, \end{aligned}
(6)

where $$\bar{Q}(t) = Q(t)\frac{R^{\alpha } (\sigma (t))}{R^{\alpha } (t)}$$).

Integrating this inequality from T to t, we have

$$w(t) \le w(T) - \int _{ T}^{ t} \biggl( \rho (s)\bar{Q}(s) - \frac{r(s)(\rho '(s))^{\alpha + 1}}{(\alpha + 1)^{\alpha + 1}\rho ^{\alpha } (s)} \biggr)\,ds.$$

From (4), we get $$\lim w(t)_{t \to \infty } = - \infty$$, this contradicts $$w(t) > 0$$.

Case (ii). If $$z(t)$$ satisfies (ii), then due to Lemma 2.3, Eq.Â (1) is oscillatory.â€ƒâ–¡

### Theorem 3.2

Assume that (2) and (3) are satisfied. If there exists a positive function$$\varphi \in C^{1}([t_{0},\infty ),(0,\infty ))$$such that, for all sufficiently large$$T \ge t_{0}$$,

$$\int _{ t_{0}}^{ \infty } \biggl[\bar{Q}(t) - \frac{\varphi ^{\alpha + 1}(t)}{r^{1 / \alpha } (t)} \biggr]\exp \biggl[(\alpha + 1) \int _{ T}^{ t} \frac{\varphi (s)}{r^{1 / \alpha } (s)}\,ds\biggr] = \infty,$$
(7)

then Eq.Â (1) is oscillatory.

### Proof

We use the counter-evidence method, suppose we have a non-oscillatory solution $$x(t)$$ of Eq.Â (1), as above, suppose that $$x(t)$$ is a positive solution of (1), by using Lemma 2.1, $$z(t)$$ satisfies one of (i) and (ii), we discuss each of the two cases separately.

Case (i). Assume that $$z(t)$$ has property (i), we obtain (5). We define a function $$V(t)$$ of a generalized Riccati transformation by

$$V(t) = \frac{r(t)(z'(t))^{\alpha }}{z^{\alpha } (t)}.$$

Then $$V(t) > 0$$, using the Yang inequality $$\frac{1}{p}a^{p} + \frac{1}{q}b^{q} \ge ab$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, similar to (6), we have

\begin{aligned} V'(t)& = \frac{(r(t)(z'(t))^{\alpha } )'}{z^{\alpha } (t)} - \frac{\alpha r(t)(z'(t))^{\alpha + 1}}{z^{\alpha + 1}(t)} \\ &\le - \bar{Q}(t) - \frac{\alpha }{r^{1 / \alpha } (t)}V^{\frac{\alpha + 1}{\alpha }} (t) \\ &= - \bigl[\bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr] - ( \alpha + 1)r^{ - \frac{1}{\alpha }} (t)\biggl[\frac{1}{\alpha + 1}\varphi ^{\alpha + 1}(t) + \frac{\alpha }{\alpha + 1}V^{\frac{\alpha + 1}{\alpha }} (t)\biggr] \\ &= - \bigl[\bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr] - ( \alpha + 1)r^{ - \frac{1}{\alpha }} (t)\varphi (t)V(t). \end{aligned}

That is,

$$V'(t) + (\alpha + 1)r^{ - \frac{1}{\alpha }} (t)\varphi (t)V(t) \le - \bigl[ \bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr].$$

We get

\begin{aligned} &\bigl[V'(t) + (\alpha + 1)r^{ - \frac{1}{\alpha }} (t)\varphi (t)V(t)\bigr]\exp [(\alpha + 1) \int _{ T}^{ t} \frac{\varphi (s)}{r^{1 / \alpha } (s)}\,ds \\ &\quad \le - \bigl[\bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr]\exp [( \alpha + 1) \int _{ T}^{ t} \frac{\varphi (s)}{r^{1 / \alpha } (s)}\,ds. \end{aligned}

That is,

\begin{aligned} &\biggl( V(t) \cdot \exp \biggl[(\alpha + 1) \int _{ T}^{ t} r^{ - \frac{1}{\alpha }} (s)\varphi (s)\,ds\biggr] \biggr)^{\prime }\\ &\quad \le - \bigl[\bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr]\exp [(\alpha + 1) \int _{ T}^{ t} \frac{\varphi (s)}{r^{1 / \alpha } (s)}\,ds. \end{aligned}

Integrating this inequality from T to t, we get

\begin{aligned} 0 &\le V(t) \cdot \exp \biggl[(\alpha + 1) \int _{ T}^{ t} r^{ - \frac{1}{\alpha }} (s)\varphi (s)\,ds\biggr]\\ &\le V(T) - \int _{ T}^{ t} \biggl( \bigl[\bar{Q}(t) - r^{ - \frac{1}{\alpha }} (t)\varphi ^{\alpha + 1}(t)\bigr]\exp [(\alpha + 1) \int _{ T}^{ t} \frac{\varphi (s)}{r^{1 / \alpha } (s)}\,ds \biggr)\,dt. \end{aligned}

Case (ii). If $$z(t)$$ satisfies (ii), then due to Lemma 2.3, Eq.Â (1) is oscillatory.â€ƒâ–¡

### Example

Consider the following equation:

$$\bigl( \bigl( x(t) - px(t - 1)' \bigr)^{\frac{1}{3}} \bigr)^{\prime } + q_{0}x^{\frac{1}{3}}(t - 2) = 0.$$
(8)

Comparing Eq.Â (8) with Eq.Â (1), let $$r(t) = 1$$, $$\alpha = \frac{1}{3}$$, $$\tau (t) = t - 1$$, $$\sigma (t) = t - 2$$, $$q(t) = q_{0} > 0$$, $$p(t) = p < 1$$ is a positive constant. Choose $$\rho (t) = t$$, $$\varphi (t) = 1$$, we now verify (3):

$$\mathop{\lim \sup}_{t \to \infty } \int _{ \tau ^{ - 1}(\sigma (t))}^{ t} \biggl( \frac{1}{r(s)} \int _{ s}^{ t} q(u)\,du \biggr)^{1 / \alpha } \,ds = \mathop{\lim \sup}_{t \to \infty } \int _{ t - 1}^{ t} q_{0}(t - s)^{3} \,ds = \frac{q_{0}}{4} > p_{0}.$$

Therefore, if $$\frac{q_{0}}{4} > p_{0}$$, obviously, the conditions of Theorem 3.1 and Theorem 3.2 are satisfied, then Eq.Â (8) is oscillatory.

Then the conditions of Theorem 3.1 and Theorem 3.2 are satisfied.

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### Acknowledgements

The authors express their sincere gratitude to the editors and referees for careful reading of the manuscript and valuable suggestions, which helped to improve the paper. This research is supported by NNSF of P.R. China (Grant No. 11361048), NSF of Yunnan Province (Grant No. 2017FH001-014), and NSF of Qujing Normal University (Grant No. ZDKC2016002), P.R. China.

### Availability of data and materials

Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

### Authorsâ€™ information

1Institute of Applied Mathematics, Qujing Normal University, Qujing, Yunnan 655011, P.R. China, professor. 2School of Information Science and Engineering, Yunnan University, Kunming, Yunnan 650091, P.R. China, doctor. 3Academy of Mathematics and systems Science, China Academy Science, Beijing, 100190, P.R. China, researcher.

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### Contributions

All three authors contributed equally to this work. All authors read and approved the final manuscript.

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Correspondence to Jun Liu.

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Liu, J., Liu, X. & Yu, Y. Oscillatory behavior of second-order nonlinear neutral differential equations. Adv Differ Equ 2020, 387 (2020). https://doi.org/10.1186/s13662-020-02606-z

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• DOI: https://doi.org/10.1186/s13662-020-02606-z

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### Keywords

• Oscillation
• Neutral differential equation
• Riccati transformation