Oscillation of solutions of third order nonlinear neutral differential equations

Abstract

The main objective of this article is to improve and complement some of the oscillation criteria published recently in the literature for third order differential equation of the form

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{\prime }+q(t)f \bigl(x \bigl(\sigma (t) \bigr) \bigr)=0,\quad t\geq t_{0}>0, $$

where $z(t)=x(t)+p(t)x(\tau (t))$ and α is a ratio of odd positive integers in the two cases $\int _{t_{0}}^{\infty }r^{\frac{-1}{\alpha } }(s)\,\mathrm {d}s<\infty$ and $\int _{t_{0}}^{\infty }r^{\frac{-1}{\alpha } }(s)\,\mathrm {d}s=\infty$. Some illustrative examples are presented.

1 Introduction

Consider the nonlinear third order differential equation

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{\prime }+q(t)f \bigl( x \bigl( \sigma ( t ) \bigr) \bigr) =0, $$

(1.1)

where $t\geq t_{0}>0$, $z(t)=x(t)+p(t)x ( \tau (t) )$, and α is a ratio of odd positive integers. We assume that the following conditions hold:

(H₁):

$r(t),p(t),q(t),\tau (t),\sigma (t)\in C ( [t_{0},\infty ) )$, $r(t)$, $q(t)$ are positive and $0\leq p(t)\leq p_{0}<\infty$;

(H₂):

$\lim_{t\rightarrow \infty }\tau (t)=\lim_{t \rightarrow \infty }\sigma (t)=\infty$, $\sigma (t)>0$, and $\tau (t)\leq t$;

(H₃):

$f(u)\in C ( \mathbb{R} )$ and there exists a positive constant k such that $f(u)/u^{\gamma }\geq k$ for all $u\neq 0$ and γ is a ratio of odd positive integers;

(H₄):

$\tau ^{\prime }(t)\geq \tau _{0}>0$ and $\tau \circ \sigma =\sigma \circ \tau$.

By a solution of (1.1), we mean a nontrivial function $x(t)\in C ( [T_{x},\infty ) )$, $T_{x}\geq t_{0}$, which has the properties $z(t)\in C^{2} ( [T_{x},\infty ) )$, $r(t)(z^{\prime \prime }(t))^{\alpha }\in C^{1} ( [T_{x},\infty ) )$ and satisfies (1.1) on $[T_{x},\infty )$. Our attention is restricted to those solutions $x(t)$ of (1.1) satisfying $\sup \{ \vert x(t) \vert :t\geq T \} >0$ for all $T\geq T_{x}$. We assume that (1.1) possesses such a solution. A solution of (1.1) is called oscillatory if it has arbitrarily large zeros on $[T_{x},\infty )$; otherwise, it is termed nonoscillatory. Equation (1.1) is said to be oscillatory if all its solutions are oscillatory.

The oscillatory behavior of solutions of various classes of nonlinear differential and dynamic equations on time scales has received much attention, we refer the reader to [1–17] and the references cited therein.

In 2012, Liu et al. [9] established new oscillation criteria for the second order Emden–Fowler equation

$$ \bigl( r(t) \bigl\vert z^{\prime }(t) \bigr\vert ^{\alpha -1}z^{ \prime }(t) \bigr) ^{\prime }+q(t) \bigl\vert x \bigl(\sigma (t) \bigr) \bigr\vert ^{\gamma -1}x \bigl(\sigma (t) \bigr)=0 $$

(1.2)

under the assumptions

$$\begin{aligned} &0\leq p(t)\leq 1, \end{aligned}$$

(1.3)

$$\begin{aligned} &r^{\prime }(t)\geq 0, \qquad \sigma ^{\prime }(t)>0, \end{aligned}$$

(1.4)

and $\alpha \geq \gamma >0$. In 2016, Wang et al. [16] studied Eq. (1.2) with condition (1.3),

$$ \int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t= \infty , $$

(1.5)

and $\sigma ^{\prime }(t)>0$ with $\alpha \geq \gamma >1$ when the condition $r^{\prime }(t)\geq 0$ is neglected. Meanwhile, Wu et al. [17] established oscillation criteria for (1.2) in the general case when $\alpha >0$ and $\gamma >0$ are constants with conditions (1.3) and (1.4). Baculíková et al. [2] considered (1.2) in the more general case when $0\leq p(t)\leq p_{0}<\infty$ with condition (1.5) and $\sigma ^{\prime }(t)\geq 0$. For the case of third order differential equations, Džurina et al. [18] obtained sufficient conditions for the oscillation of solutions of the differential equation

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{\prime }+q(t)x^{\alpha } \bigl(\sigma (t) \bigr)=0, $$

(1.6)

where

$$ 0 \leq p(t) \leq p_{0}< 1 $$

(1.7)

with condition (1.5). Meanwhile, Baculíková et al. [1] and Su et al. [19] discussed the oscillatory behavior of third order Eq. (1.6) when $r^{\prime }(t)\geq 0$, (1.7) and (1.5) hold. Also Thandapani et al. [14] studied Eq. (1.6) when (1.7) holds and

$$ \int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t< \infty . $$

(1.8)

Recently, Jiang et al. [7] established new oscillation criteria for Eq. (1.1), where $\gamma =\alpha \geq 1$ and (1.5) hold without requiring (1.4).

More recently, Graef et al. [6] discussed the special case of Eq. (1.1) in which $r=1$ and $\alpha =\gamma$.

The main goal of this paper is to establish new oscillation criteria motivated by [6, 7], and [17] for Eq. (1.1) under all cases of γ, α (i.e., $\gamma >\alpha$, $\gamma =\alpha$, and $\gamma <\alpha$), $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t<\infty$ and $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t=\infty$ without assumption (1.4). We consider the two cases when (H₄) holds or not.

In the sequel, we give the following notations:

$$\begin{aligned}& Q(t)=\min \bigl\{ q(t),q\bigl(\tau (t)\bigr) \bigr\} ,\qquad R(t)=\max \bigl\{ r(t),r\bigl(\tau (t)\bigr) \bigr\} , \\& \eta _{+}^{{\prime }}(t)=\max \bigl\{ 0,\eta ^{\prime }(t)\bigr\} ,\qquad p^{\ast }(t)=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \biggr) , \\& p_{\ast }(t)=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \frac{m_{*}(\tau ^{-1}(\tau ^{-1}(t)))}{m_{*}(\tau ^{-1}(t))} \biggr) ,\quad \mbox{and} \\& p_{\ast \ast }(t)=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \frac{m_{**}(\tau ^{-1}(\tau ^{-1}(t)))}{m_{**}(\tau ^{-1}(t))} \biggr) , \end{aligned}$$

where $\tau ^{-1}$ is the inverse of τ, $m_{*}$ and $m_{**}$ are functions to be specified later. All functional inequalities considered in this article are assumed to hold eventually, that is, they are satisfied for all t large enough.

2 Some preliminaries

We enlist some known results which will be needed. We first present the following classes of nonoscillatory (let us say positive) solutions of (1.1):

• $z(t)\in N_{I}\Leftrightarrow z^{\prime }(t)>0$, $z^{\prime \prime }(t)>0$, $( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }<0$,

• $z(t)\in N_{\mathit{II}}\Leftrightarrow z^{\prime }(t)<0$, $z^{\prime \prime }(t)>0$, $( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }<0$, and

• $z(t)\in N_{\mathit{III}}\Leftrightarrow z^{\prime }(t)>0$, $z^{\prime \prime }(t)<0$, $( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }<0$, eventually.

The following lemma comes directly from combining Lemma 1 and Lemma 2 in [13] with Lemma 3 and Lemma 4 in [20].

Lemma 2.1

Assume that$A\geq 0$and$B\geq 0$. Then

$$ ( A+B ) ^{\lambda }\leq A^{\lambda }+B^{\lambda }\leq 2^{1- \lambda } ( A+B ) ^{\lambda },\quad 0< \lambda \leq 1, $$

(2.1)

and

$$ 2^{1-\lambda } ( A+B ) ^{\lambda }\leq A^{\lambda }+B^{ \lambda } \leq ( A+B ) ^{\lambda },\quad \lambda \geq 1. $$

(2.2)

Lemma 2.2

Let$g>0$. Then

$$ g^{r}\leq rg+(1-r) \quad \textit{for }0< r\leq 1 $$

(2.3)

and

$$ g^{r}\geq rg+(1-r) \quad \textit{for }r\geq 1. $$

(2.4)

Proof

See [21, p. 28]. □

Lemma 2.3

[17]Assume that$A\geq 0$, $B>0$, $U\geq 0$, and$\lambda >0$. Then

$$ AU-BU^{1+\frac{1}{\lambda }}\leq \frac{\lambda ^{\lambda }}{ ( \lambda +1 ) ^{\lambda +1}} \frac{A^{\lambda +1}}{B^{\lambda }}. $$

(2.5)

Lemma 2.4

Assume thatxis an eventually positive solution of (1.1). If (1.5) holds, then$z(t)\in N_{I}$or$z(t)\in N_{\mathit{II}}$. While if (1.8) holds, then either$z(t)\in N_{I}$or$z(t)\in N_{\mathit{II}}$or$z(t)\in N_{\mathit{III}}$.

Proof

The proof is similar to [22, Theorem 2.1 and Theorem 2.2]. □

Lemma 2.5

([5, 23])

Let the function$f(t)$satisfy$f^{(i)}(t)>0$, $i=0,1,2,\ldots,n$, and$f^{(n+1)}(t)<0$eventually, then there exists a constant$k_{1}\in (0,1)$such that$\frac{f(t)}{f^{\prime }(t)}\geq \frac{k_{1}t}{n}$eventually.

3 Oscillation criteria in the case when (H₄) holds

In this section, we establish new oscillation criteria for Eq. (1.1) in the case when (H₄) holds.

Theorem 3.1

Assume that (H₁)–(H₄) hold. If there exists a positive function$\rho (t)\in C^{1} ( [t_{0},\infty ) )$such that

$$\begin{aligned} &\int _{t_{\ast }}^{\infty } \biggl[K\rho (s)Q(s) \biggl( \frac{\int _{t_{2}}^{\lambda _{1}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma } \\ &\quad {}- \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\lambda } \biggr]\,\mathrm {d}s=\infty , \end{aligned}$$

(3.1)

where

$$\begin{aligned}& \lambda =\min \{\alpha ,\gamma \}, \qquad m= \textstyle\begin{cases} 1, & \gamma =\alpha, \\ 0< m\leq 1, & \gamma \neq \alpha, \end{cases}\displaystyle \qquad K= \textstyle\begin{cases} \frac{k}{2^{\gamma -1}}, & \gamma >1, \\ k, & \gamma \leq 1, \end{cases}\displaystyle \end{aligned}$$

(3.2)

$$\begin{aligned}& g = \textstyle\begin{cases} 1, & \gamma \geq \alpha, \\ \frac{\gamma }{\alpha }, & \gamma < \alpha \end{cases}\displaystyle \quad \textit{and}\quad \lambda _{1}(t)= \textstyle\begin{cases} t, & \sigma (t)\geq t, \\ \sigma (t), & \sigma (t)< t \end{cases}\displaystyle \end{aligned}$$

(3.3)

holds for some constant$k>0$, sufficiently large$t_{1}\geq t_{0}$, and for some$t_{\ast }>t_{2}>t_{1}$, then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{I}$.

Proof

Assume that $x(t)$ is a positive solution of Eq. (1.1) satisfying $z(t)\in N_{I}$ for $t\geq t_{1}$. Then from (1.1) and (H₃) it follows that

$$ \bigl( r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha } \bigr) ^{\prime }=-q(t)f \bigl(x \bigl( \sigma (t) \bigr) \bigr)\leq -kq(t)x^{\gamma } \bigl(\sigma (t) \bigr)< 0. $$

(3.4)

Since $( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{ \prime }= ( r ( z^{\prime \prime } ) ^{\alpha } ) ^{\prime }(\tau (t))\tau ^{\prime }(t)$, then in view of (H₄) there exists $t_{2}\geq t_{1}$ such that

$$\begin{aligned}& \bigl( r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha } \bigr) ^{\prime }+ \frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl( \tau (t) \bigr) \bigr)^{\alpha } \bigr) ^{\prime } \\& \quad \leq -kQ(t) \bigl[ x^{\gamma } \bigl( \sigma (t) \bigr)+p_{0}^{\gamma }x^{\gamma } \bigl(\tau \bigl( \sigma (t) \bigr) \bigr) \bigr]\quad \text{for }t\geq t_{2}. \end{aligned}$$

(3.5)

In the following, we consider the two cases $\gamma >1$ and $\gamma \leq 1$. Firstly, assume that $\gamma >1$. Using (2.2) with (3.5), we get

$$\begin{aligned} & \bigl( r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha } \bigr) ^{\prime }+ \frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl( \tau (t) \bigr) \bigr)^{\alpha } \bigr) ^{\prime } \\ &\quad \leq -\frac{k}{2^{\gamma -1}}Q(t) \bigl[ x \bigl(\sigma (t) \bigr)+p_{0}x \bigl(\tau \bigl( \sigma (t) \bigr) \bigr) \bigr] ^{\gamma } \leq - \frac{k}{2^{\gamma -1}}Q(t)z^{ \gamma } \bigl(\sigma (t) \bigr). \end{aligned}$$

(3.6)

Define the functions $\omega (t)$ and $\nu (t)$ by

$$ \omega (t)=\rho (t) \frac{r(t)(z^{\prime \prime }(t))^{\alpha }}{ ( z^{\prime }(t) ) ^{\gamma }} $$

(3.7)

and

$$ \nu (t)=\rho (t) \frac{r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }},\quad t\geq t_{2}. $$

(3.8)

Then clearly $\omega (t)$ and $\nu (t)$ are positive for $t\geq t_{2}$ and satisfy

$$ \omega ^{\prime }(t)=\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)+ \rho (t)\frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}-\gamma \rho (t)r(t) \frac{ ( z^{\prime \prime }(t) ) ^{\alpha +1}}{ ( z^{\prime }(t) ) ^{\gamma +1}} $$

(3.9)

and

$$ \nu ^{\prime }(t) =\frac{\rho ^{\prime }(t)}{\rho (t)}\nu (t)+\rho (t)\frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }} - \gamma \rho (t)r \bigl(\tau (t) \bigr)\tau ^{\prime }(t) \frac{ ( z^{\prime \prime }(\tau (t)) ) ^{\alpha +1}}{ ( z^{\prime }(\tau (t)) ) ^{\gamma +1}}. $$

(3.10)

Now, we consider the two cases $\gamma \geq \alpha$ and $\gamma <\alpha$. We first assume that $\gamma \geq \alpha$. From (3.7), we have

$$ z^{\prime \prime }(t)= \bigl( z^{\prime }(t) \bigr) ^{ \frac{\gamma }{\alpha }} \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{\frac{1}{\alpha }}. $$

Substituting into (3.9), we get

$$ \omega ^{\prime }(t)=\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)+ \rho (t)\frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}-\gamma \rho (t)r(t) \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{1+\frac{1}{\alpha }} \bigl( z^{\prime }(t) \bigr) ^{\frac{\gamma }{\alpha }-1}. $$

(3.11)

But since $z^{\prime }(t)$ is positive and increasing, it follows that there exists a constant $M>0$ satisfying $z^{\prime }(t)\geq M$ and

$$ \omega ^{\prime }(t) \leq \rho _{+}^{\prime }(t)r(t) \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) -\gamma M^{\frac{\gamma }{\alpha }-1}\rho (t)r(t) \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{1+\frac{1}{\alpha }} +\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}. $$

Using inequality (2.5) with $A=\rho _{+}^{\prime }(t)r(t)$, $U=\frac{\omega (t)}{\rho (t)r(t)}$, and $B=\gamma M^{\frac{\gamma }{\alpha } -1}\rho (t)r(t)$, it follows that

$$\begin{aligned} \omega ^{\prime }(t) &\leq r(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{\alpha }{\gamma M^{\frac{\gamma }{\alpha }-1}\rho (t)} \biggr) ^{ \alpha }+\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }} \\ &\leq r(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{ \alpha +1} \biggl( \frac{1}{M^{\frac{\gamma }{\alpha }-1}\rho (t)} \biggr) ^{ \alpha }+\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}. \end{aligned}$$

(3.12)

In view of (3.8), we have

$$ z^{\prime \prime } \bigl(\tau (t) \bigr)= \bigl( z^{\prime } \bigl(\tau (t) \bigr) \bigr) ^{ \frac{\gamma }{\alpha }} \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) ^{ \frac{1}{\alpha }}. $$

Substituting into (3.10), we get

$$\begin{aligned} \nu ^{\prime }(t) &=\frac{\rho ^{\prime }(t)}{\rho (t)}\nu (t)+\rho (t)\frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }} \\ &\quad {} -\gamma \rho (t)r \bigl(\tau (t) \bigr)\tau ^{\prime }(t) \bigl( z^{ \prime } \bigl(\tau (t) \bigr) \bigr) ^{\frac{\gamma }{\alpha }-1} \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) ^{1+\frac{1}{\alpha }} \\ &\leq \rho _{+}^{\prime }(t)r \bigl(\tau (t) \bigr) \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) +\rho (t) \frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }} \\ &\quad {} -\gamma M^{\frac{\gamma }{\alpha }-1}\rho (t)r \bigl(\tau (t) \bigr)\tau ^{ \prime }(t) \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) ^{1+ \frac{1}{\alpha }}. \end{aligned}$$

Again by inequality (2.5), we get

$$ \nu ^{\prime }(t) \leq r \bigl(\tau (t) \bigr) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{1}{M^{\frac{\gamma }{\alpha }-1}\rho (t)\tau ^{\prime }(t)} \biggr) ^{\alpha } +\rho (t) \frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }}. $$

But since $z^{\prime \prime }(t)>0$ and $\tau (t) \leq t$, we obtain

$$ \nu ^{\prime }(t) \leq r \bigl(\tau (t) \bigr) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{1}{M^{\frac{\gamma }{\alpha }-1}\rho (t)\tau ^{\prime }(t)} \biggr) ^{\alpha } +\rho (t) \frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}. $$

(3.13)

Combining (3.12) and (3.13) and using (3.6), we get

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq {}&{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{z(\sigma (t))}{z^{\prime }(t)} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\alpha +1}} \biggr) R(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{ \alpha +1} \biggl( \frac{1}{M^{\frac{\gamma }{\alpha }-1}\rho (t)} \biggr) ^{ \alpha }. \end{aligned}$$

(3.14)

Now, assume that $\gamma <\alpha$. Then from (3.7) we have

$$ \frac{1}{z^{\prime }(t)}= \frac{ ( \frac{\omega (t)}{\rho (t)r(t)} ) ^{\frac{1}{\gamma }}}{ ( z^{\prime \prime }(t) ) ^{\frac{\alpha }{\gamma }}}. $$

Substituting into (3.9), we get

$$\begin{aligned} \omega ^{\prime }(t) &=\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)+ \rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}-\gamma \rho (t)r(t) \bigl( z^{ \prime \prime }(t) \bigr) ^{1-\frac{\alpha }{\gamma }} \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{1+\frac{1}{\gamma }} \\ &=\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)+\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }} \\ &\quad {} -\gamma \rho (t) \bigl( r(t) \bigr) ^{1-\frac{1}{\alpha }+ \frac{1}{\gamma }} \bigl( r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t) \bigr) ^{1- \frac{\alpha }{\gamma }} \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{1+ \frac{1}{\gamma }}. \end{aligned}$$

It is clear that $(r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t) )^{1- \frac{\alpha }{\gamma }}$ is positive and increasing, and so there exists a positive constant $m_{1}$ such that

$$\begin{aligned} \omega ^{\prime }(t) \leq {}&\rho _{+}^{\prime }(t)r(t) \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) +\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }} \\ &{}-\gamma m_{1}\rho (t) \bigl( r(t) \bigr) ^{1-\frac{1}{\alpha }+ \frac{1}{\gamma }} \biggl( \frac{\omega (t)}{\rho (t)r(t)} \biggr) ^{1+ \frac{1}{\gamma }} \end{aligned}$$

for all sufficiently large t. Using inequality (2.5), we conclude that

$$ \omega ^{\prime }(t)\leq \biggl( \frac{\rho _{+}^{\prime }(t)}{\gamma +1} \biggr) ^{\gamma +1} \biggl( \frac{r^{\frac{1}{\alpha }}(t)}{m_{1}\rho (t)} \biggr) ^{\gamma }+\rho (t) \frac{ ( r(t)(z^{\prime \prime }(t))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}. $$

(3.15)

But since from (3.8) we have

$$ \frac{1}{z^{\prime }(\tau (t))}= \frac{ ( \frac{\nu (t)}{\rho (t)r(\tau (t))} ) ^{\frac{1}{\gamma }}}{ ( z^{\prime \prime }(\tau (t)) ) ^{\frac{\alpha }{\gamma }}}, $$

then, by substituting into (3.10), we get

$$\begin{aligned} \nu ^{\prime }(t) &\leq \rho _{+}^{\prime }(t)r \bigl( \tau (t) \bigr) \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) +\rho (t) \frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(\tau (t)) ) ^{\gamma }} \\ &\quad{} -\gamma m_{1}\rho (t) \bigl( r \bigl(\tau (t) \bigr) \bigr) ^{1- \frac{1}{\alpha }+\frac{1}{\gamma }}\tau ^{\prime }(t) \biggl( \frac{\nu (t)}{\rho (t)r(\tau (t))} \biggr) ^{1+\frac{1}{\gamma }}. \end{aligned}$$

This with (2.5) leads to

$$ \nu ^{\prime }(t)\leq \biggl( \frac{\rho _{+}^{\prime }(t)}{\gamma +1} \biggr) ^{\gamma +1} \biggl( \frac{r^{\frac{1}{\alpha }}(\tau (t))}{m_{1}\rho (t)\tau ^{\prime }(t)} \biggr) ^{\gamma }+\rho (t) \frac{ ( r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha } ) ^{\prime }}{ ( z^{\prime }(t) ) ^{\gamma }}. $$

(3.16)

Combining (3.15) and (3.16), using (3.6), we get

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq {}&{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{z(\sigma (t))}{z^{\prime }(t)} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\gamma +1}} \biggr) R^{ \frac{\gamma }{\alpha }}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\gamma +1} \biggr) ^{ \gamma +1} \biggl( \frac{1}{m_{1}\rho (t)} \biggr) ^{\gamma }. \end{aligned}$$

(3.17)

Combining (3.14) and (3.17), we obtain for any α, γ ratios of odd positive integers that

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq {}&{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{z(\sigma (t))}{z^{\prime }(t)} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{\lambda }. \end{aligned}$$

(3.18)

Now, we consider the two cases $\sigma (t)< t$ and $\sigma (t)\geq t$. We start by considering the case $\sigma (t)< t$. Since $r(t)(z^{\prime \prime }(t))^{\alpha }$ is positive and decreasing, we have

$$ z^{\prime }(t)\geq z^{\prime }(t)-z^{\prime }(t_{2})= \int _{t_{2}}^{t} \frac{r^{\frac{1}{\alpha }}(s)z^{\prime \prime }(s)}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s\geq r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t) \int _{t_{2}}^{t} \frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s, $$

i.e.,

$$ \biggl( \frac{z^{\prime }(t)}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{\prime }\leq 0. $$

(3.19)

But since $\sigma (t)< t$, then it follows that

$$ \frac{z^{\prime }(\sigma (t))}{z^{\prime }(t)}\geq \frac{\int _{t_{2}}^{\sigma (t)}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}. $$

(3.20)

Now since by (3.19) we have

$$ z(t)\geq z(t)-z(t_{3})= \int _{t_{3}}^{t} \frac{z^{\prime }(s)\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u}{\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u}\,\mathrm {d}s\geq z^{\prime }(t) \frac{\int _{t_{3}}^{t}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}, $$

which means that

$$ \frac{z(t)}{z^{\prime }(t)}\geq \frac{\int _{t_{3}}^{t}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \quad \text{for }t\geq t_{3}>t_{2}, $$

(3.21)

then we have

$$ \frac{z(\sigma (t))}{z^{\prime }(\sigma (t))}\geq \frac{\int _{t_{3}}^{\sigma (t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{\sigma (t)}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}. $$

(3.22)

This with (3.20) leads to

$$ \frac{z(\sigma (t))}{z^{\prime }(t)}= \frac{z(\sigma (t))}{z^{\prime }(\sigma (t))} \frac{z^{\prime }(\sigma (t))}{z^{\prime }(t)}\geq \frac{\int _{t_{3}}^{\sigma (t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}. $$

(3.23)

Substituting into (3.18), we get

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq {}&{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{\int _{t_{3}}^{\sigma (t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{\lambda }. \end{aligned}$$

(3.24)

Now, consider the case $\sigma (t)\geq t$. Since $z(t)$ is positive and increasing, it follows from (3.18) that

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq{} &{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{z(t)}{z^{\prime }(t)} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{\lambda }. \end{aligned}$$

(3.25)

Since $( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{ \prime }<0$, we get (3.19) and consequently we arrive at (3.21). Then, substituting into (3.25), we have

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq{} &{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{\int _{t_{3}}^{t}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{\lambda }. \end{aligned}$$

(3.26)

Combining (3.24) and (3.26), we get

$$\begin{aligned} \omega ^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu ^{\prime }(t) \leq {}&{-} \frac{k}{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{\int _{t_{3}}^{\lambda _{1}(t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{\gamma } \\ &{}+ \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{\lambda }. \end{aligned}$$

Integrating from $t_{4}$ (>$t_{3}$) to t, we have

$$\begin{aligned} \omega (t_{4})+\frac{p_{0}^{\gamma }}{\tau _{0}}\nu (t_{4})\geq{} & \int _{t_{4}}^{t} \biggl[\frac{k}{2^{\gamma -1}}\rho (s)Q(s) \biggl( \frac{\int _{t_{3}}^{\lambda _{1}(s)}\int _{t_{2}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma } \\ &{} - \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\lambda +1}} \biggr) R^{g}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\lambda } \biggr]\,\mathrm {d}s, \end{aligned}$$

which contradicts (3.1). Secondly, assume that $\gamma \leq 1$. Using (2.1) with (3.5), we obtain

$$ \bigl( r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha } \bigr) ^{\prime }+ \frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl( \tau (t) \bigr) \bigr)^{\alpha } \bigr) ^{\prime } \leq -kQ(t)z^{\gamma } \bigl( \sigma (t) \bigr). $$

(3.27)

By completing the proof as the above case of $\gamma >1$, using (3.27) instead of (3.6), the proof is completed. □

Lemma 3.1

Assume that conditions (H₁)–(H₄) hold. Letxbe an eventually positive solution of Eq. (1.1) and the corresponding$z(t)$satisfies$z(t)\in N_{\mathit{II}}$. If

$$ \int _{t_{0}}^{\infty }Q(s)\,\mathrm {d}s=\infty $$

(3.28)

or

$$ \int _{t_{0}}^{\infty } \int _{t}^{\infty } \biggl[ \frac{1}{r(\tau (s))} \int _{s}^{\infty }Q(u)\,\mathrm {d}u \biggr] ^{\frac{1}{\alpha }} \,\mathrm {d}s \,\mathrm {d}t= \infty , $$

(3.29)

then$\lim_{t\rightarrow \infty }x(t)=\lim_{t \rightarrow \infty }z(t)=0$.

Proof

Assume that $x(t)$ is a positive solution of Eq. (1.1) satisfying $z(t)\in N_{\mathit{II}}$ for $t\geq t_{1}$. Going through as in the proof of Theorem 3.1, we arrive at (3.5). In the following, we consider the two cases $\gamma >1$ and $\gamma \leq 1$. Firstly, assume that $\gamma >1$. Then we have (3.6). Since $z(t)$ is positive and decreasing, we have $\lim_{t\rightarrow \infty }z(t)=l\geq 0$ exists. We claim that $l=0$. If not, then there exists $t_{3}\geq t_{2}$ such that $z(\sigma (t))>l$ for $t\geq t_{3}$. Substituting into (3.6), we get

$$ \bigl( r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha } \bigr) ^{\prime }+ \frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl( \tau (t) \bigr) \bigr)^{\alpha } \bigr) ^{\prime } \leq - \frac{kl^{\gamma }}{2^{\gamma -1}}Q(t). $$

(3.30)

Integrating (3.30) from $t_{3}$ to t and taking into account (3.28), we have

$$\begin{aligned} &r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha }+ \frac{p_{0}^{\gamma }}{\tau _{0}} r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl(\tau (t) \bigr) \bigr)^{ \alpha } \\ &\quad \leq r(t_{3}) \bigl(z^{\prime \prime }(t_{3}) \bigr)^{\alpha }+ \frac{p_{0}^{\gamma }}{\tau _{0}} r \bigl(\tau (t_{3}) \bigr) \bigl(z^{\prime \prime } \bigl(\tau (t_{3}) \bigr) \bigr)^{ \alpha } -\frac{kl^{\gamma }}{2^{\gamma -1}} \int _{t_{3}}^{t}Q(s)\,\mathrm {d}s \rightarrow -\infty \quad \text{as } t\rightarrow \infty , \end{aligned}$$

which is a contradiction. Thus $l=0$ and consequently $\lim_{t\rightarrow \infty }x(t)=0$. In the following, we obtain the same conclusion in the case when $\int _{t_{0}}^{\infty }Q(s)\,\mathrm {d}s<\infty$. Integrating (3.30) from t to ∞, we have

$$ r(t) \bigl(z^{\prime \prime }(t) \bigr)^{\alpha }+ \frac{p_{0}^{\gamma }}{\tau _{0}} r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl(\tau (t) \bigr) \bigr)^{ \alpha }\geq \frac{kl^{\gamma }}{2^{\gamma -1}} \int _{t}^{\infty }Q(s) \,\mathrm {d}s. $$

But since $\tau (t)\leq t$, then we can observe that $r(\tau (t))(z^{\prime \prime }(\tau (t)))^{\alpha }\geq r(t)(z^{ \prime \prime }(t))^{\alpha }$ and consequently we have

$$ r \bigl(\tau (t) \bigr) \bigl(z^{\prime \prime } \bigl(\tau (t) \bigr) \bigr)^{\alpha }\geq \frac{kl^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )} \int _{t}^{\infty }Q(s)\,\mathrm {d}s, $$

i.e.,

$$ z^{\prime \prime } \bigl(\tau (t) \bigr)\geq \biggl[ \frac{kl^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )} \biggr]^{\frac{1}{\alpha }} \biggl[\frac{1}{r(\tau (t))} \int _{t}^{\infty }Q(s) \,\mathrm {d}s \biggr]^{\frac{1}{\alpha }}. $$

Integrating from t to ∞ followed by integrating from $t_{3}$ to ∞, we obtain

$$ \frac{1}{\tau _{0}^{2}}z \bigl(\tau (t_{3}) \bigr)\geq \biggl[ \frac{kl^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )} \biggr]^{\frac{1}{\alpha }} \int _{t_{3}}^{\infty } \int _{t}^{\infty } \biggl[\frac{1}{r(\tau (s))} \int _{s}^{\infty }Q(u)\,\mathrm {d}u \biggr]^{ \frac{1}{\alpha }} \,\mathrm {d}s \,\mathrm {d}t, $$

which contradicts (3.29). Thus $\lim_{t\rightarrow \infty }x(t)=0$. Secondly, assume that $\gamma \leq 1$. As in the proof of Theorem 3.1, we have (3.27). By completing the proof as in the above case of $\gamma >1$, using (3.27) instead of (3.6), the proof is completed. □

Theorem 3.2

Assume that (H₁)–(H₄) hold. If

$$ \int _{t_{0}}^{\infty } \biggl[ \frac{1}{r(t)} \int _{t_{0}}^{t}Q(s) \bigl( \sigma (s) \bigr) ^{\gamma } \biggl( \int _{\sigma (s)}^{ \infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr) ^{\gamma }\,\mathrm {d}s \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}t= \infty , $$

(3.31)

then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{\mathit{III}}$.

Proof

Let $x(t)$ be an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{\mathit{III}}$ for all $t\geq t_{1}\geq t_{0}$. Since $z^{\prime \prime }(t)<0$ and $z^{\prime }(t)>0$, then by Lemma 2.5, there exist $t_{2}\geq t_{1}$ and a constant $k_{1}$ satisfying $0< k_{1}<1$ such that $z(t)\geq k_{1}tz^{\prime }(t)$ for $t\geq t_{2}$, i.e.,

$$ z \bigl(\sigma (t) \bigr)\geq k_{1}\sigma (t)z^{\prime } \bigl( \sigma (t) \bigr),\quad t\geq t_{2} \geq t_{1}. $$

(3.32)

Going through as in Theorem 3.1, we arrive at (3.5). In the following, we consider the two cases $\gamma >1$ and $\gamma \leq 1$. Firstly, assume that $\gamma >1$. Then we have (3.6), and using (3.32) we get

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{\prime }+\frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl( z^{ \prime \prime } \bigl(\tau (t) \bigr) \bigr) ^{\alpha } \bigr) ^{\prime } \leq - \frac{kk_{1}^{\gamma }}{2^{\gamma -1}}Q(t) \bigl( \sigma (t) \bigr) ^{\gamma } \bigl( z^{\prime } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma }. $$

(3.33)

But since $v(t)=-r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t)$ is positive and increasing, then there exists a constant $g_{1}>0$ such that $v(t)\geq g_{1}$ for $t\geq t_{3}\geq t_{2}$. Hence

$$ z^{\prime } \bigl(\sigma (t) \bigr)\geq \int _{\sigma (t)}^{\infty } \frac{v(s)}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s\geq g_{1} \int _{\sigma (t)}^{ \infty }\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s. $$

(3.34)

Substituting into (3.33) and integrating from $t_{3}$ to t, we get

$$\begin{aligned}& -r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha }- \frac{p_{0}^{\gamma }}{\tau _{0}}r \bigl(\tau (t) \bigr) \bigl( z^{\prime \prime } \bigl(\tau (t) \bigr) \bigr) ^{ \alpha } \\& \quad \geq \frac{kk_{1}^{\gamma }g_{1}^{\gamma }}{2^{\gamma -1}}\int _{t_{3}}^{t}Q(s) \bigl( \sigma (s) \bigr) ^{\gamma } \biggl( \int _{ \sigma (s)}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr) ^{ \gamma }\,\mathrm {d}s. \end{aligned}$$

(3.35)

But since $\tau (t)\leq t$, then we can conclude that $r(\tau (t)) ( z^{\prime \prime }(\tau (t)) ) ^{\alpha } \geq r(t) ( z^{\prime \prime }(t) ) ^{\alpha }$. Now since from (3.35) we have

$$ -r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \geq \frac{kk_{1}^{\gamma }g_{1}^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )}\int _{t_{3}}^{t}Q(s) \bigl( \sigma (s) \bigr) ^{\gamma } \biggl( \int _{ \sigma (s)}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr) ^{ \gamma }\,\mathrm {d}s, $$

i.e.,

$$ -z^{\prime \prime }(t) \geq \biggl( \frac{kk_{1}^{\gamma }g_{1}^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )} \biggr) ^{\frac{1}{\alpha }} \biggl[ \frac{1}{r(t)}\int _{t_{3}}^{t}Q(s) \bigl( \sigma (s) \bigr) ^{\gamma } \biggl( \int _{ \sigma (s)}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr) ^{ \gamma }\,\mathrm {d}s \biggr] ^{\frac{1}{\alpha }}. $$

Then integrating from $t_{4}$ ($\geq t_{3}$) to t, we get

$$ z^{\prime }(t_{4}) \geq \biggl( \frac{kk_{1}^{\gamma }g_{1}^{\gamma }}{2^{\gamma -1} (1+\frac{p_{0}^{\gamma }}{\tau _{0}} )} \biggr) ^{\frac{1}{\alpha }} \int _{t_{4}}^{t} \biggl[ \frac{1}{r(s)} \int _{t_{3}}^{s}Q(u) \bigl( \sigma (u) \bigr) ^{\gamma } \biggl( \int _{\sigma (u)}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \biggr) ^{\gamma }\,\mathrm {d}u \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}s, $$

which contradicts (3.31). Secondly, assume that $\gamma \leq 1$. As in the proof of Theorem 3.1, we arrive at (3.27), and then using (3.32) we get

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }+\frac{p_{0}^{\gamma }}{\tau _{0}} \bigl( r \bigl(\tau (t) \bigr) \bigl( z^{ \prime \prime } \bigl(\tau (t) \bigr) \bigr) ^{\alpha } \bigr) ^{\prime } \leq -kk_{1}^{ \gamma }Q(t) \bigl( \sigma (t) \bigr) ^{\gamma } \bigl( z^{\prime } \bigl( \sigma (t) \bigr) \bigr) ^{\gamma }. $$

(3.36)

Going through as in the proof of the case $\gamma >1$, using (3.36) instead of (3.33), this completes the proof. □

The following results are immediate consequences of Lemma 2.4, Lemma 3.1, Theorem 3.1, and Theorem 3.2.

Theorem 3.3

Assume that (1.8) and all the conditions of Lemma3.1, Theorem3.1, and Theorem3.2hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

Theorem 3.4

Assume that (1.5) and all the conditions of Lemma3.1and Theorem3.1hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

The following results deal with the special case $\alpha \leq 1$ and $\gamma \geq 1$ of Eq. (1.1).

Theorem 3.5

Assume that conditions (H₁)–(H₄), $\alpha \leq 1$, and$\gamma \geq 1$hold. If there exists a positive function$\rho (t)\in C^{1} ( [t_{0},\infty ) )$such that

$$ \int _{t_{\ast }}^{\infty } \biggl[ \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t) \biggl( \frac{\int _{t_{2}}^{\lambda _{1}(t)}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) -G_{1}(t) \biggr] \,\mathrm {d}t=\infty $$

(3.37)

holds for any positive constantsk, M, sufficiently large$t_{1}\geq t_{0}$, and for some$t_{\ast }>t_{2}>t_{1}$, where$\lambda _{1}(t)$is defined by (3.3) and

$$\begin{aligned} G_{1}(t) ={}&\frac{1}{4}\alpha \rho (t) \biggl[ r(t) \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+\frac{1-\alpha }{\alpha M} \biggr) ^{2}+ \frac{p_{0}^{\gamma }}{\tau _{0}^{2}}r \bigl(\tau (t) \bigr) \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+ \frac{ ( 1-\alpha ) \tau ^{\prime }(t)}{\alpha M} \biggr) ^{2} \biggr] \\ &{}+\frac{k(\gamma -1)}{2^{\gamma -1}M}\rho (t)Q(t), \end{aligned}$$

then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{I}$.

Proof

Assume that $x(t)$ is an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{I}$. As in the proof of Theorem 3.1, we arrive at (3.6). Now define the function $W(t)$ by

$$ W(t)=\rho (t) \frac{r(t) ( z^{\prime \prime }(t) ) ^{\alpha }}{z^{\prime }(t)},\quad t\geq t_{1}\geq t_{0}. $$

(3.38)

Then $W(t)>0$ for $t\geq t_{1}$ and

$$\begin{aligned} W^{\prime }(t) &=\frac{\rho ^{\prime }(t)}{\rho (t)}W(t)+\rho (t) \frac{ ( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)}- \rho (t) \frac{r(t) ( z^{\prime \prime }(t) ) ^{\alpha +1}}{ ( z^{\prime }(t) ) ^{2}} \\ &=\frac{\rho ^{\prime }(t)}{\rho (t)}W(t)+\rho (t) \frac{ ( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)}-W(t) \frac{z^{\prime \prime }(t)}{z^{\prime }(t)}. \end{aligned}$$

(3.39)

Since $z^{\prime }(t)$ and $z^{\prime \prime }(t)$ are positive, then there exist $t_{2}\geq t_{1}$ and constant $M>0$ such that $z^{\prime }(t)\geq M$ for all $t\geq t_{2}$. Now, from (3.38) and (2.3), we get

$$ \frac{z^{\prime \prime }(t)}{z^{\prime }(t)}\geq \frac{W(t)}{\alpha \rho (t)r(t)}-\frac{(1-\alpha )}{\alpha M}. $$

(3.40)

This with (3.39) yields

$$\begin{aligned} W^{\prime }(t) &\leq \frac{\rho ^{\prime }(t)}{\rho (t)}W(t)+\rho (t) \frac{ ( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)}- \frac{W^{2}(t)}{\alpha \rho (t)r(t)}+\frac{(1-\alpha )}{\alpha M}W(t) \\ &\leq \rho (t) \frac{ ( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)}+ \frac{1}{4}\alpha \rho (t)r(t) \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+\frac{(1-\alpha )}{\alpha M} \biggr) ^{2}. \end{aligned}$$

(3.41)

Now define

$$ V(t)=\rho (t) \frac{r(\tau (t)) ( z^{\prime \prime }(\tau (t)) ) ^{\alpha }}{z^{\prime }(\tau (t))}. $$

(3.42)

As we did for W, we can get

$$ V^{\prime }(t) \leq \frac{\rho ^{\prime }(t)}{\rho (t)}V(t)+\rho (t) \frac{ ( r(\tau (t)) ( z^{\prime \prime }(\tau (t)) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(\tau (t))} -\frac{\tau ^{\prime }(t)V^{2}(t)}{\alpha \rho (t)r(\tau (t))}+\frac{(1-\alpha )\tau ^{\prime }(t)}{\alpha M}V(t). $$

But since $z^{\prime }$ is increasing and $\tau (t) \leq t$, then

$$ V^{\prime }(t) \leq \rho (t) \frac{ ( r(\tau (t)) ( z^{\prime \prime }(\tau (t)) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)}+ \frac{1}{4} \frac{\alpha \rho (t)r(\tau (t))}{\tau ^{\prime }(t)} \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+ \frac{(1-\alpha )\tau ^{\prime }(t)}{\alpha M} \biggr) ^{2}. $$

(3.43)

This with (3.41) leads to

$$\begin{aligned} &W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t) \\ &\quad \leq \rho (t) \biggl[ \frac{ ( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{\prime }+\frac{p_{0}^{\gamma }}{\tau _{0}} ( r(\tau (t)) ( z^{\prime \prime }(\tau (t)) ) ^{\alpha } ) ^{\prime }}{z^{\prime }(t)} \biggr] \\ &\qquad {} +\frac{1}{4}\alpha \rho (t) \biggl[ r(t) \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+\frac{(1-\alpha )}{\alpha M} \biggr) ^{2}+ \frac{p_{0}^{\gamma }r(\tau (t))}{\tau _{0}^{2}} \biggl( \frac{\rho ^{\prime }(t)}{\rho (t)}+ \frac{(1-\alpha )\tau ^{\prime }(t)}{\alpha M} \biggr) ^{2} \biggr] . \end{aligned}$$

Thus, by (3.6) and (2.4), we get

$$ W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t)\leq - \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t) \frac{z(\sigma (t))}{z^{\prime }(t)}+G_{1}(t). $$

(3.44)

Now, we consider the two cases $\sigma (t)< t$ and $\sigma (t)\geq t$.

First assume that $\sigma (t)< t$. As in the proof of Theorem 3.1, we get (3.23). Substituting into (3.44), we have

$$ W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t)\leq - \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t) \frac{\int _{t_{3}}^{\sigma (t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}+G_{1}(t)\quad \text{for }t\geq t_{3}>t_{2}. $$

(3.45)

Secondly, assume that $\sigma (t)\geq t$. Since $z^{\prime }(t)>0$, it follows from (3.44) that

$$ W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t)\leq - \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t)\frac{z(t)}{z^{\prime }(t)}+G_{1}(t). $$

(3.46)

As in the proof of Theorem 3.1, we arrive at (3.21). Then, substituting into (3.46), we have

$$ W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t)\leq - \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t) \frac{\int _{t_{3}}^{t}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}+G_{1}(t). $$

(3.47)

This with (3.45) yields

$$ W^{\prime }(t)+\frac{p_{0}^{\gamma }}{\tau _{0}}V^{\prime }(t)\leq - \frac{k\gamma }{2^{\gamma -1}}\rho (t)Q(t) \frac{\int _{t_{3}}^{\lambda _{1}(t)}\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}+G_{1}(t). $$

Integrating from $t_{4}$ ($>t_{3}$) to t, we get

$$ W(t_{4})+\frac{p_{0}^{\gamma }}{\tau _{0}}V(t_{4})\geq \int _{t_{4}}^{t} \biggl[\frac{k\gamma }{2^{\gamma -1}}\rho (s)Q(s) \frac{\int _{t_{3}}^{\lambda _{1}(s)}\int _{t_{2}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{2}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u}-G_{1}(s) \biggr]\,\mathrm {d}s. $$

This contradicts (3.37) and completes the proof. □

Theorem 3.6

Assume that (1.8) and all the conditions of Lemma3.1, Theorem3.2, and Theorem3.5hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

Theorem 3.7

Assume that (1.5) and all the conditions of Lemma3.1and Theorem3.5hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

4 Oscillation criteria without condition (H₄)

In this section, we study the oscillation of Eq. (1.1) when either of the two conditions $0\leq p(t)\leq p_{0}<1$ or $p(t)\geq 1$, $p(t)\not \equiv 1$ holds for large t. Now, we begin by establishing new oscillation criteria for Eq. (1.1) in the case when $p(t)\geq 1$, $p(t)\not \equiv 1$ for large t with the condition $\tau (t)< t$ and $\tau (t)$ is strictly increasing.

Theorem 4.1

Assume that (H₁)–(H₃) hold, $p(t)\geq 1$, $p(t)\not \equiv 1$for sufficiently larget, $\tau (t)< t$and$\tau ^{\prime }(t)>0$. Further assume that there exists a positive function$m_{*}(t)\in C^{1} ( [t_{0},\infty ) )$such that

$$ m_{*}(t) \int _{t_{1}}^{t}\frac{\,\mathrm {d}s}{r^{\frac{1}{\alpha }}(s)}-m_{*}^{ \prime }(t) \int _{t_{2}}^{t} \int _{t_{1}}^{s} \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s\leq 0 $$

(4.1)

and$p_{\ast }(t)>0$for sufficiently larget. If there exists a positive function$\rho (t)\in C^{1} ( [t_{0},\infty ) )$such that

$$\begin{aligned} &\int _{t_{\ast }}^{\infty } \biggl[k\rho (s)q(s) \bigl( p_{\ast } \bigl( \sigma (s) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\lambda _{2}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma } \\ &\quad {}-r^{g}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\lambda +1} \biggr) ^{ \lambda +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\lambda } \biggr]\,\mathrm {d}s= \infty \end{aligned}$$

(4.2)

holds for some constant$k>0$, sufficiently large$t_{1}\geq t_{0}$, and for some$t_{\ast }>t_{2}>t_{1}$, whereλ, m, gare defined by (3.2), (3.3), and

$$ \lambda _{2}(t)= \textstyle\begin{cases} \tau ^{-1}(\sigma (t)), & \sigma (t)< \tau (t) , \\ t, & \sigma (t)\geq \tau (t), \end{cases} $$

(4.3)

then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{I}$.

Proof

Assume that $x(t)$ is an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{I}$ for $t\geq t_{1}$. From the definition of z (see also (2.2) in [6]), we have

$$\begin{aligned} x(t) &=\frac{1}{p(\tau ^{-1}(t))} \bigl( z \bigl(\tau ^{-1}(t) \bigr)-x \bigl(\tau ^{-1}(t) \bigr) \bigr) \\ &=\frac{z(\tau ^{-1}(t))}{p(\tau ^{-1}(t))}- \frac{1}{p(\tau ^{-1}(t))p(\tau ^{-1}(\tau ^{-1}(t)))} \bigl( z \bigl(\tau ^{-1} \bigl( \tau ^{-1}(t) \bigr) \bigr)-x \bigl(\tau ^{-1} \bigl(\tau ^{-1}(t) \bigr) \bigr) \bigr) \\ &\geq \frac{z(\tau ^{-1}(t))}{p(\tau ^{-1}(t))}- \frac{1}{p(\tau ^{-1}(t))p(\tau ^{-1}(\tau ^{-1}(t)))}z \bigl(\tau ^{-1} \bigl( \tau ^{-1}(t) \bigr) \bigr). \end{aligned}$$

(4.4)

Define the function $\omega (t)$ as in (3.7). Then $\omega (t)>0$ for $t\geq t_{1}$ satisfying (3.9). As in the proof of Theorem 3.1, since $( r(t) ( z^{\prime \prime }(t) ) ^{\alpha } ) ^{ \prime }<0$, we have (3.19) and then

$$ \frac{z(t)}{z^{\prime }(t)}\geq \frac{\int _{t_{2}}^{t}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \quad \text{for }t\geq t_{2}>t_{1}. $$

(4.5)

This with (4.1) yields

$$\begin{aligned} \biggl( \frac{z(t)}{m_{*}(t)} \biggr) ^{\prime } &= \frac{1}{m_{*}^{2}(t)} \bigl[ z^{\prime }(t)m_{*}(t)-z(t)m_{*}^{ \prime }(t) \bigr] \\ &\leq \frac{z(t)}{m_{*}^{2}(t)} \biggl[ \frac{m_{*}(t)\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}{\int _{t_{2}}^{t}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}-m_{*}^{\prime }(t) \biggr] \leq 0. \end{aligned}$$

This means that $\frac{z(t)}{m_{*}(t)}$ is nonincreasing. But since $\tau (t)< t$ and $\tau ^{\prime }(t)>0$, it follows that $\tau ^{-1}(t)\leq \tau ^{-1}(\tau ^{-1}(t))$, and so

$$ z \bigl(\tau ^{-1} \bigl(\tau ^{-1}(t) \bigr) \bigr)\leq \frac{m_{*}(\tau ^{-1}(\tau ^{-1}(t)))z(\tau ^{-1}(t))}{m_{*}(\tau ^{-1}(t))}. $$

(4.6)

Substituting from (4.6) into (4.4), we get

$$ x(t)\geq p_{\ast }(t)z \bigl(\tau ^{-1}(t) \bigr). $$

(4.7)

This in the view of (1.1) leads to

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kq(t) \bigl( p_{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma }z^{ \gamma } \bigl(\tau ^{-1} \bigl( \sigma (t) \bigr) \bigr). $$

(4.8)

In the following, we consider the two cases $\gamma \geq \alpha$ and $\gamma <\alpha$.

First, assume that $\gamma \geq \alpha$. As in the proof of Theorem 3.1, we have (3.12). Then, substituting from (4.8) into (3.12), we obtain

$$ \omega ^{\prime }(t) \leq r(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{1}{M^{\frac{\gamma }{\alpha }-1}\rho (t)} \biggr) ^{\alpha } -k\rho (t)q(t) \bigl( p_{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{ \gamma } \biggl( \frac{z(\tau ^{-1}(\sigma (t)))}{z^{\prime }(t)} \biggr) ^{\gamma }. $$

(4.9)

Now assume that $\gamma <\alpha$. As in the proof of Theorem 3.1, we have (3.15). Then, substituting from (4.8) into (3.15), we obtain

$$ \omega ^{\prime }(t) \leq r^{\frac{\gamma }{\alpha }}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\gamma +1} \biggr) ^{\gamma +1} \biggl( \frac{1}{m_{1}\rho (t)} \biggr) ^{\gamma } -k\rho (t)q(t) \bigl( p_{ \ast } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma } \biggl( \frac{z(\tau ^{-1}(\sigma (t)))}{z^{\prime }(t)} \biggr) ^{\gamma }. $$

(4.10)

This with (4.9) yields

$$ \omega ^{\prime }(t) \leq r^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{ \lambda } -k\rho (t)q(t) \bigl( p_{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{ \gamma } \biggl( \frac{z(\tau ^{-1}(\sigma (t)))}{z^{\prime }(t)} \biggr) ^{\gamma }. $$

(4.11)

Now, consider the two cases $\sigma (t)<\tau (t)$ and $\sigma (t)\geq \tau (t)$. First assume that $\sigma (t)<\tau (t)$. Since $\tau ^{-1}(\sigma (t))< t$ and $( \frac{z^{\prime }(t)}{\int _{t_{1}}^{t}\frac{\,\mathrm {d}s}{r^{\frac{1}{\alpha }}(s)}} ) ^{\prime }\leq 0$, then by (4.5) we have

$$ \frac{z(\tau ^{-1}(\sigma (t)))}{z^{\prime }(t)}\geq \frac{\int _{t_{2}}^{\tau ^{-1}(\sigma (t))}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s}. $$

Substituting into (4.11), we get

$$\begin{aligned} \omega ^{\prime }(t) \leq{} & r^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{ \lambda } \\ &{}-k\rho (t)q(t) \bigl( p_{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\tau ^{-1}(\sigma (t))}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{ \gamma }. \end{aligned}$$

(4.12)

Secondly, assume that $\sigma (t)\geq \tau (t)$. Hence since $z^{\prime }(t)>0$ and $\tau ^{-1}(\sigma (t))\geq t$, we have $z(\tau ^{-1}(\sigma (t)))\geq z(t)$. Thus it follows from (4.11) and (4.5) that

$$\begin{aligned} \omega ^{\prime }(t) \leq& r^{g}(t) \biggl( \frac{\rho _{+}^{\prime }(t)}{\lambda +1} \biggr) ^{\lambda +1} \biggl( \frac{1}{m\rho (t)} \biggr) ^{ \lambda } \\ &{}-k\rho (t)q(t) \bigl( p_{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{ \gamma } \biggl( \frac{\int _{t_{2}}^{t}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{1}}^{t}\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s} \biggr) ^{\gamma }. \end{aligned}$$

(4.13)

Combining (4.12) and (4.13) and then integrating from $t_{3}$ ($>t_{2}$) to t, we get

$$\begin{aligned} \omega (t_{3}) \geq{}& \int _{t_{3}}^{t} \biggl[k\rho (s)q(s) \bigl( p_{ \ast } \bigl(\sigma (s) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\lambda _{2}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{ \gamma } \\ &{}-r^{g}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\lambda +1} \biggr) ^{ \lambda +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\lambda } \biggr]\,\mathrm {d}s, \end{aligned}$$

which contradicts (4.2). This completes the proof. □

Theorem 4.2

Assume that (H₁)–(H₃) hold, $p(t)\geq 1$, $p(t)\not \equiv 1$for sufficiently larget, $\tau (t)< t$, $\tau ^{\prime }(t)>0$, and$p^{\ast }(t)>0$. If$x(t)$is an eventually positive solution of Eq. (1.1) satisfying$z(t)\in N_{\mathit{II}}$with

$$ \int _{t_{0}}^{\infty }q(s) \bigl( p^{\ast } \bigl(\sigma (s) \bigr) \bigr) ^{ \gamma }\,\mathrm {d}s=\infty $$

(4.14)

or

$$ \int _{t_{0}}^{\infty } \int _{t}^{\infty } \biggl[ \frac{1}{r(s)}\int _{s}^{\infty }q(u) \bigl( p^{\ast } \bigl(\sigma (u) \bigr) \bigr) ^{ \gamma }\,\mathrm {d}u \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}s \,\mathrm {d}t=\infty , $$

(4.15)

then$\lim_{t\rightarrow \infty }x(t)=0$.

Proof

Let $x(t)$ be an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{\mathit{II}}$ for $t\geq t_{1}$. Going through as in the proof of Theorem 4.1, we arrive at (4.4). Since $z(t)$ is decreasing and $\tau (t)< t$, then $z(\tau ^{-1}(t))\geq z(\tau ^{-1}(\tau ^{-1}(t)))$. Substituting into (4.4), we get

$$ x(t)\geq p^{\ast }(t)z \bigl(\tau ^{-1}(t) \bigr). $$

(4.16)

This with (1.1) leads to

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kq(t) \bigl( p^{\ast } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma }z^{ \gamma } \bigl(\tau ^{-1} \bigl( \sigma (t) \bigr) \bigr). $$

(4.17)

Since $z(t)>0$ and $z^{\prime }(t)<0$, then $\lim_{t\rightarrow \infty }z(t)=l\geq 0$ exists. We claim that $l=0$. If not, then there exists $t_{2}\geq t_{1}$ such that $\tau ^{-1}(\sigma (t))>t_{1}$ and $z(\tau ^{-1}(\sigma (t)))\geq l$ for $t\geq t_{2}$. Substituting into (4.17), we get

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kl^{\gamma }q(t) \bigl( p^{\ast } \bigl( \sigma (t) \bigr) \bigr) ^{ \gamma }. $$

(4.18)

Integrating from $t_{2}$ to t and taking into account (4.14), we have

$$ r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha }\leq r(t_{2}) \bigl( z^{\prime \prime }(t_{2}) \bigr) ^{\alpha } -kl^{\gamma } \int _{t_{2}}^{t}q(s) \bigl( p^{\ast } \bigl(\sigma (s) \bigr) \bigr) ^{\gamma } \,\mathrm {d}s\rightarrow -\infty \quad \text{as } t\rightarrow \infty , $$

which is a contradiction. Thus $l=0$ and $\lim_{t\rightarrow \infty }x(t)=0$. In the following, we obtain the same conclusion in the case when $\int _{t_{0}}^{\infty }q(s) ( p^{\ast }(\sigma (s)) ) ^{ \gamma }\,\mathrm {d}s<\infty$. Integrating (4.18) from t to ∞ and dividing both sides by $r(t)$, we have

$$ z^{\prime \prime }(t)\geq \bigl( kl^{\gamma } \bigr) ^{ \frac{1}{\alpha }} \biggl[ \frac{1}{r(t)} \int _{t}^{\infty }q(s) \bigl( p^{\ast } \bigl(\sigma (s) \bigr) \bigr) ^{\gamma }\,\mathrm {d}s \biggr] ^{\frac{1}{\alpha }},\quad t \geq t_{2}. $$

Integrating again from t to ∞, we obtain

$$ -z^{\prime }(t)\geq \bigl( kl^{\gamma } \bigr) ^{ \frac{1}{\alpha }} \int _{t}^{\infty } \biggl[ \frac{1}{r(s)} \int _{s}^{\infty }q(u) \bigl( p^{\ast } \bigl(\sigma (u) \bigr) \bigr) ^{\gamma }\,\mathrm {d}u \biggr] ^{ \frac{1}{\alpha }}\,\mathrm {d}s,\quad t\geq t_{3}\geq t_{2}. $$

Moreover, by integrating again from $t_{3}$ to ∞, we get

$$ z(t_{3})\geq \bigl( kl^{\gamma } \bigr) ^{\frac{1}{\alpha }} \int _{t_{3}}^{\infty } \int _{t}^{\infty } \biggl[ \frac{1}{r(s)}\int _{s}^{\infty }q(u) \bigl( p^{\ast } \bigl(\sigma (u) \bigr) \bigr) ^{ \gamma }\,\mathrm {d}u \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}s \,\mathrm {d}t, $$

which contradicts (4.15). Hence, $l=0$. So from the fact that $0< x(t)< z(t)$, it follows that $\lim_{t\rightarrow \infty }x(t)=0$. □

Theorem 4.3

Assume that (H₁)–(H₃) hold, $p(t)\geq 1$, $p(t)\not \equiv 1$for sufficiently larget, $\tau (t)< t$and$\tau ^{\prime }(t)>0$. If for some constant$k_{1}\in (0,1)$there exists a function$m_{**}(t)\in C^{1} ( [t_{0},\infty ),(0,\infty ) )$such that

$$ m_{**}(t)-k_{1}tm_{**}^{\prime }(t) \leq 0, $$

(4.19)

$p_{\ast \ast }(t)>0$for all sufficiently largetand

$$ \int _{t_{0}}^{\infty } \biggl[ \frac{1}{r(t)} \int _{t_{0}}^{t}q(s) \biggl[ \tau ^{-1} \bigl(\sigma (s) \bigr)p_{\ast \ast } \bigl(\sigma (s) \bigr) \int _{\tau ^{-1}( \sigma (s))}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(u)} \,\mathrm {d}u \biggr] ^{\gamma }\,\mathrm {d}s \biggr] ^{ \frac{1}{\alpha }}\,\mathrm {d}t=\infty , $$

(4.20)

then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{\mathit{III}}$.

Proof

Let $x(t)$ be an eventually positive solution of Eq. (1.1) such that $x(t)>0$, $x(\tau (t))>0$, and $x(\sigma (t))>0$, $z(t)$ satisfies $z(t)\in N_{\mathit{III}}$ and $\tau ^{-1}(\sigma (t))>t_{0}$ for $t\geq t_{1}\geq t_{0}$. From the definition of z, we have (4.4) as in the proof of Theorem 4.1. Since $z^{\prime \prime }(t)<0$ and $z^{\prime }(t)>0$, then by Lemma 2.5 there exists $t_{2}\geq t_{1}$ such that

$$ z(t)\geq k_{1}tz^{\prime }(t),\quad t\geq t_{2}. $$

(4.21)

This with (4.19) yields

$$\begin{aligned} \biggl( \frac{z(t)}{m_{**}(t)} \biggr) ^{\prime } &= \frac{1}{m_{**}^{2}(t)} \bigl[ m_{**}(t)z^{\prime }(t)-z(t)m_{**}^{ \prime }(t) \bigr] \\ &\leq \frac{z(t)}{k_{1}tm_{**}^{2}(t)} \bigl[ m_{**}(t)-k_{1}tm_{**}^{ \prime }(t) \bigr] \leq 0, \end{aligned}$$

and so $\frac{z(t)}{m_{**}(t)}$ is nonincreasing. Hence $z(\tau ^{-1}(\tau ^{-1}(t)))\leq \frac{m_{**}(\tau ^{-1}(\tau ^{-1}(t)))z(\tau ^{-1}(t))}{m_{**}(\tau ^{-1}(t))}$. Now, from (1.1), (4.4), and (4.21), we have

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kk_{1}^{\gamma }q(t) \bigl( \tau ^{-1} \bigl(\sigma (t) \bigr) \bigr) ^{\gamma } \bigl( p_{\ast \ast } \bigl(\sigma (t) \bigr) \bigr) ^{ \gamma } \bigl( z^{\prime } \bigl(\tau ^{-1} \bigl(\sigma (t) \bigr) \bigr) \bigr) ^{\gamma }. $$

(4.22)

But since $-r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t)$ is positive and increasing, then we have $-r^{\frac{1}{\alpha }}(t)z^{\prime \prime }(t)\geq g_{1}$ for $t\geq t_{1}$. Hence

$$ z^{\prime }(t)\geq \int _{t}^{\infty } \frac{-r^{\frac{1}{\alpha }}(s)z^{\prime \prime }(s)}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s\geq g_{1} \int _{t}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s. $$

Thus

$$ z^{\prime } \bigl(\tau ^{-1} \bigl(\sigma (t) \bigr) \bigr) \geq g_{1} \int _{\tau ^{-1}( \sigma (t))}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s. $$

(4.23)

This with (4.22) leads to

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kk_{1}^{\gamma }g_{1}^{\gamma }q(t) \bigl( \tau ^{-1} \bigl( \sigma (t) \bigr) \bigr) ^{\gamma } \bigl( p_{\ast \ast } \bigl(\sigma (t) \bigr) \bigr) ^{\gamma } \biggl( \int _{\tau ^{-1}(\sigma (t))}^{\infty } \frac{1}{r^{\frac{1}{\alpha }}(s)}\,\mathrm {d}s \biggr) ^{\gamma }. $$

Integrating from $t_{2}$ to t, we get

$$ -z^{\prime \prime }(t)\geq \bigl( kk_{1}^{\gamma }g_{1}^{\gamma } \bigr) ^{\frac{1}{\alpha }} \biggl[ \frac{1}{r(t)} \int _{t_{2}}^{t}q(s) \biggl[ \tau ^{-1} \bigl(\sigma (s) \bigr)p_{ \ast \ast } \bigl(\sigma (s) \bigr) \int _{\tau ^{-1}(\sigma (s))}^{\infty } \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr] ^{\gamma }\,\mathrm {d}s \biggr] ^{ \frac{1}{\alpha }}. $$

Integrating again from $t_{3}$ ($\geq t_{2}$) to t, we have

$$ \frac{z^{\prime }(t_{3})}{ ( kk_{1}^{\gamma }g_{1}^{\gamma } ) ^{\frac{1}{\alpha }}} \geq \int _{t_{3}}^{t} \biggl[ \frac{1}{r(s)} \int _{t_{2}}^{s}q(u) \biggl[ \tau ^{-1} \bigl(\sigma (u) \bigr)p_{\ast \ast } \bigl(\sigma (u) \bigr) \int _{\tau ^{-1}( \sigma (u))}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(v)} \,\mathrm {d}v \biggr] ^{\gamma }\,\mathrm {d}u \biggr] ^{ \frac{1}{\alpha }}\,\mathrm {d}s. $$

This contradicts (4.20) and completes the proof. □

Theorem 4.4

Assume that (1.8) and all the conditions of Theorem4.1, Theorem4.2, and Theorem4.3hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

Theorem 4.5

Assume that (1.5) and all the conditions of Theorem4.1and Theorem4.2hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

Remark 1

The assumptions concerning the existence of the two functions $m_{\ast }(t)$ and $m_{\ast \ast }(t)$ hold, for example, $\mu (t)=\xi (t)$, $\mu (t)= ( \xi (t) ) ^{\eta }$, $\mu (t)=\xi (t)e^{\xi ^{\epsilon }(t)}$, $\mu (t)= ( \xi (t) ) ^{\eta }e^{\epsilon \xi (t)}$ with

$$ \xi (t)= \textstyle\begin{cases} \int _{t_{2}}^{t}\int _{t_{1}}^{s} \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s, & \mu (t)=m_{\ast } (t), \\ t^{\frac{1}{k_{1}}}, & \mu (t)= m_{\ast \ast } (t), \end{cases}$$

$\eta \geq 1$ and $\epsilon \geq 0$, etc.

Remark 2

From Theorem 4.4 and Theorem 4.5, we can obtain more than one oscillation criterion for Eq. (1.1) in the two theorems with different choices of $m_{\ast }(t)$ and $m_{\ast \ast }(t)$ which are mentioned in Remark 1.

In the following, we discuss the oscillatory behavior of solutions of Eq. (1.1) in the case when $0\leq p(t)\leq p_{0}<1$.

Theorem 4.6

Assume that (H₁)–(H₃) hold and$0\leq p(t)\leq p_{0}<1$. If there exists a positive function$\rho (t)\in C^{1}([t_{0},\infty ))$such that

$$\begin{aligned} &\int _{t_{\ast }}^{\infty } \biggl[k\rho (s)q(s) \bigl( 1-p \bigl(\sigma (s) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\lambda _{1}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma } \\ &\quad {}-r^{g}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\lambda +1} \biggr) ^{ \lambda +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\lambda } \biggr]\,\mathrm {d}s= \infty \end{aligned}$$

(4.24)

holds for some constant$k>0$, for sufficiently large$t_{1}\geq t_{0}$, and for some$t_{\ast }>t_{2}>t_{1}$, whereλ, m, g, $\lambda _{1}(t)$are as defined by (3.2) and (3.3), then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{I}$.

Proof

Let $x(t)$ be an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{I}$. From the definition of z, we have

$$ x(t) =z(t)-p(t)x \bigl(\tau (t) \bigr) \geq \bigl(1-p(t) \bigr)z(t). $$

This with (1.1) yields

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kq(t) \bigl( 1-p \bigl(\sigma (t) \bigr) \bigr) ^{\gamma }z^{ \gamma } \bigl(\sigma (t) \bigr). $$

(4.25)

Defining $\omega (t)$ by (3.7), completing the proof as in the proof of Theorem 4.1 by applying (4.25) instead of (4.8), and considering the two cases $\sigma (t)< t$ and $\sigma (t)\geq t$ instead of the two cases $\sigma (t)<\tau (t)$ and $\sigma (t)\geq \tau (t)$, we get a contradiction to (4.24). □

Theorem 4.7

Assume that (H₁)–(H₃) hold, $0\leq p(t)\leq p_{0}<1$, and$x(t)$is an eventually positive solution of Eq. (1.1) satisfying$z(t)\in N_{\mathit{II}}$. If

$$ \int _{t_{0}}^{\infty }q(s)\,\mathrm {d}s=\infty $$

(4.26)

or

$$ \int _{t_{0}}^{\infty } \int _{t}^{\infty } \biggl[ \frac{1}{r(s)}\int _{s}^{\infty }q(u)\,\mathrm {d}u \biggr] ^{\frac{1}{\alpha }} \,\mathrm {d}s \,\mathrm {d}t=\infty , $$

(4.27)

then$\lim_{t\rightarrow \infty }x(t)=0$.

Proof

Let $x(t)$ be an eventually positive solution of Eq. (1.1) satisfying $z(t)\in N_{\mathit{II}}$ for $t\geq t_{1}\geq t_{0}$. Since $z(t)$ is positive and decreasing, we have $\lim_{t\rightarrow \infty }z(t)=l\geq 0$ exists. We claim that $l=0$. If not, then for any $\epsilon >0$ we have $l< z(t)< l+\epsilon$ eventually. Choose $0<\epsilon <\frac{l(1-p_{0})}{p_{0}}$. It is easy to verify that

$$ x(t) =z(t)-p(t)x \bigl(\tau (t) \bigr) \geq z(t)-p(t)z \bigl(\tau (t) \bigr) >l-p_{0}(l+ \epsilon )=k_{2}(l+\epsilon )>k_{2}z(t), $$

where $k_{2}=\frac{l-p_{0}(l+\epsilon )}{(l+\epsilon )}>0$. Now, it follows from (1.1) that

$$ \bigl( r(t) \bigl( z^{\prime \prime }(t) \bigr) ^{\alpha } \bigr) ^{ \prime }\leq -kk_{2}^{\gamma }q(t)z^{\gamma } \bigl(\sigma (t) \bigr)\leq -k(k_{2}l)^{ \gamma }q(t). $$

(4.28)

Going through as in the proof of Theorem 4.2 by applying (4.28) instead of (4.18), we can get a contradiction to (4.26) or (4.27). This completes the proof. □

Using a similar technique to the proof of Theorem 4.3 and using (4.25) with (4.21) instead of (4.22), we can get the following result.

Theorem 4.8

Assume that (H₁)–(H₃) hold and$0\leq p(t)\leq p_{0}<1$. If

$$ \int _{t_{0}}^{\infty } \biggl[ \frac{1}{r(t)} \int _{t_{0}}^{t}q(s) \biggl[ \sigma (s) \bigl(1-p \bigl(\sigma (s) \bigr) \bigr) \int _{\sigma (s)}^{\infty } \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr] ^{\gamma }\,\mathrm {d}s \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}t= \infty , $$

(4.29)

then there exists no positive solution$x(t)$of Eq. (1.1) satisfying$z(t)\in N_{\mathit{III}}$.

Theorem 4.9

Assume that (1.8) and all the conditions of Theorem4.6, Theorem4.7, and Theorem4.8hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

Theorem 4.10

Assume that (1.5) and all the conditions of Theorem4.6and Theorem4.7hold. Then every solution$x(t)$of Eq. (1.1) is either oscillatory or satisfies$\lim_{t\rightarrow \infty }x(t)=0$.

5 Examples

Example 1

Consider the third order differential equation

$$ \biggl( \biggl[ \biggl( x(t)+\frac{25}{4}x \biggl(\frac{t}{2} \biggr) \biggr) ^{\prime \prime } \biggr] ^{\frac{1}{3}} \biggr) ^{\prime }+\frac{3}{t}x(t)=0,\quad t\geq 1. $$

(5.1)

Here, $r(t)=1$, $p=\frac{25}{4}$, $\tau (t)=\frac{t}{2}$, $q(t)=\frac{3}{t}$, $\sigma (t)=t$, and $1=\gamma >\alpha =\frac{1}{3}$. It is clear that $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t= \infty$. Choosing $\rho (t)=\frac{1}{t}$, then we have $\rho _{+}^{\prime }(t)=0$, and

$$\begin{aligned} & \int _{t_{\ast }}^{\infty } \biggl[k\rho (s)Q(s) \biggl( \frac{\int _{t_{2}}^{s}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma }- \biggl( 1+ \frac{p_{0}^{\gamma }}{\tau _{0}^{\alpha +1}} \biggr) R(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\alpha } \biggr]\,\mathrm {d}s \\ &\quad = \int _{t_{\ast }}^{\infty }\frac{3}{s^{2}} \biggl( \frac{\int _{t_{2}}^{s}\int _{t_{1}}^{u}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\,\mathrm {d}u} \biggr) \,\mathrm {d}s \geq \int _{t_{\ast }}^{\infty } \biggl( \frac{3}{2s}- \frac{3t_{1}}{s^{2}}-\frac{3t_{2}^{2}}{2s^{3}}+\frac{3t_{1}t_{2}}{s^{3}} \biggr) \,\mathrm {d}s= \infty . \end{aligned}$$

Thus, it follows from Theorem 3.4 that every solution $x(t)$ of Eq. (5.1) is either oscillatory or satisfies $\lim_{t\rightarrow \infty }x(t)=0$. In fact, $x(t)=\frac{1}{t}$ is a solution of Eq. (5.1).

Example 2

Consider the third order differential equation

$$ \biggl( \biggl[ \biggl( x(t)+p_{0}x \biggl(t-\frac{1}{2} \biggr) \biggr) ^{\prime \prime } \biggr] ^{5} \biggr) ^{\prime }+ \biggl( t-\frac{1}{2} \biggr) ^{\frac{4}{3}}x^{\frac{1}{3}} \biggl(t-\frac{1}{2} \biggr)=0,\quad t\geq 1, p_{0}>0. $$

(5.2)

Here, $r(t)=1$, $p=p_{0}$, $\tau (t)=t-\frac{1}{2}$, $q(t)= ( t-\frac{1}{2} ) ^{\frac{4}{3}}$, $\sigma (t)=t-\frac{1}{2}$, and $\frac{1}{3}=\gamma <\alpha =5$. It is clear that $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t=\infty$. Choosing $\rho (t)=1$, we have $\rho _{+}^{\prime }(t)=0$, and

$$\begin{aligned} & \int _{t_{\ast }}^{\infty } \biggl[k\rho (s)Q(s) \biggl( \frac{\int _{t_{2}}^{\sigma (s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{ \gamma } \\ &\qquad {}- \biggl( 1+\frac{p_{0}^{\gamma }}{\tau _{0}^{\gamma +1}} \biggr) R^{ \frac{\gamma }{\alpha }}(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\gamma +1} \biggr) ^{\gamma +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\gamma } \biggr] \,\mathrm {d}s \\ &\quad = \int _{t_{\ast }}^{\infty } \biggl( \frac{1}{2} \biggr) ^{\frac{1}{3}} ( s-1 ) ^{\frac{4}{3}} \biggl( \frac{ ( s-t_{1}-\frac{1}{2} ) ^{2}-(t_{2}-t_{1})^{2}}{s-t_{1}} \biggr) ^{\frac{1}{3}}\,\mathrm {d}s \\ &\quad \geq \int _{t_{\ast }}^{\infty } \biggl( \frac{1}{2} \biggr) ^{ \frac{1}{3}} ( s-1 ) \biggl( \biggl( s-t_{1}- \frac{1}{2} \biggr) ^{2}-(t_{2}-t_{1})^{2} \biggr) ^{\frac{1}{3}}\,\mathrm {d}s \\ &\quad > \int _{t_{\ast }}^{\infty } \biggl( \frac{1}{2} \biggr) ^{\frac{1}{3}} \biggl( s- \biggl(t_{1}+ \frac{1}{2} \biggr) \biggr) \biggl( \biggl( s-t_{1}- \frac{1}{2} \biggr) ^{2}-(t_{2}-t_{1})^{2} \biggr) ^{\frac{1}{3}}\,\mathrm {d}s=\infty . \end{aligned}$$

Thus, by Theorem 3.4, it follows that every solution $x(t)$ of Eq. (5.2) is either oscillatory or satisfies $\lim_{t\rightarrow \infty }x(t)=0$.

Example 3

Consider the third order differential equation

$$ \biggl( t \biggl[ \biggl( x(t)+\frac{1}{3\sqrt{3}}x \biggl( \frac{t}{3} \biggr) \biggr) ^{ \prime \prime } \biggr] ^{\frac{1}{3}} \biggr) ^{\prime }+\lambda t^{6}x^{3} \biggl( \frac{t}{2} \biggr)=0,\quad t>1, \lambda >0. $$

(5.3)

Here, $r(t)=t$, $p=\frac{1}{3\sqrt{3}}$, $\tau (t)=\frac{t}{3}$, $q(t)=\lambda t^{6}$, $\sigma (t)=\frac{t}{2}$, and $3=\gamma >\alpha =\frac{1}{3}$. It is clear that $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(t)}\,\mathrm {d}t=\int _{1}^{\infty }\frac{1}{t^{3}}\,\mathrm {d}t=\frac{1}{2}< \infty$. Choosing $\rho (t)=\frac{1}{t^{10}}$, we have $\rho _{+}^{\prime }(t)=0$, and

$$\begin{aligned} & \int _{t_{\ast }}^{\infty } \biggl[\frac{k}{2^{\gamma -1}}\rho (s)Q(s) \biggl( \frac{\int _{t_{2}}^{\sigma (s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{\gamma } \\ &\qquad {}- \biggl( 1+ \frac{p_{0}^{\gamma }}{\tau _{0}^{\alpha +1}} \biggr) R(s) \biggl( \frac{\rho _{+}^{\prime }(s)}{\alpha +1} \biggr) ^{\alpha +1} \biggl( \frac{1}{m\rho (s)} \biggr) ^{\alpha } \biggr] \,\mathrm {d}s \\ &\quad =\frac{\lambda t_{1}^{6}}{4(3)^{6}} \int _{t_{\ast }}^{\infty } \frac{s^{6}}{s} \biggl( \frac{\frac{2}{s^{2}}+\frac{1}{2t_{1}^{2}}-\frac{\frac{1}{t_{2}}+\frac{t_{2}}{t_{1}^{2}}}{s}}{s^{2}-t_{1}^{2}} \biggr) ^{3}\,\mathrm {d}s \\ &\quad >\frac{\lambda t_{1}^{6}}{4(3)^{6}} \int _{t_{\ast }}^{\infty } \frac{ ( s^{2}-t_{1}^{2} ) ^{3}}{s} \biggl( \frac{\frac{2}{s^{2}}+\frac{1}{2t_{1}^{2}}-\frac{\frac{1}{t_{2}}+\frac{t_{2}}{t_{1}^{2}}}{s}}{s^{2}-t_{1}^{2}} \biggr) ^{3}\,\mathrm {d}s=\infty , \end{aligned}$$

and

$$\begin{aligned} & \int _{t_{0}}^{\infty } \biggl[ \frac{1}{r(t)} \int _{t_{0}}^{t}Q(s) \bigl( \sigma (s) \bigr) ^{\gamma } \biggl( \int _{\sigma (s)}^{ \infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr) ^{\gamma }\,\mathrm {d}s \biggr] ^{ \frac{1}{\alpha }}\,\mathrm {d}t \\ &\quad =\frac{\lambda ^{3}}{(3)^{18}(4)^{3}} \int _{1}^{\infty } \frac{(t^{4}-1)^{3}}{t^{3}}\,\mathrm {d}t = \frac{\lambda ^{3}}{(3)^{18}(4)^{3}} \int _{1}^{\infty } \frac{(t-1)^{3}(t+1)^{3}(t^{2}+1)^{3}}{t^{3}}\,\mathrm {d}t \\ &\quad >\frac{\lambda ^{3}}{(3)^{18}(4)^{3}} \int _{1}^{\infty }(t-1)^{3} \bigl(t^{2}+1 \bigr)^{3} \,\mathrm {d}t >\frac{\lambda ^{3}}{(3)^{18}(4)^{3}} \int _{1}^{\infty }(t-1)^{3} \,\mathrm {d}t= \infty . \end{aligned}$$

Thus, by Theorem 3.3, it follows that every solution $x(t)$ of Eq. (5.3) is either oscillatory or satisfies $\lim_{t\rightarrow \infty }x(t)=0$. We may note that, for $\lambda =\frac{\sqrt[3]{35}}{\sqrt{2^{17}}}$, we have $x(t)=\frac{1}{t^{\frac{5}{2}}}$ is a solution of Eq. (5.3).

Example 4

Consider the third order neutral delay differential equation

$$ \biggl( t^{3} \biggl( x(t)+t^{\frac{5}{3}} \frac{5t+6}{t+1}x \biggl(\frac{t}{2} \biggr) \biggr) ^{\prime \prime } \biggr) ^{\prime }+t^{9}x^{3}(t-1)=0,\quad t \geq t_{0}=2. $$

(5.4)

Here, $r(t)=t^{3}$, $p(t)=t^{\frac{5}{3}}\frac{5t+6}{t+1}$, $q(t)=t^{9}$, $\tau (t)=\frac{t}{2}$, $\sigma (t)=t-1$, $f(u)=u^{3}$, $\alpha =1$, and $\gamma =3$. It is clear that $p(t)=t^{\frac{5}{3}} [ 5+\frac{1}{t+1} ] \geq (5)(2^{\frac{5}{3}})\simeq 15.874>1$, $\tau \circ \sigma \neq \sigma \circ \tau$, $\sigma (t)\geq \tau (t)$, conditions (H₁)–(H₃), and (1.8) hold, and

$$ p \bigl( \tau ^{-1} \bigl( \tau ^{-1} ( t ) \bigr) \bigr) =\frac{20t+6}{4t+1}(4t)^{\frac{5}{3}}= \biggl[ 5+ \frac{1}{4t+1} \biggr] (4t)^{ \frac{5}{3}} >(5) (4t)^{\frac{5}{3}}>(5) (8)^{\frac{5}{3}}\simeq 160. $$

(5.5)

Let $m_{\ast }(t)=\int _{t_{2}}^{t}\int _{t_{1}}^{s} \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s$ and $m_{\ast \ast }(t)=t^{\frac{1}{k_{1}}}$. Thus

$$\begin{aligned} p_{\ast }(t) &=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \frac{\int _{t_{2}}^{\tau ^{-1}(\tau ^{-1}(t))}\int _{t_{1}}^{s} \frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s}{\int _{t_{2}}^{\tau ^{-1}(t)}\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \,\mathrm {d}s} \biggr) \\ &\geq \frac{1}{p(2t)} \biggl( 1-\frac{1}{160} \frac{12t^{2}-13t+3}{6t^{2}-13t+6} \biggr) =\frac{1}{p(2t)} \biggl( 1- \frac{1}{160} \bigl(2+\phi (t) \bigr) \biggr), \end{aligned}$$

(5.6)

where $\phi (t)=\frac{13t-9}{6t^{2}-13t+6}$. Since $\phi ^{\prime }(t)=\frac{-78t^{2}+108t-39}{(6t^{2}-13t+6)^{2}}$, which is negative for $t\geq t_{2}=3>t_{1}=2$. Thus $\phi (t)$ is positive and decreasing for $t\geq t_{2}=3$. It follows that $\phi (t)\leq \frac{10}{7}$. Thus by (5.6) we have

$$ p_{\ast }(t)\geq \frac{1}{p(2t)} \biggl( 1-\frac{1}{160} \frac{24}{7} \biggr)=\frac{137}{(140)(2t)^{\frac{5}{3}}} \frac{2t+1}{10t+6} >0 \quad \text{for }t\geq t_{2}=3. $$

By choosing $\rho (t)=\frac{1}{t^{8}}$, condition (4.2) becomes

$$\begin{aligned} &\int _{t_{\ast }}^{\infty } \biggl[k\rho (s)q(s) \bigl( p_{\ast } \bigl( \sigma (s) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\lambda _{2}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{ \gamma } \biggr]\,\mathrm {d}s \\ &\quad \geq \int _{t_{\ast }}^{\infty } \biggl[\frac{1}{s^{8}}s^{9} \biggl( \frac{137}{(140)(2^{\frac{5}{3}})}\frac{2s-1}{10s-4} \frac{1}{ ( s-1 ) ^{\frac{5}{3}}} \biggr) ^{3} \biggl( \frac{\int _{t_{2}}^{s}\int _{t_{1}}^{u}\frac{1}{v^{3}}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{u^{3}}\,\mathrm {d}u} \biggr) ^{3} \biggr] \,\mathrm {d}s \\ &\quad \geq \int _{t_{\ast }}^{\infty } \biggl[s \biggl( \frac{137}{(140)(2^{\frac{5}{3}})}\zeta (s)\frac{1}{s^{\frac{5}{3}}} \biggr) ^{3} \biggl( \frac{\int _{t_{2}}^{s} ( \frac{-1}{2u^{2}}+\frac{1}{2t_{1}^{2}} ) \,\mathrm {d}u}{\frac{-1}{2s^{2}}+\frac{1}{2t_{1}^{2}}} \biggr) ^{3} \biggr]\,\mathrm {d}s, \end{aligned}$$

(5.7)

where $\zeta (s)=\frac{2s-1}{10s-4}$. Then $\zeta ^{\prime }(s)=\frac{2}{(10s-4)^{2}}>0$, i.e., $\zeta (s)$ is positive and increasing and $\zeta (s)\geq \frac{5}{26}$ for $s\geq t_{2}=3$. Now from (5.7) we have

$$\begin{aligned} & \int _{t_{\ast }}^{\infty } \biggl[k\rho (s)q(s) \bigl( p_{\ast } \bigl( \sigma (s) \bigr) \bigr) ^{\gamma } \biggl( \frac{\int _{t_{2}}^{\lambda _{2}(s)}\int _{t_{1}}^{u}\frac{1}{r^{\frac{1}{\alpha }}(v)}\,\mathrm {d}v \,\mathrm {d}u}{\int _{t_{1}}^{s}\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u} \biggr) ^{ \gamma } \biggr]\,\mathrm {d}s \\ &\quad \geq \int _{t_{\ast }}^{\infty } \biggl[s \biggl( \frac{137}{(140)(2^{\frac{5}{3}})}\frac{5}{26}\frac{1}{s^{\frac{5}{3}}} \biggr) ^{3} \biggl( \frac{\frac{1}{2s}+\frac{s}{2t_{1}^{2}}-\frac{1}{2t_{2}}-\frac{t_{2}}{2t_{1}^{2}}}{\frac{-1}{2s^{2}}+\frac{1}{2t_{1}^{2}}} \biggr) ^{3} \biggr]\,\mathrm {d}s \\ &\quad \geq \int _{t_{\ast }}^{\infty } \biggl[ \bigl(2.082656208\times 10^{-4} \bigr) \frac{1}{s^{4}} \biggl( \frac{t_{1}^{2}s+s^{3}-\frac{t_{1}^{2}}{t_{2}}s^{2}-t_{2}s^{2}}{s^{2}} \biggr) ^{3} \biggr]\,\mathrm {d}s=\infty . \end{aligned}$$

But since by (5.5) we have

$$ p^{\ast }(t)=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \biggr) \geq \frac{159}{ ( 160 ) ( 2^{\frac{5}{3}} ) }\frac{2t+1}{10t+6}\frac{1}{t^{\frac{5}{3}}}>0, $$

then it follows that condition (4.14) reads

$$ \int _{t_{0}}^{\infty }q(u) \bigl( p^{\ast } \bigl(\sigma (u) \bigr) \bigr) ^{ \gamma }\,\mathrm {d}u \geq \int _{t_{0}}^{\infty }u^{9} \biggl( \frac{159}{ ( 160 ) ( 2^{\frac{5}{3}} ) }\frac{3}{16}\frac{1}{u^{\frac{5}{3}}} \biggr) ^{3}\,\mathrm {d}u \simeq \epsilon _{1} \int _{t_{0}}^{\infty }u^{4}\,\mathrm {d}u=\infty , $$

where $\epsilon _{1}=(0.05868968172)^{3}$. Moreover, since by using (5.5) and letting $k_{1}=\frac{1}{2}$ we have

$$\begin{aligned} p_{\ast \ast }(t) &=\frac{1}{p(\tau ^{-1}(t))} \biggl( 1- \frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \biggl[ \frac{\tau ^{-1}(\tau ^{-1}(t))}{\tau ^{-1}(t)} \biggr]^{\frac{1}{k_{1}}} \biggr) \\ &\geq \frac{1}{p(2t)} \biggl( 1-\frac{1}{160} \biggl( \frac{4t}{2t} \biggr)^{2} \biggr) =\frac{39}{(40)(2^{\frac{5}{3}})} \frac{2t+1}{10t+6}\frac{1}{t^{\frac{5}{3}}}>0, \end{aligned}$$

then condition (4.20) becomes

$$\begin{aligned} & \int _{t_{0}}^{\infty } \biggl[ \frac{1}{r(t)} \int _{t_{0}}^{t}q(s) \biggl[ \tau ^{-1} \bigl(\sigma (s) \bigr)p_{\ast \ast } \bigl(\sigma (s) \bigr) \int _{\tau ^{-1}( \sigma (s))}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(u)}\,\mathrm {d}u \biggr] ^{ \gamma }\,\mathrm {d}s \biggr] ^{\frac{1}{\alpha }}\,\mathrm {d}t \\ &\quad \geq \int _{2}^{\infty }\frac{1}{t^{3}} \int _{2}^{t}s^{9} \biggl[ \frac{1}{4(s-1)} \biggl( \frac{39}{(40)(2^{\frac{5}{3}})}\frac{2s-1}{10s-4} \frac{1}{ ( s-1 ) ^{\frac{5}{3}}} \biggr) \biggr] ^{3}\,\mathrm {d}s \,\mathrm {d}t \\ &\quad \geq \int _{2}^{\infty }\frac{1}{t^{3}} \int _{2}^{t}s^{9} \biggl[ \frac{1}{4s} \biggl( \frac{39}{(40)(2^{\frac{5}{3}})}\frac{3}{16} \frac{1}{s^{\frac{5}{3}}} \biggr) \biggr] ^{3}\,\mathrm {d}s \,\mathrm {d}t \\ &\quad \simeq \epsilon _{2} \int _{2}^{\infty } \biggl[ \frac{1}{2t}- \frac{2}{t^{3}} \biggr] \,\mathrm {d}t=\infty \text{, where }\epsilon _{2}=(0.01439558231)^{3}. \end{aligned}$$

Thus, all the conditions of Theorem 4.4 are satisfied, and so every solution $x(t)$ of Eq. (5.4) is either oscillatory or satisfies $\lim_{t\rightarrow \infty }x(t)=0$.

6 General remarks

• (1) In this paper, several new oscillation criteria for Eq. (1.1) have been presented which complement and improve the existing results introduced in the cited papers. In fact, our results are applicable in the cases either with $p(t)$ is bounded or unbounded and where the restriction $r^{\prime }(t)\geq 0$ imposed by the authors in [1, 8, 9, 14, 19], and [17] is dropped in this paper.

• (2) It is our belief that the present paper is of significance because it extends most of the cited papers which are concerned with unbounded $p(t)$ and relaxes some of their conditions. For example, Theorem 4.5 includes Theorem 2.6 and Theorem 2.9 of [15], where the author was only concerned with the special case $\alpha =\gamma$ with $\int _{t_{0}}^{\infty }r^{-\frac{1}{\alpha }}(s)\,\mathrm {d}s=\infty$, and with the restriction $\sigma (t)$ is nonincreasing. Moreover, our results in this paper extend those of [5] in the special case $r(t)=1$, $\alpha =1$, and $f(u)=u^{\gamma }$, where $\gamma \leq 1$. At the same time it extends those of [4] in the special case $p(t)=0$, $\alpha =\gamma$, with $\sigma (t)$ being strictly increasing.

• (3) Our criteria could be extended to the dynamic equation on time scales. In this case, if we consider $m_{\ast }(t)=\int _{t_{2}}^{t}\int _{t_{1}}^{s} \frac{1}{r^{\frac{1}{\alpha }}(u)}\Delta u \Delta s$ and $\int _{t_{0}}^{\infty }\frac{1}{r^{\frac{1}{\alpha }}(s)}\Delta s= \infty$, then the obtained results will be more general than those of [10], because one may note that the results of [10] are applicable only in the case $\gamma \leq \alpha$, $0\leq p(t)\leq p_{0}<1$, and $\sigma (t)$ is nondecreasing, while our results are applicable in the case $\gamma >\alpha$ and $p(t)\geq 1$.