Oscillatory behavior of third-order neutral delay differential equations with distributed deviating arguments

Abstract

The main contribution of this paper is to establish some new criteria, which ensure that every solution of third-order neutral delay differential equations with distributed deviating arguments is either oscillatory or tends to zero. The obtained theorems extend and improve several known results in the literature. Two examples are provided to illustrate the main results.

Introduction

Over the last several years, an increasing attention has been given to the oscillation theory and asymptotic behavior of various classes of second-order and high-order differential equations and dynamic equations on time scales [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]. So far, much research activity concerns the oscillation problem of the third-order (TO) neutral differential and dynamic equations [18,19,20,21,22,23,24,25,26]. Recently, the research focus has been shifted to the study of the TO differential equations (DE) with distributed deviating arguments (DDA), and some results can be found in [27,28,29,30,31,32,33,34,35].

Li et al. [3] investigated

$$ \bigl(r(t) \bigl\vert z'(t) \bigr\vert ^{\alpha -1}z'(t) \bigr)'+ \int ^{b}_{a} q(t,\xi ) \bigl\vert x\bigl[g(t, \xi )\bigr] \bigr\vert ^{\alpha -1}x\bigl[g(t,\xi )\bigr]\,d\sigma (\xi )=0, $$

where $z(t)=x(t)+p(t)x(\tau (t))$, $0\leq p(t)\leq p_{0}<\infty$ and $\alpha \geq 1$ is a constant. Under the methods proposed by Li et al. [3, 20], Jiang and Li [30] studied the following equation with DDA:

$$ \bigl(r(t) \bigl\vert z''(t) \bigr\vert ^{\alpha -1}z''(t) \bigr)'+ \int ^{b}_{a} q(t,\xi ) \bigl\vert x\bigl[g(t, \xi )\bigr] \bigr\vert ^{\alpha -1}x\bigl[g(t,\xi )\bigr]\,d\sigma (\xi )=0, $$

(1.1)

where $\alpha >0$ is a constant, and obtained several theorems for (1.1) whenever

$$ \int ^{\infty }_{t_{0}}r^{-\frac{1}{\alpha }}(t)\,dt=\infty , \quad \text{or}\quad \int ^{\infty }_{t_{0}}r^{-\frac{1}{\alpha }}(t)\,dt< \infty . $$

Furthermore, Elabbasy and Moaaz [31], and Wang et al. [32] examined a TODE with DDA under the assumption $0\leq p(t) \leq P<1$. However, the obtained oscillation theorems cannot be applied when $p(t)\geq 1$. Then Tunç [33] utilized a new technique, different from the existing methods, to give some criteria for a TODE with DDA, when $p(t)\geq 1$.

The main objective here is to establish several oscillation criteria for the TO neutral delay (ND) DE with DDA

$$ E'_{2}(t)+ \int ^{b}_{a} q(t,\xi ) \bigl\vert x\bigl(g(t, \xi )\bigr) \bigr\vert ^{\alpha _{3}-1}x\bigl(g(t, \xi )\bigr)\,d\sigma (\xi )=0, $$

(1.2)

where $t\geq t_{0}>0$,

$$\begin{aligned} &E_{2}(t) =r_{2}(t) \bigl\vert E'_{1}(t) \bigr\vert ^{\alpha _{2}-1}E'_{1}(t), \\ &E_{1}(t) =r_{1}(t) \bigl\vert z'(t) \bigr\vert ^{\alpha _{1}-1}z'(t), \\ &z(t) =x(t)+p(t)x\bigl(\tau (t)\bigr), \end{aligned}$$

(1.3)

and $\alpha _{1}$, $\alpha _{2}$, and $\alpha _{3}$ are positive constants. We assume that:

• (A1) $r_{i}(t)\in C([t_{0},\infty ),(0,\infty ))$, $\int ^{ \infty }_{t_{0}}r^{-\frac{1}{\alpha _{i}}}_{i}(t)\,dt=\infty$, $i=1, 2$;

• (A2) $p(t)\in C([t_{0},\infty ),[1,\infty ))$ with $p(t)\not \equiv 1$, and $q(t,\xi )\in C([t_{0},\infty )\times [a,b],[0,\infty ))$ with $q(t,\xi )\not \equiv 0$ eventually;

• (A3) $\tau (t)\in C^{1}([t_{0},\infty ),\mathbb{R})$, $\tau (t)\leq t$, $\tau '(t)>0$ and $\lim_{t\rightarrow \infty } \tau (t)=\infty$;

• (A4) $g(t,\xi )\in C([t_{0},\infty )\times [a,b],\mathbb{R})$ is a nondecreasing function for ξ, which satisfies $\liminf_{t\rightarrow \infty }g(t,\xi )=\infty$ for $\xi \in [a,b]$;

• (A5) $\sigma (\xi )\in C([a,b],\mathbb{R})$ is nondecreasing and the integral of (1.2) is taken in the Riemann–Stieltjies sense.

This article is organized in the following manner. Section 2 presents three lemmas to prove our results. Section 3 establishes some new oscillation criteria for (1.2). Two examples finalize this article.

Some lemmas

For simplicity, we use some notations for sufficiently large $t_{1}$ with $t_{1}\geq t_{0}$ as below:

$$\begin{aligned} &g_{1}(t) =g(t,a),\qquad g_{2}(t)=g(t,b),\qquad d_{+}(t)=\max \bigl\{ 0,d(t)\bigr\} , \\ &\delta _{1}(t,t_{1}) = \int ^{t}_{t_{1}} \biggl(\frac{1}{r_{2}(s)} \biggr)^{\frac{1}{ \alpha _{2}}}\,ds,\qquad \delta _{2}(t,t_{1})= \biggl(\frac{\delta _{1}(t,t_{1})}{r _{1}(t)} \biggr)^{\frac{1}{\alpha _{1}}}, \\ &\delta _{3}(t,t_{1}) = \int ^{t}_{t_{1}}\delta _{2}(s,t_{1})\,ds, \quad t\geq t _{1}. \end{aligned}$$

Furthermore, assume that

$$\begin{aligned} &p_{1}(t) =\frac{1}{p(\tau ^{-1}(t))} \biggl(1-\frac{1}{p(\tau ^{-1}(\tau ^{-1}(t)))} \biggr)>0, \end{aligned}$$

(2.1)

$$\begin{aligned} &p_{2}(t) =\frac{1}{p(\tau ^{-1}(t))} \biggl(1-\frac{\delta _{3}(\tau ^{-1}( \tau ^{-1}(t)),t_{1})}{p(\tau ^{-1}(\tau ^{-1}(t)))\delta _{3}(\tau ^{-1}(t),t _{1})} \biggr)>0, \\ &q_{1}(t) = \int ^{b}_{a} q(t,\xi )p^{\alpha _{3}}_{1} \bigl(g(t,\xi )\bigr)\,d\sigma (\xi ), \\ &q_{2}(t) = \int ^{b}_{a} q(t,\xi )p^{\alpha _{3}}_{2} \bigl(g(t,\xi )\bigr)\,d\sigma (\xi ). \end{aligned}$$

(2.2)

Lemma 2.1

Assume that (A1)–(A5) hold. Furthermore, let $x(t)$ be an eventually positive solution of (1.2). Then $z(t)$ satisfies either

$$ (\mathrm{I})\quad z(t)>0,\qquad z'(t)>0,\qquad E'_{1}(t)>0,\qquad E'_{2}(t) \leq 0, $$

or

$$ (\mathrm{II})\quad z(t)>0,\qquad z'(t)< 0,\qquad E'_{1}(t)>0,\qquad E'_{2}(t) \leq 0. $$

Proof

From the condition of Lemma 2.1, there exists a $t_{1}\geq t_{0}$ such that

$$ x(t)>0,\qquad x\bigl(\tau (t)\bigr)>0\quad \text{and}\quad x \bigl(g(t,\xi )\bigr)>0,\quad \xi \in [a,b], $$

(2.3)

for $t\geq t_{1}$. Then (1.2) implies that

$$ E'_{2}(t)=- \int ^{b}_{a} q(t,\xi ) \bigl\vert x\bigl(g(t, \xi )\bigr) \bigr\vert ^{\alpha _{3}-1}x\bigl(g(t, \xi )\bigr)\,d\sigma (\xi )\leq 0, $$

(2.4)

which means that $E_{2}(t)$ is nonincreasing and of constant sign, and $E'_{1}(t)$ is also of constant sign. We claim that $E'_{1}(t)>0$. To prove this, assume on the contrary that $E'_{1}(t)<0$ and there exists $M_{1}$, s.t. for $t\geq t_{2} \geq t_{1}$,

$$ E_{2}(t)\leq -M_{1}< 0. $$

Then

$$ \bigl\vert E'_{1}(t) \bigr\vert \geq \biggl(\frac{M_{1}}{r_{2}(t)} \biggr)^{ \frac{1}{\alpha _{2}}}>0. $$

(2.5)

We integrate (2.5) to get

$$ E_{1}(t)\leq E_{1}(t_{2})-M^{\frac{1}{\alpha _{2}}}_{1} \int ^{t}_{t_{2}} \biggl(\frac{1}{r_{2}(s)} \biggr)^{\frac{1}{\alpha _{2}}}\,ds. $$

Letting $t\rightarrow \infty$ and from (A1), we obtain $\lim_{t\rightarrow \infty }E_{1}(t)=-\infty$. Then there exist constants $M_{2}$ and $t_{3}\geq t_{2}$ such that

$$ E_{1}(t)\leq -M_{2}< 0,\quad t\geq t_{3}, $$

which yields that

$$ \bigl\vert z'(t) \bigr\vert \geq \biggl(\frac{M_{2}}{r_{1}(t)} \biggr)^{\frac{1}{\alpha _{1}}}>0. $$

Integrating the latter inequality from $t_{3}$ to t, we have

$$ z(t)\leq z(t_{3})-M^{\frac{1}{\alpha _{1}}}_{2} \int ^{t}_{t_{3}} \biggl(\frac{1}{r _{1}(s)} \biggr)^{\frac{1}{\alpha _{1}}}\,ds. $$

We use (A1) again to have $\lim_{t\rightarrow \infty }z(t)=-\infty$, which contradicts $z(t)>0$. Hence, $E'_{1}(t)>0$ for $t\geq t_{1}$, and $z(t)$ has properties (I) and (II). □

Lemma 2.2

Assume that (A1)–(A5) and (2.1) hold. Furthermore, suppose that $x(t)$ is an eventually positive solution of (1.2) and $z(t)$ has property (II) of Lemma 2.1. If

$$ \int ^{\infty }_{t_{0}} \biggl[\frac{1}{r_{1}(u)} \int ^{\infty }_{u} \biggl(\frac{1}{r _{2}(v)} \int ^{\infty }_{v}q_{1}(s) \,ds \biggr)^{\frac{1}{\alpha _{2}}}\,dv \biggr]^{\frac{1}{\alpha _{1}}}\,du=\infty , $$

(2.6)

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

Proof

One can see that (1.3) yields (see [33])

$$ z\bigl(\tau ^{-1}(t)\bigr)=x\bigl(\tau ^{-1}(t)\bigr)+p \bigl(\tau ^{-1}(t)\bigr)x(t), $$

which can be rewritten as

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

(2.7)

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

(2.8)

Combining (2.4) and (2.8), we get

$$\begin{aligned} E'_{2}(t) &\leq - \int ^{b}_{a} q(t,\xi )p^{\alpha _{3}}_{1} \bigl(g(t,\xi )\bigr)z ^{\alpha _{3}}\bigl(\tau ^{-1}\bigl(g(t,\xi )\bigr)\bigr)\,d\sigma (\xi ) \\ &\leq -q_{1}(t)z^{\alpha _{3}}\bigl(\tau ^{-1} \bigl(g_{2}(t)\bigr)\bigr), \end{aligned}$$

(2.9)

based on the fact that $z'(t)<0$ for $t\geq t_{1}$. Since $z(t)$ has property (II) of Lemma 2.1, one gets $\lim_{t\rightarrow \infty }z(t)=l\geq 0$. We claim that $l=0$. Indeed, if we assume on the contrary that $l>0$, then there exists $t_{2}\geq t_{1}$ s.t. $\tau ^{-1}(g_{2}(t))\geq t_{2}$ and $z(\tau ^{-1}(g_{2}(t)))\geq l$, $t\geq t_{2}$. Inequality (2.9) then yields

$$ E'_{2}(t)\leq -l^{\alpha _{3}}q_{1}(t). $$

(2.10)

We integrate (2.10) to get

$$ r_{2}(t) \bigl(E'_{1}(t) \bigr)^{\alpha _{2}}\geq l^{\alpha _{3}} \int ^{\infty }_{t} q_{1}(s)\,ds, $$

which indicates that

$$ E'_{1}(t)\geq \biggl( \frac{l^{\alpha _{3}}}{r_{2}(t)} \int ^{\infty }_{t} q _{1}(s)\,ds \biggr)^{\frac{1}{\alpha _{2}}}. $$

(2.11)

Integrating (2.11) from t to ∞, we have

$$ -E_{1}(t)\geq l^{\frac{\alpha _{3}}{\alpha _{2}}} \int ^{\infty }_{t} \biggl(\frac{1}{r_{2}(v)} \int ^{\infty }_{v} q_{1}(s)\,ds \biggr)^{\frac{1}{ \alpha _{2}}}\,dv, $$

and then

$$ \bigl\vert z'(t) \bigr\vert \geq l^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}} \biggl[ \frac{1}{r _{1}(t)} \int ^{\infty }_{t} \biggl(\frac{1}{r_{2}(v)} \int ^{\infty }_{v} q _{1}(s)\,ds \biggr)^{\frac{1}{\alpha _{2}}}\,dv \biggr]^{\frac{1}{\alpha _{1}}}, $$

since $z'(t)<0$ for $t\geq t_{2}$. Integrating the latter inequality from $t_{2}$ to ∞, we get

$$ z(t_{2})> l^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}} \int ^{\infty } _{t_{2}} \biggl[\frac{1}{r_{1}(u)} \int ^{\infty }_{u} \biggl( \frac{1}{r_{2}(v)} \int ^{\infty }_{v} q_{1}(s)\,ds \biggr)^{\frac{1}{\alpha _{2}}}\,dv \biggr]^{\frac{1}{\alpha _{1}}}\,du, $$

which contradicts (2.6). Thus, we obtain $l=0$ and $\lim_{t\rightarrow \infty }x(t)=0$, since $0< x(t)\leq z(t)$. □

Lemma 2.3

Assume that (A1)–(A5) and (2.2) hold. Furthermore, suppose that $x(t)$ is an eventually positive solution of (1.2) and $z(t)$ has property (I) of Lemma 2.1. Then for $t\geq t_{1} \geq t_{0}$,

$$ \bigl[r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}} \bigr]'+q _{2}(t)z^{\alpha _{3}} \bigl(\tau ^{-1}\bigl(g_{1}(t)\bigr)\bigr)\leq 0. $$

(2.12)

Proof

Property (I) of $z(t)$ yields

$$ E_{2}(t)=r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}}>0. $$

Applying the monotonicity of $E^{\frac{1}{\alpha _{2}}}_{2}(t)$ gives

$$\begin{aligned} r_{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}} &=r_{1}(t_{1}) \bigl(z'(t_{1}) \bigr)^{\alpha _{1}}+ \int ^{t}_{t_{1}} \frac{r^{\frac{1}{\alpha _{2}}}_{2}(s)(r_{1}(s)(z'(s))^{ \alpha _{1}})'}{r^{\frac{1}{\alpha _{2}}}_{2}(s)}\,ds \\ &\geq \delta _{1}(t,t_{1})r^{\frac{1}{\alpha _{2}}}_{2}(t) \bigl(r_{1}(t) \bigl(z'(t)\bigr)^{ \alpha _{1}} \bigr)'. \end{aligned}$$

(2.13)

It can be seen from (2.13) that

$$ \biggl(\frac{r_{1}(t)(z'(t))^{\alpha _{1}}}{\delta _{1}(t,t_{1})} \biggr)' \leq 0, $$

which, together with $r_{1}(t)(z'(t))^{\alpha _{1}}>0$, yields that $z'(t)/\delta _{2}(t,t_{1})$ is nonincreasing for $t\geq t_{1}$. Therefore,

$$\begin{aligned} z(t)=z(t_{1})+ \int ^{t}_{t_{1}}\frac{z'(s)}{\delta _{2}(s,t_{1})}\delta _{2}(s,t_{1})\,ds \geq \frac{\delta _{3}(t,t_{1})}{\delta _{2}(t,t_{1})}z'(t), \end{aligned}$$

(2.14)

which leads to

$$ \biggl(\frac{z(t)}{\delta _{3}(t,t_{1})} \biggr)'\leq 0. $$

(2.15)

Then for $t\geq t_{2}\geq t_{1}$,

$$ z\bigl(\tau ^{-1}\bigl(\tau ^{-1}(t) \bigr)\bigr)\leq \frac{\delta _{3}(\tau ^{-1}(\tau ^{-1}(t)),t _{1})z(\tau ^{-1}(t))}{\delta _{3}(\tau ^{-1}(t),t_{1})}, $$

(2.16)

due to $\tau ^{-1}(t)\leq \tau ^{-1}(\tau ^{-1}(t))$. Substituting (2.16) into (2.7), one has

$$ x(t)\geq p_{2}(t)z\bigl(\tau ^{-1}(t)\bigr), $$

which leads to

$$\begin{aligned} x\bigl(g(t,\xi )\bigr)\geq p_{2}\bigl(g(t,\xi )\bigr)z\bigl(\tau ^{-1}\bigl(g_{1}(t)\bigr)\bigr). \end{aligned}$$

(2.17)

Combining (1.2) and (2.17), we obtain (2.12). □

Main results

We respectively consider two cases $g_{1}(t)\leq \tau (t)$ and $g_{1}(t)\geq \tau (t)$ for $t\geq t_{0}$. Now, we begin with the first case.

Theorem 3.1

Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold, and $g_{1}(t)\leq \tau (t)$. If there exists $\rho (t)\in C^{1}([t _{0},\infty ),(0,\infty ))$ s.t.

$$\begin{aligned} & \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{\lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s) \gamma (s)\delta _{2}(s,t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds=\infty , \end{aligned}$$

(3.1)

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, where

$$\begin{aligned} &\lambda = \biggl(\frac{\alpha _{1}\alpha _{2}}{\alpha _{3}} \biggr)^{\alpha _{1}\alpha _{2}} \biggl( \frac{1}{\alpha _{1}\alpha _{2}+1} \biggr)^{\alpha _{1} \alpha _{2}+1}, \\ &\gamma (t) =\textstyle\begin{cases} m_{1}(\delta _{3}(t,t_{1}))^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}, \quad m_{1} \textit{ is any positive constant},& \textit{if } \alpha _{1} \alpha _{2}>\alpha _{3}, \\ m_{2}, \quad m_{2} \textit{ is any positive constant},& \textit{if } \alpha _{1}\alpha _{2}\leq \alpha _{3}, \end{cases}\displaystyle \end{aligned}$$

then every solution of (1.2) is either oscillatory or tends to zero.

Proof

Suppose that (1.2) has a nonoscillatory solution $x(t)$. We may assume that (2.3) holds for $t\geq t_{1}\geq t _{0}$. So we have that $z(t)$ is positive and satisfies the two properties for $t\geq t_{1}$.

We first consider property (I). Define $\omega (t)$ by

$$ \omega (t)=\rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(t)},\quad t\geq t_{1}. $$

(3.2)

Then $\omega (t)>0$ and

$$\begin{aligned} \omega '(t) =&\rho '(t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(t)} + \rho (t) \biggl[\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)} \\ &{}-\frac{\alpha _{3} r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}z'(t)}{z^{\alpha _{3}+1}(t)} \biggr] \end{aligned}$$

(3.3)

$$\begin{aligned} = &\frac{\rho '(t)}{\rho (t)}\omega (t) +\rho (t)\frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z ^{\alpha _{3}}(t)} \\ &{}-\alpha _{3} \rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}z'(t)}{z^{\alpha _{3}+1}(t)}. \end{aligned}$$

(3.4)

Based on (2.12), we have

$$ \frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)} \leq -q_{2}(t) \biggl( \frac{z(\tau ^{-1}(g_{1}(t)))}{z(t)} \biggr)^{\alpha _{3}}. $$

(3.5)

Since $g_{1}(t)\leq \tau (t)$, we get $\tau ^{-1}(g_{1}(t))\leq t$. Applying (2.15), we obtain

$$ \frac{z(\tau ^{-1}(g_{1}(t)))}{z(t)}\geq \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})}. $$

(3.6)

From (2.13), we have

$$ z'(t)\geq \delta _{2}(t,t_{1}) \bigl(r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}} \bigr)^{\frac{1}{\alpha _{1}\alpha _{2}}}. $$

(3.7)

We combine (3.4)–(3.7) to conclude that

$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\delta _{2}(t,t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)}z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t) \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$

(3.8)

Next, we will compute $z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)$ and consider the following two cases:

Case (i). Assume that $\alpha _{1}\alpha _{2}>\alpha _{3}$. Since $z(t)/\delta _{3}(t,t_{1})$ is nonincreasing, due to (2.15), there exist constants $h_{1}>0$ and $t_{2}\geq t_{1}$ such that

$$ \frac{z(t)}{\delta _{3}(t,t_{1})}\leq \frac{z(t_{2})}{\delta _{3}(t_{2},t _{1})}=h_{1},\quad t\geq t_{2}, $$

and

$$ z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\geq m_{1}\bigl(\delta _{3}(t,t _{1})\bigr)^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}, $$

(3.9)

where $m_{1}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{1}$.

Case (ii). Assume that $\alpha _{1}\alpha _{2}\leq \alpha _{3}$. Since $z'(t)>0$, there exists $h_{2}>0$ such that

$$ z(t)\geq z(t_{1})=h_{2},\quad t\geq t_{1}, $$

and

$$ z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\geq m_{2},\quad t\geq t _{1}, $$

(3.10)

where $m_{2}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{2}$. We combine (3.8) with (3.9) and (3.10) to have

$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$

(3.11)

By using the inequality (see [18])

$$ Bu-Au^{\frac{1}{\alpha }+1}\leq \frac{\alpha ^{\alpha }}{(\alpha +1)^{ \alpha +1}} \frac{B^{\alpha +1}}{A^{\alpha }},\quad A>0, $$

(3.12)

where

$$ B=\frac{\rho '_{+}(t)}{\rho (t)},\qquad A=\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)},\quad \alpha = \alpha _{1}\alpha _{2}, u=\omega (t), $$

we obtain

$$\begin{aligned} &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t)-\frac{\alpha _{3}\gamma (t) \delta _{2}(t,t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t) \\ &\quad \leq \biggl(\frac{\alpha _{1}\alpha _{2}}{\alpha _{3}\rho (t)\gamma (t) \delta _{2}(t,t_{1})} \biggr)^{\alpha _{1}\alpha _{2}} \biggl( \frac{ \rho '_{+}(t)}{\alpha _{1}\alpha _{2}+1} \biggr)^{\alpha _{1}\alpha _{2}+1} \\ &\quad = \frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t) \gamma (t)\delta _{2}(t,t_{1}))^{\alpha _{1}\alpha _{2}}}. \end{aligned}$$

(3.13)

We combine (3.11) and (3.13) to conclude that

$$ \omega '(t)\leq -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} +\frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t)\gamma (t)\delta _{2}(t,t_{1}))^{\alpha _{1}\alpha _{2}}}. $$

We integrate the latter inequality to make

$$\begin{aligned} & \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma (s)\delta _{2}(s,t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds \\ &\quad \leq \omega (t_{2})-\omega (t)< \omega (t_{2}), \end{aligned}$$

which contradicts (3.1).

Secondly, we investigate property (II) and deduce $\lim_{t\rightarrow \infty }x(t)=0$. □

Theorem 3.2

Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that $g_{1}(t)\leq \tau (t)$ and $\alpha _{1} \alpha _{2}=\alpha _{3}$. If there exists $\rho (t)$ s.t.

$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\rho '_{+}(s)}{(\delta _{3}(s,t_{1}))^{\alpha _{3}}} \biggr]\,ds=\infty , $$

(3.14)

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, then we get the same conclusion as in Theorem 3.1.

Proof

Suppose that (2.3) holds for $t\geq t_{1}\geq t _{0}$. Then $z(t)$ satisfies the two properties.

We first consider property (I). From (2.13) and (2.14), we have

$$\begin{aligned} r_{1}(t)z^{\alpha _{1}}(t) &\geq \biggl(\frac{\delta _{3}(t,t_{1})}{\delta _{2}(t,t_{1})} \biggr)^{\alpha _{1}}r_{1}(t) \bigl(z'(t) \bigr)^{\alpha _{1}} \\ &\geq \biggl(\frac{\delta _{3}(t,t_{1})}{\delta _{2}(t,t_{1})} \biggr)^{ \alpha _{1}} \delta _{1}(t,t_{1})r^{\frac{1}{\alpha _{2}}}_{2}(t) \bigl(r_{1}(t) \bigl(z'(t)\bigr)^{ \alpha _{1}} \bigr)', \end{aligned}$$

and

$$ z^{\alpha _{3}}(t)\geq \bigl(\delta _{3}(t,t_{1}) \bigr)^{\alpha _{3}}r_{2}(t) \bigl(\bigl(r _{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}}. $$

(3.15)

Define $\omega (t)$ by (3.2). As in the proof of Theorem 3.1, we get (3.3). Since $E_{2}(t)>0$, (3.3) indicates

$$ \omega '(t)\leq \rho '_{+}(t) \frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}}{z^{\alpha _{3}}(t)} +\rho (t)\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)}. $$

Combining the latter inequality with (3.5), (3.6), and (3.15), we see that

$$ \omega '(t)\leq \frac{\rho '_{+}(t)}{(\delta _{3}(t,t_{1}))^{\alpha _{3}}} -\rho (t)q_{2}(t) \biggl(\frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t _{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}}. $$

(3.16)

An integration of (3.16) from $t_{2}$ ($t_{2}\geq t_{1}$) to t leads to

$$ \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\rho '_{+}(s)}{( \delta _{3}(s,t_{1}))^{\alpha _{3}}} \biggr]\,ds\leq \omega (t_{2})-\omega (t)< \omega (t_{2}), $$

for all sufficiently large t, which contradicts (3.14).

Secondly, if property (II) holds, then $\lim_{t\rightarrow \infty }x(t)=0$. □

Theorem 3.3

Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that $g_{1}(t)\leq \tau (t)$ and $\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1$. If there exists $\rho (t)$ s.t.

$$\begin{aligned} &\limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{(\rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (s)\delta _{2}(s,t _{1})(\delta _{3}(s,t_{1}))^{\alpha _{3}-1}} \biggr]\,ds=\infty , \end{aligned}$$

(3.17)

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, where $\gamma (t)$ is given in Theorem 3.1, then we get the same conclusion as in Theorem 3.1.

Proof

Suppose that (2.3) holds for $t\geq t_{1}\geq t _{0}$. Then $z(t)$ satisfies the two properties.

We first consider property (I). Proceeding as in the proof of Theorem 3.1, we get (3.11), and

$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})\omega ^{2}(t)}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t). \end{aligned}$$

(3.18)

From (3.2) and (3.15), one has

$$\begin{aligned} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) &= \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) \biggl( \frac{r_{2}(t)((r_{1}(t)(z'(t))^{ \alpha _{1}})')^{\alpha _{2}}}{z^{\alpha _{3}}(t)} \biggr)^{\frac{1}{\alpha _{1}\alpha _{2}}-1} \\ &\geq \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) \bigl(\delta _{3}(t,t_{1}) \bigr)^{ \alpha _{3}-1}. \end{aligned}$$

(3.19)

We substitute (3.19) into (3.18) to see that

$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})(\delta _{3}(t,t_{1}))^{ \alpha _{3}-1}}{\rho (t)} \omega ^{2}(t), \end{aligned}$$

from which one gets

$$ \omega '(t)\leq -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} +\frac{( \rho '_{+}(t))^{2}}{4\alpha _{3}\rho (t)\gamma (t)\delta _{2}(t,t_{1})( \delta _{3}(t,t_{1}))^{\alpha _{3}-1}}, $$

by completing the square with respect to $\omega (t)$. We integrate the latter inequality from $t_{2}$ ($t_{2}\geq t_{1}$) to t to obtain

$$\begin{aligned} & \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{(\rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (s)\delta _{2}(s,t _{1})(\delta _{3}(s,t_{1}))^{\alpha _{3}-1}} \biggr]\,ds\leq \omega (t_{2}), \end{aligned}$$

for all sufficiently large t, which contradicts (3.17).

Secondly, if property (II) holds, then $\lim_{t\rightarrow \infty }x(t)=0$. □

Next, we consider $g_{1}(t)\geq \tau (t)$ for $t\geq t_{0}$.

Theorem 3.4

Assume that conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, and $g_{1}(t)\geq \tau (t)$. If there exists $\rho (t)$ s.t.

$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{ \lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma ( \tau (s))\delta _{2}(\tau (s),t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds= \infty , $$

(3.20)

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, then we get the same conclusion as in Theorem 3.1.

Proof

Suppose that (2.3) holds for $t\geq t_{1}\geq t _{0}$. Then $z(t)$ satisfies the two properties.

We first consider property (I). Define $\nu (t)$ by

$$ \nu (t)=\rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))},\quad t\geq t_{1}. $$

(3.21)

Then $\nu (t)>0$ and

$$\begin{aligned} \nu '(t) = &\rho '(t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))} +\rho (t) \biggl[\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}( \tau (t))} \\ &{}-\frac{\alpha _{3} r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}(z(\tau (t)))'}{z^{\alpha _{3}+1}(\tau (t))} \biggr] \end{aligned}$$

(3.22)

$$\begin{aligned} = &\frac{\rho '(t)}{\rho (t)}\nu (t) +\rho (t)\frac{ [r _{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(\tau (t))} \\ &{}-\alpha _{3} \rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}(z(\tau (t)))'}{z^{\alpha _{3}+1}(\tau (t))}. \end{aligned}$$

(3.23)

Since $\tau ^{-1}(g_{1}(t))\geq t\geq \tau (t)$ and $z'(t)>0$, we have

$$ \frac{z(\tau ^{-1}(g_{1}(t)))}{z(\tau (t))}\geq 1,\quad t\geq t_{1}, $$

which indicates that

$$ \frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(\tau (t))} \leq -q_{2}(t), $$

(3.24)

due to (2.12). Based on (3.7), $E'_{2}(t)\leq 0$ and $\tau (t)\leq t$, so one has $\tau (t)\geq t_{1}$ and

$$ \bigl(z\bigl(\tau (t)\bigr)\bigr)'\geq \delta _{2}\bigl(\tau (t),t_{1}\bigr) \bigl(r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{ \alpha _{1}} \bigr)'\bigr)^{\alpha _{2}} \bigr)^{\frac{1}{\alpha _{1}\alpha _{2}}}, $$

(3.25)

for $t\geq t_{2}>t_{1}$. Combining (3.9), (3.10), (3.23)–(3.25), we conclude that

$$\begin{aligned} \nu '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\nu (t)-\rho (t)q_{2}(t) \\ &{}-\frac{\alpha _{3}\gamma (\tau (t))\delta _{2}(\tau (t),t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \nu ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$

(3.26)

Using (3.12) and (3.26) with

$$ B=\frac{\rho '_{+}(t)}{\rho (t)},\qquad A=\frac{\alpha _{3}\gamma (\tau (t)) \delta _{2}(\tau (t),t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)}, $$

one gets

$$ \nu '(t)\leq -\rho (t)q_{2}(t) + \frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t)\gamma (\tau (t))\delta _{2}(\tau (t),t _{1}))^{\alpha _{1}\alpha _{2}}}. $$

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

$$ \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) -\frac{\lambda (\rho '_{+}(s))^{ \alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma (\tau (s))\delta _{2}(\tau (s),t _{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds\leq \omega (t_{2}), $$

for all sufficiently large t, which contradicts (3.20).

Secondly, if property (II) holds, then $\lim_{t\rightarrow \infty }x(t)=0$. □

Theorem 3.5

Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that $g_{1}(t)\geq \tau (t)$ and $\alpha _{1} \alpha _{2}=\alpha _{3}$. If there exists $\rho (t)$ s.t.

$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{ \rho '_{+}(s)}{(\delta _{3}(\tau (s),t_{1}))^{\alpha _{3}}} \biggr]\,ds= \infty , $$

(3.27)

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, then we get the same conclusion as in Theorem 3.1.

Proof

Suppose that (2.3) holds for $t\geq t_{1}\geq t _{0}$. Then $z(t)$ satisfies the two properties.

We first consider property (I). Proceeding as in the proof of Theorem 3.4, we get (3.22), which implies

$$ \nu '(t)\leq \rho '_{+}(t) \frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))} +\rho (t)\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}( \tau (t))}. $$

(3.28)

Applying (3.15), the monotonicity of $E_{2}(t)$ and the fact that $\tau (t)\leq t$, one has $\tau (t)\geq t_{1}$ and

$$ z^{\alpha _{3}}\bigl(\tau (t)\bigr)\geq \bigl(\delta _{3}\bigl(\tau (t),t_{1}\bigr)\bigr)^{\alpha _{3}}r _{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t) \bigr)^{\alpha _{1}}\bigr)'\bigr)^{\alpha _{2}}, $$

(3.29)

for $t\geq t_{2}>t_{1}$. Combining (3.24), (3.28), and (3.29), one gets

$$ \nu '(t)\leq \frac{\rho '_{+}(t)}{(\delta _{3}(\tau (t),t_{1}))^{\alpha _{3}}} -\rho (t)q_{2}(t). $$

(3.30)

Upon integrating (3.30) from $t_{2}$ to t, we obtain a contradiction to (3.27).

Secondly, if property (II) holds, then $\lim_{t\rightarrow \infty }x(t)=0$. □

Theorem 3.6

Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that $g_{1}(t)\geq \tau (t)$ and $\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1$. If there exists $\rho (t)$ s.t.

$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{( \rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (\tau (s))\delta _{2}( \tau (s),t_{1})(\delta _{3}(\tau (s),t_{1}))^{\alpha _{3}-1}} \biggr]\,ds= \infty , $$

for $t_{1}$ and $t_{*}$ with $t_{*}\geq t_{1}\geq t_{0}$, then we get the same conclusion as in Theorem 3.1.

We omit the proof of Theorem 3.6 here, since it is similar to that of Theorem 3.3.

Examples

The following examples are given to show the applications of Theorems 3.1 and 3.5.

Example 4.1

For $t>k_{1}\geq 1$, consider a TONDDE with DDA

$$ E'_{2}(t)+ \int ^{k_{1}+1}_{k_{1}} 10(t+\xi ) \biggl\vert x \biggl(t-k_{1}-\frac{1}{ \xi } \biggr) \biggr\vert ^{\frac{4}{3}}x \biggl(t-k_{1}-\frac{1}{\xi } \biggr)\,d\xi =0, $$

(4.1)

where

$$\begin{aligned} &E_{2}(t) = \bigl\vert E'_{1}(t) \bigr\vert ^{4} E'_{1}(t), \\ &E_{1}(t) =(t-k_{1}) \bigl\vert z'(t) \bigr\vert ^{-\frac{2}{3}}z'(t), \\ &z(t) =x(t)+\frac{4t+5}{t+1}x(t-k_{1}). \end{aligned}$$

Let $\alpha _{1}=1/3$, $\alpha _{2}=5$, $\alpha _{3}=7/3$, $a=k_{1}$, $b=k_{1}+1$, $r_{1}(t)=t-k_{1}$, $r_{2}(t)=1$, $\tau (t)=t-k_{1}$, $g(t,\xi )=t-k_{1}-1/\xi$, $\sigma (\xi )=\xi$, $p(t)=(4t+5)/(t+1)$, $q(t,\xi )=10(t+\xi )$. Choose $t_{0}=t_{1}=k_{1}$. Then we obtain $\alpha _{1}\alpha _{2}<\alpha _{3}$, $4\leq p(t)<5$,

$$\begin{aligned} &g_{1}(t) =g(t,k_{1})=t-k_{1}- \frac{1}{k_{1}}, \\ &\delta _{1}(t,t_{1}) =\delta _{1}(t,k_{1})=t-k_{1}, \\ &\delta _{2}(t,t_{1}) = \biggl(\frac{\delta _{1}(t,k_{1})}{t-k_{1}} \biggr)^{3}=1, \\ &\delta _{3}(t,t_{1}) =\delta _{3}(t,k_{1})=t-k_{1}, \\ &\delta _{3}\bigl(\tau ^{-1}(t),t_{1} \bigr) =\delta _{3}(t+k_{1},k_{1})=t, \\ &\delta _{3}\bigl(\tau ^{-1}\bigl(\tau ^{-1}(t)\bigr),t_{1}\bigr) =\delta _{3}(t+2k_{1},k_{1})=t+k _{1}, \\ &\delta _{3}\bigl(\tau ^{-1}\bigl(g_{1}(t) \bigr),t_{1}\bigr) =\delta _{3}\biggl(t- \frac{1}{k_{1}},k _{1}\biggr)=t-k_{1}- \frac{1}{k_{1}}. \end{aligned}$$

Furthermore, we deduce that

$$\begin{aligned} &p_{1}(t) >\frac{1}{5} \biggl(1-\frac{1}{4} \biggr)=\frac{3}{20}>0, \\ &p_{2}(t) >\frac{1}{5} \biggl(1-\frac{1}{4} \cdot \frac{t+k_{1}}{t} \biggr) > \frac{1}{10}>0, \\ &q_{1}(t) > \int ^{k_{1}+1}_{k_{1}}\frac{3}{20}\cdot 10(t+ \xi )\,d\xi = \frac{3}{2} \biggl(t+k_{1}+ \frac{1}{2} \biggr), \\ &q_{2}(t) > \int ^{k_{1}+1}_{k_{1}}\frac{1}{10}\cdot 10(t+ \xi )\,d\xi =t+k _{1}+\frac{1}{2}. \end{aligned}$$

It is easy to verify that

$$\begin{aligned} & \int ^{\infty }_{t_{0}} \biggl[\frac{1}{r_{1}(u)} \int ^{\infty }_{u} \biggl(\frac{1}{r_{2}(v)} \int ^{\infty }_{v}q_{1}(s) \,ds \biggr)^{\frac{1}{ \alpha _{2}}}\,dv \biggr]^{\frac{1}{\alpha _{1}}}\,du \\ &\quad > \int ^{\infty }_{k_{1}} \biggl[\frac{1}{u-k_{1}} \int ^{ \infty }_{u} \biggl( \int ^{\infty }_{v} \frac{3}{2} \biggl(s+k_{1}+ \frac{1}{2} \biggr)\,ds \biggr)^{\frac{1}{5}}\,dv \biggr]^{3}\,du \\ &\quad = \infty . \end{aligned}$$

Therefore, conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, and $g_{1}(t)\leq \tau (t)$. We choose $\rho (t)=t$ and $t_{*}=k_{1}+2$. Applying Theorem 3.1, it remains to check (3.1), where

$$ \lambda = \biggl(\frac{5}{7} \biggr)^{\frac{5}{3}} \biggl( \frac{3}{8} \biggr)^{ \frac{8}{3}}. $$

Then we get

$$\begin{aligned} & \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma (s)\delta _{2}(s,t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds \\ &\quad > \int ^{t}_{k_{1}+2} \biggl[s \biggl(s+k_{1}+ \frac{1}{2} \biggr) \biggl(\frac{s-k_{1}-\frac{1}{k_{1}}}{s-k_{1}} \biggr)^{\frac{7}{3}} -\frac{ (\frac{5}{7} )^{\frac{5}{3}} (\frac{3}{8} )^{\frac{8}{3}}}{(m _{2} s)^{\frac{5}{3}}} \biggr]\,ds \rightarrow \infty , \end{aligned}$$

as $t\rightarrow \infty$, since $\int ^{t}_{k_{1}+2}s^{-\frac{5}{3}}\,ds< \infty$. Hence, we get the same conclusion as in Theorem 3.1.

Example 4.2

For $t>k_{1}\geq 1$, consider a TONDDE with DDA

$$ E'_{2}(t)+ \int ^{k_{1}+l}_{k_{1}} \frac{40\xi }{t} \biggl\vert x \biggl(\frac{t+ \xi }{2} \biggr) \biggr\vert ^{2} x \biggl(\frac{t+\xi }{2} \biggr)\,d\xi =0, $$

(4.2)

where l is a positive integer,

$$\begin{aligned} &E_{2}(t) = \bigl\vert E'_{1}(t) \bigr\vert ^{-\frac{2}{3}} E'_{1}(t), \\ &E_{1}(t) =(t-k_{1}) \bigl\vert z'(t) \bigr\vert ^{8} z'(t), \\ &z(t) =x(t)+\frac{5t+4k_{1}}{t+k_{1}}x \biggl(\frac{t}{2} \biggr). \end{aligned}$$

Let $\alpha _{1}=9$, $\alpha _{2}=1/3$, $\alpha _{3}=3$, $a=k_{1}$, $b=k_{1}+l$, $r_{1}(t)=t-k_{1}$, $r_{2}(t)=1$, $\sigma (\xi )=\xi$,

$$ \tau (t)=\frac{t}{2},\qquad g(t,\xi )=\frac{t+\xi }{2},\qquad p(t)=\frac{5t+4k _{1}}{t+k_{1}},\qquad q(t,\xi )=\frac{40\xi }{t}. $$

Choose $t_{0}=t_{1}=k_{1}$. Then we have $\alpha _{1}\alpha _{2}=\alpha _{3}$, $4< p(t)<5$,

$$\begin{aligned} &g_{1}(t) =g(t,k_{1})=\frac{t+k_{1}}{2}, \\ &\delta _{3}\bigl(\tau ^{-1}(t),t_{1} \bigr) =\delta _{3}(2t,k_{1})=2t-k_{1}, \\ &\delta _{3}\bigl(\tau ^{-1}\bigl(\tau ^{-1}(t)\bigr),t_{1}\bigr) =\delta _{3}(4t,k_{1})=4t-k _{1}, \\ &\delta _{3}\bigl(\tau (t),t_{1}\bigr) =\delta _{3} \biggl(\frac{t}{2},k_{1} \biggr)= \frac{t}{2}-k_{1}>\frac{t}{4}, \end{aligned}$$

where $t\geq t_{2}>4k_{1}$, $\delta _{1}(t,t_{1})$, $\delta _{2}(t,t _{1})$, and $\delta _{3}(t,t_{1})$ are the same as in Example 4.1. Furthermore, we deduce that

$$\begin{aligned} &p_{1}(t) >\frac{1}{5} \biggl(1-\frac{1}{4} \biggr)=\frac{3}{20}>0, \\ &p_{2}(t) >\frac{1}{5} \biggl(1-\frac{1}{4} \cdot \frac{4t-k_{1}}{2t-k _{1}} \biggr) >\frac{1}{20}>0, \\ &q_{1}(t) > \int ^{k_{1}+1}_{k_{1}}\frac{3}{20}\cdot \frac{40\xi }{t}\,d \xi =\frac{6k_{1} l+3l^{2}}{t}, \\ &q_{2}(t) > \int ^{k_{1}+1}_{k_{1}}\frac{1}{20}\cdot \frac{40\xi }{t}\,d \xi =\frac{2k_{1} l+l^{2}}{t}. \end{aligned}$$

Clearly, (2.6) holds. Choosing $\rho (t)=t^{2}$ and $t_{*}=t _{2}$, one has

$$\begin{aligned} & \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{\rho '_{+}(s)}{(\delta _{3}(\tau (s),t_{1}))^{\alpha _{3}}} \biggr]\,ds \\ &\quad > \int ^{t}_{t_{2}} \biggl[s^{2}\cdot \frac{2k_{1} l+l^{2}}{s} -\frac{2s}{ (\frac{s}{4} )^{3}} \biggr]\,ds \rightarrow \infty , \end{aligned}$$

as $t\rightarrow \infty$, which means that (3.27) holds, and all conditions of Theorem 3.5 are satisfied. Hence, we get the same conclusion as in Theorem 3.1.