Oscillation criteria for second-order nonlinear neutral dynamic equations with distributed deviating arguments on time scales

Abstract

In this article, we establish some new oscillation criteria and give sufficient conditions to ensure that all solutions of nonlinear neutral dynamic equation of the form

$${(r(t){\left({(y(t)+p(t)y(\tau (t)))}^{\mathrm{△}}\right)}^{\gamma })}^{\mathrm{△}}+{\int }_a^{b}f(t,y(\delta (t,\xi )))\mathrm{△}\xi =0$$

are oscillatory on a time scale $\mathbb{T}$, where $\gamma ⩾1$ is a quotient of odd positive integers.

1 Introduction

The aim of this article is to develop some oscillation theorems for a second-order nonlinear neutral dynamic equation

$${(r(t){\left({(y(t)+p(t)y(\tau (t)))}^{\mathrm{△}}\right)}^{\gamma })}^{\mathrm{△}}+{\int }_a^{b}f(t,y(\delta (t,\xi )))\mathrm{△}\xi =0$$

(1)

on a time scale $\mathbb{T}$. Throughout this paper, it is assumed that $\gamma ⩾1$ is a quotient of odd positive integers, $0<a<b$, $\tau (t):\mathbb{T}\to \mathbb{T}$, is rd-continuous function such that $\tau (t)⩽t$ and $\tau (t)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$, $\delta (t,\xi ):\mathbb{T}\times [a,b]\to \mathbb{T}$ is rd-continuous function such that decreasing with respect to ξ, $\delta (t,\xi )⩽t$ for $\xi \in [a,b]$, $\delta (t,\xi )\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$, $r(t)>0$ and $0⩽p(t)<1$ are real valued rd-continuous functions defined on $\mathbb{T}$, $p(t)$ is increasing and

(H₁) ${\int }_{t_0}^{\mathrm{\infty }}{(\frac{1}{r(t)})}^{\frac{1}{\gamma }}\mathrm{△}t=\mathrm{\infty }$,

(H₂) $f:\mathbb{T}\times \mathbb{R}\to \mathbb{R}$ is a continuous function such that $uf(t,u)>0$ for all $u\ne 0$ and there exists a positive function $q(t)$ defined on $\mathbb{T}$ such that $|f(t,u)|⩾q(t)|u^{\gamma }|$.

A nontrivial function $y(t)$ is said to be a solution of (1) if $y(t)+p(t)y(\tau (t))\in C_{rd}^1[t_y,\mathrm{\infty }]$ and $r(t){({(y(t)+p(t)y(\tau (t)))}^{\mathrm{△}})}^{\gamma }\in C_{rd}^1[t_y,\mathrm{\infty }]$ for $t_y⩾t_0$ and $y(t)$ satisfies equation (1) for $t_y⩾t_0$. A solution of (1), which is nontrivial for all large t, is called oscillatory if it has no last zero. Otherwise, a solution is called nonoscillatory.

We note that if $\mathbb{T}=\mathbb{R}$, we have $\sigma (t)=t$, $\mu (t)=0$, $y^{\mathrm{△}}(t)=y^{\prime }(t)$ and, therefore, (1) becomes a second-order neutral differential equation with distributed deviating arguments

$${(r(t){\left({(y(t)+p(t)y(\tau (t)))}^{\prime }\right)}^{\gamma })}^{\prime }+{\int }_a^{b}f(t,y(\delta (t,\xi )))d\xi =0.$$

If $\mathbb{T}=\mathbb{N}$, we have $\sigma (t)=t+1$, $\mu (t)=1$, $y^{\mathrm{△}}(t)=\mathrm{△}y(t)=y(t+1)-y(t)$ and therefore (1) becomes a second-order neutral difference equation with distributed deviating arguments

$$\mathrm{△}(r(t){\left(\mathrm{△}(y(t)+p(t)y(\tau (t)))\right)}^{\gamma })+\sum _{\xi =a}^{b-1}f(t,y(\delta (t,\xi )))=0$$

and if $\mathbb{T}=h\mathbb{N}$, $h>0$, we have $\sigma (t)=t+h$, $\mu (t)=h$, $y^{\mathrm{△}}(t)={\mathrm{△}}_{h}y(t)=\frac{y(t+h)-y(t)}{h}$ and, therefore, (1) becomes a second-order neutral difference equation with distributed deviating arguments

$${\mathrm{△}}_h(r(t){\left({\mathrm{△}}_h(y(t)+p(t)y(\tau (t)))\right)}^{\gamma })+\sum _{k=\frac{a}{h}}^{\frac{b}{h}-1}f(t,y(\delta (t,kh)))h=0.$$

In recent years, there has been important research activity about the oscillatory behavior of second-order neutral differential, difference and dynamic equations. For example, Grace and Lalli [1] considered the following second-order neutral delay equation

$${(a(t){(x(t)+p(t)x(t-\tau ))}^{\prime })}^{\prime }+q(t)f(x(t-\tau ))=0,t⩾t_0$$

and Graef et al. [2] considered the nonlinear second-order neutral delay equation

$${(y(t)+p(t)y(\tau (t)))}^{″}+q(t)f(y(t-\delta ))=0,t⩾t_0.$$

Recently, Agarwal et al. [3] considered second-order nonlinear neutral delay dynamic equation

$${(r(t){\left({(y(t)+p(t)y(\tau (t)))}^{\mathrm{△}}\right)}^{\gamma })}^{\mathrm{△}}+f(t,y(t-\delta ))=0.$$

(2)

Later, Saker [4] considered (2) but he used different technique to prove his results. In [5] and [6], the authors considered the second order neutral functional dynamic equation of the form

$${(r(t){\left({(y(t)+p(t)y(\tau (t)))}^{\mathrm{△}}\right)}^{\gamma })}^{\mathrm{△}}+f(t,y(\delta (t)))=0,$$

which is more general than (2). For more papers related to oscillation of second-order nonlinear neutral delay dynamic equation on time scales, we refer the reader to [7–10]. For neutral equations with distributed deviating arguments, we refer the reader to the paper by Candan [11]. To the best of our knowledge, [12] is the only paper regarding to the distributed deviating arguments on time scales. The books [13, 14] gives time scale calculus and some applications.

2 Main results

Throughout the paper, we use the following notations for simplicity:

$$x(t)=y(t)+p(t)y(\tau (t)),x^{[1]}=r{\left(x^{\mathrm{△}}\right)}^{\gamma },x^{[2]}={\left(x^{[1]}\right)}^{\mathrm{△}}$$

(3)

and ${\theta }_1(t)=\delta (t,a)$ and ${\theta }_2(t)=\delta (t,b)$.

Theorem 2.1 Assume that (H₁) and (H₂) hold. In addition, assume that $r^{\mathrm{△}}(t)⩾0$. Then every solution of (1) oscillates if the inequality

$$x^{[2]}(t)+A(t)x^{[1]}({\theta }_1(t))⩽0,$$

(4)

where

$$A(t)=\frac{(b-a)q(t){(1-p({\theta }_1(t)))}^{\gamma }}{r({\theta }_1(t))}{\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma }$$

has no eventually positive solution.

Proof Let $y(t)$ be a nonoscillatory solution of (1), without loss of generality, we assume that $y(t)>0$ for $t⩾t_0$, then $y(\tau (t))>0$ and $y(\delta (t,\xi ))>0$ for $t⩾t_1>t_0$ and $b⩾\xi ⩾a$. In the case when $y(t)$ is negative, the proof is similar. In view of (1), (H₂) and (3)

$$x^{[2]}(t)+{\int }_a^{b}q(t)y^{\gamma }(\delta (t,\xi ))\mathrm{△}\xi ⩽0$$

(5)

for all $t⩾t_1$, and we see that $x^{[1]}(t)$ is an eventually decreasing function. We claim that $x^{[1]}(t)>0$ eventually. Assume not then there exists a $t_2⩾t_1$ such that $x^{[1]}(t_2)=c<0$, then we have $x^{[1]}(t)⩽c$ for $t⩾t_2$ and it follows that

$$x^{\mathrm{△}}(t)⩽{\left(\frac{c}{r(t)}\right)}^{1/\gamma }.$$

(6)

Now integrating (6) from $t_2$ to t and using (H₁), we obtain

$$x(t)⩽x(t_2)+c^{1/\gamma }{\int }_{t_2}^t{\left(\frac{1}{r(s)}\right)}^{1/\gamma }\mathrm{△}s\to -\mathrm{\infty }\text{as }t\to \mathrm{\infty }$$

which contradicts the fact that $x(t)>0$ for all $t⩾t_0$. Hence, $x^{[1]}(t)$ is positive. Therefore, one sees that there is a $t_2⩾t_1$ such that

$$x(t)>0,x^{\mathrm{△}}(t)>0,x^{[1]}(t)>0,x^{[2]}(t)<0,t⩾t_2.$$

(7)

For $t⩾t_3⩾t_2$, this implies that

$$y(t)⩾x(t)-p(t)x(\tau (t))⩾(1-p(t))x(t)$$

then we conclude that

$$y^{\gamma }(\delta (t,\xi ))⩾{(1-p(\delta (t,\xi )))}^{\gamma }{x}^{\gamma }(\delta (t,\xi )),t⩾t_4⩾t_3,\xi \in [a,b].$$

(8)

Multiplying (8) by $q(t)$ and integrating both sides from a to b, we have

$${\int }_a^{b}q(t)y^{\gamma }(\delta (t,\xi ))\mathrm{△}\xi ⩾{\int }_a^{b}q(t){(1-p(\delta (t,\xi )))}^{\gamma }{x}^{\gamma }(\delta (t,\xi ))\mathrm{△}\xi .$$

(9)

Substituting (9) into (5), we obtain

$$x^{[2]}(t)+{\int }_a^{b}q(t){(1-p(\delta (t,\xi )))}^{\gamma }{x}^{\gamma }(\delta (t,\xi ))\mathrm{△}\xi ⩽0.$$

(10)

On the other hand, we can verify that $x^{\mathrm{△}\mathrm{△}}(t)⩽0$ for $t⩾t_4$ and, therefore, we obtain

$$x(t)=x(t_4)+{\int }_{t_4}^t{x}^{\mathrm{△}}(s)\mathrm{△}s⩾(t-t_4)x^{\mathrm{△}}(t)⩾\frac{t}{2}{x}^{\mathrm{△}}(t),t⩾t_5⩾2t_4.$$

From the last inequality, it can be easily seen that

$$x(\delta (t,\xi ))⩾\left(\frac{\delta (t,\xi )}{2}\right)x^{\mathrm{△}}(\delta (t,\xi ))⩾\left(\frac{{\theta }_2(t)}{2}\right)x^{\mathrm{△}}(\delta (t,\xi )),t⩾t_6⩾t_5,\xi \in [a,b].$$

Substituting the last inequality into (10), we have

$$x^{[2]}(t)+{\int }_a^{b}q(t){(1-p(\delta (t,\xi )))}^{\gamma }{\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma }{\left(x^{\mathrm{△}}(\delta (t,\xi ))\right)}^{\gamma }\mathrm{△}\xi ⩽0$$

and it can be found

$$x^{[2]}(t)+(b-a)q(t){(1-p({\theta }_1(t)))}^{\gamma }{\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma }{\left(x^{\mathrm{△}}({\theta }_1(t))\right)}^{\gamma }⩽0,$$

or

$$x^{[2]}(t)+\frac{(b-a)q(t){(1-p({\theta }_1(t)))}^{\gamma }}{r({\theta }_1(t))}{\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma }{x}^{[1]}({\theta }_1(t))⩽0,$$

which is the inequality (4). As a consequence of this, we have a contradiction and therefore every solution of (1) oscillates. □

Theorem 2.2 Assume that (H₁) and (H₂) hold. In addition, assume that $r^{\mathrm{△}}(t)⩾0$, $\delta (t,\xi )$ is increasing with respect to t and that the inequality

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{{\theta }_1(t)}^{t}A(s)\mathrm{△}s>1$$

(11)

holds. Then every solution of (1) oscillates.

Proof Let $y(t)$ be a nonoscillatory solution of (1). We can proceed as in the proof of Theorem 2.1 to get (4). Integrating (4) from ${\theta }_1(t)$ to t for sufficiently large t, we have

$$\begin{array}{rcl}0& ⩾& {\int }_{{\theta }_1(t)}^t(x^{[2]}(s)+A(s)x^{[1]}({\theta }_1(s)))\mathrm{△}s\\ =& x^{[1]}(t)-x^{[1]}({\theta }_1(t))+{\int }_{{\theta }_1(t)}^{t}A(s)x^{[1]}({\theta }_1(s))\mathrm{△}s\\ ⩾& x^{[1]}(t)-x^{[1]}({\theta }_1(t))+x^{[1]}({\theta }_1(t)){\int }_{{\theta }_1(t)}^{t}A(s)\mathrm{△}s\\ =& x^{[1]}(t)+x^{[1]}({\theta }_1(t))({\int }_{{\theta }_1(t)}^{t}A(s)\mathrm{△}s-1)>0.\end{array}$$

By making use of (11), we reach to a contradiction therefore the proof is complete. □

Theorem 2.3 Assume that (H₁) and (H₂) hold. In addition, assume that $r^{\mathrm{△}}(t)⩾0$, $\delta (t,\xi )$ is increasing with respect to t and there exists a positive rd-continuous △-differentiable function $\alpha (t)$ such that

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t(\alpha (s)Q(s)-\frac{{({({\alpha }^{\mathrm{△}}(s))}_{+})}^{2}r({\theta }_2(s))}{4\gamma {(\frac{{\theta }_2(s)}{2})}^{\gamma -1}\alpha (s)})\mathrm{△}s=\mathrm{\infty },$$

(12)

where ${({\alpha }^{\mathrm{△}}(s))}_{+}=max\{0,{\alpha }^{\mathrm{△}}(s)\}$ and $Q(s)=(b-a)q(s){(1-p({\theta }_1(s)))}^{\gamma }$. Then every solution of (1) is oscillatory on $[t_0,\mathrm{\infty })$.

Proof Suppose to the contrary that $y(t)$ is nonoscillatory solution of (1). We may assume without loss of generality that $y(t)>0$ for $t⩾t_0$, then $y(\tau (t))>0$ and $y(\delta (t,\xi ))>0$ for $t⩾t_1>t_0$ and $b⩾\xi ⩾a$. Proceeding as in the proof of Theorem 2.1, we obtain (7) and the inequality (10). Using (7) and Pötzsche’s chain rule [[15], Theorem 1], we obtain

$$\begin{array}{rcl}{(x^{\gamma }(t))}^{\mathrm{△}}& =& \gamma {\int }_0^1{[x(t)+h\mu (t)x^{\mathrm{△}}(t)]}^{\gamma -1}dh{x}^{\mathrm{△}}(t)\\ ⩾& \gamma {\int }_0^1{(x(t))}^{\gamma -1}dh{x}^{\mathrm{△}}(t)=\gamma {(x(t))}^{\gamma -1}{x}^{\mathrm{△}}(t)>0.\end{array}$$

(13)

From (10) and (13), we obtain

$$x^{[2]}(t)⩽-(b-a)q(t){(1-p({\theta }_1(t)))}^{\gamma }{x}^{\gamma }({\theta }_2(t))=-Q(t)x^{\gamma }({\theta }_2(t)),t⩾t_4.$$

(14)

Define the function

$$z(t)=\alpha (t)\frac{x^{[1]}(t)}{x^{\gamma }({\theta }_2(t))},t⩾t_4.$$

(15)

It is obvious that $z(t)>0$. Taking the derivative of $z(t)$, we see that

$$\begin{array}{rcl}{z}^{\mathrm{△}}(t)& =& {\left(x^{[1]}\right)}^{\sigma }(t){\left(\frac{\alpha (t)}{x^{\gamma }({\theta }_2(t))}\right)}^{\mathrm{△}}+\frac{\alpha (t)}{x^{\gamma }({\theta }_2(t))}{x}^{[2]}(t)\\ =& \frac{\alpha (t)x^{[2]}(t)}{x^{\gamma }({\theta }_2(t))}+{\left(x^{[1]}\right)}^{\sigma }(t)\left(\frac{x^{\gamma }({\theta }_2(t)){\alpha }^{\mathrm{△}}(t)-\alpha (t){(x^{\gamma }({\theta }_2(t)))}^{\mathrm{△}}}{x^{\gamma }({\theta }_2(t)){(x^{\sigma }({\theta }_2(t)))}^{\gamma }}\right).\end{array}$$

(16)

Now using (14) in (16), we obtain

$$z^{\mathrm{△}}(t)⩽-\alpha (t)Q(t)+\frac{{\alpha }^{\mathrm{△}}(t)z^{\sigma }(t)}{{\alpha }^{\sigma }(t)}-\frac{\alpha (t){(x^{[1]})}^{\sigma }(t){(x^{\gamma }({\theta }_2(t)))}^{\mathrm{△}}}{x^{\gamma }({\theta }_2(t)){(x^{\sigma }({\theta }_2(t)))}^{\gamma }}.$$

(17)

On the other hand, as in the proof of Theorem 2.1, it can be shown that for sufficiently large $t⩾t_5$

$$x(t)⩾\left(\frac{t}{2}\right)x^{\mathrm{△}}(t),t⩾t_5⩾2t_4$$

and then

$$\gamma x^{\gamma -1}(t)⩾\gamma {\left(\frac{t}{2}\right)}^{\gamma -1}{(x^{\mathrm{△}}(t))}^{\gamma -1}$$

or

$$\gamma x^{\gamma -1}({\theta }_2(t))⩾\gamma {\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma -1}{\left(x^{\mathrm{△}}({\theta }_2(t))\right)}^{\gamma -1},t⩾t_6⩾t_5.$$

(18)

Since $x^{[2]}(t)<0$, we have

$$x^{[1]}(t)>x^{[1]}(\sigma (t)).$$

(19)

Multiplying (18) by $x^{\mathrm{△}}({\theta }_2(t))$ and using (19), it follows that

$$\begin{array}{rcl}\gamma x^{\gamma -1}({\theta }_2(t))x^{\mathrm{△}}({\theta }_2(t))& ⩾& \gamma {\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma -1}{\left(x^{\mathrm{△}}({\theta }_2(t))\right)}^{\gamma }\\ ⩾& \gamma {\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma -1}\frac{r({\theta }_2(\sigma (t)))}{r({\theta }_2(t))}{\left(x^{\mathrm{△}}\left({\theta }_2(\sigma (t))\right)\right)}^{\gamma }\\ ⩾& \gamma {\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma -1}\frac{{(x^{[1]})}^{\sigma }({\theta }_2(t))}{r({\theta }_2(t))}.\end{array}$$

(20)

From (13), for sufficiently large $t⩾t_7⩾t_6$, we have

$${\left(x^{\gamma }({\theta }_2(t))\right)}^{\mathrm{△}}⩾\gamma x^{\gamma -1}({\theta }_2(t))x^{\mathrm{△}}({\theta }_2(t)).$$

(21)

From (20) and (21), it follows that

$${\left(x^{\gamma }({\theta }_2(t))\right)}^{\mathrm{△}}⩾\gamma {\left(\frac{{\theta }_2(t)}{2}\right)}^{\gamma -1}\frac{{(x^{[1]})}^{\sigma }({\theta }_2(t))}{r({\theta }_2(t))}.$$

(22)

Substituting (22) into (17), we obtain

$$z^{\mathrm{△}}(t)⩽-\alpha (t)Q(t)+\frac{{\alpha }^{\mathrm{△}}(t)z^{\sigma }(t)}{{\alpha }^{\sigma }(t)}-\frac{\gamma {(\frac{{\theta }_2(t)}{2})}^{\gamma -1}\alpha (t)}{{({\alpha }^{\sigma }(t))}^{2}r({\theta }_2(t))}{(z^{\sigma }(t))}^2.$$

Using the fact $u-m{u}^2⩽\frac{1}{4m}$, $m>0$, we have

$$\begin{array}{rcl}{z}^{\mathrm{△}}(t)& ⩽& -\alpha (t)Q(t)+\frac{{({\alpha }^{\mathrm{△}}(t))}_{+}}{{\alpha }^{\sigma }(t)}(z^{\sigma }(t)-\frac{\gamma {(\frac{{\theta }_2(t)}{2})}^{\gamma -1}\alpha (t)}{({({\alpha }^{\mathrm{△}}(t))}_{+}){\alpha }^{\sigma }(t)r({\theta }_2(t))}{(z^{\sigma }(t))}^2)\\ ⩽& -(\alpha (t)Q(t)-\frac{{({({\alpha }^{\mathrm{△}}(t))}_{+})}^{2}r({\theta }_2(t))}{4\gamma {(\frac{{\theta }_2(t)}{2})}^{\gamma -1}\alpha (t)}).\end{array}$$

Integrating the last inequality from $t_7$ to t, we obtain

$$-z(t_7)<z(t)-z(t_7)⩽-{\int }_{t_7}^t(\alpha (s)Q(s)-\frac{{({({\alpha }^{\mathrm{△}}(s))}_{+})}^{2}r({\theta }_2(s))}{4\gamma {(\frac{{\theta }_2(s)}{2})}^{\gamma -1}\alpha (s)})\mathrm{△}s$$

or

$$z(t_7)>{\int }_{t_7}^t(\alpha (s)Q(s)-\frac{{({({\alpha }^{\mathrm{△}}(s))}_{+})}^{2}r({\theta }_2(s))}{4\gamma {(\frac{{\theta }_2(s)}{2})}^{\gamma -1}\alpha (s)})\mathrm{△}s$$

which contradicts (12). Therefore, the proof is complete. □

Theorem 2.4 Assume that (H₁) and (H₂) hold and $\sigma (t)\ne t$ for each $t\in \mathbb{T}$. Let $\alpha (t)$, $\delta (t,\xi )$, and $Q(s)$ be as defined in Theorem  2.3. If

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t(\alpha (s)Q(s)-\frac{{({({\alpha }^{\mathrm{△}}(s))}_{+})}^{2}r({\theta }_2(s))}{2^{3-\gamma }{(\mu ({\theta }_2(s)))}^{\gamma -1}\alpha (s)})\mathrm{△}s=\mathrm{\infty },$$

then every solution of (1) is oscillatory on $[t_0,\mathrm{\infty })$.

Proof Following the same lines as in the proof of Theorem 2.1, we get (7) and (10). Using the inequality,

$$x^{\gamma }-y^{\gamma }⩾2^{1-\gamma }{(x-y)}^{\gamma },\gamma ⩾1,$$

we have

$$\begin{array}{rcl}{(x^{\gamma }(t))}^{\mathrm{△}}& =& \frac{x^{\gamma }(\sigma (t))-x^{\gamma }(t)}{\mu (t)}⩾2^{1-\gamma }\frac{{(x(\sigma (t))-x(t))}^{\gamma }}{\mu (t)}\\ =& 2^{1-\gamma }{(\mu (t))}^{\gamma -1}{\left(\frac{x(\sigma (t))-x(t)}{\mu (t)}\right)}^{\gamma }=2^{1-\gamma }{(\mu (t))}^{\gamma -1}{(x^{\mathrm{△}}(t))}^{\gamma }.\end{array}$$

(23)

Now setting $z(t)$ by (15), using (17) and (23) we see that

$$z^{\mathrm{△}}(t)⩽-\alpha (t)Q(t)+\frac{{({\alpha }^{\mathrm{△}}(t))}_{+}{z}^{\sigma }(t)}{{\alpha }^{\sigma }(t)}-\frac{2^{1-\gamma }{(\mu ({\theta }_2(t)))}^{\gamma -1}\alpha (t)}{{({\alpha }^{\sigma }(t))}^{2}r({\theta }_2(t))}{(z^{\sigma }(t))}^2.$$

The remaining part of the proof is similar to that of Theorem 2.3, hence it is omitted. □

Example 2.5

Consider the following second-order neutral nonlinear dynamic equation

$${\left({\left({(y(t)+\left(\frac{t+a-1}{t+a}\right)y(\tau (t)))}^{\mathrm{△}}\right)}^{5/3}\right)}^{\mathrm{△}}+{\int }_a^b{t}^{-1/3}y(t-\xi )\mathrm{△}\xi =0,t\in \mathbb{T}$$

where $\gamma =\frac{5}{3}$, $r(t)=1$, $p(t)=(\frac{t+a-1}{t+a})$, $q(t)=t^{-1/3}$. One can verify that the conditions of Theorem 2.3 are satisfied. Note that taking $\alpha (s)=s$, we see that

$$\begin{array}{r}\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t(\alpha (s)Q(s)-\frac{{({({\alpha }^{\mathrm{△}}(s))}_{+})}^{2}r({\theta }_2(s))}{4\gamma {(\frac{{\theta }_2(s)}{2})}^{\gamma -1}\alpha (s)})\mathrm{△}s\\ =\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t((b-a)s^{-1}-\frac{1}{\frac{20}{3}{(\frac{s-b}{2})}^{2/3}s})\mathrm{△}s=\mathrm{\infty }.\end{array}$$

Therefore, (1) is oscillatory.