Oscillation criteria for second-order quasi-linear delay dynamic equations on time scales

Abstract

This paper is concerned with oscillations of the second-order delay nonlinear dynamic equation ${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\beta }(\tau (t))=0$ on a time scale , where a and q are real-valued rd-continuous positive functions on , α and β are ratios of odd positive integers, $\tau :\mathbb{T}\to \mathbb{T}$, $\tau (t)\le t$, $\mathrm{\forall }t\in \mathbb{T}$, and ${lim}_{t\to \mathrm{\infty }}\tau (t)=\mathrm{\infty }$. We establish some new sufficient conditions for this equation.

MSC:34K11, 39A10, 39A99.

1 Introduction

The study of dynamic equations on time scales, which goes back to Hilger [1], provides a rapidly expanding body of literature where the main idea is to unify, extend, and generalise concepts from continuous, discrete and quantum calculus to arbitrary time scales analysis, where a time scale is simply any nonempty closed subset of the reals.

For detailed information regarding calculus on time scales, we refer the reader to Bohner and Peterson [2, 3].

Now we consider the second-order delay nonlinear dynamic equation

$${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\beta }(\tau (t))=0$$

(1.1)

on an arbitrary time scale , where a and q are real-valued rd-continuous positive functions defined on ; α and $\beta >0$ are ratios of odd positive integers; $\tau :\mathbb{T}\to \mathbb{T}$, $\tau (t)\le t$, $\mathrm{\forall }t\in \mathbb{T}$ and ${lim}_{t\to \mathrm{\infty }}\tau (t)=\mathrm{\infty }$.

We shall also consider the case

$${\int }_{t_0}^{\mathrm{\infty }}\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}=\mathrm{\infty }.$$

(1.2)

Since we are interested in the oscillatory and asymptotic behavior of solutions near infinity, we assume that $sup\mathbb{T}=\mathrm{\infty }$, and we define the time scale interval ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$ by ${[t_0,\mathrm{\infty })}_{\mathbb{T}}:=[t_0,\mathrm{\infty })\cap \mathbb{T}$. By a solution of (1.1) we mean a nontrivial real-valued function $x\in C_{\mathrm{rd}}^1[T_x,\mathrm{\infty })$, $T_x\ge t_0$ which has the property that $a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha }\in C_{\mathrm{rd}}^1[T_x,\mathrm{\infty })$ and satisfies (1.1) on $[T_x,\mathrm{\infty })$; here $C_{\mathrm{rd}}$ is the space of rd-continuous functions. The solutions vanishing in some neighborhood of infinity will be excluded from our consideration. A solution x of (1.1) is said to be oscillatory if it is neither eventually positive nor eventually negative, otherwise it is nonoscillatory.

If $\mathbb{T}=\mathbb{R}$ then $\sigma (t)=t$, $\mu (t)=0$, $f^{\mathrm{\Delta }}(t)=f^{\prime }(t)$, ${\int }_a^{b}f(t)\mathrm{\Delta }t={\int }_a^{b}f(t)dt$ and when $\alpha =\beta$ (1.1) becomes the half-linear delay differential equation

$${(a(t){(x^{\prime }(t))}^{\alpha })}^{\prime }+q(t)x^{\alpha }(\tau (t))=0.$$

(1.3)

When $a(t)=1$ and $\alpha =1$, (1.3) becomes

$$x^{″}(t)+q(t)x(\tau (t))=0.$$

(1.4)

Ohriska [4] proved that every solution of (1.4) oscillates if

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}t{\int }_t^{\mathrm{\infty }}q(s)\left(\frac{\tau (s)}{s}\right)ds>1$$

(1.5)

holds. Agarwal et al. [5] considered (1.3) and extended the condition (1.5) and proved that if

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{t}^{\alpha }{\int }_t^{\mathrm{\infty }}q(s){\left(\frac{\tau (s)}{s}\right)}^{\alpha }ds>1,$$

(1.6)

then every solution of (1.3) oscillates.

If $\mathbb{T}=\mathbb{Z}$, then $\sigma (t)=t+1$, $\mu (t)=1$, $f^{\mathrm{\Delta }}(t)=\mathrm{\Delta }f(t)$, ${\int }_a^{b}f(t)\mathrm{\Delta }t={\sum }_{t=a}^{b-1}f(t)$, and (1.1) becomes the quasi-linear difference equation

$$\mathrm{\Delta }(a(t){(\mathrm{\Delta }x(t))}^{\alpha })+q(t)x^{\beta }(\tau (t))=0.$$

(1.7)

When $\alpha =\beta$, (1.1) becomes the half-linear delay dynamic equation which has been considered by some authors and some oscillation and nonoscillation results have been obtained [2, 3, 5–10]. As a special case of (1.1) Agarwal et al. [11] considered the second-order delay dynamic equations on time scales

$$x^{\mathrm{\Delta }\mathrm{\Delta }}(t)+q(t)x(\tau (t))=0$$

(1.8)

and established some sufficient conditions for oscillations of (1.8) when

$${\int }_{t_0}^{\mathrm{\infty }}\tau (t)q(t)\mathrm{\Delta }t=\mathrm{\infty }.$$

(1.9)

Saker [12] studied (1.8) to extend the result of Lomtatidze [13]. Recently Higgins [14] proved oscillation results for (1.8). Saker [15] examined oscillations for the half-linear dynamic equation

$${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\alpha }(t)=0$$

(1.10)

on time scales, where $\alpha >1$ is an odd positive integer and Agarwal et al. [6] and Grace et al. [10] studied oscillations for the same equation, (1.10), where $\alpha >1$ is the quotient of odd positive integers which cannot be applied when $0<\alpha \le 1$. Han et al. [16] and Hassan [17] solved this problem and improved Agarwal’s and Saker’s results. Erbe et al. [7] considered the half-linear delay dynamic equation

$${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\alpha }(\tau (t))=0$$

(1.11)

on time scales, where $\alpha >1$ is the quotient of odd positive integers and

$$a^{\mathrm{\Delta }}(t)\ge 0\text{and}{\int }_{t_0}^{\mathrm{\infty }}{\tau }^{\alpha }(t)q(t)\mathrm{\Delta }t=\mathrm{\infty }$$

(1.12)

and utilised a Riccati transformation technique and established some oscillation criteria for (1.11). Erbe et al. [8] considered the half-linear delay dynamic equation (1.11) on time scales, where $0<\alpha \le 1$ is the quotient of odd positive integers and established some sufficient conditions for oscillations when (1.12) holds. Han et al. [18] considered (1.11) and followed the proof that has been used in [15] and established some sufficient conditions for oscillations when $a^{\mathrm{\Delta }}(t)\ge 0$. For oscillations of quasi-linear dynamic equations, Grace et al. [19] considered the equation

$${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\beta }(t)=0;$$

(1.13)

and Saker and Grace [20] considered the delay equation

$${(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+q(t)x^{\beta }(\tau (t))=0$$

(1.14)

on an arbitrary time scale where $\tau (t)\le t$ or $\tau (t)\ge t$. The special case of (1.14) where $0<\alpha =\beta \le 1$ has been studied in [9] by Erbe et al.

In this paper, we establish some new sufficient conditions for oscillations of (1.1).

2 Preliminary result

For a function $f:\mathbb{T}\to \mathbb{R}$, the (delta) derivative $f^{\mathrm{\Delta }}(t)$ at $t\in \mathbb{T}$ is defined to be the number (if it exists) such that for all $ϵ>0$ there is a neighborhood U of t with

$$|f(\sigma (t))-f(s)-f^{\mathrm{\Delta }}(\sigma (t)-s)|\le ϵ|\sigma (t)-s|$$

(2.1)

for all $s\in U$. If the (delta) derivative $f^{\mathrm{\Delta }}(t)$ exists for all $t\in \mathbb{T}$, then we say that f is (delta) differentiable on . For two (delta) differentiable functions f and g, the derivative of the product fg and the quotient $f/g$ (where $g{g}^{\sigma }\ne 0$) are given as in [[2], Theorem 1.20]

$${(fg)}^{\mathrm{\Delta }}=f^{\mathrm{\Delta }}g+f^{\sigma }{g}^{\mathrm{\Delta }}=f{g}^{\mathrm{\Delta }}+f^{\mathrm{\Delta }}{g}^{\sigma },$$

(2.2)

$${\left(\frac{f}{g}\right)}^{\mathrm{\Delta }}=\frac{f^{\mathrm{\Delta }}g-f{g}^{\mathrm{\Delta }}}{g{g}^{\sigma }}$$

(2.3)

as well as of the chain rule [[2], Theorem 1.90] for the derivative of the composite function $fog$ for a continuously differentiable function $f:\mathbb{R}\to \mathbb{R}$ and a (delta) differentiable function $g:\mathbb{T}\to \mathbb{R}$

$${(fog)}^{\mathrm{\Delta }}=\left[{\int }_0^1{f}^{\prime }(g+h\mu g^{\mathrm{\Delta }})dh\right]g^{\mathrm{\Delta }}.$$

(2.4)

For $b,c\in \mathbb{T}$ and a differentiable function f, the Cauchy integral of $f^{\mathrm{\Delta }}$ is defined by

$${\int }_b^c{f}^{\mathrm{\Delta }}(t)\mathrm{\Delta }t=f(c)-f(b)$$

(2.5)

and the improper integral is defined as

$${\int }_b^{\mathrm{\infty }}f(t)\mathrm{\Delta }t=\underset{c\to \mathrm{\infty }}{lim}{\int }_b^{c}f(t)\mathrm{\Delta }t.$$

(2.6)

Lemma 2.1 If A and B are nonnegative real numbers and $\lambda >1$, then

$$A^{\lambda }-\lambda A{B}^{\lambda -1}+(\lambda -1)B^{\lambda }\ge 0,$$

where the equality holds if and only if $A=B$.

3 The main results

In this section, by employing a Riccati transformation technique, we establish oscillation criteria for (1.1). To prove our main result, we will use the formula

$${(x^{\alpha }(t))}^{\mathrm{\Delta }}=\alpha \left\{{\int }_0^1{[h{x}^{\sigma }(t)+(1-h)x(t)]}^{\alpha -1}dh\right\}{x}^{\mathrm{\Delta }}(t),$$

(3.1)

which is a simple consequence of the Pötzsche chain rule [2].

In the following theorems, we assume that (1.2) holds.

Theorem 3.1 Assume that there exists a positive nondecreasing delta differentiable function $\xi (t)$ such that, for all sufficiently large $T\ge t_0$, and for $\tau (t)>g(T)$, we have

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t[{\theta }^{\beta }(s,g(T))\xi (s)q(s)-\frac{{(\frac{\alpha }{\beta })}^{\alpha }a(s){({\xi }^{\mathrm{\Delta }}(s))}^{\alpha +1}}{{(\alpha +1)}^{\alpha +1}{(\xi (s){\delta }_1(s))}^{\alpha }}]\mathrm{\Delta }s=\mathrm{\infty },$$

(3.2)

where ${({\xi }^{\mathrm{\Delta }}(s))}_{+}:=max\{0,{\xi }^{\mathrm{\Delta }}(s)\}$,

$${\delta }_1(t)=\{\begin{array}{cc}{c}_1\mathit{\text{ is any positive constant}},\hfill & \mathit{\text{if }}\beta >\alpha ,\hfill \\ 1,\hfill & \mathit{\text{if }}\beta =\alpha ,\hfill \\ c_2{({\eta }^{\sigma }(t))}^{\frac{\alpha -\beta }{\alpha }},c_2\mathit{\text{ is any positive constant}},\hfill & \mathit{\text{if }}\beta <\alpha ,\hfill \end{array}$$

(3.3)

and

$$\theta (t,g(T))=\frac{{\int }_{g(T)}^{\tau (t)}\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}}{{\int }_{g(T)}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}},\eta (t)={\left({\int }_{t_1}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\right)}^{-1}.$$

(3.4)

Then every solution of (1.1) is oscillatory on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$.

Proof Let x be a nonoscillatory solution of (1.1) on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$. Then, without loss of generality, there is a $t_1\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, sufficiently large, such that $x(t)>0$ and $x(\tau (t))>0$ on ${[t_1,\mathrm{\infty })}_{\mathbb{T}}$. Since $a{(x^{\mathrm{\Delta }})}^{\alpha }$ is a strictly decreasing function, it is of one sign. We claim that $x^{\mathrm{\Delta }}>0$ on ${[t_1,\mathrm{\infty })}_{\mathbb{T}}$. If not, then there is a $t_2\in {[t_1,\mathrm{\infty })}_{\mathbb{T}}$ such that $x^{\mathrm{\Delta }}(t_2)<0$. Using the fact $a{(x^{\mathrm{\Delta }})}^{\alpha }$ is strictly decreasing, we get

$$x(t)\le x(t_2)+{[a(t_2){(x^{\mathrm{\Delta }}(t_2))}^{\alpha }]}^{1/\alpha }{\int }_{t_2}^t\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}\to -\mathrm{\infty },\text{as }t\to \mathrm{\infty },$$

which implies that $x(t)$ is eventually negative. This is a contradiction. Hence $x^{\mathrm{\Delta }}>0$ on ${[t_1,\mathrm{\infty })}_{\mathbb{T}}$. Consider the generalized Riccati substitution

$$w(t)=\frac{\xi (t)a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha }}{x^{\beta }(t)},$$

(3.5)

then $w(t)>0$. By the product rule and then the quotient rule, we have

$$\begin{array}{rcl}{w}^{\mathrm{\Delta }}(t)& =& \left(\frac{\xi (t)}{x^{\beta }(t)}\right){(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\mathrm{\Delta }}+{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }{\left(\frac{\xi (t)}{x^{\beta }(t)}\right)}^{\mathrm{\Delta }}\\ =& -q(t)\xi (t){\left(\frac{x(\tau (t))}{x(t)}\right)}^{\beta }+{\xi }^{\mathrm{\Delta }}(t)\frac{{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }}{x^{\beta }(\sigma (t))}\\ -\xi (t)\frac{{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }{(x^{\beta }(t))}^{\mathrm{\Delta }}}{x^{\beta }(t)x^{\beta }(\sigma (t))}.\end{array}$$

(3.6)

Using the fact that $x(t)$ is increasing and $a{(x^{\mathrm{\Delta }})}^{\alpha }$ is decreasing, we have

$$\begin{array}{rcl}x(t)-x(\tau (t))& =& {\int }_{\tau (t)}^t{\frac{(a(s){(x^{\mathrm{\Delta }}(s))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\\ \le & {\int }_{\tau (t)}^t{\frac{(a(\tau (t)){(x^{\mathrm{\Delta }}(\tau (t)))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\\ =& a^{\frac{1}{\alpha }}(\tau (t))x^{\mathrm{\Delta }}(\tau (t)){\int }_{\tau (t)}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\end{array}$$

and so

$$\frac{x(t)}{x(\tau (t))}\le 1+\frac{a^{\frac{1}{\alpha }}(\tau (t))x^{\mathrm{\Delta }}(\tau (t))}{x(\tau (t))}{\int }_{\tau (t)}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}$$

(3.7)

for $t\ge t_1$. Also, we see that

$$\begin{array}{c}\begin{array}{rl}x(\tau (t))& >x(\tau (t))-x(g(T))={\int }_{g(T)}^{\tau (t)}{\frac{(a(s){(x^{\mathrm{\Delta }}(s))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\\ \ge a^{\frac{1}{\alpha }}(\tau (t))x^{\mathrm{\Delta }}(\tau (t)){\int }_{g(T)}^{\tau (t)}\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)},\end{array}\hfill \\ \frac{a^{\frac{1}{\alpha }}(\tau (t))x^{\mathrm{\Delta }}(\tau (t))}{x(\tau (t))}\le {\left({\int }_{g(T)}^{\tau (t)}\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\right)}^{-1}.\hfill \end{array}$$

(3.8)

Therefore, (3.7) and (3.8) imply that

$$\frac{x(\tau (t))}{x(t)}\ge \frac{{\int }_{g(T)}^{\tau (t)}\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}}{{\int }_{g(T)}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}}=\theta (t,g(T)).$$

(3.9)

In view of (3.1), we get

$$\begin{array}{rcl}{(x^{\beta }(t))}^{\mathrm{\Delta }}& =& \beta x^{\mathrm{\Delta }}(t){\int }_0^1{\left[(x(t)+h\mu (t)x^{\mathrm{\Delta }}(t))\right]}^{\beta -1}dh\\ \ge & \{\begin{array}{cc}\beta {(x^{\sigma }(t))}^{\beta -1}{x}^{\mathrm{\Delta }}(t),\hfill & 0<\beta \le 1,\hfill \\ \beta {(x(t))}^{\beta -1}{x}^{\mathrm{\Delta }}(t),\hfill & \beta >1.\hfill \end{array}\end{array}$$

(3.10)

From (3.9), (3.10), the fact that $x(t)$ is an increasing function and the definition of $w(t)$, if $0<\beta \le 1$, we have

$$\begin{array}{rcl}{w}^{\mathrm{\Delta }}(t)& \le & -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\beta \xi (t)\frac{{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }}{{(x^{\sigma }(t))}^{\beta +1}}{x}^{\mathrm{\Delta }}(t)\\ =& -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\beta \xi (t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\frac{x^{\mathrm{\Delta }}(t)}{x^{\sigma }(t)},\end{array}$$

whereas, if $\beta >1$, we have

$$\begin{array}{rcl}{w}^{\mathrm{\Delta }}(t)& \le & -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\beta \xi (t)\frac{{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }}{{(x^{\sigma }(t))}^{\beta +1}}{x}^{\mathrm{\Delta }}(t)\\ =& -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\beta \xi (t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\frac{x^{\mathrm{\Delta }}(t)}{x^{\sigma }(t)}.\end{array}$$

Using the fact that $a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha }$ is strictly decreasing on ${[t_1,\mathrm{\infty })}_{\mathbb{T}}$, we find

$$\frac{x^{\mathrm{\Delta }}(t)}{x^{\sigma }(t)}\ge \frac{{({(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma })}^{\frac{1}{\alpha }}}{a^{\frac{1}{\alpha }}(t){(x^{\sigma }(t))}^{\frac{\beta }{\alpha }}}{(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}={\left(\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\right)}^{\frac{1}{\alpha }}\frac{{(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}}{a^{\frac{1}{\alpha }}(t)}.$$

Thus for $\beta >0$, we have

$$\begin{array}{rl}{w}^{\mathrm{\Delta }}(t)\le & -q(t)\xi (t){\theta }^{\beta }(t,g(T))\\ +{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\frac{\beta \xi (t)}{a^{\frac{1}{\alpha }}(t)}{\left(\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\right)}^{\frac{\alpha +1}{\alpha }}{(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}.\end{array}$$

(3.11)

We consider the following three cases:

Case (i). $\beta >\alpha$.

In this case, since $x^{\mathrm{\Delta }}(t)>0$, there exists $t_2\ge t_1$ such that $x^{\sigma }(t)\ge x(t)\ge c>0$. This implies that ${(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}\ge c_1$, where $c_{{}^1}=c^{\frac{\beta -\alpha }{\alpha }}$.

Case (ii). $\beta =\alpha$.

In this case, we see that ${(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}=1$.

Case (iii). $\beta <\alpha$.

Then there exists a positive constant $b:=a(t_1){(x^{\mathrm{\Delta }}(t_1))}^{\alpha }$. Using the decreasing of $a{(x^{\mathrm{\Delta }})}^{\alpha }$, we have

$$x^{\mathrm{\Delta }}(t)\le b^{\frac{1}{\alpha }}{a}^{\frac{-1}{\alpha }}(t)\text{for }t\ge t_1.$$

Integrating this inequality from $t_1$ to t, we have

$$x(t)\le x(t_1)+b^{\frac{1}{\alpha }}{\int }_{t_1}^t{a}^{\frac{-1}{\alpha }}(s)\mathrm{\Delta }s.$$

Thus, there exist a constant $b_1>0$ and $t_2\ge t_1$ such that

$$x(t)\le b_1{\eta }^{-1}(t)\text{for }t\ge t_2$$

and hence

$${(x^{\sigma }(t))}^{\frac{\beta -\alpha }{\alpha }}\ge c_2{\left({\eta }^{\sigma }\right)}^{\frac{\alpha -\beta }{\alpha }}(t)\text{for }t\ge t_2,$$

where $c_2=b_1^{\frac{\beta -\alpha }{\alpha }}$.

By (3.3), we get

$$w^{\mathrm{\Delta }}(t)\le -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}-\beta a^{-\frac{1}{\alpha }}(t)\xi (t){\left(\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\right)}^{\frac{\alpha +1}{\alpha }}{\delta }_1(t)$$

(3.12)

for $t\ge t_2$. Defining $A\ge 0$ and $B\ge 0$ by

$$\begin{array}{c}A={(\beta {\delta }_1(t)\xi (t))}^{\frac{\alpha }{\alpha +1}}{a}^{\frac{-1}{\alpha +1}}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)},\hfill \\ B={\left(\frac{\alpha }{\alpha +1}\right)}^{\alpha }{\left(\frac{{\xi }^{\mathrm{\Delta }}(t)}{{\xi }^{\sigma }(t)}\right)}^{\alpha }{[{(\beta {\delta }_1(t)\xi (t))}^{\frac{-\alpha }{\alpha +1}}{\xi }^{\sigma }(t)a^{\frac{1}{\alpha +1}}(t)]}^{\alpha }\hfill \end{array}$$

and using Lemma 2.1 we get

$$\beta {\delta }_1(t)\xi (t)a^{\frac{-1}{\alpha }}(t){\left(\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\right)}^{\frac{\alpha +1}{\alpha }}-{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}\ge -\frac{{(\frac{\alpha }{\beta })}^{\alpha }a(t){({\xi }^{\mathrm{\Delta }}(t))}^{\alpha +1}}{{(\alpha +1)}^{\alpha +1}{({\delta }_1(t)\xi (t))}^{\alpha }},$$

where $\lambda =\frac{\alpha +1}{\alpha }$. Therefore, by (3.12), we have

$$w^{\mathrm{\Delta }}(t)\le -q(t)\xi (t){\theta }^{\beta }(t,g(T))+\frac{{(\frac{\alpha }{\beta })}^{\alpha }a(t){({\xi }^{\mathrm{\Delta }}(t))}^{\alpha +1}}{{(\alpha +1)}^{\alpha +1}{({\delta }_1(t)\xi (t))}^{\alpha }}.$$

(3.13)

Integrating (3.13) from $t_2$ to t, we get as $t\to \mathrm{\infty }$

$$\begin{array}{rcl}w(t)& \le & w(t_2)-{\int }_{t_2}^t[q(s)\xi (s){\theta }^{\beta }(s,g(T))-\frac{{(\frac{\alpha }{\beta })}^{\alpha }a(s){({\xi }^{\mathrm{\Delta }}(s))}^{\alpha +1}}{{(\alpha +1)}^{\alpha +1}{({\delta }_1(s)\xi (s))}^{\alpha }}]\mathrm{\Delta }s\\ =& -\mathrm{\infty },\end{array}$$

which contradicts (3.5). □

Theorem 3.2 Assume that there exists a positive nondecreasing delta differentiable function $\xi (t)$ such that, for all sufficiently large $T\ge t_0$, and for $\tau (t)>g(T)$, we have

$$\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_{t_0}^t[q(s)\xi (s){\theta }^{\beta }(s,g(T))-{\delta }_2(s){\eta }^{\alpha }(s){\xi }^{\mathrm{\Delta }}(s)]\mathrm{\Delta }s=\mathrm{\infty },$$

(3.14)

where

$${\delta }_2(t)=\{\begin{array}{cc}{c}_1\mathit{\text{ is any positive constant}},\hfill & \mathit{\text{if }}\beta >\alpha ,\hfill \\ 1,\hfill & \mathit{\text{if }}\beta =\alpha ,\hfill \\ c_2{\eta }^{\beta -\alpha }(t),c_2\mathit{\text{ is any positive constant}},\hfill & \mathit{\text{if }}\beta <\alpha ,\hfill \end{array}$$

(3.15)

and θ is as in Theorem 3.1. Then (1.1) is oscillatory on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$.

Proof Let x be a nonoscillatory solution of (1.1) on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$. Then, without loss of generality, there is a $t_1\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, sufficiently large, such that $x(t)>0$ and $x(\tau (t))>0$ on $[t_1,\mathrm{\infty })$. Proceeding as in the proof of Theorem 3.1, we obtain (3.11). Therefore

$$w^{\mathrm{\Delta }}(t)\le -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{w^{\sigma }(t)}{{\xi }^{\sigma }(t)}.$$

(3.16)

Using the definition of $w(t)$, it follows from (3.16) that

$$\begin{array}{rcl}{w}^{\mathrm{\Delta }}(t)& \le & -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}^{\sigma }}{x^{\beta }(\sigma (t))}\\ \le & -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)\frac{a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha }}{x^{\beta }(t)}\\ =& -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t)a(t){\left(\frac{x^{\mathrm{\Delta }}(t)}{x(t)}\right)}^{\alpha }{x}^{\alpha -\beta }(t).\end{array}$$

(3.17)

Now, from

$$\begin{array}{rcl}x(t)& =& x(t_1)+{\int }_{t_1}^t{\frac{(a(s){(x^{\mathrm{\Delta }}(s))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\\ \ge & a^{\frac{1}{\alpha }}(t)x^{\mathrm{\Delta }}(t){\int }_{t_1}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\end{array}$$

it follows that

$${\left(\frac{x^{\mathrm{\Delta }}(t)}{x(t)}\right)}^{\alpha }\le a^{-1}(t){\left({\int }_{t_1}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\right)}^{-\alpha }=\frac{{\eta }^{\alpha }(t)}{a(t)}.$$

(3.18)

Using (3.18) in (3.17), we have

$$w^{\mathrm{\Delta }}(t)\le -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\xi }^{\mathrm{\Delta }}(t){(\eta (t))}^{\alpha }{x}^{\alpha -\beta }(t).$$

(3.19)

Next as in the proof of Theorem 3.1, we consider the Cases (i), (ii) and (iii).

Case (i). $\beta >\alpha$.

In this case, since $x^{\mathrm{\Delta }}(t)>0$, there exists $t_2\ge t_1$ such that $x^{\sigma }(t)\ge x(t)\ge c>0$. This implies that $x^{\alpha -\beta }(t)\le c_1$, where $c_{{}^{{}_1}}=c^{\alpha -\beta }$.

Case (ii). $\beta =\alpha$.

In this case, we see that $x^{\alpha -\beta }(t)=1$.

Case (iii). $\beta <\alpha$.

Proceeding as in the proof of Theorem 3.1, there exist a constant $b_1>0$ and $t_2\ge t_1$ such that

$$x(t)\le b_1{\eta }^{-1}(t)\text{for }t\ge t_2$$

and hence

$$x^{\alpha -\beta }(t)\ge c_2{\eta }^{\beta -\alpha }(t)\text{for }t\ge t_2,$$

where $c_2=b_1^{\alpha -\beta }$.

Using these three cases in (3.19) and the definition of ${\delta }_2(t)$, we get

$$w^{\mathrm{\Delta }}(t)\le -q(t)\xi (t){\theta }^{\beta }(t,g(T))+{\delta }_2(t){(\eta (t))}^{\alpha }{\xi }^{\mathrm{\Delta }}(t)$$

for $t\ge t_2$. Integrating the above inequality from $t_2$ to t, we have

$$0<w(t)\le w(t_2)-{\int }_{t_2}^t[q(s)\xi (s){\theta }^{\beta }(s,g(T))-{\delta }_2(s){\eta }^{\alpha }(s){\xi }^{\mathrm{\Delta }}(s)]\mathrm{\Delta }s$$

which gives a contradiction using (3.14). □

We next state and prove a Philos-type oscillation criterion for (1.1).

Theorem 3.3 Assume that there exist functions H and h such that for each fixed t, $H(t,s)$ and $h(t,s)$ are rd-continuous with respect to s on $D=\{(t,s),t\ge s\ge t_0\}$ such that

$$H(t,t)=0,t\ge t_0,H(t,s)>0,t>s\ge t_0$$

(3.20)

and H has a non-positive continuous Δ-partial derivative $H^{{\mathrm{\Delta }}_s}(t,s)$ with respect to the second variable, and that it satisfies

$$-H^{{\mathrm{\Delta }}_s}(t,s)-H(t,s)\frac{{\xi }^{\mathrm{\Delta }}(s)}{{\xi }^{\sigma }(s)}=\frac{h(t,s)}{{\xi }^{\sigma }(s)}{(H(t,s))}^{\frac{\alpha }{\alpha +1}}$$

(3.21)

and for all sufficiently large $T\ge t_0$, and for $\mathrm{\forall }t\ge T$, $\tau (t)>g(T)$, we have

$$\begin{array}{r}\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{1}{H(t,T)}\\ \times {\int }_T^t[q(s)\xi (s){\theta }^{\beta }(s,g(T))H(t,s)-\frac{{(\frac{\alpha }{\beta })}^{\alpha }{(h_{-}(t,s))}^{\alpha +1}a(s)}{{(\alpha +1)}^{\alpha +1}{\xi }^{\alpha }(s){({\delta }_1(s))}^{\alpha }}]\mathrm{\Delta }s\\ =\mathrm{\infty },\end{array}$$

(3.22)

where $\xi (s)$ is a positive Δ-differentiable function and $h_{-}(t,s):=max\{0,-h(t,s)\}$. Then every solution of (1.1) is oscillatory on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$.

Proof Let x be a nonoscillatory solution of (1.1) on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$. Then, without loss of generality, there is a $t_1\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, sufficiently large, such that $x(t)>0$ and $x(\tau (t))>0$ on $[t_1,\mathrm{\infty })$. Again we define $w(t)$ function as in the proof of Theorem 3.1. Then, proceeding as in the proof of Theorem 3.1, we obtain (3.12). Multiplying both sides of inequality (3.12), with t replaced by s, by $H(t,s)$ and integrating with respect to s from $t_2$ to t, we get

$$\begin{array}{r}{\int }_{t_2}^{t}H(t,s)q(s)\xi (s){\theta }^{\beta }(s,g(T))\mathrm{\Delta }s\\ \le -{\int }_{t_2}^{t}H(t,s)w^{\mathrm{\Delta }}(s)\mathrm{\Delta }s+{\int }_{t_2}^{t}H(t,s){\xi }^{\mathrm{\Delta }}(s)\frac{w^{\sigma }(s)}{{\xi }^{\sigma }(s)}\mathrm{\Delta }s\\ -\beta {\int }_{t_2}^{t}H(t,s)a^{-\frac{1}{\alpha }}(s)\xi (s){\left(\frac{w^{\sigma }(s)}{{\xi }^{\sigma }(s)}\right)}^{\frac{\alpha +1}{\alpha }}{\delta }_1(s)\mathrm{\Delta }s.\end{array}$$

(3.23)

Integrating (3.23) by parts and using (3.20) and (3.21), we obtain

$$\begin{array}{r}{\int }_{t_2}^{t}H(t,s){\theta }^{\beta }(s,g(T))q(s)\xi (s)\mathrm{\Delta }s\\ \le H(t,t_2)w(t_2)+{\int }_{t_2}^t\frac{h(t,s)}{{\xi }^{\sigma }(s)}{(H(t,s))}^{\frac{\alpha }{\alpha +1}}{w}^{\sigma }(s)\mathrm{\Delta }s\\ -\beta {\int }_{t_2}^{t}H(t,s)a^{-\frac{1}{\alpha }}(s)\xi (s){\left(\frac{w^{\sigma }(s)}{{\xi }^{\sigma }(s)}\right)}^{\frac{\alpha +1}{\alpha }}{\delta }_1(s)\mathrm{\Delta }s.\end{array}$$

(3.24)

Again, defining $A\ge 0$ and $B\ge 0$ by

$$A={\left[\frac{\beta H(t,s){\delta }_1(s)\xi (s)}{a^{1/\alpha }(s)}\right]}^{\frac{\alpha }{\alpha +1}}\frac{w^{\sigma }(s)}{{\xi }^{\sigma }(s)},B=\frac{{(h_{-}(t,s))}^{\alpha }{a}^{\frac{\alpha }{\alpha +1}}(s)}{{(\frac{\alpha +1}{\alpha })}^{\alpha }{(\beta {\delta }_1(s)\xi (s))}^{\frac{{\alpha }^2}{\alpha +1}}}$$

and using Lemma 2.1 where $\lambda =\frac{\alpha +1}{\alpha }$, we get

$$\begin{array}{r}\frac{\beta H(t,s)\xi (s){\delta }_1(s){(w^{\sigma }(s))}^{\frac{\alpha +1}{\alpha }}}{a^{1/\alpha }(s){({\xi }^{\sigma }(s))}^{\frac{\alpha +1}{\alpha }}}-\frac{h_{-}(t,s)}{{\xi }^{\sigma }(s)}{(H(t,s))}^{\frac{\alpha }{\alpha +1}}{w}^{\sigma }(s)\\ \ge \frac{\frac{1}{\alpha }{(h_{-}(t,s))}^{\alpha +1}a(s)}{{\beta }^{\alpha }{(\frac{\alpha +1}{\alpha })}^{\alpha +1}{\xi }^{\alpha }(s){({\delta }_1(s))}^{\alpha }}.\end{array}$$

(3.25)

From (3.24) and (3.25), we have

$$\frac{1}{H(t,t_2)}{\int }_{t_2}^t[H(t,s){\theta }^{\beta }(s,g(T))q(s)\xi (s)-\frac{{(\frac{\alpha }{\beta })}^{\alpha }{(h_{-}(t,s))}^{\alpha +1}a(s)}{{(\alpha +1)}^{\alpha +1}{\xi }^{\alpha }(s){({\delta }_1(s))}^{\alpha }}]\mathrm{\Delta }s\le w(t_2),$$

which contradicts assumption (3.22).

Now we introduce the following notation for $T\in {[t_1,\mathrm{\infty })}_{\mathbb{T}}$. For all sufficiently large T such that $g(T)<\tau (T)$,

$$p_{\ast }:=\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{t^{\alpha }}{a(t)}{\int }_{\sigma (t)}^{\mathrm{\infty }}{\theta }^{\beta }(s,g(T))q(s)\mathrm{\Delta }s,$$

(3.26)

$$q_{\ast }:=\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{1}{t}{\int }_T^t\frac{s^{\alpha +1}}{a(s)}{\theta }^{\beta }(s,g(T))q(s)\mathrm{\Delta }s,$$

(3.27)

$$r_{\ast }:=\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{t^{\alpha }{w}^{\sigma }(t)}{a(t)},$$

(3.28)

$$R_{\ast }:=\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{t^{\alpha }{w}^{\sigma }(t)}{a(t)}.$$

(3.29)

Assume that $l={lim\hspace{0.17em}inf}_{t\to \mathrm{\infty }}\frac{t}{\sigma (t)}$. Note that $0\le l\le 1$. We assume that

$${\int }_T^{\mathrm{\infty }}{\theta }^{\beta }(s,g(T))q(s)\mathrm{\Delta }s<\mathrm{\infty },T\in {[t_1,\mathrm{\infty })}_{\mathbb{T}}.$$

(3.30)

 □

Theorem 3.4 Let $\beta \ge \alpha$ and assume $a(t)$ is a delta differentiable function such that $a^{\mathrm{\Delta }}(t)\ge 0$ and (3.30) holds. Furthermore, assume $l>0$ and

$$p_{\ast }>\frac{{(\frac{{\alpha }^2}{\beta c_1})}^{\alpha }}{{(\alpha +1)}^{\alpha +1}{l}^{{\alpha }^2}}$$

(3.31)

or

$$p_{\ast }+q_{\ast }>{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }\frac{1}{l^{\alpha (\alpha +1)}}$$

(3.32)

for all large T. Then every solution of (1.1) is oscillatory on ${[t_0,\mathrm{\infty })}_{\mathbb{T}}$.

Proof Let x be a nonoscillatory solution of (1.1). Without loss of generality we assume that there is a $t_1\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, sufficiently large, such that $x(t)>0$ and $x(\tau (t))>0$ on $[t_1,\mathrm{\infty })$. Again we define $w(t)$ as in the proof of Theorem 3.1 by putting $\xi (t)=1$ and $\delta (t)=c_1$. Proceeding as in the proof of Theorem 3.1, we obtain from (3.12)

$$w^{\mathrm{\Delta }}(t)\le -q(t){\theta }^{\beta }(t,g(T))-\frac{\beta c_1{(w^{\sigma }(t))}^{\frac{\alpha +1}{\alpha }}}{a^{\frac{1}{\alpha }}(t)}.$$

(3.33)

First, we assume that inequality (3.31) holds. It follows from (3.5) and $a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha }$ being strictly decreasing that

$$\begin{array}{rcl}x(t)& =& x(t_0)+{\int }_{t_0}^t{\frac{(a(s){(x^{\mathrm{\Delta }}(s))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\\ \ge & x(t_0)+{\int }_{t_0}^t{\frac{(a(t){(x^{\mathrm{\Delta }}(t))}^{\alpha })}{a^{\frac{1}{\alpha }}(s)}}^{\frac{1}{\alpha }}\mathrm{\Delta }s\ge a^{\frac{1}{\alpha }}(t)x^{\mathrm{\Delta }}(t){\int }_{t_0}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\end{array}$$

and it follows that

$$a(t){\left(\frac{x^{\mathrm{\Delta }}(t)}{x(t)}\right)}^{\alpha }\le {\left({\int }_{t_0}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\right)}^{-\alpha }$$

and hence

$$w(t)=a(t){\left(\frac{x^{\mathrm{\Delta }}(t)}{x(t)}\right)}^{\alpha }{x}^{\alpha -\beta }(t)\le c_1{\left({\int }_{t_0}^t\frac{\mathrm{\Delta }s}{a^{\frac{1}{\alpha }}(s)}\right)}^{-\alpha }.$$

(3.34)

Using (1.2), we have ${lim}_{t\to \mathrm{\infty }}w(t)=0$. Integrating (3.33) from $\sigma (t)$ to ∞ and using ${lim}_{t\to \mathrm{\infty }}w(t)=0$, we have

$$w^{\sigma }(t)\ge {\int }_{\sigma (t)}^{\mathrm{\infty }}q(s){\theta }^{\beta }(s,g(T))\mathrm{\Delta }s+\beta c_1{\int }_{\sigma (t)}^{\mathrm{\infty }}\frac{{(w^{\sigma }(s))}^{\frac{\alpha +1}{\alpha }}}{a^{\frac{1}{\alpha }}(s)}\mathrm{\Delta }s.$$

(3.35)

Multiplying (3.35) by $\frac{t^{\alpha }}{a(t)}$ we get

$$\frac{t^{\alpha }}{a(t)}{w}^{\sigma }(t)\ge \frac{t^{\alpha }}{a(t)}{\int }_{\sigma (t)}^{\mathrm{\infty }}q(s){\theta }^{\beta }(s,g(T))\mathrm{\Delta }s+\beta c_1\frac{t^{\alpha }}{a(t)}{\int }_{\sigma (t)}^{\mathrm{\infty }}\frac{{(w^{\sigma }(s))}^{\frac{\alpha +1}{\alpha }}}{a^{\frac{1}{\alpha }}(s)}\mathrm{\Delta }s.$$

(3.36)

Let $0<ϵ<l$. Then by the definition of $p_{\ast }$ and $r_{\ast }$ we can pick $t_1\in {[T,\mathrm{\infty })}_{\mathbb{T}}$ sufficiently large, so we have

$$\begin{array}{rcl}\frac{t^{\alpha }{w}^{\sigma }(t)}{a(t)}& \ge & (p_{\ast }-ϵ)+\beta c_1\frac{t^{\alpha }}{a(t)}{\int }_{\sigma (t)}^{\mathrm{\infty }}\frac{s{(w^{\sigma }(s))}^{\frac{1}{\alpha }}{s}^{\alpha }{w}^{\sigma }(s)a(s)}{a^{\frac{1}{\alpha }}(s)a(s)s^{\alpha +1}}\mathrm{\Delta }s\\ \ge & (p_{\ast }-ϵ)+\beta c_1{(r_{\ast }-ϵ)}^{\frac{\alpha +1}{\alpha }}\frac{t^{\alpha }}{a(t)}{\int }_{\sigma (t)}^{\mathrm{\infty }}\frac{a(s)}{s^{\alpha +1}}\mathrm{\Delta }s\\ \ge & (p_{\ast }-ϵ)+\frac{\beta c_1}{\alpha }{(r_{\ast }-ϵ)}^{\frac{\alpha +1}{\alpha }}{t}^{\alpha }{\int }_{\sigma (t)}^{\mathrm{\infty }}\frac{\alpha \mathrm{\Delta }s}{s^{\alpha +1}};\end{array}$$

(3.37)

for $\sigma (t)\in {[t_1,\mathrm{\infty })}_{\mathbb{T}}$. Using the Pötzsche chain rule we get

$$\begin{array}{rcl}{\left(\frac{-1}{s^{\alpha }}\right)}^{\mathrm{\Delta }}& =& \alpha {\int }_0^1\frac{1}{{[s+\mu h]}^{\alpha +1}}dh\\ \le & {\int }_0^1\left(\frac{\alpha }{s^{\alpha +1}}\right)dh=\frac{\alpha }{s^{\alpha +1}}.\end{array}$$

(3.38)

Then from (3.37) and (3.38), we have $t\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}$, and

$$\frac{t^{\alpha }{w}^{\sigma }(t)}{a(t)}\ge (p_{\ast }-ϵ)+\frac{\beta c_1}{\alpha }{(r_{\ast }-ϵ)}^{\frac{\alpha +1}{\alpha }}{\left(\frac{t}{\sigma (t)}\right)}^{\alpha }.$$

Taking the lim inf of both sides as $t\to \mathrm{\infty }$ we get

$$r_{\ast }\ge p_{\ast }-ϵ+\frac{\beta c_1}{\alpha }{(r_{\ast }-ϵ)}^{\frac{\alpha +1}{\alpha }}{l}^{\alpha }.$$

Since $ϵ>0$ is arbitrary, we get

$$p_{\ast }\le r_{\ast }-\frac{\beta c_1}{\alpha }{(r_{\ast })}^{\lambda }{l}^{\alpha },$$

(3.39)

where $\lambda =\frac{\alpha +1}{\alpha }$. By using Lemma 2.1, with

$$A:={\left(\frac{\beta c_1}{\alpha }\right)}^{\frac{\alpha }{\alpha +1}}{r}_{\ast }{l}^{\frac{{\alpha }^2}{\alpha +1}}\text{and}B:=\frac{1}{{(\frac{\alpha +1}{\alpha })}^{\alpha }{(\frac{\beta c_1}{\alpha })}^{\frac{{\alpha }^2}{\alpha +1}}{l}^{\frac{{\alpha }^3}{\alpha +1}}}$$

we get

$$\frac{\beta c_1}{\alpha }{(r_{\ast })}^{\frac{\alpha +1}{\alpha }}{l}^{\alpha }-r_{\ast }\ge -\frac{{(\frac{{\alpha }^2}{\beta c_1})}^{\alpha }}{{(\alpha +1)}^{\alpha +1}{l}^{{\alpha }^2}}.$$

(3.40)

It follows from (3.39) and (3.40) that

$$p_{\ast }\le \frac{{(\frac{{\alpha }^2}{\beta c_1})}^{\alpha }}{{(\alpha +1)}^{\alpha +1}{l}^{{\alpha }^2}},$$

which contradicts (3.31). Next, we assume that (3.32) holds. Multiplying both sides of (3.33) by $\frac{t^{\alpha +1}}{a(t)}$, and integrating from T to t ($t\ge T$) we get

$${\int }_T^t\frac{s^{\alpha +1}{w}^{\mathrm{\Delta }}(s)}{a(s)}\mathrm{\Delta }s\le -{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s-\beta c_1{\int }_T^t\frac{s^{\alpha +1}{(w^{\sigma }(s))}^{\frac{\alpha +1}{\alpha }}}{a^{\frac{\alpha +1}{\alpha }}(s)}\mathrm{\Delta }s.$$

Using integration by parts, the quotient rule and applying the Pötzsche chain rule, we obtain

$$\begin{array}{rcl}\frac{t^{\alpha +1}w(t)}{a(t)}& \le & \frac{T^{\alpha +1}w(T)}{a(T)}+(\alpha +1){\int }_T^t{\frac{(\sigma (s))}{a(s)}}^{\alpha }{w}^{\sigma }(s)\mathrm{\Delta }s\\ -{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s\\ -\beta c_1{\int }_T^t{\left(\frac{s^{\alpha }{w}^{\sigma }(s)}{a(s)}\right)}^{\frac{\alpha +1}{\alpha }}\mathrm{\Delta }s.\end{array}$$

Let $0<ϵ<l$ be given. Then using the definition of l, we can assume without loss of generality that T is sufficiently large so that

$$\frac{s}{\sigma (s)}\ge l-ϵ,s\ge t.$$

It follows that

$$\sigma (s)\le Ls,s\ge T,\text{where }L:=\frac{1}{l-ϵ}>0.$$

Therefore

$$\begin{array}{rcl}\frac{t^{\alpha +1}w(t)}{a(t)}& \le & \frac{T^{\alpha +1}w(T)}{a(T)}-{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s\\ +{\int }_T^t[(\alpha +1)\frac{L^{\alpha }{s}^{\alpha }{w}^{\sigma }(s)}{a(s)}-\beta c_1{\left(\frac{s^{\alpha }{w}^{\sigma }(s)}{a(s)}\right)}^{\frac{\alpha +1}{\alpha }}]\mathrm{\Delta }s.\end{array}$$

(3.41)

It follows that

$$\begin{array}{rcl}\frac{t^{\alpha +1}w(t)}{a(t)}& \le & \frac{T^{\alpha +1}w(T)}{a(T)}-{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s\\ +{\int }_T^t[(\alpha +1)L^{\alpha }u(s)-\beta c_1{u}^{\lambda }(s)]\mathrm{\Delta }s,\end{array}$$

where $u(s)=\frac{s^{\alpha }{w}^{\sigma }(s)}{a(s)}$ and $\lambda =\frac{\alpha +1}{\alpha }$. Using the inequality of Lemma 2.1 with

$$A^{\lambda }=\beta c_1{u}^{\lambda }(s)\text{and}{B}^{\lambda -1}=\frac{\alpha }{{(\beta c_1)}^{\frac{1}{\lambda }}}{L}^{\alpha }$$

we get

$$\begin{array}{rcl}\beta c_1{u}^{\lambda }(s)-(\alpha +1)L^{\alpha }u(s)& \ge & -(\lambda -1){\left(\frac{\alpha }{{(\beta c_1)}^{\frac{1}{\lambda }}}{L}^{\alpha }\right)}^{\frac{\lambda }{\lambda -1}}\\ =& -\frac{1}{\alpha }{\left(\frac{\alpha }{{(\beta c_1)}^{\frac{1}{\lambda }}}{L}^{\alpha }\right)}^{\alpha +1}\\ =& -{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }{L}^{\alpha (\alpha +1)}.\end{array}$$

(3.42)

Then we have

$$\frac{t^{\alpha +1}w(t)}{a(t)}\le \frac{T^{\alpha +1}w(T)}{a(T)}-{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s+{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }{L}^{\alpha (\alpha +1)}(t-T).$$

Since $w^{\mathrm{\Delta }}(t)\le 0$ and dividing the last inequality by t we have

$$\frac{t^{\alpha }{w}^{\sigma }(t)}{a(t)}\le \frac{T^{\alpha +1}w(T)}{ta(T)}-\frac{1}{t}{\int }_T^t\frac{s^{\alpha +1}q(s){\theta }^{\beta }(s,g(T))}{a(s)}\mathrm{\Delta }s+{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }{L}^{\alpha (\alpha +1)}(1-\frac{T}{t}).$$

Taking the lim sup of both sides as $t\to \mathrm{\infty }$ we obtain

$$R_{\ast }\le -q_{\ast }+{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }{L}^{\alpha (\alpha +1)}.$$

Since $0<ϵ<l$ is arbitrary, we get

$$R_{\ast }\le -q_{\ast }+{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }\frac{1}{l^{\alpha (\alpha +1)}}.$$

Using (3.39), we have

$$p_{\ast }\le r_{\ast }-\frac{\beta c_1}{\alpha }{(r_{\ast })}^{\frac{\alpha +1}{\alpha }}{l}^{\alpha }\le r_{\ast }\le R_{\ast }\le -q_{\ast }+{\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }\frac{1}{l^{\alpha (\alpha +1)}}.$$

Therefore

$$p_{\ast }+q_{\ast }\le {\left(\frac{\alpha }{\beta c_1}\right)}^{\alpha }\frac{1}{l^{\alpha (\alpha +1)}},$$

which contradicts (3.32). □

4 Examples

In this section, we give some examples to illustrate our main results. To obtain the conditions for oscillations, we will use the following fact:

$${\int }_{t_0}^{\mathrm{\infty }}\frac{\mathrm{\Delta }t}{t^p}=\mathrm{\infty },0\le p\le 1,t\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}.$$

Example 4.1 Consider the second-order delay half-linear dynamic equation on times scales

$${\left(\frac{1}{{}^{{(t+\sigma (t))}^3}}{(x^{\mathrm{\Delta }}(t))}^3\right)}^{\mathrm{\Delta }}+\frac{{(t+1)}^3}{t^2}{x}^3(\sqrt{t})=0,t\in {[1,\mathrm{\infty })}_{\mathbb{T}},$$

(4.1)

where $a(t)=\frac{1}{{}^{{(t+\sigma (t))}^3}}$, $q(t)=\frac{{(t+1)}^3}{t^2}$, $\tau (t)=\sqrt{t}$ and $\alpha =\beta =3$. We see that

$${\int }_{t_0}^{\mathrm{\infty }}\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}={\int }_{t_0}^{\mathrm{\infty }}(t+\sigma (t))\mathrm{\Delta }t={\int }_{t_0}^{\mathrm{\infty }}{\left(t^2\right)}^{\mathrm{\Delta }}\mathrm{\Delta }t=\mathrm{\infty },t\in {[t_0,\mathrm{\infty })}_{\mathbb{T}}.$$

Taking $T\ge t_0$ such that $g(T)=1$, we get $\mathrm{\forall }t>t_0$, $\tau (t)>1$. Then

$$\theta (t,g(T))=\frac{{\int }_1^{t^{1/2}}\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}}{{\int }_1^t\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}}=\frac{t-1}{t^2-1}=\frac{1}{t+1}.$$

Then, taking $\xi (t)=t$ by Theorem 3.1, we have

$$\begin{array}{r}\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_T^t[{\theta }^{\alpha }(s,1)\xi (s)q(s)-\frac{a(s){({\xi }^{\mathrm{\Delta }}{(s)}_{+})}^{\alpha +1}}{{(\alpha +1)}^{\alpha +1}{\xi }^{\alpha }(s)}]\mathrm{\Delta }s\\ =\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_1^t[\frac{1}{{(s+1)}^3}\cdot s\cdot \frac{{(s+1)}^3}{s^2}-\frac{1}{{}^{4^4{s}^3{(s+\sigma (s))}^3}}]\mathrm{\Delta }s\\ \ge \underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_1^t[\frac{1}{s}-\frac{1}{{}^{4^4{s}^3{(2s)}^3}}]\mathrm{\Delta }s=\mathrm{\infty },t\in {[1,\mathrm{\infty })}_{\mathbb{T}}.\end{array}$$

Then (4.1) is oscillatory.

Example 4.2 Consider the second-order delay half-linear dynamic equation on times scales

$${\left(\frac{1}{{}^{{(t+\sigma (t))}^3}}{(x^{\mathrm{\Delta }}(t))}^3\right)}^{\mathrm{\Delta }}+\frac{{(T^2{t}^2-1)}^3}{t^2{(T^{2}t-1)}^3}{x}^3(\sqrt{t})=0,t\in {[1,\mathrm{\infty })}_{\mathbb{T}},$$

(4.2)

where T is sufficiently large, $a(t)=\frac{1}{{}^{{(t+\sigma (t))}^3}}$, $q(t)=\frac{{(T^2{t}^2-1)}^3}{t^2{(T^{2}t-1)}^3}$, $\alpha =\beta =3$ and $\tau (t)=t^{1/2}$.

If we choose $g(T)=\frac{1}{T}$, we get

$$\theta (t,\frac{1}{T})=\frac{{\int }_{\frac{1}{T}}^{t^{1/2}}\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}}{{\int }_{\frac{1}{T}}^t\frac{\mathrm{\Delta }s}{a^{1/\alpha }(s)}}=\frac{t-\frac{1}{T^2}}{t^2-\frac{1}{T^2}}=\frac{T^{2}t-1}{T^2{t}^2-1}.$$

When $\xi (t)=t$ by Theorem 3.1, we have

$$\begin{array}{r}\underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_1^t[{\left(\frac{T^{2}s-1}{T^2{s}^2-1}\right)}^3\frac{{(T^2{s}^2-1)}^3}{s^2{(T^{2}s-1)}^3}s-\frac{1}{{}^{4^4{s}^3{(s+\sigma (s))}^3}}]\mathrm{\Delta }s\\ \ge \underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\int }_1^t[\frac{1}{s}-\frac{1}{{}^{4^4{s}^3{(2s)}^3}}]\mathrm{\Delta }s=\mathrm{\infty }.\end{array}$$

Then (4.2) is oscillatory on ${[1,\mathrm{\infty })}_{\mathbb{T}}$.