Theory and Modern Applications

# Oscillation theorems for three classes of conformable fractional differential equations

## Abstract

In this paper, we consider the oscillation theory for fractional differential equations. We obtain oscillation criteria for three classes of fractional differential equations of the forms

\begin{aligned}& T_{\alpha}^{t_{0}} x(t)+\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)=0,\quad t\geqslant t_{0}, \\& T_{\alpha}^{t_{0}} \bigl(r(t) \bigl(T_{\alpha}^{t_{0}} \bigl(x(t)+p(t)x\bigl(\tau(t)\bigr)\bigr)\bigr)^{\beta}\bigr)+q(t)x^{\beta}\bigl(\sigma(t)\bigr)=0, \quad t\geqslant t_{0}, \end{aligned}

and

$$T_{\alpha}^{t_{0}}\bigl(r_{2}T_{\alpha}^{t_{0}} \bigl(r_{1}\bigl(T_{\alpha}^{t_{0}} y \bigr)^{\beta}\bigr)\bigr) (t)+p(t) \bigl(T_{\alpha}^{t_{0}} y(t)\bigr)^{\beta}+q(t)f\bigl(y\bigl(g(t)\bigr)\bigr)=0, \quad t \geqslant t_{0},$$

where $$T_{\alpha}$$ denotes the conformable differential operator of order α, $$0<\alpha\leqslant1$$.

## 1 Introduction

Fractional differential equations have been of great interest recently. Apart from diverse areas of mathematics, fractional differential equations arise in rheology, dynamical processes in self-similar and porous structures, fluid flows, electrical networks, chemical physics, and many other branches of science.

The oscillation of fractional differential equations as a new research field has received significant attention, and some interesting results have already been obtained. We refer to [1,2,3,4,5,6,7,8,9,10,11] and the references therein. The definition of the fractional-order derivative used is either the Caputo or the Riemann–Liouville fractional-order derivative involving an integral expression and the gamma function. Because of the definition, the oscillation of these types of fractional equations cannot be studied by regular methods, for example, by the Riccati transformation. It can only be studied by transforming it into an integer-order equation. In 2012, Chen et al. [4] studied the oscillation behavior of the following fractional differential equation:

$$\bigl[r(t) \bigl(D_{-}^{\alpha}\bigr)\eta(t) \bigr]'-q(t)f \biggl( \int_{t}^{\infty}(v-t)^{-\alpha}y(v)\,dv \biggr)=0 \quad \text{for } t>0,$$

where $$D_{-}^{\alpha}y$$ denotes the Liouville right-sided fractional derivative of order α,

$$\bigl(D_{-}^{\alpha}y\bigr) (t):=-\frac{1}{\varGamma(1-\alpha)} \frac{d}{dt} \int_{t}^{\infty}(v-t)^{-\alpha}y(v)\,dv \quad \text{for } t\in\mathbb{R}_{+}:=(0,\infty).$$

By the Riccati transformation the authors obtained some sufficient conditions.

Recently, Khalil et al. [12] introduced a new well-behaved definition of local fractional derivative, called the conformable fractional derivative, depending just on the basic limit definition of the derivative. This new theory is improved by Abdeljawad [13]. For recent results on conformable fractional derivatives, we refer the reader to [14,15,16,17,18,19,20,21,22,23]. This new definition satisfies formulas of the derivatives of the product and quotient of two functions and has a simpler chain rule. In addition to the definition of conformable fractional derivative, a definition of conformable fractional integral, the Rolle theorem, and the mean value theorem for conformable fractional differentiable functions were given. These properties are more conducive to the study of the oscillation of fractional-order equations.

In fact, some works in this field have shown the significance of conformable fractional derivative. For example, [24] discusses the potential conformable quantum mechanics, [25] discusses the conformable Maxwell equations, and [26, 27] show that the conformable fractional derivative models present good agreements with experimental data, but there are less oscillation results.

In the paper, we study oscillation criteria of conformable fractional differential equations. Our main goal is to generalize the oscillatory criteria in [28,29,30,31,32,33,34,35,36,37] to the conformable fractional derivative. The three equations represent three classes of equations of different orders. For example, in 2016, Akca et al. [33] studied the equation

$$x'(t)+\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)=0, \quad t\geqslant0,$$

and obtained the following:

### Theorem 1.1

Assume that $$0<\varsigma:=\liminf_{t\rightarrow\infty}\int_{\tau(t)}^{t}\sum_{i=1}^{m}p_{i}(s)\,ds\leqslant\frac{1}{e}$$ and for some $$r\in\mathbb {N}$$, we have

$$\limsup_{t\rightarrow\infty} \int^{t}_{h(t)}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta> \frac{1+\ln\lambda_{0}}{\lambda_{0}},$$

where $$h(t)=\max_{1\leqslant i\leqslant m}h_{i}(t)$$, $$h_{i}(t)=\sup_{0\leqslant s\leqslant t}\tau_{i}(s)$$, $$a_{1}(t,s):=\exp \{\int_{s}^{t}\sum_{i=1}^{m}p_{i}(\zeta )\,d\zeta \}$$, $$a_{r+1} (t,s):=\exp \{\int_{s}^{t}\sum_{i=1}^{m}p_{i}(\zeta )a_{r}(\zeta,\tau_{i}(\zeta))\,d\zeta \}$$, and $$\lambda_{0}$$ is the smaller root of the equation $$e^{\varsigma\lambda}=\lambda$$. Then the above equation oscillates.

From this we can unify the oscillation theory of integral-order and fractional-order differential equations. Through the inequality principle, iterative sequences, and the Riccati transformation method this can be extended to the conformable fractional derivatives by Lemma 2.2.

A solution x is called oscillatory if it is eventually neither positive nor negative. Otherwise, the solution is said to be nonoscillatory. An equation is oscillatory if all its solutions oscillate. In this paper, x is differentiable on $$[t_{0},\infty)$$. This paper is organized as follows. In Sect. 2, we introduce some notation and definitions on conformable fractional integrals. In Sect. 3, we present the main theorems on α-order equations. Section 4 is devoted to the oscillatory results on 2α-order equation. In Sect. 5, we demonstrate the oscillatory results for 3α-order equations. In each section, we give examples to illustrate the significance of the results.

## 2 Conformable fractional calculus

For the convenience of the reader, we give some background from fractional calculus theory. These materials can be found in the recent literature, see [12, 13, 23].

### Definition 2.1

([13]) The (left) fractional derivative of a function $$f : [ a ,\infty)\rightarrow R$$ of order $$\alpha\in(0,1]$$ starting from a is defined by

$$\bigl(T^{a}_{\alpha}f \bigr) ( t ) = \lim _{\varepsilon\rightarrow 0}\frac{f(t+\varepsilon(t-a)^{1-\alpha})-f(t)}{\varepsilon}.$$

When $$a=0$$, we write $$T_{\alpha}$$.

Note that if f is differentiable, then $$(T_{\alpha}^{a}f)(t)=(t-a)^{1-\alpha}f'(t)$$.

### Definition 2.2

([13]) The left fractional integral of order $$\alpha\in(0,1]$$ starting at a is defined by

$$\bigl(I_{\alpha}^{a}f\bigr) (t)= \int_{a}^{t}f(x)\,d\alpha(x,a)= \int _{a}^{t}(x-a)^{\alpha-1}f(x)\,dx.$$

### Definition 2.3

([13]) Let $$f:[a,\infty)\rightarrow\mathbb{R}$$ be a continuous function, and let $$\alpha\in(0,1]$$. Then, for all $$t>a$$, we have

$$T_{\alpha}^{a}I_{\alpha}^{a}f(t)=f(t).$$

### Definition 2.4

([13]) Let $$f:(a,b)\rightarrow\mathbb{R}$$ be let a differentiable function, and let $$\alpha\in(0,1]$$. Then, for all $$t>a$$, we have

$$I^{a}_{\alpha}T^{a}_{\alpha}(f) (t) =f(t)-f(a).$$

### Proposition 2.1

([13]) Let $$f:(a,\infty)\rightarrow\infty\rightarrow\mathbb{R}$$ be a twice differentiable function, and let $$0 < \alpha, \beta\leqslant 1$$ be such that $$1<\alpha+\beta\leqslant2$$. Then

$$\bigl(T_{\alpha}^{a}T_{\beta}^{a}\bigr) (t)=T_{\alpha+\beta}^{a}f(t)+(1-\beta ) (t-a)^{-\beta}T_{\alpha}^{a}f(t).$$

### Proposition 2.2

([23]) Let $$\alpha\in(0,1]$$, and let f and g be α-differentiable at a point $$t>0$$ on $$[a,\infty)$$. Then

1. (1)

$$T_{\alpha}^{a}(af+bg)=aT_{\alpha}^{a}(f)+bT_{\alpha}^{a}(g)$$ for all $$a, b\in\mathbb{R}$$,

2. (2)

$$T_{\alpha}^{a}(\lambda)=0$$ for all constant functions $$f(t)=\lambda$$,

3. (3)

$$T_{\alpha}^{a}(fg)=fT_{\alpha}^{a}(g)+gT_{\alpha}^{a}(f)$$,

4. (4)

$$T_{\alpha}^{a}(\frac{f}{g})=\frac{gT_{\alpha}^{a}(f)-fT_{\alpha}^{a}(g)}{g^{2}}$$,

5. (5)

$$T_{\alpha}^{a}(t^{n})=nt^{n-\alpha}$$ for all $$n\in\mathbb{R}$$, and

6. (6)

$$T_{\alpha}^{a}(f\circ g)(t)=f'(g(t))T_{\alpha}^{a}(g)(t)$$ for f differentiable at $$g(t)$$.

### Lemma 2.1

([13]) Let $$f, g:[a,b]\rightarrow\mathbb{R}$$ be two functions such that fg is differentiable, and let $$\alpha\in(0,1]$$. Then

$$\int^{b}_{a} f(x)T_{\alpha}^{a}(g) (x)\,d\alpha(x,a)=fg\bigg|_{a}^{b}- \int _{a}^{b}g(x)T_{\alpha}^{a}(f) (x)\,d\alpha(x,a).$$

### Lemma 2.2

Let $$f : (t_{0} , \infty) \rightarrow\mathbb{R}$$ be differentiable, and let $$\alpha\in(0,1]$$. If $$T_{\alpha}^{t_{0}}f(t)=M(t)$$, then for all $$t > s>t_{0}$$, we have

$$f(t)-f(s)=I_{\alpha}^{t_{0}}M(t).$$

### Proof

We can conclude from $$T_{\alpha}^{t_{0}}f(t)=M(t)$$ that

$$\biggl(\frac{t-t_{0}}{t-s} \biggr)^{1-\alpha}T_{\alpha}^{s}f(t)=M(t),$$

that is,

$$T_{\alpha}^{s}f(t)= \biggl(\frac{t-t_{0}}{t-s} \biggr)^{\alpha-1}M(t).$$

Then applying $$I_{\alpha}$$ to the latter from s to t, we have

$$I_{\alpha}^{s}T_{\alpha}^{s}f(t)=I_{\alpha}^{s} \biggl[ \biggl(\frac {t-t_{0}}{t-s} \biggr)^{\alpha-1}M(t) \biggr],$$

that is,

$$f(t)-f(s)=I_{\alpha}^{t_{0}}M(t).$$

The proof of Lemma 2.2 is complete. □

## 3 α-Order conformable fractional differential equations with finite nonmonotone delay arguments

In this section, we deal with the differential equations of the form

\begin{aligned}[b] T_{\alpha}^{t_{0}} x(t)+\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)=0, \quad t\geqslant t_{0}, \end{aligned}
(3.1)

where $$T_{\alpha}$$ denotes the conformable differential operator of order $$\alpha\in(0,1]$$, $$p_{i}(t)$$, $$1\leqslant i\leqslant m$$, are nonnegative functions, $$\tau_{i}(t)$$, $$1\leqslant i\leqslant m$$, are nonmonotone functions of positive real numbers such that

\begin{aligned}[b] \tau_{i}(t)\leqslant t,\qquad t\geqslant t_{0}, \qquad\lim_{t\rightarrow \infty}\tau_{i}(t)= \infty,\qquad 1\leqslant i \leqslant m. \end{aligned}

To prove our main results, we establish some fundamental results in this section.

### Lemma 3.1

Assume that $$x(t)$$ is an eventually positive solution of (3.1) and $$a_{r}(t,s)$$, $$r\in\mathbb{N}^{+}$$, is defined as

\begin{aligned} &a_{1}(t,s)=\exp \Biggl\{ \int^{t}_{s}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta \Biggr\} , \\ &a_{r+1}(t,s)=\exp \Biggl\{ \int^{t}_{s}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(\zeta,\tau_{i}(\zeta)\bigr) \Biggr\} . \end{aligned}
(3.2)

Then

\begin{aligned}[b] x(t)a_{r}(t,s) \leqslant x(s), \quad0\leqslant s\leqslant t, r\in \mathbb{N}^{+}. \end{aligned}
(3.3)

### Proof

Let $$x(t)$$ be an eventually positive solution of equation (3.1). Then there exists $$t_{1}>t_{0}$$ such that $$x(t)>0$$ and $$x(\tau_{i}(t))>0$$, $$1\leqslant i\leqslant m$$, for all $$t\geqslant t_{1}$$, so

$$T_{\alpha}^{t_{0}} x(t)=-\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)\leqslant 0, \quad t\geqslant t_{1}.$$

This means that $$x(t)$$ is monotonically decreasing, that is, $$x(\tau_{i}(t))\geqslant x(t)$$, $$1\leqslant i\leqslant m$$, and it is easy to put it into the original equation:

$$T_{\alpha}^{t_{0}} x(t)+x(t)\sum_{i=1}^{m}p_{i}(t) \leqslant0, \quad t\geqslant t_{1}.$$

Dividing this equation by $$x(t)$$, we get

$$\frac{T_{\alpha}^{t_{0}} x(t)}{x(t)}\leqslant-\sum_{i=1}^{m}p_{i}(t), \quad t\geqslant t_{1},$$

that is,

$$(t-t_{0})^{1-\alpha}\frac{ x'(t)}{x(t)}\leqslant-\sum _{i=1}^{m}p_{i}(t), \quad t \geqslant t_{1}.$$

Integrating the last inequality from s to t, $$0\leqslant s\leqslant t$$, we get

$$\ln x(\zeta)\bigg|_{s}^{t}\leqslant \int^{t}_{s} \Biggl((\zeta-t_{0})^{\alpha-1} \Biggl(-\sum_{i=1}^{m}p_{i}( \zeta)\Biggr) \Biggr)\,d\zeta,$$

that is,

$$\ln x(t)\leqslant\ln x(s)- \int^{t}_{s}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )\,d \zeta.$$

So

$$x(s)\geqslant x(t)\exp \Biggl\{ \int^{t}_{s}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta \Biggr\} ,$$

that is, estimate (3.3) is valid for $$r = 1$$. Supposing that (3.3) is established for $$r=n$$, we obtain

$$x(t)a_{n}(t,s)\leqslant x(s),$$

so

$$T_{\alpha}^{t_{0}} x(t)+\sum_{i=1}^{m}p_{i}(t)x(t)a_{n} \bigl(t,\tau_{i}(t)\bigr)\leqslant0.$$

Repeating these steps can, we obtain

$$x(s)\geqslant x(t)\exp \Biggl\{ \int^{t}_{s}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}( \zeta)a_{n}\bigl(\zeta,\tau_{i}(\zeta)\bigr)\,d\zeta \Biggr\} ,$$

that is, $$x(t)a_{n+1}(t,s)\leqslant x(s)$$. So Lemma 3.1 is proved by mathematical induction. □

### Lemma 3.2

Assume that $$x(t)$$ is an eventually positive solution of (3.1) and

\begin{aligned}[b] 0< \beta:=\liminf _{t\rightarrow\infty} \int^{t}_{h(t)}(\zeta -t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta\leqslant \frac{1}{e}, \end{aligned}
(3.4)

where

\begin{aligned}[b] h(t)=\max_{1\leqslant i\leqslant m}h_{i}(t), \qquad h_{i}(t)=\max_{0\leqslant s\leqslant t}\tau_{i}(s), \quad t\geqslant0. \end{aligned}
(3.5)

Then

\begin{aligned}[b] \gamma=\liminf_{t\rightarrow\infty} \frac{x(h(t))}{x(t)}\geqslant \lambda_{0}, \end{aligned}
(3.6)

where $$\lambda_{0}$$ is the smaller root of the equation $$\lambda=e^{\beta\lambda}$$.

### Proof

Let $$x(t)$$ be an eventually positive solution of equation (3.1). Then there exists $$t_{1}>t_{0}$$ such that $$x(t)>0$$ and $$x(\tau_{i}(t))>0$$, $$1\leqslant i \leqslant m$$, for all $$t\geqslant t_{1}$$. Thus we can conclude from (3.1) that

$$T_{\alpha}^{t_{0}} x(t)=-\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)\leqslant 0, \quad t\geqslant t_{1}.$$

This means that $$x(t)$$ is monotonically decreasing and positive.

By (3.4), for any $$\varepsilon\in(0,\beta)$$, there is $$t_{\varepsilon}$$ such that

$$\int^{t}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )\,d \zeta\geqslant\beta-\varepsilon, \quad t\geqslant t_{\varepsilon}\geqslant t_{1}.$$

We will show that

\begin{aligned}[b] \liminf_{t\rightarrow\infty} \frac{x(h(t))}{x(t)}\geqslant\lambda_{1}, \end{aligned}
(3.7)

where $$\lambda_{1}$$ is the smaller root of the equation

$$e^{(\beta-\varepsilon)\lambda}=\lambda.$$

$$\gamma=\liminf_{t\rightarrow\infty}\frac{x(h(t))}{x(t)}< \lambda_{1}.$$

Therefore

\begin{aligned}[b] e^{(\beta-\varepsilon)\gamma}>\gamma. \end{aligned}
(3.8)

Then for any $$\delta\in(0,\gamma)$$, there exists $$t_{\delta}$$ such that $$\frac{x(h(t))}{x(t)}\geqslant\gamma-\delta$$ for $$t\geqslant t_{\delta}$$. Dividing both sides of (3.1) by $$x(t)$$, we have

$$-\frac{T_{\alpha}^{t_{0}} x(t)}{x(t)}=\sum_{i=1}^{m}p_{i}(t) \frac{x(\tau _{i}(t))}{x(t)}\geqslant\sum_{i=1}^{m}p_{i}(t) \frac{x(h(t))}{x(t)} \geqslant(\gamma-\delta)\sum_{i=1}^{m}p_{i}(t).$$

Integrating the latter from $$h(t)$$ to t, we obtain

$$- \int_{h(t)}^{t}\frac{x'(s)}{x(s)}\,ds\geqslant \int_{h(t)}^{t} (s-t_{0})^{\alpha-1} \Biggl((\gamma-\delta)\sum_{i=1}^{m}p_{i}(s) \Biggr)\,ds,$$

or

$$- \int_{h(t)}^{t}\frac{x'(s)}{x(s)}\,ds\geqslant( \gamma-\delta) (\beta -\varepsilon),$$

so

$$\frac{x(h(t))}{x(t)}\geqslant e^{(\gamma-\delta)(\beta-\varepsilon)}.$$

Therefore

$$\gamma=\liminf_{t\rightarrow\infty}\frac{x(h(t))}{x(t)}\geqslant e^{(\gamma-\delta)(\beta-\varepsilon)},$$

which implies

$$\gamma\geqslant e^{(\beta-\varepsilon)\gamma},$$

which is a contradiction to hypothesis (3.8). So (3.7) is true. Since (3.7) implies (3.6), the proof of Lemma 3.2 is complete. □

### Theorem 3.1

Assume that (3.4) holds and for some r, we have

\begin{aligned}[b] \limsup_{t\rightarrow\infty} \int^{t}_{h(t)}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta> \frac{1+\ln\lambda_{0}}{\lambda_{0}}, \end{aligned}
(3.9)

where $$h(t)$$ is defined by (3.5), $$a_{r} (t,s)$$ is defined by (3.2), and $$\lambda_{0}$$ is the smaller root of the equation $$e^{\beta\lambda}=\lambda$$. Then equation (3.1) oscillates.

### Proof

If equation (3.1) has a solution $$x(t)$$, then $$-x(t)$$ is also a solution of equation (3.1), so we only consider the situation where a solution of (3.1) is eventually positive, that is, there is an integer $$t_{1}\geqslant t_{0}$$ such that $$x(t)>0$$ and $$x(\tau_{i}(t))>0$$, $$1\leqslant i\leqslant m$$, for all $$t\geqslant t_{1}$$. By (3.1) we have

$$T_{\alpha}^{t_{0}} x(t)=-\sum_{i=1}^{m}p_{i}(t)x \bigl(\tau_{i}(t)\bigr)\leqslant 0, \quad t\geqslant t_{1}.$$

It is shown that $$x(t)$$ is an eventually decreasing function.

By Lemma 3.2 inequality (3.6) holds. It can be easily seen that $$\lambda_{0}>1$$, so for any real number $$0<\varepsilon\leqslant\lambda_{0}-1$$, we have

$$\frac{x(h(t))}{x(t)}\geqslant\lambda_{0}-\varepsilon, \quad t \geqslant t_{2}\geqslant t_{1}.$$

Then there is $$t^{\ast}\in(h(t),t)$$ satisfying

\begin{aligned}[b] \frac{x(h(t))}{x(t^{\ast})}= \lambda_{0}-\varepsilon, \quad t\geqslant t_{2}. \end{aligned}
(3.10)

Then integrating from $$t^{\ast}$$ to t equation (3.1) and substituting into (3.3), we have

$$x(t)-x\bigl(t^{\ast}\bigr)+x\bigl(h(t)\bigr) \int_{t^{\ast}}^{t}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta \leqslant0.$$

Combining this with (3.10), we have

\begin{aligned}[b] \int_{t^{\ast}}^{t}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta\leqslant \frac{x(t^{\ast})}{ x(h(t))}=\frac{1}{\lambda_{0}-\varepsilon}. \end{aligned}
(3.11)

Dividing (3.1) by $$x(t)$$, substituting into (3.3), and then integrating from $$h(t)$$ to $$t^{\ast}$$, we have

$$- \int^{t^{\ast}}_{h(t)}\frac{ x'(\zeta)}{x(\zeta)}\,d\zeta \geqslant \int_{h(t)}^{t^{\ast}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \frac{x(h(t))}{x(\zeta)}a_{r}\bigl(h(t),\tau _{i}(\zeta) \bigr)\,d\zeta,$$

and because of $$T_{\alpha}^{t_{0}} x(t)<0$$, we have

\begin{aligned}[b] & (\lambda_{0}-\varepsilon) \int^{t^{\ast}}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta \\ &\quad\leqslant \int^{t^{\ast}}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \frac{x(h(t))}{x(t)}a_{r}\bigl(h(t), \tau_{i}(\zeta) \bigr)\,d\zeta \\ &\quad\leqslant- \int^{t^{\ast}}_{h(t)}\frac{ x'(\zeta)}{x(\zeta )}\,d\zeta, \end{aligned}

that is,

\begin{aligned}[b] \int^{t^{\ast}}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta \leqslant \frac{1}{ \lambda_{0}-\varepsilon}\ln\frac{x(h(t))}{x(t^{\ast})}. \end{aligned}
(3.12)

Adding (3.12) to (3.11), we get

$$\int^{t}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta\leqslant \frac{1+\ln(\lambda_{0}-\varepsilon)}{\lambda_{0}-\varepsilon}.$$

This inequality holds for all $$0<\varepsilon\leqslant\lambda_{0}-1$$, so as $$\varepsilon\rightarrow0$$, we obtain

$$\limsup_{t\rightarrow\infty} \int^{t}_{h(t)}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta \leqslant\frac{1+\ln\lambda_{0}}{\lambda_{0}}.$$

This is a contradiction to (3.9). The proof of Theorem 3.1 is complete. □

### Lemma 3.3

Assume that $$x(t)$$ is an eventually positive solution of (3.1) and that β and $$h(t)$$ are defined by (3.4) and (3.5). Then

\begin{aligned}[b] \liminf_{t\rightarrow\infty} \frac{x(t)}{x(h(t))}\geqslant\frac {1}{2}\bigl(1-\beta-\sqrt{1-2\beta- \beta^{2}}\bigr):=A(\beta). \end{aligned}
(3.13)

### Proof

Assume that $$x(t)>0$$ for $$t>T_{1}\geqslant t_{0}$$. Then there exists $$T_{2}\geqslant T_{1}$$ such that $$x(\tau_{i}(t))>0$$, $$i=1,2,\dots,m$$. In view of (3.1), $$T_{\alpha}^{t_{0}}x(t)\leqslant0$$ on $$[T_{2},\infty)$$. Clearly, (3.13) holds for $$\beta=0$$. If $$0<\beta\leqslant\frac{1}{e}$$, then for any $$\varepsilon\in(0,\beta)$$, there exists $$N_{\varepsilon}$$ such that

\begin{aligned}[b] \int^{t}_{h(t)}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )\,d \zeta>\beta-\varepsilon, \quad t>N_{\varepsilon}. \end{aligned}
(3.14)

For fixed ε, we will show that for each $$t>N_{\varepsilon}$$, there exists $$\lambda_{t}$$ such that $$h(\lambda_{t})< t<\lambda_{t}$$ and

\begin{aligned}[b] \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta=\beta-\varepsilon. \end{aligned}
(3.15)

In fact, for a given $$t>N_{\varepsilon}$$, $$f(\lambda):=\int^{\lambda}_{t}(\zeta-t_{0})^{\alpha-1}\sum_{i=1}^{m}p_{i}(\zeta)\,d\zeta$$ is continuous. Because of $$\lim_{t\rightarrow\infty}h(t)=\infty$$ and (3.14), we have $$\lim_{\lambda\rightarrow\infty}f(\lambda)>\beta-\varepsilon >0$$. Hence there exists $$\lambda_{t}>t$$ such that $$f(\lambda)=\beta-\varepsilon$$, that is, (3.15) holds. From (3.14) we have

$$\int^{\lambda_{t}}_{h(\lambda_{t})}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta>\beta-\varepsilon = \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta,$$

and therefore $$h(\lambda_{t})< t$$.

Integrating (3.1) from t ($$>T_{3}=\max\{T_{2},N_{\varepsilon}\}$$) to $$\lambda_{t}$$, we have

\begin{aligned}[b] x(t)-x(\lambda_{t}) \geqslant \int_{t}^{\lambda_{t}}(\zeta -t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)x \bigl(\tau_{i}(\zeta )\bigr)\,d\zeta. \end{aligned}
(3.16)

We see that $$h(t)\leqslant h(\zeta)\leqslant h(\lambda_{t})< t$$ for $$t\leqslant y\leqslant\lambda_{t}$$. Integrating (3.1) from $$\tau_{i}(\zeta)$$ to t, we have that for $$t\leqslant\zeta\leqslant \lambda_{t}$$,

\begin{aligned}[b] &x\bigl(\tau_{i}( \zeta)\bigr)-x(t)\\ &\quad \geqslant \int_{\tau_{i}(\zeta )}^{t}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)x \bigl(\tau_{i}(u)\bigr)\,du \\ &\quad \geqslant x\bigl(h(t)\bigr) \int_{\tau_{i}(\zeta)}^{t}(u-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}(u)\,du \\ &\quad >x\bigl(h(t)\bigr) \Biggl( \int_{h(\zeta)}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du- \int_{t}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du \Biggr) \\ &\quad >x\bigl(h(t)\bigr) \Biggl((\beta-\varepsilon)- \int_{t}^{\zeta }(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du \Biggr). \end{aligned}
(3.17)

From (3.16) and (3.17) we have

\begin{aligned}[b]x(t)&\geqslant x( \lambda_{t})+ \int_{t}^{\lambda_{t}}(\zeta -t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)x \bigl(\tau_{i}(\zeta )\bigr)\,d\zeta \\ &>x(\lambda_{t})+ \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha -1}\\ &\quad {}\times \sum_{i=1}^{m}p_{i}(\zeta) \Biggl[x(t)+x\bigl(h(t)\bigr) \Biggl((\beta-\varepsilon)- \int_{t}^{\zeta }(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du \Biggr) \Biggr] \,d\zeta \\ &=x(\lambda_{t})+x(t) (\beta-\varepsilon) \\ &\quad {}+x\bigl(h(t)\bigr) \Biggl[(\beta-\varepsilon)^{2}\\ &\quad {}- \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{t}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1} ^{m}p_{i}(u) \,du \Biggr]\,d\zeta. \end{aligned}
(3.18)

Noting the known formula

\begin{aligned}[b] & \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{t}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du\,d \zeta \\ &\quad= \int_{t}^{\lambda_{t}}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u) \int_{u}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta \,du, \end{aligned}

or

\begin{aligned}[b] & \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{t}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du\,d \zeta \\ &\quad= \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{\zeta}^{\lambda_{t}}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du\,d \zeta, \end{aligned}

we have

\begin{aligned}[b] & \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{t}^{\zeta}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du\,d \zeta \\ &\quad=\frac{1}{2} \int_{t}^{\lambda_{t}}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}(\zeta) \int_{t}^{\lambda_{t}}(u-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(u)\,du\,d \zeta \\ &\quad=\frac{1}{2} \Biggl[ \int_{t}^{\lambda_{t}}(\zeta -t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta)\,d \zeta \Biggr]^{2} \\ &\quad=\frac{1}{2}(\beta-\varepsilon)^{2}. \end{aligned}

Substituting this into (3.18), we have

\begin{aligned}[b] x(t)>x(\lambda_{t})+x(t) (\beta-\varepsilon)+\frac{1}{2}(\beta -\varepsilon)^{2}x \bigl(h(t)\bigr). \end{aligned}
(3.19)

Hence

$$\frac{x(t)}{x(h(t))}>\frac{(\beta-\varepsilon)^{2}}{2(1-\beta +\varepsilon)}:=d_{1},$$

and then

$${x(\lambda_{t})}>\frac{(\beta-\varepsilon)^{2}}{2(1-\beta +\varepsilon)}{x\bigl(h( \lambda_{t})\bigr)}=d_{1}x\bigl(h(\lambda_{t}) \bigr)\geqslant d_{1}x(t).$$

Substituting this into (3.19), we obtain

$$x(t)>x(t) (m+d_{1}-\varepsilon)+\frac{1}{2}(m- \varepsilon)^{2}x\bigl(h(t)\bigr),$$

and hence

$$\frac{x(t)}{x(h(t))}>\frac{(\beta-\varepsilon)^{2}}{2(1-\beta -d_{1}+\varepsilon)}:=d_{2}.$$

In general, we have

$$\frac{x(t)}{x(h(t))}>\frac{(\beta-\varepsilon)^{2}}{2(1-\beta -d_{n}+\varepsilon)}:=d_{n+1}, \quad n=1,2, \dots.$$

It is not difficult to see that if ε is small enough, then $$1\geqslant d_{n}>d_{n-1}$$, $$n=2,3,\dots$$ . Hence $$\lim_{n\rightarrow\infty}d_{n}=d$$ exists and satisfies

$$-2d^{2}+2d(1-\beta+\varepsilon)=(\beta-\varepsilon)^{2},$$

that is,

$$d=\frac{1-\beta+\varepsilon\pm\sqrt{1-2(\beta-\varepsilon)-(\beta -\varepsilon)^{2}}}{2}.$$

Because of $$T_{\alpha}^{t_{0}}\leqslant0$$, we have $$d<1$$. Therefore, for all large t,

$$\frac{x(t)}{x(h(t))}>\frac{1-\beta+\varepsilon-\sqrt{1-2(\beta -\varepsilon)-(\beta-\varepsilon)^{2}}}{2}.$$

Letting $$\varepsilon\rightarrow0$$, we obtain that

$$\frac{x(t)}{x(h(t))}>\frac{1-\beta-\sqrt{1-2\beta-\beta ^{2}}}{2}=A(\beta).$$

This shows that (3.13) holds. □

### Theorem 3.2

Assume (3.4) holds and that for some r, we have

\begin{aligned}[b] &\limsup_{t\rightarrow\infty} \int^{t}_{h(t)}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta\\ &\quad >1- \frac{1}{2}\bigl(1-\beta-\sqrt{1-2\beta-\beta^{2}}\bigr), \end{aligned}
(3.20)

where $$h(t)$$ is defined by (3.5), $$a_{r} (t,s)$$ is defined by (3.2), and $$\lambda_{0}$$ is the smaller root of the equation $$e^{\beta\lambda}=\lambda$$. Then equation (3.1) oscillates.

### Proof

If equation (3.1) has a solution $$x(t)$$, then $$-x(t)$$ is also a solution of equation (3.1), so we only consider the situation where a solution of (3.1) is eventually positive, that is, $$x(t)>0$$ and $$x(\tau_{i}(t))>0$$, $$1\leqslant i \leqslant m$$, for all $$t\geqslant T_{3}$$. By (3.1) we have

$$x'(t)-x\bigl(h(t)\bigr) (t-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(t) \leqslant0, \quad t\geqslant T_{3}.$$

Integrating from $$h(t)$$ to t the latter and substituting into (3.3), we have

\begin{aligned}[b] x(t)-x\bigl(h(t)\bigr)+x\bigl(h(t)\bigr) \int_{h(t)}^{t}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta \leqslant 0. \end{aligned}

Consequently,

$$\int_{h(t)}^{t}(\zeta-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\zeta )a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta\leqslant1- \frac{x(t)}{ x(h(t))},$$

which gives

$$\limsup_{t\rightarrow\infty} \int_{h(t)}^{t}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta \leqslant1-\liminf_{t\rightarrow\infty}\frac{x(t)}{x(h(t))},$$

and by (3.13) the last inequality leads to

$$\limsup_{t\rightarrow\infty} \int_{h(t)}^{t}(\zeta-t_{0})^{\alpha -1} \sum_{i=1}^{m}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr) \,d\zeta \leqslant1-\frac{1}{2}\bigl(1-\beta-\sqrt{1-2\beta-\beta^{2}} \bigr),$$

which contradicts (3.20). The proof of the theorem is complete. □

### Example 3.1

We consider the delay differential equation

\begin{aligned}[b] T_{\frac{1}{2}} x(t)+p_{1}(t)x\bigl(\tau_{1}(t)\bigr)+p_{2}(t)x \bigl(\tau_{2}(t)\bigr)=0, \quad t\geqslant0, \end{aligned}
(3.21)

where

\begin{aligned}& \tau_{1}(t)= \textstyle\begin{cases} t-1,& t\in[3k,3k+1],\\ -3t+12k+3,& t\in[3k+1,3k+2],\\ 5t-12k-3,& t\in[3k+2,3k+3], \end{cases}\displaystyle \\& k\in\mathbb{N},\quad \mbox{and} \quad \tau_{2}(t)= \tau_{1}(t)-1, \\& p_{i}(t)=\frac{1}{8}t^{\frac{1}{2}},\quad i=1,2. \end{aligned}

By (3.5) we obtain

\begin{aligned}& h_{1}(t)=\max_{0\leqslant s\leqslant t}\tau_{1}(s)= \textstyle\begin{cases} t-1,& t\in[3k,3k+1],\\ 3k,& t\in[3k+1,3k+2],\\ 5t-12k-13,& t\in[3k+2,3k+3], \end{cases}\displaystyle \\& k\in\mathbb{N},\quad \mbox{and} \quad h_{2}(t)=h_{1}(t)-1. \end{aligned}

So $$h(t)=\max_{1\leqslant i\leqslant2}\{h_{i}(t)\}=h_{1}(t)$$.

The functions $$F_{r}:\mathbb{N}\rightarrow\mathbb{R}^{+}$$ are defined as $$F_{r}(t)=\int^{t}_{h(t)}\sum_{i=1}^{m}p_{i}(\zeta)a_{r}(h(t),\tau _{i}(\zeta))\,d\zeta$$. When $$t=3k+2.6$$, $$t\in\mathbb{N}$$, for any $$r\in\mathbb{N}^{+}$$, the function $$F_{r}(t)$$ attains its maximum. In particular,

$$F_{1}(t=3k+2.6)= \int_{3k}^{3k+2.6}\zeta^{-\frac{1}{2}}\sum _{i=1}^{2}p_{i}( \zeta)a_{r}\bigl(h(t),\tau_{i}(\zeta)\bigr)\,d\zeta,$$

where

\begin{aligned}[b] a_{r}\bigl(h(t),\tau_{i}( \zeta)\bigr)&=\exp \Biggl\{ \int_{\tau_{i}(\zeta )}^{h(t)}(\xi-t_{0})^{\alpha-1} \sum_{i=1}^{m}p_{i}(\xi) \,d \xi \Biggr\} =\exp \biggl\{ \int_{\tau_{i}(\zeta)}^{h(t)}\xi^{-\frac {1}{2}} \frac{1}{4}\xi^{\frac{1}{2}}\,d\xi \biggr\} \\ &=\exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau_{i}(\zeta) \bigr) \biggr\} , \end{aligned}

so

\begin{aligned} F_{1}(t=3k+2.6)&= \int_{3k}^{3k+2.6}\zeta^{-\frac{1}{2}}\sum _{i=1}^{2}\frac{1}{8} \zeta^{\frac{1}{2}} \exp \biggl\{ \frac{1}{4}\bigl(h(t)- \tau_{i}(\zeta)\bigr) \biggr\} \,d\zeta \\ &=\frac{1}{4} \int_{3k}^{3k+2.6} \biggl(\exp \biggl\{ \frac {1}{4}\bigl(h(t)-\tau_{1}(\zeta)\bigr) \biggr\} + \exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau_{2}(\zeta)\bigr) \biggr\} \biggr)\,d\zeta \\ &=\frac{1}{4} \int_{3k}^{3k+1} \biggl(\exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau _{1}(\zeta)\bigr) \biggr\} + \exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau_{2}(\zeta)\bigr) \biggr\} \biggr)\,d\zeta \\ &\quad{}+\frac{1}{4} \int_{3k+1}^{3k+2} \biggl(\exp \biggl\{ \frac {1}{4}\bigl(h(t)-\tau_{1}(\zeta)\bigr) \biggr\} + \exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau_{2}(\zeta)\bigr) \biggr\} \biggr)\,d\zeta \\ &\quad{}+\frac{1}{4} \int_{3k+2}^{3k+2.6} \biggl(\exp \biggl\{ \frac {1}{4}\bigl(h(t)-\tau_{1}(\zeta)\bigr) \biggr\} + \exp \biggl\{ \frac{1}{4}\bigl(h(t)-\tau_{2}(\zeta)\bigr) \biggr\} \biggr)\,d\zeta \\ &\approx1.5052, \end{aligned}

and therefore

$$\limsup_{t\rightarrow\infty}F_{1}(t)\geqslant1.5052.$$

Now we see that

$$\beta=\liminf_{t\rightarrow\infty} \int^{t}_{h(t)}\zeta^{-\frac {1}{2}}\sum _{i=1}^{m}p_{i}(\zeta)\,d\zeta = \frac{1}{4}\bigl(t-h(t)\bigr)=\frac{1}{4}\leqslant \frac{1}{e}.$$

The solution of $$\lambda=e^{\beta\lambda}$$ is $$\lambda_{0}=1.435$$, so we get

\begin{aligned}& 1.5052>\frac{1+\ln\lambda_{0}}{\lambda_{0}}\thickapprox0.9485, \\& 1.5052>1>A(\beta). \end{aligned}

Therefore equation (3.21) satisfies the conditions of Theorems 3.1 and 3.2, and thus equation (3.21) oscillates.

## 4 Oscillation of 2α-order neutral conformable fractional differential equation

In this section, we deal with differential equations of the form

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl(r(t) \bigl(T_{\alpha}^{t_{0}}\bigl(x(t)+p(t)x\bigl(\tau(t) \bigr)\bigr)\bigr)^{\beta}\bigr)+q(t)x^{\beta}\bigl(\sigma(t) \bigr)=0, \quad t\geqslant t_{0}, \end{aligned}
(4.1)

where $$T_{\alpha}$$ denotes the conformable differential operator of order $$\alpha\in(0,1]$$, $$\beta\geqslant1$$ is a quotient of odd positive integers, and the functions r, p, q, τ, σ are such that $$r, p, q, \tau, \sigma\in C^{1}([t_{0},\infty),(0,\infty))$$. We also assume that, for all $$t\geqslant t_{0}$$, $$\tau(t)\leqslant t$$, $$\sigma(t)\leqslant t$$, $$T_{\alpha}^{t_{0}}\sigma(t)>0$$, $$\lim_{t\rightarrow\infty}\tau(t)=\lim_{t\rightarrow\infty }\sigma(t)=\infty$$, $$0\leqslant p(t)<1$$, $$q(t)\geqslant0$$, and q does not vanish eventually.

We further use the following notation:

\begin{aligned}& \varepsilon:=\bigl(\beta/(\beta+1)\bigr)^{\beta+1}, \qquad Q(t):=q(t) \bigl(1-p\bigl(\sigma(t)\bigr)\bigr)^{\beta}, \\& z(t)=x(t)+p(t)x\bigl(\tau(t)\bigr)< \infty, \qquad\pi(t):= \int_{t}^{\infty}(s-t_{0})^{\alpha-1}r(s)^{-1/\beta}\,ds. \end{aligned}

### Lemma 4.1

Let $$\beta\geqslant1$$ be a ratio of two odd numbers. Then

\begin{aligned} &A^{(\beta+1)/\beta}-(A-B)^{(\beta+1)/\beta}\leqslant \frac{4}{2} {B^{1/\beta}}{\beta}\bigl[(1+\beta)A-B\bigr], \quad AB\geqslant0. \\ &{-}Cv^{(\beta+1)/\beta}+Dv\leqslant \frac{\beta^{\beta}}{(\beta +1)^{\beta+1}}\frac{D^{\beta+1}}{C^{\beta}}, \quad C>0. \end{aligned}
(4.2)

### Theorem 4.1

Assume that $$\pi(t)=\int_{t}^{\infty}(s-t_{0})^{\alpha-1}r(s)^{-1/\beta }\,ds<\infty$$ and there exists a function $$\rho\in C^{1}([t_{0},\infty),(0,\infty))$$ such that

\begin{aligned}[b] \limsup_{t\rightarrow\infty}I_{\alpha}^{t_{0}} \biggl(\rho (t)Q(t)-\bigl(\sigma(t)-t_{0}\bigr)^{(1-\alpha)\beta} \frac{(T_{\alpha}^{t_{0}}\rho_{+}(t))^{\beta+1}r(\sigma(t))}{(\beta +1)^{\beta+1} \rho^{\beta(t)}(T_{\alpha}^{t_{0}}\sigma(t))^{\beta}} \biggr)=\infty. \end{aligned}
(4.3)

Suppose that there exists a function $$\delta\in C^{1}([t_{0},\infty),(0,\infty))$$ such that

\begin{aligned}[b] \limsup_{t\rightarrow\infty}I^{t_{0}}_{\alpha} \biggl[\psi(t)-\frac {\delta(t)r(t)((\varphi(t))_{+})^{\beta+1}}{ (\beta+1)^{\beta+1}}\biggr]=\infty, \end{aligned}
(4.4)

where

\begin{aligned}& \psi(t):=\delta(t) \biggl[q(t) \biggl(1-p\bigl(\sigma(t)\bigr) \frac{\pi(\tau(\sigma (t)))}{\pi(\sigma(t)}\biggr)^{\beta}+ \frac{1-\beta}{r^{1/\beta}(t)\pi^{\beta+1}(t)} \biggr], \\& p(t)< \pi(t)/\pi\bigl(\tau(t)\bigr),\qquad \varphi(t):=\frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}+ \frac {1+\beta}{r^{1/\beta}(t)\pi(t)}, \end{aligned}

and $$(\varphi(t))_{+}:=\max\{0,\varphi(t)\}$$. Then equation (4.1) oscillates.

### Proof

Let $$x(t)$$ be a nonoscillating solution of (4.1) on $$[t_{0},\infty)$$. Without loss of generality, we may assume that there exists $$t_{1}\geqslant t_{0}$$ such that $$x(t)>0$$, $$x(\tau(t))>0$$, and $$x(\sigma(t))>0$$ for all $$t\geqslant t_{1}$$. Then $$z(t)\geqslant x(t)>0$$, and since

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl\{ r(t)\bigl[T_{\alpha}^{t_{0}}\bigl(z(t)\bigr) \bigr]^{\beta}\bigr\} =-q(t)x^{\beta}\bigl(\sigma(t)\bigr) \leqslant0, \end{aligned}
(4.5)

the function $$[r(t)T_{\alpha}^{t_{0}}z(t)]^{\beta}$$ is nonincreasing for all $$t\geqslant t_{1}$$. Therefore $$T_{\alpha}^{t_{0}} z(t)$$ does not change sign eventually, that is, there exists $$t_{2}\geqslant t_{1}$$ such that either $$T_{\alpha}^{t_{0}} z(t)>0$$ or $$T_{\alpha}^{t_{0}} z(t)<0$$ for all $$t\geqslant t_{2}$$.

Case I. Assume first that $$T_{\alpha}^{t_{0}} z(t)>0$$ for all $$t\geqslant t_{2}$$. Note that $$T_{\alpha}^{t_{0}}z(t)|_{t=\sigma(t)}=T_{\alpha}^{t_{0}}(z(\sigma(t)))$$. Then

$$r(t) \bigl(T_{\alpha}^{t_{0}}\bigl(z(t)\bigr) \bigr)^{\beta}\leqslant r\bigl(\sigma(t)\bigr) \bigl(T_{\alpha}^{t_{0}}z \bigl(\sigma(t)\bigr)\bigr)^{\beta},$$

from which it follows that

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl(z\bigl(\sigma(t)\bigr)\bigr)\geqslant\bigl(T_{\alpha}^{t_{0}}\bigl(z(t)\bigr)\bigr) \biggl(\frac{r(t)}{r(\sigma(t))} \biggr)^{1/\beta}. \end{aligned}
(4.6)

Since $$x(t)\leqslant z(t)$$, we see that

\begin{aligned}[b] x(t)\geqslant\bigl[1-p(t) \bigr]z(t), \quad t\geqslant t_{2}. \end{aligned}
(4.7)

In view of (4.7) and (4.1),

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl(r(t) \bigl(T_{\alpha}^{t_{0}}\bigl(x(t)+p(t)x\bigl(\tau(t) \bigr)\bigr)\bigr)^{\beta}\bigr)+Q(t)z^{\beta}\bigl(\sigma(t) \bigr)\leqslant0, \quad t\geqslant t_{2}. \end{aligned}
(4.8)

Put

\begin{aligned}[b] w(t)=\rho(t)\frac{r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta}}{z^{\beta}(\sigma(t))}, \quad t\geqslant t_{2}. \end{aligned}
(4.9)

Clearly, $$w(t)>0$$. Applying $$T_{\alpha}^{t_{0}}$$ to (4.9) and using (4.6) and (4.8), we obtain

\begin{aligned}[b] &T_{\alpha}^{t_{0}}\bigl(w(t) \bigr)\\ &\quad =\frac{T_{\alpha}^{t_{0}}\rho_{+}(t)}{\rho (t)}w(t)+\rho(t)\frac{T_{\alpha}^{t_{0}}(r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta})}{z^{\beta}(\sigma(t))} -\rho(t) \frac{r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta}\beta z'(\sigma(t) )T_{\alpha}^{t_{0}}\sigma(t)}{z^{\beta+1}(\sigma(t))} \\ &\quad \leqslant\frac{T_{\alpha}^{t_{0}}\rho_{+}(t)}{\rho(t)}w(t)-\rho(t)Q(t) -\frac{\beta T_{\alpha}^{t_{0}}\sigma(t)}{(\rho(t)r(\sigma(t)))^{\frac{1}{\beta }}(\sigma(t)-t_{0})^{1-\alpha}}w^{\frac{\beta+1}{\beta}}(t), \end{aligned}

where $$T_{\alpha}^{t_{0}}\rho_{+}(t)=\max\{T_{\alpha}^{t_{0}}\rho (t),0\}$$. Set

$$F(v)=\frac{T_{\alpha}^{t_{0}}\rho_{+}(t)}{\rho(t)}v-\frac{\beta T_{\alpha}^{t_{0}}\sigma(t)}{(\rho(t)r(\sigma(t)))^{\frac{1}{\beta }}(\sigma(t)-t_{0})^{1-\alpha}}v^{\frac{\beta+1}{\beta}}, \quad v>0.$$

By calculation letting $$v_{0}=(\sigma(t)-t_{0})^{(1-\alpha)\beta}\frac{1}{(\beta+1)^{\beta }}\frac{(T_{\alpha}^{t_{0}}\rho_{+}(t))^{\beta}}{ \rho^{\beta-1}(t)}\frac{r(\sigma(t))}{(T_{\alpha}^{t_{0}}\sigma (t))^{\beta}}$$, we have that when

$$v=v_{0},$$

the function $$F(v)$$ attains its maximum $$F(v_{0})$$. So

$$F(v)\leqslant F(v_{0})=\bigl(\sigma(t)-t_{0} \bigr)^{(1-\alpha)\beta}\frac{(T_{\alpha}^{t_{0}}\rho_{+}(t))^{\beta+1}r(\sigma(t))}{(\beta+1)^{\beta+1} \rho^{\beta}(t)(T_{\alpha}^{t_{0}}\sigma(t))^{\beta}}.$$

Therefore

$$T_{\alpha}^{t_{0}}\bigl(w(t)\bigr)\leqslant-\rho(t)Q(t)+\bigl( \sigma (t)-t_{0}\bigr)^{(1-\alpha)\beta} \frac{(T_{\alpha}^{t_{0}}\rho_{+}(t))^{\beta+1}r(\sigma(t))}{(\beta +1)^{\beta+1} \rho^{\beta}(t)(T_{\alpha}^{t_{2}}\sigma(t))^{\beta}}.$$

Applying $$I_{\alpha}$$ to the last inequality from $$t_{0}$$ to t, we have

$$0< w(t)\leqslant w(t_{0})-I_{\alpha}^{t_{0}} \biggl( \rho(t)Q(t)-\bigl(\sigma (t)-t_{0}\bigr)^{(1-\alpha)\beta} \frac{(T_{\alpha}^{t_{0}}\rho _{+}(t))^{\beta+1}r (\sigma(t))}{(\beta+1)^{\beta+1}\rho^{\beta(t)}(T_{\alpha}^{t_{0}}\sigma(t))^{\beta}} \biggr).$$

Letting $$t\rightarrow\infty$$ in this inequality, we get a contradiction to (4.3).

Case II. Assume now that $$T_{\alpha}^{t_{0}} z(t)<0$$ for all $$t\geqslant t_{0}$$. It follows from (4.1) that $$T_{\alpha}^{t_{0}}(r(T_{\alpha}^{t_{0}} z)^{\beta})<0$$ for all $$s\geqslant t \geqslant t_{2}$$, and thus

\begin{aligned}[b] T_{\alpha}^{t_{0}} z(s)\leqslant \biggl(\frac{r(t)}{r(s)} \biggr)^{1/\beta}T_{\alpha}^{t_{0}}z(t). \end{aligned}
(4.10)

Dividing (4.10) by $$(s-t_{0})^{1-\alpha}$$ and then integrating from t to l, $$l\geqslant t\geqslant t_{2}$$, we have

\begin{aligned}[b] z(l)-z(t)\leqslant \int_{t}^{l}(s-t_{0})^{\alpha-1} \biggl\{ \biggl(\frac {r(t)}{r(s)} \biggr)^{1/\beta}T_{\alpha}^{t_{0}} \bigl(z(t)\bigr) \biggr\} \,ds& \\ =r(t) ^{1/\beta} T_{\alpha}^{t_{0}}\bigl(z(t)\bigr) \int_{t}^{l}(s-t_{0})^{\alpha-1}r(s)^{-1/\beta}\,ds. \end{aligned}

Letting $$l\rightarrow\infty$$, we get

\begin{aligned}[b] z(t)\geqslant- \pi(t)r^{1/\beta}(t)T_{\alpha}^{t_{0}}\bigl(z(t)\bigr), \end{aligned}
(4.11)

which implies that

\begin{aligned}[b] T_{\alpha}^{t_{0}} \biggl( \frac{z(t)}{\pi(t)} \biggr)&=\frac{\pi (t)T_{\alpha}^{t_{0}} z(t)-z(t)T_{\alpha}^{t_{0}}\pi(t)}{\pi^{2}(t)} =\frac{\pi(t)T_{\alpha}^{t_{0}} z(t)+z(t)r^{-1/\beta}(t))}{\pi ^{2}(t)} \\ &\geqslant\frac{\pi(t)T_{\alpha}^{t_{0}} z(t)-\pi(t)T_{\alpha}^{t_{0}} z(t)}{\pi^{2}(t)}=0. \end{aligned}

Hence we conclude that

\begin{aligned}[b] x(t)=z(t)-p(t)x\bigl(\tau(t)\bigr) \geqslant z(t)-p(t)z\bigl(\tau(t)\bigr)\geqslant \biggl(1-p(t) \frac{\pi(\tau(t))}{\pi (t)} \biggr)z(t). \end{aligned}
(4.12)

Using (4.12) in (4.5), we have

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl\{ r(t)\bigl[T_{\alpha}^{t_{0}}\bigl(z(t)\bigr) \bigr]^{\beta}\bigr\} \leqslant -q(t) \biggl(1-p\bigl(\sigma(t)\bigr) \frac{\pi(\tau(\sigma(t)))}{ \pi(\sigma(t))} \biggr)^{\beta}z^{\beta}\bigl(\sigma(t) \bigr)\leqslant0. \end{aligned}
(4.13)

Define a generalized Riccati substitution by

\begin{aligned}[b] w(t):=\delta(t) \biggl[ \frac{r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta}}{z^{\beta}(t)}+\frac{1}{\pi^{\beta}(t)} \biggr]. \end{aligned}
(4.14)

By (4.11), $$w(t)\geqslant0$$ for all $$t\geqslant t_{2}$$. Applying $$T_{\alpha}^{t_{0}}$$ to (4.14), we have

\begin{aligned}[b] T_{\alpha}^{t_{0}} w(t)&=\frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}w(t)\\ &\quad {} +\delta(t) \biggl(\frac{T_{\alpha}^{t_{0}}(r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta})}{z^{\beta}} - \frac{\beta r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta+1}}{z^{\beta +1}(t)}-\beta\pi^{-(\beta+1)}T_{\alpha}^{t_{0}} \pi(t) \biggr) \\ &=\frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}w(t)\\ &\quad {}+\delta(t)\frac {T_{\alpha}^{t_{0}}(r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta})}{z^{\beta}}-\beta\delta(t)r(t) \biggl(\frac {w(t)}{\delta(t)r(t)}-\frac{1}{r(t)\pi^{\beta}(t)} \biggr) ^{(\beta+1)/\beta} \\ &\quad{}+\frac{\beta\delta(t)}{r^{1/\beta}(t)\pi^{1+\beta}(t)}. \end{aligned}
(4.15)

Let $$A:=w(t)/(\delta(t)r(t))$$ and $$B=1/(r(t)\pi^{\beta}(t))$$. Using Lemma 4.1, we conclude that

\begin{aligned} &\biggl(\frac{w(t)}{\delta(t)r(t)}-\frac{1}{r(t)\pi^{\beta}(t)} \biggr)^{\frac{\beta+1}{\beta}} \\ &\quad \geqslant \biggl(\frac{w(t)}{\delta(t)r(t)} \biggr)^{\frac{\beta +1}{\beta}}- \frac{1}{\beta r^{1/\beta}(t)\pi(t)} \biggl[(1+\beta)\frac{w(t)}{\delta(t)r(t)}-\frac{1}{r(t)\pi^{\beta}(t)} \biggr]. \end{aligned}

On the other hand, we get by (4.13) that $$T_{\alpha}^{t_{0}}z<0$$ and from $$\sigma(t)\leqslant t$$ that

$$\frac{T_{\alpha}^{t_{0}}\{r(t)[T_{\alpha}^{t_{0}}(z(t))]^{\beta}\} }{z^{\beta}(t)}\leqslant-q(t) \biggl(1-p\bigl(\sigma(t)\bigr) \frac{\pi(\tau(\sigma(t)))}{\pi(\sigma(t))} \biggr)^{\beta}.$$

Thus (4.15) yields

\begin{aligned}[b] T_{\alpha}^{t_{0}} w(t)&= \frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}w(t)+\delta(t)\frac {T_{\alpha}^{t_{0}}(r(t)(T_{\alpha}^{t_{0}} z(t))^{\beta})}{z^{\beta}} \\ &\quad{}-\beta\delta(t)r(t) \biggl(\frac {w(t)}{\delta(t)r(t)}-\frac{1}{r(t)\pi^{\beta}(t)} \biggr) ^{(\beta+1)/\beta}+\frac{\beta\delta(t)}{r^{1/\beta}(t)\pi ^{1+\beta}(t)} \\ &\leqslant\frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}w(t)-\delta (t)q(t) \biggl(1-p\bigl(\sigma(t) \bigr) \frac{\pi(\tau(\sigma(t)))}{\pi(\sigma(t))} \biggr)^{\beta}+\frac {\beta\delta(t)}{r^{1/\beta}(t)\pi^{1+\beta}(t)} \\ &\quad{}-\beta\delta(t)r(t) \biggl( \biggl(\frac{w(t)}{\delta(t)r(t)} \biggr)^{\frac{\beta+1}{\beta}}- \frac{1}{\beta r^{1/\beta}(t)\pi(t)} \biggl[(1+\beta)\frac{w(t)}{\delta(t)r(t)} -\frac{1}{r(t)\pi^{\beta}(t)} \biggr]\biggr) \\ &=-\delta(t) \biggl[q(t) \biggl(1-p\bigl(\sigma(t)\bigr)\frac{\pi(\tau(\sigma (t)))}{\pi(\sigma(t))} \biggr)^{\beta}+\frac{1-\beta}{r^{1/\beta}(t)\pi^{\beta+1}(t)} \biggr] \\ &\quad{}+ \biggl[\frac{T_{\alpha}^{t_{0}}\delta(t)}{\delta(t)}+ \frac{1+\beta}{r^{1/\beta}(t)\pi(t)} \biggr]w(t)- \frac{\beta }{(\delta(t)r(t))^{1/\beta}}w^{(\beta+1)/\beta}(t) , \end{aligned}

that is,

\begin{aligned}[b] T_{\alpha}^{t_{0}} w(t)\leqslant-\psi(t)+\bigl(\varphi(t)\bigr)_{+}w(t)- \frac{\beta}{(\delta (t)r(t))^{1/\beta}}w^{(\beta+1)/\beta}(t). \end{aligned}
(4.16)

Denote $$C:=\beta/(\delta(t)r(t))^{1/\beta}$$, $$D:=(\varphi(t))_{+}$$, and $$v:=w(t)$$. Applying inequality (4.2), we obtain

\begin{aligned}[b] \bigl(\varphi(t) \bigr)_{+}w(t)-\frac{\beta}{(\delta(t)r(t))^{1/\beta }}w^{(\beta+1)/\beta}(t)\leqslant \frac{\delta(t)r(t) ((\varphi(t))_{+})^{\beta+1}}{(\beta+1)^{\beta+1}}. \end{aligned}
(4.17)

By (4.16) and (4.17) we have

$$T_{\alpha}^{t_{0}} w(t)\leqslant-\psi(t)+\frac{\delta(t)r(t)((\varphi(t))_{+})^{\beta +1}}{(\beta+1)^{\beta+1}}.$$

Applying $$I_{\alpha}$$ to the latter inequality from $$t_{0}$$ to t, we have

$$I_{\alpha}^{t_{0}} \biggl[\psi(t)-\frac{\delta(t)r(t)((\varphi (t))_{+})^{\beta+1}}{(\beta+1)^{\beta+1}} \biggr] \leqslant -w(t)+w(t_{0}),$$

which contradicts (4.4). Therefore (4.1) oscillates. □

### Example 4.1

We consider the equation

\begin{aligned}[b] T_{\frac{1}{2}}^{1} \biggl(t^{2}T_{\frac{1}{2}}^{1}\biggl(x(t)+p(t)x \biggl(\frac {t}{2}\biggr)\biggr)\biggr)+q(t)x(t)=0, \quad t\geqslant0, \end{aligned}
(4.18)

where $$p(t)=\frac{1}{5}$$ and $$q(t)=(2+\frac{4\sqrt{2}}{5})t$$. Let $$\rho(t)=1$$ and $$\delta(t)=1/t$$. Then we have

\begin{aligned}[b] &I_{\alpha}^{t_{0}} \biggl( \rho(t)Q(t)-\bigl(\sigma(t)-t_{0}\bigr)^{(1-\alpha )\beta} \frac{(T_{\alpha}^{t_{0}}\rho_{+}(t))^{\beta+1} r(\sigma(t))}{ (\beta+1)^{\beta+1}\rho^{\beta(t)}(T_{\alpha}^{t_{0}}\sigma (t))^{\beta}} \biggr)\\ &\quad =I_{\alpha}^{t_{0}}Q(t)=I_{\alpha}^{t_{0}} \biggl(\frac{4}{5}\biggl(2+\frac{4\sqrt{2}}{5}\biggr)t \biggr), \end{aligned}

and it is obvious that (4.3) holds. Because of $$\varphi(t)=2/\sqrt{t}$$, $$\psi(t)=(q_{0}(1-2\sqrt{2}p_{0}))/t=\frac{34}{25}$$. So

$$I^{t_{0}}_{\alpha}\biggl[\psi(t)-\frac{\delta(t)r(t)((\varphi (t))_{+})^{\beta+1}}{(\beta+1)^{\beta+1}}\biggr] =I^{t_{0}}_{\alpha}\biggl[\frac{34}{25} - \frac{\frac{1}{t}t^{2}(\frac{2}{\sqrt {t}})^{2}}{2^{2}}\biggr]=I^{t_{0}}_{\alpha} \frac{9}{25},$$

and we can conclude that condition (4.4) is satisfied. Hence by Theorem 4.1 we deduce that (4.18) oscillates.

## 5 Oscillation of 3α-order damped conformable fractional differential equation

This section deals with oscillatory behavior of all solutions of the 3α-order nonlinear delay damped equation of the form

\begin{aligned}[b] &T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} \bigl(r_{1} \bigl(T_{\alpha}^{t_{0}} y\bigr)^{\beta}\bigr)\bigr) (t)+p(t) \bigl(T_{\alpha}^{t_{0}} y(t)\bigr)^{\beta}+q(t)f \bigl(y\bigl(g(t)\bigr)\bigr)=0,\quad t\geqslant t_{0}, \end{aligned}
(5.1)

where $$0<\alpha\leqslant1$$, and $$\beta\geqslant1$$ is the ratio of positive odd integers. We further assume that the following conditions are satisfied:

(H1):

$$r_{1},r_{2},p,q\in C(I,\mathbb{R}^{+})$$, where $$I=[t_{0},\infty)$$, $$\mathbb{R}^{+}=(0,\infty)$$;

(H2):

$$g\in C^{1}(I,\mathbb{R})$$, $$T_{\alpha}^{t_{0}} g(t)\geqslant0$$ and $$g(t)\rightarrow\infty$$ as $$t\rightarrow\infty$$;

(H3):

$$f \in C(\mathbb{R},\mathbb{R})$$ is such that $$xf(x)>0$$ for $$x\neq0$$, and $$f(x)/x^{\gamma}\geqslant k>0$$, where γ is the ratio of positive odd integers.

We define

\begin{aligned}[b] &R_{1}(t,t_{0})=I_{\alpha}^{t_{0}} \frac{1}{r_{1}^{1/\beta}(t)},\qquad R_{2}(t,t_{0})=I_{\alpha}^{t_{0}} \frac{1}{r_{2}(t)},\quad \text{and} \quad R^{*}(t,t_{0})=I_{\alpha}^{t_{0}} \biggl(\frac {R_{2}(t,t_{0})}{r_{1}(t)} \biggr)^{1/\beta} \end{aligned}

for $$t_{0}\leqslant t_{1}\leqslant t\leqslant\infty$$ and assume that

\begin{aligned}[b] &R_{1}(t,t_{0}) \rightarrow\infty, \quad t\rightarrow\infty, \end{aligned}
(5.2)

and

\begin{aligned}[b] &R_{2}(t,t_{0}) \rightarrow\infty, \quad t\rightarrow\infty. \end{aligned}
(5.3)

A function y is called a solution of (5.1) if $$y, r_{1}(T_{\alpha}^{t_{0}} y)^{\beta}, r_{2}(r_{1}(T_{\alpha}^{t_{0}} y)^{\beta})\in C^{1}([t_{y},\infty),\mathbb{R})$$ and y satisfies (5.1) for $$[t_{y},\infty)$$ for some $$t_{y}\geqslant t_{0}$$.

For brevity, we define

$$\begin{gathered} L_{0}y(t)=y(t),\qquad L_{1}y(t)=r_{1}(t) \bigl(T_{\alpha}^{t_{0}}(L_{0}y)\bigr)^{\beta}(t),\\ L_{2}y(t)=r_{2}(t)T_{\alpha}^{t_{0}}(L_{1}y) (t),\qquad L_{3}y(t)=T_{\alpha}^{t_{0}}(L_{2}y) (t) \end{gathered}$$

on I. Then (5.1) can be written as

$$L_{3}y(t)+\frac{p(t)}{r_{1}(t)}L_{1}y(t)+q(t)f\bigl(y \bigl(g(t)\bigr)\bigr)=0.$$

The purpose of this section is to ensure that any solution of (5.1) oscillates when the related second-order linear ordinary fractional differential equation without delay

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl\{ r_{2}(t)T_{\alpha}^{t_{0}} z(t) \bigr\} + \frac{p(t)}{r_{1}(t)}z(t)=0 \end{aligned}
(5.4)

is nonoscillatory.

Next, we state and prove the following lemmas.

### Lemma 5.1

Let y be a nonoscillatory solution of (5.1) on I. Suppose (5.4) is nonoscillatory. Then there exists $$t_{2}\in[t_{1},\infty)$$ such that $$y(t)L_{1}y(t)>0$$ or $$y(t)L_{1}y(t)<0$$, $$t\geqslant t_{2}$$.

### Proof

Let y be a nonoscillatory solution of (5.1) on $$[t_{1},\infty)$$, say $$y(t)>0$$ and $$y(g(t))>0$$ for $$t\geqslant t_{1}\geqslant t_{0}$$. Let $$x=-L_{1}y(t)$$. By (5.1) we have

$$T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr) (t) +\frac{p(t)}{r_{1}(t)}x(t)=q(t)f\bigl(y\bigl(g(t)\bigr)\bigr)>0, \quad t\geqslant t_{1}.$$

Let $$u(t)$$ be a positive solution of (5.4), say $$u(t)>0$$ for $$t\geqslant t_{1}\geqslant t_{0}$$. If x is oscillatory, then x has consecutive zeros at a and b ($$t_{1}< a< b$$) such that $$T_{\alpha}^{t_{0}} x(a)\geqslant0$$, $$T_{\alpha}^{t_{0}} x(b)\leqslant0$$, and $$x(t)>0$$ for $$t\in(a,b)$$. Then we obtain

\begin{aligned} 0&< \int^{b}_{a} \biggl[T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr) (t)+ \frac{p(t)}{r_{1}(t)}x(t) \biggr]u(t)\,d \alpha(t,a) \\ &= \int^{b}_{a}(t-a)^{1-\alpha} \bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr)'(t) \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}u(t)\,d \alpha(t,a)+ \int _{a}^{b}\frac{p(t)}{r_{1}(t)}x(t)u(t)\,d \alpha(t,a) \\ &=r_{2}(t)T_{\alpha}^{t_{0}} x(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}u(t)\bigg|_{a}^{b}- \int _{a}^{b}\bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr) (t)T_{\alpha}^{a} \biggl[ \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha} u(t) \biggr]\,d \alpha(t,a) \\ &\quad{} + \int_{a}^{b}\frac{p(t)}{r_{1}(t)}x(t)u(t)\,d \alpha(t,a) \\ &=r_{2}(t)T_{\alpha}^{t_{0}} x(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}u(t)\bigg|_{a}^{b}+ \int_{a}^{b}\frac {p(t)}{r_{1}(t)}x(t)u(t)\,d \alpha(t,a) \\ &\quad{} - \int_{a}^{b} r_{2}(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}T_{\alpha}^{a} \biggl[ \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha} u(t) \biggr](t-a)^{1-\alpha} x'(t)\,d \alpha(t,a) \\ &=r_{2}(t)T_{\alpha}^{t_{0}} x(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}u(t)\bigg|_{a}^{b}- \biggl\{ r_{2}(t) \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha} T_{\alpha}^{a} \biggl[ \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha} u(t) \biggr] \biggr\} x(t)\bigg|^{b}_{a} \\ &\quad{} + \int_{a}^{b}T_{\alpha}^{a} \biggl\{ r_{2}(t) \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}T_{\alpha}^{a} \biggl[ \biggl(\frac{t-t_{0}}{t-a} \biggr)^{1-\alpha} u(t) \biggr] \biggr\} x(t)\,d \alpha(t,a) \\ &\quad{}+ \int_{a}^{b}\frac{p(t)}{r_{1}(t)}u(t)x(t)\,d\alpha (t,a) \\ &=r_{2}(t)T_{\alpha}^{t_{0}} x(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha}u(t)\bigg|_{a}^{b}+ \int _{a}^{b} \biggl\{ T_{\alpha}^{t_{0}} \bigl\{ r_{2}(t)T_{\alpha}^{t_{0}} u(t) \bigr\} + \frac{p(t)}{r_{1}(t)}u(t) \biggr\} x(t)\,d \alpha(t,a) \\ &=r_{2}T_{\alpha}^{t_{0}} x(t) \biggl( \frac{t-t_{0}}{t-a} \biggr)^{1-\alpha }u(t)\bigg|_{a}^{b} \leqslant0, \end{aligned}

which yields a contradiction. This completes the proof. □

### Lemma 5.2

If y is a nonoscillatory solution of (5.1) and $$y(t)L_{1}y(t)>0$$, $$t\geqslant t_{1}\geqslant t_{0}$$, then

\begin{aligned}[b] L_{1}y(t)\geqslant R_{2}(t,t_{0})L_{2}y(t) \quad\textit{for all } t\geqslant t_{1} \end{aligned}
(5.5)

and

\begin{aligned}[b] y(t)\geqslant R^{*}(t,t_{0}) (L_{2}y)^{1/\beta}(t) \quad\textit{for all } t\geqslant t_{1}. \end{aligned}
(5.6)

### Proof

If y is a nonoscillatory solution of (5.1), then $$y(t)>0$$, $$y(g(t))>0$$, and $$L_{1}y(t)>0$$ for $$t\geqslant t_{1}\geqslant t_{0}$$. It is easy to see that

$$L_{3}y(t)=-\frac{p(t)}{r_{1}(t)}L_{1}y(t)-q(t)f \bigl(y\bigl(g(t)\bigr)\bigr)\leqslant0,$$

which implies that $$L_{2}y(t)$$ is nonincreasing on $$[t_{1},\infty)$$. Applying $$I_{\alpha}$$ to $$T_{\alpha}^{t_{0}}L_{1}y(t)=\frac{L_{2}y(t)}{r_{2}(t)}$$ from $$t_{1}$$ to t and Lemma 2.2, we get

\begin{aligned}[b] L_{1}y(t)=L_{1}y(t_{1})+I_{\alpha}^{t_{0}} \biggl[\frac {L_{2}y(t)}{r_{2}(t)} \biggr]\geqslant L_{2}y(t)I_{\alpha}^{t_{0}} \frac{1}{r_{2}(t)}=L_{2}y(t)R_{2}(t,t_{0}) \quad \text{for any } t\geqslant t_{1}. \end{aligned}

Then

$$T_{\alpha}^{t_{0}} y(t)\geqslant \biggl(\frac{R_{2}(t,t_{0})}{r_{1}(t)} \biggr)^{1/\beta }(L_{2}y)^{1/\beta}(t).$$

Now, applying $$I_{\alpha}$$ to the last inequality from $$t_{1}$$ to t, we can obtain from Lemma 2.2 that

\begin{aligned}[b] y(t)&\geqslant y(t_{1})+I_{\alpha}^{t_{0}} \biggl[ \biggl(\frac {R_{2}(t,t_{0})}{r_{1}(t)} \biggr)^{1/\beta}(L_{2}y)^{1/\beta }(t) \biggr] \\ &\geqslant (L_{2}y)^{1/\beta}(t)I_{\alpha}^{t_{0}} \biggl(\frac {R_{2}(t,t_{0})}{r_{1}(t)} \biggr)^{1/\beta} =R^{*}(t,t_{0}) (L_{2}y)^{1/\beta}(t)\quad \text{for } t\geqslant t_{1}. \end{aligned}

This completes the proof. □

In the following two lemmas, we consider the second-order delay differential inequality

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} x(t)\bigr)\geqslant Q(t)x\bigl(h(t)\bigr), \quad t>t_{0}, \end{aligned}
(5.7)

where the function $$r_{2}$$ is as in (5.1), $$Q(t)\in C(I,\mathbb{R}^{+})$$, and $$h(t)\in C^{1}(I,\mathbb{R})$$ is such that $$h(t)\leqslant t$$, $$T_{\alpha}^{t_{0}} h(t)\geqslant0$$ for $$t\geqslant t_{0}$$, and $$h(t)\rightarrow\infty$$ as $$t\rightarrow\infty$$.

### Lemma 5.3

If

\begin{aligned}[b] \limsup_{t\rightarrow\infty}R_{2} \bigl(h(t),t_{0}\bigr)I_{\alpha}^{t_{0}}Q(t)>1, \end{aligned}
(5.8)

then all bounded solutions of (5.7) are oscillatory.

### Proof

Let $$x(t)$$ be a bounded nonoscillatory solution of (5.7), say $$x(t)>0$$ and $$x(h(t))>0$$ for $$t\geqslant t_{1}$$ for some $$t_{1}\geqslant t_{0}$$. By (5.7), $$r_{2}T_{\alpha}^{t_{0}} x(t)$$ is strictly increasing on $$[t_{1},\infty)$$. Hence, for any $$t_{2}\geqslant t_{1}$$, applying $$I_{\alpha}$$ from $$t_{2}$$ to t in $$T_{\alpha}^{t_{0}} x(t)=\frac{r_{2}(t)T_{\alpha}^{t_{0}} x(t)}{r_{2}(t)}$$ and Lemma 2.2 yield

\begin{aligned}[b] x(t)&=x(t_{2})+I_{\alpha}^{t_{0}} \biggl[\frac{r_{2}(t)T_{\alpha}^{t_{0}} x(t)}{r_{2}(t)} \biggr]>x(t_{2}) +r_{2}(t_{2})T_{\alpha}^{t_{0}} x(t_{2})I_{\alpha}^{t_{0}}\frac {1}{r_{2}(t)} \\ &=x(t_{2})+r_{2}(t_{2})T_{\alpha}^{t_{0}} x(t_{2})R_{2}(t,t_{0}), \end{aligned}

so $$T_{\alpha}^{t_{0}} x(t_{2})<0$$, as otherwise (5.3) would imply $$x(t)\rightarrow\infty$$ as $$t\rightarrow\infty$$, a contradiction to the boundedness of x. Altogether,

\begin{aligned}[b] x>0, T_{\alpha}^{t_{0}} x< 0,\quad \text{and} \quad T_{\alpha}^{t_{0}}\bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr)>0 \quad\text{on } [t_{1},\infty). \end{aligned}

Now, for $$v\geqslant u\geqslant t_{1}$$, repeating the previous steps, we have

\begin{aligned}[b] x(u)&>x(u)-x(v)=-I_{\alpha}^{t_{0}} \biggl[\frac{r_{2}(v)T_{\alpha}^{t_{0}} x(v)}{r_{2}(v)} \biggr] \geqslant-r_{2}(v)T_{\alpha}^{t_{0}} x(v)I_{\alpha}^{t_{0}} \frac{1}{r_{2}(v)}\\ &=-r_{2}(v)T_{\alpha}^{t_{0}} x(v)R_{2}(v,t_{0}). \end{aligned}
(5.9)

For $$t\geqslant s\geqslant t_{1}$$, setting $$u=h(s)$$ and $$v=h(t)$$ in (5.9), we get

$$x\bigl(h(s) \bigr)>-r_{2}\bigl(h(t)\bigr)T_{\alpha}^{t_{0}} x \bigl(h(t)\bigr)R_{2}\bigl(h(t),t_{0}\bigr).$$

Applying $$I_{\alpha}$$ to (5.7) from $$h(t)\geqslant t_{1}$$ to t, we obtain from Lemma 2.2 that

\begin{aligned}[b] -r_{2}\bigl(h(t)\bigr)T_{\alpha}^{t_{0}} x\bigl(h(t)\bigr)&>r_{2}(t) T_{\alpha}^{t_{0}}x(t)-r_{2} \bigl(h(t)\bigr)T_{\alpha}^{t_{0}} x\bigl(h(t)\bigr) \\ &\geqslant I_{\alpha}^{t_{0}} \bigl(Q(t)x\bigl(h(t)\bigr) \bigr) \\ &>-r_{2}\bigl(h(t)\bigr)T_{\alpha}^{t_{0}} x \bigl(h(t)\bigr)R_{2}\bigl(h(t),t_{0} \bigr)I_{\alpha}^{t_{0}}Q(t), \end{aligned}

that is,

\begin{aligned}[b] 1>R_{2}\bigl(h(t),t_{0} \bigr)I_{\alpha}^{t_{0}}Q(t). \end{aligned}

Taking lim sup as $$t\rightarrow\infty$$ on both sides of this inequality yields a contradiction to (5.8). This completes the proof. □

### Lemma 5.4

If

\begin{aligned}[b] \limsup_{t\rightarrow\infty,u\rightarrow\infty }R_{2}(u,t_{0})I_{\alpha}^{t_{0}}Q(t)>1, \end{aligned}
(5.10)

then all bounded solutions of (5.7) are oscillatory.

### Proof

Let x be a bounded nonoscillatory solution of (5.7), say $$x(t)>0$$ and $$x(h(t))>0$$ for $$t\geqslant t_{1}$$ for some $$t_{1}\geqslant t_{0}$$. As in Lemma 5.1, we obtain

\begin{aligned}[b] x>0,\qquad T_{\alpha}^{t_{0}} x< 0,\quad \text{and} \quad T_{\alpha}^{t_{0}}\bigl(r_{2}T_{\alpha}^{t_{0}} x\bigr)>0 \quad\text{on } [t_{1},\infty). \end{aligned}

Applying $$I_{\alpha}$$ to(5.7) from $$u\geqslant t_{1}$$ to t, we obtain from the previous forms that

\begin{aligned}[b] {-}r_{2}(u)T_{\alpha}^{t_{0}} x(u)>r_{2}(t)T_{\alpha}^{t_{0}} x(t)-r_{2}(u)T_{\alpha}^{t_{0}} x(u)\geqslant I_{\alpha}^{t_{0}} \bigl(Q(t)x\bigl(h(t)\bigr) \bigr)\geqslant x\bigl(h(t)\bigr)I_{\alpha}^{t_{0}}Q(t), \end{aligned}

so

\begin{aligned}[b] -T_{\alpha}^{t_{0}} x(u)> \biggl(\frac{1}{r_{2}(u)}I_{\alpha}^{t_{0}}Q(t) \biggr)x \bigl(h(t)\bigr). \end{aligned}
(5.11)

We obtain from (5.11) that

\begin{aligned}[b] x\bigl(h(t)\bigr)>x\bigl(h(t)\bigr)-x(u)\geqslant x\bigl(h(t)\bigr)I_{\alpha}^{t_{0}} \biggl[ \biggl( \frac{1}{r_{2}(u)}I_{\alpha}^{t_{0}}Q(t) \biggr) \biggr], \end{aligned}

that is,

\begin{aligned}[b] 1>R_{2}(u,t_{0})I_{\alpha}^{t_{0}}Q(t). \end{aligned}

Taking lim sup as $$u, t\rightarrow\infty$$ on both sides of this inequality yields a contradiction to (5.10). This completes the proof. □

### Theorem 5.1

Assume that (5.2) and (5.3) hold and $$\beta\geqslant \gamma$$. Suppose that there exist two functions $$m,h\in C^{1}(I, \mathbb{R})$$ such that

$$g(t)\leqslant h(t)\leqslant t,\qquad T_{\alpha}^{t_{0}} h(t) \geqslant0,\quad \textit{and} \quad m(t)>0,\quad t\in I,$$

satisfying

\begin{aligned}[b] \limsup_{t\rightarrow\infty}I_{\alpha}^{t_{0}} \biggl[km(t)q(t)-\frac {A^{2}(t)}{4B(t)} \biggr]=\infty, \end{aligned}
(5.12)

and for $$t\geqslant t_{1}$$,

$$\textstyle\begin{cases} A(t)=\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}-\frac {p(t)}{r_{1}(t)}R_{2}(t,t_{0}),\\ B(t)=c^{*}m^{-1}(t)T_{\alpha}^{t_{0}} g(t)(R^{*}(g(t),t_{0}))^{\gamma-1} (\frac{R_{2}(g(t),t_{0})}{r_{1}(g(t))} )^{1/\beta }(t-t_{0})^{\alpha-1}, \end{cases}$$
(5.13)

and that (5.8) or (5.10) holds with

$$Q(t)=ckq(t) \bigl(R_{1}\bigl(h(t),t_{0}\bigr) \bigr)^{\gamma}-\frac{p(t)}{r_{1}(t)}\geqslant 0,\quad t\geqslant t_{1},$$

with $$c,c^{*}> 0$$. Then every solution y of (5.1) and $$L_{2}y(t)$$ are oscillatory.

### Proof

Let y be a nonoscillatory solution of (5.1) on $$[t_{1},\infty)$$, $$t_{1}\geqslant t_{0}$$. We assume that $$y(t)>0$$ and $$y(g(t))>0$$ for $$t\geqslant t_{1}$$. From Lemma 5.1 we have $$L_{1}y(t)<0$$ or $$L_{1}y(t)>0$$ for $$t\geqslant t_{1}$$.

Step 1. We assume that $$L_{1}y(t) > 0$$ on $$[t_{1},\infty)$$. By (5.1) $$L_{2}y$$ is strictly decreasing. Hence, for any $$t_{2}\geqslant t_{1}$$, we have from Lemma 2.2 that

\begin{aligned}[b] L_{1}y(t)&=L_{1}y(t_{2})+I_{\alpha}^{t_{0}} \biggl[ \frac{L_{2}y(t)}{r_{2}(t)} \biggr]\leqslant L_{1}y(t_{2})+L_{2}y(t_{2})I_{\alpha}^{t_{0}} \frac{1}{r_{2}(t)}=L_{1}y(t_{2})+L_{2}y(t_{2})R_{2}(t,t_{2}). \end{aligned}

So $$L_{2}y(t_{2})>0$$ as otherwise (5.3) would imply $$L_{1}y(t)\rightarrow-\infty$$ as $$t\rightarrow\infty$$, a contradiction to the positivity of $$L_{1}y$$. Altogether, $$L_{2}y>0$$ on $$[t_{1},\infty)$$.

Define the following generalized Riccati transformation:

\begin{aligned}[b] w(t)=m(t)\frac{L_{2}y(t)}{y^{\gamma}(g(t))},\quad t \in[t_{1},\infty). \end{aligned}
(5.14)

By the product and quotient rules, α-differentiating w, we obtain

\begin{aligned}[b] T_{\alpha}^{t_{0}} w(t)&=T_{\alpha}^{t_{0}} \biggl[m(t)\frac{L_{2}y(t)}{ y^{\gamma}(g(t))} \biggr] \\ &=T_{\alpha}^{t_{0}} m(t)\frac{L_{2}y(t)}{y^{\gamma}(g(t))}\\ &\quad {}+m(t) \frac{T_{\alpha}^{t_{0}} (L_{2}y(t))y^{\gamma}(g(t)) -\gamma[y^{\gamma-1}(g(t))]y'(g(t))T_{\alpha}^{t_{0}} g(t)L_{2}y(t)}{y^{2\gamma}(g(t))} \\ &=\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)+m(t)\frac{T_{\alpha}^{t_{0}} (L_{2}y(t))}{y^{\gamma}(g(t))} -m(t)\frac{\gamma y'(g(t))T_{\alpha}^{t_{0}} g(t)L_{2}y(t)}{y^{\gamma +1}(g(t))} \\ &=\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)+\frac{T_{\alpha}^{t_{0}} (L_{2}y)(t)}{L_{2}y(t)}w(t) -\gamma T_{\alpha}^{t_{0}} g(t)\frac{ y'(g(t))}{y(g(t))}w(t). \end{aligned}

Using (5.1), (5.5), and assumption (H3) on f, we obtain

\begin{aligned}[b] &\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)+\frac{T_{\alpha}^{t_{0}} (L_{2}y)(t)}{L_{2}y(t)}w(t) \\ &\quad =\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)-\frac{\frac{p(t)}{r_{1}(t)}L_{1}y(t) +q(t)f(y(g(t)))}{L_{2}y(t)}w(t) \\ &\quad =\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)-\frac{\frac{p(t)}{r_{1}(t)}L_{1}y(t)}{L_{2}y(t)}w(t) -\frac{q(t)f(y(g(t)))}{L_{2}y(t)}w(t) \\ &\quad \leqslant\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}w(t)-\frac {p(t)}{r_{1}(t)}R_{2}(t,t_{0})w(t)-km(t)q(t) \\ &\quad = \biggl[\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}-\frac {p(t)}{r_{1}(t)}R_{2}(t,t_{0}) \biggr]w(t)-km(t)q(t) \\ &\quad =A(t)w(t)-km(t)q(t). \end{aligned}

By the definition of $$L_{1}y(t)$$ and (5.5) we obtain

\begin{aligned}[b] (t-t_{0})^{1-\alpha}\bigl(y \bigl(g(t)\bigr)\bigr)'&=T_{\alpha}^{t_{0}} y \bigl(g(t)\bigr)= \biggl(\frac {1}{r_{1}(g(t))}L_{1}y\bigl(g(t)\bigr) \biggr)^{1/\beta} \\ &\geqslant \biggl(\frac{R_{2}(g(t),t_{0})}{r_{1}(g(t))} \biggr)^{1/\beta} \bigl(L_{2}y\bigl(g(t)\bigr) \bigr)^{1/\beta} \\ &\geqslant \biggl(\frac{R_{2}(g(t),t_{0})}{r_{1}(g(t))} \biggr)^{1/\beta} \bigl(L_{2}y(t) \bigr)^{1/\beta}. \end{aligned}

Then

\begin{aligned}[b] \frac{y'(g(t))}{y(g(t))}&\geqslant(t-t_{0})^{\alpha-1} \biggl(\frac {R_{2}(g(t),t_{0})}{m(t)r_{1}(g(t))} \biggr)^{1/\beta} \frac{m^{1/\beta}(t)(L_{2}y)^{1/\beta}(t)}{y^{\gamma/\beta }(g(t))}y^{\gamma/\beta-1} \bigl(g(t)\bigr) \\ &\overset{(5.14)}{=}(t-t_{0})^{\alpha-1} \biggl(\frac {R_{2}(g(t),t_{0})}{m(t)r_{1}(g(t))} \biggr)^{1/\beta}w^{1/\beta} (t)y^{\gamma/\beta-1}\bigl(g(t)\bigr), \end{aligned}

and we obtain

\begin{aligned}[b] T_{\alpha}^{t_{0}} w(t)&\leqslant A(t)w(t)-km(t)q(t) \\ &\quad{}-\gamma T_{\alpha}^{t_{0}} g(t) (t-t_{0})^{\alpha-1} \biggl(\frac {R_{2}(g(t),t_{0})}{m(t)r_{1}(g(t))} \biggr)^{1/\beta}w^{1/\beta}(t) y^{\gamma/\beta-1}\bigl(g(t)\bigr)w(t) \\ &\leqslant A(t)w(t)-km(t)q(t)\\ &\quad {}-\gamma T_{\alpha}^{t_{0}} g(t) (t-t_{0})^{\alpha-1}w^{1/\beta+1}(t)y^{\gamma/\beta -1} \bigl(g(t)\bigr) \biggl(\frac{R_{2}(g(t),t_{0})}{m(t)r_{1}(g(t))} \biggr) ^{1/\beta}. \end{aligned}
(5.15)

Since $$L_{3}y(t)<0$$, we have $$0 < L_{2}y(t)\leqslant L_{2}y(t_{1})$$, $$L_{2}y(t_{1})=c_{1}$$ for $$t\geqslant t_{1}$$. Then

$$r_{2}(t)T_{\alpha}^{t_{0}} (L_{1}y) (t)=L_{2}y(t)\leqslant c_{1},\quad t\geqslant t_{1},$$

and thus we get from Lemma 2.2 that

\begin{aligned}[b] r_{1}(t) \bigl(T_{\alpha}^{t_{0}} y\bigr)^{\beta}(t)&=L_{1}y(t)=L_{1}y(t_{1})+I_{\alpha}^{t_{0}} \biggl[\frac {r_{2}(t)T_{\alpha}^{t_{0}}(L_{1}y(t))}{r_{2}(t)} \biggr]\leqslant L_{1}y(t_{1}) +c_{1}I_{\alpha}^{t_{0}}\frac{1}{r_{2}(t)} \\ & =L_{1}y(t_{1})+c_{1}R_{2}(t,t_{0})= \biggl[\frac {L_{1}y(t_{1})}{R_{2}(t,t_{0})}+c_{1} \biggr]R_{2}(t,t_{0}) \\ &\leqslant \biggl[\frac{L_{1}y(t_{1})}{R_{2}(t_{2},t_{0})}+c_{1} \biggr]R_{2}(t,t_{0})= \tilde{c}_{1}R_{2}(t,t_{0}) \end{aligned}

(note that $$L_{1}y(t_{1})>0$$), where

$$\tilde{c}_{1}=c_{1}+\frac{L_{1}y(t_{1})}{R_{2}(t_{2},t_{0})}.$$

Therefore, we get for all $$t\geqslant t_{2}$$ that

\begin{aligned}[b] y(t)&=y(t_{2})+I_{\alpha}^{t_{0}} \bigl[T_{\alpha}^{t_{0}} y(t) \bigr]\leqslant y(t_{2})+I_{\alpha}^{t_{0}} \biggl( \frac{\tilde {c_{1}}R_{2}(t,t_{0})}{r_{1}(t)} \biggr)^{1/\beta} \\ &=y(t_{2})+\tilde{c_{1}}^{1/\beta}R^{*}(t,t_{0}) = \biggl[\frac{y(t_{2})}{R^{*}(t,t_{0})}+\tilde{c_{1}}^{1/\beta} \biggr]R^{*}(t,t_{0}) \\ &\leqslant \biggl[\frac{y(t_{2})}{R^{*}(t_{2},t_{0})}+\tilde {c_{1}}^{1/\beta} \biggr]R^{*}(t,t_{0}) \\ &=c_{2}R^{*}(t,t_{0}) \end{aligned}

(note that $$y(t_{2})>0$$), where

$$c_{2}=\frac{y(t_{2})}{R^{*}(t_{2},t_{0})}+\tilde{c_{1}}^{1/\beta}.$$

Then we get

\begin{aligned}[b] y^{\gamma/\beta-1}\bigl(g(t)\bigr) \geqslant c_{2}^{\gamma/\beta-1}\bigl(R^{*} \bigl(g(t),t_{0}\bigr)\bigr)^{\gamma/\beta-1},\quad t\geqslant t_{2}. \end{aligned}
(5.16)

By (5.14) and (5.6) we have

\begin{aligned}[b] w(t)&=m(t)\frac{L_{2}y(t)}{y^{\gamma}(g(t))} \leqslant m(t)\frac{L_{2}y(g(t))}{y^{\gamma}(g(t))}\\ &\leqslant m(t) \bigl(R^{*} \bigl(g(t),t_{0}\bigr)\bigr)^{-\beta}y^{\beta-\gamma} \bigl(g(t)\bigr),\quad t\geqslant t_{2}. \end{aligned}
(5.17)

Using (5.16) in (5.17), we get

\begin{aligned}[b] w(t)\leqslant c_{2}^{\beta-\gamma} m(t) \bigl(R^{*}\bigl(g(t),t_{0}\bigr) \bigr)^{-\gamma }, \quad t\geqslant t_{2}. \end{aligned}

Then

\begin{aligned}[b] w^{1/\beta-1}(t)\geqslant& c_{2}^{(1/\beta-1)(\beta-\gamma)}m^{1/\beta -1}(t) \bigl(R^{*} \bigl(g(t),t_{0}\bigr)\bigr)^{-\gamma(1/\beta-1)},\quad t\geqslant t_{2}. \end{aligned}
(5.18)

Using (5.16) and (5.18) in (5.15), we get

\begin{aligned}[b] &T_{\alpha}^{t_{0}} w(t)\\ &\quad \leqslant A(t)w(t)-km(t)q(t) \\ &\qquad{} -\gamma c_{2}^{-\beta+\gamma}m^{-1}T_{\alpha}^{t_{0}} g(t) \bigl(R^{*}\bigl(g(t),t_{0}\bigr) \bigr)^{\gamma-1} \biggl(\frac {R_{2}(g(t),t_{0})}{r_{1}(g(t))} \biggr)^{1/\beta}(t-t_{0})^{\alpha -1}w^{2}(t) \\ &\quad =A(t)w(t)-km(t)q(t)-B(t)w^{2}(t) \\ &\quad =-km(t)q(t)- \biggl(\sqrt{B(t)}w(t)-\frac{A(t)}{2\sqrt{B(t)}} \biggr)^{2}+ \frac{A^{2}(t)}{ 4B(t)} \\ &\quad \leqslant-km(t)q(t)+\frac{A^{2}(t)}{4B(t)},\quad t\geqslant t_{2}, \end{aligned}
(5.19)

where $$c^{*}=\gamma c_{2}^{\gamma-\beta}$$, and A and B are as in (5.13). Applying $$I_{\alpha}$$ to (5.19) from $$t_{0}$$ to t, we get

$$I_{\alpha}^{t_{0}} \biggl[km(t)q(t)-\frac{A^{2}(t)}{4B(t)} \biggr]\leqslant w(t_{0})-w(t)\leqslant w(t_{0}),$$

Step 2. Let $$L_{1}y(t) < 0$$ on $$[t_{1},\infty)$$. We consider the function $$L_{2}y(t)$$. The case $$L_{2}y(t)\leqslant0$$ cannot hold for all large t, say $$t\geqslant t_{2}\geqslant t_{1}$$, since by double integration of

$$T_{\alpha}^{t_{0}}y(t)= \biggl(\frac{L_{1}y(t)}{r_{1}(t)} \biggr)^{1/\beta}\leqslant \biggl(\frac{L_{1}y(t_{2})}{r_{1}(t)} \biggr) ^{1/\beta},\quad t\geqslant t_{2},$$

we get from (5.2) that $$y(t) \leqslant0$$ for all large t, which is a contradiction. Thus we assume that $$y(t)>0$$, $$L_{1}y(t)<0$$, and $$L_{2}y(t)\geqslant0$$ for all large t, say $$t\geqslant t_{3}\geqslant t_{2}$$. Now, for $$v\geqslant u\geqslant t_{3}$$, we have

\begin{aligned}[b] y(u)&>y(u)-y(v)=-I_{\alpha}^{t_{0}} \biggl[\frac{r_{1}^{1/\beta }(v)T_{\alpha}^{t_{0}}y(v)}{r_{1}^{1/\beta}(v)} \biggr] \\ &\geqslant-I_{\alpha}^{t_{0}} \biggl[\frac{1}{r_{1}^{1/\beta}(v)} \biggr]r_{1}^{1/\beta}(v)T_{\alpha}^{t_{0}}y(v) \\ &=R_{1}(v,t_{0}) \bigl(-L_{1}y(v) \bigr)^{1/\beta}. \end{aligned}

Letting $$u=g(t)$$ and $$v=h(t)$$, we obtain

\begin{aligned}[b] y\bigl(g(t)\bigr)&\geqslant R_{1} \bigl(h(t),t_{0}\bigr) \bigl(-L_{1}y\bigl(h(t)\bigr) \bigr)^{1/\beta} \\ &=R_{1}\bigl(h(t),t_{0}\bigr)x\bigl(h(t)\bigr),\quad \text{for } h(t)\geqslant g(t)\geqslant t_{3}, \end{aligned}

where $$x(t)=(-L_{1}y(t))^{1/\beta}>0$$ for $$t\geqslant t_{3}$$. By (5.1), since that $$x(t)$$ is decreasing and $$g(t)\leqslant h(t)\leqslant t$$, we get

$$T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} z\bigr) (t)+\frac{p(t)}{r_{1}(t)}z\bigl(h(t)\bigr)\geqslant kq(t) \bigl(R_{1}\bigl(h(t),t_{0}\bigr) \bigr)^{\gamma}z\bigl(h(t)\bigr)z^{\gamma/\beta-1}\bigl(h(t)\bigr),$$

where $$z(t)= x^{\beta}(t)$$. Because $$z(t)$$ is decreasing and $$\beta\geqslant\gamma$$, there exists a constant $$c_{4}>0$$ such that $$z^{\gamma/\beta-1}(t)\geqslant c_{4}$$ for $$t\geqslant t_{2}$$. Then we have

\begin{aligned}[b] T_{\alpha}^{t_{0}} \bigl(r_{2}T_{\alpha}^{t_{0}} z\bigr) (t)&\geqslant kq(t) \bigl(R_{1}\bigl(h(t),g(t)\bigr)\bigr) )^{\gamma}z \bigl(h(t)\bigr)z^{\gamma/\beta -1}\bigl(h(t)\bigr)-\frac{p(t)}{r_{1}(t)}z \bigl(h(t)\bigr) \\ &\geqslant \biggl[c_{4}kq(t) \bigl(R_{1} \bigl(h(t),g(t)\bigr)\bigr)^{\gamma}-\frac {p(t)}{r_{1}(t)} \biggr]z \bigl(h(t)\bigr). \end{aligned}

Proceeding exactly as in the proofs of Lemmas 5.3 and 5.4, we arrive at the desired conclusion, thus completing the proof. □

### Example 5.1

\begin{aligned}[b] &T_{\frac{1}{2}} \bigl(T_{\frac{1}{2}}\bigl(t^{-\frac{3}{2}}T_{\frac{1}{2}} y(t)\bigr) \bigr) +t^{-\frac{5}{2}}T_{\frac{1}{2}} y(t)\\ &\quad {}+ \biggl[\frac{1}{2}(t-2)^{-2}t^{-\frac {1}{2}}+2(t-2)^{-2}t^{-1}+1 \biggr]f\bigl(y(t-2)\bigr)=0, \quad t>0, \end{aligned}
(5.20)

where $$r_{1}(t)=t^{-\frac{3}{2}}$$, $$r_{2}(t)=1$$, $$q(t)=\frac{1}{2}(t-2)^{-2}t^{-\frac{1}{2}}+2(t-2)^{-2}t^{-1}+1$$, $$p(t)=t^{-\frac{5}{2}}$$, $$g(t)=t-2$$, $$h(t)=t-2$$, $$\alpha=\frac{1}{2}$$, $$\beta=1$$, $$\gamma=1$$, $$c^{\ast}=1$$. By taking $$m(t)=1$$ we get

\begin{aligned}& R_{1}(t,t_{0})=I_{\alpha}^{t_{0}} \frac{1}{r_{1}^{1/\beta}(t)}=I_{\alpha}t^{\frac{3}{2}}=\frac{1}{2}t^{2} \rightarrow\infty \quad\text{as } t\rightarrow\infty, \\& R_{2}(t,t_{0})=I_{\alpha}^{t_{0}} \frac{1}{r_{2}(t)}=I_{\alpha}1=2 t^{\frac{1}{2}}\rightarrow\infty \quad\text{as } t\rightarrow\infty, \\& \textstyle\begin{cases} A(t)=\frac{T_{\alpha}^{t_{0}} m(t)}{m(t)}-\frac {p(t)}{r_{1}(t)}R_{2}(t,t_{0})=-t^{-1}R_{2}(t,t_{0})=-t^{-1}2t^{\frac {1}{2}}=2t^{-\frac{1}{2}},\\ B(t)=c^{*}m^{-1}(t)T_{\alpha}^{t_{0}} g(t)(R^{*}(g(t),t_{0}))^{\gamma-1} (\frac{R_{2}(g(t),t_{0})}{r_{1}(g(t))} )^{1/\beta }(t-t_{0})^{\alpha-1}=2(t-2)^{2}t^{-\frac{1}{2}}, \end{cases}\displaystyle \\& \begin{aligned}[b] I_{\alpha}^{t_{0}} \biggl[km(t)q(t)-\frac{A^{2}(t)}{4B(t)} \biggr]&=I_{\alpha}\biggl( \frac{1}{2}(t-2)^{-2}t^{-\frac{1}{2}} +2(t-2)^{-2}t^{-1}+1- \frac{4t^{-1}}{8(t-2)^{2}t^{-\frac{1}{2}}} \biggr) \\ &=I_{\alpha}\bigl(2(t-2)^{-2}t^{-1}+1 \bigr), \end{aligned} \end{aligned}

so

\begin{aligned}& \limsup_{t\rightarrow\infty}I_{\alpha}^{t_{0}} \biggl[km(t)q(t)-\frac {A^{2}(t)}{4B(t)} \biggr]=\infty. \\& \begin{aligned}[b] Q(t)&=ckq(t) \bigl(R_{1} \bigl(h(t),t_{0}\bigr)\bigr)^{\gamma}-\frac{p(t)}{r_{1}(t)} \\ &=\biggl(\frac{1}{2}(t-2)^{-2}t^{-\frac{1}{2}}+2(t-2)^{-2}t^{-1}+1 \biggr) \biggl(\frac{1}{2}(t-2)^{2}\biggr)-t^{-1} \\ &=\frac{1}{4}t^{-\frac{1}{2}}+\frac{1}{2}(t-2)^{2} \geqslant0, \end{aligned} \end{aligned}

and we obtain that

\begin{aligned}[b] I_{\alpha}^{t_{0}} Q(t)&=I_{\alpha}\frac{1}{4}t^{-\frac{1}{2}}+ \frac{1}{2}(t-2)^{2}= \int_{0}^{t} \biggl(\frac{1}{4}s^{-1}+ \frac{1}{2}(s-2)^{2} s^{-\frac{1}{2}} \biggr)\,ds \\ &= \int_{0}^{t} \biggl(\frac{1}{4}s^{-1}+ \frac{1}{2}s^{\frac {3}{2}}-2s^{\frac{1}{2}}+2s^{-\frac{1}{2}} \biggr)\,ds \\ &=\frac{1}{4}\ln t+\frac{1}{5}t^{\frac{5}{2}}- \frac{4}{3}t^{\frac{3}{2}}+4t^{\frac{1}{2}} -\frac{1}{4} \ln0. \end{aligned}

Hence

\begin{aligned}& I_{\alpha}^{t_{0}} Q(t)>I_{\alpha}^{t_{0}} Q(1)=0+\frac{1}{5}-\frac{4}{3}+4-\frac{1}{4}\ln0>1, \quad t>1, \\& R_{2}\bigl(h(t),t_{0}\bigr)>R_{2}\biggl(h \biggl(\frac{9}{4}\biggr),t_{0}\biggr)=2\biggl( \frac{9}{4}-2\biggr)^{\frac {1}{2}}=1, \quad t>\frac{9}{4}, \\& R_{2}(u,t_{0})>R_{2}\biggl( \frac{1}{4},t_{0}\biggr)=2\biggl(\frac{1}{4} \biggr)^{\frac{1}{2}}=1, \quad u>\frac{1}{4}. \end{aligned}

So

\begin{aligned}& \limsup_{t\rightarrow\infty}R_{2}\bigl(h(t),t_{0} \bigr)I_{\alpha}^{t_{0}}Q(t)>1, \\& \limsup_{t\rightarrow\infty,u\rightarrow\infty }R_{2}(u,t_{0})I_{\alpha}^{t_{0}}Q(t)>1, \end{aligned}

Then we see that (5.8) and (5.10) are clearly satisfied, and it is easy to verify that the equation

\begin{aligned}[b] T_{\frac{1}{2}} \bigl(T_{\frac{1}{2}}z(t)\bigr)+t^{-1}z(t)=0 \end{aligned}
(5.21)

is nonoscillatory, and one nonoscillatory solution of (5.21) is $$z(t)=18t^{\frac{1}{3}}$$. Then we get that equation (5.20) is oscillatory.

### Example 5.2

\begin{aligned}[b] T_{\frac{1}{2}} \bigl(t^{-\frac{1}{2}}T_{\frac{1}{2}}\bigl(t^{-\frac{1}{2}}T_{\frac{1}{2}} y(t)\bigr)\bigr)+2t^{-\frac{1}{2}}T_{\frac{1}{2}} y(t)+3y(t)=0, \quad t \geqslant0, \end{aligned}
(5.22)

where $$r_{1}(t)=r_{2}(t)=t^{-\frac{1}{2}}$$, $$p(t)=2t^{-\frac{1}{2}}$$, $$q(t)=3$$, $$k=1$$, $$g(t)=t$$, $$\alpha=\frac{1}{2}$$, $$\beta=\gamma=1$$, $$c^{\ast}=c=1$$. Letting $$m(t)=1$$ and $$h(t)=t$$, we can obtain

$$R_{2}(t,t_{0})=t, \qquad A(t)=-2t, \qquad B(t)=t, \qquad Q(t)=3t-2,$$

so all conditions except (5.12) are satisfied.

Equation (5.22) can be rewritten as

$$y'''(t)+2y'(t)+3y(t)=0.$$

It is obvious that the equation is nonoscillatory. It has a nonoscillatory solution $$x=e^{\frac{1}{2}t}\cos\frac{\sqrt{2}}{2}t$$. We can obtain that condition (5.12) indispensable.

## 6 Conclusion

In this paper, we study three kinds of different order conformable fractional equations and obtain oscillatory results of three equations. Those results unify the oscillation theory of the integral-order and fractional-order differential equations.

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## Funding

This research is supported by the Natural Science Foundation of China (61703180, 61803176), and supported by Shandong Provincial Natural Science Foundation (ZR2016AM17, ZR2017MA043).

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Correspondence to Shurong Sun.

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Feng, L., Sun, S. Oscillation theorems for three classes of conformable fractional differential equations. Adv Differ Equ 2019, 313 (2019). https://doi.org/10.1186/s13662-019-2247-5