Oscillation and Asymptotic Behavior of Third-Order Neutral Delay Differential Equations with Mixed Nonlinearities

Abstract

In the present article, we create new sufficient conditions for the oscillatory and asymptotic behavior of solutions of third-order nonlinear neutral delay differential equations with several super-linear and sub-linear terms. The results are obtained first by applying the arithmetic–geometric mean inequality along with the linearization method and then using comparison method as well as the integral averaging technique. Finally, we show the importance and novelty of the main results by applying them to special cases of the studied equation.

1. Introduction

This paper deals with third-order nonlinear neutral delay differential equations with mixed nonlinearities of the form

$$\left(a(t){\left(z″(t)\right)}^{\alpha }\right)\prime +\sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))=0,t\ge t_0\ge 0,$$

(1)

where $z(t)=x(t)+bx(t-\sigma ).$ In the sequel, the following conditions are assumed without further mention:

(H₁)

$\alpha ,{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ are ratios of odd positive integers such that ${\alpha }_1>{\alpha }_2>\dots >{\alpha }_m>\alpha >{\alpha }_{m+1}>\dots >{\alpha }_n,$ with $\alpha \ge 1,$$b\in [0,\infty ),$$b\ne 1$, and $\sigma \in [0,\infty )$ being constants;

(H₂)

$a,p_i\in C([t_0,\infty ),(0,\infty ))$ for $i=1,2,3,\dots ,n$ and

$$A(t,t_0)={\int }_{t_0}^t{a}^{-1/\alpha }(t)dt\mathrm{with}A(t,t_0)\to \infty \mathrm{as}t\to \infty ;$$

(H₂)

${\tau }_i\in C^1([t_0,\infty ),\mathbb{R})$ with ${\tau }_i(t)<t$ and ${lim}_{t\to \infty }{\tau }_i(t)=\infty$ for $i=1,2,3,\dots ,n.$

Let $\tau (t)=min\{{\tau }_1(t),{\tau }_2(t),\dots ,{\tau }_n(t)\}.$ By a solution of (1), we mean a function $x\in C([t_x,\infty ),\mathbb{R}),$$t_x=min\{t-\sigma ,\tau (t)\}$ such that $a{(z^{″})}^{\alpha }\in C\prime ([t_x,\infty ),\mathbb{R})$ and x satisfies (1) on $[t_x,\infty ).$ We consider only solutions of (1) which satisfy $sup\left\{\right|x(t)|:t\ge T\}>0$ for all $T\in [t_x,\infty )$ and tacitly assume that (1) possesses such solutions.

Definition 1. 

A solution of (1) containing an unbounded number of zeros on $[t_x,\infty )$ is said to be $oscillatory;$ otherwise, it is called $nonoscillatory.$

Definition 2. 

Equation (1) is said to be $almost$$oscillatory$ if its solutions are either oscillatory or tend to zero monotonically.

Oscillation and delay phenomena appear in different models from real-world applications; see the papers [1,2] and the references cited therein for more details. In recent years, the theory and applications of neutral type differential equations have attracted great interest since such equations are used to describe a variety of real-world problems in physics, engineering, mathematical biology, and so on; see, for example, [3,4,5]. For recent applications and general theory of these equations, the reader is referred to the monographs [6,7].

The oscillatory characteristics of third-order delay differential equations are peculiar in the sense that they may have both oscillatory and nonoscillatory solutions, or they may have only oscillatory solutions. For example, in [8], all the solutions of the third-order delay differential equation

$$x‴(t)+x(t-\pi )=0,$$

are oscillatory if $\pi e>3.$ However in [5], the third-order delay differential equation

$$x‴(t)+2x\prime (t)-x(t-{\displaystyle \frac{3\pi }{2}})=0,$$

has the oscillatory solution $x_1(t)=sint$ and a nonoscillatory solution $x_2(t)=exp(\beta t)$, where $\beta >0$ such that

$${\beta }^3+2\beta -exp\left(-{\displaystyle \frac{3\pi }{2}}\beta \right)=0.$$

Because of the abovementioned behavior of solutions of third-order differential equations, there has been great interest in establishing sufficient conditions for the oscillation or nonoscillation of solutions of different classes of differential equations of the third order; see, for example, [5,6,9,10,11,12,13,14,15,16,17,18,19,20,21,22] and the references contained therein.

Recently, in [23], the authors studied the oscillatory behavior of (1) for the case $n=1$ and ${\alpha }_1=\alpha ,$ and in [24], the authors studied the following equation:

$$\left(b_2(t){\left(\left(b_1(t){\left(z\prime (t)\right)}^{{\gamma }_1}\right)\prime \right)}^{{\gamma }_2}\right)\prime +\sum _{i=1}^m{q}_i(t)x^{{\alpha }_i}({\tau }_i(t))=0,$$

(2)

where $z(t)=x(t)+bx(t-{\tau }_0),$ and obtained some sufficient conditions which state that every solution of (2) is either oscillatory or tends to zero eventually (almost oscillatory) under the assumption

$${\int }_{t_0}^{\infty }{b}_1^{-1/{\gamma }_1}(t)dt={\int }_{t_0}^{\infty }{b}_2^{-1/{\gamma }_2}(t)dt=\infty .$$

Since the positive solution of (2) satisfies the condition

$$z\prime (t)>0,{\left(b_1(t){\left(z\prime (t)\right)}^{{\gamma }_1}\right)}^{\prime }>0,\left(b_2(t){\left(\left(b_1(t){\left(z\prime (t)\right)}^{{\gamma }_1}\right)\prime \right)}^{{\gamma }_2}\right)\prime \le 0,$$

using this, the authors infer that $z″(t)>0$ for $t\ge t_0,$ which is necessary to use Lemma 1.5.8 of [7] to obtain the main results in [24]. This is not true in general; for example, if $b_1(t)=t$ and ${\gamma }_1=1$, then we have $z\prime (t)+tz″(t)>0,$ and this may not imply that $z″(t)>0$ for $t\ge t_0.$ However, this is used in [24] to obtain the main results, and hence the results in [24] may not be correct unless they have to assume that $b_1(t)$ is either constant or monotonically decreasing. Note that the authors used the function $b_1(t)={\displaystyle \frac{1}{t}}$ in their examples, which is clearly monotonically decreasing.

Motivated by the above observations and inspired by recent works [23,24], in this study, we consider Equation (1), which is the same as (2) if ${\gamma }_1=1$ and $b_1(t)\equiv 1$, and then using the linearization method and the arithmetic–geometric inequality, we obtain some new criteria for the oscillation and asymptotic behavior of solutions of $(1)$. Here, we set ${\gamma }_1=1$ to apply the linearization technique to reduce the studied equation into a linear equation, and for the sake of simplicity assumed that $b_1(t)\equiv 1.$ This approach modified and corrected the results in [24]. Examples are provided to illustrate the importance and novelty of the main results.

2. Main Results

We begin with the following preliminary results, which will be used in the proof of the main results.

Lemma 1. 

Assume that

$${\alpha }_i>\alpha ,i=1,2,3,\dots ,mand{\alpha }_i<\alpha ,i=m+1,m+2,\dots ,n.$$

(3)

Then, an n-tuple $({\eta }_1,{\eta }_2,\dots ,{\eta }_n)$ exists with ${\eta }_i>0$ satisfying the conditions

$$\sum _{i=1}^n{\alpha }_i{\eta }_i=\alpha and\sum _{i=1}^n{\eta }_i=1.$$

(4)

Proof. 

From (4), we see that

$$\sum _{i=1}^n{\beta }_i{\eta }_i=1and\sum _{i=1}^n{\eta }_i=1$$

where ${\beta }_i={\displaystyle \frac{{\alpha }_i}{\alpha }}.$ The rest of the proof is similar to Lemma 1 of [25], and hence the details are omitted. This indicates the proof is complete. □

Lemma 2 

([7], Lemma 1.5.1). Let $h,g:[t_0,\infty )\to \mathbb{R}$ such that $h(t)=g(t)+bg(t-c),$ $t\ge t_0+max\{0,c\},$ where $b\ne 1$ and c are non-negative constants. Assume that there exists a constant $l\in \mathbb{R}$ such that ${lim}_{t\to \infty }h(t)=l.$

(i) 

If ${lim}_{t\to \infty }infg(t)=g_{*}\in \mathbb{R},$ then $l=(1+b)g_{*};$

(ii) 

If ${lim}_{t\to \infty }supg(t)=g^{*}\in \mathbb{R},$ then $l=(1+b)g^{*}.$

Note that the above lemma is applicable only when the neutral term has discrete delay and not for variable delay term, and it can be used for all values of b except $b=1.$

The assumption $A(t,t_0)\to \infty \mathrm{as}t\to \infty$ implies that Equation (1) is in canonical form, and therefore we can use Lemma 1 of [23] to obtain the following classification for eventually positive solutions of $(1)$.

Lemma 3. 

Let $x(t)$ be an eventually positive solution of Equation $(1)$. Then, there exists a sufficiently large $t_1\ge t_0$ such that, for all $t\ge t_1$, either

(I) 

$z(t)>0,$ $z\prime (t)>0,$ $z″(t)>0,$ $(a(t){(z″(t))}^{\alpha })\prime \le 0,$ or

(II) 

$z(t)>0,$  $z\prime (t)<0,$  $z″(t)>0,$  $(a(t){(z″(t))}^{\alpha })\prime \le 0.$

Lemma 4. 

Let $x(t)$ be an eventually positive solution of Equation (1) and assume that Case (II) of Lemma 3 holds. If

$${\int }_{t_0}^{\infty }\xi {\left({\displaystyle \frac{1}{a(\xi )}}{\int }_{\xi }^{\infty }Q(s)ds\right)}^{1/\alpha }d\xi =\infty ,$$

(5)

where

$$Q(t)=\prod _{i=1}^n{\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)}^{{\eta }_i},$$

(6)

with ${\eta }_i$ defined as in Lemma 1, then

$$\underset{t\to \infty }{lim}x(t)=0.$$

(7)

Proof. 

Since $z(t)>0$ and $z\prime (t)<0,$ there exists a constant $M>0$ such that ${lim}_{t\to \infty }z(t)=M\ge 0.$ We claim that $M=0.$ If not, then by using Lemma 2, we see that ${lim}_{t\to \infty }x(t)={\displaystyle \frac{M}{1+b}}>0.$ Then, there exists $t_1\ge t_0$, such that for all $t\ge t_1,$ we have

$$x({\tau }_i(t))\ge {\displaystyle \frac{M}{2(1+b)}},i=1,2,3,\dots ,n.$$

Using the last inequality, we see that

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))& \ge & \sum _{i=1}^n{p}_i(t){\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{{\alpha }_i}\hfill \\ & =& {\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{\alpha }\sum _{i=1}^n{p}_i(t){\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{{\alpha }_i-\alpha }.\hfill \end{array}$$

(8)

According to Lemma 2, there exists ${\eta }_1,{\eta }_2,\dots ,{\eta }_n$ with

$$\sum _{i=1}^n{\alpha }_i{\eta }_i-\alpha \sum _{i=1}^n{\eta }_i=0.$$

The arithmetic–geometric mean inequality (see [26]) leads to

$$\sum _{i=1}^n{\eta }_i{u}_i\ge \prod _{i=1}^n{u}_i^{{\eta }_i},\mathrm{for}\mathrm{any}{u}_i\ge 0,i=1,2,3,\dots ,n.$$

In view of the above inequality, we obtain

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{p}_i(t){\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{{\alpha }_i-\alpha }& =& \sum _{i=1}^n{\eta }_i\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right){\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{{\alpha }_i-\alpha }\hfill \\ & \ge & \prod _{i=1}^n{\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)}^{{\eta }_i}{\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{{\eta }_i({\alpha }_i-\alpha )}\hfill \\ & =& \prod _{i=1}^n{\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)}^{{\eta }_i}=Q(t).\hfill \end{array}$$

This together with (8) yields that

$$\begin{array}{c}\hfill \sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))\ge Q(t){\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{\alpha }.\end{array}$$

(9)

Combining (1) and (9), we take

$$(a(t){(z″(t))}^{\alpha })\prime +{\left({\displaystyle \frac{M}{2(1+b)}}\right)}^{\alpha }Q(t)\le 0.$$

We can further note that there exist constants $M_1$ and $M_2$ such that ${lim}_{t\to \infty }a(t){(z″(t))}^{\alpha }=M_1\ge 0$ and ${lim}_{t\to \infty }z\prime (t)=M_2\le 0.$

A method similar to that in Theorem 15 of [18] leads to the conclusion that ${lim}_{t\to \infty }x(t)=0.$ This completes the proof. □

Remark 1. 

In the proof of Lemma 4 (Theorem 1), we used the arithmetic–geometric mean inequality to reduce the sum of n non-negative terms into a single product, and this is essential to obtain the desired result. Also from Lemma 4, we conclude that every positive decreasing solution of (1) tends to zero as $t\to \infty .$

Lemma 5. 

Let $x(t)$ be a positive solution of (1) with a corresponding function $z(t)\in$ class (I) for all $t\ge t_1.$ Then,

(i) 

$z\prime (t)\ge (a^{1/\alpha }(t)z″(t))A(t,t_1),$

(ii) 

${\displaystyle \frac{z\prime (t)}{A(t,t_1)}}$ is decreasing,

(iii) 

$z(t)\ge (a^{1/\alpha }(t)z″(t))A_1(t,t_1),$

(iv) 

$z(t)\ge A_1(t,t_1){\displaystyle \frac{z\prime (t)}{A(t,t_1)}},$

(ii) 

${\displaystyle \frac{z(t)}{A_1(t,t_1)}}$ is decreasing,

where $A_1(t,t_1)={\int }_{t_1}^{t}A(s,t_1)ds.$

Proof. 

Since $z(t)\in$ class (I), we see that $a(t){(z″(t))}^{\alpha }>0$ and decreasing for all $t\ge t_1.$ Then,

$$\begin{array}{ccc}\hfill z\prime (t)\ge z\prime (t)-z\prime (t_1)& =& {\int }_{t_1}^t{\displaystyle \frac{{(a(s){(z″(s))}^{\alpha })}^{1/\alpha }}{a^{1/\alpha }(s)}}ds\hfill \\ & \ge & A(t,t_1)a^{1/\alpha }(t)z″(t),\hfill \end{array}$$

(10)

which proves (i).

Moreover,

$${\left({\displaystyle \frac{z\prime (t)}{A(t,t_1)}}\right)}^{\prime }={\displaystyle \frac{A(t,t_1)a^{1/\alpha }(t)z″(t)-z\prime (t)}{a^{1/\alpha }(t)A^2(t,t_1)}}\le 0,$$

which implies that ${\displaystyle \frac{z\prime (t)}{A(t,t_1)}}$ is decreasing.

Integrating (10) from $t_1$ to t yields and this proves ($ii$).

$$z(t)\ge A_1(t,t_1)a^{1/\alpha }(t)z″(t),$$

which proves ($iii$).

Since

$$z(t)-z(t_1)={\int }_{t_1}^{t}A(s,t_1){\displaystyle \frac{z\prime (s)}{A(s,t_1)}}ds,$$

or

$$z(t)\ge A_1(t,t_1){\displaystyle \frac{z\prime (t)}{A(t,t_1)}},$$

where we have used ($ii$), this proves ($iv$).

Finally,

$${\left({\displaystyle \frac{z(t)}{A_1(t,t_1)}}\right)}^{\prime }={\displaystyle \frac{A_1(t,t_1)z\prime (t)-A(t,t_1)z(t)}{A_1^2(t,t_1)}}\le 0$$

according to ($iv$). Hence, ${\displaystyle \frac{z(t)}{A_1(t,t_1)}}$ is decreasing. This completes the proof. □

Next, we state and prove the main theorems. In the first result, we use the linearization technique along with the arithmetic–geometric mean inequality to reduce the nonlinear Equation (1) into a first-order linear delay differential equation.

Theorem 1. 

Let condition (5) hold. If the first-order delay differential equation

$$w\prime (t)+{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }{Q}_1(t)w(\tau (t))=0,$$

(11)

where

$$Q_1(t)=Q(t)A_1^{\alpha }(\tau (t),t_2),$$

with $Q(t)$ defined as in (6), is oscillatory for all large $t_1\ge t_0$ and for some $t_2\ge t_1$, then Equation (1) is almost oscillatory.

Proof. 

Let $x(t)$ be a nonoscillatory solution of (1). Then, with no loss of generality, assume $x(t)>0,$$x(t-\sigma )>0$ and $x({\tau }_i(t))>0$ for $t\ge t_1$ for some $t_1\ge t_0.$ Then, we know from Lemma 3 that the corresponding function $z(t)>0$ for all $t\ge t_1$ and satisfies either $(I)$ or $(II)$. If $z(t)$ satisfies case $(II)$, then we know from Lemma 4 that (7) holds, and we need to consider the other case (I).

From (1), we see that

$$\begin{array}{ccc}\hfill (a(t){(z″(t))}^{\alpha })\prime & =& ({(a^{1/\alpha }(t)z″(t))}^{\alpha })\prime \hfill \\ & =& \alpha {(a^{1/\alpha }(t)z″(t))}^{\alpha -1}(a^{1/\alpha }(t)z″(t))\prime \hfill \\ & =& -\sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))\hfill \end{array}$$

and so

$$(a^{1/\alpha }(t)z″(t))\prime +{\displaystyle \frac{1}{\alpha }}{(a^{1/\alpha }(t)z″(t))}^{1-\alpha }\sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))=0.$$

(12)

Since $z\prime (t)>0$ and $z″(t)>0,$ there exists a constant $d_0$ (it is also possible that $d_0=\infty$) such that ${lim}_{t\to \infty }z\prime (t)=d_0>0.$

Consequently, according to Lemma 2, ${lim}_{t\to \infty }x\prime (t)={\displaystyle \frac{d_0}{1+b}}>0,$ and we conclude that

$$x\prime (t)>0.$$

(13)

Using (13), we see that $z(t)=x(t)+bx(t-\sigma )\le (1+b)x(t)$, that is,

$$x(t)\ge \left({\displaystyle \frac{1}{1+b}}\right)z(t).$$

(14)

In addition, we have

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))& \ge & \sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}(\tau (t))\hfill \\ & =& x^{\alpha }(\tau (t))\sum _{i=1}^n{p}_i(t)x^{{\alpha }_i-\alpha }(\tau (t)).\hfill \end{array}$$

(15)

In view of Lemma 1, there exists ${\eta }_1,{\eta }_2,\dots ,{\eta }_n$ with

$$\sum _{i=1}^n{\alpha }_i{\eta }_i-\alpha \sum _{i=1}^n{\eta }_1=0.$$

The arithmetic–geometric mean inequality (see [26]) gives

$$\sum _{i=1}^n{\eta }_i{u}_i\ge \prod _{i=1}^n{u}_i^{{\eta }_i}\mathrm{for}\mathrm{any}{u}_i\ge 0,i=1,2,\dots ,n.$$

Therefore, we have

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{p}_i(t)x^{{\alpha }_i-\alpha }(\tau (t))& =& \sum _{i=1}^n{\eta }_i\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)x^{{\alpha }_i-\alpha }(\tau (t))\hfill \\ & \ge & \prod _{i=1}^n{\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)}^{{\eta }_i}{\left(x(\tau (t))\right)}^{{\eta }_i({\alpha }_i-\alpha )}\hfill \\ & =& \prod _{i=1}^n{\left({\displaystyle \frac{p_i(t)}{{\eta }_i}}\right)}^{{\eta }_i}=Q(t).\hfill \end{array}$$

This together with (15) yields that

$$\sum _{i=1}^n{p}_i(t)x^{{\alpha }_i}({\tau }_i(t))\ge Q(t)x^{\alpha }(\tau (t)).$$

(16)

Using (14) and (16) in (12), we obtain

$$\left(a^{1/\alpha }(t)z″(t)\right)\prime +{\displaystyle \frac{1}{\alpha }}{\left(a^{1/\alpha }(t)z″(t)\right)}^{1-\alpha }Q(t){\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }{z}^{\alpha }(\tau (t))\le 0.$$

(17)

From Lemma 5(iii), we see that

$$z(\tau (t))\ge A_1(\tau (t),t_1)\left(a^{1/\alpha }(t)z″(t)\right)$$

(18)

for $t\ge t_1.$ Since $a^{1/\alpha }(t)z″(t)$ is nonincreasing and $\alpha \ge 1,$ we have

$${\left(a^{1/\alpha }(t)z″(t)\right)}^{1-\alpha }\ge {\left(a^{1/\alpha }(\tau (t))z″(\tau (t))\right)}^{1-\alpha }.$$

(19)

Using (19) in (17) yields

$$\left(a^{1/\alpha }(t)z″(t)\right)\prime +{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }{\left(a^{1/\alpha }(\tau (t))z″(t)\right)}^{1-\alpha }Q(t)z^{\alpha }(\tau (t))\le 0.$$

(20)

From (18) and (20), we observe that

$$\left(a^{1/\alpha }(t)z″(t)\right)\prime +{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }Q(t)A_1^{\alpha }(\tau (t),t_1)\left(a^{1/\alpha }(\tau (t))z″(t)\right)\le 0.$$

(21)

Let $w(t)=a^{1/\alpha }(t)z″(t)$ in (21). We see that w is a positive solution of the first-order linear delay differential inequality

$$w\prime (t)+{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }{Q}_1(t)w(\tau (t))\le 0.$$

The function w is clearly strictly decreasing for all $t\ge t_2$ and so according to Theorem 1 of [27], there exists a positive solution of Equation (11), which contradicts the fact that Equation (11) is oscillatory. The proof of the theorem is complete. □

The next result immediately follows on from Theorem 1 and (Theorem 2.11, [16]).

Corollary 1. 

Let condition (5) hold. If

$$\underset{t\to \infty }{lim}inf{\int }_{\tau (s)}^t{Q}_1(s)ds\ge {\displaystyle \frac{\alpha {(1+b)}^{\alpha }}{e}}$$

(22)

where $Q_1(t)$ is defined as in Theorem 1, then Equation (1) is almost oscillatory.

In our next theorems, we use Riccati transformation and the integral averaging technique to obtain oscillation results. We split the single integral into three integrals to obtain better conditions for the oscillation of $(1).$

Theorem 2. 

Let condition (5) hold and $\tau (t)\in C\prime ([t_0,\infty ))$ with $\tau \prime (t)>0.$ Assuming that there exists a function $\rho \in C\prime ([t_0,\infty ),(0,\infty )),$ for sufficiently large $t_1\ge t_0,$ there is a $t_2\ge t_1$ such that

$$\underset{t\to \infty }{lim}sup{\int }_{t_2}^t\left[\rho (s)Q_2(s)-{\displaystyle \frac{\alpha {(1+b)}^{\alpha }{(\rho \prime (s))}^2}{4\rho (s)A(\tau (s),t_1)\tau \prime (s)}}\right]ds=\infty ,$$

(23)

where $Q_2(t)=Q(t){(A_1(\tau (t)),t_1)}^{\alpha -1}$ with $Q(t)$ defined as in (6) and ${(\rho \prime (t))}_{+}=max\{0,\rho \prime (t)\}.$ Then, Equation (1) is almost oscillatory.

Proof. 

Let $x(t)$ be a nonoscillatory solution of $(1).$ Then, with no loss of generality, assume $x(t)>0,$$x(t-\sigma )>0$ and $x({\tau }_i(t))>0$ for $t\ge t_1$ for some $t_1\ge t_0.$ Then, from Lemma 3, we see that the corresponding function $z(t)>0$ and satisfies either case $(I)$ or case $(II)$ for all $t\ge t_1.$ If $z(t)$ satisfies case $(II)$, then from Lemma 4, we see that (7) holds, and we need to consider the other case (I). It follows from (18) and the fact that $a^{1/\alpha }(t)z″(t)$ is non-increasing that

$$\begin{array}{ccc}\hfill a^{1/\alpha }(t)z″(t)& \le & a^{1/\alpha }(\tau (t))z″(t)\le A_1{(\tau (t),t_1)}^{-1}z(\tau (t))\hfill \end{array}$$

and so

$${(a^{1/\alpha }(t)z″(t))}^{1-\alpha }\ge (A_1{(\tau (t),t_1)}^{\alpha -1}{(z(\tau (t)))}^{1-\alpha }.$$

Using this inequality in (20) yields

$$\left(a^{1/\alpha }(t)z″(t)\right)\prime +{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }{Q}_2(t)z(\tau (t))\le 0,t\ge t_1.$$

(24)

Define

$$w(t)={\displaystyle \frac{\rho (t)a^{1/\alpha }(t)z″(t)}{z(\tau (t))}}.$$

Then, $w(t)>0$, and using (24), we obtain

$$w\prime (t)\le {\displaystyle \frac{\rho \prime (t)}{\rho (t)}}w(t)-{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }\rho (t)Q_2(t)-{\displaystyle \frac{w(t)z\prime (\tau (t))\tau \prime (t)}{z(\tau (t))}}.$$

(25)

From Lemma 5(i), we see that

$$z\prime (\tau (t))\ge (a^{1/\alpha }(\tau (t))z″(\tau (t)))A(\tau (t),t_1)\ge (a^{1/\alpha }(t)z″(t))A(\tau (t),t_1).$$

Combining the last inequality with (25), we obtain

$$w\prime (t)\le {\displaystyle \frac{-1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }\rho (t)Q_2(t)+{\displaystyle \frac{\rho \prime (t)}{\rho (t)}}w(t)-{\displaystyle \frac{A(\tau (t),t_1)\tau \prime (t)}{\rho (t)}}{w}^2(t).$$

(26)

Using the inequality $Bu-A{u}^2\le {\displaystyle \frac{1}{4}}{\displaystyle \frac{B^2}{A}},$$A>0$ in (26), we have

$$w\prime (t)\le {\displaystyle \frac{-1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }\rho (t)Q_2(t)+{\displaystyle \frac{{(\rho \prime (t))}^2}{4\rho (t)A(\tau (t),t_1)\tau \prime (t)}},t\ge t_2\ge t_1.$$

Integrating the above inequality $t_2$ to $t,$ we obtain

$${\int }_{t_2}^t\left[{\displaystyle \frac{1}{\alpha }}{\left({\displaystyle \frac{1}{1+b}}\right)}^{\alpha }\rho (s)Q_2(s)-{\displaystyle \frac{{(\rho \prime (s))}^2}{4\rho (s)A(\tau (s),t_1)\tau \prime (s)}}\right]ds\le w(t_2)$$

which contradicts (23). The proof of the theorem is complete. □

Theorem 3. 

Let condition (5) hold and

$$\begin{array}{c}\hfill \underset{t\to \infty }{lim}sup\left\{{\displaystyle \frac{1}{A(\tau (t),t_{*})}}{\int }_{t_{*}}^{\tau (t)}{Q}_2(s)A(s,t_{*})A_1(\tau (s),t_{*})ds+{\int }_{\tau (t)}^t{Q}_2(s)A_1(\tau (s),t_{*})ds\right.\\ \hfill \left.+A_1(\tau (s),t_{*}){\int }_t^{\infty }{Q}_2(s)ds\right\}>\alpha {(1+b)}^{\alpha },\end{array}$$

(27)

where $Q_2(t)$ is as defined in Theorem 2. Then, Equation (1) is almost oscillatory.

Proof. 

Let $x(t)$ be a positive solution of (1) with $x(t-\sigma )>0$ and $x({\tau }_i(t))>0$ for $t\ge t_1$ for some $t_1\ge t_0.$ Then, the corresponding function $z(t)>0$ and satisfies case $(I)$ or case $(II)$ of Lemma 3 for all $t\ge t_1.$ If $z(t)$ satisfies case $(II)$, then from Lemma 4, we see that (7) holds, and therefore we need to consider the other case (I). Proceeding as in the proof of Theorem 2, we arrive at (24).

Integrating (24) from t to ∞ yields

$$\begin{array}{c}\hfill z″(t)\ge {\displaystyle \frac{1}{\alpha {(1+b)}^{\alpha }}}{\displaystyle \frac{1}{a^{1/\alpha }(t)}}{\int }_t^{\infty }{Q}_2(s)z(\tau (s))ds.\end{array}$$

Integrating again from $t_1$ to t, we obtain

$$\begin{array}{ccc}\hfill \alpha {(1+b)}^{\alpha }z\prime (t)& \ge & {\int }_{t_1}^t{\displaystyle \frac{1}{a^{1/\alpha }(u)}}{\int }_u^{\infty }{Q}_2(s)z(\tau (s))ds\hfill \\ & =& {\int }_{t_1}^t{\displaystyle \frac{1}{a^{1/\alpha }(u)}}{\int }_u^t{Q}_2(s)z(\tau (s))ds+{\int }_{t_1}^t{\displaystyle \frac{1}{a^{1/\alpha }(u)}}{\int }_t^{\infty }{Q}_2(s)z(\tau (s))ds\hfill \\ & =& {\int }_{t_1}^{t}A(u,t_1)Q_2(s)z(\tau (s))ds+A(t,t_1){\int }_t^{\infty }{Q}_2(s)z(\tau (s))ds.\hfill \end{array}$$

Employing Lemma 5($iv$), we have

$$\begin{array}{ccc}\hfill \alpha {(1+b)}^{\alpha }{\displaystyle \frac{z(t)A(t,t_1)}{A_1(t,t_1)}}& \ge & {\int }_{t_1}^{t}A(s,t_1)Q_2(s)z(\tau (s))ds+A(t,t_1){\int }_t^{\infty }{Q}_2(s)z(\tau (s))ds,\hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill \alpha {(1+b)}^{\alpha }{\displaystyle \frac{z(\tau (t))A(\tau (t),t_1)}{A_1(\tau (t),t_1)}}& \ge & {\int }_{t_1}^{\tau (t)}A(s,t_1)Q_2(s)z(\tau (s))ds+A(\tau (t),t_1){\int }_{\tau (t)}^t{Q}_2(s)z(\tau (s))ds\hfill \\ & & +A(\tau (t),t_1){\int }_t^{\infty }{Q}_2(s)z(\tau (s))ds.\hfill \end{array}$$

Taking into account the fact that $z(t)$ is increasing and ${\displaystyle \frac{z(t)}{A_1(t,t_1)}}$ is decreasing, one can verify that

$$\begin{array}{ccc}\hfill \alpha {(1+b)}^{\alpha }{\displaystyle \frac{z(\tau (t))A(\tau (t),t_1)}{A_1(\tau (t),t_1)}}& \ge & {\displaystyle \frac{z(\tau (t))}{A_1(\tau (t),t_1)}}{\int }_{t_1}^{\tau (t)}A(s,t_1)A_1(\tau (s),t_1)Q_2(s)ds\hfill \\ & & +{\displaystyle \frac{A(\tau (t),t_1)z(\tau (t))}{A_1(\tau (t),t_1)}}{\int }_{\tau (t)}^t{Q}_2(s)A_1(\tau (s),t_1)ds\hfill \\ & & +A(\tau (t),t_1)z(\tau (t)){\int }_t^{\infty }{Q}_2(s)ds,\hfill \end{array}$$

which yields

$$\begin{array}{ccc}\hfill \alpha {(1+b)}^{\alpha }& \ge & {\displaystyle \frac{1}{A(\tau (t),t_1)}}{\int }_{t_1}^{\tau (t)}A(s,t_1)A_1(\tau (s),t_1)Q_2(s)ds+{\int }_{\tau (t)}^t{Q}_2(s)A_1(\tau (s),t_1)ds\hfill \\ & & +A_1(\tau (t),t_1){\int }_t^{\infty }{Q}_2(s)ds.\hfill \end{array}$$

Taking $limsup$ as $t\to \infty$ on both sides of the last inequality, we are led to a contradiction with (27). The proof of the theorem is complete. □

Theorem 4. 

Let condition (5) hold, and

$$\underset{t\to \infty }{lim}infA(t,t_1){\int }_t^{\infty }{\displaystyle \frac{Q(s){(A_1(\tau (s),t_1))}^{\alpha }}{A(s,t_1)}}ds>{\displaystyle \frac{\alpha {(1+b)}^{\alpha }}{4}},$$

(28)

where $Q(t)$ defined as in (6). Then, Equation (1) is almost oscillatory.

Proof. 

Let $x(t)$ be a positive solution of (1) with $x(t-\sigma )>0$ and $x({\tau }_i(t))>0$ for $t\ge t_1$ for some $t_1\ge t_0.$ Then, the corresponding function $z(t)>0$ and satisfies case $(I)$ or case $(II)$ of Lemma 3 for all $t\ge t_1.$ If $z(t)$ satisfies case $(II)$, then from Lemma 4, we see that (7) holds and therefore we need to consider the other case (I). Proceeding as in the proof of Theorem 2, we arrive at (24).

From Lemma 5($ii$) and ($iv$), we have

$$\begin{array}{ccc}\hfill z(\tau (t))& \ge & A_1(\tau (t),t_1){\displaystyle \frac{z\prime (t)}{A(t,t_1)}}\hfill \end{array}$$

and using this inequality in (24), we obtain

$$\left(a^{1/\alpha }(t)z″(t)\right)\prime +{\displaystyle \frac{1}{\alpha {(1+b)}^{\alpha }}}{\displaystyle \frac{Q(t)A^{\alpha }(\tau (t),t)}{A(t,t_1)}}z\prime (t)\le 0.$$

Let $u(t)=z\prime (t).$ Then, we see that $u(t)$ is a positive solution of the inequality

$$\left(a^{1/\alpha }(t)u\prime (t)\right)\prime +{\displaystyle \frac{1}{\alpha {(1+b)}^{\alpha }}}{\displaystyle \frac{Q(t)A_1^{\alpha }(\tau (t),t)}{A(t,t_1)}}u(t)\le 0.$$

Define

$$w(t)={\displaystyle \frac{a^{1/\alpha }(t)u\prime (t)}{u(t)}},t\ge t_1.$$

Then $w(t)>0$ and satisfies

$$w(t)\ge {\displaystyle \frac{1}{\alpha {(1+b)}^{\alpha }}}{\int }_t^{\infty }{\displaystyle \frac{Q(s)A_1^{\alpha }(\tau (s),t_1)}{A(s,t_1)}}ds+{\int }_t^{\infty }{\displaystyle \frac{w^2(s)}{a^{1/\alpha }(s)}}ds.$$

(29)

Multiply the inequality (29) by $A(t,t_1)$ and letting $M={inf}_{t\ge t_1}R(t)w(t),$ we see that

$$\begin{array}{ccc}\hfill M& >& {\displaystyle \frac{1}{4}}+M^{2}R(t){\int }_t^{\infty }{\displaystyle \frac{1}{a^{1/\alpha }(s)R^2(s)}}ds\hfill \\ & =& {\displaystyle \frac{1}{4}}+M^{2}R(t){\displaystyle \frac{1}{R(t)}}ds={\displaystyle \frac{1}{4}}+M^2,\hfill \end{array}$$

which contradicts the admissible value of $M.$ The proof of the theorem is complete. □

3. Examples

In this section, we present some examples to illustrate the main results.

Example 1. 

Consider the third-order nonlinear neutral differential equation of the form

$${\left({\displaystyle \frac{1}{t}}\left(x(t)+2x(t-1)\right)″\right)}^{\prime }+{\displaystyle \frac{d_1}{t^4}}{x}^3\left({\displaystyle \frac{t}{2}}\right)+{\displaystyle \frac{d_2}{t^4}}{x}^{1/3}\left({\displaystyle \frac{t}{3}}\right)=0,t\ge 1,$$

(30)

where $d_1>0$ and $d_2>0$ are constants.

Here, $a(t)={\displaystyle \frac{1}{t}},$ $b=2,$ $\sigma =1,$ $p_1(t)={\displaystyle \frac{d_1}{t^4}},$ $p_2(t)={\displaystyle \frac{d_2}{t^4}},$ $\alpha =1,$ ${\alpha }_1=3,$ ${\alpha }_2={\displaystyle \frac{1}{3}},$ ${\tau }_1(t)={\displaystyle \frac{t}{2}},$ ${\tau }_2(t)={\displaystyle \frac{t}{3}}.$ A simple computation shows that $\tau (t)={\displaystyle \frac{t}{3}},$ ${\eta }_1={\displaystyle \frac{1}{4}},$ ${\eta }_2={\displaystyle \frac{3}{4}},$

$$Q(t)={\left(4d_1\right)}^{1/4}{\left({\displaystyle \frac{4}{3}}{d}_2\right)}^{3/4}{\displaystyle \frac{1}{t^4}},A(t,1)\approx {\displaystyle \frac{t^2}{2}}\mathrm{and}{A}_1(t,1)\approx {\displaystyle \frac{t^3}{6}}.$$

Condition (5) becomes

$${\int }_1^{\infty }\xi \left(\xi {\int }_{\xi }^{\infty }{(4d_1)}^{{\displaystyle \frac{1}{4}}}{\left({\displaystyle \frac{4}{3}}{d}_2\right)}^{{\displaystyle \frac{3}{4}}}{\displaystyle \frac{1}{s^4}}ds\right)d\xi ={(4d_1)}^{{\displaystyle \frac{1}{4}}}{\left({\displaystyle \frac{4}{3}}{d}_2\right)}^{{\displaystyle \frac{3}{4}}}{\displaystyle \frac{1}{3}}{\int }_1^{\infty }{\displaystyle \frac{1}{\xi }}d\xi =\infty ;$$

that is, Condition (5) is satisfied. Condition (22) becomes

$$\underset{t\to \infty }{lim}inf{\int }_{t/3}^t{(4d_1)}^{{\displaystyle \frac{1}{4}}}{\left({\displaystyle \frac{4}{3}}{d}_2\right)}^{{\displaystyle \frac{3}{4}}}{\displaystyle \frac{1}{162}}{\displaystyle \frac{1}{s}}ds={(4d_1)}^{{\displaystyle \frac{1}{4}}}{\left({\displaystyle \frac{4}{3}}{d}_2\right)}^{{\displaystyle \frac{3}{4}}}{\displaystyle \frac{1}{162}}ln3>{\displaystyle \frac{3}{e}};$$

that is, Condition (22) is satisfied if $d_1^{{\displaystyle \frac{1}{4}}}{d}_2^{{\displaystyle \frac{3}{4}}}>{\displaystyle \frac{243{(3)}^{3/4}}{2eln3}}.$ Thus by Corollary 1, Equation (30) is almost oscillatory if $d_1^{{\displaystyle \frac{1}{4}}}{d}_2^{{\displaystyle \frac{3}{4}}}>92.7423.$ In particular, for $d_1=d_2=93,$ we see that $93>92.7423$, and so Equation (30) is almost oscillatory.

Example 2. 

Consider the third-order nonlinear neutral delay differential equation

$$\left({\left(\left(x(t)+2x(t-1)\right)″\right)}^{{\displaystyle \frac{5}{3}}}\right)\prime +{\displaystyle \frac{d_1}{t^2}}{x}^3\left({\displaystyle \frac{t}{2}}\right)+{\displaystyle \frac{d_2}{t^2}}x\left({\displaystyle \frac{t}{3}}\right)=0,t\ge 1,$$

(31)

where $d_1>0$ and $d_2>0$ are constants.

Here, $a(t)=1,$ $b=2,$ $\sigma =1,$ $p_1(t)={\displaystyle \frac{d_1}{t^2}},$ $p_2(t)={\displaystyle \frac{d_2}{t^2}},$ $\alpha ={\displaystyle \frac{5}{3}},$ ${\alpha }_1=3,$ ${\alpha }_2=1,$ ${\tau }_1(t)={\displaystyle \frac{t}{2}},$ ${\tau }_2(t)={\displaystyle \frac{t}{3}}.$ Through a simple calculation, we see that $\tau (t)={\displaystyle \frac{t}{3}},$ ${\eta }_1={\displaystyle \frac{1}{3}},$ ${\eta }_2={\displaystyle \frac{2}{3}},$

$$Q(t)=3d_1^{1/3}{\left({\displaystyle \frac{d_2}{2}}\right)}^{2/3}{\displaystyle \frac{1}{t^2}},A(t,1)\approx t\mathrm{and}{A}_1(t,1)\approx {\displaystyle \frac{t^2}{2}}.$$

Condition (5) becomes

$${\int }_1^{\infty }\xi {\left({\int }_{\xi }^{\infty }3d_1^{1/3}{\left({\displaystyle \frac{d_2}{2}}\right)}^{2/3}{\displaystyle \frac{1}{s^2}}ds\right)}^{{\displaystyle \frac{3}{5}}}d\xi ={\left(3d_1^{1/3}{\left({\displaystyle \frac{d_2}{2}}\right)}^{2/3}\right)}^{{\displaystyle \frac{3}{5}}}{\int }_1^{\infty }{\xi }^{2/5}d\xi =\infty ;$$

that is, Condition (5) holds. Since $Q_2(t)={\displaystyle \frac{3d_1^{1/3}{\left(\frac{d_2}{2}\right)}^{2/3}}{{(18)}^{\frac{2}{3}}}}{\displaystyle \frac{1}{t^{2/3}}},$ then by choosing $\rho (t)=1,$ Condition (23) is clearly satisfied for all $d_1>0$ and $d_2>0.$ Therefore, according to Theorem 2, Equation (31) is almost oscillatory if $d_1>0$ and $d_2>0.$

Example 3. 

Consider the third-order nonlinear neutral delay differential equation

$${\left(t^{{\displaystyle \frac{1}{2}}}\left(x(t)+2x(t-1)\right)″\right)}^{\prime }+{\displaystyle \frac{d_1}{t^4}}{x}^{{\displaystyle \frac{5}{3}}}\left({\displaystyle \frac{t}{2}}\right)+{\displaystyle \frac{d_2}{t}}{x}^{{\displaystyle \frac{1}{3}}}\left({\displaystyle \frac{t}{4}}\right)=0,t\ge 1,$$

(32)

where $d_1>0$ and $d_2>0$ are constants.

Here, $a(t)=t^{{\displaystyle \frac{1}{2}}},$ $b=2,$ $\sigma =1,$ $p_1(t)={\displaystyle \frac{d_1}{t^4}},$ $p_2(t)={\displaystyle \frac{d_2}{t}},$ $\alpha =1,$ ${\alpha }_1={\displaystyle \frac{5}{3}},$ ${\alpha }_2={\displaystyle \frac{1}{3}},$ ${\tau }_1(t)={\displaystyle \frac{t}{2}},$ ${\tau }_2(t)={\displaystyle \frac{t}{4}}.$ Through a simple calculation, we see that $\tau (t)={\displaystyle \frac{t}{4}},$ ${\eta }_1={\eta }_2={\displaystyle \frac{1}{2}},$

$$Q(t)={\displaystyle \frac{2\sqrt{d_1{d}_2}}{t^{\frac{5}{2}}}},A(t,1)\approx 2t^{{\displaystyle \frac{1}{2}}}\mathrm{and}{A}_1(t,1)\approx {\displaystyle \frac{4}{3}}{t}^{{\displaystyle \frac{3}{2}}}.$$

Condition (5) becomes

$$2\sqrt{d_1{d}_2}{\int }_1^{\infty }\xi \left({\displaystyle \frac{1}{\sqrt{\xi }}}{\int }_{\xi }^{\infty }{\displaystyle \frac{1}{s^{\frac{5}{2}}}}ds\right)d\xi ={\displaystyle \frac{4}{3}}\sqrt{d_1{d}_2}{\int }_1^{\infty }{\displaystyle \frac{d\xi }{\xi }}=\infty ;$$

that is, Condition (5) holds. Condition (28) becomes

$$\underset{t\to \infty }{lim}inf{\displaystyle \frac{\sqrt{t}}{3}}{\int }_t^{\infty }{\displaystyle \frac{\sqrt{d_1{d}_2}}{s^{\frac{3}{2}}}}ds=\sqrt{d_1{d}_2}>{\displaystyle \frac{9}{8}};$$

that is, Condition (28) holds if $\sqrt{d_1{d}_2}>1.125.$ Also, Condition (27) holds if $\sqrt{d_1{d}_2}>0.81691.$ Hence, according to Theorem 4, Equation (32) is almost oscillatory if $\sqrt{d_1{d}_2}>1.125$, and the same conclusion holds according to Theorem 3 if $\sqrt{d_1{d}_2}>0.81691).$ Therefore, Theorem 3 is better than Theorem 4.

Note that using Corollary 1 of [24], we see that (32) is almost oscillatory if $\sqrt{d_1{d}_2}>2.3883203.$ So, our Theorems 3 and 4 significantly improve Corollary 1 of [24]. In particular, for $d_1=d_2=2$, we see that Equation (30) is almost oscillatory according to Theorems 3 and 4, and Corollary 1 of [24] does not imply this conclusion.

4. Conclusions

In this paper, we have obtained some new oscillation criteria by using the arithmetic–geometric mean inequality along with the linearization technique and then applying the comparison method and the integral averaging technique. The obtained results improve on those in [24], and this is illustrated via an example.The results already reported in the literature [5,11,12,13,14,15,17,18,19,20,21,23] cannot be applied to Equations (30)–(32), since the number of nonlinear terms is more than one.

Furthermore, using the technique presented in this paper, one can extend the results of this paper to a more general equation, Equation $(2),$ when the function $b_1(t)$ is positive and decreasing with ${\gamma }_1=1;$ the details are left to the reader. It is an interesting problem to obtain similar results of this paper for Equation $(2),$ and/or higher-order neutral differential equations when the condition $A(t,t_0)<\infty \mathrm{as}t\to \infty$ is satisfied. Another interesting problem is to obtain conditions under which all solutions of Equation (1) are only oscillatory.