**1. Introduction**

Equations with neutral terms are of particular significance, as they arise in many applications including systems of control, electrodynamics, mixing liquids, neutron transportation, networks and population models; see [1].

Asymptotic properties of solutions of second/third order differential equations have been subject to intensive research in the literature. This problem for differential equations with respective delays has received a grea<sup>t</sup> deal of attention in the last years; see for examples, [2–21].

This paper deals with the oscillation and asymptotic behavior of solutions of the class of third-order, nonlinear, mixed-type, neutral differential equations

$$\left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{a}\right)' + q\_1\left(t\right)f\_1\left(\mathbf{x}\left(\sigma\_1\left(t\right)\right)\right) + q\_2\left(t\right)f\_2\left(\mathbf{x}\left(\sigma\_2\left(t\right)\right)\right) = 0,\tag{1}$$

where

$$z\left(t\right) = x\left(t\right) + p\_1\left(t\right) \ge \left(\tau\_1\left(t\right)\right) + p\_2\left(t\right) \ge \left(\tau\_2\left(t\right)\right)t$$

and we will assume the following assumptions hold:



*Mathematics* **2020**, *8*, 485; doi:10.3390/math8040485  www.mdpi.com/journal/mathematics

(M5) *f*1, *f*2 ∈ *C* (<sup>R</sup>, <sup>R</sup>), *f*1 (*x*) /*xβ* ≥ *k*1 > 0 and *f*2 (*x*) /*x<sup>γ</sup>* ≥ *k*2 for *x* = 0 where *β* and *γ* are ratios of odd positive integers.

By a solution of Equation (1), we mean a non-trivial real function *x* ∈ *C* ([*tx*, <sup>∞</sup>)), *tx* ≥ *t*0, with *<sup>z</sup>*(*t*), *z* (*t*) and *<sup>r</sup>*1(*t*)(*z*(*t*))*<sup>α</sup>* being continuously differentiable for all *t* ∈ [*tx*, <sup>∞</sup>), and satisfying (1) on [*tx*, <sup>∞</sup>). A solution of Equation (1) is called oscillatory if it has arbitrary large zeros; otherwise it is called nonoscillatory. Equation (1) is said to be oscillatory if all its solutions are oscillatory.

Han et al. in [22] studied the asymptotic properties of the solutions of equation

$$\left(r\left(t\right)\left(z''\left(t\right)\right)\right)' + q\_1\left(t\right) \ge \left(\sigma\_1\left(t\right)\right) + q\_2\left(t\right) \ge \left(\sigma\_2\left(t\right)\right) = 0,\tag{2}$$

where *z* (*t*) = *x* (*t*) + *p*1 (*t*) *x* (*<sup>τ</sup>*1 (*t*)) + *p*2 (*t*) *x* (*<sup>τ</sup>*2 (*t*)).

Baculikova and Dzurina [5] studied the oscillation of the third-order equation

$$\left(r\left(t\right)\left(\mathbf{x}'\left(t\right)\right)^a\right)'' + q\left(t\right)f\left(\mathbf{x}\left(\tau\left(t\right)\right)\right) + p\left(t\right)h\left(\mathbf{x}\left(\sigma\left(t\right)\right)\right) = 0,$$

where *τ* (*t*) ≤ *t* and *σ* (*t*) ≥ *t*.

> Thandapani and Rama [23] established some oscillation theorems for equation

$$\left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)\right)^{\prime} + q\_1\left(t\right)x^{\alpha}\left(\sigma\_1\left(t\right)\right) + q\_2\left(t\right)x^{\emptyset}\left(\sigma\_2\left(t\right)\right) = 0,$$

where *z* (*t*) = *x* (*t*) + *p*1 (*t*) *x* (*<sup>τ</sup>*1 (*t*)) + *p*2 (*t*) *x* (*<sup>τ</sup>*2 (*t*)), and the authors used the Recati technique.

The aim of this paper is to discuss the asymptotic behavior of solutions of a class of third-order, nonlinear, mixed-type, neutral differential equations. We established sufficient conditions to ensure that the solution of Equation (1) is oscillatory or tended to zero. The results of this study basically generalize and improve the previous results. An illustrative example is provided.

#### **2. Auxiliary Lemmas**

In order to prove our results, we shall need the next auxiliary lemmas.

**Lemma 1.** *Assume that f* (*y*) = *Uy* − *Vy η*+1 *η* , *where U and V are constants, V* > 0 *and η is a quotient of odd positive integers. Then f imposes its maximum value on* R *at y*∗ = *Uη <sup>V</sup>*(*η*+<sup>1</sup>)*<sup>η</sup> and*

$$\max\_{y \in \mathbb{R}} f = f\left(y^\*\right) = \frac{\eta^{\eta}}{(\eta+1)^{\eta+1}} \mathcal{U}^{\eta+1} V^{-\eta}.$$

**Lemma 2** ([24])**.** *Assume that A* ≥ 0 *and B* ≥ 0. *If δ* > 1, *then*

$$(A+B)^{\delta} \le 2^{\delta-1} \left(A^{\delta} + B^{\delta}\right),$$

*Moreover, if* 0 < *δ* < 1, *then* (*A* + *B*)*<sup>δ</sup>* ≤ -*A<sup>δ</sup>* + *B<sup>δ</sup>* .

**Lemma 3** ([17])**.** *If the function y satisfies y*(*i*) > 0, *i* = 0, 1, ..., *n*, *and y*(*n*+<sup>1</sup>) < 0, *then*

$$\frac{y'(t)}{t^n/n!} \ge \frac{y'(t)}{t^{n-1}/'(n-1)!}$$

**Lemma 4** ([23])**.** *Assume that u* (*t*) > 0, *u* (*t*) > 0, *u* (*t*) > 0 *and u* (*t*) < 0 *on* (*<sup>T</sup>*, <sup>∞</sup>)*. Then,*

$$\frac{u'(t)}{u'(t)} \ge \frac{t - T}{2} \ge \frac{\mu t}{2}$$

*for t* ≥ *T and some μ* ∈ (0, 1)*.*

**Lemma 5.** *Let x be a positive solution of Equation (1). Then z has only one of the following two properties eventually:*

(**i**) *z* (*t*) > 0, *z* (*t*) > 0 *and z* (*t*) > 0; (**ii**) *z* (*t*) > 0, *z* (*t*) < 0 *and z* (*t*) > 0.

**Proof.** The proof is similar to that of Lemma 2.1 of [10] and hence the details are omitted.

**Lemma 6.** *Let x be a positive solution of Equation (1), and z has the property* (**ii**)*. If β* = *γ and*

$$\int\_{t\_0}^{\infty} \int\_{v}^{\infty} \left( \frac{1}{r(u)} \int\_{u}^{\infty} (k\_1 q\_1(s) + k\_2 q\_2(s)) \right) \mathrm{d}s \right)^{1/u} \mathrm{d}u \mathrm{d}v = \infty,\tag{3}$$

*then the solution x of Equation (1) converges to zero as t* → ∞*.*

**Proof.** Let *x* be a positive solution of Equation (1). Since *z* satisfies the property (**ii**), we ge<sup>t</sup> lim*<sup>t</sup>*→∞ *z*(*t*) = *δ* ≥ 0. Next, we will prove that *δ* = 0. Suppose that *δ* > 0, then we have for all *ε* > 0 and *t* enough large *δ* < *z*(*t*) < *δ* + *ε*. By choosing *ε* < 1−*c*1−*c*<sup>2</sup> *<sup>c</sup>*1+*c*2*δ*, we obtain

$$\begin{aligned} x(t) &=& z(t) - p\_1\left(t\right) \ge \left(\tau\_1\left(t\right)\right) - p\_2\left(t\right) \ge \left(\tau\_2\left(t\right)\right) \\ &>& \delta - \left(c\_1 + c\_2\right) z\left(\tau\_1\left(t\right)\right) \\ &>& \delta - \left(c\_1 + c\_2\right) \left(\delta + \varepsilon\right) \\ &>& L\left(\delta + \varepsilon\right) > Lz\left(t\right), \end{aligned}$$

where *L* = *<sup>δ</sup>*−(*<sup>c</sup>*1+*c*2)(*<sup>δ</sup>*+*ε*) *δ*+*ε* > 0. Thus, from (1) and (M5), we have

$$\begin{aligned} 0 &\geq \left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{a}\right)' + k\_{1}q\_{1}\left(t\right)x^{\emptyset}\left(\sigma\_{1}\left(t\right)\right) + k\_{2}q\_{2}\left(t\right)x^{\emptyset}\left(\sigma\_{2}\left(t\right)\right),\\ &\geq \left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{a}\right)' + L^{\emptyset}\left(k\_{1}q\_{1}\left(t\right) + k\_{2}q\_{2}\left(t\right)\right)z^{\emptyset}\left(\sigma\_{2}\left(t\right)\right),\end{aligned}$$

and so,

$$\left(r\left(t\right)\left(z''\left(t\right)\right)^{\alpha}\right)' \le -L^{\beta}\delta^{\beta}\left(k\_{1}q\_{1}\left(t\right) + k\_{2}q\_{2}\left(t\right)\right).$$

By integrating this inequality two times from *t* to <sup>∞</sup>, we ge<sup>t</sup>

$$-z'(t) > L^{\beta/a} \delta^{\beta/a} \int\_t^{\infty} \left( \frac{1}{r(u)} \int\_u^{\infty} (k\_1 q\_1(s) + k\_2 q\_2(s)) \right) ds \Big|^{1/a} du.$$

Integrating the last inequality from *t*1 to <sup>∞</sup>, we have

$$\varpi(t\_1) > L^{\beta/a} \delta^{\beta/a} \int\_{t\_1}^{\infty} \int\_{\upsilon}^{\infty} \left( \frac{1}{r(u)} \int\_u^{\infty} (k\_1 q\_1 \left(s\right) + k\_2 q\_2 \left(s\right)) \right) ds \right)^{1/a} du dv.$$

Thus, we are led to a contradiction with (3). Then, lim*<sup>t</sup>*→∞ *z*(*t*) = 0; moreover, the fact that *x* (*t*) ≤ *z* (*t*) implies lim*<sup>t</sup>*→∞ *x*(*t*) = 0.

#### **3. Main Results**

In this section, we will establish new oscillation criteria for solutions of the Equation (1). For the sake of convenience, we insert the next notation:

$$\begin{aligned} \mathcal{R}\_{\boldsymbol{u}}\left(t\right) &:= \int\_{\boldsymbol{u}}^{t} \frac{1}{r^{1/\boldsymbol{\alpha}}\left(s\right)} \mathrm{d}s, \\\mathcal{R}\_{\boldsymbol{u}}^{\*}\left(t\right) &:= \min\_{t \ge t\_{0}} \left\{ \mathcal{R}\_{\boldsymbol{u}}\left(t\right), \mathcal{R}\_{\boldsymbol{u}}\left(\tau\_{1}\left(t\right)\right) \right\}. \end{aligned}$$

and

$$q\_i^\*\left(t\right) := \min\_{t \ge t\_0} \left\{ q\_i\left(t\right), \ q\_i\left(\tau\_1\left(t\right)\right), \ q\_i\left(\tau\_2\left(t\right)\right) \right\}, i = 1, 2.$$

**Theorem 1.** *Assume that (M*1*)–(M*5*) and (3) hold. Let β* = *γ* ≥ *α*, *σ*1 (*t*) ≤ *τ*1 (*t*) *and σ*1 (*t*) > 0*. If there exists a positive function ρ* ∈ *C*<sup>1</sup> ([*<sup>t</sup>*0, ∞)) *such that*

$$\limsup\_{t \to \infty} \int\_{t0}^{t} \left( \Theta\_1(s) - \left( 1 + c\_1^{\frac{\beta}{\beta}} + \frac{c\_2^{\beta}}{2^{\beta - 1}} \right) \frac{1}{(a+1)^{a+1}} \frac{\left( \rho'\_+(s) \right)^{a+1} r \left( \sigma\_1(s) \right)}{\left( \rho \left( s \right) \sigma'\_1(s) \right)^{a}} \right) ds = \infty,\tag{4}$$

*where ρ*+ (*s*) = max {*ρ* (*s*), 0} *and*

$$
\Theta\_1\left(t\right) = \frac{\mu^{\alpha}\upsilon^{\beta-\alpha}}{2^{2\beta+\alpha-2}}\rho\left(t\right)\sigma\_1^{\beta}\left(t\right)\left(k\_1q\_1^\*\left(t\right) + k\_2q\_2^\*\left(t\right)\right),
$$

*then every solution of equation (1) either oscillates or tends to zero as t* → ∞*.*

**Proof.** Let *x* be non-oscillatory solution of Equation (1). Without loss of generality, we assume that *x* (*t*) > 0; then there exists a *t*1 ≥ *t*0 such that *x* (*t*) > 0, *x* (*τi* (*t*)) > 0 and *x* (*σi* (*t*)) > 0 for *t* ≥ *t*1and *i* = 1, 2. From Lemma 5, we have that *z* has the property (**i**) or the property (**ii**). From Lemma 6, if *z* (*t*) has the property (**ii**), then we obtain lim*<sup>t</sup>*→∞ *x*(*t*) = 0. Next, let *z* have the property (**i**). Using (1) and (M5), we obtain

$$\left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{a}\right)' + k\_1q\_1\left(t\right)\ge^{\beta}\left(\sigma\_1\left(t\right)\right) + k\_2q\_2\left(t\right)\ge^{\beta}\left(\sigma\_2\left(t\right)\right) \le 0.1$$

Thus, we ge<sup>t</sup>

0 ≥ *r* (*t*) *z* (*t*)*<sup>α</sup>* + *k*1*q*1 (*t*) *xβ* (*<sup>σ</sup>*1 (*t*)) + *k*2*q*2 (*t*) *xβ* (*<sup>σ</sup>*2 (*t*)) <sup>+</sup>*cβ*1 [*r* (*<sup>τ</sup>*1 (*t*)) *z* (*<sup>τ</sup>*1 (*t*))*<sup>α</sup>* + *k*1*q*1 (*<sup>τ</sup>*1 (*t*)) *xβ* (*<sup>σ</sup>*1 (*<sup>τ</sup>*1 (*t*))) + *k*2*q*2 (*<sup>τ</sup>*1 (*t*)) *xβ* (*<sup>σ</sup>*2 (*<sup>τ</sup>*1 (*t*))) + *cβ*2 2*β*−<sup>1</sup> [*r* (*<sup>τ</sup>*2 (*t*)) *z* (*<sup>τ</sup>*2 (*t*))*<sup>α</sup>* + *k*1*q*1 (*<sup>τ</sup>*2 (*t*)) *xβ* (*<sup>σ</sup>*1 (*<sup>τ</sup>*2 (*t*))) + *k*2*q*2 (*<sup>τ</sup>*2 (*t*)) *xβ* (*<sup>σ</sup>*2 (*<sup>τ</sup>*2 (*t*))) .

> That is

$$\begin{split} \left(r\left(t\right)\left(z''\left(t\right)\right)^{a}\right)' + c\_{1}^{\beta}\left(r\left(\tau\_{1}\left(t\right)\right)\left(z''\left(\tau\_{1}\left(t\right)\right)\right)^{a}\right)' + \frac{c\_{2}^{\beta}}{2^{\beta-1}}\left(r\left(\tau\_{2}\left(t\right)\right)\left(z''\left(\tau\_{2}\left(t\right)\right)\right)^{a}\right)' \\ + k\_{1}q\_{1}^{\ast}\left(t\right)\left(\mathbf{x}^{\theta}\left(\sigma\_{1}\left(t\right)\right) + c\_{1}^{\theta}\mathbf{x}^{\theta}\left(\sigma\_{1}\left(\tau\_{1}\left(t\right)\right)\right) + \frac{c\_{2}^{\beta}}{2^{\beta-1}}\mathbf{x}^{\theta}\left(\sigma\_{1}\left(\tau\_{2}\left(t\right)\right)\right)\right) \\ + k\_{2}q\_{2}^{\ast}\left(t\right)\left(\mathbf{x}^{\theta}\left(\sigma\_{2}\left(t\right)\right) + c\_{1}^{\theta}\mathbf{x}^{\theta}\left(\sigma\_{2}\left(\tau\_{1}\left(t\right)\right)\right) + \frac{c\_{2}^{\beta}}{2^{\beta-1}}\mathbf{x}^{\theta}\left(\sigma\_{2}\left(\tau\_{2}\left(t\right)\right)\right)\right) \leq 0. \end{split} \tag{5}$$

From Lemma 2, we obtain

$$\begin{split} \mathbf{z}^{\beta} \begin{pmatrix} t \\ \end{pmatrix} &\leq \begin{pmatrix} \mathbf{x} \begin{pmatrix} t \\ \end{pmatrix} + c\_{1} \begin{pmatrix} t \end{pmatrix} \mathbf{x} \begin{pmatrix} \tau\_{1} \begin{pmatrix} t \end{pmatrix} \end{pmatrix} + c\_{2} \begin{pmatrix} t \end{pmatrix} \mathbf{x} \begin{pmatrix} \tau\_{2} \begin{pmatrix} t \end{pmatrix} \end{pmatrix} \end{split} \\ &\leq \begin{array}{c} \mathbf{4}^{\beta - 1} \left( \mathbf{x}^{\beta} \begin{pmatrix} t \end{pmatrix} + c\_{1}^{\beta} \mathbf{x}^{\beta} \begin{pmatrix} \tau\_{1} \begin{pmatrix} t \end{pmatrix} \end{pmatrix} + \frac{c\_{2}^{\beta}}{2^{\beta - 1}} \mathbf{x}^{\beta} \begin{pmatrix} \tau\_{2} \begin{pmatrix} t \end{pmatrix} \end{pmatrix} \right) . \end{split} \tag{6}$$

which with (5) gives

$$\begin{split} \left(r\left(t\right)\left(z''\left(t\right)\right)^{a}\right)' + \varepsilon\_{1}^{\beta}\left(r\left(\tau\_{1}\left(t\right)\right)\left(z''\left(\tau\_{1}\left(t\right)\right)\right)^{a}\right)' + \frac{\varepsilon\_{2}^{\beta}}{2^{\beta-1}}\left(r\left(\tau\_{2}\left(t\right)\right)\left(z''\left(\tau\_{2}\left(t\right)\right)\right)^{a}\right)' \\ + \frac{k\_{1}}{4^{\beta-1}}q\_{1}^{\*}\left(t\right)z^{\beta}\left(\sigma\_{1}\left(t\right)\right) + \frac{k\_{2}}{4^{\beta-1}}q\_{2}^{\*}\left(t\right)z^{\beta}\left(\sigma\_{2}\left(t\right)\right) \leq 0. \end{split}$$

This implies that

$$\begin{aligned} \left(r\left(t\right)\left(z''\left(t\right)\right)^a\right)' + c\_1^{\beta}\left(r\left(\tau\_1\left(t\right)\right)\left(z''\left(\tau\_1\left(t\right)\right)\right)^a\right)' + \frac{c\_2^{\beta}}{2^{\beta-1}}\left(r\left(\tau\_2\left(t\right)\right)\left(z''\left(\tau\_2\left(t\right)\right)\right)^a\right)' \\ + \frac{1}{4^{\beta-1}}\left(k\_1q\_1^\*\left(t\right) + k\_2q\_2^\*\left(t\right)\right)z^{\beta}\left(\sigma\_1\left(t\right)\right) \le 0. \end{aligned} \tag{7}$$

.

Now, we define

$$
\omega\_1\left(t\right) = \rho\left(t\right) \frac{r\left(t\right)\left(z''\left(t\right)\right)^{\alpha}}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^{\alpha}}
$$

Then *ω*1 (*t*) > 0. By differentiating, we ge<sup>t</sup>

$$
\omega\_1'\left(t\right) = \frac{\rho'\left(t\right)}{\rho\left(t\right)}\omega\_1\left(t\right) + \rho\left(t\right)\frac{\left(r\left(t\right)\left(z''\left(t\right)\right)^a\right)'}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^a} - a\rho\left(t\right)\frac{r\left(t\right)\left(z''\left(t\right)\right)^a}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^{a+1}}z''\left(\sigma\_1\left(t\right)\right)\sigma\_1'\left(t\right)\dots
$$

Since *r* (*t*) (*z* (*t*))*<sup>α</sup>* < 0 and *σ*1 (*t*) < *t*, we obtain

$$r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{\alpha} \le r\left(\sigma\_1\left(t\right)\right)\left(z^{\prime\prime}\left(\sigma\_1\left(t\right)\right)\right)^{\alpha},$$

and hence

$$
\omega\_1' \left( t \right) \le \frac{\rho\_+' \left( t \right)}{\rho \left( t \right)} \omega\_1 \left( t \right) - \mathfrak{a} \frac{\sigma\_1' \left( t \right)}{\rho^{1/a} \left( t \right) r^{1/a} \left( \sigma\_1 \left( t \right) \right)} \omega\_1^{\frac{a+1}{a}} \left( t \right) + \rho \left( t \right) \frac{\left( r \left( t \right) \left( z'' \left( t \right) \right)^a \right)'}{\left( z' \left( \sigma\_1 \left( t \right) \right) \right)^a}.
$$

> Using Lemma 1 with

$$\eta = \mathfrak{a}, \; \mathcal{U} = \frac{\rho\_+^{\prime}\begin{pmatrix} t \\ \end{pmatrix}}{\rho\_-(t)}, \; V = \mathfrak{a} \frac{\sigma\_1^{\prime}\begin{pmatrix} t \\ \end{pmatrix}}{\rho^{1/\alpha}\begin{pmatrix} t \end{pmatrix} r^{1/\alpha} \begin{pmatrix} \sigma\_1\begin{pmatrix} t \end{pmatrix} \end{pmatrix}} \text{ and } y = \omega\_1.$$

we obtain

$$
\omega\_1'\left(t\right) \le \rho\left(t\right) \frac{\left(r\left(t\right)\left(z''\left(t\right)\right)^a\right)'}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^a} + \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+'\left(t\right)\right)^{a+1} r\left(\sigma\_1\left(t\right)\right)}{\left(\rho\left(t\right)\sigma\_1'\left(t\right)\right)^a}.\tag{8}
$$

Further, we define the function

$$
\omega\_2\left(t\right) = \rho\left(t\right) \frac{r\left(\tau\_1\left(t\right)\right) \left(z''\left(\tau\_1\left(t\right)\right)\right)^{\kappa}}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^{\kappa}}.
$$

Then *ω*2 (*t*) > 0. By differentiating *ω*2 and using *σ*1 (*t*) ≤ *τ*1 (*t*), we find

$$
\omega\_2'(t) \le \frac{\rho'(t)}{\rho(t)} \omega\_2\left(t\right) + \rho\left(t\right) \frac{\left(r\left(\tau\_1\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_1\left(t\right)\right)\right)^a\right)'}{\left(z^{\prime}\left(\sigma\_1\left(t\right)\right)\right)^a} - a \frac{\sigma\_1'\left(t\right)}{\rho^{1/a}\left(t\right)r^{1/a}\left(\sigma\_1\left(t\right)\right)} \omega\_2^{\frac{a+1}{a}}\left(t\right) \dots
$$

Using Lemma 1, we obtain

$$
\omega\_2' \left( t \right) \le \rho \left( t \right) \frac{\left( r \left( \tau\_1 \left( t \right) \right) \left( z'' \left( \tau\_1 \left( t \right) \right) \right)^a \right)^t}{\left( z' \left( \sigma\_1 \left( t \right) \right) \right)^a} + \frac{1}{\left( a+1 \right)^{a+1}} \frac{\left( \rho\_+' \left( t \right) \right)^{a+1} r \left( \sigma\_1 \left( t \right) \right)}{\left( \rho \left( t \right) \sigma\_1' \left( t \right) \right)^a}. \tag{9}
$$

Next, we define another function

$$
\omega\_3\left(t\right) = \rho\left(t\right) \frac{r\left(\tau\_2\left(t\right)\right) \left(z^{\prime\prime}\left(\tau\_2\left(t\right)\right)\right)^{\alpha}}{\left(z^{\prime}\left(\sigma\_1\left(t\right)\right)\right)^{\alpha}}.
$$

Thus *ω*3 (*t*) > 0. By differentiating, and similar to (9) we have

$$
\omega\_3'(t) \le \rho\left(t\right) \frac{\left(r\left(\tau\_2\left(t\right)\right)\left(z''\left(\tau\_2\left(t\right)\right)\right)^a\right)^t}{\left(z'\left(\sigma\_1\left(t\right)\right)\right)^a} + \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+'\left(t\right)\right)^{a+1} r\left(\sigma\_1\left(t\right)\right)}{\left(\rho\left(t\right)\sigma\_1'\left(t\right)\right)^a}.\tag{10}
$$

From (8)–(10), we ge<sup>t</sup>

$$\begin{split} \omega\_{1}^{\prime}\left(t\right) &+ c\_{1}^{\delta}\omega\_{2}^{\prime}\left(t\right) + \frac{c\_{2}^{\delta}}{2^{\delta-1}}\omega\_{3}^{\prime}\left(t\right) \leq \frac{\rho\left(t\right)}{\left(z^{\prime}\left(\sigma\_{1}\left(t\right)\right)\right)^{\mathfrak{a}}} \left(\left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^{\mathfrak{a}}\right)^{\mathfrak{a}}\right)^{\mathfrak{b}} + \\ &+ c\_{1}^{\delta}\left(r\left(\tau\_{1}\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_{1}\left(t\right)\right)\right)^{\mathfrak{a}}\right)^{\mathfrak{b}} + \frac{c\_{2}^{\delta}}{2^{\delta-1}}\left(r\left(\tau\_{2}\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_{2}\left(t\right)\right)\right)^{\mathfrak{a}}\right)^{\mathfrak{b}}\right)^{\mathfrak{b}} \\ &+ \left(1 + c\_{1}^{\delta} + \frac{c\_{2}^{\delta}}{2^{\delta-1}}\right)\frac{1}{\left(a+1\right)^{\mathfrak{a}+1}}\frac{\left(\rho\_{+}^{\prime}\left(t\right)\right)^{\mathfrak{a}+1}r\left(\sigma\_{1}\left(t\right)\right)}{\left(\rho\left(t\right)\sigma\_{1}^{\prime}\left(t\right)\right)^{\mathfrak{a}}}, \end{split}$$

which with (7) gives

$$
\omega\_1' \left( t \right) + c\_1^{\beta} \omega\_2' \left( t \right) + \frac{c\_2^{\beta}}{2^{\beta - 1}} \omega\_3' \left( t \right) \le -\frac{\rho \left( t \right)}{4^{\beta - 1}} \left( k\_1 q\_1^\* \left( t \right) + k\_2 q\_2^\* \left( t \right) \right) \frac{z^{\beta} \left( \sigma\_1 \left( t \right) \right)}{\left( z' \left( \sigma\_1 \left( t \right) \right) \right)^a}
$$

$$
+ \left( 1 + c\_1^{\beta} + \frac{c\_2^{\beta}}{2^{\beta - 1}} \right) \frac{1}{\left( a + 1 \right)^{a + 1}} \frac{\left( \rho\_+' \left( t \right) \right)^{a + 1} r \left( \sigma\_1 \left( t \right) \right)}{\left( \rho \left( t \right) \sigma\_1' \left( t \right) \right)^a}. \tag{11}
$$

> Using Lemma 4, we have, for some *μ* ∈ (0, <sup>1</sup>),

$$\frac{z\left(\sigma\_1\left(t\right)\right)}{z'\left(\sigma\_1\left(t\right)\right)} \ge \frac{\mu}{2}\sigma\_1\left(t\right)\dots$$

From property (**i**), we ge<sup>t</sup>

$$\begin{aligned} z\left(t\right) &= \left. z\left(t\_1\right) + \int\_{t\_1}^{t} z'\left(s\right)ds \\ &\ge \left. \left(t - t\_1\right)z'\left(t\_1\right) \right| \ge \frac{\upsilon}{2}t\_\prime \end{aligned} \tag{12}$$

.

for some *υ* > 0 and for *t* enough large. Therefore, for some *μ* ∈ (0, 1) and *υ* > 0, we find

$$\frac{z^{\beta}\left(\sigma\_{1}\left(t\right)\right)}{\left(z'\left(\sigma\_{1}\left(t\right)\right)\right)^{\alpha}} \geq \frac{\mu^{\alpha}\upsilon^{\beta-\alpha}}{2^{\alpha}}\sigma\_{1}^{\beta}\left(t\right)\,.$$

Combining the last inequality with (11), we obtain

$$\begin{aligned} \omega\_1'\left(t\right) + \mathfrak{c}\_1^{\beta}\omega\_2'\left(t\right) + \frac{\mathfrak{c}\_2^{\beta}}{2^{\beta-1}}\omega\_3'\left(t\right) &\leq -\Theta\left(t\right) \\ + \left(1 + \mathfrak{c}\_1^{\beta} + \frac{\mathfrak{c}\_2^{\beta}}{2^{\beta-1}}\right) \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+'\left(t\right)\right)^{a+1} r\left(\sigma\_1\left(t\right)\right)}{\left(\rho\left(t\right)\sigma\_1'\left(t\right)\right)^a}. \end{aligned}$$

Integrating the above inequality from *t*1 to *t* , we have

$$\begin{split} \int\_{t\_1}^{t} \left( \Theta \left( s \right) - \left( 1 + c\_1^{\beta} + \frac{c\_2^{\beta}}{2^{\beta - 1}} \right) \frac{1}{\left( a + 1 \right)^{a + 1}} \frac{\left( \rho\_+^{\prime} \left( s \right) \right)^{a + 1} r \left( \sigma\_1 \left( s \right) \right)}{\left( \rho \left( s \right) \sigma\_1^{\prime} \left( s \right) \right)^{a}} \right) ds \\ \leq \omega\_1 \left( t\_1 \right) + c\_1^{\beta} \omega\_2 \left( t\_1 \right) + \frac{c\_2^{\beta}}{2^{\beta - 1}} \omega\_3 \left( t\_1 \right) . \end{split}$$

Taking the superior limit as *t* → <sup>∞</sup>, we ge<sup>t</sup> a contradiction with (4). The proof is complete.

**Remark 1.** *In the Theorem 1, if σ*1 (*t*) ≥ *τ*1 (*t*) *and τ*1 (*t*) > 0, *then the assumption (4) is replaced by*

$$\limsup\_{t \to \infty} \int\_{t\_0}^t \left( \Theta\_1 \left( s \right) - \left( 1 + c\_1^{\theta} + \frac{c\_2^{\theta}}{2^{\theta - 1}} \right) \frac{1}{\left( a + 1 \right)^{a + 1}} \frac{\left( \rho\_+^{\prime} \left( s \right) \right)^{a + 1} r \left( \tau\_1 \left( s \right) \right)}{\left( \rho \left( s \right) \tau\_1^{\prime} \left( s \right) \right)^{a}} \right) ds = \infty.$$

**Theorem 2.** *Assume that (M*1*)–(M*5*) and (3) hold. Let β* = *γ* ≥ *α and r* (*t*) > 0*. If there exists a positive function ρ* ∈ *C*<sup>1</sup> ([*<sup>t</sup>*0, ∞)) *such that*

$$\limsup\_{t \to \infty} \int\_{t\_0}^t \left( \Theta\_2(s) - \left( 1 + c\_1^{\beta} + \frac{c\_2^{\beta}}{2^{\beta - 1}} \right) \frac{1}{(a+1)^{a+1}} \frac{(\rho'\_+(t))^{a+1}}{\left( \rho \left( t \right) R\_{t\_0}^\*(t) \right)^a} \right) ds = \infty,\tag{13}$$

*where*

$$
\Theta\_2\left(t\right) = \frac{\upsilon^{\beta-\alpha}}{2^{\beta\beta-\alpha-2}t^{2a}}\rho\left(t\right)\sigma\_1^{\beta+\alpha}\left(t\right)\left(k\_1q\_1^\*\left(t\right) + k\_2q\_2^\*\left(t\right)\right)\rho
$$

*then every solution of Equation (1) either oscillates or tends to zero as t* → ∞*.*

**Proof.** Proceeding as in the proof of Theorem 1, we have that (7) holds. Since *r* (*t*) (*z* (*t*))*<sup>α</sup>* < 0, we obtain

$$z'\left(t\right) = \left.z'\left(t\_1\right) + \int\_{t\_1}^{t} \frac{\left[r\left(s\right)\left(z''\left(s\right)\right)^a\right]^{1/a}}{r^{1/a}\left(s\right)}ds$$

$$\geq \left.\left[r\left(t\right)\left(z''\left(t\right)\right)^a\right]^{1/a}R\_{t\_1}\left(t\right)\right.\tag{14}$$

Now, we define

$$
\omega\_1\left(t\right) = \rho\left(t\right) \frac{r\left(t\right)\left(z''\left(t\right)\right)^{\alpha}}{z^{\alpha}\left(t\right)}.
$$

Then *ω*1 (*t*) > 0. By differentiating *ω*1 and using (14), we ge<sup>t</sup>

$$
\omega\_1' \left( t \right) \le \frac{\rho\_+' \left( t \right)}{\rho \left( t \right)} \omega\_1 \left( t \right) - a \frac{\mathcal{R}\_{l\_1} \left( t \right)}{\rho^{1/a} \left( t \right)} \omega\_1^{\frac{a+1}{a}} \left( t \right) + \rho \left( t \right) \frac{\left( r \left( t \right) \left( z^{\prime \prime} \left( t \right) \right)^a \right)'}{z^a \left( t \right)}.
$$

Using Lemma 1 with *η* = *α*, *U* = *ρ*+(*t*) *ρ*(*t*) , *V* = *α Rt*1 (*t*) *ρ*1/*α*(*t*) and *y* = *ω*1, we obtain

$$
\omega\_1'\left(t\right) \le \rho\left(t\right) \frac{\left(r\left(t\right)\left(z^{\prime\prime}\left(t\right)\right)^a\right)'}{z^a\left(t\right)} + \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+^{\prime}\left(t\right)\right)^{a+1}}{\left(\rho\left(t\right)R\_{\ell\_1}\left(t\right)\right)^a}.\tag{15}
$$

Next, we define a function

$$
\omega\_2\left(t\right) = \rho\left(t\right) \frac{r\left(\tau\_1\left(t\right)\right) \left(z^{\prime\prime}\left(\tau\_1\left(t\right)\right)\right)^d}{z^{\alpha}\left(t\right)}.\tag{16}
$$

Then *ω*2 (*t*) > 0. Since *z* (*t*) > 0 and *τ*1 (*t*) < *t*, we obtain *z* (*t*) > *z* (*<sup>τ</sup>*1 (*t*)). Hence, from (14), we find

$$z'(t) > \left[ r\left(\tau\_1\left(t\right)\right) \left(z''\left(\tau\_1\left(t\right)\right)\right)^a \right]^{1/a} \mathcal{R}\_{t\_1}\left(\tau\_1\left(t\right)\right). \tag{17}$$

for *t* ≥ *t*2 ≥ *t*1. By differentiating (16) and using (17), we ge<sup>t</sup>

$$
\omega\_2'\left(t\right) \le \frac{\rho'\left(t\right)}{\rho\left(t\right)}\omega\_2\left(t\right) - a\frac{\mathbb{R}\_{l\_1}\left(\tau\_1\left(t\right)\right)}{\rho^{1/a}\left(t\right)}\omega\_2^{\frac{a+1}{a}}\left(t\right) + \rho\left(t\right)\frac{\left(r\left(\tau\_1\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_1\left(t\right)\right)\right)^a\right)'}{z^a\left(t\right)}.
$$

By using Lemma 1, we obtain

$$
\omega\_2'(t) \le \rho\left(t\right) \frac{\left(r\left(\tau\_1\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_1\left(t\right)\right)\right)^a\right)'}{z^a\left(t\right)} + \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+^{\prime}\left(t\right)\right)^{a+1}}{\left(\rho\left(t\right)R\_{t\_1}\left(\tau\_1\left(t\right)\right)\right)^a}.\tag{18}
$$

Additionally, we define another function

$$
\omega\_3\left(t\right) = \rho\left(t\right) \frac{r\left(\tau\_2\left(t\right)\right) \left(z''\left(\tau\_2\left(t\right)\right)\right)^a}{z''\left(t\right)}.\tag{19}
$$

> Thus *ω*3 (*t*) > 0. Using *r* (*t*) (*z* (*t*))*<sup>α</sup>* < 0, *τ*2 (*t*) > *t* and (14), we note that

$$z'(t) > \left[ r\left(\tau\_2\left(t\right)\right) \left( z''\left(\tau\_2\left(t\right)\right) \right)^a \right]^{1/a} \mathcal{R}\_{t\_1}\left(t\right) . \tag{20}$$

By differentiating (19) and using (20) and Lemma 1, we ge<sup>t</sup>

$$
\omega\_3'(t) \le \rho\left(t\right) \frac{\left(r\left(\tau\_2\left(t\right)\right)\left(z^{\prime\prime}\left(\tau\_2\left(t\right)\right)\right)^a\right)^{\prime}}{z^a\left(t\right)} + \frac{1}{\left(a+1\right)^{a+1}} \frac{\left(\rho\_+^{\prime}\left(t\right)\right)^{a+1}}{\left(\rho\left(t\right)R\_{t\_1}\left(t\right)\right)^a}.\tag{21}
$$

From (7), (15), (18) and (21), we find

$$
\omega\_1' \left( t \right) + c\_1^\beta \omega\_2' \left( t \right) + \frac{c\_2^\beta}{2^{\beta - 1}} \omega\_3' \left( t \right) \le -\frac{\rho \left( t \right)}{4^{\beta - 1}} \left( k\_1 q\_1^\* \left( t \right) + k\_2 q\_2^\* \left( t \right) \right) \frac{z^\beta \left( \sigma\_1 \left( t \right) \right)}{z^a \left( t \right)}
$$

$$
+ \left( 1 + c\_1^\beta + \frac{c\_2^\beta}{2^{\beta - 1}} \right) \frac{1}{\left( a + 1 \right)^{a + 1}} \frac{\left( \rho\_+^\prime \left( t \right) \right)^{a + 1}}{\left( \rho \left( t \right) R\_{t\_1}^\* \left( t \right) \right)^a}. \tag{22}
$$

Using (12) and Lemma 6, we have

$$\frac{z^{\beta}\left(\sigma\_{1}\left(t\right)\right)}{z^{\alpha}\left(t\right)} \geq \frac{\upsilon^{\beta-\alpha}}{2^{\beta-\alpha}t^{2\alpha}}\sigma\_{1}^{\beta+\alpha}\left(t\right)\,.$$

As in the proof of Theorem 1, we are led to a contradiction with (13). This completes the proof.

In the following Theorems, we are concerned with the oscillation of solutions of Equation (1) when *α* = 1 and *r* (*t*) = 1.

**Theorem 3.** *Assume that (M*1*)-(M*5*) and (3) hold. Let* 0 < *β* < 1 < *γ and τ*<sup>−</sup><sup>1</sup> *i exists for i* = 1, 2*. If the inequalities*

$$y^{\prime\prime\prime}(t) + \left(\frac{k\_1}{\lambda\_1}\right)^{\lambda\_1} \left(\frac{k\_2}{4^{\gamma-1}\lambda\_2}\right)^{\lambda\_2} \frac{\left(q\_1^\*\left(t\right)\right)^{\lambda\_1} \left(q\_2^\*\left(t\right)\right)^{\lambda\_2}}{\left(1 + c\_1^{\beta} + c\_2^{\beta}\right)} y\left(\tau\_i^{-1}\left(\sigma\_{\backslash}(t)\right)\right) \le 0,\tag{23}$$

*where i*, *j* = 1, 2, *i* = *j*, *λ*1 = *γ*−1 *γ*−*β and λ*2 = 1−*β γ*−*β* , *have oscillatory solutions, then every solution of Equation (1) is oscillatory.*

**Proof.** Let *x* non-oscillatory solution of Equation (1). Without loss of generality we assume that *x* > 0; then, there exists a *t*1 ≥ *t*0 such that *x* (*t*) > 0, *x* (*τi* (*t*)) > 0 and *x* (*σi* (*t*)) > 0 for *t* ≥ *t*1and *i* = 1, 2. By Lemma 6, we ge<sup>t</sup> that *z* (*t*) > 0, *z* (*t*) > 0 and *z* (*t*) < 0. Now, we define a function

$$y\left(t\right) = z\left(t\right) + c\_1^\beta z\left(\tau\_1\left(t\right)\right) + c\_2^\beta z\left(\tau\_2\left(t\right)\right). \tag{24}$$

Thus *y* (*t*) > 0 and *y* (*t*) > 0. From (1) and (M5), we obtain

$$z^{\prime\prime\prime}(t) \le -k\_1 q\_1(t) \ge^{\beta} \left(\sigma\_1(t)\right) - k\_2 q\_2(t) \ge^{\gamma} \left(\sigma\_2(t)\right). \tag{25}$$

> Combining (24) with (25), we ge<sup>t</sup>

$$\begin{aligned} y^{\prime\prime\prime}(t) &= \begin{aligned} z^{\prime\prime\prime}(t) + c\_1^{\beta} z^{\prime\prime\prime}(\tau\_1(t)) + c\_2^{\beta} z^{\prime\prime\prime}(\tau\_2(t)) \\ &\leq -k\_1 q\_1(t) \, \mathbf{x}^{\beta} \, (\sigma\_1(t)) - k\_2 q\_2 \, (t) \, \mathbf{x}^{\gamma} \, (\sigma\_2(t)) \\ &- c\_1^{\beta} \left( -k\_1 q\_1 \, (t) \, \mathbf{x}^{\beta} \left( \sigma\_1 \left( \tau\_1(t) \right) \right) - k\_2 q\_2 \, (t) \, \mathbf{x}^{\gamma} \left( \sigma\_2 \left( \tau\_1(t) \right) \right) \right) \\ &- c\_2^{\beta} \left( -k\_1 q\_1 \, (t) \, \mathbf{x}^{\beta} \left( \sigma\_1 \left( \tau\_2 \left( t \right) \right) \right) - k\_2 q\_2 \, (t) \, \mathbf{x}^{\gamma} \left( \sigma\_2 \left( \tau\_2 \left( t \right) \right) \right) \right) \end{aligned}$$

and so,

$$\begin{split} \left(y^{\prime\prime\prime}(t)\right) &\leq \ -k\_{1}q\_{1}^{\ast}\left(t\right)\left(\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{1}\left(t\right)\right)+c\_{1}^{\left\Vert\theta}\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{1}\left(\tau\_{1}\left(t\right)\right)\right)+c\_{2}^{\left\Vert\theta}\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{1}\left(\tau\_{2}\left(t\right)\right)\right)\right)\right)\right.\\ &\left. -k\_{2}q\_{2}^{\ast}\left(t\right)\left(\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{2}\left(t\right)\right)+c\_{1}^{\left\Vert\theta}\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{2}\left(\tau\_{1}\left(t\right)\right)\right)+c\_{2}^{\left\Vert\theta}\mathbf{x}^{\left\Vert\theta}\left(\sigma\_{2}\left(\tau\_{2}\left(t\right)\right)\right)\right)\right)\right.. \end{split}\right). \end{split}$$

By Lemma 2, since *c*1 + *c*2 < 1 and *β* < 1 < *γ*, we obtain

$$\begin{aligned} &y'''(t) + k\_1 q\_1^\*\left(t\right) z^{\beta} \left(\sigma\_1\left(t\right)\right) \\ &+ k\_2 q\_2^\*\left(t\right) \left(\mathbf{x}^{\gamma} \left(\sigma\_2\left(t\right)\right) + c\_1^{\gamma} \mathbf{x}^{\gamma} \left(\sigma\_2\left(\tau\_1\left(t\right)\right)\right) + \frac{c\_2^{\gamma}}{2^{\gamma-1}} \mathbf{x}^{\gamma} \left(\sigma\_2\left(\tau\_2\left(t\right)\right)\right)\right) \le 0. \end{aligned}$$

This implies

$$y'''(t) + k\_1 q\_1^\*\left(t\right) z^{\beta} \left(\sigma\_1\left(t\right)\right) + \frac{k\_2}{4^{\gamma - 1}} q\_2^\*\left(t\right) z^{\gamma} \left(\sigma\_2\left(t\right)\right) \le 0. \tag{26}$$

,

Using Lemma 6, we have two cases for *z* (*t*). If *z* (*t*) > 0, we find

$$y'''(t) + k\_1 q\_1^\*\left(t\right) z^{\beta} \left(\sigma\_1\left(t\right)\right) + \frac{k\_2}{4^{\gamma - 1}} q\_2^\*\left(t\right) z^{\gamma} \left(\sigma\_1\left(t\right)\right) \le 0. \tag{27}$$

Using arithmetic-geometric mean inequality with *u*1 = *k*1*λ*1 *q*∗1 (*t*) *zβ* (*<sup>σ</sup>*1 (*t*)) and *u*2 = *k*2 <sup>4</sup>*γ*−1*λ*2*q*∗2 (*t*) *zγ* (*<sup>σ</sup>*1 (*t*)), we ge<sup>t</sup>

$$\begin{split} \lambda\_1 \mu\_1 + \lambda\_2 \mu\_2 &\geq \quad u\_1^{\lambda\_1} u\_2^{\lambda\_2} \\ &= \quad \left(\frac{k\_1}{\lambda\_1}\right)^{\lambda\_1} \left(\frac{k\_2}{4^{\gamma - 1} \lambda\_2}\right)^{\lambda\_2} (q\_1^\*\left(t\right))^{\lambda\_1} (q\_2^\*\left(t\right))^{\lambda\_2} z\left(\sigma\_1\left(t\right)\right). \end{split} \tag{28}$$

Since *τ*1 (*t*) < *t* < *τ*2 (*t*), we note that

$$y'(t) \le \left(1 + c\_1^{\theta} + c\_2^{\theta}\right) z\left(\tau\_2\left(t\right)\right).$$

Hence, from (28), (27) becomes

$$y^{\prime\prime\prime}(t) + \left(\frac{k\_1}{\lambda\_1}\right)^{\lambda\_1} \left(\frac{k\_2}{4^{\gamma-1}\lambda\_2}\right)^{\lambda\_2} \frac{\left(q\_1^\*\left(t\right)\right)^{\lambda\_1} \left(q\_2^\*\left(t\right)\right)^{\lambda\_2}}{\left(1 + c\_1^{\beta} + c\_2^{\beta}\right)} y\left(\tau\_2^{-1}\left(\sigma\_1\left(t\right)\right)\right) \le 0. \tag{29}$$

Then, the condition (23) implies (29) has oscillatory solution, which contradicts *y* (*t*) > 0.

> Let *z* (*t*) < 0. As in the previous case, we ge<sup>t</sup>

$$y'''(t) + \left(\frac{k\_1}{\lambda\_1}\right)^{\lambda\_1} \left(\frac{k\_2}{4^{\gamma - 1}\lambda\_2}\right)^{\lambda\_2} \frac{\left(q\_1^\*\left(t\right)\right)^{\lambda\_1} \left(q\_2^\*\left(t\right)\right)^{\lambda\_2}}{\left(1 + c\_1^{\beta} + c\_2^{\beta}\right)} y\left(\tau\_1^{-1}\left(\sigma\_2\left(t\right)\right)\right) \le 0. \tag{30}$$

Hence, the condition (23) implies (30) has oscillatory solution, which contradicts *y* (*t*) > 0. This contradiction completes the proof.

**Remark 2.** *There are numerous results concerning the oscillation of the equation*

$$y''''(t) + q\left(t\right)y\left(\sigma\left(t\right)\right) = 0,$$

*(see [2,18,20,21]), which include Hille and Nehari types, Philos type, etc.*

Assume that

$$
\pi\_l(t) = t + (-1)^i \widetilde{\tau}\_{l\prime} \,\sigma\_l(t) = t - (-1)^i \widetilde{\sigma}\_{l\prime} \tag{31}
$$

where *τ*\**i*, \**σi* are positive constants for *i* = 1, 2. It is well known (see [9]) that the differential inequalities (29) and (30) are oscillatory if

$$\liminf\_{t \to \infty} \int\_{t - (\overline{\tau}\_2 + \overline{\sigma}\_1)/3}^t (\widetilde{\tau}\_2 + \widetilde{\sigma}\_1)^2 \left( q\_1^\* \left( t \right) \right)^{\lambda\_1} \left( q\_2^\* \left( t \right) \right)^{\lambda\_2} > \frac{9}{2\varepsilon} \left( \frac{\lambda\_1}{k\_1} \right)^{\lambda\_1} \left( \frac{4^{\gamma - 1} \lambda\_2}{k\_2} \right)^{\lambda\_2} \tag{32}$$

and

$$\liminf\_{t \to \infty} \int\_{t}^{t + \overline{\tau}\_{1} + \overline{\sigma}\_{2}} (s - t)^{2} \left( q\_{1}^{\*} \left( t \right) \right)^{\lambda\_{1}} \left( q\_{2}^{\*} \left( t \right) \right)^{\lambda\_{2}} > 2 \left( \frac{\lambda\_{1}}{k\_{1}} \right)^{\lambda\_{1}} \left( \frac{4^{\gamma - 1} \lambda\_{2}}{k\_{2}} \right)^{\lambda\_{2}},\tag{33}$$

respectively. Hence, we conclude the following theorem:

**Theorem 4.** *Assume that* 0 < *β* < 1 < *γ and (31) hold. If (32) and (33) hold, then every solution of Equation (1) is oscillatory.*

**Remark 3.** *In the case where α* = 1, *r* (*t*) = 1 *and pi* (*t*) = 0*, Equation (1) becomes*

$$\left(\mathbf{x}^{\prime\prime\prime}(t) + q\_1\left(t\right)f\_1\left(\mathbf{x}\left(\sigma\_1\left(t\right)\right)\right) + q\_2\left(t\right)f\_2\left(\mathbf{x}\left(\sigma\_2\left(t\right)\right)\right) = 0. \tag{34}$$

*Baculikova and Dzurina [5] proved that every nonoscillatory solution x of (34) satisfies x* < 0*. Thus, Theorems 3 and 4 improve the results in [5].*

**Remark 4.** *A manner similar to the Theorem 3, we can study the oscillation of solutions of Equation (1) when* 0 < *γ* < 1 < *β*.

**Remark 5.** *If α* = 1, *f*1 (*x*) = *<sup>x</sup>β*, *f*2 (*x*) = *<sup>x</sup>γ*, *τ*1 (*t*) = *t* − *<sup>τ</sup>*\*1, *σ*1 (*t*) = *t* − \**σ*1, *τ*2 (*t*) = *t* + *<sup>τ</sup>*\*2, *σ*2 (*t*) = *t* + \**σ*2 *and τ*\**i*, \**σi are positive constants, then Theorem 1 extends Theorem 2.5 and 2.7 in [23].*

**Remark 6.** *The results of Theorem 3 can be extended to the third-order differential equation*

$$\left( \left( z \left( t \right) \right)^{a} \right)^{\prime\prime\prime} + q\_1 \left( t \right) f\_1 \left( \ge \left( \sigma\_1 \left( t \right) \right) \right) + q\_2 \left( t \right) f\_2 \left( \ge \left( \sigma\_2 \left( t \right) \right) \right) = 0;$$

*the details are left to the reader.*

**Example 1.** *Consider the equation*

$$\left(\mathbf{x} + \frac{1}{3}\mathbf{x}\left(\frac{1}{3}t\right) + \frac{1}{3}\mathbf{x}\left(2t\right)\right)^{\prime\prime\prime} + \frac{q\_0}{t^3}\mathbf{x}\left(\frac{1}{2}t\right) + \frac{q\_1}{t^3}\mathbf{x}\left(2t\right) = \mathbf{0},\tag{35}$$

*where q*0 > 0*. We note that α* = *β* = *γ* = 1, *r* (*t*) = 1, *p*1 (*t*) = *p*2 (*t*) = 1/3, *τ*1 (*t*) = 1/3*t, σ*1 (*t*) = 1/2*t*, *τ*2 (*t*) = *σ*2 (*t*) = 2/*t and q*∗ (*t*) = *<sup>q</sup>*0/*t*3*. Hence, it is easy to see that*

$$\int\_{t\_0}^{\infty} \frac{1}{r^{1/a} \left(s\right)} ds = \infty.$$

*Now, if we set ρ* (*s*) := *t and k*1 = *k*2 = 1*, then we have*

$$\Theta\_1\left(t\right) = \frac{q\_0}{2s}.$$

*Thus, we find*

$$\begin{split} & \limsup\_{t \to \infty} \int\_{t\_0}^t \left( \Theta\_1 \left( s \right) - \left( 1 + c\_1^{\delta} + \frac{c\_2^{\delta}}{2^{\beta - 1}} \right) \frac{1}{\left( a + 1 \right)^{a + 1}} \frac{\left( \rho\_+^{\prime} \left( s \right) \right)^{a + 1} r \left( \sigma\_1 \left( s \right) \right)}{\left( \rho \left( s \right) \sigma\_1^{\prime} \left( s \right) \right)^a} \right) ds \\ & = \limsup\_{t \to \infty} \int\_{t\_0}^t \left( \frac{q\_0}{2s} - \frac{5}{6s} \right) ds. \end{split}$$

*Thus, the conditions become*

#### *q*0 > 1.66.

*Thus, by using Theorem 1, Equation (35) is either oscillatory if q*0 > 1.66 *or tends to zero as t* → ∞*.*

**Author Contributions:** O.M. and O.B.: Writing original draft, and writing review and editing. D.C.: Formal analysis, writing review and editing, funding and supervision. All authors have read and agreed to the published version of the manuscript.

**Funding:** The authors received no direct funding for this work.

**Acknowledgments:** The authors thank the reviewers for for their useful comments, which led to the improvement of the content of the paper.

**Conflicts of Interest:** There are no competing interests for the authors.
