EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 2, 2020, 185-199 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Some new oscillation results for fourth-order neutral differential equations Osama Moaaz1, Clemente Cesarano2,∗, Ali Muhib3 1 Department of Mathematics, Faculty of Science, Mansoura University, 35516 Mansoura, Egypt 2 Section of Mathematics, International Telematic University Uninettuno, CorsoVittorio Emanuele II, 39, 00186 Roma, Italy 3 Department of Mathematics, Faculty of Education – Al-Nadirah, Ibb University, Ibb, Yemen Abstract. By employing the Riccati substitution technique, we establish new oscillation criteria for a class of fourth-order neutral differential equations. Our new criteria complement a number of existing ones. An illustrative example is provided. 2020 Mathematics Subject Classifications: 34K10, 34K11 Key Words and Phrases: Fourth-order differential equations, Neutral delay, Oscillation 1. Introduction For several decades, an increasing interest in obtaining sufficient conditions for oscil- latory and nonoscillatory behavior of different classes of differential equations has been observed; see, for instance, the monographs [1]-[5], the papers [6]-[20], and the references cited therein. Neutral differential equations are used in numerous applications in technology and natural science. For instance, they are frequently used for the study of distributed net- works containing lossless transmission lines; see Hale [22], and therefore their qualitative properties are important. In this paper, we are concerned with the oscillation of solutions of the fourth-order neutral differential equation( r (t) ( (x (t) + p (t)x (τ (t)))′′′ )α)′ + q (t)xβ (σ (t)) = 0, (1) where t ≥ t0. In this work, we assume that α and β are quotients of odd positive integers, r, p, q ∈ C[t0,∞), r (t) > 0, r′ (t) ≥ 0, q (t) > 0, 0 ≤ p (t) < p0 < ∞, τ, σ ∈ C[t0,∞), ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i2.3654 Email addresses: o moaaz@mans.edu.eg (O. Moaaz), c.cesarano@uninettunouniversity.net (C. Cesarano),muhib39@students.mans.edu.eg (A. Muhib) http://www.ejpam.com 185 c© 2020 EJPAM All rights reserved. O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 186 τ (t) ≤ t, limt→∞ τ (t) = limt→∞ σ (t) = ∞. Moreover, we study (1) under the condition that ∫ ∞ t0 1 r1/α (s) ds =∞, (2) and we define the function z (t) := x (t) + p (t)x (τ (t)) . By a solution of (1) we mean a function x ∈ C3[tx,∞), tx ≥ t0, which has the property r (t) (z′′′ (t))α ∈ C1[tx,∞), and satisfies (1) on [tx,∞). We consider only those solutions x of (1) which satisfy sup{|x (t)| : t ≥ T} > 0, for all T ≥ tx. Definition 1. A solution x of (1) is said to be non-oscillatory if it is positive or negative, ultimately; otherwise, it is said to be oscillatory. The equation itself is termed oscillatory if all its solutions oscillate. Let us briefly comment on a number of related results which motivated our study. A number of oscillation results for differential equation( r (t) ( x(n−1) (t) )α)′ + q (t) f (x (τ (t))) = 0, have been established by Baculikova et al. [16] under the conditions (2) and∫ ∞ r−1/α (t) dt <∞. Asymptotic behavior of higher-order quasilinear neutral differential equations of the form( r (t) ( z(n−1) (t) )α)′ + q (t)xβ (σ (t)) = 0 have been studied by Li and Rogovchenko [21]. Agarwal et al. [6] investigated the oscil- latory behavior of a higher-order differential equation( r (t) ( x(n−1) (t) )α)′ + q (t)xβ (τ (t)) = 0, under the condition (2). The purpose of this article is to give sufficient conditions for the oscillatory behavior of (1). under the condition that (2) In order to discuss our main results, we need the following lemmas: Lemma 1. [5]If the function x satisfies x(i) (t) > 0, i = 0, 1, ..., n, and x(n+1) (t) < 0, then x (t) tn/n! ≥ x′ (t) tn−1/ (n− 1)! . O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 187 Lemma 2. [3, Lemma 2.2.3]Let x ∈ Cn ([t0,∞) , (0,∞)) . Assume that x(n) (t) is of fixed sign and not identically zero on [t0,∞) and that there exists a t1 ≥ t0 such that x(n−1) (t)x(n) (t) ≤ 0 for all t ≥ t1. If limt→∞ x (t) 6= 0, then for every µ ∈ (0, 1) there exists tµ ≥ t1 such that x (t) ≥ µ (n− 1)! tn−1 ∣∣∣x(n−1) (t) ∣∣∣ for t ≥ tµ. Lemma 3. [23]Let x (t) be a positive and n-times differentiable function on an interval [T,∞) with its nth derivative x(n) (t) non-positive on [T,∞) and not identically zero on any interval of the form [T ′,∞) , T ′ ≥ T and x(n−1) (t)x(n) (t) ≤ 0, t ≥ tx then there exist constants θ, 0 < θ < 1 and N > 0 such that x′ (θt) ≥ Ntn−2x(n−1) (t) , for all sufficient large t. In this section we will find one condition to ensure the oscillation of solutions of (1) in the case p0 < 1. 2. One-condition theorems Lemma 4. Assume that x is an eventually positive solution of (1). Then( r (t) ( z′′′ (t) )α)′ ≤ −q (t) (1− p0)β zβ (σ (t)) . (3) Proof. Assume that x is an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. Since r′ (t) > 0, we have z (t) > 0, z′ (t) > 0, z′′′ (t) > 0, z(4) (t) < 0 and ( r (t) ( z′′′ (t) )α)′ ≤ 0, (4) for t ≥ t1. From definition of z, we get x (t) ≥ z (t)− p0x (τ (t)) ≥ z (t)− p0z (τ (t)) ≥ (1− p0) z (t) , which with (1) gives ( r (t) ( z′′′ (t) )α)′ + q (t) (1− p0)β zβ (σ (t)) ≤ 0. The proof is complete. Theorem 1. Assume that lim inf t→∞ 1 Ψ̃1 (t) ∫ ∞ t Ψ2 (s) Ψ̃ α+1 α 1 (s) ds > α (α+ 1) α+1 α , (5) O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 188 where Ψ1 (t) = q (t) (1− p0)βMβ−α (σ (t)) , Ψ2 (t) = αε σ2 (t) ζσ′ (t) r1/α (t) and Ψ̃1 (t) = ∫ ∞ t Ψ1 (s) ds. Then, (1) is oscillatory. Proof. Assume that x is an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. Using Lemma 4, we obtain that (3) holds. Define ω as follows ω (t) := r (t) (z′′′ (t))α zα (ζσ (t)) . (6) By differentiating and using (3), we obtain ω′ (t) ≤ −q (t) (1− p0)β zβ (σ (t)) . zα (ζσ (t)) − αr (t) (z′′′ (t))α z′ (ζσ (t)) ζσ′ (t) zα+1 (ζσ (t)) . From Lemma 3, we have ω′ (t) ≤ −q (t) (1− p0)β zβ−α (σ (t))− αr (t) (z′′′ (t))α εσ2 (t) z′′′ (σ (t)) ζσ′ (t) zα+1 (ζσ (t)) , which is ω′ (t) ≤ −q (t) (1− p0)β zβ−α (σ (t))− αεr (t)σ2 (t) ζσ′ (t) (z′′′ (t))α+1 zα+1 (ζσ (t)) , by using (6) we have ω′ (t) ≤ −q (t) (1− p0)β zβ−α (σ (t))− αεσ 2 (t) ζσ′ (t) r1/α (t) ω(α+1)/α (t) , (7) Since z′ (t) > 0, there exist a t2 ≥ t1 and a constant M > 0 such that z (t) > M. Then, (7), turn to ω′ (t) ≤ −q (t) (1− p0)βMβ−α (σ (t))− αεσ 2 (t) ζσ′ (t) r1/α (t) ω(α+1)/α (t) , that is, ω′ (t) + Ψ1 (t) + Ψ2 (t)ω(α+1)/α (t) ≤ 0. O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 189 Integrating the above inequality from t to l , we get ω (l)− ω (t) + ∫ l t Ψ1 (s) ds+ ∫ l t Ψ2 (s)ω α+1 α (s) ds ≤ 0. Letting l→∞ and using ω > 0 and ω′ < 0, we have ω (t) ≥ Ψ̃1 (t) + ∫ ∞ t Ψ2 (s)ω α+1 α (s) ds. This implies ω (t) Ψ̃1 (t) ≥ 1 + 1 Ψ̃1 (t) ∫ ∞ t Ψ2 (s) Ψ̃ α+1 α 1 (s) ( ω (s) Ψ̃1 (s) )α+1 α ds. (8) Let λ = inft≥T ω (t) /Ψ̃1 (t) . Then obviously λ ≥ 1. Thus, from (5) and (8) we see that λ ≥ 1 + α ( λ α+ 1 )(α+1)/α or λ α+ 1 ≥ 1 α+ 1 + α α+ 1 ( λ α+ 1 )(α+1)/α , which contradicts the admissible value of λ ≥ 1 and α > 0. Therefore, the proof is complete. In this section we will find two independent conditions to ensure the oscillation of solutions of (1) in the case p0 < 1 3. Two independent conditions theorems Here, we introduce Riccati substitutions ω (t) := r (t) (z′′′ (t))α zα (t) and w (t) := z′ (t) z (t) . (9) Also, for convenience, we denote that: R1 (t) : = αµ t2 2r1/α (t) , Q1 (t) : = q (t) (1− p0)βMβ−α 1 ( σ (t) t )3β and Q2 (t) := (1− p0)β/αMβ/α−1 2 ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) σβ (s) sβ ds )1/α du, O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 190 for some µ ∈ (0, 1) and every M1,M2 are positive constants. All functional inequalities are assumed to hold eventually, that is, they are assumed to be satisfied for all t sufficiently large. The proof of the next lemma is immediate from [23] and hence is omitted. Lemma 5. Assume that (2) holds and x is an eventually positive solution of (1). Then, (r (t) (z′′′ (t))α) ′ < 0 and there are the following two possible cases eventually: (C1) z (t) > 0, z′ (t) > 0, z′′ (t) > 0, z′′′ (t) > 0, z(4) (t) < 0, (C2) z (t) > 0, z′ (t) > 0, z′′ (t) < 0, z′′′ (t) > 0. Lemma 6. Let x be an eventually positive solution of (1) and the functions ω and w are defined as in (9). (I1) If x satisfies (C1), then ω′ (t) +Q1 (t) +R1 (t)ω α+1 α (t) ≤ 0; (10) (I2) If x satisfies (C2), then w′ (t) +Q2 (t) + w2 (t) ≤ 0. (11) Proof. Assume that x is an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. Using Lemma 4, we obtain that (3) holds. In the case (C1), by differentiating ω and using (3), we obtain ω′ (t) ≤ −q (t) (1− p0)β zβ (σ (t)) zα (t) − αr (t) (z′′′ (t))α zα+1 (t) z′ (t) . (12) From Lemma 1, we have that z (t) ≥ t 3 z′ (t) and hence z (σ (t)) z (t) ≥ σ3 (t) t3 . (13) It follows from Lemma 2 that z′ (t) ≥ µ1 2 t2z′′′ (t) , (14) for all µ1 ∈ (0, 1) and every sufficiently large t. Since z′ (t) > 0, there exist a t2 ≥ t1 and a constant M > 0 such that z (t) > M, (15) for t ≥ t2. Thus, by (12), (13), (14) and (15), we get ω′ (t) +Q1 (t) +R1 (t)ω α+1 α (t) ≤ 0. In the case (C2), integrating (3) from t to u, we obtain r (u) ( z′′′ (u) )α − r (t) ( z′′′ (t) )α ≤ −∫ u t q (s) (1− p0)β zβ (σ (s)) ds. (16) O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 191 From Lemma 1, we get that z (t) ≥ tz′ (t) and hence z (σ (t)) ≥ σ (t) t z (t) . (17) For (16), letting u→∞ and using (17), we see that r (t) ( z′′′ (t) )α ≥ (1− p0)β zβ (t) ∫ ∞ t q (s) σβ (s) sβ ds. Integrating this inequality again from t to ∞, we get z′′ (t) ≤ − (1− p0)β/α zβ/α (t) ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) σβ (s) sβ ds )1/α du, (18) for all µ2 ∈ (0, 1). By differentiating w and using (15) and (18), we find w′ (t) = z′′ (t) z (t) − ( z′ (t) z (t) )2 ≤ −w2 (t)− (1− p0)β/αM (β/α)−1 ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) σβ (s) sβ ds )1/α du, (19) hence w′ (t) +Q2 (t) + w2 (t) ≤ 0. The proof is complete. Theorem 2. Assume that lim inf t→∞ 1 Q̃1 (t) ∫ ∞ t R1 (s) Q̃ α+1 α 1 (s) ds > α (α+ 1) α+1 α (20) and lim inf t→∞ 1 Q̃2 (t) ∫ ∞ t0 Q̃2 2 (s) ds > 1 4 , (21) where Q̃1 (t) = ∫ ∞ t Q1 (s) ds and Q̃2 (t) = ∫ ∞ t Q2 (s) ds. (22) Then, (1) is oscillatory. Proof. Assume to the contrary that (1) has a nonoscillatory solution in [t0,∞). With- out loss of generality, we let x be an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. From Lemma 5 O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 192 there is two cases. For case (C1). Using Lemma 6, we obtain (10) holds. Integrating (10) from t to l , we get ω (l)− ω (t) + ∫ l t Q1 (s) ds+ ∫ l t R1 (s)ω α+1 α (s) ds ≤ 0. Letting l→∞ and using ω > 0 and ω′ < 0, we have ω (t) ≥ Q̃1 (t) + ∫ ∞ t R1 (s)ω α+1 α (s) ds. (23) This implies ω (t) Q̃1 (t) ≥ 1 + 1 Q̃1 (t) ∫ ∞ t R1 (s) Q̃ α+1 α 1 (s) ( ω (s) Q̃1 (s) )α+1 α ds. (24) Let λ = inft≥T ω (t) /Q̃1 (t) . Then obviously λ ≥ 1. Thus, from (20) and (24) we see that λ ≥ 1 + α ( λ α+ 1 )(α+1)/α or λ α+ 1 ≥ 1 α+ 1 + α α+ 1 ( λ α+ 1 )(α+1)/α , which contradicts the admissible value of λ ≥ 1 and α > 0. The proof of the case where (C2) holds is the same as that of case (C1). Therefore, the proof is complete. Define a sequence of functions {un (t)}∞n=0 and {vn (t)}∞n=0 as u0 (t) = Q̃1 (t) , and v0 (t) = Q̃2 (t) , un (t) = u0 (t) + ∫∞ t R1 (t)u (α+1)/α n−1 (s) ds, n > 1, vn (t) = v0 (t) + ∫∞ t v (α+1)/α n−1 (s) ds, n > 1, (25) where Q̃1 and Q̃2 defined as in (22). We see by induction that un (t) ≤ un+1 (t) and vn (t) ≤ vn+1 (t) for t ≥ t0, n > 1. Theorem 3. Let un (t) and vn (t) be defined as in (25). If lim sup t→∞ ( µ1t 3 6r1/α (t) )α un (t) > 1 (26) and lim sup t→∞ λtvn (t) > 1, (27) for some n, then (1)is oscillatory. O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 193 Proof. Assume to the contrary that (1) has a nonoscillatory solution in [t0,∞). With- out loss of generality, we let x be an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. From Lemma 5 there is two cases. In the case (C1), proceeding as in the proof of Lemma 6, we get that (14) holds. It follows from Lemma 2 that z (t) ≥ µ1 6 t3z′′′ (t) . (28) From definition of ω (t) and (28), we have 1 ω (t) = 1 r (t) ( z (t) z′′′ (t) )α ≥ 1 r (t) (µ1 6 t3 )α . Thus, ω (t) ( µ1t 3 6r1/α (t) )α ≤ 1. Therefore, lim sup t→∞ ω (t) ( µ1t 3 6r1/α (t) )α ≤ 1, which contradicts (26). The proof of the case where (C2) holds is the same as that of case (C1). Therefore, the proof is complete. Corollary 1. Let un (t) and vn (t) be defined as in (25). If∫ ∞ t0 Q1 (t) exp (∫ t t0 R1 (s)u1/αn (s) ds ) dt =∞ (29) and ∫ ∞ t0 Q2 (t) exp (∫ t t0 v1/αn (s) ds ) dt =∞, (30) for some n, then (1) is oscillatory. Proof. Assume to the contrary that (1) has a nonoscillatory solution in [t0,∞). With- out loss of generality, we let x be an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. From Lemma 5 there is two cases. In the case (C1), proceeding as in the proof of Theorem 2, we get that (23) holds. It follows from (23) that ω (t) ≥ u0 (t). Moreover, by induction we can also see that ω (t) ≥ un (t) for t ≥ t0, n > 1. Since the sequence {un (t)}∞n=0 monotone increasing and bounded above, it converges to u (t). Thus, by using Lebesgue’s monotone convergence theorem, we see that u (t) = lim n→∞ un (t) = ∫ ∞ t R1 (t)u(α+1)/α (s) ds+ u0 (t) O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 194 and u′ (t) = −R1 (t)u(α+1)/α (t)−Q1 (t) . (31) Since un (t) ≤ u (t), it follows from (31) that u′ (t) ≤ −R1 (t)u1/αn (t)u (t)−Q1 (t) . Hence, we get u (t) ≤ exp ( − ∫ t T R1 (s)u1/αn (s) ds )( u (T )− ∫ t T Q1 (s) exp (∫ s T R1 (u)u1/αn (u) du ) ds ) . This implies ∫ t T Q1 (s) exp (∫ s T R1 (u)u1/αn (u) du ) ds ≤ u (T ) <∞, which contradicts (29). The proof of the case where (C2) holds is the same as that of case (C1). Therefore, the proof is complete. 4. Further results Lemma 7. Assume that x is an eventually positive solution of (1) and p ( τ−1 ( τ−1 (t) )) ≥ ( τ−1 ( τ−1 (t) ) τ−1 (t) )3 . (32) Then ( r (t) ( z′′′ (t) )α)′ + q (t) p̃β (σ (t)) zβ ( τ−1 (σ (t)) ) ≤ 0, (33) where p̃ (t) :=  1 p(τ−1(t)) ( 1− (τ−1(τ−1(t))) 3 (τ−1(t))3p(τ−1(τ−1(t))) ) for case (C1) ; 1 p(τ−1(t)) ( 1− (τ−1(τ−1(t))) (τ−1(t))p(τ−1(τ−1(t))) ) for case (C2) . (34) Proof. Proceeding as in the proof of Lemma 4, we get that (4) holds. It follows from Lemma 5 that there exist two possible cases (C1) and (C2). From the definition of z (t), we see that x (t) = 1 p (τ−1 (t)) ( z ( τ−1 (t) ) − x ( τ−1 (t) )) . By repeating the same process, we find that x (t) = z ( τ−1 (t) ) p (τ−1 (t)) − 1 p (τ−1 (t)) ( z ( τ−1 ( τ−1 (t) )) p (τ−1 (τ−1 (t))) − x ( τ−1 ( τ−1 (t) )) p (τ−1 (τ−1 (t))) ) O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 195 ≥ z ( τ−1 (t) ) p (τ−1 (t)) − 1 p (τ−1 (t)) z ( τ−1 ( τ−1 (t) )) p (τ−1 (τ−1 (t))) . (35) Assume that Case (C1) holds. Proceeding as in the proof of Lemma 6, we get that (13) holds, which with the fact that τ (t) ≤ t gives z ( τ−1 ( τ−1 (t) )) ≤ ( τ−1 ( τ−1 (t) ) τ−1 (t) )3 z ( τ−1 (t) ) . (36) From (35) and (36), we find that x (t) ≥ 1 p (τ−1 (t)) ( 1− ( τ−1 ( τ−1 (t) ))3 (τ−1 (t))3 p (τ−1 (τ−1 (t))) ) z ( τ−1 (t) ) . (37) Assume that Case (C2) holds. Proceeding as in the proof of (C2) in Lemma 6, we get that (17) holds. Since τ−1 (t) ≤ τ−1 ( τ−1 (t) ) , we obtain τ−1 (t) z ( τ−1 ( τ−1 (t) )) ≤ τ−1 ( τ−1 (t) ) z ( τ−1 (t) ) . (38) From (35) and (38), we find x (t) ≥ 1 p (τ−1 (t)) ( 1− ( τ−1 ( τ−1 (t) )) (τ−1 (t)) p (τ−1 (τ−1 (t))) ) z ( τ−1 (t) ) . (39) Next, from (37) and (39), we get that x (t) ≥ p̃ (t) z ( τ−1 (t) ) , which with (1) yields (33). Therefore, the proof is complete. Lemma 8. Assume that σ (t) ≤ τ (t) , x is an eventually positive solution of (1) and the functions ω and w are defined as in (9). (I3) If x satisfies (C1), then ω′ (t) +Q3 (t) +R1 (t)ω α+1 α (t) ≤ 0; (I4) If x satisfies (C2), then w′ (t) +Q4 (t) + w2 (t) ≤ 0, where Q3 (t) = q (t) p̃β (σ (t))Mβ−α 3 ( τ−1 (σ (t)) t )3α and Q4 (t) = p̃β/α (σ (s))M (β/α)−1 4 ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) ( τ−1 (σ (s)) s )β ds )1/α du. O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 196 Proof. Assume that x is an eventually positive solution of (1). Then, there exists a t1 ≥ t0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0 for t ≥ t1. Using Lemma 7, we obtain that (33) holds. In the case (C1), by differentiating ω and using (33), we obtain ω′ (t) ≤ − q (t) p̃β (σ (t)) zβ ( τ−1 (σ (t)) ) zα (t) − αr (t) (z′′′ (t))α zα+1 (t) z′ (t) . (40) From Lemma 1, we have that z (t) ≥ t 3 z′ (t) and hence z ( τ−1 (σ (t)) ) z (t) ≥ ( τ−1 (σ (t)) )3 t3 . (41) It follows from Lemma 2 that z′ (t) ≥ µ1 2 t2z′′′ (t) , (42) for all µ1 ∈ (0, 1) and every sufficiently large t. Since z′ (t) > 0, there exist a t2 ≥ t1 and a constant M > 0 such that z (t) > M, (43) for t ≥ t2. Thus, by (40), (41), (42) and (43), we get ω′ (t) +Q3 (t) +R1 (t)ω α+1 α (t) ≤ 0. In the case (C2), integrating (33) from t to u, we obtain r (u) ( z′′′ (u) )α − r (t) ( z′′′ (t) )α ≤ −∫ u t q (s) p̃β (σ (s)) zβ ( τ−1 (σ (s)) ) ds ≤ 0. (44) From Lemma 1, we get that z (t) ≥ tz′ (t) and hence z ( τ−1 (σ (t)) ) ≥ τ−1 (σ (t)) t z (t) . (45) For (44), letting u→∞ and using (45), we see that r (t) ( z′′′ (t) )α ≥ p̃β (σ (s)) zβ (t) ∫ ∞ t q (s) ( τ−1 (σ (s)) s )β ds. Integrating this inequality again from t to ∞, we get z′′ (t) ≤ −p̃β/α (σ (s)) zβ/α (t) ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) ( τ−1 (σ (s)) s )β ds )1/α du, (46) for all µ2 ∈ (0, 1). By differentiating w and using (15) and (46), we find w′ (t) = z′′ (t) z (t) − ( z′ (t) z (t) )2 O. Moaaz, C. Cesarano, A. Muhib / Eur. J. Pure Appl. Math, 13 (2) (2020), 185-199 197 ≤ −w2 (t)− p̃β/α (σ (s))M (β/α)−1 ∫ ∞ t ( 1 r (u) ∫ ∞ u q (s) ( τ−1 (σ (s)) s )β ds )1/α du, (47) hence w′ (t) +Q4 (t) + w2 (t) ≤ 0. The proof is complete. Theorem 4. Assume that lim inf t→∞ 1 Q̃3 (t) ∫ ∞ t R1 (s) Q̃ α+1 α 3 (s) ds > α (α+ 1) α+1 α (48) and lim inf t→∞ 1 Q̃4 (t) ∫ ∞ t Q̃2 4 (s) ds > 1 4 , (49) where Q̃3 (t) = ∫ ∞ t Q3 (s) ds and Q̃4 (t) = ∫ ∞ t Q4 (s) ds. Then, (1) is oscillatory. Proof. Proceeding as in the proof of Theorem 2, Example 1. Consider the differential equation( x (t) + 16x ( t 2 ))(4) + q0 t4 x ( t 6 ) = 0. (50) We note that α = β = 1, r (t) = 1, p (t) = 16, τ (t) = t/2, σ (t) = t/6 and q (t) = q0/t 4. 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