EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 713-723 ISSN 1307-5543 – ejpam.com Published by New York Business Global Existence of nonoscillatory solutions of higher order nonlinear neutral differential equations B. Çına1, T. Candan2,∗and M. Tamer Şenel3 1 Zara Veysel Dursun School of Applied Science, Cumhuriyet University, Sivas, Turkey 2 College of Engineering and Technology, American University of the Middle East, Egaila 54200, Kuwait 3 Department of Mathematics, Faculty of Sciences, Erciyes University, Kayseri, Turkey Abstract. In this paper, an n-th order neutral nonlinear differential equation is studied. By using the Banach contraction principle, some sufficient conditions are established for the existence of nonoscillatory solutions of nonlinear n-th order neutral differential equation. An example is included to illustrate the results obtained. 2020 Mathematics Subject Classifications: 34K11, 34K40 Key Words and Phrases: Fixed point, Higher-order, Neutral differential equation, Nonoscilla- tory solution 1. Introduction This paper is concerned with nonoscillatory solutions of nonlinear n-th order neutral differential equation of the form [r(t)[x(t)− p(t)x(t− τ)](n−1)]′ + (−1)n[f1(t, x(σ1(t)))− f2(t, x(σ2(t)))− g(t)] = 0, (1) where n ≥ 2 is an integer, τ > 0, p, σi, g ∈ C([t0,∞),R), r ∈ C([t0,∞), (0,∞)) and limt→∞ σi(t) = ∞, i = 1, 2. Throughout this article, we assume that fi(t, x) ∈ C([t0,∞)×R,R) is a nondecreasing in x for i = 1, 2, xfi(t, x) > 0 for x ̸= 0, i = 1, 2, and satisfies |fi(t, x)− fi(t, y)| ≤ qi(t)|x− y| for t ∈ [t0,∞) and x, y ∈ [a, b], (2) where qi ∈ C([t0,∞), (0,∞)), i = 1, 2, and [a, b] (0 < a < b or a < b < 0) is any closed interval. Furthermore, suppose that∫ ∞ t0 ∫ s t0 sn−2 r(s) qi(u)duds < ∞, i = 1, 2, (3) ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4708 Email addresses: bengu58cina@gmail.com (B. Çına), Tuncay.Candan@aum.edu.kw (T. Candan) senel@erciyes.edu.tr (M. Tamer Şenel) https://www.ejpam.com 713 © 2023 EJPAM All rights reserved. B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 714 ∫ ∞ t0 ∫ s t0 sn−2 r(s) |fi(u, d)|duds < ∞ for some d ̸= 0, i = 1, 2, (4) and ∫ ∞ t0 ∫ s t0 sn−2 r(s) |g(u)|duds < ∞ (5) hold. Oscillation and nonoscillation phenomena appear in different models from real world applications; see, for instance, oscillatory and nonoscillatory solutions may appear in im- pulsive partial neutral differential equations from mathematical biology, we refer to the papers [11, 12, 16] where impulsive effects are modelled by external sources complementing partial differential equations involving taxis mechanisms, and arising in biomathematics. We also refer the reader to the papers [9, 14, 15] for the oscillation and asymptotic behav- ior of solutions to various classes of neutral differential equations. In particular, Zhou and Zhang [21] and Candan [4] studied existence of nonoscillatory solutions of higher order neutral differential equations of the form dn dtn [x(t) + cx(t− τ)] + (−1)n+1 [P (t)x(t− σ)−Q(t)x(t− δ)] = 0 (6) and [r(t)[x(t) + P (t)x(t− τ)](n−1)]′ + (−1)n[Q1(t)g1(x(t− σ1))−Q2(t)g2(x(t− σ2))− f(t)] = 0, (7) respectively. Later, Çına et al.[8] studied the existence of nonoscillatory solutions of non- linear second order neutral differential equation with forcing term of the form( r(t) (x(t)− p(t)x(t− τ))′ )′ + f1(t, x(σ1(t)))− f2(t, x(σ2(t))) = g(t). Motivated by the idea of [4, 8, 21], the goal of this paper is to present some sufficient conditions for the existence of nonoscillatory solutions of (1). For related studies on the existence of nonoscillatory solutions of second or higher order neutral differential and difference equations the reader is referred to the papers [3, 5–7, 17–20] and books [1, 2, 10, 13]. Let T0 = min{t1 − τ, inf t≥t1 σ1(t), inf t≥t1 σ2(t)} for t1 ≥ t0. By a solution of equation (1), we mean a function x ∈ C([T0,∞),R) in the sense that x(t)− p(t)x(t− τ) is n− 1 times continuously differentiable and r(t)(x(t)− p(t)x(t− τ))(n−1) is continuously differentiable on [t1,∞) and such that equation (1) is satisfied for t ≥ t1. As usual, a solution of (1) is said to be oscillatory if it has arbitrarily large zeros. Otherwise the solution is called nonoscillatory. 2. Main Results Theorem 1. Assume that (3)-(5) hold and 0 ≤ p(t) ≤ p < 1. Then (1) has a bounded nonoscillatory solution. B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 715 Proof. Suppose (4) holds with d > 0, the case d < 0 can be treated similarly. Let X be the set of all continuous and bounded functions on [t0,∞) with the ∥x∥ = sup t≥t0 |x(t)| < ∞ norm. Set A = {x ∈ X : N1 ≤ x(t) ≤ d, t ≥ t0}, where N1 is a positive constant such that N1 < (1− p)d. Clearly, A is a closed, bounded and convex subset of X. By (3)-(5) there exists a t1 > t0 sufficiently large such that t− τ ≥ t0, σ1(t) ≥ t0, σ2(t) ≥ t0 for t ≥ t1 and p+ 2 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ≤ θ1 < 1, i = 1, 2, (8) where θ1 is a constant, 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ≤ (1− p)d− α, (9) 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ≤ α−N1, (10) where α is a positive constant such that N1 < α < (1 − p)d. Define the operator S on A by (Sx)(t) =  α+ p(t)x(t− τ) + 1 (n−2)! ∫∞ t (s−t)n−2 r(s) ∫ s t1 [f1(u, x(σ1(u))) −f2(u, x(σ2(u)))− g(u)]duds, t ≥ t1 (Sx)(t1), t0 ≤ t ≤ t1. We can easily see that Sx is continuous. We shall show that SA ⊂ A. In fact, for every x ∈ A and t ≥ t1, due to (9), we have (Sx)(t) = α+ p(t)x(t− τ) + 1 (n− 2)! ∫ ∞ t (s− t)n−2 r(s) ∫ s t1 [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ≤ α+ pd+ 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ≤ d. Furthermore, by using (10), we obtain (Sx)(t) = α+ p(t)x(t− τ) + 1 (n− 2)! ∫ ∞ t (s− t)n−2 r(s) ∫ s t1 [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ≥ α− 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 716 ≥ N1. Thus, we proved that SA ⊂ A. Now we shall show that operator S is a contraction operator on A. In fact, for x, y ∈ A and t ≥ t1, in view of (2) and (8), we have |(Sx)(t)− (Sy)(t)| ≤ p|x(t− τ)− y(t− τ)| + 1 (n− 2)! 2∑ i=1 ∫ ∞ t (s− t)n−2 r(s) ∫ s t1 |fi(u, x(σi(u)))− fi(u, y(σi(u)))|duds ≤ p|x(t− τ)− y(t− τ)| + 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 (s− t)n−2 r(s) ∫ s t1 qi(u)|x(σi(u))− y(σi(u)))|duds ≤ ∥x− y∥ [ p+ 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ1∥x− y∥. This implies that ∥Sx− Sy∥ ≤ θ1∥x− y∥. Since θ1 < 1 by (8), it follow that S is a contraction mapping on A. By the Banach contraction mapping principle, S has a fixed point x ∈ A, which is obviously a positive solution of (1). This completes the proof. Theorem 2. Assume that (3)-(5) hold and 1 < p1 ≤ p(t) ≤ p2 < ∞. Then (1) has a bounded nonoscillatory solution. Proof. Suppose (4) holds with d > 0, the case d < 0 can be treated similarly. Let X be the set as in the proof of Theorem 1. Set A = {x ∈ X : N2 ≤ x(t) ≤ d, t ≥ t0}, where N2 is a positive constant such that p2N2 < (p1 − 1)d. It is clear that A is a closed, bounded and convex subset of X. By (3)-(5), we can choose a t1 > t0 sufficiently large such that σ1(t+ τ) ≥ t0, σ2(t+ τ) ≥ t0 for t ≥ t1 and 1 p1 [ 1 + 2 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ2 < 1, i = 1, 2, (11) where θ2 is a constant, 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ≤ α− p2N2, (12) 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ≤ (p1 − 1)d− α, (13) B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 717 where α is a positive constant such that p2N2 < α < (p1 − 1)d. Define the operator S on A by (Sx)(t) =  1 p(t+τ) [ α+ x(t+ τ)− 1 (n−2)! ∫∞ t+τ (s−t−τ)n−2 r(s) ∫ s t1+τ [f1(u, x(σ1(u))) −f2(u, x(σ2(u)))− g(u)]duds ] , t ≥ t1 (Sx)(t1), t0 ≤ t ≤ t1. Clearly, Sx is continuous. First, we shall show that SA ⊂ A. In fact, for every x ∈ A and t ≥ t1, using (13), we obtain (Sx)(t) = 1 p(t+ τ) [ α+ x(t+ τ)− 1 (n− 2)! ∫ ∞ t+τ (s− t− τ)n−2 r(s) ∫ s t1+τ [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ] ≤ 1 p1 [ α+ d+ 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ] ≤ d and taking (12) into account, we have (Sx)(t) = 1 p(t+ τ) [ α+ x(t+ τ)− 1 (n− 2)! ∫ ∞ t+τ (s− t− τ)n−2 r(s) ∫ s t1+τ [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ] ≥ 1 p(t+ τ) [ α− 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ] ≥ 1 p2 [ α− 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ] ≥ N2. Thus, we proved that SA ⊂ A. Second, we shall show that S is a contraction operator on A. In fact, for x, y ∈ A and t ≥ t1, in view of (2) and (11), we have |(Sx)(t)− (Sy)(t)| ≤ 1 p(t+ τ) [ |x(t+ τ)− y(t+ τ)| + 1 (n− 2)! 2∑ i=1 ∫ ∞ t (s− t− τ)n−2 r(s) ∫ s t1 |fi(u, x(σi(u)))− fi(u, y(σi(u)))|duds ] ≤ ∥x− y∥ p1 [ 1 + 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ2∥x− y∥. B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 718 This immediately implies that ∥Sx− Sy∥ ≤ θ2∥x− y∥. Since θ2 < 1 by (11), it follows that S is a contraction operator on A. By the Banach contraction mapping principle, S has a fixed point x ∈ A, and x is a positive solution of (1). Thus, the proof is completed. Theorem 3. Assume that (3)-(5) hold and −1 < −p ≤ p(t) ≤ 0. Then (1) has a bounded nonoscillatory solution. Proof. Suppose (4) holds with d > 0, the case d < 0 can be treated similarly. Let X be the set as in the proof of Theorem 1. Set A = {x ∈ X : N3 ≤ x(t) ≤ d, t ≥ t0}, where N3 is a positive constant such that N3 + pd < d. Clearly, A is a closed, bounded and convex subset of X. In view of (3)-(5), there exists a t1 > t0 sufficiently large such that t− τ ≥ t0, σ1(t) ≥ t0, σ2(t) ≥ t0 for t ≥ t1 and p+ 2 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ≤ θ3 < 1, i = 1, 2, (14) where θ3 is a constant, 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ≤ d− α, (15) 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ≤ α−N3 − pd, (16) where α is a positive constant such that N3 + pd < α < d. Define the operator S on A by (Sx)(t) =  α+ p(t)x(t− τ) + 1 (n−2)! ∫∞ t (s−t)n−2 r(s) ∫ s t1 [f1(u, x(σ1(u))) −f2(u, x(σ2(u)))− g(u)]duds, t ≥ t1 (Sx)(t1), t0 ≤ t ≤ t1. Clearly, Sx is continuous. First, we shall show that SA ⊂ A. For every x ∈ A and t ≥ t1, by using (15), we have (Sx)(t) = α+ p(t)x(t− τ) + 1 (n− 2)! ∫ ∞ t ∫ s t1 (s− t)n−2 r(s) [f1(u, x(σ1(u)))− f2(u, x(σ2(u)))− g(u)]duds ≤ α+ 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 719 ≤ d and applying (16), we have (Sx)(t) = α+ p(t)x(t− τ) + 1 (n− 2)! ∫ ∞ t ∫ s t1 (s− t)n−2 r(s) [f1(u, x(σ1(u)))− f2(u, x(σ2(u)))− g(u)]duds ≥ α− pd− 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ≥ N3. Hence, SA ⊂ A. Finally, we show that S is a contraction operator on A. In fact, for x, y ∈ A and t ≥ t1, using (2) and (14), we obtain |(Sx)(t)− (Sy)(t)| ≤ p|x(t− τ)− y(t− τ)| + 1 (n− 2)! 2∑ i=1 ∫ ∞ t (s− t)n−2 r(s) ∫ s t1 |fi(u, x(σi(u)))− fi(u, y(σi(u)))|duds ≤ p|x(t− τ)− y(t− τ)| + 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 (s− t)n−2 r(s) ∫ s t1 qi(u)|x(σi(u))− y(σi(u)))|duds ≤ ∥x− y∥ [ p+ 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ3∥x− y∥. This implies that ∥Sx− Sy∥ ≤ θ3∥x− y∥. Since θ3 < 1 by (14), it follows that S is a contraction operator on A. By the Banach contraction mapping principle, S has a fixed point x ∈ A, which is obviously a positive solution of (1). This completes the proof. Theorem 4. Assume that (3)-(5) hold and −∞ < −p1 ≤ p(t) ≤ −p2 < −1. Then (1) has a bounded nonoscillatory solution. Proof. Suppose (4) holds with d > 0, the case d < 0 can be treated similarly. Let X be the set as in the proof of Theorem 1. Set A = {x ∈ X : N4 ≤ x(t) ≤ d, t ≥ t0}, where N4 is a positive constant such that p1N4 + d < p2d. It is clear that A is a closed, bounded and convex subset of X. By (3)-(5), we can choose a t1 > t0 sufficiently large such that σ1(t+ τ) ≥ t0, σ2(t+ τ) ≥ t0 for t ≥ t1 and 1 p2 [ 1 + 2 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ4 < 1, i = 1, 2, (17) B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 720 where θ4 is a constant, 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ≤ p2d− α (18) and 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ≤ α− p1N4 − d, (19) where α is a positive constant such that p1N4 + d < α < p2d. Define the operator S on A by (Sx)(t) =  − 1 p(t+τ) [ α− x(t+ τ) + 1 (n−2)! ∫∞ t+τ (s−t−τ)n−2 r(s) ∫ s t1+τ [f1(u, x(σ1(u))) −f2(u, x(σ2(u)))− g(u)]duds ] , t ≥ t1 (Sx)(t1), t0 ≤ t ≤ t1. Clearly, Sx is continuous. We shall show that SA ⊂ A. For each x ∈ A and t ≥ t1, by using (18), we have (Sx)(t) = − 1 p(t+ τ) [ α− x(t+ τ) + 1 (n− 2)! ∫ ∞ t+τ ∫ s t1+τ (s− t− τ)n−2 r(s) [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ] ≤ 1 p2 [ α+ 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f1(u, d) + |g(u)|]duds ] ≤ d and applying (19), we obtain (Sx)(t) = − 1 p(t+ τ) [ α− x(t+ τ) + 1 (n− 2)! ∫ ∞ t+τ ∫ s t1+τ (s− t− τ)n−2 r(s) [f1(u, x(σ1(u))) − f2(u, x(σ2(u)))− g(u)]duds ] ≥ − 1 p(t+ τ) [ α− d− 1 (n− 2)! ∫ ∞ t1+τ ∫ s t1+τ (s− t− τ)n−2 r(s) [f2(u, d) + |g(u)|]duds ] ≥ 1 p1 [ α− d− 1 (n− 2)! ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) [f2(u, d) + |g(u)|]duds ] ≥ N4. Hence, we proved that SA ⊂ A. Now we shall show that S is a contraction operator on A. In fact, for x, y ∈ A and t ≥ t1, in view of (2) and (17), we have |(Sx)(t)− (Sy)(t)| ≤ 1 |p(t+ τ)| [ |x(t+ τ)− y(t+ τ)| B. Çına, T. Candan, M. Tamer Şenel / Eur. J. Pure Appl. Math, 16 (2) (2023), 713-723 721 + 1 (n− 2)! 2∑ i=1 ∫ ∞ t+τ (s− t− τ)n−2 r(s) ∫ s t1+τ |fi(u, x(σi(u)))− fi(u, y(σi(u)))|duds ] ≤ ∥x− y∥ p2 [ 1 + 1 (n− 2)! 2∑ i=1 ∫ ∞ t1 ∫ s t1 (s− t)n−2 r(s) qi(u)duds ] ≤ θ4∥x− y∥. This implies that ∥Sx− Sy∥ ≤ θ4∥x− y∥. Since θ4 < 1 by (17), S is a contraction operator on A. By the Banach contraction mapping principle, S has a fixed point x ∈ A, and x is a positive solution of (1). Thus, the proof is completed. Example 1. Consider the equation (et(x(t)− e−t−4x(t− 4))′′′)′ + e−t−5x(t− 5) −e−t−6x3(t− 2)− e−2t + e−4t + 8e−t = 0, t0 > 5, (20) where n = 4, r(t) = et, p(t) = e−t−4, τ = 4, σ1(t) = t−5, σ2(t) = t−2, f1(t, x) = e−t−5x, f2(t, x) = e−t−6x3 and g(t) = e−2t − e−4t − 8e−t. Thus, |f1(t, x)− f1(t, y)| = |e−t−5x− e−t−5y| = e−t−5|x− y|, where x, y ∈ [a, b], a > 0, |f2(t, x)− f2(t, y)| = |e−t−6x3 − e−t−6y3| = e−t−6|x2 + xy + y2||x− y| ≤ 3b2e−t−6|x− y|, where x, y ∈ [a, b], a > 0. Letting q1(t) = e−t−5 and q2(t) = 3b2e−t−6, then 1 (n− 2)! ∫ ∞ t0 ∫ s t0 sn−2 r(s) q1(u)duds = 1 2! ∫ ∞ t0 ∫ s t0 s2 es e−u−5duds < ∞ and 1 (n− 2)! ∫ ∞ t0 ∫ s t0 sn−2 r(s) q2(u)duds = 1 2! ∫ ∞ t0 ∫ s t0 s2 es 3b2e−u−6duds < ∞. Furthermore, 1 (n− 2)! ∫ ∞ t0 ∫ s t0 sn−2 r(s) |f1(u, d)|duds = 1 2! ∫ ∞ t0 ∫ s t0 s2 es e−u−5|d|duds < ∞, d ̸= 0, 1 (n− 2)! ∫ ∞ t0 ∫ s t0 sn−2 r(s) |f2(u, d)|duds = 1 2! ∫ ∞ t0 ∫ s t0 s2 es e−u−6|d|3duds < ∞, d ̸= 0, and 1 (n− 2)! ∫ ∞ t0 ∫ s t0 sn−2 r(s) |g(u)|duds = 1 2! ∫ ∞ t0 ∫ s t0 s2 es e−u−5|e−2u − e−4u − 8e−u|duds < ∞. We see that all conditions of Theorem 1 are satisfied. In fact, x(t) = e−t is a nonoscillatory solution of (20). REFERENCES 722 References [1] Ravi P Agarwal, Martin Bohner, and Wan-Tong Li. 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