Existence of nonoscillatory solutions of second-order nonlinear neutral differential equations with distributed deviating arguments

Some sufficient conditions are provided for the existence of nonoscillatory solutions of nonlinear second-order neutral differential equations with distributed deviating arguments. The main tool for proving our results is the Banach contraction principle. Two examples are given to illustrate the effectiveness of our results.


Introduction
The purpose of this article is to study the second-order neutral nonlinear differential equations with distributed deviating arguments of the form x(σ 2 (t, ξ )))dξ = g(t) (1) and where g ∈ C([t 0 , ∞), R), τ > 0, r ∈ C([t 0 , ∞), (0, ∞)), Throughout this study, we assume that f i ∈ C([t 0 , ∞) × R, R) is nondecreasing in the second variable, i = 1, 2, xf i (t, x) > 0 for x = 0, i = 1, 2, and satisfies hold. In recent years, there have been many studies concerning the oscillatory and nonoscillatory behaviour of neutral differential equations (see [1][2][3][4][5][6][7][8][9][10][11][12][13][14][15][16] and references cited therein). For some studies of the qualitative analysis of Volterra integro-differential equations and a variable delay system of differential equations of second order, we refer the reader to [17,18] and references cited therein. For example, in 2010, Candan and Dahiya [3] considered the existence of nonoscillatory solutions of first and second-order neutral equations of the form Candan [4] investigated nonoscillatory solutions of higher order neutral differential equations of form and in 2019, Çına et al. [8] studied the existence of nonoscillatory solutions of a nonlinear second-order neutral differential equation with forcing term Motivated by the above-mentioned studies, the aim of this paper is to give some sufficient conditions for the existence of nonoscillatory solutions of (1) and (2), more general than the latter equation, by using the Banach contraction principle. Let and such that Equation (1) is satisfied for t ≥ t 1 .
As is customary, a solution of (1) or (2) is said to be oscillatory if it has arbitrarily large zeros. Otherwise, the solution is called nonoscillatory.

Main results
We suppose throughout this paper that X is the set of all continuous and bounded functions on [t 0 , ∞) with the norm x = sup t≥t 0 |x(t)| < ∞.
where N 1 and N 2 are positive constants such that It is clear that A is a closed, bounded and convex subset of X. In view of (4)-(6), we can choose a t 1 > t 0 sufficiently large such that t − τ ≥ t 0 , σ i (t, ξ) ≥ t 0 , ξ ∈ [a i , b i ], i = 1, 2 for t ≥ t 1 and where θ 1 is a constant, where α ∈ (N 1 , (1 − p)N 2 ). Define a mapping S : A −→ X as follows: It is clear that Sx is continuous. For every x ∈ A and t ≥ t 1 , by (9), we get and taking (8) into account, we have Thus SA ⊂ A. Now we show that S is a contraction mapping on A. In fact, for x, y ∈ A and t ≥ t 1 , in view of (3) and (7), we have This implies that Sx − Sy ≤ θ 1 x − y . Since θ 1 < 1, S is a contraction mapping on A. Consequently, S has the unique fixed point x ∈ A such that Sx = x, which is obviously a positive solution of (1). This completes the proof.
Proof: Suppose (5) holds with d > 0, the case d < 0 can be treated similarly.
where N 3 and N 4 are positive constants such that It is obvious that A is a closed, bounded and convex subset of X. Because of (4)-(6), we can choose a t 1 > t 0 sufficiently large such that where θ 2 is a constant, where α ∈ (p 2 N 3 , (p 1 − 1)N 4 ). Consider the mapping S : It is obvious that Sx is continuous. For every x ∈ A and t ≥ t 1 , by (11), we have and from (12), we obtain Thus SA ⊂ A. Finally, for x, y ∈ A and t ≥ t 1 , in view of (3) and (10), we have This implies that Sx − Sy ≤ θ 2 x − y with θ 2 < 1. Hence, S is a contraction mapping on A. Consequently, S has the unique fixed point x ∈ A such that Sx = x, which is obviously a positive solution of (1). Thus, the theorem is proved.
where N 5 and N 6 are positive constants such that It is obvious that A is a closed, bounded and convex subset of X. In view of (4)-(6), we can choose a t where θ 3 is a constant, and where α ∈ (N 5 + pN 6 , N 6 ). Define a mapping S : A −→ X as follows: It is clear that Sx is continuous. For every x ∈ A and t ≥ t 1 , from (15), we have and by using (14), we have Thus SA ⊂ A. We shall show that S is a contraction mapping on A. In fact, for x, y ∈ A and t ≥ t 1 , in view of (3) and (13), we have (σ 1 (u, ξ ))) − f 1 (u, y(σ 1 (u, ξ )))|dξ du ds (σ 2 (u, ξ ))) − f 2 (u, y(σ 2 (u, ξ )))|dξ du ds 1 (u, ξ))|dξ du ds This implies that Sx − Sy ≤ θ 3 x − y . Since θ 3 < 1, S is a contraction mapping on A. Consequently, S has the unique fixed point x ∈ A such that Sx = x, which is obviously a positive solution of (1). This completes the proof. Proof: Suppose (5) holds with d > 0, the case d < 0 can be treated similarly. Let N 8 = d. Set where N 7 and N 8 are positive constants such that It is obvious that A is a closed, bounded and convex subset of X. In view of (4)-(6), there exists a t 1 > t 0 sufficiently large such that σ i (t + τ , ξ) ≥ t 0 , ξ ∈ [a i , b i ], i = 1, 2 for t ≥ t 1 and where θ 4 is a constant, where α ∈ (p 1 N 7 + N 8 , p 2 N 8 ). Consider the mapping S : (σ 1 (u, ξ ) It is obvious that Sx is continuous. For every x ∈ A and t ≥ t 1 , from (18), we have and from (17), we obtain Thus SA ⊂ A. In fact, for x, y ∈ A and t ≥ t 1 , by using (3) and (16), we have (σ 1 (u, ξ ))) − f 1 (u, y(σ 1 (u, ξ )))|dξ du ds (σ 2 (u, ξ ))) − f 2 (u, y(σ 2 (u, ξ )))|dξ du ds (σ 1 (u, ξ)) − y (σ 1 (u, ξ))|dξ du ds This shows that Sx − Sy ≤ θ 4 x − y with θ 4 < 1. This implies that S is a contraction mapping on A. Thus, S has the unique fixed point x ∈ A such that Sx = x, which is clearly a positive solution of (1). Hence the proof is complete.
where N 9 and N 10 are positive constants such that It is obvious that A is a closed, bounded and convex subset of X. Because of (4)-(6), we can take a t 1 > t 0 sufficiently large such that where θ 5 is a constant, where α ∈ (N 9 , (1 − p)N 10 ). Define a mapping S : A −→ X as follows: (σ 1 (u, ξ ) It is clear that Sx is continuous. For every x ∈ A and t ≥ t 1 , from (21), we have and by using (20), we get Thus SA ⊂ A. Now we show that S is a contraction mapping on A. In fact, for x, y ∈ A and t ≥ t 1 , from (3) and (19), we have (σ 1 (u, ξ ))) − f 1 (u, y(σ 1 (u, ξ )))|dξ du ds (σ 2 (u, ξ ))) − f 2 (u, y(σ 2 (u, ξ )))|dξ du ds 1 (u, ξ))|dξ du ds This implies that Sx − Sy ≤ θ 5 x − y . Since θ 5 < 1, S is a contraction mapping on A. Consequently, S has the unique fixed point x ∈ A such that Sx = x, which is obviously a positive solution of (2). This completes the proof. It is clear that A is a closed, bounded and convex subset of X. By (4)-(6), we can take a t 1 > t 0 sufficiently large where θ 6 is a constant, where α ∈ (pN 12 + N 11 , N 12 ). Consider the mapping S : It is obvious that Sx is continuous. For every x ∈ A and t ≥ t 1 , by (24), we have and taking (23) into account, we get Thus SA ⊂ A. Now we show that S is a contraction mapping on A. In fact, for x, y ∈ A and t ≥ t 1 , in view of (3) and (22), we have (σ 1 (u, ξ ))) − f 1 (u, y(σ 1 (u, ξ )))|dξ du ds (σ 2 (u, ξ ))) − f 2 (u, y(σ 2 (u, ξ )))|dξ du ds × |x(σ 2 (u, ξ)) − y(σ 2 (u, ξ))|dξ du ds This implies that Sx − Sy ≤ θ 6 x − y with θ 6 < 1, and S is a contraction mapping on A. Consequently, S has the unique fixed point x ∈ A such that Sx = x, which is obviously a positive solution of (2). Thus the theorem is proved.