# Existence of Nonoscillatory Solutions for System of Higher-Order Neutral Differential Equations with Distributed Deviating Arguments

Authors: Youjun Liu; Jianwen Zhang; Jurang Yan
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

In this paper, we consider the existence of nonoscillatory solutions for system of variable coefficients higher-order neutral differential equations with distributed deviating arguments. We use theBanach contraction principle to obtain new sufficient conditions for the existence of nonoscillatory solutions.


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## Body

## 1. Introduction and Preliminary

In this paper, we consider the system of higher-order neutral differential equations with distributed deviating arguments:(1)[r(t)(x(t)+P(t)x(t-θ))(n-1)]′+(-1)n∫cd‍Q1(t,τ)x(t-τ)dτ-∫ef‍Q2(t,σ)x(t-σ)dσ=0,t≥t0,(1)
wheren is a positive integer, n>1, θ>0, 0<c<d, and 0<e<f;(2)
r∈C([t0,∞),R+), r(t)>0, and P∈C([t0,∞),R);(3)
x∈Rn, Qi is continuous n×n matrix on [t0,∞), i=1,2, and matrix coefficients system of higher order neutral differential equations with distributed deviating arguments: (2)[r(t)(x(t)+Bx(t-θ))(n-1)]′+(-1)n∫cd‍Q1(t,τ)x(t-τ)dτ-∫ef‍Q2(t,σ)x(t-σ)dσ=0,t≥t0,(1)
wheren is a positive integer, n>1, θ>0, 0<c<d, and 0<e<f;(2)
r∈C([t0,∞),R+), r(t)>0, and B is nonsingular constant n×n matrix;(3)
x∈Rn, and Qi is continuous n×n matrix on [t0,∞), i=1,2.Recently there have been a lot of activities concerning the existence of nonoscillatory solutions for neutral differential equations with positive and negative coefficients. In 2013, Candan [1] has investigated existence of nonoscillatory solutions for system of higher-order nonlinear neutral differential equations:
(3)[x(t)+P(t)x(t-θ)](n)+(-1)n+1[Q1(t)x(t-σ1)-Q2(t)x(t-σ2)]=0
and matrix coefficient system of higher order neutral functional differential equation:
(4)[x(t)+Bx(t-θ)](n)+(-1)n+1[Q1(t)x(t-σ1)-Q2(t)x(t-σ2)]=0.
In 2012, Candan [2] studies higher-order nonlinear differential equation:
(5)[r(t)[x(t)+P(t)x(t-τ)](n-1)]′+(-1)n×[Q1(t)g1(x(t-σ1))kkkkkkkkkkk-Q2(t)g2(x(t-μ))-f(t)]=0.
has obtained sufficient conditions for the existence of nonoscillatory solutions. For related work, we refer the reader to the books [3–12].A solution of system of (1) and (2) is a continuous function x(t) defined on ([t1-μ,∞),Rn), for some t1>t0, such that x(t)-P(t)x(t-θ) and x(t)-Bx(t-θ) are n-1 times continuously differentiable, and r(t)(x(t)-P(t)x(t-θ))(n-1) and r(t)(x(t)-Bx(t-θ))(n-1) are continuously differentiable, and system of (1) and (2) holds for all n>1. Here, μ=max{θ,d,f}.


## 2. The Main Results

Theorem 1.
Assume that0≤P(t)≤p<1/2 and
(6)∫t∞‍sn-2r(s)∫t0s‍∥∫cd‍Qi(u,τ)dτ∥duds<∞,i=1,2.
Then, (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,M1≤∥x(t)∥≤M2,t≥t0}, where M1 and M2 are positive constants and b is a constant vector, such that pM2+M1<∥b∥<2∥b∥<M2+M1. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that(7)1(n-2)!∫t∞‍(s-t)n-2r(s)kkkkkkkk×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≤∥b∥-pM2-M1M2,t>t1,and define an operator T on A as follows:(8)(Tx)(t)={{∫t∞‍(s-t)n-2r(s)b-P(t)x(t-θ)+1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds}t≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous, for t≥t1, x∈A; by using (7), we have
(9)(Tx)(t)=∥b-P(t)x(t-θ)+1(n-2)!×∫t∞‍(s-t)n-2r(s)kkkkkk×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτkkkkkkkkkkk-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥≤∥b∥-p∥x(t-θ)∥+1(n-2)!×∥∫t∞‍(s-t)n-2r(s)kkkkk×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτkkkkkkkkkkk-∫ef‍Q2(u,σ)dσx(u-σ)]duds∥≤∥b∥-p∥x(t-θ)∥+1(n-2)!×∫t∞‍(s-t)n-2r(s)kkkk×∫t1s[∫cd‍∥Q1(u,τ)∥∥x(u-τ)∥dτ‍kkkkkkkkk+∫ef‍∥Q2(u,σ)∥∥x(u-σ)∥dσ]duds≤∥b∥-pM2+M2(n-2)!×∫t∞‍(s-t)n-2r(s)kkkkk×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥kkkkkkkkkkk+∥∫ef‍Q2(u,σ)dσ∥]duds≤M2;
and taking (7) into account, we have
(10)(Tx)(t)=∥b-P(t)x(t-θ)+1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥≥∥b∥∥px(t-θ)∥-1(n-2)!×∥∫t∞‍(s-t)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥≥∥b∥-∥px(t-θ)∥-1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)x(u-τ)dτ∥+∥∫ef‍Q2(u,σ)x(u-σ)dσ∥]duds≥∥b∥-pM2-M2(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≥M1.
These show that TA⊂A. Since A is a bounded, close, convex subset of Λ, in order to apply the contraction principle, we have to show that T is a contraction mapping on A. For all x1,x2∈A, and t≥t1,(11)|(Tx1)(t)-(Tx2)(t)|≤P(t)∥x1(t-θ)-x2(t-θ)∥+1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)∥x1(u-τ)-x2(u-τ)∥dτ∥+∥∫ef‍Q2(u,σ)∥x1(u-σ)-x2(u-σ)∥dσ∥]duds.Using (7), (12)|(Tx1)(t)-(Tx2)(t)|≤∥x1-x2∥×(p+1(n-2)!×∫t∞‍(s-t)n-2r(s)∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds)<q1∥x1-x2∥,(0<q1<1).This implies, with the sup norm, that
(13)∥Tx1-Tx2∥<q1∥x1-x2∥,(0<q1<1),
which shows that T is a contraction mapping on A, and therefore there exists a unique solution. Consequently there exists a unique solution of (1) x∈A of Tx=x. The proof is complete.Theorem 2.
Assume that2<p1≤P(t)≤p0<∞ and that (6) holds.
Then, (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,M3≤∥x(t)∥≤M4,t≥t0}, where M3 and M4 are positive constants such that p0M3+M4<∥b∥<2∥b∥≤p1M4+p0M3. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that
(14)1(n-2)!∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)σ∥]duds≤∥b∥-M4-p0M3M4,t>t1,
and define an operator T on A as follows:(15)(Tx)(t)={1P(t+θ)×{b-x(t+θ)+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s[∫cd‍Q1(u,τ)x(u-τ)dτ‍-∫ef‍Q2(u,σ)x(u-σ)dσ]duds}t≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous, for t≥t1, x∈A; by using (14), we have
(16)(Tx)(t)≤1p1{∥-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}b-x(t-θ)+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}≤1p1{-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}∥b∥+∥x(t-θ)∥+1(n-2)!×∥∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}≤1p1{-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}∥b∥+∥x(t-θ)∥+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍∥Q1(u,τ)∥∥x(u-τ)∥dτ+∫ef‍∥Q2(u,σ)∥∥x(u-σ)∥dσ]duds}≤1p1{∥b∥+M4+M4(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds}≤M4;
and taking (14) into account, we have
(17)(Tx)(t)≥1p0{∥-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}b-x(t-θ)+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}≥1p0{-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥∥b∥-∥x(t-θ)∥-1(n-2)!×∥∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥}≥1p0{-∫ef‍Q2(u,σ)x(u-σ)dσ]duds∥∥b∥-∥x(t-θ)∥-1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍∥Q1(u,τ)∥∥x(u-τ)∥dτ+∫ef‍∥Q2(u,σ)∥∥x(u-σ)∥dσ]duds}≥1p0{∥b∥-M4-M4(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds}≥M3.
These show that TA⊂A. Since A is a bounded, close, convex subset of Λ, in order to apply the contraction principle, we have to show that T is a contraction mapping on A. For all x1,x2∈A, and t≥t1,(18)|(Tx1)(t)-(Tx2)(t)|≤1p1{∥∫ef‍Q2(u,σ)∥∥x1(t+θ)-x2(t+θ)∥+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)∥x1(u-τ)-x2(u-τ)∥dτ∥+∥∫ef‍Q2(u,σ)∥x1(u-σ)-x2(u-σ)∥dσ∫ef‍Q2(u,)∥]duds},or using (14),
(19)|(Tx1)(t)-(Tx2)(t)|≤∥x1-x2∥p1×{1+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds}<∥x1-x2∥.
This implies, with the sup norm, that
(20)∥Tx1-Tx2∥<∥x1-x2∥,
which shows that T is a contraction mapping on A, and therefore there exists a unique solution. Consequently there exists a unique solution of (1) x∈A of Tx=x. The proof is complete.Theorem 3.
Assume that-1/2<p2≤P(t)≤0 and that (6) holds.
Then, (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,M5≤∥x(t)∥≤M6,t≥t0}, where M5 and M6 are positive constants such that -p2M6+M5<∥b∥<2∥b∥≤M6+M5. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that
(21)1(n-2)!∫t∞‍(s-t)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≤∥b∥+p2M6-M5M6,t>t1,
and define an operator T on A as follows(22)(Tx)(t)={b-P(t)x(t-θ)+1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]dudst≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous. Since the proof is similar to that of Theorem 1, we omit the remaining part of the proof. The proof is complete.Theorem 4.
Assume that-∞<p3≤P(t)≤p4<-2 and that (6) holds.
Then, (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,M7≤∥x(t)∥≤M8,t≥t0}, where M7 and M8 are positive constants such that -p3M7+M8<∥b∥<2∥b∥≤-p4M8-p3M7. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that
(23)1(n-2)!∫t+θ∞‍(s-t-θ)n-1r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≤∥b∥+p3M7+M8M8,t>t1,
and define an operator T on A as follows:(24)(Tx)(t)={1P(t+θ)×{+1(n-2)!b-x(t+θ)+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]duds}t≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous. The remaining part of the proof is similar to that of Theorem 2; therefore, it is omitted. The proof is complete.Theorem 5.
Assume that0<∥B∥<1/2 and that (6) holds.
Then, (2) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,N1≤∥x(t)∥≤N2,t≥t0}, where N1 and N2 are positive constants such that ∥B∥N2+N1<∥b∥<2∥b∥≤N2+N1. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that
(25)1(n-2)!∫t∞‍(s-t)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≤∥b∥-∥B∥N2-N1N2,t>t1,
and define an operator T on A as follows:(26)(Tx)(t)={b-Bx(t-θ)+1(n-2)!×∫t∞‍(s-t)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]dudst≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous. Since the proof is similar to that of Theorem 1, we omit the remaining part of the proof. The proof is complete.Theorem 6.
Assume that0<∥B-1∥<1/2 and that (6) holds.
Then, (2) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded vector functions on [t0,∞) and the sup norm. Set A={x∈Λ,N3≤∥x(t)∥≤N4,t≥t0}, where N3 and N4 are positive constants such that ∥B-1∥N4+N3<∥B-1b∥<2∥B-1b∥≤N4-N3. From (6), one can choose a t1≥t0, t1≥t0+μ, sufficiently large t≥t1 such that
(27)1(n-2)!∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥+∥∫ef‍Q2(u,σ)dσ∥]duds≤∥B-1b∥-N3-N4∥B-1∥N4∥B-1∥,t>t1,
and define an operator T on A as follows:(28)(Tx)(t)={B-1{b-x(t+θ)+1(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∫cd‍Q1(u,τ)x(u-τ)dτ-∫ef‍Q2(u,σ)x(u-σ)dσ]ds}t≥t1,(Tx)(t1)t0≤t≤t1.It is easy to see that Tx is continuous, for t≥t1, x∈A; by using (27), we have
(29)(Tx)(t)≤∥B-1b∥+∥B-1∥×{N4+N4(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥-∥∫ef‍Q2(u,σ)dσ∥]duds}≤N4,
and taking (27) into account, we have(30)(Tx)(t)≥∥B-1b∥-∥B-1∥×{∫t+θ∞‍(s-t-θ)n-2r(s)N4-N4(n-2)!×∫t+θ∞‍(s-t-θ)n-2r(s)×∫t1s‍[∥∫cd‍Q1(u,τ)dτ∥-∥∫ef‍Q2(u,σ)dσ∥]duds}≥N3.It is easy to see that Tx is continuous. The remaining part of the proof is similar to that of Theorem 2; therefore, it is omitted. The proof is complete.


## 3. Example

Example 1.
Consider high-order neutral differential equation with distributed deviating arguments:(31)(e-t(x(t)+e-1x(t-1))(3))′+23∫12‍τe-2t(311272)x(t-τ)dτ-23∫12‍σe-2t(2153103)x(t-σ)dσ=0.
Here, n=4, c=e=1, d=f=2, θ=1, r(t)=e-t, P(t)=e-1, Q1(t,τ)=(2/3)τe-2t(311/27/2), and Q2(t,σ)=(2/3)σe-2t(215/310/3).
It is easy to see that(32)0<P(t)=e-1<12,∫t∞‍ets2∫t0s‍∫12‍∥23τe-2u(2153103)dτ∥duds<∞,∫t∞‍ets2∫t0s‍∫12‍∥23σe-2u(311272)dσ∥duds<∞;
thus Theorem 1 holds. In fact, x(t)=(e-te-t) is a nonoscillatory solution of (31).Example 2.
Consider high-order neutral differential equation with distributed deviating arguments:(33)(1|sint|(x(t)+e-π(124353)x(t-π))(2))′+38∫13‍τ2e-4t(527251)x(t-τ)dτ-38∫13‍σ2e-4t(2473113)x(t-σ)dσ=0.
Here, n=3, r(t)=1/|sint|, θ=π, c=e=1, d=f=3, B=e-π(124/35/3), Q1(t,τ)=(3/8)τ2e-4t(5/27/251), and Q2(t,σ)=(3/8)σ2e-4t(247/311/3).
It is easy to see that(34)0<∥B∥=∥e-π(124353)∥<12,∫t∞‍|sins|s∫t0s‍38∫13‍∥τ2e-4u(527251)dτ∥duds<∞,∫t∞‍|sins|s∫t0s‍38∫13‍∥σ2e-4u(2473113)dσ∥duds<∞;
thus Theorem 5 holds. In fact, x(t)=(2+sint2+sint) is a nonoscillatory solution of (33).



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