# Existence of Nonoscillatory Solutions of Higher-Order Neutral Differential Equations with Distributed Coefficients and Delays

Authors: Youjun Liu; Huanhuan Zhao; Jurang Yan
Journal: Discrete Dynamics in Nature and Society
 (2016)

---

## Abstract

We consider the existence of nonoscillatory solutions of higher-order neutral differential equations with distributed coefficients. We use theB
a
n
a
c
h contraction principle to obtain new sufficient condition for the existence of nonoscillatory solutions.


---

## Body

## 1. Introduction

In this paper, we consider the higher-order nonlinear neutral differential equation with distributed deviating arguments:(1)
Φ
r
t
x
t
+
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
n
+
∫
c
d
q
1
t
,
τ
g
1
x
t
-
τ
d
τ
-
∫
e
f
q
2
t
,
σ
g
2
x
t
-
σ
d
σ
=
h
t
.Throughout this paper, the following conditions are assumed to hold:(1)
wheren
≥
1 is a positive integer, r
∈
C
t
0
,
∞
,
R
+
,
0
<
a
<
b
,
0
<
c
<
d
,
0
<
e
<
f;
(2)
p
∈
C
(
[
t
0
,
∞
)
×
[
a
,
b
]
,
R
)
,
q
1
∈
C
(
[
t
0
,
∞
)
×
[
c
,
d
]
,
R
+
)
,
q
2
∈
C
(
[
t
0
,
∞
)
×
[
e
,
f
]
,
R
+
)
,
h
∈
C
(
[
t
0
,
∞
)
,
R
);
(3)
Φ
(
u
) are continuously increasing real odd function with respect to u defined on R, and Φ
-
1
(
u
) satisfy the local L
i
p
s
c
h
i
t
z condition;
(4)
g
i
∈
C
(
R
,
R
)
,
g
i
(
u
) satisfy the local L
i
p
s
c
h
i
t
z condition and u
g
i
(
u
)
>
0
,
i
=
1,2, for u
≠
0.Recently there have been a lot of activities concerning the existence of nonoscillatory solutions for neutral differential equations with positive and negative coefficient. In 2002, Zhou and Zhang [1] studied higher-order linear neutral delay differential equation: (2)
d
n
d
t
n
x
t
+
c
x
t
-
τ
+
-
1
n
+
1
P
t
x
t
-
σ
-
Q
t
x
t
-
δ
=
0
.In 2005, the existence of nonoscillatory solutions of first-order linear neutral delay differential equations of the form(3)
d
d
t
x
t
+
P
t
x
t
-
τ
+
Q
1
t
x
t
-
σ
1
-
Q
2
t
x
t
-
σ
2
=
0was investigated by Zhang et al. [2] and, in the same year, Yu and Wang [3] studied nonoscillatory solutions of second-order nonlinear neutral differential equations of the form (4)
r
t
x
t
+
P
t
x
t
-
τ
′
′
+
Q
1
t
f
x
t
-
σ
1
-
Q
2
t
g
x
t
-
σ
2
=
0
.In 2010, Candan and Dahiya [4] studied nonoscillatory solutions of first-order and second-order nonlinear neutral differential equations with positive and negative coefficients: (5)
d
k
d
t
k
x
t
+
P
t
x
t
-
τ
+
∫
a
b
q
1
t
,
ξ
x
t
-
ξ
d
ξ
-
∫
c
d
q
2
t
,
μ
x
t
-
μ
d
μ
=
0
,
k
=
1,2
.In 2012, Candan [5] study higher-order nonlinear differential equation: (6)
r
t
x
t
+
P
t
x
t
-
τ
n
-
1
′
+
-
1
n
Q
1
t
g
1
x
t
-
σ
1
-
Q
2
t
g
2
x
t
-
μ
-
f
t
=
0
. In 2015, Liu et al. [6] study higher-order neutral differential equation with distributed deviating arguments: (7)
r
t
x
t
+
P
t
x
t
-
θ
n
+
∫
c
d
q
1
t
,
τ
g
1
x
t
-
τ
d
τ
-
∫
e
f
q
2
t
,
σ
g
2
x
t
-
σ
d
σ
=
f
t
.As can be seen from the development process of the above equations, the delay of neutral part in the discussed differential equations was all constant delays, and the main thought in the employed verification method also kept the same in which the coefficient neutral part in the neutral was all discussed in four cases, that is, (
-
∞
,
-
1
)
,
(
-
1,0
)
,
(
0,1
), and (
1
,
+
∞
), and then was verified by constructing the corresponding operator. However, the case for distributed deviating arguments is rather rare. In 2013, Candan [7] studied first-order neutral differential equation with distributed deviating arguments: (8)
x
t
-
∫
a
b
p
2
t
,
ξ
x
t
-
ξ
d
ξ
γ
′
+
∫
c
d
Q
2
t
,
ξ
G
x
t
-
ξ
d
ξ
=
0
,where γ is a ratio of odd positive integers; however the discussion only covered the condition for coefficient being 0
<
∫
a
b
‍
p
2
(
t
,
ξ
)
d
ξ
<
1, while being without the other three conditions, which might be caused by the difficulty in establishing feasible operator. In view of the above, here, in this paper, the difficulty of operator establishment was settled and sufficient condition for the existence of nonoscillatory solutions of differential equation with coefficient of ∫
a
b
‍
p
2
(
t
,
ξ
)
d
ξ in the four cases was obtained. Thus, this paper may present its theoretical value as well as practical application value. For related work, we refer the reader to the references [8–12].As usual, a solution of (1) is said to be oscillatory if it has arbitrarily large zeros. Otherwise the solution is said to be nonoscillatory.A solution of (1) is a continuous function x
(
t
) defined on [
t
1
-
μ
,
∞
), for some t
1
>
t
0, such that Φ
(
r
(
t
)
x
(
t
)
+
∫
a
b
‍
p
(
t
,
θ
)
x
(
t
-
θ
)
d
θ
) is n times continuously differentiable and (1) holds for all n
≥
1. Here, μ
=
m
a
x
{
b
,
d
,
f
}.LetL
i
,
i
=
1,2 denote the L
i
p
s
c
h
i
t
z constants of functions g
i
,
i
=
1,2 on the set A, K denote the L
i
p
s
c
h
i
t
z constants of functions Φ
-
1
(
u
), respectively, and L
=
m
a
x
{
L
1
,
L
2
}
,
β
i
=
m
a
x
x
∈
A
g
i
(
x
)
,
i
=
1,2
.


## 2. The Main Results

Theorem 1.
Assume that0
≤
∫
a
b
‍
p
(
t
,
θ
)
d
θ
≤
p
1
<
1
,
1
≤
r
(
t
)
≤
1
/
p
1, and (9)
∫
t
∞
s
n
-
1
∫
c
d
q
1
s
,
τ
d
τ
d
s
<
∞
,
∫
t
∞
s
n
-
1
∫
e
f
q
2
s
,
σ
d
σ
d
s
<
∞
,
∫
t
∞
s
n
-
1
h
s
d
s
<
∞
.Then (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded functions on [
t
0
,
∞
) and the norm be x
(
t
)
=
s
u
p
t
0
≤
t
<
+
∞
x
(
t
). Set A
=
{
x
∈
Λ
,
M
1
≤
x
(
t
)
≤
M
2
,
t
≥
t
0
}, where M
1
,
M
2 are two positive constants such that p
1
M
2
+
M
1
/
p
1
<
α
<
M
2
. From (9), one can choose t
1
≥
t
0
+
μ and sufficiently large t
≥
t
1 such that(10)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
Φ
M
2
-
Φ
α
,
(11)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≤
Φ
α
-
Φ
p
1
M
2
+
M
1
p
1
,
(12)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
≤
1
-
p
1
K
L
,and define an operator T on A as follows: (13)
T
x
t
=
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
-
h
s
d
s
,
t
≥
t
1
,
T
x
t
1
,
t
0
≤
t
≤
t
1
. It is easy to see that T is continuous, for t
≥
t
1
,
x
∈
A. By using (10), we have(14)
T
x
t
≤
1
r
t
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
h
s
d
s
≤
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
Φ
-
1
Φ
α
+
Φ
M
2
-
Φ
α
=
M
2
,and, taking (11) into account, we have(15)
T
x
t
≥
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
-
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
+
h
s
d
s
≥
p
1
-
p
1
M
2
+
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≥
p
1
-
p
1
M
2
+
Φ
-
1
Φ
α
-
Φ
α
+
Φ
p
1
M
2
+
M
1
p
1
=
M
1
.These show that T
A
⊂
A. Since A is 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 ∀
x
1
,
x
2
∈
A, and t
≥
t
1,(16)
T
x
1
t
-
T
x
2
t
≤
1
r
t
∫
a
b
p
t
,
θ
x
1
t
-
θ
-
x
2
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
d
σ
-
h
s
d
s
-
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
2
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
2
s
-
σ
d
σ
-
h
s
d
s
≤
1
r
t
p
1
x
1
-
x
2
+
K
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
-
g
1
x
2
s
-
τ
d
τ
+
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
-
g
2
x
2
s
-
σ
d
σ
d
s
.Using (12), (17)
T
x
1
t
-
T
x
2
t
≤
x
1
-
x
2
p
1
+
K
L
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
<
ρ
x
1
-
x
2
.This implies with the sup norm that (18)
T
x
1
-
T
x
2
<
ρ
x
1
-
x
2
,where ρ
=
1, which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) x
∈
A, such that T
x
=
x. The proof is complete.Theorem 2.
Assume that1
<
p
3
≤
∫
a
b
‍
p
(
t
,
θ
)
d
θ
≤
p
2
<
2
p
3
<
+
∞
,
2
p
3
≤
r
(
t
)
≤
2
p
2 and that (9) holds.
Then (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded functions on [
t
0
,
∞
) and the norm be x
(
t
)
=
s
u
p
t
0
≤
t
<
+
∞
x
(
t
). Set A
=
{
x
∈
Λ
,
M
3
≤
x
(
t
)
≤
M
4
,
t
≥
t
0
}, where M
3
,
M
4 are two positive constants such that p
2
M
4
+
2
p
2
M
3
<
α
<
2
p
3
M
4. From (9), one can choose t
1
≥
t
0
+
μ and sufficiently large t
≥
t
1 such that (19)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
Φ
2
p
3
M
4
-
Φ
α
,
(20)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≤
Φ
α
-
Φ
p
2
M
4
-
2
p
2
M
3
,
(21)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
≤
2
p
3
-
p
2
K
L
,and define an operator T on A as follows: (22)
T
x
t
=
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
-
h
s
d
s
,
t
≥
t
1
,
T
x
t
1
,
t
0
≤
t
≤
t
1
.It is easy to see that T is continuous, for t
≥
t
1
,
x
∈
A. By using (19), we have(23)
T
x
t
≤
1
r
t
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
h
s
d
s
≤
1
2
p
3
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
M
4
,and, taking (20) into account, we have(24)
T
x
t
≥
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
+
h
s
d
s
≥
1
2
p
2
-
p
2
M
4
+
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≥
M
3
.These show that T
A
⊂
A. Since A is 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 ∀
x
1
,
x
2
∈
A, and t
≥
t
1, (25)
T
x
1
t
-
T
x
2
t
≤
1
r
t
∫
a
b
p
t
,
θ
x
1
t
-
θ
-
x
2
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
d
σ
-
h
s
d
s
-
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
2
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
2
s
-
σ
d
σ
-
h
s
d
s
≤
1
r
t
p
2
x
1
-
x
2
+
K
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
-
g
1
x
2
s
-
τ
d
τ
+
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
-
g
2
x
2
s
-
σ
d
σ
d
s
.Or, using (21), (26)
T
x
1
t
-
T
x
2
t
≤
x
1
-
x
2
1
2
p
3
p
2
+
K
L
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
<
ρ
x
1
-
x
2
.This implies with the sup norm that (27)
T
x
1
-
T
x
2
<
ρ
x
1
-
x
2
,where ρ
=
1, which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) x
∈
A, such that T
x
=
x. The proof is complete.Theorem 3.
Assume that-
1
<
p
4
≤
∫
a
b
‍
p
(
t
,
θ
)
d
θ
≤
0,1
≤
r
(
t
)
≤
1
/
-
p
4 and that (9) holds.
Then (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded functions on [
t
0
,
∞
) and the norm be x
(
t
)
=
s
u
p
t
0
≤
t
<
+
∞
x
(
t
). Set A
=
{
x
∈
Λ
,
M
5
≤
x
(
t
)
≤
M
6
,
t
≥
t
0
}, where M
5
,
M
6 are two positive constants such that M
5
/
-
p
4
<
α
<
(
1
+
p
4
)
M
6
. From (9), one can choose t
1
≥
t
0
+
μ and sufficiently large t
≥
t
1 such that (28)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
Φ
1
+
p
4
M
6
-
Φ
α
,
(29)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≤
Φ
α
-
Φ
-
M
5
p
4
,
(30)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
≤
1
+
p
4
K
L
.and define an operator T on A as follows: (31)
T
x
t
=
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
-
h
s
d
s
,
t
≥
t
1
,
T
x
t
1
,
t
0
≤
t
≤
t
1
.It is easy to see that T is continuous, for t
≥
t
1
,
x
∈
A. By using (28), we have(32)
T
x
t
≤
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
h
s
d
s
≤
-
p
4
M
6
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
M
6
,and, taking (29) into account, we have(33)
T
x
t
≥
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
+
h
s
d
s
≥
-
p
4
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≥
-
p
4
Φ
-
1
Φ
α
-
Φ
α
+
Φ
-
M
5
p
4
=
M
5
.These show that T
A
⊂
A. Since A is 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 ∀
x
1
,
x
2
∈
A, and t
≥
t
1,(34)
T
x
1
t
-
T
x
2
t
≤
1
r
t
∫
a
b
p
t
,
θ
x
1
t
-
θ
-
x
2
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
d
σ
-
h
s
d
s
-
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
2
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
2
s
-
σ
d
σ
-
h
s
d
s
≤
1
r
t
-
p
4
x
1
-
x
2
+
K
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
-
g
1
x
2
s
-
τ
d
τ
+
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
-
g
2
x
2
s
-
σ
d
σ
d
s
.Or, using (30),(35)
T
x
1
t
-
T
x
2
t
≤
x
1
-
x
2
-
p
4
+
K
L
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
<
ρ
x
1
-
x
2
.This implies with the sup norm that (36)
T
x
1
-
T
x
2
<
ρ
x
1
-
x
2
,where ρ
=
1, which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) x
∈
A, such that T
x
=
x. The proof is complete.Theorem 4.
Assume that-
∞
<
2
p
5
<
p
6
≤
∫
a
b
‍
p
(
t
,
θ
)
d
θ
≤
p
5
<
-
1
,
-
2
p
5
<
r
(
t
)
<
-
2
p
6 and that (9) holds.
Then (1) has a bounded nonoscillatory solution.Proof.
LetΛ be the set of all continuous and bounded functions on [
t
0
,
∞
) and the norm be x
(
t
)
=
s
u
p
t
0
≤
t
<
+
∞
x
(
t
). Set A
=
{
x
∈
Λ
,
M
7
≤
x
(
t
)
≤
M
8
,
t
≥
t
0
}, where M
7
,
M
8 are two positive constants such that -
2
p
6
M
7
<
α
<
(
-
2
p
5
+
p
6
)
M
8. From (9), one can choose t
1
≥
t
0
+
μ and sufficiently large t
≥
t
1 such that (37)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
Φ
-
2
p
5
M
8
+
p
6
M
8
-
Φ
α
,
(38)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≤
Φ
α
-
Φ
-
2
p
6
M
7
,
(39)
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
≤
p
6
-
2
p
5
K
L
,and define an operator T on A as follows: (40)
T
x
t
=
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
-
h
s
d
s
,
t
≥
t
1
,
T
x
t
1
,
t
0
≤
t
≤
t
1
.It is easy to see that T is continuous, for t
≥
t
1
,
x
∈
A. By using (37), we have(41)
T
x
t
≤
1
r
t
-
∫
a
b
p
t
,
θ
x
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
s
-
τ
d
τ
-
h
s
d
s
≤
1
-
2
p
5
-
p
6
M
8
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
1
∫
c
d
q
1
s
,
τ
d
τ
+
h
s
d
s
≤
M
8
,and, taking (38) into account, we have(42)
T
x
t
≥
1
r
t
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
e
f
q
2
s
,
σ
g
2
x
s
-
σ
d
σ
+
h
s
d
s
≥
1
-
2
p
6
Φ
-
1
Φ
α
-
∫
t
∞
s
-
t
n
-
1
n
-
1
!
β
2
∫
e
f
q
2
s
,
σ
d
σ
+
h
s
d
s
≥
M
7
.These show that T
A
⊂
A. Since A is 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 ∀
x
1
,
x
2
∈
A, and t
≥
t
1,(43)
T
x
1
t
-
T
x
2
t
≤
1
r
t
∫
a
b
p
t
,
θ
x
1
t
-
θ
-
x
2
t
-
θ
d
θ
+
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
d
σ
-
h
s
d
s
-
Φ
-
1
Φ
α
+
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
2
s
-
τ
d
τ
-
∫
e
f
q
2
s
,
σ
g
2
x
2
s
-
σ
d
σ
-
h
s
d
s
≤
1
r
t
-
p
6
x
1
-
x
2
+
K
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
g
1
x
1
s
-
τ
-
g
1
x
2
s
-
τ
d
τ
+
∫
e
f
q
2
s
,
σ
g
2
x
1
s
-
σ
-
g
2
x
2
s
-
σ
d
σ
d
s
.Or, using (39),(44)
T
x
1
t
-
T
x
2
t
≤
x
1
-
x
2
1
-
2
p
5
-
p
6
+
K
L
∫
t
∞
s
-
t
n
-
1
n
-
1
!
∫
c
d
q
1
s
,
τ
d
τ
+
∫
e
f
q
2
s
,
σ
d
σ
d
s
<
ρ
x
1
-
x
2
.This implies with the sup norm that (45)
T
x
1
-
T
x
2
<
ρ
x
1
-
x
2
,where ρ
=
1, which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) x
∈
A, such that T
x
=
x. The proof is complete.


## 3. Remark

We note that whenn
=
1
,
r
(
t
)
≡
1
,
Φ
(
u
)
=
u
γ
,
q
2
(
t
,
σ
)
=
0, and h
(
t
)
=
0, we obtain (8). Thus, this paper great achievement was further obtained compared to that of [7].


## 4. Example

Example 1.
Consider high-order neutral differential equation with distributed deviating arguments:(46)
3
2
+
sin
⁡
t
x
t
+
∫
π
/
2
π
e
-
t
x
t
-
θ
d
θ
3
′
′
+
36
∫
π
3
π
/
2
τ
e
-
t
x
t
-
τ
d
τ
-
72
∫
π
/
2
π
σ
e
-
2
t
x
t
-
σ
d
σ
=
e
-
3
t
15
c
o
s
3
t
-
30
s
i
n
3
t
+
63
π
s
i
n
2
t
-
18
π
c
o
s
2
t
+
1
2
sin
⁡
2
t
15
cos
⁡
t
-
27
s
i
n
2
t
+
33
π
+
6
π
2
cos
⁡
t
-
7
sin
⁡
t
+
9
π
3
+
18
e
-
2
t
8
π
cos
⁡
t
+
cos
⁡
2
t
-
sin
⁡
2
t
+
4
sin
⁡
t
-
4
cos
⁡
t
-
3
π
2
+
9
e
-
t
4
π
cos
⁡
t
-
6
π
sin
⁡
t
+
5
π
2
+
2
π
.Here, n
=
2, r
(
t
)
=
3
/
2
+
sin
⁡
t, P
(
t
,
θ
)
=
e
-
t, q
1
(
t
,
τ
)
=
36
τ
e
-
t, q
2
(
t
,
σ
)
=
72
σ
e
-
2
t, Φ
(
u
)
=
u
3, g
1
(
u
)
=
g
2
(
u
)
=
u, a
=
e
=
π
/
2, b
=
c
=
f
=
π, d
=
3
π
/
2. 1
<
r
(
t
)
=
3
/
2
+
sin
⁡
t
<
3, ∫
π
/
2
π
‍
e
-
t
d
θ
=
π
/
2
e
-
t
<
1, ∫
t
0
∞
‍
∫
3
π
/
2
π
‍
36
τ
s
e
-
s
d
τ
d
s
<
∞, ∫
t
0
∞
‍
∫
π
/
2
π
‍
σ
s
e
-
2
s
d
σ
d
s
<
∞, ∫
t
0
∞
‍
e
-
3
s
|
15
c
o
s
3
s
-
30
s
i
n
3
s
+
63
π
s
i
n
2
s
-
18
π
c
o
s
2
s
+
1
/
2
sin
⁡
2
s
15
cos
⁡
s
-
27
s
i
n
2
s
+
33
π + 6
π
2
cos
⁡
s
-
7
sin
⁡
s
+
9
π
3
| + 18
e
-
2
s
|
8
π
cos
⁡
s
+
cos
⁡
2
s
-
sin
⁡
2
s
+
4
sin
⁡
s
-
4
cos
⁡
s
-
3
π
2
| + 9
e
-
s
|
4
π
cos
⁡
s
-
6
π
sin
⁡
s
+
5
π
2 + 2
π
|
d
s
<
∞
.
Then it is easy to see that all the conditions of Theorem1 are satisfied. In fact, x
(
t
)
=
2
+
sin
⁡
t is a nonoscillatory solution of (46).



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*Source: 1026726-2016-08-25.xml*