Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 39, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF NON-OSCILLATORY SOLUTIONS TO
FIRST-ORDER NEUTRAL DIFFERENTIAL EQUATIONS
TUNCAY CANDAN
Abstract. This article presents sufficient conditions for the existence of nonoscillatory solutions to first-order differential equations having both delay and
advance terms, known as mixed equations. Our main tool is the Banach contraction principle.
1. Introduction
In this article, we consider a first-order neutral differential equation
d
[x(t) + P1 (t)x(t − τ1 ) + P2 (t)x(t + τ2 )]
dt
+ Q1 (t)x(t − σ1 ) − Q2 (t)x(t + σ2 ) = 0,
(1.1)
where Pi ∈ C([t0 , ∞), R), Qi ∈ C([t0 , ∞), [0, ∞)), τi > 0 and σi ≥ 0 for i = 1, 2.
We give some new criteria for the existence of non-oscillatory solutions of (1.1).
Recently, the existence of non-oscillatory solutions of first-order neutral functional differential equations has been investigated by many authors. Yu and Wang
[16] showed that the equation
d
[x(t) + px(t − c)] + Q(t)x(t − σ) = 0, t ≥ t0
dt
has a non-oscillatory solution for p ≥ 0. Later, in 1993, Chen et al [9] studied the
same equation and they extended the results to the case p ∈ R\{−1}. Zhang et al
[17] investigated the existence of non-oscillatory solutions of the first-order neutral
delay differential equation with variable coefficients
d
[x(t) + P (t)x(t − τ )] + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0, t ≥ t0 .
dt
They obtained sufficient conditions for the existence of non-oscillatory solutions
depending on the four different ranges of P (t). In [10], existence of non-oscillatory
solutions of first-order neutral differential equations
d
[x(t) − a(t)x(t − τ )] = p(t)f (x(t − σ))
dt
was studied.
2010 Mathematics Subject Classification. 34K11, 34C10.
Key words and phrases. Neutral equations; fixed point; non-oscillatory solution.
c
2016
Texas State University.
Submitted October 14, 2015. Published January 27, 2016.
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T. CANDAN
EJDE-2016/39
On the other hand, there has been research activities about the oscillatory behavior of first and higher order neutral differential equations with advanced terms.
For instance, in [1] and [5], n-th order neutral differential equations with advanced
term of the form
[x(t) + ax(t − τ ) + bx(t + τ )](n) + δ (q(t)x(t − g) + p(t)x(t + h)) = 0
and
[x(t)+λax(t−τ )+µbx(t+τ )](n) +δ
Z
d
Z
q(t, ξ)x(t−ξ)dξ +
d
p(t, ξ)x(t+ξ)dξ = 0,
c
c
were studied, respectively.
This article was motivated by the above studies. To the best of our knowledge,
this current paper is the only paper regarding to the existence of non-oscillatory
solutions of neutral differential equation with advanced term. Some other papers for
the existence of non-oscillatory solutions of first, second and higher order neutral
functional differential and difference equations; see [13, 18, 6, 7, 8, 15] and the
references contained therein. We refer the reader to the books [14, 12, 4, 11, 2, 3]
on the subject of neutral differential equations.
Let m = max{τ1 , σ1 }. By a solution of (1.1) we mean a function x ∈ C([t1 −
m, ∞), R), for some t1 ≥ t0 , such that x(t) + P1 (t)x(t − τ1 ) + P2 (t)x(t + τ2 ) is
continuously differentiable on [t1 , ∞) and (1.1) is satisfied for t ≥ t1 .
As it is customary, a solution of (1.1) is said to be oscillatory if it has arbitrarily
large zeros. Otherwise the solution is called non-oscillatory.
The following theorem will be used to prove the theorems.
Theorem 1.1 (Banach’s Contraction Mapping Principle). A contraction mapping
on a complete metric space has exactly one fixed point.
2. Main Results
To show that an operator S satisfies the conditions for the contraction mapping
principle, we consider different cases for the ranges of the coefficients P1 (t) and
P2 (t).
Theorem 2.1. Assume that 0 ≤ P1 (t) ≤ p1 < 1, 0 ≤ P2 (t) ≤ p2 < 1 − p1 and
Z ∞
Z ∞
Q1 (s)ds < ∞,
Q2 (s)ds < ∞,
(2.1)
t0
t0
then (1.1) has a bounded non-oscillatory solution.
Proof. Because of (2.1), we can choose a t1 > t0 ,
t1 ≥ t0 + max{τ1 , σ1 }
(2.2)
sufficiently large such that
Z ∞
M2 − α
, t ≥ t1 ,
M2
t
Z ∞
α − (p1 + p2 )M2 − M1
Q2 (s)ds ≤
, t ≥ t1 ,
M2
t
Q1 (s)ds ≤
where M1 and M2 are positive constants such that
(p1 + p2 )M2 + M1 < M2
and α ∈ (p1 + p2 )M2 + M1 , M2 .
(2.3)
(2.4)
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3
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : M1 ≤ x(t) ≤ M2 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define an operator
S : Ω → Λ as follows:


P (t)x(t − τ1 ) − P2 (t)x(t + τ2 )

α −
R ∞1
+
[Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )]ds, t ≥ t1 ,
(Sx)(t) =
t


(Sx)(t ),
t0 ≤ t ≤ t1 .
1
Obviously, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.3) and (2.4), respectively, it follows that
Z ∞
Z ∞
(Sx)(t) ≤ α +
Q1 (s)x(s − σ1 )ds ≤ α + M2
Q1 (s)ds ≤ M2
t
t
and
Z ∞
(Sx)(t) ≥ α − P1 (t)x(t − τ1 ) − P2 (t)x(t + τ2 ) −
Q2 (s)x(s + σ2 )ds
t
Z ∞
≥ α − p 1 M2 − p 2 M2 − M2
Q2 (s)ds ≥ M1 .
t
This means that SΩ ⊂ Ω. To apply the contraction mapping principle, the remaining is to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and
t ≥ t1 ,
|(Sx1 )(t) − (Sx2 )(t)|
≤ P1 (t)|x1 (t − τ1 ) − x2 (t − τ1 )| + P2 (t)|x1 (t + τ2 ) − x2 (t + τ2 )|
Z ∞
+
(Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )| + Q2 (s)|x1 (s + σ2 ) − x2 (s + σ2 )|) ds
t
or
|(Sx1 )(t) − (Sx2 )(t)|
Z
≤ kx1 − x2 k p1 + p2 +
∞
(Q1 (s) + Q2 (s)) ds
t
M2 − α α − (p1 + p2 )M2 − M1 +
kx1 − x2 k
≤ p1 + p2 +
M2
M2
= λ1 kx1 − x2 k,
where λ1 = (1 −
M1
M2 ).
This implies that
kSx1 − Sx2 k ≤ λ1 kx1 − x2 k,
where the supremum norm is used. Since λ1 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.2. Assume that 0 ≤ P1 (t) ≤ p1 < 1, p1 − 1 < p2 ≤ P2 (t) ≤ 0 and
(2.1) hold, then (1.1) has a bounded non-oscillatory solution.
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T. CANDAN
EJDE-2016/39
Proof. Because of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.2)
such that
Z ∞
(1 + p2 )N2 − α
Q1 (s)ds ≤
, t ≥ t1 ,
(2.5)
N2
Zt ∞
α − p1 N2 − N1
Q2 (s)ds ≤
, t ≥ t1 ,
(2.6)
N2
t
where N1 and N2 are positive constants such that
N1 + p1 N2 < (1 + p2 )N2
and α ∈ (N1 + p1 N2 , (1 + p2 )N2 ).
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : N1 ≤ x(t) ≤ N2 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define an operator
S : Ω → Λ as follows:


P (t)x(t − τ1 ) − P2 (t)x(t + τ2 )

α −
R ∞1
(Sx)(t) = + t [Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds, t ≥ t1 ,


(Sx)(t ),
t ≤t≤t .
1
0
1
Obviously, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.5) and (2.6), respectively, it follows that
Z ∞
(Sx)(t) ≤ α − p2 N2 + N2
Q1 (s)ds ≤ N2 ,
t
Z ∞
(Sx)(t) ≥ α − p1 N2 − N2
Q2 (s)ds ≥ N1 .
t
This proves that SΩ ⊂ Ω. To apply the contraction mapping principle, it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and t ≥ t1 ,
Z ∞
|(Sx1 )(t) − (Sx2 )(t)| ≤ kx1 − x2 k p1 − p2 +
(Q1 (s) + Q2 (s)) ds
t
≤ λ2 kx1 − x2 k,
where λ2 = (1 −
N1
N2 ).
This implies
kSx1 − Sx2 k ≤ λ2 kx1 − x2 k,
where the supremum norm is used. Since λ2 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.3. Assume that 1 < p1 ≤ P1 (t) ≤ p10 < ∞, 0 ≤ P2 (t) ≤ p2 < p1 − 1
and (2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. In view of (2.1), we can choose a t1 > t0 ,
t 1 + τ1 ≥ t 0 + σ1 ,
(2.7)
sufficiently large such that
Z ∞
p 1 M4 − α
, t ≥ t1 ,
M4
t
Z ∞
α − p10 M3 − (1 + p2 )M4
Q2 (s)ds ≤
, t ≥ t1 ,
M4
t
Q1 (s)ds ≤
(2.8)
(2.9)
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EXISTENCE OF NON-OSCILLATORY SOLUTIONS
5
where M3 and M4 are positive constants such that
p10 M3 + (1 + p2 )M4 < p1 M4
and α ∈ p10 M3 + (1 + p2 )M4 , p1 M4 .
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : M3 ≤ x(t) ≤ M4 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define a mapping
S : Ω → Λ as follows:

1

 P1 (t+τ1 ) {α − x(t + τ1 ) − P2 (t + τ1 )x(t + τ1 + τ2 )

R∞
t ≥ t1 ,
(Sx)(t) = + t+τ1 [Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds},


(Sx)(t ),
t ≤t≤t .
1
0
1
Clearly, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.8) and (2.9), respectively,
it follows that
Z ∞
Z ∞
1
1
(Sx)(t) ≤
α + M4
Q1 (s)ds ≤
α + M4
Q1 (s)ds ≤ M4
P1 (t + τ1 )
p1
t
t
and
Z ∞
1
α − (1 + p2 )M4 − M4
Q2 (s)ds
P1 (t + τ1 )
t
Z ∞
1 ≥
α − (1 + p2 )M4 − M4
Q2 (s)ds ≥ M3 .
p10
t
(Sx)(t) ≥
This means that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and t ≥ t1 ,
Z ∞
1
(Q1 (s) + Q2 (s)) ds
|(Sx1 )(t) − (Sx2 )(t)| ≤ kx1 − x2 k 1 + p2 +
p1
t
≤ λ3 kx1 − x2 k,
where λ3 = (1 −
p10 M3
p1 M4 ).
This implies
kSx1 − Sx2 k ≤ λ3 kx1 − x2 k,
where the supremum norm is used. Since λ3 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.4. Assume that 1 < p1 ≤ P1 (t) ≤ p10 < ∞, 1 − p1 < p2 ≤ P2 (t) ≤ 0
and (2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. In view of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.7)
such that
Z ∞
(p1 + p2 )N4 − α
Q1 (s)ds ≤
, t ≥ t1 ,
(2.10)
N4
Zt ∞
α − p10 N3 − N4
Q2 (s)ds ≤
, t ≥ t1 ,
(2.11)
N4
t
where N3 and N4 are positive constants such that
p10 N3 + N4 < (p1 + p2 )N4
and α ∈ p10 N3 + N4 , (p1 + p2 )N4 .
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T. CANDAN
EJDE-2016/39
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : N3 ≤ x(t) ≤ N4 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define a mapping
S : Ω → Λ as follows:

1

{α − x(t + τ1 ) − P2 (t + τ1 )x(t + τ1 + τ2 )

1)
 P1R(t+τ
∞
+
[Q
t ≥ t1 ,
(Sx)(t) =
1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds},
t+τ1


(Sx)(t ),
t ≤t≤t .
1
0
1
Clearly, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.10) and (2.11), respectively,
it follows that
Z ∞
1
α − p2 N4 + N4
Q1 (s)ds
(Sx)(t) ≤
P1 (t + τ1 )
t
Z ∞
1
≤
α − p2 N4 + N4
Q1 (s)ds ≤ N4
p1
t
and
Z ∞
1
α − N4 − N4
(Sx)(t) ≥
Q2 (s)ds
P1 (t + τ1 )
t
Z ∞
1 α − N4 − N4
≥
Q2 (s)ds ≥ N3 .
p10
t
This proves that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and t ≥ t1 ,
Z ∞
1
|(Sx1 )(t) − (Sx2 )(t)| ≤ kx1 − x2 k 1 − p2 +
(Q1 (s) + Q2 (s)) ds
p1
t
≤ λ4 kx1 − x2 k,
where λ4 = (1 −
p10 N3
p1 N4 ).
This implies
kSx1 − Sx2 k ≤ λ4 kx1 − x2 k,
where the supremum norm is used. Since λ4 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.5. Assume that −1 < p1 ≤ P1 (t) ≤ 0, 0 ≤ P2 (t) ≤ p2 < 1 + p1 and
(2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. Because of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.2)
such that
Z ∞
(1 + p1 )M6 − α
, t ≥ t1 ,
(2.12)
Q1 (s)ds ≤
M6
t
and
Z ∞
α − p2 M6 − M5
Q2 (s)ds ≤
, t ≥ t1 ,
(2.13)
M6
t
where M5 and M6 are positive constants such that
M5 + p2 M6 < (1 + p1 )M6
and α ∈ (M5 + p2 M6 , (1 + p1 )M6 ) .
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7
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : M5 ≤ x(t) ≤ M6 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define an operator
S : Ω → Λ as follows:


P (t)x(t − τ1 ) − P2 (t)x(t + τ2 )

α −
R ∞1
+
[Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds, t ≥ t1 ,
(Sx)(t) =
t


(Sx)(t ),
t 0 ≤ t ≤ t1 .
1
Obviously, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.12) and (2.13), respectively, it follows that
Z ∞
(Sx)(t) ≤ α − p1 M6 + M6
Q1 (s)ds ≤ M6 ,
t
Z ∞
(Sx)(t) ≥ α − p2 M6 − M6
Q2 (s)ds ≥ M5 .
t
This proves that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω, t ≥ t1 ,
Z ∞
|(Sx1 )(t) − (Sx2 )(t)| ≤ kx1 − x2 k − p1 + p2 +
(Q1 (s) + Q2 (s)) ds
t
≤ λ5 kx1 − x2 k,
where λ5 = (1 −
M5
M6 ).
This implies
kSx1 − Sx2 k ≤ λ5 kx1 − x2 k,
where the supremum norm is used. Since λ5 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.6. Assume that −1 < p1 ≤ P1 (t) ≤ 0, −1 − p1 < p2 ≤ P2 (t) ≤ 0 and
(2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. Because of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.2)
such that
Z ∞
(1 + p1 + p2 )N6 − α
Q1 (s)ds ≤
, t ≥ t1 ,
(2.14)
N6
t
and
Z
∞
Q2 (s)ds ≤
t
α − N5
,
N6
t ≥ t1 ,
(2.15)
where N5 and N6 are positive constants such that
N5 < (1 + p1 + p2 )N6
and α ∈ (N5 , (1 + p1 + p2 )N6 ).
Let Λ be the set of continuous and bounded functions on [t0 , ∞) with the supremum
norm. Set
Ω = {x ∈ Λ : N5 ≤ x(t) ≤ N6 , t ≥ t0 }.
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T. CANDAN
EJDE-2016/39
It is clear that Ω is a bounded, closed and convex subset of Λ. Define an operator
S : Ω → Λ as follows:


P (t)x(t − τ1 ) − P2 (t)x(t + τ2 )
α −

R ∞1
(Sx)(t) = + t [Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds, t ≥ t1 ,


(Sx)(t ),
t 0 ≤ t ≤ t1 .
1
Obviously, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.14) and (2.15), respectively, it follows that
Z ∞
(Sx)(t) ≤ α − p1 N6 − p2 N6 + N6
Q1 (s)ds ≤ N6 ,
t
Z ∞
(Sx)(t) ≥ α − N6
Q2 (s)ds ≥ N5 .
t
This proves that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and t ≥ t1 ,
Z ∞
|(Sx1 )(t) − (Sx2 )(t)| ≤ kx1 − x2 k − p1 − p2 +
(Q1 (s) + Q2 (s)) ds
t
≤ λ6 kx1 − x2 k,
where λ6 = (1 −
N5
N6 ).
This implies
kSx1 − Sx2 k ≤ λ6 kx1 − x2 k,
where the supremum norm is used. Since λ6 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.7. Assume that −∞ < p10 ≤ P1 (t) ≤ p1 < −1, 0 ≤ P2 (t) ≤ p2 <
−p1 − 1 and (2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. In view of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.7)
such that
Z ∞
p 1 M7 + α
, t ≥ t1 ,
(2.16)
Q1 (s)ds ≤ 0
M8
t
and
Z ∞
(−p1 − 1 − p2 )M8 − α
, t ≥ t1 ,
(2.17)
Q2 (s)ds ≤
M8
t
where M7 and M8 are positive constants such that
−p10 M7 < (−p1 − 1 − p2 )M8
and α ∈ (−p10 M7 , (−p1 − 1 − p2 )M8 ) .
Let Λ be the set of all continuous and bounded functions on [t0 , ∞) with the supremum norm. Set
Ω = {x ∈ Λ : M7 ≤ x(t) ≤ M8 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define a mapping
S : Ω → Λ as follows:

−1

{α + x(t + τ1 ) + P2 (t + τ1 )x(t + τ1 + τ2 )

1)
 P1R(t+τ
∞
t ≥ t1
(Sx)(t) = − t+τ1 [Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds},


(Sx)(t ),
t ≤t≤t .
1
0
1
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9
Clearly, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.17) and (2.16), respectively,
it follows that
Z ∞
−1
(Sx)(t) ≤
α + M8 + p2 M8 + M8
Q2 (s)ds ≤ M8
p1
t
and
Z ∞
−1
α − M8
Q1 (s)ds ≥ M7 .
p10
t
This implies that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω and t ≥ t1 ,
Z ∞
−1
kx1 − x2 k 1 + p2 +
(Q1 (s) + Q2 (s)) ds
|(Sx1 )(t) − (Sx2 )(t)| ≤
p1
t
≤ λ7 kx1 − x2 k,
(Sx)(t) ≥
where λ7 = (1 −
p10 M7
p1 M8 ).
This implies
kSx1 − Sx2 k ≤ λ7 kx1 − x2 k,
where the supremum norm is used. Since λ7 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Theorem 2.8. Assume that −∞ < p10 ≤ P1 (t) ≤ p1 < −1, p1 +1 < p2 ≤ P2 (t) ≤ 0
and (2.1) hold, then (1.1) has a bounded non-oscillatory solution.
Proof. In view of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.7)
such that
Z ∞
p1 N7 + p2 N8 + α
Q1 (s)ds ≤ 0
, t ≥ t1 ,
(2.18)
N8
t
and
Z ∞
(−p1 − 1)N8 − α
, t ≥ t1 ,
(2.19)
Q2 (s)ds ≤
N8
t
where N7 and N8 are positive constants such that
−p10 N7 − p2 N8 < (−p1 − 1)N8
and α ∈ (−p10 N7 − p2 N8 , (−p1 − 1)N8 ).
Let Λ be the set of continuous and bounded functions on [t0 , ∞) with the supremum
norm. Set
Ω = {x ∈ Λ : N7 ≤ x(t) ≤ N8 , t ≥ t0 }.
It is clear that Ω is a bounded, closed and convex subset of Λ. Define a mapping
S : Ω → Λ as follows:
 −1

 P1R(t+τ1 ) {α + x(t + τ1 ) + P2 (t + τ1 )x(t + τ1 + τ2 )
∞
(Sx)(t) = − t+τ
[Q1 (s)x(s − σ1 ) − Q2 (s)x(s + σ2 )] ds},
t ≥ t1 ,
1


(Sx)(t1 ),
t 0 ≤ t ≤ t1 .
Clearly, Sx is continuous. For t ≥ t1 and x ∈ Ω, from (2.19) and (2.18), respectively,
it follows that
Z ∞
−1 (Sx)(t) ≤
α + N8 + N8
Q2 (s)ds ≤ N8
p1
t
and
Z
∞
−1 (Sx)(t) ≥
α + p2 N8 − N8
Q1 (s)ds ≥ N7 .
p10
t
10
T. CANDAN
EJDE-2016/39
These prove that SΩ ⊂ Ω. To apply the contraction mapping principle it remains
to show that S is a contraction mapping on Ω. Thus, if x1 , x2 ∈ Ω, t ≥ t1 ,
Z ∞
−1
|(Sx1 )(t) − (Sx2 )(t)| ≤
kx1 − x2 k 1 − p2 +
(Q1 (s) + Q2 (s)) ds
p1
t
≤ λ8 kx1 − x2 k,
where λ8 = (1 −
p10 N7
p1 N8 ).
This implies
kSx1 − Sx2 k ≤ λ8 kx1 − x2 k,
where the supremum norm is used. Since λ8 < 1, S is a contraction mapping on Ω.
Thus S has a unique fixed point which is a positive and bounded solution of (1.1).
This completes the proof.
Example 2.9. Consider the equation
h
i0
1
1
t x(t) − x(t − 2π) +
− exp(− ) x(t + 5π)
2
2
2
(2.20)
1
t
t
5π
+ exp(− )x(t − 4π) − exp(− )x(t +
) = 0, t > −2 ln(1/2)
2
2
2
2
and note that
1
t
1
t
t
1
P1 (t) = − , P2 (t) = − exp(− ), Q1 (t) = exp(− ), Q2 (t) = exp(− ).
2
2
2
2
2
2
A straightforward verification yields that the conditions of Theorem 2.5 are valid.
We note that x(t) = 2 + sin t is a non-oscillatory solution of (2.20).
Example 2.10. Consider the equation
h
1
1 3
1 i0
x(t) −
− exp(−t) x(t − 1) − exp(1/4) + exp(−t) x(t + )
exp(1) 4
4
4
(2.21)
1
1
3
+ exp(−t − 1)x(t − 1) − exp(−t + )x(t + ) = 0, t ≥
4
4
2
and note that
1 3
1 1
P1 (t) = −
− exp(−t) , P2 (t) = − exp( ) + exp(−t) ,
exp(1) 4
4 4
1
Q1 (t) = exp(−t − 1), Q2 (t) = exp(−t + ).
4
It is easy to verify that the conditions of Theorem 2.6 are valid. We note that
x(t) = 1 + exp(−t) is a non-oscillatory solution of (2.21).
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Tuncay Candan
Department of Mathematics, Faculty of Arts and Sciences, Niğde University, Niğde
51200, Turkey
E-mail address: tcandan@nigde.edu.tr