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Abstract
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.
1. Introduction
The purpose of this article is to study the second-order neutral nonlinear differential equations with distributed deviating arguments of the form
(1)
(1) and
(2)
(2) where
,
,
,
,
for
and
with
for
,
Throughout this study, we assume that is nondecreasing in the second variable,
for
,
and satisfies
(3)
(3)
where ,
and
(0<e<f or e<f<0) is any closed interval. Furthermore, suppose that
(4)
(4)
(5)
(5)
(6)
(6) hold.
In recent years, there have been many studies concerning the oscillatory and nonoscillatory behaviour of neutral differential equations (see [Citation1–16] 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 [Citation17, Citation18] and references cited therein. For example, in 2010, Candan and Dahiya [Citation3] considered the existence of nonoscillatory solutions of first and second-order neutral equations of the form
in 2016, Candan [Citation4] investigated nonoscillatory solutions of higher order neutral differential equations of form
and in 2019, Çına et al. [Citation8] 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 (Equation1
(1)
(1) ) and (Equation2
(2)
(2) ), more general than the latter equation, by using the Banach contraction principle.
Let for
. By a solution of Equation (Equation1
(1)
(1) ), we mean a function
in the sense that both
and
are continuously differentiable on
and such that Equation (Equation1
(1)
(1) ) is satisfied for
.
Let for
. By a solution of Equation (Equation2
(2)
(2) ), we mean a function
in the sense that both
and
are continuously differentiable on
and such that Equation (Equation2
(2)
(2) ) is satisfied for
.
As is customary, a solution of (Equation1(1)
(1) ) or (Equation2
(2)
(2) ) is said to be oscillatory if it has arbitrarily large zeros. Otherwise, the solution is called nonoscillatory.
2. Main results
We suppose throughout this paper that X is the set of all continuous and bounded functions on with the norm
.
Theorem 2.1
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold and
. Then (Equation1
(1)
(1) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0. A similar argument holds for d<0. Let
. Set
where
and
are positive constants such that
It is clear that A is a closed, bounded and convex subset of X. In view of (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), we can choose a
sufficiently large such that
,
,
, i = 1, 2 for
and
(7)
(7) where
is a constant,
(8)
(8) and
(9)
(9) where
Define a mapping
as follows:
It is clear that Sx is continuous. For every
and
, by (Equation9
(9)
(9) ), we get
and taking (Equation8
(8)
(8) ) into account, we have
Thus
. Now we show that S is a contraction mapping on A. In fact, for
and
, in view of (Equation3
(3)
(3) ) and (Equation7
(7)
(7) ), we have
This implies that
Since
, S is a contraction mapping on A. Consequently, S has the unique fixed point
such that
, which is obviously a positive solution of (Equation1
(1)
(1) ). This completes the proof.
Theorem 2.2
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold and
. Then (Equation1
(1)
(1) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0, the case d<0 can be treated similarly. Let
. Set
where
and
are positive constants such that
It is obvious that A is a closed, bounded and convex subset of X. Because of (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), we can choose a
sufficiently large such that
,
i = 1, 2 for
and
(10)
(10) where
is a constant,
(11)
(11) and
(12)
(12) where
Consider the mapping
defined by
It is obvious that Sx is continuous. For every
and
, by (Equation11
(11)
(11) ), we have
and from (Equation12
(12)
(12) ), we obtain
Thus
. Finally, for
and
, in view of (Equation3
(3)
(3) ) and (Equation10
(10)
(10) ), we have
This implies that
with
. Hence, S is a contraction mapping on A. Consequently, S has the unique fixed point
such that
, which is obviously a positive solution of (Equation1
(1)
(1) ). Thus, the theorem is proved.
Theorem 2.3
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold and
. Then (Equation1
(1)
(1) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0. A similar argument holds for d<0. Let
. Set
where
and
are positive constants such that
It is obvious that A is a closed, bounded and convex subset of X. In view of (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), we can choose a
sufficiently large such that
,
,
, i = 1, 2 for
and
(13)
(13) where
is a constant,
(14)
(14) and
(15)
(15) where
Define a mapping
as follows:
It is clear that Sx is continuous. For every
and
, from (Equation15
(15)
(15) ), we have
and by using (Equation14
(14)
(14) ), we have
Thus
. We shall show that S is a contraction mapping on A. In fact, for
and
, in view of (Equation3
(3)
(3) ) and (Equation13
(13)
(13) ), we have
This implies that
Since
, S is a contraction mapping on A. Consequently, S has the unique fixed point
such that
, which is obviously a positive solution of (Equation1
(1)
(1) ). This completes the proof.
Theorem 2.4
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold and
. Then (Equation1
(1)
(1) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0, the case d<0 can be treated similarly. Let
. Set
where
and
are positive constants such that
It is obvious that A is a closed, bounded and convex subset of X. In view of (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), there exists a
sufficiently large such that
,
i = 1, 2 for
and
(16)
(16) where
is a constant,
(17)
(17) and
(18)
(18) where
Consider the mapping
defined by
It is obvious that Sx is continuous. For every
and
, from (Equation18
(18)
(18) ), we have
and from (Equation17
(17)
(17) ), we obtain
Thus
. In fact, for
and
, by using (Equation3
(3)
(3) ) and (Equation16
(16)
(16) ), we have
This shows that
with
. This implies that S is a contraction mapping on A. Thus, S has the unique fixed point
such that
, which is clearly a positive solution of (Equation1
(1)
(1) ). Hence the proof is complete.
Theorem 2.5
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold,
and
. Then (Equation2
(2)
(2) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0. A similar argument holds for d<0. Let
. Set
where
and
are positive constants such that
It is obvious that A is a closed, bounded and convex subset of X. Because of (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), we can take a
sufficiently large such that
,
,
, i = 1, 2 for
and
(19)
(19) where
is a constant,
(20)
(20) and
(21)
(21) where
Define a mapping
as follows:
It is clear that Sx is continuous. For every
and
, from (Equation21
(21)
(21) ), we have
and by using (Equation20
(20)
(20) ), we get
Thus
. Now we show that S is a contraction mapping on A. In fact, for
and
, from (Equation3
(3)
(3) ) and (Equation19
(19)
(19) ), we have
This implies that
Since
, S is a contraction mapping on A. Consequently, S has the unique fixed point
such that
, which is obviously a positive solution of (Equation2
(2)
(2) ). This completes the proof.
Theorem 2.6
Assume that (Equation3(3)
(3) )–(Equation6
(6)
(6) ) hold,
and
. Then (Equation2
(2)
(2) ) has a bounded nonoscillatory solution.
Proof.
Suppose (Equation5(5)
(5) ) holds with d>0, the case d<0 can be treated similarly. Let
. Set
where
and
are positive constants such that
It is clear that A is a closed, bounded and convex subset of X. By (Equation4
(4)
(4) )–(Equation6
(6)
(6) ), we can take a
sufficiently large such that
,
,
, i = 1, 2 for
and
(22)
(22) where
is a constant,
(23)
(23) and
(24)
(24) where
Consider the mapping
defined by
It is obvious that Sx is continuous. For every
and
, by (Equation24
(24)
(24) ), we have
and taking (Equation23
(23)
(23) ) into account, we get
Thus
. Now we show that S is a contraction mapping on A. In fact, for
and
, in view of (Equation3
(3)
(3) ) and (Equation22
(22)
(22) ), we have
This implies that
with
and S is a contraction mapping on A. Consequently, S has the unique fixed point
such that
, which is obviously a positive solution of (Equation2
(2)
(2) ). Thus the theorem is proved.
Example 2.7
For t>4, consider the equation
(25)
(25) Note that
,
,
,
,
,
,
and
We can check that all the conditions of Theorem 2.1 are satisfied. We note that
is a nonoscillatory solution of (Equation25
(25)
(25) ).
Example 2.8
For t>4, consider the equation
(26)
(26) Equation (Equation26
(26)
(26) ) is a special case of (Equation2
(2)
(2) ) with
,
,
,
,
,
and
The conditions of Theorem 2.5 are clearly satisfied. It is obvious that
is a nonoscillatory solution of (Equation26
(26)
(26) ).
Disclosure statement
No potential conflict of interest was reported by the authors.
ORCID
M. Tamer Şenel http://orcid.org/0000-0003-1915-5697
T. Candan http://orcid.org/0000-0002-6603-3732
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