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J. Nonlinear Sci. Appl. 8 (2015), 884–892
Research Article
Existence of nonoscillatory solutions to second-order
nonlinear neutral difference equations
Yazhou Tiana,b , Yuanli Caia , Tongxing Lib,∗
a
School of Electronic and Information Engineering, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, P. R. China.
b
Qingdao Technological University, Feixian, Shandong 273400, P. R. China.
Communicated by Martin Bohner
Abstract
We study a class of second-order neutral delay difference equations with positive and negative coefficients
∆(rn (∆(xn + pxn−m ))) + pn f (xn−k ) − qn g(xn−l ) = 0,
n = n0 , n0 + 1, . . . ,
where p ∈ R, m, k, l, n0 ∈ N , pn , qn , rn ∈ R+ , f, g ∈ C(R, R) with xf (x) > 0 and xg(x) > 0 (x 6= 0). Some
sufficient
conditions for the
equation expressed in terms
Pstudied
P existence of a nonoscillatory solution of the
P
n
1
c
of ∞ Rn pn < ∞ and ∞ Rn qn < ∞ are obtained, where Rn =
,
All rights
s=n0 rs n ≥ n0 . 2015
reserved.
Keywords: Nonoscillatory solution, neutral delay difference equation, second-order, positive and negative
coefficients.
2010 MSC: 34B15, 34B25.
1. Introduction
This paper is concerned with a second-order neutral delay difference equation with positive and negative
coefficients
∆(rn (∆(xn + pxn−m ))) + pn f (xn−k ) − qn g(xn−l ) = 0, n = n0 , n0 + 1, . . . ,
(1.1)
where ∆ stands for the forward difference operator, ∆xn = xn+1 − xn , p ∈ R, m, k, l, n0 ∈ N , pn , qn , rn ∈
R+ , f, g ∈ C(R, R), xf (x) > 0, and xg(x) > 0 for all x 6= 0. Throughout, we suppose that the following
assumptions are satisfied.
∗
Corresponding author
Email addresses: tianyazhou369@163.com (Yazhou Tian), ylicai@mail.xjtu.edu.cn (Yuanli Cai), litongx2007@163.com
(Tongxing Li)
Received 2014-12-23
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
885
(H1 ) f and g satisfy local Lipchitz conditions, Lipchitz constants are denoted by Lf (A) and Lg (A), where
A is the domain that f and g are defined;
P
P
P
(H2 ) Rn = ns=n0 r1s , n ≥ n0 , ∞ Rs ps < ∞, and ∞ Rs qs < ∞.
In recent years, there has been an increasing interest in studying the oscillatory and nonoscillatory behavior of various classes of differential, difference, and dynamic equations; see, for instance, the monographs
[1, 2], papers [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], and the references cited therein. Candan [3] investigated a
higher-order nonlinear neutral differential equation
h
i0
r(t)(x(t) + P (t)x(t − τ ))(n−1) + (−1)n [Q1 (t)g1 (x(t − σ1 )) − Q2 (t)g2 (x(t − σ2 )) − f (t)] = 0,
(1.2)
where t ≥ t0 , n ≥ 2 is an integer, r ∈ C([t0 , ∞), R+ ), P, f ∈ C([t0 , ∞), R), Qi ∈ C([t0 , ∞), R+ ), i = 1, 2,
and gi ∈ C(R, R), i = 1, 2, satisfy the local Lipschitz condition with xgi (x) > 0, i = 1, 2 for x 6= 0.
Using the Banach contraction principle, the author obtained some sufficient conditions for the existence of
nonoscillatory solutions to (1.2). Cheng [6] studied the existence of nonoscillatory solution of a second-order
linear neutral difference equation
∆2 (xn + pxn−m ) + pn xn−k − qn xn−l = 0,
n = n0 , n0 + 1, . . . ,
(1.3)
where p ∈ R, m, k, l, n0 ∈ N , pn , qn ∈ R+ , and some other special cases of equation (1.3) were considered
by Li et al. [9] and Zhang and Zhou [13]. In particular, Cheng [6] established the following result.
P
P
Theorem 1.1 (See [6, Theorem 1]). Suppose that p 6= −1, ∞ sps < ∞, and ∞ sqs < ∞. Then equation
(1.3) has a nonoscillatory solution.
To the best of our knowledge, there are few results for second-order nonlinear difference equations with
positive and negative coefficients. Motivated by the ideas exploited in [3, 6], we obtain the global results
(with respect to p), which are some sufficient conditions for the existence of a nonoscillatory solution of
(1.1) for p 6= −1. The results obtained extend those reported in [6]. An example is considered to illustrate
the possible applications.
2. Main results
Theorem 2.1. Assume that p 6= −1 and conditions (H1 ) and (H2 ) are satisfied. Then (1.1) has a bounded
nonoscillatory solution.
Proof. The proof of Theorem 2.1 will be divided into five cases, depending on the five different ranges of the
parameter p. Let ln∞0 be the Banach space which is composed of all bounded real sequences x = {xn }∞
n=n0
with the norm ||x|| = supn≥n0 |xn |.
Case 1. p = 1. By (H1 ) and (H2 ), one can choose an n∗ ≥ n0 + max{m, k, l} sufficiently large such
that, for all n ≥ n∗ ,
∞
X
1
(Ru − Rn−1 )pu ≤ ,
α
u=n
∞
X
(Ru − Rn−1 )qu ≤
u=n
∞
X
1
,
β
(Ru − Rn−1 )(pu + qu ) < min
u=n
(2.1)
1 1
1
, +
L α β
,
where α = max1≤x≤3 {f (x)}, β = max1≤x≤3 {g(x)}, and L = max{Lf ([1, 3]), Lg ([1, 3])}.
We define a bounded, closed, and convex subset S in ln∞0 by
S = {x = {xn } ∈ ln∞0 : 1 ≤ xn ≤ 3, n ≥ n0 }.
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
886
Consider the operator T : S → ln∞0 defined by
(T x)n =




n+2jm
X
∞
X
2−
j=1 s=n+(2j−1)m



!
∞
1 X
(pu f (xu−k ) − qu g(xu−l )) ,
rs u=s
n ≥ n∗ ,
n0 ≤ n ≤ n ∗ .
(T x)n∗ ,
Clearly, T xn is a real sequence. It is not difficult to show that T is a continuous mapping on S. For every
x = {xn } ∈ S and n ≥ n∗ , we obtain


n+(2j−1)m
n+2jm
∞
∞
∞
X
X
X
X
X
1
1

(T x)n ≤ 2 +
qu g(xu−l ) +
qu g(xu−l )
rs u=s
rs u=s
j=1
=2+
=2+
s=n+(2j−1)m
∞
X
∞
X
1
r
s=n s
∞
X
qu g(xu−l ) = 2 +
u=s
s=n+(2j−2)m
∞ X
u
X
1
qu g(xu−l )
r
s
u=n s=n
(Ru − Rn−1 )qu g(xu−l ) ≤ 2 + β
u=n
∞
X
(Ru − Rn−1 )qu ≤ 3.
u=n
On the other hand, we have
(T x)n ≥ 2 −

∞
X

j=1
=2−
=2−
n+2jm
X
s=n+(2j−1)m
∞
X
∞
X
1
r
s=n s
∞
X
∞
1 X
pu f (xu−k ) +
rs u=s
pu f (xu−k ) = 2 −
u=s
X
s=n+(2j−2)m

∞
X
1
pu f (xu−k )
rs u=s
∞ X
u
X
1
pu f (xu−k )
r
u=n s=n s
(Ru − Rn−1 )pu f (xu−l ) ≥ 2 − α
u=n
n+(2j−1)m
∞
X
(Ru − Rn−1 )pu ≥ 1.
u=n
Thus, we conclude that T S ⊆ S.
Next, we prove that T is a contraction mapping on S. As a matter of fact, for every x, y ∈ S and n ≥ n∗ ,
we get
!
n+2jm
∞
∞
X
X
1 X
(pu |f (xu−k ) − f (yu−k )| + qu |g(xu−l ) − g(yu−l )|)
|T xn − T yn | ≤
rs u=s
j=1 s=n+(2j−1)m
∞
∞ X
u
∞
X
X
1 X
1
≤ L||x − y||
(pu + qu )
(pu + qu ) = L||x − y||
r
r
s=n s u=s
u=n s=n s
= L||x − y||
∞
X
(Ru − Rn−1 )(pu + qu ) = p0 ||x − y||,
u=n
which implies that
||T x − T y|| ≤ p0 ||x − y||,
P∞
where p0 = L u=n (Ru − Rn−1 )(pu + qu ). Using (2.1), we have p0 < 1, and thus T is a contraction mapping.
Consequently, T has a unique fixed x such that (T x)n = xn , that is,

n+2jm
∞
∞
X
X

1 X

2 −
[pu f (xu−k ) − qu g(xu−l )], n ≥ n∗ ,
rs u=s
xn =
j=1
s=n+(2j−1)m



(T x)n∗ , n0 ≤ n ≤ n∗ .
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
887
Furthermore, we have
xn + xn−m = 4 −
∞
X


j=1
=4−
n+2jm
X
∞
X
n+(2j−1)m
+

∞
X
 1
(pu f (xu−k ) − qu g(xu−l ))
rs u=s
s=n+(2j−2)m
s=n+(2j−1)m
∞
X
1
r
s=n s
X
(pu f (xu−k ) − qu g(xu−l )).
u=s
Therefore,
∆(rn (∆(xn + xn−m ))) + pn f (xn−k ) − qn g(xn−l ) = 0,
and xn is obviously a positive solution of (1.1). This completes the proof of Case 1.
Case 2. p ∈ (0, 1). By virtue of conditions (H1 ) and (H2 ), we can choose an n1 ≥ n0 + max{m, k, l}
sufficiently large such that
∞
X
p − (1 − N1 )
Rs ps ≤
,
α1
s=n
∞
X
s=n
∞
X
R s qs ≤
1 − p − pN1 − M1
,
β1
Rs (ps + qs ) <
s=n
(2.2)
1−p
L1
hold for all n ≥ n1 , where N1 ≥ M1 > 0, 1 − N1 < p < (1 − M1 )/(1 + N1 ), α1 = maxM1 ≤x≤N1 {f (x)},
β1 = maxM1 ≤x≤N1 {g(x)}, and L1 = max{Lf ([M1 , N1 ]), Lg ([M1 , N1 ])}. Set
A1 = {x = {xn } ∈ ln∞0 : M1 ≤ xn ≤ N1 , n ≥ n0 }.
Define an operator T : A1 → ln∞0 by

∞
X


1 − p − pxn−m + Rn−1
(ps f (xs−k ) − qs g(xs−l ))




s=n−1


n−2
X
(T x)n =

+
Rs (ps f (xs−k ) − qs g(xs−l )), n ≥ n1 ,




s=n1



(T x)n1 ,
n0 ≤ n ≤ n 1 .
For every x ∈ A1 and n ≥ n1 , we have
∞
X
(T x)n ≤ 1 − p + α1 Rn−1
s=n−1
≤ 1 − p + α1
∞
X
n−2
X
ps + α1
Rs ps
s=n1
Rs ps ≤ N1 .
s=n1
Furthermore, we get
(T x)n ≥ 1 − p − pN1 − Rn−1
≥ 1 − p − pN1 − β1
∞
X
s=n−1
∞
X
qs g(xs−l ) −
Rs qs ≥ M1 ,
s=n1
n−2
X
s=n1
Rs qs g(xs−l )
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
888
and hence T A1 ⊆ A1 .
Now, for x, y ∈ A1 and n ≥ n1 , we obtain
∞
X
|T xn − T yn | ≤ p|xn−m − yn−m | + Rn−1
ps |f (xs−k ) − f (ys−k )|
s=n−1
∞
X
+ Rn−1
qs |g(xs−l ) − g(ys−l )| +
+
Rs ps |f (xs−k ) − f (ys−k )|
s=n1
s=n−1
n−2
X
n−2
X
Rs qs |g(xs−l ) − g(ys−l )|
s=n1
≤ p||x − y|| + L1 ||x − y||
∞
X
Rs (ps + qs )
s=n1
= q̂1 ||x − y||,
where q̂1 = p + L1
P∞
s=n1
Rs (ps + qs ) < 1 due to (2.2). This immediately yields
||T x − T y|| ≤ q̂1 ||x − y||,
and so T is a contraction mapping. Consequently, T has a unique fixed x, which is obviously a positive
solution of (1.1). This completes the proof of Case 2.
Case 3. p ∈ (1, ∞). From (H1 ) and (H2 ), one can choose an n2 ≥ n0 + max{m, k, l} sufficiently large such
that
∞
X
1 − p(1 − N2 )
Rs ps ≤
,
α
2
s=n
∞
X
s=n
∞
X
Rs qs ≤
(1 − M2 )p − (1 + N2 )
,
β2
Rs (ps + qs ) <
s=n
(2.3)
p−1
L2
hold for all n ≥ n2 , where N2 ≥ M2 > 0, (1 − M2 )p > 1 + N2 , p(1 − N2 ) < 1, α2 = maxM2 ≤x≤N2 {f (x)},
β2 = maxM2 ≤x≤N2 {g(x)}, and L2 = max{Lf ([M2 , N2 ]), Lg ([M2 , N2 ])}. Set
A2 = {x = {xn } ∈ ln∞0 : M2 ≤ xn ≤ N2 , n ≥ n0 }.
Define an operator T : A2 → ln∞0 as

∞
X

 1 − 1 − 1 xn+m + 1 Rn+m−1
(ps f (xs−k ) − qs g(xs−l ))



p p
p

s=n+m−1


n+m−2
(T x)n =
1 X

+
Rs (ps f (xs−k ) − qs g(xs−l )), n ≥ n2 ,



p s=n

2



(T x)n2 , n0 ≤ n ≤ n2 .
For every x ∈ A2 and n ≥ n2 , we get
(T x)n ≤ 1 −
≤1−
1 1
+ α2 Rn+m−1
p p
∞
X
∞
X
s=n+m−1
1 1
+ α2
Rs ps ≤ N2 .
p p s=n
2
n+m−2
X
1
ps + α2
Rs ps
p
s=n
2
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
889
Furthermore, we have
1 1
1
(T x)n ≥ 1 − − N2 − β2 Rn+m−1
p p
p
≥1−
∞
X
s=n+m−1
n+m−2
X
1
qs − β2
Rs qs
p
s=n
2
∞
X
1 1
1
Rs qs ≥ M2 ,
− N2 − β2
p p
p s=n
2
and thus T A2 ⊆ A2 . Since A2 is a bounded, closed, and convex subset of ln∞0 , we have to prove that T is a
contraction mapping on A2 to apply the contraction principle.
Now, for x, y ∈ A2 and n ≥ n2 , we obtain
1
1
|T xn − T yn | ≤ |xn+m − yn+m | + Rn+m−1
p
p
1
+ Rn+m−1
p
+
≤
1
p
n+m−2
X
∞
X
∞
X
ps |f (xs−k ) − f (ys−k )|
s=n+m−1
qs |g(xs−l ) − g(ys−l )| +
s=n+m−1
n+m−2
1 X
Rs ps |f (xs−k ) − f (ys−k )|
p s=n
2
Rs qs |g(xs−l ) − g(ys−l )|
s=n2
∞
X
1
1
||x − y|| + L2 ||x − y||
Rs (ps + qs )
p
p
s=n
2
= q̂2 ||x − y||,
which yields
||T x − T y|| ≤ q̂2 ||x − y||.
From (2.3), we have q̂2 = 1/p(1 + L2 Σ∞
s=n2 Rs (ps + qs )) < 1. Therefore, T is a contraction mapping.
Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). The proof of Case 3
is complete.
Case 4. p ∈ (−1, 0). Combining (H1 ) and (H2 ), we can choose an n3 ≥ n0 + max{m, k, l} sufficiently large
such that
∞
X
(1 + p)N3 − (1 + p)
Rs ps ≤
,
α
3
s=n
∞
X
s=n
∞
X
s=n
Rs qs ≤
1 + p − M3 (1 + p)
,
β3
Rs (ps + qs ) <
(2.4)
1+p
L3
hold for all n ≥ n3 , where M3 and N3 are positive constants satisfying 0 < M3 < 1 < N3 ,
α3 = maxM3 ≤x≤N3 {f (x)}, β3 = maxM3 ≤x≤N3 {g(x)}, and L3 = max{Lf ([M3 , N3 ]), Lg ([M3 , N3 ])}. Set
A3 = {x = {xn } ∈ ln∞0 : M3 ≤ xn ≤ N3 , n ≥ n0 }.
Define an operator T : A3 → ln∞0 by

∞
X

 1 + p − pxn−m + Rn−1
(ps f (xs−k ) − qs g(xs−l ))




s=n−1


n−2
X
(T x)n =

+
Rs (ps f (xs−k ) − qs g(xs−l )), n ≥ n3 ,




s=n
3



(T x)n3 , n0 ≤ n ≤ n3 .
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
890
For every x ∈ A and n ≥ n3 , we have
∞
X
(T x)n ≤ 1 + p − pN3 + α3 Rn−1
p s + α3
∞
X
Rs ps
s=n3
s=n−1
≤ 1 + p − pN3 + α3
n−2
X
Rs ps ≤ N3 .
s=n3
Furthermore, we conclude that
(T x)n ≥ 1 + p − pM3 − Rn−1
≥ 1 + p − pM3 − β3
∞
X
qs g(xs−l ) −
n−2
X
Rs qs g(xs−l )
s=n3
s=n−1
∞
X
Rs qs ≥ M3 ,
s=n3
and thus T A3 ⊆ A3 .
Next, we prove that T is a contraction mapping on A3 . In fact, for every x, y ∈ A3 and n ≥ n3 , we have
|T xn − T yn | ≤ −p|xn−m − yn−m | + Rn−1
∞
X
ps |f (xs−k ) − f (ys−k )|
s=n−1
+ Rn−1
∞
X
qs |g(xs−l ) − g(ys−l )| +
n−2
X
Rs ps |f (xs−k ) − f (ys−k )|
s=n3
s=n−1
+
n−2
X
Rs qs |g(xs−l ) − g(ys−l )|
s=n3
≤ −p||x − y|| + L3 ||x − y||
∞
X
Rs (ps + qs )
s=n3
= q̂3 ||x − y||.
This immediately yields
||T x − T y|| ≤ q̂3 ||x − y||,
P∞
where q̂3 = −p + L3 s=n3 Rs (ps + qs ) < 1 due to (2.4), which implies that T is a contraction mapping.
Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). This completes the
proof of Case 4.
Case 5. p ∈ (−∞, −1). From (H1 ) and (H2 ), one can choose an n4 ≥ n0 + max{m, k, l} sufficiently large
such that
∞
X
−(p + 1)(N4 − 1)
Rs ps ≤
,
β
4
s=n
∞
X
s=n
∞
X
s=n
Rs qs ≤
−(1 + p)(1 − M4 )
,
α4
Rs (ps + qs ) <
(2.5)
−(p + 1)
L4
hold for all n ≥ n4 , where M4 and N4 are positive constants satisfying 0 < M4 < 1 < N4 ,
α4 = maxM4 ≤x≤N4 {f (x)}, β4 = maxM4 ≤x≤N4 {g(x)}, and L4 = max{Lf ([M4 , N4 ]), Lg ([M4 , N4 ])}. Set
A4 = {x = {xn } ∈ ln∞0 : M4 ≤ xn ≤ N4 , n ≥ n0 }.
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
891
Define an operator T : A4 → ln∞0 as

∞
X
1
1 1


(ps f (xs−k ) − qs g(xs−l ))
−
x
+
R
1
+

n+m
n+m−1


p p
p

s=n+m−1


n+m−2
(T x)n =
1 X

+
Rs (qs f (xs−k ) − qs g(xs−l )), n ≥ n4 ,



p s=n

4



(T x)n4 , n0 ≤ n ≤ n4 .
For every x ∈ A4 and n ≥ n4 , we get
1 1
1
− N4 − β4 Rn+m−1
p p
p
(T x)n ≤ 1 +
∞
X
s=n+m−1
n+m−2
X
1
qs − β4
Rs qs
p
s=n
4
∞
X
1 1
1
Rs qs ≤ N4 .
− N4 − β4
p p
p s=n
≤1+
4
Furthermore, we have
(T x)n ≥ 1 +
≥1+
1 1
1
− M4 + α4 Rn+m−1
p p
p
∞
X
s=n+m−1
n+m−2
X
1
ps + α4
Rs ps
p
s=n
4
∞
X
1 1
1
− M4 + α 4
Rs qs ≥ M4 ,
p p
p s=n
4
and so T A4 ⊆ A4 . Since A4 is a bounded, closed, and convex subset of ln∞0 , we have to prove that T is a
contraction mapping on A4 to apply the contraction principle.
Now, for x, y ∈ A4 and n ≥ n4 , we have
1
1
|T xn − T yn | ≤ − |xn+m − yn+m | − Rn+m−1
p
p
1
− Rn+m−1
p
−
1
p
n+m−2
X
∞
X
s=n+m−1
∞
X
ps |f (xs−k ) − f (ys−k )|
s=n+m−1
n+m−2
1 X
qs |g(xs−l ) − g(ys−l )| −
Rs ps |f (xs−k ) − f (ys−k )|
p s=n
4
Rs qs |g(xs−l ) − g(ys−l )|
s=n4
∞
X
1
1
≤ − ||x − y|| − L4 ||x − y||
Rs (ps + qs )
p
p
s=n
4
= q̂4 ||x − y||.
This immediately implies that
||T x − T y|| ≤ q̂4 ||x − y||.
By virtue of (2.5), we get q̂4 = 1/p(−1 − L4 Σ∞
s=n4 Rs (ps + qs )) < 1, which proves that T is a contraction
mapping. Consequently, T has a unique fixed x, which is obviously a positive solution of (1.1). This
completes the proof of Case 5. Therefore, the proof of Theorem 2.1 is complete.
Remark 2.2. One can easily see that Theorem 2.1 includes Theorem 1.1 when rn = 1 and f (u) = g(u) = u.
Y. Tian, Y. Cai, T. Li, J. Nonlinear Sci. Appl. 8 (2015), 884–892
892
3. Applications
Example 3.1. Consider a second-order difference equation
∆(rn ∆(xn + xn−1 )) + pn xn−2 − qn x3n−2 = 0,
n = 2, 3, . . . ,
(3.1)
where p = 1, rn = 1/n, f (x) = x, g(x) = x3 ,
pn =
2(n − 2)
,
(n + 2)(n + 1)n(n − 1)(2n − 3)
qn =
8(n − 2)3
.
(n + 2)(n + 1)n(n − 1)(2n − 3)3
and
It is easy to verify that
n
n
X
X
1
1
Rn =
=
s = (n + 2)(n − 1),
rs
2
s=2
∞
X
n=2
s=2
Rn pn < ∞,
and
∞
X
Rn qn < ∞.
n=2
Therefore, conditions (H1 ) and (H2 ) are satisfied. By Theorem 2.1, equation (3.1) has a bounded nonoscillatory solution. As a matter of fact, the sequence {xn } = {2 + 1/n} is a nonoscillatory solution of (3.1).
Acknowledgements
This research was supported by the NNSF of China (Grant Nos. 11171178 and 11271225) and the NSF
of Shandong Province (ZR2012AL03).
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