Asymptotic stability in totally nonlinear neutral diﬀerence equations … · 2015. 9. 11. · signiﬁcant advantage over Lyapunov’s direct method. The conditions of the former

Asymptotic stability in totally nonlinear neutraldifference equations

Abdelouaheb ArdjouniUniversity Souk Ahras, Algeria

andAhcene Djoudi

University Annaba, AlgeriaReceived : February 2015. Accepted : July 2015

Proyecciones Journal of MathematicsVol. 34, No 3, pp. 255-276, September 2015.Universidad Catolica del NorteAntofagasta - Chile

Abstract

In this paper we use fixed point method to prove asymptotic stabil-ity results of the zero solution of the totally nonlinear neutral differ-ence equation with variable delay

4x (n) = −a (n) f (x (n− τ (n))) +4g (n, x (n− τ (n))) .

An asymptotic stability theorem with a sufficient condition is proved,which improves and generalizes some results due to Raffoul (2006)[23], Yankson (2009) [27], Jin and Luo (2009) [17] and Chen (2013)[9].

Subjclass [2000] : 39A30, 39A70.

Keywords : Fixed point, Stability, Neutral difference equations,Variable delay.

256 Abdelouaheb Ardjouni and Ahcene Djoudi

1. INTRODUCTION

Certainly, the Lyapunov direct method has been, for more than 100 years,the efficient tool for the study of stability properties of ordinary, functional,partial differential and difference equations. Nevertheless, the application ofthis method to problems of stability in differential and difference equationswith delay has encountered serious difficulties if the delay is unbounded orif the equation has unbounded terms ([6],[7],[11]-[13],[15],[25]). Recently,Burton, Furumochi, Zhang, Raffoul, Islam, Yankson and others have no-ticed that some of these difficulties vanish or might be overcome by means offixed point theory (see [1],[2],[3],[6],[7],[9],[16],[17], [23],[24],[27]-[29]). Thefixed point theory does not only solve the problem on stability but has asignificant advantage over Lyapunov’s direct method. The conditions of theformer are often averages but those of the latter are usually pointwise (see[6]). Yet the stability theory of difference equations with/without delay hasbeen considered by many authors without the application of Lyapunov andfixed point methods, see the papers [4],[5],[8],[14],[19]-[22],[30].

In this paper, we consider the nonlinear neutral difference equation withvariable delay

4x (n) = −a (n) f (x (n− τ (n))) +4g (n, x (n− τ (n))) ,(1.1)

with the initial condition

x (n) = ψ (n) for n ∈ [m (n0) , n0] ∩ Z,

where ψ : [m (n0) , n0] ∩ Z→ R is a bounded sequence and for n0 ≥ 0,

m (n0) = inf {n− τ (n) , n ≥ n0} .

Throughout this paper we assume that a : Z+ → R, f : R → R andτ : Z+ → Z+ with n − τ (n) → ∞ as n → ∞. The function g (n, x) islocally Lipschitz continuous in x. That is, there is positive constant E sothat if |x| ≤ l for some positive constant l then

|g (n, x)− g (n, y)| ≤ E |x− y| .(1.2)

We also assume thatg (n, 0) = 0.(1.3)

Special cases of (1.1) have been considered and investigated by manyauthors. For example, Raffoul in [23] and Yankson in [27] studied theequation

Asymptotic stability in totally nonlinear neutral difference ... 257

4x (n) = −a (n)x (n− τ (n)) ,(1.4)

and proved the following theorems.

Theorem A (Raffoul [23]). Suppose that τ (n) = r and a (n+ r) 6= 1 andthere exists a constant α < 1 such that

n−1Xs=n−r

|a (s+ r)|+n−1Xs=0

⎛⎝|a (s+ r)|

¯¯ n−1Yk=s+1

[1− a (k + r)]

¯¯ s−1Xu=s−r

|a (u+ r)|

⎞⎠ ≤ α,

(1.5)

for all n ∈ Z+ andn−1Ys=0

[1− a (s+ r)] → 0 as n → ∞ . Then, for every

small initial sequence ψ : [−r, 0] ∩ Z→ R, the solution x (n) = x (n, 0, ψ)of (1.4) is bounded and tends to zero as n→∞.

Theorem B (Yankson [27]). Suppose that Q (n) 6= 0 for all n ∈ [n0,∞)∩Z,the inverse sequence g of n − τ (n) exists and there exists a constant α ∈(0, 1) for all n ∈ [n0,∞) ∩ Z such that

n−1Xs=n−τ(n)

|a (g (s))|+n−1Xs=n0

⎛⎝|1−Q (s)|

¯¯ n−1Yk=s+1

Q (k)

¯¯ s−1Xu=s−τ(s)

|a (g (u))|

⎞⎠ ≤ α,

(1.6)where Q (n) = 1− a (g (n)). Then the zero solution of (1.4) is asymptoti-

cally stable ifn−1Ys=n0

Q (s)→ 0 as n→∞ .

Obviously, Theorem B improves and generalizes Theorem A. On otherhand, Jin and Luo in [17] and Chen in [9] considered the generalized formof (1.4),

4x (n) = −a (n) f (x (n− τ (n))) ,(1.7)

and obtained the following theorems.

Theorem C (Jin and Luo [17]). Suppose that τ (n) = r. Let f be odd,increasing on [0, l], satisfy a Lipschitz condition, and let x− f (x) be non-decreasing on [0, l]. Suppose that |a (n)| < 1 and for each l1 ∈ (0, l] wehave


|l1 − f (l1)| supn∈Z+Pn−1

s=0 |a (s+ r)|n−1Y

k=s+1

[1− a (k + r)]

+ f (l1) supn∈Z+Pn−1

s=0 |a (s+ r)|n−1Y

k=s+1

[1− a (k + r)]Pn−1

u=s−r |a (u+ r)|

+ f (l1) supn∈Z+Pn−1

s=n−r |a (s+ r)| ≤ αl1.(1.8)

Then the zero solution of (1.7) is stable.

Theorem D (Chen [9]). Suppose that the following conditions are satisfied

(i) the function f is odd, increasing on [0, l],

(ii) f (x) and x−f (x) satisfy a Lipschitz condition with constant K onan interval [−l, l], and x− f (x) is nondecreasing on [0, l],

(iii) the inverse function g (n) of n− τ (n) exists and |a (g (n))| < 1,(iv) there exists a constant α ∈ (0, 1) for all n ∈ Z+ such that

Pn−1s=0 |a (g (s))|

n−1Yk=s+1

[1− a (g (k))] +Pn−1

s=n−r |a (g (s))|

+Pn−1

s=0 |a (g (s))|n−1Y

k=s+1

[1− a (g (k))]Pn−1

u=s−r |a (g (u))|

≤ α.

Then the zero solution of (1.7) is asymptotically stable ifn−1Yk=0

[1− a (g (k))]→ 0 as n→∞.

Obviously, Theorem D improves Theorem C.

Our purpose here is to improve Theorems A—D and extend it to inves-tigate a wide class of nonlinear neutral difference equation with variabledelay presented in (1.1). Our results are obtained with no need of furtherassumptions on the inverse of sequence n−τ (n), so that for a given boundedinitial sequence ψ a mapping P for (1.1) is constructed in such a way tomap a, carefully chosen, complete metric space Sψ into itself on which P isa contraction mapping possessing a fixed point. This procedure will enableus to establish and prove by means of the contraction mapping theoreman asymptotic stability theorem for the zero solution of (1.1) with a lessrestrictive conditions. It is important to note that, in our consideration,the neutral term 4g (n, x (n− τ (n))) of (1.1) produces nonlinearity in the


neutral term 4x (n− τ (n)). While, the neutral term in [3] enters linearly.As a consequence, we have performed an appropriate analysis which is dif-ferent from that used in [3] to construct the mapping in order to employfixed point theorems. For details on contraction mapping principle we re-fer the reader to [26] and for more on the calculus of difference equations,we refer the reader to [10] and [18]. The results presented in this paperimprove and generalize the main results in [9],[17],[23],[27].

2. Main results

For a fixed n0, we denoteD (n0) the set of bounded squences ψ : [m (n0) , n0]∩Z→ R with the norm |ψ|0 = max {|ψ (n)| : n ∈ [m (n0) , n0] ∩ Z}. For each(n0, ψ) ∈ Z+ ×D (n0), a solution of (1.1) through (n0, ψ) is a sequence x :[m (n0) ,∞)∩Z→ R such that x satisfies (1.1) on [n0,∞)∩Z and x = ψ on[m (n0) , n0]∩Z. We denote such a solution by x (n) = x (n, n0, ψ). For each(n0, ψ) ∈ Z+×D (n0), there exists a unique solution x (n) = x (n, n0, ψ) of(1.1) defined on [m (n0) ,∞) ∩ Z.

Let h : [m (n0) ,∞) ∩ Z → R be an arbitrary sequence. Rewrite (1.1)as 4x (n) = −h (n) f (x (n)) +4n

Pn−1s=n−τ(n) h (s) f (x (s))

+{h (n− τ (n))− a (n)} f (x (n− τ (n))) +4g (n, x (n− τ (n)))= −h (n)x (n) + h (n) [x (n)− f (x (n))] +4n

Pn−1s=n−τ(n) h (s) f (x (s))

+ {h (n− τ (n))− a (n)} f (x (n− τ (n))) +4g (n, x (n− τ (n))) ,(2.1)

where 4n represents that the difference is with respect to n. If we letH (n) = 1− h (n) then (2.1) is equivalent to

x (n+ 1) = H (n)x (n)+h (n) [x (n)− f (x (n))]+4n

n−1Xs=n−τ(n)

h (s) f (x (s))

+ {h (n− τ (n))− a (n)} f (x (n− τ (n))) +4g (n, x (n− τ (n))) .

(2.2)


In the process, for any sequence x, we denote

bXk=a

x (k) = 0 andbY

k=a

x (k) = 1 for any a > b.

Lemma 2.1. Suppose that H (n) 6= 0 for all n ∈ [n0,∞) ∩ Z. Then x is asolution of equation (1.1) if and only if

x(n) =nx (n0)− g (n0, x (n0 − τ (n0)))−

Pn0−1s=n0−τ(n0) h (s) f (x (s))

o×

n−1Yu=n0

H (u)

+Pn−1

s=n0 h (s)n−1Y

u=s+1

H (u) [x (s)− f (x (s))] +Pn−1

s=n−τ(n) h (s) f (x (s))

−Pn−1s=n0 {h (s)}

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) h (v) f (x (v))

+Pn−1

s=n0

n−1Yu=s+1

H (u) {h (s− τ (s))− a (s)} f (x (s− τ (s)))

+ g (n, x (n− τ (n)))−Pn−1s=n0 {h (s)}

n−1Yu=s+1

H (u) g (s, x (s− τ (s))) .

(2.3)

Proof. Let x be a solution of (1.1). By multiplying both sides of (2.2)

bynY

u=n0

[H (u)]−1 and by summing from n0 to n− 1 we obtain

n−1Xs=n0

4"s−1Yu=n0

[H (u)]−1 x (s)

#

=n−1Xs=n0

sYu=n0

[H (u)]−1 h (s) [x (s)− f (x (s))]

+n−1Xs=n0

sYu=n0

[H (u)]−14s

s−1Xv=s−τ(s)

h (v) f (x (v))


+n−1Xs=n0

sYu=n0

[H (u)]−1 {h (s− τ (s))− a (s)} f (x (s− τ (s)))

+n−1Xs=n0

sYu=n0

[H (u)]−14g (s, x (s− τ (s))) .

As a consequence, we arrive atn−1Yu=n0

[H (u)]−1 x (n)−n0−1Yu=n0

[H (u)]−1 x (n0)

=n−1Xs=n0

sYu=n0

[H (u)]−1 h (s) [x (s)− f (x (s))]

+n−1Xs=n0

sYu=n0

[H (u)]−14s

s−1Xv=s−τ(s)

hj (v) fj (x (v))

+n−1Xs=n0

sYu=n0

[H (u)]−1 {h (s− τ (s))− a (s)} f (x (s− τ (s)))

+n−1Xs=n0

sYu=n0

[H (u)]−14g (s, x (s− τ (s))) .

By dividing both sides of the above expression byn−1Yu=n0

[H (u)]−1 we get

x(n) = x (n0)n−1Yu=n0

H (u)

+n−1Xs=n0

n−1Yu=s+1

H (u)h (s) [x (s)− f (x (s))]

+n−1Xs=n0

n−1Yu=s+1

H (u)4sPs−1

v=s−τ(s) h (v) f (x (v))

+n−1Xs=n0

n−1Yu=s+1

H (u) {h (s− τ (s))− a (s)} f (x (s− τ (s)))

+n−1Xs=n0

n−1Yu=s+1

H (u)4g (s, x (s− τ (s))) .

(2.4)

By performing a summation by parts, we have


n−1Xs=n0

n−1Yu=s+1

H (u)4s

s−1Xv=s−τ(s)

h (v) f (x (v))

=n−1X

s=n−τ(n)h (s) f (x (s))−

n−1Yu=n0

H (u)n0−1X

s=n0−τ(n0)h (s) f (x (s))

−n−1Xs=n0

{h (s)}n−1Y

u=s+1

H (u)s−1X

v=s−τ(s)h (v) f (x (v)) ,

(2.5)

and

n−1Xs=n0

n−1Yu=s+1

H (u)4g (s, x (s− τ (s)))(2.6)

= −g (n0, x (n0 − τ (n0)))n−1Yu=n0

H (u) + g (n, x (n− τ (n)))

−Pn−1s=n0 {h (s)}

n−1Yu=s+1

H (u) g (s, x (s− τ (s))) .

Finally, substituting (2.5) and (2.6) into (2.4) completes the proof. 2From equation (2.3) we shall derive a fixed point mapping P for (1.1).

But the challenge here is to choose a suitable metric space of sequences onwhich the map P can be defined. Below a weighted metric on a specificspace is defined. Let C be the Banach space of real bounded sequencesϕ : [m (n0) ,∞) ∩ Z→ R with the supremum norm k.k, that is, for ϕ ∈ C,

kϕk = sup {|ϕ (n)| : n ∈ [m (n0) ,∞) ∩ Z} .

In other words, we carry out investigations in the complete metric space(C, d) where d denotes the supremum metric d (ϕ1, ϕ2) = kϕ1 − ϕ2k forϕ1, ϕ2 ∈ C. For a given initial sequence ψ : [m (n0) , n0] ∩ Z→ [−l, l] withl > 0, define the set

Slψ = {ϕ ∈ C, ϕ (n) = ψ (n) for n ∈ [m (n0) , n0] ∩ Z, |ϕ (n)| ≤ l} .

Since Slψ is a closed subset of C, the metric space

³Slψ, d

´is complete.


Definition 2.2. The zero solution of (1.1) is Lyapunov stable if for any> 0 and any integer n0 ≥ 0 there exists a δ > 0 such that |ψ (n)| ≤ δ for

n ∈ [m (n0) , n0] ∩ Z implies |x (n, n0, ψ)| ≤ for n ∈ [n0,∞) ∩ Z.

Theorem 2.3. Define a mapping P on Slψ as follows, for ϕ ∈ Sl

ψ (Pϕ) (n) =ψ (n) if n ∈ [m (n0) , n0] ∩ Z, while, for n ∈ [n0,∞) ∩ Z

(Pϕ) (n) =nψ (n0)− g (n0, ψ (n0 − τ (n0)))−

Pn0−1s=n0−τ(n0) h (s) f (ψ (s))

o×

n−1Yu=n0

H (u)

+Pn−1

s=n0 h (s)n−1Y

u=s+1

H (u) [ϕ (s)− f (ϕ (s))] +Pn−1

s=n−τ(n) h (s) f (ϕ (s))

−Pn−1s=n0 h (s)

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) h (v) f (ϕ (v))

+Pn−1

s=n0

n−1Yu=s+1

H (u) {h (s− τ (s))− a (s)} f (ϕ (s− τ (s)))

+ g(n, ϕ (n− τ (n)))−Pn−1s=n0 h (s)

n−1Yu=s+1

H (u) g (s, ϕ (s− τ (s))) .

(2.7)

Suppose that (1.2) and (1.3) hold and the following conditions are sat-isfied,

(i) the function f is odd, increasing on [0, l],(ii) f (x) and x−f (x) satisfy a Lipschitz condition with constant K on

an interval [−l, l], and x− f (x) is nondecreasing on [0, l],(iii) |h (n)| < 1 for n ∈ [m (n0) ,∞) ∩ Z and |h (n− τ (n))− a (n)| < 1,

ρE < 1 for n ∈ [n0,∞) ∩ Z, ρ > 6,(iv) there exist a constant α ∈ (0, 1) for all n ∈ [n0,∞) ∩ Z such that

f (l)

⎧⎨⎩Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) +Pn−1

s=n−τ(n) |h (s)|

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|


+Pn−1

s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)|

⎫⎬⎭+ lE

⎧⎨⎩1 +Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u)

⎫⎬⎭ ≤ αf (l) .

Then there exists δ > 0 such that for any ψ : [m n0) , n0]∩Z→ (−δ, δ),we have that P : Sl

ψ → Slψ and P is a contraction mapping with respect to

the metric defined on Slψ.

Proof. Since f is odd and satisfies the Lipshitz condition on [−l, l],f (0) = 0 and f is uniformly continuous on [−l, l]. So we can choose a δthat satisfies

δ

⎛⎝1 +E +Kn0−1X

s=n0−τ(n0)|h (s)|

⎞⎠ ≤ (1− α) l.(2.8)

Let ψ ∈ D (n0) such that |ψ (n)| ≤ δ for n ∈ [m (n0) , n0] ∩ Z. Notethat (2.8) implies δ < l since f (l) ≤ l by condition (ii). Thus, |ψ (n)| ≤ lfor n ∈ [m (n0) , n0] ∩ Z. Now we show that for such a ψ the mappingP : Sl

ψ → Slψ. Indeed, consider (2.7). For an arbitrary ϕ ∈ Sl

ψ, if followsfrom conditions (i) and (ii) that

|(Pϕ) (n)| ≤n(1 +E) kψk+Pn0−1

s=n0−τ(n0) |h (s)| |f (ψ (s))|o n−1Yu=n0

H (u)

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u) |ϕ (s)− f (ϕ (s))|

+Pn−1

s=n−τ(n) |h (s)| |f (ϕ (s))|

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| |f (ϕ (v))|

+Pn−1

s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)| |f (ϕ (s− τ (s)))|

+ |g (n, ϕ (n− τ (n)))|+Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) |g (s, ϕ (s− τ (s)))|


≤ δ³1 +E +K

Pn0−1s=n0−τ(n0) |h (s)|

´+(l − f (l))

Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) + f (l)Pn−1

s=n−τ(n) |h (s)|

+f (l)Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|

+f (l)Pn−1

s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)|

+lE

⎧⎨⎩1 +Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u)

⎫⎬⎭ ,

for n ∈ [n0,∞) ∩ Z. By applying (iv) and (2.8), we see that

|(Pϕ) (n)| ≤ δ

⎛⎝1 +E +Kn0−1X

s=n0−τ(n0)|h (s)|

⎞⎠+ (l − f (l))α+ f (l)α

≤ (1− α) l + (l − f (l))α+ f (l)α = l.

Hence, |(Pϕ) (n)| ≤ Z because |(Pϕ) (n)| = |ψ (n)| ≤ l forn ∈ [m (n0) , n0] ∩ Z. Therefore, Pϕ ∈ Sl

ψ.

Suppose that ρ > max {6, 1/K}. If we define a metric on Slψ as follows,

|ϕ− η|ρ

:= supn∈[n0,∞)∩Z

n−1Yu=n0

[1− |h (u)|] [1− |h (s− τ (u))− a (u)|] [1−E |h (u)| /K]ρK [1 + |h (u)|] [1 + |h (u− τ (u))− a (u)|] [1 +E |h (u)| /K]

× |ϕ (n)− η (n)| ,(2.9)

then³Slψ, |.|ρ

´is a complete metric space.

Next, we show that P is a contraction mapping on Slψ with respect to

the metric (2.9). For ϕ, η ∈ Slψ, we have

|(Pϕ) (n)− (Pη) (n)|


≤Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) |ϕ (s)− f (ϕ (s))− η (s) + f (η (s))|

+Pn−1

s=n−τ(n) |h (s)| |f (ϕ (s))− f (η (s))|

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| |f (ϕ (v))− f (η (v))|

+Pn−1

s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)| |f (ϕ (s− τ (s)))− f (η (s− τ (s)))|

+|g (n,ϕ (n− τ (n)))− g (n, η (n− τ (n)))|

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u) |g (s, ϕ (s− τ (s)))− g (s, η (s− τ (s)))| .

(2.10)

Let F (x) = x − f (x), then F (x) satisfies a Lipschitz condition withconstant K > 0 on an interval [−l, l]. If we multiply both sides of (2.10) by

n−1Yu=n0

[1− |h (u)|] [1− |h (u− τ (u))− a (u)|] [1−E |h (u)| /K]ρK [1 + |h (u)|] [1 + |h (u− τ (u))− a (u)|] [1 +E |h (u)| /K] ,

then the first term on the right-hand side of (2.10) becomesn−1Yu=n0

[1− |h (u)|] [1− |h (u− τ (u))− a (u)|] [1−E |h (u)| /K]ρK [1 + |h (u)|] [1 + |h (u− τ (u))− a (u)|] [1 +E |h (u)| /K]

×Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) |F (ϕ (s))− F (η (s))|

≤ KPn−1

s=n0|h(s)|[1−|h(s)|][1−|h(s−τ(s))−a(s)|][1−E|h(s)|/K]ρK[1+|h(s)|][1+|h(s−τ(s))−a(s)|][1+E|h(s)|/K]

×s−1Yu=n0

[1−|h(u)|][1−|h(u−τ(u))−a(u)|][1−E|h(u)|/K]ρK[1+|h(u)|][1+|h(u−τ(u))−a(u)|][1+E|h(u)|/K] |ϕ (s)− η (s)|

×n−1Y

u=s+1

H(u)[1−|h(u)|][1−|h(u−τ(u))−a(u)|][1−E|h(u)|/K]ρK[1+|h(u)|][1+|h(u−τ(u))−a(u)|][1+E|h(u)|/K]

≤ K 1ρK |ϕ− η|ρ

Pn−1s=n0 |h (s)|

n−1Yu=s+1

[1− |h (u)|]

≤ 1ρ |ϕ− η|ρ .


Similarly, we haven−1Yu=n0


×Pn−1s=n−τ(n) |h (s)| |f (ϕ (s))− f (η (s))|

≤ KPn−1


×s−1Yu=n0


×n−1Y

u=s+1

[1−|h(u)|][1−|h(u−τ(u))−a(u)|][1−E|h(u)|/K]ρK[1+|h(u)|][1+|h(u−τ(u))−a(u)|][1+E|h(u)|/K]


Pn−1s=n0 |h (s)|

n−1Yu=s+1

[1− |h (u)|]

≤ 1ρ |ϕ− η|ρ ,

n−1Yu=n0


×Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| |f (ϕ (v))− f (η (v))|

≤ KPn−1


×n−1Y

u=s+1


×Ps−1v=s−τ(s) |h (v)|

v−1Yu=n0

[1−|h(v)|][1−|h(v−τ(v))−a(v)|][1−E|h(v)|/K]ρK[1+|h(v)|][1+|h(v−τ(v))−a(v)|][1+E|h(v)|/K] |ϕ (v)− η (v)|

×s−1Yu=v

[1−|h(v)|][1−|h(v−τ(v))−a(v)|][1−E|h(v)|/K]ρK[1+|h(v)|][1+|h(v−τ(v))−a(v)|][1+E|h(v)|/K]


Pn−1s=n0 |h (s)|

n−1Yu=s+1

[1− |h (u)|]Ps−1v=s−τ(s) |h (v)|

×s−1Yu=v

[1− |h (v)|]

≤ 1ρ |ϕ− η|ρ ,


n−1Yu=n0


×Pn−1s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)| |

×f (ϕ (s− τ (s)))− f (η (s− τ (s)))

≤ KPn−1

s=n0|h(s−τ(s))−a(s)|[1−|h(s)|][1−|h(s−τ(s))−a(s)|][1−E|h(s)|/K]

ρK[1+|h(s)|][1+|h(s−τ(s))−a(s)|][1+E|h(s)|/K]

×s−1Yu=n0


×n−1Y

u=s+1



Pn−1s=n0 |h (s− τ (s))− a (s)|

×n−1Y

u=s+1

[1− |h (u− τ (u))− a (u)|]

≤ 1ρ |ϕ− η|ρ ,

n−1Yu=n0


× |g (n, ϕ (n− τ (n)))− g (n, η (n− τ (n)))|

≤ E |ϕ− η|ρ ≤ 1ρ |ϕ− η|ρ ,

and

n−1Yu=n0


×Pn−1s=n0 |h (s)|

n−1Yu=s+1

H (u) |g (s, ϕ (s− τ (s)))− g (s, η (s− τ (s)))|

≤ EPn−1


×s−1Yu=n0


×n−1Y

u=s+1




Pn−1s=n0 E |h (s)| /K

n−1Yu=s+1

[1−E |h (u)| /K]

≤ 1ρ |ϕ− η|ρ .

Hence, |Pϕ− Pη|ρ ≤6

ρ|ϕ− η|ρ, since ρ > 6, we have that P is a

contraction mapping on Slψ. 2

Theorem 2.4. Assume that the hypotheses of Theorem 2.3 hold. Thenthe zero solution of (1.1) is stable.

Proof. Let P be defined as in Theorem 2.3. By the contraction mappingprinciple ([26], p. 2), P has a unique fixed point in Sl

ψ, which is a solutionof (1.1) with x = ψ on [m (n0) , n0] ∩ Z.

To prove stability at n = n0, let > 0 be given, then we choose m > 0so that m < min {l, }. By considering Sm

ψ , we obtain there is a δ > 0such that kψk < δ implies that the unique solution of (1.1) with x = ψ on[m (n0) , n0] ∩ Z satisfies |x (n)| ≤ m < for all n ∈ [m (n0) ,∞) ∩ Z. Thisproves that the zero solution of (1.1) is stable. 2

Definition 2.5. The zero solution of (1.1) is asymptotically stable if it isLyapunov stable and if for any integer n0 ≥ 0 there exists a δ > 0 suchthat |ψ (n)| ≤ δ for n ∈ [m (n0) , n0]∩Z implies x (n, n0, ψ)→ 0 as n→∞.

Theorem 2.6. Assume that the hypotheses of Theorem 2.3 hold. Alsoassume that

n−1Yu=n0

H (u)→ 0 as n→∞,(2.11)

where H (u) = 1− h (u). Then the zero solution of (1.1) is asymptoticallystable.


Proof. From Theorem 2.4, the zero solution of (1.1) is stable. For agiven > 0 let ψ ∈ D (n0) such that |ψ (n)| ≤ δ for n ∈ [m (n0) , n0] ∩ Zwhere δ > 0 and define

Sψ= {ϕ ∈ C, ϕ (n) = ψ (n) for n ∈ [m (n0) , n0] ∩ Z, kϕk ≤ and

ϕ (n)→ 0 as n→∞.

Then Sψ is a complete metric space with respect to the metric (2.9).Define P : Sψ → Sψ by (2.9). From the proof of Theorem 2.3, the mappingP is a contraction and for every ϕ ∈ Sψ, kPϕk ≤ .

We next show that (Pϕ) (n)→ 0 as n→∞. There are seven terms onthe right hand side in (2.7). Denote them, respectively, by Ik, k = 1, 2, ..., 7.It is obvious that the first term I1 tends to zero as t → ∞, by condition(2.11). Therefore, the second term I2 in (2.7) satisfies

|I2| =

¯¯Pn−1

s=n0 h (s)n−1Y

u=s+1

H (u) [ϕ (s)− f (ϕ (s))]

¯¯

≤ KPn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u) |ϕ (s)|

≤ Kα < K.

Thus, I2 → 0 as n → ∞. Also, due to the conditions (1.2) and (1.3)and the facts that ϕ (n) → 0 and n − τ (n) → ∞ as n → ∞, the terms I3and I6 tend to zero, as n→∞.

Now, for a given 1 ∈ (0, ), there exists a N1 > n0 such that s ≥ N1implies |ϕ (s− τj (s))| < 1. Thus, for n ≥ N1, the term I4 in (2.7) satisfies

|I4| =

¯¯Pn−1

s=n0 h (s)n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) h (v) f (ϕ (v))

¯¯

≤PN1−1s=n0 |h (s)|

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| |f (ϕ (v))|

+Pn−1

s=N1|h (s)|

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| |f (ϕ (v))|

≤ supσ≥m(n0) |ϕ (σ)|KPN1−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|

+ 1KPn−1

s=N1|h (s)|

n−1Yu=s+1

H (u)Ps−1

v=s−τ(s) |h (v)| .

By (2.11), we can find N2 > N1 such that n ≥ N2 implies


supσ≥m(n0) |ϕ (σ)|KPN1−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|

= supσ≥m(n0) |ϕ (σ)|Kn−1Yu=N2

H (u)PN1−1

s=n0 |h (s)|N2−1Yu=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|

< 1K.Now, apply condition (iv) to have |I4| < 1K + 1Kα < 2 1K. Thus,

I4 → 0 as n → ∞. Similarly, by using (2.11), then, if n ≥ N2 then termsI5 and in I7 (2.7) satisfy

|I5| =

¯¯Pn−1

s=n0

n−1Yu=s+1

H (u) {h (s− τ (s))− a (s)} f (ϕ (s− τ (s)))

¯¯

≤ supσ≥m(n0) |ϕ (σ)|Kn−1Yu=N2

H (u)PN1−1

s=n0

N2−1Yu=s+1

H (u) |h (s− τ (s))− a (s)|

+ 1KPn−1

s=N1

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)|

≤ 1K + 1Kα < 2 1K,and

|I7| =

¯¯Pn−1

s=n0 h (s)n−1Y

u=s+1

H (u) g (s, ϕ (s− τ (s)))

¯¯

≤ supσ≥m(n0) |ϕ (σ)|En−1Yu=N2

H (u)PN1−1

s=n0 |h (s)|

×N2−1Yu=s+1

H (u) + 1EPn−1

s=N1|h (s)|

n−1Yu=s+1

H (u)

≤ 1 + 1αf(l)l <

³1 + f(l)

l

´1.

Thus, I5, I7 → 0 as n→∞. In conclusion (Pϕ) (n)→ 0 as n→∞, asrequired. Hence P maps Sψ into Sψ.

By the contraction mapping principle, P has a unique fixed point x ∈ Sψwhich solves (1.1). Therefore, the zero solution of (1.1) is asymptoticallystable. 2

Letting g (n, x) = 0, we have

Corollary 2.7. Let h : [m (n0) ,∞) ∩ Z → R be an arbitrary sequence.Suppose that the following conditions are satisfied,


(i) the function f is odd, increasing on [0, l],(ii) f (x) and x−f (x) satisfy a Lipschitz condition with constant K on

an interval [−l, l], and x− f (x) is nondecreasing on [0, l],(iii) |h (n)| < 1 for n ∈ [m (n0) ,∞) ∩ Z and |h (n− τ (n))− a (n)| < 1

for n ∈ [n0,∞) ∩ Z,(iv) there exist constants α ∈ (0, 1) for all n ∈ [n0,∞) ∩ Z such thatPn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u) +Pn−1

s=n−τ(n) |h (s)|

+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

H (u)Ps−1

v=s−τ(s) |h (v)|

+Pn−1

s=n0

n−1Yu=s+1

H (u) |h (s− τ (s))− a (s)|

≤ α.

Then the zero solution of (1.7) is asymptotically stable if

n−1Yu=n0

H (u)→ 0 as n→∞.

Remark 2.8. When τ (s) = r and h (s) = a (s+ r), Corollary 2.7 improvesTheorem C. When h (s) = a (g (s)), where g (s) is the inverse function ofs−τ (s), Corollary 2.7 reduces to Theorem D. Thus Theorem 2.6 generalizesand improves Theorems C and D.

For the special case g (n, x) = 0 and f (x) = x, we can get

Corollary 2.9. Suppose that H (n) 6= 0 for all n ∈ [n0,∞) ∩ Z and thereexists a constant α ∈ (0, 1) such that for n ∈ [n0,∞) ∩ ZPn−1

s=n−τ(n) |h (s)|+Pn−1

s=n0 |h (s)|n−1Y

u=s+1

|H (u)|Ps−1v=s−τ(s) |h (v)|

+Pn−1

s=n0

n−1Yu=s+1

|H (u)| |h (s− τ (s))− a (s)|

≤ α.

Then the zero solution of (1.4) is asymptotically stable if

n−1Yu=n0

H (u)→ 0 as n→∞.


Remark 2.10. When τ (s) = r and h (s) = a (s+ r), Corollary 2.9 reducesto Theorem A. When h (s) = a (g (s)), where g (s) is the inverse function ofs− τ (s), Corollary 2.9 reduces to Theorem B. Thus Theorem 2.6 improvesTheorems A and B.

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Abdelouaheb ArdjouniFaculty of Sciences and Technology,Department of Mathematics and Informatics,University Souk Ahras,P. O. Box 1553,Souk Ahras, 41000,Algeriae-mail: abd [email protected]

and


Ahcene DjoudiApplied Mathematics Lab,Faculty of Sciences,Department of Mathematics,University Annaba,P. O. Box 12, Annaba 23000,Algeriae-mail: [email protected]

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