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Advances in Dierence Equations Volume 2010, Article ID 461014,17pages doi:10.1155/2010/461014

Research Article

Explicit Conditions for Stability of Nonlinear Scalar Delay Impulsive Difference Equation

Bo Zheng

College of Mathematics and Information Sciences, Guangzhou University, Guangzhou 510006, China

Correspondence should be addressed to Bo Zheng,[email protected] Received 20 March 2010; Accepted 2 June 2010

Academic Editor: Leonid Shaikhet

Copyrightq2010 Bo Zheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Sufficient conditions are obtained for the uniform stability and global attractivity of the zero solution of nonlinear scalar delay impulsive difference equation, which extend and improve the known results in the literature. An example is also worked out to verify that the global attractivity condition is a sharp condition.

1. Introduction and Main Results

LetR,N, andZbe the sets of real numbers, natural numbers, and integers, respectively. For anya, b∈Z, defineZa {a, a1, . . .}andZa, b {a, a1, . . . , b}whenab.

It is well known that the theory of impulsive differential equations is emerging as an important area of investigation, since it is not only richer than the corresponding theory of differential equations without impulse effects but also represents a more natural framework for mathematical modeling of many world phenomena1 . Moreover, such equations may exhibit several real-world phenomena, such as rhythmical beating, merging of solutions, and noncontinuity of solutions. And hence ordinary differential equations and delay differential equations with impulses have been considered by many authors, and numerous papers have been published on this class of equations and good results were obtainedsee, e.g.,1–10 and references therein.

Since the behavior of discrete systems is sometimes sharply different from the behavior of the corresponding continuous systems and discrete analogs of continuous problems may yield interesting dynamical systems in their own right see11–13 , many scholars have investigated difference equations independently. However, there are few concerned with the impulsive difference equations or delay impulsive difference equationssee14–19 . On the other hand, stability is one of the major problems encountered in applications, but, to the

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best of our knowledge, very little has been done with the stability of impulsive difference equationssee15,20 . Motivated by this, the aim of this paper is devoted to studying the uniform stability and global attractivity of the zero solution of the following nonlinear scalar delay impulsive difference equation:

Δxn fn, xn, n∈Z0, n /nj, Δx

nj

Ij

x nj

, j ∈Z1, 1.1

whereΔdenotes the forward difference operator defined byΔxn xn1−xn,f : Z0− {njS → R,Sis the set of all functionsφ :Z−k,0 → Rfor somek ∈ N,and xnSis defined byxnm xnmform∈Z−k,0,Ij :R → R,and 0≤n1< n2 <· · ·<

nj< nj1<· · ·, withnj → ∞asj → ∞. By a solution of1.1, we mean a sequence{xn}of real numbers which is defined for alln∈Zn0kand satisfies1.1forn∈Zn0for some n0 ∈ Z0. It is easy to see that, for any givenn0 ∈ Z0and a given initial functionφS, there is a unique solution of1.1, denoted byxn, n0, φsuch that

x

n0m, n0, φ

φm, form∈Z−k,0. 1.2

We assume thatfn,0≡0 andIj0≡0, so thatxn≡0 is a solution of1.1, which we call the zero solution.

ForφS, define the norm ofφas φmaxφ

j:j∈Z−k,0

, 1.3

and for anyH >0, define

SH

φS:φ< H

. 1.4

Definition 1.1. The zero solution of1.1is stable, if, for anyε >0 andn0∈Z0, there exists a δδn0, >0 such thatφSδimplies that|xn, n0, φ|< εforn∈Zn0. Ifδis independent ofn0, we say that the zero solution of1.1is uniformly stable.

Definition 1.2. The zero solution of1.1is globally attractive, if every solution of1.1tends to zero asn → ∞.

A simple example of1.1is given by

Δxn Cnxnk 0, n∈Z0, n /nj, Δx

nj cjx

nj

, j ∈Z1, 1.5

where k ∈ N,C : Z0 → R, and{cj} is a sequence of real numbers. In15 , the author studied the stability of the zero solution of1.5, whereCn ≥ 0 forn ∈ Z0− {nj}and cj∈−1,0 , and obtained the following result.

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Theorem 1.3. If

n in−k

Ci

nj∈Zi−k,i−1

1cj

−1

≤ 3

2, forn∈Zk, 1.6

then the zero solution of 1.5is stable.

In this paper, we assume that there exists a positive constantHand a sequence{Pn}

of nonnegative real numbers such that

PnM

φ

≥ −f n, φ

≥ −PnM

−φ

, forn∈Z0, φ∈SH, 1.7 where max{0,maxi∈Z−k,0φi}. Furthermore, we assume that there is a sequence {bj}of positive numbers withbj≤1 such that

bjx2x

xIjx

x2, forj∈Z1, |x|< H. 1.8 The main purpose of this paper is to establish the following theorems.

Theorem 1.4. Assume that1.7and1.8hold and n

in−k

Pi

nj∈Zi−k,i−1

b−1j ≤ 3

2 1

2k1, forn∈Zk. 1.9

Then the zero solution of 1.1is uniformly stable.

Remark 1.5. Theorem 1.4generalizes and improvesTheorem 1.3greatly.

The next theorem provides a sufficient condition for every solution of1.1tends to zero as n → ∞, that is, the zero solution of1.1is globally attractive.

Theorem 1.6. Assume that1.7and1.8hold and n

in−k

Pi

nj∈Zi−k,i−1

b−1jα < 3

2 1

2k1, forn∈Zk, 1.10

and assume that, for each bounded solution{xn}, either

nlim→ ∞xn>0, n0

fn, xn −∞, forn∈Zk, 1.11

or

nlim→ ∞xn<0, n0

fn, xn ∞, forn∈Zk. 1.12

Then every solution of 1.1tends to zero asn → ∞.

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Remark 1.7. An example is worked out in Section 3 to verify that the upper bound 3/2 1/2k1in1.10is best possible, that is, the upper bound in1.10cannot be improved.

One special form of1.1is when

fn, xn

m j1

pn,jx nkn,j

, 1.13

wherepn,j ≤0, andkn,j ∈Z−k,0forn∈Z1,j ∈Z1, m. Then for anyH >0,1.7holds withPn m

j1|pn,j|. As a consequence ofTheorem 1.4, we have the following.

Corollary 1.8. Assume that1.8holds and

n in−k

m j1

pi,j

nj∈Zi−k, i−1

bj−1 ≤ 3

2 1

2k1, forn∈Zk. 1.14

Then the zero solution of the equation

Δxn m

j1

pn,jx

nkn,j

, n∈Z0, n /nj, Δx

nj Ij

x nj

, j ∈Z1,

1.15

is uniformly stable.

For the sake of convenience, throughout this paper, we will use the convention

j ni

Pn 0, wheneverji−1,

i∈A

Pi 1, wheneverAis an empty set, xn x

n, n0, φ .

1.16

2. Proofs of Main Results

Define

gjx

⎧⎪

⎪⎪

⎪⎪

⎪⎩ x

xIjx, x /0, 1

bj, x0

2.1

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forj∈Z1. Then 1≤gjx≤1/bjforj ∈Z1if|x|< H. Equation1.1can be rewritten as Δxn fn, xn, n∈Z0, n /nj,

x nj1

gj−1 x

nj

x nj

, j∈Z1. 2.2

Set

yn xn

nj∈Z0,n−1

gj x

nj

, n∈Z−k, 2.3

then2.2reduces to

Δyn fn, xn

nj∈Z0,n−1

gj x

nj

, n∈Z0, n /nj,

Δy nj

0, j∈Z1.

2.4

And it is easy to see that

|xn| ≤yn≤ |xn|

nj∈Z0,n−1

b−1j , forj∈Z1, n∈Z−k. 2.5

To prove Theorems1.4and1.6, we need the following lemma.

Lemma 2.1. Let{xn},n ∈ Zn0k,be a solution of 1.1,{yn}is defined by2.3,n ∈ Zn02k2, andBn max{|yn|:n∈Zn−3k−2, n−1}< H. If

n in−k

Pi

nj∈Zi−k,i−1

b−1jc k2

2k1, forn∈Zk 2.6

holds for somec∈k2/2k1,1 and either

yn≥0, yn> yn−1 2.7

or

yn<0, yn< yn−1, 2.8

then|yn| ≤cBn.

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Proof. We assume thatyn≥0 andyn> yn−1. The case whereyn<0 andyn<

yn−1is similar and is omitted.

It is easy to see thatLemma 2.1holds foryn 0.

Ifk0, thenc1 andPn≤2 for alln∈Z0by2.6. Thus we only need to prove that|yn| ≤Bn. Sinceyn>0,yn> yn−1, by1.7we have

−Pn−1M−xn−1≤ −fn−1, xn−1<0, 2.9

that is,

−Pn−1max{0,−xn−1} ≤ −fn−1, xn−1<0. 2.10

Soxn−1<0 which admits thatyn−1<0. Hence

ynyn−1 fn−1, xn−1

nj∈Z0,n−2

gj

x nj

Pn−1max{0,−xn−1}

nj∈Z0,n−2

gj

x nj

−Pn−1yn−1≤ −2yn−1,

2.11

which implies thatyn ≤ −yn−1, so|yn| ≤ |yn−1| ≤ Bnby the definition of Bn.

Now, assume thatyn>0,yn> yn−1,andk≥1. By1.7, we also have

−Pn−1M−xn−1≤ −fn−1, xn−1<0. 2.12

So,

max

0, max

i∈Z−k,0xn−1i

>0. 2.13

Hence, there isn1∈Znk, nsuch thatxn1−1<0 andxn≥0 for alln∈Zn1, n. So, yn1−1<0 andyn≥ 0 for alln∈Zn1, n. Forn∈Zn0, n−1, by1.7and2.4, one has

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Δyn fn, xn

nj∈Z0,n−1

gj

x nj

PnM−xn

nj∈Z0,n−1

gj x

nj

Pnmax

0, max

i∈Z−k,0−xni

nj∈Z0,n−1

gj

x nj

Pnmax

0, max

i∈Z−k,0−xni

nj∈Z0,n−1

gj

x nj

Pnmax

0, max

i∈Zn−k,n−xi

nj∈Z0,n−1

gj

x nj

Pnmax

⎧⎨

⎩0, max

i∈Zn−k,n

−yi

nj∈Z0,i−1

gj−1 x

nj

n

j∈Z0,n−1

gj x

nj

BnPn max

i∈Zn−k,nnj∈Zi,n−1

b−1j

BnPn

nj∈Zn−k,n−1

b−1j

2.14

provided thatZn0, n−1contains no impulsive points. And sinceΔyn 0 ifnmeets one of impulsive points, we always have

Δyn≤BnPn

nj∈Zn−k,n−1

bj−1ΔPnBn 2.15

forn∈Zn0, n−1, wherePn Pn

nj∈Zn−k,n−1b−1j .

By the choice ofn1, there is a real numberξ∈n1−1, n1 such that

yn1−1

yn1yn1−1

ξ−n11 0. 2.16

Next, we will show that, for anyl∈Z0, k,

−yn−lBn n

1−1

in−k

Pi−n1ξPn1−1

, forn∈Zn1−1, n−1. 2.17

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In fact, for anyl∈Z0, k,

−yn−l −yn1−1 n1−2

in−l

Δyi Δyn1−1ξ−n11 n1−2

in−l

Δyi

Bn

ξ−n11Pn1−1 n1−2

in−l

Pi

Bn n

1−1

in−l

Pi−n1ξPn1−1

Bn n

1−1

in−k

Pi−n1ξPn1−1

,

2.18

which shows that2.17holds.

Substituting2.17into2.4, it is easy to get

Δyn≤BnPn n

1−1

in−k

Pi−n1ξPn1−1

, forn∈Zn1−1, n−1. 2.19

Let

βc k2

2k1, dn

−1

nn1

Pn n1ξPn1−1. 2.20

There are two cases to consider.

Case 1d≤1. We have by2.6,2.16, and2.19

yn yn1 n−1 nn1

Δyn n1ξΔyn1−1 n

−1

nn1

Δyn

≤n1ξPn1−1Bn n

1−1

in1−k−1

Pi−n1ξPn1−1

n

−1

nn1

PnBn n

1−1

in−k

Pi−n1ξPn1−1

Bn

n1ξPn1−1

β−n1ξPn1−1

n

−1

nn1

Pn

βn

in1

Pi−n1ξPn1−1

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Bn

βdn

−1

nn1

Pnn

in1

Pi−n1ξPn1−1n

−1

nn1

Pn−n1ξ2P2n1−1

Bn

βd−1 2d2−1

2 n−1

nn1

P2n n1ξ2P2n1−1

.

2.21

Since

n−1 nn1

P2n n1ξ2P2n1−1≥ 1 nn11

n−1 nn1

Pn n1ξPn1−1 2

1

nn11d2≥ 1 k1d2,

2.22

we have

ynBn

βd−1

2d2− 1

2k1d2 Bn

βdk2 2k1d2

Bn

!

βk2 2k1

"

cBn.

2.23

Case 2d >1. In this case, there existsn2∈Zn1, n−1such that

n−1 nn2

Pn≤1,

n−1 nn2−1

Pn>1. 2.24

Therefore, we may choose a real numberη∈n2−1, n2 such that

n−1 nn2

Pn

n2η

Pn2−1 1. 2.25

Notice that

yn

Δyn1−1n1ξ n2−2

nn1

Δyn

ηn21

Δyn2−1

n2η

Δyn2−1 n

−1

nn2

Δyn

.

2.26

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So, by2.6,2.15,2.19, and2.25, we have

ynBn

Pn1−1n1ξ n2−2

nn1

Pn

ηn21

Pn2−1

× n−1

nn2

Pn n2η

Pn2−1

Bn n2η

Pn2−1

× n

1−1

in2−k−1

Pi−n1ξPn1−1

n

−1

nn2

BnPn n

1−1

in−k

Pi−n1ξPn1−1

Bn n−1

nn2

Pn n

2−1

in−k

Pi− n2η

Pn2−1

n2η

Pn2−1 n

2−1

in2−k−1

Pi− n2η

Pn2−1

Bn n−1

nn2

Pn

βn

in2

Pi− n2η

Pn2−1

n2η

Pn2−1 β

n2η

Pn2−1

Bn

! β− 1

2− 1

2k1

"

cBn.

2.27

The proof is completed by combining Cases1and2.

Proof ofTheorem 1.4. By Lemma 2.1, 1.9implies that 2.6 holds with c 1. For any ε ∈ 0, H, assume that Zn0, n0 2k1contains m mn0, kimpulsive points: n0nl1 <

nl2<· · ·< nlmn02k1. Set

β 3

2 1

2k1, δ ε

1β2k. 2.28

We will prove thatφSδimplies that

|xn|x

n, n0, φ< ε, forn∈Zn0. 2.29 To this end, we first prove that

|xn|< δ

1β2k

< ε, forn∈Zn0, n02k1. 2.30

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Ifnl1 n0, then|xnl11| |xn0 I0xn0| ≤ |xn0| < δ < . Ifnl1 > n0, then, for n∈Zn0, nl1, we claim that

|xn|< δ

1βnl1−n0 < . 2.31

In fact,

|xn01| ≤ |xn0|fn0, xn0< δPn0δ≤δ 1β

< ,

|xn02| ≤ |xn01|fn01, xn01

< δ 1β

βxn01δ 1β

βδ 1β δ

1β2

< .

2.32

In general, we can obtain2.31by induction. And so|xn| < δ1βnl1−n0< < Hfor either case withn∈Zn0, nl1. Forn∈Znl11, nl2, we have

|xnl11| ≤ |xnl1||Il1xnl1| ≤ |xnl1|< δ

1βnl1−n0 ,

|xnl12| ≤ |xnl11|fnl11, xnl11

< δ

1βnl1−n0βxnl11

δ

1βnl1−n0βδ

1βnl1−n0

δ

1βnl1−n01

< .

2.33

And so

|xn|< δ

1βnl2−n0−1

, forn∈Znl11, nl2, 2.34 In general, we have

|xn|< δ

1βnli1−n0−1, forn∈Znli1, nli1, i1,2, . . . , m−1. 2.35

Now forn∈Znlm1, n02k1,

|xnlm1| ≤ |xnlm||Ilmxnlm| ≤ |xnlm|< δ

1βnlm−n0−1,

|xnlm2| ≤ |xnlm1|fnlm1, xnlm1

< δ

1βnlm−n0−1βxnlm1

δ

1βnlm−n0−1 βδ

1βnlm−n0−1 δ

1βnlm−n0

< .

2.36

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And so

|xn|< δ

1β2k−1

, forn∈Znlm1, n02k1 2.37

which has proved that2.30holds. Now, we will prove that

|xn|< , forn∈Zn02k2. 2.38

Thus we only need to prove that

yn< , forn∈Zn02k2. 2.39

For anyn∈Zn02k2andBn< H,we claim that

ynBn. 2.40 In fact, we can assume thatyn ≥ 0. The case whereyn < 0 is similar and is omitted.

Ifynyn−1, by the definition ofBn,2.40holds. Ifyn > yn −1, then by Lemma 2.1we have that2.40holds.

Since

Bn02k2 maxyn:n∈Z−k, n02k1

< ε < H, 2.41

by2.40we have

yn02k2< ε. 2.42

By repeatedly using2.40we get that2.39holds.

Combining2.30and2.39, we find that2.29holds and the proof ofTheorem 1.4is complete.

Proof ofTheorem 1.6. In view ofTheorem 1.4, we see that the zero solution of1.1is uniformly stable. Therefore, for anyn0∈Z0, there existsδ >0 such thatφSδimplies that

|xn|x

n, n0, φ< 1

2H, forn∈Zn0. 2.43

Next, we will prove that

nlim→ ∞xn 0. 2.44

There are two cases to consider.

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Case 1. {xn}is eventually positive, that is, there existsn1 ∈Zn0such thatxn>0 for all n ∈ Zn1. Hence, by1.7and1.8, we haveΔxn ≤ 0 forn ∈Zn1k. That is,{xn}

is eventually nonincreasing and hence limn→ ∞xn ≥ 0. Thus, by1.11we see that 2.44 holds. The case when{xn}is eventually negative is similar and will be omitted.

Case 2. {xn} is oscillatory in the sense that {xn} is neither eventually positive nor eventually negative, so{yn}is also oscillatory. To prove2.44, we only need to prove that

nlim→ ∞yn 0. 2.45

By the proof ofTheorem 1.4, we have yn< 1

2H, forn∈Zn0. 2.46

Let

nlim→ ∞supyn p, lim

n→ ∞infyn q, 2.47

thenq≤0≤p. It suffices to show that

pq0. 2.48

In fact, if2.48does not hold, we assume thatp≥ −qandp >0. The case wherep <−q andq <0 is similar and is omitted.

Let 0 < ε0 < 1−c/2p, where c < 1 is given inLemma 2.1. Then there exists an integerm0∈Z2ksuch that

qε0yn0, forn∈Zm0. 2.49

Since limn→ ∞supyn p > 0 and{yn}is oscillatory, there must exist an integer m1∈Zm03k2such that

ym1> pε0, ym1> ym1−1. 2.50

ByLemma 2.1and2.49we have

pε0< ym1 ym1cBm1c 0

, forn∈Zm0. 2.51

Equations 2.49 and 2.51 imply that pε0 < cpε0, which contradicts the fact that ε0<1−c/2p; thus2.48holds and the proof is complete.

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3. Example

In this section we will give a result which guarantees that the upper bound 3/21/2k1 is best possible inTheorem 1.6.

Example 3.1. Consider the delay impulsive difference equation

Δxn Pnxnk 0, n∈Z0, n / 6k1i,

Δxni bixni, i∈Z0, ni 6k1i, 3.1

wherebi −1/i22, i ∈ Z0, k ∈ Z1, and {Pn} is a sequence of nonnegative real numbers defined by

Pn

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

0, n∈Z6k1i, k 6k1i

#Z3k2 6k1i,6k 6k1i, 1

k1, n∈Zˇk1 6k1i,2k 6k1i

#Z2k2 6k1i,3k1 6k1i,

c, n2k1 6k1i,

3.2

wherei∈Z0andc >0 is an undetermined constant. In view ofTheorem 1.6, if n

in−k

Pi

nj∈Zi−k,i−1

1bj

−1

c k k1 < 3

2 1

2k1, forn∈Zk, 3.3

or equivalently

c < k4

2k1, 3.4

then every solution of3.1tends to zero asn → ∞.

The following theorem shows that if3.4does not hold, then there is a solution of3.1 which does not tend to zero asn → ∞. This shows that the upper bound 3/21/2k1 cannot be improved.

Theorem 3.2. Assume that

ck4

2k1. 3.5

Then there exists a solution of 3.1which does not tend to zero asn → ∞.

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Proof. Let{xn}be a solution of3.1with initial condition of the form

xn , forn∈Z−k,0, 3.6

where >0 is a given constant. Then by3.1and the definition of{Pn}, we have xn 1b0ε, forn∈Z1, k1,

Δxn − 1

k11b0, forn∈Zk1,2k. 3.7

And hence

xn 1b0ε

! 2− n

k1

"

, forn∈Zk1,2k1, 3.8

x2k2 x2k1−cxk1 1b0ε

! 1 k1 −c

"

. 3.9

By virtue of3.1and3.8we get

Δxn −1b0ε 1 k1

!

2−nk k1

"

, forn∈Z2k2,3k1. 3.10

Summing up from 2k2 to 3k1 and using3.9, we have

x3k2 x2k2− 1b0 k1

3k1

n2k2

!

2−nk k1

"

1b0

! 1 k1 −c

"

−1b0 k 2k1 −1b0

!

c k−2 2k1

"

def≡ 1b01.

3.11

Furthermore, we have, by the definition of{Pn},

xn 1b01, forn∈Z3k2,6k1. 3.12

Define the sequence{i}as follows:

0, i1i

!

c k−2 2k1

"

, fori∈Z0. 3.13

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Then we have by3.5

|i1||i|

c k−2 2k1

≥ |i|, fori∈Z0, 3.14

which implies that{|i|}is a non-decreasing sequence and does not tend to zero asn → ∞.

Repeating the above argument, we find that, forn∈Z3k2i6k1, i16k 1, i∈Z0,

xn 1b01b1· · ·1bii1. 3.15

By the definition of{bi}, we know thatn

i01bi0 asn → ∞, soxn0 asn → ∞.

The proof is complete.

Acknowledgments

The author would like to express her thanks to the referees for helpful suggestions. This research is supported by Guangdong College Yumiao Project2009.

References

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