Dynamics for Nonlinear Difference Equation

doi:10.1155/2009/235691

Research Article

Dynamics for Nonlinear Difference Equation

x

_n

1

αx

n

−

k

/

β

γx

p

_n

₋

_l

Dongmei Chen,

_{Xianyi Li,}

_{and Yanqin Wang}

1_{College of Mathematics and Computational Science, Shenzhen University, Shenzhen,}

Guangdong 518060, China

2_{School of Physics & Mathematics, Jiangsu Polytechnic University, Changzhou, 213164 Jiangsu, China}

Correspondence should be addressed to Xianyi Li,[email protected]

Received 19 April 2009; Revised 19 August 2009; Accepted 9 October 2009 Recommended by Mariella Cecchi

We mainly study the global behavior of the nonlinear difference equation in the title, that is, the global asymptotical stability of zero equilibrium, the existence of unbounded solutions, the existence of period two solutions, the existence of oscillatory solutions, the existence, and asymptotic behavior of non-oscillatory solutions of the equation. Our results extend and generalize the known ones.

Copyrightq2009 Dongmei Chen et al. 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.

1. Introduction

Consider the following higher order difference equation:

xn1

αxn−k

βγxp_n−l, n0,1, . . . , 1.1

wherek, l,∈ {0,1,2, . . .}, the parametersα, β, γandp,are nonnegative real numbers and the initial conditionsx₋max{k, l}, . . . , x−1andx0are nonnegative real numbers such that

βγxp_n−l>0, ∀n≥0. 1.2

It is easy to see that if one of the parametersα, γ, pis zero, then the equation is linear. Ifβ 0, then1.1can be reduced to a linear one by the change of variablesxn eyn. So in

The change of variablesxn β/γ1/pynreduces1.1into the following equation:

yn1 ryn−k

1y_np_−l, n0,1, . . . , 1.3

whererα/β >0.

Note that y₁ 0 is always an equilibrium point of 1.3. When r > 1, 1.3 also possesses the unique positive equilibriumy₂ r−11/p.

The linearized equation of1.3about the equilibrium pointy₁0 is

zn1rzn−k, n0,1, . . . , 1.4

so, the characteristic equation of1.3about the equilibrium pointy₁0 is either, fork≥l,

λk1_{−r 0, 1.5}

or, fork < l,

λl−k_λk1_{−r0. 1.6}

The linearized equation of1.3about the positive equilibrium pointy₂ r−11/p has the form

zn1 −

pr−1

r zn−lzn−k, n0,1, . . . , 1.7

with the characteristic equation either, fork≥l,

λk1pr−1

r λk−l−10 1.8

or, fork < l,

λl1_−λl−kpr−1

r 0. 1.9

Whenp 1,k, l∈ {0,1},1.1has been investigated in1–4. Whenk1,l 2,1.1 reduces to the following form:

xn1

αxn−1

βγxp_n−2, n0,1, . . . . 1.10

On the other hand, whenk 0, p 1,1.1is just the discrete delay logistic model investigated in4,P75. Therefore, it is both theoretically and practically meaningful to study

1.1.

Our aim in this paper is to extend and generalize the work in 3. That is, we will investigate the global behavior of1.1, including the global asymptotical stability of zero equilibrium, the existence of unbounded solutions, the existence of period two solutions, the existence of oscillatory solutions, the existence and asymptotic behavior of nonoscillatory solutions of the equation. Our results extend and generalize the corresponding ones of3.

For the sake of convenience, we now present some definitions and known facts, which will be useful in the sequel.

Consider the difference equation

xn1Fxn, xn−1, . . . , xn−k, n0,1, . . . , 1.11

wherek≥1 is a positive integer, and the functionFhas continuous partial derivatives. A pointxis called an equilibrium of1.11if

xFx, . . . , x. 1.12

That is,xnxforn >0 is a solution of1.11, or equivalently,xis a fixed point ofF.

The linearized equation of1.11associated with the equilibrium pointxis

yn1

k

i0 ∂F

∂uix, . . . , xyn−i, n0,1, . . . . 1.13

We need the following lemma.

Lemma 1.1see4–6. iIf all the roots of the polynomial equation

λk1₋k

i0 ∂F

∂uix, . . . , xλ

k−i_{0 1.14}

lie in the open unit disk|λ|<1, then the equilibriumxof 1.11is locally asymptotically stable. iiIf at least one root of 1.11has absolute value greater than one, then the equilibriumxof 1.11is unstable.

For the related investigations for nonlinear difference equations, see also7–11and the references cited therein.

2. Global Asymptotic Stability of Zero Equilibrium

Lemma 2.1. The following statements are true.

aIfr <1, then the equilibrium pointy₁0of 1.3is locally asymptotically stable.

bIfr >1, then the equilibrium pointy₁of 1.3is unstable. Moreover, fork < l,1.3has a l−k-dimension local stable manifold and ak1-dimension local unstable manifold.

cIfr >1,kis odd andlis even, then the positive equilibrium pointy₂ r−11/pof 1.3

is unstable.

Proof. aWhenr < 1, it is clear from1.5and1.6that every characteristic rootλsatisfies |λ|k1_{r <1 or|λ|0, and so, byLemma 1.1i,y}

1is locally asymptotically stable.

b When r > 1, if k ≥ l, then it is clear from 1.5that every characteristic rootλ satisfies |λ|k1 _{r > 1, and so, by Lemma 1.1ii, y}

1 is unstable. If k < l, then 1.6 has l−kcharacteristic rootsλsatisfying|λ|l−k0<1, which corresponds to al−k-dimension local stable manifold of1.3, andk1 characteristic rootsλsatisfying|λ|k1 r >1, which corresponds to ak1-dimension local unstable manifold of1.3.

cIf kis odd andlis even, then, regardless ofk ≥ l ork < l, correspondingly, the characteristic equation1.8 or1.9always has one characteristic rootλlying the interval

−∞, −1. It follows fromLemma 1.1iithaty₂is unstable.

Remark 2.2. Lemma 2.1aincludes and improves3, Theorem 3.1i.Lemma 2.1bandc include and generalize3,Theorem 3.1iiandiii, respectively.

Now we state the main results in this section.

Theorem 2.3. Assume that r < 1, then the equilibrium point y₁ 0 of 1.3 is globally asymptotically stable.

Proof. We know from Lemma 2.1 that the equilibrium point y₁ 0 of 1.3 is locally asymptotically stable. It suffices to show that limn→ ∞yn 0 for any nonnegative solution

{yn}∞n−max{k,l}of1.3.

Since

0≤yn1 ryn−k

1y_np_−l ≤ryn−k≤yn−k, 2.1

{y₋jk1i}∞i0converges for anyj ∈ {0,1, . . . , k}. Let limi→ ∞y−jk1i α−j, j ∈ {0,1, . . . , k},

then

α0 rα0

1αp_−1−l, . . . , αl1−k rαl1−k

1αp_−k, αl−k rαl−k

1αp₀, . . . , α−k rα−k

1αp_−l. 2.2

Thereout, one has

that is,

lim

i→ ∞y−jk1i0, for anyj ∈ {0,1, . . . , k}, 2.4

which implies limn→ ∞yn0. The proof is over.

Remark 2.4. Theorem 2.3includes3, Theorem 3.3.

3. Existence of Eventual Period Two Solution

In this section, one studies the eventual nonnegative prime period two solutions of1.3. A solution {xn}∞n−max{k,l} of 1.3is said to be eventual prime periodic two solution if there

exists ann0 ∈ {−max{k, l},−max{k, l}1, . . .}such thatxn2 xnforn ≥ n0 andxn1/xn holds for alln≥n0.

Theorem 3.1. a Assumek is odd andl is even, then 1.3possesses eventual prime period two solutions if and only ifr1.

bAssumekis odd andlis odd, then1.3possesses eventual prime period two solutions if and only ifr >1.

cAssumekis even andlis even. Then the necessary condition for1.3to possess eventual prime period two solutions isr2_{p−2> pandr}p_>max{p/p−2p−2_r2_,p−1r2_−p}.

dAssumekis even andlis odd. Then,1.3has no eventual prime period two solutions.

Proof. aIf1.3has the eventual nonnegative prime period two solution. . . , ϕ, ψ, ϕ, ψ, . . . , then, we eventually haveϕrϕ/1ψpandψrψ/1ϕp. Hence,

ϕ1−rψp0, ψ1−rϕp0. 3.1

Ifr /1, then we can derive from3.1thatϕ 0 ifψ 0 or vice versa, which contradicts the assumption that. . . , ϕ, ψ, ϕ, ψ, . . .is the eventual prime period two solution of1.3. So, ϕψ /0. Accordingly, 1−rψp0 and 1−rϕp0, which indicate thatϕψwhenr >1 or thatϕandψdo not exist whenr <1, which are also impossible. Therefore,r 1.

Conversely, ifr 1, then choose the initial conditions such asy−ky−k2· · ·0 and y₋k1y−k3 · · ·y0 ϕ >0, or such asy−ky−k2· · ·ϕ >0 andy−k1 y−k3· · · y00. We can see by induction that. . . ,0, ϕ,0, ϕ, . . .is the prime period two solution of1.3.

bLet. . . , ϕ, ψ, ϕ, ψ, . . .be the eventual prime period two solution of1.3, then, it holds eventually thatϕrϕ/1ϕpandψ rψ/1ψp. Hence,

ϕ1−rϕp0, ψ1−rψp0. 3.2

c Assume that 1.3 has the eventual nonnegative prime period two solution . . . , ϕ, ψ, ϕ, ψ, . . . ,then eventually

ϕ rψ

1ψp, ψ rϕ

1ϕp. 3.3

Obviously,ϕ 0 impliesψ 0 or vice versa. This is impossible. Soϕψ > 0. It is easy to see from3.3thatϕandψsatisfy the equation

gy1ypp−11−r2_yp_rp_yp_{0, 3.4}

that is,ϕandψare two distinct positive roots ofgy 0. From3.4we can see thatgy 0 does not have two distinct positive roots at all whenr ≤1, alternatively,1.3does not have the nonnegative prime period two solution at all whenr ≤1. Therefore, we assumer >1 in the following.

Let 1ypxin3.4, then the equationfx xp−r2_xp−1_rp_{x−1 0, x >1,has}

at least two distinct positive roots. By simple calculation, one has

f_{x px}p−2 _x−

p−1r2 p

rp_{, fx pp−1x}p−3 _x−

p−2r2 p

. 3.5

Ifp−1r2/p≤1, we can seefx>0 for allx∈1,∞. This means thatfxis strictly increasing in the interval1,∞and hence the equation,fx xp−r2_xp−1_rp_{x−1 0, x >1,}

cannot have two distinct positive roots. So, next we considerp−1r2_{/p > 1, which implies} p >1. Denotex0 p−2r2/p. We need to discuss several cases, respectively, as follows.

Case 1. It holds thatx0 ≤1. Thenfx >0 for allx∈1,∞, hence,fxis convex. Again, f1 1−r2_{<0. So it is impossible forfxto have two distinct positive roots.}

Case 2. It holds thatx0 > 1 and fx0 rp1−p−2/pp−2rp−2 ≥ 0. Then, for x > x0, f_{x>0 and sofx> fx}

0≥0; for 1< x < x0,fx<0 and sofx> fx0≥0. At this time, one always hasfx≥fx0≥0. Thenfxcannot have two distinct positive roots.

Case 3. It holds thatx0 >1,fx0<0 andf1 p−p−1r2rp≤0. Then, for 1< x < x0, f_{x<0 and sofx< f1≤0 and hencefx}

0< fx< f1 1−r2<0, that is,fx 0 has no solutions for 1 < x < x0; forx > x0,fx > 0, that is, fxis convex forx > x0. Noticingfx0<0, it is also impossible forfxto have two distinct positive roots forx > x0.

dLet . . .,ϕ,ψ,ϕ, ψ,. . . be the eventual nonnegative prime period two solution of

1.3, then, it is eventually true that

ϕ rψ

1ϕp, ψ rϕ

1ψp. 3.6

It is easy to see thatϕ >0 andψ >0. So, we have

ϕp_1ϕpp1_rp_1−r2_ϕp_0,

ψp_1ϕpp1_rp_1−r2_ψp_0, 3.7

that is,ϕandψ are two distinct positive roots ofhx xp1xpp1rp1−r2_xp _0.

Obviously, whenr≤1, thehx 0 has no positive roots.

Now letr >1 and set 1xpy. Then the function,fy yp2_−yp1_yrp_−rp2_{, y >1,} has at least two distinct positive roots. However,fy ypp2y−p1 rp>0 for any y∈1,∞, which indicates thatfyis strictly increasing in the interval1,∞. This implies that the functionfydoes not have two distinct positive roots at all in the interval1,∞. In turn,1.3does not have the prime period two solution whenr >1.

4. Existence of Oscillatory Solution

For the oscillatory solution of1.3, we have the following results.

Theorem 4.1. Assumer > 1,kis odd andlis even. Then, there exist solutions{yn}∞n−max{k,l} of 1.3which oscillate abouty₂ r−11/pwith semicycles of length one.

Proof. We only prove the case wherek ≥ l the proof of the case wherek < lis similar and will be omitted. Choose the initial values of1.3such that

y−k, y−k2, . . . , y−1≤y₂, y−k1, y−k3, . . . , y0≥y2, 4.1

or

y−k, y−k2, . . . , y−1≥y₂ y−k1, y−k3, . . . , y0≤y2. 4.2

We will only prove the case where4.1holds. The case where4.2holds is similar and will be omitted. According to1.3, one can see that

y1 ry₋k

1yp_−l < y−k≤y2, y2 ry₋k1

1yp_1−l ≥y−k1≥y2, ..

.

yk ry−1

1yp_k−1−l < y−1≤y2, yk1 ry0

1yp_k−l ≥y0≥y2.

4.3

5. Existence of Unbounded Solution

With respect to the unbounded solutions of1.3, the following results are derived.

Theorem 5.1. Assumer >1,kis odd, andlis even, then1.3possesses unbounded solutions.

Proof. We only prove the case wherek ≥ l the proof of the case wherek < lis similar and will be omitted. Choose the initial values of1.3such that

0< y₋k, y−k2, . . . , y−1< y2, y−k1, y−k3, . . . , y0> y2. 5.1

In the following, assumej ≥ −k. From the proof ofTheorem 4.1, one can see thatyj < y₂

whenjis odd and thatyj> y₂forjeven. Together with

yjk1i1

ryjk1i

1y_kp_−ljk1i, 5.2

It is derived that

0< yjk1i1< yjk1i< y₂ forj odd,

yjk1i1 > yjk1i> y2 forj even.

5.3

So,{yjk1i}∞i0is decreasing forjodd whereas{yjk1i}∞i0is increasing forjeven. Let

lim

i→ ∞yjk1iαj, ∀j ≥ −k, 5.4

then one has

10≤αj< y2forjodd andy2 < αj ≤ ∞forjeven,

2αjαjk1i, j∈ {−k,−k1, . . .}, i∈ {0,1, . . .}.

Now, eitherαj∞for some evenjin which case the proof is complete, orαj<∞for

all evenj. We shall prove that this latter is impossible. In fact, we prove thatαj ∞for all

evenj.

Assumeαj < ∞for some evenj ≥ −k, then one has, by5.2,αj rαj/1αp_jk−l.

Noticing1, one hence further getsαjk−l y₂. Howeverjk−lis odd, according to1,

αjk−l< y₂. This is a contradiction.

Therefore, αj ∞ for any even j. Accordingly, {yjk1i1} are unbounded

subsequences of this solution{yn}of1.3for evenj. Simultaneously, for oddj, we get

αj lim

i→ ∞yjk1iilim→ ∞yk1jk1iilim→ ∞

ryjk1i

1y_kp_−ljk1i 0. 5.5

The proof is complete.

6. Existence and Asymptotic Behavior of Nonoscillatory Solution

In this section, we consider the existence and asymptotic behavior of nonoscillatory solution of1.3. Because all solutions of1.3are nonnegative, relative to the zero equilibrium point y₁, every solution of 1.3is a positive semicycle, a trivial nonoscillatory solution! Thus, it suffices to consider the positive equilibriumy₂ when studying the nonoscillatory solutions of1.3. At this time,r >1.

Firstly, we have the following results.

Theorem 6.1. Every nonoscillatory solution of 1.3with respect toy₂approachesy₂.

Proof. Let{yn}∞n−max{k,l}be any one nonoscillatory solution of1.3with respect toy2. Then, there exists ann0 ∈ {−max{k, l},−max{k, l}1, . . .}such that

yn≥y₂ forn≥n0 6.1

or

yn< y2 forn≥n0. 6.2

We only prove the case where6.1holds. The proof for the case where6.2holds is similar and will be omitted. According to6.1, forn≥n0max{k, l}, one has

yn1 ryn−k

1y_np_−l ≤yn−k. 6.3

So,{yjk1i}∞i0is decreasing forj ∈ {−max{k, l},−max{k, l}1, . . . ,−1,0}with upper bound y₂. Hence, limi→ ∞yjk1iexists and is finite. Denote

lim

i→ ∞yjk1iαj, j ∈ {−max{k, l},−max{k, l}1, . . . ,−1,0}. 6.4

Thenαj≥y₂. Taking limits on both sides of1.3, we can derive

αjy₂ forj∈ {−max{k, l},−max{k, l}1, . . . ,−1,0}, 6.5

which shows limn→ ∞yny2and completes this proof.

A problem naturally arises: are there nonoscillatory solutions of1.3? Next, we will positively answer this question. Our result is as follows.

Theorem 6.2. However 1.3 possesses asymptotic solutions with a single semicycle (positive semicycle or negative semicycle).

refer also to13. Consider a general real nonlinear difference equation of orderm≥ 1 with the form

Fxn, xn1, . . . , xnm 0, 6.6

whereF:Rm1 _{→R, n∈N}

0. Letϕnandψnbe two sequences satisfyingψn>0 andψnoϕn

asn → ∞. Thenmaybe under certain additional conditions, for any given >0, there exist a solution{xn}∞n−1of6.6and ann0 ∈Nsuch

ϕn− ψn≤xn≤ϕn ψn, n≥n0 . 6.7

Denote

X xn:ϕn− ψn≤xn≤ϕn ψn, n≥n0 , 6.8

which is called an asymptotic stripe. So, ifxn ∈X , then it is implied that there exists a real

sequenceCnsuch thatxnϕnCnψnand|Cn| ≤ forn≥n0 . We now state the inclusion theorem12.

Lemma 6.3. Let Fω0, ω1, . . . , ωm be continuously differentiable when ωi yni, for i 0,1, . . . , m, andyn∈X1. Let the partial derivatives ofFsatisfy

Fωi

yn, yn1, . . . , ynm∼Fωi

ϕn, ϕn1, . . . , ϕnm 6.9

asn → ∞uniformly inCjfor|Cj| ≤1,n≤j ≤nm, as far asFωi/≡0. Assume that there exist a sequencefn>0and constantsA0, A1, . . . , Amsuch that

Fϕn, . . . , ϕnmofn,

ψniFwi

ϕn, . . . , ϕnm∼Aifn

6.10

fori0,1, . . . , masn → ∞, and suppose there exists an integers, with0≤s≤m, such that

|A0|· · ·|As−1||As1|· · ·|Am|<|As|. 6.11

Then, for sufficiently largen, there exists a solution{xn}∞n−1of6.6satisfying6.7.

Proof ofTheorem 6.2. We only prove the case wherek ≥lthe proof of the case wherek < lis similar and will be omitted. Putxn yn−yy₂ is denoted intoyfor short. Then1.3is

transformed into

xnk1y

1xnk−lyp−rxny0, n−k,−k1, . . . . 6.12

An approximate equation of6.12is

znk1

provided thatxn → 0 asn → ∞. The general solution of6.13is

Now choosefntλn. Noting

Fϕn, ϕn1, . . . , ϕnk1

tnk1_y1tnk−l_yp_−rtn_y

tnk1_yrpyp−1_tnk−l_Ot2nk−l_−rtn_y

tn_rtk1_pr−1tk−l_−rtnk1_yOt2nk−l

tnk1_yOt2nk−l_,

6.20

we haveFϕn, ϕn1, . . . , ϕnk1 ofn.Again,

ψniFωi

ϕn, ϕn1, . . . , ϕnk1

∼Aifn, i0, k−l, k1, 6.21

where

A0−r, Ak−l pr−1tλk−l,

Ak1rtλk1.

6.22

Therefore, one has

|A1|· · ·|Ak1|pr−1tλk−lrtλk1< pr−1tk−lrtk1r|A0|. 6.23

Up to here, all conditions ofLemma 6.3withmk1 ands 0 are satisfied. Accordingly, we see that, for arbitrary ∈ 0,1and for sufficiently largen, sayn ≥ N0 ∈ N,6.12has a solution{xn}∞n−k in the stripeϕn− ψn ≤ xn ≤ ϕn ψn, n ≥ N0, whereϕnandψnare as

defined in6.16. Becauseϕn− ψn > ϕn−ψn tn−tλn > 0,xn > 0 forn≥ N0. Thus,1.3

has a solution{yn}∞n−ksatisfyingyn xny > yforn≥ N0. Since1.3is an autonomous equation,{ynN0k}

∞

n−kstill is its solution, which evidently satisfiesynN0k > yforn≥ −k. Therefore, the proof is complete.

Remark 6.4. If we takeϕn −tn in6.16, thenϕn ψn < −tntλn < 0. At this time,1.3

possesses solutions{yn}∞n−kwhich remain below its equilibrium for alln≥ −k, that is,1.3

has solutions with a single negative semicycle.

Remark 6.5. The appropriate equation6.12is just the linearized equation of1.3associated withy₂.

Acknowledgments

This work of the second author is partly supported by NNSF of ChinaGrant: 10771094and the Foundation for the Innovation Group of Shenzhen UniversityGrant: 000133. Y. Wang work is supported by School Foundation of JiangSu Polytechnic UniversityGrant: JS200801.

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