Dynamical Properties in a Fourth Order Nonlinear Difference Equation

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doi:10.1155/2010/679409

Research Article

Dynamical Properties in a Fourth-Order Nonlinear

Difference Equation

Yunxin Chen

1

_{and Xianyi Li}

2

1_{School of Mathematics and Physics, University of South China, Hengyang, Hunan 421001, China} 2_{College of Mathematics and Computational Science, Shenzhen University, Shenzhen, Guangdong 518060,}

China

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

Received 21 October 2009; Accepted 29 March 2010

Academic Editor: Jianshe S. Yu

Copyrightq2010 Y. Chen and X. Li. 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.

The rule of trajectory structure for fourth-order nonlinear diﬀerence equationxn1 xan−2

xn−3/xa_n₋2xn−31,n0,1,2, . . . ,wherea∈0,1and the initial valuesx−3, x−2, x−1, x0∈0,∞,

is described clearly out in this paper. Mainly, the lengths of positive and negative semicycles of its nontrivial solutions are found to occur periodically with prime period 15. The rule is 4,3−,1,2−,2,1−,1,1−in a period. By utilizing this rule its positive equilibrium point is verified to be globally asymptotically stable.

1. Introduction

In this paper we consider the following fourth-order nonlinear diﬀerence equation:

xn1

xa

n−2xn−3

xa

n−2xn−31, n0,1,2, . . . ,

1.1

wherea∈0,1and the initial valuesx₋3, x−2, x−1, x0∈0,∞.

Whena0,1.1becomes the trivial casexn11,n0,1, . . . .Hence, we will assume

in the sequel that 0< a <1.

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In this paper, it is of key for us to find that the lengths of positive and negative semi-cycles of nontrivial solutions of1.1occur periodically with prime period 15 with the rule 4, 3−, 1, 2−, 2, 1−, 1, 1− and in a period. With the help of this rule and utilizing the monotonicity of solution the positive equilibrium point of the equation is verified to be globally asymptotically stable.

Essentially, we derive the following results for solutions of1.1.

Theorem CL. The rule of the trajectory structure of 1.1is that all of its solutions asymptotically approach its equilibrium; furthermore, any one of its solutions is either

1eventually trivial

2nonoscillatory and eventually negative (i.e.,xn≥1) or

3strictly oscillatory with the lengths of positive and negative semi-cycles periodically successively occurring with prime period 15 and the rule to be4,3−,1,2−,2,1−,1,1−

in a period.

It follows from the results stated below that Theorem CL is true. It is easy to see that the positive equilibriumxof 1.1satisfies

x xax

xax1, 1.2

from which one can see that1.1has a unique equilibriumx1.

In the following, we state some main definitions used in this paper.

Definition 1.1. A positive semi-cycle{x_n}∞_n₋₃ of a solution of 1.1consists of a “string” of termsxl, xl1, . . . , xm, all greater than or equal to the equilibriumx, withl ≥ −3 andm ≤ ∞

such that

eitherl−3 orl >−3, xl−1< x ,

eitherm∞or m <∞, x_m1< x.

1.3

A negative semi-cycle of a solution {x_n}∞_n₋₃ of 1.1 consists of a “string” of term xl, xl1, . . . , xm, all less thanx, withl≥ −3 andm≤ ∞such that

eitherl−3 orl >−3, x_l₋1≥x ,

eitherm∞or m <∞, x_m1≥x.

1.4

The length of a semi-cycle is the number of the total terms contained in it.

Definition 1.2. A solution{xn}∞n−3 of1.1is said to be eventually trivial ifxnis eventually

equal tox1; Otherwise, the solution is said to be nontrivial. A solution{x_n}∞_n₋₃of1.1is said to be eventually positivenegativeifxnis eventually greaterlessthanx1.

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2. Three Lemmas

Before drawing a qualitatively clear picture for the solutions of1.1, we first establish three basic lemmas which will play key roles in the proof of our main results.

Lemma 2.1. A solution{xn}∞_n₋₃of 1.1is eventually trivial if and only if

x−3−1x−2−1x−1−1x0−1 0. 2.1

Proof. Suﬃciency. Assume that 2.1 holds. Then it follows from 1.1 that the following conclusions hold:

iifx−3−10, thenxn1 forn≥7;

iiifx₋2−10, thenxn1 forn≥4;

iiiifx−1−10, thenxn1 forn≥5; ivifx0−10, thenxn1 forn≥6.

Necessity. Conversely, assume that

x−3−1x−2−1x−1−1x0−1/0. 2.2

Then one can show that

xn/1 for anyn≥1. 2.3

Assume the contrary that for someN≥1,

xN1 and that xn/1 for −3≤n≤N−1. 2.4

Clearly,

1x_N x a

N−3xN−4

xa

N−3xN−41

, 2.5

which implies thatx_Na₋₃−1x_N₋4−1 0, which contradicts2.4.

Remark 2.2. Lemma2.1actually demonstrates that a solution{xn}∞_n₋₃ of1.1is eventually nontrivial if and only if

x−3−1x−2−1x−1−1x0−1/0. 2.6

Therefore, if a solution{x_n}∞_n₋₃is nontrivial, thenx_n_/1 forn≥ −3.

Lemma 2.3. Let{xn}∞_n₋₃be a nontrivial positive solution of 1.1. Then the following conclusions

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a xn1−1xn−2−1xn−3−1<0forn≥0; b xn1−xn−3xn−3−1<0forn≥0.

Proof. In view of1.1, we can see that

xn1−1−

xa n−2−1

xn−3−1

xa

n−2xn−31

, n0,1, . . . ,

xn1−xn−3

xa

n−21−xn−31xn−3

xa

n−2xn−31 , n0,1, . . .

2.7

from which inequalitiesaandbfollow. So the proof is complete.

Lemma 2.4. There exist nonoscillatory solutions of 1.1, which must be eventually negative. There

do not exist eventually positive non-oscillatory solutions of 1.1.

Proof. Consider a solution of1.1 with x₋3 < 1, x−2 < 1, x−1 < 1, and x0 < 1. We then

know from Lemma2.3athatxn < 1 forn ≥ −3. So, this solution is just a non-oscillatory solution, and furthermore, eventually negative. Suppose that there exist eventually positive nonoscillatory solutions of1.1. Then, there exists a positive integerNsuch thatx_n >1 for n≥N. Thereout, forn≥N3,xn1−1xn−2−1xn−3−1>0. This contradicts Lemma2.3a.

So, there do not exist eventually positive non-oscillatory solutions of1.1, as desired.

3. Main Results and Their Proofs

First we analyze the structure of the semi-cycles of nontrivial solutions of 1.1. Here we confine us to consider the situation of the strictly oscillatory solution of1.1.

Theorem 3.1. Let{x_n}∞_n₋₃be any strictly oscillatory solution of 1.1. Then, the lengths of positive

and negative semi-cycles of the solution periodically successively occur with prime period 15. And in a period, the rule is4,3−,1,2−,2,1−,1,1−.

Proof. By Lemma2.3a, one can see that the length of a positive semi-cycle is not larger than 4, whereas, the length of a negative semi-cycle is at most 3. Based on the strictly oscillatory character of the solution, we see, for some integerp≥0, that one of the following four cases must occur.

Case 1. xp−3>1,xp−2<1,xp−1>1,xp>1.

Case 2. x_p₋3>1,xp−2<1,xp−1>1,xp<1.

Case 3. x_p₋3>1,xp−2<1,xp−1<1,xp>1.

Case 4. xp−3>1,xp−2<1,xp−1<1,xp<1.

If Case1occurs, it follows from Lemma2.3athatx_p₋3>1,xp−2<1,xp−1>1,xp>1, xp1 > 1, xp2 > 1, xp3 < 1, xp4 < 1,xp5 < 1, xp6 > 1, xp7 < 1, xp8 < 1, xp9 > 1,

xp10 >1,xp11 < 1,xp12 > 1,xp13 < 1,xp14 >1,xp15 > 1,xp16 > 1,xp17 >1,xp18 <1,

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xp28 <1,xp29 > 1,xp30 > 1,xp31 > 1,xp32 >1,xp33 < 1,xp34 < 1,xp35 <1,xp36 >1,

xp37 < 1,xp38 < 1,xp39 > 1, xp40 > 1, xp41 < 1,. . ., which means that the rule for the

lengths of positive and negative semi-cycles of the solution of1.1to successively occur is . . . ,4,3−,1,2−,2,1−,1,1−,4,3−,1,2−,2,1−,1,1−, . . . .

If Case2happens, then Lemma2.3atells us thatxp−3 >1,xp−2 <1,xp−1 >1,xp <1, xp1 >1,xp2 >1,xp3>1,xp4 >1,xp5 <1,xp6 <1,xp7 <1,xp8 >1,xp9 <1,xp10 <1,

xp11 >1,xp12 > 1,xp13 < 1,xp14 > 1,xp15 <1,xp16 > 1,xp17 > 1,xp18 >1,xp19 >1,

xp20 <1,xp21 < 1,xp22 < 1,xp23 > 1,xp24 <1,xp25 < 1,xp26 > 1,xp27 >1,xp28 <1,

xp29 >1,xp30 < 1,xp31 > 1,xp32 > 1,xp33 >1,xp34 > 1,xp35 < 1,xp36 <1,xp37 <1,

xp38>1,xp39 <1,xp40<1,xp41>1,xp42 >1,. . .. This shows that the rule for the numbers

of terms of positive and negative semi-cycles of the solution of1.1to successively occur still is. . . ,4,3−,1,2−,2,1−,1,1−,4,3−,1,2−,2,1−,1,1−, . . . .

If Case3happens, then Lemma2.3aimplies thatxp−3>1,xp−2 <1,xp−1 <1,xp >1,

xp1 >1,xp2 <1,xp3>1,xp4 <1,xp5 >1,xp6 >1,xp7 >1,xp8 >1,xp9 <1,xp10 <1,

xp11 <1,xp12 > 1,xp13 < 1,xp14 < 1,xp15 >1,xp16 > 1,xp17 < 1,xp18 >1,xp19 <1,

xp20 >1,xp21 > 1,xp22 > 1,xp23 > 1,xp24 >1,xp25 < 1,xp26 < 1,xp27 >1,xp28 <1,

xp29 <1,xp30 > 1,xp31 > 1,xp32 < 1,xp33 >1,xp34 < 1,xp35 > 1,xp36 >1,xp37 >1,

xp38>1,xp39<1,xp40 <1,xp41 <1,xp42>1,. . . .This shows that the rule for the numbers

of terms of positive and negative semi-cycles of the solution of1.1to successively occur still is. . . ,4,3−,1,2−,2,1−,1,1−,4,3−,1,2−,2,1−,1,1−, . . . .

If Case4 happens, then it is to be seen from Lemma2.3a thatx_p₋3 > 1,xp−2 < 1,

xp−1 < 1,xp < 1,xp1 >1,xp2 < 1,xp3 < 1,xp4 >1,xp5 > 1,xp6 <1,xp7 > 1,xp8 <1,

xp9 > 1,xp10 > 1,xp11 > 1,xp12 > 1,xp13 < 1,xp14 < 1,xp15 < 1,xp16 > 1,xp17 <1,

xp18 <1,xp19 > 1,xp20 > 1,xp21 < 1,xp22 >1,xp23 < 1,xp24 > 1,xp25 >1,xp26 >1,

xp27 >1,xp28 < 1,xp29 < 1,xp30 < 1,xp31 >1,xp32 < 1,xp33 < 1,xp34 >1,xp35 >1,

xp36 <1,xp37 >1,xp38 <1,xp39 >1,xp40 >1,xp41 >1,xp42>1,. . .. This shows that the

rule for the numbers of terms of positive and negative semi-cycles of the solution of1.1to successively occur still is. . . ,4,3−,1,2−,2,1−,1,1−,4,3−,1,2−,2,1−,1,1−. . . .

Hence, the proof is complete.

Now, we present the global asymptotical stable results for1.1.

Theorem 3.2. Assume thata ∈ 0, 1. Then the unique positive equilibrium of 1.1 is globally asymptotically stable.

Proof. When a 0, 1.1 is trivial. So, we only consider the case a > 0 and prove that the positive equilibrium point xof 1.1 is both locally asymptotically stable and globally attractive. The linearized equation of1.1about the positive equilibriumx1 is

yn10·yn0·yn−10·yn−20·yn−3, n0,1, . . . . 3.1

By virtue of3, Remark 1.3.7, page 13, xis locally asymptotically stable. It remains to be verified that every positive solution{xn}∞_n₋₃of1.1converges toxasn → ∞. Namely, we want to prove that

lim

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If the initial values of the solutions satisfy2.1, that is to say, the solution is a trivial solution, then Lemma2.1says that the solution is eventually equal to 1 and of course3.2 holds.

If the solution is a nontrivial solution, then we can further divide the solution into two cases.

anon-oscillatory solution;

boscillatory solution.

If Caseahappens, then it follows from Lemma2.3that the solution is actually an eventually negative one. According to Lemma2.3b, we see thatx4n,x4n−1,x4n−2andx4n−3

are eventually increasing and bounded from the upper by the constant 1. So the limits

lim

and taking the limits on both sides of the above equalities, respectively, one may obtain

L MaL If casebhappens, the solution is strictly oscillatory.

Consider nowxn to be strictly oscillatory about the positive equilibrium pointx of

1.1. By virtue of Theorem3.1, one understands that the lengths of positive and negative semi-cycles of the solution periodically successively occur, and in a period, the rule is 4,3−,1,2−,2,1−,1,1−.

For simplicity, for some integer p ≥ 0, we denote by {xp, xp1, xp2,xp3} the

terms of a positive semi-cycle of length four, followed by {x_p4, xp5, xp6}− negative

semi-cycle with length three, then a positive semi-cycle {xp7}, a negative

semi-cycle {xp8, xp9}−, a positive semi-cycle {xp10, xp11}, a negative semi-cycle {xp12}−, a

positive semi-cycle {x_p13}, and a negative semi-cycle {xp14}−. Namely, the rule for

the lengths of negative and positive semi-cycles to occur successively can be periodi-cally expressed as follows:{xp15n, xp15n1, xp15n2, xp15n3}, {xp15n4, xp15n5, xp15n6}−, {xp15n7},{xp15n8, xp15n9}−,{xp15n10, xp15n11},{xp15n12}−,{xp15n13},{xp15n14}−,

andn0,1, . . . .

From Lemma2.3b, we may immediately obtain the following results:

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Also, the following inequalities hold: similarly proved. Noticing that 0≤a <1 and from that the observation

xp15n6

we know that the first inequality in vii holds. The other inequality in vii can be analogously proved.

Combining the above inequalities, one can derive that

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It follows from3.8that{xp15n3}∞_n0is decreasing with lower bound 1. So, the limit

lim

n→ ∞xp15n3L 3.11

exists and is finite. Accordingly, by view of3.8, we obtain

lim

n→ ∞xp15n10L, nlim→ ∞xp15n14 nlim→ ∞xp15n6 1

L. 3.12

It is easy to see from3.9that{xp15n}∞n0 is decreasing with lower bound 1. So, the

limit

lim

n→ ∞xp15nM 3.13

exists and is finite. Thereout, in light of3.9, one has

lim

n→ ∞xp15n11 M, nlim→ ∞xp15n8 nlim→ ∞xp15n4 1

M. 3.14

It follows from3.10that{x_p15n1}∞n0is decreasing with lower bound 1. So, the limit

lim

n→ ∞xp15n1 N 3.15

exists and is finite. Accordingly, by view of3.10, we obtain

lim

n→ ∞xp15n13N, nlim→ ∞xp15n9 nlim→ ∞xp15n5 1

N. 3.16

Taking the limits on both sides ofxp15n18 x_pa 15n15xp15n14/xa_p15n15xp15n14

1, one hasL Ma1/L/Ma/L1, which gives rise toL1. We further obtain fromi and3.12that

lim

n→ ∞xp15nk1, k3,6,7,10,11,14,15. 3.17

Hence,M1. Therefore,

lim

n→ ∞xp15nk1, k3,4,6,7,8,10,11,14,15. 3.18

It is easy to derive fromvthat 1> xp15n2>1/x_p15n6. Noticing that limn→ ∞xp15n61,

one can see that lim_n_{→ ∞}x_p15n21.

Similarly, taking the limits on both sides of xp15n13 xa_p15n10

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the limits on both sides of xp15n12 xa_p₁₅_n₉ xp15n8/xa_p15n9xp15n8 1, one has

lim_n_{→ ∞}x_p15n121.

Up to now, we have shown that

lim

n→ ∞xp15nk1, k1,2, . . . ,15. 3.19

So, the proof for Theorem3.2is complete.

Remark 3.3. One can see from the process of proofs stated previously that these results in this paper also hold fora1.

Acknowledgment

This work is partly supported by NNSF of Chinagrant: 10771094and the Foundation for the Innovation Group of Shenzhen Universitygrant: 000133.

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