Báo cáo hóa học: " Research Article Dynamical Properties for a Class of Fourth-Order Nonlinear Difference Equations" potx

The key is for us to find that the successive lengths of positive and negative semicycles for non-trivial solutions of this equation periodically occur with same prime period 5.. Althoug

fference Equations

Volume 2008, Article ID 678402, 13 pages

doi:10.1155/2008/678402

Research Article

Dynamical Properties for a Class of Fourth-Order Nonlinear Difference Equations

Dongsheng Li, 1, 2 Pingping Li, 1 and Xianyi Li 3

1 Ministry Education Key Laboratory of Modern Agricultural Equipment and Technology,

Jiangsu University, Jiangsu 212013, Zhenjiang, China

2 School of Economics and Management, University of South China, Hunan 421001, Hengyang, China

3 College of Mathematics and Computational Science, Shenzhen University, Shenzhen 518060,

Guangdong, China

Correspondence should be addressed to Xianyi Li, xianyili6590@163.com

Received 6 May 2007; Revised 14 August 2007; Accepted 18 September 2007

Recommended by Jianshe Yu

We consider the dynamical properties for a kind of fourth-order rational difference equations The key is for us to find that the successive lengths of positive and negative semicycles for non-trivial solutions of this equation periodically occur with same prime period 5 Although the pe-riod is same, the order for the successive lengths of positive and negative semicycles is com-pletely different The rule is , 3 , 2−, 3, 2−, 3, 2−, 3, 2−, , or , 2, 1−, 1, 1−, 2, 1−, 1, 1−, , or , 1, 4−, 1, 4−, 1, 4−, 1, 4−, By the use of the rule, the positive equilibrium point of this

equa-tion is proved to be globally asymptotically stable.

Copyright q 2008 Dongsheng Li 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 and preliminaries

Rational difference equation, as a kind of typical nonlinear difference equations, is always

a subject studied in recent years Especially, some prototypes for the development of the basic theory of the global behavior of nonlinear difference equations of order greater than one come from the results of rational difference equations For the systematical investiga-tions of this aspect, refer to the monographs1 3, the papers 4 9, and the references cited therein

Motivated by the work5 7, we consider in this paper the following fourth-order ratio-nal difference equation:

x n1 F



x n , x n−1 , x n−2 , x n−3

Gx n , x n−1 , x n−2 , x n−3 , n  0, 1, , 1.1

Fx, y, z, w  x u y v  x u z k  x u w j  y v z k  y v w j  z k w j  x u y v z k w j  1  a,

Gx, y, z, w  x u  y v  z k  w j  x u y v z k  x u y v w j  x u z k w j  y v z k w j  a, 1.2

the parametera ∈ 0, ∞, u ∈ 0, 1, v, k, j ∈ 0, ∞, and the initial values x−3, x−2, x−1, x0 ∈

0, ∞.

Mainly, by analyzing the rule for the length of semicycle to occur successively, we clearly describe out the rule for the trajectory structure of its solutions With the help of several key lemmas, we further derive the global asymptotic stability of positive equilibrium of1.1 To the best of our knowledge,1.1 has not been investigated so far; therefore, it is theoretically meaningful to study its qualitative properties

It is easy to see that the positive equilibriumx of 1.1 satisfies

x  1 a  x uv  x uk  x uj  x vk  x vj  x kj  x uvkj

a  x u  x v  x k  x j  x uvk  x uvj  x ukj  x vkj . 1.3

From this, we see that1.1 possesses a unique positive equilibrium x  1.

It is essential in this note for us to obtain the general rule for the trajectory structure of solutions of1.1 as follows

Theorem 1.1 The rule for the trajectory structure of any solution of 1.1  is as follows:

i the solution is either eventually trivial,

ii or the solution is eventually nontrivial,

1 and further, either the solution is eventually positive nonoscillatory,

2 or the solution is strictly oscillatory, and moreover, the successive lengths for

posi-tive and negaposi-tive semicycles occur periodically with prime period 5, and the rule is , 3, 2−, 3, 2−, 3, 2−, 3, 2−, , or , 2, 1−, 1, 1−, 2, 1−, 1, 1−, , or , 1, 4−,

1, 4−, 1, 4−, 1, 4−,

The positive equilibrium point of 1.1 is a global attractor of all its solutions.

It follows from the results stated in the sequel thatTheorem 1.1is true

For the corresponding concepts in this paper, see3 or the papers 5 7

2 Nontrivial solution

Theorem 2.1 A positive solution {x n}∞n−3 of 1.1 is 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, according to 1.1, we know that the following conclusions are true: ifx−3 1, x−2 1, x−1 1, or x0 1, then xn  1 for n ≥ 1.

Necessity Conversely, assume that



x−3− 1x−2− 1x−1− 1x0− 1/ 0. 2.2

Then, we can show that x n / 1 for any n ≥ 1 For the sake of contradiction, assume that for

someN ≥ 1,

Clearly,

1 xN F



x N−1 , x N−2 , x N−3 , x N−4

From this, we can know that

0 xN− 1 



x u N−1− 1x v

N−2− 1x k

N−3− 1x j N−4− 1

Gx N−1 , x N−2 , x N−3 , x N−4 , 2.5 which implies thatx N−1  1, xN−2  1, xN−3  1, or xN−4  1 This contradicts 2.3

Remark 2.2. Theorem 2.1actually demonstrates that a positive solution{xn}∞n−3of1.1 is even-tually nontrivial ifx−3− 1x−2− 1x−1− 1x0− 1 / 0 So, if a solution is a nontrivial one, then

x n / 1 for any n ≥ − 3.

3 Several key lemmas

We state several key lemmas in this section, which will be important in the proofs of the sequel DenoteN k  {k, k  1, } for any integer k.

Lemma 3.1 If the integer i ∈ N−3, then

x n1 − xi K



x n , x n−1 , x n−2 , x n−3 , x i

Gx n , x n−1 , x n−2 , x n−3  , n  0, 1, , 3.1

where

Kx, y, z, w, p

1− x u p1 y v z k  y v w j  z k w j

 a1− xix u − py v  z k  w j  y v z k w j

.

3.2

Lemma 3.2 If the integer i ∈ N−3, t  1, 2, , then

1− xn1 x1/u t



x n , x n−1 , x n−2 , x n−3 , x i

Gx n , x n−1 , x n−2 , x n−3  , n  0, 1, , 3.3

where

Mx, y, z, w, p x u − p1/u t

1 y v z k  y v w j  z k w j

 a1− p1/u t

1− x u p1/u t

y v  z k  w j  y v z k w j

Lemma 3.3 If the integer i ∈ Nư3, t  1, 2, , then

x n1 ư x1/u t



x n , x nư1 , x nư2 , x nư3 , x i

Gx n , x nư1 , x nư2 , x nư3  , n  0, 1, , 3.5

where

Nx, y, z, w, p 1ư x u p1/u t

1 y v z k  y v w j  z k w j a1ư p1/u t

x u ư p1/u t

y v  z k  w j  y v z k w j. 3.6

The results of Lemmas3.1,3.2, and3.3can be easily obtained from1.1, and so we omit their proofs here

Lemma 3.4 Let {x n}∞nư3 be a positive solution of 1.1 which is not eventually equal to 1, then the

following conclusions are valid:

a xn1 ư 1xn ư 1xnư1 ư 1xnư2 ư 1xnư3 ư 1 > 0, for n ≥ 0;

b xn1 ư xnxn ư 1 < 0, for n ≥ 0;

c xn1 ư xnư1xnư1 ư 1 < 0, for n ≥ 0;

d x n1 ư x nư2 x nư2 ư 1 < 0, for n ≥ 0;

e xn1 ư xnư3xnư3 ư 1 < 0, for n ≥ 0.

Proof First, let us investigatea According to 1.1, it follows that

x n1ư 1 



x u

nư 1x v

nư1ư 1x k

nư2ư 1x j nư3ư 1

Gx n , x nư1 , x nư2 , x nư3 , n  0, 1, 3.7 So,



x n1ư 1x u

nư 1x v

nư1ư 1x k

nư2ư 1x nư3 j ư 1> 0. 3.8 Noting thatu ∈ 0, 1, v, k, j ∈ 0, ∞, one has x u

n ư 1xn ư 1 > 0, x v

nư1 ư 1xnư1 ư 1 > 0,

x k

nư2 ư 1xnư2 ư 1 > 0, and x j nư3 ư 1xnư3 ư 1 > 0 From those, one can easily obtain the result

ofa

Second,b comes From 3.1, we obtain

x n1 ư xn K



x n , x nư1 , x nư2 , x nư3 , x n

Gx n , x nư1 , x nư2 , x nư3  , 3.9 where

Kx n , x nư1 , x nư2 , x nư3 , x n

1ư x u1

n



1 x v nư1 x k nư2  x v nư1 x nư3 j  x k

nư2 x j nư3 a1ư xn

x u

n ư x nx v

nư1  x k nư2  x j nư3  x v

nư1 x k nư2 x j nư3

1ư x u1

n



1 x v nư1 x k nư2  x v nư1 x nư3 j  x k

nư2 x j nư3 a1ư xn

1ư x1ưu

x u

n

x v nư1  x k nư2  x j nư3  x v

nư1 x k nư2 x j nư3.

3.10

Fromu ∈ 0, 1 and {x n}∞nư3, being eventually not equal to 1, one can see that



1ư x u1

n



1ư xn> 0, 1ư x1ưu

n



1ư xn≥ 0, Gx n , x nư1 , x nư2 , x nư3

> 0. 3.11 This tells us thatxn1 ư xn1 ư xn > 0, n  0, 1, That is, xn1 ư xnxn ư 1 < 0, n  0, 1,

So, the proof ofb is complete

Third, let us provec From 3.1 one has

x n1 ư xnư1 K



x n , x nư1 , x nư2 , x nư3 , x nư1

Gx n , x nư1 , x nư2 , x nư3  , 3.12 where

Kx n , x nư1 , x nư2 , x nư3 , x nư1

1ư x u

n x nư1

1 x v nư1 x k nư2  x v nư1 x j nư3  x k

nư2 x j nư3 a1ư xnư1

x u

n ư xnư1x v

nư1  x k nư2  x j nư3  x v

nư1 x k nư2 x j nư3.

3.13 From3.3, one gets

1ư xn x1/u

nư1 M



x nư1 , x nư2 , x nư3 , x nư4 , x nư1

Gx nư1 , x nư2 , x nư3 , x nư4 , 3.14

where

Mx nư1 , x nư2 , x nư3 , x nư4 , x nư1

 a1ư x1/u

nư1



x u nư1 ư x1/u nư1



1 x v nư2 x k nư3  x v nư2 x j nư4  x k

nư3 x j nư4

1ư x u1/u nư1 x v

nư2  x k nư3  x j nư4  x v

nư2 x k nư3 x j nư4

1ư x nư1 1ưu2/ux u

nư1



1 x v nư2 x k nư3  x v nư2 x nư4 j  x k

nư3 x nư4 j  a1ư x1/u

nư1



1ư x u1/u nư1 x v

nư2  x k nư3  x j nư4  x v

nư2 x k nư3 x j nư4.

3.15

According tou ∈ 0, 1 and {x n}∞nư3, being eventually not equal to 1, one arrives at



1ư x1/u nư1



1ư xnư1> 0, 1ư x u1/u nư1 1ư xnư1> 0,



From3.14, 3.15, and 3.16, we know that 1 ư xn x1/u

nư1 1 ư xnư1 > 0 So, we can get

imme-diately



1ư x u

n x nư1

From3.5, one can have

x n ư x1/u nư1 N



x nư1 , x nư2 , x nư3 , x nư4 , x nư1

Gx nư1 , x nư2 , x nư3 , x nư4 , 3.18

Nx nư1 , x nư2 , x nư3 , x nư4 , x nư1

1ư x nư1 u1/u1 x v

nư2 x k nư3  x v nư2 x nư4 j  x k

nư3 x nư4 j  a1ư x1/u

nư1



x u nư1 ư x1/u nư1



x v nư2  x k nư3  x j nư4  x v

nư2 x k nư3 x j nư4

1ư x nư1 u1/u1 x v

nư2 x k nư3  x v nư2 x nư4 j  x k

nư3 x nư4 j  a1ư x1/u

nư1



1ư x 1ưu nư12/ux u

nư1



x v nư2  x k nư3  x j nư4  x v

nư2 x k nư3 x j nư4.

3.19

From3.16, 3.18, and 3.19, we can obtain xn ư x1/u

nư1 1 ư xnư1 > 0, that is,



x u

By virtue of3.12, 3.13, 3.17, and 3.20, we see that c is true

The proofs ofd and e are similar to those of c The proof for this lemma is complete

Lemma 3.5 Let {x n}∞nư3 be a positive solution of 1.1 which is not eventually equal to 1, then xn1ư

x nư4xnư4 ư 1 < 0, for n ≥ 1.

Proof From3.1, one has

x n1 ư xnư4 K



x n , x nư1 , x nư2 , x nư3 , x nư4

Gx n , x nư1 , x nư2 , x nư3  , n  0, 1, , 3.21

where

Kx n , x nư1 , x nư2 , x nư3 , x nư4

1ư x u

n x nư4

1 x v nư1 x k nư2  x v nư1 x j nư3  x k

nư2 x j nư3 a1ư xnư4

x u

n ư x nư4x v

nư1  x k nư2  x nư3 j  x v

nư1 x k nư2 x j nư3.

3.22 From3.3, one has

1ư x nư3 x1/u4



x nư4 , x nư5 , x nư6 , x nư7 , x nư4

Gx nư4 , x nư5 , x nư6 , x nư7 , 3.23 where

Mx nư4 , x nư5 , x nư6 , x nư7 , x nư4

x u nư4 ư x1/u4

nư4



1 x v nư5 x k nư6  x v nư5 x j nư7  x k

nư6 x j nư7 a1ư x1/u4

nư4



1ư x u1/u4

nư4



x v nư5  x k nư6  x nư7 j  x v

nư5 x k nư6 x j nư7.

3.24 Noticing thatu ∈ 0, 1, we have



x u nư4 ư x1/u4

nư4



1ư xnư4 x u

nư4



1ư x 1ưu nư45/u41ư xnư4≥ 0,



1ư x1/u4

nư4



1ư xnư4> 0, 1ư x u1/u4

nư4



1ư xnư4> 0. 3.25

From3.23, 3.24, and 3.25, we know that 1 − x n−3 x1/u4

n−4 1 − x n−4  > 0 So,



1− x u n−3 x1/u3

n−4



From3.5, one can recognize that

x n−3 − x1/u4



x n−4 , x n−5 , x n−6 , x n−7 , x n−4

Gx n−4 , x n−5 , x n−6 , x n−7 , 3.27 where

Nx n−4 , x n−5 , x n−6 , x n−7 , x n−4

1− x u1/u n−4 41 x v n−5 x k n−6  x v n−5 x j n−7  x k n−6 x j n−7 a1− x1/u4

n−4



x u n−4 − x1/u4

n−4



x v n−5  x k n−6  x j n−7  x v

n−5 x k n−6 x j n−7.

3.28 From3.25, 3.27, and 3.28, we derive xn−3 − x1/u4

n−4 1 − xn−4 > 0 So,



x u n−3 − x1/u3

n−4



Equation3.5 shows that

x n−2 − x1/u3



x n−3 , x n−4 , x n−5 , x n−6 , x n−4

Gx n−3 , x n−4 , x n−5 , x n−6 , 3.30

where

Nx n−3 , x n−4 , x n−5 , x n−6 , x n−4

1−xu n−3 x1/u3

n−4



1xv n−4 x k n−5 x v n−4 x j n−6 x k

n−5 x j n−6 a1− x1/u3

n−4



x u n−3 − x1/u3

n−4



x v n−4  x k n−5  x j n−6  x v

n−4 x k n−5 x j n−6.

3.31

By using3.26, 3.29, 3.30, and 3.31, and noting that 1−x1/u3

n−4 1−xn−4 > 0 when u ∈ 0, 1,

we getxn−2 − x1/u3

n−4 1 − xn−4 > 0 Hence,



x u n−2 − x1/u2

n−4



It follows from3.3 that

1− xn−2 x1/u3



x n−3 , x n−4 , x n−5 , x n−6 , x n−4

Gx n−3 , x n−4 , x n−5 , x n−6 , 3.33

where

Mx n−3 , x n−4 , x n−5 , x n−6 , x n−4

x u n−3 − x1/u3

n−4



1 x v n−4 x k n−5  x v n−4 x j n−6  x k

n−5 x j n−6 a1− x1/u3

n−4



1− x u

n−3 x1/u3

n−4



x v n−4  x k n−5  x n−6 j  x v

n−4 x k n−5 x j n−6.

3.34

By virtue of3.26, 3.29, 3.33, and 3.34, as well as 1 − x1/u3

n−4 1 − xn−4 > 0 for u ∈ 0, 1,

one has1 − xn−2 x1/u3

n−4 1 − xn−4 > 0 Accordingly,



1− x u n−2 x1/u2

n−4



Equation3.3 instructs us that

1− xn−1 x1/u2



x n−2 , x n−3 , x n−4 , x n−5 , x n−4

Gx n−2 , x n−3 , x n−4 , x n−5 , 3.36 where

Mx n−2 , x n−3 , x n−4 , x n−5 , x n−4

x u n−2 − x1/u2

n−4



1 x v n−3 x k n−4  x v n−3 x j n−5  x k

n−4 x j n−5 a1− x1/u2

n−4



1− x u

n−2 x1/u2

n−4



x v n−3  x k n−4  x n−5 j  x v

n−3 x k n−4 x j n−5.

3.37

By virtue of 3.32, 3.35, 3.36, and 3.37, together with 1 − x1/u2

n−4 1 − x n−4  > 0 when

u ∈ 0, 1, one sees that 1 − x n−1 x1/u2

n−4 1 − xn−4 > 0 So,



1− x u n−1 x1/u n−4



From3.5, one obtains

x n−1 − x1/u2



x n−2 , x n−3 , x n−4 , x n−5 , x n−4

Gx n−2 , x n−3 , x n−4 , x n−5 , 3.39 where

Nx n−2 , x n−3 , x n−4 , x n−5 , x n−41− x u

n−2 x1/u2

n−4



1 x v n−3 x k n−4  x v n−3 x j n−5  x k

n−4 x j n−5a1− x1/u2

n−4



x u n−2 − x1/u2

n−4



x v n−3  x k n−4  x j n−5  x v

n−3 x k n−4 x n−5 j .

3.40

By virtue of 3.32, 3.35, 3.39, and 3.40, in addition to 1 − x1/u2

n−4 1 − xn−4 > 0 when

u ∈ 0, 1, one can see that x n−1 − x1/u2

n−4 1 − xn−4 > 0 Thus,



x u n−1 − x1/u n−4



From3.3, we can see that

1− xn x1/u

n−4 M



x n−1 , x n−2 , x n−3 , x n−4 , x n−4

Gx n−1 , x n−2 , x n−3 , x n−4 , 3.42 where

Mx n−1 , x n−2 , x n−3 , x n−4 , x n−4

x u n−1 − x1/u n−4



1 x v n−2 x k n−3  x v n−2 x j n−4  x k

n−3 x j n−4 a1− x1/u

n−4



1− x u

n−1 x1/u n−4



x v n−2  x k n−3  x j n−4  x v

n−2 x k n−3 x n−4 j .

3.43

Utilizing3.38, 3.41, 3.42, and 3.43, and adding 1 − x1/u

n−4 1 − xn−4 > 0 when u ∈ 0, 1,

we know that the following is true:1 − xn x1/u

n−4 1 − xn−4 > 0 So,



1− x u

n x n−4

Similar to3.44, by virtue of 3.5, 3.38, 3.41, and 1 − x1/u

n−4 1 − x n−4  > 0 when u ∈ 0, 1,

we know thatxn − x1/u

n−4 1 − xn−4 > 0 is true So,



x u

From3.21, 3.22, 3.44, and 3.45, one knows that the following is true:



This shows thatLemma 3.5is true

4 Oscillation and nonoscillation

Theorem 4.1 There exist nonoscillatory solutions of 1.1  with x−3, x−2, x−1, x0 ∈ 1, ∞, which

must be eventually positive There are not eventually negative nonoscillatory solutions of 1.1.

Proof Consider a solution of 1.1 with x−3, x−2, x−1, x0 ∈ 1, ∞ We then know from

Lemma 3.4a that xn > 1 for n ∈ N−3 So, this solution is just a nonoscillatory solution and

it is, furthermore, eventually positive

Suppose that there exist eventually negative nonoscillatory solutions of1.1 Then, there exists a positive integerN such that x n < 1 for n ≥ N Thereout, for n ≥ N  3, x n1 − 1xn− 1xn−2− 1xn−3 − 1  0 This contradictsLemma 3.4a So, there are not eventually negative nonoscillatory solutions of1.1, as desired

5 Rule of cycle length

Theorem 5.1 Let {x n}∞−3be a strictly oscillatory solution of 1.1, then the rule for the lengths of

posi-tive and negaposi-tive semicycles of this solution to occur successively is , 3, 2−, 3, 2−, 3, 2−, 3, 2−, ,

or , 2, 1−, 1, 1−, 2, 1−, 1, 1−, , or , 1, 4−, 1, 4−, 1, 4−, 1, 4−,

Proof ByLemma 3.4a, one can see that the length of a negative semicycle is at most 4, and that of a positive semicycle is at most 3 On the basis of the strictly oscillatory character of the solution, we see, for some integerp ≥ 0, that one of the following sixteen cases must occur:

1 x p > 1, x p1 > 1, x p2 > 1, x p3 > 1;

2 xp > 1, x p1 > 1, x p2 > 1, x p3 < 1;

3 xp > 1, x p1 > 1, x p2 < 1, x p3 > 1;

4 x p > 1, x p1 > 1, x p2 < 1, x p3 < 1;

5 xp > 1, x p1 < 1, x p2 > 1, x p3 > 1;

6 xp > 1, x p1 < 1, x p2 > 1, x p3 < 1;

7 xp > 1, x p1 < 1, x p2 < 1, x p3 > 1;

8 xp > 1, x p1 < 1, x p2 < 1, x p3 < 1;

9 xp < 1, x p1 > 1, x p2 > 1, x p3 > 1;

10 xp < 1, x p1 > 1, x p2 > 1, x p3 < 1;

11 xp < 1, x p1 > 1, x p2 < 1, x p3 > 1;

12 x p < 1, x p1 > 1, x p2 < 1, x p3 < 1;

13 xp < 1, x p1 < 1, x p2 > 1, x p3 > 1;

14 xp < 1, x p1 < 1, x p2 > 1, x p3 < 1;

15 xp < 1, x p1 < 1, x p2 < 1, x p3 > 1;

16 xp < 1, x p1 < 1, x p2 < 1, x p3 < 1.

If case1 occurs, of course, it will be a nonoscillatory solution of 1.1

If case2 occurs, it follows fromLemma 3.4a that xp4 < 1, x p5 > 1, x p6 > 1, x p7 >

1, x p8 < 1, x p9 < 1, x p10 > 1, x p11 > 1, x p12 > 1, x p13 < 1, x p14 < 1, x p15 > 1, x p16 >

1, x p17 > 1, x p18 < 1, x p19 < 1,

This means that the rule for the lengths of positive and negative semicycles of the solu-tion of1.1 to occur successively is

, 3, 2−, 3, 2−, 3, 2−, 3, 2−, 3, 2−, 3, 2−, 3, 2−, 3, 2− . 5.1

If case3 occurs, it follows fromLemma 3.4a that x p4 < 1, x p5 > 1, x p6 > 1, x p7 < 1,

x p8 > 1, x p9 < 14, x p10 > 1, x p11 > 1, x p12 < 1, x p13 > 1, x p14 < 1, x p15 > 1, x p16 > 1,

x p17 < 1, x p18 > 1, x p19 < 1, , which means that the rule for the lengths of positive and

negative semicycles of the solution of1.1 to occur successively is

, 2, 1−, 1, 1−, 2, 1−, 1, 1−, 2, 1−, 1, 1−, 2, 1−, 1, 1−, 5.2

If case 8 is reached, Lemma 3.4a tells us that xp4 < 1, x p5 > 1, x p6 < 1, x p7 < 1,

x p8 < 1, x p9 < 1, x p10 > 1, x p11 < 1, x p12 < 1, x p13 < 1, x p14 < 1, x p15 > 1, x p16 < 1, x p17 <

1,x p18 < 1, x p19 < 1,

This implies that the rule for the lengths of positive and negative semicycles of the solu-tion of1.1 to occur successively is

, 1, 4−, 1, 4−, 1, 4−, 1, 4−, 1, 4−, 1, 4−, 1, 4−, 1, 4−, 5.3

Moreover, the rule for the cases2.3, 3.5, 3.13, 3.15, and 3.17 is the same as that of case2.1 And cases 3.1, 3.3, and 3.15 are completely similar to case 3 except possibly for the first semicycle And cases3.16, 3.18, 3.19, and 3.20 are like case 8 with a possible exception for the first semicycle

Up to now, the proof ofTheorem 5.1is complete

6 Global asymptotic stability

First, we consider the local asymptotic stability for unique positive equilibrium pointx of 1.1

We have the following result