# Global Behavior of a Discrete Anticompetitive System in the Plane

Authors: Lin-Xia Hu
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

The main goal of this paper is to investigate the global asymptotic behavior of the difference systemxn+1=γ1yn/A1+xn,yn+1=β2xn/B2+yn,n=0,1,2,…. with γ1,β2,A1,B2∈(0,∞) and the initial condition (x0,y0)∈[0,∞)×[0,∞). We obtain some global attractivity results of this system for different values of the parameters, which answer the open problem proposed in “Rational systems in the plane, J. Difference Equ. Appl. 15 (2009), 303-323”.


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## Body

## 1. Introduction and Preliminaries

Difference equations or systems have been attracting more and more attention by authors since these models display some complicated character comparing with its analogue differential equations; see [1–11]. In [12], the authors gave a discussion on the following rational system in the plane:(1)xn+1=α1+β1xn+γ1ynA1+B1xn+C1yn,yn+1=α2+β2xn+γ1ynA2+A2xn+C2yn,n=0,1,2,⋯.System (1) contains, as special cases, a large number of equations whose dynamics have not been thoroughly understood yet and there still exists research aspect to be further studied. Several global asymptotic results for some special cases of (1) have been obtained in [13–18]. As a special anticompetitive system of (1), the difference system(2)xn+1=γ1ynA1+xn,yn+1=β2xnA2+yn,n=0,1,2,…,is considered in this paper, where the parameters β2,γ1,A1,A2∈(0,∞) and the initial value (x0, y0)∈R+2=[0,∞)×[0,∞).In [12], system is labeled as (16,16) and is concerned as an open problem (Open Problem 7) which asked for determining the boundedness of its solutions, the local stability of its equilibrium, the existence of prime period-two solutions, and the global character of system (2). To answer the open problem, the main goal of this paper is to study the dynamic behavior of system (2).As far the definition of stability and the method of linearized stability, one can see [19–21].


## 2. Linearized Stability

The equilibrium(x¯,y¯) of system (2) is the intersection of the two following curves:(3)C1:x=γ1yA1+x,C2:y=β2xA2+y.The slopes of the tangent line of the two curves at the origin (0,0) are(4)kC1=A1γ1,kC2=β2A2.Whenβ2γ1≤A1A2, kC1≥kC2, the two curves C1 and C2 have a unique intersection in R+2, that is (0,0). When β2γ1>A1A2, kC1<kC2, the curves C1 and C2 have also another intersection E+=(x¯,y¯) locating in the first quadrant, satisfying (A1+x¯)(A2+y¯)=β2γ1. So we have the following lemma.Lemma 1.
Ifβ2γ1≤A1A2, then system (2) has a unique fixed point E0=(0,0). If β2γ1>A1A2, then system (2) has another positive fixed point E+=(x¯,y¯) locating in the first quadrant, satisfying (A1+x¯)(A2+y¯)=β2γ1.The linearized system of (2) about E0 is(5)xn+1=γ1A1yn,yn+1=β2A2xn,and its characteristic equation is λ2-β2γ1/A1A2=0. So λ1,2=±β2γ1/A1A2.The linearized system of (2) about E+ is(6)xn+1=-x¯A1+x¯xn+γ1A1+x¯yn,yn+1=β2A2+y¯xn-y¯A2+y¯yn.Noticing the fact that (A1+x¯)(A2+y¯)=β2γ1, the characteristic equation associated with E+ can be written as(7)λ2+pλ+q=0,where p=A1y¯+A2x¯+2x¯y¯/β2γ1=1+x¯y¯-A1A2/β2γ1>0, q=x¯y¯-β2γ1/β2γ1<0.A simple calculation shows that(8)Δ=p2-4q=1+x¯y¯-A1A2β2γ12-4x¯y¯-β2γ1β2γ1>1+x¯y¯-A1A2β2γ12-4x¯y¯-A1A2β2γ1=1-x¯y¯-A1A2β2γ12>0.Since 1+x¯y¯-A1A2/β2γ1>0, we get x¯y¯-A1A2/β2γ1>-1, and hence 1±x¯y¯-A1A2/β2γ1<Δ<3+x¯y¯-A1A2/β2γ1. Hence,(9)0<λ1=12-1-x¯y¯-A1A2β2γ1+Δ<1,λ2=12-1-x¯y¯-A1A2β2γ1-Δ<-1.Therefore, we have the following results.Theorem 2.
(i) Assume thatβ2γ1≤A1A2. Then E0 is the only equilibrium of system (2), and it is locally stable while β2γ1<A1A2 and is nonhyperbolic while β2γ1=A1A2.
(ii) Assume thatβ2γ1>A1A2. Then E0 and E+ are equilibria of system (2) and they are all unstable. In fact, E0 is a repeller, and E+ is a saddle point.


## 3. Periodic Character

Leta=γ1/A1, b=β2/A2. Then the following statements are true.Lemma 3.
(i) The initiation value(x0,0) generates the solution(10)x0,0,0,bx0,abx0,0,…,0,b2n-1a2nx0,ab2nx0,0,…,(ii) The initiation value (0,y0) generates the solution(11)0,y0,ay0,0,0,aby0,…,a2nb2n-1y0,0,0,ab2ny0,….(iii) The initial value (x0,y0) with x0y0>0 generates the solution {(xn,yn)} with xnyn>0 for n≥0.Theorem 4.
System (2) has positive prime period-two solution if and only if β2γ1=A1A2. Further, the prime period-two solution possesses the form such that
(i)⋯,(x0,0),0,β2/A2x0,(x0,0),(0,β2/A2x0),⋯ with x0>0;
(ii)⋯,(0,y0),(γ1/A1y0,0),(0,y0),(γ1/A1y0,0),⋯ with y0>0.Proof.
Applying Lemma3 (i)-(ii) and the condition that β2γ1=A1A2, it is easy to see that the solution of system (2) which starting on either axis are all of period-two. This establishes the sufficient condition. Moreover, it is also a necessary and vital condition, as may be seen by the following argument.
Let(m,l), (M,L) be a period-two solution of system (2). Then they should satisfy the following equation:(12)M=γ1lA1+m,m=γ1LA1+M,L=β2mA2+l,l=β2MA2+L,from which it follows that(13)A1M-m=γ1l-L,A2L-l=β2m-M.
Obviously,m=M yields l=L, or vice versa, this is a contradiction. Thus m≠M and l≠L, and (13) yields β2γ1=A1A2.
Equation (12) also yields(14)mM=γ12lLA1+mA1+M,lL=β22mMA2+lA2+L,from which it follows that(15)mMlLA1+mA1+MA2+lA2+L-β22γ12=0.
Sinceβ2γ1=A1A2 and m+M>0, l+L>0, we obtain(16)A1+mA1+MA2+lA2+L-β22γ12>A1A2m+Ml+L>0,and so mMlL=0.
Ifm=0, then L=0, or vice versa. In this case, we claim that M≠0. Otherwise, there would be the fact that m=M=l=L=0 holds, a contradiction. So system (2) possesses a period-two solution(17)…,M,0,0,β2A2M,M,0,…,where M=x0>0.
Similarly, ifM=0, then l=0, or vice versa, and system (2) possesses a period-two solution(18)…,0,L,γ1A1L,0,0,L,…,where L=y0>0.
The proof is complete.


## 4. Global Attractivity

### 4.1. The Caseβ2γ1≤A1A2

Theorem 5.
Assume thatβ2γ1<A1A2. Then the unique equilibrium E0 is globally asymptotically stable.Proof.
Let{(xn,yn)} be a solution of system (2). Clearly, β2γ1<A1A2 implies that ab<1. Notice that for n>0 yield(19)x2ny2n≤ay2n-1bx2n-1≤abx2n-2y2n-2≤⋯≤abnx0y0,and(20)x2n+1y2n+1≤ay2nbx2n≤abx2n-1y2n-1≤⋯≤abnx1y1.So (xn,yn)→(0,0). Consequently, E0 is a global attractor and hence by Theorem 2 (i), the conclusion follows.Consider the mapF on R2 associated with system (2); that is,(21)Fx,y=fx,y,gx,y=γ1yA1+x,β2xA2+y,x,y≥0.Set(22)Δx=f2x,y-x=xβ2γ1A1+xA2+yA1A1+x+γ1y-1,Δy=g2x,y-y=yβ2γ1A2+yA1+xA2A2+y+β2x-1.Lemma 6.
Assume thatβ2γ1=A1A2. Then Δx<0, Δy<0 with x,y>0.Proof.
Using (22) and the condition that β2γ1=A1A2,  x,y>0, we have(23)Δx=-xyA2γ1+A12+A1x+γ1yA2+yA1A1+x+γ1y<0and(24)Δy=-xyA1β2+A22+A2y+β2xA1+xA2A2+y+β2x<0.
The proof is complete.Theorem 7.
Assume thatβ2γ1=A1A2. Then
(i) every solution of system (2) starting on either axis is of period-two;
(ii) every solution of system (2) with x0y0>0 converges to E0.Proof.
(i) It is a direct consequence of Theorem4.
(ii) Let{(xn,yn)} be a solution of system (2) with initial value x0y0>0. Then by Lemma 6, the subsequence {x2n}, {x2n+1} and {y2n}, {y2n+1} are all strictly decreasing and bounded below by zero. Notice the form of period-two solution of system (2) mentioned by Theorem 4; then there is only one result; that is, limn→∞⁡(xn,yn)=(0,0). This completes the proof.


### 4.2. The Caseβ2γ1>A1A2

LetQi for i=1,…,4 be the usual four quadrants at E+ and numbered in a counterclockwise direction, e.g., Q1={(x,y)∈R+2:x≥x¯,y≥y¯}. Then we have the following results.Lemma 8.
Assume thatβ2γ1>A1A2. Then F(Q2)⊆Q4, F(Q4)⊆Q2. Further, the regions Q2 and Q4 are invariant under the iteration by F2.Proof.
Let(x,y)∈Q2. Then x≤x¯ and y≥y¯, so(25)fx,y=γ1yA1+x≥γ1y¯A1+x¯=x¯,gx,y=β2xA2+y≤β2x¯A2+y¯=y¯,from which it follows that (F(x,y)=(f(x,y),g(x,y))∈Q4.
Further,(26)f2x,y=γ1gx,yA1+fx,y≤γ1y¯A1+x¯=x¯,g2x,y=β2fx,yA2+gx,y≥β2x¯A2+y¯=y¯,from which it follows that F2(x,y)∈Q2.
The proof forF(Q4)⊆Q2 is similar and is omitted, finishing the proof.To obtain the global attractivity of the positive equilibriumE+, we further divide R+2 into four distinct regions, which are described as follows:(i)R1={(x,y)∣x¯/γ1(A1+x)<y<β2/y¯x-A2,x>x¯},(ii)R2={(x,y)∣y≥max⁡{x¯/γ1(A1+x),β2/y¯x-A2},x>0}\{E+},(iii)R3={(x,y)∣max⁡{0,β2/y¯x-A2}<y<x¯/γ1(A1+x),0<x<x¯},(iv)R4={(x,y)∣0<y≤min⁡{x¯/γ1(A1+x),β2/y¯x-A2},x>A2y¯/β2}\{E+}.Lemma 9.
Assume thatβ2γ1>A1A2. Then
(i)F(R1)⊆Q1, F(R3)⊆Q3;
(ii)F(R2)⊆Q4, F(R4)⊆Q2.Proof.
(i) Let(x,y)∈R1. Then x¯/γ1(A1+x)<y<β2/y¯x-A2 and x>x¯; thus(27)fx,y=γ1yA1+x>γ1x¯/γ1A1+xA1+x=x¯,gx,y=β2xA2+y>β2xA2+β2/y¯x-A2=y¯,from which it follows that (f(x,y),g(x,y))∈Q1.
The proof ofF(R3)⊆Q3 is similar and is omitted.
(ii) Here we only prove thatF(R2)⊆Q4; the proof for F(R4)⊆Q2 is the same and is omitted.
Set(x,y)∈R2. Then yield y≥x¯/γ1(A1+x)>β2/y¯x-A2 if x≤x¯, and y≥β2/y¯x-A2>x¯/γ1(A1+x) if x>x¯. Hence(28)fx,y=γ1yA1+x≥γ1x¯/γ1A1+xA1+x=x¯,gx,y=β2xA2+y≤β2xA2+β2/y¯x-A2=y¯,from which it follows that F(x,y)=(f(x,y),g(x,y))∈Q4.
The proof is complete.Lemma 10.
Assume thatβ2γ1>A1A2. Then
(i)Δx<0,   Δy>0 with (x,y)∈Q2;
(ii)Δx>0,   Δy<0 with (x,y)∈Q4.Proof.
Using the equalities that(A1+x¯)(A2+y¯)=β2γ1, β2γ1(A1+x¯)/(A2+y¯)[A1(A1+x¯)+γ1y¯]=1, and β2γ1(A2+y¯)/(A1+x¯)[A2(A2+y¯)+β2x¯]=1, Δx and Δy can be rewritten as follows: (29)Δx=xβ2γ1A1+xA2+yA1A1+x+γ1y-β2γ1A1+x¯A2+y¯A1A1+x¯+γ1y¯=β2γ1xx¯A2+y¯x-x¯+A1A1+x+γ1A2+y¯+γ1yy¯-yA2+yA1A1+x+γ1y,and(30)Δy=yβ2γ1A2+yA1+xA2A2+y+β2x-β2γ1A2+y¯A1+x¯A2A2+y¯+β2x¯=β2γ1yA2A2+y+β2A1+x¯+β2xx¯-x+y¯A1+x¯y-y¯A1+xA2A2+y+β2x.
(i) In the regionQ2, yield x≤x¯  and y≥y¯; hence Δx<0, Δy>0.
(ii) In the regionQ4, yield x≥x¯  and y≤y¯; hence Δx>0, Δy<0.
The proof is complete.Lemma 11.
Assume thatβ2γ1>A1A2. Then(i)Δx<0, Δy<0 with (x,y)∈R1;(ii)Δx>0, Δy>0 with (x,y)∈R3.Proof.
Here, we only prove (i), the proof of (ii) is similar and is omitted.
(i) Let(x,y)∈R1. Then we have(31)x¯γ1A1+x<y<β2y¯x-A2and x>x¯, y>y¯. So (32)Δx=xβ2γ1A1+x-A1A1+xA2+y-γ1yA2+yA2+yA1A1+x+γ1y<xβ2γ1A1+x-A1A1+xA2+y¯-γ1x¯/γ1A1+xA2+y¯A2+yA1A1+x+γ1y=xA1+xβ2γ1-A1A2+y¯-x¯A2+y¯A2+yA1A1+x+γ1y=0.From (31), we can get that x>y¯/β2(A2+y), and so (33)Δy=yβ2γ1A2+y-A2A1+xA2+y-β2xA1+xA1+xA2A2+y+β2x<yβ2γ1A2+y-A2A1+x¯A2+y-β2y¯/β2A2+yA1+x¯A1+xA2A2+y+β2x=yA2+yβ2γ1-A2A1+x¯-y¯A1+x¯A1+xA2A2+y+β2x=0.
The proof is complete.Theorem 12.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R2 has the following character: (i)
(x2n,y2n) eventually enters the region Q2 and satisfies (x2n,y2n)→(0,∞);(ii)
(x2n+1,y2n+1) eventually enters the region Q4 and satisfies (x2n+1,y2n+1)→(∞,0).Proof.
By Lemmas9(ii) and 8, it easy to obtain that (x2n,y2n)∈Q2 and (x2n+1,y2n+1)∈Q4 for n≥0 with (x0,y0)∈R2.
In view of Lemma10 (i), we find that in the region Q2 the sequence {x2n} is strictly decreasing and bounded below by 0, and the sequence {y2n} is strictly increasing. Notice that, in this case, system (2) has no prime period-two solution and the other equilibrium point; hence (x2n,y2n)→(0,∞).
From Lemma10 (ii), we find that in the region Q4 the sequence {x2n+1} is strictly increasing, the sequence {y2n+1} is strictly decreasing and bounded below by 0. Hence y2n+1→0, and x2n+1→∞, and the proof is complete.Similarly, we have the following result.Theorem 13.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R4 has the following character: (i)
(x2n,y2n) eventually enters the region Q4 and satisfies (x2n,y2n)→(∞,0);(ii)
(x2n+1,y2n+1) eventually enters the region Q2 and satisfies (x2n+1,y2n+1)→(0,∞).Theorem 14.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R1 has stability trichotomy; that is, exactly one of the following three cases holds: (i)
(x2n,y2n) eventually enters the region Q2 satisfying (x2n,y2n)→(0,∞) and (x2n+1,y2n+1) eventually enters the region Q4 satisfying (x2n+1,y2n+1)→(∞,0);(ii)
(x2n,y2n) eventually enters the region Q4 satisfying (x2n,y2n)→(∞,0) and (x2n+1,y2n+1) eventually enters the region Q2 satisfying (x2n+1,y2n+1)→(0,∞);(iii)
it remains in the regionR1 forever and satisfies (xn,yn)→E+.Proof.
Let(x0,y0)∈R1. Then Lemma 8 implies that (x1,y1)∈Q1. Hence (x1,y1)∈Q1∩R2⊂R2, (x1,y1)∈Q1∩R4⊂R4, or (x1,y1)∈R1.
Case (a). If (x1,y1)∈R2, then Theorem 12 implies that (x2n+1,y2n+1)∈Q2, (x2n,y2n)∈Q4 for n≥1. Further, (x2n+1,y2n+1)→(0,∞), (x2n,y2n)→(∞,0).
Case (b). If (x1,y1)∈R4, then Theorem 13 implies that (x2n+1,y2n+1)∈Q4, (x2n,y2n)∈Q2 for n≥1. Further, (x2n+1,y2n+1)→(∞,0), (x2n,y2n)→(0,∞).
Case (c). If (x1,y1)∈R1 and there exists an integer N1 such that (xN1,yN1)∈R2, then byCase (a), we have (xN1+2n,yN1+2n)∈Q2, (xN1+2n+1,yN1+2n+1)∈Q4 for n≥0, and (xN1+2n,yN1+2n)→(0,∞), (xN1+2n+1,yN1+2n+1)→(∞,0).
If(x1,y1)∈R1 and there exists an integer N2 such that (xN2,yN2)∈R4, then byCase (b), we have (xN2+2n,yN2+2n)∈Q4, (xN2+2n+1,yN2+2n+1)∈Q2 for n≥0. Furthermore, (xN2+2n,yN2+2n)→(∞,0), (xN2+2n+1,yN2+2n+1)→(0,∞).
If(x1,y1)∈R1, and (xn,yn)∈R1 for all n≥1, then by Lemma 11 (i), we have that the sequences {x2n}, {x2n+1} and {y2n}, {y2n+1} are all strictly decreasing, and bounded below by x¯, and y¯, respectively. Hence they are all convergent; moreover, (xn,yn)→E+.
The proof is complete.Similarly, we can show that the orbits starting inR3 have the following character.Theorem 15.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R3 has stability trichotomy; that is, exactly one of the following three cases holds: (i)
(x2n,y2n) eventually enters the region Q2 satisfying (x2n,y2n)→(0,∞) and (x2n+1,y2n+1) eventually enters the region Q4 satisfying (x2n+1,y2n+1)→(∞,0);(ii)
(x2n,y2n) eventually enters the region Q4 satisfying (x2n,y2n)→(∞,0) and (x2n+1,y2n+1) eventually enters the region Q2 satisfying (x2n+1,y2n+1)→(0,∞);(iii)
it remains in the regionR3 forever and satisfies (xn,yn)→E+.


## 4.1. The Caseβ2γ1≤A1A2

Theorem 5.
Assume thatβ2γ1<A1A2. Then the unique equilibrium E0 is globally asymptotically stable.Proof.
Let{(xn,yn)} be a solution of system (2). Clearly, β2γ1<A1A2 implies that ab<1. Notice that for n>0 yield(19)x2ny2n≤ay2n-1bx2n-1≤abx2n-2y2n-2≤⋯≤abnx0y0,and(20)x2n+1y2n+1≤ay2nbx2n≤abx2n-1y2n-1≤⋯≤abnx1y1.So (xn,yn)→(0,0). Consequently, E0 is a global attractor and hence by Theorem 2 (i), the conclusion follows.Consider the mapF on R2 associated with system (2); that is,(21)Fx,y=fx,y,gx,y=γ1yA1+x,β2xA2+y,x,y≥0.Set(22)Δx=f2x,y-x=xβ2γ1A1+xA2+yA1A1+x+γ1y-1,Δy=g2x,y-y=yβ2γ1A2+yA1+xA2A2+y+β2x-1.Lemma 6.
Assume thatβ2γ1=A1A2. Then Δx<0, Δy<0 with x,y>0.Proof.
Using (22) and the condition that β2γ1=A1A2,  x,y>0, we have(23)Δx=-xyA2γ1+A12+A1x+γ1yA2+yA1A1+x+γ1y<0and(24)Δy=-xyA1β2+A22+A2y+β2xA1+xA2A2+y+β2x<0.
The proof is complete.Theorem 7.
Assume thatβ2γ1=A1A2. Then
(i) every solution of system (2) starting on either axis is of period-two;
(ii) every solution of system (2) with x0y0>0 converges to E0.Proof.
(i) It is a direct consequence of Theorem4.
(ii) Let{(xn,yn)} be a solution of system (2) with initial value x0y0>0. Then by Lemma 6, the subsequence {x2n}, {x2n+1} and {y2n}, {y2n+1} are all strictly decreasing and bounded below by zero. Notice the form of period-two solution of system (2) mentioned by Theorem 4; then there is only one result; that is, limn→∞⁡(xn,yn)=(0,0). This completes the proof.


## 4.2. The Caseβ2γ1>A1A2

LetQi for i=1,…,4 be the usual four quadrants at E+ and numbered in a counterclockwise direction, e.g., Q1={(x,y)∈R+2:x≥x¯,y≥y¯}. Then we have the following results.Lemma 8.
Assume thatβ2γ1>A1A2. Then F(Q2)⊆Q4, F(Q4)⊆Q2. Further, the regions Q2 and Q4 are invariant under the iteration by F2.Proof.
Let(x,y)∈Q2. Then x≤x¯ and y≥y¯, so(25)fx,y=γ1yA1+x≥γ1y¯A1+x¯=x¯,gx,y=β2xA2+y≤β2x¯A2+y¯=y¯,from which it follows that (F(x,y)=(f(x,y),g(x,y))∈Q4.
Further,(26)f2x,y=γ1gx,yA1+fx,y≤γ1y¯A1+x¯=x¯,g2x,y=β2fx,yA2+gx,y≥β2x¯A2+y¯=y¯,from which it follows that F2(x,y)∈Q2.
The proof forF(Q4)⊆Q2 is similar and is omitted, finishing the proof.To obtain the global attractivity of the positive equilibriumE+, we further divide R+2 into four distinct regions, which are described as follows:(i)R1={(x,y)∣x¯/γ1(A1+x)<y<β2/y¯x-A2,x>x¯},(ii)R2={(x,y)∣y≥max⁡{x¯/γ1(A1+x),β2/y¯x-A2},x>0}\{E+},(iii)R3={(x,y)∣max⁡{0,β2/y¯x-A2}<y<x¯/γ1(A1+x),0<x<x¯},(iv)R4={(x,y)∣0<y≤min⁡{x¯/γ1(A1+x),β2/y¯x-A2},x>A2y¯/β2}\{E+}.Lemma 9.
Assume thatβ2γ1>A1A2. Then
(i)F(R1)⊆Q1, F(R3)⊆Q3;
(ii)F(R2)⊆Q4, F(R4)⊆Q2.Proof.
(i) Let(x,y)∈R1. Then x¯/γ1(A1+x)<y<β2/y¯x-A2 and x>x¯; thus(27)fx,y=γ1yA1+x>γ1x¯/γ1A1+xA1+x=x¯,gx,y=β2xA2+y>β2xA2+β2/y¯x-A2=y¯,from which it follows that (f(x,y),g(x,y))∈Q1.
The proof ofF(R3)⊆Q3 is similar and is omitted.
(ii) Here we only prove thatF(R2)⊆Q4; the proof for F(R4)⊆Q2 is the same and is omitted.
Set(x,y)∈R2. Then yield y≥x¯/γ1(A1+x)>β2/y¯x-A2 if x≤x¯, and y≥β2/y¯x-A2>x¯/γ1(A1+x) if x>x¯. Hence(28)fx,y=γ1yA1+x≥γ1x¯/γ1A1+xA1+x=x¯,gx,y=β2xA2+y≤β2xA2+β2/y¯x-A2=y¯,from which it follows that F(x,y)=(f(x,y),g(x,y))∈Q4.
The proof is complete.Lemma 10.
Assume thatβ2γ1>A1A2. Then
(i)Δx<0,   Δy>0 with (x,y)∈Q2;
(ii)Δx>0,   Δy<0 with (x,y)∈Q4.Proof.
Using the equalities that(A1+x¯)(A2+y¯)=β2γ1, β2γ1(A1+x¯)/(A2+y¯)[A1(A1+x¯)+γ1y¯]=1, and β2γ1(A2+y¯)/(A1+x¯)[A2(A2+y¯)+β2x¯]=1, Δx and Δy can be rewritten as follows: (29)Δx=xβ2γ1A1+xA2+yA1A1+x+γ1y-β2γ1A1+x¯A2+y¯A1A1+x¯+γ1y¯=β2γ1xx¯A2+y¯x-x¯+A1A1+x+γ1A2+y¯+γ1yy¯-yA2+yA1A1+x+γ1y,and(30)Δy=yβ2γ1A2+yA1+xA2A2+y+β2x-β2γ1A2+y¯A1+x¯A2A2+y¯+β2x¯=β2γ1yA2A2+y+β2A1+x¯+β2xx¯-x+y¯A1+x¯y-y¯A1+xA2A2+y+β2x.
(i) In the regionQ2, yield x≤x¯  and y≥y¯; hence Δx<0, Δy>0.
(ii) In the regionQ4, yield x≥x¯  and y≤y¯; hence Δx>0, Δy<0.
The proof is complete.Lemma 11.
Assume thatβ2γ1>A1A2. Then(i)Δx<0, Δy<0 with (x,y)∈R1;(ii)Δx>0, Δy>0 with (x,y)∈R3.Proof.
Here, we only prove (i), the proof of (ii) is similar and is omitted.
(i) Let(x,y)∈R1. Then we have(31)x¯γ1A1+x<y<β2y¯x-A2and x>x¯, y>y¯. So (32)Δx=xβ2γ1A1+x-A1A1+xA2+y-γ1yA2+yA2+yA1A1+x+γ1y<xβ2γ1A1+x-A1A1+xA2+y¯-γ1x¯/γ1A1+xA2+y¯A2+yA1A1+x+γ1y=xA1+xβ2γ1-A1A2+y¯-x¯A2+y¯A2+yA1A1+x+γ1y=0.From (31), we can get that x>y¯/β2(A2+y), and so (33)Δy=yβ2γ1A2+y-A2A1+xA2+y-β2xA1+xA1+xA2A2+y+β2x<yβ2γ1A2+y-A2A1+x¯A2+y-β2y¯/β2A2+yA1+x¯A1+xA2A2+y+β2x=yA2+yβ2γ1-A2A1+x¯-y¯A1+x¯A1+xA2A2+y+β2x=0.
The proof is complete.Theorem 12.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R2 has the following character: (i)
(x2n,y2n) eventually enters the region Q2 and satisfies (x2n,y2n)→(0,∞);(ii)
(x2n+1,y2n+1) eventually enters the region Q4 and satisfies (x2n+1,y2n+1)→(∞,0).Proof.
By Lemmas9(ii) and 8, it easy to obtain that (x2n,y2n)∈Q2 and (x2n+1,y2n+1)∈Q4 for n≥0 with (x0,y0)∈R2.
In view of Lemma10 (i), we find that in the region Q2 the sequence {x2n} is strictly decreasing and bounded below by 0, and the sequence {y2n} is strictly increasing. Notice that, in this case, system (2) has no prime period-two solution and the other equilibrium point; hence (x2n,y2n)→(0,∞).
From Lemma10 (ii), we find that in the region Q4 the sequence {x2n+1} is strictly increasing, the sequence {y2n+1} is strictly decreasing and bounded below by 0. Hence y2n+1→0, and x2n+1→∞, and the proof is complete.Similarly, we have the following result.Theorem 13.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R4 has the following character: (i)
(x2n,y2n) eventually enters the region Q4 and satisfies (x2n,y2n)→(∞,0);(ii)
(x2n+1,y2n+1) eventually enters the region Q2 and satisfies (x2n+1,y2n+1)→(0,∞).Theorem 14.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R1 has stability trichotomy; that is, exactly one of the following three cases holds: (i)
(x2n,y2n) eventually enters the region Q2 satisfying (x2n,y2n)→(0,∞) and (x2n+1,y2n+1) eventually enters the region Q4 satisfying (x2n+1,y2n+1)→(∞,0);(ii)
(x2n,y2n) eventually enters the region Q4 satisfying (x2n,y2n)→(∞,0) and (x2n+1,y2n+1) eventually enters the region Q2 satisfying (x2n+1,y2n+1)→(0,∞);(iii)
it remains in the regionR1 forever and satisfies (xn,yn)→E+.Proof.
Let(x0,y0)∈R1. Then Lemma 8 implies that (x1,y1)∈Q1. Hence (x1,y1)∈Q1∩R2⊂R2, (x1,y1)∈Q1∩R4⊂R4, or (x1,y1)∈R1.
Case (a). If (x1,y1)∈R2, then Theorem 12 implies that (x2n+1,y2n+1)∈Q2, (x2n,y2n)∈Q4 for n≥1. Further, (x2n+1,y2n+1)→(0,∞), (x2n,y2n)→(∞,0).
Case (b). If (x1,y1)∈R4, then Theorem 13 implies that (x2n+1,y2n+1)∈Q4, (x2n,y2n)∈Q2 for n≥1. Further, (x2n+1,y2n+1)→(∞,0), (x2n,y2n)→(0,∞).
Case (c). If (x1,y1)∈R1 and there exists an integer N1 such that (xN1,yN1)∈R2, then byCase (a), we have (xN1+2n,yN1+2n)∈Q2, (xN1+2n+1,yN1+2n+1)∈Q4 for n≥0, and (xN1+2n,yN1+2n)→(0,∞), (xN1+2n+1,yN1+2n+1)→(∞,0).
If(x1,y1)∈R1 and there exists an integer N2 such that (xN2,yN2)∈R4, then byCase (b), we have (xN2+2n,yN2+2n)∈Q4, (xN2+2n+1,yN2+2n+1)∈Q2 for n≥0. Furthermore, (xN2+2n,yN2+2n)→(∞,0), (xN2+2n+1,yN2+2n+1)→(0,∞).
If(x1,y1)∈R1, and (xn,yn)∈R1 for all n≥1, then by Lemma 11 (i), we have that the sequences {x2n}, {x2n+1} and {y2n}, {y2n+1} are all strictly decreasing, and bounded below by x¯, and y¯, respectively. Hence they are all convergent; moreover, (xn,yn)→E+.
The proof is complete.Similarly, we can show that the orbits starting inR3 have the following character.Theorem 15.
Assume thatβ2γ1>A1A2. Then every solution with the initial value (x0,y0)∈R3 has stability trichotomy; that is, exactly one of the following three cases holds: (i)
(x2n,y2n) eventually enters the region Q2 satisfying (x2n,y2n)→(0,∞) and (x2n+1,y2n+1) eventually enters the region Q4 satisfying (x2n+1,y2n+1)→(∞,0);(ii)
(x2n,y2n) eventually enters the region Q4 satisfying (x2n,y2n)→(∞,0) and (x2n+1,y2n+1) eventually enters the region Q2 satisfying (x2n+1,y2n+1)→(0,∞);(iii)
it remains in the regionR3 forever and satisfies (xn,yn)→E+.



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*Source: 7825631-2019-08-01.xml*