# Global Behavior of Four Competitive Rational Systems of

Authors: M. Garić-Demirović; M. R. S. Kulenović; M. Nurkanović
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We investigate the global dynamics of solutions of four distinct competitive rational
systems of difference equations in the plane. We show that the basins of attractions of
different locally asymptotically stable equilibrium points are separated by the global stable
manifolds of either saddle points or  nonhyperbolic equilibrium points. Our results give
complete answer to Open Problem 2 posed recently by Camouzis et al. (2009).


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## Body

## 1. Introduction and Preliminaries

We consider the following open  problem (see [1, Open Problem 2]).Foreach of the following four distinct systems(1.1)(14,21),(15,21),(21,21),(21,38),determine the following:(i)
the boundedness character of its solutions,(ii)
the local stability of its equilibrium points,(iii)
the existence of prime period-two solutions,(iv)
the global character of the systems.Equation (3.4) is of the form(1.2)xn+1=β1xnA1+yn,n=0,1,…;
equation (3.5) is of the form(1.3)xn+1=β1xnB1xn+yn,n=0,1,…;
equation (3.16) is either of the form(1.4)xn+1=α1+γ1xnyn,n=0,1,…
or the form(1.5)yn+1=α2+γ2ynxn,n=0,1,…
depending on whether it appears as first or second equation in the system; equation (38) is of the form(1.6)yn+1=γ2ynA2+B2xn+yn,n=0,1,….The typical results are the following theorems. The first theorem is a combination of Theorems2.3 and 2.5 and the second theorem is Theorem 3.3.Theorem 1.1.
Consider system (14, 21) and assume thatγ2A1≠α2. If β1>A1, then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(1.7)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(1.8)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.
Assume thatβ1≤A1. Every solution {(xn,yn)} of system (14, 21), with x0>0,y0≥0, satisfies
(1.9)limn→∞xn=0,limn→∞yn=∞.Theorem 1.2.
Consider system (21, 21). There exists a set𝒞⊂ℛ which is invariant and a subset of the basin of attraction of the unique equilibrium E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(1.10)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex}.
which satisfy
(1.11)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.All considered systems are competitive systems, which we discuss next.A first-order system of difference equations(1.12)xn+1=f(xn,yn)yn+1=g(xn,yn)n=0,1,2,…,(x-1,x0)∈ℛ,
where ℛ⊂ℝ2, (f,g):ℛ→ℛ, f, g are continuous functions, is competitive if f(x,y) is nondecreasing in x and nonincreasing in y, and g(x,y) is nonincreasing in x and nondecreasing in y. If both f and g are nondecreasing in x and y, the system (1.12) is cooperative. A map T that corresponds to the system (1.12) is defined as T(x,y)=(f(x,y),g(x,y)). Competitive and cooperative maps, which are called monotone maps, are defined similarly. Strongly competitive systems of difference equations or maps are those for which the functions f and g are coordinatewise strictly monotone.Ifv=(u,v)∈ℝ2, we denote with 𝒬ℓ(v), ℓ∈{1,2,3,4}, the four quadrants in ℝ2 relative to v, that is, 𝒬1(v)={(x,y)∈ℝ2:x≥u,y≥v}, 𝒬2(v)={(x,y)∈ℝ2:x≤u,y≥v}, and so on. Define the South-East partial order ⪯se on ℝ2 by (x,y)⪯se(s,t) if and only if x≤s and y≥t. Similarly, we define the North-East partial order ⪯ne on ℝ2 by (x,y)⪯ne(s,t) if and only if x≤s and y≤t. For 𝒜⊂ℝ2 and x∈ℝ2, define the distance fromxto𝒜 as dist(x,𝒜):=inf{∥x-y∥:y∈𝒜}. By int𝒜 we denote the interior a set 𝒜.It is easy to show that a mapF is competitive if it is nondecreasing with respect to the South-East partial order, that is, if the following holds:(1.13)(x1y1)⪯se(x2y2)⇒F(x1y1)⪯seF(x2y2).Competitive systems were studied by many authors; see [2–17], and others. All known results, with the exception of [2, 3, 18], deal with hyperbolic dynamics. The results presented here are results that hold in both the hyperbolic and the nonhyperbolic case.We now state three results for competitive maps in the plane. The following definition is from [17].Definition 1.3.
Let𝒮 be a nonempty subset of ℝ2. A competitive map T:𝒮→𝒮 is said to satisfy condition (O+) if for every x, y in 𝒮, T(x)⪯neT(y) implies x⪯ney, and T is said to satisfy condition (O-) if for every x, y in 𝒮, T(x)⪯neT(y) implies y⪯nex.The following theorem was proved by DeMottoni-Schiaffino for the Poincaré map of a periodic competitive Lotka-Volterra system of differential equations. Smith generalized the proof to competitive and cooperative maps [14, 15].Theorem 1.4.
Let𝒮 be a nonempty subset of ℝ2. If T is a competitive map for which (O+) holds then for all x∈𝒮, {Tn(x)} is eventually componentwise monotone. If the orbit of x has compact closure, then it converges to a fixed point of T. If instead (O-) holds, then for all x∈𝒮, {T2n} is eventually componentwise monotone. If the orbit of x has compact closure in 𝒮, then its omega limit set is either a period-two orbit or a fixed point.The following result is from [17], with the domain of the map specialized to be the cartesian product of intervals of real numbers. It gives a sufficient condition for conditions (O+) and (O-).Theorem 1.5.
Letℛ⊂ℝ2 be the cartesian product of two intervals in ℝ. Let T:ℛ→ℛ be a C1 (continuously differentiable) competitive map. If T is injective and detJT(x)>0 for all x∈ℛ, then T satisfies (O+). If T is injective and detJT(x)<0 for all x∈ℛ, then T satisfies (O-).The next results are the modifications of [8, Theorem 4]. See [18].Theorem 1.6.
LetT be a monotone map on a closed and bounded rectangular region ℛ⊂ℝ2. Suppose that T has a unique fixed point e̅ in ℛ. Then e̅ is a global attractor T on ℛ.The following four results were proved by Kulenović and Merino [18] for competitive systems in the plane, when one of the eigenvalues of the linearized system at an equilibrium (hyperbolic or nonhyperbolic) is by absolute value smaller than 1 while the other has an arbitrary value. These results are useful for determining basins of attraction of fixed points of competitive maps.Our first result gives conditions for the existence of a global invariant curve through a fixed point (hyperbolic or not) of a competitive map that is differentiable in a neighborhood of the fixed point, when at least one of two nonzero eigenvalues of the Jacobian matrix of the map at the fixed point has absolute value less than one. A regionℛ⊂ℝ2 is rectangular if it is the cartesian product of two intervals in ℝ.Theorem 1.7.
LetT be a competitive map on a rectangular region ℛ⊂ℝ2. Let x¯∈ℛ be a fixed point of T such that Δ:=ℛ∩int(𝒬1(x¯)∪𝒬3(x¯)) is nonempty (i.e., x¯ is not the NW or SE vertex of ℛ), and T is strongly competitive on Δ. Suppose that the following statements are true. (a)
The mapT has a C1 extension to a neighborhood of x¯.(b)
The Jacobian matrix ofT at x¯ has real eigenvalues λ, μ such that 0<|λ|<μ, where |λ|<1, and the eigenspace Eλ associated with λ is not a coordinate axis.Then there exists a curve 𝒞⊂ℛ through x¯ that is invariant and a subset of the basin of attraction of x¯, such that 𝒞 is tangential to the eigenspace Eλ at x¯, and 𝒞 is the graph of a strictly increasing continuous function of the first coordinate on an interval. Any endpoints of 𝒞 in the interior of ℛ are either fixed points or minimal period-two points. In the latter case, the set of endpoints of 𝒞 is a minimal period-two orbit of T.Corollary 1.8.
IfT has no fixed point nor periodic points of minimal period-two in Δ, then the endpoints of 𝒞 belong to ∂ℛ.For maps that are strongly competitive near the fixed point, hypothesis (b) of Theorem1.7 reduces just to |λ|<1. This follows from a change of variables [17] that allows the Perron-Frobenius Theorem to be applied to give that at any point, the Jacobian matrix of a strongly competitive map has two real and distinct eigenvalues, the larger one in absolute value being positive, and that corresponding eigenvectors may be chosen to point in the direction of the second and first quadrant, respectively. Also, one can show that in such case no associated eigenvector is aligned with a coordinate axis.The following result gives a description of the global stable and unstable manifolds of a saddle point of a competitive map. The result is a modification of [8, Theorem 5].Theorem 1.9.
In addition to the hypotheses of Theorem1.7, suppose that μ>1 and that the eigenspace Eμ associated with μ is not a coordinate axis. If the curve 𝒞 of Theorem 1.7 has endpoints in ∂ℛ, then 𝒞 is the global stable manifold 𝒲s(x¯) of x¯, and the global unstable manifold 𝒲u(x¯) is a curve in ℛ that is tangential to Eμ at x¯ and such that it is the graph of a strictly decreasing function of the first coordinate on an interval. Any endpoints of 𝒲u(x¯) in ℛ are fixed points of T.The next result is useful for determining basins of attraction of fixed points of competitive maps.Theorem 1.10.
Assume the hypotheses of Theorem1.7, and let 𝒞 be the curve whose existence is guaranteed by Theorem 1.7. If the endpoints of 𝒞 belong to ∂ℛ, then 𝒞 separates ℛ into two connected components, namely,
(1.14)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
such that the following statements are true. (i)
𝒲- is invariant, and dist(Tn(x),𝒬2(x¯))→0 as n→∞ for every x∈𝒲-.(ii)
𝒲+ is invariant, and dist(Tn(x),𝒬4(x¯))→0 as n→∞ for every x∈𝒲+.If, in addition, x¯ is an interior point of ℛ and T is C2 and strongly competitive in a neighborhood of x¯, then T has no periodic points in the boundary of 𝒬1(x¯)∪𝒬3(x¯) except for x¯, and the following statements are true. (i)
For everyx∈𝒲- there exists n0∈ℕ such that Tn(x)∈int𝒬2(x¯) for n≥n0.(ii)
For everyx∈𝒲+ there exists n0∈ℕ such that Tn(x)∈int𝒬4(x¯) for n≥n0.In this paper we study the global dynamics of four rational systems of difference equations mentioned earlier, where all parameters are positive numbers and initial conditionsx0 and y0 are arbitrary nonnegative numbers. Two of these systems have a nonhyperbolic semistable equilibrium point. In general all four systems share the common feature that the global stable manifolds of either saddle points or nonhyperbolic equilibrium points serve as boundaries of basins of attraction of different local attractors or points at infinities. The techniques used here can be applied to treat number of competitive systems which appear in applications, such as Leslie-Gower competition model, see [19], or Leslie-Gower competition model with stocking, see [20], or genetic model, see [13]. An important new feature of our techniques is that they are applicable to nonhyperbolic case as well, which was shown for the first time in [18] where we have completed analysis of basic Leslie-Gower competition model from [19]. Furthermore, system (21,38) can be considered as a variant of Leslie-Gower competition model, where the first equation has been replaced by another equation, which does not allow extinction of both species. In fact, all four considered competitive systems share common feature that they do not allow the extinction of both species.


## 2. System (14,21)

Now we consider the following system of difference equations:(2.1)xn+1=β1xnA1+yn,yn+1=α2+γ2ynxn,n=0,1,…,
where the parameters A1,β1,α2, and γ2 are positive numbers and initial conditions x0>0,y0≥0.System (2.1) was considered in [1, Example 1], where it was shown that the associated map T(x,y)=(β1x/(A1+y),(α2+γ2y)/x) is injective and(2.2)detJT(x,y)=β1(A1+y)2x·(γ2A1-α2).
When γ2A1>α2, detJT(x,y)>0. Therefore, in view of Theorems 1.4 and 1.5 every solution of system (2.1) is eventually componentwise monotonic. If γ2A1<α2, then detJT(x,y)<0, and four subsequences(2.3){x2n},{x2n+1},{y2n},{y2n+1}
of every solution {(xn,yn)} of system (2.1) are eventually monotonic.Thus, ifγ2A1≠α2, the Jacobian matrix of T in (x,y) is invertible.The Jacobian matrix of the corresponding mapT(x,y) is of the form(2.4)JT(x,y)=[β1A1+y-β1x(A1+y)2-α2+γ2yx2γ2x].


### 2.1. Linearized Stability Analysis

The equilibrium points(x¯,y¯) of system (2.1) are solutions of the system of equations(2.5)x¯=β1x¯A1+y¯,y¯=α2+γ2y¯x¯,
from which we obtain(2.6)y¯=β1-A1,x¯=α2β1-A1+γ2.Lemma 2.1.
(i) Ifβ1>A1, then system (2.1) has a unique equilibrium point:
(2.7)E=(α2β1-A1+γ2,β1-A1),
which is a saddle point.
(ii) Ifβ1≤A1, then system (2.1) has no equilibrium points.Proof.
By (2.6) and (2.4) the Jacobian matrix evaluated at the equilibrium point E has the form
(2.8)JT(E)=[1-x¯β1-y¯x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E is
(2.9)λ2-pλ+q=0,
where
(2.10)p=TrJT(x¯,y¯)=1+γ2x¯>0,q=DetJT(x¯,y¯)=γ2x¯-y¯β1.
Notice that in view of (2.6) y̅/β1=1-A1/β1 and so
(2.11)1+q=1+γ2x¯-y¯β1=1+γ2x¯-1+A1β1>0.
Since p>0 and 1+q>0, we need to show(I)
p>1+q,(II)
p2-4q>0.
Indeed,
(I)
p>1+q⇔1+γ2/x¯>1+γ2/x¯-y¯/β1⇔0>-y¯/β1,which is satisfied (because β1>0 and y¯>0). Furthermore(II)
p2-4q>0⇔(1+γ2/x¯)2-4(γ2/x¯)+4(y¯/β1)>0⇔(1-γ2/x¯)2+4(y¯/β1)>0,which is satisfied.


### 2.2. Global Results

#### 2.2.1. Caseβ1>A1

Theorem 2.2.
System (2.1) has no prime period-two solutions.Proof.
System (2.1) can be reduced to the following second-order difference equation:
(2.12)yn+2=yn+1(A1+yn)(α2+γ2yn+1)β1(α2+γ2yn),
or to the following second-order difference equation:
(2.13)xn+2=β1xnxn+12(A1xn+α2-γ2A1)xn+1+γ2β1xn.
Now it is sufficient to prove that both of the difference equations (2.12) and (2.13) have no prime period-two solutions. Assume that this is not true for (2.12), that is, that
(2.14)ϕ,ψ,ϕ,ψ,…,(ϕ≠ψ)
is a prime period-two solution of (2.12). Then we have
(2.15)ϕ=ψ(A1+ϕ)(α2+γ2ψ)β1(α2+γ2ϕ),ψ=ϕ(A1+ψ)(α2+γ2ϕ)β1(α2+γ2ψ).
This implies
(2.16)β1ϕ(α2+γ2ϕ)=ψ(A1+ϕ)(α2+γ2ψ),β1ψ(α2+γ2ψ)=ϕ(A1+ψ)(α2+γ2ϕ).
By subtraction, we obtain
(2.17)β1(ϕ-ψ)[α2+γ2(ϕ+ψ)]=A1[α2(ψ-ϕ)+γ2(ψ2-ϕ2)]+ϕψγ2(ψ-ϕ),
that is,
(2.18)(ϕ-ψ)[β1(α2+γ2(ϕ+ψ))+A1(α2+γ2(ψ+ϕ))+ϕψγ2]=0,
and this implies that ϕ=ψ, which is a contradiction.
Now assume that
(2.19)χ,φ,χ,φ,…,(χ≠φ)
is a prime period-two solution of (2.13). Then we have
(2.20)χ=β1χφ2(A1χ+α2-γ2A1)φ+γ2β1χ,φ=β1φχ2(A1φ+α2-γ2A1)χ+γ2β1φ,
from which
(2.21)(χ-φ)[A1χφ+γ2β1(χ+φ)+β1χφ]=0,
and this implies that χ=φ, which is a contradiction.Theorem 2.3.
Consider system (2.1) and assume that β1>A1 and γ2A1≠α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(2.22)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(2.23)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (2.1) is strongly competitive on (0,∞)×[0,∞). In view of Theorem 2.2 we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied with ℛ=(0,∞)×[0,∞) and so the conclusion follows.Remark 2.4 (see [1]).
Ifγ2A1=α2, then system (2.1) can be decoupled as follows:
(2.24)xn+1=β1xn2A1xn+β1γ2,yn+1=1β1yn(A1+yn),n=0,1,…
and every solution of this system (depending of the choice of the initial condition (x0,y0)) is either bounded and converges to an equilibrium point or increases monotonically to infinity.


#### 2.2.2. Caseβ1≤A1

In this case system (2.1) has no equilibrium points. Now we have the following.Theorem 2.5.
Assume thatβ1≤A1 and γ2A1≠α2. Every solution {(xn,yn)} of system (2.1), with x0>0,y0≥0, satisfies
(2.25)limn→∞xn=0,limn→∞yn=∞.Proof.
Ifβ1<A1, then
(2.26)xn+1<β1A1xn⇒xn<(β1A1)nx0→0(n→∞),
which implies limn→∞xn=0.
On the other hand, ifβ1=A1, then xn+1=A1xn/(A1+yn)<xn, and we obtain that the sequence {xn}n=0∞ is strictly decreasing. Because xn>0 for all n, we see that {xn}n=0∞ is convergent and limn→∞xn=0, since otherwise, that is, limn→∞xn=a>0, the first equation of system (2.1) implies limn→∞yn=β1-A1=0 or the second equation of system (2.1) implies limn→∞yn=α2/(a-γ2)≠0, which is a contradiction, since otherwise system (2.1) would have an equilibrium point in the first quadrant.
We see that ifβ1≤A1, then every solution {(xn,yn)} of system (2.1) satisfies limn→∞xn=0.
But then the denominator in
(2.27)yn+1=α2+γ2ynxn
is, for all large n, strictly less than a constant η<γ2, which in turn implies
(2.28)yn+1>α2η+γ2ηyn,n≥N.
Iterating this inequality we obtain
(2.29)yn>α2η+γ2η(α2η)n-N+(γ2η)n-1-NyN,n≥N,
and this forces yn to infinity.The obtained results lead to the following characterization of the boundedness of solutions of system (2.1).Corollary 2.6.
Consider system (2.1) subject to the condition α2≠A1γ2. If β1>A1, then all bounded solutions converge to the unique equilibrium with the corresponding initial conditions belonging to the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start in the complement of 𝒞 are asymptotic to either (∞,0) or (0,∞). If β1≤A1, then all solutions are unbounded in the sense that {xn} is bounded and {yn} approaches ∞.


## 2.1. Linearized Stability Analysis

The equilibrium points(x¯,y¯) of system (2.1) are solutions of the system of equations(2.5)x¯=β1x¯A1+y¯,y¯=α2+γ2y¯x¯,
from which we obtain(2.6)y¯=β1-A1,x¯=α2β1-A1+γ2.Lemma 2.1.
(i) Ifβ1>A1, then system (2.1) has a unique equilibrium point:
(2.7)E=(α2β1-A1+γ2,β1-A1),
which is a saddle point.
(ii) Ifβ1≤A1, then system (2.1) has no equilibrium points.Proof.
By (2.6) and (2.4) the Jacobian matrix evaluated at the equilibrium point E has the form
(2.8)JT(E)=[1-x¯β1-y¯x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E is
(2.9)λ2-pλ+q=0,
where
(2.10)p=TrJT(x¯,y¯)=1+γ2x¯>0,q=DetJT(x¯,y¯)=γ2x¯-y¯β1.
Notice that in view of (2.6) y̅/β1=1-A1/β1 and so
(2.11)1+q=1+γ2x¯-y¯β1=1+γ2x¯-1+A1β1>0.
Since p>0 and 1+q>0, we need to show(I)
p>1+q,(II)
p2-4q>0.
Indeed,
(I)
p>1+q⇔1+γ2/x¯>1+γ2/x¯-y¯/β1⇔0>-y¯/β1,which is satisfied (because β1>0 and y¯>0). Furthermore(II)
p2-4q>0⇔(1+γ2/x¯)2-4(γ2/x¯)+4(y¯/β1)>0⇔(1-γ2/x¯)2+4(y¯/β1)>0,which is satisfied.


## 2.2. Global Results

### 2.2.1. Caseβ1>A1

Theorem 2.2.
System (2.1) has no prime period-two solutions.Proof.
System (2.1) can be reduced to the following second-order difference equation:
(2.12)yn+2=yn+1(A1+yn)(α2+γ2yn+1)β1(α2+γ2yn),
or to the following second-order difference equation:
(2.13)xn+2=β1xnxn+12(A1xn+α2-γ2A1)xn+1+γ2β1xn.
Now it is sufficient to prove that both of the difference equations (2.12) and (2.13) have no prime period-two solutions. Assume that this is not true for (2.12), that is, that
(2.14)ϕ,ψ,ϕ,ψ,…,(ϕ≠ψ)
is a prime period-two solution of (2.12). Then we have
(2.15)ϕ=ψ(A1+ϕ)(α2+γ2ψ)β1(α2+γ2ϕ),ψ=ϕ(A1+ψ)(α2+γ2ϕ)β1(α2+γ2ψ).
This implies
(2.16)β1ϕ(α2+γ2ϕ)=ψ(A1+ϕ)(α2+γ2ψ),β1ψ(α2+γ2ψ)=ϕ(A1+ψ)(α2+γ2ϕ).
By subtraction, we obtain
(2.17)β1(ϕ-ψ)[α2+γ2(ϕ+ψ)]=A1[α2(ψ-ϕ)+γ2(ψ2-ϕ2)]+ϕψγ2(ψ-ϕ),
that is,
(2.18)(ϕ-ψ)[β1(α2+γ2(ϕ+ψ))+A1(α2+γ2(ψ+ϕ))+ϕψγ2]=0,
and this implies that ϕ=ψ, which is a contradiction.
Now assume that
(2.19)χ,φ,χ,φ,…,(χ≠φ)
is a prime period-two solution of (2.13). Then we have
(2.20)χ=β1χφ2(A1χ+α2-γ2A1)φ+γ2β1χ,φ=β1φχ2(A1φ+α2-γ2A1)χ+γ2β1φ,
from which
(2.21)(χ-φ)[A1χφ+γ2β1(χ+φ)+β1χφ]=0,
and this implies that χ=φ, which is a contradiction.Theorem 2.3.
Consider system (2.1) and assume that β1>A1 and γ2A1≠α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(2.22)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(2.23)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (2.1) is strongly competitive on (0,∞)×[0,∞). In view of Theorem 2.2 we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied with ℛ=(0,∞)×[0,∞) and so the conclusion follows.Remark 2.4 (see [1]).
Ifγ2A1=α2, then system (2.1) can be decoupled as follows:
(2.24)xn+1=β1xn2A1xn+β1γ2,yn+1=1β1yn(A1+yn),n=0,1,…
and every solution of this system (depending of the choice of the initial condition (x0,y0)) is either bounded and converges to an equilibrium point or increases monotonically to infinity.


### 2.2.2. Caseβ1≤A1

In this case system (2.1) has no equilibrium points. Now we have the following.Theorem 2.5.
Assume thatβ1≤A1 and γ2A1≠α2. Every solution {(xn,yn)} of system (2.1), with x0>0,y0≥0, satisfies
(2.25)limn→∞xn=0,limn→∞yn=∞.Proof.
Ifβ1<A1, then
(2.26)xn+1<β1A1xn⇒xn<(β1A1)nx0→0(n→∞),
which implies limn→∞xn=0.
On the other hand, ifβ1=A1, then xn+1=A1xn/(A1+yn)<xn, and we obtain that the sequence {xn}n=0∞ is strictly decreasing. Because xn>0 for all n, we see that {xn}n=0∞ is convergent and limn→∞xn=0, since otherwise, that is, limn→∞xn=a>0, the first equation of system (2.1) implies limn→∞yn=β1-A1=0 or the second equation of system (2.1) implies limn→∞yn=α2/(a-γ2)≠0, which is a contradiction, since otherwise system (2.1) would have an equilibrium point in the first quadrant.
We see that ifβ1≤A1, then every solution {(xn,yn)} of system (2.1) satisfies limn→∞xn=0.
But then the denominator in
(2.27)yn+1=α2+γ2ynxn
is, for all large n, strictly less than a constant η<γ2, which in turn implies
(2.28)yn+1>α2η+γ2ηyn,n≥N.
Iterating this inequality we obtain
(2.29)yn>α2η+γ2η(α2η)n-N+(γ2η)n-1-NyN,n≥N,
and this forces yn to infinity.The obtained results lead to the following characterization of the boundedness of solutions of system (2.1).Corollary 2.6.
Consider system (2.1) subject to the condition α2≠A1γ2. If β1>A1, then all bounded solutions converge to the unique equilibrium with the corresponding initial conditions belonging to the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start in the complement of 𝒞 are asymptotic to either (∞,0) or (0,∞). If β1≤A1, then all solutions are unbounded in the sense that {xn} is bounded and {yn} approaches ∞.


## 2.2.1. Caseβ1>A1

Theorem 2.2.
System (2.1) has no prime period-two solutions.Proof.
System (2.1) can be reduced to the following second-order difference equation:
(2.12)yn+2=yn+1(A1+yn)(α2+γ2yn+1)β1(α2+γ2yn),
or to the following second-order difference equation:
(2.13)xn+2=β1xnxn+12(A1xn+α2-γ2A1)xn+1+γ2β1xn.
Now it is sufficient to prove that both of the difference equations (2.12) and (2.13) have no prime period-two solutions. Assume that this is not true for (2.12), that is, that
(2.14)ϕ,ψ,ϕ,ψ,…,(ϕ≠ψ)
is a prime period-two solution of (2.12). Then we have
(2.15)ϕ=ψ(A1+ϕ)(α2+γ2ψ)β1(α2+γ2ϕ),ψ=ϕ(A1+ψ)(α2+γ2ϕ)β1(α2+γ2ψ).
This implies
(2.16)β1ϕ(α2+γ2ϕ)=ψ(A1+ϕ)(α2+γ2ψ),β1ψ(α2+γ2ψ)=ϕ(A1+ψ)(α2+γ2ϕ).
By subtraction, we obtain
(2.17)β1(ϕ-ψ)[α2+γ2(ϕ+ψ)]=A1[α2(ψ-ϕ)+γ2(ψ2-ϕ2)]+ϕψγ2(ψ-ϕ),
that is,
(2.18)(ϕ-ψ)[β1(α2+γ2(ϕ+ψ))+A1(α2+γ2(ψ+ϕ))+ϕψγ2]=0,
and this implies that ϕ=ψ, which is a contradiction.
Now assume that
(2.19)χ,φ,χ,φ,…,(χ≠φ)
is a prime period-two solution of (2.13). Then we have
(2.20)χ=β1χφ2(A1χ+α2-γ2A1)φ+γ2β1χ,φ=β1φχ2(A1φ+α2-γ2A1)χ+γ2β1φ,
from which
(2.21)(χ-φ)[A1χφ+γ2β1(χ+φ)+β1χφ]=0,
and this implies that χ=φ, which is a contradiction.Theorem 2.3.
Consider system (2.1) and assume that β1>A1 and γ2A1≠α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(2.22)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(2.23)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (2.1) is strongly competitive on (0,∞)×[0,∞). In view of Theorem 2.2 we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied with ℛ=(0,∞)×[0,∞) and so the conclusion follows.Remark 2.4 (see [1]).
Ifγ2A1=α2, then system (2.1) can be decoupled as follows:
(2.24)xn+1=β1xn2A1xn+β1γ2,yn+1=1β1yn(A1+yn),n=0,1,…
and every solution of this system (depending of the choice of the initial condition (x0,y0)) is either bounded and converges to an equilibrium point or increases monotonically to infinity.


## 2.2.2. Caseβ1≤A1

In this case system (2.1) has no equilibrium points. Now we have the following.Theorem 2.5.
Assume thatβ1≤A1 and γ2A1≠α2. Every solution {(xn,yn)} of system (2.1), with x0>0,y0≥0, satisfies
(2.25)limn→∞xn=0,limn→∞yn=∞.Proof.
Ifβ1<A1, then
(2.26)xn+1<β1A1xn⇒xn<(β1A1)nx0→0(n→∞),
which implies limn→∞xn=0.
On the other hand, ifβ1=A1, then xn+1=A1xn/(A1+yn)<xn, and we obtain that the sequence {xn}n=0∞ is strictly decreasing. Because xn>0 for all n, we see that {xn}n=0∞ is convergent and limn→∞xn=0, since otherwise, that is, limn→∞xn=a>0, the first equation of system (2.1) implies limn→∞yn=β1-A1=0 or the second equation of system (2.1) implies limn→∞yn=α2/(a-γ2)≠0, which is a contradiction, since otherwise system (2.1) would have an equilibrium point in the first quadrant.
We see that ifβ1≤A1, then every solution {(xn,yn)} of system (2.1) satisfies limn→∞xn=0.
But then the denominator in
(2.27)yn+1=α2+γ2ynxn
is, for all large n, strictly less than a constant η<γ2, which in turn implies
(2.28)yn+1>α2η+γ2ηyn,n≥N.
Iterating this inequality we obtain
(2.29)yn>α2η+γ2η(α2η)n-N+(γ2η)n-1-NyN,n≥N,
and this forces yn to infinity.The obtained results lead to the following characterization of the boundedness of solutions of system (2.1).Corollary 2.6.
Consider system (2.1) subject to the condition α2≠A1γ2. If β1>A1, then all bounded solutions converge to the unique equilibrium with the corresponding initial conditions belonging to the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start in the complement of 𝒞 are asymptotic to either (∞,0) or (0,∞). If β1≤A1, then all solutions are unbounded in the sense that {xn} is bounded and {yn} approaches ∞.


## 3. System (21,21)

Now we consider the following system of difference equations:(3.1)xn+1=β1xn+α1yn,yn+1=α2+γ2ynxn,n=0,1,…,
where the parameters α1,β1,α2, and γ2 are positive numbers and initial conditions x0>0,y0>0.System (3.1) was considered in [1, Example 3], where it was shown that the associated map T is injective and(3.2)detJT(x,y)=-α1(α2+γ2y)-β1α2xx2y2<0,
that is, the Jacobian matrix of T in (x,y) is invertible. Therefore, in view of Theorems 1.4 and 1.5, four subsequences(3.3){x2n},{x2n+1},{y2n},{y2n+1}
of every solution {(xn,yn)} of system (3.1) are eventually monotonic.


### 3.1. Linearized Stability Analysis

Equilibrium points of system (3.1) are solutions of the system(3.4)x¯=β1x¯+α1y¯,y¯=α2+γ2y¯x¯.
Since x¯≠0 and y¯≠0, we have(3.5)y¯±=12γ2[-(α2-α1-β1γ2)±D1],
where(3.6)D1=(α2-α1-β1γ2)2+4β1γ2α2.
Since y¯-<0 and y¯+>0, system (3.1) has a unique positive equilibrium E=(x¯+,y¯+), where(3.7)x¯+=12β1(α2-α1+β1γ2+D2),
where D2=(α2-α1+β1γ2)2+4β1γ2α1.Lemma 3.1.
System (3.1) has a unique positive equilibrium point:
(3.8)E=(12β1(α2-α1+β1γ2+D2),12γ2(-(α2-α1-β1γ2)+D1)),
which is a saddle point.Proof.
The Jacobian matrix of the corresponding mapT(x,y)=((β1x+α1)/y,(α2+γ2y)/x) is of the form
(3.9)JT(x,y)=[β1y-β1x+α1y2-α2+γ2yx2γ2x].
By using (3.4) we obtain
(3.10)JT(E)=[β1y¯-x¯y¯-y¯x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E of system (3.1) is
(3.11)λ2-pλ+q=0,
where
(3.12)p=TrJT(x¯,y¯)=β1y¯+γ2x¯>0,q=DetJT(x¯,y¯)=β1γ2x¯y¯-1.
Notice that
(3.13)1+q=1+β1γ2x¯y¯-1=β1γ2x¯y¯>0.
Since p>0 and 1+q>0, we need to show(I)
p>1+q,(II)
p2-4q>0.Now, we get(I)
p>1+q⇔β1/y¯+γ2/x¯>β1γ2/x¯y¯⇔x¯y¯-α1+x¯y¯-α2>β1γ2.By using (3.4), (3.5), and (3.7) we obtain
(3.14)β1x¯+γ2y¯=2x¯y¯-α1-α2=β1γ2+12D1+12D2>β1γ2.
Furthermore(II)
p2-4q>0⇔(β1/y¯+γ2/x¯)2-4(β1γ2/x¯y¯-1)>0⇔(β1/y¯-γ2/x¯)2+4>0,which is satisfied.


### 3.2. Global Results

Theorem 3.2.
System (3.1) has no prime period-two solutions.Proof.
System (3.1) can be reduced to the following second-order difference equation:
(3.15)yn+2=ynyn+1(α2+γ2yn+1)α1yn+1+β1(α2+γ2yn),
or to the following second-order difference equation:
(3.16)xn+2=(β1xn+1+α1)xnxn+1α2xn+1+γ2(β1xn+α1).
Now it is sufficient to prove that both of the difference equations (3.15) and (3.16) have no prime period-two solutions. Assume that this is not true for (3.15), that is, that
(3.17)ϕ,ψ,ϕ,ψ,…,(ϕ≠ψ)
is a prime period-two solution of (3.15). Then we have
(3.18)ϕ=ϕψ(α2+γ2ψ)α1ψ+β1(α2+γ2ϕ),ψ=ϕψ(α2+γ2ϕ)α1ϕ+β1(α2+γ2ψ),
that is,
(3.19)ψ(α2+γ2ψ)α1ψ+β1(α2+γ2ϕ)=ϕ(α2+γ2ϕ)α1ϕ+β1(α2+γ2ψ),
from which
(3.20)(ψ-ϕ){α1ϕψγ2+β1[α22+2α2γ2(ϕ+ψ)+γ22(ϕ2+ϕψ+ψ2)]}=0,
and this implies that ϕ=ψ, which is a contradiction.
Now assume that
(3.21)χ,φ,χ,φ,…,(χ≠φ)
is a prime period-two solution of (3.16). Then we have
(3.22)χ=(β1φ+α1)χφα2φ+γ2(β1χ+α1),φ=(β1χ+α1)χφα2χ+γ2(β1φ+α1),
from which
(3.23)(χ-φ)[γ2β1(χ+φ)+γ2α1+β1χφ]=0,
and this implies that χ=φ, which is a contradiction.The global behavior system (3.1) is described by the following result.Theorem 3.3.
Consider system (3.1). There exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(3.24)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(3.25)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
In view of Theorem3.2 and the injectivity of the map T we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied with ℛ=(0,∞)×[0,∞) and so the conclusion follows.The obtained result leads to the following characterization of the boundedness of solutions of system (3.1).Corollary 3.4.
All bounded solutions of system (3.1) converge to the unique equilibrium with the corresponding initial conditions which belong to the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start in the complement of 𝒞 are asymptotic to either (∞,0) or (0,∞).


## 3.1. Linearized Stability Analysis

Equilibrium points of system (3.1) are solutions of the system(3.4)x¯=β1x¯+α1y¯,y¯=α2+γ2y¯x¯.
Since x¯≠0 and y¯≠0, we have(3.5)y¯±=12γ2[-(α2-α1-β1γ2)±D1],
where(3.6)D1=(α2-α1-β1γ2)2+4β1γ2α2.
Since y¯-<0 and y¯+>0, system (3.1) has a unique positive equilibrium E=(x¯+,y¯+), where(3.7)x¯+=12β1(α2-α1+β1γ2+D2),
where D2=(α2-α1+β1γ2)2+4β1γ2α1.Lemma 3.1.
System (3.1) has a unique positive equilibrium point:
(3.8)E=(12β1(α2-α1+β1γ2+D2),12γ2(-(α2-α1-β1γ2)+D1)),
which is a saddle point.Proof.
The Jacobian matrix of the corresponding mapT(x,y)=((β1x+α1)/y,(α2+γ2y)/x) is of the form
(3.9)JT(x,y)=[β1y-β1x+α1y2-α2+γ2yx2γ2x].
By using (3.4) we obtain
(3.10)JT(E)=[β1y¯-x¯y¯-y¯x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E of system (3.1) is
(3.11)λ2-pλ+q=0,
where
(3.12)p=TrJT(x¯,y¯)=β1y¯+γ2x¯>0,q=DetJT(x¯,y¯)=β1γ2x¯y¯-1.
Notice that
(3.13)1+q=1+β1γ2x¯y¯-1=β1γ2x¯y¯>0.
Since p>0 and 1+q>0, we need to show(I)
p>1+q,(II)
p2-4q>0.Now, we get(I)
p>1+q⇔β1/y¯+γ2/x¯>β1γ2/x¯y¯⇔x¯y¯-α1+x¯y¯-α2>β1γ2.By using (3.4), (3.5), and (3.7) we obtain
(3.14)β1x¯+γ2y¯=2x¯y¯-α1-α2=β1γ2+12D1+12D2>β1γ2.
Furthermore(II)
p2-4q>0⇔(β1/y¯+γ2/x¯)2-4(β1γ2/x¯y¯-1)>0⇔(β1/y¯-γ2/x¯)2+4>0,which is satisfied.


## 3.2. Global Results

Theorem 3.2.
System (3.1) has no prime period-two solutions.Proof.
System (3.1) can be reduced to the following second-order difference equation:
(3.15)yn+2=ynyn+1(α2+γ2yn+1)α1yn+1+β1(α2+γ2yn),
or to the following second-order difference equation:
(3.16)xn+2=(β1xn+1+α1)xnxn+1α2xn+1+γ2(β1xn+α1).
Now it is sufficient to prove that both of the difference equations (3.15) and (3.16) have no prime period-two solutions. Assume that this is not true for (3.15), that is, that
(3.17)ϕ,ψ,ϕ,ψ,…,(ϕ≠ψ)
is a prime period-two solution of (3.15). Then we have
(3.18)ϕ=ϕψ(α2+γ2ψ)α1ψ+β1(α2+γ2ϕ),ψ=ϕψ(α2+γ2ϕ)α1ϕ+β1(α2+γ2ψ),
that is,
(3.19)ψ(α2+γ2ψ)α1ψ+β1(α2+γ2ϕ)=ϕ(α2+γ2ϕ)α1ϕ+β1(α2+γ2ψ),
from which
(3.20)(ψ-ϕ){α1ϕψγ2+β1[α22+2α2γ2(ϕ+ψ)+γ22(ϕ2+ϕψ+ψ2)]}=0,
and this implies that ϕ=ψ, which is a contradiction.
Now assume that
(3.21)χ,φ,χ,φ,…,(χ≠φ)
is a prime period-two solution of (3.16). Then we have
(3.22)χ=(β1φ+α1)χφα2φ+γ2(β1χ+α1),φ=(β1χ+α1)χφα2χ+γ2(β1φ+α1),
from which
(3.23)(χ-φ)[γ2β1(χ+φ)+γ2α1+β1χφ]=0,
and this implies that χ=φ, which is a contradiction.The global behavior system (3.1) is described by the following result.Theorem 3.3.
Consider system (3.1). There exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(3.24)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(3.25)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
In view of Theorem3.2 and the injectivity of the map T we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied with ℛ=(0,∞)×[0,∞) and so the conclusion follows.The obtained result leads to the following characterization of the boundedness of solutions of system (3.1).Corollary 3.4.
All bounded solutions of system (3.1) converge to the unique equilibrium with the corresponding initial conditions which belong to the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start in the complement of 𝒞 are asymptotic to either (∞,0) or (0,∞).


## 4. System (15,21)

Now we consider the following system of difference equations:(4.1)xn+1=β1xnB1xn+yn,yn+1=α2+γ2ynxn,n=0,1,…,
where the parameters β1,B1,α2, and γ2 are positive numbers and initial conditions x0>0,y0≥0. The Jacobian matrix of the corresponding map T(x,y)=(β1x/(B1x+y),(α2+γ2y)/x) is of the form(4.2)JT(x,y)=[β1y(B1x+y)2-β1x(B1x+y)2-α2+γ2yx2γ2x].System (4.1) was considered in [1, Example 2], where it was shown that the corresponding map T is injective and(4.3)detJT(x,y)=-β1α2x(B1x+y)2<0,
that is, the Jacobian matrix of T in (x,y) is invertible. Therefore, in view of Theorems 1.4 and 1.5, four subsequences(4.4){x2n},{x2n+1},{y2n},{y2n+1}
of every solution {(xn,yn)} of system (4.1) are eventually monotonic.


### 4.1. Linearized Stability Analysis

Equilibrium points of system (4.1) are solutions of the system(4.5)x¯=β1x¯B1x¯+y¯,y¯=α2+γ2y¯x¯.
Since x¯≠0, we obtain
(4.6)x¯±=β1+γ2B1±D32B1,
where D3=(β1-B1γ2)2-4B1α2≥0.This implies that we have the following three cases for the equilibrium points.(i)
Ifβ1-B1γ2>2B1α2, then there exist two equilibrium points of system (4.1):
(4.7)E+=(β1+γ2B1+D32B1,β1-γ2B1-D32),E-=(β1+γ2B1-D32B1,β1-γ2B1+D32).(ii)
Ifβ1-B1γ2=2B1α2, then system (4.1) has a unique equilibrium point:
(4.8)E=(B1γ2+β12B1,β1-B1γ22).(iii)
Ifβ1-B1γ2≤0 or 0<β1-B1γ2<2B1α2, then system (4.1) has no equilibrium points.Next, by using (4.5) we have(4.9)JT(x¯,y¯)=[1-B1β1x¯-x¯β1B1-β1x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E=(x¯,y¯) is(4.10)λ2-pλ+q=0,
where(4.11)p=TrJT(x¯,y¯)=y¯β1+γ2x¯=1-B1x¯β1+γ2x¯,q=DetJT(x¯,y¯)=γ2x¯-B1γ2β1+B1x¯β1-1=γ2y¯β1x¯-y¯β1.
Notice that p>0.Lemma 4.1.
Ifβ1-B1γ2>2B1α2, then the equilibrium point E+ of system (4.1) is locally asymptotically stable and the equilibrium point E- is a saddle point.
Ifβ1-B1γ2=2B1α2, then the equilibrium point E of system (4.1) is nonhyperbolic.Proof.
First, assumeβ1-B1γ2>2B1α2. For the equilibrium point E+ we need to prove that
(4.12)|p|<1+q<2,
or equivalently (because p>0):(I)
p<1+q,(II)
q<1.Indeed,(I)
we have(4.13)p<1+q⇔1-B1x¯β1+γ2x¯<1+γ2x¯-B1γ2β1+B1x¯β1-1⇔1+B1γ2β1<2B1x¯β1⇔1+B1γ2β1<β1+γ2B1+D3β1⇔0<D3,which is true. Furthermore(II)
we have(4.14)q<1⇔γ2x¯-B1γ2β1+B1x¯β1-1<1⇔γ2(1x¯-B1β1)+B1x¯-β1β1<1⇔γ2(β1-B1x¯β1x¯)-β1-B1x¯β1<1⇔β1-B1x¯β1(γ2x¯-1)<1⇔y¯β1(1-α2x¯y¯-1)<1⇔-α2β1x¯<1,
which is true.
For the equilibrium pointE- we need to prove that
(4.15)|p|>|1+q|,p2-4q>0,
that is (because p>0 and 1+q>0)(I)
p>1+q,(II)
p2-4q>0.Indeed,
(4.16)1+q>0⇔1+γ2x¯-B1γ2β1+B1x¯β1-1>0⇔γ2β1-B1x¯x¯β1+B1x¯β1>0⇔γ2y¯x¯β1+B1x¯β1>0.
Now (I)
we have(4.17)p>1+q⇔1-B1x¯β1+γ2x¯>1+γ2x¯-B1γ2β1+B1x¯β1-1⇔1+B1γ2β1>2B1x¯β1⇔1+B1γ2β1>β1+γ2B1-D3β1⇔0>-D3,
which is true.
Similarly(II)
we have(4.18)p2-4q>0⇔(y¯β1+γ2x¯)2-4(γ2y¯β1x¯-y¯β1)>0⇔y¯2β12-2γ2y¯β1x¯+γ22x¯2+4y¯β1>0⇔(y¯β1-γ2x¯)2+4y¯β1>0,
which is satisfied.
Assume thatβ1-B1γ2=2B1α2.
We need to prove that
(4.19)|1+q|=|p|,
that is (because p>0 and,1+q>0),
(4.20)1+q=p.
We have
(4.21)1+q=p⇔1+B1γ2β1=2B1x¯β1⇔1+B1γ2β1=2B1β1B1γ2+β12B1.


### 4.2. Global Results

Theorem 4.2.
System (4.1) has no prime period-two solutions.Proof.
The second iterate of mapT is
(4.22)T2(x,y)=T(β1xB1x+y,α2+γ2yx)=(β1(β1x/(B1x+y))B1(β1x/(B1x+y))+(α2+γ2y)/x,α2+γ2((α2+γ2y)/x)β1x/(B1x+y))=(β12x2B1β1x2+(α2+γ2y)(B1x+y),(B1x+y)(α2x+γ2α2+γ22y)β1x2).
Period-two solution satisfies
(4.23)β12x2B1β1x2+(α2+γ2y)(B1x+y)-x=0,(B1x+y)(α2x+γ2α2+γ22y)β1x2-y=0,β12x=B1β1x2+(α2+γ2y)(B1x+y),β1yx2=(B1x+y)(α2x+α2γ2+γ22y).
From this system we have(i)
x=(1/2B1)(β1+γ2B1+(β1-B1γ2)2-4B1α2),y=(1/2B1)(β1-γ2B1-(β1-B1γ2)2-4B1α2),(ii)
x=(1/2B1)(β1+γ2B1-(β1-B1γ2)2-4B1α2),y=(1/2B1)(β1-γ2B1+(β1-B1γ2)2-4B1α2),(iii)
x=-γ2+(1/2β12)(α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1),y=-(1/2β1γ2)(α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1),(iv)
x=-γ2+(1/2β12)(α2β1+β12γ2-B1α2γ2+B1β1γ22+Δ1),y=-(1/2β1γ2)(α2β1+β12γ2-B1α2γ2+B1β1γ22+Δ1),(v)
x=0,y=-α/γ,where Δ1=(α-βγ)(αβ2+3β3γ+B2αγ2+2Bβ2γ2-B2βγ3-2Bαβγ).
In cases (i) and (ii) solutions(x,y) are equilibrium points E+ and E-, and in case (v) solution (x,y) is not in the first quadrant in the plane. It is sufficient to prove that solutions (x,y) in cases (iii) and (iv) are not in the first quadrant in the plane. Namely, if Δ1<0, x and y are not real. Supose that Δ1≥0. If α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1≥0, then y≤0. If α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1<0, then x<0 for solution in case (iii). By analogous reasoning we have that the same conclusion for case (iv) holds.Our linearized stability analysis indicates that there are three cases with different asymptotic behavior, depending on the values of parametersβ1,B1,α2, and γ2.Case 1.
β1-B1γ2>2B1α2.Case 2.
β1-B1γ2=2B1α2.Case 3.
β1-B1γ2≤0 or 0<β1-B1γ2<2B1α2.


#### 4.2.1. Global Results—Case1

Theorem 4.3.
Consider system (4.1) and assume that β1-B1γ2>2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E-. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.24)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey}𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.25)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(β1+γ2B1+D32B1,β1-γ2B1-D32)forevery(x0,y0)∈𝒲+Proof.
Clearly, system (4.1) is strongly competitive on ℛ=(0,∞)×[0,∞). In view of injectivity of T, invertibility of JT, and Theorem 4.2, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


#### 4.2.2. Global Results—Case2

Theorem 4.4.
Consider system (4.1) and assume that β1-B1γ2=2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.26)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.27)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=E=(B1γ2+β12B1,β1-B1γ22)forevery(x0,y0)∈𝒲+.Proof.
In this case system (4.1) has a unique equilibrium point E=((B1γ2+β1)/2B1,(β1-B1γ2)/2) which is nonhyperbolic. For p=q+1, the corresponding characteristic equation is of the form
(4.28)λ2-pλ+p-1=0.
This implies
(4.29)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. We will show that p<1. Indeed
(4.30)p<1⇔B1α2β1+γ2B1B1γ2+B1α2<1⇔B1γ2B1α2+B1α2+β1γ2B1<β1γ2B1+β1B1α2⇔B1γ2+B1α2<β1=B1γ2+2B1α2
which is satisfied. Thus, λ1∈(-1,0).
The eigenvector corresponding toλ1=p-1 is
(4.31)(12B1β1(β1-B1γ2)(B1γ2+β1)2)=(14B1β1B1α2(B1γ2+β1)2).
It means that all conditions of Theorems 1.7 and 1.10 are satisfied with ℛ=(0,∞)×[0,∞).
Assume that(x0,y0)∈𝒲+. Then (xn,yn)∈𝒲+ for all n, and sequences {x2n},{x2n+1},{y2n}, and {y2n+1} are monotone and bounded since xn≤β1/B1. Thus these sequences are convergent, which in view of Theorem 4.2 shows that they converge to the equilibrium point. Since E is the unique equilibrium point in 𝒲+ the statement for 𝒲+ follows. The same conclusion is obtained by using Theorem 1.6.
If(x0,y0) is in 𝒲-, by Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬2(E). Assume (without loss of generality) that (x0,y0)∈int𝒬2(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(-1,B1). Choose a value of t small enough so that E+tv∈𝒬2(E) and (x0,y0)⪯E+tv. Let us show that T(E+tv)⪯E+tv. Indeed
(4.32)T(E+tv)=(B1γ2+β12B1-t,B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t)⪯(B1γ2+β12B1-t,β1-B1γ22+B1t)
because
(4.33)B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t≥β1-B1γ22+B1t
reduces to
(4.34)4B12t2+4B1α2+2B1β1γ2≥β12+B12γ22=4B1α2+2B1β1γ2,
where the last equality follows from the condition β1-B1γ2=2B1α2.
SinceT(E+tv)⪯E+tv, it follows that {Tn(E+tv)} is a monotonically decreasing sequence in 𝒬2(E) which is bounded above by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has second coordinate which is monotone and unbounded. But (xn,yn):=Tn(x0,y0)⪯Tn(E+tv), which implies that yn→∞. From (4.1) it follows that xn→0.


#### 4.2.3. Global Results—Case3

Theorem 4.5.
Consider system (4.1) and assume that β1-B1γ2≤0 or 0<β1-B1γ2<2B1α2. Then every solution {(xn,yn)} of system (4.1) satisfies
(4.35)limn→∞xn=0,limn→∞yn=∞.Proof.
In this case system (4.1) has no equilibrium points. Consider now the following system satisfied by subsequences of the solution of system (4.1):
(4.36)x2k+1=β1x2kB1x2k+y2k,x2k+2=β1x2k+1B1x2k+1+y2k+1,y2k+1=α2+γ2y2kx2k,y2k+2=α2+γ2y2k+1x2k+1.
We know that each of the four subsequences
(4.37){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {(xn,yn)} of system (4.1) is eventually monotonic. The subsequences {x2k} and {x2k+1} are bounded (by β1/B1), which implies that they are convergent. Suppose that (a)limk→∞x2k=xE and (b)limk→∞x2k+1=xO. For the other two subsequences the following four cases are possible: (1) limk→∞y2k=yE, (2) limk→∞y2k=∞, (3) limk→∞y2k+1=yO, or (4) limk→∞y2k+1=∞.
Case1 and Case 3 imply
(4.38)xO=β1xEB1xE+yE,xE=β1xOB1xO+yO,yO=α2+γ2yExE,yE=α2+γ2yOxO
that is, system (4.1) has a period-two solution, which is a contradiction by Theorem 4.2.
Case1 and Case 4 imply
(4.39)limk→∞x2k=xE=0⇒limk→∞x2k+1=xO=0⇒limk→∞y2k+2=∞,
which is a contradiction by Case 1.
Case2 and Case 3 imply
(4.40)limk→∞x2k+1=xO=0⇒limk→∞x2k+2=xE=0⇒limk→∞y2k+1=∞,
which is a contradiction by Case 3.
Case2 and Case 4 imply
(4.41)limk→∞x2k=xE=limk→∞x2k+1=xO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (3.1).Corollary 4.6.
Consider system (4.1) and assume that β1-B1γ2≥2B1α2. All bounded solutions of system (4.1) converge to the unique equilibrium with the corresponding initial conditions which belong to region below and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start above 𝒞 are asymptotic to (0,∞).
Consider system (4.1) and assume that either β1-B1γ2≤0 or β1-B1γ2<2B1α2. Then every solution of (4.1) is asymptotic to (0,∞).


## 4.1. Linearized Stability Analysis

Equilibrium points of system (4.1) are solutions of the system(4.5)x¯=β1x¯B1x¯+y¯,y¯=α2+γ2y¯x¯.
Since x¯≠0, we obtain
(4.6)x¯±=β1+γ2B1±D32B1,
where D3=(β1-B1γ2)2-4B1α2≥0.This implies that we have the following three cases for the equilibrium points.(i)
Ifβ1-B1γ2>2B1α2, then there exist two equilibrium points of system (4.1):
(4.7)E+=(β1+γ2B1+D32B1,β1-γ2B1-D32),E-=(β1+γ2B1-D32B1,β1-γ2B1+D32).(ii)
Ifβ1-B1γ2=2B1α2, then system (4.1) has a unique equilibrium point:
(4.8)E=(B1γ2+β12B1,β1-B1γ22).(iii)
Ifβ1-B1γ2≤0 or 0<β1-B1γ2<2B1α2, then system (4.1) has no equilibrium points.Next, by using (4.5) we have(4.9)JT(x¯,y¯)=[1-B1β1x¯-x¯β1B1-β1x¯γ2x¯].
The corresponding characteristic equation evaluated at the equilibrium point E=(x¯,y¯) is(4.10)λ2-pλ+q=0,
where(4.11)p=TrJT(x¯,y¯)=y¯β1+γ2x¯=1-B1x¯β1+γ2x¯,q=DetJT(x¯,y¯)=γ2x¯-B1γ2β1+B1x¯β1-1=γ2y¯β1x¯-y¯β1.
Notice that p>0.Lemma 4.1.
Ifβ1-B1γ2>2B1α2, then the equilibrium point E+ of system (4.1) is locally asymptotically stable and the equilibrium point E- is a saddle point.
Ifβ1-B1γ2=2B1α2, then the equilibrium point E of system (4.1) is nonhyperbolic.Proof.
First, assumeβ1-B1γ2>2B1α2. For the equilibrium point E+ we need to prove that
(4.12)|p|<1+q<2,
or equivalently (because p>0):(I)
p<1+q,(II)
q<1.Indeed,(I)
we have(4.13)p<1+q⇔1-B1x¯β1+γ2x¯<1+γ2x¯-B1γ2β1+B1x¯β1-1⇔1+B1γ2β1<2B1x¯β1⇔1+B1γ2β1<β1+γ2B1+D3β1⇔0<D3,which is true. Furthermore(II)
we have(4.14)q<1⇔γ2x¯-B1γ2β1+B1x¯β1-1<1⇔γ2(1x¯-B1β1)+B1x¯-β1β1<1⇔γ2(β1-B1x¯β1x¯)-β1-B1x¯β1<1⇔β1-B1x¯β1(γ2x¯-1)<1⇔y¯β1(1-α2x¯y¯-1)<1⇔-α2β1x¯<1,
which is true.
For the equilibrium pointE- we need to prove that
(4.15)|p|>|1+q|,p2-4q>0,
that is (because p>0 and 1+q>0)(I)
p>1+q,(II)
p2-4q>0.Indeed,
(4.16)1+q>0⇔1+γ2x¯-B1γ2β1+B1x¯β1-1>0⇔γ2β1-B1x¯x¯β1+B1x¯β1>0⇔γ2y¯x¯β1+B1x¯β1>0.
Now (I)
we have(4.17)p>1+q⇔1-B1x¯β1+γ2x¯>1+γ2x¯-B1γ2β1+B1x¯β1-1⇔1+B1γ2β1>2B1x¯β1⇔1+B1γ2β1>β1+γ2B1-D3β1⇔0>-D3,
which is true.
Similarly(II)
we have(4.18)p2-4q>0⇔(y¯β1+γ2x¯)2-4(γ2y¯β1x¯-y¯β1)>0⇔y¯2β12-2γ2y¯β1x¯+γ22x¯2+4y¯β1>0⇔(y¯β1-γ2x¯)2+4y¯β1>0,
which is satisfied.
Assume thatβ1-B1γ2=2B1α2.
We need to prove that
(4.19)|1+q|=|p|,
that is (because p>0 and,1+q>0),
(4.20)1+q=p.
We have
(4.21)1+q=p⇔1+B1γ2β1=2B1x¯β1⇔1+B1γ2β1=2B1β1B1γ2+β12B1.


## 4.2. Global Results

Theorem 4.2.
System (4.1) has no prime period-two solutions.Proof.
The second iterate of mapT is
(4.22)T2(x,y)=T(β1xB1x+y,α2+γ2yx)=(β1(β1x/(B1x+y))B1(β1x/(B1x+y))+(α2+γ2y)/x,α2+γ2((α2+γ2y)/x)β1x/(B1x+y))=(β12x2B1β1x2+(α2+γ2y)(B1x+y),(B1x+y)(α2x+γ2α2+γ22y)β1x2).
Period-two solution satisfies
(4.23)β12x2B1β1x2+(α2+γ2y)(B1x+y)-x=0,(B1x+y)(α2x+γ2α2+γ22y)β1x2-y=0,β12x=B1β1x2+(α2+γ2y)(B1x+y),β1yx2=(B1x+y)(α2x+α2γ2+γ22y).
From this system we have(i)
x=(1/2B1)(β1+γ2B1+(β1-B1γ2)2-4B1α2),y=(1/2B1)(β1-γ2B1-(β1-B1γ2)2-4B1α2),(ii)
x=(1/2B1)(β1+γ2B1-(β1-B1γ2)2-4B1α2),y=(1/2B1)(β1-γ2B1+(β1-B1γ2)2-4B1α2),(iii)
x=-γ2+(1/2β12)(α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1),y=-(1/2β1γ2)(α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1),(iv)
x=-γ2+(1/2β12)(α2β1+β12γ2-B1α2γ2+B1β1γ22+Δ1),y=-(1/2β1γ2)(α2β1+β12γ2-B1α2γ2+B1β1γ22+Δ1),(v)
x=0,y=-α/γ,where Δ1=(α-βγ)(αβ2+3β3γ+B2αγ2+2Bβ2γ2-B2βγ3-2Bαβγ).
In cases (i) and (ii) solutions(x,y) are equilibrium points E+ and E-, and in case (v) solution (x,y) is not in the first quadrant in the plane. It is sufficient to prove that solutions (x,y) in cases (iii) and (iv) are not in the first quadrant in the plane. Namely, if Δ1<0, x and y are not real. Supose that Δ1≥0. If α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1≥0, then y≤0. If α2β1+β12γ2-B1α2γ2+B1β1γ22-Δ1<0, then x<0 for solution in case (iii). By analogous reasoning we have that the same conclusion for case (iv) holds.Our linearized stability analysis indicates that there are three cases with different asymptotic behavior, depending on the values of parametersβ1,B1,α2, and γ2.Case 1.
β1-B1γ2>2B1α2.Case 2.
β1-B1γ2=2B1α2.Case 3.
β1-B1γ2≤0 or 0<β1-B1γ2<2B1α2.


### 4.2.1. Global Results—Case1

Theorem 4.3.
Consider system (4.1) and assume that β1-B1γ2>2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E-. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.24)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey}𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.25)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(β1+γ2B1+D32B1,β1-γ2B1-D32)forevery(x0,y0)∈𝒲+Proof.
Clearly, system (4.1) is strongly competitive on ℛ=(0,∞)×[0,∞). In view of injectivity of T, invertibility of JT, and Theorem 4.2, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


### 4.2.2. Global Results—Case2

Theorem 4.4.
Consider system (4.1) and assume that β1-B1γ2=2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.26)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.27)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=E=(B1γ2+β12B1,β1-B1γ22)forevery(x0,y0)∈𝒲+.Proof.
In this case system (4.1) has a unique equilibrium point E=((B1γ2+β1)/2B1,(β1-B1γ2)/2) which is nonhyperbolic. For p=q+1, the corresponding characteristic equation is of the form
(4.28)λ2-pλ+p-1=0.
This implies
(4.29)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. We will show that p<1. Indeed
(4.30)p<1⇔B1α2β1+γ2B1B1γ2+B1α2<1⇔B1γ2B1α2+B1α2+β1γ2B1<β1γ2B1+β1B1α2⇔B1γ2+B1α2<β1=B1γ2+2B1α2
which is satisfied. Thus, λ1∈(-1,0).
The eigenvector corresponding toλ1=p-1 is
(4.31)(12B1β1(β1-B1γ2)(B1γ2+β1)2)=(14B1β1B1α2(B1γ2+β1)2).
It means that all conditions of Theorems 1.7 and 1.10 are satisfied with ℛ=(0,∞)×[0,∞).
Assume that(x0,y0)∈𝒲+. Then (xn,yn)∈𝒲+ for all n, and sequences {x2n},{x2n+1},{y2n}, and {y2n+1} are monotone and bounded since xn≤β1/B1. Thus these sequences are convergent, which in view of Theorem 4.2 shows that they converge to the equilibrium point. Since E is the unique equilibrium point in 𝒲+ the statement for 𝒲+ follows. The same conclusion is obtained by using Theorem 1.6.
If(x0,y0) is in 𝒲-, by Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬2(E). Assume (without loss of generality) that (x0,y0)∈int𝒬2(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(-1,B1). Choose a value of t small enough so that E+tv∈𝒬2(E) and (x0,y0)⪯E+tv. Let us show that T(E+tv)⪯E+tv. Indeed
(4.32)T(E+tv)=(B1γ2+β12B1-t,B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t)⪯(B1γ2+β12B1-t,β1-B1γ22+B1t)
because
(4.33)B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t≥β1-B1γ22+B1t
reduces to
(4.34)4B12t2+4B1α2+2B1β1γ2≥β12+B12γ22=4B1α2+2B1β1γ2,
where the last equality follows from the condition β1-B1γ2=2B1α2.
SinceT(E+tv)⪯E+tv, it follows that {Tn(E+tv)} is a monotonically decreasing sequence in 𝒬2(E) which is bounded above by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has second coordinate which is monotone and unbounded. But (xn,yn):=Tn(x0,y0)⪯Tn(E+tv), which implies that yn→∞. From (4.1) it follows that xn→0.


### 4.2.3. Global Results—Case3

Theorem 4.5.
Consider system (4.1) and assume that β1-B1γ2≤0 or 0<β1-B1γ2<2B1α2. Then every solution {(xn,yn)} of system (4.1) satisfies
(4.35)limn→∞xn=0,limn→∞yn=∞.Proof.
In this case system (4.1) has no equilibrium points. Consider now the following system satisfied by subsequences of the solution of system (4.1):
(4.36)x2k+1=β1x2kB1x2k+y2k,x2k+2=β1x2k+1B1x2k+1+y2k+1,y2k+1=α2+γ2y2kx2k,y2k+2=α2+γ2y2k+1x2k+1.
We know that each of the four subsequences
(4.37){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {(xn,yn)} of system (4.1) is eventually monotonic. The subsequences {x2k} and {x2k+1} are bounded (by β1/B1), which implies that they are convergent. Suppose that (a)limk→∞x2k=xE and (b)limk→∞x2k+1=xO. For the other two subsequences the following four cases are possible: (1) limk→∞y2k=yE, (2) limk→∞y2k=∞, (3) limk→∞y2k+1=yO, or (4) limk→∞y2k+1=∞.
Case1 and Case 3 imply
(4.38)xO=β1xEB1xE+yE,xE=β1xOB1xO+yO,yO=α2+γ2yExE,yE=α2+γ2yOxO
that is, system (4.1) has a period-two solution, which is a contradiction by Theorem 4.2.
Case1 and Case 4 imply
(4.39)limk→∞x2k=xE=0⇒limk→∞x2k+1=xO=0⇒limk→∞y2k+2=∞,
which is a contradiction by Case 1.
Case2 and Case 3 imply
(4.40)limk→∞x2k+1=xO=0⇒limk→∞x2k+2=xE=0⇒limk→∞y2k+1=∞,
which is a contradiction by Case 3.
Case2 and Case 4 imply
(4.41)limk→∞x2k=xE=limk→∞x2k+1=xO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (3.1).Corollary 4.6.
Consider system (4.1) and assume that β1-B1γ2≥2B1α2. All bounded solutions of system (4.1) converge to the unique equilibrium with the corresponding initial conditions which belong to region below and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start above 𝒞 are asymptotic to (0,∞).
Consider system (4.1) and assume that either β1-B1γ2≤0 or β1-B1γ2<2B1α2. Then every solution of (4.1) is asymptotic to (0,∞).


## 4.2.1. Global Results—Case1

Theorem 4.3.
Consider system (4.1) and assume that β1-B1γ2>2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E-. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.24)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey}𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.25)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(β1+γ2B1+D32B1,β1-γ2B1-D32)forevery(x0,y0)∈𝒲+Proof.
Clearly, system (4.1) is strongly competitive on ℛ=(0,∞)×[0,∞). In view of injectivity of T, invertibility of JT, and Theorem 4.2, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


## 4.2.2. Global Results—Case2

Theorem 4.4.
Consider system (4.1) and assume that β1-B1γ2=2B1α2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(4.26)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(4.27)limn→∞(xn,yn)=(0,∞)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=E=(B1γ2+β12B1,β1-B1γ22)forevery(x0,y0)∈𝒲+.Proof.
In this case system (4.1) has a unique equilibrium point E=((B1γ2+β1)/2B1,(β1-B1γ2)/2) which is nonhyperbolic. For p=q+1, the corresponding characteristic equation is of the form
(4.28)λ2-pλ+p-1=0.
This implies
(4.29)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. We will show that p<1. Indeed
(4.30)p<1⇔B1α2β1+γ2B1B1γ2+B1α2<1⇔B1γ2B1α2+B1α2+β1γ2B1<β1γ2B1+β1B1α2⇔B1γ2+B1α2<β1=B1γ2+2B1α2
which is satisfied. Thus, λ1∈(-1,0).
The eigenvector corresponding toλ1=p-1 is
(4.31)(12B1β1(β1-B1γ2)(B1γ2+β1)2)=(14B1β1B1α2(B1γ2+β1)2).
It means that all conditions of Theorems 1.7 and 1.10 are satisfied with ℛ=(0,∞)×[0,∞).
Assume that(x0,y0)∈𝒲+. Then (xn,yn)∈𝒲+ for all n, and sequences {x2n},{x2n+1},{y2n}, and {y2n+1} are monotone and bounded since xn≤β1/B1. Thus these sequences are convergent, which in view of Theorem 4.2 shows that they converge to the equilibrium point. Since E is the unique equilibrium point in 𝒲+ the statement for 𝒲+ follows. The same conclusion is obtained by using Theorem 1.6.
If(x0,y0) is in 𝒲-, by Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬2(E). Assume (without loss of generality) that (x0,y0)∈int𝒬2(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(-1,B1). Choose a value of t small enough so that E+tv∈𝒬2(E) and (x0,y0)⪯E+tv. Let us show that T(E+tv)⪯E+tv. Indeed
(4.32)T(E+tv)=(B1γ2+β12B1-t,B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t)⪯(B1γ2+β12B1-t,β1-B1γ22+B1t)
because
(4.33)B1(2α2+β1γ2-B1γ22+2B1γ2t)B1γ2+β1+2B1t≥β1-B1γ22+B1t
reduces to
(4.34)4B12t2+4B1α2+2B1β1γ2≥β12+B12γ22=4B1α2+2B1β1γ2,
where the last equality follows from the condition β1-B1γ2=2B1α2.
SinceT(E+tv)⪯E+tv, it follows that {Tn(E+tv)} is a monotonically decreasing sequence in 𝒬2(E) which is bounded above by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has second coordinate which is monotone and unbounded. But (xn,yn):=Tn(x0,y0)⪯Tn(E+tv), which implies that yn→∞. From (4.1) it follows that xn→0.


## 4.2.3. Global Results—Case3

Theorem 4.5.
Consider system (4.1) and assume that β1-B1γ2≤0 or 0<β1-B1γ2<2B1α2. Then every solution {(xn,yn)} of system (4.1) satisfies
(4.35)limn→∞xn=0,limn→∞yn=∞.Proof.
In this case system (4.1) has no equilibrium points. Consider now the following system satisfied by subsequences of the solution of system (4.1):
(4.36)x2k+1=β1x2kB1x2k+y2k,x2k+2=β1x2k+1B1x2k+1+y2k+1,y2k+1=α2+γ2y2kx2k,y2k+2=α2+γ2y2k+1x2k+1.
We know that each of the four subsequences
(4.37){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {(xn,yn)} of system (4.1) is eventually monotonic. The subsequences {x2k} and {x2k+1} are bounded (by β1/B1), which implies that they are convergent. Suppose that (a)limk→∞x2k=xE and (b)limk→∞x2k+1=xO. For the other two subsequences the following four cases are possible: (1) limk→∞y2k=yE, (2) limk→∞y2k=∞, (3) limk→∞y2k+1=yO, or (4) limk→∞y2k+1=∞.
Case1 and Case 3 imply
(4.38)xO=β1xEB1xE+yE,xE=β1xOB1xO+yO,yO=α2+γ2yExE,yE=α2+γ2yOxO
that is, system (4.1) has a period-two solution, which is a contradiction by Theorem 4.2.
Case1 and Case 4 imply
(4.39)limk→∞x2k=xE=0⇒limk→∞x2k+1=xO=0⇒limk→∞y2k+2=∞,
which is a contradiction by Case 1.
Case2 and Case 3 imply
(4.40)limk→∞x2k+1=xO=0⇒limk→∞x2k+2=xE=0⇒limk→∞y2k+1=∞,
which is a contradiction by Case 3.
Case2 and Case 4 imply
(4.41)limk→∞x2k=xE=limk→∞x2k+1=xO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (3.1).Corollary 4.6.
Consider system (4.1) and assume that β1-B1γ2≥2B1α2. All bounded solutions of system (4.1) converge to the unique equilibrium with the corresponding initial conditions which belong to region below and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start above 𝒞 are asymptotic to (0,∞).
Consider system (4.1) and assume that either β1-B1γ2≤0 or β1-B1γ2<2B1α2. Then every solution of (4.1) is asymptotic to (0,∞).


## 5. System (21,38)

Now we consider the following system of difference equations:(5.1)xn+1=α1+β1xnyn,yn+1=γ2ynA2+B2xn+yn,n=0,1,…,
where the parameters α1,β1,A2,B2, and γ2 are positive numbers and initial conditions x0≥0,y0>0.The Jacobian matrix of the corresponding mapT(x,y)=((α1+β1x)/y,γ2y/(A2+B2x+y)) is of the form(5.2)JT(x,y)=[β1y-α1+β1xy2-γ2B2y(A2+B2x+y)2γ2(A2+B2x)(A2+B2x+y)2].System (5.1) was considered in [1, Example 4], where it was shown that the map T is injective. In addition, when(5.3)β1A2>α1B2,
we see that(5.4)detJT(x,y)=(β1A2-α1B2)(A2+B2x+y)2y>0.
Therefore, when (5.3) holds, the Jacobian matrix of T in (x,y) is invertible and in view of Theorems 1.4 and 1.5 every solution of system (5.1) is eventually componentwise monotonic.When(5.5)β1A2<α1B2,
we see that(5.6)detJT(x,y)=(β1A2-α1B2)(A2+B2x+y)2y<0,
and the Jacobian matrix of T in (x,y) is invertible. Therefore, in view of Theorems 1.4 and 1.5, four subsequences(5.7){x2n},{x2n+1},{y2n},{y2n+1}
of every solution {(xn,yn)} of system (5.1) are eventually monotonic.


### 5.1. Linearized Stability Analysis

Equilibrium points of system (5.1) are solutions of the system(5.8)x¯=α1+β1x¯y¯,y¯=γ2y¯A2+B2x¯+y¯.
Since y¯≠0, we have(5.9)x¯±=12B2(γ2-A2-β1±D4),
where(5.10)D4=(γ2-A2-β1)2-4α1B2.
It is easy to prove that the following result holds.Lemma 5.1.
(i) Ifγ2-A2-β1>2α1B2, then system (5.1) has two equilibrium points:
(5.11)E+=(γ2-A2-β1+D42B2,γ2-A2+β1-D42),E-=(γ2-A2-β1-D42B2,γ2-A2+β1+D42).
(ii) Ifγ2-A2-β1=2α1B2, then system (5.1) has a unique equilibrium point:
(5.12)E=(γ2-A2-β12B2,γ2-A2+β12).
(iii) Ifγ2≤A2+β1 or 0<γ2-A2-β1<2α1B2, then system (5.1) has no equilibrium points.Lemma 5.2.
Ifγ2-A2-β1>2α1B2, then the equilibrium point E+ of system (5.1) is a saddle point and E- is locally asymptotically stable.
Ifγ2-A2-β1=2α1B2, then the equilibrium point E of system (5.1) is nonhyperbolic.Proof.
By using (5.8) we have
(5.13)JT(x¯,y¯)=[β1y¯-x¯y¯-B2y¯γ21-y¯γ2].
The corresponding characteristic equation evaluated at the equilibrium point E(x¯,y¯) is
(5.14)λ2-pλ+q=0,
where
(5.15)p=TrJT(x¯,y¯)=β1y¯+1-y¯γ2,q=DetJT(x¯,y¯)=β1y¯-β1γ2-B2x¯γ2.
For the equilibrium pointE+ we need to prove that
(5.16)|p|>|1+q|,p2-4q>0,
that is (because p>0 and 1+q>0),(I)
p>1+q,(II)
p2-4q>0.Indeed
(5.17)p=β1y¯+1-y¯γ2=1γ2(β1γ2y¯+γ2-y¯)=1γ2(β1γ2y¯+A2+B2x¯)>0,
which is always true, and in view of (5.8) γ2-B2x̅=y̅+A2 we obtain
(5.18)1+q>0⇔1+β1y¯-β1γ2-B2x¯γ2>0⇔γ2+β1γ2y¯-β1-B2x¯>0⇔y¯+A2+β1γ2-y¯y¯>0,
which is true because y¯<γ2.
Next, in view of (5.8) γ2-B2x̅=y̅+A2,
(I)
we have(5.19)p>1+q⇔β1y¯+1-y¯γ2>1+β1y¯-β1γ2-B2x¯γ2⇔y¯<β1+B2x¯⇔γ2-A2-B2x¯<β1+B2x¯⇔0<D4,
which is true.Similarly,(II)
we have(5.20)p2-4q>0⇔(β1y¯+1-y¯γ2)2-4β1y¯+4β1γ2+4B2x¯γ2>0⇔(β1y¯)2+2β1y¯+1-2y¯γ2β1y¯-2y¯γ2+(y¯γ2)2-4β1y¯+4β1γ2+4B2x¯γ2>0⇔(β1y¯-1+y¯γ2)2+4B2x¯γ2>0,
which is true.
For the equilibrium pointE- we need to prove that
(5.21)p<1+q<2,
or equivalently(I)
p<1+q,(II)
q<1.Indeed, (I)
we have(5.22)p<1+q⇔β1y¯+1-y¯γ2<1+β1y¯-β1γ2-B2x¯γ2⇔y¯>β1+B2x¯⇔γ2-A2-B2x¯>β1+B2x¯⇔0>-D4,
which is true, and in view of (5.8) β1/y̅=1-α1/x̅y̅(II)
we obtain(5.23)q<1⇔β1y¯-β1γ2-B2x¯γ2<1⇔1-α1x¯y¯-β1γ2-B2x¯γ2<1⇔-α1x¯y¯-β1γ2-B2x¯γ2<0,
which is true.
Assume thatγ2-A2-β1=2α1B2.
Let us prove that1+q=p. We have
(5.24)1+q=p⇔1+β1y¯-β1γ2-B2x¯γ2=β1y¯+1-y¯γ2⇔y¯=β1+B2x¯⇔2y¯=β1+γ2-A2,
which is true.


### 5.2. Global Results

Theorem 5.3.
System (5.1) has no prime period-two solutions.Proof.
The second iterate of the mapT is
(5.25)T2(x,y)=T(α1+β1xy,γ2yA2+B2x+y)=(α1+β1((α1+β1x)/y)γ2y/(A2+B2x+y),γ2(γ2y/(A2+B2x+y))A2+B2((α1+β1x)/y)+γ2y/(A2+B2x+y)),
that is,
(5.26)T2(x,y)=((A2+B2x+y)(α1y+α1β1+β12x)γ2y2γ22y2y[(A2+B2x+y)(A2y+B2α1+B2β1x)+γ2y2])
Period-two solutions satisfy
(5.27)(A2+B2x+y)(α1y+α1β1+β12x)γ2y2-x=0,γ22y2y[(A2+B2x+y)(A2y+B2α1+B2β1x)+γ2y2]-y=0.
From this system we have(i)
x=(γ2-A2-β1+D4)/2B2,y=(γ2-A2+β1-D4)/2,(ii)
x=(γ2-A2-β1-D4)/2B2,y=(γ2-A2+β1+D4)/2,(iii)
x=-A2/B2,y=0,(iv)
x=-α1/β1,y=0,(v)
x=-(1/B2)(A2+γ2)-(1/2B2β1γ2)(Λ+Δ2),y=(1/2γ2(A2+γ2))(Λ+Δ2),(vi)
x=-(1/B)(A+γ)-(1/2B2β1γ2)(Λ-Δ2),y=(1/2γ2(A2+γ2))(Λ-Δ2),Where
(5.28)Λ=A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2,Δ2=-(A2β1-B2α1+β1γ2)(2A22β12+2β12γ22-A2β13-A23β1+3β1γ23-β13γ2+A22B2α1+A2β1γ22+4A2β12γ2+B2α1β12-3A22β1γ2+B2α1γ22-2A2B2α1β1+2A2B2α1γ2-2B2α1β1γ2).
In cases (i) and (ii) solutions (x,y) are the equilibrium points E+ and E-, and in cases (iii) and (iv) solution(x,y) is not in the first quadrant in the plane. It is sufficient to prove that solutions (x,y) in cases (v) and (vi) are not in the first quadrant in the plane. Namely,  Δ2<0 implies that x and y are not real. Suppose that Δ2≥0. If
(5.29)A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2+Δ2≥0,
then y≥0 and x<0. If
(5.30)A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2+Δ2<0,
then y<0 for solution in the case (v). By analogous reasoning we have that the same conclusion for case (vi) holds.Remark 5.4 (see [1]).
When(5.31)β1A2=α1B2,
we see that
(5.32)yn+1=β1γ2B2xn+1+β1,n=0,1,…,
and so system (5.1) can be decoupled as follows:
(5.33)xn+1=(α1+β1xn)(β1+B2xn)β1γ2,yn+1=γ2yn2yn2+(A2-β1)yn+β1γ2,n=0,1,….
The solutions of first equation (depending of the choice of the initial condition x0) are either bounded and converge to a finite limit or increase monotonically to infinity. Using this and (5.32) we find the behavior of solutions of second equation.Our linearized stability analysis indicates that there are three cases with different asymptotic behavior, depending on the values of parametersβ1,B1,α2, and γ2:Case 1.
γ2-A2-β1>2α1B2.Case 2.
γ2-A2-β1=2α1B2.Case 3.
γ2≤A2+β1 or 0<γ2-A2-β1<2α1B2.


#### 5.2.1. Caseγ2−A2−β1>2α1B2

Theorem 5.5.
Consider system (5.1) and assume that γ2-A2-β1>2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E+. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(5.34)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex}.
which satisfy
(5.35)limn→∞(xn,yn)=(γ2-A2-β1-D42B2,γ2-A2+β1+D42)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (5.1) is strongly competitive on ℛ=[0,∞)×(0,∞). In view of the injectivity of T, the invertibility of JT and Theorem 5.3, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


#### 5.2.2. Caseγ2−A2−β1=2α1B2

Theorem 5.6.
Consider system (5.1) and assume that γ2-A2-β1=2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely
(5.36)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(5.37)limn→∞(xn,yn)=E=(γ2-A2-β12B2,γ2-A2+β12)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
In this case system (5.1) has a unique equilibrium point E=((γ2-A2-β1)/2B2,(γ2-A2+β1)/2) which is nonhyperbolic. By p=q+1, the corresponding characteristic equation is of the form
(5.38)λ2-pλ+p-1=0.
This implies
(5.39)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. In view of  β1/y¯=1-α1/x¯y¯, we have
(5.40)p<2⇔β1y¯+1-y¯γ2<2⇔1-α1x¯y¯+1-y¯γ2<2,
which is satisfied. Thus, |λ1|<1.
The eigenvector corresponding toλ1=p-1 is
(5.41)(1B2(γ2-A2+β1)22γ2(γ2-A2-β1))=(1B2(α1B2+β1)2γ2α1B2).
It means that all conditions of  Theorems1.7 and 1.10 are satisfied with ℛ=[0,∞)×(0,∞). In view of the fact that yn≤γ2 we obtain the conclusion of the theorem in the case (x0,y0)∈𝒲-. The same conclusion is obtained by using Theorem 1.6.
Next, assume that(x0,y0)∈𝒲+. By Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬4(E). Assume (without loss of generality) that (x0,y0)∈int𝒬4(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(1,-B2). Choose a value of t small enough so that E+tv∈𝒬4(E) and E+tv⪯(x0,y0). Let us show that E+tv⪯T(E+tv). Indeed
(5.42)T(E+tv)=(2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t),γ2-A2+β12-B2t)≽(γ2-A2-β12B2+t,γ2-A2+β12-B2t)=E+tv
because
(5.43)2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t)≥γ2-A2-β12B2+t
reduces to
(5.44)(γ2-A2-β1)2≤4B2t2+4α1B2=4B2t2+(γ2-A2-β1)2,
where the last equality follows from the condition γ2-A2-β1=2α1B2.
SinceE+tv⪯T(E+tv), it follows that {Tn(E+tv)} is a monotonically increasing sequence in 𝒬4(E) which is bounded below by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has a first coordinate which is monotone and unbounded. But Tn(E+tv)⪯(xn,yn):=Tn(x0,y0), which implies that xn→∞. From (5.1) it follows that yn→0.


#### 5.2.3. Caseγ2≤A2+β1 or 0<γ2−A2−β1<2α1B2

In this case system (5.1) has no equilibrium points.Theorem 5.7.
Consider system (5.1) and assume that γ2≤A2+β1 or 0<γ2-A2-β1<2α1B2 and β1A2≠α1B2. Then every solution {(xn,yn)} of system (4.1) satisfies
(5.45)limn→∞xn=∞,limn→∞yn=0.Proof.
(1) Assume that β1A2>α1B2. Then every solution of system (5.1) is eventually componentwise monotonic. The sequence {yn} is bounded (by γ2), which implies that it converges, that is, limn→∞yn=Y. For the sequence {xn} the following two cases are possible: (a) limn→∞xn=X,  (b) limn→∞xn=∞.
Iflimn→∞xn=X, then we obtain
(5.46)X=α1+β1XY,Y=γ2YA2+B2X+Y;
that is, (X,Y) is an equilibrium point of system (5.1), which is a contradiction.
Iflimn→∞xn=∞, then limn→∞yn=Y=0.
(2) Assume that β1A2<α1B2. Consider now the following system satisfied by subsequences of the solution of system (5.1):
(5.47)x2k+1=α1+β1x2ky2k,x2k+2=α1+β1x2k+1y2k+1,y2k+1=γ2y2kA2+B2x2k+y2k,y2k+2=γ2y2k+1A2+B2x2k+1+y2k+1.
We know that each of the four subsequences
(5.48){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {xk,yk} of system (5.1) is eventually monotonic. The subsequences {y2k} and {y2k+1} are bounded by γ2, which implies that they are convergent. Suppose that  (a) limk→∞y2k=yE and (b) limk→∞y2k+1=yO. For the other two subsequences the following four cases are possible: (1) limk→∞x2k=xE,  (2) limk→∞x2k=∞,  (3) limk→∞x2k+1=xO, or (4) limk→∞x2k+1=∞.
By similar reasoning as in the proof of Theorem4.5, we obtain
(5.49)limk→∞y2k=yE=limk→∞y2k+1=yO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (5.1).Corollary 5.8.
Consider system (5.1) and assume that β1A2≠α1B2. If γ2-A2-β1≥2B2α1, then all bounded solutions of system (5.1) converge to the unique equilibrium with the corresponding initial conditions which belong to the region above and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start below 𝒞 are asymptotic to (∞,0).
Consider system (5.1) and assume that either γ2≤A2+β1 or 0<γ2-β1-A2γ2<2B2α1. Then every solution of (5.1) is asymptotic to (∞,0).


## 5.1. Linearized Stability Analysis

Equilibrium points of system (5.1) are solutions of the system(5.8)x¯=α1+β1x¯y¯,y¯=γ2y¯A2+B2x¯+y¯.
Since y¯≠0, we have(5.9)x¯±=12B2(γ2-A2-β1±D4),
where(5.10)D4=(γ2-A2-β1)2-4α1B2.
It is easy to prove that the following result holds.Lemma 5.1.
(i) Ifγ2-A2-β1>2α1B2, then system (5.1) has two equilibrium points:
(5.11)E+=(γ2-A2-β1+D42B2,γ2-A2+β1-D42),E-=(γ2-A2-β1-D42B2,γ2-A2+β1+D42).
(ii) Ifγ2-A2-β1=2α1B2, then system (5.1) has a unique equilibrium point:
(5.12)E=(γ2-A2-β12B2,γ2-A2+β12).
(iii) Ifγ2≤A2+β1 or 0<γ2-A2-β1<2α1B2, then system (5.1) has no equilibrium points.Lemma 5.2.
Ifγ2-A2-β1>2α1B2, then the equilibrium point E+ of system (5.1) is a saddle point and E- is locally asymptotically stable.
Ifγ2-A2-β1=2α1B2, then the equilibrium point E of system (5.1) is nonhyperbolic.Proof.
By using (5.8) we have
(5.13)JT(x¯,y¯)=[β1y¯-x¯y¯-B2y¯γ21-y¯γ2].
The corresponding characteristic equation evaluated at the equilibrium point E(x¯,y¯) is
(5.14)λ2-pλ+q=0,
where
(5.15)p=TrJT(x¯,y¯)=β1y¯+1-y¯γ2,q=DetJT(x¯,y¯)=β1y¯-β1γ2-B2x¯γ2.
For the equilibrium pointE+ we need to prove that
(5.16)|p|>|1+q|,p2-4q>0,
that is (because p>0 and 1+q>0),(I)
p>1+q,(II)
p2-4q>0.Indeed
(5.17)p=β1y¯+1-y¯γ2=1γ2(β1γ2y¯+γ2-y¯)=1γ2(β1γ2y¯+A2+B2x¯)>0,
which is always true, and in view of (5.8) γ2-B2x̅=y̅+A2 we obtain
(5.18)1+q>0⇔1+β1y¯-β1γ2-B2x¯γ2>0⇔γ2+β1γ2y¯-β1-B2x¯>0⇔y¯+A2+β1γ2-y¯y¯>0,
which is true because y¯<γ2.
Next, in view of (5.8) γ2-B2x̅=y̅+A2,
(I)
we have(5.19)p>1+q⇔β1y¯+1-y¯γ2>1+β1y¯-β1γ2-B2x¯γ2⇔y¯<β1+B2x¯⇔γ2-A2-B2x¯<β1+B2x¯⇔0<D4,
which is true.Similarly,(II)
we have(5.20)p2-4q>0⇔(β1y¯+1-y¯γ2)2-4β1y¯+4β1γ2+4B2x¯γ2>0⇔(β1y¯)2+2β1y¯+1-2y¯γ2β1y¯-2y¯γ2+(y¯γ2)2-4β1y¯+4β1γ2+4B2x¯γ2>0⇔(β1y¯-1+y¯γ2)2+4B2x¯γ2>0,
which is true.
For the equilibrium pointE- we need to prove that
(5.21)p<1+q<2,
or equivalently(I)
p<1+q,(II)
q<1.Indeed, (I)
we have(5.22)p<1+q⇔β1y¯+1-y¯γ2<1+β1y¯-β1γ2-B2x¯γ2⇔y¯>β1+B2x¯⇔γ2-A2-B2x¯>β1+B2x¯⇔0>-D4,
which is true, and in view of (5.8) β1/y̅=1-α1/x̅y̅(II)
we obtain(5.23)q<1⇔β1y¯-β1γ2-B2x¯γ2<1⇔1-α1x¯y¯-β1γ2-B2x¯γ2<1⇔-α1x¯y¯-β1γ2-B2x¯γ2<0,
which is true.
Assume thatγ2-A2-β1=2α1B2.
Let us prove that1+q=p. We have
(5.24)1+q=p⇔1+β1y¯-β1γ2-B2x¯γ2=β1y¯+1-y¯γ2⇔y¯=β1+B2x¯⇔2y¯=β1+γ2-A2,
which is true.


## 5.2. Global Results

Theorem 5.3.
System (5.1) has no prime period-two solutions.Proof.
The second iterate of the mapT is
(5.25)T2(x,y)=T(α1+β1xy,γ2yA2+B2x+y)=(α1+β1((α1+β1x)/y)γ2y/(A2+B2x+y),γ2(γ2y/(A2+B2x+y))A2+B2((α1+β1x)/y)+γ2y/(A2+B2x+y)),
that is,
(5.26)T2(x,y)=((A2+B2x+y)(α1y+α1β1+β12x)γ2y2γ22y2y[(A2+B2x+y)(A2y+B2α1+B2β1x)+γ2y2])
Period-two solutions satisfy
(5.27)(A2+B2x+y)(α1y+α1β1+β12x)γ2y2-x=0,γ22y2y[(A2+B2x+y)(A2y+B2α1+B2β1x)+γ2y2]-y=0.
From this system we have(i)
x=(γ2-A2-β1+D4)/2B2,y=(γ2-A2+β1-D4)/2,(ii)
x=(γ2-A2-β1-D4)/2B2,y=(γ2-A2+β1+D4)/2,(iii)
x=-A2/B2,y=0,(iv)
x=-α1/β1,y=0,(v)
x=-(1/B2)(A2+γ2)-(1/2B2β1γ2)(Λ+Δ2),y=(1/2γ2(A2+γ2))(Λ+Δ2),(vi)
x=-(1/B)(A+γ)-(1/2B2β1γ2)(Λ-Δ2),y=(1/2γ2(A2+γ2))(Λ-Δ2),Where
(5.28)Λ=A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2,Δ2=-(A2β1-B2α1+β1γ2)(2A22β12+2β12γ22-A2β13-A23β1+3β1γ23-β13γ2+A22B2α1+A2β1γ22+4A2β12γ2+B2α1β12-3A22β1γ2+B2α1γ22-2A2B2α1β1+2A2B2α1γ2-2B2α1β1γ2).
In cases (i) and (ii) solutions (x,y) are the equilibrium points E+ and E-, and in cases (iii) and (iv) solution(x,y) is not in the first quadrant in the plane. It is sufficient to prove that solutions (x,y) in cases (v) and (vi) are not in the first quadrant in the plane. Namely,  Δ2<0 implies that x and y are not real. Suppose that Δ2≥0. If
(5.29)A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2+Δ2≥0,
then y≥0 and x<0. If
(5.30)A2β12-A22β1-β1γ22+β12γ2+A2B2α1-2A2β1γ2-B2α1β1+B2α1γ2+Δ2<0,
then y<0 for solution in the case (v). By analogous reasoning we have that the same conclusion for case (vi) holds.Remark 5.4 (see [1]).
When(5.31)β1A2=α1B2,
we see that
(5.32)yn+1=β1γ2B2xn+1+β1,n=0,1,…,
and so system (5.1) can be decoupled as follows:
(5.33)xn+1=(α1+β1xn)(β1+B2xn)β1γ2,yn+1=γ2yn2yn2+(A2-β1)yn+β1γ2,n=0,1,….
The solutions of first equation (depending of the choice of the initial condition x0) are either bounded and converge to a finite limit or increase monotonically to infinity. Using this and (5.32) we find the behavior of solutions of second equation.Our linearized stability analysis indicates that there are three cases with different asymptotic behavior, depending on the values of parametersβ1,B1,α2, and γ2:Case 1.
γ2-A2-β1>2α1B2.Case 2.
γ2-A2-β1=2α1B2.Case 3.
γ2≤A2+β1 or 0<γ2-A2-β1<2α1B2.


### 5.2.1. Caseγ2−A2−β1>2α1B2

Theorem 5.5.
Consider system (5.1) and assume that γ2-A2-β1>2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E+. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(5.34)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex}.
which satisfy
(5.35)limn→∞(xn,yn)=(γ2-A2-β1-D42B2,γ2-A2+β1+D42)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (5.1) is strongly competitive on ℛ=[0,∞)×(0,∞). In view of the injectivity of T, the invertibility of JT and Theorem 5.3, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


### 5.2.2. Caseγ2−A2−β1=2α1B2

Theorem 5.6.
Consider system (5.1) and assume that γ2-A2-β1=2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely
(5.36)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(5.37)limn→∞(xn,yn)=E=(γ2-A2-β12B2,γ2-A2+β12)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
In this case system (5.1) has a unique equilibrium point E=((γ2-A2-β1)/2B2,(γ2-A2+β1)/2) which is nonhyperbolic. By p=q+1, the corresponding characteristic equation is of the form
(5.38)λ2-pλ+p-1=0.
This implies
(5.39)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. In view of  β1/y¯=1-α1/x¯y¯, we have
(5.40)p<2⇔β1y¯+1-y¯γ2<2⇔1-α1x¯y¯+1-y¯γ2<2,
which is satisfied. Thus, |λ1|<1.
The eigenvector corresponding toλ1=p-1 is
(5.41)(1B2(γ2-A2+β1)22γ2(γ2-A2-β1))=(1B2(α1B2+β1)2γ2α1B2).
It means that all conditions of  Theorems1.7 and 1.10 are satisfied with ℛ=[0,∞)×(0,∞). In view of the fact that yn≤γ2 we obtain the conclusion of the theorem in the case (x0,y0)∈𝒲-. The same conclusion is obtained by using Theorem 1.6.
Next, assume that(x0,y0)∈𝒲+. By Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬4(E). Assume (without loss of generality) that (x0,y0)∈int𝒬4(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(1,-B2). Choose a value of t small enough so that E+tv∈𝒬4(E) and E+tv⪯(x0,y0). Let us show that E+tv⪯T(E+tv). Indeed
(5.42)T(E+tv)=(2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t),γ2-A2+β12-B2t)≽(γ2-A2-β12B2+t,γ2-A2+β12-B2t)=E+tv
because
(5.43)2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t)≥γ2-A2-β12B2+t
reduces to
(5.44)(γ2-A2-β1)2≤4B2t2+4α1B2=4B2t2+(γ2-A2-β1)2,
where the last equality follows from the condition γ2-A2-β1=2α1B2.
SinceE+tv⪯T(E+tv), it follows that {Tn(E+tv)} is a monotonically increasing sequence in 𝒬4(E) which is bounded below by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has a first coordinate which is monotone and unbounded. But Tn(E+tv)⪯(xn,yn):=Tn(x0,y0), which implies that xn→∞. From (5.1) it follows that yn→0.


### 5.2.3. Caseγ2≤A2+β1 or 0<γ2−A2−β1<2α1B2

In this case system (5.1) has no equilibrium points.Theorem 5.7.
Consider system (5.1) and assume that γ2≤A2+β1 or 0<γ2-A2-β1<2α1B2 and β1A2≠α1B2. Then every solution {(xn,yn)} of system (4.1) satisfies
(5.45)limn→∞xn=∞,limn→∞yn=0.Proof.
(1) Assume that β1A2>α1B2. Then every solution of system (5.1) is eventually componentwise monotonic. The sequence {yn} is bounded (by γ2), which implies that it converges, that is, limn→∞yn=Y. For the sequence {xn} the following two cases are possible: (a) limn→∞xn=X,  (b) limn→∞xn=∞.
Iflimn→∞xn=X, then we obtain
(5.46)X=α1+β1XY,Y=γ2YA2+B2X+Y;
that is, (X,Y) is an equilibrium point of system (5.1), which is a contradiction.
Iflimn→∞xn=∞, then limn→∞yn=Y=0.
(2) Assume that β1A2<α1B2. Consider now the following system satisfied by subsequences of the solution of system (5.1):
(5.47)x2k+1=α1+β1x2ky2k,x2k+2=α1+β1x2k+1y2k+1,y2k+1=γ2y2kA2+B2x2k+y2k,y2k+2=γ2y2k+1A2+B2x2k+1+y2k+1.
We know that each of the four subsequences
(5.48){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {xk,yk} of system (5.1) is eventually monotonic. The subsequences {y2k} and {y2k+1} are bounded by γ2, which implies that they are convergent. Suppose that  (a) limk→∞y2k=yE and (b) limk→∞y2k+1=yO. For the other two subsequences the following four cases are possible: (1) limk→∞x2k=xE,  (2) limk→∞x2k=∞,  (3) limk→∞x2k+1=xO, or (4) limk→∞x2k+1=∞.
By similar reasoning as in the proof of Theorem4.5, we obtain
(5.49)limk→∞y2k=yE=limk→∞y2k+1=yO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (5.1).Corollary 5.8.
Consider system (5.1) and assume that β1A2≠α1B2. If γ2-A2-β1≥2B2α1, then all bounded solutions of system (5.1) converge to the unique equilibrium with the corresponding initial conditions which belong to the region above and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start below 𝒞 are asymptotic to (∞,0).
Consider system (5.1) and assume that either γ2≤A2+β1 or 0<γ2-β1-A2γ2<2B2α1. Then every solution of (5.1) is asymptotic to (∞,0).


## 5.2.1. Caseγ2−A2−β1>2α1B2

Theorem 5.5.
Consider system (5.1) and assume that γ2-A2-β1>2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E+. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely,
(5.34)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex}.
which satisfy
(5.35)limn→∞(xn,yn)=(γ2-A2-β1-D42B2,γ2-A2+β1+D42)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
Clearly, system (5.1) is strongly competitive on ℛ=[0,∞)×(0,∞). In view of the injectivity of T, the invertibility of JT and Theorem 5.3, we see that all conditions of Theorems 1.7, 1.9, and 1.10 and Corollary 1.8 are satisfied and the conclusion of the theorem follows.


## 5.2.2. Caseγ2−A2−β1=2α1B2

Theorem 5.6.
Consider system (5.1) and assume that γ2-A2-β1=2α1B2 and β1A2≠α1B2. Then there exists a set 𝒞⊂ℛ which is invariant and a subset of the basin of attraction of E. The set 𝒞 is a graph of a strictly increasing continuous function of the first variable on an interval (and so is a manifold) and separates ℛ into two connected and invariant components, namely
(5.36)𝒲-:={x∈ℛ∖𝒞:∃y∈𝒞withx⪯sey},𝒲+:={x∈ℛ∖𝒞:∃y∈𝒞withy⪯sex},
which satisfy
(5.37)limn→∞(xn,yn)=E=(γ2-A2-β12B2,γ2-A2+β12)forevery(x0,y0)∈𝒲-,limn→∞(xn,yn)=(∞,0)forevery(x0,y0)∈𝒲+.Proof.
In this case system (5.1) has a unique equilibrium point E=((γ2-A2-β1)/2B2,(γ2-A2+β1)/2) which is nonhyperbolic. By p=q+1, the corresponding characteristic equation is of the form
(5.38)λ2-pλ+p-1=0.
This implies
(5.39)λ1=p-1,λ2=1,
and |λ1|<1⇔|p-1|<1⇔0<p<2.
It is obvious thatp>0. In view of  β1/y¯=1-α1/x¯y¯, we have
(5.40)p<2⇔β1y¯+1-y¯γ2<2⇔1-α1x¯y¯+1-y¯γ2<2,
which is satisfied. Thus, |λ1|<1.
The eigenvector corresponding toλ1=p-1 is
(5.41)(1B2(γ2-A2+β1)22γ2(γ2-A2-β1))=(1B2(α1B2+β1)2γ2α1B2).
It means that all conditions of  Theorems1.7 and 1.10 are satisfied with ℛ=[0,∞)×(0,∞). In view of the fact that yn≤γ2 we obtain the conclusion of the theorem in the case (x0,y0)∈𝒲-. The same conclusion is obtained by using Theorem 1.6.
Next, assume that(x0,y0)∈𝒲+. By Theorem 1.10 the orbit of (x0,y0) eventually enters 𝒬4(E). Assume (without loss of generality) that (x0,y0)∈int𝒬4(E). An eigenvector associated with the nonhyperbolic eigenvalue λ2=1 is v=(1,-B2). Choose a value of t small enough so that E+tv∈𝒬4(E) and E+tv⪯(x0,y0). Let us show that E+tv⪯T(E+tv). Indeed
(5.42)T(E+tv)=(2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t),γ2-A2+β12-B2t)≽(γ2-A2-β12B2+t,γ2-A2+β12-B2t)=E+tv
because
(5.43)2α1B2+β1(γ2-A2-β1+2B2t)B2(γ2-A2+β1-2B2t)≥γ2-A2-β12B2+t
reduces to
(5.44)(γ2-A2-β1)2≤4B2t2+4α1B2=4B2t2+(γ2-A2-β1)2,
where the last equality follows from the condition γ2-A2-β1=2α1B2.
SinceE+tv⪯T(E+tv), it follows that {Tn(E+tv)} is a monotonically increasing sequence in 𝒬4(E) which is bounded below by E. Since {Tn(E+tv)} is coordinatewise monotone and it does not converge (if it did it would have to converge to E, which is impossible), we have that Tn(E+tv) has a first coordinate which is monotone and unbounded. But Tn(E+tv)⪯(xn,yn):=Tn(x0,y0), which implies that xn→∞. From (5.1) it follows that yn→0.


## 5.2.3. Caseγ2≤A2+β1 or 0<γ2−A2−β1<2α1B2

In this case system (5.1) has no equilibrium points.Theorem 5.7.
Consider system (5.1) and assume that γ2≤A2+β1 or 0<γ2-A2-β1<2α1B2 and β1A2≠α1B2. Then every solution {(xn,yn)} of system (4.1) satisfies
(5.45)limn→∞xn=∞,limn→∞yn=0.Proof.
(1) Assume that β1A2>α1B2. Then every solution of system (5.1) is eventually componentwise monotonic. The sequence {yn} is bounded (by γ2), which implies that it converges, that is, limn→∞yn=Y. For the sequence {xn} the following two cases are possible: (a) limn→∞xn=X,  (b) limn→∞xn=∞.
Iflimn→∞xn=X, then we obtain
(5.46)X=α1+β1XY,Y=γ2YA2+B2X+Y;
that is, (X,Y) is an equilibrium point of system (5.1), which is a contradiction.
Iflimn→∞xn=∞, then limn→∞yn=Y=0.
(2) Assume that β1A2<α1B2. Consider now the following system satisfied by subsequences of the solution of system (5.1):
(5.47)x2k+1=α1+β1x2ky2k,x2k+2=α1+β1x2k+1y2k+1,y2k+1=γ2y2kA2+B2x2k+y2k,y2k+2=γ2y2k+1A2+B2x2k+1+y2k+1.
We know that each of the four subsequences
(5.48){x2k},{x2k+1},{y2k},{y2k+1}
of every solution {xk,yk} of system (5.1) is eventually monotonic. The subsequences {y2k} and {y2k+1} are bounded by γ2, which implies that they are convergent. Suppose that  (a) limk→∞y2k=yE and (b) limk→∞y2k+1=yO. For the other two subsequences the following four cases are possible: (1) limk→∞x2k=xE,  (2) limk→∞x2k=∞,  (3) limk→∞x2k+1=xO, or (4) limk→∞x2k+1=∞.
By similar reasoning as in the proof of Theorem4.5, we obtain
(5.49)limk→∞y2k=yE=limk→∞y2k+1=yO=0.The obtained results lead to the following characterization of the boundedness of solutions of system (5.1).Corollary 5.8.
Consider system (5.1) and assume that β1A2≠α1B2. If γ2-A2-β1≥2B2α1, then all bounded solutions of system (5.1) converge to the unique equilibrium with the corresponding initial conditions which belong to the region above and on the graph of a continuous increasing function 𝒞 in the plane of initial conditions. All solutions that start below 𝒞 are asymptotic to (∞,0).
Consider system (5.1) and assume that either γ2≤A2+β1 or 0<γ2-β1-A2γ2<2B2α1. Then every solution of (5.1) is asymptotic to (∞,0).



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*Source: 153058-2009-11-19.xml*