Open Access

Local dynamics and global attractivity of a certain second-order quadratic fractional difference equation

Advances in Difference Equations20142014:68

https://doi.org/10.1186/1687-1847-2014-68

Received: 24 December 2013

Accepted: 5 February 2014

Published: 18 February 2014

Abstract

We investigate the local and global character of the equilibrium and the local stability of the period-two solution of the difference equation x n + 1 = β x n x n 1 + γ x n 1 2 + δ x n B x n x n 1 + C x n 1 2 + D x n where the parameters β, γ, δ, B, C, D are nonnegative numbers which satisfy B + C + D > 0 and the initial conditions x 1 and x 0 are arbitrary nonnegative numbers such that B x n x n 1 + C x n 1 2 + D x n > 0 for all n 0 .

MSC:39A10, 39A11, 39A30.

Keywords

boundednessdifference equationglobal attractivitylocal stabilityperiod-two solutions

1 Introduction and preliminaries

In this paper we study the global dynamics of the following rational difference equation:
x n + 1 = β x n x n 1 + γ x n 1 2 + δ x n B x n x n 1 + C x n 1 2 + D x n , n = 0 , 1 , 2 , ,
(1)

where the parameters β, γ, δ, B, C, D are nonnegative numbers which satisfy B + C + D > 0 and the initial conditions x 1 and x 0 are arbitrary nonnegative numbers such that B x n x n 1 + C x n 1 2 + D x n > 0 for all n 0 .

Equation (1), which has been studied in [13], is a special case of a general second-order quadratic fractional equation of the form
x n + 1 = A x n 2 + B x n x n 1 + C x n 1 2 + D x n + E x n 1 + F a x n 2 + b x n x n 1 + c x n 1 2 + d x n + e x n 1 + f , n = 0 , 1 ,
(2)

with nonnegative parameters and initial conditions such that A + B + C > 0 , a + b + c + d + e + f > 0 and a x n 2 + b x n x n 1 + c x n 1 2 + d x n + e x n 1 + f > 0 , n = 0 , 1 ,  . Several global asymptotic results for some special cases of (2) were obtained in [47].

The change of variable x n = 1 / u n transforms (1) into the difference equation
u n + 1 = D u n 1 2 + C u n + B u n 1 δ u n 1 2 + γ u n + β u n 1 , n = 0 , 1 , ,
(3)

where we assume that δ + β + γ > 0 and that the nonnegative initial conditions u 1 , u 0 are such that δ u n 1 2 + γ u n + β u n 1 > 0 for all n 0 . Thus the results of this paper extend to (3).

The first systematic study of global dynamics of a special quadratic fractional case of (2) where A = C = D = a = c = d = 0 was performed in [1, 2]. The dynamics of some related quadratic fractional difference equations was considered in the papers [47]. In this paper we will perform the local stability analysis of the unique equilibrium and the period-two solution and we will give the necessary and sufficient conditions for the equilibrium to be locally asymptotically stable, a saddle point, a repeller or a non-hyperbolic equilibrium. The local stability analysis indicates that some possible dynamics scenarios for (1) include period-doubling bifurcations and Naimark-Sacker bifurcation and global attractivity of the equilibrium, see [8, 9]. This means that the techniques we used in [3, 4, 914] are applicable. We will also obtain the global asymptotic stability results for (1). As we have seen in [11] an efficient way of studying the dynamics of (1) is considering the dynamics of 49 special cases of (1) which are obtained when one or more coefficients are set to zero. Based on our results in [11], it is difficult to prove global asymptotic stability results of the unique equilibrium even for linear fractional difference equations; there are still two remaining cases one needs to study to prove the general conjecture that the local stability of the unique equilibrium implies the global stability.

Some interesting special cases of (1), which were thoroughly studied in [11], are the following equations.
  1. (1)
    The Beverton-Holt difference equation when γ = δ = C = 0 :
    x n + 1 = β x n 1 B x n 1 + D , n = 0 , 1 , ,
     
which represents the basic discrete model in population dynamics, see [15].
  1. (2)
    The Riccati difference equation when γ = C = 0 :
    x n + 1 = β x n 1 + δ B x n 1 + D , n = 0 , 1 , .
     
  2. (3)
    The difference equation studied in [11, 16, 17], when δ = D = 0 :
    x n + 1 = β x n + γ x n 1 B x n + C x n 1 , n = 0 , 1 , ,
    (4)
     

which represents the discretization of the differential equation model in biochemical networks, see [18].

The global attractivity results obtained specifically for the complicated cases of (2) are the following theorems [19].

Theorem 1 Assume that (2) has the unique equilibrium x ¯ . If the following condition holds:
( | A a x ¯ | + | B b x ¯ | + | C c x ¯ | ) ( U + x ¯ ) + | D d x ¯ | + | E e x ¯ | ( a + b + c ) L 2 + ( d + e ) L + f < 1 ,

where L and U are lower and upper bounds of all solutions of (2) and L + f > 0 , then x ¯ is globally asymptotically stable.

Theorem 2 Assume that (2) has the unique equilibrium x ¯ in the interval [ m , M ] , where m = min { x ¯ , x 1 , x 0 } and M = max { x ¯ , x 1 , x 0 } are lower and upper bounds of a specific solution of (2) and m + f > 0 . If the following condition holds:
( | A a x ¯ | + | B b x ¯ | + | C c x ¯ | ) ( M + x ¯ ) + | D d x ¯ | + | E e x ¯ | < ( a + b + c ) m 2 + ( d + e ) m + f ,

then x ¯ is globally asymptotically stable on the interval [ m , M ] .

In the case of (1) Theorems 1 and 2 give the following special results.

Corollary 1 If the following condition holds:
( | β B x ¯ | + | γ C x ¯ | ) ( U + x ¯ ) + | δ D x ¯ | ( B + C ) L 2 + D L < 1 ,
(5)

where L > 0 and U are lower and upper bounds of all solutions of (1), then x ¯ is globally asymptotically stable.

Corollary 2 If the following condition holds:
( | β B x ¯ | + | γ C x ¯ | ) ( M + x ¯ ) + | δ D x ¯ | < ( B + C ) m 2 + D m ,
(6)

where m = min { x ¯ , x 1 , x 0 } > 0 and M = max { x ¯ , x 1 , x 0 } are lower and upper bounds of a specific solution of (1), then the unique equilibrium x ¯ is globally asymptotically stable on the interval [ m , M ] .

In this paper we present the local stability analysis for the unique equilibrium and the period-two solutions of (1) and then we apply Corollaries 1 and 2 to some special cases of (1) to obtain global asymptotic stability results for those equations. The obtained results will give the regions of the parametric space where the unique positive equilibrium of (1) is globally asymptotically stable. In an upcoming manuscript we will give more precisely the dynamics in some special cases of (1) such as the case where the right-hand side of (1) is decreasing in x n and increasing in x n 1 ; here the theory of monotone maps can be applied to give the global dynamics. The application of the monotone map theory requires precise information on the local stability of the equilibrium solutions and the period-two solutions which will be given in this paper. See [20, 21] for an application of the monotone maps techniques to some competitive systems of linear fractional difference equations. These results will give the parameter regions where a global period-doubling bifurcation takes place, see [9]. The special cases of (1) where the unique equilibrium changes its stability character from the local stability to repeller are cases where the Naimark-Sacker bifurcation occurs, see [8, 22], and these cases will be treated in an upcoming manuscript. Following the approach from [11], we divide (1) into 49 special cases of types ( k , m ) where k (resp. m) denotes the number of positive parameters in the numerator (resp. denominator). We summarize information as regards the stability of both the equilibrium solution and the period-two solution as well as the monotonic character of the right-hand side of the special cases of types ( 1 , 2 ) , ( 2 , 1 ) , ( 1 , 3 ) , ( 3 , 1 ) and ( 2 , 2 ) of (1) in Tables 1-5. We did not include the cases of the type ( 1 , 1 ) , which are well known from [11] as well as 7 cases of types ( 3 , 2 ) , ( 2 , 3 ) , and ( 3 , 3 ) for which global stability will be given in Section 6. Using the techniques established in [2325] one can determine the rate of convergence for all regions of parameters for which we established convergence.
Table 1

Equations of type ( 1 , 2 )

Equation

Equilibrium point

Stability of equilibrium point

Period-two solution and stability

Partial derivatives

x n + 1 = β x n 1 x n C x n 1 2 + x n

x ¯ = β 1 C for β>1

no eq. point for β ≤ 1

LAS for β>1

no period-two solution

f u = C v 3 β ( C v 2 + u ) 2

f v = u β ( u C v 2 ) ( C v 2 + u ) 2

x n + 1 = γ x n 1 2 C x n 1 2 + x n

x ¯ = γ 1 C for γ>1

no eq. point for γ ≤ 1

LAS for γ>3

a saddle point for 1<γ<3

a non-hyp. eq. for γ = 3

{0,γ/C}-LAS

{ γ + 1 ( γ 3 ) ( γ + 1 ) 2 C , γ + 1 + ( γ 3 ) ( γ + 1 ) 2 C }

a saddle point for γ>3

a non-hyp. eq. for γ = 3

f u = γ v 2 ( C v 2 + u ) 2

f v = 2 γ u v ( C v 2 + u ) 2

x n + 1 = γ x n 1 2 B x n 1 x n + x n

x ¯ = γ 1 B for γ>1

no eq. point for γ ≤ 1

a saddle point for γ>1

no period-two solution

f u = γ v 2 u 2 ( B v + 1 )

f v = γ v ( B v + 2 ) u ( B v + 1 ) 2

x n + 1 = δ x n B x n x n 1 + x n 1 2

x ¯ = δ B + 1

a repeller for δ>0, B>0

no period-two solution

f u = δ ( B u + v ) 2

f v = u δ ( B u + 2 v ) ( B u v + v 2 ) 2

x n + 1 = δ x n C x n 1 2 + x n

x ¯ = 4 C δ + 1 1 2 C

LAS for <2

a repeller for >2

a non-hyp. eq. for  = 2

no period-two solution

f u = C v 2 δ ( C v 2 + u ) 2

f v = 2 C u v δ ( C v 2 + u ) 2

Table 2

Equations of type ( 2 , 1 )

Equation

Equilibrium point

Stability of equilibrium point

Period-two solution and stability

Partial derivatives

x n + 1 = β x n x n 1 + γ x n 1 2 x n

x ¯ = t , t>0 for β + γ = 1

no eq. point for β + γ ≠ 1

a non-hyp. for β + γ = 1

no minimal period-two solution

f u = v 2 γ u 2

f v = 2 v γ u + β

x n + 1 = β x n 1 x n + δ x n x n 1 2

x ¯ = 1 2 ( β 2 + 4 δ + β )

a repeller for β,δ>0

no minimal period-two solution

f u = v β + δ v 2

f v = u ( v β + 2 δ ) v 3

x n + 1 = γ x n 1 2 + δ x n x n 1 x n

x ¯ = 1 2 ( γ 2 + 4 δ + γ )

LAS for 4 δ > 3 γ 2

a saddle point for 4 δ < 3 γ 2

a non-hyp. for 4 δ = 3 γ 2

{ γ δ + δ 2 ( 4 δ 3 γ 2 ) 2 ( γ 2 δ ) , γ δ δ 2 ( 4 δ 3 γ 2 ) 2 ( γ 2 δ ) }

exists for 3 γ 2 < 4 δ < 4 γ 2

a saddle point for  3 γ 2 < 4 δ < 4 γ 2

f u = v γ u 2

f v = v 2 γ u δ u v 2

x n + 1 = γ x n 1 2 + δ x n x n 1 2

x ¯ = 1 2 ( γ 2 + 4 δ + γ )

LAS for δ < 2 γ 2

a repeller for δ > 2 γ 2

a non-hyp. for δ = 2 γ 2

possible Naimark-Sacker bifurcation

f u = δ v 2

f v = 2 u δ v 3

x n + 1 = γ x n 1 2 + δ x n x n

x ¯ = δ γ 1 for γ<1

LAS for 3γ<1

a saddle point for 3γ>1

a non-hyp. for 3γ = 1

{ϕ,ψ} for 3γ<1

ϕ = ( γ + 1 ) ( 1 3 γ ) δ 2 + γ δ + δ 2 γ ( γ + 1 )

ψ = ( γ + 1 ) ( 1 3 γ ) δ 2 + γ δ + δ 2 γ ( γ + 1 )

a saddle point for 3γ<1

f u = v 2 γ u 2

f v = 2 v γ u

Table 3

Equations of type ( 2 , 2 )

Equation

Equilibrium point

Stability of equilibrium point

Period-two solution and stability

Partial derivatives

x n + 1 = β x n x n 1 + γ x n 1 2 C x n 1 2 + x n

x ¯ = β + γ 1 C for β + γ>1

no eq. point for β + γ ≤ 1

LAS for (3β + γ>3γ ≤ 3)γ>3

a saddle point for β<1β + γ>13β + γ<3

a non-hyp. eq. for γ<33β + γ = 3

{ ϕ 1 , ψ 1 } = { 0 , γ / C } -LAS for β<1

saddle for β>1; non-hyp. for β = 1

{ ϕ 2 , ψ 2 } exists for β<1, 3β + γ>3

a saddle point for β<1, 3β + γ>3

ψ 1 = ( γ + 1 β ) + ( γ + 1 β ) ( 3 β + γ 3 ) 2 c

ϕ 2 = ( γ + 1 β ) ( γ + 1 β ) ( 3 β + γ 3 ) 2 c

f u = v 2 ( C v β γ ) ( C v 2 + u ) 2

f v = u ( v ( 2 γ C v β ) + u β ) ( C v 2 + u ) 2

x n + 1 = β x n x n 1 + γ x n 1 2 B x n 1 x n + x n

x ¯ = β + γ 1 B for β + γ>1

LAS for β>γ + 1

a saddle for 1 − γ<β<γ + 1

a non-hyp. for β = γ + 1

{ϕ,ψ} for ϕ,ψ>0

for β = γ + 1

f u = v 2 γ u 2 ( B v + 1 )

f v = v γ ( B v + 2 ) + u β u ( B v + 1 ) 2

x n + 1 = β x n 1 x n + δ x n B x n x n 1 + x n 1 2

x ¯ = 4 B δ + β 2 + 4 δ + β 2 ( B + 1 )

LAS for B β 2 > δ

a repeller for B β 2 < δ

a non-hyp. for B β 2 = δ

no minimal period-two solution

f u = v β + δ ( B u + v ) 2

f v = u ( B u δ + v 2 β + 2 v δ ) v 2 ( B u + v ) 2

x n + 1 = β x n 1 x n + δ x n C x n 1 2 + x n

x ¯ = 4 C δ + ( β 1 ) 2 + β 1 2 C

a LAS for <2D(1 + β)

a repeller for >2D(1 + β)

a non-hyp. eq. for  = 2D(1 + β)

no minimal period-two solution

f u = C v 2 ( v β + δ ) ( C v 2 + u ) 2

f v = u ( u β C v ( v β + 2 δ ) ) ( C v 2 + u ) 2

x n + 1 = γ x n 1 2 + δ x n B x n 1 x n + x n 1 2

x ¯ = 4 B δ + γ 2 + 4 δ + γ 2 ( B + 1 )

LAS for 3 ( B 1 ) γ 2 ( B + 2 ) 2 < 4 δ or δ < 2 ( 2 B + 1 ) γ 2

saddle point for B > 1 δ < 3 ( B 1 ) γ 2 4 ( B + 2 ) 2

non-hyp. eq. for ( 4 b + 2 ) γ 2 = δ or 4 ( B + 2 ) 2 δ = 3 ( B 1 ) γ 2

repeller for δ > 2 ( 2 B + 1 ) γ 2

possible Naimark-Sacker bifurcation

f u = δ B v γ ( B u + v ) 2

f v = u ( B v 2 γ 2 v δ B u δ ) v 2 ( B u + v ) 2

x n + 1 = γ x n 1 2 + δ x n B x n x n 1 + x n

x ¯ = 4 B δ + ( 1 γ ) 2 + γ 1 2 B

LAS for δ > ( γ + 1 ) ( 3 γ 1 ) 4 B

a saddle point for δ < ( γ + 1 ) ( 3 γ 1 ) 4 B

a non-hyperbolic eq. for δ = ( γ + 1 ) ( 3 γ 1 ) 4 B

{ϕ,ψ} exists for ( γ + 1 ) ( 3 γ 1 ) 4 B < δ < γ ( γ + 1 ) B

ϕ = δ ( 4 B δ γ ( 3 γ + 2 ) + 1 + γ + 1 ) 2 ( B δ + γ 2 + γ )

ψ = δ ( 4 B δ γ ( 3 γ + 2 ) + 1 + γ + 1 ) 2 ( B δ + γ 2 + γ )

a saddle point for ( γ + 1 ) ( 3 γ 1 ) 4 B < δ < γ ( γ + 1 ) B

f u = v 2 γ u 2 ( B v + 1 )

f v = v γ ( B v + 2 ) B u δ u ( B v + 1 ) 2

x n + 1 = γ x n 1 2 + δ x n C x n 1 2 + x n

x ¯ = 4 C δ + ( 1 γ ) 2 + γ 1 2 C

LAS δ < ( γ + 2 ) ( 2 γ + 1 ) C and ( 3 γ ) ( 3 γ 1 ) 16 C < δ

a repeller for δ > ( γ + 2 ) ( 2 γ + 1 ) C

a saddle for ( 3 γ ) ( 3 γ 1 ) 16 C > δ

a non-hyp. eq. for ( 3 γ ) ( 3 γ 1 ) 16 C = δ or δ = ( γ + 2 ) ( 2 γ + 1 ) C

possible Naimark-Sacker bifurcation

f u = v 2 ( C δ γ ) ( C v 2 + u ) 2

f v = 2 u v ( γ C δ ) ( C v 2 + u ) 2

Table 4

Equations of type ( 3 , 1 )

Equation

Equilibrium point

Stability of equilibrium point

Period-two solution and stability

Partial derivatives

x n + 1 = β x n x n 1 + γ x n 1 2 + δ x n x n 1 x n

x ¯ = ( β + γ ) 2 + 4 δ + β + γ 2

LAS for β>γ or β γ β 2 + 2 β γ + 4 δ > 3 γ 2

a saddle point for β < γ β 2 + 2 β γ + 4 δ < 3 γ 2

a non-hyp. eq. for β < γ β 2 + 2 β γ + 4 δ = 3 γ 2

no minimal period-two sol.

f u = v γ u 2

f v = v 2 γ u δ u v 2

x n + 1 = β x n x n 1 + γ x n 1 2 + δ x n x n 1 2

x ¯ = ( β + γ ) 2 + 4 δ + β + γ 2

LAS for δ<γ(β + 2γ)

a repeller for δ>γ(β + 2γ)

a non-hyp. eq. for δ = γ(β + 2γ)

possible Naimark-Sacker bifurcation

f u = v β + δ v 2

f v = u v β 2 u δ v 3

x n + 1 = β x n x n 1 + γ x n 1 2 + δ x n x n

x ¯ = δ 1 β γ

LAS for β>γ or β + 3γ<1

a saddle for β + 3γ>1

a non-hyperbolic eq. for β + 3γ = 1

{ϕ,ψ} exists for β + 3γ<1

ϕ = δ ( γ β + 1 + ( β γ 1 ) ( β + 3 γ 1 ) ) 2 γ ( β + γ + 1 )

ψ = δ ( γ β + 1 ( β γ 1 ) ( β + 3 γ 1 ) ) 2 γ ( β + γ + 1 )

a saddle point for β + 3γ<1

f u = v 2 γ u 2

f v = u β + 2 v γ u

Table 5

Equations of type ( 1 , 3 )

Equation

Equilibrium point

Stability of equilibrium point

Period-two solution and stability

Partial derivatives

x n + 1 = x n 1 x n B x n x n 1 + C x n 1 2 + D x n

x ¯ = 1 D B + C exists for D<1

LAS for D<1

no minimal period-two sol.

f u = C v 3 ( v ( B u + C v ) + D u ) 2

f v = D u 2 C u v 2 ( B u v + C v 2 + D u ) 2

x n + 1 = x n 1 2 B x n x n 1 + C x n 1 2 + D x n

x ¯ = 1 D B + C exists for D<1

LAS for D < C B B + 3 C

a saddle for D > C B B + 3 C

a non-hyp. for D = C B B + 3 C

{ ϕ 1 , ψ 1 } = { 0 , 1 / C } -LAS

{ ϕ 2 , ψ 2 } exists for D < C B B + 3 C

ϕ 2 = D + 1 + Δ B C 2 C

ϕ 2 = D + 1 Δ B C 2 C

Δ = (D + 1)(B − C)(BD + B + 3CD − C)

a saddle point for D < C B B + 3 C

f u = v 2 ( B v + D ) ( v ( B u + C v ) + D u ) 2

f v = B u v 2 + 2 D u v ( B u v + C v 2 + D u ) 2

x n + 1 = x n B x n x n 1 + C x n 1 2 + D x n

x ¯ = 4 B + 4 C + D 2 D 2 ( B + C )

LAS for B > C 2 2 C D 2 D 2

a repeller for B < C 2 2 C D 2 D 2

a non-hyp. eq. for B = C 2 2 C D 2 D 2

no minimal period-two sol.

f u = C v 2 ( v ( B u + C v ) + D u ) 2

f v = u ( B u + 2 C v ) ( B u v + C v 2 + D u ) 2

Some special cases of (1) have very interesting dynamics such as x n + 1 = β x n 1 x n C x n 1 2 + x n given in Table 1 where, in the case β 1 , every solution converges to 0 although 0 is out of range of this equation. Another interesting example is the equation x n + 1 = γ x n 1 2 C x n 1 2 + x n from Table 1, where, in the case γ 1 , every solution converges to 0 or to the unique period-two solution. It is interesting to notice that 0 is out of the range of this equation. Another interesting example is the equation x n + 1 = β x n x n 1 + γ x n 1 2 x n from Table 2, which has the property that if β 1 every solution approaches ∞. None of these dynamics scenarios were possible in the case of the linear fractional difference equation, which is also a special case of (2) and which was studied in great detail in [11].

2 Local stability of the positive equilibrium

In this section we investigate the equilibrium points of Eq. (1) where β , γ , δ , A , B , C [ 0 , ) , β + γ + δ , A + B + C ( 0 , ) and where the initial conditions x 1 and x 0 are arbitrary nonnegative real numbers such B x n x n 1 + C x n 1 2 + D x n > 0 for all n 0 .

In view of the above restriction on the initial conditions of (1), the equilibrium points of (1) are positive solutions of the equation
x ¯ = β x ¯ 2 + γ x ¯ 2 + δ x ¯ B x ¯ 2 + C x ¯ 2 + D x ¯ ,
(7)
or equivalently
x ¯ 2 ( B + C ) + x ¯ ( D β γ ) δ = 0 .
(8)
When
δ = 0 , B + C > 0 and β + γ > D
the unique positive equilibrium of (1) is given by
x ¯ = β + γ D B + C .
When
δ > 0 , B + C = 0 and D > β + γ
the unique positive equilibrium of (1) is given by
x ¯ = δ D β γ .
Finally when
δ > 0 and B + C > 0
the only equilibrium point of (1) is the positive solution
x ¯ = 4 δ ( B + C ) + ( D β γ ) 2 D + β + γ 2 ( B + C )

of the quadratic equation (8).

In summary, it is interesting to observe that when (1) has a positive equilibrium x ¯ , then x ¯ is unique and it satisfies (7) and (8). This observation simplifies the investigation of the local stability of the positive equilibrium of (1).

Next, we investigate the stability of the positive equilibrium of (1). Set
f ( u , v ) = u v β + u δ + v 2 γ B u v + C v 2 + D u
and observe that
f u ( u , v ) = v 2 ( C ( v β + δ ) γ ( B v + D ) ) ( v ( B u + C v ) + D u ) 2
and
f v ( u , v ) = u ( B u δ + B v 2 γ C v ( v β + 2 δ ) + D ( u β + 2 v γ ) ) ( v ( B u + C v ) + D u ) 2 .
If x ¯ denotes an equilibrium point of (1), then the linearized equation associated with (1) about the equilibrium point x ¯ is
z n + 1 = p z n + q z n 1 ,
where
p = f u ( x ¯ , x ¯ ) and q = f v ( x ¯ , x ¯ ) .
Theorem 3 Assume that
δ = 0 , B + C > 0 and β + γ > D .
Then the unique equilibrium point
x ¯ = β + γ D B + C
of (1) is
  1. (i)

    locally asymptotically stable if C ( 3 D + 3 β + γ ) > B ( D β + γ ) ;

     
  2. (ii)

    a saddle point if C ( 3 D + 3 β + γ ) < B ( D β + γ ) ;

     
  3. (iii)

    a non-hyperbolic equilibrium if C ( 3 D + 3 β + γ ) = B ( D β + γ ) or ( B = γ = D = 0 β > 0 C > 0 ) .

     
Proof It is easy to see that
p = f u ( x ¯ , x ¯ ) = C ( β D ) B γ ( B + C ) ( β + γ ) and q = f v ( x ¯ , x ¯ ) = B ( D + γ ) + C ( 2 D β ) ( B + C ) ( β + γ ) .
Then the proof follows from Theorem 1.1.1 in [11] and the fact that
1 p q = β + γ D β + γ > 0 , p q + 1 = C ( 3 D + 3 β + γ ) B ( D β + γ ) ( B + C ) ( β + γ )
and
q + 1 = B ( D + β + 2 γ ) + C ( 2 D + γ ) ( B + C ) ( β + γ ) 0 .

 □

Theorem 4 Assume that
δ > 0 , B + C = 0 and D > β + γ .
Then the unique equilibrium point
x ¯ = δ D β γ
of (1) is
  1. (i)

    locally asymptotically stable if D > β + 3 γ ;

     
  2. (ii)

    a saddle point if β + γ < D < β + 3 γ ;

     
  3. (iii)

    a non-hyperbolic if D = β + 3 γ .

     
Proof It is easy to see that
p = f u ( x ¯ , x ¯ ) = γ D and q = f v ( x ¯ , x ¯ ) = β + 2 γ D .
Then the proof follows from Theorem 1.1.1 in [11] and the fact that
1 p q = D β γ β + γ , p q + 1 = D β 3 γ D , q + 1 = D + β + 2 γ D .

 □

As we previously mentioned if
δ > 0 and B + C > 0
the only equilibrium point of (1) is the positive solution
x ¯ = 4 δ ( B + C ) + ( D β γ ) 2 D + β + γ 2 ( B + C )

of the quadratic equation (8).

By using the identity
x ¯ 2 ( B + C ) = x ¯ ( β + γ D ) + δ
one can see that
p = f u ( x ¯ , x ¯ ) = x ¯ ( C β B γ ) + C δ D γ ( x ¯ ( B + C ) + D ) 2 = x ¯ ( C β B γ ) + C δ D γ 2 D x ¯ ( B + C ) + x ¯ 2 ( B + C ) 2 + D 2 = x ¯ ( C β B γ ) + C δ D γ x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 ,
(9)
q = f v ( x ¯ , x ¯ ) = x ¯ ( B γ C β ) δ ( B + 2 C ) + D ( β + 2 γ ) ( x ¯ ( B + C ) + D ) 2 = x ¯ ( B γ C β ) δ ( B + 2 C ) + D ( β + 2 γ ) 2 D x ¯ ( B + C ) + x ¯ 2 ( B + C ) 2 + D 2 = x ¯ ( B γ C β ) δ ( B + 2 C ) + D ( β + 2 γ ) x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 ,
(10)
and
p q + 1 = x ¯ ( B ( D + β γ ) + C ( D + 3 β + γ ) ) + 2 δ ( B + 2 C ) + D 2 D ( β + 3 γ ) x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 , 1 p q = x ¯ ( B + C ) ( D + β + γ ) + 2 δ ( B + C ) + D 2 D ( β + γ ) x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 , q + 1 = x ¯ ( B ( D + β + 2 γ ) + C ( D + γ ) ) C δ + D ( D + β + 2 γ ) x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 , q 1 = x ¯ ( B γ C β ) δ ( B + 2 C ) + D ( β + 2 γ ) x ¯ ( B + C ) ( D + β + γ ) + δ ( B + C ) + D 2 .
(11)
Let
ρ 1 = D ( D + β + 3 γ ) 2 δ ( B + 2 C ) B ( D + β γ ) + C ( D + 3 β + γ ) , ρ 2 = D ( D + β + γ ) 2 δ ( B + C ) ( B + C ) ( D + β + γ ) , ρ 3 = C δ D ( D + β + 2 γ ) B ( D + β + 2 γ ) + C ( D + γ ) .
(12)
Now if we set
F ( u ) = u 2 ( B + C ) + u ( D β γ ) δ

it is clear that F ( x ¯ ) = 0 and that x ¯ > ρ if and only if F ( ρ ) < 0 while x ¯ < σ if and only if F ( σ ) > 0 for some ρ , σ [ 0 , ) .

A straightforward computation gives
F ( ρ 1 ) = ( δ ( B + C ) D ( β + γ ) ) × ( ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) + 4 δ ( B + 2 C ) 2 ) ( B D + B β B γ + C D + 3 C β + C γ ) 2 , F ( ρ 2 ) = ( δ ( B + C ) D ( β + γ ) ) ( 4 δ ( B + C ) + ( D + β + γ ) 2 ) ( B + C ) ( D + β + γ ) 2 , F ( ρ 3 ) = ( D ( β + γ ) δ ( B + C ) ) ( B ( D + β + 2 γ ) 2 + C ( ( 2 D + γ ) ( D + β + 2 γ ) C δ ) ) ( B D + B β + 2 B γ + C D + C γ ) 2 .
Lemma 1 Let p and q be partial derivatives given by (9) and (10). Assume that
δ > 0 , B + C > 0 .
  1. (a)

    Then 1 p q > 0 is true for all values of parameters.

     
  2. (b)
    Then p q + 1 > 0 if and only if
    δ > ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 .
     
  3. (c)
    Then q + 1 > 0 if and only if
    C = 0
     
or
C > 0 and δ < ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 .
Proof (a) The inequality 1 q p > 0 is equivalent to
D ( D + β + γ ) 2 δ ( B + C ) 0 or ( D ( D + β + γ ) 2 δ ( B + C ) > 0  and  F ( ρ 2 ) < 0 ) ,
which is equivalent to
C D ( D + β + γ ) 2 B δ 2 δ or ( C < D ( D + β + γ ) 2 B δ 2 δ  and  C < B δ + D β + D γ δ ) .
Since
B δ + D β + D γ δ D ( D + β + γ ) 2 B δ 2 δ = D 2 + D β + D γ 2 δ 0
we find that 1 q p > 0 is always true.
  1. (b)

    There are three cases to consider.

     
  2. (i)
    Assume B ( D + β γ ) + C ( D + 3 β + γ ) > 0 . Then p q + 1 > 0 is equivalent to x ¯ > ρ 1 . One can see that
    F ( ρ 1 ) < 0
     
if and only if
δ ( ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 , D ( β + γ ) B + C ) ,
since
D ( β + γ ) B + C + ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 = ( B ( D + β γ ) + C ( D + 3 β + γ ) ) ( B ( D + β + 3 γ ) + C ( 3 D + β + 3 γ ) ) 4 ( B + C ) ( B + 2 C ) 2 > 0 .
From (12), we have p q + 1 > 0 if and only if
δ D ( D β 3 γ ) 2 ( B + 2 C ) or ( δ < D ( D β 3 γ ) 2 ( B + 2 C )  and  F ( ρ 1 ) < 0 ) ,
which is equivalent to
δ D ( D β 3 γ ) 2 ( B + 2 C ) or δ ( ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 , D ( D β 3 γ ) 2 ( B + 2 C ) ) ,
(13)
since
D ( D β 3 γ ) 2 ( B + 2 C ) ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 = ( D β 3 γ ) ( B ( D + β γ ) + C ( D + 3 β + γ ) ) 4 ( B + 2 C ) 2 < 0
(14)
and
D ( D β 3 γ ) 2 ( B + 2 C ) + D ( β + γ ) B + C = D ( B ( D + β γ ) + C ( D + 3 β + γ ) ) 2 ( B + C ) ( B + 2 C ) .
Statement (13) is equivalent to
δ > ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 ,
from which the proof follows.
  1. (ii)
    Assume B ( D + β γ ) + C ( D + 3 β + γ ) < 0 . Then p q + 1 > 0 if and only if x ¯ < ρ 1 . It is easy to see that
    F ( ρ 1 ) > 0
     
if and only if
δ < D ( β + γ ) B + C or δ > ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 ,
which implies that p q + 1 > 0 if and only if
δ > D ( D β 3 γ ) 2 ( B + 2 C ) and F ( ρ 1 ) > 0
which is equivalent to
δ > D ( D β 3 γ ) 2 ( B + 2 C ) and δ > ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 ,
(15)
since
D ( D β 3 γ ) 2 ( B + 2 C ) + D ( β + γ ) B + C = D ( B ( D + β γ ) + C ( D + 3 β + γ ) ) 2 ( B + C ) ( B + 2 C ) < 0 .
In view of the left-hand side of (14) we see that (15) is equivalent to
δ > ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 ,
from which the proof follows.
  1. (iii)

    If B ( D + β γ ) + C ( D + 3 β + γ ) = 0 , then the proof follows from (11).

     
  2. (c)
    The inequality q > 1 is equivalent to
    C δ D ( D + β + 2 γ ) 0 or ( C δ D ( D + β + 2 γ ) > 0  and  F ( ρ 3 ) < 0 ) ,
    (16)
     
which is true for C = 0 . If C > 0 , then (16) is equivalent to
δ D ( D + β + 2 γ ) C
(17)
or
δ > D ( D + β + 2 γ ) C and δ ( D ( β + γ ) B + C , ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 ) ,
(18)
since
( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 D ( β + γ ) B + C = ( B ( D + β + 2 γ ) + C ( D + γ ) ) ( B ( D + β + 2 γ ) + C ( 2 D + β + 2 γ ) ) C 2 ( B + C ) > 0 .
It is easy to see that
D ( D + β + 2 γ ) C ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 = ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( D + γ ) ) C 2 0
and
D ( β + γ ) B + C D ( D + β + 2 γ ) C = D ( B ( D + β + 2 γ ) + C ( D + γ ) ) C ( B + C ) 0 ,
from which it follows that (18) is equivalent to
δ ( D ( D + β + 2 γ ) C , ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 ) .
(19)
Since C > 0 , in view of (19) we find that (17) and (18) are equivalent to
δ < ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 .

 □

Theorem 5 Assume
δ > 0 , B + C > 0 .
Then the unique equilibrium point
x ¯ = 4 δ ( B + C ) + ( D β γ ) 2 D + β + γ 2 ( B + C )
of (1) is
  1. (i)

    locally asymptotically stable if and only if any of the following holds:

     
  2. (a)
    C > 0
     
and
( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 < δ < ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2
  1. (b)
    C = 0 and δ > ( D β 3 γ ) ( D β + γ ) 4 B ;
     
  1. (ii)
    a repeller if and only if the following holds:
    C > 0 and δ > ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 ;
     
  2. (iii)
    a saddle point if and only if the following holds:
    δ < ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2 ;
     
  3. (iv)

    a non-hyperbolic equilibrium if and only if any of the following holds:

     
  4. (a)
    δ = ( D β 3 γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 4 ( B + 2 C ) 2
     
  5. (b)
    C > 0 and δ = ( D + β + 2 γ ) ( B ( D + β + 2 γ ) + C ( 2 D + γ ) ) C 2 .
     

Proof The proof follows from Theorem 1.1.1 in [11] and Lemma 1. □

3 Existence of period-two solutions

Assume that { ϕ , ψ } is a minimal period-two solution of (1). Then
ϕ = f ( ψ , ϕ ) and ψ = f ( ϕ , ψ ) with  ψ , ϕ [ 0 , )  and  ϕ ψ ,
which is equivalent to
ϕ = β ϕ ψ + γ ϕ 2 + δ ψ B ϕ ψ + C ϕ 2 + D ψ and ψ = β ϕ ψ + γ ψ 2 + δ ϕ B ϕ ψ + C ψ 2 + D ϕ ,
from which it immediately follows that
ϕ ( B ϕ ψ + C ϕ 2 + D ψ ) = β ϕ ψ + γ ϕ 2 + δ ψ
(20)
and
ψ ( B ϕ ψ + C ψ 2 + D ϕ ) = β ϕ ψ + γ ψ 2 + δ ϕ .
(21)
Lemma 2 Equation (1) has a minimal period-two solution { ϕ , ψ } with ϕ ψ = 0 if and only if the following holds:
  1. (i)

    δ = 0 , γ > 0 and C > 0 , then { ϕ , ψ } = { 0 , γ / C } is the minimal period-two solution.

     
  2. (ii)

    δ = 0 , γ = 0 and C = 0 , then { ϕ , ψ } = { 0 , ψ } , with ψ 0 is a minimal period-two solution.

     
Proof If ϕ = 0 , then (20) and (21) are equivalent to
ψ δ = 0 and ψ 2 ( C ψ γ ) = 0 ,

from which the proof follows. □

Assume that ϕ ψ 0 . Subtracting equations (20) and (21) we get
( ϕ ψ ) ( ϕ ψ ( B + C ) + C ( ϕ 2 + ψ 2 ) γ ( ϕ + ψ ) + δ ) = 0 .
(22)
Dividing (20) by ϕ and (20) by ψ and subtracting them we get
( ϕ ψ ) ( ϕ ψ ( C ( ϕ + ψ ) D + β γ ) + δ ( ϕ + ψ ) ) ϕ ψ = 0 .
If we set
ϕ + ψ = x and ϕ ψ = y ,
where x , y > 0 , then ϕ and ψ are positive and different solutions of the quadratic equation
t 2 x t + y = 0 .
(23)

In addition to the conditions x , y > 0 it is necessary that x 2 4 y > 0 .

From (22) and (23) we get the system
{ y ( C x D + β γ ) + x δ = 0 , y ( B C ) + C x 2 x γ + δ = 0 .
(24)
Theorem 6 For (1) the following holds:
  1. (i)

    If γ = 0 then (1) has no a minimal period-two solution.

     
  2. (ii)
    If C = 0 , γ > 0 and δ > 0 then (1) has the minimal period-two solution { ϕ , ψ } where
    ϕ = δ ( D β + γ + 4 B δ + ( D β 3 γ ) ( D β + γ ) ) 2 γ ( D β + γ ) 2 B δ ,
    (25)
     
ψ = δ ( D β + γ 4 B δ + ( D β 3 γ ) ( D β + γ ) ) 2 γ ( D β + γ ) 2 B δ
(26)
if and only if
4 B δ + ( D β 3 γ ) ( D β + γ ) > 0 and γ ( D β + γ ) B δ > 0 .
  1. (iii)
    If δ = 0 , γ > 0 and C > 0 then (1) has two minimal period-two solutions { 0 , γ / C } and { ϕ , ψ } where
    ϕ = ( B C ) ( D β + γ ) + D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C ) , ψ = ( B C ) ( D β + γ ) D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C ) ,
     
if and only if
( ( B + 3 C ) ( D β ) + γ ( B C ) ) ( B C ) > 0 and D β + γ > 0 ( B C ) ( D β ) < 0 .
  1. (iv)

    If δ = C = 0 , γ > 0 then (1) has no minimal period-two solution.

     
Proof (i) The proof follows from (22), since ϕ , ψ 0 and ϕ ψ .
  1. (ii)
    Assume that C = 0 and δ > 0 . By using (24) we see that x and y satisfy the following equations:
    y ( D + β γ ) + x δ = 0 B y = ( x γ δ ) .
    (27)
     
Assume that γ ( D β + γ ) B δ 0 . The solution of system (24) is given by
x = δ ( D β + γ ) γ ( D β + γ ) B δ , y = δ 2 γ ( D β + γ ) B δ ,
and x , y > 0 x 2 4 y > 0 if and only if
4 B δ + ( D β 3 γ ) ( D β + γ ) > 0 γ ( D β + γ ) B δ > 0 ,
since
x 2 4 y = δ 2 ( D β + γ ) 2 ( γ ( D β + γ ) B δ ) 2 4 δ 2 γ ( D β + γ ) B δ = δ 2 ( 4 B δ + ( D β 3 γ ) ( D β + γ ) ) ( γ ( D β + γ ) B δ ) 2 .
In this case the equation
t 2 δ ( D β + γ ) γ ( D β + γ ) B δ t + δ 2 γ ( D β + γ ) B δ = 0
has positive distinct solutions which are given by
t ± = δ ( D β + γ ± 4 B δ + ( D β 3 γ ) ( D β + γ ) ) 2 γ ( D β + γ ) 2 B δ .
If γ ( D β + γ ) B δ = 0 , it is easy to see that system (27) has no solutions from which follows that (1) has no minimal period-two solution.
  1. (iii)
    Assume that δ = 0 , γ > 0 and C > 0 . By using (24) we find that x and y satisfy the following equations:
    { y ( C x D + β γ ) = 0 , y ( B C ) + C x 2 x γ = 0 .
    (28)
     
Assume that y 0 and B C . The solution of system (28) is given by
x = D β + γ C and y = ( D β ) ( D β + γ ) C ( B C ) ,
and x , y > 0 x 2 4 y > 0 if and only if
( ( B + 3 C ) ( D β ) + γ ( B C ) ) ( B C ) > 0 D β + γ > 0 ( B C ) ( D β ) < 0 ,
since
x 2 4 y = ( D β + γ ) 2 C 2 + 4 ( D β ) ( D β + γ ) C ( B C ) = ( D β + γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) C 2 ( B C ) .
In this case the equation
t 2 D β + γ C t ( D β ) ( D β + γ ) C ( B C ) = 0
has positive distinct solutions which are given by
t ± = ( B C ) ( D β + γ ) ± D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C ) .

If y = 0 , then from (28) we have x = γ C , which implies that { γ C , 0 } is the minimal period-two solution.

If y 0 and B = C , then the rest of the proof follows from Lemma 2.
  1. (iv)

    The proof follows from the proof of Lemma 2. □

     
Theorem 7 Assume that C = B > 0 , δ > 0 and γ > 0 . Let
x ± = γ ± γ 2 4 C δ 2 C , y ± = δ ( ( γ ± γ 2 4 C δ ) ( D β + γ ) 2 C δ ) 2 C ( C δ + ( D β ) ( D β + γ ) ) .
Then for (1) the following holds:
  1. (i)
    If
    γ 2 > 4 c δ and D > 3 2 ( γ 2 4 C δ + γ ) + β
     
then (1) has two minimal period-two solutions { ϕ + , ψ + } and { ϕ , ψ } , where ϕ + and ψ + are solutions of equation t 2 x + t + y + = 0 and ϕ and ψ are solutions of equation t 2 x t + y = 0 .
  1. (ii)
    If
    γ 2 > 4 c δ and 3 2 ( γ γ 2 4 C δ ) + β < D 3 2 ( γ + γ 2 4 C δ ) + β
     
then (1) has one minimal period-two solution { ϕ + , ψ + } where ϕ + and ψ + are solutions of equation t 2 x + t + y + = 0 .
  1. (iii)

    In all other cases (1) has no minimal period-two solution.

     
Proof It is clear that ( x ± , y ± ) are solutions of system (24). Then minimal period-two solutions are solutions of the equation t 2 x + t + y + = 0 if x + , y + > 0 x + 2 4 y + > 0 and the equation t 2 x t + y = 0 if x , y > 0 x 2 4 y > 0 . Let
Δ = D 2 ( γ 2 2 C δ ) + D ( 4 C β δ 6 C γ δ 2 β γ 2 + γ 3 ) + β 2 ( γ 2 2 C δ ) β ( γ 3 6 C γ δ ) + 3 C δ ( 2 C δ γ 2 )
and
Γ = D ( 4 C δ 2 β γ + γ 2 ) + 4 C β δ 3 C γ δ + D 2 γ + β 2 γ β γ 2 .
One can show that the following identities hold:
x ± 4 y ± = Δ ± Γ γ 2 4 c δ C δ + ( D β ) ( D β + γ ) , Δ 2 Γ 2 ( γ 2 4 c δ ) = 4 C 2 δ 2 ( C δ + ( D β ) ( D β + γ ) ) ( 9 C δ + ( D β ) ( D β 3 γ ) ) ,

from which the proof follows. □

Assume now that B C and C > 0 . Solving the second equation of system (24) for y we get
y = C x 2 + x γ δ B C .
(29)
Substituting (29) in the first equation of system (24) we see that x satisfies the following equation:
C 2 x 3 + C x 2 ( D β + 2 γ ) + x ( B δ 2 C δ D γ + β γ γ 2 ) + δ ( D β + γ ) = 0 .
(30)
In a similar way one can show that y satisfies the following equation:
C 2 y 3 ( C B ) + C y 2 ( 2 B δ + C δ D 2 + 2 D β D γ β 2 + β γ ) + y δ ( B δ C δ + D γ β γ + γ 2 ) δ 3 = 0 .
Let
a = C 2 , b = C ( D β + 2 γ ) , c = ( B δ 2 C δ D γ + β γ γ 2 ) , d = δ 3 .
The solutions of (30) are given by
x 1 = S + T b 3 a , x 2 = S + T 2 b 3 a + i 3 2 ( S T ) , x 3 = S + T 2 b 3 a i 3 2 ( S T ) ,
where
S = R + Q 3 + R 2 3 and T = R Q 3 + R 2 3
and
Q = 3 a c b 2 9 a 2 , R = 9 a b c 27 a 2 d 2 b 3 54 a 3 .
Then the solutions of the system (24) are given by
( x i , y i ) = C x i 2 + x i γ δ B C , i = 1 , 2 , 3 .
If x i , y i > 0 and x i 2 4 y i > 0 , then (1) has minimal period-two solutions given by
{ ϕ i = x i x i 2 4 y i 2 , ψ i = x i + x i 2 4 y i 2 } for  i = 1 , 2 , 3 .
Let g be a function given by
g ( x ) = C 2 x 6 ( C B ) C x 5 ( C B ) ( D β + 2 γ ) + x 4 ( δ ( B 2 B C + C 2 ) + γ ( C B ) ( D β + γ ) + C ( D β ) ( D β + γ ) ) + x 3 ( δ ( B ( 2 D 2 β + γ ) 2 C ( D β + γ ) ) + γ ( β D ) ( D β + γ ) ) + x 2 δ ( 2 B δ + C δ + ( D β + γ ) 2 ) x δ 2 ( 2 D 2 β + γ ) + δ 3 .
Eliminating ϕ and ψ from (20) and (21) implies that if { ϕ , ψ } is a minimal period-two solution; then
g ( ϕ ) = 0 and g ( ψ ) = 0 with  ϕ ψ ,
from which it follows that
g ( t ) = C 2 ( C B ) ( t 2 x 1 t + y 1 ) ( t 2 x 2 t + y 2 ) ( t 2 x 3 t + y 3 )

for B C , since { ϕ i , ψ i } are distinct roots of equation t 2 x i t + y i = 0 .

4 Local stability of period-two solutions

Let { ϕ , ψ } be a minimal period-two solution of (1).

Set
u n = x n 1 and v n = x n for  n = 0 , 1 ,
and write (1) in the equivalent form
u n + 1 = v n , v n + 1 = β v n u n + γ u n 2 + δ v n B u n v n + C u n 2 + D v n
for n = 0 , 1 ,  . Let T be the function defined by
T ( u , v ) = ( v , β v u + γ u 2 + δ v B u v + C u 2 + D v ) .
Then { ϕ , ψ } is a fixed point of T 2 , the second iterate of T, and
T 2 ( u , v ) = ( G ( u , v ) , H ( u , v ) ) ,
where H ( u , v ) = G ( v , G ( u , v ) ) and
G ( u , v ) = u 2 γ + u v β + v δ B u v + C u 2 + D v , H ( u , v ) = u v ( v ( B v γ + β 2 ) + β δ ) + u 2 γ ( v ( C v + β ) + δ ) + v ( D v 2 γ + δ ( v β + δ ) ) C v 3 ( B u + D ) + ( B v + D ) ( u 2 γ + u v β + v δ ) + C 2 u 2 v 2 .
By definition
J T 2 ( ϕ , ψ ) = ( G u ( ϕ , ψ ) G v ( ϕ , ψ ) H u ( ϕ , ψ ) H v ( ϕ , ψ ) ) .
Theorem 8 If C = 0 , γ > 0 and δ > 0 , then (1) has the minimal period-two solution { ϕ , ψ } where ϕ and ψ are given by (25) and (26) if and only if
4 B δ + ( D β 3 γ ) ( D β + γ ) > 0 γ ( D β + γ ) B δ > 0 .

In this case the minimal period-two solution { ϕ , ψ } is a saddle point.

Proof The existence of the minimal period-two solution follows from Theorem 6. Now, we prove that the minimal period-two solution is a saddle point. Since G ( ϕ , ψ ) = ϕ we have
δ = ϕ ( ψ ( B ϕ + D β ) γ ϕ ) ψ .
(31)
Using (31) and the fact C = 0 we see that the Jacobian matrix of T 2 at the point { ϕ , ψ } is given by
J T 2 ( ϕ , ψ ) = ( 2 γ ϕ B ψ ϕ + β ψ ( D + B ϕ ) ψ γ ϕ 2 ( D + B ϕ ) ψ 2 γ ψ ( 2 γ ϕ + B ψ ϕ β ψ ) ϕ 2 ( D + B ϕ ) ( D + B ψ ) ϕ ψ ( D + B ψ ) γ 2 + ( D + B ϕ ) ( 2 D ψ 2 + B ( ϕ 3 + ψ 3 ) ) γ ϕ ( D + B ϕ ) 2 ( B ϕ β ) ψ ϕ ( D + B ϕ ) ψ ( D + B ψ ) 2 ) .
(32)
The determinant of the Jacobian matrix (32) is given by
det J T 2 ( ϕ , ψ ) = ( B ϕ ψ β ψ 2 γ ϕ ) ( ϕ ψ ( B ϕ + D ) ( B ϕ β ) γ ( B ( ϕ 3 + ψ 3 ) + 2 D ψ 2 ) ) ϕ ψ 2 ( B ϕ + D ) ( B ψ + D ) 2 .
(33)
The trace of the Jacobian matrix (32) is given by
tr J T 2 ( ϕ , ψ ) = ϕ ψ ( B ϕ + D ) 2 ( B ϕ β ) + γ 2 ϕ ψ ( B ψ + D ) + γ ( B ϕ + D ) ( B ( ϕ 3 + ψ 3 ) + 2 D ψ 2 ) ϕ ( B ψ + D ) 2 B ϕ ψ + β ψ + 2 γ ϕ ψ ( B ϕ + D ) .
(34)
Substituting (25) and (26) into (33) and (34) we find that the determinant of the Jacobian matrix (32) is given by
det J T 2 ( ϕ , ψ ) = B 2 δ 2 B δ ( D ( β + 2 γ ) + γ ( 6 γ 5 β ) ) + γ ( D β + γ ) ( 2 D β β 2 2 β γ + 4 γ 2 ) B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ ) ,
and the trace of the Jacobian matrix (32) is given by
tr J T 2 ( ϕ , ψ ) = 2 B 2 δ 2 B δ ( D 2 5 D γ + β 2 + γ ( β γ ) ) + γ ( 2 D 2 2 D γ + γ 2 ) ( D β + γ ) B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ ) .
The period-two solution { ϕ , ψ } is a saddle point if and only if
| tr J T 2 ( ϕ , ψ ) | > | 1 + det J T 2 ( ϕ , ψ ) | .
One can see that
1 + det J T 2 ( ϕ , ψ ) tr J T 2 ( ϕ , ψ ) = ( γ ( D β + γ ) B δ ) ( 4 B δ + ( D β 3 γ ) ( D β + γ ) ) B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ )
and
1 + det J T 2 ( ϕ , ψ ) + tr J T 2 ( ϕ , ψ ) = γ ( D β + γ ) ( 3 D 2 2 γ ( D + β ) + 2 D β β 2 + 5 γ 2 ) B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ ) B δ ( 4 γ ( D + β ) + ( D + β ) 2 + 5 γ 2 ) B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ ) .
Since
γ ( D β + γ ) B δ > 0
(35)
we have
β < B δ + D γ + γ 2 γ .
Since
B 2 δ 2 + B D γ δ + D 3 γ + D 2 γ 2 D ( B δ + D γ ) B δ + D γ + γ 2 γ = B 2 δ 2 ( D + γ ) D γ ( B δ + D γ ) 0
we have
B 2 δ 2 + B D γ δ + D 3 γ + D 2 γ 2 D ( B δ + D γ ) > β ,
which implies
B 2 δ 2 + B D δ ( γ β ) + D 2 γ ( D β + γ ) > 0 .
Hence, we prove that 1 + det J T 2 ( ϕ , ψ ) tr J T 2 ( ϕ , ψ ) < 0 . From (35) we have
B < γ ( D β + γ ) δ .
Let h ( x ) = x 2 4 γ x + 5 γ 2 . Since the discriminant of h is negative we have h ( x ) > 0 for x R , which implies
( D + β ) 2 4 γ ( D + β ) + 5 γ 2 > 0 .
Since
γ ( D β + γ ) ( 3 D 2 2 γ ( D + β ) + 2 D β β 2 + 5 γ 2 ) δ ( 4 γ ( D + β ) + ( D + β ) 2 + 5 γ 2 ) γ ( D β + γ ) δ = 2 γ ( D + β ) ( D β + γ ) 2 δ ( 4 γ ( D + β ) + ( D + β ) 2 + 5 γ 2 ) 0 ,
we have
γ ( D β + γ ) ( 3 D 2 2 γ ( D + β ) + 2 D β β 2 + 5 γ 2 ) δ ( 4 γ ( D + β ) + ( D + β ) 2 + 5 γ 2 ) > B .
(36)
Inequality (36) is equivalent to
γ ( D β + γ ) ( 3 D 2 2 γ ( D + β ) + 2 D β β 2 + 5 γ 2 ) B δ ( 4 γ ( D + β ) + ( D + β ) 2 + 5 γ 2 ) > 0 ,
from which it follows that
1 + det J T 2 ( ϕ , ψ ) + tr J T 2 ( ϕ , ψ ) > 0 .

 □

Theorem 9 Assume δ = 0 , γ > 0 and C > 0 . Then (1) has the minimal period-two solution { ϕ , ψ } where
ϕ = ( B C ) ( D β + γ ) + D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C ) , ψ = ( B C ) ( D β + γ ) D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C )
if and only if
( ( B + 3 C ) ( D β ) + γ ( B C ) ) ( B C ) > 0 D β + γ > 0 ( B C ) ( D β ) < 0 .
The minimal period-two solution { ϕ , ψ } is
  1. (i)
    locally asymptotically stable if
    D β + γ > 0 D < β ( B + 3 C ) ( D β ) + γ ( B C ) > 0 ,
     
  2. (ii)
    a saddle point if
    γ + 3 β 3 D > 0 D > β ( B + 3 C ) ( D β ) + γ ( B C ) < 0 .
     
Proof The existence of the minimal period-two solution follows from Theorem 6. Now, we investigate the stability of { ϕ , ψ } . The Jacobian matrix of T 2 at the point { ϕ , ψ } is given by
J T 2 ( ϕ , ψ ) = ( e f g h ) ,
(37)
where
e = ψ ( C β ϕ 2 + γ ( 2 D + B ϕ ) ϕ + D β ψ ) ( C ϕ 2 + ( D + B ϕ ) ψ ) 2 , f = ϕ 2 ( C β ϕ γ ( D + B ϕ ) ) ( C ϕ 2 + ( D + B ϕ ) ψ ) 2 , g = ψ 3 ( C β ϕ 2 + γ ( 2 D + B ϕ ) ϕ + D β ψ ) ( C β ψ γ ( D + B ψ ) ) ( C ( D + B ϕ ) ψ 3 + C 2 ϕ 2 ψ 2 + ϕ ( D + B ψ ) ( γ ϕ + β ψ ) ) 2 , h = ( ϕ ( β 2 ψ 2 ( D β ϕ C ψ 2 ( B ϕ + D ) ) + γ 2 ϕ ( C ϕ 2 ψ ( B ψ + 2 D ) + ψ 2 ( B ϕ + D ) ( 2 B ψ + 3 D ) + D β ϕ 2 ) ) ) / ( ( C ψ 3 ( B ϕ + D ) + ϕ ( B ψ + D ) ( β ψ + γ ϕ ) + C 2 ϕ 2 ψ 2 ) 2 ) + β γ ϕ ψ ( ψ ( 2 ψ ( B ϕ + D ) ( D C ϕ ) + B ψ 2 ( B ϕ + D ) + C ϕ 2 ( D C ϕ ) ) + 2 D β ϕ 2 ) ( C ψ 3 ( B ϕ + D ) + ϕ ( B ψ + D ) ( β ψ + γ ϕ ) + C 2 ϕ 2 ψ 2 ) 2 .
The determinant of the Jacobian matrix (37) is given by
det J T 2 ( ϕ , ψ ) = e h g f = ϕ ψ ( β ψ + γ ϕ ) ( γ ϕ ( B ϕ + 2 D ) + C β ϕ 2 D β ψ ) ( ψ ( B ϕ + D ) + C ϕ 2 ) 2 × β ψ ( C ψ ( ψ ( B ϕ + D ) + C ϕ 2 ) D β ϕ ) ( C ψ 3 ( B ϕ + D ) + ϕ ( B ψ + D ) ( β ψ + γ ϕ ) + C 2 ϕ 2 ψ 2 ) 2 γ ( ψ ( B ψ + 2 D ) ( ψ ( B ϕ + D ) + C ϕ 2 ) + D β ϕ 2 ) ( C ψ 3 ( B ϕ + D ) + ϕ ( B ψ + D ) ( β ψ + γ ϕ ) + C 2 ϕ 2 ψ 2 ) 2 .
By using
ϕ = ( B C ) ( D β + γ ) + D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C ) , ψ = ( B C ) ( D β + γ ) D β + γ ( B C ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) 2 C ( B C )
we find that
det J T 2 ( ϕ , ψ ) = ( D β ) ( B 2 D ( D β + γ ) + B C ( 2 D β ) ( D γ ) C 2 ( β 2 D ) 2 ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) ,
(38)
tr J T 2 ( ϕ , ψ ) = e + h = ( D β ) ( B 2 β ( β D ) + B C β ( β 3 D ) + C 2 ( D 2 + 2 D β β 2 ) ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) + γ ( D β ) ( B 2 ( D β ) + B C ( 2 D β ) + C 2 ( β 2 D ) ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) + γ 2 ( B C ) ( B D + C ( β 2 D ) ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) .
(39)
From (38) and (39) it follows that
1 + det J T 2 ( ϕ , ψ ) tr J T 2 ( ϕ , ψ ) = ( D β ) ( D β + γ ) ( ( B + 3 C ) ( D β ) + γ ( B C ) ) γ 2 ( C D B β ) + β γ ( B + C ) ( β D ) + C β 2 ( D β ) , 1 + det J T 2 ( ϕ , ψ ) + tr J T 2 ( ϕ , ψ ) = ( β D ) ( D 2 ( B 2 + 2 B C 5 C 2 ) β 2 ( B 2 + C 2 ) + 2 C D β ( B + C ) ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) × γ 2 ( B C ) ( B ( D + β ) + C ( β 3 D ) ) 2 C γ ( D β ) ( B ( β 2 D ) + C D ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) , 1 det J T 2 ( ϕ , ψ ) = γ ( D β ) ( B 2 ( D + β ) + B C ( β 2 D ) C 2 β ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) + ( D β ) 2 ( B 2 D + B C ( 2 D + β ) 4 C 2 D ) + γ 2 ( B C ) ( B β C D ) ( B C ) ( γ 2 ( B β C D ) + β γ ( B + C ) ( D β ) + C β 2 ( β D ) ) .

The rest of the proof follows from Lemma 4. □

Lemma 3 Assume that D β + γ > 0 .
  1. (1)

    If D < β then 4 D 2 5 D β + β 2 + 5 β γ + γ 2 > 0 .

     
  2. (2)

    2 D 2 D ( β + 2 γ ) 3 β 2 5 β γ γ 2 < 0 .

     
Proof
  1. (1)
    It is sufficient to prove that the inequality 4 D 2 5 D β + β 2 + 5 β γ > 0 holds for D β + γ > 0 and D < β . Since
    4 D 2 + 5 D β β 2 5 β ( β D ) = 2 ( 2 D 3 β ) ( D β ) 5 β < 0
     
we find that
4 D 2 + 5 D β β 2 5 β < β D < γ
which implies
4 D 2 + 5 D β β 2 5 β γ = 4 D 2 + 5 D β β 2 5 β γ 5 β < 0 ,
from which the inequality follows.
  1. (2)
    In view of the assumption of the lemma we have
    2 D 2 D ( β + 2 γ ) 3 β 2 5 β γ γ 2 = 2 D ( D γ ) D β 3 β 2 5 β γ γ 2 < D β 3 β 2 5 β γ γ 2 < ( β + γ ) β 3 β 2 5 β γ γ 2 = 2 β 2 4 β γ γ 2 < 0 ,
     

from which the proof follows.

 □

Lemma 4 If C > 0 , γ > 0 and
( ( B + 3 C ) ( D β