# Global Asymptotic Stability and Naimark-Sacker Bifurcation of Certain Mix Monotone Difference Equation

Authors: M. R. S. Kulenović; S. Moranjkić; M. Nurkanović; Z. Nurkanović
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

We investigate the global asymptotic stability of the following second order rational difference equation of the formxn+1=Bxnxn-1+F/bxnxn-1+cxn-12,n=0,1,…, where the parameters B, F, b, and c and initial conditions x-1 and x0 are positive real numbers. The map associated with this equation is always decreasing in the second variable and can be either increasing or decreasing in the first variable depending on the parametric space. In some cases, we prove that local asymptotic stability of the unique equilibrium point implies global asymptotic stability. Also, we show that considered equation exhibits the Naimark-Sacker bifurcation resulting in the existence of the locally stable periodic solution of unknown period.


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## Body

## 1. Introduction and Preliminaries

In this paper, we investigate the local and global dynamics of the following difference equation:(1)xn+1=Bxnxn-1+Fbxnxn-1+cxn-12n=0,1,…where the parameters B,F,b,c are positive real numbers and initial conditions x-1 and x0 are arbitrary positive real numbers. Equation (1) is the special case of a general second order quadratic fractional equation of the form(2)xn+1=Axn2+Bxnxn-1+Cxn-12+Dxn+Exn-1+Faxn2+bxnxn-1+cxn-12+dxn+exn-1+f,n=0,1,…,with nonnegative parameters and initial conditions such that A+B+C>0, a+b+c+d+e+f>0 and axn2+bxnxn-1+cxn-12+dxn+exn-1+f>0,n=0,1,…. Several global asymptotic results for some special cases of Equation (2) were obtained in [1–11]. Also, Equation (1) is a special case of the equation(3)xn+1=Bxnxn-1+Cxn-12+Fbxnxn-1+cxn-12+f,n=0,1,…,with positive parameters and nonnegative initial conditions x-1, x0. Local and global dynamics of Equation (3) was investigated in [12].The special case of Equation (3) when B=C=0, i.e.,(4)xn+1=Fbxnxn-1+cxn-12+f,n=0,1,…was studied in [8]. The authors performed the local stability analysis of the unique equilibrium point and gave the necessary and sufficient conditions for the equilibrium to be locally asymptotically stable, a repeller or nonhyperbolic equilibrium. Also, it was shown that Equation (4) exhibits the Naimark-Sacker bifurcation.The special case of Equation (3) (when B=F=0 and C=1) is the following equation:(5)xn+1=xn-12bxnxn-1+cxn-12+f,n=0,1,…,where the parameters b, c, and f are nonnegative numbers with condition b+c>0, f≠0 and the initial conditions x-1, x0 arbitrary nonnegative numbers such that x-1+x0>0. Equation (5) is a perturbed Sigmoid Beverton-Holt difference equation and it was studied in [9]. The special case of Equation (5) for b=0 is the well-known Thomson equation(6)xn+1=xn-12cxn-12+f,n=0,1,…,where the parameters c and f are positive numbers and the initial conditions x-1, x0 are arbitrary nonnegative numbers, is used in the modelling of fish population [13].The dynamics of (6) is very interesting and follows from the dynamics of related equation(7)xn+1=xn2cxn2+f,n=0,1,….Indeed (6) is delayed version of (7) and so it exhibits the existence of period-two solutions.Two interesting special cases of Equation (2) are the following difference equations:(8)xn+1=α+γxn-1Bxn+Dxnxn-1+xn-1,n=0,1,…,studied in [14], and(9)xn+1=xnxn-1+αxn+βxn-1axnxn-1+bxn-1,n=0,1,…,studied in [5]. In both equations, (8) and (9), the associated map changes its monotonicity with respect to its variable.In this paper, in some cases when the associated map changes its monotonicity with respect to the first variable in an invariant interval, we will use Theorems1 and 2 below in order to obtain the convergence results. However, if F=Fg=B/b3c, we would not be able to use this method, so we will use the semicycle analysis; see [15] to show that each of the following four sequences x4kk=1∞, x4k+1k=0∞, x4k+2k=0∞, x4k+3k=0∞ converges to the unique equilibrium point.Also, we will show that Equation (1) exhibits the Naimark-Sacker bifurcation resulting in the existence of the locally stable periodic solution of unknown period.Note that the problem of determining invariant intervals in the case when the associated map changes its monotonicity with respect to its variable has been considered in [17, 18].In this paper, we will use the following well-known results, Theorem 2.22, in [16], and Theorem 1.4.7 in [19].Theorem 1.
Let[a,b] be a compact interval of real numbers and assume that f:[a,b]×[a,b]→[a,b] is a continuous function satisfying the following properties:(a)
f(x,y) is nondecreasing in x∈[a,b] for each y∈[a,b], and f(x,y) is nonincreasing in y∈[a,b] for each x∈[a,b];(b)
If(m,M)∈[a,b]×[a,b] is a solution of the system(10)fm,M=mandfM,m=M,thenm=M.
Then(11)xn+1=fxn,xn-1,n=0,1,⋯has a unique equilibrium x¯∈[a,b] and every solution of Equation (11) converges to x¯.Theorem 2.
Let[a,b] be an interval of real numbers and assume that f:[a,b]×[a,b]→[a,b] is a continuous function satisfying the following properties:(a)
f(x,y) is nonincreasing in both variables(b)
If(m,M)∈[a,b]×[a,b] is a solution of the system(12)fm,m=MandfM,M=m,thenm=M.
Then, (11) has a unique equilibrium x¯∈[a,b] and every solution of Equation (11) converges to x¯.Remark 3.
As is shown in [20] the unique equilibrium in Theorems 1 and 2 is globally asymptotically stable.The rest of this paper is organized as follows. The second section presents the local stability of the unique positive equilibrium solution and the nonexistence of the minimal period-two solution. The third section gives global dynamics in certain regions of the parametric space. The results and techniques depend on monotonic character of the transition functionf(x,y) which is either decreasing in both arguments or increasing in first and decreasing in second argument. In simpler situations Theorems 1 and 2 are sufficient to prove global stability of the unique equilibrium. In more complicated situations we use the semicycle analysis, which is extensively used in [15, 19] for many linear fractional equations, to prove that every solution has four convergent subsequences, which leads to the conclusion that every solution converges to period-four solution. In some parts of parametric space we prove that there is no minimal period-four solution and so every solution converges to the equilibrium, while in other parts of parametric space we prove that the period-four solution exists. The semicycle analysis presented here uses innovative techniques based on analysis of systems of polynomial equations which coefficients depend on four parameters. Finally in the region of parameters complementary to the one where the period-four solution exists we prove that the Naimark-Sacker bifurcation takes place which produces locally stable periodic solution. All numerical simulations indicate that the equilibrium solution is globally asymptotically stable whenever it is locally asymptotically stable and that the dynamics is chaotic whenever the equilibrium is repeller. An interesting feature of Equation (1) is that it gives an example of second order difference equation with period-four solution for which period-two solution does not exist. The global dynamics of Equation (11) when the transition function f(x,y) is either increasing in both arguments or decreasing in the first and increasing in the second argument is fairly simple as every solution {xn} breaks into two eventually monotonic subsequences {x2n} and {x2n+1}; see [21–23]. The global dynamics of Equation (11) when the transition function f(x,y) is either decreasing in both arguments or increasing in the first and decreasing in the second argument could be quite complicated ranging from global asymptotic stability of the equilibrium, see [19, 21, 22, 24–26] to conservative and nonconservative chaos, see [3, 19, 26]. Interesting applications can be found in [27].


## 2. Linearized Stability

In this section, we present the local stability of the unique positive equilibrium of Equation (1) and the nonexistence of the minimal period-two solution of Equation (1).In view of the above restriction on the initial conditions of Equation (1), the equilibrium points of Equation (1) are the positive solutions of the equation(13)x¯=Bx¯2+Fb+cx¯2,or equivalently(14)b+cx¯3-Bx¯2-F=0.Equation (1) has the unique positive solution x¯ given as(15)x¯=B3b+c+23B23b+cΔ+Δ2-4B63+Δ+Δ2-4B63343,where(16)Δ=2B3+27Fb+c2.Now, we investigate the stability of the positive equilibrium of Equation (1). Set(17)fu,v=Buv+Fbuv+cv2=Buv+Fvbu+cv,and observe that (18)fu′=Bcv2-bFvbu+cv2,fv′=-Bcuv2+2cFv+bFuv2bu+cv2<0.The linearized equation associated with Equation (1) about the equilibrium point x¯ is(19)zn+1=szn+tzn-1where(20)s=fu′x¯,x¯andt=fv′x¯,x¯.Theorem 4.
LetF0=B/c3b. The unique equilibrium point x¯ of Equation (1) given by (15) is(i)
locally asymptotically stable ifF<F0,(ii)
a repeller ifF>F0,(iii)
a nonhyperbolic point of elliptic type ifF=F0.Proof.
In view of(21)s=fu′x¯,x¯=Bcx¯2-bFx¯3b+c2=cb+c-Fb+cx¯3,t=fv′x¯,x¯=-Bcx¯2+2cF+bFx¯3b+c2=-s-2Fb+cx¯3=-cb+c-Fb+cx¯3<0,we have that(22)s+t=-2Fx¯3b+c<0and(23)s2-1-t2=Bcx¯2-bFx¯3b+c22-1+Bcx¯2+2cF+bFx¯3b+c22=Fb-Bcx¯22x¯6b+c4-b2x¯3+2bcx¯3+Fb+c2x¯3+Bcx¯2+2Fc2x¯6b+c4=-2F+bx¯3+cx¯3b2x¯3+2bcx¯3+c2x¯3+2Bcx¯2+2Fcx¯6b+c3<0,and so |s|<|1-t|.
Also, we have(24)1-t=1+Bcx¯2+2cF+bFx¯3b+c2>0⇒t<1.Since |s|<|1-t|, the equilibrium point x¯ will be nonhyperbolic if t=-1 and |s|<2. From t=-1 we obtain(25)-cb+c-Fb+cx¯3=-1⇔x¯=Fb3,and by using (14), we have(26)b+cFb33-BFb32-F=0⇔F=F0=Bc3b.Now,(27)s=cb+c-Fb+cF/b=-b-cb+cand the characteristic equation of (19) is of the form (28)λ2+b-cb+cλ+1=0,from which (29)λ1,2=c-b±ib+3c3b+c2b+candλ1,2=1,that is, x¯ is nonhyperbolic equilibrium point. Let us denote (30)gx=b+cx3-Bx2-F.Then,(31)gx¯nh=0⇔F=F0,andg0=-F=-F0.The condition |s|<|1-t|=1-t is always satisfied. Hence, it holds: the equilibrium solution x¯ is locally asymptotically stable if(32)t<-1⇔-cb+c-Fb+cx-3<-1⇔x-<Fb3=x-nh,i.e., F<F0 and a repeller if t>-1, which is equivalent with x¯>F/b3=x¯nh, i.e., F>F0. See Figure 1.Figure 1
IfF<F0, then x¯<x¯nh, i.e., x¯ is LAS, and if F>F0, then x¯>x¯nh, i.e., x¯ is repeller.Lemma 5.
Equation (1) has no minimal period-two solution.Proof.
Otherwise Equation (1) has a minimal period-two solution …x,y,x,y,… which satisfies (33)x=Byx+Fbyx+cx2,y=Bxy+Fbxy+cy2.Then,(34)bx2y+cx3=Bxy+F,bxy2+cy3=Bxy+F,which yields(35)x-ybxy+cx2+xy+y2=0,which implies x=y. So, there is no a minimal period-two solution.


## 3. Global Results

In this section, we prove several global attractivity results in the parts of parametric space.We notice that the functionf(u,v) is always decreasing with respect to the second variable and can be either decreasing or increasing with respect to the first variable, depending on the sign of the nominator of fu′. Therefore, (36)fu′=0⇔v=bFBc,and the function f(u,v) is nonincreasing in both variables if v≤bF/Bc, and nondecreasing with respect to the first variable and nonincreasing with respect to the second variable if v≥bF/Bc. Since(37)fbFBc,bFBc=Bb,if we denote Fg=B/b3c, we can have three possible cases:(38)Bb>bFBc⇔F<Fg,Bb=bFBc⇔F=Fg,Bb<bFBc⇔F>Fg.As we have been seen, the nature of the local stability of the equilibrium point depends on the parameter F0, so we distinguish the following scenarios:(1)
Fg≤F0,(2)
Fg>F0.Case 1 (Fg≤F0).
Notice first thatFg<F0 implies c<b and that Fg=F0 implies c=b. Now, we observe three subcases.
(a) F < F g ≤ F 0. If F<Fg≤F0, the function f(u,v) is nondecreasing with respect to the first variable and nonincreasing with respect to the second variable on the invariant interval of Equation (1) which is given by (39)L,U=bFBc,Bb,i.e., it holds(40)f:bFBc,Bb2→bFBc,Bb.Indeed, since(41)maxx,y∈L,U2⁡fx,y=fU,Landminx,y∈L,U2⁡fx,y=fL,Uwe have that(42)fU,L=fBb,bFBc=F+B2/b1/BFb/cB1/BFb/c+1/BFb=Bb=U,and (43)fL,U≥L⇔fbFBc,Bb=F+B2/b1/BFb/cB1/BFb/c+B2/b2c≥bFBc⇔F+B2b1BFbc≥BbFBc+B2b2cbFBc⇔Fb2c+bcB2bFBc≥b3F+B2c2bFBc⇔Fb2c-b≥c-bcB2bFBcwhich is true for c≤b and F<Fg.
Also, sinceF<B3/b3c=F, we obtain (44)gbFBcgBb=-b+cB3c-Fb3F-cbF/Bc3b3c<0.This means that the equilibrium point x¯ belongs to the invariant interval [L,U].Theorem 6.
IfF<Fg≤F0, then the equilibrium point x¯ is globally asymptotically stable.Proof.
The system of algebraic equations(45)fm,M=m,fM,m=M,is reduced to the system(46)F+BMm=mcM2+bmM,F+BMm=Mcm2+Mbm,which yields(47)Mmb-cM-m=0.Since c≠b, then it implies that m=M=x¯. Now, by using Theorems 1 and 4, the conclusion follows.For some numerical values of parameters we give a visual evidence for Theorem6 which indicates that in the case when F<Fg<F0, the corresponding orbit converges very quickly (see Figure 2(a)), and in the case when F<Fg=F0, the corresponding orbit converges significantly slower (see Figure 2(b)).Figure 2
The orbit for (a)b=2, c=1, B=4, F=7, Fg=8, F0=128, and (x0,x-1)=(0.3,2.8) and (b) b=0.5, c=0.5, B=2, F=30, Fg=F0=32, and (x0,x-1)=(1.1,1) generated by Dynamica 4 [16].
(a)
F < F g < F 0 (b)
F < F g = F 0(b) F g < F < F 0Lemma 7.
IfF<Fd=4B3/b+c2, then the system of algebraic equations(48)fm,m=MandfM,M=m,has the unique solution m,M=x¯,x¯Proof.
From (48) we have that(49)Bm2+F=Mm2b+c,BM2+F=mM2b+c,that is, (50)Mbm2+cm2-Bm2-mbM2+cM2+BM2=0M-mBm+M-b+cmM=0,from which m=M=x¯ or Bm+M-b+cmM=0.
IfBm+M-b+cmM=0, then m=BM/-B+Mb+c. Since M=Bm2+F/m2(b+c)>B/b+c (see the first equation of system (49)), then m>0. After substituting m in the second equation of system (49), we get (51)BM2+F=M2b+c2BM-B+Mb+c⇔B2M2+Fb+cM+BF=0,from which we have that (52)M1,2=12B2Fb+c±F2b+c2-4B3F>0.Straightforward calculation show that m1=M2 and m2=M1. Notice that the solution m2,M2 is exactly the same as the solution M1,m1, and that system (48) has a unique solution m=M=x¯ if F≤4B3/(b+c)2=Fd.Theorem 8.
IfFg<F≤Fd<F0, where Fd=4B3/(b+c)2, then the equilibrium x¯ is globally asymptotically stable.Proof.
IfFg<F<B3/bc, then(53)f:Bb,bFBc2→Bb,bFBc,which means that the interval L,U=B/b,bF/Bc is an invariant interval.
Indeed, since the functionf is nonincreasing in both variables on the invariant interval, then(54)maxx,y∈L,U2⁡fx,y=fL,Landminx,y∈L,U2⁡fx,y=fU,U,and we obtain that(55)fU,U=fbFBc,bFBc=Fb/c+F1/BFb2/c+1/BFb=Bb=L,and(56)fL,L≤U⇔fBb,Bb=B3+Fb2B2b+c≤bFBc⇔B3+Fb2B2b+c2-bFBc<0⇔B3c-Fb3B3-Fbc<0⇔Fg=B3b3c<F<B3bc.Hence,(57)fBb,Bb=Fb2+B3B2b+c∈Bb,bFBc,ifc<bandFg=B3b3c<F<B3bc.The following calculation will show that Fd<B3/bc. Indeed, (58)4B3b+c2<B3bc⇔-B3b-c2bcb+c2<0,which is true.
Also, sincegbF/BcgB/b<0, it means that the equilibrium point x¯ belongs to the invariant interval [L,U].
Now, by using Lemma7, Theorems 2 and 4, we get the conclusion that the equilibrium x¯ is globally asymptotically stable.For some numerical values of parameters we give a visual evidence for Theorem8. See Figure 3.Figure 3
The orbit and the phase portrait forb=2, c=1, B=4, F=28, Fg=8, F0=128, and (x0,x-1)=(0.3,2.8) generated by Dynamica 4 [16].Lemma 9.
Assume thatF≠Fg.(i)
Ifxn-1<bF/Bc, then xn+1>B/b.(ii)
Ifxn-1>bF/Bc, then xn+1<B/b.(iii)
Ifxn-1=bF/Bc, then xn+1=B/b.Proof.
Since the map associated with the right-hand side of Equation (1) is always decreasing in the second variable, we have that(59)xn-1<>=bFBc⇒xn+1=Bxnxn-1+Fbxnxn-1+cxn-12><=BxnbF/Bc+FbxnbF/Bc+bF/B=Bb.Note that under assumption of Lemma9, the following inequality holds:(60)min⁡bFBc,Bb<x¯<max⁡bFBc,Bb.(c)  F=Fg<F0. By substituting parameter F=Fg=B/b3c=E3c, where x¯=E=B/b, in Equation (1), we obtain(61)xn+1=Bxnxn-1+B3/b3cbxnxn-1+cxn-12=B/bxnxn-1+B/b3c/bxnxn-1+c/bxn-12=Exnxn-1+E3c/bxnxn-1+c/bxn-12.Lemma 10.
(i) Assume thatF=Fg<F0, i.e., c<b. Then Equation (61) does not possess a minimal period-four solution.
(ii) Assume thatF=Fg=F0, i.e., c=b. Then Equation (61) has the minimal period-four solutions of the form(62)…,E,E2t,E,t,E,E2t,E,t,…where E=x¯=B/b(=B/c) and t>0 is an arbitrary number depending on initial conditions x0 and x-1.Proof.
Suppose that Equation (61) has a minimal period-four solution …x,y,z,t,x,y,z,t,…; then it holds(63)x=Etz+E3c/btz+c/bz2,y=Ext+E3c/bxt+c/bt2,z=Exy+E3c/bxy+c/bx2,t=Eyz+E3c/byz+c/by2,where E=x¯. By eliminating x and y we obtain(64)zz-EWt,z=0,tt-EUt,z=0,where the functions W(t,z) and U(t,z) can be written in the polynomial form as(65)Wt,z=β4z4+β3z3+β2z2+β1z+β0,(66)Ut,z=α5z5+α4z4+α3z3+α2z2+α1z+α0,where(67)β4=c3tEb2E-c2t,β3=cbc3E4-c4t2E2+b2c2t4-b2c2t2E2+2b3ct2E2-3bc3t3E+b4t3E,β2=b5t4E-c5t2E3+b3c2t5-bc4t3E2+2b2c3tE4-2b2c3t4E+b4ct3E2-b2c3t2E3+2b3c2t2E3-2b3c2t3E2-bc4tE4+b4ct4E,β1=bctE3-c3E2+2b3t2-c3t2+bc2E2+b2ct2-3bc2tE+b2ctE,β0=bc2tE5b2t-c2E,α5=bc4tE>0,α4=c2b2t2cE+bt-c3E3+b3-c3tE2+b2-c2ct2E,α3=bbc3E4+b4t3E-2c4tE3-2c4t3E-2c4t2E2+b3ct4-bc3t2E2+b2c2t3E+b3ct2E2,α2=bctE-2c3E3+b2cE3+2b3tE2+b3t2E-bc2t2E-bc2t3-bc2tE2-b2ct2E,α1=b2c2tzE4bE+2tb-c>0,α0=b2c3tE6>0.Since z≠0 and t≠0, from system (64), we obtain the following four cases:(1)
The system(68)z-E=0,t-E=0,impliesz=t=E, and by using (63), we get x=y=E.(2)
The system(69)z-E=0,Ut,z=0,impliesz=E and(70)Ut,E=E3b-cb+cct2+tEb+c+cE2cE+bt2>0ifc<b. If b=c, then U(t,z)=0 is satisfied for every t>0, and by using system (63), it follows that the periodic solution of the minimal period four is of the form (62).(3)
The system(71)Wt,z=0,t-E=0,impliest=E and(72)WE,z=E3b-cb+ccz2+zEb+c+cE2c2z2+bzEb+c+bcE2>0,so the conclusion is the same as in the previous case.(4)
The system(73)Wt,z=0,Ut,z=0,demands more detailed analysis.
(a) Assume thatb>c. Then we can write b=c+ξ, ξ>0. Consider the polynomials Wt,z and Ut,z as polynomials in one variable t:(74)Wt,z=a5t5+a4t4+a3t3+a2t2+a1t+a0,Ut,z=b4t4+b3t3+b2t2+b1t+b0,where(75)a5=c2z2c+ξ3,a4=z2c+ξ2c3z+ξ3E+4cξ2E+5c2ξE,a3=2c5zEE2-zE-z2+c4zξE10E2-3zE+z2+3c3zξ2E5E2+2z2+c2zξ3E2zE+9E2+4z2+czξ4EzE+2E2+z2,a2=c5EE4-2zE3-z4+c4ξE23E3-3zE2+4z2E+4z3+c3ξ2E23E3+5z2E+5z3+c2ξ3E2z+EE2+2z2,a1=c3E2c+ξcz4-E4+z4ξ+cz2E2+2z2ξE2+zξE3,a0=c4z3E4c+ξ,b4=cz2c+ξ2zc+ξ2-c2E,b3=z2Ec+ξ3zc+ξ2-c2E+cz2ξE22c+ξc+ξ2+c2z3c+ξξ2E+2cξE+zc+ξ2-c2E,b2=c5z2EE-z2+2c4zEξE3+3zE2+2z3+c3zEξ24E3+11zE2+5z2E+2z3+2c2zE2ξ34zE+E2+2z2+cz2E2ξ4z+2E,b1=c5EE-zz+EE3+zE2-z2+c4ξEz2E3+2E3-z3E2+3zE4+3z4E+z5+c3ξ2E23z2E2+E4+3zE3+3z4+c2zξ3E2E3+zE2+z3,b0=c3z3E3Ec+ξ2-c2z.If z∈c2E/c+ξ2,E, then bi⩾0, for i=0,1,4 and b2>0,b3>0, so we have that(76)Ut,z>0forz∈c2Ec+ξ2,E.Since Wt,z and Ut,z are polynomials of the fifth and fourth degrees, respectively, the resultant of these polynomials is the determinant of the ninth degree:(77)restW,U=deta5a4a3a2a1a00000a5a4a3a2a1a00000a5a4a3a2a1a00000a5a4a3a2a1a0b4b3b2b1b000000b4b3b2b1b000000b4b3b2b1b000000b4b3b2b1b000000b4b3b2b1b0,from which we obtain(78)restW,U=z-E4Υ1zΥ2zΥ3zΥ4zΦz,E,c+ξ,c,where(79)Υ1z=ξz2+ξEz-cE2,Υ2z=c2z2+ξbEz-c+ξcE22,Υ3z=c+ξ9c17z9E10z+E62c+ξz2+cz+EE>0,Υ4z=c2z2+z2c+ξbE+c+ξcE2>0,Φz,E,c+ξ,c=c4z4+c3z3+c2z2+c1z+c0,c4=8c22cξ+ξ2+2c2c+ξ7,c3=-2E2cξ+ξ2+2c2-55c2ξ4-50c3ξ3+2c4ξ2-4ξ6-24cξ5+32c5ξ+16c6c+ξ3,c2=96c11E2+512c10ξE2+1208c9ξ2E2+1512c8ξ3E2+768c7ξ4E2-644c6ξ5E2-1600c5ξ6E2-1542c4ξ7E2-903c3ξ8E2-338c2ξ9E2-76cξ10E2-8ξ11E2,c1=-2cE32cξ+ξ2+2c231c2ξ4+48c3ξ3+62c4ξ2+4ξ6+16cξ5+48c5ξ+16c6c+ξ2,c0=16c3E4c+ξ8.If the equation restW,U=0 has solutions for variable z, then they are the common roots of both equations in system (73) for a fixed value of t. One of these positive roots is z1=E, but for z=E and t>0 system (73) has no solutions since Ut,E>0, see (76). Therefore, in this case, Equation (61) has no minimal period-four solution.
The positive solution of the equationΥ1z=0 is(80)z2=E-ξ+ξξ+4c2ξ.We will show later that z2 can not be a component of any positive solutions of system (73).
The positive solution of the equationΥ2z=0 is(81)z3=E-c+ξξ+2c+ξc+ξξ2-cξ+2c22c2,and z3∈c2E/c+ξ2,E. Namely,(82)z3<E⇔4c2ξ2>0,which is true, and(83)c2c+ξ2E<z3⇔c2Ec+ξ2-E-c+ξξ+2c+ξc+ξξ2-cξ+2c22c2<0⇔-4c3ξ4cξ3+7c3ξ+7c2ξ2+4c4+ξ4<0,which is also true. So, system (73) has no solutions since Ut,E>0, see (76).
Now, we prove that the eventually positive roots of the equationΦz,E,c+ξ,c=0 for variable z can belong only to the interval c2E/c+ξ2,E and that every z∈c2E/c+ξ2,E can not be a component of any positive solution of system (73). First, we prove that Φz,E,c+ξ,c>0 for z>E. Let z=E+γ, γ>0; then we obtain(84)ΦE+γ,E,c+ξ,c=16c11γ4+128c10γ2ξγ2+E2+γE+8c9ξ257γ4+16E4+142γ2E2+48γE3+111γ3E+8c8ξ3119γ4+76E4+522γ2E2+256γE3+349γ3E+8c7ξ4161γ4+162E4+1101γ2E2+627γE3+657γ3E+4c6ξ5294γ4+402E4+3013γ2E2+1839γE3+1646γ3E+2c5ξ6364γ4+636E4+5653γ2E2+3550γE3+2879γ3E+2c4ξ7148γ4+330E4+3708γ2E2+2333γE3+1789γ3E+c3ξ872γ4+221E4+3375γ2E2+2082γE3+1570γ3E+2c2ξ9γ+E4γ3+22E3+281γE2+231γ2E+4cξ10Eγ+E22γ2+E2+25γE+8γξ11Eγ+E2,i.e., ΦE+γ,E,c+ξ,c>0. It means that the function Φz,E,c+ξ,c eventually has the positive roots in the interval 0,E. Since we already considered the case when z=E, now we investigate the existence of the positive roots of the equation Φz,E,b,c=0 for 0<z<E. As we have seen, Ut,z>0 for c2E/c+ξ2<z<E and t>0, so system (73) has no solution in the case when the equation Φz,E,c+ξ,c=0 has the positive roots in the interval c2E/c+ξ2,E. This implies that Equation (61) has no minimal period-four solution whenever any root of equation restW,U=0 lies in the interval c2E/c+ξ2,E.
Now, we prove that the equationΦz,E,c+ξ,c=0 has no root for variable z if z∈0,c2E/c+ξ2 and ξ>0. It is easy to see the following: (85)Φ0=Φ0,E,c+ξ,c=16c3E4c+ξ8>0.For Φ1=Φc2E/c+ξ2,E,c+ξ,c we have(86)Φ1=c3ξ2E472cξ7+184c7ξ+278c2ξ6+618c3ξ5+877c4ξ4+820c5ξ3+500c6ξ2+32c8+8ξ82c+ξ2c+ξ4>0,and(87)Φ0-Φ1=c3E42cξ+2c2+ξ25cξ2+4c2ξ+2c3+2ξ326cξ6+36c6ξ+76c2ξ5+135c3ξ4+154c4ξ3+104c5ξ2+4c7+4ξ7c+ξ4>0.Now, it is sufficient to prove dΦz,E,b,c/dz<0 in 0,c2E/c+ξ2,ξ>0. Since(88)dΦz,E,c+ξ,cdz=2c+ξ2+c2Pz,E,c+ξ,c,where(89)Pz,E,c+ξ,c=16c9z-E3+16c8ξz-E-8zE+5E2+7z2+6c7ξ2-29E3+62zE2-73z2E+56z3+4c6ξ3-55E3+70zE2-51z2E+140z3+c5ξ4-189E3-82zE2+501z2E+560z3+c4ξ5-126E3-380zE2+1011z2E+336z3+c3ξ6-67E3-379zE2+873z2E+112z3+c2ξ7-24E3-202zE2+417z2E+16z3+4cξ8E-15zE-E2+27z2+4zξ9E3z-2E,then for z=c2E/kc+ξ2, ξ>0, k>1 we obtain(90)Pc2Ekc+ξ2,E,c+ξ,c=cE3Ωc,ξ,kk3c+ξ2<0,with(91)Ωc,ξ,k=-16c10k-13-16c9ξk-16kk-1+k2+3-2c8ξ251k-186k2+175k3-24-8c7ξ3k-120k+55k2+2+26k3-c6ξ4k82k+488k2+315k2-1-c5ξ5k380k+487k2+237k2-1-c4ξ6k379k+424k2+84k2-1-2c3ξ7k101k+136k2+6k2-1-c2ξ8k2119k+60-8cξ9k24k+1-4ξ10k3<0.Similarly, now we will consider Wt,z and Ut,z as polynomials in the variable t (with the coefficients αi, βj, i∈0,1,…,5, j∈0,1,…,4). The resultant of these polynomials is (92)reszW,U=t-E2t+E4c+ξt-cE6Λ1tΛ2tΦt,E,c+ξ,c,where(93)Λ1t=c+ξ4c18t4E152c+ξt2+cEt+EcE+c+ξt4>0,Λ2t=ξt2+ξEt-cE2.If the equation reszW,U=0 has solutions for variable t, then they are the common roots of both equations in system (73) for a fixed value of z. One of these positive roots is t1=E, and for t=E and z>0 system (73) has no solutions since (94)WE,z,c+ξ,c=ξE32c+ξcz2+zE2c+ξ+cE2c2z2+z2c+ξ+cEc+ξE>0for ξ>0. It means that Equation (61) has no minimal period-four solution.
Similarly, for the second roott2=cE/c+ξ, ξ>0 we obtain U~=UcE/c+ξ,z,c+ξ,c,(95)U~=c3E2z+E2c4z+Ez-E2+c3ξ3E3+3z2E+2z3+c2ξ23E3+4zE2+6z2E+z3+cξ3E2z+E2+ξ4zEz+Ec+ξ2>0,where ξ>0. It means that Equation (61) has no minimal period-four solution in this case. As we have already seen, the equation Φt,E,c+ξ,c=0 has eventually positive roots only in the interval t∈c2E/c+ξ2,E, ξ>0. Then,(96)β4=c3tEE-tc2+ξ2E+2cξE>0,β3=c5E-tE3-t3+tE2+t2E+c4ξ4t2E2+E4+t3E+2t4+c3ξ2t2t+Et+5E+2c2ξ3t2E2t+E+ct3ξ4E>0,β2=c5tt-E2t+E2+c4tξ3E2E2-t2+4tE3+5t3E+3t4+c3tξ22E4+5tE3+14t3E+3t4+c2t2ξ32E3+2tE2+14t2E+t3+ct3ξ4E6t+E+t4ξ5E>0,β1=ctE3c+ξcξc+ξt2+ct2+cE2+c+ξtcEξ+2c+ξ2t-c2E>0,β0=c2tE5c+ξc+ξ2t-c2E>0.This implies Wt,z>0 for c2E/c+ξ2<t<E and z>0, so system (73) has no solution in the case when the equation Φt,E,c+ξ,c=0 has the positive roots in the interval c2E/c+ξ2,E, which further means that Equation (61) has no minimal period-four solution.
Also, the positive solution of the equationΛ2t=0 is(97)t3=E-ξ+ξξ+4c2ξ.Note that t3=z2. Now, we prove that z2,z2 can not be solution of system (73). Indeed, suppose the opposite, i.e.,(98)Uz2,z2,c+ξ,c=0,Wz2,z2,c+ξ,c=0,⇕c10+2c9ξ+8c8ξ2+26c7ξ3+45c6ξ4+43c5ξ5+13c4ξ6-27c3ξ7-35c2ξ8-15cξ9-2ξ10=0,c10-2c9ξ+27c8ξ2+51c7ξ3-30c6ξ4-165c5ξ5-214c4ξ6-148c3ξ7-58c2ξ8-12cξ9-ξ10=0,⇕c=0,ξ=0,which is a contradiction with the assumption that c>0 and ξ>0.
Consequently system (73) does not have positive solutions when b>c.
(b) Assume thatb=c. Then, system (73)(99)Wt,z,c,c=0,Ut,z,c,c=0,is of the form (100)t-Et4z2+t3z+Ez2-t2z3E+z2E2-2zE3-tz4E+z3E2+z2E3-E5-z3E3=0=0,z-Et4z2+t3z+Ez2-t2z2E2-z3E-t-z4E-z3E2+z2E3+2zE4+E5-z3E3=0,and combining those equations, we have the following four cases:(i)
(101)t-E=0,z-E=0,and the solution in this case ist=z=E,(ii)
(102)t-E=0,t4z2+t3z+Ez2-t2z2E2-z3E-t-z4E-z3E2+z2E3+2zE4+E5-z3E3=0,and substitutingt by E we obtain(103)E2z-Ez+E3=0from which we get that the solution ist=z=E,(iii)
(104)t4z2+t3z+Ez2-t2z3E+z2E2-2zE3-tz4E+z3E2+z2E3-E5-z3E3=0,z-E=0,i.e.,(105)E2t-Et+E3=0,and the solution ist=z=E,(iv)
(106)t4z2+t3z+Ez2-t2z3E+z2E2-2zE3-tz4E+z3E2+z2E3-E5-z3E3=0,t4z2+t3z+Ez2-t2z2E2-z3E-t-z4E-z3E2+z2E3+2zE4+E5-z3E3=0.By subtracting we get(107)-2tEz-Ez+EzE+E2+tz+z2=0,i.e.,(108)z=E.Hence, the solution ist=z=E.This means thatE,E is a solution of system (73) and that Equation (61) does not possess a minimal period-four solution.
Thus, ifF=Fg<F0, then Equation (61) does not possess a minimal period-four solution. Consequently if F=Fg=F0, then Equation (61) has the minimal period-four solutions of the form (62).Theorem 11.
Assume thatF=Fg=B/b3c<F0. Then, the unique equilibrium point x¯=B/b of Equation (61) is globally asymptotically stable. Also, every solution of Equation (61) oscillates about the equilibrium point x¯ with semicycles of length two.Proof.
Notice that(109)xn+1-Bb=BcB+bxn-1b2bxn+cxn-1xn-1Bb-xn-1,i.e., xn+1 and xn-1 are from the different sides of the equilibrium point (see also Lemma 9, when bF/Bc=B/b). Also, that means xn+1 and xn+5 are always from the same side of the equilibrium point x¯=B/b. Since(110)xn+4-xn=Bxn+3xn+2+B3/b3cbxn+3xn+2+cxn+22-xn=Hb3bxn+3xn+2+cxn+22,where H=Bb3xn+3xn+2+B2c-b3bxn+3xn+2+cxn+22xn is a linear function in variable xn, it can be seen that H=0⇔xn+4=xn=B/b because Equation (61) has no period-two solutions nor period-four solutions (and it holds that xn=B/b⇒xn+2=B/b⇒xn+4=B/b, see Lemma 9). Also,(111)xn>Bb⇒H<0⇒xn>xn+4>Bb,n∈N,xn<Bb⇒H>0⇒xn<xn+4<Bb,n∈N,which means that every sequence x4kk=1∞, x4k+1k=0∞, x4k+2k=0∞, x4k+3k=0∞ is monotone and bounded. That implies that each of the sequences is convergent. Since, by Lemmas 5 and 10, Equation (61) has neither minimal period-two nor period-four solutions, it holds (112)limk→∞⁡x4k=limk→∞⁡x4k+1=limk→∞⁡x4k+2=limk→∞⁡x4k+3=x¯,which implies that equilibrium x¯ is an attractor and by using Theorem 4, which completes the proof of the theorem.For some numerical values of parameters we give a visual evidence for Theorem11. See Figure 4.Figure 4
The orbit and the phase portrait forb=2, c=1, B=4, F=8, Fg=8, F0=128, and (x0,x-1)=(0.3,2.8) generated by Dynamica 4 [16].Remark 12.
One can see from Theorems6, 8, and 11 that the equilibrium point x¯ is globally asymptotically stable for all values of parameter F such that 0<F≤Fd, where Fg<F0, i.e., c<b (see Figure 5(a)) and for all values of parameter F such that 0<F<Fg=F0=Fd, i.e., c=b (see Figure 5(b)).
(d)  F=Fg=F0. Since Fg=F0 implies c=b, Equation (61) is of the form(113)xn+1=Bb3b2xnxn-1+B2xnxn+xn-1.In this case, by using Lemma 10, we see that Equation (61) has minimal period-four solutions of the form (62). Based on our many numerical simulations and the proof of Theorem 11, we believe that the following conjectures are true.Figure 5
Visual representation of local and global asymptotic stability of Equation (1) when (a) Fg<Fd<F0, i.e., c<b, and (b) F0=Fg, i.e., c=b.
(a) (b)Conjecture 13.
IfF=Fg=F0 (that is b=c), then every solution of Equation (61) converges to some period-four solution of the form (62) or to the equilibrium point x¯.For some numerical values of parameters we give a visual evidence for this case. See Figures6 and 7.Figure 6
The orbit and phase portraits forb=1, c=1, B=2, and F=Fg=F0=8 generated by Dynamica 4 [16].
(a)
( x 0 , x 1 ) = (1.5, 1) (b)
( x 0 , x 1 ) = (5.9, 5)Figure 7
Bifurcation diagram in(F,x) plane for b=1, c=1, and B=2, generated by Dynamica 4 [16].Case 2 (F<F0<Fg).
We give a visual evidence for some numerical values of parameters which indicates very interesting behaviour and verifies our suspicion that the equilibrium pointx¯ is globally asymptotically stable in this case also. See Figures 8 and 9.Figure 8
The orbit and phase portrait forc=2,b=1,B=4, F=7, F0=8, Fg=128, and (x0,x-1)=(0.3,2.8) generated byDynamica 4 [16].Figure 9
The orbit and phase portrait forc=2,b=1,B=4, F=3, F0=8, Fg=128, and (x0,x-1)=(0.3,2.8) generated byDynamica 4 [16].Conjecture 14.
IfF<F0<Fg (that is, b<c), then the equilibrium point x¯ of Equation (1) is globally asymptotically stable.


## 4. Naimark-Sacker Bifurcation forb≠c

In this section, we consider bifurcation of a fixed point of map associated with Equation (1) in the case where the eigenvalues are complex conjugates and of unit module. We use the following standard version of the Naimark-Sacker result, see [28, 29]Theorem 15 (Naimark-Sacker or Poincare-Andronov-Hopf Bifurcation for maps).
Let(114)F:R×R2→R2;λ,x→Fλ,xbe a C4 map depending on real parameter λ satisfying the following conditions:(i)
F(λ,0)=0 for λ near some fixed λ0;(ii)
DF(λ,0) has two nonreal eigenvalues μ(λ) and μ(λ)¯ for λ near λ0 with |μ(λ0)|=1;(iii)
d/dλ|μ(λ)|=d(λ0)≠0 at λ=λ0;(iv)
μk(λ0)≠1 for k=1,2,3,4.
Then there is a smoothλ-dependent change of coordinate bringing F into the form(115)Fλ,x=Fλ,x+Ox5and there are smooth function a(λ),b(λ), and ω(λ) so that in polar coordinates the function F(λ,x) is given by(116)F:rθ→μλr-aλr3θ+ωλ+bλr2.If a(λ0)>0, then there is a neighborhood U of the origin and a δ>0 such that for |λ-λ0|<δ and x0∈U, then ω-limit set of x0 is the origin if λ<λ0 and belongs to a closed invariant C1 curve Γ(λ) encircling the origin if λ>λ0. Furthermore, Γ(λ0)=0.
Ifa(λ0)<0, then there is a neighborhood U of the origin and a δ>0 such that for |λ-λ0|<δ and x0∈U, then α-limit set of x0 is the origin if λ>λ0 and belongs to a closed invariant C1 curve Γ(λ) encircling the origin if λ<λ0. Furthermore, Γ(λ0)=0.Consider a general mapFλ,x that has a fixed point at the origin with complex eigenvalues μ(λ)=α(λ)+iβ(λ) and μ(λ)¯=α(λ)-iβ(λ) satisfying α(λ)2+β(λ)2=1 and β(λ)≠0. By putting the linear part of such a map into Jordan Canonical form, we may assume F to have the following form near the origin:(117)Fλ,x=αλ-βλβλαλx1x2+g1λ,x1,x2g2λ,x1,x2.Then the coefficient a(λ0) of the cubic term in Equation (116) in polar coordinate is equal to(118)aλ0=Re1-2μλ0μλ0¯21-μλ0ξ11ξ20+12ξ112+ξ022-Reμλ0¯ξ21,where(119)ξ20=18∂2g10,0∂x12-∂2g10,0∂x22+2∂2g20,0∂x1∂x2+i∂2g20,0∂x12-∂2g20,0∂x22-2∂2g10,0∂x1∂x2,(120)ξ11=14∂2g10,0∂x12+∂2g10,0∂x22+i∂2g20,0∂x12+∂2g20,0∂x22,(121)ξ02=18∂2g10,0∂x12-∂2g10,0∂x22-2∂2g20,0∂x1∂x2+i∂2g20,0∂x12-∂2g20,0∂x22+2∂2g10,0∂x1∂x2,and(122)ξ21=116∂3g1∂x13+∂3g1∂x1∂x22+∂3g2∂x12∂x2+∂3g2∂x23+i∂3g2∂x13+∂3g2∂x1∂x22-∂3g1∂x12∂x2-∂3g1∂x23.Theorem 16.
Assume thatb,c,B>0, F0=B/c3b, and x¯=B/c.(i)
Ifk2c<b<c or c<b<k3c, where k2 and k3 are positive solutions of the equation 3k3-9k2-3k+1=0, then there is a neighborhood U of the equilibrium point x¯ and ρ>0 such that for F-F0<ρ and x0,x-1∈U then ω-limit set of solution of Equation (1), with initial condition x0,x-1 is the equilibrium point x¯ if F<F0 and belongs to a closed invariant C1 curve ΓF0 encircling the equilibrium point x¯ if F>F0. Furthermore, ΓF0=0.(ii)
If0<b<k2c or k3c<b<+∞, then there is a neighborhood U of the equilibrium point x¯ and a ρ>0 such that for F-F0<ρ and x0,x-1∈U then α-limit set of x0,x-1 is the equilibrium point x¯ if F>F0 and belongs to a closed invariant C1 curve ΓF0 encircling the equilibrium point x¯ if F<F0. Furthermore, ΓF0=0.Proof.
See Figures10 and 11 for visual illustration. In order to apply Theorem 15 we make a change of variable(123)yn=xn-x¯⇒xn=yn+x¯,yn+1=Byn+x¯yn-1+x¯+Fbyn+x¯yn-1+x¯+cyn-1+x¯2-x¯.Set (124)un=yn-1andvn=ynforn=0,1,…,then(125)un+1=vn,vn+1=Bvn+x¯un+x¯+Fbvn+x¯un+x¯+cun+x¯2-x¯.Let us define the function (126)Kuv=vBv+x¯u+x¯+Fbv+x¯u+x¯+cu+x¯2-x¯.Then Ku,v has the unique fixed point 0,0. The Jacobian matrix of Ku,v is given by(127)JKuv=01-bFv+x¯+cu+x¯2F+Bu+x¯v+x¯u+x¯2cu+x¯+bv+x¯2-bF+Bcu+x¯2u+x¯cu+x¯+bv+x¯2and its value at the zero equilibrium is(128)J0=JK00=01-bF+2cF+Bcx¯2x¯3b+c2-bF+Bcx¯2x¯3b+c2,i.e.,(129)J0=JK00=01-cb+c-Fx¯3b+ccb+c-Fx¯3b+c.The eigenvalues μ(F), μ(F)¯, using (128), are(130)μF=-bF+Bcx¯2±i4b+c2x¯3bF+2cF+Bcx¯2-bF-Bcx¯222b+c2x¯3because (131)bF-Bcx¯22-4b+c2x¯3bF+2cF+Bcx¯2=bF-Bcx¯22-4b+cBx¯2+FbF+2cF+Bcx¯2=-8F2c+3F2b2+3B2c2x4+12F2bc+4BFb2x2+12BFc2x2+4B2bcx4+18BFbcx22<0.Then(132)Kuv=01-cb+c-Fx¯3b+ccb+c-Fx¯3b+cuv+f1F,u,vf2F,u,v,where (133)f1F,u,v=0,f2F,u,v=Bv+x¯u+x¯+Fbv+x¯u+x¯+cu+x¯2+cx¯3+Fx¯3b+cu+F-cx¯3x¯3b+cv-x¯.Denote F0=B/c3b. For F=F0 we have x¯=F0/b3=B/c. The eigenvalues of J0 are μF0 and μF0¯ where (134)μF0=c-b+ib+3c3b+c2b+candμF0=1.The eigenvectors corresponding to μF0 and μF0¯ are vF0 and vF0¯ where (135)vF0=c-b+ib+3c3b+c2b+c,1.Further, (136)μF0=1,μ2F0=-c2+6bc+b22b+c2-ic-bb+3c3b+c2b+c2,μ3F0=b-cb2+4bc+c2b+c3+i2bcb+3c3b+cb+c3,μ4F0=-b4-4bc3-4b3c-26b2c2+c42b+c4-ib-cb2+6bc+c2b+3c3b+c2b+c4,and μkF0≠1 for k=1,2,3,4 for c>0, b>0, and b≠c. For F=F0 and x¯=B/c(137)Kuv=01-1-b-cb+cuv+h1u,vh2u,v,and(138)h1u,v=f1F0,u,v=0,h2u,v=f2F0,u,v=BB2b+B2c+Bc2u+Bc2v+c3uvcB+cuc2u+Bb+Bc+bcv+u+b-cb+cv-Bc.Hence, for F=F0 system (125) is equivalent to (139)un+1vn+1=01-1-b-cb+cunvn+h1un,vnh2un,vn.Let unvn=Pξnηn, where(140)P=c-b2b+cb+3c3b+c2b+c10,P-1=012b+cb+3c3b+c-c-bb+3c3b+c.Then system (125) is equivalent to its normal form (141)ξn+1ηn+1=c-b2b+c-b+3c3b+c2b+cb+3c3b+c2b+cc-b2b+cξnηn+P-1HPξnηnwhere(142)Huv≔h1u,vh2u,v.Let(143)Guv=g1u,vg2u,v=P-1HPuv.By the straightforward calculation we obtain that (144)g1u,v=-Bc+b-cb+cu+c-bu+b+3c3b+cv2b+c+Ωu,v,where (145)Ωu,v=2Bb+cκ1cκ2κ3,κ1=2B2b2+2B2c2+c4u2+4B2bc+3Bc3u-bc3u2+Bc2vb+3c3b+c+c3uvb+3c3b+c+Bbc2u,κ2=c2u+2Bb+2Bc+cvb+3c3b+c-bcu,κ3=2Bb2+2Bc2+c3u+c2vb+3c3b+c+bc2u+2b2cu+4Bbc,and (146)g2u,v=b-cb+3c3b+cg1u,v.Further,(147)∂2g10,0∂u2=cb-c3b2+c22Bb+c3,∂2g10,0∂u∂v=bc2b+3c3b+cBb+c3,∂2g10,0∂v2=c3b+cb+3c2Bb+c2,∂3g10,0∂u3=-3c2b-c5b4+4b3c+6b2c2+c44B2b+c5,∂3g10,0∂u2∂v=c2-3b4-8b3c+2b2c2-8bc3+c4b+3c3b+c4B2b+c5,∂3g10,0∂u∂v2=-c23b+cb+3c-b3+3b2c+9bc2+c34B2b+c5,∂3g10,0∂v3=-3c23b+cb+3cb+3c3b+c4B2b+c3.Now, by using (118), (119), (120), (121), and (122) we obtain(148)ξ11=c7bc2+5b2c+3b3+c34Bb+c31+ib-cb+3c3b+c,ξ20=-c24Bb+c23b+c+i7b2+2bc-c2b+3c3b+c,ξ02=c24Bb+c3-5b2+2bc+c2+i-b3+11b2c+5bc2+c3b+3c3b+c,ξ21=c28B2b+c4-3b4+2b3c+8b2c2-c4+i3b5+35b4c+60b3c2+74b2c3+33bc4+3c5b+3c3b+c,1-2μF0μF0¯21-μF0=-8bc+b2+3c22b+c2+i10bc+3b2+3c25b2-c223b+cb+c2,ξ11ξ20=-c33b3+5b2c+7bc2+c316B2b+c52b3+19b2c+11bc2+c33b2+10bc+3c2+i25b2-c23b2+10bc+3c2,Re1-2μF0μF0¯21-μF0ξ11ξ20=c313b2+12bc+3c23b3+5b2c+7bc2+c38B2b+c43b+cb+3c,12ξ112=c23b3+5b2c+7bc2+c328B2b+c43b+cb+3c,ξ022=19b3c4+15b2c5+5bc6+c74B23b+cb+3cb+c3,ReμF0¯ξ21=c23b4+14b3c+19b2c2+14bc3+2c48B2b+c4and finally,(149)aF0=-bc33b3-9b2c-3bc2+c38B2b+c43b+c.If we substitute b with kc we obtain(150)3b3-9b2c-3bc2+c3=3kc3-9kc2c-3kcc2+c3=c33k3-9k2-3k+1.So,(151)aF0=0⇔3k3-9k2-3k+1=0.Solutions, determined numerically, are k1≈-0.48445, k2≈0.21014, and k3≈3.2743. Since b>0 and c>0 it must be k>0. Now,(152)aF0>0forb=kc,k∈k2,1∪1,k3,aF0<0forb=kc,k∈0,k2∪k3,+∞.Further,(153)μF=-bF+Bcx¯2±i4b+c2x¯3bF+2cF+Bcx¯2-bF-Bcx¯222b+c2x¯3and μ(F)μ(F)¯=bF+2cF+Bcx2/(b+c)2x3, i.e., |μ(F)|=bF+2cF+Bcx2/(b+c)2x3. By differentiating the equilibrium equation (154)b+cx3-Bx2-F=0with respect to F and solving for x′(F) we obtain x′(F)=1/3(b+c)x2-2Bx, i.e., x(F0)=F0/b3=B/c. (155)x′F0=13b+cB/c2-2BB/c=c2B23b+c.Now, (156)dμFdF=12bF+2cF+Bcx2/b+c2x3b+2c+2Bcxx′b+c2x3-3bF+2cF+Bcx2x′b+c2x4.By substituting x′(F) in the above expression and considering the fact that |μ(F0)|=1, we obtain(157)dμFdFF0=12b+2c+2BcB/cc2/B23b+cb+c2B/c3-3c2/B23b+cB/c,i.e.,(158)dμFdFF0=c42B3b+c3b+c>0and that completes the proof of the theorem.Figure 10
Bifurcation diagrams in(F,x) plane for b=0.5, c=1, and B=2, generated by Dynamica 4 [16].Figure 11
Phase portraits whenc=1,b=0.5,B=2, x-1=x0=2.1 (green), and x-1=x0=6.8 (blue), generated byDynamica 4 [16].
(a)
F = 3 < F 0 (b)
F = 5 > F 0



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