Proof.
Assume that there exist distinct nonnegative real numbers Φ and Ψ, such that
(27)…,Φ,Ψ,Φ,Ψ,…
is a prime period-two solution of (3); there are two cases to be considered.
Case 1 (k is even). In this case Φ and Ψ satisfy
(28)Φ=r+pΨ+Ψz+qΨ+Ψ, Ψ=r+pΦ+Φz+qΦ+Φ.
Furthermore,
(29)zΦ+(q+1)ΨΦ=r+(p+1)Ψ,(30)zΨ+(q+1)ΦΨ=r+(p+1)Φ.
Subtracting (30) from (29), we have
(31)z(Φ-Ψ)=(p+1)(Ψ-Φ),
so
(32)(Φ-Ψ)(z+p+1)=0 ⟹Φ=Ψ or z+p=-1.
This contradicts the hypothesis that Φ and Ψ are distinct nonnegative real numbers. Also, z+p=-1 contradicts the hypothesis that z and p are positive real numbers.
Case 2 (k is odd). (a) If
(33)p+z≥1,
then in this case Φ and Ψ satisfy
(34)Φ=r+pΨ+Φz+qΨ+Φ, Ψ=r+pΦ+Ψz+qΦ+Ψ.
Furthermore,
(35)Φ(z+qΨ+Φ)=r+pΨ+Φ,(36)Ψ(z+qΦ+Ψ)=r+pΦ+Ψ.
Subtracting (35) from (36), we have
(37)(Φ+Ψ)=(1-z-p).
But, p+z≥1, this implies that Φ+Ψ≤0 which contradicts the hypothesis that Φ,Ψ are distinct positive real numbers.
(b) If
(38)p+z<1,
then in this case Φ and Ψ satisfy
(39)Φ=r+pΨ+Φz+qΨ+Φ,(40)Ψ=r+pΦ+Ψz+qΦ+Ψ.
Moreover,
(41)Φ(z+qΨ+Φ)=r+pΨ+Φ,(42)Ψ(z+qΦ+Ψ)=r+pΦ+Ψ.
Subtracting (41) from (42), we have
(43)(Φ+Ψ)=(1-z-p).
Furthermore, one adds (41) to (42), makes use of (43), and then does some elementary algebraic manipulation; we have
(44)ΦΨ=p(1-z-p)+rq-1.
Equation (44) leads to the following conclusion:
(45)q>1,
that follows from the facts that
(46)ΦΨ>0, 1-p-z>0.
Notice that when q=1 then adding (41) to (42) gives 2r+2p(1-p-z)=0, which is impossible.
Construct the quadratic equation
(47)t2-(1-z-p)t+p(1-z-p)+rq-1=0, q>1.
So Φ and Ψ are the positive and distinct solutions of the above quadratic equation, that is,
(48)t=(1-z-p)±(1-z-p)2-4((p(1-z-p)+r)/(q-1))2, q>1.
Proof.
To investigate the local stability of the two cycles
(49)…,Φ,Ψ,Φ,Ψ,…
we first vectorize (3) by introducing the following change of variables:
(50)z1(n)=yn-k,z2(n)=yn-k+1,z3(n)=yn-k+2,⋮ zk(n)=yn-1,zk+1(n)=yn,
and write (3) in the equivalent form:
(51)(z1(n+1)z2(n+1)⋮zk(n+1)zk+1(n+1))=T(z1(n)z2(n)⋮zk(n)zk+1(n)), n=0,1,2,…,
where
(52)T(z1z2⋮zkzk+1)=(z2z3⋮zk+1r+pzk+1+z1z+qzk+1+z1).
Now Φ and Ψ generate a period-two solution of (3) only if
(53)(ΦΨ⋮ΦΨ)
is a fixed point of T2, the second iterate of T. Furthermore,
(54)T2(z1z2⋮zkzk+1)=(f1(z1,z2,…,zk+1)f2(z1,z2,…,zk+1)⋮fk(z1,z2,…,zk+1)fk+1(z1,z2,…,zk+1)),
where
(55)f1(z1,z2,…,zk+1)=z3,f2(z1,z2,…,zk+1)=z4, ⋮ fk(z1,z2,…,zk+1)=r+pzk+1+z1z+qzk+1+z1fk+1(z1,z2,…,zk+1) =r+p((r+pzk+1+z1)/(z+qzk+1+z1))+z2z+q((r+pzk+1+z1)/(z+qzk+1+z1))+z2.
The prime period-two solution of (3) is asymptotically stable if the eigenvalues of the Jacobian matrix JT2, evaluated at (ΦΨ⋮ΦΨ) lie inside the unit disk.
We have(56)JT2(ΦΨ⋮ΦΨ)=(0010...00001...000001..0................0000...11-Φz+qΨ+Φ000...p-qΦz+qΨ+Φ(p-qΨ)(1-Φ)(z+qΨ+Φ)(z+qΦ+Ψ)1-Ψz+qΦ+Ψ00...(p-qΦ)(p-qΨ)(z+qΨ+Φ)(z+qΦ+Ψ)).
Now let P(λ)=det(JT2-λI) be the characteristic polynomial of JT2. Then, by the Laplace expansion in the (k+1) row, (57)P(λ)=|-λ010⋯⋯00-λ01⋯⋯000-λ01⋯0⋮⋮⋮⋮⋮⋱⋮0000-λ⋯11-Φz+qΨ+Φ000⋯-λp-qΦz+qΨ+Φ(p-qΨ)(1-Φ)(z+qΨ+Φ)(z+qΦ+Ψ)1-Ψz+qΦ+Ψ0000(p-qΨ)(p-qΦ)(z+qΨ+Φ)(z+qΦ+Ψ)-λ|=-(p-qΨ)(1-Φ)(z+qΨ+Φ)(z+qΦ+Ψ)|010⋯⋯0-λ01⋯⋯00-λ01⋯0⋮⋮⋮⋮⋱⋮000-λ⋯1000⋯-λp-qΦz+qΨ+Φ|︸Ak+1-Ψz+qΦ+Ψ|-λ10⋯⋯0001⋯⋯00-λ01⋯0⋮⋮⋮⋮⋱⋮000-λ⋯11-Φz+qΨ+Φ00⋯-λp-qΦz+qΨ+Φ|︸Bk+((p-qΨ)(p-qΦ)(z+qΨ+Φ)(z+qΦ+Ψ)-λ)|-λ010⋯⋯00-λ01⋯⋯000-λ01⋯⋮⋮⋮⋮⋮⋱10000-λ⋯01-Φz+qΨ+Φ000⋯⋯-λ|.︸Ck
However; by Lemma 5, Corollary 6, and the fact that k is odd,
(58)Ak=p-qΦz+qΨ+ΦDk-1+λDk-2=(p-qΦz+qΨ+Φ)λ(k-1)/2,Bk=-λAk-1+(1-Φz+qΨ+Φ)=-λ[p-qΦz+qΨ+ΦDk-2+λDk-3]+(1-Φz+qΨ+Φ)=-λ(k+1)/2+(1-Φz+qΨ+Φ),Ck=(-λ)k+(1-Φz+qΨ+Φ)Dk-1=-λk+(1-Φz+qΨ+Φ)λ(k-1)/2.
Therefore,
(59)P(λ)=-(p-qΨ)(1-Φ)(z+qΨ+Φ)(z+qΦ+Ψ)×(p-qΦz+qΨ+Φ)λ(k-1)/2+(1-Ψz+qΦ+Ψ)[-λ(k+1)/2+(1-Φz+qΨ+Φ)]+((p-qΨ)(p-qΦ)(z+qΨ+Φ)(z+qΦ+Ψ)-λ)×[-λk+(1-Φz+qΨ+Φ)λ(k-1)/2]=λk+1-(p-qΨ)(p-qΦ)(z+qΨ+Φ)(z+qΦ+Ψ)λk-(1-Ψz+qΦ+Ψ+1-Φz+qΨ+Φ)λ(k+1)/2+(1-Ψ)(1-Φ)(z+qΦ+Ψ)(z+qΨ+Φ).
Hence, the characteristic polynomial is given by
(60)f(λ)=λk+1-Qλk-Lλ(k+1)/2+μ=0,
where
(61)Q=(p-qΦ)(p-qΨ)(z+qΦ+Ψ)(z+qΨ+Φ),L=1-Φz+qΨ+Φ︸a+1-Ψz+qΦ+Ψ︸b,μ=(1-Φ)(1-Ψ)(z+qΨ+Φ)(z+qΦ+Ψ)=ab.
Assume that 0<Φ<Ψ. Then, by (39),
(62)1=(r/Φ)+p(Ψ/Φ)+1z+qΨ+Φ>1z+qΨ+Φ.
Hence,
(63)z+qΨ+Φ>1.
Similarly, we observe that
(64)z+qΦ+Ψ>1.
Furthermore, since p+z<1, (43) implies the sum of Φ, Ψ is less than 1 and, a fortiori, each is less than 1. Indeed, we have
(65)0<Φ<min{Ψ,12}<1.
With that in mind, it is clear that
(66)0<a, b<1.
In addition, with understanding that
(67)(Φ+Ψ)=(1-z-p)>0, ΦΨ=p(1-z-p)+rq-1
and the fact that
(68)q>1,
we have to establish
(69)Q>0,(70)Q+L<1+μ.
First, we will establish inequality (69). To this end, observe that inequality (69) is equivalent to
(71)(p-qΦ)(p-qΨ)(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is true if and only if
(72)p2-pq(Φ+Ψ)+q2ΦΨ(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is true if and only if
(73)p2-pq(1-p-z)+q2((p(1-p-z)+r)/(q-1))(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is true if and only if
(74)p2(q-1)-pq(q-1)(1-p-z)+pq2(1-p-z)+rq2(q-1)(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is true if and only if
(75)p2(q-1)-pq(1-p-z)[q-1-q]+rq2(q-1)(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is true if and only if
(76)p2(q-1)+pq(1-p-z)+rq2(q-1)(z+qΦ+Ψ)(z+qΨ+Φ)>0,
which is clearly satisfied.
Next we will establish inequality (70). Observe that inequality (70) is equivalent to
(77)1-Φz+qΨ+Φ+1-Ψz+qΦ+Ψ +(p-qΦ)(p-qΨ)-(1-Φ)(1-Ψ)(z+qΦ+Ψ)(z+qΨ+Φ)<1,
which is true if and only if
(78)1-Φz+qΨ+Φ+1-Ψz+qΦ+Ψ +p2-1+(q2-1)ΦΨ+(1-pq)(Φ+Ψ)(z+qΦ+Ψ)(z+qΨ+Φ)<1,
which is true if and only if
(79)1-Φz+qΨ+Φ+1-Ψz+qΦ+Ψ +(p2-1+(q-1)(q+1) ×((p(1-p-z)+r)/(q-1)) +(1-pq)(1-p-z){p2}) ×((z+qΦ+Ψ)(z+qΨ+Φ))-1<1,
which is true if and only if
(80)1-Φz+qΨ+Φ+1-Ψz+qΦ+Ψ +p2-1+r(q+1)+(1-p-z)(p+1)(z+qΦ+Ψ)(z+qΨ+Φ)<1,
which is true if and only if
(81)1-Φz+qΨ+Φ+1-Ψz+qΦ+Ψ +r(q+1)-z(p+1)(z+qΦ+Ψ)(z+qΨ+Φ)<1,
which is true if and only if
(82)(1-Φ)(z+qΦ+Ψ) +(1-Ψ)(z+qΨ+Φ) +r(q+1)-z(p+1) <(z+qΨ+Φ)(z+qΦ+Ψ).
Now observe that the righthand side of (82) is
(83)I=(z+qΨ+Φ)(z+qΦ+Ψ)=z2+z(q+1)(Ψ+Φ)+q(Ψ2+Φ2)+(1+q2)ΨΦ=z2+z(q+1)(Ψ+Φ)+q((Ψ+Φ)2-2ΨΦ)+(1+q2)ΨΦ =z2+z(q+1)(Ψ+Φ)+q(Ψ+Φ)2+(q-1)2ΨΦ=z2+z(q+1)(1-p-z)+q(1-p-z)2+(q-1)2(p(1-p-z)+rq-1)=z2+r(q-1)+(1-p-z)×(z(q+1)+q(1-p-z)+p(q-1))=z2+r(q-1)+(1-p-z)(q-p+z)=r(q-1)-z(p-1)-(p-q)(1-p-z)=r(q-1)+z(1-q)+q(1-p)+p(1-p)=(z-r)(1-q)+(p-q)(p-1)=z-r-qz+qr+p2-p-qp+q.
The lefthand side of (82) is
(84)II=(1-Φ)(z+qΦ+Ψ)+(1-Ψ)(z+qΨ+Φ)+r(q+1)-z(p+1)=2z+q(Φ+Ψ)+(Φ+Ψ)-z(Φ+Ψ)-q(Ψ2+Φ2)-2ΦΨ+r(q+1)-z(p+1)=2z+(q+1-z)(Φ+Ψ)-q(Ψ2+Φ2)-2ΦΨ+r(q+1)-z(p+1)=2z+(q+1-z)(Φ+Ψ)-q[(Ψ+Φ)2-2ΦΨ]-2ΦΨ+r(q+1)-z(p+1)=2z+(q+1-z)(Φ+Ψ)-q(Ψ+Φ)2+2(q-1)ΦΨ+r(q+1)-z(p+1)=2z+(q+1-z)(Φ+Ψ)-q(Ψ+Φ)2+2(q-1)(p(1-p-z) +rq-1)+r(q+1)-z(p+1)=2z+2r+(Φ+Ψ)[1-z+q-q(Ψ+Φ)+2p]+r(q+1)-z(p+1)=2z+2r+(1-p-z)(1-z+q-q(1-p-z)+2p)+r(q+1)-z(p+1)=2z+2r+(1-p-z)(1-z+qp+qz+2p)+r(q+1)-z(p+1)=3r+1+p-z+rq+pq+zq-2pz-2pzq+z2-qz2-qp2-2p2-2pz.
Hence, inequality (70) is true if and only if
(85)3r+1+p-z+rq+pq+zq-2pz -2pzq+z2-qz2-qp2-2p2-2pz <z-r-qz+qr+p2-p-qp+q
or equivalently
(86) 4r<q-qp-qz-1-3p+z-pq+qp2+qzp+p +3p2-zp-zq+qzp+qz2+z+3pz-z2⟺4r<(1-p-z)[q-qp-qz-1-3p+z]⟺4r<(1-p-z)[q(1-p-z)-(1+3p-z)] ⟺r<(1-p-z)[q(1-p-z)-(1+3p-z)]4,
which is clearly satisfied (condition (25)).
Now, by applying Theorem 4 we shall show that the zeros of f in (60) lie in the open unit disk |λ|<1. To do so, suppose to the contrary that f has a zero λ such that |λ|≥1. Then, by the triangle inequality,
(87)(|λ|(k+1)/2-a)(|λ|(k+1)/2-b) ≤|(λ(k+1)/2-a)(λ(k+1)/2-b)|=Q|λ|k.
Thus,
(88)f(|λ|)=|λ|k+1-Q|λ|k-L|λ|(k+1)/2+μ≤0.
However, by the Descartes’ Rule of Signs f has either two or no positive zeros. Furthermore,
(89)f(0)=μ>0,f(a(k+1)/2)<0,f(1)=1+μ-Q-L>0,
and so, by the Intermediate Value Theorem, f(x) has two positive zeros in the open interval (0,1). Moreover, since f(1)>0, we conclude that f(x)>0 for all x≥1 which contradicts inequality (88).
The proof is complete.