# Existence of a Period-Two Solution in Linearizable Difference Equations

Authors: E. J. Janowski; M. R. S. Kulenović
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

Consider the difference equationxn+1=f(xn,…,xn−k),n=0,1,…, where k∈{1,2,…} and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it can be rewritten as the nonautonomous linear equation xn+l=∑i=1−lkgixn−i,   n=0,1,…, where l,k∈{1,2,…} and the functions gi:ℝk+l→ℝ. We give some necessary and sufficient conditions for the equation to have a minimal period-two solution when l=1.


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## Body

## 1. Introduction

Consider the difference equation(1)xn+1=f(xn,…,xn-k),n=0,1,…,
where k∈{1,2,…} and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it has the linearization of the form
(2)xn+l=∑i=1-lk‍gixn-i,n=0,1,…,
where k,l∈{1,2,…} and the functions gi:ℝk+l→ℝ. By “(1) has the linearization (2)” we mean that (1) can be rewritten as the nonautonomous linear equation (2); see [1, 2].The importance of a period-two solution is well known in the case of first order difference equations of the form of (1) with k=0, where the periods of the solutions appear in the well-known Sharkovsky ordering starting with period two. As a consequence of the results on Sharkovsky ordering the nonexistence of the period-two solution implies the nonexistence of periodic solutions of any period; see [3–5].In the case of second order difference equations the following result has been obtained in [6].Theorem 1.
LetI⊆R and let f∈C[I×I,I] be a function which either increases in both variables or decreases in the first variable and increases in the second variable. Then for every solution of (1) with k=1 the subsequences {x2n}n=0∞ and {x2n+1}n=-1∞ of even and odd terms of the solution do exactly one of the following.(i)
Eventually they are both monotonically increasing.(ii)
Eventually they are both monotonically decreasing.(iii)
One of them is monotonically increasing and the other is monotonically decreasing.As a consequence of Theorem1 every bounded solution of (1) with k=1 approaches either an equilibrium solution, a period-two solution, or a finite point at the boundary, and every unbounded solution is asymptotic to the point at infinity in a monotonic way. In view of Theorem 1 the results on the nonexistence of period-two solutions are as important as the results on the existence of these solutions. The importance of the existence or nonexistence of period-two solutions is also clear from the fact that one of the two most common local bifurcations for second order monotone autonomous difference equations is period-doubling bifurcation; see [3, 4, 7–9] for related results. Also the only known global bifurcation for second order monotone autonomous difference equations is period-doubling bifurcation [10]. See [11–13] for related results. The nonexistence results for periodic solutions which are the discrete analogue of the Bendixson’s nonexistence result for periodic solutions of differential equations have been obtained in [14, 15].The results obtained in this paper are applicable to both autonomous and nonautonomous difference equations as the coefficientsgi in (2) are in general functions of n and xn-i,i=0,1,…. Some of our examples will reflect this situation. The method of finding period-two solutions in the autonomous case consists of finding the fixed points of the second iterate of the corresponding map. However, in the nonautonomous case this method does not work and the results which will be presented in this paper can be used to find period-two solutions.Some interesting points of our results can be demonstrated by the following example.Example 2.
The period-two solution{Φ,Ψ}, Φ≠Ψ of the difference equation
(3)xn+1=anxn+bnxn-1,n=0,1,…,
where {an} and {bn} are two real sequences, satisfies
(4)Φ=anΨ+bnΦ,Ψ=anΦ+bnΨ,n=0,1,…,
which implies
(5)bn-an=1,n=0,1,….
Conversely, if condition (5) holds, then any possible period-two solution {Φ,Ψ}, Φ≠Ψ of (3) must satisfy
(6)an(Φ+Ψ)=0,n=0,1,….
If aN≠0 for some N=0,1,… then Ψ=-Φ and (3) has an infinite number of period-two solutions of the form {Φ,-Φ}, Φ≠0. If an=0 for every n=0,1,… then bn=1 for every n=0,1,… in which case every nonequilibrium solution of (3) is a period-two solution.
Thus, condition (5) is a necessary and sufficient condition for the existence of a period-two solution. This condition is clearly satisfied if, for instance,
(7)an=1n+1-1,bn=1n+1,n=0,1,…,
or if {an} and {bn} are period-two sequences which satisfy
(8)b2n=1,b2n+1=b,a2n=0,a2n+1=b-1,n=0,1,…,
or if {an} and {bn} are period-k sequences which satisfy
(9)bn+i-an+i=1,i=0,1,…,k-1,n=0,1,….
An example of a nonautonomous nonlinear difference equation for which one can find a period-two solution is the following equation
(10)xn+1=(cn+xn-2+xn-4-1)xn+(cn+xn-2+xn-4)xn-1,n=0,1,….
The quadratic second order difference equation
(11)xn+1=Axn2+Cxn-12+Dxn+Exn-1,hn=0,1,…,
where A,C,D,andE are constants, can be linearized as
(12)xn+1=(Axn+D)xn+(Cxn-1+E)xn-1,hiiiin=0,1,…,
which is of the form of (3) where an=Axn+D,bn=Cxn-1+E and condition (5) becomes the first order linear difference equation
(13)Cxn-1+E-Axn-D=1,n=0,1,….
Since an≠0,n≥0 then the period-two solution of (11) has the form {x-1,-x-1}, x-1≠0.


## 2. The Constant Case

In this section we consider the case when the sums of the even indexed functionsgi and the odd indexed functions gi are both constants.The following simple result will be a useful technical tool.Lemma 3.
Suppose that (1) has the linearization (2). Let
(14)α=∑i∈{1-l,…,k:ieven}‍gi,β=∑i∈{1-l,…,k:iodd}‍gi,n=0,1,….
Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,  andΦ=x2n-1 for n≥0. (a)
Forl∈{1,3,…},
(15)ifniseven,thenΦ(1-β)=Ψα;(16)ifnisodd,thenΨ(1-β)=Φα.(b)
Forl∈{2,4,…},
(17)ifniseven,thenΨ(1-α)=Φβ;(18)ifnisodd,thenΦ(1-α)=Ψβ.Proof.
By pluggingΨ=x2n and Φ=x2n-1 for n≥0 in (2) and assuming that l is odd we obtain immediately that when n is even (15) holds, while in the case when n is odd (16) holds. Similarly assuming that l is even we obtain immediately that in the case when n is even (17) holds, while when n is odd (18) holds.Theorem 4.
Suppose that (1) has the linearization (2) with l=1 and that α,β are given by (14). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0 if and only if the following hold: (a)
ifn is even, then Φ(1-β)=Ψα;(b)
ifn is odd, then Ψ(1-β)=Φα.Proof.
The necessary part of the proof follows from part (a) of Lemma3.
For the proof of the sufficient part choose the initial conditionsΨ=x0= x-2=⋯,Φ=x-1=x-3=⋯ and x0≠x-1. Setting n=0,1,… we get that, by part (a),
(19)x1=αΨ+βΦ=Φ,x2=αΦ+βΨ=Ψ,x3=αΨ+βΦ=Φ,x4=αΦ+βΨ=Ψ.
By using induction we get that Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0.Theorem 5.
Suppose that (1) has the linearization (2) and that α,β are given by (14). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0. (1)
Ifl∈{1,3,…}, then
(a)
β=1 if and only if α=0;(b)
ifβ≠1, then β-α=1 and Ψ=-Φ.(2)
Ifl∈{2,4,…}, then
(a)
α=1 if and only if β=0;(b)
ifα≠1, then α-β=1 and Ψ=-Φ.Proof.
The proof is as follows.(1)
Letl∈{1,3,…}. In view of Lemma 3 part (a) the identities (15) and (16) are satisfied.
(a)
Assume thatβ=1. Then by (15) Ψα=0. Hence either Ψ=0 or α=0. If Ψ=0, then by (16) Φα=0, and since Φ≠0, we have α=0.Assume thatα=0. Then by (15) Φ(1-β)=0. Hence either Φ=0 or β=1. If Φ=0, then by (16) Ψ(1-β)=0, and since Ψ≠0, we obtain that β=1.(b)
Assume thatβ≠1. Then by (15) and (16) Φ2(1-β)=ΦΨα=Ψ2(1-β). Thus Φ2=Ψ2 and so Ψ=-Φ. By (15) Φ(1-β)=-Φα, and since Φ≠0, then β-α=1.(2)
Letl∈{2,4,…}. In view of Lemma 3 part (b) the identities (17) and (18) are satisfied.
(a)
Assume thatα=1. Then by (17) Φβ=0. Hence either Φ=0 or β=0. If Φ=0, then by (18) Ψβ=0, and since Ψ≠0, we have β=0.Conversely, assume thatβ=0. Then by (17) Ψ(1-α)=0. Hence either Ψ=0 or α=1. If Ψ=0, then by (18) Φ(1-α)=0, and since Φ≠0, we obtain that α=1.(b)
Assume thatα≠1. Then by (17) and (18) Φ2(1-α)=ΦΨβ=Ψ2(1-α). Thus Φ2=Ψ2 and so Ψ=-Φ. By (17) -Φ(1-α)=Φβ, and since Φ≠0, then α-β=1.Theorem 6.
Suppose that (1) has the linearization (2) with l=1 and that α,β are given by (14). Then (1) has a minimal period-two solution if and only if
(20)β-α=1.Proof.
If (1) has a minimal period-two solution then by Theorem 5 part (1) the necessary condition follows.
Conversely, assume thatβ=1. Then α=0. Choose the initial conditions
(21)x0=x-2=⋯,x-1=x-3=⋯,x0≠x-1.
Then for n=0 we get x1=αx0+βx-1=x-1 and for n=1 we get x2=αx-1+βx0=x0, which shows that {xn} is a minimal period-two solution.
Now suppose thatβ≠1. Choose the initial conditions
(22)x0=x-2=⋯,x-1=x-3=⋯,x0=-x-1.
Then for n=0 we get x1=αx0+βx-1=(β-α)x-1=x-1 and for n=1 we get x2=αx-1+βx0=(β-α)x0=x0, which shows that {xn} is a minimal period-two solution.An immediate consequence of Theorem6 is this result.Corollary 7.
(a) Ifβ=1 and α=0, then the minimal period-two solution of (1) is …,Ψ,Φ,….
(b) Ifβ≠1 and β-α=1, then the minimal period-two solution of (1) is either …,-Φ,Φ,… or …,-Ψ,Ψ,….


## 3. The Nonconstant Case

In this section we consider the case when the sums of the even indexed functionsgi and the odd indexed functions gi are both nonconstants.Theorem 8.
Suppose that (1) has the linearization (2) with l=1. Let
(23)An=∑i∈{1-l,…,k;ieven}‍gi,Bn=∑i∈{1-l,…,k;iodd}‍gi,n=0,1,….
Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0 if and only if Bn-An=1,n=0,1,….Proof.
The proof follows from the same reasoning as in Example2.Note that it is possible for (3) to have a minimal period-two solution other than {x-1,-x-1},x-1≠0, when an≠0 for all n≥0 (see Example 12). In order to handle the cases not covered by Theorem 8 we establish the following results.The following simple result will be a useful technical tool.Lemma 9.
Suppose that (1) has the linearization (2) and that An,Bn are given by (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0. Then for n=0,1,…,(a)
ifl∈{1,3,…}, then
(24)Φ(1-B2n)=ΨA2n,(25)Ψ(1-B2n+1)=ΦA2n+1;(b)
ifl∈{2,4,…}, then
(26)Ψ(1-A2n)=ΦB2n,(27)Φ(1-A2n+1)=ΨB2n+1.Proof.
Assume thatl is odd. By plugging Ψ=x2n,Φ=x2n-1 for n≥0 in (2) and setting n=0,1,… we obtain immediately that
(28)xl=A0Ψ+B0Φ=Φ,xl+1=A1Φ+B1Ψ=Ψ,xl+2=A2Ψ+B2Φ=Φ,xl+3=A3Φ+B3Ψ=Ψ.
Now simple induction completes the proof of (24) and (25) and so of part (a). The proof of part (b) is similar.Theorem 10.
Suppose that (1) has the linearization (2) with l=1 and that An,Bn are given by (23). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0 if and only if (24) and (25) hold.Proof.
The necessary part of the proof follows from part (a) of Lemma9.
For the proof of sufficient part choose the initial conditionsΨ=x0=x-2=⋯,Φ=x-1=x-3=⋯, and x0≠x-1.(1)
AssumeΦ,Ψ≠0. Setting n=0,1,… we get that
(29)by(24)x1=A0Ψ+B0Φ=Φ,by(25)x2=A1Φ+B1Ψ=Ψ,by(24)x3=A2Ψ+B2Φ=Φ,by(25)x4=A3Φ+B3Ψ=Ψ.
By using induction we get that Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0.(2)
AssumeΨ=0. We obtain from (24) and (25) that B2k=1 and A2k+1=0 for k=0,1,…,   which implies that x2n=0 and x2n+1=Φ,n=0,1,….(3)
AssumeΦ=0. We obtain from (24) and (25) that A2k=0 and B2k+1=1 for k=0,1,…,  which implies that x2n+1=0 and x2n=Ψ,n=0,1,….When (1) has been embedded into a higher order equation, the following results can be used to establish either the nonexistence or the necessary conditions for existence of a minimal period-two solution.Theorem 11.
Letl∈{1,3,…}. Suppose that (1) has the linearization (2) and that An,Bn are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ, Ψ=x2n,andΦ=x2n-1  for n≥0.(1)
Let eitherBn≠1 or An≠0  for n≥0; then
(a)
Φ/Ψ=A0/(1-B0)=A2n/(1-B2n),n=0,1,…,(b)
Ψ/Φ=A1/(1-B1)=A2n+1/(1-B2n+1),n=0,1,….(2)
LetBn=1 for n≥0.
(a)
IfA2n=A2n+1,n=0,1,…,  then A2n=A2n+1=0,n=0,1,….(b)
IfA2n≠A2n+1,n=0,1,…,  then the following are true:
(i)
A2n≠0 if and only if A2n+1=0 for n=0,1,…;(ii)
A2n=0 if and only if A2n+1≠0 for n=0,1,…;(iii)
eitherΨ=0 or Φ=0.(3)
LetAn=0 for n≥0.
(a)
IfB2n=B2n+1,n=0,1,…,  then B2n=B2n+1=1,n=0,1,….(b)
IfB2n≠B2n+1,n=0,1,…,  then the following are true:
(i)
B2n≠1 if and only if B2n+1=1 for n=0,1,…;(ii)
B2n=1 if and only if B2n+1≠1 for n=0,1,…;(iii)
eitherΨ=0 or Φ=0.Proof.
The proof is as follows.(1)
Assume thatBn≠1 for n≥0. Then Φ,Ψ≠0. Otherwise, suppose that Ψ=0. Then Φ≠0 and in view of (24) Φ(1-B2n)=0. This implies B2n=1 for n≥0, which is a contradiction. Now, suppose that Φ=0. Then Ψ≠0 and in view of (25) Ψ(1-B2n+1)=0. This implies B2n+1=1 for n≥0, which is a contradiction.ThusΦ,Ψ≠0, which in view of Lemma 9, implies An≠0 for n≥0.
(a)
By (24) Φ/Ψ=A0/(1-B0)=A2n/(1-B2n),n=0,1,….(b)
By (25) Ψ/Φ=A1/(1-B1)=A2n+1/(1-B2n+1),n=0,1,….Next, assume thatAn≠0 for n≥0. Then Φ,Ψ≠0. Otherwise, suppose that Ψ=0. Then Φ≠0 and in view of (25) ΦA2n+1=0. This implies A2n+1=0 for n≥0, which is a contradiction. Now, suppose that Φ=0. Then Ψ≠0 and in view of (24) ΨA2n=0. This implies A2n=0 for n≥0, which is a contradiction.ThusΦ,Ψ≠0, which in view of Lemma 9, implies Bn≠1 for n≥0.The rest of the proof is similar to the first part of the proof and will be omitted.(2)
In view of (24) and (25) the condition Bn=1 for n≥0 implies A2nΨ=A2n+1Φ=0 for n=0,1,….
(a)
Assume thatA2n=A2n+1,n=0,1,….IfA2n≠0 for some n≥0, then A2n+1≠0 for this n and so Φ=Ψ=0, which is a contradiction. Thus A2n=0 for n≥0, and so A2n=A2n+1=0,n=0,1,….(b)
Assume thatA2n≠A2n+1 for n≥0.
(i)
IfA2n+1=0 for n≥0, then A2n≠0 for n≥0. Now suppose that A2n≠0 for n≥0; then from A2nΨ=A2n+1Φ=0 we get that Ψ=ΦA2n+1/A2n=0 for n≥0.SinceΨ=0, then Φ≠0. Thus A2n+1=0 for n≥0.(ii)
The proof is similar to part (i) and will be omitted.(iii)
By (24) A2nΨ=0 for n≥0. Now suppose that A2n≠0 for n≥0, then Ψ=0.IfA2n=0 for n≥0; then A2n+1≠0 for n≥0. In view of (25) A2n+1Φ=0 for n≥0 and so Φ=0.(3)
In view of (24) and (25) condition An=0 for n≥0 implies (1-B2n)Φ=(1-B2n+1)Ψ=0 for n=0,1,….
(a)
Assume thatB2n=B2n+1,n=0,1,….IfB2n≠1 for some n≥0, then B2n+1≠1 for this n and so Φ=Ψ=0, which is a contradiction. Thus B2n=1 for n=0,1,… and consequently B2n=B2n+1=1,n=0,1,….(b)
Assume thatB2n≠B2n+1 for n≥0.
(i)
IfB2n+1=1 for n≥0, then B2n≠1 for n≥0. Now suppose that B2n≠1 for n≥0; then from (1-B2n)Φ=(1-B2n+1)Ψ=0 we get that Φ=Ψ(1-B2n+1)/(1-B2n)=0 for n≥0. Since Φ=0, then Ψ≠0. Thus B2n+1=1 for n≥0.(ii)
The proof is similar to part (i) and will be omitted.(iii)
By (24) (1-B2n)Φ=0 for n≥0. Now suppose that B2n≠1 for n≥0, then Φ=0. If B2n=1 for n≥0, then B2n+1≠1 for n≥0. In view of (25) Ψ(1-B2n+1)=0 for n≥0 and so Ψ=0.(i)
IfB2n+1=0 for n≥0; then B2n≠0 for n≥0. Now suppose that B2n≠0 for n≥0; then from B2nΦ=B2n+1Ψ=0 we get that Φ=ΨB2n+1/B2n=0 for n≥0. Since Φ=0, then Ψ≠0. Thus B2n+1=0 for n≥0.(ii)
The proof is similar to the proof of part (i).(iii)
By (26) B2nΦ=0 for n≥0. If B2n≠0 for n≥0, then Φ=0. Now if B2n=0 for n≥0, then B2n+1≠0 for n≥0. In view of (27) ΨB2n+1=0 for n≥0 and so Ψ=0.Example 12.
The difference equation(30)xn+1=cnn+1xn+(1-1n+1)xn-1,n=0,1,…,
where {cn} is a period-two sequence such that c2n=γ≠0,1, c2n+1=1/γ, has an infinite number of period-two solutions of the form {x-1,x-1/γ},x-1≠0, which can be seen by immediate checking.
This equation is an illustration of Theorem11 part 1. In this case An={γ,1/2γ,γ/3,1/4γ,γ/5,…} and Bn={0,1/2,2/3,3/4,4/5,…}.Theorem 13.
Letl∈{2,4,…}. Suppose that (1) has the linearization (2) and An,Bn are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,  Ψ=x2n,andΦ=x2n-1 for n≥0. (1)
Let eitherAn≠1 or Bn≠0 for n≥0; then
(a)
Ψ/Φ=B0/(1-A0)=B2n/(1-A2n),n=0,1,…,(b)
Φ/Ψ=B1/(1-A1)=B2n+1/(1-A2n+1),n=0,1,….(2)
LetAn=1 for n≥0.
(a)
IfB2n=B2n+1,n=0,1,…,  then B2n=B2n+1=0,n≥0.(b)
IfB2n≠B2n+1,n=0,1,…, then the following are true:
(i)
B2n≠0 if and only if B2n+1=0 for n=0,1,…;(ii)
B2n=0 if and only if B2n+1≠0 for n=0,1,…;(iii)
eitherΨ=0 or Φ=0.(3)
LetBn=0 for n≥0.
(a)
IfA2n=A2n+1,n=0,1,…, then A2n=A2n+1=1,n≥0.(b)
IfA2n≠A2n+1,n=0,1,…, then the following are true:
(i)
A2n≠1 if and only if A2n+1=1 for n=0,1,…;(ii)
A2n=1 if and only if A2n+1≠1 for n=0,1,…;(iii)
eitherΨ=0 or Φ=0.Proof.
The proof is as follows.(1)
Assume thatAn≠1 for n≥0. Then ΦΨ≠0. Otherwise, if Φ=0 then Ψ≠0. By (26) Ψ(1-A2n)=0 and so A2n=1 for n≥0, which is a contradiction. In the other case, if Ψ=0 then Φ≠0. By (27) Φ(1-A2n+1)=0 and so A2n+1=1 for n≥0, which is a contradiction. Hence Bn≠0 for n≥0.
(a)
By (26) Ψ/Φ=B0/(1-A0)=B2n/(1-A2n) for n≥0.(b)
By (27) Φ/Ψ=B1/(1-A1)=B2n+1/(1-A2n+1) for n≥0.Now assume thatBn≠0 for n≥0. Then ΦΨ≠0. Otherwise, if Ψ=0 then Φ≠0. By (26) ΦB2n=0 and so B2n=0 for n≥0, which is a contradiction. In the other case, if Φ=0 then Ψ≠0. By (27) ΨB2n+1=0 and so B2n+1=0 for n≥0, which is a contradiction. Hence An≠1 for n≥0 and the proof follows similarly to the previous part.(2)
In view of (26) and (27) condition An=1 for n≥0 implies B2nΦ=B2n+1Ψ=0 for n=0,1,….
(a)
Assume thatB2n=B2n+1,n=0,1,….IfB2n≠0 for some n≥0, then B2n+1≠0 for this n and so Φ=Ψ=0, which is a contradiction. Hence B2n=0 for n≥0. Thus B2n=B2n+1=0,n=0,1,….(b)
Assume thatB2n≠B2n+1 for n≥0.
(i)
IfB2n+1=0 for n≥0; then B2n≠0 for n≥0. Now suppose that B2n≠0 for n≥0; then from B2nΦ=B2n+1Ψ=0 we get that Φ=ΨB2n+1/B2n=0 for n≥0. Since Φ=0, then Ψ≠0. Thus B2n+1=0 for n≥0.(ii)
The proof is similar to the proof of part (i).(iii)
By (26) B2nΦ=0 for n≥0. If B2n≠0 for n≥0, then Φ=0. Now if B2n=0 for n≥0, then B2n+1≠0 for n≥0. In view of (27) ΨB2n+1=0 for n≥0 and so Ψ=0.(3)
In view of (26) and (27) condition Bn=0 for n≥0 implies (1-A2n)Ψ=(1-A2n+1)Φ=0 for n=0,1,….
(a)
Assume thatA2n=A2n+1,n=0,1,….IfA2n≠1 for some n≥0, then A2n+1≠1 for this n and so Φ=Ψ=0, which is a contradiction. Hence A2n=1 for n≥0. Thus A2n=A2n+1=1,n=0,1,….(b)
Assume thatA2n≠A2n+1 for n≥0.
(i)
IfA2n+1=1 for n≥0, then A2n≠1 for n≥0. Now assume that A2n≠1 for n≥0; then from (1-A2n)Ψ=(1-A2n+1)Φ=0 we get that Ψ=Φ(1-A2n+1)/(1-A2n)=0 for n≥0. Since Ψ=0, then Φ≠0. Thus A2n+1=1 for n≥0.(ii)
The proof is similar to the proof of part (i).(iii)
By (26) (1-A2n)Ψ=0 for n≥0. If A2n≠1 for n≥0, then Ψ=0. Now if A2n=1 for n≥0, then A2n+1≠1 for n≥0. In view of (27) (1-A2n+1)Φ=0 for n≥0 and so Φ=0.Corollary 14.
Suppose that (1) has the linearization (2) and that An,Bn are given by (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0. (1)
Letl∈{1,3,…}.
(a)
An≠0 if and only if Bn≠1,n=0,1,….(b)
IfA0/(1-B0)=A1/(1-B1) and either Bn≠1 or An≠0 for n=0,1,…,  then
(i)
B2n-A2n=1 and B2n+1-A2n+1=1,n=0,1,…;(ii)
Ψ=-Φ.(2)
Letl∈{2,4,…}.
(a)
An≠1 if and only if Bn≠0,n=0,1,….(b)
IfB0/(1-A0)=B1/(1-A1) and either An≠1 or Bn≠0 for n=0,1,…,  then
(i)
A2n-B2n=1 and A2n+1-B2n+1=1,n=0,1,…;(ii)
Ψ=-Φ.Proof.
Part (1a) follows from the proof of Theorem11 part (1). Part (1b) follows from the proof of Theorem 11 part (1) and A0/(1-B0)=A1/(1-B1), which implies Φ/Ψ=Ψ/Φ. Hence Φ2=Ψ2 and so Φ=-Ψ. Thus -1=A2n/(1-B2n)=A2n+1/(1-B2n+1),n≥0 which implies B2n-A2n=1 and B2n+1-A2n+1=1 for n≥0.
Part (2a) follows from the proof of Theorem13 part (1) and the proof of part (2b) follows in a similar way as the proof of part (1b).The next result gives a necessary and sufficient condition for the existence of a minimal period-two solution in the special case whenl=1 in (2).Theorem 15.
Suppose that (1) has the linearization (2) with l=1 and that An,Bn are as in (23). Then (1) has a minimal period-two solution if and only if one of the following holds. (1)
All the conditions of Theorem11 part (1) are satisfied. Furthermore the period-two solution is …,A0x0/(1-B0),A1x-1/(1-B1),….(2)
Bn=1 for n≥0 and one of the following conditions is satisfied for n≥0:
(a)
A2n=A2n+1=0,(b)
A2n=0 and A2n+1≠0,(c)
A2n≠0 and A2n+1=0.(3)
An=0 for n≥0 and one of the following conditions is satisfied for n≥0:
(a)
B2n=B2n+1=1,(b)
B2n=1 and B2n+1≠1,(c)
B2n≠1 and B2n+1=1.Proof.
The necessary part follows from Theorem11. We will now prove the sufficient part.(1)
Assume that conditions (a) and (b) of Theorem11 part (1) are satisfied. Then by (a) x0=(1-B0)x-1/A0=(1-B2n)x-1/A2n for n≥0 and by (b) x-1=(1-B1)x0/ A1=(1-B2n+1)x0/A2n+1 for n≥0. Choose the initial conditions x0=x-2=⋯,  x-1=x-3=⋯, and x0≠x-1. Then by using these equalities we get
(31)x1=A0x0+B0x-1=(1-B0)x-1+B0x-1=x-1,x2=A1x-1+B1x0=(1-B1)x0+B1x0=x0,x3=A2x0+B2x-1=(1-B2)x-1+B2x-1=x-1,x4=A3x-1+B3x0=(1-B3)x0+B3x0=x0.
Simple induction completes the proof.(2)
Next, suppose thatBn=1 for n≥0.
(a)
Assume thatA2n=A2n+1=0 for n≥0. By choosing the initial conditions x0=x-2=⋯,x-1=x-3=⋯, and x0≠x-1, an immediate calculation shows that {x-1,x0} is a minimal period-two solution.(b)
Assume thatA2n=0 and A2n+1≠0 for n≥0. By choosing the initial conditions x0=x-2=⋯,x-1=x-3=⋯=0, and x0≠0, we obtain
(32)x1=A0x0+B0x-1=0,x2=A1x1+B1x0=x0,x3=A2x2+B2x1=0,x4=A3x3+B3x2=x0,
and straightforward induction shows that {0,x0} is the minimal period-two solution.(c)
Assume thatA2n≠0 and A2n+1=0 for n≥0. Choose the initial conditions x0=x-2=⋯=0,x-1=x-3=⋯, and x-1≠0. By straightforward induction we obtain that {x-1,0} is the minimal period-two solution.(3)
Next, suppose thatAn=0 for n≥0.
(a)
Assume thatB2n=B2n+1=1 for n≥0. By choosing the initial conditions x0 = x-2=⋯,x-1=x-3=⋯, and x0≠x-1, an immediate calculation shows that {x-1,x0} is the minimal period-two solution.(b)
Assume thatB2n=1 and B2n+1≠1 for n≥0. By choosing the initial conditions x0=x-2=⋯=0,x-1=x-3=⋯, and x-1≠0, we obtain
(33)x1=A0x0+B0x-1=x-1,x2=A1x1+B1x0=0,x3=A2x2+B2x1=x-1,x4=A3x3+B3x2=0,
and straightforward induction shows that {x-1,0} is the minimal period-two solution.(c)
Assume thatB2n≠1 and B2n+1=1 for n≥0. By choosing the initial conditions x0=x-2=⋯,x-1=x-3=⋯=0, and x0≠0, and using a straightforward induction we obtain that {0,x0} is the minimal period-two solution.Example 16.
The difference equation(34)xn+1=Bxnxn-1dxn+exn-1,n=0,1,…,
where B,d,e>0, has a minimal period-two solutions of the form {x-1,-x-1},x-1≠0, if and only if B=d-e and B≠d.
This equation is an illustration of Theorem15. The linearization of (34) gives
(35)xn+1=g1xn-1,n≥0,
where g1=Bxn/(dxn+exn-1),n≥0. Since An=0 for all n≥0, then from Theorem 15 part (3a) a period-two solution exists if and only if
(36)g1=Bn=Bxndxn+exn-1=1,∀n=0,1,….
Observe that if B=d, then exn-1=0 for n≥0, which is a contradiction.
Equation (36) gives the first order equation
(37)xn=eB-dxn-1,n=0,1,…
which has a period-two solution if and only if B=d-e and x0=-x-1 as Theorem 15 part (1) is satisfied.
Therefore, (34) has a period-two solution of the form {x-1,-x-1},x-1≠0 if and only if B=d-e,B≠d.
This example can be extended to a more general equation of the form(38)xn+1=Bxnxn-1xn-2a0xn-1xn-2+a1xnxn-2+a2xnxn-1,iiiiiin=0,1,…,
where B,ai>0,i=0,1,2. Similar reasoning gives the necessary and sufficient conditions for the existence of a period-two solution to be
(39)B=a1-(a0+a2),B≠a1.
In this case there is an infinite number of period-two solutions of the form {x-1,-x-1}, x-1≠0.Corollary 17.
Suppose that (1) has the linearization (2) with l=1 and that An,Bn are as in (23). Assume that A0/(1-B0)=A1/(1-B1) and either Bn≠1 or An≠0 for n≥0. Then (1) has a minimal period-two solution if and only if B2n+1-A2n+1=1andB2n-A2n=1 for n≥0.Proof.
The necessary part follows from Corollary14 part (1b).
For the proof of sufficient part choose the initial conditionsx0=x-2=⋯,x-1=x-3=⋯, and x0=-x-1. Setting n=0,1,… we obtain
(40)x1=A0x0+B0x-1=-A0x-1+B0x-1=x-1,x2=A1x1+B1x0=A1x-1-B1x-1=-x-1,x3=A2x2+B2x1=-A2x-1+B2x-1=x-1,x4=A3x3+B3x2=A3x-1-B3x-1=-x-1.
Simple induction completes the proof.So far we have considered the cases whenl∈{1,3,…} and Bn is either equal or not equal to one for all n≥0. But what happens when l∈{1,3,…} and Bn=1 for some n’s and Bn≠1 for other n’s or when An=0 for some n’s and An≠0 for other n’s? We will now investigate these cases along with the cases when l∈{2,4,…} and Bn=0 for some n’s and Bn≠0 for other n’s or when An=1 for some n’s and An≠1 for other n’s.Theorem 18.
Letl∈{1,3,…}. Suppose that (1) has the linearization (2) and that An,Bn are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0. (1)
LetΦΨ≠0. Then
(a)
for someN∈{0,1,…}BN=1 if and only if AN=0;(b)
if someN are even and BN≠1, then AN=(Φ/Ψ)(1-BN);(c)
if someN are odd and BN≠1, then AN=(Ψ/Φ)(1-BN).(2)
IfΨ=0, then for all n≥0B2n=1 and A2n+1=0.(3)
IfΦ=0, then for all n≥0B2n+1=1 and A2n=0.Proof.
The proof is as follows.(1)
The result follows from Lemma9 part (a).(2)
SinceΨ=0, then Φ≠0 and the result follows from Lemma 9 part (a).(3)
SinceΦ=0, then Ψ≠0 and the result follows from Lemma 9 part (a).Remark 19.
Note that in part(2) of Theorem 18  B2n+1∈ℝ and A2n∈ℝ for all n≥0. Similarly in part (3) of Theorem 18  B2n∈ℝ and A2n+1∈ℝ for all n≥0.Theorem 20.
Letl∈{2,4,…}. Suppose that (1) has the linearization (2) and that An,Bn are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0. (1)
LetΦΨ≠0. Then
(a)
for someN∈{0,1,…}AN=1 if and only if BN=0;(b)
if someN are even and AN≠1, then BN=(Ψ/Φ)(1-AN);(c)
if someN are odd and AN≠1, then BN=(Φ/Ψ)(1-AN).(2)
IfΨ=0, then for all n≥0B2n=0 and A2n+1=1.(3)
IfΦ=0, then for all n≥0B2n+1=0 and A2n=1.Proof.
The results follow from Lemma9 part (b).Theorem 21.
Suppose that (1) has the linearization (2) with l=1 and that An,Bn are as in (23). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=x2n,andΦ=x2n-1 for n≥0 if and only if one of the following holds: (1)
ΦΨ≠0 and for each n≥0 either
(a)
Bn=1 and An=0;(b)
n is even, Bn≠1, and An=(Φ/Ψ)(1-Bn);(c)
n is odd, Bn≠1, and An=(Ψ/Φ)(1-Bn).(2)
IfΨ=0, then for all n≥0B2n=1 and A2n+1=0.(3)
IfΦ=0, then for all n≥0B2n+1=1 and A2n=0.Proof.
The necessary part follows from Theorem18. The sufficient part follows by direct calculation of the period-two solution which satisfies specific initial conditions in a similar way as in the proof of Theorem 15.An illustration of Theorem20 is the following example.Example 22.
The difference equation(41)xn+1=sin(xn-1)n+1xn+xn-1exn,n=0,1,…,
has an infinite number of period-two solutions of the form {0,x0},x0≠0, which can be seen by immediate checking. In view of Theorem 20 part (3)  A2n=sin(x2n-1)/(2n+1) and B2n+1=ex2n+1=1 for n=0,1,… when Φ=0.



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