# Periodic Solutions to a Third-Order Conditional Difference Equation over the Integers

Authors: Li He; Wanping Liu
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

This paper studies a third-order conditional difference equation which is a generalization from the literature. We investigate this equation by transforming it into a first-order system. Finally it is proved that the equation has no period-two (or three) integer solutions. Besides, its all period-four (and five) integer solutions are derived under appropriate rational parameters.


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## Body

## 1. Introduction

As mentioned in [1], difference equations appear naturally as a discrete analogue and as a numerical solution of differential and delay differential equations having applications in various scientific branches, such as biology, ecology, physics, economy, technics, and sociology. The stability, asymptotic behavior, and periodic property of solutions to difference equations had been widely investigated, such as [2–14]. Recently, the study of max-type difference equation attracted a considerable attention, for example, [7, 11, 15–25]. Although max-type difference equations are relatively simple in form, it is unfortunately extremely difficult to understand thoroughly the behavior of their solutions. The max operator arises naturally in certain models in automatic control theory. On the other hand, there exists another kind of difference equations called conditional difference equations, which also have simple forms, but it is difficult to understand clearly the behavior of their solutions.From [2, 5], we know that the following conditional difference equation
(1.1)an={an-1+an-22,if2∣(an-1+an-2),an-1+an-2,otherwise,
with positive initial integers a0,a1, has the property that each positive integer solution (an) is either stationary or unbounded.Later, Ladas [26] gave the conjecture (also mentioned in [5]) that all solutions to the conditional difference equation(1.2)an={an-1+an-23,if3∣(an-1+an-2),an-1+an-2,otherwise,
are unbounded except for certain obvious periodic solutions, such as the solutions 1,1,2,1,1,2,… or 7,14,7,7,14,7,…. For such kind of conditional difference equations, any solution which is not eventually periodic must be unbounded; hence, the only problem is to classify the periodic solutions. However, this problem seems extremely difficult.Equations such as (1.1) and (1.2) have many other generalizations, see [6, 26]. For instance, Clark [3] studied periodic solutions to an=⌈can-1⌉-an-2 for various real c.Greene and Niedzielski [5] considered a generalization of (1.1) and (1.2) and studied the following conditional difference equation:(1.3)an={r(an-1+an-2),ifr(an-1+an-2)∈Z,an-1+an-2,otherwise,
where r is some fixed rational number. However, they pointed out that it was also very hard to characterize the periodic solutions to (1.3). Hence, they addressed a different, easier question, and then did some research.Motivated by the above works, in this paper, we study the following conditional third-order difference equation, which is another generalization of (1.1), (1.2), and (1.3):(1.4)yn={r(yn-1+yn-2+yn-3),ifr(yn-1+yn-2+yn-3)∈Z,yn-1+yn-2+yn-3,otherwise,
where r is some appropriate rational number. We study this equation by transforming it into a first-order system. It is eventually proved that the equation has no period-two (or three) positive integer solutions. Besides, its all period-four (and five) positive integer solutions are derived under appropriate rational parameters. The main results are presented in Section 4.


## 2. Auxiliary Results

For the convenience of investigation, higher-order difference equations are usually converted into first- or lower-order difference equation systems. As we all know, second-order difference equations can be transformed into first-order difference equation system. First, we define a mappingℱ:ℤ3→ℤ3 such that(2.1)F(x)={Dx,ifr(x+y+z)∈Z,Cx,otherwise,
where x=(x,y,z)T∈ℤ3, r is a rational parameter, and(2.2)D=(010001rrr),C=(010001111).Applying the defined mappingℱ(x), the third-order difference equation (1.4) can be converted into a corresponding first-order difference equation system as follows. Let xn=(xn,yn,zn)T∈ℤ3, n∈ℕ0 and xn=ℱn(x0), n∈ℕ, then we have the following first-order difference equation system:(2.3)xn+1=F(xn)={Dxn,ifr(xn+yn+zn)∈Z,Cxn,otherwise,
where x0=(y0,y1,y2)T, y0,y1,y2 are initial integers in (1.4). Note that the condition r(xn+yn+zn)∈ℤ can be replaced by the equivalent matrix product condition r((1,1,1)xn)∈ℤ.For instance, whenr=1/5, then the periodic solution (1,1,1,3,1,1,1,3,…) to (1.4) corresponds to the following periodic solution to system (2.3):(2.4)(111),(113),(131),(311),(111),….
Given an initial vector x0, each subsequent xn can be got from x0 via a formula xn=Lnx0, where Ln is an appropriate product of n matrices, each of which is C or D. Hence, for the above example, if x0=(1,1,1)T, then x1=Cx0, x2=Dx1=DCx0, x3=D2Cx0, x4=D3Cx0, and so forth. Note that the matrices multiply x0 from right to left and that x4=x0, so D3Cx0=x0. We have the following obvious result.Lemma 2.1.
If system (2.3) has a periodic solution with period k and x is a vector in that periodic solution, then there exists a corresponding matrix L, which is a product of k matrices, each of which is C or D, such that x is an eigenvector of L with eigenvalue 1.The proof of Lemma2.1 is simple; here we point out that the converse of Lemma 2.1 is not true. The problem is that the sum of the entries in some xi may be divisible by the denominator of r at a time, when multiplication by C is called for.Next, we present the following linear algebra facts, for example, [27, Chapter  7], (similar results presented in [5]), followed by the form most convenient to us.Lemma 2.2.
(i) Ifv is an eigenvector with eigenvalue λ for a matrix L, then v is an eigenvector with eigenvalue λk for Lk.
(ii) The characteristic polynomial for a3×3 matrix L has the form
(2.5)fL(λ)=det⁡(λI-L)=λ3-tr⁡(L)λ2+(det⁡(L1,1)+det⁡(L2,2)+det⁡(L3,3))λ-det⁡(L),
where tr(L) is the trace of L.
(iii) The characteristic polynomials forLM and ML are identical for any n×n matrices L and M. In particular, tr(LM)=tr(ML).
(iv) The trace is linear. That is,tr(αL+βM)=αtr(L)+βtr(M), for any n×n matrices L,M and scalars α,β.
(v) Every matrix satisfies its characteristic polynomial. In particular, for a3×3 matrices L, one has
(2.6)L3-tr⁡(L)L2+(det⁡(L1,1)+det⁡(L2,2)+det⁡(L3,3))L-det⁡(L)I=0.Let the sequences(Hn)n≥0, (Jn)n≥0 be two solutions to the difference equation(2.7)yn=yn-1+yn-2+yn-3,n≥3
with initial values H0=H1=0, H2=1 and J0=J2=1, J1=0, respectively.Then, we get the following result about the matrixC in (2.2).Lemma 2.3.
For the matrixC  in (2.2), one has
(2.8)Cn=(Hn-1Jn-1HnHnJnHn+1Hn+1Jn+1Hn+2),
for n∈ℕ.Proof.
This result can be proved by induction. By (2.2) and the definitions of the sequences (Hn)n≥0, (Jn)n≥0, we have
(2.9)C=(010001111)=(H0J0H1H1J1H2H2J2H3).
Therefore, (2.8) holds for n=1. Now, we assume that (2.8) holds for n=ω, ω∈ℕ. In the following, it suffices to prove that (2.8) holds for n=ω+1. By (2.2) and the associative law of matrix multiplication, we have
(2.10)Cω+1=CCω=(010001111)(Hω-1Jω-1HωHωJωHω+1Hω+1Jω+1Hω+2)=(HωJωHω+1Hω+1Jω+1Hω+2Hω+2Jω+2Hω+3).
Therefore, (2.8) holds for k=ω+1. The proof is complete.Given a3×3 matrix L=N1N2⋯Nk, k∈ℕ, where each Nj is either C or D, by Lemma 2.2 we define the following polynomial:(2.11)PL(r)=-det⁡(I-L)=-1+tr⁡(L)+det⁡(L)-(det⁡(L1,1)+det⁡(L2,2)+det⁡(L3,3)),
where I is the identity matrix, and seek L for which 𝒫L(r) has rational zeros.


## 3. Some Other Properties

In this section, several properties of𝒫L(r) defined in (2.11) are derived. They are used to derive restrictions on values of r that allow periodic solutions to (1.4).Lemma 3.1.
For the polynomial𝒫L(r)  in (2.11),  if 𝒫L(r)=0,  then  𝒫Lk(r)=0,  for each  k∈ℕ.Proof.
Since𝒫Lk(r)=-det(I-Lk)=-det(I-L)det(Lk-1+Lk-1+⋯+I), then 𝒫Lk(r)=𝒫L(r)q(r) for some polynomial
(3.1)q(r)=det⁡(Lk-1+Lk-1+⋯+I),
and the result follows.Lemma 3.2.
IfC=(010001111),E=(010001xxx), then tr(Cn1ECn2E⋯CnkE) is a polynomial in x of degree k with nonnegative integer coefficients, with leading coefficient ∏j=1kHnj+3, where nj∈ℕ.Proof.
By Lemma2.3, we get that
(3.2)CnE=(xHnxHn+Hn-1xHn+Jn-1xHn+1xHn+1+HnxHn+1+JnxHn+2xHn+2+Hn+1xHn+2+Jn+1)=x(HnHnHnHn+1Hn+1Hn+1Hn+2Hn+2Hn+2)+(0Hn-1Jn-10HnJn0Hn+1Jn+1)=x(Hn,Hn+1,Hn+2)T1+Fn,
where 1 represents the vector (1,1,1), and
(3.3)Fn=(0Hn-1Jn-10HnJn0Hn+1Jn+1).
Hence,
(3.4)Cn1ECn2E⋯CnkE=(x(Hn1,Hn1+1,Hn1+2)T1+Fn1)⋯(x(Hnk,Hnk+1,Hnk+2)T1+Fnk)=xk(Hn1,Hn1+1,Hn1+2)T1(Hn2,Hn2+1,Hn2+2)T1⋯(Hnk,Hnk+1,Hnk+2)T1+⋯+Fn1Fn2⋯Fnk=xk∏j=1k(Hnj,Hnj+1,Hnj+2)T1+⋯+∏j=1kFnj.
To obtain the leading coefficient, we may induct on the simple calculation
(3.5)((a,b,c)T1)((d,e,f)T1)=(d+e+f)((a,b,c)T1)
to show that the trace of the product is the product of the traces, such as the following. Easily, the trace of each individual matrix is Hnj+3,j=1,2,…,k. By (2.7) and (3.5), we have that
(3.6)∏j=1k(Hnj,Hnj+1,Hnj+2)T1=(Hnk+Hnk+1+Hnk+2)∏j=1k-1(Hnj,Hnj+1,Hnj+2)T1=Hnk+3∏j=1k-1(Hnj,Hnj+1,Hnj+2)T1=⋯=Hnk+3Hnk-1+3⋯Hn2+3(Hn1,Hn1+1,Hn1+2)T1=∏j=2kHnj+3(Hn1,Hn1+1,Hn1+2)T1.
Thus the leading coefficient is
(3.7)∏j=2kHnj+3(Hn1+Hn1+1+Hn1+2)=∏j=1kHnj+3.
The proof is complete.


## 4. Periodic Solutions

In this section, we prove that (1.4) has no periodic solution with prime period two or three and derive all periodic solutions to (1.4) with prime period four and five.Theorem 4.1.
Suppose thatr is a rational number and r∉ℤ, then (1.4) has no periodic solutions with prime period two.Proof.
Suppose thatx∈ℤ3 be a vector of a period-two solution to system (2.3), then, for some matrix L2, we have x=L2x; here the matrix L2 has four possible cases C2, CD, DC and D2.
TakeL2=D2, for example, through some calculations, we get that
(4.1)D2=(001rrrr2r(r+1)r(r+1)),PD2(r)=3r2+2r-1.
By solving the equation𝒫D2(r)=3r2+2r-1=0, we get two real roots r1=1/3 and r2=-1. Through similar calculations, we can get Table 1.
The only value ofr which may lead to period-two solutions to system (2.3) is r=1/3 since r∉ℤ. Hence, by solving the following matrix equation (with r=1/3)
(4.2)(D2-I)x=0,
we get its all integer solutions x=(t,t,t)T, t∈ℤ. Obviously, the solutions are equilibrium points which contradicts the assumption. The proof is complete.Table 1L2𝒫L2(r)Roots of𝒫L2(r)=0D23r2+2r-11/3,-1DC4r0CD4r0C24∖Theorem 4.2.
Suppose thatr is a rational number and r∉ℤ, then (1.4) has no periodic solutions with prime period three.Proof.
Assume thatx∈ℤ3 be a vector of a period-three solution to system (2.3), then we get x=L3x where L3 is a matrix product of three matrices, each of which is C or D.
Obviously,L3 has eight possible cases which can be divided into four categories such as three Cs, two Cs and one D, one C and two Ds, and three Ds. By (2.2) and through certain calculations, the matrices C2D,CDC,DC2 are similar, and; thus, they generate the same polynomials 𝒫L3(r), so do the matrices CD2,DCD,D2C.
In the following, we takeL3=D3, for example. Through some calculations, we have
(4.3)D3=(rrrr2r(r+1)r(r+1)r2(r+1)r2(r+2)r(r2+2r+1)),PD3(r)=3r-1.
By solving the equation 𝒫D3(r)=3r-1=0, we get the only root r=1/3. Through similar calculations, we can obtain Table 2.
The only value ofr which may lead to period-three solutions to system (2.3) is r=1/3 since r∉ℤ. Hence, by solving the following matrix equation (with r=1/3)
(4.4)(D3-I)x=0,
we get its all integer solutions x=(t,t,t)T,t∈ℤ. Obviously, the solutions are equilibrium points which contradicts the assumption. The proof is complete.Table 2L3𝒫L3(r)Roots of𝒫L3(r)=0D33r-11/31D2Csr+1−11C2Ds2r0C32∖In the following, denote by𝒮(k,r) the set of initial values (y0,y1,y2) which lead to period-k solutions to (1.4) with the parameter r.Theorem 4.3.
Suppose thatr is a rational number and r∉ℤ, then (1.4) has periodic solutions with prime period four if and only if r=-1/3 or 1/5.  Moreover, 𝒮(4,-1/3) =  {(t,t,t),(-t,t,t),(t,-t,t),(t,t,-t)∣t∈ℤ,3∤t}, and 𝒮(4,1/5) = {(t,t,3t),(t,3t,t),(3t,t,t),(t,t,t)∣t∈ℤ,5∤t}.Proof.
Letx∈ℤ3 be a vector of a period-four solution to system (2.3), then we get x=L4x, where L4 is an appropriate product of four matrices, each of which is C or D. Through similar calculation to those in Theorems 4.1 and 4.2, Table 3 is derived. The only values of r which possibly lead to period-four solutions to system (2.3) are r=1/3,-1/3, or 1/5 because of r∉ℤ.
Case 1 (r=1/3).
By solving the following matrix equation(4.5)(D4-I)x=0,
we get its all integer solutions x=(t,t,t)T,t∈ℤ. Obviously, the solutions are equilibrium points.Case 2 (r=-1/3).
In this case, the matrixL4 has four possible cases C3D,C2DC,CDC2, and DC3.
By solving the matrix equation(4.6)(C3D-I)x=0,
we get its all integer solutions x=(t,t,t)T,t∈ℤ. On the condition of that 3∤t, then we can verify that the initial vector x0=(t,t,t)T leads to a period-four solution to system (2.3), such as the following:
(4.7)(ttt),(tt-t),(t-tt),(-ttt),(ttt),….
Similarly, for the following three matrix equations
(4.8)(C2DC-I)x=0,(CDC2-I)x=0,(DC3-I)x=0,
we derive its all integer solutions x=(-t,t,t)T,x=(t,-t,t)T,x=(t,t,-t)T,t∈ℤ, respectively.
Note that, on the condition3∤t, the initial vectors x0=(-t,t,t)T,(t,-t,t)T, or (t,t,-t)T also lead to period-four solutions to system (2.3).Case 3 (r=1/5).
In this case, the matrixL4 also has four possible cases CD3,DCD2,D2CD, and D3C. By solving the matrix equation
(4.9)(CD3-I)x=0,
we get its all integer solutions x=(t,t,3t)T, t∈ℤ. On the condition of that 5∤t, then we can verify that the initial vector x0=(t,t,3t)T leads to a period-four solution to system (2.3), such as the following:
(4.10)(tt3t),(t3tt),(3ttt),(ttt),(tt3t),….
Obviously, on the condition 5∤t, the initial vectors x0=(t,3t,t)T, (3t,t,t)T, or (t,t,t)T (which are integer solutions to appropriate matrix equations corresponding to DCD2, D2CD, D3C, resp.) also lead to period-four solutions to system (2.3). The proof is complete.Table 3L4𝒫L4(r)Roots of𝒫L4(r)=0D43r4+6r2+8r3-11⁄3, −1(triple)3Ds1C5r3+9r2+3r-11⁄5, −1(double)2Ds2Cs8r2+8r-1(double)1D3Cs12r+4-1/3C416∖To illustrate the results, we give two orbits (see the following Figure1) of period-four integer solutions to (1.4) of the particular cases r=-1/3, t=8 and r=1/5, t=7 in Theorem 4.3.Equation (1.4). (a) r=-1/3. Initial values y0=8,y1=8, y2=8(t=8). (b) r=1/5. Initial values y0=7, y1=7, y2=21(t=7).
(a)(b)Theorem 4.4.
Suppose thatr is a rational number and r∉ℤ, then (1.4) has periodic solutions with prime period five if and only if r=-7/15.  Moreover, 𝒮(5,-7/15) = {(5t,t,9t),(t,9t,-7t),(9t,-7t,3t),(-7t,3t,5t),(3t,5t,t)∣t∈ℤ,3∤t,5∤t}.Proof.
Letx∈ℤ3 be a vector of a period-five solution to system (2.3), then we get x=L5x where L5 is an appropriate product of five matrices, each of which is C or D. Through similar calculation to those in Theorems 4.1 and 4.2, Table 4 is obtained. Apparently, the only values of r which possibly lead to period-three solutions to system (2.3) are r=1/3 or -7/15 because of r∉ℤ.
Case 1 (r=1/3).
By solving the matrix equation(4.11)(D5-I)x=0,
we get its all integer solutions x=(t,t,t)T, t∈ℤ. Obviously, the solutions are equilibrium points.Case 2 (r=-7/15).
In this case, the matrixL5 also has five possible cases C4D,C3DC,C2DC2,CDC3, and DC4. Solve the matrix equation
(4.12)(C4D-I)x=0,
we get its all integer solutions x=(5t,t,9t)T, t∈ℤ. On the conditions that 3∤t,5∤t, then we can verify that the initial vector x0=(5t,t,9t)T leads to a period-four solution to system (2.3), such as the following:
(4.13)(5tt9t),(t9t-7t),(9t-7t3t),(-7t3t5t),(3t5tt),(5tt9t),….
Obviously, on the conditions3∤t,5∤t, the initial vectors x0=(t,9t,-7t)T, (9t,-7t,3t)T, (-7t,3t,5t)T, or (3t,5t,t)T (which are integer solutions to appropriate matrix equations corresponding to DC4, CDC3, C2DC2, and C3DC, resp.) also lead to period-four solutions to system (2.3). The proof is complete.Table 4L5𝒫L5(r)Rational roots of𝒫L5(r)=0D53r5+5r4+10r3+5r2-11/34Ds1C4r4+8r3+9r2+2r-1∖3Ds2Cs6r3+10r2+7r-1∖2Ds3Cs9r2+12r+1∖1D4Cs15r+7-7/15C522∖To illustrate the results, we give two orbits (see Figure2) of period-five integer solutions to (1.4) of the particular cases t=2,11 in Theorem 4.4.Equation (1.4) r=-7/15. (a): Initial values y0=10, y1=2, y2=18(t=2); (b): Initial values y0=55, y1=11, y2=99(t=11).
(a)(b)



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