# Periodic Solutions of Certain Differential Equations with Piecewise Constant Argument

Authors: James Guyker
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

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## Abstract

Existence criteria are derived for the eventually periodic solutions of a class of differential equations with piecewise constant argument whose solutions at consecutive integers satisfy nonlinear recurrence relations. The proof characterizes the initial values of periodic solutions in terms of the coefficients of the resulting difference equations. Sufficient conditions for the unboundedness, boundedness, and symmetry of general solutions also follow from the corresponding properties of the difference
equations.


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## Body

## 1. Introduction

Since the seminal works of Shah and Wiener [1] and Cooke and Wiener [2], differential equations with piecewise constant arguments of the form(1)x′t=ft,xt,xt,where f is continuous and [·] is the greatest integer function, have been treated widely in the literature and applied to certain biomedical models (see [3–7] and references therein). Continuity of the solutions of these equations implies recurrence relations for the values of solutions at consecutive integers. Therefore, there is a natural interplay between properties of these differential equations and properties of difference equations.In this paper, we consider a class of equations of the above form but wheref is discontinuous: the chaotic and eventually periodic behavior and symmetry of solutions of initial-value problems of the form(2)x′t=Axt+Bxt+CFxt,x0=x0on [0,∞) are determined, where A, B, and C are constants; F:R→R is defined by(3)Fx≡0,ifx≤λ1,ifx>λfor some positive number λ, and the sequence x(n)(n=0,1,…) satisfies a nonlinear difference equation. As in the case when f is continuous, by a solution of (2), we mean a function x(t) that is defined on [0,∞) with these properties:(1)
x(t) is continuous on [0,∞).(2)
The derivativex′(t) exists at each point t in [0,∞), with the possible exception of the points [t] in [0,∞), where one-sided derivatives exist.(3)
x(t) satisfies (2) on [n,n+1) for each nonnegative integer n.Specifically, since(4)ddtxte-At=Bxte-At+CFxte-At,it follows that if n is a positive integer and n-1≤t′<n, then(5)xt′e-At′-xn-1e-An-1=Bxn-1+CFxn-1∫n-1t′e-Atdt.Letting t′ increase to n, we have the difference equation(6)xn=a∗xn-1-b∗Fxn-1,where x(0)=x0,(7)a∗≡eA+BeA-1A,ifA≠01+B,ifA=0,b∗≡-CeA-1A,ifA≠0-C,ifA=0.Therefore, the unique solution to (2) is(8)xt=eAt-t+B∫tteAt-sdsxt+C∫tteAt-sdsFxt,where (9)∫tteAt-sds=eAt-t-1A,ifA≠0t-t,ifA=0.Moreover,x(t) satisfies (1)–(3) above since xt is continuous on [0,∞) with left derivatives(10)x′n-=Axn+Bxn-1+CFxn-1at integers n>0, with right derivatives(11)x′n+=A+Bxn+CFxnat integers n≥0, and more generally with derivatives at nonintegral t given by(12)x′t=A+Bxt+CFxteAt-t.We recall that the solutionx=x(t) to (2) isoscillatory if it has arbitrarily large zeros [3]. Accordingly, a sequence xn(n=0,1,…) isoscillatory if there exists a subsequence xnk such that xnkxnk+1≤0 for all k. Moreover, astationary state of xn is a term xi such that xi=xi+1 [8]. The following holds for solutions of (2).Proposition 1.
Letx=x(t) be given by (8), where x(n)(n=0,1,…) satisfies (6).(1)
Ifa∗≠1 and x(i) is a stationary state of x(n), then either x=x(t)=0 for all t≥i or x=x(t)=b∗/(a∗-1) for all t≥i.(2)
Leta∗>1.
(a)
Ifx=x(t) is oscillatory, then x(n) is oscillatory with stationary state 0.(b)
Ifx(n) diverges to ∞, then limt→∞x(t)=∞.(c)
Ifx(n) diverges to -∞, then limt→∞x(t)=-∞.Proof.
(1) Leta∗≠1 and let x(i)=x(i+1). Suppose first that x(i)≤λ. Then x(i+1)=a∗x(i)=x(i) so x(i)=x(i+1)=0 since a∗≠1. By induction, x(i+j)=0<λ and F(x(i+j))=0 for all integers j≥0; and, by (8), x(t)=0 for all t≥i.
On the other hand, assumex(i)>λ. Then x(i+1)=a∗x(i)-b∗=x(i) so x(i)=x(i+1)=b∗/(a∗-1)>λ since a∗≠1. Thus x(i+2)=a∗x(i+1)-b∗=b∗/(a∗-1)>λ and, by induction, x(i+j)=b∗/(a∗-1) for all integers j≥0. Since b∗/(a∗-1)=-C/(A+B) and F(b∗/(a∗-1))=1, it follows from (8) that x(t)=b∗/(a∗-1) for all t≥i.
(2) Leta∗>1. Then B>-A and in (8) we have(13)eAt-t+B∫tteAt-sds>1.
(a) Suppose by way of contradiction thatx=x(t) is oscillatory but 0 is not a stationary state of x(n). It follows that x(n-1)≠0 for all positive integers n.
Assume first thatx(n-1)<0 for some integer n≥1. Then F(x(n-1))=0 and x(n)=a∗x(n-1)<0<λ. By induction, x([t])<0 and F(x([t]))=0 for all t≥n-1.
Letn-1≤[t]<t<[t]+1. By (8) and (13), x(t)<0 for all t≥n-1; and thus x=x(t) is not oscillatory in this case, a contradiction.
Therefore, assume thatx(n-1)>0 for all positive integers n. Let n be a positive integer. We show that x(t)>0 for all t in the interval (n-1,n): note that x′(t)=x′(n-1+)eA(t-[t]).
Suppose thatx′(n-1+)=0. Then x′(t)=0 and(i)
ifA≠0, then, by (2),(14)xt=-Bxn-1+CFxn-1A=xn-1>0for allt in (n-1,n);(ii)
ifA=0, then, by (2), Bx(n-1)+CF(x(n-1))=0 and, by (8),(15)xt=1+Bt-txn-1+Ct-tFxn-1=xn-1>0for allt in (n-1,n).
Ifx′(n-1+)>0, then x(t) is strictly increasing on (n-1,n) and x(n-1)>0, so x(t)>0 for all t in (n-1,n) by continuity.
Ifx′(n-1+)<0, then x(t) is strictly decreasing on (n-1,n), but x(t) is still positive on (n-1,n) by continuity since x(n-1) and x(n) are both positive.
Therefore,x(t)>0 for all t in [n-1,n] (and hence for all t). Thus x=x(t) is not oscillatory in this case, contrary to our assumption.
(b) Suppose thatlimn→∞x(n)=∞. There exists N such that x(n)>λ for all integers n≥N. Let [t]≥N. Then F(x([t]))=1 and, by (8) and (13),(16)xt≥xt+C∫tteAt-sds.
Thus, ifC≥0, then x(t)≥x([t]). And if C<0, then x(t)≥x([t])-b∗ since(17)0≤eAt-t-1A≤eA-1Awhen A≠0 and t-[t]<1 otherwise. Therefore, limt→∞x(t)=∞.
(c) Suppose thatlimn→∞x(n)=-∞. There exists N such that x(n)≤0 for all integers n≥N. Let [t]≥N. Then F(x([t]))=0 and, by (8) and (13), x(t)≤x([t]). Hence, limt→∞x(t)=-∞.Our main results characterize eventually periodic solutionsx=x(t) of (2). Since x(t) is generally not differentiable at integers t=n but is always differentiable between integers, we restrict our attention to integral periods (see [3, 6]). In this case, by (8), we have the following:x=x(t)is eventually periodic with positive integral periodpif and only ifx(n)(n=0,1,…)is eventually periodic with periodp.Equation (2) is similar to recent models related to neural networks ([8–12]). We treat a generalized version of (6) as follows.For real numbersa, b, x0, and λ>0, define, for n≥1,(18)xn=axn-1-bFxn-1,where λ and F are given by (3).Remark 2.
Equation (18) is the difference equation (6) of a differential equation (2) with x(0)≡x0, λ≡λ, and A being arbitrary: if A=0, choose B≡a-1 and C≡-b. And if A≠0, then let B≡(a-eA)A/(eA-1) and C≡-bA/(eA-1). The resulting solution x=x(t) given by (8) satisfies (2) with a∗≡a and b∗≡b.The next result shows that we may henceforth assume thata>1 and b>0 in (18).Proposition 3.
Letxn be defined by (18) for real a, b, x0, and λ>0, and let x=x(t) satisfy (8), where a∗ and b∗ are given in (6). In the following, j is an arbitrary positive integer.
(1) (a) Suppose that0≤a≤1. If xk≤λ, for some k, then xk+j=ajxk. On the other hand, if xk>λ, for all k, then(19)xj=ajx0-ba-1+ba-1,ifa≠1x0-jb,ifa=1b≤0.
(b) In particular, assume that0≤a∗≤1. If x(k)≤λ, for some nonnegative integer k, then, for all t′ in [k,k+1), one has that x(t′+j)=a∗jx(t′). On the other hand, if x(k)>λ, for all nonnegative integers k, then, for all t′ in [0,1),(20)xt′+j=a∗jxt′-b∗a∗-1+b∗a∗-1,ifa∗≠1xt′-jb∗,ifa∗=1b∗≤0.
(2) (a) Suppose thata>1 and b≤0. If x0≤0, then xj=ajx0. If x0>0, then there exists k≥1 such that xk+j=aj(xk-b/(a-1))+b/(a-1).
(b) In particular, assume thata∗>1 and b∗≤0. If x(0)≤0, then x(t′+j)=a∗jx(t′) for all t′ in [0,1). If x(0)>0, then there exists a positive integer k such that x(t′+j)=a∗j(x(t′)-b∗/(a∗-1))+b∗/(a∗-1) for all t′ in [k,k+1).
(3) (a) Suppose thata<0. Either xn is oscillatory or there exists k≥0 such that xk+j=aj(xk-b/(a-1))+b/(a-1).
(b) In particular, assume thata∗<0. Either x=x(t) is oscillatory or there exists an integer k≥0 such that x(t′+j)=a∗j(x(t′)-b∗/(a∗-1))+b∗/(a∗-1) for all t′ in [k,k+1).Proof.
(1a) Suppose that0≤a≤1. If xk≤λ, for some k, then either 0≤xk+1=axk≤xk≤λ or xk+1=axk≤0<λ. Hence, the desired form of xk+j follows by induction in this case.
Assume thatxk>λ for all k. Thus, if a≠1, then(21)xj=axj-1-b=a2xj-2-a2-1a-1b=⋯=ajx0-ba-1+ba-1.And if a=1, then xj=xj-1-b=xj-2-2b=⋯=x0-jb. (Since xj>λ>0, for all j, it follows that b≤0 in this case.)
(1b) Suppose that0≤a∗≤1. Let x(k)≤λ for some integer k≥0. By (a), x(k+j)=a∗jx(k)≤λ and F(x(k+j))=0. Let t′ be in [k,k+1). Then k+j≤t′+j<k+j+1 and, by (8), (22)xt′+j=eAt′+j-t′+j+B∫t′+jt′+jeAt′+j-sdsxk+j=a∗jeAt′-t′+B∫t′t′eAt′-sdsxk=a∗jxt′.
Next, assume thatx(k)>λ and thus F(x(k))=1 for all integers k≥0. Suppose first that a∗≠1 and 0≤t′<1. By (a), x(j)=a∗j(x(0)-b∗/(a∗-1))+b∗/(a∗-1). Since j≤t′+j<j+1 and b∗/(a∗-1)=-C/(A+B), (23)xt′+j=a∗jeAt′-t′+B∫t′t′eAt′-sds·x0-b∗a∗-1+b∗a∗-1as in the proof of Proposition 1(1). Therefore, x(t′+j)=a∗j(x(t′)-b∗/(a∗-1))+b∗/(a∗-1).
Finally suppose thata∗=1. Thus B=-A and, by (a), x(j)=x(0)-jb∗>λ and b∗≤0. As above, for 0≤t′<1, we have that j≤t′+j<j+1 and(24)xt′+j=xj+C∫t′t′eAt′-sds=xt′-jb∗.
(2a) Suppose thata>1 and b≤0. If x0≤0 (<λ), then clearly xj=ajx0. Assume that x0>0. There exists a positive integer k such that xk-1=ak-1x0≤λ<xk=akx0. Thus, xk+1=ak+1x0-b=axk-b>xk>λ since a>1, xk>0, and b≤0. As in the proof of (1a), xk+j=ajxk-((aj-1)/(a-1))b (>λ).
(2b) Suppose thata∗>1 and b∗≤0. If x(0)≤0, then x(j)=a∗jx(0)<λ by (2a); and hence the result follows as in the proof of (1b).
Assume thatx(0)>0. By (2a), there exists a positive integer k such that x(k)>λ and x(k+j)=a∗j(x(k)-b∗/(a∗-1))+b∗/(a∗-1)>λ. Let k≤t′<k+1 so that k+j≤t′+j<k+j+1. Then, by (8), (25)xt′+j=a∗jeAt′-t′+B∫t′t′eAt′-sds·xk-b∗a∗-1+b∗a∗-1=a∗jxt′-b∗a∗-1+b∗a∗-1as in the proof of (1b).
(3a) Suppose thata<0. One of the following must be true:(i)
For every integerk≥0, there exists an integer nk≥k such that xnk≤λ.(ii)
There exists an integerk≥0 such that xn>λ for all n≥k.
Assume that (i) holds. Ifxj=0, for some j, then xn is oscillatory with stationary state 0; thus we further assume that xn has no zero terms. Therefore, there exists a subsequence of xn with alternating signs.
We may choosexn0>0 as follows. If x0>0, then let xn0≡x0. If x0<0 (<λ), then let xn0≡x1=ax0>0.
Next choosexn1<0, where n1>n0: by (i), there exists n1′≥n0 such that xn1′≤λ. If xn1′<0, then n1′>n0 so let n1≡n1′. If xn1′>0, then xn1′+1=axn1′<0 and n1′+1>n0; thus, in this case, let n1≡n1′+1.
Sincexn1<0<λ, we have xn2≡xn1+1=axn1>0 and n2>n1. By induction, a sequence xnk is constructed such that nk+1>nk and xnkxnk+1<0 for all k. Thus xn is oscillatory.
On the other hand, if (ii) holds, then, as in the proof of (1a), there existsk≥0 such that xk+j=aj(xk-b/(a-1))+b/(a-1).
(3b) Suppose thata∗<0. By (3a), either x(n) is oscillatory (and hence x=x(t) is oscillatory) or x(k+j)=a∗j(x(k)-b∗/(a∗-1))+b∗/(a∗-1)>λ (with x(k)>λ) for some integer k≥0. The desired result follows in the latter case as in the proof of (2b).Thus we assume thata∗>1 and b∗>0; and therefore Proposition 1 applies to the resulting solution x=x(t) of (2). It will follow from the third section that if λ is outside the interval (b∗/a∗2,b∗/(a∗-1)-b∗/a∗2), then x(t) is either eventually constant or unbounded. Moreover, if a∗≥2 and λ is in [b∗/(a∗(a∗-1)),b∗/a∗], then, by the fourth section, all initial values x(0) are derived such that x(t) is eventually periodic; and more generally, for any λ>0, the eventually periodic solutions of (2) are the bounded solutions with these initializations x(0).Remark 4.
For integersa, b, x0, and λ>0, periodic solutions of difference equation (18) were used in [13] to determine the real eigenvalues of certain arbitrarily large, sparse matrices.


## 2. Unbounded Solutions

Letxn satisfy (18), where a>1 and b>0. If xn is unbounded, then xn is eventually geometric: we define(26)f-i,j≡ajxiwheneverxi≤0,f+i,j≡ajxi-ba-1+ba-1whenxi≥ba-1.The following result shows that we may assume(27)0<x0<ba-1,λ<ba-1since otherwise there exists i≥0 such that xi+j=f±(i,j) for all j.Lemma 5.
Assume thatxn is defined by (18), where a>1 and b>0.(1)
Ifxn≤λ, for all n≥i≥0, then xi+j=f-(i,j) for all j≥0. In particular, if xi≤0, for some i, then xi+j=f-(i,j) for all j.(2)
Ifxn>λ, for all n≥i≥0, then xi+j=f+(i,j) for all j≥0. In particular, if xi≥b/(a-1)>λ, for some i, then xi+j=f+(i,j) for all j.(3)
Ifxi>λ≥b/(a-1), for some i, then xi+j=f+(i,j) for all j≥0. In particular, if either 0<xi≤b/(a-1)≤λ or b/(a-1)≤xi≤λ, for some i, then there exists k>i such that xk+j=f+(k,j) for all j.Proof.
(1) Assume thatxn≤λ for all n≥i. Then xi+j=ajxi for all j≥0. If xi>0, then, since a>1, it follows that xn is not bounded above which contradicts our hypothesis. Thus xi≤0 and xi+j=f-(i,j).
Ifxi≤0 (<λ), for some i, then clearly xn≤λ for all n≥i.
(2) Suppose thatxn>λ for all n≥i. Then(28)xi+1=axi-b=axi-ba-1+ba-1,xi+2=axi+1-b=a2xi-ba-1+ba-1and, by induction, for all j,(29)xi+j=ajxi-ba-1+ba-1.If xi<b/(a-1), then, since a>1, xn is not bounded below which is contrary to our hypothesis. Thus xi≥b/(a-1) and xi+j=f+(i,j) for all j.
Assume thatxi≥b/(a-1)>λ for some i. Then xi=f+(i,0)>λ since xi≥b/(a-1), and(30)xi+1=axi-b≥aba-1-b=ba-1>λ,where xi+1=f+(i,1) as above.
Similarly, by induction,(31)xi+j=axi+j-1-b≥ba-1>λand xi+j=f+(i,j) for all j.
(3) Assume thatxi>λ≥b/(a-1) for some i. Then xi=f+(i,0)>λ and (32)xi+1=axi-b>aλ-a-1λ=λ,where xi+1=f+(i,1). By induction, xi+j=axi+j-1-b>λ and xi+j=f+(i,j) for all j.
Suppose that0<xi≤b/(a-1)≤λ for some i. There exists k>i such that(33)xk-1=ak-1-ixi≤λ<xk=ak-ixi.Thus, by the general case, xk+j=f+(k,j) for all j.
Similarly, ifb/(a-1)≤xi≤λ, then there is k>i such that xk+j=f+(k,j) for all j.The next result is central to our analysis.Lemma 6.
Letxn be given by (18), where a>1, b>0, and 0<λ,x0<b/(a-1).(1)
Ifλ≤b/a(a-1), then xn≤b/(a-1) for all n. In particular, if λ<b/a(a-1), then xn<b/(a-1) for all n.(2)
Ifb/a(a-1)≤x0≤λ, then x1+j=f+(1,j) for all j≥0. In this case, x0=b/a(a-1) if and only if x1=b/(a-1).(3)
Ifλ≥b/a, then xn>0 for all n.(4)
Ifb/a≥x0>λ, then x1+j=f-(1,j) for all j≥0. In this case, x0=b/a if and only if x1=0.Proof.
(1) Assumeλ≤b/a(a-1) and x0<b/(a-1).
Ifx0≤λ, then x1=ax0≤aλ≤b/(a-1). And if x0>λ, then x1=ax0-b<a(b/(a-1))-b=b/(a-1). Thus x1≤b/(a-1) in both cases, and, similarly, by induction, xn≤b/(a-1) for all n.
Ifλ<b/a(a-1), then xn<b/(a-1) for all n by the above argument.
(2) Suppose thatb/a(a-1)≤x0≤λ. Then x1=ax0≥b/(a-1)>λ and x1+j=f+(1,j) for all j by Lemma 5(2).
Ifb/a(a-1)=x0≤λ, then x1=ax0=b/(a-1). Conversely, if b/a(a-1)≤x0≤λ and x1=b/(a-1), then x1=ax0 so x0=b/a(a-1).
(3) Assumeλ≥b/a. If x0≤λ, then x1=ax0>0. And if x0>λ, then x1=ax0-b>aλ-b≥0. Thus x1>0 and, by induction, xn>0 for all n.
(4) Suppose thatb/a≥x0>λ. Then x1=ax0-b≤a(b/a)-b=0 and x1+j=f-(1,j) for all j by Lemma 5(1).
Clearly,x0=b/a if and only if x1=0 in this case.Remark 7.
Letxn be defined as in Lemma 6. As in the proof of Lemma 5(3), if x0≤λ, then there is a unique positive integer i such that ai-1x0≤λ<aix0. In this case, xk=akx0 for k=1,…,i; and if xi≤b/a, then xi+1+j=ajxi+1 for all j by Lemma 6(4) since xi>λ.
Similarly, ifx0>λ, then there is a unique positive integer i such that(34)0<ai-1ba-1-x0<ba-1-λ≤aiba-1-x0.In this case, writing x0=b/(a-1)-(b/(a-1)-x0)>λ, we have that(35)xk=axk-1-b=ba-1-akba-1-x0>λfor k=1,…,i; and if xi≥b/a(a-1), then(36)xi+1+j=ba-1-ajba-1-xi+1for all j by Lemma 6(2) since xi≤λ.Example 8.
Leta≥2 and let b>0 and assume that b/a(a-1)≤λ≤b/a. Let x0 be the weighted average(37)x0≡2a-1b/a2+b/aa-12a.Then x0<λ and(38)x1=ax0=2a-1a-1+ab2a2a-1>ba≥λ.Thus(39)x2=ax1-b=b2aa-1<baa-1≤λand x3=ax2=b/2(a-1) is the midpoint of [b/a(a-1),b/a].
Ifx3≤λ, then x4+j=f+(4,j) for all j by Lemma 6(2). And if x3>λ, then x4+j=f-(4,j) for all j by Lemma 6(4).


## 3. Bounded Solutions

By Lemma5, solutions xn of (18) such that a>1, b>0, and 0<λ,x0<b/(a-1) are bounded only when 0≤xn≤b/(a-1) for all n. In particular, by Lemma 6, if a≤2 and b/a≤λ≤b/a(a-1), then xn is bounded. It is possible that xn is bounded but not eventually periodic.Example 9.
Assume that1<a=p/q<2, where p and q are odd and even integers, respectively; b>0 and b/a≤λ≤b/a(a-1). If x0≡(b/(a-1))(r/s), where r and s are odd and even integers, respectively, such that r<s, then the solution xn is bounded but not eventually periodic: by Lemma 6, 0<xn≤b/(a-1) for all n. Note that(40)ax0=ba-1prqs,ax0-b=ba-1pr-p-qsqsso that x1=(b/(a-1))(p1/qs), where p1 is odd and qs is even.
Similarly, by induction, for alln≥1, xn=(b/(a-1))(pn/qns), where pn is odd and qns is even. It follows that xn is not eventually periodic since if xm=xm+n for some m≥0 and n≥1, then pm+n=qnpm is even.We now classify the types of solutions that may be bounded. Our results will be stated in terms of the decomposition of(0,b/(a-1)) into the disjoint union of the intervals I1=(0,b/a2],  I2=(b/a2,min⁡{b/a,b/a(a-1)}),  I3=[min⁡b/a,b/a(a-1),max⁡{b/a,b/a(a-1)}],  I4=(max⁡{b/a,b/a(a-1)},b/(a-1)-b/a2), and I5=[b/(a-1)-b/a2,b/(a-1)).Definition 10.
Leta>1, b>0, and 0<λ,x0<b/(a-1). A solution xn defined by (18) is of(i)
type A ifλ and xk are in I2 for some k such that xk≤λ,(ii)
type B ifλ and xk are in I4 for some k such that xk>λ,(iii)
type C ifλ is in I3 and either a≥2 such that xk is in I2∪I4 for some k or a<2.The unbounded solution in Example8 is of type C.Letxn satisfy (18) as in Definition 10. Since λ<b/(a-1), we have that c is a stationary state of xn if and only if xi=c=0 or xi=c=b/(a-1) for some i, in which case xi+j=c for all j≥0: if xi=0 (<λ) or xi=b/(a-1) (>λ), then clearly xi+1=xi=c. Conversely, suppose that xi+1=xi=c for some i. If c≤λ, then c=xi+1=axi=ac so c=0 since a>1. And if c>λ, then c=xi+1=axi-b=ac-b so c=b/(a-1).The solutionxn iseventually periodic if there are integers m≥0 and p≥1 such that xm=xm+p (and thus, by (18), xm+j=xm+p+j for all j≥0). The following eventually periodic solution is either of type A or type C.Example 11.
Leta>1 and let b>0 and let p≥2 be an integer such that ap≥a/(a-1). (If a≥2, then p≥2 is arbitrary.) Choose λ such that(41)ba2≤b/a21-1/ap=bap-2ap-1≤λ≤baand, for any integer m≥0, let(42)x0≡bamap-1.Then xm=xm+p: x0≤bap-2/(ap-1)≤λ so x1=b/(am-1ap-1). Similarly, xm=amx0=b/(ap-1)  ≤  bap-2 /  (ap-1)≤λ and, for k=1,…,p-2,xm+k=bak  /  (ap-1)≤bap-2  /  (ap-1)≤λ. Thus xm+(p-1)=bap-1/(ap-1)>b/a≥λ and(43)xm+p=axm+p-1-b=bap-1=xm.
Note that ifa≥2, then, since b/a2≤xm+p-2≤b/a(a-1)≤b/a, it follows that xn is of type A whenever λ is in I2 (and xm+p-2≤λ) and is of type C when λ is in I3. If a<2, then xn is of type A since xm+p-2 and λ are in I2 and xm+p-2≤λ. The following slight modification is eventually periodic of type C when a<2.Example 12.
Let1<a<2 and let b>0 and let p≥2 be an integer such that(44)ap>max⁡aa-1,12-a,and let λ in I3 satisfy(45)ba2<bap-2ap-1<ba<λ<bap-1ap-1<baa-1.Define (46)x0≡baqamap-1for integers m≥0 and q such that 0≤q≤p-2. Then x0≤bap-2/(ap-1)<λ and, as in Example 11,(47)xm=baqap-1≤bap-2ap-1<λ,xm+p-q-2=ap-q-2xm=bap-2ap-1<λ.Therefore, xm+p-q-1=bap-1/(ap-1)>λ, (48)xm+p-q=axm+p-q-1-b=bap-1<λ,xm+p=aqxm+p-q=baqap-1=xm.Note that, for an elementx0 in I1, there exists an integer k≥2 such that(49)bak+1<x0≤bak.Similarly, for x0 in I5, there exists an integer k≥2 such that(50)ba-1-bak≤x0<ba-1-bak+1.Inequalities (49) and (50) will be used repeatedly in the next result.In seeking bounded solutions without stationary states as in Examples11 and 12, we may further assume that λ is in I2∪I3∪I4 by the next result.Theorem 13.
Letxn be a solution of (18) such that a>1, b>0, and 0<λ,x0<b/(a-1). Then x0 fits one and only one of the following cases.(1)
Letλ be in I1.
(a)
Ifx0≤λ and k≥2 is given by (49), then xi+j=f-(i,j) for all j≥0, where i is the first integer in [2,k] such that ai-1x0>λ; that is, xi-1>λ. In this case, if λ=b/ak, then i=2; and x2<0 when k>2.(b)
Ifλ<x0≤b/a, then x1+j=f-(1,j) for all j≥0.(2)
Letλ be in I2.
(a)
Ifx0 is in I1 and k≥2 is given by (49), then xk-1=ak-1x0 and one has the following.
(i)
Ifb/a2<xk-1≤λ, then xn is of type A.(ii)
Ifλ<xk-1≤b/a, then xk+j=f-(k,j) for all j.(b)
(i)
Ifb/a2<x0≤λ, then xn is of type A.(ii)
Ifλ<x0≤b/a, then x1+j=f-(1,j) for all j.(3)
Letλ be in I1∪I2.
(a)
Letx0≠b/a be in I3 and let ϕ=(1+5)/2 be the golden ratio with the property that ϕ(ϕ-1)=1.
(i)
Ifϕ≤a<2, then either (1) or (2) holds for the sequence xn(n≥1) with initial term x1=ax0-b.(ii)
Ifa<ϕ and r is the least positive integer such that ar(a-1)≥1, then there exists an integer s in [1,r] such that xs=b/(a-1)-as(b/(a-1)-x0) is in I1∪I2∪{b/a}; and therefore (1) or (2) applies to xn(n≥s).(b)
Letx0 be in I4.
(i)
Ifa≥2, then x1=ax0-b is in I1∪I2 and xn(n≥1) satisfies either (1) or (2).(ii)
Ifa<2, then x1=ax0-b≠b/a(a-1) is in I1∪I2∪I3 and xn(n≥1) satisfies (1), (2), or (3a).(c)
Letx0 be in I5 and let k≥2 be given by (50). Then xk=b/(a-1)-ak(b/(a-1)-x0) and one has the following.
(i)
Ifa≥2, then either xk≤0 (and xk+j=f-(k,j), for all j), or xk is in I1∪I2 and xn(n≥k) satisfies (1) or (2).(ii)
Ifa<2, then xk≠b/a(a-1) is in I1∪I2∪I3 and xn(n≥k) satisfies (1), (2), or (3a).(4)
Letλ be in I3. By Definition 10, if either a<2 or a≥2 and I2∪I4 contains a term of the sequence xn, then xn is of type C. Therefore, suppose that a≥2 and x0 is in I1∪I3∪I5.
(a)
Letx0 be in I1 and let k≥2 be given by (49). Then xk-1=ak-1x0 is in I2∪I3. If xk-1 is in I2, then xn is of type C. Let xk-1 be in I3. Then
(i)
xk+j=f+(k,j) for all j≥0 whenever b/a(a-1)≤xk-1≤λ;(ii)
xk+j=f-(k,j) for all j≥0 when λ<xk-1≤b/a.(b)
Letx0 be in I3.
(i)
Ifx0≤λ, then x1+j=f+(1,j) for all j.(ii)
Ifx0>λ, then x1+j=f-(1,j) for all j.(c)
Letx0 be in I5 and let k≥2 be given by (50). Then xk-1=b/(a-1)-ak-1(b/(a-1)-x0) is in I3∪I4. Thus, if xk-1 is in I4, then xn is of type C. Let xk-1 be in I3. Then
(i)
xk+j=f+(k,j) for all j≥0 whenever b/a(a-1)≤xk-1≤λ;(ii)
xk+j=f-(k,j) for all j≥0 when λ<xk-1≤b/a.(5)
Letλ be in I4.
(a)
Ifx0 is in I5 and k≥2 is given by (50), then xk-1=b/(a-1)-ak-1(b/(a-1)-x0) and one has the following.
(i)
Ifb/a(a-1)≤xk-1≤λ, then xk+j=f+(k,j) for all j.(ii)
Ifλ<xk-1<b/(a-1)-b/a2, then xn is of type B.(b)
(i)
Ifb/a(a-1)≤x0≤λ, then x1+j=f+(1,j) for all j.(ii)
Ifλ<x0<b/(a-1)-b/a2, then xn is of type B.(6)
Letλ be in I5.
(a)
Ifx0>λ and k≥2 is given by (50), then xi+j=f+(i,j) for all j≥0, where i is the first integer in [2,k] such that ai-1(b/(a-1)-x0)≥b/(a-1)-λ; that is, xi-1≤λ. In this case, if λ=b/(a-1)-b/ak, then i=2 and x2>b/(a-1).(b)
Ifb/a(a-1)≤x0≤λ, then x1+j=f+(1,j) for all j≥0.(7)
Letλ be in I4∪I5.
(a)
Letx0≠b/a(a-1) be in I3.
(i)
Ifϕ≤a<2, then either (5) or (6) holds for the sequence xn(n≥1) with initial term x1=ax0.(ii)
Ifa<ϕ and r is the least positive integer such that ar(a-1)≥1, then there exists an integer s in [1,r] such that xs=asx0 is in {b/a(a-1)}∪I4∪I5 and thus (5) or (6) applies to xn(n≥s).(b)
Letx0 be in I2.
(i)
Ifa≥2, then x1=ax0 is in I4∪I5 and xn(n≥1) satisfies either (5) or (6).(ii)
Ifa<2, then x1=ax0≠b/a is in I3∪I4∪I5 and xn(n≥1) satisfies (5), (6), or (7a).(c)
Letx0 be in I1 and let k≥2 be given by (49). Then xk=akx0 and one has the following.
(i)
Ifa≥2, then either xk≥b/(a-1) (hence xk+j=f+(k,j) for all j) or xk is in I4∪I5 and xn(n≥k) satisfies (5) or (6).(ii)
Ifa<2, then xk≠b/a is in I3∪I4∪I5 and xn(n≥k) satisfies (5), (6), or (7a).Proof.
(1a) Assume0<x0≤λ≤b/a2. By (49), there exists an integer k≥2 such that(51)bak+1<x0≤bak≤ba2,λ≤ba2<ak-1x0.Therefore, as in Remark 7, there is a smallest integer i in [2,k] such that(52)xi-2=ai-2x0≤λ<xi-1=ai-1x0.It follows that(53)xi=axi-1-b=aix0-b≤akx0-b≤b-b=0and xi+j=f-(i,j) for all j by Lemma 5(1).
Suppose thatλ=b/ak. Then λ/a<x0≤λ and λ<x1=ax0 so i=2. And if k>2, then (54)x2=ax1-b=a2x0-b<0sincex0≤bak<ba2.
(1b) Lemma6(4).
(2) Letλ be in I2. By (49), if x0 is in I1, then(55)bak+1<x0≤bak≤ba2<λ.Therefore, b/a2<xk-1=ak-1x0 and (2) follows from Definition 10 and Lemma 6(4).
Note that, for anyλ in I1∪I2, (1) and (2) cover the following cases:(i)
x0 is in I1∪I2∪I3 whenever a≥2.(ii)
x0 is in I1∪I2∪{b/a} when a<2.
(3) Letλ be in I1∪I2.
(a) Suppose thatx0≠b/a is in I3 and a<2. Then λ<b/a<x0≤b/a(a-1) and(56)0=aba-b<x1=ax0-b≤ba-1-b<baa-1.(i)
Suppose thatϕ≤a<2. Then a(a-1)≥1 so b/(a-1)-b≤b/a since b/a(a-1)≤b. Thus, by (56), x1 is in I1∪I2∪{b/a} and (1) or (2) applies to xn (n≥1).(ii)
On the other hand, assume thata<ϕ so that a(a-1)<1. By (56), x1 is in I1∪I2∪I3. If x1 is in I1∪I2∪{b/a}, then (1) or (2) applies to xn (n≥1). And if x1≠b/a is in I3, then, since x1≤b/(a-1)-b by (56), the above argument shows(57)0<x2=ax1-b≤ba-1-ab<baa-1andx2 is in I1∪I2∪I3.
Note that, in the latter case, ifx2≠b/a is in I3, then λ<b/a<x2≤b/(a-1)-ab so a2(a-1)<1.
It follows by induction that, for everyj≥1, either xj is in I1∪I2∪{b/a} or xj≠b/a is in I3 and aj(a-1)<1. Since a>1, there are least positive integers r and s≤r such that ar(a-1)≥1 and xs is in I1∪I2∪{b/a}. Since xi>λ, for i=0,…,s-1, the form of xs follows from Remark 7.
(b) Letx0 be in I4.(i)
Leta≥2. Then λ<b/a<x0<b/(a-1)-b/a2,(58)0=aba-b<x1=ax0-b<aba-1-ba2-b=baa-1,andx1 is in I1∪I2.(ii)
Leta<2. Then λ<b/a(a-1)<x0<b/(a-1)-b/a2,(59)0<abaa-1-b<x1=ax0-b<aba-1-ba2-b=baa-1,andx1≠b/a(a-1) is in I1∪I2∪I3.
(c) Letx0 be in I5 and let k≥2 be given by (50). Then(60)λ<ba-1-ba2≤ba-1-bak≤x0<ba-1-bak+1so that, for j=1,…,k-2,(61)λ<ba-1-ba2≤ba-1-bak-j≤xj=axj-1-b<ba-1-bak-j+1.Hence,(62)λ<baa-1=ba-1-ba≤xk-1=axk-2-b<ba-1-ba2,ba-1-b≤xk=axk-1-b<ba-1-ba=baa-1,where b/(a-1)-b>0 when a<2. Therefore, (3c) follows as above.
Moreover, by Remark7, xk=b/(a-1)-ak(b/(a-1)-x0).
(4) Assume thatλ is in I3.
(a) Letx0 be in I1 and a≥2. Then b/a(a-1)≤λ≤b/a and, as in the proof of (2), if k is given by (49), then b/a2<xk-1=ak-1x0≤b/a. Thus (4a) follows from Definition 10 and Lemma 6((2), (4)).
(b) Letx0 be in I3 and a≥2 so that b/a(a-1)≤x0≤b/a. Hence, (i) and (ii) follow from Lemma 6((2), (4)).
(c) Letx0 be in I5 and a≥2. As in the proof of (3c),(63)baa-1≤xk-1=ba-1-ak-1ba-1-x0<ba-1-ba2.Therefore, (4c) follows from Definition 10 and Lemma 6((2), (4)).
(5) Assume thatλ is in I4.
(a) Letx0 be in I5. As above, (63) holds for xk-1. Moreover,(64)max⁡ba,baa-1<λ<ba-1-ba2.Thus (5a) follows from Lemma 6(2) and Definition 10.
(b) A direct consequence of Lemma6(2) and Definition 10.
(6) Letλ be in I5.
(a) Assume thatb/(a-1)-b/a2≤λ<x0<b/(a-1) and let k≥2 be the integer satisfying (50):(65)ba-1-bak≤x0<ba-1-bak+1.Thus(66)ba-1-ak-1ba-1-x0<ba-1-ba2≤λand, by Remark 7, there is a smallest integer i in [2,k] such that xi-1≤λ and(67)xi-1=ba-1-ai-1ba-1-x0≤λ<xi-2=ba-1-ai-2ba-1-x0.It follows that(68)xi=axi-1=aba-1-aiba-1-x0≥ba-1if and only if ai(b/(a-1)-x0)≤b. But, by (50), ai(b/(a-1)-x0)≤b/ak-i≤b. Therefore, xi≥b/(a-1)>λ and xi+j=f+(i,j) for all j by Lemma 5(2).
Suppose thatλ=b/(a-1)-b/ak. Then (69)ba-1-ba2≤ba-1-bak=λ<x0<ba-1-bak+1,baa-1≤ba-1-bak-1<x1=ax0-b<ba-1-bak=λ.Thus i=2 and x2=ax1>a(b/a(a-1))=b/(a-1).
(b) Follows immediately from Lemma6(2).
Note that, for anyλ in I4∪I5, (5) and (6) cover the following cases:(i)
x0 is in I3∪I4∪I5 whenever a≥2.(ii)
x0 is in {b/a(a-1)}∪I4∪I5 when a<2.
(7) Letλ be in I4∪I5.
(a) Leta<2 and let x0≠b/a(a-1) be in I3. Then b/a≤x0<b/a(a-1)<λ and(70)ba<b=aba≤x1=ax0<abaa-1=ba-1.(i)
Assume thatϕ≤a<2. Then a(a-1)≥1 so b/a(a-1)≤b and therefore, by (70), x1 is in {b/a(a-1)}∪I4∪I5. Hence, (5) or (6) applies to xn (n≥1).(ii)
Suppose thata<ϕ so that a(a-1)<1. By (70), x1 is in I3∪I4∪I5. If x1 is in {b/a(a-1)}∪I4∪I5, then (5) or (6) applies to xn (n≥1).
Assumex1≠b/a(a-1) is in I3. Then, by (70), b≤x1<b/a(a-1)<λ and(71)ba<b<ab≤x2=ax1<ba-1.Hence, x2 is in I3∪I4∪I5. Note that if x2≠b/a(a-1) is in I3, then ab≤x2<b/a(a-1)<λ and a2(a-1)<1.
Continuing by induction, for everyj≥1, either xj is in {b/a(a-1)}∪I4∪I5 or xj≠b/a(a-1) is in I3 and aj(a-1)<1. Therefore, since a>1, if r is the least positive integer such that ar(a-1)≥1, then there exists an integer s in [1,r] such that xs=asx0 is in {b/a(a-1)}∪I4∪I5.
(b) Leta>1 and let x0 be in I2. Then b/a2<x0<b/a(a-1)<λ and b/a<x1=ax0<b/(a-1). Therefore, x1 is in I4∪I5 when a≥2; and x1≠b/a is in I3∪I4∪I5 otherwise.
(c) Letx0 be in I1 and let k≥2 be given by (49). Then, for j=0,1,…,k-2,(72)bak+1-j<xj=ajx0≤bak-j≤ba2<λ.Moreover, b/a2<xk-1=ak-1x0≤b/a<λ and(73)ba<xk=akx0≤b.(i)
Ifa≥2, then, by (73), either xk≥b/(a-1) or xk is in I4∪I5.(ii)
Ifa<2, then b<b/(a-1) and therefore, by (73), xk≠b/a is in I3∪I4∪I5.


## 4. Periodic Solutions

The following sets are basic components of any eventually periodic solution of the difference equation (18).Definition 14.
LetP-1(a)≡{0} and, for k≥0, let Pk(a) be the set of all polynomials in a with degree at most k and with all coefficients either 0 or 1.A consequence of the next result is that ifxm=xm+p, then there are 2m+p possibilities for x0.Lemma 15.
Letxn be defined by (18) with a>1, b>0, and 0<λ,x0<b/(a-1). If xm=xm+p, for some integers m≥0 and p≥1, then there exist polynomials pp-1(a) in Pp-1(a) and qm-1(a) in Pm-1(a) such that(74)x0=bpp-1a+ap-1qm-1aamap-1.Proof.
Note thatP0(a)={0,1} and, for k≥1,(75)Pka=∑j=0kp0k-jaaj:p0k-ja∈P0a∀j.We solve the equation xm=xm+p for general initial value x0.
The possible expressions forxm+1 may be written as follows:(76)xm+1=axm-p00ab,p00a∈P0a.Similarly,(77)xm+2=axm+1-p01ab=a2xm-p01a+ap00ab,where p0(1)(a)∈P0(a).
Continuing in this manner, we have(78)xm+p=apxm-pp-1ab,where pp-1(a)=∑j=0p-1‍p0(p-1-j)(a)aj is an arbitrary element of Pp-1(a).
In particular,(79)xm=amx0-qm-1ab,qm-1a∈Pm-1a.Therefore, xm=xm+p is equivalent to(80)xm=pp-1abap-1=amx0-qm-1aband thus the desired form (74) for x0 follows.Note that the initial value of the unbounded solution of Example8 is not of form (74).Our main results are converses of Lemma15. If x0 is defined by (74), then(81)0≤x0≤bap-1/a-1+ap-1am-1/a-1amap-1=ba-1.Furthermore, we have the following refinement.Lemma 16.
Leta>1 and let b>0 and suppose that x0 satisfies (74) for integers m≥0 and p≥1 and polynomials pp-1(a) in Pp-1(a) and qm-1(a) in Pm-1(a).
(1) If(82)deg⁡qm-1a<m-1whenm>1,qm-1a=0whenm=1,deg⁡pp-1a<p-1whenm=0,p>1,pp-1a=0whenm=0,p=1,then0≤x0≤b/a(a-1).
(2) If(83)deg⁡qm-1a=m-1whenm>1,qm-1a=1whenm=1,deg⁡pp-1a=p-1whenm=0,p>1,pp-1a=1whenm=0,p=1,thenb/a≤x0≤b/(a-1).
Moreover, the converses of (1) and (2) hold wheneverx0 is not in I3.Proof.
Suppose thatm>1 and deg⁡qm-1(a)<m-1. Then 0≤qm-1(a)≤(am-1-1)/(a-1) and 0≤pp-1(a)≤(ap-1)/(a-1) and thus, by (74),(84)0≤x0≤bap-1/a-1+ap-1am-1-1/a-1amap-1=ba-1.
Similarly, ifm>1 and deg⁡qm-1(a)=m-1, then qm-1(a)≥am-1 and, by (74) and (81),(85)ba-1≥x0≥bpp-1a+ap-1am-1amap-1≥ba.
Suppose thatm=1. Then Pm-1(a)={0,1}. If qm-1(a)=0, then, by (74),(86)0≤x0=baa-1pp-1aap-1/a-1≤baa-1.And if qm-1(a)=1, then(87)ba-1≥x0=bpp-1a+ap-1aap-1≥ba.
Assume next thatm=0 and p>1. Then Pm-1(a)={0}. If deg⁡pp-1(a)<p-1, then pp-1(a)≤(ap-1-1)/(a-1) and(88)0≤x0≤bpp-1aap-1≤ba-1ap-1-1ap-1<baa-1since a>1. And if deg⁡pp-1(a)=p-1, then pp-1(a)≥ap-1 and(89)ba-1≥x0=bpp-1aap-1≥bap-1ap-1>ba.
Finally assume thatm=0 and p=1. Then Pm-1(a)={0}. If pp-1(a)=0, then x0=0. And if pp-1(a)=1, then x0=b/(a-1).
Conversely, suppose that0≤x0≤b/a(a-1) and  m>1 but x0 is not in I3. Either deg⁡qm-1(a)<m-1 or deg⁡qm-1(a)=m-1. If deg⁡qm-1(a)=m-1, then b/a≤x0≤b/(a-1) by (2). Thus, in this case, if a<2, then x0 is in I3; and if a≥2, then a=2 and x0 is again in I3. Hence, deg⁡qm-1(a)<m-1.
The other converses follow similarly.The converses in Lemma16 may fail when x0 is in I3: in Example 12, we have a<2, xm+p-q-1=bap-1/(ap-1) is in I3, 0≤xm+p-q-1≤b/a(a-1), and xm+p-q-1 satisfies (74) with m=0 and p>1 but deg⁡pp-1(a)=p-1.Ifλ is in I1∪I5, then, by Theorem 13, xn either has a stationary state or is unbounded. Using Lemma 16, we may extend this result to the other cases of λ.Theorem 17.
Letxn be a solution of (18) such that a≥2, b>0, and x0 satisfies (74) for some integers m≥0 and p≥1 and some polynomials pp-1(a) in Pp-1(a) and qm-1(a) in Pm-1(a).(1)
Ifλ is in I2, then either xm=xm+p or there exists a positive integer i≤m+p such that xi+j=f-(i,j) for all j≥0.(2)
Letλ be in I3. If λ≠b/a or pp-1(a)≠0, then xm=xm+p. On the other hand, if λ=b/a and pp-1(a)=0, then either x0=0 or there exists a positive integer i≤m such that xi+j=f+(i,j) for all j≥0, where xi=b.(3)
Ifλ is in I4, then either xm=xm+p or there exists a positive integer i≤m+p such that xi+j=f+(i,j) for all j≥0.Proof.
(2) Letλ be in I3 and suppose that either λ≠b/a or pp-1(a)≠0. We first verify that, for m≥1,(90)xm=bpp-1aap-1.
Suppose thatqm-1(a)=0. Then(91)x0=ba-1pp-1aam1+a+⋯+ap-1≤ba-11am≤ba-11a≤λsince a≥2 and λ is in I3. Thus x1=ax0 and similarly xj=axj-1≤λ for j=1,…,m-1. Hence, xm=amx0 satisfies (90).
Next assumeqm-1(a)≠0 and m=1. Since a≥2 and λ is in I3, it follows that(92)x0=bpp-1a+ap-1aap-1≥ba≥λ.Additionally, since λ<b/a or pp-1(a)>0, at least one of the inequalities is strict and x0>λ. Thus x1=ax0-b satisfies (90).
Hence, assume thatqm-1(a)≠0 and m>1. There exists a nonempty subset Λm-1 of {0,1,…,m-1} such that(93)x0=bpp-1aamap-1+qm-1aam,where qm-1(a)=∑k∈Λm-1‍ak. Let i≡min⁡Λm-1 and j≡max⁡Λm-1=deg⁡qm-1(a). We will show that(94)xm-i=bpp-1aaiap-1and therefore (90) will hold as in the initial case qm-1(a)=0.
We begin by showing that, form>1,(95)x0>λiffj=m-1.
Letx0>λ. If a=2, then b/a(a-1)=λ=b/a<x0 and therefore x0 is not in I3. On the other hand, if a>2, then b/a(a-1)<b/a and, by Lemma 16, x0 is again not in I3.
Sinceb/a(a-1)≤λ≤b/a, x0>λ, and x0 is not in I3, it follows that b/a<x0≤b/(a-1) and, by Lemma 16, j=m-1.
Conversely, letj=m-1. Then qm-1(a)≥am-1 and(96)x0≥bpp-1aamap-1+am-1am≥ba≥λ.Since λ<b/a or pp-1(a)>0, at least one of the latter inequalities is strict and thus x0>λ. Hence, (95) holds.
Next considerx1, where qm-1(a)≠0 and x0 is given by (93). By (95),(97)x1=bpp-1aam-1ap-1+qm-2aam-1,where qm-2(a)=∑k∈Λm-2‍ak such that(98)Λm-2≡Λm-1,ifj<m-1Λm-1∖j,ifj=m-1,which is a subset of Λm-1∩{0,…,m-2}. In addition, since i≤j≤m-1, it follows that(99)qm-2a=0iff1=m-i,in which case (94) holds for x1.
Suppose that1<m-i. Then qm-2(a)≠0, i is in Λm-2, and x1 is of form (93) with m replaced by m-1. By (95),(100)x1>λiffmax⁡Λm-2=m-2.Moreover, as above, (94) holds for x2 if 2=m-i.
By induction, forr=0,1,…,m-i-1, we may assume(101)xr=bpp-1aam-rap-1+qm-r-1aam-r,where qm-r-1(a)=∑k∈Λm-r-1‍ak≠0 such that(102)Λm-r-1≡Λm-r,ifmax⁡Λm-r<m-rΛm-r∖max⁡Λm-r,ifmax⁡Λm-r=m-r,which is a subset of Λm-r∩{0,…,m-r-1}, and, by (95),(103)xr>λiffmax⁡Λm-r-1=m-r-1.
Moreover, afterr≡m-i steps, we conclude that qm-r-1(a)=0 and xm-i satisfies (94). Thus (90) follows as in the case qm-1(a)=0.
Finally, we verify thatxm=xm+p, where xm is given by (90). If p=1, then xm is 0 or b/(a-1); so xm=xm+p. Thus assume p>1. In this case, we prove that(104)xm≤λiffdeg⁡pp-1a<p-1.
Letxm≤λ. Suppose first that xm is in I3. Since a≥2, we have b/a(a-1)≤b/a; and, by Lemma 16, b/a(a-1)=xm=λ=b/a and a=2. Hence, by (90), pp-1(a)=(ap-1)/a, which is impossible since a=2 and pp-1(a) is a positive integer but ap-1 and a are relatively prime.
Thus assumexm is not in I3. Then xm is in I1∪I2 since λ is in I3 and xm≤λ. Hence, by Lemma 16, deg⁡pp-1(a)<p-1 since p>1.
Conversely, letdeg⁡pp-1(a)<p-1 (and p>1). Then 0≤xm≤b/a(a-1)≤λ by Lemma 16 and therefore xm≤λ. Thus (104) follows.
As in (93), by (90),(105)xm=b∑k∈Λp-1akap-1for some nonempty subset Λp-1 of {0,…,p-1}. By (104),(106)xm+1=b∑k∈Λp-1Tp-1akap-1,where Tp-1 is the mapping on the set {a0,a1,…,ap-1} defined by(107)Tp-1ak≡ak+1,ifk<p-1a0,ifk=p-1.
Similarly, forr=1,2,…,(108)xm+r=b∑k∈Λp-1Tp-1rakap-1and since Tp-1p(ak)=ak, for all k, it follows that xm=xm+p.
We now turn to the second part of (2). Assume thata≥2, λ=b/a, and x0=bqm-1(a)/am>0 and thus m≥1. We prove the following:(109)Ifx0=bQjaak≠0whereQjais inPjasuch that0≤j=deg⁡Qja<k≤m,then eitherx0=ba=λorx1=bQj′aak-1≠0whereQj′ais inPj′aand0≤j′=deg⁡Qj′a<k-1≤m.
Suppose thatx0=bQj(a)/ak≠0, where Qj(a) satisfies the above hypotheses. Consider two cases for j.
(i) Assumej≤k-2. Then x0=bQj(a)/ak<λ=b/a: Qj(a)≤1+a+⋯+aj=(aj+1-1)/(a-1)<ak-1 since(110)ak-1+aj+1-1≤ak-1+ak-1-1=2ak-1-1≤aak-1-1<ak.Hence, (109) holds since x1=ax0=bQj(a)/ak-1≠0, where 0≤j=deg⁡Qj(a)≤k-2<k-1≤m so let j′=j.
(ii) Suppose thatj=k-1. If Qj(a)=aj, then x0=b/a=λ so assume that Qj(a)>aj since j=deg⁡Qj(a). In this case, (109) also holds since x0=bQj(a)/ak>λ=b/a and hence x1=ax0-b=bQj′(a)/ak-1≠0, where Qj′(a)≡Qj(a)-aj and 0≤j′≡deg⁡Qj′(a)≤k-2<k-1≤m.
Thus, starting withx0=bqm-1(a)/am=bQj(a)/ak≠0, Qj(a)∈Pj(a), 0≤j=deg⁡Qj(a)<k=m, and, applying (109) to x1,x2,…, we conclude that xi-1=b/a=λ for some i in [1,m]. Therefore,(111)xi=axi-1=b≥ba-1>ba=λand xi+j=f+(i,j) for all j≥0 by Lemma 5(2).
(1) Letλ be in I2. The condition “xi+j=f-(i,j) for allj≥0” is equivalent to “xi≤0.” Thus we further assume that xi>0 for i=1,…,m+p and show that xm=xm+p following the outline of the proof of (2).
We first verify thatxm satisfies (90) for m≥1. Suppose that qm-1(a)=0. Then, since a≥2 and m≥1,(112)x0=ba-1pp-1aamap-1/a-1≤ba-11am≤ba-11a≤ba.If x0>λ, then x1≤0 by Lemma 6(4). Therefore, x0≤λ and(113)x1=ax0=bpp-1aam-1ap-1.Similarly, xj≤λ for j=1,…,m-1 and hence xm=amx0 satisfies (90).
Suppose next thatqm-1(a)≠0 and m=1. Since λ is in I2,(114)x0=bpp-1a+ap-1aap-1≥ba>λand x1=ax0-b satisfies (90).
Thus assume thatqm-1(a)≠0 and m>1. Then x0 is given by (93), where qm-1(a)=∑k∈Λm-1‍ak for some nonempty subset Λm-1 of {0,1,…,m-1}. We will show that (94) holds, where i≡min⁡Λm-1, and therefore (90) will hold as in the case qm-1(a)=0.
Letj≡max⁡Λm-1=deg⁡qm-1(a). We begin by showing (95) for m>1.
Letx0>λ. If x0 is in (λ,b/a(a-1))∪I3, then x1≤0 by Lemma 6(4). Since x1>0 by assumption and x0>λ, it follows that x0 is in I4∪I5; and therefore j=m-1 by Lemma 16 since m>1.
Conversely, letj=m-1. By Lemma 16,(115)λ<baa-1≤ba≤x0≤ba-1and x0>λ. Thus (95) holds.
Equation (94) and consequently (90) follow from (95) as in the proof of (2).
Finally, we verify thatxm=xm+p, where xm is given by (90). We may assume p>1 since otherwise xm is 0 or b/(a-1), and therefore xm=xm+p. We start by proving (104): since λ is in I2 and xm+1>0 by assumption, it follows by Lemma 6(4) that xm is not in (λ,b/a(a-1))∪I3.
Letxm≤λ. Then xm is in I1∪I2 so deg⁡pp-1(a)<p-1 by Lemma 16 since xm is not in I3 and p>1.
Conversely, assume thatdeg⁡pp-1(a)<p-1. By Lemma 16, 0≤xm≤b/a(a-1), but xm is not in (λ,b/a(a-1)]. Therefore xm≤λ and (104) holds.
By (90) and (104), the desired result xm=xm+p follows as in the proof of (2).
(3) Letλ be in I4. The condition “xi+j=f+(i,j) for all j≥0” is equivalent to “xi≥b/(a-1).” Thus we assume that xi<b/(a-1) for i=1,…,m+p and show that xm=xm+p following the outline of the proof of (2).
We first verify thatxm satisfies (90) for m≥1. Suppose that qm-1(a)=0. Then, since a≥2, m≥1, and λ is in I4,(116)x0=ba-1pp-1aamap-1/a-1≤ba-11am≤ba-11a≤ba<λ.Thus x1=ax0 and similarly xj≤λ for j=1,…,m-1. Hence, xm=amx0 satisfies (90).
Next, we assumeqm-1(a)≠0 and m=1. Then(117)x0=bpp-1a+ap-1aap-1≥ba≥baa-1.If x0≤λ, then x1≥b/(a-1) by Lemma 6(2), contrary to our hypothesis. Thus x0>λ and x1=ax0-b satisfies (90).
Therefore assume thatqm-1(a)≠0 and m>1. Then x0 is given by (93), where qm-1(a)=∑k∈Λm-1‍ak for some nonempty subset Λm-1 of {0,1,…,m-1}. We will show that (94) holds, where i≡min⁡Λm-1, and hence (90) will follow as in the case qm-1(a)=0.
Letj≡maxΛm-1=deg⁡qm-1(a). We begin by showing (95) for m>1: if x0 is in I3∪(b/a,λ), then x1≥b/(a-1) by Lemma 6(2). Thus x0 is not in I3∪(b/a,λ) by our hypothesis.
Letx0>λ. Then x0 is in I4∪I5,  m>1, and hence j=m-1 by Lemma 16 since x0 is not in I3.
Conversely, letj=m-1 (and m>1). By Lemma 16, b/a(a-1)≤b/a≤x0≤b/(a-1). Since x0 is not in (b/a,λ), we have that x0>λ; and (95) holds.
Equation (94) and consequently (90) follow from (95) as in the proof of (2).
Finally, we verify thatxm=xm+p, where xm is given by (90). We may assume p>1 since otherwise xm is 0 or b/(a-1), and therefore xm=xm+p. We start by proving (104).
Letxm≤λ. Then xm is not in I3∪(b/a,λ) by Lemma 6(2) since xm+1<b/(a-1). Thus xm is in I1∪I2 and therefore deg⁡pp-1(a)<p-1 by Lemma 16 since xm is not in I3 and p>1.
Conversely, ifdeg⁡pp-1(a)<p-1 (and p>1), then 0≤xm≤b/a(a-1)<λ by Lemma 16, and (104) follows.
Finally,xm=xm+p by (90) and (104) as in the proof of (2).The following two examples illustrate symmetry between type C solutions about the pointb/2(a-1) that will be generalized in the next section. In particular, type B solutions will be shown to be reflections of type A solutions.Example 18.
Leta≥2 and let b>0 and let λ be in I3. For positive integer k, consider the weighted average(118)x0≡b/a2+akb/aa-11+akin I2 (that converges to b/a(a-1) as k tends to infinity). Then(119)x0=bakak-1/a-1+a2k-1a2a2k-1so x2=x2+2k by Theorem 17(2).Example 19.
For integerk≥1, redefine x0 in Example 18 to be the weighted average(120)x0≡b/a-1-b/a2+akb/a1+akin I4 (that converges to b/a as k→∞). Then x2=x2+2k as above since(121)x0=bak-1/a-1+a2k-1aa2a2k-1in form (74). Note that, in this case, b/(a-1)-x0 is the initial value in Example 18.Theorem17 may be applied to cryptography.Example 20.
Arbitrary vectorsv=v0,…,vp-1 of zeros and ones may be encoded by selecting any real numbers a≥2, b>0, and λ in I3 and defining(122)x0≡b∑viaiap-1in form (74). By Theorem 17(2), the “cipher-set” {a,b,λ,x0(computed value)} may then be decoded for v as follows: if x0=0, then v is a zero vector of indeterminate length. If x0=b/(a-1), then v is a vector of all ones of indeterminate length.
Assumex0≠0,b/(a-1). Compute enough terms of the periodic sequence xn to determine its period p. Then p is the length of v and the proof of Theorem 17(2) shows that(123)vi=1iffxp-1-i>λi=0,…,p-1.The following eventually periodic solutionxn with x0 of form (74) is of type A.Example 21.
Leta≥2 and let b>0 and, for integers m≥1 and p>2, let x0 be the weighted average(124)x0≡bap-2/ap-1+am-1-1b/aa-1am-1in I2. Then x0 is of form (74) with pp-1(a)=ap-1 and qm-1(a)=(am-1-1)/(a-1). Choose λ in I2 such that x0≤λ. Then xm=bap-1/(ap-1): (125)x1=ax0=bap-1am-1ap-1+bam-1am-1-1a-1,so the claim is true for m=1.
Assumem>1. We prove by induction that(126)xi=bap-1am-iap-1+bam-iam-i-1a-1>ba>λfor i=1,…,m-1. By the form of x1 above,(127)x1>bam-1am-1-1a-1≥ba>λsince am-1≥a; so (126) holds for i=1.
Suppose that (126) is true for some i<m-1. Then (128)xi+1=axi-b=bap-1am-i+1ap-1+bam-i+1am-i+1-1a-1≥bam-i+1am-i+1-1a-1≥ba>λas in the case i=1 since m>i+1. Thus (126) holds for i=1,…,m-1.
In particular,xm=axm-1-b=bap-1/(ap-1)>b/a>λ.
Finallyxm+p=xm:   xm+1=axm-b=b/(ap-1)≤b/a2<λ since p>2, and, by inducton, xm+i=bai-1/(ap-1)≤b/a2<λ for i=1,…,p-2. Hence, xm+(p-1)=axm+(p-2)=bap-2/(ap-1)<x0≤λ by the form of x0 and the choice of λ. Therefore, xm+p=bap-1/(ap-1)=xm.Unfortunately, not all type A solutionsxn with x0 of form (74) are bounded.Example 22.
Leta≥2 and let b>0 and(129)ba2<x0≡ba2+1aa3-1<λ≡baa-1a2+1<baa-1.Then x0 is of form (74) with m=1, p=3, pp-1(a)=a2+1, and qm-1(a)=0. Moreover,(130)x1=ax0=ba2+1a3-1>ba>λ,x2=ax1-b=ba+1a3-1>λ,x3=ax2-b=ba3-1/a-1-a3a3-1<0since a≥2. Therefore, x3+j=f-(3,j) for all j≥0.Suppose thatx0 is defined by (74). By Theorems 13 and 17, if a≥2 and xn is bounded, then xn is eventually periodic. In fact, if a=2 and λ is in I3, then xm=xm+p by treating the cases pp-1(a)≠0 and pp-1(a)=0 separately. However, if a<2 and λ is in I3, then xn is bounded (Lemma 6) but may not be eventually periodic.Example 23.
Assume that1<a=r/s<ϕ, where r and s are odd and even integers, respectively; b>0 and x0≡ba/(a2-1). Then x0 is of form (74) with m=0, p=2, and pp-1(a)=a; and x0 is in I3 since a(a-1)<1. Choose λ in I3 such that x0≤λ. Then λ≠b/a and pp-1(a)≠0, but xn is not eventually periodic: with r0≡r and r1≡r2, we have as in Example 9, for n≥0,(131)xn=ba-1rnsnr+s,where rn is odd. If xn is eventually periodic, then xu=xu+v for some u≥0 and v≥1; and therefore ru+v=svru is both even and odd.


## 5. Symmetric Solutions

Theorems13 and 17 indicate symmetry about the midpoint of (0,b/(a-1)) between pairs of solutions of (18). Moreover, if x0 satisfies (74), then so does b/(a-1)-x0: if(132)x0=bpp-1a+ap-1qm-1aamap-1,then(133)ba-1-x0=bap-1/a-1-pp-1a+ap-1am-1/a-1-qm-1aamap-1.Therefore, in view of Theorem 17, a natural question is, if xn and yn are given by (18), where y0≡b/(a-1)-x0, does it follow that yn=b/(a-1)-xn for all n? We show affirmative answers in general for at least two choices of λ, the first of which requires that λ is not a term of xn.Theorem 24.
Letxn satisfy difference equation (18) with respect to a>1, b>0, 0<λ,x0<b/(a-1), and Fλ≡F; and let yn be defined by (18) in terms of a≡a, b≡b, λy≡b/(a-1)-λ, y0≡b/(a-1)-x0, and the corresponding F denoted by Fλy. Then(134)y1=ba-1-x1,ifx0≠λaba-1-x1,ifx0=λ.Moreover, if x0=λ and λ≥b/a(a-1), then x1+j=f+(1,j) for all j≥0.Proof.
Suppose first thatx0>λ. Then x1=ax0-b=b/(a-1)-a(b/(a-1)-x0) and y0=b/(a-1)-x0<b/(a-1)-λ=λy so that y1=ay0=b/(a-1)-x1.
Next, assume thatx0≤λ. Then x1=ax0. If x0<λ, then y0=b/(a-1)-x0>b/(a-1)-λ=λy and therefore y1=ay0-b=b/(a-1)-ax0=b/(a-1)-x1. And if x0=λ, then y0=b/(a-1)-λ=λy and y1=ay0=a(b/(a-1))-x1.
Finally, ifx0=λ≥b/a(a-1), then x1+j=f+(1,j) by Lemma 6(2).For the solutionsxn in the following examples (with the conditions imposed), yn=b/(a-1)-xn for all n by Theorem 24 since xk≠λ for all k: Examples 8 (if λ≠b/2(a-1)), 12, 21 (if x0<λ), and 22.Letxn and yn be given as in Theorem 24 and suppose further that xk≠λ for all k. By Remark 2, xn=x(n), where x=x(t) satisfies(135)x′t=Axt+Bxt+CFλxtfor constants A, B, and C such that b/(a-1)=-C/(A+B). Therefore, yn=y(n), where y(t)≡b/(a-1)-x(t) is the symmetric solution about the line x=b/2(a-1) of the differential equation (136)y′t=Ayt+Byt+CFλyytwith y(0)=b/(a-1)-x0; and Fλy(y([t]))=1-Fλ(x([t])) since x([t])≠λ, and hence x([t])≤λ if and only if y([t])>λy.For type C solutions, whereλ≠b/a, there may be another symmetric solution obtained by reflecting x0 about b/2(a-1).Theorem 25.
Letxn satisfy (18) with respect to a>2, b>0, and 0<x0<b/(a-1) such that either(1)
b/a(a-1)≤λ<b/2(a-1) and xn is bounded belowor(2)
b/2(a-1)≤λ<b/a and xn is bounded above.
Ifzn satisfies (18) in terms of a≡a, b≡b, λ≡λ, and z0≡b/(a-1)-x0, then zn=b/(a-1)-xn for all n.Proof.
Suppose that (1) holds. Ifx0≤λ, then x1=ax0 and z0=b/(a-1)-x0≥b/(a-1)-λ>b/(a-1)-b/2(a-1)=b/2(a-1)>λ. Thus, z1=az0-b=b/(a-1)-ax0=b/(a-1)-x1 in this case.
Next, assume thatx0>λ. Then x1=ax0-b=b/(a-1)-a(b/(a-1)-x0)=b/(a-1)-az0. We claim that z0≤λ. By way of contradiction, assume that z0>λ≥b/a(a-1). Then x0=b/(a-1)-z0<b/(a-1)-λ≤b/(a-1)-b/a(a-1)=b/a. Hence, x1=ax0-b<a(b/a)-b=0 and xn is not bounded below by Lemma 5(1), contrary to (1). Therefore z0≤λ and z1=az0=b/(a-1)-x1.
It follows in either case thatz1=b/(a-1)-x1; and, by induction, zn=b/(a-1)-xn for all n.
Suppose now that (2) holds. Ifx0>λ, then x1=ax0-b and z0=b/(a-1)-x0<b/(a-1)-λ≤b/(a-1)-b/2(a-1)=b/2(a-1)≤λ. Thus z1=az0=ab/(a-1)-ax0=ab/(a-1)-(x1+b)=b/(a-1)-x1 in this case.
Finally assume thatx0≤λ. We claim that z0>λ. Suppose otherwise that z0≤λ. Then x0=b/(a-1)-z0≥b/(a-1)-λ>b/(a-1)-b/a=b/a(a-1) and x1=ax0>a(b/a(a-1))=b/(a-1); so xn is not bounded above by Lemma 5(2), contradicting (2). Therefore z0>λ and z1=az0-b=b/(a-1)-ax0=b/(a-1)-x1.
Hence,z1=b/(a-1)-x1 in both cases; and, by induction, zn=b/(a-1)-xn for all n.Note that Theorem25 may fail when λ=b/a: let a>2 and let b>0 and let x0≡b/a=λ. Then x1=ax0=b and z0=b/(a-1)-x0=b/a(a-1)<λ. Thus z1=az0=b/(a-1)≠b/(a-1)-x1=b/(a-1)-b.The solutionxn in Example 11 satisfies yn=b/(a-1)-xn for all n whenever λ≠bap-2/(ap-1) by Theorem 24 since λ is thus not a term of xn. Furthermore, in this example, if a>2 and λ≠b/a is in I3, then zn=b/(a-1)-xn for all n by Theorem 25 since xn is eventually periodic and therefore bounded.Letxn and zn be given as in Theorem 25. By Remark 2, xn=x(n), where x=x(t) satisfies (2) for constants A, B, and C such that b/(a-1)=-C/(A+B). Therefore, zn=z(n), where z(t)≡b/(a-1)-x(t) is the symmetric solution about x=b/2(a-1) of the same differential equation(137)z′t=Azt+Bzt+CFztbut with z(0)=b/(a-1)-x0, since, by the proof of Theorem 25, F(z([t]))=1-F(x([t])) (i.e., x([t])≤λ if and only if z([t])>λ).Ifx0 is given by (74), then b/(a-1)-x0 is given by (133) and Theorem 17 may be applied to Theorems 24 and 25.Corollary 26.
Letxn satisfy (18), where a≥2, b>0, λ is in I3, and x0 is given by (74) for some integers m≥0 and p≥1 and some polynomials pp-1(a) in Pp-1(a) and qm-1(a) in Pm-1(a).(1)
Ifλ≠b/a(a-1) or pp-1(a)≠(ap-1)/(a-1) or a=2, then ym=ym+p. Moreover, if xn is bounded with no stationary states, then yn=b/(a-1)-xn for all n. In particular, if pp-1(a)≠0,(ap-1)/(a-1), then yn=b/(a-1)-xn for all n.(2)
Ifλ≠b/a or pp-1(a)≠(ap-1)/(a-1), then zm=zm+p. Moreover, if a>2, λ≠b/a, and xn is bounded, then zn=b/(a-1)-xn for all n.Proof.
Suppose thata≥2, b>0, λ is in I3, and x0 is given by (74). The first lines of (1) and (2) follow directly from (133) and Theorem 17(2).
(1) Assumexn is bounded with no stationary states. Then x0≠0,b/(a-1) and 0<x0<b/(a-1) by (81). By Theorem 13, xk is not in I3 so xk≠λ for all k. Thus yn=b/(a-1)-xn by Theorem 24.
In particular, suppose thatpp-1(a)≠0,(ap-1)/(a-1). Then xn is eventually periodic, and hence bounded, by Theorem 17(2); and, by the proof of this theorem, the only possible stationary states are when(i)
qm-1(a)=0 and x0=bpp-1(a)/am(ap-1)=0 (≤b/a(a-1)) or(ii)
xm=bpp-1(a)/(ap-1) is 0 or b/(a-1),
that is, whenpp-1(a) is 0 or (ap-1)/(a-1), which are ruled out. Thus, by the previous case, yn=b/(a-1)-xn for all n.
(2) Next, assume thata>2, λ≠b/a, and xn is bounded. By (81), 0≤x0≤b/(a-1). If x0=0, then xn=0 for all n; and hence z0=b/(a-1) and thus zn=b/(a-1)=b/(a-1)-xn for all n. Similarly, if x0=b/(a-1), then zn=b/(a-1)-xn for all n.
Therefore, we may assume0<x0<b/(a-1) and the desired result is now immediate from Theorem 25.Example18 (revisited). In this example, a≥2, b>0, λ is in I3, and x0 is of form (74) with m=2, p=2k≥2, pp-1(a)=ak((ak-1)/(a-1)), and qm-1(a)=1. Moreover, x2=x2+2k. By Corollary 26, yn=b/(a-1)-xn for all n and z2=z2+2k. Furthermore, if a>2 and λ≠b/a, then zn=b/(a-1)-xn for all n.Example 27.
Leta≥2 and let b>0 and let λ be in I3. For positive integer k, consider the weighted average(138)x0≡b/a+akb/a-1-b/a21+akin I4 (that converges to b/(a-1)-b/a2 as k tends to infinity). Then x0 is of form (74) with m=1, p=2k≥2, pp-1(a)=ak-1((ak-1)/(a-1)), and qm-1(a)=1 so x1=x1+2k by Theorem 17(2). Therefore, as above,(i)
yn=b/(a-1)-xn for all n;(ii)
z1=z1+2k;(iii)
ifa>2 and λ≠b/a, then zn=b/(a-1)-xn for all n.
Moreover,(139)ba-1-x0=b/aa-1+akb/a21+ak=bak-1/a-1ak-ak-1+1+a2k-10aa2k-1.



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*Source: 828952-2015-07-15.xml*