# Nonnegative Periodic Solutions of a Three-Term Recurrence Relation Depending on Two Real Parameters

Authors: Yen Chih Chang; Sui Sun Cheng; Wei Chang Yeh
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

Simple dynamic systems representing time varying states of interconnected neurons may exhibit extremely complex behaviors when bifurcation parameters are switched from one set of values to another. In this paper, motivated by simulation results, we examine the steady states of one such system with bang-bang control and two real parameters. We found that nonnegative and negative periodic states are of special interests since these states are solutions of linear nonhomogeneous three-term recurrence relations. Although the standard approach to analyse such recurrence relations is the method of finding the general solutions by means of variation of parameters, we find novel alternate geometric methods that offer the tracking of solution trajectories in the plane. By means of this geometric approach, we are then able, without much tedious computation, to completely characterize the nonnegative and negative periodic solutions in terms of the bifurcation parameters.


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## Body

## 1. Introduction

Simple dynamic systems representing time varying states of interconnected compartments or “neurons” may exhibit extremely complex behaviors when bifurcation parameters are switched from one set of values to another. An example has been given in several of our previous studies [1] and a slightly modified model of which is described as follows. Let n neurons be placed on the vertices of a regular n-gon and let the time dependent state values of the neurons be denoted by uit, for i=1,2,…,n. Suppose the rate of change u2′t is determined by the resultant effect of a constant multiple of u2t, plus a magnified on-off (or bang bang) state dependent control mechanism as well as its two near neighbors u1t, u3t in the form(1)u2′t=u1t+ρu2t+u2t-μHu2t. Then assuming “uniformity” among the neuron interactions, we have(2)ui′t=ui-1t+ρuit+ui+1t-μHuit,i=1,2,…,n,where u0(t) is identified with un-1(t) while un+1(t) with u1(t) for compatibility reasons. Here Hx is the on-off Heaviside step function defined by(3)Hx=1ifx≥0,-1ifx<0.When the real parametersρ and μ are switched among different real values, it is expected (and verified by simulations) that complex dynamic behaviors will be more than abundant. Some of these behaviors can be explained (see [1–3]) but some not (at least to the best of our knowledge). Similar models of piecewise constant dynamic systems which exhibit similar behaviors with parameters can be found in many recent investigations; see for examples [4–10] and the references therein.In this note, we discuss the steady state solutions of the above dynamic system (i.e., those that satisfyuk(t)≡φk for all t and for each k=1,2,…, or n). Then we will face the existence problem of periodic solutions of the following three-term recurrence relation:(4)φk+ρφk+1+φk+2=μHφk+1,k∈Z≔…,-1,0,1,….In [1–3], we are lucky to obtain complete information about the periodic solutions of (4) when μ=1 and ρ=1, 0, or -1 by breaking the solutions into two sequences, a companion and an error (more specifically, for a solution φ=φkk∈N of (4) with μ=1 and ρ∈-1,0,1, the companion sequence is λ=λkk∈N which is an even integral sequence so that λk=2(1/2)φk, where 2x/2 is the greatest even integer that is less than or equal to x∈R, and the error sequence ϵ=ϵkk∈N is defined by ϵk=φk-λk,k∈Z).Yet, for other values ofμ and ρ, simulations show complex periodic behaviors beyond our present comprehension, except when the periodic solutions are also “nonnegative” (or “negative”). Such exceptional results, when examined more closely, can be explained.In this paper, we will devote ourselves to explaining the behaviors of these nonnegative (or negative) and periodic solutions of (7).First, a real sequenceφ=φkk∈Z is a solution of (4) if it renders (4) into an identity after substitution. It is nonnegative (or negative) if all its terms are nonnegative (respectively negative). If ζ=ζkk∈Z is a nonnegative solution of (4), then clearly ζ satisfies(5)ζm+ρζm+1+ζm+2=μ,m∈Z,while if ξ is a negative solution of (4), then(6)ξn+ρξn+1+ξn+2=-μ,n∈Z.Hence it seems that we are back to the usual nonhomogeneous linear second-order difference equations studied in elementary theory of difference equations. Since μ∈R in (5) and (6), we can further restrict ourselves in the sequel to the following difference equation:(7)φk+ρφk+1+φk+2=μ,μ∈R,k∈Z,with arbitrary μ∈R. By means of the techniques of general solutions plus the method of variation of parameters, it may be argued that the existence of nonnegative (respectively negative) periodic solutions can be handled completely. Such an assertion may be true in theory, but the general solution here is ρ- as well as μ-dependent and hence actual attempts lead to many complications.That said, in this paper, we will handle our equation from a novel approach and the crux of which is based on representing each pair of two consecutive terms(φk,φk+1) of a solution φ=φkk∈Z as a point in the plane and invent a geometric method to track the movements of these points. First, the totality of such pairs is called the orbit of this solution. More precisely, let φ=φkk∈Z be a solution of (4). We define the orbit of φ by(8)Oφ≔xk,yk∈R2∣xk=φk,yk=φk+1,k∈Zand the “positive” orbit by(9)O+φ≔xk,yk∈R2∣xk=φk,yk=φk+1,k∈N,N≔0,1,2,….As examples, let μ=1 and let ζ and ξ be solutions of (4) with ρ=1 or 0, respectively. The orbits Oζ with ζ0,ζ1=9,1 and Oξ with ξ0,ξ1=15/2,1/2 are depicted in Figures 1 and 2, respectively. These figures clearly suggest that ζ is “28-periodic,” that ξ is “30-periodic,” and that there are accompanying “distinctive” features which can be exploited further.Figure 1
The orbit ofζ with ρ=1.Figure 2
The orbit ofξ with ρ=0.Several sets will be encountered in the ensuing discussions and we denote them as follows:(i)
R≔-∞,∞ is the entire set of real numbers.(ii)
Z≔…,-1,0,1,…is the set of integers(iii)
Z+≔1,2,3,…is the set of positive integers.(iv)
N≔0,1,2,…is the set of nonnegative integers.(v)
Q≔±m/n∣m∈N,n∈Z+is the set of rational numbers.Before we enter into discussions on the necessary and sufficient conditions for a solution of (7) to be nonnegative and periodic, we take a note that any solution φ=φkk∈Z of our three-term equation (7) is uniquely determined by two of its consecutive terms. Furthermore, in view of the fact that (7) can be written as(10)φm-1+φm+1=μ-ρφm,m∈Z,as well as(11)φm+1+ρφm+2+φm+3=μ,m∈Z,it is both “symmetric” (or “reflection invariant”) and “translation invariant.” More precisely, let φjj∈Z be the solution of (7) that satisfies φ0=α and φ1=β. Then the solution ξjj∈Z of (7) determined by ξ0=α and ξ-1=β, called the reflection of φ and denoted by ξ=Uφ, will satisfy(12)ξj=φ-j,j∈Z,while the solution ζjj∈Z of (7) determined by ζk=α and ζk+1=β for any k∈Z, called the k-translation of φ and denoted by ζ=Ekφ, will satisfy(13)ζj=φj-k,k∈Z.In view of these invariances, to study (7), we may simply concentrate our attention on the positive orbits of its solutions! Indeed, let ψ=ψkk∈Z be the solution of (7) with ψ0,ψ1=σ,τ. We investigate ψ0,ψ1,ψ2,… by observing the positive orbit O+ψ defined by σ,τ. On the other hand, we analyze ψ1,ψ0,ψ-1,ψ-2,… by studying the positive orbit O+UE-1ψ. Note that for all ψn,ψn+1∈Oψ, where n<0, ψn+1,ψn=Qn, where Qn∈O+UE-1ψ. In the sequel, if no doubt arises, we frequently let φ=φkk∈Z be a solution of (7) which is defined by φ0,φ1=α,β and we denote(14)O⊥+φ=xk,yk=vk+1,uk+1∣uk+1,vk+1∈O+UE-1φ,k∈N so that(15)Oφ=O+φ∪O⊥+φ.Proposition 1.
Letφ=φkk∈Z be a solution of (7) defined by φ0,φ1=α,β and O+φ=xk,ykkk∈N be the “positive” orbit of φ. Then ω is a period of φ if and only if xω,yω=α,β. Furthermore, if limk→∞xk=∞, then φ is aperiodic and if limk→∞xkyk<0, then φ is neither a nonnegative nor a negative solution.The statements of Proposition1 can be verified directly from the definitions, as well as the invariant properties of (7), and, thus, the proof is omitted here.


## 2. Some Basic Results of (7)

Periods and the prime periods of sequences are defined as usual. Furthermore, letφ=φkk∈Z be a nonnegative solution of (7). Note that for some distinct m,n∈N, if φm,φm+1=φn,φn+1, then m-n is a period of φ and the least one ω among the periods is the least or prime period of φ (we also say that φ is ω-periodic). A constant solution of (7) is 1-periodic. We first find the necessary and sufficient conditions for φ to be a constant solution.Proposition 2.
Letφ=φmm∈Z be a solution of (7) with φ0,φ1=α,β.(i)
Supposeρ≠-2. Then φ is a constant solution if and only if α=β=μ/(2+ρ).(ii)
Supposeρ=-2. Then φ is a constant solution if and only if μ=0 and α=β.Proof.
Supposeρ≠-2. If φ is a constant solution of (7), then α=β=φ2 and(16)φ2=α=μ-ρφ1-φ0=μ-ρβ-α=μ-ρα-α so that(17)α2+ρ=μ. Accordingly, α=μ/(2+ρ)=β. The converse can be verified by direct iteration. Next, suppose ρ=-2. If φ is a constant solution of (7), then α=β and by (7)(18)φ2=μ+2β-α=μ+α=αwhich leads to μ=0. Conversely, if μ=0 and α=β, then by iteration φ is clearly a constant solution.
The proof is complete.Corollary 3.
Letφ=φkk∈Z be a solution of (7) with φ0=φ1=γ. Let γ=μ/2+ρ where ρ≠-2. Then φ is 1-periodic. φ is a nonnegative solution if and only if γ≥0; and φ is a negative solution if and only if γ<0.The results can be obtained from Proposition2 directly and, hence, we omit the proof.In view of Proposition2, it is easy to see that the orbit Oφ of a constant solution is just the set containing the only point μ/(2+ρ),μ/(2+ρ).Proposition 4.
Letφ=φkk∈Z be a solution of (5) with φ0,φ1=α,β. Then φ is 2-periodic if and only if ρ=2, α+β=μ/2, and α≠β.Proof.
Supposeφ is 2-periodic. Then φ2=α as well as φ3=β and by (5), φ2=μ-ρβ-α=α and φ3=μ-ρα-β=β which lead us to 2α+ρβ=μ and ρα+2β=μ, respectively. Accordingly, we have 2+ρα+β=2μ and 2-ρα-β=0. Note that since φ is 2-periodic, we can be sure that α≠β. Hence, if ρ≠2, then α-β=0 which is a contradiction; if ρ=2, then α+β=μ/2 and α≠β. The converse may be checked by direct substitution into (5) and this completes the proof.Corollary 5.
Letφ=φkk∈Z be a solution of (7) with ρ=2 and φ0,φ1=α,β. Suppose α+β=μ/2. Then φ is periodic with period 2.  φ is nonnegative if and only if α,β,μ≥0; and φ is negative if and only if α,β,μ<0.Proof.
Supposeα+β=μ/2. First of all, if α=β=μ/4, then, by Proposition 2, φ is 1-periodic; otherwise, by Proposition 4, φ is 2-periodic. Hence, φ is periodic with period 2. If φ is a nonnegative solution, then α,β≥0 which implies μ≥0. Furthermore, since φ is periodic with period 2, we can see that φk≥0 for all k∈Z. If α,β,μ≥0, then α+β=μ/2 is satisfied and φ is a nonnegative solution. The case where φ is a negative solution can be handled similarly and this completes the proof.In this paper, since we are interested in the periodicity of the solutions of (4) which are also nonnegative (or negative), the following results will be useful.Proposition 6.
Letφ=φkk∈Z be a nonnull solution of (7). If μ≥0 and ρ≥0, then φ cannot be a nonpositive solution; if μ≤0 and ρ≤0, then φ cannot be a nonnegative solution.Proof.
Supposeμ≥0 and ρ≥0. If φ is nonpositive, then 0≥φ2=-ρφ1-φ0+μ≥0 and 0≥φ3=-ρφ2-φ1+μ≥0. Hence φ2=φ3=0, which implies φ is null, contrary to our assumption.
Supposeμ≤0 and ρ≤0. If φ is nonnegative, then 0≤φ2=-ρφ1-φ0+μ≤0 and 0≤φ3=-ρφ2-φ1+μ≤0 so that φ2=φ3=0. A contradiction is arrived and this completes the proof.Proposition 7.
φ is a solution of (5) if and only if -φ is a solution of (6).Proof.
Suppose thatφ=φmm∈Z is a solution of (5) with φ0,φ1=α,β. By (5), we have φ2=μ-ρβ-α. Let ψ=-φ such that ψ0,ψ1=-α,-β. Accordingly, ψ2=-φ2 and we see that(19)ψ2=-μ-ρβ-α=-μ+ρβ+α=-μ-ρψ1-ψ0which satisfies (6). By induction, it follows that ψ is a solution of (6) and this completes the proof.Next, we discuss nonconstant solutions of (5) (although 2-periodic solutions are discussed in Proposition 4, we may include them in the following discussions). The behavior of solution φ of (7) with least period 1 (or 2) was quite easy to analyze directly, yet it is difficult to conduct similar analyses for φ with larger least period (or φ is aperiodic) by similar manner. Hence, we will investigate the behavior of φ from a new perspective.


## 3. The Orbits of Solutions of (7)

We will need the following quadratic function:(20)Γμ,ρx,y=x2+ρxy+y2-μx-μy,wherex,y∈R2,and its f-level curve Γμ,ρ,f is defined by(21)Γμ,ρ,f=x,y∈R2∣Γμ,ρx,y=f,f∈R.Since Γμ,ρ is a quadratic function, its level curves are plane conic sections. The properties of these conic sections are well known. First, we see that the x=y line is one of the principal axes of Γμ,ρ,f and if ρ≠-2; then μ/(2+ρ),μ/(2+ρ) is the center of Γμ,ρ,f and the other principal axis is the x+y=2μ/(2+ρ) line. In particular (see, e.g., [11]) Γμ,ρ,f can be classified by the value of its discriminant ρ2-4 and the values (ρ-2)μ2 as well as f:(1)
Ifρ>2, then Γμ,ρ,f is a hyperbola (see the green and red curves in Figures 3 and 4) when f≠-μ2/2+ρ or a degenerate hyperbola consisting of two intersecting lines when f=-μ2/2+ρ (see the blue lines in Figures 3 and 4).(2)
Ifρ<2, then Γμ,ρ,f is an ellipse (see Figure 5) when f≠-μ2/2+ρ or a degenerate point ellipse when f=-μ2/2+ρ.(3)
Ifρ=2, then Γμ,ρ,f is a parabola when ρ=-2 or a degenerate parabola consisting of a single or two distinct parallel lines when ρ=2 (see the green and blue lines in Figure 6).Figure 3
The hyperbolae withρ>2.Figure 4
The hyperbolae withρ<-2.Figure 5
The ellipses with-2<ρ<2.Figure 6
The parabolae withρ=2.Unless indicated by the adjective “degenerate,” a conic section is meant to be nondegenerate. The conic sectionsΓμ,ρ,f are symmetric with respect to the x=y line. This property is a simple consequence of the fact that Γρ,fx,y is a symmetric function in x and y. Hence, the x=y line is one of the principal axes of the conic sections.To discuss the principal axes further, we define the two variable functions:(22)Πx,y=x-yand the associated line Π¯=(x,y)∈R2∣Π(x,y)=0. Also, for ρ≠-2, we let(23)Π⊥x,y=x+y-2μ2+ρand Π¯⊥=(x,y)∈R2∣Π⊥(x,y)=0. Furthermore, let f=Γμ,ρα,β, where α,β∈R2 and δ=ρ2-4 for ρ>2. We define the following two variable functions:(24)Lμ,ρ,f+x,y≔y+12ρ+δx-μ21+δ2+ρifρ>2,x+y-12μ+f+14μ2ifρ=2,x-y+fifρ=-2,as well as(25)Lμ,ρ,f-x,y≔y+12ρ-δx-μ21-δ2+ρifρ>2,x+y-12μ-f+14μ2ifρ=2,x-y-fifρ=-2,with L¯μ,ρ,f+=(x,y)∈R2∣Lμ,ρ,f+x,y=0 and L¯μ,ρ,f-=(x,y)∈R2∣Lμ,ρ,f-x,y=0. Note that in view of (24) and (25), Lμ,ρ,f+x,y as well as Lμ,ρ,f-x,y are independent from f when ρ>2.LetS⊂R2. Define(26)S′=x,y∈R2∣y,x∈S,S′′=x,y∈R2∣-y,-x∈S. Then S′ and S are symmetric with respect to the x=y line, while S′′ and S are symmetric with respect to the x+y=0 line. The boundary of the plane set S will be denoted by ∂S.The conic sections will be of great help in the analysis of solutions of (4). The most significant result is the following.Theorem 8.
Letφ=φkk∈Z be a solution of (5). If f=Γμ,ρφm,φm+1 for some m∈Z, then Oφ⊆Γμ,ρ,f. Furthermore, if α,β∈Γμ,ρ,f, where f=Γμ,ρα,β, then -α,-β∈Γμ,ρ,f′′.Proof.
Without loss of generality, we letm=0 and φ0,φ1=α,β. Let f=Γμ,ρα,β. Then α,β∈Γμ,ρ,f. By (21), we see that(27)α2+ραβ+β2-μα-μβ-f=0 and by (7), it follows φ1,φ2=β,μ-ρβ-α so that (28)Γμ,ρβ,μ-ρβ-α=β2+βμ-ρβ-αρ+μ-ρβ-α2-β-μ-ρβ-α=α2+ραβ+β2-μα-μβ=f,which leads to φ1,φ2∈Γμ,ρ,f. By induction, for any n∈N, φn,φn+1∈Γμ,ρ,f which implies O+φ⊆Γμ,ρ,f. Next, we consider φn,φn+1, where n∈Z∖N. Let ψ=ψkk∈Z=UE-1φ such that ψ0,ψ1=β,α and φn=ψ-n+1. Since Γμ,ρ,f is symmetric with respect to the x=y line, β,α∈Γμ,ρ,f and by the previous discussions, we can be sure that O+ψ⊂Γμ,ρ,f which implies Qn′∈Γμ,ρ,f, where Qn∈O+ψ. Accordingly, O⊥+φ⊆Γμ,ρ,f and by (15), it follows that Oφ⊆Γμ,ρ,f. Furthermore, by the definitions, if α,β∈Γμ,ρ,f, where f=Γμ,ρα,β, then -β,-α∈Γμ,ρ,f′′ and since Γμ,ρ,f′′ itself is symmetric with respect to the x=y line, it is clear -α,-β∈Γμ,ρ,f′′ as desired. The proof is complete.Letφ=φkk∈Z be a solution of (5). Then φ is uniquely determined by two consecutive terms φm and φm+1. Although we can calculate the next term φm+2 directly, the conic sections allow us to easily “track” the movements of the points from (φm,φm+1) to (φm+1,φm+2). Indeed, by Theorem 8, we see that Oφ⊆Γμ,ρ,f, where f=Γμ,ρφm,φm+1. Then by means of this conic section, we may plot the corresponding orbit as follows.Tracking Procedure. Input α,β∈R2.(1)
PlotΓμ,ρ,f where f=Γμ,ρα,β on the plane.(2)
Takek=0 and start at the point Pk=xk,yk=α,β.(3)
LetPk′=yk,xk be the point on Γμ,ρ,f which is symmetric to Pk with respect to the x=y line.(4)
Draw a vertical lineLk through Pk′.(5)
The lineLk and the conic section Γμ,ρ,f can intersect at Pk′ and possibly at another point Pk⊥. If Pk′=Pk⊥, then let Pk+1=Pk′; otherwise, take Pk+1=Pk⊥.(6)
IfPk+1=P0, then stop; otherwise, let Pk=Pk+1 and go to step (3).Note that by the Principal Axes Theorem, the level curveΓμ,ρ,f is defined by setting Γμ,ρ=f, which is a polynomial of degree 2. The conic section has one principal on the x=y line which implies any vertical line through Pk′ meets Γμ,ρ,f at least once and at most twice through two points Pk′ or Pk⊥. If Pk′=Pk⊥, then Pk′ is on the ρx+2y=μ line and this can be verified directly by (7). In view of theTracking Procedure, we can generate each Pk=xk,yk as (φk,φk+1) by initializing P0=x0,y0=φ0,φ1 and it is not difficult to see that the above algorithm can yield the “positive” orbit O+φ; on the other hand, the “negative” orbit of Oφ can be plotted by the “positive” orbit O⊥+φ which is generated by theTracking Procedure with input φ1,φ0. In Figures 7 and 8, we illustrate the orbits of Pk on ellipses; Figures 9 and 10 depict the orbits on a parabola and a degenerate parabola correspondingly; and in Figures 11 and 12, we show the orbits on a degenerate hyperbola and a hyperbola, respectively.Figure 7
The orbit onΓ-1,2,4,42.Figure 8
The orbit onΓ-1,-2cos⁡2π/9,-1/4.Figure 9
The orbit onΓ-1,-2,4.Figure 10
The orbit onΓ-0,2,4.Figure 11
The orbit onΓ-5,3,-5.Figure 12
The orbit onΓ-1,3,1/5.As we have shown previously, thef-level curve Γμ,ρ,f of Γμ,ρ is a conic section and it can be classified into hyperbola, ellipse, or parabola (or the degenerate curves). Let φ=φkk∈Z be a solution of (7) defined by φ0,φ1=α,β. By means of theTracking Procedure, we can easily analyze the asymptotic behavior of φk when k→∞. In the sequel, we will discuss the sequence xk,ykk∈N which is generated by theTracking Procedure with input x0,y0=α,β depending on ρ>2, ρ=2, and ρ<2.We first consider the case whereρ>2. By the previous discussions, the points Pk∈xk,ykk∈N generated by theTracking Procedure with input x0,y0 move on Γμ,ρ,f where f=Γμ,ρx0,y0 and Γμ,ρ,f is a hyperbola (or a degenerate one made up of two lines). Since Γμ,ρ,f is a hyperbola, we first locate the asymptotes of Γμ,ρ,f. Note that Pc=μ/(2+ρ),μ/(2+ρ) is the center of Γμ,ρ,f. In view of (21), we rewrite the equation Γμ,ρ,f as(29)y±x=-12ρx-μ±14ρx-μ2-x2+μx+f, and suppose y=m±x+n± are the oblique asymptotes of Γμ,ρ,f. Then we have (30)m±=limx→∞+⁡y±xx=limx→∞+⁡-12ρ-μx±14ρ-μx2-1+μx+fx2=-12ρ±12ρ2-4.As for n±, since lines y=m±x+n± intersect at Pc, we apply Pc to the lines and obtain n±=μ-μm±/2+ρ. Accordingly, we see that L¯μ,ρ,f+ and L¯μ,ρ,f- can be expressed by y-m-x-n-=0 as well as y-m+x-n+=0, respectively. For ρ>2, one relation between Pk and Lμ,ρ,f+ as well as Lμ,ρ,f- defined in (24) and (25), respectively, is revealed by(31)Lμ,ρ,f+yk,μ-ρyk-xk=-Lμ,ρ,f-yk,xkas well as(32)Lμ,ρ,f-yk,μ-ρyk-xk=-Lμ,ρ,f+yk,xk.Moreover, with the oblique asymptotes L¯μ,ρ,f+ and L¯μ,ρ,f-, we may illustrate the asymptotic behaviors of Γμ,ρ,f where ρ>2.Theorem 9.
Supposeρ>2 and let α,β∈R2 such that f=Γμ,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then Γμ,ρ,f is a hyperbola (or a degenerate one made up of two lines L¯μ,ρ,f- and L¯μ,ρ,f+) and we have the following.(i)
SupposeΓμ,ρ,f is a degenerate hyperbola. If α,β∈L¯μ,ρ,f+, then xm,ym∈L¯μ,ρ,f+ for all m∈N; if α,β∈L¯μ,ρ,f-, then xn,yn∈L¯μ,ρ,f- for all n∈N.(ii)
SupposeΓμ,ρ,f is a hyperbola. If ρ≥0, then xm,xm+1 and xm+1,xm+2 are sitting on different branches for all m∈N; if ρ<0, then xn,xn+1 and xn+1,xn+2 lie on the same branch for all n∈N.Proof.
Note thatx0=α and x1=y0=β. By Theorem 8 and (7), we see that y1=μ-ρy0-x0=μ-ρβ-α=γ. Suppose α,β∈L¯μ,ρ,f+ or α,β∈L¯μ,ρ,f-. In the former case, L¯μ,ρ,f+α,β=L¯μ,ρ,f-β,α=0 and by (31), as well as (32), L¯μ,ρ,f+β,γ=-L¯μ,ρ,f-β,α=0 which implies x1,y1=β,γ∈L¯μ,ρ,f+; in the latter case, by similar arguments, we may see that L¯μ,ρ,f-β,γ=0. Note that if α,β∈L¯μ,ρ,f+∩L¯μ,ρ,f-, then α,β=μ/(2+ρ),μ/(2+ρ) and by theTracking Procedure, x1,y1=μ/(2+ρ),μ/(2+ρ) which implies xi,yi=μ/(2+ρ),μ/(2+ρ) for all i∈N. Suppose α,β∉L¯μ,ρ,f+∪L¯μ,ρ,f-. Then Γμ,ρ,f is a hyperbola and α,β is on one branch of Γμ,ρ,f. By theTracking Procedure, we may see that if ρ≥0, then P1 “jumps” to another branch (see Figure 3), while if ρ<0, then P1 will stay on the same branch (see Figure 4). By induction, we may see that (ii) holds. This completes our proof.In view of Theorem9, we may follow the movements of xk,yk generated by theTracking Procedure when ρ>2. This enables us to see the asymptotic behavior of xk,yk when k→∞ in Theorem 10. Before introducing Theorem 10, we denote the distance between two points P1 and P2 on R2 by P1-P2 and the distance from a point P to the line L by dP,L.Theorem 10.
Supposeρ>2 and let α,β∈R2 such that f=Γμ,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with x0,y0=α,β. Then Γμ,ρ,f is a hyperbola (or a degenerate one made up of two lines L¯μ,ρ,f+ and L¯μ,ρ,f-) and we have the following.(i)
Supposeα,β∈L¯μ,ρ,f+∪L¯μ,ρ,f-. If α,β=μ/(2+ρ),μ/(2+ρ), then xk,yk=α,β for all k∈N; if α,β∈L¯μ,ρ,f+ and α,β≠μ/(2+ρ),μ/(2+ρ), then(33)limk→∞⁡dxk,yk,L¯μ,ρ,f-=∞ifρ>20ifρ<-2;and ifα,β∈L¯μ,ρ,f- and α,β≠μ/(2+ρ),μ/(2+ρ), then(34)limk→∞⁡dxk,yk,L¯μ,ρ,f+=0ifρ>2∞ifρ<-2.(ii)
Ifρ>2 and Γμ,ρ,f is a hyperbola, then limk→∞dxk,yk,L¯μ,ρ,f+=0; and if ρ<-2 and Γμ,ρ,f is a hyperbola, then limk→∞dxk,yk,L¯μ,ρ,f-=0.Proof.
For the sake of convenience, we letPc=μ/(2+ρ),μ/(2+ρ) and Pk=xk,yk so that P0=α,β and f=Γμ,ρα,β. Let y1=γ=μ-ρβ-α. By Proposition 2, if P0=Pc, then φ is a constant solution of (7) which implies Pkk∈N=Pc. Now, we suppose P0≠Pc. Since ρ>2, Γμ,ρ,fis a hyperbola (or a degenerate one). Hence, we discuss the two cases where P0 is on a degenerate hyperbola or a hyperbola.
SupposeP0∈L¯μ,ρ,f+∪L¯μ,ρ,f- and P0≠Pc. Then by Theorem 9, P0′ and P1′ are on the same line. In view of (24) and (25), we let m+=ρ+δ,  m-=ρ-δ, Δ+=μ+δ/(2+ρ), and Δ+=μ-δ/(2+ρ). For the sake of convenience, we also let(35)κ=1+m-21+m+2.Since Γμ,ρ,f is symmetric with respect to the x=y line, we see that if ρ>2, then κ<1; if ρ<-2, then κ>1. Suppose ρ>2. If P0∈L¯μ,ρ,f- and P0≠Pc, then P0′∈L¯μ,ρ,f+ and P1∈L¯μ,ρ,f- so that α+m+β+Δ+=0 and γ+m-β+Δ-=0, respectively. Also, by the symmetry of Γμ,ρ,f with respect to the x=y line, we can be sure that dP0,L¯μ,ρ,f+=dP0′,L¯μ,ρ,f- so that(36)dP0,L¯μ,ρ,f+=α+m-β+Δ-1+m-2=α-γ1+m-2.On the other hand, we calculate(37)dP1,L¯μ,ρ,f+=γ+m+β+Δ+1+m+2=γ-α1+m+2;and by (35) and (36), as well as (37), it follows that(38)dP1,L¯μ,ρ,f+=κdP0,L¯μ,ρ,f+. By induction, we have dPi,L¯μ,ρ,f+=κidP0,L¯μ,ρ,f+, where κ<1 and i∈N. Accordingly, (34) holds. The case where ρ<-2 and P0∈L¯μ,ρ,f-∖Pc can be handled similarly with dPj,L¯μ,ρ,f+=κ-jdP0,L¯μ,ρ,f+, where κ>1 and j∈N. By similar arguments, (33) is also true.
SupposeP0∉L¯μ,ρ,f+∪L¯μ,ρ,f-. We first consider the case where ρ>2. Then κ<1 and by (24)(39)dP1,L¯μ,ρ,f+=Lμ,ρ,f+β,γ1+m+2=-Lμ,ρ,f-β,α1+m+2=-Lμ,ρ,f-β,α1+m-21+m-21+m+2=κdP0′,L¯μ,ρ,f-=κdP0,L¯μ,ρ,f+.By induction, it follows that dPi,L¯μ,ρ,f+=κidP0,L¯μ,ρ,f+, where i∈N. Consequently, case (ii) is true when ρ>2 and by similar arguments, we also see that ρ<-2 holds with dPj,L¯μ,ρ,f-=κ-jdP0,L¯μ,ρ,f-, where j∈N and κ>1.
The proof is complete.Corollary 11.
Supposeρ>2 and let α,β∈R2 such that f=Γμ,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then Γμ,ρ,f is a hyperbola (or a degenerate one made up of two lines L¯μ,ρ,f+ and L¯μ,ρ,f-). Suppose α,β≠μ/(2+ρ),μ/(2+ρ). If dα,β,L¯μ,ρ,f+≤dα,β,L¯μ,ρ,f-, where ρ>2, then xkxk+1<0 and xk→∞ when k→∞; if dα,β,L¯μ,ρ,f-≤dα,β,L¯μ,ρ,f+, where ρ<-2, then xkxk+1>0 and xk→∞ when k→∞.The argument of Corollary11 can be verified directly from Theorem 10 based on the facts that the slope of L¯μ,ρ,f- is characterized by -ρ+δ<0 and the slope of L¯μ,ρ,f+ is characterized by -ρ-δ>0, where ρ>2 and ρ<-2, respectively, and, thus, the proof is omitted.In Theorem10, we showed the asymptotic behaviors of Pkk∈N generated by theTracking Procedure. We illustrate the movements of Pkk∈N when k→∞ in Figures 13, 14, and 15 where the arrows indicate the moving directions of Pk as k→∞.Figure 13
The asymptotic behavior of{Pk} plotted byTracking Procedure with ρ>2.Figure 14
The asymptotic behavior of{Pk} plotted byTracking Procedure with ρ<-2.Figure 15
The asymptotic behavior of{Pk} plotted byTracking Procedure with ρ=-2.Next, we consider the asymptotic behaviors ofPkk∈N with ρ=2 or ρ=-2.Theorem 12.
Supposeρ=2 and let α,β∈R2 such that f=Γμ,2α,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then f≥-μ2/4 and Γμ,ρ,f is a degenerate parabola made up of L¯μ,2,-μ2/4+ and L¯μ,2,-μ2/4-. If f=-μ2/4, then L¯μ,2,-μ2/4+=L¯μ,2,-μ2/4- and xk,yk=xk+2,yk+2 for all k∈N; otherwise, xk,yk as well as xk+1,yk+1 are not on the same line. Furthermore, if f>-μ2/4, then dxk,yk,Π¯→∞ and xkyk<0 when k→∞.Proof.
For the sake of convenience, we letPc=μ/(2+ρ),μ/(2+ρ) and Pk=xk,yk. First of all, we see that(40)Γμ,2x,y=x2+2xy+y2-μx-μy=x+y2-μx+y=x+y-μ22-14μ2, which implies f≥-μ2/4. Furthermore,(41)Γμ,2,f=x+y-12μ-f+14μ2x+y-12μ+f+14μ2=0,which leads to Γμ,2,f being two parallel lines. Consequently, if f=-μ2/4, then L¯μ,2,-μ2/4+=L¯μ,2,-μ2/4-; otherwise, L¯μ,2,-μ2/4+ and L¯μ,2,-μ2/4- are two distinct parallel lines. Suppose f=-μ2/4. Then P0∈L¯μ,2,-μ2/4+=L¯μ,2,-μ2/4-. In view of theTracking Procedure, it is found that if P0=Pc, then P1=P0 which implies Pm=P0 for all m∈N; otherwise, P1′=P1⊥=P2=P0 which leads to Pn=Pn+2 for all n∈N. Next, suppose f>-μ2/4.By theTracking Procedure, we can directly observe (see Figure 10) that for an arbitrary n∈N, if xn,yn∈L¯μ,2,f+, then xn+1,yn+1∈L¯μ,2,f- and vice versa. Now, we discuss the asymptotic behaviors of xk,ykk∈N. By theTracking Procedure, we first observe that(42)dPm+1,Π=Pm,Π¯+dL¯μ,2,f+,L¯μ,2,f-ifdPm+1,Π¯>dPm,Π¯,Pm,Π¯-dL¯μ,2,f+,L¯μ,2,f-ifdPm+1,Π¯<dPm,Π¯,and clearly, the converses are also true. Note that dPn,Π≠dPn+1,Π for any n∈N. Furthermore, by theTracking Procedure again, if dP1,Π¯>dP0,Π¯, then for all k∈N(43)dPk,Π¯=dP0,Π¯+kdL¯μ,2,f+,L¯μ,2,f-;otherwise, we see that for some j∈N,(44)dPj,Π¯=dP0,Π¯-jdL¯μ,2,f+,L¯μ,2,f-.In view of (44), since dPj,Π¯≥0,  dP0,Π¯≥0, and dL¯μ,2,f+,L¯μ,2,f- is a positive constant, there is i∈N such that dPi,Π¯≥0 as well as dPi+1,Π¯<0 and, thus, by the previous discussions, dPi+1,Π¯>dPi,Π¯. Accordingly, by the previous discussions again, dPk,Π¯→∞ when k→∞ and since the slopes of L¯μ,2,f+ and L¯μ,2,f- are -1, xkyk<0 as k→∞. The proof is complete.Next, we consider the case whereρ=-2.Theorem 13.
Supposeρ=-2 and let α,β∈R2 such that f=Γμ,-2α,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then we have the following.(i)
Supposeμ=0. Then f≥0 and Γ0,-2,f is a degenerate parabola made up of two parallel lines L¯0,-2,f+ and L¯0,-2,f-. If f=0, then L¯0,-2,f+=L¯0,-2,f-=Π¯ and α=β such that xk,yk=α,α for all k∈N; otherwise, xk,yk and xk+1,yk+1 are on the same line and dxk,yk,P0→∞ when k→∞.(ii)
Supposeμ≠0. Then Γμ,-2,f is a parabola and dxk,yk,Π¯→∞ when k→∞.Proof.
Supposeμ=0. By (20), Γ0,-2x,y=x2-2xy+y2=x-y2 and, thus, f=Γ0,-2α,β=α-β2≥0. Accordingly, Γ0,-2,f is a degenerate parabola which is made up of two parallel lines L¯0,2,f+ and L¯0,2,f- as defined in (24) and (25), respectively. More precisely, if α=β, then f=0 which leads to L¯0,2,0+=L¯0,2,0-=Π¯; if α≠β, then f>0 which implies L¯0,2,f+ and L¯0,2,f- are two distinct parallel lines. By theTracking Procedure, if α=β, then xk,yk=α,α for all k∈N; if α≠β, then xk,yk and xk+1,yk+1 are on the same line. Furthermore, if α≠β, then we have Δ=dL¯0,-2,f+,L¯0,-2,f-=2f and by theTracking Procedure again, dxk,yk,α,β=kΔtan⁡π/4=kΔ, where k∈N. Accordingly, dxk,yk,P0→∞ when k→∞. Next, suppose μ≠0. Then Γμ,-2,f is parabola. By (7), for k∈N, we have yk=xk+1 such that xk+2-xk+1=μ+xk+1-xk and thus(45)dxk,yk,Π¯=μ+2xk+1-xk-xk+12=μ+xk+1-xk2, which leads to(46)dxk,yk,Π¯=12kμ+β-α.Hence, dxk,yk,Π¯→∞ when k→∞. Note that, if μ>0, then Γμ,-2,f is concave upward (see Figure 15), while if μ<0, then Γμ,-2,f is concave downward. Accordingly, (ii) is true and this completes the proof.Corollary 14.
Supposeρ=-2 and let α,β∈R2 such that f=Γμ,-2α,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then we have the following.(i)
Supposeμ=0. Then f≥0 and Γ0,-2,f is a degenerate parabola with two parallel lines, L¯0,-2,f+ and L¯0,-2,f-. If α=β, then f=0 such that L¯0,-2,f+=L¯0,-2,f-=Π¯ and xk=α for all k∈N; otherwise, f>0 such that L¯0,-2,f+ as well as L¯0,-2,f- are two distinct parallel lines and(47)xk⟶∞ifα,β∈L¯0,-2,f+,-∞ifα,β∈L¯0,-2,f-.(ii)
Supposeμ≠0. Then Γμ,-2,f is a parabola. If μ>0, then Γμ,-2,f is concave upward and xk→∞ when k→∞; otherwise, Γμ,-2,f is concave downward and xk→-∞ when k→∞.The results stated in Corollary14 can be derived directly from Theorem 13 by observing the asymptotic behavior of the points xk,yk.Letxk,ykk∈N be generated by theTracking Procedure with inputs x0,y0=α,β and f=Γμ,ρα,β. In Theorems 10, 12, and 13, we have shown the asymptotic behaviors of φ with ρ≥2. Next, we consider the case where ρ<2 and α,β≠μ/(2+ρ),μ/(2+ρ). Suppose ρ<2. Then the conic section Γμ,ρ,f is an ellipse which implies xk is both bounded above and below. Moreover, by the Principal Axes Theorem, the nondegenerate Γμ,ρ,f is an ellipse which is rotated by ϑ∈-π/4,π/4 and translated by xc,yc=μ/(2+ρ),μ/(2+ρ) from a standard ellipse with major axis a as well as minor axis b. Thus, Γμ,ρ,f where f≠-μ2/2+ρ2 can be expressed parametrically by(48)xθyθ=xcyc+cos⁡ϑ-sin⁡ϑsin⁡ϑcos⁡ϑacos⁡θbsin⁡θ,where the polar angle θ∈0,2π. For the sake of convenience, we let(49)xθ,yθ=xc+Acos⁡θ-Bsin⁡θ,yc+Acos⁡θ+Bsin⁡θifϑ=π4,xc+Acos⁡θ+Bsin⁡θ,yc-Acos⁡θ+Bsin⁡θifϑ=-π4,where A,B=(2/2)a,(2/2)b. By the previous discussions, if α=β=μ/(2+ρ), then Γμ,ρ,-μ2/2+ρ where f=Γμ,ρμ/2+ρ,μ/2+ρ is a point ellipse and in view of (49), we may arrive at the following result.Theorem 15.
Supposeρ<2 and α,β≠μ/(2+ρ),μ/(2+ρ) such that f=Γμ,ρα,β. Let xθk,yθkk∈N be generated by theTracking Procedure with input α,β=xθ0,yθ0. For any m∈N,(50)θm+1-θm=arccos⁡-ρ2.Proof.
In view of (49), we first consider the case where ϑ=π/4. By (49), it follows that(51)xk=xc+Acos⁡θk-Bsin⁡θk,xk+1=yc+Acos⁡θk+Bsin⁡θk,as well as(52)xk+1=xc+Acos⁡θk+1-Bsin⁡θk+1,xk+2=yc+Acos⁡θk+1+Bsin⁡θk+1.Note that xc=yc=μ/2+ρ. By (7) and (52), we also have (53)xk+2=μ-ρxk+1-xk=μ-ρyc+Acos⁡θk+1-Bsin⁡θk+1-xc+Acos⁡θk-Bsin⁡θk=yc+A-ρcos⁡θk+1-cos⁡θk+Bsin⁡θk+ρsin⁡θk+1. In view of (52), with respect to xk+2, it follows that(54)-ρcos⁡θk+1-cos⁡θk2+sin⁡θk+ρsin⁡θk+12=1, which leads to(55)cos⁡θk+1cos⁡θk+sin⁡θk+1sin⁡θk=cos⁡θk+1-θk=-ρ2. The case where ϑ=-π/4 can be handled similarly. Hence, (50) is true and this completes the proof.By Theorem15, we can find the necessary and sufficient condition for a solution φ of (7) with ρ<2 to be a periodic sequence.Corollary 16.
Supposeρ<2 and α,β≠μ/(2+ρ),μ/(2+ρ) such that f=Γμ,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then xk,ykk∈N is periodic if and only if(56)1πarccos⁡-ρ2∈Q.Furthermore, suppose π-1arccos-ρ/2=m/n∈Q such that m/n is irreducible and 0<m<n. If m is even, then xk,ykk∈N is n-periodic; otherwise, xk,ykk∈N is 2n-periodic.Proof.
Note that ifρ≠-2 and α,β=μ/(2+ρ),μ/(2+ρ), then by theTracking Procedure and Proposition 1, x1,y1=α,β which implies xk,yk=α,β for all k∈N. Suppose ρ<2 and α,β≠μ/(2+ρ),μ/(2+ρ). By (49), we have(57)xk,yk=xc+Acos⁡θk-Bcos⁡θk,yc+Acos⁡θk+Bsin⁡θk. By Theorem 15, for any k∈N, θΔ=θk+1-θk is constant such that cos⁡θΔ=-ρ/2. Hence θk=θ0+kθΔ. If xk,ykk∈N is periodic with period n∈Z+, then we see that there is some m∈Z+ such that θn=θ0+2mπ. Accordingly, nθΔ=2mπ which implies(58)θΔπ=1πarccos⁡-ρ2=mn∈Q.On the other hand, if (56) holds, then, by the previous discussions and (58), xk,ykk∈N is periodic with period 2n. Note that by Propositions 2 and 4, if xk,ykk∈N is ω-periodic with α,β≠μ/(2+ρ),μ/(2+ρ), then ω>2. Next, suppose r∈Q, where r=π-1arccos⁡-ρ/2. Since 2jπ where j∈Z are periods of cos⁡x, we can first restrict ourselves to 0<r<2, where r≠π. Also, if π<x<2π, then cos⁡x=cos⁡-x=cos⁡2π-x and this allows us to further concentrate on 0<r<1. Note that cos⁡x is bijective on 0,π. Hence, for each ρ such that r∈Q, where 0<r<1, there is a unique pair m,n such that r=m/n, where m and n are coprime. In view of (58), it is easy to see that if m is even, then n is a period of xk,ykk∈N; if m is odd, then xk,ykk∈N is periodic with period 2n. In the former case, we claim that xk,ykk∈N is q-periodic. To this end, if q′<q is a period of xk,ykk∈N, then, by the previous discussions, there is p′∈Z+ such that p′ as well q′ are coprime and r=p′/q′. Hence, p′/q′=p/q which is not possible. The latter case can be handled similarly and this completes the proof.In view of Corollary16, we have found the necessary and sufficient condition for xk,ykk∈N, which is generated by theTracking Procedure with input x0,y0≠μ/(2+ρ),μ/(2+ρ) such that f=Γμ,ρx0,y0 and ρ<2, to be periodic. Let ρ=-1, 0, or 1; for instance, by (58), we take m/n=1/3, 1/2, or 2/3, correspondingly, so that xk,ykk∈N is 6-, 4-, or 3-periodic, respectively. For another example, by (58) again, if m/n=2/5 or 4/5, then xk,ykk∈N is 5-periodic where ρ=(1-5)/2 or ρ=(1+5)/2, respectively.For the analysis of nonnegative or negative solutions of (7) with ρ<2, we need one more result.Theorem 17.
Letr∉Q. Suppose ϕk≡2πkr mod 2π where k∈N. Then ϕk∣k∈N is dense in 0,2π.In view of Theorem17, let ηk≡kr mod 1, where k∈N. Then we see that ηk is dense in 0,1 (see the details in [12] in pages 280 and 281). Suppose s∈0,2π. Since s/2π∈0,1, there exists a subsequence {ηkj}j∈N of ηkk∈N such that ηkj=s/2π, when j→∞, from which it follows that(59)limj→∞⁡2πηkj=s. Accordingly, ϕk is dense in 0,2π as desired.Recall that by the Principal Axes Theorem,Π¯ and Π¯⊥ are the principal axes of Γμ,ρ,f, where ρ≠-2 and f=Γμ,ρα,β, α,β∈R2. For -2<ρ<0 and α,β∈R2 such that f=Γμ,ρα,β, one property of xk,ykk∈N which is generated by theTracking Procedure with input x0,y0=α,β≠μ/(2+ρ),μ/(2+ρ) is revealed as follows.Corollary 18.
Suppose-2<ρ<0 and α,β≠μ/(2+ρ),μ/(2+ρ) such that f=Γμ,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. The set xm,ym∣m∈N cannot lie entirely in one of the half planes divided by Π¯⊥.Proof.
Since-2<ρ<0 and α,β≠μ/(2+ρ),μ/(2+ρ), Γμ,ρ,f is an ellipse. By Corollary 16 and Propositions 2 and 4, if π-1arccos⁡-ρ/2∈Q, then φ is ω-periodic, where ω≥3; otherwise, φ is aperiodic. Suppose π-1arccos⁡-ρ/2∈Q. Then xk,ykk∈N is ω-periodic, where ω≥3 such that xω,yω=α,β. First of all, by theTracking Procedure, we see that xk,yk moves on Γμ,ρ,f in a clockwise manner. Secondly, Γμ,ρ,f is symmetric with respect to x=y. Hence, without loss of generality, we suppose Πα,β≥0. Since xk,yk move on Γμ,ρ,f in a clockwise manner; by theTracking Procedure, we can be sure that for some i∈0,1,2,…,ω-1, HΠ⊥xi,yiHΠ⊥α,β<0. Note that if HΠ⊥α,β=0, then we may investigate the orbit O+E-1φ or OE1φ so that HΠ⊥β,μ-ρβ-α≠0 or HΠ⊥μ-ρα-β,α≠0, respectively. Suppose π-1arccos⁡-ρ/2∉Q. Without loss of generality, we suppose Π⊥α,β<0 and s,t∈Γμ,ρ,f such that Π⊥s,t>0. Then by Theorem 17, there is a subsequence xki,ykii∈N such that limi→∞xki,yki=s,t as required and this completes the proof.The previous discussions about the asymptotic behaviors of the solution of (7) help us to understand the periodicity of φ. To further analyze the positivity of solutions of (7), we require one specific solution χρμ=χkρμk∈Z with μ≠0,  ρ<2, and χ0ρμ,χ1ρμ=μ/2,0. We note that f=Γμ,ρμ/2,0=-μ2/4. Also, some properties of χρμ are useful for the ensuing analysis and they are as follows.Proposition 19.
Forμ,ρ∈R, the lines y=0 and x=0 intercept with Γμ,ρ,-μ2/4 at μ/2,0 and 0,μ/2, respectively. Furthermore, if ρ<2 and μ≠0, then Γμ,ρ,-μ2/4 is a nondegenerate conic section and the lines y=0 and x=0 are tangents of Γμ,ρ,-μ2/4 at the points μ/2,0 and 0,μ/2, respectively.Proof.
First of all, we show thatΓμ,ρ,-μ2/4 meets the y=0 line at μ/2,0. To this end, note that(60)Γμ,ρ,-μ2/4x,0=x2-μx+14μ2=x-12μ2=0and, thus, x=μ/2 is the unique solution to Γμ,ρ,-μ2/4 and y=0. Accordingly, the y=0 line intercept with Γμ,ρ,-μ2/4 at μ/2,0. By the previous discussions, if ρ=2, then Γμ,2,-μ2/4 is a degenerate parabola made up of the x+y-(1/2)μ=0 line; if ρ=-2 and μ=0, then Γ0,-2,0 is a degenerate parabola made up of the x+y=0 line. Note that the Γμ,-2,-μ2/4 is a parabola with μ≠0. Accordingly, if ρ<2 and μ≠0, then Γμ,ρ,-μ2/4 is not a degenerate conic section which implies Γμ,-2,-μ2/4 is tangent to x=0 and y=0 at 0,μ/2 as well as μ/2,0, respectively. The proof is complete.We illustrate Proposition19 by several plots of Γ1,ρ,-μ2/4, namely, ellipses with 0<ρ<2 plotted in Figure 16, ellipses with -2<ρ<0 plotted in Figure 17, and hyperbola with ρ<-2 plotted in Figure 18.Figure 16
TheΓ1,ρ,-1/4, where 0<ρ<2.Figure 17
TheΓ1,ρ,-1/4, where -2<ρ<0.Figure 18
The plot ofΓ1,ρ,-1/4, where ρ=-3.In the following section, we establish the relation between the nonnegative (or negative) solution of (7) and a special solution χρμ=χkρμk∈N in the geometric setting.


## 4. A Special Positive Orbit

Letχρμ=χρμk∈Z be the solution of (7) with χ0ρμ,χ1ρμ=μ/2,0. By (12) and (13), we see that Uχρμ=E-2χρμ which implies χkρμ=χ-k+2ρμ for all m∈Z∖N. Hence, it is found that O⊥+χρμ⊂O+χρμ and by (15), we have(61)Oχρμ=O+χρμ∪xk,yk=vk,uk∣uk,vk∈O+χρμ,k∈N.Note that (61) allows us to obtain Oχρμ from plotting O+χρμ only. In the sequel, we focus on the the “positive” orbit of O+χρμ when the orbit Oχρμ is discussed. By theTracking Procedure with χ0ρμ,χ1ρμ=μ/2,0, we can generate the positive orbit O+χρμ. Next, we establish the connection between the nonnegative solutions and the specific solution χρμ.Let(62)Φμ,ρ=⋃Oξ∣ξis a nonnegative solution of7,Ωμ,ρ=⋃Oξ∣ξis a negative solution of7. Let φ=φkk∈Z of (7) with φm,φm+1=α,β for some m∈Z. Then by the definitions of Φμ,ρ and Ωμ,ρ, we can be sure that(1)
φ is a nonnegative solution if and only if α,β∈Φμ,ρ;(2)
φ is a negative solution if and only if α,β∈Ωμ,ρ.Now, it is our objective to findΦμ,ρ and Ωμ,ρ and in several special cases where ρ<2 and μ>0, we will show that Φμ,ρ and Ωμ,ρ can be expressed as the intersections of certain half planes in R2.Proposition 20.
Letφ=φkk∈Z and c=ckk∈Z be solutions of (7) with φ0,φ1=α,β and c0,c1=0,0. Let a=akk∈Z and b=bkk∈Z be solutions of (7) with μ=0, with a0,a1=0,1, and with b0,b1=1,0. Then we have(63)φk=bkα+akβ+ck.Proof.
For the sake of convenience, we supposek∈N. First of all, if a0,b0,c0=0,1,0 and a1,b1,c1=1,0,0, then φ0,φ1=α,β which implies (63) is true when k≤1. Next, we show that (63) holds for k≥2. To this end, by (7), it follows that(64)φk+2=μ-ρφk+1-φk,(65)=μ-ρbk+1α+ak+1β+ck+1-bkα+akβ+ck(66)=-ρbk+1-bkα+-ρak+1-akβ+μ-ρck+1-ck(67)=bk+2α+ak+2β+ck+2,as required. Note that in view of (66), as well as (67), we have(68)ak+2=-ρak+1-ak,bk+2=-ρbk+1-bk,ck+2=μ-ρck+1-ck.Next, we consider the case where k∈Z∖N. Let ψ=UE-1φ such that ψ0,ψ1=β,α. Thus, we may reformulate (63) for ψm, where m∈N as(69)ψm=akα+bkβ+ck.Furthermore, in view of (68), UE-1a=b and UE-1c=c which implies (63) is true and this completes the proof.Withak,bk,ck defined in Proposition 20, we define the 2-variable function(70)hμ,ρ,kx,y=akx+bky+ckand the set(71)h¯μ,ρ,kx,y=x,y∈R2∣akx+bky+ck=0.Then we construct a set which is an intersection of half planes:(72)Hμ,ρ=⋂k∈Zx,y∈R2∣hμ,ρ,kx,y≥0.We claim that(73)Φμ,ρ=Hμ,ρ.Let φ=φkk∈Z be a solution of (7). Note that by Corollaries 3 and 5 and the definition of (72), we see that Φμ,ρ is nonempty which implies Hμ,ρ≠∅. If φm,φm+1∈Φμ,ρ, for some m∈Z, then, by the definition, φ is nonnegative which implies φn≥0 for all n∈Z. Since each hμ,ρ,k is characterized by φk, in view of (72), we see that φn,φn+1∈Hμ,ρ. On the other hand, if φi,φi+1∉Φμ,ρ, then φ is not nonnegative which implies there is some j∈Z such that φj<0. Accordingly, we have hμ,ρ,jx,y<0 and by (72), it follows that Hμ,ρ=∅ which is not possible.Theoretically, we can calculatehμ,ρ,kx,y for all k∈Z to determine (73) explicitly. Suppose φ=φkk∈Z is an ω-periodic solution of (7) with given μ and ρ. Then we only have to calculate hμ,ρ,kx,y, where k∈0,1,2,…,ω-1, to determine Φμ,ρ (or Ωμ,ρ). Even though this is feasible for us, it is still difficult to illustrate Φμ,ρ explicitly on R2 when ω is sufficiently large. In the sequel, we find Φμ,ρ, where ρ<2 and μ≠0, by one specific Oχρμ⊆Γμ,ρ,-μ2/4. Note that the cases where Φμ,ρ, with ρ≥2 or μ=0, can be handled separately. By means of investigating the specific Oχρμ and Γμ,ρ,-μ2/4 where ρ<2 and μ≠0, Φμ,ρ can be found in an explicit way. Recall that if ρ<2 and μ≠0, then the x-axis and the y-axis are tangent to Γμ,ρ,-μ2/4 at 0,μ/2 as well as μ/2,0, respectively. Let χρμ=χkρμk∈Z be a solution of (7) with χ0ρμ,χ1ρμ=μ/2,0. The tangent line l¯μ,ρ,kx,y of Γμ,ρ,-μ2/4 at xk,yk is given by lμ,ρ,kx,y=0, where(74)lμ,ρ,kx,y=2xk+ρyk-μx+ρxk+2yk-μy+12μ2-μxk-μyk.In addition, we also let l¯μ,ρ,k′ be the line so that l¯μ,ρ,k′ and l¯μ,ρ,k are symmetric with respect to the x=y line. Now, we establish the connection between Hμ,ρ and the set of intersection of half planes divided by l¯μ,ρ,kx,y.Theorem 21.
Givenρ,μ∈R, where ρ<2 and μ>0, then we have(75)Hμ,ρ=⋂k∈Nx,y∈R2∣lμ,ρ,kx,y≤0∪x,y∈R2∣lμ,ρ,k′x,y≤0.Proof.
First of all, we consider the case wherek≥0 and we show that hμ,ρ,kx,y=rlμ,ρ,kx,y, where r=μρ-2/2<0. To this end, we show the following results,(76)rak=2xk+ρyk-μ,rbk=ρxk+2yk-μ,rck=12μ2-μxk-μyk,are true for all k∈Z. Let xk,ykk∈N be O+χρμ with x0,y0=χ0ρμ,χ1ρμ=μ/2,0. Since a0,b0,c0=0,1,0and a1,b1,c1=1,0,0, by (70) and (74), it follows that (75) holds for m≤1. Now, we suppose that for some j∈N such that (75) holds for j and j+1. Then we have seen that for i∈j,j+1(77)rai,bi,ci=2xi+ρyi-μ,ρxi+2yi-μ,12μ2-xi-yi.For cj+2, by (68), (76), and (7), it follows that (78)rcj+2=rμ-ρcj+1-cj=rμ-ρrcj+1-rcj=rμ-ρ12μ2-μxj+1-μyj+1-12μ2-μxj-μyj=12ρμ2-μ2-12ρμ2-12μ2+μρxj+1+xj+μρyj+1+yj=12μ2-μμ-ρxj+1-xj-μμ-ρyj+1-yj=12μ2-μxj+2-μyj+2, as required. Next we consider aj+2 and bj+2. By (68), (77), and (7) again, in the former case,(79)raj+2=r-ρaj+1-aj=-ρraj+1-raj=-ρ2xj+1+ρyj+1-μ-2xj+ρyj-μ=-2ρxj+1+xj-μ-ρρyj+1+yj-μ-μ=2xj+2+ρyj+2-μ, as desired; the latter case can be handled similarly. By induction, we see that for all m∈N, hμ,ρ,kx,y=rlμ,ρ,kx,y is true. Now, we consider the case where k<0. By the reflection and translation invariances, we transform ψ=UE-1φ in Proposition 20 and ζ=UE-1χρμ and by similarly applied arguments for k≥0, we see that hμ,ρ,kx,y=rlμ,ρ,kx,y, for all k<0. Note that Uχρμ=E-2χρμ which implies lμ,ρ,kx,y=lμ,ρ,-k+2′x,y, where k<0. Accordingly, if k≥0, then hμ,ρ,kx,y=rlμ,ρ,kx,y; if k<0, then hμ,ρ,kx,y=rlμ,ρ,kx,y=rlμ,ρ,-k+2′x,y. Finally, since r<0, we see that (75) holds and this completes the proof.Corollary 22.
Letφ=φkk∈Z be a solution of (7) with φm,φm+1=σ,τ, for some m∈Z. If σ,τ∈H¯μ,ρ, then there is some n∈Z such that φn=0.Proof.
By the definitions ofHμ,ρ in (75), if σ,τ∈H¯μ,ρ, then σ,τ is on the tangent line l¯jx,y at χjρμ,χj+1ρμ, for some j∈Z. By (75) again, without loss of generality, we may further suppose that σ,τ∈l¯ix,y, for some i∈N. For the sake of convenience, we let δ=μ-ρτ-σ and χiρμ,χi+1ρμ=s,t such that χi+2ρμ=μ-ρt-s=u. Since σ,τ∈l¯ix,y, by (74), we have(80)liσ,τ=ρs+2t-μτ+2s+ρt-μσ-μs-μt+12μ2=0. Next, we show that τ,δ∈l¯i+1x,y. To this end, we first have(81)li+1x,y=ρt+2u-μy+2t+ρu-μx-μt-μu+12μ2.We apply x,y=τ,δ=τ,μ-ρτ-σ and u=μ-ρt-s to (81) and by simplifying, it is found that li+1τ,δ=liσ,τ=0 which implies τ,δ∈l¯i+1x,y as desired. Since l¯0 and l¯1 are the y=0 and x=0 lines, respectively, by the previous discussions, there is some n∈Z such that φn=0 and this completes the proof.In this section, we introduced the definitions ofΦμ,ρ (or Ωμ,ρ) so that the necessary and sufficient conditions for a solution φ of (7) to be nonnegative (or negative) can be obtained, respectively. Also, in Theorem 21, we discussed the specific solution χρμ=χkρμk∈Z of (7) with χ0ρμ,χ1ρμ=μ/2,0 and the intersection of the closed half planes generated by the tangent lines at χkρμ,χk+1ρμ, for all k∈Z. For the case where ρ<2 and μ>0, we will show that some Φμ,ρ can be expressed explicitly in terms of Hμ,ρ in the sequel. Note that by Proposition 7 and Theorem 8, some Φμ,ρ and Ωμ,ρ with ρ<2 and μ<0 can be derived from the results of the cases where ρ<2 and μ>0. Hence, with all the results derived in the previous discussions together, we are ready to discuss the necessary and sufficient conditions for φ to be nonnegative (or negative).


## 5. Nonnegative and Negative Solutions whenρ>2

Letφ=φkk∈Z be a solution of (7) defined by φ0,φ1=α,β. In the last section, we have completed the discussions about the periodicity of φ and the asymptotic behaviors of O+φ which is generated by theTracking Procedure with input α,β. Note that φk,φk+1=xk,yk=xk,xk+1, where xk,ykk∈N is generated by theTracking Procedure with input x0,y0=α,β. To facilitate the ensuing discussions, we define two sets Θμ,ρ and Υμ,ρ on R2 where(82)Θμ,ρ=x,y∈R2∣L0,ρ,f+x,y≥0,L0,ρ,f-x,y≤0,x≥0,y≥0,Υμ,ρ=x,y∈R2∣L0,ρ,f+x,y≤0,L0,ρ,f-x,y≥0,x<0,y<0.Note that Θμ,ρ is defined by four closed half planes and Υμ,ρ is defined by two open as well as two closed half planes. Also note that by the definitions in (24) and (25), L0,ρ,f+x,y and L0,ρ,f-x,y are both independent of f. Now, we are able to apply these results to derive necessary and sufficient conditions for φ to be nonnegative (or negative).Theorem 23.
Letφ=φkk∈Z be a solution of (7) with ρ>2. Suppose φm,φm+1≠μ/(2+ρ),μ/(2+ρ), for some m∈Z. Then φ is neither a nonnegative solution nor a negative solution.Proof.
Without loss of generality, we letm=0. Let φ0,φ1=α,β such that f=Γμ,ρα,β. Let O+φ=xk,ykk∈N and O⊥+φ=sk,tkk∈N be generated by theTracking Procedure with inputs x0,y0=α,β and s0,t0=β,α, respectively. Then Γμ,ρ,f is a hyperbola (or a degenerate one made up of two lines L¯μ,ρ,f+ and L¯μ,ρ,f-) such that O+φ,O⊥+φ⊂Γμ,ρ,f. For the sake of convenience, we let Pc=μ/(2+ρ),μ/(2+ρ). If α,β=Pc, then, by Proposition 2, xk,yk=sk,tk=Pc, for all k∈N. Suppose α,β≠Pc. By the previous discussions, we first consider the case where α,β∈L¯μ,ρ,f+∪L¯μ,ρ,f-∖Pc. If α,β∈L¯μ,ρ,f+∖Pc, then dα,β,L¯μ,ρ,f+≤dα,β,L¯μ,ρ,f- and by Corollary 11, it is found that xkxk+1=φkφk+1<0 when k→∞; if α,β∈L¯μ,ρ,f-, then we take β,α∈L¯μ,ρ,f+ and we see that β,α∈O⊥+φ so that sksk+1<0 when k→∞as required. Next, suppose Γμ,ρ,f is a hyperbola. Then by Corollary 11 again, xkxk+1=φkφk+1<0 when k→∞. Accordingly, if α,β≠Pc, then φ is neither a nonnegative nor a negative solution and this completes the proof.Next, we consider the case whereρ<-2 and μ∈R. This case is a little more complicated than the one described in Theorem 23, and we give the result of μ=0 first.Theorem 24.
Letφ=φkk∈Z be a solution of (7) with ρ<-2 and μ=0. Suppose φm,φm+1=α,β, for some m∈Z such that f=Γμ,ρα,β. Then φ is a nonnegative solution if and only if α,β∈Θ0,ρ; and φ is a negative solution if and only if α,β∈Υ0,ρ.Proof.
Note that in view of (82), Θ0,ρ is symmetric with respect to the x=y line and Θ0,ρ∖0,0 and Υ0,ρ are symmetric with respect to the x+y=0 line. Let xk,ykk∈N and sk,tkk∈N be generated by theTracking Procedure with inputs x0,y0=α,β and s0,t0=β,α, respectively. Here, Γ0,ρ,f, where f=Γ0,ρα,β, is a hyperbola (or a degenerate one made up of two lines L0,ρ,f+ and L0,ρ,f-). Note that Θ0,ρ is a cone with two rays L0,ρ,f+ and L0,ρ,f- such that u,v≥0, where u,v∈Θ0,ρ. First of all, we show that Φ0,ρ=Θ0,ρ. To this end, we investigate xk,yk and sk,yk directly, and by Theorem 10, it is found that xk,yk,sk,tk≥0 for all k∈N if and only if x0,y0∈Θ0,ρ. Accordingly, we have Φ0,ρ=Θ0,ρ and the argument about Υ0,ρ can be handled similarly. Note that both Θ0,ρ and Θ0,ρ are symmetric with respect to the x=y line. Hence, the cases where k∈Z∖N can be handled by the invariances of translation and reflection. The proof is complete.Next, we discuss the case whereρ<-2 and μ≠0.Theorem 25.
Letφ=φkk∈Z be a solution of (7) with ρ<-2 and μ≠0. Suppose φm,φm+1=α,β for some m∈Z such that f=Γμ,ρα,β.(i)
Supposeμ>0. Then φ is a nonnegative solution if and only if α,β∈Hμ,ρ which is the intersection of the closed half planes defined in (75). Furthermore, φ is a negative solution if and only if α,β∈Υμ,ρ.(ii)
Supposeμ<0.  φ is a nonnegative solution if and only if α,β∈Θμ,ρ; and φ is a negative solution if and only if α,β∈H-μ,ρ′′∖∂H-μ,ρ′′, where H-μ,ρ is the intersection of the closed half planes defined in (75).Proof.
Without loss of generality, we letm=0 and Pc=μ/(2+ρ),μ/(2+ρ). Since f=Γμ,ρα,β, Γμ,ρ,f is a hyperbola (or a degenerate one, two lines L¯μ,ρ,f+ and L¯μ,ρ,f+). Let O+φ=xk,ykk∈N and O⊥+=sk,tkk∈N be generated by theTracking Procedure with inputs x0,y0=α,β and s0,t0=β,α, respectively. Suppose μ>0. Then μ/2+ρ<0. By Theorem 21, it follows that Φμ,ρ=Hμ,ρ as desired. Next, by Corollary 11, if α,β∈Υμ,ρ, then xk,sk<0, for all k∈N. Conversely, if xk,sk<0 for all k∈N, then, by Theorem 10 and Corollary 11, α,β∈Υμ,ρ. Now, we consider the case where μ<0. From the previous discussions, Proposition 7 and Theorem 8, we can be sure that Φμ,ρ=Θμ,ρ. Next, we claim that Ωμ,ρ=H-μ,ρ′′∖∂H-μ,ρ′′, where H-μ,ρ is given by (75). From the previous discussions and Corollary 22, we can be sure that φ is a positive solution if and only if H-μ,ρ∖∂H-μ,ρ. Thus, from Proposition 7 and Theorem 8, it follows that Ωμ,ρ=H-μ,ρ′′∖∂H-μ,ρ′′ and this completes the proof.Corollary 26.
Letφ=φkk∈Z be a solution of (7) with ρ>2.  φ is a nonnegative and periodic solution if and only if μ≥0 and φm,φm+1=μ/(2+ρ),μ/(2+ρ) for some m∈Z; φ is a negative and periodic solution if and only if μ<0 and φn,φn+1=μ/(2+ρ),μ/(2+ρ) for some n∈Z.From Proposition2, Theorem 10, Corollary 11, and Theorems 23, 24, and 25, we can easily see that the statement of Corollary 26 holds.This section shows that bothΦμ,ρ and Ωμ,ρ are nonempty and if φ is not a constant solution of (7) with μ>2, then φ is not periodic. Next, we will discuss Φμ,ρ and Ωμ,ρ where ρ=2.


## 6. Nonnegative and Negative Solutions whenρ=2

Letφ=φkk∈Z be a solution of (7) with ρ=2 and φ0,φ1=α,β. Let f=Γμ,ρα,β. Recall that Oφ is on the conic section Γμ,ρ,f which is a parabola or degenerate parabola that is made up of either one line or two distinct parallel lines. In this section, we now discuss the nonnegative and negative solutions of (7) by investigating ρ=2 and ρ=-2.Theorem 27.
Letφ=φkk∈Z be a solution of (7) with ρ=2. Suppose φm,φm+1=α,β for some m∈Z.(i)
Supposeμ≥0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if(83)α,β∈x,y∈L¯μ,2,-μ2/4+∣x,y≥0.(ii)
Supposeμ<0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if(84)α,β∈x,y∈L¯μ,2,-μ2/4-∣x,y<0.Proof.
From the previous discussions,Γμ,2,f, where f=Γμ,2α,β≥-μ2/4, is a degenerate parabola made up of two lines, L¯μ,2,f+ and L¯μ,2,f- defined by (24) and (25), respectively. Moreover, if f=-μ2/4, then L¯μ,2,f+=L¯μ,2,f-; otherwise, L¯μ,2,f+ and L¯μ,2,f- are two distinct parallel lines. We first consider the case where μ≥0. By Theorem 12 and theTracking Procedure, if α,β≥0 such that f=Γμ,2α,β=-μ2/4, then φ is periodic with period 2 and φ2≥0 which implies φ is a nonnegative solution. Conversely, if φ is a nonnegative solution, then α,β≥0 and by Theorem 12, f=-μ2/4 which implies α,β∈L¯μ,2,-μ2/4+=L¯μ,2,-μ2/4-. Accordingly, we see that(85)Φμ,2=x,y∈L¯μ,2,-μ2/4+∣x,y≥0. On the other hand, by Theorem 12 again, we can be sure that φ cannot be a nonpositive solution which implies Ωμ,ρ is empty. Next, we suppose μ<0. From the previous discussions and Proposition 7, if φ is a nonnegative solution then -φ is a solution of (7) with -μ>0 which is not possible. Also, L¯μ,2,-μ2/4+ and L¯μ,2,-μ2/4- are symmetric with respect to line x+y=0 and from Proposition 7 and Theorem 8, it follows that(86)Ωμ,ρ=x,y∈L¯μ,2,-μ2/4-∣x,y<0 and this completes the proof.Theorem 28.
Letφ=φkk∈Z be a solution of (7) with ρ=-2. Suppose φm,φm+1=α,β for some m∈Z.(i)
Supposeμ=0. Then φ is a nonnegative solution if and only if α,β≥0 and α,β∈Π¯ which is the x=y line; furthermore, φ is negative solution if and only if α,β∈Π¯ and α,β<0.(ii)
Supposeμ>0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if α,β∈Hμ,-2 which is the intersection of the closed half planes defined by (75).(iii)
Supposeμ<0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if(87)α,β∈H-μ,-2′′∖∂H-μ,-2′′,whereH-μ,-2 is the intersection of the closed half planes defined by (75).Proof.
Recall thatΠ¯ is the line x=y which is one principal axis of Γμ,ρ,f, where f=Γμ,ρα,β. Suppose μ=0. By Theorem 13, if α,β∉Π¯, then φ is neither a nonnegative nor a nonpositive solution. If α,β∈Π¯, then φ is 1-periodic and we see that (i) holds by direct verification. Suppose μ>0. Then by Theorem 21, we can be sure that Φμ,-2=Hμ,-2 as required and Hμ,-2 is depicted in Figure 19. For the case of (iii), from Proposition 7 and Theorem 8, it follows that (87) is true and this completes the proof.Figure 19
The setΦ1, −2.In view of Theorem28. we have shown that Φμ,-2=Hμ,-2 which is the intersection of the half planes defined by the tangent lines at all the points of Oχρμ on the parabola Γμ,-2,-μ2/4. Figure 19 illustrates Φ1,-2 where the red-dashed curve is the parabola Γ1,-2,-1/4. The discussions of the cases where ρ≥2 are done and move forward to analyze the last cases where ρ<2.


## 7. Nonnegative and Negative Solutions whenρ<2

In the previous discussions, we have analyzed the cases whereρ≥2 and derived the necessary and sufficient conditions for a solution φ to be nonnegative (or negative). Let φ=φkk∈Z be a solution of (7). From our analyses of the case where ρ≥2, the periodicity of φ has no significant role in the analysis (albeit φ may be periodic with least period ω≤2); yet the periodicity of φ is of great importance when discussing the case where ρ<2. In Corollary 16, we illustrated the necessary and sufficient conditions for φ to be a periodic solution of (7) with ρ<2 and we will discuss the case where μ=0,  μ>0, and μ<0 in the sequel.Theorem 29.
Letφ=φkk∈Z be a solution of (7) with ρ<2 and μ=0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if φ is the null solution.Proof.
Letφ0,φ1=α,β such that f=Γ0,ρα,β. Let xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Suppose α,β=0,0. From the previous discussions, f=Γ0,ρ0,0=0 which implies Γ0,ρ,0 is a point ellipse. On the other hand, by Proposition 2, φ0,φ1=0,0 which leads to φk=0, for all k∈N and, thus, φ is a nonnegative solution with least period 1. Now, we suppose α,β≠0,0. Then Γ0,ρ,f is an ellipse and we show that φ is neither a nonnegative nor a negative solution. To this end, we consider 0≤ρ<2 and -2<ρ<0. Suppose 0≤ρ<2. By Proposition 6, φ cannot be a nonpositive solution which implies φ is not negative. If φ is a nonnegative solution, then by Proposition 7, -φ is a nonpositive solution which is a contradiction. The case where -2<ρ<0 can be handled similarly and this completes the proof.Next, we consider the corresponding results whenρ<2 and μ≠0.Theorem 30.
Letφ=φkk∈Z be a solution of (7) with ρ<2 and μ≠0. Suppose π-1arccos⁡-ρ/2∉Q and let φm,φm+1=α,β for some m∈Z.(i)
Supposeμ>0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if(88)α,β∈x,y∈R2∣Γμ,ρx,y≤14μ2,which is the interior of the ellipseΓμ,ρ,-μ2/4.(ii)
Supposeμ<0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if α,β∈Λ-μ,ρ′′, where(89)Λ-μ,ρ=x,y∈R2∣Γ-μ,ρx,y≤14μ2∖Oχρμ.Proof.
Without loss of generality, we letm=0. Also, for the sake of convenience, let Pc=μ/(2+ρ),μ/(2+ρ) and r=π-1arccos⁡-ρ/2. Also, let O+φ=xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Then we consider two cases where μ>0 and μ<0. Here, if α,β=Pc, then, by the previous discussions, Γμ,ρ,-μ2/2+ρ is a point ellipse so that xk,yk=Pc, for all k∈N. Note that Γμ,ρ,-μ2/4 is tangent to lines x=0 and y=0 at 0,μ/2 and μ/2,0, respectively. Suppose μ>0. We first show that φ cannot be a nonpositive solution. If ρ≥0, then, by Proposition 6, φ is not a nonpositive solution. If φ is nonpositive with ρ<0, then, by the previous discussions, α,β≠Pc and by Corollary 18, there is some m∈N such that xm≥0 or ym≥0. Furthermore, if α=β=0, then, by (7), φ2=μ>0. Accordingly, φ is not a nonpositive solution when μ>0.
Now, we show(90)Φμ,ρ=x,y∈R2∣Γμ,ρx,y≤14μ2,where ρ>0. Note that s,t≥0, where s,t∈x,y∈R2∣Γμ,ρx,y≤(1/4)μ2. If (88) holds, then, by Theorem 8, we can be sure that Γμ,ρ,f⊆Φμ,ρ, where f=Γμ,ρα,β. Accordingly, φm≥0 for all m∈Z. On the other hand, suppose (88) does not hold. Since r∉Q, by Theorem 17, there is some i∈N such that l¯iα,β>0 (or some j∈N such that l¯j′α,β>0) which implies φ is not nonnegative. Hence, (90) is true. Next, suppose μ<0. First of all, by the previous discussions and Corollary 22, we see that φ is a positive solution if and only if α,β∈Λ-μ,ρ. Thus, from Proposition 7 and Theorem 8, it follows that Ωμ,ρ=Λ-μ,ρ′′ as required.
The proof is complete.Theorems30 and 17 and Proposition 7 help us to analyze the necessary and sufficient conditions for an aperiodic solution φ of (7) to be nonnegative (or negative). For the case when φ is periodic, we have to resort to the help of the specific solution χρμ of (7).Theorem 31.
Letφ=φkk∈Z be a solution of (7) with ρ<2 and μ≠0. Suppose π-1arccos⁡-ρ/2∈Q and φm,φm+1=α,β, for some m∈Z. Then φ is periodic and we have the following.(i)
Supposeμ>0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if α,β∈Hμ,ρ which is the intersection of the closed half planes defined by (75).(ii)
Supposeμ<0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if(91)α,β∈H-μ,ρ′′∖∂H-μ,ρ′′,whereH-μ,ρ is the intersection of the closed half planes defined by (75).Proof.
Without loss of generality, we letm=0. Also, let Oφ=xk,ykk∈N be generated by theTracking Procedure with input x0,y0=α,β. Suppose μ>0. From the previous discussions, if α,β=μ/(2+ρ),μ/(2+ρ), then φ is a nonnegative solution with least period 1. If α,β≠μ/(2+ρ),μ/(2+ρ), then, by Corollary 16, φ is periodic and Γμ,ρ,f is an ellipse, where f=Γμ,ρα,β. Also, by Theorem 21, we can be sure that Φμ,ρ=Hμ,ρ as desired. Next, we show that φ is not a nonpositive solution. First of all, by Proposition 6, φ is not nonpositive when ρ≥0. Secondly, suppose φ is a nonpositive solution of (7) with ρ<0. Then by Corollary 18, there is some m∈N such that xm or ym≥0. If xm=ym=0, then, by (7), xm+1=μ>0 which is a contradiction. Accordingly, φ is not a nonpositive solution.
Supposeμ<0. By Corollary 22, Proposition 7, and Theorem 8, it follows that (91) holds. Also, if φ is a nonnegative solution of (7) with μ<0, then, by Proposition 7, -φ is a solution of (7) with -μ>0 which is not possible and this completes the proof.Corollary 32.
Letφ=φkk∈Z be a solution of (7) with ρ<2 and μ>0. Suppose r=π-1arccos⁡-ρ/2=n/m∈Q such that 0<r<1 and r is irreducible. Then Φμ,ρ is a closed polygon with ω sides which is circumscribed around the ellipse Γμ,ρ,-μ2/4 and ω is either 2n or n depending on whether m is even or odd, respectively.Proof.
In view of Theorem21, Φμ,ρ=Hμ,ρ which is defined by the tangent lines of Γμ,ρ,-μ2/4. Furthermore, all the lines are tangent to Γμ,ρ,-μ2/4 at the points which are generated by a specific solution χρμ of (7) with χρμ0,χρμ1=μ/2,0. Since π-1arccos⁡-ρ/2∈Q, by Corollary 16, χρμ is ω-periodic where ω is dependent on n and the parity of m. Note that ω>2. Accordingly, we see that Φμ,ρ is defined by ω distinct lines. To see Φμ,ρ is a closed polygon with ω sides which is circumscribed around Γμ,ρ,-μ2/4, it is sufficient to show that Φμ,ρ is closed. If ρ=0, then χρμ is 4-periodic such that(92)Φμ,ρ⊂x,y∈R2∣0≤x≤μ,0≤y≤μ; if 0<ρ<2, then, by (70), hμ,ρ,2x,y=μ-ρx-y such that(93)Φμ,ρ⊂x,y∈R2∣μ-ρx-y≥0,x≥0,y≥0. Accordingly, Φμ,ρ is closed, where 0≤ρ<2. Suppose -2<ρ<0. To see Φμ,ρ is closed, it is sufficient to show that there is some i∈N such that Π⊥χρμi,χρμi+1>0. Since Π⊥χρμ0,χρμ1<0, in view of Corollary 18, there is some j∈N such that Π⊥χρμj,χρμj+1>0 as required and this completes the proof.Corollary32 gives an explicit description of Φμ,ρ, where ρ<0, and two examples for the cases ρ,μ=1,1 and ρ,μ=-1,1 are in Figures 20 and 21 respectively.Figure 20
The setΦ1,1.Figure 21
The setΦ1,-1.


## 8. Conclusions

In the previous sections, we have analyzed the nonnegative (or negative) solutionφ=φkk∈Z of (7) by investigating the pairs φm,φm+1 for all m∈Z which are on a conic section (or degenerate one). Specifically, the orbit Oφ is the collection of φm,φm+1 for all m∈Z which is a subset of a conic section and by theTracking Procedure, φn+1,φn+2 is uniquely determined by giving φn,φn+1. By means of this geometric method, many qualitative properties of  (7) can be obtained. In particular, we obtain the necessary and sufficient conditions for φ to be a nonnegative (or negative) solution.Now, we summarize the necessary and sufficient conditions forφ to be a nonnegative (or negative) periodic solution of (7) as follows (recall that for a set S⊂R2, S′′ is the set on R2 such that S and S′′ are symmetric with respect to the x+y=0 line).(1)
Supposeρ>2.  φ is a nonnegative and periodic solution if and only if ρμ≥0, and for some i∈Z, φi,φi+1=μ/(2+ρ),μ/(2+ρ). Furthermore, such φ must satisfy φj=μ/2+ρ for all j∈Z (and hence is 1-periodic).(2)
Supposeρ<2.  φ is a nonnegative and periodic solution if and only if μ>0(94)1πarccos⁡-12ρ=mn∈Q,(wherem/n is irreducible and 0<m<n) and for some i∈Z, φi,φi+1∈Hμ,ρ which, by Corollary 32, is a closed ω-side polygon where ω is either n or 2n depending on whether m is even or odd, respectively. Furthermore, such φ must either satisfy φj,φj+1=μ/(2+ρ),μ/(2+ρ) for all j∈Z (and hence is 1-periodic) or φ is ω-periodic.(3)
Supposeρ=2. If ρ=2, then φ is a nonnegative and periodic solution if and only if μ≥0, and for some m∈Z, φm,φm+1≥0 and φm,φm+1 is on the x+y=μ/4 line; if ρ=-2, then φ is a nonnegative and periodic solution if and only if μ=0, and for some n∈Z, φn,φn+1≥0, and φn,φn+1 lie on the x=y line. Furthermore, such φ is either 1-periodic if φm=φm+1 (and 2-periodic if φm≠φm+1) when ρ=2 or 1-periodic when ρ=-2.As for the negative periodic solutions, we have the following.(1)
Supposeρ>2.  φ is a negative and periodic solution if and only if ρμ<0 and for some i∈Z, φi,φi+1=μ/(2+ρ),μ/(2+ρ). Furthermore, such φ must satisfy φj=μ/2+ρ for all j∈Z (and hence is 1-periodic).(2)
Supposeρ<2.  φ is a negative and periodic solution if and only if μ<0(95)1πarccos⁡-12ρ=mn∈Q,(wherem/n is irreducible and 0<m<n) and for some i∈Z, φi,φi+1∈H-μ,ρ′′∖∂H-μ,ρ′′, where by Corollary 32, H-μ,ρ is a closed ω-side polygon and ω is either n or 2n depending on whether m is even or odd, respectively. Furthermore, such φ must either satisfy φj,φj+1=μ/(2+ρ),μ/(2+ρ) for all j∈Z (i.e., φ is 1-periodic) or φ is ω-periodic.(3)
Supposeρ=2. If ρ=2, then φ is a negative and periodic solution if and only if μ<0, and for some m∈Z, φm,φm+1<0 and φm,φm+1 is on the x+y=μ/4 line; if ρ=-2, then φ is a negative and periodic solution if and only if μ=0, and for some n∈Z, φn,φn+1<0 and φn,φn+1 lie on the x=y line. Furthermore, such φ is either 1-periodic if φm=φm+1 (and 2-periodic if φm≠φm+1) when ρ=2 or 1-periodic when ρ=-2.With the help of theTracking Procedure, the behaviors of the solution φ of (7) can be visualized (instead of investigating each term of φ by the method of general solutions). By means of our geometric method, we may save lots of calculation and most properties of φ which may be obtained by the general solutions are also available from our geometric discoveries.In [1–3], we conducted complete analyses on the periodic solutions (7) with μ=1 and ρ∈-1,0,1. By means of breaking a solution ψ into two subsequences, the companion sequence and the error sequence, we successfully show that both the companion sequence and error sequence are periodic so that ψ is periodic. Note that for ρ=-1, 0, and 1, we have arccos-ρ/2=π/3, π/2, and 2π/3, respectively.The nice thing here is that by means of our geometric method, not only can we track the orbit of a solutionφ on R2 easily but also we may obtain the conditions for φ to be nonnegative (or negative) quickly (especially for the case where ρ<2 and φ is periodic). By [1–3], we have observed the importance of nonnegative solutions of (7) with ρ=0, 1, and -1, respectively. However, we also realize the difficulties of analyzing the solution ζ of (7) with ρ∉-1,0,1 since the breaking of ζ into companion and error sequences is not applicable for all the solutions anymore.Finally, in this paper we have only touched upon one class of steady state solutions of our original neural network. Much work has to be done before the dynamic behaviors of its solutions can be fully understood. Our contributions here, however, are that elementary geometric methods can be used to explain some of the complex phenomena obtained from simulations (see, e.g., Figures1 and 2) and these methods are also expected to be applied in other piecewise smooth dynamic systems.



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