Convergence, Periodicity and Bifurcations for the Two-parameter Absolute-Difference Equation

Journal of Difference Equations and pplications ISSN: 123-6198 (Print) 1563-512 (Online) Journal homepage: http://www.tandfonline.com/loi/gdea2 Convergence, Periodicity and Bifurcations for the Two-parameter bsolute-difference Equation C.M. Kent & H. Sedaghat To cite this article: C.M. Kent & H. Sedaghat (24) Convergence, Periodicity and Bifurcations for the Two-parameter bsolute-difference Equation, Journal of Difference Equations and pplications, 1:9, 817-841, DOI: 1.18/12361941168549 To link to this article: http://dx.doi.org/1.18/12361941168549 Published online: 22 ug 26. Submit your article to this journal rticle views: 42 View related articles Citing articles: 4 View citing articles Full Terms & Conditions of access and use can be found at http://www.tandfonline.com/action/journalinformation?journalcode=gdea2 Download by: [VCU Libraries Serials] Date: 2 February 217, t: 12:43

Journal of Difference Equations and pplications, Vol. 1, No. 9, 1 ugust 24, pp. 817 841 Convergence, Periodicity and Bifurcations for the Two-parameter bsolute-difference Equation C.M. KENT and H. SEDGHT* Department of Mathematics, Virginia Commonwealth University, Richmond, V 23284-214, US (Received 21 December 23; In final form 11 February 24) The two-parameter absolute-difference equation x nþ1 ¼jax n 2 bx n21 j is studied. Based on the parameter values a, b and a pair of initial values, we consider the existence and bifurcations of solutions having one or more of the following properties: (i) unbounded, (ii) convergent (to zero or to a positive constant), (iii) monotonic, (iv) periodic and (v) non-periodic oscillatory. The semiconjugate first order equation satisfied by the ratios {x n =x n21 } is used to significant advantage for points ða; bþ in certain regions of the parameter plane. Some open problems and conjectures are presented. Keywords: Global asymptotic stability; Oscillatory; Bifurcations; Ratios; Semiconjugate; The golden mean MS Subject Classification Numbers: 3911; 391 Consider the second-order difference equation x nþ1 ¼jax n 2 bx n21 j; a; b $ ; n ¼ ; 1; 2;...: ð1þ n equation such as this may appear implicitly in smooth difference equations (or difference relations) that are in the form of e.g. quadratic polynomials. In such cases, the solutions of the quadratic equation include the solutions of Eq. (1) so statements can be made with regard to the existence and stability of the solutions for the more general, smooth equations based on the solutions of Eq. (1). We may assume that the initial values x 21, x in Eq. (1) are non-negative and for non-triviality, at least one is positive. Dividing both sides of Eq. (1) by x n we obtain a ratios equation x nþ1 x n ¼ a 2 bx n21 ; x n which can be written as r nþ1 ¼ a 2 b r ; n ¼ ; 1; 2;...; ð2þ n *Corresponding author. E-mail: hsedagha@vcu.edu Journal of Difference Equations and pplications ISSN 123-6198 print/issn 1563-512 online q 24 Taylor & Francis Ltd http://www.tandf.co.uk/journals DOI: 1.18/12361941168549

818 C.M. KENT ND H. SEDGHT where we define r n ¼ x n =x n21 for every n $ : We may think of Eq. (2) as the recursion r nþ1 ¼ f ðr n Þ where f is the piecewise smooth mapping f ðrþ ¼ a 2 b r ; r. : In this format, solutions {r n } of Eq. (2) can be written as r n ¼ f n ðr Þ for n $ 1: Note that Eq. (2) is a first order equation and a one-dimensional semicongjugate of Eq. (1) with the ratio x/y as a link map; see Ref. [6]. lso see Refs. [2,3] for an application of ratio links. dditional background material for this paper may be found in Refs. [1,6]. The special case where a ¼ b ¼ 1 in Eqs. (1) and (2) is studied in Ref. [9] where the following is proved: Theorem numbers. Let a ¼ b ¼ 1 in Eq. (1) and let Q þ denote the set of all non-negative rational (a) If x =x 21 Ó Q þ then the corresponding solution {x n } of Eq. (1) converges to zero. (b) If x =x 21 [ Q þ or x 21 ¼ then the corresponding solution {x n } of Eq. (1) has period 3 eventually and for all large n it has the form {; a; a; ; a; a; ;...}; where a. : (c) Equation (2) has a p-periodic solution {r 1 ;...; r p } for every p 3 given by r 1 ¼ 1 þ p ffiffi pffiffiffi 5 5 2 1 ; r 2 ¼ pffiffi ð p ¼ 2Þ; 2 5 þ 1 r 1 ¼ 1 h qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i 2 y p24 þ y 2 p24 þ 4y p24y p21 ; r k ¼ y k24r 1 2 y k22 ; 2 # k # p; ð p $ 4Þ; y k23 2 y k25 r 1 where y n is the n-th Fibonacci number; i:e: y nþ1 ¼ y n þ y n21 for n $22 where we define y 23 ¼ 21; y 22 ¼ 1: (d) The mapping f has a scrambled set S; hence, if {x n } is a solution of Eq. (1) with initial value ratio x =x 21 [ S; then the sequence {x n =x n21 } of consecutive ratios is chaotic. lthough Eq. (1) is a rather simple equation, Theorem suggests that for some values of the parameters a, b it exhibits interesting dynamics. In this paper, we study the asymptotic behavior of Eq. (1) for different parameter values a; b $ : For comparison, it is interesting to note that Eq. (1) relates to certain other simple equations such as u nþ1 ¼ au n þ bu n21 v nþ1 ¼ max{av n ; bv n21 } w nþ1 ¼ min{aw n ; bw n21 } ð3þ through the relations ja 2 bj ¼2 max{a; b} 2 ða þ bþ ¼max{a; b} 2 min{a; b}: In contrast to Eq. (1), each of the equations in (3) produces solutions with simple behaviors for all parameter values a, b. Some indication of the major difference between these equations and Eq. (1) may be had by looking at their semiconjugate factors through ratio links: f u ðrþ ¼a þ b r ; f vðrþ ¼max a; b r ; f w ðrþ ¼min a; b : r

TWO-PRMETER BSOLUTE-DIFFERENCE EQN 819 It is evident from a quick examination of these mappings that only the factor f which was defined for Eq. (1) is capable of generating complex behavior on the half-line. Indeed, an important feature of f is its nonsmooth critical point at b/a whose pre-image is the origin, namely, the point of discontinuity of f. This source of complexity is lacking in the factors f u, f v and f w all of which are non-increasing maps. THE UNIT SQURE a; b # 1 We first consider the parameter values ða; bþ in the unit square. The function f (or equivalently, its vectorization V f ) in Lemma 1 below is said to be a weak contraction on the set a ; see Ref. [6] for a general theory of weak contractions and weak expansions as well as a proof of the lemma. Lemma 1 Let f : R m! R be continuous and let x be an isolated fixed point of x nþ1 ¼ f ðx n ; x n21 ;...; x n2m Þ: Let V f ðu 1 ;...; u m Þ¼ðf ðu 1 ;...; u m Þ; u 1 ;...; u m21 Þ and for a [ ð; 1Þ define the set a ¼ fðu 1 ;...; u m Þ : j f ðu 1 ;...; u m Þ 2 xj g # a max{ju 1 2 xj;...; ju m 2 xj}: If S is a subset of a such that V f ðs), S and ðx;...; xþ [ S; then ðx;...; xþ is asymptotically ðin fact, exponentiallyþ stable relative to S. Corollary 1 If a; b, 1 then the origin is globally asymptotically stable for Eq. (1). Proof Define f ðx; yþ ¼jax 2 byj and note that for x; y [ ½; 1Þ 2 we have f ðx; yþ # max{ax; by} # max{a; b}max{x; y}: Hence f is a weak contraction on ½; 1Þ 2 and Lemma 1 applies. The proof is completed upon recalling that for each solution {x n } of Eq. (1) x n $ for all n $ 1: Lemma 2 Let, a; b # 1 and either a ¼ 1 or b ¼ 1: For each solution {x n } of Eq. (1) if x k. x k21. for some k $ then x n, x k for all n. k: Proof First, let b ¼ 1 and, a # 1: If ax k. x k21 then x kþ1 ¼ ax k 2 x k21, ax k # x k : ð4þ From this we infer that ax kþ1, ax k # x k So x kþ2 ¼ x k 2 ax kþ1, x k : ð5þ Now from Eqs. (4) and (5) it follows that x kþ3 # max{ax kþ2 ; x kþ1 }, x k : The last step is easily extended by induction to all n. k: Next, assume that ax k # x k21 : Then x kþ1 ¼ x k21 2 ax k, x k21, x k and it follows that ax kþ1, ax k # x k : This again implies Eq. (5) and we argue as before.

82 C.M. KENT ND H. SEDGHT Now, assume that a ¼ 1 and, b # 1: Since x k. x k21 $ bx k21. by hypothesis, it follows that, x kþ1 ¼ x k 2 bx k21, x k : If x kþ1 $ bx k ; then x kþ2 ¼ x kþ1 2 bx k, x kþ1 so that x kþ3 # max{x kþ2 ; bx kþ1 }, x k : Inductively, x n. x k for n. k: If x kþ1, bx k then x kþ2 ¼ bx k 2 x kþ1, bx k # x k ; which implies x kþ3, x k and by induction the proof is completed. Theorem 1 For all ða; bþ [ ½; 1Š 2 except at the three boundary points ð1; 1Þ; ð1; Þ and ð; 1Þ; the origin is globally asymptotically stable for Eq. (1). Proof In light of Corollary 1 it remains only to consider the boundaries where either a ¼ 1 or b ¼ 1: For such points ða; bþ; first assume that, ab, 1 and let {x n } be a solution of Eq. (1). Either x n, x n21 for all n in which case x n converges to zero monotonically, or there is k 1 $ such that x k1. x k1 21. : In the latter case, Lemma 2 implies that x n, x k1 for all n. k 1 : If the sequence {x n } is not eventually decreasing, then there is an increasing sequence k i of positive integers such that and for i ¼ 1; 2; 3;... x k1. x k2.. x ki. x n, x ki if k i, n # k iþ1 : These facts imply that x n! as n!1: Stability of the origin follows from Lemma 2 also since x n # max{x ; x 21 } for all n $ 1: It remains for us to examine the three boundary points mentioned in the statement of the theorem. t ða; bþ ¼ð1; 1Þ Theorem shows that the origin is not globally attracting. t ða; bþ ¼ð1; Þ; Eq. (1) reduces to the trivial equation x nþ1 ¼ x n whose solutions are the constants x.tða; bþ ¼ð; 1Þ each solution of x nþ1 ¼ x n21 trivially has period 2: {x 21 ; x }: Since for a ¼ ð1þ reduces to x nþ1 ¼ bx n21 and for b ¼ it reduces to x nþ1 ¼ ax n ; it is evident that along the rays from the origin through ð; 1Þ and through ð1; Þ all solutions of Eq. (1) are unbounded when b. 1 or a. 1 respectively. Therefore, the points ð; 1Þ and ð1; Þ are bifurcation points on their respective rays. The same is true for the ray through ð1; 1Þ; though this is much harder to verify. In the remainder of this paper we consider the behavior of solutions of Eq. (1) when ða; bþ is outside the unit square. THE REGION b # a 2 =4 The semiconjugate factor map f on the half-line can be used to further explore the behavior of solutions in both convergent and non-convergent cases. The mapping f always has a fixed point r through its left half f L ðrþ ¼ b r 2 a;, r # b a :

TWO-PRMETER BSOLUTE-DIFFERENCE EQN 821 Solving f L ðrþ ¼r gives the value of r as r ¼ 1 hpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i a 2 2 þ 4b 2 a : It may be noted that r is an unstable fixed point for all positive values of a, b, since f L ðrþ b ¼ r 2 ¼ b b 2 ar. 1: The right half of f, namely, f R ðrþ ¼a 2 b r ; r $ b a can have up to two fixed points if the equation f R ðrþ ¼r possesses real solutions. Solving the latter equation yields two values r 1 ¼ 1 h 2 a 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 2 4b i; r 2 ¼ 1 h 2 a þ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i a 2 2 4b both of which are real if and only if a 2 $ 4b: Stated differently, f has up to two additional fixed points for all points ða; bþ on or below the parabola b ¼ a 2 =4 in the parameter plane. On this parabola, r 1 ¼ r 2 ¼ a 2 and the single fixed point of f is semistable. In the sub-parabolic region b, a 2 =4; b a, r 1, r 2 with r 1 unstable and r 2 asymptotically stable as may be readily verified. The tangent bifurcation of the factor map f that produces r 1 ; r 2 when a parameter point ða; bþ crosses the parabola b ¼ a 2 =4 corresponds to significant behavioral changes in the solutions of Eq. (1) as well. In this section, we examine the region below the parabola, beginning with the following lemma whose straighforward proof is omitted. Lemma 3 Let b # a 2 =4: (a) r 2 # a with equality holding if and only if b ¼ : (b) Let a, 2: Then r 1, 1; also r 2, 1 if and only if b. a 2 1: Further, r 2 ¼ 1 if b ¼ a 2 1: (c) Let a $ 2: Then r 2 $ 1; also r 1 # 1 if and only if b # a 2 1: Further, r 1 ¼ 1 if b ¼ a 2 1: In Lemma 3 notice that the line b ¼ a 2 1 is tangent to the parabola b ¼ a 2 =4 at a ¼ 2: Lemma 4 Let b # a 2 =4 and let {x n } be a solution of Eq. (1) such that x k $ r 1 x k21 for some integer k $ : (a) If b a 2 =4 then there are real constants c 1, c 2 such that x n ¼ c 1 ðr 1 Þn þ c 2 ðr 2 Þn ; n $ k: ð6þ

822 C.M. KENT ND H. SEDGHT (b) If b ¼ a 2 =4 then there are real constants c 1 ; c 2 such that with r 1 ¼ r 2 ¼ a=2: x n ¼ðc 1 þ c 2 nþða=2þ n ; n $ k ð7þ Proof By hypothesis, r k ¼ x k =x k21 $ r 1 : If r n ¼ f ðr n21 Þ for n. k; then since r n $ r 1 for n $ k; we may write x nþ1 x n ¼ r nþ1 ¼ f R ðr n Þ¼a 2 bx n21 x n i.e. x nþ1 ¼ ax n 2 bx n21 for n $ k: The eigenvalues of the preceding linear equation are none other than r 1 and r 2 so Eq. (6) or Eq. (7) holds as appropriate for n $ k and suitable constants c 1 ; c 2. Theorem 2 Let a, 2: (a) If a 2 1, b # a 2 =4 then every nontrivial solution of Eq. (1) converges to zero eventually monotonically. (b) If b ¼ a 2 1 then every solution of Eq. (1) converges to a non-negative constant eventually monotonically. (c) If b, a 2 1 with a. 1 then any solution {x n } of Eq. (1) for which x k. r 1 x k21 for some integer k $ is unbounded and stricly increasing eventually. On the other hand, if a solution of Eq. (1) is bounded, then it is strictly decreasing to zero; such solutions do exist. Proof (a) First, let {x n } be a positive solution of Eq. (1), i.e. x n. for n $21: Then the sequence {r n } where r n ¼ f ðr n21 Þ and r ¼ x =x 21 is well defined for all n.ifx n, r 1 x n21 for all n $ then x n! as n!1because r 1, 1 by Lemma 3. Otherwise, x k $ r 1 x k21 for some k $ so by Lemma 4, Eq. (6) or Eq. (7) holds as appropriate, and since r 2, 1 the proof is completed. Next, suppose that x m ¼ for some m $21: Then x mþ2 ¼ ax mþ1 and therefore, r mþ2 ¼ x mþ2 ¼ a: x mþ1 By Lemma 3, r 2 # a so r mþ2 $ r 2. r 1 : Letting k ¼ m þ 2 and applying the above argument once more we obtain Eq. (6) or Eq. (7) and complete the proof. (b) Here r 2 ¼ 1 and r 1, 1; the proof is similar to that for Part (a). (c) With r 2. 1. r 1 ; if {x n} is a solution with r k ¼ x k =x k21. r 1 for some k, then apply Lemma 4 with Eq. (6) to show that {x n } is unbounded and eventually increasing. This argument also shows that if {x n } is bounded, then it has to be strictly decreasing with x n # r 1 x n21 for all n $ : It is clear in this case that x n! as n!1: Finally, to verify that the latter type of solution does exist, let x =x 21 ¼ r belong to either of the sets B ¼ [1 i¼ f 2i ðrþ; B 1 ¼ [1 i¼ f 2i ðr 1 Þ

of backward iterates of r or r 1 respectively under the inverse map f 21. Then r n ¼ r, 1 or r n ¼ r 1, 1 for all large n and x n converges to zero exponentially. Remark 1 The region in the parameter plane that is mentioned in Theorem 2(a) contains points outside the unit square. In addition to being globally attracting in this region, it is also not hard to see that the origin is stable as well, using the facts that r 2, 1 and that the iterates f take a fixed number (i.e. independent of x, x 21 ) to drop below 1 on their way to r 2 : Relating to Theorem 2(b), if b ¼ a ^ 1 then as might be expected the origin is not the unique fixed point of Eq. (1). Lemma 5 (a) r $ 1 if and only if b $ a þ 1: lso r p ¼ 1 if b ¼ a þ 1: (b) If r $ 1 and b # a 2 =4 then a $ 2 þ ffiffi 8 < 4:828: p Note that the line b ¼ a þ 1 intersects the parabola at a ¼ 2 þ ffiffi 8 : Further, on both lines b ¼ a ^ 1; (1) is degenerate in the sense that if x 21 ¼ x then x n ¼ x for all n $ : In particular, the origin is not the only fixed point of Eq. (1). Theorem 3 Let a $ 2 and b # a 2 =4: (a) ny solution {x n } of Eq. (1) for which x k. r 1 x k21 for some integer k $ is unbounded and strictly increasing eventually. (b) If b, a þ 1; then there are solutions of Eq. (1) that converge to zero eventually monotonically, and if b ¼ a ^ 1; then there are also solutions that are eventually constant and positive. Proof TWO-PRMETER BSOLUTE-DIFFERENCE EQN 823 (a) This is proved similarly to Theorem 2ðcÞ, the only difference being that here it is possible that r 1 $ 1: (b) If b, a þ 1; then by Lemma 5, r, 1soifx =x 21 ¼ r [ B where B is the set of backward iterates defined in the proof of Theorem 2(c), then x n! eventually monotonically. If b ¼ a þ 1; then r ¼ 1sor [ B implies that r n ¼ 1 for all large n and thus x n is eventually constant. For b ¼ a 2 1 it is the case that r 1 ¼ 1 so letting r [ B 1 we obtain an eventually constant solution. THE REGION b > a 2 =4 In this region, the absence of fixed points such as r 1 and r 2 mandates the use of different tools and methods. In particular, periodic solutions present themselves as interesting substitutes. We observe that a positive, p-periodic solution {x n } of Eq. (1) induces a periodic ratio sequence {x n /x n21 } of the same period that satisfies the identity Q p i¼1 x i=x i21 ¼ 1: It follows that positive, period-p solutions of Eq. (1) correspond in a one-to-one fashion to the p-cycles {r n } of Eq. (2) with the property that Q p i¼1 r i ¼ 1: We begin with the next lemma, which shows in particular that unlike the fixed points r 1 and r 2 ; we can always expect to find twocycles for Eq. (2), even when b. a 2 =4:

824 C.M. KENT ND H. SEDGHT Lemma 6 (a) For fixed a; b. ; Eq. (2) has a unique two-cycle {r 1 ; r 2 } where r 1 ¼ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 a 4 þ 4b 2 2 a 2 ; r 2 ¼ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2a a þ a 4 þ 4b 2 2 a 2 : ð8þ 2a (b) For fixed a 2. b. ; Eq. (2) has a unique three-cycle {r 1 ; r 2 ; r 3 } where r 2 ¼ f L ðr 1 Þ, b a ; r 3 ¼ f L ðr 2 Þ. b a ; r 1 ¼ f R ðr 3 Þ, b a and r 1 is given by r 1 ¼ aða p 2 þ 3bÞ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 ða 2 þ bþ 2 2 4b 3 2ða 2 : ð9þ þ bþ Proof (a) If {r 1 ; r 2 } is a two-cycle of Eq. (2) with r 1, r 2 ; then it is the case that r 1, b=a, r 2 : This is true because f R is increasing and also f 2 LðrÞ ¼r implies that r ¼ r: Hence both of r 1 and r 2 cannot be on one side of b=a. Now setting r 2 ¼ f L ðr 1 Þ and r 1 ¼ f R ðr 2 Þ gives r 1 ¼ a 2 b f L ðr 1 Þ ¼ a 2 br 1 ¼ ab 2 ða 2 þ bþr 1 : b 2 ar 1 b 2 ar 1 Solving the above equation for r 1 we obtain the value in Eq. (8) as the only solution in the interval ð; b=aþ: Then r 2 ¼ f L ðr 1 Þ is the other value Eq. (8). (b) We calculate the three-cycle indicated in the statement of the lemma similarly to Part(a), although the algebraic manipulations are more extensive. The unique value for r 1 [ ð; b=aþ in the form (9) is defined only when the expression under the square root is non-negative; i.e. r 1 exists if and only if aða 2 þ bþ $ 2b 3=2 : This inequality can be written as a cubic polynomial inequality in a p a 3 þ ab 2 2b ffiffi b $ : Noting that a ¼ p ffiffi b is a root of the polynomial on the left hand side, we have a factorisation p ða 2 ffiffi p b Þða 2 þ a ffiffi b þ 2bÞ $ : Since the quadratic factor p is positive for a; b. we conclude that r 1 (hence also the threecycle) exist when a $ ffiffi b or equivalently, a 2 $ b: If the equality holds, then we calculate from Eq. (9) r 1 ¼ a ¼ b a ; where the last equality is equivalent to b ¼ a 2 : However, b=a maps to zero and cannot be a periodic point of f; therefore, the strict inequality a 2. b must hold.

In Lemma 6, we point out that b. because it is necessary that r 1. in both Eqs. (8) and (9). We need one more lemma before presenting our main theorem on periodicity. The next lemma concerns solutions that contain zeros, i.e. they pass through the origin repeatedly. Note that such solutions cannot be represented by the mapping f. Lemma 7 ssume that b. and let {x n } be a solution of Eq. (1) such that x k ¼ for some k $ and x k21. : (a) If x kþ2 ¼ ; then a ¼ and x kþ2n21 ¼ b n x k21 for n ¼ 1; 2; 3;... (b) If x kþ3 ¼ ; then a 2 ¼ b and x kþn ¼ a nþ1 x k21 for n 3j; j ¼ 1; 2; 3;... Proof (a) We have x kþ1 ¼ bx k21. and ¼ x kþ2 ¼ ax kþ1 : Thus a ¼ and the stated formula is easily established by induction on n. (b) Since x kþ3 ¼jax kþ2 2 bx kþ1 j¼; it follows that ax kþ2 ¼ bx kþ1 : In particular, a and x kþ2 : In fact, so TWO-PRMETER BSOLUTE-DIFFERENCE EQN 825 x kþ2 ¼ ax kþ1 ¼ abx k21 ; ¼ x kþ3 ¼ bja 2 2 bjx k21 and it follows that b ¼ a 2 : We may now write x kþ1 ¼ a 2 x k21 and x kþ2 ¼ a 3 x k21 : Further, x kþ4 ¼ bx kþ2 ¼ a 5 x k21 ; x kþ5 ¼ ax kþ4 ¼ a 6 x k21 : Induction on n now completes the proof. Theorem 4 (a) Equation (1) has a period-2 solution if and only if b 2 2 a 2 ¼ 1 or equivalently b ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 þ 1 for a $ : ð1þ Further, if a. ; then the period-2 solutions are positive and confined to the pair of lines y ¼ r 1 x and y ¼ r 2 x in phase space, where r 1,r 2 are given by r 1 ¼ b 2 1 a ; r 2 ¼ b þ 1 : ð11þ a On the other hand, the only period-2 solutions of Eq. (1) that pass through the origin occur at a ¼ where b ¼ 1: (b) Equation (1) has a period-3 solution if and only if a 3 þ ab 2 b 3 ¼ 1; a $ 1: ð12þ Further, if a. 1; then the period-3 solutions are positive and confined to the three lines y ¼ r i x in phase space where for i ¼ 1; 2; 3; r i are given by r 1 ¼ ab þ 1 a 2 þ b ; r 2 ¼ b 2 2 a ab þ 1 ; r 3 ¼ b þ a 2 b 2 2 a : ð13þ

826 C.M. KENT ND H. SEDGHT Proof On the other hand, the only period-3 solutions of Eq. (1) that pass through the origin occur at a ¼ 1 where b ¼ 1. (a) First, let us assume that a. and use Eq. (8) to compute r 1 r 2 ¼ 1 hpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i a 2 4 þ 4b 2 2 a 2 : ð14þ Setting r 1 r 2 ¼ 1 and simplifying, we obtain the quadratic curve in Eq. (1). Next, using Eq. (1) we may simplify the formulas in Eq. (8) to obtain the values in Eq. (11). For parameter values on the quadratic curve (1), if x =x 21 equals either r 1 or r 2 then f ðx =x 21 Þ¼r 2 and f ðx 1 =x Þ¼r 1 or conversely, so the corresponding solution {x n } of Eq. (1) with period 2 is confined to the lines y ¼ r 1 x and y ¼ r 2 x in phase space. Now consider the case a ¼ in Eq. (1). lthough r 1 ; r 2 are not defined, it is clear that for ða; bþ ¼ð; 1Þ every solution of Eq. (1) has period 2 and by Lemma 7(a) these include all the period-2 solutions that pass through the origin. Since every positive period-2 solution of Eq. (1) induces a two-cycle in the ratios, it follows from Lemmas 6 and 7 that there are no other period-2 solutions of Eq. (1) than the ones already mentioned. (b) With r 1 given by Eq. (9) and r 2 ¼ f L ðr 1 Þ¼ b 2 ar 1 ; r 3 ¼ f L ðr 2 Þ¼ ða 2 þ bþr 1 2 ab r 1 b 2 ar 1 we see that r 1 r 2 r 3 ¼ða 2 þ bþr 1 2 ab ¼ 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 aða 2 þ bþ 2 a 2 ða 2 þ bþ 2 2 4b 3 : ð15þ Setting r 1 r 2 r 3 ¼ 1 and re-arranging terms, gives the equation qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 ða 2 þ bþ 2 2 4b 3 ¼ aða 2 þ bþ 2 2; ð16þ which can hold only when (i) aða 2 þ bþ 2 2 $ or equivalently, b $ 2=a 2 a 2 and (ii) b, a 2 so that the square root is real and r 1. : The curve 2=a 2 a 2 is strictly decreasing and intersects the parabola b ¼ a 2 at a ¼ 1: It follows that Eq. (16) holds for a. 1: Now, if we square both sides of Eq. (16) and simplify, we get the cubic equation in (12). This equation in turn may be used to simplify the values r 1 ; r 2 ; r 3 in Lemma 6(b) to get the values in Eq. (13). It follows from Lemma 6 that positive period-3 solutions of Eq. (1) exist if and only if the parameters a; b satisfy Eq. (12) with a. 1 and in this case the solution in the phase space is confined to the three straight lines mentioned in the statement of the theorem. Next, consider a ¼ 1 in Eq. (12). Here, also b ¼ 1, so Lemma 7(b) shows that all period-3 solutions that pass through the origin are accounted for. The proof is now complete. Remarks 2 (a) s seen earlier, for ða; bþ ¼ð1; 1Þ the period 3 solutions of Eq. (1) pass through the origin and are not positive. Hence, they cannot correspond to three-cycles of Eq. (2), a fact that is consistent with Theorem (c). On the other hand, note that as ða; bþ

TWO-PRMETER BSOLUTE-DIFFERENCE EQN 827 approaches ð1; 1Þ along the curve (12) the quantities r 1, r 2, r 3 in Eq. (13) have the following limits lim ða;bþ! ð1;1þ r 1 ¼ 1; lim r 2 ¼ ; lim r 3 ¼ 1: ða;bþ! ð1;1þ ða;bþ! ð1;1þ These correspond to, respectively, the line y ¼ x; the x-axis and the y-axis in the phase plane of Eq. (1) which are indeed lines containing the period-3 solutions of Theorem (b) in the phase plane. Therefore, the non-positive period-3 solutions of Eq. (1) may be interpreted as limiting values of the positive period-3 solutions as ða; bþ approaches ð1; 1Þ along the curve (14). lso see Corollary 3 below. (b) No period-2 or period-3 solution of Eq. (1) is stable, whether asymptotically or structurally. Theorem 4 shows the structural instability; as for asymptotic instability, recall that since a p-cycle {r 1 ;...; r p } of Eq. (2) does not contain the minimum point of f, itisunstable if that is, if 1, Yp jf ðr i Þj ¼ Yp b r 2 i¼1 i¼1 i Y p i¼1 r i, b p=2 : ¼ b p Q p r 2 i i¼1 If p ¼ 2 and a. ; then r 1 r 2, ða 2 þ 2b 2 a 2 Þ=2 ¼ b by Eq. (14), so the positive twoperiodic solutions are not asymptotically stable. The same conclusion clearly holds when a ¼ and b ¼ 1: For p ¼ 3 and a. 1 we have from Eq. (15) that r 1 r 2 r 3, b 3=2 if and only if qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aða 2 þ bþ 2 a 2 ða 2 þ bþ 2 2 4b 3, 2b 3=2 : The preceding inequality reduces to 2b 3=2, aða 2 þ bþ which is true if r 1 is real. Hence, positive period-3 solutions of Eq. (1) are not asymptotically stable. lso, Theorem shows that the period-3 solutions passing through the origin are unstable, although rather ironically, these are the only solutions that appear in computer simulations because of their rationality! We can now state the following result concerning the bifurcations of solutions of Eq. (1) at the point ða; bþ ¼ð1; 1Þ: Corollary 2 Every neighborhood of ð1; 1Þ in the parameter plane contains a pair ða; bþ for which one of the following is true: (a) Every solution of Eq. (1) converges to zero; (b) There are unbounded solutions of Eq. (1) that are equal to zero infinitely often; (c) There are positive solutions of Eq. (1) that are 3-periodic (hence also bounded but not convergent). Proof (a) For each ða; bþ in the open unit square every solution converges to zero by Theorem 1. (b) On the parabola b ¼ a 2 which contains ð1; 1Þ; Lemma 7 shows that the desired solutions exist for every value a. 1: In fact, Lemma 7 shows that as ða; bþ passes ð1; 1Þ on

828 C.M. KENT ND H. SEDGHT the curve b ¼ a 2 ; solutions of Eq. (1) change qualitatively from bounded and convergent to unbounded and non-convergent, with an uneasy compromise taking place at ð1; 1Þ: (c) For each ða; bþ on the cubic curve a 3 þ ab 2 b 3 ¼ 1 which contains ð1; 1Þ; Theorem 4 establishes the existence of positive period-3 solutions for every value a. 1: s the preceding results show, periodic ratios are useful in answering certain questions about the dynamics of Eq. (1). The next result is another example of this usage. Theorem 5 Let b. a 2 =4: (a) If b # ða 2 þ aþ=ða þ 2Þ then there are solutions of Eq. (1) that converge to zero eventually monotonically. p (b) If ða 2 þ aþ=ða þ 2Þ, b, ffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 þ 1 then for all a $ there are solutions of Eq. (1) that converge p to zero in an oscillatory fashion. (c) If b. ffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 þ 1 then for all a $ there are solutions of Eq. (1) that are oscillatory and unbounded. However, if b, a þ 1; then there are also solutions that converge to zero monotonically (eventually monotonically if b, 2a 2 ). (d) If b ¼ a; then for all a. there are solutions of Eq. (1) that converge to zero in an oscillatory fashion as well as solutions that converge to zero eventually monotonically. Proof (a) and (b) Setting r 2 # 1 in Eq. (8) and simplifying gives b # a 2 þ a ð17þ a þ 2 with p equality holding if and only if r 2 ¼ 1: Using Eq. (1) we find that r 1 r 2, 1 if and only if b, ffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi a 2 þ 1: Further, for a. ; we have a 2 þ 1. a whereas ða 2 þ aþ=ða þ 2Þ, a: It follows that the region in Part (b) is nonempty and in there, r 1 r 2, 1 but r 2. 1: Hence, if e.g. x =x 21 ¼ r 1 then the ratio sequence is 2-periodic. In this case, x 1 ¼ r 2 x. x while x 2 ¼ r 1 r 2 x, x : Inductively, for all n $ 1; it follows that x 2n21. r 2 r 1 x 2n21 ¼ r 2 x 2n ¼ x 2nþ1. x 2n. r 1 r 2 x 2n ¼ x 2nþ2 : It is clear that {x n } in this case converges to zero in an oscillatory fashion. This proves statement (b); we also note that the region of the parameter plane in this case contains a part of the unit square. On the other hand, if Eq. (17) holds, then both r 2 and r 1 r 2 are less than unity and the solution {x n } with x =x 21 ¼ r 1 converges to zero monotonically. This concludes the proof of Part (a). (c) In this case r 1 r 2. 1, so solutions of Eq. (1) with x =x 21 ¼ r 1 are unbounded and they oscillate because r 1, 1: In fact, this last inequality holds provided that b. ða 2 2 aþ=ð2a 2 1Þ which is true because pffiffiffiffiffiffiffiffiffiffiffiffiffi ða 2 2 aþ=ð2a 2 1Þ ¼a½ða 2 1Þ=ð2a 2 1ÞŠ, a, a 2 þ 1, b: If b, a þ 1; then by Lemma 5, r, 1 so if x ¼ rx 21 then x n ¼ rx n21 for p all n and {x n } converges to zero monotonically. If also b, 2a 2 (possible for a. ½ð1 þ ffiffiffiffiffi 17 Þ=8Š 1=2 Þ then we find that r, a so that f 21 ðr) is not empty. In this case, we can find solutions {x n } of Eq. (1) where there is k $ such that x k ¼ rx k21 and thus {x n } converges to zero eventually monotonically. Note that these conclusions extend similar ones made in Theorem 3(b) for the region b # a 2 =4:

TWO-PRMETER BSOLUTE-DIFFERENCE EQN 829 ðdþ The line b ¼ a in the parameter space lies within the region in Part ðbþ so there are solutions of Eq. (1) in this case that converge to zero in an oscillatory fashion. Further, b, a þ 1 so r, 1 and if {x n } is a solution of Eq. (1) with x =x 21 in the set B of backward iterates of r under f, then x n ¼ rx n21 for all sufficiently large n. Clearly, such {x n } converges to zero eventually monotonically. If we raise the lower limit in Theorem 5(c) to b. a þ 1 the following stronger conclusion is obtained. Theorem 6 Let b. a 2 =4: If b. a þ 1; then every solution of Eq. (1) is unbounded. If a solution {x n } has the property that x k # x k21 for some k $ then {x n } is unbounded and oscillatory. There are also unbounded solutions of Eq. (1) that are monotonically increasing. Proof Since the case a ¼ is discussed in Lemma 7, we assume from here on that a. : Define c ¼ b 2 a and note that c. 1: Let {x n } be a non-zero solution of Eq. (1) where there is k $ such that x k # x k21 : Then x kþ1 ¼jbx k21 2 ax k j¼jcx k21 2 aðx k 2 x k21 Þj $ cx k21 $ cx k. x k : ð18þ Notice that a decrease is always followed by an increase in such a way that the postdecrease high value x kþ1 $ cx k21 with x k21 representing the pre-decrease high. Next, x kþ2 ¼jcx k 2 aðx kþ1 2 x k Þj and here are two possible cases: If x kþ2 ¼ cx k 2 aðx kþ1 2 x k Þ ð19þ then clearly x kþ2, cx k # cx k21 # x kþ1 which implies that x kþ3 ¼jcx kþ1 2 aðx kþ2 2 x kþ1 Þj. cx kþ1. x kþ1. x kþ2 : Therefore, the situation in Eq. (18) is repeated with the new low and high values x kþ2 and x kþ3. If Eq. (19) does not hold, then its negative holds for possibly more than one index value: x kþj ¼ aðx kþj21 2 x kþj22 Þ 2 cx kþj22 ¼ ax kþj21 2 bx kþj22 ; j ¼ 2; 3; 4...: ð2þ Equation (2) is linear with complex eigenvalues since b. a 2 =4: If l ^ are these eigenvalues, then jl ^j p ¼ ffiffiffi b. 1ifb. a þ 1: It follows that there is an integer jmax ða; bþ $ 2 such that x kþjmax þ1 # x kþjmax and once again a situation analogous to that in Eq. (19) is obtained. We have shown that after at most a finite number of terms (determined only by the values of a, b) there will be a drop in the value of x n for all n, i.e. {x n } is oscillatory. Further, if {n m } where m ¼ ; 1; 2;...is the sequence of indicies at which drops in x n occur, then x 21þnm $ x 1þnm21 ; so from the preceding arguments and Eq. (18) it may be concluded that x 1þnm $ cx 21þnm $ cx 1þnm21 : Therefore, inductively x 1þnm $ cx 1þnm21. c 2 x 1þnm22.. c m x 1þn from which we may conclude that {x n } is unbounded. Finally, if b. a þ 1 then any strictly increasing solution of Eq. (1) must approach infinity, since the origin is the only possible fixed point of Eq. (1). Such strictly increasing solutions do exist; for example, since r. 1 by Lemma 5, a solution {x n } with x =x 21 ¼ r converges to infinity monotonically.

83 C.M. KENT ND H. SEDGHT We note here that if a þ 1, b # a 2 =4; then by Theorem 3 the solutions of Eq. (1) are generally unbounded, approaching infinity eventually monotonically. The special role of period 3 in the Li-Yorke Theorem in Ref. [4] gives the following result on the complex nature of the mapping f. The significant consequence of the following corollary and its extensions is that regardless of whether a solution {x n } of Eq. (1) converges or not, it may oscillate in a complicated way. Corollary 3 Let b # a 2 : Then the mapping f is chaotic in the sense of [4]; i.e. it has an uncountable set S called a scrambled set with the following properties: (i) S contains no periodic points of f and f ðsþ, S; (ii) For every x; y [ S and x y; lim supkf k ðxþ 2 f k ð yþk. ; k!1 lim infkf k ðxþ 2 f k ð yþk ¼ : k!1 (iii) For every x [ S and periodic y, lim supkf k ðxþ 2 f k ð yþk. : k!1 Proof Let ½; 1Š be the one-point compactification of the closed half-line ½; 1Þ and extend f continuously to ½; 1Š as follows: f*ðþ ¼1; f*ð1þ ¼a; f*ðrþ ¼f ðrþ if r [ ð; 1Þ: Then f* defines a continuous dynamical system on the compact interval ½; 1Š: Now, if b, a 2 then by Lemma 6(b) f has a three-cycle, which is also clearly a three-cycle for f*. lso if b ¼ a 2 then since a ¼ b a f! f*!1 f*! a it follows that f* again has a three-cycle, namely, {a; ; 1}: In either case, since ½; 1Š is homeomorphic to ½; 1Š; by the Li-Yorke theorem f* has a scrambled set S*, ½; 1Š: Define " # S ¼ S* 2 {1} < [1 f 2n ðþ : Since < 1 n¼ f 2n ðþ is countable for each n, it follows that S is an uncountable subset of ð; 1Þ > S* which satisfies conditions (i) (iii) above because n¼ fj S ¼ f*j S : Hence, S is a scrambled set for f as required. With the aid of a more general result on chaos from Ref. [5] Corollary 3 can be extended to the region b # 2a 2 where f may not have a period-3 point. Indeed, it can be shown that f has a snap-back repeller (in the weak or non-smooth sense see Refs. [5,7]) if and only if b, 2a 2 and also when b ¼ 2a 2 then f* has a snap-back repeller. We do not work out the routine details of this extension here, but refer to [7,9] for similar problems.

THE LINE b 5 1 : PERIODICITY By Lemma 7 and Theorem 4 in the preceding section, periodic solutions of Eq. (1) occur when ða; bþ ¼ð; 1Þ; ð1; 1Þ: These points are both located on the part of the line b ¼ 1 that lies inside the region b. a 2 =4 in the parameter plane. In this section, we find that this is not a coincidence, but part of a larger picture. Let us begin with the next result, which extends Lemma 7 to periods 4,5 and provides additional information with respect to periodic solutions that pass through the origin. Lemma 8 ssume that a; b. and let {x n } be a solution of Eq. (1) such that x k ¼ and x k21. for some k $ : (a) If x kþ4 ¼ then a 2 ¼ 2b and TWO-PRMETER BSOLUTE-DIFFERENCE EQN 831 x kþ4nþj ¼ b 2 x kþ4ðn21þþj ; j ¼ ; 1; 2; 3; n $ 1: ð21þ p In particular, {x n } is 4-periodic if and only if b ¼ 1 and a ¼ ffiffi pffiffi 2 : (b) If x kþ5 ¼ then a 2 ¼ gb; a 2 ¼ b=g or a 2 ¼ g 2 b where g ¼ð 5 þ 1Þ=2 is the golden mean. If a 2 ¼ gb then x kþ5nþj ¼ ab 2 g 2 x kþ5ðn21þþj; j ¼ ; 1; 2; 3; 4; n $ 1: ð22þ In particular, {x n } is 5-periodic if and only if ab 2 =g 2 ¼ 1; i.e. b ¼ g 3=5 a ¼ g 4=5 : If a 2 ¼ b=g then and x kþ5nþj ¼ ab 2 g x kþ5ðn21þþj; j ¼ ; 1; 2; 3; 4; n $ 1: ð23þ In particular, {x n } is 5-periodic if and only if ab 2 =g ¼ 1; i.e. b ¼ g 3=5 a ¼ g 21=5 : If a 2 ¼ g 2 b then and x kþ5nþj ¼ ab 2 g x kþ5ðn21þþj; j ¼ ; 1; 2; 3; 4; n $ 1: ð24þ In particular, {x n } is 5-periodic if and only if ab 2 =g ¼ 1; i.e. b ¼ 1 and a ¼ g: Proof In general, x kþ1 ¼ bx k21. and x kþ2 ¼ ax kþ1. : (a) From ¼ x kþ4 ¼jax kþ3 2 bx kþ2 j we get ax kþ3 ¼ bx kþ2 ¼ abx kþ1 i.e. x kþ3 ¼ bx kþ1 : But x kþ3 ¼jax kþ2 2 bx kþ1 j¼ja 2 2 bjx kþ1 so ja 2 2 bj ¼b: For positive a; b this equation is equivalent to a 2 ¼ 2b: Further, we calculate x kþ5 ¼ bx kþ3 ¼ b 2 x kþ1 ; x kþ6 ¼ ax kþ5 ¼ abx kþ3 ¼ b 2 x kþ2 :

832 C.M. KENT ND H. SEDGHT These in turn imply that x kþ7 ¼jab 2 x kþ2 2 b 3 x kþ1 j¼b 2 x kþ3 ; x kþ8 ¼jab 2 x kþ3 2 b 3 x kþ2 j¼b 2 x kþ4 ¼ : Now simple induction on n proves Eq. (21). The statement about 4-periodic solutions is now obvious. (b) x kþ3 ¼jax kþ2 2 bx kþ1 j¼ja 2 2 bjx kþ1 and x kþ4 ¼jax kþ3 2 bx kþ2 j¼jaja 2 2 bj 2 abjx kþ1 : Since ¼ x kþ5 ¼jax kþ4 2 bx kþ3 j we have ax kþ4 ¼ bx kþ3 ; i.e. a 2 jja 2 2 bj 2 bj ¼bja 2 2 bj: For positive a; b this equation is equivalent to one of three possible forms: b 2 2 a 2 b 2 a 4 ¼ ; a 2, b b 2 þ a 2 b 2 a 4 ¼ ; b, a 2, 2b b 2 2 3a 2 b þ a 4 ¼ ; a 2. 2b: For a 2, b solving the corresponding quadratic equation for b in terms of a gives a single positive solution pffiffiffi 5 þ 1 b ¼ a 2 ¼ ga 2 or a 2 ¼ 1 2 g b: For the range b, a 2, 2b solving the corresponding quadratic equation also yields a single positive solution pffiffiffi 5 2 1 b ¼ a 2 ¼ a 2 or a 2 ¼ gb 2 g and for the last range, we find two positive solutions only one of which is acceptable for a 2. 2b; namely, b ¼ 3 2 p ffiffi 5 a 2 ¼ a 2 2 g 2 or a 2 ¼ g 2 b: First, let a 2 ¼ gb: Then x kþ6 ¼ bx kþ4 ¼ abjja 2 2 bj 2 bjx kþ1 ¼ ab 2 g 2 x kþ1; x kþ7 ¼ ax kþ6 ¼ a 2 b 2 g 2 x kþ1 ¼ ab 2 g 2 x kþ2; x kþ8 ¼jax kþ7 2 bx kþ6 j¼ ab 2 g 2 jax kþ2 2 bx kþ1 j¼ ab 2 g 2 x kþ3; x kþ9 ¼jax kþ8 2 bx kþ7 j¼ ab 2 g 2 jax kþ3 2 bx kþ2 j¼ ab 2 g 2 x kþ4; x kþ1 ¼jax kþ9 2 bx kþ8 j¼ ab 2 g 2 jax kþ4 2 bx kþ3 j¼ ab 2 g 2 x kþ5 ¼ :

TWO-PRMETER BSOLUTE-DIFFERENCE EQN 833 Now induction on n proves Eq. (22). The assertion about periodicity follows immediately. The proofs of Eqs. (23) and (24) being similar to the above, are omitted. For period-5 solutions of Eq. (1) that pass through the origin, Lemma 8 shows that they do not all occur at the parameter value b ¼ 1; 5 is evidently the least period with this property. However, for b ¼ 1 Lemmas 7 and 8 do show that eventually periodic solutions of Eq. (1) of periods 2 5 exist that pass through the origin. In fact, if a p denotes the value of a at which a period-p solution of Eq. (1) occurs with b ¼ 1; then ¼ a 2, a 3 ¼ 1, a 4, a 5, 2: In particular, a p seems to be an increasing function of p and periods greater than 3 occur when 1, a, 2: In the remainder of this section, we see why this happens on the line b ¼ 1 and further, we establish the remarkable fact that periodic solutions of Eq. (1) for all p. 3 exist when 1, a, 2 and b ¼ 1: We proceed by introducing a linear equation corresponding to Eq. (1) that plays an important role in the proof of the existence of periodic solutions ( z nþ1 ¼ az n 2 z n21 ; for n ¼ ; 1;... ð25þ z 21 ; z [ ð; 1Þ; a [ ð1; 2Þ In particular, in the sequel we consider Eq. (25) with initial conditions Next, let g 1 ; g 2 ;...: ½1; 2Š! R be recursively defined by z 21 ¼ 1 and z ¼ a: ð26þ g nþ1 ðxþ ¼xg n ðxþ 2 g n21 ðxþ; n ¼ ; 1;...; ð27þ where g 21 ðxþ ¼1 and g ðxþ ¼x: ð28þ Then, we define two more collections of functions. For n ¼ ; 1;...; let G n : D n! R be defined by G n ðxþ ¼ g nðxþ g n21 ðxþ ; ð29þ where D n ¼ {x [ ½1; 2Š : g k ðxþ ; # k # n}: ð3þ For n ¼ ; 1;... and D n (x) as defined in Eq. (3), let H n : D n! R be defined by H n ðxþ ¼G n ðxþ 2 1 ; n ¼ ; 1;...: ð31þ x Observe that the particular solution {z n } 1 n¼21 of Eqs. (25) and (26) is also the sequence of values {g n ðaþ} 1 n¼21 : Lemma 9 Consider the sequence of polynomials {g n ðxþ} 1 n¼21 ; as defined by Eqs. (27) and (28), and the sequence of rational functions, {G n ðxþ} 1 n¼21 ; together with their sequence of domains {D n } 1 n¼21 as defined by Eqs. (29) and (3). Then{g nðxþ} 1 n¼21 and

834 C.M. KENT ND H. SEDGHT {G n ðxþ} 1 n¼21 satisfy the following properties: (P1) For all n $ ; G n is strictly increasing on its domain, D n. (P2) If, for some N [ {1; 2;...}; there exists a ðnþ [ ½1; 2Š such that g N ða ðnþ Þ¼; then aðnþ is the unique zero of g N ðxþ in the interval ½1; 2Š and g N ðxþ. for all x [ a ðnþ ; 2Š: (P3) For every n $ 1; there exists a ðnþ [ ½1; 2Š such that g n ða ðnþ Þ¼: Furthermore, 1 ¼ a ð1þ, a ð2þ,, 2: Proof We first establish Property (P1). For each integer n $ ; let PðnÞ be the proposition that G n ðxþ, G n ð yþ; for x; y [ D n and x, y: We will show by induction that PðnÞ is true for all n $ : Clearly, PðÞ is true, where, for x; y [ D ¼½1; 2Š and x, y; G ðxþ ¼x, y ¼ G ð yþ: Next suppose that n $ is an integer such that P(n) is true. We will show that Pðn þ 1Þ is true. Now, let x; y [ D nþ1 and x, y: Then x; y [ D n ; since D nþ1, D n by Eq. (3). Thus, we have G n ðxþ, G n ð yþ and by assumption, G n ðxþ, G n ð yþ: We then can write x 2 y,, 1 G n ðxþ, y 2 1 G n ð yþ : It then follows that x 2 1 G n ðxþ, y 2 1 G n ð yþ ; which, in turn, implies that xg n ðxþ 2 g n21 ðxþ, yg nð yþ 2 g n21 ð yþ g n ðxþ g n ð yþ by Eq. (29). From Eq. (27), we then have g nþ1 ðxþ g n ðxþ, g nþ1ð yþ g n ð yþ ; and, so, from Eq. (29) G nþ1 ðxþ, G nþ1 ð yþ: Therefore, Pðn þ 1Þ is true and Property (P1) is established. To prove Property (P2), suppose that for some N [ {1; 2;...}; there exists a ðnþ ; bðnþ [ ½1; 2Š such that g N ða ðnþ Þ¼ and g Nðb ðnþ Þ¼: Without loss of generality, we assume that a ðnþ # b ðnþ Then, from Eq. (27), we have and g N g N a ðnþ b ðnþ ¼ a ðnþ ¼ b ðnþ g N21 a ðnþ 2 gn22 a ðnþ ¼ g N21 b ðnþ 2 gn22 b ðnþ ¼ :

From these and the fact that a ðnþ # b ðnþ it follows that g N21 a ðnþ $ g N21 b ðnþ ; while a ðnþ g N22 a ðnþ On the other hand, note that TWO-PRMETER BSOLUTE-DIFFERENCE EQN 835 g N22 b ðnþ # b ðnþ : ð32þ g N21 ðxþ g N22 ðxþ ¼ G N21ðxÞ; and G N21 is strictly increasing on D N21, by Property (P1). Hence, in Eq. (32), we must have a ðnþ ¼ b ðnþ ; and, so, g N (x) has a unique zero in the interval ½1; 2Š: Moreover, we claim that g N ðxþ. for all x [ ða ðnþ ; 2Š: For we have g Nð2Þ. by the observation before this lemma and the fact that the particular solution of Eqs. (25) and (26) with a ¼ 2 is z n ¼ 2 þ n; n ¼ 21; ; 2;...: ð33þ Hence, Property (P2) is established. Given Properties (P1) and (P2), we can now show that Property (P3) is true. For each integer n $ 1; let P(n) be the proposition that there exists a ðnþ ; aðnþ1þ [ ½1; 2Š such that for i ¼ n; n þ 1; g i ða ðiþ Þ¼ and aðiþ is a unique zero of g i(x) in the interval [1,2] and 1 # a ðnþ, a ðnþ1þ, 2: We show by induction that P(n) is true for all n $ 1: Clearly, P(1) is true, where, in the interval [1,2], g 1 ðxþ ¼x 2 2 1 has the unique zero a ð1þ ¼ 1; g 2 ðxþ ¼x 3 2 2x has the unique zero a ð2þ p ¼ ffiffi 2 ; and a ð1þ ¼ 1, a ð2þ p ¼ ffiffi 2 : Next suppose that n $ 1 is an integer such that P(n) is true. We will show that P(n þ 1) is true. By assumption, we have that, in the interval [1,2], g n (x) has the unique zero a ðnþ ; g nþ1(x) has the unique zero a ðnþ1þ ; and a ðnþ, a ðnþ1þ : From Property (P2), we also have that g n (x), g nþ1 ðxþ. for all x [ ða ðnþ1þ ; 2Š: Thus, for some a ðnþ1þ, a # 2; g n ðxþ. g nþ1 ðxþ for all x [ ða ðnþ1þ ; aš: On the other hand, by the observation before this lemma and Eq. (33), we have g n ð2þ, g nþ1 ð2þ: Therefore, since g n (x) and g nþ1 (x) are continuous, there exists ~a [ ða ðnþ1þ ; 2Þ such that g n ð~aþ ¼g nþ1 ð~aþ: ð34þ So, we have the following: (i) G nþ1 ða ðnþ1þ Þ¼; by Eq. (29). (ii) G nþ1 ð~aþ ¼1; by Eqs. (29) and (34). (iii) G nþ1 (x) is defined and continuous on the interval ½a ðnþ1þ ; 2Š; by Eq. (3) and Property (P2), where a ðnþ is the unique zero of g n (x) and a ðnþ, a ðnþ1þ : Therefore, from Statements (i), (ii) and (iii), Eqs. (3) and (31), and the fact that 1, a ðnþ1þ, ~a, 2; we have the following: (i) H nþ1 a ðnþ1þ, : (ii) H nþ1 ð~aþ. ; by Eqs. (29) and (34). (iii) H nþ1 (x) is defined and continuous on the interval ½a ðnþ1þ ; 2Š.

836 C.M. KENT ND H. SEDGHT Hence, by the intermediate value theorem, there exists a [ ða ðnþ1þ ; ~aþ such that H nþ1 ðaþ ¼ which, in turn, implies that g nþ2 ðaþ ¼ ð35þ by Eq. (27). Let a ðnþ2þ ¼ a: Then, by Eq. (35) and Property (P2), a ðnþ2þ is the unique zero of g nþ2 ðaþ ¼ and 1 # a ðnþ1þ, a ðnþ2þ, 2: So, P(n þ 1) is true and Property (P3) is established. Lemma 1 Consider the sequence of polynomials {g n ðxþ} 1 n¼1 ; as defined by Eqs. (27) and (28), together with the sequence of their unique zeros in the interval [1,2], {a ðnþ }1 n¼1 : Then lim n!1 a ðnþ ¼ 2: Proof From Lemma 9, Property (P3), and the fact that a ð2þ p ¼ ffiffi 2 for g2 ðxþ ¼x 3 2 2x; we infer that (i) n lim!1 a ðnþ exists; (ii) a p ðnþ ffiffiffi # n lim!1 a ðnþ for all n $21; (iii) 2, n lim!1 a ðnþ # 2: Let lim a ðnþ n!1 ¼ L p ffiffi and for the sake of contradiction, assume that L [ ð 2 ; 2Þ: Observe that the particular solution {u n } 1 n¼21 of the initial value problem ( u nþ1 ¼ Lu n 2 u n21 ; for n ¼ ; 1;...; u 21 ¼ 1; u ¼ L; ð36þ is also the sequence of values {g n ðlþ} 1 n¼21 : lso observe that the particular solution of Eq. (36) is given by u n ¼ L cos nu þ L2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin nu; n $21; 4 2 L 2 where pffiffiffiffiffiffiffiffi pffiffi p ffiffi pffiffiffiffiffiffiffiffi S 42L (i) 2 L [ ð; 3 Þ for L [ ð 2 ; 2Þ, ð1; 2Þ; so that u ¼ arctan 42L 2 L [ ; p 3 p; 4p 3 (ii) L; p L2 ffiffiffiffiffiffiffiffi 22. 42L 2 p ffiffi since L [ ð 2 ; 2Þ: Therefore, g n ðlþ ¼Lcos nu þ L2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin nu; n $21 4 2 L 2 and, if for some N $ 1; we have Nu [ ½p; 3p=2Š; then g N ðlþ, : Indeed, we have the following, given that u [ ð; p=3þ < ðp; 4p=3Þ: (i) If u [ p; 4p 3 ; then for N ¼ 1; we have gn ðlþ, : (ii) If, u, p p 6 ; then there exists k $ 2 such that 3ðkþ1Þ, u, p 3k ; which, in turn, implies that for N ¼ 3ðk þ 1Þ; we have g N ðlþ, : (iii) If p 6 # u, 2p 9 ; then for N ¼ 6; we have g NðLÞ, :

(iv) If 2p 9 # u, 5p 18 ; then for N ¼ 5; we have g NðLÞ, : (v) If 5p 18 # u, p 3 ; then for N ¼ 4; we have g NðLÞ, : Hence, there exists N $ 1 such that g N ðlþ, ; a result which contradicts Property (P2) of Lemma 9 since a ðnþ # L for all n $21: p ffiffi Therefore, our original assumption that L [ ð 2 ; 2Þ is false and we have L ¼ 2: Note that Eq. (25) can be written as TWO-PRMETER BSOLUTE-DIFFERENCE EQN 837 z n21 ¼ az n 2 z nþ1 ð37þ and that with initial conditions z 21, z, we can determine not only the future terms, z 1 ; z 2 ;...; but the past terms, z 22 ; z 23 ;...: We will refer to Eq. (37) as the backwards-intime version of Eq. (25). We also note that for a difference equation x nþ1 ¼ f ðx n ; x n21 Þ; n ¼ ; 1;...; ð38þ and each integer N $21; we refer to {x n } N n¼21 as the partial solution of Eq. (38). We now make several observations with respect to Eq. (25) and its backwards-in-time version. lthough these observations might be considered trivial in nature, we list them explicitly in order to facilitate any references made to them in the proof of the main result of this paper in the next section. Remark 3 Let {z n } 1 n¼21 be a solution of Eq. (25). Let N $ ; and consider the equation ( w mþ1 ¼ aw m 2 w m21 ; for m ¼ ; 1;...; w 21 ¼ z N ; w ¼ z N21 ; a [ ð1; 2Þ ð39þ Then the solution, {w m } 1 m¼21 ; of Eq. (39) is equivalent to the solution, {z n} 21 n¼n ; of the backwards-in-time version of Eq. (25), ( z n22 ¼ az n21 2 z n ; for n ¼ N; N 2 1;... z N ¼ w 21 ; z N21 ¼ w ; a [ ð1; 2Þ In particular, the partial solution {w m } N m¼21 of Eq. (39) is such that {w m } N m¼21 ¼ z N; z N21 ;...; z ; z 21 ; ð4þ where w m ¼ z N2m21 ; for m ¼ 21; ;...; N 2 1; N: ð41þ Remark 4 Let {~z n } 1 n¼21 be a particular solution of Eq. (25) with initial conditions z 21 and ~z. Let {z n } 1 n¼21 be another particular solution of Eq. (25) with z 21 ¼ l~z 21 and z ¼ l~z : Then it is easy to show that z n ¼ l~z n ; for all n $21:

838 Remark 5 C.M. KENT ND H. SEDGHT Let {x n } 1 n¼21 be a solution of Eq. (1), and suppose there exists N $ 1 such that ax n 2 x n21 $ ; for n ¼ 21; ;...; N: Let {z n } 1 n¼21 be a solution of Eq. (25) with z 21 ¼ x 21 and z ¼ x and let {w m } 1 m¼21 be a solution of Eq. (39) in which we have w 21 ¼ z N and w ¼ z N21 : Then x n ¼ z n ¼ w N2n21 ; for n ¼ 21; ;...; N: Remark 6 Let {z n } 1 n¼21 be a solution of Eq. (25) and let {w m} 1 m¼21 be a solution of w mþ1 ¼ aw m 2 w m21 ; for m ¼ ; 1;... with w 21 ¼ z 21 and w ¼ z : Then {w m } 1 m¼21 ¼ {z n} 1 n¼21 : We now state the main result of this section. Theorem 7 Let b ¼ 1 in Eq. (1). There exists a strictly increasing sequence of values {a p } 1 p¼3 with a3 ¼ 1 and lim a p!1 p ¼ 2 such that for each p ¼ 3; 4;...; the particular solution {x n } 1 n¼21 of Eq. (1) with b ¼ 1; a ¼ a p and initial values x 21 ¼ 1; x ¼ a p ; is periodic with period p. Proof The proof is a consequence of the following two lemmas. Lemma 11 Let ~a [ ½1; 2Þ and let {x n } 1 n¼21 be a particular solution of Eq. (1) with a ¼ ~a and x 21 ¼ 1 and x ¼ ~a: Suppose there exists N $ such that x Nþ1 ¼ and x n for n ¼ 21; ; 1;...; N: Then {x n } 1 n¼21 is periodic with period N þ 3: Proof Let ~a [ ½1; 2Þ and let {x n } 1 n¼21 be a particular solution of Eq. (1) with a ¼ ~a and x 21 ¼ 1 and x ¼ ~a : Suppose there exists N $ such that x Nþ1 ¼ and x n for n ¼ 21; ; 1;...; N: Then, from Lemma 9 and the observation preceding it, and from Property (P2), we have ~ax n 2 x n21. ; for n ¼ ; 1;...; N 2 1: ð42þ Therefore, if we let {z n } 1 n¼21 be a particular solution of Eqs. (25) and (26), then it follows from Remark 6 that x 21 ¼ z 21 ¼ 1; x ¼ z ¼ ~a; x 1 ¼ z 1 ;...; x N ¼ z N ; x Nþ1 ¼ z Nþ1 ¼ : ð43þ

From Eq. (43) and the fact that x Nþ1 ¼ ; we have x N21 ¼ ~ax N 2 x Nþ1 ¼ ~ax N : Let x N ¼ l; where l. since x N. : Then x N ¼ l and x N21 ¼ l~a; l. : ð44þ Thus, from Eqs. (43) and (44), we have z N ¼ l and z N21 ¼ l~a; l. : ð45þ Now, let {w n } 1 n¼21 be the particular solution of w mþ1 ¼ aw m 2 w m21 ; for m ¼ ; 1;... with w 21 ¼ l and w ¼ l~a. Then we have the following: 1. From Eq. (45) and Remark 3, TWO-PRMETER BSOLUTE-DIFFERENCE EQN 839 w 21 ¼ z N ; w ¼ z N21 ;...; w N21 ¼ z ; w N ¼ z 21 : 2. On the other hand, from Eq. (45) and Remarks 4 and 6 we have w 21 ¼ lz 21 ; w ¼ lz ;...; w N21 ¼ lz N21 ; w N ¼ lz N : ð46þ ð47þ Therefore, from Eqs. (43), (45), (46) and (47), we have 1 ¼ x 21 ¼ z 21 ¼ w N ¼ lz N ¼ l 2 : Hence, l ¼ 1; and this, in turn, implies that x N21 ¼ ~a and x N ¼ 1: Therefore, x Nþ2 ¼j~ax Nþ1 2 x N j¼j~a 2 1j ¼1; x Nþ3 ¼j~ax Nþ2 2 x Nþ1 j¼j~a 12 j ¼~a: Clearly, then, {x n } 1 n¼21 is periodic with period N þ 3: Lemma 12 Let b ¼ 1 in Eq. (1). There exists a strictly increasing sequence of values {a p } 1 p¼3 with a 3 ¼ 1 and p lim a!1 p ¼ 2; such that for each p ¼ 3; 4;...; the particular solution {x n } 1 n¼21 of Eq. (1) with b ¼ 1 and initial values x 21 ¼ 1; x ¼ a p ; is characterized by the following: (i) x p ¼ : (ii) x n for n ¼ 21; ; 1;...; p 2 1: (iii) a p x n 2 x n21. for n ¼ 21; ; 1;...; p 2 2: Proof The result follows from Lemma 9, Properties (P2) and (P3) and Lemma 1. OPEN PROBLEMS ND CONJECTURES It is fitting to close this paper with a section on what remains to be done, which is considerable. The preceding results clearly indicate that the two-parameter absolute difference equation (1) is a simple equation exhibiting complex behavior over ranges that