Monotone and oscillatory solutions of a rational difference equation containing quadratic terms

Monotone and oscillatory solutions of a rational difference equation containing quadratic terms M. Dehghan, C.M. Kent, R. Mazrooei-Sebdani N.L. Ortiz, H. Sedaghat * Department of Mathematics, Virginia Commonwealth University, Richmond, Virginia, 23284-2014, USA; Department of Applied Mathematics, Amirkabir University of Technology, Tehran, Iran Abstract We show that the second order rational difference equation x n+1 = Ax2 n + Bx nx n 1 + Cx 2 n 1 αx n + βx n 1 has several qualitatively different types of positive solutions. Depending on the nonnegative parameter values A, B, C, α, β, all solutions may converge to 0, or they may all be unbounded. There are parameter values for which both cases can occur depending on the initial values. We discuss cases where converging solutions are monotonic and cases where they are not. Although the above equation has no isolated fixed points, it can have periodic solutions with period 2 (and no other periodic solutions). A semiconjugate relation facilitates derivations of many of these results by providing a link to a rational first order equation. Keywords. Rational, quadratic, monotonic, non-monotonic, period 2, semiconjugate AMS Subject Classification. 39A10, 39A11 1 Introduction The second order difference equation x n+1 = Ax2 n + Bx n x n 1 + Cx 2 n 1 αx n + βx n 1 (1) is an example of a rational equation with a quadratic numerator and a linear denominator, or a QLR equation for short. Rational difference equations having both a linear numerator and 0 *Corresponding author. Email: hsedagha@vcu.edu 1

a linear denominator have been studied extensively by Gerry Ladas and colleagues; see e.g. [6-9] which include extensive lists of references. This extensive research makes a strong case for studying the behavior of rational difference equations and their work provides a great deal of information about the behavior of rational equations with linear terms. However, rational equations containing quadratic terms in the numerator or the denominator have not yet been systematically studied; a few papers that we are aware of include [1-3] and [5] for second order equations. In [2] we considered QLR difference equations in some detail and found that they exhibit a typically broader range of behavior than equations having only linear terms. Therefore, understanding the nature of solutions of QLR equations adds significantly to our general understanding of the remarkable class of rational difference equations. In this paper, we study the positive solutions of (1). Clearly, if the parameters (or coefficients) satisfy the conditions α > 0, A, B, C, β 0 with A + B + C > 0 (2) then each solution {x n } of (1) with initial values x 0, x 1 (0, ) (3) is a positive solution. Although the weaker inequality α + β > 0 is sufficient for positive solutions, we do not consider the case α = 0 in this paper and assume that (2) and (3) both hold without further explicit mention. We also assume generally that β + C > 0 so that (1) is not reduced to a linear equation. Equation (1), subject to (2) and (3), is essentially different from QLR equations considered in [2] or in most other studies because (1) has no isolated fixed points. Therefore, typical methods of analysis that utilize fixed points (e.g., linearization, semicycle analysis) cannot be used. By way of comparison, the linear-over-linear analog of (1), i.e. the constants-free equation x n+1 = Ax n + Bx n 1 αx n + βx n 1, A, B, α, β 0, A + B, α + β > 0. (4) has a unique, isolated, positive fixed point x = (A + B)/(α + β). Thus (4) is fully amenable to linearization and semicycle analysis; see [9] for a detailed study of the solutions of (4) using these methods. For Equation (1), we observe that it is homogeneous of degree 1 (see [11]) and thus has a semiconjugate factorization that ties it to a first order rational equation. This feature enables us to show the existence of several qualitatively different types of positive solutions for (1). Depending on the parameter values, we show that all solutions of (1) may converge to 0, or they may all be unbounded. There are parameter values for which both cases occur depending on the initial values. We discuss cases where converging solutions are monotonic 2

and cases where they are not. We also show that even without fixed points, (1) can have periodic solutions with period 2 that are the only periodic solutions of (1) and they occur for almost all (but not all) initial values. Finally, we note that our results for the special case C = 0 solve Open Problem 6.10.1(b) in [9] for non-negative coefficients. 2 The ratio mapping For background concepts used in this and subsequent sections we refer to the texts [4, 7, 10]. We may write Equation (1) as x n+1 x n = Ax2 n + Bx n x n 1 + Cx 2 n 1 αx 2 n + βx nx n 1 = A(x n/x n 1 ) 2 + B(x n /x n 1 ) + C α(x n /x n 1 ) 2 + β(x n /x n 1 ) Now if we define the ratios r n = x n /x n 1 for n 0 then the following first order, rational difference equation is satisfied by the sequence of ratios: r n+1 = Ar2 n + Br n + C αr 2 n + βr n. (5) Under conditions (2) {r n } is a positive solution of (5) if r 0 = x 0 /x 1 > 0; this is guaranteed by (3). For such {r n }, we note that x n = r n x n 1 = r n r n 1 x n 2 = = r n r n 1 r 1 x 0. (6) Hence each positive solution of (1) can be obtained from a given positive solution of (5) in the form (6). Equations (5) and (1) are semiconjugates; see [10]. Also see [11] for a generalization of the ratios idea to homogeneous difference equations on groups. Aside from being of different orders, Equations (5) and (1) are different in two additional respects: (5) does not have the symmetric form of (1) and further, it contains quadratic terms in both the numerator and the denominator. We exploit such differences to clarify the different roles played by the various parameters in Equation (1). Another important difference between (5) and (1) is the fact that (5) has an isolated fixed point in (0, ). Lemma 1. Let B + C > 0 or A > β. (a) Equation (5) has a unique fixed point r (0, ). (b) r = 1 if and only if A + B + C = α + β. (c) For r 1, ( r 1)(A + B + C α β) > 0; i.e., r < 1 (respectively, r > 1) if and only if A + B + C < α + β (respectively, A + B + C > α + β). 3

Proof. (a) Fixed points of (5) in (0, ) are positive solutions of the equation Ar 2 + Br + C αr 2 + βr = r. This equation is equivalent to the polynomial equation φ(r). = αr 3 + (A β)r 2 + Br + C = 0. First, assume that C > 0. Since α > 0 and φ(0) = C > 0 it is clear that φ has at least one positive root r. Further, there is precisely one sign change in φ regardless of the sign of A β, so by the Descartes rule of signs [12], φ has at most one positive root. It follows that r is unique. Next, if C = 0 but B > 0 then the non-zero roots of φ are the same as the roots of the quadratic equation αr 2 (A β)r B = 0 i.e., A β (A β) 2 + 4αB < 0 < A β + (A β) 2 + 4αB. 2α 2α Thus, once again there is a unique positive root. Finally, if B = C = 0 then by hypothesis A > β and the only non-zero root of φ is (A β)/α, which is positive and unique as such. (b) φ(1) = A + B + C α β = 0 so r = 1. (c) To show that r 1 has the same sign as A+B+C α β, from the equation φ( r) = 0 we obtain A β + B α α r + C = r. (7) α r 2 Suppose first that r < 1. Then (7) gives 0 > r 1 A β α + B α + C α 1 = A + B + C α β. α This chain of inequalities is valid if and only if A + B + C < α + β. Similarly, if r > 1 then from (7) it follows that 0 < r 1 A β α + B α + C α 1 = A + B + C α β α which holds if and only if A + B + C > α + β. The proof is now complete. 4

It is convenient in what follows to define the ratio mapping g(r) = Ar2 + Br + C αr 2 + βr so that Equation (5) can be stated as r n+1 = g(r n ). Lemma 1 thus shows that the continuous map g : (0, ) (0, ) has a unique fixed point r. It is an interesting fact that although the precise value of r is not generally easy to calculate, its position relative to 1 is easily determined from Lemma 1. This fact is important in the study of solutions of (1). Remark. (The invariant ray) Suppose that the hypotheses of Lemma 1 hold. Then the ray {(x, rx) : x (0, )}, or rx for short, is an invariant set of (1) in the state space (0, ) 2 since if (x 1, x 0 ) is a point on this ray so that x 0 = rx 1 then r 0 = r and thus x 1 x 0 = r 1 = r = x 1 = rx 0 i.e., (x 0, x 1 ) is on rx. By induction, the state space orbit (x n 1, x n ) is on the invariant ray for all n. Now if r < 1 then every orbit of (1) starting on rx will converge monotonically to zero on rx since by (6) x n = ( r) n x 0. (8) This inequality also shows that if r > 1 then every orbit in rx goes to infinity monotonically and if r = 1 then every orbit in rx is stationary (a point). The invariant ray rx is analogous to a fixed point for (1), in the sense that by taking the quotient of (0, ) 2 modulo rx, Equation (1) is transformed into a topological conjugate of (5), and the ray rx into the point r, on the space of rays through the origin (see [10]). Monotonic behavior on the invariant ray may or may not be representative of other solutions. In most cases in the next section, the behavior on the invariant ray is in fact representative of all solutions but in Section 4 this is not the case. 3 Monotone solutions Let {x n } be a positive solution of (1). We say that {x n } converges to 0 eventually monotonically if {x n } is a decreasing sequence for all n greater than some positive integer k and has limit 0. We also say that {x n } converges to eventually monotonically if {1/x n } converges to 0 eventually monotonically. The next result is essential for determining when all solutions of (1) are eventually monotonic. Its proof also provides information that we use in the next section on periodic and other non-monotonic solutions. 5

Lemma 2. (a) Let B + C > 0. The fixed point r of g is globally asymptotically stable on (0, ) if and only if αc (β + A)B + 2A C(β + A). (9) with the inequality strict if A = 0. (b) If B = C = 0 and A > β then the positive fixed point r = (A β)/α is globally asymptotically stable on (0, ). Proof. (a) We show that the sign of the function g 2 (r) r = g(g(r)) r is opposite to r r for r > 0, r r (Theorem 2.1.2 in [10]). g 2 (r) r = ( ) Ar 2 2 ( ) + Br + C Ar 2 + Br + C A + B + C αr 2 + βr αr 2 + βr ( ) Ar 2 2 ( ) r + Br + C Ar 2 + Br + C α + β αr 2 + βr αr 2 + βr = A (Ar2 + Br + C) 2 + B (Ar 2 + Br + C)(αr 2 + βr) + C (αr 2 + βr) 2 [ α (Ar 2 + Br + C) 2 + β (Ar 2 + Br + C)(αr 2 + ] βr) r α(ar 2 + Br + C) 2 + β (Ar 2 + Br + C)(αr 2 + βr) α (Ar 2 + Br + C) 2 + β (Ar 2 + Br + C)(αr 2 + βr) Combining the fractions and simplifying the numerator gives: g 2 (r) r = [ αa (A + β)r 5 + ( A 3 + Cα 2 ABα Aβ 2 Bαβ ) r 4 + ( 2A 2 B + ABβ + Cαβ 2ACα Bβ 2) r 3 + ( 2A 2 C + AB 2 + B 2 β BCα ) r 2 + ( 2ABC + BCβ C 2 α ) r + AC] [ α ( Ar 2 + Br + C )2 + β ( Ar 2 + Br + C )( αr 2 + βr )]. Let P(r) be the quintic polynomial in the numerator of g 2 (r) r. Since the denominator of g 2 (r) r is positive for r > 0, the sign of P(r) is the same as the sign of g 2 (r) r for r > 0. Next, we divide P(r) by the cubic polynomial φ(r) that determines the fixed points of g in Lemma 1. The polynomials are divisible and the quotient is the quadratic polynomial ψ(r) = (A + β)ar 2 (αc AB βb)r + AC. (10) Observe that if ψ(r) > 0 for r > 0 then g 2 (r) r has the same sign as φ(r) on (0, ). Since under the given hypotheses by Lemma 1 φ(r) has a unique positive root at r, φ(0) 0 6

and also φ(r) as r, it must be that the sign of φ(r) is opposite of the sign of r r. Thus to complete the proof we find conditions implying ψ(r) > 0. We consider two cases: First, if A = 0, then ψ(r) > 0 for r > 0 if and only if αc < (A + β)b = βb. (11) Since A = 0 implies that B + C > 0, (9) with strict inequality reduces to (11). Now assume that A > 0. Then the roots of ψ can be found explicitly as αc AB βb ± (αc AB βb) 2 4(A + β)a 2 C ρ ± =. (12) 2(A + β)a Note that ψ(r) > 0 for r > 0 if and only if ψ has no real and positive roots, i.e., if and only if ρ ± are either complex or they are real and non-positive. First, ρ ± are complex if and only if or equivalently, αc (A + β)b < 2A C(A + β). (β + A)B 2A C(β + A) < αc < (β + A)B + 2A C(β + A). (13) Also, since B + C > 0 it follows that ρ ± are real and non-positive if and only if αc (A + β)b 2A C(A + β) and αc < (A + β)b or equivalently, αc (β + A)B 2A C(β + A) (14) Inequalities (13) and (14) together, i.e., one or the other holding, are equivalent to (9) with the strict inequality. Finally, to account for possible equality, if αc (A + β)b = 2A C(A + β) and αc > (A + β)b we obtain and ρ = ρ + = αc (A + β)b = 2A C(A + β) (15) αc (A + β)b 2(A + β)a = 2A C(A + β) 2(A + β)a = C A + β > 0. 7

Notice that ( ) C φ A + β = αc A + β = C A + β αc + (A + β)b A + β (A β)c + + B A + β C A + β + 2AC A + β = 0 C A + β + C where the value 0 is obtained by using (15) again. Therefore, the unique positive root of ψ is the same as the unique root of φ, i.e., the fixed point r = C/(A + β). Since ψ is a quadratic polynomial this value of r gives its minimum value of 0, so ψ(r) > 0 if r r. Hence once again the sign of φ determines the sign of g 2 (r) r and since (15) is the same as (9) with equality, the proof of (a) is complete. (b) In this case ψ(r) = (A + β)ar 2 > 0 for r > 0 so the conclusion follows easily from Lemma 1 and the arguments in the proof of (a). Theorem 1. (a) Assume that either B + C > 0 and (9) holds, or B = C = 0 and A > β. (i) If A + B + C < α + β then every positive solution of (1) converges to 0 eventually monotonically. (ii) If A + B + C > α + β then every positive solution of (1) converges to eventually monotonically. (b) Let B = C = 0 and A β. Then every positive solution of (1) converges to 0 eventually monotonically. Proof. (a),(i): By Lemma 1 there is a fixed point r (0, 1) for g which is globally attracting by Lemma 2. Hence there is k 1 such that r n < 1 for all n > k and (6) implies that x n is decreasing to zero if n > k. (a),(ii): The argument is similar to that for (a),(i) except that now r > 1 so that r n > 1 for all sufficiently large n. (b): In this case φ(r) = αr 3 (β A)r 2 < 0 for r > 0 so that g(r) < r (in particular, g has no positive fixed points). Thus r n 0 as n and (6) implies that x n is (eventually) decreasing to zero. Setting C = 0 in Theorem 1 gives the next result concerning Open Problem 6.10.1 in [9]; also see Corollaries 3 and 5 below. Corollary 1. Let β, A + B > 0 in the following difference equation: x n+1 = x n ( Axn + Bx n 1 αx n + βx n 1 8 ), x 1, x 0 > 0. (16)

(a) Every positive solution of (16) converges to 0 eventually monotonically if either of the following conditions holds: (i) A + B < α + β with A > β if B = 0; (ii) B = 0 and A β. (b) Every positive solution of (16) converges to eventually monotonically if A + B > α + β. 1. Setting A = 0 in Theorem 1 gives the next result, which may be compared with Corollary Corollary 2. Let B, β > 0 and βb > αc in the following difference equation: x n+1 = x n 1 ( Bxn + Cx n 1 αx n + βx n 1 ), x 1, x 0 > 0. (17) (a) Every positive solution of (17) converges to 0 eventually monotonically if B + C < α + β. (b) Every positive solution of (17) converges to eventually monotonically if B + C > α + β. Example. By considering a special case, we show that if the reverse of the inequality in Corollary 2 holds then (17) has solutions that are not eventually monotonic. Consider x n+1 = x2 n 1 x n (18) which is a special case of (17) with B = β = 0 and C = α. The ratios equation for (18) is r n+1 = 1 r 2 n whose explicit solution is easily obtained as r n = r ( 2)n 0. This solution and (6) give the explicit solution of (18) as x n = r 2+( 2)2 + +( 2) n 1 0 x 0 = x 0 r 2[( 2)n 1]/3 0 = x ( 2)n+1 +1/3 0 /x 1. It is clear from this that for all x 0, x 1 > 0, we have x 2n 1 0 and x 2n as n ; i.e. {x n } is unbounded but does not converge to. Remark. The preceding example indicates the occurrence of a type of behavior for the solutions of (1) that is different from other types of behavior that we have discussed so far. We will not discuss these types of solutions any further because the methods needed to analyze them diverge significantly from the methodology of this paper. 9

4 Non-monotonic and periodic solutions In this section we discuss bounded solutions, periodic solutions and solutions that converge to 0 or in a non-monotonic way depending on how A+B+C compares to α+β. We recall that under the hypotheses of Lemma 1 the behavior on the invariant ray is still monotonic so monotonic solutions can coexist with non-monotonic ones. We start by discussing the special case A+B+C = α+β. Under the conditions of Lemma 1 this equality implies the existence of a fixed point r = 1 for g. When this fixed point is globally attracting, it is generally difficult to reach any conclusions about the asymptotic behavior of solutions of (1) by using relation (6). So we use the next result instead; it is also a result of general interest. Lemma 3. Let y n be a given sequence of real numbers. If there exists a sequence {p n } of positive real numbers such that y n+1 y n p n y n y n 1, n = 0, 1,... (19) and lim n p n = p < 1, then y n converges to a finite limit. Proof. (a) Since lim n p n = p < 1, there exists a constant 0 < ρ < 1, and k 1 such that 0 < p n ρ for every n k. Thus (19) reduces to the following: From this we obtain y n+1 y n ρ y n y n 1, n = k, k + 1,... y n+1 y n ρ y n y n 1 ρ 2 y n 1 y n 2 ρ n k+1 y k y k 1, Now, if m > n > k then y n y m y n y n+1 + y n+1 y n+2 + + y m 1 y m y n y n+1 + ρ y n y n+1 + + ρ m n+1 y n y n+1 1 1 ρ y n y n+1 ρn k+1 1 ρ y k y k 1. It follows that y n is a Cauchy sequence, hence convergent. Lemma 4. Let A + B + C = α + β with B + C > 0. Then the following conditions are equivalent: 10

(i) C < A + β; (ii) α < 2A + B; (iii) αc < (β + A)B + 2A (β + A)C. Proof. (i) (ii): C < A + β A + B + C < 2A + B + β α < 2A + B. (ii)= (iii): If C > 0 then α < B + 2A = αc < BC + 2A ( C ) 2 < B(A + β) + 2A (A + β)c. If C = 0 then B > 0 so that αc = 0 < B(A + β) = B(A + β) + 2A (A + β)c. (iii)= (ii): We show that the negation of (ii) implies the negation of (iii). Since the negation of (ii), i.e. 2A + B α has already been shown equivalent to the negation of (i), i.e. A + β C, we have B(A + β) + 2A (A + β)c BC + 2AC = (2A + B)C αc which is the negation of (iii), as required. Theorem 2. Let A + B + C = α + β and B + C > 0. If α < 2A + B, then every positive solution of (1) converges to a finite limit. Proof. Lemma 1 implies that g has a fixed point at 1. By Lemma 4, (9) holds and thus 1 is globally attracting by Lemma 2(a). Next, use the hypotheses to write Therefore, x n+1 x n = (A α)x2 n + (B β)x n x n 1 + Cx 2 n 1 αx n + βx n 1 = [(α A)x n + Cx n 1 ](x n 1 x n ) αx n + βx n 1. x n+1 x n α A x n + Cx n 1 αx n + βx n 1 x n x n 1. Since 1 is globally attracting we have lim n (x n /x n 1 ) = 1. Therefore, ( ) xn α A + C α A x n + Cx n 1 x n 1 α A + C lim = lim ( ) =. n αx n + βx n 1 n xn α + β α + β 11 x n 1

Now, α A + C α + β < 1 α A < α + β C = A + B B < α < 2A + B which is true by hypothesis. Thus by Lemma 3 x n converges to a finite limit and the proof is complete. Corollary 3. (a) In Equation (16) assume that A + B = α + β, B > 0 and α < 2A + B. Then every positive solution of (16) converges to a finite limit. (b) In Equation (17) assume that B + C = α + β and α < B. Then every positive solution of (17) converges to a finite limit. So far we have not considered whether the positive solutions of (1) under the conditions of Theorem 2 are monotone or not. If the mapping g is decreasing at the fixed point 1, i.e., if g (1) < 0, then the attracting nature of 1 means that the ratios sequence r n oscillates about 1. Thus the sequence x n cannot be eventually monotonic. Lemma 5. The mapping g is decreasing on (0, ) if βa αb and C > 0 or if βa < αb and C 0. Proof. The derivative of g is g (r) = (2Ar + B)(αr2 + βr) (2αr + β)(ar 2 + Br + C) (αr 2 + βr) 2. So g (r) < 0 when the numerator is negative. Multiplying terms in the numerator and rearranging them gives the requirement (βa αb)r 2 2αCr βc < 0. (20) This last inequality is clearly true for r > 0 under the stated hypotheses and the proof is complete. Remark. (Non-monotonicity) In Theorem 2 or Corollary 3 solutions may or may not be monotonic. For instance, in Corollary 3(b) the conditions of Lemma 5 are satisfied so solutions are not eventually monotonic; they approach their limits in an oscillatory fashion (unless, of course, x 0 = x 1 ). However, in Corollary 3(a) if βa > αb then from (20) it is clear that the right hand side is positive (given that C = 0) so that g is increasing for r > 0; thus the converging solutions are eventually monotonic in this case. For more details see Theorem 3 below; also see Corollary 4 which can be compared with Theorem 2. Lemma 6. Assume that C > 0. Then: 12

(a) g has a unique pair of distinct, positive, period-2 points, namely, ρ ± in (12), if and only if αc > (β + A)B + 2A C(β + A). (21) (b) ρ < 1 < ρ + if and only if αc > (β + A)B + (A + C + β)a = AC + (A + β)(a + B). (22) (c) The product of the period-2 points is given by ρ ρ + = C A + β. (23) Proof. (a) The positive roots of the mapping ψ in (10) are the non-fixed point roots of g 2 (r); hence, these roots of ψ are the periodic points of g. As seen in the proof of Lemma 2, distinct real roots of ψ, i.e. the numbers ρ ±, exist and are positive if and only if (21) holds. (b) From (12) we find that ρ < 1 (αc AB βb) 2 4(A + β)a 2 C > αc (2A + B)(A + β) ρ + > 1 (αc AB βb) 2 4(A + β)a 2 C > [αc (2A + B)(A + β)]. Therefore, ρ < 1 < ρ + if and only if (αc AB βb) 2 4(A + β)a 2 C > αc (2A + B)(A + β) [αc (A + β)b] 2 4(A + β)a 2 C > {[αc (A + β)b] 2A(A + β)} 2 AC > αc + (A + β)b + (A + β)a The last expression is equivalent to (22). As might be expected from a comparison of (21) and (22) it is easy to verify that indeed (A + C + β)a 2A C(β + A) with equality holding if and only if A + β = C. (c) The relation (23) may be verified by direct multiplication or more quickly, by writing to see, using (10), that (A + β)aρ ρ + = AC. ψ(r) = (A + β)a(r ρ )(r ρ + ) 13

Lemma 7. Assume that A, C > 0, βa αb and (21) holds. Then the 2-cycle {ρ, ρ + } attracts all orbits of g in (0, ) except for r. Proof. With the given hypotheses Lemma 5 implies that g is decreasing on (0, ) and Lemma 6 implies that {ρ, ρ + } is its unique, positive 2-cycle. Note that g 2 = g g is increasing on (0, ) since g is decreasing. Also Lemmas 1 and 2 imply that g has a unique positive fixed point r that is unstable. Now let r 0 > 0 and r 0 r. Without loss of generality, we may assume that r 0 < r; for if r 0 > r, then r 1 = g(r 0 ) < r and we may start with r 1. Note that There are three possible cases: (i) r 2 = r 0 ; in this case r 4 = g 2 (r 2 ) = g 2 (r 0 ) = r 2 = r 0 r 2 = g 2 (r 0 ) < g 2 ( r) = r. (24) so by induction, r 2n = r 0 for n 1. Also, r 2n+1 = g(r 2n ) = g(r 0 ) = r 1 for n 1. It follows that {r 0, r 1 } is a positive 2-cycle of g and must therefore coincide with {ρ, ρ + } by uniqueness. (ii) r 2 > r 0 ; now using (24) and various inequalities: r 4 = g 2 (r 2 ) > g 2 (r 0 ) = r 2 > r 0 r 4 = g 2 (r 2 ) < g 2 ( r) = r. Hence, r 2 < r 4 < r. By induction, r 2n 2 < r 2n < r for n 1. This fact implies also that so we have r 2n 1 = g(r 2n 2 ) > g(r 2n ) = r 2n+1 > r, n 1 r 0 < r 2 < r 4 < < r < < r 5 < r 3 < r 1. Define λ = lim n r 2n and µ = lim n r 2n+1 and note that λ < r < µ. The numbers λ, µ are distinct from r because r is repelling and cannot attract any orbit of g (see Lemma 2.1.5 in [HS]). Further ( ) µ = lim g(r 2n) = g lim r 2n = g(λ), n n λ = lim n g(r 2n 1) = g ( lim n r 2n 1 ) = g(µ). Therefore, {λ, µ} is a 2-cycle of g which by uniqueness must coincide with {ρ, ρ + }. 14

(iii) r 2 < r 0 ; in this case, reasoning as in (ii) with inequalities reversed, we find that < r 4 < r 2 < r 0 < r < r 1 < r 3 < r 5 < Define λ and µ as in (ii) so that µ = g(λ) and λ = g(µ). Since g is decreasing and lim r g(r) = A/α > 0 it follows that g(r) > A/α for r > 0. Thus r n = g(r n 1 ) > A ( ) A α = r n+1 = g(r n ) < g, n 1. α Therefore, r n is bounded so µ < and thus λ = g(µ) > 0. Once again, {λ, µ} is a 2-cycle of g which by uniqueness must coincide with {ρ, ρ + }. This completes the proof. Before presenting the next result, a few definitions and observations are made. We say that a solution {x n } of (1) converges to 0 in an oscillatory fashion if {x n } converges to 0 but not eventually monotonically; i.e., lim n x n = 0 but {x n } is not an eventually decreasing sequence. We also say that {x n } converges to in an oscillatory fashion if {1/x n } converges to 0 in an oscillatory fashion; thus lim n x n = but {x n } is not an eventually increasing sequence. If {ρ 1, ρ 2,..., ρ k } is one cycle of a positive periodic solution of (5) with period k, then defining x 1 = ρ 1 x 0 gives x 2 = r 2 x 1 = ρ 2 ρ 1 x 0,, x k = ρ k ρ 2 ρ 1 x 0. Thus if the product ρ k ρ 2 ρ 1 = 1 then x k = x 0, x k+1 = x 1,..., x 2k = x k ; i.e., {x 1,..., x k } is one cycle of a solution of (1) with period k. Conversely, if {x 1,..., x k } is one cycle of a positive solution of (1) with period k then with ρ i = x i /x i 1 for i = 1,..., k we get ρ k+1 = x k+1 x k = x 1 x 0 = ρ 1, ρ k ρ k 1 ρ 1 = x k x k 1 x k 1 x k 2 x1 x 0 = x k x 0 = 1. Theorem 3. Let A, C > 0, βa αb and assume that (22) holds. (a) Let A + B + C < α + β; (i) If C = A+β then every solution of (1) not on the invariant ray rx converges to a periodic solution with period 2. (ii) If C > A+β then every solution of (1) not on the invariant ray rx converges to in an oscillatory fashion. (iii) If C < A + β and (A + B + C)(A + β) < (α + β)c then every solution of (1) not on the invariant ray rx converges to 0 in an oscillatory fashion. (iv) Solutions on the invariant ray rx (i.e., x 0 = rx 1 ) converge to 0 monotonically. 15

(b) If A+B+C α+β then every solution of (1) not on the invariant ray rx converges to in an oscillatory fashion. If A+B+C = α+β then solutions on the invariant ray rx are stationary or constant solutions. If A + B + C > α + β then solutions on the invariant ray rx converge to eventually monotonically. Proof. (a),(i): By Lemmas 6 and 7, g has a globally attracing, positive 2-cycle {ρ, ρ + } with ρ ρ + = 1. Thus by (6) and the remarks preceding this theorem, solutions of (1) with x 0 /x 1 r converge to a period-2 solution (the two limit points depend on the initial values). (a),(ii): As in (a),(i) g has a globally attracing, positive 2-cycle {ρ, ρ + } with ρ ρ + > 1. For sufficiently large n 1, the even and odd terms r 2n and r 2n 1 are close to ρ, ρ +. Without loss of generality assume that r 2n 1 ρ and r 2n ρ +. Then there is k 1 and 1 < γ < ρ ρ + such that r 2n 1 < 1 < r 2n and r 2n 1 r 2n γ > 1 for all n k. Thus x 2n = r 2n x 2n 1 > x 2n 1 and x 2n+1 = r 2n+1 x 2n < x 2n for n k; i.e., {x n } is an eventually oscillatory solution of (1). Further, for n > k n n x 2n = x 2k 1 (r 2i r 2i 1 ) > x 2k 1 γ n k and x 2n+1 = x 2k (r 2i+1 r 2i ) > x 2k γ n k. i=k Therefore, both x 2n, x 2n+1 as n as required. (a),(iii): We first need to consider a consequence of inequality (22); i.e., A + B + C < αc AC A + β + C = (α + β)c A + β i=k = (α + β)ρ ρ +. This inequality is stronger than the hypothesis A + B + C α + β if ρ ρ + < 1. If this stronger inequality holds, then by modifying the argument used to prove (b) appropriately (e.g., reversing the obvious inequalities and using ρ ρ + < δ < 1 instead of γ) shows that our claim is true. (a),(iv): This is clear by our earlier observations about the invariant ray. (b): As in the proof of (a),(iii) from (22) we obtain α + β A + B + C < (α + β)c A + β = C > A + β. The rest of the argument goes as in the proofs of (a),(iii) and (iv), thus completing the proof. The next result shows what happens when the inequality in Theorem 2 is reversed. 16

Corollary 4. Assume that A, C > 0, βa αb and A+B+C = α+β. If α > 2A+B then every solution of (1) not on the diagonal (i.e., x 0 x 1 ) converges to in an oscillatory fashion. Proof. Since α = 2A + B implies C = A + β, from Lemma 4 it follows that C > A + β. Also multiplying the inequality by C gives αc > 2AC + BC = AC + (A + B)C > AC + (A + B)(A + β). Thus (22) holds and we apply Theorem 3(b) to complete the proof. We note that the hypotheses of Theorem 3 are not satisfied if C = 0. The next result shows that this is not a deficiency of Theorem 3. Corollary 5. Equation (16) has no positive periodic solutions. Proof. Note that the function ψ in (10) takes the form ψ(r) = (A + β)ar 2 + (A + β)br which has no positive roots; in fact, ψ(r) > 0 for r > 0. Hence the only zeros of g 2 (r) r are the fixed points of g and there are no points of period 2. Now if (16) has a periodic solution then g must also have periodic points. Since g is continuous, by the Shrakovski ordering (see, e.g. [10]) g has to have points of period 2, which is a contradiction. References. [1] Camouzis, E., Ladas, G., Northshield, S. and Rodriguez, I.W., On the rational recursive sequence x n+1 = βx 2 n/(1 + x 2 n 1), Adv. Difference Eqs. Computers and Math. Appl. (1994) 37-43. [2] Dehghan, M., Kent, C.M., Mazrooei-Sebdani, R., Ortiz, N.L. and Sedaghat, H., Dynamics of rational difference equations containing quadratic terms, J. Difference Eqs. and Appl. (to appear) [3] DeVault, R., Galminas, L., Janowski, E.J., and Ladas, G., On the recursive sequence x n+1 = A/x n + B/x n 1,J. Difference Eqs. and Appl., 6 (2000) 121-125. [4] Elaydi, S.N., An Introduction to Difference Equations (2nd ed.) Springer, New York, 1999. [5] Grove, E.A., Janowski, E.J., Kent, C.M. and Ladas, G., On the rational recursive sequence x n+1 = (αx n + β)/(γx n + δ)x n 1, Comm. Appl. Nonlin. Analysis, 1 (1994) 61-72. [6] Grove, E.A. and Ladas, G., Periodicities in Nonlinear Difference Equations, Chapman & Hall/CRC, Boca Raton, 2005. [7] Kocic, V.L. and Ladas, G., Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Kluwer, Dordrecht, 1993. 17

[8] Kulenovic, M.R.S. and Ladas, G., Open problems and conjectures: On period two solutions of x n+1 = (α + βx n + γx n 1 )/(A + Bx n + Cx n 1 ), J. Difference Eqs. and Appl., 6 (2000) 641-646. [9] Kulenovic, M.R.S. and Ladas, G., Dynamics of Second Order Rational Difference Equations with Open Problems and Conjectures, Chapman and Hall/CRC, Boca Raton, 2002. [10] Sedaghat, H., Nonlinear Difference Equations: Theory with Applications to Social Science Models, Kluwer, Dordrecht, 2003. [11] Sedaghat, H., A note: All homogeneous second order difference equations of degree one have semiconjugate factorization, J. Difference Eqs. and Appl., 13 (2007) 453-456. [12] Weisstein, E.W., CRC Concise Encyclopedia of Mathematics (2nd ed.) Chapman & Hall/CRC, Boca Raton, 2003. 18