Unboundedness Results for Rational Difference Equations

Transcription

1 University of Rhode Island Open Access Dissertations 203 Unboundedness Results for Rational Difference Equations Gabriel Lugo University of Rhode Island, Follow this and additional works at: Terms of Use All rights reserved under copyright Recommended Citation Lugo, Gabriel, "Unboundedness Results for Rational Difference Equations" 203 Open Access Dissertations Paper 7 This Dissertation is brought to you for free and open access by DigitalCommons@URI It has been accepted for inclusion in Open Access Dissertations by an authorized administrator of DigitalCommons@URI For more information, please contact digitalcommons@etaluriedu

2 UNBOUNDEDNESS RESULTS FOR RATIONAL DIFFERENCE EQUATIONS BY GABRIEL LUGO A DISSERTATION SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY IN MATHEMATICS UNIVERSITY OF RHODE ISLAND 203

3 DOCTOR OF PHILOSOPHY DISSERTATION OF GABRIEL LUGO APPROVED: Dissertation Committee: Major Professor Gerasimos Ladas Edward A Grove George Tsiatas Nasser H Zawia DEAN OF THE GRADUATE SCHOOL UNIVERSITY OF RHODE ISLAND 203

4 ABSTRACT We present a collection of techniques for demonstrating the existence of unbounded solutions We then use these techniques to determine the boundedness character of rational dierence equations and systems of rational dierence equations We study the rational dierence equation x n = + x n Cx n 2 + x n 3, n N Particularly, we show that for nonnegative and C, whenever C = 0 and C + > 0, unbounded solutions exist for some choice of nonnegative initial conditions Moreover, we study the rational dierence equation x n = + βx n + x n 2 x n 3, n N Particularly, we show that whenever 0 < β < 3 and [0, ], unbounded solutions exist for some choice of nonnegative initial conditions Following these two results, we then present some new results regarding the boundedness character of the k th order rational dierence equation x n = + k i= β ix n i A + k j= B jx n j, n N When applied to the general fourth order rational dierence equation, these results prove the existence of unbounded solutions for 49 special cases of the fourth order rational dierence equation, where the boundedness character has not been established yet This resolves 49 conjectures posed by E Camouzis and G Ladas Finally, we study k th order systems of two rational dierence equations x n = + k i= β ix n i + k i= γ iy n i A + k j= B jx n j + k j= C, n N, jy n j

5 y n = p + k i= δ ix n i + k i= ɛ iy n i q + k j= D jx n j + k j= E, n N jy n j In particular, we assume non-negative parameters and non-negative initial conditions We develop several approaches, which allow us to prove that unbounded solutions exist for certain initial conditions in a range of the parameters

6 ACKNOWLEDGMENTS The rst person I would like to acknowledge is my advisor Professor Ladas I'd like to thank him for his help, support, and knowledge I secondly would like to thank my main research collaborator Frank Palladino I'd like to thank him for his brilliant ideas and knowledge Other professors at URI who I have I gotten signicant help, knowledge and support from are professors Eaton, Grove, Kulenovi, Ladas, Merino and Pakula Other people I'd like to thank for their support are my friend Gailia Rutan and my brother Jeann Lugo iv

7 DEDICATION I dedicate this to the poor v

8 PREFACE This dissertation is prepared in accordance with the University of Rhode Island Manuscript Plan option The main body of the dissertation consists of three manuscripts which have been written in a form that is suitable for publication in peer reviewed international mathematical journals Manuscript, Unboundedness for some classes of rational dierence equations, has been published in the International Journal of Dierence Equations 52009, Manuscript 2, Unboundedness results for fourth order rational dierence equations, has been accepted in the Journal of Dierence Equations and Applications Manuscript 3, Unboundedness results for systems, has been published in the Central European Journal of Mathematics 72009, The references for each individual manuscript are located at the end of the manuscript before the start of the next manuscript vi

9 TABLE OF CONTENTS ABSTRACT ii ACKNOWLEDGMENTS DEDICATION PREFACE iv v vi TABLE OF CONTENTS vii MANUSCRIPT Unboundedness for some classes of rational dierence equations Introduction 2 2 Todd's Equation 3 3 Special Case # Special Case # Conclusion 8 2 Unboundedness results for fourth order rational dierence equations 20 2 Introduction 2 22 Some General Unboundedness Results Unboundedness by Iteration The equation x n = +x n 3 Bx n +x n Conclusion Appendix A 37 3 Unboundedness results for systems 42 vii

10 Page 3 Introduction Unboundedness Results Involving Modulo Classes Adapting an unboundedness result to systems Some Examples for Rational Systems in the Plane Conclusion 67 viii

11 MANUSCRIPT Unboundedness for some classes of rational dierence equations by Gabriel Lugo and Frank J Palladino Publication Status: Published G Lugo and FJ Palladino, Unboundedness for some classes of rational dierence equations, Int J Dierence Equ 42009, Mathematics Subject Classication: 39A0,39A Keywords: dierence equation; periodic convergence; unbounded solutions; boundedness character

12 2 Abstract We study the rational dierence equation x n = + x n Cx n 2 + x n 3, n N Particularly, we show that for nonnegative and C, whenever C = 0 and C + > 0, unbounded solutions exist for some choice of nonnegative initial conditions Moreover, we study the rational dierence equation x n = + βx n + x n 2 x n 3, n N Particularly, we show that whenever 0 < β < 3 and [0, ], unbounded solutions exist for some choice of nonnegative initial conditions Introduction In Ref [], Camouzis and Ladas devote a chapter to the study of unbounded solutions for the k th order rational dierence equation with nonnegative parameters and nonnegative initial conditions x n = + k i= β ix n i A + k j= B jx n j, n N In the introduction of said chapter, the authors of Ref [] pose ve conjectures regarding the boundedness character of ve dierent special cases of the third order rational dierence equation Particularly, we are referring to the special cases #28, #44, #56, #70, and #20 These are the only remaining cases of third order for which the boundedness character has not been established First, we study special cases #56 and #20 x n = + βx n + x n 2 x n 3, n N Using a standard induction technique, we show that whenever 0 < β < 3 and [0, ], unbounded solutions exist for some choice of nonnegative initial conditions

13 3 We then study special cases #44 and #28 x n = + x n Cx n 2 + x n 3, n N We show that for nonnegative and C, whenever C = 0 and C + > 0, unbounded solutions exist for some choice of nonnegative initial conditions The proof is presented in two special cases The case where > 0 and the case where C > 0 2 Todd's Equation Consider the third order rational dierence equation x n = + βx n + x n 2 x n 3, n N 2 There have been signicant results concerning the case where β = In this case, the equation is generally referred to by the cognomen Todd's equation and possesses the invariant: + x n + x n + x n 2 + xn + + = constant x n x n 2 For more information regarding Todd's equation see Refs [5-7] In the following theorem we show that whenever 0 < β < 3 and [0, ], the dierence equation 2 has unbounded solutions for some choice of nonnegative initial conditions Theorem Suppose 0 < β < 3 and [0, ], then the dierence equation 2 has unbounded solutions for some choice of nonnegative initial conditions Proof Choose initial conditions so that min x 0, x 2 > max β, x β

14 4 We shall rst prove by induction that for all j N, min x 2j, x 2j 2 > max β, x 2j 22 β The initial conditions provide the base case Assume the following holds for some j N, min x 2j 2, x 2j 4 > max β, x 2j 3 β Since βx 2j 2 > x 2j 3, x 2j 2 > β, βx 2j 2 >, and x 2j 4 > β > 3, we see that x 2j = + βx 2j 2 + x 2j 3 x 2j 4 < 3βx 2j 2 x 2j 4 < βx 2j 2 Thus we have shown x 2j 2 > max β, x 2j β Since βx 2j 4 > x 2j 3 and 0 < β <, we have 3 Also x 2j = + βx 2j + x 2j 2 x 2j 3 > x 2j 2 x 2j 3 > x 2j 2 βx 2j 4 > 3x 2j 2 x 2j 4 = 3βx 2j 2 βx 2j 4 > x 2j β Thus This completes the induction x 2j > x 2j 2 x 2j 3 > x 2j 2 βx 2j 2 = β min x 2j, x 2j 2 > max β, x 2j β Using Equation 22 we now prove that x 8η > x 8η 8 9β 2 for all η N: x 8η = + βx 8η + x 8η 2 > x 8η 2 = x 8η 3 x 8η 3 x 8η 6 + βx8η 3 + x 8η 4 > + βx 8η 4 + x 8η 5 x 8η 5 x 8η 6 x 8η 6 x 8η 8 = 3βx 8η 5 3β + βx 8η 6 + x 8η 7 > x 8η 6x 8η 8 9β 2 x 8η 6 x8η 6 3βx 8η 4 x8η 4 = x 8η 5 = x 8η 8 9β 2 Since 0 < β < 3, 9β2 < Thus, we have a subsequence of our solution which diverges to Hence, the solution is unbounded

15 5 3 Special Case #44 We now study special case #44 x n = + x n x n 3, n N 3 Particularly, we show that whenever > 0, Equation 3 has unbounded solutions for some initial conditions The following lemma provides a useful technique for constructing divergent subsequences of solutions for rational dierence equations Lemma Let {x n } n= be a sequence in [0, Suppose that there exists D > and hypotheses H,, H k so that for all n N there exists p n N so that the following holds Whenever x n i satises H i for all i {,, k}, then x n+pn i satises H i for all i {,, k} and x n+pn Dx n Further assume that for some N N, x N i satises H i for all i {,, k} and x N > 0 Then {x n } n= is unbounded Particularly, {x zm } m= is a subsequence of {x n} n= which diverges to, where z m = z m + p zm and z 0 = N Proof Let z m = z m + p zm and z 0 = N Using induction, we prove that given m N the following holds x zm D m x N and x zm i satises H i for all i {,, k} By assumption, x N i satises H i for all i {,, k} and x N D 0 x N This provides the base case Assume x zm i satises H i for all i {,, k} and x zm D m x N Using our earlier assumption, this implies that there exists p zm so that x zm +p zm i satises H i for all i {,, k} and x zm +p zm Dx zm DD m x N = D m x N So we have shown that x zm D m x N for all m N Hence, the subsequence {x zm } m= of {x n } n= clearly diverges to, since D > The above argument merely simplies the following arguments by removing a

16 6 somewhat onerous construction Theorem 2 If > 0, then Equation 3 has unbounded solutions for some initial conditions Proof We choose initial conditions so that 2 5 x 0 > max, , x 3 > max 2, + 2 2, x 2 > 2 We show that there exists D = 4 3 so that for all n N there exists p n {7, 8} so that the following holds Whenever 2 5 x n > max, , x 3 n 2 > max Then we have 2 5 x n+pn > max First assume 2 5 x n > max, x n+pn 3 > 2,, , , x n+pn 2 > max x n+p n 4 x n 3, x n 3 > 2 2, + 2 2, 2, x n 2 > max, , x n 3 > 2 Since x n, x n 2, x n 3 > 0, we may write η = log 2 x n, l = log 2 x n 2, and ρ = log 2 x n 3 Hence 2 η = x n, 2 l = x n 2, and 2 ρ = x n 3 We use such representations for ease of computations First we see that x n = + x n x n 3 = x n 3 + x n x n 3 = 2 ρ + 2η ρ ; 32 x n+ = + x n = x n 2 x n l 2 + ρ 2η ρ 2 = l 2 + l 2 + ; 33 l+ρ 2l+ρ η x n+2 = + x n+ = x n x n 2 + η 2 + l 2 + l+ρ 2 l+ρ η 2 η = 2 + η 2 + η+l 2 + η+l+ρ 2 ; l+ρ 34

17 x n+3 = + x n+2 = 2ρ x n x n η 2 + η 2 + η+l ρ 35 η+l+ρ 2 l+ρ + 2 η We will make use of these identities later We prove the result in two cases Let us rst assume l + ρ η We show that if this inequality is satised for some n N, then p n = 7 First, we prove that x n+pn 3 = x n+4 > Notice that 2 From Equation 33 we see that x n+ = x n+4 = + x n+3 x n+ > x n+ + + = 2 l 2 l+ρ 2 l+ρ η 2 l + 2 l ρ + 2 η l ρ We now use the assumption l + ρ η This assumption implies that 2 η l ρ 2 0 = Earlier we assumed that 2 ρ < 2 Moreover, from our assumptions, 7 2 l > = so 2 l < 2 2 and 2 l < 2 Using these inequalities we obtain the following 2 l + 2 l ρ + 2 η l ρ > So we have shown x n+pn 3 = x n+4 > 2 We now prove that x n+pn 2 = x n+5 > max = 2 x n+5 = + x n+4 x n+2 > x n+4 x n+2 > > 2 2 2, Notice that 2x n+2 Since l+ρ η, 2 l+ρ 2 η Moreover, as we have recently shown, 2 l > 2, similarly 2 η > 2 5 So l > > 0 and η > 5 > 0 So from Equation 34, Hence, x n+2 = 2 η + x n+5 > 2x n+2 > 2 + η+l 2 + η+l+ρ 2 < l+ρ 2 + η 2 + η 2 + η 2 = η 2 η 2 3+ = 2 η 2 4 > max , η

18 8 So, x n+5 > 2 4 max 3 +, > = > When, 2 < + 2 so = max, Thus, the only remaining case is when < In this case we have the following: x n+5 > 2 4 max 3 +, So we have shown x n+pn 2 = x n+5 > max We now prove that x n+pn = x n > > , x n > max 3, First assume, Notice that max + 2 5, η ρ 37 x n+6 = + x n+5 x n+3 > x n+5 x n+3 = + x n+4 x n+2 x n+3 > x n+4 x n+2 x n+3 = + x n+3 x n+2 x n+3 x n+ > x n+2 x n+ We use Equation 34, our induction assumption, our assumption 37, and the fact that 2 ρ < 2 to obtain x n+ = 2 + 2η ρ + < l 2l+ρ 2 l 2 + max + 2 5, +25 l 2 + l max 2, Notice that if, max + 2 5, +25 max , = + 2 <

19 9 Also if <, max + 2 5, +25 max , = So, < < < x n+ < 2 l + 2 l < = < 3 + Now, using the inequality we have just shown and Equation 36, we have x n+6 > Thus we have shown > x n+ x n η > η = x n+pn = x n+6 4 x n x n > max 3, when 37 holds Now assume the opposite inequality in 37 Using Equation 35 and Equation 36, we have the following: So, Since < 2ρ + 2 η + > x n+3 = 2ρ 2 ρ x n+2 η + 2 η ρ 2 η + 2 η x n+6 > x n+5 x n+3 > + x n+4 x n+3 x n+2 x n+ > 2 η > < 2 ρ η η > x n+4 x n+3 x n+2 x n+3 x n+2 2 ρ η η xn+2 x n+ > > 3 +, we see that 3+ 2 η < and using Equation 33, we get x n+6 > 2 ρ η + x n+2 x n+ = 2 ρ η + x n+2 2 l + 2 l+ρ + 2 l+ρ η

20 0 Distributing the 2 ρ η, we have x n+6 > + x n l+η ρ 2 l+η + 2 l 38 Now let us assume Then we have x n+6 > In this case + x n l+η ρ 2 l+η + 2 l = 2 l + x n+2 2 η ρ + 2 η l > max, , so x n+6 > 2 + x n η ρ + 2 η Since we assumed the reversed inequality in 37, we have that 2 η ρ > 2 5 > Furthermore, we know from earlier that 2 η > 2 5 > Using this information, we obtain x n+6 > Now we use Equation 36 and we obtain 2 + > 2 + x n x 2 η ρ 2 η n+2 x n+6 > x n > x n = 2 9 x n > 4 x n 3 We now prove the case when < Here we continue from Equation 38 with the following: x n+6 > In this case so we have + x n l+η ρ 2 l+η + 2 l = 2 2 l > max, , x n+6 > 2 + x n η ρ + 2 η 2 l + x n+2 2 η ρ + 2 η +

21 Since we assumed the reverse inequality in 37, we have that 2 η ρ > 2 5 > Furthermore, we know from earlier that 2 η > 2 5 > Using this information, we obtain x n+6 > 2 + x n η ρ + 2 η > x n+2 Now we use Equation 36 and the assumption < to obtain x n+6 > Thus, we have shown x n > 2 24 x n > x n+pn = x n+6 4 x n x n > max 3, when the opposite inequality of 37 holds Therefore, we have nished the case where l + ρ η We now consider the case l+ρ < η We show that if this inequality is satised for some n N, then p n = 8 First we prove that x n+pn 3 = x n+5 > Notice that 2 since our assumptions have changed, Equations 36 and 38 no longer hold We will now make a new analogue for Equation 36, namely the forthcoming Equation 39 Since l + ρ < η we have 2 l+ρ < 2 η Moreover, since 2 l > 2 and 2 η > 2 5 we have l > 0 and η > 0 So from Equation 6, x n+2 = 2 η + So we have Notice that 2 + η+l 2 + η+l+ρ 2 < l+ρ 2 + l+ρ 2 + l+ρ = 39 l+ρ 2l+ρ 2l+ρ x n+5 = + x n+4 x n+2 > x n+2 > 2l+ρ l > max 3 +, = max max 2 + 2, , >

22 2 So x n+5 > 2 ρ > 2 We now prove that 2 x n+pn 2 = x n+6 > max, Notice that from Equation 35, we have x n+3 = Since x n+5 > 2 ρ, we get 2ρ 2 ρ x n+2 < 2 ρ η + x η + 2 η n+2 30 x n+6 > x n+5 x n+3 > 2 η ρ x n+5 + x n+2 > 2 η + x n+2 3 We assume + and we use Equation 39 We know that 2ρ > 2 and 2 l > So, x n+2 < 3 + < l+ρ +2 So, since +, x n+6 > 2 η + x n+2 > 2 = < < η > max , > = > We now assume > + and we use Equation 39 We know that 2ρ > 2 and 2 l > So, x n+2 < 3 + < l+ρ +2 2 = < < We now use Equation 3 and our assumption > + to obtain the following: x n+6 > 2 η + x n+2 > 2 η > max , > = 25 2 > 2 2 >

23 3 Thus we have shown that Now we prove Notice that 2 x n+pn 2 = x n+6 > max, x n+pn = x n x n > max 3, x n+7 = + x n+6 x n+4 > x n+6 x n+4 = + x n+5 x n+4 x n+3 > x n+5 x n+4 x n+3 = + x n+4 x n+2 x n+3 x n+4 > Using Equations 39 and 30, we have x n+7 > x n+2 x n+3 > 2 l+ρ 3 + x n+3 > 2 l+η x n+2 x n+2 x n+3 Earlier we demonstrated that x n+2 < 2 8 Furthermore, we have assumed that 2 l > Thus, x n+7 > 2 l+η x n+2 > 2 l+η > η = x n > 4 x n 3 3 Hence x n+pn = x n+7 4 xn > max 25, Now we apply Lemma, 3 3 and then the proof is done 4 Special Case #28 We now study special case #28 x n = x n Cx n 2 + x n 3, n N 4 Particularly, we show that whenever C > 0, the dierence equation 4 has unbounded solutions for some initial conditions

24 4 Theorem 3 If C > 0, then the dierence equation 4 has unbounded solutions for some initial conditions Proof We choose initial conditions so that x 0 > max 000C + 3, 000C + 3, C x > max 0, 0 C, C 3 00C + 3, 00C 2 + C, C 3 We show that there exists D = 2 so that for all n N there exists p n {7, 8} so that the following holds: Whenever x n > max 000C + 3, then we have 000C + 3, C x n 2 > max x n+pn > max 000C + 3, First assume 0, 0 C,, C 3 000C + 3, C 00C + 3, 00C 2 + C, C 3 00C + 3, 00C 2 + C, C 3 x n+pn 2 > max 0, 0 C,, x C 3 n+pn 2x n x n > max 000C + 3, 000C + 3, C x n 2 > max 0, 0 C, C 3 Using algebra, we immediately obtain the following: x n = x n Cx n 2 + x n 3 < x n+ = x n = x n Cx n 2 < x n 0 ; x n+ = x n Cx n + x n 2 < x n Cx n 2 + x n 3 < x n 00C + 3, 00C 2 + C, C 3 x n Cx n + x n 2 < x n 000C ; < Cx n x n < x n Cx n 2 0C ; x n x n 2 < x n 0 ;

25 5 x n+2 = x n+ Cx n + x n < x n+ x n < x n+ 00C ; x n+2 = So we get the following inequalities: Using Equation 43 we get x n+ Cx n + x n < x n+ x n < x n x n+2 < 00x n+ < 0x n < x n ; C 3 x n+2 < 0000C 2 x n+ < 0Cx n < x n 43 x n+2 x n+3 = Cx n+ + x n < 2x n 44 We use Equations 42 and 43 to get C Cx n+3 + x n+2 = x n+2 + Cx n+ + x n < x n+2 + < x n+2 + x n+ 000x n+2 = x n = x n Cx n + x n Cx n + x n < + Cx n < < In short, Using Equations 43 and 44, we have x n+5 = x n+5 = Furthermore we have x n+4 = x n+4 Cx n+3 + x n+2 = Cx n+2 + x n+ Cx n+3 + x n+2 < 45 x n+3 Cx n+2 + x n+ < x n+3 x n+ = Cx n+2 + x n+ C + x n+2 x n+3 > 2x n+ x n+3 Cx n+3 + x n+2 C + 2x n 46 Cx n+ + x n Cx n + x n < 47 x n x n Using Equations 46 and 47, we obtain x n+5 x n+6 = > C Cx n+4 + x n+3 2x n+ C + 2x n x n x n + x n+3

26 6 xn = C 2x n+ C + 2x n x n + x n x n+3 Cxn + x n 2 Cx n 2 + x n 3 = C 2C + 4x n x n + x n x n+3 Cxn + x n 2 2 Cx n 2 + x n 3 = C 2C + 4x n x n+ + x n x n+3 Cx n + x n 2 Using Equations 42 and 43, we see x n x n+3 Cx n +x n 2 = x n x n+2 Cx n + x n 2 Cx n+ + x n = x n x n+ Cx n + x n 2 Cx n+ + x n Cx n + x n = x n x n Cx n+ + x n Cx n + x n < < C 3 x n 2 < CCx n + x n 2 Cx n 2 + x n 3 x n = C x n+ Using this in the prior inequality, we get Using this fact, we have x n+7 = x n+6 Cx n+5 + x n+4 = x n+6 > x n Cx n + x n 2 4C 2 + 8Cx n 48 x n+6 = Cxn+4 Cx n+3 +x n+2 + x n+4 xn+6 x n+4 x n+4 = C Cx n+3 + x n+2 Cx n+4 + x n+3 x n+4 + = > C + Cx n+3 + x n+2 Cx n+4 + x n+3 Cx n+4 + x n+3 We now use Equations 42 and 43 to show x n+3 = < x n+2 Cx n+ + x n = C + Cx n+3 +x n+2 Cx n+3 +x n+2 x n+ Cx n+ + x n Cx n + x n x n+ Cx n Cx n+ + 5x n+ Cx n + x n < 50C + 5x n C +

27 C x nx n + x n 4C+4 < 4C + 4x n We now use this fact and Equation 47 to obtain x n+7 > = C + x n C x n + 4C C + Suppose x n 4CC + we will show that in this case p n = 7 Using Equation 46, we have Cx n + x n 2 2x n = x n+pn 2 = x n+5 > > C + 2x n 2x n+ Cx n 2x n C + 2x n = C + 2x n x n 8C + C + 8CC + > x n 00C + 3 > max Also using Equation 48, we have x n+pn = x n+6 > x n Cx n + x n 2 4C 2 + 8Cx n > = x 2 n 4C + 84CC + > Cx n 8CC + C + 8CC + 0, 0 C, C 3 Cx 2 n 4C 2 + 8C4CC + x 2 n 50C 2 + C > 2x n Now suppose x n > 4CC + We will show that in this case p n = 8 Using Equation 48, we have x n+pn 2 = x n+6 > x n Cx n + x n 2 > x n Cx n + x n 2 4C 2 + 8Cx n 9Cx n = Cx n + x n 2 Cx n 2 + x n 3 9C Also using Equation 49, we have x n+pn = x n+7 > > x n C + 4CC+ 4C+ > x n 2 > max x n C x n + 4C+ C + = 2x n C + Hence, after application of Lemma, the proof is complete 0, 0 C, C 3

28 8 5 Conclusion Theorem establishes the boundedness character of special cases #56 and #20 in a range of their parameters Theorems 2 and 3 establish the boundedness character of the special cases #44 and #28 respectively Further work should focus on expanding the range for which boundedness character of special cases #56 and #20 is known and resolving Conjecture 30 in Ref [] There remains only one special case for which Conjecture 30 has not yet been established This is special case #70 It is worthwhile to note that special case #70 is part of the period-six trichotomy conjecture The resolution of the period-six trichotomy conjecture will immediately resolve Conjecture 30 in Ref [] See Ref [2] for more details regarding the period-six trichotomy conjecture We restate, for the convenience of the reader, the period-six trichotomy conjecture Conjecture Assume that, C [0, Then the following period-six trichotomy result is true for the rational equation x n = + x n Cx n 2 + x n 3, n N 5 a Every solution of Equation 5 converges to its positive equilibrium if and only if C 2 > b Every solution of Equation 5 converges to a not necessarily prime periodsix solution of Equation 5 if and only if C 2 = c Equation 5 has unbounded solutions if and only if C 2 < For more on boundedness character see Refs [-4] List of References [] E Camouzis and G Ladas, Dynamics of Third-Order Rational Dierence Equations with Open Problems and Conjectures, Chapman & Hall/CRC Press, Boca Raton, 2007

29 9 [2] E Camouzis and G Ladas, Three trichotomy conjectures, J Dierence Equ Appl 82002, [3] E Camouzis, G Ladas, F Palladino, and EP Quinn, On the Boundedness Character of Rational Equations, Part, J Dierence Equ Appl 22006, [4] EA Grove and G Ladas, Periodicities in Nonlinear Dierence Equations, Chapman & Hall/CRC Press, Boca Raton, 2005 [5] EA Grove and G Ladas, Periodicity in nonlinear dierence equations, Revista Cubo May2002, [6] VL Kocic and G Ladas, Global Behavior of Nonlinear Dierence Equations of Higher Order with Applications, Kluwer Academic Publishers, Dordrecht, 993 [7] MRS Kulenovic and G Ladas, Dynamics of Second Order Rational Dierence Equations With Open Problems and Conjectures, Chapman & Hall/CRC Press, 200

30 20 MANUSCRIPT 2 Unboundedness results for fourth order rational dierence equations by Gabriel Lugo and Frank J Palladino Publication Status: Accepted for Publication G Lugo and FJ Palladino, Unboundedness results for fourth order rational difference equations, J Dierence Equ Appl accepted 99 Mathematics Subject Classication: 39A0,39A Keywords: dierence equation; unboundedness; boundedness character; fourth order equations

31 2 Abstract We present some new results regarding the boundedness character of the k th order rational dierence equation x n = + k i= β ix n i A + k j= B jx n j, n N When applied to the general fourth order rational dierence equation, these results prove the existence of unbounded solutions for 49 special cases of the fourth order rational dierence equation, where the boundedness character has not been established yet This resolves 49 conjectures posed by E Camouzis and G Ladas 2 Introduction In Ref [], E Camouzis and G Ladas posed numerous conjectures regarding the boundedness character of third and fourth order rational dierence equations Later on, the same authors published [2], where they proved that there exist unbounded solutions for 60 additional special cases, whose boundedness character had not been established at that time After this, there remained 49 special cases of third and fourth order, for which E Camouzis and G Ladas had conjectured that there exist unbounded solutions and the conjecture remained open These cases were listed in Ref [2] in Appendix A Later on, [5] was published In Ref [5], the authors resolved the conjectures 28, 44, 56, and 20, listed in Appendix A of Ref [2] Thus, there remains only one special case of third order for which the boundedness character has not been established yet This is special case #70, which is the following equation, x n = + x n Cx n 2 + x n 3, n N In the process of proving a general periodic trichotomy result, the author of Ref

32 22 [8] resolved the conjectures 296, 578, 586, 608, 60, 66, and 68 Furthermore, the conjectures 584, 609, 6, 67, and 69 were resolved in Ref [4] Thus, there remain 49 6 = 33 special cases of third and fourth order, for which E Camouzis and G Ladas have conjectured that there exist unbounded solutions and the conjecture has not been established yet When applied to fourth order rational dierence equations, the results in this manuscript establish 49 out of the 33 remaining conjectures from Appendix A of Ref [2] Theorem 4 establishes the conjectures 620, 62, 622, 623, 624, 625, 632, 633, 634, 635, 636, 637, 638, 639, 864, 865, 872, 873, 874, 875, 876, 877, 878, 879, 880, 88, 888, 889, 890, 89, 892, 893, 894, and 895 Theorem 5 establishes the conjectures 64, 65, 626, 627, 630, 63, 866, 867, 870, 87, 882, 883, 886, and 887 Theorem 6 establishes conjecture 585 So, there now remain = 84 special cases of third and fourth order, for which E Camouzis and G Ladas have conjectured that there exist unbounded solutions and the conjecture has not been established yet We include a list of the remaining cases in Appendix A The results in this paper can be considered a generalization of the results in Refs [2-5,7-8] To be more specic, the idea for Theorem 4 came from attempts to combine the methods used in Ref [2] with the modulo class techniques used in Ref [8] The idea for Theorem 5 came from attempts to combine the methods used in Ref [2] with the technique of iteration developed in Refs [3] and [7] Furthermore, Theorem 6 is a direct generalization of methods used in Refs [4] and [5] 22 Some General Unboundedness Results In Ref [8], techniques for proving unboundedness involving modular arithmetic on the indices are introduced These techniques are expanded upon in Ref [4] and partially extended to systems in Ref [6] Here, we extend these ideas further in order to solve some conjectures in Ref [2]

33 23 We rst introduce a condition which allows us to construct unbounded solutions, namely Condition Before doing so, let us rst introduce some notation Let us dene the following sets of indices : I β = {i {, 2,, k} β i > 0} and I B = {j {, 2,, k} B j > 0} These sets are used extensively in Ref [9] when referring to the k th order rational dierence equation Similarly, we shall make extensive use of this notation In the following proof, we use nite subsets of the set {,, k} as indexing sets in sums For example, we may write i I B I β β i In the case where I B I β = {, 2, 3} then i I B I β β i = β +β 2 +β 3 Let us point out the notational convention that if I B I β =, then i I B I β β i = 0 The notation is similar for all such sums indexed in this way Condition We say that Condition is satised if, for some p N, p gcd I β \ I B We also must have disjoint sets B, L {0,, p } with B and with the following properties For all b B, {b j mod p : j I B } L 2 For all l L, there exists j I B so that l j mod p B We now present Theorem 4, which makes use of Condition In the remainder of this section, we will verify Condition for a number of special cases of the fourth order rational dierence equation, thereby conrming several conjectures in Refs [2] and [] Theorem 4 Consider the k th order rational dierence equation, x n = + k i= β ix n i A + k j= B jx n j, n N 22

34 24 Assume nonnegative parameters and nonnegative initial conditions Further assume that 4 i I B I β β i k j= B j min j IB B j < i I β \I B β i A, and that Condition is satised for Equation 22 Then unbounded solutions of Equation 22 exist for some initial conditions Proof By assumption, we may choose p N and B, L {0,, p } so that Condition is satised Choose initial conditions x m, where m {0,, k }, so that the following holds If m mod p B, then 4 k j= x m > B j min j IB B j i I β \I B β i A + 2 i I β \I B β i min j IB B j If m mod p L, then x m < i I β \I B β i A 2 k j= B j Also, assume x m > 0 for all m {0,, k } Under this choice of initial conditions, our solution {x n } has the following properties a x n > whenever n mod p B 4 k j= B j min j IB B j i I β \I B β i A + 2 i I β \I B β i min j IB B j b whenever n mod p L x n < i I β \I B β i A 2 k j= B j c x n > 0 for all n N We prove this using induction on n, our initial conditions provide the base case Assume that the statement is true for all n N We show the statement

35 25 for n = N This induction proof has three cases Let us begin by assuming N mod p B Condition tells us that in this case, {N j mod p : j I B } L Hence, x N j < i I β \I B β i A 2 k j= B j for all j I B Hence, we have the following: x N = + k i= β ix N i A + k j= B jx N j A + k j= B j = i I β \I B β i A + i I β \I β i A B 2 To complete this case, notice that since min x N i i I β \I B 0 4 i I B I β β i k j= B j min j IB B j we have that A < i I β \I B β i Thus, i I β \I B β i i I β \I β i A B 2 k j= B j min x N i i I β \I B = 2 i I β \I B β i A + min x N i i I β \I B β i i I β \I B < i I β \I B β i A, So, this gives us 2 i I β \I B β i A + i I β \I B β i > x N > min x N i i I β \I B Since p gcdi β \ I B, N mod p = N i mod p for all i I β \ I B Thus, for all i I β \ I B, x N i > 4 k j= B j min j IB B j i I β \I B β i A + 2 i I β \I B β i min j IB B j Thus, x N > This nishes case a 4 k j= B j min j IB B j i I β \I B β i A + 2 i I β \I B β i min j IB B j

36 26 We now assume N mod p L Since p gcdi β \ I B, N mod p = N i mod p for all i I β \ I B Hence x N i < i I β \I B β i A 2 k j= B j for all i I β \ I B Condition 2 guarantees that there exists j I B so that 4 k j= x N j > B j min j IB B j i I β \I B β i A + 2 i I β \I B β i min j IB B j Hence, we have the following: x N = + k i= β ix N i A + k j= B jx N j < min j IB B j = + i I β \I B β i i I β \I B β i A 2 k j= B j 4 k j= B j min j IB B j + 2 i I β \I β i A B + i I β \I B β i 2 i I β \I B β i A 2 k j= B j i I β \I β i B min j IB B j 2 k j= B j + + i I β \I β i A i I β \I B β i B + i I B I β β i min j IB B j i I B I β β i min j IB B j = 2 k j= B j + i I β \I β i A i I β \I B β i B 2 k j= 2 B j + i I β \I β i A i I β \I B β i B = i I β \I B β i A 4 k j= B j Since we have assumed that we get that 4 i I B I β β i k j= B j min j IB B j x N i I β \I B β i A 4 k j= B j + i I β \I B β i A 2 k j= B + j + i I B I β β i min j IB B j < i I β \I B β i A, i I B I β β i min j IB B j i I B I β β i min j IB B j < i I β \I B β i A 2 k j= B j

37 27 This completes case b It is clear that if x n > 0 for n < N, then x N > 0 so case c is trivial We now use the facts we obtained from our induction to prove that a particular subsequence is unbounded Take b B We now show that {x mp+b } m= diverges to We explained earlier that x mp+b 2 i I β \I B β i A + min x mp+b i i I β \I B β i i I β \I B This implies that the following inequality holds for C = 2 i I β \I B β i A+ i I β \I B β i > x mp+b C min i {,, k p } x mp+b ip This is a dierence inequality which holds for the subsequence {x mp+b } for m k We now rename this subsequence and apply the methods used in [7] We set z m = x mp+b for m N As we have just shown, {z m } satises the following dierence inequality, z m C min i {,, k p } z m i, m k Using the results of Ref [7], particularly Theorem 6, we have that for m k, min z m,, z m k min y p,, y m k m k k p Where {y m } m=0 is a solution of the dierence equation, y m = Cy m, m N 222 With y 0 = min z k,, z k k Clearly every positive solution diverges to p for the simple dierence equation 222, since C > Hence, using the inequality we have obtained, {z m } m= diverges to Hence, with given initial conditions, there is a subsequence of our solution {x n } n=, namely {x mp+b } m=, which diverges

38 28 to Hence, our solution {x n } n= is unbounded So, we have exhibited an unbounded solution under our current assumptions Now, let us apply Theorem 4 to resolve some conjectures in Ref [2] Corollary Consider the 4 th order rational dierence equation, x n = + βx n + γx n 2 + δx n 3 + ɛx n 4 A + Bx n + Cx n 2 + Ex n 4, n N 223 Assume nonnegative initial conditions and assume that B, C, E, δ > 0 and that all other parameters are allowed to take on arbitrary nonnegative values Further choose δ large enough so that A + 4β + γ + ɛb + C + E minb, C, E < δ Then unbounded solutions of Equation 223 exist for some initial conditions Proof We check the hypotheses of Theorem 4 and then apply Theorem 4 Since we have chosen δ large enough so that A + 4β + γ + ɛb + C + E minb, C, E < δ, we need only check that Condition is satised Notice that for all choices of parameters we have allowed here 3 I β and I β \ I B = {3} This is not accidental In this case, we let p = 3, B = {}, and L = {0, 2} Notice that B L = and notice the following : mod 3 = 0 L, 2 mod 3 = 2 L, 4 mod 3 = 0 L, 0 2 mod 3 = B, 2 mod 3 = B Thus Condition is satised and so Theorem 4 applies and unbounded solutions of Equation 223 exist for some initial conditions

39 29 Corollary establishes the conjectures 620, 62, 622, 623, 632, 633, 634, 635, 636, 637, 638, 639, 872, 873, 874, 875, 876, 877, 878, 879, 888, 889, 890, 89, 892, 893, 894, and 895 Corollary 2 Consider the 4 th order rational dierence equation, x n = + γx n 2 + δx n 3 + ɛx n 4 A + Cx n 2 + Ex n 4, n N 224 Assume nonnegative initial conditions and assume that C, E, δ > 0 and that all other parameters are allowed to take on arbitrary nonnegative values Further choose δ large enough so that A + 4γ + ɛc + E minc, E < δ Then unbounded solutions of Equation 224 exist for some initial conditions Proof We check the hypotheses of Theorem 4 and then apply Theorem 4 Since we have chosen δ large enough so that A + 4γ + ɛc + E minc, E < δ, we need only check that Condition is satised Notice that for all choices of parameters we have allowed here 3 I β and I β \ I B = {3} This is not accidental In this case, we let p = 3, B = {}, and L = {0, 2} Notice that B L = and notice the following : 2 mod 3 = 2 L, 4 mod 3 = 0 L, 0 2 mod 3 = B, 2 4 mod 3 = B Thus, Condition is satised and so Theorem 4 applies and unbounded solutions of Equation 224 exist for some initial conditions

40 30 Corollary 2 establishes the conjectures 624,625,864,865,880, and 88 Theorem 4 resolves Open Problem 6 in Ref [2] Theorem 4 generalizes Theorem 6 in Ref [2] through the use of modulo class techniques, such as those used in Ref [8] 23 Unboundedness by Iteration The technique of iteration has been a very useful tool for proving that every solution is bounded in many special cases of the k th order rational dierence equation See Refs [3] and [7] for a discussion on the technique of boundedness by iteration In the following theorem, we use the technique of iteration in order to obtain bounds for certain subsequences of the solutions which we study Obtaining bounds for these subsequences is critical in order to create an unbounded solution For this reason, we label this technique unboundedness by iteration Theorem 5 Consider the following fourth order rational dierence equation x n = + βx n + γx n 2 + δx n 3 + ɛx n 4 A + Cx n 2 + Ex n 4, n N, Suppose that all of the following conditions hold i A, C, E > 0, ii δ > 2A + C + E, iii + γ + ɛ + + β + β2 + βγ + βɛ + βδ minc,e C E A A 2 AC AC AE AE Then, under a proper choice of initial conditions, lim n x 3n+ = So, the above dierence equation has unbounded solutions Proof We choose initial conditions in such a way that for n <, x n if n mod 3, and x n δ if n mod 3 This provides the base case in the following induction proof Suppose that for all n < N, x n if n mod 3, and x n δ if n mod 3 We shall then prove that x N if N mod 3, and x N δ if

41 3 N mod 3 The proof is divided into two cases The case N mod 3 and the case N mod 3 Consider the case where N mod 3, then x N = + βx N + γx N 2 + δx N 3 + ɛx N 4 A + Cx N 2 + Ex N 4 δx N 3 A + Cx N 2 + Ex N 4 Now since N mod 3, N 2 mod 3 and N 4 mod 3 Thus, by our induction hypothesis, x N 2, x N 4 Thus, we have x N δx N 3 A + C + E 2x N 3 Since N 3 mod 3 by our induction hypothesis we get, x N 2x N 3 δ Now consider the case where N mod 3, then x N = + βx N + γx N 2 + δx N 3 + ɛx N 4 A + Cx N 2 + Ex N 4 δx N 3 βx N + + γ A + Cx N 2 + Ex N 4 A + Cx N 2 + Ex N 4 C + ɛ E + A Since N mod 3, either N 2 mod 3 or N 4 mod 3 Thus, by our induction hypothesis, either x N 2 δ or x N 4 δ Further, since N mod 3, N 3 mod 3 so x n 3 These facts combine to give us x N δ minc, Eδ + βx N + γ A + Cx N 2 + Ex N 4 C + ɛ E + A To complete the proof, we iterate the x N term In other words, we substitute in a copy of our dierence equation one step back for this term Notice that since N mod 3 in this case and N N, we have that N 2 in this case and so this is permissable Once we iterate, we get x N minc, E + γ C + ɛ E + A + β + βx N 2 + γx N 3 + δx N 4 + ɛx N 5 A + Cx N 2 + Ex N 4 A + Cx N 3 + Ex N 5

42 32 minc, E + γ C + ɛ E + A + β A + β2 2 AC + βγ AC + βɛ AE + βδ AE This completes the induction proof Notice that along the way we proved that, for all n mod 3, x n 2x n 3 Thus lim n x 3n+ = This concludes our proof of unboundedness Notice that the only parameters which are required to be positive for the given equation in Theorem 5 are A, C, E, δ > 0 B and D must be zero as shown above The rest are allowed to take on arbitrary nonnegative values Thus, Theorem 5 establishes that there exist unbounded solutions for some choice of parameters and some choice of initial conditions for the special cases: 64, 65, 626, 627, 630, 63, 866, 867, 870, 87, 882, 883, 886, and The equation x n = +x n 3 Bx n +x n 4 For the following dierence equation x n = + x n 3 Bx n + x n 4, n N, we show that whenever B > 2 8 and < B 3 unbounded solutions exist for some choice of nonnegative initial conditions Our proof will establish the Conjecture 585 in Ref [2] We make use of the argument structure presented in Lemma of Ref [5] The following lemma is an adaptation to be used for our case here Lemma 2 Let {x n } n= be a sequence in [0, Suppose that there exists D > and hypotheses H,, H k so that for all n N there exists p n N so that the following holds Whenever x n i satises H i for all i {,, k}, then x n+pn i satises H i for all i {,, k} and x n+pn 3 Dx n 3 Further assume that for some N N, x N i satises H i for all i {,, k} and x N 3 > 0 Then {x n } n=

43 33 is unbounded Particularly {x zm 3} m= is a subsequence of {x n} n= which diverges to, where z m = z m + p zm and z 0 = N Proof Recall from the assumptions that for all n N there exists p n N so that the following holds Whenever x n i satises H i for all i {,, k}, then x n+pn i satises H i for all i {,, k} and x n+pn 3 Dx n 3 This p n may not necessarily be unique for any given n N If there is more than one value that can act as p n for a given n, then take the smallest such value and call that value p n Let z m = z m + p zm and z 0 = N, this is well dened from the prior explanation Using induction, we prove that given m N the following holds x zm 3 D m x N 3 and x zm i satises H i for all i {,, k} By assumption, x N i satises H i for all i {,, k} and x N 3 D 0 x N 3 This provides the base case Assume x zm i satises H i for all i {,, k} and x zm 3 D m x N 3 Using our earlier assumption, this implies that there exists p zm so that x zm +p zm i satises H i for all i {,, k} and x zm +p zm 3 Dx zm 3 DD m x N 3 = D m x N 3 So we have shown that x zm 3 D m x N 3 for all m N Hence, the subsequence {x zm 3} m= of {x n } n= clearly diverges to, since D > Theorem 6 Consider the fourth order rational dierence equation, x n = + x n 3 Bx n + x n 4, n N 24 Suppose B > 2 8 and < B 3, then Equation 24 has unbounded solutions for some initial conditions Proof We choose initial conditions so that x 2 > B, x 3 < 4, and one of the following holds

44 34 x 0 < 4B and x < B, 2 4B x 0 2x 2 and x < 2 B 2 x 2, 3 x 0 > 2x 2 and x < 2 B 2 x 2 We show that there exists D = 2 so that for all n N, there exists p n {2, 3, 5} so that the following holds Whenever x n 3 > B, x n 4 < 4, and one of the following holds x n < 4B and x n 2 < B, 2 4B x n 2x n 3 and x n 2 < 2 B 2 x n 3, 3 x n > 2x n 3 and x n 2 < 2 B 2 x n 3 Then, we have x n+pn 3 > Dx n 3 > B, x n+pn 4 < 4, and one of the following holds x n+pn < 4B and x n+p n 2 < B, 2 4B x n+p n 2x n+pn 3 and x n+pn 2 < 2 B 2 x n+pn 3, 3 x n+pn > 2x n+pn 3 and x n+pn 2 < 2 B 2 x n+pn 3 First assume x n < 4B, x n 2 < B, x n 3 > B, x n 4 < 4

45 35 In this case, p n = 3 Since B > 2 8, we have Since x n 4 < and x 4 n <, we have 4B x n+pn 3 = x n = Since x n 2 < B and < B 3, x n+pn 2 = x n+ = x n+pn 4 = x n < 4B < 4 + x n 3 Bx n + x n 4 + x n 2 Bx n + x n 3 x n 3 2 max Bx n, x n 4 > 2x n 3 > B B 3 + B Bx n < 2 B 2 x n < 2 B 3 < B Hence, regardless of the value of x n+pn, one of our requirements is satised If x n+pn < 4B, then requirement is satised If 4B x n+p n 2x n+pn 3, then requirement 2 is satised If x n+pn > 2x n+pn 3, then requirement 3 is satised Next assume 4B x n 2x n 3, x n 2 < In this case, p n = 5 Since B > 2 8 and < B 3, we have x n+pn 4 = x n+ = 2 B 2 x n 3, x n 3 > B, x n 4 < 4 + x n 2 Bx n + x n 3 < B 3 x n 3 + x n 2 x n 3 Since x n 2 < 2 B 2 x n 3 < B 3 x n B 2 x 2 n 3 and B > 2 8, we have < 4 B 3 x n 3 < 4 Also notice that, x n+pn 3 = x n+2 = > + x n Bx n+ + x n 2 x n 2 max Bx n+, x n 2 x n 2 max 4 B 2 x n 3, 2 B 2 x n 3 B2 x n B > 2x n 3 > B + x n x n+pn 2 = x n+3 = Bx n+2 + x n

46 < Bx n+2 + Bx n+2 + B 2 x n+2 x n + x n 3 Bx n+2 + x n Bx n + x n 4 x n 3 Bx n+2 + x n Bx n + x n 4 < + Bx n+2 x n 3 Bx n+2 Bx n + x n 4 < 2 5 x n 3 + Bx n+2 B 2 x n 3 Bx n + x n 4 < + 25 < Bx n+2 B 3 x n B B < 2 B Notice that since B > 2 8, < B 3, x n 3 > B, x n 2x n 3, and x n+ < 4 B 3 x n 3, x n+pn = x n+4 = + x n+ < 4 + Bx n+3 + x n x n B 3 x n 3 Bx n+3 + x n < 4 + x n B 3 x n 3 x n Bx n + x n 4 x n 3 + 4Bx n + 4x n 4 B 3 x 2 n 3 < 2Bx n B 3 x n 3 + 8Bx n 3 + B 3 x 2 n 3 < 6Bx n B 3 x n 3 < 6 B B 3 x n 3 < 4B Hence, requirement is satised in this case Finally, assume x n > 2x n 3, x n 2 < In this case, p n = 2 Immediately, we have Also by assumption, x n+pn 4 = x n 2 < 2 B 2 x n 3, x n 3 > B, x n 4 < 4 2 B 2 x n 3 < 4 x n+pn 3 = x n > 2x n 3 > B Furthermore, since x n > 2x n 3 and < B 3, x n+pn 2 = x n = Furthermore, + x n 3 Bx n + x n 4 < Bx n + x n 3 Bx n < 2B 5 + 2B < B 36 x n+pn = x n+ = + x n 2 Bx n + x n 3 < x n 3 + x n 2 x n 3 < B 3 x n B 2 x 2 n 3 < 4B Hence, requirement is satised in this case, so after application of Lemma 2 the proof is complete

47 37 25 Conclusion Our work here resolves more than one third of the outstanding conjectures regarding the existence of unbounded solutions for 4 th order rational dierence equations However, there remain 84 special cases of third and fourth order, for which E Camouzis and G Ladas have conjectured that there exist unbounded solutions and the conjecture has not been established yet We include a list of the remaining cases in the attached Appendix A One special case of particular interest is special case #70, since it is the only remaining third order rational dierence equation whose boundedness character is yet to be determined Special case #70 is the following equation, x n = + x n Cx n 2 + x n 3, n N The technique used here in Theorem 6 and originally developed in Ref [5] is a new approach that we have found useful for tackling some of the particularly thorny cases Perhaps this type of approach will conquer the special case #70 For the most recent ideas regarding boundedness character, the reader should look to Refs [,5,8] Much of our work here is built o of the ideas in these papers For readers wishing to expand into higher order rational dierence equations, for example fth order and beyond, Conjecture of Ref [8] may be of interest We feel that further work in this direction would be best focused on resolving any of the conjectures we have just mentioned 26 Appendix A In this appendix, we present the remaining 84 special cases of order less than or equal to four, for each of which E Camouzis and G Ladas have conjectured that the equation has unbounded solutions in some range of the parameters and for some initial conditions

48 38 #70 : x n+ = + x n /Cx n + x n 2 #292 : x n+ = βx n + ɛx n 3 /Cx n #293 : x n+ = + βx n + ɛx n 3 /Cx n #294 : x n+ = βx n + ɛx n 3 /A + Cx n #295 : x n+ = + βx n + ɛx n 3 /A + Cx n #297 : x n+ = + ɛx n 3 /Bx n + Cx n #308 : x n+ = βx n + γx n + ɛx n 3 /Cx n #309 : x n+ = + βx n + γx n + ɛx n 3 /Cx n #30 : x n+ = βx n + γx n + ɛx n 3 /A + Cx n #3 : x n+ = + βx n + γx n + ɛx n 3 /A + Cx n #328 : x n+ = δx n 2 + ɛx n 3 /Bx n #329 : x n+ = + δx n 2 + ɛx n 3 /Bx n #330 : x n+ = δx n 2 + ɛx n 3 /A + Bx n #33 : x n+ = + δx n 2 + ɛx n 3 /A + Bx n #332 : x n+ = βx n + δx n 2 + ɛx n 3 /Bx n #333 : x n+ = + βx n + δx n 2 + ɛx n 3 /Bx n #334 : x n+ = βx n + δx n 2 + ɛx n 3 /A + Bx n #335 : x n+ = + βx n + δx n 2 + ɛx n 3 /A + Bx n #352 : x n+ = δx n 2 + ɛx n 3 /Cx n #353 : x n+ = + δx n 2 + ɛx n 3 /Cx n #354 : x n+ = δx n 2 + ɛx n 3 /A + Cx n #355 : x n+ = + δx n 2 + ɛx n 3 /A + Cx n #356 : x n+ = βx n + δx n 2 + ɛx n 3 /Cx n #357 : x n+ = + βx n + δx n 2 + ɛx n 3 /Cx n #358 : x n+ = βx n + δx n 2 + ɛx n 3 /A + Cx n

49 39 #359 : x n+ = + βx n + δx n 2 + ɛx n 3 /A + Cx n #368 : x n+ = γx n + δx n 2 + ɛx n 3 /Cx n #369 : x n+ = + γx n + δx n 2 + ɛx n 3 /Cx n #370 : x n+ = γx n + δx n 2 + ɛx n 3 /A + Cx n #37 : x n+ = + γx n + δx n 2 + ɛx n 3 /A + Cx n #372 : x n+ = βx n + γx n + δx n 2 + ɛx n 3 /Cx n #373 : x n+ = + βx n + γx n + δx n 2 + ɛx n 3 /Cx n #374 : x n+ = βx n + γx n + δx n 2 + ɛx n 3 /A + Cx n #375 : x n+ = + βx n + γx n + δx n 2 + ɛx n 3 /A + Cx n #404 : x n+ = βx n + γx n + ɛx n 3 /Dx n 2 #405 : x n+ = + βx n + γx n + ɛx n 3 /Dx n 2 #406 : x n+ = βx n + γx n + ɛx n 3 /A + Dx n 2 #407 : x n+ = + βx n + γx n + ɛx n 3 /A + Dx n 2 #47 : x n+ = + ɛx n 3 /Cx n + Dx n 2 #420 : x n+ = βx n + ɛx n 3 /Cx n + Dx n 2 #42 : x n+ = + βx n + ɛx n 3 /Cx n + Dx n 2 #57 : x n+ = + βx n /Ex n 3 #532 : x n+ = βx n + γx n /Ex n 3 #533 : x n+ = + βx n + γx n /Ex n 3 #537 : x n+ = + γx n /Bx n + Ex n 3 #548 : x n+ = βx n /Cx n + Ex n 3 #549 : x n+ = + βx n /Cx n + Ex n 3 #564 : x n+ = βx n + γx n /Cx n + Ex n 3 #577 : x n+ = + δx n 2 /Ex n 3 #580 : x n+ = βx n + δx n 2 /Ex n 3 #58 : x n+ = + βx n + δx n 2 /Ex n 3

50 40 #592 : x n+ = γx n + δx n 2 /Ex n 3 #593 : x n+ = + γx n + δx n 2 /Ex n 3 #594 : x n+ = γx n + δx n 2 /A + Ex n 3 #595 : x n+ = + γx n + δx n 2 /A + Ex n 3 #596 : x n+ = βx n + γx n + δx n 2 /Ex n 3 #597 : x n+ = + βx n + γx n + δx n 2 /Ex n 3 #598 : x n+ = βx n + γx n + δx n 2 /A + Ex n 3 #599 : x n+ = + βx n + γx n + δx n 2 /A + Ex n 3 #600 : x n+ = γx n + δx n 2 /Bx n + Ex n 3 #60 : x n+ = + γx n + δx n 2 /Bx n + Ex n 3 #62 : x n+ = βx n + δx n 2 /Cx n + Ex n 3 #63 : x n+ = + βx n + δx n 2 /Cx n + Ex n 3 #628 : x n+ = βx n + γx n + δx n 2 /Cx n + Ex n 3 #629 : x n+ = + βx n + γx n + δx n 2 /Cx n + Ex n 3 #644 : x n+ = βx n /Dx n 2 + Ex n 3 #645 : x n+ = + βx n /Dx n 2 + Ex n 3 #660 : x n+ = βx n + γx n /Dx n 2 + Ex n 3 #66 : x n+ = + βx n + γx n /Dx n 2 + Ex n 3 #676 : x n+ = βx n /Cx n + Dx n 2 + Ex n 3 #677 : x n+ = + βx n /Cx n + Dx n 2 + Ex n 3 #692 : x n+ = βx n + γx n /Cx n + Dx n 2 + Ex n 3 #848 : x n+ = γx n + δx n 2 + ɛx n 3 /Ex n 3 #849 : x n+ = + γx n + δx n 2 + ɛx n 3 /Ex n 3 #850 : x n+ = γx n + δx n 2 + ɛx n 3 /A + Ex n 3 #85 : x n+ = + γx n + δx n 2 + ɛx n 3 /A + Ex n 3 #852 : x n+ = βx n + γx n + δx n 2 + ɛx n 3 /Ex n 3

51 4 #853 : x n+ = + βx n + γx n + δx n 2 + ɛx n 3 /Ex n 3 #854 : x n+ = βx n + γx n + δx n 2 + ɛx n 3 /A + Ex n 3 #855 : x n+ = + βx n + γx n + δx n 2 + ɛx n 3 /A + Ex n 3 #868 : x n+ = βx n + δx n 2 + ɛx n 3 /Cx n + Ex n 3 #869 : x n+ = + βx n + δx n 2 + ɛx n 3 /Cx n + Ex n 3 #884 : x n+ = βx n + γx n + δx n 2 + ɛx n 3 /Cx n + Ex n 3 #885 : x n+ = + βx n + γx n + δx n 2 + ɛx n 3 /Cx n + Ex n 3 List of References [] E Camouzis and G Ladas, Dynamics of Third-Order Rational Dierence Equations with Open Problems and Conjectures, Chapman & Hall/CRC Press, Boca Raton, 2007 [2] E Camouzis and G Ladas, On third-order rational dierence equations, part 2, J Dierence Equ Appl 42008, [3] E Camouzis, G Ladas, F Palladino, and EP Quinn, On the Boundedness Character of Rational Equations, Part, J Dierence Equ Appl 22006, [4] G Lugo, On the existence of unbounded solutions for some rational equations, Involve 22009, [5] G Lugo and FJ Palladino, Unboundedness for some classes of rational difference equations, Int J Dierence Equ 42009, 97-3 [6] G Lugo and FJ Palladino, Unboundedness results for systems, Cent Eur J Math 72009, [7] FJ Palladino, Dierence inequalities, comparison tests, and some consequences, Involve 2008, 9-00 [8] FJ Palladino, On periodic trichotomies, J Dierence Equ Appl 52009, [9] FJ Palladino, On the Characterization of Rational Dierence Equations, J Dierence Equ Appl 52009,