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Hindawi Publishing Corporation Advances in Difference Equations Volume 2009, Article ID 463169, 11 pages doi:10.1155/2009/463169 Research Article On Boundedness of Solutions of the Difference Equation xn pxn qxn−1 / xn for q > p > Hongjian Xi,1, Taixiang Sun,1 Weiyong Yu,1 and Jinfeng Zhao1 Department of Mathematics, Guangxi University, Nanning, Guangxi 530004, China Department of Mathematics, Guangxi College of Finance and Economics, Nanning, Guangxi 530003, China Correspondence should be addressed to Taixiang Sun, stx1963@163.com Received February 2009; Revised 19 April 2009; Accepted June 2009 Recommended by Agacik Zafer We study the boundedness of the difference equation xn pxn qxn−1 / xn , n 0, 1, , where q > p > and the initial values x−1 , x0 ∈ 0, ∞ We show that the solution {xn }∞ −1 of n this equation converges to x q p − if xn ≥ x or xn ≤ x for all n ≥ −1; otherwise {xn }∞ −1 is n unbounded Besides, we obtain the set of all initial values x−1 , x0 ∈ 0, ∞ × 0, ∞ such that the positive solutions {xn }∞ −1 of this equation are bounded, which answers the open problem 6.10.12 n proposed by Kulenovi´ and Ladas 2002 c Copyright q 2009 Hongjian Xi et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Introduction In this paper, we study the following difference equation: xn pxn qxn−1 , xn n 0, 1, , 1.1 where p, q ∈ 0, ∞ with q > p and the initial values x−1 , x0 ∈ 0, ∞ The global behavior of 1.1 for the case p q < is certainly folklore It can be found, for example, in see also a precise result in The global stability of 1.1 for the case p q follows from the main result in see also Lemma in Stevi´ ’s paper Some generalizations of Copson’s result can be found, c for example, in papers 5–8 Some more sophisticated results, such as finding the asymptotic behavior of solutions of 1.1 for the case p q even when p can be found, for Advances in Difference Equations example, in papers see also 8–11 Some other properties of 1.1 have been also treated in The case q p was treated for the first time by Stevi´ ’s in paper 12 The main trick c from 12 has been later used with a success for many times; see, for example, 13–15 Some existing results for 1.1 are summarized as follows 16 Theorem A If p q ≤ 1, then the zero equilibrium of 1.1 is globally asymptotically stable If q 1, then the equilibrium x p of 1.1 is globally asymptotically stable If < q < p, then every positive solution of 1.1 converges to the positive equilibrium x p q − If q p, then every positive solution of 1.1 converges to a period-two solution If q > p, then 1.1 has unbounded solutions In 16 , Kulenovi´ and Ladas proposed the following open problem c Open problem B (see Open problem 6.10.12of [16]) Assume that q ∈ 1, ∞ a Find the set B of all initial conditions x−1 , x0 ∈ 0, ∞ × 0, ∞ such that the solutions {xn }∞ −1 of 1.1 are bounded n b Let x−1 , x0 ∈ B Investigate the asymptotic behavior of {xn }∞ −1 n In this paper, we will obtain the following results: let p, q ∈ 0, ∞ with q > p, and let {xn }∞ −1 be a positive solution of 1.1 with the initial values x−1 , x0 ∈ 0, ∞ × 0, ∞ n If xn ≥ x for all n ≥ −1 or xn ≤ x for all n ≥ −1 , then {xn }∞ −1 converges to x q p − n Otherwise {xn }∞ −1 is unbounded n For closely related results see 17–34 Some Definitions and Lemmas In this section, let q > p > and x q p − be the positive equilibrium of 1.1 Write D 0, ∞ × 0, ∞ and define f : D → D by, for all x, y ∈ D, f x, y y, py qx y It is easy to see that if {xn }∞ −1 is a solution of 1.1 , then f n x−1 , x0 n Let A1 0, x × 0, x , A3 0, x × x, ∞ , A2 A4 2.1 xn−1 , xn for any n ≥ x, ∞ × x, ∞ , x, ∞ × 0, x , R0 {x} × 0, x , L0 {x} × x, ∞ , R1 0, x × {x}, L1 x, ∞ × {x} 2.2 Advances in Difference Equations Then D ∪4 Ai ∪ L0 ∪ L1 ∪ R0 ∪ R1 ∪ { x, x } The proof of Lemma 2.1 is quite similar to i that of Lemma in 35 and hence is omitted Lemma 2.1 The following statements are true f : D → f D is a homeomorphism f L1 L0 and f L0 ⊂ A4 f R1 {x} × p, x and f R0 ⊂ A3 f A3 ⊂ A4 and f A4 ⊂ A3 A2 ∪ L1 ⊂ f A2 ⊂ A2 ∪ L1 ∪ A4 and A1 ∪ R1 ⊂ f A1 ⊂ A1 ∪ R1 ∪ A3 p > 1, and let {xn }∞ −1 be a positive solution of 1.1 n Lemma 2.2 Let q > 1 If limn → ∞ x2n a ∈ 0, ∞ and a / p, then limn → If limn → ∞ x2n−1 ∞ x2n b ∈ 0, ∞ and b / p, then limn → a b ∞ x2n x x Proof We show only because the proof of follows from by using the change yn xn−1 and the fact that is autonomous Since limn → ∞ x2n a ∈ 0, ∞ and a / p, by 1.1 we have lim x2n n→ ∞ lim n→ ∞ q−1 a a−p qx2n − x2n x2n − p 2.3 Also it follows from 1.1 that a lim x2n n→ ∞ from which we have a qx2n−1 − x2n ∞ x2n − p lim n→ x and limn → ∞ x2n a q−1 a q−1 a−p a−p , 2.4 x This completes the proof Lemma 2.3 Let q > p > 1, and let {xn }∞ −1 be a positive solution of 1.1 with the initial values n x−1 , x0 ∈ A4 If there exists some n ≥ such that x2n−1 ≥ x2n , then x2n ≥ x2n Proof Since x−1 , x0 ∈ A4 , it follows from Lemma 2.1 that x2n−1 , x2n ∈ A4 for any n ≥ Without loss of generality we may assume that n 0, that is, x−1 ≥ x1 Now we show x0 ≥ x2 Suppose for the sake of contradiction that x0 < x2 , then x−1 ≥ x1 x0 < x2 px0 qx−1 , x0 px1 qx0 x1 2.5 2.6 By 2.5 we have x0 ≥ x−1 q − , x−1 − p 2.7 Advances in Difference Equations and by 2.6 we get q − − qx−1 x0 p2 q − − p x0 pqx−1 > 2.8 Claim If x−1 ≥ x, then p2 q − − qx−1 − q − − p pqx−1 ≥ 2.9 Proof of Claim Let g x p2 q − − qx − q − − p pqx x ≥ x , then we have g x qx − p2 − q − 4pq q − − p 2q ≥ 2q q − p2 p 1−q 2q q − q − p − p p 2.10 p > Since x−1 ≥ x, it follows p2 q − − qx−1 ≥ q2 qp − 2q q2 − 2q qp 2 − q − − p pqx−1 − p2 − p2 − q − − p qp q 2qp q2 − 2q − q − 2q q2 − 2q 2 − p2 − pq 1−p p−1 − p2 2.11 pq ≥ This completes the proof of Claim By 2.8 , we have x0 > λ1 qx−1 − p2 − q qx−1 − p2 − q q−1−p − 4pq q − − p x−1 2.12 Advances in Difference Equations or x0 < λ2 qx−1 − p2 − q − qx−1 − p2 − q − 4pq q − − p x−1 q−1−p 2.13 Claim We have λ1 ≥ x, 2.14 x−1 q − x−1 − p 2.15 λ2 ≤ Proof of Claim Since q q p − − p2 − q q2 − p2 − 2q q − 4pq q − − p p q−1 − qp 2.16 p−1 q−1−p − q q p − − p2 − q , we have λ1 qx−1 − p2 − q qx−1 − p2 − q − 4pq q − − p x−1 q−1−p ≥ qx − p2 − q qx − p2 − q − 4pq q − − p x q−1−p q q p − − p2 − q q q p − − p2 − q − 4pq q − − p p q−1 q−1−p ≥ q p−1 x 2.17 The proof of 2.14 is completed Now we show 2.15 Let h x pq x − p − x−p q−1 qx − p2 − q q−1 q − − p x 2.18 Advances in Difference Equations Note that 2pq − 2q q − < 0; it follows that if x ≥ x, then h x qx − p2 − q 2pq x − p − q−1 ≤ 2pq q − − q−1 2pq − q − p2 q−1 q p p q q−1 x−p − q−1 q−1−p q2 − p − q < 0, 2.19 which implies that h x is decreasing for x ≥ x Since x−1 ≥ x and pq q − − q−1 q−1 q−1 hx q−1−p q q q p−1 p − − p2 − q 2.20 0, it follows that pq x−1 − p h x−1 q−1 − x−1 − p q − qx−1 − p2 − q q − − p x−1 ≤ h x 2.21 Thus q−1 qx−1 − p2 − q ≥ 4p2 q2 x−1 − p q−1 2 − 4pq q − − p x−1 − 4pq x−1 − p q − qx−1 − p2 − q qx−1 − p2 − q 2.22 This implies that q−1 qx−1 − p2 − q ≥ 2pq x−1 − p − q − − 4pq q − − p x−1 2.23 qx−1 − p2 − q Finally we have x−1 q − ≥ x−1 − p q − − p pqx−1 q−1−p 1 qx−1 − p2 − q qx−1 − p2 − q − 1 qx−1 − p2 − q q−1−p qx−1 − p2 − q 2 − 4pq q − − p x−1 − 4pq q − − p x−1 λ2 2.24 The proof of 2.15 is completed Advances in Difference Equations Note that x0 < x since x−1 , x0 ∈ A4 By 2.12 , 2.13 , 2.14 , and 2.15 , we see x0 < x−1 q − / x−1 − p , which contradicts to 2.7 The proof of Lemma 2.3 is completed Main Results In this section, we investigate the boundedness of solutions of 1.1 Let q > p > 1, and let {xn }∞ −1 be a positive solution of 1.1 with the initial values x−1 , x0 ∈ 0, ∞ × 0, ∞ , n then we see that xn − x xn − x < for some n ≥ −1 or xn ≥ x for all n ≥ −1 or xn ≤ x for all n ≥ −1 Theorem 3.1 Let q > p > 1, and let {xn }∞ −1 be a positive solution of 1.1 such that xn ≥ x for n all n ≥ −1 or xn ≤ x for all n ≥ −1, then {xn }∞ −1 converges to x q p − n Proof Case < xn ≤ x for any n ≥ −1 If < x2n ≤ q − for some n, then x2n − x2n−1 px2n qx2n−1 − x2n−1 − x2n−1 x2n > x2n 3.1 If q − < x2n ≤ x for some n, then px2n px ≥ x2n − q x − q which implies that px2n ≥ x2n−1 x2n − q x2n − x2n−1 px2n x ≥ x2n−1 , and qx2n−1 − x2n−1 − x2n−1 x2n ≥ x2n Thus x ≥ x2n ≥ x2n−1 for any n ≥ In similar fashion, we can show x ≥ x2n n ≥ Let limn → ∞ x2n a and limn → ∞ x2n b, then a which implies a b x pb 3.2 qa , b b pa qb , a 3.3 ≥ x2n for any 3.4 Advances in Difference Equations Case xn ≥ x p q − for any n ≥ −1 Since f x, y decreasing in y, it follows that for any n ≥ −1, xn py qx / y x > p/q is pxn qxn xn 3.5 px qxn ≤ ≤ xn x In similar fashion, we can show that limn → proof ∞ x2n limn → ∞ x2n x This completes the Lemma 3.2 see 20, Theorem Let I be a set, and let F : I × I → I be a function F u, v which decreases in u and increases in v, then for every positive solution {xn }n∞ of equation xn −1 F xn , xn−1 , {x2n }∞ and {x2n−1 }∞ exactly one of the following n n They are both monotonically increasing They are both monotonically decreasing Eventually, one of them is monotonically increasing, and the other is monotonically decreasing Remark 3.3 Using arguments similar to ones in the proof of Lemma 3.2, Stevi´ proved c Theorem in 25 Beside this, this trick have been used by Stevi´ in 18, 28, 29 c Theorem 3.4 Let q > p > 1, and let {xn }∞ −1 be a positive solution of 1.1 such that xn n x xn − x < for some n ≥ −1, then {xn }∞ −1 is unbounded n − Proof We may assume without loss of generality that x0 − x x−1 − x < and x−1 , x0 ∈ A4 the proof for x−1 , x0 ∈ A3 is similar From Lemma 2.1 we see x2n−1 , x2n ∈ A4 for all n ≥ 0.If {x2n }∞ is eventually increasing, then it follows from Lemma 2.3 that {x2n−1 }∞ is n n eventually increasing Thus limn → ∞ x2n−1 b > x and limn → ∞ x2n a ≤ x, it follows from Lemma 2.2 that b ∞ If {x2n }∞ is not eventually increasing, then there exists some N ≥ such that n x2N ≥ x2N px2N qx2N , x2N 3.6 from which we obtain x2N ≥ px2N / x2N − q ≥ p, since x2N ≥ x p q − and q > Since f y, x py qx / y p qx − p / y x ≥ p, y ≥ p is increasing in x and is decreasing in y, we have that x2n ≥ p for any n ≥ N It follows from Lemma 3.2 that {x2n }∞ is eventually decreasing Thus limn → ∞ x2n a < x and limn → ∞ x2n−1 b ≥ x n It follows from Lemma 2.2 that b ∞ This completes the proof By Theorems 3.1 and 3.4 we have the following Corollary 3.5 Let q > p > 1, and let {xn }∞ −1 be a positive bounded solution of 1.1 , then n xn−1 ≥ xn ≥ x for all n ≥ or x ≥ xn ≥ xn−1 for all n ≥ Advances in Difference Equations Now one can find out the set of all initial values x−1 , x0 ∈ 0, ∞ × 0, ∞ such that the positive solutions of 1.1 are bounded Let P0 A2 , Q0 A1 For any n ≥ 1, let Pn f −1 Pn−1 , It follows from Lemma 2.1 that P1 Qn f −1 Qn−1 f −1 P0 ⊂ P0 , Q1 Pn ⊂ Pn−1 , Qn ⊂ Qn−1 3.7 f −1 Q0 ⊂ Q0 , which implies 3.8 for any n ≥ Let S be the set of all initial values x−1 , x0 ∈ 0, ∞ × 0, ∞ such that the positive solutions {xn }∞ −1 of 1.1 are bounded Then we have the following theorem n Theorem 3.6 S ∞ n Qn ∪ ∞ n Pn ⊂ A1 ∪ A2 ∪ { x, x } Proof Let {xn }∞ −1 be a positive solution of 1.1 with the initial values x−1 , x0 ∈ S n xn−1 , xn ∈ A1 for any n ≥ 0, which implies If x−1 , x0 ∈ ∞ Qn , then f n x−1 , x0 n xn ≤ x for any n ≥ −1 It follows from Theorem 3.1 that limn → ∞ xn x xn−1 , xn ∈ A2 , which implies xn ≥ x for any If x−1 , x0 ∈ ∞ Pn , then f n x−1 , x0 n n ≥ −1 It follows from Theorem 3.1 that limn → ∞ xn x Now assume that {xn }∞ −1 is a positive solution of 1.1 with the initial values n x−1 , x0 ∈ D − S If x−1 , x0 ∈ A3 A4 L0 L1 R0 R1 , then it follows from Lemma 2.1 that x1 , x2 ∈ { x, y : x − x y − x < 0}, which along with Theorem 3.4 implies f x−1 , x0 that {xn } is unbounded If x−1 , x0 ∈ A2 − ∞ Pn , then there exists n ≥ such that x−1 , x0 ∈ Pn − Pn n xn−1 , xn ∈ A2 − f −1 A2 By Lemma 2.1, we obtain f −n A2 − f −n−1 A2 Thus f n x−1 , x0 n n x−1 , x0 ∈ L1 A4 and f x−1 , x0 xn , xn ∈ A4 , which along with Theorem 3.4 f implies 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the two cycle of the difference equation xn xn−1 / p qxn xn−1 ,” Journal of Difference Equations and Applications, vol 13, no 10, pp 945–952, 2007  ... 4pq q − − p p q? ? ?1 − qp 2 .16 p? ? ?1 q? ? ?1? ? ?p − q q p − − p2 − q , we have ? ?1 qx? ?1 − p2 − q qx? ?1 − p2 − q − 4pq q − − p x? ?1 q? ? ?1? ? ?p ≥ qx − p2 − q qx − p2 − q − 4pq q − − p x q? ? ?1? ? ?p q q p − − p2 − q q q p. .. x qx − p2 − q − 4pq q − − p 2q ≥ 2q q − p2 p 1? ? ?q 2q q − q − p − p p 2 .10 p > Since x? ?1 ≥ x, it follows p2 q − − qx? ?1 ≥ q2 qp − 2q q2 − 2q qp 2 − q − − p pqx? ?1 − p2 − p2 − q − − p qp q 2qp q2 ... − q Finally we have x? ?1 q − ≥ x? ?1 − p q − − p pqx? ?1 q? ? ?1? ? ?p 1 qx? ?1 − p2 − q qx? ?1 − p2 − q − 1 qx? ?1 − p2 − q q? ?1? ? ?p qx? ?1 − p2 − q 2 − 4pq q − − p x? ?1 − 4pq q − − p x? ?1 λ2 2.24 The proof of 2 .15