RESEARC H Open Access Positive periodic solution of higher-order functional difference equation Mei-Lan Tang and Xin-Ge Liu * * Correspondence: liuxgliuhua@163.com School of Mathematical Science and Computing Technology, Central South University Changsha, Hunan 410083, China Abstract Based on a fixed point theorem in a cone, a new sufficient condition for the existence of a positive periodic solution to a class of higher-order functional difference equations is established in this article. The result obtained in this article is different from the existing results in previous literature. Mathematic Subject Classification 2000: 34k13; MSC 39A70. Keywords: positive periodic solution, fixed point theorem, cone, existence 1 Introduction The existence of positive periodic solutions of discrete mathematical models such as the disc rete model of blood cell productio n and the single-species discrete periodic popula- tion model has been studied extensively in recent years (see [1-8], for example). Most of these discrete mathematical models are first-order functional difference equations. Rela- tively, few articles focused on the existence of positive periodic solutions of higher-order functional difference equations. In 2010, Wang and Chen [9] have studi ed the existence of positive periodic solutions for the following general higher-order functional difference equation x(n + m + k) − ax(n + m) − bx(n + k)+abx(n)=f(n, x(n − τ (n))) (1) where a ≠ 1, b ≠ 1 are positive constants, τ: Z ® Z and τ(n + ω)=τ(n), f(n + ω, u)= f(n, u) for any u Î R, ω, m, k Î N where N denotes the set of positive integers. Based on fixed point theorem in a cone [10,11 ], some new sufficient conditions on the exis- tence of positive perio dic solutions to the higher-order functional difference equation (1) are obtained. However, the main results in [9] require that a should be positive constant, l should satisfy conditio n l = ω where l = ω (m, ω) and (m, ω)arethegreatest common divisor of m and ω. In fact, in most cases, m and ω do not satisfy such severe constraint l = ω. In general, l ≤ ω. I n this article, we consider the following higher- order functional difference equation x(n + m + k) − a(n + m)x(n + m) − bx(n + k)+a(n)bx(n)=f (n, x(n − τ (n))) (2) Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 © 2011 Tang and Liu; licensee Springer. This is an Open Access a rticle distributed under the terms of the Creative Common s Attribution License (http://creativecommons.org/lice nses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. where b ≠ 1 is positive constant, a: Z ® R + with a(n) ≠ 1anda(n + ω)=a(n), τ: Z ® Z and τ(n + ω)=τ(n), f(n + ω, u)=f(n, u) for any u Î R, k, ω, m Î N where N denotes the set of positive integers. The purpose of this article is to consider the existence of positive periodic solution of higher -order functional differen ce equatio n (2), we will remove the constrains on a and l in [9]. We will replace constant a in [9] with function a(n). At same time, we will remove the unreasonable assumption l = w. Based on a fixed point theorem in a cone, a new sufficient condition is es tablished for the existence of positive periodic solutions for higher-order functional difference equation. 2 Some preparation Let X be the set of all real ω periodic sequences, then X is a Banach space with the maximum norm ||x|| =max n∈[0,ω−1] |x(n)| . Lemma 1 (Deimling [10]) LetXbeaBanachspaceandKbeaconeinX.Suppose Ω 1 and Ω 2 are open subsets of X such that 0 ∈ 1 ⊂ ¯ 1 ⊂ 2 and suppose that : K ∩ ( ¯ 2 \ 1 ) → K is a completely continuous operator such that (i) ||Fu|| ≤ ||u|| for u Î K ∩ ∂Ω 1 and there exists ψ Î K\{0} such that x ≠ Fx + lψ for x Î K ∩ ∂Ω 2 and l >0;or (ii) ||Fu|| ≤ ||u|| for u Î K ∩ ∂Ω 2 and there exists ψ Î K\{0} such that x ≠ Fx + lψ for x Î K ∩ ∂Ω 1 and l >0. Then, F has a fixed point in K ∩ ( ¯ 2 \ 1 ) . Let d Î N. Consider the equation x(n + d)=cx(n)+γ (n) (3) where g Î X. Set (d, ω) as the greatest common divisor of d and ω, p = ω/(d, ω). Lemma 2 [9]Assume that 0<c ≠ 1, then (3) has a unique periodic solution x(n)=[c −p − 1] −1 p i=1 c −i γ (n +(i − 1)d). Let y(n)=x(n + k)-a(n)x(n), ¯ a =max 1≤n≤ω a(n), a − = min 1≤n≤ω a(n) , then (2) can be rewrit- ten as x(n + k)=a − x(n)+y(n)+[a(n) − a − ]x(n), y(n + m)=by(n)+f (n, x(n − τ (n))). (4) Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 2 of 8 Let h = ω (k, ω) , l = ω (m, ω) .Assumethatx Î X solution of (2), then y Î X.From Lemma 2, we have x(n)=[a − −h − 1] −1 h i=1 a − −1 {y(n +(i − 1)k)+[a(n +(i − 1)k) − a − ]x(n +(i − 1)k)}, y(n)=[b −l − 1] −1 l i=1 b −i f (n +(i − 1)m, x(n +(i − 1)m − τ (n +(i − 1)m))). If f(n, x(n - τ(n))) ≥ 0 and 0 <b < 1, then y(n) ≥ 0. We introduce the following conditions: (H) 0 <a(n)<1,0<b <1,h = ω and f: R × (0, +∞) ® [0, +∞) is continuous. Define the operator T by (Tx)(n)= a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m))) + a − h (1 − a − h ) h i=1 a − −i [a(n +(i − 1)k) − a − ]x(n +(i − 1)k). Define the cone by K = {x ∈ X, x(n) ≥ δ||x||} where δ = a − h b l (1 − a − h )(1 − b l )/ω . Lemma 3 Assume that (H) holds and 0<r 1 <r 2 , then T : ¯ K r 2 \K r 1 → K is completely continuous, where K r ={x Î K:||x|| <r} and ¯ K r = {x ∈ K : ||x|| ≤ r} . Proof Since 0 <a(n) < 1, then 0 < a − < 1 .Notingthat0<b < 1 and f(n, x(n - τ(n))) ≥ 0, we hav e y(n) ≥ 0. So (Tx)(n) ≥ 0on[0,ω -1].Sinceτ(n + ω)=τ(n ) and f(n + ω, u) = f(n, u) for any u >0,(Tx)(n + ω)=(Tx)(n) for x Î X. Since l = ω (m, ω) ≤ ω we have l j=1 f (n +(j − 1)m, x(n +(j − 1)m − τ (n +(j − 1)m))) ≤ ω j=1 f (j, x(j − τ (j))). On the other hand, from (H), h = ω (k, ω) = ω , we have h i=1 f (n +(i − 1)k, x(n +(i − 1)k − τ (n +(i − 1)k))) = ω i=1 f (i, x(i − τ (i))) Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 3 of 8 and h i=1 [a(n +(i − 1)k) − a − ]x(n +(i − 1)k)= ω i=1 [a(i) − a − ]x(i). For any x ∈ ¯ K r 2 \K r 1 , (Tx)(n)= a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m))) + a − h (1 − a − h ) h i=1 a − −i [a(n +(i − 1)k) − a − ]x(n +(i − 1)k) ≤ a − h b l (1 − a − h )(1 − b l ) a − −h b −l h i=1 l j=1 {f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m)))} + a − h (1 − a − h ) a − −h h i=1 [a(n +(i − 1)k) − a − ]x(n +(i − 1)k) = a − h b l (1 − a − h )(1 − b l ) a − −h b −l l j=1 h i=1 {f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m)))} + a − h (1 − a − h ) a − −h ω i=1 (a(i) − a − )x(i) ≤ 1 (1 − a − h )(1 − b l ) ω j=1 ω i=1 f (i, x(i − τ (i))) + 1 (1 − a − h ) ω i=1 (a(i) − a − )x(i) ≤ ω (1 − a − h )(1 − b l ) ω i=1 f (i, x(i, τ (i))) + 1 (1 − a − h) ω i=1 (a(i) − a − )x(i) ≤ ω (1 − a − h )(1 − b l ) ω i=1 {f (i, x(i, τ (i))) + (a(i) − a − )x(i)}. So ||Tx|| ≤ ω (1 − a − h )(1 − b l ) ω i=1 {f (i, x(i − τ (i))) + (a(i) − a − )x(i)}. (5) Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 4 of 8 At the same time (Tx)(n) ≥ a − h b l (1 − a − h )(1 − b l ) a − −1 b −1 h i=1 l j=1 {f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m)))} + a − h (1 − a − h ) a −1 h i=1 [a(n +(i − 1)k) − a − ]x(n +(i − 1)k) = a − h b l (1 − a − h )(1 − b l ) a − −1 b −1 l j=1 h i=1 {f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ (n +(i − 1)k +(j − 1)m)))} + a − h (1 − a − h ) a − −1 ω i=1 (a(i) − a − )x(i) ≥ a − h (1 − a − h ) lb l (1 − b l ) ω i=1 f (i, x(i − τ (i))) + a − h (1 − a − h ) ω i=1 (a(i) − a − )x(i) ≥ a − h (1 − a − h ) ω i=1 [ b l (1 − b l ) f (i, x(i, τ (i))) + (a(i) − a − )x(i)] ≥ a − h ω i=1 [b l f (i, x(i − τ (i))) + (a(i) − a − )x(i)] ≥ a − h b l ω i=1 [f (i, x(i − τ (i))) + (a(i) − a − )x(i)]. We have (Tx)(n) ≥ δ||Tx||. (6) Thus T : ¯ K r 2 \K r 1 → K is well defined. Since X is finite-dimensional Banach space, one can easily show that T is completely continuous. This completes the proof. We can easily obtain the following result. Lemma 4 The fixed point of T in K is a positive periodic solution of (2). 3 Main result Let ϕ(s)=max{f (n, u), n ∈ [0, ω − 1], u ∈ [δs, s]} ψ(s) = min f (n, u) u , n ∈ [0, ω − 1], u ∈ [δs, s] Let ¯ a =max 1≤n≤ω a(n), a − = min 1≤n≤ω a(n) . Theorem 1 Assume that (H) holds and there exist two positive constants a, b with a ≠ b such that Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 5 of 8 ϕ(α) ≤ ( ¯ a − 1)(b − 1)α, ψ(β) ≥ (a − − 1)(b − 1) (7) Then (2) has at least one positive ω-periodic solution x with min{a, b} ≤ ||x|| ≤ max {a, b}. Proof Without loss of generality, we assume that (H) holds, a <b. Obviously, 0 < ¯ a < 1, 0 < a − < 1 . We claim that: (i) ||Tx|| ≤ ||x||, x Î ∂K a , (ii) x ≠ Tx + l ·1,∀x Î ∂K b ,1Î K and l >0. From (7), we have that f (n, x) ≤ ( ¯ a − 1)(b − 1)α, ∀0 ≤ n ≤ ω − 1, ∀δα ≤ x ≤ α, (8) f (n, x) ≥ (a − − 1)(b − 1)x, ∀0 ≤ n ≤ ω − 1, ∀δβ ≤ x ≤ β. (9) In order to prove (i), let x Î ∂K a , then ||x|| = a and δa ≤ x(n) ≤ a for 0 ≤ n ≤ ω -1. So (Tx)(n)= a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j f (n +(i − 1)k +(j − 1)m, x(n +(i − 1)k +(j − 1)m − τ ( n +(i − 1)k +(j − 1)m))) + a − h (1 − a − h ) h i=1 a − −i [a(n +(i − 1)k) − a − ]x(n +(i − 1)k) ≤ a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j {( ¯ a − 1)(b − 1)α} + a − h (1 − a − h h i=1 a − −i [ ¯ a − a − ]||x|| ≤ ⎧ ⎨ ⎩ b l (1 − b l ) (1 − b) l j=1 b −j ⎫ ⎬ ⎭ a − h (1 − a − h ) h i=1 a − −i {(1 − ¯ a)α} + a − h (1 − a − h ) h i=1 a − −i [ ¯ a − a − ]α = a − h (1 − a − h ) h i=1 a − −i [1 − a − ]α = α. It follows that ||Tx|| ≤ ||x||, x ∈ ∂K α . (10) Next, let ψ =1Î K in Lemma 1, we prove (ii). If not, there exists u o Î ∂ K b and l o > 0 such that u 0 =(Tu 0 )(n)+λ 0 . (11) Since u o Î ∂ K b , then ||u o || = b and δb ≤ u o (n) ≤ b. Put u o (n) = min{u o (i)|0 ≤ i ≤ ω - 1} for some n Î[0, ω - 1]. Noting that u o (n) > 0 and 0 < a − < 1 , we have Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 6 of 8 u 0 (n)=(Tu 0 )(n)+λ 0 = a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j f (n +(i − 1)k +(j − 1) m, u 0 (n +(i − 1)k +(j − 1)m − τ(n +(i − 1)k +(j − 1)m))) + a − h (1 − a − h ) h i=1 a − −i [a(n +(i − 1)k) − a − ]u 0 (n +(i − 1)k)+λ 0 ≥ a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j {f (n +(i − 1)k +(j − 1)m, u 0 (n +(i − 1)k +(j − 1)m − τ(n +(i − 1)k +(j − 1)m)))} + λ 0 ≥ a − h b l (1 − a − h )(1 − b l ) h i=1 a − −i l j=1 b −j (a − − 1)(b − 1)u 0 (n +(i − 1)k +(j − 1)m − τ(n +(i − 1)k +(j − 1)m)) + λ 0 ≥ u 0 (n)+λ 0 which implies that u o (n)>u o (n), a contradiction. Therefore, by Lemma 1, T has a fixed point x Î K b \K a .Furthermore,a ≤ ||x|| ≤ b and x(n) ≥ δa, which means that x is one positive periodic solution of (2). The proof is completed. 4 Example Now, an example is given to demonstrate our result. Example 1 Consider the difference equation x(n + m + k) − a(n + m)x(n + m) − bx(n + k)+a(n)bx(n)=f (n, x(n − τ (n))) (12) where b =1/2,m =3,k =5,ω =6,τ: Z ® Z and τ(n + ω)=τ(n), a: Z ® R + with a(n)= 1 2 + 1 16 cos nπ 3 , f (n, u)=(1− 7 16 )(1 − 1 2 )u 3 [1 + 1 2 (−1) n cos πu 3 ] . Obviously, a(n + ω)=a(n +6)=a(n), f(n + ω, u)=f(n +6,u)=f(n, u) for any u Î R. h = ω (k, ω) = 6 (5, 6) =6,l = ω m, ω = 6 (3, 6) =2. ¯ a =max 1≤n≤ω a(n)= 9 16 , a − = min 1≤n≤ω a(n)= 7 16 , δ = 7 16 6 1 2 2 1 − 7 16 6 1 − 1 2 2 /6 . Let α = 1 2 , then ϕ(α)=ϕ 1 2 ≤ 1 − 7 16 1 − 1 2 1 2 3 1+ 1 2 = 1 − 7 16 1 − 1 2 1 2 2 3 4 < 9 16 1 2 1 − 1 2 1 2 < 1 − 9 16 1 − 1 2 1 2 . (13) So ϕ(α) ≤ ( ¯ a − 1)(b − 1)α . Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 7 of 8 Let β = 2 δ .Ifu Î [δb,b], then u ≥ 2. Furthermore, ψ(β) ≥ 1 − 7 16 1 − 1 2 2 3 2 1 − 1 2 =2 1 − 7 16 1 − 1 2 > 1 − 7 16 1 − 1 2 . (14) So ψ(β) ≥ (a − − 1)(b − 1) . By Theorem 1 in this article, (12) has at least one positive 6-periodic solution. Acknowledgements The authors would like to thank the reviewers for their valuable comments and constructive suggestions. This study was partly supported by the ZNDXQYYJJH under grant no. 2010QZZD015, Hunan Scientific Plan under grant no. 2011FJ6037, NSFC under grant no. 61070190 and NFSS under grant no. 10BJL020. Authors’ contributions All authors contributed equally to the manuscript and read and approved the final draft. Competing interests The authors declare that they have no competing interests. Received: 19 May 2011 Accepted: 21 November 2011 Published: 21 November 2011 References 1. 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Submit your manuscript to a journal and benefi t from: 7 Convenient online submission 7 Rigorous peer review 7 Immediate publication on acceptance 7 Open access: articles freely available online 7 High visibility within the fi eld 7 Retaining the copyright to your article Submit your next manuscript at 7 springeropen.com Tang and Liu Advances in Difference Equations 2011, 2011:56 http://www.advancesindifferenceequations.com/content/2011/1/56 Page 8 of 8 . on the existence of positive periodic solutions of higher-order functional difference equations. In 2010, Wang and Chen [9] have studi ed the existence of positive periodic solutions for the. 39A70. Keywords: positive periodic solution, fixed point theorem, cone, existence 1 Introduction The existence of positive periodic solutions of discrete mathematical models such as the disc rete model of. es tablished for the existence of positive periodic solutions for higher-order functional difference equation. 2 Some preparation Let X be the set of all real ω periodic sequences, then X is a