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# Positive periodic solution of higher-order functional difference equation

## 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.

## 1 Introduction

The existence of positive periodic solutions of discrete mathematical models such as the discrete model of blood cell production and the single-species discrete periodic population model has been studied extensively in recent years (see [18], for example). Most of these discrete mathematical models are first-order functional difference equations. Relatively, few articles focused on the existence of positive periodic solutions of higher-order functional difference equations. In 2010, Wang and Chen [9] have studied the existence of positive periodic solutions for the following general higher-order functional difference equation

$x ( n + m + k ) - a x ( n + m ) - b x ( n + k ) + a b x ( n ) = f ( n , x ( n - τ ( n ) ) )$
(1)

where a ≠ 1, b ≠ 1 are positive constants, τ: ZZ and τ(n + ω) = τ(n), f(n + ω, u) = f(n, u) for any u R, ω, m, k 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 existence of positive periodic 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 condition l = ω where $l= ω ( m , ω )$ and (m, ω) are the greatest common divisor of m and ω. In fact, in most cases, m and ω do not satisfy such severe constraint l = ω. In general, lω. In this article, we consider the following higher-order functional difference equation

$x ( n + m + k ) - a ( n + m ) x ( n + m ) - b x ( n + k ) + a ( n ) b x ( n ) = f ( n , x ( n - τ ( n ) ) )$
(2)

where b ≠ 1 is positive constant, a: ZR + with a(n) ≠ 1 and a(n + ω) = a(n), τ: ZZ and τ(n + ω) = τ(n), f(n + ω, u) = f(n, u) for any u R, k, ω, m 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 difference equation (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 established 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]) Let X be a Banach space and K be a cone in X. Suppose Ω 1 and Ω 2 are open subsets of X such that $0 ∈ Ω 1 ⊂ Ω ̄ 1 ⊂ Ω 2$ and suppose that

$Φ : K ∩ ( Ω ̄ 2 \ Ω 1 ) → K$

is a completely continuous operator such that

(i) ||Φu|| ≤ ||u|| for u K∂Ω 1 and there exists ψ K\{0} such that xΦx + λψ for x K∂Ω 2 and λ > 0; or

(ii) ||Φu|| ≤ ||u|| for u K∂Ω 2 and there exists ψ K\{0} such that xΦx + λψ for x K∂Ω 1 and λ > 0.

Then, Φ has a fixed point in $K∩ ( Ω ̄ 2 \ Ω 1 )$.

Let d N. Consider the equation

$x ( n + d ) = c x ( n ) + γ ( n )$
(3)

where γ 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

$x ( n ) = [ c - p - 1 ] - 1 ∑ i = 1 p c - i γ ( n + ( i - 1 ) d ) .$

Let y(n) = x(n + k) - a(n)x(n), $ā= max 1 ≤ n ≤ ω a ( n ) , a - = min 1 ≤ n ≤ ω a ( n )$, then (2) can be rewritten as

$x ( n + k ) = a - x ( n ) + y ( n ) + [ a ( n ) - a - ] x ( n ) , y ( n + m ) = b y ( n ) + f ( n , x ( n - τ ( n ) ) ) .$
(4)

Let $h= ω ( k , ω ) ,l= ω ( m , ω )$. Assume that x X solution of (2), then y X. From Lemma 2, we have

$x ( n ) = [ a - - h - 1 ] - 1 ∑ i = 1 h a - - 1 { y ( n + ( i - 1 ) k ) + [ a ( n + ( i - 1 ) k ) - a - ] x ( n + ( i - 1 ) k ) } , y ( n ) = [ b - l - 1 ] - 1 ∑ i = 1 l b - i f ( n + ( i - 1 ) m , x ( n + ( i - 1 ) m - τ ( n + ( i - 1 ) m ) ) ) .$

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

$( T x ) ( n ) = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ a ( n + ( i − 1 ) k ) − a ¯ ] x ( n + ( i − 1 ) k ) .$

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 K: ||x|| < r} and $K ̄ r = { x ∈ K : | | x | | ≤ r }$.

Proof Since 0 < a(n) < 1, then $0< a - <1$. Noting that 0 < b < 1 and f(n, x(n - τ(n))) ≥ 0, we have y(n) ≥ 0. So (Tx)(n) ≥ 0 on [0, ω - 1]. Since τ(n + ω) = τ(n) and f(n + ω, u) = f(n, u) for any u > 0, (Tx)(n + ω) = (Tx)(n) for x X. Since $l= ω ( m , ω ) ≤ω$ we have

$∑ j = 1 l f ( n + ( j - 1 ) m , x ( n + ( j - 1 ) m - τ ( n + ( j - 1 ) m ) ) ) ≤ ∑ j = 1 ω f ( j , x ( j - τ ( j ) ) ) .$

On the other hand, from (H), $h= ω ( k , ω ) =ω$, we have

$∑ i = 1 h f ( n + ( i - 1 ) k , x ( n + ( i - 1 ) k - τ ( n + ( i - 1 ) k ) ) ) = ∑ i = 1 ω f ( i , x ( i - τ ( i ) ) )$

and

$∑ i = 1 h [ a ( n + ( i - 1 ) k ) - a - ] x ( n + ( i - 1 ) k ) = ∑ i = 1 ω [ a ( i ) - a - ] x ( i ) .$

For any $x∈ K ̄ r 2 \ K r 1$,

$( T x ) ( n ) = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ a ( n + ( i − 1 ) k ) − a ¯ ] x ( n + ( i − 1 ) k ) ≤ a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) a ¯ − h b − l ∑ i = 1 h ∑ j = 1 l { f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) } + a ¯ h ( 1 − a ¯ h ) a ¯ − h ∑ i = 1 h [ a ( n + ( i − 1 ) k ) − a ¯ ] x ( n + ( i − 1 ) k ) = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) a ¯ − h b − l ∑ j = 1 l ∑ i = 1 h { f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) } + a ¯ h ( 1 − a ¯ h ) a ¯ − h ∑ i = 1 ω ( a ( i ) − a ¯ ) x ( i ) ≤ 1 ( 1 − a ¯ h ) ( 1 − b l ) ∑ j = 1 ω ∑ i = 1 ω f ( i , x ( i − τ ( i ) ) ) + 1 ( 1 − a ¯ h ) ∑ i = 1 ω ( a ( i ) − a ¯ ) x ( i ) ≤ ω ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 ω f ( i , x ( i , τ ( i ) ) ) + 1 ( 1 − a ¯ h ) ∑ i = 1 ω ( a ( i ) − a ¯ ) x ( i ) ≤ ω ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 ω { f ( i , x ( i , τ ( i ) ) ) + ( a ( i ) − a ¯ ) x ( i ) } .$

So

$| | T x | | ≤ ω ( 1 - a - h ) ( 1 - b l ) ∑ i = 1 ω { f ( i , x ( i - τ ( i ) ) ) + ( a ( i ) - a - ) x ( i ) } .$
(5)

At the same time

$( T x ) ( n ) ≥ a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) a ¯ − 1 b − 1 ∑ i = 1 h ∑ j = 1 l { f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) } + a ¯ h ( 1 − a ¯ h ) a − 1 ∑ i = 1 h [ a ( n + ( i − 1 ) k ) − a ¯ ] x ( n + ( i − 1 ) k ) = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) a ¯ − 1 b − 1 ∑ j = 1 l ∑ i = 1 h { f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) } + a ¯ h ( 1 − a ¯ h ) a ¯ − 1 ∑ i = 1 ω ( a ( i ) − a ¯ ) x ( i ) ≥ a ¯ h ( 1 − a ¯ h ) l b l ( 1 − b l ) ∑ i = 1 ω f ( i , x ( i − τ ( i ) ) ) + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 ω ( a ( i ) − a ¯ ) x ( i ) ≥ a ¯ h ( 1 − a ¯ h ) ∑ i = 1 ω [ b l ( 1 − b l ) f ( i , x ( i , τ ( i ) ) ) + ( a ( i ) − a ¯ ) x ( i ) ] ≥ a ¯ h ∑ i = 1 ω [ b l f ( i , x ( i − τ ( i ) ) ) + ( a ( i ) − a ¯ ) x ( i ) ] ≥ a ¯ h b l ∑ i = 1 ω [ f ( i , x ( i − τ ( i ) ) ) + ( a ( i ) − a ¯ ) x ( i ) ] .$

We have

$( T x ) ( n ) ≥ δ | | T x | | .$
(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

$φ ( s ) = max { f ( n , u ) , n ∈ [ 0 , ω - 1 ] , u ∈ [ δ s , s ] } ψ ( s ) = min f ( n , u ) u , n ∈ [ 0 , ω - 1 ] , u ∈ [ δ s , s ]$

Let $ā = max 1 ≤ n ≤ ω a ( n ) , a - = min 1 ≤ n ≤ ω a ( n )$.

Theorem 1 Assume that (H) holds and there exist two positive constants α, β with αβ such that

$φ ( α ) ≤ ( ā - 1 ) ( b - 1 ) α , ψ ( β ) ≥ ( a - - 1 ) ( b - 1 )$
(7)

Then (2) has at least one positive ω-periodic solution x with min{α, β} ≤ ||x|| ≤ max{α, β}.

Proof Without loss of generality, we assume that (H) holds, α < β. Obviously, $0<ā<1,0< a - <1$. We claim that:

1. (i)

||Tx|| ≤ ||x||, x ∂K α ,

2. (ii)

xTx + λ · 1, x ∂K β , 1 K and λ > 0.

From (7), we have that

$f ( n , x ) ≤ ( ā - 1 ) ( b - 1 ) α , ∀ 0 ≤ n ≤ ω - 1 , ∀ δ α ≤ x ≤ α ,$
(8)
$f ( n , x ) ≥ ( a ¯ − 1 ) ( b − 1 ) x , ∀ 0 ≤ n ≤ ω − 1 , ∀ δ α ≤ x ≤ β .$
(9)

In order to prove (i), let x ∂K α , then ||x|| = α and δαx(n) ≤ α for 0 ≤ nω - 1. So

$( T x ) ( n ) = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j f ( n + ( i − 1 ) k + ( j − 1 ) m , x ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ a ( n + ( i − 1 ) k ) − a ¯ ] x ( n + ( i − 1 ) k ) ≤ a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j { ( a ¯ − 1 ) ( b − 1 ) α } + a ¯ h ( 1 − a ¯ h ∑ i = 1 h a ¯ − i [ a ¯ − a ¯ ] | | x | | ≤ { b l ( 1 − b l ) ( 1 − b ) ∑ j = 1 l b − j } a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i { ( 1 − a ¯ ) α } + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ a ¯ − a ¯ ] α = a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ 1 − a ¯ ] α = α .$

It follows that

$| | T x | | ≤ | | x | | , x ∈ ∂ K α .$
(10)

Next, let ψ = 1 K in Lemma 1, we prove (ii). If not, there exists u o ∂K β and λ o > 0 such that

$u 0 = ( T u 0 ) ( n ) + λ 0 .$
(11)

Since u o ∂K β , then ||u o|| = β and δβu o(n) ≤ β. Put u o(n) = min{u o(i)|0 ≤ iω - 1} for some n [0, ω - 1]. Noting that u o(n) > 0 and $0< a - <1$, we have

$u 0 ( n ) = ( T u 0 ) ( n ) + λ 0 = a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j f ( n + ( i − 1 ) k + ( j − 1 ) m , u 0 ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) + a ¯ h ( 1 − a ¯ h ) ∑ i = 1 h a ¯ − i [ a ( n + ( i − 1 ) k ) − a ¯ ] u 0 ( n + ( i − 1 ) k ) + λ 0 ≥ a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − j { f ( n + ( i − 1 ) k + ( j − 1 ) m , u 0 ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) ) } + λ 0 ≥ a ¯ h b l ( 1 − a ¯ h ) ( 1 − b l ) ∑ i = 1 h a ¯ − i ∑ j = 1 l b − 1 ( a ¯ − 1 ) ( b − 1 ) u 0 ( n + ( i − 1 ) k + ( j − 1 ) m − τ ( n + ( i − 1 ) k + ( j − 1 ) m ) ) + λ 0 ≥ u 0 ( n ) + λ 0$

which implies that u o(n) > u o(n), a contradiction.

Therefore, by Lemma 1, T has a fixed point x K β \K α . Furthermore, α ≤ ||x|| ≤ β and x(n) ≥ δα, 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

$x ( n + m + k ) - a ( n + m ) x ( n + m ) - b x ( n + k ) + a ( n ) b x ( n ) = f ( n , x ( n - τ ( n ) ) )$
(12)

where b = 1/2, m = 3, k = 5, ω = 6, τ: ZZ and τ(n + ω) = τ(n), a: ZR + with $a ( n ) = 1 2 + 1 16 cos n π 3 ,f ( n , u ) = ( 1 - 7 16 ) ( 1 - 1 2 ) u 3 [ 1 + 1 2 ( - 1 ) n c o s π u 3 ]$.

Obviously, a(n + ω) = a(n + 6) = a(n), f(n + ω, u) = f(n + 6, u) = f(n, u) for any u R. $h= ω ( k , ω ) = 6 ( 5 , 6 ) =6,l= ω m , ω = 6 ( 3 , 6 ) =2.ā= max 1 ≤ n ≤ ω a ( n ) = 9 16 , a - = min 1 ≤ n ≤ ω a ( n ) = 7 16 ,δ= 7 16 6 1 2 2 1 - 7 16 6 1 - 1 2 2 ∕6$.

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 $φ ( α ) ≤ ( ā - 1 ) ( b - 1 ) α$.

Let $β= 2 δ$. If u [δβ,β], 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.

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## 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.

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Correspondence to Xin-Ge Liu.

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All authors contributed equally to the manuscript and read and approved the final draft.

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Tang, M., Liu, X. Positive periodic solution of higher-order functional difference equation. Adv Differ Equ 2011, 56 (2011). https://doi.org/10.1186/1687-1847-2011-56