Theory and Modern Applications

# Periodic solutions for a class of higher-order difference equations

## Abstract

In this article, we discuss the existence of periodic solutions for the higher-order difference equation

$x\left(n+k\right)=g\left(x\left(n\right)\right)-f\left(n,x\left(n-\tau \left(n\right)\right).$

We show the existence of periodic solutions by using Schauder's fixed point theorem, and illustrate three examples.

MSC 2010: 39A10; 39A12.

## 1 Introduction and main results

Let denote the set of the real numbers, the integers and the positive integers. In this article, we investigate the existence of periodic solutions of the following high-order functional difference equation

$x\left(n+k\right)=g\left(x\left(n\right)\right)-f\left(n,x\left(n-\tau \left(n\right)\right),\phantom{\rule{1em}{0ex}}n\in ℤ,$
(1.1)

where k N, τ : and τ (n + ω) = τ (n), f (n + ω, u) = f (n, u) for any (n, u) × , ω .

Difference equations have attracted the interest of many researchers in the last 20 years since they provided a natural description of several discrete models, in which the periodic solution problem is always a important topic, and the reader can consult [17] and the references therein. There are many good results about existence of periodic solutions for first-order functional difference equations [812]. Only a few article have been published on the same problem for higher-order functional difference equations. Recently, using coincidence degree theory, Liu [13] studied the second-order nonlinear functional difference equation

${\Delta }^{2}x\left(n-1\right)=f\left(n,x\left(n-{\tau }_{1}\left(n\right)\right),x\left(n-{\tau }_{2}\left(n\right)\right),\dots ,x\left(n-{\tau }_{m}\left(n\right)\right)\right),$
(1.2)

and obtain sufficient conditions for the existence of at least one periodic solution of equation (1.2). By using fixed point theorem in a cone, Wang and Chen [14] discussed the following higher-order functional difference equation

$x\left(n+m+k\right)-ax\left(n+m\right)-bx\left(n+k\right)+abx\left(n\right)=f\left(n,x\left(n-\tau \left(n\right)\right)\right),$
(1.3)

where a ≠ 1, b ≠ 1 are positive constants, τ : and τ(n+ω) = τ(n), ω, m, k , and obtained existence theorem for single and multiple positive periodic solutions of (1.3).

Our aim of this article is to study the existence of periodic solutions for the higher-order difference equations (1.1) using the well-known Schauder's fixed point theorem. Our results extend the known results in the literature.

The main results of this article are following sufficient conditions which guarantee the existence of a periodic solution for (1.1).

Theorem 1.1. Assume that there exist constants m < M, r > 0 such that g C 1[m, M] with rg'(u) ≤ 1 for any u [m, M] and f (n, u): × [m, M] → is continuous in u,

$g\left(M\right)-M\le f\left(n,u\right)\le g\left(m\right)-m$
(1.4)

for any (n, u) × [m, M], then (1.1) has at least one ω-periodic solution x with mxM.

Theorem 1.2. Assume that there exist constants m < M such that g C 1[m, M] with g'(u) ≥ 1 for any u [m, M] and f (n, u): × [m, M] → is continuous in u,

$g\left(m\right)-m\le f\left(n,u\right)\le g\left(M\right)-M$
(1.5)

for any (n, u) × [m, M], then (1.1) has at least one ω-periodic solution x with mxM.

## 2 Some examples

In this section, we present three examples to illustrate our conclusions.

Example 2.1. Consider the difference equation

$x\left(n+k\right)=ax\left(n\right)+q\left(n\right)\sqrt[3]{x\left(n-\tau \left(n\right)\right)},$
(2.1)
$x\left(n+k\right)=bx\left(n\right)-q\left(n\right)\sqrt[3]{x\left(n-\tau \left(n\right)\right)},$
(2.2)

where k , 0 < a < 1, b > 1, q is one ω-periodic function with q(n) > 0 for all n [1, ω] and τ : and τ (n + ω) = τ (n).

Let m > 0 be sufficiently small and M > 0 sufficiently large. It is easy to check that

$\begin{array}{c}\left(a-1\right)M\le -q\left(n\right)\sqrt[3]{u}\le \left(a-1\right)m,\\ \left(b-1\right)m\le q\left(n\right)\sqrt[3]{u}\le \left(b-1\right)M\end{array}$

for n and u [m.M]. By Theorem 1.1 (Theorem 1.2), Equation (2.1) (or (2.2)) has at least one positive ω-periodic solution x with m x M. When k = 1, this conclusion about (2.1) and (2.2) can been obtained from the results in [15]. Our result holds for all k .

Remark 1 Consider the difference equations

$x\left(n+k\right)=ax\left(n\right)+q\left(n\right)f\left(x\left(n-\tau \left(n\right)\right)\right),$
(2.3)
$x\left(n+k\right)=bx\left(n\right)-q\left(n\right)f\left(x\left(n-\tau \left(n\right)\right)\right),$
(2.4)

where k , 0 < a < 1, b > 1, q is one ω-periodic function with q(n) > 0 for all n [1, ω], τ : and τ(n + ω) = τ (n) and f : (0, + ∞) → (0, + ∞) is continuous.

The following result generalizes the conclusion of Example 2.1.

Proposition 2.1 Assume that f 0 = + ∞ and f , = 0, here

${f}_{0}=\underset{u\to 0+}{lim}\frac{f\left(u\right)}{u},\phantom{\rule{1em}{0ex}}{f}_{\infty }=\underset{u\to \infty }{lim}\frac{f\left(u\right)}{u},$

then (2.3) or (2.4) has at least one positive ω-periodic solution.

Proof Here, we only consider (2.3). From f 0 = + ∞ and f = 0, we obtain that there exist 0 < ρ 1 < ρ 2 such that

$f\left(u\right)\ge \frac{1-a}{minq\left(n\right)}u,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0

Let A = min q(n) min {f (u): ρ 1 ≤ u ≤ ρ 2} and B = max q(n) max{f (u): ρ 1 ≤ u ≤ ρ 2}. Choosing θ (0, 1) such that

$\frac{A}{1-a}\ge \theta {\rho }_{1},\phantom{\rule{1em}{0ex}}\frac{B}{1-a}\le {\theta }^{-1}{\rho }_{2},$

we obtain that

$\begin{array}{c}f\left(u\right)\ge \frac{1-a}{minq\left(n\right)}u\ge \frac{\theta \left(1-a\right){\rho }_{1}}{minq\left(n\right)},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\theta {\rho }_{1}\le u\le {\rho }_{1},\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(u\right)\le \frac{{\theta }^{-1}\left(1-a\right){\rho }_{2}}{maxq\left(n\right)},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{\rho }_{2}\le u\le {\theta }^{-1}{\rho }_{2},\\ \phantom{\rule{1em}{0ex}}A\le q\left(n\right)f\left(u\right)\le B,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\forall n\in ℤ,{\rho }_{1}\le u\le {\rho }_{2}.\end{array}$

Using the above three inequalities, we have

$\left(1-a\right)\theta {\rho }_{1}\le q\left(n\right)f\left(u\right)\le \left(1-a\right){\theta }^{-1}{\rho }_{2},\phantom{\rule{1em}{0ex}}\forall n\in ℤ,\theta {\rho }_{1}\le u\le {\theta }^{-1}{\rho }_{2}.$

By Theorem 1.1, Equation (2.3) has at least one positive ω-periodic solution x with θ ρ 1 ≤ x ≤ θ -1 ρ 2. □

Example 2.2. Consider the difference equation

$x\left(n+k\right)=-\frac{1}{{x}^{\alpha }\left(n\right)}+q\left(n\right),$
(2.5)

where k , α > 0, q is one ω-periodic function.

We claim that there is a λ > 0 such that (2.5) has at least two positive ω-periodic solutions for min q(n) > λ.

In fact, g(x) = - x -α. Let $0 be sufficiently small and $b>\sqrt[\alpha +1]{\alpha }$ be sufficiently large, then

$\begin{array}{c}\frac{\alpha }{{b}^{\alpha +1}}\le {g}^{\prime }\left(x\right)=\frac{\alpha }{{x}^{\alpha +1}}\le 1,\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[\sqrt[\alpha +1]{\alpha },b\right],\\ {g}^{\prime }\left(x\right)=\frac{\alpha }{{x}^{\alpha +1}}\ge 1,\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[a,\sqrt[\alpha +1]{\alpha }\right].\end{array}$

If the following conditions are fulfilled

$-\frac{1}{{b}^{\alpha }}-b\le -q\left(n\right)\le -\frac{1}{\sqrt[\alpha +1]{{\alpha }^{\alpha }}}-\sqrt[\alpha +1]{\alpha },\forall n\in ℤ,$
(2.6)
$-\frac{1}{{a}^{\alpha }}-a\le -q\left(n\right)\le -\frac{1}{\sqrt[\alpha +1]{{\alpha }^{\alpha }}}-\sqrt[\alpha +1]{\alpha },\forall n\in ℤ,$
(2.7)

then (2.5) has at least one periodic solution $\left[a,\sqrt[\alpha +1]{\alpha }\right]$ and $\left[\sqrt[\alpha +1]{\alpha },b\right]$ respectively. When min q(n) is sufficiently large, the conditions (2.6) and (2.7) are satisfied.

Example 2.3. Consider the difference equation

$x\left(n+k\right)={x}^{3}\left(n\right)-2x\left(n\right)-q\left(n\right){x}^{2}\left(n-\tau \left(n\right)\right),$
(2.8)

where k , q is one ω-periodic function with q(n) > 0 for all n [1, ω], τ : and τ (n + ω) = τ (n).

Let m = 1, M > 3 + max q(n) and g(u) = u 3 - 2u, f (n, u) = q(n)u 2. It is easy to check that g'(u) ≥ 1 for u [m, M], and

$g\left(m\right)-m=-2

By Theorem 1.2, Equation (2.8) has at least one positive ω-periodic solution x with mxM.

Remark 2 Consider the difference equation

$x\left(n+k\right)=g\left(x\left(n\right)\right)-q\left(n\right)f\left(x\left(n-\tau \left(n\right)\right)\right),$
(2.9)

where k , q is one ω-periodic function with q(n) > 0 for all n [1, ω], τ : ' → ' and τ(n + ω) = τ(n) and f : (0, +∞) → (0, +∞) is continuous.

Proposition 2.2 Assume that there exists a > 0 such that g C 1([a, +∞), R) with g'(u) ≥ 1 for u > a, f(u) ≥ (g(a) - a)/min q(n) for ua. Further suppose that

$\underset{u\to +\infty }{lim}\frac{g\left(u\right)-u}{f\left(u\right)}>maxq\left(n\right),\phantom{\rule{1em}{0ex}}\underset{u\to +\infty }{lim}\left(g\left(u\right)-u\right)=+\infty .$

Then (2.9) has at least one positive ω-periodic solution.

Proof There exist ρ > 0 such that

$g\left(u\right)-u\ge f\left(u\right)maxq\left(n\right),\phantom{\rule{1em}{0ex}}u\ge \rho .$

Let A = min q(n) min{f(u): auρ} and B = max q(n) max{f(u): auρ}. Since lim u→+∞(g(u) - u) = +∞ and g'(u) ≥ 1 for u > a, there is M > ρ such that g(M) - M > B and

$f\left(u\right)maxq\left(n\right)\le g\left(u\right)-u\le g\left(M\right)-M,\phantom{\rule{1em}{0ex}}\rho \le u\le M.$

Thus, (2.9) has at least one ω-periodic solution x with axM. □

3 Proof

Let X be the set of all real ω-periodic sequences. When endowed with the maximum norm ||x|| = max n[0, ω-1]|x(n)|, X is a Banach space.

Let k and 0 < c ≠ 1, and consider the equation

$x\left(n+k\right)=cx\left(n\right)+\gamma \left(n\right),$
(3.1)

where γ X. Set (k, ω) is the greatest common divisor of k and ω, h = ω/(k, ω). We obtain that if x X satisfies (3.1), then

$\begin{array}{c}{c}^{-1}x\left(n+k\right)-x\left(n\right)={c}^{-1}\gamma \left(n\right),\\ {c}^{-2}x\left(n+2k\right)-{c}^{-1}x\left(n+k\right)={c}^{-2}\gamma \left(n+k\right),\\ \cdots \cdots \cdots \cdots \\ {c}^{-p}x\left(n+hk\right)-{c}^{1-p}x\left(n+\left(h-1\right)k\right)={c}^{-p}\gamma \left(n+\left(h-1\right)k\right).\end{array}$

By summing the above equations and using periodicity of x, we obtain the following result.

Lemma 3.1. Assume that 0 < c ≠ 1, then (3.1) has a unique periodic solution

$x\left(n\right)={\left({c}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{c}^{-i}\gamma \left(n+\left(i-1\right)k\right).$

The following well-known Schauder's fixed point theorem is crucial in our arguments.

Lemma 3.2. [16]Let X be a Banach space with D X closed and convex. Assume that T : DD is a completely continuous map, then T has a fixed point in D.

Now, we rewrite (1.1) as

$x\left(n+k\right)=px\left(n\right)+\left[g\left(x\left(n\right)\right)-f\left(n,x\left(n-\tau \left(n\right)\right)-px\left(n\right)\right],$
(3.2)

where p > 0 is a constant which is determined later. By Lemma 3.1, if x is a periodic solution of (1.1), x satisfies

$x\left(n\right)={\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right),$

where h = ω/(k, ω), the mapping H p is defined as

$\left({H}_{p}x\right)\left(n\right)=g\left(x\left(n\right)\right)-px\left(n\right)-f\left(n,\phantom{\rule{2.77695pt}{0ex}}x\left(n-\tau \left(n\right)\right),\phantom{\rule{2.77695pt}{0ex}}x\in X.$

Define a mapping T p in X by

$\left({T}_{p}x\right)\left(n\right)={\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right),\phantom{\rule{1em}{0ex}}x\in X.$

Clearly, the fixed point of T p in X is a periodic solution of (1.1).

Proof of Theorem 1.1 Let p = r and Ω = {x X : mx(n) ≤ M for n '}, then Ω is a closed and convex set. If r = 1, then g(u) = u on [m, M]. It is easy to check that any constant c [m, M] is a periodic solution of (1.1). Set r < 1. Now we show that T r satisfies all conditions of Lemma 3.2. Noting that the function g(u) - ru is nondecreasing in [m, M], we have for any x Ω,

$g\left(m\right)-rm\le g\left(x\left(n\right)\right)-rx\left(n\right)\le g\left(M\right)-rM,\phantom{\rule{1em}{0ex}}\forall n\in ℤ.$

Let (1.4) be fulfilled. For any x Ω and n ,

$\begin{array}{c}\left({H}_{r}x\right)\left(n\right)=g\left(x\left(n\right)\right)-px\left(n\right)-f\left(n,x\left(n-\tau \left(n\right)\\ \le g\left(M\right)-rM-\left(g\left(M\right)-M\right)\\ =\left(1-r\right)M,\\ \left({H}_{r}x\right)\left(n\right)=g\left(x\left(n\right)\right)-px\left(n\right)-f\left(n,x\left(n-\tau \left(n\right)\\ \ge g\left(m\right)-rm-\left(g\left(m\right)-m\right)\\ =\left(1-r\right)m.\end{array}$

Hence, for any x Ω and n ,

$\begin{array}{ll}\hfill \left({T}_{r}x\right)\left(n\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}{\left({r}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{r}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right)\\ \le \phantom{\rule{1em}{0ex}}{\left({r}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{r}^{-i}\left(1-r\right)M=M,\\ \hfill \left({T}_{r}x\right)\left(n\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}{\left({r}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{r}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right)\\ \ge \phantom{\rule{1em}{0ex}}{\left({r}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{r}^{-i}\left(1-r\right)m=m.\end{array}$

Hence, T r (Ω) Ω.

Since X is finite-dimensional and g(u), f(n, u) are continuous in u, one easily show that T r is completely continuous in Ω. Therefore, T r has a fixed point x Ω by Lemma 3.2, which is a ω = periodic solution of (1.1). The proof is complete. □

Proof of Theorem 1.2 Since g C 1[m, M], max{g'(u): muM} exists and max{g'(u): muM} ≥ 1. Let p = max{g'(u): muM}. If p = 1, then g(u) ≡ u on [m, M]. It is easy to check that any constant c [m, M] is a periodic solution of (1.1). Next, we assume that p > 1. Set Ω = {x X : mx(n) ≤ M for n }. Noting that the function g(u) - pu is nonincreasing in [m, M], we have for any x Ω,

$g\left(M\right)-pM\le g\left(x\left(n\right)\right)-px\left(n\right)\le g\left(m\right)-pm,\phantom{\rule{1em}{0ex}}\forall n\in ℤ.$

For any x Ω and n ,

$\begin{array}{c}\left({H}_{p}x\right)\left(n\right)=g\left(x\left(n\right)\right)-px\left(n\right)-f\left(n,x\left(n-\tau \left(n\right)\\ \le g\left(m\right)-pm-\left(g\left(m\right)-m\right)\\ =\left(1-p\right)m,\\ \left({H}_{p}x\right)\left(n\right)=g\left(x\left(n\right)\right)-px\left(n\right)-f\left(n,x\left(n-\tau \left(n\right)\\ \ge g\left(M\right)-pM-\left(g\left(M\right)-M\right)\\ =\left(1-p\right)M.\end{array}$

Hence, for any x Ω and n ,

$\begin{array}{ll}\hfill \left({T}_{p}x\right)\left(n\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}{\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right)\\ \ge \phantom{\rule{1em}{0ex}}{\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left(1-p\right)m=m,\\ \hfill \left({T}_{p}x\right)\left(n\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}{\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left({H}_{p}x\right)\left(n+\left(i-1\right)k\right)\\ \le \phantom{\rule{1em}{0ex}}{\left({p}^{-h}-1\right)}^{-1}\sum _{i=1}^{h}{p}^{-i}\left(1-p\right)M=M.\end{array}$

Hence, T p (Ω) Ω. T p has a fixed point x Ω. The proof is complete. □

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## Acknowledgements

The authors would like to thank the referee for the comments which help to improve the article. The study was supported by the NNSF of China (10871063) and Scientific Research Fund of Hunan Provincial Education Department (10B017).

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Zhu, H., Wang, W. Periodic solutions for a class of higher-order difference equations. Adv Differ Equ 2011, 66 (2011). https://doi.org/10.1186/1687-1847-2011-66