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 [1–8], 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

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

(1)

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 ∈ 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 $l=\frac{\omega }{\left(m,\omega \right)}$ 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, l ≤ ω. In this article, we consider the following higher-order functional difference equation

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

(2)

where b ≠ 1 is positive constant, a: Z → R ₊ with a(n) ≠ 1 and a(n + ω) = a(n), τ: Z → Z 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 $\left|\right|x\left|\right|=\underset{n\in \left[0,\omega -1\right]}{max}\left|x\left(n\right)\right|$.

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 $0\in {\Omega }_1\subset {\bar{\Omega }}_1\subset {\Omega }_2$ and suppose that

$$\Phi :K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)\to K$$

is a completely continuous operator such that

(i) ||Φu|| ≤ ||u|| for u ∈ K ∩ ∂Ω ₁ and there exists ψ ∈ K\{0} such that x ≠ Φx + 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 x ≠ Φx + λψ for x ∈ K ∩ ∂Ω ₁ and λ > 0.

Then, Φ has a fixed point in $K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$.

Let d ∈ N. Consider the equation

$$x\left(n+d\right)=cx\left(n\right)+\gamma \left(n\right)$$

(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

$$x\left(n\right)={\left[c^{-p}-1\right]}^{-1}\sum _{i=1}^p{c}^{-i}\gamma \left(n+\left(i-1\right)d\right).$$

Let y(n) = x(n + k) - a(n)x(n), $ā=\underset{1\le n\le \omega }{max}a\left(n\right),\underset{-}{a}=\underset{1\le n\le \omega }{min}a\left(n\right)$, then (2) can be rewritten as

$$\left\{\begin{array}{c}\hfill x\left(n+k\right)=\underset{-}{a}x\left(n\right)+y\left(n\right)+\left[a\left(n\right)-\underset{-}{a}\right]x\left(n\right),\hfill \\ \hfill y\left(n+m\right)=by\left(n\right)+f\left(n,x\left(n-\tau \left(n\right)\right)\right).\hfill \end{array}\right.$$

(4)

Let $h=\frac{\omega }{\left(k,\omega \right)},l=\frac{\omega }{\left(m,\omega \right)}$. Assume that x ∈ X solution of (2), then y ∈ X. From Lemma 2, we have

$$\begin{array}{c}x\left(n\right)={\left[{\underset{-}{a}}^{-h}-1\right]}^{-1}\sum _{i=1}^h{\underset{-}{a}}^{-1}\left\{y\left(n+\left(i-1\right)k\right)+\left[a\left(n+\left(i-1\right)k\right)-\underset{-}{a}\right]x\left(n+\left(i-1\right)k\right)\right\},\\ y\left(n\right)={\left[b^{-l}-1\right]}^{-1}\sum _{i=1}^l{b}^{-i}f\left(n+\left(i-1\right)m,x\left(n+\left(i-1\right)m-\tau \left(n+\left(i-1\right)m\right)\right)\right).\end{array}$$

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

$$\begin{array}{c}(Tx)(n)=\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-j}}f(n+(i-1)k+(j-1)m,\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}[a(n+(i-1)k)-\underset{¯}{a}]x(n+(i-1)k).\end{array}$$

Define the cone by

$$K=\left\{x\in X,x\left(n\right)\ge \delta \left|\right|x\left|\right|\right\}$$

where $\delta ={\underset{-}{a}}^h{b}^l\left(1-{\underset{-}{a}}^h\right)\left(1-b^l\right)∕\omega$.

Lemma 3 Assume that (H) holds and 0 < r ₁ < r ₂, then $T:{\bar{K}}_{r_2}\backslash K_{r_1}\to K$ is completely continuous, where K _r = {x ∈ K: ||x|| < r} and ${\bar{K}}_r=\left\{x\in K:\left|\right|x\left|\right|\le r\right\}$.

Proof Since 0 < a(n) < 1, then $0<\underset{-}{a}<1$. 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 $l=\frac{\omega }{\left(m,\omega \right)}\le \omega$ we have

$$\sum _{j=1}^{l}f\left(n+\left(j-1\right)m,x\left(n+\left(j-1\right)m-\tau \left(n+\left(j-1\right)m\right)\right)\right)\le \sum _{j=1}^{\omega }f\left(j,x\left(j-\tau \left(j\right)\right)\right).$$

On the other hand, from (H), $h=\frac{\omega }{\left(k,\omega \right)}=\omega$, we have

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

and

$$\sum _{i=1}^h\left[a\left(n+\left(i-1\right)k\right)-\underset{-}{a}\right]x\left(n+\left(i-1\right)k\right)=\sum _{i=1}^{\omega }\left[a\left(i\right)-\underset{-}{a}\right]x\left(i\right).$$

For any $x\in {\bar{K}}_{r_2}\backslash K_{r_1}$,

$$\begin{array}{l}(Tx)(n)=\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-j}}f(n+(i-1)k+(j-1)m,\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}[a(n+(i-1)k)-\underset{¯}{a}]x(n+(i-1)k)\\ \le \frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\underset{¯}{a}}^{-h}{b}^{-l}{\displaystyle \sum _{i=1}^h{\displaystyle \sum _{j=1}^l\{f(n+(i-1)k+(j-1)m,}}\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\underset{¯}{a}}^{-h}{\displaystyle \sum _{i=1}^h[a(n+(i-1)k)-\underset{¯}{a}]x(n+(i-1)k)}\\ =\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\underset{¯}{a}}^{-h}{b}^{-l}{\displaystyle \sum _{j=1}^l{\displaystyle \sum _{i=1}^h\{f(n+(i-1)k+(j-1)m,}}\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\underset{¯}{a}}^{-h}{\displaystyle \sum _{i=1}^{\omega }(a(i)-\underset{¯}{a})x(i)}\\ \le \frac{1}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{j=1}^{\omega }{\displaystyle \sum _{i=1}^{\omega }f(i,x(i-\tau (i)))}}\\ +\frac{1}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^{\omega }(a(i)-\underset{¯}{a})x(i)}\\ \le \frac{\omega }{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^{\omega }f(i,x(i,\tau (i)))}\\ +\frac{1}{(1-\underset{¯}{a}h)}{\displaystyle \sum _{i=1}^{\omega }(a(i)-\underset{¯}{a})x(i)}\\ \le \frac{\omega }{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^{\omega }\{f(i,x(i,\tau (i)))+(a(i)-\underset{¯}{a})x(i)\}.}\end{array}$$

So

$$\left|\right|Tx\left|\right|\le \frac{\omega }{\left(1-{\underset{-}{a}}^h\right)\left(1-b^l\right)}\sum _{i=1}^{\omega }\left\{f\left(i,x\left(i-\tau \left(i\right)\right)\right)+\left(a\left(i\right)-\underset{-}{a}\right)x\left(i\right)\right\}.$$

(5)

At the same time

$$\begin{array}{l}(Tx)(n)\ge \frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\underset{¯}{a}}^{-1}{b}^{-1}{\displaystyle \sum _{i=1}^h{\displaystyle \sum _{j=1}^l\{f(n+(i-1)k+(j-1)m,}}\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{a}^{-1}{\displaystyle \sum _{i=1}^h[a(n+(i-1)k)-\underset{¯}{a}]x(n+(i-1)k)}\\ =\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\underset{¯}{a}}^{-1}{b}^{-1}{\displaystyle \sum _{j=1}^l{\displaystyle \sum _{i=1}^h\{f(n+(i-1)k+(j-1)m,}}\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\underset{¯}{a}}^{-1}{\displaystyle \sum _{i=1}^{\omega }(a(i)-\underset{¯}{a})x(i)}\\ \ge \frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}\frac{l{b}^l}{(1-b^l)}{\displaystyle \sum _{i=1}^{\omega }f(i,x(i-\tau (i)))}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^{\omega }(a(i)-\underset{¯}{a})x(i)}\\ \ge \frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^{\omega }[\frac{b^l}{(1-b^l)}f(i,x(i,\tau (i)))+(a(i)-\underset{¯}{a})x(i)]}\\ \ge {\underset{¯}{a}}^h{\displaystyle \sum _{i=1}^{\omega }[b^{l}f(i,x(i-\tau (i)))+(a(i)-\underset{¯}{a})x(i)]}\\ \ge {\underset{¯}{a}}^h{b}^l{\displaystyle \sum _{i=1}^{\omega }[f(i,x(i-\tau (i)))+(a(i)-\underset{¯}{a})x(i)]}.\end{array}$$

We have

$$\left(Tx\right)\left(n\right)\ge \delta \left|\right|Tx\left|\right|.$$

(6)

Thus $T:{\bar{K}}_{r_2}\backslash K_{r_1}\to K$ 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

$$\begin{array}{c}\phi \left(s\right)=max\left\{f\left(n,u\right),n\in \left[0,\omega -1\right],u\in \left[\delta s,s\right]\right\}\\ \psi \left(s\right)=min\left\{\frac{f\left(n,u\right)}{u},n\in \left[0,\omega -1\right],u\in \left[\delta s,s\right]\right\}\end{array}$$

Let $ā=\underset{1\le n\le \omega }{max}a\left(n\right),\underset{-}{a}=\underset{1\le n\le \omega }{min}a\left(n\right)$.

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

$$\phi \left(\alpha \right)\le \left(ā-1\right)\left(b-1\right)\alpha ,\psi \left(\beta \right)\ge \left(\underset{-}{a}-1\right)\left(b-1\right)$$

(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, $0<ā<1,0<\underset{-}{a}<1$. We claim that:

• (i) ||Tx|| ≤ ||x||, x ∈ ∂K _α ,

• (ii) x ≠ Tx + λ · 1, ∀x ∈ ∂K _β , 1 ∈ K and λ > 0.

From (7), we have that

$$f\left(n,x\right)\le \left(ā-1\right)\left(b-1\right)\alpha ,\forall 0\le n\le \omega -1,\forall \delta \alpha \le x\le \alpha ,$$

(8)

$$f(n,x)\ge (\underset{¯}{a}-1)(b-1)x,\forall 0\le n\le \omega -1,\forall \delta \alpha \le x\le \beta .$$

(9)

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

$$\begin{array}{l}(Tx)(n)=\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-j}}f(n+(i-1)k+(j-1)m,\\ x(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}[a(n+(i-1)k)-\underset{¯}{a}]x(n+(i-1)k)\\ \le \frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}{\displaystyle \sum _{j=1}^l{b}^{-j}}\{(\overline{a}-1)(b-1)\alpha \}}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}[\overline{a}-\underset{¯}{a}]\left|\right|x\left|\right|}\\ \le \left\{\frac{b^l}{(1-b^l)}(1-b){\displaystyle \sum _{j=1}^l{b}^{-j}}\right\}\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}\{(1-\overline{a})\alpha \}\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}[\overline{a}-\underset{¯}{a}]\alpha \\ =\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}[1-\underset{¯}{a}}]\alpha \\ =\alpha .\end{array}$$

It follows that

$$\left|\right|Tx\left|\right|\le \left|\right|x\left|\right|,x\in \partial K_{\alpha }.$$

(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

$$u_0=\left(T{u}_0\right)\left(n\right)+{\lambda }_0.$$

(11)

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

$$\begin{array}{l}{u}_0(n)=(T{u}_0)(n)+{\lambda }_0\\ =\frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-j}}f(n+(i-1)k+(j-1)m,\\ u_0(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\\ +\frac{{\underset{¯}{a}}^h}{(1-{\underset{¯}{a}}^h)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}[a(n+(i-1)k)-\underset{¯}{a}]u_0(n+(i-1)k)+{\lambda }_0\\ \ge \frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-j}}\{f(n+(i-1)k+(j-1)m,\\ u_0(n+(i-1)k+(j-1)m-\tau (n+(i-1)k+(j-1)m)))\}+{\lambda }_0\\ \ge \frac{{\underset{¯}{a}}^h{b}^l}{(1-{\underset{¯}{a}}^h)(1-b^l)}{\displaystyle \sum _{i=1}^h{\underset{¯}{a}}^{-i}}{\displaystyle \sum _{j=1}^l{b}^{-1}(\underset{¯}{a}-1)(b-1)u_0(n+(i-1)k}\\ +(j-1)m-\tau (n+(i-1)k+(j-1)m))+{\lambda }_0\\ \ge u_0(n)+{\lambda }_0\end{array}$$

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

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

(12)

where b = 1/2, m = 3, k = 5, ω = 6, τ: Z → Z and τ(n + ω) = τ(n), a: Z → R ₊ with $a\left(n\right)=\frac{1}{2}+\frac{1}{16}cos\frac{n\pi }{3},f\left(n,u\right)=\left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right)u^3\left[1+\frac{1}{2}{\left(-1\right)}^{n}cos\frac{\pi u}{3}\right]$.

Obviously, a(n + ω) = a(n + 6) = a(n), f(n + ω, u) = f(n + 6, u) = f(n, u) for any u ∈ R. $h=\frac{\omega }{\left(k,\omega \right)}=\frac{6}{\left(5,6\right)}=6,l=\frac{\omega }{m,\omega }=\frac{6}{\left(3,6\right)}=2.ā=\underset{1\le n\le \omega }{max}a\left(n\right)=\frac{9}{16},\underset{-}{a}=\underset{1\le n\le \omega }{min}a\left(n\right)=\frac{7}{16},\delta ={\left(\frac{7}{16}\right)}^6{\left(\frac{1}{2}\right)}^2\left[1-{\left(\frac{7}{16}\right)}^6\right]\left[1-{\left(\frac{1}{2}\right)}^2\right]∕6$.

Let $\alpha =\frac{1}{2}$, then

$$\begin{array}{c}\phi (\alpha )=\phi \left(\frac{1}{2}\right)\\ \le \left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right){\left(\frac{1}{2}\right)}^3\left[1+\frac{1}{2}\right]\\ =\left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right){\left(\frac{1}{2}\right)}^2\frac{3}{4}\\ <\left(\frac{9}{16}\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\frac{1}{2}\\ <\left(1-\frac{9}{16}\right)\left(1-\frac{1}{2}\right)\frac{1}{2}.\end{array}$$

(13)

So $\phi \left(\alpha \right)\le \left(ā-1\right)\left(b-1\right)\alpha$.

Let $\beta =\frac{2}{\delta }$. If u ∈ [δβ,β], then u ≥ 2. Furthermore,

$$\begin{array}{c}\psi (\beta )\ge \left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right)\left(\frac{2^3}{2}\right)\left[1-\frac{1}{2}\right]\\ =2\left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right)\\ >\left(1-\frac{7}{16}\right)\left(1-\frac{1}{2}\right).\end{array}$$

(14)

So $\psi \left(\beta \right)\ge \left(\underset{-}{a}-1\right)\left(b-1\right)$.

By Theorem 1 in this article, (12) has at least one positive 6-periodic solution.