0 Introduction

Fractional linear maps have been studied extensively^[1-3]. A special form of Ricatti map, so-called Beverton-Holt map

$ h(x)=\frac{\mu K x}{K+(\mu-1) x} $

is deeply concerned. This map originated from an investigation of the reaction of species to periodically changing living environment, where μ>0 denotes intrinsic growth rate and K>0 expresses carrying capacity of the environment.

The Beverton-Holt equation models density dependent growth which shows compensation as opposed to over-compensation. This equation takes the following form:

$ x_{n+1}=\frac{\mu K x_n}{K+(\mu-1) x_n}, x_0>0, \mu>0, K>0 $

(1)

Obviously, x=0 is a fixed point of Eq.(1). It is globally asymptotical stable if μ∈(0, 1). x=K is another fixed point, and is globally asymptotical stable for μ>1.

It is known that living environment surrounding the species influences the population dramatically. In Ref. [4], authors considered a situation where carrying capacity K fluctuates periodically with a minimal integer period p as a result of seasonally varying environment. Then Eq. (1) becomes

$ x_{n+1}=\frac{\mu K_n x_n}{K_n+(\mu-1) x_n}, x_0>0, K_n>0 $

(2)

where K_n+p=K_p, n=0, 1, 2, ….

Let

$ f_i(x)=\frac{\mu K_i x}{K_i+(\mu-1) x}, i=0, 1, \cdots $

then a (nonautonomous) difference system with period p is obtained.

$ x_{n+1}=f_n\left(x_n\right), f_{n+p}(x)=f_n(x), n=0, 1, \cdots $

(3)

In Ref.[5], Cushing and Henson made two conjectures for p-periodic Beverton-Holt Eq. (2) when μ>1 and p≥2:

Conjecture (Conj.) 1  A positive p-periodic solution $\left\{\bar{x}_0, \quad \bar{x}_1, \cdots, \quad \bar{x}_{p-1}\right\}$ exists, which globally attracts all positive solutions.

Conjecture (Conj.) 2  This p-periodic solution is attenuant, that is

$ a v\left(\bar{x}_n\right)=\frac{1}{p} \sum\limits_{i=0}^{p-1} \bar{x}_i<a v\left(K_n\right)=\frac{1}{p} \sum\limits_{i=0}^{p-1} K_i $

The correctness of Conj.2 means that a varying habitat harms the population. That is to say, the average population is less in a periodically oscillating habitat than it is in a constant habitat with the same average^[6].

Elaydi and Sacker proved these two conjectures in Refs. [7] and [8]. There are other creative works relating the average $a v\left(\bar{x}_n\right)$ of periodic solutions and the average av(K_n) of carrying capacities. For example, Kocic^[9] considered the above two conjectures for a more general case of bounded K_n. In Ref. [10], Kon proved Conj.2 for some systems including periodically forced Beverton-Holt Eq. (2).

In Ref. [11], Haskell et al. studied the attenuation of Beverton-Holt equation when μ=μ_n is also p-periodic. They gave a condition on μ_n and K_n to make the second conjecture true.

Refs.[12] and [13] proved the existence and globally asymptotically stability of periodic orbit of period r for periodic nonautonomous difference equations via the concept of skew-product dynamical systems.

In present work, we investigate the Beverton-Holt equation for the case of changing μ_n and K_n periodically with same period p. Some sufficient conditions are obtained to guarantee the attenuation of periodic solution for p-periodic Beverton-Holt equation. Our proof about the attenuation is different from that in Ref. [11].

1 Average of Periodic Cycle

As mentioned above, it is already known that Conj. 1 and Conj. 2 are correct for constant μ and periodically changing K_n. However, the situation is drastically different if μ is also p-periodic. That is to say, the intrinsic growth rate of the species also changes periodically with time.

Consider Eq. (2) with p-periodic μ_n,

$ x_{n+1}=f_n\left(x_n\right)=\frac{\mu_n K_n x_n}{K_n+(\mu-1) x_n}, n=0, 1, \cdots $

(4)

where μ_n and K_n have the same minimal period p≥2 satisfying μ_n>1, K_n>0.

The existence and globally asymptotically stability of p-cycle $\left\{\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}\right\}$ was proved in Ref. [14] for Eq. (4), while p may not be the minimal period.

Ref.[14] extends Conj.2 to Eq. (4). The following inequality is obtained:

$ a v\left(\tilde{x}_n\right)<\frac{\mu_{\max }}{\mu_{\min }} \cdot \frac{\mu_{\max }-1}{\mu_{\min }-1} \cdot a v\left(K_n\right) $

where $\mu_{\max }=\max \left\{\mu_n\right\}, \mu_{\min }=\min \left\{\mu_n\right\}$.

As for the average of p-cycle{$\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}$}, there are two more results.

Proposition 1  If K_i≠K_i+1 for at least one i∈{0, 1, …, p-1}, hen for any p-periodic solution $\left\{\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}\right\}$ of difference Eq. (4) with μ_n>1, K_n>0, there is

$ K_{\min }<a v\left(\tilde{x}_n\right)<K_{\max } $

where $K_{\min }=\min\limits_{0 \leqslant n \leqslant p-1}\left\{\mu_n\right\}, K_{\max }=\max\limits_{0 \leqslant n \leqslant p-1}\left\{\mu_n\right\}$.

Proof  Define

$ f_i(x)=\frac{\mu_i K_i x}{K_i+\left(\mu_i-1\right) x}, i=0, 1, \cdots, p-1 $

and F(x)=f_p-1°f_p-2°…°f₁°f₀(x). Clearly, $\tilde{x}_0$ is a fixed point of F(x). F(x) can be obtained by p-1 iterations inductively:

$ F(x)=\frac{\mu_{p-1} \cdots \mu_0 K_{p-1} \cdots K_0}{K_{p-1} \cdots K_0+E_{p-1} x} \cdot x $

where E_i satisfies the linear difference equation below:

$ \begin{gathered} E_i=K_i E_{i-1}+\left(\mu_i-1\right) \mu_{i-1} \cdots \mu_0 K_{i-1} \cdots K_0 \\ E_0=\mu_0-1, i=0, 1, \cdots, p-1 \end{gathered} $

So

$ E_{p-1}=K_{p-1} \cdots K_0 \cdot \sum\limits_{i=0}^{p-1} \frac{\left(\mu_i-1\right) \mu_{i-1} \cdots \mu_0}{K_i} $

Fixed point $\tilde{x}_0$ of F(x) is obtained by solving equation F(x)=x:

$ \tilde{x}_0=\frac{K_{p-1} \cdots K_0\left(\mu_{p-1} \cdots \mu_0-1\right)}{E_{p-1}}=\frac{1}{\frac{r_{p-1}}{K_{p-1}}+\cdots+\frac{r_0}{K_0}} $

where coefficients r_i are defined as follows:

$ r_i=\left\{\begin{array}{l} \frac{\mu_0-1}{\mu_{p-1} \cdots \mu_0-1}, i=0 \\ \frac{\left(\mu_i-1\right) \mu_{i-1} \cdots \mu_0}{\mu_{p-1} \mu_{p-2} \cdots \mu_0-1}, i=1, 2, \cdots, p-1, \end{array}\right. $

(5)

where $\sum\limits_{i=0}^{p-1} r_i=1$.

If K_i≠K_i+1 for at least one i∈{0, 1, …, p-1}, there is

$ K_{\min }=\frac{1}{\frac{r_{p-1}}{K_{\min }}+\cdots+\frac{r_0}{K_{\min }}}<\tilde{x}_0<\frac{1}{\frac{r_{p-1}}{K_{\max }}+\cdots+\frac{r_0}{K_{\max }}}=K_{\max } $

(6)

Since $\tilde{x}_1=f\left(\tilde{x}_0\right)$, so

$ \tilde{x}_1=f_0{ }^{\circ} {f_{p-1}} ^{\circ} {f_{p-2}} ^{\circ} \cdots ^{\circ} f_1\left(\tilde{x}_1\right)=F_1\left(\widetilde{x_1}\right) $

where F₁(x₁)=f₀°f_p-1°f_p-2°…°f₁(x₁). By the symmetry, there is

$ \tilde{x}_1=\frac{K_0 K_{p-1} \cdots K_1\left(\mu_0 \mu_{p-1} \cdots \mu_1-1\right)}{E_{p-1}^1}=\frac{1}{\frac{r_{p-1}^1}{K_0}+\cdots+\frac{r_0^1}{K_1}} $

where

$ \begin{gathered} E_{p-1}^1=K_0 K_{p-1} \cdots K_1 \cdot \sum\limits_{i=0}^{p-1} \frac{\left(\mu_{i+1}-1\right) \mu_i \cdots \mu_1}{K_{i+1}}, \\ K_p=K_0, \mu_p=\mu_0 \end{gathered} $

and

$ r_i^1=\left\{\begin{array}{l} \frac{\mu_1-1}{\mu_0 \mu_{p-1} \cdots \mu_1-1}, i=0 \\ \frac{\left(\mu_{i+1}-1\right) \mu_i \cdots \mu_1}{\mu_0 \mu_{p-1} \cdots \mu_1-1}, i=1, 2, \cdots, p-1 \end{array}\right. $

(7)

where $\sum\limits_{i=0}^{p-1} r_i^1=1, \mu_p=\mu_0$.

Similar to Eq. (6), there is

$ \begin{aligned} K_{\min }=& \frac{1}{\frac{r_{p-1}^1}{K_{\min }}+\cdots+\frac{r_0^1}{K_{\min }}}<\tilde{x}_1<\\ & \frac{1}{\frac{r_{p-1}^1}{K_{\max }}+\cdots+\frac{r_0^1}{K_{\max }}}=K_{\max } \end{aligned} $

(8)

By the same way, there is

$ K_{\min }<\tilde{x}_i<K_{\max }, i=2, 3, \cdots, p-1 $

(9)

because $\tilde{x}_i={f_{i-1}} ^{\circ} \cdots{ }^{\circ} {f_0}^{\circ} {f_{p-1}} ^{\circ} \cdots ^{\circ} f_i\left(\tilde{x}_i\right)$.

This means $K_{\min }<\tilde{x}_i<K_{\max }(i=0, 1, \cdots, p-1)$ and the proof is completed.

Proposition 2  If K_i≠K_i+1 for at least one i∈{0, 1, …, p-1}, then for any p-periodic solution of difference Eq. (4) with μ_n>1, K_n>0, there is

$ a v\left(\tilde{x}_n\right)<\frac{\mu_{\max }}{\mu_{\min }-1} \cdot a v\left(K_n\right) $

Proof  The proposition is correct because

$ a v\left(\tilde{x}_n\right)=\frac{1}{p} \sum\limits_{i=0}^{p-1} \tilde{x}_i=\frac{1}{p} \sum\limits_{i=0}^{p-1} \tilde{x}_{i+1}= \\ \frac{1}{p} \sum\limits_{i=0}^{p-1} \frac{\mu_i K_i \tilde{x}_i}{K_i+\left(\mu_i-1\right) \tilde{x}_i}= \\ \frac{1}{p} \sum\limits_{i=0}^{p-1} \frac{\frac{\mu_i K_i}{\mu_i-1} \cdot\left(\frac{\mu_i-1}{K_i} \tilde{x}\right)}{1+\frac{\mu_i-1}{K_i} \tilde{x}_i}< \\ \frac{1}{p} \sum\limits_{i=0}^{p-1} \frac{\mu_i K_i}{\mu_i-1}<\frac{\mu_{\max }}{\mu_{\min }-1} \cdot a v\left(K_n\right) $

The following theorem summarizes the above conclusions.

Theorem 1  For any p-periodic solution of difference Eq. (4) with μ_n>1, K_n>0 and K_i≠K_i+1 for at least one i∈{0, 1, …, p-1}, the following inequalities hold:

$ \begin{aligned} &K_{\min }<a v\left(\tilde{x}_n\right)< \\ &\min \left\{K_{\max }, \frac{\mu_{\max }}{\mu_{\min }-1} \cdot a v\left(K_n\right), \frac{\mu_{\max }}{\mu_{\min }} \cdot \frac{\mu_{\max }-1}{\mu_{\min }-1} \cdot a v\left(K_n\right)\right\} \end{aligned} $

Theorem 1 shows a concrete range of $a v\left(\tilde{x}_n\right)$ for the periodic solutions: $\left\{\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}\right\}$ of Eq. (4) for any p-periodic with μ_n>1 and K_n>0.

2 Results

In this section, an extension of Conj. 2 to Eq. (4) is discussed. The following example illustrates that Conj. 2 may not hold for some p-periodic μ_n and K_n.

Example 1  Taking p=2, μ₀=2, μ₁=3, K₀=3, K₁=4 in Eq. (4), 2-periodic solution {15/4, 10/3} and av(K_n)=85/24>av(K_n)=7/2 can be easily obtained.

Two theorems about the attenuation of periodic solution of Eq. (4) for periodically changing μ_i and K_i are proved in this section. Our proofs are more direct and different from those in Ref. [11]. Firstly, a lemma is proved which will be used below.

Lemma 1  The following inequality is true for a_i>0, x_i>0, i=1, 2, …n,

$ \begin{gathered} \frac{x_1 x_2 \cdots x_n}{a_1 x_2 \cdots x_n+\cdots+a_n x_1 x_2 \cdots x_{n-1}} \leqslant \\ a_1 x_1+a_2 x_2+\cdots+a_n x_n \end{gathered} $

where $\sum\limits_{i=1}^n a_i=1$. "=" holds if and only if x₁=x₂=…=x_n.

Proof  This lemma can be proved easily by applying Jessen's inequality to the convex function h(x)=x^－1 defined on (0, +∞).

Next the conditions on Eq.(4) is explored to prove Conj. 2 is true.

Recalling the proof of Proposition 1, $\tilde{x}_0$ can be rewritten as

$ \tilde{x}_0=\frac{K_{p-1} K_{p-2} \cdots K_0}{r_{p-1} K_{p-2} \cdots K_0+\cdots+r_0 K_{p-1} \cdots K_1} $

where all r_i are given by Eq. (5). With Lemma 1, there is

$ \tilde{x}_0<r_{p-1} K_{p-1}+\cdots+r_0 K_0 $

(10)

By $f\left(\tilde{x}_0\right)=\tilde{x}_1$, there is

$ \widetilde{x}_1=\frac{K_{p-1} K_{p-2} \cdots K_0}{\frac{r_{p-1}}{\mu_0 \prod\limits_{j \neq p-1}} K_j+\cdots} \rightarrow \\ \;\;\;\;\;\leftarrow \frac{1}{+\frac{r_1}{\mu_0} \prod\limits_{j \neq 1} K_j+\mu_{p-1} \cdots \mu_1 r_0 \prod\limits_{j \neq 0} K_j} $

Since

$ \frac{r_{p-1}}{\mu_0}+\cdots+\frac{r_1}{\mu_0}+\mu_{p-1} \cdots \mu_1 r_0=1 $

the following inequality is obtained:

$ \tilde{x}_1<\frac{r_{p-1}}{\mu_0} K_{p-1}+\cdots+\frac{r_1}{\mu_0} K_1+\mu_{p-1} \cdots \mu_1 r_0 K_0 $

(11)

In the same way, there is

$ \begin{aligned} &\tilde{x}_i=\frac{K_{p-1} K_{p-2} \cdots K_0}{\frac{1}{\mu_{i-1} \cdots \mu_0} \cdot \sum\limits_{t=i}^{p-1} r_t \prod\limits_{j \neq t} K_j+\cdots} \rightarrow\\ &\longleftarrow \frac{1}{+\mu_{p-1} \cdots \mu_i \cdot \sum\limits_{t=0}^{i-1} r_t \prod\limits_{j \neq t} K_j} \text {, }\\ &i=2, \cdots, p-1 . \end{aligned} $

It is easy to check

$ \frac{1}{\mu_{i-1} \cdots \mu_0} \cdot \sum\limits_{t=i}^{p-1} r_t+\cdots+\mu_{p-1} \cdots \mu_i \cdot \sum\limits_{t=0}^{p-1} r_t=1 $

Therefore

$ \begin{gathered} \tilde{x}_i<\frac{1}{\mu_{i-1} \cdots \mu_0} \cdot \sum\limits_{i=i}^{p-1} r_t K_t+\mu_{p-1} \cdots \mu_i \cdot \sum\limits_{t=0}^{i-1} r_t K_t, \\ i=2, \cdots, p-1 \end{gathered} $

(12)

For $\widetilde{x}_{p-1}$, there is

$ \begin{aligned} &\widetilde{x}_{p-1}=\frac{K_{p-1} K_{p-2} \cdots K_0}{\frac{r_{p-1}}{\mu_{p-2} \cdots \mu_{0 j \neq p-1}} \prod\limits_j K_j+\cdots} \rightarrow\\ &\leftarrow \frac{1}{+\mu_{p-1}\left(r_{p-2} \prod\limits_{j \neq p-2} K_j+\cdots+r_0 \prod\limits_{j \neq 0} K_j\right)} \end{aligned} $

By equation

$ \frac{r_{p-1}}{\mu_{p-2} \cdots \mu_0}+\mu_{p-1} r_{p-2}+\cdots+\mu_{p-1} r_0=1 $

and Lemma 1 again, the following expression is obtained:

$ \begin{aligned} \tilde{x}_{p-1}<& \frac{r_{p-1}}{\mu_{p-2} \cdots \mu_0} K_{p-1}+\mu_{p-1} r_{p-2} K_{p-2}+\cdots+\\ & \mu_{p-1} r_0 K_0 \end{aligned} $

(13)

Adding up inequalities from (10) to (13) yields

$ \begin{aligned} &\tilde{x}_0+\cdots+\tilde{x}_{p-1}<\left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{p-2} \cdots \mu_0}\right) r_{p-1} K_{p-1}+ \\ &\quad\left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{p-3} \cdots \mu_0}+\mu_{p-1}\right) r_{p-2} K_{p-2}+\cdots+ \\ &\quad\left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{i-1} \cdots \mu_0}+\mu_{p-1} \cdots \mu_{i+1}+\cdots+\right. \\ &\left.\quad \mu_{p-1}\right) r_i K_i+\cdots+\left(1+\mu_{p-1} \cdots \mu_1+\mu_{p-1} \cdots \mu_2+\right. \\ &\left.\quad \cdots+\mu_{p-1}\right) r_0 K_0=K_{p-1}+K_{p-2}+\cdots+K_0+ \\ &\sum\limits_{i=0}^{p-1} s_i K_i \end{aligned} $

That is

$ \tilde{x}_0+\cdots+\tilde{x}_{p-1}<K_{p-1}+K_{p-2}+\cdots+K_0+\sum\limits_{i=0}^{p-1} s_i K_i $

(14)

where

$ \begin{aligned} &s_i= \\ &\left\{\begin{array}{l} \left(1+\mu_{p-1} \cdots \mu_1+\mu_{p-1} \cdots \mu_2+\cdots+\mu_{p-1}\right) r_0-1, i=0 \\ \left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{p-2} \cdots \mu_0}\right) r_{p-1}-1, i=p-1 \\ \left(1+\cdots+\frac{1}{\mu_{i-1} \cdots \mu_0}+\cdots+\mu_{p-1}\right) r_i-1, i=1, 2, \cdots, p-2 \end{array}\right. \end{aligned} $

(15)

Or

$ \begin{aligned} s_0=& \frac{\mu_0-1}{\prod\limits_{j=0}^{p-1} \mu_j-1} \cdot\left(\mu_{p-1} \cdots \mu_2+\cdots+\mu_{p-1}\right)+\\ & \frac{\mu_0-\prod\limits_{j \neq 0} \mu_j}{\prod\limits_{j=0}^{p-1} \mu_j-1} \end{aligned} $

(16a)

$ \begin{aligned} s_{p-1}=& \frac{\mu_p-1}{\prod\limits_{j=0}^{p-1} \mu_j-1} \cdot\left(\mu_{p-2} \cdots \mu_1+\cdots+\mu_{p-2}\right)+\\ & \frac{\mu_{p-1}-\prod\limits_{j \neq p-1} \mu_j}{\prod\limits_{j=0}^{p-1} \mu_j-1} \end{aligned} $

(16b)

$ \begin{aligned} s_i=& \frac{\mu_i-1}{\prod\limits_{j=0}^{p-1} \mu_j-1} \cdot\left(\mu_{i-1} \cdots \mu_0 \cdot \mu_{p-1} \cdots \mu_{i+2}+\cdots+\right.\\ &\left.\mu_{i-1} \cdots \mu_0 \cdot \mu_{p-1}\right)+\frac{\mu_i-1}{\prod\limits_{j=0}^{p-1} \mu_j-1} \cdot \end{aligned} \\ \;\;\; \begin{aligned} &\left(\mu_{i-1} \cdots \mu_0+\cdots+\mu_{i-1}\right)+ \\ &\frac{\mu_i-\prod\limits_{j \neq i} \mu_j}{\prod\limits_{j=0}^{p-1} \mu_j-1} \end{aligned} $

(16c)

in which i=1, 2, …, p-2.

Define $\overrightarrow{\boldsymbol{s}}=\left(s_0, s_1, \cdots, s_{p-1}\right), \overrightarrow{\boldsymbol{K}}=\left(K_0, K_1, \cdots, K_{p-1}\right)$. By inequality (14), the following theorem is immediately obtained.

Theorem 2  Suppose μ_n>1, K_n>0, and $\left\{\tilde{x}_n\right\}_{n=0}^{p-1}=\left\{\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}\right\}$ is a p-periodic solution of Eq. (4), then $a v\left(\tilde{x}_n\right)<a v\left(K_n\right)$ when $\overrightarrow{\boldsymbol{s}} \cdot \overrightarrow{\boldsymbol{K}} \leqslant 0$.

Theorem 2 provides a sufficient condition on s_n and K_n to guarantee the attenuation of periodic solution, i.e., $a v\left(\tilde{x}_n\right)<a v\left(K_n\right)$.

Now, s_i is closely studied. According to Eq. (15) and $\sum\limits_{i=0}^{p-1} r_i=1$, there is

$ \begin{aligned} &\sum\limits_{i=0}^{p-1} s_i=-p+\left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{p-2} \cdots \mu_0}\right) r_{p-1}+\\ &\left(1+\frac{1}{\mu_0}+\cdots+\frac{1}{\mu_{p-3} \cdots \mu_0}+\mu_{p-1}\right) r_{p-2}+\cdots+\\ &\left(1+\mu_{p-1} \cdots \mu_1+\mu_{p-1} \cdots \mu_2+\cdots+\mu_{p-1}\right) r_0=\\ &-p+\sum\limits_{i=0}^{p-1} r_i+\frac{r_{p-1}+\cdots+r_1+r_0 \prod\limits_{i=0}^{p-1} \mu_i}{\mu_0}+\\ &\frac{r_{p-1}+\cdots+r_2+\left(r_1+r_0\right) \prod\limits_{i=0}^{p-1} \mu_i}{\mu_1 \mu_0}+\cdots+\\ &\frac{r_{p-1}+\left(r_{p-2}+\cdots+r_0\right) \prod\limits_{i=0}^{p-1} \mu_i}{\mu_{p-2} \cdots \mu_1 \mu_0}=0 \end{aligned} $

or

$ s_0+s_1+\cdots+s_{p-1}=0 $

(17)

The following lemma shows how the signs of s_i change with μ₀, μ₁, …, μ_p-1.

Lemma 2   Let μ_n>1, K_n>0 be both p-periodic, and s_i be defined by Eq.(15). If p-periodic sequence {μ_n} satisfies μ₀≤μ₁≤…≤μ_p-1, then the following statements are true.

(S₁) s₀ < 0, s_p-1>0;

(S₂) If s_i-1≥0, then s_i>0 for i=2, …, p-1.

Proof  (S₁) can be proved easily. By Eq. (15), there is

$ s_0=\frac{\sum\limits_{j=2}^{p-1} \mu_{p-1} \mu_{p-2} \cdots \mu_j\left(\mu_0-\mu_{j-1}\right)+\left(\mu_0-\mu_{p-1}\right)}{\prod\limits_{j=0}^{p-1} \mu_j-1} \\ s_{p-1}=\frac{\sum\limits_{j=1}^{p-2} \mu_{p-2} \mu_{p-3} \cdots \mu_j\left(\mu_{p-1}-\mu_{j-1}\right)+}{\prod\limits_{j=0}^{p-1} \mu_j-1} \rightarrow \\ \longleftarrow \frac{\left(\mu_{p-1}-\mu_{p-2}\right)}{1} $

Obviously, s₀ < 0 and s_p-1>0 because of μ₀=μ_min and μ_p-1=μ_max.

(S₂) is proved next. With Eq. (15) again, s_i is rewritten as follows:

$ \begin{aligned} s_i=& \frac{1}{\prod\limits_{j=0}^{p-1} \mu_j-1} \cdot\left[\mu_{i-1} \mu_{i-2} \cdots \mu_0 \mu_{p-1} \cdots \mu_{i+2}\left(\mu_i-\mu_{i+1}\right)+\right.\\ & \cdots+\mu_{i-1} \mu_{i-2} \cdots \mu_0 \mu_{p-1}\left(\mu_i-\mu_{p-2}\right)+\mu_{i-1} \mu_{i-2} \cdots \\ & \mu_0\left(\mu_i-\mu_{p-1}\right)+\mu_{i-1} \mu_{i-2} \cdots \mu_1\left(\mu_i-\mu_0\right)+\mu_{i-1} \mu_{i-2} \cdots \\ &\left.\mu_2\left(\mu_i-\mu_1\right)+\cdots+\mu_{i-1}\left(\mu_i-\mu_{i-2}\right)+\left(\mu_i-\mu_{i-1}\right)\right] \end{aligned} $

(18)

Suppose s_i≤0 and s_i-1≥0 for some i∈ 2, 3, …, p-1 (p>3). Then the following two inequalities are obtained:

$ \left\{\begin{array}{l} \mu_{p-1} \leqslant \frac{A}{\mu_{i-2} \mu_{i-3} \cdots \mu_0 B} \\ \mu_{p-1} \geqslant \frac{C}{\mu_{i-2} \mu_{i-3} \cdots \mu_0 D} \end{array}\right. $

(19)

where

$ \begin{aligned} A=& \mu_{i-2} \mu_{i-3} \cdots \mu_0 \mu_{i-1}+\sum\limits_{j=1}^{i-2} \mu_{i-2} \mu_{i-3} \cdots \\ & \mu_j\left(\mu_{i-1}-\mu_{j-1}\right)+\left(\mu_{i-1}-\mu_{i-2}\right) \end{aligned} $

(20a)

$ \begin{aligned} B=& \sum\limits_{j=i+1}^{p-2} \mu_{p-2} \mu_{p-3} \cdots \mu_j\left(\mu_{j-1}-\mu_{i-1}\right)+\\ &\left(\mu_{p-2}-\mu_{i-1}\right)+1 \end{aligned} $

(20b)

$ \begin{aligned} C=& \mu_{i-2} \mu_{i-3} \cdots \mu_0 \mu_i+\sum\limits_{j=1}^{i-2} \mu_{i-2} \mu_{i-3} \cdots \\ & \mu_j\left(\mu_i-\mu_{j-1}\right)+\left(\mu_i-\mu_{i-2}\right)+\frac{\mu_i-\mu_{i-1}}{\mu_{i-1}} \end{aligned} $

(20c)

$ D=\sum\limits_{j=i+2}^{p-2} \mu_{p-2} \mu_{p-3} \cdots \mu_j\left(\mu_{j-1}-\mu_i\right)+\left(\mu_{p-2}-\mu_i\right)+1 $

(20d)

The following inequality is obtained by Eq.(19):

$ \frac{C}{D} \leqslant \frac{A}{B} $

(21)

On the other hand, there is C>A>0 and B>D> 0 by assumptions on μ_i. This contradiction means that s_i>0 whenever s_i-1≥0. So (S₂) of lemma holds.

Now the main result of this paper is given.

Theorem 3  Let μ_n>1, K_n>0 and $\left\{\tilde{x}_n\right\}_{n=0}^{p-1}$=$\left\{\tilde{x}_0, \tilde{x}_1, \cdots, \tilde{x}_{p-1}\right\}$ be a p-periodic solution of Eq. (4). This periodic solution is attenuant if the following two conditions are satisfied: μ₀≤μ₁≤…≤μ_p-1 and K₀≥K₁≥…≥K_p-1.

Proof  By Lemma 2, it is known that s₀ < 0, s_p-1>0 and s_i≥0 for some integer N(i=N, N+1, …, p-1). That is

$ \begin{gathered} s_0<0, s_1<0, \cdots, s_N \geqslant 0 \\ s_{N+1}>0, \cdots, s_{p-1}>0 \end{gathered} $

By Eq. (17), there is

$ \begin{aligned} \overrightarrow{\boldsymbol{s}} \cdot \overrightarrow{\boldsymbol{K}}=& s_0 K_0+s_1 K_1+\cdots+s_{p-2} K_{p-2}+\left(-s_0-s_1-\cdots\right.\\ &\left.-s_{p-2}\right) K_{p-1}=s_0\left(K_0-K_{p-1}\right)+\\ & s_1\left(K_1-K_{p-1}\right)+\cdots+s_{p-2}\left(K_{p-2}-K_{p-1}\right)<\\ & s_0\left(K_N-K_{p-1}\right)+s_1\left(K_N-K_{p-1}\right)+\cdots+\\ & s_{p-2}\left(K_N-K_{p-1}\right)=-s_{p-1}\left(K_N-K_{p-1}\right) \leqslant 0 \end{aligned} $

The conclusion is proved by using Theorem 2.

3 Conclusions

In this paper, we investigated the properties of periodic solutions of non-autonomous Beverton-Holt equation with μ_n and K_n both varying periodically. $a v\left(\widetilde{x}_n\right)$ is related to av(K_n), K_min and K_max in Section 1. Example 1 indicates that Conj. 2 may be wrong for general system (4). A sufficient condition is found in Section 2 which proves that the second conjecture is true. Our proof is simpler than that given in Ref. [11].