Let u(t)=p(\tau t). Then (1.1) can be rewritten as

\dot{u}(t)=\tau ku(t)[f(u(t-1))-c].

(2.1)

Assume Eq. (2.1) has a positive equilibrium point {u}^{\ast}. Then {u}^{\ast} satisfies

f\left({u}^{\ast}\right)=c.

If we employ the nonstandard finite-difference scheme (1.2) to Eq. (2.1) and choose the function Φ as

\mathrm{\Phi}(h)=\frac{1-{e}^{-\tau kh}}{\tau k},

it yields the delay difference equation

{u}_{n+1}={u}_{n}+(1-{e}^{-\tau kh}){u}_{n}[f({u}_{n-m})-c].

(2.2)

Note that {u}^{\ast} also is the unique positive equilibrium of (2.2). Set {y}_{n}={u}_{n}-{u}^{\ast}, then it follows that

{y}_{n+1}={y}_{n}+(1-{e}^{-\tau kh})({y}_{n}+{u}^{\ast})[f({y}_{n-m}+{u}^{\ast})-c].

(2.3)

By introducing a new variable {Y}_{n}={({y}_{n},{y}_{n-1},\dots ,{y}_{n-m})}^{T}, we can rewrite (2.3) in the form

{Y}_{n+1}=F({Y}_{n},\tau ),

(2.4)

where F={({F}_{0},{F}_{1},\dots ,{F}_{m})}^{T}, and

{F}_{j}=\{\begin{array}{ll}{y}_{n}+(1-{e}^{-\tau kh})({y}_{n}+{u}^{\ast})[f({y}_{n-m}+{u}^{\ast})-c],& j=0\\ {y}_{n-j+1},& 1\le j\le m.\end{array}

(2.5)

Clearly the origin is an equilibrium of (2.4), and the linear part of (2.4) is

where

A=\left(\begin{array}{ccccc}1& 0& \cdots & 0& (1-{e}^{-\tau kh}){u}^{\ast}{f}^{\prime}({u}^{\ast})\\ 1& 0& \cdots & 0& 0\\ 0& 1& \cdots & 0& 0\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0& 0& \vdots & 1& 0\end{array}\right).

The characteristic equation of *A* is given by

a(\lambda ):={\lambda}^{m+1}-{\lambda}^{m}-(1-{e}^{-\tau kh}){u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right)=0.

(2.6)

It is well known that the stability of the zero equilibrium solution of (2.4) depends on the distribution of zeros of the roots of (2.6). In this paper, we employ the results from Zhang *et al.* [9] and He *et al.* [10] to analyze the distribution of zeros of characteristic Eq. (2.6). In order to prove the existence of the local Neimark-Sacker bifurcation at equilibrium, we need some lemmas as follows.

**Lemma 2.1** *There exists a* \overline{\tau}>0 *such that for* 0<\tau <\overline{\tau} *all roots of* (2.6) *have modulus less than one*.

*Proof* When \tau =0, (2.6) becomes

{\lambda}^{m+1}-{\lambda}^{m}=0.

The equation has, at \tau =0, an m-fold root \lambda =0 and a simple root \lambda =1.

Consider the root \lambda (\tau ) such that \lambda (0)=1. This root depends continuously on *τ* and is a differential function of *τ*. From (2.6), we have

\frac{d\lambda}{d\tau}=\frac{kh{e}^{-\tau kh}{u}^{\ast}{f}^{\prime}({u}^{\ast})}{(m+1){\lambda}^{m}-m{\lambda}^{m-1}}

(2.7)

and

\frac{d\overline{\lambda}}{d\tau}=\frac{kh{e}^{-\tau kh}{u}^{\ast}{f}^{\prime}({u}^{\ast})}{(m+1){\overline{\lambda}}^{m}-m{\overline{\lambda}}^{m-1}}.

(2.8)

Noticing f(\cdot ) is a non-negative continuous, strictly decreasing function, we have

\frac{d{|\lambda |}^{2}}{d\tau}{|}_{\tau =0,\lambda =1}=[\lambda \frac{d\overline{\lambda}}{d\tau}+\overline{\lambda}\frac{d\lambda}{d\tau}]{|}_{\tau =0,\lambda =1}=2kh{u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right)<0.

So, with the increase of \tau >0, *λ* cannot cross \lambda =1. Consequently, all roots of Eq. (2.6) lie in the unit circle for sufficiently small positive \tau >0, and the existence of the \overline{\tau} follows. □

A Neimark-Sacker bifurcation occurs when a complex conjugate pair of eigenvalues of *A* cross the unit circle as *τ* varies. We have to find values of *τ* such that there are roots on the unit circle. Denote the roots on the unit circle by {e}^{i{\omega}^{\ast}}, {\omega}^{\ast}\in (-\pi ,\pi ). Since we are dealing with complex roots of a real polynomial, we only need to look for {\omega}^{\ast}\in (0,\pi ).

**Lemma 2.2** *There exists an increasing sequence of values of the time delay parameter* \tau ={\tau}_{j}, j=0,1,2,\dots ,[\frac{m-1}{2}] *satisfying*

\{\begin{array}{l}cos{\omega}_{j}=1-\frac{1}{2}{[(1-{e}^{-{\tau}_{j}kh}){u}^{\ast}{f}^{\prime}({u}^{\ast})]}^{2},\\ {\tau}_{j}=-\frac{1}{kh}ln(1+\frac{sin{\omega}_{j}}{{u}^{\ast}{f}^{\prime}({u}^{\ast})sinm{\omega}_{j}}),\\ h=\frac{1}{m},\end{array}

(2.9)

*where* {\omega}_{j}\in (\frac{2j\pi}{m},\frac{(2j+1)\pi}{m}), j=0,1,2,\dots ,[\frac{m-1}{2}].

*Proof* Denote the roots of Eq. (2.6) on the unit circle by {e}^{i{\omega}^{\ast}}, {\omega}^{\ast}\in (0,\pi ). Then

{e}^{i{\omega}^{\ast}}-1-(1-{e}^{-{\tau}_{j}kh}){u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right){e}^{-im{\omega}^{\ast}}=0.

(2.10)

Separating the real part and the imaginary part from Eq. (2.10), there are

cos{\omega}^{\ast}-(1-{e}^{-{\tau}^{\ast}kh}){u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right)cosm{\omega}^{\ast}=1

(2.11)

and

sin{\omega}^{\ast}+(1-{e}^{-{\tau}^{\ast}kh}){u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right)sinm{\omega}^{\ast}=0.

(2.12)

So,

cos{\omega}^{\ast}=1-\frac{1}{2}{\left[(1-{e}^{-{\tau}^{\ast}kh}){u}^{\ast}{f}^{\prime}\left({u}^{\ast}\right)\right]}^{2}.

(2.13)

Then the roots {e}^{i{\omega}^{\ast}} of (2.6) satisfy Eqs. (2.10)-(2.13). From (2.12) we get

{\tau}^{\ast}=-\frac{1}{kh}ln(1+\frac{sin{\omega}^{\ast}}{{u}^{\ast}{f}^{\prime}({u}^{\ast})sinm{\omega}^{\ast}}).

(2.14)

Substituting (2.14) into (2.11), we have

sin(m+1){\omega}^{\ast}=sinm{\omega}^{\ast}.

(2.15)

Then Eq. (2.15) has a unique solution {\omega}^{\ast} in every interval (\frac{2j\pi}{m},\frac{(2j+1)\pi}{m}), j=0,1,2,\dots ,[\frac{m-1}{2}], we set

{\omega}_{j}:={\omega}^{\ast}\in (\frac{2j\pi}{m},\frac{(2j+1)\pi}{m}),\phantom{\rule{1em}{0ex}}j=0,1,2,\dots ,\left[\frac{m-1}{2}\right].

(2.16)

From (2.14), we have

{\tau}_{j}=-\frac{1}{kh}ln(1+\frac{sin{\omega}_{j}}{{u}^{\ast}{f}^{\prime}({u}^{\ast})sinm{\omega}_{j}}).

(2.17)

This completes the proof. □

**Lemma 2.3** *Let* {\lambda}_{j}(\tau )={r}_{j}(\tau ){e}^{i{\omega}_{j}(\tau )} *be a root of* (2.6) *near* \tau ={\tau}_{j} *satisfying* {r}_{j}({\tau}_{j})=1 *and* {\omega}_{j}({\tau}_{j})={\omega}_{j}. *Then*

\frac{d{r}_{j}^{2}(\tau )}{d\tau}{|}_{\tau ={\tau}_{j},\omega ={\omega}_{j}}>0.

*Proof* From (2.11) and (2.12), we obtain that

cosm{\omega}_{j}=\frac{cos{\omega}_{j}-1}{(1-{e}^{-{\tau}_{j}kh}){u}^{\ast}{f}^{\prime}({u}^{\ast})},

(2.18)

sinm{\omega}_{j}=\frac{-sin{\omega}^{\ast}}{(1-{e}^{-{\tau}_{j}kh}){u}^{\ast}{f}^{\prime}({u}^{\ast})}.

(2.19)

It is easy to see that

cos(m+1){\omega}_{j}=cosm{\omega}_{j}cos{\omega}_{j}-sinm{\omega}_{j}sin{\omega}_{j}=\frac{1-cos{\omega}_{j}}{(1-{e}^{-{\tau}_{j}kh}){u}^{\ast}{f}^{\prime}({u}^{\ast})}.

(2.20)

From (2.7), (2.8) and using (2.18)-(2.20), we have

\begin{array}{rl}\frac{d{r}_{j}^{2}}{d\tau}{|}_{\tau ={\tau}_{j},\omega ={\omega}_{j}}& =\frac{d{|{\lambda}_{j}|}^{2}}{d\tau}{|}_{\tau ={\tau}_{j},\omega ={\omega}_{j}}=[\lambda \frac{d\overline{\lambda}}{d\tau}+\overline{\lambda}\frac{d\lambda}{d\tau}]{|}_{\tau ={\tau}_{j},\omega ={\omega}_{j}}\\ =\frac{2kh{e}^{-{\tau}_{j}kh}(2m+1)(1-cos{\omega}_{j})}{(1-{e}^{-{\tau}_{j}kh}){|(m+1){e}^{im{\omega}_{j}}-m{e}^{i(m-1){\omega}_{j}}|}^{2}}>0.\end{array}

This completes the proof. □

Lemmas 2.1-2.3 immediately lead to the stability of the zero equilibrium of Eq. (2.3), and equivalently, of the positive equilibrium {u}^{\ast} of Eq. (2.2). So, we have the following results on stability and bifurcation in system (2.2).

**Theorem 2.1** *There exists a sequence of values of the time*-*delay parameter* 0<{\tau}_{0}<{\tau}_{1}<\cdots <{\tau}_{[\frac{m-1}{2}]} *such that the positive equilibrium* {u}^{\ast} *of Eq*. (2.2) *is asymptotically stable for* \tau \in [0,{\tau}_{0}) *and unstable for* \tau >{\tau}_{0}. *Equation* (2.2) *undergoes a Neimark*-*Sacker bifurcation at the positive equilibrium* {u}^{\ast} *when* \tau ={\tau}_{j}, j=0,1,2,\dots ,[\frac{m-1}{2}], *where* {\tau}_{j} *satisfies* (2.13).