In this section, we mainly focus on the flip bifurcation and Neimark-Sacker bifurcation of the positive fixed point \((x^{*}, y^{*})\). We choose the parameter *d* as a bifurcation parameter for analyzing the flip bifurcation and Neimark-Sacker bifurcation of \((x^{*}, y^{*})\) by using the center manifold theorem and bifurcation theory of [23, 24].

First, we have the following result on the flip bifurcation of system (3).

### Theorem 3.1

*System* (3) *undergoes a flip bifurcation at*
\((x^{*},y^{*})\)
*if the following conditions are satisfied*: \(c>0\), \(a>3\), \(a\neq\frac{9+5c}{1+c}\), *and*
\(d=\frac{a(1+c)(3+c)}{3+a-c+ac}\). *Moreover*, *if*
\(3< a<\frac{9+5c}{1+c}\), *then period*-2 *points that bifurcate from this fixed point are unstable*.

### Proof

If \(d^{*}=\frac{a(1+c)(3+c)}{3+a-c+ac}\), then the eigenvalues of the fixed point \((x^{*},y^{*})\) are \(\lambda_{1} = -1\) and \(\lambda_{2} = \frac{6-a+4c-ac}{3+c}\). The condition \(|\lambda_{2}|\neq1\) leads to \(a\neq3,\frac{9+5c}{1+c}\). In addition, note that the existence of the positive fixed point is assured by the relation \(d>\frac{a(1+c)}{a-1}\) (\(a>1\)), so we get \(a>3\). Hence, we further assume that \(a>3\) and \(a\neq\frac{9+5c}{1+c}\).

Let \(u=x-x^{*}\), \(v=y-y^{*}\), and \(\bar{d}=d-d^{*}\). We consider the parameter *d̄* as a new and dependent variable. Then the map (3) becomes

$$ \left ( \textstyle\begin{array}{@{}c@{}} u\\ \bar{d}\\ v \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} \frac{2c-a(1+c)}{3+c}&0&\frac{c-a(1+c)-3}{a(3+c)}\\ 0&1&0\\ \frac{2a(a-3)(1+c)}{3+a-c+ac}&\frac{2(a-3)(3-c+a+ac)}{a(3+c)^{2}}&1 \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} u\\ \bar{d}\\ v \end{array}\displaystyle \right )+ \left ( \textstyle\begin{array}{@{}c@{}} f_{1}(u,\bar{d},v)\\ 0\\ f_{2}(u,\bar{d},v) \end{array}\displaystyle \right ), $$

(10)

where

$$\begin{aligned}& f_{1}(u,\bar{d},v)=-au^{2}-uv , \\& f_{2}(u,\bar{d},v)=\frac{a(1+c)(3+c)}{3+a-c+ac}uv+\frac{2(a-3)}{3+c}u\bar {d}+ \frac{3+a-c+ac}{a(3+c)}v\bar{d}+uv\bar{d} . \end{aligned}$$

Let

$$T=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} -\frac{3+a-c+ac}{a(a-3)(1+c)}&-\frac{1}{a}&-\frac{3+a-c+ac}{2a(3+c)}\\ 0&\frac{a(1+c)(3+c)^{2}}{(3+a-c+ac)^{2}}&0\\ 1&1&1 \end{array}\displaystyle \right ), $$

and use the translation \(\Bigl ( {\scriptsize\begin{matrix}{} u\cr \bar{d}\cr v \end{matrix}} \Bigr )=T \Bigl ( {\scriptsize\begin{matrix}{} X\cr \mu\cr Y \end{matrix}} \Bigr )\). Then the map (10) becomes

$$ \left ( \textstyle\begin{array}{@{}c@{}} X\\ \mu\\ Y \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} -1&0&0\\ 0&1&0\\ 0&0&\frac{6+4c-a(1+c)}{3+c} \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} X\\ \mu\\ Y \end{array}\displaystyle \right )+ \left ( \textstyle\begin{array}{@{}c@{}} F_{1}(X,\mu,Y)\\ 0\\ F_{2}(X,\mu,Y) \end{array}\displaystyle \right ), $$

(11)

where

$$\begin{aligned}& F_{1}(X,\mu,Y)= - \frac {(3+c)(9-2c-3c^{2}+a(1+c)^{2})X^{2}}{(a-3)(1+c)(a-9-5c+ac)}-\frac {(3+c)(3+a-c+ac)XY}{2(a-9-5c+ac)} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(a-3)(a+c)(1+c)^{2}Y^{2}}{2(3+c)(a-9-5c+ac)}-\frac {2(a-3)^{2}(1+c)^{2}(3+c)\mu^{2}}{(a-9-5c+ac)(3+a-c+ac)^{2}} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(3+c)^{2}X\mu}{a-9-5c+ac}-\frac{(a-3)(1+c)^{2}(9+c-a(5+c))Y\mu }{2(27+6c-5c^{2}+6a(1+c)^{2}-a^{2}(1+c)^{2})} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(a-3)(1+c)^{2}(3+c)Y^{2}\mu}{2(a-9-5c+ac)(3+a-c+ac)}-\frac {(1+c)(3+c)XY\mu}{2(a-9-5c+ac)} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(1+c)(3+c)^{2}X^{2}\mu}{(a-9-5c+ac)(3+a-c+ac)}-\frac {2(1+c)(3+c)^{2}(a-2c+ac)X\mu^{2}}{(a-9-5c+ac)(3+a-c+ac)^{2}} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(a-3)(1+c)^{2}(3+c)(9+a+c+ac)Y\mu^{2}}{2(a-9-5c+ac)(3+a-c+ac)^{2}} \\& \hphantom{F_{1}(X,\mu,Y)={}}{}- \frac{(a-3)(1+c)^{2}(3+c)^{2}\mu^{3}}{(a-9-5c+ac)(3+a-c+ac)^{2}}, \\& F_{2}(X,\mu,Y) = \frac{2(3+c)^{3}X^{2}}{(a-3)(1+c)(a-9-5c+ac)}+\frac {(c(3-c)+a^{2}(1+c)+a(3+c^{2}))X Y}{(a-3)(a-9-5c+ac)} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{(1+c)(27+21c+2c^{2}-6a(1+c)+a^{2}(1+c))Y^{2}}{2(3+c)(a-9-5c+ac)} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{2(3+c)^{2}(a(5+c)-9-c)X\mu}{(a-3)(-27-6c+5c^{2}-6a(1+c)^{2}+a^{2}(1+c)^{2})} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{4(a-3)(1+c)(3+c)^{2}\mu^{2}}{(a-9-5c+ac)(3+a-c+ac)^{2}}+\frac {(1+c)(a+c)Y\mu}{a-9-5c+ac} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{(1+c)(3+c)^{2}Y^{2}\mu}{(a-9-5c+ac)(3+a-c+ac)}+\frac{(3+c)^{2}XY\mu }{(a-3)(a-9-5c+ac)} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{2(3+c)^{3}X^{2}\mu}{(a-3)(a-9-5c+ac)(3+a-c+ac)} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{4(3+c)^{3}(a-2c+ac)X\mu^{2}}{(a-3)(a-9-5c+ac)(3+a-c+ac)^{2}} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{(1+c)(3+c)^{2}(9+a+c+ac)Y\mu^{2}}{(a-9-5c+ac)(3+a-c+ac)^{2}} \\& \hphantom{F_{2}(X,\mu,Y) ={}}{}+ \frac{2(1+c)(3+c)^{3}\mu^{3}}{(a-9-5c+ac)(3+a-c+ac)^{2}}. \end{aligned}$$

By the center manifold theorem we know that the stability of \((X,Y)=(0,0)\) near \(\mu=0\) can be determined by studying a one-parameter family of maps on a center manifold, which can be represented as follows:

$$W^{c}(0)= \bigl\{ (X,\mu,Y)\in R^{3}| Y=h^{*}(X,\mu), h^{*}(0,0)=0, Dh^{*}(0,0)=0 \bigr\} . $$

Assume that

$$h^{*}(X,\mu)=\alpha X^{2}+\beta X \mu+\gamma\mu^{2}+O \bigl( \bigl(\vert X\vert +|\mu| \bigr)^{3} \bigr). $$

By approximate computation for the center manifold, we obtain

$$\begin{aligned}& \alpha=\frac{2(3+c)^{4}}{(a-3)^{2}(1+c)^{2}(a-9-5c+ac)} , \\& \beta=\frac{2(3+c)^{3}(5a-9-c+ac)}{(a-3)(3+a-c+ac)(a-9-5c+ac)^{2}} , \\& \gamma=\frac{4(3+c)^{3}}{(a-9-5c+ac)(3+a-c+ac)^{2}} . \end{aligned}$$

Thus, the map restricted to the center manifold is given by

$$\begin{aligned} \tilde{F}:X \rightarrow&-X+h_{1}X^{2}+h_{2}X \mu+h_{3}\mu^{2} +h_{4}X^{3}+h_{5}X^{2} \mu +h_{6}X\mu^{2}+h_{7}\mu^{3} \\ &{}+O \bigl( \bigl(\vert X\vert +|\mu| \bigr)^{4} \bigr), \end{aligned}$$

where

$$\begin{aligned}& h_{1}=-\frac{(3+c)(9-2c-3c^{2}+a(1+c)^{2})}{(a-3)(1+c)(a-9-5c+ac)} , \\& h_{2}=-\frac{(3+c)^{2}}{a-9-5c+ac} , \\& h_{3}=-\frac{2(a-3)^{2}(1+c)^{2}(3+c)}{(a-9-5c+ac)(3+a-c+ac)^{2}} , \\& h_{4}=-\frac{(3+c)^{5}(3+a-c+ac)}{(a-3)^{2}(1+c)^{2}(a-9-5c+ac)^{2}} , \\& h_{5}=-\frac {(1+c)(3+c)^{2}(-81-36c-5c^{2}-2c^{3}+a^{2}(1+c)^{2}+2a(36+25c+6c^{2}+c^{3}))}{(3+a-c+ac)(a-9-5c+ac)^{3}} , \\& h_{6}=(3+c)^{2}\bigl(243+135c+186c^{2}+234c^{3}+67c^{4}-c^{5}-2a^{3}(1+c)^{4} \\& \hphantom{h_{6}={}}{}+a(1+c)^{2}\bigl(216-57a-6c-7ac-44c^{2}+9ac^{2}+2c^{3}-ac^{3} \bigr)\bigr) \\& \hphantom{h_{6}={}}{}/(3+a-c+ac)^{2}(a-9-5c+ac)^{3} , \\& h_{7}=-\frac {(a-3)^{2}(3+4c+c^{2})^{2}(27+10c-c^{2}+a(1+c)^{2})}{(a-9-5c+ac)^{2}(3+a-c+ac)^{3}} . \end{aligned}$$

If the map (11) undergoes a flip bifurcation, then it must satisfy the following conditions:

$$\alpha_{1}= \biggl[\frac{\partial F}{\partial\mu}\cdot\frac{\partial^{2} F}{\partial X^{2}}+2 \frac{\partial^{2} F}{\partial X\, \partial\mu} \biggr]\bigg|_{(0,0)}\neq0 $$

and

$$\alpha_{2}= \biggl[\frac{1}{2}\cdot \biggl(\frac{\partial^{2} F}{\partial X^{2}} \biggr)^{2}+\frac {1}{3}\cdot\frac{\partial^{3} F}{\partial X^{3}} \biggr]\bigg|_{(0,0)}\neq0. $$

By a simple calculation we obtain

$$\alpha_{1}=-\frac{2(3+c)^{2}}{a-9-5c+ac}\neq0 \quad \text{for } c>0, a>3 \text{ and } a\neq\frac{9+5c}{1+c} $$

and

$$\alpha_{2}=\frac {2(3+c)^{2}(a(1+c)^{3}-2c(c^{2}-5))}{(a-3)^{2}(1+c)^{2}(a-9-5c+ac)}\neq0\quad \text{for } c>0, a>3 \text{ and } a\neq\frac{9+5c}{1+c}. $$

It is easy to check that if \(3< a<\frac{9+5c}{1+c}\), then \(|\lambda _{2}|<1\) and \(\alpha_{2}<0\). Thus, period-2 points that bifurcate from this fixed point are unstable.

This completes the proof of Theorem 3.1. □

For Neimark-Sacker bifurcation, we have the following theorem.

### Theorem 3.2

*System* (3) *undergoes a Neimark*-*Sacker bifurcation at the fixed point*
\((x^{*}, y^{*})\)
*if the following conditions are satisfied*: \(c>0\), \(1< a<9\), \(a\neq\frac{5+3c}{1+c}, \frac {7+4c}{1+c}\), *and*
\(d=\bar{d}^{*}=\frac{a(2+c)}{a-1}\). *Moreover*, \(k < 0\), *and thus an attracting invariant closed curve bifurcates from the fixed point for*
\(d>\bar{d}^{*}\).

### Proof

The characteristic equation associated with the linearized system (3) at the fixed point \((x^{*}(d),y^{*}(d))\) is given by

$$ \lambda^{2}+p(d)\lambda+q(d)=0. $$

(12)

The eigenvalues of the characteristic equation (12) are given as

$$\lambda_{1,2}(d)= \frac{-p(d)\pm\sqrt{p(d)^{2}-4q(d)}}{2}, $$

where \(p(d)= c-a+(2a-d)x^{*}+y^{*}\) and \(q(d)= -ac+a(2c+d)x^{*}-2ad{x^{*}}^{2}+cy^{*}\).

The eigenvalues \(\lambda_{1,2}\) are complex conjugates for \(p(d)^{2}-4q(d)< 0\), which leads to

$$ d>\frac{a(1+c)+\sqrt{a^{2}(c+c^{2})+a^{3}(1+c)}}{2(a-1)}. $$

(13)

Let

$$ \bar{d}^{*}=\frac{a(2+c)}{a-1}\quad \text{for } 1< a< 9. $$

(14)

We get \(q(d)=1\) and \(\lambda_{1,2}=\frac{5-a+3c-ac}{2(2+c)}\pm\frac {i\sqrt{(1-a)(1+c)(a-9-5c+ac)}}{2(2+c)}=\rho\pm i\omega \). Under condition (14), we have

$$\bigl\vert \lambda_{1,2}(d) \bigr\vert = \bigl(q(d) \bigr)^{\frac{1}{2}}\quad \text{and}\quad d_{1}=\frac {d|\lambda_{1,2}(d)|}{dd}\bigg|_{d=\bar{d}^{*}}= \frac {(a-1)^{2}(1+c)}{2a(2+c)}\neq0. $$

In addition, if \(p(\bar{d}^{*})\neq0, 1\), which leads to

$$a\neq\frac{5+3c}{1+c}\quad \text{and}\quad a\neq\frac{7+4c}{1+c}, $$

then we obtain that \(\lambda_{1,2}^{n}(\bar{d}^{*})\neq1\) (\(n = 1, 2, 3, 4\)).

Letting \(u=x-x^{*}\) and \(v=y-y^{*}\), the map (3) becomes

$$ \left ( \textstyle\begin{array}{@{}c@{}} u\\ v \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} \frac{3+2c-a(1+c)}{2+c}&-\frac{(a-1)(1+c)}{a(2+c)}\\ a&1 \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} u\\ v \end{array}\displaystyle \right )+ \left ( \textstyle\begin{array}{@{}c@{}} f_{1}(u,v)\\ f_{2}(u,u) \end{array}\displaystyle \right ), $$

(15)

where \(f_{1}(u,v)=-au^{2}-uv\) and \(f_{2}(u,v)=\frac{a(2+c)}{a-1}uv\).

Let

$$T=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} -\frac{\sqrt{(1-a)(1+c)(a-9-5c+ac)}}{2a(2+c)}&\frac{1+c-a-ac}{2a(2+c)}\\ 0&1 \end{array}\displaystyle \right ) $$

and use the translation \(\bigl ( {\scriptsize\begin{matrix}{} u\cr v \end{matrix}} \bigr )=T \bigl ( {\scriptsize\begin{matrix}{} X\cr Y \end{matrix}} \bigr )\). Then the map (15) becomes

$$ \left ( \textstyle\begin{array}{@{}c@{}} X\\ Y \end{array}\displaystyle \right )\mapsto \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} \rho&-\omega\\ \omega&\rho \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{@{}c@{}} X\\ Y \end{array}\displaystyle \right )+ \left ( \textstyle\begin{array}{@{}c@{}} F_{1}(X,Y)\\ F_{2}(X,Y) \end{array}\displaystyle \right ), $$

(16)

where

$$\begin{aligned}& F_{1}(X,Y)=\frac{\sqrt{(1-a)(1+c)(a-9-5c+ac)}}{2(2+c)}X^{2}-\frac {4+c-c^{2}-2a(1+c)}{2(2+c)}XY \\& \hphantom{F_{1}(X,Y)={}}{}- \frac{(a-1)(1+c)(a-3+ac+c^{2})}{2(2+c)\sqrt {(1-a)(1+c)(a-9-5c+ac)}}Y^{2}, \\& F_{2}(X,Y)=-\frac{\sqrt{(1-a)(1+c)(a-9-5c+ac)}}{2(a-1)}XY-\frac{1+c}{2}Y^{2}. \end{aligned}$$

Notice that (16) is exactly in the form on the center manifold in which the coefficient *k* [23] is given by

$$k=-\operatorname{Re} \biggl[\frac{(1-2\lambda)\bar{\lambda}^{2}}{1-\lambda}\xi_{11}\xi _{20} \biggr]-\frac{1}{2}|\xi_{11}|^{2}-| \xi_{02}|^{2}+\operatorname{Re}(\bar{\lambda}\xi _{21}), $$

where

$$\begin{aligned}& \xi_{20}=\frac{1}{8} \bigl[(F_{1XX}-F_{1YY}+2F_{2XY})+i(F_{2XX}-F_{2YY}-2F_{1XY}) \bigr] , \\& \xi_{11}=\frac{1}{4} \bigl[(F_{1XX}+F_{1YY})+i(F_{2XX}+F_{2YY}) \bigr] , \\& \xi_{02}=\frac{1}{8} \bigl[(F_{1XX}-F_{1YY}-2F_{2XY})+i(F_{2XX}-F_{2YY}+2F_{1XY}) \bigr] , \\& \xi_{21}=\frac {1}{16} \bigl[(F_{1XXX}+F_{1XYY}+F_{2XXY}+F_{2YYY})+i(F_{2XXX}+F_{2XYY}-F_{1XXY}-F_{1YYY}) \bigr] . \end{aligned}$$

Thus, a complex calculation gives

$$k=-\frac{A}{8(a-1)(2+c)^{3}}< 0 \quad \text{for } 1< a< 9, $$

where \(A=3+14c+14c^{2}+4c^{3}+a^{4}(1+c)^{3}+a^{3}(1+c)^{2}(c^{2}-3c-6)-3a^{2}(1+c)^{2}(c^{2}-c-4)+a(6+19c+35c^{2}+31c^{3}+11c^{4}+c^{5})\).

Thus the fixed point \((X,Y)=(0,0)\) is a Neimark-Sacker bifurcation point for the map (16). This completes the proof. □