In this section, we will investigate the existence of the Neimark–Sacker bifurcation for the model (5) and present the conditions and direction of the Neimark–Sacker bifurcation [18–25].

A Neimark–Sacker bifurcation occurs at a bifurcation point if and only if system (5) satisfies the following conditions: eigenvalue assignment, transversality and nonresonance condition [26, 27]. The following lemma gives the eigenvalue assignment condition for the Neimark–Sacker bifurcation.

### Lemma 3.1

([28])

*A pair of complex conjugate roots of*
\(p(\lambda )=\lambda^{2}+p_{1}\lambda+p_{0}\)
*lie on the unit circle if and only if*

$$\begin{aligned} &(\mathrm{a})\quad p(1)=1+p_{1}+p_{0}>0, \\ &(\mathrm{b})\quad p(-1)=1-p_{1}+p_{0}>0, \\ &(\mathrm{c})\quad D_{1}^{-}=1-p_{0}=0. \end{aligned}$$

### Proposition 3.2

(Eigenvalue assignment)

*Suppose that*
\(-v+abH^{\ast}+H^{\ast}mv>0\), \(2v-abH^{\ast}+H^{\ast}kv-H^{\ast }mv>0\)
*and*
\(-1+bH^{\ast}k>0\). *If*
\(\max \{R_{1},R_{2}\}< R=R_{3\text{ }}\)*then the eigenvalue assignment condition of the Neimark–Sacker bifurcation in Lemma *3.1
*holds*.

### Proof

From the condition \(R>R_{1}\) we have \(p(1)>0\). On the other hand, the conditions \(-v+abH^{\ast}+H^{\ast}mv>0\), \(2v-abH^{\ast}+H^{\ast }kv-H^{\ast}mv>0\) and \(R>R_{2}\) lead to \(p(-1)>0\). Solving the equation \(D_{1}^{-}=1-p_{0}=0\) with the fact \(-1+bH^{\ast}k>0\), we get \(R=R_{3}\). This completes the proof. □

Now we can analyze the transversality and nonresonance condition for the Neimark–Sacker bifurcation. It is easy to see that the Jacobian matrix of the system (5) has the eigenvalues

$$ \lambda_{1,2}(R)=\alpha\mp i\beta, $$

(23)

where

$$ \alpha=\frac{e^{H^{\ast}m}H^{\ast}v^{b}k+ae^{H^{\ast }m}H^{2}v^{b}k+Rv-abH^{\ast}R-H^{\ast}mRv}{2Rv} $$

and

$$ \beta=\frac{\sqrt{-4H^{\ast}vk(e^{H^{\ast}m}H^{\ast}v^{b}mv+aH^{\ast }be^{H^{\ast}m}v^{b}-Rv)R-(e^{H^{\ast}m}H^{\ast}v^{1+b}k-(-v+aH^{\ast }b+H^{\ast}mv)R)^{2}}}{2Rv}. $$

For

$$ R=\frac{be^{H^{\ast}m}H^{\ast2}v^{-1+b}k(ab+mv)}{-1+bH^{\ast}k}, $$

these eigenvalues become

$$ \lambda_{1,2}(R)=\alpha_{1}\mp i \beta_{1}, $$

where

$$\begin{aligned} &\alpha_{1}=\frac{1}{2}\biggl(1+b\biggl(-1+ \frac{1}{v}\biggr)-H^{\ast}m+\frac {v(-1+H^{\ast}k)}{H^{\ast}(mv+ab)}\biggr), \\ & \beta_{1}=\frac{\sqrt{\mu^{2}-\eta^{2}}}{H^{\ast}(mv+ab)}, \\ &\mu=2H^{\ast}v\bigl(m+a\bigl(b+H^{\ast}m\bigr) \bigr) \end{aligned}$$

and

$$\begin{aligned} \eta={}&1+H^{\ast}\bigl(-k+a(1+v)+b\bigl(-v+abH^{\ast} \bigr)-H^{\ast }k(1+v)\bigr)-m\\ &{}+aH^{\ast }(-1-v+2bv)m+H^{\ast}m^{2}v^{2}. \end{aligned}$$

It is easy to see that

$$ \bigl\vert \lambda_{1,2}(R) \bigr\vert = \vert \alpha_{1}\mp i\beta _{1} \vert =1. $$

On the other hand, the transversality condition leads to

$$\begin{aligned} \frac{d\vert\lambda_{i}(R)\vert}{dR}\Big\vert _{R=R_{3}} =\frac{e^{-H^{\ast}m}(1+aH^{\ast })^{1-b}(-1+H^{\ast }k)^{2}}{2H^{\ast2}k(m+a(b+H^{\ast}m))} \neq0,\quad i=1,2. \end{aligned}$$

From the nonresonance condition \(trJ(R_{3})=-p_{1}\neq0,-1\), we have

$$ R\neq\frac{e^{H^{\ast}m}H^{\ast}v^{1+b}k}{-v+abH^{\ast}+H^{\ast }mv},\qquad R\neq \frac{e^{H^{\ast}m}H^{\ast}v^{1+b}k}{-1-v+abH^{\ast }+H^{\ast}mv}, $$

which leads to

$$ \lambda_{i}^{k}(R_{3})\neq1 \quad\text{for }k=1,2,3,4. $$

### Theorem 3.3

*Suppose that*
\((H^{\ast},P^{\ast})\)
*is the positive equilibrium point of the system* (5). *If Proposition* (3.2) *holds*, \(R\neq\frac{e^{H^{\ast}m}H^{\ast} v^{1+b}k}{-v+abH^{\ast}+H^{\ast }mv}\), \(R\neq\frac{e^{H^{\ast}m}H^{\ast} v^{1+b}k}{-1-v+abH^{\ast}+H^{\ast}mv}\), *and*
\(a(0)<0\) (*respectively*
\(a(0)>0\)), *then the Neimark–Sacker bifurcation of the system* (5) *at*
\(R=R_{3}\)
*is supercritical* (*respectively*, *subcritical*) *and there exists a unique closed invariant curve bifurcation from*
\((H^{\ast},P^{\ast})\)
*for*
\(R=R_{3}\), *which is asymptotically stable* (*respectively*, *unstable*).

### Example 3.4

Let us take \(m=1,a=1.1,b=1.15\), \(k=2.2\) and initial condition \((H_{0},P_{0})=(0.5,0.6)\). From solutions of the system (7) and (16), the positive equilibrium point and bifurcation point of the system (5) are obtained: \((H^{\ast},P^{\ast})=(0.76664,0.525223)\) and \(R_{3}=13.81043\), respectively.

In this situation it is easy to check that

$$ \bigl\vert \lambda_{1,2}(r) \bigr\vert = \vert 0.119179\pm 0.992873i \vert =1 $$

and

$$ \frac{d \vert \lambda_{i}(R) \vert }{dR}\Big\vert _{R=R_{3}} =0.0248583\neq0\quad\text{and}\quad\lambda _{i}^{k}(R_{3})\neq1\quad\text{for }k=1,2,3,4. $$

To compute the coefficients of the normal form, we convert the origin of the coordinates to equilibrium point \((H^{\ast},P^{\ast})\) by the change of variables,

$$\begin{aligned} \textstyle\begin{cases} H=0.76664+x_{1}, \\ P=0.525223+x_{2}. \end{cases}\displaystyle \end{aligned}$$

(24)

This transforms the system (5) into

$$\begin{aligned} \textstyle\begin{cases} x_{1}(n+1)=\frac{ 13.81043e^{-0.76664-x_{1}-2.2(0.525223.+x_{2})}(0.76664+x_{1})}{ (1+1.1(0.76664+x_{1})^{1.15}}, \\ x_{2}(n+1)=(1-e^{-2.2(0.525223+x_{2})})(0.76664+x_{1}). \end{cases}\displaystyle \end{aligned}$$

(25)

This system can be written as

$$ X_{n+1}=JX_{n}+\frac{1}{2}B(X_{n},X_{n})+ \frac{1}{6}C(X_{n},X_{n},X_{n})+O \bigl(X_{n}^{4}\bigr). $$

(26)

Now, the Jacobian matrix of the discrete dynamical system (25) at the equilibrium point is

J({R}_{3})=\left[\begin{array}{cc}-0.2927602& -1.6866079\\ 0.6850969& 0.5311179\end{array}\right]

(27)

and the multilinear functions *B* and *C* are defined by

$$ B_{i}(x,y)=\sum_{j,k=1}^{2} \frac{\partial^{2}X_{i}(\varepsilon,0)}{\partial\varepsilon_{j}\partial\varepsilon_{k}}\Big|_{\varepsilon =0}x_{j}y_{k},\quad i=1,2 $$

and

$$ C_{i}(x,y,z)=\sum_{j,k,l=1}^{2} \frac{\partial^{3}X_{i}(\varepsilon ,0)}{\partial\varepsilon_{j}\partial\varepsilon_{k}\partial\varepsilon _{l}}\Big|_{\varepsilon=0}x_{j}y_{k}z_{l},\quad i=1,2. $$

Let \(q\in C^{2}\) be an eigenvector of \(J(R_{3})\) corresponding to the eigenvalue \(\lambda_{1}(R_{3})\) such that \(\lambda _{1}(R_{3})q=e^{i\theta _{0}}q\) and let \(p\in C^{2}\) be an eigenvector of the transposed matrix \(J^{T}(R_{3})\) corresponding to its eigenvalue \(\overline{\lambda_{1}(R_{3})} \) such that \(J^{T}(R_{3})p=e^{-i\theta_{0}}p\). By direct calculation, we have

$$ q\sim(-0.8432789,\text{ }0.2059666+0.4964278i)^{T} $$

and

$$ p\sim(0.2059666+0.4964278i,\text{ }0.8432789)^{T}. $$

These values satisfy

$$ J(R_{3})q=e^{1.451333i}q $$

and

$$ J^{T}(R_{3})p=e^{-1.451333i}p. $$

To obtain the normalization \(\langle p,q \rangle =1\), we can take the normalized vectors as

$$ q=(-0.8432789,0.2059666+0.4964278i)^{T} $$

and

$$ p=(-0.592919+0.246002,0.0000026+1.0072021i)^{T}. $$

Now we form \(x=zq+\overline{z}\overline{q}\). In this way, the system (25) can be transformed for sufficiently small \(\vert R \vert \) into the following form:

$$ z\rightarrow\lambda_{1}(R)z+g(z,\overline{z},R), $$

where \(\lambda_{1}(R)\) can be written as \(\lambda_{1}(R)=(1+\varphi (r))e^{i\theta(R)}\) (where \(\varphi(R)\) is a smooth function with \(\varphi(R_{3})=0\) and *g* is a complex-valued smooth function). The Taylor expression of g with respect \((z,\overline{z})=(0,0)\) is

$$ g(z,\overline{z,}r)=\sum_{k+1\geq2} \frac {1}{k!1!}g_{kl}(R)z^{k}\overline{z}^{l}, $$

where

$$\begin{aligned} &g_{20}(R_{3})=\bigl\langle p,B(q,q)\bigr\rangle =0.181094+0.267089i, \\ &g_{11}(R_{3})=\bigl\langle p,B(q,\overline{q})\bigr\rangle =-0.132395+0.527414i, \\ &g_{21}(R_{3})=\bigl\langle p,C(q,q,\overline{q})\bigr\rangle =1.56056-0.084732i, \\ &g_{02}(R_{3})=\bigl\langle p,B(\overline{q}, \overline{q})\bigr\rangle =1.72284+0.527423i. \end{aligned}$$

Now, the coefficient \(a(0)\), which determines the direction of the appearance of the invariant curve in a generic system exhibiting a Neimark–Sacker bifurcation, can be computed via

$$ a(0)=\operatorname{Re}\biggl[\frac{e^{-i\theta_{0}}g_{21}}{2}\biggr]-\operatorname{Re}\biggl[ \frac{(1-2e^{i\theta _{0}})e^{-2i\theta_{0}}}{2(1-e^{i\theta_{0}})}g_{20}g_{11}\biggr]-\frac {1}{2} \vert g_{11} \vert ^{2}-\frac{1}{4} \vert g_{02} \vert ^{2}. $$

(28)

From Eq. (28), the critical real part is obtained: \(a(0)=-1.03383\). Therefore, a supercritical Neimark–Sacker bifurcation occurs at \(R_{3}=13.81043\) (Figs. 4 and 5).

### 3.1 Bifurcation analysis of model (4)

In this section, we give stability conditions of coexistence equilibrium points of the model (6) and show that the model (6) undergoes a Neimark–Sacker bifurcation. Also, we obtain the bifurcation diagrams and phase portraits.

### Theorem 3.5

*Suppose that*
\(-1-H^{\ast}+bH^{\ast}>0\), \(2+2H^{\ast }-bH^{\ast}+H^{\ast}k+H^{\ast2}k>0\)
*and*
\(-1+H^{\ast}k>0\). *Assume that*

$$\begin{aligned} &R_{11}=\frac{(1+H^{\ast})^{b}(1+H^{\ast}+bH^{\ast})k}{b+k+H^{\ast}k} , \end{aligned}$$

(29)

$$\begin{aligned} &R_{12}=\frac{H^{\ast}(1+H^{\ast})^{b}(-1-H^{\ast}+bH^{\ast})k}{2+2H^{\ast}-bH^{\ast}+H^{\ast}k+H^{\ast2}k} , \end{aligned}$$

(30)

$$\begin{aligned} &R_{13}=\frac{bH^{\ast2}(1+H^{\ast})^{-1+b}k}{-1+H^{\ast}k}. \end{aligned}$$

(31)

*The positive equilibrium point*
\(( H^{\ast},P^{\ast} ) \)
*of the system* (6) *is locally asymptotically stable if and only if*
\(\max \{R_{11},R_{12}\}< R< R_{13}\).

### Example 3.6

For the parameter values \(b=1.15,k=2.2,R=2\), the positive equilibrium point \((H^{\ast},P^{\ast})\) of the system (6) is calculated as \((H^{\ast},P^{\ast})=(0.10076,0.506785)\). Using these parameter values and equilibrium point, we obtain \(R_{11}=1.64979\), \(R_{12}=-0.40156\), and \(R_{13}=6.01248\). From Theorem 3.5, the stability region is obtained: \(1.64979< R<6.01248\). As a result, for the above parameter values, the equilibrium point \((H^{\ast},P^{\ast })=(0.10076,0.506785)\) of the system (6) is locally asymptotically stable (Fig. 6).

### Proposition 3.7

(Eigenvalue assignment)

*Suppose that*
\(-1-H^{\ast }+bH^{\ast}>0\), \(2+2H^{\ast}-bH^{\ast}+H^{\ast}k+H^{\ast2}k>0\)
*and*
\(-1+H^{\ast}k>0\). *If*
\(\max\{R_{11},R_{12}\}< R=R_{13}\)
*then the eigenvalue assignment condition of the Neimark–Sacker bifurcation holds*.

### Theorem 3.8

*Suppose that*
\((H^{\ast},P^{\ast})\)
*is positive equilibrium point of the system* (6). *If the Proposition* (3.7) *holds*, \(R\neq\frac{H^{\ast}(1+H^{\ast})^{1+b}k}{-1-H^{\ast}+bH^{\ast}}\), \(R\neq\frac{H^{\ast}(1+H^{\ast})^{1+b}k}{-2-2H+bH}\)
*and*
\(a(0)<0\) (*respectively*
\(a(0)>0\)), *then the Neimark–Sacker bifurcation of the system* (6) *at*
\(R=R_{13}\)
*is supercritical* (*respectively*, *subcritical*) *and there exists a unique closed invariant curve bifurcation from*
\((H^{\ast},P^{\ast})\)
*for*
\(R=R_{13}\), *which is asymptotically stable* (*respectively*, *unstable*).

### Example 3.9

For the parameters \(b=1.15,k=2.2\) and initial condition \((H_{0},P_{0})=(0.5,0.6)\), the equilibrium point \((H^{\ast},P^{\ast })=(0.59904,0.263645)\) and bifurcation point \(R_{13}=3.06434\) are obtained. In this case, the norm of the eigenvalues is \(\vert \lambda _{1,2}(r) \vert = \vert {-}0.346474\pm0.938059i \vert =1\) (Fig. 7).