From the above discussion and analysis, we find that the structure of basins of some periodic cycles of system (7) is quite complicated. For that, we set \(q_{1,t+1}=\acute{q}_{1}\) and \(q_{2,t+1}=\acute{q}_{2}\) in system (7), which means the time evolution is obtained by the iteration \(T:(q_{1},q_{2})\rightarrow (\acute{q}_{1},\acute{q}_{2})\) given by

$$ T: \textstyle\begin{cases} \acute{q}_{1}=q_{1}(t)+\alpha q_{1}(t) ( \frac{a+q_{2}}{ ( a+Q ) ^{2}}-c_{1} ) , \\ \acute{q}_{2}=\sqrt{\frac{a+q_{1}(t)}{c_{2}}}-q_{1}(t)-a.\end{cases} $$

(13)

System (13) is a noninvertible system because for a given point \((\acute{q}_{1},\acute{q}_{2})\in \mathbb{R} ^{2}\) it may have up to four real rank-1 preimages; these preimages can be obtained by solving algebraically system (13) with respect to the variables \(q_{1}\) and \(q_{2}\). Furthermore, the phase plane contains these points in different regions. Therefore, the phase plane may be divided into several regions represented by \(Z_{i}\), where *i* refers to the rank-1 preimages number in each region. This requires to calculate both *LC* and \(LC_{-1}\). The later is determined by vanishing the determinant of the Jacobian of system (7) as follows:

$$ \frac{kq_{1}(a-q_{1}+q_{2}) ( 1-2\sqrt{c_{2}(a+q_{1})} ) }{ ( a+q_{1}+q_{2} ) ^{3}\sqrt{c_{2}(a+q_{1})}}=0. $$

(14)

This means that \(LC_{-1}\) is represented by the union of the following:

$$ LC_{-1a}:q_{1}=0, \qquad LC_{-1b}:q_{2}=q_{1}-a, \qquad LC_{-1c}:q_{1}=\frac{1-4ac_{2}}{4c_{2}} . $$

(15)

Then the critical curve \(LC=T(LC_{-1})\) becomes

$$ LC_{a}:q_{2}=\sqrt{\frac{a}{c_{2}}}-a, \qquad LC_{b}:q_{2}^{2}= \frac{1}{4c_{2}}(a+q_{1}), \qquad LC_{c}:q_{2}=\frac{1}{4c_{2}} . $$

(16)

It is clear in Fig. 4 that at this set of parameter values the critical curve *LC* divides the phase plane of period-2 cycle into two different regions: \(Z_{0}\) and \(Z_{2}\). Therefore, system (7) is noninvertible. The region \(Z_{0}\) does not have rank-1 preimages, while the region \(Z_{2}\) possesses two real rank-1 preimages. Because of the complexity, to get an analytical expressions for those preimages, we can obtain real preimages for the points \(( \acute{q}_{1},0 ) \in \mathbb{R} ^{2}\) and \((0,\acute{q}_{2})\in \mathbb{R} ^{2}\) as follows.

### Proposition 2

*The real preimages of the point*
\(( 0,\acute{q}_{2} ) \in \mathbb{R} ^{2}\)
*belong to the same invariant axis*
\(q_{1}=0\)
*or lie on the curve*

$$ \begin{gathered} q_{2}=-a+\frac{\alpha }{2 ( \alpha c_{1}-1 ) }-q_{1}\pm \sqrt {q_{1}^{2}+ \frac{\alpha ( 2+\alpha -2ac_{1} ) }{2 ( \alpha c_{1}-1 ) ^{2}}};\\ \alpha c_{1}>1,\qquad q_{1}\leq \frac{\alpha }{4 ( \alpha c_{1}-1 ) }. \end{gathered} $$

(17)

### Proof

Substituting \(( 0,\acute{q}_{2} ) \) in (13), we get

$$ \textstyle\begin{cases} 0=q_{1}+\alpha q_{1} ( \frac{a+q_{2}}{ ( a+Q ) ^{2}}-c_{1} ) , \\ \acute{q}_{2}=\sqrt{\frac{a+q_{1}}{c_{2}}}-q_{1}-a. \end{cases} $$

(18)

The first equation of system (18) is satisfied provided that

$$ q_{1}=0\quad \text{or}\quad 0=1-\alpha c_{1}+ \frac{\alpha ( a+q_{2} ) }{ ( a+q_{1}+q_{2} ) ^{2}}. $$

Suppose that \(0=1-\alpha c_{1}+ \frac{\alpha ( a+q_{2} ) }{ ( a+q_{1}+q_{2} ) ^{2}}\) holds. That restricts \(\alpha c_{1}>1\) for the term \(( a+q_{1}+q_{2} ) ^{2}= \frac{\alpha ( a+q_{2} ) }{\alpha c_{1}-1}\) to be positive. Solving this term with respect to \(q_{2}\) completes the proof. □

### Proposition 3

*The two real preimages of the point*
\(( \acute{q}_{1},0 ) \in \mathbb{R} ^{2}\)
*are*

$$ \begin{gathered} \biggl( -a, \frac{a ( \alpha \pm \sqrt{\alpha ^{2}(1+ac_{1})-4\alpha (a+\acute{q}_{1})} ) }{2 ( a\alpha c_{1}-a-\acute{q}_{1} ) } \biggr) , \\ \acute{q}_{1}< \frac{\alpha }{4}(1+4ac_{1})-a\end{gathered} $$

*or*

$$\begin{aligned}& \biggl( \frac{1-ac_{2}}{c_{2}},\Psi \biggr) , \\& \begin{aligned} \Psi &= \frac{2-2c_{2}(a+\acute{q}_{1})-\alpha ( 1+ac_{2} ) ( 2c_{1}-c_{2} ) }{c_{2} [ c_{2}(a+\acute{q}_{1})+\alpha c_{1}(1-ac_{2})-1 ] } \\ &\quad {}\pm \frac{(1-ac_{2})}{ [ 1-c_{2}(a+\acute{q}_{1})-\alpha c_{1}(1-ac_{2}) ] }\sqrt{ \biggl[ 1- \frac{4\alpha c_{1}}{c_{2}}(1-ac_{2})-\frac{4}{c_{2}} \bigl( 1+c_{2}(a+\acute{q}_{1}) \bigr) \biggr] }, \end{aligned} \\& \acute{q}_{1}< \frac{\alpha }{4}+ \frac{(1-ac_{2})(1-\alpha c_{1})}{c_{2}}. \end{aligned}$$

### Proof

Substituting \(( \acute{q}_{1},0 ) \) in (13), we get

$$ \begin{gathered} \acute{q}_{1}=q_{1}+\alpha q_{1} \biggl( \frac{a+q_{2}}{ ( a+Q ) ^{2}}-c_{1} \biggr) , \\ 0=\sqrt{\frac{a+q_{1}}{c_{2}}}-q_{1}-a.\end{gathered} $$

(19)

The second equation of (19) gives \(q_{1}=-a\) or \(q_{1}=\frac{1-ac_{2}}{c_{2}}\). So we have the following cases. □

Case 1: Substituting \(q_{1}=-a\) in the first equation of (19) with simple calculations gives

$$ ( a\alpha c_{1}-a-\acute{q}_{1} ) q_{2}^{2}-a \alpha q_{2}-a^{2} \alpha =0. $$

(20)

Solving algebraically (20) with respect to \(q_{2}\), we get the first part of Proposition 3 proved.

Case 2: Substituting \(q_{1}=\frac{1-ac_{2}}{c_{2}}\) in the first equation of (19) and solving it as in case 1 complete the proof.

Another important aspect of system (13) we should mention here is that it is undefined at the line \(q_{2}=-a-q_{1}\). This makes it belonging to the family of maps cited in [23–25] and defined in the form \((\acute{q}_{1},\acute{q}_{2})= ( \frac{N(q_{1},q_{2})}{D(q_{1},q_{2})},G(q_{1},q_{2}) ) \). Our system (13) is defined for all \((q_{1},q_{2})\in \mathbb{R} ^{2}\) in the phase plane except at the point \(( q_{1},-a-q_{1} ) \) that makes \(D(q_{1},q_{2})=0\) and its preimages too. Furthermore, system (13) is in the form \(0/0\) because \(N(q_{1},q_{2})=0\) and \(D(q_{1},q_{2})=0\) at the point \((0,-a)\). The point \((0,-a)\) is called a focal point. But the line \(q_{2}=-a-q_{1}\) makes \(D(q_{1},q_{2})=0\) only and forms a set of non-definition points denoted by \(\mathcal{L}_{S}= \{ (q_{1},q_{2})\in \mathbb{R} ^{2}:q_{2}=-a-a_{1} \} \) and its set of prefocal points gets \(\mathcal{L}_{Q}:q_{2}=-a\). More information about the properties of this point and its lobes is well discussed in [5].