# A Simple Predator-Prey Population Model with Rich Dynamics
## 3 Preliminary Results  
### Existence of Equilibria  
In this section, we explore the existence of all nonnegative equilibria. First, the origin \((0,0)\) is still a trivial equilibrium and the predator-free equilibrium \((K,0)\) exists. Moreover, it is easy to see that there exists no positive equilibrium in region \(D\) if \(n_{0}\leq 1\). We thus assume \(n_{0}>1\) hereafter, and present our results of the existence of positive equilibria as follows. **Theorem 1**: _System Equation (3) has a positive coexistence equilibrium \(E^{*}(N^{*},P^{*})\), in the subregion of \(D\) with \(0<P\leq P_{0}\),_  
\[N^{*}=\frac{K}{\hat{n}},\quad P^{*}=\frac{r}{a}\left(1-\frac{1}{\hat{n}}\right) \tag{7}\]  
_if and only if \(P_{0}\geq\frac{r}{a}\) and \(\hat{n}>1\), or \(P_{0}<\frac{r}{a}\) and \(1<\hat{n}\leq\frac{r}{r-aP_{0}}\), where \(\frac{r}{a}\) represents the maximum predator density for which the prey population can establish itself from a small initial population and \(\hat{n}\) is the net reproductive number of the predator under harvesting defined by_  
\[\hat{n}=\frac{caK}{d+m} \tag{8}\]  
In the subregion of \(D\) with \(0<P\leq P_{0}\), a positive equilibrium of Equation (3) satisfies  
\[ \begin{cases}rN^{*}\left(1-\frac{N^{*}}{K}\right)-aN^{*}P^{*}=0\\ caN^{*}P^{*}-(d+m)P^{*}=0\end{cases}  \tag{9}\]  
Then it follows that  
\[N^{*}=\frac{d+m}{ca}=\frac{K}{\hat{n}}\]  
and then  
\[P^{*}=\frac{r}{a}\left(1-\frac{1}{\hat{n}}\right)>0\]  
if \(\hat{n}>1\). In the mean time, it follows from  
\[P^{*}=\frac{r}{a}\left(1-\frac{1}{\hat{n}}\right)\leq P_{0}\]  
and then  
\[\frac{1}{\hat{n}}\geq\frac{r-aP_{0}}{r}\]that there exists a positive equilibrium in this subregion if  
\[\frac{r-aP_{0}}{r}\leq 0\]  
or if \(\frac{r-aP_{0}}{r}>0\) and  
\[\hat{n}\leq\frac{1}{1-\frac{aP_{0}}{r}}=\frac{r}{r-aP_{0}} \tag{10}\]  
\(\square\)  
To investigate the existence of positive equilibria in the subregion of \(D\) with \(P>P_{0}\), we first give the relation of roots and coefficients for a quadratic equation. Quadratic equation \((x-A)(B-x)=C\), with constants \(A\), \(B\), and \(C\) positive, has two positive roots  
\[x_{1}=\frac{A+B-\sqrt{(B-A)^{2}-4C}}{2}<x_{2}=\frac{A+B+\sqrt{(A-B)^{2}-4C}}{2}\]  
a unique positive root  
\[x=\frac{A+B}{2}\]  
or no positive root, if  
\[C<\frac{(B-A)^{2}}{4},\;C=\frac{(B-A)^{2}}{4},\;\text{or}\;C>\frac{(B-A)^{2}}{4}\]  
respectively. The results for the existence of positive equilibria of Equation (3) in the subregion of \(D\) with \(P>P_{0}\) are provided as follows. **Theorem 2**. : _We assume \(P_{0}<\frac{r}{a}\) and define_  
\[\hat{P}=\frac{r(n_{0}-1)}{2 an_{0}},\;h_{1}=\frac{crK}{4}\left(1-\frac{1}{n_{0} }\right)^{2},\;h_{2}=\frac{rm}{a}\left(1-\frac{1}{\hat{n}}\right) \tag{11}\]  
_System Equation (3), in the subregion of \(D\) with \(P>P_{0}\), has_  
1. _No positive equilibrium if_ \[h>h_{1},\quad\text{or}\quad \begin{cases}P_{0}>\hat{P}\\ h_{2}\leq h<h_{1}\end{cases} \]
2. _A unique positive equilibrium with_ \[N=\frac{K(1+n_{0})}{2n_{0}},\quad P=\frac{r(K-N)}{aK}=\frac{r(n_{0}-1)}{2 an_ {0}}\] (12) _if_ \[h=h_{1},\quad \begin{cases}P_{0}>\hat{P},\\ h<h_{1}\text{ and }h<h_{2}\end{cases} \quad\text{or}\quad \begin{cases}P_{0}<\hat{P},\\ h<h_{1}\text{ and }h\leq h_{2}\end{cases} \]
3. _Two positive equilibria_ \(E_{i}(N_{i},P_{i})\)_,_ \(i=1,2\)_, where_ \[N_{1} =\frac{crK(1+n_{0})-n_{0}\sqrt{4 crK\left(h_{1}-h\right)}}{2 crn_{0}}\] (13) \[<\frac{crK(1+n_{0})+n_{0}\sqrt{4 crK\left(h_{1}-h\right)}}{2 crn_{0}}=N _{2}\]_and_  
\[P_{2}=\frac{r(K-N_{2})}{aK}<P_{1}=\frac{r(K-N_{1})}{aK}\]  
_if \(h<h_{1}\) and_  
\[ \begin{cases}P_{0}<\hat{P}\\ \hat{n}<1\end{cases} \qquad or\quad \begin{cases}P_{0}<\hat{P}\\ h_{2}<h\end{cases} \]  
Proof.