# A Simple Predator-Prey Population Model with Rich Dynamics
## 3 Preliminary Results
### Existence of Equilibria
: In the subregion of \(D\) with \(P>P_{0}\), a positive equilibrium satisfies  
\[rN(1-\frac{N}{K})-aNP =0 \tag{14a}\] \[caNP-dP-h =0 \tag{14b}\]  
It follows from Equation (14a) that  
\[P=\frac{r(K-N)}{aK}\]  
Substituting it into Equation (14b) yields  
\[\left(N-\frac{K}{n_{0}}\right)(K-N)=\frac{K}{cr}h \tag{15}\]  
It follows from the relation of roots and coefficients for the quadratic equation \((x-A)(B-x)=c\) shown above that equation Equation (15) has no, unique, or two positive solutions if \(h>h_{1}\), \(h=h_{1}\), or \(h<h_{1}\). To have \(P_{i}>P_{0}\), \(i=1,2\), we need  
\[P_{i}=\frac{r(K-N_{i})}{aK}>P_{0}\]  
that is,  
\[N_{i}<K-\frac{aK}{r}P_{0}=\frac{aK}{r}\left(\frac{r}{a}-P_{0}\right),\;\;i=1,2 \tag{16}\]  
Thus, if \(P_{0}\geq r/a\), there is no positive equilibrium of Equation (15) in the subregion of \(D\) with \(P>P_{0}\). We assume \(P_{0}<r/a\). Suppose there are two positive solutions, \(N_{1}<N_{2}\), to Equation (15) and hence \(P_{2}<P_{1}\). Then \(P_{2}>P_{0}\) if and only if  
\[N_{2}<K-\frac{aK}{r}P_{0} \tag{17}\]  
Substituting \(N_{2}\) in Equation (13) into Equation (17) yields  
\[n_{0}\sqrt{4 crK\left(h_{1}-h\right)}<crK(n_{0}-1)-2 caKn_{0}P_{0}\]  
that is,  
\[\frac{\sqrt{4 crK}}{2 caK}\sqrt{(h_{1}-h)}<\frac{r(n_{0}-1)}{2 an_{0}}-P_{0}=\hat {P}-P_{0} \tag{18}\]  
Then if \(P_{0}\geq\hat{P}\), we have \(P_{2}\leq P_{0}\). Assume \(P_{0}<\hat{P}\). Squaring both sides of Equation (18) yields  
\[\frac{r}{ca^{2}K}(h_{1}-h)<\frac{r^{2}(n_{0}-1)^{2}}{4a^{2}n_{0}^{2}}-\frac{r (n_{0}-1)}{an_{0}}P_{0}+P_{0}^{2} \tag{19}\]It follows from the definition of \(h_{1}\) that  
\[caNP-dP-h =0 \tag{14b}\]0  
Then Equation (19) becomes  
\[ \begin{split} 0&<h^{2}-\frac{rm}{a}\left(\frac{n_{0}-1}{n _{0}}-\frac{m}{caK}\right)h=\left(h-\frac{rm}{a}\left(1-\frac{d+m}{caK}\right) \right)h\\ &=\ \left(h-\frac{rm}{a}\left(1-\frac{1}{\hat{n}}\right)\right)h\end{split}  \tag{20}\]  
If \(\hat{n}<1\), we have \(P_{2}>P_{0}\). Otherwise, Equation (20) is equivalent to  
\[0<(h-h_{2})h\]  
Thus, if \(h_{2}<h<h_{1}\), \(P_{2}>P_{0}\), and if \(h\leq h_{2}\), \(P_{2}<P_{0}\). We now consider \(P_{1}>P_{0}\), that is,  
\[N_{1}<\frac{aK}{r}\left(\frac{r}{a}-P_{0}\right) \tag{21}\]  
Substituting \(N_{1}\) in Equation (13) into Equation (21), we have  
\[\frac{\sqrt{4 crK}}{2 caK}\sqrt{(h_{1}-h)}>P_{0}-\hat{P} \tag{22}\]  
If \(P_{0}\leq\hat{P}\), inequality Equation (21) is satisfied which implies \(P_{1}>P_{0}\). Suppose \(P_{0}>\hat{P}\). Similarly as above, inequality Equation (22) is equivalent to  
\[0>(h-h_{2})h\]  
Thus, if \(h<h_{2}\), \(P_{1}>P_{0}\), and if \(h\geq h_{2}\), \(P_{1}<P_{0}\). By putting all together, the proof is completed. **Remark 1**: System Equation (1) with linear predator harvest strategy leads to the predator extinct if the net reproductive number \(\hat{n}<1\). However, by Theorem 2 (\(c\)), we find that for system Equation (1) with switched predator harvest strategy; that is, when the density of predator is below harvest level \(P_{0}\), the linear harvesting rate is applied to the system, whereas when the density of predators is higher than harvest level, the system adopts nonzero constant harvesting rate, even if \(\hat{n}<1\), the prey and predator may coexist.