## 2 Materials and Methods
### Mathematical Analysis
#### 2.3.2 Boundedness
The threshold condition establishes a relationship between the parameters of the model system and the minimum and maximum values achievable by the variables within the varying quantities. Simply put, it delineates the range of resources available and subsequently determines the minimum and maximum number of possible livestock that the system can sustain, facilitating the coexistence of both the plant and livestock populations without risking extinction. To uphold this equilibrium, we have the option to either introduce more livestock or reduce their numbers (via harvesting) within the system. Note that \(P^{*}_{\min}\leq P^{*}(t)\leq P^{*}_{max}\), so that \(\mathcal{R}^{m}_{0}\leq\mathcal{R}_{0}\leq\mathcal{R}^{M}_{0}\). The threshold value, \(\mathcal{R}_{0}\), helps us to specify the non-extinction condition for the livestock population and hence can be considered as the effective reproduction number of the livestock population. That means to guarantee coexistence of both populations, we need this ratio to be always greater than unity, that is, \(\mathcal{R}_{0}>1\). **Theorem 4**. : _Suppose that \(\mathcal{R}_{0}>1\). For \(\epsilon>0\), \(n_{0}>b+P^{*}\), if \(P^{*}(0)>\frac{n_{0}}{(c\alpha\gamma e^{-r^{*}(t)\gamma t)}}\) where \(\gamma=\big{(}\frac{1}{2(d+\mu)}\big{)}\big{[}1+\ln\big{(}\frac{L^{*}(0)}{d+ \mu}\big{)}\big{]}\) is a positive constant then \(L^{*}(t)\
rightarrow 0\) as \(t\rightarrow\infty\)._  
Proof. : Suppose \(L^{*}(0)\geq d+\mu\). Then, since \(d+\mu>0\), let us define \(\epsilon\) as a very small positive number, \(n_{0}\) to be a sufficiently large number such that \(n_{0}>b+P^{*}_{\max}\), and also let \(\gamma=\big{(}\frac{1}{2(d+\mu)}\big{)}\big{[}1+\ln\big{(}\frac{L^{*}(0)}{d+ \mu}\big{)}\big{]}\). Then, \(c\alpha\gamma e^{-r^{*}(t)\gamma}>0\). Let us choose \(P^{*}(0)\) so that \(P^{*}(0)>\frac{n_{0}}{(c\alpha\gamma e^{-r^{*}(t)\gamma t)}}\). When using the equation for \(\frac{dP}{dt}\) and Lemma 3, we have  
\[\frac{dP}{dt}>0,\]  
which implies that  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]0  
Now for \(\epsilon>0\), and because of the choice of \(n_{0}\), we have \(\frac{P^{*}}{b+P^{*}}>\frac{p^{*}}{n_{0}}\); using our choice of \(P^{*}(0)>\frac{n_{0}}{(c\alpha\gamma e^{-r^{*}(t)\gamma t)}}\), we obtain  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]1  
Hence,  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]2  
Moreover, using the equation for \(\frac{dL}{dt}\) and Lemma 3 since \(\epsilon>0\), and \(n_{0}>b+P^{*}_{\max}\), \(\frac{P^{*}_{\max}}{b+P^{*}_{\max}}>\frac{P^{*}_{\max}}{n_{0}}\), we have  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]3  
Then, using Equation (19) we obtain  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]4But then \(L(t)\geq L^{*}(0)e^{\left(\frac{1}{\gamma}-d-u\right)t}\) for \(t\geq\gamma\). Hence, \(L^{*}(\gamma)\geq L^{*}(0)e^{(1-\gamma(d-\mu))}\). Again from the definition of \(\gamma\), we can clearly see that  
\[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]5 \[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]6 \[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]7  
Hence \(L^{*}(\gamma)>(d+\mu)\). This implies that the livestock population will not go to extinction, that is, \(L^{*}(t)\
rightarrow 0\) for \(t\rightarrow\infty\). Theorem 4 tells us that coexistence is possible in our system, but it does not tell us what happens to the size of the population as time goes on. The answer to this is found in Theorem 3.3 from [9]. This can also be represented by simulation, as shown in \[\frac{caP(t)}{b+P(t)}>d+\mu. \tag{12}\]8 This figure delineates the range of resources (forage) available and subsequently determines the minimum and maximum number of livestock that the system can sustain, facilitating the coexistence of both the plant and livestock populations without risking extinction. This is referred to as a stable limit cycle. To uphold this equilibrium, we have the option to either introduce more livestock or reduce their numbers (via harvesting) within the system.