# The Impact of Intermittent Androgen Suppression Therapy in Prostate Cancer Modeling
## 4 Dynamics of the New Model
Thus, substitute \(x_{2}^{*}=x_{1}^{*}/y^{*}\) in Equation (20) to obtain  
\[Q^{*}=\frac{mA^{*}x_{1}^{*}(1+\frac{1}{y^{*}})+(\mu_{1}q_{1}+\mu_{2}q_{2}\frac {1}{y^{*}})x_{1}^{*}}{mx_{1}^{*}(1+\frac{1}{y^{*}})+(\mu_{1}+\mu_{2}\frac{1}{ y^{*}})x_{1}^{*}}=\frac{mA^{*}(1+\frac{1}{y^{*}})+(\mu_{1}q_{1}+\mu_{2}q_{2} \frac{1}{y^{*}})}{m(1+\frac{1}{y^{*}})+(\mu_{1}+\mu_{2}\frac{1}{y^{*}})}.\]  
This shows that \(Q^{*}\) is unique. Finally, from Equation (14), \(x_{1}^{*}=\frac{A_{1}(Q^{*})}{\delta_{11}}\). It follows that \(x_{1}^{*}\) and \(x_{2}^{*}\) are also unique. Since \(\lim_{t\to\infty}A(t)=\frac{\gamma_{2}}{\gamma_{1}}+A_{0}=:A^{*}\), the existence of a unique positive equilibrium follows. The third proposition shows that when the patient is on treatment, under biologically reasonable conditions, the boundary equilibrium with the extinction of AD cells is the only equilibrium of the system. In this case, it establishes a sufficient and necessary condition for the boundary equilibrium to be globally stable. Additionally, it provides sufficient conditions for both cancer subpopulations to eventually become extinct. **Proposition 3**. : _In the case of \(u=0\), if_  
\[\frac{m\frac{\gamma_{1}}{\gamma_{2}}+\max\{\mu_{1}q_{1},\mu_{2}q_{2}\}}{m+\min \{\mu_{1},\mu_{2}\}}<q_{1},\]  
_then the system does not have a positive steady-state and \(x_{1}\) becomes extinct as \(t\to\infty\). Furthermore, \(Q(t)\) approaches \(Q_{*}=\frac{mA_{*}+\mu_{1}q_{1}}{m+\mu_{1}}\), where \(A_{*}=\frac{\gamma_{2}}{\gamma_{1}}\)._  
_In addition, if_  
\[\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}>0,\]  
_then \(x_{2}\) approaches \(\frac{\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}} {\delta_{2}}\) and \(E_{1}=(0,\frac{\mu_{2}(1-q_{2}/Q_{*})-D_{2}(Q_{*})}{\delta_{22}},Q_{*},A_{*})\) is globally stable. On the other hand, if either_  
\[\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}>0,\]  
_or,_  
\[\frac{m\frac{\gamma_{1}}{\gamma_{2}}+\max\{\mu_{1}q_{1},\mu_{2}q_{2}\}}{m+ \min\{\mu_{1},\mu_{2}\}}\leq q_{2},\]  
_then both \(x_{1}\) and \(x_{2}\) become extinct in the limit of \(t\to\infty\)._  
Proof. : Similar to the previous proof, we establish that in the limit of \(t\to\infty\), \(A\to\frac{\gamma_{2}}{\gamma_{1}}\) and  
\[f(y)=\left(\lambda(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y^{2}+\left(A_{2 }(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y-A_{1}(Q^{*})=0. \tag{16}\]0  
Suppose \(\overline{Q}<q_{1}\). This implies that \(\lim_{t\to\infty}x_{1}^{\prime}(t)<-(D_{1}(Q)+\delta_{1}x_{1}+\lambda(Q))x_{1}<0\). Thus there is no positive steady-state for \(x_{1}(t)\) and \(\lim_{t\to\infty}x_{1}(t)=0\). Then in the limit of \(x_{1}\to 0\) and \(A\to\frac{\gamma_{2}}{\gamma_{1}}\),  
\[f(y)=\left(\lambda(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y^{2}+\left(A_{2 }(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y-A_{1}(Q^{*})=0. \tag{16}\]1This implies that \(\lim_{t\to\infty}Q(t)=\frac{m\frac{\eta_{2}^{2}+\mu_{2}q_{2}}{\Omega+\mu_{2}}}{m+ \mu_{2}}=Q_{*}\). Furthermore, in the limit of \(t\to\infty\),  
\[f(y)=\left(\lambda(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y^{2}+\left(A_{2 }(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y-A_{1}(Q^{*})=0. \tag{16}\]2  
If \(\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}>0\), then \(x_{2}\) approaches \(\frac{\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}} {\delta_{2}\frac{Q_{*}+R_{2}}{Q_{*}+R_{2}}}\) and \(E_{1}\) is globally stable. On the other hand, if \(\mu_{2}\left(1-\frac{q_{2}}{Q_{*}}\right)-d_{2}\frac{R_{2}}{Q_{*}+R_{2}}<0\), then \(x_{2}\) becomes extinct. Alternatively, if \(\overline{Q}<q_{2}\), then \(\lim_{t\to\infty}x_{2}^{\prime}(t)<-(D_{2}(Q)+\delta_{1}x_{2})x_{2}<0\). This implies \(\lim_{t\to\infty}x_{2}(t)=0\).