# The Impact of Intermittent Androgen Suppression Therapy in Prostate Cancer Modeling
## 4 Dynamics of the New Model
Additionally, it establishes sufficient conditions for the existence of a unique positive steady-state. **Proposition 2**. : _In the case of \(u=1\), assume \(q_{1}>q_{2}\), then \(E_{1}=\left(0,\frac{\mu_{2}(1-q_{2}/Q_{*})-D_{2}(Q_{*})}{\delta_{22}},Q_{*},A_{*}\right)\), where \(Q_{*}=\frac{mA_{*}+\mu_{1}q_{1}}{m+\mu_{1}}\) and \(A_{*}=\left(\gamma_{2}+\gamma_{1}A_{0}\right)/\gamma_{1}\), is the only equilibrium state where the AD subpopulation becomes extinct. Furthermore, suppose conditions (1) and (2) hold:_  
1. \(\frac{mA_{0}}{m+\max\{\mu_{1},\mu_{2}\}}>q_{1}\)__
2. \(\frac{mA_{0}+\min\{\mu_{1}q_{1},\mu_{2}q_{2}\}}{m+\max\{\mu_{1},\mu_{2}\}}> \frac{1}{2}\left(\Delta+\sqrt{\left(\Delta\right)^{2}-4\max\{R_{1},K\}}\right) >0\)_,_  
_where \(\Delta=q_{1}+\left(1+\frac{d_{1}c}{\mu_{1}}\right)\max\{R_{1},K\}\). Then the system has a unique positive steady-state \((x_{1}^{*},x_{2}^{*},Q^{*},A^{*})\)._  
Proof. : For \(u=1\) (off-treatment interval), we first consider the boundary equilibria of the system. Since \((x_{1},x_{2})=(0,0)\) is undefined for the system and \(x_{2*}=0\) implies \(x_{1*}=0\), the only equilibrium involving an extinction of a cancer subpopulation takes place when \(x_{1*}=0\). In this case, solving \(x_{2}^{\prime}=0\) yields \(x_{2*}=\frac{\mu_{2}(1-q_{2}/Q_{*})-D_{3}(Q_{*})}{\delta_{22}}\). Additionally, solving \(Q^{\prime}=0\) and \(A^{\prime}=0\) yields \(Q_{*}=\frac{mA_{*}+\mu_{1}q_{1}}{m+\mu_{1}}\) and \(A_{*}=(\gamma_{2}+\gamma_{1}A_{0})\backslash\gamma_{1}\). For the positive steady state, suppose that \(x_{1}^{*},x_{2}^{*},Q^{*},A^{*}>0\). Then solving for \(x_{1}^{\prime}=0\) and \(x_{2}^{\prime}=0\) gives:  
\[ \begin{cases}0&=A_{1}x_{1}^{*}-\delta_{11}(x_{1}^{*})^{2}\\ 0&=A_{2}x_{2}^{*}-\delta_{22}(x_{2}^{*})^{2}+\lambda(Q^{*})x_{1}^{*}\end{cases}  \Rightarrow \begin{cases}\delta_{11}(x_{1}^{*})^{2}&=A_{1}x_{1}^{*}\\ \delta_{22}(x_{2}^{*})^{2}&=A_{2}x_{2}^{*}+\lambda(Q^{*})x_{1}^{*}\end{cases}  \tag{14}\]  
where \(A_{i}\) are the coefficients of \(x_{i}^{*}\), e.g., \(A_{1}=\mu_{1}(1-\frac{q_{1}}{Q^{*}})-D_{1}(Q^{*})-\lambda(Q^{*})\). Dividing the previous two equations to obtain  
\[\frac{\delta_{11}}{\delta_{22}}\left(\frac{x_{1}^{*}}{x_{2}^{*}}\right)^{2}= \frac{A_{1}\left(\frac{x_{1}^{*}}{x_{2}^{*}}\right)}{A_{2}+\lambda(Q^{*}) \left(\frac{x_{1}^{*}}{x_{2}^{*}}\right)} \tag{15}\]  
Set \(y=\left(\frac{x_{1}^{*}}{x_{2}^{*}}\right)\), then  
\[f(y)=\left(\lambda(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y^{2}+\left(A_{2 }(Q^{*})\frac{\delta_{11}}{\delta_{22}}\right)y-A_{1}(Q^{*})=0. \tag{16}\]  
Note that \(\lambda(Q^{*})=c\frac{K}{Q^{*}+K}>0\), so if \(A_{1}>0\), then by [PERSON]'s rule of sign, the system has a unique solution \(y\), regardless of the sign for \(A_{2}\). The uniqueness of \(y\) implies that the ratio \(\frac{x_{1}^{*}}{x_{2}^{*}}\) is constant. We proceed to establish the positivity of \(A_{1}\). Explicitly, \(A_{1}\) takes the form  
\[A_{1}(Q^{*}) =\mu_{1}\left(1-\frac{q_{1}}{Q^{*}}\right)-D_{1}(Q^{*})-\lambda(Q^ {*})\] \[=\mu_{1}\left(1-\frac{q_{1}}{Q^{*}}\right)-d_{1}\frac{R_{1}}{Q^{* }+R_{1}}-c\frac{K}{Q^{*}+K}\]  
Note that, since the treatment is off, biologically \(Q(t)>>q_{1}\), which allows strong growth of the AD cell subpopulation. We will establish later that there is indeed a lower bound of \(Q(t)\) satisfying \(\liminf Q(t)>q_{1}\), under reasonable biological constraints. For the moment, we assume such lower bound exists. Then  
\[Q^{*}=\frac{mA^{*}+\frac{\mu_{1}q_{1}x_{1}^{*}+\mu_{2}q_{2}x_{2}^{*}}{x_{1}^{*}+x_ {2}^{*}}}{m+\frac{\mu_{1}x_{1}^{*}+\mu_{2}x_{2}^{*}}{x_{1}^{*}+x_{2}^{*}}}=\frac{mA ^{*}(x_{1}^{*}+x_{2}^{*})+\mu_{1}q_{1}x_{1}^{*}+\mu_{2}q_{2}x_{2}^{*}}{m(x_{1}^ {*}+x_{2}^{*})+\mu_{1}x_{1}^{*}+\mu_{2}x_{2}^{*}}>0. \tag{20}\]  
Finally, to show uniqueness, recall that the ratio of \(x_{1}^{*}\) and \(x_{2}^{*}\) is a fixed constant, \(y^{*}\).