## 3 System Properties for Constant Concentrations
### Reduction to the Uncontrolled System and Basic Dynamical System Properties
Our aim in the following is thus to determine the asymptotic behavior of the trajectories of the system. We start with some basic properties. The positive orthant  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]0  
is positively invariant for the dynamics. This is because the planes \(Q=0\) and \(E=0\) are invariant under Equations (6) and (8) and \(\dot{P}\geq k_{P}Q\) whenever \(P=0\). Thus, starting at a positive initial condition \((Q_{0},P_{0},E_{0})\), it follows that the solutions remain positive for all times. For the long-term behavior of the system, the equilibrium solutions in the closure of \(\mathbb{P}\), \(\bar{\mathbb{P}}=\mathrm{clos}(\mathbb{P})\), also matter. Recall that the system is defined and continuous on \(\bar{\mathbb{P}}\) due to the use of the limits as \(P\to 0\) and \(E\to 0\) in place of \(P=0\) and \(E=0\), respectively. The vector field defining the \(P\) and \(E\) dynamics is not continuously differentiable at \(P=0\) or \(E=0\), but these values are repelling and thus this does not become an issue. The equilibrium solution \(E_{*}=0\) is repelling: there exists a positive threshold \(E_{\Delta}<E_{\mathrm{ss}}\) such that \(\frac{dE}{dt}\) is positive on \((0,E_{\Delta}]\). In particular, once \(E_{\mathrm{ss}}>E(\tau)\geq E_{\Delta}\), then \(E(t)\geq E_{\Delta}\) for all \(t\geq\tau\). Furthermore, for \(E(0)<E_{\mathrm{ss}}\), \(E\) will remain below \(E_{\mathrm{ss}}\). The terms in the last parentheses in Equation (8) are bounded between \(1\) and \(1+P_{\mathrm{max},2}\) and thus, as \(E\to 0\), the Gompertzian growth dominates the dynamics. Specifically, let  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]1  
then \(E_{\Delta}\leq E_{\mathrm{ss}}\) and for \(E<E_{\Delta}\) we have that \(\frac{dE}{dt}>0\). Furthermore, for \(E=E_{\mathrm{ss}}\), Equation (8) reduces to \(\frac{dE}{dt}=-\delta_{E}\left(1+\frac{P_{\mathrm{max},2}P}{1+P}\right)E<0\) and thus the values of \(E\) cannot reach the value \(E_{\mathrm{ss}}\) if they start below \(E_{\mathrm{ss}}\). The equilibrium solution \(P_{*}\equiv 0\) is repelling: there exists a positive threshold \(P_{\Delta}<P_{\mathrm{ss}}\) such that \(\frac{dP}{dt}\) is positive on \((0,P_{\Delta}]\). In particular, once \(P_{\mathrm{ss}}>P(\tau)\geq P_{\Delta}\), we have \(P(t)\geq P_{\Delta}\) for all \(t\geq\tau\). For values of \(E\) less than \(E_{\mathrm{ss}}\), we have that  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]2  
for all times. Choosing \(P_{\Delta}\) as  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]3the result follows: for \(P<P_{\Delta}\) we have that  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]4  
This proves the result. Corollary 1: _The equilibrium solutions \(E_{*}\equiv 0\) and \(P_{*}\equiv 0\) are unstable._  
Note, however, that \(P\) is not necessarily bounded. For, with \(P=P_{\text{ss}}\), Equation (7) becomes  
\[\hat{r}_{P} =\left(1-\frac{U1_{\max,1}u_{1}}{U1C_{50}+u_{1}}\right)\left(1- \frac{U2_{\max,2}u_{2}}{U2C_{50}+u_{2}}\right)r_{P},\]5  
and thus, if \(Q\) is large enough, this term will be positive. Hence, if \(Q\) grows exponentially, \(P\) will diverge to \(+\infty\). Lemma 3: _If \(Q\) increases exponentially with time, then \(\lim_{t\to\infty}P(t)=+\infty\)._  
Proof: We need to show that for every positive value \(\hat{P}\) there exists a time \(\hat{T}\) so that \(P(t)\geq\hat{P}\) for all \(t\geq\hat{T}\). We first remark that \(P\) is unbounded.