## 3 System Properties for Constant Concentrations
### Dynamics on the Plane \(Q=0\)  
The plane \(Q=0\) is invariant under the dynamics and can have regions that are repelling or attractive. We first analyze the reduced dynamical system in this boundary stratum of \(\mathbb{P}\), i.e., consider the equations\[\frac{dP}{dt} =\left[r_{P}\ln\left(\frac{P_{\text{ss}}}{P}\right)-\delta_{P}\left(1 +\frac{E_{\text{max},2}E}{1+E}\right)\right]P, \tag{9}\] \[\frac{dE}{dt} =\left[s_{E}\left(1+\frac{P_{\text{max},1}P}{1+P}\right)\ln\left( \frac{E_{\text{ss}}}{E}\right)-\delta_{E}\left(1+\frac{P_{\text{max},2}P}{1+P} \right)\right]E\] (10) \[=\left[s_{E}\ln\left(\frac{E_{\text{ss}}}{E}\right)-\delta_{E} \frac{1+P+P_{\text{max},2}P}{1+P+P_{\text{max},1}P}\right]\left(1+\frac{P_{ \text{max},1}P}{1+P}\right)E. \tag{11}\]  
Let \(\mathbb{P}_{0}\) denote the open rectangle  
\[\mathbb{P}_{0}=\left\{(P,E):0<P<P_{\text{ss}},\ 0<E<E_{\text{ss}}\right\}\]  
and denote by \(\bar{\mathbb{P}}_{0}\) its closure, \(\bar{\mathbb{P}}_{0}=\left\{(P,E):0\leq P\leq P_{\text{ss}},\ 0\leq E\leq E_{\text{ss}}\right\}\). For \(Q\equiv 0\) the variable \(P\) is bounded above by \(P_{\text{ss}}\) and therefore the compact set \(\bar{\mathbb{P}}_{0}\) is positively invariant under Equations (9) and (11). The dynamical system has the following _trivial equilibrium solutions_ in the boundary of \(\bar{\mathbb{P}}_{0}\): \((0,0)\), \((P_{*},0)\) with  
\[P_{*}=P_{\text{ss}}\exp\left(-\frac{\delta_{P}}{r_{P}}\right)<P_{\text{ss}}\]  
and \((0,E_{*})\) with \(E_{*}\) given by  
\[E_{*}=E_{\text{ss}}\exp\left(-\frac{\delta_{E}}{s_{E}}\right)<E_{\text{ss}}.\]  
In view of Lemmas 1 and 2 these solutions are unstable. While the origin has two unstable modes, the equilibrium points \((P_{*},0)\) and \((0,E_{*})\) are saddles with the respective axes forming the stable manifolds and the unstable modes entering the interior of \(\mathbb{P}_{0}\). It is clear from this that there needs to exist at least one more equilibrium point \((P_{*},E_{*})\) in \(\mathbb{P}_{0}\). **Lemma 4**. : _There are no periodic orbits in \(\mathbb{P}_{0}\)._  
**Proof. ** Changing variables to \(\bar{P}=\ln P\) and \(\tilde{E}=\ln E\), the dynamics transforms into  
\[\frac{d\tilde{P}}{dt} =r_{P}\left(\ln P_{\text{ss}}-\tilde{P}\right)-\delta_{P}\left(1 +E_{\text{max},2}\frac{e^{\tilde{E}}}{1+e^{\tilde{E}}}\right),\] \[\frac{d\tilde{E}}{dt} =s_{E}\left(1+P_{\text{max},1}\frac{e^{\tilde{P}}}{1+e^{\tilde{ P}}}\right)\left(\ln E_{\text{ss}}-\tilde{E}\right)-\delta_{E}\left(1+P_{\text{ max},2}\frac{e^{\tilde{P}}}{1+e^{\tilde{P}}}\right).\]  
The divergence of this vector field is given by  
\[-r_{P}-s_{E}\left(1+P_{\text{max},1}\frac{e^{\tilde{P}}}{1+e^{\tilde{P}}} \right)<0\]  
and thus the result follows from [PERSON]'s negative criterion because of the monotonicity of the logarithm function. The relations defining equilibrium points inside \(\mathbb{P}_{0}\) are  
\[P_{*}=P_{\text{ss}}\exp\left(-\frac{\delta_{P}}{r_{P}}\left(1+E_{\text{max},2} \frac{E_{*}}{1+E_{*}}\right)\right) \tag{12}\]  
and  
\[\frac{dE}{dt} =\left[s_{E}\left(1+\frac{P_{\text{max},1}P}{1+P}\right)\ln\left( \frac{E_{\text{ss}}}{E}\right)-\delta_{E}\left(1+\frac{P_{\text{max},2}P}{1+P} \right)\right]E\]0or, equivalently,  
\[\frac{dE}{dt} =\left[s_{E}\left(1+\frac{P_{\text{max},1}P}{1+P}\right)\ln\left( \frac{E_{\text{ss}}}{E}\right)-\delta_{E}\left(1+\frac{P_{\text{max},2}P}{1+P} \right)\right]E\]1  
Define  
\[\frac{dE}{dt} =\left[s_{E}\left(1+\frac{P_{\text{max},1}P}{1+P}\right)\ln\left( \frac{E_{\text{ss}}}{E}\right)-\delta_{E}\left(1+\frac{P_{\text{max},2}P}{1+P} \right)\right]E\]2  
and  
\[\frac{dE}{dt} =\left[s_{E}\left(1+\frac{P_{\text{max},1}P}{1+P}\right)\ln\left( \frac{E_{\text{ss}}}{E}\right)-\delta_{E}\left(1+\frac{P_{\text{max},2}P}{1+P} \right)\right]E\]3  
Then equilibrium values \(P_{*}\) are fixed points of the function \(\Phi\), \(P=\Phi(P)\), in the interval \([0,P_{\text{ss}}]\). Since \(\Phi(0)>0\), \(\Phi(P_{\text{ss}})<P_{\text{ss}}\), and \(\Phi\) is continuous in \(P\), it follows that there exists at least one solution.