# Understanding Neutrophil Dynamics during COVID-19 Infection
## 3 Results  
### Asymptotic Analysis Results  
To determine the equilibrium solutions of model Equation (3), we assume that the initial tissue damage stimulus has been eliminated, and hence set \(\alpha=0\). **Proposition 1**. : _Model Equation (3) has a single equilibrium \(S_{1}\) when_  
\[\alpha_{c}<2\beta_{a}\gamma_{c}\frac{1}{k_{a}}\sqrt{\frac{\alpha_{m}}{\gamma_ {m}}\frac{\phi}{\gamma_{c}}k_{a}+1} \tag{5}\]  
_and three equilibria \(S_{1},S_{2}\), and \(S_{3}\) when_  
\[\alpha_{c}>2\beta_{a}\gamma_{c}\frac{1}{k_{a}}\sqrt{\frac{\alpha_{m}}{\gamma_ {m}}\frac{\phi}{\gamma_{c}}k_{a}+1}. \tag{6}\]  
Proof. : We set \(\frac{d\bar{X}}{dt}=0\) for \(\bar{X}=\{\bar{n},\bar{m},\bar{c},\bar{g},\bar{e}\}\) in Equation (3) and solve for components of vector \(\bar{X}\) in terms of \(\bar{a}\). This yields the following expressions \begin{table}
\begin{tabular}{c c c} \hline
**Parameter** & **Description** & **Mild** & **Severe** \\ \hline \(\alpha_{e}\) & CD8 T-cells expansion rate & \(1.2\times 10^{-2}\) & \(1.6\times 10^{-8}\) \\ \(\phi\) & Rate at which macrophages engulf neutrophils & \(4.1\times 10^{-3}\) & \(3.3\times 10^{-4}\) \\ \(\
u\) & Rate of neutrophil apoptosis & \(0.35\) & \(0.46\) \\ \(\alpha_{c}\) & Neutrophil recruitment rate & \(3.44\) & \(4.28\) \\ ssq & Sum of squares & \(2.84\) & \(4.14\) \\ \hline \end{tabular}
\end{table} 
Table 2: Best parameter estimates from fitting model Equation (3) to data from mild and severe COVID-19 cases. \[ \begin{split}\bar{n}&=\frac{\bar{a}^{3}(\gamma_{\alpha} \gamma_{g}\alpha_{m}\phi k_{a})+\bar{a}\big{(}\bar{a}^{2}+\beta_{a}^{2}\big{)} \gamma_{c}\gamma_{m}\gamma_{g}\gamma_{a}}{\bar{a}^{3}(\gamma_{a}\alpha_{m}vk_{g }\phi k_{a})+(\bar{a}^{2}+\beta_{a}^{2})\gamma_{c}\gamma_{m}v_{g}},\\ \bar{m}&=\frac{\alpha_{m}}{\gamma_{m}}\frac{\gamma_{a }k_{a}}{\gamma_{c}}\frac{\bar{a}^{2}}{(\bar{a}^{2}+\beta_{a}^{2})},\\ \bar{c}&=\frac{\gamma_{a}k_{a}}{\gamma_{c}}\frac{ \bar{a}^{2}}{(\bar{a}^{2}+\beta_{a}^{2})},\\ \bar{g}&=\frac{s_{g}}{\gamma_{g}}+\frac{\alpha_{m}} {\gamma_{m}}\frac{\gamma_{a}k_{g}\phi k_{a}}{\gamma_{c}\gamma_{g}}\frac{\bar{ a}^{3}}{(\bar{a}^{2}+\beta_{a}^{2})},\\ \bar{e}&=\frac{s_{e}}{\gamma_{e}}+\frac{\bar{a}^{2 }(\alpha_{e}\alpha_{m}\gamma_{a}\gamma_{g}k_{a})}{\bar{a}^{3}(\gamma_{e}\alpha_ {m}\beta_{e}\gamma_{a}k_{g}\phi k_{a})+(\bar{a}^{2}+\beta_{a}^{2})\gamma_{e} \gamma_{c}\gamma_{m}(s_{g}\beta_{e}+\gamma_{g})}.\end{split}  \tag{7}\]  
Next, we solve for \(\bar{a}\) by substituting \(\bar{c}\) and \(\bar{m}\) into  
\[\frac{d\bar{a}}{dt}=\alpha_{c}\bar{c}-(\gamma_{a}\bar{a}+\phi\bar{m}\bar{a})=0.\]  
Simplifying yields  
\[0=\bar{a}\bigg{[}\bar{a}^{2}(\gamma_{a}\gamma_{c}\gamma_{m}+k_{a}\phi\alpha_{m }\gamma_{a})-\bar{a}(\alpha_{c}k_{a}\gamma_{a}\gamma_{m})+\beta_{a}^{2}\gamma_ {a}\gamma_{c}\gamma_{m}\bigg{]}, \tag{8}\]  
which has one solution  
\[\bar{a}_{1}=0, \tag{9}\]  
when  
\[\alpha_{c}<2\beta_{a}\gamma_{c}\frac{1}{k_{a}}\sqrt{\frac{\alpha_{m}}{\gamma _{m}}\frac{\phi}{\gamma_{c}}k_{a}+1},\]  
and three solutions  
\[ \begin{split}\bar{a}_{1}&=0,\\ \bar{a}_{2,3}&=\frac{\alpha_{c}\gamma_{m}k_{a}\pm \sqrt{\gamma_{m}}\sqrt{\alpha_{c}^{2}\gamma_{m}k_{a}^{2}-4\beta_{a}^{2}\gamma_ {c}^{2}\gamma_{m}-4\alpha_{m}\beta_{a}^{2}\gamma_{c}\phi k_{a}}}{2(\gamma_{c} \gamma_{m}+\alpha_{m}\phi k_{a})},\end{split}  \tag{10}\]  
when  
\[\alpha_{c}>2\beta_{a}\gamma_{c}\frac{1}{k_{a}}\sqrt{\frac{\alpha_{m}}{\gamma _{m}}\frac{\phi}{\gamma_{c}}k_{a}+1}. \tag{11}\]  
Since \(\{\bar{n},\bar{m},\bar{c},\bar{g},\bar{e}\}\) are positive when \(\bar{a}\) is positive, we obtain that a single equilibrium solution called _no-inflammation_ equilibrium exists when condition Equation (5) holds,  
\[S_{1}=(0,0,0,0,\frac{s_{g}}{\gamma_{g}},\frac{s_{e}}{\gamma_{e}}).\]  
Conversely, three equilibria solutions exist when condition Equation (6) holds. They are \(S_{1}\) (as before) and  
\[S_{2,3}=(\bar{n},\bar{m},\bar{a},\bar{c},\bar{g},\bar{e})_{2,3}\]  
given by Equations (7) and (10) and called _chronic inflammation_ equilibria. This concludes our proof.