Equilibrium analysis

Let us for the sake of simplicity rearrange the equation shown in the previous section:

\[\begin{eqnarray} \frac{dB}{dt}&=& \underbrace{r_{max}H''_{eDAR}B}_{growth} - \underbrace{dBP}_{infection} \\ \frac{dP}{dt}&=& \underbrace{c\big(1-P_L\big)\mu_pI}_{\text{lytic burst}} + \underbrace{c\mu_iL}_{\text{induct growth}} - \underbrace{mP}_{decay} -\underbrace{dBP}_{infection} \\ \frac{dI}{dt}&=& \underbrace{dBP}_{infection} - \underbrace{\mu_p I}_{lysogenic+burst} \\ \frac{dL}{dt}&=& \underbrace{(r_{max}H''_{eDAR} - \mu_i)L }_{\text{effective growth}} + \underbrace{\mu_p P_LI}_{\text{new lysogens}} \end{eqnarray}\]

We set all equations equal to zero to obtain the equilibrium conditions. The equilibrium concentrations are:

\[\begin{eqnarray} B^{*}&=&\frac{\mu_p}{r H''} I^{*} \\ P^{*}&=&\frac{r H''}{d} \\ I^{*}&=&\frac{\mu_i - r H''}{\mu_p P_L} L^*\\ L^{*}&=&\frac{m \mu_p P_L r H''}{d[c(\mu_i - r H'' (1 - \mu_p P_L) + \mu_p (r H'' - \mu_i)]} \end{eqnarray}\]

The derivations section provides more detain on how to obtain these concentrations.