Maximum value of modified Hill functions

The modified Hill function reads:

\[\begin{equation*} H'(eDAR)=\frac{(eDAR - eDAR_{min})^{n}}{K^n + \alpha (eDAR - eDAR_{min})^{n}} \, . \end{equation*}\]

With

\[\begin{equation*} \alpha=\frac{(eDAR_{max} - eDAR_{min})^{n} - 2(1 - eDAR_{min})^{n}}{(eDAR_{max}- eDAR_{min})\ ^{n} - (1 - eDAR_{min})^{n}} \end{equation*}\]

and

\[\begin{equation*} K=(2-\alpha)^{1/n}(1-eDAR_{min}) \end{equation*}\]

Here, we prove that \(H'(eDAR=eDAR_{max})=1\). For the sake of simplicity let us introduce some compact notation:

\[\begin{eqnarray*} eDAR - eDAR_{min}&=&\Delta \\ eDAR_{max} - eDAR_{min}&=&\Delta_{tot} \\ 1 - eDAR_{min}&=&\Delta_{1} \end{eqnarray*}\]

Then we have that:

\[\begin{equation*} K^n=\frac{\Delta^n_{tot} \Delta^n_{1} }{\Delta^n_{tot} - \Delta^n_{1}} \end{equation*}\]

And the modified Hill function reads now:

\[\begin{equation*} H'(eDAR)=\frac{\Delta^n}{\frac{\Delta^n_{tot} \Delta^n_1 + (\Delta^n_{tot} - 2 \Delta^n_1) \Delta^n}{\Delta^n_{tot} - \Delta^n_1}}= \frac{(\Delta^n_{tot} - \Delta^n_1)\Delta^n}{\Delta^n_{tot} \Delta^n_1 + (\Delta^n_{tot} - 2 \Delta^n_1) \Delta^n} \end{equation*}\]

If \(eDAR=eDAR_{max}\), then \(\Delta=\Delta_{tot}\) which gives \(H'(eDAR=eDAR_{max})=1\).

Alternatively, if \(eDAR=1\), then \(\Delta=Delta_1\), which gives \(H'(eDAR=1)=0.5\).

Equilibrium analysis

The equilibrium analysis is done by taking the ODEs and setting them to zero:

\[\begin{eqnarray} \frac{dB}{dt}&=&0; \underbrace{r_{max}H''_{eDAR}B}_{growth} - \underbrace{dBP}_{infection}& = &0 \\ \frac{dP}{dt}&=&0; \underbrace{c\big(1-P_L\big)\mu_pI}_{\text{lytic burst}} + \underbrace{c\mu_iL}_{\text{induct growth}} - \underbrace{mP}_{decay} - \underbrace{dBP}_{infection} &=& 0 \\ \frac{dI}{dt}&=&0; \underbrace{dBP}_{infection} - \underbrace{\mu_p I}_{lysogenic+burst}& =& 0 \\ \frac{dL}{dt}&=&0; \underbrace{(r_{max}H''_{eDAR} - \mu_i)L }_{\text{effective growth}} + \underbrace{\mu_p P_LI}_{\text{new lysogens}} &=&0 \label{eq:equilibrium} \end{eqnarray}\]

From Eq. \ref{eq:equilibrium} .(1) we get:

\[\begin{equation} P^{*}=\frac{rH''}{d} \label{eq:phage_eq} \end{equation}\]

From Eq. \ref{eq:equilibrium} .(3) we have:

\[\begin{equation} dBP=\mu_p I \label{eq:infected_sensitive} \end{equation}\]

Substituting Eq. \ref{eq:infected_sensitive} in Eq. \ref{eq:equilibrium} .(2):

\[\begin{equation} I^{*}=\frac{1}{c(1 - \mu_p P_L) - \mu_p} \big[ m P^{*} - c \mu_i L^*\big] \label{eq:lysogens_1} \end{equation}\]

From Eq. \ref{eq:equilibrium} .(4):

\[\begin{equation} I^{*}=\frac{\mu_i - rH''}{\mu_p P_L} L^* \label{eq:lysogens_2} \end{equation}\]

Putting Eq. \ref{eq:lysogens_1} and Eq. \ref{eq:lysogens_2} together and rearranging terms, we get the equilibrium concentration of lysogens:

\[\begin{equation} L^{*}=\frac{m \mu_p P_L r H''}{d[c(\mu_i - r H'' (1 - \mu_p P_L) + \mu_p (r H'' - \mu_i)]} \label{eq:lysogens_eq} \end{equation}\]

Getting the equilibrium concentration of infected bacteria \(I^*\) from Eq. \ref{eq:lysogens_eq} is straightforward:

\[\begin{equation} I^{*}=\frac{\mu_i - r H''}{\mu_p P_L} L^* . \label{eq:infected_eq} \end{equation}\]

Similarly, for the equilibrium concentration of sensitive bacteria \(B^*\):

\[\begin{equation} B^{*}=\frac{\mu_p}{r H''} I^* . \end{equation} \label{eq:sensitive_eq}\]