In the model considered here the population is divided into susceptibles (S), infectives (I), isolated or quarantined individuals (Q), and recovered individuals (R), for whom permanent immunity is assumed. Let N denote the total population i.e. $N=S+I+Q+R$, and let $A = S + I + R$ denote the active (nonisolated) individuals. The model takes the form:
$$\begin{align} \frac{dS}{dt}&=\mu N-\mu S− \sigma S \frac{I}{A}\\ \frac{dI}{dt}&=-(\mu + \gamma)I+\sigma S \frac{I}{A}\\ \frac{dQ}{dt}&=-(\mu+\xi)Q+\gamma I\\ \frac{dR}{dt}&=−\mu R+\xi Q \\ A&=S+I+R \end{align}$$
All newborns are assumed to be susceptible. $\mu$ is the per capita mortality rate, $\sigma$ is the per capita infection rate of an average susceptible individual provided everybody else is infected, $\gamma$ is the rate at which individuals leave the infective class, and $\xi$ is the rate at which individuals leave the isolated class; all are positive constants.
Question: Rescale the model by: $\tau=\sigma t$, $u=S/A$, $y=I/A$, $q=Q/A$, $z=R/A$. Rearrange your new model as follows:
$$\begin{align} \dot{y}&=y(1-\nu -\theta -y-z+\theta y-(\nu +\zeta)q)\\ \dot{q}&=(1+q)(\theta y-(\nu+\zeta)q) \\ \dot{z}&=\zeta q-\nu z +z(\theta y-(\nu+\zeta)q). \end{align}$$
Express the new parameters in terms of the old parameters. Check that all the new parameters and variables are dimensionless.
The new parameters are $\nu$, $\theta$, and $\zeta$. The answer should be $\nu=\frac{\mu}{\sigma}$, $\theta=\frac{\gamma}{\sigma}$, and $\zeta=\frac{\xi}{\sigma}$. I am unsure on how to rescale the original model. It has 4 equations, but the new model only has 3 equations. Can someone explain how I should go about rescaling the original model?
What I did not understand was that $$\begin{align} \frac{dI}{dt}&=\frac{dI}{d\tau}\frac{d\tau}{dt}\\ \frac{dI}{dt}&=\dot{I}\sigma \end{align}$$
Since $I=Ay$ we have $\frac{dI}{dt}=(\dot{A}y+A\dot{y})\sigma=\left(\frac{1}{\sigma}\frac{dA}{dt}+A\dot{y}\right)\sigma=\frac{dA}{dt}+A\dot{y}\sigma$.
Using the fact that A=S+I+R, we have $\frac{dA}{dt}=\mu Q-\gamma I+\xi Q$, so $\frac{dI}{dt}=(\mu Q-\gamma I+\xi Q)y+A\dot{y}\sigma$.
From our original model, $\frac{dI}{dt}=-(\mu+\gamma)I+\sigma S\frac{I}{A}$, so by equating our two equations and then solving for $\dot{y}$ we have
$$-(\mu+\gamma)I+\sigma S\frac{I}{A}=(\mu Q-\gamma I+\xi Q)y+A\dot{y}\sigma$$
$$\dot{y}=\frac{1}{A\sigma}\left(-(\mu+\gamma)I+\sigma S\frac{I}{A}-(\mu Q-\gamma I+\xi Q)y\right)$$
$$=-\left(\frac{\mu}{\sigma}+\frac{\gamma}{\sigma}\right)\frac{I}{A}+\frac{S}{A}\frac{I}{A}-\frac{\mu}{\sigma}\frac{Q}{A}y+\frac{\gamma}{\sigma}\frac{I}{A}y-\frac{\xi}{\sigma}\frac{Q}{A}y$$
Using the proper substitutions, we have $$\dot{y}=-\left(\frac{\mu}{\sigma}+\frac{\gamma}{\sigma}\right)y+uy-\frac{\mu}{\sigma}qy+\frac{\gamma}{\sigma}y^2-\frac{\xi}{\sigma}qy$$
By factoring, rearranging and using the fact that u=1-y-z, we have $$\dot{y}=y\left(1-\frac{\mu}{\sigma}-\frac{\gamma}{\sigma}-y-z+\frac{\gamma}{\sigma}y-\left(\frac{\mu}{\sigma}+\frac{\xi}{\sigma}\right)q\right)$$
and can easily see that $\nu=\frac{\mu}{\sigma},\theta=\frac{\gamma}{\sigma}, \zeta=\frac{\xi}{\sigma}$