Certainly, if that would help. The basic form is:
\frac{dT}{dt} = 1 - \theta_{12} T I \\
\frac{dI}{dt} = \theta_{12} T I - E I \\
\frac{dE}{dt} = \theta_3 - \theta_4 E \frac{Q^n}{1+Q^n} \\
\frac{dQ}{dt} = \theta_5 I \\
T(0) = \theta_6 e^{-\theta_{12}} \\
I(0) = \theta_6 \left(1-e^{-\theta_{12}} \right) \\
E(0) = 0 \\
Q(0) = 0.
The variable I is the problem variable that goes negative (or becomes very close to zero) and its log is the main observable.