 Slow and fast limits of a delayed stem-cell model

For resting-cell recruitment $g(u)=bu/(1+u^n)$, a positive constant equilibrium exists precisely when $\mu b>1$, with $u_{\ast }=(\mu b-1{)}^{1/n}$. On slow time $T=ϵt$, expand the delayed value as $u(T-ϵ)=u-ϵu_T+O({ϵ}^2)$. Substitution gives $(1+g^{\prime }(u))u_T=\mu g(u)-u$ to leading order. This reduction requires smooth slow motion and $1+g^{\prime }(u)$ bounded away from zero.

On the original fast time, only the explicit order-$ϵ$ terms are dropped, leaving $u^{\prime }=g(u(t-1))-g(u(t))$. The unit delay remains order one and cannot be Taylor-expanded on that scale. For the biological two-compartment derivation, a fixed cell-cycle time $\tau$ and proliferating death hazard ${\gamma }_d$ give survival $e^{-{\gamma }_d\tau }$ and daughter-cell gain twice the surviving exit rate. Scaling gives $b={\beta }_0\tau$, $ϵ=\delta \tau$ and $1+ϵ\mu =2e^{-{\gamma }_d\tau }$. Thus small resting-cell loss over one cycle creates the slow-fast structure.