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Resting cells enter proliferation at rate $\beta (N)N$. A proliferating cell takes a fixed cycle time $\tau$ to divide and survives that cycle with probability $e^{-\gamma \tau }$ under a constant death hazard $\gamma$. Thus exit from the proliferating compartment at time $t$ is $e^{-\gamma \tau }\beta (N(t-\tau ))N(t-\tau )$, and it returns twice as many daughter cells to the resting population. Resting-cell loss at rate $\delta N$ is separate.

More explicitly, the number currently in the proliferating stage isfor a consistent initial history. Differentiation gives the stated $P$ equation. Counting resting-cell entry, loss and daughter-cell return gives its $N$ equation. Negative feedback in $\beta (N)$ suppresses recruitment when the resting population is high; its Hill exponent gives a steep response around the concentration scale $\theta$.

Set $\boxed{u=N/\theta ,\hat{t}=t/\tau ,u_1=u(\hat{t}-1)}$ and defineThen $1+ϵ\mu =2e^{-\gamma \tau }$, so the resting-cell equation becomes exactlyThe nonzero steady state solves $u=\mu g(u)$, or $1=\mu b/(1+u^n)$ for positive $u$. The latter right side decreases strictly from $\mu b$ to zero. Thus when $\mu b>1$ there is exactly one positive equilibrium,Zero is also an equilibrium; negative populations are outside the biological domain. The same result is visible by intersecting the line $u$ with the hump-shaped $\mu g(u)$.

Put $d_g=g^{\prime }(u_{\ast })=b[1+(1-n)u_{\ast }^n]/(1+u_{\ast }^n{)}^2$. Linearization and a mode $e^{\sigma \hat{t}}$ giveHere ${\gamma }_{\mathrm{ch}}$ is the coefficient called $\gamma$ in the source's characteristic equation; it is not the proliferating-cell death hazard. Any characteristic root with positive real part makes the equilibrium linearly unstable. A nonzero imaginary crossing satisfies $\alpha +{\gamma }_{\mathrm{ch}}\cos \omega =0$ and $\omega -{\gamma }_{\mathrm{ch}}\sin \omega =0$; these are its Hopf conditions, subject to nondegeneracy.

On slow time $T=ϵ\hat{t}$, the unit delay becomes $u_1=u(T-ϵ)=u-ϵu_T+O({ϵ}^2)$. Taylor expansion gives $g(u_1)-g(u)=-ϵg^{\prime }(u)u_T+O({ϵ}^2)$. Dividing the equation by $ϵ$ therefore yieldsThis approximation requires slow smooth variation and a denominator bounded away from zero; a vanishing denominator marks failure of this slow reduction. On the fast time $\hat{t}=O(1)$, omit only the explicit order-$ϵ$ terms, retaining the finite delay:Replacing the delayed argument by the present value on that fast scale would incorrectly erase the dynamics. These are the slow and fast limits of a delayed stem-cell model.