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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.

The decreasing production term represents negative feedback: low red-cell abundance stimulates production, whereas high abundance suppresses it. The linear loss term models removal at constant per-cell rate. SetThen $u^{\prime }=f(u)-\delta u$, with $f(u)=1/(1+u^n)$. On the physical half-line $u\ge 0$, $f$ decreases from one to zero while $\delta u$ increases from zero to infinity. Their unique intersection $u_{\ast }>0$ is the steady state. Its linear derivative is $f^{\prime }(u_{\ast })-\delta <0$, so it is asymptotically stable.

For small $\delta$, the intersection lies at large $u$, where $f\sim u^{-n}$; for large $\delta$ it lies at small $u$, where $f\sim 1$. HenceImplicit differentiation of $f(u_{\ast })=\delta u_{\ast }$ gives $d{u}_{\ast }/d\delta =u_{\ast }/[f^{\prime }(u_{\ast })-\delta ]<0$. The plot joins these two decreasing asymptotic regimes.

The delay $t_0$ represents the time between feedback sensing and the appearance of newly produced mature cells. In dimensionless time it is $d=F_0{t}_0/\theta$, so the equation becomesIt requires a nonnegative initial history on $[-d,0]$, not just one initial value. Its constant equilibrium is unchanged. Put $a=-f^{\prime }(u_{\ast })=n{u}_{\ast }^{n-1}/(1+u_{\ast }^n{)}^2>0$. A small perturbation obeys $v^{\prime }=-\delta v-av(\tau -d)$, whose characteristic equation isThere is no real root with $\sigma \ge 0$, because all three terms would have positive sum; zero is never a root either. Starting from the stable zero-delay case, loss of linear stability can therefore occur only when a complex conjugate pair crosses the imaginary axis at a nonzero frequency, the linear signature of a Hopf bifurcation.

For completeness, substituting $\sigma =i\omega$ gives $\delta +a\cos (\omega d)=0$ and $\omega =a\sin (\omega d)$. Such a crossing requires $a>\delta$, withThe first crossing is transversal: $\mathrm{Re}(d\sigma /dd)={\omega }^2/[(1+\delta d{)}^2+{\omega }^2{d}^2]>0$. If $a\le \delta$, the magnitude relation rules out roots in the closed right half-plane for every delay. A nondegenerate nonlinear Hopf conclusion additionally uses the usual bifurcation hypotheses. This is delayed negative feedback in red-cell production.

Glacier sliding is motion of the ice base relative to its bed, distinct from internal ice deformation. Near a melting bed, basal water, pressure-melting/refreezing around obstacles and creep past bed roughness can permit motion; a cold frozen bed can suppress it. A simple sliding law is increasing in basal traction, for example $u_b\propto {\tau }_b^m$ on the positive-traction branch, with resistance also affected in richer models by roughness and basal water pressure.

Slow ice motion has negligible inertial acceleration. In a long shallow glacier, leading vertical stress is hydrostatic and longitudinal stress gradients are neglected relative to transverse shear gradients. Resolving gravity downslope and normally yields the two stated balances. This approximates a two-dimensional slab/valley centreline; lateral drag, rapidly varying geometry and strong longitudinal-stress regions need additional terms.

With a stress-free surface,Put $K=\rho g\sin \alpha$ and $D=1-\cot \alpha h_x$, assuming positive driving shear for the stated power law. Integrating from the bed givesThe surface speed is $u_s=u_b+A(KD{)}^n{h}^{n+1}/(n+1)$. Mass conservation, with $a$ measured as ice-equivalent thickness accumulation, is $\boxed{h_t+q_x=a}$.

Choose $h=H\hat{h}$, $x=l\hat{x}$, $u_b=U{\hat{u}}_b$ and $t=(l/U)\hat{t}$, with $U=(KH/R{)}^m$. Scale accumulation with $HU/l$. The dimensionless basal law and parameters areThe resulting flux is exactly $h{u}_b+\delta (1-\mu h_x{)}^n{h}^{n+2}/(n+2)$.

When internal deformation is small and slope corrections are uniformly small, $u_b\simeq h^m$ andFor a steady accumulation profile and prescribed upstream flux $q_{\mathrm{in}}$, $\boxed{h_{\ast }(x)=[q_{\mathrm{in}}+{\int }_0^{x}a(\xi )d\xi {]}^{1/(m+1)}}$. For constant $a$ and zero upstream flux this is $(ax{)}^{1/(m+1)}$ on the region of positive flux.

A small thickness disturbance $\zeta$ satisfies ${\zeta }_t+[c(x)\zeta {]}_x=0$ with $c=(m+1)h_{\ast }^m$. Hence it travels down-glacier along $dx/dt=c>0$, with amplitude changing as $d\zeta /dt=-c^{\prime }\zeta$. Since surface speed is $u_s\simeq u_b=h_{\ast }^m$ in this sliding-dominated limit,The thickness wave therefore outruns individual ice particles. This is the sliding-dominated glacier kinematic wave.