Resolve gravity into components $g\sin \alpha$ down the bed and $-g\cos \alpha$ normal to it. For slow incompressible flow, inertia is negligible and Cauchy momentum equation reduces to the divergence of the stress tensor plus the gravitational force density being zero. Write the dominant shear component as $\tau ={\sigma }_{xz}$ and the leading normal stresses as $-p$.

The shallow-ice approximation assumes a small thickness-to-length ratio. Vertical variation of the shear is retained in the along-bed equation, while longitudinal deviatoric-stress gradients and the shear contribution to the normal equation are higher order. The leading balances are consequentlyThe second is a hydrostatic pressure balance; the first balances gravitational driving and the surface-induced pressure gradient against vertical shear. This approximation requires shallow geometry and sufficiently slow flow, not a Newtonian constitutive equation for ice.

The water mass conservation equation inherited from the earlier model is $ϵh_t+(hu{)}_x=0$. It is essential: without it the printed displayed equations are not closed. Take the slow bed branch $ϵ=0$, and perturb the uniform state by $h=1+\hat{h}{e}^{\lambda t+ikx}$, $u=1+U{e}^{\lambda t+ikx}$, $c=1+C{e}^{\lambda t+ikx}$, $s=S{e}^{\lambda t+ikx}$ and $\eta =1+\hat{\eta }{e}^{\lambda t+ikx}$. For $k\ne 0$, mass conservation gives $\hat{h}=-U$ and $\hat{\eta }=S-U$.

Put $m=E^{\ast \prime }(1)+1$; the extra $1$ comes from linearizing $c/u$. The linearized momentum and suspended-sediment equations areEliminating $U,C$ gives the dispersion relationAway from $F=1$, for fixed nonzero $k$ and negligible $\delta$,Thus growth requires $m/(F^2-1)>0$. For $F>1$, any $E^{\ast \prime }(1)>-1$ gives an unstable branch; for $F<1$ growth would require $E^{\ast \prime }(1)<-1$. A physically increasing erosion function therefore gives the supercritical instability in this simplified model, rather than proving growth of ordinary subcritical dunes. Retaining drag giveswhich shows its stabilizing correction and the need to reconsider the limit near critical flow.

The negative bed-to-surface ratio for subcritical flow means bed crests lie beneath surface troughs: the dune phase relation. Its positive sign for supercritical flow gives aligned crests: an antidune. In the unstable supercritical case the imaginary part gives upstream migration, since $-\mathrm{Im}\lambda /k=-m/[(F^2-1)(1+k^2)]$. The figure shows these two phase relations. They identify the bedforms but do not by themselves establish growth. The qualitative limitation of the suspended-sediment mechanism is also discussed in the official ^[ref].

Let the original area be $A_0>0$ and the increased upstream area be $A_1>A_0$. At $ϵ=0$, the conservation law flux $q(A)=A^{5/4}$ is strictly convex. Characteristics behind the increase travel faster than those ahead, forming an admissible shock wave. Rankine-Hugoniot conditions give its speedAt $x=1$ the idealized hydrograph jumps from $Q_0=A_0^{5/4}$ to $Q_1=A_1^{5/4}$ at $t=1/c_s$. With positive $ϵ$, the diffusion term spreads the rise over a finite interval. The hydrograph is rounded, with an earlier small response and a later approach to the new discharge, rather than an instantaneous jump.

The physical cause is the longitudinal pressure gradient: a local increase in depth influences neighbouring sections and redistributes water before pure downstream transport would deliver a discontinuity. For a small increase, linearization about $A_0$ has speed $c=\frac{5}{4}{A}_0^{1/4}$ and diffusivity $D=ϵA_0^{3/4}/4$. The smooth curve in the figure is the exact half-line step response for this linearized advection-diffusion equation; it illustrates the shape without claiming to solve a large nonlinear jump exactly.

First neglect inertia relative to the $O(ϵ)$ pressure correction, requiring $F\ll 1$ for smooth fields on these scales. The positive velocity isMultiplying by area gives the discharge $Q=A^{5/4}-(ϵ/4)A^{3/4}{A}_x+O({ϵ}^2)$. HenceThe leading equation is a nonlinear kinematic wave equation. The pressure correction has positive diffusion coefficient for $A>0$ and smooths steep gradients. Dropped terms are $O({ϵ}^2)$ and inertial terms $O(ϵF)$ on the assumed scales; sufficiently thin fronts require a separate balance rather than an unrestricted Taylor expansion.

Write $R=k_R\sqrt{A}$, $\bar{h}=k_H\sqrt{A}$, where the given geometry givesTake a Chézy drag law $\tau =\rho c_{d}u|u|$ and restrict to downstream flow $u>0$. Choose an area scale $A_0$, length scale $L$, and speed scale $U_0$ satisfyingThis makes gravitational driving and drag equal in the reference uniform steady flow. Dividing the printed momentum equation by $gS$ givesMass conservation retains $A_t+(Au{)}_x=0$. In this notation $F$ is the square of the velocity-to-gravity-wave-speed ratio based on the chosen depth scale; authors sometimes instead denote its square root by Froude number. If the actual triangular pressure head from the preceding solution is used, replace $k_H$ by $2k_H$ in both parameter definitions. The displayed dimensionless form remains the same, with consistently redefined parameters.

Assume constant mass density, a prismatic channel, approximately uniform cross-sectional velocity $u$, a hydrostatic pressure distribution, small bed slope $S$, and no lateral inflow. Integrating mass conservation over a slice of length $dx$ gives $\rho A_{t}dx=\rho Au{|}_x-\rho Au{|}_{x+dx}$, hence $A_t+(Au{)}_x=0$.

Let $I(A)={\int }_A(H-z)dA$ be the hydrostatic first moment, where $H$ is the actual free-surface depth. The downstream momentum balance isUsing $A_t+(Au{)}_x=0$ reduces its left side to $A(u_t+u{u}_x)+g{I}_x$. For a fixed prismatic cross-section, $dI/dH=A$, so $I_x/A=H_x$. Therefore the exact depth-averaged result isHere hydraulic radius converts perimeter shear stress into force per unit fluid volume. The printed notation $\bar{h}$ yields the requested equation if it denotes this hydraulic pressure head, whose derivative equals $H_x$.

The geometrical definition printed in the next part needs care. Triangular-channel hydrostatic pressure head gives the factor explicitly. For the stated symmetric triangular notch, the actual depth is $H=(\ell /2)\sin \theta$, whereas its printed $\bar{h}=(\ell /4)\sin \theta$ is $H/2$. Indeed $A=H^2\cot \theta$ and $I=AH/3$, giving $I_x/A=H_x=2{\bar{h}}_x$. Thus the equation with precisely the printed geometrical $\bar{h}$ has pressure term $-2g{\bar{h}}_x$. A first-principles derivation cannot remove that factor. The subsequent intended scaling can be carried out algebraically using the printed convention, as below, or with the pressure scale doubled for the actual triangular channel.

Initially $a=1/2$ and $T=T_{+}/T_0>1$, so $\epsilon T_t=1-T^4<0$. On the fast time $s=t/\epsilon$, albedo changes by only $O(\epsilon )$ andThus temperature rapidly falls towards $1$. At the same time $a_t=a^{\ast }(T)-1/2<0$ while $T>1$, which seeds selection of the warm branch. Afterwards the approximate slow manifold is $T^4=2(1-a)$: decreasing albedo makes temperature rise slowly back towards $T_{+}/T_0$. The preceding figure separates these two time scales; a single time axis would compress the initial fall into a nearly vertical segment.

In the sharp-transition regime, after selection of the warm branch $a^{\ast }(T)\simeq 1/4$. Solving the linear ordinary differential equation with the leading matched value $a(0^{+})=1/2$ givesThese are outer solutions: their initial temperature is $1$, so the fast cooling layer is required to satisfy the actual initial temperature.

There is a useful qualification to the two-limit argument. For fixed finite $\mu >2$, put $x=1-2a$ on the slow manifold. Then $x_t=-x+\frac{1}{2}\tanh (\mu x)$, and $x=0$ is an unstable equilibrium with growth rate $\mu /2-1$. The small positive $O(\epsilon )$ seed from the fast layer can therefore produce a delay of order $\log (1/\epsilon )/(\mu /2-1)$. The simple exponential approximation assumes that the sharp albedo transition is sufficiently fast to make this delay negligible on the displayed slow scale. It is not a uniform approximation for every fixed $\mu >2$ as $\epsilon \to 0$.

Use dimensional quantities $T_d,t_d$ and set $T_d=T_{0}T$, $t_d=t_{0}t$. Dividing the energy balance by $\sigma \gamma T_0^4$ and the albedo equation by its time scale givesThus $\epsilon$ is the ratio of the atmospheric thermal response time to the much longer albedo response time. Converting kilometres to metres and years to seconds gives $C=10^7\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, $T_0\simeq 266.3\mathrm{K}$, $t_0\simeq 3.156\times 10^{10}\mathrm{s}$, andThe thermal time is about half a year, compared with a millennium for albedo adjustment. This is a singular perturbation: setting $\epsilon =0$ removes the independent initial condition for temperature.

Set $x=T^4/T_0^4-1$. Since $\sigma \gamma T_0^4=Q/8$, the steady state energy balance is $2(1-a_0)=1+x$. Substitution givesThe two sides have slopes $\mu$ and $2$ at the origin. For $x>0$, $\tanh (\mu x)$ is strictly concave, its initial slope is $\mu$, and it remains below $1$. Consequently no positive root exists for $\mu \le 2$, while for $\mu >2$ there is exactly one positive root $x_{+}\in (0,1/2)$. Oddness supplies the corresponding negative root. At $\mu =2$ the only root is still zero. Hence multiple steady temperatures occur precisely when $\mu >{\mu }_c=2$.

The largest root corresponds monotonically to the largest positive absolute temperature:For large $\mu$, the positive root tends to $1/2$, so $a_{+}\simeq 1/4$ and $T_{+}/T_0\simeq (3/2{)}^{1/4}$. The negative branch has larger albedo and lower temperature.

The heat capacity per unit horizontal area is $C=\rho c_{p}d$: mass density times specific heat capacity times atmospheric depth. Thus $C\dot{T}$ is the rate of stored thermal energy per unit area. The solar constant $Q$ is the incident radiative flux on a surface normal to sunlight; the ratio of the Earth's intercepting disk area to its spherical area is $1/4$. The absorbed fraction is $1-a$, where albedo $a$ is the reflected fraction. The outgoing radiative flux is $\sigma \gamma T^4$, a Stefan–Boltzmann law with effective radiation factor $\gamma$ incorporating the simplified greenhouse effect. Every term therefore has units of power per area.

The albedo curve decreases smoothly, passes through $(T_0,1/2)$, and approaches $1/4$ at high temperature. At $T=0$ it is $(2+\tanh \mu )/4$, close to $3/4$ when $\mu$ is large. Its derivative isAn ice-covered planet reflects more sunlight than exposed land or ocean. Warming removes reflective ice, lowers albedo, and increases absorbed radiative flux: this is ice-albedo feedback. The parameter $\mu$ controls the sharpness of this transition; the formula is an idealized global response, rather than a local melting law.

The positive-part law $\mathcal{V}(P)=G[P-P_0{]}_{+}$ is a threshold-linear approximation to the ventilatory response. Below the apnoeic threshold $P_0$, this model assigns zero ventilation; above it, each pressure increment produces a proportional ventilation increase, with gain $G>0$. Taking the positive part ensures a nonnegative ventilation rate rather than the unphysical negative rate that unrestricted linear extrapolation would give. It is a local idealization of the response, not a claim that actual physiology lacks other respiratory drives or saturates nowhere. In the delayed model this response is evaluated at the delayed pressure.

Chemoreceptors respond to carbon-dioxide levels, but feedback is delayed by blood transport to sensing sites, neural processing and the response of respiratory muscles. A fixed positive delay $\tau$ approximates their combined effect. The ventilation at the present instant therefore depends on the earlier pressure, $\dot{V}(t)=\mathcal{V}(P(t-\tau ))$, not just on $P(t)$. This delayed negative feedback can overcorrect after the pressure has already changed, providing a mechanism for oscillatory breathing. The delay equation requires a pressure history over an interval of length $\tau$, rather than a single initial value.

In delayed carbon-dioxide ventilation feedback, $P$ is the effective carbon-dioxide partial pressure of the lumped body compartment. Its stored carbon-dioxide amount is proportional to $P$, with effective capacitance $K$, so $KdP/dT$ is its accumulation rate. Metabolism produces carbon dioxide at rate $M$. Ventilation removes it at a rate proportional to partial pressure times ventilation, $P\dot{V}$, with gas-conversion constants absorbed into the model coefficients. The equation is therefore a one-compartment carbon-dioxide balance: production minus respiratory removal equals accumulation. It idealizes mixing and treats metabolic production as constant over the time scale of interest.

In a relaxation cycle, cytoplasmic calcium rises slowly on the left attracting branch while total calcium is accumulated, then rises rapidly during store discharge along a constant-$w$ fast path to the right branch. It subsequently decreases on the right slow branch as extrusion dominates input, and finally drops rapidly during store uptake back to the left branch. The result is a repeated pulse with slow sections joined by sharp transitions, rather than a sinusoid except near a small-amplitude Hopf bifurcation.

For explicit time labelling, the compatible scaling $u=c/K_1$, $v=c_s/K_2$ gives $\tau =kt$, $\gamma =K_2/K_1$, $ϵ=k{K}_2/V_2$, $\mu =r/(k{K}_1)$, $\beta =V_1/V_2$ and $\delta =k_s{K}_2/V_2$. Hence an interval $\Delta \tau$ corresponds to physical duration $\Delta t=\Delta \tau /k$. Slow-branch times can be computed by integratingin the direction of motion; fast-jump durations are omitted at leading order. The plot uses that quadrature, with vertical lines representing the limiting rapid transitions. Finite $ϵ$ rounds those lines and changes the precise amplitude and period, especially near folds.

For Hopf thresholds in a two-pool calcium model, with the unique equilibrium parametrized by $u=\mu$, locate changes of stability by solving the two equationsIf there are two simple roots ${\mu }_{-}<{\mu }_{+}$ with positive trace between them, the equilibrium is a repeller throughout that interval, since its determinant stays positive. At a nondegenerate crossing a Hopf bifurcation is possible, with linear frequency $[-f_v/ϵ{]}^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the trace calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V^{\prime }=-1/\gamma$.

Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v^{\prime }\le (\beta -\delta v)/ϵ$ bounds $v$, and the $u$ equation then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen forward invariant. For an initial point distinct from the repelling unique equilibrium, its omega-limit set cannot contain that equilibrium or a boundary equilibrium. The Poincaré-Bendixson theorem supplies a periodic orbit. In the strong separation regime it follows the two attracting branches and jumps between them, a relaxation oscillation.

The interval requires the maximum of $f_v-\gamma f_u$ along the critical manifold to exceed $ϵ$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta =0.1$, $\delta =10^{-5}$, $ϵ=5\times 10^{-4}$ and $\gamma =0.1$ give approximately ${\mu }_{-}=0.22618$ and ${\mu }_{+}=0.37350$. At $\mu =0.3$ the trace is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.

At every equilibrium the store balance gives $f=0$, and adding $\gamma$ times that balance to the cytoplasmic one gives $u=\mu$. The unique-root result therefore gives a single positive equilibrium $(\mu ,V(\mu ))$. Its Jacobian iswith derivatives evaluated at that equilibrium. At small $\mu$, $f_u>0$ and the trace is negative; sufficiently large $\mu$ also has $f_u>0$, since $J^{\prime }$ is of order $u^{-3}$ and $K^{\prime }$ of order $u^{-5}$. Thus both extreme regimes have a linearly stable unique equilibrium.

The fast variable is not $v$ alone. Using total-calcium fast fibres, introduce total scaled calcium $w=u+\gamma v$. Then $\dot{w}=\mu -u$, while on fast time $\sigma =\tau /ϵ$ the leading equations give $du/d\sigma =-\gamma f$, $dv/d\sigma =f$ and $w$ constant. Fast paths are therefore straight lines of slope $-1/\gamma$ in the $(u,v)$ plane. Attracting portions of the critical manifold satisfy $f_v-\gamma f_u<0$, rather than merely $f_v<0$. Along $v=V(u)$, slow motion obeysStarting with large $u$ and $v=0$, $f>0$ drives a rapid move upward and left, loading the store until the trajectory approaches the right attracting branch. For large $\mu$ in the right stable regime, slow motion on that branch carries it toward $u=\mu$; the direction reverses appropriately if the starting value lies below that equilibrium. For sufficiently small $\mu$, slow motion travels left along the right branch until its fold, then a fast approximately constant-$w$ jump carries the trajectory to the left attracting branch. Subsequent slow motion reaches the small-$u$ equilibrium. At a fold the slow approximation is singular, so the full equations resolve the transition. At very small $u$ an additional loss of time-scale separation is possible because the rates themselves vanish; the fast/slow drawing describes the separated portions, not a uniform estimate arbitrarily near the origin.