With constant lake surface area, raising its level by $dh$ changes its volume by $S_{L}dh$. The symbol $M_L$ in the proposed balance is an effective filling rate measured as lake-level rise per unit time, so the volume inflow is $S_L{M}_L$. Volume conservation givesThus the lake rises when its inflow exceeds channel discharge and falls in the opposite case. If an inflow were instead specified directly as a volume rate $I_L$, its contribution would be $I_L/S_L$; keeping these units distinct is necessary for the interpretation of $M_L$.

For Röthlisberger channel drainage, $S$ is the water-filled cross-sectional area of the subglacial channel, $Q$ its volume discharge, $m$ the mass of ice melted per unit channel length per unit time, and $N=p_i-p_w$ the effective pressure, namely ice overburden pressure minus water pressure. The first balance converts melt mass to area growth through division by ice density ${\rho }_i$, and subtracts material creep closure $KS{N}^3$. The cubic pressure dependence comes from the adopted ice-flow law; $K$ includes its material and geometric factors.

The second relation balances the downslope hydraulic driving gradient ${\rho }_{w}g\sin \alpha$ against the adopted frictional pressure loss. It is an instantaneous discharge-area law, giving $Q=(\sin \alpha /f{)}^{1/2}{S}^{4/3}$ for positive downstream flow. The final balance equates latent heat $mL$ needed to melt the channel walls with the gravitational power dissipated per unit length, ${\rho }_{w}g\sin \alpha Q$. It assumes that this dissipated power is available for melting rather than substantial sensible heating or other energy losses. The model thus couples melt opening, pressure-driven material creep closure and hydraulic discharge.

In physical concentration coordinates, the two nullclines areSince $K_S<K_M$, the substrate nullcline lies above the complex nullcline for $S>0$. Starting at $(S_0,0)$, the complex rises and substrate falls; with $E_0/S_0\ll 1$, this fast segment is nearly vertical when the axes use their respective concentration scales. The trajectory then follows the complex quasi-steady manifold toward the origin. On its slow descending portion it is slightly above the complex nullcline so that $\dot{C}<0$; the leading outer curve itself sets that derivative to zero only at zeroth order.

The fast scale resolves the rapid complex buildup that is compressed into a thin initial layer on the substrate-depletion scale. The outer scale follows depletion over a much longer time and cannot satisfy $c(0)=0$ by itself.

In the overlap $1\ll {\tau }_f\ll {ϵ}^{-1}$, the inner solution tends to $(s,c)=(1,1/(1+K))$. The outer solution as ${\tau }_s\downarrow 0$ has the same limit. Subtracting this common part produces a leading composite approximation,It satisfies the leading initial values and tends to the outer quasi-steady solution after the initial layer. Corrections of order $ϵ$ describe the small substrate loss during complex buildup.

Use the longer scale $T_s=(k_1{E}_0{)}^{-1}=T_f/ϵ$ and ${\tau }_s=t/T_s=ϵ{\tau }_f$. The equations becomeAt leading order the fast complex variable has equilibrated to the quasi-steady manifold. ThereforeMatching from the initial layer gives $s(0)=1$ for this outer problem; its implicit solution is $\boxed{s+K\log s=1-\lambda {\tau }_s}$. The dimensional substrate loss is the Michaelis-Menten rate with maximum rate $k_2{E}_0$ and Michaelis constant $(k_{-1}+k_2)/k_1$.

Set $T_f=(k_1{S}_0{)}^{-1}$, $ϵ=E_0/S_0$, $K=(k_{-1}+k_2)/(k_1{S}_0)$ and $\lambda =k_2/(k_1{S}_0)$. On fast time ${\tau }_f=t/T_f$,For ${\tau }_f=O(1)$, leading substrate remains $s=1$. The initial complex buildup therefore solves $c^{\prime }=1-(1+K)c$, $c(0)=0$, givingTime belongs inside the exponential. Its placement outside the exponential bracket in the printed expression is a typographical error; the derived form satisfies both the kinetic equation and zero initial complex concentration.

Mass action givesThus $E+C=E_0$ and $S+C+P=S_0$ are conserved. Eliminating free enzyme yields the stated two-equation system.

With $s=S/S_0$, $c=C/E_0$ and $\tau =t/T$, substitution givesThe dimensionless initial data are $\boxed{s(0)=1,c(0)=0}$. The parameter named $\delta$ here is an enzyme kinetic coefficient and is unrelated to the loss parameter used in the later blood-cell question.

For an assay begun by mixing free enzyme and substrate, take $\boxed{[S](0)=S_0>0,[E](0)=E_0>0,[C](0)=[P](0)=0}$. The complex has not yet formed and no product has been made. The later small-enzyme limit assumes $E_0\ll S_0$. Other preincubation protocols could require nonzero initial complex, but these initial conditions are the ones used by the stated inner solution.

Write $u=1+\tilde{u}$, $c=1+\tilde{c}$ and let $a_E=E^{\ast \prime }(1)>0$. To first order $\tilde{s}=(1-F^2)\tilde{u}$, and the other balances give ${\tilde{c}}_x=a_E\tilde{u}-\tilde{c}=-{\tilde{s}}_t$. Seek modes $e^{\sigma t+ikx}$. ThenThus every nonzero wavenumber grows when $F>1$, while it decays when $F<1$. The zero mode is neutral in this reduced model. The exceptional $F=1$ limit makes the algebraic depth-bed relation degenerate and requires neglected effects. If the entrainment derivative vanishes at the reference state, this first-order instability calculation is itself degenerate.

For $F>1$, the growth rate tends to the finite positive limit $a_E/(F^2-1)$ as $|k|\to \infty$. It does not diverge with wavenumber, although this reduced theory selects no finite fastest scale and should not be extrapolated beyond its long-wave regime. The phase speed isThe unstable bedforms migrate upstream against the water flow and represent antidunes arising from coupled erosion, deposition and supercritical free-surface flow. This is the relaxation-controlled antidune instability.

The reduced mass equation gives $hu=1$ after normalizing the reference discharge. The reduced momentum equation integrates to $\eta +F^2{u}^2/2=1+F^2/2$ with the reference free-surface level one. Since $h=1/u$,Also $hu=1$ reduces the concentration equation toWith $E^{\ast }(1)=1$, a steady uniform reference state is $\boxed{u=h=c=\eta =1,s=0}$. The integration constants above specify this reference normalization; a different imposed discharge or head changes them.

Choose reference depth $h_0$, velocity $u_0=\sqrt{gS{h}_0/f}$ and concentration $c_0=E(u_0)$. Set the adaptation length $L=h_0{u}_0/v_s$ and the slow bed-evolution time $t_b={\rho }_s(1-\varphi )h_0/(v_s{c}_0)$. Scale both bed and surface elevation with $h_0$. ThenThese choices balance sediment advection with settling, bed mass change with exchange, and reference drag with downslope gravity. Direct substitution givesHere $E^{\ast }(1)=1$ by construction. The small parameter $ϵ$ is the suspended solid mass relative to bed solid mass at the chosen scales, explaining the separation between fast fluid motion and slow morphology.

The depth $h$ is free-surface height $\eta$ minus bed elevation $s$. The first differential equation conserves water volume. Momentum balances local and advective acceleration against the free-surface pressure gradient, mean gravitational bed-slope forcing and the quadratic drag $f{u}^2/h$.

The concentration $c$ is suspended sediment mass per water volume. Multiplication of its material rate by depth gives the sediment change per bed area. Settling at speed $v_s$ removes mass at rate $v_{s}c$, while entrainment supplies $v_{s}E(u)$; increasing equilibrium concentration $E(u)$ represents stronger erosive capacity at greater speed. The same exchange changes the bed mass per area, whose solid volume fraction is $1-\varphi$ and grain density is ${\rho }_s$. Hence net suspended gain equals $-{\rho }_s(1-\varphi )s_t$: erosion lowers the bed and deposition raises it.

One derives these balances by depth integration with hydrostatic pressure and a sectionally averaged velocity, followed by closures for drag and sediment entrainment/settling. The model neglects additional bedload transport and uses the dilute-suspension approximation to keep the water-volume balance independent of sediment volume.

The physically meaningful extension requires $h\ge 0$: negative net input removes water where wet and cannot keep removing nonexistent surface water where dry. Keeping the algebraic source $E(t)<0$ unchanged in a dry zone would instead produce negative depths, so the drying-front description implicitly includes this availability condition.

Use the same $H(t)$, increasing until $t^{\ast }$ and decreasing afterwards. While $H(t)>0$, there is one earlier wetting time ${\tau }_d(t)\in [0,t^{\ast }]$ such that $H({\tau }_d(t))=H(t)$. The boundary-emitted characteristic with this label has just exhausted its water. Its position is the drying front:The integrand is nonnegative because $H(r)\ge H(t)$ between these two equal-height times. At $t=t^{\ast }$ the interval collapses and $s_d=0$; as net input becomes negative the front moves downstream. Differentiating gives $s_d^{\prime }=-(10/9)E(t){\int }_{{\tau }_d(t)}^t[H(r)-H(t){]}^{-1/3}dr>0$, where endpoint singularities are integrable away from the turning-time limit.

The wet solution ahead of the front still follows unexhausted characteristics. With $S_0(t)=\frac{5}{3}{\int }_0^{t}H(r{)}^{2/3}dr$,where $0\le \tau <{\tau }_d(t)$ is determined by the same characteristic-position integral as in (iv). If cumulative loss reaches $H(t)=0$, all remaining initial-line water has also vanished and the surface is dry thereafter while $E<0$. If the total later loss is smaller, the formula applies for all later times with $H(t)>0$. This is the drying front in a rainfall-driven Manning kinematic wave.

Let $H(t)={\int }_0^{t}E(r)dr$. For positive rainfall excess it is strictly increasing. The characteristic equations areA characteristic starting on the initial line at position $s_0\ge 0$ has $h=H(t)$ and $s=s_0+S_0(t)$, where $S_0(t)=\frac{5}{3}{\int }_0^{t}H(r{)}^{2/3}dr$. It supplies the downstream region $s\ge S_0(t)$.

A characteristic emitted at the upstream boundary at time $\tau$ hasFor fixed $t$, this position decreases strictly from $S_0(t)$ to zero as $\tau$ increases from zero to $t$. Thus it uniquely determines $\tau =\tau (s,t)$ in the remaining region. The complete solution isThe formulas agree at the separator, satisfy both zero data and cannot form a characteristic crossing while $E>0$. At the upstream endpoint $\tau =t$, height is zero.