For the three-region calcium nullcline, defineFor $u>0$, $f(u,0)>0$, $f\to -\infty$ as $v\to \infty$, and $f_v=-2K(u)v/(1+v^2{)}^2-\delta <0$. Thus there is exactly one positive root $v=V(u)$ for every positive $u$. The curve starts at the origin and initially satisfies $v\sim \beta u^2/\delta$.

For $0<\beta <1$, let $u_c>0$ solve $J(u_c)=K(u_c)$ and put $K_c=K(u_c)$, $J_c^{\prime }=J^{\prime }(u_c)$, $K_c^{\prime }=K^{\prime }(u_c)$. Explicitly, with $z_c=u_c^2$,In particular $u_c\sim \sqrt{\beta }$ and $K_c^{\prime }-J_c^{\prime }>0$ for small $\beta$.

On the large-$v$ branch away from the crossing, replace $v^2/(1+v^2)$ by one. This gives $v\sim [J(u)-K(u)]/\delta$ where this is large and positive. Near $u_c$ retain the next term $1-v^{-2}$ and linearize $J-K$:Finally where the leak is negligible compared with the other terms, $v^2/(1+v^2)\simeq J(u)/K(u)$, so $v\simeq [J/(K-J){]}^{1/2}$ on the side $J<K$. These are overlapping asymptotic descriptions with different validity ranges, not three exact disconnected curves. The curve rises to a large peak, falls sharply near $u_c$, then approaches a small finite level as $u\to \infty$. At leading small-$\delta$ order that last level is $[\beta /(1-\beta ){]}^{1/2}$.

The cytoplasmic concentration $c$ gains a constant influx $r$ from outside the cell and loses calcium through a first-order extrusion term $kc$. The pump flux $J_{+}$ moves calcium from cytoplasm into the sarcoplasmic reticulum, saturating with cytoplasmic concentration through a Hill function with scale $K_1$, maximal rate $V_1$ and exponent $n$. The release flux $J_{-}$ returns stored calcium to the cytoplasm. It requires both calcium in the store and calcium-dependent activation of release; its two saturating factors have scales $K_2,K_3$ and exponents $m,p$. This represents calcium-induced calcium release. The leak $k_s{c}_s$ is passive return from the store.

Every internal transfer appears with opposite signs in the two balances, so $d(c+c_s)/dt=r-kc$ in the model's compatible volume normalization. Only influx and extrusion change the total. The nonlinear pump and activated release can redistribute this total rapidly and generate oscillations even with constant external input.

Replace only the free-substrate equations byThe other four equations remain as above. The carrier invariant remains unchanged because no protein is added or removed. The two external substrate terms cancel in the total-substrate balance, so the second invariant also remains unchanged. This cancellation describes a balanced source and sink, even though there is a nonzero throughput. A constant sink is physically meaningful only while substrate concentrations remain nonnegative; it is not a state-dependent removal law valid after an interior reservoir is exhausted.

Adding the two complex and two free-carrier equations cancels binding and conformational exchange, givingEach free or bound carrier contains one carrier protein, and the model neither synthesizes nor degrades it. Adding the two free-substrate and two bound-substrate equations likewise givesBinding stores substrate temporarily and loaded conformational exchange relocates it; neither changes total substrate. The two invariants count distinct conserved chemical moieties. The constants $c_0$ and $s_0$ should not be interchanged: protein and substrate need not be present in equal amounts.

In alternating-access carrier transport, the four reversible reaction pairs are $S_i+C_i\rightleftharpoons P_i$, $S_e+C_e\rightleftharpoons P_e$, $C_i\rightleftharpoons C_e$ and $P_i\rightleftharpoons P_e$. Binding and unbinding use $k_{+},k_{-}$; both conformational pairs use $k$ in each direction. Define the net binding rates $B_i=k_{+}{s}_i{c}_i-k_{-}{p}_i$ and $B_e=k_{+}{s}_e{c}_e-k_{-}{p}_e$. Mass action and the change of each species in each reaction giveFor example binding consumes one free interior substrate and one free interior carrier, and produces one interior complex; loaded-carrier transformation then moves that bound substrate across the membrane. A bare-carrier transformation alone does not transport substrate. The concentration convention implicitly uses compatible compartment-volume normalization, as required for the stated unweighted conservation laws.

For quasistatic bed stability with suspended sediment, in the leading approximation, $(hu{)}_x=0$ and ${\eta }_x=0$. For spatially varying perturbations at fixed unit water flux and fixed surface level, write $u=1+U$, $h=1+H$, $c=1+C$, $s=s_{\ast }+S_1$. The constraints give $H=-U$ and $S_1=-H=U$. Linearizing the suspended and bed equations yieldsFor a Fourier mode $e^{\sigma t+i\kappa x}$, the first relation gives $C=4U/(1+i\kappa )$; substitution into the second givesThus every nonzero spatial frequency has negative growth rate, independently of $\beta$. The parameter changes propagation, not damping. The spatially uniform mode has $\sigma =0$ and is neutral, representing a change among uniform reference states compatible with the leading constraints. Hence there is no growing linear mode in this approximation; the claim is stability, not strict decay of every homogeneous perturbation. Restoring the small discarded terms or changing boundary conditions requires a separate calculation.

Take an arbitrary reference depth $h_0>0$, and select $u_0=(gS{h}_0/f{)}^{1/2}$ so the uniform reference flow balances drag and slope. Work with positive downstream velocity, so $u_{\ast }=\sqrt{f}u$ in the dimensional constitutive laws. Introduce the erosion and bedload scalesMatching erosion with deposition, suspended advection with erosion, and bed storage with erosion gives respectivelyUse $h_0$ for the scales of $\eta$ and $s$ too, so $h=\eta -s$ survives unchanged. After dividing the balances by their matched scales, the parameters areFor example the momentum balance divided by $g{h}_0/x_0$ has inertial prefactor $F^2$, surface-gradient term $-{\eta }_x$, and equal slope/drag prefactors $S{x}_0/h_0=\delta$, because $f{u}_0^2=gS{h}_0$. The suspended balance has source $u^3-c/u$, and the bed balance has its negative with unit bed-time coefficient by the definition of $t_0$. These substitutions give the entire stated dimensionless system; $q_b=u^3$ is dimensionless bedload, with its dimensional scale already absorbed into $\beta$.

Here $h$ is water depth, $u$ the depth-averaged downstream speed, $s$ bed elevation relative to the reference slope, and $\eta =s+h$ free-surface elevation relative to that reference. The constants ${\rho }_w,{\rho }_s$ are water and grain densities, $S$ is the reference bed slope, $\tau$ bed shear stress, $\varphi$ bed porosity, $c$ suspended sediment mass concentration, $q_b$ bedload solid volume flux per unit width, and $v_E,v_D$ erosion and deposition velocities. The settling speed is denoted by $v_s$ in the constitutive laws.

Depth integration of incompressible water conservation gives the depth-storage plus discharge-divergence balance. Depth-averaged momentum has local and advective acceleration on the left, and gravity driven by the actual surface gradient together with bed drag on the right. Hydrostatic pressure and a shallow-water approximation underlie this Saint-Venant description.

Suspended sediment conservation first has the conservative form $(hc{)}_t+(huc{)}_x={\rho }_s(v_E-v_D)$; the water balance reduces it to the material-derivative form printed in the source. Solid bed volume per unit horizontal area is $(1-\varphi )s$. Its storage plus bedload divergence equals deposition minus erosion, giving the Exner bed balance. Adding the suspended and bed balances after converting bed volume to grain mass cancels exchange terms: erosion and deposition transfer sediment between the two reservoirs rather than create it.

Taylor expansion gives $Q(A_{\mathrm{eq}}+a)=Q(A_{\mathrm{eq}})+c{A}_{\mathrm{eq}}^{m}a+O(a^2)$. Subtract the equilibrium equation and write $V(s)=c{A}_{\mathrm{eq}}^m(s)$ to obtain $a_t+(Va{)}_s=0$. Set $b=Va$. Since $V$ is time independent, $b_t+V{b}_s=0$. The travel coordinate for a conservative transport perturbation $\xi (s)={\int }_0^{s}V(r{)}^{-1}dr$ transforms this into $b_t+b_{\xi }=0$, and thereforeThe conserved quantity along a characteristic is the flux perturbation $Va$, explaining the diminishing area amplitude as $V$ grows.

Differentiate the equilibrium relation to get $A_{\mathrm{eq}}^{\prime }=M/(c{A}_{\mathrm{eq}}^m)=M/V$. Thus the arrival time is explicitlyFor zero upstream flux the integral remains finite, giving the limiting formula with its second term zero. However $V(0)=0$ then makes the displayed amplitude formula singular at the head; a uniformly small arbitrary area disturbance imposed exactly there is outside this linearization. A positive upstream flux avoids that degeneracy.

At equilibrium $Q_s=M$, so $Q=Q_{\mathrm{in}}+Ms$ for an upstream boundary flux $Q_{\mathrm{in}}$. HenceWithout a specified upstream condition this integration constant must be retained. On a region of positive area, the characteristic speed is $Q^{\prime }(A)=c{A}^m>0$, increasing downstream along the equilibrium. Small disturbances move downstream, spread in space as this speed increases, and decrease in area amplitude in the manner derived next. They are transported away from a fixed location; the model has no diffusion giving instantaneous smoothing. Finite-amplitude compressive disturbances can steepen into shocks because the flux is convex, $Q^{\prime \prime }(A)=cm{A}^{m-1}>0$, so the small-disturbance picture is not a global no-shock assertion.

To obtain a power-law river kinematic wave, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow force balance per unit river length is $\rho gAI={\tau }_b{P}_b$. With a Chézy friction law ${\tau }_b=\rho f{u}^2$, it gives $u=(gI/f{)}^{1/2}(A/P_b{)}^{1/2}$. Thus the hydraulic radius determines the power of area.

For a channel family with $P_b=P_0{A}^q$, where $q<1$, we have $u=C{A}^{(1-q)/2}$, with $C=(gI/(f{P}_0){)}^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. ThenFor example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the friction balance, not a universal arbitrary value fixed by Chezy's law alone.

The cross-sectional area $A(s,t)$ multiplied by the mean downstream speed $u(s,t)$ is the volume flux $Q=Au$. The term $A_t$ is local storage change per unit river length, $(Au{)}_s$ is net downstream export per unit length, and $M>0$ is lateral volume input per unit length per unit time, for example distributed tributary or runoff supply.

For a fixed reach $[s_1,s_2]$, incompressibility and constant density giveApply the fundamental theorem of calculus and localize the resulting integral identity to obtain $A_t+Q_s=M$. The density cancels because every term transports the same water; no sediment source is included in this balance.

The characteristic roots satisfyThey are a nonreal conjugate pair precisely when ${\mu }^{1/2}/9-16{\mu }^{7/6}<0$, or ${\mu }^{2/3}>1/144$. ThereforeIn lake-coupled Röthlisberger drainage instability, above this value the real part is ${\mu }^{1/4}/6>0$, so nonzero linear perturbations are sinusoidal oscillations with an exponentially growing envelope. Their angular frequency is $[4{\mu }^{7/6}-{\mu }^{1/2}/36{]}^{1/2}$.

Physically the feedback represents repeated lake filling and pressure increase, channel enlargement by melting, rapid drainage, pressure reduction and material creep closure. Large drainage events are glacial lake outburst floods, or jökulhlaups. The unbounded amplitude is a conclusion of the linearized equations, not a proof of infinite physical discharge or an indefinitely growing nonlinear solution. Once perturbations are comparable with the equilibrium scales, nonlinear dynamics and constraints such as lake depletion must be retained. At the threshold the two real roots coincide; below it the instability is nonoscillatory.

For $\mu >0$, the positive equilibrium isSet $S=S_{\ast }+\epsilon S_1$ and $N=N_{\ast }+\epsilon N_1$ and retain first-order terms. The linear system isDifferentiating its first row and using the second givesIn particular the trace is positive and determinant is $4{\mu }^{7/6}>0$. The steady drainage state is linearly unstable for every positive supply, whether its instability is oscillatory or nonoscillatory.

Write dimensional variables as their scales times dimensionless variables, with positive scales $S_0,Q_0,m_0,N_0,t_0$. For compactness defineA consistent explicit choice isThese identities give $t_0{m}_0/({\rho }_i{S}_0)=1$, $K{t}_0{N}_0^3=1$, $f{Q}_0^2/(\sin \alpha S_0^{8/3})=1$ and ${\rho }_{w}g{t}_0{Q}_0/(S_L{N}_0)=1$. Substitution therefore leaves the unit coefficients in all four balances, and the sole supply parameter is $\mu =M_L{S}_L/Q_0$. Expanding its scales yieldsAfter dropping scale hats, eliminate the instantaneous quantities to obtain the useful two-dimensional systemThese scales assume a positive slope, positive friction coefficient and nonzero lake area; they nondimensionalize exactly the hydraulic law printed in the source.

Hydrostatic pressure at the entrance is $p_w=p_{\mathrm{ref}}+{\rho }_{w}gh$, with a constant reference pressure. Therefore $N=p_i-p_{\mathrm{ref}}-{\rho }_{w}gh$, and constant overburden pressure givesA filling lake raises water pressure and lowers effective pressure, reducing material creep closure. A draining lake has the reverse effect. These signs supply the pressure feedback in the coupled drainage model.