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For positive $q,K,g^{\prime }$ in a steady depositing gravity current, the function $K(h)=q^2/h+g^{\prime }{h}^2/2$ has a unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, where $\mathrm{Fr}=1$ and $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. When $K>K_c$ there are two positive depth roots: $h_{-}<h_c$ is supercritical flow and $h_{+}>h_c$ is subcritical flow. Differentiating the momentum invariant and using $g_x^{\prime }=-v_s{g}^{\prime }/q$ givesThe supercritical depth decreases towards $q^2/K$ as $g^{\prime }\downarrow 0$, whereas the subcritical depth increases like $\sqrt{2K/g^{\prime }}$. Because $K_c$ decreases downstream, an initially strict branch never reaches a finite-distance critical point. A critical inlet instead gives a singular depth derivative, so a smooth critical continuation is not justified by this model.

A dune is a sediment bedform. In the elementary subcritical flow description of river dunes, bed undulations and free-surface undulations are out of phase. This phase relation identifies a hydraulic response; instability additionally requires a positive real part of the relevant dispersion relation. Contrast the in-phase surface of an antidune.

A hydraulic bore is a propagating abrupt depth change described as a moving hydraulic jump. In the shock frame, an admissible rightward through-flow goes from supercritical flow to subcritical flow, with the appropriate characteristic curves entering the jump.

A hydraulic control is a critical section where a shallow water flow passes smoothly between subcritical flow and supercritical flow. One long-wave characteristic speed relative to the bed vanishes there. Regularity supplies an extra relation between discharge, depth and channel geometry, allowing the control to set the discharge or upstream state. A critical depth alone is not sufficient for a regular transcritical solution: the channel forcing must also be compatible there.

A slowly varying, steady, inviscid channel flow has constant discharge $Q=uh$ per unit width and the displayed Bernoulli equation. The bed-height function of depth has its maximum at $h_c=(Q^2/g{)}^{1/3}$, where the Froude number is one. At a nondegenerate critical crest, a differentiable profile changes between the subcritical flow and supercritical flow branches. With subcritical upstream squared Froude number $F<1$ and upstream depth $h_1$, the shock-free supercritical downstream depth is $h_1(F+\sqrt{F^2+8F})/4$.

A hydraulic jump connects shallow supercritical flow to deeper subcritical flow while conserving discharge and momentum and dissipating mechanical energy. Its Rankine-Hugoniot condition uses the channel's actual area and integrated hydrostatic pressure, which depend on cross-sectional geometry.

In a stationary rectangular-channel hydraulic jump, orient the discharge $q=hu>0$ from side 1 to side 2. Momentum conservation gives $q^2=g{h}_1{h}_2(h_1+h_2)/2$ for unequal depths. The specific mechanical energy is $B=u^2/2+gh$, soDividing by $g$ gives the displayed loss of head, measured as a length. The mechanical energy flux lost is $\rho q(B_1-B_2)$, so a dissipative jump must have $h_2>h_1$. With $R=h_2/h_1>1$, the squared Froude numbers are $R(1+R)/2>1$ upstream and $(1+R)/(2R^2)<1$ downstream. Thus the shallow water equations connect supercritical flow to subcritical flow through the admissible jump.

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 $B$ be the constant canal width, large compared with water depth. Then $A\simeq Bh$ and $l\simeq B$, so friction is $f{u}^2/h$. Choose a reference uniform water depth $H$, normal-flow speed $U$ satisfying $f{U}^2/H=gS$, and scalesThe continuity equation becomes ${\hat{h}}_{\hat{t}}+(\hat{h}\hat{u}{)}_{\hat{s}}=0$. Dividing momentum by $gS$ and dropping hats givesThe Froude number compares velocity to long gravity-wave speed. At the reference state, $F<1$ is subcritical flow, $F=1$ critical flow, and $F>1$ supercritical flow. In dimensional variables the wave characteristics travel at $u\pm \sqrt{gh}$: one can propagate upstream in subcritical flow, while both move downstream in supercritical flow. Away from the reference state the local Froude number is $F|u|/\sqrt{h}$.

For a one-layer shallow water flow the Froude number is $F=u/\sqrt{g^{\prime }h}$. Supercritical flow has $F>1$, so both characteristic speeds $u\pm \sqrt{g^{\prime }h}$ point downstream. Subcritical flow has $F<1$, allowing information to travel upstream. At hydraulic control the flow passes through $F=1$ at a geometry-induced critical section; that condition fixes discharge in relation to upstream head when the exit is free.

For escaping light air, reflect the usual lower-layer picture vertically: a lowered doorway roof acts as an inverted broad-crested weir. The lower return flow can be ignored dynamically only while it has a sufficiently large available area and small velocity. If the warm interface lies far below the doorway top, the cool inflow occupies a substantial dynamically active fraction of the opening; its inertia and pressure cannot be replaced by a motionless deep ambient. A two-layer exchange model is then required. If the warm layer reaches below the doorway bottom, the assumed separate lower replacement path is lost entirely.

Outside a freely controlled doorway the outgoing current becomes supercritical and then rises rapidly. That upturn violates the small-vertical-acceleration shallow water approximation, but it does not alter the upstream discharge as long as it does not submerge or back-pressure the control. Information cannot return through the supercritical outflow to the critical section. A blocked conservatory or a drowned doorway is different and can move the control upstream.

At $V=0$, the particle budget gives $\varphi =1$, so $g^{\prime }=g_0^{\prime }$ is constant. The primitive momentum conservation equation integrates toIntroduce $q=Q/Q_0=1-Wx/Q_0$, $y=h/h_0$ and $\epsilon ={\mathrm{Fr}}_0^2=Q_0^2/(g_0^{\prime }{h}_0^3)\ll 1$. The exact relation on the subcritical flow branch isExpanding uniformly for $0\le q\le 1$ givesThe depth increases slightly because horizontal kinetic energy decreases as fluid is removed. In fact the depth equation has $h^{\prime }\ge 0$ on this branch, andThus the small-Froude number approximation becomes better downstream, including as the flux vanishes. At leading order $u=(Q_0-Wx)/h_0$.

With $W=0$ and $V>0$, $Q=Q_0$ and $\varphi =e^{-kx}$, where $k=V/Q_0$. No carrier momentum is withdrawn; hence the integrated momentum flux is exactly constant:On an initially small-Froude number branch, the pressure term dominates. At leading orderA slightly more accurate large-distance amplitude is $\sqrt{2K/g_0^{\prime }}$, so $h\sim \sqrt{2K/g_0^{\prime }}{e}^{kx/2}$. The Froude number satisfies ${\mathrm{Fr}}^2\sim {\mathrm{Fr}}_0^2{e}^{-kx/2}$ and remains small. The growing depth can eventually violate confinement or the shallow water assumption; this does not make the asymptotic momentum calculation inconsistent within its intended regime.

On an initially large-Froude number branch, horizontal momentum flux dominates instead. The leading depth is constant, $\boxed{h\sim h_0}$, while the decaying reduced gravity makes the Froude number still larger. To resolve the small change, put $\delta ={\mathrm{Fr}}_0^{-2}\ll 1$ and $y=h/h_0$:soThe depth decreases slightly and tends exactly to $h_{\infty }=Q_0^2/K=h_0/(1+\delta /2)$. Its Froude number grows like ${\mathrm{Fr}}_0{e}^{kx/2}$. Both approximations agree with the exact depth equation $h^{\prime }=kh/[2(1-{\mathrm{Fr}}^2)]$: the subcritical flow deepens, while the supercritical flow thins.

The shallow-water approximation requires the depth to be small compared with the horizontal scale of variation, small vertical accelerations and nearly hydrostatic pressure. Take cross-sectionally uniform lower-layer velocity and a sufficiently deep, effectively stationary upper layer. Under the Boussinesq approximation, $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0$ is the reduced gravity. Conservation of lower-layer volume gives constant discharge $Q=bhu$.

Dividing the Bernoulli equation by $g^{\prime }$ gives the shallow-water specific energyAt fixed $Q,b$, the energy curve has its minimum at the critical depth $h_c=(Q^2/(g^{\prime }{b}^2){)}^{1/3}$, where $E_c=3h_c/2$. Above this minimum there are two depth branches: deeper subcritical flow has $F<1$, and shallower supercritical flow has $F>1$.

The two long-wave speeds relative to the bed are $u\pm \sqrt{g^{\prime }h}$. In subcritical flow one family can propagate upstream and one downstream; in supercritical flow both travel downstream for $u>0$. At hydraulic control, $u-\sqrt{g^{\prime }h}=0$, and a smooth transcritical solution must also satisfy the geometric regularity condition. The flow may remain entirely on either branch, cross smoothly through the control from subcritical to supercritical, or take the reverse smooth branch under appropriate boundary conditions. A downstream hydraulic jump can return supercritical flow to the subcritical branch while losing energy.

The sketch distinguishes smooth branch changes at a critical section from a jump: a jump conserves mass and momentum, not the original Bernoulli head.

Assume a steady, incompressible, inviscid flow with a nearly uniform velocity $u(x)$ across each water section. The channel varies slowly enough for vertical accelerations to be negligible, pressure to be hydrostatic and the water surface to be nearly horizontal across the width. Atmospheric pressure is uniform; neglect surface tension, wall friction and energy losses. The hydraulic equations apply away from jumps and short nonhydrostatic regions.

The Euler equation for horizontal momentum, with hydrostatic pressure, givesIntegrating gives a constant Bernoulli equation head $\mathcal{B}=H+h+u^2/(2g)$. Conservation of mass gives the constant volume discharge $Q=bhu$. Thus the shallow-water specific energy, measured above the local bed, isDefine the Froude number by $F=u/\sqrt{gh}$ for the positive flow direction. ThenAt fixed discharge and width, energy tends to infinity at both zero and infinite depth. It has a unique minimum atIndeed $E_{hh}=3Q^2/(g{b}^2{h}^4)>0$. Above the minimum there are two depth branches: the shallower one is supercritical flow and the deeper one is subcritical flow.

The local long-wave propagation speeds from the shallow water equations are $u\pm \sqrt{gh}$. In subcritical positive flow, one wave family travels upstream and one downstream. In supercritical flow both travel downstream. At critical flow the upstream-directed speed is zero, so the flow can provide hydraulic control separating upstream adjustment from downstream conditions. This statement concerns long waves in the slowly varying hydraulic approximation, rather than all short-wave disturbances.

For a smooth positive discharge on the flat bed, differentiate the constant specific energy. With $q(x)=Q/b(x)$,At a regular critical point, $F=1$ and $q>0$, so $q_x=0$. Since $Q$ is constant, this requires $b_x=0$. The given width has $b_x=2b_{0}x/L^2$, proving that a regular hydraulic control can occur only at $x=0$.

The throat is the unique width minimum and hence the maximum of the critical-energy envelopeA steady positive-depth flow must have its conserved energy at least as large as every local minimum energy. To be controlled smoothly at the throat it must satisfyEquivalently the controlled discharge at a prescribed head is $Q=b_0\sqrt{g}(2\mathcal{B}/3{)}^{3/2}$. If the head is larger than this critical value at fixed $Q$, the two depth branches remain separate everywhere and the flow need not be controlled. If it is smaller, no smooth steady solution carrying that discharge through the throat exists; upstream head or discharge must adjust.

Boundary conditions must also select the appropriate connected branches. A local expansion about the throat givesAt the controlled head, $h_x(0)=\pm \sqrt{2/3}{h}_c/L$. The usual accelerating positive-discharge connection uses the negative slope: deep subcritical flow upstream becomes shallow supercritical flow downstream. The throat is a maximum of the critical-head envelope, as required for a real smooth transcritical connection in hydraulic control in a variable-width channel.

Assume the shallow-water approximation: the horizontal scale is long compared with the depth, the hydrostatic approximation applies, and the cross-section-averaged velocity is approximately uniform. Neglect viscosity and drag for this part. Volume conservation gives the constant discharge $Q=bhu$. The steady horizontal momentum equation isso its integral is the Bernoulli equation head $\mathcal{H}=H+h+u^2/(2g)$. The shallow-water specific energy measured above the bed is thereforeThe Froude number compares the flow velocity with the long-wave speed $\sqrt{gh}$:At fixed $Q,b$, the shallow-water specific energy tends to infinity as $h$ tends to zero or infinity. Its unique minimum occurs at the critical depthThe two depth branches above that minimum are subcritical flow, $F<1$ with $h>h_c$, and supercritical flow, $F>1$ with $h<h_c$. The long-wave characteristic speeds relative to the bed are $u\pm \sqrt{gh}$. Thus a subcritical flow can receive information from downstream, whereas both characteristic curves travel downstream in a positive supercritical flow.

A hydraulic control is a critical section where one characteristic speed vanishes and a smooth flow can pass between the two branches. Regularity there supplies a condition relating $Q$, the depth and the geometry; together with reservoir or inlet data, it can determine the discharge. Merely finding $F=1$ is not enough: a finite depth gradient also requires the geometric forcing to vanish in the critical momentum equation.

The upstream flow is subcritical flow, with $h_1>h_c$. On this branch $dD/dh<0$, so the depth decreases as the bed rises; the velocity $Q/h$ increases. Since the crest height is below $D^{\ast }$, the depth remains above $h_c$ and never becomes critical. Over the falling side of the hump the depth increases again to $h_1$ downstream. The free surface falls over the hump: $d(h+D)/dD=-{\mathrm{Fr}}^2/(1-{\mathrm{Fr}}^2)<0$ on the subcritical branch.

Let states $1$ and $2$ be just left and right of a hydraulic bore moving at speed $s$, and let $[f]=f_2-f_1$. Integrating the conservative prismatic-channel shallow water equations across it gives the Rankine-Hugoniot conditionEquivalently the discharge in the shock frame is common on the two sides,Thus, for unequal positive depths,Momentum is conserved through the bore in this ideal model, but mechanical energy is dissipated; an energy-conservation jump must not be imposed as well.

For a stationary bore with rightward flow, $s=0$, $A_1{u}_1=A_2{u}_2>0$, and the momentum jump must still hold. A compressive hydraulic jump has $h_2>h_1$ andThe $u-c$ characteristic curves approach the stationary bore from both sides: upstream flow is supercritical flow, while downstream flow is subcritical flow using the local wave speed $c_i=\sqrt{2g{h}_i/3}$. The $u+c$ family travels downstream on both sides. These characteristic directions accompany, rather than replace, the two stationary conservation jumps.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

Put $R=h_2/h_1>1$. The squared Froude numbers follow directly from the discharge relation:Thus the jump connects supercritical flow to subcritical flow. Long surface gravity waves have speed $\sqrt{gh}$ relative to the fluid, so the characteristic speeds in the stationary frame are $u\pm \sqrt{gh}$. Upstream both point with the flow; downstream one points against it. The jump changes whether long-wave information can propagate upstream through the local flow.