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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.

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 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].

Write $e=E^{\prime }(1)>0$, $q_1=q^{\prime }(1)>0$ and introduce perturbations $(H,U,C,S)e^{\lambda t+ikx}$ to water depth, velocity, concentration and bed height. Take the leading quasi-steady approximation $ϵ=0$ for the fast water and suspension response, retaining the small friction parameter $\delta$. For $k\ne 0$, the linearized equations areHere surface height is $H+S$, and linearization of $1-u^2/h$ gives $H-2U=-3U$. Eliminating $H,C,U$ gives the dispersion relationIn particular, putting $B=F^2-1\ne 0$ and expanding at large $|k|$,Thus the temporal growth rate tends to a finite constant, while the phase velocity isFor supercritical flow $F>1$, the waves propagate upstream. They grow if $\beta e(F^2-1)>3\delta q_1$; at leading order in small $\delta$, this holds for fixed positive $\beta ,e$ away from critical flow. They are antidunes, upstream-migrating bedforms coupled to the free surface. Indeed $H=-U$ and $S\simeq -(F^2-1)U$, so surface height $H+S\simeq -F^{2}U$ is in phase with the bed perturbation in this regime.

For clarity, neglecting $\delta$ completely gives the exact leading-order expressionsThese also identify the missing qualification in the source: the unstable upstream branch requires $F>1$. For $F<1$ this leading model gives decay and downstream propagation. The near-critical case $F^2-1\simeq 0$ needs a different balance; it is not covered by the expansion. Small nonzero $ϵ$ perturbs the slow bedform branch and adds fast water modes, so the stated result concerns the leading slow morphodynamic branch. This is the high-wavenumber antidune dispersion calculation.

In a reduced sediment-bed model with normalized discharge one, water momentum gives $s=1-u^{-1}+F^2(1-u^2)/2$, while concentration adaptation and bed exchange give $c_x=E(u)-c=-s_t$. Linearizing about $u=c=1,s=0$ yields $\tilde{s}=(1-F^2)\tilde{u}$ and $(1+ik)\tilde{c}=E^{\prime }(1)\tilde{u}$. Elimination proves the displayed dispersion relation.

For $E^{\prime }(1)>0$, the growth rate is $E^{\prime }(1)k^2/[(F^2-1)(1+k^2)]$, positive exactly on the supercritical side $F>1$ for nonzero $k$. The phase speed is $-E^{\prime }(1)/[(F^2-1)(1+k^2)]$, so unstable bedforms migrate upstream and are antidunes. At high wavenumber growth approaches a finite plateau rather than becoming unbounded. This closure selects no finite fastest wavelength; short-scale selection needs additional physics. The critical limit $F=1$ and a vanishing entrainment derivative are degenerate cases outside the displayed conclusion.