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Substitution of $e^{i(kx-\omega t)}$ givesThe phase velocity is negative for all nonzero wavenumbers, so crests travel left. The group velocity is negative for $|k|<\ell$, zero at $|k|=\ell$, and positive for $|k|>\ell$. Its minimum is $-\beta /{\ell }^2$ at zero and maximum $\beta /(8{\ell }^2)$ at $|k|=\sqrt{3}\ell$; it tends to zero from above at infinity. The odd dispersion relation has extrema at $k=\pm \ell$, with $\omega (\ell )=-\beta /(2\ell )$.

The Fourier solution is $\phi (x,t)={\int }_{\mathbb{R}}A(k)e^{i[kx-\omega (k)t]}dk$. The stationary-phase statements require enough amplitude regularity and decay, for example $A$ smooth and rapidly decreasing; realness and evenness alone do not guarantee a convergent integral or a stationary-phase expansion. For $x=Vt$, set ${\Phi }_V(k)=kV-\omega (k)$ and solve ${\omega }^{\prime }(k)=V$ implicitly.

For an axisymmetric vertical mode of a shearing sheet in this Boussinesq approximation, the perturbations have no horizontal gradients, so background-shear advection vanishes. The linear Coriolis acceleration and perturbation advection of the background shear givewhile thermal conduction and incompressibility giveHere $\beta$ is the thermal diffusion rate of a Fourier mode, not the cooling exponent in Question 2. Since $k\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $P^{\prime }=0$ for this nonzero-wavenumber amplitude.

The three remaining amplitudes satisfyA nontrivial solution requires the determinant to vanish, giving the convective overstability dispersion relationNo division by $s$ or $s+\beta$ was needed, so the thermal energy mode of a shearing sheet is retained. The notation $N^2$ denotes a signed buoyancy frequency squared and can be negative for an adverse stratification; it must not be restricted to the square of a real positive frequency.

Treat $n^2=N^2/{\Omega }^2$ as a signed small parameter, with $|n^2|\ll 1$, as printed in the original PDF. At $n^2=0$, the dispersion relation factors asThus the two epicyclic motion modes have $s_0=\pm i\Omega$, while the thermal energy mode of a shearing sheet hasIt decays on the thermal diffusion time ${\beta }^{-1}$. For completeness, its next correction is $s_E=-\beta +\beta N^2/({\beta }^2+{\Omega }^2)+O(\Omega n^4)$ at fixed $\beta /\Omega$.

For either epicyclic motion root, the implicit function theorem applied to the dispersion relation givesHenceSince $\beta >0$, the oscillations undergo overstability precisely whenThe convective overstability growth rate tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta /({\beta }^2+{\Omega }^2)$ shows that its maximum occurs at $\beta =\Omega$, corresponding to $k^2=\Omega /\xi$. ThereforeThis is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The convective overstability arises when an adverse radial specific entropy gradient couples to epicyclic motion with a finite thermal conduction lag.

Let $q=\alpha h$, $d=\alpha (1-h)$, and use the four regional amplitudes above. Continuity of $\hat{w}$ at the two jumps givesPressure continuity, using vorticity-jump matching for an inviscid shear flow, gives the other two equations:These are the required homogeneous four equations. As a useful reduction, put $X=\tanh q$, $Y=\tanh d$, $H=q(1+XY)$ and $J=q(X+Y)$. Eliminating $A,D$ leavesThe vanishing determinant gives $c^2=(H-Y)(J-XY)/(HJ)$, which expands to the printed dispersion relation.

For fixed $h\in (0,1)$, $X=\alpha h+O({\alpha }^3)$ and $Y=\alpha (1-h)+O({\alpha }^3)$. The factored dispersion relation therefore giveswhere the latter follows from $X,Y\to 1$ and $c^2\sim (1-1/(2\alpha h){)}^2$. A real negative $c^2$ yields a growing branch $c=i\sqrt{-c^2}$ for positive $k$. Thus $h<1/2$ gives instability at sufficiently small nonzero wavenumber.

For the converse, monotonicity toward the limiting value one would suffice, as permitted in the question. In fact a direct sign argument avoids this extra assumption. With $q=\alpha h>0$, $X=\tanh q<q$, so $J-XY=qX+(q-X)Y>0$. If $h\ge 1/2$, then $q\ge d=\alpha (1-h)$ and $Y=\tanh d<d\le q$. Hence $H-Y=q(1+XY)-Y>0$, so $c^2>0$ for every $\alpha >0$. We obtain the exact instability threshold for a bounded piecewise-linear shear layer:At ${\zeta }_L=1/2$ the long-wave limiting value is zero, but every finite positive wavenumber has $c^2>0$.

Take $x$ in the injection direction and $y$ along the flat interface. The specified $U$ must be a Darcy velocity, so the pore-fluid velocity and unperturbed interface speed are $U/\varphi$. Write the perturbed interface as $x=X(t)+\eta (y,t)$, with $\dot{X}=U/\varphi$, and choose the normal from fluid 1 into fluid 2. The background pressure gradients from Darcy law are ${\partial }_x{p}_i^{(0)}=-{\mu }_{i}U/k$.

By incompressibility and uniform permeability of a porous medium, pressure perturbations are harmonic functions. A transverse Fourier mode with positive wavenumber $\alpha$ has the decaying formsThe kinematic boundary condition and equal normal Darcy flux implyWithout capillarity the pressure is continuous at the displaced interface. Its linear perturbation is thereforeConsequently the planar viscous-fingering dispersion relation reduces toThis is the Saffman–Taylor instability: less viscous fluid advances preferentially through protrusions. Without a short-wave regularizing mechanism the idealized rate has no finite maximum.

For the prescribed apparent capillary pressure, the chosen normal is $\mathbf{n}=(1,-{\eta }_y)/\sqrt{1+{\eta }_y^2}$, so $\nabla \cdot \mathbf{n}=-{\eta }_{yy}+O({\eta }^2)={\alpha }^2\eta +O({\eta }^2)$. Subtract the flat-front pressure jump $\gamma (1+\beta X){\kappa }_0$. Linearization of the jump at $x=X+\eta$ givesThere is no first-order product of the surface-tension perturbation and the perturbed macroscopic curvature. ThusUsing the capillary number convention $\mathrm{Ca}={\mu }_{2}U/\gamma$ and ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$, this isIf $x$ is measured from the unperturbed front and $\gamma$ denotes its local apparent tension, set $X=0$ to obtain the usual fixed-coefficient expression. For a physically fixed spatial gradient, $X=Ut/\varphi$ and the displayed rate is an instantaneous, generally time-dependent growth rate; the amplitude solves $\dot{a}=\sigma (\alpha ;X(t))a$, rather than one constant exponential law for the entire displacement. The parameter $\beta$ has dimensions of inverse length; the absolute tension gradient is $\gamma \beta$.

Assume $1+\beta X>0$, as required for positive apparent surface tension, and define $A=({\mu }_2-{\mu }_1)U-k\gamma \beta {\kappa }_0$. If $A>0$, unstable modes satisfy $0<{\alpha }^2<A/[k\gamma (1+\beta X)]$. Differentiating the cubic dispersion relation gives the fastest-growing modeWhen $A\le 0$, every nonzero mode decays, with a neutral zero-wavenumber translation in the infinite-domain limit. For $\beta {\kappa }_0>0$, surface-tension-gradient stabilization of viscous fingering suppresses the instability entirely at positive injection speeds satisfyingIf $\beta {\kappa }_0\le 0$, no positive injection speed eliminates every long-wave unstable mode in an unbounded interface. The prescribed wetting variation is treated as an effective normal pressure law for porous-media flow.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Eliminate the transport amplitudes in the linear stability analysis. For a general bed shear response $\mathcal{T}(k)$, the dispersion relation isThis is the most specific result available without adding the missing hydrodynamic closure. With $\mathcal{T}(k)={\tau }_{0}k(A+iB)$, setMultiplication by $1-ix$ gives the growth rate and migration velocityThe packed-bed volume fraction cancels. For a real physical closure the negative wavenumber response is the complex conjugate; the displayed $k>0$ formulas cover the independent normal modes.

If $A>0$, $B>0$ are treated as constants, the upstream shear stress phase lead promotes growth, while the inclined-bed sediment threshold reduces it and the finite saturation length delays sediment transport. There is an unstable band precisely when $b>0$: $0<k<b/(a{L}_{\mathrm{sat}})$. Within this band $c>0$, so the growing bedforms migrate downstream. The fastest normal mode satisfiesThis last maximization assumes constant $A,B$; if the hydrodynamic coefficients vary with wavenumber, their derivatives also enter the selection condition. When $b\le 0$ and $a>0$, all nonzero normal modes decay.

Just above the sediment entrainment threshold,Thus the common regime $B<1/{\mu }_s$ is stable near onset: the slope penalty overwhelms the hydrodynamic phase lead. This conclusion is conditional on that inequality, not universal. For constant $B>0$, instability begins only when ${\tau }_0>{\tau }_{\mathrm{th},0}/({\mu }_{s}B)$ as well as ${\tau }_0>{\tau }_{\mathrm{th},0}$. The transport sensitivity $C=\chi \gamma {\Delta }^{\gamma -1}$ vanishes at onset for $\gamma >1$, is finite for $\gamma =1$, and is singular for $0<\gamma <1$. Exactly at threshold, the positive-part transport law needs separate treatment; the above linearization with perturbations small compared with $\Delta$ is unavailable.

Far above the sediment entrainment threshold, $b\simeq {\tau }_{0}B$. If $A,B>0$, the unstable cutoff tends to $B/(A{L}_{\mathrm{sat}})$, and the dominant wavelength is set by the saturation length. Both growth rate and migration velocity scale with $C{\tau }_0\sim \chi \gamma {\tau }_0^{\gamma }$, apart from their length factors. Their numerical values and any detailed dependence on wavenumber still require a flow closure.

For example, prescribing uniform shear stress independently of the bed gives $\mathcal{T}=0$ andThis equally admissible closure illustrates why instability cannot be asserted from the printed sediment equations alone.

The plots use constant $a>0$ and three illustrative ratios $b/a$. They show how the sign of the shear phase lead minus the slope correction determines stability; they are not numerical predictions for unspecified flow conditions.

Let a slowly varying wave have phase $S(\mathbf{x},t)$, wavevector $\mathbf{k}=\nabla S$, and angular frequency $\omega =-{\partial }_{t}S$. The local dispersion relation is the Hamilton-Jacobi equation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf{x}}={\nabla }_{\mathbf{k}}\Omega$, gives the Hamiltonian ray-tracing equationsHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf{x}}$ is the group velocity.

For $\Omega =kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin \psi =(k_x^2+k_y^2{)}^{1/2}$ and $k=\omega /c$. Hence Snell's law is

For the ray launched with $\mathbf{k}=k(\cos \varphi ,0,\sin \varphi )$ at $z=0$, this constant is $|\cos \varphi |/c_0$. Thereforewhich proves

For $0<\varphi <\pi /2$, write $a=\cos \varphi$. Since the ray and wavevector are parallel,Near the upper turning point, put $z=z_m-\zeta$. Expansion givesIf $x=x_m$ at the turn, integration yields the local ray pathThe ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\varphi$ decreases $z_m$, while $\varphi \downarrow 0$ sends the turning level outward.

Melting raises permeability of a porous medium. An advancing protrusion therefore offers a less resistive path to hot liquid, attracting larger Darcy flux and receiving more heat. The resulting faster melting reinforces the protrusion: this is a reactive infiltration instability. A permeability decrease would reverse this feedback, while zero contrast gives no growth in this ideal model.

For positive contrast, the growth rate is proportional to transverse wavenumber, so the model has no finite fastest-growing wavelength. The latent-to-sensible heat ratio slows the front but supplies no short-wave cutoff. Finite thermal transport and a resolved melting zone can introduce a length scale; pore geometry eventually limits the continuum description. If interface curvature changes the equilibrium melting temperature through the Gibbs--Thomson relation, that can also oppose short-wave corrugations. The modified dispersion relation must include such effects before a preferred wavelength can be calculated; a length scale alone does not guarantee that every added mechanism selects one.

The linear dispersion relation for a mode $e^{i(kx-\omega t)}$ is $5{\omega }^2=k^4+4$. The proposed modes share phase velocity $c$, so $5c^2{k}_j^2=k_j^4+4$. A quadratic wave interaction generates the second harmonic and the difference harmonic. With only two positive wavenumbers, phase matching for a quadratic wave interaction requires $k_2=2k_1$. Equating their phase velocities givesand thereforeThis is a two-to-one resonance of dispersive waves. Its amplitude changes accumulate on $t=O({ϵ}^{-1})$. Choose the slow time $T=ϵt/5$; this harmless constant rescaling makes the later amplitude formulas simple. With $\xi =x-t$, the leading real field is $A_0{e}^{i\xi }+B_0{e}^{2i\xi }+{\overline{A}}_0{e}^{-i\xi }+{\overline{B}}_0{e}^{-2i\xi }$.

For the fundamental, the order-$ϵ$ cross derivative in $5{\psi }_{tt}$ is $-2i(A_0{)}_T{e}^{i\xi }$, while the fundamental coefficient in ${\psi }_0{\psi }_{0x}$ is $i{\overline{A}}_0{B}_0$. For the second harmonic the corresponding terms are $-4i(B_0{)}_T{e}^{2i\xi }$ and $i{A}_0^2$. The solvability condition in the method of multiple scales removes the resonant forcing and givesThe nonresonant third and fourth harmonics enter the correction. They do not change these leading amplitude equations.

Write $A_0=R{e}^{i\theta }$, $B_0=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. Taking real and imaginary parts gives the explosive two-to-one amplitude system in polar form:The polar phases are used where their corresponding amplitudes are nonzero. These equations implyFor the second identity, differentiate using ${\delta }_T=(R^2/(4P)+P)\sin \delta$: the terms proportional to $\sin \delta \cos \delta$ cancel. Thereforeare first integrals, and in particular

For constant phases with $\varphi =2\theta +\pi$, these real equations reduce to $R_T=RP/2$ and $P_T=R^2/4$. The initial data give $R^2-2P^2=8$, so the Riccati equation for $P$ is $2P_T=P^2+4$. Integrating and using $P(0)=2$ givesThe tangent and secant function have a pole at $T=\pi /4$, corresponding to $t=5\pi /(4ϵ)$. Both modes grow through their locked resonant interaction. This is formal finite-time blowup of the reduced amplitude equations; the weakly nonlinear expansion loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.

Take $U_0,N_0,k>0$ and assume an initially propagating mode, $k<N_0/U_0$. Put $s=N_0/U_0$ and $\delta (z)=1+\gamma z$. Neglecting $U^{\prime \prime }/U$ as stipulated, the Scorer parameter and stationary internal-wave WKB solution areThe positive root $m>0$ selects upward group velocity for the negative intrinsic frequency branch $\hat{\omega }=-kU$. The WKB approximation needs a slowly varying background relative to the local vertical wavelength, in particular $|m^{\prime }|\ll m^2$ and $|m^{\prime \prime }|\ll |m{|}^3$, away from a turning level or singular mean profile.

The local dispersion relation is $\omega =kU-Nk/(k^2+m^2{)}^{1/2}=0$. Its intrinsic and laboratory group velocities are, with $K=(k^2+m^2{)}^{1/2}$,Thus intrinsic energy propagation is upstream and upward, while laboratory energy propagation is downstream and upward, parallel to $(k,m)$. Constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m$ and normal spacing $2\pi /K$; the horizontal spacing remains $2\pi /k$.

If $\gamma <0$, $m$ grows with height. The wavevector and laboratory energy ray become more vertical; phase lines become flatter and closer together. For finite nonzero $N$, the heightis a critical level of an internal gravity wave: $U\to 0$ and the vertical wavenumber diverges. The laboratory group velocity tends to zero. Importantly, $|m^{\prime }|/m^2\to |\gamma |/s$, so the formal WKB condition need not deteriorate before this level if that ratio is small. Nevertheless the stationary equation is singular at the level, the phase has no finite limit, and the horizontal perturbation velocity grows as $m^{1/2}$; the inviscid linear wave cannot be continued uniformly through it without additional physics.

If $\gamma >0$, $m$ decreases to zero atThe wavevector and laboratory ray become horizontal, the phase lines become vertical, and the normal spacing increases. Above $z_t$, $m^2<0$ and the disturbance is an evanescent wave. Here $|m^{\prime }|/m^2$ diverges, so the WKB approximation fails in a turning region and must be replaced by a local connection solution.

For a progressive plane wave, writeThe momentum equation gives ${\rho }_0\omega U=kP$, while the dispersion relation is $\omega =c_{0}k$. Therefore the pressure and longitudinal velocity amplitudes obeyThe constant $Z$ is the acoustic impedance.