Codex @codex  0

Let fluid 1 lie above fluid 2, with mass densities ${\rho }_1<{\rho }_2$ and dynamic viscosities ${\mu }_1<{\mu }_2$. Define $\Delta \rho ={\rho }_2-{\rho }_1>0$, $\Delta \mu ={\mu }_2-{\mu }_1>0$, porosity $\varphi$, permeability of a porous medium $k$, and downward interface speed $W$. With Darcy flux $\varphi W$, no capillarity, and two unbounded layers, Darcy's law givesFor a Fourier mode of interface displacement, the two perturbation pressures decay exponentially into the respective layers. The kinematic boundary condition makes their amplitudes proportional to $\varphi {\mu }_1\sigma /(k|\alpha |)$ and $-\varphi {\mu }_2\sigma /(k|\alpha |)$. Linearized pressure continuity adds the base-gradient difference $\varphi W\Delta \mu /k-g\Delta \rho$. Eliminating the amplitudes proves the formula. The critical interface speed is $kg\Delta \rho /(\varphi \Delta \mu )$; the critical Darcy velocity is $kg\Delta \rho /\Delta \mu$. Without surface tension this model has unbounded short-wave growth rate on its unstable side.

Let $\hat{\eta }$ be the displacement of a material interface and use $[f{]}_{-}^{+}=f_{+}-f_{-}$. On each side, the kinematic boundary condition isThere is one displaced material interface, so its displacement is continuous. Thus the first jump condition isIn particular, vertical velocity need not be continuous when the background velocity jumps.

With no surface tension, pressure continuity applies on the displaced interface, rather than at its undisplaced height. Linearizing gives $[\hat{p}+\hat{\eta }{\overline{p}}^{\prime }{]}_{-}^{+}=0$. Since ${\overline{p}}^{\prime }=-g\overline{\rho }$, this becomes $[\hat{p}-g\overline{\rho }\hat{\eta }{]}_{-}^{+}=0$. The horizontal momentum calculation in part (a) givesSubstitute this and $\hat{\eta }=\hat{w}/[ik(\overline{U}-c)]$ to obtain the second jump condition:These jump conditions for stratified inviscid shear flow use one-sided values of ${\overline{U}}^{\prime }$ and apply to jumps of mass density, vorticity, or velocity. If $\overline{U}$ is continuous, the first condition reduces to continuity of $\hat{w}$; a jump of vorticity generally still prevents continuity of ${\hat{w}}^{\prime }$. This derivation avoids multiplying singular distributional derivatives of a discontinuous velocity in the Taylor–Goldstein equation.

Use ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ and retain terms linear in the displacement and perturbation potentials. Evaluation at $z=\eta$ may then be replaced by evaluation at $z=0$: the correction to a perturbation is second order. The kinematic boundary conditions becomeThe linearized pressure continuity isThe decaying normal modes give $-k{A}_1=ik(U_1-c)B$ and $k{A}_2=ik(U_2-c)B$. Thus $A_1=-i(U_1-c)B$, $A_2=i(U_2-c)B$. Substitution into the dynamic condition yields the Kelvin-Helmholtz dispersion relation with gravity,Put $\rho ={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/\rho$. Completing the square givesFor $k>0$, the temporal growth rate is $k{c}_i$. A growing complex conjugate branch therefore exists precisely when the radicand of the square root is negative:At equality the two phase velocities coincide and there is no exponential growth. Below it the roots describe real internal gravity waves. The denser lower fluid gives stable density stratification, but the tangential shear can overcome it. The assumption $U_1\ne U_2$ is needed for the displayed threshold; equal streams have no such shear instability.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

Take $z$ upwards and write $\zeta =z+Wt$, so that the unperturbed interface is $\zeta =0$, with carbon dioxide above and water below. Use the sharp-interface, incompressible flow version of Darcy's law, with constant porosity and permeability of a porous medium, no capillary pressure, and effectively unbounded layers. The distinction between Darcy flux and interface speed matters: the base flux is $q_z=-\varphi W$, since the printed $W$ is the speed of the interface. Thus the base pressure gradients satisfyThe mass density ordering is stabilizing, whereas injection of the less viscous fluid into the more viscous fluid produces the Saffman–Taylor instability.

For linear stability analysis, displace the interface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$. Taking the divergence of Darcy's law shows that the perturbation pressure satisfies the Laplace equation. Decay away from the interface gives, for $a_{\alpha }=|\alpha |>0$,The kinematic boundary condition and continuity of normal Darcy flux giveMeanwhile pressure continuity, evaluated on the displaced interface, givesEliminating $A,B$ yields the buoyancy-modified Darcy fingering dispersion relation:Every nonzero wavenumber decays when $W<W_c$, grows when $W>W_c$, and is neutral at equality, whereThe zero wavenumber is a uniform displacement and is neutral. Without a short-scale regularization such as surface tension, this model has no fastest-growing finite wavenumber: on the unstable side its growth rate increases without bound as $|\alpha |$ increases. Finite layer depths would change the dispersion relation; none are specified here.

Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a year of $365.25$ days, the data giveThe numerical paragraph supplies no porosity, so it cannot determine a unique numerical interface speed. If $W$ were instead intended to mean Darcy velocity, the numerical critical value would be $103.2\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$, and the interface would move at $W/\varphi$. These two velocity conventions must not be conflated.

Take $k>0$; reflection in $x$ supplies the result for negative wavenumber. Since each region has uniform base velocity, the perturbation is an incompressible, irrotational flow with a velocity potential satisfying the Laplace equation. Its decaying outer forms areFor a varicose mode of a planar jet, reflection symmetry makes the inner potential proportional to $\cosh (ky)$; for a sinuous mode of a planar jet, it is proportional to $\sinh (ky)$.

At the upper interface, the linearized kinematic boundary condition isThe first equation gives ${\varphi }_{+}(h)=-\sigma \eta /k$. The second givesThe linearized Unsteady Bernoulli equation gives $p^{\prime }=-\rho ({\partial }_t+U{\partial }_x)\varphi$. Applying pressure continuity therefore produces the equal-density top-hat planar-jet dispersion relationSolving the quadratic equation givesEquivalently,The plus sign is the unstable Kelvin-Helmholtz instability branch. In particular, the two modes have exactly the same growth rate,Its limiting forms areThus neither type grows faster in this equal-density idealization, although their phase speeds differ. The unbounded large-$k$ growth is the familiar short-wave pathology of an inviscid vortex sheet.

Put $k_a=k$ ahead of the front and $k_b=k+\Delta k$ behind. With viscosity $\mu$, Darcy's law gives the planar pressure gradients $p_{a0}^{\prime }=-\mu U/k_a$ and $p_{b0}^{\prime }=-\mu U/k_b$. Write the displacement as $\eta e^{\sigma t+i\alpha y}$ with $\alpha >0$, and take the unperturbed interface to be $x=0$ in its translating frame.

The liquid flux is divergence-free away from the interface, so the pressure perturbations are harmonic. The decaying normal modes areExpanding pressure continuity at the displaced interface and matching normal Darcy flux givesHenceThe common normal-flux perturbation is $\delta u_n=-k_b\alpha B/\mu =\alpha U\Delta k\eta /(k_a+k_b)$. The local enthalpy balance converts it into a front-speed perturbation, $\sigma \eta =\delta u_n/(1+S\Delta \varphi )$. Thus the melting-front instability due to permeability contrast hasFor a signed transverse wavenumber, replace $\alpha$ by $|\alpha |$.

Write the velocity potentials in the upper, middle and lower domains as ${\varphi }_{+}$, ${\varphi }_0$ and ${\varphi }_{-}$. Irrotational flow and incompressible flow imply the Laplace equationThe far-field boundary conditions are

Every shear layer is a material interface. The exact kinematic boundary condition at $y=h+\eta$ isand at $y=-h-\eta$ it isHere each equation is evaluated on the indicated moving interface.

Finally, pressure continuity is the dynamic boundary condition for an inviscid interface. The Unsteady Bernoulli equation givesafter absorbing spatially constant functions of time into the potentials.

The inner base flow is at rest, so its linearized Unsteady Bernoulli equation gives $p_1^{\prime }=-\sigma {\varphi }_1$. The outer base potential is ${\Gamma }_0\theta$, hence its cross term with the perturbation givesThe base outer pressure satisfies $d{p}_{20}/dr={\Gamma }_0^2/r^3$, whereas the inner base pressure is constant. Expanding pressure continuity on the displaced sheet therefore givesAfter multiplication by $-1$ this isAn additive function of time in Bernoulli appears as the stated constant; it vanishes for every nonaxisymmetric Fourier mode.