 Solution

Let the unperturbed interface rise at pore velocity $U/\varphi$, and write its displacement asIn each fluid, Darcy's law and incompressible flow implyThe decaying pressure perturbations are proportional to $e^{-\alpha |z|}$. Continuity of normal velocity, the kinematic boundary condition $\varphi {\eta }_t=u_n^{\prime }$, and continuity of pressure giveThus a less mobile displaced fluid, ${\mu }_2>{\mu }_1$, and a denser fluid above a lighter one both drive the Saffman–Taylor instability.

For immiscible fluids, the Young–Laplace equation adds the pressure jump $-\gamma {\eta }_{xx}$. The dispersion relation becomeswhereIf $A>0$, the unstable band is $0<\alpha <\sqrt{A/C}$ and differentiation gives

Define the signed characteristic buoyancy velocitySince ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$ and $Ca={\mu }_{2}U/\gamma$,This expression applies when the quantity in the final parentheses is positive.

Every growth curve starts at the origin. For equal densities its initial slope is proportional to $M$; for ${\rho }_2>{\rho }_1$ buoyancy shifts that slope upward, while for ${\rho }_2<{\rho }_1$ it shifts it downward. When $A>0$, the curve rises to one positive maximum and then crosses zero before its stabilizing $-{\alpha }^3$ capillary tail. When $A\le 0$, every nonzero wavenumber decays.

For a prescribed nonzero wavenumber, neutral stability requiresorIn the quasistatic limit $U\to 0$, viscosity contrast disappears and this reduces to the capillary Rayleigh-Taylor instability threshold ${\rho }_2-{\rho }_1=\gamma {\alpha }^2/g$. Without surface tension, neutral stability in that limit simply requires equal densities.