= Solution
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 $p_{a0}'=-\mu U/k_a$ and $p_{b0}'=-\mu U/k_b$. 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> are
$$
\delta p_b=B e^{\alpha x}e^{\sigma t+i\alpha y},\qquad \delta p_a=A e^{-\alpha x}e^{\sigma t+i\alpha y}.
$$
Expanding <pressure continuity> at the displaced interface and matching normal <Darcy flux> gives
$$
B-A=\mu U\left(\frac1{k_b}-\frac1{k_a}\right)\eta=-\frac{\mu U\Delta k}{k_ak_b}\eta,\qquad -k_bB=k_aA.
$$
Hence
$$
B=-\frac{\mu U\Delta k}{k_b(k_a+k_b)}\eta,\qquad A=\frac{\mu U\Delta k}{k_a(k_a+k_b)}\eta.
$$
The 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, $\sigma\eta=\delta u_n/(1+S\Delta\phi)$. Thus the <melting-front instability due to permeability contrast> has
$$
\boxed{\sigma=\frac{\alpha U}{1+S\Delta\phi}\frac{\Delta k}{2k+\Delta k}.}
$$
For a signed transverse <wavenumber>, replace $\alpha$ by $|\alpha|$.
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