= Solution
Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,
$$
d=h+(\eta_2-\eta_1)e^{i\alpha z+\sigma t}.
$$
The <lubrication theory> flux down the vertical surface, with a <no-slip boundary condition> at the ice and a <stress-free boundary condition> at the water-air interface, is
$$
q=\frac{d^3}{3\nu}
\left(g-\frac1\rho p_z\right).
$$
For the unperturbed film $p_z=0$, so
$$
q=\frac{gh^3}{3\nu},
\qquad
\boxed{h=\left(\frac{3\nu q}{g}\right)^{1/3}}.
$$
The linearized <Young–Laplace equation> gives the capillary-pressure perturbation
$$
p'=\gamma\alpha^2\eta_2e^{i\alpha z+\sigma t},
\qquad
p'_z=i\gamma\alpha^3\eta_2e^{i\alpha z+\sigma t}.
$$
Because the prescribed <volume flux> is uniform, its first-order perturbation must vanish. <Linearization> of the flux law gives
$$
0=\frac{gh^2}{\nu}(\eta_2-\eta_1)
-\frac{ih^3\gamma\alpha^3}{3\rho\nu}\eta_2.
$$
Thus, with $\Gamma=\gamma/(3\rho g)$,
$$
\eta_2(1-i\Gamma\alpha^3h)=\eta_1,
$$
and hence
$$
\boxed{
\eta_2=\frac{\eta_1}
{1-i\Gamma\alpha^3h}}.
$$
The complex amplitude ratio records the phase shift caused by <surface tension> in the <long-wave approximation>.
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