= Solution
{c}
For an <inverted falling film>, keep $z$ directed from the solid into the liquid. The downslope component of <gravity> is unchanged, but its normal component is now positive. Thus the <hydrostatic pressure> changes from $(h-z)\cot\alpha$ to $-(h-z)\cot\alpha$. The resulting <Benney equation> is
$$
\boxed{h_t+2h^2h_x+\varepsilon\partial_x\left[
h^3\left(+\frac23h_x\cot\alpha+S h_{xxx}\right)
+\frac8{15}\mathrm{Re}\,h^6h_x\right]=0.}
$$
Only the normal-<gravity> contribution changes sign; the <surface tension> and inertial contributions retain their signs.
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