= Solution
{c}
Use an <asymptotic expansion> at fixed $z$, evaluating the <boundary conditions> at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives $\hat u_{0,zz}=0$. The wall condition and the surface condition $\bar u_z(h)+\hat u_{0,z}(h)=0$ imply
$$
\boxed{\hat u_0=2(h-1)z,\qquad
u^{(0)}=\bar u+\hat u_0=2hz-z^2.}
$$
Here $u^{(0)}$ denotes the total leading <velocity>, to distinguish it from the dimensional scale $U_0$. The <continuity equation> and wall impermeability then give
$$
w^{(0)}=-z^2h_x.
$$
The transverse momentum equation gives $\hat p_{0,z}=0$. The <normal stress> condition therefore yields
$$
\hat p_0=(h-1)\cot\alpha-\frac32S h_{xx},\qquad
p^{(0)}=(h-z)\cot\alpha-\frac32S h_{xx}.
$$
The leading <volume flux per unit width> is $q^{(0)}=\int_0^hu^{(0)}\,dz=2h^3/3$. Consequently <mass conservation> gives $h_t=-2h^2h_x+O(\varepsilon)$. This relation eliminates $h_t$ inside the first inertial correction:
$$
\begin{aligned}
u^{(0)}_t+u^{(0)}u^{(0)}_x+w^{(0)}u^{(0)}_z
&=2zh_t+(2hz-z^2)(2zh_x)-z^2h_x(2h-2z)\\
&=2hz(z-2h)h_x+O(\varepsilon).
\end{aligned}
$$
Set $G=2h_x\cot\alpha-3S h_{xxx}$. The $O(\varepsilon)$ downslope equation and <boundary conditions> are
$$
\hat u_{1,zz}=G+2\mathrm{Re}\,h h_x(z^2-2hz),\qquad
\hat u_1(0)=0,\quad \hat u_{1,z}(h)=0.
$$
Two <integrations> give
$$
\boxed{\hat u_1=G\left(\frac{z^2}{2}-hz\right)
+\mathrm{Re}\,h_x\left(\frac{hz^4}{6}-\frac{2h^2z^3}{3}+\frac{4h^4z}{3}\right).}
$$
The <pressure> correction $\hat p_1$ is not needed for this calculation because its downslope <pressure gradient> enters only at $O(\varepsilon^2)$.
Integrating the total <velocity> across the film gives
$$
\begin{aligned}
q&=\int_0^h(\bar u+\hat u_0+\varepsilon\hat u_1)\,dz+O(\varepsilon^2)\\
&=\frac23h^3+\varepsilon\left[
-\frac{h^3}{3}G+\frac8{15}\mathrm{Re}\,h^6h_x\right]+O(\varepsilon^2).
\end{aligned}
$$
Indeed the three inertial <polynomial> terms integrate to $\mathrm{Re}\,h^6h_x(1/30-1/6+2/3)=8\mathrm{Re}\,h^6h_x/15$. Integrating the <continuity equation> and using the <kinematic boundary condition> gives $h_t+q_x=0$. Hence the <Benney equation> is
$$
\boxed{h_t+2h^2h_x+\varepsilon\partial_x\left[
h^3\left(-\frac23h_x\cot\alpha+S h_{xxx}\right)
+\frac8{15}\mathrm{Re}\,h^6h_x\right]=0.}
$$
Normal <gravity> smooths height variations on the upper side of the slope, while the inertial correction can amplify them. <Surface tension> suppresses sufficiently short <wavelengths>.
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