= Solution
{c}
Define the aspect ratio and the thickness-based <Reynolds number> by
$$
\boxed{\varepsilon=\frac{h_0}{L}\ll1,\qquad
\mathrm{Re}=\frac{\rho U_0h_0}{\mu}.}
$$
Use the <nondimensionalization>
$$
x=Lx_*,\quad z=h_0z_*,\quad t=\frac L{U_0}t_*,\quad
u=U_0u_*,\quad w=\varepsilon U_0w_*,\quad
p=p_a+\rho gh_0\sin\alpha\,p_*.
$$
The small transverse <velocity> scale follows from <incompressibility>. Dropping stars, the base fields become $\bar u=2z-z^2$ and $\bar p=(1-z)\cot\alpha$. Before subtracting the base state, the dimensionless <Navier-Stokes equation> reads
$$
\begin{aligned}
u_x+w_z&=0,\\
\mathrm{Re}\,\varepsilon(u_t+uu_x+wu_z)
&=-2\varepsilon p_x+u_{zz}+\varepsilon^2u_{xx}+2,\\
\mathrm{Re}\,\varepsilon^2(w_t+uw_x+ww_z)
&=-2p_z+\varepsilon(w_{zz}+\varepsilon^2w_{xx})-2\cot\alpha.
\end{aligned}
$$
For example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0h_0^2/(\mu L)=\mathrm{Re}\,\varepsilon$ and the <pressure gradient> coefficient $2\varepsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller <velocity> scale; this gives the displayed $\mathrm{Re}\,\varepsilon^2$ coefficient.
Now substitute $u=\bar u+\hat u$, $w=\hat w$, $p=\bar p+\hat p$. Since $\bar u_{zz}+2=0$, $\bar p_z=-\cot\alpha$, and the base fields have no $x$ or $t$ dependence, subtraction gives
$$
\boxed{\begin{aligned}
\hat u_x+\hat w_z&=0,\\
\mathrm{Re}\,\varepsilon[\hat u_t+(\bar u+\hat u)\hat u_x+\hat w(\bar u_z+\hat u_z)]
&=-2\varepsilon\hat p_x+\hat u_{zz}+\varepsilon^2\hat u_{xx},\\
\mathrm{Re}\,\varepsilon^2[\hat w_t+(\bar u+\hat u)\hat w_x+\hat w\hat w_z]
&=-2\hat p_z+\varepsilon(\hat w_{zz}+\varepsilon^2\hat w_{xx}).
\end{aligned}}
$$
These are nonlinear equations for the deviations; small <amplitude> has not been assumed. The later <asymptotic expansion> takes $\mathrm{Re}=O(1)$ as $\varepsilon\to0$, or more generally requires the retained inertial correction to remain small.
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