= Solution
{c}
The flat plate moves tangentially and the curved plate is stationary. The <no-slip boundary condition> and <no-penetration boundary condition> give
$$
\boxed{u=-U,\quad w=0\quad(z=0),\qquad
u=0,\quad w=0\quad(z=h(x)).}
$$
At the curved plate, impermeability can also be written $w=u h_x$; together with zero tangential <velocity>, this gives both components zero. The native PDF writes $v'=0$ in the scaled wall conditions although the normal component is defined as $w'$. We use $w'$ consistently.
The <parabolic lubrication gap> changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is therefore
$$
\boxed{L=\sqrt{Rh_0},\qquad \delta=\frac{h_0}{L}=\sqrt{\frac{h_0}{R}}\ll1.}
$$
The proposed normal <velocity> scale $h_0U/L$ then makes the two terms of the <continuity equation> comparable. Balancing the horizontal <pressure gradient> against the transverse viscous term in the <Navier-Stokes equation> requires $P/L=\mu U/h_0^2$, so
$$
\boxed{P=\frac{\mu UL}{h_0^2}.}
$$
Here <pressure> is measured from a common reference value, which is chosen at the outer fluid in part (b).
For clarity, define the dimensionless <material derivative>
$$
\mathcal D'=\partial_{t'}+u'\partial_{x'}+w'\partial_{z'},
\qquad \mathrm{Re}=\frac{\rho LU}{\mu}.
$$
Substitution of all the scalings gives the exact dimensionless equations
$$
\begin{aligned}
u'_{x'}+w'_{z'}&=0,\\
\delta^2\mathrm{Re}\,\mathcal D'u'&=-p'_{x'}+u'_{z'z'}+\delta^2u'_{x'x'},\\
\delta^4\mathrm{Re}\,\mathcal D'w'&=-p'_{z'}+\delta^2w'_{z'z'}+\delta^4w'_{x'x'}.
\end{aligned}
$$
For example, the factor in the last left side is obtained by dividing the vertical <acceleration> scale $\rho h_0U^2/L^2$ by the vertical <pressure gradient> scale $P/h_0$.
Taking $\delta\ll1$ and $\delta^2\mathrm{Re}\ll1$ leaves the <lubrication approximation>
$$
\boxed{u'_{x'}+w'_{z'}=0,\qquad
0=-p'_{x'}+u'_{z'z'},\qquad 0=-p'_{z'}.}
$$
The scaled gap and <boundary conditions> are
$$
\boxed{h'(x')=1+\frac{x'^2}{2},\qquad
(u',w')=(-1,0)\ (z'=0),\qquad
(u',w')=(0,0)\ (z'=h'(x')).}
$$
The <Reynolds number> itself need not be small: it is $\delta^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.
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