= Solution
{c}
Write $H(x')=h'(x')$ and $p'_x=dp'/dx'$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x'$. The vertical <pressure gradient> is zero, and integrating $u'_{z'z'}=p'_x$ twice with the wall <boundary conditions> gives the local <Couette-Poiseuille flow>
$$
\boxed{u'(x',z')=-1+\frac{z'}{H}
+\frac{p'_x}{2}z'(z'-H).}
$$
The first two terms are <Couette flow> due to the moving plate; the last is <plane Poiseuille flow> due to the local <pressure gradient>. Its dimensionless <volume flux per unit width> is
$$
q(x')=\int_0^{H(x')}u'(x',z')\,dz'
=-\frac H2-\frac{H^3}{12}\frac{dp'}{dx'}.
$$
Use the <Leibniz integral rule> and <incompressibility> to differentiate this <integral>:
$$
\frac{dq}{dx'}=\int_0^H u'_{x'}\,dz'+u'(x',H)\frac{dH}{dx'}
=-w'(x',H)+w'(x',0)=0.
$$
Thus \b[the flux is independent of position], and
$$
\boxed{\frac d{dx'}\left(\frac{H^3}{12}\frac{dp'}{dx'}+\frac H2\right)=0.}
$$
The normal <velocity> can then be recovered from <incompressibility>:
$$
w'(x',z')=-\int_0^{z'}u'_{x'}(x',s)\,ds.
$$
It is zero on the lower wall by construction and zero on the upper wall by the constant-<volume flux> relation. This supplies the remaining component of the leading <velocity field>.
The endpoint <pressure> conditions are matching conditions, not an assertion that the <lubrication approximation> remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous <pressure> variation is of size $\mu U/R$, whose ratio to the inner scale is
$$
\frac{\mu U/R}{P}=\frac{h_0^2}{RL}=\delta^3\ll1.
$$
Even an outer inertial <pressure> scale $\rho U^2$ has ratio
$$
\frac{\rho U^2}{P}=\delta^2\mathrm{Re}\ll1.
$$
Consequently the gap <pressure> is much larger than either outer variation. Matching to fluid with a common ambient <pressure> at both ends, and taking that value as zero <gauge pressure>, gives
$$
\boxed{p'(x')\longrightarrow0\quad(x'\to\pm\infty).}
$$
There is an <overlap region> $1\ll|x'|\ll\delta^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed <pressure> difference comparable with $P$ would change these conditions and the resulting <volume flux>; equal outer <pressure> is the physical assumption being used here.
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