= Solution
Let $X=x_{\rm dim}/L$, $Y=y/h_0$, scale horizontal and vertical velocities by $U$ and $Uh_0/L$, and define $P=(p_{\rm dim}-p_0)h_0^2/(\mu UL)$. The small <aspect ratio> makes $P$ independent of $Y$ at leading order. The second small parameter, $Uh_0^2/(\nu L)$, makes <inertia> negligible compared with transverse viscous stress. Thus the <lubrication theory>[lubrication equations] are $u_{YY}=P_X$ and $u_X+V_Y=0$.
<No-slip boundary condition>[No slip] gives $u(X,0)=1$, $u(X,h)=0$, so
$$
\boxed{u=1-\frac Yh+\frac{P_X}{2}Y(Y-h),\qquad q=\int_0^hu\,dY=\frac h2-\frac{h^3P_X}{12}.}
$$
<No-penetration boundary condition>[No penetration] at the two solid walls and continuity make $q$ constant. Therefore, writing the <dimensionless> variables again as $x,p$,
$$
\boxed{\frac d{dx}(h^3p'-6h)=0,\qquad p(0)=p(1)=0.}
$$
The <pressure> scale is set by balancing the longitudinal <pressure gradient> with shear across the thin gap, not by the inertial dynamic-<pressure> scale.
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