= Rotating roller under a nearly flat free surface
{title2=$Q=4Ub/3$}
A rotating roller with top speed $U$ and gap $d(x)=b+x^2/(2a)$ drives a shear-free <lubrication approximation> flow. No slip below and zero shear above give $u=U+p_x(y^2-d^2)/(2\mu)$ and $Q=Ud-d^3p_x/(3\mu)$. The <pressure recovery condition in lubrication flow> then gives $Q=U\int d^{-2}dx/\int d^{-3}dx=4Ub/3$. With $s=x/\sqrt{2ab}$, the excess pressure is $-\mu U\sqrt{2a}\,b^{-3/2}s/(1+s^2)^2$. Gravity supports it through the small surface displacement $h=(p-p_0)/(\rho g)$; the flat leading geometry is consistent when $\mu|U|\sqrt a/(\rho g b^{5/2})\ll1$.
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