Rotating roller under a nearly flat free surface ($Q=4Ub/3$)

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 $Q=Ud-d^3{p}_x/(3\mu )$. 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 $\mu |U|\sqrt{a}/(\rho g{b}^{5/2})\ll 1$.