 Solution

In the laboratory frame the lower wall has velocity zero and the cylinder surface has velocity $U$. Solving $\mu u_{yy}=p_X$ with these no-slip conditions givesThe relative velocity in the cylinder frame is $u-U$. Its flux is constant by incompressibility:Equal pressures at the two infinities require $\int p_{X}dX=0$. Put $I_j=\int h^{-j}dX$; then $q=-(U/2)I_2/I_3$. With $X=\ell s$ and $h=h_0(1+s^2)$, the given integrals yield $I_2/I_3=4h_0/3$. Hence $q=-2U{h}_0/3$ andIndeed $p=2\mu UX/h^2$ has this derivative and tends to ambient pressure at both ends. The distinction between the constant comoving flux and the varying laboratory flux is necessary because the gap translates.