= 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 gives
$$
\boxed{u(X,y)=\frac{Uy}{h}+\frac{p_X}{2\mu}y(y-h).}
$$
The relative <velocity> in the cylinder frame is $u-U$. Its flux is constant by incompressibility:
$$
q=-\frac{Uh}{2}-\frac{h^3p_X}{12\mu}.
$$
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 $q=-2Uh_0/3$ and
$$
\boxed{p_X=\frac{2\mu U(4h_0-3h)}{h^3}.}
$$
Indeed $p=2\mu UX/h^2$ 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.
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