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In the lubrication regime the thin region has longitudinal scale $\ell \sim \sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell /b\sim \mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.

Use a frame translating with the cylinder. The wall moves with velocity $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, soContinuity makes $q$ constant. Equal ambient pressures at both infinities give $\int p_{x}dx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. ThereforeThe vertical velocity, with the wall condition $w(0)=0$, follows explicitly from incompressibility:It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.

The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}s$, its integral isForce balance equates this wall force to the required cylinder force at leading order; the pressure and viscous stresses on remote boundaries make no leading contribution. This is the cylinder translating parallel to a wall in a thin gap result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.