Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 39D (Fluid Dynamics)

The steady Stokes equations are linear and contain no inertial term. If every imposed boundary velocity and body force is reversed, then the velocity and pressure departure from any uniform reference pressure reverse sign. Thus every fluid particle retraces its path in the reversed experiment. This is the principle of kinematic reversibility.

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.

At vertical displacement $x$ from the axis, the gap isFor $|x|\ll a$,soThe narrow-gap length scale is therefore $(a{h}_0{)}^{1/2}$.

Take $x$ upward, let $y$ measure distance from the wall, and work in the cylinder frame. The wall at $y=0$ moves upward with speed $V$. Choose the sign of $\Omega$ so that the near-cylinder surface at $y=h$ moves upward with speed $a\Omega$. Lubrication theory givesThusThe vertical volume flux per unit cylinder length is independent of $x$ and equalsThereforeBecause $p(+\infty )=p(-\infty )$, its integral over $x$ vanishes. Withthe quoted integrals giveIt follows that

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder isSubstitution of the pressure gradient givesA torque-free cylinder requires ${\int }_{-\infty }^{\infty }{\tau }_{c}dx=0$. Also,Using $Q=2h_0(V+a\Omega )/3$, the torque condition reduces toHence