Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 74 / 1

For a fixed fluid domain, consider an incompressible Newtonian Stokes flow with prescribed velocities on its boundaries and prescribed far-field behaviour. Conservative body forces, such as gravity, may be absorbed into the pressure. Among all sufficiently regular divergence-free trial velocity fields with exactly those boundary velocities and far-field behaviour, the Stokes solution minimizes the viscous dissipationTo see this Minimum-dissipation theorem for Stokes flow, write a trial field as $\mathbf{v}=\mathbf{u}+\mathbf{w}$, with $\nabla \cdot \mathbf{w}=0$ and homogeneous velocity boundary data. The Newtonian fluid stress tensor satisfies $\nabla \cdot \mathbf{\sigma }=0$, so integration by parts givesThereforeEquality requires the difference to be a rigid motion, eliminated by the fixed wall and far-field data. The theorem varies the velocity field in one fixed geometry; it does not minimize over particle positions. If a nonconservative body force is prescribed, the appropriate variational functional also includes its work, rather than being the dissipation alone.

Neglect tube-end effects and use the localized flow with no imposed throughflow, as implied by the sphere being the only cause of motion. Extend the fluid velocity through the sphere by its rigid-body velocity. This extended field is continuous across the surface by the no-slip boundary condition and divergence-free both inside and outside. The divergence theorem therefore makes its total flux through a complete tube section independent of height. In sections far from the sphere the fluid is quiescent, so this total flux is zero.

At height $z$ cutting the sphere, let $A_s(z)$ be the area of its circular slice and take $z$ upward. The translation contributes $V_z{A}_s$. The vertical component of $\mathbf{\Omega }\times (\mathbf{x}-{\mathbf{x}}_c)$ integrates to zero over that circle because its first moments about the centre vanish. Hence the zero total flux for localized tube Stokes flow giveswhere $V_z<0$ for a falling sphere. Thus the net fluid flux is upwards in every section intersecting the sphere's interior, even though some fluid locally moves downwards. At a tangential section $A_s=0$ and the flux vanishes. The zero-total-flux condition is important: an imposed throughflow or appreciable open-tube end effects would change this argument.

The Minimum-dissipation theorem for Stokes flow compares fields with fixed geometry and boundary velocities. Moving the sphere changes the geometry, while sedimentation prescribes its effective weight rather than its velocity. It therefore supplies no rule for migrating toward a position of least dissipation. Indeed, at fixed force $W$, the power is $WV$, and a lower drag resistance gives a larger settling speed and larger power, rather than a smaller power. This is a fixed-force comparison of minimum viscous dissipation issue.

The proposed migration is itself false in ideal Stokes flow. Reflect the uniform vertical tube in a horizontal plane through the sphere centre. The sphere's lateral position is unchanged, but the applied axial force reverses. Reflection leaves a transverse translational velocity unchanged. By Linearity of Stokes flow, however, reversing the force reverses that transverse velocity. It must therefore be zero. This proves no lateral migration of a single sphere in a uniform tube in Stokes flow:Rotation and a wall-dependent vertical speed are allowed. Inertial lift, particle deformation or tube-end effects belong to different problems and do not follow from the dissipation theorem.

Let downward velocity be positive in this part, and let $W>0$ be the dense sphere's weight minus buoyancy. For the original centreline configuration, write its six-component translation-and-rotation velocity as $q$, and its diagonal positive hydrodynamic resistance matrix as $R_0$. Its single-sphere dissipation isWith only axial force and no torque, the original velocity is $V_0=W/R_{zz}$.

Now add the neutral sphere. In its actual Stokes flow, extend the velocity rigidly through the new sphere's interior. This extension is a valid incompressible trial field for the domain without that sphere, with exactly the dense sphere's actual six boundary velocities; its strain and dissipation inside the inserted ball are zero. The Minimum-dissipation theorem for Stokes flow and extra dissipation due to a rigid inclusion therefore giveThe neutral sphere has zero net external force and torque, and the dense sphere has only the axial force $W$. The dissipation equals total external power, so $D=W{V}_z$. Positivity gives $V_z>0$, and the comparison yields $V_z\le W/R_{zz}$ without assuming that either sphere translates only vertically or that either angular velocity vanishes.

For a finite nonoverlapping placement, the first inequality is strict. Equality would make the extended field the old Stokes solution and thus make that solution rigid throughout the added ball. Analyticity and connectedness then force its strain to vanish throughout the original fluid domain; the stationary tube wall and quiescent far field would force the whole motion to vanish, contradicting $V_z>0$. Hence the force-free inclusion reduces axial mobility of a centred settling sphere:The difference tends to zero as the added sphere recedes infinitely far away. The lateral and rotational contributions are nonnegative in the resistance comparison, which is precisely why they cannot invalidate the bound.

Let $\delta \ll a$ be the minimum gap as the spheres approach, and $w=-\dot{\delta }>0$ their normal relative approach speed. For smooth equal-radius spheres the gap has local form $\delta +O(s^2/a)$, so the thin-gap region has lateral scale $\ell \sim (a\delta {)}^{1/2}$. Incompressibility drives a radial escape speed of order $w\ell /\delta$. The lubrication pressure therefore scales asThe available gravitational forcing remains finite. Thus lubrication resistance gives $w=O(W\delta /(\mu a^2))$: the normal approach becomes proportionally slower as the gap closes. In particular, for a bounded coefficient $C$, $\dot{\delta }\ge -C\delta$, so $\delta (t)\ge \delta (t_1)e^{-C(t-t_1)}>0$ at every finite time. Equivalently the estimated time to reach zero contains ${\int }_0^{{\delta }_1}d\delta /\delta$, which diverges. This is how lubrication prevents finite-time collision of smooth spheres.

The small nonzero initial lateral offset breaks the exactly collinear configuration; the spheres can deflect sideways and pass with a positive gap rather than touch. The condition $R>2a$ leaves space for a side-by-side passage. Smooth surfaces, continuum hydrodynamics and the no-slip boundary condition are essential to the noncontact conclusion; the argument is not a model of surface roughness or molecular contact.

Choose time $t_c$ when the passing sphere centres are at the same height, and let $S$ reflect positions in that horizontal plane. For each instantaneous configuration, Stokes flow determines particle velocities uniquely and linearly from the fixed external forces. Time reversal reverses all velocities and the dense sphere's applied force; reflecting in the horizontal plane reverses that force again. The neutral sphere remains force-free under both operations. Thus the reflected time-reversed motion obeys the same original forcing.

At $t_c$, reflection fixes both particle centres. Since the spheres have no orientational shape variables and there is no contact, uniqueness of the particle-trajectory initial-value problem makes the reflected backward motion coincide with the forward motion:This reflection symmetry of a two-sphere passing trajectory fixes the transverse coordinates and reverses the relative axial separation. Consequently, at matching large separations above and below, both spheres return to their original distances from the centreline. Their intermediate lateral excursions are undone. This uses kinematic reversibility of Stokes flow, geometric reflection symmetry and the noncontact result, not a dissipation-minimizing trajectory. It assumes the spheres pass one another; a perfectly collinear approach would not supply the side-by-side symmetry event.

When both spheres have the same effective downward weight, the second sphere adds a force-driven disturbance instead of being merely a force-free rigid inclusion. The power now contains both $W{V}_{1z}$ and $W{V}_{2z}$, so the one-forced-sphere bound from part (iv) no longer applies.

For $a\ll R$, place the spheres in a tandem configuration with $a\ll s\ll R$. Locally the flow around the second sphere is dominated by its downward Stokeslet. Its axial incident velocity at the first sphere is downward, so the first sphere falls faster to leading order. By contrast, the zero total flux for localized tube Stokes flow requires compensating upward flow away from the downward-moving particle core in sections cutting the sphere. There are therefore positions in the return flow where the disturbance acts upwards.

One can choose the second sphere in such an upward-flow region generated by the first. The Lorentz reciprocal theorem makes the axial cross coefficients of the hydrodynamic mobility matrix equal, so the downward force on the second sphere then produces an upward contribution to the first sphere's velocity. With small well-separated spheres, this force-driven interaction dominates the higher-order self-resistance change from simply inserting the other sphere. In mobility notation, with downward positive,and the cross coefficient can have either sign in the tube. ThusThe distinction is the sign of the incident flow, not whether adding a second geometrical obstacle lowers a fixed-velocity dissipation.