Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 (Slow Viscous Flow)

In a cross-section normal to the cylinder axes, measure $y$ from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at $y=w+o(w)$, so the leading cross-sectional area isThe area of the small meniscus cap is $O(w^4/a^2)$ and is lower order. At its upper end the gap has width $h(w)=w^2/a$. A tangent semicircle therefore has radius $w^2/(2a)$ and curvature

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed $y$, lubrication theory gives a planar Poiseuille flow through a gap of width $h(y)$, with axial flux per unit $y$Integrating across the cusp,The continuity equation $A_t+Q_z=0$ now gives

Set $q=w^3$. Since $w^4{w}_z=\frac{1}{5}(q^{5/3}{)}_z$, this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution $q=t^{-3/8}F(z/t^{3/8})$. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is $z_N=(C/B{)}^{1/2}{t}^{3/8}$. Usingin the volume constraint gives $C^2=8aV\sqrt{B}/\pi$. Therefore

For vertical cylinders at equilibrium, hydrostatic pressure gives $p-p_{\mathrm{atm}}=-\rho gz$. Balancing this with the capillary pressure $-2\gamma a/w^2$ gives the large-height profile

More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the surface tension there, so the resulting Marangoni stress drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is chemophoresis.

The bulk concentration obeys the advection-diffusion equationAt $r=a$,equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.

With advection neglected, $\Gamma$ is harmonic. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is ${\partial }_r\Gamma =0$ at $r=a$. This gives $B=a^3/2$, and hence

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.

The traction is a first spherical harmonic, so the decaying, force-free Stokes flow has no Stokeslet and is generated by the indicated Papkovich–Neuber representation. Comparing the supplied tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.

Near the closest point, the cylinder–plane gap isFor translation parallel to the cylinder axis, the leading flow is Couette flow. Its shear traction is $\mu U/h$, soHence $A_{yy}=2\pi {\epsilon }^{-1/2}\mu L$.

For transverse translation, the local Couette-Poiseuille flow in a thin gap and mass conservation give the Reynolds lubrication equation. Writing $x=a\sqrt{\epsilon }X$ and using pressure recovery at both ends determines its integration constant. The pressure and viscous contributions to the horizontal traction reduce towhere the final term is the direct Couette shear contribution. Therefore

Treat each short torus segment as a straight cylinder. For translation with velocity $U{\mathbf{e}}_x$, its components normal and tangent to the circular centreline are $U\cos \theta$ and $-U\sin \theta$. Integrating the local resistance per unit length around $ds=Rd\theta$ gives

Rotation at angular velocity $\Omega {\mathbf{e}}_z$ gives the everywhere tangential speed $\Omega R$. Only the axial-cylinder resistance $2\pi \mu {\epsilon }^{-1/2}$ contributes. Multiplying the force by its moment arm $R$ and integrating around the centreline gives

In the torus frame the two planes translate with velocity $U{\mathbf{e}}_x$. Away from the $O(a)$ neighborhood of the torus, the depth-averaged Hele–Shaw flow between planes separated by $2a$ isNegligible leakage imposes ${\overline{u}}_r=0$ at $r=R$. The harmonic pressure that decays at infinity in the exterior and the regular harmonic pressure in the interior are thereforeup to a common constant. The interior velocity is zero, while the exterior flow is the uniform stream diverted around a circular obstacle. In plan view the inside has high pressure on the $+x$ side and low pressure on the $-x$ side; the immediately adjacent exterior has the opposite signs, producing the pressure jump across the torus.

The jump at $r=R$ isIntegrating it over the projected vertical area $2aRd\theta$ gives the global pressure resistanceThere are two narrow gaps, so their local resistance is twice the one-plane result from part b:Consequently the local gap resistance dominates when $a\ll R\ll a{\epsilon }^{-1/2}$, whereas the global Hele–Shaw pressure resistance dominates when

To interpret the upper bound, the pressure jump has scale $\Delta p\sim \mu UR/a^2$. Each narrow gap has thickness $a\epsilon$ and streamwise lubrication length $a\sqrt{\epsilon }$. Its pressure-driven leakage flux per unit centreline length therefore scales asThe blocked Hele–Shaw flux has scale $Ua$. Leakage is negligible precisely when $UR{\epsilon }^{5/2}\ll Ua$, or