Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 (Slow Viscous Flow)

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

Put $S(\mathbf{x})=\mathbf{x}\cdot \mathbf{A}\cdot \mathbf{x}$. Since $\mathbf{u}=-S{r}^{-5}\mathbf{x}$ and $\nabla \times \mathbf{x}=0$, the product rule givesFor the symmetric stresslet tensor $\mathbf{A}$,The final term is parallel to $\mathbf{x}$ and drops out of the cross product, leaving

The Linearity of Stokes flow makes every velocity linear in ${\mathbf{G}}_1$. The only available isotropic polar vector built from the axial vector ${\mathbf{G}}_1$ and separation vector $\mathbf{R}$ is ${\mathbf{G}}_1\times \mathbf{R}$. Dimensional analysis then givesAn angular velocity is axial, so the two independent isotropic possibilities are ${\mathbf{G}}_1$ and $({\mathbf{G}}_1\cdot \mathbf{R})\mathbf{R}$. Hencefor dimensionless scalar functions ${\beta }_i,{\gamma }_i$.

Write $\mathbf{\Omega }={\mathbf{G}}_1/(8\pi \mu a^3)$. The incident rotlet of sphere 1 at sphere 2 isand it is harmonic away from sphere 1. Since sphere 2 is force-free, Faxén's first law therefore givesThe vorticity of the rotlet isBecause ${\mathbf{G}}_2=0$, Faxén's rotational law gives

The symmetric rate-of-strain tensor of the incident rotlet at sphere 2 isAfter translation and rotation have matched the uniform and antisymmetric parts of the incident flow, the leading perturbation from sphere 2 is the stresslet part of the supplied straining-sphere solution:Part b gives its vorticity asAt the centre of sphere 1, $\mathbf{x}=-\mathbf{R}$, soApplying Faxén's rotational law to sphere 1 produces half this ambient vorticity and proves

To hold sphere 2 fixed against the incident angular velocity from part ii, its applied couple must generate the bare rotationThe resulting rotlet advects the force-free sphere 1. Since the field is harmonic, Faxén's first law gives

The returned rotlet has velocity $O(\Omega a^6/R^5)$ and rate of strain $O(\Omega a^6/R^6)$ at sphere 1. That strain induces a stresslet of strength $O(\mu \Omega a^9/R^6)$, whose velocity at sphere 2 is $O(\Omega a^9/R^8)$. This method of reflections for Stokes flow gives the stated order of the next correction to ${\mathbf{U}}_2$.

Write the perturbation as $\eta e^{ikx+st}$ and use the same factor for all velocity and pressure amplitudes. At the upper interface, the linearized kinematic boundary condition, zero tangential traction, and normal-stress balance areThe first term in the normal stress is the linearization of the attractive disjoining pressure $V/h^3$, while the second is the stabilizing capillary pressure. Symmetry about $y=0$ makes $u$ even and $v$ odd, and supplies the lower-interface conditions.

For a two-dimensional Fourier mode, the Papkovich–Neuber representation is equivalently expressed by the odd Biharmonic stream function for planar Stokes flowThe tangential-stress condition at $y=h_0$, with $K=k{h}_0$, giveswhereas the kinematic condition gives $iB\sinh K=s\eta$. The associated normal traction isEquating this with the linearized interfacial traction yields the dispersion relationThis is the Van der Waals rupture instability of a viscous sheet.

For $\Gamma =0$, the growth rate is positive, starts from $3V/(4\mu h_0^3)$ as $K\to 0$, and decreases to zero like $3V/(2\mu h_0^{3}K)$ as $K\to \infty$. For $\Gamma =1$, it has the same long-wave limit, vanishes at $K=1$, and is negative for $K>1$: surface tension damps wavelengths shorter than the cutoff. Long waves feel the attractive interaction but require coherent flow over a large distance; at short wavelengths viscous resistance suppresses the clean-film instability, while capillarity adds direct decay. A finite film size, fluid inertia, surrounding-fluid stresses, gravity, surface viscosity, and failure of the continuum disjoining pressure law can shift the observable most unstable wavelength.

During a growing thin spot, interfacial flow stretches the surface and dilutes its surfactant, thereby increasing the local surface tension above ${\gamma }_0$. Adjacent less-stretched regions retain more surfactant and lower tension. The resulting surface-tension gradient pulls toward the thin spot and opposes the outward flow that drives thinning. This is surfactant stabilization of film rupture; in the strong limit the surfaces behave almost as immobile boundaries.

For strong surfactant and $\Gamma \gg 1$, instability requires $K=O({\Gamma }^{-1/2})$. The Taylor expansionreduces the supplied relation toIt is maximal atThus the characteristic rupture time is $s_{\max }^{-1}=4\mu {\gamma }_0{h}_0^5/(3V^2)$. Surfactant greatly extends the life of a soap bubble while its film is moderately thick, but the $h_0^{-5}$ growth-rate dependence predicts rapid final rupture after drainage has made the film sufficiently thin.

Reflect the configuration in a plane perpendicular to the tube axis. The geometry and radial displacement are unchanged, whereas the imposed pressure gradient and every axial velocity reverse. By kinematic reversibility of Stokes flow, a radial migration velocity would also have to reverse; the spatial reflection leaves that component unchanged. Uniqueness of Stokes flow therefore forces it to vanish. Reflection in the meridional plane through the two axes similarly excludes azimuthal drift, so $c$ remains constant.

Apply axial momentum balance to the fluid between the remote sections $x_1$ and $x_2$. The pressure forces on the end discs, the integrated tube-wall shear, and the force exerted by the sphere are the only resultant axial forces. The sphere is force-free, so its contribution vanishes andThis is the wall-shear and pressure-drop balance in a tube.

The nearly occluding sphere in a cylindrical tube has gap thickness $H=O(ϵa)$ and axial length $L=O(\sqrt{aH})=O(a\sqrt{ϵ})$. In the gap, lubrication theory therefore givesFar ahead of and behind the sphere, the length and width scales are both $a$, so

In the sphere frame the tube wall at $y=0$ moves with velocity $-U$ and the sphere surface at $y=h$ is stationary. The local Couette-Poiseuille flow in a thin gap isIts axial volume flux per unit circumferential width isDifferentiating the profile and eliminating $p_x$ gives the two wall stressesThe continuity equation integrated across the gap says that changes of $q$ along $x$ are balanced by circumferential flux divergence. Circumferential variations occur on scale $a$, much longer than the axial scale $a\sqrt{ϵ}$, so $q=q(\theta )$ is independent of $x$ at leading order.

Put $h=h_0(1+{\xi }^2)$ with $x=\sqrt{2a{h}_0}\xi$. The leading $O({ϵ}^{-3/2})$ pressure jump must vanish in the global balance from part ii. Equivalently, the pressure recovery condition in lubrication flow givesUsing $I_2=\pi /2$ and $I_3=3\pi /8$,

The total flux through the gap is consequentlyFar from the sphere, translation of the tube contributes $-\pi a^{2}U$, while Hagen-Poiseuille flow contributes $\pi a^4{p}_x/(8\mu )$. Equating these fluxes gives

At the next order, substitute $q=-2U{h}_0/3$ into the tube-wall shear and integrate through the gap:The balance in part ii then gives the leading pressure drop

Finally, the shear on the sphere integrates tobecause $I_1=\pi$ and $I_2=\pi /2$. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.