Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 (Slow Viscous Flow)

Put $\delta =H/L\ll 1$. In a thin gap, the streamwise Stokes flow balance isso $P\sim \mu UL/H^2$. The characteristic shear stress is $\tau \sim \mu U/H$, and thereforeThis is the pressure-dominance scaling of lubrication theory.

Let the centre displacement be $\lambda \Delta$ in the downward direction. To first order in $\Delta /a$, projecting this displacement onto the radial direction at polar angle $\theta$ changes the concentric gap $\Delta$ by $-\lambda \Delta \cos \theta$. Hence

Axisymmetric mass conservation in the spherical gap givesSince $h_t=-V\cos \theta$ and regularity requires $q\sin \theta =0$ at $\theta =0$, integration gives

The leading pressure-driven lubrication flux isSubstitution of $h$ and $q$ givesand therefore

Take downward as the positive vertical direction. The constant pressure contributes no resultant, while the pressure force on the inner sphere is opposite its outward normal. ThusWith $t=\lambda \cos \theta$ and the supplied integral,The sign is upward for $V>0$, so this is a drag force. As $\lambda \to 0$, the bracket is $4{\lambda }^3/3+O({\lambda }^5)$ and $F_z\to -8\pi \mu a^{4}V/{\Delta }^3$.

Rotation about the vertical axis gives the inner surface the azimuthal speed $\Omega a\sin \theta$. The leading Couette flow shear traction is opposing and has magnitudeIts moment arm about the vertical axis is $a\sin \theta$, and $dS=2\pi a^2\sin \theta d\theta$. ConsequentlyPutting $t=\lambda \cos \theta$ and using the supplied integral givesThis viscous shear torque opposes the rotation. Its concentric limit is $-8\pi \mu \Omega a^4/(3\Delta )$, agreeing with the thin-gap limit of Torque in rotational Stokes flow between concentric spheres.

A rotation by $\pi$ about the vertical symmetry axis maps a horizontal angular velocity ${\mathbf{\Omega }}^{\prime }$ to $-{\mathbf{\Omega }}^{\prime }$ but leaves a possible vertical force unchanged. Linearity and uniqueness of Stokes flow therefore require that vertical force to equal its own negative, so it vanishes. A horizontal force and a horizontal couple are allowed by the same symmetry.

Write $\lambda =1-ϵ$ and examine the lower pole, where $\theta \ll 1$. Since $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$,Thus $h=O(ϵ\Delta )$ for $\theta =O({ϵ}^{1/2})$, a circular patch of radius

In the broad region, the surface speed is $O({\Omega }^{\prime }a)$, the shear is $O(\mu {\Omega }^{\prime }a/\Delta )$, the area is $O(a^2)$, and the moment arm is $O(a)$. HenceIn the narrow patch, the shear rises to $O(\mu {\Omega }^{\prime }a/(ϵ\Delta ))$, while its area falls to $O(ϵa^2)$; the moment arm remains $O(a)$. Therefore

Dropping the moment arm gives the shear-force scale. Pressure produces the same horizontal order after multiplication by the small surface slope. Both regions consequently contribute

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

The body is invariant under the reflections and half-turns that preserve its $z$-axis. A polar vector force $F{\mathbf{e}}_z$ can therefore produce only the polar velocity $U_z{\mathbf{e}}_z$; every rotation is excluded because angular velocity is a pseudovector. Hence only $U_z$ is nonzero when $\mathbf{F}=(0,0,F)$ and $\mathbf{G}=0$.

For $\mathbf{F}=(F,0,0)$, the surviving symmetry-allowed components arewhile $U_y,U_z,{\Omega }_x,{\Omega }_z$ vanish. The coupling between translation in $x$ and rotation about $y$ is permitted because the two horizontal rods lie at opposite vertical offsets.

For pure translation, integrate the slender-body force density over each rod and take its moment about $O$. The three coordinate directions giveThuswhen $\mathbf{\Omega }=0$.

LetThe symmetry proved in part (a) determines the force generated by rotation from the translation-generated couple. Combining this with the given rotational resistance gives the complete matrix

The zero-couple equations in the two coupled horizontal blocks areand ${\Omega }_z=0$. The force equations then becomeThereforeand

A vector fixed in space has body-frame derivative of a space-fixed vectorSubstitution of part (d) gives

If $\mathbf{\Omega }$ is constant and nonzero, then $k_x$ and $k_y$ are constant and not both zero. Their evolution equations force $k_z=0$, and the last equation then gives $k_x^2=k_y^2$. Hence

For $\mathbf{k}=(1,1,0)$, both the translational velocity and the angular velocity are parallel to the fixed vertical force, with the latter oppositely directed. The body therefore falls on a straight vertical line while spinning steadily about that line. Its body $z$-axis remains horizontal, and its $x$- and $y$-axes remain at $45^{\circ }$ to the vertical.

For the initial condition $(1,0,0)$, symmetry preserves $k_y=0$. With $c=\gamma /L$ the exact body-frame solution isThe body rotates about its $y$-axis, its fall path bends slightly because the horizontal and axial mobilities differ, and $\mathbf{k}\to (0,0,-1)$. Thus the body $z$-axis becomes vertical and the angular velocity tends to zero; asymptotically it falls without rotating.

For an axisymmetric surface $r=a(z,t)$, twice the mean curvature with the outward-normal convention isWriting $a=a_0+\eta$ and retaining linear terms gives

At the unperturbed boundary $r=a_0$, the linearized kinematic boundary condition, tangential-stress condition, and Young–Laplace equation areFor a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, ${\sigma }_{rz}(a_0)=0$, and

The Papkovich–Neuber representation of body-force-free Stokes flow iswhere $\mathbf{\Phi }$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential givesUsing the modified Bessel function identity $(x{I}_1{)}^{\prime }=x{I}_0$, the tangential stress simplifies toThe stress-free condition at $r=a$ therefore yields

The expression in braces in the first equation is the net axial force. The term $-\pi a^2\gamma /a$ is the compressive force from the capillary pressure $\gamma /a$, $3\pi \mu a^2{w}_z$ is the Newtonian extensional tension with Trouton ratio three, and $2\pi a\gamma$ is the axial pull of surface tension around the circumference. Its $z$-derivative vanishes because axial force is conserved.

The final equation is conservation of an insoluble surfactant. The surface divergence $w_z+u/a$ is the sum of axial and circumferential extension rates. Positive surface divergence increases interfacial area and dilutes $C$; negative divergence concentrates it. The absence of a diffusion term expresses the assumption of negligible surface diffusion.

Linearizing the area and surfactant equations givesThusand hencewhere $\psi$ is fixed by the initial data.

Linearizing the displayed net axial force and using ${\gamma }^{\prime }=-A{C}^{\prime }$ givesEliminating $w_z$ and $C^{\prime }$ therefore gives directlyThus the three displayed evolution equations implyThe target formula printed later in the paper contains an additional factor of $\pi$ in both denominators. That factor does not follow from the displayed equations because every term in the axial-force balance contains the same factor $\pi$. If the target formula is adopted as the intended normalization, its corresponding value is $s=({\gamma }_0-A{C}_0)/(6\pi \mu a_0)$.

Initially $\eta =0$ and $C^{\prime }>0$, so $\psi >0$ and ${\eta }_t<0$: a surfactant-rich, low-tension region begins to neck as neighboring higher tension pulls fluid away. The accompanying axial extension dilutes the surfactant.

If $A<{\gamma }_0/C_0$, then $s>0$. The Rayleigh–Plateau instability overwhelms the weak surface-elastic response: the necking perturbation grows in the linear model while the original concentration excess is diluted and eventually changes sign.

If $A>{\gamma }_0/C_0$, then $s<0$. Strong surface elasticity arrests the disturbance atThe concentration perturbation becomes negative, raising the local surface tension until its axial force balances that of the wider regions. This is stabilization by a surfactant-induced Marangoni stress.

The condition $a\gamma =a_0{\gamma }_0$ makes the leading extensional axial force uniform, but it does not make the capillary pressure uniform. Sincea nonuniform radius gives a nonuniform pressure. Its axial gradient must drive flow, so the prediction $w=0$ is inconsistent.

For $a=a_0+\eta$ varying over the axial scale $L$,An axial Hagen-Poiseuille flow in a cylinder has speed scaleCross-sectional mass conservation gives ${\eta }_t\sim a_{0}w/L$, and consequently