Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 (Slow Viscous Flow)

The force-free Stokes flow equations arePut $p={\nabla }^2\Pi$. Then ${\nabla }^2(\mu \mathbf{u}-\nabla \Pi )=0$, so writeIncompressibility requires ${\nabla }^2\Pi =-\nabla \cdot \mathbf{\Phi }$. Since ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$ for harmonic $\mathbf{\Phi }$, takeThis gives the Papkovich–Neuber representation

For a rotating sphere the boundary data are toroidal, tangent to every concentric sphere, linear in $\mathbf{\Omega }$, and decay at infinity. The harmonic vector fieldhas precisely these symmetries; it is harmonic because its components are derivatives of $1/r$, and $\mathbf{x}\cdot \mathbf{\Phi }=\nabla \cdot \mathbf{\Phi }=0$. HenceThe first pressure argument is $p=-\nabla \cdot \mathbf{\Phi }=0$. Independently, this velocity is harmonic, so the Stokes momentum equation gives $\nabla p=0$; matching the ambient pressure sets that constant to zero.

At $r=a$,The surface traction is $\mathbf{\sigma }\mathbf{n}=-3\mu \mathbf{\Omega }\times \mathbf{n}$. Its moment gives the standard rotational resistance

At leading order the particle is a sphere. A pure applied couple produces no translation, while torque balance with the rotating sphere in Stokes flow givesWith $mathbfn$ directed from the particle into the fluid, the exact conditions on the true surface $S$ areEvaluate no slip at $mathbf{x}_s=(a+\epsilon f)\mathbf{n}$ and expand about $mathbfx=a\mathbf{n}$. The order-$\epsilon$ terms give

The divergence of the Newtonian stress vanishes in the surrounding fluid, and its symmetry makes the divergence of angular-momentum flux vanish as well. The total force and torque may therefore be evaluated on any homologous enclosing surface, in particular the fixed reference sphere. Since the applied force is zero and the applied couple is fixed independently of $\epsilon$,Using a fixed enclosing sphere is also why no separate terms involving the shape $f$ and leading stress ${\mathbf{\sigma }}_0$ appear.

Apply the Lorentz reciprocal theorem for Stokes flow to ${\mathbf{u}}_1$ and the test flow around a sphere rotating with arbitrary $\hat{\mathbf{\Omega }}$. On $r=a$,The reciprocal integral containing $\hat{\mathbf{u}}\cdot {\mathbf{\sigma }}_1\mathbf{n}$ vanishes by the first-order torque condition. Since ${\partial }_r{\mathbf{u}}_0=-2{\mathbf{\Omega }}_0\times \mathbf{n}$, the first-order boundary velocity isThe translational term integrates to zero. Therefore, for every $\hat{\mathbf{\Omega }}$,NowFor $f=a(\mathbf{n}\cdot \mathbf{Dn})$, tracelessness of $mathbfD$ and the stated fourth-moment identity giveIt follows thatA centered ellipsoid has inversion symmetry. An applied axial couple is unchanged under inversion, whereas a translational velocity is reversed, so uniqueness of Stokes flow forces

At leading order in the small axial slope, the outer flow is locally the two-dimensional radial incompressible flowbecause $(r{u}_r{)}_r=0$ and the kinematic boundary condition is $u_r(a)=a_t$. This field is harmonic as a vector field away from the axis, so the exterior pressure is spatially constant and may be set to zero. Its radial normal stress at the interface isNeglecting the $O(a_z)$ tangential corrections, axial curvature, and the small internal viscous normal stress, the Young–Laplace equation with cylindrical curvature $1/a$ givesThus

Interpret the stated ansatz asMultiplying the pressure relation by $a$ gives $2\mu a_t=Pa-\gamma$. Equating its constant and sinusoidal parts yieldsThe volume per wavelength is proportional to the mean of $a^2$, namely $f^2+g^2$. HenceChoosingthen gives exactly

For $g\ll f\simeq a_0$, the disturbance amplitude obeysso its long-wave linear growth rate isThe minimum radius is $a_{\min }=f-\sqrt{2}g$, andSince the trajectory follows the circle $f^2+g^2={\alpha }_0$ toward increasing $g$, it reaches $f=\sqrt{2}g$ and hence $a_{\min }=0$. Thus the nonlinear disturbance narrows monotonically to pinch-off within this approximation.

When internal pressure gradients matter, axial lubrication flow in the low-viscosity interior has volume fluxConservation of cross-sectional area, $(\pi a^2{)}_t+Q_z=0$, and the normal-stress expression for $P$ give

Let $\tau =t^{\ast }-t$. Balancing $2\mu a_t/a$ with $\gamma /a$ gives $a\sim \tau$, so $p=1$. Balancing either pressure contribution after two axial derivatives with $a{a}_t$ then gives $q=1$. DefineAt fixed $z$, $a_t=C(\zeta A^{\prime }-A)$, andSubstitution leaves the parameter-free third-order equationThus the similarity exponents are $\boxed{p=q=1}$; boundary and matching conditions would select a particular profile $A$.

Let $y$ measure distance normal to the plane. The normal momentum balance and capillary pressure condition giveThe downslope lubrication equation isApply no slip $u(0)=-U$ and zero tangential stress $u_y(h)=0$. Integration gives the fluxThe thin-film equation $h_t+q_x=0$ is therefore

Long-wave information near a uniform film propagates with the kinematic speedWhen $c>0$, disturbances travel from $x=-\infty$ into the domain, so $h\to h_0$ is an admissible upstream boundary condition. When $c<0$, information travels toward $x=-\infty$ from the pool, so the same condition cannot independently be imposed there.

For a stationary vertical plane, set $U=0$, $\sin \theta =1$, and $\cos \theta =0$. One integration of the steady equation, using $h\to h_0$ and vanishing derivatives far above the pool, givesDefineThen

Put $H=1+y$ and linearize to obtain $y^{\prime \prime \prime }=-3y$. The two characteristic roots with positive real part are $3^{1/3}{e}^{\pm i\pi /3}$. These are the modes that decay as $X\to -\infty$, soThe oscillations grow as one moves from the uniform film toward the pool.

In a static meniscus, hydrostatic pressure variation $\rho g\ell$ balances capillary pressure $\gamma /\ell$, giving the capillary lengthThe small parameter satisfiesIt is reasonable to match to a nearly static meniscus because the bulk pool has negligible thin-film viscous resistance, while its local capillary-hydrostatic shape approaches the vertical wall almost tangentially.

Since $h=h_{0}H$ and $x=h_{0}X/ϵ$, dimensional curvature isMatching it to $(2\rho g/\gamma {)}^{1/2}=\sqrt{2}/\ell$ gives

The dimensionless equation is $H^3{H}^{\prime \prime \prime }=1-H^3$. At a crest, $H\gg 1$, soThe imposed far-field flux is negligible there; the dominant physical balance is between local gravity-driven drainage and the capillary-pressure gradient.

At a trough, $H\ll 1$, soLocal gravity-driven flux is negligible; the capillary-pressure gradient drives the fixed flux through the trough against viscous resistance. These alternating balances generate the strongly nonlinear capillary waves.

Write $\delta =ϵ/2$. In the final trough, useThe trough equation $H^3{H}^{\prime \prime \prime }=1$ requires $C^4/L^3=1$, so $L=C^{4/3}$. Matching the limiting curvature $H^{\prime \prime }\to {\delta }^{-1/2}$ to ${\varphi }^{\prime \prime }\to 1$ requiresThusand the final trough isIf $\varphi (\xi )\sim -A\xi$ as $\xi \to -\infty$, its slope on the crest side is

Integrating ${\psi }^{\prime \prime \prime }=-1$ and using $\psi (0)={\psi }^{\prime }(0)=0$ givesThe conditions at ${\eta }_0>0$ imply $C={\eta }_0/3$ and $-{\eta }_0^2/6=-1$. Hence

Let the final crest have $H=B\psi ((X-X_c)/L)$. The crest balance $H^{\prime \prime \prime }=-1$ gives $B=L^3$. Matching its terminal slope to the final-trough slope from part (e) givesThereforeAt the end matching to the penultimate trough,

The trough scaling from part (e) says that a trough matched to curvature $K$ has thickness scale $K^{-3/5}$. With $K\asymp {\delta }^{-1/20}$, the penultimate trough therefore hasup to numerical constants. The exponent $3/100$ is extremely small, so attaining a clean asymptotic separation would require unrealistically tiny $ϵ$. Finite geometry, nonzero outer effects, molecular forces, and eventual rupture can intervene before experiments display many members of the predicted wave hierarchy.