Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 / 1

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