Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 66 / 1

Work in the frame in which the distant fluid is stationary, and take the pressure at infinity as zero. In the Unscaled Papkovich–Neuber representation, choose the harmonic potentialsBoth are harmonic away from the sphere's centre: $1/r$ is harmonic there and $\mathbf{V}\cdot \mathbf{x}/r^3=-\mathbf{V}\cdot \nabla (1/r)$. Substitution into $\mathbf{u}=\nabla (\mathbf{x}\cdot \mathbf{\Phi }+\chi )-2\mathbf{\Phi }$, $p=2\mu \nabla \cdot \mathbf{\Phi }$, gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. At $r=a$ the coefficients of $\mathbf{I}$ sum to one and those of $\mathbf{nn}$ sum to zero, so the no-slip boundary condition is satisfied. The field decays at infinity. The Papkovich–Neuber representation enforces the Stokes equation and incompressibility, and Uniqueness of Stokes flow identifies the physical solution.

Differentiate radially at fixed direction $\mathbf{n}$:On the surface, multiplying the specified Newtonian fluid stress tensor by the outward normal from the solid givesThus the surface traction in translating-sphere Stokes flow is uniform and the fluid's force on the sphere isThis is the Stokes drag law; an external force of the opposite sign maintains the motion.

Let $\mathbf{F}$ denote the applied force, and take the normal ${\mathbf{n}}_S$ outward from the solid. At leading order the Stokes drag law and the torque on a rotating sphere in Stokes flow giveThe exact no-slip boundary condition and the exact resultant-load conditions areThe stress condition specifies the total force and torque, not a pointwise prescribed traction.

For the boundary perturbation of a nearly spherical particle, evaluate the no-slip condition at $\mathbf{x}=(a+\epsilon f)\mathbf{n}$. Taylor expansion givesThere is no $f{\mathbf{\Omega }}_0\times \mathbf{n}$ term because ${\mathbf{\Omega }}_0=0$. Therefore

To expand the integral conditions correctly, use surface independence of Stokes force and torque integrals. The Stokes equation implies $\nabla \cdot \mathbf{\sigma }=0$, and symmetry of the stress tensor implies that the angular-momentum flux is divergence-free as well. Evaluate the exact force and torque on a fixed sphere of radius $R$ enclosing the perturbed particle. There are then no geometric terms: the $O(\epsilon )$ integrals of ${\mathbf{\sigma }}_1$ on that fixed sphere vanish because the external force and torque have no $O(\epsilon )$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the divergence theorem transfers those integrals back to $r=a$. HenceIf the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of ${\mathbf{\sigma }}_0$, the changed normal and the changed area do appear individually. Their sum vanishes by the same surface-independence identity. Thus their absence in the final equations is a cancellation, not permission to omit geometric terms independently.

Write the first-order surface velocity as ${\mathbf{u}}_1={\mathbf{U}}_1+{\mathbf{\Omega }}_1\times \mathbf{x}+\mathbf{g}$, where the previous derivative givesAs a test field, choose a sphere translating with arbitrary velocity $\hat{\mathbf{V}}$. Its surface velocity is $\hat{\mathbf{V}}$ and its traction is $\hat{\mathbf{\sigma }}\mathbf{n}=-3\mu \hat{\mathbf{V}}/(2a)$. The Lorentz reciprocal theorem for Stokes flow givesTerms at infinity vanish because both velocities decay. Since $\int \mathbf{n}dS=0$, the rotation term has zero surface average. The arbitrary test velocity therefore gives the first-order mobility of a nearly spherical particle:

For the ellipsoidal perturbation, the second spherical moment gives $\int fdS=(4\pi a^3/3)\mathrm{tr}\mathbf{D}=0$. The supplied fourth moment gives, componentwise,Substitution yieldswhere the last notation uses symmetry of the tensor.

One can also use a rotating-sphere test with surface traction $-3\mu \hat{\mathbf{\Omega }}\times \mathbf{n}$. Its reciprocal integral, together with zero first-order torque, givesHere $f(-\mathbf{n})=f(\mathbf{n})$, so the last integrand is odd and its integral vanishes. Thus ${\mathbf{\Omega }}_1=0$. Inversion symmetry forbids a translation-to-rotation coupling for this centred ellipsoidal perturbation.