Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 342 / 4

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Let $n$ point from the sphere into the exterior fluid, so the surface position relative to the centre is $r=an$. The physical surface velocity isFree rotation means the actual body is torque-free: the total hydrodynamic torque $T={\int }_{S}an\times (\sigma \cdot n)dS$ vanishes. Use the auxiliary rotating sphere in Stokes flow with arbitrary $\hat{\Omega }$, boundary velocity $\hat{u}=\hat{\Omega }\times an$ and the given traction $\hat{\sigma }\cdot n=-3\mu \hat{\Omega }\times n$. The auxiliary flow is the decaying exterior rotation of a sphere, not a rigid rotation of the entire infinite fluid.

The left-hand cross-work in the Lorentz reciprocal theorem for Stokes flow isThe other cross-work becomesThe spherical integral identities areUsing $(\Omega \times n)\cdot (\hat{\Omega }\times n)=\Omega \cdot \hat{\Omega }-(\Omega \cdot n)(\hat{\Omega }\cdot n)$, the rotational contribution is $-8\pi \mu a^3\Omega \cdot \hat{\Omega }$. The slip contribution is $-3\mu \hat{\Omega }\cdot {\int }_{S}n\times u^{\prime }dS$. Since $\hat{\Omega }$ is arbitrary, the torque-free rotation of a spherical squirmer satisfiesThe viscosity cancels, as expected for a kinematically prescribed slip in a linear viscous problem. If the freely moving sphere also has a translation $U$, its added cross-work is zero because the auxiliary rotating sphere has zero total force, ${\int }_S\hat{\sigma }\cdot ndS=0$. Thus translation does not alter this rotational formula. Without the torque-free assumption an external-torque contribution would have to be retained.

Write the angular dependence as $f(\varphi )={\sum }_{n\ge 1}{a}_n\sin n\varphi +{\sum }_{n\ge 0}{b}_n\cos n\varphi$, so $u^{\prime }=\sin \theta f(\varphi )e_{\varphi }$. In the spherical coordinate system, $n=e_r$ and $e_r\times e_{\varphi }=-e_{\theta }$, withThus the integral in the torque-free rotation of a spherical squirmer has componentsThe vanishing follows from north-south cancellation, including the potentially relevant first azimuthal harmonics. In the $z$ direction, orthogonality of complex exponentials givesConsequently every rotational component is determined:This assumes an integrable prescribed angular function, with the Fourier series convergent sufficiently to justify the surface integrals; a finite Fourier sum is enough for the construction below. Other harmonics can generate exterior flows even though they contribute no net rotation.

For flow-free rotation of a spherical squirmer, choose $b_0\ne 0$ and set every $a_n$ and every $b_n$ with $n\ge 1$ to zero. Then $u^{\prime }=b_0\sin \theta e_{\varphi }$, while the rigid-rotation velocity isThe total laboratory-frame boundary velocity is therefore zero everywhere. With fluid at rest at infinity, Uniqueness of Stokes flow givesA constant pressure supplies zero resultant force and torque, so this is consistent with free motion. The imposed surface actuation counter-rotates relative to the material sphere; no claim is being made that ordinary no-slip rigid rotation produces zero flow.