For two zero-body-force <Stokes flows> $(\mathbf u,\boldsymbol\sigma)$ and $(\widehat{\mathbf u},\widehat{\boldsymbol\sigma})$ in the same domain, the <Lorentz reciprocal theorem for Stokes flow> states
$$
\int_{\partial D}\mathbf u\mathbin\cdot\widehat{\boldsymbol\sigma}\mathbf n\,dS
=\int_{\partial D}\widehat{\mathbf u}\mathbin\cdot\boldsymbol\sigma\mathbf n\,dS.
$$
Apply it first with the auxiliary translating-sphere solution and then with the auxiliary rotating-sphere solution. The swimmer is <force-free> and <torque-free>, while the auxiliary surface tractions are known. The resulting <surface slip velocity> formulas are
$$
\boxed{
\mathbf V=-\frac1{4\pi a^2}\int_{r=a}\mathbf u_s\,dS,
\qquad
\boldsymbol\omega=-\frac3{8\pi a^4}\int_{r=a}\mathbf x\times\mathbf u_s\,dS.}
$$

On $r=a$, the first part of the prescribed slip is
$$
\frac{(\mathbf A\times\mathbf x)\times\mathbf x}{a^2}
=\frac{\mathbf x(\mathbf A\mathbin\cdot\mathbf x)}{a^2}-\mathbf A.
$$
Its surface average is $-2\mathbf A/3$, whereas the $\mathbf B$ term has zero average by <odd function>[oddness]. Thus
$$
\boxed{\mathbf V=\frac23\mathbf A.}
$$
The $\mathbf A$ term contributes no rotation. For the other term, the isotropic second and fourth surface moments give
$$
\int_{r=a}\mathbf x\times\mathbf u_s\,dS
=\frac{8\pi a^3}{15}|\mathbf B|^2\mathbf B,
$$
and hence
$$
\boxed{\boldsymbol\omega=-\frac{|\mathbf B|^2}{5a}\mathbf B.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 329 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-329/1/b/solution