Use the <Papkovich–Neuber representation> in the form
$$
\mu\mathbf u=\boldsymbol\Phi-\frac12\nabla(\mathbf x\mathbin\cdot\boldsymbol\Phi)+\nabla\chi,
\qquad
p=-\nabla\mathbin\cdot\boldsymbol\Phi,
$$
where $\boldsymbol\Phi$ and $\chi$ are <harmonic functions>. A torque is a <pseudovector>[axial vector], so rotational covariance and decay select the harmonic vector field
$$
\boldsymbol\Phi=\frac{\mathbf G\times\mathbf x}{8\pi r^3},
\qquad \chi=0.
$$
Here $\mathbf x\cdot\boldsymbol\Phi=0$ and $\nabla\cdot\boldsymbol\Phi=0$, so
$$
\boxed{\mathbf u(\mathbf x)=\frac{\mathbf G\times\mathbf x}{8\pi\mu r^3},\qquad p=0.}
$$
This <rotlet> equals the <rotating sphere in Stokes flow>
$$
\mathbf u=\frac{a^3}{r^3}\boldsymbol\Omega\times\mathbf x
$$
when $\boldsymbol\Omega=\mathbf G/(8\pi\mu a^3)$. It satisfies the <no-slip boundary condition> on $r=a$, decays at infinity, and its <Newtonian fluid stress tensor> transmits the applied couple $\mathbf G$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-329/1/a/solution