= Solution
The force-free <Stokes flow> equations are
$$
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad \nabla\cdot\mathbf u=0.
$$
Put $p=\nabla^2\Pi$. Then $\nabla^2(\mu\mathbf u-\nabla\Pi)=0$, so write
$$
\mu\mathbf u=\nabla\Pi+\boldsymbol\Phi,
\qquad \nabla^2\boldsymbol\Phi=0.
$$
Incompressibility requires $\nabla^2\Pi=-\nabla\cdot\boldsymbol\Phi$. Since $\nabla^2(\mathbf x\cdot\boldsymbol\Phi)=2\nabla\cdot\boldsymbol\Phi$ for harmonic $\boldsymbol\Phi$, take
$$
\Pi=-\frac12\mathbf x\cdot\boldsymbol\Phi+\chi,
\qquad \nabla^2\chi=0.
$$
This gives the <Papkovich–Neuber representation>
$$
\boxed{\mu\mathbf u=\boldsymbol\Phi-\frac12\nabla(\mathbf x\cdot\boldsymbol\Phi)+\nabla\chi},
\qquad
\boxed{p=-\nabla\cdot\boldsymbol\Phi}.
$$
For a rotating sphere the boundary data are toroidal, tangent to every concentric sphere, linear in $\boldsymbol\Omega$, and decay at infinity. The harmonic vector field
$$
\boldsymbol\Phi=\mu a^3\frac{\boldsymbol\Omega\times\mathbf x}{r^3},
\qquad \chi=0,
$$
has precisely these symmetries; it is harmonic because its components are derivatives of $1/r$, and $\mathbf x\cdot\boldsymbol\Phi=\nabla\cdot\boldsymbol\Phi=0$. Hence
$$
\boxed{\mathbf u=\frac{a^3}{r^3}\boldsymbol\Omega\times\mathbf x},
\qquad \boxed{p=0}.
$$
The first pressure argument is $p=-\nabla\cdot\boldsymbol\Phi=0$. Independently, this velocity is harmonic, so the Stokes momentum equation gives $\nabla p=0$; matching the ambient pressure sets that constant to zero.
At $r=a$,
$$
\boxed{\frac{\partial\mathbf u}{\partial r}=-2\boldsymbol\Omega\times\mathbf n=-\frac2a\boldsymbol\Omega\times\mathbf x}.
$$
The surface traction is $\boldsymbol\sigma\mathbf n=-3\mu\boldsymbol\Omega\times\mathbf n$. Its moment gives the standard rotational resistance
$$
\boxed{\mathbf G_{\rm fluid}=
\int_{r=a}\mathbf x\times(\boldsymbol\sigma\mathbf n)\,dS
=-8\pi\mu a^3\boldsymbol\Omega}.
$$
Back to article page
