= Solution
The <Papkovich–Neuber representation> writes a homogeneous incompressible <Stokes flow> in terms of a harmonic vector field $\boldsymbol\Phi$ and harmonic scalar $\chi$:
$$
2\mu\mathbf u
=\nabla(\mathbf x\mathbin\cdot\boldsymbol\Phi+\chi)
-2\boldsymbol\Phi,
\qquad
p=\nabla\mathbin\cdot\boldsymbol\Phi,
$$
with $\nabla^2\boldsymbol\Phi=0$ and $\nabla^2\chi=0$. The representation satisfies incompressibility because $\nabla^2(\mathbf x\mathbin\cdot\boldsymbol\Phi)=2\nabla\mathbin\cdot\boldsymbol\Phi$, and substitution then verifies the <Stokes equation>.
For a sphere translating with constant vector velocity $\mathbf U$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:
$$
\boldsymbol\Phi=-\frac{3\mu a}{2r}\mathbf U,
\qquad
\chi=\frac{\mu a^3}{2}\frac{\mathbf U\mathbin\cdot\mathbf x}{r^3}.
$$
Substitution gives the <translating sphere in Stokes flow>
$$
\boxed{
\mathbf u(\mathbf x)=
\frac{3a}{4r}\left(\mathbf I+
\frac{\mathbf x\mathbf x}{r^2}\right)\mathbf U
+\frac{a^3}{4r^3}\left(\mathbf I-3
\frac{\mathbf x\mathbf x}{r^2}\right)\mathbf U},
$$
$$
\boxed{p(\mathbf x)=
\frac{3\mu a}{2r^3}\mathbf U\mathbin\cdot\mathbf x}.
$$
At $r=a$ the radial tensor terms cancel and $\mathbf u=\mathbf U$, while $\mathbf u\to0$ as $r\to\infty$, so the <no-slip boundary condition> and far-field condition hold. The resulting traction integrates to the <Stokes drag law> $6\pi\mu a\mathbf U$ in magnitude.
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