= Solution
A convenient normalization of the <Papkovich–Neuber representation> uses a harmonic vector $\mathbf h$ and harmonic scalar $\phi$:
$$
\boxed{\mathbf u=\mathbf h-\tfrac12\nabla(\mathbf x\cdot\mathbf h+\phi),
\qquad p=-\mu\nabla\cdot\mathbf h.}
$$
Indeed, $\Delta(\mathbf x\cdot\mathbf h)=2\nabla\cdot\mathbf h$, so $\nabla\cdot\mathbf u=0$ and $\mu\Delta\mathbf u=\nabla p$. This is a rescaling of the usual harmonic-potential representation of <Stokes flow>.
Measure $\mathbf x$ from the sphere center, set $r=|\mathbf x|$ and $\mathbf n=\mathbf x/r$. A translational vector harmonic $\mathbf U/r$ provides the decaying $1/r$ <force> field, while the scalar dipole $(\mathbf U\cdot\mathbf x)/r^3$ adjusts the surface <velocity> without changing that leading far field. For rotation, $(\boldsymbol\Omega\times\mathbf x)/r^3$ is harmonic, divergence-free and perpendicular to $\mathbf x$. Thus try
$$
\mathbf h=A\frac{\mathbf U}{r}+B\frac{\boldsymbol\Omega\times\mathbf x}{r^3},
\qquad\phi=C\frac{\mathbf U\cdot\mathbf x}{r^3}.
$$
The translational <velocity> is $\frac A{2r}(I+\mathbf n\mathbf n)\mathbf U-\frac C{2r^3}(I-3\mathbf n\mathbf n)\mathbf U$. Matching its independent tangential and radial components at $r=a$ gives $A=3a/2$, $C=-a^3/2$; matching rotation gives $B=a^3$. Consequently
$$
\boxed{\mathbf u=\left[\frac{3a}{4r}(I+\mathbf n\mathbf n)
+\frac{a^3}{4r^3}(I-3\mathbf n\mathbf n)\right]\mathbf U
+\frac{a^3}{r^3}\boldsymbol\Omega\times\mathbf x,\qquad
p=\frac{3\mu a}{2r^3}\mathbf U\cdot\mathbf x.}
$$
This superposes the <translating sphere in Stokes flow> and the <rotating sphere in Stokes flow>. It is exactly $\mathbf U+\boldsymbol\Omega\times\mathbf x$ at the sphere and tends to zero at infinity; the additive ambient <pressure> is set to zero.
Back to article page
