 Solution

A convenient normalization of the Papkovich–Neuber representation uses a harmonic vector $\mathbf{h}$ and harmonic scalar $\varphi$: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, $(\mathbf{\Omega }\times \mathbf{x})/r^3$ is harmonic, divergence-free and perpendicular to $\mathbf{x}$. Thus tryThe translational velocity is $\frac{A}{2r}(I+\mathbf{nn})\mathbf{U}-\frac{C}{2r^3}(I-3\mathbf{nn})\mathbf{U}$. Matching its independent tangential and radial components at $r=a$ gives $A=3a/2$, $C=-a^3/2$; matching rotation gives $B=a^3$. ConsequentlyThis superposes the translating sphere in Stokes flow and the rotating sphere in Stokes flow. It is exactly $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at the sphere and tends to zero at infinity; the additive ambient pressure is set to zero.