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The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and a harmonic scalar $\chi$ asFor translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf{U}$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\mathbf{\Omega }\times \mathbf{x}/r^3$. Matching the no-slip boundary condition $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at $r=a$ gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.

When $\mathbf{U}=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

The force-free Stokes flow equations arePut $p={\nabla }^2\Pi$. Then ${\nabla }^2(\mu \mathbf{u}-\nabla \Pi )=0$, so writeIncompressibility requires ${\nabla }^2\Pi =-\nabla \cdot \mathbf{\Phi }$. Since ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$ for harmonic $\mathbf{\Phi }$, takeThis gives the Papkovich–Neuber representation

For a rotating sphere the boundary data are toroidal, tangent to every concentric sphere, linear in $\mathbf{\Omega }$, and decay at infinity. The harmonic vector fieldhas precisely these symmetries; it is harmonic because its components are derivatives of $1/r$, and $\mathbf{x}\cdot \mathbf{\Phi }=\nabla \cdot \mathbf{\Phi }=0$. HenceThe first pressure argument is $p=-\nabla \cdot \mathbf{\Phi }=0$. Independently, this velocity is harmonic, so the Stokes momentum equation gives $\nabla p=0$; matching the ambient pressure sets that constant to zero.

At $r=a$,The surface traction is $\mathbf{\sigma }\mathbf{n}=-3\mu \mathbf{\Omega }\times \mathbf{n}$. Its moment gives the standard rotational resistance