 Solution

Write $D={\partial }_n{A}_n$ and use repeated-index summation. Because every $A_i$ is a harmonic function,andThus $\mu {\nabla }^2{u}_i-{\partial }_{i}p=0$, the incompressible Stokes equation. MoreoverThe antisymmetric first two terms cancel in the sum with ${\partial }_i{u}_j$, giving the Newtonian fluid Cauchy stress tensorFor $A_i=V_i(1-a/(2r))$, harmonic outside the origin,At $r=a$ this gives $\mathbf{u}=\frac{1}{2}[\mathbf{V}-(\mathbf{V}\cdot \mathbf{n})\mathbf{n}]$, where $\mathbf{n}=\mathbf{x}/a$, so the normal velocity is zero. AlsoHence on the surface,a purely normal traction, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$, so the fluid's force on the sphere isThis is the clean-bubble Stokes drag in the far-field velocity $\mathbf{V}$.