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At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equationsTaking the divergence gives ${\nabla }^{2}p=0$. Taking the curl and writing $\mathbf{\omega }=\nabla \times \mathbf{u}$ givesThus both pressure and vorticity are harmonic in the fluid.

In the laboratory frame, the boundary conditions areand, on $r=a$,The first is impermeability of the solid surface and the second is viscous no slip. Together they say $\mathbf{u}=\mathbf{U}$ on the sphere.

For the stated solution, putThen $u_j=A{U}_j+Bs{x}_j$, andwhere