= Solution
At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equations
$$
-\nabla p+\mu\nabla^2\mathbf u=0,
\qquad
\nabla\cdot\mathbf u=0.
$$
Taking the divergence gives $\nabla^2p=0$. Taking the curl and writing $\boldsymbol\omega=\nabla\times\mathbf u$ gives
$$
\mu\nabla^2\boldsymbol\omega=0.
$$
Thus both pressure and vorticity are harmonic in the fluid.
In the laboratory frame, the boundary conditions are
$$
\mathbf u\to0\quad(r\to\infty),
$$
and, on $r=a$,
$$
\mathbf u\cdot\mathbf n=\mathbf U\cdot\mathbf n,
\qquad
(\mathbf I-\mathbf n\mathbf n^T)\mathbf u
=(\mathbf I-\mathbf n\mathbf n^T)\mathbf U.
$$
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, put
$$
A(r)=\frac{3a}{4r}+\frac{a^3}{4r^3},
\qquad
B(r)=\frac{3a}{4r^3}-\frac{3a^3}{4r^5},
\qquad
s=\mathbf U\cdot\mathbf x.
$$
Then $u_j=AU_j+Bs x_j$, and
$$
\boxed{
\frac{\partial u_j}{\partial x_i}
=\frac{A'}r x_iU_j
+\frac{B'}r s x_ix_j
+B U_ix_j+B s\delta_{ij},}
$$
where
$$
A'=-\frac{3a}{4r^2}-\frac{3a^3}{4r^4},
\qquad
B'=-\frac{9a}{4r^4}+\frac{15a^3}{4r^6}.
$$
Back to article page
