= Solution
When $\mathbf B=0$, part (b) gives $\mathbf V=2\mathbf A/3$ and $\boldsymbol\omega=0$. The total boundary velocity is therefore
$$
\mathbf V+\mathbf u_s
=-\frac13\mathbf A+\frac{\mathbf x(\mathbf A\mathbin\cdot\mathbf x)}{a^2}.
$$
The decaying <velocity potential>
$$
\phi=-\frac{a^3}{3}\frac{\mathbf A\mathbin\cdot\mathbf x}{r^3}
$$
is harmonic for $r>a$, and
$$
\boxed{\mathbf u=\nabla\phi
=-\frac{a^3}{3}\nabla\left(\frac{\mathbf A\mathbin\cdot\mathbf x}{r^3}\right)}
$$
has exactly this value at $r=a$. It is a force-free potential-dipole field and decays as $r^{-3}$.
When $\mathbf A=0$, define the degree-three <harmonic polynomial>
$$
H_3(\mathbf x)=(\mathbf B\mathbin\cdot\mathbf x)^3
-\frac35|\mathbf B|^2r^2(\mathbf B\mathbin\cdot\mathbf x).
$$
An appropriate decaying harmonic potential is
$$
\Psi=C\frac{H_3(\mathbf x)}{r^7},
\qquad
\mathbf u=\nabla\Psi\times\mathbf x.
$$
Indeed, the tangential boundary value is proportional to
$$
\left[\frac{(\mathbf B\mathbin\cdot\mathbf x)^2}{a^3}
-\frac{|\mathbf B|^2}{5a}\right]\mathbf B\times\mathbf x,
$$
which combines the prescribed slip with the rigid rotation found in part (b). Since $H_3/r^7=O(r^{-4})$ and multiplication of its gradient by $\mathbf x$ preserves that order, the exterior velocity decays as
$$
\boxed{|\mathbf u|=O(r^{-4}).}
$$
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