= Solution
Let the sphere have radius $a$; set $a=1$ for the unit sphere implicit in the question. Since
$$
\boldsymbol\Phi=\frac{\alpha\mathbf U}{r},
\qquad
\chi=\beta\mathbf U\mathbin{\cdot}\nabla\frac1r
=-\beta\frac{\mathbf U\mathbin{\cdot}\mathbf x}{r^3},
$$
one finds
$$
\mathbf u
=\left(-\frac\alpha r-\frac\beta{r^3}\right)\mathbf U
+\left(-\frac\alpha{r^3}+\frac{3\beta}{r^5}\right)
(\mathbf U\mathbin{\cdot}\mathbf x)\mathbf x.
$$
On $r=a$, the <no-slip boundary condition> requires $\mathbf u=\mathbf U$ for every surface normal. The coefficient of
$(\mathbf U\mathbin{\cdot}\mathbf n)\mathbf n$ must vanish, while the coefficient of $\mathbf U$ must equal one:
$$
-\frac\alpha a+\frac{3\beta}{a^3}=0,
\qquad
-\frac\alpha a-\frac\beta{a^3}=1.
$$
Solving,
$$
\boxed{\alpha=-\frac{3a}{4},\qquad
\beta=-\frac{a^3}{4}.}
$$
For a unit sphere,
$$
\boxed{\alpha=-\frac34,\qquad\beta=-\frac14.}
$$
Substitution reproduces the standard <translating sphere in Stokes flow>.
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