= Solution
Put $s=a^2/r$ and define the image term $G(r,\theta)=(r/a)\Psi(s,\theta)$. The key step is to verify that inversion preserves the <Stokes streamfunction equation for irrotational flow>. Direct differentiation gives
$$
G_r=\frac1a(\Psi-s\Psi_s),\qquad
G_{rr}=\frac{s^2}{ar}\Psi_{ss},\qquad
G_{\theta\theta}-\cot\theta\,G_\theta
=\frac r a(\Psi_{\theta\theta}-\cot\theta\,\Psi_\theta),
$$
where the quantities on the right are evaluated at $(s,\theta)$. Hence
$$
\mathcal L_rG=\frac r a\left[s^2\Psi_{ss}+\Psi_{\theta\theta}-\cot\theta\,\Psi_\theta\right]
=\frac r a\mathcal L_s\Psi=0.
$$
By <linearity>, $\widetilde\Psi=\Psi-G$ obeys the same equation and describes an <incompressible flow> and an <irrotational flow> wherever the original exterior flow is regular.
On the <sphere>, $s=a$ and the two terms cancel identically:
$$
\widetilde\Psi(a,\theta)=0,\qquad
\boxed{\widetilde u_r(a,\theta)=\frac{\partial_\theta\widetilde\Psi(a,\theta)}{a^2\sin\theta}=0.}
$$
Thus the <sphere> is a stream surface and satisfies the <no-penetration boundary condition>. Tangential slip is allowed in this <potential flow> model. For every finite $r>a$, the image argument satisfies $0<s<a$, where $\Psi$ has no singularities. The prefactor $r/a$ is regular there. Therefore the construction creates \b[no new finite exterior singularities]; any original exterior singularities remain those of the prescribed incident flow. This is the <sphere theorem for a Stokes streamfunction>.
For the far-field claim, the stated bound gives
$$
|G(r,\theta)|\leq\frac r a c\frac{a^2}r=ca
\qquad(r\text{ sufficiently large}).
$$
One must compare <velocities>, rather than require the correction to the <Stokes streamfunction> to tend to zero. Smoothness at the original origin, together with $\Psi(0,\theta)=0$, gives $\Psi_s=O(1)$ and $\Psi_\theta=O(s)$ there. Thus $G_r=O(r^{-1})$ and $G_\theta=O(1)$, and the correction has
$$
\delta u_r=-\frac{G_\theta}{r^2\sin\theta}=O(r^{-2}),\qquad
\delta u_\theta=\frac{G_r}{r\sin\theta}=O(r^{-2}).
$$
These estimates hold away from the coordinate poles, with the regular-axis limits supplied by the original smooth physical flow. Consequently
$$
\boxed{\widetilde{\mathbf u}-\mathbf u\longrightarrow\mathbf0\quad(r\to\infty).}
$$
The far-field <velocity field> is unchanged. For a fully smooth incident <velocity field> at the origin, the usual regular expansion is stronger and makes the sphere disturbance decay as $r^{-3}$.
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