= Solution
The far-field derivative fixes the growing $n=1$ cosine coefficient to $U$ and removes all other growing modes. The possible logarithmic term has derivative $D_0/a$ at the wall, so the zero wall derivative removes it. For each decaying Fourier mode, the wall condition relates its coefficient to its growing partner: the $n=1$ cosine partner has coefficient $Ua^2$, and every other partner vanishes. Thus the <potential flow around a circular cylinder> has
$$
\boxed{\phi(r,\theta)=U\left(r+\frac{a^2}{r}\right)\cos\theta+C}.
$$
Indeed, $\phi_r=U(1-a^2/r^2)\cos\theta$, which is zero at $r=a$ and tends to $U\cos\theta$ at infinity. The additive constant does not change the velocity. The <equipotential curves> satisfy $(r+a^2/r)\cos\theta=\text{constant}$. They approach vertical straight lines far away and meet the cylinder radially wherever the tangential velocity is nonzero. The two stagnation points are at $\theta=0,\pi$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-4-cylinder-potentials.png]
{title=Equipotential curves outside a circular cylinder in uniform potential flow}
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