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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 $U{a}^2$, and every other partner vanishes. Thus the potential flow around a circular cylinder hasIndeed, ${\varphi }_r=U(1-a^2/r^2)\cos \theta$, 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 $\theta =0,\pi$.