= Solution
Use instantaneous <polar coordinates> centered on the moving <circular cylinder>, but measure the fluid <velocity> in the laboratory frame. Impermeability equates the normal components of body and fluid <velocity>:
$$
(\mathbf u-U\mathbf e_x)\cdot\mathbf e_r=0\quad(r=a).
$$
Since $\mathbf e_x\cdot\mathbf e_r=\cos\theta$ and $u_r=\phi_r$, the <no-penetration boundary condition> is
$$
\boxed{\phi_r(a,\theta)=U\cos\theta.}
$$
The <velocity potential> obeys the exterior <Laplace equation>, by <incompressibility>. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\phi=C\cos\theta/r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-Ua^2$. The zero <circulation> condition rules out an angular potential term, and the integrated boundary condition gives zero net <volume flux>, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. Hence
$$
\boxed{\phi=-\frac{Ua^2}{r}\cos\theta.}
$$
Using the <streamfunction in polar coordinates>, $u_r=r^{-1}\psi_\theta$, $u_\theta=-\psi_r$, gives
$$
\boxed{\psi=\frac{Ua^2}{r}\sin\theta,\qquad
u_r=\frac{Ua^2}{r^2}\cos\theta,\quad u_\theta=\frac{Ua^2}{r^2}\sin\theta.}
$$
One may also obtain both functions as the real and imaginary parts of $w(z)=-Ua^2/z$, a <two-dimensional potential doublet>. The <velocity> decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the <added mass of a circular cylinder>.
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