= Solution
{c}
Use $w=\phi+i\psi$ for the <complex potential of a two-dimensional potential flow>, so its <complex velocity> is $q=w'=u-iv$. The incident <uniform flow> and <line source> have
$$
f(z)=Uz+\log(z-b).
$$
The coefficient of the <complex logarithm> is one because a <line source> of strength $Q$ has <complex potential of a two-dimensional potential flow> $(Q/(2\pi))\log(z-b)$, and here $Q=2\pi$. Applying the <Milne-Thomson circle theorem> to the unit <circle> gives
$$
w(z)=f(z)+\overline{f(1/\bar z)}
=Uz+\log(z-b)+\frac Uz+\log(1/z-b).
$$
Since $1/z-b=-b(z-1/b)/z$, locally consistent branches of the <complex logarithm> give
$$
\boxed{w(z)=Uz+\frac Uz+\log(z-b)
+\log(z-1/b)-\log z+C}.
$$
Changes of branch only alter the constant in a local expression; $w'$ is single valued away from its singularities. The added terms represent a <two-dimensional potential doublet> at the origin, an image <line source> at $1/b$, and a compensating <line sink> at the origin. The image <line source> and <line sink> have equal and opposite strengths, so no net <volume flux> enters the <cylinder>.
On $|z|=1$, the <Milne-Thomson circle theorem> makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the <streamfunction> is constant there, establishing the <no-penetration boundary condition>. All <line source> strengths are real, and the <two-dimensional potential doublet> has zero <circulation>, so the <circulation> around the <cylinder> is zero. There is no additional <line vortex> term.
Back to article page
