= Force on a unit cylinder due to a uniform flow and an external source
{title2=$F_x=2\pi\rho\left[\frac{m^2}{b(b^2-1)}-\frac{Um}{b^2}\right]$}
In a planar <potential flow>, superpose a positive-$x$ <uniform flow> of speed $U$ and a <line source> of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular <cylinder> with zero <circulation>. The <Milne-Thomson circle theorem> gives the <complex potential of a two-dimensional potential flow>
$$
w=U(z+z^{-1})+m[\log(z-b)+\log(z-b^{-1})-\log z].
$$
The displayed horizontal <force per unit length> acts on the <cylinder>, and $F_y=0$ by <reflection symmetry>. To prove it, let $q=w'$ be the <complex velocity>. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite <residues> of $q^2$ is $2Um$. At the external <pole> $b$, write $q=m/(z-b)+H(z)$; then
$$
\operatorname{Res}_{b}q^2=2mH(b),\qquad
H(b)=U-\frac U{b^2}+\frac{m}{b(b^2-1)}.
$$
The <residue theorem> and the counterclockwise <Blasius force theorem> give $F_x-iF_y=2\pi\rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.
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