= Solution
Start from the uniform-flow <complex potential of a two-dimensional potential flow> $Uz$. <Milne-Thomson's circle theorem> adds the image doublet $Ua^2/z$, yielding constant <stream function> on $|z|=a$. Add a <line vortex> with positive counterclockwise <circulation> $\Gamma$:
$$
\boxed{w(z)=U\left(z+\frac{a^2}{z}\right)-\frac{i\Gamma}{2\pi}\log z}.
$$
Indeed, on $z=ae^{i\theta}$ the uniform-flow and doublet terms sum to the real number $2Ua\cos\theta$, while the vortex has imaginary part $-\Gamma\log a/(2\pi)$, also constant. Thus the cylinder is a <streamline> and satisfies the <impermeability condition>. The vortex velocity is $\Gamma/(2\pi r)$ in the counterclockwise direction. Although the potential is multivalued when $\Gamma\ne0$, its velocity is single-valued.
Differentiate and locate the coefficient of $z^{-1}$ in the square:
$$
w'=U-\frac{i\Gamma}{2\pi z}-\frac{Ua^2}{z^2},\qquad\operatorname{Res}_{z=0}(w')^2=-\frac{iU\Gamma}{\pi}.
$$
For the counterclockwise contour in the printed <Blasius force theorem>, the <residue theorem> gives $\oint_C(w')^2dz=2U\Gamma$. Consequently
$$
F_x-iF_y=i\rho U\Gamma,\qquad\boxed{F_x=0,\quad F_y=-\rho U\Gamma}.
$$
These are force components per unit cylinder length; with the stated <circulation> convention the lift is downward for positive $U\Gamma$.
Back to article page
