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The Kutta condition selects the circulation of an idealized potential flow past a body with a sharp trailing edge by requiring finite limiting velocity at that edge. For the Joukowski map $z=\zeta +a^2/\zeta$, the trailing edge $z=2a$ corresponds to a simple zero of $dz/d\zeta$ at $\zeta =a$. If $W$ is the complex potential in the circle plane, then $dw/dz=W^{\prime }(\zeta )/(1-a^2/{\zeta }^2)$ is finite there only if $W^{\prime }(a)=0$. For a uniform flow of speed $U$ at angle $\alpha$ and a point vortex of circulation $\Gamma$, this gives $\Gamma =-4\pi Ua\sin \alpha$, and cancellation of the simple zeros gives the tangential trailing-edge velocity $U\cos \alpha$. The opposite endpoint can retain a leading-edge singularity.

On $\zeta =a{e}^{i\theta }$, the Joukowski map gives $z=2a\cos \theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidatesOn $D=\mathbb{C}\setminus [-2a,2a]$, choose the analytic square root $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to 1$ at infinity. One explicit definition iswhere the square root on the right is the principal square root. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required holomorphic function. Then $g(z)=(z+s(z))/2\sim z$.

The two quadratic roots have product $a^2$. For $z\notin S$, neither root lies on $|\zeta |=a$, since that circle maps into $S$. Consequently exactly one root lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, thenwhose second factor cannot vanish. Thus the Joukowski map is a conformal map from $|\zeta |>a$ onto $D$, and $g$ is conformal throughout $D$. The two endpoints are branch points and are excluded.

The complex potential in the circle plane is the sum of the uniform flow term $U{e}^{-i\alpha }\zeta$, its circle-image term $U{a}^2{e}^{i\alpha }/\zeta$, and the point vortex term $-i\Gamma \log \zeta /(2\pi )$. On $\zeta =a{e}^{i\theta }$, the first two sum to $2Ua\cos (\theta -\alpha )$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma \log a/(2\pi )$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required uniform flow; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.

Conformal transport therefore givesIn the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The complex velocity isAt the trailing edge $z=2a$, corresponding to $\zeta =a$, the denominator has a simple zero. Finite velocity requiresso the Kutta condition isWith this choice, cancellation of the common zero givesThus the limiting velocity has zero transverse component and is tangent to the plate.