Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 8C / Solution

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.