Past exam of the options II course of the University of Oxford / 2013 / K3 / Solution

Orient $\partial B$ counterclockwise, and define positive circulation by that orientation. The outward normal from the body satisfies $nds=(dy,-dx)$. Pressure acts inward, soThe steady Bernoulli equation for this constant-density irrotational flow is $p+\rho |v{|}^2/2=C$. The constant has zero integral around a closed curve. With complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, let $W=w^{\prime }=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline{W}d\overline{z}$, and $|W{|}^{2}d\overline{z}=W^{2}dz$. Substitution proves Blasius force theorem:At infinity the Laurent expansion has the formThere is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma (U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the residue theorem gives $\oint W^{2}dz=2\Gamma (U-iV)$. HenceIts component along $(U,V)$ is zero: the drag is zero. The lift magnitude is $\boxed{\rho |\Gamma |\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho \Gamma \sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the Kutta–Joukowski theorem with its orientation made explicit.