Blasius force theorem ($X-iY=(i\rho /2){\oint }_C(W^{\prime }{)}^{2}dz$)

For a steady two-dimensional incompressible flow and irrotational flow around an impermeable body, the displayed contour formula gives the force on the body per unit span. Here $W^{\prime }=u-iv$ is the complex velocity, $\rho$ is the constant mass density, and $C$ runs counterclockwise around the body, excluding other fluid singularities. To verify the sign, Bernoulli's principle gives $p=p_0-\rho |q{|}^2/2$. On a counterclockwise boundary, the complex pressure force is $d(X-iY)=-ipd\bar{z}$. Tangency gives $qdz=\bar{q}d\bar{z}$, and therefore $|q{|}^{2}d\bar{z}=q^{2}dz$. Integrating removes the constant $p_0$ term and gives the formula. The theorem is named after Heinrich Blasius.

The Milne-Thomson circle theorem places an image dipole inside a cylinder at the inverse point of an external dipole. In the residue of the squared complex velocity, only the cross term between the analytic external velocity and the image double pole contributes. The product of their conjugate dipole strengths is independent of orientation, so the cylinder force is radial toward the external dipole.

For a two-dimensional steady ideal flow outside a body, let $q=w^{\prime }$ be the complex velocity. Suppose $q$ is holomorphic in the exterior and $q^2$ has a Laurent series near infinity, with coefficient $a_{-1}$ of $z^{-1}$. Orient the body contour counterclockwise and assume the usual impermeability and Bernoulli hypotheses of the Blasius force theorem. Deforming the contour to a large circle gives $\oint q^{2}dz=2\pi i{a}_{-1}$. Substitution in Blasius' formula proves $F_x-i{F}_y=-\pi \rho a_{-1}$.

In particular, $q(z)=q_{\infty }+O(z^{-2})$ makes the force zero: its square has no reciprocal term. This criterion handles branch-cut obstacles by applying the contour in their analytic exterior, rather than separately assigning forces to endpoint singularities.

Place a held point vortex of circulation $\Gamma$ at real position $b>a$ outside a circular cylinder $|z|=a$, and prescribe circulation $-\Gamma$ about the cylinder. An opposite image at $c=a^2/b$ gives complex velocity $q(z)=-i\Gamma (1/(z-b)-1/(z-c))/(2\pi )$. On the positively oriented cylinder contour, the only pole of $q^2$ inside is at $c$, with residue $-{\Gamma }^2/(2{\pi }^2(b-c))$. The residue theorem and Blasius force theorem give $F_x-i{F}_y=\rho {\Gamma }^2/(2\pi (b-c))$. Thus $F_y=0$ and $F_x=\rho {\Gamma }^{2}b/(2\pi (b^2-a^2))>0$: the force on the cylinder points toward the vortex. The specified cylinder circulation is part of the hypothesis; a zero-circulation cylinder has an additional central image term.

For steady incompressible flow and irrotational flow of constant mass density, with no body force, this formula gives the hydrodynamic torque about the origin per unit span on an impermeable body. Orient its boundary counterclockwise, and let positive torque be counterclockwise. The proof uses $M=\mathrm{Re}{\oint }_{C}pzd\bar{z}$, the Bernoulli equation $p=P_{\ast }-\rho |w^{\prime }{|}^2/2$, and the tangency identity $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. The real part of the constant-pressure term vanishes because $\mathrm{Re}(zd\bar{z})=d|z{|}^2/2$. The full complex integral of $zd\bar{z}$ need not vanish.

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 flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ 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 $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.