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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.

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 $b=(m+\sqrt{m^2+4U^2})/(2U)$.

For a two-dimensional potential doublet of fixed complex strength $M$ outside a stationary circle of radius $a$, the Milne-Thomson circle theorem gives the image contribution $\bar{M}z/(a^2-\bar{b}z)$. If the singularity is stipulated to follow the regular fluid velocity at its position, differentiating this term and using the complex velocity convention gives the displayed motion law. Thus $\mathrm{Im}(b\bar{M})$ is constant: the trajectory is a straight line parallel to $M$. With $M=m{e}^{i\alpha }$ and initial location $b_0>a$ real, its polar coordinates satisfy $r\sin (\theta -\alpha )=-b_0\sin \alpha$. This statement assumes fixed orientation; the translational rule alone supplies no equation for orientation evolution.

In a planar incompressible flow, a line source injects volume flux $Q$ per unit length perpendicular to the plane. Its radial velocity is $u_r=Q/(2\pi r)$, since integrating the outward velocity around a circle gives $2\pi r{u}_r=Q$. Its complex potential of a two-dimensional potential flow is $w=Q\log (z-b)/(2\pi )$ and its complex velocity is $w^{\prime }=Q/[2\pi (z-b)]$. Here $Q>0$ means outward flow; a line sink has negative $Q$. Away from $b$, this is an irrotational flow as well as an incompressible flow.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

Take positive $\Gamma$ to mean counterclockwise circulation. A line vortex at ${\zeta }_0$ has complex potential of a two-dimensional potential flow $\Gamma \log (\zeta -{\zeta }_0)/(2\pi i)$. The method of images imposes the no-penetration boundary condition on the real axis by placing an image line vortex of strength $-\Gamma$ at ${\bar{\zeta }}_0$. ThusFor real $\zeta$, the numerator and denominator have equal modulus, so the streamfunction isThe real axis is therefore a streamline. The image line vortex lies outside the mapped fluid region, while the complex velocitydecays as $|\zeta |\to \infty$. This selects the flow due to the line vortex and the boundary, without superposing an imposed background velocity field.

A free point vortex is advected by the regular part of the velocity field at its position. Its own singular complex velocity $\Gamma /[2\pi i(z-c)]$ must first be subtracted. The image expression therefore givesEquivalently, $\dot{c}$ is the complex conjugate of the boxed expression. This is the instantaneous velocity, evaluated with the current value of $c$.

To see why mapping the image velocity alone would miss a term, expand the mapped physical singularity:Thus the same point-vortex drift under a conformal map can be writtenThe finite term involving $g^{\prime \prime }$ is part of the physical regular velocity field.

If $c=id$ with $d>1$, direct substitution into the image expression givesThe bracket is strictly positive, and the result is real. Hence a positive line vortex initially moves horizontally to the right, a negative line vortex moves to the left, and the initial vertical velocity is zero. As $d\to \infty$, the leading term $\Gamma /(4\pi d)$ is the familiar drift of a point vortex above a straight wall.

The physical complex potential of a two-dimensional potential flow is $w(z)=W(g(z))$. Applying the chain rule gives the complex velocityThe physical velocity is $(\mathrm{Re}{w}^{\prime },-\mathrm{Im}{w}^{\prime })$. A useful equivalent expression follows fromFactoring the image expression in the same way makes the $-1/z$ terms cancel, leavingThis represents the physical line vortex at $c$, an image line vortex of the same strength at $c^{-1}$, and image point vortices of opposite strength at $\bar{c}$ and ${\bar{c}}^{-1}$. All three image locations are outside the fluid region. Both boundary pieces inherit the constant streamfunction of the real axis in the mapped complex plane.

For $q=W^{\prime }=u-iv$, the Blasius force theorem states that the force per unit length exerted by a steady incompressible flow and irrotational flow on an impermeable body iswhere $C$ runs counterclockwise around the body and excludes external flow singularities. The sign corresponds to the force on the body.

Let $c=a^2/b$ and $A=m{a}^2{e}^{-i\alpha }/b^2$. The complex velocity may be writtenOnly the image point $c$ lies inside the circular contour. The term $A^2/(z-c{)}^4$ has zero residue; $H^2$ is regular there. In the cross term, the linear term of the Taylor expansion of $H$ suppliesThe orientation factors have canceled. The residue theorem now givesSince $b>a$, $X>0$: the obstacle is pulled toward the doublet, independently of $\alpha$. This is a force per unit length perpendicular to the plane of flow.

Keep the prescribed orientation $\alpha$ and strength $m$ fixed, and now let the doublet position $b(t)$ be complex. The Milne-Thomson circle theorem applies instantaneously withWhen taking its spatial derivative, $b(t)$ is a parameter. Removing the doublet's own singular term leaves the regular complex velocity at its position:Because $q=u-iv$, the stipulated image-induced motion of a planar doublet obeys $\dot{b}=\overline{q_{\mathrm{reg}}(b)}$, not $q_{\mathrm{reg}}(b)$ itself. HenceIts direction is constant. Indeed $\mathrm{Im}(b{e}^{-i\alpha })$ is constant and, since $b(0)=b_0>a$ is real, equals $-b_0\sin \alpha$. Writing $b=r{e}^{i\theta }$ in polar coordinates gives the path equationwhere the quotient is used when $\sin \alpha \ne 0$ and the branch through $(r,\theta )=(b_0,0)$ with $r>a$ is selected. It is the straight line through $b_0$ parallel to $e^{i\alpha }$. An unambiguous parametrization, also covering the degenerate polar cases, isIf $\sin \alpha =0$, the path is radial along the positive real axis, with $\theta =0$ and changing $r$; it cannot be represented as a single-valued $r(\theta )$. The sign of $m$ selects which direction along the line is followed. These paths apply only while the singularity remains outside the obstacle; the model prescribes no orientation evolution.

For the simple shear flow, the vertical vorticity isFor $\beta \ne 0$, this is not an irrotational flow. A complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, analytic in the flow region, has complex velocity $w^{\prime }=v_x-i{v}_y$. The Cauchy-Riemann equations imply both incompressibility and zero vorticity, so this shear has no such complex potential. It does have a Cartesian streamfunction, $\psi =\beta y^2/2$. In the degenerate case $\beta =0$, the resting fluid has a constant complex potential.

Using $w=\varphi +i\psi$ and the complex velocity convention $w^{\prime }=v_x-i{v}_y$,The streamlines are $xy=\psi /(2\beta )$, a family of rectangular hyperbolas whose asymptotes are the coordinate axes. The origin is a stagnation point. Fluid moves inward along the $y$ axis and outward along the $x$ axis, with the same directions approached by the neighbouring streamlines. This is a planar extensional incompressible flow, illustrated in the left panel below.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.

Let $\zeta =g(z)$ be a conformal map, and let a point vortex of circulation $\Gamma$ lie at $z=c$. Suppose the mapped complex potential of a two-dimensional potential flow iswhere $R$ is a holomorphic function near $g(c)$ and represents the other point vortices, boundaries, and any imposed flow. The physical complex velocity is $w^{\prime }=g^{\prime }{W}^{\prime }$. The point vortex follows its regular velocity, found by removing $\Gamma /[2\pi i(z-c)]$ before taking the limit. A Taylor expansion proveswhich gives the title formula. The complex conjugate occurs because the complex velocity convention is $w^{\prime }=u-iv$, whereas $\dot{c}=u+iv$.

For a mapped upper half-plane with no imposed flow, the method of images gives $R(\zeta )=-\Gamma \log (\zeta -\overline{g(c)})/(2\pi i)$, henceThe first term is essential: excluding a mapped singularity and then transforming its regular velocity does not in general equal subtracting the physical singularity after transformation.

A two-dimensional potential doublet is a localized potential flow singularity with complex potential of a two-dimensional potential flow $W=M/(z-b)$, where $M=m{e}^{i\alpha }$ specifies strength and orientation. It is the limiting field of an equal point source and sink as their separation tends to zero with strength times separation fixed, with the sign absorbed in $M$. Its complex velocity is $-M/(z-b{)}^2$, so the velocity decays as inverse distance squared. The total volumetric flow rate around the singularity is zero. Its two-dimensional decay differs from that of the three-dimensional source dipole.