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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.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

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.

Use $W=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so $W^{\prime }(z)=u-iv$. The Milne-Thomson circle theorem says that if $f(z)$ is an incident complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenmakes $|z|=a$ a streamline. Indeed, on that circle $a^2/\bar{z}=z$, so $W=f+\bar{f}$ is real and the stream function is constant. This imposes the no-penetration boundary condition.

For the two-dimensional potential doublet, put $M=m{e}^{i\alpha }$ and $f=M/(z-b)$, where $b>a$ is real. The theorem givesUp to an additive constant, this is equivalentlyThe method of images has placed the image two-dimensional potential doublet at $a^2/b$, inside the obstacle. We use the solution with zero added circulation and no imposed far-field flow; the original doublet is the only singularity in the liquid.

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.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.