Past exam of the options II course of the University of Oxford / 2008 printed variant / K (Fluid Dynamics and Waves)

The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.

Reflection in the real axis replaces a dipole at $a$ with an image at $\bar{a}$ of moment $\bar{m}$. Reflection in the imaginary axis places the image at $-\bar{a}$ with moment $-\bar{m}$. Applying both reflections gives the image at $-a$ with moment $-m$. ThusFor example, on the real axis the first two terms are conjugates and the last two are conjugates. On the imaginary axis the first and third terms are conjugates, as are the second and fourth. The potential is real on each wall, so both walls are streamlines. The only singularity in the open first quadrant is the original dipole.

Reflection in the line $y=-x$ is $Rz=\overline{iz}=-i\bar{z}$. The printed expression has two conjugation bars: $w(z)+\overline{w(\overline{iz})}$. On the line $Rz=z$, this is $w(z)+\overline{w(z)}$, which is real; hence its stream function vanishes there. Applying this reflection to the dipole givesConsequentlyThe image lies at $-i\bar{a}$, with moment $-i\bar{m}$, on the opposite side of the wall.

For each case omit the dipole's own singular field when evaluating the surrounding complex velocity. With the convention $W^{\prime }=u-iv$ this gives $d\bar{a}/dt=W_{\mathrm{image}}^{\prime }(a)$, which is the conjugated equation requested in the source. In case (a), differentiating $\bar{m}z/(1-\bar{a}z)$ givesFor case (b), differentiate the three image terms:For case (c), differentiation givesThe physical equation for $da/dt$ is the complex conjugate of the displayed equation. These expressions apply away from the walls, where the point-dipole model is nonsingular.

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.