Past exam of the fluids and waves course of the University of Oxford / 2023 / 2

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.

The complex potential of a two-dimensional potential flow givesAt a stagnation point, $q=0$. For $m\ne 0$, taking the two square-root alternatives givesThereforeThe denominators are nonzero because $b>a$. Moreoverso both stagnation points satisfy $|z_{\sigma }|=a$. They are distinct: equality would require $2e^{-i\alpha }(a^2-b^2)=0$. Neither is a singularity of $q$. Conversely, the numerator of $q$ is a quadratic polynomial with nonzero leading coefficient, so these are all its finite zeros. Thus there are exactly two boundary stagnation points and none in the open fluid region. If $m=0$, the flow instead vanishes everywhere, which is the degenerate exception to the assertion.

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.