Past exam of the options II course of the University of Oxford / 2012 / K (Fluid Dynamics and Waves)

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem:

The Milne-Thomson circle theorem states that if $f$ is an analytic complex potential with no singularity on or inside $|z|=a$, thenmakes $|z|=a$ a streamline while retaining the original exterior singularities and adding reflected singularities inside the circle. On the circle the two terms are conjugates, so the resulting streamfunction is zero there.

Put $A=m{e}^{i\alpha }$ and $c=a^2/b$. Applying the formula to the exterior planar potential doublet givesEquivalently, up to an irrelevant constant,The image planar potential doublet is inside the obstacle at $c<a$; the only physical fluid singularity is the prescribed one at $b$.

The complex velocity obtained from part (b) isFor a nonzero planar potential doublet, $B/A=(a^2/b^2)e^{-2i\alpha }$. Its zeros therefore satisfyand henceThe denominators are nonzero because $a/b<1$. For either sign,so $|z_{\pm }|=a$. They are distinct because $t_{+}\ne t_{-}$ and $t\mapsto (c-bt)/(1-t)$ is injective, its determinant being $c-b\ne 0$. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.

In the analytic continuation inside the circle, only the image location $c=a^2/b$ is singular; $b$ lies outside. Squaring the velocity givesThe last term has no residue, and the middle term has residuebecause $AB=m^2{a}^2/b^2$ and $c-b=-(b^2-a^2)/b$. The residue theorem and part (a) therefore yieldThus $F_y=0$ and $F_x>0$. The force points towards the planar potential doublet and is independent of its orientation, since the factors $e^{i\alpha }$ and $e^{-i\alpha }$ cancel in $AB$.

A material interface obeys $D(y-\eta (x,t))/Dt=0$. In each fluid this gives ${\varphi }_{jy}={\eta }_t+(U_j+{\varphi }_{jx}){\eta }_x$ at $y=\eta$, where $U_1=U$ and $U_2=V$. Discarding products of perturbations and evaluating at the undisturbed interface givesFor an inviscid irrotational fluid, unsteady Bernoulli is ${\Phi }_t+\frac{1}{2}|\nabla \Phi {|}^2+p/\rho +gy=C(t)$, with total potential $\Phi =U_{j}x+{\varphi }_j$. Subtract the undisturbed relation and linearize at the moving interface. The pressure perturbation there is $-{\rho }_j({\varphi }_{jt}+U_j{\varphi }_{jx}+g\eta )$. With no surface tension the two pressures agree, givingUse complex amplitudes and take real parts at the end. For $k>0$, solutions of Laplace's equation decaying away from the interface have the formPut ${\Omega }_1=\omega -Uk$ and ${\Omega }_2=\omega -Vk$. The kinematic conditions imply $B_1=-i{\Omega }_{1}A/k$ and $B_2=i{\Omega }_{2}A/k$. Substituting into the pressure condition givesA nonzero amplitude therefore requiresTo determine stability, let $R={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_{1}U+{\rho }_{2}V)/R$. Completing the square yieldsFor $U\ne V$, the right-hand side is negative exactly whenThen the two frequencies have opposite nonzero imaginary parts. The root with positive imaginary part gives exponential growth through $e^{-i\omega t}$. In terms of wavelength $\lambda =2\pi /k$, the unstable range isThis is the short-wavelength Kelvin-Helmholtz instability for a density-stable interface without surface tension. Equality is the neutral threshold; if $U=V$, the gravitational term is positive and this shear instability is absent.