Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 1

The steady radial equation first gives the base-pressure gradientFor the stated normal mode, putRetaining terms linear in the disturbance in the Euler equations for an inviscid fluid givestogether with incompressible flowThe $-2\Omega v^{\prime }$ and $(2\Omega +r{\Omega }^{\prime })u^{\prime }$ terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.

The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation givesEliminating $w^{\prime }$ with incompressibility and then $p^{\prime }$ with the axial equation yieldswhere the Rayleigh discriminant isImpermeability at the two solid walls gives $u^{\prime }(r_i)=u^{\prime }(r_o)=0$.

Multiply the equation by $r\overline{u^{\prime }}$, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtainsThe denominator is positive. Therefore $\Phi \ge 0$ throughout the annulus excludes positive real ${\sigma }^2$ and gives centrifugal stability. Since $r^4{\Omega }^2=(r^2\Omega {)}^2$ is the square of the specific angular momentum, Rayleigh's circulation criterion isAn outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

The displaced sheet is the material surface $r=R_0+\xi (\theta ,t)$. Its kinematic boundary condition equates the radial velocity on each side to the material velocity of the sheet. To linear order,For a mode $e^{\sigma t+im\theta }$ these become

The inner base flow is at rest, so its linearized Unsteady Bernoulli equation gives $p_1^{\prime }=-\sigma {\varphi }_1$. The outer base potential is ${\Gamma }_0\theta$, hence its cross term with the perturbation givesThe base outer pressure satisfies $d{p}_{20}/dr={\Gamma }_0^2/r^3$, whereas the inner base pressure is constant. Expanding pressure continuity on the displaced sheet therefore givesAfter multiplication by $-1$ this isAn additive function of time in Bernoulli appears as the stated constant; it vanishes for every nonaxisymmetric Fourier mode.

Away from the cylindrical vortex sheet, the disturbance is both incompressible and irrotational, so its potential is harmonic. Regularity at the axis and decay at infinity selectfor a positive integer $m$. The kinematic conditions give their values on the sheet:Substitution into the dynamical condition gives the dispersion relationThusFor every $m\ge 3$, one root has positive real part. The cylindrical sheet is therefore subject to a Kelvin-Helmholtz instability.