Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 61 / 2

The boundary of a steady Kida vortex must be material. Put $E=x^2/b^2+y^2/a^2$. Requiring $\mathbf{u}\cdot \nabla E=0$ on the ellipse givesThe relative vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=-\beta -\alpha =-S+{\zeta }_0$, so the supplied vorticity relation impliesCombining these independent conditions yieldsFor the usual positive orbital shear both are positive, the flow circulates clockwise and its streamlines have the same aspect ratio as the patch. The velocity coefficients in this question are unrelated to the alpha viscosity parameter in Question 1.

The steady incompressible shearing sheet equation isThe final term follows from the local tidal potential $-\Omega S{x}^2$. For the given linear velocity, the advective acceleration is $-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$, while the Coriolis acceleration is $2\Omega \beta x{\mathbf{e}}_x+2\Omega \alpha y{\mathbf{e}}_y$. Equating components givesThe cross derivatives vanish, so the pressure Hessian of an elliptical shearing-sheet vortex integrates toThe additive constant is fixed by exterior matching or the pressure reference. No zero-pressure free-boundary condition is imposed on an incompressible vortex patch: it is embedded in a surrounding fluid.

A pressureless test particle feels tidal gravity, Coriolis acceleration and linear drag toward the prescribed gas velocity. Its planar equations areThere is no gas-pressure force on this test particle. For $(x,y)=(X,Y)e^{\lambda t}$, the coefficient matrix isNontrivial solutions require a zero determinant. Expansion givesEach root has a nonzero modal vector in the matrix kernel; real motions are obtained from conjugate pairs. Repeated roots may also supply polynomially weighted exponential solutions. Negative real parts make the centre linearly asymptotically attracting for trajectories governed by this interior linear flow.

Take every finite positive drag coefficient $\gamma >0$ and defineThe coefficients of the characteristic polynomial are $c_3=2\gamma$, $c_2=K+{\gamma }^2$, $c_1=2L\gamma$ and $c_0=P{\gamma }^2$. The nontrivial Routh-Hurwitz criterion reduces exactly toThe printed strict chain $K>L>P>0$ makes every coefficient positive and both bracket terms positive, so it proves attraction for every positive drag coefficient.

There is, however, a boundary error in the asserted necessity. To have the bracket positive for every $\gamma >0$, its constant and slope as a function of ${\gamma }^2$ must be nonnegative, and cannot both vanish. Together with coefficient positivity this gives the exact drag-polynomial stability for all positive stopping rates:Necessity follows by taking arbitrarily small and arbitrarily large positive $\gamma$; if one bracket coefficient is negative, the inequality fails in the corresponding limit. Sufficiency follows because at least one coefficient is strictly positive. Thus $K=L>P$ and $K>L=P$ still give strict decay at each finite positive $\gamma$. They need not give a uniform stability margin in the zero- or infinite-drag limit.

In a Keplerian shearing sheet, $S=3\Omega /2$ andThe differences areHence $r>3$ establishes the requested strict chain and trapping conclusion. At $r=3$, however, $K=L={\Omega }^2$ and $P=9{\Omega }^2/16$, soAll coefficients remain positive. This is an explicit counterexample within the stated vortex model: $r=3$ also attracts particles for every finite positive drag coefficient, so the literal strict “if and only if” is false. The exact Keplerian all-positive-drag condition is $r\ge 3$. If $\gamma =0$ is included, the polynomial has zero and purely imaginary roots and there is no asymptotic attraction; “all drag coefficients” must mean positive drag. The local attraction claim presumes the particle remains in the patch, as is assured for sufficiently small perturbations of its centre.