Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 (Hydrodynamic Stability)

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.

Linearization about $w=0$ removes the quadratic term and givesThe three homogeneous boundary conditions select the vertical normal modesTake a horizontal Laplacian eigenfunction satisfyingOn $W_{j}f$, the full Laplacian has eigenvalue $-(a^2+j^2{\pi }^2)$. The growth rate is consequentlyNeutrality occurs atFor fixed nonzero $a$, this increases strictly with $j$, so the first unstable vertical mode is $j=1$ and

At criticality choose the fundamental modeIts quadratic self-interaction satisfiesThe critical linear operator has eigenvalues $-64{\pi }^6$ on $\sin (2\pi z)$ and $-60(a^2+{\pi }^2{)}^3$ on $\cos (2ax)\sin (2\pi z)$. The slaved second-order correction is thereforewhere

Introduce a slow time $\tau$ and project the next-order equation onto the fundamental mode. Detuning $R-R_c$ contributes $a^2(R-R_c)A$, while the interaction of $w_1$ with $w_2$ has fundamental componentThe solvability condition is the Landau amplitude equation

SetThe amplitude equation is $\dot{A}=\mu A-g{A}^3$. Its equilibria arefor every $R$, and, when $R>R_c$,For $R<R_c$, the zero branch is stable and every sufficiently small amplitude decays to zero. At $R=R_c$ it loses stability. For $R>R_c$, zero is unstable and the two nonzero branches are stable; positive initial amplitudes approach $A_{+}$ and negative ones approach $A_{-}$. The bifurcation diagram is therefore a supercritical pitchfork.

A matrix is non-normal when it does not commute with its adjoint matrix:Nonorthogonal decaying eigenmodes can interfere constructively and produce transient growth. For example,has two negative eigenvalues but is non-normal. Starting from $x(0)=(0,1{)}^T$ givesAt $t=\log 2$, its squared Euclidean norm is $2.5^2+0.25^2>1=|x(0){|}^2$. The energy grows transiently even though both eigenmodes eventually decay.

Because $L$ is diagonalizable, writewhere every diagonal entry of $\Lambda$ has negative real part. Define the equivalent normThenfor every $t\ge 0$. Equivalently, this norm comes from the positive-definite inner-product matrix $H=V^{-†}{V}^{-1}$. Thus stable eigenvalues always admit a norm with no growth, even though the standard Euclidean norm may show transient amplification.

The first component obeys ${\dot{x}}_1={\lambda }_1{x}_1$, so $x_1(t)=e^{{\lambda }_{1}t}{x}_1(0)$. Variation of constants inthen givesHence the matrix exponential is

WriteThe extremal squared amplification factors are the largest and smallest singular values of $A$ squared, equivalently the eigenvalues $G$ ofSince its trace is $a^2+b^2+c^2$ and its determinant is $a^2{c}^2$, the requested quadratic is

As ${\lambda }_1\to {\lambda }_2=\lambda$, L'Hôpital's rule gives $b\to t{e}^{\lambda t}$, soThe quadratic becomesFor $t\ge 0$, its maximum root is

For $-1/2<\lambda <0$, logarithmic differentiation givesThe unique maximizing time is thereforeWriting $\alpha =-\lambda >0$ and $q=\sqrt{1-4{\alpha }^2}$, the maximum isConsequently, as $\lambda \to 0^{-}$,At $\lambda =0$ exactly there is no finite maximizing time: the Jordan-block shear produces unbounded quadratic growth. The formulas describe how the optimal transient moves to later times and becomes larger as the damping tends to zero.