Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 (Hydrodynamic Stability)

Write the Rayleigh equation for inviscid shear flow asMultiply by ${\varphi }^{\ast }$, integrate, and use integration by parts:Its imaginary part isFor an unstable mode $c_i>0$, the integral can vanish only if $U^{\prime \prime }$ changes sign somewhere in the flow. Thus the velocity profile must have an inflection point. This is Rayleigh's inflection-point theorem; it is necessary, not sufficient, for inviscid instability.

Put $\varphi =(U-c)F$. The Rayleigh equation for inviscid shear flow becomesMultiplication by $F^{\ast }$ and integration givesFor $c_i>0$, its real and imaginary parts implyThus $c_r$ is a weighted mean of $U$ and $c_i^2$ is its weighted variance. If $U_{-}\le U\le U_{+}$, the sharp bounded-variable variance estimate givesCompleting the square proves Howard's semicircle theorem:

Away from $z=0,\pm 1$, the base profile is linear or constant, so $U^{\prime \prime }=0$ and ${\varphi }^{\prime \prime }-{\alpha }^2\varphi =0$. Across a corner $z=z_0$, integration of the Rayleigh equation gives the jump conditionFor the even mode, at $z=0$ this givesFor $0<z<1$, therefore,For $z>1$, decay requires $\varphi \propto e^{-\alpha (z-1)}$. The jump at $z=1$, where $[U^{\prime }]=1$, givesSubstitution and elementary simplification yield

The coefficients are real, so instability occurs when the quadratic discriminant is negative. At $\alpha =1$ it iswhereas at $\alpha =2$, using $e^{-4}\simeq 1/55$, it is positive. By continuity there is a first threshold $1<{\alpha }_s<2$ at which the discriminant vanishes. Hence one conjugate root has $c_i>0$ for

The energy isThusThis quadratic form can be negative for positive $c_1,c_2$ precisely whenThe decaying vectors must therefore be sufficiently nonorthogonal and oppositely directed. This is transient growth from non-normal modes.

With $c_2=1$ and $\mu =c_1/c_2$, maximizing $E^{\prime }(0)$ givesso

Dot the momentum equation with $\mathbf{u}$, multiply the temperature equation by $\sigma \mathrm{Ra}\theta$, and integrate over $V$. The pressure term vanishes by incompressibility, while skew-symmetry of incompressible transport removes both nonlinear advection terms. Integration by parts and the boundary conditions givewhereFor $\mathrm{Ra}=0$, this reduces toso viscosity monotonically dissipates kinetic energy.

Vary the integral on the right-hand side, imposing $\nabla \cdot \mathbf{u}=0$ with multiplier $2p$. After integration by parts, independent variations of $\mathbf{u}$, $\theta$, and $p$ giveDot the first equation with $\mathbf{u}$ and the second with $\sigma \mathrm{Ra}\theta$, then integrate. Their sum says exactly that the energy-production functional is zero. Hence any nonzero stationary point is a perturbation whose energy initially neither grows nor decays.

The linear normal-mode problem isAt $\lambda =0$ it is exactly the variational Euler--Lagrange system from part ii. Therefore the energy-stability threshold coincides with the stated linear neutral threshold ${\mathrm{Ra}}_{\mathrm{crit}}$. Below it the quadratic production functional is negative for every nonzero perturbation, so there is no transient energy growth; at the threshold a neutral perturbation exists, and above it the leading real eigenvalue produces exponential growth.

After the pressure enforces incompressibility, let $L$ denote the displayed linear operator. Integration by parts givesThe pressure terms vanish by incompressibility and the boundary conditions. The expression is symmetric under $i\leftrightarrow j$, soThus $L$ is a self-adjoint operator in the energy inner product and hence a normal operator. Its orthogonal eigenmodes cannot generate non-normal transient growth.

Substitute the stated scales and divide the momentum equation by $2\Omega V$. The ratio of inertial to Coriolis acceleration is the Rossby numberwhile the dimensionless buoyancy coefficient isThusThe dimensionless $B$ is the Burgers number for this rotating stratified flow.

Forboth material derivatives vanish because the fields are independent of $x$. Since $\hat{\mathbf{z}}\times \mathbf{U}=z\hat{\mathbf{y}}$, steady momentum balance requiresThese derivatives are compatible and integrate toThe cross-stream temperature gradient and vertical stratification are in thermal-wind balance with the vertical shear.

Let $D={\partial }_t+z{\partial }_x$. Retaining terms linear in the primed fields givesThe terms ${\mathrm{Ro}}^2{w}^{\prime }$ and $-v^{\prime }/B$ respectively arise from perturbation advection of the velocity and temperature gradients in the basic state.

Differentiate the $y$-momentum equation with respect to $x$, subtract the $y$ derivative of the $x$-momentum equation, and use incompressibility. With vertical perturbation vorticityone obtainsTherefore

At leading order as $\mathrm{Ro}\to 0$, horizontal geostrophic balance and vertical hydrostatic balance giveThe temperature equation givesMeanwhile ${\zeta }^{\prime }=p_{xx}^{\prime }+p_{yy}^{\prime }$. The leading vertical-vorticity equation is $D{\zeta }^{\prime }=w_z^{\prime }$. Since derivatives in $z$ do not commute with $D$,The $p_{xz}^{\prime }$ terms therefore cancel, leaving the conserved three-dimensional quasi-geostrophic potential vorticityNo normal flow at $y=0,1$ means $v^{\prime }=p_x^{\prime }=0$. At $z=0,1$, $w^{\prime }=0$, so

InsertThe side-wall condition is automatic, and the potential-vorticity equation givesDefineThenThe top and bottom conditions becomeSetting the determinant of these two homogeneous equations for $A,C$ to zero and simplifying gives

For every $\lambda >0$, $\tanh \lambda <\lambda$, soThe radicand in part f is therefore negative precisely when its second factor is negative. The two wave speeds are then complex conjugates, one with positive imaginary part and exponential growth. Hence