Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 (Hydrodynamic Stability)

With ${\mathbf{u}}_0=0$ and ${\theta }_0=-z$, the incompressibility condition and temperature equation hold because ${\nabla }^{2}z=0$. The momentum equation is satisfied by the hydrostatic pressurebecause $\nabla p_0=\sigma Ra{\theta }_0\hat{\mathbf{z}}$. Thus this is the conductive basic state.

Write $\mathbf{u}={\mathbf{u}}^{\prime }$ and $\theta =-z+{\theta }^{\prime }$. Dropping quadratic perturbation terms gives the Linearized Boussinesq equationsandThe fixed temperatures give ${\theta }^{\prime }=0$ at $z=0,1$. Impermeable stress-free boundary conditions give

Apply $\hat{\mathbf{z}}\cdot \nabla \times$ to the linear momentum equation. The pressure and buoyancy terms vanish, while incompressibility gives $\hat{\mathbf{z}}\cdot \nabla \times (\hat{\mathbf{z}}\times {\mathbf{u}}^{\prime })=-DW$. HenceApplying $\hat{\mathbf{z}}\cdot \nabla \times \nabla \times$ and using the identity supplied in the question gives

For a normal mode proportional to $e^{\mu t+ikx}$, put ${\hat{D}}^2=D^2-k^2$. The three scalar equations becomeandMultiplying through by the two scalar operators and eliminating $\hat{\omega },\hat{\theta }$ yields

At stationary onset, $\mu =0$. The lowest stress-free vertical mode is $\hat{W}=\sin \pi z$, for which $D^2=-{\pi }^2$ and ${\hat{D}}^2=-({\pi }^2+k^2)$. Substitution givesThe rotation term is positive, so rotation raises the critical Rayleigh number and is stabilizing.

Put $x=k^2$ and $T=\lambda /\sigma$. Differentiating $Ra$ gives the exact stationarity equationWhen $T\gg 1$, the optimum has $x\gg {\pi }^2$, so $2x^3\sim {\pi }^2{T}^2$. Therefore

Let $r=Ra-R{a}_c$. The amplitude equation is $\dot{A}=A(r-A^2)$. For $r<0$, $A=0$ is the sole equilibrium and every solution tends monotonically to it. At $r=0$, the origin remains attracting but only algebraically. For $r>0$, the origin is unstable and the two equilibriaare stable: positive initial data tend to $A_{+}$, negative initial data tend to $A_{-}$, and $A(0)=0$ remains zero. A plot of $A(t)$ therefore shows a pitchfork bifurcation normal form at $Ra=R{a}_c$.

The linearized vorticity equation for a inviscid parallel shear flow isSubstituting $\psi =\varphi (y)e^{ik(x-ct)}$ gives Rayleigh's equation

No normal flow at a rigid wall gives $\boxed{\varphi (\pm 1)=0}$. At a free surface $y=1+\eta$, the linearized kinematic boundary condition isso $(U-c)\eta =-\varphi$. The linearized tangential momentum equation gives the pressure amplitude $p=(U-c){\varphi }^{\prime }-U^{\prime }\varphi$. Constant surface pressure therefore requires

For $U=y$, Rayleigh's equation reduces to ${\varphi }^{\prime \prime }-k^2\varphi =0$, so $\varphi =A\cosh ky+B\sinh ky$. Applying the free-surface condition at $y=\pm 1$ and setting the determinant to zero givesFor $k>0$, $k-\tanh k>0$, so instability occurs exactly when $c^2<0$, orThus $0<k<k_c$, where the unique positive cutoff satisfies $\boxed{k_c\tanh k_c=1}$.

Linearization about zero and expansion in the Fourier sine series give, for the $n$th mode,Every mode is oscillatory exactly when the smallest coefficient, at $n=1$, is nonnegative. Since $f^{\prime }(0)<0$,

Write $T=\epsilon t$ and expandAt order ${\epsilon }^2$, the assumptions $f^{\prime \prime }(0)=0$ and the orthogonality normalization leave a homogeneous equation for $u_2$ with no forcing, hence $\boxed{u_2=0}$.

At order ${\epsilon }^3$, project the equation onto the null mode $\sin z$. Sincethe Fredholm solvability condition obtained directly from the definitions printed in the question isThe paper asks for $A/f^{\prime }(0{)}^2$ in place of $f^{\prime }(0{)}^{2}A$, but that coefficient is incompatible with $R_c=-1/f^{\prime }(0)$ and ${\epsilon }^2=R-R_c$: differentiating $1/R$ at $R_c$ gives $-1/R_c^2=-f^{\prime }(0{)}^2$. Thus the displayed target appears to contain a reciprocal typo. It would agree with the expansion only under a correspondingly rescaled definition of the small parameter.

The imposed orthogonality of every $u_j$, $j\ge 2$, removes the freedom to transfer a multiple of $\sin z$ between $A$ and the higher-order terms.

The eigenvalues of $L$ areSince $\beta <1$, the origin is linearly stable exactly when the determinant is positive:or $|\beta |<\sqrt{1+{\epsilon }^2}$.

Direct multiplication shows $L{L}^t\ne L^{t}L$ for $\beta \ne 0$, so $L$ is a non-normal matrix. MeanwhileThe symmetric part has eigenvalues $-\epsilon \pm |\beta |$, so instantaneous growth is possible exactly when $|\beta |>\epsilon$. Under the intended $0\le \beta <1$ regime this is $\boxed{\beta >\epsilon }$.

For $\epsilon =0$, $L^2=-(1-{\beta }^2)I=-{\Gamma }^{2}I$. The matrix exponential therefore gives

The maximum of $E(t)/E(0)$ is the square of the largest singular value of $A(t)$, hence the largest eigenvalue $\lambda$ of $A(t{)}^{t}A(t)$. Its determinant is one and its trace givesFor $0<\beta <1$, this is largest when $|\sin \Gamma t|=1$, namely $\Gamma t=\pi /2$ modulo $\pi$. ThenAt such a time $A(t)$ is off diagonal, and the maximizing initial condition is $\boxed{x(0)=0}$ with $y(0)\ne 0$. For negative $\beta$, the axes interchange and the formula uses $|\beta |$.

When $\epsilon =0$, trajectories conserveso they are closed ellipses around the origin. Ordinary energy measures circular radius rather than this conserved elliptical radius. Starting on the short-energy axis and rotating to the long-energy axis produces the transient amplification from part d; the state later returns, so the growth is transient despite neutral eigenvalues.