Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 77 (Convection)

Use the layer depth $d$, the thermal diffusion time $d^2/\kappa$, velocity $\kappa /d$, and the imposed temperature difference $\Delta T$ as scales. Let $\theta$ be the temperature departure from the conductive profile, $\mathbf{u}$ the velocity, $w=\mathbf{u}\cdot \hat{\mathbf{z}}$, and let pressure absorb hydrostatic terms. The dimensionless Boussinesq equations for Rayleigh-Bénard convection areThe Prandtl number and Rayleigh number are $\sigma =\nu /\kappa$ and $R=g{\alpha }_T\Delta T{d}^3/(\nu \kappa )$, with kinematic viscosity $\nu$, thermal diffusivity $\kappa$ and coefficient of thermal expansion ${\alpha }_T$. At $z=0,1$, the stress-free boundary condition and perfectly conducting thermal boundary condition give $w=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and $\theta =0$.

For the fundamental vertical normal mode, take $w=W\sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$ and $\theta =\vartheta \sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Here $k=|\mathbf{k}|>0$ is the horizontal wavenumber. With $s=k^2+{\pi }^2$, projection onto incompressible flow eliminates pressure and givesThus the stress-free convection growth-rate polynomial is $(\lambda +\sigma s)(\lambda +s)=\sigma R{k}^2/s$.

For the infinite-Prandtl number limit, the finite thermal growth rate isThe other, viscously damped root tends to $-\sigma s$. For $\sigma =1$, both roots are explicit:The plus root determines instability. A higher vertical harmonic replaces ${\pi }^2$ by $n^2{\pi }^2$; the fundamental is the most favorable one.

For the large-$R$ maximum in the infinite-Prandtl number limit, put $x=k^2$ and $p={\pi }^2$. Differentiating gives $d{\lambda }_{\infty }/dx=R(p-x)/(p+x{)}^3-1$. The unique maximum for sufficiently large $R$ therefore satisfies $R(p-x)=(p+x{)}^3$, whenceThis is fastest growth in infinite-Prandtl convection. The large-$R$ expansion is taken after the infinite-Prandtl number limit; at finite $\sigma$, neglecting inertia additionally requires $|\lambda |\ll \sigma s$. It selects the fastest growing mode far above onset, rather than the neutral-curve minimum $k^2={\pi }^2/2$.

The Lorenz model of thermosolutal convection is a Galerkin method truncation retaining a roll velocity and the leading temperature and concentration modes. The variable $a$ measures the roll velocity; $b$ is the temperature mode correlated with the vertical motion, while $c$ is the horizontally averaged temperature-profile distortion generated by that motion. The corresponding concentration modes are $d$ and $e$. The symbol $d$ in this model is a solute amplitude, not the dimensional layer depth. The terms $ab$ and $ad$ change the mean scalar profiles; $a(1-c)$ and $a(1-e)$ express advection of the remaining background gradients.

The positive parameters are the Prandtl number $\sigma =\nu /\kappa$, the ratio $\tau ={\kappa }_s/\kappa$ of solutal diffusivity to thermal diffusivity, and the geometric vertical-mode decay ratio $\varpi$. In the usual fundamental-roll truncation with time normalized to the thermal roll-mode decay rate, $\varpi =4{\pi }^2/(k^2+{\pi }^2)$; treating it as an independently specified positive truncation parameter leaves the arguments below unchanged.

The parameter $r$ is the destabilizing thermal Rayleigh number divided by the free-slip critical-mode value, and $r_s>0$ measures stabilizing solutal buoyancy in the same thermal-diffusive normalization. For example, using a positive bottom-to-top stabilizing solute difference $\Delta S$ and density coefficient ${\beta }_S$,If a solutal Rayleigh number is instead defined with ${\kappa }_s$ in its denominator, the coefficient $r_s$ here is $\tau$ times that normalized number. Thermal buoyancy drives the roll; solutal buoyancy opposes it and diffuses at a different rate. This is the essential mechanism of thermosolutal convection.

Set the time derivatives to zero. The scalar equations giveFor the nonzero convection branch, division of the velocity equation by $a$ givesIts zero-amplitude limit is $r_e=1+r_s/\tau$. Differentiation yieldsFor $0<\tau <1$, the sharp condition for a subcritical bifurcation at the steady onset is $r_s>{\tau }^3/(1-{\tau }^2)$. The stronger bound printed in the question is sufficient, since $(1-\tau {)}^2<1-{\tau }^2$. Under it the branch initially runs toward smaller $r$, and the unique saddle-node bifurcation hasIt is a genuine positive-amplitude minimum: $r^{\prime \prime }(x)=2r_s\tau (1-{\tau }^2)/({\tau }^2+x{)}^3>0$. Put $h=\sqrt{r_s\tau }$ and $j=\sqrt{1-{\tau }^2}$. At the minimum, $1+x_{\ast }=j(j+h)$ and ${\tau }^2+x_{\ast }=hj$, soThese results describe the steady fold of a thermosolutal Lorenz model. They do not by themselves assert stability against every oscillatory disturbance.

Take $\beta \ge 0$ without loss of generality, since its definition fixes only ${\beta }^2$. With $r_s={\beta }^2\tau$, the steady-branch minimum expands as $r_{\min }=1+2\beta \tau +O({\tau }^2)$. For fixed positive $\sigma$, the source's supplied approximation therefore givesThus the requested inequality follows from the supplied approximation, with equality at first order when $\beta =\delta$. If negative $\beta$ is retained, replace it by $|\beta |$ throughout.

There is an independent source-consistency issue with calling that supplied expression the oscillatory threshold of the displayed model. Linearizing about conduction, the $c,e$ modes decay separately and the $a,b,d$ characteristic polynomial isWriting it as ${\lambda }^3+A{\lambda }^2+B\lambda +C$, a nonzero imaginary pair requires $C=AB$ and $B>0$, by the Routh-Hurwitz stability criterion. Direct algebra gives the oscillatory threshold of a thermosolutal Lorenz modelConsequently, with $\delta =\sigma /(1+\sigma )$,The two factors of $\delta$ in the supplied threshold are interchanged. For example $\sigma =1$, $\beta =2$, $\tau =10^{-3}$ gives a genuine marginal imaginary pair at $r_H=1.004003$, whereas the supplied first-order expression gives $1.0085$. The intended comparison remains true after correction:This comparison applies to a physical Hopf bifurcation when ${\omega }_H^2>0$ and to an interior steady fold when $a_{\ast }^2>0$. For fixed nonzero $\beta$ and fixed $\sigma >0$, both conditions hold for sufficiently small positive $\tau$. At a vanishing squared difference, neglected terms must be examined before drawing a higher-order conclusion.

For the real-coefficient cubic-quintic amplitude equation, the trivial pattern $A=0$ exists for every $\mu$. A nonzero spatially uniform pattern has arbitrary constant phase and intensity $q=|A{|}^2>0$ satisfying $\mu +\alpha q-q^2=0$. ThusNonzero patterns exist for $\mu \ge -{\alpha }^2/4$. The upper branch $q_{+}$ exists throughout that range; a distinct positive lower branch $q_{-}$ exists only for $-{\alpha }^2/4<\mu <0$ and reaches zero at $\mu =0$.

For a homogeneous perturbation, write $A=R{e}^{i\varphi }$. Then $\dot{R}=R(\mu +\alpha R^2-R^4)$ and $\dot{\varphi }=0$. The radial eigenvalue at a nonzero pattern iswhile the phase eigenvalue is zero. Therefore the upper branch is stable modulo its arbitrary phase for $\mu >-{\alpha }^2/4$; the lower branch is unstable. The phase-neutral direction prevents asymptotic attraction to one particular complex amplitude, but permits orbital stability of the circle of equivalent patterns.

Conduction is asymptotically stable for $\mu <0$ and unstable for $\mu >0$. At $\mu =0$, the positive cubic term destabilizes arbitrarily small nonzero amplitudes, even though its linear eigenvalue vanishes. At $\mu =-{\alpha }^2/4$, the merged nonzero branch is semistable from above only: $\dot{R}=-R(R^2-\alpha /2{)}^2$, so amplitudes just below it eventually decrease toward zero. Thus the bistable interval is $-{\alpha }^2/4<\mu <0$, with the endpoints treated separately rather than declared stable merely because an eigenvalue is zero.

For the intended nontrivial interface, take the limiting amplitude at the patterned end to be $R_{-}>0$; the identically zero solution is otherwise a trivial exception. Splitting the stationary real-coefficient cubic-quintic amplitude equation into real and imaginary parts givesThe imaginary equation implies $R^2{\varphi }^{\prime }=J$, a constant. The radial first integral isA regular solution has finite $E$. As $R\to 0$, the nonnegative $J^2/(2R^2)$ term would diverge unless $J=0$. Hence ${\varphi }^{\prime }=0$ wherever the nontrivial front has positive amplitude. This conclusion does not assume in advance that the patterned end has no phase gradient.

For a heteroclinic orbit connecting two constant states, $R^{\prime }$ tends to zero at both ends. Thus $E=V(0)=0$, while the nonzero end must obey both $\mu +\alpha q-q^2=0$ and $V(\sqrt{q})=0$. Eliminating $\mu$ givesThis is the stationary front of a cubic-quintic amplitude equation: a unique equal-potential parameter in the bistable interval.

Existence, not just a necessary condition, follows from the first integral. At ${\mu }_M$, $V(R)=-R^2(R^2-q{)}^2/6$. The descending solution obeys $R^{\prime }=-R(q-R^2)/\sqrt{3}$, which integrates toIt connects the stable upper pattern at $x\to -\infty$ to conduction at $x\to +\infty$, with arbitrary translation $x_0$ and constant phase ${\varphi }_0$.

Assume $\alpha >0$, so a finite nonzero critical wavenumber exists. The linear growth rate of a Fourier mode is $\lambda (k)=-\alpha +\mu k^2-k^4$. Maximizing over $k$ and setting that maximum to zero givesSet $K=k_c^2=\sqrt{\alpha }$. The resonant three-wave triad has wavevectors satisfy ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$ and ${\mathbf{k}}_i\cdot {\mathbf{k}}_j=-K/2$ for $i\ne j$. With the slow time and amplitude scaling, every growth, quadratic and cubic term first enters at order ${ϵ}^3$. Projection onto each critical mode removes the nonresonant correction ${\Theta }_3$ by the method of multiple scales solvability condition.

The parameter displacement contributes ${\mu }_{2}KA$ to the first mode. For the quadratic term, use $\nabla \cdot ({\Theta }_1\nabla {\Theta }_1)=\frac{1}{2}{\nabla }^2({\Theta }_1^2)$. The resonant coefficient of ${\Theta }_1^2$ at ${\mathbf{k}}_1$ is $2\bar{B}\bar{C}$, so its contribution is $+\gamma K\bar{B}\bar{C}$.

For the cubic gradient term, an ordered triple of modes with wavevectors $\mathbf{p},\mathbf{q},\mathbf{r}$ summing to ${\mathbf{k}}_1$ contributes $(\mathbf{p}\cdot \mathbf{q})({\mathbf{k}}_1\cdot \mathbf{r})$ times its amplitude product. The three self-interaction permutations yield $-3K^2|A{|}^{2}A$. The six permutations of $({\mathbf{k}}_1,{\mathbf{k}}_2,-{\mathbf{k}}_2)$ yield $-2K^2-4({\mathbf{k}}_1\cdot {\mathbf{k}}_2{)}^2=-3K^2$, and likewise for the third mode. Therefore the hexagonal convection amplitude equations areFor a nonzero convection roll, $B=C=0$ and $|A{|}^2={\mu }_2/(3K)$, requiring ${\mu }_2>0$. Choose its phase so $A=a>0$. The radial perturbation decays, but a perturbation in the two other orientations givesThe eigenvalues are $\pm |\gamma |Ka$. ThusThis is roll instability from broken up-down symmetry. A growing homogeneous amplitude perturbation already proves instability, without a sideband calculation. The source's slightly asymmetric setting means nonzero quadratic coupling. If $\gamma =0$ is allowed literally, the displayed leading equations instead have a continuous sphere of constant-total-intensity states and neutral orientation perturbations. They then establish marginality, not strict growing-mode instability; higher-order effects would decide further selection. Likewise $\alpha =0$ would invalidate the assumed finite-$k_c$ triad scaling.

Use the same length, thermal-time, velocity and temperature scales as in thermal convection. Write the imposed magnetic field as $B_0\hat{\mathbf{z}}$ and measure its perturbation in units of $B_0$. The physical linear magnetohydrodynamic momentum equation contains magnetic tension $(B_0^2/{\mu }_0){\partial }_z\mathbf{b}$; the gradient part of the Lorentz force density is absorbed into magnetohydrodynamic total pressure.

Divide the momentum equation by the viscous force scale ${\rho }_0\nu \kappa /d^3$. Its acceleration coefficient is $\kappa /\nu =1/\sigma$ and its magnetic coefficient isHere $Q$ is the Chandrasekhar number, comparing magnetic and viscous-diffusive effects, and $\zeta$ is the ratio of magnetic diffusivity to thermal diffusivity. The induction scale is $B_0\kappa /d^2$, so linearization givesComplete the linear system with $\dot{\theta }=w+{\nabla }^2\theta$, $\nabla \cdot \mathbf{u}=0$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{b}=0$. The omitted advective and perturbed-field tension terms are quadratic in the disturbances. Magnetic tension resists bending of the vertical field, while diffusion allows that bending to relax.

Interpret perfect conductivity here as the perfectly conducting thermal boundary condition $\theta =0$; impose the stated magnetic verticality separately. Together with the stress-free boundary condition, admissible vertical profiles areeach multiplied by $e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Solenoidality requires $i\mathbf{k}\cdot {\mathbf{U}}_h=-\pi W$ and $i\mathbf{k}\cdot {\mathbf{B}}_h=\pi Z$. The resistive induction equation preserves both profile choices. In particular ${\mathbf{b}}_h=0$ at each boundary and ${\partial }_z{b}_z=0$ there.

Set $s=k^2+{\pi }^2$ and $p={\pi }^2$. The pressure-projected vertical momentum, thermal and vertical induction equations reduce toTheir determinant gives the vertical-field magnetoconvection dispersion relationIt is best kept as a polynomial rather than canceling a factor that might vanish at a special eigenvalue. For $\lambda =0$,For an oscillatory marginal mode $\lambda =i\omega$ with $\omega \ne 0$, write the polynomial as ${\lambda }^3+a_1{\lambda }^2+a_2\lambda +a_3$. Its coefficients areSeparating real and imaginary parts gives $a_3=a_1{a}_2$ and ${\omega }^2=a_2>0$. Solving yieldsTherefore oscillatory marginality in vertical-field magnetoconvection requiresThe equality has zero frequency and is not an oscillatory marginal mode. For some positive horizontal wavenumber to satisfy this condition, it is necessary and sufficient that $Q>\zeta (1+\sigma ){\pi }^2/[\sigma (1-\zeta )]$, although the associated threshold must still be minimized over admissible modes. When ${\omega }^2>0$, $R^{(o)}<R^{(e)}$ at the same wavenumber; this also follows from $R^{(e)}-R^{(o)}=(\sigma +1+\zeta )s{\omega }^2/(\sigma \zeta k^2)$.

Both neutral curves can be written with $x=k^2$ and $p={\pi }^2$ asFor steady convection, $(C,D)=(1,1)$; for oscillatory convection,The oscillatory result is physically relevant only with positive frequency; for fixed $0<\zeta <1$ and fixed positive $\sigma$, that holds at the minimizing wavenumber for sufficiently large $Q$.

Differentiation gives the exact minimization conditionIts left side is strictly increasing for $x>0$, so the minimum is unique. Define $h=[D{p}^{2}Q/(2C){]}^{1/3}$. An asymptotic expansion givesSubstituting back, the leading magnetic term is order $Q$, while the two balanced terms $C{x}^2$ and $DQ{p}^2/x$ sum to $3C{h}^2$. Hence, through order $Q^{2/3}$,In particular,while the same boxed general formulas with $(C_o,D_o)$ give the oscillatory minimum and critical wavenumber. Thus strong-field wavenumber selection in magnetoconvection has $k\propto Q^{1/6}$ for both types, with different parameter-dependent constants.

For the exact steady result at arbitrary $Q\ge 0$, the stationarity equation with $C=D=1$ lets us eliminate $Q$ from $R^{(e)}$. It givesAt $Q=0$, $x=p/2$ and $R_0=27p^2/4$. ConsequentlyThis exact critical Rayleigh relation for vertical-field magnetoconvection includes the nonmagnetic minimum, with $R_c\ge R_0$ and $k_c^2\ge {\pi }^2/2$.