Measure $z$ upward from the heat exchanger and let the steady ice front be at $z=h>0$. In the exchanger frame, salt in the liquid satisfies the advection-diffusion equationThe decaying solution and the prescribed total salt mass aresoThe linear liquidus condition fixes the interface temperature asA solid layer between the exchanger and the interface can therefore exist only if

Write $\vartheta =T-T_{\infty }$. Heat advection, conduction, and environmental loss giveIts characteristic exponents areBelow the exchanger and above the ice front, boundedness givesIn $0<z<h$, the ice temperature iswhereSubstitution of these fields into the Stefan conditiongives one scalar equation for the steady height $h$, which can be solved numerically. The heat flux may jump at $z=0$ because the exchanger supplies the required localized cooling.

The local equilibrium freezing temperature ahead of the front isBecause $T_l(h)=T_L(h)$, constitutional supercooling begins when the actual liquid-temperature gradient at the interface is smaller than the liquidus gradient:Using the solutions above, this is

At criticality the inequality is an equality. If $F/k\ll V^2/{\kappa }^2$, then $q\sim V/\kappa$ and, for $\kappa >D$,The critical curve is proportional to $S^{-1}$. If $F/k\gg V^2/{\kappa }^2$, put $q\sim \sqrt{F/k}$ to obtainThis curve behaves as $S^{-1/2}$ for small $S$ and approaches $D\sqrt{F/k}$ for large $S$. Supercooling occurs above the corresponding critical curve, where solute rejection steepens the liquidus faster than heat transport raises the actual temperature.

LetThe Neumann solution of the Stefan problem in the ice and substrate isandContinuity of heat flux at the contact gives, with $r=\sqrt{k_s/k}$,and therefore

At the ice–water interface, the water is isothermal at $T_m$. The Stefan condition $\rho L\dot{h}=k{T}_{i,z}(h^{-})$ givesEliminating $T_0$ yields the implicit equationwhich determines $\lambda$ and hence $h(t)=2\lambda \sqrt{\kappa t}$.

If $k_s\gg k$, then $r\gg 1$, $T_0\sim T_s$, andThe highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase Stefan problem.

If $k_s\ll k$, then $r\ll 1$, $T_0\sim T_m$, and $\lambda \sim r/(\sqrt{\pi }S)$. ConsequentlyHere heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.

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.

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 $|\beta |$.

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

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 $\boxed{\beta >\epsilon }$.

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

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 $Ra=R{a}_c$.

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

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.

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

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

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

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.

In the torus frame the two planes translate with velocity $U{\mathbf{e}}_x$. Away from the $O(a)$ neighborhood of the torus, the depth-averaged Hele–Shaw flow between planes separated by $2a$ isNegligible leakage imposes ${\overline{u}}_r=0$ at $r=R$. The harmonic pressure that decays at infinity in the exterior and the regular harmonic pressure in the interior are thereforeup to a common constant. The interior velocity is zero, while the exterior flow is the uniform stream diverted around a circular obstacle. In plan view the inside has high pressure on the $+x$ side and low pressure on the $-x$ side; the immediately adjacent exterior has the opposite signs, producing the pressure jump across the torus.

The jump at $r=R$ isIntegrating it over the projected vertical area $2aRd\theta$ gives the global pressure resistanceThere are two narrow gaps, so their local resistance is twice the one-plane result from part b:Consequently the local gap resistance dominates when $a\ll R\ll a{\epsilon }^{-1/2}$, whereas the global Hele–Shaw pressure resistance dominates when

To interpret the upper bound, the pressure jump has scale $\Delta p\sim \mu UR/a^2$. Each narrow gap has thickness $a\epsilon$ and streamwise lubrication length $a\sqrt{\epsilon }$. Its pressure-driven leakage flux per unit centreline length therefore scales asThe blocked Hele–Shaw flux has scale $Ua$. Leakage is negligible precisely when $UR{\epsilon }^{5/2}\ll Ua$, or