Codex @codex  0

Let the basic temperature be $\overline{T}=T_{\ast }-\beta z$ with $\beta >0$, the density be ${\rho }_0$, gravitational acceleration be $-g\hat{\mathbf{z}}$, and thermal expansion coefficient be $a_T$. Denote kinematic viscosity, thermal diffusivity and magnetic diffusivity by $\nu ,\kappa ,\eta$. Write the dimensional magnetic perturbation as ${\mathbf{b}}_d$, the temperature perturbation as ${\theta }_d$, and use constant permeability ${\mu }_0$. Linearizing the Boussinesq equations and the resistive induction equation about the static state givesThe Lorentz force identity is $(\nabla \times {\mathbf{b}}_d)\times B_0\mathbf{c}=B_0(\mathbf{c}\cdot \nabla ){\mathbf{b}}_d-\nabla (B_0\mathbf{c}\cdot {\mathbf{b}}_d)$. Its gradient term is absorbed into $P_d$, the perturbation magnetohydrodynamic total pressure. The temperature source follows from ${\mathbf{u}}_d\cdot \nabla \overline{T}=-\beta u_{d,z}$.

Choose any reference length $d$, since no vertical boundaries are prescribed. Scale length by $d$, time by $d^2/\kappa$, velocity by $\kappa /d$, temperature perturbation by $\beta d$, magnetic perturbation by $B_0$, and total pressure by ${\rho }_0\nu \kappa /d^2$. The dimensionless parameters areThey are the Prandtl number, magnetic-to-thermal diffusivity ratio, Rayleigh number, and Chandrasekhar number, respectively. Division by the chosen scales givesIn particular $Q\zeta =B_0^2{d}^2/({\mu }_0{\rho }_0\nu \kappa )$, so the diffusivity convention agrees with the momentum coefficient.

For a plane wave put $\mathbf{l}=(k,0,m)$, $q=|\mathbf{l}{|}^2=k^2+m^2$, and $d_{\parallel }=\mathbf{c}\cdot \mathbf{l}=k\sin \alpha +m\cos \alpha$. The induction and temperature equations imply $\hat{\mathbf{b}}=i{d}_{\parallel }\hat{\mathbf{u}}/(s+\zeta q)$ and $\hat{\theta }=\hat{w}/(s+q)$. The projection $P=I-\mathbf{l}{\mathbf{l}}^{\mathsf{T}}/q$ removes pressure. Because $(P\hat{\mathbf{z}}{)}_z=k^2/q$, its vertical momentum equation isThus the oblique-field plane-wave magnetoconvection dispersion relation isIf out-of-plane velocity is also allowed, its decoupled magnetic branch satisfies $(s/\sigma +q)(s+\zeta q)+Q\zeta d_{\parallel }^2=0$; it has no buoyant driving. At stationary marginality, for $k\ne 0$,These are stationary thresholds; an oscillatory marginal mode, if present for particular diffusivity ratios, is a different onset calculation.

For $m=1$, write $R_{mathrmstat}=F(k)+QG(k)$, where $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. Direct differentiation gives $F^{\prime }(k)=-2k^{-3}+6k+4k^3$, so its two minima are $\pm 1/\sqrt{2}$, with $F=27/4$ and $F^{\prime \prime }=36$. For $Q>0$ the negative sign has the smaller magnetic penalty: at every positive $k$, $G(-k)<G(k)$ because both sine and cosine of the inclination are positive. The wavenumber selection in oblique-field magnetoconvection therefore starts from $a=-1/\sqrt{2}$. Expanding $F^{\prime }(k_c)+Q{G}^{\prime }(k_c)=0$ gives $k_c=a+C_{1}Q+O(Q^2)$, where $C_1=-G^{\prime }(a)/F^{\prime \prime }(a)$. Taylor expansion of the value then givesThis proves the small-field forms, including the order of the correction.

For large $Q$, $G$ is nonnegative and vanishes only at $b=-\cot \alpha$. The candidate $k=b$ has finite threshold $F(b)$, so a global minimizer must satisfy $QG(k_c)\le F(b)$ and hence tend to $b$. Here $G^{\prime \prime }(b)=2(1+b^2){\sin }^2\alpha /b^2>0$. Expanding the stationarity condition at $b$ yieldsThe value expansion isThe leading coefficient follows from $1+{\cot }^2\alpha ={\csc }^2\alpha$. The expansion holds at fixed $0<\alpha <\pi /2$.

Finally choose $\cot \alpha =1/\sqrt{2}$. Then the nonmagnetic minimizing wavevector also has $d_{\parallel }=0$, and its magnetic penalty vanishes. Since $F(k)\ge 27/4$ and $QG(k)\ge 0$, this is the global minimum for every nonnegative field strength:This field-independent stationary threshold in oblique magnetoconvection has a simple physical explanation: the disturbance is constant along the imposed field, so it does not bend field lines.

Let the layer depth be $d$, its imposed temperature drop $\Delta T$, and its dimensional vertical field $B_0\hat{\mathbf{z}}$. Use length $d$, thermal time $d^2/\kappa$, velocity $\kappa /d$, temperature $\Delta T$ and field $B_0$ as units. Write the total dimensionless field as $\hat{\mathbf{z}}+\mathbf{b}$, and temperature as the conductive profile plus $\theta$. The Boussinesq equations coupled to the resistive induction equation areThe dimensionless parameters in this normalization areHere ${\alpha }_T$ is the thermal expansion coefficient, $\nu$ is kinematic viscosity, $\kappa$ is thermal diffusivity, and $\eta$ is magnetic diffusivity. The first parameter is the Rayleigh number and $\sigma$ is the Prandtl number.

The displayed dispersion relation fixes the magnetic normalization: its $Q$ must be the thermal-diffusion magnetic-field parameter above. The conventional Chandrasekhar number is instead $Q_{\mathrm{Ch}}=B_0^2{d}^2/({\mu }_0\rho \nu \eta )$, so $Q=\zeta Q_{\mathrm{Ch}}$. If $Q$ is reserved for that conventional definition, the magnetic term in the printed polynomial requires an extra factor $\zeta$. We keep $Q$ for the printed polynomial throughout the following calculation.

Linearize about the conductive state. Absorbing linear magnetic pressure into $p$, the magnetic force becomes $\sigma Q{\partial }_z\mathbf{b}$. The pressure-eliminated vertical momentum equation isThe normal mode implicit in the given relation has $w,\theta \propto \sin \pi z$ and $b_z\propto \cos \pi z$, as for stress-free, fixed-temperature boundaries with the corresponding vertical-field magnetic conditions. Put $x=k^2$, $p_0={\pi }^2$ and $m=x+p_0={\beta }^2$. Taking growth proportional to $e^{st+i\mathbf{k}\cdot {\mathbf{x}}_h}$ gives the amplitude equationsTheir determinant isThis is the stated vertical-field magnetoconvection dispersion relation after accounting for its field-strength convention. The determinant derivation remains valid even at values where an intermediate division by $s+m$ or $s+\zeta m$ would not be allowed.

For steady marginality set $s=0$. One obtainsFor oscillatory marginality write $s=i\omega$ with $\omega \ne 0$. Dividing the polynomial by $m$ gives a real cubic $s^3+A{s}^2+Bs+C_0$, whereIts real and imaginary parts imply ${\omega }^2=B$ and $C_0=AB$. Solving for the oscillatory convection threshold givesandA physical oscillatory branch therefore requires $0<\zeta <1$ and positive frequency squared. In particular an algebraic continuation of $R_o$ into ${\omega }^2<0$ is not an oscillatory onset.

The steady-Hopf merger in vertical-field magnetoconvection occurs at zero frequency:Substitution verifies $R_s=R_o$. Equivalently the merged horizontal wavenumber satisfiesA finite-wavenumber intersection exists only when this expression is positive. At the intersection the growth polynomial has a double zero root: the nonzero-frequency neutral branch ends there.

For the requested merger at a magnetoconvection neutral-curve minimum, first note that a curve of the form $R=[A_{\ast }{m}^3+C_{\ast }m]/x$ has derivative zero exactly whenIts unique positive minimum has $x>p_0/2$, because the left side increases there from zero to infinity. At this minimum $R_c=2A_{\ast }{m}^3/p_0$.

Merger at the steady minimum. For $R_s$, the stationarity condition is $m_s^2(2x_s-p_0)=Q{p}_0^2/\zeta$. Combining it with the merger equation givesThe corresponding critical Rayleigh number is $R_{s,c}=2(x_s+p_0{)}^3/p_0$. For fixed positive $\sigma$ and large $Q$, set $\zeta =1-{\delta }_s$ and retain the leading powers. ThenMore generally, before taking $\zeta \to 1$, the steady minimum obeys $k_s^2\sim [Q{\pi }^4/(2\zeta ){]}^{1/3}$.

Merger at the oscillatory minimum. Define $A_o=(\sigma +\zeta )(1+\zeta )/\sigma$. The stationarity condition for $R_o$ isCombining it with the zero-frequency relation instead givesHere $R_{o,c}=2A_o(x_o+p_0{)}^3/p_0$. For fixed $\sigma >0$ and $Q\gg 1$, setting $\zeta =1-{\delta }_o$ givesThese two merger conditions are distinct parameter surfaces, not simultaneous assertions about the same minimum. Both have the strong-field wavenumber selection in magnetoconvection scaling $k=O(Q^{1/6})$.

If the first subleading Rayleigh corrections are desired, let $X_s=[Q{p}_0^2/(2\zeta ){]}^{1/3}$ and $X_o=[\sigma Q{p}_0^2/(2(1+\sigma )(1+\zeta )){]}^{1/3}$. With parameters bounded in the above regime,These retain the $O(Q^{2/3})$ corrections, including the effect of the $O(Q^{-1/3})$ diffusivity deficit at a merger.

Use the flat-surface Boussinesq approximation, with small thermal expansion, constant properties, negligible surface deformation, and constant surface tension. The top is impermeable and shear-free, while the solid walls are no-slip. Temperature-dependent surface tension would introduce a separate Marangoni stress, which is not included in this buoyancy-driven model. The stated time scale requires $\kappa$ to mean thermal diffusivity; if the material property given is a dimensional thermal conductivity $k$, use $\kappa =k/({\rho }_0{c}_p)$.

Take $z$ positive downwards and define the nondimensional streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. Velocity is scaled by $\kappa /L$, streamfunction by $\kappa$, and temperature is measured from the cold-end surface value. Define the Prandtl number and horizontal-length Rayleigh number bywhere ${\alpha }_T$ is the thermal expansion coefficient. Put $J(\psi ,f)={\psi }_z{f}_x-{\psi }_x{f}_z$, and let ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the curl of the Boussinesq equations eliminates pressure and gives the full nonlinear vorticity-temperature systemThe buoyancy sign follows from the downward coordinate: warmer fluid has body acceleration $-g{\alpha }_T\Delta TT{\mathbf{e}}_z$.

Choose the same streamfunction constant, zero, on the connected boundary. The mechanical and thermal boundary conditions areThe upper stress-free boundary condition reduces to ${\psi }_{zz}=0$, since $\psi =0$ there already gives ${\psi }_{xx}=0$. An arbitrary additive constant in $T$ changes only the reference hydrostatic pressure.

In steady infinite-Prandtl-number convection, momentum inertia disappears but thermal advection remains:For a shallow lake, with fixed $\mathrm{Ra}$ as $ϵ\to 0$, the leading interior temperature is $T_0=x$. This is an outer solution: it does not satisfy the insulated sidewall derivative right at $x=0,1$. The leading global conductive solution, which resolves those end regions, isIt is harmonic, has $T_c(x,0)=x$, and satisfies the insulated sides and bottom. Away from sidewall layers of horizontal width $O(ϵ)$ it reduces to $x$ with exponentially small end corrections. These layers also turn the flow and enforce the sidewall no-slip boundary condition.

In the shallow interior the dominant momentum derivatives are vertical, so ${\psi }_{0,zzzz}=-\mathrm{Ra}$. Integrating with the surface and bottom conditions gives the shallow horizontal convection with an insulated bottom solution, with $y=z/ϵ$:Surface flow is toward the colder end, with a deeper return flow. The depth-integrated volume flux is zero because ${\psi }_0(ϵ)={\psi }_0(0)=0$. The bulk expressions apply away from the turning regions, not unchanged on the vertical walls. A useful small-advection condition in the end layers is $\mathrm{Ra}{ϵ}^4\ll 1$, automatic for fixed Rayleigh number in the stated thin limit.

The heat flux in these units is $\mathbf{q}=(uT-T_x,wT-T_z)$. For the leading bulk temperature, the horizontal transport through a vertical section per unit out-of-plane width isThus leading advective horizontal transport vanishes: both layers have the same temperature at a given $x$, and their opposing volume fluxes cancel. The leading conductive transport is from the hot end toward the cold end. In dimensional form it is $-{\rho }_0{c}_p\kappa \Delta Tϵ=-k\Delta Tϵ$ per unit width. Heat enters through the warmer portion of the maintained upper surface and leaves through its colder portion, with the conductive end adjustments particularly important near the two insulated sidewalls. It cannot enter through the insulated sides or bottom.

For the advection correction in shallow horizontal convection, set $T=x+\Theta (z)$ in the bulk. Since $w_0=0$ and $T_x=1$, the steady heat equation gives ${\Theta }_{zz}=u_0={\psi }_{0,z}$. The conditions $\Theta (0)=0$ and ${\Theta }_z(ϵ)=0$ imply ${\Theta }_z={\psi }_0$ and henceFor example the bottom bulk temperature is $T(x,ϵ)=x-\mathrm{Ra}{ϵ}^5/320$. The correction makes the returning deeper fluid slightly colder. Its horizontal advective heat transport isThus the transport including this bulk correction is $\boxed{{\mathcal{H}}_x=-ϵ-19{\mathrm{Ra}}^2{ϵ}^9/1451520}$, both terms directed toward the colder end.

Vertical transport must distinguish local flux from net transport. In the leading interior $w_0=T_{0,z}=0$, whereas the temperature correction gives a local downward conductive flux $q_z=-{\Theta }_z=-{\psi }_0>0$. The end circulation gives an upward advective transport: with impermeable sidewalls and $T_0\simeq x$, integration by parts gives ${\int }_0^{1}w{T}_{0}dx=-{\int }_0^{1}x{\psi }_{x}dx={\int }_0^1\psi dx\simeq {\psi }_0(z)<0$. It compensates that downward conduction. More generally, steady heat conservation and insulated sides give $d[{\int }_0^1{q}_{z}dx]/dz=0$, and the insulated impermeable bottom sets the constant to zero. ThereforeThere is no net vertical heat transport through the lake, despite its local recirculating heat flux and its nonzero horizontal heat transport.

Let $a(t)$ be the particle radius and $\Delta T=T_s-T_{\infty }>0$ the supercooling. Use spherical symmetry, constant thermal properties, equal phase densities $\rho$, and negligible fluid advection. Let $\mathcal{L}$ be latent heat per unit mass, $c_p$ the liquid specific heat, $k_{\ell }$ its thermal conductivity and $\kappa =k_{\ell }/(\rho c_p)$ its thermal diffusivity. The exterior temperature satisfies the radial heat equationwith $T(r,0)=T_{\infty }$ for $r>a_0$, $a(0)=a_0$ and $T(r,t)\to T_{\infty }$ as $r\to \infty$.

The Gibbs-Thomson effect gives curvature-induced melting-temperature depression. For a spherical interface writewhere $\gamma$ is solid-liquid interfacial energy per area and the factor two is the sphere's curvature $2/a$. Here $T_s$ in the thermodynamic formula is an absolute temperature. The notation $\Gamma$ also permits a given measured capillary coefficient without requiring its microscopic derivation.

In the one-phase Stefan problem, the solid has no internal heat flux and the Stefan condition isThe sign follows from heat balance: positive outward liquid heat flux releases latent heat as the solid grows. If the solid's thermal storage is retained, its temperature $T^{(s)}$ instead satisfies its own radial heat equation, ${\partial }_r{T}^{(s)}(0,t)=0$, and $T^{(s)}(a^{-},t)=T(a^{+},t)$. The two-phase condition is $\rho \mathcal{L}\dot{a}=k_s{T}_r^{(s)}(a^{-},t)-k_{\ell }{T}_r(a^{+},t)$. It reduces to the one-phase condition in the quasistatic limit because a regular harmonic temperature inside a sphere is spatially constant. An initial solid temperature profile is additional data for a fully time-dependent two-phase problem.

Two reciprocal Stefan number conventions are common. The large parameter appropriate to the requested slow-interface limit is the latent-to-sensible heat ratioIt measures the energy required for phase change relative to cooling or heating by the supercooling. The often-used sensible-to-latent definition is $\mathrm{Ste}=c_p\Delta T/\mathcal{L}=1/S$. Thus “large Stefan number” here means large $S$, not large $\mathrm{Ste}$ in that other convention.

The thermal diffusion time for scale $a_0$ is $a_0^2/\kappa$, whereas the interface time is $S{a}_0^2/\kappa$. More explicitly, with $R=r/a_0$, $\theta =(T-T_{\infty })/\Delta T$ and slow time $\tau =\kappa t/(S{a}_0^2)$, the liquid equation isFor $S\gg 1$, temperature therefore equilibrates before the radius changes appreciably. The quasistatic spherical Stefan problem with capillarity has a harmonic exterior temperature, fixed by its interface and far-field values:The small initial diffusion adjustment is omitted at this order. Inserting $T_r(a)=-\Delta T/a+\Gamma /a^2$ into the Stefan condition givesThis is the critical radius for a supercooled spherical seed. For $a_0>a_c$, the interface is warmer than the liquid and loses heat, driving growth. For $0<a_0<a_c$, its curvature-depressed interface temperature is colder than the liquid, so heat enters and the solid melts. At $a=a_c$, the uniform temperature $T_{\infty }$ supplies an equilibrium radius. It is unstable, since the radius-velocity derivative there is $\kappa /(S{a}_c^2)>0$. For a growing particle with $a\gg a_c$, $a^2$ increases asymptotically as $2\kappa t/S$.

For a subcritical particle the radius decreases monotonically and remains below $a_0$. HenceIt cannot remain positive beyond $S{a}_0^3/[3\kappa (a_c-a_0)]$, proving finite-time extinction of a subcritical spherical seed. The exact extinction time in the reduced model is obtained by separating variables:With ${\eta }_0=a_0/a_c\in (0,1)$, this isNear disappearance, the capillary term dominates and $\dot{a}\sim -\kappa a_c/(S{a}^2)$, so $\boxed{a^3\sim 3\kappa a_c(t_e-t)/S}$. For $a_0\ll a_c$, the lifetime is $t_e\sim S{a}_0^3/(3\kappa a_c)$; it diverges logarithmically as $a_0\to a_c^{-}$, consistently with the unstable equilibrium. The threshold and lifetime above are conclusions of the specified large-$S$ quasistatic continuum model, rather than an assertion that its approximation is uniform through every finite-$S$ shrinking-interface transient.

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.

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.

Let $K_w$ be the water thermal conductivity, $L_m$ the latent heat of fusion per unit ice mass, and ${\rho }_i$ the ice density. Define $v_m>0$ as the normal ice-retreat speed that enlarges the cavity. With the water thermal boundary layer at $T_s$ on its warm side and $T_m$ at the melting interface, the heat supply per unit interface area is approximatelyThe ice is specified to be uniformly at $T_m$, so no leading sensible-heat flux into colder ice must be subtracted. The Stefan condition is therefore ${\rho }_i{L}_m{v}_m=q_T$, givingEquivalently $K_w={\rho }_w{c}_p{\kappa }_T$ expresses the heat flux in terms of water thermal diffusivity. The imposed temperatures and constant boundary-layer thickness make this melt rate uniform and constant over the wetted interface. Melting adds latent energy demand to the flow; for an isolated finite pulse, maintaining $T_s$ indefinitely would require heat replenishment. The constant-temperature model is consequently an imposed closure, not a prediction of a permanently hot finite water volume.

Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.