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For a stationary, uniformly permeable skeleton, Darcy law gives $\mathbf{u}=-(\Pi /\mu )\nabla p$. Constant-density conservation of mass gives $\nabla \cdot \mathbf{u}=0$, hence the pressure is harmonic:With uniform gravitational acceleration, the same argument applies to pressure after subtracting its hydrostatic part. The spherical calculation below neglects body-force variation and uses the excess pressure relative to the far field.

Set $ϵ=1-{\rho }_s/\rho >0$ and $\beta =\varphi ϵ$. When a shell freezes, its pore mass decreases at rate $4\pi a^2\varphi (\rho -{\rho }_s)\dot{a}$; the difference must leave as liquid. Since Darcy velocity is volume discharge per total area, not pore-fluid speed, the interfacial and radial balances giveThe positive sign means outward expulsion by the expansion on freezing. Integrating Darcy law with $p(\infty )=0$ givesThere is no liquid Darcy velocity in the ice-filled region.

To state a closed Stefan problem, let $k_{-},k_{+}$ be the effective thermal conductivities of the frozen and unfrozen composites and $C_{-},C_{+}$ their volumetric heat capacities. If the matrix mass density is ${\rho }_m$, the common specific heat gives $C_{-}=c_p[(1-\varphi ){\rho }_m+\varphi {\rho }_s]$ and $C_{+}=c_p[(1-\varphi ){\rho }_m+\varphi \rho ]$. These material coefficients are not numerically specified. With $\mathcal{L}=\varphi {\rho }_{s}L$, the thermal equations areThe exterior equation retains advected sensible heat. A more detailed mechanical-energy model adds the Darcy dissipation source $\mu u^2/\Pi$; its smallness at leading order is checked below. At the origin require ${\partial }_r{T}_{-}(0)=0$, and at infinity require $T_{+}\to T_{\infty }$, $p\to 0$. At the phase boundary impose temperature continuity, phase equilibrium and the Stefan condition:These equations, the radial liquid-mass balance and Darcy law, together with an initial radius and compatible initial temperature fields, specify the moving-boundary model. No curvature correction is imposed. A vanishing initial seed corresponds to the leading growth law below with $a(0)=0$.

Write $\Delta =T_m-T_{\infty }$ and $S=L/(c_p\Delta )$, the latent-to-sensible heat ratio. A conductive interfacial balance gives $a\dot{a}\lesssim k_{+}\Delta /\mathcal{L}$. On length scale $a$, the ratio of thermal diffusion time to growth time is consequently bounded byIt is small for $S\gg 1$ with fixed porosity and comparable constituent densities and thermal properties. Very small porosity requires the sharper condition $C_{\pm }\Delta /(\varphi {\rho }_{s}L)\ll 1$; the bare heat-capacity ratio alone should not be used uniformly as $\varphi \to 0$. The advective-to-conductive ratio isEven if mechanical heating is retained, its size relative to latent-heat release is $p_i\beta /\mathcal{L}=(T_m-T_i)/T_m\le \Delta /T_m$. For fixed water properties this is also small as $S\to \infty$, consistently with the linearized Clausius-Clapeyron relation. Thus unsteadiness, thermal advection and mechanical heating are higher-order effects. The local leading temperature fields satisfy Laplace's equation. The approximation describes radii of order $a$; the distant transient diffusion tail supplies the far-field matching and need not be globally steady.

Regularity at zero and the interface/far-field values giveThe Stefan condition becomes $\mathcal{L}a\dot{a}=k_{+}(T_i-T_{\infty })$. Introduce the dimensionless pressure-feedback factorCombining this heat balance with pressure-dependent phase equilibrium yields the growth and interface values:This is pressure-limited spherical freezing in a porous medium. The interface values are constant during this leading solution, although $p(r,t)$ changes at a fixed exterior radius.

As permeability of a porous medium decreases, $\Lambda$ increases: the interface pressure rises toward the finite value $\Delta /\gamma$, while the interface temperature falls toward $T_{\infty }$. Hydraulic resistance suppresses freezing through pressure-induced freezing-point depression. In the limit $\Pi \to 0$, $a\dot{a}\sim \Pi \Delta /(\gamma \mu \beta )$; a zero-seed radius grows as $\sqrt{\Pi t}$ and tends to zero at fixed time. A finite seed instead hardly changes. At exactly zero permeability an incompressible rigid model cannot expel the expansion volume, so sustained growth stops; deformation or compressibility would require a different model.

Choose $x=0$ at the contact line, with $x>0$ pointing into the mountains and the plate moving in the positive $x$ direction at speed $U>0$. Let $z=0$ be the plate. For viscous mountain spreading over an underthrust plate, the lubrication approximation assumes slow horizontal variation, hydrostatic pressure $p=\rho g(h-z)$, and $\mu u_{zz}=p_x=\rho g{h}_x\equiv G$. Impose no slip at the plate and zero tangential traction at the free surface. Without sediment,The horizontal mass flux is $\rho q$. For a steady mountain range with no mountain material entering across its contact line, $q=0$. Therefore $h_x=3\mu U/(\rho g{h}^2)$, and the vanishing height at the bare contact gives the mountain profileThe large slope extremely close to $x=0$ is an inner region where the lubrication approximation ceases to be quantitatively valid.

Now let sediment occupy $0<z<d$ and mountains $d<z<h$, with $H=h-d$ the mountain-material thickness and $M={\mu }_s/\mu$. Equal densities make $G=\rho g{h}_x$ the same in both layers. Continuity of velocity and shear stress at $z=d$, no slip at $z=0$ and zero shear at $z=h$ give the shear stress $G(z-h)$ throughout. Thus the two velocity profiles areTheir volume fluxes areMultiplying by $\rho$ gives the separate mass fluxes. The coupled conservation of mass equations areThis is the sediment-lubricated viscous mountain model. It includes shear-driven transport by the plate and pressure-driven gravitational spreading.

In steady state the mountain-material flux is zero, while the upstream uniform sediment layer has $q_s=U{d}_{\infty }$. For $H>0$ this givesEliminating $G$ proves the algebraic thickness relation:At the contact $H\to 0^{+}$, the last coefficient must vanish. The nonzero physical solution is consequentlyThe condition comes from the limiting zero flux per unit mountain thickness, not merely the automatic vanishing of $q_m$ at $H=0$. The surface slope anywhere under the mountains isLower sediment viscosity flattens the near-contact surface, while its contact height remains $3d_{\infty }$. Far inside, where $H\gg d/M$, the mountains recover the bare-layer slope scaling. The formulas determine the mountain profile by the algebraic relation plus this first-order equation. For example, put $\delta =d/d_{\infty }$ and $w=H/d$; the physical root satisfies $2M(\delta -1)w^2+3(\delta -2)w+\delta -3=0$, taking $w\ge 0$ for $1<\delta \le 3$. This selects the branch smoothly connected to the contact.

In front of the contact, $x<0$, there is only sediment, so $h=d$ andDefine $\ell =\rho g{d}_{\infty }^3/(3{\mu }_{s}U)$ and $P(\delta )={\delta }^3/3+{\delta }^2/2+\delta +\ln (\delta -1)$. Integrating from the contact yields the upstream basin profile of a sediment-lubricated mountain:Equivalently, $d(x)=d_{\infty }{P}^{-1}(P(3)+x/\ell )$ on $1<\delta \le 3$. The inverse is uniquely defined because $P^{\prime }(\delta )={\delta }^3/(\delta -1)>0$; an elementary explicit inverse is unnecessary. This profile tends to $d_{\infty }$ as $x\to -\infty$, approaches $3d_{\infty }$ at the contact and has the same surface slope there as the mountain branch. Its far-field excess decays exponentially on length scale $\ell$, which increases as the sediment viscosity decreases.

In the laboratory frame the lower wall has velocity zero and the cylinder surface has velocity $U$. Solving $\mu u_{yy}=p_X$ with these no-slip conditions givesThe relative velocity in the cylinder frame is $u-U$. Its flux is constant by incompressibility:Equal pressures at the two infinities require $\int p_{X}dX=0$. Put $I_j=\int h^{-j}dX$; then $q=-(U/2)I_2/I_3$. With $X=\ell s$ and $h=h_0(1+s^2)$, the given integrals yield $I_2/I_3=4h_0/3$. Hence $q=-2U{h}_0/3$ andIndeed $p=2\mu UX/h^2$ has this derivative and tends to ambient pressure at both ends. The distinction between the constant comoving flux and the varying laboratory flux is necessary because the gap translates.

Reflection about the closest point leaves the gap geometry unchanged and reverses the direction of translation. The linear Stokes equations make the excess pressure change sign when the imposed velocity changes sign. Combining reflection with this linearity shows that pressure is odd in $X$. Its leading normal force on the stationary plane therefore cancels between the two sides, giving ${\mathbf{e}}_y\cdot \mathbf{F}=0$ in the lubrication limit. Normal viscous stresses are of smaller order in the aspect ratio. This uses symmetry only, rather than a pressure calculation.

Near the closest point, in $X=x-Ut$, the gap is $h(X)\simeq h_0+X^2/(2a)$. The horizontal length scale is $\ell =\sqrt{2a{h}_0}$, up to an inessential constant. Thus the aspect ratio $h_0/\ell \ll 1$ follows from $h_0\ll a$. The lubrication approximation assumes an incompressible Newtonian fluid, no slip on the surfaces, slow variation of the gap, and negligible inertia relative to transverse viscous stress:This is the reduced Reynolds-number condition; it is weaker than requiring a Reynolds number based on $\ell$ to be small. In the cylinder frame the geometry and flow are steady. Pressure is uniform across the narrow gap to leading order, and equal ambient pressures far to either side close the model. The calculation assumes a full liquid film; cavitation would change that pressure condition.

The fluid's tangential traction on the lower stationary surface is $\tau =\mu u_y{|}_{y=0}$ in the positive $x$ direction. Using the pressure gradient found above givesThe drag on a plane beneath a translating cylinder follows by integrating along the wall:per unit cylinder-axis length. This is the force on the lower surface, so its direction follows the cylinder's translation.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.

At long wavelengths $F_e=-q+2q^2+O(q^3)$ and $F_o=q-2q^2/3+O(q^3)$: a thickness or column-density disturbance has a gravitational instability, whereas a bending disturbance is restored. At short wavelengths both approach $q-1/2$, so both surface branches are stable. The slab equilibrium is therefore linearly unstable overall, because it admits the long-wavelength even modes despite the stability of its odd sector.