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Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.

Work in the frame in which the distant fluid is stationary, and take the pressure at infinity as zero. In the Unscaled Papkovich–Neuber representation, choose the harmonic potentialsBoth are harmonic away from the sphere's centre: $1/r$ is harmonic there and $\mathbf{V}\cdot \mathbf{x}/r^3=-\mathbf{V}\cdot \nabla (1/r)$. Substitution into $\mathbf{u}=\nabla (\mathbf{x}\cdot \mathbf{\Phi }+\chi )-2\mathbf{\Phi }$, $p=2\mu \nabla \cdot \mathbf{\Phi }$, gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. At $r=a$ the coefficients of $\mathbf{I}$ sum to one and those of $\mathbf{nn}$ sum to zero, so the no-slip boundary condition is satisfied. The field decays at infinity. The Papkovich–Neuber representation enforces the Stokes equation and incompressibility, and Uniqueness of Stokes flow identifies the physical solution.

Differentiate radially at fixed direction $\mathbf{n}$:On the surface, multiplying the specified Newtonian fluid stress tensor by the outward normal from the solid givesThus the surface traction in translating-sphere Stokes flow is uniform and the fluid's force on the sphere isThis is the Stokes drag law; an external force of the opposite sign maintains the motion.

First estimate transverse relaxation without assuming a thin cross-section. With both depth and width of order $h_0$, a transverse surface-height variation of order $h_0$ gives a pressure gradient of order $\rho g$. The Stokes equation then gives transverse horizontal and vertical velocities of order $g{h}_0^2/\nu$. Moving the surface through distance $h_0$ takesFor longitudinal variations over $L\gg h_0$, the hydrostatic pressure gradient is of order $\rho g{h}_0/L$, so the longitudinal velocity is $u\sim g{h}_0^3/(\nu L)$. Incompressibility gives a vertical velocity $w\sim h_{0}u/L$, henceThus cross-channel variations relax rapidly compared with the longitudinal evolution. These are viscous estimates, requiring inertial relaxation to be negligible; the slow-flow regime in particular excludes a large transverse inertial response. After that initial relaxation the free surface is nearly horizontal across the channel, so its height is $h_0(x,t)$ and its local vertical depth is $h_0-|y|\cot \alpha$.

Retain the full cross-sectional geometry for a fixed $0<\alpha <\pi /2$; no small-angle approximation is used. With $|h_{0x}|\ll 1$, the leading hydrostatic pressure is $p=p_a+\rho g(h_0-z)$. Axial viscous derivatives are small relative to cross-sectional ones. The axial Stokes equation is thereforeIntroduce $Y=y/h_0$, $Z=z/h_0$, and the triangular domainLet the dimensionless cross-sectional Poisson equation beThe two lower sloping sides impose the no-slip boundary condition; the upper horizontal side imposes the stress-free boundary condition. Symmetry would additionally give ${\varphi }_Y=0$ at $Y=0$ if only half the domain were used. The full domain needs no extra midline boundary condition. Define $Q(\alpha )={\int }_{{\mathcal{D}}_{\alpha }}\varphi dYdZ$. ThenThe cross-sectional area is $A=h_0^2\tan \alpha$. Conservation of mass, $A_t+q_x=0$, gives the V-shaped channel gravity current equationIn particular, neither replacing this triangular Poisson problem by a vertically parabolic film nor taking an extreme angle is necessary.

Put $C=gQ(\alpha )/(\nu \tan \alpha )$. For a symmetric fixed-volume similarity solution, write $h_0=t^{-b}f(x{t}^{-d})$. Volume conservation gives $d=2b$, and balancing $(h_0^2{)}_t=C(h_0^4{h}_{0x}{)}_x$ gives $2b+1=5b+2d$. Hence $b=1/7$, $d=2/7$. Integrating the similarity equation once, with zero flux at the centre, yieldsIntegrating again and imposing $h_0=0$ at $x=\pm X(t)$ givesThe constant-volume similarity in a V-shaped channel has a compact support. If $H(t)=h_0(0,t)$, its volume isEliminating $H$ therefore givesThis describes the large-time spreading from a localized release; the point-release profile is singular at $t=0$, and the idealized front has a narrow region where its divergent slope violates the long-wave approximation. Neither feature changes the leading bulk similarity law.

At a fixed nonzero $x^{\ast }$, the depth is initially zero until $X(t)=|x^{\ast }|$. After arrival it rises and then falls. Because $X^2\propto t^{4/7}$, differentiation of $h_0(x^{\ast },t{)}^3=3(X^2-x^{\ast 2})/(7Ct)$ givesThus its unique maximum occurs at $X^2=7x^{\ast 2}/3$; thereafter the depth tends to zero as $t^{-1/7}$. If $t_a$ is the arrival time, $t_{\max }/t_a=(7/3{)}^{7/4}$. At the singular release point $x^{\ast }=0$, the point-source similarity profile instead decreases from the start; a finite initial release regularizes its early behaviour.

The right-hand sketch shows the triangular bulk at half its previous maximum depth, with thin draining films on the previously wetted portions of both walls. These residual wall films in a receding gravity current are not captured by a perfectly dry-wall triangular section. They drain rapidly compared with bulk longitudinal spreading. For an order-one angle, a wall film of thickness $b$ and wetted length $\ell$ has speed $O(g{b}^2/\nu )$ by falling film flow; its drainage time gives $b\sim (\nu \ell /(gt){)}^{1/2}$. At a fixed nonzero station the previous maximum wetted height is finite, so film area decays as $t^{-1/2}$, faster than the bulk area $h_0^2\sim t^{-2/7}$. In stations whose arrival and peak times are of order the observation time, the typical residual film fraction is $O(\sqrt{{\tau }_y/t})\to 0$. Therefore the films modify the detailed receding cross-section but do not hold a leading-order fraction of the volume or invalidate the large-time nose law. Microscopic contact-line physics is beyond this ideal viscous model.

Let the longitudinal variation scale be $\ell \gg h$. The stress-free boundary condition gives $\mu (u_z+w_x)=0$ at each face. Incompressibility implies $w=O(hu/\ell )$, so $w_x$ is smaller than a putative $u_z=O(u/h)$ by $O(h^2/{\ell }^2)$. The leading Stokes equation and the two shear-free conditions therefore make $u$ independent of $z$; its thickness-dependent correction is smaller-order. Symmetry gives $w=0$ at $z=0$, and incompressibility integrates toVertical momentum makes the leading normal stress independent of $z$. With the normal directed from liquid into gas, the upper-face curvature is $-h_{xx}/2$ to leading order. The Young–Laplace equation givesThe Newtonian fluid stress tensor also gives ${\sigma }_{zz}=-p+2\mu w_z=-p-2\mu u_x$. Hence $p=p_a-\gamma h_{xx}/2-2\mu u_x$ andThe factor four is the planar extension factor: both the longitudinal viscous stress and the pressure adjustment from transverse contraction contribute.

For force balance on the illustrated slice, the vertical cuts give $(h{\sigma }_{xx}{)}_x\delta x$. Pressure on the sloping broad faces gives $p_a{h}_x\delta x$ horizontally. Each of the two interfaces exerts a tangential surface tension pull at each cut, with horizontal component $\gamma \cos \theta$, where $\tan \theta =h_x/2$. Thus the horizontal force balance is the derivative of the effective tensionto the retained small-slope order. The ambient-pressure contributions cancel in this excess tension. Differentiating cancels the two $h_x{h}_{xx}$ terms, leaving the planar viscous-sheet stretching equationsThe upper-face kinematic condition is $w(x,h/2,t)=(h_t+u{h}_x)/2$. Combining it with $w=-z{u}_x$ gives the second equation,which is conservation of mass per unit transverse width.

For the two bubbles, let $p_g$ be the common gas pressure and $p_e$ the surrounding liquid pressure away from the sheet. A cylindrical bubble has one nonzero principal curvature, so the Young–Laplace equation gives $p_g-p_e\simeq \gamma /a$. In the nearly flat sheet the normal stress is $-p_g$, giving liquid pressure $p_{\mathrm{film}}=p_g-2\mu u_x$. The extensional correction is small compared with $\gamma /a$ because the transition scale will satisfy $\sqrt{a{h}_0}/L\ll 1$. Thus $p_{\mathrm{film}}\simeq p_g>p_e$, and this capillary pressure difference drives liquid out through both ends.

For a spatially uniform flat sheet, $h_x=h_{xxx}=0$, so the longitudinal equation gives $u_{xx}=0$. Reflection symmetry at $x=0$ selectsHere $U$ is the outward speed at the positive end. The mass equation then gives $h_t=-Uh/L$, independent of $x$, so initially uniform thickness remains uniform. Writing it as $h_0(t)$,The end transitions determine $U$, as derived in the following parts.

Take $p$ as excess pressure relative to the fluid far from the gap, and let ${\dot{h}}_0<0$ for approach. The lower surface of the sphere is $h_0+a-\sqrt{a^2-r^2}$ above the plane. In the dominant region $r=O(\sqrt{a{h}_0})\ll a$, this is the parabolic lubrication gapIts slope is $O(\sqrt{h_0/a})\ll 1$, justifying lubrication theory. Both rigid surfaces have zero radial velocity, so the no-slip boundary condition and radial Stokes equation give the radial volume flux per unit circumferenceAxisymmetric conservation of mass gives $h_t+r^{-1}(rq{)}_r=0$. Symmetry at $r=0$ therefore yields $q=-r{\dot{h}}_0/2$, hence $p_r=6\mu {\dot{h}}_{0}r/h^3$. Matching to $p\to 0$ outside the pressure-bearing region givesExtending the parabolic profile to infinity in this integral is a leading-order matching device; the main pressure and force come from $r=O(\sqrt{a{h}_0})$.

The radial scale $\sqrt{a\hat{h}}$ balances the two contributions to the gap. The time scale $\hat{h}/\hat{v}$ is the time for a typical gap change, and the lubrication pressure scale $\mu a\hat{v}/{\hat{h}}^2$ follows either from the formula above or from $p\sim \mu \hat{v}{\ell }^2/{\hat{h}}^3$ with ${\ell }^2=a\hat{h}$. The elastic surface's upward displacement is $-\eta p$, so the actual gap is increased by $+\eta p$. Thus the dimensionless gap in weakly compliant sphere-plane squeeze flow isA dot now means differentiation with respect to $T$. The dimensionless Reynolds lubrication equation isEven without fluid inertia, the elastic surface has a pressure-dependent normal velocity, so ${\mathcal{H}}_T$ includes $\epsilon P_T$; omitting it would wrongly discard the acceleration term below.

Write $d=H_0+R^2/2$ and abbreviate $H=H_0$, $v={\dot{H}}_0$, $b={\ddot{H}}_0$ during the calculation. The rigid term isAt first order the Reynolds equation givesBecause $RdR=dd$, one integration givesSet $\xi =R^2/(2H)$, so $d=H(1+\xi )$. Dividing and changing variables yieldsThe first three terms are derivatives of $I_3,I_4,I_5$, respectively. Integrating from infinity to enforce $P_1\to 0$ givesThe rational expansion of $I_n$ follows by differentiating $\log [\xi /(1+\xi )]+{\sum }_{j=1}^{n-1}[j(1+\xi {)}^j{]}^{-1}$, obtaining $1/[\xi (1+\xi {)}^n]$, and fixing its value to zero at infinity.

Near $\xi =0$ every $I_n$ has the same singular term $\log \xi$. Its coefficient must vanish to keep the pressure regular, soEquivalently, the integrated radial flux above must vanish at $R=0$; otherwise $P_{1R}$ would have a $1/R$ singularity. Using $I_4=I_3+[3(1+\xi {)}^3{]}^{-1}$ and $I_5=I_4+[4(1+\xi {)}^4{]}^{-1}$ cancels all logarithms and gives the simplified regular expression

The upward hydrodynamic force has scale $\mu a^2\hat{v}/\hat{h}$. Its leading contribution is pressure integrated over horizontal projected area; vertical shear forces are smaller in the lubrication limit. Thus $F=2\pi {\int }_0^{\infty }(P_0+\epsilon P_1+\cdots )RdR$. Use $RdR=H_{0}d\xi$ and ${\int }_0^{\infty }(1+\xi {)}^{-n}d\xi =1/(n-1)$. The two contributions areTherefore

For weight-controlled sedimentation, let the fixed effective load be $F=6\pi A$, $A>0$, and neglect particle inertia as well as fluid inertia. At leading order ${\dot{H}}_0=-A{H}_0$ and ${\ddot{H}}_0=A^2{H}_0$. The first-order bracket evaluated on this trajectory is $-(7/5)A^2/H_0^2$. Solving the force balance perturbatively therefore givesHence weak compliance initially increases the sedimentation speed in the regime where the perturbation is valid. The squeeze pressure indents the coating away from the sphere, widens the flow passage and lowers resistance. This is a load-controlled conclusion on the slow sedimenting branch, not a claim about an arbitrary externally imposed acceleration at startup.

The load-controlled breakdown of weak squeeze-flow compliance occurs when the relative correction $O(\epsilon A/H_0^2)$ becomes order one:The central rigid pressure has size $P_0(0)\sim A/H_0$, so the relative elastic gap change is also $\epsilon P_0(0)/H_0\sim \epsilon A/H_0^2$. Thus breakdown means that elastic deflection is comparable with the nominal undeformed gap. The regular expansion about a rigid plane then ceases to be uniform and the coupled elastic-lubrication problem must be retained. It does not by itself imply contact, fluid inertia or failure of the thin-gap geometry.

Write $ϵ=1-{\rho }_s/\rho >0$ and $\Delta v=1/{\rho }_s-1/\rho =ϵ/{\rho }_s$. When the ice sphere grows by volume $4\pi a^2\dot{a}dt$, it consumes liquid volume $({\rho }_s/\rho )4\pi a^2\dot{a}dt$. The difference is expelled from the fixed cavity. Thus mass conservation fixes the outward Darcy flux in the surrounding porous medium:The Darcy law $u_r=-(\Pi /\mu ){\partial }_{r}p$ and $p(\infty )=p_m$ implyMatching at the cavity wall, whose internal pressure is uniform, givesThe solid and liquid pressures are equal when the Gibbs--Thomson relation is neglected; nevertheless the coexistence temperature changes with their common pressure because their specific volumes differ. Along coexistence, equality of chemical potentials givesHence the common-pressure melting-point slope is $d{T}_m/dp=-T_m\Delta v/L$. To first order in the pressure change,The $T_m$ multiplying this coefficient is absolute temperature, approximately $273.15\mathrm{K}$ for water; a temperature difference may equally be measured in kelvin or degrees Celsius. Substituting the numerical value $0$ from the Celsius scale into a thermodynamic denominator would be incorrect.

In the quasi-stationary thermal conduction approximation, neglect advective heat transport and use the stated effective conductivity $k$ outside the sphere. The radial Laplace equation gives the bounded solid temperature and the liquid/porous-region temperature:This treats the exterior as a uniform conducting medium, as required by the printed reduction; distinct cavity and porous-medium conductivities would give an additional thermal resistance. There is no solid-side heat flux in the bounded quasi-stationary solution. The Stefan condition therefore becomesEliminate $T_i$ and $p$ to obtainFor $x=a/R$, $t=t_0\tau$ and $t_0=R^2{\rho }_{s}L/(k\Delta T)$, this isThe pressure follows either from the Darcy flux or the Stefan condition:In particular $T_i-T_{\infty }=\Delta TK/(x+K)$: rising pressure progressively reduces the available undercooling. This is spherical freezing inside a porous cavity, whose fixed hydraulic outlet radius $R$ distinguishes it from freezing directly through a porous matrix.

For the idealized zero-radius start, integration givesA finite seed replaces the right side by its difference from the initial value. The reduction ends when the sphere fills the cavity.

For $K\gg 1$, hydraulic drainage is easy and the pressure feedback is small throughout the cavity. The leading expressions areThis is ordinary conduction-controlled spherical solidification, with a completion time asymptotic to $t_0/2$.

For $K\ll 1$, distinguish the initial crossover from the pressure-dominated regime. At $x\ll K$, equivalently $\tau \ll K^2$, the same square-root growth applies and $p-p_m\simeq p^{\ast }x/K$. At $K\ll x\le 1$, equivalently $K^2\ll \tau \lesssim 1/K$,The interface approaches $T_{\infty }$, the pressure approaches the finite coexistence value $p_m+p^{\ast }$, and growth becomes hydraulically limited. The leading completion time is $t_0/(3K)$. The cube-root approximation is not valid at the instant of nucleation, where the initial square-root regime must be retained.

For a comoving fluid volume, $V\propto a^3$ and $E=\rho c^{2}V$. The first-law relation $dE=-PdV$ gives $c^2(Vd\rho +\rho dV)=-PdV$. Since $\dot{V}/V=3\dot{a}/a$, the cosmological perfect-fluid continuity equation isFor the Newtonian energy argument, assume negligible pressure, spherical homogeneity, Newtonian gravity and constant enclosed mass. A test particle at physical radius $r=a\chi$ has constant specific energyDivide by $a^2{\chi }^2/2$ and set $k=-2ϵ/(c^2{\chi }^2)$ to obtain the Friedmann equationHomogeneity makes $k$ independent of the chosen comoving shell. In the relativistic interpretation it specifies the spatial-curvature sign: positive, zero or negative $k$ gives closed spherical, flat or open hyperbolic spatial geometry, with a conventional rescaling allowing $k=1,0,-1$. Newtonian mechanics itself derives the energy constant, not the spatial metric. When pressure is appreciable, the general relativistic equations still have the displayed Friedmann form, but the constant-mass Newtonian test-particle derivation cannot be applied without that qualification.

Let ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_P$. The Maxwell relations giveSetting $dS=0$ gives $(\partial T/\partial P{)}_S=TV{\alpha }_V/C_P$, hence ${\kappa }_S={\kappa }_T-TV{\alpha }_V^2/C_P$. Also $dS=(C_V/T)dT+(\partial P/\partial T{)}_{V}dV$, and the volume differential gives $(\partial P/\partial T{)}_V={\alpha }_V/{\kappa }_T$. Evaluating along constant pressure therefore yields $C_P-C_V=TV{\alpha }_V^2/{\kappa }_T$. Eliminate ${\alpha }_V$ to obtainThese derivatives are taken at fixed particle number, in a thermodynamically regular stable phase.