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Define $g(\mathbf{x},h,t)dh$ as the fraction of horizontal ocean area covered by sea ice of thickness in $[h,h+dh]$. Allow a Dirac delta mass at $h=0$ for open water, so the whole-area distribution has $\int gdh=1$; its first moment $H=\int hgdh$ is ice volume per ocean area, rather than mean thickness conditional on ice being present. Let $f=dh/dt$ be thermodynamic growth, $\mathbf{v}$ the horizontal ice velocity and $\psi$ the mechanical redistribution of sea ice.

Conservation in a horizontal cell and a thickness interval gives horizontal flux $\mathbf{v}g$ and thickness-space flux $fg$. For $h_1<h_2$, their contribution is $-\nabla \cdot {\int }_{h_1}^{h_2}\mathbf{v}gdh-[fg{]}_{h_1}^{h_2}$; adding ${\int }_{h_1}^{h_2}\psi dh$ and shrinking the interval yieldsThis is the sea-ice thickness distribution equation. Its distributional formulation must include creation and removal of open water when characteristics reach or leave $h=0$. On a positive-thickness smooth portion it is equivalentlyIf a nonconservative form uses a redistribution function $\tilde{\psi }$, then $\tilde{\psi }=\psi -g\nabla \cdot \mathbf{v}$; the two definitions must not be interchanged.

With the open-water boundary flux accounted for consistently, whole-area normalization requires $\int \psi dh=\nabla \cdot \mathbf{v}$. Mechanical rearrangement preserves ice volume locally, so $\int h\psi dh=0$. The volume moment consequently obeyswhere formation of new sea ice from the open-water category is included in the thermal term. Compression raises local volume per area through convergence, even though ridging does not create ice mass.

The three processes have distinct effects on the curve. Pure advection imports the distributions found upstream, so a fixed location may acquire a different mode or thick tail. Along an incompressible material trajectory, advection alone does not alter its distribution. Convergence and divergence also change area fractions through the horizontal flux, with the redistribution and open-water terms ensuring normalization.

Thermodynamic growth moves probability along thickness characteristics $\dot{h}=f(h,t)$, with $g(h,t)dh=g(h_0,0)d{h}_0$ when mechanics and horizontal divergence vanish. For example, winter conductive growth $f=\kappa /h$ gives $h^2=h_0^2+2\kappa t$ andThin ice grows faster, so the mode moves to larger thickness and neighboring thin-ice characteristics converge, sharpening the level-ice peak. Summer ablation moves classes left and creates open water when ice disappears; if melt rate depends on thickness, the shape changes as well.

Mechanical redistribution of sea ice removes area from thin or level ice and populates thick rafted or classes containing sea-ice pressure ridges, broadening the upper tail. It can also create sea-ice leads. A simple volume-conserving example, stacking a fraction $r$ of an initial distribution into double-thickness ice, isThe stacked component has area $r/2$ and unchanged volume $r\int h{g}_0(h)dh$; the freed area $r/2$ becomes open water. Thus mechanical thickening is not merely a translation of the entire curve.

The figure uses illustrative distributions, not measured data. The advection panel compares local ice with incoming thicker ice; the growth panel applies the thickness-characteristic formula; the mechanical panel uses $r=0.4$ and shows its open-water mass separately from the continuous density.

For the thick tail, take the common-shape sea-ice pressure ridge idealization to mean triangular profile sections with a common effective side slope $s=\tan \delta$, while peak thicknesses vary. Let $n(H)dH$ count sea-ice pressure ridge peaks per track length with maximum thickness in $[H,H+dH]$. Above a level-ice baseline $h_0$, a sea-ice pressure ridge with $H\ge h$ contributes two flank lengths $dh/s$ at thickness $h$; it contributes nothing at $h>H$. For a representative transect, summing these lengths gives the triangular ridge occupation identitySuppose $n(H)=B{e}^{-\lambda H}$ for $H\ge h_0$. ThenThus an exponential ridge-draft model, expressed in the same vertical coordinate as the profile, generates an exponential occupation tail with the same decay parameter. The peak-count distribution and the area-weighted thickness distribution are different distributions, so their amplitudes carry essential information.

If $\mu ={\int }_{h_0}^{\infty }n(H)dH$ is ridge line density and $p={\int }_{h_0}^{\infty }{g}_{\mathrm{tail}}(h)dh$ is the tail's area fraction, the same result becomesThe decay rate alone does not fix slope: sea-ice pressure ridge frequency or tail amplitude is also needed. If side slopes vary independently of peak height, replace $1/s$ by $⟨1/s⟩$, giving an effective harmonic-mean slope. Measured along-track slopes differ from slopes normal to a ridge if the transect is oblique. Overlap, irregular shape and porosity limit this simple geometric model. Literally identical complete ridge cross-sections would have one height, so “common shape” here necessarily permits different ridge sizes.

Sea-ice pressure ridge spacings are positive and broadly distributed. A spatial Poisson process gives an exponential distribution of gaps, a useful simple approximation. Observed sea-ice pressure ridge spacings, however, often fit a shifted lognormal distribution better, with few very small spacings because of ridge geometry, identification rules and finite beam width. This differs from the successful exponential model for peak drafts. Observed ridge-spacing and draft distributions establish this distinction.

Sea-ice lead-width distributions contain very many narrow openings and a broad tail. A power law often describes widths over a resolved range better than one negative exponential; its exponents and apparent cutoffs depend on the lead definition, resolution and count-versus-area weighting. Sea-ice lead spacings may be approximately exponential over an intermediate range, but clustering produces excess close pairs and large gaps, so there is no universal independent-spacing law. Both behaviors were found in Greenland Sea and Eurasian Basin lead statistics. Advection imports thickness classes, thermodynamics transports them in $h$, and mechanics exchanges thin-ice area for ridged ice and leads; common-slope exponential ridge populations reproduce the exponential thick tail.

For the proposed exponential profile, put $\ell =u\tau /\varphi$. Then $h_x=h/\ell$ and $(h{h}_x{)}_x=2h^2/{\ell }^2$. Thus the exact ratio of the discarded spreading term to the retained advective term isThis is a quantitative validity condition for the advection limit of an inclined porous current. For a specified small tolerance $\eta >0$, at a fixed $x$ it is enough that the inlet characteristic has arrived andIf the logarithm is negative, arrival itself already gives this tolerance. A large-distance estimate comparing the overall current length $ct$ with a height-induced pressure length $h\cot \theta$ is $t\gg \varphi h\cot \theta /u$, but the local gradient criterion above is the sharper test for this particular injection history.

At the advective nose, $x=ut/\varphi$, the exponential factor is one. Therefore large time alone is insufficient for uniform validity: the outer profile also needs $2\varphi h_0\cot \theta /(u\tau )\ll 1$, and its abrupt nose needs an inner spreading layer. This limitation follows by direct substitution, not from a missing constant in the solution.

Indeed, with $\xi =x-ut/\varphi$, the full equation becomes $h_t=D(h{h}_{\xi }{)}_{\xi }$, $D=u\cot \theta /\varphi$. Once a finite-volume injection pulse is effectively over, nonlinear diffusion continues to broaden it. Dimensional balances give width ${\ell }_d\sim (DMt{)}^{1/3}$ and height $M/{\ell }_d$, where $M=\int hd\xi$ is the conserved area. A fixed exponential translating profile cannot be the uniform infinite-time limit of that equation. The printed formula is consequently an outer advection approximation with the displayed smallness condition.

Let $X(t)$ be the advancing nose into previously unoccupied pores. In the advective limit the interior flux is $uh$, while the exterior current thickness and flux are zero. Applying the Rankine-Hugoniot condition to the storage $\varphi h$ givesThis is the leading-edge kinematic boundary condition. The outer depth immediately behind this nose is $h(X(t{)}^{-},t)=h_0$, whereas ahead it is zero. A first-order advection model allows this jump; imposing zero depth on its interior trace would contradict its characteristic solution.

For the full inclined porous gravity current, the nose is instead resolved by a continuous spreading layer with $h(X,t)=0$. Taking the flux divided by depth at that interface gives $\dot{X}=(u/\varphi )(1-\cot \theta h_x{|}_{X^{-}})$, when this one-sided derivative exists. Its leading outer value is $u/\varphi$ only when the nose correction is negligible on the scale of the approximation. The zero-depth condition belongs to that resolved layer, not to the discontinuous outer profile itself.

Choose $x$ upslope and $z$ normally downwards from the impermeable roof, so the buoyant layer occupies $0<z<h(x,t)$. Use the hydrostatic approximation and neglect the ambient return-flow pressure correction of relative order $h/H$. Pressure continuity at the interface gives the pressure inside the buoyant layer asThe along-slope Darcy velocity is thereforeDefine $u=kg\Delta \rho \sin \theta /\mu$. Integrating over depth gives the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. Local fluid-volume conservation, including porosity, is $\varphi h_t+F_x=0$. Hence the inclined porous gravity current satisfiesThe first term is upslope advection driven by the mass density difference; the second is nonlinear hydrostatic spreading. The Darcy velocity $u$ and the corresponding pore velocity $u/\varphi$ must be distinguished.

Here $k$ is turbulent kinetic energy per unit mass, $k=\frac{1}{2}\overline{u^{\prime 2}+v^{\prime 2}+w^{\prime 2}}$, rather than the permeability of a porous medium used in the porous-flow questions. The K-epsilon turbulence model represents the high-Reynolds-number transport of this energy and its turbulent kinetic energy dissipation rate $ϵ$.

In the $k$ equation, $d[({\nu }_T/{\sigma }_k)k_y]/dy$ is the divergence of modeled turbulent diffusion, $\mathcal{P}$ is mean-shear production, and $-ϵ$ is molecular conversion of fluctuation energy into heat. Statistical stationarity and fully developed flow have removed storage and streamwise advection from this reduction.

In the $ϵ$ equation, $d[({\nu }_T/{\sigma }_{ϵ}){ϵ}_y]/dy$ transports dissipation intensity; $C_{ϵ1}\mathcal{P}ϵ/k$ models its generation associated with energy production; and $-C_{ϵ2}{ϵ}^2/k$ models its destruction. The time scale $k/ϵ$ supplies the source-rate dimension. The $ϵ$ equation is a closure, not another energy-conservation equation. Finally, ${\nu }_T=C_{\mu }{k}^2/ϵ$ has dynamic viscosity dimensions and parametrizes Reynolds stress transport; ${\sigma }_k,{\sigma }_{ϵ}$ set the relative diffusion strengths, and the remaining constants calibrate the closure.

The free surface is a material surface, so its kinematic boundary condition isWith constant atmospheric pressure $p_a$ and no surface tension, the dynamic boundary condition for an inviscid interface follows from Unsteady Bernoulli equation:Write $\varphi =Ux+\phi$, where $\phi$ and $\eta$ are small. Choose the time-dependent potential gauge so the undisturbed right-hand side is $U^2/2$. Retaining first-order terms, evaluated at $y=0$, givesThe incompressibility condition also gives Laplace's equation for $\phi$.

Set $\Omega =\omega -kU$. Substitution of the specified normal mode into the two linearized boundary conditions yields $-i\Omega a=ikb$ and $\Omega b+ga=0$. Eliminating $b$ gives the dispersion relationThe decay factor $e^{ky}$ in a fluid extending to $y=-\infty$ uses $k>0$. With signed wavenumber the decaying factor is $e^{|k|y}$ and the right-hand side is $g|k|$. The frequency shift is the advection of a deep-water gravity wave by the uniform current.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

The imposed pressure gradient and Darcy's law give a linear velocity profile. Use $U$ as the mean pore velocity in the contaminant advection-diffusion equation; thenThe last conditions express zero transverse solute flux under the standard closed-layer interpretation. Transverse exchange through the layer boundaries would change this cell problem and requires additional boundary data. If the prescribed mean speed is a Darcy velocity, its corresponding pore velocity is that speed divided by porosity.

The given parameter is the ratio of transverse mixing time to travel time:Thus transverse dispersion mixes the contaminant across the layer many times during its passage along the long layer. Different fluid speeds still matter: repeated exchange between slow and fast portions produces Taylor dispersion, rather than transport solely at each parcel's original speed. The averaged description applies after $t\gg H^2/D_T$ and on longitudinal scales large compared with the distance travelled during that mixing time.

Write $C=\overline{c}$ and, to leading correction order, $c=C+\chi (y)C_x+\cdots$, with $\overline{\chi }=0$. Taking $C_t\simeq -U{C}_x$ in the local equation gives the cell problemWith $\eta =y/H$, its solution isAveraging the local advection-diffusion equation now yields $C_t+U{C}_x=[D_L-\overline{(u-U)\chi }]C_{xx}$. An integration by parts in the cell problem gives $-\overline{(u-U)\chi }=D_T\overline{({\chi }^{\prime }{)}^2}$. Since ${\int }_0^1({\eta }^2-\eta {)}^{2}d\eta =1/30$, the effective equation isThis is Taylor dispersion in a linear porous-layer velocity profile. For a localized pulse away from the inlet and outlet, its mean position advances at $U$ and its longitudinal variance grows as $2D_{\mathrm{eff}}t$.

For the inlet step, use the semi-infinite inlet approximation $x>0$, $C(0,t)=1$, $C(x,0)=0$ and $C\to 0$ as $x\to \infty$ at fixed $t$. The resulting constant-concentration inlet solution isOne derivation is to Laplace transform in time: $\tilde{C}(x,p)=p^{-1}\exp [(U-\sqrt{U^2+4D_{\mathrm{eff}}p})x/(2D_{\mathrm{eff}})]$. Inverting gives the two complementary error functions. Their sum is one at $x=0$; for fixed $x>0$ both terms vanish as $t\downarrow 0$, and substitution verifies the averaged equation. Retaining only the first term gives the familiar moving error-function front far from the inlet when longitudinal advection dominates diffusion, but does not satisfy the inlet condition exactly.

For a literal finite layer $0<x<L$, a downstream boundary condition is also needed once the outlet influences the solution. The PDF's step-inlet request specifies only the initially clean region $x>0$, so the formula above is the semi-infinite/long-layer solution, not a claim of a unique finite-interval solution with unspecified outlet data.

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.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

For a normal mode, ${\partial }_t\mapsto -i\omega$, ${\partial }_x\mapsto ik$, and ${\partial }_x^2\mapsto -k^2$. Substitution into the linear complex Ginzburg-Landau equation givesTherefore the dispersion relation isFor real $k$, its temporal growth rate is ${\omega }_i=\mu -k^2$, with unique maximum $\mu$ at $k=0$. Its real angular frequency is $Uk+c_d{k}^2$, displaying the separate advection and wave dispersion terms.

In the linear complex Ginzburg-Landau equation, take $U,c_d,\mu$ real. $U$ gives advection, carrying the envelope downstream at speed $U$. $c_d$ gives wave dispersion: the imaginary part of the second-derivative coefficient changes wave phase without directly dissipating a real-wavenumber Fourier mode. $\mu$ gives local linear production or damping: a spatially uniform perturbation grows for $\mu >0$ and decays for $\mu <0$.

The real part of the diffusion coefficient is the scaled value one. It smooths the envelope and damps a real-wavenumber Fourier mode at rate $k^2$. Thus the linear complex Ginzburg-Landau equation models competing advection, smoothing through a diffusion equation, wave dispersion and local growth, rather than nonlinear saturation.

Choose the positive $x$ direction along the background flow, so $U>0$; reverse $x$ if necessary. Let $z$ measure depth below the impermeable roof, and set $\Delta \rho ={\rho }_w-{\rho }_c>0$. The deep ambient water has $p_w=p_0(x)+{\rho }_{w}gz$. Continuity of pressure at $z=h$ and hydrostatic pressure inside the buoyant CO2 giveHere $k$ is permeability of a porous medium and $\mu$ is dynamic viscosity. Since $h/H\ll 1$, the ambient return-flow correction to the imposed pressure gradient is higher order. The small aspect ratio supplies the hydrostatic pressure approximation; capillary trapping, dissolution, compressibility and dispersion are omitted in this sharp-interface model.

The depth-integrated Darcy flux is $q=Uh-K_{b}h{h}_x$. The actual stored fluid volume is $\varphi h$ per unit horizontal area, where $\varphi$ is the aquifer porosity. Thus the porous gravity current with background flow satisfieswith $u=U/\varphi$, $a=K_b/\varphi$ and $J=Q/\varphi$. Both $Q$ and the later $V$ denote actual injected volume per unit transverse width. Away from injection, the equation is an advected porous medium equation. At each finite front impose $h=0$ and vanishing outgoing volume flux per unit width. At the injection point $h$ is continuous and $q(0^{+})-q(0^{-})=Q$. For an initially empty aquifer, $\varphi \int hdx=Qt$. The advective speed is $U/\varphi$, not $U$.

For a constant-flux porous gravity current, write characteristic depth $H_c$ and extent $L_c$ temporarily. At early times volume conservation gives $H_c{L}_c\sim Jt$; gravity spreading gives $L_c^2/t\sim a{H}_c$. HenceBackground advection becomes comparable when $L_c\sim ut$, so the transition scales areFor $t\ll t_c$, the leading similarity solution is symmetric:For $0<\eta <{\eta }_N$, its model ordinary differential equation and boundary conditions areChoose the nonnegative finite-front branch, reflect evenly in $x$, and require $F{F}^{\prime }\to 0$ at the front. The factor $1/2$ divides the injection equally between the two halves. Integration of the ordinary differential equation gives $2{\int }_0^{{\eta }_N}Fd\eta =1$; the limiting front slope is $F^{\prime }({\eta }_N^{-})=-2{\eta }_N/3$. This boundary-value problem determines the initially symmetric shape without assuming an unjustified elementary formula. Independent numerical integration gives ${\eta }_N\simeq 1.17619$ in this normalization.

For $t\gg t_c$, upstream storage approaches a finite steady value. With no upstream leakage, $q=0$ for $x<0$, giving $h_x=U/K_b$ wherever $h>0$. Downstream the uniform interior carries $q=Q$, soThe upstream volume flux vanishes even though local Darcy velocity contributions from background flow and buoyancy cancel. The upstream stored volume is $\varphi h_{\mathrm{int}}|x_{-}|/2$, a finite constant. The downstream plateau therefore extends a distance $ut+O({\ell }_c)$, and its front advances at $u$ to leading order.

The diffusive nose of an advected porous gravity current has widthIn a frame translating at $u$, balance the time derivative against the nonlinear gravity diffusion. With $\eta =(x-x_c(t))/\sqrt{a{h}_{\mathrm{int}}t}$ and $x_c(t)=ut+O({\ell }_c)$, the leading transition is $h=h_{\mathrm{int}}F(\eta )$, withIts finite front has $F\sim ({\eta }_N/2)({\eta }_N-\eta )$ as $\eta \uparrow {\eta }_N$. Independent numerical integration of this boundary-value problem gives ${\eta }_N\simeq 1.23849$, so a more accurate leading nose position is $x_N\simeq x_c+1.23849\sqrt{a{h}_{\mathrm{int}}t}$ and its local shape is $F\simeq 0.619247(1.23849-\eta )$. The entire transition has no elementary closed form. A useful explicit, approximate shape is the volume-preserving linear rampThis gives $x_N\simeq x_c+\sqrt{a{h}_{\mathrm{int}}t}$. Its ramp width is $2\sqrt{a{h}_{\mathrm{int}}t}$, and its edge obeys the exact kinematic boundary condition ${\dot{x}}_N=u-a{h}_x(x_N^{-},t)$ to the displayed order. It also has the same area as a sharp step at $x_c$. This is an approximate ramp closure, not an exact solution of the nonlinear nose ordinary differential equation. A constant phase shift $x_c=ut-|x_{-}|/2$ balances the finite upstream storage within this ramp approximation; the full solution's subleading matching may alter constant offsets.