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After recharge stops, the finite interval sets the horizontal scale. Apply the separable draining profile for power-law diffusion to each limiting equationWith $\xi =x/L$ and virtual age $\tau =t+t_0$, the similarity solution isThe positive profile fixes $c_m={\lim }_{\xi \downarrow 0}{F}_m^m{F}_m^{\prime }={\int }_0^1{F}_{m}d\xi$. The discharge follows either from the boundary Darcy flux or integrated mass conservation:In the deep regime this givesIn the final shallow regime,The deep profile contains the thin outlet layer in a deep unconfined aquifer described above. The whole aquifer eventually becomes shallow, and its nonzero basal permeability controls the ultimate ${\tau }^{-2}$ discharge.

Estimating the mobility crossover by a typical height $2/\beta$ givesUsing the deep profile's height at the divide makes this ${\tau }_{\mathrm{cross}}\simeq \varphi \beta L^2{F}_2(1{)}^2/(4u_b)$. These are regime-dependent similarity approximations: the virtual origins are set by matching to the initial drainage and crossover, and a deep regime occurs only if the aquifer is initially sufficiently deep.

With both ice accumulation and ice ablation removed, the equation is $h_t+x^{-1}(xD{h}^3{)}_x=0$ and the conserved volume is $V_0=2\pi \cos \alpha M$, where $M={\int }_0^{x_N}xhdx$.

If the thickness and extent scales are $H(t)$ and $X(t)$, mass conservation gives $H{X}^2\sim M$, while the flow equation gives $H/t\sim D{H}^3/X$. Therefore $X\propto t^{1/5}$ and $H\propto t^{-2/5}$. Set $h={\tau }^{-2/5}f(\xi )$, $\xi =x{\tau }^{-1/5}$, with a possible virtual time origin $\tau =t+t_0$. The similarity solution satisfiesIntegration and regular zero total flux at the apex give $D\xi f^3={\xi }^{2}f/5$, hence the positive profile iswith dry bed beyond the front. Volume normalization givesso the volume-conserving conical ice-current similarity solution hasThe terminus has finite thickness and is a shock wave in the gravity-only kinematic wave equation. The Rankine-Hugoniot condition gives ${\dot{x}}_N=q_N/h_N=D{h}_N^2=x_N/(5\tau )$, exactly agreeing with the similarity extent. The characteristic speed behind the front is $3D{h}_N^2$, larger than its speed, while the dry-bed characteristic speed is zero, so the front is compressive and gives an entropy solution.

This solution describes the long-time spreading, rather than exactly matching the earlier steady profile at the instant snowfall stops. The initial transient can be described by the characteristic transformation for conical ice drainage: with starting point $s$, put $K(s)=s^{1/3}{h}_s(s)$; thenwhere characteristics remain smooth. Subsequent crossings are resolved by the same conservation and entropy conditions. Restoring the neglected local pressure gradient would smooth the idealized front.

Use the axisymmetric streamfunction conventionFor the similarity solution $\psi =\nu zg(\eta )$, $\eta =r/z$, regularity on the axis requires $g(0)=g^{\prime }(0)=0$, while matching to still ambient fluid requires $g^{\prime }(\eta )/\eta \to 0$ as $\eta \to \infty$. The given equation has the first integralThe axis conditions give $C=0$, and separation yieldsfor some $a>0$.

The axial velocity isNormalizing by the conserved momentum flux givesso $a=3M/(32{\nu }^2)$. The similarity solution for a laminar round jet is therefore

The specific angular momentum of a circular Keplerian orbit is $\sqrt{G{M}_{\ast }r}$. Therefore the disk angular momentum iswhereThe absence of an external or inner-boundary torque makes $J$, and hence $B$, constant.

From $\bar{\nu }=A{r}^{9/2}{\Sigma }^2$,The combination $A{B}^{2}t$ has dimension $L^{5/2}$. Dimensional analysis therefore gives

For the supplied similarity solution, put $\tau =t/t_0$ and $u=\sqrt{r/R(t)}$. Then $r=R{u}^2$ and $dr=2Rudu$. The total mass isSince $R^{1/2}=R_0^{1/2}{\tau }^{1/5}$,

Before the river disturbance reaches the divide, use the similarity solutionSubstitution gives the nonlinear boundary-value problemIt can be solved numerically. If $C={\lim }_{\eta \downarrow 0}H{H}^{\prime }>0$, then the riverward flux and near-river profile areThe negative sign means flow toward decreasing $x$; the discharge into the river is $-q(0,t)$.

The final state is displacement ventilation: fresh air of density ${\rho }_0$ forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height $h$. Steady volume conservation requires the plume volume flux there to equal the imposed ventilation flux,

Below the interface the ambient is uniform, so $B_0=F_0/{\rho }_0$ is constant. Put $J=M/{\rho }_0$. The triangular-profile wall line plume equations reduce toThe pure plume conditions $V,J\to 0$ at the radiator select the similarity solutionConsequently the displacement-ventilation interface height isThe upper-layer reduced gravity follows from its steady buoyancy balance as $g_u^{\prime }=B_0/A=F_0/({\rho }_{0}A)$. The two-layer solution applies when the calculated interface satisfies $0<h<H$.

Seek a similarity solution in the dimensionally completed formwhere the fixed scales absorb $A,D$, and $Q_0$. Its inlet volume flux scales asso the prescribed source requiresThe time derivative, horizontal lubrication theory term, and pressure-dependent Darcy drainage term scale respectively asEquating the three exponents givesSolving these linear equations together with the inlet condition yields the similarity exponents of a draining gravity currentEquivalently,and, up to constant dimensional scales,

For $m=4,n=2$, the equation is the one-dimensional porous medium equation $S_{\tau }=(S^3{)}_{xx}$. Use a similarity solution $S={\tau }^{-1/4}f(\xi )$ and $\xi =x{\tau }^{-1/4}$. ThenThe no-mass-flux condition at $x=0$ sets the integration constant to zero:With $f(0)=1$,ConsequentlyIts edge is $R=12r_0{\tau }^{1/2}\propto t^{1/2}$. Moreover,is time-independent. This agrees with part (b), since $2-m+2n=2$.

Choose the stream function conventionwhich satisfies continuity identically. Introduceso thatTake the similarity solutionThenandA direct differentiation shows that the terms containing $\eta f^{\prime }{f}^{\prime \prime }$ cancel:Also,while the delta scaling givesThe definitions of $D$ and $U_0$ implyAfter cancelling the common dimensional factor, the similarity equation for a two-dimensional laminar plume isThus

For $n,A>0$, the one-dimensional porous medium equation $h_t=A(h^n{h}_x{)}_x$ has a volume-preserving similarity solution $h=[c_n(R^2-x^2)/(At){]}_{+}^{1/n}$, where $c_n=n/[2(n+2)]$ and $R\propto t^{1/(n+2)}$. On a reflecting half-line with conserved area $M$, $R=[M(At/c_n{)}^{1/n}/I_n{]}^{n/(n+2)}$, where $I_n=\frac{1}{2}B(\frac{1}{2},1+1/n)$. For a whole symmetric release use half its total area as $M$. The Beta function fixes normalization, while finite-front zero volume flux fixes the edge condition.

For $N^2=K{z}^{\beta }$ and positive pure-power fluxes, the similarity solution exponents are $q=3+\beta /2$, $m=4+\beta$ and $f=4+3\beta /2$. Increasing volume and momentum and decreasing positive buoyancy flux require $-4<\beta <-8/3$. At fixed nonzero $K$ the endpoint coefficient limits are singular.

For the planar porous medium equation $n_t=[D_0/(3n_0^2)]\Delta (n^3)$, the mass-$Q$ point-source similarity solution is $n=n_0{\lambda }^{-2}\sqrt{[1-r^2/(r_0^2{\lambda }^2){]}_{+}}$, with $\lambda =(6D_{0}t/r_0^2{)}^{1/6}$ and $r_0^2=3Q/(2\pi n_0)$. Its front grows as $t^{1/6}$ and its central density falls as $t^{-1/3}$. The mass conservation integral uses the planar area element $2\pi rdr$. Although the density slope diverges at the front, its diffusion flux vanishes there, giving a compactly supported weak solution with initial measure $Q{\delta }^{(2)}$.

For pi-normalized top-hat fluxes, $b=Q/\sqrt{M}$. The pure similarity solution therefore has $b=(6\alpha /5)(z+z_0)$, so the radius grows linearly with slope $6\alpha /5$. The corresponding plume virtual origin is $-z_0$ and a finite source has $z_0=5b_s/(6\alpha )$. This law translates geometric plume-contact criteria into heights.

For external pressure $\Delta \rho gh$ and edge condition $h{\sigma }_{rr}=-\Delta \rho g{h}^2/2$, the fixed-volume similarity solution is a uniformly thick disc:Conservation of mass fixes the similarity velocity. The radial equation then forces a differentiable thickness profile to be constant, and the edge stress fixes that constant. The singular point-release solution describes the spreading scaling; a finite initial uniform disc corresponds to shifting time.

The source-type similarity solution of $h_t=D(h^2{h}_x{)}_x$ with whole-line conserved area $\mathcal{V}$ is $h=H(t)[1-x^2/L(t{)}^2{]}_{+}^{1/2}$, where $L=2(\mathcal{V}/\pi {)}^{1/2}(Dt{)}^{1/4}$ and $H=(\mathcal{V}/\pi {)}^{1/2}(Dt{)}^{-1/4}$. Its volume flux vanishes at the finite front, while the velocity tends to $\dot{L}=L/(4t)$. A reflecting half-line uses twice its prescribed area in these formulas.

The unsteady Boussinesq triangular-profile line-plume balances admit a buoyant similarity solution $Q=\alpha {\rho }_0{z}^2/\tau$, $M=(2/3)\alpha {\rho }_0{z}^3/{\tau }^2$ and $F=(2/3)\alpha {\rho }_0{z}^3/{\tau }^3$, with $\tau =t+t_{\ast }>0$. It reconstructs $b=\alpha z$, $W=z/\tau$ and reduced gravity $z/{\tau }^2$. Origins can be shifted, but the zero-time unshifted solution is singular. Source and initial data are required to select the physical member and region of validity.

If a draining gravity current is supplied with volume flux proportional to $t^{\alpha }$, a first-kind similarity solution hasprovidedThese relations follow by balancing the time derivative, horizontal lubrication theory flux, vertical drainage, and imposed inlet flux.

A laminar column can contract through nearly hydrostatic equilibrium when its cooling time greatly exceeds its vertical dynamical timescale of a disk. For cooling $A{T}^{1+\beta }$ and fixed column mass, a similarity solution has $\rho \propto {\eta }^{-1}$, $T\propto {\eta }^2$, $P\propto \eta$, and vertical velocity $u_z=\dot{\eta }z/\eta$. The coefficient $C$ also depends on the initial mass density normalization.

For a saline Stefan problem with common thermal properties, $ϵ=\sqrt{D/\kappa }\ll 1$, and fixed latent-to-sensible heat ratio, set $T_0=-m{C}_0$, ${\theta }_{-}=T_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }-T_0>0$. A finite-$\lambda$ similarity solution has $T_i=(T_{\infty }+T_{-\infty })/2+O(ϵ)$. At the onset of constitutional supercooling, $\lambda =O(ϵ)$ and the error in this mean is $O({ϵ}^2)$. The salt-rejection function for a saline Stefan front then gives $m\lambda C_i=({\theta }_{-}-{\theta }_{+})/(2\sqrt{\pi })+O({ϵ}^2)$, whereas $\sqrt{Dt}{T}_{l,x}(a^{+})=ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Comparing the actual temperature gradient to the liquidus gradient gives the displayed onset, with supercooling on the side of greater ${\theta }_{-}$. For a freezing front at $a\ge 0$ and $D<\kappa$, the ratio of these two gradients decreases into the liquid, so their interface comparison also decides whether a supercooled interval occurs anywhere ahead. The formula is an asymptotic criterion; finite-$ϵ$ onset requires the full Stefan condition and salt balance.

Let $x$ be slant distance from an upward apex, and let the surface slope down at angle $\alpha$ to the horizontal. Axisymmetric lubrication theory givesFor conserved volume $V$ and $x_N^3\gg V/\sin \alpha$, the leading similarity solution has $h=\sqrt{x/(5K\sin \alpha t)}$ in $0<x<x_N$, withThe outer front is a discontinuity in this reduced conservation law; the hydrostatic-gradient term and the full local flow resolve the front and apex regions where the outer approximation fails.