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The complementary error function avoids subtracting a quantity close to one when describing a small Gaussian integral tail. For real $x>0$, $\mathrm{erfc}x=(2/\sqrt{\pi }){\int }_x^{\infty }{e}^{-t^2}dt$; analytic continuation defines it for complex arguments.

The Faddeeva function is an entire function built from the complementary error function. Its integral representation for $\mathrm{Im}z>0$ is $w(z)=(i\pi {)}^{-1}{\int }_{-\infty }^{\infty }{e}^{-t^2}/(t-z)dt$, as normalized in NIST's integral representation. Its reflection identity is $w(-z)=2e^{-z^2}-w(z)$, directly from the odd function property of the error function.

The normalized Gaussian mass of an interval isIt is the convolution of an interval indicator with the heat kernel, and has an error function representation. As the variance tends to zero it converges to the interval indicator away from its endpoints, with value one half at an endpoint. This is a concrete approximate identity.

For $z=2n+i\mu \sqrt{n}+O(1)$, the factor $t^{2n}{e}^{z(1-t)}$ has saddle scale $t=1+u/\sqrt{n}$ and becomes $e^{-u^2-i\mu u}$. If the amplitude has a simple pole there, its leading local integral contains $e^{-u^2-i\mu u}/u$. Differentiating its Cauchy principal value with respect to $\mu$ reduces it to a Gaussian integral, producing an error function. The nonsingular amplitude $(1+t{)}^{-1}$ gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$.

In ${\epsilon }^2{y}^{\prime \prime }+a(x)y^{\prime }+\cdots =f$, a simple zero $a(x_{\ast })=0$, $a^{\prime }(x_{\ast })\ne 0$, balances diffusion and advection on width $\epsilon$. With $a(x)\simeq 2(x-x_{\ast })$, the leading homogeneous inner equation is $Y^{\prime \prime }+2z{Y}^{\prime }=0$, whose bounded solutions are constants and the error function. Thus distinct order-one outer solutions can match across an interior transition. Reversing the diffusion sign makes the nonconstant homogeneous solution grow at both ends, changing the matching structure.

The Neumann solution places a planar phase boundary at $h(t)=2\lambda \sqrt{\kappa t}$. Similarity reduction of the heat equation gives temperature profiles in terms of the error function, while the Stefan condition determines $\lambda$ implicitly.

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 $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Let $h=\log \psi$, with $\psi >0$, and set $f=2\alpha h_{\theta }$. Direct differentiation givesThe heat equation ${\psi }_Z=\alpha {\psi }_{\theta \theta }$ implies $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require $\alpha >0$.

Take $U,L>0$ for the stated Burgers N-wave. Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\alpha )$ and normalizing the exterior value to one givesThis function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yieldsPut $a=1+UZ$. Completing the square givesChanging the interior integration variable to $\eta =a\phi$ changes its limits to $\pm La$ and the Gaussian function width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the Cole-Hopf solution for a Burgers N-wave isHere $I_{\alpha }$ is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed $w>0$, the Gaussian function concentrates at $\phi =\theta$ as $\alpha \downarrow 0$. Consequentlyaway from the endpoints; at $\theta =\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an approximate identity argument, not a uniform step approximation across the endpoints.

For fixed $Z>0$ and $|\theta |\ll L$, both interval masses tend to one, and the exponentially large positive $\hat{\psi }$ dominates the exterior correction. Its logarithmic derivative therefore givesFor $|\theta |\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta )\simeq 0}$ in the specified far exterior.

An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\hat{\psi }$ gives the sharper inviscid Burgers N-wave fronts $|\theta |=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the vanishing-viscosity limit isInside $|\theta |<La$, this follows from the sign of $L^2-{\theta }^2/a$ in the exponential. Outside $La$, the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is $UL/(2\sqrt{a})$, equal both to the derivative of $L\sqrt{a}$ and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.

Write $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. The first correction in the inner expansion satisfies ${\mathcal{L}}_0{Y}_1=2zE$. Direct differentiation verifies that $zE+bg$ is a particular solution. Adding the constant and error function homogeneous solutions gives $Y_1=zE+bg+A+BE$.

The left boundary condition forces $A=-b$. Re-expanding the outer expansion at $x=sz$ givesTherefore $Y_1\sim z$, with no constant term in its large-$z$ overlap. Since $E\to 1$ and $g\to 0$, this sets $B=b$. Thus the required half-order correction isIt vanishes at zero and has the correct linear overlap. Keeping this $O(\sqrt{\epsilon })$ term is essential even though the outer solution has no correction of that order.

For the matched asymptotic expansion, first determine the outer expansion. Setting $\epsilon =0$ away from $x=0$ gives $y_0^{\prime }=y_0$, and the right boundary condition sets $y_0=e^x$. At first order,soThere is no outer $\sqrt{\epsilon }$ term: its homogeneous coefficient would be forced to zero by the right boundary condition. The logarithmic overlap forces a switchback term in the inner expansion.

Near zero, balance $\epsilon y^{\prime \prime }$ against $2x{y}^{\prime }$ to get the square-root boundary layer at a vanishing drift:For fixed $z$ expand $Y=Y_0+s{Y}_1+s^2{Y}_2+\cdots$, allowing logarithmic dependence on $s$ in $Y_2$. With ${\mathcal{L}}_0={\partial }_z^2+2z{\partial }_z$, the leading equation is ${\mathcal{L}}_0{Y}_0=0$. Its bounded solutions are a constant plus an error function. The left boundary condition and the overlap value one fixThe following two solution sections complete the half-order and first-order matching; their headings retain the scaffold's organization of the supplied hints.

Interpret $P(k)$ in the variance integral as the unsmoothed cosmological density power spectrum of $\delta$. A linear smoothing window gives $P_R(k)=P(k)|\tilde{W}(kR){|}^2$ for the smoothed field; the PDF's description of $P(k)$ as already belonging to ${\delta }_R$ would otherwise count the window twice.

For a scale-free matter power spectrum $P(k)=A{k}^{n_{\mathrm{eff}}}$, substitute $q=kR$ to obtainWhenever the dimensionless integral is finite and nonzero, this proves the scale-free smoothed density varianceThe printed condition $n_{\mathrm{eff}}<3$ is insufficient by itself. For a normalized window with $\tilde{W}(0)=1$, infrared convergence requires $n_{\mathrm{eff}}>-3$. Ultraviolet convergence depends on the filter: a spherical top-hat window function requires $n_{\mathrm{eff}}<1$, a Gaussian filter suppresses every ultraviolet power, and a sharp-k smoothing filter has finite Fourier support. In particular, the monotone relation $R\to \infty \Rightarrow S={\sigma }_R^2\to 0$ requires $n_{\mathrm{eff}}>-3$ in the scale-free model.

At a single variance $S_{\ast }$, the Gaussian random field has ${\delta }_{S_{\ast }}\sim N(0,S_{\ast })$. Thus its endpoint tail isHere $\mathrm{erfc}$ is the complementary error function.

For the crossing probability, the key assumption is independent, symmetric increments as the smoothing scale changes. A sharp-k smoothing filter, $\tilde{W}(kR)=1_{k<1/R}$, supplies this property: decreasing $R$ adds independent Fourier shells of the underlying Gaussian field. Parameterizing their accumulated variance by $S$ gives a Brownian motion with covariance $\min (S,S^{\prime })$. An arbitrary real-space window gives correlated increments and does not justify this argument.

Reflect a trajectory after its first hit of ${\delta }_c$. The Strong Markov property and symmetric future increments preserve its probability, and reflection maps an endpoint below the barrier to one above it. Every endpoint above the barrier has already crossed; the reflected paths give an equally probable set that crossed but ended below. The Brownian reflection principle therefore yields the excursion-set description of halo formation:Its derivative is the Brownian first-passage time densityThe factor of two resolves trajectories that crossed a larger-scale barrier but finish below it on the chosen smaller scale.

Define the halo peak height $\nu ={\delta }_c/{\sigma }_R$. The cumulative probability $\mathcal{P}(>M)$ is a dimensionless mass fraction. With $V_M=M/\bar{\rho }$, its conversion to halo number density givesConsequently the Press-Schechter halo mass function isIt is positive when ${\sigma }_R$ decreases with mass. A chosen mass-radius convention $M\propto R^3$ gives $d\log {\sigma }_R/d\log M=-(n_{\mathrm{eff}}+3)/6$. For a sharp Fourier filter that mass assignment needs a convention, since its real-space kernel is not a localized top-hat volume.

For warm dark matter, the cutoff introduces a new length and invalidates the pure power-law variance scaling near that length. With the same sharp-k filter and $n_{\mathrm{eff}}>-3$, the variance is exactlyThe Brownian walk runs only up to $S_{\max }$ and then stops. The cutoff does not destroy independence of the Fourier shells that are still present. Thus the first-crossing density and derivative mass-function formula remain valid on the invertible part of $S(M)$, using the actual cutoff variance. For $R<k_{\mathrm{WDM}}^{-1}$, $d{\sigma }_R/dM=0$ and this idealized sharp-filter model has no new crossings or new low-mass haloes. Its total collapsed fraction isThis is halo first crossing with a finite variance cutoff: some trajectories never cross, so one must not force the collapsed fraction to one. The uncut power-law abundance cannot be extrapolated to small masses. For other windows, a variance plateau still occurs but the Markov/reflection derivation is not exact; the same mass function is then a modelling approximation, rather than a result established by this calculation.

Adopt the Boussinesq approximation, hydrostatic approximation, traditional approximation, and inviscid quasi-geostrophic approximation. Assume stable constant $N_0^2>0$, nonzero Coriolis parameter $f_0$, a beta plane, small Rossby number, $\beta L/|f_0|$ of that small order, and small displacement of the background stratification. Vertical advection of the buoyancy perturbation is then higher order. Define the quasi-geostrophic streamfunction byThe buoyancy and vertical vorticity equations at the retained order areDifferentiate the first in $z$. Since $J({\psi }_z,{\psi }_z)=0$, its material-derivative term becomes $f_0{D}_g{\psi }_{zz}$. Elimination of $w_z$ gives diabatically forced quasi-geostrophic potential vorticity:Thus it is the height gradient of heating, rather than heating alone, that produces potential vorticity at this order.

For the steady response linearized about rest, assume $\beta \ne 0$. The quasi-geostrophic potential-vorticity equation reduces to $\beta {\psi }_x=f_0{R}_z/N_0^2$. Choosing the horizontally uniform pressure gauge so that $\psi \to 0$ in the east gives the steady quasi-geostrophic response to localized heatingSetUsing the error function to integrate the Gaussian function givesThese velocities vanish as $x\to +\infty$. A depth-dependent but horizontally uniform addition to $\psi$ is dynamically irrelevant for the requested horizontal velocity. At $z=0$, streamlines are contours of $e^{-y^2/L^2}I(x)$. For $f_0,\beta ,R_0>0$, flow approaches the heating region from the west on its northern side, turns southward, and returns westward on its southern side. The contours are open, not closed gyres: $I$ tends to a nonzero constant in the west.

For $-H<z<0$, $R_z<0$ and the forcing decreases potential vorticity; steady balance requires $\beta v_g=f_0{R}_z/N_0^2<0$, giving southward flow. Below $z=-H$, $R_z>0$ and the meridional flow reverses. A constant heating rate independent of $z$ would have no such interior potential vorticity source.

For the nonlinear diagnostic quasi-geostrophic omega equation, write $\zeta ={\nabla }_h^2\psi$. Apply ${\nabla }_h^2$ to buoyancy evolution and $f_0{\partial }_z$ to vorticity evolution:Subtracting eliminates the pressure tendency and yieldsThe beta term is linear and must remain when the nonlinear terms are neglected. For the steady weak-forcing solution, $f_0\beta {\psi }_{xz}=f_0^2{R}_{zz}/N_0^2$, soThis also follows immediately from the steady linear buoyancy equation. It satisfies $w=0$ at $z=0$, decay at depth and horizontal infinity, and the linear quasi-geostrophic omega equation; with these homogeneous conditions the elliptic problem has no extra decaying homogeneous solution. The rigid-lid interpretation is an additional boundary approximation for this diagnostic response.

At $x=0$, the displayed $v_g$ and $w$ provide the requested meridional/vertical arrows. For positive heating there is upwelling concentrated near $y=0$, maximal in depth at $z=-H$, with northward flow below that depth and southward flow above. The vertical circulation tends to zero far from the forcing. A western zonal wake persists, but its meridional and vertical components vanish as $x\to -\infty$. Horizontal ageostrophic flow supplies the divergence associated with the upwelling; the projected $(v_g,w)$ slice is not itself a two-dimensional incompressible flow and need not have closed streamlines.

The steady calculation fixes an interior circulation, not its attainability from every possible initial and boundary state. For example, a rigid lid with initially uniform boundary buoyancy conserves that buoyancy at linear order because $R(0)=w(0)=0$. The formal steady profile instead generally has nonuniform $b^{\prime }(0)=f_0{\psi }_z(0)$. Such additional initial boundary data require a time-dependent adjustment and cannot simply be imposed on this particular steady solution. No boundary buoyancy data are supplied for the requested steady problem.

The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

An integrating factor explains the second hint and the first-order inner calculation. Formultiplication by $e^{z^2}$ gives $(e^{z^2}{Y}^{\prime }{)}^{\prime }=e^{z^2}(2a_{1}z+a_2+a_3\mathrm{erf}z)$. The term $a_{1}z$ already produces $2a_{1}z$, and integrating the remaining terms twice gives $a_2{E}_0+a_3{E}_1$, in terms of the Gaussian drift primitives. The homogeneous solutions are a constant and the error function. HenceFor the piecewise forcing in the inner problem, $F$ replaces $a_{1}z$ and provides the continuously matched particular solution for $2zH(-z)$. The coefficients $2c$ and $2d$ then account for the $2Y_0$ forcing.

Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.