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A two-phase diffusion model with constant properties has similarity solutions built from the complementary error function in each half-line. Equilibrium gives $T_a=-m{C}_a$ and $C_s(a)=k_D{C}_a$. The Stefan condition and interfacial solute balance determine $\lambda$ and $C_a$; finite diffusion on both sides must be retained when constructing the narrow transition near $\lambda =0$.

The complementary error function describes the Gaussian integral area remaining beyond the endpoint. For a real $x$,It is an entire function for complex arguments when defined by $1-\mathrm{erf}(z)$. On the real line it decreases from two to zero, with $\mathrm{erfc}(0)=1$ and $\mathrm{erfc}(-x)=2-\mathrm{erfc}(x)$.^[ref]

For example, $\mathrm{erfc}(1)\approx 0.157299$ is a sizeable tail, whereas $\mathrm{erfc}(8)\approx 1.12243\times 10^{-29}$ is extremely small. In the tested floating-point arithmetic, a direct evaluation of $1-\mathrm{erf}(8)$ gives zero because the computed error function has already rounded to one. Evaluating the complementary error function directly preserves this tail. It is the useful quantity in an upper-tail normal distribution probability and in the decaying side of an error-function diffusion front.

For large positive real $x$, the leading asymptotic expansion isThe first corrections are $1-1/(2x^2)+3/(4x^4)-\cdots$ multiplying the leading term. This is an asymptotic expansion, not an assertion that summing arbitrarily many terms converges at a fixed $x$.^[ref] The complementary error function tail bound provides finite-$x$ inequalities instead.

The half-line solution of $C_t+U{C}_x=D{C}_{xx}$ with initially zero concentration, unit inlet concentration for positive time, and decay at infinity at fixed time. Both complementary error functions are needed to satisfy the inlet exactly. A Laplace transform gives $\tilde{C}=p^{-1}{e}^{(U-\sqrt{U^2+4Dp})x/(2D)}$. A finite layer requires an additional downstream boundary condition.

For a passive scalar on a half-line, impose constant total solute flux at its inlet, with initially zero concentration and constant positive advection speed $U$ and mass diffusivity $\mathcal{D}$. A Laplace transform in time givesWith $\xi =Ux/\mathcal{D}$, $\vartheta =U^{2}t/\mathcal{D}$ and $z_{\pm }=(\xi \pm \vartheta )/(2\sqrt{\vartheta })$, its inverse isThe complementary error function describes the smeared front. This is a flux boundary condition, not the constant-concentration inlet solution. At large axial Péclet number, both have the leading advancing front $\frac{1}{2}\mathrm{erfc}(z_{-})$.

For stationary liquid below a salt-free solid and an upward melting interface $a(t)=2A\sqrt{Dt}$, the complementary error function salt profile and flux balance $D{C}_z(a)=-\dot{a}{C}_i$ give the displayed concentration ratio. For $A>0$ it lies below one. At large $A$ it behaves as $e^{-A^2}/(2\sqrt{\pi }A)$.

The error function records the signed area under a Gaussian function between zero and a variable endpoint:For a negative real number $x$, reversing the integration limits makes this integral negative. The normalization uses the Gaussian integral: the full area of $e^{-t^2}$ is $\sqrt{\pi }$, so its positive half contributes one after multiplication by $2/\sqrt{\pi }$. The same definition extends to a complex number $z$ along any path from zero to $z$, giving an entire function.^[ref]

For example, $\mathrm{erf}(0)=0$, $\mathrm{erf}(1)\approx 0.842701$, and $\mathrm{erf}(-1)\approx -0.842701$. A finite positive endpoint captures only part of the positive Gaussian integral; $\mathrm{erf}(x)$ approaches one as $x\to +\infty$ but never reaches it at a finite real $x$.

The useful idea is to name a recurring integral once instead of carrying it through every calculation. The error function and the normal distribution express probabilities and quantiles through this area, while an error-function diffusion front describes how diffusion smooths a sharp jump. The complementary error function retains the small area left in the tail.

The fundamental theorem of calculus givesThus the real-axis error function is strictly increasing, with values in $(-1,1)$. Its integrand is an even function, so the error function is an odd function, $\mathrm{erf}(-x)=-\mathrm{erf}(x)$. The derivative also gives ${\mathrm{erf}}^{\prime \prime }(x)=-4x{e}^{-x^2}/\sqrt{\pi }$: the curve is concave for $x>0$ and convex for $x<0$. These real-axis properties do not imply boundedness throughout the complex plane; the existing quadrant asymptotics of the error function describe a different behavior in other directions.

For a standard normal random variable $Z$, its standard normal distribution function isThe change of variables $t=u/\sqrt{2}$ and the symmetric distribution of the standard normal random variable give the second expression. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable then givesIn particular, the intervals within one, two, and three standard deviations have calculated probabilities about $0.682689$, $0.954500$, and $0.997300$. These values depend on a normal distribution; they are not universal coverage guarantees for an arbitrary random variable. This is the direct connection between a Gaussian function area and the familiar central intervals.

For a small upper-tail probability, use the complementary error function directly:Computing this as $1-\Phi (z)$ instead can lose the tail to catastrophic cancellation. The Gaussian tail bound gives another useful estimate without requiring an exact error function value.

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 Gaussian integral gives the exact area under a Gaussian function. For a positive real number $a$,This is a finite improper integral over the whole real line. For example, $I(1)=\sqrt{\pi }\approx 1.772454$ and $I(4)=\sqrt{\pi }/2\approx 0.886227$. Increasing $a$ from one to four narrows the curve and halves its area, while its height at zero stays one. The change of variables $u=\sqrt{a}x$ explains the factor $a^{-1/2}$.

The useful consequence is an exact normalization. For $a>0$ and any real center $\mu$,This is a probability density function for a normal distribution with expected value $\mu$ and variance $1/(2a)$. A narrower probability density function must be taller to keep total probability one. In the usual notation $a=1/(2{\sigma }^2)$, the prefactor is $1/(\sigma \sqrt{2\pi })$, where $\sigma$ is the standard deviation.

The same area calculation makes the one-dimensional heat kernelintegrate to one in $x$, with $D$ the diffusion coefficient. A spreading Gaussian function can therefore preserve total mass while its height falls. This links the Gaussian integral with both the normal distribution and the diffusion equation.

Finite intervals lead to the error function, rather than this single constant:The complementary tail is described by the complementary error function. These definitions and their real-axis limits are given in the NIST reference. ^[ref]

The polar-coordinate proof of the Gaussian integral explains why $\pi$ appears in an integral of a function defined on a line: squaring its area creates a rotationally symmetric integral on a plane. Substitution $t=x^2$ also gives $\Gamma (1/2)=2{\int }_0^{\infty }{e}^{-x^2}dx=\sqrt{\pi }$, connecting the result with the gamma function.

For the reduced quantum harmonic oscillator, $V(q)=q^2/2$ and the ground-state energy is $E_0=1/2$. Thus the classical turning points satisfy $q^2/2=1/2$, giving $q=\pm 1$. Normalizing the wavefunction gives ${\varphi }_0(q)={\pi }^{-1/4}{e}^{-q^2/2}$, because ${\int }_{-\infty }^{\infty }{e}^{-q^2}dq=\sqrt{\pi }$. The Born rule therefore gives the harmonic-oscillator forbidden-region probabilityThe factor two includes both tails. Using the exact complementary error function gives $\mathrm{erfc}(1)=0.1572992$. A stationary real wavefunction has zero probability current, so this result describes forbidden-region occupancy, not continuous loss of the bound particle.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

Write $H=L/c_p$, $k=\rho c_p\kappa$ and $ϵ=\sqrt{D/\kappa }$. The requested model neglects density differences and uses the same thermal transport coefficients in the interfacial balances. Let $C_b=C(b(t),t)$ and $T_b=T_m-m{C}_b$. With no imposed heat gradient in the fresh liquid, its temperature remains $T_m$. The prescribed linear temperature in the ice isThe salt and temperature fields in the lower liquid satisfy diffusion equations, $C_t=D{C}_{zz}$ and $T_t=\kappa T_{zz}$. For constant interfacial values and the proposed similarity solution, their decaying solutions areThese follow by substituting a variable $z/(2\sqrt{\chi t})$ into the diffusion equation, which gives $U^{\prime \prime }+2\eta U^{\prime }=0$; integrating gives the complementary error function. Their values at $b$ and at infinity give the required boundary data.

For the upward-moving fresh-water/ice boundary, the Stefan condition is $\rho L\dot{a}=k{T}_{i,z}$: negative $\dot{a}$ means fresh water freezes. At the lower boundary the signed heat balance isSalt cannot enter the pure ice. With no bulk liquid motion its interfacial conservation law isThe derivatives and interface velocities areDefine $F(x)=\sqrt{\pi }x{e}^{x^2}\mathrm{erfc}x$. Substitution into all three balances givesTogether with $T_b=T_m-m{C}_b$, these determine the interfacial constants. In particular $C_b=C_0/[1-F({\lambda }_b)]$.

The physical branch has ${\lambda }_a<{\lambda }_b<0$. Since $F$ is negative at negative arguments, the salt relation gives $0<C_b<C_0$: lower-surface melting dilutes the brine. Combining the two heat balances eliminates the ice gradient and givesThus $T_m>T_b>T_{\infty }>T_L(C_0)$. The interface is at local liquidus equilibrium with diluted brine, while the undiluted far-field brine remains above its own liquidus. The similarity model describes a layer starting with zero thickness, or the long-time limit when an initial finite thickness no longer sets the dominant scale.

For the small-diffusivity-ratio limit, put $A=ϵ{\lambda }_a$ and $\Delta =T_m-T_{\infty }$, keeping $S=H/\Delta$ and the imposed concentrations fixed. Seek $A=O(1)$ and ${\lambda }_b=O(1)$. Then $F(ϵ{\lambda }_b)\sim \sqrt{\pi }ϵ{\lambda }_b$ and ${\lambda }_b-{\lambda }_a\sim -{\lambda }_a$. The upper heat balance givesIn the lower balance the two terms of order ${ϵ}^{-1}$ must cancel, givingConsequentlyand $T_b-T_{\infty }\sim -\sqrt{\pi }HA$. Adding the two temperature differences yields $2A^2-\sqrt{\pi }A=1/S$. The negative root isThe resulting $C_b=(T_m-T_b)/m$ lies strictly between zero and $C_0$. The salt relation therefore determines a finite negative ${\lambda }_b$, confirming its $O(1)$ scaling. The top interface travels on the thermal scale $\sqrt{\kappa t}$, whereas the lower melting interface travels on the shorter solutal scale $\sqrt{Dt}$. This is the brine-ice sandwich with disparate diffusion scales; $S$ is the latent-to-sensible heat ratio, the reciprocal of the usual sensible-to-latent Stefan number convention.

The sketches show constant fresh-water temperature, a declining linear ice temperature, and a much broader thermal than solutal boundary layer in the brine. Concentration is zero in freshwater and pure ice, jumps to $C_b$ across the lower interface, and rises to $C_0$. Accordingly the local liquidus is $T_m$ on the salt-free side, jumps to $T_b$ and falls rapidly to $T_L(C_0)$. In the phase diagram, the brine starts on the liquidus at $(C_b,T_b)$ and ends above it at $(C_0,T_{\infty })$; the pure ice has concentration zero, and the interface tie-line joins it to the diluted brine.

Physically, the cold saline liquid permits freshwater to freeze above, but salt lowers the liquidus and allows the existing ice to melt into brine below even though that brine is colder than $T_m$. Dilution raises the lower interfacial freezing temperature. Heat released by upper freezing supplies the lower melting and the conductive loss to the colder deep brine. With $|{\lambda }_a|\gg |{\lambda }_b|$, upper freezing exceeds lower melting and the ice layer thickens as $b-a\propto \sqrt{t}$.

Use $\ell =L/c_p$, $ϵ=\sqrt{D_s/D_l}$ and $\delta =\sqrt{D_l/\kappa }$, with $0<k_D<1$. Equal reference densities and constant material properties are implicit in these diffusion balances. Put $a(t)=2\lambda \sqrt{D_{l}t}$ and $T_a=-m{C}_a$. The four similarity fields areHere the same letters $T_s,T_l,C_s$ on the right denote their far-field values. Substitution into the heat and solute diffusion equations verifies the fields; their interface values and far-field limits verify every boundary condition. The concentration has an equilibrium jump from $k_D{C}_a$ to $C_a$, whereas temperature is continuous.

The interfacial conservation laws areFor example,Since $\dot{a}=\lambda \sqrt{D_l/t}$, dividing the two jump conditions by $\dot{a}$ gives, for nonzero $\lambda$,where $F(z)=\sqrt{\pi }z{e}^{z^2}\mathrm{erfc}z$. At $\lambda =0$, use the undivided flux conditions or the continuous limit, rather than divide by zero. The liquidus relation and the segregation coefficient also give the solidus $T_S(C)=-mC/k_D$.

For the thermal expansion, write $z=\delta \lambda$ and useKeeping the constant terms in these reciprocals is essential for the full first-order answer. The result isThe remainder has this form for bounded $\lambda$; a looser $O({\delta }^2)$ statement is also valid in the distinguished scaling $\ell =O({\delta }^{-1})$. Dropping the temperature-difference term while claiming first-order accuracy for fixed $\ell$ would be incorrect.

Solving the solute balance algebraically givesFor fixed positive $\lambda$, $F(-\lambda /ϵ)\to -\infty$, and $C_a\to 0$, with a positive exponentially small correction. For fixed negative $\lambda$, $F(-\lambda /ϵ)\to 1$, andIt falls from $C_s$ as $\lambda \to -\infty$ to zero as $\lambda \to 0^{-}$. The outer positive branch is zero. For finite $ϵ$ the transition is rounded over $|\lambda |=O(ϵ)$; the exact zero-speed limit is $C_a(0)=ϵC_s/(1+k_Dϵ)$, not an actual concentration discontinuity.

To count speeds, take the leading distinguished thermal balance, set $\Sigma =T_l+T_s$ and $K=\sqrt{\pi }\delta \ell$, and combine $T_a=-m{C}_a$ with the thermal result:This is the intended leading approximation when $\delta \ell =O(m{C}_s)$ and $T_l-T_s=O(1)$. The full first-order expression instead replaces $K$ by $\sqrt{\pi }\delta [\ell +2(T_l-T_s)/\pi ]$. Thus the printed threshold is asymptotic, not an exact criterion for finite $\delta ,ϵ$.

For $r=-\lambda >0$, put $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$. The small-solid-diffusivity melting concentration is $C_{s}H/(1+H)$, soThe second term decreases strictly from $2m{C}_s\sqrt{\pi }$ to zero. One can check the strict decrease without guessing its shape. Set $A(r)=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$, so $A^{\prime }=2rA+2$, $A\ge \sqrt{\pi }+2r$, and $H=rA$. Direct differentiation givesThe bracket increases with $A$ on $r\ge 0$. Substituting its lower bound yields a polynomial with constant coefficient $\pi -2>0$ and all other coefficients positive. Hence $[H^{\prime }/(1+H{)}^2{]}^{\prime }<0$, as claimed.

If $\delta \ell >2m{C}_s$, then $K>2m{C}_s\sqrt{\pi }$ and $\Sigma (r)$ increases from zero to infinity. The freezing branch has $\Sigma =-K\lambda <0$ and is also monotone. Together these give one speed for each imposed sum within the approximation. If $0<\delta \ell <2m{C}_s$, the melting branch initially decreases, has exactly one minimum and then increases to infinity. For sums between that minimum and zero there are two melting speeds plus one freezing speed: three solutions. At the lower fold there are two distinct speeds, one being a double solution. At zero the limiting outer branches join; finite solid diffusivity rounds the join into a nearby fold. Away from the fold values the other sums have one solution. At the threshold itself the initial slope vanishes; it is the transition between these cases.

For $\Sigma >0$, the solution is on the melting branch $\lambda <0$. Its thermal layers have scale $\sqrt{\kappa t}=\sqrt{D_{l}t}/\delta$, much larger than the liquid-solute scale $\sqrt{D_{l}t}$. Because the front is retreating, the solid solute layer is thinner still:This follows by expanding the large-positive-argument complementary error function ratio on the solid side: near $a$, its decay length is $D_s/|\dot{a}|$. Using simply $\sqrt{D_{s}t}$ would miss the retreating-front compression of that layer.

Across this narrow layer, solid concentration drops from almost $C_s$ to $k_D{C}_a$, while temperature stays almost $T_a$. At the interface $T_a=T_S(k_D{C}_a)$. Moving slightly into the solid increases its concentration toward $C_s$, lowering its solidus to $-m{C}_s/k_D$, whereas the temperature has barely changed. Since $C_a<C_s$ and $k_D<1$, $T_a=-m{C}_a>-m{C}_s/k_D$. Thus an adjacent region can satisfy $T>T_S(C)$: constitutional superheating. Its existence is especially clear when far-field solid is initially stable but heat diffuses across the narrow composition layer.

The phase-diagram path goes from $(C_s,T_s)$ through a broad thermal rise, then nearly horizontally toward $(k_D{C}_a,T_a)$ across the thin solid diffusion layer. At the interface it traverses the isothermal equilibrium tie line to $(C_a,T_a)$. On the liquid side concentration falls to zero over the liquid-solute layer, while the remaining thermal adjustment to $T_l$ takes place over the much wider heat layer. The path of the depleted solid approaches the solidus from its superheated side. That region should melt at additional sites or develop a two-phase mushy layer; the original single-boundary similarity construction is then a mathematical branch rather than a physically stable description of the entire sample.

Watson lemma expands a Laplace integral by integrating the small-$t$ expansion term by term. More precisely, if $f(t)\sim {\sum }_{j\ge 0}{a}_j{t}^{\lambda +j-1}$ with $\mathrm{Re}\lambda >0$, and the tail satisfies bounds permitting this localization, thenFor ordinary powers $f(t)\sim \sum c_j{t}^j$, the coefficients are $c_{j}j!$. For the Gaussian weight, put $s=t^2$ separately on the positive and negative half-lines. The integral becomes a Laplace integral with amplitude $[f(\sqrt{s})+f(-\sqrt{s})]/(2\sqrt{s})$. Odd terms cancel and Watson lemma givesThese assertions concern asymptotic expansions, not necessarily convergent series.

For the complex integral in the lower half-plane, the identity $(z-t{)}^{-1}=i{\int }_0^{\infty }{e}^{-i(z-t)s}ds$ and the Gaussian Fourier transform giveThe last expression, initially obtained by completing the square, is entire and therefore gives the requested analytic continuation. Expanding the denominator formally, with odd Gaussian moments zero, gives the algebraic expansionHere $(-1)!!=1$. The complementary error function expansion shows $I(z)\sim \mathcal{A}(z)$ on closed subsectors of $-5\pi /4<\arg z<\pi /4$, using an unwrapped argument centered on the lower half-plane.

In the upper half-plane, the directly evaluated real-contour integral is $J(z)=-i\pi e^{-z^2}\mathrm{erfc}(-iz)$. Since $\mathrm{erfc}(w)+\mathrm{erfc}(-w)=2$, continuation from below givesThe upper expansion for $J$ holds on closed subsectors of $-\pi /4<\arg z<5\pi /4$. The added term is the residue acquired by the continued contour passing the pole; it is exponentially dominant in the upper sector $\pi /4<\arg z<3\pi /4$ and exponentially small near either real ray. On the real axis the continued value is the principal-value integral plus $i\pi e^{-x^2}$.

To make the Stokes phenomenon convention explicit, the exponential switches its asymptotic multiplier across the real rays $\arg z=0,\pi$; these are the Stokes switching lines. The diagonal rays $\arg z=\pi /4+j\pi /2$ separate exponential growth from decay and are often called anti-Stokes lines, or critical boundaries when regions are classified by dominance. Thus both sets of critical directions and the exact jump are specified, independently of which terminology is used. This is the Gaussian Cauchy transform and its Stokes jump.

Far from the inner edge of a steady Keplerian accretion disk, the inner angular-momentum flux is negligible compared with $\dot{M}j$. Since $\Omega \propto R^{-3/2}$, the viscous torque in an accretion disk becomes $\mathcal{G}=3\pi \nu \Sigma j$. The steady balance $\mathcal{G}\simeq \dot{M}j$ therefore gives $\nu \Sigma =\dot{M}/(3\pi )$. With the stated surface density of a disk,Steady conservation of mass gives $\dot{M}=-2\pi R\Sigma v_R$, and consequentlyBoth relations use the far-inner-edge approximation, rather than extrapolating the stress-free boundary profile to the inner edge.

For the trace passive scalar, put $\sigma =\Sigma C$. Substitute this into the contaminant conservation equation and expand its advective flux:The braces multiplying $C$ vanish by background conservation of mass. Hence the concentration satisfiesThe small contaminant fraction ensures it does not alter the ambient disk dynamics. The surface-density model already averages vertically; an axisymmetric background and ring-like injection give $C=C(R,t)$. More generally, rapid differential orbital motion winds azimuthal structure and facilitates mixing, but radial-only dependence remains a symmetry or mixing assumption, not a property of every initial injection.

For the axisymmetric advection-diffusion equation, cylindrical geometry givesHere $k=\zeta \nu$ and $k\Sigma =\zeta {\nu }_0{\Sigma }_0$ are independent of radius. ThusLet $Y=\ln (R/R_{\ast })$, with any fixed reference length $R_{\ast }$. Since $C_Y=R{C}_R$ and $C_{YY}=R^2{C}_{RR}+R{C}_R$, this is the tracer diffusion in a power-law accretion disk equationAssume $\zeta >0$, ${\nu }_0>0$, and write $v=3{\nu }_0/2$, $D=\zeta {\nu }_0$. The change $X=Y+vt$ turns this into the heat equation $C_t=D{C}_{XX}$. The radial Dirac delta function must be transformed with its Jacobian:Convolution with the Gaussian heat kernel givesThis full-line logarithmic solution extends the far-edge power-law model to $0<R<\infty$. An actual finite inner boundary would require its own boundary condition and modified kernel.

The fraction must be computed with the contaminant mass measure, rather than with $CdR$. IndeedThe total contaminant mass is the conserved quantity $M_c=2\pi {\Sigma }_0{C}_0/R_0$. Its normalized logarithmic radius is therefore a normal distribution with mean $Y_0-vt$ and variance $2Dt$. Integrating this tail gives, for $r>0$,The complementary error function accounts for the one-time mass fraction outside the specified radius.

For an outward target $r_1>1$, put $\ell =\ln r_1>0$. Since $\mathrm{erfc}$ is decreasing, maximize $F$ by minimizing $z(t)=(\ell +vt)/\sqrt{4Dt}$. Its derivative has the sign of $vt-\ell$, so the unique maximum occurs at $t_{\ast }=\ell /v=2\ln r_1/(3{\nu }_0)$. At this time $z(t_{\ast })=\sqrt{v\ell /D}$. Thus the printed expression is correctly derived as the maximum instantaneous exterior fraction: