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Saturation equation. Introduce intrinsic permeability of a porous medium $K$ and porosity $\varphi$. The relative porous permeabilities depend on the wetting fluid saturation. Define the phase mobilities and fractional flowUse the conventional capillary pressure $p_c=p_{nw}-p_w$, generally decreasing as wetting saturation increases. Neglect gravity in this horizontal model. The two Darcy fluxes are $q_w=-K{\lambda }_w{p}_{w,x}$ and $q_n=-K{\lambda }_n(p_{w,x}+p_c^{\prime }(s)s_x)$. Eliminate the common pressure gradient using $q_w+q_n=Q$ to obtainThe wetting-phase mass conservation law is consequentlyFor $p_c^{\prime }\le 0$, $D_c\ge 0$ and the capillary pressure provides nonlinear diffusion. The initially uniform saturation is $s_0$. Pure wetting-fluid injection prescribes $q_w(0,t)=Q$ and $q_n(0,t)=0$; in the zero-capillarity, zero-residual-nonwetting idealization this corresponds to inlet saturation one. With residual nonwetting fluid, replace one by the appropriate maximum accessible saturation.

Shock formation and its speed. Neglect capillarity first. Smooth saturation values travel along characteristic curves at speed $(Q/\varphi )F^{\prime }(s)$. When the trailing values have larger characteristic speeds than the values ahead, the curves intersect, and the Buckley-Leverett equation requires a shock selected as an entropy solution. This is typical for the increasing-convex part of a physical fractional-flow curve; shocks are not inevitable for every possible $F$.

Integrating mass conservation across a shock from upstream $s_s$ to downstream $s_0$ gives the Rankine-Hugoniot conditionIf an upstream rarefaction wave joins the shock smoothly in similarity coordinates, its terminal characteristic speed equals $V_s$. Therefore the saturation at that join obeys the fractional-flow tangent constructionThis is the printed equality. It is a tangent-selection condition, not the dimensional speed by itself; the speed includes $Q/\varphi$. An arbitrary shock between prescribed constant states satisfies the jump condition but need not satisfy this additional tangency relation.

Positive capillary diffusion spreads a jump into a transition layer. For a travelling profile $s(x-V_{s}t)$, integration givesThe endpoint states still satisfy the same Rankine-Hugoniot condition. Thus small capillarity primarily gives the front a finite thickness rather than changing the selected limiting speed; vanishing phase mobility at an endpoint can make the regularization degenerate.

The specified reciprocal fractional flow. On its stated branch, $F^{\prime }=1/s^2$ and $F^{\prime \prime }=-2/s^3<0$. If $F$ extends continuously to the initial saturation and the inlet is at one, the decreasing saturation from behind to ahead makes characteristic speeds increase forwards: there is a rarefaction wave, not a compressive shock. Write $a=Q/\varphi$ and $\xi =x/t$. The self-similar solution isThe fan connects continuously to the initial state and broadens linearly in time. This interpretation requires $F(s_0)=2-1/s_0$. A nonnegative physical fractional flow on the entire interval $[s_0,1]$ also requires $s_0\ge 1/2$.

The PDF specifies the formula only for $s>s_0$ and earlier calls $s_0$ residual. If that means an immobile wetting phase with $F(s_0)=0$, the endpoint is an extra constraint and the continuous-branch answer applies without conflict only at $s_0=1/2$. For $0<s_0<1/2$, the printed branch would give negative fractional flow for $s_0<s<1/2$, so it cannot be the complete physical constitutive law. With an additional admissible extension, for example $F=0$ up to $1/2$ and the given positive branch above it, a compound fan–shock is possible. The tangent condition then givesThe fan $s=\sqrt{a/\xi }$ ends at $\xi =V_s$, followed by a jump to $s_0$. This is a conditional physical extension, not data supplied by the question. For $s_0>1/2$ an immobile residual endpoint is likewise incompatible with a continuous version of the stated branch. These distinctions identify what is, and is not, determined by the printed constitutive assumption.

The Inviscid Burgers equation here has the negative quadratic flux $F(f)=-f^2/2$. By the method of characteristics, $df/dZ=0$ and $d\theta /dZ=-f$, so a characteristic beginning at ${\theta }_0$ satisfiesThis is a classical solution until characteristic curves intersect. Across a shock wave, the Rankine-Hugoniot condition givesIt uses the one-sided limiting values and agrees with the weak-shock expression.

For $U>0$, the right-state characteristic curves travel left with speed $-U$ while the left-state characteristic curves have speed zero. They converge into a compressive shock wave with speed $-U/2$. The entropy solution isThe characteristic speeds satisfy $0>-U/2>-U$, so both sides enter the shock.

For $U<0$, right-state characteristic curves travel right with speed $-U>0$, leaving a gap. A steep continuous transition fills that gap with a rarefaction wave. The entropy solution isThe middle value is fixed by $\theta /Z=F^{\prime }(f)=-f$. A discontinuity joining the two states would satisfy the jump equation but violate entropy admissibility. If $U=0$, the solution is identically zero.

Let $h=\log \psi$ and $f=2\alpha h_{\theta }$. Direct substitution into the viscous Burgers equation givesIf ${\psi }_Z=\alpha {\psi }_{\theta \theta }$, then $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, proving the Cole-Hopf transformation. A function of $Z$ inside the parentheses can be removed by rescaling $\psi$ by a time-dependent factor, which leaves $f$ unchanged.

For the forward heat equation and the displayed Gaussian kernel we require $\alpha >0$ and $Z>0$. The algebraic transformation works for nonzero $\alpha$, but the printed convolution is not a forward solution for negative $\alpha$: its Gaussian then grows and the step-data integral diverges.

Integrating the initial logarithmic derivative fixes a convenient positive initial heat datum,Split the heat kernel convolution at zero and complete the square in the positive half. With $h_{\ast }=U(2\theta +UZ)/(4\alpha )$, putThen $\psi =(A_0+e^{h_{\ast }}{C}_0)/\sqrt{4\pi \alpha Z}$. On differentiation, the moving-limit terms cancel because $e^{h_{\ast }}{e}^{-(\theta +UZ{)}^2/(4\alpha Z)}=e^{-{\theta }^2/(4\alpha Z)}$. HenceThis is the viscous Burgers step solution with negative flux. Both integrals can be written as $\sqrt{\pi \alpha Z}$ times a complementary error function.

For fixed $Z>0$, the Gaussian-tail asymptotics give $f\to U$ as $\theta \to +\infty$ and $f\to 0$ as $\theta \to -\infty$. For example $J{e}^{-h_{\ast }}$ tends to zero on the right with a Gaussian factor $e^{-(\theta +UZ{)}^2/(4\alpha Z)}$, while on the left it diverges with a Gaussian factor $e^{{\theta }^2/(4\alpha Z)}$. This holds for either sign of $U$; for $U=0$ the solution is already zero.

The Gaussian tail is strictly decreasing in its lower limit, so $J=1$ occurs exactly at $\theta =-UZ/2$. There $h_{\ast }=0$, and $f=U/2$. For $U>0$, the two lower limits are both far into the negative tail in the mature shock region, so $J\simeq 1$ throughout its thin transition. The solution is then approximately the traveling viscous frontcentred at the inviscid shock position, with thickness of order $\alpha /U$.

In the vanishing-viscosity limit, for $U>0$ the solution tends to the compressive entropy solution of part (a), away from its shock. For $U<0$, it tends instead to the rarefaction wave. To see the latter explicitly inside $0<\theta <-UZ$, both tails have positive lower limits, and their leading asymptotics give $J{e}^{-h_{\ast }}\to (-\theta -UZ)/\theta$. Thus $f\to -\theta /Z$ in the fan, with the constant states outside. Although $J=1$ still marks the fan midpoint, it is not approximately one throughout the expanding fan; replacing it by one there would create an inadmissible compressive-front approximation.

The release produces a rarefaction wave expanding into the dry region. Every incoming $C_{-}$ characteristic connects to the undisturbed pool, where $J_{-}=0$, so throughout the simple wave $u=2(c-c_0)$. The outgoing $C_{+}$ rays emanate from $(0,0)$ and have $x/t=u+c=3c-2c_0$. Solving givesThe depth is $h=c^2/g$. At the right edge these formulas match the stationary pool; at the left edge $c=0$ and the fluid front travels at $-2c_0$. Beyond it the bed is dry, so a fluid velocity need not be assigned. Values at $t=0$ are understood as one-sided limits of the fan.

The first wall interaction occurs at $t_0=a/c_0$. The leftmost reflected signal follows a $C_{-}$ ray through $(a,t_0)$. Until a point has encountered this signal it remains in the original fan. In that fan its speed is $u-c=x/(3t)-4c_0/3$. ThusMatching at $(a,t_0)$ gives the reflected frontConsequently the same rarefaction formula remains valid for $-2c_{0}t\le x\le x_f(t)$ when $t\ge t_0$. Between $x_f(t)$ and the wall, information from the wall changes $J_{-}$ and the simple-wave relation no longer holds. In particular continuing the old formula to $x=a$ would give a negative velocity for $t>t_0$, violating the no-penetration condition $u(a,t)=0$. The reflected region must accommodate both characteristic families; it is not another untouched part of the original fan.

A bounded weak solution satisfiesfor every compactly supported $C^1$ test function, of either sign, allowed to meet $t=0$. Across a moving jump $x=s(t)$ with distinct left/right constant states, the distributional delta coefficient vanishes exactly when the Rankine-Hugoniot condition holds:It is required on every interface of a piecewise constant solution; the initial trace must also agree with the data.

For nonuniqueness respecting the paper's global bound on $f^{\prime }$, use the bounded-derivative Burgers flux $f(z)=z^2/2$ on $|z|\le 1$ and $f(z)=|z|-1/2$ outside. It is $C^1$ and $|f^{\prime }|\le 1$. Take data $-1$ for $x<0$, $1$ for $x>0$. A stationary jump obeys the condition because $f(-1)=f(1)$. Another weak solution is the rarefaction waveThe fan solves the equation inside each smooth region, is continuous at its edges, and has the same local $L^1$ initial trace. Thus weak solutions are not unique. The stationary expansion shock violates the entropy inequality, whereas the fan is admissible; using the unmodified quadratic flux alone would miss the printed global bounded-derivative hypothesis.

The undisturbed gas has $u=0$, $c=c_0$, and hence $R_{-}=0$. The disturbance from the receding piston is a right-moving simple wave, so this invariant remains zero throughout it:

Let $s$ be the time at which the relevant $C_{+}$ characteristic leaves the piston. The boundary condition is $u={\dot{X}}_p(s)$ there. Both $u$ and $c$ are constant along this characteristic, whose speed isHence $s=s(x,t)$ is determined implicitly by the receding-piston rarefaction wave equationWithin $X_p(t)<x<c_{0}t$, the solution is thereforeFor $x\ge c_{0}t$, the gas remains undisturbed: $u=0$ and $\rho ={\rho }_0$. The speed bound keeps $c$ and $\rho$ positive, and monotonic increase of $|{\dot{X}}_p|$ makes this an expanding rarefaction wave rather than a characteristic-crossing compression.

Now the mass per unit radius satisfies the nonlinear scalar conservation lawIts characteristic curves have speedBecause $v_m<0$ and $|q^{\prime }|$ increases with $M_r$, the high central part of a localized pulse travels inward faster than its low-density edges. On the inner, rising side, faster high-density characteristics catch slower low-density ones, causing characteristic crossing and an inward-facing shock wave. On the outer, falling side, the characteristics separate and form a rarefaction wave. The initial bump therefore develops a sharp inner front and a broad outer tail while moving toward the star.

For downward settling speed $u_s(1-\varphi )$ in upward coordinate $z$, particle flux is $j=-u_s\varphi (1-\varphi )$. The concentration-characteristic speed is $j^{\prime }=u_s(2\varphi -1)$ and differs from particle speed. Since $j^{\prime \prime }=2u_s>0$, decreasing concentration with height creates a compressive shock, while increasing concentration creates a rarefaction wave.

A Riemann problem for a one-dimensional system of conservation laws ${\partial }_{t}U+{\partial }_{x}f(U)=0$ has two constant initial states, $U_L$ for $x<0$ and $U_R$ for $x>0$. The data and equation are invariant under $(x,t)\mapsto (\lambda x,\lambda t)$. When the physically admissible entropy solution is unique, this invariance makes it a self-similar solution depending on $x/t$. For ideal-gas compressible flow, the characteristic speeds are $u-c,u,u+c$. The two acoustic families carry shock waves or rarefaction waves, while the middle family carries a contact discontinuity. Sufficiently strong separating velocities can instead create a vacuum region.

Two positive uniform ideal-gas states can be joined by one nontrivial rarefaction wave precisely when their specific entropy agrees, $p_L/{\rho }_L^{\gamma }=p_R/{\rho }_R^{\gamma }$, and for one $\sigma \in \{\pm 1\}$ they satisfy $u_R-u_L=2\sigma (c_R-c_L)/(\gamma -1)$ with $\sigma (c_R-c_L)>0$. These equations are the constant-entropy and Riemann invariant conditions of the self-similar ideal-gas rarefaction. They are sufficient because its affine $u,c$ profiles match the two states at ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and ${\xi }_R>{\xi }_L$. The left family has $c_R<c_L$, while the right family has $c_R>c_L$. The inequality $u_R>u_L$ alone does not impose the two matching equalities.

For a flux that is a differentiable convex function with unique minimum at $s_{\ast }$, define $f^{+}(u)=f(u)-f(s_{\ast })$ for $u>s_{\ast }$ and zero otherwise, and $f^{-}(u)=f(u)-f(s_{\ast })$ for $u<s_{\ast }$ and zero otherwise. The Engquist-Osher flux is $F(a,b)=f(s_{\ast })+f^{+}(a)+f^{-}(b)$. Thus transsonic rarefaction wave states $a\le s_{\ast }\le b$ give $F=f(s_{\ast })$, while reversed transsonic states give $F=f(a)+f(b)-f(s_{\ast })$. This follows by integrating the positive and negative parts of $f^{\prime }$.

In a prismatic triangular channel with wetted area proportional to $h^2$, the Riemann invariants are $u\pm 4c$, with $c^2=gh/2$. Releasing water initially at rest into a dry channel produces a rarefaction wave with $u+4c=4c_0$ and $x/t=u-c$. Hence $c=(4c_0-x/t)/5$, $u=4(c_0+x/t)/5$ and the displayed depth profile for $-c_{0}t<x<4c_{0}t$. The dry front advances at $4c_0$ before finite-reservoir or varying-channel effects become important.

For positive diffusivity, the Cole-Hopf transformation with $f=2\alpha {\partial }_{\theta }\log \psi$ converts the negative-flux equation into the heat equation. Step data give a ratio of two Gaussian-tail integrals $J$. At $\theta =-UZ/2$, $J=1$ and $f=U/2$. Its vanishing-viscosity limit selects the shock or rarefaction wave in the Burgers Riemann problem with negative flux, depending on the sign of $U$.

A travelling wave of the viscous Burgers equation can connect constant states $u_L>u_R$ through a smooth decreasing layer. Require $u(x,t)=U(z)$, $z=x-st$, $U(-\infty )=u_L$, $U(+\infty )=u_R$, and $U^{\prime }\to 0$ at both ends. Substitution and one integration giveThe speed follows by evaluating the integration constant at both end states. It is precisely the Rankine-Hugoniot condition for conservation law flux $f(u)=u^2/2$. Choosing the center $U(0)=s$ gives the displayed hyperbolic tangent profile. Its characteristic width scales as $\nu /(u_L-u_R)$; a smaller positive $\nu$ makes a thinner smooth layer.

For an original worked example in dimensionless quantities, take $u_L=2$, $u_R=0$ and $s=1$:This is an exact classical solution on $\mathbb{R}$ with its own smooth initial condition $u(x,0)=1-\tanh (x/(2\nu ))$ and limiting states $2,0$. It is a family of moving profiles, not the finite-$\nu$ solution starting from an exact discontinuous step. For that different initial condition, the Cole-Hopf transformation supplies the evolution. At $\nu =1/2$, this front starts from the same $1-\tanh x$ profile as the Burgers characteristic breaking example: positive diffusion maintains a smooth moving layer while the inviscid profile breaks at $t=1$.

To check the partial differential equation explicitly, set $w=(x-t)/(2\nu )$ and $S=1-{\tanh }^{2}w$. ThenThe independent Rankine-Hugoniot condition check gives $s=(0-2)/(0-2)=1$, using $f(2)=2$ and $f(0)=0$. At fixed $x\ne t$, the vanishing viscosity approximation tends to the entropy shock $2$ for $x<t$ and $0$ for $x>t$. At the center the smooth profiles stay at $1$; changing the value on the discontinuity does not change the limiting weak solution.

For $u_L<u_R$, the same integrated ordinary differential equation has $U^{\prime }<0$ between the two states when $\nu >0$, so it cannot supply an increasing connecting front. The inviscid admissible expansion is the rarefaction wave of the Burgers Riemann problem with positive flux.