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Put $c_0=\sqrt{g^{\prime }{h}_0/2}$. The undisturbed reservoir has $u=0$ and Riemann invariants $u\pm 4c=\pm 4c_0$. The right-going characteristic curves supply the invariant $u+4c=4c_0$ throughout the release. The spreading disturbance is the other family, whose speed is $u-c$. In its rarefaction wave let $\xi =x/t=u-c$. ThenAt the reservoir edge $u=0,c=c_0$, so the back of the rarefaction wave travels at $-c_0$. At the dry edge $c=0$ the Saint-Venant dry-front condition givesThus the fan occupies $-c_{0}t\le x\le 4c_{0}t$, while $x<-c_{0}t$ remains undisturbed. Its full length grows at $5c_0$; there is no finite-depth uniform shelf between the fan and the dry front. The optional interior depth is $h=h_0[(4c_0-x/t)/(5c_0){]}^2$.

The fan's minus-family characteristic curves are rays $x=\xi t$. For the plus family in the fan, $dx/dt=u+c=(8c_0+3x/t)/5$; integration gives $x=4c_{0}t+C{t}^{3/5}$, with constants selected at entry from the reservoir. These curved paths transport the constant plus Riemann invariant.

Put the broken dam at $x=0$ and let $c_0=\sqrt{gH/3}$. A dry bed is assumed downstream. The left-family rarefaction wave joins the undisturbed reservoir $h=H,u=0$ to the dry front. The other Riemann invariant is fixed at $u+6c=6c_0$. Within the fan its characteristic speed equals $\xi =x/t=u-c$, soThus the dry-bed dam break with cubic wetted area has the fieldsVelocity is not a meaningful fluid quantity in the dry region; its wet-side limiting value is $u_f=6c_0=2\sqrt{3gH}$. The upstream head travels at $-c_0$, while the front travels at the stated constant speed in this inviscid model.

Bottom drag slows the flow and changes the dry-tip structure, where the depth is small and bed stress per unit fluid mass becomes especially important. It can replace the inviscid zero-depth high-speed tail by a drag-controlled nose. An exact altered profile or front speed requires a specified drag law; the inviscid speed should not be extrapolated to a thin real irrigation-channel front without that qualification.

Before surface-wave breaking or reflected waves return, the shallow water equations are $h_t+(hu{)}_x=0$ and $u_t+u{u}_x+g{h}_x=0$. Set $c=\sqrt{gh}$ and $c_0=\sqrt{gH}$. Their Riemann invariants are $u\pm 2c$ along $dx/dt=u\pm c$. An outgoing simple wave entering resting water has $u-2c=-2c_0$, so the imposed boundary velocity givesThis formula requires $c_0+u/2>0$ and an appropriate inflow/outflow characteristic curve count. In particular $U<2c_0/3$ ensures that only the outgoing family enters the domain throughout the cycle; stronger forcing can change the boundary problem.

A characteristic curve emitted at time $s$ carries constant $u=U\sin \omega s$ and $h=h(0,s)$ and followsIncreasing boundary velocity produces faster following characteristic curves, steepening the advancing side of a crest; decreasing boundary velocity spreads the curves and produces a rarefaction wave. A trough has reduced depth and smaller propagation speed. Once the compressive curves intersect, the single-valued simple wave must be replaced by an admissible hydraulic bore, with mass conservation and momentum conservation across it and loss of mechanical energy. In a very weak or dispersive wavetrain an undular bore can replace the ideal discontinuity.

For a bore of speed $s_b$ entering undisturbed depth $H$, the Rankine-Hugoniot condition gives $s_b[h]=[hu]$ and $s_b[hu]=[h{u}^2+g{h}^2/2]$. If $h_2>H$ is its depth just behind, elimination of the behind-bore velocity yields $s_b^2=g{h}_2(h_2+H)/(2H)$ and $u_2=s_b(h_2-H)/h_2$. Thus the newly formed front travels faster than $c_0$, carries a depth jump, and dissipates energy; the behind-bore state must satisfy these balances rather than continuing intersecting smooth curves. The final sketch is schematic and does not claim a solved post-breaking boundary-value problem.

There is a numerical defect in the printed leading-edge claim. The first leading edge is the $s=0$ curve $x=c_{0}t$. The envelope condition ${\partial }_{s}x=0$ givesAt the leading edge,This follows directly from the specified boundary acceleration and the exact Riemann invariant. The printed coefficient $4/9$ is incompatible with these data: at that distance the leading-edge characteristic curves have not yet intersected. For the first compression interval, $0<\omega s<\pi /2$, one has $v^{\prime \prime }<0$. Consequently $d(s+v/v^{\prime })/ds=2-v{v}^{\prime \prime }/v^{\prime 2}>0$, so the first envelope occurs at $s=0$. This boundary-driven shallow-water wave steepening calculation also establishes that no later characteristic in the initial compression forms a bore earlier.

Without settling, $b=b_0=G{\varphi }_0$ and $c_0=\sqrt{b_{0}H}$. The left-running rarefaction wave has the unchanged Riemann invariant $u+2c=2c_0$ and similarity coordinate $\xi =x/t=u-c$. HenceMatching to the Froude number condition $u_f=F{c}_f$ givesThe full outer profiles are $(h,u)=(H,0)$ for $x/t<-c_0$; the displayed fan for $-c_0<x/t<{\xi }_s$; and $(h,u)=(h_f,u_f)$ for ${\xi }_s<x/t<u_f$, whereAhead of $x_f$ there is only ambient fluid, so the particle-laden layer has zero depth. The shelf width is $(u_f-{\xi }_s)t=c_{f}t>0$ for every finite positive $F$. It disappears only in the formal $F\to \infty$ dry-front limit, where $u_f\to 2c_0$. For $F<1$, the fan/shelf junction lies to the left of the initial barrier; the profiles remain consistent. This is the finite-Froude lock-release rarefaction, valid while the reservoir can be treated as semi-infinite.

The undisturbed reservoir remains at $h=H$, $u=0$, while the particle budget gives $\varphi ={\varphi }_0{e}^{-V_{s}t/H}$. The edge of the rarefaction wave advancing into that reservoir travels with the local left-running characteristic speed, $-\sqrt{b_{0}H}{e}^{-V_{s}t/(2H)}$. ThusThis is the disturbance head into the initially resting reservoir, also called the leading edge of the rarefaction wave; it reduces to $-c_{0}t$ as $V_s\to 0$. The rightmost fan/shelf junction is the rarefaction tail, not this reservoir edge. Its trajectory in the full sedimenting system cannot be obtained by retaining the constant-buoyancy Riemann invariants.

For the requested near-front approximation, hold $h_f=4H/(F+2{)}^2$ fixed and take the shelf's concentration initially to be ${\varphi }_0$. The locally uniform shelf approximation gives $b_f=b_0{e}^{-V_{s}t/h_f}$, soThis tends to $u_{f}t$ at small settling and predicts a finite limiting front location within the fixed-depth closure. It is not an exact solution of the spatially varying variable-buoyancy shallow water equations: different material histories and resulting buoyancy gradients eventually invalidate a perfectly uniform shelf. No exact finite-time stoppage is claimed.

The Engquist-Osher method is an upwind finite volume method. For cell values $U_j^n$, mesh width $h$ and time step $k$, put $r=k/h$ and use the conservative updatewhere the numerical flux isConsistency of a numerical method follows from $F(u,u)=f(u)$. Positive characteristic speeds use the left state, and negative speeds use the right state. The flux difference telescopes when summed over cells, giving a discrete conservation law. The method is first order in space and time for smooth solutions; its conservation law form is also appropriate when shock waves develop.

Let $s_{\ast }$ be the unique sonic point of a scalar flux, $f^{\prime }(s_{\ast })=0$. Convexity implies $f^{\prime }<0$ to its left and $f^{\prime }>0$ to its right. Put $f_{\ast }=f(s_{\ast })$ and splitThen $f=f_{\ast }+f^{+}+f^{-}$, $(f^{+}{)}^{\prime }\ge 0$, $(f^{-}{)}^{\prime }\le 0$, andThis is the sonic-point splitting of a convex Engquist-Osher flux. If both states are above $s_{\ast }$, $F=f(a)$; if both are below it, $F=f(b)$. Across a transsonic rarefaction wave, $a\le s_{\ast }\le b$, the flux is the minimum $f_{\ast }$. In the opposite crossing it is $f(a)+f(b)-f_{\ast }$, which in general differs from the exact Godunov numerical flux. This distinction does not affect the monotonicity proof.

Here stability requires the usual Courant–Friedrichs–Lewy condition. Let the initial values lie in $[m,M]$, and chooseA differentiable convex function has a continuous derivative on this interval, so the maximum is finite. Write the three-input update as $H(a,b,c)$. Differentiating givesThus the update $T$ preserves componentwise order. Constants are fixed, since $H(c,c,c)=c$, so comparison with the constants $m,M$ proves $m\le U_j^n\le M$ for every time level. In particular the same CFL condition remains applicable. This is a mesh-independent maximum-principle stability bound.

A stronger stability statement is contraction in discrete $L^1$. Consider two solutions $U,V$ in the same invariant interval, on a periodic mesh or on the infinite lattice with summable differences. Let $Q=U\vee V$ be their componentwise maximum. Monotonicity gives $TQ\ge TU,TV$, henceThe discrete conservation law for the difference gives the equality: on a finite periodic mesh the flux sum cancels exactly, and on the infinite lattice it follows by truncation and the fact that the flux is a Lipschitz function on $[m,M]$. Interchanging $U,V$ and adding provesThis is L1 contraction of a monotone conservative scheme. Since $T$ commutes with the cell shift, applying this contraction to $U$ and its shift gives ${\sum }_j|U_{j+1}^{n+1}-U_j^{n+1}|\le {\sum }_j|U_{j+1}^n-U_j^n|$. The scheme is therefore a total variation diminishing scheme as well.

The same order argument proves a discrete entropy inequality, explaining why this stable discretization selects entropy solutions rather than nonphysical expansion shock waves. For a constant $c\in [m,M]$, setBecause $T(U\vee c)\ge TU,c$ and $T(U\wedge c)\le TU,c$, their componentwise difference is at least $|TU-c|$. Subtracting their updates givesFor constants outside $[m,M]$, the entropy inequality is an equality by the invariant bound and the conservative update. At equal states $Q_c(u,u)=\mathrm{sgn}(u-c)[f(u)-f(c)]$, the Kruzhkov entropy flux. The flux is a Lipschitz function on the invariant interval: writing $L=\max |f^{\prime }|$, the conservative update gives $h{\sum }_j|U_j^{n+1}-U_j^n|\le 2kL{\sum }_j|U_{j+1}^n-U_j^n|$. This time-translation bound, the invariant bound and the total variation estimate give local space-time compactness for bounded-variation initial conditions as the mesh is refined; the conservative update passes to the weak conservation law, and the displayed inequality passes to its Kruzhkov entropy inequality. This identifies the limit as the entropy solution.

Finally the time-step qualification is essential. For the convex Inviscid Burgers equation flux $f(u)=u^2/2$, whose unique sonic point of a scalar flux is zero, linearize about a positive constant $U$. The perturbation update becomes $v_j^{n+1}=(1-rU)v_j^n+rU{v}_{j-1}^n$. At Fourier frequency $\pi$ its multiplier is $1-2rU$, with modulus greater than one if $rU>1$. Wave packets near this phase give linear instability even for summable Cauchy perturbations. Thus the assumptions on $f$ imply the stability results above under the CFL condition, not for arbitrary $k/h$. This is CFL necessity for explicit Engquist-Osher stability.

For isentropic flow of a perfect gas, $p=K{\rho }^{\gamma }$ with $\gamma >1$ and $c^2=dp/d\rho$. The mass and momentum equations implyAdding and subtracting the appropriate multiples therefore provesThese are the Riemann invariants for one-dimensional isentropic flow.

In the piston-driven rarefaction wave, the incoming minus invariant has its initial value zero. Thus $c=c_0+(\gamma -1)u/2$. At the piston $u_p=-ft$, giving $c_p=c_0-(\gamma -1)ft/2$. A plus characteristic emitted at time $\tau$ has constant $u=-f\tau$ and $c=c_0-(\gamma -1)f\tau /2$, so its straight trajectory isThe first characteristic is $x=c_{0}t$, ahead of which the gas is undisturbed. Later characteristics have decreasing propagation speed and spread into a rarefaction. More explicitly ${\partial }_{\tau }x=-c_0-(\gamma +1)ft/2+\gamma f\tau <0$ for $\tau \le t<t_v$, with equality only at the vacuum endpoint. They never intersect, so no compressive shock forms.

Since $p/p_0=(c/c_0{)}^{2\gamma /(\gamma -1)}$, the vacuum formation at an accelerating withdrawing piston givesAt $t_v$, density, sound speed and pressure at the gas edge vanish, and the piston velocity is $\boxed{\dot{X}(t_v)=-2c_0/(\gamma -1)}$. Its speed is the positive magnitude $2c_0/(\gamma -1)$. Afterwards the gas edge continues at this limiting velocity, while the piston keeps accelerating away; a vacuum gap of width $f(t-t_v{)}^2/2$ opens and the gas pressure on the piston stays zero.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

Let $c_L=(\gamma p_L/{\rho }_L{)}^{1/2}$ and $c_R=(\gamma p_R/{\rho }_R{)}^{1/2}$, with positive pressures and densities. A nontrivial single rarefaction wave has constant specific entropy and constant Riemann invariant, so for one family label $\sigma =\pm 1$ the necessary matching conditions areThe final inequality says that the fan opens rather than compresses. Its edge speeds are ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and their difference is $\sigma (c_R-c_L)(\gamma +1)/(\gamma -1)>0$. The profiles in part (a) then interpolate the two end states and supply a sufficient construction. For the left family, $c_R<c_L$, hence ${\rho }_R<{\rho }_L$ and $p_R<p_L$; for the right family all three inequalities reverse. In both cases $u_R>u_L$. These are the single-rarefaction matching conditions for an ideal-gas Riemann problem.

For a concrete nontrivial example, take $\gamma =5/3$ andThe adiabatic sound speeds are $c_L=1$ and $c_R=1/2$, with common $K=3/5$ and left-family invariant $u+3c=3$. The fan edges are ${\xi }_L=-1$ and ${\xi }_R=1$. The complete self-similar solution is uniform at these stated left and right values outside the fan, while inside $-1<\xi <1$,It matches both states continuously, has $u^{\prime }=3/4=2/(\gamma +1)$, and has constant specific entropy. For fixed $x\ne 0$, its $t\downarrow 0$ limit is the required initial step.

For zero viscosity the Inviscid Burgers equation is $q_X-q{q}_{\theta }=0$. Along a characteristic curve, $dq/dX=0$ and $d\theta /dX=-q$. Labelling a characteristic by its initial position $\eta$ givesWhere $f$ is differentiable, $q_{\theta }=f^{\prime }(\eta )/(1-X{f}^{\prime }(\eta ))$. The first characteristic crossing occurs at $X=1/\max f^{\prime }$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the conservation law $q_X+{\partial }_{\theta }(-q^2/2)=0$ and replace the overturning part by an entropy-admissible shock wave. The Rankine-Hugoniot conditions give its trajectoryThe states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.

For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The Burgers Riemann problem with negative flux is solved byIt is a shock, not a rarefaction wave; the minus sign in the nonlinear flux is decisive.

For positive viscosity, substitute $q=2ϵ{\partial }_{\theta }\log \psi$ into the viscous Burgers equation. Writing $w=\log \psi$ givesAn $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The Cole-Hopf transformation therefore gives ${\psi }_X=ϵ{\psi }_{\theta \theta }$. Normalize its initial value continuously at zero:Split the heat kernel integral at zero. Completing the square in the positive-half integral yieldsWhen differentiating, the Gaussian terms from the two complementary error functions cancel, becauseThus the exact viscous Burgers step solution with negative flux isFor $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow 0$. At $\theta =-UX/2$ it equals $U/2$. Its vanishing-viscosity limit gives the entropy shock above. At large $U\sqrt{X/ϵ}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta +UX/2)/(2ϵ)}]$, with thickness of order $ϵ/U$.

A failed dam launches a rarefaction wave back into the reservoir and an advancing current into the initially dry downstream channel. Take the failed dam as $x=0$, initial depth $h_0$ and zero velocity for $x<0$, and a dry bed for $x>0$. Locally assume a flat, prismatic triangular section, $S=C{h}^2$. The early flow is governed byIts Riemann invariants are $u\pm 4c$, since $\int (g/c)dh=4c$. In the left-running simple wave, $u+4c=4c_0$, where $c_0=\sqrt{g{h}_0/2}$, and the expanding fan has $x/t=u-c$. Solving givesThus the triangular-channel dry-bed dam break profile isThe reservoir-side edge moves at $-c_0$, and the dry front moves atVelocity is zero in the undisturbed reservoir and has the fan value above where water is present. This approximation applies before the rarefaction reaches the surviving dam and before channel variation, friction or finite-volume reflections substantially modify the flow. If water is already present downstream, a bore replaces the dry-bed front and requires a different matching problem.

The exterior characteristics give $u=0$ for $x<0$ and $u=1$ for $x>t$. In the gap put $\xi =x/t$ and $u=f(\xi )$. The Inviscid Burgers equation becomes $(f-\xi )f^{\prime }=0$. Matching the increasing states selects the rarefaction wave $f(\xi )=\xi$, soIt is continuous and piecewise smooth at positive times and satisfies the entropy condition. A discontinuous shock joining the same increasing states would be nonentropic.

For the early release, approximate the intact left half by water at rest at depth $H$, and the right half by a dry bed. Put $c_0=\sqrt{2gH/3}$. The parabolic-channel dry-bed dam break is a centered rarefaction wave of the $u-c$ family, with the other Riemann invariant fixed at $u+3c=3c_0$. Inside the fan, $\xi =x/t=u-c$, soHenceThe left edge moves into the reservoir at $-c_0$ and the dry front advances at $3c_0$. At the failed dam $h=9H/16$ and $u=c=3c_0/4$. There is initially no finite-depth hydraulic bore into the dry bed: the depth falls continuously to zero.

On the $x$–$t$ diagram, the $u-c$ characteristic curves inside the fan are straight rays from the origin. The $u+c$ curves obey $dx/dt=3c_0/2+x/(2t)$, so they have form $x=3c_{0}t+C\sqrt{t}$. The solution describes the period before the finite lake ends and their reflected signals matter. Immediately after failure, near the dam and very near the thin nose, strong vertical acceleration can invalidate the shallow-water approximation. The outer fan becomes appropriate when the horizontal flow scale is large compared with its depth.

For the negative-flux Inviscid Burgers equation, the method of characteristics givesThe characteristic curve starting at ${\theta }_0$ therefore has $\theta ={\theta }_0-f_0({\theta }_0)z$, and henceFor smooth data this describes a single-valued classical solution as long as the characteristic flow map is invertible. Its Jacobian is $1-z{f}_0^{\prime }({\theta }_0)$; after characteristic crossing, one must instead select a weak solution with the appropriate entropy solution condition.

The conservation form is $f_z+G(f{)}_{\theta }=0$, with conservation law flux $G(f)=-f^2/2$. Integrating this scalar conservation law across a moving discontinuity, or differentiating its Heaviside representation as a distribution, yields the Rankine-Hugoniot conditionFor distinct one-sided limits it simplifies toThis jump-speed relation is exact for the Burgers conservation law; an additional entropy condition is needed to distinguish a physical compressive shock wave from an expansion discontinuity.

The step gives the Burgers Riemann problem with negative flux. If $U>0$, characteristic curves from the left have speed zero, while those from the right have speed $-U$: they converge. The entropy shock wave has speed $-U/2$ and thusCharacteristics enter this shock from both sides, since $0>-U/2>-U$.

If $U<0$, the right-hand speed $-U$ is positive and the two families separate. A smooth steep approximation to the initial step spreads into a rarefaction wave. In the fan, the self-similar characteristic curve relation is $-f=\theta /z$, givingThe endpoint values match continuously. The discontinuity moving at $-U/2$ would satisfy the jump condition even for $U<0$, but its characteristic curves leave the discontinuity and it fails entropy admissibility. For $U=0$ the solution is identically zero.

Set $q=\log \psi$ and use the sign-appropriate Cole-Hopf transformation $f=2\epsilon q_{\theta }$. Direct differentiation givesThus the heat equation ${\psi }_z=\epsilon {\psi }_{\theta \theta }$ implies the viscous Burgers equation with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\epsilon$ where $\psi \ne 0$.

For a forward dissipative initial-value solution, take $\epsilon >0$ and $z>0$. This is essential for the supplied heat kernel: at $\epsilon <0$, $z>0$, its real Gaussian integral diverges even for $\psi (0,\varphi )=1$. More generally that kernel requires $\epsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\epsilon \ne 0$ does not justify a forward Gaussian function formula.

Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\epsilon )$ gives a positive continuous initial factor, normalized asIts heat kernel convolution converges for both signs of $U$, since a Gaussian function dominates the one-sided exponential. DefineThe left half-line contributes $J(\theta )$; completing the square on the right half-line gives $e^{S}J(-\theta -Uz)$. ThusWhen differentiating, the two moving-endpoint Gaussian function terms cancel, becauseOnly ${\partial }_{\theta }{e}^S=(U/(2\epsilon ))e^S$ remains in the logarithmic derivative. The viscous Burgers step solution with negative flux is thereforeThe denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta =0$ are the required step.

For the spatial limits, the Gaussian function tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to +\infty$, the complementary error function asymptotic givesAt $\theta \to +\infty$, $J(\theta )e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to 0$. At $\theta \to -\infty$, use the identity above to obtainwhile $J(\theta )$ approaches its full Gaussian integral. Thusfor both signs of $U$.

Since $J$ is strictly decreasing, $\alpha =1$ holds exactly when $\theta =-\theta -Uz$. At this point $S=0$, so the midpoint symmetry of a viscous Burgers step givesMore generally the exact symmetry is $f(z,-Uz-\theta )=U-f(z,\theta )$. For $U>0$, the midpoint follows the inviscid shock trajectory. For $U<0$, it is instead the center of an expanding fan, not a shock.

The vanishing viscosity approximation makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta =-Uz/2$ both tail integrals tend to the full Gaussian integral, so $\alpha \to 1$ andThe layer has width $O(\epsilon /U)$ and tends to the entropy shock, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.

For $U<0$, inside $0<\theta <-Uz$ both Gaussian function tails have large positive lower limits. Their asymptotics giveOutside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the rarefaction wave, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding entropy solution construction, while a negative diffusivity would not furnish this dissipative selection.

Write the negative-flux Inviscid Burgers equation as $f_Z+{\partial }_{\theta }F(f)=0$, with $F(f)=-f^2/2$. Along a characteristic curve,so a characteristic starting at ${\theta }_0$ has $\theta ={\theta }_0-f_0({\theta }_0)Z$ andThis formula is a classical solution only while the method of characteristics is one-to-one; after characteristic crossing one must select an entropy solution.