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Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.

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.

Use the jump convention $[f]=f_1-f_2$, consistent with region 1 occupying $x>Vt$. The one-dimensional Lighthill stress tensor is $T=\rho u^2+p-c_0^2\rho$, so $T=T_2+[T]H(x-Vt)$. Both bulk stresses are constant. The distributional jump formula for a Heaviside product therefore givesThe Rankine-Hugoniot condition for conservation of mass is $[\rho u]=V[\rho ]$, while conservation of momentum gives $[\rho u^2+p]=V[\rho u]$. Hence a physical moving shock wave also satisfiesThis sign can be checked directly. Subtracting a constant reference density does not affect derivatives, and ${\rho }^{\prime }=\text{constant}+[\rho ]H(x-Vt)$. Thus ${\rho }_{tt}^{\prime }=V^2[\rho ]{\delta }^{\prime }(x-Vt)$ and ${\rho }_{xx}^{\prime }=[\rho ]{\delta }^{\prime }(x-Vt)$, exactly reproducing the coefficient above.

Let $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$, so the Lighthill stress tensor is $T_{ij}^{-}+D_{ij}H(S)$. Write $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. The distributional derivative of the Heaviside step function gives ${\partial }_{i}H(S)=S_i\delta (S)$. Applying the product rule a second time yields the surface sources of a discontinuous Lighthill stress tensor:Repeated indices are summed. The first line gives the two bulk acoustic quadrupole distributions. The second line consists of additional surface delta distributions and their derivatives, all supported on the shock wave $S=0$. For a symmetric Lighthill stress tensor, the first two coefficients of $\delta (S)$ combine to $2D_{ij,i}{S}_j$ after relabelling indices.

A discontinuity therefore adds both a surface-delta source and a surface-delta-derivative source. The level function need not be a signed distance: retaining its derivatives makes the expression invariant under a smooth reparametrization of the same surface. If the two bulk fields are only continuously differentiable, their second derivatives in the first line are read weakly; no classical second derivative is being assumed. The coefficients multiplying ${\delta }^{\prime }(S)$ are kept as extensions, not replaced prematurely by traces.

In the energy-conserving blast regime, neglect ambient pressure, radiative losses, ejecta mass and gravity compared with the explosion. The cold ambient ideal gas has no sound speed scale, and its power-law mass density has no preferred radius. The point explosion introduces only its energy $E$ and elapsed time $t$. This scale-free setting supports a self-similar blast wave; extra physics would limit the time interval over which that description applies.

The mass swept up to radius $R$ isIt is finite because $\beta <3$. With no energy loss, the kinetic energy and internal energy both scale as $C{R}^{3-\beta }{\dot{R}}^2$. Therefore $R^{3-\beta }{\dot{R}}^2$ is constant and integration givesThe dimensionless coefficient $A$ depends on the adiabatic exponent and ambient exponent through the similarity profiles. Indeed, $[E/C]=L^{5-\beta }{T}^{-2}$, which independently verifies the scaling. The shock wave decelerates for $0<\beta <3$, since its time exponent lies between $2/5$ and $1$.

Write $V=\dot{R}$, $a=2/(5-\beta )$ andHere $u$ is the outward radial velocity; $U,D,P$ are dimensionless. The relationsare sufficient for the reduction. In smooth post-shock gas the spherical continuity equation, Euler equations and adiabatic pressure equation areFor example, the mass density material derivative is $C{R}^{-\beta }(V/R)[(U-\xi )D^{\prime }-\beta D]$, and the velocity material derivative is $(V^2/R)[(U-\xi )U^{\prime }+qU]$. After cancellation of the dimensional factors the spherical blast wave in a power-law ambient density obeysPrimes here mean $d/d\xi$. In the pressure equation the scale derivative $2q-\beta =-3$ explains the constant term. Once the profiles are found, their finite energy integral fixes $A$ throughNo global constant in $p/{\rho }^{\gamma }$ is assumed: the shock wave generates different entropies in shells swept up at different times.

Assume $\gamma >1$. The upstream ideal gas is at rest in the laboratory frame and has mass density ${\rho }_1=C{R}^{-\beta }$. In the shock frame its incident speed is $V$. The strong-shock limit of the Rankine-Hugoniot conditions for a perfect gas gives compression $\chi =(\gamma +1)/(\gamma -1)$ and downstream shock-frame speed $V/\chi$. Transforming back to the laboratory frame gives $u_2=V(1-1/\chi )=2V/(\gamma +1)$.

The normal momentum jump yieldswhere the cold upstream pressure has been neglected. Thus the boundary values in the normalization of part (b) areThe velocity jump uses the relative speed $V-u_2$, not the laboratory speed $u_2$. This distinction is essential to all three shock wave values.

Put $U=b\xi$ and $D=d\xi$. The shock wave conditions fix $b=2/(\gamma +1)$ and $d=(\gamma +1)/(\gamma -1)$. The continuity equation then reduces toso a linear mass density profile requiresMomentum requires $P^{\prime }=-db(b-1+q){\xi }^2$. At this exponent $q=-2(\gamma -1)/(\gamma +1)$, hence integration gives $P=2{\xi }^3/(\gamma +1)+P_0$. Substitution in the pressure equation makes the cubic coefficient vanish because $(\gamma +1)b=2$, while its constant term requires $(3\gamma b-3)P_0=0$. For $\gamma >1$ this implies $P_0=0$. Therefore all three equations and all shock wave conditions hold for the linear-profile spherical blast wave,The prescribed range $0<\beta <3$ restricts this family to $1<\gamma <7$.

For $\gamma =5/3$ one has $\beta =2$, $a=2/3$, $U=3\xi /4$, $D=4\xi$ and $P=3{\xi }^3/4$. The energy integral isConsequently $E=(3\pi /2)CR{\dot{R}}^2=(2\pi /3)C{R}^3/t^2$, which fixes the radius and dimensional fields completely:The kinetic energy and internal energy integrands are each $9{\xi }^5/8$, so each accounts for half the explosion energy. Integrating the mass density also gives the correct swept-up mass $4\pi CR$.

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.

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 $ϵ/U$.

In the shock rest frame the upstream and downstream speed magnitudes are $U$ and $U-V$. Denote the downstream density by ${\rho }_2$, and write $v_i=1/{\rho }_i$ for specific volume. The Rankine-Hugoniot conditions for a perfect gas express conservation of mass, momentum and total energy:These follow by balancing the respective fluxes across a stationary infinitesimal control volume around the shock wave. The mass relation also gives $V=j(v_1-v_2)$, and momentum gives $j^2(v_1-v_2)=p_2-p_1$.

Eliminate the kinetic-energy difference in the energy condition using momentum:Substitute the given perfect-gas internal energy and collect the coefficients of $v_1,v_2$:HenceFinally $V^2=j^2(v_1-v_2{)}^2=(p_2-p_1)(v_1-v_2)$ givesFor the advancing piston choose the compressive branch $p_2>p_1$, $U>V>0$; the algebraic squared relation alone does not specify that branch.

Let $e=\rho u^2/2+p/(\gamma -1)$ be the total gas energy density, with $\gamma >1$. The one-dimensional continuity and momentum equations areUsing continuity to put the momentum equation into conservative form givesThe internal energy density $e_{\mathrm{int}}=p/(\gamma -1)$ satisfiesby the adiabatic pressure equation. Multiplying the momentum equation by $u$ and using continuity givesAdding these equations combines the pressure work into $-{\partial }_z(pu)$. The conservation law variables and conservation law fluxes are thereforeThe three rows express conservation of mass, momentum and total energy. Their integral conservation laws remain meaningful across a shock wave, where the differential pressure and velocity equations cannot be applied pointwise.

Integrating each conservation law across a vanishingly thin interval around the stationary shock wave leaves equal conservation law fluxes on its two sides. HenceThese are the Rankine-Hugoniot conditions for a perfect gas. Signed velocities may both be negative when the material travels from positive to negative $z$; no sign change is needed in the conservation law fluxes.

Put $r={\rho }_2/{\rho }_1=u_1/u_2$, $P=p_2/p_1$ and $m=M_1^2={\rho }_1{u}_1^2/(\gamma p_1)$. Momentum conservation givesDivide the energy condition by the nonzero mass flux. Equality of kinetic energy plus specific enthalpy givesSubstituting $P$ and multiplying by $2(\gamma -1)r^2/\gamma$ yieldsThe factor $r-1$ is the continuous, no-shock solution. On the nontrivial normal shock wave branch,An admissible compressive gas shock wave has $M_1>1$, so $r>1$ and $P>1$. The algebraic jump equations alone also allow a reversed expansive discontinuity; the entropy production in a perfect-gas shock excludes that branch. At $M_1=1$ the nontrivial branch joins the continuous solution.

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.

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.

For a discontinuity with left and right values $f_{-}$ and $f_{+}$, the Rankine-Hugoniot condition follows by integrating conservation across a moving small interval:For a nonzero jump, factor the difference of squares to obtainThe negative flux makes increasing jumps compressive. Decreasing jumps spread into rarefaction waves; they cannot be retained as nonphysical expansion shock waves.