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A conservation law flux $F(U)$ transports the conserved density $U$ across a surface. In one dimension $U_t+{\partial }_{x}F(U)=0$; a moving discontinuity satisfies $s[U]=[F]$. For a stationary shock wave, each conserved flux has the same value on both sides.

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.