= Solution
Integrating each <conservation law> across a vanishingly thin interval around the stationary <shock wave> leaves equal <conservation law fluxes> on its two sides. Hence
$$
\boxed{\rho_2u_2=\rho_1u_1,\quad
p_2+\rho_2u_2^2=p_1+\rho_1u_1^2,\quad
u_2\left(\frac{\rho_2u_2^2}{2}+\frac{\gamma p_2}{\gamma-1}\right)=u_1\left(\frac{\rho_1u_1^2}{2}+\frac{\gamma p_1}{\gamma-1}\right).}
$$
These 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 $m=M_1^2=\rho_1u_1^2/(\gamma p_1)$. Momentum conservation gives
$$
P=1+\gamma m\left(1-\frac1r\right).
$$
Divide the energy condition by the nonzero mass flux. Equality of kinetic energy plus <specific enthalpy> gives
$$
\frac{\gamma m}{2}+\frac{\gamma}{\gamma-1}
=\frac{\gamma m}{2r^2}+\frac{\gamma P}{(\gamma-1)r}.
$$
Substituting $P$ and multiplying by $2(\gamma-1)r^2/\gamma$ yields
$$
(r-1)\left[\bigl((\gamma-1)m+2\bigr)r-(\gamma+1)m\right]=0.
$$
The factor $r-1$ is the continuous, no-shock solution. On the nontrivial <normal shock wave> branch,
$$
\boxed{\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2},\qquad
\frac{p_2}{p_1}=\frac{2\gamma M_1^2-(\gamma-1)}{\gamma+1}.}
$$
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.
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