 Solution

Integrate any conservation equation from part (a) through a vanishingly thin control volume moving with shock speed $s$. Its singular terms require $[F]=s[Q]$. The stationary shock has $s=0$, so all five fluxes are continuous. Substitution gives the Rankine-Hugoniot conditions for a perfect gas:Let the common mass flux be $j={\rho }_1{v}_1={\rho }_2{v}_2>0$. Dividing the transverse conditions by $j$ proves $u_{y2}=u_{y1}$ and $u_{z2}=u_{z1}$. Dividing the energy condition by $j$ shows that $u^2/2+w$ is unchanged. Thus tangential kinetic energies cancel when solving for the normal shock compression. No global equality of the entropy parameter $p/{\rho }^{\gamma }$ across the shock has been assumed: physical dissipative shocks increase that parameter.