 Solution

Set $a=v_2/v_1$. Mass and normal momentum conservation implyAfter cancelling the unchanged tangential kinetic energies, the energy condition is $v_1^2/2+\gamma p_1/[(\gamma -1){\rho }_1]=a^2{v}_1^2/2+\gamma a{p}_2/[(\gamma -1){\rho }_1]$. With $v_{s1}^2=\gamma p_1/{\rho }_1$ and normal Mach number $\mathcal{M}=v_1/v_{s1}$, rearrangement factors it asThe first root is the no-shock solution. The nontrivial perfect-gas shock jump conditions therefore giveIts density and pressure ratios arePhysical admissibility selects a compressive, entropy-increasing shock. This can be verified directly: with $X={\mathcal{M}}^2$, define $K_i=p_i/{\rho }_i^{\gamma }$. The derivative of $\ln (K_2/K_1)$ after inserting these ratios iswhere pressures are positive, and $K_2/K_1=1$ at $X=1$. Thus the entropy increases precisely on the branch $X>1$; the positive-pressure expansion branch $X<1$ decreases entropy. Since the normal speed is positive,The endpoint one is the zero-strength limit, not a finite shock. The downstream normal Mach number obeys ${\mathcal{M}}_2^2=[(\gamma -1){\mathcal{M}}^2+2]/[2\gamma {\mathcal{M}}^2-(\gamma -1)]<1$.