= Solution
Set $a=v_2/v_1$. Mass and normal <momentum conservation> imply
$$
\rho_2=\rho_1/a,\qquad p_2=p_1+\rho_1v_1^2(1-a).
$$
After cancelling the unchanged tangential kinetic energies, the energy condition is $v_1^2/2+\gamma p_1/[(\gamma-1)\rho_1]=a^2v_1^2/2+\gamma a p_2/[(\gamma-1)\rho_1]$. With $v_{s1}^2=\gamma p_1/\rho_1$ and normal <Mach number> $\mathcal M=v_1/v_{s1}$, rearrangement factors it as
$$
(1-a)\left[(\gamma-1)+\frac2{\mathcal M^2}-(\gamma+1)a\right]=0.
$$
The first root is the no-shock solution. The nontrivial <perfect-gas shock jump conditions> therefore give
$$
\boxed{\frac{u_{x2}}{u_{x1}}=
\frac{(\gamma-1)\mathcal M^2+2}{(\gamma+1)\mathcal M^2}.}
$$
Its <density> and <pressure> ratios are
$$
\frac{\rho_2}{\rho_1}=\frac{(\gamma+1)\mathcal M^2}{(\gamma-1)\mathcal M^2+2},\qquad
\frac{p_2}{p_1}=\frac{2\gamma\mathcal M^2-(\gamma-1)}{\gamma+1}.
$$
Physical 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 is
$$
\frac{d}{dX}\ln\frac{K_2}{K_1}
=\frac{2\gamma(\gamma-1)(X-1)^2}
{[2\gamma X-(\gamma-1)]X[(\gamma-1)X+2]}>0\quad(X\ne1),
$$
where 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,
$$
\boxed{1<\mathcal M<\infty.}
$$
The endpoint one is the zero-strength limit, not a finite shock. The downstream normal <Mach number> obeys $\mathcal M_2^2=[(\gamma-1)\mathcal M^2+2]/[2\gamma\mathcal M^2-(\gamma-1)]<1$.
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