Across a stationary <normal shock wave>, the <Rankine-Hugoniot conditions for a perfect gas> conserve
$$
j=\rho_1u_1=\rho_2u_2,\qquad
p_1+\rho_1u_1^2=p_2+\rho_2u_2^2,
\qquad
C_B=\frac{u^2}{2}+\frac{c_s^2}{\gamma-1}.
$$
Momentum conservation and $1/\rho=u/j$ give
$$
p_2-p_1=j(u_1-u_2).
$$
Direct elimination of the two pressures and densities from these three jump conditions gives
$$
u_1u_2=\frac{2(\gamma-1)}{\gamma+1}C_B.
$$
Part c identifies the right side with the square of the critical speed, so the <Prandtl shock relation> is
$$
\boxed{u_1u_2=c_{\rm cr}^2}.
$$
It maps the unique upstream supersonic state on a given Bernoulli streamline to its downstream subsonic state.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 2 / d / Solution

{c}
{title2=$u_1u_2=c_{\rm cr}^2$}

For a stationary normal shock in a polytropic perfect gas, the upstream and downstream normal speeds obey $u_1u_2=c_{\rm cr}^2$, where $c_{\rm cr}$ is the sonic speed on their common Bernoulli streamline.


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