 Solution

Use the jump convention $[f]=f_1-f_2$, consistent with region 1 occupying $x>Vt$. The one-dimensional Lighthill stress tensor is $T=\rho u^2+p-c_0^2\rho$, so $T=T_2+[T]H(x-Vt)$. Both bulk stresses are constant. The distributional jump formula for a Heaviside product therefore givesThe Rankine-Hugoniot condition for conservation of mass is $[\rho u]=V[\rho ]$, while conservation of momentum gives $[\rho u^2+p]=V[\rho u]$. Hence a physical moving shock wave also satisfiesThis sign can be checked directly. Subtracting a constant reference density does not affect derivatives, and ${\rho }^{\prime }=\text{constant}+[\rho ]H(x-Vt)$. Thus ${\rho }_{tt}^{\prime }=V^2[\rho ]{\delta }^{\prime }(x-Vt)$ and ${\rho }_{xx}^{\prime }=[\rho ]{\delta }^{\prime }(x-Vt)$, exactly reproducing the coefficient above.