 Solution

For an isentropic perfect gas, $p=K{\rho }^{\gamma }$ and $c^2=dp/d\rho$. The one-dimensional Euler equations give $D\rho =-\rho u_x$ and $Du=-c^2{\rho }_x/\rho$, where $D={\partial }_t+u{\partial }_x$. Since $dc/d\rho =(\gamma -1)c/(2\rho )$, these becomeAdding the appropriate $\pm c{\partial }_x$ terms showsThese are the Riemann invariants. A simple wave is a solution in which one invariant is spatially and temporally constant, so all local state variables depend on the other one. For a right-moving wave connected to rest, the minus invariant is zero and $c=c_0+(\gamma -1)u/2$. The plus characteristic then has speed $u+c=c_0+(\gamma +1)u/2$, givingFor the supplied damped equation put $b=(\gamma +1)/2$. Along $\dot{x}=c_0+bu$, $\dot{u}=-\alpha u$, so an initial label $x_0$ givesThis solves the initial-value problem while the label map remains invertible. Its derivative is $1+(b/\alpha )(1-e^{-\alpha t})v^{\prime }(x_0)$. Nonnegative slopes cannot cause crossing; for negative slopes, the condition $\alpha >b{\sup }_{v^{\prime }<0}|v^{\prime }|$ keeps that derivative strictly positive. Hence the stated damping condition prevents shock formation at every finite time. Equality also avoids a finite-time crossing, although the derivative may tend to zero as time tends to infinity. The strict inequality in the question is sufficient rather than necessary.