= 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 become
$$
Du=-\frac{2c}{\gamma-1}c_x,\qquad Dc=-\frac{\gamma-1}{2}c u_x.
$$
Adding the appropriate $\pm c\partial_x$ terms shows
$$
\boxed{(\partial_t+(u\pm c)\partial_x)\left[u\pm\frac{2(c-c_0)}{\gamma-1}\right]=0.}
$$
These 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 $u+c=c_0+(\gamma+1)u/2$, giving
$$
\boxed{u_t+\left(c_0+\frac{\gamma+1}{2}u\right)u_x=0.}
$$
For the supplied damped equation put $b=(\gamma+1)/2$. Along $\dot x=c_0+bu$, $\dot u=-\alpha u$, so an initial label $x_0$ gives
$$
\boxed{u=v(x_0)e^{-\alpha t},\qquad
x-c_0t=x_0+\frac b\alpha(1-e^{-\alpha t})v(x_0).}
$$
This solves the initial-value problem while the label map remains invertible. Its derivative is $1+(b/\alpha)(1-e^{-\alpha t})v'(x_0)$. Nonnegative slopes cannot cause crossing; for negative slopes, the condition $\alpha>b\sup_{v'<0}|v'|$ keeps that derivative strictly positive. Hence \b[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.
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