 Solution

For isentropic gas with $P=K{\rho }^{\gamma }$, $\gamma >1$, the sound speed obeys $c^2=P^{\prime }(\rho )$. The one-dimensional mass and momentum equations areSince $c\propto {\rho }^{(\gamma -1)/2}$, these implyAdding and subtracting yields the Riemann invariantsA simple wave has one invariant constant throughout. For a right-moving wave into a resting state with sound speed $c_0$, $R_{-}=0$, so $c=c_0+(\gamma -1)u/2$ and the right characteristic speed is $c_0+bu$, $\boxed{b=(\gamma +1)/2}$.

For the damped scalar equation, characteristics from $\xi$ satisfyAssume smooth bounded initial data with bounded derivative, and the stated maximal compressive slope exists. ThenFor the damping threshold for a simple wave, if $m={\max }_{\xi }[-u_0^{\prime }(\xi )]>0$, put $\boxed{{\alpha }_c=bm}$. For $0<\alpha <{\alpha }_c$, a gradient blow-up first occurs atFor $\alpha >{\alpha }_c$, $J$ stays uniformly positive. At $\alpha ={\alpha }_c$ it still satisfies $J\ge e^{-\alpha t}>0$ for every finite time. At a point attaining the minimum slope, $u_x=u_0^{\prime }$ for all time, rather than blowing up. Hence the precise conclusion is no finite-time shock occurs when $\alpha \ge {\alpha }_c$, not the printed strict “if and only if $\alpha >{\alpha }_c$”. Equality only makes the limiting characteristic map degenerate as $t\to \infty$. The profile $u_0(\xi )=-\sin \xi$ directly verifies this endpoint distinction. If there is no compressive slope, take ${\alpha }_c=0$; no finite-time shock occurs for positive damping. A nonattained maximal slope is interpreted with a supremum and the corresponding infimum of breakdown times. The gas interpretation also requires $c_0+(\gamma -1)u/2>0$.