 Solution

For one-dimensional isentropic compressible flow, continuity and momentum balance areSet $Q(\rho )={\int }_{{\rho }_0}^{\rho }c(s)ds/s$. Combining the equations gives the Riemann invariantsFor the power-law equation of state, $c=c_0(\rho /{\rho }_0{)}^{(\gamma -1)/2}$ and direct integration gives $Q=2(c-c_0)/(\gamma -1)$.

In the expansion into the region left of the undisturbed gas, the invariant $u-Q$ remains zero. Hence $c=c_0+(\gamma -1)u/2$. In the centred simple wave, the other characteristic speed equals $\xi =x/t=u+c$, givingWhen the gas follows the piston, $u=-V$ at the piston and $c_p=c_0-(\gamma -1)V/2$. This requires $V<2c_0/(\gamma -1)$ for positive density. The three gas regions areAt the threshold the uniform region collapses. If $V$ exceeds the threshold, a vacuum gap separates the piston from the gas, whose free edge moves at $-2c_0/(\gamma -1)$.

A particle initially at $x_0>0$ stays at rest until the fan head arrives at $t_0=x_0/c_0$. Afterwards it lies in the fan and satisfiesMultiplication by $t^{-2/(\gamma +1)}$ and integration give $x=C{t}^{2/(\gamma +1)}-2c_{0}t/(\gamma -1)$. Matching $x(t_0)=x_0$ determines $C$, yieldingSince $2/(\gamma +1)<1$, differentiating provesThus each fixed particle approaches the velocity of the vacuum edge.