 Solution

For a one-dimensional isentropic perfect gas, continuity and momentum are $D\rho =-\rho u_x$ and $Du=-c^2{\rho }_x/\rho$, where $D={\partial }_t+u{\partial }_x$ and $c^2=\partial P/\partial \rho$. Since $P=K{\rho }^{\gamma }$, differentiation givesThus applying $D+c{\partial }_x$ to $u+2(c-c_0)/(\gamma -1)$ cancels both terms:The complementary invariant $u-2(c-c_0)/(\gamma -1)$ is transported along $dx/dt=u-c$. In the right-running piston simple wave, those characteristics carry the undisturbed value zero, giving $c=c_0+(\gamma -1)u/2$.

A right-running characteristic emitted by the piston at time $\tau$ has $u=f\tau$ and speed $u+c=c_0+(\gamma +1)f\tau /2$, constant along that characteristic. It starts at $x=f{\tau }^2/2$, soNeighboring characteristics first intersect when $\partial x/\partial \tau =0$. This derivative is $\frac{\gamma +1}{2}ft-c_0-\gamma f\tau$, so the envelope time isFor $\gamma >1$, it is smallest at $\tau =0$. Consequently the first gradient catastrophe and ensuing shock form atThis uses the smooth isentropic simple-wave solution up to its first characteristic crossing, not beyond the newly formed shock.