Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 39A / b

The undisturbed gas has $u=0$, $c=c_0$, and hence $R_{-}=0$. The disturbance from the receding piston is a right-moving simple wave, so this invariant remains zero throughout it:

Let $s$ be the time at which the relevant $C_{+}$ characteristic leaves the piston. The boundary condition is $u={\dot{X}}_p(s)$ there. Both $u$ and $c$ are constant along this characteristic, whose speed isHence $s=s(x,t)$ is determined implicitly by the receding-piston rarefaction wave equationWithin $X_p(t)<x<c_{0}t$, the solution is thereforeFor $x\ge c_{0}t$, the gas remains undisturbed: $u=0$ and $\rho ={\rho }_0$. The speed bound keeps $c$ and $\rho$ positive, and monotonic increase of $|{\dot{X}}_p|$ makes this an expanding rarefaction wave rather than a characteristic-crossing compression.

At $t=t_c$, the piston velocity reachesThe simple-wave relation then gives $c=0$ and $\rho =0$. For later times the formal continuation would require negative sound speed, so the gas can no longer follow the piston. By vacuum formation behind a receding piston, a vacuum region opens between $X_p(t)$ and the gas edgeThe earlier rarefaction solution remains to the right of $x_v(t)$, while $X_p(t)<x<x_v(t)$ is vacuum.

HereWrite $A=(\gamma +1)/2$. Substituting $x=(c_0-a)t$ into the $C_{+}$ characteristic equation givesFor $0<a<A$ and $0<\alpha \ll 1$, the relevant emission time is exponentially small. The terms containing an explicit factor $s$ are therefore beyond all algebraic orders, leavingThus the singular small-power piston emission time isAlthough $s\to 0$, the limiting state is nontrivial:which is the centered-rarefaction state on the ray $x/t=c_0-a$ when $\alpha =0$.