 Solution

For $U<0$ the same positive algebraic root remains real and satisfies $0<V<c_0$, because $V^2-c_0^2=(\gamma +1)UV/2<0$. The jump relations then give a density decrease, ${\rho }_1/{\rho }_0=V/(V-U)<1$, and a pressure decrease. Where both states remain thermodynamically positive, this is an expansion shock: its upstream shock-frame flow is subsonic and its entropy change has the forbidden sign. At sufficiently large negative $U$ the formula even predicts nonpositive pressure, an additional reason it cannot represent a physical gas state.

The observed outgoing wave is instead a continuous isentropic rarefaction wave. Its plus-family rays spread from the initial piston corner rather than coalescing. The fan head enters the undisturbed gas at $x=c_{0}t$; for moderate recession the tail is $x=(U+c_1)t$, with the uniform gas next to the piston moving at $U$ and $c_1=c_0+(\gamma -1)U/2>0$. The appropriate Riemann invariant is $u-2c/(\gamma -1)=-2c_0/(\gamma -1)$, so inside the fan, with $\xi =x/t$,These expressions show the spreading characteristics in the right panel of the preceding figure. If $U\le -2c_0/(\gamma -1)$, the gas cannot follow the piston with positive sound speed; a vacuum gap forms, bounded by the limiting rarefaction wave edge. Thus the fan description also identifies the strong-recession limit.