= Solution
For <isentropic flow> of a perfect gas, $p=K\rho^\gamma$ with $\gamma>1$ and $c^2=dp/d\rho$. The mass and momentum equations imply
$$
c_t+u c_x+\frac{\gamma-1}{2}cu_x=0,\qquad u_t+uu_x+\frac{2c}{\gamma-1}c_x=0.
$$
Adding and subtracting the appropriate multiples therefore proves
$$
\boxed{\left[\partial_t+(u\pm c)\partial_x\right]\left[u\pm\frac{2(c-c_0)}{\gamma-1}\right]=0.}
$$
These are the <Riemann invariants for one-dimensional isentropic flow>.
In the piston-driven <rarefaction wave>, the incoming minus invariant has its initial value zero. Thus $c=c_0+(\gamma-1)u/2$. At the piston $u_p=-ft$, giving $c_p=c_0-(\gamma-1)ft/2$. A plus characteristic emitted at time $\tau$ has constant $u=-f\tau$ and $c=c_0-(\gamma-1)f\tau/2$, so its straight trajectory is
$$
x(t;\tau)=-\frac12f\tau^2+\left[c_0-\frac{\gamma+1}{2}f\tau\right](t-\tau).
$$
The first characteristic is $x=c_0t$, ahead of which the gas is undisturbed. Later characteristics have decreasing propagation speed and spread into a rarefaction. More explicitly $\partial_\tau x=-c_0-(\gamma+1)ft/2+\gamma f\tau<0$ for $\tau\le t<t_v$, with equality only at the vacuum endpoint. They never intersect, so no compressive shock forms.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-2-piston-characteristics.png]
{title=Withdrawing piston, outgoing rarefaction characteristics, incoming characteristics and the vacuum gap}
Since $p/p_0=(c/c_0)^{2\gamma/(\gamma-1)}$, the <vacuum formation at an accelerating withdrawing piston> gives
$$
\boxed{p_p(t)=p_0(1-t/t_v)^{2\gamma/(\gamma-1)}\quad(0\le t<t_v),\qquad t_v=\frac{2c_0}{f(\gamma-1)}.}
$$
At $t_v$, density, <sound speed> and pressure at the gas edge vanish, and the piston velocity is $\boxed{\dot X(t_v)=-2c_0/(\gamma-1)}$. Its speed is the positive magnitude $2c_0/(\gamma-1)$. Afterwards the gas edge continues at this limiting velocity, while the piston keeps accelerating away; a vacuum gap of width $f(t-t_v)^2/2$ opens and the gas pressure on the piston stays zero.
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