= Solution
For smooth <isentropic flow> with $p=K\rho^\gamma$, $\gamma>1$, the continuity and momentum equations give
$$
u_t+u u_x+\frac{2c}{\gamma-1}c_x=0,\qquad
c_t+u c_x+\frac{\gamma-1}{2}c u_x=0,
$$
where $c^2=dp/d\rho$. Adding and subtracting these equations shows
$$
\boxed{[\partial_t+(u\pm c)\partial_x]\left[u\pm\frac{2(c-c_0)}{\gamma-1}\right]=0.}
$$
These are the <Riemann invariants> on the two characteristic families.
Before any shock, the incoming $C_-$ characteristics originate in the initially uniform gas, so their invariant is zero. Thus $c=c_0+(\gamma-1)u/2$ throughout the disturbed <simple wave>. A $C_+$ characteristic leaving the piston at time $\tau$ has constant $u=\dot X(\tau)$ and speed $u+c=c_0+(\gamma+1)\dot X(\tau)/2$. Therefore
$$
\boxed{u(x,t)=\dot X(\tau),\qquad x=X(\tau)+\left[c_0+\frac{\gamma+1}{2}\dot X(\tau)\right](t-\tau).}
$$
This parametrization applies for $0\le\tau\le t$ in $X(t)\le x\le c_0t$ as long as the characteristics remain ordered. The gas for $x\ge c_0t$ is still at rest. The values $\tau=t$ and $\tau=0$ give the piston and leading disturbance respectively.
Characteristic crossing first occurs when the derivative of the displayed $x(\tau,t)$ with respect to $\tau$ vanishes. With $U(\tau)=\dot X(\tau)$ this gives
$$
t_s(\tau)=\tau+\frac{c_0+(\gamma-1)U(\tau)/2}{(\gamma+1)U'(\tau)/2}.
$$
For the printed $X=2c_0t^3/(3T^2)$, $U=2c_0\tau^2/T^2$, so
$$
t_s(\tau)=\frac{(3\gamma+1)\tau+T^2/\tau}{2(\gamma+1)}.
$$
Its minimum over $\tau>0$ occurs at $\tau_*=T/\sqrt{3\gamma+1}$ and satisfies $t_s>\tau_*$. Hence \b[the first shock forms at]
$$
\boxed{t_{\rm shock}=\frac{T\sqrt{3\gamma+1}}{\gamma+1}.}
$$
The characteristic formula is restricted to earlier times; crossing afterwards must be replaced by a shock satisfying the conservation laws.
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