= Solution
Here
$$
X_p(s)=-\frac{s^{1+\alpha}}{1+\alpha},
\qquad
\dot X_p(s)=-s^\alpha.
$$
Write $A=(\gamma+1)/2$. Substituting $x=(c_0-a)t$ into the $C_+$ characteristic equation gives
$$
at=c_0s+At\,s^\alpha
-\left(A-\frac1{1+\alpha}\right)s^{1+\alpha}.
$$
For $0<a<A$ and $0<\alpha\ll1$, the relevant emission time is exponentially small. The terms containing an explicit factor $s$ are therefore beyond all algebraic orders, leaving
$$
s^\alpha\sim\frac aA.
$$
Thus the <singular small-power piston emission time> is
$$
\boxed{s(x,t)\sim
\left(\frac aA\right)^{1/\alpha}
=\left(\frac{2a}{\gamma+1}\right)^{1/\alpha}.}
$$
Although $s\to0$, the limiting state is nontrivial:
$$
u\sim-\frac{2a}{\gamma+1},
\qquad
c\sim c_0-\frac{\gamma-1}{\gamma+1}a,
$$
which is the centered-rarefaction state on the ray $x/t=c_0-a$ when $\alpha=0$.
Back to article page
