= Solution
<Taylor expansion> gives $Q(A_{\rm eq}+a)=Q(A_{\rm eq})+cA_{\rm eq}^m a+O(a^2)$. Subtract the <equilibrium> <equation> and write $V(s)=cA_{\rm eq}^m(s)$ to obtain $a_t+(Va)_s=0$. Set $b=Va$. Since $V$ is <time> independent, $b_t+Vb_s=0$. The <travel coordinate for a conservative transport perturbation> $\xi(s)=\int_0^sV(r)^{-1}\,dr$ transforms this into $b_t+b_\xi=0$, and therefore
$$
\boxed{a(s,t)=\frac{\phi(\xi(s)-t)}{cA_{\rm eq}^m(s)}.}
$$
The conserved quantity along a <characteristic> is the <flux> perturbation $Va$, explaining the diminishing <area> <amplitude> as $V$ grows.
Differentiate the <equilibrium> relation to get $A_{\rm eq}'=M/(cA_{\rm eq}^m)=M/V$. Thus the arrival <time> is explicitly
$$
\boxed{t_l=\xi(l)=\frac{A_{\rm eq}(l)-A_{\rm eq}(0)}M
=\frac1M\left(\frac{m+1}{c}\right)^{1/(m+1)}\left[(Q_{\rm in}+Ml)^{1/(m+1)}-Q_{\rm in}^{1/(m+1)}\right].}
$$
For zero upstream <flux> the <integral> remains finite, giving the limiting formula with its second term zero. However $V(0)=0$ then makes the displayed <amplitude> formula singular at the head; a uniformly small arbitrary <area> disturbance imposed exactly there is outside this <linear stability analysis>[linearization]. A positive upstream <flux> avoids that degeneracy.
Back to article page
