 Solution

Taylor expansion gives $Q(A_{\mathrm{eq}}+a)=Q(A_{\mathrm{eq}})+c{A}_{\mathrm{eq}}^{m}a+O(a^2)$. Subtract the equilibrium equation and write $V(s)=c{A}_{\mathrm{eq}}^m(s)$ to obtain $a_t+(Va{)}_s=0$. Set $b=Va$. Since $V$ is time independent, $b_t+V{b}_s=0$. The travel coordinate for a conservative transport perturbation $\xi (s)={\int }_0^{s}V(r{)}^{-1}dr$ transforms this into $b_t+b_{\xi }=0$, and thereforeThe conserved quantity along a characteristic is the flux perturbation $Va$, explaining the diminishing area amplitude as $V$ grows.

Differentiate the equilibrium relation to get $A_{\mathrm{eq}}^{\prime }=M/(c{A}_{\mathrm{eq}}^m)=M/V$. Thus the arrival time is explicitlyFor 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 linearization. A positive upstream flux avoids that degeneracy.