= Solution
Use the explicit printed width law $b(z)=\alpha z^2$. Its wetted area and hydrostatic pressure integral are
$$
A(h)=\int_0^h b(z)dz=\frac{\alpha h^3}{3},\qquad
\mathcal P(h)=g\int_0^h(h-z)b(z)dz=\frac{g\alpha h^4}{12}.
$$
Assuming hydrostatic pressure, a horizontal prismatic bed and cross-sectionally uniform velocity, <volume conservation> and <momentum conservation> give the <prismatic-channel shallow water equations>
$$
A_t+(Au)_x=0,\qquad (Au)_t+(Au^2+\mathcal P)_x=0.
$$
Equivalently,
$$
h_t+uh_x+\frac h3u_x=0,\qquad u_t+uu_x+gh_x=0.
$$
The coefficient matrix in variables $(h,u)$ is $\begin{pmatrix}u&h/3\\g&u\end{pmatrix}$. Its eigenvalues and the corresponding gravity-wave speed are
$$
\boxed{\lambda_\pm=\frac{dx}{dt}=u\pm c,\qquad c=\sqrt{\frac{gh}{3}}.}
$$
For $h>0$ these eigenvalues are distinct and real, establishing strict hyperbolicity in the <hyperbolic partial differential equation> sense. Since $dc/dh=c/(2h)$, substitution in the two equations shows
$$
\boxed{J_\pm=u\pm6c\text{ is constant along }\frac{dx}{dt}=u\pm c.}
$$
For example $(\partial_t+(u+c)\partial_x)(u+6c)=0$. This provides both the slopes and the conserved characteristic quantities. At a dry bed $h=0$, strict hyperbolicity degenerates and the front is interpreted as the limit from positive depth.
The descriptive word “parabolic” alone would normally suggest $b\propto\sqrt z$ and different coefficients. The explicit $z^2$ width in the PDF determines the calculation here; it must not be silently replaced by that other geometry.
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