= Solution
Treat the channel as prismatic, with horizontal bottom datum $z=0$, uniform cross-sectional velocity, and surface height $h(x,t)$. The area and depth-integrated <hydrostatic pressure> force divided by density are
$$
A(h)=\int_0^h z^{1/2}\,dz=\frac23h^{3/2},\qquad
P(h)=g\int_0^h(h-z)z^{1/2}\,dz=\frac4{15}gh^{5/2}.
$$
The <prismatic-channel shallow water equations> are
$$
A_t+(Au)_x=0,\qquad (Au)_t+(Au^2+P)_x=0.
$$
Since $P'(h)=gA(h)$, they become
$$
\boxed{h_t+uh_x+\frac23h\,u_x=0,\qquad
u_t+uu_x+gh_x=0,\qquad c^2=\frac{gA}{A'}=\frac23gh.}
$$
Their two <characteristic speeds> are $u\pm c$. Using $dc/dh=c/(2h)$ gives the <Riemann invariants>
$$
\boxed{\frac{dx}{dt}=u\pm c,\qquad \frac{d}{dt}(u\pm3c)=0.}
$$
The factor $3$ replaces the familiar $2$ for a rectangular channel. It comes from the depth-dependent width, not from a different gravitational acceleration.
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