Prismatic-channel shallow water equations ($A_t+(Au{)}_x=0,(Au{)}_t+(A{u}^2+P{)}_x=0$)

In a channel whose transverse width $b(z)$ does not depend on downstream position, the wetted area is $A(h)={\int }_0^{h}b(z)dz$ and the integrated hydrostatic pressure force divided by density is $P(h)=g{\int }_0^h(h-z)b(z)dz$. With cross-sectionally uniform velocity, volume conservation and momentum conservation give the displayed shallow water equations. Since $P^{\prime }(h)=gA(h)$, the two characteristic speeds are $u\pm c$ with $c^2=gA/A^{\prime }$. For $b(z)=C{z}^p$, $c^2=gh/(p+1)$ and the Riemann invariants are $u\pm 2(p+1)c$.

A channel with width $b(z)=\alpha z^2$ has wetted area $A=\alpha h^3/3$. Its inviscid shallow water equations have characteristic speeds $u\pm c$ and the displayed Riemann invariants. Releasing a reservoir of depth $H$ at rest into a dry channel gives a left-family rarefaction wave: $c=(6c_0-x/t)/7$, $u=6(c_0+x/t)/7$, $h/H=[(6-x/(c_{0}t))/7{]}^2$ between $x=-c_{0}t$ and $x=6c_{0}t$, where $c_0=\sqrt{gH/3}$. This width law differs from a geometrically parabolic trough, whose width is proportional to $\sqrt{z}$.

An impermeable wall at $x=-\ell$ first receives the upstream-moving head at $t_1$. The leading downstream characteristic curve of the reflected adjustment obeys $\dot{x}=u+c=(12c_0+5x/t)/7$ in the unmodified rarefaction wave, giving the displayed curve. It stays behind the inviscid dry tip $x_f=6c_{0}t$ at every finite time. The finite reservoir drains and its upstream depth decreases, while only the region behind $x_r$ has yet received the wall signal. Bottom drag or other tip physics can change the front law.

In a prismatic triangular channel with wetted area proportional to $h^2$, the Riemann invariants are $u\pm 4c$, with $c^2=gh/2$. Releasing water initially at rest into a dry channel produces a rarefaction wave with $u+4c=4c_0$ and $x/t=u-c$. Hence $c=(4c_0-x/t)/5$, $u=4(c_0+x/t)/5$ and the displayed depth profile for $-c_{0}t<x<4c_{0}t$. The dry front advances at $4c_0$ before finite-reservoir or varying-channel effects become important.

For water initially at rest at depth $H$ on one side of a dam and a dry bed on the other, a channel with $b(z)=C\sqrt{z}$ has $c_0=\sqrt{2gH/3}$. The centered rarefaction wave has $u+3c=3c_0$ and $x/t=u-c$. Thus $c=(3c_0-x/t)/4$ and $u=3(c_0+x/t)/4$ within $-c_{0}t<x<3c_{0}t$. The reservoir remains undisturbed to the left and the bed dry to the right. This is an outer shallow-water approximation before finite-reservoir boundary signals return.