Parabolic-channel dry-bed dam break ($h/H=(3-x/(c_{0}t){)}^2/16$)

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.