= Parabolic-channel dry-bed dam break
{title2=$h/H=(3-x/(c_0t))^2/16$}
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 $-c_0t<x<3c_0t$. 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.
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