I am looking for some help understanding if there is a difference in these two forms of the shallow water equations:
$\partial_t(\eta u)+\partial_x((\eta u)^2+\frac{1}{2}g\eta^2)=0$ (Momentum)
$\partial_t \eta+\partial_x(u\eta)=0$ (Area Conservation)
Simplifying the Momentum equation with the Area Conservation equation yields:
$\partial_tu+\partial_x(u^2/2+g\eta)=0$
$\partial_t \eta+\partial_x(u\eta)=0$.
where $u$ is the depth averaged horizontal velocity, $\eta$ is the depth, and $g$ is a constant.
Are the solutions to these two sets of equations the same? I am especially confused since they have different fluxes which would lead to different Rankine-Hugoniot conditions. Does anyone have a clear explanation or resource for how to think about this?
The two systems have the same differentiable solutions, as you prove by manipulating one system into the other; this argument relies on the functions being differentiable.
For solutions with shocks, they do not have the same solutions, because as you say, the jump conditions are different for the two systems.
The first system ought to be taken as the more important one, because it is intended as a model of a physical process, and the derivation uses fundamental physical reasoning, namely conservation of mass and conservation of momentum (or Newton's law, whichever you prefer to call it). The second system is less physical because there is no such thing as conservation of velocity. I prefer to think of the unknown functions in the first system as the physically conserved things, say $(\eta,m)$, depth and momentum $m=u\eta$.
By the way your first equation should have only one place where $\eta$ is squared.
Interestingly, it seems that both systems agree about the speed of a very small jump in $(\eta,u)$, approximately $\sqrt{g\eta}$. I might be wrong about this, but don't have time right now to recheck.