= Solution
Vertical hydrostatic balance gives $p=\rho g(h-z)$; depth-independent horizontal <momentum> and depth-integrated incompressibility yield the two shallow-water equations. Curling <momentum> and combining with mass conservation gives $D[(\zeta+f)/h]/Dt=0$. Dot <momentum> with $\rho h u$ and use continuity to obtain $E_t+\nabla\cdot J=0$ with $E=\rho h|u|^2/2+\rho gh^2/2$ and $J=u(E+\rho gh^2/2)$. It is kinetic plus gravitational <potential energy>. A no-normal-flow boundary kills the flux and conserves its <integral>; a localized nonuniform height has excess positive energy relative to the uniform rest state with the same mass, so it cannot decay to that state without dissipation.
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