= Solution
Write $g'=g(\rho_f-\rho_a)/\rho_f>0$. The <Saint-Venant dry-front condition> describes a liquid advancing over a dry bed, or into an ambient whose inertia is negligible, as when $\rho_a/\rho_f\ll1$. The edge has $h_f=0$; its <velocity> is obtained by continuing the appropriate <shallow water> <Riemann invariant> to zero depth. There is no independent universal finite-depth head <Froude number>. In the triangular geometry a release of a resting layer gives $u_f=2\sqrt{2g'h_0}$, whereas a rectangular release gives $2\sqrt{g'h_0}$.
The <Benjamin deep-ambient front condition> includes the inertia of an ambient that must turn around a finite-depth <gravity current> head. In the ideal deep-ambient <Boussinesq approximation>, $\rho_a/\rho_f$ is close to one and
$$
\boxed{u_f=F\sqrt{g'h_f},\qquad F=\sqrt2.}
$$
Here $h_f$ is the current depth just behind its head, $u_f$ its speed relative to the far ambient, and $F$ the ideal head <Froude number>. The ambient is much deeper than $h_f$; a finite upper-layer depth changes the closure. The lower density must exceed the ambient density for a bottom current. A non-Boussinesq head requires an appropriate density-ratio correction rather than using $F=\sqrt2$ at every ratio. The Saint-Venant and Benjamin models therefore describe different front physics, even though both use the same long-wave equations in the interior.
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