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For a one-sided fixed-volume gravity current in a unit-width channel with a closed rear wall, use front closure $\dot{L}=F\sqrt{g^{\prime }{h}_f}$ with $0<F<2$. An exact late-time similarity solution has $u=\dot{L}x/L$, $h=h_f[1-F^2/4+(F^2/4)(x/L{)}^2]$ and $h_{f}L=V/(1-F^2/6)$. Substituting into the shallow water equations proves the radius law. The Benjamin deep-ambient front condition has $F=\sqrt{2}$. The virtual time origin $t_v$ is fixed by matching to the finite release, not by volume alone.

The volume alone does not specify the early release geometry. For a concrete sketch, use a rectangular lock initially occupying $0<x<{\ell }_0$ with depth $h_0=V/{\ell }_0$, at rest, and remove its gate at $t=0$. Let $c_0=\sqrt{g^{\prime }{h}_0}$ and impose a gravity-current front condition $u_f=F\sqrt{g^{\prime }{h}_f}$, with fixed $F$. The Benjamin deep-ambient front condition gives $F=\sqrt{2}$ for the usual deep-ambient idealization.

A rarefaction wave propagates back from the release. Before information returns from the closed wall, its invariant is $u+2c=2c_0$. With $s=(x-{\ell }_0)/t$, the fan has $u=2(c_0+s)/3$, $c=(2c_0-s)/3$. The front-adjacent shelf hasThe fan extends from $s=-c_0$ to $s=u_f-c_f$; a uniform shelf connects its tail to the finite-depth head at $x={\ell }_0+u_{f}t$. A small nonhydrostatic head displaces ambient fluid; the hydrostatic interior model does not resolve its overturning structure.

The fan head first reaches the rear wall at $t_r={\ell }_0/c_0$. The wall condition is $u=0$, and the returned positive characteristic curve begins to alter the interior. In the still-undisturbed fan it solves $dx/dt=(4c_0+(x-{\ell }_0)/t)/3$, henceIt reaches the fan's uniform shelf at $t_{\ast }=t_r[(F+2)/2{]}^{3/2}$, then propagates with speed $u_f+c_f$ to the head at $t_{\mathrm{hit}}=2t_{\ast }$. These times refer to this particular initial lock and front closure. After that first arrival, the finite supply influences the front: the constant-speed slumping stage gives way to deceleration and eventually the finite-volume inertial gravity current with $L\propto t^{2/3}$, depth proportional to $t^{-2/3}$ and speed proportional to $t^{-1/3}$. The initial fan and reflected signal are shown below; later characteristic trajectories change as the current adjusts.

For the unit-width channel put $\eta =x/L(t)$, $h=h_{f}H(\eta )$ and $u=\dot{L}U(\eta )$. Fixed volume gives $V=h_{f}L{\int }_0^{1}Hd\eta$, soThe closed rear wall has $U(0)=0$. Substituting into volume conservation gives $-H-\eta H^{\prime }+(HU{)}^{\prime }=0$, or $[H(U-\eta ){]}^{\prime }=0$. The wall sets this constant to zero. Wherever $H>0$, $U(\eta )=\eta$. Thus the velocity is $u=(\dot{L}/L)x$ and its material acceleration is $\ddot{L}\eta$.

Use a constant-Froude gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }{h}_f}$, with the front depth normalized by $H(1)=1$. Differentiating its square and using the volume relation gives $\ddot{L}=-{\dot{L}}^2/(2L)$. The Euler equation now reduces toIntegrate from $\eta =1$ to obtainA filled rear-wall solution needs $0<F<2$; $F=2$ is the limiting zero rear-depth profile. Integrating the front equation gives the complete similarity solutionHere $u$ is the horizontal velocity; the virtual origin $t_v$ accounts for matching to the finite initial release. With the Benjamin deep-ambient front condition, $F=\sqrt{2}$ and $A_F=2/3$, this simplifies toThe volume, wall condition and finite front depth can all be checked directly. Choosing another physically appropriate nose Froude number changes the coefficient and profile; the interior equations alone cannot choose it.

Write $g^{\prime }=g({\rho }_f-{\rho }_a)/{\rho }_f>0$. The Saint-Venant dry-front condition describes a liquid advancing over a dry bed, or into an ambient whose inertia is negligible, as when ${\rho }_a/{\rho }_f\ll 1$. 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 $u_f=2\sqrt{2g^{\prime }{h}_0}$, whereas a rectangular release gives $2\sqrt{g^{\prime }{h}_0}$.

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 andHere $h_f$ 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 $F=\sqrt{2}$ 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.

Use a gravity-current box model: approximate the current by uniform depth $h(t)$ and length $L(t)$, with constant reduced gravity, no deposition and no fluid entrainment. The triangular cross-sectional area is $h^2/2$, so the fixed released volume isThe Benjamin deep-ambient front condition closes the integral model:For a finite initial length $L_0$, direct integration givesThe ideal point-release limit takes $L_0=0$ and has $L\propto t^{4/5}$, $h\propto t^{-2/5}$. Its divergent initial depth is outside the shallow water approximation, so a real finite reservoir regularizes the early stage. This is the finite-volume triangular-channel current integral law; it does not assert a spatially uniform exact solution of the local equations.