= Solution
For the unit-width channel put $\eta=x/L(t)$, $h=h_fH(\eta)$ and $u=\dot L U(\eta)$. Fixed volume gives $V=h_fL\int_0^1H\,d\eta$, so
$$
\boxed{L\dot h_f=-h_f\dot L.}
$$
The closed rear wall has $U(0)=0$. Substituting into <volume conservation> gives $-H-\eta H'+(HU)'=0$, or $[H(U-\eta)]'=0$. The wall sets this constant to zero. Wherever $H>0$, \b[$U(\eta)=\eta$]. 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> $\dot L=F\sqrt{g'h_f}$, 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 to
$$
H'=-\frac{L\ddot L}{g'h_f}\eta=\frac{F^2}{2}\eta.
$$
Integrate from $\eta=1$ to obtain
$$
H=1-\frac{F^2}{4}+\frac{F^2}{4}\eta^2,\qquad
A_F=\int_0^1H\,d\eta=1-\frac{F^2}{6},\qquad h_f=\frac{V}{A_FL}.
$$
A 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 solution>
$$
\boxed{L^3=\frac{9F^2g'V}{4A_F}(t-t_v)^2,\quad
h(x,t)=\frac{V}{A_FL}\left(1-\frac{F^2}{4}+\frac{F^2x^2}{4L^2}\right),\quad
u(x,t)=\frac{2x}{3(t-t_v)}.}
$$
Here $u$ is the horizontal <velocity>; the virtual origin $t_v$ accounts for matching to the finite initial release. With the <Benjamin deep-ambient front condition>, $F=\sqrt2$ and $A_F=2/3$, this simplifies to
$$
\boxed{L^3=\frac{27}{4}g'V(t-t_v)^2,\qquad
h=\frac{L^2+x^2}{9g'(t-t_v)^2},\qquad u=\frac{2x}{3(t-t_v)}.}
$$
The 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.
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