Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 43 / 1 / c

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.