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For a one-sided draining gravity current with injected volume per unit width $V_0(t/\tau {)}^3$, defineThe similarity solution satisfieswhere $K=k[g^3{\tau }^3/({\nu }^3{V}_0){]}^{2/5}$. The inlet condition is $-H(0{)}^3{H}^{\prime }(0)=9$, the integral of $H+\varphi L$ is one, and both depths vanish at the advancing front. The endpoint ${\xi }_N$ gives $x_N={\xi }_{N}D{t}^2$ and depends on $K$ and $\varphi$.

For $\alpha =2$, part iii gives $a=1$, $b=2$, and $\beta =1$. PutSubstitution into the nonlinear partial differential equation gives the ordinary differential equationwith boundary conditions

To find the leading edge, set $y={\eta }_N-\eta$ and suppose $F\sim C{y}^p$. The two singular terms in the ordinary differential equation have orders $y^{4p-2}$ and $y^{p-1}$. Their exponents agree only when $p=1/3$. Their leading coefficients then satisfywhereas $F$ and the constant drainage $D$ are lower-order terms. HenceThe draining gravity current therefore has a one-third-power leading edge and its flux vanishes there.

If a draining gravity current is supplied with volume flux proportional to $t^{\alpha }$, a first-kind similarity solution hasprovidedThese relations follow by balancing the time derivative, horizontal lubrication theory flux, vertical drainage, and imposed inlet flux.