= Solution
For $\alpha=2$, part iii gives $a=1$, $b=2$, and $\beta=1$. Put
$$
H=tF(\eta),
\qquad
\eta=\frac{x}{t^2},
\qquad
x_N(t)=\eta_Nt^2.
$$
Substitution into the nonlinear <partial differential equation> gives the <ordinary differential equation>
$$
\boxed{
A(F^3F')'+2\eta F'-F-D=0}
$$
with <boundary conditions>
$$
\boxed{-AF(0)^3F'(0)=Q_0,
\qquad F(\eta_N)=0,
\qquad F^3F'\longrightarrow0
\ \text{as }\eta\uparrow\eta_N}.
$$
To find the leading edge, set $y=\eta_N-\eta$ and suppose $F\sim Cy^p$. 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 satisfy
$$
\frac{AC^4}{9}-\frac{2\eta_NC}{3}=0,
$$
whereas $F$ and the constant drainage $D$ are lower-order terms. Hence
$$
\boxed{
F(\eta)\sim
\left(\frac{6\eta_N}{A}\right)^{1/3}
(\eta_N-\eta)^{1/3}}
\qquad(\eta\uparrow\eta_N).
$$
The <draining gravity current> therefore has a one-third-power leading edge and its flux vanishes there.
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