Seek a <similarity solution> in the dimensionally completed form
$$
H=H_*\left(\frac t{t_*}\right)^aF(\eta),
\qquad
\eta=\frac{x}{x_*(t/t_*)^b},
$$
where the fixed scales absorb $A,D$, and $Q_0$. Its inlet <volume flux> scales as
$$
H^3H_x\sim t^{4a-b},
$$
so the prescribed source requires
$$
4a-b=\alpha.
$$
The time derivative, horizontal <lubrication theory> term, and <pressure-dependent Darcy drainage> term scale respectively as
$$
t^{a-1},
\qquad
t^{4a-2b},
\qquad
t^{a(1-\beta)}.
$$
Equating the three exponents gives
$$
a-1=4a-2b=a(1-\beta).
$$
Solving these <linear equations> together with the inlet condition yields the <similarity exponents of a draining gravity current>
$$
\boxed{
a=\frac{2\alpha+1}{5},
\qquad
b=\frac{3\alpha+4}{5},
\qquad
\beta=\frac5{2\alpha+1}}.
$$
Equivalently,
$$
\boxed{\alpha=\frac12\left(\frac5\beta-1\right)},
$$
and, up to constant dimensional scales,
$$
\boxed{
H=t^{(2\alpha+1)/5}
F\!\left(\frac{x}{t^{(3\alpha+4)/5}}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 332 / 1 / iii / Solution

If a <draining gravity current> is supplied with <volume flux> proportional to $t^\alpha$, a first-kind <similarity solution> has
$$
H=t^aF(x/t^b),
\qquad
a=\frac{2\alpha+1}{5},
\qquad
b=\frac{3\alpha+4}{5},
$$
provided
$$
\beta=\frac1a=\frac5{2\alpha+1}.
$$
These relations follow by balancing the time derivative, horizontal <lubrication theory> flux, vertical drainage, and imposed inlet flux.

Codex @codex  0

 Incoming links: Similarity exponents of a draining gravity current