Let $K=k\Delta\rho g/\mu$. <Hydrostatic pressure> continuity at the lower boundary of the light current gives
$$
p(x,z,t)=p_H-\rho g(H-z)+\Delta\rho g\,[h(x,t)-z]
$$
inside the current, so $p_x=\Delta\rho g h_x$. <Darcy law> gives the depth-integrated horizontal flux per unit transverse width
$$
q=-Kh h_x.
$$
Thus the <porous gravity current> satisfies
$$
\boxed{\phi h_t+q_x=0,
\qquad q=-\frac{k\Delta\rho g}{\mu}hh_x}
$$
away from the fracture. The boundary and front conditions are
$$
q(0,t)=Q,
\quad h(x_N,t)=0,
\quad q(x_N,t)=0.
$$
At $x=L$, the pressure excess at the base of the fracture is $\Delta\rho g h_L$. Taking upward leakage as positive,
$$
Q_l=\frac{W\alpha k\Delta\rho g}{\mu b}h_L,
\qquad
\boxed{q(L^+,t)=q(L^-,t)-Q_l.}
$$
This jump is the local <mass conservation> law for a <leaky porous gravity current>.

At late times, $q=Q$ to leading order on $0<x<L$, while $Q_l\simeq Q$. Integrating $q=-K(h^2)_x/2$ gives
$$
\boxed{
h(x)=\left[h_0^2-(h_0^2-h_L^2)\frac{x}{L}\right]^{1/2}.}
$$
Leakage balance and the pressure-driven drop determine
$$
\boxed{h_L=\frac{\mu bQ}{W\alpha k\Delta\rho g},}
\qquad
\boxed{h_0=\left(h_L^2+\frac{2\mu QL}{k\Delta\rho g}\right)^{1/2}.}
$$

In the far field, $h(L,t)\simeq h_L$ and
$$
\phi h_t=K(hh_x)_x.
$$
Balancing the two sides with $h=O(h_L)$ gives
$$
\boxed{x_N-L=O\left[\left(\frac{k\Delta\rho g h_L}{\phi\mu}t\right)^{1/2}\right].}
$$
More precisely, set
$$
h=h_L f(\eta),
\qquad
\eta=\frac{x-L}{(Kh_Lt/\phi)^{1/2}}.
$$
The <self-similar solution> is determined by
$$
(ff')'+\frac\eta2f'=0,
\qquad
f(0)=1,
\quad f(\eta_N)=0,
\quad ff'(\eta_N)=0,
$$
and $x_N=L+\eta_N(Kh_Lt/\phi)^{1/2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 332 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-332/2/solution