Use a well-mixed <gravity-current box model> of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard <gravity-current front condition> are
$$
V_0\dot\phi=-w_dL\phi,
\qquad
\dot L=\operatorname{Fr}\sqrt{g'(\phi)h}
=\operatorname{Fr}\sqrt{\frac{gV_0}{\rho_0L}(R_1\phi+R_2\phi^2)}.
$$
These two <ordinary differential equations> are the required integral model.

First suppose $R_1>0$ and put
$$
a=\frac{R_2}{R_1},
\qquad
C=\frac{\operatorname{Fr}V_0}{w_d}
\sqrt{\frac{gV_0R_1}{\rho_0}}.
$$
Eliminating time gives
$$
L^{3/2}\frac{dL}{d\phi}
=-C\sqrt{\frac{1+a\phi}{\phi}}.
$$
With
$$
I(f)=\sqrt{f(1+af)}+\frac1{\sqrt a}\operatorname{arsinh}\sqrt{af},
$$
integration from $(L,\phi)=(L_0,1)$ yields
$$
\boxed{L^{5/2}=L_0^{5/2}+\frac52C\,[I(1)-I(\phi)].}
$$
As $t\to\infty$, the <concentration> tends to zero and the <runout length of a gravity current> is
$$
\boxed{L_\infty=
\left\{L_0^{5/2}+\frac52C\left[
\sqrt{1+a}+\frac1{\sqrt a}\operatorname{arsinh}\sqrt a
\right]\right\}^{2/5}.}
$$
The formula has a regular $a\to0$ limit. If $R_1=0<R_2$, direct integration instead gives
$$
L^{5/2}=L_0^{5/2}+\frac52\frac{\operatorname{Fr}V_0}{w_d}
\sqrt{\frac{gV_0R_2}{\rho_0}}(1-\phi),
$$
whose value at $\phi=0$ gives $L_\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-345/2/c/solution