= Solution
Take $x$ increasing upslope and let $u$ be the superficial <Darcy velocity>, so the mobile discharge is $uh$. Its pore transport speed is $u/\phi$. This interpretation is required by the factors of <porosity> in the printed equations; a literal pore-speed convention would instead replace $u$ there by $\phi u$.
For <capillary residual trapping>, let $M(x,t)=\max_{0\leq\tau\leq t}h(x,\tau)$ be the maximum invaded thickness, including the initial state. The stored carbon-dioxide volume per plan area is
$$
\phi h+\phi s(M-h)=\phi[(1-s)h+sM].
$$
During advance, $M=h$ and new pores fill with mobile <carbon dioxide>; during recession, $M$ is fixed and the newly vacated pores retain saturation $s$. Conservation of mobile plus trapped fluid is
$$
\partial_t\{\phi[(1-s)h+sM]\}+\partial_x(uh)=0.
$$
Consequently
$$
\boxed{h_t+v_Ah_x=0\quad(h_t>0),\qquad
h_t+v_Rh_x=0\quad(h_t<0),\quad
v_A=\frac u\phi,\quad v_R=\frac{u}{\phi(1-s)}}.
$$
Assume $a,L,u,\phi>0$ and $0<s<1$. The two speeds differ because the advancing front fills a whole pore volume while the receding tail removes only its mobile fraction.
The <method of characteristics> keeps height constant on $x=x_0+v_A t$ on the leading face and $x=x_0+v_Rt$ on the trailing face. Matching the two linear profiles gives the <triangular current with capillary retention>:
$$
\boxed{h(x,t)=a\max\left\{0,\min\{x-v_Rt,\ 2L+v_At-x\}\right\}}.
$$
Before extinction, the rear, front, crest position and crest height are
$$
x_R=v_Rt,\quad x_F=2L+v_At,\quad
x_m=L+\frac{v_R+v_A}{2}t,\quad
h_m=a\left(L-\frac{v_R-v_A}{2}t\right).
$$
Every positive height has one trailing and one leading characteristic; their intersections trace the crest. The mobile current disappears at
$$
\boxed{t_* =\frac{2L}{v_R-v_A}=\frac{2\phi(1-s)L}{us},\qquad x_* =\frac{2L}{s}}.
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-69-capillary-retention.png]
{title=Advancing and receding characteristics, shrinking mobile profiles, and the final capillary trapping envelope}
{height=840}
The final trapped region is the <maximum-invasion envelope of a retained current>. For $0<x<L$, the initial height is already the maximum, $M_\infty=ax$. For $L<x<x_*$, the maximum occurs when the crest passes $x$, at $t_m=2(x-L)/(v_R+v_A)$. Substituting into the trailing profile gives
$$
\boxed{M_\infty(x)=\begin{cases}
ax,&0<x<L,\\
\dfrac{a(2L-sx)}{2-s},&L<x<2L/s,\\
0,&\text{otherwise}.
\end{cases}}
$$
If $z$ is distance into the aquifer measured normally from its upper boundary, trapped <carbon dioxide> occupies $0<z<M_\infty(x)$ at residual pore saturation $s$. The occupied pore volume per unit transverse width is
$$
\boxed{\phi s\int M_\infty(x)\,dx=\phi aL^2},
$$
exactly the initial mobile volume. This is an independent mass-conservation check of both the envelope and extinction distance. In the limit $s=0$, the current translates without shrinking and nothing is trapped; the finite-extinction formula is not used in that limit.
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