 Solution

Take $x$ increasing upslope and let $u$ be the superficial Darcy velocity, so the mobile discharge is $uh$. Its pore transport speed is $u/\varphi$. This interpretation is required by the factors of porosity in the printed equations; a literal pore-speed convention would instead replace $u$ there by $\varphi u$.

For capillary residual trapping, let $M(x,t)={\max }_{0\le \tau \le t}h(x,\tau )$ be the maximum invaded thickness, including the initial state. The stored carbon-dioxide volume per plan area isDuring 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 isConsequentlyAssume $a,L,u,\varphi >0$ 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 $x=x_0+v_{R}t$ on the trailing face. Matching the two linear profiles gives the triangular current with capillary retention:Before extinction, the rear, front, crest position and crest height areEvery positive height has one trailing and one leading characteristic; their intersections trace the crest. The mobile current disappears at

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 $L<x<x_{\ast }$, the maximum occurs when the crest passes $x$, at $t_m=2(x-L)/(v_R+v_A)$. Substituting into the trailing profile givesIf $z$ 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 isexactly 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.