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For an impulsive leaky porous gravity current, write $s=(1-e^{-\Omega t})/\Omega$ and $h=(1-\Omega s)[R^2-x^2{]}_{+}/(6Ds)$, where $R=C{s}^{1/3}$. At a fixed lateral location, $h_t=0$ when $x^2=R^2(1+2\Omega s)/3$. The maximum invaded thickness is the envelope $x_{\ast }(s)=C{s}^{1/3}\sqrt{(1+2\Omega s)/3}$, $h_{\ast }(s)=C^2(1-\Omega s{)}^2/(9D{s}^{1/3})$. The source solution is singular at the initial line release; finite source size and aquifer depth limit its near-source validity.

Let $K=k\Delta \rho g/\mu$. Hydrostatic pressure continuity at the lower boundary of the light current givesinside the current, so $p_x=\Delta \rho g{h}_x$. Darcy law gives the depth-integrated horizontal flux per unit transverse widthThus the porous gravity current satisfiesaway from the fracture. The boundary and front conditions areAt $x=L$, the pressure excess at the base of the fracture is $\Delta \rho g{h}_L$. Taking upward leakage as positive,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 $q=-K(h^2{)}_x/2$ givesLeakage balance and the pressure-driven drop determine

In the far field, $h(L,t)\simeq h_L$ andBalancing the two sides with $h=O(h_L)$ givesMore precisely, setThe self-similar solution is determined byand $x_N=L+{\eta }_N(K{h}_{L}t/\varphi {)}^{1/2}$.