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Even when most of an unconfined aquifer with depth-dependent permeability is deep, its height vanishes at an absorbing outlet. Near that outlet, finite outward volume flux per unit width $Q$ gives $u_{b}h{h}_x\simeq Q$ and hence $h\simeq (2Qx/u_b{)}^{1/2}$. The high-depth outer approximation instead has $h\simeq [6Qx/(u_b\beta ){]}^{1/3}$. Their crossover lies at height of order $1/\beta$ and distance of order $u_b/({\beta }^{2}Q)$; the thin inner boundary layer transmits the same leading discharge.

Use the Dupuit approximation: the saturated region is shallow enough that hydrostatic pressure is $p=\rho g(h-z)$ and flow is predominantly horizontal. Darcy's law then gives the horizontal Darcy velocity $u_x=-u_b(1+\beta z)h_x$, where $u_b=k_0\rho g/\mu$. Integrating through the saturated depth gives the volume flux per unit widthThe stored water volume per unit area is $\varphi h$, so mass conservation yields the unconfined aquifer with depth-dependent permeability equationDefine the positive river discharge by $Q(t)=-q(0,t)$. A useful check on all subsequent results is the integrated mass conservation law