 Solution

Because $\zeta <\eta \le h(x)$, the lower layer is saturated along the entire barrier. Both layers have the same horizontal head gradient but different Darcy velocities. Integrating their conductivity over the actual cross-section gives the transmissivityThe total discharge is $Q=-\mathcal{T}(h)h_x$. Hence $QL={\int }_{\eta }^H\mathcal{T}(h)dh$, givingThis is a parallel-layer conductivity integral, not a series-resistance average: the layers share the horizontal pressure drop. It reduces to the previous result when $k_1=k_2$. The assumption $\zeta <\eta$ is important; if the downstream free surface cuts the lower layer, the transmissivity and its head integration must be split into different saturation regimes.