= Solution
{c}
Let $H\ll L$ be the characteristic saturated thickness and let $\delta=H/L$. Scale <pressure> by $\rho gH$. The horizontal part of <Darcy's law> gives the scale
$$
u\sim\frac{k\rho g}{\mu}\frac HL=K\delta,\qquad
K=\frac{k\rho g}{\mu}.
$$
By <incompressibility>, $w\sim\delta u\sim K\delta^2$. The vertical part of <Darcy's law> is
$$
w=-\frac{k}{\mu}(p_z+\rho g).
$$
Its left side is smaller than the scale $K$ by $\delta^2$. Therefore $p_z+\rho g=O(\rho g\delta^2)$, and the leading <pressure> is <hydrostatic pressure>:
$$
\boxed{p=p_a+\rho g[h(x,t)-z]+O(\rho gH\delta^2).}
$$
Here the <water table> has <pressure> $p_a$. This is the <Dupuit approximation>, valid away from the small region near an outlet where the surface slope need not remain small.
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