= Dipole similarity solution of a draining porous current
{title2=$h\sim t^{-1/2},\quad x_f\sim t^{1/4}$}
The <conserved first moment of a draining porous current> fixes $\alpha+2\beta=0$ in a <self-similar solution> $h=a t^\alpha f(x/(dt^\beta))$. The <Boussinesq equation for an unconfined aquifer> supplies $\alpha=2\beta-1$, giving $\alpha=-1/2$, $\beta=1/4$. Normalize $\kappa a=d^2$ and the nose at $\eta=1$. Then $-f/2-\eta f'/4=(ff')'$ is solved by
$$
f(\eta)=\frac{\sqrt\eta-\eta^2}{6}\quad(0<\eta<1),\qquad
D=\frac{ad^2}{40},\quad d=(40\kappa D)^{1/4},\quad a=d^2/\kappa.
$$
The square-root outlet has finite drainage flux and the nose has zero flux. This is a particular similarity solution and its scaling family, not an exact description of arbitrary initial data.
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