Conserved first moment of a draining porous current ($D={\int }_0^{\infty }xhdx$)

For $h_t=\kappa (h{h}_x{)}_x$ with a drained boundary $h(0,t)=0$ and a finite current vanishing at its nose, integration by parts gives $\dot{D}=-\kappa [h^2{]}_0^{\infty }/2=0$. The outlet can have finite volume flux because its contribution to the first-moment flux is multiplied by $x=0$. A positive prescribed boundary height instead gives $\dot{D}=\kappa h(0,t{)}^2/2$, so drainage is essential to this invariant.

The conserved first moment of a draining porous current fixes $\alpha +2\beta =0$ in a self-similar solution $h=a{t}^{\alpha }f(x/(d{t}^{\beta }))$. 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 $-f/2-\eta f^{\prime }/4=(f{f}^{\prime }{)}^{\prime }$ is solved byThe 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.

For the dipole similarity solution of a draining porous current, $V=\varphi \int hdx=\varphi ad{t}^{-1/4}/18$. Thus $\dot{V}/V=-1/(4t)$. A shifted similarity origin replaces $t$ by $t+t_0$. Neither a positive rate nor a rate $-1/(2t)$ is compatible with this two-dimensional first-moment-conserving profile.