Consider a <nonlinear diffusion equation> with uniform supply, $\phi h_t=D_m(h^m h_x)_x+R$, on $x>0$, with an absorbing boundary $h(0,t)=0$ and initially $h=0$. Far from the boundary, $h=Rt/\phi$. Balancing the <time derivative>, supply, and diffusion gives
$$
h=\frac{Rt}{\phi}F(\eta),\qquad \eta=\frac{x}{\ell(t)},\qquad \ell(t)=\left[\frac{D_mR^m t^{m+1}}{\phi^{m+1}}\right]^{1/2}.
$$
The <similarity solution> satisfies
$$
(F^mF')'+1-F+\frac{m+1}{2}\eta F'=0,\qquad F(0)=0,\qquad F(\infty)=1.
$$
The outward boundary <volume flux per unit width> is $Q=R\ell c_m$, where $c_m=\lim_{\eta\downarrow0}F^mF'$. Integrating the equation gives
$$
c_m=\frac{m+3}{2}\int_0^\infty(1-F)\,d\eta.
$$
Thus the growing region of depleted storage fixes the discharge prefactor, and $Q\propto t^{(m+1)/2}$.

Forced filling similarity for power-law diffusion

Assume $B$ is <continuously differentiable>. Apply the <chain rule> to each component with $z_j=tx_j$:
$$
\partial_{x_j}F_i=t\partial_{z_j}B_i(tx),\qquad
\partial_tF_i=x_j\partial_{z_j}B_i(tx).
$$
Contracting the first identity with $x_j$ proves
$$
\boxed{(x\cdot\nabla)F=t\,\partial_tF.}
$$
Here $\nabla$ differentiates with respect to $x$ at fixed $t$, while the <time derivative> holds $x$ fixed. The equality follows directly even at $t=0$; no division by $t$ is needed. It holds componentwise for the <vector field>, so it is not restricted to <scalar> functions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 3 / 9B / a / Solution

{title2=$\partial_t$}

A time derivative is the <partial derivative> of an evolving field with respect to time while holding its spatial coordinates fixed. For a quantity depending only on time, it is an ordinary <derivative>.

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