= Solution
When sliding dominates and the whole cap is below the <snowline>, its thickness obeys
$$
h_t=\beta(hh_x)_x-\alpha.
$$
With $\alpha=0$, this is a one-dimensional <porous medium equation>. Its mass-preserving <Barenblatt solution> is
$$
\boxed{
h_{ss}(x,t)=
\left[A t^{-1/3}-\frac{x^2}{6\beta t}\right]_+}.
$$
Within its support,
$$
(h_{ss})_{xx}=-\frac1{3\beta t}.
$$
Now set $h=h_{ss}+ct$. Since $h_{ss}$ solves the unforced equation,
$$
h_t-\beta(hh_x)_x
=c-\beta ct(h_{ss})_{xx}
=\frac43c.
$$
Matching the uniform ablation term $-\alpha$ gives
$$
\boxed{c=-\frac{3\alpha}{4}},
\qquad
\boxed{
h(x,t)=
\left[A t^{-1/3}-\frac{x^2}{6\beta t}
-\frac{3\alpha t}{4}\right]_+}.
$$
The similarity solution has a virtual origin. Writing $s=t+t_0$ gives the exact decaying family
$$
h(x,t)=
\left[
C s^{-1/3}-\frac{x^2}{6\beta s}
-\frac{3\alpha s}{4}
\right]_+.
$$
If its initial centre thickness is $h^*$, then $C=(h^*+3\alpha t_0/4)t_0^{1/3}$ and the exact extinction time is
$$
t_{\rm melt}
=t_0\left[
\left(1+\frac{4h^*}{3\alpha t_0}\right)^{3/4}-1
\right].
$$
Its value depends on the initial cap width through the virtual age $t_0$, but its dimensional scale is unambiguously
$$
\boxed{t_{\rm melt}=O\left(\frac{h^*}{\alpha}\right)}.
$$
The additive melting correction alone has the corresponding time $4h^*/(3\alpha)$.
Solved by gpt-5.6-sol high.
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