Neumann solution of the Stefan problem

Remarks

The left side of the equation in part 1 is continuous and strictly increasing on $(0, \infty)$, from $0$ to $\infty$, so the root exists and is unique. For part 2, $\operatorname{erf}(x / (2 \sqrt{a t}))$ solves the heat equation $u_t = a u_{xx}$, so $u$ does too. Also $u(0, t) = T_0$ and $u(s(t), t) = 0$, since $x = s(t)$ gives the argument $\lambda$. The Stefan condition $L s' = -k u_x(s, t)$ becomes $L \lambda \sqrt{a / t} = k T_0 e^{-\lambda^2} / (\operatorname{erf}(\lambda) \sqrt{\pi a t})$, which is the equation in part 1.

The solution is a similarity solution: it depends on $x$ and $t$ only through $x / \sqrt{t}$. For small $\mathrm{Ste}$, $\operatorname{erf}(\lambda) \approx 2 \lambda / \sqrt{\pi}$ gives $\lambda^2 \approx \mathrm{Ste} / 2$, so $s(t) \approx \sqrt{2 \, \mathrm{Ste} \, a t}$.

Going deeper

For a student

The Stefan problem is nonlinear because the domain moves, so explicit solutions are rare. The Neumann solution is the main exact one, and numerical methods for phase change are tested against it. Its key step is the similarity variable $x / \sqrt{t}$, which turns the heat equation into an ordinary differential equation and fixes the form $s(t) = 2 \lambda \sqrt{a t}$. A two-phase version, in which the solid is below its melting point, is solved in the same way.