Asymptotics Neumann Solution Stefan problem

Question

Consider the one-phase Stefan problem, defined on the moving domain $[0,s(t)]$ where the temperature inside the domain is determined by the heat equation $$T_t=T_{xx},\qquad 0<x<s(t),$$ subject to the boundary conditions $$T(0,t)=-1,\quad T(s(t),t)=0,$$ and the moving interface is determined by the Stefan condition $$\beta\dot s=T_x|_{x=s(t)},$$ where $\beta$ is the Stefan number. This is one of the rare configurations where the problem has an analytical solution, known as the Neumann solution:$$T(x,t)=\frac{\text{erf}(x/2\sqrt{t})}{\text{erf}(\lambda)}-1,\qquad s(t)=2\lambda\sqrt{t},$$ where $\lambda$ satisfies the transcendental equation \begin{equation} \beta\sqrt{\pi}\lambda e^{\lambda^2}\text{erf}(\lambda)=1. \end{equation} For large Stefan numbers we have $\lambda\sim\beta^{-1}$, but what happens in the case where $\beta\ll1$? The Stefan problem itself is much more complicated to study in this case, but I am only interested in seeing what happens to the solution of the transcendental equation above. Any hint on how to start?

Answer 1

Let's rearrange the equation to

$$ \sqrt{\pi}\lambda e^{\lambda^2}\operatorname{erf}(\lambda) = \frac{1}{\beta}. $$

Note that the function $f(\lambda) = \lambda e^{\lambda^2} \operatorname{erf}(\lambda)$ is strictly increasing for $\lambda \geq 0$ with $f(0) = 0$ and $f(\infty) = \infty$, so this equation has a single positive solution $\lambda$ for any $\beta > 0$. Further,

$$ \lim_{\beta \to \infty} \lambda = 0 \qquad \text{and} \qquad \lim_{\beta \to 0^+} \lambda = \infty. $$

So, when $\beta$ is large $\lambda$ is small, and for small $\lambda$ we have the leading order approximation

$$ \frac{1}{\beta} = \sqrt{\pi}\lambda \cdot e^{\lambda^2} \cdot \operatorname{erf}(\lambda) \sim \sqrt{\pi} \lambda \cdot 1 \cdot \frac{2\lambda}{\sqrt{\pi}} = 2\lambda^2, $$

which imples that

$$ \lambda \sim \frac{1}{\sqrt{2\beta}} \qquad \text{as } \beta \to \infty. \tag{$*$} $$

When $\beta$ is small $\lambda$ is large, leading to the approximation

$$ \frac{1}{\beta} = \sqrt{\pi}\lambda e^{\lambda^2} \cdot \operatorname{erf}(\lambda) \sim \sqrt{\pi}\lambda e^{\lambda^2} \cdot 1 = \sqrt{\pi}\lambda e^{\lambda^2}. $$

Taking logarithms, this yields

$$ \log \frac{1}{\beta} = \lambda^2 + \log \sqrt{\pi}\lambda + o(1) \sim \lambda^2, $$

and thus

$$ \lambda \sim \sqrt{\log \frac{1}{\beta}} \qquad \text{as } \beta \to 0^+. \tag{$**$} $$