Stefan problem as a variational inequality

Example

Take the one-dimensional one-phase Stefan problem in units with thermal diffusivity and conductivity equal to $1$ and latent heat $L > 0$. The liquid occupies $0 < x < s(t)$, with $s(0) = 0$ and $s$ increasing. The end $x = 0$ is heated, with a given temperature $u(0, t) > 0$, and the temperature $u$ satisfies: $$\partial_t u = \partial_x^2 u \ \text{ for } 0 < x < s(t), \qquad u(s(t), t) = 0, \qquad L s'(t) = -\partial_x u(s(t), t).$$ Extend $u$ by $0$ to the solid region $x > s(t)$, and define the integrated temperature $$w(x, t) = \int_0^t u(x, \tau) \, d\tau.$$ Then $w$ solves the obstacle-type system $$w \ge 0, \qquad \partial_t w - \partial_x^2 w + L \ge 0, \qquad w \, \big(\partial_t w - \partial_x^2 w + L\big) = 0.$$

Check in the liquid: fix $x > 0$ and let $\theta(x)$ be the time with $s(\theta(x)) = x$, so $u(x, \tau) = 0$ for $\tau \le \theta(x)$. Then $w(x, t) = \int_{\theta(x)}^t u(x, \tau) \, d\tau$ for $t > \theta(x)$. Differentiating twice in $x$ and using $u(x, \theta(x)) = 0$ gives $\partial_x^2 w = \int_{\theta(x)}^t \partial_x^2 u \, d\tau - \partial_x u(x, \theta(x)) \, \theta'(x)$. The Stefan condition and $s'(\theta(x)) \, \theta'(x) = 1$ give $-\partial_x u(x, \theta(x)) \, \theta'(x) = L$. The heat equation turns the integral into $u(x, t) = \partial_t w$, so $\partial_t w - \partial_x^2 w = -L$ where $w > 0$. In the solid, $w = 0$, so $\partial_t w - \partial_x^2 w + L = L > 0$.