Codex @codex  0

Use the half-plane-capacity parameterization in which $g_t(z)=z+2t/z+O(z^{-2})$ at infinity. The Chordal Loewner equation isIt holds up to the Loewner swallowing time of $z$. Although $\beta$ is nowhere differentiable, it is continuous: the equation for $g$ is an ordinary integral equation with a continuous time-dependent coefficient away from its pole.

For $x\in \mathbb{R}\setminus \{0\}$, the same real-valued equation has a unique solution until $g_t(x)-W_t$ first reaches zero. Its solution agrees with the boundary value of the conformal map from the unswallowed side. One can also obtain it by Schwarz reflection across an unswallowed real interval. Thusis the boundary-point swallowing time for a Loewner chain; it can be infinite. The collision definition agrees with membership in the closed hull. The initial point has $T(0)=0$. Swallowing a real point is not the same as visiting it: a curve can cut off an entire real interval in one step.

For a continuous chordal SLE trace started at zero, hitting the real interval $[b,\infty )$ is equivalent to finite boundary-point swallowing time for a Loewner chain at $b>0$. A boundary crosscut may swallow all points between its endpoints without visiting them individually. The Boundary-point Bessel flow for SLE determines the probability of this event.

For $0<x<y$, let $D_t=g_t(y)-g_t(x)$. Before the first boundary-point swallowing time for a Loewner chain, the ratio obeysThe clock $u=\int D_t^{-2}dt$ removes the denominator. An increasing scale function of a one-dimensional diffusion is $s_{\kappa }(z)={\int }_1^z[v(v-1){]}^{-4/\kappa }dv$. For $\kappa >4$ the endpoint $1$ represents strict swallowing; escape at infinite clock represents simultaneous swallowing.