Loewner driving function

The Loewner driving function $U_t$ is the real boundary point at which the mapped-out hull grows. Scaling the hulls by $a^{-1}$ and time by $a^2$ transforms it to $a^{-1}{U}_{a^{2}t}$.

Suppose the mapped increments $K_{s,t}$ have real-centred enclosing radii bounded by a nondecreasing modulus $\omega (t-s)\to 0$. Nested closures define the unique growing real point ${\xi }_s$. For $v\in K_{t,t+h}$, its preimage $z=g_{K_{s,t}}^{-1}(v)$ belongs to $K_{s,t+h}$. The uniform displacement estimate for a mapping-out function gives $|v-z|\le C\omega (t-s)$, while $|z-{\xi }_s|\le 2\omega (t-s+h)$. Letting $h\downarrow 0$ with $h\le t-s$ proves the displayed estimate. It proves uniform continuity of the Loewner transform on each bounded time interval without evaluating a conformal map on a rough boundary.

The domains of a capacity-parametrized Loewner chain determine their hydrodynamically normalized mapping-out functions of compact H-hulls. At a surviving point, the Chordal Loewner equation gives $U_t=g_t(z)-2/{\partial }_t{g}_t(z)$. For a continuous driver and $t>0$, the left derivative uses only the hull past. High points $in$, $n\in \mathbb{N}$, survive up to any fixed finite time for sufficiently large $n$, making this a countable local reconstruction of the driver and its past filtration.

After time $t$, map remaining domains by $g_t-U_t$. The new mapping-out function at elapsed time $s$ is ${\hat{g}}_s(z)=g_{t+s}(g_t^{-1}(z+U_t))-U_t$. Direct differentiation of the Chordal Loewner equation proves that its driver is $U_{t+s}-U_t$, and its half-plane capacity is $2s$. Defining transformed hulls as complements of transformed domains avoids ambiguity at the old hull boundary.

For the Chordal Loewner equation driven by $U_t=a\sqrt{t}$, the hull is a straight slit whose endpoint is proportional to $\sqrt{t}$. Put $r_{\pm }=(a\pm \sqrt{a^2+16})/2$ and $\alpha =-r_{-}/(r_{+}-r_{-})\in (0,1)$. Its inverse map isusing upper-half-plane branches. The interval between the two real roots maps onto the two sides of a segment of argument $\pi \alpha$.

For a real $x$ before its swallowing time,It lies in $(0,1]$ and decreases with time.