For $x>0$, the <Chordal Loewner equation> and $U_t=\sqrt\kappa B_t$ give
$$
dV_t^x=\frac2{V_t^x}dt-\sqrt\kappa\,dB_t.
$$
Put $D_t=V_t^r-V_t^1$. The common Brownian term cancels, so
$$
dD_t=-\frac{2D_t}{V_t^rV_t^1}dt.
$$
The <Itô formula> applied to $Z_t=\log D_t-\log V_t^1$ gives
$$
dZ_t
=\frac{\sqrt\kappa}{V_t^1}dB_t
+\left[
\frac{\kappa-4}{2(V_t^1)^2}
-\frac2{V_t^rV_t^1}
\right]dt.
$$
Since
$$
\frac{V_t^1}{V_t^r}=\frac1{1+e^{Z_t}},
$$
the clock $q(u)=\int_0^u(V_s^1)^{-2}ds$ and its inverse $\sigma$ turn the local-martingale term into Brownian motion by the <Dambis-Dubins-Schwarz theorem>. Therefore
$$
d\widetilde Z_t
=\sqrt\kappa\,dW_t
+\left(
\frac{\kappa-4}{2}
-\frac2{1+e^{\widetilde Z_t}}
\right)dt,
\qquad
\widetilde Z_0=\log(r-1).
$$
This is the <SLE boundary-point logarithmic separation diffusion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 203 / 2 / c / i / Solution

{c}

For every $x\in\mathbb R\setminus\{0\}$, the probability that a chordal $\operatorname{SLE}_\kappa$ trace in $(\mathbb H,0,\infty)$ passes through $x$ is zero. The <SLE boundary-point logarithmic separation diffusion> shows that the chance for $x$ to be swallowed strictly before a nearby point tends to zero as that point approaches $x$; <Scaling invariance of SLE> and reflection then handle every nonzero $x$.

SLE does not hit a fixed nonzero boundary point

{c}

For two boundary points $1<r$, let $V_t^x=g_t(x)-U_t$ and set $Z_t=\log(V_t^r-V_t^1)-\log V_t^1$. After the clock $q(u)=\int_0^u(V_s^1)^{-2}ds$, the <Dambis-Dubins-Schwarz theorem> gives
$$
d\widetilde Z_t=\sqrt\kappa\,dW_t+\left(\frac{\kappa-4}{2}-\frac2{1+e^{\widetilde Z_t}}\right)dt.
$$
This one-dimensional <diffusion process> compares the swallowing times of nearby boundary points.

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 Incoming links: SLE boundary-point logarithmic separation diffusion