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

For $x>0$, the Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ givePut $D_t=V_t^r-V_t^1$. The common Brownian term cancels, soThe Itô formula applied to $Z_t=\log D_t-\log V_t^1$ givesSincethe 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. ThereforeThis is the SLE boundary-point logarithmic separation diffusion.

It is enough by Scaling invariance of SLE and reflection in the imaginary axis to treat $x=1$. Take $r=1+ϵ$. The drifttends to $(\kappa -8)/2<0$ as $z\to -\infty$. Choose $L<0$ and $a<0$ such that $b(z)\le a$ for $z\le L$. Before $\tilde{Z}$ reaches $L$,The infinite-horizon crossing probability for Brownian motion with negative drift shows that this process has a finite running maximum almost surely, so

Order preservation for the Loewner flow gives ${\tau }_1\le {\tau }_{1+ϵ}$. On $\{{\tau }_1<{\tau }_{1+ϵ}\}$, one has $V_t^1\to 0$ while $V_t^{1+ϵ}$ remains positive, so ${\tilde{Z}}_t\to +\infty$ and in particular $T_L<\infty$. HenceIf the trace itself hit the fixed boundary point $1$, then $1$ would be the right endpoint of the swallowed interval and ${\tau }_1<{\tau }_{1+ϵ}$ for every $ϵ>0$. Its probability is therefore zero. Scaling and reflection prove that SLE does not hit a fixed nonzero boundary point for every $x\in \mathbb{R}\setminus \{0\}$.