If $g_t$ is driven by $U_t=\sqrt\kappa B_t$, then the mapping-out functions of
$$
\widetilde\gamma(t)=r\gamma(t/r^2)
$$
are
$$
\widetilde g_t(z)=r g_{t/r^2}(z/r),
$$
and their driver is $\widetilde U_t=rU_{t/r^2}$. By <Brownian scaling>, $(rB_{t/r^2})_{t\geq0}$ is Brownian. Thus $\widetilde U$ has the same law as $U$, and uniqueness of the <Chordal Loewner equation> proves
$$
\boxed{(r\gamma(t/r^2))_{t\geq0}\overset d=(\gamma(t))_{t\geq0}.}
$$
This is the <Scaling invariance of SLE>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-203/2/a/solution