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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-203/2/c/i/solution