For $t<T_x$, differentiate the Loewner equation with respect to $x$:
$$
\partial_tg_t'(x)
=-\frac{2g_t'(x)}{(g_t(x)-U_t)^2},
\qquad g_0'(x)=1.
$$
Therefore
$$
g_t'(x)=
\exp\left(-\int_0^t
\frac{2\,ds}{(g_s(x)-U_s)^2}\right).
$$
The exponent is nonpositive and decreases with $t$, so $0<g_t'(x)\leq1$ and $g_t'(x)$ is decreasing before $x$ is swallowed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-203/3/d/solution