Since $\partial_t\sqrt g=\sqrt g(\partial_sW-kU)$,
$$
\partial_tS(u,t)
=\int_0^u\sqrt g(\partial_sW-kU)\,du'
=\int_0^{S(u,t)}(\partial_sW-kU)\,ds.
$$
Thus
$$
\boxed{
\partial_tS
=W(S,t)-W(0,t)-\int_0^S kU\,ds
}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 3 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-344/3/b/ii/solution