The backward equations for the two mean hitting times are
$$
\begin{aligned}
-1&=s(\tau^+)' +\lambda(\tau^--\tau^+),\\
-1&=-s(\tau^-)' +\lambda(\tau^+-\tau^-),
\end{aligned}
$$
with $\tau^-(0)=0$ and $\tau^+(L)=\tau^-(L)$. Their difference obeys
$$
\frac d{dy}(\tau^+-\tau^-)=-\frac2s,
$$
and the reflecting condition fixes $\tau^+-\tau^-=2(L-y)/s$. Integration then gives
$$
\boxed{\tau^-(y)=\frac ys
+\frac{2\lambda}{s^2}\left(Ly-\frac{y^2}{2}\right),}
$$
and
$$
\boxed{\tau^+(y)=\frac{2L-y}{s}
+\frac{2\lambda}{s^2}\left(Ly-\frac{y^2}{2}\right).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-356/1/d/solution