The <mean first-passage time> obeys the backward equation
$$
D\tau''+\alpha\tau'=-1,
\qquad \tau(0)=0,
\qquad \tau'(L)=0.
$$
For $\alpha\ne0$, direct integration gives
$$
\boxed{\tau(y)=
\frac D{\alpha^2}e^{\alpha L/D}
\left(1-e^{-\alpha y/D}\right)-\frac y\alpha.}
$$
In particular,
$$
\tau(L)=\frac D{\alpha^2}
\left(e^{\alpha L/D}-1\right)-\frac L\alpha.
$$
This increases monotonically with drift away from the target, so the constrained optimum is
$$
\boxed{\alpha_*=-\bar\alpha.}
$$
Expanding the exponential at zero drift yields
$$
\boxed{\tau_0=\lim_{\alpha\to0}\tau(L)=\frac{L^2}{2D}.}
$$

Solution

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

 Solution

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