## Info

Show that the deterministic return time Td(L) to reach ^/e, given by the solution of dx/dt =—x, with x(0) = L, is Td(L) = log(L) — log (\/e). Thus, conclude that the deterministic time to return from initial point L to the vicinity of the origin scales as log(L=/e).

Our second calculation is not much more complicated. Suppose that we allow T(x) to denote the mean time to escape from the interval [— L, L], given that X(0) = x. We know that T(x) satisfies the equation

with the boundary conditions T(— L) = T(L) = 0. The solution of Eq. (8.48) with these boundary conditions is not too difficult, but it is very cumbersome and hard to learn from. Let us think a bit more about the situation. First note that Eq. (8.48) is symmetrical about x = 0, because if we set y = —x, we obtain the same differential equation and same boundary conditions. Second, think about what happens to the stochastic process when it returns, ever so momentarily to X = 0: at that point there is no deterministic component to the dynamics and the mean of the fluctuations is 0 as well. In other words, we could think of the process at X = 0 being reflected rather than continuing through to negative values. Thus, we can equivalently consider Eq. (8.48) with the boundary condition T(L) = 0 and Tx|x=0 = 0, i.e. reflection at the origin.

Show that the solution of Eq. (8.48) satisfying the boundary conditions T(L) = 0 and Tx|x=0 = 0 is

expl 7

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