By Sidney Redner
Preface; Errata; 1. First-passage basics; 2. First passage in an period; three. Semi-infinite procedure; four. Illustrations of first passage in uncomplicated geometries; five. Fractal and nonfractal networks; 6. platforms with round symmetry; 7. Wedge domain names; eight. purposes to uncomplicated reactions; References; Index
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Additional resources for First-passage process
Then the probability for a random walk that starts at this input point to escape, that is, never return to its starting point, is simply proportional to the network conductance G. It is amazing that a subtle feature of random walks is directly related to currents and voltages in a resistor network! One appeal of this connection is that network conductances can be computed easily. In one dimension, the conductance of an inﬁnitely long chain of identical resistors is clearly zero. Thus Pescape = 0 or, equivalently, Preturn = 1; that is, a random walk in one dimension is recurrent.
4, this integral has two fundamentally different ∞ behaviors, depending on whether P(0, t) dt diverges or converges. In the 22 First-Passage Fundamentals former case, we apply the last step in Eq. 2) where the dimension-dependent prefactor Ad is of the order of 1 and does not play any role in the asymptotic behavior. ∞ For d > 2, the integral P(0, t) dt converges and we apply Eq. 8) to compute the asymptotic behavior of P(0, 1) − P(0, z). By deﬁnition, F(0, 1) = t F(0, t) = 1 − [P(0, 1)]−1 . Further, t F(0, t) is just the eventual probability R that a random walk reaches the origin, so that P(0, 1) = (1 − R)−1 .
This notation allows us to write the Green’s function in the entire domain as a single expression. With practice, one should be able to immediately write down this nearly complete form of the Green’s function for these simple boundary-value problems. We determine the remaining constant by integrating Eq. 6) over an inﬁnitesimal interval that includes x0 . 2. Time-Dependent Formulation 45 discontinuity in the ﬁrst derivative of the Green’s function at x = x 0 to be c (x, s) x=x 0+ − c (x, s) =− x=x 0− 1 .