problemset4

# Problemset4

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Unformatted text preview: −vx0 /2D 0 t x0 � 4πDq 3 2 e−x0 /4Dq e−v 2 q/4D dq. Using the substitution u2 = x2 /4Dt and the P´clet number Pe = v x0 /2D, we ﬁnd e �∞ 2 2 2 2 e−u −Pe /4u du S (t) = 1 − √ e−Pe √ π x/ 4Dt � � �� √ 1 x0 Pe 4Dt = 1− 1 − erf √ + 2 x0 2 4Dt � � �� √ e−2Pe x0 Pe 4Dt + 1 − erf √ − . 2 2 x0 4Dt � � �� √ 1 −Pe−|Pe| |Pe| 4Dt ∼ 1− e 2 − erfc . 2 x0 2 As t → ∞, we get two diﬀerent behaviors for S (t), depending on the sign of v : � 1 − e−2Pe for Pe > 0 S (t) ∼ x −Pe2 Dt/x2 0 √0 e for Pe ≤ 0 Pe π Dt � 1 − −vx0 /D for v > 0 �e ∼ 4D −v 2 t/4D for v ≤ 0. e πv 2 t From these expressions, we see that if v > 0 then there is a probability of e−vx0 /D of eventual ﬁrst passage. M. Z. Bazant – 18.366 Random Walks and Diﬀusion – Problem Set 4 Solutions 1.3 3 Minimum ﬁrst passage time Let the random variables for the ﬁrst passage times be T1 , T2 , . . . , TN . Since the walkers are inde­ pendent, we know that P (min{T1 , T2 , . . . , TN } > t) = P (T1 > t, T2 > t, . . . , TN > t) = P (T1 > t)P (T2 > t) . . . P (TN > t) = S (t)N and hence the PDF of the minimum ﬁrst passage time is given by d S (t)N = f (t)N S (t)N −1 dt where f (t) and S (t) are explicitly given in the previous sections. pn (t) = − 2 2.1 First passage for anomalous walks Unbiased Cauchy walk Appendix A provides a simple C++ code to simulate ﬁrst passage times for the Cauchy walk. For a large number of trials, it was found that the standard C++ math rand() function was inadequate, and that slight biases in the probabilities around n = 30 could be seen. A second code, listed in appendix B, was therefore written, making use of the more advanced random number generation routines found in the GNU Scientiﬁc Library (GSL) [1]. The GSL code was run with 2 × 1010 trials for the case of d = 0.0. Wal...
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## This note was uploaded on 01/23/2014 for the course MATH 18.366 taught by Professor Martinbazant during the Fall '06 term at MIT.

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