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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.
 Fall '06
 MartinBazant

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