problemset4

problemset4 - Solutions to Problem Set 4 Chris H Rycroft 1...

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Solutions to Problem Set 4 Chris H. Rycroft November 10, 2006 1 First passage for biased diffusion 1.1 The first passage time to the origin The PDF ρ ( x, t ) of a continuous diffusion process with drift velocity v and diffusivity D satisfies a Fokker-Planck equation ρ t + x = xx . For this problem, we are interested in solving in the domain x > 0. Walkers which reach x = 0 achieve first passage and are removed, so we make use of the boundary condition ρ (0 , t ) = 0. Since the walker starts at x = x 0 , our initial condition is ρ ( x, 0) = δ ( x x 0 ). Without the boundary, we would just get a solution of the form 1 4 πDt e ( x x 0 vt ) 2 / 4 Dt . For this problem, we make use of the image method, introducing another term starting at x = x 0 with magnitude A . Our PDF is therefore ρ ( x, t ) = 1 e ( x x 0 vt ) 2 / 4 Dt + Ae ( x + x 0 vt ) 2 / 4 Dt . 4 πDt This trivially satisfies the Fokker-Planck equation, and we wish to choose A so that our boundary condition is satisfied. Setting x = 0 gives ρ (0 , t ) = 1 e ( x 0 vt ) 2 / 4 Dt + Ae ( x 0 vt ) 2 / 4 Dt 4 πDt 2 ( x + v 2 t 2 ) / 4 Dt = e 0 4 πDt e x 0 v/ 2 D + Ae x 0 v/ 2 D . If A = e x 0 v/D then our boundary condition is satisfied, and hence 4 1 πDt e ( x x 0 vt ) 2 / 4 Dt e x 0 v/D e ( x + x 0 vt ) 2 / 4 Dt ρ ( x, t ) = . Solutions to problems 1 and 3 based on sections of A Guide to-First Passage Processes by Sidney Redner (2001). 1

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| �� M. Z. Bazant 18.366 Random Walks and Diffusion Problem Set 4 Solutions 2 By evaluating the probability current at x = 0, we find that the first passage probability density is given by f ( t ) = + x x =0 = D x x 0 vt e ( x x 0 vt ) 2 / 4 Dt + x + x 0 vt e x 0 v/D e ( x + x 0 vt ) 2 / 4 Dt 4 πDt 2 Dt 2 Dt x =0 = 1 x 0 e ( x 0 + vt ) 2 / 4 Dt + x 0 x 0 v/D ( x 0 + vt ) 2 / 4 Dt e e 4 πt 2 DT 2 DT = x 0 e ( x 0 + vt ) 2 / 4 Dt . 4 πDt 3 1.2 The survival probability By integrating the f ( t ), we find that the survival probability is t S ( t ) = 1 f ( q ) dq 0 t = 1 e vx 0 / 2 D x 0 e x 2 0 / 4 Dq e v 2 q/ 4 D dq. 0 4 πDq 3 Using the substitution u 2 = x 2 / 4 Dt and the eclet number Pe = vx 0 / 2 D , we find 2 2 S ( t ) = 1 − √ 2 π e Pe x/ 4 Dt e u Pe 2 / 4 u du 1 x 0 Pe 4 Dt �� = 1 2 1 erf 4 Dt + 2 x 0 e 2 Pe x 0 Pe 4 Dt �� + 2 1 erf 4 Dt 2 x 0 . 1 e Pe −| Pe | 2 erfc | Pe | 4 Dt �� . 1 2 2 x 0 As t → ∞ , we get two different behaviors for S ( t ), depending on the sign of v : 1 e 2 Pe for Pe > 0 S ( t ) Pe x 0 πDt e Pe 2 Dt/x 2 0 for Pe 0 1 e vx 0 /D for v > 0 4 D 2 t/ 4 D e v for v 0. πv 2 t From these expressions, we see that if v > 0 then there is a probability of e vx 0 /D of eventual first passage.
M. Z. Bazant 18.366 Random Walks and Diffusion Problem Set 4 Solutions 3 1.3 Minimum first passage time Let the random variables for the first passage times be T 1 , T 2 , . . . , T N . Since the walkers are inde- pendent, we know that P (min { T 1 , T 2 , . . . , T N } > t ) = P ( T 1 > t, T 2 > t, . . . , T N > t ) = P ( T 1 > t ) P ( T 2 > t ) . . . P

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