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18.366 Random Walks and Diﬀusion, Spring 2005 10.95/18.325 Mathematical Modeling of Electrochemical Systems, Spring 2009 Escape from a Symmetric Trap Notes by MIT Student (and MZB) Problem Statement. (Exam 2, 18.366, 2005) Consider a diﬀusing particle which feels a conser­ vative force, f (x) = −φ" (x), in a smooth, symmetric potential, φ(x) = φ(−x), causing a drift velocity, v(x) = bf (x), where b = D/kT is the mobility and D is the diﬀusion constant. If the particle starts at the origin, then PDF of the position, P (x, t), satisﬁes the Fokker-Planck equation, ∂P ∂ ∂2P + (v(x)P (x, t)) = D 2 , ∂t ∂x ∂x with P (x, 0) = δ(x). Suppose that the potential has a minimum φ = 0 at x = 0 with φ"" (0) = K0 > 0 and two equal maxima φ = E > 0 at x = ±x1 with φ"" (x1 ) = −K1 < 0. Let τ be the mean ﬁrst passage time to reach one of the barriers at x = ±x1 (and then escape from the well with probability 1/2). 1. Derive the general formula τ= x1 1 D x dx eφ(x)/kT 0 dy e−φ(y)/kT 0 2. In the low temperature limit, kT /E → 0, calculate the leading-order asymptotics of the escape rate, R = 1/2τ ∼ R0 (T ), using the saddle-point method. Verify the classical result of Kramers: R0 (T ) ∝ e−E/kT . 3. Calculate the ﬁrst correction to the Kramers escape rate: R(T ) ∼ R0 (T ) 1 + a kT E . SOLUTION 1. Mean escape time. We have: D ∂ ∂P ∂2P =D 2 + φ" (x)P (x, t) kT ∂x ∂t ∂x where P (x, t) = p(x, t|0, 0) within initial condition at the bottom of the well, P (x, 0) = δ(x), and absorbing boundary conditions at the exit points, P (±x1 , t) = 0. We can write: ∂P = Lx P ∂t (1) where the spatial operator Lx can be simpliﬁed using an integrating factor: Lx = D ∂ −φ(x)/kT ∂ e eφ(x)/kT ∂x ∂x (2) The probability S(t) of realization which have started at x = 0 and which have not yet reached x = ±x1 up to time t is given by: x1 S(t) = x1 p(x, t|0, 0)dx = −x1 P (x, t)dx −x1 1 The distribution function f (t) for the ﬁrst passage time is then given by: Z x1 ∂P ∂ ∂S f (t) = 1 − S(t) = − =− dx ∂t ∂t −x1 ∂t The mean escape time is then given by1 : Z ∞ Z x1 τ= tf (t)dt = g1 (x)dx Z with g1 (x) = − −x1 0 ∞ t 0 ∂P dt ∂t Performing an integration by parts (assuming that P (x, t) decays quickly enough in time that limt→∞ tP (x, t) = 0) gives: Z ∞ g1 (x) = P (x, t)dt 0 By applying the operator Lx on both sides of this relation, we get: Z ∞ Z ∞ ∂P Lx P (x, t)dt = dt = −P (x, 0) = −δ(x) Lx g1 (x) = ∂t 0 0 where we have used (1). Using the expression (2) for Lx , it is easy to solve: Z Z y e−φ(x)/kT x1 φ(y)/kT g1 (x) = e δ(z)dz dy D x 0 Now we can express the mean escape time: Z x1 Z τ= g1 (x)dx = 2 −x1 1 = D x1 g1 (x)dx 0 x1 Z e −φ(x)/kT 0 x1 Z eφ(y)/kT dy dx x Switch the order of integration, to get ﬁnally: Z Z x 1 x1 τ= dx eφ(x)/kT dy e−φ(y)/kT D 0 0 2. Kramers Mean Escape Rate. We use the saddle-point asymptotics (just Laplace’s method on the real axis, in this case) to evaluate the integrals as kT → 0. Factors of 1/2 arise since the maximum and minimum occur at the endpoints. Z x 1 2πkT −φ(0)/kT πkT e−φ(y)/kT dy ∼ e = "" 2 φ (0) 2K0 0 So that: Z x1 dx eφ(x)/kT 0 x Z dy e−φ(y)/kT ∼ 0 πkT 2K0 x1 Z eφ(x)/kT dx 0 with: Z 0 x1 eφ(x)/kT dx ∼ 1 2 − 2πkT φ(x1 )/kT e = φ"" (x1 ) πkT E/kT e 2K1 Escape occurs with probability 1/2 from either of 2 activation barriers, so √ 11 1 2D K0 K1 −E/kT R=2 = ∼ R0 (T ) = e ∝ e−E/kT 2τ τ πkT 1 The function g1 (x) from 10.95 Lecture 11 is denoted g0 (x) in 18.366 Lecture 18 from 2005, and some of this derivation can also be found in both sets of scribe notes. 2 3. First Correction to the Kramers Escape Rate. In the next derivation, we will use the following relation: Z +∞ Z +∞ b2 x6 −ax2 −ax2 +bx3 +cx4 3 4 e dx ∼ 1 + bx + cx + e dx 2 −∞ −∞ r π 3 c 15 b2 + = 1+ a 4 a2 16 a3 Using saddle-point asymptotics with the previous formula: s (3) 2 ! Z x (4) (0) φ (0) 1 2πkT kT φ 5kT e−φ(y)/kT dy ∼ 1− e−φ(0)/kT 2 + 00 2 φ (0) 8 φ00 (0) 24 φ00 (0) 3 0 Since the well is symmetric, φ(3) (0) = 0, we end up with: r Z x πkT kT M0 −φ(y)/kT 1− e dy ∼ 2K0 8 K02 0 with M0 = φ(4) (0). The same way: s (3) 2 ! Z x1 (4) (x ) φ (x ) 2πkT kT φ 5 k T 1 1 1 eφ(x)/kT dx ∼ − 00 1+ eφ(x1 )/kT − 2 φ (x1 ) 8 φ00 (x1 ) 2 24 φ00 (x1 ) 3 0 that we write: Z x1 r φ(x)/kT e dx ∼ 0 π kT kT M1 5kT L21 E/kT 1+ − e 2K1 8 K12 24 K13 with L1 = φ(3) (x1 ) and M1 = φ(4) (x1 ). We write then: kT M1 5kT L21 1 k T M0 1+ − τ∼ 1− 8 K12 24 K13 R0 (T ) 8 K02 1 kT M1 M0 5 L21 ∼ 1+ − 2 − R0 (T ) 8 K12 3 K13 K0 that is: kT R(T ) ∼ R0 (T ) 1 − 8 with: K0 = φ00 (0) K1 = −φ00 (x1 ) M1 M0 5 L12 − 2 − 3 K13 K12 K0 L1 = φ(3) (x1 ) 3 M0 = φ(4) (0) M1 = φ(4) (x1 ) MIT OpenCourseWare http://ocw.mit.edu 10.626 Electrochemical Energy Systems Spring 2014 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.