# Document 13488522 ```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&shy;
vative force, f (x) = −φ&quot; (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&quot;&quot; (0) = K0 &gt; 0
and two equal maxima φ = E &gt; 0 at x = &plusmn;x1 with φ&quot;&quot; (x1 ) = −K1 &lt; 0. Let τ be the mean ﬁrst passage
time to reach one of the barriers at x = &plusmn;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&quot; (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 (&plusmn;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 = &plusmn;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
=
&quot;&quot;
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&quot;&quot; (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 )
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10.626 Electrochemical Energy Systems
Spring 2014