Average velocity and effective diffusion of a Brownian particle driven by a constant force over a static periodic potential
Hongyun Wang Department of Applied Mathematics and Statistics, Jack Baskin School of Engineering, University of California, Santa Cruz, CA 95064, USA Email:
[email protected]
Abstract In this manuscript, we consider the case where a Brownian particle is subject to a static periodic potential and is driven by a constant force. We derive analytic formulas for the average velocity and the effective diffusion.
Mathematical formulation We consider the one-dimensional stochastic motion of a particle subject to a periodic potential and a constant driving force. In particular, we are interested in the effective diffusion coefficient of the particle, and how the periodic potential and the driving force affect the effective diffusion coefficient. The probability density of the particle is governed by the Fokker-Planck equation:
∂ (φ ′ − f ) ∂ ρ ∂ρ =D ρ+ ∂ x kB T ∂x ∂t
(1)
where ρ( x, t) is the probability density that the particle is at position x at time t, D is the diffusion constant of the particle, φ ( x ) is the periodic potential with period = L, f is the constant driving force, kB is the Boltzmann constant, and T is the absolute temperature. The probability density ρ( x, t) satisfies the normalizing condition
1
∞
∫ ρ( x, t)dx = 1
−∞
Consider the transform x = L x˜ , t =
L2 ˜ t D
L2 t˜ The function ρ˜ ( x˜, t˜ ) = L ρ L x˜, satisfies D ∞
∂ ˜ ∂ ρ˜ ∂ ρ˜ = φ ′ − f˜ ρ˜ + , ∂ x˜ ∂ t˜ ∂ x˜
(
)
∫ ρ˜ ( x˜, t˜)dx˜ = 1
(2)
−∞
φ ( L x˜ ) fL where φ˜ ( x˜ ) = is a periodic function of x˜ with period = 1 and f˜ = is a constant. kB T kB T Thus, we only need to consider equation (1) for L = 1 , D = 1 , kB T = 1
The effective diffusion coefficient The mean and the variance of the particle position at time t are
X ( t) =
∞
∫ x ρ( x, t) dx
−∞
var {X ( t)} =
( X ( t) −
X ( t)
)
2
∞
=
∫ ( x − X (t) ) ρ( x, t) dx 2
−∞
It can be shown that for large t, both the mean and the variance increase approximately linearly. The average velocity and the effective diffusion coefficient of the particle are defined as def
V = lim t →∞
X ( t) t
def
Deff = lim t →∞
var {X ( t)} 2t
The effective drag coefficient can be defined as def
ζ eff =
f V 2
Let us introduce functions
ρk ( x, t) =
∞
∑
j =−∞
j k ρ( j + x, t)
(3)
In appendix A, we derive that 1
1 lim ∫ ρ1 ( x, t) dx = V t →∞ t 0
(4)
1 1 1 x t dx x t dx lim ρ ( , ) − ρ ( , ) 2 ∫ 1 t →∞ 2 t ∫ 0 0
2
= Deff
Below we give the formulas for the average velocity and effective diffusion. The detailed derivations are given in Appendices B, C and D. The steady state of ρ0 ( x, t) is
u0 ( x ) =
w0 ( x ) 1
∫ w ( x )dx 0
0
where 1
w 0 ( x ) = ∫ exp(φ ( x + s) − φ ( x ) − f s) ds 0
The average velocity V, the effective diffusion Deff, and the effective drag coefficient ζeff are given by (see Appendices B, C and D for derivations)
V=
1 − exp(− f )
∫
1
0
w 0 ( x ) dx
2 ∫0 (w0 ( x )) ∫0 exp(φ ( x ) − φ ( x − s) − f s)dsdx 1
Deff =
ζ eff =
,
1
1 w ( x ) dx ∫0 0
3
1 f w ( x ) dx 1 − exp(− f ) ∫0 0
3
(5)
Results and discussion Below we discuss the behaviors of effective diffusion and effective drag coefficient for several cases (the detailed derivation is in Appendix E). Case A:
f is small
V=
f 1+ a0
a 1 f 1 − + O( f 2 ) a0 2
1 a ζ eff = a0 1 + f − 1 + O( f 2 ) 2 a0 Deff =
1 1+ a0
a 1 f 1 − + O( f 2 ) a0 2
(
)
ζ eff Deff = 1 + O( f 2 ) where
1 1 a0 = ∫ exp(−φ ( x )) dx ∫ exp(φ ( s)) ds 0 0 1 1
a1 = ∫ ∫ exp(−φ ( x ) + φ ( x + s)) sds dx 0 0
If
a1 1 ≠ (this happens when the potential is asymmetric), then the minimum of Deff is attained a0 2
at non-zero value of f, and min Deff
ζ ,
Deff > D
and
In both Case A and Case B, we see that the non-dimensional Einstein relation ζ eff Deff = 1 is approximately satisfied. Case A:
ζ eff Deff = 1 + O( f 2 )
for small f
Case B:
1 ζ eff Deff = 1 + O 2 f
for large f
− A φ( x) = A
Case C:
0 ≤ x < 0.5 0.5 ≤ x < 1
Appendix A In this appendix, we derive equations (4) and (5): 1
1 ρ1 ( x, t) dx = V t →∞ t ∫ 0
lim
1 1 1 lim ∫ ρ2 ( x, t) dx − ∫ ρ1 ( x, t) dx t →∞ 2 t 0 0
We first write 1
2
= Deff
1
∫ ρ ( x, t) dx as 0
1
1
∞
1
∫ ρ ( x, t) dx = ∫ ∑ ( j + x )ρ( j + x, t) dx − ∫ x ρ ( x, t) dx ≡ X (t) 1
0
0 j =−∞
0
0
1
T1 = ∫ x ρ0 ( x, t) dx satisfies 0 ≤ T1 ≤ 1. 0
5
− T1
1
[
]
1 1 lim ∫ ρ1 ( x, t) dx = lim X ( t) − T1 = V t →∞ t t →∞ t 0
==>
Then we have 2
1 1 ∫ ρ2 ( x, t) dx − ∫ ρ1( x, t) dx = ∫ 0 0 0 1
1
=∫
2
1 j − ∫ ρ1 ( s, t) ds ρ( j + x, t) dx ∑ j =−∞ 0 ∞
∞
∑ {[ j + x − X (t) ] + [T − x ]} ρ( j + x, t) dx 2
1
0 j =−∞
1
= var {X ( t)} + 2 ∫
∞
∑ [ j + x − X (t) ] ⋅ [T − x ]ρ( j + x, t) dx 1
0 j =−∞
∞
1
∑ [T − x ] ρ( j + x, t) dx ≡ var{X (t)} + T
+∫
2
2
1
+ T3
0 j =−∞
Using the Cauchy-Schwarz inequality, we see that T2 is bounded by 1
T2 ≤ 2 ∫
∞
∑
j + x − X ( t) ρ( x + j, t) dx
0 j =−∞
∞
1
≤2
∫ ∑ [ j + x − X (t) ] ρ( x + j, t) dx = 2 2
0 j =−∞
var {X ( t)}
which leads to 1 1 1 x t dx x t dx lim ρ ( , ) − ρ ( , ) 2 1 ∫ t →∞ 2 t ∫ 0 0
2
1 = lim var {X ( t)} + T2 + T3 = Deff t →∞ 2t
[
Appendix B In this appendix, we derive the results for ρ0 ( x, t) : •
ρ0 ( x, t)
converges to a steady state u0 ( x ) given by
u0 ( x ) =
1 w ( x) M0 0
6
]
where 1
w 0 ( x ) = ∫ exp(φ ( x + s) − φ ( x ) − f s) ds , 0
•
1
M 0 = ∫ w 0 ( x ) dx 0
The steady state flux of u0 ( x ) is J 0 = −((φ ′ − f ) u0 ( x ) + u0′ ( x )) =
1 (1 − exp(− f )) M0
The boundary condition for ρ0 ( x, t) is ρ0 ( x + 1, t) = ρ0 ( x, t) . So ρ0 ( x, t) is governed by ∂ ρ0 ∂ ∂ ρ0 = (φ ′ − f ) ρ0 + ∂x ∂t ∂ x 1 ρ ( x + 1, t) = ρ ( x, t) , ρ0 ( x, t) dx = 1 0 0 ∫ 0
Let u0 ( x ) be the steady state solution of (6). u0 ( x ) satisfies
(φ ′ − f )u0 + u0′ = − J 0 ==>
[exp(φ ( x ) − f x ) u ( x )]′ = − J
==>
x +1 J0 u0 ( x ) = exp(−φ ( x ) + f x ) c1 + exp(φ ( s) − f s) ds ∫ 1 − exp(− f ) x
0
0
exp(φ ( x ) − f x )
The boundary condition: u0 ( x + 1) = u0 ( x ) implies c1 = 0 . Thus, we get u0 ( x ) =
J0 w ( x) 1 − exp(− f ) 0
where 1
w 0 ( x ) = ∫ exp(φ ( x + s) − φ ( x ) − f s) ds . 0
Applying the normalizing condition: J0 =
1
∫ u ( x )dx = 1, we have 0
0
1 (1 − exp(− f )) , M0
7
(6)
where 1
M 0 = ∫ w 0 ( x ) dx 0
Now we show that lim ρ0 ( x, t) = u0 ( x ) t →∞
We consider r0 ( x, t) =
ρ0 ( x, t) − 1. r0 ( x, t) satisfies u0 ( x )
∂ r0 ∂ r0 ∂ = u0 − J 0 r0 + u0 ∂x ∂t ∂ x 1 r ( x + 1, t) = r ( x, t) , u0 ( x ) r0 ( x, t) dx = 0 0 0 ∫ 0
To show lim r0 ( x, t) = 0 , we consider the time evolution of t →∞
(7)
∫ u ( x )(r ( x, t)) 1
0
0
0
2
dx .
1 1 1 ∂r ∂ ∂ r0 d 2 dx r0 − J 0 r0 + u0 0 dx = u r dx u r = 2 2 0 0 0 0 ∫ ∫ ∫ ∂x ∂x ∂t dt 0 0 0 1 ∂ r0 ( x, t) ∂ r0 ∂ r0 = −2 ∫ − J 0 r0 + u0 dx = −2 ∫ u0 dx ≤ 0 ∂x ∂x ∂x 0 0 2
1
The quantity
1
∫ ur 0
2
0 0
dx is non-negative and non-increasing. Hence, we obtain
∂ r ( x, t) lim ∫ u0 0 dx = 0 t →∞ ∂x 0 2
1
1
==>
lim ∫ t →∞
0
∂ r0 ( x, t) dx = 0 ∂x
Using the constraint
1
∫ u ( x )r ( x, t)dx = 0 , we write r ( x, t) as 0
0
0
0
x ξ ∂ r0 ( s, t) ∂ r ( s, t) ds ds u0 (ξ ) dξ + ∫ 0 r0 ( x, t) = − ∫ ∫ ∂x ∂x 00 0 1
Thus, we have lim r0 ( x, t) = 0 t →∞
8
lim ρ0 ( x, t) = u0 ( x )
==>
t →∞
For simplicity, we start with the initial probability density
u ( x ) 0 ≤ x ≤ 1 ρ( x,0) = 0 otherwise 0 With this initial probability density, we have
ρ0 ( x, t) = u0 ( x )
and
ρ1 ( x, 0) ≡ 0
Appendix C In this appendix, we derive the results for ρ1 ( x, t) : •
p1 ( x, t) = ρ1 ( x, t) − u0 ( x ) J 0 t
converges to a steady state u1 ( x ) given by
x M1 w ( x) 1 u1 ( x ) = 3 + u0 ( x ) − u0 ( x ) ∫ u0 ( s) ds − 1 3 M0 J 0 M0 J 0 2 0
where 1
w1 ( x ) = ∫ ( w 0 ( x + s)) exp(φ ( x + s) − φ ( x ) − f s) ds , 2
0
1
M1 = ∫ w1 ( x ) dx 0
•
The steady state flux of u1 ( x ) is x
M J −((φ ′ − f ) u1 + u1′) = 13 + 0 − J 0 ∫ u0 ( s) ds 2 M0 0 The boundary condition for ρ1 ( x, t) is ρ1 ( x + 1, t) = ρ1 ( x, t) − u0 ( x ) . So ρ1 ( x, t) is governed by
∂ ρ1 ∂ ∂ ρ1 = (φ ′ − f ) ρ1 + ∂x ∂t ∂ x ρ ( x + 1, t) = ρ ( x, t) − u ( x ) 1 0 1
(8)
9
Integrating (8), we obtain x =1
1 ∂ρ d ρ1dx = (φ ′ − f ) ρ1 + 1 = −((φ ′ − g) u0 (0) + u0′ (0)) = J 0 ∫ ∂ x x =0 dt 0
(9)
This shows that ρ1 ( x, t) does not converge to a steady state.
p1 ( x, t) = ρ1 ( x, t) − u0 ( x ) J 0 t is governed by ∂ p1 ∂ ∂ p1 + J 0 u0 = (φ ′ − f ) p1 + ∂x ∂x ∂t 1 p ( x + 1, t) = p ( x, t) − u ( x ) , p1 ( x, t) dx = 0 1 1 0 ∫ 0
(10)
Let u1 ( x ) be the stead state solution of (10). u1 ( x ) satisfies
′ J 0 u0 = [(φ ′ − f ) u1 + u1′] u1 ( x + 1) = u1 ( x ) − u0 ( x ) ,
1
∫ u ( x )dx = 0 0
1
x
==>
(φ ′ − f )u1 + u1′ = −c1 + J 0 ∫ u0 (s)ds 0
==>
x exp(φ ( x ) − f x ) u1 ( x ) ′ = exp(φ ( x ) − f x ) −c1 + J 0 ∫ u0 ( s) ds 0
[
Using exp(φ ( x ) − f x ) =
]
−1 exp(φ ( x ) − f x ) u0 ( x ) ′ , we obtain J0
[
]
[
]
x
c u1 ( x ) = 1 u0 ( x ) − exp(−φ ( x ) + f x ) ∫ exp(φ ( x ) − f x ) u0 ( x ) ′ ∫ u0 ( s) ds dx J0 0 x
c 2 = 1 u0 ( x ) − u0 ( x ) ∫ u0 ( s) ds + exp(−φ ( x ) + f x ) ∫ exp(φ ( x ) − f x )( u0 ( x )) dx J0 0 x exp(−φ ( x ) + f x ) x +1 c1 2 exp(φ ( s) − f s)( u0 ( s)) ds + c 2 = u0 ( x ) − u0 ( x ) ∫ u0 ( s) ds − ∫ J0 1 − exp(− f ) x 0
The condition u1 ( x + 1) = u1 ( x ) − u0 ( x ) implies c 2 = 0 . Thus, we obtain
10
x
c 1 w ( x) u1 ( x ) = 1 u0 ( x ) − u0 ( x ) ∫ u0 ( s) ds − J0 M03J 0 1 0 where 1
w1 ( x ) = ∫ ( w 0 ( x + s)) exp(φ ( x + s) − φ ( x ) − f s) ds 2
0
The condition
∫
1
0
u1 ( x ) dx = 0 yields c1 =
M1 J 0 + M03 2
1
where M1 = ∫ w1 ( x ) dx 0
Thus, the steady state flux of u1 ( x ) is
−((φ ′ − f ) u1 + u1′) =
x
M1 J 0 + − J 0 ∫ u0 ( s) ds M03 2 0
Now we show that lim p1 ( x, t) = u1 ( x ) t →∞
We consider r1 ( x, t) =
p1 ( x, t) u1 ( x ) − . r ( x, t) is governed by u0 ( x ) u0 ( x ) 1
∂ r1 ∂ r1 ∂ = u0 − J 0 r1 + u0 ∂x ∂t ∂ x 1 r ( x + 1, t) = r ( x, t) , u0 ( x ) r1 ( x, t) dx = 0 1 1 ∫ 0
This is the same system as (7) Thus, we have lim r1 ( x, t) = 0 t →∞
==>
lim p1 ( x, t) = u1 ( x ) t →∞
Appendix D In this appendix, we derive the effective diffusion. The boundary condition for ρ2 ( x, t) is ρ2 ( x + 1, t) = ρ2 ( x, t) − 2 ρ1 ( x, t) + u0 ( x ) . So ρ2 ( x, t) is governed by
11
∂ ρ2 ∂ ∂ ρ2 = (φ ′ − f ) ρ2 + ∂x ∂t ∂ x ρ ( x + 1, t) = ρ ( x, t) − 2 ρ ( x, t) + u ( x ) 2 1 0 2
(11)
Integrating (11) and using ρ2 ( x + 1, t) = ρ2 ( x, t) − 2 ρ1 ( x, t) + u0 ( x ) and ρ1 ( x, t) = p1 ( x, t) + u0 ( x ) J 0 t , we obtain x =1
1 ∂ρ ∂ρ d ρ2 dx = (φ ′ − f ) ρ2 + 2 = −2 (φ ′ − f ) ρ1 + 1 − J0 ∫ ∂ x x =0 ∂ x x =0 dt 0
∂p = −2 (φ ′ − f ) p1 + 1 + 2J 02t − J 0 ∂ x x =0
(
)
Using the steady state flux of u1 ( x ) , we get 1 M ∂u d lim ∫ ρ2 dx = −2 (φ ′ − f ) u1 + 1 + 2 J 0 2 t − J 0 = 2 13 + 2 J 0 2 t t →∞ dt ∂ x x =0 M0 0
(
1 M d = lim ∫ ρ2 dx − 2 J 0 2 t = 2 13 M0 t →∞ dt 0
==>
1 1 d lim ∫ ρ2 dx − ∫ ρ1dx t →∞ dt 0 0
==>
1 1 1 Deff = lim ∫ ρ2 dx − ∫ ρ1dx t →∞ 2 t 0 0
2
)
2
M = 13 M 0
The average velocity is
V = J0 =
1 − exp(− f ) M0
The effective drag coefficient is
ζ eff =
f f M0 = V 1 − exp(− f )
Notice that 1
1 1
M1 = ∫ w1 ( x ) dx = ∫ ∫ ( w 0 ( x + s)) exp(φ ( x + s) − φ ( x ) − f s) dsdx 2
0
0 0
12
1 1
= ∫ ∫ ( w 0 ( x )) exp(φ ( x ) − φ ( x − s) − f s) dsdx 2
0 0
Appendix E In this appendix, we derive the asymptotic behaviors of the effective diffusion and the effective drag coefficient for several cases Case A:
f is small
f 1 − exp(− f ) = f 1 − + O( f 2 ) 2 f f = 1 + + O( f 2 ) 1 − exp(− f ) 2 1
w 0 ( x ) = ∫ exp(φ ( x + s) − φ ( x ) − f s) ds 0
1
[
]
= exp(−φ ( x )) ∫ exp(φ ( x + s)) 1 − f s + O( f 2 ) ds 0
1
==>
a1
∫ w ( x ) dx = a 1 − f a 0
0
0
0
+ O( f 2 )
1 1 where a0 = ∫ exp(−φ ( x )) dx ∫ exp(φ ( s)) ds 0 0 1 1
a1 = ∫ ∫ exp(−φ ( x ) + φ ( x + s)) sdsdx 0 0
==>
V=
1 − exp(− f ) 1
∫ w ( x ) dx 0
ζ eff =
0
=
a1 1 f 2 1 + f − + O( f ) a0 a0 2
1
∫ w ( x ) dx = a 1 + f 1 − a + O( f ) 1 − exp(− f ) 2 a f
0
0
1
0
0
13
2
1
1
∫ (w ( x )) ∫ exp(φ ( x ) − φ ( x − s) − f s)dsdx 2
0
0
0
1
= ∫e
−φ ( x )
0
2 1 1 φ ( s) 1 φ ( s) 1 φ ( s) 1 φ ( x + s) −φ ( x − s) sds ∫ e ds − 2 f ∫ e ds∫ e sds dx ∫ e ds − f ∫ e 0 0 0 0 0
1 = a0 2 − f 2 a0 a1 + a0 2 + O( f 2 ) = a0 2 1 − 2
1 w ( x ) dx ∫ 0 0
−3
=
a1 1 + O( f 2 ) 3 1 + 3 f a0 a0
∫ (w ( x )) ∫ exp(φ ( x ) − φ ( x − s) − f s)dsdx = 1 1 + f a = a a w ( x ) dx ∫ 1
Deff
==>
2a 1 f 1 + + O( f 2 ) a0 2
0
2 1
0
1
0
3
1
0
0
0
0
1 − + O( f 2 ) 2
ζ eff Deff = 1 + O( f 2 ) If the potential φ ( x ) is an even function, then 1 1
1 1
0 0
0 0
a1 = ∫ ∫ exp(φ ( x + s) − φ ( x )) sdsdx = ∫ ∫ exp(φ ( x + 1 − s) − φ ( x ))(1 − s) dsdx 1 1
1 1
0 0
0 0
= ∫ ∫ exp(φ (− x − s) − φ (− x ))(1 − s) dsdx = ∫ ∫ exp(φ ( x + s) − φ ( x ))(1 − s) dsdx = a0 − a1
==>
a1 1 = a0 2
In this case, we have
[
]
ζ eff = a0 1 + O( f 2 ) Deff =
Case B:
[
]
1 1 + O( f 2 ) a0
f is large
14
s f w0 ( x ) = f ∫ exp(φ ( x + s) − φ ( x ) − f s) ds = ∫ expφ x + − φ ( x ) − s ds f 0 0 f
1
(
f
)
s2 s 2 = ∫ φ ′ ( x ) + (φ ′ ( x )) + φ ′′ ( x ) 2 exp(− s) ds f f 2 0
=1+
(
1
==>
f ∫ w 0 ( x ) dx = 1 + 0
==>
)
1 1 1 2 φ ′ ( x ) + 2 (φ ′ ( x )) + φ ′′ ( x ) + O 3 f f f 1 1 2 2 ∫ (φ ′ ( x )) dx + O f 0 f 3 1
1 − exp(− f )
1 1 1 2 V= 1 = f 1 − 2 ∫ (φ ′ ( x )) dx + O 3 f ∫ w0 ( x ) dx f 0 0
ζ eff =
f
1
∫ w ( x ) dx = 1 + 0
0
1 − exp(− f )
1 1 2 2 ∫ (φ ′ ( x )) dx + O f 0 f 3 1
1 s f ∫ exp(φ ( x ) − φ ( x − s) − f s) ds = ∫ expφ ( x ) − φ x − − s ds f 0 0 f
(
f
)
s2 s 2 exp(− s) ds = ∫ φ ′ ( x ) + (φ ′ ( x )) − φ ′′ ( x ) f 2 f 2 0
=1+
(
1
==>
f
3
)
1 1 1 2 φ ′ ( x ) + 2 (φ ′ ( x )) − φ ′′ ( x ) + O 3 f f f 1
∫ (w ( x )) ∫ exp(φ ( x ) − φ ( x − s) − f s)dsdx 2
0
0
0
(
)dx
3 1 2 = ∫ 1 + φ ′ ( x ) + 2 6(φ ′ ( x )) + φ ′′ ( x ) f f 0 1
1 6 2 = 1 + 2 ∫ (φ ′ ( x )) dx + O 3 f 0 f 1
15
2 ∫0 (w0 ( x )) ∫0 exp(φ ( x ) − φ ( x − s) − f s)dsdx 1
==>
Deff =
1
1 w ( x ) dx ∫0 0
3
=1+
1 3 2 2 ∫ (φ ′ ( x )) dx + O f 0 f 3
ζ eff Deff = 1 +
1 4 2 2 ∫ (φ ′ ( x )) dx + O f 0 f 3
1
1
Acknowledgements This work was partially supported by NSF.
16
