Smoluchowski approximation is completely invalid in this case. When

Barnes and Silverman reproduced the graph, it got turned over, reversing

the direction of time. However, the over-all appearance of the curves is

very much the same and in fact this stochastic process is invariant under

time reversal. Are there any beats in this graph, and should there be?

Fig. 4b on p. 242 of Kappler (¬g. 5a on p. 169 of Barnes and Silverman)

BROWNIAN MOTION IN A FORCE FIELD 57

represents an intermediate case.

Figure 2a

Figure 2b

Figure 2c

We illustrate crudely the two cases (Fig. 2). Fig. 2a is the highly

overdamped case, a Markov process. Fig. 2b is the underdamped case,

not a Markov process. Fig. 2c illustrates a case that does not occur.

(The only repository for memory is in the velocity, so over-all sinusoidal

behavior implies local smoothness of the curve.)

One has the feeling with some of Kappler™s curves that one can occa-

sionally see where an exceptionally energetic gas molecule gave the mirror

a kick. This is not true. Even at the lowest pressure used, an enormous

number of collisions takes place per period, and the irregularities in the

curves are due to chance ¬‚uctuations in the sum of enormous numbers of

individually negligible events.

58 CHAPTER 10

It is not correct to think simply that the jiggles in a Brownian trajec-

tory are due to kicks from molecules. Brownian motion is unbelievably

gentle. Each collision has an entirely negligible e¬ect on the position of

the Brownian particle, and it is only ¬‚uctuations in the accumulation of

an enormous number of very slight changes in the particle™s velocity that

give the trajectory its irregular appearance.

The experimental results lend credence to the statement that the

Smoluchowski approximation is valid when the friction is large (β large).

A theoretical proof does not seem to be in the literature. Ornstein and

Uhlenbeck [22] show only that if a harmonically bound particle starts

at x0 at time 0 with a Maxwellian-distributed velocity, the mean and

variance of x(t) are approximately the mean and variance of the Smolu-

β ’1 . We shall examine the

chowski theory provided β 2ω and ∆t

Smoluchowski approximation in the case of a general external force, and

prove a result that says that it is in a very strong sense the limiting case

of the Ornstein-Uhlenbeck theory for large friction.

Consider the equations (10.1) of the Ornstein-Uhlenbeck theory. Let w

be a Wiener process with di¬usion coe¬cient D (variance parameter 2D)

as in the Einstein-Smoluchowski theory. Then if we set B = βw the

process B has the correct variance parameter 2β 2 D for the Ornstein-

Uhlenbeck theory. The idea of the Smoluchowski approximation is that

the relaxation time β ’1 is negligibly small but that the di¬usion coe¬cient

D = kT /mβ and the velocity K/β are of signi¬cant size. Let us therefore

de¬ne b(x, t) (having the dimensions of a velocity) by

K(x, t)

b(x, t) =

β

and study the solution of (10.1) as β ’ ∞ with b and D ¬xed. The

equations (10.1) become

dx(t) = v(t)dt

dv(t) = ’βv(t)dt + βb x(t), t dt + βdw(t).

Let x(t) = x(β, t) be the solution of these equations with x(0) = x0 ,

v(0) = v0 . We will show that as β ’ ∞, x(t) converges to the solution

y(t) of the Smoluchowski equation

dy(t) = ’b y(t), t dt + dw(t)

with y(0) = x0 . For simplicity, we treat the case that b is independent

of the time, although the theorem and its proof remain valid for the case

BROWNIAN MOTION IN A FORCE FIELD 59

that b is continuous and, for t in compact sets, satis¬es a uniform Lips-

chitz condition in x.

THEOREM 10.1 Let b : ‚ ’ ‚ satisfy a global Lipschitz condition and

let w be a Wiener process on ‚ . Let x, v be the solution of the coupled

equations

dx(t) = v(t)dt; x(0) = x0 , (10.2)

dv(t) = ’βv(t)dt + βb x(t), t dt + βdw(t); v(0) = v0 . (10.3)

Let y be the solution of

dy(t) = b y(t) dt + dw(t); y(0) = x0 . (10.4)

For all v0 , with probability one

lim x(t) = y(t),

β’∞

uniformly for t in compact subintervals of [0, ∞).

Proof. Let κ be the Lipschitz constant of b, so that |b(x1 ) ’ b(x2 )| ¤

κ|x1 ’ x2 | for all x1 , x2 in ‚ . Let

1

tn = n

2κ

for n = 0, 1, 2, . . . . Consider the equations on [tn , tn+1 ]. By (10.2),

t

x(t) = x(tn ) + v(s) ds, (10.5)

tn

and by (10.3),

t t

v(t) = v(tn ) ’ β b x(s) ds + β[w(t) ’ w(tn )], (10.6)

v(s) ds + β

tn tn

or equivalently,

t t

v(tn ) v(t)

’ b x(s) ds + w(t) ’ w(tn ).

v(s) ds = + (10.7)

β β

tn tn

60 CHAPTER 10

By (10.5) and (10.7),

t

v(tn ) v(t)

’ b x(s) ds + w(t) ’ w(tn ).

x(t) = x(tn ) + + (10.8)

β β tn

By (10.4),

t

b y(s) ds + w(t) ’ w(tn ),

y(t) = y(tn ) + (10.9)

tn

so that by (10.8) and (10.9),

v(tn ) v(t)

x(t) ’ y(t) = x(tn ) ’ y(tn ) + ’

β β

(10.10)

t

[b x(s) ’ b y(s) ]ds

+

tn

The integral in (10.10) is bounded in absolute value by

(t ’ tn )κ |x(s) ’ y(s)|,

sup

tn ¤s¤tn+1

1

and (t ’ tn )κ ¤ for tn ¤ t ¤ tn+1 , so that

2

v(s)

|x(t) ’ y(t)| ¤ |x(tn ) ’ y(tn )| + 2 sup

β

tn ¤s¤tn+1

(10.11)

1

sup |x(s) ’ y(s)|

+

2 tn ¤s¤tn+1

for tn ¤ t ¤ tn+1 . Since this is true for all such t, we can take the

supremum of the left hand side and combine it with the last term on the

right hand side. We ¬nd

v(s)

|x(t) ’ y(t)| ¤ 2|x(tn ) ’ y(tn )| + 4

sup sup . (10.12)

β

tn ¤t¤tn+1 tn ¤s¤tn+1

Suppose we can prove that

v(s)

’0

sup (10.13)

β

tn ¤s¤tn+1

with probability one as β ’ ∞, for all n. Let

|x(t) ’ y(t)|.

ξn = sup

tn ¤t¤tn+1

BROWNIAN MOTION IN A FORCE FIELD 61

Since x(t0 ) ’ y(t0 ) = x0 ’ x0 = 0, by (10.12) and (10.13), ξ1 ’ 0 as

β ’ ∞. By induction, it follows from (10.12) and (10.13) that ξn ’ 0

for all n, which is what we wish to prove. Therefore, we need only prove

(10.13).

If we regard the x(t) as being known, (10.3) is an inhomogeneous

linear equation for v, so that by Theorem 8.2,

t

’β(t’tn )

e’β(t’s) b x(s) ds

v(t) = e v(tn ) + β

tn

(10.14)

t

e’β(t’s) dw(s).

+β

tn

Now consider (10.8). Since

¤ b x(tn ) + κ|x(s) ’ x(tn )|,

b x(s) (10.15)

it follows from (10.8) that

v(tn ) v(t)

|x(t) ’ x(tn )| ¤ + (t ’ tn ) b x(tn )

+

β β (10.16)

+ (t ’ tn )κ sup |x(s) ’ x(tn )| + |w(t) ’ w(tn )|.

tn ¤s¤tn+1

Remembering that (t ’ tn ) ¤ 1/2κ for tn ¤ t ¤ tn+1 , we ¬nd as before

that

v(t)

|x(t) ’ x(tn )| ¤ 4

sup sup

β

tn ¤t¤tn+1 tn ¤s¤tn+1

(10.17)

1

|w(t) ’ w(tn )|.

+ b x(tn ) +2 sup

κ tn ¤t¤tn+1

From (10.15) and (10.17), we can bound b x(s) for tn ¤ s ¤ tn+1 :

¤ 2 b x(tn )

sup b x(s)

tn ¤s¤tn+1

(10.18)

v(t)

+ 2κ sup |w(t) ’ w(tn )|.

+ 4κ sup

β

tn ¤t¤tn+1 tn ¤t¤tn+1

If we apply this to (10.14) and observe that

t

e’β(t’s) ds ¤ 1,

β

tn

62 CHAPTER 10

we obtain

|v(t)| ¤ |v(tn )| + 2 b x(tn )

sup

tn ¤t¤tn+1

v(t)

+ 2κ sup |w(t) ’ w(tn )|

+ 4κ sup (10.19)

β

tn ¤t¤tn+1 tn ¤t¤tn+1

t

e’β(t’s) dw(s) .

+ sup β