Transcript L4_stochastics

Stochastic models
of chemical kinetics
4. Random walks and diffusion
Karl Pearson, 1905, ‘Nature’
The Problem of the Random Walk
Can any of your readers refer me to a work wherein I should find a solution
of the following problem, or failing the knowledge of any existing solution
provide me with an original one? I should be extremely grateful for aid in
the matter.
A man starts from a point O and walks l yards in a straight line; he then turns
through any angle whatever and walks another l yards in a second straight
line. He repeats this process n times. I require the probability that after
these n stretches he is at a distance between r and r + dr from his starting
point, O.
The problem is one of considerable interest, but I have only succeeded in
obtaining an integrated solution for two stretches. I think, however, that a
solution ought to be found, if only in the form of a series in powers of 1/n,
when n is large.
KARL PEARSON. The Gables, East Ilsley, Berks.
Rayleigh, 1905, ‘Nature’
The Problem of the Random Walk
This problem, proposed by Prof. Karl Pearson in the current
number of Nature, is the same as that of the composition of n isoperiodic vibrations of unit amplitude and of phases distributed at
random, considered in Phil. Mag., x., p. 73, 1880; xlvii., p. 246,
1899; ("Scientific Papers," i., p. 491, iv., p. 370). If n be very great,
the probability sought is
2
e
n
r2

n
rdr
RAYLEIGH. Terling Place, July 29.
A random step
Step size l.
Direction determined by flipping a fair coin:
p (l )  1/ 2, p(l )  1/ 2
s  random position after one step
E[ s ]  lp (l )  ( l ) p (l )  0
Var ( s )  E[ s 2 ]  ( E[ s ]) 2  l 2
A random walk
A sequence of n independent steps, each of size l.
n
x   si  position after n steps
i 1
 n 
E  x   E   si   nE  s   0
 i 1 
average displacement is zero.
Var  x   nVar ( s )  nl 2
variance scales linearly with number of steps.
Distribution function for position (1)
E  x  0
Var  x   nl 2  mean and variance.
We want the entire distribution.
Use the Central Limit Theorem to approximate:
n
s
x i 1
When n  1:

n
n
i
Var ( s ) 

N  E[ s ],

n


2


x
l
2
E[ s ]  0 Var ( s )  l  ~ N  0, 
n
 n
Distribution function for position (2)
n
s
x i 1
n  1:

n
n
 l2 
Var ( s ) 

N  E[ s],
  N  0, 
n 

 n
i

l
2
N  n  0, n 
n

2
x
x2


1
2
2 nl 2
e
  N  0, nl  
2
2 nl

Convert number of steps to time
t
n  , where  is time between jumps.

p ( x, t ) 
1
2 nl 2

e

x2
2 nl 2

1
2
l2

e
x2
2t
l2

t

 p( x, t )dx  1

p l 2  2 p
p ( x, t ) satisfies

with p ( x, t  0)   ( x)
2
t 2 x
Connection to concentration (1)
Consider N independent particles that start at x  0.
M (t )  number of particles in (x, x  dx) :
Probability to find a single particle in (x, x  dx) is
a Bernoulli random variable, with probability of
success given by p ( x, t ).
Number of particles in (x, x  dx) is Binomial:
M (t ) ~ Bin( N , p ( x, t ))  E  M (t )   N  p ( x, t )
Connection to concentration (2)
E  M (t )   N  p ( x, t )  concentration C ( x, t )  E  M (t ) 
Since C ( x, t ) is just scaled p ( x, t ), it satisfies the same equation:
C l 2  2C

t 2 x 2
Relate this to conservation of mass:
l2
D
2
C l 2  2C
J



t 2 x 2
x
l 2 C
J 
 Flux related to the gradient of concentration
2 x
Fick’s Law
l 2 C
J 
2 x
l2
D
2
C
J  D
x
Nothing but probability!
Retrace the steps of derivation
•
•
•
•
•
•
•
Single random step (Bernoulli)
Position after n steps (Normal)
Number of steps, time → p(x,t)
p(x,t) satisfies diffusion equation
This was done without deriving equation itself
Let’s approach the problem differently
Derive equation, then get p(x,t) as its solution
Differential equation for p(x,t)
Particle jumps on a lattice. Jump size l , jump frequency  .
Jumps left and right.
Particle never stays in one place.
p ( x, t )  probability that a particle is in  x  l , x  l 
1
1
p ( x, t   )  p ( x  l , t )  p ( x  l , t )
2
2
The Law of Total Probability
Differential equation for p(x,t)
1
1
p ( x, t   )  p ( x  l , t )  p ( x  l , t )
2
2
Taylor expand
p
1
1 p 1 l 2  2 p
p ( x, t )  
 ... 
p ( x, t )  l 

2
t
2
2 x 2 2 x
1
1 p 1 l 2  2 p
 p ( x, t )  l 
 ...
2
2
2 x 2 2 x
p l 2  2 p

t 2 x 2

l2
D
2
( Einstein,1905)
Average displacement
p
2 p
 D 2 , p (, t )  0, p ( x, t  0)   ( x)
t
x


 p
2 p 
x   xp ( x, t )dx   x 
 D 2 dx
t
x 

 
d x
 0  x  x(t  0)  0
dt
Note that this did not require having the explicit expression for p( x, t )
Mean square displacement

x
2

 p
2 p 
  x p ( x, t )dx   x 
 D 2 dx
x 
 t


d x2
dt
2
2

2 p
D x
dx 
2

x

2
 2 p   p 
D x
  2 x dx  
 x   x 




D  2 xp ( x, t )    2 p( x, t )dx   2 D



d x2
dt
 2 D  x 2  2 Dt
Diffusivity can be obtained by analyzing statistical properties
of trajectories. Very useful in experiments and simulations!
Rayleigh, 1905, ‘Nature’
The Problem of the Random Walk
This problem, proposed by Prof. Karl Pearson in the current number of Nature, is
the same as that of the composition of n iso-periodic vibrations of unit
amplitude and of phases distributed at random, considered in Phil. Mag., x., p.
73, 1880; xlvii., p. 246, 1899; ("Scientific Papers," i., p. 491, iv., p. 370). If n be
very great, the probability sought is
2
e
n
r2

n
rdr  you should be able to derive this!
RAYLEIGH. Terling Place, July 29.
Pearson’s response, 1905
The Problem of the Random Walk
I have to thank several correspondents for assistance in this matter. Mr. G. J.
Bennett finds that my case of n=2 can really be solved by elliptic integrals,
and, of course, Lord Rayleigh's solution for n very large is most valuable, and
may very probably suffice for the purposes I have immediately in view.
I ought to have known it, but my reading of late years has drifted into other
channels, and one does not expect to find the first stage in a biometric
problem provided in a memoir on sound.
The lesson of Lord Rayleigh's solution is that in open country the most
probable place to find a drunken man who is at all capable of keeping on his
feet is somewhere near his starting point!
KARL PEARSON
Questions
• Reaction and diffusion
– What happens when particle reacts?
– Distribution of reaction points, etc.
• Fluctuations in concentration
– N particles, starting at the same time
– Statistics of particle numbers at a point?
• More complex microscopic mechanisms
– Asymmetric jumps
– Distribution of jumping times, etc