Diffusion and FokkerPlanck equations' - PowerPoint PPT Presentation

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Diffusion and FokkerPlanck equations'

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Title: Diffusion and FokkerPlanck equations'


1
Diffusion and Fokker-Planck equations. Stochastic
processes are closely related to some partial
differential equations which are important in
physics applications. We shall see later that
very similar equations occur in financial
problems - an important reason why the financial
world wants people with expertise in mathematics
and physics.
2
We consider first a collection of particles lying
on a line and all following the same basic
Wiener process
Now, let ?(x,t) be the density of particles on
the line and consider the following question. How
does ? evolve in time as a result of the random
motion of the particles? The equation for the
change in ??over a time interval ?t is
where P(?x,?t) is the probability that a particle
will move a distance ?x in time ?t. The integral
just gives the average number of particles which
will end up at x.
3
Now expand in a Taylor series
Now
since P is a probability density function,
since this is the average jump, and
the variance over an interval ?t.
4
So, we obtain
This is correct to order ?t so by letting this
interval become small we conclude that the
density obeys the equation
This is a diffusion equation and we have related
the diffusion coefficient D to the properties of
the underlying random motion of particles. This
relation is due to Einstein.
5
The solution of the diffusion equation on an
infinite interval, given the initial density is
As t increases, the exponential function becomes
broader and gives the smearing out of the initial
shape expected of a diffusion process. The
diagram on the next slide illustrates this for a
square initial profile and D1. The density is
plotted at times 0, 0.25, 0.5..
6
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7
The solution is easily understood if we think of
the group of particles initially in the range s -
sds. The number of such particles is
and at time t they will have spread out into a
Gaussian distribution with variance ?2t2Dt. They
will thus have a density
Integrating over s gives us the solution quoted
earlier. This shows that we get equivalent
results from a macroscopic picture where we
consider a continuous density of the substance
and describe the evolution of the density by
diffusion, and in a microscopic picture where we
think of the random motion of individual
particles.
8
Now let us consider a more general random
process. Historically this was first done for
Brownian motion and the resulting equation is
called Fokker-Planck equation. We consider this
process for the sake of illustration, though the
methods apply to any situation described by
a similar stochastic DE. Suppose the velocity of
a particle undergoing Brownian motion
is described by the stochastic DE
where A is a friction term, which we take to be
velocity dependent. W is a standard Wiener
process, so the standard deviation of the random
change of velocity in time dt is Bvdt. Now let
f(v) be the velocity distribution function, ie
the probability density function for the
distribution of velocities.
9
Following the same procedure as we used to obtain
the diffusion equation we have
Notice one important difference here. Since the
stochastic process contains coefficients which
depend on v we need to introduce a v dependence
into the probability of a jump ?v in time ?t.
Note also that we evaluate this probability using
the value of the velocity at the beginning of the
jump. This is consistent with the assumption of a
Markov process in which the probability of the
change during any interval just depends on the
state at the beginning of that interval.
10
Now expand in a Taylor series as before to get
Note how the evaluation of the probability at the
beginning of the jump puts P inside the
derivatives.
11
The first term under the integral just gives
f(v,t). The second involves
and the third
These are all valid to lowest order in ?t so if
we let this tend to zero we get the Fokker-Planck
equation
12
Note that that the initial probability
distribution must satisfy
The form of the equation then implies that
so the distribution function remains properly
normalised. This would not be the case had we not
invoked the Markov property to get the
coefficients inside the derivatives.
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