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3. Fokker Planck equation.
LAST TIME DISCUSSED
MALTHUS-VERHULST PROBLEM
Birthrate: $b(n)=\beta n$
Death rate: $d(n)=\alpha n +\frac{1}{\Omega}n(n-1)$
Made a system size expansion

\begin{displaymath}n=\Omega\phi(t)+\sqrt{\Omega} x\end{displaymath}

Found that $\phi$ satisfied a macroscopic rate equation

\begin{displaymath}\frac{d\phi}{dt}=(\beta-\alpha)\phi-\phi^2\end{displaymath}

If $\beta>\alpha,\; \phi_s=(\beta-\alpha)$ stable steady state
An initial fluctuation x0 evolved so that

\begin{displaymath}\langle x(t)\rangle=x_0e^{-(\beta-\alpha)t}\end{displaymath}


\begin{displaymath}\sigma(t)^2=\langle x(t)^2\rangle -\langle x(t)\rangle^2=\beta(1-e^{-2(\beta-\alpha)t})\end{displaymath}



FOKKER PLANCK EQUATION FOR
FLUCTUATIONS ABOUT MACROSCOPIC STEADY STATE

\begin{displaymath}\frac{\partial\Pi(x,t)}{\partial t}=(\beta-\alpha)\frac{\part...
...al x}x\Pi+\beta(\beta-\alpha)\frac{\partial^2}{\partial x^2}\Pi\end{displaymath}

Find time independent distribution $\Pi(x)$ for long times

\begin{displaymath}\frac{d}{dx}[x\Pi+\beta\frac{d\Pi}{dx}]=0\end{displaymath}

Get

\begin{displaymath}x\Pi+\beta\frac{d\Pi}{dx}=const\end{displaymath}

Requirement that $\Pi \geq 0,\; -\infty <x<\infty$ gives const=0
Normalization condition $\int dx\Pi=1$ gives

\begin{displaymath}\Pi(x)=\frac{1}{\sqrt{2\pi\beta}}\exp \frac{-x^2}{2\beta}\end{displaymath}

Gaussian distribution!

TIME EVOLUTION OF FLUCTUATIONS

Can also solve time dependent equation
All necessary information in P(x,t|x0,0) (Notation $\Rightarrow, \;\Pi(A\vert B)$ is the conditional probability for A given B) Can show (e.g. by Fourier methods)


\begin{displaymath}P(x,t\vert x_0,0)=\frac
{\exp
{-\frac
{(x-e^{-(\beta-\al...
...\alpha)t})}}
}
{\sqrt{2\pi\beta(1-e^{-2(\beta-\alpha)t})}}
\end{displaymath}

Probability distribution Gaussian with
evolving variance and mean


\fbox{\parbox{8cm}{
\begin{displaymath}\Pi(x,t)=\frac{1}{\sqrt{2\pi\sigma(t)^2}}\exp \frac{-(x-\langle x(t)\rangle)^2}{2\sigma(t)^2}\end{displaymath}
}}

Variance and mean same as before



\begin{figure}
\epsfysize=300pt
\epsffile{evfluct.eps}
\end{figure}


So far have argued only from an example
More general form of FOKKER-PLANCK equation

\begin{displaymath}\frac{\partial P(x,t)}{\partial t}=-\frac{\partial}{\partial x}A(x)P(x,t)+\frac{1}{2}\frac{\partial^2}{\partial x^2}B(x)P(x,t)\end{displaymath}

Also called Smoluchowski equation, or
second Kolmogorov equation.
Structure of equation:

\fbox{\parbox{7cm}{
\begin{displaymath}\frac{\partial P(x,t)}{\partial t}=-\fra...
...A(x)P(x,t)-\frac{1}{2}\frac{\partial}{\partial x}B(x)P(x,t)\end{displaymath}
}}

J=Probabiliy current
A(x)P(x,t)=drift term or transport term
$\frac{1}{2}\frac{\partial}{\partial x}B(x)P(x,t)=$ diffusion term

ANALOGY WITH
MACROSCOPIC CURRENTS
Suppose n(x)= concentration of conserved quantity e.g. particle number

\begin{displaymath}\int n(x)dx=const\end{displaymath}

$\frac{\partial n}{\partial t}=-\nabla\cdot$j
j= particle current
Current from constitutive relation e.g.
j$=\mu$f $ n-\nabla(Dn)$
$\mu$=mobility given by $\langle$v $\rangle =\mu$ f
f=force on particles (e.g. mg or eE.)
D= diffusion constant
In Fokker-Planck equation deal with probability density rather than particle density!

LINEAR PROCESSES

By definition
Fokker-Planck equation linear in P
Fokker-Planck equations commonly referred to as linear if A(x) linear and B(x)=const. Stationary processes governed by linear Fokker Planck equation

\begin{displaymath}\frac{\partial P(x,t)}{\partial t}=-\frac{\partial}{\partial x}[A_0+A_1x]P+\frac{B}{2}\frac{\partial^2P}{\partial x^2}\end{displaymath}

called Ornstein-Uhlenbeck processes. Stationarity requires A1<0. Time dependent Fokker-Planck equation can be solved explicitly for linear processes (Malthus-Verhulst problem example). Many numerical methods available for nonlinear processes!

STEADY STATE SOLUTION
(when exist)
can be found for arbitrary $A(x), \;B(x)$
In steady state probability current is zero
for unbounded system

\begin{displaymath}0=A(x)P_s(x)-\frac{1}{2}\frac{d}{d x}B(x)P_s(x)\end{displaymath}

with solution

\fbox{\parbox{7cm}{
\begin{displaymath}P_s(x)=\frac{const}{B(x)}\exp\left[2\int^x\frac{A(y)}{B(y)}dy\right]\end{displaymath}
}}

PLANCK'S DERIVATION
Continuous version of master equation

\begin{displaymath}\frac{\partial P(q,t)}{\partial t}=\int dq'[W(q\vert q')P(q',t)-W(q'\vert q)P(q,t)]\end{displaymath}

Define r=q-q' and w(q',r)=W(q|q')

\begin{displaymath}\frac{\partial P(q,t)}{\partial t}=\int drw(q-r,r)P(q-r,t)-P(q,t)\int dr w(q,-r)\end{displaymath}

Expand $\int dr w(q-r,r)P(q-r,t)$ w.r.t. r in first argument

\begin{displaymath}\int dr w(q,r)P(q,t)-\frac{\partial}{\partial q}\int drrw(q,r)P(q,t)\end{displaymath}


\begin{displaymath}+
\frac{1}{2}\frac{\partial^2}{\partial q^2}\int dr r^2w(q,r)P(q,t)+\cdots\end{displaymath}

Next, define the jump moments

\begin{displaymath}a_n=\int dr r^n w(q,r)\end{displaymath}

Get Fokker-Planck equation to 2nd order!

\fbox{\parbox{9cm}{
\begin{displaymath}\frac{\partial P(q,t)}{\partial t}=-\fra...
...)]+\frac{1}{2}\frac{\partial^2}{\partial q^2}[a_2(q)
P(q,t)]\end{displaymath}}}

ADDED BONUS OF DERIVATION:

Physical interpretation of A(q), B(q) in terms of jump moments.

Why stop at second order? Include all terms in expansion! Kramers-Moyal expansion:

\fbox{\parbox{8cm}{
\begin{displaymath}\frac{\partial P(q,t)}{\partial t}=\sum_...
...c{(-1)^n}{n!} \frac{\partial^n}{\partial q^n}[a_n(q)P(q,t)]\end{displaymath}
}}

Pawula's theorem $\Rightarrow$ if break off at finite order beyond 2nd, P(q,t) no longer positive
definite! Expansion to all order equivalent to master
equation. System size expansion $\Rightarrow$ higher than 2nd order jump moments higher order contributions in $\Omega^{-1/2}$ (van Kampen).

SUMMARY Click here for 4. Brownian particles Click here for Return to title page

 
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Next: About this document ...
Birger Bergersen
1998-09-18