8. Transformations and blow torches

LAST TIME:

• Discussed absorbing boundary conditions.
• Artificial boundary $\Rightarrow P(S,t)=0$ on boundary S
• Natural boundaries $\Rightarrow P(S,t)$ "well behaved".
• Examples: Diffusion near "cliff". Genetic drift.
• Derived Kramers' escape rate:

where

$$\alpha=\frac{1}{kT}\frac{d^2 U(x)}{dx}\;At\; potenial\; minimum$$

$$\beta=-\frac{1}{kT}\frac{d^2 U(x)}{dx}\;At\; potenial\; maximum$$

HETEROGENEOUS DIFFUSION
Consider Fokker-Planck equation on form

$$\frac{\partial P(x,t)}{\partial t}=-\frac{\partial}{\partial ... ...)P(x,t)-\frac{1}{2}\frac{\partial}{\partial x}B(x)P(x,t)\right]$$

Coordinate transformation

$$x\Rightarrow y=y(x);\;\;y'=\frac{dy}{dx}$$

$$P(x,t)=\frac{dy}{dx}\Pi(y,t)=y'\Pi(y,t)$$

a(y)=A(x);b(y)=B(x)

$$\frac{\partial\Pi(y,t)}{\partial t}=-\frac{\partial}{\partial... ...-\frac{y'}{2}\frac{\partial}{\partial y}b(y)y'\Pi(y,t) \right]$$

Using

$$y'\frac{dy'}{dy}=\frac{d^2y}{dx^2}=y''$$

$$y'\frac{db}{dy}=B'$$

we find

$$\frac{\partial\Pi}{\partial t}=-\frac{\partial}{\partial y} ... ...']\Pi(y,t)+\frac{1}{2}(y')^2b\frac{\partial^2\Pi}{\partial y^2}$$

Choose transformation

$$\frac{1}{2}(y')^2b(y)=\frac{1}{2}(\frac{dy}{dx})^2B(x)=D=const.$$

$$y=\sqrt{2D}\int^x\frac{dz}{\sqrt{B(z)}}$$

where lower limit of integral is arbitrary.
Get Fokker-Planck equation on form

$$\frac{\partial\Pi(y,t)}{\partial t}=-\frac{\partial f\Pi}{\partial y}+D\frac{\partial^2\Pi}{\partial y^2}$$

where we have a new "effective" force

$$f(y)=y'(a-\frac{B'}{2})-\frac{1}{2}by''$$

Problem with state dependent diffusion has been transformed into problem with constant diffusion but new drift term.
Physical example: Thermoelectric effect!

Example PERIODIC FORCE AND TEMPERATURE

$$A(x)=-\frac{dU(x)}{dx}=-\eta\sin x$$

$$B(x)=2\xi(1 +\alpha\sin x)$$

Fokker-Planck equation

$$\frac{\partial P(x,t)}{\partial t}=\eta\frac{\partial}{\parti... ...n (x) P+\xi\frac{\partial^2 }{\partial x^2}(1 +\alpha \sin x)P$$

Coordinate transformation:
Choose $D=\xi$

$$y(x)=\int^x\frac{dz}{\sqrt{1+\alpha\sin z}}$$

$$y'=\frac{dy}{dx}=\frac{1}{\sqrt{1+\alpha \sin x}}$$

$$y''=-\frac{\alpha\cos x}{2(1+\alpha\sin x)^{3/2}}$$

$$B'=2\xi\alpha\cos x$$

Effective force

$$f(y)=y'(a-\frac{B'}{2})-\frac{1}{2}by''\equiv F(x)$$

$$=\frac{-\eta \sin x-\frac{\xi\alpha}{2}\cos x}{\sqrt{1+\alpha\sin x}}$$

The associated potential is

$$U=-\int^ydz f(z)=\int^xdx\;y'F(x)$$

$$=\int^x\frac{\eta\sin x +\frac{\xi\alpha}{2}\cos x}{1+\alpha\sin x}$$

We are interested in the case $\alpha<<1$

$$U\approx\int dx[\eta\sin x-\alpha(\eta\sin^2x-\frac{\xi}{2}\cos x)]$$

$$=-\eta\cos x+\frac{\alpha}{2}[\eta(-x+\sin x\cos x)-\xi\sin x]+const.$$

For small $\alpha$ the potential has minima at

$$x=n 2\pi-\frac{\xi\alpha}{2\eta};\;\;n=0,\pm 1,\pm 2\cdots$$

The maxima occur at

$$x=(2n+1)\pi+\frac{\xi\alpha}{2\eta};\;\;n=0,\pm 1,\pm 2\cdots$$

The potential difference when a particle jumps over a barrier to the left or right is

$$\Delta U_L=\eta(2 +\frac{\alpha\pi}{2});\;\;\Delta U_R=\eta(2 -\frac{\alpha\pi}{2})$$

The Kramers escape rates are then

$$r_{L,R}=\eta\exp[-\frac{\eta}{\xi}(2 \pm\frac{\alpha\pi}{2})]$$

The net velocity of a particle in the potential is thus



BLOWTORCHES The previous example is an illustration of Landauer's blowtorch paradox.
The Kramers' escape rate formula depends only on the behavior of the potential near maxima and minima.
This interpretation can be misleading!
Blowtorch applied to "irrelevant" part of path effectively lowers potential maximum.

Some people believe similar considerations apply to markets.
Sometimes seemingly "irrelevant" events affects volatility!
Such events called "sunspots". SUNSPOTS MATTER!

TRANSFORMATION TO SCHRÖDINGER EQUATION
Fokker-Planck equation generally not self-adjoint.
Eigenfunctions associated with reflecting or
absorbing boundaries need not be orthogonal.
Consider Fokker-Planck equation

$$\frac{\partial P(x,t)}{\partial t}= -\frac{\partial f(x)P}{\partial x}+\xi\frac{\partial^2P}{\partial x^2}$$

Introduce new variable $\Psi$

$$P(x,t)=e^{-\lambda t}\exp(-\frac{U(x)}{2\xi})\Psi(x)$$

where

$$f=-\frac{dU}{dx}$$

Find that $\Psi$ satisfies Schrödinger type equation

$$-\xi\frac{d^2\Psi}{dx^2}+\left[\frac{1}{4\xi}(\frac{dU}{dx})^2 -\frac{1}{2}\frac{d^2 U}{dx^2}\right]\Psi=\lambda\Psi$$

with the expression inside the square bracket playing the role of a potential.

SUMMARY

• We have introduced two useful transformation of the Fokker-Planck equation.
• The first transformation eliminates a heterogeneous diffusion term.
• The second transformation changes the Fokker-Planck equation into a self-adjoint Schrödinger type equation.
• We have discussed in detail and example of variable diffusion constant.
• The result of discussion resolved the Blowtorch paradox.
  
  Click here for 9.Langevin approach
  Click here for Return to title page