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Variable substitution in Langevin equation and Fokker-Planck equation

  1. Jun 27, 2012 #1
    Dear all,
    I have a question about the variable substitution in Langevin equation and Fokker-Planck equation and this has bothered me a lot. The general Langevin equation is:
    $$\frac{dx}{dt}=u(x)+\sqrt{2 D(x)}\eta(t)$$
    and the corresponding Fokker-Planck equation is thus:
    $$\frac{\partial \rho(x)}{\partial t}=-\frac{\partial}{\partial x}\left[u(x)\rho(x)\right]+\frac{\partial^2}{\partial x^2}\left[D(x)\rho(x)\right]$$
    which means the stationary distribution of x should satisfy
    $$u(x)\rho(x)=\frac{\partial}{\partial x}\left[D(x)\rho(x)\right]$$
    However, problem emerges when I want to use a variable substitution y(x), since the Langevin equation becomes
    $$\frac{dy}{dt}=u(x)y'(x)+\sqrt{2 D(x)}y'(x)\eta(t)$$
    which the corresponding F-P equation
    $$\frac{\partial \rho(y)}{\partial t}=-\frac{\partial}{\partial y}\left[u(x)y'\rho(y)\right]+\frac{\partial^2}{\partial y^2}\left[D(x)y'^2\rho(y)\right]$$
    and the stationary distribution of y is thus
    $$u(x)y'\rho(y)=\frac{\partial}{\partial y}\left[D(x)y'^2\rho(y)\right]$$
    Considering
    $$\rho(x)dx=\rho(y)dy \Rightarrow \rho(x)=\rho(y)y'$$
    we can rewrite the stationary ρ(y) equation before as
    $$u(x)\rho(x)=\frac{\partial}{\partial y}\left[D(x)y'\rho(x)\right]=\frac{\partial}{\partial x}\left[D(x)y'\rho(x)\right]x'(y)=\frac{1}{y'}\frac{\partial}{\partial x}\left[D(x)y'\rho(x)\right]$$
    which is not equal to the stationary ρ(x) derived before. Is there anything wrong with my derivation? Can anyone help me to figure this out? :frown::frown::frown::frown::frown::frown::frown:
    I have posted the same question in classical physics forum a few days ago but not one replies me. I hope people here can show me where I am going wrong or why this result happens. Thanks so much! :smile::smile::smile:
     
  2. jcsd
  3. Jun 27, 2012 #2

    Mute

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    Homework Helper

    You have to be careful when doing changes of variables in a stochastic differential equation. There are two main interpretations of stochastic calculus that physicists like to use: the Stratonovich interpretation and the Ito interpretation. The nature of your problem dictates which one to use, and things like changing variables are not the same in both (I'm afraid I don't remember the guidelines for which one you should use in which case). In Stratonovich, the change of variables follows the usual chain rule, while in Ito an additional term is produced. See http://en.wikipedia.org/wiki/Ito_Calculus#It.C5.8D_calculus_for_physicists

    I believe the distinction is very important when you have multiplicative noise, as you do in your equation. (Also, I have been told that the distinction is only important when you have delta-function or singular noise correlations. If the noise correlations are given by some smoothly decaying function C(t-t'), then apparently the distinction between the two interpretations is unimportant).

    So, I think the problem may be that you should be using the Ito calculus and the additional term produced by the change of variables.
     
  4. Jun 28, 2012 #3
    Yes I think you get the point. It seems that Ito calculus uses a different method for changing the variables. I am really unfamiliar with this area before so your link really helps. Thanks a lot!
     
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