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Fokker Planck Solution Biased Random Walk

  1. Dec 15, 2013 #1
    This is part b) of an assignment question. In part a) we were asked to derive the Fokker Planck relation for the biased random walk. The answer is:

    dP/dt = -vdP/dx + D d2P/dx2

    Where the first term is the drift term due to the biased motion and the second term is the diffusion term.

    Then in part b) we were asked to solve this differential equation for P(x,t). I already have the written out solution to this question however I don't understand one of the steps the lecturer has taken.
    First of all we define a new frame of reference y=x-vt in an attempt to get rid of the drift term (as we will then simply have the non-biased fokker-planck relation which we can solve). We then set out to rewrite the partial derivatives from the above equation in terms of the new variable y. Here the lecturer's solution emphasises that when deriving with respect to time, y is not a constant (y=y(t)) and x is considered constant. He writes the following:

    dP(y,t)/dt |x=const=dP(y,t)/dt|y=x-vt

    =dP(y,t)/dt + dy/dt * dP(y,t)/dy

    I'm not sure why he has taken that last step - where have the two terms come from? It seems to me like he is adding the derivative twice (as the second term in the last line is just the dP/dt term again expanded using the chain rule).

    If anyone can explain why he's done it this way then I will understand the rest of the question okay.

    Thanks!
     
  2. jcsd
  3. Dec 18, 2013 #2
    It's just the chain rule:
    [tex]\frac{d}{dt} f(y(t), z(t)) = \frac{\partial f}{\partial y} \frac{dy}{dt}+\frac{\partial f}{\partial z} \frac{dz}{dt}[/tex]

    Where your function has the form [itex]P = P(y(t), t)[/itex] <-- has two t dependent terms
    To make this clearer let y(t) just as is and put z(t)=t, So that your [itex]P(y(t), z(t))[/itex],
    So by the chain rule
    [tex]\frac{d}{dt} P(y(t), z(t)) = \frac{\partial P}{\partial y} \frac{dy}{dt}+\frac{\partial P}{\partial z} \frac{dz}{dt}[/tex]
    but remember [itex]z[/itex] was just [itex]t[/itex], so [itex]dz/dt=1[/itex] and you get
    [tex]\frac{d}{dt} P(y(t), z(t)) = \frac{\partial P}{\partial y} \frac{dy}{dt}+\frac{\partial P}{\partial t}[/tex]
    The Last piece is just a bit of notational detritus since [itex]P[/itex] is actually a function of [itex]x[/itex]
    too you write the derivative on the left as partials
    [tex]\frac{\partial }{\partial t} P(y(x,t), t) = \frac{\partial P}{\partial y} \frac{dy}{dt}+\frac{\partial P}{\partial t}[/tex]
    but you have to remember that [itex]\frac{\partial }{\partial t}[/itex] has different meanings on each side
    of the equation. on the left it is holding x constant while on the right it is holding y constant.
     
  4. Dec 18, 2013 #3
    That makes perfect sense now, thanks a lot!
     
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