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Fokker-Planck P(y,t)?

  1. Apr 5, 2013 #1
    I am Reading in a Book of Stochastic Processes.

    I understood the Derivation of the Fokker-Planck equation from the master equation.
    The Result is (the FPE):
    $$
    \frac{\partial P(y,t)}{\partial t}
    =
    - \frac{\partial}{\partial y}
    { \lbrace {a_{1}(y)P} \rbrace }
    +
    \frac{1}{2}
    \frac{\partial ^{2} }{\partial ^{2} y}
    {\lbrace {a_{2}(y)P} \rbrace}
    $$

    Than the author recommits to the FPE, which he introduced at the beginning of the chapter.
    He says, both are equal.

    $$
    \frac{\partial P(y,t)}{\partial t}
    =
    - \frac{\partial}{\partial y}
    A(y)P
    +
    \frac{1}{2}
    \frac{\partial ^{2} y}{\partial ^{2}}
    B(y)P
    $$

    I don't understand why they should be equal.
    I think that they are just equal, wenn [itex]\frac{\partial P(y,t)}{\partial y} = 0[/itex]. But why sould it be zero/ P=const ?
     
  2. jcsd
  3. Oct 24, 2013 #2
    I know this is an old post, but what is the author of the book you are reading?

    I am not sure if I can answer your question.
     
  4. Oct 24, 2013 #3
    I know this is an old post, but what is the author of the book you are reading?

    I am not sure if I can answer your question.
     
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