Fokker-Planck P(y,t)?

1. Apr 5, 2013

Abigale

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 $\frac{\partial P(y,t)}{\partial y} = 0$. But why sould it be zero/ P=const ?

2. Oct 24, 2013

X89codered89X

I know this is an old post, but what is the author of the book you are reading?