- #1
Abigale
- 53
- 0
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 ?
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 ?