# Particle in a 1-d potential/ Fokker-Planck(FPE)

1. Apr 6, 2013

### Abigale

Hi guys,
I have a Problem.
I regard a brownian particle in a potential $U(z)=gz$ with $f=-\frac{d}{dz}U$ and it is connected to a heat reservoir.
The particle is in the interval $z>0$.

I regard the equilibrium-state.
So I got the FPE.

$$\frac{\partial}{\partial t} P_{1}(z,t) = -\frac{\partial}{\partial z}[\Gamma~f(z)P_{1}(z,t)] + \frac{D}{2} \frac{\partial^{2}}{\partial z^{2}} P_{1}(z,t) =0$$

$$\Rightarrow -\Gamma~f(z)P_{1}(z,t) + \frac{D}{2} \frac{\partial}{\partial z} P_{1}(z,t) =0$$

The solution is:
$$P_{eq}(z,t)=\frac{2\Gamma g}{D}\exp{\lbrace\frac{-2\Gamma g z}{D} \rbrace}$$

Please check if the following callculations are correct?

Now I regard the situation:

At a certain time $t_{0}$ opens a door, which increases the first volume.
The volume ranges now from $-z_{0}$ to $+ \infty$. ($z_{0}>0$)
But I am just interested in the equilibrium state of this situtaion.

I think the FPE must have the same form like the first case. So again:

$$\frac{\partial}{\partial t} P_{2}(z,t) = -\frac{\partial}{\partial z}[\Gamma~f(z)P_{2}(z,t)] + \frac{D}{2} \frac{\partial^{2}}{\partial z^{2}} P_{2}(z,t) =0$$

$$\Rightarrow -\Gamma~f(z)P_{2}(z,t) + \frac{D}{2} \frac{\partial}{\partial z} P_{2}(z,t) =0$$

$$\Rightarrow \Gamma~gP_{2}(z,t) + \frac{D}{2} \frac{\partial}{\partial z} P_{2}(z,t) =0$$

$$\Rightarrow \Gamma~g dz = - \frac{D}{2} \frac{1}{P_{2}(z,t)} d(P_{2}(z,t))$$
$$P_{2}=C\exp{\lbrace \frac{-2\Gamma g z}{D} \rbrace}$$

Now for normalisation:
$$\int \limits_{-z_{0}}^ {+ \infty} P_{2}(z,t) ~dz =1$$

Can this be true? Is the range of integration, the only thing i have to change for this situation?

THX
and sorry for the long letter.
Bye
Abby