The derivation of the backward fokker planck equation

[science_boy]

Homework Statement

Starting with the F.P. equation, where the probability has been defined in terms of the stationary probability and another probability function, Q(x,t), i am trying to derive the backward F.P. equation in terms of Q(x,t).

Homework Equations

The F.P. equation provided: [d(P(x,t)]/dt=-{d[f(x)P(x,t)]/dx}+D{[d^2P(x,t)]/dx^2}
where f(x) is the applied force

P(x,t) = Ps(x)Q(x,t), where Ps(x) is the stationary distribution function of the F.P. eqn.

I know the end result, ie the backward F.P. eqn is:

[d(Q(x,t)]/dt=f(x){d[Q(x,t)]/dx}+D{[d^2Q(x,t)]/dx^2}

In my attempt i have made use of equations:

Fick’s law: J(x,t)=-D*del[n(x,t)]

conservation of mass

and the diffusion equation:

d[n(x,t)]/dt=D*del^2[(n(x,t)]

The Attempt at a Solution

So far i have managed to gain the backward F.P. eqn. However i made use of this equation:

d[Ps(x)f(x)]/dx-D[d^2(Ps(x))/dx^2]=0 (eq.1)

Whilst familiar, i cannot seem to obtain this equation form. Instead of just stating it, i really need to understand where it comes from. Through research i seem to believe it comes from Fick's law with an external force. Where making use of Fick's law along with the conservation of mass law i can obtain a similar equation. This is the diffusion equation equalling to zero. Still, i am unconvinced of that method to gain (eq.1).

Does anyone recognise Eq.1. or can you seem to get a more conclusive derivation of Eq.1 than me! Any advice would be much appreciated.

Thanks

Science_boy

 

[mark726]

Dear Science_boy,

Thank you for your post. I can see that you have put a lot of effort into your attempt to derive the backward F.P. equation. It is important to understand the derivation of equations in order to fully grasp their meaning and significance.

I would suggest starting with the definition of the stationary distribution function, Ps(x), which is the solution to the F.P. equation when the time derivative is equal to zero. Therefore, we have:

0 = -{d[f(x)Ps(x)]/dx}+D{[d^2Ps(x)]/dx^2}

Rearranging this equation, we get:

{d[f(x)Ps(x)]/dx} = D{[d^2Ps(x)]/dx^2}

Now, let's substitute P(x,t) = Ps(x)Q(x,t) into the F.P. equation:

[d(P(x,t)]/dt = -{d[f(x)P(x,t)]/dx}+D{[d^2P(x,t)]/dx^2}

Substituting in the expression for P(x,t), we get:

[d(Ps(x)Q(x,t))/dt = -{d[f(x)Ps(x)Q(x,t)]/dx}+D{[d^2(Ps(x)Q(x,t))]/dx^2}

Expanding the derivatives, we get:

[dPs(x)/dt * Q(x,t) + Ps(x) * dQ(x,t)/dt] = -{d[f(x)Ps(x)]/dx * Q(x,t) + f(x)Ps(x) * dQ(x,t)/dx} + D{[d^2Ps(x)]/dx^2 * Q(x,t) + 2dPs(x)/dx * dQ(x,t)/dx + Ps(x) * d^2Q(x,t)/dx^2}

Using the definition of Ps(x), we can rewrite the first and fourth terms on the right-hand side of the equation as zero. Therefore, we are left with:

[dPs(x)/dt * Q(x,t) + Ps(x) * dQ(x,t)/dt] = -{d[f(x)Ps(x)]/dx * Q(x,t) + f(x)Ps(x) * dQ(x,t)/dx} + D{2dPs(x)/dx * dQ(x,t)/dx}

Now, let