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With the derivation of the Fokker-Planck equation we get:

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

We then interpret the first term as the drift with $A(x,t)$ as drift velocity and the second term as diffusion with $B(x,t)$ as $2D$ with $D$ as the diffusion coefficient. The second term looks similar to the diffusion equation, so I understand this part of the interpretation. What is the reason for the drift interpretation?

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    $\begingroup$ Do you understand the derivation you link to? I feel like if you really understood it you would know why this is a drift term. $\endgroup$ Jun 5, 2019 at 14:46
  • $\begingroup$ In the lecture we did derive the Fokker Planck equation from the master equation doing the Kramers-Moyal expansion. I would say I understand that mathematical derivation but the interpretation then seemed rather sudden. The link above was rather for the concept itself than for the derivation, because the lecture took a different approach and I wanted to understand that one and not the derivation on Wikipedia. That was an unfavorable place for the link. $\endgroup$
    – PhylomatX
    Jun 6, 2019 at 4:55
  • $\begingroup$ I guess it's just weird that you keep calling it an interpretation when it's actually a drift term. It's like saying that in the expression for kinetic energy $K=\frac12mv^2$ one can interpret $v$ to be a velocity $\endgroup$ Jun 6, 2019 at 4:57
  • $\begingroup$ Sure, as I said, I'm not sure how to see that it is a drift term when I'm doing the expansion, so there is definitely something which I don't understand. I think I have to reformulate the question then. $\endgroup$
    – PhylomatX
    Jun 6, 2019 at 5:00

2 Answers 2

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That term doesn't just appear - it comes from introducing a drift term in the first step of the derivation from your link.

$${\displaystyle dX_{t}=\mu (X_{t},t)\,dt+\sigma (X_{t},t)\,dW_{t}}$$

The $\mu$ is a mean drift velocity which leads to a $dx = v*dt$ relationship in that equation, while the second term models diffusion. So to answer your question, we aren't interpreting it once we get that final result; we know it is a drift term from our initial assumptions in writing that first step down.

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    $\begingroup$ The link was misplaced. I want to understand the derivation doing the Kramers-Moyal expansion. The derivation on Wikipedia is rather short and I haven't learned anything about SDEs yet. I'll post a new reformulated question. $\endgroup$
    – PhylomatX
    Jun 6, 2019 at 5:18
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As already stated in alex1stef2's answer, $A$ comes from the literal drift term in the SDE of the process.

But, let's look at this another way and assume you were only given the FP-equation without knowledge of its derivation.

If you start the dynamics with an initial deterministic distribution $P(z,0)=\delta(z-x0)$ and set $B\equiv 0$, you'll find that $ P(z,t) = \delta(z-x(t)) $ where $x(t)$ is the solution of the ordinary differential equation $$ \dot{x} = A(x) $$ i.e. the deterministic equation of motion with initial value $x_0$.

In the absence of the diffusion term, the distribution stays deterministic and moves according to the drift prescribed by $A$.

See also: Gardiner, Stochastic Methods for a derivation and discussion.


Addendum: Plugging in $\delta(z-x(t))$ on either side of the Fokker-Planck-equation yields

1.) $$ \partial_t \delta(z-x(t)) = -\partial_z\delta(z-x(t))\frac{dx}{dt} = -\partial_z\delta(z-x(t))A(x(t)) $$ 2.) $$-\partial_z\left[A(z)\delta(z-x(t))\right] = -\partial_z\left[A(x(t))\delta(z-x(t))\right] = -A(x(t))\partial_z\delta(z-x(t)) $$

The first equality in 2.) is due to the delta-function. Check this if you are unsure.

Note that I have renamed $x$ to $z$ from the orginal answer to avoid ambiguity.

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  • $\begingroup$ That approach of just looking at the equation is nice, but I don't understand how you get to your $\dot{x} =A(x)$ formula in detail. I guess I'll have to get the Gardiner book somewhere. $\endgroup$
    – PhylomatX
    Jun 6, 2019 at 5:22
  • $\begingroup$ @JoKli Okay, I'll expand on that. But I cannot recommend Gardiner's book highly enough. IMHO it's excellent both as a reference and for learning. $\endgroup$
    – Nephente
    Jun 6, 2019 at 6:51

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