# Why do we interpret the first term of the Fokker-Planck equation as drift?

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?

• 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. Commented Jun 5, 2019 at 14:46
• 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. Commented Jun 6, 2019 at 4:55
• 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 Commented Jun 6, 2019 at 4:57
• 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. Commented Jun 6, 2019 at 5:00

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.

• 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. Commented Jun 6, 2019 at 5:18

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 $$A$$.

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.) $$-\partial_z\left[A(z)\delta(z-x(t))\right] = -\partial_z\left[A(x(t))\delta(z-x(t))\right] = -A(x(t))\partial_z\delta(z-x(t))$$
Note that I have renamed $$x$$ to $$z$$ from the orginal answer to avoid ambiguity.
• 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. Commented Jun 6, 2019 at 5:22