Schrodinger equation in term of Fokker-Planck equation

Question

From Wikipedia on the Fokker-Planck equation:

$$\tag{1}\frac{\partial }{\partial t}f\left( x^{\prime },t\right) ~=~\int_{-\infty}^\infty dx\left( \left[ D_{1}\left( x,t\right) \frac{\partial }{\partial x}+D_2 \left( x,t\right) \frac{\partial^2}{\partial x^2}\right] \delta\left( x^{\prime}-x\right) \right) f\left( x,t\right).\qquad $$

Integrate over a time interval $\varepsilon$,

$$f\left( x^\prime ,t+\varepsilon \right) $$ $$~=~\int_{-\infty }^\infty \, dx\left(\left( 1+\varepsilon \left[ D_{1}\left(x,t\right) \frac{\partial }{\partial x}+D_{2}\left( x,t\right) \frac{\partial^{2}}{\partial x^{2}}\right]\right) \delta \left( x^\prime - x\right) \right) f\left( x,t\right)$$ $$\tag{2}+O\left( \varepsilon ^{2}\right).\qquad $$

OK, but Fokker-Planck equation for one dimension is usually

$$\tag{0} \frac{\partial}{\partial t}f(x,t) = -\frac{\partial}{\partial x}\left[\mu(x,t)f(x,t)\right] + \frac{\partial^2}{\partial x^2}\left[ D(x,t)f(x,t)\right].$$

I was not able to understand how one gets from the original equation (0) to the above (1) and how does the first equation (1) lead to the second equation (2). Can anyone explain this?

Answer 1

Hints:

$\underline{(0) \Rightarrow (1)}$: Don't try to accomplish everything at once. Do it slowly in as many steps as you need to be sure that you are calculating correctly and understand everything. The trick is to integrate by part. Be very careful to keep track of what depends on $x$ and what depends on $x^{\prime}$.

$\underline{(1) \Rightarrow (2)}$: Use Taylor series

$$f( x^{\prime} ,t+\varepsilon ) ~=~f( x^{\prime} ,t) +\varepsilon\frac{\partial }{\partial t}f\left( x^{\prime },t\right) +{\cal O}\left( \varepsilon ^{2}\right).$$

Answer 2

The Wikipedia article's author points out that the equations are “formally equivalent”. I think this means that both, Schrodinger and Fokker-Planck, equations describe the evolution of a function over time. Then, as it is done in quantum mechanics with Feynman path integrals, we may write the partial differential equation in terms of a path integral and talk about “propagating” the initial state through time.

Here is how to step from (0) to (1).

First, observe this property of the delta-function: $$ \int_{-\infty}^{\infty}f\left(x\right)\delta^{\left(n\right)}\left(x'-x\right)dx=\left(-1\right)^{n}f^{\left(n\right)}\left(x'\right) $$

where the superscript $\left(n\right)$ indicates the nth derivative. You can show this by integration by parts.

On the right hand side of equation (0), there are first and second $x$ derivatives of $f(x,t)$ and the derivatives of $\mu\left(x,t\right)$ and $D\left(x,t\right)$ can be lumped into the $D_{1}\left(x,t\right)$ and $D_{2}\left(x,t\right)$ of equation (1). $$ \frac{\partial}{\partial t}f(x,t)=\left(-D_{1}\left(x,t\right)\frac{\partial}{\partial x}+D_{2}\left(x,t\right)\frac{\partial^{2}}{\partial x^{2}}\right)f\left(x,t\right) $$

I am not sure what happens to the term $\left(\frac{\partial^{2}D\left(x,t\right)}{\partial x^{2}}-\frac{\partial\mu\left(x,t\right)}{\partial x}\right)f\left(x,t\right)$ . Maybe these space derivatives can be ignored, if we assume that the functions $\mu\left(x,t\right)$ and $D\left(x,t\right)$ do not vary significatlly with $x$.

Now, the delta-function derivatives can be used to write the terms on right hand side as:

$$ -D_{1}\left(x',t\right)\frac{\partial}{\partial x'}f\left(x',t\right)=-\int_{-\infty}^{\infty}D_{1}\left(x,t\right)\frac{\partial f\left(x,t\right)}{\partial x}\delta\left(x'-x\right)dx=\int_{-\infty}^{\infty}D_{1}\left(x,t\right)\frac{\partial\delta\left(x'-x\right)}{\partial x}f\left(x,t\right)dx $$

and similarly for the second term. Putting it together we get equation (1).

To get to equation (2) follow Qmechanic's hint and Taylor expand around $t$.

Answer 3

Going from equation (0) to (1) is basically by writing the adjoint of the linear operator in the Fokker Planck equation (with a standard $L^2$ inner product). So the first equation is basically the following statements- $$ \dfrac{\partial f(x,t)}{\partial t}=\mathcal{L}_xf\text{ where }\mathcal{L}_x\equiv-\dfrac{\partial}{\partial x}D_1(x,t)+\dfrac{\partial^2}{\partial x^2}D_2(x,t) $$ and given (for all real valued functions) $$ \langle f,g\rangle=\int_{\mathbb{R}}\mathrm{d}xf(x)g(x) $$ then the adjoint ($\mathcal{L}_x^\dagger$) of $\mathcal{L}_x$ is defined as $$ \langle\mathcal{L}f,g\rangle=\langle f,\mathcal{L}_x^\dagger g\rangle $$ which implies $$ \mathcal{L}_x^\dagger=D_1(x,t)\dfrac{\partial}{\partial x}+D_2(x,t)\dfrac{\partial^2}{\partial x^2} $$