I need your help, could you please explain me the sentence

"The diffusion equation is the Fokker-Planck equation for the Brownian motion".

I have tried to use some assumption and transform a multidimensional Fokker-Planck equation into the diffusion equation. Could you please explain me how to do that?

  • 1
    $\begingroup$ What's unclear about the sentence, exactly? It's essentially saying that you can represent an SDE as a PDE. $\endgroup$ – alarge Oct 25 '18 at 6:41
  • $\begingroup$ I have not clear what are the assumption that made this possible: in particolare why the mixed derivatives in the diffusion term disappear $\endgroup$ – user502940 Oct 25 '18 at 11:30
  • $\begingroup$ It is more an algebraic problem, of how I can obtain, using some assumption like drift coffiecient equal to zero, the diffusion equation from the FP equation $\endgroup$ – user502940 Oct 25 '18 at 11:52
  • $\begingroup$ There are no extra assumptions. If you're looking for a derivation of FP, Google will find a host of resources and as it's part of any textbook on the topic, it's not probably very useful to copy a derivation here. I also don't quite understand what you mean by mixed derivatives disappearing, as per the Wikipedia page, multidimensional FP certainly has mixed derivatives. In any case, your question does not seem to have anything to do with physics and you will probably have better luck getting your question answered at the Maths Stack Exchange. $\endgroup$ – alarge Oct 26 '18 at 11:28

TL;DR we make the physical assumption that $\mathbf{D}$ is diagonal and independent of position.

The Fokker-Planck equation Wikipedia page states the equivalence of the two key equations.

The stochastic differential equation $$ d\mathbf{X}_t = \boldsymbol{\mu}(\mathbf{X}_t,t)\, dt + \boldsymbol{\sigma}(\mathbf{X}_t,t)\, d\mathbf{W}_t $$ is essentially the Langevin equation (without inertia) describing the Brownian motion of a set of particles whose positions are represented by the vector $\mathbf{X}_t \equiv \{X_{i,t}\}$ at time $t$. The drift term $\boldsymbol{\mu}\equiv\{\mu_i\}$ may be related to external and inter-particle forces, and to the particle mobilities (hence to diffusion coefficients). The term $d\mathbf{W}_t$ represents a Wiener process and the prefactor $\boldsymbol{\sigma}$ is a matrix, also related to diffusion (see below).

The Fokker-Planck equation is $$ \frac{\partial p(\mathbf{x},t)}{\partial t} = - \sum_i \frac{\partial}{\partial x_i} \left[ \mu_i(\mathbf{x},t)p(\mathbf{x},t) \right] + \sum_i \sum_j \frac{\partial^2}{\partial x_i\partial x_j} \left[ D_{ij}(\mathbf{x},t)p(\mathbf{x},t) \right] $$ In the absence of external forces and interparticle forces, the drift terms $\boldsymbol{\mu}\equiv\{\mu_i\}$ are zero, but it is not essential to make this assumption. The diffusion tensor may be expressed $$ D_{ij}=\frac{1}{2}\sum_k \sigma_{ik} \sigma_{jk} $$

The link between those two equations can be found in many places, but my impression is that you are just worried about how the Fokker-Planck equation becomes a simple diffusion equation. This is because we usually make the physical assumptions that the system is homogeneous and that there are no hydrodynamic interactions between the particles. So we take the diffusion tensor to be

  1. diagonal, so $D_{ij}=0$ if $i\neq j$;
  2. position-independent, so $\partial D_{ii}/\partial x_i=0$

Once this is done, the diffusive term becomes simply $$ \sum_i D_{ii}\frac{\partial^2}{\partial x_i^2} p(\mathbf{x},t) $$ and (if we have no drift term) the Fokker-Planck equation factorizes into separate diffusion equations for each of the particles, involving the usual $D_{ii}\nabla_i^2 p_i$ diffusive term for each. Along with this, we get independent Langevin equations. If we allow inhomogeneity by letting the diffusion tensor depend on positions, but still assume that it is diagonal, then we get a slightly more complicated version of the diffusion equation. In the presence of drift terms, particularly interparticle forces, then the equations are coupled through these terms.

If we don't make those assumptions, then the full form of the Fokker-Planck equation essentially is the diffusion equation for the system.

We do not always make those assumptions. In treatments of Brownian motion with hydrodynamic interactions, the averaged effects of the hydrodynamics are often represented approximately through a diffusion tensor which depends on the separation vectors of all pairs of particles. Examples are the Oseen tensor or the Rotne-Prager tensor (which is always positive-definite, and hence preferable, because of the relation to $\boldsymbol{\sigma}$). In each case, for $N$ particles in 3D, so $\mathbf{x}\equiv\{\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N\}$ is a vector of length $3N$ consisting of $N$ position vectors, the diffusion tensor takes the form of an $N\times N$ array of individual $3\times3$ matrices. The matrices along the diagonal take the usual diagonal form, while the matrices corresponding to the coupling between particles are functions of the vector between them. Here, as an example, is a typical off-diagonal term in the Oseen tensor $$ \mathbf{D}_{ab}^{(3\times3)}= \frac{k_B T}{8\pi\eta r_{ab}} \bigl(\boldsymbol{1} + \hat{\mathbf{r}}_{ab} \hat{\mathbf{r}}_{ab}\bigr) \qquad a\neq b $$ where $a$ and $b$ label the particles, $r_{ab}$ is the distance between them, $\hat{\mathbf{r}}_{ab}$ the unit vector pointing between them, $k_B$ is Boltzmann's constant and $\eta$ the viscosity of the surrounding fluid. Then the cross-terms in the diffusion equation, and in the Langevin equation, quite properly, do not go away.


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