# More intuition in understanding diffusion

This is a short conceptual question: (I think I'm confusing some very key concepts, here's an attempt to try and resolve it)

• Systems whose dynamics are governed by diffusion, for example a particle in a liquid, are subject to constant random forces (e.g. from molecular collisions with the liquid). Say we choose a simplistic model for this random process, namely a Gaussian process, then essentially, with no additional considerations, the particle's position evolution in time follows a Markov process. Furthermore, we know that the diffusion equation is first order in time, which implies(?) that knowing the position at some time $t=t_0$ determines the value of the position in succeeding times $t.$ (similar to Hamilton's equations of motion)
• But if the assumed dynamics is inherently random in the above, how can the governing differential equation be first order in time? (in other words, shouldn't the randomness imply unpredictability in the determination of position for the future of the system?)
• Does my mistake lie in trying to associate the position of the particle with the diffusion equation? whereas one usually relates the diffusive coefficient only with the root mean squared of the position in time and not the actual position: $\langle x^2 \rangle \propto Dt$.

Your confusion comes from the fact that you are confusing two different ways of representing the random motion. These two ways go by the names stochastic differential equation and fokker-planck equation

To establish a base case to reference later in the answer, let's first discuss what happens when there is no randomness. In this case, your differential equation looks like this:

$\dot{x} = 0.$

This is a first order ordinary differential equation and for each initial condition on $x$, this differential equation determines a unique trajectory (the trajectory where the particle just sits where it started).

# Stochastic differential equation

Now lets see what happens if we had randomness. There are two ways of represent the randomness. As I said, one way is called a stochastic differential equation. The equation is usually written this way:

$$\dot{x} = \eta(t)\tag{SDE}$$

Here $\eta$ has a random value drawn from a gaussian distribution with mean $0$ and standard deviation $\sigma$ at each moment in time. The values of $\eta$ at each point in time are independent of eachother. People often represent the last two facts about $\eta$ by the equations

$$\langle \eta(t) \rangle = 0$$

and

$$\langle \eta(t) \eta(s)\rangle = \sigma^2 \delta(t-s).$$

Notice that (SDE) is a first order differential equation, so given an initial condition and each realization of $\eta$, there is a unique solution. However, since $\eta$ is a random variable there will actually be many different possible solutions, even though the differential equation is first order. (I think this last point is crucial to resolving your confusion.) Thus this differential equation really gives you a probability distribution for $x$ at each time $t$.

# Fokker-planck equation

The other way of representing the randomness is to represent your knowledge about the particle position as a distribution $\rho$ right from the start, and to give a ordinary, totally deterministic differential equation for $\rho$. For general stochastic systems, this is called the fokker-planck equation. In the case of diffusion, the fokker-planck equation is just the diffusion equation:

$$\dot{\rho} = D \nabla^2 \rho$$

Integrating this differential equation, you can get the probability distribution $\rho$ as a function of time.

# Summary

So what we have found is that you can either get a stochastic differential equation for a postion, or a deterministic differential equation for a distribution, but you can't frame it as a deterministic differential equation for the position because the position needs to depend on the random steps it takes. So although both differential equations are first order, an initial condition still does not give you a unique solution. The randomness messes it up.

• A very succinct exposure, thanks a lot. Now I have a very consistent picture in mind, seen from both approaches. So is the simplified SDE you took as example, simply a continuous markov process? Where the future only depends on the current state? (thus $\delta(t-s)$ appearing in the correlator, I guess this becomes time dependent e.g. for self avoiding walks of the particle or some other constraint, like boundaries and so on.) Would you happen to have any recomnendations for getting started on SDE's? – user929304 Dec 6 '15 at 19:37
• I think your statements are correct. As for as recommendations, I don't really know much about SDE's myself. I guess I would look for books on math.se if I were you. Here are some links I found. math.stackexchange.com/q/1206722/75297, math.stackexchange.com/q/1028502/75297 – Brian Moths Dec 6 '15 at 20:02

Yes (to your last question). Statistics can be very accurate at predicting the distribution of a large number of samples, even though each sample is random within the laws that govern it. When you see a cloud of dye spreading it in a beaker of water, you are seeing the distribution.

• Very interesting. So my mistake was that I should have thought of the distribution of positions, and that the diffusuion equation is a diff. eq. for the distribution (and not single particle trajectories), right? – user929304 Dec 6 '15 at 9:03
• Correct. Each particle performs a random walk. For a smell in a room, you don't know where an individual molecule goes, but you know whether enough of them get into someone's nose that they can detect it. – Spirko Dec 6 '15 at 13:56