Fokker–Planck equations and probability transport
Derive the forward density equation from the diffusion generator, conserve probability through flux, and calculate an Ornstein–Uhlenbeck equilibrium.
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The question for today
What you’ll understand
- Distinguish the generator acting on observables from its adjoint acting on densities.
- Write the correct drift and diffusion signs in the forward equation.
- Check normalization, flux conditions, and a stationary Gaussian density.
A diffusion can be described by random paths or by the evolving probability of finding those paths in different regions. The Fokker–Planck equation is the second description. It does not track a particular particle, and its density is not a quantum amplitude. Drift transports probability while noise spreads it. Deriving the equation from Itô's formula fixes the signs and factors, and expressing it as a conservation law makes boundary conditions visible. A stationary solution then requires probability balance as well as a plausible-looking curve.
Average a test observable before differentiating a density
For a time-homogeneous Itô diffusion dX=b(X)dt+sigma(X)dB, apply Itô's formula to a smooth compactly supported test function f. Under suitable integrability, the stochastic integral has mean zero. The expectation of f therefore changes according to the generator Lf=b f′+(sigma²/2)f″. This weak formulation remains meaningful in settings where a classical smooth density has not yet been established. Existence of a density and enough regularity to differentiate it are separate analytical questions, not consequences of writing the SDE symbolically.
When a sufficiently smooth density p exists, write the expectation as the integral of f against p. Move derivatives from f to the coefficients times p using integration by parts. The first derivative gains a minus sign; the second derivative retains a plus sign. Boundary terms vanish for compactly supported probes, while global probability conservation still requires appropriate boundary or tail conditions. The resulting forward operator is the adjoint L* acting on distributions or densities, rather than the generator L acting on observables.
Flux states what crosses a boundary
Define probability flux J=bp−one-half the spatial derivative of sigma squared times p. Then the forward equation says that density decreases where more probability leaves than enters. Integrating across an interval gives the difference between incoming and outgoing boundary fluxes. Reflecting boundaries impose zero flux. Absorbing boundaries permit loss of probability from the interior; tracking that lost mass as absorption restores a complete probabilistic description. The boundary rule is part of the model and cannot be recovered from the interior differential equation alone.
For constant noise sigma=√(2D), pure Brownian diffusion has equation pₜ=D pₓₓ. Its fundamental Gaussian solution has variance 2Dt, which checks the coefficient. Confusing sigma with D introduces a factor-of-two error. For state-dependent noise, the diffusion term is one-half the second derivative of sigma²p, not merely one-half sigma² times pₓₓ. Derivatives of the coefficient matter. A change from Itô to Stratonovich notation also changes the corresponding drift unless the appropriate conversion is made.
Stationarity balances inward drift and outward diffusion
For the Ornstein–Uhlenbeck process dX=−kappa X dt+√(2D)dB, with positive kappa and D, the restoring drift pulls toward zero while diffusion spreads probability. On the real line, a normalizable zero-flux stationary density solves D p′=−kappa x p. Integrating the logarithmic derivative produces a Gaussian with variance D/kappa. The negative exponent is essential: changing its sign would yield a nonintegrable function that grows in the tails rather than a probability density.
Starting from a deterministic x₀, the mean decays as x₀e^(−kappa t), while variance grows from zero toward D/kappa. This separates relaxation of the center from spreading around it. Stationarity means the full distribution remains unchanged, not that sample paths stop fluctuating. Zero flux characterizes this equilibrium on the line with natural decay conditions, but it is not a universal requirement for every stationary system: a process on a circle can carry a nonzero constant circulating current while retaining a stationary density.
Work it through
Relaxation with an exactly checkable density
Consider with X₀=2. Find its stationary variance and its mean and variance at .
Match √(2D)=2, giving D=2, and identify κ=2. The stationary variance is D/κ=1, so the invariant density is the standard normal density, despite the process starting at 2.
At the specified time, . The mean is . The variance is . Thus the transient law is N(1,3/4).
Check stationary zero flux using . Then . The cancellation tests both the drift sign and the diffusion coefficient in the forward equation.
The invariant variance is 1. At time (log 2)/2 the transient mean is 1 and variance is 3/4; the stationary standard normal density has zero probability flux.
The living notebook
Explore the model
Interactive lab · fields
A drift field is only one part of probability evolution
Set rotation Ω to zero and source strength q below zero to inspect an inward-pointing softened two-dimensional vector field. Reverse q to see outward drift, then add rotation. This illustrates directional transport, but its q r/(r²+ε) field is not the linear Ornstein–Uhlenbeck drift, and the lab contains no density, diffusion, boundary flux, or Fokker–Planck solver.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA drift diagram can suggest transport direction, while density evolution additionally depends on diffusion and boundary conditions.
Pause & reason
Check your understanding
Your turn
Try it yourself
For dX=−3X dt+√6 dB, find the invariant variance and check whether increasing both the restoring rate and diffusion coefficient by the same factor changes that variance.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Identify D from the squared noise amplitude divided by two, then form D/κ.
Compare with the solution
Here κ=3 and D=6/2=3, so the stationary variance is 1. Multiplying both κ and D by a common positive factor leaves D/κ unchanged, preserving the invariant density while speeding up time evolution. The mean relaxation time 1/κ decreases, showing that an unchanged equilibrium does not imply unchanged dynamical behavior.
Keep the idea
Take it with you
- The forward density operator is the adjoint of the diffusion generator, with signs fixed by integration by parts.
- Probability conservation depends on flux and boundary conditions as well as the interior equation.
- For an Ornstein–Uhlenbeck process, stationary variance is D/κ when noise amplitude is √(2D).
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