Deriving Stationary Densities from the Fokker-Planck Equation

From Probability Flow to a Stationary Density

A stationary density is a probability density that does not change with time under the diffusion. The Fokker–Planck equation describes how a density evolves; at stationarity, it says that the probability current is constant in space. Setting that current to zero gives a first-order differential equation whose solution is the familiar one-dimensional density formula. The zero-current condition is an additional choice: stationarity alone implies constant current, not necessarily zero current.

Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Consider a nonexplosive solution, taking values in an open interval \(I\subseteq\mathbb{R}\), of \(dX_t=b(X_t)\,dt+\sigma(X_t)\,dB_t\). Here \(b\) is the drift coefficient, \(\sigma\) is the diffusion coefficient, and \(a(x)=\sigma(x)^2\). Assume for the density derivation that \(b\) and \(\sigma\) are locally regular enough for the displayed derivatives and integrals to exist, and that a classical density \(p(t,x)\) exists and has the derivatives used below. In the formula for a zero-current density, assume \(a(x)>0\) and \(b,a\) are continuous on \(I\).

Every Itô integral here is understood with an adapted, progressively measurable integrand in \(\mathcal{L}^2\) on the time interval in question: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For the test-function calculation below, \(H_s=\varphi'(X_s)\sigma(X_s)\); if \(\varphi\) is smooth with compact support and \(\sigma\) is locally bounded, this integrand is bounded and hence satisfies that condition.

Deriving the Stationary Probability-Current Equation

For a smooth compactly supported test function \(\varphi\), Itô’s formula has \(f'(x)=\varphi'(x)\) and \(f''(x)=\varphi''(x)\), and gives

The stochastic integral has mean zero: its integrand is progressively measurable and square-integrable, so the Itô integral is a square-integrable martingale. Taking expectations and using the density of \(X_t\) yields the weak form of the Fokker–Planck equation. Integration by parts, with no boundary terms because \(\varphi\) has compact support in \(I\), gives the classical equation when \(p\) is sufficiently smooth.

Define the probability current \(J(t,x)\) as the net density flow to the right: drift transports density at rate \(bp\), while diffusion contributes the derivative term below. Then the Fokker–Planck equation is a conservation law.

Proof. Stationarity makes \(\partial_t p=0\). Substituting into the conservation equation gives \(\partial_xJ(x)=0\) throughout \(I\). A differentiable function with zero derivative on a connected interval is constant there, so \(J(x)=j\) for some constant \(j\). This argument does not imply \(j=0\); that value requires a zero-current condition, often motivated by the boundary behavior of a particular model. \(\square\)

Solving the Zero-Current Equation

Now impose \(J(x)=0\). This is the stationary probability-current equation with zero flux, and it reduces to an ordinary differential equation. Fix any reference point \(x_0\in I\).

Proof. The equation \(J=0\) means \(bp-\tfrac12(ap)'=0\), or \((ap)'=(2b/a)(ap)\). Set \(q=ap\), which is positive for any positive solution because \(a>0\). Dividing by \(q\) gives \(q'/q=2b/a\). Integrating from \(x_0\) to \(x\) gives \(q(x)=q(x_0)\exp(\int_{x_0}^{x}2b(y)/a(y)\,dy)\). Dividing by \(a(x)\) proves the stated form with a positive multiplicative constant. Conversely, differentiating that expression gives \((ap)'=(2b/a)(ap)=2bp\), so its current is zero. The identically zero solution is also a solution of the differential equation, but is not a nonzero density. Finally, a positive multiple can have integral one exactly when the displayed function has finite positive integral; choosing its reciprocal as the constant gives total mass one. This proves both directions of the claim. \(\square\)

The formula is local to the interval where \(a>0\); it does not settle how the process behaves at endpoints or whether the candidate density is invariant for the SDE. Those questions require additional boundary and process-level conditions. Also, if a stationary solution has \(J=j\ne0\), it solves a different inhomogeneous first-order equation, so the zero-current formula does not describe every stationary solution.

Worked Densities and Direct Current Checks

When the Formula Does Not Give a Probability Density

The formula produces a positive solution of an ordinary differential equation wherever its assumptions hold, but a stationary probability density must also have finite integral. For example, if \(b(x)=\mu\) and \(\sigma(x)=s>0\) on \(\mathbb{R}\), the zero-current solution is proportional to \(\exp(2\mu x/s^2)\). If \(\mu=0\), this is constant and has infinite integral on \(\mathbb{R}\). If \(\mu>0\), it grows as \(x\to\infty\); if \(\mu<0\), it grows as \(x\to-\infty\). None of these cases is normalizable on the whole real line. A formal solution is not automatically an invariant probability density.

A common error is to infer that every stationary density must have zero current. The stationary equation only says \(J'=0\). For instance, replacing that conclusion by \(J=0\) silently discards the possible constant \(j\ne0\). A second error is to use the exponential factor alone and omit \(1/a(x)\). The zero-current equation is \((ap)'=(2b/a)(ap)\), so it is \(ap\), not \(p\), that satisfies the simple logarithmic derivative equation. Both distinctions matter when diffusion varies with position.