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For a sufficiently regular unit-mass solution, the Ornstein-Uhlenbeck Fokker-Planck equation is ${\partial }_{t}f=\nabla \cdot (f\nabla \log (f/\gamma ))$. Integration by parts proves the displayed identity. Consequently relative entropy decreases, and its dissipation is relative Fisher information.

Multiply the Fokker-Planck equation by $B^n$ and integrate over $B>0$ and all $\sigma$, assuming the required boundary terms vanish. Integration by parts givesThe mixed moment is not determined by $\mathbb{E}[B^n]$: the current strain and the accumulated magnetic field are correlated. Zero mean strain does not imply $\mathbb{E}[\sigma B^n]=0$, so this is an unclosed moment equation.

Keep the strain dependence by defining $P_n(\sigma ,t)={\int }_0^{\infty }{B}^{n}P(B,\sigma ,t)dB$. The magnetic drift term satisfiesThe strain derivatives commute with the $B$ integral. HenceThis weighted density obeys a tilted Ornstein-Uhlenbeck Fokker-Planck equation. Its integral over $\sigma$ is the desired moment, while retaining $\sigma$ makes the evolution closed.

For the normalized standard Gaussian density,The two terms cancel, so the Ornstein-Uhlenbeck Fokker-Planck equation has stationary density for a Fokker-Planck equationEquivalently its Fokker-Planck probability current $-\nabla \gamma -v\gamma$ vanishes identically. The normalization follows from the Gaussian integral in each coordinate.

A probability density left unchanged by the Fokker-Planck equation ${\partial }_{t}f=L^{\ast }f$. Equivalently its Fokker-Planck probability current has zero divergence; the current itself need not vanish. For the Ornstein-Uhlenbeck Fokker-Planck equation, the normalized standard Gaussian density has zero current and is stationary.