# Fokker-Planck equation with time-dependent potential

Consider a Fokker-Planck (FP) equation where the advection term is a function of time, i.e. \begin{align} \frac{\partial P ( x , t )}{\partial t} = -\nabla \cdot \left[ -\mu \, P \, \nabla U (x,t) - D \nabla P \right]. \qquad\qquad ({\rm I}) \end{align} Q1 Are there general steady-state distributions (ie $$\partial_t P = 0$$) associated with this FP?

if in similarity with the equilibrium case, we set the probability current to zero, we obtain \begin{align} P (x,t) \propto \exp( - \mu U (x,t) / D ), \qquad\qquad ({\rm II}) \end{align} which is time-dependent, and therefore does not satisfy the FP equation.
However, I guess there could exist certain regimes that it can approximate the real solution (for example if $$U$$ varies slowly with time).

Q2 Under what conditions $$({\rm II}$$) could approximate the solution to $$({\rm I})$$? (note that a physical justification could also help).

Q3 In case the system (approximately) reaches the distribution given in $$({\rm II})$$, what sets the corresponding time-scale?‌ To clarify, I am trying to understand whether this would be a diffusive scale such as $$L^2/D$$ where $$L$$ is a typical length-scale in the system, or it would be set by the time dependence of the potential $$U$$.

Q1 As you have shown yourself, this equation does not have a steady-state distribution: if we set $$\partial_t P = 0$$, i.e., if we assume that the solution is time-independent, we still obtain a solution that depends on time, contradicting our assumption.
Q2 and Q3 In some situations one could indeed approximate the solution using form (II). The conditions can be obtained by substituting this form to the original equation and demanding that the residual term is small. One could consider, for example, a situation of approaching this quasi-equilibrium by using conjecture $$P(x,t) = C\exp\left[-\mu U(x,t)/D\right] + p(x,t),$$ where $$p(x,t)$$ describes the deviation from the quasi-equilibrium. One could then study whether this perturbation have enough time to dissipate on the time scale of potential $$U(x,t)$$, which will depend on the diffusion coefficient, the spatial scale of $$U(x,t)$$, and other spatial scales in the system (e.g., those set by the boundary conditions).
• Well, in principle you could have $\nabla\cdot\mathbf{J} = 0$ without $\mathbf{J}=0$. This could correspond to a steady-state solution (rather than an equilibrium solution where $\mathbf{J}$ is identical to zero). – SaMaSo Oct 10 '20 at 15:44
• in case it was not already clear, $\mathbf{J} = - \mu P \nabla U -D \nabla P$ is what I call probability current. – SaMaSo Oct 10 '20 at 15:46