Is there a classical solution to the following Fokker-Planck equation?:

$$\frac{\partial}{\partial t} P(x,t) = -\frac{\partial}{\partial x} (x\,P(x,t)),$$ $$ P(x,0) = \delta(x).$$

I know that if the initial condition was taken as $x(0)=x_0$ or equivalently $P(x,0)=\delta(x-x_0)$ with $x_0 \neq 0$, then there exists a deterministic solution given by $$x(t) = x_0 \, e^{t}.$$

The solution is as follows,

Take $g(x,t) = x\,P(x,t)$. Then $g$ satisfies the following PDE,

$$ \frac{\partial}{\partial t} g(x,t) = -x \,\frac{\partial}{\partial x} g(x,t),$$ $$ g(x,0)= \delta(x-x_0).$$ The solution to the above PDE is deterministic:

$$x(t) = x_0\,e^t.$$

But what if the initial value was zero, i.e., $x_0 =0$?.

My guess is that, at the beginning, the particle fluctuates, but once it reaches some point x with $|x|>0$, it grows exponentially as the above solution for the non-zero initial condition. So, $P(x,t)$ is the convex combination of a Gaussian distribution and a delta distribution(?). How can we put these together to have an exact solution?

  • $\begingroup$ I am not sure this is a Fokker-Planck equation. The $x$ should be inside the derivative operator, a minus is missing, and the case is degenerate (there is no diffusion) $\endgroup$
    – Snoop
    Aug 4, 2022 at 21:06
  • $\begingroup$ This exact case is very simple, the particle is just still forever. With arbitrarily small diffusion you would have a different picture. Also, as was correctly stated already, this is not a Fokker-Planck type equation even for a deterministic flow, since it cannot be put into divergence form. $\endgroup$
    – Ian
    Aug 4, 2022 at 21:09
  • $\begingroup$ @Snoop You're right. I jumped from FP to a PDE equation fast. I have edited the question. There is no diffusion term. It vanishes in the limit: $\delta t=\delta x$ and then $\delta t \to 0$. $\endgroup$
    – Dan
    Aug 4, 2022 at 21:29
  • $\begingroup$ @Ian I have edited the question. Thanks $\endgroup$
    – Dan
    Aug 4, 2022 at 21:37

1 Answer 1


The distributional solution of this pde with initial data $\delta(x_0 - x)$ is $\delta(x - x_0 e^t)$, for all $x_0$ including $x = 0$. The particle just sits at $x_0 = 0$.


The above answer is for the equation $P_t = xP_x, \, P(x,0) = \delta(x - x_0)$.

For $P_t = -(xP)_x = - xP_x - P, \, P(x,0) = \delta(x - x_0)$, the distributional solution is $e^{-t} \delta(x - x_0e^{-t})$. That is if $P(x,0) = \varphi(x)$, then $P(x,t) = e^{-t} \varphi(xe^{-t})$. Note that $\int_\mathbb{R}P(x,t) \, dx$ is constant.

So for $P(x,0) = \delta(x - x_0)$, we obtain the distributional solution $e^{-t} \delta(x)$.

  • $\begingroup$ Thank you, I have edited the question. I was missing something. Not sure how the edited question is different from the original one. But now the deterministic flow grows exponentially. $\endgroup$
    – Dan
    Aug 4, 2022 at 21:33
  • $\begingroup$ Thank you for the answer! $\endgroup$
    – Dan
    Aug 4, 2022 at 23:11
  • $\begingroup$ How can this be right when the integral isn't constant? $\endgroup$
    – Ian
    Aug 5, 2022 at 3:51
  • $\begingroup$ @Ian the solution is actually $e^{-t}\delta(xe^{-t})$ not $e^{-t}\delta(x)$ (probably there is a typo in the answer but please correct me if I'm wrong) and the integral of $e^{-t}\delta(xe^{-t})$ is 1 based on the scaling property of Dirac delta. $\endgroup$
    – Dan
    Aug 5, 2022 at 5:01
  • $\begingroup$ Sure, then it is fine, though that is just a weird way of writing $\delta(x)$ again. $\endgroup$
    – Ian
    Aug 5, 2022 at 11:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .