# Is there a classical solution to this Fokker-Planck equation?

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?

• 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) Aug 4, 2022 at 21:06
• 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.
– Ian
Aug 4, 2022 at 21:09
• @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 $\delta t \to 0$.
– Dan
Aug 4, 2022 at 21:29
• @Ian I have edited the question. Thanks
– Dan
Aug 4, 2022 at 21:37

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 $$x_0 = 0$$.

Edit

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 $$e^{-t} \delta(x)$$.

• 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.
– Dan
Aug 4, 2022 at 21:33
• Thank you for the answer!
– Dan
Aug 4, 2022 at 23:11
• How can this be right when the integral isn't constant?
– Ian
Aug 5, 2022 at 3:51
• @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.
– Dan
Aug 5, 2022 at 5:01
• Sure, then it is fine, though that is just a weird way of writing $\delta(x)$ again.
– Ian
Aug 5, 2022 at 11:43