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