# A variant to the Fokker–Planck equation

Consider the PDE of $$p(t,x)\ge 0$$ given as

$$\partial_t p = \frac{\partial^2_{xx}p}{(1+m(t))^2} - \partial_x p,\quad \forall t,~x \in (0,\infty)$$

with initial and boundary conditions $$p(0,\cdot)=\rho$$, $$p(\cdot,0)=0$$ and $$p(\cdot,\infty)=0$$, where $$\rho:(0,\infty)\to\mathbb R_+$$ is a probability density and

$$m(t)=\int_0^{\infty}p(t,x)dx.$$

This PDE originates from Fokker–Planck equation for very degenerate diffusion processes and $$p(t,\cdot)$$ stands for a flow of sub-probabilities whose total mass is decreasing, i.e. $$0\le m\le 1$$ is decreasing on $$\mathbb R_+$$. Could we expect an explicit formula for $$p$$, or derive the ODE satisfied by $$m$$?

This is not an answer, while I believe it is more suitable to write here instead of posting a related question. Let me know if it is not acceptable.

Set for $$t\ge 0$$

$$M(t):=\int_0^t(1+m(s))^2ds.$$

Then $$M:\mathbb R_+\to\mathbb R_+$$ is continuous (even differentiable) and strictly increasing. Define $$q: \Omega\to\mathbb R_+$$ by $$q(t,x):=p\big(M(t), x+M(t)\big)$$, where $$\Omega:=\{(t,x)\in\mathbb R^2: t\ge 0, x\ge -M(t)\}$$. A straightforward computation yields

$$\begin{eqnarray} \partial_t q &=& \partial^2_{xx} q,\quad \forall t>0, x>-M(t),~~~~~~~~~~~~(1) \\ M(t) &=& \int_0^t (1+m(s))^2ds,\quad \forall t\ge 0~~~~~~~~~~~~ ~(2) \\ q(0,x)&=&\rho(x),\quad \forall x\ge 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (3)\\ q(t,-M(t))&=& 0,\quad \forall t>0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (4)\\ \partial_x q\big(t,-M(t)\big)&=&-m'(t)\big(1+m(t)\big)^2, \quad \forall t>0~~~~~~~~~~~~(5) \end{eqnarray}$$

Here $$(1-5)$$ is quit similar to the system arising in Lemma 2.1 of Classical Solutions for a nonlinear Fokker-Planck equation arising in Computational Neuroscience. So my question can be reformulated as whether $$(1-5)$$ is well posed?

PS : (5) follows from the integration of $$p$$ on $$\mathbb R_+$$ with respect to $$x$$:

$$\begin{eqnarray} m'(t)=\partial_t\left(\int_0^{\infty}p(t,x)dx\right)& =&\int_0^{\infty}\partial_tp(t,x)dx \\ &=& \frac{1}{(1+m(t))^2}\int_0^{\infty}\partial^2_{xx}p(t,x)dx - \int_0^{\infty}\partial_{x}p(t,x)dx \\ &=& \frac{1}{(1+m(t))^2}\partial_{x}p(t,x)|_0^{\infty} \\ &=& -\frac{1}{(1+m(t))^2}\partial_{x}p(t,0). \end{eqnarray}$$

• Could you specify why $(5)$ holds?
– user128095
Dec 2, 2021 at 12:14
• @Neymar Please see the reasoning below $(1-5)$ Dec 2, 2021 at 12:21