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


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 $m$?

Any answer, comments or references are highly appreciated.


1 Answer 1


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 $t\ge 0$


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}

  • $\begingroup$ Could you specify why $(5)$ holds? $\endgroup$
    – user128095
    Dec 2, 2021 at 12:14
  • $\begingroup$ @Neymar Please see the reasoning below $(1-5)$ $\endgroup$
    – GJC20
    Dec 2, 2021 at 12:21

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