Consider


$$X\_t=X\_0 + \int\_0^t b(s)ds+ \int\_0^t \sigma(s)dW\_s,\quad \forall t\ge 0,$$


where $X\_0\ge 0$ is a random variable of density $\rho$, $(W\_t)\_{t\ge 0}$ is an independent Brownian motion and $b,\sigma$ are measurable function "as nice as possible". Set $\tau:=\inf\{t\ge 0: X\_t\le 0\}$ and $Y\_t:=X\_{t\wedge \tau}$ for $t\ge 0$. Denote by $p(t,\cdot)$ be the density of $Y\_t$ restricted on $(0,\infty)$, i.e.


$$\mathbb E[f(Y\_t){\bf 1}\_{\{Y\_t>0\}}] = \int\_0^{\infty}f(x)p(t,x)dx,\quad \mbox{for all continuous and bounded function } f: \mathbb R\to\mathbb R.$$


It is known that $p$ satisfies the Fokker–Planck equation as below :


\begin{eqnarray}
\partial\_t p &=& \frac{\sigma(t)^2}{2}\partial^2\_{xx} p - b(t)\partial\_x p,\quad \forall t>0, x>0 \\
p(0,x) &=& \rho(x),\quad \forall x\ge 0 \\
p(t,0) &=& 0,\quad \forall t>0. 
\end{eqnarray}


From the probabilistic point of view, it is straightforward to see $p\ge 0$ and the function


$$m(t):=\int\_0^{\infty}p(t,x)dx \in [0,1]$$


is non-increasing. From the PDE point of view, why the above Fokker–Planck equation together with the initial and boundary conditions implies $p\ge 0$ and $m$ is non-increasing?


Any answer, comments or references are appreciated.


PS : When $b\equiv 0$ and $\sigma\equiv 1$, one has in view of the reflection principle,


$$p(t,x)=\int\_0^{\infty}\frac{1}{\sqrt{2\pi t}}\left(\exp\left(-\frac{(x-y)^2}{2t}\right)-\exp\left(-\frac{(x+y)^2}{2t}\right)\right)\rho(y)dy$$


which clearly satisfies the desired properties. For general $b,\sigma$, do we still have the closed-form expression of $p$?



PDE interpretation of some properties of the solution to Fokker–Planck equations
Differentiating $m(t)$ and using the equation leads to
$$
m'(t):=\int\_0^{\infty}\left(
\frac{\sigma^2(t)}{2}\partial^2\_{xx} p(x,t) - b(t)\partial\_x p(x,t)\right)\,dx=
-\frac{\sigma(t)^2}{2}\partial\_{x} p(0,t)<0.
$$
The last inequality follows from the Zaremba-Giraud theorem for the sign of the solution's normal derivative at the boundary. See, for example, Nazarov A.I., [*A Centennial of the Zaremba–Hopf–Oleinik Lemma*](https://arxiv.org/abs/1101.0164).


For $b\equiv0$
$$
p(t,x)=\int\_0^{\infty}\frac{1}{\sqrt{2\pi \int\_0^t{\sigma^2(\tau)}\,d\tau}}\left(\exp\left(-\frac{(x-y)^2}{2\int\_0^t{\sigma^2(\tau)}\,d\tau}\right)-\exp\left(-\frac{(x+y)^2}{2\int\_0^t{\sigma^2(\tau)}\,d\tau}\right)\right)\rho(y)dy.
$$
For the general case the Green's function of the first BVP cannot be expressed via elementary functions.


