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 $p$?