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 $p$?
Best Answer
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.
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.