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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 $p$?

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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.

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  • $\begingroup$ Thank you so much Andrew for the answer. I will accept your solution as soon as I read the paper that you suggested. For the case where $b>0$ is constant, can we expect some "nice" expression for $p$? Here "good" does not mean as explicit as your formula above, but as explicit as possible $\endgroup$
    – GJC20
    Dec 31, 2021 at 18:19
  • $\begingroup$ Happy New Year! $\endgroup$
    – GJC20
    Dec 31, 2021 at 18:19
  • $\begingroup$ Further, why the positivity of $p$ is ensured by the PDE? $\endgroup$
    – GJC20
    Jan 1 at 13:29
  • $\begingroup$ @GJC20 Positivity follows from $\rho\ge0$, $\rho>0$ somewhere and the maximum principle. Or, in this case, from the formula for $p$ in your post. $\endgroup$
    – Andrew
    Jan 1 at 15:17
  • $\begingroup$ Thanks for the quick reply. I mean for the general case where $b,\sigma$ that are not constant. Could you please provide the reference of the maximum principle that can be applied here? Also, for the case where $b$ is constant, can we have some "nice" expression for $p$? $\endgroup$
    – GJC20
    Jan 1 at 16:42

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