# PDE interpretation of some properties of the solution to Fokker–Planck equations

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?

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 $$p$$?

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.
• 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 Dec 31, 2021 at 18:19
• Further, why the positivity of $p$ is ensured by the PDE? Jan 1, 2022 at 13:29
• @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. Jan 1, 2022 at 15:17
• 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 $p$? Jan 1, 2022 at 16:42