Consider the Laplace equation in $\mathcal{R}^3$ \begin{equation} \Delta u = f, ~~~\lim_{x\to \infty} u(x) = 0. \end{equation} Here we assume $f$ is a smooth, compactly supported function. Of course, $u$ can be explicitly solved with the Green function. I am considering if we can use a stochastic process (like Brownian motion) to express its solution? I know the solution can be expressed as an expectation of a stochastic process (with a stopping time) if we consider a bounded domain with a certain boundary condition. In this case, the boundary seems to play a quite important role. Can we have a similar result for the whole space case?

  • $\begingroup$ In these notes (James Nolen, PDEs and Diffusion Processes), it appears what you are looking for is treated in Chapter 4. $\endgroup$ – Willie Wong Feb 3 at 1:05
  • $\begingroup$ Thank you for the note @WillieWong! I found in Theorem 4.1.3 there is a further condition for the representation: the Poisson equation should be like $\sum_{i,j}1/2a_{ij}u_{x_ix_j} + \sum_jb_j(x)u_{x_j} - c(x)u = f(x)$, with $c(x) > c_0 > 0.$ It seems the strict positiveness of $c(x)$ is required for the representation. Do we have to use this condition if we simply consider the Laplace equation? $\endgroup$ – Jacob Lu Feb 3 at 2:04

If $f(x) / (1 + |x|)$ is integrable, then the solution $u$ is equal to the Newtonian potential of $f$: $$ -u(x) = \frac{1}{4\pi} \int_{\mathbb R^3} \frac{f(y)}{|x - y|} \, dy . $$ And the Newtonian potential kernel is the occupation density of the Brownian motion: $$ \frac{1}{4\pi} \, \frac{1}{|x|} = \int_0^\infty \frac{1}{(4 \pi t)^{3/2}} \, e^{-|x|^2 / (4 t)} dt . $$ Thus, $$ -u(x) = \mathbb E^x \int_0^\infty f(B_t) dt , $$ where $B_t$ is the 3-D Brownian motion (with covariance matrix $2 \operatorname{Id}$).

  • $\begingroup$ Thanks a lot! What if we consider a slightly more general equation $b(x)\cdot\nabla u + \Delta u = f$ with a suitable vector field $b(x)$? Can we get a similar conclusion in this situation? Do we need more structure condition for $b$ to make the argument true (for example, $b(x)$ is a divergence free vector field). $\endgroup$ – Jacob Lu Feb 3 at 19:44
  • $\begingroup$ If there is a Markov process $X_t$ with infinitesimal generator $L = b(x) \cdot \nabla + \delta$, then its potential operator $U$, defined by $U f(x) = \mathbb E^x \int_0^\infty f(X_t) dt$, is roughly the inverse of $-L$. Detailed statements require extra care, and this belongs to the field of probabilistic potential theory. Off the top of my head, I do not have a good reference, but, especially if you are willing to read more about it, you can try Dynkin's two-volume book Markov processes. Although very old, it is still a very good read. (Of course, there are more excellent resources.) $\endgroup$ – Mateusz Kwaśnicki Feb 3 at 21:00
  • $\begingroup$ Many thanks for your suggestions! They are really helpful! $\endgroup$ – Jacob Lu Feb 7 at 2:02

consider its boundary condition $lim_{x\to{\infty}}u(x)=0$, to express its inhomogeneity as a hyperbolic term, by inverse function theorem. thus not adequate to define a fast variation function. since stochastic process require a nonzero term under variation two, to make this wave be more frequently, even to make its vibration not ever disappearing.

then define an Ito calculus to represent its vibration, $df=\frac{\partial{f}}{\partial{t}}dt+\frac{\partial{f}}{\partial{x}}dB_t+\frac{1}{2}\frac{\partial^2{f}}{\partial{x}^2}d(B_t)^2=(\frac{\partial{f}}{\partial{t}}+\frac{1}{2}\frac{\partial^2{f}}{\partial{x}^2})dt+\frac{\partial{f}}{\partial{x}}dB_t$, action as a centralized source at both positive and negative side, between $y=(u-\epsilon)t$ and $y=(u+\epsilon)t$ , spanned at a distance of $\Delta{s}$, represented by its non-convergent part.

then define a mandatory function by orthogonal condition, such that $\delta(x-x_0)+c\phi_n{(x)}$. to handle its inhomogeneity boundary by generalized green function. satisfied space symmetry and time reciprocity, and the condition of compact support and smoothness in variation theory.

for example, since $\partial{f}$ is both $t$ and $B_t$'s derivation, are equivalent under the scalar derivation $dB_t$ at order two. so this time's green integral can be expressed as $\int[G(x,t;x',0)$$\psi{(x')}-\phi{(x')}$$\frac{\partial{G(x,t;x',t')}}{\partial{t'}}\vert_{t'=0}]dx'$, correspond to the reciprocal of time.

next, to express its space symmetry as a green integral be similar to the form :

$a^2{\int{[v(t')}}$$\frac{\partial{G}} {\partial{x'}}\vert_{x'=l}-\mu{(t')}$$\frac{\partial{G}}{\partial{x'}}\vert_{x'=0}]dt'$

is it helpful? thank you.


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