Green's function of the Ornstein-Uhlenbeck operator The Ornstein-Uhlenbeck operator $L$ is given by
$$
Lu = \Delta u- \frac{1}{2}x\cdot \nabla u.
$$
Is there a known closed form expression of the Green's function of $L$ on $\mathbb R^d$ (for $d\geq 2$ or at least for $d=2$) ?
Any references?
Thanks a lot!
 A: Denote by $H_k(x)$ the $k$-th  Hermite polynomial in one variable,
$$H_k(x) =\delta^k 1,$$
where $\delta f(x)=xf(x)-f'(x)$, $\newcommand{\bR}{\mathbb{R}}$ $\forall f\in C^\infty(\bR)$. $\newcommand{\bx}{\boldsymbol{x}}$  For $\bx=(x_1,\dotsc,x_d)\in\bR^d$ and $\newcommand{\bZ}{\mathbb{Z}}$$\alpha\in\bZ^d_{\geq 0}$  we set
$$ H_\alpha(\bx):=H_{\alpha_1}(x_1)\cdots H_{\alpha_d}(x_d). $$ 
Denote by $\Gamma$ the standard Gaussian measure on $\bR^d$,
$$ \Gamma(d\bx)=\frac{1}{(2\pi)^{d/2}} e^{-\frac{1}{2}\Vert\bx\Vert^2} d\bx. $$
We have
$$\int_{\bR^d} H_\alpha(\bx)^2\Gamma(\bx)=\alpha!:=\prod_j \alpha_j!, $$
$$ L H_\alpha = -|\alpha|\, H_{\alpha},\;\;\alpha=\sum_j\alpha_j. $$
Moreover, the linear span of the set of  polynomials $H_\alpha(\bx)$, $\alpha\in\bZ^d_{\geq 0}$, is dense in the Hilbert space $L^2\bigl(\,\bR^d, \Gamma(d\bx)\,\bigr)$.   Thus  any $f\in L^2(\bR^d,\Gamma)$ has an orthogonal decomposition
$$ f(\bx)=\sum_\alpha\frac{f_\alpha}{\alpha!} H_\alpha(\bx),\;\;f_\alpha=\int_{\bR^d} f(\bx) H_\alpha(\bx)\Gamma(d\bx). $$
The range of $L$ is the codimension $1$ subspace of $L^2(\bR^d,\Gamma)$ consisting of functions $f$ such that $f_0=0$.  Consider the bounded operator
$$G:L^2(\bR^d,\Gamma)\to L^2(\bR^d,\Gamma), $$
given by
$$ G[f](\bx)= -\sum_{\alpha\neq 0}\frac{1}{|\alpha|} \frac{f_\alpha}{\alpha!} H_\alpha(\bx). $$
Then $G[f]$  belongs to the domain of $L$  for any $f\in L^2(\bR^d,\Gamma)$ and
$$ LG [f]=f-f_0. $$
The operator $G$ is an integral operator and its  integral kernel has the form $\newcommand{\by}{\boldsymbol{y}}$
$$K_G(\bx,\by)=-\sum_{\alpha\neq 0}\frac{1}{|\alpha|\cdot \alpha!} H_\alpha(\bx) H_\alpha(\by). $$
This means that
$$G[f](\bx)=\int_{\bR^d} K_G(\bx,\by) f(\by) \Gamma(d\bx). $$
You can take $K_G$ as your Green's function.  You can simplify the description of $K_G$ a bit by using Mehler's formula
$$\sum_{k\geq 0}H_k(x)H_k(y)\frac{r^k}{k!}=\frac{1}{\sqrt{1-r^2}}  \exp\left(-\frac{(rx)^2-2rxy+(ry)^2}{2(1-r^2)}\right).
 $$
For more details you can check Malliavin's book Integration and Probability.
A: Not a major in mathematics, forgive me if it sounds somewhat stupid.
You want to find the Green's function to a Linear operator $L$ on $\mathbb{R}^d$, but it seems that you need to state what boundary conditions you would like to impose on the Green's function $G(x,x^{\prime})$. For example, if $L$ is the Laplacian, and you want the Green's function on $[0,L]^n$, you need to state the boundary conditions you would like to impose on the faces of the hypercube.
Can we assume spherical symmetry and translation symmetry? If we can, then write $u(r,\theta, \phi) = u(r)$ and we would like to solve $L u = \delta(\vec{x})$. The exact expression for $\delta(\vec{x})$ depends on the number of dimensions. In three dimensions ${1 \over r^2} \partial_r r^2 \partial_r u(r) - {1 \over 2} r \partial_r u(r) = {1 \over r^2} \delta(r)$. When $r > 0$, we just need to solve ${1 \over r^2} \partial_r r^2 \partial_r u(r) - {1 \over 2} r \partial_r u(r) = 0$. It seems that we will have two independent solutions. A person should examine whether there's a solution that gives a $\delta$-function at $r=0$. If there is, then a solution exists.
