Consider the following PDE on $\Omega\subset \mathbb{R}^n$ for $n\geq 2:$
\begin{align}
\Delta u - x\cdot \nabla u &= f(x),\text{ in } \Omega\\
u&=0 \text{ on }\partial \Omega
\end{align}


Are there any explicit expressions for a kernel $K$ such that,
$$u(x)=\int\_{\Omega} K(x,y)f(y)dy$$ when $\Omega=\mathbb{R}^n$ or $\Omega=B(0,1)$?



Kernel for an equation involving the Ornstein-Uhlenbeck operator
There is a trick that reduces the equation $u\_t=Lu$, $L=\Delta-x \nabla$ to the heat equation $u\_t-\Delta$. It is genuinely parabolic and gives the parabolic kernel in the whole space, from which the elliptic kernel can be deduced by integrating in time.
If $v\_t(t,x)=\Delta v(t,x)-x\nabla v (t,x)$ with $v(0,x)=f(x)$, then $u(t,x)=e^{-2t}v(t, e^tx)$ solves $u\_t(t,x)=\Delta u(t,x)-2u(t,x)$ with $u(0,x)=f(x)$.
It does not work in an a ball where probably an expansion in spherical harmonics can give the result for the elliptic case directly.


EDIT. Sorry for the mistake, but the equation for $u$ is wrong. Let us do it in two steps. First put $u(t,x)=v(t, e^t x)$. Then $u\_t(t,x)=e^{-2t}\Delta u(t,x)$ with $u(0,x)=f(x)$, which is a simple non-autonomous heat equation. Then, setting $$u(t,x)=w(\frac {1-e^{-2t}}{2}, x)$$ we have $w\_t=\Delta w$ with $w(0,x)=f(x)$.


Therefore the final transformation is $v(t,x)=w(\frac {1-e^{-2t}}{2}, e^{-t}x)$.


