# Kernel for an equation involving the Ornstein-Uhlenbeck operator

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 $$\Omega=B(0,1)$$?

• If $\Omega$ is generic an explicit formula cannot be found, as for $\Delta$. Jun 28, 2022 at 16:58
• I am interested in the case when the domain is the whole space or a unit ball for instance. Jun 29, 2022 at 8:38
• there is also a way of writing $\Delta u - \nabla \gamma(x) \cdot \nabla u = f(x)$ in divergence form... or you can write an energy for this. Which might be helpful for certain things. Jul 1, 2022 at 19:05

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 $$v(t,x)=w(\frac {1-e^{-2t}}{2}, e^{-t}x)$$.