2
$\begingroup$

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

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ If $\Omega$ is generic an explicit formula cannot be found, as for $\Delta$. $\endgroup$
    – Giorgio Metafune
    Commented Jun 28, 2022 at 16:58
  • $\begingroup$ I am interested in the case when the domain is the whole space or a unit ball for instance. $\endgroup$
    – Student
    Commented Jun 29, 2022 at 8:38
  • $\begingroup$ 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. $\endgroup$
    – Math604
    Commented Jul 1, 2022 at 19:05

1 Answer 1

Reset to default
3
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Thanks a lot for sharing this trick, it is indeed very interesting! $\endgroup$
    – Student
    Commented Jun 29, 2022 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .