1
$\begingroup$

consider a domain $U\subset \mathbb{R}^n$ which is not bounded and denote its boundary by $\partial U$. (The boundary I'm confronted with is actually not very irregular. Think of a smooth manifold with corners.). Denote the heat kernel for the domain $U$ by $K_{U}(t,x,y)$ and the heat kernel for $\mathbb{R}^n$ by $K(t,x,y)$.

Now I have read that one can always write $K_U(t,x,y)=K(t,x,y)-H(t,x,y)$, where $H(t,x,y)$ is the continuous solution of the following equations for all $y\in U$ fixed:

$(\partial_t-\Delta)H(t,x,y)=0, x\in U, t>0$

$H(0,x,y)=0, x\in U, t=0,$

$H(t,x,y)=K(t,x,y), x\in\partial U, t>0$.

Unfortunately I'm not very acquainted with boundary value problems of this type. So if such a function $H(t,x,y)$ exists, then it should become small at infinity, i.e. $\forall \epsilon>0 \exists K\subset U$ compact, s.t. $\sup_{(x,t)\in U-K} H(t,x,y)<\epsilon$. How can I show this?

Best wishes

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

If such $H$ exists, the $H(t,x,y)\le K(t,x,y)$ due to the maximum principle.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Hello Andrew. Which maximum principle do you refer to? Usually the maximum principle is proven only for bounded domains. In my case the domain is unfortunately unbounded. $\endgroup$
    – Jop Kop
    Feb 19, 2015 at 18:08
  • $\begingroup$ Yes, it doesn't directly work. One has to assume some conditions at infinity. Here is another argument. Denote $B_R=\{|x|<R\}$ a ball in $\mathbb R^n$. Suppose that there exists a monotone sequence $R_n\to\infty$ s.t. $U\cap B_{R_{n}}$ is regular enough and let $H_n$ be the corresponding solutions. Then $H_n\le K$ and for fixed values of parameters $(x,y,t)$ sequence $H_n$ is increasing. In such case there exists a pointwise limit of $H_n$, which will have all the required properties. $\endgroup$
    – Andrew
    Feb 20, 2015 at 15:15
  • $\begingroup$ Thank you for this wonderful argument! I suppose the pointwise limit still satisfies the heat equation, but I'm not sure why this is the case. Is this easy to see? $\endgroup$
    – Jop Kop
    Feb 21, 2015 at 6:28
  • $\begingroup$ In this question math.stackexchange.com/questions/266785/… it is proved for harmonic functions. The same argument holds for solutions of the heat equation. $\endgroup$
    – Andrew
    Feb 21, 2015 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.