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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

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

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  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented Feb 21, 2015 at 6:57

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