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 to 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 $K_U(t,x,y)=K(t,x,y)-H(t,x,y)$, wher $H(t,x,y)$ 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

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