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It is known that the Laplacian operator with Dirichlet boundary condition in $L^2(\Omega),$ (with $\Omega$ being a open subset of $R^n$) generates a $C_0-$semigroup in $L^2(\Omega)$). Moreover, in the case of the whole space $R^n$, the semigroup can be expressed by the means of the Gauss–Weierstrass kernel. I m wondering if this is also true for a bounded open subset $\Omega$ of $R^n$.

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If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $\Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is, $$ -\Delta \varphi_n =\lambda_n \varphi_n, \;\;\; \|\varphi_n\|_{L^2(\Omega)}=1,\\ \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots, $$ and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^{t\Delta}f$, which is the solution of $\psi_{t}=\Delta \psi$ can be expressed in terms of the eigenfunctions of $-\Delta$, $$ H(t)f = \sum_{n=1}^{\infty}e^{-\sqrt{\lambda_n}t}\langle f,\varphi_n\rangle\varphi_n. $$ There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(\Omega)$.

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  • $\begingroup$ In fact I m interested with the continuity of the mapping $t\to \|S(t)x\|_{L^\infty(\Omega)}, $ and I didn't succeed to do it using this formula. $\endgroup$ – Rabat Mar 17 at 16:54
  • $\begingroup$ @Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions. $\endgroup$ – DisintegratingByParts Mar 17 at 17:13

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