Definition Laplace operator on $H^1_0(\Omega)$

Question

Let $\Omega \subseteq \mathbb{R}^n$ be a bounded open set. Is the Laplace operador defined in the space $H^2(\Omega)$ as a sum of second weak derivatives? How is the Laplace operador defined in the space $H^1_0(\Omega)$?. I ask that because I'm reading about eigenvalues of Dirichlet Laplace problem and for study that in the book is considered the Laplace operator $\Delta: H_0^1(\Omega) \to L^2(\Omega)$

Answer 1

There are two common ways how the Laplace operator $\Delta$. The first case is what you suggested, where we have $\Delta : H^2(\Omega)\to L^2(\Omega)$.

But it is also possible to define $\Delta$ on $H_0^1(\Omega)$. However, it is wrong to consider $\Delta : H_0^1(\Omega)\to L^2(\Omega)$, and the book contains probably a typo here. Instead, it is possible to define $H_0^1(\Omega) \to H^{-1}(\Omega)$, where $H^{-1}$ is the dual space of $H_0^1(\Omega)$.

The definition can be done using the duality product as follows: $$ \langle -\Delta u , v \rangle_{H^{-1}(\Omega) \times H_0^1(\Omega)} = \int_\Omega \nabla v^\top \nabla u \,\mathrm dx \quad \forall u,v \in H_0^1(\Omega). $$

Answer 2

Suppose that $f \in C^\infty_0(\Omega)$ and $g \in H_0^1(\Omega)$. Then Green's identity gives you $$ \int_\Omega (\Delta f) g \, dx = - \int_\Omega Df \cdot Dg \, dx.$$

If you have only that $f \in H_0^1(\Omega)$, you could say $\Delta f$ exists in the weak sense if there exists $h \in L^2(\Omega)$ satisfying $$\int_\Omega h g \, dx = - \int_\Omega Df \cdot Dg \, dx$$ in which case you write $\Delta f = h$. The operator $\Delta : H_0^1(\Omega) \to L^2(\Omega)$ takes $f$ to $h$.