Understanding the minimal and maximal closed extensions of Laplace operator

Question

Let's consider the following setting. $\Omega$ is a connected, riemmanian manifold with boundary (although i think there is no harm on thinking it is a bounded, close region of $\mathbb{R}^n$).

One can define the Laplace(-Beltrami) operator $-\Delta$ on smooth functions as usual, and the symmetric operator $-\Delta_0 = -\Delta|_{C_0^\infty(\Omega)}$ given by its restriction to compact supported smooth functions. Consider its minimal and maximal closed extensions defined as follows. The minimal closed extension $\Delta_\text{min}$ is the closure of $\Delta_0$ with respect to the Sobolev norm $\Vert \cdot \Vert_{H^2(\Omega)}$, that is $$ \mathcal{D}(\Delta_\text{min}) = \overline{C_0^\infty(\Omega)}^{\Vert \cdot \Vert_{H^2(\Omega)}} = H^2_0(\Omega).$$ The maximal closed extension is the adjoint of the minimal, $\Delta_\text{max} = \Delta_\text{min}^\dagger$. I belive this is like saying $$ \mathcal{D}(\Delta_\text{max}) = \{u \in L^2(\Omega) : \Delta u \in L^2(\Omega)\}, $$ where $\Delta u$ must be understood in the weak sense.

That's what I would say I understand. I've been told that $$ \mathcal{D}(\Delta_\text{max}) = H^1_0 \cap H_2 \oplus \mathrm{ker}\, (\Delta_\text{max}), $$ but I can't see even how to start proving that, which is what lead me to ask you for any help or reference on this.

Also I've heard that $\mathcal{D}(\Delta_\text{max})$ is greater than $H^2(\Omega)$ in the general case, but I don't really understand why. As I see it, the maximal closed extension can be seen as a map between $H^2(\Omega)$ and its dual, and thus $$ \Delta_\text{max} u\colon v \in H^2(\Omega) \mapsto \Delta_\text{max} u(v) = \langle \Delta_\text{max} u, v \rangle_{L^2(\Omega)}. $$ I can't see how $\Delta_\text{max} u$ can be identified (in the weak sense) with a function in $L^2(\Omega)$ if $u \notin H^2(\Omega)$.

Again, any hint or reference would be appreciated.

Answer 1

I'm still struggling with the second part of the answer, but for the first part I was told how to prove it. It follows from the following result which might be known:

Theorem. Let $H$ be a Hilbert space and $A, A^\dagger$ denote a closed symmetric operator on $H$ and its adjoint respectively. Suppose that $A_\lambda \subset A^\dagger$ is a closed extension of $A$ (not necessarily self-adjoint) such that $\lambda$ is in the resolvent set of $A_\lambda$. Then $$ \mathcal{D}(A^\dagger) = \mathcal{D}(A_\lambda) \oplus \mathrm{ker}\,(A^\dagger - \lambda). $$

Proof. It is clear that $\mathcal{D}(A^\dagger) \supset \mathcal{D}(A_\lambda) \oplus \mathrm{ker}\, (A^\dagger - \lambda)$, so we need to prove the other inclusion, $\mathcal{D}(A^\dagger) \subset\mathcal{D}(A_\lambda) \oplus \mathrm{ker}\, (A^\dagger - \lambda)$. Take $u \in \mathcal{D}(A^\dagger)$ and define $$ u_\lambda = (A_\lambda -\lambda)^{-1} (A^\dagger - \lambda) u. $$ It is well defined since $u \in \mathcal{D}(A^\dagger)$ and $\lambda \in \rho(A_\lambda)$. Moreover, $(A_\lambda -\lambda)^{-1} \colon H \to \mathcal{D}(A_\lambda)$, so $u_\lambda \in \mathcal{D}(A_\lambda)$. Now, define $u_0 = u - u_\lambda$. Since $A^\dagger \supset A_\lambda$, $$ \begin{alignedat}{2} (A^\dagger - \lambda) u_0 &= (A^\dagger - \lambda)u - (A^\dagger - \lambda)u_\lambda \\ &= (A^\dagger - \lambda) u - (A_\lambda- \lambda) u_\lambda \\ &= (A^\dagger - \lambda) u - (A_\lambda- \lambda) (A_\lambda -\lambda)^{-1} (A^\dagger - \lambda) u \\ &= (A^\dagger - \lambda) u - (A^\dagger - \lambda) u = 0. \end{alignedat}$$ That is, $u_0 \in \mathrm{ker}\,(A^\dagger - \lambda)$. To complete the proof we still need to show that $\mathcal{D}(A_\lambda) \cap \mathrm{ker}\,(A^\dagger - \lambda) = \{0\}$ so that we actually have a direct sum. To see this, consider $u \in \mathcal{D}(A_\lambda) \cap \mathrm{ker}\,(A^\dagger - \lambda)$; therefore $$ 0 = (A^\dagger - \lambda) u = (A_\lambda - \lambda) u, $$ where we have used again that $A^\dagger \supset A_\lambda$. Since $\lambda \in \rho(A_\lambda)$, $A_\lambda - \lambda$ is injective and thus the previous equation implies $u = 0$. $\square$

As a corollary, taking $A = \Delta_\text{min}$, $A^\dagger = \Delta_\text{min}$ and $A_0 = \Delta_\text{Dirichlet}$ (which has $0 \in \rho(A_0)$) it follows that $$ \mathcal{D}(\Delta_\text{max}) = H^2 \cap H^1_0 \oplus \mathrm{ker}\,(\Delta_\text{max}). $$

Answer 2

The maximal domain is in general a beast. For the sake of simplicity, I will assume that $\Omega$ is compact (it will not matter if it is a domain in Euclidean space, or a manifold with boundary). The trace map $H^2(\Omega)\to H^{3/2}(\partial \Omega)$ is continuous but there are plenty of elements in $\mathrm{ker}(\Delta_{\rm max})$ whose trace belong to $H^{-1/2}(\partial \Omega)$ but not $H^{3/2}(\partial \Omega)$ which tells us that $\mathrm{Dom}(\Delta_{\rm max})$ is much bigger than $H^2(\Omega)$. Below I give an example where the trace map $\mathrm{ker}(\Delta_{\rm max})\to H^{-1/2}(\partial \Omega)$ is surjective.

Here is an example. Take $\Omega$ to be the unit disc in the plane. Then any element $u$ of $\mathrm{ker}(\Delta_{\rm max})$ is smooth in the interior of $\Omega$ and can be written in polar coordinates as $$u(r,\theta)=\sum_{n\in \mathbb{Z}} a_n r^{|n|}\mathrm{e}^{in\theta}.$$ An expression of this type a priori defines a harmonic distribution on $\Omega$ under very mild assumptions on the sequence $(a_n)_{n\in \mathbb{Z}}$, so to get some control we need to compute $L^2$-norms. We have that $$\|u\|_{L^2(\Omega)}^2=2\pi \sum_{n\in \mathbb{Z}} \frac{|a_n|^2}{2|n|+1}.$$ Since $u$ is harmonic, then whenever $u\in L^2(\Omega)$ we have that $u\in \mathrm{ker}(\Delta_{\rm max})$ and the norm of $u$ equals $\|u\|_{L^2(\Omega)}$. If we write $g=\gamma_0(u)$ for the boundary restriction, then $$g(\theta)=\sum_{n\in \mathbb{Z}} a_n \mathrm{e}^{in\theta}.$$ In particular, we have that $$\|u\|_{L^2(\Omega)}\sim \|g\|_{H^{-1/2}(\partial \Omega)}.$$ Here $\sim$ denotes that the two are comparable up to a universal constant. We can consider the construction of $u$ as a map $H^{-1/2}(\partial \Omega)\to \mathrm{ker}(\Delta_{\rm max})$ which by the norm estimate above is a Banach space isomorphism and splits the trace mapping $\mathrm{ker}(\Delta_{\rm max})\to H^{-1/2}(\partial \Omega)$. In this example we therefore have that the trace map $\mathrm{ker}(\Delta_{\rm max})\to H^{-1/2}(\partial \Omega)$ is even an isomorphism of Banach spaces.