The Stokes Dirichlet-to-Neumann operator

Abstract

Let $\Omega \subset \mathbb {R}^d$ be a bounded open connected set with Lipschitz boundary. Let $A^N$ and $A^D$ be the Stokes Neumann operator and Stokes Dirichlet operator on $\Omega$, respectively. We study the associated Stokes version of the Dirichlet-to-Neumann operator and show a Krein formula which relates these three Stokes version operators. We also prove a Stokes version of the Friedlander inequalities, which relates the Dirichlet eigenvalues and the Neumann eigenvalues.

A. Appendix

We conclude with a converse of Lemma 2.12 for $C^2$-domains.

Proposition A.1

Let $\Omega \subset \mathbb {R}^d$ be a bounded open connected set with $C^2$-boundary. Let $\alpha \in (-1,1]$ and $\lambda \in \mathbb {R}$. Let ${{\mathcal {N}}}_\lambda$ be the Dirichlet-to-Neumann graph with parameter $\alpha$ as in Sect. 2. Further let $A^D$ be the Stokes Dirichlet operator on $\Omega$. Then ${{\mathcal {N}}}_\lambda$ is an operator if and only if $\lambda \in \mathbb {R}\setminus \sigma (A^D)$.

Proof

By Lemma 2.12 it remains to show that $\textrm{mult}({{\mathcal {N}}}_\lambda ) \ne \{ 0 \}$ for all $\lambda \in \sigma (A^D)$. Let $\lambda \in \sigma (A^D)$. Since $A^D$ has a point spectrum, there exists a $u \in D(A^D)$ such that $A^D u = \lambda \, u$ and $u \ne 0$. By Proposition 2.4 there exists a $\pi \in L_2(\Omega )$ such that $- \Delta u + \nabla \pi = \lambda \, u$ in $H^{-1}(\Omega ,\mathbb {C}^d)$. Then [29] Theorem III.2.1.1(e) implies that $u \in H^2(\Omega ,\mathbb {C}^d)$ and $\nabla \pi \in L_2(\Omega ,\mathbb {C}^d)$. Hence $\pi \in H^1(\Omega )$ by [19] Corollary 1.1.11. Therefore Lemma 2.5 implies that $\partial _\nu (u,\pi ) \in L_2(\partial \Omega , \mathbb {C}^d)$. Adding a suitable constant to $\pi$ one may assume that $\partial _\nu (u,\pi ) \in L_{2,0}(\partial \Omega , \mathbb {C}^d)$.

Now suppose that $\partial _\nu (u,\pi ) = 0$. Then $u \in D(A^N)$, the Stokes Dirichlet operator, by Proposition 2.10. Moreover, $A^N u = \lambda \, u$. By the verification of Condition (I) in the proof of Theorem 3.11 one obtains that $u = 0$, which is a contradiction. So $\partial _\nu (u,\pi ) \in L_{2,0}(\partial \Omega , \mathbb {C}^d) {\setminus } \{ 0 \}$. Using (2) one deduces that

$$\begin{aligned} (\partial _\nu (u,\pi ), {\textrm{Tr}\,\,}v)_{L_{2,0}(\partial \Omega , \mathbb {C}^d)} = \mathfrak {a}(u,v) - \int _\Omega \lambda \, u \cdot {\overline{v}} = \mathfrak {b}_\lambda (u,v) \end{aligned}$$

for all $v \in V$, where $\mathfrak {a}$ and $\mathfrak {b}_\lambda$ are as in Sect. 2. Hence $({\textrm{Tr}\,\,}u, \partial _\nu (u,\pi )) \in {{\mathcal {N}}}_\lambda$. But ${\textrm{Tr}\,\,}u = 0$ since $u \in D(A^D)$. Therefore $0 \ne \partial _\nu (u,\pi ) \in \textrm{mult}({{\mathcal {N}}}_\lambda )$. $\square$