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# Biharmonic Steklov problems with Neumann boundary conditions and spectral inequalities on differential forms

Rodolphe Abou Assali

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Rodolphe Abou Assali. Biharmonic Steklov problems with Neumann boundary conditions and spectral inequalities on differential forms. 2025. hal-05133032

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# Biharmonic Steklov problems with Neumann boundary conditions and spectral inequalities on differential forms

Rodolphe Abou Assali*¹,²

¹Lebanese University, Faculty of Sciences II, Department of Mathematics, P.O. Box 90656 Fanar-Matn, Lebanon

²Université de Lorraine, CNRS, IECL, F-54000 Nancy, France

July 2, 2025

**Abstract.** We introduce three biharmonic Steklov problems on differential forms with Neumann boundary conditions and show that they are elliptic. We prove the existence of a discrete spectrum for each of those problems and give associated variational characterizations for their eigenvalues. We establish eigenvalue estimates known as Kuttler-Sigillito inequalities, relating the eigenvalues of these problems with the eigenvalues of the Steklov, Dirichlet and Neumann standard problems as well as the biharmonic Steklov problem with Dirichlet boundary conditions on differential forms.

*Mathematics Subject Classification* (2020): 35A15, 35G15, 35J40, 35P15, 58C40, 58J32, 58J50

*Keywords:* Riemannian manifolds with boundary, biharmonic Steklov operator, Neumann boundary conditions, discrete spectrum, eigenvalue estimates

# 1 Introduction

## 1.1 Historical background

Spectral theory is a vast field that intersects with several aspects of mathematics, such as analysis, geometry, and topology. The applications of spectral theory are closely related to physics. The Dirichlet and Neumann problems are the primary examples of spectral problems [4, 27], allowing the study of the properties of eigenfunctions and eigenvalues on geometric domains. Their applications are directly relevant to the study of vibration modes and the diffusion of sound and heat in domains with boundaries. For instance, given a planar domain ("drum"), how does its geometry influence its vibrations or the evolution of temperature within it? Mathematically, this corresponds to studying the spectrum of a differential operator, which in our case is the Laplacian under certain boundary conditions. The Steklov problem is another example of spectral problems, introduced by the Russian mathematician Vladimir Andreevich Steklov in a talk given in 1895, see [18].

In this paper we work on biharmonic Steklov operators. The study of these differential operators originated from the theory of elastic oscillating membranes and was mathematically examined for the first time on planar domains by J. Kuttler and V. Sigillito [16, 17] and L.E. Payne [22]. The latter gave some isoperimetric inequalities for the first eigenvalue and eigenvalue estimates were given in

<footnote>*rodolphe.abou-assali@univ-lorraine.fr, rodolphe.abouassali.1@ul.edu.lb</footnote>
[20, 21]. The problem also appeared in [3], where some results about the first Steklov eigenvalue of the biharmonic operator in bounded domains are given. There are two distinct but related natural spectral problems, depending on the boundary conditions. Let (M, g) be a compact Riemannian manifold with a smooth boundary and let $\nu$ be the inward unit normal along the boundary $\partial M$. The scalar biharmonic Steklov problem with Dirichlet boundary conditions is defined as:

$$ \left\{ \begin{array}{ll} \Delta^2 u = 0 & \text{on } M \\ u = 0 & \text{on } \partial M \\ \Delta u - q \partial_\nu u = 0 & \text{on } \partial M, \end{array} \right. \qquad (1) $$

and the scalar biharmonic Steklov problem with Neumann boundary conditions is:

$$ \left\{ \begin{array}{ll} \Delta^2 u = 0 & \text{on } M \\ \partial_\nu u = 0 & \text{on } \partial M \\ \partial_\nu(\Delta u) - \ell u = 0 & \text{on } \partial M, \end{array} \right. \qquad (2) $$

where $u$ is a smooth function on $M$, and $\Delta$ is the (positive) scalar Laplacian operator. In both problems, the real numbers $q$ and $\ell$ are the eigenvalues of the operator. Following the original studies [16, 17] on planar domains, which in particular showed the existence of a discrete spectrum consisting of real eigenvalues of finite multiplicity for problems (1) and (2), these problems were more recently studied on higher-dimensional Euclidean domains in [8]. It was subsequently shown by S. Raulot and A. Savo in [23] that the biharmonic Steklov-Dirichlet operator on forms of maximal degree is related to the Serrin problem [26]. The Serrin problem is the following boundary problem: given $M$ subset of $\mathbb{R}^n$, we look for a solution $u$ such that, for some real constant $c$ it satisfies the overdetermined problem

$$ \left\{ \begin{array}{ll} \Delta u = 1 & \text{on } M \\ u = 0 & \text{on } \partial M \\ \partial_\nu u|_{\partial M} = c & \text{on } \partial M. \end{array} \right. \qquad (3) $$

Actually $u$ describes the velocity of an incompressible viscous fluid in a straight pipe with a circular section.

More recently, A. Hassannezhad and A. Siffert [11] generalized some of the estimates of [16, 17] for both scalar biharmonic Steklov operators on arbitrary Riemannian manifolds and showed that other quantities appear in their estimates that depend on the curvature of the manifold. In the same context, S. Kouzayha established in her PhD thesis [15] an inequality that links the first eigenvalue of the scalar biharmonic Steklov-Dirichlet problem with those of the Dirichlet and Robin Laplacians.

In a recent work [6], the biharmonic Steklov problem with Dirichlet boundary conditions (1) is generalized to differential forms of arbitrary degree. Appropriate boundary conditions are considered to make the corresponding problem well-posed. The smallest eigenvalue is characterized, and various estimates for this eigenvalue are obtained, some of which involve eigenvalues for other problems such as Dirichlet, Neumann, Robin, and Steklov. Independently, other inequalities concerning the eigenvalues of these latter problems are established.

In this work we extend the biharmonic Steklov problem with Neumann boundary condition (2) to differential forms of arbitrary degree. This extension is not unique, as there are several possible choices of boundary conditions. We provide the variational characterization of the eigenvalues for three possible choices of such extensions. This allows us to generalize Kuttler-Sigillito spectral inequalities, which relate eigenvalues of different problems on a bounded planar domain [17]. Some of these inequalities on functions were also established by A. Hassannezhad and A. Siffert on compact Riemannian manifolds with $C^2$ boundary.

From now on, we will refer to the Biharmonic Steklov problem with Dirichlet boundary condition as BSD and the Biharmonic Steklov problem with Neumann boundary condition as BSN.
## 1.2 Biharmonic Steklov problems on functions

In this section, we introduce the BSD and BSN operators and review on the existence of a discrete spectrum for both operators for the scalar case on Riemannian manifolds as in [11]. Here and from now on, we define the scalar Laplacian for all functions $u$ on a Riemannian manifold $(M, g)$ by $\Delta u = -\operatorname{tr}(\operatorname{Hess}(u))$. The bi-Laplacian will be defined as the square of the Laplacian.

**Definition 1.1.** We define the BSD operator denoted by $BSD^{func}$, as follows:

$$BSD^{func}: L^2(\partial M) \to L^2(\partial M) \\ f \mapsto \Delta u$$

where $u$ is related to $f$ by: $\Delta^2 u = 0$ on $M$, $u = 0$ on $\partial M$, and $\partial_\nu u = f$ on $\partial M$.

For the existence of such $u$ see for instance Theorem 2.6 below.

We now recall the BSD problem on a compact Riemannian manifold for the scalar case.

**Theorem 1.2.** [8, Paragraph 2] Let $(M^n, g)$ be a compact Riemannian manifold of dimension $n \ge 2$ with smooth boundary. The operator $BSD^{func}$ is elliptic positive and self adjoint. It has a discrete spectrum consisting of an unbounded monotonously non-decreasing sequence of positive real eigenvalues of finite multiplicities $(q_k)_{k \ge 1}$. In other words, for every $k$ there exists a finite dimensional space of smooth solutions $u$ to the following boundary problem:

$$\left\{
\begin{array}{ll}
\Delta^2 u = 0 & \text{on } M \\
u = 0 & \text{on } \partial M \\
\Delta u - q_k \partial_\nu u = 0 & \text{on } \partial M.
\end{array}
\right.
\qquad (4)$$

**Definition 1.3.** Similarly we define the BSN operator denoted by $\Delta_{BSN}$, as follows:

$$BSN^{func}: L^2(\partial M) \to L^2(\partial M) \\ f \mapsto -\partial_\nu \Delta u$$

such that $u$ is related to $f$ by: $\Delta^2 u = 0$ on $M$, $\partial_\nu u = 0$ on $\partial M$, and $u = f$ on $\partial M$.

For the existence of such $u$ see for instance Theorem 2.6 below. We now recall the BSN problem on a compact Riemannian manifold for the scalar case.

**Theorem 1.4.** [11, Theorem 4.8] Let $(M^n, g)$ be a compact Riemannian manifold of dimension $n \ge 2$ with smooth boundary. The operator $BSN^{func}$ is elliptic non-negative and self adjoint. It has a discrete spectrum consisting of an unbounded monotonously non-decreasing sequence of non-negative real eigenvalues of finite multiplicities $(\ell_k)_{k \ge 0}$. In other words, for every $k$ there exists a finite dimensional space of smooth solutions $u$ to the following boundary problem:

$$\left\{
\begin{array}{ll}
\Delta^2 u = 0 & \text{on } M \\
\partial_\nu u = 0 & \text{on } \partial M \\
\partial_\nu \Delta u + \ell_k u = 0 & \text{on } \partial M.
\end{array}
\right.
\qquad (5)$$

Moreover, $\ell_0 = 0$ is the eigenvalue associated with the constant eigenfunction.

The proof of Theorem 1.4 was given briefly in [11]. In fact, it follows from Lemma 1.15 that Theorem 1.2 and Theorem 1.4 are particular cases of Theorem 1.10 which we prove later on in this paper.
## 1.3 Kuttler-Sigillito inequalities

After obtaining well posed BSD and BSN problems and finding their variational characterizations, inequalities connecting the eigenvalues of different problems were established. These are known under the name of Kuttler-Sigillito inequalities. These inequalities were introduced in [17] on a subset of $\mathbb{R}^2$ and were partly generalized in [11] to the manifold setting. We provide these inequalities from [11, 17], that we extend for differential forms later on. We denote by $\lambda_k$, $\mu_k$, $\sigma_k$, $q_k$, and $\ell_k$ the $k^{th}$ eigenvalues of the following problems, respectively: Dirichlet, Neumann, Steklov, BSD, and BSN.

**Note that we always index the eigenvalues of non-negative operators so that $k = 1$ corresponds to the first positive eigenvalue.**

**Theorem 1.5.** [11, Theorem 1.1] Let $(M^n, g)$ be a compact Riemannian manifold of dimension $n \ge 2$ with a $C^2$ boundary. For all integer $k \ge 1$, we have:

$$ \mu_k \sigma_1 \le \ell_k \quad \text{and} \quad \mu_1 \sigma_k \le \ell_k. $$

**Theorem 1.6.** [17, Table 1] On a bounded domain $M$ of the plane with piecewise $C^1$ boundary $\partial M$, the following inequalities hold:

(i) $q_1 \sigma_1^2 \le \ell_1,$

(ii) $\mu_1^{-1} \le \lambda_1^{-1} + (q_1 \ell_1)^{-\frac{1}{2}},$

(iii) $\mu_1^{-1} \le \lambda_1^{-1} + (q_1 \sigma_1)^{-1}.$

## 1.4 BSD on differential forms

We have seen above the BSD problem, defined for functions on a domain of $\mathbb{R}^2$ and on a compact Riemannian manifold with smooth boundary. This problem has been generalized for differential forms of arbitrary degree by F. El Chami, N. Ginoux, G. Habib, and O. Makhoul in [6]. Motivated by the extension of Serrin problem (3) to differential forms, they defined the boundary problem (6). They showed the existence of a discrete spectrum and provided classical and alternative characterizations for the first eigenvalue of this problem. We cite some of their results.

Let $(M^n, g)$ be a compact Riemannian manifold of dimension $n \ge 2$ with smooth boundary $\partial M$.

**Notation 1.7.** We denote by $\nu$ the inward unit normal along the boundary $\partial M$, by $d$ the exterior differential operator, by $\delta$ the codifferential operator, by $\iota: \partial M \to M$ the inclusion map, and by $\lceil$ the interior product. We write $\Delta = d\delta + \delta d$ the Hodge-de-Rham Laplacian on differential forms (see Section 2.1.1).

By [6, Theorem 2.2] the boundary problem

$$ (BSD) \quad \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \omega = 0 & \text{on } \partial M \\ \nu \lceil \Delta \omega + q \iota^* \delta \omega = 0 & \text{on } \partial M \\ \iota^* \Delta \omega - q \nu \lceil d \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (6) $$

on $p$-forms, has a discrete spectrum consisting of an unbounded monotonously non-decreasing sequence of positive eigenvalues of finite multiplicities $(q_{j,p})_{j \ge 1}$.
**Theorem 1.8.** [6, Theorem 2.6] The first eigenvalue $q_{1,p}$ of BSD (6) can be characterized as follows:

$$ q_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\nu \rfloor d\omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2} \mid \omega \in \Omega^p(M), \omega|_{\partial M} = 0 \text{ and } \nabla_{\nu} \omega \neq 0 \right\} \quad (7) $$

$$ = \inf \left\{ \frac{\| \omega \|_{L^2(\partial M)}^2}{\| \omega \|_{L^2(M)}^2} \mid \omega \in \Omega^p(M) \setminus \{0\}, \Delta \omega = 0 \text{ on } M \right\}. \quad (8) $$

Both infima are indeed minima, where the infimum in (7) is attained by a $q_{1,p}$-eigenform for BSD1 (6) and (8) is attained by $\Delta \omega$, where $\omega$ is a $q_{1,p}$-eigenform for BSD1 (6).

## 1.5 BSN on differential forms

To establish inequalities between the eigenvalues for differential forms, as was done in the scalar case, we need to find a suitable BSN problem for differential forms, inspired by the work done in [6]. This first following problem is the natural analogue of the BSD problem presented in [6], adapted to Neumann boundary conditions.

Let us first define the space that will be the kernel of many boundary problems we are interested in.

**Definition 1.9.** The absolute de Rham cohomology of order p is the space

$$ H_A^p(M) = \{ \omega \in \Omega^p(M) \mid d\omega = 0 \text{ on } M, \delta\omega = 0 \text{ on } M, \nu \rfloor \omega = 0 \text{ on } \partial M \}. \quad (9) $$

It has finite dimension, see [23, Section 2].

**Theorem 1.10.** The following boundary problem

$$ (BSN1) \qquad \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \nu \rfloor \omega = 0 & \text{on } \partial M \\ \nu \rfloor d\omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta \omega + \ell_i^* \delta \omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta d\omega + \ell_i^* \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (10) $$

for differential p-forms, has a discrete spectrum consisting of an unbounded non-decreasing sequence of positive real eigenvalues with finite multiplicities $(\ell_i)_{i \ge 1}$ and possibly $\ell_0 = 0$. Moreover, its kernel is $H_A^p(M)$.

It happens that we could not use this version of the BSN, which we will refer to as BSN1, to establish Kuttler-Sigillito inequalities due to the non-suitable boundary conditions. For that reason, we have first introduced another version of the BSN, called BSN2, which can be used to establish the equivalent of Theorem 1.5.

**Theorem 1.11.** The following boundary problem

$$ (BSN2) \qquad \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \nu \rfloor \omega = 0 & \text{on } \partial M \\ \nu \rfloor d\omega = 0 & \text{on } \partial M \\ \ell_i^* \delta \omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta d\omega + \ell_i^* \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (11) $$

for differential p-forms, has a discrete spectrum consisting of an unbounded non-decreasing sequence of positive real eigenvalues with finite multiplicities $(\ell_i)_{i \ge 1}$ and possibly $\ell_0 = 0$. Moreover, its kernel is $H_A^p(M)$.
The BSN2 version allows us to establish the analogous of Theorem 1.5, but not the analogous of all the inequalities from Theorem 1.6 for differential forms. That is why we need to introduce a third BSN, called BSN3, that will improve the inequality of Theorem 1.5 for BSN2 and will allow us to prove the rest of the inequalities of Theorem 1.6.

**Theorem 1.12.** The following boundary problem

$$ (BSN3) \quad \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \nu \rfloor \omega = 0 & \text{on } \partial M \\ \nu \rfloor d\omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta \omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta d\omega + l^* \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (12) $$

for differential p-forms, has a discrete spectrum consisting of an unbounded non-decreasing sequence of positive real eigenvalues with finite multiplicities ($l_i$)$_{i \ge 1}$ and possibly $l_0 = 0$. Moreover, its kernel is $H_A^p(M)$.

The restriction to dimension $n \ge 2$ is needed, as shown by the following lemma.

**Lemma 1.13.** The spectrum of BSN3 on functions $f : [0, 1] \to \mathbb{R}$ on the interval $M = [0, 1]$ is reduced to $\{0, 24\}$.

*Proof.* We get the spectrum by simply solving the system

$$ \left\{ \begin{array}{ll} f^{(4)} = 0 & \text{on } M \\ f'(0) = 0 & \text{on } \partial M \\ f'(1) = 0 & \text{on } \partial M \\ -f^{(3)}(0) + lf(0) = 0 & \text{on } \partial M \\ f^{(3)}(1) + lf(1) = 0 & \text{on } \partial M. \end{array} \right. $$

□

The spectrum of BSD on function (4) on an interval $[-1, 1]$ is also reduced to $\{1, 3\}$, see [8, Remark 1.4].

The following Lemma will be shown in Lemma 6.1 later on.

**Lemma 1.14.** Using the notations defined previously, for all $p \in \{0, \dots, n\}$ and all $k \ge 0$, we have the following inequalities:

$$ \ell_{k,p} \le l_{k,p} \le \mathbf{1}_{k,p}. $$

Throughout this paper, we fix the degree $p \in \{0, \dots, n\}$. To simplify the notation, for all $k \ge 1$ we will write $\ell_k, \mathbf{l}_k$ and $l_k$ or simply $\ell, \mathbf{l}$ and $l$ instead of $\ell_{k,p}, \mathbf{l}_{k,p}$ and $l_{k,p}$.

**Lemma 1.15.** For $p = 0$, the three problems BSN1, BSN2 and BSN3 reduce to the classical BSN problem for scalar functions (5).

For $p = n$, the BSN1 reduce to the scalar BSD (4), the BSN2 and BSN3 are trivial.

*Proof.* It is immediate that the three BSN problems for $p = 0$ reduce to the BSN problem for functions.

For $p = n$ we have $\omega = f \operatorname{Vol} g$ with $f$ a real function. Then in the BSN1, BSN2 and BSN3 the second and last boundary conditions are trivial. The remaining conditions in the BSN1 give us the scalar BSD conditions. In BSN2 and BSN3, we have $\Delta f = 0$ on $M$ and $f|_{\partial M} = 0$, using an integration by parts, we get that $f = 0$ and so $\omega = 0$. Hence these problems are trivial. □

That is why for all the results related to BSN2 and BSN3 we consider $p \in \{0, \dots, n-1\}$.
## 1.6 Kuttler-Sigillito inequalities for differential forms

Now that we have the BSN problems, we can establish the Kuttler-Sigillito inequalities. We will cite them here and we will prove them later on. Let us write respectively $\lambda_k$, $\mu_k$ and $\sigma_k$ the $k^{th}$ positive eigenvalue of the Dirichlet, Neumann and Steklov problems on differential forms (see Section 2.3). For BSN2 and BSN3, we have the following results on differential forms:

**Theorem 1.16.** For $k \ge 1$, we have the following inequalities between the eigenvalues of the Steklov, Dirichlet, Neumann, BSD, BSN2 and BSN3 problems:

$$ \mu_k \sigma_1 \le l_k \le l_k, \qquad (13) $$

$$ \mu_1 \sigma_k \le l_k \le l_k. \qquad (14) $$

The inequality is strict for $k = 1$, i.e., $\mu_1 \sigma_1 < l_1$.

$$ q_1 \sigma_1^2 < l_1 \le l_1. \qquad (15) $$

$$ \mu_1^{-1} < \lambda_1^{-1} + (q_1 l_1)^{-\frac{1}{2}}. \qquad (16) $$

$$ \mu_1^{-1} < \lambda_1^{-1} + (q_1 \sigma_1)^{-1}. \qquad (17) $$

**Remark 1.17.** Inequalities (15), (16) and (17) are new on manifolds, even for $p = 0$, i.e., for scalar BSD and BSN problems.

**Remark 1.18.** As mentioned earlier, the inequalities of Theorem 1.16 are a priori not valid for BSN1. Moreover, inequalities (16) and (17) may not hold either for BSN2. Later, we will provide some insights by presenting parts of the proof to illustrate why they may not hold.

**Remark 1.19.** Kuttler and Sigillito [17] established other spectral inequalities that involve geometric controls of the boundary. These were partially extended to scalar functions on manifolds in [11]. We will study such inequalities in a forthcoming paper.

**Outline of the Article:** In Section 2 of this article, we provide a review of concepts related to differential forms, elliptic operators, and general theorem for the existence of solutions of boundary problems. We also recall spectral problems involving differential forms, particularly the Dirichlet, Neumann, Steklov, and biharmonic Steklov problems, while presenting their variational characterizations of eigenvalues. In Sections 3, 4, and 5, we respectively study the spectra BSN1, BSN2 and BSN3 for differential forms. Finally, in Section 6, we establish generalizations of the Kuttler-Sigillito inequalities.

**Acknowledgment:** I would like to express my sincere gratitude to my PhD supervisors, Nicolas Ginoux, Georges Habib and Samuel Tapie, for their invaluable help, guidance, support and ideas since the beginning of my PhD. I am deeply grateful for the time they have dedicated to me throughout this journey. I would also like to thank the IECL and the SAFAR scholarship for funding my PhD and for making this collaboration possible, along with everyone else who contributed to it. I also acknowledge the support of the Grant ANR-24-CE40-0702 ORBISCAR.

# 2 Analytical overview

In this section, we provide a review of some geometric and analytic tools that we frequently use.
## 2.1 Differential forms

We will give some definitions and results without proofs, some of which are from [19, Chapter 5], [25, Chapter 1] and [27, Chapter 1, Section 13].

**Notation 2.1.** For a differential form $\alpha \in \Omega^p(M)$, we denote by $\phi^*\alpha$ its pullback by a diffeomorphism $\phi$, and by $X \rfloor \alpha$ its interior product by a vector field $X$.

### 2.1.1 Differential forms on manifolds

Let $M$ be a smooth compact manifold of dimension $n \ge 2$ with smooth boundary. A differential form of degree $p$ on $M$ is a smooth section of the bundle $\Lambda^p T^*M$ and the set of $p$-differential forms on $M$ is denoted $\Omega^p(M)$.

We denote by $\iota: \partial M \to M$ the inclusion map. Thus, for any $\omega \in \partial M$, $\iota^*\omega$ is the tangential component of $\omega|_{\partial M}$, and $\nu \rfloor \omega$ is its normal component.

Let $M$ be an oriented Riemannian manifold. Recall that the Hodge operator is a linear map denoted by $\star \star$, defined as $\star: \Lambda^p M^* \to \Lambda^{n-p} M^*$ such that if $(\epsilon_i)_{i=1}^n$ is a positively oriented orthonormal basis of $T^*M$, then

$$ \star(\epsilon_{i_1} \wedge \cdots \wedge \epsilon_{i_p}) = \epsilon_{j_1} \wedge \cdots \wedge \epsilon_{j_{n-p}} $$

where, $(j_1, \dots, j_{n-p})$ is chosen such that $(\epsilon_{i_1}, \dots, \epsilon_{i_p}, \epsilon_{j_1}, \dots, \epsilon_{j_{n-p}})$ is a positively oriented orthonormal basis. As a consequence we have $\star 1 = \epsilon_1 \wedge \cdots \wedge \epsilon_n$, $\star(\epsilon_1 \wedge \cdots \wedge \epsilon_n) = 1$ and for a $p$-form $\omega$ we get $\star \star \omega = (-1)^{p(n-p)} \omega$.

For differential forms, we denote by $\langle \cdot, \cdot \rangle$ the natural inner product induced by the Riemannian metric, and we define $(\omega, \omega') := \int_M \langle\omega, \omega'\rangle d\mu_g$. The codifferential operator, denoted $\delta$, is a map $\delta: \Omega^p(M) \to \Omega^{p-1}(M)$. It is the formal adjoint of the exterior derivative $d$, i.e., for all $\alpha \in \Omega^p(M)$ and $\beta \in \Omega^{p+1}(M)$ with support in $M \setminus \partial M$, we have $(d\alpha, \beta) = (\alpha, \delta\beta)$. It turns out that, for every $p$-form $\omega$,

$$ \delta \omega = (-1)^{n(p+1)+1} \star d \star \omega. $$

We denote by $\nabla$ the Levi-Civita connection. For a vector field $X$, we write $X^\flat$ for the differential 1-form such that for all vector field $Y$ we have $X^\flat = g(X, Y)$.

Given a local orthonormal basis $(e_i)_{i=1}^n$, we have the local expressions of $d\omega = \sum_i e_i^\flat \wedge \nabla_{e_i} \omega$ and $\delta\omega = -\sum_i e_i \rfloor \nabla_{e_i} \omega$, see for instance [9, Proposition 2.61]. The Hodge-de-Rham Laplacian is the operator denoted $\Delta$ from $\Omega^p(M)$ to itself, defined for all $\omega \in \Omega^p(M)$ by

$$ \Delta \omega = d\delta \omega + \delta d\omega. $$

A $p$-differential form $\omega$ is said to be harmonic if $\Delta \omega = 0$.

We have the following properties

1. The Hodge-de-Rham Laplacian is formally self-adjoint, i.e., for all $\alpha, \beta \in \Omega^p(M)$ with support in $M \setminus \partial M$, we have $(\Delta\alpha, \beta) = (\alpha, \Delta\beta)$.

2. $\Delta\star = \star\Delta$.

We will frequently use the following integration by parts formulas

**Proposition 2.2.** For $\omega \in \Omega^p(M)$ and $\omega' \in \Omega^{p+1}(M)$, we have

$$ \int_M \langle d\omega, \omega' \rangle d\mu_g = \int_M \langle \omega, \delta\omega' \rangle d\mu_g - \int_{\partial M} \langle \iota^* \omega, \nu \rfloor \omega' \rangle d\mu_g. \quad (18) $$
Moreover, for any $\omega, \omega' \in \Omega^p(M)$, we have

$$ 
\begin{aligned}
& \int_M (\langle \Delta \omega, \omega' \rangle - \langle \omega, \Delta \omega' \rangle) d\mu_g \\
& = \int_{\partial M} (\langle \nu_\Box d\omega, \iota^* \omega' \rangle - \langle \iota^* \omega, \nu_\Box d\omega' \rangle + \langle \nu_\Box \omega, \iota^* \delta \omega' \rangle - \langle \iota^* \delta \omega, \nu_\Box \omega' \rangle) d\mu_g.
\end{aligned} 
\qquad (19) $$

Eventually, for all $\omega, \omega' \in \Omega^p(M)$ we have the following identity

$$ 
\begin{aligned}
\int_M \langle \Delta \omega, \omega' \rangle d\mu_g &= \int_M \langle d\omega, d\omega' \rangle d\mu_g + \int_M \langle \delta \omega, \delta \omega' \rangle d\mu_g \\
&\quad + \int_{\partial M} \langle \nu_\Box d\omega, \iota^* \omega' \rangle d\mu_g - \int_{\partial M} \langle \iota^* \delta \omega, \nu_\Box \omega' \rangle d\mu_g.
\end{aligned} 
\qquad (20) $$

*Proof.* Equations (18) and (19) are shown in [27, p.182]. Equation (20) is shown in [5, Equation 30] and [27, p.421]. □

Replacing $\omega$ with $\Delta \omega$ in (19), we obtain:

$$ 
\begin{aligned}
\int_M \langle \Delta^2 \omega, \omega' \rangle d\mu_g &= \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g \\
&\quad + \int_{\partial M} \left( \langle \nu_\Box d\Delta \omega, \iota^* \omega' \rangle - \langle \iota^* \Delta \omega, \nu_\Box d\omega' \rangle + \langle \nu_\Box \Delta \omega, \iota^* \delta \omega' \rangle - \langle \iota^* \delta \Delta \omega, \nu_\Box \omega' \rangle \right) d\mu_g.
\end{aligned} 
\qquad (21) $$

In addition, for $\omega = \omega'$ in (20) we obtain the following formula:

$$ 
\begin{aligned}
\int_M \langle \Delta \omega, \omega \rangle d\mu_g &= \int_M (|d\omega|^2 + |\delta\omega|^2) d\mu_g + \int_{\partial M} \langle \nu_\Box d\omega, \iota^* \omega \rangle d\mu_g \\
&\quad - \int_{\partial M} \langle \iota^* \delta\omega, \nu_\Box \omega \rangle d\mu_g.
\end{aligned} 
\qquad (22) $$

Moreover, applying (21) to $\int_M \langle \Delta^2 \omega, \omega' \rangle d\mu_g - \int_M \langle \omega, \Delta^2 \omega' \rangle d\mu_g$ gives us the following formula:

$$ 
\begin{aligned}
& \int_M \langle \Delta^2 \omega, \omega' \rangle d\mu_g - \int_M \langle \omega, \Delta^2 \omega' \rangle d\mu_g \\
&= \int_{\partial M} \langle \nu_\Box d\Delta \omega, \iota^* \omega' \rangle d\mu_g - \int_{\partial M} \langle \iota^* \omega, \nu_\Box d\Delta \omega' \rangle d\mu_g \\
&\quad - \int_{\partial M} \langle \iota^* \delta \Delta \omega, \nu_\Box \omega' \rangle d\mu_g + \int_{\partial M} \langle \nu_\Box \omega, \iota^* \delta \Delta \omega' \rangle d\mu_g \\
&\quad + \int_{\partial M} \langle \nu_\Box \Delta \omega, \iota^* \delta \omega' \rangle d\mu_g - \int_{\partial M} \langle \iota^* \delta \omega, \nu_\Box \Delta \omega' \rangle d\mu_g \\
&\quad - \int_{\partial M} \langle \iota^* \Delta \omega, \nu_\Box d\omega' \rangle d\mu_g + \int_{\partial M} \langle \nu_\Box d\omega, \iota^* \Delta \omega' \rangle d\mu_g.
\end{aligned} 
\qquad (23) $$

Note that in [27], the identities involve $\omega$ on the boundary rather than $\iota^* \omega$, as in the above proposition. This is equivalent since for all $\alpha \in \Omega^p(M)$, $\beta \in \Omega^{p+1}(M)$ we have $\langle \alpha, \nu_\Box \beta \rangle = \langle \iota^* \alpha, \nu_\Box \beta \rangle$ on $\partial M$.

## 2.2 Elliptic operators and existence of solutions

### 2.2.1 Sobolev spaces

We still denote by $\nabla$ the Levi-Civita connection on the tangent bundle. Let $E \to M$ be a smooth vector bundle equipped with a bundle norm. We write $\Gamma(E)$ for smooth sections of $E$. For $1 \le p < \infty$ and $m \in \mathbb{N}$. We will use the following Sobolev spaces:
• $L^k(M; E) = \{\omega \in \Gamma(E) \mid \int_M |\omega|^k d\mu_g < +\infty\} / \{\omega \in \Gamma(E) \mid \omega = 0 \text{ a.e.}\}$,

• $W^{m,k}(M; E) = \{\omega \in L^k(M) \mid \forall 1 \le j \le m, \nabla^j \omega \in L^k(M; E^{(j)})\}, \text{where } E^{(j)} = (TM)^{\otimes j} \otimes E,$

• $H^m(M; E) = W^{m,2}(M),$

• $H_0^1(M; E) = \{\omega \in H^1(M) \mid \omega|_{\partial M} = 0\},$

• $H_0^2(M; E) = \{\omega \in H^2(M) \mid \omega|_{\partial M} = 0 \text{ and } (\nabla \omega)|_{\partial M} = 0\}.$

In the sequel, the vector bundle $E$ will be one of the $\Lambda^p T^*M$ and we will omit it in the notations by writing $L^p(M) = L^p(M; E)$ and similarly for other spaces.

### 2.2.2 Elliptic operators

**Definition 2.3.** [27, Chapter 2, Section 10, p.183] Let $E = \Lambda^p T^*M$ and $A : \Gamma(E) \to \Gamma(E)$ a linear differential operator of order $k \ge 1$.

Its principal symbol $\sigma_A(\xi) : E_x \to E_x$ is given by $\sigma_A(\xi) = \frac{1}{k!} [\cdots [[A, f], f] \cdots f]$, where for all $x \in M$ and $\xi \in T_x^*M$, $f$ is a smooth map such that $d_x f = \xi$.

The operator $A$ is elliptic if and only if for all $x \in M$ and $\xi \in T_x^*M \setminus \{0\}$, its principal symbol is invertible, i.e., $\det(\sigma_A(\xi)) \ne 0$.

It follows from [27, p.177] that if $A$ and $B$ are differential operators then for all $\xi \in T^*M$, $\sigma_{A \circ B}(\xi) = \sigma_A(\xi) \circ \sigma_B(\xi)$.

**Examples 2.1.** The Hodge-de-Rham Laplacian and bi-Laplacian are elliptic, and their principal symbols are given by:

$$ \sigma_{\Delta}(\xi) = -|\xi|^2 \quad \text{and} \quad \sigma_{\Delta^2}(\xi) = |\xi|^4, $$

where $\xi \in T_x^*M$.

For the proof, see [27, p.182].

### 2.2.3 Elliptic boundary problems in the sense of Shapiro-Lopatinskij

The aim of this paragraph is to give the definition of the ellipticity of a differential boundary problem, in the sense of Shapiro-Lopatinskij, as presented in [25, Paragraph 1.6]. Let us consider the following boundary problem:

$$ \left\{ \begin{array}{ll} A \omega = 0 & \text{in } M, \\ B_1 \omega = 0 & \text{on } \partial M, \\ B_2 \omega = 0 & \text{on } \partial M, \\ \vdots & \vdots \\ B_p \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (24) $$

where $A : \Gamma(E) \to \Gamma(E)$ is a linear differential operator of order $k$, and for all $1 \le j \le p$ $B_j : \Gamma(E) \to \Gamma(E_j)$ is a linear differential operator. Here $E \to M$ and $(E_j \to \partial M)_{1 \le j \le p}$ are vector bundles.

The principal symbol

$$ \begin{array}{rcl} \sigma_A : T_x^*M & \to & \text{End}(E_x) \\ \xi & \mapsto & \sigma_A(\xi) \end{array} $$
has locally a polynomial homogeneous expression of order $k$ in the coordinates of $\xi$. Therefore, it extends to generalized covectors with values in the algebra of differential operators on $C^\infty(\mathbb{R}, E_x \otimes \mathbb{C})$, i.e., $\tilde{\xi} \in T_x^*M \otimes (\mathbb{C} \oplus \mathbb{C} \cdot \partial_t)$, where the multiplication in $\mathbb{C} \oplus \mathbb{C} \cdot \partial_t$ is given by $(a_1+b_1\partial_t)(a_2+b_2\partial_t) = a_1a_2 + (a_1b_2 + a_2b_1)\partial_t + b_1b_2\partial_t^2$.

**Definition 2.4.** [12, Definition 20.1.1][25, Definition 1.6.1] The problem (24) is said to be elliptic in the sense of Shapiro-Lopatinskij if and only if:

* A is elliptic in the sense of Definition 2.3, i.e., $\sigma_A(\xi)$ is invertible for all $\xi \in T^*M \setminus \{0\}$.

* For all $v \in T_x^*M \setminus \{0\}$, the map $\Phi$

$$ \Phi: M_{A,v}^{+} \to \bigoplus_{j=1}^{k} E_j \\ y \mapsto (\sigma_{B_1}(-iv + \partial_t v^b)y, \sigma_{B_2}(-iv + \partial_t v^b)y, \dots, \sigma_{B_k}(-iv + \partial_t v^b)y) \quad (0) $$

is an isomorphism, where $-iv + \partial_t v^b \in T_x^*M \otimes (\mathbb{C} \oplus \mathbb{C} \cdot \partial_t)$ for $x \in \partial M$ and

$$ M_{A,v}^{+} := \{ \text{bounded solutions } y = y(t) \text{ on } \mathbb{R}_{+} \text{ of the ODE } \sigma_A(-iv + \partial_t v^b)y = 0 \}. \quad (25) $$

**Lemma 2.5.** In the case of the bi-Laplacian, the space $M_{\Delta^2,v}^+$ is given by:

$$ M_{\Delta^2,v}^{+} = \operatorname{Span} \left( e^{-|v|t} (at + b) \cdot \omega_0 \mid a, b \in \mathbb{R}, \omega_0 \in \Lambda^p T_x^* M_{|\partial M} \right) $$

and its dimension is $2(\begin{pmatrix} n \\ p \end{pmatrix})$.

*Proof.* Let $\xi \in T^*M$, $v \in T^*\partial M$, and $\nu \in TM_{|\partial M}$. For $A = \Delta^2$ in (25) we have

$$ M_{\Delta^2,v}^{+} := \{\text{bounded solutions on } \mathbb{R}_{+} \text{ of the ODE } \sigma_{\Delta^2}(-iv + \partial_t v^b)y = 0\}, $$

where $\sigma_{\Delta^2}(\xi) = |\xi|^4$. Thus,

$$ \begin{aligned} \sigma_{\Delta^2}(-iv + \partial_t v^b) &= |-iv + \partial_t v|^4 \\ &= (|-iv + \partial_t v|^2)^2 \\ &= ((-iv + \partial_t v, -iv + \partial_t v))^2 \\ &= ((-i)^2 \langle v, v \rangle + \langle \partial_t v, \partial_t v \rangle - 2 \langle \underbrace{iv}_{\in T^*\partial M}, \underbrace{\partial_t v}_{\perp T^*\partial M} \rangle)^2. \end{aligned} $$

Therefore,

$$ \sigma_{\Delta^2}(-iv + \partial_t v^b) = (-|v|^2 + (\partial_t)^2 \underbrace{|v|^2}_{1})^2 = ((\partial_t)^2 - |v|^2)^2 = \partial_t^4 - 2|v|^2\partial_t^2 + |v|^4. $$

Now solving the equation $y = y(t) : \mathbb{R}_{+} \to \Lambda^p T^*M$,

$$ (\sigma_{\Delta^2}(-iv + \partial_t v^b)y)(t) = 0 $$

is equivalent to solving

$$ (\partial_t^4 - 2|v|^2\partial_t^2 + |v|^4)y = 0. $$

Thus, the equation becomes $y^{(4)}(t) - 2|v|^2y''(t) + |v|^4y(t) = 0$, which is a linear ODE with constant coefficients. The general solution is given by:

$$ y(t) = (a_1t + b_1)e^{|v|t} \omega_0 + (a_2t + b_2)e^{-|v|t} \omega'_0, \quad a_1, b_1, a_2, b_2 \in \mathbb{R} \text{ and } \omega_0, \omega'_0 \in \Lambda^p T_x^* M_{|\partial M}. $$
Since we are looking for a bounded solution, we require that $|y(t)| \le M$ for all $t \ge 0$, which implies $a_1 = b_1 = 0$, hence

$$y(t) = (at + b)e^{-|v|t} \omega_0, \quad a, b \in \mathbb{R} \text{ and } \omega_0 \in \Lambda^p T_x^* M|_{\partial M}.$$

Consequently,

$$\mathcal{M}_{\Delta^2, v}^{+} = \operatorname{Span} \left( e^{-|v|t}(at + b) \cdot \omega_0 \mid a, b \in \mathbb{R}, \omega_0 \in \Lambda^p T_x^* M|_{\partial M} \right).$$

To complete the proof of the lemma, $\mathcal{M}_{\Delta^2, v}^{+}$ depends on two parameters $a$ and $b$. Moreover, $\omega_0 \in \Lambda^p T_x^* M$, which is a space of dimension $\binom{n}{p}$. Thus, we have

$$\dim(\mathcal{M}_{\Delta^2, v}^{+}) = 2\binom{n}{p}.$$

□

### 2.2.4 Existence of solutions of boundary problems

In this section, we present some results from [14, Chapter 2], which will be useful for our work. We denote by $\rho(x)$ a smooth function in $M$, that is positive in $M \setminus \partial M$, vanishes on the boundary $\partial M$, and whose normal derivative is non-zero on the boundary.

As in [14, Chapter 2 - Definition 6.1], the Hilbert space $\Xi^s(M)$ is defined for $s \in \mathbb{N} \cup \{0\}$, as:

$$\Xi^s(M) := \{u \mid \rho^j \nabla^j u \in L^2(M), \forall 0 \le j \le s\},$$

equipped with the norm:

$$\|u\|_{\Xi^s(M)} = \sum_{j \le s} \|\rho^j \nabla^j u\|_{L^2(M)}.$$

The space $\Xi^s(M)$ does not depend on the choice of $\rho$. We note that $\Xi^0(M) = L^2(M)$ and $H^s(M) \subset \Xi^s(M) \subset L^2(M)$. In addition $(\Xi^s(M))' = \Xi^{-s}(M)$ represents the dual of $\Xi^s(M)$.

As in [14, Chapter 2 - Definition 6.2], we define the Hilbert space $\Xi^s(M)$ for $s \in \mathbb{R}_+^*$, $s = k + \theta$ with $k \in \mathbb{Z}_+$ and $0 < \theta < 1$, by $\Xi^s(M) = [\Xi^{k+1}(M), \Xi^k(M)]_{1-\theta}$. For the definition of $[X, Y]_\theta$ where $X, Y$ are two Hilbert spaces and $0 \le \theta \le 1$, see [14, Chapter 1 - Definition 2.1].

The following is an interpretation of [14, Chapter 2 - Theorem 2.1] in our context. On $\Gamma(E)$, we say that the problems

$$\left\{
\begin{array}{ll}
Au = 0 & \text{on } M \\
B_1u = 0 & \text{on } \partial M \\
\vdots & \vdots \\
B_ku = 0 & \text{on } \partial M
\end{array}
\right.
\quad \text{and} \quad
\left\{
\begin{array}{ll}
A^*u = 0 & \text{on } M \\
C_1u = 0 & \text{on } \partial M \\
\vdots & \vdots \\
C_ku = 0 & \text{on } \partial M
\end{array}
\right.$$

are dual, if

1. $A$ is a linear elliptic differential operator of order $m$ and $A^*$ is the formal adjoint of $A$, i.e., for all $u, v \in \Gamma(E)$ with compact support in $M \setminus \partial M$, $\int_M \langle Au, v \rangle d\mu_g = \int_M \langle u, A^*v \rangle d\mu_g$.

2. There exist operators $(T_1, \dots, T_k)$, $(S_1, \dots, S_k)$ such that $\forall i \in \{1, \dots, k\}$, $\deg(T_i) = m - 1 - \deg(B_i)$ and $\deg(S_i) = m - 1 - \deg(C_i)$.
3. For all $u, v \in \Gamma(E)$, we have

$$ \int_M \langle Au, v \rangle d\mu_g - \int_M \langle u, A^*v \rangle d\mu_g = \sum_{i=1}^{k} \int_{\partial M} \langle S_i u, C_i v \rangle d\mu_g - \sum_{i=1}^{k} \int_{\partial M} \langle B_i u, T_i v \rangle d\mu_g. \quad (26) $$

In this paper, we take $A = \Delta$ or $A = \Delta^2$, both are formally self adjoint: $A = A^*$. According to A and the boundary conditions which we consider, the operators $B_j$, $C_j$, $S_j$, and $T_j$ vary in each situation due to the integration by parts formulas presented in Proposition 2.2. For example in the proof of Proposition 3.20 we have $C_1 \omega = -\nu_\perp d\omega$, $C_2 \omega = -\nu_\perp \omega$, $B_1 \omega = \nu_\perp \omega$, $B_2 \omega = \nu_\perp d\omega$, $T_1 \omega = -\iota^* \delta \omega$, $T_2 \omega = -\iota^* \omega$, $S_1 \omega = \iota^* \omega$ and $S_2 \omega = \iota^* \delta \omega$.

Assuming that the problems are dual, let $\mathcal{N}$ and $\mathcal{N}^*$ be the spaces defined by: $\mathcal{N} := \{u \in \Gamma(E) \mid Au = 0, B_1u = 0, \dots, B_ku = 0\}$ and $\mathcal{N}^* := \{u \in \Gamma(E) \mid A^*u = 0, C_1u = 0, \dots, C_ku = 0\}$.

**Theorem 2.6.** [14, Chapter 2 - Theorems 5.3 and 5.4] Under the following assumptions:

i. M is a compact manifold with smooth boundary $\partial M$.

ii. $A: \Gamma(E) \to \Gamma(E)$ is a differential operator of order m.

iii. The problem (24) is elliptic in the sense of Shapiro-Lopatinskij,

Then for any $s \ge m$, the following map

$$ \mathcal{P}: H^s(M)/\mathcal{N} \to \{H^{s-m}(M) \times \prod_{j=1}^k H^{s-m_j-\frac{1}{2}}(\partial M)\} \\ [\omega] \mapsto (A\omega, B_1\omega, \dots, B_k\omega) $$

is a topological and algebraic isomorphism onto the space

$$ \mathrm{Im}(\mathcal{P}) = \left\{ \begin{array}{l} (f; g_1, \dots, g_k) \in H^{s-m}(M) \times \prod_{j=1}^{k} H^{s-m_j-\frac{1}{2}}(\partial M) \mid \\ \qquad \int_M \langle f, v \rangle d\mu_g + \sum_{j=1}^{k} \int_{\partial M} \langle g_j, T_j v \rangle d\mu_g = 0 \quad \text{for all } v \in \mathcal{N}^* \end{array} \right\}. $$

In particular, solutions of $(A\omega, B_1\omega, B_2\omega, \dots, B_k\omega) = (0, 0, \dots, 0)$ are in all $H^s(M)$ with $s \ge m$, i.e, are smooth.

Following [14, Equations 6.24 and 7.21], for all $s < m$, we define the following space

$$ D_A^s(M) = \{u \mid u \in H^s(M) \mid Au \in \Xi^{-m+s}(M)\}. $$

**Theorem 2.7.** [14, Chapter 2 - Theorem 6.6] Under the same hypotheses as in Theorem 2.6, for all $s \in [-\infty, 0]$ such that $s - \frac{1}{2} \notin \mathbb{Z}$, the following map:

$$ \mathcal{P}: D_A^s(M)/\mathcal{N} \to \Xi^{s-m}(M) \times \prod_{j=1}^k H^{s-m_j-\frac{1}{2}}(\partial M) \\ [\omega] \mapsto (A\omega, B_1\omega, \dots, B_k\omega) $$
is an algebraic and a topological isomorphism onto the space Im($\mathcal{P}$) that is given by

$$ \mathrm{Im}(\mathcal{P}) = \left\{ (f; g_1, \dots, g_k) \in \Xi^{s-m}(M) \times \prod_{j=1}^{k} H^{s-m_j-\frac{1}{2}}(\partial M) \mid \int_M \langle f, v \rangle d\mu_g + \sum_{j=1}^{k} \int_{\partial M} \langle g_j, T_j v \rangle d\mu_g = 0 \text{ for all } v \in \mathcal{N}^* \right\}. \quad (27) $$

**Theorem 2.8.** [14, Chapter 2 - Theorem 7.4] Under the same hypotheses as in Theorem 2.6, for all $0 < \mathbf{s} < \mathbf{m}$, the following map:

$$ \mathcal{P}: D_A^s(M)/\mathcal{N} \to \Xi^{s-m}(M) \times \prod_{j=1}^k H^{s-m_j-\frac{1}{2}}(\partial M) \\ [\omega] \mapsto (A\omega, B_1\omega, \dots, B_k\omega) $$

is an algebraic and topological isomorphism onto the space Im($\mathcal{P}$) (27).

## 2.3 Variational characterizations of classical eigenvalue problems

In this section, we will recall the classical problems on differential forms and their variational characterization, as we will use them later to establish the Kuttler-Sigillito inequalities.

### 2.3.1 Classical eigenvalue problems on differential forms

**Definition 2.9.** Let $(M^n, g)$ be a compact connected Riemannian manifold with boundary $\partial M$. The Dirichlet problem for differential forms is defined as follows:

$$ \left\{ \begin{array}{ll} \Delta \omega = \lambda \omega & \text{on } M \\ \omega = 0 & \text{on } \partial M, \end{array} \right. \quad (28) $$

where $\lambda \in \mathbb{R}$. The Neumann problem for differential forms (with absolute boundary conditions) is defined by

$$ \left\{ \begin{array}{ll} \Delta \omega = \mu \omega & \text{on } M \\ \nu \rfloor \omega = 0 & \text{on } \partial M \\ \nu \lfloor d\omega = 0 & \text{on } \partial M, \end{array} \right. \quad (29) $$

where $\mu \in \mathbb{R}$.

For $p \in \{0, \dots, n\}$, the kernel of the Dirichlet problem is trivial and the kernel of the Neumann problem is $H_A^p(M)$ defined in (9). The spectrum of the Laplacian for the Dirichlet and Neumann problems on forms are discrete, consisting respectively of eigenvalues

$$ 0 < \lambda_{1,p} \le \lambda_{2,p} \le \dots \quad \text{and} \quad 0 \le \mu_{1,p} \le \mu_{2,p} \le \dots. $$

Now, let $p \in \{0, \dots, n\}$. For every $p$-form $\omega$ on the boundary, according to [23] (see also Lemma 3.22 below) there exists a unique $p$-form $\hat{\omega}$ on $M$ solving the following problem:

$$ \left\{ \begin{array}{ll} \Delta \hat{\omega} = 0 & \text{on } M \\ \iota^* \hat{\omega} = \omega & \text{on } \partial M \\ \nu \rfloor \hat{\omega} = 0 & \text{on } \partial M. \end{array} \right. \quad (30) $$

The differential form $\hat{\omega}$ is called the tangential harmonic extension of $\omega$.
**Definition 2.10.** The Steklov operator $T^{[p]} : \Lambda^p(\partial M) \to \Lambda^p(\partial M)$ is defined by $T^{[p]}(\omega) = -\nu_{\lvert\lvert} d\hat{\omega}$.

It is proved in [23, Section 2], that the operator $T^{[p]}$ is an elliptic, self-adjoint and pseudodifferential operator with a discrete spectrum consisting of eigenvalues

$$0 \le \sigma_{1,p} \le \sigma_{2,p} \le \dots$$

Its kernel is also $H_A^p(M)$.

### 2.3.2 Variational characterizations

In this section, we provide some of the variational characterizations of the eigenvalues of the previous problems on differential forms, derived from the min-max principle (see [4, Chapter 1]), that we will use later on. Several of these expressions can also be found in [5, 10, 13, 23]. The proofs primarily follow the same lines as those used for the scalar case and we omit them here.

Recall that $H^s(M)$ and $H_0^2(M)$ have been defined in Section 2.2.1. Let us set

$$H_N^1(M) := \{\omega \in H^1(M) \mid \nu_{\lvert\lvert} \omega = 0\}$$

and

$$H_{N,1}^2(M) := \{\omega \in H^2(M) \mid \nu_{\lvert\lvert} \omega = 0 \text{ and } \nu_{\lvert\lvert} d\omega = 0\}.$$

**Proposition 2.11.** The first eigenvalue $\lambda_{1,p}$ of the problem (28) is given by:

$$\lambda_{1,p} = \inf \left\{ \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(M)}^2} \mid \omega \in H_0^1(M) \right\}. \quad (31)$$

**Proposition 2.12.** The $k^{th}$ eigenvalue $\lambda_{k,p}$ of the problem (28) is given by

$$\lambda_{k,p} = \inf_{V \subset H_0^1(M) \atop \dim(V)=k} \sup_{0 \neq \omega \in V} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(M)}^2} \quad (32)$$

$$= \inf_{V \subset H^2(M) \cap H_0^1(M) \atop \dim(V)=k} \sup_{0 \neq \nabla \omega \in V} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \quad (33)$$

**Proposition 2.13.** The first non-zero eigenvalue $\mu_{1,p}$ of the problem (29) is given by

$$\mu_{1,p} = \inf \left\{ \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(M)}^2} \mid \omega \in H_N^1(M) \text{ and } \omega \perp_{L^2(M)} H_A^p(M) \right\}. \quad (34)$$

**Proposition 2.14.** An alternative non-zero variational characterization of $\mu_{1,p}$ for the problem (29) is given by

$$\mu_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \mid \omega \in H_{N,1}^2(M) \setminus H_A^p(M) \right\}. \quad (35)$$

**Proposition 2.15.** The $k^{th}$ non-zero eigenvalue $\mu_{k,p}$ of the problem (29) is given by

$$\mu_{k,p} = \inf_{V \subset H_N^1(M) \atop \dim(V)=k+\dim H_A^p(M)} \sup_{0 \neq \omega \in V} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(M)}^2} \quad (36)$$

$$= \inf_{V \subset H_{N,1}^2(M)/H_A^p(M) \atop \dim(V)=k} \sup_{0 \neq \omega \in V/H_A^p(M)} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \quad (37)$$
**Proposition 2.16.** [23, Theorem 11] The first non-zero eigenvalue of the Steklov problem on differential forms is given by

$$ \sigma_{1,p} = \inf \left\{ \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \mid \omega \in H_N^1(M) \text{ and } \omega \perp_{L^2(\partial M)} H_A^p(M) \right\}. \quad (38) $$

**Proposition 2.17.** An alternative variational characterization of the first non-zero eigenvalue of the Steklov problem on differential forms is given by

$$ \sigma_{1,p} = \inf \left\{ \frac{\|\nu_\perp d\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \mid \omega \in \Omega^p(M), \Delta\omega = 0 \text{ on } M, \nu_\perp\omega = 0 \text{ on } \partial M, \omega \perp_{L^2(M)} H_A^p(M) \right\} \quad (39) $$

$$ = \inf \left\{ \frac{\|\nu_\perp d\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \mid \omega \in \Omega^p(M) \setminus H_A^p(M), \Delta\omega = 0 \text{ on } M, \nu_\perp\omega = 0 \text{ on } \partial M \right\} \quad (40) $$

**Proposition 2.18.** The $k^{th}$ non-zero eigenvalue $\sigma_{k,p}$ of the problem (30) is given by

$$ \sigma_{k,p} = \inf_{V \subset \tilde{H}_N^1(M) \atop \dim(V)=k} \sup_{0 \neq \omega \in V} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \quad (41) $$

$$ = \inf_{V \subset \mathcal{A} \atop \dim(V)=k} \sup_{0 \neq \omega \in V} \frac{\|\nu_\perp d\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}, \quad (42) $$

where $\tilde{H}_N^1(M) := \{\omega \in H_N^1(M) \mid \omega \perp_{L^2(\partial M)} H_A^p(M)\}$ and $\mathcal{A} := \{\omega \in H^2(M) \mid \nu_\perp\omega = 0, \Delta\omega = 0\}$.

# 3 Spectral properties of BSN1

We first consider the problem BSN1 (10), which is the analogue of (6) given in [6] by adapting its boundary conditions to the Neumann ones.

## 3.1 Ellipticity of BSN1

We begin with showing the ellipticity of the problem BSN1 (10).

**Proposition 3.1.** Let $(M^n, g)$ be a compact Riemannian manifold with a smooth boundary $\partial M$. The BSN1 (10) is elliptic in the sense of Shapiro-Lopatinskij.

*Proof.* Let $B_1, B_2, B_3$ et $B_4$ be the following maps

$$ B_1 : \Omega^p(M) \to \Gamma(E_1 = \Lambda^{p-1}T^*\partial M) \ ; \quad B_2 : \Omega^p(M) \to \Gamma(E_2 = \Lambda^p T^*\partial M) \\ \omega \mapsto \nu_\perp\omega \qquad \omega \mapsto \nu_\perp d\omega $$

$$ B_3 : \Omega^p(M) \to \Gamma(E_3 = \Lambda^p T^* \partial M) \ ; \quad B_4 : \Omega^p(M) \to \Gamma(E_4 = \Lambda^{p-1} T^* \partial M) \\ \omega \mapsto \nu_\perp \Delta d\omega + \ell t^* \omega \qquad \omega \mapsto \nu_\perp \Delta \omega + \ell t^* \delta \omega $$

$$ B_3 : \Omega^p(M) \to \Gamma(E_3 = \Lambda^p T^* \partial M) \ ; \quad B_4 : \Omega^p(M) \to \Gamma(E_4 = \Lambda^{p-1} T^* \partial M) \\ \omega \mapsto \nu_\perp \Delta d\omega + \ell t^* \omega \qquad \omega \mapsto \nu_\perp \Delta \omega + \ell t^* \delta \omega $$

$$ B_3 : \Omega^p(M) \to \Gamma(E_3 = \Lambda^p T^* \partial M) \ ; \quad B_4 : \Omega^p(M) \to \Gamma(E_4 = \Lambda^{p-1} T^* \partial M) \\ \omega \mapsto \nu_\perp \Delta d\omega + \ell t^* \omega \qquad \omega \mapsto \nu_\perp \Delta \omega + \ell t^* \delta \omega $$

where $\ell$ is a fixed real number. We note that $E_1$ and $E_4$ have rank $\binom{n-1}{p-1}$ and that $E_2$ and $E_3$ have rank $\binom{n-1}{p}$. Therefore, the rank of $\bigoplus_{j=1}^4 E_j$ is $2(\binom{n-1}{p} + \binom{n-1}{p-1}) = 2\binom{n}{p} = \dim \mathcal{M}_{\Delta^2, \nu}^+$. We have to compute the principal symbols of the $B_j$ for $1 \le j \le 4$.
Let us fix $\xi \in T^*M$ and let $f: M \to \mathbb{R}$ be such that $d_x f = \xi$. The operator $B_1$ is of order 0, so its principal symbol is $B_1$ itself: $\sigma_{B_1}(\xi) \omega = \nu_{\lceil} \omega$.

The operator $B_2$ is of order 1, recall that $\xi$ has both a normal and a tangential component, i.e., $\xi = \xi_{\nu} \nu^{\flat} + \iota^* \xi$ for all point in $\partial M$. By definition,

$$
\begin{align*}
\sigma_{B_2}(\xi) \omega &= [B_2, f] \omega \\
&= B_2(f \omega) - f B_2(\omega) \\
&= \nu_{\lceil} d(f \omega) - f \nu_{\lceil} d \omega \\
&= \nu_{\lceil} f d \omega + \nu_{\lceil} (df \wedge \omega) - f \nu_{\lceil} d \omega \\
&= \nu_{\lceil} (df \wedge \omega) \\
&= (\nu_{\lceil} df) \omega - df \wedge (\nu_{\lceil} \omega) \\
&= (\nu_{\lceil} \xi) \omega - \xi \wedge (\nu_{\lceil} \omega) \\
&= \xi_{\nu} \omega - \xi_{\nu} \nu \wedge (\nu_{\lceil} \omega) - \iota^* \xi \wedge (\nu_{\lceil} \omega) \\
&= \xi_{\nu} (\omega - \nu \wedge (\nu_{\lceil} \omega)) - \iota^* \xi \wedge (\nu_{\lceil} \omega) \\
&= \xi_{\nu} \iota^* \omega - \iota^* \xi \wedge (\nu_{\lceil} \omega).
\end{align*}
$$

The operator $B_3$ is of order 3 and $\sigma_{B_3}(\xi) \omega = (\sigma_{\nu_{\lceil}} \circ \sigma_{\Delta} \circ \sigma_d)(\xi) \omega$. We have $\sigma_d(\xi) \omega = \xi \wedge \omega$, see [27, p.181].

Therefore $\sigma_{\Delta \circ d} = -|\xi|^2 \xi \wedge \omega$. Since $\nu_{\lceil}$ is of order 0, we get $\sigma_{B_3}(\xi) \omega = -|\xi|^2 \nu_{\lceil}(\xi \wedge \omega)$.

Finally, $B_4$ is of order 2 and $\sigma_{B_4}(\xi) \omega = \sigma_{\nu_{\lceil}} \circ \sigma_{\Delta}(\xi) \omega = -\nu_{\lceil} |\xi|^2 \omega = -|\xi|^2 \nu_{\lceil} \omega$.

We now need to evaluate all of the principal symbols for $\xi = (-iv + \partial_t \nu^{\flat})$ with $v \in T^*M \setminus \{0\}$ and apply it to $y(t) = e^{-|v|t}(at + b)\omega_0$. For $\sigma_{B_1}$ we get

$$
\begin{align*}
\sigma_{B_1}(-iv + \partial_t \nu^{\flat})(y) &= \nu_{\lceil} y \\
&= \nu_{\lceil}(at + b)e^{-|v|t} \omega_0 \\
&= (at + b)e^{-|v|t} \nu_{\lceil} \omega_0,
\end{align*}
$$

then $\sigma_{B_1}(-iv + \partial_t \nu^{\flat})(y)(0) = b\nu_{\lceil} \omega_0$.

Similarly for $\sigma_{B_2}$,

$$
\begin{align*}
\sigma_{B_2}(-iv + \partial_t \nu^{\flat})(y) &= \partial_t \iota^* y - \iota^*(-iv + \partial_t) \wedge (\nu_{\lceil} y) \\
&= \partial_t \iota^* (at + b)e^{-|v|t} \omega_0 + \iota^*(iv) \wedge (\nu_{\lceil} (at + b)e^{-|v|t} \omega_0) \\
&= \iota^* a e^{-|v|t} \omega_0 - \iota^*(at + b)|v| e^{-|v|t} \omega_0 \\
&\quad + i(at + b)e^{-|v|t} \iota^* v \wedge (\nu_{\lceil} \omega_0).
\end{align*}
$$

Thus

$$
\begin{align*}
\sigma_{B_2}(-iv + \partial_t \nu^{\flat})(y)(0) &= a \iota^* \omega_0 - b|v| \iota^* \omega_0 + i b \iota^* v \wedge (\nu_{\lceil} \omega_0) \\
&= \iota^* \omega_0 (a - b|v|) + i b \iota^* v \wedge (\nu_{\lceil} \omega_0).
\end{align*}
$$
Next for $\sigma_{B_3}$ we have

$$
\begin{align*}
\sigma_{B_3}(-iv + \partial_t v^\flat)(y) &= -|-iv + \partial_t |^2 v_\flat ((-iv + \partial_t) \wedge y) \\
&= -|-iv + \partial_t v|^2 v_\flat ((-iv + \partial_t) \wedge (at + b)e^{-|v|t} \cdot \omega_0) \\
&= -\langle -iv + \partial_t v, -iv + \partial_t v \rangle. \\
&\quad [v_\flat (-i(at + b)e^{-|v|t} v \wedge \omega_0 + \partial_t (at + b)e^{-|v|t} v \wedge \omega_0)] \\
&= (|v|^2 - \partial_t^2) [-i(at + b)e^{-|v|t} \underbrace{(v_\flat v)}_0 \cdot \omega_0 - (v \wedge v)_\flat \omega_0) \\
&\quad + (be^{-|v|t} - |v|(at + b)e^{-|v|t})v_\flat (v \wedge \omega_0)] \\
&= (|v|^2 - \partial_t^2)e^{-|v|t} [i(at + b)v \wedge (v_\flat \omega_0) + b - |v|(at + b)v_\flat (v \wedge \omega_0)] \\
&= |v|^2 e^{-|v|t} [i(at + b)v \wedge (v_\flat \omega_0) + b - |v|(at + b)v_\flat (v \wedge \omega_0)] \\
&\quad - |v|^2 e^{-|v|t} [i(at + b)v \wedge (v_\flat \omega_0) + b - |v|(at + b)v_\flat (v \wedge \omega_0)] \\
&\quad + 2|v|e^{-|v|t} ((iav \wedge (v_\flat \omega_0) - |v|a v_\flat (v \wedge \omega_0)) \\
&= 2|v|e^{-|v|t} ((iav \wedge (v_\flat \omega_0) - |v|a v_\flat (v \wedge \omega_0)).
\end{align*}
$$

Noting that $v_\flat(\nu \wedge \omega_0) = (v_\flat \nu) \omega_0 - \nu \wedge (v_\flat \omega_0) = \omega_0 - \nu \wedge (v_\flat \omega_0) = \iota^* \omega_0$, we obtain that

$$
\sigma_{B_3}(-iv + \partial_t v^{\flat})(y) = 2|v|e^{-|v|t}a \left( iv \wedge (v_\flat \omega_0) - |v|\iota^* \omega_0 \right),
$$

and thus

$$
\sigma_{B_3}(-iv + \partial_t v^{\flat})(y)(0) = 2|v|a \left( iv \wedge (v_\flat \omega_0) - |v|\iota^* \omega_0 \right).
$$

Finally for $\sigma_{B_4}$,

$$
\begin{align*}
\sigma_{B_4}(-iv + \partial_t v^\flat)(y) &= -v_\flat |\xi|^2 y \\
&= -v_\flat |-iv + \partial_t v|^2 (at + b)e^{-|v|t} \omega_0 \\
&= -v_\flat (\partial_t^2 - |v|^2)(at + b)e^{-|v|t} \omega_0 \\
&= |v|^2 (at + b)e^{-|v|t} v_\flat \omega_0 - v_\flat (-2a|v|e^{-|v|t} + (at + b)|v|^2 e^{-|v|t}) \omega_0 \\
&= 2a|v|e^{-|v|t} v_\flat \omega_0,
\end{align*}
$$

we conclude $\sigma_{B_4}(-iv + \partial_t v^\flat)(y)(0) = 2a|v|\nu_\flat \omega_0$. Based on the calculations above, we obtain the
following map:

$$
\Phi : \mathcal{M}_{\Delta^2, v}^{+} \to \bigoplus_{j=1}^{4} E_j
$$
$$
y \mapsto \Phi(y)
$$

such that

$$
\Phi(y) = (b\nu_\flat\omega_0, (a - b|v|)\iota^*\omega_0 + ib\iota^*v \wedge (\nu_\flat\omega_0), 2|v|(iav \wedge (\nu_\flat\omega_0) - |v|a\iota^*\omega_0), 2a|v|\nu_\flat\omega_0).
$$

To end the proof, we must show that $\Phi$ is an isomorphism. Since both source and target spaces
have the same finite dimension, it is enough to show that $\Phi$ is injective. Suppose that $\Phi(y) = 0$,
then

$$
\left\{
\begin{array}{l@{\qquad}l}
b\nu_{\flat}\omega_0 & = 0 \quad (1) \\
(a - b|v|)\iota^*\omega_0 + ib\iota^*v \wedge (\nu_{\flat}\omega_0) & = 0 \quad (2) \\
2|v|(iav \wedge (\nu_{\flat}\omega_0) - |v|a\iota^*\omega_0) & = 0 \quad (3) \\
2a|v|\nu_{\flat}\omega_0 & = 0 \quad (4)
\end{array}
\right.
$$
If a and b are zero, then y is zero. If a or b are non-zero, then from (1) and (4), we deduce that $\nu \rfloor \omega_0 = 0$. Substituting this result into (2) and (3), we find that $\iota^* \omega_0 = 0$. Consequently, $\omega_0 = 0$, which implies that $y = 0$. Hence, $\Phi$ is injective and the problem (10) is elliptic.

□

## 3.2 Discrete spectrum of BSN1

The aim of this paragraph is to establish the following theorem. The approach of the proof is mainly inspired by the work in [6, 8].

**Theorem 3.2.** Let $(M^n, g)$ be a compact Riemannian manifold with a smooth boundary $\partial M$ and let $\nu$ be the unit inward normal vector to the boundary. Then the BSN1 (10) for differential $p$-forms, has a discrete spectrum consisting of an unbounded non-decreasing sequence of non-negative real eigenvalues with finite multiplicities $(\ell_i)_{i \ge 0}$ in the form

$$0 < \ell_1 \le \ell_2 \le \ell_3 \le \dots$$

Its kernel is given by $H_A^p(M)$ (9). Moreover, all its eigenforms are smooth.

**Lemma 3.3.** For $f \in \Omega^p(M)$, $\omega_1 \in \Gamma(\Lambda^{p-1}T^*\partial M)$, $\omega_2 \in \Gamma(\Lambda^p T^*\partial M)$, $\omega_3 \in \Omega^p(\partial M)$, and $\omega_4 \in \Omega^{p-1}(\partial M)$ the boundary problem

$$(\text{biLap}_1) \left\{ \begin{array}{ll} \Delta^2 \omega & = f \quad \text{on } M \\ \nu \rfloor \omega & = \omega_1 \quad \text{on } \partial M \\ \nu \rfloor d\omega & = \omega_2 \quad \text{on } \partial M \\ \iota^* \omega & = \omega_3 \quad \text{on } \partial M \\ \iota^* \delta \omega & = \omega_4 \quad \text{on } \partial M. \end{array} \right. \qquad (43)$$

has a unique solution. Moreover, if $f, \omega_1, \dots, \omega_n$ are smooth, then $\omega$ is also smooth.

*Proof.* Let us first show that the problem $(\text{biLap}_1)$ is elliptic in the sense of Shapiro-Lopatinskij. As in section Section 2.2.3, we denote by $B_1, B_2, B_3$, and $B_4$ the following maps:

$$B_1 : \Omega^p(M) \longrightarrow \Gamma(E_1 = \Lambda^{p-1}T^*\partial M) ; \quad B_2 : \Omega^p(M) \longrightarrow \Gamma(E_2 = \Lambda^p T^*\partial M) \\ \omega \longmapsto \nu \rfloor \omega \qquad \omega \longmapsto \nu \rfloor d\omega$$

$$B_2 : \Omega^p(M) \longrightarrow \Gamma(E_2 = \Lambda^p T^* \partial M) \\ \omega \longmapsto \nu \rfloor d \omega$$

$$B_3 : \Omega^p(M) \longrightarrow \Gamma(E_3 = \Lambda^p T^* \partial M) ; \quad B_4 : \Omega^p(M) \longrightarrow \Gamma(E_4 = \Lambda^{p-1} T^* \partial M) \\ \omega \longmapsto \iota^* \omega \qquad \omega \longmapsto \iota^* \delta \omega$$

$$B_4 : \Omega^p(M) \longrightarrow \Gamma(E_4 = \Lambda^{p-1} T^* \partial M) \\ \omega \longmapsto \iota^* \delta \omega$$

Note that $\dim \mathcal{M}_{\Delta^2, \nu}^+ = \dim \bigoplus_{j=1}^4 E_j = 2(\binom{n}{p})$. Similarly to Section 3.1, we compute the principal symbols of the $B_i$ applied to $y = e^{-|v|t}(at + b)\omega_0$ for some $\omega_0 \in \Lambda^p T^* M|_{\partial M}$. We find

$$\sigma_{B_1}(-iv + \partial_t \nu^b)(y)(0) = b \nu \rfloor \omega_0,$$

$$\sigma_{B_2}(-iv + \partial_t \nu^b)(y)(0) = \iota^* \omega_0(a - b|v|) + ib \iota^* v \wedge (\nu \rfloor \omega_0),$$

$$\sigma_{B_3}(-iv + \partial_t \nu^b)(y)(0) = b \iota^* \omega_0,$$

$$\sigma_{B_4}(-iv + \partial_t \nu^b)(y)(0) = -a \nu \rfloor \omega_0 + |v|b \nu \rfloor \omega_0 + ivb \rfloor \iota^* \omega_0.$$

Therefore, for all $y = e^{-|v|t}(at + b) \in \mathcal{M}_{\Delta^2, \nu}^+$, the map $\Phi$ of Definition 2.4 is given by

$$\Phi(y) = (b\nu \rfloor \omega_0, (a - b|v|)\iota^* \omega_0 + ib\iota^* v \wedge (\nu \rfloor \omega_0), b\iota^* \omega_0, (b|v| - a)\nu \rfloor \omega_0 + bv \rfloor \iota^* \omega_0).$$
Let us show that $\Phi$ is injective. Suppose $\Phi(y) = 0$, then

$$ 
\left\{ 
\begin{array}{lclcl}
b\nu\lrcorner\omega_0 & = & 0 & \\
(a - b|v|)\iota^*\omega_0 + ib\iota^*v \wedge (\nu\lrcorner\omega_0) & = & 0 & \\
b\iota^*\omega_0 & = & 0 & \\
(b|v| - a)\nu\lrcorner\omega_0 + bv\lrcorner\iota^*\omega_0 & = & 0 & \\
\end{array}
\right.
\implies
\left\{ 
\begin{array}{lclcl}
b\nu\lrcorner\omega_0 & = & 0 & \\
(a - b|v|)\iota^*\omega_0 & = & 0 & \\
b\iota^*\omega_0 & = & 0 & \\
-(a - b|v|)\nu\lrcorner\omega_0 & = & 0. & \\
\end{array}
\right.
$$

Hence, $(a - b|v|)\omega_0 = 0$ and $b\omega_0 = 0$, so $y = 0$. Thus, $\Phi$ is injective and consequently it is an isomorphism. Therefore $(biLap_1)$ is elliptic.

Performing an integration by parts (23) we get

$$ 
\begin{aligned}
(\Delta^2 \omega, \omega')_{L^2(M)} - (\omega, \Delta^2 \omega')_{L^2(M)}
&= (\nu\lrcorner d\Delta \omega, \iota^* \omega')_{L^2(\partial M)} - (\iota^* \omega, \nu\lrcorner d\Delta \omega')_{L^2(\partial M)} \\
&\quad - (\iota^* \delta\Delta \omega, \nu\lrcorner \omega')_{L^2(\partial M)} + (\nu\lrcorner \omega, \iota^* \delta\Delta \omega')_{L^2(\partial M)} \\
&\quad + (\nu\lrcorner \Delta \omega, \iota^* \delta \omega')_{L^2(\partial M)} - (\iota^* \delta \omega, \nu\lrcorner \Delta \omega')_{L^2(\partial M)} \\
&\quad - (\iota^* \Delta \omega, \nu\lrcorner d \omega')_{L^2(\partial M)} + (\nu\lrcorner d \omega, \iota^* \Delta \omega')_{L^2(\partial M)}.
\end{aligned}
$$

Using the notations given in Section 2.2.4 $T_1\omega = -\iota^*\Delta\omega$, $T_2\omega = -\iota^*\delta\Delta\omega$, $T_3\omega = \nu\lrcorner\Delta\omega$, and $T_4\omega = \nu\lrcorner d\Delta\omega$. Similarly, $C_1\omega = \iota^*\omega$, $C_2\omega = -\nu\lrcorner\omega$, $C_3\omega = \iota^*\delta\omega$, and $C_4\omega = -\nu\lrcorner d\omega$.

This implies that $N^* = 0$ using an integration by parts (19) and that every harmonic differential form that vanishes on the boundary also vanishes identically on $M$ (see [1]). The kernel $N$ of problem (43) is also trivial.

So by Theorem 2.6 the map

$$ 
\begin{array}{rcl}
\mathcal{P}_{biLap_1} : H^s(M) & \to & H^{s-4}(M) \oplus H^{s-\frac{1}{2}}(\partial M) \oplus H^{s-\frac{3}{2}}(\partial M) \oplus H^{s-\frac{1}{2}}(\partial M) \oplus H^{s-\frac{3}{2}}(\partial M) \\
[\omega] & \mapsto & (\Delta^2 \omega, \nu\lrcorner \omega, \nu\lrcorner d\omega, \iota^* \omega, \iota^* \delta \omega)
\end{array}
$$

is an isomorphism on its image that is given for all $s \ge 4$ by

$$ \mathrm{Im}(\mathcal{P}_{biLap_1}) = H^{s-4}(M) \times H^{s-\frac{3}{2}}(\partial M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M) \times H^{s-\frac{1}{2}}(\partial M). $$

Which concludes the proof of the lemma.

<figure><img src="image_3.png" /></figure>

By analogy to Definition 1.3, we define the BSN1 operator denoted by $\mathrm{BSN1}^{\mathrm{form}}$ as:

$$ 
\begin{array}{rcl}
\mathrm{BSN1}^{\mathrm{form}} : H^{\frac{7}{2}}(\partial M) \times H^{\frac{5}{2}}(\partial M) & \to & H^{\frac{3}{2}}(\partial M) \times H^{\frac{1}{2}}(\partial M) \\
(\omega_1, \omega_2) & \mapsto & (-\nu\lrcorner\Delta\hat{\omega}, -\nu\lrcorner d\hat{\omega})
\end{array}
$$

where $\hat{\omega} \in H^4(M)$ is the solution of

$$ 
\begin{array}{l}
\left( \begin{array}{lclcl}
\Delta^2 \hat{\omega} & = & 0 & \text{on } M & \\
\nu\lrcorner \hat{\omega} & = & 0 & \text{on } \partial M & \\
\nu\lrcorner d \hat{\omega} & = & 0 & \text{on } \partial M & \\
\iota^* \hat{\omega} & = & \omega_3 & \text{on } \partial M & \\
\iota^* \delta \hat{\omega} & = & \omega_4 & \text{on } \partial M, & \\
\end{array} \right) \\
\text{(biLap}_1) \\
\end{array}
\qquad (44)
$$

which is a particular case of the problem (43). In the particular case of (43) where $f = 0$, $\omega_1 = 0$ and $\omega_2 = 0$ we have $\Delta^2 \hat{\omega} = 0$ on $M$, $\nu\lrcorner \hat{\omega} = 0$ and $\nu\lrcorner d\hat{\omega} = 0$ on $\partial M$, $\iota^* \hat{\omega} = \omega_3$, and $\iota^* \delta \hat{\omega} = \omega_4$. Note that the eigenforms of this problem are the same as those of the BSN1 (10).
Proof of Theorem 3.2. We define the following spaces:

$$H_{N,1}^2(M) := \{\omega \in H^2(M) \mid \nu \rfloor d\omega = 0 \text{ and } \nu \rfloor \omega = 0 \text{ on } \partial M\},$$

$$\check{H}_{N,1}^2(M) := \{\omega \in H_{N,1}^2(M) \mid \omega \perp_{L^2(\partial M)} H_A^p(M)\},$$

$$Z_1 := \{\omega \in \Omega^p(M) \cap H_{N,1}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M, \} \subset H_{N,1}^2(M)$$

and

$$\check{Z}_1 := \{\omega \in Z_1 \mid \omega \perp_{L^2(\partial M)} H_A^p(M)\} = Z_1 \cap \check{H}_{N,1}^2(M).$$

For any $\ell$-eigenform of BSN1 $\omega$ and $\omega' \in Z_1$, we have from equation (21) the following identity:

$$\int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g = \ell \int_{\partial M} \left( \langle \iota^* \omega, \iota^* \omega' \rangle + \langle \iota^* \delta \omega, \iota^* \delta \omega' \rangle \right) d\mu_g. \quad (45)$$

Let us define for all $\omega, \omega' \in \Omega^p(M)$ the following bilinear forms:

$$(\omega, \omega')_{W_1} := \int_{\partial M} \langle \iota^* \omega, \iota^* \omega' \rangle d\mu_g + \int_{\partial M} \langle \iota^* \delta \omega, \iota^* \delta \omega' \rangle d\mu_g$$

and

$$(\omega, \omega')_V := \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g.$$

**Lemma 3.4.** The forms $(\cdot, \cdot)_{W_1}$ and $(\cdot, \cdot)_V$ are scalar products on $Z_1$ and $\check{H}_{N,1}^2(M)$, respectively.

*Proof.* It is easy to verify that the forms defined above are bilinear, symmetric, and non-negative. Let us show that they are positive definite. For $\omega \in Z_1$, if $(\omega, \omega)_{W_1} = 0$, then $\int_{\partial M} (|\iota^* \omega|^2 + |\iota^* \delta \omega|^2) d\mu_g = 0$, which implies that $\iota^* \omega$ and $\iota^* \delta \omega$ are zero on $\partial M$. Substituting into (21) with $\omega \in Z_1$, we obtain $\Delta \omega = 0$ on $M$. Since $\omega|_{\partial M} = 0$ we conclude that $\omega = 0$ on $M$ according to [1].

Similarly, for $\omega \in \check{H}_{N,1}^2(M)$, if $(\omega, \omega)_V = 0$ then $\Delta \omega = 0$ on $M$. Thus, (22) gives that $d\omega$ and $\delta \omega$ are zero, and since $\nu \rfloor \omega = 0$ it follows that $\omega \in H_A^p(M)$. Moreover, since $\omega \in \check{Z}_1$ this implies that $\int_{\partial M} \langle \iota^* \omega, \iota^* \omega \rangle d\mu_g = 0$ by the definition of elements of $\check{Z}_1$. Therefore, $\int_{\partial M} |\iota^* \omega|^2 d\mu_g = 0$, which means that $\iota^* \omega = 0$. As a result, $\omega|_{\partial M} = 0$, and because $\omega$ is harmonic, we can conclude again that $\omega = 0$ on $M$ according to [1]. $\square$

We denote by $W_1$ the completion of $Z_1$ with respect to $(\cdot, \cdot)_{W_1}$, and by $V_1$ the completion of $\check{Z}_1$ with respect to $(\cdot, \cdot)_V$.

From (19) and (21) by taking $\omega \in Z_1$ and $\omega' = \omega$, we get

$$\int_M |\Delta \omega|^2 d\mu_g + \int_{\partial M} \left( \langle \nu \rfloor d\Delta \omega, \iota^* \omega \rangle + \langle \nu \rfloor \Delta \omega, \iota^* \delta \omega \rangle \right) d\mu_g = 0. \quad (46)$$

**Lemma 3.5.** The kernel of BSN1 coincides with $H_A^p(M)$. Moreover, if $\omega$ is a solution of BSN1 with $\ell \neq 0$, then $\omega \perp_{L^2(\partial M)} H_A^p(M)$ and $\ell > 0$.

*Proof.* Let $\omega$ be a solution of BSN1 with $\ell = 0$. By integrating by parts using equation (21), where we take $\omega = \omega'$ we obtain:

$$\int_M \langle \underbrace{\Delta^2 \omega}_{0}, \omega \rangle d\mu_g = \int_M \langle \Delta \omega, \Delta \omega \rangle d\mu_g + \int_{\partial M} \left( \langle \underbrace{\nu \rfloor d\Delta \omega}_{0}, \iota^* \omega \rangle - \langle \iota^* \Delta \omega, \underbrace{\nu \rfloor d\omega}_{0} \rangle + \langle \underbrace{\nu \rfloor \Delta \omega}_{0}, \iota^* \delta \omega \rangle - \langle \iota^* \delta \Delta \omega, \underbrace{\nu \rfloor \omega}_{0} \rangle \right) d\mu_g,$$
thus $\Delta \omega = 0$. Substituting this into (22), we have

$$ \int_M \langle \underbrace{\Delta \omega, \omega}_{0} \rangle d\mu_g = \int_M (|d\omega|^2 + |\delta\omega|^2) d\mu_g + \int_{\partial M} \langle \underbrace{\nu \rfloor d\omega, \iota^* \omega}_{0} \rangle d\mu_g - \int_{\partial M} \langle \iota^* \delta\omega, \underbrace{\nu \rfloor \omega}_{0} \rangle d\mu_g. $$

This implies that $d\omega = 0$ and $\delta\omega = 0$, so $\omega$ lies in $H_A^p(M)$.

Conversely, for $\omega \in H_A^p(M)$, it is a solution of BSN1 for $\ell = 0$ by definition of $H_A^p(M)$ (9). Now if $\omega$ is a solution of the BSN1, $\ell \neq 0$ and $\omega \in Z_1$, we have by (45) for all $\alpha \in H_A^p(M)$ that

$$ 0 = \int_M \langle \Delta \omega, \Delta \alpha \rangle d\mu_g = \ell \int_{\partial M} \langle \iota^* \omega, \iota^* \alpha \rangle d\mu_g. $$

Since $\ell \neq 0$ we get $\omega \perp L^2(\partial M) = H_A^p(M)$, therefore $\omega \in \check{Z}_1 \subset V_1$. Moreover, still by (45) we get $\int_M |\Delta \omega|^2 d\mu_g = \ell \int_{\partial M} (|\iota^* \omega|^2 + |\iota^* \delta \omega|^2) d\mu_g$. Therefore $\ell > 0$, which concludes the proof. $\square$

**Lemma 3.6.** The norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$ are equivalent on $\check{H}_{N,1}^2(M)$. In particular they are equivalent in $\check{Z}_1$ and $V_1 \subset \check{H}_{N,1}^2(M)$.

*Proof.* As $\Delta$ is a second-order differential operator, there exists $c > 0$ such that for all $\omega \in \check{Z}_1$, $\|\omega\|_V \le c\|\omega\|_{H^2(M)}$. It remains to show that for all $\omega \in \check{Z}_1$, $\|\omega\|_{H^2(M)} \le c\|\omega\|_V$. To do this, let us first show that the space $H_{N,1}^2(M)$ is closed in $H^2(M)$. Indeed, it is since $H_{N,1}^2(M)$ is the kernel of the map

$$ J: H^2(M) \to L^2(\partial M) \oplus L^2(\partial M) \\ \omega \mapsto (\nu \rfloor d\omega, \nu \rfloor \omega), $$

which is continuous by the trace theorem.

By definition, we have

$$ H_{N,1}^2(M) = \check{H}_{N,1}^2(M) \oplus H_A^p(M). $$

We define the space

$$ \check{L}^2(M) := \{ \omega \in L^2(M) \mid \forall \alpha \in H_A^p(M), \int_M \langle \alpha, \omega \rangle d\mu_g = 0 \}, $$

and we prove the following lemma.

**Lemma 3.7.** The following map $\Delta_{|\check{H}_{N,1}^2(M)} : \check{H}_{N,1}^2(M) \to \check{L}^2(M)$, is an isomorphism.

*Proof.* This map is well-defined. Indeed, doing an integration by parts (20) we get $\Delta \omega \in \check{L}^2(M)$ for all $\omega \in \check{H}_{N,1}^2(M)$. Now let $\omega \in \check{H}_{N,1}^2(M)$ such that $\Delta \omega = 0$, performing an integration by parts (20) we obtain that $d\omega$ and $\delta\omega$ are zero. Additionally, since $\nu \rfloor \omega = 0$, we conclude that $\omega \in H_A^p(M)$. Thus, $\iota^* \omega = 0$ as $\omega \in H_A^p(M)$ and $\omega \perp L^2(\partial M) = H_A^p(M)$. Therefore, $\omega = 0$ by [1], so $\ker \Delta_{|\check{H}_{N,1}^2(M)} = \{0\}$. Hence, the operator $\Delta_{|\check{H}_{N,1}^2(M)} : \check{H}_{N,1}^2(M) \to \check{L}^2(M)$ is injective.

In fact, it is an isomorphism, as follows from Theorem 2.6, which states that for $\mathcal{N} = H_A^p(M)$:

$$ \begin{aligned} \mathcal{P}_{\Delta_{Neu}} : H_{N,1}^2(M)/H_A^p(M) &\to \check{L}^2(M) \oplus H^3(\partial M) \oplus H^1(\partial M) \\ [\omega] &\mapsto (\Delta \omega, \nu \rfloor \omega, \nu \rfloor d\omega) \end{aligned} $$
is a topological isomorphism onto its image, which is given by

$$ \mathrm{Im}(\mathcal{P}_{\Delta_{Neu}}) = \left\{ (\hat{\omega}, \omega_1, \omega_2) \in L^2(M) \times H^{-\frac{1}{2}}(\partial M) \times H^{-\frac{3}{2}}(\partial M) \mid \forall \omega' \in \mathcal{N}^*, \right. \\ \left. (\hat{\omega}, \omega') + \sum_{j=1}^{2} (\omega_j, T_j \omega') = 0 \right\}, $$

where $(\cdot, \cdot)$ is the duality product and $T_j$ is the operator defined in Section 2.2.4.

Note that $\check{L}^2(M) \oplus \{0\} \oplus \{0\} \subset \mathrm{Im} \, \mathcal{P}$. We have in this case

$$ \mathcal{N}^* = \{ \omega \in \Omega^p(M) \mid \Delta \omega = 0, C_1 \omega = 0 \text{ and } C_2 \omega = 0 \} = H_A^p(M), \quad (47) $$

since $C_1 \omega = \nu \rfloor d\omega$ and $C_2 \omega = \nu \rfloor \omega$ by Section 2.2.4. In fact, $(\hat{\omega}, \omega_1, \omega_2)$ belongs to $\mathrm{Im}(\mathcal{P})$ if and only if for all $\alpha \in H_A^p(M)$ we have $(\hat{\omega}, \alpha) - (\omega_1, \iota^*\delta\alpha) - (\omega_2, \iota^*\alpha) = 0$, i.e., $(\Delta \omega, \alpha) = (\omega_2, \alpha)$. Thus,

$(\hat{\omega}, 0, 0) \in \mathrm{Im}(\mathcal{P})$ if and only if for all $\alpha \in H_A^p(M)$, $(\hat{\omega}, \alpha)_{L^2(M)} = 0$. Therefore, $\hat{\omega} \in H_A^p(M)^{\perp(\cdot, \cdot)_{L^2(M)}}$. We deduce that the set $\check{L}^2(M) \times \{0\} \times \{0\}$ is included in $\mathrm{Im}(\mathcal{P})$. $\square$

Now let

$$ \mathcal{L} := (\mathcal{P}_{\Delta_{Neu}})_{|\check{L}^2(M) \oplus \{0\} \oplus \{0\}}^{-1} : \check{L}^2(M) \oplus \{0\} \oplus \{0\} \to H_{N,1}^2(M)/H_A^p(M) \\ (\hat{\omega}, 0, 0) \mapsto [\omega] $$

be the continuous map such that $\Delta \omega = \hat{\omega}$ on $M$, $\nu \rfloor \omega = 0$ and $\nu \rfloor d\omega = 0$ on $\partial M$. By continuity of $\mathcal{L}$, for all $\hat{\omega} \in \check{L}^2(M)$, there exists $c > 0$ such that

$$ \|\mathcal{L}(\hat{\omega})\|_{H^2(M)} \le c \|\hat{\omega}\|_{L^2(M)}, $$

therefore, for all $\omega \in \check{H}_{N,1}^2(M)$, there exists $c' > 0$ such that

$$ \|\omega\|_{H^2(M)} \le c' \|\Delta \omega\|_{L^2(M)}. $$

Since $Z_1 \subset H_{N,1}^2(M)$, we have $\check{Z}_1 \subset \check{H}_{N,1}^2(M)$, so there exists $c' > 0$ such that for all $\omega \in \check{Z}_1$,

$$ \|\omega\|_{H^2(M)} \le c' \|\Delta \omega\|_{L^2(M)} = c' \|\omega\|_V. $$

Thus, the norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$ are equivalent on $\check{H}_{N,1}^2(M)$ and therefore on $\check{Z}_1$ and $V_1$ by density, which completes the proof. $\square$

**Lemma 3.8.** There exists $c > 0$ such that $\|\cdot\|_{W_1} \le c\|\cdot\|_V$ on $\check{Z}_1$.

*Proof.* Recall that $\|\omega\|_{H^1(\partial M)}^2 = \|\omega\|_{L^2(\partial M)}^2 + \|\nabla \omega\|_{L^2(\partial M)}^2$. For all $\omega \in \check{Z}_1$, we have by definition $\|\omega\|_{W_1}^2 = \|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2$.

Moreover,

$$ \|\iota^* \omega\|_{L^2(\partial M)}^2 = \|\omega\|_{L^2(\partial M)}^2 \le \|\omega\|_{H^1(\partial M)}^2 $$

and

$$ \|\iota^* \delta \omega\|_{L^2(\partial M)}^2 \le c \|\nabla \omega\|_{L^2(\partial M)}^2 \le c \|\omega\|_{H^1(\partial M)}^2. $$

Furthermore, by continuity of the trace operator $T: H^2(M) \to H^1(\partial M)$, there exists $c > 0$ such that $\|\omega\|_{W_1}^2 \le c\|\omega\|_{H^1(\partial M)}^2 \le c\|\omega\|_{H^2(M)}^2$.

Finally, using the equivalence of the norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$, we obtain $\|\omega\|_{W_1}^2 \le c\|\omega\|_V^2$ for some constant $c$. $\square$
From Lemma 3.8, and since by definition $\tilde{Z}_1$ is dense in both $V_1$ and $W_1$, the inclusion map $i_{W_1}: \tilde{Z}_1 \to W_1$, which is linear and continuous, extends to another continuous "inclusion" map $I_1$ denoted as

$$ \begin{array}{ccc} I_1: V_1 & \to & W_1 \\ & \omega & \mapsto \omega \end{array} \qquad (48) $$

such that $i_{W_1} = I_1 \circ i_{V_1}$, where $i_{V_1}: \tilde{Z}_1 \to V_1$.

**Lemma 3.9.** *The map $I_1$ is compact.*

*Proof.* Let $J_1$ be the following map:

$$ \begin{array}{ccc} J_1: Z_1 & \to & L^2(\partial M) \oplus L^2(\partial M) \\ & \omega & \mapsto (\iota^* \omega, \iota^* \delta \omega) \end{array} $$

Its extension to $W_1$ exists and is an isometry by definition of $\|\cdot\|_{W_1}$. We prove the compactness of $I_1$ via the compactness of the map $J_1 \circ I_1$:

$$ \begin{array}{ccc} J_1 \circ I_1: V_1 & \xrightarrow{I_1} & W_1 \\ & \omega & \mapsto \omega \end{array} \qquad \begin{array}{ccc} & \xrightarrow{J_1} & L^2(\partial M) \oplus L^2(\partial M) \\ & & (\iota^* \omega, \iota^* \delta \omega) \end{array} $$

First, note that $V_1$ is included in $H^2(M)$ by the equivalence of norms. According to Lemma 3.6, and since $H^2(M) \subset H^1(M)$, the map $C: V_1 \to H^1(M) \oplus H^1(M)$, $\omega \mapsto (\omega, \delta \omega)$ is continuous. Moreover, we have that $H^1(M) \subset H^{1/2}(\partial M)$ due to the trace theorem, and the inclusion map $H^{1/2}(\partial M) \subset L^2(\partial M)$ is compact by the Rellich-Kondrachov theorem. Hence, $J_1 \circ I_1$ can be represented as follows:

$$ \begin{array}{ccc} V_1 & \to & H^1(M) \oplus H^1(M) \\ & \omega & \mapsto (\omega, \delta \omega) \end{array} \qquad \begin{array}{ccc} & \to & H^{1/2}(\partial M) \oplus H^{1/2}(\partial M) \\ & & (\iota^* \omega, \iota^* \delta \omega) \end{array} \qquad \begin{array}{ccc} & \to & L^2(\partial M) \oplus L^2(\partial M) \\ & & (\iota^* \omega, \iota^* \delta \omega) \end{array} $$

Thus, $J_1 \circ I_1$ is compact as it is composed of continuous maps, one of which is compact. To deduce that $I_1$ is compact, we require the following general lemma.

**Lemma 3.10.** Let $E, E'$ and $E''$ be three Banach spaces. Let $F: E \to E''$ be a compact map, $G: E' \to E''$ an isometry and $H: E \to E'$ a map such that $F = G \circ H$. Then $H$ is compact.

*Proof.* Let $(v_n)_n$ be a bounded sequence in $E$. By the compactness of $F$, there exists a subsequence $(v_{k_n})_n$ such that $F(v_{k_n}) = G(H(v_{k_n}))$ converges in $E''$. Since $G: E' \to E''$ is an isometry then $G(E')$ is a complete set in $E''$, consequently, $G(E')$ is closed in $E''$. Thus, the sequence $G(H(v_{k_n}))$ converges to $G(x)$ for some $x \in E'$. Due to the isometry property of $G$, we have $H(v_{k_n}) \to x$ in $E'$. Therefore, we can deduce that the map $H$ is compact. $\square$

Consequently, $I_1$ is compact according to Lemma 3.10. $\square$

We provide the definition of a weak solution adapted to our situation.

**Definition 3.11.** For $f \in L^2(M)$, a weak solution $\omega$ of $\Delta^2 \omega = f$ on $M$ is a form $\omega \in H_{N,1}^2(M)$ such that

$$ (\Delta \omega, \Delta \omega')_{L^2(M)} = (f, \omega')_{L^2(M)}, \quad \forall \omega' \in H_0^2(M). \qquad (49) $$

A general definition of the weak solution in an open subset of $\mathbb{R}^n$ is given in [7, Sections 6.1.1 and 6.1.2].
**Lemma 3.12.** We have the following equality

$$V_1 = \{\omega \in \tilde{H}_{N,1}^2(M) \mid \Delta^2 \omega = 0 \text{ weakly in } M\}.$$

*Proof.* We know by Lemma 3.6 that $\tilde{Z}_1 \subset \tilde{H}_{N,1}^2(M) \subset H_{N,1}^2(M)$. Let us fix $\omega \in V_1$. There exists a sequence $(\omega_m)_m$ in $\tilde{Z}_1$ such that $\|\omega_m - \omega\|_V \xrightarrow{m \to \infty} 0$. By the equivalence of the norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$, on $V_1$ the sequence $(\omega_m)_m$ converges in $H^2(M)$ to $\omega$. Hence, $(\Delta \omega_m)_m$ converges in $L^2(M)$ to $\Delta \omega$. For $\omega' \in H_0^2(M)$, $\omega_m \in \tilde{Z}_1$ and replacing $\omega$ by $\omega_m$ in (21), we obtain $(\Delta \omega_m, \Delta \omega')_{L^2(M)} = 0$. Since $(\Delta \omega_m)_m$ converges in $L^2(M)$ to $\Delta \omega$, it follows that $(\Delta \omega, \Delta \omega')_{L^2(M)} = 0$ for all $\omega' \in H_0^2(M)$. Thus, $\omega \in H_{N,1}^2(M)$ and $\Delta^2 \omega = 0$ weakly in $M$, proving the first inclusion.

Conversely, let $\omega \in \tilde{H}_{N,1}^2(M)$ such that $\Delta^2 \omega = 0$ weakly in $M$. Since the norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$ are equivalent on $\tilde{H}_{N,1}^2(M)$, we get that $\Delta^2 \omega = 0$ sur $M$. Hence the equality. $\square$

**Lemma 3.13.** The map $I_1$ is injective.

*Proof.* The injectivity of $J_1 \circ I_1$ implies the injectivity of $I_1$. Let $\omega \in V_1$ be such that $J_1 \circ I_1(\omega) = 0$ i.e., $\iota^* \omega = 0$ and $\iota^* \delta \omega = 0$ on $\partial M$. Since, $\omega|_{\partial M} = 0$ and $\nu \rfloor d\omega = 0$ ($\omega \in Z_1$), its differential $d\omega = \nu^\flat \wedge (\nu \rfloor d\omega) + \iota^* d\omega$ is zero on the boundary. Furthermore, $\delta \omega = 0$ is zero on the boundary as $\iota^* \delta \omega = 0$ and $\nu \rfloor \delta \omega = -\delta(\nu \rfloor \omega) = 0$. Consequently, $d\omega = \nu^\flat \wedge \nabla_\nu \omega = 0$ and $\delta \omega = -\nu \rfloor \nabla_\nu \omega = 0$ along the boundary. From the above, we have $\nu \rfloor \underbrace{(\nu^\flat \wedge \nabla_\nu \omega)}_{0} = \underbrace{\nu^\flat(\nu) \nabla_\nu \omega}_{1} - \underbrace{\nu^\flat \wedge (\nu \rfloor \nabla_\nu \omega)}_{0} = 0$, so $\nabla_\nu \omega = 0$, implying $\nabla \omega|_{\partial M} = 0$. Therefore, $\omega \in H_0^2(M)$, and we can replace $\omega$ by $\omega'$ in (49) to obtain $\Delta \omega = 0$. Given that $\omega|_{\partial M} = 0$, we have $\omega = 0$ by [1]. We conclude that $J_1 \circ I_1$ is injective and consequently $I_1$ is also. $\square$

To complete the proof of Theorem 3.2, we introduce the operator $K_1 : V_1 \to V_1$ defined as follows:

$$K_1 := D_{V_1}^{-1} \circ {}^t I_1 \circ D_{W_1} \circ I_1 \qquad (50)$$

where $D_{V_1} : V_1 \to V_1'$ and $D_{W_1} : W_1 \to W_1'$ are natural dual isomorphisms defined as follows: $D_{V_1}(\omega) := (., \omega)_V$ and $D_{W_1}(\omega) := (., \omega)_{W_1}$ for all $\omega$ in $V_1$ and $W_1$ respectively, and ${}^t I_1(\theta) = \theta \circ I_1$ for all $\theta$ in $W_1'$. Note that, $K_1$ is characterized for all $\omega$ and $\omega'$ in $V_1$ by:

$$(K_1 \omega, \omega')_V = ({}^t I_1 \circ D_{W_1} \circ I_1(\omega))(\omega') = (D_{W_1} \circ I_1(\omega))(I_1(\omega')) = (I_1(\omega), I_1(\omega'))_{W_1} = (\omega, \omega')_{W_1}, \quad (51)$$

since $V_1 \subset W_1$. We observe that the operator $K_1$ is self-adjoint and non-negative. Moreover, since $I_1$ is injective, $\ker I_1 = \{0\}$, thus $\ker K_1 = \{0\}$, so $K_1$ is positive definite. The compactness of $K_1$ follows from the fact that it is the composition of continuous mappings with $I_1$, which is compact. Finally, since $K_1$ is compact, self-adjoint, and positive definite on $V_1$, which is a Hilbert space, there exists a sequence of real, positive, and non-increasing eigenvalues $(\chi_{i,1})_{i \ge 1}$ that tend to zero, with finite-dimensional eigenspaces, as per [24, p.431]. The eigenspaces form a dense space in $V_1$, and the $(\omega_i)_i$ form a Hilbert basis. To conclude the proof of Theorem 3.2 let us show the following lemma.

**Lemma 3.14.** A form $\omega$ solves (10) for $\ell \in \mathbb{R}^*$ if and only if,

$$K_1 \omega = \frac{1}{\ell} \omega. \qquad (52)$$

*Proof.* Let us fix $\ell \in \mathbb{R}^*$ and a smooth solution $\omega$ of the BSN1 (10). Using (21) $\omega' \in \tilde{Z}_1$, we get

$$(\Delta \omega, \Delta \omega')_{L^2(M)} = \ell((\iota^* \omega, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \omega, \iota^* \delta \omega')_{L^2(\partial M)}).$$
By density of $\tilde{Z}_1$ in $W_1$, we get for all $\omega'$ in $V_1$, $(\omega, \omega')_V = \ell(\omega, \omega')_{W_1}$. Note that $\ell > 0$ by Lemma 3.5. Thus, by the definition of $K_1$, $(\omega, \omega')_V = \ell(K_1 \omega, \omega')_V$ for all $\omega' \in V_1$, so $K_1 \omega = \frac{1}{\ell} \omega$. Therefore, $\omega$ is an eigenform of $K_1$ associated with the eigenvalue $\chi = \frac{1}{\ell}$.

Conversely, let $\omega_i \in V_1$ be a $\chi_{i,1}$-eigenform for $K_1$. Since $\omega_i \in V_1$, we have by Lemma 3.12 that $\Delta^2 \omega_i = 0$ weakly on $M$ with $\nu \rfloor \omega_i = 0$ and $\nu \rfloor d\omega_i = 0$. Let us show that $\nu \rfloor \Delta d\omega_i = -\ell t^* \omega_i$ and $\nu \rfloor \Delta \omega_i = -\ell t^* \delta \omega_i$ on $\partial M$ with $\ell = \frac{1}{\chi_{i,1}}$. For any $\omega \in \tilde{Z}_1$, we have by definition:

$$ 
\begin{aligned}
\chi_{i,1}(\Delta \omega_i, \Delta \omega)_{L^2(M)} &= \chi_{i,1}(\omega_i, \omega)_V \\
&= (K \omega_i, \omega)_V \\
&= (I_1(\omega_i), I_1(\omega))_{W_1} \\
&= (\omega_i, \omega)_{W_1} \\
&= (t^* \omega_i, t^* \omega)_{L^2(\partial M)} + (t^* \delta \omega_i, t^* \delta \omega)_{L^2(\partial M)}.
\end{aligned} 
\quad (53) $$

**Lemma 3.15.** For all $\omega \in V_1$, $\omega' \in H_{N,1}^2(M)$, we have

$$ \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g + \int_{\partial M} \langle \nu \rfloor d\Delta \omega, t^* \omega' \rangle d\mu_g + \int_{\partial M} \langle \nu \rfloor \Delta \omega, t^* \delta \omega' \rangle d\mu_g = 0. $$

*Proof.* Let $\omega' \in H_{N,1}^2(M)$ and $(\omega_m)_m \in H_{N,1}^2(M)$ be a sequence that converges to $\omega$ in $V_1$. Using (21), we obtain:

$$ \int_M \langle \Delta \omega_m, \Delta \omega' \rangle d\mu_g + \int_{\partial M} \langle \nu \rfloor d\Delta \omega_m, t^* \omega' \rangle d\mu_g + \int_{\partial M} \langle \nu \rfloor \Delta \omega_m, t^* \delta \omega' \rangle d\mu_g = 0. \quad (54) $$

Now, $V_1 \subset H^2(M)$, and by the equivalence of the norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$, the sequence $(\omega_m)_m$ converges in $H^2(M)$ to $\omega$. Hence, $(\Delta \omega_m)_m$ converges in $L^2(M)$ to $\Delta \omega$, $(\nu \rfloor d\Delta \omega_m)_m$ converges in $H^{-\frac{3}{2}}(\partial M)$ to $\nu \rfloor d\Delta \omega$, and $(\nu \rfloor \Delta \omega_m)_m$ converges in $H^{-\frac{1}{2}}(\partial M)$ to $\nu \rfloor \Delta \omega$. Furthermore, the bilinear forms in (54) are well-defined and continuous because

$$ \langle \underbrace{\nu \rfloor d\Delta \omega_m}_{\in H^{-\frac{3}{2}}(\partial M)}, \underbrace{t^* \omega'}_{\in H^{\frac{3}{2}}(\partial M)} \rangle \quad \text{and} \quad \langle \underbrace{\nu \rfloor \Delta \omega_m}_{\in H^{-\frac{1}{2}}(\partial M)}, \underbrace{t^* \delta \omega'}_{\in H^{\frac{1}{2}}(\partial M)} \rangle, $$

see [2, Definition 1.7.3]. Thus, the result follows. $\square$

Thus, for all $\omega \in Z_1$, we have

$$ (\Delta \omega_i, \Delta \omega)_{L^2(M)} + (\nu \rfloor d\Delta \omega_i, t^* \omega)_{L^2(\partial M)} + (\nu \rfloor \Delta \omega_i, t^* d\omega)_{L^2(\partial M)} = 0. $$

Using (53), we obtain:

$$ \left( \frac{1}{\chi_i} t^* \omega_i + \nu \rfloor d \Delta \omega_i, t^* \omega \right)_{L^2(\partial M)} + \left( \frac{1}{\chi_i} t^* \delta \omega_i + \nu \rfloor \Delta \omega_i, t^* \delta \omega \right)_{L^2(\partial M)} = 0. \quad (55) $$

Consider the following map:

$$ O_1 : Z_1 \longrightarrow \Omega^p(\partial M) \oplus \Omega^{p-1}(\partial M) \\ \omega \longmapsto (t^* \omega, t^* \delta \omega). $$

It is continuous with respect to the norms $\|\cdot\|_V$ and $\|\cdot\|_{L^2(\partial M)}$ and is injective since it is the restriction of $J_1 \circ I_1$ on $Z_1$, which is continuous and injective by Lemma 3.13.

**Lemma 3.16.** The map $O_1$ is surjective.
Proof. This follows from Lemma 3.3. □

We know by (54) that for all $\omega \in Z_1$,

$$(\nu \rfloor \Delta \omega_i + \ell \ell^* \delta \omega_i, \ell^* \delta \omega)_{L^2(\partial M)} + (\nu \rfloor d \Delta \omega_i + \ell \ell^* \omega_i, \ell^* \omega)_{L^2(\partial M)} = 0.$$

Now using Lemma 3.16, we have that $O_1$ is surjective. Consequently, for all $\alpha, \beta \in \Omega^p(\partial M)$, we get

$$(\nu \rfloor \Delta \omega_i + \ell \ell^* \delta \omega_i, \alpha)_{L^2(\partial M)} + (\nu \rfloor d \Delta \omega_i + \ell \ell^* \omega_i, \beta)_{L^2(\partial M)} = 0.$$

By duality it follows that

$$(\nu \rfloor \Delta \omega_i + \ell \ell^* \delta \omega_i, \nu \rfloor d \Delta \omega_i + \ell \ell^* \omega_i) = (0, 0)$$

holds in $H^{-\frac{1}{2}}(\partial M) \times H^{-\frac{3}{2}}(\partial M)$. Now by Theorem 2.6, $\omega_i$ is hence smooth and is an eigenform of the problem (10) associated with the eigenvalue $\ell_i = \frac{1}{\chi_i}$. □

This completes the proof of Theorem 3.2. □

## 3.3 Variational characterization of eigenvalues

In this section, we will provide variational characterizations of the eigenvalues of the BSN1 problem.

**Theorem 3.17.** A variational characterization of the first strictly positive eigenvalue of BSN1 is given by

$$\ell_{1,p} = \min \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2} \mid \omega \in \tilde{H}_{N,1}^2(M), \ell^* \omega \neq 0 \text{ or } \ell^* \delta \omega \neq 0 \text{ on } \partial M \right\}. \quad (56)$$

This infimum is attained if and only if $\omega$ is an $\ell_{1,p}$-eigenform for BSN1.

*Proof.* By applying (46) to $\omega_{1,p}$, a $\ell_{1,p}$-eigenform, we obtain

$$\int_M |\Delta \omega_{1,p}|^2 d\mu_g = \ell_{1,p} \int_{\partial M} \left( \langle \ell^* \omega_{1,p}, \ell^* \omega_{1,p} \rangle + \langle \ell^* \delta \omega_{1,p}, \ell^* \delta \omega_{1,p} \rangle \right) d\mu_g.$$

Therefore,

$$\ell_{1,p} = \frac{\|\Delta \omega_{1,p}\|_{L^2(M)}^2}{\|\ell^* \omega_{1,p}\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega_{1,p}\|_{L^2(\partial M)}^2}.$$

Let $\omega \in V_1$. By the proof of Theorem 3.2, $\omega = \sum_i (\omega, \omega_i)_V \omega_i$, where $(\omega_i)_i$ is a Hilbertian orthonormal basis of $V_1$ consisted of eigenforms for the operator $K_1$, whose eigenvalues $(\chi_i)_{i \ge 1}$ satisfy $\chi_i = \frac{1}{\ell_{i,p}}$. We have hence

$$\|\Delta \omega\|_{L^2(M)}^2 = \|\omega\|_V^2 = \sum_i (\omega, \omega_i)_V^2.$$
Using (52), we compute:

$$
\begin{align*}
\|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2 &= \|\omega\|_{W_1}^2 \\
&= (K_1 \omega, \omega)_V \\
&= \left( K_1 \sum_i (\omega_i, \omega)_V \omega_i, \sum_j (\omega_j, \omega)_V \omega_j \right)_V \\
&= \sum_i \chi_i(\omega_i, \omega)_V^2 \\
&\le \chi_1 \sum_i (\omega_i, \omega)_V^2 \\
&= \frac{1}{\ell_{1,p}} \|\omega\|_V^2 \\
&= \frac{1}{\ell_{1,p}} \|\Delta \omega\|_{L^2(M)}^2.
\end{align*}
$$

Thus,

$$
\ell_{1,p} \le \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2},
$$

and the inequality is strict unless $\omega$ is a $\chi_1$-eigenform of $K_1$, i.e., $\omega$ is an $\ell_{1,p}$-eigenform of BSN1 (10). We have shown that,

$$
\ell_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2} \mid \omega \in V_1 \setminus \{0\} \right\}.
$$

Let us now show that the same characterization holds for all $\omega \in \tilde{H}_{N,1}^2(M)$ that are non-zero on the boundary. For this, we require the following lemma:

**Lemma 3.18.** On $\tilde{H}_{N,1}^2(M)$, we have the following orthogonal decomposition with respect to $(\cdot, \cdot)_V$:

$$
\tilde{H}_{N,1}^2(M) = V_1 \oplus H_0^2(M).
$$

*Proof.* Let $\omega' \in \tilde{H}_{N,1}^2(M)$ and $\omega \in \check{Z}_1$. From (21), we have:

$$
\int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g = - \int_{\partial M} \left( \langle \nu \rfloor d\Delta \omega, \iota^* \omega' \rangle + \langle \nu \rfloor \Delta \omega, \iota^* \delta \omega' \rangle \right) d\mu_g.
$$

Hence, for $\omega' \in H_0^2(M)$, we have $(\omega, \omega')_V = 0$, which implies $H_0^2(M) \subset \check{Z}_1^\perp$. By density of $\check{Z}_1$ in $V_1$, we have $\check{Z}_1^\perp = V_1^\perp$.

Conversely, let $\omega' \in V_1^\perp \subset \tilde{H}_{N,1}^2(M)$, then $(\omega_i, \omega')_V = 0$ for all $i$. Then, by equation (46), we have:

$$
(\nu \rfloor d \Delta \omega_i, \iota^* \omega')_{L^2(\partial M)} + (\nu \rfloor \Delta \omega_i, \iota^* \delta \omega')_{L^2(\partial M)} = 0
$$

for all $i$. Hence,

$$
(\iota^* \omega_i, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \omega_i, \iota^* \delta \omega')_{L^2(\partial M)} = 0. \quad (57)
$$

The goal now is to show that $\omega' \in H_0^2(M)$. Since $\omega' \in \tilde{H}_{N,1}^2(M)$, we will see that it is enough to show that $\iota^* \omega' = 0$ and $\iota^* \delta \omega' = 0$.

Recall that the map $O_1$ is surjective according to Lemma 3.16, bounded with respect to $\|\cdot\|_V$ and $\|\cdot\|_{L^2(\partial M)}$, and that the $\omega_i$'s generate a dense space in $V_1$. Consequently, for any $\epsilon > 0$, there exists $\omega_1 \in \Omega^p(\partial M)$ and $\omega_2 \in \Omega^{p-1}(\partial M)$ such that $\|\iota^* \omega' - \omega_1\|_{L^2(\partial M)} < \epsilon$ and $\|\iota^* \delta \omega' - \omega_2\|_{L^2(\partial M)} < \epsilon$.
Thus, there exists a unique $\omega_{\epsilon} \in \Omega^p(M)$ such that $\Delta^2 \omega_{\epsilon} = 0$, $\nu \rfloor \omega_{\epsilon} = 0$, $\nu \rfloor d \omega_{\epsilon} = 0$, $\iota^* \omega_{\epsilon} = \omega_1$, and $\iota^* \delta \omega_{\epsilon} = \omega_2$. Therefore, for every $\epsilon > 0$, there exists $\omega_{\epsilon} \in \Omega^p(M)$ such that $\Delta^2 \omega_{\epsilon} = 0$, $\nu \rfloor \omega_{\epsilon} = 0$, $\nu \rfloor d \omega_{\epsilon} = 0$, $\|\iota^* \omega_{\epsilon} - \iota^* \omega'\|_{L^2(\partial M)} < \epsilon$, and $\|\iota^* \delta \omega_{\epsilon} - \iota^* \delta \omega'\|_{L^2(\partial M)} < \epsilon$. We can then decompose $\omega_{\epsilon} \in Z_1$ as follows: $\omega_{\epsilon} = \tilde{\omega}_{\epsilon} + \alpha_{\epsilon}$, where $\tilde{\omega}_{\epsilon} \in \check{Z}_1$ and $\alpha_{\epsilon} \in H_A^p(M)$, the decomposition is w.r.t $(\cdot, \cdot)_{W_1}$. Since we have (57) and the map $O_1$ is continuous, we obtain the following:

$$ \sum_i (\tilde{\omega}_\epsilon, \omega_i)_V [ (\iota^* \omega_i, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \omega_i, \iota^* \delta \omega')_{L^2(\partial M)} ] = \left( \sum_i (\tilde{\omega}_\epsilon, \omega_i)_V \iota^* \omega_i, \iota^* \omega' \right)_{L^2(\partial M)} + \left( \sum_i (\tilde{\omega}_\epsilon, \omega_i)_V \iota^* \delta \omega_i, \iota^* \delta \omega' \right)_{L^2(\partial M)} = 0. $$

Again by the continuity of the map $O_1$, we have $(\iota^* \tilde{\omega}_\epsilon, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \tilde{\omega}_\epsilon, \iota^* \delta \omega')_{L^2(\partial M)} = 0$. Given that for $\alpha_\epsilon \in H_A^p(M)$ and $\omega' \in \check{H}_{N,1}^2(M)$, we have:

$$ (\iota^* \alpha_\epsilon, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \alpha_\epsilon, \iota^* \delta \omega')_{L^2(\partial M)} = 0, $$

therefore:

$$ (\iota^* \omega_\epsilon, \iota^* \omega')_{L^2(\partial M)} + (\iota^* \delta \omega_\epsilon, \iota^* \delta \omega')_{L^2(\partial M)} = 0. $$

By taking $\epsilon$ to 0, we obtain $\|\iota^* \omega'\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega'\|_{L^2(\partial M)}^2 = 0$, which implies that $\iota^* \omega' = 0$ and $\iota^* \delta \omega' = 0$. We then have: $\omega'_{|\partial M} = 0$, $d\omega'_{|\partial M} = \nu^b \wedge \nabla_\nu \omega' = 0$ and $\delta \omega'_{|\partial M} = -\nu \rfloor \nabla_\nu \omega' = 0$. Thus, by the proof of Lemma 3.13, we have $\nabla_\nu \omega' = 0$ and $\nabla_X \omega' = 0$ for all $X \in T\partial M$. Therefore, $\nabla_X \omega' = 0$ for all $X \in TM_{|\partial M}$, so $(\nabla \omega)_{|\partial M} = 0$, and hence $\omega' \in H_0^2(M) = V^\perp$. $\square$

We can then perform the following decomposition for $\omega \in \Omega^p(M)$ such that $\nu \rfloor d\omega = 0$ and $\nu \rfloor \omega = 0$, $\omega = \omega_{V_1} + \hat{\omega}$, with $\omega_{V_1} \in V_1$ and $\hat{\omega} \in H_0^2(M)$. We write

$$ \|\Delta \omega\|_{L^2(M)}^2 = \|\omega\|_V^2 = \|\omega_{V_1}\|_V^2 + \|\hat{\omega}\|_V^2 \quad (58) $$

and

$$ \|\iota^* \delta \omega\|_{L^2(\partial M)}^2 + \|\iota^* \omega\|_{L^2(\partial M)}^2 = \|\iota^* \delta \omega_{V_1}\|_{L^2(\partial M)}^2 + \|\iota^* \omega_{V_1}\|_{L^2(\partial M)}^2 \quad (59) $$

since $\iota^* \delta \hat{\omega} = 0$ and $\iota^* \hat{\omega} = 0$. Thus, using (58) and (59), we get

$$ \begin{aligned} \|\Delta \omega\|_{L^2(M)}^2 &= \|\Delta \omega_{V_1}\|_{L^2(M)}^2 + \|\Delta \hat{\omega}\|_{L^2(M)}^2 \\ &\geq \|\Delta \omega_{V_1}\|_{L^2(M)}^2 \\ &\geq \ell_{1,p} (\|\iota^* \delta \omega_{V_1}\|_{L^2(\partial M)}^2 + \|\iota^* \omega_{V_1}\|_{L^2(\partial M)}^2) \\ &= \ell_{1,p} (\|\iota^* \delta \omega\|_{L^2(\partial M)}^2 + \|\iota^* \omega\|_{L^2(\partial M)}^2). \end{aligned} \quad (60) $$

We conclude finally that

$$ \ell_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2} \mid \omega \in \check{H}_{N,1}^2(M), \iota^* \omega \neq 0 \text{ or } \iota^* \delta \omega \neq 0 \text{ on } \partial M \right\} $$

and the infimum is only reached for $\ell_{1,p}$-eigenforms of BSN1. $\square$

We provide below an alternative variational characterization of the first eigenvalue of the BSN1.

**Theorem 3.19.** Let us define

$$ \ell'_{1,p} := \inf \left\{ \frac{\|\nu \rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu \rfloor d \omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} \mid \omega \in H^2(M), \omega \perp_{L^2(M)} H_A^p(M) \text{ and } \Delta \omega = 0 \text{ on } M \right\}. \quad (61) $$

Then we have $\ell_{1,p} = \ell'_{1,p}$.
**Proposition 3.20.** The infimum $l'_{1,p}$ defined in Theorem 3.19, is a positive minimum.

*Proof.* We prove first the following lemma.

**Lemma 3.21.** The boundary problem

$$ (\Delta_{Neu}) \left\{ \begin{array}{ll} \Delta \omega = \omega_0 & \text{on } M \\ \nu \rfloor \omega = \omega_1 & \text{on } \partial M \\ \nu \lfloor d\omega = \omega_2 & \text{on } \partial M. \end{array} \right. \qquad (62) $$

has a unique solution $\omega$ if and only if $\omega_0, \omega_1$ and $\omega_2$ lie in $\text{Im}(\mathcal{P}_{\Delta_{Neu}})$. In this case the set of solutions is $\{\omega + H_A^p(M)\}$. This is equivalent to say that the map

$$ \mathcal{P}_{\Delta_{Neu}} : H^s(M)/\mathcal{N} \longrightarrow H^{s-2}(M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M) \\ [\omega] \longmapsto (\Delta \omega, \nu \rfloor \omega, \nu \lfloor d\omega) $$

is a topological isomorphism onto its image which is

$$ \text{Im}(\mathcal{P}_{\Delta_{Neu}}) = \{ (\omega_0, \omega_1, \omega_2) \in H^{s-2}(M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M) \mid \forall \omega' \in \mathcal{N}^*, (\omega_0, \omega') + (\omega_1, \underbrace{\omega'}_0) + (\omega_2, \iota^* \omega') \} $$

here $\mathcal{N}$ and $\mathcal{N}^*$ are equal to $H_A^p(M)$.

*Proof. Step 1:* We start by showing the ellipticity of the problem (62) in the sense of Shapiro-Lopatinskij.

In the same way as in Section 3.1, let us show that the map

$$ \Phi : \mathcal{M}_{\Delta,v}^{+} \longrightarrow E_1 \oplus E_2 \\ y \longmapsto (\sigma_{B_1}(-iv + \partial_t \nu^b)(y), \sigma_{B_2}(-iv + \partial_t \nu^b)(y)) \quad (0) $$

is an isomorphism where in this case

$$ \mathcal{M}_{\Delta,v}^{+} = \text{Span} (ae^{-|v|t} \cdot \omega_0 \mid a \in \mathbb{R}, \omega_0 \in \Lambda^p T_x^* M|_{\partial M}) $$

and has dimension $\binom{n}{p}$. Let $B_1$ and $B_2$ be the following maps:

$$ B_1 : \Omega^p(M) \longrightarrow \Gamma(E_1 = \Lambda^{p-1}T^*\partial M) \quad ; \quad B_2 : \Omega^p(M) \longrightarrow \Gamma(E_2 = \Lambda^p T^*\partial M) \\ \hat{\omega} \longmapsto \nu \rfloor \hat{\omega} \qquad \hat{\omega} \longmapsto \nu \lfloor d\hat{\omega}. $$

We calculate the principal symbols of $B_1$ and $B_2$ for $\xi = (-iv+\partial_t \nu^b)$ and applied to $y$. Let $y \in \mathcal{M}_{\Delta,v}^+$, we obtain $\sigma_{B_1}(-iv+\partial_t \nu^b)(y)(0) = a\nu \rfloor \omega_0$ and $\sigma_{B_2}(-iv+\partial_t \nu^b)(y)(0) = -a|v|\iota^* \omega_0 + i\alpha \iota^* v \wedge (\nu \rfloor \omega_0)$. Thus, the map $\Phi$ is

$$ \Phi : \mathcal{M}_{\Delta,v}^{+} \longrightarrow E_1 \oplus E_2 \\ y \longmapsto (a\nu \rfloor \omega_0, -a|v|\iota^* \omega_0 + i\alpha \iota^* v \wedge (\nu \rfloor \omega_0)). $$

Suppose that $\Phi(y) = 0$, then either $a = 0$, or

$$ \left\{ \begin{array}{ll} a\nu \rfloor \omega_0 & = 0 \\ -a|v|\iota^* \omega_0 + i\alpha \iota^* v \wedge (\nu \rfloor \omega_0) & = 0 \end{array} \right. \implies \left\{ \begin{array}{ll} \nu \rfloor \omega_0 & = 0 \\ \iota^* \omega_0 & = 0 \end{array} \right. $$

Therefore, in both cases $\omega_0 = 0$, so $y = 0$, and thus $\Phi$ is injective.
**Step 2:** By Theorem 2.6 we have hence that $\mathcal{P}_{\Delta_{Neu}}$ is a topological isomorphism on its image so it is invertible and its inverse is continuous. In this case $\mathcal{N} = \mathcal{N}^* = H_A^p(M)$ (see Section 2.2.4 for their definitions). Indeed, it is clear by the reformulation of the formula (19) as it is presented in (26), for $\omega$ and $\omega'$ solutions of (62) we have:

$$ 
\begin{aligned}
& \int_M (\langle \Delta \omega, \omega' \rangle - \langle \omega, \Delta \omega' \rangle) d\mu_g \\
& = \int_{\partial M} \left( \langle \nu_{\perp} \omega, \iota^* \delta \omega' \rangle + \langle \nu_{\perp} d\omega, \iota^* \omega' \rangle - \langle \iota^* \omega, \nu_{\perp} d\omega' \rangle - \langle \iota^* \delta \omega, \nu_{\perp} \omega' \rangle \right) d\mu_g.
\end{aligned}
$$

Then $(\omega_0, \omega_1, \omega_2) \in \mathrm{Im}(\mathcal{P}_{\Delta_{Neu}})$ if and only if for every $\alpha \in H_A^p(M)$, we have

$$ (\omega_0, \alpha) - (\omega_1, \underbrace{\iota^* \delta \alpha}_{0}) - (\omega_2, \iota^* \alpha) = 0 $$

which implies

$$ (\Delta \omega, \alpha) = (\omega_2, \iota^* \alpha). $$

Thus, $(0, \omega_1, \omega_2) \in \mathrm{Im}(\mathcal{P}_{\Delta_{Neu}})$ if and only if for every $\alpha \in H_A^p(M)$, $\int_{\partial M} \langle \omega_2, \iota^* \alpha \rangle d\mu_g = 0$. Therefore, we must choose $\omega_2 \in H_A^{p \perp L^2(\partial M)}(M) := \{\theta \in H^{-\frac{3}{2}}(\partial M) | \forall \alpha \in H_A^p(M), (\theta, \iota^* \alpha) = 0\}$ to get the result. Consequently, we have the inclusion

$$ \{0\} \times H^{-\frac{1}{2}}(\partial M) \times H_A^{p \perp L^2(\partial M)}(M) \subset \mathrm{Im}(\mathcal{P}_{\Delta_{Neu}}). $$

□

**Lemma 3.22.** For all $(\omega_1, \omega_2) \in L^2(\partial M) \times L^2(\partial M)$ there exists a unique $\omega \in H^2(M)$ such that in the weak sense we have

$$ 
\left\{
\begin{array}{ll}
\Delta \omega = 0 & \text{on } M \\
\nu_{\perp} \omega = \omega_1 & \text{on } \partial M \\
\nu_{\perp} d\omega = \omega_2 & \text{on } \partial M \\
\omega \perp_{L^2(M)} H_A^p(M). &
\end{array}
\right.
\qquad (63) $$

Moreover,

$$ 
\begin{array}{rcl}
E : L^2(\partial M) \oplus L^2(\partial M) & \to & \check{L}^2(M) \\
(\omega_1, \omega_2) & \mapsto & \omega
\end{array}
$$

is compact.

*Proof.* We rely on an argument of [8, p.318]. The map $E$ is well defined by Lemma 3.21. The goal for now is to show that it is continuous. To prove this, we express the composition of the map $E$ as follows:

$$ 
\begin{array}{rcl}
L^2(\partial M) \oplus L^2(\partial M) & \xrightarrow{Q} & H^{-\frac{1}{2}}(\partial M) \oplus H^{-\frac{3}{2}}(\partial M) \\
(\omega_1, \omega_2) & \mapsto & (\omega_1, \omega_2) \\
& & \xrightarrow{(\mathcal{P}_{\Delta_{Neu}})^{-1}} D_{\Delta}^0(M)/\mathcal{N} \\
& & \xrightarrow{[\omega]} \check{L}^2(M) \\
& & \xmapsto{\pi} \check{L}^2(M) \\
& & \xmapsto{\omega} \omega
\end{array}
$$

Here $Q$ is the inclusion map and $\mathcal{P}_{\Delta_{Neu}}$ is given by Lemma 3.21. By the Rellich-Kondrachov theorem, the map $H^{\frac{1}{2}}(\partial M) \to L^2(\partial M)$ is compact. Thus, by duality, we obtain that $L^2(\partial M) \to H^{-\frac{1}{2}}(\partial M)$ is a compact inclusion. In addition, since $H^{\frac{3}{2}}(M) \subset H^{\frac{1}{2}}(M)$ is a continuous inclusion, we have that $L^2(\partial M) \to H^{-\frac{3}{2}}(\partial M)$ is compact. Therefore, the application $Q$ is compact. Since by Lemma 3.21, $\mathcal{N} = H_A^p(M)$, the map $\pi$ is continuous which shows the continuity and compactness of $E$. □
Let us come back to the proof of Proposition 3.20. Let $(\omega_1, \omega_2) \in L^2(\partial M) \times L^2(\partial M)$ and let $\omega_0 = E(\omega_1, \omega_2)$ be the harmonic extension given by Lemma 3.22. Since $E$ is continuous, there exists a constant $c > 0$ such that

$$ \begin{aligned} \|\omega_0\|_{L^2(M)}^2 &= \|E(\omega)\|_{L^2(M)}^2 \le c (\|\omega_1\|_{L^2(\partial M)}^2 + \|\omega_2\|_{L^2(\partial M)}^2) \\ &= c (\|\nu_\perp \omega_0\|_{L^2(\partial M)}^2 + \|\nu_\perp d\hat{\omega}\|_{L^2(\partial M)}^2). \end{aligned} $$

Therefore, $\ell'_{1,p} \ge \frac{1}{c}$ is positive.

Now let $\mathbb{S}$ be the unit sphere,

$$ \mathbb{S} = \{(\omega_1, \omega_2) \in L^2(\partial M) \oplus H_A^p(M)^{\perp L^2(\partial M)} \mid \|\omega_1\|_{L^2(\partial M)}^2 + \|\omega_2\|_{L^2(\partial M)}^2 = 1\}. $$

Since $E$ is compact, $E(\mathbb{S})$ is relatively compact in $\check{L}^2(M)$, and thus there exists $\tilde{\omega} \in \overline{E(\mathbb{S})} \subset \check{L}^2(M)$ such that $\|\tilde{\omega}\|_{L^2(M)} = \sup\{\|E\omega\|_{L^2(M)} \mid \omega \in \mathbb{S}\}$. Therefore, there exists a subsequence $(\omega_m)_m \in \mathbb{S}$ such that $E\omega_m \xrightarrow[m \to \infty]{L^2(M)} \tilde{\omega}$. Then, $\Delta(E\omega_m) \xrightarrow[m \to \infty]{H^{-2}(M)} \Delta\tilde{\omega}$. Moreover, since $E(\omega_m) \in H^2(M)$ and $\Delta(E\omega_m) = 0 \in L^2(M)$ we get $\Delta(E\omega_m) \xrightarrow[m \to \infty]{L^2(M)} 0 = \Delta\tilde{\omega}$. Using the inequality

$$ \|\omega\|_{H^2(M)} \le c (\|\Delta\omega\|_{L^2(M)} + \|\omega\|_{L^2(M)}) $$

given in [2, Theorem 1.6.12], we obtain for some positive constant $c$:

$$ \|E \omega_m - \tilde{\omega}\|_{H^2(M)} \le c (\|\Delta(E \omega_m - \tilde{\omega})\|_{L^2(M)} + \|E \omega_m - \tilde{\omega}\|_{L^2(M)}), $$

so in the limit, $E\omega_m \xrightarrow[m \to \infty]{H^2(M)} \tilde{\omega}$. This implies that $\nu_\perp E(\omega_m) \xrightarrow[m \to \infty]{H^3(\partial M)} \nu_\perp \tilde{\omega}$, and $\nu_\perp dE(\omega_m) \xrightarrow[m \to \infty]{H^1(\partial M)} \nu_\perp d\tilde{\omega}$. Since $\|\nu_\perp E(\omega_m)\|_{L^2(\partial M)}^2 + \|\nu_\perp dE(\omega_m)\|_{L^2(\partial M)}^2 = 1$, we conclude that in the limit, $\|\nu_\perp \tilde{\omega}\|_{L^2(\partial M)}^2 + \|\nu_\perp d\tilde{\omega}\|_{L^2(\partial M)}^2 = 1$. Thus, $\tilde{\omega} \in E(\mathbb{S})$, and since $\|\tilde{\omega}\|_{L^2(M)} = \max\{\|E\omega\|_{L^2(M)} \mid \omega \in \mathbb{S}\}$, we have $\|\tilde{\omega}\|_{L^2(M)} = \frac{1}{\ell'_{1,p}}$. Therefore, $\ell'_{1,p}$ is a minimum, which completes the proof. $\square$

**Proof of Theorem 3.19.** Let $\omega$ be a $\ell_{1,p}$-eigenform of BSN1. Without loss of generality, we can assume that

$$ \|\ell^* \omega\|_{L^2(\partial M)} + \|\ell^* \delta \omega\|_{L^2(\partial M)} = 1. $$

Consider the following problem

$$ \left\{ \begin{array}{ll} \Delta \hat{\omega} & = 0 & \text{on } M \\ \nu_\perp \hat{\omega} & = \ell_{1,p} \ell^* \delta \omega & \text{on } \partial M \\ \nu_\perp d \hat{\omega} & = \ell_{1,p} \ell^* \omega & \text{on } \partial M, \end{array} \right. \qquad (64) $$

which is a particular case of (62). Since $\ell^* \omega \perp_{L^2(\partial M)} H_A^p(M)$ by Lemma 3.22 and there exists a unique solution $\hat{\omega} \perp_{L^2(M)} H_A^p(M)$. Given that $\mathcal{N}^* = H_A^p(M)$, we have by Theorem 2.6 the map

$$ \mathcal{P}_{\Delta_{Neu}} : H^2(M)/H_A^p(M) \to L^2(M) \times H^3(\partial M) \times H^1(\partial M) \\ [\omega] \mapsto (\Delta \omega, \nu_\perp \omega, \nu_\perp d \omega) $$

is an isomorphism on

$$ \begin{aligned} \mathrm{Im}(\mathcal{P}_{\Delta_{Neu}}) = & \{(\omega_0, \omega_1, \omega_2) \mid \forall \omega' \in H_A^p(M), (\omega_0, \omega') + (\omega_1, \ell^* \delta \omega') + (\omega_2, \ell^* \omega') = 0\} \\ & \supset \{(\omega_0, \omega_1, \omega_2) \mid \forall \omega' \in H_A^p(M), (0, \omega') + \ell_{1,p}(\underbrace{\ell^* \delta \omega, \ell^* \delta \omega'}_{0} + \ell_{1,p}(\underbrace{\ell^* \omega, \ell^* \omega'}_{0})) = 0\}, \end{aligned} $$
where $(\cdot, \cdot)$ is the duality product. Therefore $(0, \ell_{1,p} \ell^* \delta \omega, \ell_{1,p} \ell^* \omega)$ belongs to $\mathrm{Im}(\mathcal{P}_{\Delta_{Neu}})$. Let $\hat{\omega} \in \Omega^p(M)$ be a solution of (64) and $\hat{\omega}_1 := \mathrm{proj}_{\perp L^2(M)}(\hat{\omega}) \in H_A^p(M)$. Then $\hat{\omega}_2 := \hat{\omega} - \hat{\omega}_1$ which will be a solution of (64) such that $\hat{\omega}_2$ is $L^2(M)$-orthogonal to $H_A^p(M)$.

Next, using (20) for $\omega \in Z_1$, we obtain

$$ \int_M \langle \hat{\omega}_2, \Delta \omega \rangle d\mu_g + \ell_{1,p} \int_{\partial M} \left( \langle \ell^* \omega, \ell^* \omega \rangle + \langle \ell^* \delta \omega, \ell^* \delta \omega \rangle \right) d\mu_g = 0, $$

hence

$$ \int_M \langle \hat{\omega}_2, \Delta \omega \rangle d\mu_g + \ell_{1,p} \underbrace{\int_{\partial M} (|\ell^* \omega|^2 + |\ell^* \delta \omega|^2) d\mu_g}_{=1} = \int_M \langle \hat{\omega}_2, \Delta \omega \rangle d\mu_g + \ell_{1,p} = 0. $$

Using the Cauchy-Schwarz inequality, we get

$$ \ell_{1,p} \le \|\Delta \omega\|_{L^2(M)} \|\hat{\omega}_2\|_{L^2(M)}. \qquad (65) $$

Since $\omega$ is an eigenform of $K_1$ with eigenvalue $\frac{1}{\ell_{1,p}}$, we have

$$ \|\omega\|_{W_1}^2 = (K_1 \omega, \omega)_V = \frac{1}{\ell_{1,p}} \|\omega\|_V^2 = \frac{1}{\ell_{1,p}} \|\Delta \omega\|_{L^2(M)}^2 $$

which implies $\|\Delta \omega\|_{L^2(M)}^2 = \ell_{1,p}$ hence $\|\Delta \omega\|_{L^2(M)} = \ell_{1,p}^{\frac{1}{2}}$, consequently by (65) we get $\ell_{1,p} \le \|\hat{\omega}_2\|_{L^2(M)}^2$. Thus

$$ \frac{\|\nu \rfloor \hat{\omega}_2\|_{L^2(\partial M)}^2 + \|\nu \rfloor d\hat{\omega}_2\|_{L^2(\partial M)}^2}{\|\hat{\omega}_2\|_{L^2(M)}^2} = \frac{\ell_{1,p}^2 \|\ell^* \omega\|_{L^2(\partial M)}^2 + \ell_{1,p}^2 \|\ell^* \delta \omega\|_{L^2(\partial M)}^2}{\|\hat{\omega}_2\|_{L^2(M)}^2} \\ \le \frac{\ell_{1,p}^2 (\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2)}{\ell_{1,p}} = \ell_{1,p}. $$

Finally

$$ \inf_{\substack{\omega \perp L^2(M) \\ \Delta \omega = 0}} \frac{\|\nu \rfloor \hat{\omega}_2\|_{L^2(\partial M)}^2 + \|\nu \rfloor d\hat{\omega}_2\|_{L^2(\partial M)}^2}{\|\hat{\omega}_2\|_{L^2(M)}^2} \le \ell_{1,p}. $$

Conversely, let $\omega \in \Omega^p(M) \setminus \{0\}$ be a minimizer of $\ell_{1,p}'$ as given in Proposition 3.20, i.e., $\omega$ is $L^2(M)$-orthogonal to $H_A^p(M)$, $\Delta \omega = 0$ on $M$, and

$$ \|\nu \rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu \rfloor d \omega\|_{L^2(\partial M)}^2 = \ell_{1,p}' \|\omega\|_{L^2(M)}^2. $$

Again, using Lemma 3.22, for $\omega_0 = \omega$, $\omega_1 = 0$ and $\omega_2 = 0$, the following problem which is a particular case of (62) has a solution since $\omega \perp_{L^2(M)} H_A^p(M)$:

$$ \left\{ \begin{array}{ll} \Delta \hat{\omega} & = \omega \quad \text{on } M \\ \nu \rfloor \hat{\omega} & = 0 \quad \text{on } \partial M \\ \nu \rfloor d \hat{\omega} & = 0 \quad \text{on } \partial M. \end{array} \right. \qquad (66) $$

Indeed, using integration by parts (19) we get

$$ \int_M \langle \omega, \alpha \rangle d\mu_g = \int_M \langle \Delta \hat{\omega}, \alpha \rangle d\mu_g \\ = \int_M \langle \hat{\omega}, \underbrace{\Delta \alpha}_{0} \rangle d\mu_g + \int_{\partial M} \left( \langle \underbrace{\nu \rfloor d\hat{\omega}}_{0}, \ell^* \alpha \rangle - \langle \ell^* \hat{\omega}, \underbrace{\nu \rfloor d\alpha}_{0} \rangle \right. \\ \left. + \langle \underbrace{\nu \rfloor \hat{\omega}}_{0}, \ell^* \delta \alpha \rangle - \langle \ell^* \delta \hat{\omega}, \underbrace{\nu \rfloor \alpha}_{0} \rangle \right) d\mu_g, $$
then $\int_M \langle \omega, \alpha \rangle d\mu_g = 0$ hence, $\omega \perp L^2(M) H_A^p(M)$. Therefore, there exists a such solution $\hat{\omega}$.

We have the following decomposition: $\hat{\omega} = \hat{\omega}_1 + \hat{\omega}_2$, where $\hat{\omega}_1 \in H_A^p(M)$ and $\hat{\omega}_2 \in H_A^p(M)^{\perp L^2(\partial M)}$.
Using the integration by parts (19), we get

$$ \begin{aligned} \|\omega\|_{L^2(M)}^2 &= (\Delta\hat{\omega}_2, \omega)_{L^2(M)} \\ &= -(\iota^*\hat{\omega}_2, \nu_\rfloor d\omega)_{L^2(\partial M)} - (\iota^*\delta\hat{\omega}_2, \nu_\rfloor \omega)_{L^2(\partial M)}. \end{aligned} \quad (67) $$

Note that

$$ \ell_{1,p} \left( \|\iota^* \hat{\omega}_2\|_{L^2(\partial M)}^2 + \|\iota^* \delta \hat{\omega}_2\|_{L^2(\partial M)}^2 \right) \le \|\omega\|_{L^2(M)}^2. \quad (68) $$

Using Cauchy-Schwarz inequality in (67), we obtain

$$ \begin{aligned} \|\omega\|_{L^2(M)}^2 &\le \left( \|\iota^* \hat{\omega}_2\|_{L^2(\partial M)}^2 + \|\iota^* \delta \hat{\omega}_2\|_{L^2(\partial M)}^2 \right)^{\frac{1}{2}} \left( \|\nu_\rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu_\rfloor d \omega\|_{L^2(\partial M)}^2 \right)^{\frac{1}{2}} \\ &\le \frac{\|\omega\|_{L^2(M)} \left( \|\nu_\rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu_\rfloor d \omega\|_{L^2(\partial M)}^2 \right)^{\frac{1}{2}}}{\ell_{1,p}^{\frac{1}{2}}}. \end{aligned} $$

Thus,

$$ \ell_{1,p}^{\frac{1}{2}} \le \frac{\left( \|\nu_\rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu_\rfloor d \omega\|_{L^2(\partial M)}^2 \right)^{\frac{1}{2}}}{\|\omega\|_{L^2(M)}}. $$

Hence, the desired characterization follows.

<figure><img src="image_4.png" /></figure>

**Corollary 3.23.** Let $\omega$ be any harmonic $p$-form which is $L^2(M)$-orthogonal to $H_A^p(M)$. Then the equality

$$ \ell_{1,p} = \frac{\|\nu_\rfloor \omega\|_{L^2(\partial M)}^2 + \|\nu_\rfloor d \omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} $$

holds if and only if there exists a $\ell_{1,p}$-eigenform $\hat{\omega}$ for BSN1 such that $\Delta\hat{\omega} = \omega$.

*Proof.* Inequality (68) is strict when $\hat{\omega}_2$ is an eigenform for $K_1$ with eigenvalue $\frac{1}{\ell_{1,p}}$, i.e., a solution of BSN1 associated with $\ell_{1,p}$. $\square$

<figure><img src="image_5.png" /></figure>

**Theorem 3.24.** For every $k \in \mathbb{N}$, we have the following characterization for the $k^{th}$ positive eigenvalue of BSN1:

$$ \ell_{k,p} = \inf_{\substack{V_k \subset V_1 \\ \dim V_k = k}} \sup_{0 \ne \omega \in V_k} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2 + \|\iota^* \delta \omega\|_{L^2(\partial M)}^2}. $$

We will rely on the following immediate Lemma:

**Lemma 3.25.** Let $H$ be a Hilbert space, and let $V$ and $E_{k-1}$ be subspaces of $H$ of dimension $k$ and $k-1$ respectively, where $k \ge 1$. Then $V \cap E_{k-1}^\perp \ne \{0\}$.

*Proof of Theorem 3.24.* Let $(\omega_i)_{i \ge 1}$ be a Hilbert basis of $(V_1, (\cdot, \cdot)_V)$, with $\omega_i$ being $\ell_{i,p}$-eigenforms. According to Lemma 3.25, for $V_k \subset V_1$ such that $\dim V_k = k$, there exists a non-zero $\omega \in V_k$ such that $(\omega, \omega_i)_V = 0$ for all $i = 1, \dots, k-1$. Thus, $\omega = \sum_{i \ge k} (\omega, \omega_i)_V \omega_i$ and we have:

$$ \|\omega\|_V^2 = \sum_{i \ge k} (\omega, \omega_i)_V^2, $$
and

$$ \|\omega\|_{W_1}^2 = (K \omega, \omega)_V = \sum_{i \ge k} \frac{1}{\ell_{i,p}} (\omega, \omega_i)_V^2 \le \frac{1}{\ell_{k,p}} \sum_{i \ge 1} (\omega, \omega_i)_V^2. $$

Thus, we conclude that $\ell_{k,p} \le \frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2}$. Therefore, $\ell_{k,p} \le \sup_{0 \ne \omega \in \mathcal{V}_k} \frac{\|\omega\|_{W_1}^2}{\|\omega\|_{W_1}^2}$. This implies that

$$ \ell_{k,p} \le \inf_{\substack{\mathcal{V}_k \subset V_1 \\ \dim \mathcal{V}_k = k}} \sup_{0 \ne \omega \in \mathcal{V}_k} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2}. $$

To show the second inequality, we define $\bar{V}_k = \text{Span}(\omega_1, \dots, \omega_k) \subset V_1$. For $\omega \in \bar{V}_k$, we have

$$ \|\omega\|_V^2 = \sum_{i=1}^k (\omega, \omega_i)_V^2, $$

and

$$ \|\omega\|_{W_1}^2 = (K \omega, \omega)_V = \sum_{i=1}^k \frac{1}{\ell_{i,p}} (\omega, \omega_i)_V^2 \ge \frac{1}{\ell_{k,p}} \sum_{i=1}^k (\omega, \omega_i)_V^2. $$

Thus, $\frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2} \le \ell_{k,p}$, therefore, $\sup_{0 \ne \omega \in \mathcal{V}_k} \frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2} \le \ell_{k,p}$. This concludes the proof. $\square$

**Theorem 3.26.** For all $k \in \mathbb{N}$, we have the following characterization of the $k^{th}$ positive eigenvalue of BSN1:

$$ \ell_{k,p} = \inf_{\substack{\mathcal{V}_k \subset \tilde{H}_{N,1}^2(M) \\ \dim(\mathcal{V}_k/H_0^2(M) \cap \mathcal{V}_k) = k}} \sup_{\omega \in \mathcal{V}_k \setminus H_0^2(M)} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2}. $$

*Proof.* Recall that by Lemma 3.18 we have the following orthogonal decomposition $\tilde{H}_{N,1}^2(M) = V \oplus H_0^2(M)$ with respect to the inner product $(\cdot, \cdot)_V$. Let $\mathcal{V}_k \subset \tilde{H}_{N,1}^2(M)$ such that $\dim(\mathcal{V}_k/H_0^2(M) \cap \mathcal{V}_k) = k$, and let $p_{V_1} : \tilde{H}_{N,1}^2(M) \to V_1$ be the orthogonal projection with respect to the inner product $(\cdot, \cdot)_V$. We define $\mathcal{U} := p_{V_1}(\mathcal{V}_k)$ hence $\mathcal{U} \subset V_1$ and $\dim \mathcal{U} = k$. Then, by Theorem 3.24, we have $\ell_{k,p} \le \sup_{0 \ne u \in \mathcal{U}} \frac{\|u\|_V^2}{\|u\|_{W_1}^2}$. Now, let $\omega \in \mathcal{V}_k \setminus H_0^2(M)$, we decompose $\omega$ as follows: $\omega = p_{V_1}(\omega) + \omega - p_{V_1}(\omega)$ where $p_{V_1}(\omega) \in \mathcal{U} \subset V_1$ and $\omega - p_{V_1}(\omega) \in V_1^{\perp(\cdot, \cdot)_V} \subset \mathcal{U}^{\perp(\cdot, \cdot)_V}$. Thus, we have $\|\omega\|_V^2 = \|p_{V_1}(\omega)\|_V^2 + \|\omega - p_{V_1}(\omega)\|_V^2$ and since $\omega - p_{V_1}(\omega) \in H_0^2(M)$, we get $\|\omega - p_{V_1}(\omega)\|_{W_1}^2 = 0$ so $\|\omega\|_{W_1}^2 = \|p_{V_1}(\omega)\|_{W_1}^2 + \|\omega - p_{V_1}(\omega)\|_{W_1}^2$.

We now have:

$$ \frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2} = \frac{\|p_{V_1}(\omega)\|_V^2 + \|\omega - p_{V_1}(\omega)\|_V^2}{\|p_{V_1}(\omega)\|_{W_1}^2} \ge \frac{\|p_{V_1}(\omega)\|_V^2}{\|p_{V_1}(\omega)\|_{W_1}^2}. $$

Therefore,

$$ \sup_{0 \ne \omega \in \mathcal{V}_k \setminus H_0^2(M)} \frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2} \ge \sup_{0 \ne \omega \in \mathcal{V}_k \setminus H_0^2(M)} \frac{\|p_{V_1}(\omega)\|_V^2}{\|p_{V_1}(\omega)\|_{W_1}^2} = \sup_{u \in p_{V_1}(\mathcal{V}_k)} \frac{\|u\|_V^2}{\|u\|_{W_1}^2} \ge \ell_{k,p}. $$

This implies that for all $k \in \mathbb{N}$, we have

$$ \ell_{k,p} \le \inf_{\substack{\mathcal{V}_k \subset \tilde{H}_{N,1}^2(M) \\ \dim(\mathcal{V}_k/H_0^2(M) \cap \mathcal{V}_k) = k}} \sup_{u \in \mathcal{V}_k \setminus H_0^2(M)} \frac{\|u\|_V^2}{\|u\|_{W_1}^2}. $$
Finally, let $\bar{V}_k = \text{Span}(\omega_1, \dots, \omega_k) \subset V_1$. Since $\bar{V}_k$ is a subspace of $V_1$ orthogonal to $H_0^2(M)$, we have $\bar{V}_k \cap H_0^2(M) = \{0\}$. Therefore,

$$ \inf_{V_k \subset \dot{H}_{N,1}^2(M)} \sup_{\omega \in V_k \setminus H_0^2(M)} \frac{\|\omega\|_V^2}{\|\omega\|_{W_1}^2} \le \sup_{u \in V_k \setminus H_0^2(M)} \frac{\|u\|_V^2}{\|u\|_{W_1}^2} = \ell_{k,p}. $$
$$ \dim(V_k/H_0^2(M) \cap V_k) = k $$

This concludes the proof.

<figure><img src="image_6.png" /></figure>

**Remark 3.27.** For every $k \in \mathbb{N}$, the following characterization holds for the $k^{th}$ positive eigenvalue of BSN1:

$$ \ell_{k,p} = \inf_{\substack{\omega \in V_1 \\ \omega \perp W_1 \text{Span}(\omega_1, \dots, \omega_{k-1})}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2}. $$

## 3.4 No Kuttler-Sigillito inequalities using BSN1

Let us describe the problem to establish the Kuttler-Sigillito inequality $\mu_1\sigma_1 \le \ell_1$ based on this BSN1. Regardless of the conditions imposed in the characterizations, for any $\omega \in \Omega^p(M)$, we have

$$ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \times \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2} \\ \ge \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\ell^* \omega\|_{L^2(\partial M)}^2 + \|\ell^* \delta \omega\|_{L^2(\partial M)}^2}. \quad (69) $$

Thus, using the characterizations (35), (38) and (56) does not yield Kuttler-Sigillito inequalities. We can also try using the characterizations (34), (40), and (61). So we get

$$ \mu_1 \sigma_1 = \inf_{\substack{\omega \perp L^2(M) H_A^p(M) \\ \nu \lrcorner \omega = 0}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(M)}^2} \times \inf_{\substack{\omega \notin H_A^p(M) \\ \nu \lrcorner \omega = 0 \\ \Delta \omega = 0}} \frac{\|\nu \lrcorner d\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \\ \le \inf_{\substack{\omega \perp L^2(M) H_A^p(M) \\ \nu \lrcorner \omega = 0 \\ \Delta \omega = 0}} \frac{\|\nu \lrcorner d\omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2}. $$

Because of the boundary condition $\nu \lrcorner \omega = 0$ the infimum above is a priori greater than

$$ \ell_1 = \inf_{\substack{\omega \perp L^2(M) H_A^p(M) \\ \Delta \omega = 0}} \frac{\|\nu \lrcorner \omega\|_{L^2(\partial M)}^2 + \|\nu \lrcorner d\omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2}, $$

which also prevents us from concluding.

In a forthcoming paper, we will see that on some examples the inequality $\mu_1\sigma_1\ell_1$ can indeed fail.
# 4 Spectral properties of BSN2

## 4.1 Formulation of the boundary problem

To establish the inequalities (13), and (14), it follows from (56) and (69) that it is enough to have a BSN on differential forms, whose first non-zero eigenvalue satisfies

$$ \mathbf{l}_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \check{H}_{N,1}^2(M), \iota^* \delta \omega = 0 \text{ and } \iota^* \omega \neq 0 \text{ on } \partial M \right\}. $$

We start by defining the following spaces:

$$ \begin{gathered} H_{N,2}^2(M) := \{\omega \in H^2(M) \mid \nu \rfloor d\omega = 0, \nu \rfloor \omega = 0, \iota^* \delta \omega = 0 \text{ on } \partial M\}, \\ \check{H}_{N,2}^2(M) := \{\omega \in H^2(M) \mid \nu \rfloor d\omega = 0, \nu \rfloor \omega = 0, \iota^* \delta \omega = 0 \text{ on } \partial M \text{ and } \omega \perp_{L^2(\partial M)} H_A^p(M)\} \subset \check{H}_{N,1}^2(M). \end{gathered} \tag{70} $$

$$ Z_2 := \{\omega \in \Omega^p(M) \cap H_{N,2}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M\} \subset Z_1 \subset H_{N,1}^2(M), $$

and

$$ \check{Z}_2 := \{\omega \in Z_2 \mid \omega \perp_{L^2(\partial M)} H_A^p(M)\} = Z_2 \cap \check{H}_{N,1}^2(M) = Z_2 \cap \check{Z}_1. $$

We compute the critical points of the following functional: $Q(\omega) := \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2}$ on $\check{H}_{N,2}^2(M)$, which will be the eigenforms of the new BSN2 problem that we will determine. The critical points of the functional $Q(\omega)$ are those $\omega$ such that $d_\omega Q$ is zero, i.e, $\omega \in \check{H}_{N,2}^2(M)$ is a critical point of $Q(\omega)$ if and only if for all $\omega' \in \check{Z}_2 \subset \check{H}_{N,2}^2(M)$, we have

$$ (\Delta \omega, \Delta \omega')_{L^2(M)} \|\iota^* \omega\|_{L^2(\partial M)}^2 = (\iota^* \omega, \iota^* \omega')_{L^2(\partial M)} \|\Delta \omega\|_{L^2(M)}^2. \tag{71} $$

Using an integration by parts (21) in (71), we get

$$ (\Delta^2 \omega, \omega')_{L^2(M)} - (\nu \rfloor d\Delta \omega, \iota^* \omega')_{L^2(\partial M)} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2} (\iota^* \omega, \iota^* \omega')_{L^2(\partial M)}. $$

Denoting $\mathbf{l} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2}$, it follows that

$$ (\Delta^2 \omega, \omega')_{L^2(M)} = (\nu \rfloor d\Delta \omega + \mathbf{l} \iota^* \omega, \iota^* \omega')_{L^2(\partial M)}. \tag{72} $$

Let us now show that $\Delta^2 \omega = 0$ on $M$. We define the sets $D_\epsilon := \{x \in M \mid d(x, \partial M) \ge \epsilon\}$ and $D_{2\epsilon} := \{x \in M \mid d(x, \partial M) \ge 2\epsilon\}$. There exists $\chi_\epsilon \in C^\infty(M, [0, 1])$ such that $\chi_\epsilon|_{D_{2\epsilon}} = 1$ and $\chi_\epsilon|(M \setminus D_\epsilon) = 0$. Let $\omega_\epsilon := \chi_\epsilon \Delta^2 \omega \in \Omega^p(M)$ then $\omega_\epsilon|_{(M \setminus D_\epsilon)} = 0$ thus, $\nu \rfloor \omega_\epsilon$, $\nu \rfloor d\omega_\epsilon$, $\iota^* \delta \omega_\epsilon$ and $\iota^* \omega_\epsilon$ vanish on $\partial M$ since $\omega_\epsilon$ vanishes on a neighbourhood of $\partial M$. Then $\int_M \langle \Delta^2 \omega, \chi_\epsilon \Delta^2 \omega \rangle d\mu_g = 0 = \int_M \chi_\epsilon |\Delta^2 \omega|^2 d\mu_g$, hence $\chi_\epsilon \Delta^2 \omega = 0$. Thus, $\chi_\epsilon \Delta^2 \omega = 0$, so $\Delta^2 \omega = 0$ on $D_\epsilon$, and letting $\epsilon$ tend to 0, we conclude that $\Delta^2 \omega = 0$ on $M$.

<figure><img src="image_7.png" /></figure>
We show now that (72) is valid for all $\omega \in \tilde{H}_{N,2}^2(M)$, $\omega' \in H_A^p(M)$. Then, using (21) for such $\omega$ and $\omega'$, we get

$$ 
\begin{aligned}
\int_M \langle \Delta^2 \omega, \omega' \rangle d\mu_g &= \int_{\partial M} \langle \nu \rfloor d\Delta \omega, \iota^* \omega' \rangle d\mu_g \\
&= \int_{\partial M} \langle \nu \rfloor d\Delta \omega + \mathbf{l}^* \omega, \iota^* \omega' \rangle d\mu_g \\
&\quad - \int_{\partial M} \langle \mathbf{l}^* \omega, \iota^* \omega' \rangle d\mu_g
\end{aligned}
$$

noting that here $(\iota^* \omega, \iota^* \omega')_{L^2(\partial M)} = 0$. From the above, we deduce that (72) is true for all $\omega' \in H_{N,2}^2(M)$. By Lemma 3.3, there exists a unique $\omega' \in \Omega^p(M)$ such that

$$ 
\left\{
\begin{array}{ll}
\Delta^2 \omega' = 0 & \text{on } M, \\
\nu \rfloor \omega' = 0 & \text{on } \partial M, \\
\nu \rfloor d \omega' = 0 & \text{on } \partial M, \\
\iota^* \delta \omega' = 0 & \text{on } \partial M, \\
\iota^* \omega' = \nu \rfloor d \Delta \omega + \mathbf{l}^* \omega & \text{on } \partial M.
\end{array}
\right.
$$

Since $\omega' \in H_{N,2}^2(M)$, then by (72) we get $(\nu \rfloor d\Delta \omega + \mathbf{l}^* \omega, \iota^* \omega')_{L^2(\partial M)} = 0$, hence $\nu \rfloor d\Delta \omega + \mathbf{l}^* \omega = 0$.

Therefore, $\omega$ is a smooth critical point of $Q$ if and only if $\omega$ satisfies:

$$ 
\left\{
\begin{array}{ll}
\Delta^2 \omega = 0 & \text{on } M \\
\nu \rfloor \omega = 0 & \text{on } \partial M \\
\nu \rfloor d \omega = 0 & \text{on } \partial M \\
\iota^* \delta \omega = 0 & \text{on } \partial M \\
\nu \rfloor d \Delta \omega + \mathbf{l}^* \omega = 0 & \text{on } \partial M,
\end{array}
\right.
$$

for $l = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2}$. This problem will be called BSN2 from now on.

## 4.2 Ellipticity of BSN2

**Theorem 4.1.** The BSN2 problem (11) is elliptic in the sense of Shapiro-Lopatinskij.

*Proof.* We follow the method used in Section 3.1 to show that BSN1 is elliptic. Let us compute the principal symbols of $B_1$, $B_2$, $B_3$ and $B_4$ respectively, applied to $y = e^{-|v|t}(at + b) \cdot \omega_0 \in \mathcal{M}_{\Delta^2,v}^+$ where $B_1 \omega = \nu \rfloor \omega$, $B_2 \omega = \nu \rfloor d \omega$, $B_3 \omega = \nu \rfloor d \Delta \omega$, and $B_4 \omega = \iota^* \delta \omega$, we get

$$ 
\begin{aligned}
\sigma_{B_1}(-iv + \partial_t \nu^b)(y)(0) &= b\nu \rfloor \omega_0, \\
\sigma_{B_2}(-iv + \partial_t \nu^b)(y)(0) &= \iota^* \omega_0(a - b|v|) + ib\iota^* v \wedge (\nu \rfloor \omega_0), \\
\sigma_{B_3}(-iv + \partial_t \nu^b)(y)(0) &= 2|v|a(iv \wedge (\nu \rfloor \omega_0) - |v|\iota^* \omega_0), \\
\sigma_{B_4}(-iv + \partial_t \nu^b)(y)(0) &= -a\nu \rfloor \omega_0 + |v|b\nu \rfloor \omega_0 + ivb \rfloor \iota^* \omega_0.
\end{aligned}
$$

Let us show that the corresponding map

$$ \Phi : \mathcal{M}_{\Delta^2, v}^{+} \to \bigoplus_{j=1}^{4} E_j $$
is injective. Assume that $\Phi(y) = 0$. Then:

$$ 
\left\{ 
\begin{array}{ll}
b\nu\upharpoonright\omega_0 & = 0 \quad (1) \\
\iota^*\omega_0(a - b|v|) + ib\iota^*v \wedge (\nu\upharpoonright\omega_0) & = 0 \quad (2) \\
2|v|a(iv \wedge (\nu\upharpoonright\omega_0) - |v|\iota^*\omega_0) & = 0 \quad (3) \\
-a\nu\upharpoonright\omega_0 + |v|b\nu\upharpoonright\omega_0 + ivb\upharpoonright\iota^*\omega_0 & = 0 \quad (4)
\end{array}
\right.
$$

If $a$ and $b$ are zero then $y = 0$. If $a = 0$ and $b$ non-zero, we get from (1) that $\nu\upharpoonright\omega_0 = 0$. Replacing in (2) and (4), we get $\iota^*\omega_0 = 0$. If $a$ non-zero and $b = 0$, from (2) and (4) we get that $\iota^*\omega_0 = 0$ and $\nu\upharpoonright\omega_0 = 0$ respectively. In both cases we conclude that $\omega_0 = 0$, hence $y = 0$, which implies that $\Phi$ is injective. Therefore, the problem (11) is elliptic. $\square$

## 4.3 Discrete spectrum of BSN2

Let us show that BSN2 has a discrete spectrum of eigenvalues.

**Theorem 4.2.** Let $(M^n, g)$ be a compact Riemannian manifold with a smooth boundary $\partial M$. The BSN2 problem (11) for differential $p$-forms has a discrete spectrum consisting of an unbounded non-decreasing sequence of positive real eigenvalues with finite multiplicities $(l_i)_{i \ge 1}$ and possibly $l_0 = 0$. Its kernel is given by $H_A^p(M)$ (see (9)).

The proof of this theorem is similar to that of Theorem 3.2.

*Proof.* The bilinear forms

$$ (\omega, \omega')_V := \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g $$

and

$$ (\omega, \omega')_{W_2} := \int_{\partial M} \langle \iota^* \omega, \iota^* \omega' \rangle d\mu_g. $$

are inner products on $\tilde{Z}_1$ and $Z_2$ respectively since $(\omega, \omega')_{W_1}$ and $(\omega, \omega')_V$ are inner products on $Z_1$ and $\tilde{Z}_1$ (see Lemma 3.4). Note that for all $\omega, \omega' \in Z_2$ we have $(\omega, \omega')_{W_2} = (\omega, \omega')_{W_1}$.

We denote by $W_2$ the completion of $Z_2$ with respect to $(\omega, \omega')_{W_2}$, and by $V_2$ the completion of $\tilde{Z}_2$ with respect to $(\omega, \omega')_V$.

The norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$ are equivalent on $\tilde{Z}_2$ since they are on $\tilde{Z}_1$ and $\tilde{Z}_2 \subset \tilde{Z}_1$ (see Lemma 3.6). Furthermore, for the same reason, there exists a positive constant $c$ such that $\|\cdot\|_{W_2} \le c\|\cdot\|_V$ on $\tilde{Z}_2$ (see Lemma 3.8). We can then say that $\tilde{Z}_2$ is dense in $V_2$ and $W_2$.

The inclusion map $i_{W_2}: \tilde{Z}_2 \to W_2$ can be extended to another continuous map

$$ 
\begin{array}{rcl}
I_2 & = & I_{1|V_2}: V_2 \longrightarrow W_2 \\
    &   & \omega \longmapsto \omega
\end{array}
$$

The map $I_2$ is compact since $I_1$ is compact and $W_2$ is closed (see Lemma 3.9). The map $I_2$ is injective since $I_1$ is injective (see Lemma 3.13).

Let the operator $K_2: V_2 \to V_2$ such that

$$ K_2 := D_{V_2}^{-1} \circ {}^t I_2 \circ D_{W_2} \circ I_2, $$

where

$$ 
\begin{array}{rcl}
D_{V_2}: V_2 & \to & V_2' \quad \text{and} \quad D_{W_2}: W_2 \to W_2' \\
\omega & \mapsto & (\cdot, \omega)_V \qquad \omega \mapsto (\cdot, \omega)_{W_2}
\end{array}
$$
In addition, as in (51), $K_2$ is characterized by

$$ (K_2 \omega, \omega')_V = (\omega, \omega')_{W_2}. $$

The operator $K_2$ is compact, self-adjoint, and positive definite in $V_2$. Therefore, there exists a sequence of eigenvalues $(\chi_{i,2})_{i \ge 1}$, which are real, positive, and decreasing, going to 0 with finite-dimensional eigenspaces, according to [24, p.431]. The eigenforms $\omega_i$ form a Hilbert basis of $V_2$. To conclude the proof, we show the following Lemma.

**Lemma 4.3.** A form $\omega$ is a solution of BSN2 for $l \in \mathbb{R}^*$ if and only if $K_2 \omega = \frac{1}{l} \omega$.

*Proof.* Let $l \in \mathbb{R}^*$ such that $\omega$ is a solution of BSN2, integrating by parts (21) for $\omega' \in \check{Z}_2$, we have:

$$ (\Delta \omega, \Delta \omega')_{L^2(M)} = \mathbf{I}(\iota^* \omega, \iota^* \omega')_{L^2(\partial M)}. $$

Since $\check{Z}_2$ is dense in $W_2$, we have for all $\omega' \in V_2$, $(\omega, \omega')_V = \mathbf{I}(\omega, \omega')_{W_2}$. Then by definition of $K_2$ we get $(\omega, \omega')_V = \mathbf{I}(K_2 \omega, \omega')_V$ for all $\omega' \in V_2$. Hence $K_2 \omega = \frac{1}{l} \omega$, consequently $\omega$ is an eigenform for $K_2$ associated to $\frac{1}{l} = \chi$.

Conversely, let $\omega_i$ be a $\chi$-eigenform in $V_2$ for $K_2$. Since $V_2 \subset V_1$, we have by Lemma 3.12 that $\Delta^2 \omega_i = 0$ weakly on $M$, $\nu \rfloor \omega_i = 0$, $\nu \rfloor d\omega_i = 0$ and $\iota^* \delta \omega_i = 0$ on $\partial M$. It remains to show that $\nu \rfloor \Delta d\omega_i = -\mathbf{I} \iota^* \omega_i$. For all $\omega \in \check{Z}_2$, we have by definition

$$ \begin{aligned} \chi(\Delta \omega_i, \Delta \omega)_{L^2(M)} &= \chi(\omega_i, \omega)_V \\ &= (K_2 \omega_i, \omega)_V \\ &= (I_2(\omega_i), I_2(\omega))_{W_2} \\ &= (\omega_i, \omega)_{W_2} \\ &= (\iota^* \omega_i, \iota^* \omega)_{L^2(\partial M)}. \end{aligned} \quad (73) $$

Then by integrating by parts (21), for all $\omega_i \in V_2$, $\omega \in Z_2$, as in Lemma 3.14, we get:

$$ (\Delta \omega_i, \Delta \omega)_{L^2(M)} + (\underbrace{\nu \rfloor d\Delta \omega_i}_{\in H^{-\frac{3}{2}}(\partial M)}, \underbrace{\iota^* \omega}_{\in H^{\frac{3}{2}}(\partial M)})_{L^2(\partial M)} = 0. $$

Using (73), we get

$$ \frac{1}{\chi}(\iota^* \omega_i, \iota^* \omega)_{L^2(\partial M)} + (\nu \rfloor d\Delta \omega_i, \iota^* \omega)_{L^2(\partial M)} = 0. $$

Finally we get

$$ \left( \frac{1}{\chi} \iota^* \omega_i + \nu \rfloor d \Delta \omega_i, \iota^* \omega \right)_{L^2(\partial M)} = 0. $$

Moreover, it follows from Lemma 3.3 that the map

$$ \begin{aligned} O_2 : Z_2 &\to \Omega^p(\partial M) \\ \omega &\mapsto \iota^* \omega \end{aligned} $$

is surjective since $O_1$ is surjective (see Lemma 3.16). This implies that $\nu \rfloor d\Delta \omega_i + \iota^* \omega_i = 0$ in $H^{-\frac{3}{2}}(\partial M)$. By the definition of $Z_2$, the restriction $O_2|_{\check{Z}_2} : \check{Z}_2 \to \Omega^p(M) \cap H_A^p(M)^{\perp L^2(\partial M)}$ is also surjective. Hence, by Theorem 2.6, $\omega_i \in \Omega^p(M)$, therefore $\omega_i$ is a solution of BSN2.

□

Which concludes the proof of Theorem 4.2.

□
## 4.4 Variational characterization of eigenvalues

To establish Kuttler-Sigillito inequalities, we will present in this section the variational characterizations associated with the eigenvalues of problem BSN2. The proofs of these characterizations are done in the same way as those of problem BSN1.

**Theorem 4.4.** A variational characterization of the first strictly positive eigenvalue of problem (11) is given by

$$l_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \check{H}_{N,2}^2(M), l^* \omega \neq 0 \text{ on } \partial M \right\}. \quad (74)$$

*Proof.* The proof is identical to that of Theorem 3.17. □

**Theorem 4.5.** For any $k \in \mathbb{N}$, the following characterization of the $k^{th}$ strictly positive eigenvalue holds:

$$l_{k,p} = \inf_{\substack{\nu_k \subset \check{H}_{N,2}^2(M) \\ \dim(\nu_k/(H_0^2(M) \cap \nu_k)) = k}} \sup_{\omega \in \nu_k \setminus H_0^2(M)} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2}.$$

*Proof.* The proof is identical to that of Theorem 3.26. □

# 5 Spectral properties of BSN3

As we will see in Section 6, BSN2 is enough to show Kuttler-Sigillito inequalities (13) and (14) but not to show the rest of the inequalities of Theorem 1.16. A new BSN3 is introduced with the aim of improving what was done using BSN2 and establishing the inequalities that remain.

## 5.1 Formulation of the boundary problem

In this section, we mainly use the same working method as in Section 4.1. The goal is to obtain the following variational characterization:

$$l_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \check{H}_{N,1}^2(M) \right\},$$

i.e., removing the boundary condition $l^* \delta \omega = 0$.

To do this, we compute again the critical points of the following functional: $Q(\omega) := \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2}$ on $\check{H}_{N,1}^2(M)$. For all $\omega, \omega'$ satisfying the above conditions, we have $(\Delta \omega, \Delta \omega')_{L^2(M)} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2} (l^* \omega, l^* \omega')_{L^2(\partial M)}$. Denoting $l = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2}$ and using integration by parts (21), we obtain

$$(\Delta^2 \omega, \omega')_{L^2(M)} - (\nu \rfloor d\Delta \omega, l^* \omega)_{L^2(\partial M)} - (\nu \rfloor \Delta \omega, l^* \delta \omega)_{L^2(\partial M)} = l(l^* \omega, l^* \omega')_{L^2(\partial M)}.$$

It implies that

$$(\Delta^2 \omega, \omega')_{L^2(M)} = (\nu \rfloor d\Delta \omega + l l^* \omega, l^* \omega')_{L^2(\partial M)} + (\nu \rfloor \Delta \omega, l^* \delta \omega')_{L^2(\partial M)},$$

for all $\omega'$ such that $\nu \rfloor \omega'$ and $\nu \rfloor d \omega'$ vanish. By following the same steps as in Section 4.1, we show that $\Delta^2 \omega = 0$ on M. Again by Lemma 3.3 there exists $\omega' \in H_{N,1}^2(M)$ such that
$$
\left\{
\begin{array}{ll}
\Delta^2 \omega' = 0 & \text{on } M \\
\nu \lrcorner \omega' = 0 & \text{on } \partial M \\
\nu \lrcorner d \omega' = 0 & \text{on } \partial M \\
\iota^* \delta \omega' = \nu \lrcorner \Delta \omega & \text{on } \partial M \\
\iota^* \omega = \nu \lrcorner d \Delta \omega + l \iota^* \omega & \text{on } \partial M.
\end{array}
\right.
$$

We conclude in the same way we did for BSN2. The critical points of $\mathcal{Q}$ are the eigenforms $\omega$ for
the following BSN3 problem:

$$
\left\{
\begin{array}{ll}
\Delta^2 \omega = 0 & \text{on } M \\
\nu \lrcorner \omega = 0 & \text{on } \partial M \\
\nu \lrcorner d \omega = 0 & \text{on } \partial M \\
\nu \lrcorner \Delta \omega = 0 & \text{on } \partial M \\
\nu \lrcorner d \Delta \omega + l \iota^* \omega = 0 & \text{on } \partial M.
\end{array}
\right.
$$

5.2 Ellipticity of BSN3

**Theorem 5.1.** The BSN3 problem (12) is elliptic in the sense of Shapiro-Lopatinskij.

*Proof.* Since its principal symbols are identical to those of the BSN1 (10), the problem BSN3 (12) is also elliptic in the sense of Shapiro-Lopatinskij.

5.3 Discrete spectrum of BSN3

**Theorem 5.2.** Let $(M^n, g)$ be a compact Riemannian manifold with a smooth $\partial M$ and let $\nu$ be the unit inward normal vector to the boundary. Then the BSN3 (12) for differential $p$-forms, has a discrete spectrum consisting of an unbounded non-decreasing sequence of positive real eigenvalues with finite multiplicities $(l_i)_{i \ge 0}$. Its kernel is given by $H_A^p(M)$ (9).

**Lemma 5.3.** The problem

$$
(biLap_2) \left\{
\begin{array}{lll}
\Delta^2 \omega & = f & \text{on } M \\
\nu \lrcorner \omega & = \omega_1 & \text{on } \partial M \\
\nu \lrcorner d \omega & = \omega_2 & \text{on } \partial M \\
\iota^* \omega & = \omega_3 & \text{on } \partial M \\
\nu \lrcorner \Delta \omega & = \omega_4 & \text{on } \partial M.
\end{array}
\right.
$$

admits a unique solution $\omega$ for all $(f, \omega_1, \omega_2, \omega_3, \omega_4) \in H^{s-4}(M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M)$ where $s \ge 4$, in particular for $f=0$, $\omega_1=0$, $\omega_2=0$, $\iota^*\omega = \omega_3 \perp L^2(\partial M)$, $H_A^p(M)$ and $\nu\lrcorner\Delta\omega = \omega_4$.

*Proof.* Indeed, by **Theorem 2.6**, the problem above admits a unique solution. It is elliptic since by
doing the same calculations of principal symbols as in **Section 3.1**, we obtain that the corresponding
map

$$
\Phi : \mathcal{M}_{\Delta^2, \nu}^{+} \to \bigoplus_{j=1}^{4} E_j
$$
$$
y \mapsto (b \nu \lrcorner \omega_0, (a - b|v|) \iota^* \omega_0 + ib \iota^* v \wedge (\nu \lrcorner \omega_0), b \iota^* \omega_0, 2a|v| \nu \lrcorner \omega_0)
$$
is an isomorphism.

The operator of this problem is injective since its kernel $\mathcal{N}$ is trivial, and this is shown by doing an integration by parts:

$$ \int_M \langle \underbrace{\Delta^2 \omega}_{0}, \omega \rangle d\mu_g = \int_M \langle \Delta \omega, \Delta \omega \rangle d\mu_g \\ + \int_{\partial M} \left( \langle \nu \rfloor d\Delta \omega, \underbrace{\iota^* \omega}_{0} \rangle - \langle \iota^* \Delta \omega, \nu \rfloor d\omega \rangle + \langle \nu \rfloor \Delta \omega, \iota^* \delta \omega \rangle - \langle \iota^* \delta \Delta \omega, \nu \rfloor \omega \rangle \right) d\mu_g. $$

Thus, $\Delta \omega = 0$ and since $\omega|_{\partial M} = 0$ we get by [1] that $\omega = 0$.

To evaluate $\mathcal{N}^*$, we use Theorem 2.6, and we denote by $B_0 \omega = \nu \rfloor d\omega$, $B_1 \omega = \nu \rfloor \omega$, $B_2 \omega = \nu \rfloor \Delta \omega$ and $B_3 \omega = \iota^* \omega$ and integrate by parts we get

$$ \begin{aligned} (\Delta^2 \omega, \omega')_{L^2(M)} - (\omega, \Delta^2 \omega')_{L^2(M)} & = (\nu \rfloor d\Delta \omega, \iota^* \omega')_{L^2(\partial M)} - (\iota^* \omega, \nu \rfloor d\Delta \omega')_{L^2(\partial M)} \\ & \quad - (\iota^* \delta \Delta \omega, \nu \rfloor \omega')_{L^2(\partial M)} + (\nu \rfloor \omega, \iota^* \delta \Delta \omega')_{L^2(\partial M)} \\ & \quad + (\nu \rfloor \Delta \omega, \iota^* \delta \omega')_{L^2(\partial M)} - (\iota^* \delta \omega, \nu \rfloor \Delta \omega')_{L^2(\partial M)} \\ & \quad - (\iota^* \Delta \omega, \nu \rfloor d\omega')_{L^2(\partial M)} + (\nu \rfloor d\omega, \iota^* \Delta \omega')_{L^2(\partial M)}. \end{aligned} $$

We note, based on the above, that $T_0 \omega = -\iota^* \Delta \omega$, $T_1 \omega = -\iota^* \delta \Delta \omega$, $T_2 \omega = -\iota^* \delta \omega$ and $T_3 \omega = \nu \rfloor d\Delta \omega$ as well as $C_0 \omega = \iota^* \omega$, $C_1 \omega = -\nu \rfloor \omega$, $C_2 \omega = -\nu \rfloor \Delta \omega$ and $C_3 \omega = -\nu \rfloor d\omega$, which implies that $\mathcal{N}^* = 0$. So by Theorem 2.6 the map

$$ \mathcal{P}_{biLap_2}: H^s(M) \longrightarrow H^{s-4}(M) \oplus H^{s-\frac{1}{2}}(\partial M) \oplus H^{s-\frac{1}{2}}(\partial M) \oplus H^{s-\frac{3}{2}}(\partial M) \oplus H^{s-\frac{3}{2}}(\partial M) \\ [\omega] \longmapsto (\Delta^2 \omega, \nu \rfloor \omega, \nu \rfloor d\omega, \iota^* \omega, \nu \rfloor \Delta \omega) $$

is an isomorphism on its image that is given for all $s \ge 4$ by:

$$ \mathrm{Im}(\mathcal{P}_{biLap_2}) = H^{s-4}(M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M) \times H^{s-\frac{1}{2}}(\partial M) \times H^{s-\frac{3}{2}}(\partial M). $$

In particular, by Theorem 2.6 (biLap$_2$) admits a unique solution. $\square$

The proof of Theorem 5.2 is done in a very similar way to that of both Theorem 3.2 and Theorem 4.2.

**Proof of Theorem 5.2.** We begin by defining the following spaces:

$$ Z_3 := \{ \omega \in \Omega^p(M) \cap H_{N,3}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M \} \subset Z_1, $$

$$ \check{Z}_3 := \{ \omega \in Z_3 \mid \omega \perp_{L^2(\partial M)} H_A^p(M) \} \subset \check{Z}_1, $$

$$ H_{N,3}^2(M) := \{ \omega \in H^2(M) \mid \nu \rfloor \omega = 0, \nu \rfloor d\omega = 0, \text{ and } \nu \rfloor \Delta \omega = 0 \text{ on } \partial M \} $$

and

$$ \check{H}_{N,3}^2(M) := \{ \omega \in H^2(M) \mid \nu \rfloor d\omega = 0, \nu \rfloor \omega = 0, \nu \rfloor \Delta \omega = 0 \text{ on } \partial M \text{ and } \omega \perp_{L^2(\partial M)} H_A^p(M) \}. $$

The bilinear forms

$$ (\omega, \omega')_V := \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g $$

and

$$ (\omega, \omega')_{W_3} := \int_{\partial M} \langle \iota^* \omega, \iota^* \omega' \rangle d\mu_g. $$

are inner products on $\check{Z}_3$ and $Z_3$ respectively since $(\omega, \omega')_{W_1}$ and $(\omega, \omega')_V$ are inner products on $Z_1$ and $\check{Z}_1$ (see Lemma 3.4).
We denote by $W_3$ the completion of $Z_3$ with respect to $(\omega, \omega')_{W_3}$, and by $V_3$ the completion of $\tilde{Z}_3$ with respect to $(\omega, \omega')_V$.

The norms $\|\cdot\|_V$ and $\|\cdot\|_{H^2(M)}$ are equivalent on $\tilde{Z}_3$ since they are on $\tilde{Z}_1$ and $\tilde{Z}_3 \subset \tilde{Z}_1$ (see Lemma 3.6). Furthermore, for the same reason, there exists a positive constant $c$ such that $\|\cdot\|_{W_3} \le c\|\cdot\|_V$ on $\tilde{Z}_3$ (see Lemma 3.8). We can then say that $\tilde{Z}_3$ is dense in $V_3$ and $W_3$. The inclusion map $i_{W_3}: \tilde{Z}_3 \to W_3$ can be extended to another continuous map

$$I_3 = I_{1|V_3} : V_3 \to W_3 \\ \omega \mapsto \omega$$

The map $I_3$ is compact since $I_1$ is compact and $W_3$ is closed (see Lemma 3.9). The map $I_3$ is injective since $I_1$ is injective (see Lemma 3.13).

Let the operator $K_3: V_3 \to V_3$ such that

$$K_3 := D_{V_3}^{-1} \circ {}^t I_3 \circ D_{W_3} \circ I_3,$$

where

$$D_{V_3}: V_3 \to V'_3 \quad \text{and} \quad D_{W_3}: W_3 \to W'_3 \\ \omega \mapsto (\cdot, \omega)_V \qquad \omega \mapsto (\cdot, \omega)_{W_3}$$

In addition, as in (51), $K_3$ is characterized by

$$(K_3 \omega, \omega')_V = (\omega, \omega')_{W_3}.$$

The operator $K_3$ is compact, self-adjoint, and positive definite in $V_3$. Therefore, there exists a sequence of eigenvalues $(\chi_{i,3})_{i \ge 1}$, which are real, positive, and decreasing, going to 0 with finite-dimensional eigenspaces, according to [24, p.431]. The eigenforms $\omega_i$ form a Hilbert basis of $V_3$. To conclude the proof, we show the following Lemma.

**Lemma 5.4.** A form $\omega$ is a solution of BSN3 for $l \in \mathbb{R}^*$ if and only if $K_3 \omega = \frac{1}{l} \omega$.

*Proof.* Let $l \in \mathbb{R}^*$ such that $\omega$ is a solution of BSN3, integrating by parts (21) for $\omega' \in \tilde{Z}_3$, we have:

$$(\Delta \omega, \Delta \omega')_{L^2(M)} = l(\iota^* \omega, \iota^* \omega')_{L^2(\partial M)}.$$

Since $\tilde{Z}_3$ is dense in $W_3$, we have for all $\omega' \in V_3$, $(\omega, \omega')_V = l(\omega, \omega')_{W_3}$. Then by definition of $K_3$ we get $(\omega, \omega')_V = l(K_3 \omega, \omega')_V$ for all $\omega' \in V_3$. Hence $K_3 \omega = \frac{1}{l} \omega$, consequently $\omega$ is an eigenform for $K_3$ associated to $\frac{1}{l} = \chi$.

Conversely, let $\omega_i$ be a $\chi$-eigenform in $V_3$ for $K_3$. Since $V_3 \subset V_1$, we have by Lemma 3.12 that $\Delta^2 \omega_i = 0$ weakly on $M$, $\nu \rfloor \omega_i = 0$, $\nu \rfloor d\omega_i = 0$ and $\nu \rfloor \Delta \omega_i = 0$ on $\partial M$. It remains to show that $\nu \rfloor \Delta d\omega_i = -l^* \omega_i$. For all $\omega \in \tilde{Z}_3$, we have by definition

$$\begin{aligned} \chi(\Delta \omega_i, \Delta \omega)_{L^2(M)} &= \chi(\omega_i, \omega)_V \\ &= (K_3 \omega_i, \omega)_V \\ &= (I_3(\omega_i), I_3(\omega))_{W_3} \\ &= (\omega_i, \omega)_{W_3} \\ &= (\iota^* \omega_i, \iota^* \omega)_{L^2(\partial M)}. \end{aligned} \qquad (75)$$

Then by integrating by parts (21), for all $\omega_i \in V_3$, $\omega \in Z_3$, as in Lemma 3.14, we get:

$$(\Delta \omega_i, \Delta \omega)_{L^2(M)} + (\underbrace{\nu \rfloor d \Delta \omega_i}_{\in H^{-\frac{3}{2}}(\partial M)}, \underbrace{\iota^* \omega}_{\in H^{\frac{3}{2}}(\partial M)})_{L^2(\partial M)} = 0.$$
Using (75), we get

$$ \frac{1}{\chi}(\iota^* \omega_i, \iota^* \omega)_{L^2(\partial M)} + (\nu \rfloor d\Delta \omega_i, \iota^* \omega)_{L^2(\partial M)} = 0. $$

Finally we get

$$ \left( \frac{1}{\chi} \iota^* \omega_i + \nu \rfloor d \Delta \omega_i, \iota^* \omega \right)_{L^2(\partial M)} = 0. $$

Moreover, by Lemma 5.3, for all $\omega_0 \in H^{\frac{3}{2}}(\partial M)$, there exists $\omega \in H^2(M)$ such that

$$ \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M, \\ \nu \rfloor \omega = 0 & \text{on } \partial M, \\ \nu \rfloor d\omega = 0 & \text{on } \partial M, \\ \iota^* \omega = \omega_0 & \text{on } \partial M, \\ \nu \rfloor \Delta \omega = 0 & \text{on } \partial M. \end{array} \right. $$

Therefore, the map

$$ O_3 : Z_3 \to \Omega^p(\partial M) \\ \omega \mapsto \iota^* \omega $$

is surjective since $O_1$ is surjective (see Lemma 3.16). The map

$$ O_{3|\tilde{Z}_3} : \tilde{Z}_3 \to \Omega^p(\partial M) \cap H_A^p(M)^{\perp L^2(\partial M)} $$

is also surjective. This implies that $\nu \rfloor d\Delta \omega_i + \iota^* \omega_i = 0$ in $H^{-\frac{3}{2}}(\partial M)$. Hence, by Theorem 2.6, $\omega_i \in \Omega^p(M)$, therefore $\omega_i$ is a solution of BSN3.

□

Which concludes the proof of Theorem 4.2.

□

## 5.4 Variational characterization of eigenvalues

In this section, we will provide the variational characterizations associated with the eigenvalues of BSN3, which will be used to establish the Kuttler-Sigillito inequalities of Section 1.6.

**Theorem 5.5.** A variational characterization of the first strictly positive eigenvalue of problem (12) is given by

$$ l_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \tilde{H}_{N,3}^2(M) \text{ and } \iota^* \omega \neq 0 \text{ on } \partial M \right\}. \quad (76) $$

It is a minimum, attained only by eigenforms of BSN3.

*Proof.* The proof of this theorem is done in the same way as the one of Theorem 3.17.

□

Actually the condition $\nu \rfloor \Delta \omega$ can be dropped in (76): we have the following stronger theorem.

**Theorem 5.6.** A variational characterization of the first strictly positive eigenvalue of problem (12) is given by

$$ l_{1,p} = \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \tilde{H}_{N,1}^2(M) \text{ and } \iota^* \omega \neq 0 \text{ on } \partial M \right\}. \quad (77) $$
Proof. We recall that the operator $K_3: V_3 \to V_3$ is self-adjoint, compact with a discrete spectrum consisting of the eigenvalues $\frac{1}{l_{i,p}}$. Thus,

$$ \frac{1}{l_{1,p}} = \sup_{\omega \in V_3 \setminus \{0\}} \frac{(K_3 \omega, \omega)_V}{\|\omega\|_V^2} = \sup_{\omega \in V_3 \setminus \{0\}} \frac{\|l^* \omega\|_{L^2(\partial M)}^2}{\|\Delta \omega\|_{L^2(M)}^2} $$

so that

$$ \begin{aligned} l_{1,p} &= \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2}, \omega \in V_3 \right\} \\ &= \inf \left\{ \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|l^* \omega\|_{L^2(\partial M)}^2} \mid \omega \in \Omega^p(M), \nu \rfloor \omega = 0, \nu \lfloor d\omega = 0, \nu \rfloor \Delta \omega = 0 \text{ on } \partial M \text{ and } \omega \perp_{L^2(\partial M)} H_A^p(M) \right\}. \end{aligned} $$

We want to remove the condition $\nu \rfloor \Delta \omega = 0$. To do so, we establish the following decomposition:

$$ \tilde{H}_{N,3}^2(M) = V_3 \oplus_{\perp(\cdot, \cdot)_V} H_0^1(M) \cap \tilde{H}_{N,3}^2(M). $$

Note that the inner product $(\cdot, \cdot)_V$ is well-defined on $\tilde{H}_{N,3}^2(M) \subset \tilde{H}_{N,1}^2(M)$. Let $\omega \in \tilde{Z}_3, \omega' \in \tilde{H}_{N,3}^2(M)$, then by performing an integration by parts (21), we obtain

$$ \int_M \langle \Delta \omega, \Delta \omega' \rangle d\mu_g = - \int_{\partial M} \langle \nu \rfloor d\Delta \omega, l^* \omega' \rangle d\mu_g, $$

so that for $\omega' \in H_0^1(M) \cap \tilde{H}_{N,3}^2(M)$, we have $(\omega, \omega')_V = 0$, thus $H_0^1(M) \cap \tilde{H}_{N,3}^2(M) \subset \tilde{Z}_3^\perp = V_3^\perp$.

Conversely, let $\omega' \in V_3^\perp \subset \tilde{H}_{N,3}^2(M)$, then $(\omega_i, \omega')_V = 0$ so that $(\nu \rfloor d\Delta \omega_i, l^* \omega')_{L^2(\partial M)} = 0$, thus $-l_i(l^* \omega_i, l^* \omega')_{L^2(\partial M)} = 0$, and consequently, $(l^* \omega_i, l^* \omega')_{L^2(\partial M)} = 0$.

The goal now is to show that if $\omega' \in V_3^\perp \subset \tilde{H}_{N,3}^2(M)$ and satisfies

$$ \begin{cases} \nu \rfloor d \omega' = 0 & \text{on } \partial M \\ \nu \rfloor \omega' = 0 & \text{on } \partial M \\ \nu \rfloor \Delta \omega' = 0 & \text{on } \partial M, \end{cases} $$

then $l^* \omega' = 0$.

Now, the operator $\Delta^2$ with the boundary conditions of the problem (78) is an invertible operator and thus has a trivial kernel (see Lemma 5.3). Therefore, the problem

$$ \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \nu \rfloor d \omega = 0 & \text{on } \partial M \\ \nu \rfloor \omega = 0 & \text{on } \partial M \\ \nu \rfloor \Delta \omega = 0 & \text{on } \partial M \\ l^* \omega = 0 & \text{on } \partial M, \end{array} \right. \qquad (78) $$

which is a particular case of $(biLap_2)$ (5.3) admits $\omega = 0$ as unique solution. We have shown in Section 5.2 that the operator $O_3: Z_3 \to \Omega^p(M)$ is surjective, so for every strictly positive $\epsilon$, there exists $\omega \in \Omega^p(\partial M)$ such that $\|l^* \omega' - l^* \omega\|_{L^2(\partial M)} < \epsilon$. Therefore, there exists $\omega_\epsilon \in \Omega^p(M)$ such that

$$ \left\{ \begin{array}{ll} \Delta^2 \omega_\epsilon = 0 & \text{on } M \\ \nu \rfloor d \omega_\epsilon = 0 & \text{on } \partial M \\ \nu \rfloor \omega_\epsilon = 0 & \text{on } \partial M \\ \nu \rfloor \Delta \omega_\epsilon = 0 & \text{on } \partial M \\ l^* \omega_\epsilon = l^* \omega & \text{on } \partial M, \end{array} \right. $$
which implies that

$$
\left\{
\begin{array}{ll}
\Delta^2 \omega_{\epsilon} = 0 & \text{on } M \\
\nu \rfloor d\omega_{\epsilon} = 0 & \text{on } \partial M \\
\nu \rfloor \omega_{\epsilon} = 0 & \text{on } \partial M \\
\nu \rfloor \Delta \omega_{\epsilon} = 0 & \text{on } \partial M \\
\| \iota^* \omega_{\epsilon} - \iota^* \omega' \|_{L^2(\partial M)} < \epsilon.
\end{array}
\right.
$$

We then decompose the $\omega_\epsilon \in Z_3$ as follows: $\omega_\epsilon = \check{\omega}_\epsilon + \alpha_\epsilon$ such that $\check{\omega}_\epsilon \in \check{Z}_3$ and $\alpha_\epsilon \in H_A^p(M)$. By the continuity of the trace map and having $\|\cdot\|_V \sim \|\cdot\|_{H^2(M)}$ (see the proof of Theorem 5.2) we get the continuity of the following maps:

$$
\begin{array}{rcl}
V_3 & \to & H^{\frac{3}{2}}(\partial M) \to L^2(\partial M) \\
\omega & \mapsto & \iota^* \omega \mapsto \iota^* \omega
\end{array}
$$

Now given that $(\iota^* \omega_i, \iota^* \omega')_{L^2(\partial M)} = 0$, we have by the continuity of the map above

$$
(\iota^* \check{\omega}_\epsilon, \iota^* \omega')_{L^2(\partial M)} = \sum_i (\check{\omega}_\epsilon, \omega_i)_V (\iota^* \omega_i, \iota^* \omega')_{L^2(\partial M)} = 0.
$$

Moreover, since for $\alpha_\epsilon \in H_A^p(M)$ and $\omega' \in \check{H}_{N,3}^2(M)$ we have $(\iota^*\alpha_\epsilon, \iota^*\omega')_{L^2(\partial M)} = 0$, this implies that $(\iota^*\omega_\epsilon, \iota^*\omega')_{L^2(\partial M)} = 0$. Letting $\epsilon$ tend to 0, we obtain $\|\iota^*\omega'\|_{L^2(\partial M)}^2 = 0$, hence $\iota^*\omega' = 0$. Since we already have $\nu \rfloor \omega' = 0$, it follows that $\omega'_{|_{\partial M}} = 0$, so $\omega' \in H_0^1(M)$. Finally, we conclude that

$$
\check{H}_{N,3}^2(M) = V_3 \oplus_{\perp(\cdot, \cdot)_V} H_0^1(M) \cap \check{H}_{N,3}^2(M).
$$

Let us now show that the infimum in (77) exists and is a minimum. We know that

$$
\begin{align*}
\check{H}_{N,1}^2(M) &= \left\{ \omega \in H^2(M) \mid \nu \rfloor \omega = 0, \nu \rfloor d\omega = 0 \text{ on } \partial M, \omega \perp_{L^2(\partial M)} H_A^2(M) \right\} \\
&= \check{H}_{N,3}^2(M) \oplus (\check{H}_{N,3}^2(M))^{\perp V}.
\end{align*}
$$

Since $\|\cdot\|_V \sim \|\cdot\|_{H^2(M)}$ on $\check{H}_{N,1}^2(M)$ and we have $\|\cdot\|_{W_3} \le \|\cdot\|_{W_1} \le c\|\cdot\|_{H^{\frac{3}{2}}(M)}$ on $Z_1$ and so on $Z_3$, and the embedding $H^2(M) \hookrightarrow H^{\frac{3}{2}}(M)$ is compact, thus, the map

$$
\begin{array}{rcl}
\check{H}_{N,1}^2(M) \setminus \{0\} & \to & \mathbb{R} \\
\omega & \mapsto & \frac{\|\omega\|_{W_3}^2}{\|\omega\|_V^2}
\end{array}
$$

admits a maximum on $\bar{B}_1^{\check{H}_{N,1}^2(M)}(0)$. That is, there exists $\hat{\omega} \in B_1 \subset \check{H}_{N,1}^2(M)$ such that $\frac{\|\hat{\omega}\|_{W_3}^2}{\|\omega\|_V^2} = \sup\left\{\frac{\|\hat{\omega}\|_{W_3}^2}{\|\omega\|_V^2} \text{ on } B_1\right\}$.

To conclude, consider the following functional: $Q(\omega) := \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2}$ and let $\omega \in \check{H}_{N,3}^2(M)$ such that $\omega = u + v$ where $u \in V_3$ and $v \in H_0^1(M)$. Given that $\iota^* v \in H_0^1(M)$, we have then:

$$
Q(\omega) = \frac{\|\Delta u + \Delta v\|_{L^2(M)}^2}{\|\iota^* u\|_{L^2(\partial M)}^2} = \frac{\|u + v\|_V^2}{\|\iota^* u\|_{L^2(\partial M)}^2} = \frac{\|u\|_V^2 + \|v\|_V^2}{\|\iota^* u\|_{L^2(\partial M)}^2} \ge \frac{\|u\|_V^2}{\|\iota^* u\|_{L^2(\partial M)}^2}.
$$

So we can conclude that

$$
\inf\{Q(\omega) \mid \omega \in V_3\} = l_{1,p} = \inf\{Q(\omega) \mid \omega \in \check{H}_{N,3}^2(M)\}
$$
is attained.

Next, let $\omega \in \check{H}_{N,1}^2(M)$ be such that $\omega = u' + v'$ where $u' \in \check{H}_{N,3}^2(M)$ and $v' \in (\check{H}_{N,3}^2(M))^{\perp v}$. Then,

$$Q(\omega) = \frac{\|\Delta u' + \Delta v'\|_{L^2(M)}^2}{\|\iota^* u' + \iota^* v'\|_{L^2(\partial M)}^2} = \frac{\|\Delta u'\|_{L^2(M)}^2 + \|\Delta v'\|_{L^2(M)}^2}{\|\iota^* u'\|_{L^2(\partial M)}^2 + \|\iota^* v'\|_{L^2(\partial M)}^2},$$

because of:

$$\|\iota^* u' + \iota^* v'\|_{L^2(\partial M)}^2 = \|\iota^* u'\|_{L^2(\partial M)}^2 + \|\iota^* v'\|_{L^2(\partial M)}^2 + 2 \int_{\partial M} \langle \iota^* u', \iota^* v' \rangle d\mu_g.$$

It is enough to show that

$$\int_{\partial M} \langle \iota^* u', \iota^* v' \rangle d\mu_g = 0 \quad \text{for } u' \in V_3,$$

because it is obvious for $u' \in H_0^1(M)$. We decompose $u' \in V_3$ as an orthogonal sum over the eigenforms of $u' = \sum_{i \in I} \alpha_i l_i \omega_i$, where $\omega_i$ is an $l_i$-eigenform. Doing an integration by parts and using the fact that $\iota^*$ is continuous on $V_3$, we get

$$\int_{\partial M} \langle \iota^* u', \iota^* v' \rangle d\mu_g = \sum_{i \in I} \alpha_i \int_{\partial M} \langle l_1 \iota^* \omega_i, \iota^* v' \rangle d\mu_g.$$

For a fixed $i \in I$, we have

$$\begin{aligned} \int_{\partial M} \langle l_i, \iota^* \omega_i, \iota^* v' \rangle d\mu_g &= - \int_{\partial M} \langle \nu_{\lceil} d\Delta \omega_i, \iota^* v' \rangle d\mu_g \\ &= \int_M \underbrace{\langle \Delta \omega_i, \Delta v' \rangle}_{0} d\mu_g + \int_M \underbrace{\langle \Delta^2 \omega_i, v' \rangle}_{0} d\mu_g \\ &\quad + \int_{\partial M} \underbrace{\langle \nu_{\lceil} \Delta \omega_i, \iota^* \delta v' \rangle}_{0} d\mu_g - \int_{\partial M} \langle \iota^* \delta \omega_i, \underbrace{\nu_{\lceil} v' \rangle}_{0} \rangle d\mu_g = 0. \end{aligned}$$

Thus $\int_{\partial M} \langle \iota^* u', \iota^* v' \rangle d\mu_g = 0$. Let us next show that the closure of $\iota^* V_3$ is equal to $\check{L}^2(\partial M) = L^2(\partial M) \cap H_A^{p\perp}(M)$. Hence, by Lemma 5.3, $(\check{H}_{N,3}^2(M))^{\perp \check{H}_{N,1}^2(M)} \subset \{\omega \in \check{H}_{N,1}^2(M), \iota^* \omega = 0\}$ and thus $Q(\omega) \ge Q(u')$. Finally,

$$\inf\{Q(\omega), \omega \in \check{H}_{N,1}^2(M)\} = \inf\{Q(\omega), \omega \in V_3\}.$$

This completes the proof.

<figure><img src="image_8.png" /></figure>

**Theorem 5.7.** For all $k \in \mathbb{N}$, we have the following characterization of the $k^{th}$ positive eigenvalue:

$$l_{k,p} = \inf_{\substack{V_k \subset \check{H}_{N,1}^2(M) \\ \dim(V_k/(H_0^1(M) \cap V_k)) = k}} \sup_{\omega \in V_k \setminus H_0^1(M)} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\iota^* \omega\|_{L^2(\partial M)}^2}. \quad (79)$$

*Proof.* The proof is done in the same way of the one of Theorem 3.26.

<figure><img src="image_9.png" /></figure>

**Theorem 5.8.** Let us define

$$l_{1,p}^{\prime} := \inf \left\{ \frac{\|\nu_{\lceil} d \omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} \mid \omega \in \Omega^p(M), \Delta \omega = 0 \text{ on } M, \nu_{\lceil} \omega = 0 \text{ on } \partial M \text{ and } \omega \perp_{L^2(M)} H_A^p(M) \right\}$$

and

$$\tilde{l}_{1,p} := \inf \left\{ \frac{\|\nu_{\lceil} d \omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} \mid \omega \in \Omega^p(M), \Delta \omega = 0 \text{ on } M, \nu_{\lceil} \omega = 0 \text{ on } \partial M \text{ and } \omega \perp_{L^2(\partial M)} H_A^p(M) \right\}. \quad (80)$$

We have then $l_{1,p}^{\prime} = l_{1,p} = \tilde{l}_{1,p}$.
**Lemma 5.9.** The characterizations $l'_{1,p}$ and $\tilde{l}_{1,p}$ given in Theorem 5.8 are positive minima.

*Proof.* The proof is an adaptation of Proposition 3.20, using the BSN3 adapted boundary conditions. □

*Proof of Theorem 5.8.* We need to show that $l'_{1,p} = l_{1,p}$. Let $\omega$ be an $l_1$-eigenform in $\tilde{Z}_3$. By taking $\|\iota^* \omega\|_{L^2(\partial M)} = 1$ in (76), we obtain $\|\Delta \omega\|_{L^2(M)}^2 = l_1$. Let $\Delta \omega = \hat{\omega}$, then we have the following elliptic problem which is a particular case of (62)

$$ \left\{ \begin{array}{ll} \Delta \hat{\omega} = 0 & \text{on } M \\ \nu \rfloor d\hat{\omega} = -l_1 \iota^* \omega & \text{on } \partial M \\ \nu \rfloor \hat{\omega} = 0 & \text{on } \partial M. \end{array} \right. \qquad (81) $$

By Lemma 3.21 there exists a solution $\hat{\omega} \in \Omega^p(M)$, we denote $\hat{\omega}' = \text{proj}_{\perp L^2(\partial M)}^{H_A^p(M)}(\hat{\omega})$. Using integration by parts (19) since $\omega$ is an $l_1$-eigenform ($\omega \in \tilde{Z}_3$), we obtain

$$ \int_M \langle \hat{\omega}', \Delta \omega \rangle d\mu_g + \int_{\partial M} \langle \nu \rfloor d\hat{\omega}', \iota^* \omega \rangle d\mu_g = 0. $$

Thus,

$$ \int_M \langle \hat{\omega}', \Delta \omega \rangle d\mu_g - l_1 \underbrace{\int_{\partial M} \langle \iota^* \omega, \iota^* \omega \rangle d\mu_g}_{1} = 0. $$

By Cauchy-Shwarz inequality, we have

$$ \|\hat{\omega}'\|_{L^2(M)} \cdot \underbrace{\|\Delta \omega\|_{L^2(M)}}_{l_1^{\frac{1}{2}}} \geq \int_M \langle \hat{\omega}', \Delta \omega \rangle = l_1, $$

$$ \text{so } l_1 \leq \|\hat{\omega}'\|_{L^2(M)}^2. $$

Therefore,

$$ \frac{\|\nu \rfloor d\hat{\omega}'\|_{L^2(M)}^2}{\|\hat{\omega}'\|_{L^2(M)}^2} = l_1^2 \frac{\|\iota^* \omega\|_{L^2(\partial M)}^2}{\|\hat{\omega}'\|_{L^2(M)}^2} \leq l_1^2 \frac{\|\iota^* \omega\|_{L^2(\partial M)}^2}{l_1} = l_1. $$

Since

$$ \frac{\|\nu \rfloor d\hat{\omega}\|_{L^2(\partial M)}^2}{\|\hat{\omega}\|_{L^2(M)}^2} \leq \frac{\|\nu \rfloor d\hat{\omega}'\|_{L^2(\partial M)}^2}{\|\hat{\omega}'\|_{L^2(M)}^2} \leq l_1 $$

we have then

$$ \tilde{l}_1 \leq l_1 $$

and

$$ \inf \left\{ \frac{\|\nu \rfloor d\hat{\omega}\|_{L^2(\partial M)}^2}{\|\hat{\omega}\|_{L^2(M)}^2} \mid \Delta \omega = 0 \text{ on } M, \nu \rfloor \omega = 0, \text{ on } \partial M \text{ and } \omega \perp L^2(M) \right\} \leq l_1. $$

Before concluding the proof, we establish the following lemma:

**Lemma 5.10.** Let $\hat{\omega} = \Delta \omega$ where $\omega$ is an $l_1$-eigenform for BSN3, then $\hat{\omega} \perp_{L^2(M)} H_A^p(M)$.

*Proof.* Let $\hat{\omega} = \Delta \omega$ where $\omega$ is an $l_1$-eigenform for BSN3 and $\alpha \in H_A^p(M)$. From (21), we obtain

$$ \int_M \langle \hat{\omega}, \alpha \rangle d\mu_g = \int_M \langle \Delta \omega, \alpha \rangle d\mu_g = -l_1 \int_{\partial M} \langle \iota^* \omega, \iota^* \alpha \rangle d\mu_g = 0. $$

□
Returning to the proof, conversely, let $\omega \in \Omega^p(M)\setminus\{0\}$ such that $\omega \perp_{L^2(M)} H_A^p(M)$, $\Delta\omega = 0$ on $M$, $\nu\rfloor\omega = 0$ on $\partial M$ and $\|\nu\rfloor d\omega\|_{L^2(M)}^2 = l_1'\|\omega\|_{L^2(M)}^2$ where $l_1'$ is a positive minimum. Let $\hat{\omega}$ be a solution of the problem

$$ \begin{cases} \Delta\hat{\omega} = \omega & \text{on } M \\ \nu\rfloor d\hat{\omega} = 0 & \text{on } \partial M \\ \nu\rfloor\hat{\omega} = 0 & \text{on } \partial M, \end{cases} $$

given by Lemma 3.21.

Again from an integration by parts (19), we have

$$ \|\omega\|_{L^2(M)}^2 = \int_M \langle \Delta\hat{\omega}, \omega \rangle d\mu_g = - \int_{\partial M} \langle \iota^*\hat{\omega}, \nu\rfloor d\omega \rangle d\mu_g, $$

and using the Cauchy-Schwarz inequality, we obtain

$$ \|\omega\|_{L^2(M)}^2 \le \|\iota^*\hat{\omega}\|_{L^2(\partial M)} (l_1')^{\frac{1}{2}} \|\omega\|_{L^2(M)}. $$

Thus,

$$ \|\omega\|_{L^2(M)} \le \|\iota^*\hat{\omega}\|_{L^2(\partial M)} (l_1')^{\frac{1}{2}}, $$

so

$$ l_1 \le \frac{\|\Delta\hat{\omega}\|_{L^2(M)}^2}{\|\iota^*\hat{\omega}\|_{L^2(\partial M)}^2} \le l_1', $$

which implies that $l_1 \le l_1'$, thus we obtain the equality.

<figure><img src="image_10.png" /></figure>

# 6 Spectral inequalities for BSN2 and BSN3

After having stated the three BSN problems and given the characterization of their eigenvalues, we will now present a generalization of the Kuttler-Sigillito inequalities [11, 17] on differential forms.

**Lemma 6.1.** Using the notations defined previously, we have the following inequalities:

$$ \ell_{k,p} \le l_{k,p} \le \mathbf{1}_{k,p}. $$

*Proof.* For all $k \ge 1$, it follows immediately from Theorem 3.26, Theorem 4.5 and Theorem 5.7. $\square$

**Theorem 6.2.** Let $(M^n, g)$ be a compact Riemannian manifold with smooth boundary $\partial M$. The following inequality holds between the first non-zero eigenvalues of the Steklov, Neumann, BSN2 and BSN3 problems on differential forms:

$$ \mu_1\sigma_1 < l_1 \le \mathbf{1}_1. $$
Proof. Using the following variational characterizations (35), (38), and (76), we obtain:

$$
\begin{align*}
\mu_1 \sigma_1 &= \inf_{\substack{\omega \notin H_A^p(M) \\ \nu \downarrow \omega = 0 \\ \nu \downarrow d \omega = 0}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \times \inf_{\substack{\nu \downarrow \omega = 0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \\
&\le \inf_{\substack{\nu \downarrow \omega = 0 \\ \nu \downarrow d \omega = 0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \times \inf_{\substack{\nu \downarrow \omega = 0 \\ \nu \downarrow d \omega = 0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \\
&\le \inf_{\substack{\nu \downarrow \omega = 0 \\ \nu \downarrow d \omega = 0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} = l_1 \le l_1.
\end{align*}
$$

We then have the inequality in the broad sense. To show that it is strict, we prove the following lemma:

**Lemma 6.3.** If $\mu_1\sigma_1 = l_1$, then there exists $\omega$ an $l_1$-eigenform for problem (12) such that $\omega$ is also a $\sigma_1$-eigenform for Steklov problem (30) and $\omega - \text{proj}_{L^2(M)}^{H_A^p(M)}(\omega)$ is a $\mu_1$-eigenform for Neumann problem (29).

*Proof.* Let $\omega \in \check{H}_{N,3}^2(M)$ be an $l_1$-eigenform for BSN3 (12). We know that

$$
\frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \geq \sigma_1
$$

and the quotient

$$
\frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2}
$$

is invariant by translation by an element of $H_A^p(M)$. Therefore, for $\tilde{\omega} = \omega - \text{proj}_{L^2(M)}^{H_A^p(M)}(\omega)$, we have

$$
\frac{\|\Delta \tilde{\omega}\|_{L^2(M)}^2}{\|d\tilde{\omega}\|_{L^2(M)}^2 + \|\delta\tilde{\omega}\|_{L^2(M)}^2} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \geq \mu_1
$$

which implies that

$$
\mu_1 \sigma_1 \le \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \times \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} = l_1 = \mu_1 \sigma_1.
$$

Thus,

$$
\mu_1 = \frac{\|\Delta \tilde{\omega}\|_{L^2(M)}^2}{\|d\tilde{\omega}\|_{L^2(M)}^2 + \|\delta\tilde{\omega}\|_{L^2(M)}^2} = \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}
$$

and

$$
\sigma_1 = \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2}.
$$
Hence, $\tilde{\omega}$ and $\omega$ are solutions of the following problems:

$$ 
\left\{ 
\begin{array}{ll} 
\Delta \tilde{\omega} = \mu_1 \tilde{\omega} & \text{on } M \\ 
\nu \rfloor \tilde{\omega} = 0 & \text{on } \partial M \\ 
\nu \rfloor d \tilde{\omega} = 0 & \text{on } \partial M 
\end{array} 
\right. 
; 
\left\{ 
\begin{array}{ll} 
\Delta^2 \omega = 0 & \text{on } M \\ 
\nu \rfloor \omega = 0 & \text{on } \partial M \\ 
\nu \rfloor d \omega = 0 & \text{on } \partial M \\ 
\nu \rfloor \Delta \omega = 0 & \text{on } \partial M \\ 
\nu \rfloor d \Delta \omega + l_1 t^* \omega = 0 & \text{on } \partial M 
\end{array} 
\right. 
; 
\left\{ 
\begin{array}{ll} 
\Delta \omega = 0 & \text{on } M \\ 
-\nu \rfloor d \omega = \sigma_1 t^* \omega & \text{on } \partial M \\ 
\nu \rfloor \omega = 0 & \text{on } \partial M 
\end{array} 
\right. 
$$

<figure><img src="image_11.png" /></figure>

Back to the proof of Theorem 6.2. Since $\nu \rfloor d \omega = 0$, it follows that $t^* \omega$ is zero due to the conditions of the Steklov problem. Moreover, as $\nu \rfloor \omega = 0$, we get that $\omega|_{\partial M} = 0$, which implies together with $\Delta \omega = 0$ that $\omega = 0$ by [1]. This is a contradiction, because $\omega$ is an eigenform and cannot be zero. Consequently, the inequality is strict. Now using Lemma 6.1 we get $\mu_1 \sigma_1 < l_1 \le l_1$.

<figure><img src="image_12.png" /></figure>

The following theorem is a generalization of Theorem 6.2 that is given in [11, Theorem 1.1] and [17, Inequality I].

**Theorem 6.4.** Let $(M^n, g)$ be a compact Riemannian manifold with a smooth boundary $\partial M$. The following inequalities hold for $k \ge 2$, between the eigenvalues of the Steklov, Neumann, BSN2 and BSN3 problems for differential forms:

$$ \mu_k \sigma_1 \le l_k \le \mathbf{1}_k, \qquad (82) $$

and

$$ \mu_1 \sigma_k \le l_k \le \mathbf{1}_k. \qquad (83) $$

*Proof.* We begin by proving (82). Let $\mathcal{V} = \text{Span}(\omega_1, \dots, \omega_k)$, where the $\omega_i$ are $l_i$-eigenforms, note that $\mathcal{V}$ has dimension $k$. Let $\tilde{\mathcal{V}} = H_A^p(M) \oplus \mathcal{V}$ such that $\dim \tilde{\mathcal{V}} = k + \dim H_A^p(M)$. For all $\omega \in \mathcal{V}$, we have $\omega \perp L^2(M) H_A^p(M)$. Since $\omega$ satisfies the boundary conditions of the Neumann problem, we can use the characterization (36), and deduce that

$$ 
\begin{align*}
\mu_k &\le \sup_{\omega \in \tilde{\mathcal{V}}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \\
&= \sup_{\omega \in \mathcal{V}} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \times \frac{\|\omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \\
&\le l_k \sup_{\omega \in \mathcal{V}} \frac{\|\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2} \\
&= l_k \left( \inf_{\omega \in \mathcal{V}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta \omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \right)^{-1} \\
&\le \frac{l_k}{\sigma_1},
\end{align*}
$$

from which we deduce using Lemma 6.1 that $\mu_k \sigma_1 \le l_k \le \mathbf{1}_k$.
Let us next show (83). Using the same definitions above of $\mathcal{V}$ and $\tilde{\mathcal{V}}$, we have

$$
\begin{align*}
\sigma_k &\le \sup_{\omega \in \mathcal{V}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\omega\|_{L^2(\partial M)}^2} \\
&= \sup_{\substack{\tilde{\omega} \in \tilde{\mathcal{V}} \\ \tilde{\omega}|_{\partial M} \ne 0}} \frac{\|d\tilde{\omega}\|_{L^2(M)}^2 + \|\delta\tilde{\omega}\|_{L^2(M)}^2}{\|t^*\tilde{\omega}\|_{L^2(\partial M)}^2} \\
&\le \sup_{\substack{\omega \in \mathcal{V} \\ \omega|_{\partial M} \ne 0}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|t^*\omega\|_{L^2(\partial M)}^2} \times \frac{\|\Delta\omega\|_{L^2(M)}^2}{\|\Delta\omega\|_{L^2(M)}^2} \\
&\le l_k \sup_{\omega \in \mathcal{V}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|\Delta\omega\|_{L^2(M)}^2} \\
&= l_k \left( \inf_{\omega \in \mathcal{V}} \frac{\|\Delta\omega\|_{L^2(M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \right)^{-1} \\
&\le \frac{l_k}{\mu_1},
\end{align*}
$$

we conclude that $\mu_1\sigma_k \le l_k \le l_k$ using Lemma 6.1. $\square$

**Remark 6.5.** Unlike the case where $k = 1$, the generalized inequalities may not be strict because having

$$l_{k,p} = \sup_{\omega \in \mathcal{V}_k / \mathcal{V}_k \cap H_0^2(M)} \frac{\|\Delta \omega\|_{L^2(M)}^2}{\|t^* \omega\|_{L^2(\partial M)}^2}$$

for $\mathcal{V}_k \subset \tilde{H}_{N,3}^2(M)$ and $\dim(\mathcal{V}_k / \mathcal{V}_k \cap H_0^2) = k$, does not imply that $\mathcal{V} = \text{Span}\{\omega_1, \dots, \omega_k\}$.

The following theorem is a generalization of [17, Inequality II].

**Theorem 6.6.** Let $(M^n, g)$ be a compact Riemannian manifold with smooth boundary $\partial M$. The following inequality holds between the eigenvalues of the Steklov, BSD, BSN2 and BSN3 problems for differential forms:

$$q_1 \sigma_1^2 < l_1 \le l_1. \qquad (84)$$

*Proof.* Using the characterizations (38) and (39) of the classical Steklov problem, we have:

$$
\begin{align*}
\sigma_1^2 \le \inf_{\substack{\nu \rfloor \omega=0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2}{\|t^*\omega\|_{L^2(\partial M)}^2} \times \inf_{\substack{\nu \rfloor \omega=0 \\ \Delta \omega=0}} \frac{\|\nu \rfloor d\omega\|_{L^2(\partial M)}^2}{\|d\omega\|_{L^2(M)}^2 + \|\delta\omega\|_{L^2(M)}^2} \\
\le \inf_{\substack{\nu \rfloor \omega=0 \\ \Delta \omega=0}} \frac{\|\nu \rfloor d\omega\|_{L^2(\partial M)}^2}{\|t^*\omega\|_{L^2(\partial M)}^2}.
\end{align*}
$$

Multiplying by $q_1$ and using $\tilde{l}_1$ given in (80), we obtain:

$$q_1 \sigma_1^2 \le \inf_{\Delta \omega=0} \frac{\|t^* \omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} \times \inf_{\substack{\nu \rfloor \omega=0 \\ \omega \perp L^2(\partial M) H_A^p(M)}} \frac{\|\nu \rfloor d \omega\|_{L^2(\partial M)}^2}{\|t^* \omega\|_{L^2(\partial M)}^2}$$
$$ \leq \inf_{\substack{\nu \lrcorner \omega = 0 \\ \Delta \omega = 0 \\ \omega \perp L^2(\partial M)} \omega \perp L^2(\partial M)} \frac{\|\nu \lrcorner d\omega\|_{L^2(\partial M)}^2}{\|\omega\|_{L^2(M)}^2} \leq \tilde{l}_1 = l_1. $$

Let us next show that the inequality is strict. We first show the following lemma:

**Lemma 6.7.** If $q_1\sigma_1^2 = l_1$, then there exists $\omega \in \Omega^p(M)$ eigenform associated to both $l_1$ and $\sigma_1$.

*Proof.* Let $\omega$ be an $l_1$-eigenform for BSN3, in the same way as in the proof of Lemma 6.3, we show that $\omega$ is also a $\sigma_1$-eigenform so it satisfies the following problems:

$$ \left\{ \begin{array}{ll} \Delta^2 \omega = 0 & \text{on } M \\ \nu \lrcorner \omega = 0 & \text{on } \partial M \\ \nu \lrcorner d\omega = 0 & \text{on } \partial M \\ \nu \lrcorner \Delta \omega = 0 & \text{on } \partial M \\ \nu \lrcorner d\Delta \omega + l\iota^* \omega = 0 & \text{on } \partial M \end{array} \right. ; \left\{ \begin{array}{ll} \Delta \omega = 0 & \text{on } M \\ \nu \lrcorner d\omega = -\sigma_1 \iota^* \omega & \text{on } \partial M \\ \nu \lrcorner \omega = 0 & \text{on } \partial M. \end{array} \right. $$

<figure><img src="image_13.png" /></figure>

Let us finish the proof of Theorem 6.6. In this case, $\nu \lrcorner \omega = 0$ and $\iota^* \omega = 0$, so $\omega|_{\partial M} = 0$. Moreover, since $\Delta \omega = 0$, we have $\omega = 0$ on $M$ according to [1]. This leads to a contradiction, and the inequality is strict. Now, using Lemma 6.1, we get $q_1\sigma_1^2 < l_1 \leq l_1$.

<figure><img src="image_14.png" /></figure>

The following is a generalization of [17, Inequality III].

**Theorem 6.8.** Let $(M^n, g)$ be a compact Riemannian manifold with smooth boundary $\partial M$. The following inequality holds between the eigenvalues of the Dirichlet, Neumann, BSD and BSN3 problems for differential forms:

$$ \mu_1^{-1} < \lambda_1^{-1} + (q_1 l_1)^{-\frac{1}{2}}. \qquad (85) $$

*Proof.* Let $\omega_1 \in \Omega^p(M)$ such that $\nu \lrcorner d\omega_1 = 0$, $\nu \lrcorner \omega_1 = 0$, and $\omega_1 \perp L^2(\partial M)$ $H_A^p(M)$. Let $v \in \Omega^p(M)$ satisfy the following conditions:

$$ \left\{ \begin{array}{ll} \Delta v = \Delta \omega_1 & \text{on } M \\ v|_{\partial M} = 0, & \end{array} \right. $$

and which has a solution due to Theorem 2.6. Let $u \in \Omega^p(M)$ such that $u = \omega_1 - v$, so we have $\Delta u = \Delta \omega_1 - \Delta v = 0$. So we decompose $\omega_1 = u + v$ such that $\Delta u = 0$ and $v|_{\partial M} = 0$. On the one hand, we have

$$ \|u\|_{L^2(M)} \leq (q_1^{-1})^{\frac{1}{2}} \|u\|_{L^2(\partial M)}. $$

Since $v|_{\partial M} = 0$, we get

$$ \|u\|_{L^2(M)} \leq (q_1^{-1})^{\frac{1}{2}} \|\omega_1\|_{L^2(\partial M)}. \qquad (86) $$

On the other hand, by multiplying (32) and (33) we get $\|v\|_{L^2(M)} \leq \lambda_1^{-2} \|\Delta v\|_{L^2(M)}$. Now $\Delta v = \Delta \omega_1$ as $\Delta u = 0$, so that

$$ \|v\|_{L^2(M)} \leq \lambda_1^{-1} \|\Delta \omega_1\|_{L^2(M)}. \qquad (87) $$
Thus, using (86) and (87) we get

$$
\begin{align*}
\|\omega_1\|_{L^2(M)} &\le \|u\|_{L^2(M)} + \|v\|_{L^2(M)} \\
&\le (q_1^{-1})^{\frac{1}{2}} \|\omega_1\|_{L^2(\partial M)} + (\lambda_1^{-1}) \|\Delta \omega_1\|_{L^2(M)} \\
&\le (q_1^{-1})^{\frac{1}{2}} (l_1^{-1})^{\frac{1}{2}} \|\Delta \omega_1\|_{L^2(M)} + \lambda_1^{-1} \|\Delta \omega_1\|_{L^2(M)} \\
&\le (\lambda_1^{-1} + q_1^{-\frac{1}{2}} l_1^{-\frac{1}{2}}) \|\Delta \omega_1\|_{L^2(M)}. \tag{88}
\end{align*}
$$

Now let $\omega$ be a $\mu_1$-eigenform for the Neumann problem (29), so we have $\|\Delta \omega\|_{L^2(M)} = \mu_1 \|\omega\|_{L^2(M)}$.
We decompose $\omega$ as follows, $\omega = \omega_0 + \omega_1$ such that $\omega_0 \in H_A^p(M)$ and $\omega_1 \perp L^2(\partial M)$, $\omega_0$, then
$\Delta \omega = \Delta \omega_1$. We perform an integration by parts (20), we obtain $(\Delta \omega_1, \omega_0)_{L^2(M)} = 0$. As we said
before, $\Delta \omega = \Delta \omega_1$, then $\Delta \omega_1 = \mu_1(\omega_1 + \omega_0)$, so

$$
0 = (\Delta \omega_1, \omega_0)_{L^2(M)} = \mu_1(\omega_1, \omega_0)_{L^2(M)} + \mu_1 \|\omega_0\|_{L^2(M)}^2.
$$

Hence we get $(\omega_1, \omega_0)_{L^2(M)} = -\|\omega_0\|_{L^2(M)}^2$. Using the calculations above, we have

$$
\begin{align*}
\|\omega\|_{L^2(M)} &= \|\omega_0\|_{L^2(M)} + \|\omega_1\|_{L^2(M)} + 2(\omega_0, \omega_1)_{L^2(M)} \\
&= \|\omega_0\|_{L^2(M)} + \|\omega_1\|_{L^2(M)} - 2\|\omega_0\|_{L^2(M)} \\
&= \|\omega_1\|_{L^2(M)} - \|\omega_0\|_{L^2(M)} \\
&\le \|\omega_1\|_{L^2(M)}. \tag{89}
\end{align*}
$$

Going back to (88) and using (89) and that $\Delta \omega = \Delta \omega_1$, we get

$$
\|\omega\|_{L^2(M)} \le \|\omega_1\|_{L^2(M)} \le \|\Delta \omega_1\|_{L^2(M)} [(q_1 l_1)^{-\frac{1}{2}} + \lambda_1^{-1}] = \|\Delta \omega\|_{L^2(M)} [(q_1 l_1)^{-\frac{1}{2}} + \lambda_1^{-1}].
$$

Which implies finally that

$$
\mu_1^{-1} \le \lambda_1^{-1} + (q_1 l_1)^{-\frac{1}{2}}
$$

since $\|\Delta \omega\|_{L^2(M)} = \mu_1 \|\omega\|_{L^2(M)}$. Let us now show that the inequality is strict. Then

$$
\|\omega\|_{L^2(M)}^2 = (\|u\|_{L^2(M)} + \|v\|_{L^2(M)})^2,
$$

so

$$
\|u\|_{L^2(M)}^2 + \|v\|_{L^2(M)}^2 + 2(u, v)_{L^2(M)} = \|u\|_{L^2(M)}^2 + \|v\|_{L^2(M)}^2 + 2\|u\|_{L^2(M)}\|v\|_{L^2(M)},
$$

thus $u$ and $v$ are collinear, either $v = ku$ or $u = kv$, where $k \in \mathbb{R}$. However, $\omega$ is an $\mu_1$-eigenform,
so it satisfies the following conditions:

$$
\begin{cases}
\Delta \omega = \mu_1 \omega & \text{on } M \\
\nu \perp \omega = 0 & \text{on } \partial M \\
\nu \perp d\omega = 0 & \text{on } \partial M,
\end{cases}
$$

thus $k\Delta v = \Delta u = 0$, contradiction. When $u = kv$ there are two cases: the first case is if $k = 0$,
then $u = 0$, so $\omega|_{\partial M} = 0$ and $\omega$ is then associated to $\lambda_1$, which implies $\mu_1 = \lambda_1$, hence $(l_1 q_1)^{-1} = 0$,
a contradiction. The second case is if $k \neq 0$, then $\Delta v = 0$, so $\Delta \omega = 0$, and hence $\omega = 0$, a
contradiction. Therefore, the inequality (85) is strict. $\square$

The following is a generalization of [17, Inequality IV].

Theorem 6.9. Let $(M^n, g)$ be a compact Riemannian manifold with smooth boundary $\partial M$. The following inequality holds between the eigenvalues of the Dirichlet, Neumann, Steklov, and BSD problems for differential forms:

$$
\mu_1^{-1} < \lambda_1^{-1} + (q_1 \sigma_1)^{-1}. \qquad (90)
$$
Proof. Using the inequality (84), we have

$$ \sigma_1 \le l_1^{\frac{1}{2}} q_1^{-\frac{1}{2}} $$

and by multiplying then by $q_1$, we obtain

$$ q_1 \sigma_1 \le l_1^{\frac{1}{2}} q_1^{\frac{1}{2}}. $$

Finally, by inserting the previous inequality into (85), we get

$$ \mu_1^{-1} \le \lambda_1^{-1} + (q_1 \sigma_1)^{-1}. $$

Since the two inequalities (84) and (85) are strict, the inequality (90) is also strict.

□

## A List of all spaces

In this Appendix we will list all the functional spaces we use throughout the paper.

* $H_A^p(M) = \{\omega \in \Omega^p(M) \mid d\omega = 0 \text{ on } M, \delta\omega = 0 \text{ on } M, \nu\rfloor\omega = 0 \text{ on } \partial M\}$.

* $L^k(M) = \{\omega \in \Gamma(E) \mid \int_M |\omega|^k d\mu_g < +\infty\} / \{\omega \in \Gamma(E) \mid \omega = 0 \text{ a.e.}\}$.

* $\check{L}^2(M) := \{\omega \in L^2(M) \mid \forall \alpha \in H_A^p(M), \int_M \langle \alpha, \omega \rangle d\mu_g = 0\}$.

* $W^{m,k}(M) = \{\omega \in L^k(M) \mid \forall 1 \le j \le m, \nabla^j \omega \in L^k(M; E^{(j)})\}$, where $E^{(j)} = (TM)^{\otimes j} \otimes E$.

* $H^m(M) = W^{m,2}(M)$.

* $H_0^1(M) = \{\omega \in H^1(M) \mid \omega|_{\partial M} = 0\}$.

* $H_0^2(M) = \{\omega \in H^2(M) \mid \omega|_{\partial M} = 0 \text{ and } (\nabla\omega)|_{\partial M} = 0\}$.

* $M_{A,v}^+ := \{\text{bounded solutions } y = y(t) \text{ on } \mathbb{R}_+ \text{ of the ODE } \sigma_A(-iv + \partial_t v^b)y = 0\}$.

* $\Xi^s(M) := \{u \mid \rho^j \nabla^j u \in L^2(M), \forall 0 \le j \le s\}$, for $s = 0, 1, \dots$

* $\mathcal{N} := \{u \in \Gamma(E) \mid Au = 0, B_1u = 0, \dots, B_ku = 0\}$.

* $\mathcal{N}^* := \{u \in \Gamma(E) \mid A^*u = 0, C_1u = 0, \dots, C_ku = 0\}$.

* $A := \{\omega \in H^2(M) \mid \nu\rfloor\omega = 0, \Delta\omega = 0\}$.

* $H_N^1(M) := \{\omega \in H^1(M) \mid \nu\rfloor\omega = 0\}$.

* $\check{H}_N^1(M) := \{\omega \in H^1(M) \mid \nu\rfloor\omega = 0, \omega \perp L^2(\partial M) H_A^p(M)\}$.

* $H_{N,1}^2(M) := \{\omega \in H^2(M) \mid \nu\rfloor d\omega = 0 \text{ and } \nu\rfloor\omega = 0 \text{ on } \partial M\}$.

* $\check{H}_{N,1}^2(M) := \{\omega \in H_{N,1}^2(M) \mid \omega \perp L^2(\partial M) H_A^p(M)\}$.

* $H_{N,2}^2(M) := \{\omega \in H^2(M) \mid \nu\rfloor d\omega = 0, \nu\rfloor\omega = 0 \text{ and } \iota^*\delta\omega = 0 \text{ on } \partial M\}$.

* $\check{H}_{N,2}^2(M) := \{\omega \in H^2(M) \mid \nu\rfloor d\omega = 0, \nu\rfloor\omega = 0, \iota^*\delta\omega = 0 \text{ on } \partial M \text{ and } \omega \perp L^2(\partial M) H_A^p(M)\}$.

* $H_{N,3}^2(M) := \{\omega \in H^2(M) \mid \nu\rfloor\omega = 0, \nu\rfloor d\omega = 0, \text{ and } \nu\rfloor\Delta\omega = 0 \text{ on } \partial M\}$.
• $\tilde{H}_{N,3}^2(M) := \{\omega \in H^2(M) \mid \nu\rfloor d\omega = 0, \nu\rfloor \omega = 0, \nu\rfloor \Delta \omega = 0 \text{ on } \partial M \text{ and } \omega \perp L^2(\partial M) H_A^p(M)\}$.

• $Z_1 := \{\omega \in \Omega^p(M) \cap H_{N,1}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M\} \subset H_{N,1}^2(M)$.

• $\check{Z}_1 := \{\omega \in Z_1 \mid \omega \perp L^2(\partial M) H_A^p(M)\} = Z_1 \cap \check{H}_{N,1}^2(M)$.

• $Z_2 := \{\omega \in \Omega^p(M) \cap H_{N,2}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M\} \subset Z_1 \subset H_{N,1}^2(M)$.

• $\check{Z}_2 := \{\omega \in Z_2 \mid \omega \perp L^2(\partial M) H_A^p(M)\} = Z_2 \cap \check{H}_{N,1}^2(M) = Z_2 \cap \check{Z}_1$.

• $Z_3 := \{\omega \in \Omega^p(M) \cap H_{N,3}^2(M) \mid \Delta^2 \omega = 0 \text{ on } M\}$.

• $\check{Z}_3 := \{\omega \in Z_3 \mid \omega \perp L^2(\partial M) H_A^p(M)\}$.

• $V_1 = \{\omega \in \check{H}_{N,1}^2(M) \mid \Delta^2 \omega = 0 \text{ weakly in } M\}$.

• $\mathcal{H} := \{\omega \in \Omega^p(M) \mid \nu\rfloor \omega = 0, \nu\rfloor d\omega = 0, \iota^* \omega = 0 \text{ and } \iota^* \delta \omega = 0 \text{ on } \partial M\}$.

• $\mathcal{S} = \{(\omega_1, \omega_2) \in L^2(\partial M) \oplus H_A^p(M)^{\perp L^2(\partial M)} \mid \|\omega_1\|_{L^2(\partial M)}^2 + \|\omega_2\|_{L^2(\partial M)}^2 = 1\}$.

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