1 Introduction

Let (M, g) be a smooth compact connected Riemannian manifold of dimension \(n \ge 2\) with smooth boundary \(\Sigma \). The Steklov problem on (M, g) consists of finding the real numbers \(\sigma \) and the harmonic functions \(f: M \longrightarrow {\mathbb {R}}\) such that \(\partial _\nu f=\sigma f\) on \(\Sigma \), where \(\nu \) denotes the outward normal on \(\Sigma \). Such a \(\sigma \) is called a Steklov eigenvalue of (M, g). It is well known that the Steklov spectrum forms a discrete sequence \(0= \sigma _0(M,g) < \sigma _1(M,g) \le \sigma _2(M,g) \le \cdots \nearrow \infty \). Each eigenvalue is repeated with its multiplicity, which is finite. If the context is clear, then we simply write \(\sigma _k(M)\) for \(\sigma _k(M,g)\).

It is known [3, Thm. 1.1] that for any connected compact manifold (M, g) of dimension \(n \ge 3\), there exists a family \((g_\varepsilon )\) of Riemannian metrics conformal to g which coincide with g on the boundary of M, such that

$$\begin{aligned} \sigma _1(M, g_\varepsilon ) \underset{\varepsilon \rightarrow 0}{\longrightarrow }\ \infty . \end{aligned}$$

Therefore, to obtain upper bounds for the Steklov eigenvalues, it is necessary to study manifolds that satisfy certain additional constraints. We refer to [6] for an overview of the current state-of-the-art on geometric upper bounds for the Steklov eigenvalues.

Recently, authors investigated the Steklov problem on manifolds of revolution [9, 10, 12, 13]. A natural constraint for the manifolds is that they are (hyper)surfaces of revolution in Euclidean space. Some work has already been done on these kinds of manifolds, see for example [4, 5]. We refer to [4, Sect. 3.1] for a review about what these manifolds are, and consider a particular case in this paper that we define below (see Definition 1).

This work led to the discovery of lower and upper bounds for the Steklov eigenvalues of a hypersurface of revolution. We begin by recalling some recent results.

We first consider results for hypersurfaces of revolution with one boundary component that is isometric to \({\mathbb {S}}^{n-1}\). In dimension \(n=2\), it is proved in [4, Prop. 1.10] that each surface of revolution \(M \subset {\mathbb {R}}^3\) with boundary \({\mathbb {S}}^1 \subset {\mathbb {R}}^2 \times \{0\}\) is Steklov isospectral to the unit disk. In dimension \(n \ge 3\), many bounds were given. It is proved that each hypersurface of revolution \(M \subset {\mathbb {R}}^{n+1}\) with one boundary component isometric to \({\mathbb {S}}^{n-1}\) satisfies \(\sigma _k(M) \ge \sigma _k({\mathbb {B}}^n)\), where \({\mathbb {B}}^n\) is the Euclidean ball and equality holds if and only if \(M= {\mathbb {B}}^n \times \{0\}\), see [4, Thm. 1.8]. In [5, Thm. 1], the authors show the following upper bound: if \(M \subset {\mathbb {R}}^{n+1}\) is a hypersurface of revolution with one boundary component isometric to \({\mathbb {S}}^{n-1}\), then for each \(k \ge 1\), we have

$$\begin{aligned} \sigma _{(k)}(M) < k+n-2, \end{aligned}$$

where \(\sigma _{(k)}(M)\) is the kth distinct Steklov eigenvalue of M. Although there exists no equality case within the collection of hypersurfaces of revolution, this upper bound is sharp. Indeed, for each \(\varepsilon >0\) and each \(k\ge 1\), there exists a hypersurface of revolution \(M_\varepsilon \) such that \(\sigma _{(k)}(M_\varepsilon ) > k+n-2-\varepsilon \).

These results concern hypersurfaces of revolution that have one boundary component isometric to \({\mathbb {S}}^{n-1}\). Therefore, the goal of this paper is to investigate the Steklov problem on a hypersurface of revolution with two boundary components. As was already done in [4] and in [5], we will consider hypersurfaces with boundary components isometric to \({\mathbb {S}}^{n-1}\). We begin by defining the context.

Definition 1

An n-dimensional compact hypersurface of revolution (M, g) in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) is the warped product \(M=[0, L] \times {\mathbb {S}}^{n-1}\) endowed with the Riemannian metric

$$\begin{aligned} g(r,p) = dr^2 + h^2(r)g_0(p), \end{aligned}$$

where \((r,p) \in [0, L] \times {\mathbb {S}}^{n-1}\), \(g_0\) is the canonical metric of the \((n-1)\)-sphere of radius one and \(h: [0, L] \longrightarrow {\mathbb {R}}_+^*\) is a smooth function which satisfies:

  1. (1)

    \(|h'(r)| \le 1\) for all \(r \in [0, L]\);

  2. (2)

    \(h(0)=h(L)=1\).

Assumption (1) comes from the fact that (M, g) is a hypersurface in Euclidean space \({\mathbb {R}}^{n+1}\), see [4, Sect. 3.1] for more details. Assumption (2) implies that each component of the boundary is isometric to \({\mathbb {S}}^{n-1}\), as commented in Fig. 1.

We now make some remarks on the terminology used throughout this paper. If \(M = [0, L] \times {\mathbb {S}}^{n-1}\) and \(h: [0, L] \longrightarrow {\mathbb {R}}_+^*\) satisfies the properties above, we say that M is a hypersurface of revolution, we say that \(g(r, p) = dr^2 + h^2(r)g_0(p)\) is a metric of revolution on M induced by h and we call the number L the meridian length of M.

Fig. 1
figure 1

Since \(h(0)=h(L)=1\), the boundary of M consists of two copies of \({\mathbb {S}}^{n-1}\)

Full size image

Some lower bounds have already been obtained is this case. Indeed, [4, Thm. 1.11] states that if \(M \subset {\mathbb {R}}^{n+1}\), \(n \ge 3\), is a hypersurface of revolution (in the sense of Definition 1), and \(L>2\) is the meridian length of M, then for each \(k \ge 1\),

$$\begin{aligned} \sigma _k(M) \ge \sigma _k({\mathbb {B}}^n \sqcup {\mathbb {B}}^n). \end{aligned}$$

Moreover, this inequality is sharp. In the case \(0 < L \le 2\), a lower bound is also obtained:

$$\begin{aligned} \sigma _k(M) \ge \left( 1- \frac{L}{2}\right) ^{n-1} \sigma _k(C_L, dr^2 + g_0), \end{aligned}$$

However, this inequality does not appear to be sharp.

In this paper, we will look for upper bounds for the Steklov eigenvalues of hypersurfaces of revolution. First, we recall that there exists a bound \(B_n^k(L)\) such that for all metrics of revolution g on M, we have \(\sigma _k(M, g) < B_n^k(L)\). Indeed, Proposition 3.3 of [4] states that if \(M = [0, L] \times {\mathbb {S}}^{n-1}\) is a hypersurface of revolution, then we have

$$\begin{aligned} \sigma _k(M) \le \left( 1+ \frac{L}{2}\right) ^{n-1} \sigma _k(C_L, dr^2 + g_0). \end{aligned}$$

As such, a natural question is the following:

figure a

Our investigations show that the answer is negative. Indeed, a sharp upper bound \(B_n^k(L)\) exists, but no metric of revolution on \(M=[0,L]\times {\mathbb {S}}^{n-1}\) achieves the equality case. However, there exists a non-smooth metric \(g^*\), that we will call a degenerated maximizing metric, which maximizes the kth Steklov eigenvalue, for each \(k \in {\mathbb {N}}\). This metric is non-smooth, therefore \(g^*\) is not a metric of revolution on M in the sense of Definition 1. Endowed with this metric, \((M,g^*)\) can be seen as two annuli glued together; we provide more information about this degenerated maximizing metric \(g^*\) and the geometric representation of \((M, g^*)\) in Sect. 3.

We state our first result:

Theorem 2

Let \((M=[0,L] \times {\mathbb {S}}^{n-1}, g_1)\) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) and meridian length L. We suppose \(n \ge 3\). Then there exists a metric of revolution \(g_2\) on M such that for each \(k \ge 1\),

$$\begin{aligned} \sigma _k(M,g_1) < \sigma _k(M, g_2). \end{aligned}$$

This result implies that among all metrics of revolution on M, none maximizes the kth non zero Steklov eigenvalue. Nevertheless, given any metric of revolution \(g_1\) on M, we can iterate Theorem 2 to generate a sequence of metrics \((g_i)_{i=1}^\infty \) on M. This sequence converges to a unique non-smooth metric \(g^*\) on M, which is quite simple (see Sect. 3) and which maximizes the kth Steklov eigenvalue. That is why we call \(g^*\) the degenerated maximizing metric. Hence, as we search for the optimal bounds \(B_n^k(L)\), we must use information contained in \(g^*\).

We start by studying the case \(k=1\). We fix \(n \ge 3\) and \(L >0\) and search for a sharp upper bound \(B_n(L)\) for \(\sigma _1(M,g)\). In this case, we are able to calculate an expression for \(B_n(L)\):

Theorem 3

Let \((M= [0, L] \times {\mathbb {S}}^{n-1}, g)\) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) and dimension \(n \ge 3\). Then the first non trivial Steklov eigenvalue \(\sigma _1(M,g)\) is bounded above, by a bound that depends only on the dimension n and the meridian length L of M:

$$\begin{aligned} \sigma _1(M,g) < B_n(L) := \min \left\{ \frac{(n-2)\left( 1+L/2 \right) ^{n-2}}{\left( 1+L/2 \right) ^{n-2}-1}, \frac{(n-1)\left( \left( 1+L/2\right) ^n -1 \right) }{\left( 1+ L/2 \right) ^n +n-1}\right\} . \end{aligned}$$

Moreover, this bound is sharp: for each \(\varepsilon >0\), there exists a metric of revolution \(g_\varepsilon \) on M such that \(\sigma _1(M, g_\varepsilon ) >B_n(L)- \varepsilon \).

We have the following asymptotic behaviour:

$$\begin{aligned} B_n(L)&\underset{L \rightarrow \infty }{\longrightarrow }\ n-2 \\ B_n(L)&\underset{L \rightarrow 0}{\longrightarrow }\ 0, \end{aligned}$$

see Fig. 4.

We also study the function \(L \longmapsto B_n(L)\). This allows us to find a sharp upper bound \(B_n\) such that for all meridian lengths \(L>0\) and metrics of revolution g on M, we have \(\sigma _1(M,g) < B_n\):

Corollary 4

Let \(n \ge 3\). Then there exists a bound \(B_n < \infty \) such that for all hypersurfaces of revolution (M, g) in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\), we have

$$\begin{aligned} \sigma _1(M, g) < B_n := \frac{(n-2)\left( 1+\frac{L_1}{2}\right) ^{n-2}}{\left( 1+\frac{L_1}{2}\right) ^{n-2}-1}, \end{aligned}$$

where \(L_1\) is the unique real positive solution of the equation

$$\begin{aligned} \left( 1+L/2\right) ^{2n-2}-(n-1)\left( 1+L/2\right) ^n-(n-1)^2\left( 1+L/2\right) ^{n-2}+n-1=0. \end{aligned}$$

Moreover, this bound is sharp: for each \(\varepsilon >0\), there exists a hypersurface of revolution with two boundary components each isometric to a unit sphere \((M_\varepsilon , g_\varepsilon )\) such that \(\sigma _1(M_\varepsilon , g_\varepsilon ) > B_n - \varepsilon \).

We say that \(L_1\) is a critical length associated with \(k=1\), see Definition 8.

Proposition 5

Let \(n \ge 3\), and let \(L_1 = L_1(n)\) be the critical length associated with \(k =1\). Then we have:

$$\begin{aligned} \lim _{n\rightarrow \infty } L_1(n) = 0 \quad \text{ and } \quad \lim _{n\rightarrow \infty } B_n = \infty . \end{aligned}$$

Note that the behaviour of \(L_1\) is surprising since we know that when n is fixed, then \(L \ll 1\) implies \(\sigma _1(M,g) \ll 1\). Indeed, by [4, Prop. 3.3], we have

$$\begin{aligned} \sigma _1(M) \le \left( 1+ \frac{L}{2} \right) ^{n-1} \sigma _1(C_L) \underset{L \rightarrow 0}{\longrightarrow }\ 0. \end{aligned}$$

Now that we have provided information about sharp upper bounds for \(\sigma _1(M, g)\), it is natural to wonder what kind of stability properties the hypersurfaces of revolution possess. A first interesting question is the following:

figure b

The answer to this question is positive. Indeed we will prove that if L is not close to \(L_1\), then \(\sigma _1(M, g)\) is not close to \(B_n\). Additionally, given the information that \(\sigma _1(M, g)\) is \(\delta \)-close to \(B_n\), we will show that the distance between L and \(L_1\) is less than \(\delta \), up to a constant of proportionality which depends only on the dimension n.

Theorem 6

Let \(M=[0, L] \times {\mathbb {S}}^{n-1}\), with \(L >0\) and \(n \ge 3\). We suppose \(L \ne L_1\). Then there exists a constant \(C(n, L) >0\) such that for all metrics of revolution g on M, we have

$$\begin{aligned} B_n - \sigma _1(M, g) \ge C(n, L). \end{aligned}$$

Moreover, there exists a constant \(C(n) > 0\) such that for all \(0< \delta < \frac{B_n - (n-2)}{2}\), we have

$$\begin{aligned} |B_n - \sigma _1(M, g) |< \delta \implies |L_1 -L| < C(n) \cdot \delta . \end{aligned}$$

We also consider the following question about stability properties:

figure c

We prove that if g is not close to \(g^*\), then \(\sigma _1(M,g)\) is not close to \(B_n(L)\).

For this purpose, given \(m \in [1, 1+L/2)\), we define

$$\begin{aligned} {\mathcal {M}}_m :=&\{\text{ metrics } \text{ of } \text{ revolution } g \text{ on } M \\&\text{ induced } \text{ by } \text{ a } \text{ function } h \text{ such } \text{ that } \max _{r \in [0, L]} \{h(r)\} \le m \}. \end{aligned}$$

The collection \({\mathcal {M}}_m\) can be thought of the set of all metrics of revolution that are not close to the degenerated maximizing metric \(g^*\), where the qualitative appreciation of the word "close" is given by the parameter m. The larger m is, the closer to \(g^*\) the metrics in \({\mathcal {M}}_m\) can be.

We get the following result:

Theorem 7

Let \((M=[0, L] \times {\mathbb {S}}^{n-1}, g)\) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) and dimension \(n \ge 3\). Let \(m \in [1, 1+L/2)\) and \({\mathcal {M}}_m\) as above. Then there exists a constant \(C(n, L, m) >0\) such that for all \(g \in {\mathcal {M}}_m\), we have

$$\begin{aligned} B_n(L)- \sigma _1(M, g) \ge C(n, L, m). \end{aligned}$$

These results solve the case \(k=1\). Therefore, it would be interesting to find the same kind of results for any \(k \ge 1\). After having calculated sharp upper bounds for some higher values of k in Sects. 6.1 and 6.2, we will see that in order to get an expression for \(B_n^k(L)\), we need to distinguish between many cases. As such, giving a general formula for \(B_n^k(L)\) or \(B_n^k:= \sup _{L \in {\mathbb {R}}_+^*} \{B_n^k(L)\}\) via this method seems difficult. We discuss this in Remark 20.

Definition 8

We say that \(L_k \in {\mathbb {R}}_+^*\) is a finite critical length associated with k if we have \(B_n^k = B_n^k(L_k)\). We say that k has a critical length at infinity if it satisfies \(B_n^k= \lim _{L\rightarrow \infty } B_n^k(L)\).

These lengths are critical in the following sense: if \(L_k \in {\mathbb {R}}_+^*\) is a finite critical length for a certain \(k \in {\mathbb {N}}\) and if we write \(g^*\) the degenerated maximizing metric on \(M_k=[0, L_k]\times {\mathbb {S}}^{n-1}\), then

$$\begin{aligned} B_n^k= \sigma _k(M_k,g^*). \end{aligned}$$

Given \(n \ge 3\), there exist some k which have a finite critical length associated with them. Indeed, thanks to Corollary 4, we know that \(k=1\) has this property. Moreover, we know that there exist some k which have a critical length at infinity, see Sect. 6.1.

Since we want to study upper bounds for the Steklov eigenvalues, it is then natural to ask what qualitative and quantitative information we can provide about these critical lengths.

We get the following result:

Theorem 9

Let \(n \ge 3\). Then there exist infinitely many \(k \in {\mathbb {N}}\) which have a finite critical length associated with them. Moreover, if we call \((k_i)_{i=1}^\infty \subset {\mathbb {N}}\) the increasing sequence of such k and if we call \((L_i)_{i=1}^\infty \) the associated sequence of finite critical lengths, then we have

$$\begin{aligned} \lim _{i\rightarrow \infty }L_i=0. \end{aligned}$$

The existence of finite critical lengths is something surprising when we compare with what happens in the case of hypersurfaces of revolution with one boundary component. Indeed, using our vocabulary, we can state that in the case of hypersurfaces of revolution with one boundary component, each \(k \in {\mathbb {N}}\) has a critical length at infinity, see [5, Prop. 7]. Nevertheless, in our case, Theorem 9 guarantees that there exist infinitely many \(k \in {\mathbb {N}}\) which have a finite critical length associated with them. Moreover, we will show in Sect. 6.1 that there exist some k which have a critical length at infinity. However, we do not know if there are infinitely many of them. This consideration leads to the following open question (Question 22):

figure d

Plan of the paper. In Sect. 2, we recall the variational characterizations of the Steklov eigenvalues before giving the expression of eigenfunctions on hypersurfaces of revolution, and we introduce the notion of mixed Steklov–Dirichlet and Steklov–Neumann problems and state some propositions about them. We will then have enough information to prove Theorem 2 in Sect. 3. This will allow us to prove Theorem 3, Corollary 4 and Proposition 5 in Sect. 4. Then we prove the stability properties of hypersurfaces of revolution, i.e Theorem 6 and Theorem 7 in Sect. 5. We continue by performing some calculation for sharp upper bounds for higher eigenvalues in Sect. 6. We conclude by proving Theorem 9 in Sect. 7.

2 Variational characterization of the Steklov eigenvalues and mixed problems

We state some general facts about Steklov eigenfunctions and define the mixed Steklov–Dirichlet and Steklov–Neumann problems.

2.1 Variational characterization of the Steklov eigenvalues

Let (M, g) be a Riemannian manifold with smooth boundary \(\Sigma \). Then we can characterize the kth Steklov eigenvalue of M by the following formula:

$$\begin{aligned} \sigma _k(M, g) = \min \left\{ R_g(f) : f \in H^1(M), \; f \perp _\Sigma f_0, f_1, \ldots , f_{k-1} \right\} , \end{aligned}$$
(1)

where

$$\begin{aligned} R_g(f) = \frac{\int _M |\nabla f|^2 dV_g}{\int _\Sigma |f|^2 dV_\Sigma } \end{aligned}$$

is called the Rayleigh quotient and

$$\begin{aligned} f \perp _\Sigma f_i \iff \int _\Sigma f f_i dV_\Sigma =0. \end{aligned}$$

Another way to characterize the kth eigenvalue of M is given by the Min-Max principle:

$$\begin{aligned} \sigma _k(M,g) = \min _{E \in {\mathcal {H}}_{k+1}(M)} \max _{0 \ne f \in E} R_g(f), \end{aligned}$$
(2)

where \({\mathcal {H}}_{k+1}\) is the set of all \((k+1)\)-dimensional subspaces in the Sobolev space \(H^1(M)\).

We state now a proposition that provides us with information about the expression of the Steklov eigenfunctions of a hypersurface of revolution.

We denote by \(0 =\lambda _0 < \lambda _1 \le \lambda _2 \le \cdots \nearrow \infty \) the spectrum of the Laplacian on \(({\mathbb {S}}^{n-1}, g_0)\) and we consider \((S_j)_{j=0}^\infty \) an orthonormal basis of eigenfunctions associated to \((\lambda _j)_{j=0}^\infty \).

Proposition 10

Let (M, g) be a hypersurface of revolution as in Definition 1. Then each eigenfunction on M can be written as \(f_k(r, p) = u_l(r)S_j(p)\), where \(u_l\) is a smooth function on [0, L].

This property is well known for warped product manifolds (and thus for our case of hypersurfaces of revolution) and it is used often, see for example [7, Remark 1.1], [8, Lemma 3], [11, Prop. 3.16] or [12, Prop. 9].

2.2 Mixed problems and their variational characterizations

Let \((N, \partial N)\) be a smooth compact connected Riemannian manifold and \(A \subset N\) be a domain which satisfies \(\partial N \subset \partial A\). We suppose that \(\partial A\) is smooth and we call \(\partial _{int}A\) the intersection of \(\partial A\) with the interior of N.

Definition 11

The Steklov–Dirichlet problem on A is the eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta f =0&{}\quad \text{ in } A \\ \partial _\nu f = \sigma f &{}\quad \text{ on } \partial N \\ f =0 &{}\quad \text{ on } \partial _{int}A. \end{array} \right. \end{aligned}$$

It is well known that this mixed problem possesses solutions that form a discrete sequence

$$\begin{aligned} 0 < \sigma _0^D(A) \le \sigma _1^D(A) \le \cdots \nearrow \infty . \end{aligned}$$

The variational characterization of the kth Steklov–Dirichlet eigenvalue is the following:

$$\begin{aligned} \sigma _k^D(A) = \min _{E \in {\mathcal {H}}_{k+1,0}(A)} \max _{0 \ne f \in E} \frac{\int _A |\nabla f|^2 dV_A}{\int _\Sigma |f|^2 dV_\Sigma }, \end{aligned}$$

where \({\mathcal {H}}_{k+1, 0}\) is the set of all \((k+1)\)-dimensional subspaces in the Sobolev space

$$\begin{aligned} H_0^1(A)= \{f \in H^1(A) : f=0 \text{ on } \partial _{int}A \}. \end{aligned}$$

Definition 12

The Steklov–Neumann problem on A is the eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta f =0&{}\quad \text{ in } A \\ \partial _\nu f = \sigma f &{}\quad \text{ on } \partial N \\ \partial _\nu f = 0 &{}\quad \text{ on } \partial _{int}A. \end{array} \right. \end{aligned}$$

It is well known that this mixed problem possesses solutions that form a discrete sequence

$$\begin{aligned} 0 = \sigma _0^N(A) \le \sigma _1^N(A) \le \cdots \nearrow \infty . \end{aligned}$$

The variational characterization of the kth Steklov–Neumann eigenvalue is the following:

$$\begin{aligned} \sigma _k^N(A) = \min _{E \in {\mathcal {H}}_{k+1}(A)} \max _{0 \ne f \in E} \frac{\int _A |\nabla f|^2 dV_A}{\int _\Sigma |f|^2 dV_\Sigma }, \end{aligned}$$

where \({\mathcal {H}}_{k+1}\) is the set of all \((k+1)\)-dimensional subspaces in the Sobolev space \(H^1(A)\).

2.3 Mixed problems on annular domains

Let \({\mathbb {B}}_1\) and \({\mathbb {B}}_R\) be the balls in \({\mathbb {R}}^n\), \(n \ge 3\), with radius 1 and \(R >1\) respectively centered at the origin. The annulus \(A_R\) is defined as follows: \(A_R = {\mathbb {B}}_R \backslash {\overline{{\mathbb {B}}}}_1\). We say that this annulus is of inner radius 1 and outer radius R. This particular kind of domain shall be useful in this paper.

For such domains, it is possible to compute \(\sigma _{(k)}^D(A_R)\) explicitly, which is the (k)th eigenvalue of the Steklov–Dirichlet problem on \(A_R\), counted without multiplicity.

We state here Proposition 4 of [5]:

Proposition 13

For \(A_R\) as above, consider the Steklov–Dirichlet problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta f = 0 &{}\quad \text{ in } A_R \\ \partial _\nu f = \sigma f &{}\quad \text{ on } \partial {\mathbb {B}}_1 \\ f = 0 &{} \quad \text{ on } \partial {\mathbb {B}}_R. \end{array} \right. \end{aligned}$$

Then, for \(k \ge 0\), the (k)th eigenvalue (counted without multiplicity) of this problem is

$$\begin{aligned} \sigma _{(k)}^D(A_R) = \frac{(k+n-2)R^{2k+n-2}+k}{R^{2k+n-2}-1}. \end{aligned}$$

By [5, Prop. 4], it is possible to get the expression of the eigenfunctions of the Steklov–Dirichlet problem on an annular domain.

Lemma 14

Each eigenfunction \(\varphi _l\) of the Steklov–Dirichlet problem on the annulus \(A_{R}\) can be expressed as \(\varphi _l(r,p)=\alpha _l(r)S_l(p)\), where \(S_l\) is an eigenfunction for the \(l^{th}\) harmonic of the sphere \({\mathbb {S}}^{n-1}\).

It is possible to compute \(\sigma _{(k)}^N(A_R)\) explicitly, which is the (k)th eigenvalue of the Steklov–Neumann problem on \(A_R\), counted without multiplicity.

We state now Proposition 5 of [5]:

Proposition 15

For \(A_R\) as above, consider the Steklov–Neumann problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta f = 0 &{}\quad \text{ in } A_R \\ \partial _\nu f = \sigma f &{}\quad \text{ on } \partial {\mathbb {B}}_1 \\ \partial _\nu f = 0 &{}\quad \text{ on } \partial {\mathbb {B}}_R. \end{array} \right. \end{aligned}$$

Then, for \(k \ge 0\), the (k)th eigenvalue (counted without multiplicity) of this problem is

$$\begin{aligned} \sigma _{(k)}^N(A_R) = k \frac{(k+n-2)(R^{2k+n-2}-1)}{kR^{2k+n-2}+k+n-2}. \end{aligned}$$

In the same manner as before, we have the following expression for the Steklov–Neumann eigenvalues, see [5, Prop. 5].

Lemma 16

Each eigenfunction \(\phi _l\) of the Steklov–Neumann problem on the annulus \(A_{R}\) can be expressed as \(\phi _l(r,p)=\beta _l(r)S_l(p)\), where \(S_l\) is an eigenfunction for the \(l^{th}\) harmonic of the sphere \({\mathbb {S}}^{n-1}\).

3 The degenerated maximizing metric

A particular case of hypersurfaces of revolution is the following: let \(M=[0, L] \times {\mathbb {S}}^{n-1}\) be endowed with a metric of revolution \(g(r,p)=dr^2+h^2(r)g_0(p)\). Let us suppose that there exists \(\varepsilon >0\) such that \(h(r) = 1+r\) on \([0, \varepsilon ]\). Let us consider the connected component of the boundary \({\mathbb {S}}_0\) associated with h(0). Then the \(\varepsilon \)-neighborhood of \({\mathbb {S}}_0\) is an annulus with inner radius 1 and outer radius \(1+\varepsilon \) (Fig. 2).

Fig. 2
figure 2

On \([0, \varepsilon ]\), we have \(h(r)=1+r\) and on \([L-\varepsilon , L]\), we have \(h(r) = -r+L+1\). This implies that the \(\varepsilon \)-neighborhood of the boundary consists of two disjoint copies of an annulus with inner radius 1 and outer radius \(1+\varepsilon \)

Full size image

This particular case is the key idea that we use to prove Theorem 2. We prove it now.

Proof

We write \(g_1(r,p)=dr^2+h_1^2(r)g_0(p)\). Because \(h_1\) is smooth and \(|h_1'| \le 1\), we have \(h_1(r) < 1+\frac{L}{2}\) for all \(r \in [0, L]\). Since \(h_1\) is continuous and [0, L] is compact, \(h_1\) reaches its maximum on [0, L]. We call

$$\begin{aligned} m := \max _{r \in [0, L]} \{h_1(r)\}. \end{aligned}$$

Notice that \(1 \le m < 1+\frac{L}{2}\).

We define a smooth function \(h_2: [0, L] \longrightarrow {\mathbb {R}}\) by

$$\begin{aligned} h_2(r)= \left\{ \begin{array}{lll} 1+r &{}\quad \text{ if } &{} 0 \le r \le m-1 \\ 1+L-r &{}\quad \text{ if } &{} L-m+1 \le r \le L. \end{array} \right. \end{aligned}$$

For \(r\in (m-1, L-m+1)\), we only require that \(h_2(r) > m\), that \(h_2(L/2)= \frac{1+L/2+m}{2}\) and that

$$\begin{aligned} g_2(r,p) := dr^2 + h_2^2(r)g_0(p) \end{aligned}$$

is a symmetric metric of revolution on M, i.e for all \(r \in [0, L]\), we have \(h_2(r)= h_2(L-r)\). Note that we have \(h_2 \ge h_1\) and that for \(r \in (m-1,L-m+1)\) we have \(h_2(r) > h_1(r)\).

Besides, for f a smooth function on M, we have

$$\begin{aligned} R_{g_1}(f) = \frac{\int _M |\nabla f|_{g_1}^2 dV_{g_1}}{\int _\Sigma |f|^2dV_\Sigma } = \frac{\int _M \left( (\partial _r f)^2 + \frac{1}{h_1^2}|{\tilde{\nabla }} f|_{g_0}^2 \right) h_1^{n-1}dV_{g_0}dr}{\int _\Sigma |f|^2dV_\Sigma } \end{aligned}$$

and

$$\begin{aligned} R_{g_2}(f) = \frac{\int _M |\nabla f|_{g_2}^2 dV_{g_2}}{\int _\Sigma |f|^2dV_\Sigma } = \frac{\int _M \left( (\partial _r f)^2 + \frac{1}{h_2^2}|{\tilde{\nabla }} f|_{g_0}^2 \right) h_2^{n-1}dV_{g_0}dr}{\int _\Sigma |f|^2dV_\Sigma }, \end{aligned}$$

where \({\tilde{\nabla }} f\) is the gradient of f seen as a function of p.

Since \(n\ge 3\), for all functions \(f \in H^1(M)\), we have \(R_{g_1}(f)\le R_{g_2}(f)\). Using the Min-Max principle, we can conclude that for all \(k \ge 1\), we have \(\sigma _k(M, g_1) \le \sigma _k(M, g_2)\). However, here we want to show a strict inequality.

Because of the existence of a continuum of points r for which \(h_1(r) < h_2(r)\), if \(\partial _r f\) does not vanish on any interval, then the inequality is strict.

Let \(k\ge 1\) be an integer. Let \(E_{k+1}:=Span(f_{0,2}, \ldots , f_{k,2})\), where \(f_{i,2}\) is a Steklov eigenfunction associated with \(\sigma _i(M,g_2)\). We can choose these functions such that for all \(i = 0, \ldots , k\), we have

$$\begin{aligned} \int _\Sigma (f_{i,2})^2 dV_\Sigma = 1, \end{aligned}$$

and hence

$$\begin{aligned} \int _M |\nabla f_{i,2}|_{g_2}^2dV_{g_2} = \sigma _i(M,g_2). \end{aligned}$$

Let \(f^{*} = \sum _{i=0}^k a_i f_{i,2} \in E_{k+1}\) be such that \(\max _{f \in E_{k+1}} R_{g_1}(f) = R_{g_1}(f^*)\).

We now consider two cases:

  1. 1.

    Let us suppose \(f^* = a_k f_{k,2}\) with \(a_k \ne 0\), i.e \(f^*\) is an eigenfunction associated with \(\sigma _k(M, g_2)\). Then by Proposition 10, we have \(f^*(r,p) = u_j(r) S_j(p)\). Moreover, using [5, Prop. 2], we know that \(u_j\) is a non trivial solution of the ODE

    $$\begin{aligned} \frac{1}{h^{n-1}} \frac{d}{dr} \left( h^{n-1} \frac{d}{dr}u_j \right) - \frac{1}{h_2^2} \lambda _j u_j = 0. \end{aligned}$$
    1. (a)

      If \(\lambda _j =0\), which means \(S_j = S_0 = const\), then \(u_j\) cannot be locally constant. Indeed, otherwise \(f^*\) would be locally constant, but since \(f^*\) is harmonic, this implies that \(f^*\) is constant, see [1]. That is not the case because \(k \ge 1\).

    2. (b)

      If \(\lambda _j \ne 0\), then \(u_j\) cannot be locally constant, otherwise the ODE is not satisfied.

    Hence \(u_j\) is not locally constant and then \(\partial _r f^*\) does not vanish on any interval. Therefore, using the Min-Max principle (2), we have

    $$\begin{aligned} \sigma _k(M, g_1) \le \max _{f \in E_{k+1}} R_{g_1}(f) = R_{g_1}(f^*) < R_{g_2}(f^*) = \sigma _k(M, g_2). \end{aligned}$$
  2. 2.

    Let us suppose \(f^* = \sum _{i=0}^k a_i f_{i,2}\) such that there exists \(0 \le i <k\) such that \(a_i \ne 0\).

    Then by the Min-Max principle (2), we have

    $$\begin{aligned} \sigma _k(M,g_1)&\le \max _{f \in E_{k+1}}R_{g_1}(f) = R_{g_1}(f^*) \le R_{g_2}(f^*) \\&= \frac{\int _M \sum _{i=0}^k a_i^2 |\nabla f_{i,2}|^2dV_{g_2}}{\int _\Sigma (\sum _{i=0}^k a_i f_{i,2})^2 dV_\Sigma } \\&= \frac{\sum _{i=0}^k a_i^2 \sigma _i(M, g_2)}{\sum _{i=0}^k a_i^2 } \quad \text{ since } \int _\Sigma f_{i,2} f_{j,2} dV_\Sigma = \delta _{ij} \\&< \sigma _k(M,g_2). \end{aligned}$$

In both cases, we have

$$\begin{aligned} \sigma _k(M, g_1) < \sigma _k(M, g_2). \end{aligned}$$

\(\square \)

Remark 17

We never used the assumption that \(g_2\) is a symmetric metric of revolution on M in the previous proof. However, it will be useful in the proofs of the theorems that follow.

The process that constructs the metric \(g_2\) from \(g_1\) can then be repeated to create a third metric \(g_3\), and so on. This generates a sequence of metrics \((g_i)\), obtained from a sequence of functions \((h_i)\) (Fig. 3).

Fig. 3
figure 3

On the left, \(M=[0,L]\times {\mathbb {S}}^{n-1}\) is endowed with a metric \(g_i\) of the sequence. On the right, \(M=[0,L]\times {\mathbb {S}}^{n-1}\) is endowed with another metric \(g_j\) of the sequence, \(j >i\)

Full size image

The sequence \((h_i)\) uniformly converges to the function

$$\begin{aligned} h^* : [0, L]&\longrightarrow {\mathbb {R}}\\ r&\longmapsto \left\{ \begin{array}{lll} 1+r &{}\quad \text{ if } &{} 0 \le r \le L/2 \\ 1+L-r &{}\quad \text{ if } &{} L/2 \le r \le L. \end{array} \right. \end{aligned}$$

This function is not smooth. Hence \((M, g^*)\), where \(g^* = dr^2 + h^{*2}(r)g_0\), is not a hypersurface of revolution in the sense of Definition 1. In the limit, \((M, g^*)\) can be seen as the gluing of two copies of an annulus of inner radius 1 and outer radius \(1+L/2\). The metric \(g^*\) is therefore a maximizing metric, but is degenerated since it is induced by the function \(h^*\) which is non-smooth. That is why, as already mentioned, we call \(g^*\) the degenerated maximizing metric on M.

4 The first non trivial eigenvalue

In this section, we prove Theorem 3. The idea consists of comparing \(\sigma _1(M, g)\) with the Rayleigh quotient of a test function that is obtained from an eigenfunction for a mixed problem (Steklov–Dirichlet or Steklov–Neumann) introduced in Sect. 2.2. Then, to show that the upper bound \(B_n(L)\) given is sharp, we take a metric of revolution \(g_\varepsilon \) on M that is close to the degenerated maximizing metric \(g^*\) and show that \(\sigma _1(M, g_\varepsilon )\) is close to \(B_n(L)\).

Proof

Let \((M= [0, L] \times {\mathbb {S}}^{n-1}, g)\) be a hypersurface of revolution, where \(L>0\) is the meridian length of M. We recall that the boundary \(\Sigma \) of M consists of two disjoint copies of \({\mathbb {S}}^{n-1}\). We want to find a sharp upper bound \(B_n(L)\) for \(\sigma _1(M, g)\).

We consider \(A_{1+L/2}\) the annulus of inner radius 1 and outer radius \(1+L/2\). Let \(\varphi _0\) be an eigenfunction for the first eigenvalue of the Steklov–Dirichlet problem on \(A_{1+L/2}\), i.e.

$$\begin{aligned} \sigma _0^D(A_{1+L/2}) =\frac{\int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \varphi _0)^2 + \frac{1}{(1+r)^2} | {\tilde{\nabla }}\varphi _0|^2 \right) (1+r)^{n-1} dV_{g_0}dr}{\int _{{\mathbb {S}}^{n-1}} \varphi _0^2(0,p) dV_{g_0}}. \end{aligned}$$

We define a new function

$$\begin{aligned} \tilde{\varphi _0} : [0,L] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\nonumber \\ (r, p)&\longmapsto \left\{ \begin{array}{ll} \varphi _0(r,p) &{}\quad \text{ if } 0 \le r \le L/2 \\ - \varphi _0(L-r,p) &{}\quad \text{ if } L/2 \le r \le L. \end{array} \right. \end{aligned}$$
(3)

The function \(\tilde{\varphi _0}\) is continuous and we can check that

$$\begin{aligned} \int _\Sigma \tilde{\varphi _0}(r,p) dV_\Sigma&= \int _{{\mathbb {S}}^{n-1}} \tilde{\varphi _0}(0,p) dV_{g_0} + \int _{{\mathbb {S}}^{n-1}} \tilde{\varphi _0}(L,p) dV_{g_0} \\&= \int _{{\mathbb {S}}^{n-1}} \varphi _0(0,p) dV_{g_0} - \int _{{\mathbb {S}}^{n-1}} \varphi _0(0,p) dV_{g_0} \\&= 0. \end{aligned}$$

Hence, thanks to formula (1), the function \(\tilde{\varphi _0}\) can be used as a test function for \(\sigma _1(M,g)\).

We have

$$\begin{aligned} \sigma _1(M, g)&\le R_g(\tilde{\varphi _0}) \nonumber \\&< R_{{\tilde{g}}}(\tilde{\varphi _0}) \quad \text{ where } {\tilde{g}}= dr^2 + {\tilde{h}}^2g_0 \text{ comes } \text{ from } \text{ Theorem } 2 \nonumber \\&= \frac{\int _0^L \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \tilde{\varphi _0})^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \tilde{\varphi _0}|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{\int _\Sigma \tilde{\varphi _0}^2(0,p) dV_{\Sigma }} \nonumber \\&= \frac{2 \times \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \tilde{\varphi _0})^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \tilde{\varphi _0}|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{2 \times \int _{{\mathbb {S}}^{n-1}} \tilde{\varphi _0}^2(0,p) dV_{g_0}} \quad \text{ since } {\tilde{g}} \text{ is } \text{ symmetric } \nonumber \\&= \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \varphi _0)^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \varphi _0|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} \varphi _0^2(0,p) dV_{g_0}} \nonumber \\&< \frac{\int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \varphi _0)^2 + \frac{1}{(1+r)^2} | {\tilde{\nabla }}\varphi _0|^2 \right) (1+r)^{n-1} dV_{g_0}dr}{\int _{{\mathbb {S}}^{n-1}} \varphi _0^2(0,p) dV_{g_0}} \nonumber \\&= \sigma _0^D(A_{1+L/2}), \end{aligned}$$
(4)

where the second strict inequality comes from the existence of a continuum of points \(r \in [0, L/2]\) such that \({\tilde{h}}(r) < 1+r\).

If \(\phi _1\) is an eigenfunction for the first non trivial eigenvalue of the Steklov–Neumann problem on \(A_{1+L/2}\), i.e

$$\begin{aligned} \sigma _1^N(A_{1+L/2}) = \frac{\int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \phi _1)^2 + \frac{1}{(1+r)^2} | {\tilde{\nabla }} \phi _1 |^2 \right) (1+r)^{n-1} dV_{g_0}dr}{\int _{{\mathbb {S}}^{n-1}} \phi _1^2(0,p) dV_{g_0}}, \end{aligned}$$

then we define a new function

$$\begin{aligned} \tilde{\phi _1} : [0, L] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\\ (r,p)&\longmapsto \left\{ \begin{array}{ll} \phi _1(r,p) &{}\quad \text{ if } 0 \le r \le L/2 \\ \phi _1(L-r,p) &{}\quad \text{ if } L/2 \le r \le L. \end{array} \right. \end{aligned}$$

The function \(\tilde{\phi _1}\) is continuous and we can check that

$$\begin{aligned} \int _\Sigma \tilde{\phi _1}(r,p) dV_\Sigma&= \int _{{\mathbb {S}}^{n-1}} \tilde{\phi _1}(0,p) + \int _{{\mathbb {S}}^{n-1}} \tilde{\phi _1}(L,p) \\&= 0 + 0 \\&= 0, \end{aligned}$$

hence we can use it as a test function for \(\sigma _1(M,g)\). The same calculations as in (4) show that

$$\begin{aligned} \sigma _1(M,g) < \sigma _1^N(A_{1+L/2}). \end{aligned}$$
(5)

Putting Inequality (4) and Inequality (5) together, we get

$$\begin{aligned} \sigma _1(M,g) < B_n(L) :&= \min \left\{ \sigma _0^D(A_{1+L/2}), \; \sigma _1^N(A_{1+L/2}) \right\} \nonumber \\&= \min \left\{ \frac{(n-2)\left( 1+L/2 \right) ^{n-2}}{\left( 1+L/2 \right) ^{n-2}-1}, \frac{(n-1)\left( \left( 1+L/2\right) ^n -1 \right) }{\left( 1+ L/2 \right) ^n +n-1}\right\} . \end{aligned}$$
(6)

We will now prove that the bound \(B_n(L)\) is sharp. This means that for each \(\varepsilon >0\), there exists a metric of revolution \(g_\varepsilon \) on M such that \(\sigma _1(M, g_\varepsilon ) > B_n(L)-\varepsilon \).

Let \(\varepsilon >0\). Let \(M=[0, L] \times {\mathbb {S}}^{n-1}\) and let \(g_\varepsilon (r,p) = dr^2+h_\varepsilon ^2(r)g_0(p)\) be a metric of revolution on M such that:

  1. 1.

    The function \(h_\varepsilon \) is symmetric: for all \(r \in [0, L]\), we have \(h_\varepsilon (r)=h_\varepsilon (L-r)\);

  2. 2.

    For all \(r \in [0, L/2- \delta ]\), we have \(h_\varepsilon (r)=(1+r)\), with \(\delta \) small enough to guarantee that for all \(r \in [0, L/2]\), we have

    $$\begin{aligned} \max \{ (1+r)^{n-3} - h_\varepsilon (r)^{n-3}, (1+r)^{n-1}-h_\varepsilon (r)^{n-1}\} < \frac{\varepsilon }{B_n(L)} =: \varepsilon ^*. \end{aligned}$$

    Geometrically, this means that \((M, g_\varepsilon )\) looks like two copies of an annulus joined by a smooth curve, see Fig. 3.

Let \(f_1\) be an eigenfunction for \(\sigma _1(M, g_\varepsilon )\). Because \((M, g_\varepsilon )\) is symmetric, then we can choose \(f_1\) symmetric or anti-symmetric, which means that for all \(r \in [0, L]\) and \(p \in {\mathbb {S}}^{n-1}\), we have \(|f_1(r,p)|=|f_1(L-r,p)|\).

Moreover, it results from the calculations in (4) that for any symmetric or anti-symmetric function f, we have

$$\begin{aligned} R_{g_\varepsilon }(f) = \frac{ \int _0^{{L}/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r f)^2 + \frac{1}{h_\varepsilon (r)^2} | {\tilde{\nabla }} f|^2 \right) h_\varepsilon (r)^{n-1} dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} f^2(0,p) dV_{g_0}}. \end{aligned}$$

We will compare

$$\begin{aligned} R_{g_\varepsilon }(f_1) = \frac{ \int _0^{{L}/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r f_1)^2 + \frac{1}{h_\varepsilon (r)^2} | {\tilde{\nabla }} f_1|^2 \right) h_\varepsilon (r)^{n-1} dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}} \end{aligned}$$

with

$$\begin{aligned} R_{A_{1+L/2}}(f_1) = \frac{ \int _0^{{L}/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r f_1)^2 + \frac{1}{(1+r)^2} | {\tilde{\nabla }} f_1|^2 \right) (1+r)^{n-1} dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}}. \end{aligned}$$

If we call \(S:= R_{A_{1+L/2}}(f_1) - R_{g_\varepsilon }(f_1)\), we have

$$\begin{aligned} S&= \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} (\partial _r f_1)^2 \left( (1+r)^{n-1} - h_\varepsilon (r)^{n-1} \right) + | {\tilde{\nabla }} f_1|^2 \left( (1+r)^{n-3} -h_\varepsilon (r)^{n-3} \right) dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}} \\&< \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} ((\partial _r f_1)^2 \cdot \varepsilon ^* + | {\tilde{\nabla }} f_1|^2 \cdot \varepsilon ^* ) dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}} \\&= \varepsilon ^* \cdot \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} ((\partial _r f_1)^2 + | {\tilde{\nabla }} f_1|^2 ) dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}} \\&< \varepsilon ^* \cdot \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} ((\partial _r f_1)^2 h_\varepsilon (r)^{n-1} + | {\tilde{\nabla }} f_1|^2 h_\varepsilon (r)^{n-3} ) dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} (f_1)^2(0,p) dV_{g_0}} \quad \text{ since } h_\varepsilon \ge 1 \\&= \varepsilon ^* \cdot \sigma _1(M, g_\varepsilon ) \quad \text{ since } f_1 \text{ is } \text{ an } \text{ eigenfunction } \\&< \varepsilon ^* \cdot B_n(L) \\&= \varepsilon . \end{aligned}$$

Hence, we have

$$\begin{aligned} R_{A_{1+L/2}}(f_1) < \sigma _1(M,g_\varepsilon ) + \varepsilon . \end{aligned}$$
(7)

We now have two cases:

  1. 1.

    \(f_1\) can be written as \(f_1(r,p)=u_0(r) S_0(p)\), where \(S_0\) is a trivial harmonic function of the sphere, i.e \(S_0\) is constant (we can choose \(S_0 \equiv 1/\text{Vol }({\mathbb {S}}^{n-1})\)), and \(u_0\) is smooth. Hence \(f_1\) is constant on \(\{0\} \times {\mathbb {S}}^{n-1}\),

    $$\begin{aligned} \int _{\{0\} \times {\mathbb {S}}^{n-1}} f_1(r,p) dV_{g_0} = u_0(0) \ne 0. \end{aligned}$$

    Moreover, since \(|f_1(r, p)| = |f_1(L-r,p)|\) for all \(r \in [0, L]\) and since

    $$\begin{aligned} \int _{\Sigma } f_1(r,p) dV_\Sigma =0, \end{aligned}$$

    we have

    $$\begin{aligned} f_1\left( \frac{L}{2}, p\right) = 0. \end{aligned}$$

    Therefore, we can use \(f_{1{\vert _{[0, L/2] \times {\mathbb {S}}^{n-1}}}}\) as a test function for \(\sigma _0^D(A_{1+L/2})\), and we can state

    $$\begin{aligned} \sigma _0^D(A_{1+L/2}) \le R_{A_{1+L/2}}(f_1). \end{aligned}$$
  2. 2.

    \(f_1\) can be written as \(f_1(r,p) = u_1(r) S_1(p)\), where \(S_1\) is a non constant harmonic function of the sphere associated with the first non zero eigenvalue and \(u_1\) is smooth. Hence

    $$\begin{aligned} \int _{\{0\} \times {\mathbb {S}}^{n-1}} f_1(r,p) dV_{g_0} = 0. \end{aligned}$$

    Moreover, we have \(u_1(L/2) > 0\).

    Added with the fact that \(|f_1(r, p)| = |f_1(L-r,p)|\) for all \(r \in [0, L]\) and because \(f_1\) is smooth, we can conclude

    $$\begin{aligned} \partial _r f_1\left( \frac{L}{2}, p\right) = 0. \end{aligned}$$

    Therefore, we can use \(f_{1{\vert _{[0, L/2] \times {\mathbb {S}}^{n-1}}}}\) as a test function for \(\sigma _1^N(A_{1+L/2})\) and we can state

    $$\begin{aligned} \sigma _1^N(A_{1+L/2}) \le R_{A_{1+L/2}}(f_1). \end{aligned}$$

But we defined \(B_n(L)\) as

$$\begin{aligned} B_n(L) = \min \{\sigma _0^D(A_{1+L/2}), \sigma _1^N(A_{1+L/2}) \}. \end{aligned}$$

Hence we have

$$\begin{aligned} B_n(L) \le R_{A_{1+L/2}}(f_1) {\mathop {<}\limits ^{(7)}} \sigma _1(M,g_\varepsilon ) + \varepsilon \end{aligned}$$

and then

$$\begin{aligned} \sigma _1(M,g_\varepsilon ) > B_n(L) - \varepsilon . \end{aligned}$$

\(\square \)

From this result we can prove Corollary 4.

Proof

By Theorem 3, the inequality (6) holds which is

$$\begin{aligned} \sigma _1(M,g) < \min \left\{ \frac{(n-2)\left( 1+L/2 \right) ^{n-2}}{\left( 1+L/2 \right) ^{n-2}-1}, \frac{(n-1)\left( \left( 1+L/2\right) ^n -1 \right) }{\left( 1+ L/2 \right) ^n +n-1}\right\} . \end{aligned}$$

We consider the two functions

$$\begin{aligned} L&\longmapsto \frac{(n-2)\left( 1+L/2 \right) ^{n-2}}{\left( 1+L/2 \right) ^{n-2}-1} = \sigma _0^D(A_{1+L/2}) \\ L&\longmapsto \frac{(n-1)\left( \left( 1+L/2\right) ^n -1 \right) }{\left( 1+ L/2 \right) ^n +n-1} = \sigma _1^N(A_{1+L/2}). \end{aligned}$$
Fig. 4
figure 4

Representation of the case \(n = 5\). The decreasing smooth curve is \(L \longmapsto \sigma _0^D(A_{1+L/2})\) while the increasing smooth curve is \(L \longmapsto \sigma _1^N(A_{1+L/2})\). The solid curve is the bound \(B_5(L)\) given by Theorem 3

Full size image

We can show that \(L \longmapsto \sigma _0^D(A_{1+L/2})\) is strictly decreasing with L (Fig. 4). Indeed, let \(L' > L\) and let \(\varphi _0\) be an eigenfunction for \(\sigma _0^D(A_{1+L/2})\). We consider

$$\begin{aligned} {\bar{\varphi }}_0 : \left[ 0, \frac{L'}{2}\right] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\end{aligned}$$

the extension by 0 of \(\varphi _0\) to the annulus \(A_{1+L'/2}\). We get

$$\begin{aligned} \sigma _0^D(A_{1+L'/2}) < R_{A_{1+L/2}}({\bar{\varphi }}_0) = R_{A_{1+L'/2}}(\varphi _0) = \sigma _0^D(A_{1+L/2}), \end{aligned}$$

where the strict inequality comes from the fact that \({\bar{\varphi }}_0\) is not an eigenfunction associated with \(\sigma _0^D(A_{1+L'/2})\). Indeed, if we suppose that \({\bar{\varphi }}_0\) is an eigenfunction for \(\sigma _0^D(A_{1+L'/2})\), then it is harmonic in \(A_{1+L'/2}\) (since it satisfies the Steklov–Dirichlet problem), and since \({\bar{\varphi }}_0\) vanishes on the open set \(A_{1+L'/2} \backslash A_{1+L/2}\), then by [1] \({\bar{\varphi }}_0\) is constant, which is a contradiction.

In the same way, we can show that \(L \longmapsto \sigma _1^N(A_{1+L/2})\) is strictly increasing with L (Fig. 4). Indeed, let \(L' > L\) and let \(\phi _1\) be an eigenfunction for \(\sigma _1^N(A_{1+L'/2})\). We consider

$$\begin{aligned} {\bar{\phi }}_1 : \left[ 0, \frac{L}{2}\right] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\end{aligned}$$

the restriction of \(\phi _1\) to the annulus \(A_{1+L/2}\). We get

$$\begin{aligned} \sigma _1^N(A_{1+L/2}) \le R_{A_{1+L/2}}({\bar{\phi }}_1) < R_{A_{1+L'/2}}(\phi _1) = \sigma _1^N(A_{1+L'/2}). \end{aligned}$$

Hence the bound we gave possesses a maximum depending only on the dimension n, given by

$$\begin{aligned} \sigma _1(M,g) < B_n := \frac{(n-2)\left( 1+L_1/2 \right) ^{n-2}}{\left( 1+L_1/2 \right) ^{n-2}-1} \end{aligned}$$

where \(L_1\) is the unique positive solution of the equation

$$\begin{aligned} \left( 1+L/2\right) ^{2n-2}-(n-1)\left( 1+L/2\right) ^n-(n-1)^2\left( 1+L/2)\right) ^{n-2}+n-1=0. \end{aligned}$$

In order to prove that this bound is sharp, let \(\varepsilon >0\). We define \(M_\varepsilon := [0, L_1]\times {\mathbb {S}}^{n-1}\). Theorem 3 guarantees that there exists a metric of revolution \(g_\varepsilon \) on \(M_\varepsilon \) such that \(\sigma _1(M_\varepsilon , g_\varepsilon ) > B_n(L_1) - \varepsilon = B_n - \varepsilon \), which ends the proof. \(\square \)

We continue by proving Proposition 5.

Proof

We know that there exists a unique positive value of L, that we call \(L_1=L_1(n)\), such that the equality

$$\begin{aligned} \left( 1+L/2\right) ^{2n-2}-(n-1)\left( 1+L/2\right) ^n-(n-1)^2\left( 1+L/2)\right) ^{n-2}+n-1=0 \end{aligned}$$

holds. To ease notation, we substitute \((1+L/2)\) by R and we can state that there is a unique value of \(R \in (1, \infty )\) such that the equality

$$\begin{aligned} R^{2n-2}-(n-1)R^n-(n-1)^2R^{n-2}+n-1=0 \end{aligned}$$

holds. This equation is equivalent to

$$\begin{aligned} R^{n-2}\left( R^{n}-(n-1)R^2-(n-1)^2\right) +n-1=0, \end{aligned}$$

and we call \(R_1=R_1(n)\) its unique solution in \((1, \infty )\). We prove that \(R_1(n) \underset{n \rightarrow \infty }{\longrightarrow }\ 1\).

We call

$$\begin{aligned} \psi _n(R) := R^{n}-(n-1)R^2-(n-1)^2 \end{aligned}$$

and

$$\begin{aligned} \Psi _n(R) := R^{n-2}\left( R^{n}-(n-1)R^2-(n-1)^2\right) +n-1. \end{aligned}$$

Then, for \(R_1\) to be such that \(\Psi _n(R_1) =0\), it is necessary that \(\psi _n(R_1)<0\).

Thus,

$$\begin{aligned} R_1^{n}&< (n-1)R_1^2 + (n-1)^2 \\&< (n-1)^2(R_1^2 +1) \\&< (n-1)^2 \cdot 2R_1^2 \hspace{2 cm} \text{ since } R_1 > 1 \\&< (n-1)^3 R_1^2 \hspace{2 cm} \text{ since } n-1 \ge 2. \end{aligned}$$

Therefore,

$$\begin{aligned} n \ln (R_1) < 3 \ln (n-1) + 2 \ln (R_1) \end{aligned}$$

so

$$\begin{aligned} \ln (R_1) < \frac{3 \ln (n-1)}{n-2} \end{aligned}$$

and

$$\begin{aligned} R_1 < e^{\frac{3 \ln (n-1)}{n-2}}. \end{aligned}$$

As we substituted \((1+L/2)\) by R, and we can state that

$$\begin{aligned} L_1(n)&< 2 \left( e^{\frac{3 \ln (n-1)}{n-2}} -1 \right) . \end{aligned}$$

Therefore, since \(\frac{3 \ln (n-1)}{n-2} \underset{n \rightarrow \infty }{\longrightarrow }\ 0\), we have

$$\begin{aligned} L_1(n) \underset{n \rightarrow \infty }{ \longrightarrow }\ 0. \end{aligned}$$

Moreover, we have

$$\begin{aligned} n-1> B_n > n-2 \underset{n \rightarrow \infty }{\longrightarrow }\ \infty . \end{aligned}$$

\(\square \)

5 Stability properties of hypersurfaces of revolution

The goal of this section is to prove Theorems 6 and 7, which show some stability properties of the hypersurfaces we are studying in this paper. For Theorem 6, the key idea is to choose \(L \ne L_1\) and compare \(\sigma _1(M = [0, L] \times {\mathbb {S}}^{n-1}, g)\) with the first non trivial eigenvalue of M when endowed with the degenerated maximizing metric, namely \(B_n(L)\). For the case of Theorem 7, the strategy consists of showing that among all metrics of revolution that are not close (in a sense properly defined) to the degenerated maximizing metric, none of them induces a first non trivial eigenvalue that is close to \(B_n(L)\). We prove these theorems now.

5.1 Proof of Theorem 6

Recall that here we suppose \(L \ne L_1\).

Proof

Let g be any metric of revolution on \(M= [0, L] \times {\mathbb {S}}^{n-1}\). Then we have

$$\begin{aligned} \sigma _1(M, g) < B_n(L), \end{aligned}$$

where \(B_n(L)\) is given by Theorem 3.

We define \(C(n, L):= B_n - B_n(L)\), which is strictly positive since we assumed \(L \ne L_1\). Then we have

$$\begin{aligned} B_n - \sigma _1(M, g) \ge B_n - B_n(L) = C(n, L). \end{aligned}$$

Let \(0< \delta < \frac{B_n-(n-2)}{2}\), and let us suppose \(|B_n - \sigma _1(M, g)| < \delta \). Therefore, we have \(|B_n - \sigma _1(M, g^*)| < \delta \), where we wrote \(g^*\) the degenerated maximizing metric on M. We consider two cases:

  1. 1.

    We suppose \(L_1 < L\). In this case, we have \(B_n(L) = \sigma _0^D(A_{1+L/2}) = \frac{(n-2)(1+L/2)^{n-2}}{(1+L/2)^{n-2}-1}\). We write

    $$\begin{aligned} R := 1+L/2 \quad \text{ and }\quad \sigma _1(R) := \frac{(n-2)R^{n-2}}{R^{n-2}-1}. \end{aligned}$$

    Hence we have \(|B_n - \sigma _1(R)| < \delta \implies R \in [R_1, R_\delta ]\), where \(R_1 = 1+L_1/2\) and \(R_\delta \) is defined by \(\sigma _1(R_\delta ) = B_n - \delta \). Note that \(R_\delta \) exists since we assumed \(\delta < B_n -(n-2)\). We can calculate that

    $$\begin{aligned} R_\delta = \left( \frac{B_n - \delta }{B_n - (n-2) - \delta }\right) ^{\frac{1}{n-2}} \quad \text{ and }\quad R_1 = \left( \frac{B_n}{B_n - (n-2)} \right) ^{\frac{1}{n-2}}. \end{aligned}$$

    Thus, we have

    $$\begin{aligned} |R_1 - R| \le R_\delta - R_1 = \left( \frac{B_n - \delta }{B_n - (n-2) - \delta }\right) ^{\frac{1}{n-2}} - \left( \frac{B_n}{B_n - (n-2)} \right) ^{\frac{1}{n-2}}. \end{aligned}$$

    To estimate this expression, we use the identity \(x^{n-2} - y^{n-2} = (x-y)(x^{n-3} + x^{n-4}y + \cdots + xy^{n-4} + y^{n-3})\), with \(x = R_\delta \) and \(y = R_1\). On the one hand, we can compute that

    $$\begin{aligned} R_\delta ^{n-2} - R_1^{n-2} = \frac{(n-2)\delta }{(B_n-(n-2)-\delta )(B_n-(n-2))} \le \frac{2(n-2)\delta }{(B_n - (n-2))^2}, \end{aligned}$$

    where the inequality comes from the assumption \(\delta < \frac{B_n - (n-2)}{2}\). On the other hand, we can compute that

    $$\begin{aligned} R_\delta ^{n-3} + R_\delta ^{n-4}R_1 + \cdots + R_\delta R_1^{n-4} + R_1^{n-3} \ge (n-2) \cdot \left( \frac{B_n}{B_n -(n-2)}\right) ^{\frac{n-3}{n-2}}. \end{aligned}$$

    Therefore,

    $$\begin{aligned} R_\delta - R_1 \le \frac{2/(B_n-(n-2))^2}{\left( B_n/(B_n-(n-2))\right) ^{\frac{n-3}{n-2}}} \cdot \delta := C_1(n) \cdot \delta . \end{aligned}$$

    Since we wrote \(R = 1+L/2\), we can conclude that, for \(L_1 < L\) and \(0< \delta < \frac{B_n -(n-2)}{2}\), we have

    $$\begin{aligned} B_n - \sigma _1(M, g)< \delta \implies L - L_1 < 2C_1(n) \cdot \delta . \end{aligned}$$
  2. 2.

    Now we suppose \(L < L_1\) and we do a similar calculation, this time with \(B_n(L) = \sigma _1^N(A_{1+L/2}) = \frac{(n-1)((1+L/2)^n-1)}{(1+L/2)^n+n-1}\). We obtain a constant

    $$\begin{aligned} C_2(n) := \frac{ (n-2)^2/(n-1-B_n)^2 }{ n \left( ((n-1)B_n+1)/(n-1-B_n) \right) ^{ \frac{1}{n} }} \end{aligned}$$

    such that

    $$\begin{aligned} B_n -\sigma _1(M, g) < \delta \implies |L_1 -L| \le 2C_2(n) \cdot \delta . \end{aligned}$$

Defining \(C(n):= 2 \cdot \max \{ C_1(n), C_2(n)\}\) concludes the proof. \(\square \)

5.2 Proof of Theorem 7

Recall that we fixed \(m \in [1, 1+L/2)\) and that we defined \({\mathcal {M}}_m:= \{\)metrics of revolution g induced by a function h such that \(\max _{r \in [0, L]} \{h(r)\} \le m\}\).

Proof

Let \(g \in {\mathcal {M}}_m\), and let \(h: [0, L] \longrightarrow {\mathbb {R}}_+^*\) be the function which induces g. We define a new function \(h_m: [0, L] \longrightarrow {\mathbb {R}}_+^*\) as follows:

$$\begin{aligned} h_m(r) = \left\{ \begin{array}{ll} 1+r &{} \quad \text{ if } 0 \le r \le m-1 \\ m &{} \quad \text{ if } m-1 \le r \le L -m+1 \\ 1+L-r &{}\quad \text{ if } L-m+1 \le r \le L. \end{array} \right. \end{aligned}$$

We call \(g_m\) the metric induced by \(h_m\). Notice that \(g_m\) is not a metric of revolution in the sense of Definition 1 since \(h_m\) is not smooth (Fig. 5).

Fig. 5
figure 5

Since \(g \in {\mathcal {M}}_m\), the function h which induces g satisfies \(h \le h_m\)

Full size image

In the same spirit as in Sect. 3, for any smooth function f on M, we have

$$\begin{aligned} R_{g}(f) = \frac{\int _M \left( (\partial _r f)^2 + \frac{1}{h^2}|{\tilde{\nabla }} f|_{g_0}^2 \right) h^{n-1}dV_{g_0}dr}{\int _\Sigma |f|^2dV_\Sigma } \end{aligned}$$

and

$$\begin{aligned} R_{g_m}(f) = \frac{\int _M \left( (\partial _r f)^2 + \frac{1}{h_m^2}|{\tilde{\nabla }} f|_{g_0}^2 \right) h_m^{n-1}dV_{g_0}dr}{\int _\Sigma |f|^2dV_\Sigma }. \end{aligned}$$

Therefore, since \(n \ge 3\) and \(h \le h_m\), we have

$$\begin{aligned} \sigma _1(M, g) \le \sigma _1(M, g_m). \end{aligned}$$

We can now consider a new function \({\tilde{h}}_m\), obtained from \(h_m\) by smoothing out the two non-smooth points, with \({\tilde{h}}_m\) satisfying:

  1. 1.

    For all \(r \in [0, L]\), we have \(h_m(r) \le {\tilde{h}}_m(r)\);

  2. 2.

    The metric \({\tilde{g}}_m\) induced by \({\tilde{h}}_m\) is a metric of revolution in the sense of Definition 1.

Remark that since \(h_m \le {\tilde{h}}_m\), we have \(\sigma _1(M, g_m) \le \sigma _1(M, {\tilde{g}}_m)\).

We define \(C(n, L, m):= B_n(L) - \sigma _1(M, {\tilde{g}}_m)\), which is strictly positive by Theorem 3. Then we have

$$\begin{aligned} B_n(L)- \sigma _1(M, g) \ge B_n(L)- \sigma _1(M, {\tilde{g}}_m) = C(n, L, m). \end{aligned}$$

\(\square \)

6 Upper bounds for higher Steklov eigenvalues

In this section, we want to compute some sharp upper bound for higher Steklov eigenvalues of hypersurfaces of revolution. Therefore, we will have to deal with the multiplicity of the eigenvalues. We write \(\lambda _{(k)}, \sigma _{(k)}, \sigma _{(k)}^D, \sigma _{(k)}^N\) for the (k)th eigenvalue counted without multiplicity.

Before we can state and prove our results, we first recall some known properties of the multiplicities of the eigenvalues under consideration.

Given a hypersurface of revolution \((M=[0,L] \times {\mathbb {S}}^{n-1}, g)\), we want to provide information about the multiplicity of the Steklov eigenvalues of (M, g).

For the classical Laplacian problem \(\Delta S = \lambda S\) on \(({\mathbb {S}}^{n-1}, g_0)\), we know [2, pp. 160–162] that the set of eigenvalues is \(\{ \lambda _{(k)} = k(n+k-2): k \ge 0 \}\), where the multiplicity \(m_0\) of \(\lambda _{(0)}=0\) is 1 and the multiplicity of \(\lambda _{(k)}\) is

$$\begin{aligned} m_k:=\frac{(n+k-3)(n+k-4) \ldots n(n-1)}{k!}(n+2k-2). \end{aligned}$$
(8)

As such, given \(k \ge 0\), there exist \(m_k\) independent functions \(S_k^1, \ldots , S_k^{m_k}\) such that \(\Delta S_k^i = \lambda _{(k)} S_k^i, \; i=1, \ldots , m_k\).

Given \(k \ge 0\), there are \(m_k\) independent Steklov–Dirichlet eigenfunctions associated with the eigenvalue \(\sigma _{(k)}^D(A_{1+L/2})\), that can be written \(\varphi _k^i(r,p) = \alpha _{k}(r)S_{k}^i(p), \; i = 1, \ldots , m_k\). For the Steklov–Neumann case, the eigenfunctions associated with \(\sigma _{(k)}^N(A_{1+L/2})\) can be written \(\phi _k^i(r,p) = \beta _{k}(r)S_k^i(p), \; i = 1, \ldots , m_k\). Indeed, for each of these problems, the multiplicity of the (k)th eigenvalue is exactly \(m_k\), see, for example, [5, Prop. 3].

6.1 Upper bound for \(\sigma _{2}(M, g), \ldots , \sigma _{m_1}(M, g)\)

In this section, we prove the following theorem:

Theorem 18

Let \((M = [0, L] \times {\mathbb {S}}^{n-1}, g)\) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) and dimension \(n \ge 3\). Let \(m_1\) be the multiplicity of the first non trivial eigenvalue of the classical Laplacian problem on \(({\mathbb {S}}^{n-1}, g_0)\). Then we have

$$\begin{aligned} \sigma _2(M, g) = \cdots = \sigma _{m_1}(M, g) < B_n^2(L) = \cdots = B_n^{m_1}(L) = : \sigma _{(1)}^N(A_{1+L/2}). \end{aligned}$$

Moreover, this bound is sharp: for all \(\varepsilon >0\) there exists a metric of revolution \(g_\varepsilon \) on M such that

$$\begin{aligned} \sigma _2(M, g_\varepsilon ) = \cdots = \sigma _{m_1}(M, g_\varepsilon ) > \sigma _{(1)}^N(A_{1+L/2}) - \varepsilon . \end{aligned}$$

Proof

We consider two cases.

  1. 1.

    Let \(M=[0,L] \times {\mathbb {S}}^{n-1}\), with \(L\le L_1\). We write \(f_1^1\) an eigenfunction associated with \(\sigma _1(M,g)\). Since \(L \le L_1\), we have \(B_n(L)=\sigma _1^N(A_{1+L/2})\) and therefore \(f_1^1(r,p)=u_1(r)S_1^1(p)\).

    We consider now a new function denoted \(f_1^2\) given by \( f_1^2(r,p)=u_1(r)S_1^2(p). \) We can check that

    $$\begin{aligned} \int _\Sigma f_1^2(r,p)dV\Sigma =0 \quad \text{ and } \quad \int _\Sigma f_1^1(r,p)f_1^2(r,p)dV\Sigma =0. \end{aligned}$$

    Moreover, we have

    $$\begin{aligned} \sigma _1(M, g)=R_g(f_1^1) = R_g(f_1^2). \end{aligned}$$

    In the same way, we write

    $$\begin{aligned} f_1^i(r,p) = u_1(r) S_1^i(p), \quad i=1, \ldots , m_1 \end{aligned}$$

    and we can conclude

    $$\begin{aligned} \sigma _1(M,g) = \sigma _2(M,g) = \cdots = \sigma _{m_1}(M, g). \end{aligned}$$

    Therefore, we already have a sharp upper bound for these eigenvalues, which is given by \(\sigma _1^N(A_{1+L/2})\).

  2. 2.

    Let \(M=[0,L] \times {\mathbb {S}}^{n-1}\), with \(L>L_1\). We call \(f_1\) an eigenfunction associated with \(\sigma _1(M,g)\). Since \(L > L_1\), we have \(B_n(L)=\sigma _0^D(A_{1+L/2})\). Therefore \(f_1(r,p)=u_0(r)S_0(p)\).

    We write now \(f_2^1(r,p) = u_2(r)S_1^1(p)\) an eigenfunction associated with \(\sigma _2(M,g)\). As before, we then consider \(m_1\) functions denoted \(f_2^i(r,p) = u_2(r)S_1^i(p), i= 1, \ldots , m_1\) and we get

    $$\begin{aligned} \sigma _2(m,g) = \cdots = \sigma _{m_1+1}(M,g). \end{aligned}$$

    We consider a function \(\phi _1(r,p) = \beta _1(r)S_1(p)\) associated with \(\sigma _{(1)}^N(A_{1+L/2})\). In the same spirit as before, we define a function

    $$\begin{aligned} \tilde{\phi _1} : [0, L] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\\ (r,p)&\longmapsto \left\{ \begin{array}{ll} \phi _1(r,p) &{}\quad \text{ if } 0 \le r \le L/2 \\ \phi _1(L-r,p) &{}\quad \text{ if } L/2 \le r \le L. \end{array} \right. \end{aligned}$$

    We can check that the function \(\tilde{\phi _1}\) is continuous and that \(\int _\Sigma \tilde{\phi _1}dV_\Sigma =0\). Moreover, it is immediate that \(\int _\Sigma \tilde{\phi _1} f_1 dV_\Sigma = 0\). Hence we can use \(\tilde{\phi _1}\) as a test function for \(\sigma _2(M, g)\) and as we did before, we can see that

    $$\begin{aligned} \sigma _2(M, g)&\le R_g(\tilde{\phi _1}) \\&< R_{{\tilde{g}}}(\tilde{\phi _1}) \text{ where } {\tilde{g}} \text{ comes } \text{ from } \text{ Theorem } 2 \\&= \frac{\int _0^L \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \tilde{\phi _1})^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \tilde{\phi _1}|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{\int _\Sigma \tilde{\phi _1}^2(0,p) dV_{g_0}} \\&= \frac{2 \times \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \tilde{\phi _1})^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \tilde{\phi _1}|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{2 \times \int _{{\mathbb {S}}^{n-1}} \tilde{\phi _1}^2(0,p) dV_{g_0}} \\&= \frac{ \int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \phi _1)^2 + \frac{1}{{\tilde{h}}(r)^2} | {\tilde{\nabla }} \phi _1|^2 \right) {\tilde{h}}(r)^{n-1} dV_{g_0}dr}{ \int _{{\mathbb {S}}^{n-1}} \phi _1^2(0,p) dV_{g_0}} \\&< \frac{\int _0^{L/2} \int _{{\mathbb {S}}^{n-1}} \left( (\partial _r \phi _1)^2 + \frac{1}{(1+r)^2} | {\tilde{\nabla }}\phi _1|^2 \right) (1+r)^{n-1} dV_{g_0}dr}{\int _{{\mathbb {S}}^{n-1}} \phi _1^2(0,p) dV_{g_0}} \\&= \sigma _{(1)}^N(A_{1+L/2}). \end{aligned}$$

Therefore, regardless of the value of \(L>0\), we have

$$\begin{aligned} \sigma _2(M,g) = \cdots = \sigma _{m_1}(M,g) < B_n^2(L) = \cdots = B_n^{m_1}(L) := \sigma _{(1)}^N(A_{1+L/2}). \end{aligned}$$

Moreover, this bound is sharp: for all \(\varepsilon >0\), there exists a metric \(g_\varepsilon \) on \(M=[0,L]\times {\mathbb {S}}^{n-1}\) such that \(\sigma _2(M, g_\varepsilon ) = \cdots = \sigma _{m_1}(M,g_\varepsilon ) > \sigma _{(1)}^N(A_{1+L/2}) - \varepsilon \). Indeed, as before it is sufficient to choose the metric \(g_\varepsilon = dr^2 + h_\varepsilon ^2g_0\), with the function \(h_\varepsilon \) such that

  1. 1.

    \(h_\varepsilon \) is symmetric;

  2. 2.

    For all \(r \in [0, L/2- \delta ]\), we have \(h_\varepsilon (r)=1+r\), with \(\delta \) small enough.

The proof of sharpness goes as in the proof of Theorem 3. \(\square \)

The upper bound we gave, namely \( \sigma _{(1)}^N(A_{1+L/2})\), depends on the dimension of M and the meridian length L of M. It is easy to see that \(\sigma _{(1)}^N(A_{1+L/2})\), which is strictly increasing, satisfies

$$\begin{aligned} \sigma _{(1)}^N(A_{1+L/2}) = \frac{(n-1)\left( \left( 1+L/2\right) ^n -1 \right) }{\left( 1+ L/2 \right) ^n +n-1} \underset{L \rightarrow \infty }{\longrightarrow }\ n-1. \end{aligned}$$

Therefore, we have got a bound that depends only on the dimension n of M. Given a hypersurface of revolution (M, g) with two boundary components, we have

$$\begin{aligned} \sigma _2(M,g) = \cdots = \sigma _{m_1}(M,g) < B_n^2 = \cdots = B_n^{m_1} := n-1. \end{aligned}$$

Moreover, this bound is sharp, in the sense that for all \(\varepsilon >0\), there exists a hypersurface of revolution \((M_\varepsilon , g_\varepsilon )\) such that \(\sigma _2(M_\varepsilon ,g_\varepsilon ) = \cdots = \sigma _{m_1}(M_\varepsilon ,g_\varepsilon ) > n-1 - \varepsilon \). Indeed, we can choose \(L_\varepsilon \) large enough for \(\sigma _1^N(A_{1+L_\varepsilon /2})\) to be \(\frac{\varepsilon }{2}\)-close to \(n-1\), and then define \(M_\varepsilon := [0,L_\varepsilon ] \times {\mathbb {S}}^{n-1}\). Now we can put a metric \(g_\varepsilon \) on \(M_\varepsilon \) such that \(\sigma _2(M_\varepsilon , g_\varepsilon ) = \cdots = \sigma _{m_1}(M_\varepsilon , g_\varepsilon ) > \sigma _1^N(A_{1+L_\varepsilon /2}) - \frac{\varepsilon }{2}\), and we are done.

Our calculations showed that the eigenvalues \(k = 2, \ldots , m_1\) have a critical length at infinity.

6.2 Upper bound for \(\sigma _{m_1+1}(M,g)\)

Now we are interested in the next eigenvalue, namely \(\sigma _{m_1+1}(M, g)\). For that reason, we define a new special meridian length \(L_2\): it is the unique solution of the equation \(\sigma _0^D(A_{1+L/2}) = \sigma _{(2)}^N(A_{1+L/2})\). We remark that we have \(L_2<L_1\). Indeed, for all \(L >0\), we have \(\sigma _{(2)}^N(A_{1+L/2}) > \sigma _{(1)}^N(A_{1+L/2})\), and the function \(L \longmapsto \sigma _0^D(A_{1+L/2})\) is strictly decreasing. Hence, comparing the intersection of the curves gives \(L_2 < L_1\). We prove the following theorem:

Theorem 19

Let \((M = [0, L] \times {\mathbb {S}}^{n-1}, g)\) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to \({\mathbb {S}}^{n-1}\) and dimension \(n \ge 3\). Let \(m_1\) be the multiplicity of the first non trivial eigenvalue of the classical Laplacian problem on \(({\mathbb {S}}^{n-1}, g_0)\). Then we have

$$\begin{aligned} \sigma _{m_1+1}(M,g)< B_n^{m_1+1}(L) := \left\{ \begin{array}{ll} \sigma _{(2)}^N(A_{1+L/2}) &{}\quad \text{ if } L \le L_2 \\ \sigma _{(0)}^D(A_{1+L/2}) &{}\quad \text{ if } L_2< L \le L_1 \\ \sigma _{(1)}^N(A_{1+L/2}) &{}\quad \text{ if } L_1 < L. \end{array} \right. \end{aligned}$$

Moreover, this bound is sharp: for all \(\varepsilon >0\), there exists a metric of revolution \(g_\varepsilon \) on M such that

$$\begin{aligned} \sigma _{m_1+1}(M, g_\varepsilon ) > B_n^{m_1+1}(L) - \varepsilon . \end{aligned}$$

A plot of the function \(L \longmapsto B_n^{m_1+1}(L)\) can be useful to visualize the sharp upper bound, see Fig. 6.

Proof

Now we have to distinguish three cases.

  1. 1.

    Let \(M=[0,L]\times {\mathbb {S}}^{n-1}\), with \(L \le L_2\). We call \(f_1^1(r,p) = u_1(r)S_1^1(p), \ldots , f_1^{m_1}(r,p)= u_1(r)S_1^{m_1}(p)\) the Steklov eigenfunctions associated with \(\sigma _{(1)}(M,g)= \sigma _1(M,g) = \cdots = \sigma _{m_1}(M,g)\).

    There exists an eigenfunction \(\phi _2(r,p) = \beta _2(r) S_2(p)\) associated with \(\sigma _{(2)}^N(M,g)= \sigma _{m_1+1}^N(M,g)\). We define a new function

    $$\begin{aligned} \tilde{\phi _2} : [0, L] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\\ (r,p)&\longmapsto \left\{ \begin{array}{ll} \phi _2(r,p) &{}\quad \text{ if } 0 \le r \le L/2 \\ \phi _2(L-r,p) &{}\quad \text{ if } L/2 \le r \le L. \end{array} \right. \end{aligned}$$

    This function is continuous, satisfies \(\int _\Sigma \tilde{\phi _2}dV_\Sigma =0\) and we can check that for all \(i=1, \ldots , m_1\),

    $$\begin{aligned} \int _\Sigma \tilde{\phi _2} f_1^idV_\Sigma =0. \end{aligned}$$

    Hence we can use \(\tilde{\phi _2}\) as a test function for \(\sigma _{m_1+1}(M,g)\). The same kind of calculations as in Inequality (5) show that we have

    $$\begin{aligned} \sigma _{m_1+1}(M, g) < \sigma _{(2)}^N(A_{1+L/2}), \end{aligned}$$

    which is a sharp upper bound.

  2. 2.

    Let \(M=[0,L]\times {\mathbb {S}}^{n-1}\), with \(L_2 < L \le L_1\). We call \(f_1^1(r,p) = u_1(r)S_1^1(p), \ldots , f_1^{m_1}(r,p)= u_1(r)S_1^{m_1}(p)\) the Steklov eigenfunctions associated with \(\sigma _{(1)}(M,g)= \sigma _1(M,g) = \cdots = \sigma _{m_1}(M,g)\).

    There exists an eigenfunction \(\varphi _0(r,p) = \alpha _0(r) S_0(p)\) associated with \(\sigma _{0}^D(M,g)\). We use the function \(\tilde{\varphi _0}\) we defined before, namely

    $$\begin{aligned} \tilde{\varphi _0} : [0,L] \times {\mathbb {S}}^{n-1}&\longrightarrow {\mathbb {R}}\\ (r, p)&\longmapsto \left\{ \begin{array}{ll} \varphi _0(r,p) &{}\quad \text{ if } 0 \le r \le L/2 \\ - \varphi _0(L-r,p) &{}\quad \text{ if } L/2 \le r \le L. \end{array} \right. \end{aligned}$$

    We already saw that \(\tilde{\varphi _0}\) is continuous, that \(\int _\Sigma \tilde{\varphi _0}dV_\Sigma =0\) and we can check that for all \(i = 1, \ldots , m_1\),

    $$\begin{aligned} \int _\Sigma \tilde{\varphi _0}f_1^i dV_\Sigma =0. \end{aligned}$$

    Using \(\tilde{\varphi _0}\) as a test function, we get

    $$\begin{aligned} \sigma _{m_1+1}(M,g) < \sigma _{(0)}^D(A_{1+L/2}), \end{aligned}$$

    which is a sharp upper bound.

  3. 3.

    Let \(M=[0,L]\times {\mathbb {S}}^{n-1}\), with \(L_1 \le L\). Then \(\sigma _1(M, g) < \sigma _2(M, g) = \cdots = \sigma _{m_1+1}(M, g)\). We already dealt with this case in the proof of Theorem 18 and we saw that

    $$\begin{aligned} \sigma _{m_1+1}(M, g) < \sigma _{(1)}^N(A_{1+L/2}), \end{aligned}$$

    which is a sharp upper bound.

Therefore, given a hypersurface of revolution \((M=[0, L] \times {\mathbb {S}}^{n-1}, g)\), we have a sharp upper bound for \(\sigma _{m_1+1}(M,g)\), depending on n and L, given by

$$\begin{aligned} \sigma _{m_1+1}(M,g)< B_n^{m_1+1}(L) := \left\{ \begin{array}{ll} \sigma _{(2)}^N(A_{1+L/2}) &{}\quad \text{ if } L \le L_2 \\ \sigma _{(0)}^D(A_{1+L/2}) &{}\quad \text{ if } L_2< L \le L_1 \\ \sigma _{(1)}^N(A_{1+L/2}) &{}\quad \text{ if } L_1 < L. \end{array} \right. \end{aligned}$$

The proof of sharpness goes as in the proof of Theorem 3. \(\square \)

Fig. 6
figure 6

Representation of the case \(n = 5\). The solid curve is the bound given in Theorem 19

Full size image

From this, one can once again look for a sharp upper bound for \(\sigma _{m_1+1}(M,g)\) that depends only on the dimension n of M. This bound is given by

$$\begin{aligned} \sigma _{m_1+1}(M,g) < B_n^{m_1+1}&:= \max \left\{ \sigma _0^D(A_{1+{L_2}/2}), n-1\right\} \nonumber \\&= \left\{ \begin{array}{ll} \sigma _0^D(A_{1+{L_2}/2}) &{}\quad \text{ if } 3 \le n \le 6 \\ n-1 &{}\quad \text{ if } 7 \le n. \end{array} \right. \end{aligned}$$
(9)

A proof of (9) is given in Appendix A.

Therefore, the eigenvalue \(k = m_1+1\) possesses a finite critical length if \(3 \le n \le 6\), and it has a critical length at infinity if \(7 \le n\).

Remark 20

It is then tempting to search for an expression for \(B_n^k:= \sup _{L \in {\mathbb {R}}_+^*} \{B_n^k(L)\}\) for any n and k; but it seems to be hard to give an explicit formula for it. Indeed, as Sects. 6.1 and 6.2 suggest, the function \( L \longmapsto B_n^k(L)\) is hard to determine and can be either smooth (as in Sect. 6.1 for instance) or piecewise smooth (as Sect. 6.2 for instance). In the second case, there are possibly many irregular points that we have to consider. Moreover, depending on the value of n and k:

  1. 1.

    Either k has a finite critical length, i.e \(B_n^k =B_n^k(L_k)\) for a certain \(L_k \in {\mathbb {R}}_+^*\). That is for instance the case of \(\sigma _1(M,g)\) or \(\sigma _{m_1+1}\) if \(n=3, 4, 5 \text{ or } 6\);

  2. 2.

    Or k has a critical length at infinity, i.e \(B_n^k = \lim _{L \rightarrow \infty } B_n^k(L)\). That is for instance the case of \(\sigma _2(M,g), \ldots , \sigma _{m_1}(M,g)\).

Furthermore, we will prove in Sect. 7 that for all \(n \ge 3\), there are infinitely many k that have a finite critical length associated to them. In all these cases, the function \( L \longmapsto B_n^k(L)\) is piecewise smooth.

7 Critical lengths of hypersurfaces of revolution

We recall that given \(n \ge 3\), we are interested in giving information about the set of finite critical lengths. We want to prove Theorem 9, i.e that there are infinitely many k such that \(B_n^k = B_n^k(L_k)\) for a certain finite \(L_k \in {\mathbb {R}}_+^*\), and that the sequence of critical lengths converges to 0.

Proof

As before, for \(j \ge 0\), we denote by \(m_j\) the number given by the formula (8), which is the multiplicity of \(\sigma _{(j)}^D(A_R)\) as well as the multiplicity of \(\sigma _{(j)}^N(A_R)\). Let \(i \ge 2\) be an integer. We claim that for all k such that

$$\begin{aligned} m_0 + \sum _{j=1}^{i-1} 2m_j + m_i < k \le m_0 + \sum _{j=1}^{i} 2m_j, \end{aligned}$$
(10)

we have

$$\begin{aligned} B_n^k = B_n^k(L_k) \end{aligned}$$

for a certain \(L_k \in {\mathbb {R}}_+^*\).

Indeed, let k satisfy (10). Then, because of the asymptotic behaviour of the functions \(L \longmapsto \sigma _{(j)}^D(A_{1+L/2})\) and \(L \longmapsto \sigma _{(j)}^N(A_{1+L/2})\) as \(L \rightarrow \infty \), there exists \(C >0\) such that for all \(L > C\), we have \(B_n^k(L) = \sigma _{(i)}^D(A_{1+L/2})\). But we can compute that

$$\begin{aligned} \frac{\partial }{\partial L} \sigma _{(i)}^D(A_{1+L/2}) = - \frac{4(L+2)(2i+n-2)^2(1+L/2)^{2i+n}}{\left( 4(1+L/2)^{2i+n}-L^2-4L-4 \right) ^2} < 0, \end{aligned}$$

which means that the function \(L \longmapsto \sigma _{(i)}^D(A_{1+L/2})\) is strictly decreasing. Hence, for \(L> L' > C\), we have \(B_n^k(L) < B_n^k(L')\). Therefore, for such a k, we have

$$\begin{aligned} B_n^k= B_n^k(L_k) \end{aligned}$$

with \(L_k\) finite, that is \(L_k\) is a finite critical length associated to k.

Then, defining \(k_1:= 1\) and for each \(i \ge 2\), defining \(k_i:= m_0 + \sum _{j=1}^{i} 2m_j\), we get a sequence \((k_i)_{i=1}^\infty \) such that

$$\begin{aligned} B_n^{k_i} = B_n^{k_i}(L_i) \end{aligned}$$

for a certain \(L_i \in {\mathbb {R}}_+^*\) finite.

Now we want to prove that the sequence of finite critical lengths \((L_i)_{i=1}^\infty \) converges to 0.

Let \(i \ge 1\). We know that \(L_i\) has to be a solution of the equation \(\sigma _{(j_1)}^D(A_{1+L/2}) = \sigma _{(j_2)}^N(A_{1+L/2})\) for a certain ordered pair \((j_1, j_2) \in {\mathbb {N}}^2\). As said before, for \(L >C\), we have \(B_n^{k_i}(L) = \sigma _{(i)}^D(A_{1+L/2})\), hence \(L_i \le L_i^*\), where \(L_i^*\) is the unique solution of the equation

$$\begin{aligned} \sigma _{(i)}^D(A_{1+L/2}) = \sigma _{(i+1)}^N(A_{1+L/2}). \end{aligned}$$

Therefore, in order to prove that \(L_i \underset{i \rightarrow \infty }{\longrightarrow }\ 0\), we prove that \(L_i^* \underset{i \rightarrow \infty }{\longrightarrow }\ 0\).

Using Propositions 13 and 15, making some calculations and substituting \((1+L/2)\) by R, we can see that solving

$$\begin{aligned} \sigma _{(i)}^D(A_{1+L/2}) = \sigma _{(i+1)}^N(A_{1+L/2}) \end{aligned}$$

is equivalent to finding the unique value \(R_i \in (1, \infty )\) which solves the equation

$$\begin{aligned} (i+1)R^{2i+n-2}\left( R^{2i+n} - R^2(2i+n-1)- \frac{(i+n-1)(2i+n-1)}{i+1} \right) +(i+n-1)=0. \end{aligned}$$

We call

$$\begin{aligned} \underbrace{(i+1)R^{2i+n-2} \left( \underbrace{ R^{2i+n} - R^2(2i+n-1)- \frac{(i+n-1)(2i+n-1)}{i+1} }_{=: \psi _i(R)} \right) +(i+n-1)}_{=: \Psi _i(R)}. \end{aligned}$$

Because \( (i+1)R^{2i+n-2} >0\) and \((i+n-1) >0\), then for \(R_i\) to be the solution of the equation \(\Psi _i(R) =0\), it is necessary that \(\psi _i(R_i) <0\).

Then we have

$$\begin{aligned} R^{2i +n}&< R^2(2i+n-1) + \frac{(i+n-1)(2i+n-1)}{i+1} \\&< R^2 \left( (2i+n-1) + \frac{(i+n-1)(2i+n-1)}{i+1} \right) \\&= R^2 \left( \frac{(2i+n-1)(2i+n)}{i+1} \right) \\&< R^2(2i+n)^2. \end{aligned}$$

Therefore we have

$$\begin{aligned} \ln (R) < \frac{2\ln (2i+n)}{2i+n-2} \end{aligned}$$

and thus

$$\begin{aligned} R < e^{\frac{2\ln (2i+n)}{2i+n-2} } \end{aligned}$$

Remember that we substituted \((1+L/2)\) by R, and then the unique solution of the equation

$$\begin{aligned} \sigma _{(i)}^D(A_{1+L/2}) = \sigma _{(i+1)}^N(A_{1+L/2}) \end{aligned}$$

is a value \(L_i^*\) which satisfies

$$\begin{aligned} 0< L_i^* < 2 \left( e^{\frac{2\ln (2i+n)}{2i+n-2} }-1 \right) . \end{aligned}$$

Therefore, since \(\frac{2 \ln (2i +n)}{2i +n -2} \underset{i \rightarrow \infty }{\longrightarrow }\ 0\), we have

$$\begin{aligned} L_i < L_i^* \underset{i \rightarrow \infty }{\longrightarrow }\ 0. \end{aligned}$$

In particular, for each \(\delta >0\) there exists \(k_0 \in {\mathbb {N}}\) such that for each \(k >k_0\) which has a finite critical length \(L_k\), then \(L_k < \delta \). \(\square \)

Remark 21

The condition given by (10) is sufficient, but is not a necessary one. Indeed, \(k=1\) does not meet condition (10) but we have \(B_n^1 = B_n^1(L_1)\), where \(L_1\) is given by Corollary 4. This consideration naturally leads to the following open question:

Question 22

Given \(n \ge 3\), are there finitely or infinitely many \(k \in {\mathbb {N}}\) such that k has a critical length at infinity?