Sharp upper bounds for Steklov eigenvalues of a hypersurface of revolution with two boundary components in Euclidean space

We investigate the question of sharp upper bounds for the Steklov eigenvalues of a hypersurface of revolution of the Euclidean space with two boundary components isometric to two copies of $\mathbb{S}^{n-1}$. For the case of the first non zero Steklov eigenvalue, we give a sharp upper bound $B_n(L)$ (that depends only on the dimension $n \ge 3$ and the meridian length $L>0$) which is reached by a degenerated metric $g^*$, that we compute explicitly. We also give a sharp upper bound $B_n$ which depends only on $n$. Our method also permits us to prove some stability properties of these upper bounds.


Introduction
Let (M, g) be a smooth compact connected Riemannian manifold of dimension n ≥ 2 with smooth boundary Σ.The Steklov problem on (M, g) consists of finding the real numbers σ and the harmonic functions f : M −→ R such that ∂ ν f = σf on Σ, where ν denotes the outward normal on Σ.Such a σ is called a Steklov eigenvalue of (M, g).It is well known that the Steklov spectrum forms a discrete sequence 0 = σ 0 (M, g) < σ 1 (M, g) ≤ σ 2 (M, g) ≤ . . .↗ ∞.Each eigenvalue is repeated with its multiplicity, which is finite.If the context is clear, then we simply write σ k (M ) for σ k (M, g).
It is known [3,Thm. 1.1] that for any connected compact manifold (M, g) of dimension n ≥ 3, there exists a family (g ε ) of Riemannian metrics conformal to g which coincide with g on the boundary of M , such that 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 S n−1 .In dimension n = 2, it is proved in [4,Prop. 1.10] that each surface of revolution M ⊂ R 3 with boundary S 1 ⊂ R 2 × {0} is Steklov isospectral to the unit disk.In dimension n ≥ 3, many bounds were given.It is proved that each hypersurface of revolution M ⊂ R n+1 with one boundary component isometric to S n−1 satisfies σ k (M ) ≥ σ k (B n ), where B n is the Euclidean ball and equality holds if and only if M = B n × {0}, see [4,Thm. 1.8].In [5,Thm. 1], the authors show the following upper bound: if M ⊂ R n+1 is a hypersurface of revolution with one boundary component isometric to S n−1 , then for each k ≥ 1, we have where σ (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 ε > 0 and each k ≥ 1, there exists a hypersurface of revolution M ε such that σ (k) (M ε ) > k + n − 2 − ε.
These results concern hypersurfaces of revolution that have one boundary component isometric to 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 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 S n−1 is the warped product M = [0, L] × S n−1 endowed with the Riemannian metric g(r, p) = dr 2 + h 2 (r)g 0 (p), where (r, p) ∈ [0, L] × S n−1 , g 0 is the canonical metric of the (n − 1)-sphere of radius one and h : [0, L] −→ R * + is a smooth function which satisfies: (1) |h ′ (r)| ≤ 1 for all r ∈ [0, L]; (2) h(0) = h(L) = 1.
Assumption (1) comes from the fact that (M, g) is a hypersurface in Euclidean space R n+1 , see [4,Sect. 3.1] for more details.Assumption (2) implies that each component of the boundary is isometric to S n−1 , as commented in Fig. 1.
We now make some remarks on the terminology used throughout this paper.If M = [0, L] × S n−1 and h : [0, L] −→ 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 .Some lower bounds have already been obtained is this case.Indeed, [4,Thm. 1.11] states that if M ⊂ R n+1 , n ≥ 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 ≥ 1, Moreover, this inequality is sharp.In the case 0 < L ≤ 2, a lower bound is also obtained: 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 k n (L) such that for all metrics of revolution g on M , we have σ k (M, g) < B k n (L).Indeed, Proposition 3.3 of [4] states that if M = [0, L] × S n−1 is a hypersurface of revolution, then we have As such, a natural question is the following: Given the dimension n ≥ 3 and the meridian length L of M , does a metric of revolution g * on M exist, such that σ k (M, g) ≤ σ k (M, g * ) for all metrics of revolution g on M ?
Our investigations show that the answer is negative.Indeed, a sharp upper bound B k n (L) exists, but no metric of revolution on M = [0, L]×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 ∈ 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]×S n−1 , g 1 ) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to S n−1 and meridian length L. We suppose n ≥ 3. Then there exists a metric of revolution g 2 on M such that for each k ≥ 1, 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 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 k n (L), we must use information contained in g * .
We start by studying the case k = 1.We fix n ≥ 3 and L > 0 and search for a sharp upper bound B n (L) for σ 1 (M, g).In this case, we are able to calculate an expression for B n (L): g) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to S n−1 and dimension n ≥ 3. Then the first non trivial Steklov eigenvalue σ 1 (M, g) is bounded above, by a bound that depends only on the dimension n and the meridian length L of M : Moreover, this bound is sharp: for each ε > 0, there exists a metric of revolution We have the following asymptotic behaviour: see Fig. 4.
We also study the function L −→ 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 σ 1 (M, g) < B n : Then there exists a bound B n < ∞ such that for all hypersurfaces of revolution (M, g) in Euclidean space with two boundary components each isometric to S n−1 , we have where L 1 is the unique real positive solution of the equation Moreover, this bound is sharp: for each ε > 0, there exists a hypersurface of revolution with two boundary components each isometric to a unit sphere We say that L 1 is a critical length associated with k = 1, see Definition 8.
Proposition 5. Let n ≥ 3, and let L 1 = L 1 (n) be the critical length associated with k = 1.
Then we have: Note that the behaviour of L 1 is surprising since we know that when n is fixed, then L ≪ 1 implies σ 1 (M, g) ≪ 1.Indeed, by [4,Prop. 3.3], we have Now that we have provided information about sharp upper bounds for σ 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: Given the information that σ 1 (M = [0, L] × S n−1 , g) is close to the sharp upper bound B n , can we conclude that the meridian length L of M is close to the critical length L 1 ?
The answer to this question is positive.Indeed we will prove that if L is not close to L 1 , then σ 1 (M, g) is not close to B n .Additionally, given the information that σ 1 (M, g) is δ-close to B n , we will show that the distance between L and L 1 is less than δ, up to a constant of proportionality which depends only on the dimension n.
Theorem 6.Let M = [0, L] × S n−1 , with L > 0 and n ≥ 3. We suppose L ̸ = L 1 .Then there exists a constant C(n, L) > 0 such that for all metrics of revolution g on M , we have Moreover, there exists a constant C(n) > 0 such that for all 0 < δ < Bn−(n−2)

2
, we have We also consider the following question about stability properties: Given the information that σ 1 (M, g) is close to the sharp upper bound B n (L), can we conclude that the metric of revolution g is close (in a sense that is defined below) to the degenerated maximizing metric g * ?
We prove that if g is not close to g * , then σ 1 (M, g) is not close to B n (L).
For this purpose, given m ∈ [1, 1 + L/2), we define M m :={metrics of revolution g on M induced by a function h such that max The collection 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 M m can be.
We get the following result: ) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to S n−1 and dimension n ≥ 3. Let m ∈ [1, 1 + L/2) and M m as above.Then there exists a constant C(n, L, m) > 0 such that for all g ∈ M m , we have These results solve the case k = 1.Therefore, it would be interesting to find the same kind of results for any k ≥ 1.After having calculated sharp upper bounds for some higher values of k in Sect.6.1 and 6.2, we will see that in order to get an expression for B k n (L), we need to distinguish between many cases.As such, giving a general formula for B k n (L) or B k n := sup L∈R * + {B k n (L)} via this method seems difficult.We discuss this in Remark 20.
Definition 8. We say that These lengths are critical in the following sense: if L k ∈ R * + is a finite critical length for a certain k ∈ N and if we write g * the degenerated maximizing metric on Given n ≥ 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 ≥ 3. Then there exist infinitely many k ∈ N which have a finite critical length associated with them.Moreover, if we call i=1 the associated sequence of finite critical lengths, then we have 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 ∈ N has a critical length at infinity, see [5,Prop. 7].Nevertheless, in our case, Theorem 9 guarantees that there exist infinitely many k ∈ 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): Given n ≥ 3, are there finitely or infinitely many k ∈ N such that k has a critical length at infinity?
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.
Acknowledgment.I would like to warmly thank my thesis supervisor Bruno Colbois for offering me the opportunity to work on this topic, and for his precious help which enabled me to solve the difficulties encountered.I also want to thank several of my friends and colleagues, Antoine Bourquin, Laura Grave De Peralta and Flavio Salizzoni, for various discussions we had, their help and advice.Moreover, I would like to thank the anonymous referee for their careful proofreading and their comments which have improved the paper greatly.

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.

Variational characterization of the Steklov eigenvalues
Let (M, g) be a Riemannian manifold with smooth boundary Σ.Then we can characterize the kth Steklov eigenvalue of M by the following formula: where Another way to characterize the kth eigenvalue of M is given by the Min-Max principle: where 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.
Proposition 10.Let (M, g) be a hypersurface of revolution as in Definition 1. Then each eigenfunction on M can be written as 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 [

Mixed problems and their variational characterizations
Let (N, ∂N ) be a smooth compact connected Riemannian manifold and A ⊂ N be a domain which satisfies ∂N ⊂ ∂A.We suppose that ∂A is smooth and we call ∂ int A the intersection of ∂A with the interior of N .
Definition 11.The Steklov-Dirichlet problem on A is the eigenvalue problem It is well known that this mixed problem possesses solutions that form a discrete sequence The variational characterization of the kth Steklov-Dirichlet eigenvalue is the following: , where H k+1,0 is the set of all (k + 1)-dimensional subspaces in the Sobolev space Definition 12.The Steklov-Neumann problem on A is the eigenvalue problem It is well known that this mixed problem possesses solutions that form a discrete sequence The variational characterization of the kth Steklov-Neumann eigenvalue is the following: , where H k+1 is the set of all (k + 1)-dimensional subspaces in the Sobolev space H 1 (A).

Mixed problems on annular domains
Let B 1 and B R be the balls in R n , n ≥ 3, with radius 1 and R > 1 respectively centered at the origin.The annulus A R is defined as follows: A R = B R \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 σ D (k) (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]: Then, for k ≥ 0, the (k)th eigenvalue (counted without multiplicity) of this problem is 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 φ l of the Steklov-Dirichlet problem on the annulus A R can be expressed as φ l (r, p) = α l (r)S l (p), where S l is an eigenfunction for the l th harmonic of the sphere S n−1 .
It is possible to compute σ N (k) (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 Then, for k ≥ 0, the (k)th eigenvalue (counted without multiplicity) of this problem is In the same manner as before, we have the following expression for the Steklov-Neumann eigenvalues, see [5,Prop. 5].
Lemma 16.Each eigenfunction ϕ l of the Steklov-Neumann problem on the annulus A R can be expressed as ϕ l (r, p) = β l (r)S l (p), where S l is an eigenfunction for the l th harmonic of the sphere S n−1 .

The degenerated maximizing metric
A particular case of hypersurfaces of revolution is the following: let M = [0, L] × S n−1 be endowed with a metric of revolution g(r, p) = dr  This implies that the ε-neighborhood of the boundary consists of two disjoint copies of an annulus with inner radius 1 and outer radius 1 + ε.
This particular case is the key idea that we use to prove Theorem 2. We prove it now.
Besides, for f a smooth function on M , we have where ∇f is the gradient of f seen as a function of p.
Using the Min-Max principle, we can conclude that for all k ≥ 1, we have σ k (M, g 1 ) ≤ σ 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 ∂ r f does not vanish on any interval, then the inequality is strict.
Let k ≥ 1 be an integer.Let E k+1 := Span(f 0,2 , . . ., f k,2 ), where f i,2 is a Steklov eigenfunction associated with σ i (M, g 2 ).We can choose these functions such that for all i = 0, . . ., k, we have and hence Let ).We now consider two cases: 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 (a) If λ 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 ≥ 1.
(b) If λ j ̸ = 0, then u j cannot be locally constant, otherwise the ODE is not satisfied.
Hence u j is not locally constant and then ∂ r f * does not vanish on any interval.Therefore, using the Min-Max principle (2), we have 2. Let us suppose f * = k i=0 a i f i,2 such that there exists 0 ≤ i < k such that a i ̸ = 0. Then by the Min-Max principle (2), we have In both cases, we have 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 ).The sequence (h i ) uniformly converges to the function 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 .

The first non trivial eigenvalue
In this Section, we prove Theorem 3. The idea consists of comparing σ 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 Section 2.2.Then, to show that the upper bound B n (L) given is sharp, we take a metric of revolution g ε on M that is close to the degenerated maximizing metric g * and show that σ 1 (M, g ε ) is close to B n (L).
Proof.Let (M = [0, L] × S n−1 , g) be a hypersurface of revolution, where L > 0 is the meridian length of M .We recall that the boundary Σ of M consists of two disjoint copies of S n−1 .We want to find a sharp upper bound B n (L) for σ 1 (M, g).
We consider A 1+L/2 the annulus of inner radius 1 and outer radius 1 + L/2.Let φ 0 be an eigenfunction for the first eigenvalue of the Steklov-Dirichlet problem on A 1+L/2 , i.e.
We define a new function The function φ0 is continuous and we can check that Hence, thanks to formula (1), the function φ0 can be used as a test function for σ 1 (M, g).We have where the second strict inequality comes from the existence of a continuum of points r ∈ [0, L/2] such that h(r) < 1 + r.
If ϕ 1 is an eigenfunction for the first non trivial eigenvalue of the Steklov-Neumann problem on A 1+L/2 , i.e The function φ1 is continuous and we can check that hence we can use it as a test function for σ 1 (M, g).The same calculations as in (4) show that Putting Inequality (4) and Inequality ( 5) together, we get We will now prove that the bound B n (L) is sharp.This means that for each ε > 0, there exists a metric of revolution g ε on M such that σ 1 (M, g ε ) > B n (L) − ε.
Let f 1 be an eigenfunction for σ 1 (M, g ε ).Because (M, g ε ) is symmetric, then we can choose f 1 symmetric or anti-symmetric, which means that for all r ∈ [0, L] and p ∈ S n−1 , we have Moreover, it results from the calculations in (4) that for any symmetric or anti-symmetric function f , we have We will compare Hence, we have We now have two cases: 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 ≡ 1/Vol(S n−1 )), and u 0 is smooth.
we have Therefore, we can use f 1| [0,L/2]×S n−1 as a test function for σ D 0 (A 1+L/2 ), and we can state 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 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 ∈ [0, L] and because f 1 is smooth, we can conclude Therefore, we can use f 1| [0,L/2]×S n−1 as a test function for σ N 1 (A 1+L/2 ) and we can state But we defined B n (L) as Hence we have < σ 1 (M, g ε ) + ε and then From this result we can prove Corollary 4.
Proof.By Theorem 3, the inequality (6) holds which is We consider the two functions We can show that L −→ σ D 0 (A 1+L/2 ) is strictly decreasing with L. Indeed, let L ′ > L and let φ 0 be an eigenfunction for σ D 0 (A 1+L/2 ).We consider the extension by 0 of φ 0 to the annulus A 1+L ′ /2 .We get where the strict inequality comes from the fact that φ0 is not an eigenfunction associated with σ D 0 (A 1+L ′ /2 ).Indeed, if we suppose that φ0 is an eigenfunction for σ D 0 (A 1+L ′ /2 ), then it is harmonic in A 1+L ′ /2 (since it satisfies the Steklov-Dirichlet problem), and since φ0 vanishes on the open set A 1+L ′ /2 \A 1+L/2 , then by [1] φ0 is constant, which is a contradiction.
In the same way, we can show that L −→ σ N 1 (A 1+L/2 ) is strictly increasing with L. Indeed, let L ′ > L and let ϕ 1 be an eigenfunction for σ N 1 (A 1+L ′ /2 ).We consider the restriction of ϕ 1 to the annulus A 1+L/2 .We get Hence the bound we gave possesses a maximum depending only on the dimension n, given by where L 1 is the unique positive solution of the equation In order to prove that this bound is sharp, let ε > 0. We define M ε := [0, L 1 ] × S n−1 .Theorem 3 guarantees that there exists a metric of revolution 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 holds.To ease notation, we substitute (1 + L/2) by R and we can state that there is a unique value of R ∈ (1, ∞) such that the equality holds.This equation is equivalent to and we call We call Then, for R 1 to be such that Ψ n (R 1 ) = 0, it is necessary that ψ n (R 1 ) < 0.
Thus, As we substituted (1 + L/2) by R, and we can state that Therefore, since 3 ln(n−1) Moreover, we have

Stability properties of hypersurfaces of revolution
The goal of this section is to prove Theorem 6 and Theorem 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 ̸ = L 1 and compare σ 1 (M = [0, L] × 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.

Proof of Theorem 6
Recall that here we suppose L ̸ = L 1 .
Proof.Let g be any metric of revolution on M = [0, L] × S n−1 .Then we have 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 ̸ = L 1 .
Then we have

2
, and let us suppose |B n − σ 1 (M, g)| < δ.Therefore, we have |B n − σ 1 (M, g * )| < δ, where we wrote g * the degenerated maximizing metric on M .We consider two cases: 1. We suppose L 1 < L. In this case, we have Hence we have We can calculate that Thus, we have Therefore, since n ≥ 3 and h ≤ h m , we have We can now consider a new function hm , obtained from h m by smoothing out the two non-smooth points, with hm satisfying: 1.For all r ∈ [0, L], we have h m (r) ≤ hm (r); 2. The metric gm induced by hm is a metric of revolution in the sense of Definition 1.
We define C(n, L, m) := B n (L)−σ 1 (M, gm ), which is strictly positive by Theorem 3. Then we have

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 λ (k) , σ (k) , σ D (k) , σ N (k) 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] × S n−1 , g), we want to provide information about the multiplicity of the Steklov eigenvalues of (M, g).
6.1 Upper bound for σ 2 (M, g), . . ., σ m 1 (M, g) In this section, we prove the following theorem: Theorem 18.Let (M = [0, L] × S n−1 , g) be a hypersurface of revolution in Euclidean space with two boundary components each isometric to S n−1 and dimension n ≥ 3. Let m 1 be the multiplicity of the first non trivial eigenvalue of the classical Laplacian problem on (S n−1 , g 0 ).Then we have Moreover, this bound is sharp: for all ε > 0 there exists a metric of revolution g ε on M such that Proof.We consider two cases.
In the same way, we write and we can conclude Therefore, we already have a sharp upper bound for these eigenvalues, which is given by σ N 1 (A 1+L/2 ).
The proof of sharpness goes as in the proof of Theorem 3. Remark 20.It is then tempting to search for an expression for B k n := sup L∈R * + {B k n (L)} for any n and k; but it seems to be hard to give an explicit formula for it.Indeed, as Sect.6.1 and 6.2 suggest, the function L −→ B k n (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. Either k has a finite critical length, i.e B k n = B k n (L k ) for a certain L k ∈ R * + .That is for instance the case of σ 1 (M, g) or σ m 1 +1 if n = 3, 4, 5 or 6; 2. Or k has a critical length at infinity, i.e B k n = lim L→∞ B k n (L).That is for instance the case of σ 2 (M, g), . . ., σ m 1 (M, g).
Furthermore, we will prove in Sect.7 that for all n ≥ 3, there are infinitely many k that have a finite critical length associated to them.In all these cases, the function L −→ B k n (L) is piecewise smooth.

Critical lengths of hypersurfaces of revolution
We recall that given n ≥ 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 k n = B k n (L k ) for a certain finite L k ∈ R * + , and that the sequence of critical lengths converges to 0.
Proof.As before, for j ≥ 0, we denote by m j the number given by the formula (8), which is the multiplicity of σ D (j) (A R ) as well as the multiplicity of σ N (j) (A R ).Let i ≥ 2 be an integer.We claim that for all k such that we have Hence, we solve the equation σ D (0) (A 1+L D /2 ) = n − 1, i.e we find the unique L D > 0 such that We find Similarly, solving the equation We have to find for which values of n we have L D < L N and vice versa.This leads to the inequality n(n + 1) 2 We analyze the function f : [9, ∞) −→ R, x −→ 1 2 x 14 11 .We have f (9) > 8, and f ′ (x) = 14 22 x 3 11 .We can compute that f ′ (9) > 1 and since f ′′ (x) = 42 242 x −9 11 > 0 for all x ∈ [9, ∞), we can conclude f ′ (x) > 1 for all x ∈ [9, ∞).Hence, f (x) > x − 1 for all x ∈ [9, ∞).
Therefore, for all integers n ≥ 9, we have > n − 1 and then L D < L N .We can compute the cases n = 3, . . ., 8 and we can conclude that Therefore, we have

Figure 1 :
Figure 1: Since h(0) = h(L) = 1, the boundary of M consists of two copies of S n−1 .
2 + h 2 (r)g 0 (p).Let us suppose that there exists ε > 0 such that h(r) = 1 + r on [0, ε].Let us consider the connected component of the boundary S 0 associated with h(0).Then the ε-neighborhood of S 0 is an annulus with inner radius 1 and outer radius 1 + ε.

Figure 2 :
Figure 2: On [0, ε], we have h(r) = 1 + r and on [L − ε, L], we have h(r) = −r + L + 1.This implies that the ε-neighborhood of the boundary consists of two disjoint copies of an annulus with inner radius 1 and outer radius 1 + ε.

Figure 3 :
Figure 3: On the left, M = [0, L] × S n−1 is endowed with a metric g i of the sequence.On the right, M = [0, L] × S n−1 is endowed with another metric g j of the sequence, j > i.

Figure 4 :
Figure 4: Representation of the case n = 5.The decreasing smooth curve is L −→ σ D 0 (A 1+L/2 ) while the increasing smooth curve is L −→ σ N 1 (A 1+L/2 ).The solid curve is the bound B 5 (L) given by Theorem 3.

Figure 5 :
Figure 5: Since g ∈ M m , the function h which induces g satisfies h ≤ h m .

Figure 6 :
Figure 6: Representation of the case n = 5.The solid curve is the bound given in Theorem 19.