Eigenvalues on Riemannian Manifolds

Publicações Matemáticas

Changyu Xia UnB

o 29 Colóquio Brasileiro de Matemática

Impresso no Brasil / Printed in Brazil

Capa: Noni Geiger / Sérgio R. Vaz

29o Colóquio Brasileiro de Matemática

• Análise em Fractais – Milton Jara • Asymptotic Models for Surface and Internal Waves - Jean-Claude Saut • Bilhares: Aspectos Físicos e Matemáticos - Alberto Saa e Renato de Sá Teles • Controle Ótimo: Uma Introdução na Forma de Problemas e Soluções - Alex L. de Castro • Eigenvalues on Riemannian Manifolds - Changyu Xia • Equações Algébricas e a Teoria de Galois - Rodrigo Gondim, Maria Eulalia de Moraes Melo e Francesco Russo • Ergodic Optimization, Zero Temperature Limits and the Max-Plus Algebra - Alexandre Baraviera, Renaud Leplaideur e Artur Lopes • Expansive Measures - Carlos A. Morales e Víctor F. Sirvent • Funções de Operador e o Estudo do Espectro - Augusto Armando de Castro Júnior • Introdução à Geometria Finsler - Umberto L. Hryniewicz e Pedro A. S. Salomão • Introdução aos Métodos de Crivos em Teoria dos Números - Júlio Andrade • Otimização de Médias sobre Grafos Orientados - Eduardo Garibaldi e João Tiago Assunção Gomes

Distribuição: IMPA Estrada Dona Castorina, 110 22460-320 Rio de Janeiro, RJ E-mail: [email protected] ISBN: 978-85-244-0354-5 http://www.impa.br

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Contents

1 Eigenvalue problems on Riemannian manifolds 1 1.1 Introduction ...... 1 1.2 Some estimates for the ﬁrst eigenvalue of the Laplacian 5

2 Isoperimetric inequalities for eigenvalues 14 2.1 Introduction ...... 14 2.2 The Faber-Krahn Inequality ...... 16 2.3 The Szeg¨o-Weinberger Inequality ...... 17 2.4 The Ashbaugh-Benguria Theorem ...... 19 2.5 The Hersch Theorem ...... 23

3 Universal Inequalities for Eigenvalues 29 3.1 Introduction ...... 29 3.2 Eigenvalues of the Clamped Plate Problem . . . . 34 3.3 Eigenvalues of the Polyharmonic Operator . . . 46 3.4 Eigenvalues of the Buckling Problem ...... 57

4 Pólya Conjecture and Related Results 73 4.1 Introduction ...... 73 4.2 The Kröger’sTheorem ...... 77 4.3 A generalized Pólya conjecture by Cheng-Yang . . 81 4.4 Another generalized Pólya conjecture ...... 85

5 The Steklov eigenvalue problems 94 5.1 Introduction ...... 94 5.2 Estimates for the Steklov eigenvalues ...... 95

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4 CONTENTS

Bibliography 103

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Chapter 1

Eigenvalue problems on Riemannian manifolds

1.1 Introduction

Let (M, g) be an n-dimensional Reimannian manifold with boundary (possibly empty). The most important operator on M is the Lapla- n cian ∆. In local coordinate system xi i=1, the Laplacian is given by { }

1 n ∂ ∂ ∆ = √Ggij √ ∂x ∂x G i,j=1 i j X ij 1 ∂ ∂ where (g ) is the inverse matrix (g )− , g = g( , ) are the ij ij ∂xi ∂xj coeﬃcients of the Riemannian metric in the local coordinates, and G = det(gij ). In local coordinates, the Riemannian measure dv on (M, g) is given by

dv = √Gdx 1...dxn.

Let φ C∞(M) and set ∈

φ 2 = φ 2 + φ 2. || ||1 |∇ | | | ZM ZM 1

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2 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

Here and in the future, the integrations on M are always taken with 2 respect to the Riemannian measure on M. Let us denote by H1 (M) o 2 and H1 (M) the completion of C∞(M) and C0∞(M) with respect to o 2 2 . The theory of Sobolev spaces tells us that H1 (M) = H1 (M) when|| || M is complete. Our purpose is to study some eigenvalue problems associated to the Laplacian operator on a compact manifold M. When ∂M = , we consider the closed eigenvalue problem: ∅ ∆u + λu = 0. (1.1)

When ∂M = , we are interested in the following eigenvalue problems. 6 ∅

The Dirichlet problem: • ∆u = λu in M, (1.2) u = 0. |∂M

The Neumannn problem: • ∆u = λu in M, (1.3) ∂u = 0, ∂ν ∂M

where ν is the unit outward normal to ∂M. The clamped plate problem: • ∆2u = λu in M, (1.4) u = ∂u = 0, |∂M ∂ν ∂M The buckling problem: • ∆2u = λ∆u in M, − (1.5) u = ∂u = 0, |∂M ∂ν ∂M The eigenvalue problem of poly-harmonic operator: • ( ∆)lu = λu in M, − ∂u− ∂l−1u (1.6) u ∂M = ∂ν = = l−1 = 0, l 2. ( | ∂M ··· ∂ν ∂M ≥

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[SEC. 1.1: INTRODUCTION 3

The buckling problem of arbitrary order: • ( ∆)lu = λ∆u in M, − ∂u− ∂l−1u (1.7) u ∂M = ∂ν = = l−1 = 0, l 2. ( | ∂M ··· ∂ν ∂M ≥

The Steklov problem of second order : • ∆u = 0 in M, (1.8) ∂u = λu on ∂M. ∂ν The Steklov problem of fourth order: • ∆2u = 0 in M, (1.9) u = ∆u λ ∂u = 0 on ∂M. − ∂ν

Let us denote by λ1 the ﬁrst non-zero eigenvalue of the above problems. We can arrange the eigenvalues of these problems as follows:

0 < λ λ + . 1 ≤ 2 ≤ · · · → ∞ For many reasons in Mathematics and Physics, it is important to obtain nice estimates for the λ′s. We will concentrate our attention on this problem. Let us list some basic facts in this direction.

Theorem 1.1 (Weyl’s asymptotic formula, [97]). In each of the eigenvalue problems (1.1), (1.2), (1.3), let N(λ) be the number of eigenvalues, counted with multiplicity, λ. Then ≤ N(λ) ω M λn/2/(2π)n (1.10) ∼ n| | n as λ , where ωn is the volume of the unit ball in R and M is the volume→ ∞ of M. In particular, | |

λn/2 (2π)n/ω k/ M (1.11) k ∼ { n} | | as λ + . → ∞ There are similar asymptotic formulas for the other eigenvalue problems above (Cf. [1], [79], [80]). Deﬁne a space H as follows:

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4 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

For the closed eigenvalue problem (1.1),

H = f H2(M) f = 0 . (1.12) ∈ 1 ZM

For the Dirichlet eigenvalue problem (1.2), o 2 H = H1 (M). (1.13) For the Neumann eigenvalue problem (1.3),

H = f H2(M) f = 0 . (1.14) ∈ 1 ZM

A fundamental fool in the theory of eigenvalues is the Mini-Max principle. We can ﬁnd a countable orthonormal basis fi , fi C∞(M) for the problems (1.1), (1.2) and (1.3) such that{ } ∈

f 2 M |∇ | λ1 = inf f 2 f H , R M | ∈ f 2 (1.15) n MR |∇ | o λi = inf f 2 f H, M ffj = 0, j = 1, , i 1 . R M | ∈ ··· − n o In particular, weR have the R

Poincar´einequality:

f 2 λ f 2, f H. (1.16) |∇ | ≥ 1 ∀ ∈ ZM ZM

For other eigenvalue problems above, similar mini-max principles also hold.

Theorem 1.2 (The Co-Area formula, [17]). Let M be a compact Riemannian manifold with boundary, f H1(M). Then for any non-negative function g on M, ∈

∞ g g = dσ (1.17) M f=σ f ! Z Z−∞ Z{ } |∇ |

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[SEC. 1.2: SOME ESTIMATES FOR THE FIRST EIGENVALUE OF THE LAPLACIAN 5 1.2 Some estimates for the ﬁrst eigenvalue of the Laplacian

In this section, we will prove some estimates for the ﬁrst eigenvalue of the Laplacian.

Theorem 1.3 ([73]). Let M be an n-dimensional complete Rie- mannian manifold with Ricci curvature RicM n 1. Then the first non-zero eigenvalue of the closed eigenvalue≥ problem− (1.1) of M satisfies λ (M) n. 1 ≥ The proof of Theorem 1.3 can be carried out by substituting a first eigenfunction into the Bochner formula and integrating on M the resulted equality (see the proof of theorem 1.6 below). An important classical result about eigenvalue is the following

Theorem 1.4 (Cheng’s Comparison Theorem, [18]). Let M be an n-dimensional complete Riemannian manifold with Ricci curvature satisfying RicM (n 1)c and let BR(p) be an open geodesic ball of radius R around≥ a− point p in M, where R < π/√c , when c > 0. Then the ﬁrst eigenvalue of the Dirichlet problem (1.2) of BR(p) satisﬁes

λ (B (p)) λ (B (c)), (1.18) 1 R ≤ 1 R

with equality holding if and only if BR(p) is isometric to BR(c), where BR(c) is a geodesic ball of radius R in a complete simply connected Riemannian manifold of constant curvature c and of dimension n.

An immediate application of Cheng’s eigenvalue comparison theorem is a rigidity theorem for compact manifolds of positive Ricci curvature.

Theorem 1.5 (The Maximal Diameter Theorem, [18]). Let M be an n-dimensional complete Riemannian manifold with Ricci curvature RicM n 1. If the diameter of M satisﬁes d(M) π, then M is isometric≥ to− an n-dimensional unit sphere. ≥

Proof. Take two points p, q M so that d(p, q) π; then B (p) B (q) = . Let f and g∈be the ﬁrst eigenfunctions≥ corre- π/2 ∩ π/2 ∅

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6 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

sponding to the ﬁrst Dirichlet eigenvalues of Bπ/2(p) and Bπ/2(q), respectively. We extend f and g on the whole M by setting f M B (p) = | \ π/2 g M B (q) = 0 and take two non-zero constants a and b such that | \ π/2

(af + bg) = 0 ZM Observe that the ﬁrst Dirichlet eigenvalue of an n-dimensional unit hemisphere is n. The mini-max principle and Cheng’s comparison theorem then imply that

n λ (M) ≤ 1 (af + bg) 2 M |∇ | ≤ (af + bg)2 R M a2 f 2 + b2 g 2 R B 2(p) B 2(q) = π/ |∇ | π/ |∇ | a2 f 2 + b2 g2 R Bπ/2(p) RBπ/2(q) 2 2 2 2 a λ1(BR π/2(p)) B (p)Rf + b λ1(Bπ/2(q)) B (q) g = π/2 π/2 a2 f 2 + b2 g2 RBπ/2(p) Bπ/2(q) R 2 2 2 2 na B (p) f R+ nb B (q) gR π/2 π/2 = n. ≤ a2 f 2 + b2 g2 RBπ/2(p) BRπ/2(q) R R We conclude from the equality case of the mini-max principle and Cheng’s comparison theorem that each of Bπ/2(p) and Bπ/2(q) is isometric to the n-dimensional unit hemisphere and

M = B (p) B (q) π/2 ∪ π/2 Consequently, M is isometric to a unit n-sphere.

The maximal diameter theorem can be also used to prove the Obata theorem below.

Theorem 1.6 ([76]). Let M be an n-dimensional complete Rie- mannian manifold with Ricci curvature RicM n 1. If the ﬁrst non-zero eigenvalue of the closed eigenvalue problem≥ − (1.1) of M is n, then M is isometric to a unit n-sphere.

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[SEC. 1.2: SOME ESTIMATES FOR THE FIRST EIGENVALUE OF THE LAPLACIAN 7

Proof. Let f be a ﬁrst eigenfunction corresponding to the ﬁrst eigenvalue n of M. From the Bochner formula, we get

1 ∆ f 2 = 2f 2 + f, (∆f) + Ric( f, f) (1.19) 2 |∇ | |∇ | h∇ ∇ i ∇ ∇ (∆f)2 n f 2 + (n 1) f 2 = nf 2 f 2. ≥ n − |∇ | − |∇ | − |∇ |

2 2 Integrating on M and noticing M (nf f ) = 0, we conclude that the inequalities in 1.19 should take equality− |∇ | sign. Thus, we have R 1 1 1 ∆( f 2 + f 2) = ∆ f 2 + ∆f 2 2 |∇ | 2 |∇ | 2 = nf 2 f 2 + f∆f + f 2 = 0 − |∇ | |∇ | and so f 2 + f 2 is a constant. Without lose of generality, we can assume|∇ that| f 2 + f 2 = 1 and so |∇ | f |∇ | = 1. 1 f 2 − Let p and q be points of Mp such that f(p) = f(q) = 1 and take a unit speed minimizing geodesic γ : [0, l] M from− p to−q. Integrating the above equation along γ, we obtain →

f 1 dt l = ds = |∇ | ds = π. 2 2 γ γ 1 f ≥ 1 √1 t Z Z − Z− − It then follows from thep maximal diameter theorem that M is isometric to an unit n-sphere.

Remark 1.1. Let M n be a compact Riemannian manifold with Ricci curvature RicM n 1 and nonempty boundary. If the mean curvature of ∂M is nonnegative,≥ − then the first Dirichlet eigenvalue n of M satisfies λ1 n with equality holding if and only if M is isometric to an n-dimensional≥ unit hemisphere [82]. Similarly, if the boundary of M is convex, then the first non-zero Neumann eigenvalue n of M must satisfy λ1 n with equality holding if and only if M is isometric to an n-dimensional≥ unit hemisphere [34, 100].

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8 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

We now prove another rigidity theorem using the techniques of eigenvalues.

Theorem 1.7 ([101]). Let M be an n-dimensional complete Rie- mannian manifold with Ricci curvature RicM n 1 and let N be a closed minimal hypersurface which divides M ≥into− two disjoint open domains Ω1 and Ω2. If there exists a point p M such that d(p, N), the distance from p to N, is no less than π/2,∈ then the pair (M,N) is n n 1 n isometric to the pair (S (1), S − (1), being S (1) the unit n-sphere.

Proof. Assume without lose of generality that p Ω1. We know from d(p, N) π/2 that B (p) Ω . It then follows∈ from the do- ≥ π/2 ⊂ 1 main monotonicity [17] that the ﬁrst Dirichlet eigenvalues of Bπ/2(p) and Ω1 satisfy

λ B (p) λ (Ω ). (1.20) 1 π/2 ≥ 1 1 On the other hand, Cheng’s comparison theorem tells us that

λ B (p) n (1.21) 1 π/2 ≤ and Reilly’s estimate implies that λ (Ω ) n. Thus, the inequalities 1 1 ≥ in (1.20) and (1.21) should be equalities. Consequently, Bπ/2(p) = Ω1 is isometric to an n-dimensional unit hemisphere and so N = ∂Ω1 = n 1 S − (1) is totally geodesic. It then follows from a result of [39] that Ω2 is also isometric to an n-dimensional unit hemisphere.

Let λ1 be the least nontrivial eigenvalue of an n-dimensional compact manifold M and let φ be the corresponding eigenfunction. By multiplying with a constant it is possible to assume that

a 1 = inf φ; a + 1 = sup φ − M M where 0 a(φ) < 1 is the median of φ. Suppose≤ that M n is a compact manifold without boundary of nonnegative Ricci curvature and of diameter d. Li-Yau [72] showed that the ﬁrst nontrivial eigenvalue satisﬁes

π2 λ 1 ≥ (1 + a)d2

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[SEC. 1.2: SOME ESTIMATES FOR THE FIRST EIGENVALUE OF THE LAPLACIAN 9

and conjectured that

π2 λ . (1.22) 1 ≥ d2 Li-Yau’s conjecture was proved by Zhong and Yang in [103]. Let us provide a proof of (1.22) given by Li in [71].

Lemma 1.1. The function 2 z(u) = arcsin(u) + u 1 u2 u π − − p deﬁned on [-1, 1] satisﬁes

2 uz′ + z′′(1 u ) + u = 0; (1.23) −

2 z′ 2zz′′ + z′ 0; (1.24) − ≥

2z uz′ + 1 0; (1.25) − ≥ and

1 u2 2 z . (1.26) − ≥ | | Proof. Diﬀerentiating yields

4 2 4u z′ = 1 u 1, z′′ = − . π − − π√1 u2 p − Thus (1.23) is satisﬁed. To see (1.24), we note that

2 4 4 2 2 z′ 2zz′′ + z′ = 1 u + u arcsin u (1 + u ) . − π√1 u2 π − − − p Since the right hand side is an even function, it suﬃces to check that 4 1 u2 + u arcsin u (1 + u2) 0 π − − ≥ p

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10 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

on [0, 1]. It is easy to see that d 4 4 1 u2 + u arcsin u (1 + u2) = arcsin u 2u du π − − π − p which is nonpositive on [0, 1]. Hence 4 1 u2 + u arcsin u (1 + u2) π − − 4p 1 u2 + u arcsin u (1 + u2) ≥ π − − u=1 p = 0.

Inequality (1.25) follows easily because 4 2z uz′ + 1 = arcsin u + 1 u 0. − π − ≥ To see (1.26), let us consider the cases 1 u 0 and 0 u 1 separately. It is clearly that the inequality− ≤ is valid≤ at -1,≤ 0 and≤ 1. Setting 4 f(u) = 1 u2 arcsin u + u 12 u2 + u; − − π − p then

4 2 f ′ = 2u (2 1 u ) +2, − − π − p 8u f ′′ = 2 + , − π√1 u2 − and 8 f ′′′ = . π1( u2)3/2 − When 1 u 0, f ′′ 0. Hence f(u) min f( 1), f(0) = 0. For − ≤ ≤ ≤ ≥ { − } the case 0 u 1, f ′′′ 0. Thus ≤ ≤ ≥ 8 f ′ max f ′(0), f ′(1) = max 2 , 0 = 0. ≤ { } − π

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[SEC. 1.2: SOME ESTIMATES FOR THE FIRST EIGENVALUE OF THE LAPLACIAN 11

Therefore f(u) f(1) which proves (1.26). ≥ Lemma 1.2. Suppose M is a compact manifold without boundary of nonnegative Ricci curvature. Assume that a nontrivial eigenfunction φ corresponding to the eigenvalue λ is normalized so that for 0 a < 1, a + 1 = sup φ and a 1 = inf φ. If u = φ a, then ≤ M − M − 2 λ(1 u2) + 2aλz(u) (1.27) |∇| ≤ − where 2 z(u) = arcsin u + u 1 u2 u. (1.28) π − − p

Proof. We need only to prove an estimate similar to (1.27) for u = ǫ(φ a) where 0 < ǫ < 1. The lemma will follow by letting ǫ 0. By− the deﬁnition of u; we have → ∆u = λ(u + ǫa) − with ǫ u ǫ. We may assume a > 0. Consider the function − ≤ ≤ Q = u 2 c(1 u2) 2aλz(u), |∇ | − − −

We can choose c large enough so that supM Q = 0. The lemma follows if c λ ; for a sequence of ǫ 1, hence we may assume that c > λ. ≤ → Let the maximizing point of Q be x0. We claim that u(x0) > 0 since otherwise u(x ) = 0 and |∇ | ∇ 0 0 = Q(x ) = c(1 u2)(x ) 2aλz(x ) (c aλ)(1 ǫ2) 0 − − 0 − 0 ≤ − − − by (1.26), which is a contradiction. Diﬀerentiating in the ei direction gives 1 Q = u u + cuu aλz′u . (1.29) 2 i j ji i − i

We can assume at x0 that u(x0) = u(x0) and using Qi = 0, we have |∇ |

2 2 u u u = (cu aλz′) . (1.30) ji ji ≥ 11 −

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12 [CAP. 1: EIGENVALUE PROBLEMS ON RIEMANNIAN MANIFOLDS

Diﬀerentiating again, using the commutation formula, Q(x0) = 0, (1.26), (1.29), and (1.30), we get

1 0 ∆Q(x ) (1.31) ≥ 2 0 = ujiuji + uj(∆u)j + Ric( u, u) + (c aλz′′) + (cu aλz′)∆u 2 ∇ ∇ −2 − (cu aλz′) + (c λ aλz′′)(c(1 u ) + 2aλz) ≥ − − − − λ(cu aλz′)(u + ǫa) − − 2 2 2 2 = acλ((1 u )z′′ + uz′ + ǫu) + a λ ( 2zz′′ + z′ + ǫz′) − − − +aλ(c λ)( uz′ + 2z + 1) + (c λ)(c aλ). − − − − However by (1.23), (1.24), and (1.25), we conclude that

2 2 0 acλ(1 ǫ)u a λ (1 ǫ)z′ + (c λ)(c aλ) (1.32) ≥ − − − − − 4 acλ(1 ǫ) a2λ2(1 ǫ) 1 + (c λ)(c aλ) ≥ − − − − π − − − (c + λ)λ(1 ǫ) + (c λ)2. ≥ − − − This implies that

2 + (1 ǫ) + (1 ǫ)(9 ǫ) c λ − − − . ≤ ( p2 ) Taking ǫ 0 one gets the desired estimate. → Theorem 1.8. ([103]) Suppose M is a compact manifold without boundary whose Ricci curvature is nonnegative. Let a u 0 be the ≥ median of a normalized first eigenfunction with a + 1 = supM φ and a 1 = infM φ; and let d be the diameter. Then the first non-zero eigenvalue− of M satisfies

6 π 4 d2λ π2 + 1 a2 π2(1 + 0.02a2). (1.33) 1 ≥ π 2 − ≥

Proof. Let u = φ a and let γ be the shortest geodesic from the minimizing point of u−to the maximizing point with length at most

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[SEC. 1.2: SOME ESTIMATES FOR THE FIRST EIGENVALUE OF THE LAPLACIAN 13

d. Integrating the gradient estimate (1.27) along this segment with respect to arc-length and using oddness, we have

dλ1/2 ds ≥ Zγ u ds |∇ | 2 ≥ γ √12 u + az Z − 1 1 1 + du 2 2 ≥ 0 √12 u + az √1z u 2a Z − − − 1 1 3a2z2 2 + du ≥ √1 u2 1 u2 Z0 − − 2 1 z π + 3a2 2 ≥ 0 √1 u Z − 3a2 π 4 = π + 1 . π2 2 − Remark 1.2. It has been shown by Hang-Wang [40] that if the equality holds in (1.33) then M is isometric to a circle.

Remark 1.3. Let M n be a compact manifold with smooth boundary and nonnegative Ricci curvature. Suppose that the second fundamental form of M is nonnegative. Then the ﬁrst nontrivial eigenvalue of the Laplacian with Neumann boundary conditions also satisﬁes the inequality (1.27). The proof runs the same as Lemma 1.1 except that the possibility of the maximum of the test function Q at the boundary must be handled. In fact, the boundary convexity assumption implies that the maximum of Q cannot occur on the boundary.

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Chapter 2

Isoperimetric inequalities for eigenvalues

2.1 Introduction

In this chapter, we will prove some isoperimetric inequalities for eigenvalues on manifolds which have always been important problems in geometric analysis. Owing to the limitation on the materials, we only select some of the results in the area. For more interesting results, we refer to [3] , [8], [17] and the references therein. The isoperimetric inequalities to be proved are : the Faber-Krahn inequality for the first eigenvalue of the Dirichlet eigenvalue; the Szegö-Weinberger inequality for the first nontrivial Neumann eigenvalue; the Hersch theorem for the first closed eigenvalue on a compact Riemannian surface of genus zero; the Ashbaugh-Benguria theorem; etc. For the conve- nience of later use, we recall now the notion of spherically symmetric rearrangement. Suppose that f is a bounded measurable function on the bounded measurable set Ω Rn. Consider the distribution ⊂ function µf (t) defined by

µ (t) = x Ω f(x) > t (2.1) f |{ ∈ || | }| 14

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[SEC. 2.1: INTRODUCTION 15

where denotes Lebesgue measure. The distribution function can be viewed| · | as a function from [0, ) to [0, Ω ] and is nonincreasing. ∞ | | The decreasing rearrangement f ∗ of f, is the inverse of µf and is deﬁned by

f ∗(s) = inf t 0 µ (t) < s . (2.2) { ≥ | f } It is a nonincreasing function on [0, Ω ]. For a bounded measurable n | | set Ω R , its spherical rearrangement Ω∗ is deﬁned as the ball centered⊂ at the origin having the same measure as Ω. The spherically ⋆ (symmetric) decreasing rearrangement f :Ω∗ R is deﬁned by → ⋆ n ⋆ f (x) = f ∗(C x ) for x Ω (2.3) n| | ∈ n/2 n Rn where Cn = π /Γ 2 + 1 is the volume of the unt ball in . An important fact we will use is that Ω 2 | | 2 ⋆ 2 f = (f ∗(s)) ds = (f ) . (2.4) ∗ ZΩ Z0 ZΩ 1 It is known that for any function f in the Sobolev space H0 (Ω), ⋆ 1 f H (Ω∗) and ∈ 0

f ⋆ 2 f 2. (2.5) ∗ |∇ | ≤ |∇ | ZΩ ZΩ For two nonnegative measurable functions f and g on Ω we have

fg f ⋆g⋆. (2.6) ≤ ∗ ZΩ ZΩ Let us recall the notion of spherically (symmetric) increasing rearrangement, which we denote by a lower ⋆. The deﬁnition is almost identical to that of spherically decreasing rearrangement, except that g⋆ should be radially increasing (in the weak sense) on Ω∗. In this case, we have

⋆ fg f g⋆. (2.7) ≥ ∗ ZΩ ZΩ

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16 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES 2.2 The Faber-Krahn Inequality

In this section, we will prove the Faber-Krahn inequality which is one of the oldest isoperimetric inequalities for an eigenvalue.

Theorem 2.1 (Faber-Krahn [36],[61]). For a bounded domain Ω Rn, the ﬁrst Dirichlet eigenvalue satisﬁes ⊂

λ (Ω) λ (Ω∗) (2.8) 1 ≥ 1

with equality if and only if Ω = Ω∗.

Proof. Let u1 be a ﬁrst Dirichlet eigenfunction for Ω. We have from (2.4), (2.5) and the mini-max principle that

2 u1 λ (Ω) = Ω |∇ | (2.9) 1 u2 R Ω 1 2 u1 = ΩR|∇ | (u )2 R Ω∗ 1∗ u 2 RΩ∗ |∇ 1| ≥ (u )2 R Ω∗ 1∗ λ (Ω∗). ≥ R1 For the characterization of the case of equality, we refer to [56]. The Faber-Krahn inequality is valid for more general manifolds. Let M be an n-dimensional complete Riemannian manifold and for a ﬁxed κ R, let Mκ be the complete simply connected n-dimensional space form∈ of constant sectional curvature κ. To each bounded domain Ω in M, associate the geodesic ball D in Mκ satisfying

Ω = D . (2.10) | | | | If κ > 0 then only consider those Ω for which Ω < M . | | | κ| Theorem 2.2. If, for all such Ω in M, equality (2.10) implies the isoperimetric inequality

∂Ω ∂D , (2.11) | | ≥ | |

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[SEC. 2.3: THE SZEGO¨-WEINBERGER INEQUALITY 17

with equality in (2.11) if and only if Ω is isometric to D, then we also have, for every bounded domain Ω in M, that equality (2.10) implies the inequality for the ﬁrst Dirichlet eigenvalue

λ (Ω) λ (D), (2.12) 1 ≥ 1 with equality holding if and only if Ω is isometric to D.

For a proof of Theorem 2.2, we refer to [17].

2.3 The Szeg¨o-Weinberger Inequality

In this section, we prove the Szeg¨o-Weinberger inequality which is a counterpart to the ﬁrst non-zero Neumann eigenvalue of the Faber- Krahn inequality.

Theorem 2.3 ([96]). Let Ω be a bounded domain in Rn. Then the ﬁrst non-zero Neumann eigenvalue of Ω satisﬁes

λ (Ω) λ (Ω∗) (2.13) 1 ≤ 1

with equality holding if and only if Ω = Ω∗.

Proof. Let R be the radius of Ω∗ and let g be the solution of the equation

n 1 n 1 g′′ + − g′ −2 g + λ (Ω∗)g = 0 r − r 1 (2.14) g(0) = 0, g′(R) = 0

By a topological argument, we can take as trial functions Pi, such that P = 0 for i = 1, , n, with Ω i ··· R x P (x) = h(r) i , i r Rn where the xi′ s are Cartesian coordinates, x = (x1, , xn) , r = x , and ··· ∈ | | g(r) for 0 r R h(r) = g(R) for r≤ R.≤ ≥

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18 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

Observe that by an appropriate choice of sign, g(r) is increasing on [0,R] and hence that h is everywhere nondecreasing for r 0. By ≥ substituting our trial functions Pi into the mini-max inequality for λ1, we ﬁnd

λ (Ω) P 2 P 2. 1 i ≤ |∇ i| ZΩ ZΩ Summing this in i for 1 i n, we arrive at ≤ ≤ n 2 Pi λ (Ω) Ω i=1 |∇ | (2.15) 1 ≤ n P 2 R PΩ i=1 i 2 n 1 2 h′(r) + −2 h(r) = ΩR P r h(r)2 R Ω A(r) = Ω R h(r)2 RΩ where R

2 n 1 2 A(r) = h′(r) + − h(r) . (2.16) r2 A(r) is easily seen to be decreasing for 0 r R by differentiating and using the differential equation (2.14).≤ One≤ finds

2 3 A′(r) = 2(λ (Ω∗)hh′ + (n 1)(rh′ h) /r ) < 0, 0 < r < R.(2.17) − 1 − − Also, A(r) = (n 1)g(R)2/r2 for r R shows that A is decreasing for r > R. Since−A is continuous for≥ all r 0, it is also decreasing there. Observe that ≥

A(r) A(r) (2.18) ≤ ∗ ZΩ ZΩ since the volumes integrated over are the same in both cases, while in passing from the left to right hand sides we are exchanging integrating over Ω Ω∗ for integrating over Ω∗ Ω which are sets of equal volume. Since A\ is (strictly) decreasing this\ clearly increases the value of the integral unless Ω = Ω∗, when equality obtains. Similarly we ﬁnd that

h(r)2 h(r)2 (2.19) ≥ ∗ ZΩ ZΩ

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[SEC. 2.4: THE ASHBAUGH-BENGURIA THEOREM 19

since h is nondecreasing. Thus we arrive at

Ω∗ A(r) λ (Ω) = λ (Ω∗), (2.20) 1 ≤ h(r)2 1 RΩ∗ R since each Pi is precisely a Neumann eigenfunction of ∆ with eigenvalue λ1(BR) for the domain BR = Ω∗. This completes the proof of the Szeg¨o-Weinberger inequality, including the characterization of the case of equality.

2.4 The Ashbaugh-Benguria Theorem

In this section we consider the sharp upper bound for λ2/λ1 for the Dirichlet eigenvalue problem proved by Ashbaugh-Benguria. In 1955 and 1956, Payne, P´olya and Weinberger [77], [78], proved that

λ 2 3 for Ω R2 λ1 ≤ ⊂ and conjectured that

λ λ j2 2 2 = 1,1 λ ≤ λ j2 1 1 disk 0,1

with equality if and only if Ω is a disk and where jp,k denotes the th k positive zero of the Bessel function Jp(t). For general dimension n 2, the analogous statements are ≥ λ 4 2 1 + for Ω Rn, λ1 ≤ n ⊂ and the PPW conjecture

λ λ j2 2 2 = n/2,1 , (2.21) λ ≤ λ j2 1 1 n ball n/2 1,1 − −

with equality if and only if Ω is an n-ball. This important conjecture was proved by Ashbaugh-Benguria (see [5], [6], [7]).

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20 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

We proceed now with the proof of (2.21). Let us start from the variational principle for λ2

2 Ω φ λ2(Ω) = min |∇ | , (2.22) 1 2 φ H (Ω),0=φ u1 φ ∈ 0 6 ⊥ R Ω which, by integration by parts, leads to R

λ (Ω) λ (Ω) (2.23) 2 − 1 P 2u2 Ω |∇ | 1 , P u H1(Ω), P u2 = 0,P = 0. ≤ P 2u2 1 ∈ 0 1 6 R Ω 1 ZΩ To get the isoperimetricR result out of (2.23), one must make very special choices of the function P , in particular, choices for which (2.23) is an equality if Ω is a ball. Thus we shall use n trial functions P = P , such that P u2 = 0 for i = 1, , n where i Ω i 1 ··· x R P = g(r) i (2.24) i r and f(r) = “right” radial function on B for 0 r R, g(r) = R (2.25) f(R) for r R. ≤ ≤ ≥ The right R in this case turns out to be the unique R such that λ1(BR) = λ1(Ω). Substituting Pi into (2.23) and summing on i, we ﬁnd B(r)u2 λ (Ω) λ (Ω) Ω 1 (2.26) 2 − 1 ≤ f(r)2u2 RΩ 1 where R

2 n 1 2 B(r) = f ′(r) + − f(r) . (2.27) r2

Now the equation (2.26) does not depend on the Pi′s and so we are in a position to deﬁne the function f. The idea is to take f as a properly quotient of Bessel functions so that the equality occur if Ω is a ball in Rn. This motivates the choice of :

f(r) = w(γr), (2.28)

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[SEC. 2.4: THE ASHBAUGH-BENGURIA THEOREM 21

where

jn/2(βx) , if 0 x < 1, w(x) = jn/2−1(αx) ≤ (2.29) ( w(1) limx 1 w(x), if x 1, ≡ → ≥

with α = jn/2 1,1, β = jn/2,1 and γ = √λ1/α. − Lemma 2.1. The equality occurs in (2.26) when Ω is a ball with λ1 as the ﬁrst Dirichlet eigenvalue and f is given by (2.28).

n Proof. If S1 is a closed ball of R of appropriate radius centered in the origin in which the problem

∆z = λz in S , 1 (2.30) z =− 0. |∂S1

has λ1 as the ﬁrst eigenvalue, then

S = x Rn; x α/ λ1 = /γ . 1 { ∈ | | ≤ 1 } p ˜ β2 The second eigenvalue of the problem (2.30) is λ2 = α2 λ1. The ﬁrst eigenfunction of S1 is

1 n/2 z(x) = c x − jn/2 1( λ1 x ), | | − | | p and the eigenfunctions corresponding to λ˜2 are:

1 n/2 xi f (x) = c x − j ( λ˜ x ) , i = 1, , n, i | | n/2 2| | x ··· q | | where c is a non-zero constant. Let

˜ jn/2 √λ2r , if 0 r < 1/γ, Q(r) = jn/2−1(√λ1r) ≤ (2.31)  ˜ jn/2 √λ2r  limr 1/γ , if r 1/γ, → jn/2−1(√λ1r) ≥ Observe that Q(r) = w(γr) = g(r) and let x x Q (x) = Q( x ) i = g(r) i , i | | x x | | | |

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22 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

then

Q z2 = 0, i = 1, , n i ··· ZS1

and Qiz are eigenfunctions of λ˜2. Thus, we have

2 (Qiz) ˜ S1 |∇ | λ2 = 2 , i = 1, , n. (Qiz) ··· R S1 Summing over i and simplifying,R we get

2 2 f (r) 2 (f (r)) + (n 1) 2 z S1 ′ r λ˜ λ = − . (2.32) 2 − 1 f 2(r)z2 R S1 This completes the proof of LemmaR 3.1. Substituting (2.28) into (2.26), we get

2 λ1 B(γr)u λ λ Ω 1 (2.33) 2 − 1 ≤ w2(γr)u2 ΩR 1 where R 2 2 w (x) B(x) = (w′(x)) + (n 1) . (2.34) − x2 From the deﬁnition of w and the properties of Bessel functions one can prove that w(t) is nondecreasing and B(t) is non-increasing. There- fore, we have

2 2 2 B(γr)u1 B(γr)∗(u1∗) B(γr)(u1∗) (2.35) ≤ ∗ ≤ ∗ ZΩ ZΩ ZΩ and

2 2 2 2 w(γr) u1 w(γr) (u1∗) w(γr)(u1∗) (2.36) ≥ ∗ ∗ ≤ ∗ ZΩ ZΩ ZΩ In order to continue the proof, we need a result of Chiti: If c is chosen so that

2 2 2 u1 = u1 = z , (2.37) ∗ ZΩ ZΩ ZS1

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[SEC. 2.5: THE HERSCH THEOREM 23

then

2 2 f(r)(u1∗) f(r)z , (2.38) ∗ ≥ ZΩ ZS1 if f is increasing, and the reverse inequality if f is decreasing. It follows from (2.38) and the monotonicity properties of B and w that

2 2 B(γr)(u1∗) B(γr)z (2.39) ∗ ≤ ZΩ ZS1 and

2 2 2 w(γr)(u1∗) w(γr) z . (2.40) ∗ ≥ ZΩ ZS1 Combining (2.33), (2.35), (2.36), (2.39), (2.40), and using the deﬁni- tion of z, we ﬁnally get

2 λ1 B(γr)z λ λ λ S1 = 1 (β2 α2). (2.41) 2 − 1 ≤ αr 2 w2(γ )z2 α2 − RS1 From here the inequalityR

2 λ2 jn/2,1 2 (2.42) λ1 ≤ jn/2 1,1 − follows immediately. Also, it is clear from the proof that the equality holding in (2.42) if and only if Ω is a ball.

2.5 The Hersch Theorem

In 1974, Hersch proved an isoperimetric inequality for the ﬁrst nontrivial eigenvalue on the 2-dimensional sphere S2.

Theorem 2.5 ([48]). For any metric on S2, the ﬁrst non-trivial eigenvalue satisﬁes 8π λ . (2.43) 1 ≤ A(S2)

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24 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

Proof. For any metric ds˜2 on S2, we can construct a conformal map S2 2 S2 2 2 S2 φ :( , ds˜ ) ( , ds0), here ds0 denotes the standard metric on . From the mini-max→ principle, we have 2 2 f dv˜ S |∇ | λ1 = inf 2 , (2.44) 2 fdv˜=0 2 f dv˜ S R S R where dv˜ is the area element withR respect to ds˜2. Take the coordi- i S2 2 i nate functions x (i = 1, 2, 3) on ( , ds0); then x φ, i = 1, 2, 3, are functions on (S2, ds˜2). ◦ Observe that φ is a conformal map and that in the case of surfaces, the Dirichlet integral of a function is a conformal invariant. Thus we have 8π (xi φ) 2dv˜ = xi 2dv = xi∆xi = 2 (xi)2 = . S2 |∇ ◦ | S2 |∇ | − S2 S2 3 Z Z Z Z Since 3 Area(S2) = dv˜ = (xi φ)2dv,˜ S2 S2 ◦ i=1 Z X Z there exists at least one i such that Area(S2) (xi φ)2dv˜ . S2 ◦ ≥ 3 Z i Also, we can choose φ satisfying 2 x φdv˜ = 0 [85]. Thus S ◦ i 2 2 (x φR) dv˜ 8π S |∇ ◦ | λ1 i 2 2 . (2.45) ≤ 2 (x φ) dv˜ ≤ A(S ) R S ◦ For the discussion of equalityR case, we refer to [48].

Remark 2.1. S2 is a Riemann surface of genus g = 0. For Riemannian surface Σg of genus g > 0, Yang-Yau obtained a similar result.

Theorem 2.6 ([99]). For any metric on Σg, the ﬁrst eigenvalue satisﬁes 8π(1 + g) λ (2.46) 1 ≤ Σ | g|

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[SEC. 2.5: THE HERSCH THEOREM 25

Remark 2.2. Hersch’s theorem can’t be generalized directly to the case of higher dimensions [86]. That is, one can’t expect that

λ Vol(M)2/n C, 1 ≤

with a constant depending only on n. It must depend also on other geometric invariants of M. Here is an interesting application of Hersch’s theorem.

Theorem 2.7 ([19]). Suppose that M is homeomorphic to S2 and φ1, φ2, φ3 are three ﬁrst eigenfunctions such that their square sum is a constant. Then M is actually isometric to a sphere with constant sectional curvature.

Proof. The assumption of Theorem 2.7 says that

∆φi + λ1(M)φi = 0, i = 1, 2, 3, 3 φ2 = c, c is a constant. i=1 i P Thus,

3 3 3 0 = ∆ φ2 = 2 φ 2 + 2 φ ∆φ i |∇ i| i i i=1 ! i=1 i=1 X X X 3 3 = 2 φ 2 2λ (M) φ2 |∇ i| − 1 i i=1 i=1 X X which gives

3 φ 2 = cλ (M). (2.47) |∇ i| 1 i=1 X Taking the Laplacian of both sides of (2.47) and using the Bochner

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26 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

formula, we get

1 3 0 = ∆ φ 2 (2.48) 2 |∇ i| i=1 X 3 3 3 = 2φ 2 + φ (∆φ ) + Ric( φ , φ ) |∇ i| ∇ i · ∇ i ∇ i ∇ i i=1 i=1 i=1 X X X 3 3 3 = 2φ 2 λ (M)2 φ2 + K φ 2 |∇ i| − 1 i |∇ i| i=1 i=1 i=1 X X X 3 ∆φ 2 | i| cλ (M)2 + Kcλ (M) ≥ 2 − 1 1 i=1 X = λ (M)2/2 + Kcλ (M), − 1 1 where K is the sectional curvature of M. Thus we have

λ (M) 2K. (2.49) 1 ≥ Integrating (2.49) and using the Gauss-Bonnet formula, we have

λ (M) area(M) 8π. (2.50) 1 × ≥ Combining (2.50) and Hersch’s theorem we know that M is a 2- sphere.

We have an isoperimetric upper bound for the ﬁrst eigenvalue of the Laplacian of a closed (compact without boundary) hypersurface embedded in Rn.

Theorem 2.8 ([92]). Let M be a connected closed hypersurface embedded in Rn(n 3). Let Ω be the region bounded by M. Denote by V and A the volume≥ of Ω and the area of M, respectively. Then the ﬁrst non-zero eigenvalue λ1 of the Laplacian acting on functions on M satisﬁes

(n 1)A ω 1/n λ − n . (2.51) 1 ≤ nV V with equality holding if and only if M is an (n 1)-sphere. −

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[SEC. 2.5: THE HERSCH THEOREM 27

Proof. Let us denote by x1, , xn, the coordinate functions on Rn. By choosing the coordinates··· origin properly, we can assume that

x = 0, i = 1, , n. i ··· ZM Since M cannot be contained in any hyperplane, each xi is not a constant function, i = 1, , n. It follows from the Poincar´einequality that for each ﬁxed i ···1, , n ∈ { ··· } λ x2 x 2, 1 i ≤ |∇ i| ZM ZM with equality holding if and only if ∆xi = λ1xi. Summing over i from 1 to n, we get −

n n λ x2 x 2 = (n 1) = (n 1)A, (2.52) 1 i ≤ |∇ i| − − M i=1 M i=1 M Z X Z X Z with equality if and only if

∆x = λ x , i 1, , n . (2.53) i − 1 i ∀ ∈ { ··· } Take a ball B in Rn of radius R centered at the origin so that vol(B) = V ; then

V 1/n R = . ω n By using the weighted isoperimetric inequality proved in [14], we have

n n x2 x2 (2.54) i ≥ i M i=1 ∂B i=1 Z X Z X = area(∂B) R2 · V 1/n = nV . ω n Substituting (2.54) into (2.52), one gets (2.51). If the equality holds in (2.51), then the inequalities (2.52) and (2.54) must take equality sign.

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28 [CAP. 2: ISOPERIMETRIC INEQUALITIES FOR EIGENVALUES

It follows that the position vector x = (x1, , xn) when restricted on M satisﬁes ···

∆x =: (∆x , , ∆x ) = λ (x , , x ). 1 ··· n − 1 1 ··· n Also, it is well known that

∆x = (n 1)−→H, − where −→H is the mean curvature vector of M. Consider now the function g = x 2 : M R. Observing that −→H is normal to M, we infer that | | → 2(n 1) wf = 2 w, x = − w, −→H = 0, w χ(M). h i − λ1 h i ∀ ∈ Thus f is a constant function and so M is a hypersphere.

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Chapter 3

Universal Inequalities for Eigenvalues

3.1 Introduction

Payne, P´olya and Weinberger proved that the Dirichlet eigenvalues of the Laplacian for Ω R2 satisfy the bound [77], [77]. ⊂

2 k λ λ λ , k = 1, 2, (3.1) k+1 − k ≤ k i ··· i=1 X This result easily extends to Ω Rn as ⊂

4 k λ λ λ , k = 1, 2, (3.2) k+1 − k ≤ kn i ··· i=1 X Many interesting generalizations of (1.3) have been done during the past years, e. g., in [3], [4], [9], [20], [21], [22], [23], [25], [26], [27], [29], [30], [33], [41], [42], [43], [44], [45], [46], [47], [50], [52], [68], [69], [90], [98]. In 1991, Yang [98] proved the following much stronger result:

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30 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Theorem 3.1. The Dirichlet eigenvalues of the Laplacian of Ω Rn satisfy the inequality ⊂ k 4 (λ λ ) λ 1 + λ 0, for k = 1, 2, . (3.3) k+1 − i k+1 − n i ≤ ··· i=1 X The inequality (3.3), as observed by Yang himself, and as later proved, e. g., in [3], [4], [9], is the strongest of the classical inequalities that are derived following the scheme devised by Payne-P´olya- Weinberger. Yang’s inequality provided a marked improvement for eigenvalues of large index. Recently, some Yang type inequalities on eigenvalues of the clamped plate problem, the buckling problem, the polyharmonic operator and some other type eigenvalue problems have been proved. This chapter is devoted to prove some of the universal inequalities in this subarea. Since the method in proving Yang’s inequality has been widely generalized in obtaining various universal inequalities for eigenvalues, we end this section by proving Yang’s inequality.

Proof of Theorem 3.1. Let uk be the orthonormal eigenfunction th corresponding to the k eigenvalue λk, i.e. uk satisﬁes

∆uk = λkuk, in Ω u =− 0, (3.4)  k|∂Ω  Ω uiuj = δij. Rn Let x1, , xn be the R standard coordinate functions in . For any ﬁxed p =··· 1, , n, put g = x and deﬁne φ by ··· p i k φ = gu a u , a = gu u = a . (3.5) i i − ij j ij i j ji j=1 Ω X Z It is easy to see that

φ u = 0, for i, j = 1, , k. (3.6) i j ··· ZΩ Letting

b = u g u , ij j∇ · ∇ i ZΩ

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[SEC. 3.1: INTRODUCTION 31

from Green’s formula, we derive

λ a = g( ∆u )u = 2b + λ a j ij − j i − ij i ij ZΩ and so 2b = (λ λ )a . (3.7) ij i − j ij Since k ∆φ = λ gu + 2 g u + λ a u , i − i i ∇ · ∇ i j ij j j=1 X we have

φ 2 = λ φ2 2 φ g u . (3.8) |∇ i| i i − i∇ · ∇ i ZΩ ZΩ ZΩ On the other hand, from the deﬁnition of φi, (3.5) and (3.6), we derive

2 φ g u (3.9) − i∇ · ∇ i ZΩ k = 2 g g u u + 2 a u g u − ∇ · i∇ i ij j∇ · ∇ i Ω j=1 Ω Z X Z k = 1 + (λ λ )a2 . i − j ij j=1 X From the mini-max principle, we obtain k (λ λ ) φ2 1 + (λ λ )a2 . (3.10) k+1 − i i ≤ i − j ij Ω j=1 Z X 2 Multiplying (3.9) by (λk+1 λi) and taking sum on i from 1 to k, we obtain − k k (λ λ )2 + (λ λ )(λ λ )2a2 (3.11) k+1 − i i − j k+1 − i ij i=1 i,j=1 X X k = 2 (λ λ )2 φ g u . − k+1 − i i∇ · ∇ i i=1 Ω X Z

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32 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

By a = a , b = b , we have ij ji ij − ji k 2 (λ λ )2 φ g u (3.12) − k+1 − i i∇ · ∇ i i=1 Ω X Z k k = (λ λ )2 4 (λ λ )b2 w. k+1 − i − k+1 − i ij ≡ i=1 i=1 X X 2 Multiplying (3.10) by (λk+1 λi) and taking sum on i from 1 to k, we infer −

k (λ λ )3 φ2 k+1 − i i i=1 Ω X Z k k (λ λ )2 4 (λ λ )b2 = w. ≤ k+1 − i − k+1 − i ij i=1 i=1 X X

From Ω uiφj = 0 for all i, j = 1, , k, we have, for arbitrary constants d , ··· R ij w2 k 2 = 2 (λ λ )2 φ g u − k+1 − i i∇ · ∇ i i=1 Ω ! X Z k 4 (λ λ )3φ2 ≤ k+1 − i i i=1 ZΩ X 2 k k (λ λ )1/2 g u d u ×  k+1 − i ∇ · ∇ i − ij j i=1 Ω j=1 X Z X   2 k k 4w (λ λ ) g u 2 + d u ≤  k+1 − i |∇ · ∇ i|  ij j i=1 Ω j=1 X Z X    k 2 d (λ λ )1/2u g u . − ij k+1 − i j∇ · ∇ i j=1 X 

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[SEC. 3.1: INTRODUCTION 33

Then we have k ∂u 2 w 4 (λ λ ) i ≤ k+1 − i ∂x i=1 Ω p X Z k k +4 2 (λ λ )1/2b + d2 . − k+1 − i ij ij i,j=1 i,j=1 X X   Putting d = (λ λ )1/2b , we obtain ij k+1 − i ij k ∂u 2 k w 4 (λ λ ) i 4 (λ λ )b2 (3.13) ≤ k+1 − i ∂x − k+1 − i ij i=1 Ω p i,j=1 X Z X and so we infer k k ∂u 2 (λ λ )2 4 (λ λ ) i . (3.14) k+1 − i ≤ k+1 − i ∂x i=1 i=1 Ω p X X Z Summing over p, we obtain

k 4 k (λ λ )2 (λ λ ) u 2 (3.15) k+1 − i ≤ n k+1 − i |∇ i| i=1 i=1 Ω X X Z 4 k = (λ λ )λ . n k+1 − i i i=1 X

Yang’s inequality has been generalized to bounded domains in complete submanifolds in Euclidean space. That is, we have

Theorem 3.2 ([20], [41]). Let Ω be a bounded domain in an n-dimensional complete Riemannian manifold M n isometrically immersed in RN . Then the Dirichlet eigenvalues of the Laplacian of Ω satisfy the inequality

k 4 k n2 (λ λ )2 (λ λ )(λ + H 2), (3.16) k+1 − k ≤ n k+1 − k i 4 || || i=1 i=1 X X where H is the mean curvature vector ﬁeld of M n and H 2 = sup H 2. || || Ω | |

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34 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES 3.2 Eigenvalues of the Clamped Plate Problem

Let us generalize Yang’s method to prove universal inequalities for eigenvalues of the clamped plate problem on Riemannian manifolds.

Theorem 3.2 ([94]). Let M be an n-dimensional complete Riemannian manifold and let Ω be a bounded domain with smooth boundary in M. Denote by ν the outward unit normal of ∂Ω and let λi the i-th eigenvalue of the problem:

∆2u = λu in Ω, (3.17) ( u = ∂u = 0. |∂Ω ∂ν ∂Ω

i) If M is isometrically immersed in Rm with mean curvature vector H, then

k (λ λ )2 (3.18) k+1 − i i=1 X 1/2 1 k (λ λ )2 n2H2 + (2n + 4)λ1/2 ≤ n k+1 − i 0 i (i=1 ) X k 1/2 (λ λ ) n2H2 + 4λ1/2 , × k+1 − i 0 i (i=1 ) X

where H = sup H . 0 Ω | | ii) If there exists a function φ :Ω R and a constant A0 such that →

φ = 1, ∆φ A , on Ω, (3.19) |∇ | | | ≤ 0

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 35

then

k (λ λ )2 (3.20) k+1 − i i=1 X k 1/2 (λ λ )2 A2 + 4A λ1/4 + 6λ1/2 ≤ k+1 − i 0 0 i i (i=1 ) X 1/2 k 2 (λ λ ) 2λ1/4 + A . × k+1 − i i 0 (i=1 ) X

iii) If there exists a function ψ :Ω R and a constant B0 such that →

ψ = 1, ∆ψ = B , on Ω, (3.21) |∇ | 0 then

k k 1/2 (λ λ )2 (λ λ )2(6λ1/2 B2) (3.22) k+1 − i ≤ k+1 − i i − 0 i=1 (i=1 ) X X k 1/2 (λ λ ) 4λ1/2 B2 . × k+1 − i i − 0 (i=1 ) X m iv) If Ω admits an eigenmap f = (f1, f2, , fm+1):Ω S (1) corresponding to an eigenvalue µ, that is, ··· →

m+1 ∆f = µf , α = 1, , m + 1, f 2 = 1, α − α ··· α α=1 X then

k k 1/2 (λ λ )2 (λ λ )2 6λ1/2 + µ (3.23) k+1 − i ≤ k+1 − i i i=1 (i=1 ) X X k 1/2 (λ λ ) 4λ1/2 + µ , × k+1 − i i (i=1 ) X

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36 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

where Sm(1) is the unit m-sphere.

Theorem 3.2 can be deduced from a general result.

Lemma 3.1 ([89]). Let λ , i = 1, , be the i-th eigenvelue of the i ··· problem (3.17) and ui the orthonormal eigenfunction corresponding to λi, that is,

2 ∆ ui = λiui in Ω, u = ∂ui = 0, (3.24)  i|∂Ω ∂ν ∂Ω u u = δ , i, j = 1, 2, .  M i j ij ∀ ···

Then for any smoothR function h :Ω R and any δ > 0, we have → k (λ λ )2 u2 h 2 (3.25) k+1 − i i |∇ | i=1 Ω X Z k δ (λ λ )2 u2(∆h)2 2u h 2∆u ≤ k+1 − i { i − i|∇ | i i=1 Ω X Z +4(( h u )2 + u ∆h h u ) ∇ · ∇ i i ∇ · ∇ i } k (λ λ ) u ∆h 2 + k+1 − i h u + i . δ ∇ · ∇ i 2 i=1 Ω X Z

Proof of Lemma 3.1. For i = 1, , k, consider the functions φ :Ω R given by ··· i → k φ = hu r u , i i − ij j j=1 X where

rij = huiuj. ZΩ

∂φi Since φi ∂Ω = ∂ν = 0 and | ∂Ω

u φ = 0, i, j = 1, , k, j i ∀ ··· ZΩ

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 37

it follows from the mini-max inequality that

2 φi∆ φi λ Ω . (3.26) k+1 ≤ φ2 R Ω i R We have

2 φi∆ φi (3.27) ZΩ k = φ ∆2(hu ) r λ u i  i − ij j j Ω j=1 Z X 2  = φi∆ (hui) ZΩ k = λ φ 2 r s i|| i|| − ij ij j=1 X + hu (∆(u ∆h) + 2∆( h u ) + 2 h (∆u ) + ∆h∆u ), i i ∇ · ∇ i ∇ · ∇ i i ZΩ

where φ 2 = φ2 and || i|| Ω i R s = u (∆(u ∆h) + 2∆( h u ) + 2 h (∆u ) + ∆h∆u ). ij j i ∇ · ∇ i ∇ · ∇ i i ZΩ

2 Multiplying the equation ∆ ui = λiui by huj, we have

2 huj∆ ui = λihuiuj. (3.28)

Changing the roles of i and j, one gets

2 hui∆ uj = λjhuiuj. (3.29)

Subtracting (3.28) from (3.29) and integrating the resulted equation

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38 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

on Ω, we get

(λ λ )r (3.30) j − i ij = (hu ∆2u hu ∆2u ) i j − j i ZΩ = (∆(hu )∆u ∆(hu )∆u ) i j − j i ZΩ = ((u ∆h + 2 h u )∆u (u ∆h + 2 h u )∆u ) i ∇ · ∇ i j − j ∇ · ∇ j i ZΩ = u (∆(u ∆h) + 2∆( h u ) + ∆h∆u + 2 (∆u ) h) j i ∇ · ∇ i i ∇ i · ∇ ZΩ = sij,

Observe that

hu (∆(u ∆h) + 2∆( h u ) + 2 h (∆u ) + ∆h∆u ) i i ∇ · ∇ i ∇ · ∇ i i ZΩ = (u2(∆h)2 + 4( h u 2 + u ∆h h u ) 2u h 2∆u ). i |∇ · ∇ i| i ∇ · ∇ i − i|∇ | i ZΩ

It follows from (3.26), (3.27) and (3.30) that

(λ λ ) φ 2 (3.31) k+1 − i || i|| (u2(∆h)2 + 4( h u 2 + u ∆h h u ) 2u h 2∆u ) ≤ i |∇ · ∇ i| i ∇ · ∇ i − i|∇ | i ZΩ k + (λ λ )r2 . i − j ij j=1 X

Set

u ∆h t = u h u + i ; (3.32) ij j ∇ · ∇ i 2 ZΩ

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 39

then tij + tji = 0 and

u ∆h ( 2)φ h u + i (3.33) − i ∇ · ∇ i 2 ZΩ k = ( 2hu h u u2h∆h) + 2 r t − i∇ · ∇ i − i ij ij Ω j=1 Z X k = u2 h 2 + 2 r t . i |∇ | ij ij Ω j=1 Z X

2 Multiplying (3.33) by (λk+1 λi) and using the Schwarz inequality, we get −

k (λ λ )2 u2 h 2 + 2 r t (3.34) k+1 − i  i |∇ | ij ij Ω j=1 Z X  u ∆h = (λ λ )2 ( 2)φ h u + i k+1 − i − i ∇ · ∇ i 2 ZM u ∆h k = (λ λ )2 ( 2)φ h u + i t u k+1 − i − i  ∇ · ∇ i 2 − ij j M j=1 Z X 3 2   δ(λk+1 λi) φi ≤ − || || 2 (λ λ ) u ∆h k + k+1 − i h u + i t u δ ∇ · ∇ i 2 − ij j ZM j=1 X 3 2 = δ(λk+1 λi) φi − || || (λ λ ) u ∆h 2 k + k+1 − i h u + i t2 δ  ∇ · ∇ i 2 − ij Ω j=1 Z X  

Substituting (3.31) into (3.34) and summing over i from 1 to k and

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40 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

noticing r = r , t = t , we get ij ji ij − ji

k k (λ λ )2 u2 h 2 2 (λ λ )(λ λ )r t k+1 − i i |∇ | − k+1 − i i − j ij ij i=1 Ω i,j=1 X Z X k (λ λ )2δ (u2(∆h)2 + 4(( h u )2 + u ∆h h u ) ≤ k+1 − i i ∇ · ∇ i i ∇ · ∇ i i=1 Ω X Z k (λ λ ) u ∆h 2 2u h 2∆u ) + k+1 − i ( h u )2 + i − i|∇ | i δ ∇ · ∇ i 2 i=1 Ω X Z k k (λ λ ) (λ λ )δ(λ λ )2r2 k+1 − i t2 , − k+1 − i i − j ij − δ ij i,j=1 i,j=1 X X which implies (3.25).

Proof of Theorem 3.2. Let u ∞ be the orthonormal eigenfunc- { i}i=1 tions corresponding to the eigenvalues λ ∞ of the problem (3.17). { i}i=1

i) Let xα, α = 1, , m, be the standard coordinate functions of m ··· R . Taking h = xα in (3.25) and summing over α, we have

k m (λ λ )2 u2 x 2 (3.35) k+1 − i i |∇ α| i=1 α=1 Ω X X Z k+1 m δ (λ λ )2 (u2(∆x )2 + 4(( x u )2 ≤ k+1 − i i α ∇ α · ∇ i i=1 α=1 Ω X X Z +u ∆x x u ) 2u x 2∆u ) i α∇ α · ∇ i − i|∇ α| i k (λ λ ) m u ∆x 2 + k+1 − i x u + i α , δ ∇ α · ∇ i 2 i=1 α=1 Ω X X Z Since M is isometrically immersed in Rm, we have

m x 2 = n |∇ α| α=1 X

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 41

which implies that

m u2 x 2 = n (3.36) i |∇ α| α=1 Ω X Z Also, we have

∆(x , , x ) (∆x , , ∆x ) = nH, (3.37) 1 ··· m ≡ 1 ··· m

m m ( x u )2 = ( u (x ))2 = u 2 (3.38) ∇ α · ∇ i ∇ i α |∇ i| α=1 α=1 X X and m m ∆x x u = ∆x u (x ) = nH u = 0. (3.39) α∇ α · ∇ i α∇ i α · ∇ i α=1 α=1 X X Substituting (3.36)-(3.39) into (3.35), we get

k n (λ λ )2 (3.40) k+1 − i i=1 X k δ (λ λ )2 (n2u2 H 2 + 4 u 2 2nu ∆u ) ≤ k+1 − i i | | |∇ i| − i i i=1 Ω X Z k (λ λ ) n2u2 H 2 + k+1 − i u 2 + i | | δ |∇ i| 4 i=1 Ω X Z k δ (λ λ )2(n2H2 + (2n + 4)λ1/2) ≤ k+1 − i 0 i i=1 X k (λ λ ) n2H2 + k+1 − i λ1/2 + 0 . δ i 4 i=1 X Here in the last inequality, we have used the fact that H H and | | ≤ 0 1/2 1/2 u 2 = u ∆u u2 (∆u )2 = λ1/2.(3.41) |∇ i| − i i ≤ i i i ZΩ ZΩ ZΩ ZΩ

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42 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Taking

2 2 1/2 k 1/2 n H0 i=1(λk+1 λi) λi + 4 δ = − ,  k P(λ λ )2(n2H 2 + (2n + 4)λ1/2)  i=1 k+1 − i 0 i  one gets (3.18).P  ii) Substituting h = φ into (3.25) and using (3.19) and the Schwarz inequality, we get

k (λ λ )2 (3.42) k+1 − i i=1 X k δ (λ λ )2 u2(∆φ)2 + 4(( φ u )2 + u ∆φ φ u ) ≤ k+1 − i { i ∇ · ∇ i i ∇ · ∇ i i=1 Ω X Z k (λ λ ) u ∆φ 2 2u ∆u + k+1 − i φ u + i − i i} δ ∇ · ∇ i 2 i=1 Ω X Z k δ (λ λ )2 (A2u2 + 4( u 2 + A u u ) 2u ∆u ) ≤ k+1 − i 0 i |∇ i| 0| i||∇ i| − i i i=1 Ω X Z k (λ λ ) A2u2 + k+1 − i u 2 + A u u + 0 i . δ |∇ i| 0| i||∇ i| 4 i=1 Ω X Z Substituting (3.41) and

1/2 1/2 u u u2 u 2 λ1/4 | i||∇ i| ≤ i |∇ i| ≤ i ZΩ ZΩ ZΩ into (3.42), we get

k (λ λ )2 k+1 − i i=1 X k δ (λ λ )2(A2 + 4A λ1/4 + 6λ1/2) ≤ k+1 − i 0 0 i i i=1 X k (λ λ ) A 2 + k+1 − i λ1/4 + 0 . δ i 2 i=1 X

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 43

Taking

2 1/2 k 1/4 A0 i=1(λk+1 λi) λi + 2 δ = − ,  k 1/4 1/2  P(λ λ )2(A2 + 4A λ + 6λ )  i=1 k+1 − i 0 0 i i  P we obtain (3.20). 

iii) Introducing h = ψ into (3.25) and using (3.21), we have

k (λ λ )2 k+1 − i i=1 X k δ (λ λ )2 (B2u2 + 4( u 2 + B u ψ u ) 2u ∆u ) ≤ k+1 − i 0 i |∇ i| 0 i∇ · ∇ i − i i i=1 Ω X Z k (λ λ ) B2u2 + k+1 − i u 2 + B u ψ u + 0 i δ |∇ i| 0 i∇ · ∇ i 4 i=1 Ω X Z k k (λ λ ) B2 δ (λ λ )2(6λ1/2 B2) + k+1 − i λ1/2 0 , ≤ k+1 − i i − 0 δ i − 4 i=1 i=1 X X where in the last inequality, we have used the fact that

1 B2 u ψ, u = u2∆ψ = 0 . ih∇ ∇ ii −2 i − 2 ZΩ ZΩ Taking 2 1/2 k 1/2 B0 i=1(λk+1 λi) λi 4 δ = − − ,  Pk (λ λ )2(6λ1/2 B2)  i=1 k+1 − i i − 0  we obtain (3.22).P  iv) Taking the Laplacian of the equation

m+1 2 fα = 1 α=1 X

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44 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

and using the fact that

∆f = µf , η = 1, , m + 1, η − η ··· we have

m+1 f 2 = µ. |∇ η| η=1 X

It then follows by taking h = fη in (3.25) and summing over η that

k µ (λ λ )2 k+1 − i i=1 X k m+1 δ (λ λ )2 µ2u2 + 4 ( f u )2 2µu ∆u ≤ k+1 − i i ∇ α · ∇ i − i i i=1 Ω α=1 ! X Z X k (λ λ ) m+1 µ2u2 + k+1 − i ( f u )2 + i δ ∇ α · ∇ i 4 i=1 Ω α=1 ! X Z X k δ (λ λ )2(µ2 + 6µλ1/2) ≤ k+1 − i i i=1 X k (λ λ ) µ2 + k+1 − i µλ1/2 + . δ i 4 i=1 X We get (3.23) by taking

1/2 k 1/2 µ i=1(λk+1 λi) λi + 4 δ = − .  Pk (λ λ )2 6λ1/2 + µ   i=1 k+1 − i i  P 

Here are some examples of manifolds supporting the functions on the whole manifolds as stated in items ii)-v) of Theorem 1.1.

Example 3.1. Let M be an n-dimensional Hadamard manifold with Ricci curvature satisfying Ric (n 1)c2, c 0 and let M ≥ − − ≥

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[SEC. 3.2: EIGENVALUES OF THE CLAMPED PLATE PROBLEM 45

γ : [0, + ) M be a geodesic ray, namely a unit speed geodesic with d(γ∞(s), γ→(t)) = t s for any t > s > 0. The Busemann function − bγ corresponding to γ deﬁned by

bγ (x) = lim (d(x, γ(t)) t) t + → ∞ − satisﬁes bγ 1(Cf. [10], [49]). Also, it follows from Theorem 3.5 in |∇ | ≡ 2 [84] that ∆bγ (n 1)c on M. Thus any Hadamard manifold with Ricci curvature| | ≤ bounded− below supports functions satisfying (3.19).

2 Example 3.2. Let (N, dsN ) be a complete Riemannian manifold and define a Riemannian metric on M = R N by × 2 2 2 2 dsM = dt + η (t)dsN , (3.43) where η is a positive smooth function defined on R with η(0) = 1. 2 The manifold (M, dsM ) is called a warped product and denoted by M = R η N. It is known that M is a complete Riemannian manifold. × t Set η = e− and consider the warped product M = R e−t N. Define ψ : M R by ψ(x, t) = t. One can show that × → ψ = 1, ∆ψ = 1 n. (3.44) |∇ | − That is, a warped product manifold M = R e−t N admits functions satisfying (3.21). × Let Hn be the n-dimensional hyperbolic space with constant curvature 1. Using the upper half-space model, Hn is given by − Rn = (x , x , , x ) x > 0 (3.45) + { 1 2 ··· n | n } with metric 2 2 2 dx1 + + dxn ds = ···2 (3.46) xn n 1 One can check that the map Φ : R −t R − given by ×e Φ(t, x) = (x, et) is an isometry. Therefore, Hn admits a warped product model, Hn = n 1 R −t R − . ×e Example 2.3. Any compact homogeneous Riemannian manifold admits eigenmaps to some unit sphere for the first positive eigenvalue of the Laplacian [70].

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46 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES 3.3 Eigenvalues of the Polyharmonic Operator

The method of proving universal bounds for eigenvalues of the clamped plate problem can be generalized to the eigenvalue problem of polyharmonic operators.

Theorem 3.3 ([55]). Let M be an n-dimensional compact Rie- mannian manifold with boundary ∂M (possibly empty) Let l be a positive integer and let λ , i = 1, , be the i-th eigenvalue of the i ··· problem (1.7) and ui be the orthonormal eigenfunction corresponding to λi, that is, l ( ∆) ui = λiui in M, − l−1 ∂ui ∂ ui ui ∂M = ∂ν = = l−1 = 0, (3.47)  | ∂M ··· ∂ν ∂M   M uiuj = δij , for any i, j = 1, 2, . ··· l+2 l+1 Then for anyR function h C (M) C (∂M) and any positive integer k, we have ∈ ∩ k (λ λ )2 u2 h 2 (3.48) k+1 − i i |∇ | i=1 M X Z k δ (λ λ )2 hu ( ∆)l(hu ) λ hu ) ≤ k+1 − i i − i − i i i=1 ZM X k (λ λ ) u ∆h 2 + k+1 − i h, u + i , δ h∇ ∇ ii 2 i=1 X

where δ is any positive constant.

Proof. For i = 1, , k, consider the functions φi : M R given by ··· → k φ = hu r u , (3.49) i i − ij j j=1 X where

rij = huiuj. (3.50) ZM

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[SEC. 3.3: EIGENVALUES OF THE POLYHARMONIC OPERATOR 47

Since ∂φ ∂l 1φ φ = i = = − i = 0 i|∂M ∂ν ··· ∂νl 1 ∂M − ∂M

and

u φ = 0, i, j = 1, , k, j i ∀ ··· ZM it follows from the mini-max inequality that

2 λk+1 φi (3.51) ZM φ ( ∆)lφ ≤ i − i ZM = λ φ 2 + φ ( ∆)lφ λ hu i|| i|| i − i − i i ZM = λ φ 2 + φ ( ∆)l(hu ) λ hu i|| i|| i − i − i i ZM k = λ φ 2 + hu ( ∆)l(hu ) λ hu r s , i|| i|| i − i − i i − ij ij ZM j=1 X where

s = ( ∆)l(hu ) λ hu u . ij − i − i i j ZM Notice that if u Cl+2(M) Cl+1(∂M) satisﬁes ∈ ∩ ∂u ∂l 1u u = = = − = 0, (3.52) |∂M ∂ν ··· ∂νl 1 ∂M − ∂M

then

u ∂M = u ∂M = ∆u ∂M = (∆u) ∂M = | ∇m| 1 | m∇1 | ··· = ∆ − u = (∆ − u) = 0, when l = 2m ∂M ∇ ∂M

and m 1 u = u = ∆u = (∆u) = = ∆ − u |∂M ∇ |∂M |∂M ∇ |∂M ··· ∂M m 1 m = (∆ − u) = ∆ u = 0, when l = 2m + 1. ∇ ∂M |∂M

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48 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

We can then use integration by parts to conclude that

u ( ∆)l(hu ) = hu ( ∆)l(u ) = λ r , j − i i − j j ij ZM ZM which gives

s = (λ λ )r . (3.53) ij j − i ij

Set

p (h) = ( ∆)l(hu ) λ hu ; i − i − i i

then we have from (3.51) and (3.53) that

(λ λ ) φ 2 φ p (h) (3.54) k+1 − i || i|| ≤ i i ZM k = hu p (h) + (λ λ )r2 . i i i − j ij M j=1 Z X Set

u ∆h t = u h u + i ; (3.55) ij j ∇ · ∇ i 2 ZM

then tij + tji = 0 and

u ∆h k ( 2)φ h u + i = w + 2 r t , (3.56) − i ∇ · ∇ i 2 i ij ij M j=1 Z X where

w = ( hu2∆h 2hu h u ). (3.57) i − i − i∇ · ∇ i ZM

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[SEC. 3.3: EIGENVALUES OF THE POLYHARMONIC OPERATOR 49

2 Multiplying (3.56) by (λk+1 λi) and using the Schwarz inequality and (3.54), we get −

k (λ λ )2 w + 2 r t k+1 − i  i ij ij j=1 X   u ∆h k = (λ λ )2 ( 2)φ h u + i t u k+1 − i − i  ∇ · ∇ i 2 − ij j M j=1 Z X 3 2   δ(λk+1 λi) φi ≤ − || || 2 (λ λ ) u ∆h k + k+1 − i h u + i t u δ ∇ · ∇ i 2 − ij j ZM j=1 X 3 2 = δ(λk+1 λi) φi − || || (λ λ ) u ∆h 2 k + k+1 − i h u + i t2 δ  ∇ · ∇ i 2 − ij j=1 X   k δ(λ λ )2 hu p (h) + (λ λ )r2 ≤ k+1 − i  i i i − j ij M j=1 Z X   (λ λ ) u ∆h 2 k + k+1 − i h u + i t2 . δ  ∇ · ∇ i 2 − ij j=1 X   Summing over i and noticing rij = rji, tij = tji, we infer − k k (λ λ )2w 2 (λ λ )(λ λ )r t k+1 − i i − k+1 − i i − j ij ij i=1 i,j=1 X X k δ (λ λ )2 hu p (h) ≤ k+1 − i i i i=1 M X Z k (λ λ ) u ∆h 2 + k+1 − i h, u + i δ h∇ ∇ ii 2 i=1 X k k (λ λ ) (λ λ )δ (λ λ )2r2 k+1 − i t2 . − k+1 − i i − j ij − δ ij i,j=1 i,j=1 X X

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50 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Using (3.48), we can obtain universal inequalities for eigenvalues of the problem (1.7) when M is a bounded domain in Rn or Sn(1).

Theorem 3.4 ([55]). Let Ω be a bounded domain in Rn and Denote by λi the i-th eigenvelue of the eigenvalue problem:

( ∆)lu = λu in Ω, − ∂u ∂l−1u (3.58) u ∂Ω = ∂ν = = l−1 = 0. ( | ∂Ω · ·· ∂ν ∂Ω

Then we have k (λ λ )2 (3.59) k+1 − i i=1 X 1/2 1/2 k 4l(n + 2l 2) 2 (l 1)/l − (λ λ ) λ − ≤ n2 k+1 − i i i=1 ! X k 1/2 (λ λ ) λ1/l × k+1 − i i i=1 ! X

Proof. Let x1, x2, , xn be the standard Euclidean coordinate n ··· functions of R . Let ui be the i-th orthonormal eigenfunction corresponding to the eigenvalue λi of the problem (3.58), i = 1, ; then ···

l l l 1 ( ∆) (x u ) = λ x u + 2l( 1) x (∆ − u ) (3.60) − α i i α i − ∇ α · ∇ i

Taking h = xα in (3.48), we infer for any δ > 0 that

k (λ λ )2 k+1 − i i=1 X k 2 l l 1 δ (λ λ ) 2l( 1) x u x (∆ − u ) ≤ k+1 − i − α i∇ α · ∇ i i=1 Ω X Z 1 k + (λ λ ) x u 2. δ k+1 − i ||∇ α · ∇ i|| i=1 X

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[SEC. 3.3: EIGENVALUES OF THE POLYHARMONIC OPERATOR 51

Summing over α, we have

k n (λ λ )2 (3.61) k+1 − i i=1 X k n 2 l l 1 2lδ (λ λ ) ( 1) x u x (∆ − u ) ≤ k+1 − i − α i∇ α · ∇ i i=1 α=1 Ω X X Z 1 k + (λ λ ) u 2. δ k+1 − i |∇ i| i=1 Ω X Z By induction, we infer

u ( ∆)ku λk/l, k = 1, , l. (3.62) i − i ≤ i ··· ZΩ Since

l 1 l 2 l 1 ∆ − (x u ) = 2(l 1) x (∆ − u ) + x ∆ − u , α i − ∇ α · ∇ i α i we have

l 1 x u x (∆ − u ) (3.63) α i∇ α · ∇ i ZΩ l 1 = x u ∆ − x u α i ∇ α · ∇ i ZΩ l 1 = ∆ − (x u ) x u α i ∇ α · ∇ i ZΩ l 2 l 1 = 2(l 1) x (∆ − u ) + x ∆ − u x u . − ∇ α · ∇ i α i ∇ α · ∇ i ZΩ On the other hand,

l 1 x u x (∆ − u ) (3.64) α i∇ α · ∇ i ZΩ l 1 = ∆ − u div(x u x ) − i α i∇ α ZΩ l 1 2 = ∆ − u ( x u + x x u ). − i |∇ α| i α∇ α · ∇ i ZΩ

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52 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Combining (3.63) and (3.64), we obtain

l 1 x u x (∆ − u ) (3.65) α i∇ α · ∇ i ZΩ l 2 1 l 1 2 = (l 1) x (∆ − u ) x u ∆ − u x u − ∇ α · ∇ i ∇ α · ∇ i − 2 i|∇ α| i ZM Observe that

n l l 1 ( 1) x u x (∆ − u ) (3.66) − α i∇ α · ∇ i α=1 Ω X Z l l 2 n l 1 = ( 1) (l 1) (∆ − ui) ui ui∆ − ui Ω − − ∇ · ∇ − 2 Z n o n l 1 = l 1 + u ( ∆) − u − 2 i − i ZΩ n (l 1)/l l 1 + λ − . ≤ − 2 i Substituting (3.62) and (3.66) into (3.61), one gets

k n (λ λ )2 (3.67) k+1 − i i=1 X k k 2 (l 1)/l 1 1/l l(n + 2l 2)δ (λ λ ) λ − + (λ λ ) λ . ≤ − k+1 − i i δ k+1 − i i i=1 i=1 X X Taking

1/2 k (λ λ ) λ1/l δ = i=1 k+1 i i , k − 2 (l 1)/l (l2(n + l 2) (λ λ ) λ − ) −P i=1 k+1 − i i we get (3.59). P

Let l be a positive integer and for p = 0, 1, 2, ..., deﬁne the poly- nomials Fp(t) inductively by

F0(t) = 1,F1(t) = t n, − 2 Fp(t) = (2t 2)Fp 1(t) (t + 2t n(n 2))Fp 2(t), p = 2, . − − − − − − ···

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(3.68) Set

l l 1 l 1 l Fl(t) = t al 1t − + + ( 1) − a1t + ( n) . (3.69) − − ··· − −

Theorem 3.4 ([55]). Let λi be the i-th eigenvalue of the eigenvalue problem: ( ∆)lu = λu in Ω, − ∂u ∂l−1u u ∂Ω = ∂ν = = l−1 = 0, ( | ∂Ω · ·· ∂ν ∂Ω

Sn where Ω is a compact domain in (1). Then we have k (λ λ )2 (3.70) k+1 − i i=1 X k 1/2 1 l−1 1 2 + l + l + (λk+1 λi) al 1λi + + a1 λi + a0 ≤ n − − · ·· (i=1 ) X k 1/2 (λ λ ) n2 + 4λ1/2 , × k+1 − i i (i=1 ) X where a+ = max 0, a . j { j }

Proof. As before, let x1, x2, , xn+1 be the standard coordinate functions of Rn+1; then ···

n+1 Sn(1) = (x , . . . , x ) Rn+1; x2 = 1 . { 1 n+1 ∈ α } α=1 X It is well known that ∆x = nx , α = 1, , n + 1. (3.71) α − α ··· n+1 2 Taking the Laplacian of the equation α=1 xα = 1 and using (3.71), we get P n+1 x 2 = n. (3.72) |∇ α| α=1 X

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Let ui be the i-th orthonormal eigenfunction corresponding to the eigenvalue λi, i = 1, 2, . For any δ > 0, by taking h = xα in (3.48), we have ···

k (λ λ )2 u2 x 2 k+1 − i i |∇ α| i=1 Ω X Z k+1 δ (λ λ )2 x u (( ∆)l(x u ) λ x u ) ≤ k+1 − i α i − α i − i α i i=1 Ω X Z 1 k u ∆x 2 + (λ λ ) x u + i α δ k+1 − i ∇ α · ∇ i 2 i=1 X

Taking sum on α from 1 to n + 1 and using (3.72), we get

k n (λ λ )2 (3.73) k+1 − i i=1 X k+1 n+1 δ (λ λ )2 x u (( ∆)l(x u ) λ x u ) ≤ k+1 − i α i − α i − i α i i=1 α=1 Ω X X Z 1 k n+1 u ∆x 2 + (λ λ ) x u + i α . δ k+1 − i ∇ α · ∇ i 2 i=1 α=1 X X

It is easy to see that

n+1 u ∆x 2 x u + i α (3.74) ∇ α · ∇ i 2 α=1 X n +1 n2u2x2 = ( x u )2 n x u u x + i α ∇ α · ∇ i − ∇ α · ∇ i i α 4 Ω α=1 Z X n2 = + u 2 4 |∇ i| ZΩ n2 + λ1/l. ≤ 4 i For any smooth functions f, g on Ω, we have from the Bochner for-

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mula that ∆( f g) (3.75) ∇ · ∇ = 2 2f 2g + f (∆g) + g (∆f) ∇ · ∇ ∇ · ∇ ∇ · ∇ +2(n 1) f g, − ∇ · ∇ where n 2f 2g = 2f(e , e ) 2g(e , e ), ∇ · ∇ ∇ s t ∇ s t s,t=1 X being e , , e orthonormal vector ﬁelds locally deﬁned on Ω. Since 1 ··· n 2x = x I, ∇ α − α we infer from (3.75) by taking f = xα that ∆( x g) (3.76) ∇ α · ∇ = 2x ∆g + x (∆g) + (n 2) x g − α ∇ α · ∇ − ∇ α · ∇ = 2x ∆g + x ((∆ + (n 2))g). − α ∇ α · ∇ − For each q = 0, 1, , thanks to (3.71) and (3.76), there are polyno- ··· mials Bq and Cq of degrees less than or equal to q such that ∆q(x g) = x B (∆)g + 2 x (C (∆)g). (3.77) α α q ∇ α · ∇ q It is obvious that B = 1,B = t n, C = 0,C = 1. (3.78) 0 1 − 0 1 It follows from (3.71), (3.76) and (3.77) that

q q 1 ∆ (xαg) = ∆(∆ − (xαg)) (3.79)

= ∆(xαBq 1(∆)g + 2 xα (Cq 1(∆)g)) − ∇ · ∇ − = xα((∆ n)Bq 1(∆) 4∆Cq 1(∆))g − − − − +2 xα ((Bq 1(∆) + (∆ + (n 2))Cq 1(∆))g). ∇ · ∇ − − − Thus, for any q = 2, , we have ··· Bq(∆) = (∆ n)Bq 1(∆) 4∆Cq 1(∆), (3.80) − − − − Cq(∆) = Bq 1(∆) + (∆ + (n 2))Cq 1(∆). (3.81) − − −

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56 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Consequently,

Bq(∆) (3.82)

= (2∆ 2)Bq 1(∆) (∆ + n 2)Bq 1(∆) 4∆Cq 1(∆) − − − − − 2 − − = (2∆ 2)Bq 1(∆) (∆ + 2∆ n(n 2))Bq 2(∆) − − − − − − +4∆[Bq 2(∆) + (∆ + n 2)Cq 2(∆) Cq 1(∆)] − − − 2 − − = (2∆ 2)Bq 1(∆) (∆ + 2∆ n(n 2))Bq 2(∆), q = 2, . − − − − − − ···

Since (3.78) and (3.82) hold, we know that Bq = Fq, q = 0, 1, . It follows from (3.77) and the divergence theorem that∀ ···

x u (( ∆)l(x u ) λ x u ) (3.83) α i − α i − i α i ZΩ = x u ( 1)l (x B (∆)u + 2 x (C (∆)u )) λ x u α i − α l i ∇ α · ∇ l i − i α i ZΩ l l l 1 l = xαui ( 1) xα(∆ al 1∆ − + + ( n) )ui − − − ··· − ZΩ +2 x (C (∆) u )) λ x u ) ∇ α · ∇ l i − i α i l l 1 l = ( 1) xαui xα( al 1∆ − + + ( n) )ui + 2 xα (Cl(∆)ui) − − − ··· − ∇ · ∇ ZΩ

Summing on α, one has

x u (( ∆)l(x u ) λ x u ) (3.84) α i − α i − i α i ZΩ l l 1 l = ui( 1) ( al 1∆ − + + ( n) a0)ui − − − ··· − ZΩ l 1 l 2 = al 1 ui( ∆) − ui + + a1 ui( ∆)ui + n ui − − ··· − ZΩ ZΩ ZΩ + (l 1)/l + 1/l l al 1λi − + + a1 λi + n . ≤ − ···

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Substituting (3.74) and (3.84) into (3.73), we infer

k n (λ λ )2 k+1 − i i=1 X k 2 + (l 1)/l + 1/l l δ (λk+1 λi) (al 1λi − + + a1 λi + n ) ≤ − − ··· i=1 X 1 k n2 + (λ λ ) λ1/l + . δ k+1 − i i 4 i=1 X Taking

1/2 k 1/l n2 i=1(λk+1 λi) λi + 4 δ = − ,  k 2 + (l 1)/l + 1/l l  i=1(λk+1P λi) (al 1λi − + + a1 λi + n )  − − ···  we getP (3.70). 

3.4 Eigenvalues of the Buckling Problem

Let Ω Rn and consider the problem ⊂ ∆2u = λ∆u, − (3.85) u = ∂u = 0 |∂Ω ∂ν ∂Ω

which is used to describe the critical buckling load of a clamped plate subjected to a uniform compressive force around its boundary. Payne, P´olya and Weinberger [77] proved λ /λ < 3 for Ω R2. 2 1 ⊂ For Ω Rn this reads ⊂ λ2/λ1 < 1 + 4/n. Subsequently Hile and Yeh [51] reconsidered this problem obtaining the improved bound

2 λ2 n + 8n + 20 Rn 2 for Ω . λ1 ≤ ( n + 2) ⊂

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58 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Ashbaugh [3] proved :

n λ (n + 4)λ . (3.86) i+1 ≤ 1 i=1 X This inequality has been improved to the following form [54]:

n 4(λ λ ) λ + 2 − 1 (n + 4)λ . i+1 n4 + ≤ 1 i=1 X Cheng and Yang introduced a new method to construct trial functions for the problem (3.85) and obtained the following universal inequality [28]:

k 4(n + 2) k (λ λ )2 (λ λ )λ . (3.87) k+1 − i ≤ n2 k+1 − i i i=1 i=1 X X It has been proved in [88] that for the problem (1.5) if M is a bounded connected domain in an n-dimensional unit sphere, then the following inequality holds

k 2 (λ λ )2 (3.88) k+1 − i i=1 X k δ2(λ (n 2)) (λ λ )2 δλ + i − − ≤ k+1 − i i 4(λ δ + n 2) i=1 i X − 1 k (n 2)2 + (λ λ ) λ + − , δ k+1 − i i 4 i=1 X where δ is any positive constant. The inequality (3.87) has been generalized to eigenvalues of buckling problem of arbitrary orders. That is, we have

Theorem 3.5 ([54]). Let l 2 and let λi be the i-th eigenvalue of the following eigenvalue problem:≥ ( ∆)lu = λ∆u in Ω, − ∂u − ∂l−1u (3.89) u ∂Ω = ∂ν = = l−1 = 0. ( | ∂Ω · ·· ∂ν ∂Ω

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where Ω is a bounded domain with smooth boundary in Rn. Then for k = 1, , we have ··· k (λ λ )2 (3.90) k+1 − i i=1 X 1/2 2 1/2 k 2(2l + (n 4)l + 2 n) 2 (l 2)/(l 1) − − (λ λ ) λ − − ≤ n k+1 − i i (i=1 ) X k 1/2 1/(l 1) (λ λ )λ − . × k+1 − i i (i=1 ) X

Before proving theorem 3.5, let us recall a method of constructing trial functions developed by Cheng-Yang (Cf. [28], [54]). Let M be an n-dimensional complete submanifold in an m-dimensional Euclidean space Rm. Denote by the canonical metric on Rm as well as that induced on M. Let ∆· and be the Laplacian and the gradient operator of M, respectively. Let∇ Ω be a bounded connected domain of M with smooth boundary ∂Ω and let ν be the outward unit normal vector ﬁeld of ∂Ω. For functions f and g on Ω, the Dirichlet inner product (f, g)D of f and g is given by

(f, g) = f g. D ∇ · ∇ ZΩ The Dirichlet norm of a function f is deﬁned by

1/2 f = (f, f) 1/2 = f 2 . || ||D { D} |∇ | ZΩ Consider the eigenvalue problem

( ∆)lu = λ∆u in Ω, − ∂u − ∂l−1u (3.91) u ∂Ω = ∂ν = = l−1 = 0. ( | ∂Ω · ·· ∂ν ∂Ω

Let 0 < λ λ λ , 1 ≤ 2 ≤ 3 ≤ · · ·

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denote the successive eigenvalues, where each eigenvalue is repeated according to its multiplicity. Let ui be the i-th orthonormal eigenfunction of the problem (3.91) corresponding to the eigenvalue λ , i = 1, 2, , that is, u satisﬁes i ··· i l ( ∆) ui = λi∆ui in Ω, − − l−1 ∂ui ∂ u ui ∂Ω = ∂ν = l−1 = 0, (3.92)  | ∂Ω · ·· ∂ν ∂Ω  (u , u ) = u , u = δ , i, j. i j D Ωh∇ i ∇ ji ij ∀

For k = 1, , l, let k denoteR the k-th covariant derivative operator on M, deﬁned··· in the∇ usual weak sense via an integration by parts formula. For a function f on Ω, the squared norm of kf is deﬁned as ∇ n 2 2 kf = kf(e , , e ) , (3.93) ∇ ∇ i1 ··· ik i1, ,i =1 ···Xk

where e1, , en are orthonormal vector fields locally defined on Ω. ··· 2 Define the Sobolev space Hl (Ω) by H2(Ω) = f : f, f , , lf L2(Ω) . l { |∇ | ··· ∇ ∈ } Then H2(Ω) is a Hilbert space with respect to the norm : l || · ||l,2 l 1/2 k 2 f l,2 = f . (3.94) || || Ω |∇ | !! Z kX=0 2 2 Consider the subspace Hl,D(Ω) of Hl (Ω) defined by ∂f ∂l 1u H2 (Ω) = f H2(Ω) : f = = − = 0 . l,D ∈ l |∂Ω ∂ν · ·· ∂νl 1 ∂Ω − ∂Ω

l 2 The operator ( ∆) defines a self-adjoint operator acting on Hl,D(Ω) with discrete eigenvalues− 0 < λ λ for the buckling 1 ≤ · · · ≤ k ≤ · · · problem (3.91) and the eigenfunctions ui i∞=1 defined in (3.92) form { } 2 a complete orthonormal basis for the Hilbert space Hl,D(Ω). If φ 2 ∈ Hl,D(Ω) satisfies (φ, uj)D = 0, j = 1, 2, , k, then the Rayleigh- Ritz inequality tells us that ∀ ···

λ φ 2 φ( ∆)lφ. (3.95) k+1|| ||D ≤ − ZΩ

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For vector-valued functions F = (f1, f2, , fm),G = (g1, g2, , gm): Ω Rm, we deﬁne an inner product (F,G··· ) by ··· → m (F,G) = fαgα. Ω α=1 Z X The norm of F is given by

m 1/2 F = (F,F )1/2 = f 2 . || || α ( Ω α=1 ) Z X 2 Let H1(Ω) be the Hilbert space of vector-valued functions given by

2 H1(Ω) = F = (f , , f ):Ω Rm; f , f L2(Ω), α = 1, , m 1 ··· m → α |∇ α| ∈ ··· with norm m 1/2 F = F 2 + f 2 . || ||1 || || |∇ α| Ω α=1 ! Z X Observe that a vector ﬁeld on Ω can be regarded as a vector-valued Rm 2 2 function from Ω to . Let H1,D(Ω) Hl (Ω) be a subspace of 2 ⊂ Hl (Ω) spanned by the vector-valued functions ui i∞=1, which form a complete orthonormal basis of H2 (Ω). For{∇ any }f H2 (Ω), we 1,D ∈ l,D have f H2 (Ω) and for any X H2 (Ω), there exists a function ∇ ∈ 1,D ∈ 1,D f H2 (Ω) such that X = f. ∈ l,D ∇ Proof of Theorem 3.5. With notations as above, we consider now the special case that Ω is a bounded domain in Rn. Let us decompose the vector-valued functions x u as α∇ i x u = h + W , (3.96) α∇ i ∇ αi αi where h Hl (Ω), h is the projection of x u in H2 (Ω) αi ∈ 2,D ∇ αi α∇ i 1,D and W H2 (Ω). Thus we have αi ⊥ 1,D

W = 0, W u = 0, u H2 (Ω) (3.97) αi|∂Ω αi · ∇ ∀ ∈ l,D ZΩ

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and from the discussions in [28] and [88] we know that

div Wαi = 0. (3.98)

For each α = 1, , n, i = 1, , k, consider the functions φαi : Ω R, given by ··· ··· →

k φ = h a u , (3.99) αi αi − αij j j=1 X where

a = x u u = a . (3.100) αij α∇ i · ∇ j αji ZΩ We have

∂φ ∂l 1φ φ = αi = − αi = 0, (3.101) αi|∂Ω ∂ν · ·· ∂νl 1 ∂Ω − ∂Ω

(φ , u ) = φ u = 0, j = 1, , k. (3.102) αi j D ∇ αi · ∇ j ∀ ··· ZΩ It follows from the Rayleigh-Ritz inequality that

λ φ 2 (3.103) k+1 |∇ αi| ZΩ φ ( ∆)lφ , α = 1, , n, i = 1, , k. ≤ αi − αi ··· ··· ZD It is easy to see that

k l l l 1 ( ∆) φ = ( 1) ∆ − (u + x ∆u ) + a λ ∆u − αi − i,α α i αij j j j=1 X

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and so

φ ( ∆)lφ (3.104) αi − αi ZΩ l l 1 = φ ( 1) ∆ − (u + x ∆u ) αi − i,α α i ZΩ k l l 1 l = h ( 1) ∆ − (u + x ∆u ) a u ( ∆) h αi − i,α α i − αij j − αi Ω j=1 Ω Z X Z k l l 2 l = ∆h ( 1) ∆ − (u + x ∆u ) a h ( ∆) u αi − i,α α i − αij αi − j Ω j=1 Ω Z X Z k l l 2 l 2 = ∆h ( 1) ((∆ − u ) + ∆ − (x ∆u )) + λ a h ∆u αi − i ,α α i j αij αi j Ω j=1 Ω Z X Z l l 2 l 2 = ( 1) (u + x ∆u )((∆ − u ) + ∆ − (x ∆u )) − i,α α i i ,α α i ZΩ k λ a h , u − j αij h∇ αi ∇ ji j=1 Ω X Z l l 2 l 1 = ( 1) (u + x ∆u )((2l 3)(∆ − u ) + x ∆ − u )) − i,α α i − i ,α α i ZΩ k λ a h , u − j αij h∇ αi ∇ ji j=1 Ω X Z l l 2 l 2 = ( 1) ((2l 3) u (∆ − u ) + x ∆u (∆ − u ) ) − − { i,α i ,α α i i ,α ZΩ k l 1 2 l 1 2 +u x ∆ − u + x ∆u ∆ − u λ a i,α α i α i i} − j αij j=1 X

Since

l 1 l 2 l 1 ∆ − (x u ) = 2(l 1)(∆ − u ) + x ∆ − u , α i − i ,α α i

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we have

l 1 l 1 xαui(∆ − ui),α = xαui∆ − ui,α (3.105) ZΩ ZΩ l 1 = ∆ − (xαui)ui,α ZΩ l 2 l 1 = 2(l 1)(∆ − u ) + x ∆ − u u . − i ,α α i i,α ZΩ On the other hand, it holds

l 1 l 1 x u (∆ − u ) = ∆ − u (u + x u ). (3.106) α i i ,α − i i α i,α ZΩ ZΩ Combining (3.105) and (3.106), we obtain

l 1 xαui(∆ − ui),α (3.107) ZΩ l 2 1 l 1 = (l 1)(∆ − u ) u u ∆ − u − i ,α i,α − 2 i i ZM Hence

l 1 l 1 l 1 x u ∆ − u = u (∆ − u + x (∆ − u ) ) (3.108) α i,α i − i i α i ,α ZΩ ZΩ l 2 1 l 1 = (l 1)(∆ − u ) u + u ∆ − u − − i ,α i,α 2 i i ZM and consequently, we have

l 2 l 2 xα∆ui(∆ − ui),α = xα∆ui∆ − ui,α (3.109) ZΩ ZΩ l 2 = ∆ − (xα∆ui)ui,α ZΩ l 2 l 1 = u 2(l 2)(∆ − u ) + x ∆ − u i,α − i ,α α i ZΩ l 2 1 l 1 = (l 3)(∆ − u ) u u ∆ − u . − i ,α i,α − 2 i i ZΩ

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Also, one has

2 uixα∆ui (3.110) ZΩ = x2 u 2 2 x u u − α|∇ i| − α i i,α ZΩ ZΩ = x2 u 2 + u2, − α|∇ i| i ZΩ ZΩ

2 l 1 xα∆ui∆ − ui (3.111) ZΩ 2 l 1 = ui∆(xα∆ − ui) ZΩ l 1 2 l l 1 = ui(2∆ − ui + xα∆ ui + 4xα(∆ − ui),α) ZΩ l 1 l 1 2 l 1 = u (2∆ − u + ( 1) − λ x ∆u + 4x (∆ − u ) ). i i − i α i α i ,α ZΩ Combining (3.107), (3.110) and (3.111), we get

2 l 1 l 2 x ∆u ∆ − u = 4(l 1) (∆ − u ) u (3.112) α i i − i ,α i,α ZΩ ZΩ l 1 2 2 2 +( 1) − λ x u + u . − i − α|∇ i| i ZΩ ZΩ Substituting (3.109), (3.111) and (3.112) into (3.104), one gets

φ ( ∆)lφ (3.113) αi − αi ZΩ l l 1 2 l 2 = ( 1) ( l + 1)u ∆ − u + (2l 4l + 3)(∆ − u ) u − − i i − i ,α i,α ZΩ k +λ x2 u 2 u2 λ a2 . i α|∇ i| − i − j αij Ω Ω j=1 Z Z X It is easy to see that

x u 2 = h 2 + W 2 (3.114) || α∇ i|| ||∇ αi|| || αi||

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and k h 2 = φ 2 + a2 , (3.115) ||∇ αi|| ||∇ αi|| αij j=1 X Combining (3.103), (3.113), (3.114) and (3.115), we infer (λ λ ) φ 2 (3.116) k+1 − i ||∇ αi|| l l 1 2 l 2 ( 1) ( l + 1)u ∆ − u + (2l 4l + 3)(∆ − u ) u ≤ − − i i − i ,α i,α ZΩ k λ ( u 2 W 2) + (λ λ )a2 , − i || i|| − || αi|| i − j αij j=1 X 2 Observe that (xαui) = ui xα + xα ui H1,D(Ω). For Aαi = (x u h ),∇ we have ∇ ∇ ∈ ∇ α i − αi u x = A W (3.117) i∇ α αi − αi and so u 2 = u x 2 = W 2 + A 2. || i|| || i∇ α|| || αi|| || αi|| Because of ( u ,W ) = 0, it follows that ∇ i,α αi 2 u 2 = 2 A u || i,α|| − αi · ∇ i,α ZΩ 1/(l 1) 2 1 2 λ − A + u ≤ i || αi|| 1/(l 1) ||∇ i,α|| λi − which gives

l−2 l−3 λ A 2 2λ l−1 u 2 + λ l−1 u 2 (3.118) − i|| αi|| ≤ − i || i,α|| i ||∇ i,α|| Introducing (3.118) into (3.116), we get (λ λ ) φ 2 (3.119) k+1 − i ||∇ αi|| l l 1 2 l 2 ( 1) ( l + 1)u ∆ − u + (2l 4l + 3)(∆ − u ) u ≤ − − i i − i ,α i,α ZΩ (l 2)/(l 1) 2 (l 3)/(l 1) 2 2λ − − u + λ − − u − i || i,α|| i ||∇ i,α|| k + (λ λ )a2 , i − j αij j=1 X

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[SEC. 3.4: EIGENVALUES OF THE BUCKLING PROBLEM 67

Since

2 x u u − α∇ i · ∇ i,α ZΩ 2 = 2 ui,α + 2 xαui,α∆ui ZΩ ZΩ 2 = 2 ui,α + 2 ui∆(xαui,α) ZΩ ZΩ = 2 u2 + 2 u x (∆u ) + 4 u x u i,α i α i ,α i∇ α · ∇ i,α ZΩ ZΩ ZΩ = 2 u2 2 ∆u (u + x u ) 4 u div(u x ) i,α − i i α i,α − i,α i∇ α ZΩ ZΩ ZΩ = 2 u2 + 2 2 x u ∆u 4 u2 i,α − α i,α i − i,α ZΩ ZΩ ZΩ = 2 u2 + 2 + 2 u (x u ) − i,α ∇ i · ∇ α i,α ZΩ ZΩ = 2 + 2 x u u , α∇ i · ∇ i,α ZΩ we have

2 x u u = 1. (3.120) − α∇ i · ∇ i,α ZΩ Set

d = u u ; αij ∇ i,α · ∇ j ZΩ then d = d and we get αij − αji

1 = 2 x u u (3.121) − α∇ i · ∇ i,α ZΩ = 2 h u − ∇ αi · ∇ i,α ZΩ k = 2 φ u 2 a d . − ∇ αi · ∇ i,α − αij αij Ω j=1 Z X

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68 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Thus, we have

k (λ λ )2 1 + 2 a d (3.122) k+1 − i  αij αij j=1 X   k = (λ λ )2 2 φ , u d u k+1 − i − ∇ αi ∇ i,α − αij ∇ j j=1 X    1 k δ(λ λ )3 φ 2 + (λ λ ) u 2 d2 , ≤ k+1 − i ||∇ αi|| δ k+1 − i ||∇ i,α|| − αij j=1 X  

where δ is any positive constant. Substituting (3.119) into (3.122), we get

k (λ λ )2 1 + 2 a d k+1 − i  αij αij j=1 X   2 l l 1 δ(λ λ ) ( 1) ( l + 1)u ∆ − u ≤ k+1 − i − { − i i ZΩ 2 l 2 +(2l 4l + 3)(∆ − u ) u − i ,α i,α} l−2 l−3 k 2λ l−1 u 2 + λ l−1 u 2 + (λ λ )a2 − i || i,α|| i ||∇ i,α|| i − j αij  j=1 X  1 k + (λ λ ) u 2 d2 , δ k+1 − i ||∇ i,α|| − αij j=1 X  

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[SEC. 3.4: EIGENVALUES OF THE BUCKLING PROBLEM 69

Summing on i from 1 to k, we infer

k k (λ λ )2 2 (λ λ )(λ λ )a d k+1 − i − k+1 − i i − j αij αij i=1 i,j=1 X X k 2 l l 1 δ (λ λ ) ( 1) ( l + 1)u ∆ − u ≤ k+1 − i − { − i i i=1 ZΩ X2 l 2 +(2l 4l + 3)(∆ − u ) u − i ,α i,α} (l 2)/(l 1) 2 (l 3)/(l 1) 2 2λ − − u + λ − − u − i || i,α|| i ||∇ i,α|| k (λ λ )(λ λ )2a2 − k+1 − i i − j αij i,j=1 X  1 k k + (λ λ ) u 2 (λ λ )d2 , δ  k+1 − i ||∇ i,α|| − k+1 − i αij  i=1 i,j=1 X X  

which gives

k (λ λ )2 k+1 − i i=1 X k 2 l l 1 δ (λ λ ) ( 1) ( l + 1)u ∆ − u ≤ k+1 − i − − i i i=1 ZΩ X2 l 2 +(2l 4l + 3)(∆ − u ) u − i ,α i,α l−2 l−3 2λ l−1 u 2 + λ l−1 u 2 − i || i,α|| i ||∇ i,α|| 1 k + (λ λ ) u 2, δ k+1 − i ||∇ i,α|| i=1 X

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70 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Taking sum for α from 1 to n, we get

k n (λ λ )2 k+1 − i i=1 X k 2 l l 1 δ (λ λ ) ( 1) n( l + 1)u ∆ − u ≤ k+1 − i − − i i i=1 ZΩ X2 l 2 +(2l 4l + 3) (∆ − u ) u − ∇ i · ∇ i n l−2 l−3 2λ l−1 + λ l−1 u 2 − i i ||∇ i,α|| α=1 ! X 1 k n + (λ λ ) u 2 δ k+1 − i ||∇ i,α|| i=1 α=1 X X k l−2 l−3 n = δ (λ λ )2 2λ l−1 + λ l−1 u 2 k+1 − i − i i ||∇ i,α|| i=1 α=1 X X 2 l 1 +(2l + (n 4)l + 3 n) u ( ∆) − u − − i − i ZΩ 1 k n + (λ λ ) u 2. δ k+1 − i ||∇ i,α|| i=1 α=1 X X But

k k u 2 = u ∆u ||∇ i,α|| − i,α i,α α=1 Ω α=1 X Z X k = u (∆u ) − i,α i ,α Ω α=1 Z X k

= ui,αα∆ui Ω α=1 Z X 2 = (∆ui) ZΩ 2 = ui∆ ui, ZΩ

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[SEC. 3.4: EIGENVALUES OF THE BUCKLING PROBLEM 71

2 ∂ ui where ui,αα = 2 . Therefore, ∂xα

k n (λ λ )2 k+1 − i i=1 X k l−2 l−3 δ (λ λ )2 2λ l−1 + λ l−1 u ∆2u ≤ k+1 − i − i i i i i=1 Ω X Z 2 l 1 +(2l + (n 4)l + 3 n) u ( ∆) − u − − i − i ZΩ 1 k + (λ λ ) u ∆2u . δ k+1 − i i i i=1 Ω X Z Observe that

l−2 1 l 1 l−1 2 l−1 u ( ∆) − u λ , u ∆ u λ . i − i ≤ i i i ≤ i ZΩ ZΩ Thus we have

k n (λ λ )2 k+1 − i i=1 X k l−2 δ(2l2 + (n 4)l + 2 n) (λ λ )2λ l−1 ≤ − − k+1 − i i i=1 X k 1 1 + (λ λ )λ l−1 . δ k+1 − i i i=1 X Taking

1 1/2 k (λ λ )λ l−1 i=1 k+1 − i i δ = , P l−2 1/2 (2l2 + (n 4)l + 2 n) k (λ λ )2λ l−1 − − i=1 k+1 − i i P we get (3.90).

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72 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Recently, Cheng-Yang have strengthened the inequality (3.87) to the following form:

k 4 n + 4 k (λ λ )2 3 (λ λ )λ . (3.123) k+1 − i ≤ n2 k+1 − i i i=1 i=1 X X The inequality (3.90) has been improved by Cheng-Qi-Wang-Xia very recently in [24]:

k n (λ λ )2 (3.124) k+1 − i i=1 X k 14 8 l−2 δ (λ λ )2 2l2 + n l + n λ l−1 ≤ i k+1 − i − 3 3 − i i=1 X k 1 1 + (λ λ )λ l−1 , δ k+1 − i i i=1 i X k where δi i=1 is any positive non-increasing monotone sequence. For{ eigenvalues} of the buckling problem on spherical domains, the inequality (3.88) has also been improved (Cf. [28])) and generalized to buckling problem of arbitrary orders (Cf. [54], [24]).

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Chapter 4

P´olya Conjecture and Related Results

4.1 Introduction

Let Ω Rn be a bounded open set and consider the eigenvalue problem⊂ of the Dirichlet Laplacian on Ω: ∆u + λu = 0 in Ω, (4.1) u = 0. |∂Ω In 1960, P´olya [81] showed that for any “plane covering domain” Ω in R2 (those that tile R2) the Weyl asymptotic relation (1.) is in fact a one-sided inequality (his proof also works for Rn-covering domains) and conjectured, for any bounded domain Ω Rn, the inequality ⊂ k 2/n (2π)2 λk C(n) k, with C(n) = 2 . (4.2) ≥ Ω ∀ n | | ωn P´olya’s conjecture has been a central problem about eigenvalues and many important developments have been made during the past years. In 1982, Li-Yau [72] showed the lower bound

k nkC(n) k 2/n λ (4.3) i ≥ n2 + Ω i=1 X | | 73

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74 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

which yields an individual lower bound on λk in the form nC(n) k 2/n λ . (4.4) k ≥ n2 + Ω | | Similar bounds for eigenvalues with Neumann boundary conditions have been proved in [62], [63] and [65]. It was pointed out in [66] that (4.3) also follows from an earlier result by Berezin [11] by the Legendre transformation. The inequality (4.3) has been improved by Melas:

Theorem 4.1 ([75]). For any bounded domain Ω Rn and any k 1, the eigenvalues of the problem (4.1) satisfy the⊂ inequality ≥ k nkC(n) k 2/n Ω λ + d(n)k | | (4.5) i ≥ n2 + Ω IΩ)( i=1 X | | where the constant dn depends only on the dimension and I(Ω) = 2 mina Rn x a dx is the “moment of inertia” of Ω. ∈ Ω | − |

Proof.R Fix a k 1 and let u1, , uk be an orthonormal set of eigenfunctions of≥ (4.1) corresponding··· to the set of eigenvalues λ , , λ . We consider the Fourier transform of each eigenfunction 1 ··· k n/2 ix ξ fj(ξ) =u ˆj(ξ) = (2π)− uj(x)e · dx. (4.6) ZΩ ¿From Plancherel’s Theorem, we know that f1, , fk is an orthonormal set in Rn. Bessel’s inequality implies that··· for every ξ Rn ∈ k 2 n ix ξ n f (ξ) (2π)− e · dx = (2π)− Ω (4.7) | j | ≤ | | | | j=1 Ω X Z and k 2 n ix ξ n f (ξ) (2π)− ixe · dx = (2π)− I(Ω). (4.8) |∇ j | ≤ | | j=1 Ω X Z Since u = 0 it is easy to see that j|∂Ω 2 2 2 ξ fj(ξ) dξ = uj = λj. (4.9) Rn | | | | |∇ | Z ZΩ

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[SEC. 4.1: INTRODUCTION 75

Setting

k F (ξ) = f (ξ) 2; (4.10) | j | j=1 X n then 0 F (ξ) (2π)− Ω , ≤ ≤ | | k k F (ξ) 2( f (ξ) 2)1/2( f (ξ) 2)1/2 (4.11) |∇ | ≤ | j | |∇ j | j=1 j=1 X X n n 2(2π)− Ω)Ω I( , ξ R , ≤ | | ∀ ∈ p

F (ξ)dξ = k (4.12) Rn Z and

k 2 ξ F (ξ)dξ = λj. (4.13) Rn | | j=1 Z X

Let F ∗(ξ) = φ( ξ ) denote the decreasing radial rearrangement of F . By approximating| | F we may assume that the function φ : [0; + ) n ∞ → [0, (2π)− Ω ] is absolutely continuous. Setting µ(t) = F ∗ > t = F > t |the| co-area formula (1.17) implies that |{ }| |{ }| (2π)−n Ω | | 1 µ(t) = F − dσsds. (4.14) t F =s |∇ | Z Z{ } n Observe that F ∗ is radial and so µ(φ(s)) = F ∗ > φ(s) = ωns n 1 |{ }| which gives nωns − = µ′(φ(s))φ′(s) a.e. It follows from (4.11), (4.14) and the isoperimetric inequality that

1 µ′(φ(s)) = F − dσs (4.15) − F =φ(s) |∇ | Z{ } 1 ρ− F = φ(s) ≥ 1|{ n 1 }| ρ− nω s − ≥ n

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76 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

and so for almost every s

ρ φ′(s) 0, (4.16) − ≤ ≤ where

n ρ = 2(2π)− Ω)Ω I( . | | From (4.12) and (4.13), we have p

k = F (ξ)dξ = F ∗(ξ)dξ (4.17) Rn Rn Z Z ∞ n 1 = nωn s − φ(s)ds Z0 and k 2 λj = xi F (ξ)dξ (4.18) Rn | | j=1 X Z 2 xi F ∗(ξ)dξ ≥ Rn | | Z ∞ n+1 = nωn s φ(s)ds. Z0 We need an elementary lemma.

Lemma 4.1([75]) Let n 1, ρ, A > 0 and ψ : [0, + ) [0, + ) be decreasing (and absolutely≥ continuous) such that ∞ → ∞

∞ n 1 ρ φ′(s) 0, s − ψ(s)ds = A. (4.19) − ≤ ≤ Z0 Then 2 ∞ n+1 1 n+2 2 Aψ(0) s ψ(s)ds (nA) n ψ(0)− n + . (4.20) ≥ n2 + 6( n + 2)ρ2 Z0 1 Applying Lemma 4.1 to the function φ with A = (nωn)− k, ρ = n 2(2π)− Ω I(Ω) we get in view of (4.12) and (4.13) that | | 2 p 2 n+2 2 n n ckφ(0) λ ωn− k n φ(0)− n + (4.21) j ≥ n2 + ( n + 2)ρ2 j=1 X

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[SEC. 4.2: THE KRO¨GER’S THEOREM 77

1 where c is any constant such that 0 < c < 6 . Observe that 0 < n n φ(0) (2π)− Ω and that if R is such that ωnR = Ω , then I(Ω) ≤ | | n+2 | | ≥ x 2dx = nωnR and so B(R) | | n2 + R 2 n+2 n n n +1 ρ 2(2π)− ωn− Ω n (4.22) ≥ n2 + | | r 1 n+1 n n (2π)− ωn− Ω n . ≥ | | 1 2 n Let us choose c independent of n satisfying c < (2π) ωn− . It is easy to see that the function 2 2 n+2 2 n n ckt g(t) = ω− k n t− t + n2 + n ( n + 2)ρ2

n n is decreasing on (0, (2π)− Ω ]. We can replace φ(0) by (2π)− Ω in (4.21) which gives the inequality| | (4.5). | |

4.2 The Kr¨oger’sTheorem

th Let λk be the k eigenvalue for the Neumann boundary value problem with respect to the Laplace operator on a bounded domain with piecewise smooth boundary in Rn. Pólysconjecture states that 2 k n λk Cn Ω . With respect to this conjecture, Krögerproved the ≤ | | following result. Theorem 4.2 ([62]). The first k + 1 Neumann eigenvalues of a a bounded domain Ω with piecewise smooth boundary in Rn satisfy the inequality

n(k + 1) 1/n r = 2π , αn 1 Ω − | |

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78 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

one gets

Corollary 4.1. Under the assumptions of the Theorem the following inequality holds for every k:

2 k n n 2 1 − n+2 λj (2π) αn 1 Ω k n (4.25) ≤ n2 + n − | | j=1 X

k Proof of Theorem 4.2. Let φj j=1 be the set of orthonormal { } n eigenfunctions for the eigenvalues λ1, ..., λk. Fix z R and consider R C iy z ∈ hz : given by hz(y) = e · . Letting aj = Ω hz(x)uj(x)dx; → iz y 2 then the projection of h (y) e · into the subspace of L (Ω) z ≡ R spanned by φ1, ..., φk can be written as

hz(y) span u k = hz(x)uj(x)dx uj(y) (4.26) | { i}i=1 j=1 Ω X Z k = ajuj(y). j=1 X It is clear that the function g (y) h (y) k a u (y) is orthog- z ≡ z − j=1 j j onal to uj, j = 1, , k. Thus we have ··· P 2 gz(y) dy λ Ω |∇ | . (4.27) k+1 ≤ g2dy R Ω z Elementary computation sows thatR

k g (y) 2dy = z 2 Ω a 2λ (4.28) |∇ z | | | | | − | j| j Ω j=1 Z X and

k g2dy = Ω a 2. (4.29) z | | − | j| Ω j=1 Z X

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[SEC. 4.2: THE KRO¨GER’S THEOREM 79

1/n Assuming that r > 2π nk ; then α 1 Ω n− | | 2 gz(y) dy dz λ Br Ω |∇ | (4.30) k+1 2 ≤ gz(y) dy dz R BrR Ω Since R R

g (y) 2dy dz (4.31) |∇ z | ZBr ZΩ k = z 2 Ω dz λ a 2dz | | | | − j| j| Br Br j=1 Z Z X n+2 k r αn 1 Ω 2 = − | | λ a dz n2 + − j | j| j=1 Br X Z and

2 (gz(y)) dy dz (4.32) ZBr ZΩ k = Ω dz a 2dz | | − | j| Br Br j=1 Z Z X n k r αn 1 Ω 2 = − | | a dz, n − | j| j=1 Br X Z we get

n iy z uˆj(z) = (2π)− 2 e · uj(y)dy Rn Z and we know from the Plancherel Theorem that u 2(z)dz = Rn | j| R b

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80 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

u 2(y)dy. Hence we have Rn | j| R 2 2 hzujdy dz hzujdy dz ≤ Rn ZBr ZΩ Z ZΩ

n 2 = (2π) uj(z ) dz Rn | | Z n 2 = (2π) ubj(y) dy Rn | | Z = (2π)n.

Let us prove by induction that

n+2 r αn−1 Ω n k n2 + | | (2π) j=1 λj λk+1 n − . (4.34) r α 1 Ω ≤ n− | | (2π)nk n − P Suppose that (4.34) is true for k 1, that is − n+2 r αn−1 Ω n k 1 − n2 + | | (2π) j=1 λj λk n − . r α 1 Ω ≤ n− | | (2π)n(k 1) n − P− We then have that

n+2 r αn−1 Ω n k n2 + | | (2π) j=1 λj A λk n − = , r α 1 Ω ≤ n− | | (2π)nk B n − P where

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[SEC. 4.3: A GENERALIZED PO´LYA CONJECTURE BY CHENG-YANG 81

we get

A + k λ C λ l=1 l l k+1 ≤ k B +P l=1 Cl k A + λPk l=1 Cl ≤ k B + Pl=1 Cl A k A + BP l=1 Cl ≤ k B + Pl=1 Cl A = . P B This completes the proof of Theorem 4.2.

4.3 A generalized P´olya conjecture by Cheng-Yang

In [31], Cheng-Yang investigated eigenvalues of the Dirichlet Lapla- cian on a bounded domain in an n-dimensional complete Riemannian manifold M and proposed a generalized P´olya conjecture.

Cheng-Yang’s Conjecture ([31]). Let Ω be a bounded domain in an n-dimensional complete Riemannian manifold M. Then, there exists a constant c(M, Ω), which only depends on M and Ω such that eigenvalue λi’s of the eigenvalue problem

∆u = λu in Ω, (4.35) u =− 0. |∂Ω satisfy

k 2 1 n 4π 2 n λi + c(M, Ω) 2 k , k = 1, , (4.36) k ≥ n2 + n ··· i=1 (ωn Ω ) X | |

2 4π 2 n λi + c(M, Ω) 2 k , k = 1, , (4.37) ≥ (ω Ω ) n ··· n| |

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82 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

Remark 4.1. Cheng-Yang believed that in the above conjecture, Sn n2 when M is the unit sphere (1), c(M, Ω) = 4 , when M is the hyper- 2 Hn (n 1) bolic space ( 1), c(M, Ω) = −4 and when M is a complete minimal submanifold− in RN , c(M,− Ω) = 0. Cheng-Yang obtained a partial solution to the above conjecture.

Theorem 4.3 ([31]). Let Ω be a bounded domain in an n-dimensional complete Riemannian manifold M. Then, there exists a constant 2 H > 0, which only depends on M and Ω such that eigenvalue λi′ s of the problem (4.35) satisfy

k 2 1 n 4π 2 2 n λi + H0 2 k , k = 1, , (4.38) k ≥ n ··· i=1 ( n + 2)(n + 4) (ωn Ω ) X | | p Sn 2 n2 Moreover, when M is the unit sphere (1), one can take H0 = 4 and when M is a complete minimal submanifold in RN , one can take 2 H0 = 0. A crucial result in the proof of Theorem 4.3 is the following

Lemma 4.2 ([25]). Let µ1 µk+1 be any non-negative real numbers satisfying ≤ · · · ≤

k 4 k (µ µ )2 µ (µ µ ). (4.39) k+1 − i ≤ t i k+1 − i i=1 i=1 X X Deﬁne

1 k 1 k 2 G = µ ,T = µ2,F = 1 + G2 T . (4.40) k k i k k i k t k − k i=1 i=1 X X Then, we have

4 k + 1 t F C(t, k) F , (4.41) k+1 ≤ k k

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[SEC. 4.3: A GENERALIZED PO´LYA CONJECTURE BY CHENG-YANG 83

where t is any positive real number and

4 1 k t 1 + 2 1 + 4 C(t, k) = 1 t t < 1. (4.42) − 3t k1 + ( k + 1)3

Proof of Theorem 4.3. From Nashs theorem, we know that M can be isometricly immersed into a Euclidean space RN , that is, there exists an isometric immersion:

φ : M RN . (4.43) → Thus, M can be seen as a complete submanifold isometricly immersed into RN . We denote by H the mean curvature of the immersion φ. From (3.16), we have | |

k 4 k n2 (λ λ )2 (λ λ )(λ + sup H 2). (4.44) k+1 − i ≤ n k+1 − i i 4 | | i=1 i=1 Ω X X Since eigenvalues are invariants of isometries, the above inequality holds for any isometric immersion from M into a Euclidean space. Let us deﬁne

Φ = φ; φ is an isometric immersion from M into a Euclidean space . { } Putting

2 H0 = inf sup H ; φ Φ | | ∈ Ω then

k 4 k n2 (λ λ )2 (λ λ )(λ + H2). (4.45) k+1 − i ≤ n k+1 − i i 4 0 i=1 i=1 X X n2 2 Letting µi = λi + 4 H0 ; then

k 4 k (µ µ )2 µ (µ µ ). (4.46) k+1 − i ≤ n i k+1 − i i=1 i=1 X X

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84 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

From theorem 2.1 with t = n of [25], we have

1 1 k + 1 n k + 1 n F C(n, k) F F . k+1 ≤ k k ≤ k k Therefore, we infer

Fk+1 Fk 4 4 . ( k + 1) n ≤ k n

For any positive integers l and k, we have

Fk+l Fk 4 4 . (k + l) n ≤ k n

From the Weyl’s asymptotic formula (1.10)

2 λl 4π lim 2 = 2 , l l n (ω Ω ) n →∞ n| | we get

1 l λi 2 λl i=1 n 4π lim 2 = 2 l l n n2 + (ω Ω ) n →∞ P n| | and

1 l λ2 4 λl i=1 i n 16π lim 4 = 4 . l l n n4 + (ω Ω ) n →∞ P n| | Hence

4 Fk+l 2n 16π lim 4 = 4 . l (k + l) n ( n + 2)(n + 4) (ω Ω ) n →∞ n| | According to (4.46), we have, for any positive integer k,

4 Fk 2n 16π 4 4 . (4.47) k n ≥ ( n + 2)(n + 4) (ω Ω ) n n| |

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[SEC. 4.4: ANOTHER GENERALIZED PO´LYA CONJECTURE 85

Since 2 2 F = 1 + G2 T G2 , k n k − k ≤ n k we get

2 4 2 Gk Fk 2n 16π 4 4 4 n k n ≥ k n ≥ ( n + 2)(n + 4) (ω Ω ) n n| | which implies (4.38). In order to ﬁnish the proof of Theorem 4.3, we need only to observe that Sn(1) can be seen as a compact hypersurface in Rn+1 with mean curvature 1 and that a complete minimal submanifold in RN has mean curvature H = 0. | | 4.4 Another generalized P´olya conjecture

Let Ω be a bounded domain in Rn and let L be the elliptic operator of order 2t deﬁned by

t m Lu = am r( ∆) u, u C∞(Ω) − − ∈ m=r+1 X

where r 0 is an integer, am r’s are constants with am r 0, r+1 ≥ − − ≥ ≤ m t, at r = 1, t a fixed positive integer. Consider the following eigenvalue≤ − problem about L which is important in the study of various branches of mathematics, such as differential equations, differential geometry and mathematical physics:

Lu = λ( ∆)ru, u C (Ω), ∞ (4.48) (∂/∂ν)j u − = 0, j∈= 0, 1, 2, , t 1. |∂Ω ··· − Let

0 < λ λ λ . (4.49) 1,r ≤ 2,r ≤ · · · ≤ k,r ≤ · · · → ∞ be the eigenvalues of the problem (4.48). In [58], the following conjecture was proposed :

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86 [CAP. 4: POLYA´ CONJECTURE AND RELATED RESULTS

Generalized P´olya Conjecture. The eigenvalues λk,r, k = 1, 2, , of the eigenvalue problem (1.7) of the operator L satisﬁes the inequality···

t r 2m/n − k λ a Cm . (4.50) k,r ≥ m Ω m=1 X | |

With respect to the above generalized P´olya conjecture, Ku-Ku- Tang showed in [58] that if r is even, then

t r 2m/n − na k λ m Cm , k = 1, 2, . (4.51) k,r ≥ n2 + m Ω ··· m=1 X | | This section provides comparison theorems between the k-th eigenvalues of the problem (4.1) and that of the problem (4.48) which shows that if the P´olya conjecture (4.2) is true then so is the generalized P´olya conjecture (4.50).

Theorem 4.3 ([93]). Let M be an n( 2)-dimensional compact Riemannian manifold with boundary. Denote≥ by ∆ the Laplacian operator on M and let L be the elliptic operator given by

t m Lu = am r( ∆) u, u C∞(M), (4.52) − − ∈ m=r+1 X where r 0 is an integer, am r’s are constants with am r 0, r + ≥ − − ≥ 1 m t, at r = 1, t a ﬁxed positive integer. Consider the following≤ eigenvalue≤ − problems: Lu = λ( ∆)ru, u C (M), ∞ (4.53) (∂/∂ν)j u − = 0, j∈= 0, 1, 2, , t 1, |∂M ··· −

( ∆)r+1u = Λ( ∆)ru, u C (M), ∞ (4.54) − (∂/∂ν)j u − = 0, j∈= 0, 1, 2, , r, |∂M ···

∆u = λu in M, (4.55) u− = 0. |∂M

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[SEC. 4.4: ANOTHER GENERALIZED PO´LYA CONJECTURE 87

Denote by 0 < λ λ (4.56) 1,r ≤ 2,r ≤ · · · → ∞ 0 < Λ Λ (4.57) 1,r ≤ 2,r ≤ · · · → ∞ and 0 < λ < λ (4.58) 1 2 ≤ · · · → ∞ the successive eigenvalues for (4.53), (4.54) and (4.55), respectively. Here each eigenvalue is repeated according to its multiplicity. Then for any k = 1, 2, , we have ··· 2 t (r+1) t r λk,r a1Λk,r + a2Λk,r + + at (r+1)Λk,r− + Λk,r− (4.59) ≥ ··· − and Λ λ . (4.60) k,r ≥ k

Remark 1.1. If M is a bounded domain in Rn, we know from the inequality

2 t (r+1) t r λk,r a1λk + a2λk + + at (r+1)λk− + λk− (4.61) ≥ ··· − which is a combination of (4.59) and (4.60) that if the P´olya conjecture (4.2) is true then so is the generalized P´olya conjecture (4.50).

k Proof of Theorem 4.3. Let ui i=1 be a set of orthonormal eigenfunctions of the problem (4.53){ corresponding} to λ k , that is, { i,r}i=1 r Lui = λi,r( ∆) ui, in M, −t−1 ∂ ui ui ∂M = = ∂νt−1 = 0,  | ··· ∂M  u ( ∆)ru = δ , i, j = 1, , k. M i − j ij ··· k Similarly, let Rvi i=1 be a set of orthonormal eigenfunctions of the problem (4.54){ corresponding} to Λ k , that is, { i,r}i=1 r+1 r ( ∆) vi = Λi,r( ∆) vi in M, − − r ∂ vi vi ∂M = = ∂νr = 0,  | ··· ∂M  v ( ∆)rv = δ , i, j = 1, , k. M i − j ij ···

 R

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Let w = k α u = 0 be such that j=1 j j 6 P w( ∆)rv = 0, j = 1, , k 1. (4.62) − j ∀ ··· − ZM Such an element w exists because αj 1 j k is a non-trivial solution of a system of (k 1)-linear{ equations| ≤ ≤ } − k α u ( ∆)rv = 0, 1 i k 1, (4.63) j j − i ≤ ≤ − j=1 M X Z in k unknowns. Notice that if u C∞(M) satisﬁes ∈ ∂u ∂t 1u u = = = − = 0, |∂M ∂ν ··· ∂νt 1 ∂M − ∂M

then

u ∂M = u ∂M = ∆u ∂M = (∆u) ∂M = = | ∇p | 1 | p ∇1 | ··· = ∆ − u = (∆ − u) = 0, when t = 2p ∂M ∇ ∂M

and k 1 u = u = ∆u = (∆u) = = ∆ − u |∂M ∇ |∂M |∂M ∇ |∂M ··· ∂M p 1 p = (∆ − u) = ∆ u = 0, when t = 2 p + 1. ∇ ∂M |∂M r Observe that M w( ∆) w = 0. In fact, from divergence theorem, we have − 6 R r/2 2 r M (∆ w) , if r is even, w( ∆) w = (r 1)/2 2 − (∆ − w) , if r is odd. ZM ( M ∇ R r R r/2 Thus, if M w( ∆) w = 0, then ∆ w = 0 when r is even and (r 1)/2 − ∆ − w = 0 when r is odd. It then follows from the maximum R r/2 1 principle for harmonic functions that ∆ − w = 0 when r is even and (r 1)/2 1 ∆ − − w = 0 when r is odd. Continuing this process, we conclude r that w = 0. This is a contradiction. Thus M w( ∆) w = 0. Let us assume without loss of generality that − 6 R w( ∆)rw = 1. (4.64) − ZM

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Hence we infer from the Rayleigh-Ritz inequality that

Λ w( ∆)r+1w (4.65) k,r ≤ − ZM

We claim that for any j = 1, , t r, ··· −

j+1 j w( ∆)r+jw w( ∆)r+j+1w . (4.66) − ≤ − ZM ZM

Let us ﬁrst prove that (4.66) holds when j = 1. In fact, if r = 2h is even, then one deduces from the divergence theorem and the H¨older’s inequality that

2 w( ∆)r+1w − ZM 2 = ( ∆)hw( ∆)h+1w − − ZM (( ∆)hw)2 (( ∆)h+1w)2 ≤ − − ZM ZM = w( ∆)r+2w. − ZM

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On the other hand, if r = 2h + 1 is odd, then

2 w( ∆)r+1w − ZM 2 = (( ∆)h+1w)( ∆)(( ∆)hw) − − − ZM 2 = (( ∆)h+1w) (( ∆)hw) ∇ − ∇ − ZM (( ∆)h+1w 2 (( ∆)hw) 2 ≤ |∇ − | |∇ − | ZM ZM = ( ∆)hw( ∆)h+1w ( ∆)h+1w( ∆)h+2w − − − − ZM ZM = w( ∆)rw w( ∆)2h+3w − − ZM ZM = w( ∆)r+2w. − ZM Thus (4.66) holds when j = 1. Suppose now that (4.66) holds for j 1, that is − j j 1 r+j 1 r+j − w( ∆) − w w( ∆) w . (4.67) − ≤ − ZM ZM When r + j is even, we have

w( ∆)r+jw (4.68) − ZM (r+j)/2 1 (r+j)/2 = ∆ − w∆ ∆ w M Z (r+j)/2 1 (r+j)/2 = ∆ − ui ∆ ui − M ∇ ∇ Z 2 1/2 2 1/2 (r+j)/2 1 (r+j)/2 ∆ − w ∆ w ≤ M ∇ M ∇ Z Z 1/2 1/2 r+j 1 r+j+1 = w( ∆) − w w( ∆) w , − − ZM ZM

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On the other hand, when r + j is odd,

w( ∆)r+j w (4.69) − ZM (r+j 1)/2 (r+j+1)/2 = ( ∆) − w( ∆) w − − ZM 2 1/2 2 1/2 (r+j 1)/2 (r+j+1)/2 ( ∆) − w ( ∆) w ≤ − − ZM ZM 1/2 1/2 r+j 1 r+j+1 = w( ∆) − w w( ∆) w . − − ZM ZM Thus we always have

w( ∆)r+j w (4.70) − ZM 1/2 1/2 r+j 1 r+j+1 ( ∆) − w w( ∆) w . ≤ − − ZM ZM Combining (4.67) and (4.70), we know that (4.66) is true for j. Using (4.67) repeatedly, we get

1/s w( ∆)r+1w w( ∆)r+sw , s = 1, , t (r + 1). − ≤ − ··· − ZΩ ZM which, combining with (4.65), gives

Λs w( ∆)r+sw, s = 1, 2, , t (r + 1). k,r ≤ − ··· − ZM Thus we have

2 t (r+1) t r a1Λk,r + a2Λk,r + + at (r+1)Λk,r− + Λk,r− ··· − w(a ( ∆)r+1 + a ( ∆)r+2 + + ( ∆)t)w ≤ 1 − 2 − ··· − ZM k = wLw = ηiηj uiLuj M i,j=1 M Z X Z k k = η η u λ ( ∆)ru = η2λ λ , i j i j,r − j i i,r ≤ k,r i,j=1 M i=1 X Z X

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where in the last equality, we have used the fact that

k η2 = w( ∆)rw = 1. i − i=1 X Z This proves (4.59). In order to prove (4.60), let us take a set of orthonormal eigen- k k functions zi i=1 of the problem (4.55) corresponding to λi i=1, that is, { } { }

∆zi = λizi in M, z =− 0, (4.71)  i ∂M | z z = δ , i, j = 1, , k.  M i j ij ··· k  R Let ξ = j=1 βjvj be such that P ξ2 = 1 and ξz = 0, j = 1, , k 1. (4.72) j ∀ ··· − ZM ZM It follows from the Rayleigh-Ritz inequality that

λ ξ( ∆ξ) (4.73) k ≤ − ZM Using the same arguments as in the proof of (4.59), we have

j+1 j ξ( ∆)j ξ ξ( ∆)j+1ξ , j = 1, , r. (4.74) − ≤ − ··· ZM ZM Thus we have

1 1 r r k λ ξ( ∆)rξ = β2 (4.75) k ≤ −  j  M j=1 Z X   On the other hand, taking j = r in (4.73), we get

r+1 r ξ( ∆)rξ ξ( ∆)r+1ξ , (4.76) − ≤ − ZM ZM

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which implies that

r k r1+ β2 ξ( ∆)r+1ξ (4.77) j ≤ − j=1 ZM X r k r1+ = β β v ( ∆)r+1v  i j i − j M i,j=1 Z X   r k r1+ = β β v Λ ( ∆)rv  i j i j,r − j M i,j=1 Z X  r  k r1+ = Λ β2  j,r j  j=1 X   r k r1+ r Λ r1+ β2 . ≤ k,r  j  j=1 X   Thus we have

k β2 Λr . (4.78) j ≤ k,r j=1 X Combining (4.75) and (4.78), one gets (4.60).

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Chapter 5

The Steklov eigenvalue problems

5.1 Introduction

Let M be an n-dimensional compact Riemannian manifold with boundary. The Stekloﬀ problem is to ﬁnd a solution of the equation

∆u = 0 in M, (5.1) ∂u = pu on ∂M, ∂ν where p is a real number. This problem was first introduced by Steklov for bounded domains in the plane in [87]. His motivation came from physics. The function u represents the steady state temperature on M such that the flux on the boundary is proportional to the temperature. Problem (5.1) is also important in conductivity and harmonic analysis as it was initially studied by Calderón(Cf. [15]). This connection arises because the set of eigenvalues for the Steklov problem is the same as the set of eigenvalues of the well- known Dirichlet-Neumann map. This map associates to each function u defined on the boundary ∂M, the normal derivative of the harmonic function on M with boundary data u. The Steklov eigenvalue problem has appeared in quite a few physical fields, such as fluid mechanics, electromagnetism, elasticity, etc., and received in-

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creasing attention [53], [64]. It has applications, for instance, in the investigation of surface waves [12], the analysis of stability of me- chanical oscillators immersed in a viscous fluid [32], and the study of the vibration modes of a structure in contact with an incompressible fluid [13]. Numerical methods have been developed for this problem. For instance, its optimal error estimates of linear finite element approximations have been obtained in [2]. Interesting estimates for eigenvalues of the Steklov problem have been obtained some of which will be introduced in this chapter.

5.2 Estimates for the Steklov eigenvalues

In this section, we prove some estimates for the Steklov eigenvalues. Let us recall firstly the Reilly’s formula. Let M be n-dimensional compact manifold M with boundary ∂M. We denote by , the Riemannian metric on M as well as that induced on ∂M. Leth i and ∆ be the connection and the Laplacian on M, respectively.∇ Let ν be the unit outward normal vector of ∂M. The shape operator of ∂M is given by S(X) = X ν and the second fundamental form of ∂M is defined as II(X,Y∇) = S(X),Y , here X,Y T ∂M. The eigenvalues of S are called theh principali curvatures of∈∂M and the 1 mean curvature H of ∂M is given by H = n 1 tr S, here tr S denotes the trace of S. For a smooth function f defined− on an n-dimensional compact manifold M with boundary ∂M, the following identity holds ∂f if u = ∂ν , z = f ∂M and Ric denotes the Ricci tencor of M, then ∂M | (see [82], p. 46):

(∆f)2 2f 2 Ric( f, f) (5.2) − |∇ | − ∇ ∇ ZM = ((n 1)Hu + 2∆I z)u + I (, z z) . − ∇ ∇ Z∂M Here 2f is the Hessian of f; ∆ and represent the Laplacian and the∇ gradient on ∂M with respect to∇ the induced metric on ∂M, respectively.

Theorem 5.1 ([35]). Let M be an n-dimensional compact connected Riemannian manifold with non-negative Ricci curvature and

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boundary. Assume that the principal curvatures of ∂M are bounded from below by a positive constant c. Then the first non-zero eigen- c value of the problem (5.1) satisfies p1 > 2 . Proof. Let f be the first eigenfunction of the Steklov problem ∂u (5.1). Setting z = f ∂M , u = ∂ν ∂M and using Reilly’s formula we find after integration| by parts that

0 > 2f 2 2 z u + c z 2. − |∇ | ≥ − ∇ · ∇ |∇ | ZM Z∂M Z∂M

0 > 2p z 2 + c z 2. (5.3) − 1 |∇ | |∇ | Z∂M Z∂M Since z = constant, otherwise f=constant on M which is a contradiction,6 we have

z 2 > 0. (5.4) |∇ | Z∂M

Thus (5.3) implies that p1 > c/2. In view of Theorem 5.1, Escobar conjectured that under the same conditions as in Theorem 5.1, p c. 1 ≥ Theorem 5.2. Let the conditions be as in Theorem 5.1. Then we have i) The non-zero eigenvalue of the Laplacian of ∂M satisﬁes

λ (n 1)c2 (5.5) 1 ≥ − with equality holding if and only if M is isometric to a ball of radius 1/c in Rn([102]). ii) The non-zero eigenvalue of the Steklov problem (5.1) satisﬁes

√λ p 1 λ + λ) (n 1 c2 (5.6) 1 ≤ ()n 1 c 1 1 − − − p p Moreover, the equality holds in (5.6) if and only if M is isometric to a ball of radius 1/c in Rn ([89]).

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Proof. Let z be an eigenfunction corresponding to the ﬁrst nonzero eigenvalue λ1 of the Laplacian of ∂M:

∆z + λ1z = 0.

Let f in C∞(M) be the solution of the Dirichlet problem ∆f = 0 in M, (5.7) f = z. |∂M It then follows from (5.2) and the nonnegativity of the Ricci curvature of M that

0 (∆f)2 2f 2 Ric( f, f) (5.8) ≥ − |∇ | − ∇ ∇ ZM = ((n 1)Hh + 2∆I z)h + I (, z z) − ∇ ∇ Z∂M ( 2( λ z)u + (n 1)cu2 + c z 2) ≥ − − 1 − |∇ | Z∂M (2λ zu + (n 1)cu2 + cλ z2) ≥ 1 − 1 Z∂M λ z 2 λ2 (n 1)c u + 1 + cλ 1 z2 ≥ − ()n 1 c 1 − ()n 1 c Z∂M ( − − ) λ2 cλ 1 z2 . ≥ 1 − ()n 1 c Z∂M − Thus we have λ2 cλ 1 0 1 − ()n 1 c ≤ − or λ (n 1)c2. 1 ≥ − 1 If M is isometric to an n-dimensional Euclidean ball of radius c , it is 2 well known that λ1(M) = (n 1)c . Now we assume conversely that 2 − λ1(M) = (n 1)c . In this case the inequalities in (5.8) must take equality sign.− In particular, we have λ z 2f = 0,H = c, u = 1 = cz. (5.9) ∇ −()n 1 c − −

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From 2f = 0, we know that f 2 is a constant and is not zero since f∇is not a constant. Without|∇ loss| of generality, we can assume f 2 = 1. Thus for any point p M, we have |∇ | ∈ 1 = f 2(p) = z2 (p) + u2(p). (5.10) |∇ | |∇ | It follows from (5.10) by integration and u = cz, that − A(∂M) = ( z 2 + u2) (5.11) |∇ | Z∂M 2 2 = (λ1z + u ) Z∂M = (nu2). Z∂M On the other hand, from 1 ∆(f 2) = f 2 + f∆f = 1 (5.12) 2 |∇ | and the divergence theorem we have 1 u2 V (M) = ∆(f 2) = zu = , 2 − c ZM Z∂M Z∂M which, combining with (5.11), gives 1 A(∂M) H = c = . (5.13) n · V (M) It then follows from a result in [83] that M is isometric to a ball in n 2 R . Since λ1(∂M) = (n 1)c , the radius of M is easily seen to be 1 − c . This proves item i) of Theorem 5.1. Now let us prove (5.6). Let f and z be as in the proof of item i). We have from the Rayleigh-Ritz inequality that (Cf. [59]) h2 p ∂M (5.14) 1 ≤ f 2 RM |∇ | and R f 2 p M |∇ | , (5.15) 1 ≤ z2 R ∂M R

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which gives

h2 p2 ∂M . (5.16) 1 ≤ z2 R∂M It then follows by substituting Rf into the Reilly’s formula that

0 (∆f)2 2f 2 Ric( f, f) (5.17) ≥ − |∇ | − ∇ ∇ ZM (n 1)c h2 2λ hz + c z 2 ≥ − − 1 |∇ | Z∂M Z∂M Z∂M 2 2 = (n 1)c h 2λ1 hz + cλ1 z − ∂M − ∂M ∂M Z Z 1 Z 1 2 2 (n 1)c h2 2λ h2 z2 + cλ z2. ≥ − − 1 1 Z∂M Z∂M Z∂M Z∂M Hence, we have

1 1 2 √λ 2 h2 1 λ + λ) (n 1 c2 z2 . ≤ ()n 1 c 1 1 − − ∂M ∂M Z − p p Z Hence √λ p 1 λ + λ) (n 1 c2 . 1 ≤ ()n 1 c 1 1 − − − p p Assume now that

√λ p = 1 λ + λ) (n 1 c2 . 1 ()n 1 c 1 1 − − − p p Then the inequalities in (5.17) should take equality sign. We infer therefore

2f = 0,H = c (5.18) ∇ and

√λ h = 1 λ + λ) (n 1 c2 z. (5.19) ()n 1 c 1 1 − − − p p

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n 1 Take a local orthonormal ﬁelds ei i=1− tangent to ∂M. We infer from (5.18) and (5.19) that { }

n 1 − 0 = 2f(e , e ) = ∆( z + n 1)Hh (5.20) ∇ i i − i=1 X √λ = λ z + (n 1)c 1 λ + λ) (n 1 c2 z, − 1 − · ()n 1 c 1 1 − − − p p 2 which gives λ1 = (n 1)c and so M is isometric to an n-dimensional − 1 Euclidean ball of radius c . We consider now a fourth order Steklov eigenvalue problem on an n-dimensional compact connected Riemannian manifold (M, , ) given by h i ∆2u = 0 in M, (5.21) ∂u u = ∆u q = 0 on ∂M, − ∂ν

wher q is a real number. Let q1 be the ﬁrst non-zero eigenvalue of the problem (5.21). As pointed by Kuttler [63], q1 is the sharp constant for a priori estimates for the Laplace equation ∆v = 0 in M, v = g on ∂M, (5.22) where g L2(∂M). It has∈ been proven by Payne that if Ω R2 is a bounded convex domain with smooth boundary then q ⊂(Ω) 2ρ with equality 1 ≥ 0 holding if and only if Ω is a disk, where ρ0 is the minimum geodesic curvature of ∂Ω. This Payne’s theorem has been extended to higher dimensional Euclidean domains by Ferrero, Gazzola and Weth [38].

Theorem 5.3 ([91]). Let (M, , ) be an n( 2)-dimensional compact connected Riemannian manifoldh i with boundary≥ ∂M and nonnegative Ricci curvature. Assume that the mean curvature of M is bounded below by a positive constant c. Let q1 be the ﬁrst eigenvalue of the following Stekloﬀ eigenvalue problem : ∆2u = 0 in M, (5.23) ∂u u = ∆u q = 0 on ∂M. − ∂ν

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[SEC. 5.2: ESTIMATES FOR THE STEKLOV EIGENVALUES 101

Then q1 nc with equality holding if and only if M is isometric to a ≥ 1 Rn ball of radius c in .

Proof. Let w be an eigenfunction corresponding to the ﬁrst eigenvalue q1 of the problem (5.23), that is

∆2w = 0 in M, (5.24) ∂w w = ∆w q = 0 on ∂M. − 1 ∂ν

∂w Set η = ∂ν ∂M ; then

2 (∆w) q = M . (5.25) 1 η2 R ∂M Substituting w into Reilly’s formula,R we have

(∆w)2 2w 2 (5.26) − |∇ | ZM = Ric( w, w) + (n 1)Hη2 ∇ ∇ − ZM Z∂M (n 1)c η2. ≥ − Z∂M The Schwarz inequality implies that 1 2w 2 (∆w)2 (5.27) |∇ | ≥ n

2 ∆w with equality holding if and only if w = n , . Combining (5.25)- (5.27), we have q nc. This completes∇ the proofh i of the ﬁrst part of 1 ≥ Theorem 1.2. Assume now that q1 = nc. In this case, the inequalities (5.26) and (5.27) must take equality sign. In particular, we have ∆w 2w = , . (5.28) ∇ n h i

Take an orthornormal frame e1, , en 1, en on M such that when { ··· −2 } restricted to ∂M, en = ν. From 0 = w(ei, en), i = 1, , n 1, and w = 0, we conclude that η =∇ρ = const. and so··· ∆w − = |∂M |∂M

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q1η = ncρ is also a constant. Since (5.26) takes equality sign and η is constant, we infer that H c. Also, we conclude from the fact that ∆w is a harmonic function≡ on M and the maximum principle that ∆w is constant on M. Suppose without loss of generality that ∆w = 1 and so we have 1 2w = , . (5.29) ∇ nh i It then follows by deriving (5.29) covariantly that 3w = 0 and from the Ricci identity, ∇

R(X,Y ) w = 0, (5.30) ∇ for any X,Y tangent vector to M, where R is the curvature tensor of M. From the maximum principle w attains its minimum at some point x0 in the interior of M. From (5.29) it follows that 1 ∂ w = r , (5.31) ∇ n ∂r

where r is the distance function to x0. Using (5.30), (5.31), Cartan’s theorem and w = 0, we conclude that M is an Euclidean ball |∂M whose center is x0, and f is given by 1 w(x) = ( x x 2 b2) n | − 0| − in M, b being the radius of the ball. Since the mean curvature of ∂M 1 is c, the radius of the ball is c . For more recent developments about Steklov eigenvalues, we refer to [37, 74, 95] and the references therein.

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