Eigenvalues on Riemannian Manifolds

Publicaes Matemticas

Eigenvalues on Riemannian Manifolds

Changyu Xia UnB

29o Colquio Brasileiro de Matemtica

Copyright 2013 by Changyu Xia

Impresso no Brasil / Printed in Brazil

Capa: Noni Geiger / Srgio R. Vaz

29o Colquio Brasileiro de Matemtica

Anlise em Fractais Milton Jara Asymptotic Models for Surface and Internal Waves - Jean-Claude Saut Bilhares: Aspectos Fsicos e Matemticos - Alberto Saa e Renato de S

Teles Controle timo: Uma Introduo na Forma de Problemas e Solues -

Alex L. de Castro Eigenvalues on Riemannian Manifolds - Changyu Xia

Equaes Algbricas 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 Vctor F. Sirvent Funes de Operador e o Estudo do Espectro - Augusto Armando de

Castro Jnior Introduo Geometria Finsler - Umberto L. Hryniewicz e Pedro A. S.

Salomo Introduo aos Mtodos de Crivos em Teoria dos Nmeros - Jlio

Andrade Otimizao de Mdias sobre Grafos Orientados - Eduardo Garibaldi e

Joo Tiago Assuno Gomes ISBN: 978-85-244-0354-5

Distribuio: IMPA Estrada Dona Castorina, 110 22460-320 Rio de Janeiro, RJ E-mail: [email protected] http://www.impa.br

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Contents

1 Eigenvalue problems on Riemannian manifolds 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Some estimates for the first eigenvalue of the Laplacian 5

2 Isoperimetric inequalities for eigenvalues 142.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 142.2 The Faber-Krahn Inequality . . . . . . . . . . . . . . . 162.3 The Szego-Weinberger Inequality . . . . . . . . . . . . 172.4 The Ashbaugh-Benguria Theorem . . . . . . . . . . . 192.5 The Hersch Theorem . . . . . . . . . . . . . . . . . . . 23

3 Universal Inequalities for Eigenvalues 293.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Eigenvalues of the Clamped Plate Problem . . . . 343.3 Eigenvalues of the Polyharmonic Operator . . . 463.4 Eigenvalues of the Buckling Problem . . . . . . . . . . 57

4 Polya Conjecture and Related Results 734.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 734.2 The Krogers Theorem . . . . . . . . . . . . . . . . . . 774.3 A generalized Polya conjecture by Cheng-Yang . . 814.4 Another generalized Polya conjecture . . . . . . . . . . 85

5 The Steklov eigenvalue problems 945.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 945.2 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-cian . In local coordinate system {xi}ni=1, the Laplacian is givenby

=1G

n

i,j=1

xi

(Ggij

xj

)

where (gij) is the inverse matrix (gij)1, gij = g(

xi

, xj ) are the

coefficients of the Riemannian metric in the local coordinates, andG = det(gij). In local coordinates, the Riemannian measure dv on(M, g) is given by

dv =Gdx1...dxn.

Let C(M) and set

||||21 =

M

||2 +

M

||2.

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 withrespect to the Riemannian measure on M . Let us denote by H21 (M)

ando

H21 (M) the completion of C(M) and C0 (M) with respect to

|| ||. The theory of Sobolev spaces tells us that H21 (M) =o

H21 (M)when M is complete. Our purpose is to study some eigenvalue prob-lems associated to the Laplacian operator on a compact manifold M .When M = , we consider the closed eigenvalue problem:

u+ u = 0. (1.1)

When M 6= , we are interested in the following eigenvalue prob-lems.

The Dirichlet problem:{

u = u in M,u|M = 0. (1.2)

The Neumannn problem:{

u = u in M,u

M

= 0,(1.3)

where is the unit outward normal to M .

The clamped plate problem:{

2u = u in M,u|M = u

M

= 0,(1.4)

The buckling problem:{

2u = u in M,u|M = u

M

= 0,(1.5)

The eigenvalue problem of poly-harmonic operator:{

()lu = u in M,u|M = u

M

= = l1ul1

M

= 0, l 2. (1.6)


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[SEC. 1.1: INTRODUCTION 3

The buckling problem of arbitrary order:{

()lu = u in M,u|M = u

M

= = l1ul1

M

= 0, l 2. (1.7)

The Steklov problem of second order:{

u = 0 in M,u = u on M.

(1.8)

The Steklov problem of fourth order:{

2u = 0 in M,u = u u = 0 on M.

(1.9)

Let us denote by 1 the first non-zero eigenvalue of the above prob-lems. We can arrange the eigenvalues of these problems as follows:

0 < 1 2 +.

For many reasons in Mathematics and Physics, it is important toobtain nice estimates for the s. We will concentrate our attentionon this problem. Let us list some basic facts in this direction.

Theorem 1.1 (Weyls asymptotic formula, [97]). In each of theeigenvalue problems (1.1), (1.2), (1.3), let N() be the number ofeigenvalues, counted with multiplicity, . Then

N() n|M |n/2/(2)n (1.10)

as , where n is the volume of the unit ball in Rn and |M | isthe volume of M . In particular,

n/2k {(2)n/n}k/|M | (1.11)

as +.

There are similar asymptotic formulas for the other eigenvalueproblems above (Cf. [1], [79], [80]).

Define a space H as follows:


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For the closed eigenvalue problem (1.1),

H =

{f H21 (M)

M

f = 0

}. (1.12)

For the Dirichlet eigenvalue problem (1.2),

H =o

H21 (M). (1.13)

For the Neumann eigenvalue problem (1.3),

H =

{f H21 (M)

M

f = 0

}. (1.14)

A fundamental fool in the theory of eigenvalues is the

Mini-Max principle. We can find a countable orthonormal ba-sis {fi}, fi C(M) for the problems (1.1), (1.2) and (1.3) suchthat

1 = inf{

M|f |2

Mf2

| f H},

i = inf{

M|f |2

Mf2

| f H,

Mffj = 0, j = 1, , i 1

}.

(1.15)

In particular, we have the

Poincare inequality:

M

|f |2 1

M

f2, f H. (1.16)

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

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

M

g =

(

{f=}

g

|f |

)d (1.17)


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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 eigenvalueof the Laplacian.

Theorem 1.3 ([73]). Let M be an n-dimensional complete Rie-mannian manifold with Ricci curvature RicM n 1. Then thefirst non-zero eigenvalue of the closed eigenvalue problem (1.1) of Msatisfies 1(M) n.

The proof of Theorem 1.3 can be carried out by substituting afirst eigenfunction into the Bochner formula and integrating on Mthe resulted equality (see the proof of theorem 1.6 below).

An important classical result about eigenvalue is the following

Theorem 1.4 (Chengs Comparison Theorem, [18]). Let M bean n-dimensional complete Riemannian manifold with Ricci curva-ture satisfying RicM (n 1)c and let BR(p) be an open geodesicball of radius R around a point p in M , where R < /

c, when

c > 0. Then the first eigenvalue of the Dirichlet problem (1.2) ofBR(p) satisfies

1(BR(p)) 1(BR(c)), (1.18)

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

An immediate application of Chengs eigenvalue comparison the-orem is a rigidity theorem for compact manifolds of positive Riccicurvature.

Theorem 1.5 (The Maximal Diameter Theorem, [18]). Let Mbe an n-dimensional complete Riemannian manifold with Ricci cur-vature RicM n 1. If the diameter of M satisfies d(M) , thenM is isometric to an n-dimensional unit sphere.

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


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

M

(af + bg) = 0

Observe that the first Dirichlet eigenvalue of an n-dimensional unithemisphere is n. The mini-max principle and Chengs comparisontheorem then imply that

n 1(M)

M|(af + bg)|2

M(af + bg)2

=a2

B/2(p)|f |2 + b2

B/2(q)

|g|2

a2

B/2(p)f2 + b2

B/2(q)

g2

=a21(B/2(p))

B/2(p)

f2 + b21(B/2(q))

B/2(q)g2

a2

B/2(p)f2 + b2

B/2(q)

g2

na2

B/2(p)

f2 + nb2

B/2(q)g2

a2

B/2(p)f2 + b2

B/2(q)

g2= n.

We conclude from the equality case of the mini-max principle andChengs comparison theorem that each of B/2(p) and B/2(q) isisometric to the n-dimensional unit hemisphere and

M = B/2(p) B/2(q)

Consequently, M is isometric to a unit n-sphere.

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

Theorem 1.6 ([76]). Let M be an n-dimensional complete Rie-mannian manifold with Ricci curvature RicM n 1. If the firstnon-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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Proof. Let f be a first eigenfunction corresponding to the firsteigenvalue n of M . From the Bochner formula, we get

1

2|f |2 = |2f |2 + f,(f) + Ric(f,f) (1.19)

(f)2

n n|f |2 + (n 1)|f |2 = nf2 |f |2.

Integrating on M and noticing

M(nf2 |f |2) = 0, we conclude

that the inequalities in 1.19 should take equality sign. Thus, we have

1

2(|f |2 + f2) = 1

2|f |2 + 1

2f2

= nf2 |f |2 + ff + |f |2 = 0

and so |f |2 + f2 is a constant. Without lose of generality, we canassume that |f |2 + f2 = 1 and so

|f |1 f2

= 1.

Let p and q be points of M such that f(p) = f(q) = 1 and take aunit speed minimizing geodesic : [0, l] M from p to q. Integratingthe above equation along , we obtain

l =

ds =

|f |1 f2

ds 1

1

dt1 t2

= .

It then follows from the maximal diameter theorem that M is iso-metric to an unit n-sphere.

Remark 1.1. Let Mn be a compact Riemannian manifold withRicci curvature RicM n 1 and nonempty boundary. If the meancurvature of M is nonnegative, then the first Dirichlet eigenvalueof M satisfies 1 n with equality holding if and only if Mn isisometric to an n-dimensional unit hemisphere [82]. Similarly, if theboundary ofM is convex, then the first non-zero Neumann eigenvalueof M must satisfy 1 n with equality holding if and only if Mn isisometric to an n-dimensional unit hemisphere [34, 100].


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We now prove another rigidity theorem using the techniques ofeigenvalues.

Theorem 1.7 ([101]). Let M be an n-dimensional complete Rie-mannian manifold with Ricci curvature RicM n 1 and let N be aclosed minimal hypersurface which divides M into two disjoint opendomains 1 and 2. If there exists a point p M such that d(p,N),the distance from p to N , is no less than /2, then the pair (M,N) isisometric to the pair (Sn(1),Sn1(1), being Sn(1) the unit n-sphere.

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

1(B/2(p)

) 1(1). (1.20)

On the other hand, Chengs comparison theorem tells us that

1(B/2(p)

) n (1.21)

and Reillys estimate implies that 1(1) n. Thus, the inequalitiesin (1.20) and (1.21) should be equalities. Consequently, B/2(p) = 1is isometric to an n-dimensional unit hemisphere and so N = 1 =S

n1(1) is totally geodesic. It then follows from a result of [39] that2 is also isometric to an n-dimensional unit hemisphere.

Let 1 be the least nontrivial eigenvalue of an n-dimensional com-pact manifold M and let be the corresponding eigenfunction. Bymultiplying with a constant it is possible to assume that

a 1 = infM; a+ 1 = sup

M

where 0 a() < 1 is the median of .Suppose that Mn is a compact manifold without boundary of

nonnegative Ricci curvature and of diameter d. Li-Yau [72] showedthat the first nontrivial eigenvalue satisfies

1 2

(1 + a)d2


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and conjectured that

1 2

d2. (1.22)

Li-Yaus conjecture was proved by Zhong and Yang in [103]. Let usprovide a proof of (1.22) given by Li in [71].

Lemma 1.1. The function

z(u) =2

(arcsin(u) + u

1 u2

) u

defined on [-1, 1] satisfies

uz + z(1 u2) + u = 0; (1.23)

z2 2zz + z 0; (1.24)

2z uz + 1 0; (1.25)

and

1 u2 2|z|. (1.26)

Proof. Differentiating yields

z =4

1 u2 1, z = 4u

1 u2.

Thus (1.23) is satisfied.To see (1.24), we note that

z2 2zz + z = 4

1 u2{

4

(1 u2 + u arcsinu

) (1 + u2)

}.

Since the right hand side is an even function, it suffices to check that

4

(1 u2 + u arcsinu

) (1 + u2) 0

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on [0, 1]. It is easy to see that

d

du

{4

(1 u2 + u arcsinu

) (1 + u2)

}=

4

arcsinu 2u

which is nonpositive on [0, 1]. Hence

4

(1 u2 + u arcsinu

) (1 + u2)

[

4

(1 u2 + u arcsinu

) (1 + u2)

]u=1

= 0.

Inequality (1.25) follows easily because

2z uz + 1 = 4

arcsinu+ 1 u 0.

To see (1.26), let us consider the cases 1 u 0 and 0 u 1separately. It is clearly that the inequality is valid at -1, 0 and 1.Setting

f(u) = 1 u2 4

(arcsinu+ u

1 u2

)+ 2u;

then

f = 2u 4

(2

1 u2) + 2,

f = 2 + 8u

1 u2,

and

f =8

(1 u2)3/2 .

When 1 u 0, f 0. Hence f(u) min{f(1), f(0)} = 0. Forthe case 0 u 1, f 0. Thus

f max{f (0), f (1)} = max{

2 8, 0

}= 0.


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Therefore f(u) f(1) which proves (1.26).

Lemma 1.2. Suppose M is a compact manifold without boundaryof nonnegative Ricci curvature. Assume that a nontrivial eigenfunc-tion corresponding to the eigenvalue is normalized so that for0 a < 1, a+ 1 = supM and a 1 = infM . If u = a, then

||2 (1 u2) + 2az(u) (1.27)

where

z(u) =2

(arcsinu+ u

1 u2

) u. (1.28)

Proof. We need only to prove an estimate similar to (1.27) foru = ( a) where 0 < < 1. The lemma will follow by letting 0. By the definition of u; we have

u = (u+ a)

with u . We may assume a > 0. Consider the function

Q = |u|2 c(1 u2) 2az(u),

We can choose c large enough so that supM Q = 0. The lemmafollows if c ; for a sequence of 1, hence we may assume thatc > .

Let the maximizing point of Q be x0. We claim that |u(x0)| > 0since otherwise u(x0) = 0 and

0 = Q(x0) = c(1 u2)(x0) 2az(x0) (c a)(1 2)

by (1.26), which is a contradiction.Differentiating in the ei direction gives

1

2Qi = ujuji + cuui azui. (1.29)

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

ujiuji u211 = (cu az)2. (1.30)


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

0 12Q(x0) (1.31)

= ujiuji + uj(u)j + Ric(u,u) + (c az) + (cu az)u (cu az)2 + (c az)(c(1 u2) + 2az)

(cu az)(u+ a)= ac((1 u2)z + uz + u) + a22(2zz + z2 + z)

+a(c )(uz + 2z + 1) + (c )(c a).

However by (1.23), (1.24), and (1.25), we conclude that

0 ac(1 )u a22(1 )z + (c )(c a) (1.32)

ac(1 ) a22(1 )(

4

1)

+ (c )(c a)

(c+ )(1 ) + (c )2.

This implies that

c {

2 + (1 ) +

(1 )(9 )2

}.

Taking 0 one gets the desired estimate.

Theorem 1.8. ([103]) Suppose M is a compact manifold withoutboundary whose Ricci curvature is nonnegative. Let a u 0 be themedian of a normalized first eigenfunction with a+ 1 = supM anda 1 = infM ; and let d be the diameter. Then the first non-zeroeigenvalue of M satisfies

d21 2 +6

(2 1)4a2 2(1 + 0.02a2). (1.33)

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


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

d1/2

ds

|u|ds1 u2 + 2az

1

0

{1

1 u2 + 2az+

11 u2 2az

}du

1

0

11 u2

{2 +

3a2z2

1 u2}du

+ 3a2( 1

0

z1 u2

)2

= +3a2

2

(2 1)4.

Remark 1.2. It has been shown by Hang-Wang [40] that if theequality holds in (1.33) then M is isometric to a circle.

Remark 1.3. Let Mn be a compact manifold with smoothboundary and nonnegative Ricci curvature. Suppose that the sec-ond fundamental form of M is nonnegative. Then the first nontrivialeigenvalue of the Laplacian with Neumann boundary conditions alsosatisfies the inequality (1.27). The proof runs the same as Lemma1.1 except that the possibility of the maximum of the test functionQ at the boundary must be handled. In fact, the boundary convex-ity assumption implies that the maximum of Q cannot occur on theboundary.


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Chapter 2

Isoperimetric

inequalities for

eigenvalues

2.1 Introduction

In this chapter, we will prove some isoperimetric inequalities for eigen-values on manifolds which have always been important problems ingeometric analysis. Owing to the limitation on the materials, we onlyselect some of the results in the area. For more interesting results,we refer to [3] , [8], [17] and the references therein. The isoperimetricinequalities to be proved are : the Faber-Krahn inequality for the firsteigenvalue of the Dirichlet eigenvalue; the Szego-Weinberger inequal-ity for the first nontrivial Neumann eigenvalue; the Hersch theoremfor the first closed eigenvalue on a compact Riemannian surface ofgenus zero; the Ashbaugh-Benguria theorem; etc. For the conve-nience of later use, we recall now the notion of spherically symmetricrearrangement. Suppose that f is a bounded measurable functionon the bounded measurable set Rn. Consider the distributionfunction f (t) defined by

f (t) = |{x ||f(x)| > t}| (2.1)

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where | | denotes Lebesgue measure. The distribution function canbe viewed as a function from [0,) to [0, ||] and is nonincreasing.The decreasing rearrangement f of f , is the inverse of f and isdefined by

f(s) = inf{t 0|f (t) < s}. (2.2)

It is a nonincreasing function on [0, ||]. For a bounded measurableset Rn, its spherical rearrangement is defined as the ballcentered at the origin having the same measure as . The spherically(symmetric) decreasing rearrangement f : R is defined by

f(x) = f(Cn|x|n) for x (2.3)

where Cn = n/2/

(n2 + 1

)is the volume of the unt ball in Rn. An

important fact we will use is that

f2 =

||

0

(f(s))2ds =

(f)2. (2.4)

It is known that for any function f in the Sobolev space H10 (),f H10 () and

|f|2

|f |2. (2.5)

For two nonnegative measurable functions f and g on we have

fg

fg. (2.6)

Let us recall the notion of spherically (symmetric) increasing rear-rangement, which we denote by a lower . The definition is almostidentical to that of spherically decreasing rearrangement, except thatg should be radially increasing (in the weak sense) on

. In thiscase, we have

fg

fg. (2.7)


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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 oneof the oldest isoperimetric inequalities for an eigenvalue.

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

1() 1() (2.8)

with equality if and only if = .

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

1() =

|u1|2u21

(2.9)

=

|u1|2

(u1)

2

|u1|2

(u1)2

1().

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 afixed R, let M be the complete simply connected n-dimensionalspace form of constant sectional curvature . To each bounded do-main in M , associate the geodesic ball D in M satisfying

|| = |D|. (2.10)

If > 0 then only consider those for which || < |M|.

Theorem 2.2. If, for all such in M , equality (2.10) impliesthe isoperimetric inequality

|| |D|, (2.11)


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[SEC. 2.3: THE SZEGO-WEINBERGER INEQUALITY 17

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

1() 1(D), (2.12)

with equality holding if and only if is isometric to D.

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

2.3 The Szego-Weinberger Inequality

In this section, we prove the Szego-Weinberger inequality which is acounterpart to the first non-zero Neumann eigenvalue of the Faber-Krahn inequality.

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

1() 1() (2.13)

with equality holding if and only if = .

Proof. Let R be the radius of and let g be the solution of theequation

{g + n1r g

n1r2 g + 1()g = 0g(0) = 0, g(R) = 0

(2.14)

By a topological argument, we can take as trial functions Pi, suchthat

Pi = 0 for i = 1, , n, with

Pi(x) = h(r)xir,

where the xis are Cartesian coordinates, x = (x1, , xn) Rn, r =|x|, and

h(r) =

{g(r) for 0 r Rg(R) for r R.


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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. Bysubstituting our trial functions Pi into the mini-max inequality for1, we find

1()

P 2i

|Pi|2.

Summing this in i for 1 i n, we arrive at

1()

ni=1 |Pi|2

ni=1 P

2i

(2.15)

=

(h(r)2 + n1r2 h(r)

2)

h(r)2

=

A(r)

h(r)2

where

A(r) = h(r)2 +n 1r2

h(r)2. (2.16)

A(r) is easily seen to be decreasing for 0 r R by differentiatingand using the differential equation (2.14). One finds

A(r) = 2(1()hh + (n 1)(rh h)2/r3) < 0, 0 < r < R.(2.17)

Also, A(r) = (n 1)g(R)2/r2 for r R shows that A is decreasingfor r > R. Since A is continuous for all r 0, it is also decreasingthere. Observe that

A(r)

A(r) (2.18)

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

h(r)2

h(r)2 (2.19)


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[SEC. 2.4: THE ASHBAUGH-BENGURIA THEOREM 19

since h is nondecreasing. Thus we arrive at

1() A(r)

h(r)2

= 1(), (2.20)

since each Pi is precisely a Neumann eigenfunction of with eigen-value 1(BR) for the domain BR =

. This completes the proofof the Szego-Weinberger inequality, including the characterization ofthe case of equality.

2.4 The Ashbaugh-Benguria Theorem

In this section we consider the sharp upper bound for 2/1 for theDirichlet eigenvalue problem proved by Ashbaugh-Benguria. In 1955and 1956, Payne, Polya and Weinberger [77], [78], proved that

21

3 for R2

and conjectured that

21

21

disk

=j21,1j20,1

with equality if and only if is a disk and where jp,k denotes thekth positive zero of the Bessel function Jp(t). For general dimensionn 2, the analogous statements are

21

1 + 4n

for Rn,

and the PPW conjecture

21

21

nball

=j2n/2,1

j2n/21,1, (2.21)

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


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We proceed now with the proof of (2.21). Let us start from thevariational principle for 2

2() = minH10 (),06=u1

||22

, (2.22)

which, by integration by parts, leads to

2() 1() (2.23)

|P |2u21P 2u21

, Pu1 H10 (),

Pu21 = 0, P 6= 0.

To get the isoperimetric result out of (2.23), one must make veryspecial choices of the function P , in particular, choices for which(2.23) is an equality if is a ball. Thus we shall use n trial functionsP = Pi, such that

Piu

21 = 0 for i = 1, , n where

Pi = g(r)xir

(2.24)

and

g(r) =

{f(r) = right radial function on BR for 0 r R,f(R) for r R. (2.25)

The right R in this case turns out to be the unique R such that1(BR) = 1(). Substituting Pi into (2.23) and summing on i, wefind

2() 1() B(r)u21

f(r)2u21

(2.26)

where

B(r) = f (r)2 +n 1r2

f(r)2. (2.27)

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

f(r) = w(r), (2.28)


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[SEC. 2.4: THE ASHBAUGH-BENGURIA THEOREM 21

where

w(x) =

{jn/2(x)

jn/21(x), if 0 x < 1,

w(1) limx1 w(x), if x 1,(2.29)

with = jn/21,1, = jn/2,1 and =1/.

Lemma 2.1. The equality occurs in (2.26) when is a ball with1 as the first Dirichlet eigenvalue and f is given by (2.28).

Proof. If S1 is a closed ball of Rn of appropriate radius centered

in the origin in which the problem{

z = z in S1,z|S1 = 0.

(2.30)

has 1 as the first eigenvalue, then

S1 = {x Rn; |x| /1 = 1/}.

The second eigenvalue of the problem (2.30) is 2 =2

21. The firsteigenfunction of S1 is

z(x) = c|x|1n/2jn/21(1|x|),

and the eigenfunctions corresponding to 2 are:

fi(x) = c|x|1n/2jn/2(2|x|)

xi|x| , i = 1, , n,

where c is a non-zero constant.Let

Q(r) =

jn/2

(2r

)

jn/21(

1r), if 0 r < 1/,

limr1/jn/2

(2r

)

jn/21(

1r), if r 1/,

(2.31)

Observe that Q(r) = w(r) = g(r) and let

Qi(x) = Q(|x|)xi|x| = g(r)

xi|x| ,


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then

S1

Qiz2 = 0, i = 1, , n

and Qiz are eigenfunctions of 2. Thus, we have

2 =

S1

|(Qiz)|2S1

(Qiz)2, i = 1, , n.

Summing over i and simplifying, we get

2 1 =

S1

((f (r))2 + (n 1) f

2(r)r2

)z2

S1f2(r)z2

. (2.32)

This completes the proof of Lemma 3.1.Substituting (2.28) into (2.26), we get

2 1 1B(r)u21

w2(r)u21

(2.33)

where

B(x) = (w(x))2 + (n 1)w2(x)

x2. (2.34)

From the definition of w and the properties of Bessel functions one canprove that w(t) is nondecreasing and B(t) is non-increasing. There-fore, we have

B(r)u21

B(r)(u1)

2

B(r)(u1)

2 (2.35)

and

w(r)2u21

w(r)(u

1)

2

w(r)(u1)

2 (2.36)

In order to continue the proof, we need a result of Chiti: If c is chosenso that

u21 =

u21 =

S1

z2, (2.37)


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[SEC. 2.5: THE HERSCH THEOREM 23

then

f(r)(u1)

2

S1

f(r)z2, (2.38)

if f is increasing, and the reverse inequality if f is decreasing. Itfollows from (2.38) and the monotonicity properties of B and w that

B(r)(u1)

2

S1

B(r)z2 (2.39)

and

w(r)(u1)

2

S1

w(r)2z2. (2.40)

Combining (2.33), (2.35), (2.36), (2.39), (2.40), and using the defini-tion of z, we finally get

2 1 1

S1B(r)z2

2

S1w2(r)z2

=12

(2 2). (2.41)

From here the inequality

21

j2n/2,1

j2n/21,1(2.42)

follows immediately. Also, it is clear from the proof that the equalityholding in (2.42) if and only if is a ball.

2.5 The Hersch Theorem

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

Theorem 2.5 ([48]). For any metric on S2, the first non-trivialeigenvalue satisfies

1 8

A(S2). (2.43)


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Proof. For any metric ds2 on S2, we can construct a conformal map : (S2, ds2) (S2, ds20), here ds20 denotes the standard metric on S2.From the mini-max principle, we have

1 = infS2

fdv=0

S2|f |2dv

S2f2dv

, (2.44)

where dv is the area element with respect to ds2. Take the coordi-nate functions xi(i = 1, 2, 3) on (S2, ds20); then x

i , i = 1, 2, 3, arefunctions on (S2, ds2).

Observe that is a conformal map and that in the case of surfaces,the Dirichlet integral of a function is a conformal invariant. Thus wehave

S2

|(xi )|2dv =

S2

|xi|2dv =

S2

xixi = 2

S2

(xi)2 =8

3.

Since

Area(S2) =

S2

dv =

3

i=1

S2

(xi )2dv,

there exists at least one i such that

S2

(xi )2dv Area(S2)

3.

Also, we can choose satisfying

S2xi dv = 0 [85]. Thus

1

S2|(xi )|2dv

S2(xi )2dv

8

A(S2). (2.45)

For the discussion of equality case, we refer to [48].

Remark 2.1. S2 is a Riemann surface of genus g = 0. ForRiemannian surface g of genus g > 0, Yang-Yau obtained a similarresult.

Theorem 2.6 ([99]). For any metric on g, the first eigenvaluesatisfies

1 8(1 + g)

|g|(2.46)


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Remark 2.2. Herschs theorem cant be generalized directly tothe case of higher dimensions [86]. That is, one cant expect that

1Vol(M)2/n C,

with a constant depending only on n. It must depend also on othergeometric invariants of M .

Here is an interesting application of Herschs theorem.

Theorem 2.7 ([19]). Suppose that M is homeomorphic to S2 and1, 2, 3 are three first eigenfunctions such that their square sum isa constant. Then M is actually isometric to a sphere with constantsectional curvature.

Proof. The assumption of Theorem 2.7 says that

{i + 1(M)i = 0, i = 1, 2, 3,3

i=1 2i = c, c is a constant.

Thus,

0 =

(3

i=1

2i

)= 2

3

i=1

|i|2 + 23

i=1

ii

= 2

3

i=1

|i|2 21(M)3

i=1

2i

which gives

3

i=1

|i|2 = c1(M). (2.47)

Taking the Laplacian of both sides of (2.47) and using the Bochner


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formula, we get

0 =1

2

3

i=1

|i|2 (2.48)

=3

i=1

|2i|2 +3

i=1

i (i) +3

i=1

Ric(i,i)

=

3

i=1

|2i|2 1(M)23

i=1

2i +K

3

i=1

|i|2

3

i=1

|i|22

c1(M)2 +Kc1(M)

= 1(M)2/2 +Kc1(M),

where K is the sectional curvature of M . Thus we have

1(M) 2K. (2.49)

Integrating (2.49) and using the Gauss-Bonnet formula, we have

1(M) area(M) 8. (2.50)

Combining (2.50) and Herschs theorem we know that M is a 2-sphere.

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

Theorem 2.8 ([92]). Let M be a connected closed hypersurfaceembedded in Rn(n 3). Let be the region bounded by M . Denoteby V and A the volume of and the area of M , respectively. Thenthe first non-zero eigenvalue 1 of the Laplacian acting on functionson M satisfies

1 (n 1)AnV

(nV

)1/n. (2.51)

with equality holding if and only if M is an (n 1)-sphere.


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Proof. Let us denote by x1, , xn, the coordinate functions onR

n. By choosing the coordinates origin properly, we can assume that

M

xi = 0, i = 1, , n.

Since M cannot be contained in any hyperplane, each xi is not a con-stant function, i = 1, , n. It follows from the Poincare inequalitythat for each fixed i {1, , n}

1

M

x2i

M

|xi|2,

with equality holding if and only if xi = 1xi.Summing over i from 1 to n, we get

1

M

n

i=1

x2i

M

n

i=1

|xi|2 =

M

(n 1) = (n 1)A, (2.52)

with equality if and only if

xi = 1xi, i {1, , n}. (2.53)

Take a ball B in Rn of radius R centered at the origin so that vol(B) =V ; then

R =

(V

n

)1/n.

By using the weighted isoperimetric inequality proved in [14], we have

M

n

i=1

x2i

B

n

i=1

x2i (2.54)

= area(B) R2

= nV

(V

n

)1/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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It follows that the position vector x = (x1, , xn) when restrictedon M satisfies

x =: (x1, ,xn) = 1(x1, , xn).

Also, it is well known that

x = (n 1)H,

whereH is the mean curvature vector of M . Consider now the func-

tion g = |x|2 : M R. Observing that H is normal to M , we inferthat

wf = 2w, x = 2(n 1)1

w,H = 0, w (M).

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, Polya and Weinberger proved that the Dirichlet eigenvaluesof the Laplacian for R2 satisfy the bound [77], [77].

k+1 k 2

k

k

i=1

i, k = 1, 2, (3.1)

This result easily extends to Rn as

k+1 k 4

kn

k

i=1

i, k = 1, 2, (3.2)

Many interesting generalizations of (1.3) have been done during thepast 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:

29


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30 [CAP. 3: UNIVERSAL INEQUALITIES FOR EIGENVALUES

Theorem 3.1. The Dirichlet eigenvalues of the Laplacian of Rn satisfy the inequality

k

i=1

(k+1 i)(k+1

(1 +

4

n

)i

) 0, for k = 1, 2, . (3.3)

The inequality (3.3), as observed by Yang himself, and as laterproved, e. g., in [3], [4], [9], is the strongest of the classical inequal-ities that are derived following the scheme devised by Payne-Polya-Weinberger. Yangs inequality provided a marked improvement foreigenvalues of large index. Recently, some Yang type inequalities oneigenvalues of the clamped plate problem, the buckling problem, thepolyharmonic operator and some other type eigenvalue problems havebeen proved. This chapter is devoted to prove some of the univer-sal inequalities in this subarea. Since the method in proving Yangsinequality has been widely generalized in obtaining various universalinequalities for eigenvalues, we end this section by proving Yangsinequality.

Proof of Theorem 3.1. Let uk be the orthonormal eigenfunctioncorresponding to the kth eigenvalue k, i.e. uk satisfies

uk = kuk, in uk| = 0,uiuj = ij .

(3.4)

Let x1, , xn be the standard coordinate functions in Rn. For anyfixed p = 1, , n, put g = xp and define i by

i = gui k

j=1

aijuj , aij =

guiuj = aji. (3.5)

It is easy to see that

iuj = 0, for i, j = 1, , k. (3.6)

Letting

bij =

ujg ui,


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from Greens formula, we derive

jaij =

g(uj)ui = 2bij + iaij

and so

2bij = (i j)aij . (3.7)Since

i = igui + 2g ui +k

j=1

jaijuj ,

we have

|i|2 = i

2i 2

ig ui. (3.8)

On the other hand, from the definition of i, (3.5) and (3.6), wederive

2

ig ui (3.9)

= 2

gg uiui + 2k

j=1

aij

ujg ui

= 1 +

k

j=1

(i j)a2ij .

From the mini-max principle, we obtain

(k+1 i)

2i 1 +k

j=1

(i j)a2ij . (3.10)

Multiplying (3.9) by (k+1 i)2 and taking sum on i from 1 to k,we obtain

k

i=1

(k+1 i)2 +k

i,j=1

(i j)(k+1 i)2a2ij (3.11)

= 2k

i=1

(k+1 i)2

ig ui.


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By aij = aji, bij = bji, we have

2k

i=1

(k+1 i)2

ig ui (3.12)

=

k

i=1

(k+1 i)2 4k

i=1

(k+1 i)b2ij w.

Multiplying (3.10) by (k+1 i)2 and taking sum on i from 1 to k,we infer

k

i=1

(k+1 i)3

2i

k

i=1

(k+1 i)2 4k

i=1

(k+1 i)b2ij = w.

Fromuij = 0 for all i, j = 1, , k, we have, for arbitrary con-

stants dij ,

w2

=

(2

k

i=1

(k+1 i)2

ig ui)2

4k

i=1

(k+1 i)32i

k

i=1

(k+1 i)1/2g ui k

j=1

dijuj

2

4wk

i=1

(k+1 i)|g ui|2 +

k

j=1

dijuj

2

2k

j=1

dij(k+1 i)1/2ujg ui

.


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Then we have

w 4k

i=1

(k+1 i)(uixp

)2

+4

2k

i,j=1

(k+1 i)1/2bij +k

i,j=1

d2ij

.

Putting dij = (k+1 i)1/2bij , we obtain

w 4k

i=1

(k+1 i)

(uixp

)2 4

k

i,j=1

(k+1 i)b2ij (3.13)

and so we infer

k

i=1

(k+1 i)2 4k

i=1

(k+1 i)

(uixp

)2. (3.14)

Summing over p, we obtain

k

i=1

(k+1 i)2 4

n

k

i=1

(k+1 i)

|ui|2 (3.15)

=4

n

k

i=1

(k+1 i)i.

Yangs inequality has been generalized to bounded domains incomplete submanifolds in Euclidean space. That is, we have

Theorem 3.2 ([20], [41]). Let be a bounded domain in ann-dimensional complete Riemannian manifold Mn isometrically im-mersed in RN . Then the Dirichlet eigenvalues of the Laplacian of satisfy the inequality

k

i=1

(k+1 k)2 4

n

k

i=1

(k+1 k)(i +n2

4||H||2), (3.16)

where H is the mean curvature vector field of Mn and ||H||2 =sup |H|2.


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3.2 Eigenvalues of the Clamped Plate

Problem

Let us generalize Yangs method to prove universal inequalities foreigenvalues of the clamped plate problem on Riemannian manifolds.

Theorem 3.2 ([94]). Let M be an n-dimensional completeRiemannian manifold and let be a bounded domain with smoothboundary in M . Denote by the outward unit normal of and leti the i-th eigenvalue of the problem:

{2u = u in ,

u| = u

= 0.(3.17)

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

k

i=1

(k+1 i)2 (3.18)

1n

{k

i=1

(k+1 i)2(n2H20 + (2n+ 4)

1/2i

)}1/2

{

k

i=1

(k+1 i)(n2H20 + 4

1/2i

)}1/2,

where H0 = sup |H|.ii) If there exists a function : R and a constant A0 such

that

|| = 1, || A0, on , (3.19)


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

then

k

i=1

(k+1 i)2 (3.20)

{

k

i=1

(k+1 i)2(A20 + 4A0

1/4i + 6

1/2i

)}1/2

{

k

i=1

(k+1 i)(2

1/4i +A0

)2}1/2

.

iii) If there exists a function : R and a constant B0 suchthat

|| = 1, = B0, on , (3.21)

then

k

i=1

(k+1 i)2 {

k

i=1

(k+1 i)2(61/2i B20)}1/2

(3.22)

{

k

i=1

(k+1 i)(4

1/2i B20

)}1/2.

iv) If admits an eigenmap f = (f1, f2, , fm+1) : Sm(1)corresponding to an eigenvalue , that is,

f = f, = 1, ,m+ 1,m+1

=1

f2 = 1,

then

k

i=1

(k+1 i)2 {

k

i=1

(k+1 i)2(6

1/2i +

)}1/2(3.23)

{

k

i=1

(k+1 i)(4

1/2i +

)}1/2,


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

Theorem 3.2 can be deduced from a general result.

Lemma 3.1 ([89]). Let i, i = 1, , be the i-th eigenvelue of theproblem (3.17) and ui the orthonormal eigenfunction correspondingto i, that is,

2ui = iui in ,

ui| = ui

= 0,Muiuj = ij , i, j = 1, 2, .

(3.24)

Then for any smooth function h : R and any > 0, we havek

i=1

(k+1 i)2

u2i |h|2 (3.25)

k

i=1

(k+1 i)2

{u2i (h)2 2ui|h|2ui

+4((h ui)2 + uihh ui)}

+k

i=1

(k+1 i)

(h ui +

uih

2

)2.

Proof of Lemma 3.1. For i = 1, , k, consider the functionsi : R given by

i = hui k

j=1

rijuj ,

where

rij =

huiuj .

Since i| = i

= 0 and

uji = 0, i, j = 1, , k,


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it follows from the mini-max inequality that

k+1 i

2i2i

. (3.26)

We have

i2i (3.27)

=

i

2(hui) k

j=1

rijjuj

=

i2(hui)

= i||i||2 k

j=1

rijsij

+

hui((uih) + 2(h ui) + 2h (ui) + hui),

where ||i||2 =2i and

sij =

uj((uih) + 2(h ui) + 2h (ui) + hui).

Multiplying the equation 2ui = iui by huj , we have

huj2ui = ihuiuj . (3.28)

Changing the roles of i and j, one gets

hui2uj = jhuiuj . (3.29)

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


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on , we get

(j i)rij (3.30)

=

(hui2uj huj2ui)

=

((hui)uj (huj)ui)

=

((uih+ 2h ui)uj (ujh+ 2h uj)ui)

=

uj((uih) + 2(h ui) + hui + 2(ui) h)= sij ,

Observe that

hui((uih) + 2(h ui) + 2h (ui) + hui)

=

(u2i (h)2 + 4(|h ui|2 + uihh ui) 2ui|h|2ui).

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

(k+1 i)||i||2 (3.31)

(u2i (h)2 + 4(|h ui|2 + uihh ui) 2ui|h|2ui)

+

k

j=1

(i j)r2ij .

Set

tij =

uj

(h ui +

uih

2

); (3.32)


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then tij + tji = 0 and

(2)i(h ui +

uih

2

)(3.33)

=

(2huih ui u2ihh) + 2k

j=1

rijtij

=

u2i |h|2 + 2k

j=1

rijtij .

Multiplying (3.33) by (k+1 i)2 and using the Schwarz inequality,we get

(k+1 i)2

u2i |h|2 + 2k

j=1

rijtij

(3.34)

= (k+1 i)2

M

(2)i(h ui +

uih

2

)

= (k+1 i)2

M

(2)i

(h ui +

uih

2

)

k

j=1

tijuj

(k+1 i)3||i||2

+(k+1 i)

M

h ui +

uih

2

k

j=1

tijuj

2

= (k+1 i)3||i||2

+(k+1 i)

(h ui +

uih

2

)2

k

j=1

t2ij

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


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noticing rij = rji, tij = tji, we get

k

i=1

(k+1 i)2

u2i |h|2 2k

i,j=1

(k+1 i)(i j)rijtij

k

i=1

(k+1 i)2

(u2i (h)2 + 4((h ui)2 + uihh ui)

2ui|h|2ui) +k

i=1

(k+1 i)

((h ui)2 +

uih

2

)2

k

i,j=1

(k+1 i)(i j)2r2ij k

i,j=1

(k+1 i)

t2ij ,

which implies (3.25).

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

i) Let x, = 1, ,m, be the standard coordinate functions ofR

m. Taking h = x in (3.25) and summing over , we have

k

i=1

(k+1 i)2m

=1

u2i |x|2 (3.35)

k+1

i=1

(k+1 i)2m

=1

(u2i (x)2 + 4((x ui)2

+uixx ui) 2ui|x|2ui)

+

k

i=1

(k+1 i)

m

=1

(x ui +

uix2

)2,

Since M is isometrically immersed in Rm, we have

m

=1

|x|2 = n


i

i

i

i

i

i

i

i


which implies that

m

=1

u2i |x|2 = n (3.36)

Also, we have

(x1, , xm) (x1, ,xm) = nH, (3.37)

m

=1

(x ui)2 =m

=1

(ui(x))2 = |ui|2 (3.38)

and

m

=1

xx ui =m

=1

xui(x) = nH ui = 0. (3.39)

Substituting (3.36)-(3.39) into (3.35), we get

n

k

i=1

(k+1 i)2 (3.40)

k

i=1

(k+1 i)2

(n2u2i |H|2 + 4|ui|2 2nuiui)

+

k

i=1

(k+1 i)

(|ui|2 +

n2u2i |H|24

)

k

i=1

(k+1 i)2(n2H20 + (2n+ 4)1/2i )

+

k

i=1

(k+1 i)

(

1/2i +

n2H204

).

Here in the last inequality, we have used the fact that |H| H0 and

|ui|2 =

uiui (

u2i

)1/2(

(ui)2

)1/2=

1/2i .(3.41)


i

i

i

i

i

i

i

i


Taking

=

ki=1(k+1 i)

(

1/2i +

n2H204

)

ki=1(k+1 i)2(n2H20 + (2n+ 4)

1/2i )

1/2

,

one gets (3.18).ii) Substituting h = into (3.25) and using (3.19) and the Schwarz

inequality, we get

k

i=1

(k+1 i)2 (3.42)

k

i=1

(k+1 i)2

{u2i ()2 + 4(( ui)2 + ui ui)

2uiui} +k

i=1

(k+1 i)

( ui +

ui

2

)2

k

i=1

(k+1 i)2

(A20u2i + 4(|ui|2 +A0|ui||ui|) 2uiui)

+

k

i=1

(k+1 i)

(|ui|2 +A0|ui||ui| +

A20u2i

4

).

Substituting (3.41) and

|ui||ui| (

u2i

)1/2(

|ui|2)1/2

1/4i

into (3.42), we get

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2(A20 + 4A01/4i + 61/2i )

+

k

i=1

(k+1 i)

(

1/4i +

A02

)2.


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i

i

i

i

i

i

i


Taking

=

ki=1(k+1 i)

(

1/4i +

A02

)2

ki=1(k+1 i)2(A20 + 4A0

1/4i + 6

1/2i )

1/2

,

we obtain (3.20).

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

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2

(B20u2i + 4(|ui|2 +B0ui ui) 2uiui)

+

k

i=1

(k+1 i)

(|ui|2 +B0ui ui +

B20u2i

4

)

k

i=1

(k+1 i)2(61/2i B20) +k

i=1

(k+1 i)

(

1/2i

B204

),

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

ui,ui = 1

2

u2i = B202.

Taking

=

ki=1(k+1 i)

(

1/2i

B204

)

ki=1(k+1 i)2(6

1/2i B20)

1/2

,

we obtain (3.22).

iv) Taking the Laplacian of the equation

m+1

=1

f2 = 1


i

i

i

i

i

i

i

i


and using the fact that

f = f, = 1, ,m+ 1,

we have

m+1

=1

|f|2 = .

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

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2

(2u2i + 4

m+1

=1

(f ui)2 2uiui)

+

k

i=1

(k+1 i)

(m+1

=1

(f ui)2 +2u2i

4

)

k

i=1

(k+1 i)2(2 + 61/2i )

+

k

i=1

(k+1 i)

(

1/2i +

2

4

).

We get (3.23) by taking

=

ki=1(k+1 i)

(

1/2i +

4

)

ki=1(k+1 i)2

(6

1/2i +

)

1/2

.

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

Example 3.1. Let M be an n-dimensional Hadamard manifoldwith Ricci curvature satisfying RicM (n 1)c2, c 0 and let


i

i

i

i

i

i

i

i


: [0,+) M be a geodesic ray, namely a unit speed geodesicwith d((s), (t)) = t s for any t > s > 0. The Busemann functionb corresponding to defined by

b(x) = limt+

(d(x, (t)) t)

satisfies |b | 1(Cf. [10], [49]). Also, it follows from Theorem 3.5 in[84] that |b | (n1)c2 on M . Thus any Hadamard manifold withRicci curvature bounded below supports functions satisfying (3.19).

Example 3.2. Let (N, ds2N ) be a complete Riemannian manifoldand define a Riemannian metric on M = R N by

ds2M = dt2 + 2(t)ds2N , (3.43)

where is a positive smooth function defined on R with (0) = 1.The manifold (M,ds2M ) is called a warped product and denoted byM = RN . It is known thatM is a complete Riemannian manifold.

Set = et and consider the warped product M = R et N .Define : M R by (x, t) = t. One can show that

|| = 1, = 1 n. (3.44)That is, a warped product manifold M = Ret N admits functionssatisfying (3.21).

Let Hn be the n-dimensional hyperbolic space with constant cur-vature 1. Using the upper half-space model, Hn is given by

Rn+ = {(x1, x2, , xn)|xn > 0} (3.45)

with metric

ds2 =dx21 + + dx2n

x2n(3.46)

One can check that the map : R et Rn1 given by(t, x) = (x, et)

is an isometry. Therefore, Hn admits a warped product model, Hn =R et Rn1.

Example 2.3. Any compact homogeneous Riemannian manifoldadmits eigenmaps to some unit sphere for the first positive eigenvalueof the Laplacian [70].


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3.3 Eigenvalues of the Polyharmonic

Operator

The method of proving universal bounds for eigenvalues of the clampedplate problem can be generalized to the eigenvalue problem of poly-harmonic operators.

Theorem 3.3 ([55]). Let M be an n-dimensional compact Rie-mannian manifold with boundary M (possibly empty) Let l be apositive integer and let i, i = 1, , be the i-th eigenvalue of theproblem (1.7) and ui be the orthonormal eigenfunction correspondingto i, that is,

()lui = iui in M,ui|M = ui

M

= = l1uil1

M

= 0,

Muiuj = ij , for any i, j = 1, 2, .

(3.47)

Then for any function h Cl+2(M) Cl+1(M) and any positiveinteger k, we have

k

i=1

(k+1 i)2

M

u2i |h|2 (3.48)

k

i=1

(k+1 i)2

M

hui(()l(hui) ihui)

)

+

k

i=1

(k+1 i)

h,ui +uih

2

2

,

where is any positive constant.

Proof. For i = 1, , k, consider the functions i : M R givenby

i = hui k

j=1

rijuj , (3.49)

where

rij =

M

huiuj . (3.50)


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i

i

i

i

[SEC. 3.3: EIGENVALUES OF THE POLYHARMONIC OPERATOR 47

Since

i|M =i

M

= = l1il1

M

= 0

and

M

uji = 0, i, j = 1, , k,

it follows from the mini-max inequality that

k+1

M

2i (3.51)

M

i()li

= i||i||2 +

M

i(()li ihui

)

= i||i||2 +

M

i(()l(hui) ihui

)

= i||i||2 +

M

hui(()l(hui) ihui

)

k

j=1

rijsij ,

where

sij =

M

(()l(hui) ihui

)uj .

Notice that if u Cl+2(M) Cl+1(M) satisfies

u|M =u

M

= = l1u

l1

M

= 0, (3.52)

then

u|M = u|M = u|M = (u)|M = = m1u

M

= (m1u)M

= 0, when l = 2m

and

u|M = u|M = u|M = (u)|M = = m1uM

= (m1u)M

= mu|M = 0, when l = 2m+ 1.


i

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i

i

i

i

i


We can then use integration by parts to conclude that

M

uj()l(hui) =

M

hui()l(uj) = jrij ,

which gives

sij = (j i)rij . (3.53)

Set

pi(h) = ()l(hui) ihui;

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

(k+1 i)||i||2

M

ipi(h) (3.54)

=

M

huipi(h) +k

j=1

(i j)r2ij .

Set

tij =

M

uj

(h ui +

uih

2

); (3.55)

then tij + tji = 0 and

M

(2)i(h ui +

uih

2

)= wi + 2

k

j=1

rijtij , (3.56)

where

wi =

M

(hu2i h 2huih ui). (3.57)


i

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i

i

i

i

i


Multiplying (3.56) by (k+1 i)2 and using the Schwarz inequalityand (3.54), we get

(k+1 i)2

wi + 2k

j=1

rijtij

= (k+1 i)2

M

(2)i

(h ui +

uih

2

)

k

j=1

tijuj

(k+1 i)3||i||2

+(k+1 i)

M

h ui +

uih

2

k

j=1

tijuj

2

= (k+1 i)3||i||2

+(k+1 i)

h ui +uih

2

2

k

j=1

t2ij

(k+1 i)2

M

huipi(h) +k

j=1

(i j)r2ij

+(k+1 i)

h ui +uih

2

2

k

j=1

t2ij

.

Summing over i and noticing rij = rji, tij = tji, we inferk

i=1

(k+1 i)2wi 2k

i,j=1

(k+1 i)(i j)rijtij

k

i=1

(k+1 i)2

M

huipi(h)

+

k

i=1

(k+1 i)

h,ui +uih

2

2

k

i,j=1

(k+1 i)(i j)2r2ij k

i,j=1

(k+1 i)

t2ij .


i

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i

i

i

i

i


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

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

{()lu = u in ,u| = u

= = l1ul1

= 0.(3.58)

Then we have

k

i=1

(k+1 i)2 (3.59)

(

4l(n+ 2l 2)n2

)1/2( k

i=1

(k+1 i)2 (l1)/li

)1/2

(

k

i=1

(k+1 i)1/li

)1/2

Proof. Let x1, x2, , xn be the standard Euclidean coordinatefunctions of Rn. Let ui be the i-th orthonormal eigenfunction cor-responding to the eigenvalue i of the problem (3.58), i = 1, ;then

()l(xui) = ixui + 2l(1)lx (l1ui) (3.60)

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

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2

2l(1)lxuix (l1ui)

+1

k

i=1

(k+1 i) ||x ui||2.


i

i

i

i

i

i

i

i


Summing over , we have

n

k

i=1

(k+1 i)2 (3.61)

2lk

i=1

(k+1 i)2n

=1

(1)lxuix (l1ui)

+1

k

i=1

(k+1 i)

|ui|2.

By induction, we infer

ui()kui k/li , k = 1, , l. (3.62)

Since

l1(xui) = 2(l 1)x (l2ui) + xl1ui,

we have

xuix (l1ui) (3.63)

=

xuil1x ui

=

l1(xui)x ui

=

(2(l 1)x (l2ui) + xl1ui

)x ui.

On the other hand,

xuix (l1ui) (3.64)

=

l1ui div(xuix)

=

l1ui(|x|2ui + xx ui).


i

i

i

i

i

i

i

i


Combining (3.63) and (3.64), we obtain

xuix (l1ui) (3.65)

=

M

{(l 1)x (l2ui)x ui

1

2l1ui|x|2ui

}

Observe that

n

=1

(1)lxuix (l1ui) (3.66)

=

(1)l{

(l 1)(l2ui) ui n

2ui

l1ui}

=(l 1 + n

2

)

ui()l1ui

(l 1 + n

2

)

(l1)/li .

Substituting (3.62) and (3.66) into (3.61), one gets

n

k

i=1

(k+1 i)2 (3.67)

l(n+ 2l 2)k

i=1

(k+1 i)2 (l1)/li +1

k

i=1

(k+1 i)1/li .

Taking

=

{ ki=1 (k+1 i)

1/li

l(n+ 2l 2)ki=1 (k+1 i)2

(l1)/li

}1/2,

we get (3.59).

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,Fp(t) = (2t 2)Fp1(t) (t2 + 2t n(n 2))Fp2(t), p = 2, .


i

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i

i

i

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(3.68)

Set

Fl(t) = tl al1tl1 + + (1)l1a1t+ (n)l. (3.69)

Theorem 3.4 ([55]). Let i be the i-th eigenvalue of the eigen-value problem:

{()lu = u in ,u| = u

= = l1ul1

= 0,

where is a compact domain in Sn(1). Then we have

k

i=1

(k+1 i)2 (3.70)

1n

{k

i=1

(k+1 i)2(a+l1i

l1l + + a+1 i

1l + a+0

)}1/2

{

k

i=1

(k+1 i)(n2 + 4

1/2i

)}1/2,

where a+j = max{0, aj}.

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

Sn(1) = {(x1, . . . , xn+1) Rn+1;

n+1

=1

x2 = 1}.

It is well known that

x = nx, = 1, , n+ 1. (3.71)

Taking the Laplacian of the equationn+1

=1 x2 = 1 and using (3.71),

we get

n+1

=1

|x|2 = n. (3.72)


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

k

i=1

(k+1 i)2

u2i |x|2

k+1

i=1

(k+1 i)2

xui(()l(xui) ixui)

+1

k

i=1

(k+1 i)

x ui +uix

2

2

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

n

k

i=1

(k+1 i)2 (3.73)

k+1

i=1

(k+1 i)2n+1

=1

xui(()l(xui) ixui)

+1

k

i=1

(k+1 i)n+1

=1

x ui +uix

2

2

.


n+1

=1

x ui +uix

2

2

(3.74)

=

n+1

=1

((x ui)2 nx uiuix +

n2u2ix2

4

)

=n2

4+

|ui|2

n2

4+

1/li .

For any smooth functions f, g on , we have from the Bochner for-


i

i

i

i

i

i

i

i


mula that

(f g) (3.75)= 22f 2g + f (g) + g (f)

+2(n 1)f g,

where

2f 2g =n

s,t=1

2f(es, et)2g(es, et),

being e1, , en orthonormal vector fields locally defined on . Since

2x = xI,

we infer from (3.75) by taking f = x that

(x g) (3.76)= 2xg + x (g) + (n 2)x g= 2xg + x (( + (n 2))g).

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(xg) = xBq()g + 2x (Cq()g). (3.77)

It is obvious that

B0 = 1, B1 = t n, C0 = 0, C1 = 1. (3.78)

It follows from (3.71), (3.76) and (3.77) that

q(xg) = (q1(xg)) (3.79)

= (xBq1()g + 2x (Cq1()g))= x(( n)Bq1() 4Cq1())g

+2x ((Bq1() + ( + (n 2))Cq1())g).

Thus, for any q = 2, , we have

Bq() = ( n)Bq1() 4Cq1(), (3.80)Cq() = Bq1() + ( + (n 2))Cq1(). (3.81)


i

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Consequently,

Bq() (3.82)

= (2 2)Bq1() ( + n 2)Bq1() 4Cq1()= (2 2)Bq1() (2 + 2 n(n 2))Bq2()

+4[Bq2() + ( + n 2)Cq2() Cq1()]= (2 2)Bq1() (2 + 2 n(n 2))Bq2(), 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

xui(()l(xui) ixui) (3.83)

=

xui((1)l (xBl()ui + 2x (Cl()ui)) ixui

)

=

xui((1)l

(x(

l al1l1 + + (n)l)ui+2x (Cl()ui)) ixui)

=

(1)lxui(x(al1l1 + + (n)l)ui + 2x (Cl()ui)

)

Summing on , one has

xui(()l(xui) ixui) (3.84)

=

ui(1)l(al1l1 + + (n)la0)ui

= al1

ui()l1ui + + a1

ui()ui + nl

u2i

a+l1(l1)/li + + a+1

1/li + n

l.


i

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[SEC. 3.4: EIGENVALUES OF THE BUCKLING PROBLEM 57

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

n

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2(a+l1(l1)/li + + a+1

1/li + n

l)

+1

k

i=1

(k+1 i)(

1/li +

n2

4

).

Taking

=

ki=1(k+1 i)

(

1/li +

n2

4

)

ki=1(k+1 i)2(a+l1

(l1)/li + + a+1

1/li + n

l)

1/2

,

we get (3.70).

3.4 Eigenvalues of the Buckling Problem

Let Rn and consider the problem{

2u = u,u| = u

= 0(3.85)

which is used to describe the critical buckling load of a clamped platesubjected to a uniform compressive force around its boundary.

Payne, Polya and Weinberger [77] proved

2/1 < 3 for R2.

For Rn this reads

2/1 < 1 + 4/n.

Subsequently Hile and Yeh [51] reconsidered this problem obtainingthe improved bound

21

n2 + 8n+ 20

(n+ 2)2for Rn.


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Ashbaugh [3] proved :

n

i=1

i+1 (n+ 4)1. (3.86)

This inequality has been improved to the following form [54]:

n

i=1

i+1 +4(2 1)n+ 4

(n+ 4)1.

Cheng and Yang introduced a new method to construct trial functionsfor the problem (3.85) and obtained the following universal inequality[28]:

k

i=1

(k+1 i)2 4(n+ 2)

n2

k

i=1

(k+1 i)i. (3.87)

It has been proved in [88] that for the problem (1.5) ifM is a boundedconnected domain in an n-dimensional unit sphere, then the followinginequality holds

2k

i=1

(k+1 i)2 (3.88)

k

i=1

(k+1 i)2(i +

2(i (n 2))4(i + n 2)

)

+1

k

i=1

(k+1 i)(i +

(n 2)24

),

where is any positive constant.The inequality (3.87) has been generalized to eigenvalues of buck-

ling problem of arbitrary orders. That is, we have

Theorem 3.5 ([54]). Let l 2 and let i be the i-th eigenvalueof the following eigenvalue problem:

{()lu = u in ,u| = u

= = l1ul1

= 0.(3.89)


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where is a bounded domain with smooth boundary in Rn. Then fork = 1, , we have

k

i=1

(k+1 i)2 (3.90)

2(2l2 + (n 4)l + 2 n)1/2

n

{k

i=1

(k+1 i)2(l2)/(l1)i

}1/2

{

k

i=1

(k+1 i)1/(l1)i

}1/2.

Before proving theorem 3.5, let us recall a method of constructingtrial functions developed by Cheng-Yang (Cf. [28], [54]). LetM be ann-dimensional complete submanifold in an m-dimensional Euclideanspace Rm. Denote by the canonical metric on Rm as well as thatinduced on M . Let and be the Laplacian and the gradientoperator of M , respectively. Let be a bounded connected domainof M with smooth boundary and let be the outward unit normalvector field of . For functions f and g on , the Dirichlet innerproduct (f, g)D of f and g is given by

(f, g)D =

f g.

The Dirichlet norm of a function f is defined by

||f ||D = {(f, f)D}1/2 =(

|f |2)1/2

.

Consider the eigenvalue problem

{()lu = u in ,u| = u

= = l1ul1

= 0.(3.91)

Let

0 < 1 2 3 ,


i

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i

i

i

i

i

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denote the successive eigenvalues, where each eigenvalue is repeatedaccording to its multiplicity.

Let ui be the i-th orthonormal eigenfunction of the problem (3.91)corresponding to the eigenvalue i, i = 1, 2, , that is, ui satisfies

()lui = iui in ,ui| = ui

= l1ul1

= 0,

(ui, uj)D =ui,uj = ij , i, j.

(3.92)

For k = 1, , l, let k denote the k-th covariant derivative operatoron M , defined in the usual weak sense via an integration by partsformula. For a function f on , the squared norm of kf is definedas

kf2 =

n

i1, ,ik=1

(kf(ei1 , , eik)

)2, (3.93)

where e1, , en are orthonormal vector fields locally defined on .Define the Sobolev space H2l () by

H2l () = {f : f, |f |, ,lf

L2()}.Then H2l () is a Hilbert space with respect to the norm || ||l,2:

||f ||l,2 =(

(l

k=0

|kf |2))1/2

. (3.94)

Consider the subspace H2l,D() of H2l () defined by

H2l,D() =

{f H2l () : f | =

f

= l1u

l1

= 0

}.

The operator ()l defines a self-adjoint operator acting on H2l,D()with discrete eigenvalues 0 < 1 k for the bucklingproblem (3.91) and the eigenfunctions {ui}i=1 defined in (3.92) forma complete orthonormal basis for the Hilbert space H2l,D(). If H2l,D() satisfies (, uj)D = 0, j = 1, 2, , k, then the Rayleigh-Ritz inequality tells us that

k+1||||2D

()l. (3.95)


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For vector-valued functions F = (f1, f2, , fm), G = (g1, g2, , gm) : Rm, we define an inner product (F,G) by

(F,G) =

m

=1

fg.

The norm of F is given by

||F || = (F, F )1/2 ={

m

=1

f2

}1/2.

Let H21() be the Hilbert space of vector-valued functions given by

H21()

={F = (f1, , fm) : Rm; f, |f| L2(), = 1, ,m

}

with norm

||F ||1 =(||F ||2 +

m

=1

|f|2)1/2

.

Observe that a vector field on can be regarded as a vector-valuedfunction from to Rm. Let H21,D() H2l () be a subspace ofH2l () spanned by the vector-valued functions {ui}i=1, which forma complete orthonormal basis of H21,D(). For any f H2l,D(), wehave f H21,D() and for anyX H21,D(), there exists a functionf H2l,D() such that X = f .

Proof of Theorem 3.5. With notations as above, we consider nowthe special case that is a bounded domain in Rn. Let us decomposethe vector-valued functions xui as

xui = hi +Wi, (3.96)

where hi H l2,D(), hi is the projection of xui in H21,D()and Wi H21,D(). Thus we have

Wi| = 0,

Wi u = 0, u H2l,D() (3.97)


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

div Wi = 0. (3.98)

For each = 1, , n, i = 1, , k, consider the functions i : R, given by

i = hi k

j=1

aijuj , (3.99)

where

aij =

xui uj = aji. (3.100)

We have

i| =i

= l1il1

= 0, (3.101)

(i, uj)D =

i uj = 0, j = 1, , k. (3.102)

It follows from the Rayleigh-Ritz inequality that

k+1

|i|2 (3.103)

D

i()li, = 1, , n, i = 1, , k.


()li = (1)ll1 (ui, + xui) +k

j=1

aijjuj


i

i

i

i

i

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i

i


and so

i()li (3.104)

=

i(1)ll1 (ui, + xui)

=

hi(1)ll1 (ui, + xui) k

j=1

aij

uj()lhi

=

hi(1)ll2 (ui, + xui) k

j=1

aij

hi()luj

=

hi(1)l((l2ui), + l2(xui)) +k

j=1

jaij

hiuj

=

(1)l(ui, + xui)((l2ui), + l2(xui))

k

j=1

jaij

hi,uj

=

(1)l(ui, + xui)((2l 3)(l2ui), + xl1ui))

k

j=1

jaij

hi,uj

=

(1)l((2l 3){ui,(l2ui), + xui(l2ui),)

+ui,xl1ui + x

2ui

l1ui} k

j=1

ja2ij

Since

l1(xui) = 2(l 1)(l2ui), + xl1ui,


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we have

xui(l1ui), =

xuil1ui, (3.105)

=

l1(xui)ui,

=

(2(l 1)(l2ui), + xl1ui

)ui,.

On the other hand, it holds

xui(l1ui), =

l1ui(ui + xui,). (3.106)

Combining (3.105) and (3.106), we obtain

xui(l1ui), (3.107)

=

M

{(l 1)(l2ui),ui,

1

2ui

l1ui

}

Hence

xui,l1ui =

ui(l1ui + x(

l1ui),) (3.108)

=

M

{(l 1)(l2ui),ui, +

1

2ui

l1ui

}

and consequently, we have

xui(l2ui), =

xuil2ui, (3.109)

=

l2(xui)ui,

=

ui,(2(l 2)(l2ui), + xl1ui

)

=

{(l 3)(l2ui),ui,

1

2ui

l1ui

}.


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Also, one has

uix2ui (3.110)

=

x2|ui|2 2

xuiui,

=

x2|ui|2 +

u2i ,

x2uil1ui (3.111)

=

ui(x2

l1ui)

=

ui(2l1ui + x

2

lui + 4x(l1ui),)

=

ui(2l1ui + (1)l1ix2ui + 4x(l1ui),).

Combining (3.107), (3.110) and (3.111), we get

x2uil1ui = 4(l 1)

(l2ui),ui, (3.112)

+(1)l1i{

x2|ui|2 +

u2i

}.

Substituting (3.109), (3.111) and (3.112) into (3.104), one gets

i()li (3.113)

=

(1)l{(l + 1)uil1ui + (2l2 4l + 3)(l2ui),ui,

}

+i

{

x2|ui|2

u2i

}

k

j=1

ja2ij .


||xui||2 = ||hi||2 + ||Wi||2 (3.114)


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and

||hi||2 = ||i||2 +k

j=1

a2ij , (3.115)

Combining (3.103), (3.113), (3.114) and (3.115), we infer

(k+1 i)||i||2 (3.116)

(1)l{(l + 1)uil1ui + (2l2 4l + 3)(l2ui),ui,

}

i(||ui||2 ||Wi||2) +k

j=1

(i j)a2ij ,

Observe that (xui) = uix + xui H21,D(). For Ai =(xui hi), we have

uix = Ai Wi (3.117)and so

||ui||2 = ||uix||2 = ||Wi||2 + ||Ai||2.Because of (ui,,Wi) = 0, it follows that

2||ui,||2 = 2

Ai ui,

1/(l1)i ||Ai||2 +1

1/(l1)i

||ui,||2

which gives

i||Ai||2 2l2l1

i ||ui,||2 + l3l1

i ||ui,||2 (3.118)Introducing (3.118) into (3.116), we get

(k+1 i)||i||2 (3.119)

(1)l{(l + 1)uil1ui + (2l2 4l + 3)(l2ui),ui,

}

2(l2)/(l1)i ||ui,||2 + (l3)/(l1)i ||ui,||2

+

k

j=1

(i j)a2ij ,


i

i

i

i

i

i

i

i


Since

2

xui ui,

= 2

u2i, + 2

xui,ui

= 2

u2i, + 2

ui(xui,)

= 2

u2i, + 2

uix(ui), + 4

uix ui,

= 2

u2i, 2

ui(ui + xui,) 4

ui,div(uix)

= 2

u2i, + 2 2

xui,ui 4

u2i,

= 2

u2i, + 2 + 2

ui (xui,)

= 2 + 2

xui ui,,

we have

2

xui ui, = 1. (3.120)

Set

dij =

ui, uj ;

then dij = dji and we get

1 = 2

xui ui, (3.121)

= 2

hi ui,

= 2

i ui, 2k

j=1

aijdij .


i

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i


Thus, we have

(k+1 i)2

1 + 2k

j=1

aijdij

(3.122)

= (k+1 i)2

2i,

ui, k

j=1

dijuj

(k+1 i)3||i||2 +1

(k+1 i)

||ui,||2 k

j=1

d2ij

,

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

(k+1 i)2

1 + 2k

j=1

aijdij

(k+1 i)2(

(1)l{(l + 1)uil1ui

+(2l2 4l + 3)(l2ui),ui,}

2l2l1

i ||ui,||2 + l3l1

i ||ui,||2 +k

j=1

(i j)a2ij

+1

(k+1 i)

||ui,||2 k

j=1

d2ij

,


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Summing on i from 1 to k, we infer

k

i=1

(k+1 i)2 2k

i,j=1

(k+1 i)(i j)aijdij

(

k

i=1

(k+1 i)2(

(1)l{(l + 1)uil1ui

+(2l2 4l + 3)(l2ui),ui,}2(l2)/(l1)i ||ui,||2 +

(l3)/(l1)i ||ui,||2

)

k

i,j=1

(k+1 i)(i j)2a2ij

+1

k

i=1

(k+1 i)||ui,||2 k

i,j=1

(k+1 i)d2ij

,

which gives

k

i=1

(k+1 i)2

k

i=1

(k+1 i)2(

(1)l{(l + 1)uil1ui

+(2l2 4l + 3)(l2ui),ui,}

2l2l1

i ||ui,||2 + l3l1

i ||ui,||2)

+1

k

i=1

(k+1 i)||ui,||2,


i

i

i

i

i

i

i

i


Taking sum for from 1 to n, we get

nk

i=1

(k+1 i)2

k

i=1

(k+1 i)2(

(1)l{n(l + 1)uil1ui

+(2l2 4l + 3)(l2ui) ui}

2l2l1

i + l3l1

i

n

=1

||ui,||2)

+1

k

i=1

(k+1 i)n

=1

||ui,||2

= k

i=1

(k+1 i)2(2

l2l1

i + l3l1

i

n

=1

||ui,||2

+(2l2 + (n 4)l + 3 n)

ui()l1ui)

+1

k

i=1

(k+1 i)n

=1

||ui,||2.

But

k

=1

||ui,||2 =

k

=1

ui,ui,

=

k

=1

ui,(ui),

=

k

=1

ui,ui

=

(ui)2

=

ui2ui,


i

i

i

i

i

i

i

i


where ui, =2uix2

. Therefore,

nk

i=1

(k+1 i)2

k

i=1

(k+1 i)2(2

l2l1

i + l3l1

i

ui2ui

+(2l2 + (n 4)l + 3 n)

ui()l1ui)

+1

k

i=1

(k+1 i)

ui2ui.

Observe that

ui()l1ui l2l1

i ,

ui2ui

1l1

i .

Thus we have

n

k

i=1

(k+1 i)2

(2l2 + (n 4)l + 2 n)k

i=1

(k+1 i)2l2l1

i

+1

k

i=1

(k+1 i)1

l1

i .

Taking

=

{ki=1(k+1 i)

1l1

i

}1/2

{(2l2 + (n 4)l + 2 n)ki=1(k+1 i)2

l2l1

i

}1/2 ,

we get (3.90).


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

k

i=1

(k+1 i)2 4(n+ 43

)

n2

k

i=1

(k+1 i)i. (3.123)

The inequality (3.90) has been improved by Cheng-Qi-Wang-Xia veryrecently in [24]:

n

k

i=1

(k+1 i)2 (3.124)

k

i=1

i(k+1 i)2(

2l2 +

(n 14

3

)l +

8

3 n

)

l2l1

i

+

k

i=1

1

i(k+1 i)

1l1

i ,

where {i}ki=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 generalizedto buckling problem of arbitrary orders (Cf. [54], [24]).


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Chapter 4

Polya Conjecture and

Related Results

4.1 Introduction

Let Rn be a bounded open set and consider the eigenvalueproblem of the Dirichlet Laplacian on :

{u+ u = 0 in ,u| = 0.

(4.1)

In 1960, Polya [81] showed that for any plane covering domain in R2 (those that tile R2) the Weyl asymptotic relation (1.) is in facta one-sided inequality (his proof also works for Rn-covering domains)and conjectured, for any bounded domain Rn, the inequality

k C(n)(k

||

)2/nk, with C(n) = (2)

2

2nn

. (4.2)

Polyas conjecture has been a central problem about eigenvalues andmany important developments have been made during the past years.In 1982, Li-Yau [72] showed the lower bound

k

i=1

i nkC(n)

n+ 2

(k

||

)2/n(4.3)

73


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74 [CAP. 4: POLYA CONJECTURE AND RELATED RESULTS

which yields an individual lower bound on k in the form

k nC(n)

n+ 2

(k

||

)2/n. (4.4)

Similar bounds for eigenvalues with Neumann boundary conditionshave 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 theLegendre transformation. The inequality (4.3) has been improved byMelas:

Theorem 4.1 ([75]). For any bounded domain Rn and anyk 1, the eigenvalues of the problem (4.1) satisfy the inequality

k

i=1

i nkC(n)

n+ 2

(k

||

)2/n+ d(n)k

||I()

(4.5)

where the constant dn depends only on the dimension and I() =minaRn

|x a|2dx is the moment of inertia of .

Proof. Fix a k 1 and let u1, , uk be an orthonormal setof eigenfunctions of (4.1) corresponding to the set of eigenvalues1, , k. We consider the Fourier transform of each eigenfunction

fj() = uj() = (2)n/2

uj(x)eixdx. (4.6)

From Plancherels Theorem, we know that f1, , fk is an orthonor-mal set in Rn. Bessels inequality implies that for every Rn

k

j=1

|fj()|2 (2)n

|eix|dx = (2)n|| (4.7)

and

k

j=1

|fj()|2 (2)n

|ixeix|dx = (2)nI(). (4.8)

Since uj | = 0 it is easy to see that

Rn

||2|fj()|2d =

|uj |2 = j . (4.9)


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Setting

F () =

k

j=1

|fj()|2; (4.10)

then 0 F () (2)n||,

|F ()| 2(k

j=1

|fj()|2)1/2(k

j=1

|fj()|2)1/2 (4.11)

2(2)n

||I(), Rn,

Rn

F ()d = k (4.12)

and

Rn

||2F ()d =k

j=1

j . (4.13)

Let F () = (||) denote the decreasing radial rearrangement of F .By approximating F we may assume that the function : [0;+) [0, (2)n||] is absolutely continuous. Setting (t) = |{F > t}| =|{F > t}| the co-area formula (1.17) implies that

(t) =

(2)n||

t

{F=s}|F |1dsds. (4.14)

Observe that F is radial and so ((s)) = |{F > (s)}| = nsnwhich gives nns

n1 = ((s))(s) a.e. It follows from (4.11),(4.14) and the isoperimetric inequality that

((s)) =

{F=(s)}|F |1ds (4.15)

1|{F = (s)}| 1nnsn1


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76 [CAP. 4: POLYA CONJECTURE AND RELATED RESULTS

and so for almost every s

(s) 0, (4.16)

where

= 2(2)n||I().

From (4.12) and (4.13), we have

k =

Rn

F ()d =

Rn

F ()d (4.17)

= nn

0

sn1(s)ds

and

k

j=1

j =

Rn

|xi|2F ()d (4.18)

Rn

|xi|2F ()d

= nn

0

sn+1(s)ds.

We need an elementary lemma.

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

(s) 0,

0

sn1(s)ds = A. (4.19)

Then

0

sn+1(s)ds 1n+ 2

(nA)n+2

n (0)2n +

A(0)2

6(n+ 2)2. (4.20)

Applying Lemma 4.1 to the function with A = (nn)1k, =

2(2)n||I() we get in view of (4.12) and (4.13) that

j=1

j n

n+ 2 2nn k

n+2n (0)

2n +

ck(0)2

(n+ 2)2(4.21)


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[SEC. 4.2: THE KROGERS THEOREM 77

where c is any constant such that 0 < c < 16 . Obs

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Eigenvalues on Riemannian Manifolds