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Differential Geometry and Geometric Analysis Osaka City University

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Estimates for eigenvalues on completeRiemannian manifolds

Qing-Ming Cheng

Saga University

March 13-15, 2011

Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 1 / 64

Contents

Contents

Introduction

Lower bounds for eigenvaluesUpper bounds for eigenvalues


Contents

Contents

Introduction

Lower bounds for eigenvalues

Upper bounds for eigenvalues


Contents

Contents

IntroductionLower bounds for eigenvalues

Upper bounds for eigenvalues


Contents

Contents

IntroductionLower bounds for eigenvaluesUpper bounds for eigenvalues


1. Introduction

1. Introduction

Let M be an n-dimensional complete Riemannianmanifold, Ω a bounded domain with a piecewisesmooth boundary ∂Ω in M. We consider

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a Dirichlet eigenvalue problem of Laplacian

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∆u = −λu, in Ω,u = 0, on ∂Ω,

(1.1)

where ∆ denotes the Laplacian on M.

This Dirichlet eigenvalue problem (1.1) is also called afixed membrane problem.


1. Introduction

It is well known that the spectrum of this eigenvalueproblem (1.1) is real and discrete.

0 < λ1 < λ2 ≤ λ3 ≤ · · · → ∞.

where each λi has finite multiplicity which is repeatedaccording to its multiplicity.


1. Introduction

Furthermore, the following Weyl’s asymptotic formulaholds:

λk ∼4π2

(ωnvolΩ) 2n

k2n , k → ∞,

where ωn is the volume of the unit ball in Rn. From theformula, it is not difficult to infer

1k

k∑i=1

λi ∼n

n + 24π2

(ωnvolΩ) 2n

k2n , k → ∞.


2. Lower bounds for eigenvalues Conjecture of Polya

2. Lower bounds for eigenvalues

For a bounded domain in the Euclidean space Rn,Polya (1961) proved, for k = 1, 2, · · · ,

λk ≥4π2

(ωnvolΩ) 2n

k2n ,

if Ω is a tiling domain in Rn.



Conjecture of Polya

Furthermore, he conjectured the following:

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Conjecture of Polya

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If Ω is a bounded domain in Rn, then eigenvalue λk ofthe Dirichlet eigenvalue problem (1.1) satisfies, fork = 1, 2, · · · ,

λk ≥4π2

(ωnvolΩ) 2n

k2n .



For this conjecture of Polya, there are manymathematicians to study it. For examples, Berezin,Lieb, Li and Yau, Levine and Protter, Laptev, Melas andso on. The following is the result of Li and Yau.

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Theorem 2.1 (Li and Yau, Comm. Math. Phys. 1983).

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If Ω is a bounded domain in Rn, then eigenvalue λk ofthe Dirichlet eigenvalue problem (1.1) satisfies, fork = 1, 2, · · · ,

1k

k∑i=1

λi ≥n

n + 24π2

(ωnvolΩ) 2n

k2n ,



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Remark

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According to the Weyl’s asymptotic formula, we knowthat the result of Li and Yau is optimal in the sense ofaverage. From this formula, we have, for k = 1, 2, · · · ,

λk ≥n

n + 24π2

(ωnvolΩ) 2n

k2n ,

which gives a partial solution for the conjecture of Polyawith a factor

nn + 2

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In order to prove this theorem, main methods, whichare used by Li and Yau, are the following:

the Fourier transforma lemma of Hormander




the Fourier transform

a lemma of Hormander




the Fourier transforma lemma of Hormander



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Lemma of Homander

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If f is a function defined in Rn satisfying

0 ≤ f ≤ a1,

∫Rn|z|2 f (z)dz ≤ a2

then, we have∫Rn

f (z)dz ≤(a1ωn

) 2n+2

an

n+2

2

(n + 2n

) nn+2.

where a1 and a2 are constant.



Proof of the theorem of Li and Yau

Let ui be an eigenfunction corresponding to eigenvalueλi such that ui becomes an orthonormal basis ofL2(Ω). By defining a function

ϕ(x, y) = ∑k

i=1 ui(x)ui(y), (x, y) ∈ Ω × Ω0, the other,

andf (z) =

∫Rn|ϕ(z, y)|2dy,

where ϕ(z, y) is the Fourier transform of ϕ(x, y) in x,



we have

0 ≤ f ≤ (2π)−nvolΩ,∫Rn

f (z)dz = k,

∫Rn|z|2 f (z)dz =

k∑i=1

λi.

Thus, from the lemma of Hormander, proof of thetheorem of Li and Yau is finished.



we have0 ≤ f ≤ (2π)−nvolΩ,

∫Rn

f (z)dz = k,

∫Rn|z|2 f (z)dz =

k∑i=1

λi.




we have0 ≤ f ≤ (2π)−nvolΩ,∫

Rnf (z)dz = k,

∫Rn|z|2 f (z)dz =

k∑i=1

λi.



2. Lower bounds for eigenvalues A generalized conjecture of Polya

A generalized conjecture of Polya

For a complete Riemannian manifold M, eigenvalues ofthe Dirichlet eigenvalue problem (1.1) also satisfy theWeyl’s asymptotic formula. Hence, it is natural to try toobtain a lower bound for eigenvalues.

In fact, for a complete Riemannian manifold, by makinguse of the Sobolev constant, Li (1980), Chavel-Feldman(1981) proved

λk ≥

c(n, s)

k1

n−1

(volΩ) 2n

, n > 2

c(n, s)k

13

volΩ, n = 2



A generalized conjecture of Polya

For a complete Riemannian manifold M, eigenvalues ofthe Dirichlet eigenvalue problem (1.1) also satisfy theWeyl’s asymptotic formula. Hence, it is natural to try toobtain a lower bound for eigenvalues.In fact, for a complete Riemannian manifold, by makinguse of the Sobolev constant, Li (1980), Chavel-Feldman(1981) proved

λk ≥

c(n, s)

k1

n−1

(volΩ) 2n

, n > 2

c(n, s)k

13

volΩ, n = 2



They has applied this result to prove the uniformconvergences of the Heat Kernel.

Since the order of k is not optimal, one wants to askwhether it is possible to get lower bounds with theoptimal order of k. Furthermore, one would like to askthe following:

Whether is it possible for one to consider the sameproblem as the conjecture of Polya for a completeRiemannian manifold other than Rn?First of all, a difficulty which we will encounter isthat there is no the Fourier transform for acomplete Riemannian manifold.





Whether is it possible for one to consider the sameproblem as the conjecture of Polya for a completeRiemannian manifold other than Rn?

First of all, a difficulty which we will encounter isthat there is no the Fourier transform for acomplete Riemannian manifold.





Whether is it possible for one to consider the sameproblem as the conjecture of Polya for a completeRiemannian manifold other than Rn?First of all, a difficulty which we will encounter isthat there is no the Fourier transform for acomplete Riemannian manifold.



In order to derive the result of Li and Yau, thelemma of Hormander plays an important role. Butthere is no this kind of lemma for a completeRiemannian manifold.

In order to consider the same problem as the conjectureof Polya, we need to come over the following problems:

What method will we use to replace the Fouriertransform?What method will we use to replace the lemma ofHormander?





What method will we use to replace the Fouriertransform?

What method will we use to replace the lemma ofHormander?





What method will we use to replace the Fouriertransform?What method will we use to replace the lemma ofHormander?



Therefore, our purpose is to study the lower boundswith the optimal order of k for eigenvalues of theDirichlet eigenvalue problem of Laplacian on a boundeddomain in complete Riemannian manifolds.

First of all, we will propose the following version of theconjecture of Polya for complete Riemannian manifolds.



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The generalized conjecture of Polya

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Let Ω be a bounded domain in an n-dimensionalcomplete Riemannian manifold M. Then, there exists aconstant c(M,Ω), which only depends on M and Ωsuch that eigenvalues of the Dirichlet eigenvalueproblem (1.1) satisfy, for k = 1, 2, · · · ,

λk + c(M,Ω) ≥ 4π2

(ωnvolΩ) 2n

k2n ,

1k

k∑i=1

λi + c(M,Ω) ≥ nn + 2

4π2

(ωnvolΩ) 2n

k2n .



For this conjecture, we have proved

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Theorem 2.2 (Cheng and Yang, J. Diff. Eqns., 2009)

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Let Ω be a bounded domain in an n-dimensionalcomplete Riemannian manifold M. Then, there exists aconstant H2

0, which only depends on M and Ω such

that eigenvalues of the Dirichlet eigenvalue problem(1.1) satisfy, for k = 1, 2, · · · ,

1k

k∑i=1

λi +n2

4H2

0 ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n ,

λk +n2

4H2

0 ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n .



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Corollary 2.1

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Let Ω be a domain in the n-dimensional unit sphereSn(1). Then, eigenvalues of the Dirichlet eigenvalueproblem (1.1) satisfy, for k = 1, 2, · · · ,

1k

k∑i=1

λi +n2

4≥ n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n ,

λk +n2

4≥ n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n ,



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Corollary 2.2

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Let Ω be a bounded domain in an n-dimensionalcomplete minimal submanifold M in a Euclidean spaceRN. Then, eigenvalues of the Dirichlet eigenvalueproblem (1.1) satisfy, for k = 1, 2, · · · ,

1k

k∑i=1

λi ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n .

λk ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n .



In order to prove our theorem, we need to come overthe problems, which we have mentioned:

What method will we use to replace the Fouriertransform?The answer is universal inequalities foreigenvalues.What method will we use to replace the lemma ofHormander?The answer is a recursion formula of Cheng andYang.




What method will we use to replace the Fouriertransform?

The answer is universal inequalities foreigenvalues.What method will we use to replace the lemma ofHormander?The answer is a recursion formula of Cheng andYang.




What method will we use to replace the Fouriertransform?The answer is universal inequalities foreigenvalues.

What method will we use to replace the lemma ofHormander?The answer is a recursion formula of Cheng andYang.




What method will we use to replace the Fouriertransform?The answer is universal inequalities foreigenvalues.What method will we use to replace the lemma ofHormander?

The answer is a recursion formula of Cheng andYang.




What method will we use to replace the Fouriertransform?The answer is universal inequalities foreigenvalues.What method will we use to replace the lemma ofHormander?The answer is a recursion formula of Cheng andYang.


2. Lower bounds for eigenvalues The recursion formula of Cheng and Yang

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A recursion formula (Cheng and Yang, Math. Ann.2007)

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Let µ1 ≤ µ2 ≤ · · · ≤ µk+1 be any non-negative realnumbers satisfying

k∑i=1

(µk+1 − µi)2 ≤ 4t

k∑i=1

µi(µk+1 − µi).

Define

Gk =1k

k∑i=1

µi, Tk =1k

k∑i=1

µ2i ,

Fk =

(1 +

2t

)G2

k− Tk.



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Then, we have the following recursion formula

Fk+1 ≤ C(t, k)(

k + 1k

)4t Fk,

where t is any positive real number and

C(t, k) = 1 − 13t

(k

k + 1

)4t

(1 +

2t

) (1 +

4t

)(k + 1)3

< 1.



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Remark

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From this recursion formula, we can obtain, for anyℓ, k

Fk+ℓ

(k + ℓ) 4t

≤Fk

k 4t

By making use of the recursion formula of Chengand Yang, we can not only derive lower bounds foreigenvalues, but also derive upper bounds foreigenvalues.


2. Lower bounds for eigenvalues Universal inequalities for eigenvalues

Universal inequalities for eigenvalues

The case of a Euclidean space

First of all, we consider universal inequalities foreigenvalues of the Dirichlet eigenvalue problem (1.1)for a bounded domain Ω in an n-dimensional Euclideanspace.The investigation of universal inequalities foreigenvalues of the Dirichlet eigenvalue problem (1.1)was initiated by Payne, Polya and Weinberger (1955and 1956). They proved

λk+1 − λk ≤4

nk

k∑i=1

λi.



Although this result of Payne, Polya and Weinbergerhas been extended by many mathematicians in severalway, there are two main contributions due to Hile andProtter (1980) and Yang (1991).

In fact, Hile and Protter (1980) improved this result ofPayne, Polya and Weinberger to

k∑i=1

λi

λk+1 − λi≥ nk

4.

Yang (1991) proved a very sharp inequality:

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)λi.



Although this result of Payne, Polya and Weinbergerhas been extended by many mathematicians in severalway, there are two main contributions due to Hile andProtter (1980) and Yang (1991).In fact, Hile and Protter (1980) improved this result ofPayne, Polya and Weinberger to

k∑i=1

λi

λk+1 − λi≥ nk

4.

Yang (1991) proved a very sharp inequality:

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)λi.



The case of a unit sphere

For a domain Ω in an n-dimensional unit sphere,universal inequalities for eigenvalues of the Dirichleteigenvalue problem (1.1) has been studied by Chengand Yang. We have proved

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Theorem 2.3 (Cheng and Yang, Math Ann. 2005).

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Let Ω be a domain in an n-dimensional unit sphere.Eigenvalue λi’s of the Dirichlet eigenvalue problem (1.1)satisfy

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)(λi +n2

4).



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Remark

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The above inequality is best possible since thisinequality does not depend on the domain Ω and whenΩ tends to the unit sphere, this inequality becomes anequality for all of k.



The case of a hyperbolic space

When M is Hn(−1), although many mathematicianswant to derive a universal inequality for eigenvalues,there are no any results on universal inequalities foreigenvalues of the Dirichlet eigenvalue problem (1.1)excepting n = 2.

If n = 2, by making use of estimates for eigenvalues ofthe eigenvalue problem of the Schrodinger like operatorwith a weight, Harrell and Michel (Comm. PDEs, 1994)and Ashbaugh (2002) have obtained several results. Infact, if n = 2, the Laplacian on H2(−1) is like to theLaplacian on R2 with a weight. But, when n > 2, thisproperty does not hold again.Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 30 / 64


For a bounded domain in Hn(−1), main reason, thatone could not derive a universal inequality, is that onecan not find an appropriate trial function.

Recently, Cheng and Yang find an appropriate trialfunction for Hn(−1). Hence, we can derive a universalinequality for eigenvalues of the Dirichlet eigenvalueproblem (1.1), that is, we prove the following:

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Theorem 2.4 (Cheng and Yang, J. Diff. Eqns, 2009)

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For a bounded domain Ω in Hn(−1), eigenvalue λi’s ofthe Dirichlet eigenvalue problem (1.1) satisfy

k∑i=1

(λk+1 − λi)2 ≤4k∑

i=1

(λk+1 − λi)(λi −(n − 1)2

4).



For a bounded domain in Hn(−1), main reason, thatone could not derive a universal inequality, is that onecan not find an appropriate trial function.Recently, Cheng and Yang find an appropriate trialfunction for Hn(−1). Hence, we can derive a universalinequality for eigenvalues of the Dirichlet eigenvalueproblem (1.1), that is, we prove the following:

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Theorem 2.4 (Cheng and Yang, J. Diff. Eqns, 2009)

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For a bounded domain Ω in Hn(−1), eigenvalue λi’s ofthe Dirichlet eigenvalue problem (1.1) satisfy

k∑i=1

(λk+1 − λi)2 ≤4k∑

i=1

(λk+1 − λi)(λi −(n − 1)2

4).



An application of theorem 2.4

We consider an application of our universal inequality.Let Ω be an n-disk of radius r > 0 in Hn(−1). McKean(J. Diff. Geom. 1970) proved that the first eigenvalueλ1(r) of the Dirichlet eigenvalue problem (1.1) satisfies

λ1(r) >(n − 1)2

4,

limr→∞λ1(r) =

(n − 1)2

4.



From domain monotonicity of eigenvalues, we have, forany bounded domain Ω in Hn(−1),

λ1(Ω) >(n − 1)2

4,

limΩ→Hn(−1)λ1(Ω) =(n − 1)2

4.

It is obvious that, for any k > 1,

λk(Ω) > λ1(Ω) >(n − 1)2

4.

It would be interesting to study behaviors of λk(Ω), fork ≥ 2, when Ω tends to Hn(−1). As an application ofour universal inequality, we can prove the following:Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 33 / 64


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Proposition 2.1 (Cheng and Yang, J. Diff. Eqns,2009)

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Let Ω be a bounded domain in Hn(−1). Then,eigenvalue λk(Ω), for any k, of the Dirichlet eigenvalueproblem (1.1) satisfies

limΩ→Hn(−1)λk(Ω) =(n − 1)2

4.

In order to prove this result, we need to investigateupper bounds for eigenvalues.



The general case

For a general n-dimensional complete Riemannianmanifold, we want to obtain universal inequalities foreigenvalues.

In the cases of a Euclidean space and a unit sphere,one can make use of the coordinate functions toconstruct the trial functions. Thus, one can deriveuniversal inequalities for eigenvalues according to theRayleigh-Ritz inequality.

But for the hyperbolic space, it has been very difficult toconstruct an appropriate trial function.



For a general complete Riemannian manifold, it iscertainly very difficult to construct an appropriate trialfunction.

Fortunately, we have Nash’s theorem.

By making use of Nash’s theorem, we successfullyconstruct trial functions which satisfy good properties.

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Nash’s Theorem

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Each complete Riemannian manifold can beisometrically immersed in a Euclidean space.



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Theorem 2.5 (Chen and Cheng, J. Math. Soc.Japan, 2008)

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that eigenvalues λ j of the Dirichlet eigenvalue problem(1.1) satisfy, for any k,

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)(λi +n2

4H2

0).

We should remark that El Soufi, Harrell and Ilias (Trans.Amer. Math. Soc. 2009) have also proved a similarresult, independently.



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Theorem 2.5 (Chen and Cheng, J. Math. Soc.Japan, 2008)

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that eigenvalues λ j of the Dirichlet eigenvalue problem(1.1) satisfy, for any k,

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)(λi +n2

4H2

0).

We should remark that El Soufi, Harrell and Ilias (Trans.Amer. Math. Soc. 2009) have also proved a similarresult, independently.Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 37 / 64


In particular, when M is a complete minimalsubmanifold in RN, we have

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Corollary 2.3

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Let Ω be a bounded domain in an n-dimensionalcomplete minimal submanifold Mn in RN. Then, wehave

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)λi.



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Remark

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our universal inequality is the same as one of Yangfor the case of Rn.

There exist many complete minimal submanifoldsin RN.The universal inequality for eigenvalues of Yangdoes not only holds for bounded domains in Rn,but also for bounded domains in any completeminimal submanifold in RN.



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Remark

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our universal inequality is the same as one of Yangfor the case of Rn.There exist many complete minimal submanifoldsin RN.

The universal inequality for eigenvalues of Yangdoes not only holds for bounded domains in Rn,but also for bounded domains in any completeminimal submanifold in RN.



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Remark

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our universal inequality is the same as one of Yangfor the case of Rn.There exist many complete minimal submanifoldsin RN.The universal inequality for eigenvalues of Yangdoes not only holds for bounded domains in Rn,but also for bounded domains in any completeminimal submanifold in RN.



A general inequality

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Proposition 2.2

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Let M be a complete Riemannian manifold and Ω abounded domain with a piecewise smooth boundary∂Ω. For any function f ∈ C3(Ω) ∩ C2(∂Ω) eigenvaluesλi of the Dirichlet eigenvalue problem (1.1) satisfy

k∑i=1

(λk+1−λi)2∥ui∇ f∥2 ≤k∑

i=1

(λk+1−λi)∥2∇ f ·∇ui+ui∆ f∥2

where ∥ f∥2 =∫

M f 2 and ∇ f · ∇ui = g(∇ f,∇ui) and uiis an orthonormal eigenfunction corresponding to λi.



Proof. Assume that ui is an orthonormal eigenfunctioncorresponding to the ith eigenvalue λi, i.e. ui satisfies

∆ui = −λiui, in Ω,ui|∂Ω = 0,∫Ω

uiu j = δi j.

Defining ϕi, ai j and bi j, for i, j = 1, · · · , k, by

ai j =

∫M

fuiu j, bi j =

∫M

u j(∇ui · ∇ f +12

ui∆ f ),

ϕi = fui −k∑

j=1

ai ju j.



we have

ai j = a ji,

∫Mϕiu j = 0, for j = 1, 2, · · · , k.

Thus, we have, from Rayleigh-Ritz formula,

λk+1∥ϕi∥2 ≤∫

M|∇ϕi|2.

By a direct computation, we infer

(λk+1 − λi)∥ϕi∥2 ≤ −∫

Mϕi(2∇ f · ∇ui + ui∆ f ) ≡ wi,



On the other hand, we have

wi = −∫

Mϕi(2∇ f · ∇ui + ui∆ f − 2

k∑j=1

bi ju j).

From Schwarz inequality, we obtain

(λk+1 − λi)w2i

≤ (λk+1 − λi)∥ϕi∥2∥2∇ f · ∇ui + ui∆ f − 2k∑

j=1

bi ju j∥2

≤ wi∥2∇ f · ∇ui + ui∆ f − 2k∑

j=1

bi ju j∥2.



Hence,

(λk+1 − λi)wi ≤ ∥2∇ f · ∇ui + ui∆ f∥2 − 4k∑

j=1

b2i j.

Multiplying the above inequality by (λk+1 − λi) andtaking sum on i from 1 to k, we have

k∑i=1

(λk+1 − λi)2wi ≤ −4k∑

i, j=1

(λk+1 − λi)b2i j

+

k∑i=1

(λk+1 − λi)(∥2∇ f · ∇ui + ui∆ f∥2.



Since2bi j = (λi − λ j)ai j

and

wi = ∥ui∇ f∥2 +

k∑j=1

(λi − λ j)a2i j,

we obtain

k∑i=1

(λk+1 − λi)2∥ui∇ f∥2

≤k∑

i=1

(λk+1 − λi)∥2∇ f · ∇ui + ui∆ f∥2.



Proof of theorem 2.5

Let M be an n-dimensional complete Riemannianmanifold with metric g. From Nash’s theorem, thereexists an isometrical immersion from M into RN. Forany point P in M, assuming that y with components yαis the position vector of P in RN, we have, for anyfunction u ∈ C∞(M),

N∑α=1

g(∇yα,∇yα) = n,N∑α=1

(∆yα)2 = n2|H|2

N∑α=1

∆yα∇yα = 0,N∑α=1

g(∇yα,∇u)2 = |∇u|2,



where H is the mean curvature vector field of thisimmersion.Letting f = yα in the proposition 2.2 and taking sum onα from 1 to N, we obtain

k∑i=1

(λk+1 − λi)2 ≤ 4n

k∑i=1

(λk+1 − λi)(λi +n2

4supΩ

|H|2).

Since eigenvalues of the Dirichlet eigenvalue problemof the Laplacian are invariant under isometrictransformations. Defining H2

0= infφ∈Φ supΩ |H|2, where

Φ is the set of all isometric immersions from M intoEuclidean spaces, we prove our result.



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Theorem 2.2 (Cheng and Yang, J. Diff. Eqns., 2009)

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that eigenvalues of the Dirichlet eigenvalue problem(1.1) satisfy, for k = 1, 2, · · · ,

1k

k∑i=1

λi +n2

4H2

0 ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n ,

λk +n2

4H2

0 ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n .



Proof of the theorem 2.2

From the theorem 2.5, we have

k∑i=1

(λk+1 − λi)2 ≤4n

k∑i=1

(λk+1 − λi)(λi +n2

4H2

0).

Letting µi = λi +n2

4H2

0, we have

k∑i=1

(µk+1 − µi)2 ≤4n

k∑i=1

(µk+1 − µi)µi.



From the recursion formula of Cheng and Yang witht = n, we infer

Fk+ℓ

(k + ℓ) 4n

≤Fk

k 4n

.

According to the Weyl’s asymptotic formula, we derive,for any positive integer k,

Fk

k 4n

≥ 2n(n + 2)(n + 4)

16π4

(ωnvolΩ) 4n

.



SinceFk ≤

2n

G2k,

we infer

1k

k∑i=1

λi +n2

4H2

0 ≥n√

(n + 2)(n + 4)

4π2

(ωnvolΩ) 2n

k2n .


3. Upper bounds for eigenvalues


First, for a bounded domain in the Euclidean space Rn,according to the partial solution of the conjecture ofPolya due to Li and Yau, we have, for k = 1, 2, · · · ,

λk ≥n

n + 24π2

(ωnvolΩ) 2n

k2n

and from the Weyl’s asymptotic formula, we know

λk ∼4π2

(ωnvolΩ) 2n

k2n .

Hence, it is very important to obtain upper bounds foreigenvalues with the optimal order of k.Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 52 / 64


But, it is very hard to obtain an upper bound foreigenvalues with the optimal order of k.

In order to obtain an upper bound for eigenvalues withthe optimal order of k, we have proved

.

A recursion formula (Cheng and Yang, Math. Ann.2007)

.

.

.

. ..

.

.

Let µ1 ≤ µ2 ≤ · · · ≤ µk+1 be any non-negative realnumbers satisfying

k∑i=1

(µk+1 − µi)2 ≤ 4t

k∑i=1

µi(µk+1 − µi).



.

.

. ..

.

.

Define

Gk =1k

k∑i=1

µi, Tk =1k

k∑i=1

µ2i ,

Fk =

(1 +

2t

)G2

k− Tk.

Then, we have the following recursion formula

Fk+1 ≤ C(t, k)(

k + 1k

)4t Fk,

where t > 0 is any positive number.



By making use of the recursion formula, we haveproved

.

Theorem 3.1 (Cheng and Yang, Math. Ann. 2007)

.

.

.

. ..

.

.

For a bounded domain Ω ⊂ Rn, eigenvalues of theDirichlet eigenvalue problem (1.1) satisfy, for any k,

λk+1 ≤ C0(n)k2nλ1,

whereC0(n) ≤ 1 +

2.6n

if k > 1.

our upper bound is best possible in the sense of orderof k and it is universal inequality.



By making use of the recursion formula, we haveproved

.

Theorem 3.1 (Cheng and Yang, Math. Ann. 2007)

.

.

.

. ..

.

.

For a bounded domain Ω ⊂ Rn, eigenvalues of theDirichlet eigenvalue problem (1.1) satisfy, for any k,

λk+1 ≤ C0(n)k2nλ1,

whereC0(n) ≤ 1 +

2.6n

if k > 1.

our upper bound is best possible in the sense of orderof k and it is universal inequality.Qing-Ming Cheng (Saga University) Estimates for eigenvalues March 13-15, 2011 55 / 64


For a complete Riemannian manifold M, Chen andCheng have obtained the following

.

Theorem 3.2 (Chen and Cheng, 2008)

.

.

.

. ..

.

.

For a bounded domain Ω in an n-dimensional completeRiemannian manifold, there exists a constant H2

0such

that eigenvalues of the Dirichlet eigenvalue problem(1.1)satisfy

λk+1 ≤ C0(n)k2n (λ1 +

n2

4H2

0),


3. Upper bounds for eigenvalues A Conjecture of PPW

Next, we consider the lower order eigenvalues of theDirichlet eigenvalue problem of the Laplacian.

For a bounded domain Ω ⊂ Rn, Payne, Polya andWeinberger (1956) conjectured the following:

.

Conjecture of PPW

.

.

.

. ..

.

.

For a bounded domain Ω ⊂ Rn, eigenvalues of theDirichlet eigenvalue problem (1.1) satisfy

(1)λ2

λ1≤λ2

λ1|n−ball,

(2)λ2 + λ3 + · · · + λn+1

λ1≤ nλ2

λ1|n−ball,



For the conjecture (1) of Payne, Polya and Weinberger,many mathematicians studied it. For examples, Payne,Polya and Weinberger, Brands, de Vries, Chiti, Hile andProtter, Marcellini and so on. Finally, Ashbaugh andBenguria (1992, Ann. of Math.) solved this conjecture.

For conjecture (2) of Payne, Polya and Weinberger,when n = 2, Payne, Polya and Weinberger provedλ2 + λ3

λ1≤ 6. Brands improved on the bound onto

λ2 + λ3

λ1≤ 3 +

√7. Furthermore, Hile and Protter

obtainedλ2 + λ3

λ1≤ 5.622.




For conjecture (2) of Payne, Polya and Weinberger,

when n = 2, Payne, Polya and Weinberger provedλ2 + λ3


λ2 + λ3

λ1≤ 3 +


obtainedλ2 + λ3

λ1≤ 5.622.




For conjecture (2) of Payne, Polya and Weinberger,when n = 2, Payne, Polya and Weinberger provedλ2 + λ3


λ2 + λ3

λ1≤ 3 +


obtainedλ2 + λ3

λ1≤ 5.622.



Marcellini provedλ2 + λ3

λ1≤ (15 +

√345)/6. Very

recently, Chen and Zheng (J. Diff. Eqns., 2011) have

provedλ2 + λ3

λ1≤ 5.3507.

For a general dimension n ≥ 2, Ashbaugh andBenguria proved

λ2 + λ3 + · · · + λn+1

λ1≤ n + 4.

Furthermore, Ashbaugh and Benguria (1994) (cf. Hileand Protter ) improved the above result to

λ2 + λ3 + · · · + λn+1

λ1≤ n + 3 +

λ1

λ2.



For a general complete Riemannian manifold Mn, Chenand Cheng (J. Math. Soc. Japan 2008) have provedthat there exists a non-negative constant H0 such that

λ2 + λ3 + · · · + λn+1

λ1≤ n + 4 +

n2H20

λ1.

Recently, by making use of the fact that eigenfunctionsform an orthonormal basis of L2(Ω) in place of theRayleigh-Ritz formula, Cheng and Qi have improvedthe above result to the following:



.

Theorem 3.3 (Cheng and Qi)

.

.

.

. ..

.

.

Let Mn be an n-dimensional Riemannian manifold,Ω ⊂ Mn a bounded domain with a piecewise smoothboundary ∂Ω. Then, the lower order eigenvalues of theDirichlet eigenvalue problem of the Laplacian satisfy

λ2 + λ3 + · · · + λn+1

λ1

≤ n +

√√(n2H20

λ1+ 4

)Q(H0, λ1, λ2)

where H0 is a non-negative constant depending on Mand Ω only, and



Q(H0, λ1, λ2) =(2 −

λ1

λ2)n2H2

0

λ1+ 3 +

λ1

λ2

2

+

√√(3 + λ1

λ2+

n2H20

λ2

)2+ 4(1 −

λ1

λ2)n2H2

0

λ2

2.

It is easy to prove

Q(H0, λ1, λ2) <n2H2

0

λ1+ 4



In particular, when Mn is a complete minimalsubmanifold in the Euclidean space RN, we have

.

Corollary 3.1

.

.

.

. ..

.

.

Let Ω be a bounded domain in an n-dimensionalcomplete minimal submanifold Mn in RN. Then, wehave

λ2 + λ3 + · · · + λn+1

λ1≤ n + 2

√3 +λ1

λ2.



Thank you!


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