Yamabe-type Equations on Complete, Noncompact Manifolds

Progress in Mathematics Volume 302 Series Editors Hyman Bass Joseph Oesterlé Yuri Tschinkel Alan Weinstein Paolo Mas...

Author: Paolo Mastrolia | Marco Rigoli | Alberto G Setti

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Progress in Mathematics Volume 302

Series Editors Hyman Bass Joseph Oesterlé Yuri Tschinkel Alan Weinstein

Paolo Mastrolia Marco Rigoli Alberto G. Setti

Yamabe-type Equations on Complete, Noncompact Manifolds

Paolo Mastrolia Dipartimento di Matematica Università degli Studi die Milano Milano, Italy

Marco Rigoli Dipartimento di Matematica Università degli Studi die Milano Milano, Italy

Alberto G. Setti Dipartimento di Fisica e Matematica Università dell’Insubria Como, Italy

ISBN 978-3-0348-0375-5 ISBN 978-3-0348-0376-2 (eBook) DOI 10.1007/978-3-0348-0376-2 Springer Basel Heidelberg New York Dordrecht London Library of Congress Control Number: 2012944419 Mathematics Subject Classification (2010): Primary: 53C21, 58-02; Secondary: 35J60, 35B45, 58J50 © Springer Basel 2012 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper

Springer Basel AG is part of Springer Science+Business Media (www.birkhauser-science.com)

Contents Introduction 1

. . . . . . . . . . . . .

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7 7 8 10 12 14 16 18 18 21 22 28 31 32

2 Pointwise conformal metrics 2.1 The Yamabe equation . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 The derivation of the Yamabe equation . . . . . . . . . 2.1.2 The Kazdan-Warner obstruction . . . . . . . . . . . . . 2.1.3 The Weyl and Cotton tensors . . . . . . . . . . . . . . . 2.2 Some applications in the compact case . . . . . . . . . . . . . . 2.2.1 A rigidity result of Obata . . . . . . . . . . . . . . . . . 2.2.2 A result by M. F. Bidaut-Véron and L. Véron . . . . . . 2.2.3 A version of Theorem 2.12 on manifolds with boundary 2.2.4 A rigidity result of Escobar . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

37 37 37 40 43 49 49 57 62 67

General nonexistence results 3.1 Some spectral considerations . . . . . . . . . . . . . . . . . . . . . 3.1.1 The main nonexistence result . . . . . . . . . . . . . . . . . 3.2 The endpoint case K = −1 and the Poisson equation . . . . . . . .

73 74 78 93

3

Some Riemannian Geometry 1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Moving frames and the first structure equations . 1.1.2 Covariant derivative of tensor fields . . . . . . . . 1.1.3 Meaning of the first structure equations . . . . . 1.1.4 Curvature: the second structure equations . . . . 1.1.5 Einstein manifolds and Schur’s Theorem . . . . . 1.2 Comparison theorems . . . . . . . . . . . . . . . . . . . 1.2.1 Ricci identities . . . . . . . . . . . . . . . . . . . 1.2.2 Cut locus and regularity of the distance function 1.2.3 The Laplacian comparison theorem . . . . . . . . 1.2.4 The Bishop-Gromov comparison theorem . . . . 1.2.5 The Hessian comparison theorem . . . . . . . . . 1.3 Some formulas for immersed submanifolds . . . . . . . .

1

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

v

vi

Contents 3.3

4

5

6

A refined version of Theorem 3.2 . . . . . . . . . . . . . . . . . . .

A priori estimates 4.1 Estimates from below . . . . . . . . . . . . . 4.2 Estimates from above . . . . . . . . . . . . . 4.3 Sharpness of the previous results . . . . . . . 4.4 Some further estimates . . . . . . . . . . . . . 4.5 Nonexistence results for the Yamabe problem

. . . . .

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. . . . .

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. . . . .

. . . . .

. . . . .

Uniqueness 5.1 A sharp integral condition . . . . . . . . . . . . . . . . . . . . . 5.2 A remark on the asymptotic behaviour of solutions: examples Rm and Hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Uniqueness via the weak maximum principle . . . . . . . . . . 5.3.1 A useful form of the weak maximum principle . . . . . . 5.3.2 A comparison result . . . . . . . . . . . . . . . . . . . . 5.3.3 Uniqueness of ground states . . . . . . . . . . . . . . . . 5.4 Some geometric applications and further uniqueness . . . . . . 5.4.1 Conformal diffeomorphisms . . . . . . . . . . . . . . . . 5.4.2 Uniqueness for the Yamabe problem . . . . . . . . . . . 5.4.3 An L∞ a priori estimate . . . . . . . . . . . . . . . . .

. . . . .

. . . . .

. . in . . . . . . . . . . . . . . . . . .

Existence 6.1 A general procedure . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Another comparison result . . . . . . . . . . . . . . . . . . . 6.1.2 More basic spectral theory and a result of Li, Tam and Yang 6.1.3 Two useful lemmas . . . . . . . . . . . . . . . . . . . . . . . 6.1.4 Existence of a maximal solution . . . . . . . . . . . . . . . . 6.2 Subsolutions and existence . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Existence with λL 1 (M ) < 0 . . . . . . . . . . . . . . . . . . . 6.2.2 λL (M ) < 0: some sufficient conditions . . . . . . . . . . . . 1 6.2.3 A more general case . . . . . . . . . . . . . . . . . . . . . . 6.3 Global sub- and supersolutions . . . . . . . . . . . . . . . . . . . . 6.4 The case of the Yamabe problem . . . . . . . . . . . . . . . . . . . 6.5 Appendix: the Monotone Iteration Scheme . . . . . . . . . . . . . .

7 Some special cases 7.1 A nonexistence result . . . . . . . . . . . . . . . . 7.1.1 A Rellich-Pohozaev formula . . . . . . . . 7.1.2 A nonexistence result for hyperbolic space 7.1.3 An integral obstruction . . . . . . . . . . 7.2 Special symmetries and existence . . . . . . . . . 7.3 The case of Euclidean space and further results . 7.3.1 A linear comparison result . . . . . . . . .

. . . . . . .

. . . . . . .

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. . . . . . .

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98 105 105 111 115 117 121 127 127 130 132 133 140 143 146 146 148 149 157 158 158 158 162 165 166 166 170 177 180 188 191 197 197 207 211 219 221 227 227

Contents

vii

7.3.2 7.3.3

Back to Corollary 5.8 . . . . . . . . . . . . . . . . . . . . . 229 The Euclidean space . . . . . . . . . . . . . . . . . . . . . . 231

Bibliography

239

List of Symbols

247

Index

253

Introduction This book originates from a graduate course given at the University of Milan in 2007. Our goal is twofold: first, to present a self-contained introduction to the geometric and analytic aspects of the Yamabe problem on a complete noncompact Riemannian manifold, treating existence, nonexistence, uniqueness and a priori estimates of the solutions. Secondly, we intend to describe in a way accessible to the nonspecialist a range of methods and techniques that can be successfully applied to more general nonlinear equations which arise in applications. The classical Yamabe problem concerns the possibility of pointwise conformally deforming a metric of scalar curvature S(x) on the manifold M to a new metric with prescribed scalar curvature K(x). In the case where K is constant it is a natural higher dimensional generalization of the Poincaré–K¨ obe Uniformization Theorem for Riemann surfaces and can be seen as a way to select a privileged metric on the manifold. If , is the original metric of the Riemannian manifold M and we denote with , = ϕ2 , , ϕ > 0, a conformally deformed metric, then the two scalar curvatures S(x) and S(x) are related by the equation = S(x) − 2(m − 1) ϕ2 S(x)

Δϕ |∇ϕ|2 − (m − 1)(m − 4) 2 ϕ ϕ

(see equation (2.7) in Chapter 2), where Laplacian, gradient, and norm are those of the metric , . In the case where the dimension m of the manifold is greater than or equal to three, it is useful to set 2

ϕ = u m−2 so that the above equation takes the form m+2 m−1 m−2 Su = Su − 4 Δu. m−2

Thus the Yamabe problem amounts to finding a positive solution u of the familiar Yamabe equation m+2 (1) cm Δu − Su + Ku m−2 = 0,

1

2

Introduction

where cm = 4 m−1 m−2 and K = S, the prescribed scalar curvature of the conformally deformed metric. If M is compact and K is constant, after an initial attempted solution by H. Yamabe [Yam60], the problem was solved thanks to efforts of N. Trudinger [Tru68], T. Aubin [Aub76] and R. Schoen [Sch84] (see the nice survey paper by J.M. Lee and T.H. Parker, [LP87], for a complete and self-contained treatment). The solution was obtained using variational methods, and one of the m+2 main analytic difficulties stems from the fact that m−2 is the critical exponent for 2m

the Sobolev embedding W 1,2 → L m−2 . A natural generalization of the classical Yamabe problem is the case where K is nonconstant and/or M is noncompact. In this direction we mention the pioneering work of J. L. Kazdan and F. W. Warner, [KW74a], [KW74b], [KW75a], [KW75b]. It should also be mentioned that even the classical Yamabe problem of deforming the metric to one of constant scalar curvature in the noncompact setting is in general not solvable, as first shown by Z. R. Jin, [Jin88]. The Yamabe problem for noncompact manifolds with variable prescribed curvature is the subject of the present monograph. Indeed, we describe methods which allows us to consider the more general Yamabe-type equations (resp. inequalities) of the form Δu + a(x)u − b(x)uσ = 0 (resp. ≥ 0) (2) where σ > 1, and we study nonexistence, a priori estimates, uniqueness and existence. Equations of the form (2) and still more general differential inequalities of the form uΔu + a(x)u2 − b(x)uσ+1 ≥ −A|∇u|2 (3) arise in complex analysis (e.g. in the study of the structure of complete Kähler manifolds, [LY90], [Li90] and [LR96], in the Schwarz Lemma for the ratio of volume elements of Kähler manifolds of the same dimension, [Gri76], in the study of pluriharmonic functions on a K¨ ahler manifold, [PRS08]), in the study of harmonic maps with bounded dilation ([EL78] and [PRS08] Chapter 8), in the classification of locally conformally flat manifolds ([PRS07]), in the study of Yang-Mills fields, and in population dynamics, to quote only a few examples. Existence and nonexistence of positive solutions of (2) clearly depend on the geometry of the underlying manifold, typically encoded by curvature or volume growth of geodesic balls, on properties of the coefficients (typically the relative signs of the coefficients a(x) and b(x)) and their asymptotic behavior and on the mutual interplay of the two. This interplay can be taken into account in terms of the relative asymptotic behavior of the coefficients versus the geometry at infinity of the manifold or, at a deeper level, in terms of spectral properties of Schr¨ odinger operators naturally associated to the equation. From the geometrical interpretation of the equation, it is natural to expect it will be easier to have existence when a and b are “close” enough, for instance they have the same sign, while it will be more difficult to have existence (and therefore

Introduction

3

it will be easier to prove nonexistence) when a and b are farther apart, for instance when they have opposite sign. This expectation is confirmed by both the existence and the nonexistence results that we will describe. The geometry of the manifold also plays a natural role in the uniqueness results as well in the a priori estimates on the solutions. The latter have a particular geometric interest since they are responsible for the completeness/noncompleteness of the deformed metric. As mentioned above, we use a variety of techniques adapted to the geometric situation at hand in which the lack of compactness and of symmetry and homogeneity prevents the use of more standard tools typical of compact situations or of the Euclidean setting. In particular, for existence we will essentially use the method of sub-super solutions, [Ama76], [Sat73]. Nonexistence will be obtained using Liouville-type results which in turn are obtained using either integral formulas or a method based on the coupling of the supposed solution of the Yamabe-type inequality with that of an appropriate Schrödinger-type inequality associated to it, in a manner reminiscent of the classical generalized maximum principle. Uniqueness will be obtained using variants of the weak maximum principle (see, e.g., [PRS05b]) and of clever integration by parts arguments. Finally, a priori estimates will be typically obtained using an elaboration of the old idea of the proof of the Schwarz’s Lemma by L. H. Ahlfors, [Ahl38], which is at the heart of the maximum principle. The book is divided into seven chapters. In the first chapter we give a quick review of Riemannian geometry using the method of moving frames. While we assume basic knowledge of Riemannian geometry, several computations will be carried out in full detail in order to acquaint the reader with notation and formalism. We concentrate on derivation of the symmetry properties of the curvature tensors together with a number of other identities that will be repeatedly used in the sequel. In particular, we will describe the commutation rules for covariant derivatives up to fourth order. Then we describe comparison results for the Laplacian of the Riemannian distance function, and for the volume of geodesic balls in terms of lower bounds for the Ricci curvature. We point out that our treatment, which follows that of [PRS05b], does not use Jacobi fields. In Chapter 2 we first derive equations for the change of curvature tensors under a conformal change of the metric and introduce the Yamabe equation. As a side product of our computations we obtain decomposition of the Riemann curvature tensor in its irreducible components and we exhibit the conformal invariance of the Weyl tensor. Then, we briefly consider the case where M is compact to illustrate the interplay between geometry and analysis, with a few illuminating examples such as the Kazdan-Warner obstruction, a result of Obata on Einstein manifolds and a far-reaching “generalization” due to Véron-Véron, through which we prove further results of Escobar. Along the way we give a detailed proof, which inspires to Petersen’s treatise [Pet06a], of a famous rigidity result of Obata. The goal is also to give some geometrical feeling on the subject that will enable us to

4

Introduction

proceed with the noncompact case: the case of the rest of our investigation. The core of the monograph begins with Chapter 3, devoted to nonexistence results. As mentioned above, since our methods apply to more general situations which have a wide range of applications, we consider in fact differential inequalities of the form (2) and (3). We describe several nonexistence results; in most of them we assume that u satisfies suitable integrability conditions, that b(x) is nonnegative and that there exists a positive solution ϕ to the differential inequality Δϕ + Ha(x)ϕ ≤ −K

|∇ϕ|2 ϕ

with H, K parameters satisfying H > 0, K > −1. Note that in the special case where K = 0 the latter condition amounts to the fact that the bottom of the spectrum of the operator −Δ − Ha(x) is nonnegative. Since −Δ is a nonnegative operator, the condition is trivially satisfied if a(x) ≤ 0 on M and may be interpreted as a measure of smallness in a spectral sense of the positive part of a(x). This agrees with the heuristic intuition on the effect of the relative signs of a(x) and b(x) on the existence of solutions. The existence of the positive function ϕ enters the proof in two different ways. In Theorem 3.2 one uses the functions ϕ to obtain an integral inequality involving u and its gradient from which one concludes that u is constant, and therefore necessarily identically zero. In a second group of results, the function ϕ is combined with the solution u to give rise to a diffusion-type differential inequality for which we prove a Liouville theorem. This yields the desired triviality. We also show that when σ is greater than or equal to the critical exponent (m + 2)/(m − 2), then, by performing an appropriate change of the metric and of the solution, the nonexistence results can be improved to allow even some controlled negativity of the coefficient b(x). Chapter 4 is devoted to establishing a priori upper and lower estimates for the asymptotic behavior of solutions of the differential inequalities Δu + a(x)u − b(x)uσ ≥ 0, resp. Δu + a(x)u − b(x)uσ ≤ 0, under assumptions on a(x) and b(x) related to an assumed radial lower bound for the Ricci curvature. As briefly mentioned above, the results are obtained by applying Alhfors’s old idea, namely, one considers an auxiliary function defined in terms of the solution u which by construction attains an extremum, and applies the usual maximum principle. Clearly, the heart of the method consists in finding the best auxiliary function for the problem at hand. We exhibit examples showing that our estimates are essentially sharp. Some further estimates, which cannot be obtained with the previous method, are provided by direct comparison with the aid of the maximum principle (see section 4.3). The chapter ends with some nonexistence results for the Yamabe problem, which complement those described in Chapter 3. In Chapter 5 we discuss some uniqueness results for positive solutions of Yamabe-type equations (2). The first, the very general Theorem 5.1, states that

Introduction

5

if the coefficient b(x) is nonnegative and not identically zero, then two solutions whose difference is L2 -integrable are necessarily the same. Although very general, it is sharp, and, remarkably, the assumption on the L2 -integrability cannot be replaced by an Lp condition with p > 2. The result is obtained by means of a clever elementary integral inequality. The second result, Theorem 5.2, follows by a comparison argument which relies on a version of the weak maximum principle (see Theorem 5.3) which is interesting in its own. While in most of the results available in the literature, uniqueness is obtained by requiring that solutions have a rather precisely determined asymptotic behavior, our result applies to solutions whose behavior at infinity is specified in a much less stringent manner, see (5.13); moreover, the conclusion is reached assuming only conditions on the volume growth of the manifold. Counterexamples show the sharpness of each result. The chapter ends with a geometric application to the group of conformal diffeomorphisms of a complete manifold and to the uniqueness of solutions of the geometric Yamabe problem. Chapter 6 deals with existence results for Yamabe-type equations (2) on the complete, noncompact, Riemannian manifold M . The main tool is the monotone iteration scheme in various forms, and we give a rather detailed description of it in the appendix at the end of the chapter. The application of the scheme in this context goes back to W. M. Ni, [Ni82], in the Euclidean setting and to P. Aviles and R. C. McOwen, [AM85] and [AM88], for noncompact manifolds. After having introduced some preliminary material on spectral theory, and a useful comparison result, the main body of the chapter is then devoted to the construction of (global and local) super- and subsolutions for the problem. In general terms, supersolutions are obtained under assumptions on the sign of b(x) and of the first eigenvalue of L = Δ + a(x) on appropriate domains. Because of the combination of signs of the coefficients, subsolutions are harder to find. We give a number of sufficient conditions which ensure that such subsolutions do exist: among them the spectral condition λL 1 (M ) < 0, for which we provide a new sufficient condition contained in Theorem 6.11. Furthermore, we mention Theorem 6.15, in which existence is guaranteed under a very week growth condition on b(x), and also Theorem 6.16, where a further weakining on the condition on the sign of b(x) is balanced by the necessity of imposing a constant negative lower bound on the Ricci curvature. We explicitly note that the assumptions of our existence theorems match those of the nonexistence results in the previous chapters. In the last chapter, Chapter 7, we consider some particular cases where the symmetry of the geometry allows one to use special techniques and to obtain stronger results. Typically this happens in Euclidean and Hyperbolic spaces, and more generally in the case of models (in the sense of R. Greene and H. Wu, [GW79]), or manifolds with special symmetry. The specific feature of models which make the analysis more precise is that the Laplacian of the distance function from the origin is given explicitly, as opposed to the case of a general manifold where only upper and lower bounds may be obtained under suitable curvature assumptions, by means of the the Laplacian and Hessian comparison theorems,

6

Introduction

and where the possible presence of the cut locus raises additional difficulties. We describe refined techniques adapted to the situation at hand and obtain results that, as a by-product, show the degree of sharpness of the general theory and methods we have developed dealing with generic complete Riemannian manifolds. It seems worth remarking that, in the specific case of Hyperbolic space, we provide a nonexistence result with the aid of a Rellich-Pohozaev type formula (see Theorem 7.7) and, even more, in Proposition 7.9 we introduce an integral obstruction to the existence of a conformal deformation which is of a different nature with respect to the Kazdan-Warner condition. Many of the results presented in this monograph have been obtained over the years by the authors jointly with many collaborators. To all of them we wish to extend our thanks and appreciation. In particular we are indebted to S. Pigola and M. Rimoldi who provided us with the proof of Theorem 2.10 in Chapter 2.

Chapter 1

Some Riemannian Geometry In this chapter we give a quick review of Riemannian geometry using the moving frame formalism. While we assume basic knowledge of Riemannian geometry, several computations will be carried out in full detail in order to acquaint the reader with notation and formalism. After having introduced frame and coframes, we will describe connection and curvature in terms of the connection and curvature forms. Symmetry properties of the curvature tensors will be described in detail and we will derive a number of identities that will be repeatedly used in the sequel. In particular, we will obtain the commutation rules for covariant derivatives up to fourth order. Along the way, we will introduce Einstein manifolds and prove Schur’s lemma. Then we introduce basic results for the Riemannian distance function from a fixed reference point o ∈ M , and discuss briefly the cut locus and some of its properties. We will then describe comparison results for the Laplacian of the Riemannian distance function and for the volume of geodesic balls in terms of lower bounds for the Ricci curvature. We point out that our treatment, which follows that of [PRS05b], does not use Jacobi fields. The chapter ends with a brief section on the geometry of immersed submanifolds to fix notation and terminology used in the second part of Chapter 2.

1.1

Preliminaries

Let (M, , ) be a Riemannian manifold of dimension m with metric , . The aim of this section is to fix notation and to describe the essential facts of the geometry of ´ Cartan’s formalism. The usefulness of this approach will become (M, , ) using E. apparent in the sequel, and it will prove to be particularly effective in Chapter 2 in the derivation of the formulae which express the change of the Riemannian, Ricci and scalar curvature under a conformal change of the metric.

P. Mastrolia et al., Yamabe-type Equations on Complete, Noncompact Manifolds, Progress in Mathematics 302, DOI 10.1007/978-3-0348-0376-2_1, © Springer Basel 2012

7

8

1.1.1

Chapter 1. Some Riemannian Geometry

Moving frames and the first structure equations

Let p ∈ M and (U, ϕ) a local chart such that U p, with coordinate functions x1 , . . . , xm , m = dim (M ) . If q is a generic point in U we have, at q, , = , ij dxi ⊗ dxj ,

(1.1)

where dxi denotes the differential of the function xi and , ij are the (local) ∂ ∂ components of the metric, defined by , ij = ∂x i , ∂xj . In relation (1.1), and throughout the book, we adopt the Einstein summation convention for repeated indices. Applying in q the Gram-Schmidt orthonormalization process we can find linear combinations of the 1-form dxk , which we will call θi , such that , = δij θi ⊗ θj ,

(1.2)

where δij is the Kronecker symbol. Since, as q varies in U , the previous process gives rise to coefficients that are C ∞ -functions of q, the set of 1-forms θi , i = 1, . . . , m, define an orthonormal system on U for the metric , , i.e., a (local) orthonormal (o.n.) coframe. We will sometimes write , =

m

(θi )2

i=1

instead of (1.2). We also define the (local) dual orthonormal frame {ei }, i = 1, . . . , m, as the set of m (local) vector fields satisfying, on the open set U , θj (ei ) = δij

(1.3)

(where δij is just a suggestive way of writing the Kronecker symbol, reflecting the position of the indexes in the pairing of θi and ej ). We have the following Proposition 1.1. Let θi be a local o.n. coframe on M , defined on an open set U ; then there exist unique 1-forms i θj , i, j = 1, . . . , m on U such that, ∀ i, j = 1, . . . , m, dθi = −θji ∧ θj ,

(1.4)

θij

(1.5)

θji +

= 0.

The forms θji are called the Levi-Civita connection forms associated to the o.n. i coframe θ . Remark. Equations (1.4) are classically known as the first structure equations.

1.1. Preliminaries

9

Proof. Let us assume the existence of the forms θji satisfying (1.5) and (1.4) and determine their expression. Of course θji = aijk θk for some aijk ∈ C ∞ (U ) (where C ∞ (U ) denotes the set of smooth functions defined on the open set U ), and (1.5) is equivalent to aijk + ajik = 0.

(1.6)

The 2-forms dθi can be written, for some (unique) coefficients bijk ∈ C ∞ (U ), as dθi =

1 i j b θ ∧ θk , 2 jk

bijk + bikj = 0.

Since (1.4) must hold we have: 1 1 i j b θ ∧ θk = −aijk θk ∧ θj = aijk θj ∧ θk = (aijk − aikj )θj ∧ θk . 2 jk 2 It follows that bijk = aijk − aikj .

(1.7)

Cyclic permutations of the indices i, j, k and use of (1.6) and (1.7) yield bkij = akij − akji = −aikj + ajki ;

(1.8)

bjki

(1.9)

=

ajki

−

ajik

=

ajki

+

aijk .

Adding (1.7) and (1.9) and subtracting (1.8) we get aijk =

1 i (b − bkij + bjki ). 2 jk

(1.10)

The previous relation determines the expression of the forms θji and also proves uniqueness. Now define 1 θji = (bijk − bkij + bjki )θk , (1.11) 2 where the bijk ’s satisfy bijk + bikj = 0. It is clear that aijk =

1 1 j bik − bkji + bikj = − bijk − bkij + bjki = −aijk , 2 2

thus (1.6) is met, and then the θji defined in (1.11) satisfy (1.5); it is also immediate to verify that they satisfy (1.4).

10


1.1.2

Covariant derivative of tensor fields

The Levi-Civita connection forms are the starting point to define a covariant derivative of tensor fields in the following way. Following standard notation we denote with Tp M the tangent space at p ∈ M and with Tp∗ M the cotangent space at p. We recall that a tensor field T of type (r, s) is a law that assigns to all points p ∈ M a multilinear map Tp : Tp∗ M × · · · × Tp∗ M × Tp M × · · · × Tp M → R s times

r times

with the usual differentiability request with respect to the variable p (see e.g. [Lee03]). The set of tensor fields of type (r, s) will be denoted by Trs (M ). Let us begin considering the case of a vector field X on M , i.e., a tensor of type (1, 0). We denote with X(M ) the set of all (smooth) vector fields on M . Let θi a local o.n. coframe and {ei } the dual frame. Definition 1.2. The covariant derivative of the vector field X, ∇X, is the tensor field of type (1, 1) ∇X defined in the following way: if X = X i ei , ∇X = (dX i ) ⊗ ei + X i ∇ei , having defined ∇ei = θij ⊗ ej . Setting Xki θk = dX i + X j θji , ∇X can be then written as ∇X = (dX i + X j θji ) ⊗ ei = Xki θk ⊗ ei , and Xki is said to be the covariant derivative of the coefficient X i . If Y ∈ X(M ) we define the covariant derivative of X in the direction of Y as the vector field ∇Y X = ∇X(Y ), which in components reads as ∇Y X = Xki θk (Y )ei = Xki Y k ei . Definition 1.3. The divergence of the vector field X ∈ X(M ) is the trace of ∇X, that is, (1.12) div X = tr (∇X) = ∇ei X, ei = Xii . Analogously, for a 1-form ω (i.e., a tensor of type (0, 1)), we have:

1.1. Preliminaries

11

Definition 1.4. The covariant derivative of the 1-form ω, ∇ω, is the tensor field of type (0, 2) defined in the following way: if ω = ωi θi , ∇ω = (dωi ) ⊗ θi + ωi ∇θi , with ∇θi = −θji ⊗ θj . Note that, setting ωik θk = dωi − ωj θij , it follows that ∇ω = ωik θk ⊗ θi . If Y ∈ X(M ) we define the covariant derivative of ω in the direction of Y as the 1-form ∇Y ω = ∇ω(Y ), which in components reads as ∇Y ω = ωik θk (Y )θi = ωik Y k θi . For a 0-form f ∈ C ∞ (M ) we set ∇f = df

(exterior differential of f ).

We point out that this notation may give rise to some ambiguity; indeed, in the literature (and also in this book) ∇f often denotes the gradient of f , i.e., the vector field dual to the 1-form df : in this case, using standard notation (see for instance [Lee97]), we can write ∇f = (df ) , where is the sharp map from the ∗ cotangent bundle T M to the tangent bundle T M defined by (df ) , Y = ∇f, Y = df (Y ) = Y (f ), for all Y ∈ X(M ). Note that, in components, setting df = fj θj for some smooth i coefficients fj , we have (∇f ) = δ ij (df )j = δ ij fj = fi (that is, in an orthonormal frame, differential and of a function have the same coefficients with respect gradient to the (dual) basis θi and {ei }). Finally, ∇ can be extended in a natural way to a generic tensor T , in order to define a connection on each tensor bundle Trs (M ): this extension of ∇ satisfies the Leibniz rule and some other nice properties, like the commutativity with the trace on any pair of indices (see again [Lee97]). Although the covariant derivative has been introduced by means of ilocally defined objects, it is possible to show (using the transformation laws of θ and i θj as the coframe changes) that the new tensor field thus obtained is globally defined. Remark. One can verify that the previous definition matches the “canonical” one usually given in terms of the Koszul formalism (see for example [Lee97], [Pet06a]). Indeed, as we will see before long, the operator ∇ coincides precisely with the Levi-Civita connection associated to the metric , of M .

12

1.1.3


Meaning of the first structure equations

We now want to discuss the geometric meaning of condition (1.5), namely θji + θij = 0. To this purpose let us compute the covariant derivative of the metric tensor , . Using the Leibniz rule, and recalling that for everytangent vector Xp = X(p) ∈ Tp M (with p in the domain of the o.n. coframe θi ) it holds that ∇Xp θi = −θji (Xp )θj , we have ∇Xp , = ∇Xp (δij θi θj ) = δij (∇Xp θi ⊗ θj + θi ⊗ ∇Xp θj ) = δij (−θki (Xp )θk ⊗ θj − θkj (Xp )θi ⊗ θk ) = −θki (Xp )θk ⊗ θi − θki (Xp )θi ⊗ θk = −(θki + θik )(Xp )θk ⊗ θi , and therefore ∇ , = 0 if and only if θji + θij = 0, i.e., condition (1.5) is equivalent to the parallelism of the metric (in other words: ∀ X, Y, Z ∈ X(M ), X Y, Z = ∇X Y, Z + Y, ∇X Z). On the other hand, the first structure equations (1.4) tell us that the metric is torsion-free. Indeed, let X and Y be two vector fields on M and [X, Y ] their Lie bracket, defined by [X, Y ](f ) = X(Y (f )) − Y (X(f )),

∀f ∈ C ∞ (M ).

(1.13)

We claim that condition [X, Y ] = ∇X Y − ∇Y X

∀ X, Y ∈ X(M )

(1.14)

is equivalent to the validity of (1.4). Note that the left-hand side of (1.14) is independent of the choice of a metric on M . Since the torsion of a generic connection ∇ on M is the (0, 2) tensor field Tor(X, Y ) = ∇X Y − ∇Y X − [X, Y ], this justifies the expression “torsion-free” used above. To prove the equivalence, recall that the exterior differential of a 1-form ω is intrinsically defined by dω(X, Y ) = X(ω(Y )) − Y (ω(X)) − ω([X, Y ]); moreover, as a consequence of the definition of covariant derivative, (∇X ω)(Y ) = X(ω(Y )) − ω(∇X Y ), so that

X θi (Y ) − θi (∇X Y ) = (∇X θi )(Y ) = −θji (X)θj (Y ),

(1.15)

1.1. Preliminaries

13

that is,

X θi (Y ) + θji (X)θj (Y ) = θi (∇X Y ). Then we compute dθi + θji ∧ θj (X, Y ), that is dθi (X, Y ) + θji ∧ θj (X, Y ) = X(θi (Y )) − Y (θi (X)) − θi ([X, Y ]) + θji (X)θj (Y ) − θji (Y )θj (X) = X(θi (Y )) + θji (X)θj (Y )) − Y (θi (X)) − θji (Y )θj (X)) − θi ([X, Y ]) = θi (∇X Y − ∇Y X − [X, Y ]), and the claim follows. Remark. By the fundamental theorem of Riemannian geometry (see for instance [Lee97] or [Pet06b]), we deduce that the connection ∇ coincides, as we said previously, with the Levi-Civita connection of the metric , .

We now define the Lie derivative of Y in the direction of X to be LX Y = [X, Y ], so that condition (1.14) can be written in the form LX Y = ∇X Y − ∇Y X.

(1.16)

LX f = X(f )

(1.17)

(LX ω)(Y ) = LX (ω(Y )) − ω(LX Y ),

(1.18)

Setting also ∞

for f ∈ C (M ), and

if ω is a 1-form, we can extend LX to a generic tensor field requiring R-linearity and the validity of the Leibniz rule (see also [Lee03], [Pet06a]). Using (1.15), we compute the Lie derivative of the metric in the direction of X, LX , (note that this latter has to be a covariant tensor of order 2, that is, a (0, 2)-tensor): (LX , )(Y, Z) = ((LX θi ) ⊗ θi + θi ⊗ (LX θi ))(Y, Z) = θi (Z)(LX θi )(Y ) + θi (Y )(LX θi )(Z) = θi (Z) LX (θi (Y )) − θi (LX Y ) + θi (Y ) LX (θi (Z)) − θi (LX Z) = θi (Z)X(θi (Y )) − θi (Z)θi (∇X Y − ∇Y X) + θi (Y )X(θi (Z)) − θi (Y )θi (∇X Z − ∇Z X) = θi (Z)(∇X θi )(Y ) + θi (Y )(∇X θi )(Z) + θi (Z)θi (∇Y X) + θi (Y )θi (∇Z X) = (∇X θi ⊗ θi + θi ⊗ ∇X θi )(Y, Z) + ∇Y X, Z + Y, ∇Z X = (∇X , )(Y, Z) + ∇Y X, Z + Y, ∇Z X = ∇Y X, Z + Y, ∇Z X ,

14


where in the last equality we have used the fact that the metric is parallel with respect to the Levi-Civita connection. Thus, we have proved the useful identity (LX , )(Y, Z) = ∇Y X, Z + Y, ∇Z X

(1.19)

for all X, Y, Z ∈ X(M ), which will be repeatedly used in the sequel. Note that equation (1.19) in components reads as (LX , )ij = ∇ei X, ej + ei , ∇ej X = Xij + Xji . (1.20) It can be proved that the Lie derivative of Y in the direction of X has the following geometric meaning (see e.g. [Lee03]): (ϕ−t )∗ Yϕt (p) − Yp d (ϕ−t )∗ Yϕt (p) = lim , (LX Y )p = t→0 dt t=0 t where ϕt is the local flow generated by X and (ϕt )∗ is the push-forward. The analogous applies to LX h, with h a generic tensor field (see also Chapter 2 for the special case of LX , ).

1.1.4

Curvature: the second structure equations

i We i now consider the second structure equations. Let θ be a local o.n. frame and θj the corresponding Levi-Civita connection forms. The curvature forms Θij associated to the coframe are defined through the second structure equations, dθji = −θki ∧ θjk + Θij .

(1.21)

Obviously the Θij are 2-forms. Since, according to (1.5), θji + θij = 0, it follows immediately that the Θij ’s satisfy the same antisymmetry condition Θij + Θji = 0.

(1.22)

Using the basis {θi ∧ θj }, 1 ≤ i < j ≤ m, of the space of skew-symmetric 2-forms 2 (U ) on the open set U , we may write Θij =

1 i k i Rjkt θk ∧ θt Rjkt θ ∧ θt = 2

(1.23)

k 0 in (0, T ) and ψ ϕ ≤ , ψ ≥ ϕ on (0, T ). ϕ ψ

(1.69)

Proof. Since ψ (0) > 0, ψ > 0 in a neighborhood of 0. We observe in passing that if G is assumed to be nonnegative, then, integrating (1.68) ii), we have r ψ (r) ≥ ψ (0) + G(s)ψ(s) ds, 0

so that ψ is positive in the interval where ψ ≥ 0, and we conclude that, in fact, ψ > 0 on (0, +∞). In the general case, where no assumption is made on the sign of G, we let β = sup {t : ψ > 0 in (0, t)}; τ = min {β, T }. The function ψ ϕ − ψϕ ∈ C 0 ([0, +∞)) vanishes in r = 0, and it satisfies (ψ ϕ − ψϕ ) = ψ ϕ − ψϕ ≥ 0 in (0, τ ). Thus, ψ ϕ − ψϕ ≥ 0 on [0, τ ), and, dividing through by ϕψ, we deduce that ψ ϕ ≥ in (0, τ ). ψ ϕ

1.2. Comparison theorems

25

Integrating between ε and r, 0 < ε < r < τ , yields ϕ(ε) ψ(r) ψ(ε)

ϕ(r) ≤ and, since lim

ε→0+

ϕ(ε) ϕ (0) = ≤ 1, ψ(ε) ψ (0)

we conclude that in fact ϕ(r) ≤ ψ(r) in [0, τ ). Since ϕ > 0 in (0, T ) by assumption, this in turn forces τ = T , for, otherwise, τ = β < T and we would have ϕ(β) > 0, while, by continuity, ψ(β) = 0, a contradiction. Lemma 1.10. Let G ∈ C 0 ([0, +∞)) and let gi ∈ C 1 ((0, Ti )), i = 1, 2 be solutions of the Riccati differential inequalities i) g1 +

g12 − αG ≤ 0; α

ii) g2 +

g22 − αG ≥ 0 α

(1.70)

satisfying the condition gi (t) =

α + O(1) as t → 0+ t

(1.71)

for some α > 0. Then T1 ≤ T2 and g1 (t) ≤ g2 (t) in (0, T1 ). Proof. Since gi = α−1 gi satisfy the conditions in the statement with α = 1, without loss of generality we assume α = 1. Observe that the functions gi (s) − 1s are bounded and integrable in a neighborhood of s = 0, thus we define ϕi ∈ C 2 ((0, Ti )) ∩ C 1 ([0, Ti )) on [0, Ti ), by setting ϕi (t) = te

t 0

(gi (s)− 1s ) ds

.

Then ϕi (0) = 0, ϕi > 0 on (0, Ti ) and straightforward computations show that ϕi (t) = gi (t)ϕi (t), and

ϕi (0) = 1

ϕ1 ≤ Gϕ1

on (0, T1 );

ϕ2

on (0, T2 ).

≥ Gϕ2

An application of Lemma 1.9 shows that T1 ≤ T2 and g1 = (0, T1 ), as required.

ϕ1 ϕ1

≤

ϕ2 ϕ2

= g2 on

We are now ready to prove the next Laplacian comparison theorem, which is a simplified (but sufficient for our purposes) version of that appearing in [PRS08]:

26


Theorem 1.11. Let (M, , ) be a complete manifold of dimension m ≥ 2. Having fixed a reference point o ∈ M , let r(x) = distM (x, o). Assume that the radial Ricci curvature Ric(∇r, ∇r) of M satisfies Ric(∇r, ∇r) ≥ −(m − 1)G(r)

(1.72)

for some nonnegative function G ∈ C 0 ([0, +∞)). Let h ∈ C 2 ([0, +∞)) be a solution of the problem h − Gh ≥ 0, (1.73) h(0) = 0, h (0) = 1. Then the inequality Δr(x) ≤ (m − 1)

h (r(x)) h(r(x))

(1.74)

holds pointwise on M \({o} ∪ cut(o)) and weakly on all of M . Proof. Fix any x ∈ M \({o} ∪ cut(o)) and let γ : [0, l] → M be a minimizing geodesic from o to x parameterized by arclength. We then arrive to (1.67), where the differential inequality is in (0, l]. Since g = (m − 1) hh satisfies g (t) +

g(t)2 ≥ (m − 1)G(t) on (0, +∞) m−1

(1.75)

and (1.71) with α = m − 1, an application of Lemma 1.10 to (1.67) and (1.75) gives h (t) ϕ(t) ≤ (m − 1) in (0, l]. h(t) Thus, in particular, since γ(l) = x and r(x) = l, Δr(x) ≤ (m − 1)

h (r(x)) , h(r(x))

showing the validity of (1.74) pointwise within the cut locus. It remains to show the validity of (1.74) weakly in all of M , which is guaranteed by the following lemma. Lemma 1.12. Set Do = M \ cut(o) and suppose that Δr ≤ α(r) pointwise on Do \{o},

(1.76)

for some α ∈ C 0 ((0, +∞)). Let v ∈ C 2 (R) be nonnegative and set u(x) = v(r(x)) on M . Suppose either i) v ≤ 0

or

ii) v ≥ 0.

(1.77)

Then we respectively have i) Δu ≥ v (r) + α(r)v (r); weakly on M .

ii) Δu ≤ v (r) + α(r)v (r)

(1.78)


27

Proof. Let Eo be the maximal star-shaped domain in To M on which expo is a diffeomorphism onto its image Do , so that we have cut(o) = ∂(expo (Eo )). Since Eo is a star-shaped domain, we can exhaust Eo by a family {Eon } of relatively compact, star-shaped domains with smooth boundary. We set Don = expo (Eon ) so that n Do ⊂ Don+1 and Don = Do . n

The fact that each Eon is star-shaped implies ∂r = ∇r, νn > 0 on ∂Don , ∂νn

(1.79)

where νn denotes the outward unit normal to ∂Don . Now we assume the validity of (1.77) i). Since r ∈ C ∞ (Don \{o}), computing we get Δu ≥ v + α(r)v

pointwise on Don \{o}.

(1.80)

Let 0 ≤ ϕ ∈ C0∞ (M ), where C0∞ (M ) denotes the set of smooth functions with compact support on M . We claim that, ∀ n, uΔϕ ≥ (v + α(r)v )ϕ + εn , Don

Don

where εn → 0 as n → +∞. Since M = Do ∪ cut(o) and cut(o) has measure 0, inequality (1.78) i) will follow by letting n → +∞. To prove the claim we fix δ > 0 small and we apply the second Green formula (see e.g. [GT01]) on Don \Bδ (o) to obtain ∂u ∂ϕ ϕ , (1.81) uΔϕ = ϕΔu − −u ∂νn ∂νn Don \Bδ (o) Don \Bδ (o) ∂Don \∂Bδ (o) where νn is the outward unit normal to ∂Don \∂Bδ (o). We note that, according to (1.77) i) and (1.79), ∂u ∂r = v (r) ≤ 0 on ∂Don . ∂νn ∂νn Using this, (1.79) and (1.81) we obtain uΔϕ ≥ (v + α(r)v )ϕ + εn + Iδ , Don

Don

with

∂ϕ , ∂νn [uΔϕ − (v + α(r)v )ϕ] − εn =

Iδ =

u

∂Don

Bδ (o)

! ∂Bδ (o)

" ∂ϕ ∂u u −ϕ , ∂r ∂r

28


∂· means ∇r, ∇·. Clearly, Iδ → 0 as δ ↓ 0+ ; on the other where the notation ∂r ∞ hand, since ϕ ∈ C0 (M ) and cut(o) has measure 0, using the divergence and Lebesgue theorems we see that, as n → +∞, εn = div(u∇ϕ) → div(u∇ϕ) = div(u∇ϕ) = 0. Don

Do

M

This proves the claim and the validity of (1.78) i). The case (1.77) ii) and (1.78) ii) can be dealt with in a similar way. Remark. We note that, for the above proofs to work, it is not necessary that (1.72) holds on the entire M : instead, for instance, if (1.72) is valid on BR (o), then (1.74) holds on BR (o)\({o} ∪ cut(o)) and weakly on BR (o).

1.2.4

The Bishop-Gromov comparison theorem

We now show how to get from the previous results a (somewhat generalized) version of what is known in the literature as the Bishop-Gromov comparison theorem (see also [PRS08]). We denote with vol BR (o) and vol ∂BR (o) the volume of the geodesic ball BR (o) and of its boundary ∂BR (o), respectively. Theorem 1.13. Let (M, , ) be a complete, m-dimensional Riemannian manifold satisfying Ric (∇r, ∇r) ≥ −(m − 1)G(r) on M (1.82) for some G ∈ C 0 ([0, +∞)), G ≥ 0, where r(x) = dist(x, o). Let h ∈ C 2 ([0, +∞)) be the nonnegative solution of the problem h − G(t)h = 0, (1.83) h(0) = 0, h (0) = 1. Then, for almost every R > 0, the function R →

vol ∂BR (o) h(R)m−1

(1.84)

is nonincreasing, and vol ∂BR (o) ≤ ωm h(R)m−1 ,

(1.85)

where ωm is the volume of the unit sphere in Rm . Moreover, vol BR (o) R → # R h(t)m−1 dt 0 is a nonincreasing function on (0, +∞).

(1.86)


29

Since it will be used in the proof of Theorem 1.13, and also in the next chapters, we first recall the useful co-area formula. We denote with W 1,1 (M ) the Sobolev space consisting of functions in L1 (M ) with (weak) gradient also in L1 (M ). We also denote with Aft the t-level set (t ∈ R) of a function f on M , i.e., Aft = {x ∈ M |f (x) = t}. Following Schoen and Yau (see [SY94], p.89) we can state the following Proposition 1.14. Let M be a compact Riemannian manifold with boundary and f ∈ W 1,1 (M ). For any nonnegative measurable function g on M the following formula holds: % +∞ $ g dt. (1.87) g= −∞ M Aft |∇f | For a proof see the classical [Fed69]. Note that, in particular, if f (x) = r(x) = distM x, o, equation (1.87) becomes

D

$

%

g= M

g dt, 0

(1.88)

∂Bt (o)

where D = supM r(x). Proof of Theorem 1.13. In case o is a pole of M (see [GW79] and also the proof of Theorem 6.8 in Chapter 6) one integrates the divergence of the radial vector field X = h(r(x))−m+1 ∇r on concentric balls BR (o), and uses the divergence and Laplacian comparison theorems. However, in general, objects are nonsmooth and inequalities are intended in the sense of distributions. Therefore, we have to take some extra care. The Laplacian comparison theorem asserts that Δr(x) ≤ (m − 1)

h (r(x)) h(r(x))

(1.89)

pointwise on the open, star-shaped, full measured set M \ cut(o) and weakly on all of M . Thus, ∀ 0 ≤ ϕ ∈ Lip0 (M ), −

∇r, ∇ϕ ≤ (m − 1)

h (r(x)) ϕ. h(r(x))

(1.90)

∀ ε > 0, consider the radial cut-off function ϕε (x) = ρε (r(x))h(r(x))−m+1 ,

(1.91)

30


where ρε is the piecewise linear function ⎧ ⎪ 0, if ⎪ ⎪ ⎪ t−r ⎪ ⎪ if ⎨ ε , ρε (t) = 1, if ⎪ ⎪ R−t ⎪ if ⎪ ε , ⎪ ⎪ ⎩0, if Note that

t ∈ [0, r) t ∈ [r, r + ε) t ∈ [r + ε, R − ε) t ∈ [R − ε, R) t ∈ [R, +∞).

(1.92)

' χR−ε, R χr, r+ε h (r(x)) − + − (m − 1) ρε h(r(x))−m+1 ∇r, ε ε h(r(x))

&

∇ϕε =

for almost all x ∈ M , where χs, t is the characteristic function of the annulus Bt (o)\Bs (o). Therefore, using ϕε into (1.90) and simplifying, we get 1 1 −m+1 h(r(x)) ≤ h(r(x))−m+1 . ε BR (o)\BR−ε (o) ε Br+ε (o)\Br (o) Using the co-area formula (1.87)) we deduce that 1 R 1 r+ε −m+1 vol(∂Bt (o)) h(t) ≤ vol(∂Bt (o)) h(t)−m+1 ε R−ε ε r and, letting ε ↓ 0,

vol(∂Br (o)) vol(∂BR (o)) ≤ h(R)m−1 h(r)m−1

(1.93)

for almost all 0 < r < R. Letting r → 0 and recalling that h(r) ∼ r and vol(∂Br ) ∼ ωm rm−1 as r → 0 (which can be deduced, for instance, integrating equation (1.66) on a geodesic ball and using the divergence theorem and Gauss’ lemma), we conclude that, for almost any R > 0, vol ∂BR (o) ≤ ωm h(R)m−1 . To prove the second statement we note that, as observed in [CGT82], for general real valued functions f (t) ≥ 0, g(t) > 0, if t →

f (t) g(t)

is decreasing, then t →

t

0t 0

f g

is

decreasing. Indeed, since f /g is decreasing, if 0 < r < R, r R r R r R r R f R f (r) r f f g= g g≥ g g≥ g g = g f g r g(r) 0 g 0 r 0 r 0 r 0 r whence r f 0

R 0

r

g= 0

r

f 0

r

g+ 0

R

g≥

f r

r 0

r

f 0

r

g+ 0

R

g r

r

f=

R

g 0

f. 0


31

In particular, applying this observation to (1.93) and using the co-area formula (1.87) we deduce that vol Br (o) r → # r h(t)m−1 dt 0

is decreasing, concluding the proof.

We conclude with the following analytical result whose proof can be found in [PRS08]: Proposition 1.15. Assume h is a solution of h − H 2 (1 + r2 )δ/2 h = 0, h(0) = 0, h (0) = 1 where H > 0 and δ ≥ −2. Set H, ( H = 1 1 + 4H 2 ), 2 (1 +

if δ > −2 if δ = −2.

Then,

h (r) ≤ H rδ/2 (1 + o(1)) as r → +∞. h Moreover, there exists a constant C > 0 such that for r > 1,

⎧ 2H 1+δ/2 ⎪ exp (1 + r) , if δ ≥ 0 ⎪ 2+δ ⎨

2H −δ/4 1+δ/2 h(r) ≤ C r , if − 2 < δ < 0 exp 2+δ r ⎪ ⎪ ⎩ H r , if δ = −2.

1.2.5

The Hessian comparison theorem

For the sake of completeness we recall here the following Hessian comparison theorem; however, since it will only be used a very few times in the sequel, we only give its statement (the interested reader can find the proof in [PRS08]). Recall that the radial sectional curvature Krad of a manifold is the sectional curvature of a 2-plane containing ∇r. Theorem 1.16. Let (M, , ) be a complete manifold of dimension m. Having fixed a reference point o ∈ M , let r(x) = distM (x, o), and let Do = M \ cut(o) be the domain of the normal geodesic coordinates centered at o. Given a smooth even function G on R, let h be the solution of the Cauchy problem h − Gh = 0, (1.94) h(0) = 0, h (0) = 1,

32


and let I = [0, r0 ) ⊆ [0, +∞) be the maximal interval where h is positive. If the radial sectional curvature of M satisfies Krad ≥ −G(r(x)) then Hess(r) ≤

on Br0 (o),

h (r(x)) { , − dr ⊗ dr} h(r(x))

(1.95)

(1.96)

on Do \ {o} ∪ Br0 (o) in the sense of quadratic forms. On the other hand, if Krad ≤ −G(r(x)) then Hess(r) ≥

on Br0 (o),

h (r(x)) { , − dr ⊗ dr}. h(r(x))

(1.97)

(1.98)

Remark. By taking traces in Theorem 1.16 we immediately obtain the corresponding estimates for Δr. However, as we have seen in Theorem 1.11, the estimate from above for the Laplacian of the distance function holds under the weaker assumption that the radial Ricci curvature (and not the full radial sectional curvature) is bounded from below by −(m − 1)G(r(x)). Furthermore the estimate in this latter case can be extended, in weak form, to the entire manifold. This is not the case for the above estimates on Hess(r).

1.3 Some formulas for immersed submanifolds Let (N, , ) and M be respectively a Riemannian manifold and a manifold of dimensions n and m, with m ≤ n. Let f : M → N be an immersion and let f ∗ , be the metric induced on M by f , where f ∗ denotes the pull-back (note that f ∗ , is indeed a metric since f is an immersion). If g is a given metric on M and f : M → N is an immersion we will say that f is an isometric immersion if g = f ∗ , . To simplify notation we use the symbol , on M to denote the induced metric; more generally, from now on we shall omit the pull-back notation, being clear from the context where forms or tensors are considered. We fix the following indexes convention: 1 ≤ i, j, k, . . . ≤ m,

m + 1 ≤ α, β, γ, . . . ≤ n,

1 ≤ a, b, c, . . . ≤ n.

Definition 1.17. Given the isometric immersion f : M → (N, , ), a Darboux coframe along f is a local o.n. coframe θi on N such that θα = 0

on M.

(1.99)

In particular, for a Darboux coframe along f we have , =

m i 2 θ i=1

on M.

(1.100)

1.3. Some formulas for immersed submanifolds

33

The dual {ea } of a Darboux coframe is called a Darboux frame along f and condition (1.99) is equivalent to say that the vectors {ei } (locally) span (the image of) T M in T N , while {eα } are orthogonal to T M (and span in fact T M ⊥ , the normal bundle that will be defined afterward). The existence of Darboux coframes along f can, of course, be proved analytically but the above geometric meaning is “evidence” of their existence. We let {θba } be the Levi-Civita connection forms relative to {θa } on N . Then θji on M are the Levi-Civita connection forms of the induced metric. Indeed, obviously, θji + θij = 0

on M.

(1.101)

Furthermore, from the first structure equations on N and by (1.99), dθi = −θji ∧ θj .

(1.102)

We now recall the following elementary but useful Lemma 1.18 (Cartan’s lemma). Let U ⊂ M be an open set of the Riemannian manifold (M, , ). Let {θi } be a local basis of T ∗ U , and assume that a set of 1-forms {ωJi } on U , where J is any set of indices, satisfies i ωJi ∧ θi = 0. Then, there exist smooth functions biJ,k on U such that ωJi = biJ,k θk

and

biJ,k = bkJ,i ,

that is, the matrix B = (biJ,k )ik is an n × n symmetric matrix of 1-forms. Proof. We can write ωJi as ωJi = biJ,k θk for some smooth functions biJ,k on U . Then from i ωJi ∧ θi = 0 we deduce biJ,k θk ∧ θi = (biJ,k − bkJ,i )θk ∧ θi , 0= i,k

i 0. By contradiction, suppose c = 1. Without loss of generality, we can assume c > 1 (indeed, in case c < 1 it suffices to consider ϕ−1 ). Now, for every n ∈ N, let ϕ(n) : M → M be the n-th iterate of ϕ. Then

∗ ϕ(n) , = c2n , , proving that, for any fixed p = q ∈ M ,

c2n d (p, q) = d ϕ(n) (p) , ϕ(n) (q) ≤ diam (M ) < +∞. Since c > 1, taking the limit as n → +∞, we obtain the desired contradiction.

40

Chapter 2. Pointwise conformal metrics We are now ready to prove the following

Theorem 2.2. Let (M, , ) be a compact manifold of dimension m ≥ 3 (for the ease of exposition) with scalar curvature s(x) ≤ 0. Let ψ : M → M be a conformal diffeomorphism with scalar curvature S(x). Thus ψ is an isometry if and only if S(x) = kS(x) for some k ∈ (0, +∞). Proof. If ψ is an isometry, clearly k = 1. Vice versa, to simplify notation let m+2 1−σ for some a ∈ (0, +∞). Since ψ is conformal cm = 4 m−1 m−2 , σ = m−2 and k = a and m ≥ 3, from equation (2.9) we deduce cm Δu = S(x) u − a1−σ uσ (2.12) with u > 0, u ∈ C ∞ (M ) such that 4

ψ ∗ , = u m−2 , . Define the vector field

!

" u σ−1 W = cm − 1 ∇u; a

computing its divergence, using (2.12) and the divergence theorem we have

!

"2 u σ−1 1 u σ−1 2 − 1 S(x)u = cm (σ − 1) |∇u| . a u a M M

Since S(x) ≤ 0 it follows that |∇u| ≡ 0 and u ≡ a, that is ψ : M → M is a homothety. The result now follows from Lemma 2.1. We shall come back to this kind of problems again in Corollary 2.9 below and later in the complete noncompact case.

2.1.2

The Kazdan-Warner obstruction

Let now T denote the traceless Ricci tensor, that is (see equation (1.38)) T = Ric −

S , . m

T enables us to immediately find an obstruction to the existence of a conformally deformed metric as in (2.8) or (2.10) with assigned scalar curvature S(x). We shall consider the case m ≥ 3 so that we shall provide a necessary condition for the existence of a positive solution on M of equation (2.9). Indeed, with respect to a local orthonormal coframe θi we have Tij = Rij −

S δij . m

2.1. The Yamabe equation

41

On the other hand, from (1.36), Rli,i =

1 Sl . 2

Thus, tracing the covariant derivative Tij,k = Rij,k −

Sk δij m

with respect to j and k yields Sk m−2 (2.13) δik = Si . m 2m Now, let X be a vector field on M and consider the vector field W associated to the 1-form T (X, ·) using the duality induced by the metric, namely, Tik,k = Rik,k −

W = T (X, ·) = X i Tij ej ,

Tij = Tij , (2.14) j where {ej } is the orthonormal frame dual to the coframe θ . Since the covariant derivative satisfies the Leibniz rule, the divergence of W is given by div W = Wkk = Xki Tik + X i Tik,k .

(2.15)

On the other hand, using the symmetry of Tik and (1.20), 1 1 Tik Xki = Xki + Xik Tik = (LX , )ik Tik . 2 2 It follows that m−2 1 X(S). (2.16) div W = (LX , )ik Tik + 2 2m We now recall that a vector field X is said to be conformal if it generates a local 1-parameter group ϕt of local conformal diffeomorphisms, that is (ϕ∗t , )p = ρt (p) , p ,

p∈M

for some positive function ρt . Using the formula which expresses the Lie derivative LX as a derivative along the flow generated by X, namely, ϕ∗t , − , , t→0 t

LX , = lim

(see e.g. [Lee03], p. 472 ff.) one proves that X is conformal if and only if LX , = λ , for some function λ = λ(p), which, using (1.19), is easily seen to be equal to 2 m (div X). Substituting into (2.16) and using the fact that T is traceless, we deduce that m−2 div W = X(S) (2.17) 2m whenever X is a conformal vector field on M . Thus we have the next result which is known as the Kazdan-Warner obstruction (see [KW74a] and [BE87]):

42

Chapter 2. Pointwise conformal metrics

Theorem 2.3. Let (M, , ) be a compact manifold of dimension m ≥ 3 and , Then, for each a metric on M conformally related to , with scalar curvature S. conformal vector field X on (M, , ) we have

+ = 0. dvol X(S)

(2.18)

M

Proof. Since X is conformal with respect to the metric , , it generates a flow of local diffeomorphisms which are conformal transformations for the metric , and therefore also for the conformally related metric , . It follows that X is conformal also in the metric , . Considering the analogous (2.17) in the metric , and integrating with the aid of the divergence theorem we obtain (2.18). Remark. As a further application of the method of a moving frame, we give here an explicit formula for LX , in terms of LX , and , , from which it can be again deduced that a vector field X conformal w.r.t. the metric , is conformal also in the metric , . Indeed we have the following Lemma 2.4. Let X ∈ X(M ) be a vector field on the Riemannian manifold (M, , ), and let , = ϕ2 , , ϕ > 0, be a conformally deformed metric. Then LX , = ϕ2 LX , + 2ϕ X, ∇ϕ , .

(2.19)

Proof. We choose a local o.n. coframe θi with associated frame {ei }. First we observe that (2.2) implies that ei = ϕ−1 ei ; then, with the notation of section 1.1.2 we have i ei X = X i ei = X (2.20) and ∇X = Xki θk ⊗ ei ,

=X ki θk ⊗ ei ∇X

(2.21)

i θk = (dX i + X j θi ). From (2.20) we deduce with Xki θk = (dX i + X j θji ) and X j k i = ϕX i , X

(2.22)

while a computation using (2.21) and (2.3) shows that ki = Xki + ϕ−1 X i ϕk + X j ϕj δik − ϕi X k , X

(2.23)

ki + X ki = Xki + Xki + 2ϕ−1 X j ϕj δik . X

(2.24)

which implies

Equation (2.19) now follows easily from (2.24) and (1.20).


43

Example. The m-dimensional standard sphere Sm , m ≥ 3, can be realized, outside the North pole N and the South pole S, as the model manifold (see also Chapter 4, section 4.3)

2 (0, π) × Sm−1 , dr ⊗ dr + (sin r) dθ2 , where dθ2 denotes the standard metric on Sm−1 . Using this isometric representation, a conformal vector field X on Sm is given by XN = XS = 0.

X = sin r∇r,

Now, consider any smooth, radial function s (r (x)) on Sm . Then, by the co-area formula (1.87), Sm

,

d s + X, ∇r dvol dr Sm π d s + dr sin r dvol = 0 ∂Br dr π d s + (∂Br ) . sin r vol = dr 0

+ = X( s) dvol

Accordingly, if s (r) is monotonic and nonconstant, we deduce Sm

+ = 0. X( s) dvol

It follows from Theorem 2.3 that any nonconstant monotonic function s (r (x)) cannot be the scalar curvature of a pointwise conformal deformation of the canonical metric of Sm .

2.1.3

The Weyl and Cotton tensors

We now establish a relation between − S T = Ric , m

and

T = Ric −

S , , m

(2.25)

that will be used below in the proof of Theorem 2.8. Suppose, once again, that (2.1) holds, that is, , = ϕ2 , . Using (2.6) and (2.7), after some computations we obtain

44


− S T = Ric , m ' & 1 1 Hess(ϕ) − 2 2 dϕ ⊗ dϕ = T − (m − 2) ϕ ϕ & ' (2.26) m − 2 Δϕ |∇ϕ|2 , + −2 2 m ϕ ϕ & ' Δ(ϕ−1 ) = T + (m − 2)ϕ Hess(ϕ−1 ) − , . m Next, we observe that using (2.4) and (2.5) we are able to detect a part of the curvature tensor which is naturally invariant with respect to a conformal change of the metric. Indeed, from (2.5) we have & " ' ! ϕjt ϕt ϕj jt − (m − 3) ϕl ϕl + ϕll δtj , (m − 2) = Rjt − ϕ2 R −2 2 ϕ ϕ ϕ2 ϕ and inserting into (2.4) gives !

" 1 j j 2 i i i Rik δt − Rjk δt + Rjt δk − Rit δk ϕ Rjkt − m−2

1 i − = Rjkt Rik δtj − Rjk δti + Rjt δki − Rit δkj m−2 & '

Δϕ 1 |∇ϕ|2 i j 2 + δk δt − δti δkj . + (m − 4) 2 m−2 ϕ ϕ On the other hand, by (2.7) 2 and we obtain

Δϕ 1 |∇ϕ|2 + (m − 4) 2 = − ϕ2 S − S ϕ ϕ (m − 1)

!

1 j jk δti + R jt δki − R it δ j Rik δt − R k m−2 .

S + δki δtj − δti δkj (m − 1)(m − 2)

1 i − = Rjkt Rik δtj − Rjk δti + Rjt δki − Rit δkj m−2

S δki δtj − δti δkj . + (m − 1)(m − 2) 2

ϕ

i jkt R −

It follows that the (1, 3)-tensor, called the Weyl tensor, defined as !

1 W = Riem − Rik δtj − Rjk δti + Rjt δki − Rit δkj m−2

" S i j i j − δ δ − δt δk θ k ⊗ θ t ⊗ θ j ⊗ e i (m − 1)(m − 2) k t

(2.27)


45

is invariant under a conformal change of the metric. Remark. It is worth noting that the corresponding (0, 4)-version of W is not conformally invariant. Taking covariant derivatives we obtain 1 (Rik,t δjs − Ris,t δjk + Rjs,t δik − Rjk,t δis ) m−2 St (δik δjs − δis δjk ), + (m − 1)(m − 2)

i i Wjks,t = Rjks,t −

t so that taking the divergence with respect to the first index, that is, Wjks,t , using (1.35) and (1.36) we get

1 1 1 1 Rtk,t δjs + Rts,t δjk − Rjs,k + Rjk,s m−2 m−2 m−2 m−2 Sk Ss + δsj − δjk (m − 1)(m − 2) (m − 1)(m − 2) 1 1 = −Rjk,s + Rjs,k + Rjk,s − Rjs,k m−2 m−2 1 1 1 1 1 1 Sk δjs − Ss δjk + − − m−2 m−1 2 m−2 m−1 2 1−m+2 −1 + m − 2 = Rjk,s + Rjs,k m−2 m−2 1 1 3−m 3−m + Sk δjs − Ss δjk m − 2 2(m − 1) m − 2 2(m − 1) m−3 = Cjsk , m−2

t t Wjks,t = Rjks,t −

where Cjsk are the components of the Cotton tensor C, i.e., Cjsk = Rjs,k − Rjk,s +

1 (Ss δjk − Sk δjs ). 2(m − 1)

Note that, if m = 3, then div W ≡ 0; moreover, it is possible to prove that if m = 3, then W ≡ 0 (see for instance [Eis49]). The Cotton tensor can also be interpreted in the following way. Let A = Ric −

S , 2(m − 1)

be the Schouten tensor of components Aij = Rij −

S δij . 2(m − 1)

(2.28)

46


Clearly A is symmetric, hence taking covariant derivatives Aij,k = Aji,k , but for the last two indices one immediately verifies that Aij,k − Aik,j = Cijk . Hence we can think of the Cotton tensor as the obstruction for the Schouten tensor to be Codazzi; in other words, A is a Codazzi tensor if and only if the Cotton tensor C is identically null. Although we shall not make use of A in what follows, we recall that it enables to write, for m ≥ 3, the decomposition of the curvature tensor: denoting again with W the (0, 4) version of the Weyl tensor we have 1 R = W+ A , , m−2 where is the Kulkarni-Nomizu product; for the definition of this latter and more information on the Schouten tensor we refer to A. Besse’s treatise [Bes08]. The importance of the Weyl and the Cotton tensors will be emphasized in Theorem 2.6 below, which is a classical result due to Weyl (case m ≥ 4) and to Schouten (case m = 3). We first recall the following Definition 2.5. A Riemannian manifold (M, , ) of dimension m ≥ 2 is said to be locally conformally flat if, for every p ∈ M , there exist an open set U p and a function ϕ ∈ C ∞ (U ), ϕ > 0, such that the manifold (U, ϕ2 , ) is flat. Remark. Recall that, by the classical Riemann theorem (see for instance [Spi79]), the flat manifold (U, ϕ2 , ) is locally isometric to Rm . It follows that M is locally conformally flat if each point x ∈ M has a coordinate chart (U, ξ) such that ξ ∗ , Rm is pointwise conformally related to , . Namely, ξ : U → Rm is a conformal imbedding. Obviuously, flat spaces are locally conformally flat. Here are some more interesting examples of both conformally flat and nonconformally flat manifolds. Example. Every 2-dimensional Riemannian manifold is locally conformally flat. This fact is known as “the existence of isothermic coordinates” on any smooth surface and was established by Korn and Lichtenstein, [Kor14], [Lic16], assuming that the Riemannian metric at hand has Hölder continuous coefficients. For a conceptually easier proof we refer the reader to the paper [Che55] by Chern. The case of real analytic Riemannian metrics goes back to Gauss. Some regularity condition on the coefficients of the metric is needed as shown by Hartman and Wintner, [HW53]. Now, since every smooth manifold supports a smooth Riemannian metric, using isothermic coordinates one concludes that any orientable smooth surface possesses an underlying complex structure.


47

Example. The standard sphere Sm and the standard hyperbolic space Hm are locally conformally flat. As for the hyperbolic space, we are identifying Hm with either its Poincaré or its half-space model. Example. An important class of m-dimensional, locally conformally flat manifolds are those admitting a conformal immersion into the standard sphere Sm . In general, such an immersion does not exist. For instance, consider the flat (hence conformally flat) torus Tm . By standard topological arguments, if Tm were immersed into Sm , the immersion would be a covering map. Therefore, the fundamental group of Tm would inject into the fundamental group of Sm , and this is clearly impossible. A general obstruction to the existence of conformal immersions into spheres is represented by the value of the Yamabe invariant of the manifold. Let (M, , ) be a Riemannian manifold of dimension m ≥ 3 with scalar curvature S(x). The Yamabe invariant of M is the real constant # 2 m−2 |∇v| + 4(m−1) S (x) v 2 M Y (M ) = inf . #

m−2 v∈C0∞ (M )\{0} 2m m m−2 v M Note that, in case S (x) ≡ 0, Y (M ) reduces to the ordinary Sobolev constant. By a result of Schoen and Yau, if M has a conformal immersion into Sm , then Y (M ) = Y (Sm ) > 0; see Chapter 6 in [SY94]. By way of example, it is a simple matter to verify that the Riemannian product M = Tm−1 ×R satisfies Y (M ) = 0. Indeed, it is a complete, flat manifold with sub-linear volume growth. Example. Suppose that m − k ≥ 2. Then, it can be shown that the Riemannian product Sm−k−1 × Hk+1 is conformally diffeomorphic to Sm \Sk , where Sk ⊂ Sm is an equatorial k-sphere. In particular, Sm−k−1 × Hk+1 is locally conformally flat. Example. If M has constant sectional curvature, then the Riemannian products M × R and M × S1 are locally conformally flat. Example. In general, the Riemannian product of locally conformally flat manifolds is not locally conformally flat. For instance, the product of standard spheres Sm × Sm , m ≥ 2, is an example of a compact, simply connected, non-locally conformally flat manifold. A direct verification is possible. However, note that this follows directly from Kuiper’s Theorem 2.7 below. Indeed, if Sm × Sm were locally conformally flat, by Theorem 2.7 it would be conformally diffeomorphic to the standard sphere S2m . In particular, the m-th de Rham cohomology groups would satisfy 2m m m HdR (Sm × Sm ) HdR S , but this is impossible. Indeed, for instance, a Mayer-Vietoris argument (see for m example [Lee11]) shows that HdR S2m = 0. On the other hand, by the K¨ unneth formula, (see [GHV72]) m−k m k m HdR (Sm × Sm ) ⊕m (Sm ) = 0, k=0 HdR (S ) ⊗ HdR m 0 (Sm ) = HdR (Sm ) = R. because, by Poincaré duality, HdR

48


One of the most important results in Riemann surfaces theory is the Riemann-K¨ obe uniformization theorem (see for instance [For91]), according to which every simply connected Riemann surface is bi-holomorphic either to the complex plane C, to the open unit disk Bm ⊂ C or to the Riemann sphere S2 . Note that a bi-holomorphism is a (orientation preserving) conformal diffeomorphism. Therefore, recalling the existence theorem for isothermic coordinates alluded to above, we conclude e.g. that every 2-dimensional, compact, simply connected (hence orientable) Riemannian manifold is conformally diffeomorphic to S2 . In general, Riemannian manifolds of dimension m ≥ 3 do not enjoy any such uniformization property. An obstruction is given in the following (see [Eis49], [Bes08], [SY94]) Theorem 2.6. Let (M, , ) be a Riemannian manifold, dim M = m ≥ 3. A necessary and sufficient condition for (M, , ) to be locally conformally flat is that C ≡ 0 if m = 3, W ≡ 0 if m ≥ 4. A complete classification of locally conformally flat manifolds is unknown. In general, in higher dimensions, uniformization does require curvature restrictions. However, in the compact setting, we have the following seminal result due to N. Kuiper, [Kui49]. Theorem 2.7. Let (M, , ) be a compact, simply connected, locally conformally flat manifold of dimension m ≥ 3. Then, M is conformally diffeomorphic to the standard sphere Sm . Proof. The idea of the proof is as follows. Details can be found in Chapter 6 of Schoen-Yau’s book [SY94]. Since M is locally conformally flat, every point x ∈ M has a neighborhood Uα conformally imbedded into Rm , hence into Sm . Let ξα be such a conformal imbedding. Consider the couple (Uα , ξα ). If (Uβ , ξβ ) is a second local conformal imbedding, the transition function ξα ◦ ξβ−1 is a local conformal automorphism of Sm . The classical Liouville theorem then shows that there exists a global conformal transformation ψ(α,β) ∈ Conf (Sm ) such that ψ(α,β) = ξα ◦ ξβ−1 , on ξβ (Uα ∩ Uβ ) . Note that, with the obvious meaning of the symbols, the following cycle conditions are satisfied: ψ(α,β) ◦ ψ(β,α) = id, ψ(α,β) ◦ ψ(β,γ) ◦ ψ(γ,α) = id.

(2.29)

We are now in the position to define a conformal immersion Φ : M → Sm . Indeed, let x0 ∈ M and let (Uα0 , ξα0 ) be a fixed conformal imbedding with x0 ∈ Uα0 . For every x ∈ M let γ be a chosen path from γ (0) = x0 to γ (1) = x which is covered

2.2. Some applications in the compact case

49

by a finite chain of elements (Uα0 , ξα0 ) , . . . , (Uαn , ξαn ) such that Uαj ∩ Uαj+1 = ∅. We define Φ (x) = ξα0 (x) , on Uα0 , and Φ (x) = ψ(a0 ,α1 ) ◦ · · · ◦ ψ(αn−1 ,αn ) ◦ ξn (x) , on Uαn . Since M is simply connected, using the cycle conditions (2.29) together with a monodromy argument shows that Φ is well defined and, by construction, Φ is a conformal immersion. It remains to show that Φ is a diffeomorphism. This follows from standard topological arguments. Indeed, since M is compact and Sm is connected, Φ is a covering map. But Sm is simply connected, hence Φ is a bijection.

2.2 Some applications in the compact case 2.2.1

A rigidity result of Obata

We now prove a rigidity result for compact Einstein manifolds due to M. Obata, [Oba62a]. Its proof relies on the transformation laws (2.26) and on a rigidity result for complete Riemannian manifolds supporting nontrivial solutions of certain differential inequalities; see Theorem 2.10 below. Theorem 2.8. Let (M, , ) be a compact, Einstein manifold of dimension m ≥ 3 and let , be a pointwise conformal deformation of , . If (M, , ) has constant scalar curvature S, then (M, , ) is Einstein. Furthermore, if (M, , ) is not m 2 conformally diffeomorphic to the standard sphere S , then , = c , , for some constant c > 0. Remark. This result has been generalized by J. Escobar, [Esc90], to the case where M has a nonempty boundary ∂M = ∅. In this situation, one also requires that the inclusion ı : ∂M → (M, , ) is totally geodesic and that ı : ∂M → (M, , ) is minimal. Thus, in the conclusion of the theorem, the standard sphere Sm is replaced by the standard hemisphere Sm + . Escobar’s proof follows closely Obata’s original argument. We shall limit ourselves to proving Theorem 2.8, since in any case Escobar’s theorem will result as a consequence of Theorem 2.16 of section 2.2.3. Before proving the theorem we point out the following simple, interesting, consequence, to be compared with Theorem 2.2 . Later on, in the setting of complete, non-Einstein manifolds, we shall give a version of the next result assuming that the conformal diffeomorphism at hand preserves the scalar curvature (see Theorem 5.9). Corollary 2.9. Let (M, , ) be a compact Einstein manifold of dimension m ≥ 3 which is not conformally diffeomorphic to Sm . Let ϕ ∈ Conf (M ) be a conformal

50


diffeomorphism such that ϕ∗ , has constant scalar curvature. Then ϕ ∈ Iso (M ), that is, it is a Riemannian isometry. Proof. By Theorem 2.8, ϕ∗ , = c2 , , for some constant c > 0. Therefore, the result follows from Lemma 2.1. Now we give the Proof of Theorem 2.8. For the sake of convenience, we set , = ϕ−2 , and denote with quantities that refer to the metric , . According to (2.26), / ϕ−1 −1 Δ − T = T + (m − 2) ϕ Hess ϕ , . m Since , is Einstein we have T ≡ 0, and ) *we deduce that, in components with , , respect to a local orthonormal coframe θj for −Tij = (m − 2) ϕ

ϕ

−1

ij

/ ϕ−1 Δ − δij . m

(2.30)

Since T is a traceless tensor, 2 ϕ−1 T = − (m − 2) ϕ−1 ij Tij . ,

On the other hand, according to equation (2.13), m−2 Si Tik,k = 2m and since S is constant, we obtain Tik,k = 0. (2.31) ) * + by Thus, if we let { ej } be the dual frame of θj , and define the vector field W the formula + = ϕ−1 Tij ej , W (2.32) i −1 −1 i = ϕ where d ϕ θ , a computation that uses (2.31) shows that i +W + = ϕ−1 Tij + ϕ−1 Tik,k = ϕ−1 Tij , div ij i ij and we conclude that 2 +W +. ϕ−1 T = − (m − 2) div ,


51

Integrating on M, , and using the divergence theorem we deduce that

M

2 + = 0. ϕ−1 T dvol ,

Accordingly, T ≡ 0 and M, , is an Einstein manifold. Suppose now that the m-dimensional manifold (M, , ) is not conformally 4 diffeomorphic to Sm , m ≥ 3. Having set as usual , = u m−2 , , we shall prove that u is identically equal to a positive constant c2 . Let S be the the scalar curvature of , , which is constant since (M, , ) is Einstein. We first claim that either S = S = 0 or S · S > 0. To see this, we recall that S and S are related by the Yamabe equation m+2 m−2 cm Δu − Su + Su = 0,

cm = 4

m−1 . m−2

Suppose that S ≤ 0 and S ≥ 0. Then, Δu ≥ 0, and since (M, , ) is compact we deduce that u is a positive constant. Therefore, inserting this information into the Yamabe equation, we conclude that 4 m−2 0 ≥ S = Su ≥ 0,

proving that S = S = 0, and the claim follows. The same conclusion holds if we assume instead that S ≥ 0 and S ≤ 0. Thus, we need to consider three possible cases. First case: S = S = 0. Using the Yamabe equation once more gives Δu = 0 and, therefore, u is a positive constant. Second case: S < 0 and S < 0. Up to rescaling , by a positive constant, we can assume that S = S < 0. Let u (x0 ) = max u, M

u (x1 ) = min u. M

Then, by the usual maximum principle, Δu (x0 ) ≤ 0 which implies, according to the Yamabe equation,

4 Su (x0 ) 1 − u m−2 (x0 ) ≤ 0. As a consequence, u (x) ≤ u (x0 ) ≤ 1. Similarly, since u achieves its minimum at x1 we have Δu (x1 ) ≥ 0. This latter, in turn, implies 1 ≤ u (x1 ) ≤ u (x) ,

52


and, therefore, u ≡ 1. 4 Third case: S > 0 and S > 0. Let us observe that, since , = u m−2 , and both the metrics are Einstein, setting 2

v = u− m−2 , by (2.26) we have Δv , = 0 on M. m We define the vector field X = ∇v so that div X = Δv and Hess(v) −

(2.33)

1 LX , = Hess(v). 2 From the above, we deduce 1 div X LX , = , , 2 m that is, X is a conformal vector field. We now show, using again the moving frame formalism, that since M is Einstein, f = div X = Δv satisfies the equation Hess(f ) +

S f , = 0, m (m − 1)

(2.34)

where, we recall, s > 0. First we observe that the previous equation can be rewritten as S Hess(Δv) + Δv , = 0, m (m − 1) or, in components, (Δv)kt = viikt = −Δv

S δkt ; m (m − 1)

then we note that, by (2.33), we have vij =

vtt δij , m

(2.35)

vijk =

vttk δij m

(2.36)

vijkl =

vttkl δij . m

(2.37)

from which we deduce and By (1.54) applied to v we deduce

viikl = (Δv)kl = vikil + vsl Rsiik + vs Rsiik, l = vkiil − vsl Rsk − vs Rsk, l = vkiil − vsl Rsk ,


53

since Rsk,l = 0. Then, using (2.35), (2.36) and (2.37) we obtain (Δv)kl = vkiil −

vii vssil δki vii δsl Rsk = − Rkl , m m m

which easily implies (2.34). In case f is constant, we get f = Δv = 0 and, since M is compact, v itself must be constant. This implies that also u is constant, as desired. Finally, in case f is nonconstant, we can apply the next result to conclude that (M, , ) is isometric 2 to the sphere Sm k2 of constant curvature k = S/m (m − 1). But this implies that (M, , ) is conformally diffeomorphic to the standard sphere Sm , against our initial assumption. The following result is again due to Obata, [Oba62a] (for the analogous statement for Sm + , the upper hemisphere, see Theorem 2.14 below); see also [PR]. Theorem 2.10. Let (M, , ) be a complete Riemannian manifold of dimension m. Suppose that there exists a nonconstant, smooth function f : M → R such that, for some constant k > 0, Hess(f ) + k 2 f , = 0.

(2.38)

Then (M, , ) is isometric to the sphere Skm2 of constant sectional curvature k 2 . Proof. First, we show that f has a critical point. Indeed, by contradiction, suppose that ∇f = 0 on M . Consider the smooth vector field X = ∇f / |∇f | on M . Since |X| ∈ L∞ (M ) and (M, , ) is geodesically complete, then X is complete (see e.g. [Lee03], Chapter 12). Let γ : R → M be an integral curve of X, namely Xγ(s) = γ˙ (s), for every s ∈ R. A direct computation that uses (2.38) shows that, for every vector field Y , Dγ˙ γ, ˙ Y=

1 1 Hess (f ) (γ, ˙ Y)− Hess (f ) (γ, ˙ γ) ˙ γ, ˙ Y = 0. |∇f | |∇f |

Therefore, γ is a unit speed geodesic of M . Evaluating (2.38) along γ we deduce that the function y (s) = f ◦ γ (s) satisfies y = −k 2 y which is oscillatory. Let s0 ∈ R be such that y (s0 ) = 0. Then, recalling that γ is an integral curve of X, we conclude 0 = y (s0 ) = ∇f (γ (s0 )) , γ˙ (s0 ) = |∇f (γ (s0 ))| = 0, a contradiction. Let o ∈ M be a critical point of f and set r (x) = d (x, o). Note that f (o) = 0 for otherwise, having fixed any unit speed geodesic γ issuing from o, we would have

54


that y (s) = f ◦ γ (s) solves the Cauchy problem ⎧ ⎨ y = −k 2 y, y (0) = 0, ⎩ y (0) = 0, and, hence, y (s) ≡ 0. Since this would be true for every geodesic γ we should conclude that f ≡ 0, a contradiction. Thus, without loss of generality, we assume that f (o) = 1. We claim that, for every x ∈ M , it holds that f (x) = cos (kr (x)) .

(2.39)

Indeed, consider a unit speed, minimizing geodesic γ : [0, r (x)] → M from γ (0) = o to γ (r (x)) = x. As noted above, the smooth function y (s) = f ◦ γ (s) is the solution of the Cauchy problem ⎧ ⎨ y = −k 2 y, y (0) = 1, ⎩ y (0) = 0. Therefore, y (s) = cos (ks) . Evaluating at s = r (x) we conclude the validity of (2.39). Now, observe that the function cos (ks) is strictly decreasing on (0, π/k). It follows from (2.39) that r (x) = k −1 arccos (f (x)) is smooth on the geodesic ball Bπ/k (o) \ {o}. Applying the Bishop density result of Chapter 1 we therefore conclude that cut (o) ∩ Bπ/k (o) = ∅.

(2.40)

Therefore, the exponential map expo : Bm π/k (0) ⊂ To M → Bπ/k (o) is a diffeomorphism. Let us introduce geodesic polar coordinates (r, θ) on To M . Furthermore, let us consider a local orthonormal coframe {θα } on Sm−1 αwithαdual frame {Eα }. Thus, the standard metric of Sm−1 is written as dθ2 = θ ⊗ θ . We extend both {θα } and {Eα } radially. Then, by Gauss’ lemma, , = dr ⊗ dr + σαβ (r, θ) θα ⊗ θβ . Furthermore, since , is infinitesimally Euclidean and the standard metric of Rm ≈ To M is written as , Rm = dr ⊗ dr + r2 δαβ θα ⊗ θβ , we have the further condition σαβ (r, θ) = δαβ r2 + o r2 , as r 0.

(2.41)


55

Now we use the fact that L∇r , = 2Hess (r) , on Bπ/k (o) \ {o} where L∇r is the Lie derivative in the radial direction ∇r. Thus, by the definition of Lie derivative we deduce that 2 Hess(r)(Eα , Eβ ) = L∇r , (Eα , Eβ ) = ∇r(σαβ ). Since ∇r = −

(2.42)

∇f , |∇f |

the Hessian of the distance function r can be expressed as , ∇f , Eβ Hess (r) (Eα , Eβ ) = − DEα |∇f | 1 Hess (f ) (Eα , Eβ ) =− |∇f | k 2 cos (kr) Eα , Eβ = k sin (kr) = k cot (kr) σαβ . Inserting into (2.42), and recalling (2.41) we obtain that σαβ are the (unique) solutions of the asymptotic Cauchy problems & ∂r σαβ = 2k cot (kr) σαβ Bπ/k (o) \ {o} , , on σαβ (r, θ) = δαβ r2 + o r2 , as r 0. Integrating, finally gives σαβ = k −2 sin2 (kr) δαβ . We have thus established that, in polar coordinates of Bm π/k (0) \ {o} = (0, π/k) × m−1 S ⊂ To M , it holds that , = dr ⊗ dr + k −2 sin2 (kr) dθ2 . Since

(0, π/k) × Sm−1 , dr ⊗ dr + k −2 sin2 (kr) dθ2

(2.43)

is isometric to the m-dimensional 2-punctured sphere Sm k2 \ {2 points} of constant curvature k 2 , we conclude that the geodesic ball Bπ/k (o) ⊂ M is isometric to Sm k2 \ {point}. To complete the proof it suffices to show that Bπ/k (o) = M \ {point} .

(2.44)

56


Indeed, suppose we have already proved this fact. Then, by Seifert-Van Kampen’s theorem, (see [Lee11]) M is simply connected. Moreover, M \ {point} has constant curvature k 2 because it is isometric to Skm2 \ {point}. By continuity, M itself has constant curvature k 2 . Therefore, the Hopf classification theorem (see [Pet06b]) tells us that M must be isometric to Skm2 , as desired. The proof of (2.44) combines Morse theoretic and cut-locus arguments. First of all, we observe that ∂Bπ/k (o) is made up entirely by nondegenerate critical points of f. Therefore, by Morse’s lemma, (see [Mil63], [Pet06b]) ∂Bπ/k (o) is a discrete, compact (hence finite) set. Indeed, for every x ∈ ∂Bπ/k (o), by continuity we have |∇f | (x) = −k sin (π) = 0, and, by assumption,

Hess (f ) (x) = −k 2 f (x) , x ,

where, f (x) = cos (π) = −1. In particular, Hess(f ) is (strictly) negative definite on ∂Bπ/k (o), as claimed. To conclude, we note that ∂Bπ/k (o) is connected. In fact,

∂Bπ/k (o) = expo ∂Bm (0) . (2.45) π/k To see this, we recall that

expo ∂Bm (0) ∩ T M \c (o) = ∂Bπ/k (o) ∩ M \cut (o) , o π/k

(2.46)

where c (o) ⊂ To M is the tangential cut-locus of o ∈ M . Let v¯ ∈ ∂Bm π/k (0) ∩ c (o). By definition, x ¯ = expo v¯ ∈ cut (o). Since, by (2.40), Bπ/k (o) ⊂ M \cut (o) , and, on a generic complete Riemannian manifold,

(0) ⊆ Bπ/k (o), expo Bm π/k we must conclude that x ¯∈ / Bπ/k (o), that is,

expo ∂Bm π/k (0) ∩ c (o) ⊆ ∂Bπ/k (o) ∩ cut (o) . Conversely, let x ∈ ∂Bπ/k (o) ∩ cut (o). By definition, there is a unit speed geodesic γ (t) = expo (vt) : [0, +∞) → M from γ (0) = o to γ (π/k) = x that does not minimize distances past π/k. Then, vπ/k ∈ ∂Bm π/k (0) ∩ c (o) and expo (vπ/k) = x. Summarizing,

expo ∂Bm (2.47) π/k (0) ∩ c (o) = ∂Bπ/k (o) ∩ cut (o) . From (2.46) and (2.47) we conclude the validity of (2.45).


2.2.2

57

A result by M. F. Bidaut-Véron and L. Véron

+ defined in (2.32), with Tij given in (2.30), will suggest to us The vector field W how to proceed to provide a proof of Theorem 2.12 below. First we give an intrinsic definition: by direct manipulation of the quantities involved we have & ' (ϕ−1 )kk −1 −1 + W = (ϕ )i −(ϕ )ij + ej δij (m − 2)ϕ m m−2 = ϕ(ϕ−1 )kk (ϕ−1 )i ei − (m − 2)ϕ(ϕ−1 )ij (ϕ−1 )i ej m −1 m − 2 −1 −1 ϕ−1 ∇ϕ = ,· ϕ(Δϕ )∇ϕ − (m − 2)ϕHess m / −1 −1 Δϕ −1 − Hess ϕ−1 ∇ϕ ∇ϕ = (m − 2)ϕ , m / −1 Δϕ 1 −1 2 −1 ∇ϕ − ∇∇ϕ . = (m − 2)ϕ m 2 ,

We set the following general Definition 2.11. Given u ∈ C 2 (M ), u > 0, the vector field & ' Δu α 1 2 Z=u ∇|∇u| − ∇u , α ∈ R, 2 m

(2.48)

is called an Obata type vector field. + above is an Obata vector field (modulo the multiplicative constant Thus W 2 − m) with respect to the metric , . Furthermore, the function involving ϕ satisfies equation (2.7) with the roles of , and , interchanged, that is ϕ2 S = S − 2(m − 1)

Δϕ − (m − 1)(m − 4) ϕ

2 ∇ϕ

,

ϕ2

.

This, in turn, implies 2 −1 = − 1 Δϕ + 2 ∇ϕ Δϕ ϕ2 ϕ3 ,

=

S m S ϕ− ϕ−1 + 2(m − 1) 2(m − 1) 2

=−

2 ∇ϕ

,

ϕ3

S S m ϕ−1 + (ϕ−1 )−1 + 2(m − 1) 2(m − 1) 2

−1 2 ∇ϕ

,

ϕ−1

.

58


Therefore, u = ϕ−1 is a positive solution of the differential equation 2 $ % ∇u m S S ,

−1 = u − u . Δu + 2 u 2(m − 1) S We have thus obtained the further suggestion to consider the vector field Z defined in (2.48), with u a positive solution of a differential equation of the type Δu = (β + 1)

|∇u|2 1 1+β(1−σ) − λu , β ∈ R \ {0}, σ > 1, λ ∈ R. (2.49) + u u β

Note that, if Δu = −f (u, |∇u|),

f = f (u, t) ∈ C 1 (R2 ),

a straightforward computation which exploits the Bochner-Weitzenb¨ ock formula (1.48) gives ' & (Δu)2 div Z = uα | Hess(u)|2 − m & ' α 1 − m ∂f + uα−1 f (u, |∇u|) + u (u, |∇u|) |∇u|2 m m ∂u + uα Ric(∇u, ∇u) + αuα−1 Hess(u)(∇u, ∇u) 1 − m α ∂f + u (u, |∇u|)|∇u|−1 Hess(u)(∇u, ∇u). m ∂t Hence, if f has the special form f (u, |∇u|) = δ

|∇u|2 + g(u), u

δ ∈ R,

the previous formula becomes ' & (Δu)2 α 2 div Z = u | Hess(u)| − m & ' α 1 |∇u|2 1−m α−1 +u g(u) + (α + m − 1)δ + ug (u) |∇u|2 m m u m + uα Ric(∇u, ∇u) ' & 1−m α−1 Hess(u)(∇u, ∇u). α + 2δ +u m To get rid of the last term containing Hess(u)(∇u, ∇u) we observe that, given γ ∈ R, div γuβ |∇u|2 ∇u = 2γuβ Hess(u)(∇u, ∇u) + γuβ |∇u|2 Δu + βγuβ−1 |∇u|4 .

2.2. Some applications in the compact case We choose β = α − 1, γ =

(m−1)δ m

V =Z+ we have

−

59

α 2

so that, for the vector field m−1 α α−1 u |∇u|2 ∇u, δ− m 2

(2.50)

' (Δu)2 div V = u | Hess(u)| − m & ' m−1 α−1 α(m + 2) − 2(m − 1)δ +u g(u) − ug (u) |∇u|2 2m m uα−2 + 3mαδ − mα(α − 1) − 2δ 2 (m − 1) |∇u|4 2m + uα Ric(∇u, ∇u). &

α

2

Now, if δ = −(β + 1) and g(u) = βλ u − β1 u1+β(1−σ) from the previous relation we obtain ' & (Δu)2 (2.51) div V = uα | Hess(u)|2 − m 1 − [2(m − 1)βσ + (m + 2)α]uα+β(1−σ) |∇u|2 2mβ 1 − mα2 + (2 + 3β)mα + 2(m − 1)(β + 1)2 uα−2 |∇u|4 2m & ' λ 1 2 2m Ric(∇u, ∇u) + [α(m + 2) + 2(m − 1)β]|∇u| uα . + 2m β We are now ready to prove the following beautiful result first due to Bidaut-Véron and Véron, [BVV91]. Theorem 2.12. Let (M, , ) be a compact manifold of dimension m ≥ 2 and Ricci curvature satisfying Ric ≥ k > 0. Let ϕ be a positive solution of Δϕ − λϕ + ϕσ = 0

on M

(2.52)

for some constant σ > 1, λ > 0. Then ϕ is constant provided (i) m = 2 and λ ≤ 2k/ (σ − 1) ; (ii) m ≥ 3,

λ ≤ mk/(m − 1)(σ − 1),

σ ≤ (m + 2) / (m − 2) and

(A) either at least one of the last two inequalities is strict or (B) (M, , ) has constant scalar curvature S and it is not isometric to Skm2 , the sphere of constant curvature k 2 = s/(m − 1)m.

60


Proof. We set u = ϕ−1/β , β = 0. Then u is positive and satisfies equation (2.49), hence (2.51) holds. With the aid of the divergence theorem we obtain & ' 1 2 4 α 2 0= 2mu |Hess(u)| − (Δu) − A uα−2 |∇u| (2.53) m M M ) * 2 2 + uα 2m Ric (∇u, ∇u) + D|∇u| − B uα−β(σ−1) |∇u| , M

M

where, for ease of notation, we have set A = mα2 + (3β + 2)mα + 2(m − 1)(1 + β)2 , 1 λ B = [2(m − 1)βσ + (m + 2)α], D = [(m + 2)α + 2(m − 1)β]. β β Next we observe that, by Newton’s inequality 2

|Hess(u)| ≥

1 2 (Δu) , m

the first integral on the right-hand side of the above is nonnegative. The idea of the proof is to find, under the conditions listed in (i) and (ii), β = 0 and α ∈ R, such that D A ≤ 0, B ≤ 0, and Ric + ≥0 (2.54) 2m and at least one of the above inequalities is strict. Once this is achieved, then (2.53) implies that ∇u ≡ 0; thus u and therefore ϕ are constant. Let y = 1 + 1/β, δ = −α/β, so that y, δ ∈ R, y = 1. Rewriting A, B and D in terms of y and δ, the inequalities to be established become (a) (b) (c)

2 2 2 m−1 m y − 2δy + δ − δ ≤ 0, m−1 2σ m+2 ≤ δ,

m 2 m+2 Ric ≥ λ δ − 2 m−1 m+2 , ,

(2.55)

with at least one strict inequality. If either (i) or (ii) holds, then, for m ≥ 2, m−1 m−1 m , (2.56) Ric ≥ λ 2 σ−2 2 m+2 m+2 m+2 and setting δ = 2σ

m−1 > 0, m+2

inequalities (2.55) (b) and (c) are satisfied. In order to find a value y = 1 which satisfies (2.55) (a) with strict inequality, it suffices that the quadratic polynomial in y on the left-hand side has two distinct solutions, which, taking into account our choice of σ, in turn amounts to the validity of the inequality (m+1)−(m−2)σ > 0.


61

This inequality being always trivially satisfied if m = 2, it remains to analyze the case where m > 2, σ = (m + 2) / (m − 2). We assume that (M, , ) has constant scalar curvature S and show that if ϕ is a positive nonconstant solution of (2.49), then (M, , ) is isometric to the sphere Sm k2 of constant sectional curvature k 2 = S/(m−1)m. Our assumption that ϕ is not constant implies that u is positive, and nonconstant; by the condition on σ, equality holds in (2.55) (b) and (a) with m y = m−2 , so that A = B = 0. Since the third and the fourth summands in (2.53) vanish, while the integrands in the first and third integral are nonnegative, they must vanish identically. In particular we must have 2

|Hess(u)| =

(Δu)2 , on M m

and therefore, by the equality case in the Cauchy-Schwarz inequality, Hess(u) =

Δu , , on M. m

Now we proceed as in the proof of Theorem 2.8. We set X = ∇u and we observe that f = div X satisfies Hess(f ) +

S f , = 0. m(m − 1)

Note that f is not constant, for otherwise the above would imply Sf ≡ 0 and since S ≥ mk > 0 by the assumption on the Ricci curvature, we would conclude that f = Δu = 0 so that u is constant on M . Obata’s Theorem 2.10 implies the conclusion.

Remark. 1. Equation (2.52) is, of course, a normalized version of the more general case Δϕ − λϕ + μϕσ = 0 on M, (2.57) with λ, μ > 0 and ϕ > 0. In this situation the required condition on λ depends on μ. 2. Similarly to what happened in the proof of Theorem 2.8, if we consider equation (2.57) with λ ≤ 0, μ ≥ 0, then Δϕ ≤ 0 on the compact manifold (M, , ), and hence ϕ is a positive constant by the maximum principle. For λ ≥ 0, μ ≤ 0, then Δϕ − λϕ ≥ 0 and we obtain the same conclusion. Finally, for λ ≤ 0, μ ≤ 0, Δϕ + μϕσ−1 ϕ ≤ 0 and again ϕ is constant. 3. The above observation also shows that Theorem 2.8 is a consequence of Theorem 2.12 for m ≥ 3. Indeed, if S is the scalar curvature of (M, , ) and 4 = 0, with S the constant , = ϕ m−2 , , ϕ > 0, then cm Δϕ − Sϕ + Sϕ scalar curvature of , . However, the Einstein condition is not required for (M, , ) in Theorem 2.12.

62


2.2.3

A version of Theorem 2.12 on manifolds with boundary

We shall now give a version of Theorem 2.12 in case M is compact with nonempty boundary ∂M . For simplicity we consider the case that M is oriented. In the next result we explicate the boundary term in applying the divergence theorem to (2.51). Lemma 2.13. Let (M, , ) be an m-dimensional oriented compact Riemannian manifold with boundary ∂M . Let II and h denote respectively the second fundamental tensor and the mean curvature of the embedding ∂M → M in the direction of the outward unit normal vector field ν. Let u ∈ C 3 (M ) and let u = u|∂M . If V is the vector field defined in (2.50), that is & ' Δu m−1 α α−1 2 α 1 u V =u |∇u|2 ∇u, ∇|∇u| − ∇u − (β + 1) + 2 m m 2 we have

div V =

2m M

∂u ∂ν ∂M ∂u 2 [2(m − 1)(β + 1) + 3mα] uα−1 |∇ u| − ∂ν ∂M 3 α−1 ∂u [2(m − 1)(β + 1) + mα] u (2.58) − ∂ν ∂M ∂u 2 + 2m uα II(∇ u, ∇ u), ν . + 2(m − 1)h uα ∂ν ∂M 2 uα [(m − 1) Hess(u)(ν, ν) − (m + 1)Δ u]

Proof. Let {e1 , . . . , em−1 , em = ν} be a Darboux frame along ∂M → M . Set hij = II(ei , ej ), ν , so that h=

1 hkk , m−1

where 1 ≤ i, j, k ≤ m − 1. Then, with obvious meaning of the notation we have u i = ui ,

um =

∂u = ∇u, ν . ∂ν

It therefore follows that ⎧ ⎨u = u + um hij , ij ij ∂u ⎩ + = umi − u j hij . ∂ν i

2 Hence, using the identity ∇|∇u| , X = 2 Hess(u)(X, ∇u), we get


V, ν =

2m ∂M

−2 uα Δu

∂M

−

63

∂u + 2m uα Hess(u)(∇u, ν) ∂ν

[2(m − 1)(β + 1) + mα] uα−1 |∇u|

2 ∂u

∂ν ∂u 2 uα [(m − 1) Hess(u)(ν, ν) − Δ u] = ∂ν ∂M ∂u 2 2(m − 1)h uα + ∂ν ∂M 0 ∂u 2 1 ∂u 2 u| + − [2(m − 1)(β + 1) + mα] uα−1 |∇ ∂ν ∂ν ∂M 2 3 + ∂u 2m uα II(∇ u, ∇ u), ν + 2m uα ∇ , ∇ + u . ∂ν ∂M ∂M

We use the first Green formula (see e.g. [GT01]) to deal with the last integrand of the above formula, that is, 2 3 + ∂u ∂u 2 ∂u α . ∇ , ∇ 2m u 2m uα Δ u u| u =− + 2mα uα−1 |∇ ∂ν ∂ν ∂ν ∂M ∂M Formula (2.58) now follows at once.

In the next result we shall make use of the following extension of Obata’s theorem 2.10 due to Escobar [Esc90]. Its proof is similar to that of Theorem 2.10 and it will therefore be omitted. Theorem 2.14. Let (M, , ) be an m-dimensional Riemannian compact manifold with boundary ∂M . Assume that there exists a nonconstant function f on M satisfying & Hess(f ) + kf , = 0 on M , (2.59) ∂f ∂ν ≡ 0 on ∂M √

with k a positive constant. Then (M, , ) is isometric to Sm k , the upper + hemisphere of radius k −1/2 with the induced metric from Rm+1 . We are now ready for the next Lemma 2.15. Let (M, , ) be an m-dimensional, m ≥ 3, compact manifold with boundary ∂M such that the embedding ∂M → M is totally geodesic. Assume that (M, , ) is Einstein but not Ricci flat. Let v be a nonconstant function satisfying & Hess(v) − Δv m , = 0 on M , (2.60) ∂v ≡ 0 on ∂M ∂ν with ν the outward unit normalto ∂M . Then the scalar curvature S is positive 4 m(m−1) m . and (M, , ) is isometric to S+ S

64


Proof. Let X = ∇v. Then, because of the first equation of (2.60) X is a conformal vector field and hence, since S is constant, f = divX = Δv satisfies (2.34), that is Hess(f ) +

S f , m(m − 1)

on M .

(2.61)

We shall now show that

∂f ≡ 0 on M. (2.62) ∂ν Towards this end we observe that, with respect to a Darboux frame {ei , em = ν}, with 1 ≤ i ≤ m − 1 along the embedding ∂M → M , from conformality of X we have 2 Xim + Xmi = (div X)δim = 0. (2.63) m Thus differentiating in the ek direction, 1 ≤ k ≤ m − 1, Ximk + Xmik = 0.

(2.64)

∂v ≡ 0 on ∂M because of (2.60), thus Xmik = 0 on ∂M for Now Xm = X, ν = ∂ν 1 ≤ i, k ≤ m − 1. From (2.64) we then deduce

Ximk ≡ 0

on ∂M .

(2.65)

Now, since X is conformal Xab + Xba =

2 (divX)δab , m

1 ≤ a, b ≤ m

and for 1 ≤ k ≤ m − 1 Xkk =

divX m

(no sum over k).

Hence

1 (div X)m m From the Ricci identities (1.47), Xkkm =

(no sum over k).

Xkkm − Xkmk = Xs Rskkm

(2.66)

(no sum over k).

Thus using (2.65) m−1

Xkkm = −Xt Rtm

on ∂M .

k=1

But (M, , ) is Einstein and therefore Rsm = 0 for 1 ≤ s ≤ m − 1. Furthermore Xm = 0, hence m−1 Xkkm = 0. (2.67) k=1


65

Using (2.66) and (2.67) we find m−1 m Xkkm = 0 (divX)m = m−1

on ∂M

k=1

S that is, the validity of (2.62). Thus (2.60) is satisfied by f = Δv with k = m(m−1) and the conclusion follows at once from Theorem 2.14 once we prove that f is nonconstant and S > 0. We reason by contradiction. If f were constant, by (2.61) since S = 0 we would have f ≡ 0 so that X = ∇v would be a Killing field and 2 2 ∂v Δv ≡ 0. Since div(v∇v) = vΔv + |∇v| = |∇v| and ∂ν ≡ 0 the divergence theorem yields 2 |∇v| ≡ 0, M

that is, v is constant contradicting the assumption of the lemma. Thus f is nonconstant. Next, to show that S > 0 we trace (2.61) to obtain S Δf + m(m−1) f = 0 on M , ∂f ∂ν ≡ 0 on ∂M . From the divergence theorem S ∂f 2 2 0= f div(f ∇f ) = f Δf + |∇f | = − f2 = + |∇f | ∂ν m(m − 1) ∂M M M M and thus

# 2 |∇f | # > 0. S = m(m − 1) M f2 M

We are now ready to prove Theorem 2.16. Let (M, , ) be a compact manifold of dimension m ≥ 2 with boundary ∂M . Assume that the embedding ∂M → M is totally geodesic and that Ric ≥ k > 0. Let σ > 1, λ > 0 and ϕ a positive solution of & Δϕ − λϕ + ϕσ = 0 ∂ϕ ∂ν ≡ 0 on ∂M

(2.68)

on M ,

with ν the outward unit normal to ∂M . Then ϕ is constant provided (i) m = 2 and λ ≤ (ii) m ≥ 3, λ ≤

2k σ−1 ;

k m m−1 σ−1 ,

σ≤

m+2 m−2

and

(A) either at least one of the last two inequalities is strict or

(2.69)

66

Chapter 2. Pointwise conformal metrics (B) (M, , ) is Einstein and it is not isometric to an upper hemisphere of constant curvature.

Proof. We set u = ϕ−1/β with β = 0. Then u satisfies (2.49) and ∂u ∂ν ≡ 0 on ∂M . Using (2.51), (2.58), ∂u ≡ 0 and the fact that ∂M → M is totally geodesic we get ∂ν again (2.53). Now the proof proceeds exactly as in Theorem 2.12 up to Hess(u) = Since diction.

∂u ∂ν

Δu , m

on M.

≡ 0 on ∂M we can now apply Lemma 2.15 and we reach a contra

Remark. As it is apparent from the proof of Lemma 2.15, we may substitute the assumption that (M, , ) is Einstein in (B) with (M, , ) has constant scalar curvature and Ric(ν, X) = 0 for every X ∈ T ∂M ⊥ ⊆ T M on ∂M . Remark. 1. An observation similar to the Remark on page 61 holds also here. Point 1. applies verbatim to the more general equation Δϕ − λϕ + μϕσ = 0

on M

(2.70)

with λ, μ > 0 and ϕ > 0. 2. If λ and μ have different sign, that is, λ ≤ 0 and μ ≥ 0 or λ ≥ 0 and μ ≤ 0, ϕ > 0, then from (2.69) ∂ϕ 0= Δϕ = λϕ − μϕσ , = ∂ν ∂M M M so that

ϕ λ − μϕσ−1 ≡ 0.

From this latter we easily deduce λ = μ = 0 and ϕ is constant from (2.69) and compactness of M , without requiring any further assumption. 3. Finally, if λ, μ < 0 we have Δϕ − λϕ = −μϕσ ≥ 0. Suppose first that ϕ assumes its maximum at a point x0 ∈ ∂M . Then, by the boundary point Lemma (see [PW67] for more details) one has ∂ϕ ∂ν (x0 ) > 0, contradicting ∂ϕ (x ) ≤ 0 on ∂M . Hence ϕ has to assume its maximum at 0 ∂ν x0 ∈ M \ ∂M . Similarly ϕ attains its minimum at x1 ∈ M \ ∂M . Then 0 ≥ Δϕ(x0 ) = λϕ(x0 ) − μϕσ (x0 ), and therefore

μ σ−1 (x0 ) ≤ 1. ϕ λ


67

Analogously, since Δϕ(x1 ) ≥ 0 one gets μ σ−1 (x1 ) ≥ 1. ϕ λ Since μλ > 0 and ϕ is positive we conclude that ϕ(x0 ) = ϕ(x1 ) and therefore ϕ is constant. Remark. In Theorem 2.16 we can substitute the assumption that the embedding of the boundary in M is totally geodesic with that of a convex boundary (see for instance [PRS03a]). It seems also worth mentioning that using this result Ilias obtained some sharp Sobolev constants on M (see [Ili96] and [PRS03a], Theorem 3.5). With the above observation and with the aid of Lemma 2.19, from Theorem 2.16 we deduce the following result of Escobar (see [Esc90], Theorem 4.1): Theorem 2.17. Let (M, , ) be a compact Einstein manifold of dimension m ≥ 4 3 and with boundary ∂M totally geodesic in (M, , ). Let , = ϕ m−2 be a conformal change of metric with constant scalar curvature S and with ∂M → M, , minimal. Then , is Einstein and if (M, , ) is not isometric to an upper hemisphere of constant curvature, ϕ is constant, that is, the conformal change is an homothety.

2.2.4

A rigidity result of Escobar

In this section we prove a nice rigidity result due to Escobar, [Esc90]. Towards this aim we need some preliminary facts contained in the next two lemmas. Lemma 2.18. Let (M, , ) be a manifold with boundary, constant scalar curvature S and trace-free Ricci tensor T . Let ν be the outward unit normal to i : ∂M → M and let X be a conformal vector field on M such that, for some ψ ∈ C ∞ (∂M ), X|∂M = ψν. Then ψT (ν, nu) = 0.

(2.71)

∂M

Proof. We let W = T (X, ) as in (2.14) so that, according to (2.15), div W = Xi,k Tik + Xi Tik,k . Since X is conformal, LX , = or, in other words,

2 div X , , m

Xik + Xki =

2 div X δik . m

(2.72)

68


Hence Xik Tik

1 = (Xik + Xki )Tik = 2

1 div X Tkk = 0, m

since T is trace-free. Thus using (2.72) and the divergence theorem, T (X, ν) = (div T )(X). ∂M

M

However, since the scalar curvature S is constant, from (2.13) div T ≡ 0 and from the above we immediately deduce (2.71). Remark. For m = dim M = 2 there is no need to assume S constant, see equation (2.13). Lemma 2.19. Let (M, , ) be a manifold with boundary ∂M and , = ϕ2 , ∞ for some ϕ > 0, ϕ ∈ C (M ). Let ν be the unit outward normal of ∂M → (M, , ) and hij the coefficient of the second fundamental form

in the direction of ν. Set hij for the analogous quantities for ∂M → M, , with respect to ν = ϕ−1 ν. Then, for 1 ≤ i, j ≤ m − 1, ∂ϕ hij = ϕ−1 hij + ϕ−2 δij , ∂ν

(2.73)

and for the mean curvature ∂log ϕ . (2.74) h = ϕ−1 h + ∂ν

In particular, if ∂M → (M, , ) and ∂M → M, , are both minimal, then ∂ϕ ∂ν

≡ 0 on ∂M .

Proof. A simple computation. Indeed, with the notation of Chapter 1 and section 2.1.1 we let θi be a Darboux frame along the inclusion map ı : ∂M → (M, , ). Thus θm = 0 on ∂M and hij are defined by the requirement θim = hij θj ,

1 ≤ i, j, . . . ≤ m − 1.

, . Similarly, θi = ϕθi is a Darboux frame along the inclusion ı : ∂M → M, Thus θm = 0 and hij θj , 1 ≤ i, j, . . . ≤ m − 1. θm = i

hij we recall from (2.3) that To relate hij with ϕm i ϕi θim = θim + θm − θ, ϕ ϕ

dϕ = ϕj θj + ϕm θm .

2.2. Some applications in the compact case Hence

69

ϕm i j ϕ hij θj = hij θj − δ θ ϕ j

and it follows that

hij = ϕ−1 hij − ϕ−2 ϕm δij ,

from which (2.73), and thus (2.74), follow immediately.

Lemma 2.20. Let (M, g0 ) be a compact Einstein manifold such that ∂M → (M, g0 ) is totally geodesic and let g = ϕ−2 g0 , ϕ > 0, ϕ ∈ C ∞ (M ) be a conformally related metric with constant scalar curvature S and such that ∂M → (M, g) has constant mean curvature h with respect to the outward unit (with respect to g) normal ν. Let T be the trace-free Ricci tensor of g. Then 2 ϕ−1 |T | = −(m − 2) hϕ−1 T (ν, ν) (2.75) ∂M

M

(where of course we are integrating with respect to the volume element of g on M and ∂M ). Proof. We set g for g0 so that g = ϕ2 g. From (2.26) −1 / −1 Δ ϕ Tij = Tij + (m − 2)ϕ ϕ − δij , ij m and since g by assumption is Einstein we get / −1 Δ ϕ−1 −Tij = (m − 2)ϕ ϕ − δij . ij m The fact that T is traceless yields 2 −|T | = (m − 2)ϕ ϕ−1 ij Tij . Integrating the above on (M, g) we obtain 2 ϕ−1 |T | = (m − 2) − M

M

ϕ−1

T . ij ij

(2.76)

Again from (2.13), since (M, g) has constant scalar curvature Tik,k = 0.

(2.77)

Similarly to what we did in the proof of Lemma 2.18 we consider the vector field W = T ∇ϕ−1 .

70


Taking its divergence in the metric g we have (see (2.72)) div W = ϕ−1 ij Tij + ϕ−1 i Tik,k . Hence, using (2.77) and the divergence theorem, −1 T ∇ ϕ−1 , ν = T . ϕ ij ij ∂M

(2.78)

M

We need now to write the left-hand side of (2.78) appropriately. First we show that S if Y ∈ T ∂M , then T (Y, ν) = 0. Since T = Ric − m , , this is clearly equivalent to showing that Ric(M,g) (Y, ν) = 0 (2.79) (and this is what we expect if we want to show that g is Einstein, see Theorem 2.21). By assumption ∂M → (M, g) is totally geodesic and therefore it follows from Lemma 2.19 that, since g and g are conformally related, ∂M → (M, g) is totally umbilical. Moreover, from (2.74) 0 = h + ϕ−1

∂ϕ , ∂ν

(2.80)

with h the constant mean curvature of ∂M → (M, g). In other words ϕh = −

∂ϕ . ∂ν

In this case we can rewrite (2.73) of Lemma 2.19, since ∂M → (M, g) is totally geodesic, as hij = hδij for 1 ≤ i, j, . . . , ≤ m − 1. Since h is constant, from the above we deduce hijk = 0. On the other hand, from the Codazzi equations, m . hijk − hikj = −Rijk

Tracing with respect to i and k we deduce Rmj = 0, where Rij are the components of the Ricci tensor of (M, g). This proves (2.79). It follows that (2.78) becomes −1 ∂ ϕ−1 T . ϕ T (ν, ν) = ij ij ∂ν ∂M M


71

On the other hand, we can rewrite (2.80) as ∂ ϕ−1 = ϕ−1 h. ∂ν Substituting into the above we get ϕ−1 hT (ν, ν) = ∂M

ϕ−1

M

T . ij ij

Hence (2.75) follows immediately from (2.81) and (2.76).

(2.81)

We are now ready to prove Escobar’s result ([Esc90]). Theorem 2.21. Let Bm be the (open) unit ball in Rm and Bm the closed unit ball of Rm ; let g be a metric on Bm conformally related to the Euclidean metric , . Assume that S, the scalar curvature of Bm , g , is constant and that ∂Bm → Bm , g has constant mean curvature h with respect to the outward unit (w.r.t. g) normal ν. Then Bm , g has constant sectional curvature and it is therefore isometric to a geodesic ball in a space form with an appropriate radius depending on h. Proof. We shall prove that Bm , g has constant sectional curvature: the remaining ´ Cartan (see e.g. [Car88]). We part of the conclusion is a well-known result of E. introduce on Bm the auxiliary metric g0 =

4 1 + |x|

2

2 , .

¯ , g¯ denoting the stanNote that g0 is obtained from the upper hemisphere Sm +,g dard metric induced on Sm by the inclusion Sm → Rm+1 , by stereographic projection π from the South pole π : Bm → Sm + % $ 2 2x 1 − |x| π : x → 2, 2 . 1 + |x| 1 + |x| It follows that Bm , g0 is Einstein and ∂Bm → Bm , g0 is totally geodesic. Since the position vector field X is conformal on Bm , , , it is conformal on Bm , g0 and, since g0 |∂Bm = , ∂Bm , X|∂Bm is the outward unit normal with respect to g0 . Since g is conformally related to , , it is conformally related to g0 . Let g = ϕ−2 g0 , ϕ > 0, ϕ ∈ C ∞ Bm . First we show that g is Einstein. Indeed, we are in the assumptions of Lemma 2.20 so that, with the same notation, we have the validity of (2.75). On the other hand, on ∂Bm , X = ϕ−1 ν

72


where ν is the outward unit normal with respect to g. From Lemma 2.18, ϕ−1 T (ν, ν) = 0, Bm

and constancy of h and (2.75) give

2

ϕ−1 |T | ≡ 0,

M

that is, T ≡ 0 on Bm , g or, in other words, g is an Einstein metric. If m = 2 we are done; for m ≥ 3, to complete the proof we consider the decomposition of the i curvature tensor Rjkl given in (2.27) that we rewrite as 1 (Tik δjl − Til δjk + Tjl δik − Tjk δil ) m−2 S (δik δjl − δil δjk ) + m(m − 1)

i i Rjkl = Wjkl +

with respect to a local o.n. coframe for g. But g is conformally related to the Euclidean metric and therefore it is conformally flat. Furthermore, g is Einstein and hence T ≡ 0 Thus the above formula becomes i Rjkl =

S (δik δjl − δil δjk ) m(m − 1)

and Bm , g has constant sectional curvature.

Chapter 3

General nonexistence results The aim of this chapter is to prove a number of very general nonexistence results for Yamabe-type inequalities of the form Δu + a(x)u − b(x)uσ ≥ 0 on complete, noncompact, Riemannian manifolds. Loosely speaking, triviality of the solutions is obtained under assumptions focusing on the nonnegativity of the spectral radius for the related Schrödinger operator and Lp -properties of the solutions themselves. The spectral condition is exploited and generalized via the existence on M of a positive solution ϕ of a differential inequality of the form Δϕ + Ha(x)ϕ ≤ −K

|∇ϕ|2 ϕ

with H, K parameters satisfying H > 0, K > −1. The function ϕ is used in two different ways: in Theorem 3.2 one uses it to obtain an integral inequality involving u and its gradient from which one concludes that u is constant, and therefore necessarily identically zero; in a second group of results, we combine ϕ with the supposed solution u to give rise to a diffusion-type differential inequality for which we prove a Liouville theorem. Note that the limiting case K = −1 amounts to providing a solution on M of the Poisson equation Δψ + a(x) = 0 and it will be considered in Section 3.2 below. In the last section we give a refined version of Theorem 3.2 by using a conformal transformation of the metric in m+2 case σ ≥ m−2 (see Proposition 3.10 and Theorem 3.11). As expected, the above geometrical limitation on σ plays an essential role with respect to the conformal transformation.


73

74

3.1

Chapter 3. General nonexistence results

Some spectral considerations

First we recall some definitions and notation. From now on (M, , ) will always denote a complete, noncompact, connected Riemannian manifold. Let a(x) ∈ C 0 (M ) and for Ω ⊂ M , a bounded domain, define # |∇ϕ|2 − Ha(x)ϕ2 LH Ω # λ1 (Ω) = inf (3.1) ϕ∈C0∞ (Ω),ϕ ≡0 ϕ2 Ω so that LH denotes the operator LH = Δ + Ha(x),

H ∈ (0, +∞).

(3.2)

H If Ω has sufficiently regular boundary, λL 1 (Ω) is achieved (see, e.g., [Dav95], [GT01]) by the nonzero solutions of the Dirichlet problem H Δu + Ha(x)u + λL 1 (Ω)u = 0 on Ω, (3.3) u≡0 on ∂Ω.

We then define the first eigenvalue of LH on M as LH H λL 1 (M ) = inf λ1 (Ω), Ω

(3.4)

where Ω runs over all bounded domains in M . Observe that, due to the monoH tonicity of λL with respect to Ω, that is 1 Ω1 ⊆ Ω2

implies

LH H λL 1 (Ω1 ) ≥ λ1 (Ω2 ),

(3.5)

we have LH H λL 1 (M ) = lim λ1 (BR ). R→+∞

◦

(3.6)

Note that in case Ω2 \Ω1 = ∅, that is the interior of Ω2 \Ω1 is nonempty, then the inequality in (3.5) becomes strict. Of course (3.5) is a trivial consequence of the variational characterization (3.1), while to prove the last statement we can proceed in an elementary way by using Picone’s identity, (3.7) below, as we are now going to show (see for instance [Swa75]). Let u, v ∈ C 2 (Ω), Ω ⊂ M a bounded domain, and suppose v = 0 on Ω. A direct computation yields , 2 u 2 u 2 u 2 − 0 ≤ v 2 ∇ , ∇v . (3.7) = = |∇u| − ∇ ∇v ∇u v v v In particular, ∇u − uv ∇v ≡ 0 on Ω if and only if u = Cv for some C ∈ R. Let now u and v be eigenfunctions on Ω1 and Ω2 respectively (Ω1 ⊆ Ω2 ), LH H with eigenvalues λL 1 (Ω1 ) and λ1 (Ω2 ) . Thus H Δu + Ha(x)u + λL 1 (Ω1 )u = 0 on Ω1 , (3.8) u≡0 on ∂Ω1 ,

3.1. Some spectral considerations

75

H Δv + Ha(x)v + λL 1 (Ω2 )v = 0 on Ω2 , v≡0 on ∂Ω2 .

(3.9)

Note that we can suppose v > 0 on Ω2 . Observing that, by the divergence theorem, ∀ f, g ∈ C 2 (Ω), f ∇g, ν = f LH g + ∇f, ∇g − f Ha(x)g, ∂Ω

Ω

Ω

Ω

with ν the outward unit normal to ∂Ω, integrating (3.7) on Ω1 and using (3.8) and (3.9) gives - , 2 u u2 2 2 , ∇v = ∇ 0≤ |∇u| − |∇u| + Ha(x)u2 LH v − v v Ω1 Ω1 Ω1 Ω1 Ω1 0 1 2 LH LH LH 2 2 = |∇u| − Ha(x)u − λ1 (Ω2 )u = λ1 (Ω1 ) − λ1 (Ω2 ) u2 . Ω1

Ω1

Ω1

◦

LH H We now reason by contradiction by assuming Ω2 \Ω1 = ∅ and λL 1 (Ω1 ) = λ1 (Ω2 ). From the above inequalities it follows that, on the connected component of Ω1 , u = Cv for some C ∈ R (depending on the component). Choose one of the ◦ 1 , with ∂ Ω 1 ∩ Ω2 = ∅ (this is possible since Ω2 \Ω1 = ∅). Since component, say Ω 1 ∩ Ω2 , contradiction. u ≡ 0 on ∂Ω1 , we have that v ≡ 0 on ∂ Ω

For H = 1 we simply write L instead of L1 . It often happens, in geometrical problems, that the coefficient a(x) of L is related to the Ricci curvature of M : precisely it is a lower bound for the radial Ricci curvature Ric (∇r, ∇r) once we fix an origin o ∈ M , that is, Ric(∇r, ∇r) ≥ −(m − 1)A(r)

(3.10)

and a(x) ≥ A(r(x)) on M . We observe that the condition a(x) ≤ 0, or Ric(∇r, ∇r) ≥ 0 if equation (3.10) holds, implies (but it is not implied by) H λL 1 (M ) ≥ 0,

(3.11)

and therefore a request of the type of this latter may be interpreted as a weakening of the assumption of nonnegative Ricci curvature. It is also worth observing that (3.11) may be satisfied even if the entire sectional curvature is strictly negative. For 2 instance, on Hm −H 2 (the hyperbolic space of constant sectional curvature −H < 0) ∞ let a(x) = A(r(x)) with A(t) ∈ C ([0, +∞)), positive, nonincreasing and such that, for some ε > 0, 2 H 2 coth(Ht) on [ε, +∞), = (m−1) 4 (3.12) A(t) (m−1)2 ≤ H 2 coth(Ht) on [0, ε). 4

76


Then, the problem sinhm−1 (Ht)α + A(t) sinhm−1 (Ht)α = 0, H > 0, on [0, +∞), α (0), α(0) > 0 has a solution α ∈ C 2 ([0, +∞)) satisfying (see Proposition 7.16) α(t) ≥ C t e−

m−1 2 Ht

on [0, +∞)

for some appropriate constant C > 0. It follows that u(x) = α(r(x)) is a positive, C 2 -solution on M of Δu + a(x)u = 0

(3.13)

and, by a result of Fisher-Colbrie and Schoen (see [FCS80], [PRS05c]), λL 1 (M ) ≥ 0.

(3.14)

We also note that, as shown in the aforementioned papers, the reverse implication is true, that is, the validity of (3.14) implies the existence of a positive solution of equation (3.13). Thus (3.11) is a genuine relaxation of the assumption Ric ≥ 0 when the coefficient of the linear term in L provides a lower bound for Ric. We now relate (3.11) with the assumption on the existence of a positive solution ϕ of Δϕ + Ha(x)ϕ ≤ −K

|∇ϕ|2 , ϕ

K > −1, H > 0.

(3.15)

As already remarked, (3.11) yields the validity of (3.15) for some ϕ ∈ C ∞ (M ), ϕ = 0,with K = 0 and even with the equality sign. Thus if (3.11) holds, then (3.15) holds with K ≤ 0. On the other hand, assume that ϕ is a positive C 2 -solution of (3.15). Let ψ ∈ C0∞ (M ) and consider the vector field ψ 2 ∇ log ϕ. We compute its divergence and we use (3.15), Schwarz’s and Young’s inequalities to obtain |∇ϕ|2 ψ ψ2 Δϕ − div(ψ 2 ∇ log ϕ) = 2 ∇ψ, ∇ϕ + ϕ ϕ ϕ 2 |∇ϕ|2 ψ2 ψ −Ha(x)ϕ − (K + 1) + 2 |∇ϕ|2 + |∇ψ|2 . ≤ ϕ ϕ ϕ Thus, integrating,

|∇ψ|2 − Ha(x)ψ 2 ≥ K M

ψ2 M

|∇ϕ|2 , ϕ2


77

and from the variational characterization of the bottom of the spectrum we conclude that H if K ≥ 0, then λL 1 (M ) ≥ 0. The following simple proposition (see [BRS98]) motivates the nonexistence results of the next section. Proposition 3.1. Let (M, , ) be a complete manifold. a(x), b(x) ∈ C 0 (M ) and suppose that i) b(x) ≥ 0; ii) b(x) ≡ 0. (3.16) Assume that for some σ ∈ R there exists a positive C 2 -solution u ∈ L2 (M ) of the differential inequality Δu + a(x)u − b(x)uσ ≥ 0. (3.17) Then λL 1 (M ) < 0, with L = Δ + a(x). Proof. We reason by contradiction and assume that λL 1 (M ) ≥ 0. We fix a geodesic ball BR and we consider a smooth cut-off function ϕ such that 0 ≤ ϕ ≤ 1 and ϕ ≡ 1 on BR ,

supp ϕ ⊂ BR+1 ,

|∇ϕ| ≤ 2,

(3.18)

where supp ϕ denotes the support of the function ϕ, i. e. the closure of the set where ϕ = 0. Let u be a positive solution of (3.17) and note that, according to the variational characterization of λL 1 (BR+1 ), we have # 2 |∇(uϕ)| − a(x)u2 ϕ2 BR+1 L # . (3.19) λ1 (BR+1 ) ≤ u 2 ϕ2 BR+1 Furthermore λL 1 (M ) ≥ 0 and the monotonicity property of eigenvalues yields λL (B ) ≥ 0. Next we consider the vector field W = uϕ2 ∇u. Using the difR+1 1 ferential inequality (3.17) and computing we obtain 2

2

div W ≥ −a(x)u2 ϕ2 + b(x)uσ+1 ϕ2 + |∇(uϕ)| − u2 |∇ϕ| . Hence, by (3.19) and the divergence theorem, 2 L 2 2 2 u ϕ − u |∇ϕ| + 0 ≥ λ1 (BR+1 ) BR+1

BR+1

b(x)uσ+1 ϕ2 . BR+1

Rearranging and using (3.16) i) and (3.18)we obtain 2 σ+1 λL (B ) u + b(x)u ≤ 4 R+1 1 BR+1

BR+1

u2 . BR+1 \BR

2

Now let R → +∞ and use u ∈ L (M ) to finally have 2 λL (M ) u + b(x)uσ+1 ≤ 0. 1 M

This contradicts

λL 1 (M )

≥ 0 and (3.16).

M

78

3.1.1


The main nonexistence result

The next theorem, which is the main nonexistence result of the chapter, has been proved in [BRS98]. Theorem 3.2. Let σ > 1 and a(x), b(x) ∈ C 0 (M ) satisfy b(x) > 0 on M ;

a(x) ≤ Cb(x) for r(x) 1

(3.20)

and some constant C > 0. Given α > 0 and K > −1, suppose that, for some H≥

(1 + α)2 4α(K + 1)

(3.21)

there exists ϕ > 0 satisfying Δϕ + Ha(x)ϕ ≤ −K

|∇ϕ|2 ϕ

on M.

(3.22)

Then, the differential inequality Δu + a(x)u − b(x)uσ ≥ 0

(3.23)

has no bounded, nonnegative, nonidentically null C 2 -solutions u satisfying ⎧ ⎨i) a(x)u1+α ∈ L1 (M ); )# *−1 (3.24) ⎩ii) u1+α ∈ L1 (+∞). ∂Br Remark. Note that if a(x) is bounded, then (3.24) i), ii) are satisfied whenever u1+α ∈ L1 (M ). Remark. We recall that a sufficient condition for the validity of (3.24) ii) is #

r ∈ L1 (+∞) 1+α u Br

(3.25)

(see [RS01], Proposition 1.3). Proof. We reason by contradiction and we assume that u ∈ C 2 (M ) is a bounded, nonnegative, non-identically null solution of (3.23) satisfying (3.24). Because of (3.20) we can suppose that a(x) ≤ Cb(x)

on M

> 0. We define for some appropriate constant C & ' 1 1 σ−1 A = max sup u, C ,1 2 M


79

and we set v(x) = Then, supM v ≤

1 2

1 u(x). 2A

and v satisfies Δv + a(x)v ≥ ¯b(x)v σ ,

(3.26)

with

¯b(x) = (2A)σ−1 b(x) ≥ Cb(x) ≥ a(x) on M. Thus, it suffices to show that, if a(x) ≤ ¯b(x) on M , then (3.26) has no nonnegative, nontrivial solutions u satisfying (3.24) and the further requirement iii) sup u < 1. M

We proceed by contradiction and we divide the argument in four steps. Step 1. We claim that there exists a sequence {rk } ↑ +∞ such that lim |uα ∇u, ∇r | = 0. k→+∞

(3.27)

∂Brk

Indeed, let ε > 0. Define Wε = (u + ε)α ∇u: compute its divergence using (3.26) and apply the divergence theorem on the geodesic ball Br to obtain (u + ε)α ∇u, ∇r − α (u + ε)α−1 |∇u|2 (3.28) ∂Br Br (¯b(x)uσ − a(x)u)(u + ε)α ≥ a(x)(uσ − u)(u + ε)α ≥ Br

Br

where, in the last inequality, we have been using ¯b(x) ≥ a(x). Letting ε ↓ 0+ , the last term on the right-hand side of (3.28) converges to a(x)(uσ − u)uα Br

and the first integral on the left-hand side tends to uα ∇u, ∇r . ∂Br

Thus, by B. Levi’s monotone convergence theorem, (u + ε)α−1 |∇u|2 −→ uα−1 |∇u|2 < +∞ Br

Br

as ε ↓ 0. In particular uα−1 |∇u|2 ∈ L1loc (M ). Moreover, uα ∇u, ∇r − α uα−1 |∇u|2 ≥ a(x)(uσ − u)uα . ∂Br

Br

Br

80


Now, since u < 1, σ > 1, (3.24) i) implies that a(x)(uσ+α − u1+α ) ∈ L1 (M ) and, letting r → +∞, we obtain & ' lim inf uα ∇u, ∇r − α uα−1 |∇u|2 = B > −∞. r→+∞

∂Br

Br

We define

uα−1 |∇u|2 .

G(r) = Br

Since uα−1 |∇u|2 ∈ L1loc (M ), it follows from the co-area formula (1.87) that G(r) is absolutely continuous and that G (r) = uα−1 |∇u|2 ∂Br

is defined almost everywhere and is locally L1 . Assume by contradiction that G(r) → +∞ as r → +∞; then, for large enough r > R, α α u ∇u, ∇r ≥ uα−1 |∇u|2 , 2 Br ∂Br and therefore, by the Hölder inequality, )α 2

*2 G(r)

&

'2

≤

u ∇u, ∇r

≤ G (r)

α

∂Br

uα+1 . ∂Br

We can also suppose to have chosen R so large that G(r) > 0 for r ≥ R. From the above we obtain & '−1 G (r) α2 α+1 u ≥ G2 (r) 4 ∂Br and integrating over [R, r), we get 1 1 α2 − + ≥ G(r) G(R) 4

r

&

'−1 u

R

α+1

dt.

∂Bt

Whence, letting r → +∞, we contradict assumption (3.24) ii). Hence uα−1 |∇u|2 ∈ L1 (M ) and therefore uα−1 |∇u|2 ∈ L1 ((0, +∞)). ∂Br

It follows that there exists a sequence {rk } ↑ +∞ such that $ %$ % uα−1 |∇u|2

uα+1

lim

k→+∞

∂Brk

∂Brk

= 0.


81

By the Hölder inequality, / 12

|uα ∇u, ∇r ≤ ∂Brk

/ 12 uα−1 |∇u|2

uα+1 ∂Brk

→ 0 as k → +∞,

∂Brk

which implies the claim (3.27). Step 2. Let ϕ > 0 be a solution of (3.22), i.e., Δϕ + Ha(x)ϕ ≤ −K We claim that

|∇ϕ|2 . ϕ

uα+1 ϕ−2 |∇ϕ|2 ∈ L1 (M ).

Indeed, let V = uα+1 ϕ−1 ∇ϕ; since K > −1 we can choose θ > 0 such that 1 K + 1 − 2θ > 0. Using (3.22), a computation yields div V = (α + 1)uα ϕ−1 ∇ϕ, ∇u + uα+1 ϕ−1 Δϕ − uα+1 ϕ−2 |∇ϕ|2 (3.29) 1 ≤ 2θ(α + 1)2 uα−1 |∇u|2 + uα+1 ϕ−2 |∇ϕ|2 − Ha(x)uα+1 2θ − (K + 1)uα+1 ϕ−2 |∇ϕ|2 1 2 α−1 2 α+1 = 2θ(α + 1) u |∇u| − Ha(x)u + − K − 1 uα+1 ϕ−2 |∇ϕ|2 , 2θ and applying the divergence theorem 1 α+1 −1 u ϕ ∇ϕ, ∇r + K + 1 − uα+1 ϕ−2 |∇ϕ|2 2θ ∂Br Br α+1 2 ≤ −H a(x)u + 2θ(α + 1) uα−1 |∇u|2 ≤ B ∈ R Br

Br

where, in the last inequality, we have used the fact that a(x)uα+1 , uα−1 ∈ L1 (M ). Setting for the ease of notation 1 F (r) = K + 1 − uα+1 ϕ−2 |∇ϕ|2 , 2θ Br the above inequality can be written in the form uα+1 ϕ−1 ∇ϕ, ∇r ≤ B − F (r). ∂Br

Now assume by contradiction that F (r) → +∞ as r → +∞. Thus, for each r ≥ R sufficiently large, 1 uα+1 ϕ−1 ∇ϕ, ∇r ≤ − F (r). 2 ∂Br

82


On the other hand, by the Schwarz inequality, uα+1 ϕ−1 ∇ϕ, ∇r F (r) ≤ −2 ∂Br

& ≤

u

α+1

'1/2 & 4

∂Br

that is,

u

α+1

ϕ

−2

|∇ϕ|

2

'1/2 ,

∂Br

'1/2

& F (r) ≤

uα+1

1/2

{8F (r)}

.

∂Br

Thus, squaring, 1 F (r) ≥ F (r)2 8

&

'−1 u

α+1

r ≥ R.

,

∂Br

Integrating this latter over [R, r] gives F (R)

−1

− F (r)

−1

1 ≥ 8

r

&

'−1 u

R

α+1

dt,

∂Bt

thus, letting r → +∞, we contradict (3.24) ii). Step 3. We note that

uα+1 ϕ−1 ∇ϕ, ∇r = 0.

lim

r→+∞

∂Br

Indeed, because of Step 2, there exists a sequence {rk } ↑ +∞ such that / / α+1 α+1 −2 2 lim =0 u u ϕ |∇ϕ| k→+∞

∂Brk

and therefore,

∂Brk

/2 u

α+1

ϕ

−1

∂Brk

/

≤

u ∂Brk

∇ϕ, ∇r /

α+1

u

α+1

ϕ

−2

|∇ϕ|

2

→0

as k → +∞.

∂Brk

It remains to show that the desired limit exists. According to (3.22), (3.29) and the divergence theorem, we have α+1 −1 u ϕ ∇ϕ, ∇r = (α + 1) uα ϕ−1 ∇ϕ, ∇u (3.30) ∂Br Br uα+1 a(x) − uα+1 ϕ−2 |∇ϕ|2 . −H Br

Br


83

Thus, we are reduced to showing that each of the summands on the right-hand side of (3.30) has a finite limit, as r → +∞. This easily follows from assumption (3.24) i), and from the following fact from Step 1: α −1 u ϕ ∇ϕ, ∇u ≤ 2uα−1 |∇u|2 + 1 uα+1 ϕ−2 |∇ϕ|2 ∈ L1 (M ). 2 Step 4. Fix ε > 0 so small that supM u + ε < 1, and define the vector field Zε = H −1

uα+1 (u + ε)α ∇u. ∇ϕ − ϕ 1 − (u + ε)σ−1

Then, using (3.22), (3.26) and ¯b(x) ≥ a(x), we have div Zε = H −1 (α + 1)uα ϕ−1 ∇ϕ, ∇u + H −1 uα+1 ϕ−1 Δϕ (u + ε)α−1 |∇u|2 1 − (u + ε)σ−1 (u + ε)α+σ−2 (u + ε)α 2 − (σ − 1) |∇u| − Δu 2 1 − (u + ε)σ−1 [1 − (u + ε)σ−1 ]

− H −1 uα+1 ϕ−2 |∇ϕ|2 − α

≤ H −1 (α + 1)uα ϕ−1 ∇ϕ, ∇u − H −1 (K + 1)uα+1 ϕ−2 |∇ϕ|2 (u + ε)α−1 (u + ε)α+σ−2 2 |∇u|2 − (σ − 1) 2 |∇u| σ−1 σ−1 1 − (u + ε) [1 − (u + ε) ] (u + ε)α ¯b(x)uσ − a(x)u ≤ − 1 − (u + ε)σ−1 − a(x)uα+1 − α

≤ H −1 (α + 1)ϕ−1 uα ∇ϕ, ∇u − H −1 (K + 1)uα+1 ϕ−2 |∇ϕ|2 −α +

(u + εα−1 (u + ε)α+σ−2 2 2 |∇u| − (σ − 1) 2 |∇u| 1 − (u + ε)σ−1 [1 − (u + ε)σ−1 ]

(u + ε)α a(x)(1 − uσ−1 )u − a(x)uα+1 . 1 − (u + ε)σ−1

Integrating over Br and applying the divergence theorem gives (u + ε)α −1 α+1 −1 u ϕ ∇ϕ, ∇r − ∇u, ∇r ≤ I + II + III, H σ−1 ∂Br ∂Br 1 − (u + ε) where we have set I= H −1 (α + 1)ϕ−1 uα ∇ϕ, ∇u − (K + 1)uα+1 ϕ−2 |∇ϕ|2 ;

Br

(u + ε)α−1 (u + ε)α+σ−2 2 |∇u|2 − (σ − 1) 2 |∇u| ; σ−1 σ−1 1 − (u + ε) [1 − (u + ε) ] Br ' & α (u + ε) 1 − uσ−1 u . a(x) uα+1 − III = 1 − (u + ε)σ−1 Br II =

−α

84


Letting ε ↓ 0+ and applying the dominated convergence theorem, the left-hand side of the above converges to uα H −1 ∇u, ∇r uα+1 ϕ−1 ∇ϕ, ∇r − σ−1 ∂Br ∂Br 1 − u while

II → −

α Br

uα−1 uα+σ−2 2 2 |∇u| + (σ − 1) 2 |∇u| 1 − uσ−1 [1 − uσ−1 ]

and finally III → 0. Then, using H −1

(α + 1)2 ≤ α, 4(K + 1)

we obtain H −1

uα uα+1 ϕ−1 ∇ϕ, ∇r − ∇u, ∇r σ−1 ∂Br ∂Br 1 − u & (α + 1)2 α−1 −1 ≤ −H (K + 1)ϕ−2 uα+1 |∇ϕ|2 + |∇u|2 u 4(K + 1) Br ' −1 α − (α + 1) ∇ϕ, ∇u ϕ u / 2 uα−1 uα+σ−2 −1 (α + 1) α−1 α |∇u|2 + (σ − 1) − u 2 −H σ−1 σ−1 1 − u 4(K + 1) [1 − u ] Br 2 u√ −1 α−1 α + 1 ≤ −H u 2√K + 1 ∇u − ϕ K + 1∇ϕ Br σ−1 1 + σ−1 2 α −u −α u(α−1)+(σ−1) 2 |∇u| . (1 − uσ−1 ) Br

Now, letting r → +∞ along the sequence {rk } constructed in Step 1 and using Step 3 yields 0≤H

−1

u Br

2 α+1 u√ 2√K + 1 ∇u − ϕ K + 1∇ϕ σ−1 1 + σ−1 α −u −α uα+σ−2 |∇u|2 ≤ 0. σ−1 )2 (1 − u Br

α−1

Let now A = ∅ be a connected component of the set {x ∈ M : u(x) > 0}.


85

The above forces ∇u = 0 on A and thus u = c1 > 0 on A. In turn this implies ∇ϕ = 0 on A and ϕ = c2 > 0 on A. But, on all of M , hence on A, Δϕ + Ha(x)ϕ ≤ 0, then Ha(x)c2 ≤ 0. Because a(x)c1 − b(x)cσ1 ≥ 0 implies a(x)c1 ≥ 0, it follows that a(x) = 0 on A and therefore, since 0 = Δu ≥ −a(x)u + ¯b(x)uσ = ¯b(x)cσ1

on A

we have ¯b(x) = 0 on A contradicting (3.20).

Before commenting on Theorem 3.2 we prove the next result (see [PRS10]) in the same spirit. Theorem 3.3. Let a(x), b(x) ∈ C 0 (M ) and assume that b(x) ≥ 0. Let H > 0, K > −1, A ∈ R be constants satisfying max {0, A} ≤ H(K + 1) − 1,

(3.31)

and suppose that there exists a positive C 2 -solution ϕ of the differential inequality Δϕ + Ha(x)ϕ ≤ −K

|∇ϕ|2 ϕ

on M.

(3.32)

Then the differential inequality uΔu + a(x)u2 − b(x)uσ+1 ≥ −A|∇u|2 , σ ≥ 1,

(3.33)

has no nonnegative C 2 -solutions u on M satisfying supp u ∩ {x ∈ M : b(x) > 0} = ∅ and

& ϕ

β+1 H (2−p)

'−1 u

p(β+1)

∈ L1 (+∞)

(3.34)

(3.35)

∂Br

for some p > 1 and max {0, A} ≤ β ≤ H(K + 1) − 1. Remark. Note that if p = 2, then the nonintegrability assumption (3.35) involves u alone and reduces to & '−1 2(β+1) u ∈ L1 (+∞). (3.36) ∂Br

86


Proof. Let u ≥ 0 be a solution of (3.33) on M satisfying (3.34) and (3.35). Fix ε > 0, α ∈ R and set v = ϕ−α (u2 + ε)

β+1 2

.

A straightforward computation that uses (3.32) and (3.33) yields |∇ϕ|2 + (β + 1)(u2 + ε)β b(x)uσ+1 v div(ϕ2α ∇v) ≥ α(K − α + 1)(u2 + ε)β+1 ϕ2 " ! u2 2 β+1 αH − (β + 1) 2 + a(x)(u + ε) u +ε " ! u2 + (β + 1)(u2 + ε)β 1 − A + (β − 1) 2 |∇u|2 . u +ε We choose α = H −1 (β + 1), so that our assumptions on β, H and K give 0 < α ≤ K + 1. Therefore α(K − α + 1) ≥ 0, and using b(x) ≥ 0 and β + 1 ≥ 0 we deduce that v div(ϕ2α ∇v) ≥ ε(β + 1)a(x)(u2 + ε)β (3.37) ! " 2 u |∇u|2 . + (β + 1)(u2 + ε)β 1 − A + (β − 1) 2 u +ε Let r(t) ∈ C 1 (R) and s(t) ∈ C 0 (R) satisfy the conditions ii) r(v) + vr (v) ≥ s(v) > 0 on [0, +∞)

i) r(v) ≥ 0,

(3.38)

and consider the vector field Z = vr(v)ϕ2α . For fixed t and δ > 0 let also ψδ be the Lipschitz function defined by ⎧ ⎪ ⎨1, ψδ (x) =

t+δ−r(x) , δ ⎪

⎩

0,

if r(x) ≤ t, if t < r(x) < t + δ, if r(x) ≥ t + δ.

Using (3.37), (3.38) and the definition of ψδ we compute div(ψδ ) = ψδ div Z + ∇ψδ , Z

' " u2 2 ≥ (β + 1)r(v)(u + ε) εa(x) + 1 − A + (β − 1) 2 |∇u| χBt u +ε 1 + s(v)ϕ2α |∇v|2 χBt + ∇r, Z χB¯t+δ \Bt δ 2

β

&

!


87

whence, integrating and using the divergence theorem and Cauchy-Schwarz inequality we obtain " & ! ' u2 2 β 2 |∇u| (β + 1)r(v)(u + ε) ε r(x) + 1 − A + (β − 1) 2 u +ε Bt 1 s(v)ϕ2α |∇v|2 ≤ |Z|. + δ Bt+δ \Bt Bt By H¨ older’s inequality the integral on the right-hand side is bounded above by / 12 / 12 2 1 r(v) 1 ϕ2α ϕ2α s(v)|∇v|2 . v2 δ Bt+δ \Bt s(v) δ Bt+δ \Bt Inserting this into the above inequality, letting δ ↓ 0+ and using the co-area formula (1.87) we deduce that

" ' u2 2 |∇u| (β + 1)r(v)(u + ε) ε a(x) + 1 − A + (β − 1) 2 u +ε Bt ' 12 & 2 2α 1 2α 2 2α r(v) 2 +s(v)ϕ |∇v| ≤ ϕ ϕ s(v)|∇v|2 2 . v s(v) 2

&

!

β

(3.39)

As ε ↓ 0, v = vε → v0 = ϕ−α uβ+1 , therefore, using the dominated convergence theorem in (3.39) we get s(v0 )ϕ2α |∇v0 |2 + (β + 1)(β − A) r(v0 )u2β |∇u|2 (3.40) Bt

&

≤

ϕ2α

2

r(v0 ) 2 v s(v0 ) 0

' 12

Bt

ϕ2α s(v0 )|∇v0 |

Define

1 2 2

.

(3.41)

ϕ2α s(v0 )|∇v0 |2 ;

h(t) = Bt

by the co-area formula (1.87) h is Lipschitz and ϕ2α s(v0 )|∇v0 |2 , h (t) = ∂Bt

and noting that the coefficient of the second integral on the left-hand side of (3.40) is nonnegative by the conditions imposed on β, we obtain & h(t) ≤

ϕ ∂Bt

2α r(v0 )

2

s(v0 )

v02

' 21

1

[h (t)] 2 .

(3.42)

88


Our aim is to show that under assumption (3.35) v0 is constant. We reason by contradiction, then there exists R0 such that h(t) > 0 ∀ t ≥ R0 and therefore the right-hand side of (3.42) is positive for t ≥ R0 . Dividing through by h(t), squaring and integrating the resulting differential inequality between R and r with R0 ≤ R ≤ r yields '−1 r& 2 −1 −1 −1 2α r(v0 ) 2 ϕ dt. (3.43) h(R) ≥ h(R) − h(r) ≥ v s(v0 ) 0 R We choose a sequence f of functions rn (t) =

1 t + n 2

p−2 2

, sn (t) = min {p − 1, 1} rn (t) ∀ n ∈ N, p > 1.

Since condition (3.38) holds for all n, so does (3.43), hence, letting n → +∞ and using the Lebesgue and monotone convergence theorems we deduce that there exists c > 0 depending only on p such that & '−1 '−1 r & v0p−2 ϕ2α |∇v0 |2 ≥C ϕ2α v0p dt. (3.44) BR

R

∂Bt

Now, recalling that α = (β + 1)/H and the definition of v0 , we have ϕ2α v0p = ϕ(2−p)(β+1)/H u2(β+1) and the required contradiction is obtained by letting r → +∞ and using assumption (3.35). Thus v0 is constant, and we deduce that there exists a constant C ≥ 0 such that uH = Cϕ. Since u is not identically zero by (3.34), C > 0 and u is strictly positive on M . We insert the expression of ϕ in terms of u in (3.32), divide by CHuH−2 and subtract the result from (3.33) to obtain [A − H(K + 1)]|∇u|2 ≥ b(x)uσ+1 . Since the coefficient of |∇u|2 is nonpositive, by (3.31) we conclude that b(x)uσ+1 ≤ 0 which contradicts (3.34).

Remark. Observe that the above proof actually shows that if A < H(K + 1) − 1, then ∇u = 0, so that u, and therefore ϕ, are necessarily constant. It follows from (3.32) and (3.33) that 0 ≥ a(x)u ≥ b(x)uσ+1 ≥ 0, so that, if a(x) does not vanish identically, then u ≡ 0 without any strict positivity assumption on b(x). On the other hand, if A = H(K + 1) − 1, then the conclusion depends on the fact that b(x) is positive somewhere.


89

Remark. We also note that if u is assumed to be strictly positive, then the conclusion of Theorem 3.3 holds, with a much easier proof, if we assume that max {−1, A} ≤ H(K + 1) − 1 and that β > −1, β ≥ A, β ≤ H(K + 1) − 1. We observe that in both Theorems 3.2 and 3.3 we require some integrability type condition for the positive (nonnegative) solution u of (3.23) Δu + a(x)u ≥ b(x)uσ ,

(3.23)

respectively (3.24) ii), that is &

'−1 u

α+1

∈ L1 (+∞)

for some α > 0

∂Br

and (3.36), that is &

u2(1+β)

'−1

∈ L1 (+∞)

for some max{0, A} ≤ β ≤ H(K + 1) − 1.

∂Br

Sometimes this type of information is already contained in the differential inequality. This is the content of the next Proposition 3.4. Let (M, , ) be a complete Riemannian manifold and let a(x), b(x) ∈ C 0 (M ) with b(x) > 0 on M . Assume that u ≥ 0 is a C 2 -solution of the differential inequality uΔu + a(x)u2 − b(x)uσ+1 ≥ −A|∇u|2

(3.45)

for A ≤ 1 and σ > 1. Then, for every p ≥ 1, p > A + 2 there exist constants C1 , C2 > 0 which depend only on p, σ and R0 > 0 sufficiently large, such that, ∀R ≥ R0 , b(x)u

p+σ−2

≤ C1 R

BR

−2 p+σ−2 σ−1

b(x)

p−1 − σ−1

+ C2

B2R

B2R

a+ (x) b(x)

p−1 σ−1

a+ (x), (3.46)

where a+ denotes the positive part of a. Proof. First we observe that we may suppose that u ≡ 0, for otherwise there is nothing to prove. Thus, there exists R0 > 0 such that u ≡ 0 on BR , ∀ R ≥ R0 . Next, ∀ R ≥ R0 let ψ = ψR : M → [0, 1] be a smooth cut-off function such that ψ ≡ 1 on BR ,

ψ ≡ 0 on M \B2R ,

|∇ψ| ≤

p−1 C p+σ−2 on B2R ψ R

(3.47)

90


for some C > 0 which depends only on σ and p. Note that this is possible since the exponent p−1 < 1. p+σ−2 Having fixed ε > 0, we let W be the vector field defined by W = ψ 2 (u + ε)p−3 u∇u; a computation that uses (3.45) yields & ' u |∇u|2 div W ≥ ψ 2 (u + ε)p−3 −a(x)u2 + b(x)uσ+1 + 1 − A − (p − 3) u+ε + 2ψ(u + ε)p−3 u ∇u, ∇ψ . We estimate the last term on the right-hand side using Cauchy-Schwarz inequality and Young inequality 2ab ≤ λa2 + λ−1 b2 with λ = p − 2 − A > 0 to obtain 1 div W ≥ ψ 2 (u + ε)p−3 −a(x)u2 + b(x)uσ+1 − u(u + ε)p−2 |∇ψ|2 . p−2−A We integrate the above inequality, apply the divergence theorem, rearrange, let ε → 0+ and use the dominated convergence theorem in this order, to deduce that 1 b(x)ψ 2 up+σ−2 ≤ up−1 |∇ψ|2 + ψ 2 a+ (x)up−1 . (3.48) p − 2 − A B2R B2R B2R If p = 1, the conclusion follows immediately using (3.47). If p > 1, we denote by I and II the two integrals on the right-hand side of (3.48), and use Hölder inequality with conjugate exponents p+σ−2 >1 p−1

and

p+σ−2 σ−1

and the assumption that b(x) > 0 to estimate p−1 & σ−1 ' p+σ−2 ' p+σ−2 & p−1 p−1 −2 σ−1 − σ−1 2 p+σ−2 2 p+σ−2 b(x)ψ u ψ b(x) |∇ψ| σ−1 I≤ B2R

and

&

B2R

2 p+σ−2

II ≤

p−1 & ' p+σ−2

2

b(x)ψ u

ψ a+ (x)

B2R

p+σ−2 σ−1

b(x)

p−1 − σ−1

σ−1 ' p+σ−2

.

B2R

Inserting into (3.48), noting that the integral on the left-hand side is strictly positive by the choice of R, and simplifying, we obtain ! " & p−1 p−1 1 −2 σ−1 − σ−1 2 p+σ−2 2 p+σ−2 b(x)ψ u ≤ ψ b(x) |∇ψ| σ−1 p − 2 − A B2R B2R p+σ−2 " σ−1 / σ−1 ! ψ 2 a+ (x)

+ B2R

p+σ−2 σ−1

p−1

b(x)− σ−1

p+σ−2

.


91

The required conclusion follows again using (3.47) and the elementary inequality (a + b)τ ≤ 2τ (aτ + bτ ) valid for a, b, τ ≥ 0. For the sake of illustration let us consider the case b(x) ≥

b(x) > 0,

C r(x)μ

for r(x) 1

(3.49)

and some constants C > 0, μ ≥ 0. Then, from (3.46) we have up+σ−2 ≤ C1 Rμ−2

p+σ−2 p−1 σ−1 + σ−1 μ

BR

In particular, if

p+σ−2

vol(B2R ) + C2 B2R

a+ (x) b(x)

a+ (x) σ−1 a+ (x). (3.50) b(x)

is bounded above and a+ (x) ∈ L1 (M ), μ−2

up+σ−2 ≤ C2 + C1 R σ−1 (p+σ−2) vol(B2R ).

(3.51)

B2R

We observe now that (3.24) or (3.36) are implied by r

#

Br

and from (3.51)

uξ

∈ L1 (+∞),

&

'−1 u

p+σ−2

ξ≥0

∈ L1 (+∞)

∂Br

in case μ−2

C1 r σ−1 (p+σ−2) vol(B2R ) ≤ Cr2 log r

for r 1,

in other words if μ−2

vol(B2r ) ≤ Cr2− σ−1 (p+σ−2) log r

for r 1.

This is a slow growth for vol(Br ). This somehow forces us to find a better way to guarantee integrability and this will be achieved with upper a priori estimates which however are obtained under lower bound type restrictions on the Ricci tensor. We shall return to this in Chapter 4. Using Proposition 3.4 we obtain the following Theorem 3.5. Let (M, , ) be a complete Riemannian manifold, and let a(x), b(x)∈ C 0 (M ) where b(x) > 0 on M and b(x) ≥

C r(x)μ

(3.52)

92


for r(x) 1 and for some constants C > 0 and 0 ≤ μ ≤ 2. Assume that a+ (x) a+ (x) = O r2−μ log r as r → +∞, < +∞ and (ii) (i) sup b(x) M Br

(3.53)

and that, for some H ≥ 1, the operator LH = Δ + Ha(x) satisfies H λL 1 (M ) ≥ 0.

(3.54)

Finally, let A and σ be such that A ≤ 1, A < H − 1, 1 < σ ≤ 2H + 1 and σ < 2H − A and assume that

2H volBr = O r2+(2−μ) σ−1 log r as r → +∞. (3.55) Then the only nonnegative C 2 solution u of the differential inequality uΔu + a(x)u2 − b(x)uσ+1 ≥ −A|∇u|2

(3.56)

is u ≡ 0. Proof. If we set p = 2H + 2 − σ, the conditions imposed on the parameters imply that p satisfies the assumptions listed in the statement of Proposition 3.4. The lemma and condition (3.53) (i) show that there exist constants Ci > 0 such that 4H 2H b(x)u2H ≤ C1 r− σ−1 b(x)1− σ−1 + C2 a+ (x) (3.57) Br

B2r

B2r

for r > 0 sufficiently large. We use condition (3.52) to estimate from below the integral on the left-hand side. On the other hand, since σ < 2H + 1, we may again use condition (3.52) to estimate from above the first integral on the right-hand side, and (3.53) (ii) to estimate from above the second integral, and deduce that, for r sufficiently large, 2H u2H ≤ C r(μ−2) σ−1 volB2r + r2 log r , Br

whence, using the volume growth condition (3.57) we conclude that u2H ≤ Cr2 log r for r 1. Br

This immediately implies that, for r large, #

1 r ≥C ∈ L1 (+∞), 2H r log r u Br

which in turn yields (see, e.g., [RS01] Proposition 1.3) #

1 ∈ L1 (+∞). u2H ∂Br

We may therefore apply Theorem 3.3 with K = 0 and β = H − 1 to deduce that supp u = ∅, that is, u ≡ 0.

3.2. The endpoint case K = −1 and the Poisson equation

93

Remark. The argument used in the proof shows that the condition that ab+ is bounded above may be removed provided we replace (3.57) with

2H 2H a+σ−1 = O r2−μ σ−1 log r as r → +∞. (3.58) Br

Note that since the integral on the left-hand side is a nondecreasing function of r this also imposes the further restriction μ ≤ (σ − 1)/H, with corresponding restrictions being imposed on the range of the other parameters.

3.2 The endpoint case K = −1 and the Poisson equation Theorem 3.3 does not cover the “endpoint” case where K = −1 in (3.22), which we are going to consider presently. We therefore assume that there exists a positive solution ϕ of |∇ϕ|2 Δϕ + Ha(x)ϕ ≤ , H > 0. (3.59) ϕ If u is a C 2 solution of (3.33) with σ ≥ 0, we define v = ϕ−γ u, γ ≥ 0. A computation that uses (3.59), (3.33) and Young inequality yields vΔv = (γH − 1)a(x)ϕ−2γ u2 + b(x)ϕ−2γ uσ+1 + γ 2 ϕ−2γ−2 u2 |∇ϕ|2 − Aϕ−2γ |∇u|2 − 2γϕ−2γ−1 u∇u, ∇ϕ ≥ (γH − 1)a(x)ϕ−2γ u2 + b(x)ϕ−2γ uσ+1 1 + γ 2 (1 − )ϕ−2γ−2 u2 |∇ϕ|2 − (A + ε)ϕ−2γ |∇u|2 . ε Choosing γ = 1/H and ε = −A, the right-hand side reduces to b(x)ϕ−2/H uσ+1 +

1 1 (1 + )ϕ−2/H−2 u2 |∇ϕ|2 , H2 A

and we easily deduce that, if A ≤ −1, then the function v satisfies Δv ≥ b(x)ϕ(σ−1)/H v σ .

(3.60)

Using (3.60) we obtain the following version of Theorem 3.3. Theorem 3.6. Let a(x), b(x) ∈ C 0 (M ), with b(x) ≥ 0, and assume that ϕ is a positive C 2 solution of (3.59) satisfying ϕ(x) ≥ Cr(x)1/δ

(3.61)

for r(x) 1, and some constants C > 0 and δ > 0. Then the differential inequality uΔu + a(x)u2 − b(x)uσ+1 ≥ −A|∇u|2 ,

94


with σ ≥ 0 and A ≤ −1, has no nonnegative C 2 solution satisfying supp u ∩ {x : M : b(x) > 0} = ∅

(3.62)

rδp ∈ L1 (+∞) p u ∂Br

(3.63)

and # for some p > 1. Proof. According to (3.60) above, the function v = ϕ−1/H u is sub-harmonic. Further, (3.61) and (3.63) imply that

−1 vp ∈ L1 (+ + ∞). ∂Br

An application of Theorem B in [RS01] shows that v is constant. The conclusion now follows as in the proof of Theorem 3.3. Note that, even in the case of Theorem 3.6, if A < −1, then the conclusion can be strengthened to assert that every nonnegative solution of (3.6) vanishes identically, unless a(x) = b(x) ≡ 0. In applying Theorem 3.6 it is of course crucial to be able to find positive solutions of (3.59) satisfying the asymptotic lower bound (3.61). By contrast, in order to apply Theorem 3.3 one needs a positive solution of (3.22), whose existence, in typical applications like the one exemplified by Theorem 3.5 above, is guaranteed by means of assumptions on the spectrum of a suitable operator. Observe now that if v is a solution of the Poisson equation Δv = a(x), then the function ϕ = e−v is a positive solution of Δϕ + a(x)ϕ =

|∇ϕ|2 , ϕ

and furthermore, an upper bound for v yields a lower bound for ϕ. The Poisson equation on complete Riemannian manifolds has been extensively studied using heat kernel techniques to obtain bounds on the Green kernel. To illustrate an application of Theorem 3.6, we consider the elementary case where the positive part of a(x) is integrable. We first recall that a manifold is said to be parabolic if every positive bounded-above subharmonic function is constant, and nonparabolic otherwise (see [Gri99], [PRS05b]); we also recall that the definition of non-parabolicity is equivalent to the existence of a positive Green kernel (see again [Gri99]). Then we have the following lemma (see, e.g., the proof of Theorem 3.2 in [NST01]).


95

Lemma 3.7. Let (M, , )) be a complete, nonparabolic manifold, and let ρ ∈ C 0,α (M ) ∩ L1 (M ), 0 ≤ α < 1, be a nonnegative function. Then, there exists a solution v ∈ C 2 of the Poisson equation Δv = ρ satisfying v ≤ 0. Proof. Let G(x, y) be the Green kernel, i.e., the minimal positive fundamental solution of the Laplacian, which exists by the assumption that M is nonparabolic. ∞ The # Green kernel is symmetric and, if ψ ∈ Cc (M ), then the function u(x) = − M G(x, y)ψ(y) is smooth and satisfies Δu = ψ. # We claim that if ρ ∈ C 0 (M )∩L1 (M ), then the function v = − M G(x, y)ρ(y) is well defined and locally bounded. Assuming the claim, for every ψ ∈ C0∞ (M ) we have vΔψ = − ψ(x) ΔG(x, y)ρ(y) M M M ρ(y) G(x, y)Δψ(x) = ρ(y)ψ(y), =− M

M

M

so that v satisfies the Poisson equation in distributional sense, and therefore, by standard elliptic regularity (see [Aub98]), Theorem 3.55), it is a classical solution. Clearly, v is nonpositive. To prove the claim, fix R > 0 and for every x ∈ BR we write G(x, y)ρ(y) = G(x, y)ρ(y) + G(x, y)ρ(y). (3.64) M

B2R

M \B2R

Since G(x, y) is locally integrable uniformly for x ∈ BR , the first integral on the right-hand side is bounded above by a constant independent of x ∈ BR . On the other hand, by the local Harnack inequality there exists a constant C independent of x ∈ BR and such that G(x, y) ≤ CG(o, y)

for every y ∈ M \ B2R .

Moreover, sup G(o, y) < +∞. M \B2R

Indeed, let Ωn be an exhaustion of M by open sets containing o and with smooth boundary and let Gn by the Green kernel of Ωn and recall that, by the standard construction of the Green kernel G(x, y), Gn (x, y) → G(x, y) locally uniformly in M \ {x}. Let C > sup∂B2R G(o, y), then, for every sufficiently large n we have C > G(o, y) ≥ Gn (o, y) for y ∈ ∂B2R and clearly C > Gn (o, y) = 0 for y ∈ ∂Ωn . Thus, by the comparison principle, C > Gn (o, y) in Ω \ B2R , whence, letting

96


n → +∞, G(o, y) ≤ C for y ∈ M \ B2R . It follows that there exists a constant C independent of x ∈ BR and y ∈ M \ B2R such that G(x, y) ≤ C . Since ρ is integrable, this implies that the second integral on the right-hand side of (3.64) is also bounded independently of x ∈ BR , as required to complete the proof of the claim. Corollary 3.8. Let (M, , ) be a complete, nonparabolic manifold, let the functions a(x) ∈ C 0,α (M ), and b(x) ∈ C 0 (M ) satisfy b(x) > 0 and a+ (x) ∈ L1 (M ),

(3.65)

and suppose that for some constants σ > 1, A ∈ (−∞, −1] and μ, p, q satisfying q > max{1, 3 − σ}, we have

0≤μ≤2

σ−1 , σ+q−2

p>

q+σ−2 , σ−1

p−1 a+ (x)p = O r[2(σ−1)−μ(q+σ−2)] q−1 as r → +∞ Br

σ+q−2 as r → +∞ volBr = O r2+(2−μ) σ−1 b(x) ≥

C r(x)μ

for r(x) 1.

(3.66) (3.67) (3.68)

Then there are no nonnegative, nonidentically zero C 2 (M ) solutions of the differential inequality uΔu + a(x)u2 ≥ b(x)uσ+1 − A|∇u|2

on M.

(3.69)

Proof. Since a+ is integrable, by Lemma 3.7 and the preceding discussion there exists a solution ϕ ≥ 1 of |∇ϕ|2 Δϕ + a+ ϕ = , ϕ and ϕ is a solution of the differential inequality (3.59). Now, let u be a nonnegative solution of (3.69). Noting that q > 1, and A ≤ −1 imply q > A + 2, applying Proposition 3.4 and using the lower bound (3.68) for b(x) we obtain q+σ−2 q+σ−2 σ+q−2 uq+σ−2 ≤ C1 r(μ−2) σ−1 volB2r + C2 rμ σ−1 a+ (x) σ−1 . (3.70) Br

B2r

We claim that (3.66) implies

σ+q−1 σ+q−2 a+ (x) σ−1 = O r2−μ σ−1 B2r

as r → +∞,

(3.71)


97

which, together with the volume growth assumption (3.67) yields uq+σ−2 = O r2 as r → +∞. Br

As in the proof of Theorem 3.5, it follows (see, e.g., [RS01] Proposition 1.3) that u satisfies condition (3.63) with δ = 0 and exponent q + σ − 2 which is greater than 1 by the conditions on q. Thus, Theorem 3.6 (with a(x) replaced by a+ (x)) applies and u vanishes identically. To conclude it remains to prove the claim. To this end, we set p = (q + σ − 2)/(σ − 1), and apply Hölder inequality with conjugate exponents (p − 1)/(p − p ) and (p − 1)/(p − 1) to estimate p−p p −1 a+ (x)p = a+ (x) p−1 +p p−1 Br

Br

≤

a+ (x) B2r

p−p p−1

a+ (x)

p

−1

pp−1

B2r

p −1 q+σ−2 = O r[2(σ−1)−μ(q+σ−2)] q−1 = O r2−μ σ−1 ,

as required.

Remark. Assume that b(x) satisfies the condition stated in the corollary, with μ < σ − 1, and that conditions (3.66) and (3.67) are replaced by

μ 2 as r → +∞, a+ (x) σ−1 = O r2[1− σ−1 ] Br

μ−2 as r → +∞. volBr = O r2[1− σ−1 ] It follows from (3.70) above with q + σ − 2 = 2, that every nonnegative solution of Δu + a(x)u − b(x)uσ = 0 satisfies

u2 ≤ Cr2 log r,

(3.72)

(3.73)

Br

and the same estimate is clearly satisfied by the difference of two solutions. An application of Theorem 4.1 in [BRS98] shows that (3.72) has at most one positive solution. We remark in this respect that if we replace u2 in (3.73) with up with p > 2, then the conclusion of Theorem 4.1 in [BRS98] fails, as the example described on pages 214-215 therein shows (see also Theorem 5.1 below). Clearly, whenever a solution to the Poisson equation is available one can obtain a version of Corollary 3.8 above. According to a recent result by O. Munteanu and N. Sesum, [MS10], if (M, , ) has Ricci curvature bounded from below by a

98


negative constant, and its Laplace operator has positive bottom of the spectrum, and if ρ satisfies the decay condition |ρ(x)| ≤

C (1 + r(x))γ

for some γ > 1, then the Poisson equation Δv = ρ(x) has a solution satisfying |v(x)| ≤ AeBr(x) for some A, B > 0. Applying this result, and arguing as in the proof of Corollary 3.8 one obtains the following Corollary 3.9. Let (M, , ) be a complete manifold with Ricci curvature bounded from below by a negative constant, and positive bottom of the spectrum. Let a(x) ∈ C 0,α (M ), and 0 < b(x) ∈ C 0 (M ) and suppose that for some constants σ > 1, A ∈ (−∞, −1], μ > 0, p ≥ 1, γ ∈ (1, 2] and λ > 0 satisfying p > A + 2,

0≤μ≤2

σ−1 , σ+p−2

γ

p+σ−2 p+σ−2 ≤ λ ≤ 2 − (μ − γ) , σ−1 σ−1

we have a+ (x) ≤

C , (1 + r(x))γ

b(x) ≥

C and volBr = O rλ . μ (1 + r(x))

Then there are no nonnegative, nonidentically zero C 2 (M ) solutions of the differential inequality (3.69) on M .

3.3 A refined version of Theorem 3.2 We shall now relax the assumption b(x) > 0 on M in Theorem 3.2. This is related to geometry and first we need to derive the transformation law of the LaplaceBeltrami operator under a conformal change of the metric. We go back to Chapter 2 and we suppose to have a conformal change of the type , = ϕ2 , ,

ϕ > 0, ϕ ∈ C ∞ (M ),

(3.74)

with the notation in Chapter 2, θi = ϕθi

i = 1, . . . , m = dim M,

(3.75)

dϕ = ϕt θt ,

(3.76)

and, having set

3.3. A refined version of Theorem 3.2

99

ϕj i ϕi j θji = θji + θ − θ . ϕ ϕ Let u ∈ C 2 (M ). Then

(3.77)

j ϕθj = uj θj , du = u j θj = u

that is, u j = uj ϕ−1 .

(3.78)

Furthermore, using (3.75) and (3.77), (3.76) ϕj k ϕk j t k −1 −1 k uj − u k θj = d(uj ϕ ) − (uk ϕ ) θj + u jt θ = d θ − θ ϕ ϕ = ϕ−1 duj − uj ϕ−2 ϕt θt − ϕ−1 uk θjk − ϕ−2 uk ϕj θk + ϕ−2 uk ϕk θj

= ϕ−1 ujt θt − ϕ−2 uj ϕt + ut ϕj − δtj uk ϕk θt 0

1 = ϕ−2 ujt − ϕ−1 uj ϕt + ut ϕj − δtj uk ϕk θt . It follows that u jt =

1 1 ujt − 3 uj ϕt + ut ϕj − δtj uk ϕk . 2 ϕ ϕ

(3.79)

Hence, 1 1 Hess(u) = Hess(u) − (du ⊗ dϕ + dϕ ⊗ du) + ∇u, ∇ϕ , ϕ ϕ and

= 1 Δu + m − 2 ∇u, ∇ϕ . Δu ϕ2 ϕ3

(3.80)

(3.81)

Note the simple form that equation (3.81) takes if m = 2: in this case, harmonic functions do depend only on the conformal class of the metric. Remark. Formula (3.81) shall also be used in Step 3 of the proof of Theorem 6.8. Proposition 3.10. Let (M, , ) be a complete Riemannian manifold of dimension m ≥ 3 and let b(x) ∈ C 0 (M ) be such that b(x) > 0

in M \BR

(3.82)

for some R > 0. Let v be a positive solution of Δv + a(x)v − b(x)v σ ≥ 0 with a(x) ∈ C 0 (M ) and σ ≥

m+2 m−2 .

(3.83)

Then, there exists ε = ε(v) > 0 such that, if

b(x) ≥ −ε

on M,

(3.84)

100


then, there exist b(x) ∈ C 0 (M ) and C > 0 satisfying i) b(x) > 0 in M,

ii) b(x) = C 2 b(x) in M \BR+1

(3.85)

and a positive solution u on M of m+2 (m − 2)(σ − 1) a(x)u − b(x)u m−2 ≥ 0. 4 Furthermore, there exists C1 > 0 such that

Δu +

u(x) = C1 v(x)

(m−2)(σ−1) 4

for r(x) 1.

(3.86)

(3.87)

and we consider the conformally related metric , = Proof. We set ϕ = v 2 ϕ , . Let ψ be a solution of the Dirichlet problem = δ on BR+2 , Δψ (3.88) ψ≡0 on ∂BR+2 σ−1 2

for some constant δ > 0. We define β(x) = ξ(x)ψ(x) + β0 > 0 on M

(3.89)

where β0 > inf ψ is a constant that we will specify in the sequel and ξ(x) ∈ C ∞ (M ), ξ : M → [0, 1] is a cut-off function such that ξ(x) ≡ 1

on BR ,

ξ(x) ≡ 0 on M \BR+1 .

(3.90)

Finally we set u(x) = ϕ(x)

m−2 2

β(x).

(3.91)

Using the transformation law (3.81) we now compute Δu. To simplify the writing we set m−2 σ−1 a= , b= , (3.92) 2 2 and we observe that − (m − 2)ϕ−1 ∇β, ∇ϕ , Δβ = ϕ2 Δβ Δϕ = bv

b−1

Δϕ + b(b − 1)v

b−2

2

|∇v| .

(3.93) (3.94)

We then have Δu = βΔϕa + ϕa Δβ + 2aϕa−1 ∇ϕ, ∇β − (m − 2)ϕa−1 ∇β, ∇ϕ = βaϕa−1 Δϕ + βa(a − 1)ϕa−2 |∇ϕ|2 + ϕa+2 Δβ + 2aϕa−1 ∇β, ∇ϕ = βabϕa−1 v b−1 Δv + βabϕa−1 (b − 1)v b−2 |∇v| + βa(a − 1)ϕa−2 |∇ϕ|2 + (2a − m + 2)ϕa−1 ∇β, ∇ϕ + ϕa+2 Δβ ≥ −βabϕa−1 v b a(x) + βabϕa−1 v b−1+σ b(x) + ab(b − 1)βϕa−2 v 2b−2 |∇v| + (2a − m + 2)ϕa−1 ∇β, ∇ϕ + ϕa+2 Δβ.


101

Therefore, + abβϕa−1 v b−1+σ b(x) Δu + aba(x)u ≥ ϕa+2 Δβ + ab[b − 1 + b(a − 1)]βϕa−2 v 2b−2 |∇v|

(3.95) (3.96)

+ (2a − m + 2)ϕ ∇β, ∇ϕ 0 1 a+2 σ−1−2b Δβ + abβv b(x) + ab(ab − 1)βϕa−2 v 2b−2 |∇v|2 =ϕ a−1

+ (2a − m + 2)ϕa−1 ∇β, ∇ϕ . Using (3.92) we get Δu +

(m−2)(σ−1) a(x)u 4

The choice σ ≥

m+2 m−2

) m+2 0 1* m+2 + (m−2)(σ−1) βb(x) ≥ u m−2 β − m−2 Δβ 4 0 1 (m−2)(σ−1) (m−2)(σ−1) (m−2)(σ−1) 4 + − 1 βv |∇ log v|2 . 4 4

implies (m − 2)(σ − 1) − 1 ≥ 0. 4

Hence, having set ! " m+2 b(x) = β − m−2 + (m − 2)(σ − 1) βb(x) , Δβ 4 (3.95) implies the validity of (3.86). We also observe that, because of (3.89) and the choice of ξ(x), b(x) = C 2 b(x) on M \BR+1 , for some constant C > 0. Furthermore, since inf

B R+2 \BR

b(x) > 0

we can choose β0 sufficiently large so that + (m − 2)(σ − 1) βb(x) > 0 on B R+2 \BR . Δβ 4

(3.97)

Finally, (3.89), (3.88), (3.84) imply that (3.97) holds on BR provided (m − 2)(σ − 1) (!ψ!∞ + β0 )ε < δ. 4 This proves that b(x) > 0 on M . (3.87) follows from (3.89) and (3.91).

102


Remark. If in (3.83) we have the equality sign this does not hold in general in m+2 (3.86) unless σ = m−2 . Note that we can also choose β0 ≥ 1 so that u = β0 v is m+2 large for σ = m−2 . Hence supM v = v ∗ > 1 implies supM u = u∗ > 1. This is used in Theorem 5.9. Using Proposition 3.10 and Theorem 3.2 we obtain the following refined version. Theorem 3.11. Let a(x), b(x) ∈ C 0 (M ) satisfy i) b(x) ≥ 0 on M ; ii) b(x) > 0 for r(x) 1; iii) a(x) ≤ Cb(x) for r(x) 1. Let σ ≥

m+2 m−2 ,

α > 0 and assume the existence of ϕ > 0 satisfying Lσ,α ϕ ≤ −K

|∇ϕ| ϕ

2

2

on M for K > −1, with Lσ,α = Δ + (m−2)(σ−1)(1+α) a(x). Then, the differential 16α inequality Δu + a(x)u − b(x)uσ ≥ 0 (3.98) has no positive bounded solutions on M satisfying a(x)u and

& u

(m−2)(σ−1)(1+α) 4

(m−2)(σ−1)(1+α) 4

∈ L1 (M )

'−1

∈ L1 (+∞).

(3.99)

(3.100)

∂BR

Remark. Theorem 3.11 will be used in the proof of Theorem 4.8 below. We would like to end the chapter with a further general nonexistence result which will be useful for the Yamabe problem once we will have obtained the a priori estimates from above in the next chapter; see Theorem 4.10. First we recall how the definition of λ1 (Ω) is extended in case Ω is not necessarily an open set. Thus, let S be an arbitrary bounded subset of M . We set L λL 1 (S) = sup λ1 (Ω),

where the supremum is taken over all open bounded sets with smooth boundary Ω such that S ⊂ Ω. Note that, by definition, if S = ∅, then λL 1 (S) = +∞. Finally, if S is an unbounded subset of M , we define L λL 1 (S) = inf λ1 (D ∩ S),

where the infimum is taken over all bounded open sets D with smooth boundary. Note that the condition that λL 1 (S) > 0 means that the set S is small in suitable spectral sense. We are now ready to prove the following


103

Theorem 3.12. Let a(x), b(x) ∈ C 0 (M ) satisfy b(x) ≥ 0

on M

and λL 1 (supp a+ ) > 0

(3.101)

with L = Δ + a(x). Then the differential equation Δu + a(x)u − b(x)uσ = 0,

σ > 1,

has no nontrivial nonnegative ground states. Remark. Recall that a ground state for the above equation is an entire solution u satisfying lim u(x) = 0. x→+∞

Proof. Let u be a nonnegative ground state. If u is not identically null it attains its absolute positive maximum at a point xo ∈ M . We claim that a(xo ) ≥ 0. Indeed, otherwise a(xo ) < 0 and 0 ≥ Δu(xo ) = −a(xo )u(xo ) + b(xo )u(xo )σ ≥ −a(xo )u(xo ) > 0, contradiction. Therefore, xo ∈ supp a+ . Using (3.101) we can find a sufficiently small bounded domain Ω xo such that λL 1 (Ω) > 0. Let ϕ be the positive corresponding eigenfunction, that is, ϕ > 0 on Ω, Δϕ + a(x)ϕ + λL 1 (Ω)ϕ = 0, ϕ=0 on ∂Ω. Thus, in particular, Δϕ + a(x)ϕ = −λL 1 (Ω)ϕ ≤ 0.

(3.102)

Next, we fix 0 < γ < u(xo ) sufficiently close to u(xo ) such that the connected component containing xo , Ωo , of {x ∈ M : u(x) > γ} is inside Ω. Note that u has an absolute maximum in Ωo and Δu + a(x)u = b(x)uσ ≥ 0 on Ωo .

(3.103)

Since ϕ satisfies (3.102) and ϕ > 0 on Ωo , by the generalized maximum principle (see [PW67], Section 5) u = Cϕ for some positive constant C. From (3.102) and (3.103) we have σ −λL on Ωo . 1 (Ω)u = b(x)u Thus at xo , contradiction.

0 ≤ b(xo )u(xo )σ−1 = −λL 1 (Ω) < 0,

Remark. If a+ (x) ≡ 0, supp a+ = ∅ and λL 1 (∅) = +∞. Thus in case a(x) ≤ 0 we have the validity of the conclusions of the theorem.

Chapter 4

A priori estimates In this chapter we determine a priori estimates on the behavior at infinity of positive solutions of the equation Δu + a(x)u − b(x)uσ = 0,

σ>1

(4.1)

on M under assumptions on a(x) and b(x) related to the geometrical requirement Ric ≥ −(m − 1)H 2 (1 + r(x)2 ) 2 δ

on M

(4.2)

for some H > 0 and δ ∈ R. The estimates we shall provide are quite sharp and naturally divide into two types: from below and from above. In both cases the method we use in establishing their validity resembles the old idea in Ahlfors’ proof of the Schwarz lemma ([Ahl38]). Some further estimates, which cannot be obtained with the previous method, are provided by direct comparison with the aid of the maximum principle (see section 4.3). The chapter ends with some nonexistence result for the Yamabe problem, which complement those described in Chapter 3.

4.1 Estimates from below We begin with an estimate from below for positive supersolutions (see [RZ07], Theorem 2.1). Theorem 4.1. Let (M, , ) be a complete manifold with Ricci tensor satisfying (4.2); let a(x), b(x) ∈ C 0 (M ) with b(x) > 0 on M \BR (o) for some R > 0 and lim inf r(x)→+∞

a(x) >0 r(x)α

(4.3)

with α > max{−2, δ}. Let ψ(t) be a positive, nondecreasing function defined in a neighbourhood of infinity such that, for some ε ∈ (0, 1),


105

106

Chapter 4. A priori estimates ψ(t) = O ψ

t 1+ε

and assume that lim inf r(x)→+∞

as t → +∞

(4.4)

a(x) ψ(r(x)) > 0. b(x)

(4.5)

Then, any positive solution u ∈ C 2 (M ) of Δu + a(x)u − b(x)uσ ≤ 0,

σ > 1,

satisfies

on M

(4.6)

1

u(x) ≥ Cψ(r(x))− σ−1

(4.7)

for r(x) 1 and some constant C > 0. Proof. We fix q ∈ M , with r(q) 1 and we set ρ(x) = dist(x, q). Fix T > 0 and consider on BT (q) the function F (x) =

u(x) [T 2

− ρ2 (x)]

ξ

for some ξ > 1. Note that, as x → ∂BT (q), F (x) → +∞, while F (x) > 0 on BT (q). Thus F attains a positive absolute minimum at x ¯ ∈ BT (q). Using a trick of Calabi, [Cal57], which enables us to suppose that ρ is smooth near x ¯, we deduce i) ∇ log F (¯ x) = 0,

ii) Δ log F (¯ x) ≥ 0.

(4.8)

Using (4.8) i) a computation yields ∇u ρ∇ρ (¯ x), (¯ x) = −2ξ 2 u T − ρ2

(4.9)

while from (4.8) ii) and (4.9) we have at x ¯, Δu Δρ2 ρ2 + ξ ≥ 0. − 4ξ(ξ − 1) 2 u [T − ρ2 ]2 T 2 − ρ2

(4.10)

We want to estimate Δρ2 ; since M is complete, we have Ric ≥ −(m − 1)Z 2

on BT (q)

(4.11)

for some constant Z > 0. Therefore, the Laplacian Comparison Theorem and a computation using Gauss’ lemma imply Δρ2 ≤ 2[m + (m − 1)Zρ]

on BT (q).

(4.12)

Inserting (4.6) and (4.12) into (4.10) we have at x ¯, u≥b

1 − σ−1

&

ρ2 m + (m − 1)Zρ a + 4ξ(ξ − 1) 2 − 2ξ 2 2 [T − ρ ] T 2 − ρ2

1 ' σ−1

.

4.1. Estimates from below

107

Since x ¯ is the minimum of F on BT (q) we then deduce x)]ξ u(y) ≥ [T 2 − ρ2 (y)]ξ u(¯ x) [T 2 − ρ2 (¯ 1

x)− σ−1 ≥ [T 2 − ρ2 (y)]ξ b(¯ ' 1 & x) ρ2 (¯ m + (m − 1)Zρ(¯ x) σ−1 × a(¯ x) + 4ξ(ξ − 1) 2 − 2ξ [T − ρ2 (¯ x)]2 T 2 − ρ2 (¯ x) for each y ∈ BT (q). In particular, for y = q, !

a(¯ x) u(q) ≥ b(¯ x)

1 & " σ−1

4ξ(ξ − 1) x) 2ξ m + (m − 1)Zρ(¯ ρ2 (¯ x) 1+ − a(¯ x) [T 2 − ρ2 (¯ x)]2 a(¯ x) T 2 − ρ2 (¯ x)

1 ' σ−1

. (4.13)

We set f (t) =

2ξ (m − 1)Zt 1 2ξ(ξ − 1) t2 − + 2 2 2 2 a(¯ x) [T − t ] a(¯ x ) T 2 − t2

and g(t) =

2ξ 1 2ξ(ξ − 1) t2 m − + 2 2 2 2 2 a(¯ x) [T − t ] a(¯ x ) T − t2

on [0, T ) and, considering the parabola (note ξ > 1) w=

2ξ(ξ − 1) 2 2ξ(m − 1)Z 1 y − y+ , a(¯ x) a(¯ x) 2

y ∈ [0, +∞),

we deduce that f (t) attains on [0, T ) its minimum value 1 1 ξ (m − 1)2 Z 2 . f¯ = − 2 2ξ−1 a(¯ x) As for g(t), we have g(0) =

1 2

−

2ξ m a(¯ x) T 2

and limt→T − g(t) = +∞; furthermore

4ξ t {(ξ − 1 − m)t2 + (ξ − 1 + m)T 2 } 2 a(¯ x) [T − t2 ]3 8 t3 ≥ ξ(ξ − 1) ≥ 0 a(¯ x) [T 2 − t2 ]3

g (t) =

on [0, T ) because of our choice of ξ. It follows that g(t) attains on [0, T ) its minimum value 1 2ξ m g¯ = g(0) = − . 2 a(¯ x) T 2 Going back to (4.13) we obtain !

a(¯ x) u(q) ≥ b(¯ x)

1 & " σ−1

ξ (m − 1)2 Z 2 2ξ m 1− − 2(ξ − 1) a(¯ x) a(¯ x) T 2

1 ' σ−1

.

(4.14)

108

Chapter 4. A priori estimates

We now choose T = εr(q) and we observe that, since ∀ x ∈ BT (q), r(q) − T ≤ r(x) ≤ r(q) + T , with our choice of T , using (4.3) we have on BT (q), a(x) ≥ C(ε)r(q)α , (4.15) δ Ric ≥ −(m − 1)H 2 1 + W (ε)2 r(q)2 2 for some constants C(ε), W (ε) > 0, depending only on ε > 0 and on the sign of α and δ respectively. Therefore, ξ (m − 1)2 Z 2 2ξ m (4.16) − 2(ξ − 1) a(¯ x) a(¯ x) T 2 ! " δ2 1 2mξ 1 ξ(m − 1)2 H 2 2 r(q)δ−α − r(q)−2−α . ≥1− + W (ε) 2(ξ − 1)C(ε) r2 (q) C(ε) ε2

Λ(¯ x)σ−1 = 1 −

Thus, using the assumption α > max{−2, δ} we can suppose to have chosen R > 0 sufficiently large such that, ∀ q with r(q) > R, we have Λ(¯ x)σ−1 ≥ 12 . Hence u(q) ≥ and in turn u(q)ψ(r(q))

1 σ−1

1 5 . σ−1 1 a(x) ψ(r(q)) min ≥ . 2 BT (q) b(x)

5

We claim that lim inf r(q)→+∞

! " 1 x) σ−1 1 a(¯ 2 b(¯ x)

. a(x) >0 ψ(r(q)) min BT (q) b(x)

(4.17)

(4.18)

so that (4.7) follows at once from (4.17). To prove the claim suppose that (4.18) is false, then there exists a sequence {yn } in M , with r(yn ) → +∞, such that . 5 a(x) = 0, (4.19) ψ(r(yn )) min lim r(yn )→+∞ BTn (yn ) b(x) where Tn = εr(yn ). Let {zn } ∈ BTn (yn ) realize the minimum, that is a(x) a(zn ) = . b(x) b(zn ) BTn (yn ) min

Fix η > 0 and choose n sufficiently large so that, from (4.19), we have η a(zn ) < . b(zn ) ψ(r(yn ))

(4.20)

4.1. Estimates from below

109

Since zn ∈ BTn (yn ), r(zn ) 1+ε and therefore, since ψ is nondecreasing and (4.4) holds we have r(yn ) ≥

a(zn ) Cη η ≤ < r(zn ) b(zn ) ψ(r(z n )) ψ( 1+ε )

(4.21)

for some constant C > 0 independent of n. In other words a(zn ) ψ(r(zn )) < Cη b(zn ) for n 1. Since η > 0 was chosen arbitrarily this contradicts (4.5) and completes the proof of the theorem. Remark. If we assume ψ(t) nonincreasing, the conclusion of Theorem 4.1 holds substituting the requirement (4.4) with t as t → +∞, (4.22) ψ(t) = O ψ 1−ε for some ε ∈ (0, 1). Thus Theorem 4.1 is valid in particular with ψ(t) = Ctβ (log t)γ for t 1, β, γ ∈ R, C > 0 and ε ∈ (0, 1). Remark. Clearly, (4.4) or (4.22) are not satisfied if ψ(t) is of exponential type, for instance ψ(t) = Ceβt , C, β > 0. In this case condition (4.4) has to be substituted with ψ(t) = O(ψ(t − T0 )) as t → +∞ (4.23) for some T0 > 0. The result continues to hold with the requirement α > max{0, δ}

(4.24)

but the proof has to be modified as follows. Following the argument of Theorem 4.1 we arrive at (4.14). Now we choose T = T0 , then (4.15) holds with C(ε), W (ε) substituted with C(T0 ), W (T0 ), positive constants depending only on T0 and the sign of α and δ respectively. We then proceed, under the modified assumption (4.3) with α > max{0, δ}, directly to 1 5 . σ−1 1 a(x) 1 u(q)ψ(r(q)) σ−1 ≥ ψ(r(q)) min . (4.25) 2 BT0 (q) b(x) The remainder of the proof is the same as in Theorem 4.1 using (4.23) instead of (4.4). Clearly if ψ(t) is nonincreasing, then (4.23) has to be substituted with ψ(t) = O(ψ(t + T0 )) for some T0 > 0.

as t → +∞

(4.26)

110


Remark. In case α = δ the proof of Theorem 4.1 can be modified to obtain the same conclusion, provided a further condition on H is satisfied. Indeed, if case (4.4) or (4.22) holds, assume that the Ricci tensor of M satisfies (4.2) with δ > −2, and having set A = lim inf r(x)→+∞

a(x) > 0, r(x)α

suppose that H satisfies A 1 0 < H2 < 2 (m − 1)2

1−ε 1+ε

|δ| .

(4.27)

Under these assumptions, following the proof of Theorem 4.1, we arrive at (4.16) which takes the form ! " δ2 1 ξ 2mξ 1 (m − 1)2 H 2 2 Λ(¯ x)σ−1 ≥ 1 − − r(q)−2−δ + W (ε) 2(ξ − 1) C(ε) r2 (q) C(ε) ε2 where C(ε) and W (ε) are given by C(ε) = A(1 − (sgn(δ))ε)δ W (ε) = (1 + (sgn(δ))ε)

(4.28)

and sgn(δ) is the signum of δ. Using (4.27) and δ > −2 we can choose ξ > 1 and R > 0 both sufficiently large that, for some η ∈ (0, 1), r(q) > R gives ! " δ2 1 ξ (m − 1)2 H 2 2 0 because of (4.24) and α = δ, we require T0 δ (1 − R ) A 0 0 < H2 < T 2 (m − 1) (1 + R0 )δ 0

for R0 > T0 and the theorem is still valid. Remark. We underline that, unfortunately, in the geometric case of Yamabe’s equation, for m ≥ 3 say, under assumption (4.2) one has a(x) = −c−1 m S(x) ≤ hence (4.27) cannot be satisfied.

δ/2 m(m − 2) 2 , H 1 + r(x)2 4

4.2. Estimates from above

111

Remark. It is worth observing that the proof of Theorem 4.1 gives no uniform positive lower bounds on compact domains for the positive solutions of (4.1). As we shall see this contrasts with the case of the estimates from above.

4.2

Estimates from above

We now derive an estimate from above for nonnegative subsolutions in case δ ≥ −2. We recall that a subsolution (resp. a supersolution) of a Yamabe-type equation is a solution of the differential inequality Δu + a(x)u − b(x)uσ ≥ 0

(resp. ≤ 0).

First we need the following (see [RRV97], [PRS05b], Lemma 2.6) Lemma 4.2. Let a(x), b(x) ∈ C 0 (M ), σ > 1, 0 < T < T and Ω ⊂⊂ BT (q) ⊂ M . Assume that b(x) > 0 on BT (q). Then, there exists a constant C > 0 such that any nonnegative C 2 -solutions u on BT (q) of Δu + a(x)u − b(x)uσ ≥ 0

(4.29)

sup u ≤ C.

(4.30)

satisfies Ω

Proof. We let ρ(x) = dist(x, q) and, on the compact ball BT (q) we consider the continuous function 2 F (x) = T 2 − ρ(x)2 σ−1 u(x) (4.31) where u is any nonnegative C 2 -solution of (4.29). Note that F (x)|∂BT (q) ≡ 0. If u ≡ 0 on BT (q), there is nothing to prove; otherwise F has a positive absolute maximum at some point x ¯ ∈ BT (q). In particular, u(¯ x) > 0. Again, using Calabi’s trick, [Cal57], we can assume that ρ is smooth near x ¯. Then i) ∇ log F (¯ x) = 0;

ii) Δ log F (¯ x ≤ 0.

(4.32)

From (4.32) i) we obtain ∇u 4 ρ∇ρ = u σ − 1 T 2 − ρ2

at x ¯;

while from (4.32) ii) and (4.33) we deduce 2 Δu 2 8 ρ2 Δρ2 0≥ − − +1 2 2 u σ−1 σ−1 σ − 1 T − ρ2 (T 2 − ρ2 ) Again,

Ric ≥ −(m − 1)A2

on BT (q)

(4.33)

at x ¯.

(4.34)

(4.35)

112


for some constant A ≥ 0, so that, by the Laplacian comparison theorem, Δρ2 ≤ 2[m + (m − 1)Aρ]

on BT (q).

Inserting (4.29) and (4.36) into (4.34) we obtain that, at x ¯, & ' 1 1 8(σ − 1) 4 m + (m − 1)Aρ σ−1 ρ2 + , u ≤ b− σ−1 a+ + (σ − 1)2 (T 2 − ρ2 )2 σ−1 T 2 − ρ2

(4.36)

(4.37)

where a+ is the positive part of a. Next, we recall the elementary inequality (v + w)δ ≤ C 2 (v δ + wδ ),

v, w ≥ 0

(4.38)

valid for any fixed δ > 0 provided C = C(δ) > 0 is sufficiently large. We use (4.38) 1 in (4.37) with δ = σ−1 to get, at x ¯, u≤C

2

a+ max BT (q) b

1 / σ−1

+C

2

1 /− σ−1 &

min b BT (q)

ρ2 m + (m − 1)Aρ + 2 (T − ρ2 )2 T 2 − ρ2

1 ' σ−1

whence 5 F (¯ x) ≤ C

2

1 1 . / σ−1 /− σ−1 2 1 a+ 4 2 T max min b +C T + AT 3 σ−1 . BT (q) b BT (q)

x) and therefore On the other hand, ∀y ∈ BT(q) , we have F (y) ≤ F (¯ 1 / σ−1 1 / σ−1 T4 a+ u(y) ≤ C max (4.39) 2 [T 2 − ρ2 (y)] BT (q) b 1 /− σ−1 1 / σ−1 T2 AT 3 2 +C min b . 2 + 2 [T 2 − ρ2 (y)] [T 2 − ρ2 (y)] BT (q)

2

Therefore, using the fact that ρ(y) ≤ T < T , from (4.39) we immediately obtain (4.30). We are now ready to prove the desired estimate (see [RRV97] and [PRS05b], Proposition 2.7). Proposition 4.3. Let a(x), b(x) ∈ C 0 (M ) and let σ > 1. Suppose that the Ricci tensor satisfies δ (4.40) Ric ≥ −(m − 1)B 2 1 + r(x)2 2 for some B > 0, δ ≥ −2. Let ω, ψ be nondecreasing positive functions defined in a neighbourhood of +∞ with the following properties: there exists ε ∈ (0, 1) for which t t = O(ψ(t)); ii) ω = O(ω(t)) as t → +∞. (4.41) i) ψ 1−ε 1−ε

4.2. Estimates from above

113

Assume that, for some R0 > 0 and ∀ x : r(x) ≥ R0 , we have a+ (x) 1 ≤ ; b(x) ω(r(x))

i)

ii) b(x) ≥ ψ(r(x))r(x) 2 −1 . δ

(4.42)

If u is a nonnegative C 2 -solution on M of Δu + a(x)u − b(x)uσ ≥ 0,

1 1 u(x) = O ω(r(x))− σ−1 + ψ(r(x))− σ−1

then

(4.43) as r(x) → +∞.

(4.44)

Proof. Let q ∈ M \B2R0 (o) and set ρ(x) = dist(x, q). We fix a radius T > 0 so small as to insure that BT (q) ⊂ M \B2R0 (o). Because of (4.40) we have 0 1δ 2 2 Ric ≥ −(m − 1)B 2 1 + (r(q) + sgn(δ)T )

on BT (q).

(4.45)

In (4.39) of Lemma 4.2 we choose y = q so that ρ(q) = 0 and 1δ 0 2 4 . A = B 1 + (r(q) + sgn(δ)T )

(4.46)

In this way, we get 2

u(q) ≤ C

+C

1 / σ−1 a+ max BT (q) b 1 /− σ−1 &

2

min b BT (q)

(4.47) 1δ 1 B0 2 4 + 1 + (r(q) + sgn(δ)T ) T2 T

1 ' σ−1

.

Set T = ε[r(q) − R0 ] in (4.47), and note that BT (q) ⊂ M \BR0 (o). Since δ ≥ −2 and r(q) ≥ 2R0 , it is easy to see that there exists a constant C > 0 independent of q such that &

1δ 1 B0 2 4 + 1 + (r(q) + sgn(δ)T ) 2 T T

1 ' σ−1

1

≤ C 2 r(q)( 2 −1) σ−1 . δ

(4.48)

Now, recalling that b(x) satisfies (4.42) ii) and that ψ is nondecreasing, we obtain min b ≥ C 2 r(q) 2 −1 ψ(r(q) − T ) ≥ C 2 ψ((1 − ε)r(q)) r(q) 2 −1 δ

BT (q)

for some constant C = C(R0 ) > 0. On the other hand, (4.42) gives a+ 1 1 ≤ ≤ . b ω((1 − ε)r(q) + εR ) ω((1 − ε)r(q)) 0 BT (q) max

δ

114


Inserting these inequalities into (4.47) and using (4.41) we obtain ) * 1 1 u(q) ≤ C 2 ω((1 − ε)r(q))− σ−1 + ψ((1 − ε)r(q))− σ−1 ) * 1 1 ≤ C 2 ω(r(q))− σ−1 + ψ(r(q))− σ−1

as required. As a special case we have the following result (see [RZ07], Theorem 2.3) Theorem 4.4. Let (M, , ) be a complete manifold with Ricci tensor satisfying Ric ≥ −(m − 1)H 2 (1 + r2 ) 2 δ

for some constant H > 0 and δ ≥ −2. Let a(x), b(x) ∈ C 0 (M ) and satisfying a(x) ≤ Ar(x)α , b(x) ≥ Br(x)β ,

δ − 1, 2 δ β ≤1− +α 2

α≥

(4.49) (4.50)

for r(x) 1 and some constants A, B > 0. Then any nonnegative solution u ∈ C 2 (M ) of Δu + a(x)u − b(x)uσ ≥ 0, σ > 1 on M (4.51) satisfies β−α

u(x) ≤ Cr(x)− σ−1

(4.52)

for r(x) 1 and some constant C > 0. As in the case of Theorem 4.1, there are a number of versions of Proposition 4.3. We describe here (without proof) one of them that can be deduced from Proposition 4.1 of [RRV97]. Theorem 4.5. Let (M, , ) be a complete manifold with Ricci tensor satisfying Ric ≥ −(m − 1)H 2 (1 + r2 ) 2 δ

for some constants H > 0 and δ ≥ −2. Let a(x), b(x) ∈ C 0 (M ) and let ψ(t) be a nondecreasing positive function in a neighbourhood of +∞ such that there exists T0 > 0 for which ψ(t + T0 ) = O(ψ(t)) as t → +∞. (4.53) Assume, depending on the sign of δ, that either ⎧ b(x) ⎪ ⎪ i) for − 2 ≤ δ ≤ 0, lim sup a+ (x) < +∞ and lim inf =A>0 ⎪ ⎪ ⎪ ψ(r(x)) r(x)→+∞ r(x)→+∞ ⎨ or ⎪ ⎪ a+ (x) b(x) ⎪ ⎪ ⎪ lim inf = A > 0. ⎩ii) for δ > 0, lim sup r(x)δ < +∞ and r(x)→+∞ r(x)δ ψ(r(x)) r(x)→+∞

(4.54)

4.3. Sharpness of the previous results

115

If u is a nonnegative C 2 -solution on M of Δu + a(x)u − b(x)uσ ≥ 0, then

1 u(x) = O ψ(r(x))− σ−1

σ > 1,

as r(x) → +∞.

(4.55)

Furthermore, in case A = +∞ in (4.54), the conclusion (4.55) improves to

1 as r(x) → +∞. (4.56) u(x) = o ψ(r(x))− σ−1

4.3 Sharpness of the previous results We now show by way of examples that the estimates we obtained are sharp. We begin with Theorem 4.1. Let g(r) ∈ C ∞ ([0, +∞]), g(r) > 0 for r > 0 be such that r on [0, 12 ], r g(r) = 2 δ 1 4 e m−1 0 (1+s ) ds on [1, +∞) for some δ ≥ −2. We define the model manifold Mg = Rm in the sense of Greene and Wu, [GW79], with metric on Mg \ {0} = (0, +∞) × Sm−1 given by , = dr2 + g(r)2 dθ2 where dθ2 is the canonical metric on the unit sphere Sm−1 . Note that, since g(r) = r on [0, 12 ], , can be smoothly extended to all of Mg . The Ricci curvature in the radial direction is given by Ric(∇r, ∇r) = −(m − 1)

g (r) g(r)

while, in the direction orthogonal to ∇r determined by the unit vector ν, Ric(ν, ν) = −

(m − 2) g (r) (1 − g (r)2 ) − . 2 g(r) g(r)

Thus, a simple computation shows that there exists an appropriate H > 0 such that δ Ric ≥ −(m − 1)H 2 (1 + r2 ) 2 on Mg . Next, we consider the function α−β

w(r) = (μ + P (r)) σ−1 > 0 on [0, +∞), where μ > 0, β ≥ α > δ and P ∈ C 2 ([0, +∞)) satisfies P (r) = r, for r 1, P (r) ≥ 0 on [0, +∞), P (0) = 0.

116


We define Hw (r) = w

−σ

&

g (r) w + (m − 1) w + A(1 + r)α w g(r)

'

for some A > 0 constant. Computing we have, &! P (r) (μ + P (r))β−α (1 + r)−β Hw (r) = β−α (1 + r) (μ + P (r))(1 + r)α α−β−σ+1 P (r) + σ−1 (μ + P (r))2 (1 + r)α " ' g (r) P (r) + A . +(m − 1) g(r) (μ + P (r))(1 + r)α Next, we note that (m − 1)

δ g (r) = (1 + r2 ) 4 g(r)

on [1, +∞), and therefore δ P (r) P (r) g (r) = (1 + r2 ) 4 (1 + r)−α on [1, +∞). α g(r) (μ + P (r))(1 + r) (μ + P (r)) (4.57) Furthermore, the left-hand side of (4.57) is bounded near zero since P (0) = 0. It follows that, up to choosing μ 1, since P (r) = r for r 1, we can choose B > 0 such that Hw (r) ≤ B(1 + r)β

(m − 1)

on [0, +∞). Recalling the definition of Hw (r), we then have w + (m − 1)

g (r) w + A(1 + r)α w − B(1 + r)β wσ ≤ 0. g(r)

(4.58)

Setting u(x) = w(r(x)), a(x) = A(1 + r(x))α and b(x) = B(1 + r(x))β > 0 on Mg we then deduce Δu + a(x)u − b(x)uσ ≤ 0. Moreover, a(x) A = (1 + r(x))α−β b(x) B on Mg and we can therefore choose ψ(t) = tβ−α . Since β ≥ α, ψ is nondecreasing, satisfies (4.4), (4.5) and according to Theorem 4.1, α−β

u(x) ≥ Cr(x) σ−1

(4.59) α−β

for r(x) 1 and some constant C > 0. Since u(x) = (μ + r(x)) σ−1 for r(x) 1, estimate (4.7) cannot be improved.

4.4. Some further estimates

117

In the above example, let, as before, δ ≥ −2. Choose α > −2 and A > 0. Then, for every β ∈ R we can find μ 1 so that (1 + r)−β Hw (r) ≥ B

∀ r > 0,

and for some B > 0. Therefore, u(x) = w(r(x)) satisfies the inequality Δu + a(x)u − b(x)uσ ≥ 0

on Mg .

(4.60)

Again, we observe that α−β

u(x) = (μ + r(x)) σ−1

on Mg \ B1 .

Thus, fixing δ > −2 and choosing α ≥ δ/2 − 1 and β ≤ 1 − δ/2 + α, applying Theorem 4.4 to the differential inequality (4.60) we have α−β

u(x) ≤ Cr(x) σ−1

for some constant C > 0 and r(x) >> 1, showing that the upper bound provided by Theorem 4.4 cannot be improved.

4.4 Some further estimates We now complement the analysis of the previous section by considering the situation where a(x) lim inf > −∞. (4.61) r(x)→+∞ r(x)α The technique used in the proof of Theorem 4.1 cannot be applied but it turns out that in this case an estimate similar to (4.7), depending directly on Ricci, can be obtained in an elementary way with the aid of the maximum principle. In fact we have (see [RZ07], Theorem 2.4) Theorem 4.6. Let (M, , ) be a complete manifold with radial Ricci curvature satisfying δ/2 Ric (∇r, ∇r) ≥ −(m − 1)H 2 1 + r(x)2 on M for some H > 0, δ > −2 and let a(x), b(x) ∈ C 0 (M ) and satisfy (4.61) and b(x)

lim sup r(x)→+∞

for some α < δ, β ≤ δ, γ >

r(x)β e

2H 2+δ (m

δ+2 (σ−1)γ 1+r(x) 2

< +∞

(4.62)

− 1), σ > 1. Then, any positive solution of

Δu + a(x)u − b(x)uσ ≤ 0

on M

(4.63)

satisfies lim inf

u(x)

δ+2 r(x)→+∞ e−γr(x) 2

> 0.

(4.64)

118


Proof. First of all, using Proposition 1.15, the Laplacian comparison theorem and the lower bound on Ricci, we deduce Δr ≤ (m − 1)Hrδ/2 (1 + o(1))

as r → +∞.

(4.65)

Next, we choose R > 0 sufficiently large such that, using (4.61) and (4.62), a(x) ≥ −Ar(x) , b(x) ≤ Br(x) e α

β

δ+2 (σ−1)γ 1+r(x) 2

(4.66)

on M \BR for some A, B > 0. We let 0 < ξ < minBR u(x). To simplify the writing, set θ = 2δ + 1 and define θ v(x) = ξe−γ (1+r(x) ) − u(x).

Note that, since γ > 0, by our choice of ξ we have v(x) ≤ ξ − u(x) < 0

on BR .

(4.67)

We now reason by contradiction and suppose that lim inf r(x)→+∞

u(x) = 0. e−γ(1+r(x)θ )

This means that there exists a sequence {xn } ⊂ M , r(xn ) → +∞, such that u(xn ) θ −γ(1+r(x n) ) e

→0

as n → +∞.

(4.68)

Since γ, ξ > 0 and u(x) > 0, v ∗ = supM v(x) < +∞. Furthermore, because of (4.68), ' & u(xn ) −γ (1+r(xn )θ ) (4.69) ξ − −γ(1+r(x )θ ) > 0 v(xn ) = e n e for n sufficiently large. Thus v ∗ > 0. Finally, γ, θ > 0 force that v ∗ has to be attained at some point x ¯ ∈ M . Let Ω be the level set Ω = {x ∈ M : v(x) > 0}; because of (4.67), Ω ⊂ M \ BR and θ u(x) < ξe−γ (1+r(x) )

on Ω.

(4.70)

On Ω we compute 1 0 θ θ Δv = ξ −γθe−γ (1+r(x) ) rθ−1 Δr + γθ γθr2θ−2 + (1 − θ)rθ−2 e−γ (1+r(x) ) − Δu so that, using (4.70), (4.65), (4.66) and (4.63),

4.4. Some further estimates

119

) 0 1 θ δ δ δ Δv ≥ ξr 2 −1+θ e−γ (1+r(x) ) γθ γθrθ1 − 2 + (1 − θ)r−1− 2 − (m − 1)H(1 + o(1)) * δ δ − Arα−θ+1− 2 − ξ σ−1 Brβ−θ+1− 2 & " ! δ+2 −γ 1+r(x) 2 δ+2 δ+2 δ −1− δ δ 2 − (m − 1)H(1 + o(1)) = ξr e γ γ− r 2 2 2 − A rα−δ − ξ σ−1 Brβ−δ . Note that, because of our assumptions on α, β, γ, δ we can choose R sufficiently large and ξ > 0 sufficiently small that Δv > 0

on Ω

contradicting the fact that x ¯ ∈ Ω.

The case δ = −2 is, loosely speaking, border line between Euclidean and hyperbolic geometry. As seen in Proposition 1.15 the bound −1 Ric (∇r, ∇r) ≥ −(m − 1)H 2 1 + r2

on M,

H>0

(4.71)

as r → +∞.

(4.72)

implies the estimate Δr ≤ (m − 1)r

−1 1

+

√

1 + 4H 2 (1 + o(1)) 2

Proceeding in a way similar to that of the argument of Theorem 4.6 we can prove the following (see [RZ07], Theorem 2.5) Theorem 4.7. Let (M, , ) be a complete manifold with radial Ricci curvature satisfying (4.71). Let a(x), b(x) ∈ C 0 (M ) satisfy (4.61) and b(x) < +∞ β r(x)→+∞ r(x) lim sup

(4.73)

with α < −2, β ≤ γ(σ − 1) − 2 and γ > (m − 1)

1+

√

1 + 4H 2 − 1. 2

(4.74)

Then any positive solution u of Δu + a(x)u − b(x)uσ ≤ 0

on M,

σ>1

satisfies u(x) > 0. r(x)→+∞ r(x)−γ lim inf

(4.75)

120


Both estimates of Theorems 4.6 and 4.7 are quite sharp with respect to the range of the exponent γ. To simplify computations let us consider (4.75). Towards this aim, for a chosen R0 > 1, let Mg be the model with g(r) ∈ C ∞ (M ) positive on [0, +∞) and such that ⎧ ⎨r on 0, 12 , √ g(r) = (4.76) 1 ⎩r 2 1+ 1+4H 2 on [R , +∞) 0

> 0. Since Ric(∇r, ∇r) = −(m − 1) g (r) we have for some H g(r) 2 (1 + r−2 )(1 + r2 )−1 2 r−2 = −(m − 1)H Ric(∇r, ∇r) = −(m − 1)H 2 (1 + R−2 ) so that on Mg \ BR0 . We let H 2 = H 0

−1 Ric(∇r, ∇r) ≥ −(m − 1)H 2 1 + r2

(4.77)

on Mg \ BR0 . It is not hard to show that g(r) can be defined on way that (4.77) holds on all of Mg . Now we set

−1 ( m − 1 γ= 1 + 1 + 4H 2 2 and we choose A, B > 0 sufficiently large that

1

2 , R0

− γ −1 g (r) 1 + r2 2 [0,R0 ] g(r) γ − γ α 2 − 2 −1 ≤ γ 1 + R0 + 1 + R02 2 A inf (1 + r)

γ(γ + 2)R02 − γ(m − 1) inf r

σ−1

+ βξ0

σγ 2 − 2

1 + R0

in such a

(4.78)

[0,R0 ]

inf (1 + r)

β

[0,R0 ]

for some fixed ξ0 > 0 and α, β ∈ R. We set α

a(x) = −A(1 + r(x)) ,

b(x) = B(1 + r(x))

β

and we define, with ξ ≥ ξ0 , − γ v(x) = ξ 1 + r(x)2 2 .

(4.79)

Then, a simple computation shows that on Mg , ) − γ −1 − γ −2 Δv + a(x)v − b(x)v σ = ξ γ(γ + 2)r2 (x) 1 + r2 (x) 2 − γ 1 + r2 (x) 2 − γ −1 g (r(x)) r(x) 1 + r2 (x) 2 − g(r(x)) − γ − A(1 + r(x)α ) 1 + r2 (x) 2 − − σγ * β − Bξ σ−1 (1 + r(x)) 1 + r2 (x) 2 . − γ(m − 1)

(4.80)

4.5. Nonexistence results for the Yamabe problem

121

Thus, since ξ ≥ ξ0 , (4.78) and inspection of (4.80) show that Δv + a(x)v − b(x)v σ ≤ 0

(4.81)

on BR0 . Next, we note that on M \ BR0 , (m − 1)

( 1 m − 1 g (r) = 1 + 1 + 4H 2 = (1 + γ − ε) g(r) 2r r

(4.82)

for some ε = ε(R0 ) with ε → 0 as R0 → +∞. Rearranging the terms in (4.80) and using (4.82) we have Δv + a(x)v − b(x)v σ & − γ2 −2 2 2 εγ − γ(γ + 2 − ε)r(x)−2 = ξr (x) 1 + r (x)

(4.83)

2 1 r2 (x) β 2− γ2 (σ−1) ' 1 1 σ−1 β+2−γ(σ−1) 1+ 1+ 2 . − Bξ r(x) r(x) r (x)

− Ar(x)α+2 1 +

1 r(x)

α

1+

If β = (σ − 1)γ − 2 we can choose ξ ≥ ξ0 sufficiently large that the above yields the validity of (4.81) on M \ BR0 . Note that, in this example, the range of α plays no role.

4.5 Nonexistence results for the Yamabe problem In this section we apply the nonexistence results of Chapter 3 and the a priori estimates from above of Chapter 4 to give nonexistence results in the geometrical case of the Yamabe equation; similar results can, of course, be obtained for Yamabetype equations on the complete, noncompact, manifold (M, , ). We begin with the next result, that was first proved in [BRS98]. Theorem 4.8. Let (M, , ) be a complete Riemannian manifold with scalar curvature S(x). Assume Ric ≥ −(m − 1)B 2 (4.84) for some B > 0 and that there exists a ϕ > 0 solution of Lϕ ≤ −K

|∇ϕ| ϕ

2

(4.85)

∞ for some K > −1, with L = Δ − m−1 4m S(x), m = dim M ≥ 3. Let K(x) ∈ C (M ) satisfy K(x) ≤ 0 on M, (4.86) 2 K(x) ≤ −Cr(x) m−1 +γ e2Br(x) for r(x) 1

122


and some constants C, γ > 0. Then the metric , cannot be pointwise conformally deformed to a metric , of scalar curvature K(x). Remark. Condition (4.86) is sharp only in the term e2Br . Indeed, for the Yamabe problem on hyperbolic space Hm −B 2 , the second of (4.86) can be relaxed to K(x) ≤ −C

e2Br r(x) log r(x)

for r(x) 1

(see Theorem 7.3 and the related Remark 2; see also the Remark after Theorem 6.15). 4 Proof. First of all we recall that if , = u m−2 , , then u > 0 has to satisfy m+2

cm Δu − S(x)u + K(x)u m−2 = 0, cm =

4(m − 1) , m−2

so that, with our notation for Yamabe-type equations, a(x) = −

m−2 S(x), 4(m − 1)

b(x) = −

m−2 K(x). 4(m − 1)

We want to apply Theorem 3.11 with the choices σ=

m+2 , m−2

α=

m . m−2

Indeed, in this case Lσ,α = Δ +

(m − 1)2 m−1 a(x) = Δ − S(x) = L, m(m − 2) 4m

L

so that (4.85) gives λ1 σ,α (M ) ≥ 0. From (4.86) we have b(x) ≥ 0 on M and b(x) > 0 for r(x) 1. We now have to show the existence of C > 0 such that a(x) ≤ Cb(x) Now, (4.84) implies sup a(x) ≤ M

for r(x) 1. m−2 mB 2 4

(4.87)

(4.88)

so that, since b(x) > 0 for r(x) 1, (4.87) is satisfied for C sufficiently large. This also shows that in order to verify (3.99) and (3.100) it is enough to show that m−1

u2 m−2 ∈ L1 (M ). Towards this aim we apply Theorem 4.5 with the choice δ = 0 and 2

ψ(t) = t m−1 +γ e2Bt

(4.89)

4.5. Nonexistence results for the Yamabe problem

123

for some γ > 0 as in (4.86). Then assumptions (4.54) i) and ii) are satisfied because of (4.88) and (4.86). It follows that 1 m−2

u(x) ≤ Cr(x)− 2 m−1 −

m−2 4 γ

e−

m−2 2 Br(x)

(4.90)

for some constant C > 0 and r(x) 1. By Bishop’s volume comparison theorem (see Chapter 1) and (4.90) m−1 m−1 m−1 u(x)2 m−2 ≤ Cr−1− 2 γ e−(m−1)Br e(m−1)Br = Cr−1− 2 γ ∂Br

and since γ > 0 it follows that

m−1

u(x)2 m−2 ∈ L1 (+∞),

∂Br

that is, (4.89) is satisfied. Suppose now that we are able to solve Yamabe’s equation m+2

cm Δu − S(x)u + K(x)u m−2 = 0

on M, u > 0

(4.91)

4 on the complete manifold (M, , ); we then ask if the new metric , = u m−2 , , m = dim M ≥ 3 is still complete. The a priori estimates of Chapter 4 give us an answer, in the negative, with the following

Theorem 4.9. Let (M, , ) be a complete Riemannian manifold of dimension m ≥ 3 and scalar curvature S(x). Suppose that Ric ≥ −(m − 1)H 2 (1 + r(x)) 2 δ

(4.92)

for some H > 0, δ ≥ −2. Let K(x) ∈ C ∞ (M ) and assume that K(x) ≤ −K 2 r(x)δ+2 [log r(x)]

2(1+γ)

for r(x) 1

(4.93)

and some constants K, γ > 0. Then any metric conformal to , and with scalar curvature K(x) is incomplete. K(x) Proof. As above we let a(x) = − S(x) cm , b(x) = − cm . We apply Proposition 4.3 with the choices

ω(t) = Ct2 (log t)2(1+γ) , ψ(t) = Ct3+ 2 (log t)2(1+γ) δ

for t 1 and some C > 0. We note that ω and ψ satisfy (4.41) i), ii) ∀ ε ∈ (0, 1). Furthermore, due to (4.92) and (4.93), we respectively have b(x) ≥

δ K2 2(1+γ) r(x)δ+2 [log r(x)] ≥ ψ(r(x))r(x) 2 −1 Cm

for r(x) 1,

124

Chapter 4. A priori estimates a+ (x) 1 ≤ b(x) ω(r(x))

for r(x) 1

provided C > 0 is chosen appropriately. An application of Proposition 4.3 yields % $ − m−2 δ +1 1+γ − m−2 m−2 ] 2 r(x) 2 ( 2 ) 1+γ − 2 +[r(x)(log r(x)) u(x) = O r(x)(log r(x)) as r(x) → +∞. Since δ ≥ −2, from this latter we deduce that − m−2 2 for r(x) 1 u(x) ≤ C r(x)(log r(x))1+γ

(4.94)

4 and some C > 0. To show that the metric , = u m−2 , is not complete, we reason as follows. Let {xn } be a sequence diverging in M , that is, {xn } is definitely outside any fixed compact set in M . Since (M, , ) is complete, r(xn ) → +∞ as n → +∞; in particular we can suppose that {xn } ⊂ M \BR on which (4.94) holds. For n fixed, by the Hopf-Rinow theorem (see e.g. [Lee97]) there exists γn , a unit speed geodesic from o to xn , realizing the distance r(xn ). Thus if t is the arclength parameter of γn , then r(γn (t)) = t.

For r(x) = dist (o, x) we have ,

r(xn ) ≤

r(xn ) 0 2

≤ C m−2 =C

2

u m−2 (γn (t)) dt

2 m−2

r(xn )

0 r(xn )

r(γn (t))(log r(γn (t)))1+γ t(log t)

1+γ −1

0

dt ≤ C

−1

2 m−2

dt

+∞ 0

t(log t)1+γ

−1

6 dt ≤ C

6 an absolute constant. Thus, with C 6 r(xn )) ≤ C

∀ n ∈ N,

and the metric , cannot be complete since {xn } is a divergent sequence in M. In the next result we show that even a slow decay to 0 of K(x) at infinity under some additional condition implies nonexistence. This will be an application of Theorems 3.12 and 4.4. Theorem 4.10. Let (M, , ) be a complete Riemannian manifold of dimension m ≥ 3 and scalar curvature S(x). Suppose that Ric ≥ −(m − 1)H 2 (1 + r(x)2 ) 2 δ

4.5. Nonexistence results for the Yamabe problem for some H > 0 and −2 ≤ δ < 2. For L = Δ −

125

1 cm S(x)

assume that

λL 1 (supp S− ) > 0,

(4.95)

where S− is the negative part of S. Let K(x) ∈ C ∞ (M ) be such that K(x) ≤ 0 on M and K(x) ≤ −K2 r(x)β (4.96) for r(x) 1 and some K > 0, δ < β ≤ 1 + 2δ . Then the metric , cannot be pointwise conformally deformed to a metric , of scalar curvature K(x). Proof. We reason by contradiction and suppose that , exists. Then we have a solution u > 0 of the differential equation m+2

Δu + a(x)u − b(x)u m−2

on M

with, as we saw in Theorem 4.9, a(x) = −

S(x) , cm

b(x) = −

K(x) . cm

The assumption on the Ricci tensor yields a(x) ≤ Ar(x)δ

for r(x) 1

and an appropriate constant A > 0, while from (4.96) we deduce b(x) ≥ Br(x)β . Note that, due to the choice of the range of the parameters δ and β, the assumptions of Theorem 4.4 are satisfied and we conclude that u(x) ≤ Cr(x)− 4 (β−δ) . m

Then u(x) → 0 as x → ∞. Using (4.95) and K(x) ≤ 0 on M we apply Theorem 3.12 to obtain a contradiction to the existence of u.

Chapter 5

Uniqueness The aim of this chapter is to prove some uniqueness results for positive solutions of Yamabe-type equations. Our first theorem in this direction depends only on the sign of the coefficient b(x) of the nonlinear term and, loosely speaking, on an L2 type estimate of the distance, at infinity, of the two solutions under consideration. It is worth observing that this very general result is sharp and that the L2 -type condition cannot be substituted with a corresponding Lp condition with p > 2. Our second result, Theorem 5.2, is obtained via a comparison result whose proof requires a version of the weak maximum principle (see Theorem 5.3) which is interesting in its own. The strength of the theorem lies both in the fact that it requires that the solutions approach one another at infinity in the weak sense of condition (5.13) below, and in the fact that the remaining assumptions are very general. Counterexamples are given after each result to prove sharpness. We end the chapter with a geometric application to the group of conformal diffeomorphisms of a complete manifold and to uniqueness for the Yamabe problem.

5.1

A sharp integral condition

The first theorem gives a condition for two positive solutions of Δu + a(x)u − b(x)uσ = 0,

σ > 1,

on M

(5.1)

to coincide. The condition is expressed in integral form and, as we shall see, is quite sharp. Theorem 5.1. Let a(x), b(x) ∈ C 0 (M ) and assume that i) b(x) ≥ 0 on M ;

ii) b(x) ≡ 0 on M.


(5.2)

127

128

Chapter 5. Uniqueness

Let u, v ∈ C 2 (M ) be positive solutions of (5.1) and suppose that &

(u − v)2

'−1

∈ L1 (+∞).

(5.3)

∂Br

Then u ≡ v on M . Note that condition (5.3) is implied by u − v ∈ L2 (M ) or even (u − v)2 = o(R2 ) as R → +∞. BR

Proof. Applying the divergence theorem on a geodesic ball BR of radius R to the vector field v

, W = (v 2 − u2 )∇ log u using (5.1) and rearranging, we obtain 2 u 2 v W, ∇r − ∇u − ∇v − ∇v − ∇u v u ∂BR BR BR b(x)(v 2 − u2 )(v σ−1 − uσ−1 ). =

(5.4)

BR

Since the right-hand side is a nonnegative and nondecreasing function of R it tends to a nonnegative limit B ≤ +∞ as R → +∞. Next by Gauss’ lemma and Schwarz’s inequality

∇v ∇u ≤ |v − u| ∇v − v ∇u + ∇u − u ∇v |W, ∇r| ≤ |v − u||v + u| − v u u v and therefore & '2 & W, ∇r ≤ 2 ∂BR

(v − u) ∂BR

2

'&

' 2 u 2 v ∇v − ∇u + ∇u − ∇v . u v ∂BR (5.5)

We claim that

2 u 2 v ∇v − ∇u + ∇u − ∇v ∈ L1 (M ). u v Indeed, assume by contradiction that 2 u 2 v G(R) = ∇v − ∇u + ∇u − ∇v → +∞ as R → +∞. u v BR Since

W, ∇r − G(R) → B ≥ 0 as R → +∞, ∂BR

(5.6)

5.1. A sharp integral condition ¯ sufficiently large, ∀R ≥ R

129

W, ∇r ≥ ∂BR

1 G(R) ≥ 0. 2

Therefore, using (5.5), ' '2 & '2 & & 1 2 W, ∇r ≤ 2G (R) (v − u) . G(R) ≤ 2 ∂BR ∂BR Arguing exactly as in Step 1 of the proof of Theorem 3.2 we contradict (5.3) and conclude that (5.6) holds. Now, because of (5.3) and (5.6) there exists an increasing sequence {rk } ↑ +∞ such that / / 2 2 u v =0 lim (u − v)2 ∇v − ∇u + ∇u − ∇v k→+∞ u v ∂Br ∂Br k

k

and then, using (5.5) we conclude that lim W, ∇r = 0. k→+∞

∂Brk

Next, we evaluate (5.4) along the sequence {rk } and let k → +∞ to obtain " ! 2 u 2 v 2 2 σ−1 σ−1 b(x)(v − u )(v −u )+ ∇v − ∇u + ∇u − ∇v = 0. (5.7) u v M M Due to the assumptions on b(x), ∇u − uv ∇v = 0 on M , so that u = Av for some constant A > 0; substituting into (5.7) yields (1 − A2 ) 1 − Aσ−1 b(x)v σ+1 = 0 M

and since v > 0, b(x) ≥ 0, b(x) ≡ 0 this forces A = 1, that is, u ≡ v on M .

Remark. Despite its generality, assumption (5.3) of Theorem 5.1 is quite sharp. Indeed suppose (M, , ) is Euclidean space Rm with its canonical metric; let a(x) and b(x) be nonnegative continuous functions satisfying (m − 2)2 −2 |x| , 4 (m−2)(σ+1) 2 |x| b(x) ≤ (log |x|)σ+1 (log log |x|)(log log log |x|)2 a(x) ≤

with equality holding for sufficiently large |x|. We shall prove in the next section that the equation Δu + a(x)u − b(x)uσ = 0,

σ>1

on Rm

(5.8)

130


has a family of positive solutions uα (α > 0) satisfying uα (0) = α

uα (x) ∼ |x|−

and

m−2 2

log |x|

as |x| → +∞.

If α1 = α2 , then uα1 and uα2 are distinct solutions for which p

|uα1 − uα2 | rm−1 ≤ Crm−1− for r = |x| 1. Thus p

|uα1 − uα2 | ≤ Crm−1−

m−2 2 p

m−2 2 p

(log r)p

(log r)p .

∂Br

Hence, for p > 2,

& |uα1 − uα2 |

p

'−1

∈ L1 (+∞),

∂Br

while, for p = 2, (5.3) just falls short of being met. Note that this same example shows that the case p = 2 is special for the validity of the theorem.

5.2

A remark on the asymptotic behaviour of solutions: examples in Rm and Hm

Let b(x) ≥ 0 and let u and v be positive solutions of the equation Δu − b(x)uσ = 0,

σ > 1 on (M, , )

(5.9)

such that lim (u(x) − v(x)) = 0.

x→∞

(5.10)

Then, a simple application of the maximum principle shows that u and v coincide. Indeed, setting w = u − v, we have 2

Δ(w2 ) = 2wΔw + 2|∇w| ≥ 2wΔw = 2(u − v)b(x)(uσ − v σ ) ≥ 0 so that w2 is a nonnegative subharmonic function on M which tends to zero at infinity. Thus it attains a nonnegative maximum and it is therefore identically zero by the maximum principle (see, [CN92]). This suggests that two positive solutions u and v of Δu + a(x)u − b(x)uσ = 0,

σ > 1 on (M, , )

may coincide if they have the “same” behaviour at infinity. The following example shows that, in general, it may be that a simple limit condition like (5.10) is not enough to conclude. Let B be the unit ball in Rm and, for a > 1, define βa (r) =

m−2 1 (a2 − r2 )− 2 2 m(m − 2)a

r ∈ [0, 1), m ≥ 3.

5.2. Asymptotic behaviour of solutions: examples in Rm and Hm

131

A simple computation shows that β satisfies m+2 m−2 β = β , r ∈ [0, 1), βa + m−1 a a r 1 βa (0) = m(m−2)a , β m a (0) = 0, so that βa (|x|) is a family of positive radial solutions of (5.9) on B with b(x) ≡ 1, m+2 σ = m−2 . Operating the change of variable r(t) =

t (1 + tm−2 )

1/(m−2)

we obtain a solution αa (t) = βa (r(t)) of m+2 m−2 , t ∈ [0, +∞) αa + m−1 t αa = b(t)αa 1 αa (0) = m(m−2)am , αa (0) = 0, with

m−1 b(t) = 1 + tm−2 −2 m−2 .

Thus αa gives rise to a family of positive radial solutions ua (x) of m+2 −2 m−1 m−2 Δua − 1 + |x|m−2 uam−2 = 0

for which lim

|x|→+∞

ua (x) =

on Rm

m−2 1 (a2 − 1)− 2 . 2 m(m − 2)a

It is clear that two solutions ua1 and ua2 of the family coincide if and only if lim

|x|→+∞

ua1 =

lim

|x|→+∞

ua2 ,

that is, a1 = a2 . Note that the geometrical meaning of these solutions is obvious: Rm can be conformally deformed to a complete Riemannian manifold of negative scalar curvature sufficiently close to zero. Thus, in this case, uniqueness is expressed by (5.10). However, we may consider the same family of solutions on the hyperbolic space Hm of constant negative curvature −1 by identifying it with Rm with metric, in polar coordinates, ds2 = dr2 + (sinh r)2 dθ2 . In this case the situation changes completely: indeed, let γa (t) = βa

t tanh 2

5

2 1 − tanh2

.− m−2 2 t 2

.

Then, ⎧ ⎨γ + (m − 1) coth(t) γ + a a ⎩γa (0) = 0, γa (0) = (m−2) 2

2

m(m−2) γa 4 1 m(m−2)am

m+2

= γam−2

on [0, +∞),

132


and − m−2 "− m−2 ! 2 2 1 2 ρ(x) 2 ρ(x) 2 a 2 cosh − tanh , wa (x) = m(m − 2)a2 2 2 with ρ(x) = distHm (x, o), is a family of positive solutions of Δwa +

m+2 m(m − 2) wa − wam−2 = 0 4

on Hm .

We observe that m−2

wa (x) ∼

m−2 m−2 2− 2 (a2 − 1)− 2 e− 2 ρ(x) m(m − 2)a2

as ρ(x) → +∞.

Thus, wa (r) → 0

as r(x) → +∞

independently of a > 1 and with the same speed. In the next section we show that uniqueness of solutions converging to zero is indeed special.

5.3 Uniqueness via the weak maximum principle In this section we prove a number of uniqueness results for nonnegative solutions of equation (5.1). We begin with the following (see [RZ07]). Theorem 5.2. Let (M, , ) be a complete manifold, a(x), b(x) ∈ C 0 (M ), σ > 1, τ ≥ 0, β + τ (σ − 1) > −2 and suppose that b(x) > 0 on M and i) b(x) ≥ Br(x)β for r(x) 1;

ii) sup M

a− (x) r(x)τ (1−σ) < +∞. b(x)

(5.11)

Suppose that u, v ∈ C 2 (M ) are two nonnegative solutions of Δu + a(x)u − b(x)uσ = 0 satisfying

on M

(5.12)

C −1 r(x)τ ≤ u(x), v(x) ≤ Cr(x)τ

(5.13)

for r(x) 1 and some constant C > 0. If lim inf r→+∞

log vol Br 2+β+τ (σ−1) r

< +∞,

(5.14)

then u ≡ v on M . Theorem 5.2 will follow immediately from a corresponding comparison result whose proof is based on the next weak maximum principle at infinity.

5.3. Uniqueness via the weak maximum principle

5.3.1

133

A useful form of the weak maximum principle

The next result, which is a version of what is known in literature as the weak maximum principle at infinity (see [RSV05], [PRS05b] and [MRS10] for a more general version), turns out to be quite useful in many contexts, both geometrical and analytical. It is worth observing that the function u under consideration is not necessarily bounded but is only required to satisfy some growth control (expressed in (5.17)). The other main assumption of the theorem is a control on the volume of geodesic balls (see (5.16)). Theorem 5.3. Let (M, , ) be a complete manifold, and let σ, μ ∈ R, be such that i) σ ≥ 0; Assume that lim inf r→+∞

ii) 2 − σ − μ > 0.

(5.15)

log vol Br = d0 < +∞. r2−σ−μ

(5.16)

Let u ∈ C 2 (M ) and suppose that u(x) < +∞. σ r(x)→+∞ r(x)

u 6 = lim sup

(5.17)

Then, given γ ∈ R such that Ωγ = {x ∈ M : u(x) > γ} = ∅, we have μ

inf [1 + r(x)] Δu ≤ d0 max {6 u, 0} C(σ, η) Ωγ

with

⎧ ⎪ ⎨0 C(σ, μ) = (2 − σ − μ)2 ⎪ ⎩ σ(2 − σ − μ)

if σ = 0; if σ > 0, μ + 2(σ − 1) < 0; if σ > 0, μ + 2(σ − 1) ≥ 0.

(5.18)

(5.19)

Proof. We begin by observing that if ν is any constant and we set uν = u + ν, then Δuν = Δu and {x ∈ M : u(x) > γ} = Ωγ = {x ∈ M : uν (x) > γ + ν}. Furthermore, if σ > 0, u 6=u 6ν so, in order to estimate μ

inf [1 + r] Δu Ωγ

we may replace u by a suitable translate. Next, fix b > max {6 u, 0}; then, there exists a constant ν such that uν < b on M and uν (x0 ) > 0 for some x0 ∈ M. (5.20) (1 + r)σ

134


We will assume that a constant ν has been selected in such a way that (5.20) holds. In accordance with the observation made at the beginning of the proof, we are going to replace u with uν , and for ease of notation we will even suppress the subscript ν. We want to show (5.18) and we may suppose, without loss of generality, that γ ≥ 0. Next, let μ

K = inf [1 + r] Δu. Ωγ

Clearly, the theorem amounts to showing that K is bounded above by the righthand side of (5.19). If K ≤ 0 there is nothing to prove; so let us assume K > 0, in which case u is nonconstant on any connected component of Ωγ , and μ

[1 + r] Δu ≥ K > 0

on Ωγ .

(5.21)

We fix θ ∈ 12 , 1 and we choose R0 > 0 large enough that |∇u| ≡ 0 on the nonempty open set BR0 ∩ Ωγ . Given R > R0 , let ϕ ∈ C ∞ (M ) be a cut-off function such that 0 ≤ ϕ ≤ 1,

ϕ ≡ 1 on BθR , ϕ ≡ 0 on M \BR , |∇ϕ| ≤

c R(1 − θ)

(5.22)

for some absolute constant c > 0. Let also λ ∈ C 1 (R) and F (v, r) ∈ C 1 (R2 ) be such that 0 ≤ λ ≤ 1; λ = 0 on (−∞, γ]; λ > 0, λ ≥ 0 on (γ, +∞)

(5.23)

and F (v, r) > 0;

∂F (v, r) < 0 on [0, +∞) × [0, +∞). ∂v

(5.24)

Finally, let W be the C 1 vector field defined on Ωγ by W = ϕ2 λ(u)F (v, r)∇u

(5.25)

v = α(1 + r)σ − u

(5.26)

where v is given by and α > b is a constant. Note that, according to (5.20), v > 0 on Ωγ , and, in fact, since u > γ > 0 on Ωγ , we have (α − b)(1 + r)σ ≤ v ≤ α(1 + r)σ

on Ωγ .

(5.27)

We note that W vanishes on ∂(Ωγ ∩ BR ) and it extends to a C 1 vector field on the whole of M by setting it equal to 0 in the complement of Ωγ ∩ BR . A computation that uses the properties of ϕ, λ and F , the definition (5.26) of v, inequality (5.21) and Cauchy-Schwarz inequality yields


135

div W = ϕ2 λ(u)F (v, r)Δu + 2ϕλ(u)F (v, r) ∇ϕ, ∇u ∂F + ϕ2 λ (u)F (v, r)|∇u|2 + ϕ2 λ(u) (v, r) ∇v, ∇u ∂v ∂F + ϕ2 λ(u) (v, r) ∇u, ∇r ∂r ≥ ϕ2 λ(u)F (v, r)(1 + r)−μ K − 2ϕλ(u)F (v, r)|∇ϕ||∇u| ∂F + ϕ2 λ(u) (v, r) ∇u, ασ(1 + r)σ−1 ∇r − ∇u ∂v ∂F 2 + ϕ λ(u) (v, r) ∇u, ∇r ∂r ≥ −2ϕλ(u)F (v, r)|∇ϕ||∇u|+ ∂F 2 + ϕ λ(u) (v, r) ∂v . / 5 ∂F F (v, r) −μ 2 σ−1 ∂r ∂F K(1 + r) + |∇u| + ∂F − ασ(1 + r) × ∇u, ∇r , ∂v

∂v

that is, ∂F div W ≥ −2ϕλ(u)F (v, r)|∇ϕ||∇u| + ϕ2 λ(u) (v, r)B(∇u, r) ∂v

(5.28)

with . 5 ∂F F (v, r) −μ σ−1 ∂r ∇u, ∇r . B(∇u, r) = |∇u| + ∂F K(1 + r) + ∂F − ασ(1 + r) 2

∂v

∂v

(5.29) For notational convenience we set η = μ + 2(σ − 1) so that by (5.15), σ − η > 0, and we divide the proof into three cases. Case I. σ > 0, η < 0. We let F (v, r) = e−qv(1+r)

−η

where q > 0 is a constant that will be specified later. An elementary computation which uses the estimate for v given in (5.27) shows that ∂F ∂r (v, r) 0 ≥ ∂F − ασ(1 + r)σ−1 = ηv(1 + r)−1 − ασ(1 + r)σ−1 ≥ −α(σ − η)(1 + r)σ−1 (v, r) ∂v (5.30) and F (v, r) 1 ∂F = (1 + r)η . (5.31) (v, r) q ∂v

136


Inserting (5.30) and (5.31) into (5.29) and using Cauchy-Schwarz inequality we deduce B(∇u, r) ≥ |∇u|2 +

K (1 + r)2(σ−1) − α(σ − η)(1 + r)σ−1 |∇u|. q

(5.32)

At this point we need to estimate the right-hand side of (5.32) so as to have B(∇u, r) ≥ Λ|∇u|2 ,

(5.33)

with a constant Λ > 0 independent of ∇u and r. A simple computation shows that if we choose 4K q< (5.34) 2, [α(σ − η)] then (5.33) holds for every 2

0 d0 there exists a divergent sequence Rk ↑ +∞ with R1 > 2R0 such that log vol(BRk ) ≤ dRkσ−η .

(5.37)

Noting that θRk > 21 Rk > R0 , we apply (5.36) with R = Rk , use the bound for |∇ϕ| and the fact that λ ≤ 1, to get 2 qλ E= λ(u)F (v, r)|∇u|2 (5.38) 2 Ωγ ∩BR0 −2 ≤ (1 + θRk )η [(1 − θ)Rk ] F (v, r). (5.39) Ωγ ∩(Brk \BθRk )


137

Now we observe that E > 0, since ∇u ≡ 0 on Ωγ ∩ BR0 . On the other hand, using the bound (5.27) for v and the expression of F we get F (v, r) ≤ e−q(α−b)(1+θRk )

σ−η

on Ωγ ∩ (BRk \BθRk ).

Inserting this into the right-hand side of (5.38), using (5.37) and rearranging we obtain the inequality σ−η

0 < E ≤ CRkη−2 edRk

−q(α−b)(1+θRk )σ−η

(5.40)

where C > 0 is a constant independent of k. In order for this inequality to hold for every k we must have d ≥ (α − b)qθσ−η , whence, letting θ tend to 1, d ≥ (α − b)q. We set α = tb, insert the definition (5.34) of q in the above inequality, solve with respect to K, and then let τ → 1 to obtain K≤d

b(σ − η)2 t2 , 4 t−1

whence minimizing with respect to t > 1 and letting d → d0 and b → max {6 u, 0}, we obtain K ≤ d0 max u 6, 0(σ − η)2 . (5.41) Case II. σ = 0 (and therefore η < 0 by (5.15)). We proceed exactly as in the previous case and conclude that for any C 2 function u bounded above (5.41) holds. But, as noted earlier, ∀ ν, lim inf (1 + r)μ Δu = Ωγ

inf

{x:uν (x)>γ+ν}

(1 + r)μ Δun u.

(5.42)

Thus we may choose ν in such a way that u 6ν ≤ 0, apply (5.41) to uν to conclude that the right-hand side of (5.42) is nonpositive and then deduce that K ≤ 0 as required. Case III. σ > 0, η ≥ 0. We choose F (v, r) = F (v) = e−qv

σ−η σ

where q > 0 is a constant to be specified later. Since σ − η > 0 by assumption, a computation gives ∂F σ − η −η (v, r) = −q v σ F (v) < 0, ∂v σ

138


while clearly ∂F ≡ 0. ∂r Using the above expression, recalling that v ≥ (α−b)(1+r)σ on Ωγ , and proceeding as in Case I, we estimate B(∇u, r) ≥ |∇u|2 − ασ(1 + r)σ−1 |∇u| +

Kσ(α − b4/σ (1 + r)2(σ−1) . q(σ − η)

It is now a simple matter to observe that, ∀ r ≥ 0 fixed, the right-hand side of the above inequality is bounded from below by Λ|n∇u|2 provided Λ≤1−

qα2 σ 2 (σ − η) . 4Kσ(α − b)η/σ

(5.43)

Since the right-hand side of the above inequality is independent of r, for every such Λ we have B(∇u, r) ≥ Λ|∇u|2 . In particular, if τ ∈ (0, 1) and we choose q=

4τ Kσ(α − b)η/σ α2 σ 2 (σ − η)

and Λ = 1 − τ,

(5.44)

then Λ is positive and satisfies (5.43). Substituting into (5.28) and using the expression of ∂F ∂v we deduce that div W ≥ −2ϕ2 λ(u)F (v)|∇ϕ||∇u|Λq

σ−η 2 ϕ λ(u)v η/σ |∇u|2 . σ

We now proceed as in Case I, repeating, with minor adaptations, the arguments that lead to (5.40), to conclude that −2 0<E= λ(u)F (v)|∇u|2 ≤ C(1 + Rk )η [(1 − θ)Rk ] Ωγ ∩BR0

×

F (v, r) Ωγ ∩(BRk \BθRk )

where C > 0 is a constant independent of k and θ. Using the inequality F (v) ≤ e−q(α−b)

σ−η (1+θRk )σ−η σ

valid on Ωγ ∩ (BRk \BθRk ) and the volume growth estimate (5.38), we deduce that 0 < E ≤ C(α − b)−q (1 + θRk )−qσ (1 + Rk )σ Rkd−2


139

with C > 0 independent of k. This forces σ + d − 2 ≥ qσ, and letting d ↓ d0 , σ + d0 − 2 ≥ qσ. Thus, if σ+d0 −2 < 2 we get a contradiction, on K ≤ 0. Otherwise, if σ+d0 −2 ≥ 0, substituting the value of q, solving for K, letting α = tb, t > 1, with τ ↑ 1 and b ↓ max {6 u, 0}, we obtain K≤

1 t2 max {6 u, 0}σ(σ + d0 − 2) ; 4 t−1

whence, minimizing over t > 1, K ≤ max {6 u, 0}σ(σ + d0 − 2). This completes the proof if σ > 0. The case σ = 0 is dealt with as in Case II. We note, for future use, the following corollary (see [PRS05b], Corollary 4.4): Corollary 5.4. Let (M, , ) be a complete manifold, and let b(x) ∈ C 0 (M ) satisfy b(x) ≥

C (1 + r(x))μ

for some μ ∈ R. Let u ∈ C 2 (M ) be a solution of the differential inequality Δu ≥ b(x)f (u)

(5.45)

and suppose that, for some 0 ≤ σ < 2 − μ, either σ > 0, lim inf t→∞ f (t) > 0 and u(x) = o (r(x)σ ) as r(x) → +∞, or σ = 0 and u is bounded above. If lim inf r→∞

vol Br < +∞, r2−σ−μ

then u∗ = sup u < +∞ and f (u∗ ) ≤ 0. Proof. Observe first of all that under the stated assumptions it follows from Theorem 5.3 that, for every γ < u∗ , inf b(x)−1 Δu ≤ C −1 inf (1 + r(x))μ Δu ≤ 0. Ωγ

Ωγ

Now, if u were unbounded, then Ωγ would be nonempty for every γ and it would follow from (5.45) that inf f (t) ≤ 0, (5.46) t>γ

contradicting the assumption that f is bounded away from zero at infinity. Thus u is necessarily bounded above and u∗ < +∞. Arguing as above shows that (5.45) holds for every γ < u∗, and therefore f (u∗ ) ≤ 0.

140

5.3.2


A comparison result

We are now ready to prove the desired comparison result (see [RZ07] and also [MR10] for the case of a diffusion-type operator), from which the proof of Theorem 5.2 follows at once. Theorem 5.5. Let a(x), b(x) ∈ C 0 (M ), σ > 1, τ ≥ 0, β + τ (σ − 1) > −2 and suppose that b(x) > 0 on M and i) b(x) ≥ Br(x)β , B > 0 for r(x) 1;

ii) sup M

a− (x) r(x)τ (1−σ) < +∞. (5.47) b(x)

Let u, v ∈ C 2 (M ) be nonnegative solutions of Δu + a(x)u − b(x)uσ ≥ 0 ≥ Δv + a(x)v − b(x)v σ

(5.48)

on M , satisfying i) v(x) ≥ C1 r(x)τ ;

ii) u(x) ≤ C2 r(x)τ

(5.49)

for r(x) 1 and some positive constants C1 , C2 . If lim inf r→+∞

log vol Br < +∞, r2+β+τ (σ−1)

(5.50)

then u(x) ≤ v(x) on M . Proof. First of all, let u(x) ≡ 0, otherwise there is nothing to prove. Next, we observe by (5.48), (5.49) and the strong maximum principle that v(x) > 0 on M . This fact, u(x) ≡ 0, and (5.49) i), ii) imply that ξ = sup M

u(x) v(x)

(5.51)

satisfies 0 < ξ < +∞. If ξ ≤ 1, then u(x) ≤ v(x) on M . Let us assume, by contradiction, ξ > 1 and define ϕ = u − ξv. Note that ϕ ≤ 0 on M . We claim sup r(x)−τ ϕ(x) = 0.

(5.52)

M

Indeed, let {xn } ⊂ M be a sequence realizing ξ; then & ' u(xn ) −ξ . r(xn )−τ ϕ(xn ) = r(xn )−τ v(xn ) v(xn )

(5.53)

5.3. Uniqueness via the weak maximum principle Now observe that

141

r(xn )−τ v(xn )

is bounded because otherwise (5.49) ii) would imply ξ = 0. Then, it follows from (5.53) that r(xn )−τ ϕ(xn ) → 0 as n → +∞ proving (5.52). We now use (5.48) to obtain Δϕ ≥ −a(x)ϕ + b(x)(uσ − (ξv)σ ) + b(x)v σ ξ(ξ σ−1 − 1). (5.54) We write (uσ − (ξv)σ )(x) = h(x)ϕ(x)

where h(x) =

σu(x)σ−1 σ u(x)−ξv(x)

# u(x) ξv(x)

(5.55)

if u(x) = ξv(x), t

σ−1

dt

if u(x) = ξv(x),

is continuous and nonnegative on M . Fix the level set Ω1 = {x ∈ M : ϕ(x) > −1}, and note that (1 + r(x))−τ v(x) is bounded above on Ω1 . Indeed, using (5.49) ii), for every y ∈ Ω1 we have ξ

v(y) (u − ϕ)(y) 1 2 + 2 + 1. = ≤C ≤C (1 + r(y))τ (1 + r(y))τ (1 + r(y))τ

Thus there exists a constant C3 > 0 such that v(x) ≤ C3 (1 + r(x))τ

on Ω1 .

(5.56)

Now, from the mean value theorem for integrals, for some y ∈ (ξv(x), u(x)) or y ∈ (u(x), ξv(x)), σ (u(x) − ξv(x))y σ−1 u(x) − ξv(x) ≤ σ uσ−1 (x) + (ξv(x))σ−1

h(x) =

≤ C(1 + r(x))τ (σ−1)

(5.57)

on Ω1

because of (5.49) ii) and (5.56). Now we note that, since b(x) > 0 on M we can rewrite (5.47) i) as + r(x))β on M, b(x) ≥ B(1 (5.58) > 0. Using b(x) > 0 and (5.55), from (5.54) and ϕ ≤ 0 we for some appropriate B deduce 1 a− (x) Δϕ ≥ + h(x) ϕ(x) + v σ (x)ξ(ξ σ−1 − 1) b(x) b(x) and therefore, from ξ > 1, (5.47) ii), (5.57), (5.49) i) and ϕ ≤ 0, (1 + r(x))−στ

1 Δϕ ≥ C(1 + r(x))−τ ϕ(x) + Dξ(ξ σ−1 − 1) b(x)

on Ω1

142


for some appropriate constants C, D > 0. Next, we choose 0 < ε < 1 sufficiently small so that 1 C(1 + r(x))−τ ϕ(x) ≥ − Dξ(ξ σ−1 − 1) (5.59) 2 on Ωε = {x ∈ M : ϕ(x) > −ε} ⊂ Ω1 . Note that this is possible since τ ≥ 0. Then, on Ωε , Δϕ ≥ 0 so that (5.58) implies (1 + r(x))−β−στ Δϕ ≥ (1 + r(x))−στ and thus inf (1 + r)−β−στ Δϕ ≥ Ωε

1 Δϕ b(x)

1 Dξ(ξ σ−1 − 1) > 0 2

(5.60)

since ξ > 1. This fact, together with (5.50), contradicts Theorem 5.3.

Combining the a priori estimates given in Theorems 4.1 and 4.4, the volume growth estimates of Chapter 1 and Theorem 5.2, we obtain Theorem 5.6. Let (M, , ) be a complete manifold with Ricci tensor satisfying Ric ≥ −(m − 1)H 2 (1 + r2 )δ/2

(5.61)

for some H > 0, δ ≥ −2. Let a(x), b(x) ∈ C 0 (M ) with b(x) > 0 on M and assume that, for some A, B > 0, A−1 r(x)α ≤ a(x) ≤ Ar(x)α , B

−1

r(x) ≤ b(x) ≤ Br(x) β

(5.62)

β

(5.63)

for r(x) 1 and with α > δ, α ≥ max β, β + 2δ − 1 . Then, there exists at most one nonnegative, nontrivial solution u ∈ C 2 (M ) of Δu + a(x)u − b(x)uσ = 0,

σ > 1,

on M.

(5.64)

Remark. The geometric case of Yamabe’s equation for which α ≥ δ will be dealt with in Section 5.3. Proof. First observe, by the maximum principle, that if u ≥ 0 and u ≡ 0 is a solution of (5.64), then u > 0 on M . Thus if u ≡ 0, according to the a priori estimates of Theorems 4.1 and 4.4 we have (note that α ≥ 2δ − 1 because α > δ ≥ −2, while β ≤ 1 − 2δ + α because of our assumptions) β−α

β−α

C −1 r(x)− σ−1 ≤ u(x) ≤ Cr(x)− σ−1

(5.65)

for some constant C > 0 and r(x) 1. To apply Theorem 5.2, since a− (x) ≡ 0 for r(x) 1 because of (5.62) we only need to have α ≥ β and α > −2. This


143

latter is guaranteed by α > δ, while the first is satisfied by assumption. It remains to check that (5.14) holds, which in this case becomes lim inf r→+∞

log vol Br < +∞. r2+α

(5.66)

Towards this aim, we need to estimate vol Br from above using assumption (5.61). Using Proposition 1.15, we see that if δ ≥ −2, then log vol Br ≤ Cr1+ 2 δ

as r → +∞

for some constant C > 0, and (5.66) is satisfied provided α≥ If δ = −2, then

vol Br ≤ C

r

s

1+

√

1+4H 2 2

δ − 1. 2

(m−1)

0

(5.67)

ds # r

1+

√

1+4H 2 2

(m−1)+1

,

and (5.66) is satisfied provided α > −2.

(5.68)

Note that α > δ implies (5.67) and (5.68).

5.3.3

Uniqueness of ground states

Our aim is now to prove uniqueness of positive solutions converging to zero at infinity, or ground states (see also Chapter 3, section 3.2). As the above examples show we have to impose some extra condition. Surprisingly, this condition is again related to λL 1 (M ) ≥ 0, as we shall see below. We begin with the following general result proved in [RRV97]; we denote with W 1,2 (M ) the Sobolev space of functions which are in L2 (M ) and have L2 weak gradient. Theorem 5.7. Let (M, , ) be a complete manifold and let a(x), b(x) ∈ C 0 (M ), b(x) ≥ 0, b(x) ≡ 0. Let u1 and u2 be positive solutions of Δu + a(x)u − b(x)uσ = 0.

(5.69)

1,2 If there exists ϕ ∈ C 0 (M ) ∩ Wloc (M ), ϕ > 0 and such that

Δϕ + a(x)ϕ − σb(x)uσ−1 ϕ≤0 i

(5.70)

in the weak sense on M for i = 1, 2, and u1 (x) − u2 (x) = o(ϕ(x)) then u1 ≡ u2 .

as r(x) → +∞,

(5.71)

144


Proof. We divide the proof in two steps. Step I. For ε > 0 we set v = vε = u1 + εϕ. We claim that H(x) = Δv + a(x)v − b(x)v σ ≤ 0

(5.72)

in the weak sense on M . Indeed we find H(x) = Δu1 + a(x)u1 − b(x)(u1 + εϕ)σ + εΔϕ + εa(x)ϕ.

(5.73)

From (5.69) we get ! " 1 H(x) = ε Δϕ + a(x)ϕ − b(x)((u1 + εϕ)σ − uσ1 ) . ε ϕ, by convexity we have Since (u1 + εϕ)σ − uσ1 ≥ εσuσ−1 1 H(x) ≤ ε Δϕ + a(x)ϕ − σb(x)uσ−1 ϕ 1

(5.74)

(5.75)

which implies (5.72) from (5.70). Step II. We claim that u1 ≥ u2 .

(5.76)

Δv Δu2 − − v σ−1 ≥ 0 − b(x) uσ−1 2 u2 v

(5.77)

From (5.69) and (5.72) we have

weakly on M . Since (u22 − v 2 )+ ∈ C 0 (M ) ∩ W 1,2 (M ) and compactly supported, by (5.71)), it is an admissible test function and we have Δu2 Δv 2 2 − b(x) uσ−1 − v σ−1 (u22 − v 2 )+ ≥ 0. (5.78) u2 − v + − 2 u v 2 M M Suppose now that Ω = {x ∈ M : u2 (x) > v(x)} = ∅. Then, (5.78) yields 2 2 v u2 − v σ−1 (u22 − v 2 ) ≤ 0 (5.79) ∇u2 − ∇v + ∇v − ∇u2 + b(x) uσ−1 2 v u 2 Ω Ω and since b(x) ≥ 0, b(x) ≡ 0 we conclude, as in Theorem 5.1, that u2 ≤ u1 + εϕ

∀ ε > 0,

that is, (5.76). Similarly, u1 ≤ u2 and therefore u1 ≡ u2 .


145

Corollary 5.8. Let (M, , ) be a complete manifold and let a(x), b(x) ∈ C 0 (M ), b(x) ≥ 0, b(x) ≡ 0. Suppose ψ is a positive solution of Δψ + a(x)ψ ≤ 0

on M.

(5.80)

If u1 and u2 are positive solutions of (5.69) on M satisfying u1 (x) − u2 (x) = o(ψ(x))

as r(x) → +∞,

(5.81)

then u1 ≡ u2 . Proof. It is enough to show that ψ satisfies (5.70) of Theorem 5.7. But this is obvious, since b(x) ≥ 0, ui > 0 and (5.80) imply 0 ≥ Δψ + a(x)ψ ≥ Δψ + a(x)ψ − σb(x)uσ−1 ψ, i

i = 1, 2.

Remark. By a result of Fisher-Colbrie and Schoen, [FCS80], condition (5.80) is equivalent to λL 1 (M ) ≥ 0 with L = Δ + a(x) (see the discussion at the beginning of Chapter 3). We observe that if 2−m t ρ(t) = C cosh > 0 on [0, +∞), 2 then the function ψ(x) = ρ(r(x)) is a radial solution of

m(m − 2) ψ=0 4 on the m-dimensional hyperbolic space Hm , of constant negative curvature −1 (m ≥ 3). In this case condition (5.81) becomes ΔHm ψ +

(u1 (x) − u2 (x))e

m−2 2 r(x)

→0

as r(x) → +∞.

In the case of the example in section 5.2 we have 5 . m−2 m−2 m−2 m−2 (a21 − 1)− 2 2− 2 (a22 − 1)− 2 r(x) 2 (wa1 (x) − wa2 (x))e ∼ − m(m − 2) a21 a22 and this converges to zero if and only if a1 = a2 . This shows that (5.81) cannot be relaxed to O(ψ(x)) as r(x) → +∞. It is clear that the applicability of Corollary 5.8 depends on the knowledge of the asymptotic behaviour of the solution ψ of (5.80). We shall come back to this later, in section 7.3.2.

146


5.4 Some geometric applications and further uniqueness We apply the uniqueness results obtained so far to the problem of characterizing isometries into the group of conformal diffeomorphisms of the complete manifold M . We then go back to uniqueness for the Yamabe problem; this will be achieved via an L∞ a priori estimate of independent interest.

5.4.1

Conformal diffeomorphisms

Let (M, , ) be a Riemannian manifold with scalar curvature S(x). To simplify the writing we consider the case m = dim M ≥ 3, but one could treat the case m = 2 with few changes. Suppose that ϕ : M → M is a conformal diffeomorphism, that is, ϕ is a diffeomorphism such that, for some u ∈ C ∞ (M ), u > 0, 4

ϕ∗ , = u m−2 ,

(5.82)

(u is usually called the stretching factor of the diffeomorphism). We shall express this fact by writing ϕ ∈ Conf(M ), where Conf(M ) denotes the group of conformal diffeomorphisms. If u ≡ 1, then ϕ is an isometry. If we denote by Iso(M ) the group of isometries, then, clearly, Iso(M ) ⊆ Conf(M ). Note that if ϕ is an isometry, then it preserves the scalar curvature (in fact the sectional curvature). It is thus meaningful to investigate when a conformal diffeomorphism preserving the scalar curvature is an isometry. If ϕ preserves scalar curvature, then its stretching factor u is a positive solution of the (special) Yamabe equation * ) 4 m−2 Δu = − (5.83) S(x)u u m−2 − 1 . 4(m − 1) Since the inverse diffeomorphism ϕ−1 also preserves the scalar curvature and 4

(ϕ−1 )∗ , = u− m−2 , , we conclude that proving that ϕ is an isometry amounts to showing that u ≤ 1. After this preparation we prove the next (see [RRV94a]) Theorem 5.9. Let (M, , ) be a complete manifold with Ricci tensor satisfying Ric ≥ −(m − 1)H 2 (1 + r(x)2 )δ/2

on M

(5.84)

for some H > 0, δ ≥ −2. Suppose furthermore that the scalar curvature S(x) satisfies S(x) ≤ 0 on M and S(x) ≤ −

d2 (1 + r(x))μ

for r(x) 1

(5.85)

5.4. Some geometric applications and further uniqueness

147

and some d > 0 and μ < 1 − 2δ . Then any conformal transformation of (M, , ) which preserves the scalar curvature is an isometry. m−2 Proof. Let a(x) = b(x) = − 4(m−1) S(x). In Proposition 4.3 we chose ω(t) = C1 > 0 constant and δ δ ψ(t) = C2 t1− 2 −μ , μ ≤ 1 − . 2

Then, it follows from (5.83) and (5.85) that the solution u > 0 of (5.83) satisfies u∗ = sup u < +∞. M

According to the discussion above, in order to show that ϕ is an isometry it is enough to prove that u∗ ≤ 1. We observe that from Theorem 3.10 and the Remark thereafter we can suppose S(x) < 0 on M without loss of generality. Now we reason by contradiction and we suppose u∗ > 1. We fix n ∈ N sufficiently large such that 1 < u∗ − n1 < u∗ and we set ' 1 . x ∈ M : u(x) > u − n

& Ωn =

∗

Fix R > 0 sufficiently large that (5.85) holds on M \B R . Then, using (5.85) and the fact that u∗ − n1 > 1 we have (1 + r(x))μ Δu ≥

4 m − 2 2 m−2 −1 ≥C >0 d u u 4(m − 1)

on Ωn ∩ (M \B R )

for some constant C while (1 + r(x))μ Δu ≥ C > 0

on Ωn ∩ B R

since S(x) < 0 on M . Now, if lim inf r→+∞

log vol Br < +∞ r2−μ

(5.86)

we can apply Theorem 5.3 to obtain the desired contradiction. Proceeding as in the proof of Theorem 5.6 we see that (5.84) implies (5.86) in the assumptions that μ < 1 − 2δ . Remark. In case δ > −2 the conclusion in the above theorem is reached under the milder assumption μ ≤ 1 − 2δ . Indeed, as shown in [RRV94a], Corollary 1, the results holds for ν ≤ 1 − 2δ in the whole range δ ≥ −2. Similarly to what we did above we can use the uniqueness result contained in Theorem 5.1 to obtain the following version:

148


Theorem 5.10. Let (M, , ) be a complete manifold of dimension m ≥ 3 and scalar curvature S(x) satisfying i) S(x) ≤ 0 on M ;

ii) S(x) ≡ 0 on M.

Let ϕ : M → M be a conformal diffeomorphism whose stretching factor u satisfies & (u − 1)

2

'−1

∈ L1 (+∞).

∂Br

Then ϕ is an isometry.

5.4.2

Uniqueness for the Yamabe problem

As we pointed out in Section 4.1 , see the remark after the proof of Theorem 4.1, the lower estimates on u obtained there do not apply to solutions of the Yamabe problem since condition (4.27) is not satisfied under the usual assumption (4.2). This implies that Theorem 5.6 cannot be applied to deduce uniqueness for the Yamabe problem. However, the next result where no assumptions on the Ricci tensor are required does hold (see [RZ07], Theorem 3.4). Theorem 5.11. Let (M, , ) be a complete manifold of dimension m ≥ 3 and scalar curvature S(x) satisfying sup S(x) < +∞.

(5.87)

M

Let K(x) ∈ C ∞ (M ), K(x) < 0 on M and suppose that K(x) ≤ −

B r(x)β

for r(x) 1,

(5.88)

some constant B > 0, β < 2 and inf M

S(x) > −∞. K(x)

(5.89)

log vol Br < +∞. r2−β

(5.90)

Assume lim inf r→+∞

4

Then, there exists at most one conformal deformation to a new metric g = u m−2 , with scalar curvature K(x) and such that u∗ = inf M u > 0. Proof. Suppose such a metric exist. Thus u > 0 satisfies m+2

cm Δu − S(x)u + K(x)u m−2 = 0


149

on M . We claim that (5.87), (5.88), (5.89) and (5.90) imply u∗ = sup u < +∞.

(5.91)

M

Postponing for the moment the proof of the claim, we have 0 < u∗ ≤ u ≤ u∗ < +∞. A direct application of Theorem 5.5 with τ = 0 gives uniqueness of u.

It remains to show that u is bounded above under the sole assumption of the volume growth condition (5.90): this is precisely the content of the next remarkable result.

5.4.3

An L∞ a priori estimate

We prove here the following result of very general interest, which was obtained in [PRS03b]. Theorem 5.12. Let (M, , ) be a complete manifold and a(x), b(x) ∈ C 0 (M ). Assume sup a+ (x) < +∞ (5.92) M

and that b(x) satisfy b(x) > 0

on M,

b(x) ≥

B r(x)β

(5.93)

for r(x) 1, B > 0 and some 0 ≤ β < 2. Assume further that a+ (x) ≤E b(x)

on M.

(5.94)

Let u ∈ C 1 (M ) be a nonnegative solution of Δu + a(x)u − b(x)uσ ≥ 0

on M

(5.95)

with σ > 1. If lim inf r→+∞

log vol Br < +∞, r2−β

(5.96)

then 1

u(x) ≤ E σ−1

on M.

To prove Theorem 5.12, we first need two technical results.

(5.97)

150


Lemma 5.13. Let a(x), b(x) ∈ C 0 (M ), a(x) = a+ (x) − a− (x), with a± ≥ 0, b(x) > 0, and assume that, for some E > 0 we have a+ (x) ≤E b(x)

on M.

(5.98)

Assume that u ∈ C 2 (M ) and γ > 0 are such that Ωγ = {x ∈ M : u(x) > γ} = ∅

(5.99)

and that u satisfies Δu ≥ b(x)uσ − a(x)u +

D 2 |∇u| , u

(5.100)

on Ωγ , for some constants D ∈ R and σ > 1. Let λ : R → [0, +∞) be a C 1 , nondecreasing function such that l(t) = 0 for t ≤ γ. Then, there exists R > 0 sufficiently large, and a constant C > 0 such that, for every r > R, and for every α > max 1 − D, 1, &

b(x)λ(u) ≤ Br

C 1 (α + σ − 1)2 + σ−1 2 γ r inf B2r b D + α − 1 1

E γ σ−1

' α+σ−1 σ−1 b(x)λ(u). B2r

(5.101)

Proof. Let R > 0 large enough that BR ∩ Ωγ = ∅, and fix ζ > 1 such that 2+

1+δ 1 − 1 > 0. σ−1 ζ

(5.102)

We choose a C ∞ cut-off function ψ : M → [0, 1] such that, for every r ≥ R, i) ψ ≡ 1 on Br ;

ii) ψ ≡ 0 on M \ B2r ;

iii) |∇ψ| ≤

C0 1/ζ ψ , r

(5.103)

for some constant C0 = C0 (ζ) > 0. Note that this is possible since ζ > 1. Finally, we fix α and consider the vector field W defined by W = ψ 2(α+σ−1) λ(u)uα−1 ∇u. Note that the properties of λ and ψ imply that u vanishes off B2r ∩ Ωγ . A computation that uses (5.100), λ ≥ 0, α ≥ 1 − D and the Cauchy–Schwarz inequality, yields div W ≥ −2(α + σ − 1)ψ 2(α+σ−1)−1 λ(u)uα−1 |∇u||∇ψ| 0 1 2 + λ(u)ψ 2(α+σ−1) b(x)uα+σ−1 − a(x)uα + ((D + α − 1))uα−2 |∇u| .


151

Recalling that W is compactly supported, the divergence theorem yields ψ 2(α+σ−1) λ(u)b(x)uα+σ−1 − ψ 2(α+σ−1) λ(u)a(x)uα 2 ≤ −(D + α − 1) ψ 2(α+σ−1) λ(u)uα−2 |∇u| + 2(α + σ − 1) ψ 2(α+σ−1)−1 λ(u)uα−1 |∇u||∇ψ|. (5.104) p p

Now we apply the inequality ab ≤ ε pa + conjugate exponents, with the choices p = q = 2,

bq εq q ,

valid for a, b ≥ 0, ε > 0, with p, q

ε = (2(D + α − 1))

1/2

to the second integral on the right-hand side of (5.104) to obtain 2(α + σ − 1) ψ 2(α+σ−1)−1 λ(u)uα−1 |∇u||∇ψ| 2 ≤ (D + α − 1) ψ 2(α+σ−1) λ(u)uα−2 |∇u| (α + σ − 1)2 2 ψ 2(α+σ−1)−2 λ(u)uα |∇ψ| . + D+α−1 Inserting this into (5.104) and using λ ≥ 0 gives ψ 2(α+σ−1) λ(u)b(x)uα+σ−1 ≤ ψ 2(α+σ−1) λ(u)a+ (x)uα (5.105)

2 (α + σ − 1)2 ψ 2(α+σ−1)−2(1−1/ζ) λ(u)uα ψ −1/ζ |∇ψ| . + D+α−1 For ease of notation, we denote by I and II the integrals on the right-hand side. We use (5.98), multiply and divide by b(x), and use H¨ older inequality with conjugate exponents p and q to be chosen later, to estimate I≤E

ψ

2(α+σ−1)

1/p λ(u)b(x)u

αp

ψ

2(α+σ−1)

1/q λ(u)b(x)

.

(5.106)

Similarly, using (5.103) iii), we see that II ≤

C02 r2

ψ 2(α+σ−1) λ(u)b(x)uαp ×

ψ

1/p

2(α+σ−1)+2q(1/ζ−1)

λ(u)b(x)

1−q

1/q ,

(5.107)

152


where C0 = C0 (ζ) is the constant appearing in (5.103) iii). In the above inequalities, it is assumed that p and q are chosen in such a way that 2(α + σ − 1) + 2q(1/ζ − 1) > 0 is positive. To proceed we need to estimate the powers of u in the above integrals. We choose p in such a way that αp is equal to the exponent of u in the integral on the left-hand side of (5.105), namely α + σ − 1. This implies p=

α+σ−1 , α

q=

α+σ−1 . σ−1

Note that, since σ > 1, we have p, q > 1, and that, by our choice of ζ, ! " 2 1 1 = (α + σ − 1) 2 + > 0. 2(α + σ − 1) + 2q ζ −1 σ−1 ζ −1 Since ψ ≤ 1, and ψ = 0 off B2r , we have I≤E

ψ 2(α+σ−1) λ(u)b(x)uα+σ−1

α α+σ−1

×

λ(u)b(x)

σ−1 α+σ−1 ,

(5.108)

B2r

and II ≤

C02 γ δ−1 r2 inf B2r b

ψ 2(α+σ−1) λ(u)b(x)u(α+σ−1)

α α+σ−1

σ−1 α+σ−1

×

λ(u)b(x)

.

(5.109)

B2r

Substituting (5.108) and (5.109) into (5.105), and rearranging yield

ψ 2(α+σ−1) λ(u)b(x)uα+σ−1 ≤

1 C (α + σ − 1)2 E+ 2 r inf B2r b D + α − 1

α+σ−1 σ−1 λ(u)b(x) B2r

with C = C02 C1 . We estimate from below the integral on the left-hand side using ψ = 1 on Br , and λ(u) = 0 if u ≤ γ, and arrive at

λ(u)b(x) ≤ Br

E γ σ−1

C γ 1−σ (α + σ − 1)2 + 2 r inf B2r b D + α − 1

α+σ−1 σ−1 λ(u)b(x), B2r

which is the required conclusion.

Lemma 5.14. Let G, Q : [R, +∞) → [0, +∞) be nondecreasing functions such that, for some constants 0 < Λ < 1 and B, θ > 0, G(r) ≤ ΛBr

θ

/Q(2r)

G(2r)

∀r ≥ R.

(5.110)


153

Then there exists a constant S = S(θ) > 0 such that, for every r ≥ 2R, Q(r) Q(r) log G(r) ≥ θ log G(R) + SB log(1/Λ). rθ r

(5.111)

Proof. Let r0 = R and rk = 2k r0 . Then, for every r ≥ 2r0 , there exists k such that rk ≤ r ≤ rk+1 . Applying inequality (5.110) k-times, we obtain G(r0 ) ≤ ΛB

k−1 j=0

rjθ /Q(2rj )

G(rk ).

(5.112)

Using the definition of rk and the fact that Q is nondecreasing we estimate k−1 j=0

rjθ /Q(2rj ) ≥

k−1 θ rk+1 r0θ rθ 1 − 2−kθ −θ 2jθ = 2 ≥S θ Q(2rk−1 ) j=0 Q(2rk−1 ) 2 − 1 Q(r)

with S = 2−θ /(2θ − 1). Substituting into (5.112), and recalling that 0 < Λ < 1 and that G is nondecreasing, we conclude that G(r0 ) ≤ ΛBSr

θ

/Q(r)

G(r),

whence (5.111) follows by taking logarithms. We are now ready for the

Proof of Theorem 5.12. First of all, we note that since !a+ !∞ < +∞, we may assume that !b!∞ < +∞. We let R, λ and γ be as in Lemma 5.13, and set, for r ≥ R, G(r) = λ(u)b(x). Br

Applying Lemma 5.13 with D = 0, the lower bound b(x) ≥ B/r(x)β , and the inequality (α + σ − 1)2 ≤ 23 α α−1 valid for α ≥ max{σ − 1, 2}, we deduce that there exists a constant C3 depending only on σ such that, for every r ≥ R, γ < u∗ and α > max{σ − 1, 2}, G(r) ≤

E γ σ−1

C3 α + σ−1 2−β γ r

α+σ−1 σ−1 G(2r).

(5.113)

Now, assume by contradiction that u∗ = +∞, so that Ωγ = ∅ for every γ > 0. Fix 0 < ρ < 1 and let γ0 > 0 be such that, for every γ ≥ γ0 , E ≤ 1 − 2ρ. γ σ−1

(5.114)

154


Next, we choose α of the form α = α(r) = ρ

γ σ−1 2−β r , C3

and observe that, since β < 2, there exists R1 ≥ R such that α(r) satisfies the condition stated before (5.113) for every r ≥ R1 and C3 α E + σ−1 2−β α ≤ 1 − ρ. γ σ−1 γ r Setting Λ = 1 − ρ, (5.113) gives G(r) ≤ ΛBγ

σ−1 2−β

r

G(2r)

∀r ≥ R1 ,

(5.115)

where B > 0 is a constant depending only on ρ and σ. We apply Lemma 5.14 to deduce that there exists a constant S such that, for every r ≥ R1 , 1 1 log b(x)λ(u) ≥ 2−β log b(x)λ(u) + SBγ σ−1 log (1/Λ). (5.116) r2−β r Br BR 1 To reach the required contradiction, we choose λ satisfying sup λ = 1/ supM b, so that b(x)λ(u) ≤ 1, and let r → +∞ in (5.116). Since 2 − β > 0, we conclude that 1 1 lim inf 2−β log vol Br ≥ lim inf 2−β log b(x)λ(u) ≥ SBγ σ−1 log (1/Λ). r→+∞ r r→+∞ r Br Since σ > 1, by taking γ sufficiently large, this contradicts (5.96). To complete the proof of the theorem we observe that the differential equation (5.95) yields Δu ≥ b(x)u(uσ−1 − E), whence the required conclusion follows applying Corollary 5.4 with f (u) = u · (uσ−1 − E). Remark. One might wonder whether a corresponding result on u∗ holds. This is the case under further restrictions on the nonlinearity of the equation, see for instance [PRS05a]. However, this does not apply to Yamabe’s equation, so that the assumption that u∗ > 0 in the statement of Theorem 5.11 is necessary. See also the final remark of Section 4.1. We conclude this section with a last application of Theorem 5.12. Towards this goal, let (M, g) be a complete Riemannian manifold with scalar curvature S(x), and, as usual, let r(x) be the Riemannian distance function from a fixed reference point o. Assume that ϕ : (M, g) → (N, h) is a conformal immersion, so that, assuming it will simplify matters, m = dim(M ) ≥ 3, 4

ϕ∗ h = u m−2 g

(5.117)


155

for some positive function u ∈ C ∞ (M ). Let also K(x) be the scalar curvature of the conformal metric ϕ∗ h. Note that if u ≤ 1 the length of curves in M with respect to the conformal metric ϕ∗ h is less than or equal to their lengths with respect to the original metric g, and therefore the Riemannian distance induced by ϕ∗ h is smaller than or equal to the original distance function. In this situation we say that ϕ is weakly distance decreasing . Theorem 5.15. Assume that inf S(x) > −∞ M

and that K(x) < 0, K(x) ≤ S(x) on M

and

K(x) ≤ −

A r(x)β

for r(x) >> 1, and for some constants A > 0 and β < 2. If lim inf r→∞

log (vol Br ) < +∞, r2−β

then ϕ is weakly distance decreasing. Proof. Recall that the stretching factor u of ϕ∗ h with respect to g satisfies the Yamabe equation m+2

cm Δu − S(x)u + K(x)u m−2 = 0

on M.

K(x) Setting S(x) = − S(x) cm and b(x) == − cm , the conditions imposed on S(x) and K(x) imply that all the assumptions of Theorem 5.12 are satisfied with E = 1. Hence u ≤ 1 and ϕ is weakly distance decreasing.

Remark. Theorem 5.15 has its roots in the work started, in the compact case, by A. Lichnerowicz, [Lic58], Obata, [Oba62b] (see Section 2.1.1), and K. Yano and T. Nagano, [YN59], and extended to the noncompact case by Yau, [Yau73] where it was assumed that K(x) = S(x) ≤ −ε < 0 and that the sectional curvature of (M, g) was bounded from below. We also remark that some negativity is necessary for the conclusion of Theorem 5.15 to hold, as the following elementary example shows: let (M, g) be a manifold of dimension m ≥ 3 with nonnegative scalar curvature S(x), and for 2 every fixed a > 1 let (N, h) = (M, a2 g). Then K(x) = a− m−2 S(x) ≤ S(x) since a > 1 and the identity map is a conformal diffeomorphism of (M, g) onto (N, h) which is not weakly distance decreasing.

Chapter 6

Existence In this chapter we provide existence results for nonnegative solutions of the Yamabe-type equation Δu + a(x)u − b(x)uσ = 0, σ > 1

(6.1)

on the complete, noncompact, manifold M . Existence is obtained by various versions of the monotone iteration scheme (see the appendix at the end of the chapter) and we consequently concentrate on the construction of (global and local) superand subsolutions for the problem. The first are obtained with some assumptions on the sign of b(x) and on that of the first eigenvalue of L = Δ + a(x) on appropriate domains. In the situation at hand, subsolutions are generally harder to find and in what follows we give a number of sufficient conditions that allow their construction. We mention in this respect Theorem 6.15, in which existence is guaranteed under a very weak growth condition on b(x), and also Theorem 6.16, where a further weakening the condition on the sign of b(x) is balanced by the necessity of imposing a constant negative lower bound on the Ricci curvature. We note that the assumptions of our existence theorems match those of the nonexistence results in the previous chapters. Existence can be in particular guaranteed for the Yamabe problem, but we limit ourselves to explicitly state only one of the possible consequences at the end of the chapter, in Theorem 6.18, leaving the derivation of further results to the interested reader. The chapter begins with some introductory material, in particular a useful comparison result and some facts from basic spectral theory; then we prove a result of Li, Tam and Yang (see [LTY98]) concerning the nonnegativity of the first eigenvalue of Schrödinger operators. These steps are used to outline a general procedure that reduces the existence problem to that of the existence of a positive subsolution; next, we provide different sufficient conditions for this latter. We then deal with a case where the existence of a supersolution is no longer a consequence of a spectral assumption and depends on a rather delicate construction. Finally, we


157

158

Chapter 6. Existence

consider specifically the case of the geometric Yamabe equation. The chapter ends with an appendix in which we describe the method of sub- and supersolutions, and where we also prove that the maximum of two subsolutions (or the minimum of two supersolutions) is again a subsolution (resp. supersolution).

6.1

A general procedure

The aim of this section is to describe the general procedure mentioned above. The various steps towards this goal, achieved in Theorem 6.16, are explained in the next subsections. The next proposition is a useful comparison result that shall be used throughout the chapter.

6.1.1

Another comparison result

Proposition 6.1. Let D ⊂ M be a bounded open set with boundary ∂D. Assume ¯ and that b(x) is nonnegative. Let u, v ∈ C 0 (D) ¯ ∩ C 2 (D) be a(x), b(x) ∈ C 0 (D), solutions on D of Δu + a(x)u − b(x)uσ ≥ 0 (6.2) and Δv + a(x)v − b(x)v σ ≤ 0

(6.3)

with u ≥ 0, v > 0 and σ ≥ 1. If u ≤ v on ∂D, then u ≤ v on D. Proof. The proof is modeled on that of the generalized maximum principle (see for example [PW67]). Set w = u/v. A computation shows that Δw ≥ b(x) uσ−1 − v σ−1 w − 2 ∇w, ∇ log v . Assume by contradiction that u > v somewhere in D, so that for a sufficiently small ε > 0, Ωε = {x ∈ D : w(x) > 1 + ε} = ∅. Then Ωε ⊂ D, and since u > v on Ωε , σ ≥ 1 and b(x) is nonnegative, Δw + 2 ∇w, ∇ log v ≥ 0

on Ωε .

Since w = 1 + ε on ∂Ωε , by the maximum principle w ≤ 1 + ε on Ωε , and this yields the required contradiction.

6.1.2

More basic spectral theory and a result of Li, Tam and Yang

In this section we collect some other simple results (see also Chapter 3) from the spectral theory of Schr¨ odinger operators . For more details, we refer to [PRS08] and references therein.

6.1. A general procedure

159

Let a(x) ∈ C 0 (M ) and L = Δ+a(x). As we already know, if Ω is a nonempty open set, the first Dirichlet eigenvalue λL 1 (Ω) is variationally defined by means of the formula & ' 1,2 2 2 2 λL (Ω) = inf |∇ϕ| − a(x)ϕ : ϕ ∈ W (Ω), ϕ = 1 , 1 0 Ω

Ω

and, if Ω is bounded and both Ω and a are sufficiently regular, the infimum is attained by a unique normalized eigenfunction v defined on Ω and satisfying Δv + a(x)v + λL on Ω; 1 (Ω)v = 0, v > 0 on Ω, v ≡ 0 on ∂Ω. We extend the above definition to an arbitrary bounded subset Υ of M by setting L λL 1 (Υ) = sup λ1 (Ω),

(6.4)

where the supremum is taken over all open bounded sets with smooth boundary Ω such that Υ ⊂ Ω (see also Chapter 3). Note that, by definition, if Υ = ∅, then λL 1 (S) = +∞. Finally, if Υ is an unbounded subset of M , we define L λL 1 (Υ) = inf λ1 (D ∩ Υ),

where the infimum is taken over all bounded open sets D with smooth boundary. Note that if {Dn } is an increasing sequence of open sets with smooth boundaries which exhausts M , then, by domain monotonicity, L λL 1 (Υ) = lim λ1 (Dn ∩ Υ). n→+∞

Since the first Dirichlet eigenvalue of the Laplacian of a ball Br grows like r−2 as r → 0 (see for instance Chavel [Cha84]), λL 1 (Br ) > 0 provided r is sufficiently small, and one may think that the condition λL 1 (Υ) > 0 expresses the fact that Υ is small in a spectral sense. This notion of smallness is appropriate for our purposes. Indeed, P. Li, L. F. Tam and D. Yang, [LTY98], have established the following relationship between the first eigenvalue of B0 = {x ∈ M : b(x) = 0}

(6.5)

and the existence of a nontrivial supersolution of (6.1). We provide here a mildly improved proof of the original result. Theorem 6.2. Let a(x), b(x) ∈ C 0,α (M ), 0 < α ≤ 1, and suppose that b(x) ≥ 0 on M and that the set B0 defined in (6.5) is bounded. Let Ω be a bounded open domain containing B0 . If Δu + a(x)u − b(x)uσ = 0,

σ>1

(6.6)

L has a positive supersolution on Ω, then λL 1 (B0 ) ≥ 0. Conversely, if λ1 (B0 ) > 0, then (6.6) has a positive supersolution on Ω.

160


Proof. Suppose (6.6) has a positive supersolution u on Ω and, by contradiction, assume that λL 1 (B0 ) = −a for some constant a > 0. By the definition of λL 1 (B0 ) we can find a sequence of open sets Ωi , i ∈ N, with smooth boundaries such that, ∀ i ∈ N, Ωi ⊂ Ω,

+∞ 7

Ωi+1 ⊂ Ωi ,

Ωi = B 0

i=0

and λi = λL 1 (Ωi ) → −a as i → +∞ monotonically. Corresponding to the eigenvalues λi there is a sequence of positive eigenfunctions vi such that Δvi + a(x)vi + λi vi = 0 on Ωi , (6.7) vi > 0 on Ωi , vi = 0 on ∂Ωi . Note that, since vi > 0 on Ωi , ∇vi , ν ≤ 0

on ∂Ωi ,

(6.8)

where ν is the outward unit normal to ∂Ωi . By assumption, u > 0 satisfies Δu + a(x)u ≤ b(x)uσ

on Ω;

(6.9)

using the second Green’s formula and (6.7), (6.8), (6.9) we have 0≥ u ∇vi , ν = u ∇vi , ν − vi ∇u, ν = uΔvi − vi Δu ∂Ωi ∂Ωi Ωi −a(x)uvi − λi uvi + a(x)uvi − b(x)vi uσ = − vi u λi + b(x)uσ−1 , ≥ Ωi

Ωi

that is,

Ωi

vi u(λi + b(x)uσ−1 ) ≥ 0.

(6.10)

Now, but b(x) ≡ 0 on

λi + b(x)uσ−1 ≤ −a + b(x)uσ−1 ,

8 i

Ωi = B0 , hence, for i sufficiently large, λi + b(x)uσ−1 < 0

on Ω1

contradicting (6.10). Conversely, suppose λL 1 (B0 ) > 0. Let D and D be bounded open domains such that B0 ⊂⊂ D ⊂⊂ D ⊂⊂ Ω


161

and λL 1 (D) > 0. Let u1 be a positive solution of Δu1 + a(x)u1 + λL 1 (D)u1 = 0, u1 = 0,

on D; on ∂D.

(6.11)

Since b(x) > 0 on M \B0 , and Ω\D ⊂⊂ M \B0 , β = inf b > 0. Ω\D

We claim that a sufficiently large positive constant u2 is a supersolution of (6.6) on Ω\D . Towards this end let α = sup a(x), Ω\D

and note that α < +∞ since Ω is bounded. Then we have Δu2 + a(x)u2 − b(x)uσ2 = u2 (a(x) − b(x)uσ−1 ) ≤ u2 (α − βuσ−1 )≤0 2 2 on Ω\D provided

1 σ−1 α u2 ≥ . β

Now let ψ ∈ C ∞ (M ) be a cut-off function such that 0 ≤ ψ ≤ 1, ψ ≡ 1 on D , supp ψ ⊂ D. Fix any constant γ > 0 and define u = γ(ψu1 + (1 − ψ)u2 ).

(6.12)

Since b(x) ≥ 0 and λL 1 (D) > 0, on D , we have

Δu + a(x)u − b(x)uσ = Δγu1 + a(x)γu1 − b(x)(γu1 )σ σ−1 = − λL γu1 < 0 1 (D) + b(x)(γu1 ) independently of the value of γ > 0. Moreover, in Ω\D, Δu + a(x)u − b(x)uσ = Δγu2 + a(x)γu2 − b(x)(γu2 )σ = γ a(x)u2 − b(x)γ σ−1 uσ2 . Now, for γ ≥ 1,

b(x)γ σ−1 ≥ b(x),

thus, on Ω\D, Δu + a(x)u − b(x)uσ ≤ γ(Δu2 + a(x)u2 − b(x)uσ2 ) ≤ 0 because of our choice of u2 . It remains to analyze the situation on D\D . On this set (Δ + a(x))(ψu1 + (1 − ψ)u2 ) ≤ C (6.13)

162


for any constant C > 0 sufficiently large. Now σ

inf b(x)(ψu1 + (1 − ψ)u2 ) > 0

D\D

and we can therefore choose C > 0 sufficiently large that (6.13) holds together with σ b(x)(ψu1 + (1 − ψ)u2 ) > C −1 (6.14) on D\D . On this set we then have Δu + a(x)u − b(x)uσ = γ(Δ + a(x))(ψu1 + (1 − ψ)u2 ) σ − b(x)γ σ (ψu1 + (1 − ψ)u2 ) ≤ γC − γ σ C −1 = γ C − γ σ−1 C −1 ≤ 0 2

provided γ ≥ C σ−1 . Thus u is a supersolution of (6.6) on Ω.

6.1.3

Two useful lemmas

In the following lemmas we proceed to construct a solution of (6.6). Lemma 6.3. Let a(x), b(x) ∈ C 0,α (M ), 0 < α < 1, and suppose that b(x) ≥ 0 on M and that B0 is bounded. Let Ω ⊃ B0 be a bounded open domain. Having set L = Δ + a(x), assume that λL (6.15) 1 (B0 ) > 0. Then, for every n ∈ (0, +∞), there exists a solution of the problem on Ω, σ > 1, Δu + a(x)u − b(x)uσ = 0, u > 0 u=n on ∂Ω.

(6.16)

Proof. By the definition of λL 1 (B0 ) and (6.15) there exists an open domain D with ∞ smooth boundary such that B0 ⊂ D ⊂⊂ Ω and λL 1 (D) > 0. Let ψ ∈ C (M ) be a cut-off function such that 0 ≤ ψ ≤ 1 and ψ≡1

ψ ≡ 0 on M \Ω. Δ Fix N ≥ max supΩ |a(x)| + 1, λΔ 1 (M \Ω) + 1 , where λ1 (M \Ω) ≥ 0 is the bottom of the spectrum of the Laplacian on M \ Ω. Define on D,

a ¯(x) = ψ(x)a(x) + N (1 − ψ(x)) ¯ =Δ+a and consider the operator L ¯(x). Since a ¯(x) ≡ a(x) on D, ¯

L λL 1 (B0 ) = λ1 (B0 ) > 0.

(6.17)


163 ¯

Furthermore, since N ≥ λ1 (M \ Ω) + 1, we have λL 1 (M \ Ω) ≤ −1 and it follows that there exists R > 0 sufficiently large such that Ω ⊂ BR (o)

¯

λL 1 (BR (o)) < 0.

and

¯ on BR (o) relative to the eigenvalue Let ϕ be the normalized eigenfunction of L ¯ λL (B (o)), so that R 1 ¯ + λL¯ (BR (o))ϕ = 0 on BR (o), Lϕ 1

ϕ=0

on ∂BR (o)

and ||ϕ||L2 (BR (o)) = 1. We fix γ > 0 sufficiently small that

¯

|∇ϕ|2 − a ¯(x)ϕ2 + γb(x)ϕ2 = λL 1 (BR (o)) + γ BR

b(x)ϕ2 < 0, BR

= Δ+a showing that the operator L ¯(x) − γb(x) satisfies λL 1 (BR (o)) < 0. Let ψ (B (o)). Then ψ satisfies be a positive eigenfunction corresponding to λL R 1 ¯ − γb(x)ψ = −λL (BR (o))ψ ≥ 0 on BR (o), Lψ 1 ψ≡0 on ∂BR (o). If we choose

$ 0 0 on BR (o), Δv− + a v− ≡ 0 on ∂BR (o). ¯

On the other hand, since λL 1 (B0 ) > 0, by Theorem 6.2 there exists v+ > 0 on BR (o) satisfying Δv+ + a ¯(x)v+ − b(x)v+ σ ≤ 0, v+ > 0 on BR (o), v+ ≥ 0 on ∂BR (o). Thus, by the monotone iteration scheme (see section 6.5), we find a solution w of the problem Δw + a ¯(x)w + −b(x)wσ = 0, w > 0 on BR (o), w≡0 on ∂BR (o).

164


¯(x) ≥ a(x) on Ω, Note that inf ∂Ω w > 0 since Ω ⊂ BR (o). Thus, recalling that a it is easily verified that if α > is sufficiently large, then the function w+ = αw satisfies Δw+ + a(x)w+ − b(x)w+ σ ≤ 0, w+ > 0 on Ω, w+ ≥ n on ∂Ω. Finally, since w− ≡ 0 is a subsolution of the problem, by the monotone iteration scheme (see section 6.5) we deduce the existence of a solution u ≥ 0 of (6.16). To conclude, note that since u satisfies Δu + a(x) − b(x)uσ−1 u = 0 it follows from the strong maximum principle on page 35 of [GT01] that u > 0. Now we produce a solution blowing up at the boundary of Ω. Lemma 6.4. In the assumption of Lemma 6.3 there exists a solution of the problem Δu + a(x)u − b(x)uσ = 0, u > 0 on Ω, σ > 1, (6.18) u = +∞ on ∂Ω. Proof. By standard regularity theory (see [GT01]) it is enough to show that the sequence {un }, with un solution of (6.16), is bounded on any compact subset K of Ω. If K ⊂ Ω\Bo , then we can find a finite covering of balls Bi for K, i = 1, . . . , t such that b(x) > 0 on each Bi . Applying Lemma 4.2 we deduce the existence of a constant C1 > 0 such that un (x) ≤ C1

∀ x ∈ K, ∀ n ∈ N.

(6.19)

It remains to find an upper bound of un on a neighbourhood of Bo . Towards this aim, for η > 0 we let Nη = {x ∈ M : d(x, Bo ) < η} where η > 0 is small enough that Nη ⊂ Ω. Furthermore, by the definition of L λL 1 (Bo ) and the fact that λ1 (Bo ) > 0, we can also suppose to have chosen η so small that λL 1 (Nη ) > 0. Now ∂Nη/2 is closed and bounded, hence compact by the completeness of M , therefore (6.19) holds on ∂Nη/2 for some constant C2 > 0. Let ϕ be a positive eigenfunction corresponding to λL 1 (Nη ). Then, there exists a positive constant μ such that μϕ > C2 on ∂Nη/2 . On Nη/2 we have Δ(μϕ) + a(x)(μϕ) = −λL 1 (Nη )(μϕ) < 0 while, by (6.16), Δun + a(x)un = b(x)uσn ≥ 0. It follows from the generalized maximum principle, [PW67] Theorem 2.10, that un the function μϕ cannot attain an interior positive maximum unless it is constant, and since, un ≤ C2 < μϕ on ∂Nη/2 ,


165

we conclude that un ≤ μϕ ≤ C3 on Nη/2 with C3 independent of n. This proves the lemma.

6.1.4

Existence of a maximal solution

With the aid of the previous lemma we are ready to prove the following general result; we denote with C 0,α (M ) the space of locally Hölder continuous functions on M with exponent α. 0,α Theorem 6.5. Let a(x), b(x) ∈ Cloc (M ) for some 0 < α ≤ 1. Assume that b(x) ≥ 0 and it is strictly positive outside a compact set, and that

B0 = {x ∈ M : b(x) = 0} 1,2 0 satisfies λL 1 (B0 ) > 0 with L = Δ + a(x). If u− ∈ C (M ) ∩ Hloc (M ), u− ≥ 0, u− ≡ 0, is a global subsolution of

Δu + a(x)u − b(x)uσ = 0, σ > 1,

(6.20)

on M , then (6.20) has a maximal positive C 2 -solution . Proof. We fix an exhausting sequence {Dk } ⊂ M by open domains with smooth boundaries such that B0 ⊂ Dk ⊂ Dk ⊂ Dk+1 ∀ k, and for every k we denote by u∞ k the solution of the problem Δu + a(x)u − b(x)uσ = 0 on Dk , u = +∞ on ∂Dk obtained by applying Lemma 6.4. It follows from Proposition 6.1 that ∞ u− ≤ u∞ k+1 ≤ uk

¯ k. on D

(6.21)

Thus {u∞ k } converges monotonically to a function u which solves (6.20), and because of (6.21) u ≥ u− ≥ 0, u− ≡ 0. Thus u ≥ 0 and u ≡ 0. By the maximum principle (see [GT01], page 35) it follows that u > 0 on M . Let now u1 > 0 be a second solution of (6.20) on M . Again by Proposition 6.1, u1 ≤ u∞ k on Dk ∀ k, and therefore u1 ≤ u, as required to show that u is a maximal positive solution. The next result is an immediate consequence of the above reasoning and it will be used in the proof of Theorem 6.8. Theorem 6.6. Let a(x), b(x) ∈ C ∞ (M ) and assume b(x) > 0. If u− ≥ 0, u− ≡ 0 is a subsolution of Δu + a(x)u − b(x)uσ = 0,

σ>1

on M \ BR0 ,

then (6.22) has a maximal positive smooth solution u ≥ u− on M \ BR0 .

(6.22)

166


6.2 Subsolutions and existence In this section we give sufficient conditions to guarantee the existence of a positive subsolution of equation (6.20). An application of Theorem 6.5 then yields the existence of a maximal solution.

6.2.1

Existence with λL1 (M ) < 0

The first assumption we consider to provide an entire (i.e., global), nonnegative and nontrivial subsolution of (6.20) is the spectral condition (6.23) below. 0,α Theorem 6.7. Let a(x), b(x) ∈ Cloc (M ) for some 0 < α ≤ 1. Assume that b(x) ≥ 0, b(x) > 0 outside a compact set, and that λL 1 (B0 ) > 0 where B0 = {x ∈ M : b(x) = 0} and L = Δ + a(x). Furthermore assume

λL 1 (M ) < 0.

(6.23)

Δu + a(x)u − b(x)uσ = 0, σ > 1

(6.24)

Then the equation possesses a minimal and a maximal (possibly coinciding) positive solutions. Proof. According to Theorem 6.5 it is enough to prove that there exists a minimal solution. Since λL 1 (M ) < 0, we can find a relatively compact domain with smooth boundary Ω0 ⊃ B0 such that λL 1 (Ω0 ) < 0. Arguing exactly as in the proof of Lemma 6.3, we see that if γ > 0 is sufficiently small, then

λL 1 (Ω0 ) < 0, = Δ + a(x) − γb(x) so that, if ψ is a positive eigenfunction corresponding where L L to λ1 (Ω0 ), then Δψ + a(x)ψ ≥ γb(x)ψ on Ω0 , ψ≡0 on ∂Ω0 −1 1 and, choosing μ ≤ γ σ−1 supΩ0 ψ , the function v− defined by v− = μψ satisfies Δv− + a(x)v− ≥ b(x)v− σ on Ω0 , (6.25) v− ≡ 0 on ∂Ω0 . Since λ1 (B0 ) > 0, Theorem 6.2 guarantees the existence of a supersolution v+ > 0 to (6.25), and possibly multiplying v+ by a large positive constant we can also suppose that v+ ≥ v− on Ω0 . By the monotone iteration scheme (see section 6.5), we deduce the existence of a positive solution u0 ∈ C 2 (Ω0 ) of Δu0 + a(x)u0 − b(x)uσ0 on Ω0 , (6.26) u0 = 0 on ∂Ω0 .

6.2. Subsolutions and existence

167

Now let Ω1 ⊃ Ω0 ⊃ Ω0 . By domain monotonicity L λL 1 (Ω1 ) ≤ λ1 (Ω0 ) < 0

and the above procedure yields the existence of a positive solution u1 of (6.26) on Ω1 . Since ∂Ω0 ⊂ Ω1 , u0 = 0 < u1 on ∂Ω0 , it follows by Proposition 6.1 that u0 ≤ u1 on Ω0 . Choosing an increasing exhaustion of M by relatively compact domains +∞ with smooth boundaries {Ωi }i=0 , the above procedure produces a sequence {ui } of solutions of (6.26) on Ωi satisfying ui ≤ ui+1

on Ωi .

Using the procedure of Lemma 6.4 we see that {ui } is uniformly bounded on Ωk for i ≥ k + 1; therefore ui converges to u > 0, solution of (6.24). The argument used to prove the monotonicity of {ui } shows that if u is any other solution of (6.24), then u ≥ ui on Ωi ∀ i ≥ 0. Thus u ≥ u and u is the required positive minimal solution of (6.24). In the same vein we prove the following theorem (see [BRS98]). Theorem 6.8. Let (M, , ) be a complete manifold of dimension m ≥ 3 satisfying δ/2 Ric ∇r, ∇r ≥ −(m − 1)H 2 1 + r2

on M,

(6.27)

for some constants H > 0 and δ ≥ −2. Let a(x), b(x) ∈ C ∞ (M ), b(x) > 0 on M and suppose that, for some constants γ ≤ 0, μ < 1 − 2δ , a(x) >0 r(x)→+∞ r(x)−μ lim inf

(6.28)

and b(x)

lim sup

4γ

r(x)→+∞

Finally, given 1 < σ ≤

m+2 m−2 ,

r(x)−μ− m−2

let Lσ = Δ +

< +∞.

(m−2)(σ−1) a(x) 4

(6.29) and assume that

σ λL 1 (M ) < 0.

(6.30)

Δu + a(x)u − b(x)uσ = 0

(6.31)

Then equation has a positive solution u ∈ C ∞ (M ) on M satisfying lim inf r(x)→+∞

u(x) 4γ

r(x) (m−2)(σ−1)

> 0.

(6.32)

168


Remark. Two comments are in order. First of all, the theorem establishes the existence of a solution to (6.31) satisfying an explicit lower bound at infinity. In this respect it is worth noticing that, if δ > −2 and μ < −δ, then any positive solution u of (6.31) satisfies the estimate (6.32) by the a priori estimates of Theorem 4.1. However in case δ ≤ μ < 1 − 2δ , which is allowed since δ > −2, the lower bound (6.32) does not follow from Theorem 4.1, and in this case one has to construct the appropriate subsolution as in the proof of Theorem 6.8. The second observation concerns the validity of the spectral assumption (6.30). Note that in general, if 0 ≤ σ1 ≤ σ2 , then, by the variational characL L terization of the bottom of the spectrum, λ1 σ1 (M ) ≥ σσ12 λ1 σ2 (M ) (see the proof ∞ of Theorem 2 in [FCS80]). Indeed, for every φ ∈ C0 (M ), σ1 σ1 |∇ψ|2 − σ1 a(x)ψ 2 = |∇ψ|2 − σ2 a(x)ψ 2 ≥ |∇ψ|2 − σ2 a(x)ψ 2 σ σ 2 2 M M M and the claim follows taking the infimum over all such ψ. Thus assumption (6.30) implies the condition λL 1 (M ) ≥ 0 considered in Theorem 6.7. Proof. Again according to Theorem 6.5 it suffices to find a nonnegative subsolution of (6.31) satisfying (6.32). We divide the proof into several steps. For the sake of simplicity we assume that the M has a pole o, i.e., a point such that exp : To M → M is a diffeomorphism (see for instance [dC92], [GW79]). The general case is dealt with noting that, applying Lemma 1.12, the radial functions which are shown to be pointwise super-, respectively subsolutions within the cut locus, are in fact global super-, respectively subsolutions in the weak sense, and this is all that is needed to carry out the argument. Step 1: reduction to the case where σ =

m+2 m−2 .

1,2 (M ) which It suffices to show that there is a positive function v− ∈ C 0 (M ) ∩ Hloc satisfies weakly on M ,

Lσ v− −

m+2 (m − 2)(σ − 1) b(x)v−m−2 ≥ 0 4

(6.33)

and v− (x) ≥ Cr(x)γ

for r(x) 1

(6.34)

a straightforward compuand some constant C > 0. Indeed, since 1 < σ ≤ tation shows that if v− satisfies (6.33), then the function u− defined by m+2 m−2 ,

4

u− = v− (m−2)(σ−1) satisfies Δu− + a(x)u− − b(x)uσ− 4 4 4 2 ≥ − 1 v− (m−2)(σ−1) −2 |∇v− | ≥ 0 (m − 2)(σ − 1) (m − 2)(σ − 1)


169

weakly on M , and therefore u− is a subsolution of (6.31) satisfying (6.32). Step 2: construction of a subsolution outside a compact set. Consider w(x) = r(x)γ = β(r(x)) for r(x) ≥ 1. Clearly w(x) satisfies the lower bound (6.34). We claim that there exist δ0 > 0 and R0 sufficiently large such that, ∀ 0 < δ ≤ δ0 , the function v6 = δw satisfies (6.33) weakly on M \ BR0 . Indeed, from (6.27), the Laplacian comparison > H of Proposition 1.15, there theorem and Proposition 1.15, having fixed H sufficiently large such that exists R0 = R0 (H) δ/2 Δr(x) ≤ (m − 1)Hr(x)

on M \ BR0 .

Using (6.28) and (6.29) we can also suppose to have chosen R0 so large that a(x) ≥ A1 r(x)−μ

4γ

b(x) ≤ A2 r(x)−μ− m−2

and

(6.35)

on M \BR0 for some constants A1 , A2 > 0. Since β ≤ 0 it follows that, on M \BR0 , (m − 2)(σ − 1) a(x)w 4 (m − 2)(σ − 1) a(x)β = β + (Δr)β + 4

Lσ w = Δw +

γ−1+δ/2 + ≥ γ(γ − 1)rγ−2 + (m − 1)Hγr

(m − 2)(σ − 1) A1 rγ−μ. 4

Thus, in case μ < 1 − 2δ we deduce, up to choosing R0 sufficiently large, that there exists a constant C > 0 such that Lσ w ≥ Crγ−μ

on M \ BR0 .

To conclude note that, having set m+2

bw = w− m−2 Lσ w, we have

4γ

bw ≥ Cr−μ− m−2 Since

on M \ BR0 . 4

bδw = δ − m−2 bw , it follows from (6.35) that, if δ is sufficiently small, bδw ≥ b(x)

(m − 2)(σ − 1) 4

on M \ BR0

(6.36)

170


as required to show that there exists δo sufficiently small that for every δ < δo , v6 solves (6.33) weakly on M \ BR0 , and satisfies (6.34). Step 3: construction of a weak subsolution v− on M . σ Since by assumption λL 1 (M ) < 0, we can find a relatively compact set with smooth σ boundary Ω ⊃ BR0 such that λL 1 (Ω) < 0. Let ϕ be a corresponding eigenfunction, so that Δϕ + (m+2)(σ−1) a(x)ϕ = −λϕ, ϕ > 0 in Ω, 4 ϕ = 0 on ∂Ω.

Arguing as above, it follows that if η > 0 is sufficiently small, then ϕη = ηϕ is a subsolution of (6.33) on Ω. Finally let δ < δ0 be small enough that the function v6 < ϕη in BR0 + \ BR0 . Note that since v6 > 0 on ∂Ω there exists an open set D with D ⊂ Ω such that v6 > ϕη in Ω \ D. We claim that the function v− defined by ⎧ ⎪ in BR0 + , ⎨ ϕη v− = max {6 v , ϕη } in Ω \ BR0 , ⎪ ⎩ v6 in M \ D is the required subsolution. Note first of all that v− is well defined by construction, and it is clearly a subsolution in BR0 + and in M \ D. On the other hand, v− is a subsolution in Ω \ BR0 as maximum of two subsolutions, by Theorem 6.20.

6.2.2 λL1 (M ) < 0: some sufficient conditions The previous Theorems 6.7 and 6.8 point out the relevance of condition (6.23) in guaranteeing the existence of positive solutions to (6.24). It is therefore interesting to find sufficient conditions in order that (6.23) holds. We begin with the following elementary result. Proposition 6.9. Let (M, , ) be a complete manifold, a(x) ∈ C 0 (M ), and L = Δ + a(x). Set v(r) = vol(∂Br ) and denote by 1 a ¯(r) = a(x) (6.37) vol(∂Br ) ∂Br the spherical mean of a(x). Let A(r) ≤ a ¯(r) and suppose that α is a solution of the problem (v(r)α ) + v(r)A(r)α ≥ 0, (6.38) α(0) = α0 > 0, α (0) = 0, and has a first zero at r = T > 0. Then λL 1 (M ) < 0.

(6.39)


171

Proof. We consider the geodesic ball BT (o) and define ϕ(x) = α(r(x)). Then, using the co-area formula (1.87) and (6.38) and integrating by parts we compute |∇ϕ|2 − a(x)ϕ2 = |∇ϕ|2 − a ¯(r)ϕ2 BT BT ≤ |∇ϕ|2 − A(r)ϕ2

BT T

= 0

[(α )2 v(r) dr − A(r)α2 v(r)] dr

=−

T 0

[(v(r)α ) + A(r)αv(r)]α dr ≤ 0.

By the Rayleigh characterization λL 1 (BT ) ≤ 0 and by domain monotonicity we get (6.39). Recalling that the solution of an ordinary differential equation is said to be oscillating if is has an infinite number of zeroes, we observe that, in order to show that a solution α of (6.38) has a first zero at T > 0, any oscillation criterion would do. For instance: Proposition 6.10. Let α be any solution of (6.38) with equality on [0, +∞) and assume that v1 ∈ L1 (+∞). Then α is oscillating provided !

+∞

lim inf A(r) v(r) r→+∞

r

ds v(s)

"2 >

1 . 4

(6.40)

Proof. We perform the standard change of variables

+∞

t = K(r) = r

ds v(s)

−1 .

(6.41)

Then, K : (0, +∞) → (0, +∞) is strictly increasing and setting γ(t) = tα K −1 (t) a direct computation shows that γ satisfies γ + p(t)γ = 0 with p(t) =

v 2 (K −1 (t)A(K −1 (t)) . t4

(6.42)

172


An application of Cauchy’s theorem, together with the definition of K and (6.40) shows that +∞ 1 lim inf t p(s) ds > . t→+∞ 4 t It follows from the Hille-Nehari oscillation theorem, see [Swa68] Chapter 2, that α is oscillating. Remark. We note that in the above oscillating result A(r) has necessarily to be positive in a neighborhood of +∞, but neither here nor in Proposition 6.9 is A(r) required to have a definite sign. Suppose now we have an estimate of the type Δr ≤ (m − 1)

h (r), h

h > 0 on (0, +∞)

which would typically follow from an appropriate lower bound for the Ricci curvature. Consider a solution of hm−1 (r)β + hm−1 (r)A(r)β = 0, β (0) = 0, β(0) = β0 > 0 and let T be the first zero of β. We set ψ(x) = β(r(x)); then, if A(r) ≤ a ¯(r), ¯(r)β ≥ β + Δrβ + A(r)β. Δψ + a ¯(r(x))ψ = β + Δrβ + a Therefore, if β ≤ 0 (which is necessarily the case if A(r) ≥ 0 satisfies Δψ + a ¯(r(x))ψ ≥ 0. Since ψ is radial, using again the co-area formula, |∇ψ|2 − a(x)ψ 2 = |∇ψ|2 − a ¯ψ 2 = − BT

BT

on [0, T )), then ψ

ψ(Δψ + a ¯ψ) ≤ 0 BT

and we conclude that λL 1 (BT ) ≤ 0. Note that in this case v(r) ≤ Chm−1 (r). We conclude this short account by quoting some new results in oscillation theory obtained in [BMR09] and [MRV], which we refer to for a complete treatment. With the notation of Proposition 6.9, let v(r) = vol(∂Br ) and let a ¯(r) be the


173

spherical mean of the potential a. Suppose that there exists an oscillating solution z ∈ Liploc of the problem ¯(t)v(t)z(t) = 0 on (0, +∞), (v(t)z (t)) + a (6.43) z (t) = O(1) as t 0+ , z(0+ ) = z0 > 0, and let 0 < t2 < t3 be two consecutive zeros, that is, z(t2 ) = z(t3 ) = 0 and z = 0 in (t2 , t3 ). Similarly to what we did in Proposition 6.9, we consider the function ϕz : M → R defined as z(r(x)) r(x) ∈ [t2 , t3 ] ϕz (x) = 0 r(x) ∈ [0, +∞) \ [t2 , t3 ]. Then, L

λ1

#

¯t ≤ B t3 \ B 2

¯t Bt 3 \ B 2

# t3 =−

t2

2

|∇ϕz | − # ¯t Bt 3 \ B 2

#

¯t Bt 3 \ B 2 2 ϕz

aϕ2z

# t3 =

t2

¯vz z (vz ) + a = 0. # t3 vz 2 t2

#t v(z )2 − t23 a ¯vz 2 # t3 vz 2 t2

Therefore, by the domain monotonicity of eigenvalues, the oscillation of the solution z implies λL 1 (M \ BR ) < 0,

for all R ≥ 0.

(6.44)

Recall now that the index of L is defined as the number of negative eigenvalues of −L. Hence condition (6.44), together with a result by D. Fisher-Colbrie [FC85] shows that L has infinite index, that is indL (M ) = sup {indL (Ω)} = +∞. Ω⊂⊂M

In fact the conclusion follows without appealing to [FC85] by noting that taking pairs of consecutive zeros of z the corresponding functions ϕz constructed above are L2 -orthogonal and this is enough to conclude that indL (M ) = +∞. Now fix a constant B ≥ 0 and define, for every t1 , t2 ∈ [0, +∞], V (t1 , t2 ) = e2B

t2 t1

ds v(s)

.

(6.45)

Then we have the following ([MRV], Theorem 14) Theorem 6.11. Let V and B as in definition (6.45) and suppose that the spherical mean a ¯(r) of a(x) satisfies a ¯(r)v 2 (r) ≥ −B 2 for all r > 0.

174


i) If there exist 0 < a < b such that 2B a(x)dx > V (b,+∞) 2B Bb \Ba V (b,+∞)−1

if v −1 ∈ / L1 (+∞), −1 if v ∈ L1 (+∞),

then λL 1 (M ) < 0. ii) L is unstable at infinity, i.e., λL 1 (M \ BR ) < 0 for every R > 0, provided either v ∈ L1 (+∞) and, for some R > 0, / ∞ ds > 1, a(x)dx lim sup v(s) t→∞ t Bt \BR or v ∈ / L1 (+∞) and lim

t→∞

/

sup

t≤q1 2B.

Bq2 \Bq1

In particular, in the assumption ii) L has infinite index. Theorem 6.11 is a consequence of Theorems 6.12 and 6.14 below which are proved in [MRV] (see also Proposition 1.2 and Theorem 1.4 in [BMR09]). Theorem 6.12. Let v(t) and a ¯(t) ∈ L∞ loc ([0, +∞)) be such that v(t) ≥ 0,

1 ∈ L∞ loc ((0, +∞)), v(t)

1 ∈ / L1 (0+ ), v

and a ¯(t) ≥ −

lim v(t) = 0

t→0+

B2 v(t)2

(6.46)

(6.47)

for some real constant B ≥ 0. Let z(t) ∈ Liploc ([0, +∞)) be a solution of problem (6.43). If z(t) = 0 for all t ∈ (0, +∞), then for every 0 ≤ a < b, b 2B if v1 ∈ / L1 (+∞), a ¯(s)v(s)ds ≤ V (b,+∞) 1 2B V (b,+∞)−1 if v ∈ L1 (+∞). a In order to prove Theorem 6.12 let us recall the following Riccati comparison result ([MRV], Lemma 18; compare also with Lemma 1.10 in Chapter 1): Lemma 6.13. Let G, v ∈ C 0 ([0, +∞)), v > 0, and qi ∈ AC(T, Ti ), i = 1, 2, be solutions of the Riccati differential inequalities q1 (t) ≥ G(t) +

1 2 q (t), v(t) 1

q2 (t) ≤ G(t) +

1 2 q (t) v(t) 2

(6.48)


175

a.e. in (T, Ti ) satisfying q1 (T ) = q2 (T ).

(6.49)

Then T1 ≤ T2 and q1 (t) ≥ q2 (t) in [T, T1 ). Conversely, if qi ∈ AC(Ti , t¯), i = 1, 2, are solutions of (6.48) a.e. in (Ti , T ) satisfying q1 (T ) = q2 (T ), then T1 ≥ T2 and q1 (t) ≤ q2 (t) in (T1 , T ]. A proof of the lemma is a minor modification of that of Corollary 2.2 in [PRS08]. We are now ready for the Proof of Theorem 6.12. Since, by assumption, z(t) > 0 on (0, +∞), the function y(t) = −v(t)

z (t) z(t)

is well defined on (0, +∞) and it is locally Lipschitz, as can be immediately seen by integrating (6.43) once. Furthermore, y satisfies y =

y2 +a ¯(t)v(t) v(t)

(6.50)

so that, according to (6.47), y ≥

y2 − B 2 v(t)

on (0, +∞).

(6.51)

First we consider the case v1 ∈ L1 (+∞). Since, by (6.46) we also have v1 ∈ L1 (0+ ), it follows that given any fixed α ∈ (0, +∞) there exists tα > 0 such that tα ds 1 = log α. (6.52) v(s) 2B 1 With the notation introduced in (6.45) define yα (t) = B

α + V (1, t) α − V (1, t)

and observe that, because of (6.52), i) yα t+ α = −∞; Furthermore yα satisfies yα =

ii) yα t− α = +∞. yα2 − B 2 v(t)

(6.53)

(6.54)

(6.55)

where defined. We claim that −B ≤ y(t) ≤ B

on (0, +∞).

(6.56)

176


By contradiction assume that there exists T ∈ (0, +∞) such that y(T ) = Y > B, the case Y < −B being similar. Let Y +B V (1, T ) Y −B

(6.57)

yα (T ) = Y = y(T )

(6.58)

α= and observe that

and that yα will thus be defined on [T, T2 ) for some maximal T2 ≤ +∞. Because of (6.51), (6.55) and (6.58) we can apply the Riccati comparison lemma above 2 with the choices q1 = y, q2 = yα , G = − Bv2 and T = T . Noting that y is defined on [T, +∞) and applying the first part of the lemma we have that yα is defined on [T, +∞). However, from (6.57) and (6.52) we have tα > T and we obtain a contradiction using (6.54) ii) (in case Y < −B we would have obtained a contradiction via (6.54) i)); thus we have the validity of (6.56). From (6.50) we obtain y ≥ a ¯(t)v(t). (6.59) Fix any 0 ≤ a < b as in the statement of the theorem and integrate (6.59) on [a, b]. Using (6.56) we have

b

a ¯(s)v(s) ds ≤ y(b) − y(a) ≤ 2B.

(6.60)

a

Suppose now only for

1 v

∈ L1 (+∞). In this case, there exists tα > 0 such that (6.52) holds α ∈ (0, V (1, +∞)).

(6.61)

We claim that −B ≤ y(t) ≤ B

V (t, +∞) + 1 V (t, +∞) − 1

on (0, +∞).

(6.62)

As above, we only prove the right-hand side inequality, the left-hand inequality being proved in a similar way. Assume by contradiction that there exists T ∈ (0, +∞) such that V (T, +∞) + 1 y(T ) = Y > B . V (T, +∞) − 1 Note that, since V (1, +∞) > 1, if we let α=

Y +C , Y −C


177

then, for C > 0 sufficiently small, α < V (1, +∞). It follows that the function yα = C

α + V (1, t) α − V (1, t)

satisfies the Riccati equation (6.55), yα (T ) = Y and there exists tα > T such that yα (t− α ) = +∞. Applying the Riccati comparison lemma and arguing as above we obtain a contradiction which proves that y(t) satisfies the required upper estimate. Integrating (6.59) as before we get

b

a ¯(s)v(s) ds ≤ a

2BV (b, +∞) . V (b, +∞) − 1

(6.63)

The estimate of Theorem 6.12 now follows from (6.60) and (6.63).

The proof of the next result can be obtained by iterating the technique of the proof above; for details we refer to the original proof of Theorem 10 in [MRV]. Theorem 6.14. Let v, a ¯ and z be defined as in Theorem 6.12. Then z is oscillating provided either v1 ∈ L1 (+∞) and &

∞

a ¯(s)v(s)ds

lim sup t→∞

t R

t

ds v(s)

' >1

for some R > 0, or v −1 ∈ / L1 (+∞) and & ' q2 sup a ¯(s)v(s)ds > 2B. lim t→∞

t≤q1 0 on M and b(x) ≥ 0. Assume also that the Ricci tensor satisfies δ/2 Ric ≥ −(m − 1)H 2 1 + r(x)2 on M, (6.66) for some δ ≥ −2 and H > 0. Let B0 = {x ∈ M : b(x) = 0} and assume that λL 1 (B0 ) > 0

(6.67)

178


and θ

ii) b(x) ≤ Br(x)β eDr(x) , for r 1, (6.68) for some α > 2δ − 1, 0 < θ < min 1 + α − 2δ , 1 + α2 , and A, B, D > 0, β ∈ R. Then there exists a positive maximal solution u on M of i) a(x) ≥ Ar(x)α ;

Δu + a(x)u − b(x)uσ = 0, σ > 1.

(6.69)

Remark. Consider the case of the hyperbolic m-dimensional space Hm −H 2 of constant negative curvature −H 2 . In this case the parameter α must satisfy α > −1, so that a behavior of a(x) ≥ 0 of the type Ar(x)−1+ε ≤ a(x) ≤ C(1 + r(x))

−1+2ε

for r(x) 1 and some ε > 0 is admissible. In this case, for a(x) sufficiently small, λL 1 (M ) ≥ 0, with L = Δ + a(x). This shows that Theorem 6.15 is not contained in Theorem 6.7 when λL 1 (M ) < 0. Remark. The upper bound for θ is sharp in the sense that, in the case of equality, m+2 existence may fail; see for instance Theorem 4.8 where σ = m−2 , α = 0, δ = 0 and 2 θ = 1, and D = 2B, β = m−1 + γ, γ > 0. Proof of Theorem 6.15. According to Theorem 6.5 we only need to construct a global subsolution u− ≥ 0, u− ≡ 0. We observe first of all that we may assume that a(x) ≥ A(1 + r(x))α

b(x) ≤ B(1 + r(x))β eDr(x)

θ

on M . Because of (6.66), Proposition 1.15 and Theorem 1.11 imply that δ/2 Δr ≤ Hr and r ≥ R. On the other hand, by the asymptotic behavior for some constant H of the metric coefficients in geodesic polar coordinates we have Δr =

m−1 + o(1) as r → 0+, r

and there exists a constant C such that 1 Δr ≤ C + rδ/2 weakly on M. r It follows that, if w(r) is a nonincreasing positive C 2 function with w (0) = 0 satisfying 1 w + C (6.70) + rδ/2 w (r) + A(1 + r)α w − B(1 + r)β wσ ≥ 0, r


179

then the function u− (x) = w(r(x)) is the required (weak) subsolution. We look for a function w of the form w(r) = (μ + P (r))ξ with P (r) ≥ 0, P (0) = 0, P (r) ≥ 0 and P (r) = eDr for r ≥ R, where μ > 0 is a constant to be chosen later, D > 0 and θ are the constants in (6.68) and 1 ξ < 1−σ < 0. Letting, for ease of notation θ

1 Hw = w + C( + rδ/2 )w (r) + A(1 + r)α w, r we claim that, for an appropriate choice of μ > 0, we have Hw (r) > 0 and

θ Hw (r) ≥ Crβ eDr w(r)σ

on [0, +∞)

for some C > 0 and r 1,

(6.71)

(6.72)

It follows that if E is a sufficiently small positive constant, then Ew satisfies (6.70). Indeed, with obvious notation, Hw (r) HEw (r) = E 1−σ (Ew(r))σ (w(r))σ and using (6.71) and (6.72) we may choose E small enough that θ HEw (r) ≥ B(1 + r)β eDr σ (Ew(r))

on [0, +∞). We compute w (r) = ξ(μ + P (r))ξ−1 P (r)

w (r)

= ξ(ξ − 1)(μ + P (r))ξ−2 (P (r))2 + ξ(μ + P (r))ξ−2 P (r) and inserting this in the expression of Hw it is easy to see that we may choose μ large enough that Hw > 0 on any chosen interval [0, R]. On the other hand, using the expression of P (r) for r large, a straightforward computation shows that

θ ξ(1−θ) Hw (r) ∼ Arα μ + eDr , σ w(r) 1 < 0 and D, θ > 0 easily imply (6.72). Thus we so that the conditions ξ < 1−σ may first find R such that (6.72) holds in [R, +∞) with a constant C > 0. Next, if necessary we may increase μ in order that (6.71) holds on [0, R]. It is clear that (6.72) will continue to hold with a possibly smaller C.

180


Remark. We note that if we replace the upper bound on the coefficient b(x), then the same procedure works with different choices of w(r). For instance, if we assume that b(x) ≤ B(1 + r)β on M, then the same arguments show that one may find a subsolution of the form u− (x) = w(r(x)) with w(r) = (μ + r2 )−γ provided β + α ≤ 2γ(σ − 1). Clearly the solution u supplied by the theorem satisfies the lower bound u(x) ≥ C(μ + r2 )−γ . In applications to the Yamabe problem this is relevant in connection to the completeness of the conformal metric.

6.3

Global sub- and supersolutions

We now give another version of Theorem 6.15, where it is no longer assumed that b is nonnegative. This will require the explicit construction of a global supersolution, and the procedure will prove to be rather delicate. The weakening of the assumptions on b is also reflected in the stronger curvature conditions and an upper bound imposed on the coefficient a. Theorem 6.16. Let (M, , ) be a complete manifold of dimension m ≥ 4 satisfying Ric ≥ −(m − 1)H 2 for some constant H > 0. Let a(x), b(x) ∈ C 0,μ (M ), 0 < μ ≤ 1, σ ≥ suppose that, for some R0 > 0 we have

(6.73) m+2 m−2

and

2

a(x) ≤

(m − 1) H 2 m−2 σ−1

b(x) > 0

on M \ BR0 ,

on M \ BR0 ,

Δ+ m−2 4 (σ−1)(a(x))

λ1

(BR0 ) > 0,

a(x) ≥ A(1 + r(x))α b(x) ≤ Cr(x)β eDr(x)

θ

(6.74) (6.75) (6.76)

on M,

(6.77)

for r(x) 1

(6.78)

and for some constants A, C, D > 0, β ∈ R, −1 < α ≤ 0, θ < 1 + α. Then, there exists η > 0 such that, if b(x) ≥ −η on M , the equation Δu + a(x)u − b(x)uσ = 0 admits a positive solution u on M .

(6.79)

6.3. Global sub- and supersolutions

181

Proof. As usual, it suffices to construct sub- and supersolutions v− and v+ in Liploc such that 0 < v− ≤ v+ . The construction of the subsolution is done as in the proof of Theorem 6.15. Note that in this respect allowing b to take negative values only helps matters. We recall that the function v− constructed in Theorem 6.15 is radial and satisfies lim v− (r) = 0. r→+∞

Recalling that for every E < 1, Ev− is still a subsolution shows that the condition v+ ≥ v− can always be satisfied whenever v+ > 0 on M

and

lim v+ (r(x)) = +∞.

r+∞

We therefore concentrate on the construction of the supersolution. We first m+2 reduce the analysis to the case where σ = m−2 . Let a(x) = (m−2)(σ−1) a(x) and 4 b = (m−2)(σ−1) b(x). Then a simple computation shows that, if u+ is a positive 4 supersolution of m+2 Δu + a(x)u − b(x)u m−2 = 0 (6.80) on M , then

4

v+ = u+(m−2)(σ−1) is a positive supersolution of (6.79) provided σ ≥ (6.76), we have Δ+ a(x) (BR0 ) > 0. λ1 Furthermore,

(6.81) m+2 m−2 .

Note that, because of (6.82)

b(x) > 0

(6.83) on M \ BR0 where B is the appropriate and if b(x) ≥ −B on M for some B > 0, then b ≥ −B multiple of B. We now proceed to construct a supersolution of (6.80) in BR0 and in M \ BR0 , and then glue them together to form a global supersolution. We divide the argument, which is adapted from the proof of Theorem 0.1 in [RRV97], into several steps. Step 1. Construction of a supersolution of (6.80) inside BR0 . If there exists x0 ∈ BR0 such that a(x0 ) > 0,

(6.84)

it follows by results of J. Escobar (see [Esc87], Theorems 3.2 and 4.2), that there exists a solution u > 0 of the problem m+2 Δu + a(x)u + u m−2 = 0 on BR0 , (6.85) u=0 on ∂BR0 , u > 0 on BR0 .

182


On the other hand, if a ≤ 0 on BR0 , then, using the fact that the λΔ 1 (BR0 ) >, we a1 can choose a small enough constant a1 > 0 such that λΔ+ (BR0 ), and apply the 1 above argument to obtain a solution u > 0 of problem (6.85) with a1 instead of a(x). Then m+2 m+2 Δu + a(x)u + u m−2 ≤ Δu + a1 (x)u + u m−2 = 0 and we conclude that (6.85) always admits a positive supersolution u vanishing on ∂BR0 . Step 2. Construction of a radial supersolution on M \ BR0 . Using the Ricci curvature assumption (6.73) and the Laplacian comparison theorem we have Δr ≤ (m − 1)H coth(Hr) on M. (6.86) Moreover, it follows from (6.74) that a≤6 a=

(m − 1)2 2 H 4

on M \ BR0 .

(6.87)

Finally, we choose a positive, decreasing function 6b(t) on [0, +∞) of the form 6b(t) = e−P (t)

(6.88)

with P ∈ C 2 ([0, +∞)) satisfying P (0) = P (0) = 0, P ≥ 0, P ≥ 0, P (t) → +∞ as t → +∞ and such that

b ≥ 6b(r(x))

(6.89)

on M \ BR0 .

(6.90)

We look for a supersolution of (6.80) on M \ BR0 of the form v(x) = β(r(x)) with β : [0, +∞) → R, where β solves the problem m+2 β + (m − 1) coth(Hr)β + 6 aβ − 6b(r)β m−2 ≤ 0, (6.91) β (0) = 0, β ≥ 0, β(r) > 0. Furthermore, we require lim inf β(r) > 0.

(6.92)

r→+∞

Indeed, if β solves (6.91), it follows from (6.86), (6.87) and (6.88) that v is positive and satisfies m+2 m+2 Δv + a(x)v − b(x)v m−2 = β + Δrβ + a(x)β − b(x)β m−2

≤ β + (m − 1)H coth(Hr)β + 6 aβ − 6b(r)β m−2 ≤ 0 m+2

on M \ BR0 and lim inf v(x) > 0. r→+∞

(6.93)


183

The previous computation makes it clear that the requirement β ≥ 0 in (6.91) is vital for our approach to work. As we shall see this will be a consequence of the definition (6.88) of 6b . To solve (6.91) we choose an increasing divergent sequence {Tn } ⊂ (0, +∞). For a given n we look for a positive solution on [0, Tn ] of m+2 β + (m − 1) coth(Hr)β + 6 a(r)β − 6b(r)β m−2 = 0, (6.94) β (0) = 0, β(Tn ) = i ∈ N. To motivate the argument that follows assume that w > 0 is a solution of the problem m+2 (6.95) Δw + a ˇw − ˇbw m−2 = 0 on an open set Ω of a manifold (N, g), and let ϕ > 0 be a C 2 solution of Δϕ + a ˇ ϕ + λ1 ϕ = 0

on Ω.

(6.96)

If we define the conformally related metric 4

g = ϕ m−2 g,

(6.97)

and use the transformation law (3.81) and equations (6.95), (6.96), an elementary computation shows that the function

satisfies

u = ϕ−1 w,

(6.98)

m+2 4 − λ1 ϕ− m−2 Δu u − ˇbu m−2 = 0 on Ω,

(6.99)

denotes the Laplace-Beltrami operator of g. Hence, if u is a positive where Δ solution of (6.99), then w = ϕu is a positive solution of (6.95). Note now that (6.94) is precisely the expression of the equation Δ Hm

u+6 au − 6bu m−2 = 0 m+2

−H 2

2 on the m-dimensional hyperbolic space Hm −H 2 of constant curvature −H when u(x) = β(r(x)) is radial. We also observe that the first eigenvalue of the Dirichlet Laplacian on the ball BR in Hm −H 2 satisfies Δ Hm

λ1

−H 2

(BR ) ≥

(m − 1)2 2 H [coth(HR)]2 . 4

Indeed, ΔHm

−H 2

r = (m − 1)H coth(Hr)

(6.100)

184


and a standard application of the divergence theorem shows that if φ is a smooth function with compact support in BR , then 2 (m − 1)H coth(HR) φ ≤ φ2 ΔHm 2 r −H

= −2

φ∇φ, ∇r ≤ 2

φ

2

1/2 |∇φ|

and the claim follows. It follows that for some λ1 > 0 we have a solution ϕ of aϕ + λ1 ϕ = 0 on BTn+1 , ΔHm 2 + 6 −H

ϕ=0

on ∂BTn+1 , ϕ > 0 on BTn+1

2

1/2 ,

(6.101)

and the radial symmetry of the problem implies that the eigenfunction ϕ is radial. The above discussion shows that in order to find a solution of (6.94) we may let (N, g) be the hyperbolic space Hm −H 2 and look for a radial solution of

m+2 4 − λ1 ϕ− m−2 Δu u − 6bu m−2 = 0 on BTn , u > 0, u ≡ ϕ(Ti n ) on ∂BTn .

(6.102)

Since λ1 > 0, the functional associated to (6.102) is strictly convex and unbounded at infinity. Thus problem (6.102) admits a radial solution u. It follows that problem (6.94) admits a radial solution βn,i > 0 on [0, Tn ]. In order to determine the sign of βn,i we proceed as follows: we consider the function v− (r) = Λe

m−2 4 P (r)

, Λ=

m−1 H 2

m−2 2 .

Using (6.88) and the definition of 6 a we compute / 2 m+2 4 (m − 1) m−2 6 av− − 6b(r)v− = v− H 2 − Λ m−2 ≡ 0. 4 With the aid of (6.89) we see that v− solves ⎧ m+2 ⎨ m av− − 6b(r)v−m−2 ≥ 0 on BTn , (in fact on Hm ΔH 2 v − + 6 −H 2 ) −H ⎩v− (x) < i on ∂BT n

provided i is sufficiently large. On the other hand, having set z(x) = βn,i (r(x)),

(6.103)

(6.104)


185

because of (6.94) we have m+2 ΔHm 2 z + 6 az − 6b(r)z m−2 = 0

on BTn ,

z(x) = i

on ∂BTn .

−H

(6.105)

Applying Proposition 6.1, βn,i (r) ≥ v− (r)

on [0, Tn ].

(6.106)

We fix r0 ∈ [0, Tn ] and we consider the function m+2 y(t) = 6b(r0 )t m−2 − 6 at

on (0, +∞).

We have

4 m + 26 1 a ≥ y(t). b(r0 )t m−2 − 6 m−2 t We let t0 = v− (r0 ). Then, y(t0 ) = 0 because of (6.103); thus y (t0 ) ≥ 0. On the other hand 4 m+2 6 m−2 −1 > 0 y (t) = 4 2 b(r0 )t (m − 2)

y (t) =

and we deduce that

y (t) ≥ 0 ∀ t ≥ t0 .

Thus, the function y is nondecreasing on [t0 , +∞). By (6.106) βn,i (r0 ) ≥ v− (r0 ) = t0 It follows that

and y(t0 ) = 0.

m+2 6b(r0 )βn,i (r0 ) m−2 −6 aβn,i (r0 ) ≥ 0.

Since r0 ∈ [0, Tn ] was chosen arbitrarily, it follows that βn,i + (m − 1)H coth(Hr)β ≥ 0 on [0, Tn ], βn,i (0) = 0. Integration of the above immediately yields βn,i (r) ≥ 0

on [0, Tn ].

(6.107)

Now let i → ∞. Using Proposition 6.1 we see that the sequence of solutions {βn,i } of (6.94) is nondecreasing. Furthermore, it is uniformly bounded on compact subsets of [0, Tn ) because of Lemma 4.2. Thus, βn,i converges as i → +∞ to a solution βn of the problem m+2 βn + (m − 1)H coth(Hr)βn + 6 aβn − 6b(r)βnm−2 = 0 on [0, Tn ), βn (0) = 0, βn (r) ≥ 0, βn (r) > 0, , βn (r) → +∞ as r → Tn−

186


and, moreover, βn (r) ≥ v− (r).

(6.108)

A second application of Proposition 6.1 shows that the sequence {βn } is decreasing and therefore it converges to a solution β of m+2 β + (m − 1)H coth(Hr)β + 6 aβ − 6b(r)β m−2 = 0 on [0, +∞), (6.109) β (0) = 0, β ≥ 0, β(r) > 0. Furthermore, because of (6.108), which is independent of n, β(r) ≥ v− (r) → +∞ as r → +∞, so that (6.92) is certainly satisfied. Thus v(x) = β(r(x)) satisfies (6.93) and m+2 Δv + a(x)v − b(x)v m−2 ≤ 0 on M \ BR0 , (6.110) v(x) > 0, so is the required radial supersolution on M \ BR0 . Step 3. Gluing the supersolutions. 1,2 First, given w ∈ C 0 (M ) ∩ Wloc , for σ =

m+2 m−2

to simplify the writing we define

bw = w−σ [Δw + a(x)w]

(6.111)

in the weak sense. Note that for E > 0 constant bEw = E 1−σ bw .

(6.112) 1,2 Hloc

such that Next, we let u + be a positive function in C 0 (M ) ∩ u(x) on BR0 −ε , u + (x) = v(x) on M \ BR0

(6.113)

for some ε > 0 sufficiently small and where u(x) has been defined in (6.85) while v(x) has been defined in (6.110). We then set u+ (x). u+ (x) = E

(6.114)

Having chosen ε > 0 sufficiently small we can suppose by (6.75) that b(x) > 0 on M \ BR0 −ε . Next we note that w is a supersolution of (6.80) if and only if bw (x) ≤ b(x) on M . Using (6.112) and (6.113) we have: i) On M \ BR0 , bu+ (x) = E 1−σ bu+ (x) = E 1−σ bv (x) ≤ E 1−σb(x) ≤ b(x); therefore u is a supersolution of (6.80) on M \ BR0 ∀ E ≥ 1, since b(x) > 0 on M \ BR0 .


187

ii) On BR0 + \ B R0 −ε , bu+ (x) = E 1−σ bu+ (x) ≤ E 1−σ

sup B R0 \BR0 −ε

bu+ (x) ≤

inf

B R0 \BR0 −ε

b(x) ≤ b(x),

provided E ≥ E0 ≥ 1 is sufficiently large. This is possible since b(x) > 0 on B R0 \ BR0 −ε . iii) On BR0 −ε , since u is a supersolution of (6.85) we have bu+ (x) = E 1−σ bu+ (x) = E 1−σ bu (x) =≤ −E 1−σ . To obtain a supersolution on BR0 −ε we need −E 1−σ ≤ b(x).

(6.115)

Thus u+ is a subsolution if and only if b ≥ −E 1−σ where E is determined in ii) that is b(x) ≥ −η where η = cm E 1−σ . If this condition is satisfied, we have therefore found the required positive supersolution u+ to (6.79) with the property that lim r(x)→+∞

u+ (x) = +∞.

Remark. It should be noted that the upper bound on b(x) affects only the behavior of the subsolution, and, as observed after the proof of Theorem 6.15, different upper bounds for b(x) give rise to subsolutions with different asymptotic behaviour, and therefore to different asymptotic estimates for the solution. See the remark after the proof of Theorem 6.18 More generally, the positivity of a(x) together with (6.77) and (6.78) may by replaced by any other set guaranteeing the existence of a subsolution of (6.79). For instance we can assume that Δ+a(x)

λ1

(M ) < 0

to construct a subsolution v− on a sufficiently large ball BR as in the proof of Theorem 6.7. In this case v− satisfies σ ≥ 0, v− > 0 on BR , Δv− + a(x)v− − b+ (x)v− v− ≡ 0 on ∂BR .

Therefore u− (x) =

Dv− (x) 0

on BR , on M \ BR

yields the desired subsolution for D ∈ (0, 1] sufficiently small.

188

6.4


The case of the Yamabe problem

It is clear that we can specialize the conclusions of the previous theorems to the case of the Yamabe equation m+2

cm Δu − S(x)u + K(x)u m−2 = 0 −1 K(x). We will state such a simply replacing a(x) = −c−1 m S(x) and b(x) = −c result explicitly only in the case of Theorem 6.16, leaving the other cases to the interested reader. It should also be pointed out that, in application of the theorems of previous sections to the Yamabe equation, it is often possible to replace the condition that S(x) be everywhere negative with the assumption that it is nonpositive and strictly negative off a compact set. This is based on the following proposition (compare with Proposition 3.10).

Proposition 6.17. Let Ω0 ⊂⊂ Ω1 ⊂ (M, , ) be relatively compact domains with smooth boundaries. Assume that the scalar curvature S(x) verifies S(x) ≤ c2 h(x)

on M \ Ω0

(6.116)

for some c > 0, h ∈ C 0 (M ) nonpositive, h(x) < 0 on Ω1 \ Ω0 . Then there exists η > 0 such that, if S(x) ≤ η on M, (6.117) then there exists a complete conformal metric , homothetic to , on M \ Ω1 , whose scalar curvature S(x) satisfies S(x) ≤ c2 h(x)

on M,

(6.118)

for some c > 0. Proof. We elaborate some ideas of [AM88]. It is clear that, if , is a metric conformal to , and homothetic to , on M \ Ω1 , then (6.116) implies (6.118) on M \ Ω1 for some suitable c > 0. Therefore, up to modifying c, it is enough to show that , can be chosen so that S(x) < 0 on Ω1 . We fix a constant δ > 0 and solve the Dirichlet problem cm Δψ = δ on Ω1 , (6.119) ψ=0 on ∂Ω1 . Next, we choose a domain Ω2 such that Ω0 ⊂⊂ Ω2 ⊂⊂ Ω1 and a smooth cut-off function ξ : M → [0, 1] such that ξ ≡ 1 on Ω0

and

ξ ≡ 0 on M \ Ω2 .

6.4. The case of the Yamabe problem

189

We define ϕ = ξψ on M ; now we choose a positive constant ϕ0 so that ϕ(x) + ϕ0 > 0 on M and S(x)(ϕ(x) + ϕ0 ) − cm Δϕ ≤ −

(6.120)

δ 2

on Ω1

(6.121)

(note that, in order to achieve (6.121), we use h(x) < 0 on Ω1 \ Ω0 and S(x) ≤ η). Finally, we set 4 , = (ϕ(x) + ϕ0 ) m−2 , . Then S(x) is given by − S(x) = (ϕ(x) + ϕ0 ) m−2 [S(x)(ϕ(x) + ϕ0 ) − cm Δϕ(x)]. m+2

Now (6.120) and (6.121) imply S(x) < 0 on Ω1 .

We now give the announced version of Theorem 6.16 for the Yamabe equation. It is a variation of Theorem 0.1 in [RRV97]. Theorem 6.18. Let (M, , ) be a complete manifold of dimension m ≥ 3, scalar curvature S(x), sectional curvature SectM and let K(x) ∈ C ∞ (M ). Let R > 0 and suppose that the ball BR centered at o does not intersect the cut locus of o. Suppose that i) Ric ≥ −(m − 1)H 2 2

ii) SectM ≤ −A

on M, (6.122)

on BR , 2

iii) S(x) ≤ −m(m − 1)C (1 + r(x)]

α

on M,

for some 0 ≤ A, C ≤ H and −1 < α ≤ 0 and 0 < A ≤ B such that 1 2 A2 [coth(AR)]2 . H < 1+ m(m − 2) Assume also that K(x) ≥ −C1 r(x)β eDr(x)

θ

(6.123)

(6.124)

on M \ BR for some constants C1 , D > 0, β ∈ R, and θ < 1 + α. Then, there exists η > 0 such that, if K(x) ≤ η on M, the metric , can be pointwise conformally deformed to a new metric of scalar curvature K(x). Proof. We only need to verify that the assumptions of Theorem 6.16 are satisfied −1 with a(x) = −c−1 K(x). First of all, the bound on the Ricci m S(x) and b(x) = −c curvature implies s(x) ≥ −m(m − 1)H 2 ,

190


so that, recalling that cm = 4 m−1 m−2 , −c−1 m s(x) ≤

m(m − 2) 2 (m − 1)2 2 H ≤ H , 4 4

(6.125)

and condition (6.74) in the statement of Theorem 6.16 is automatically satisfied. Observe next that the sectional curvature condition and the fact that the ball BR does not intersect the cut locus of o implies, by the Laplacian comparison theorem, that (6.126) Δr ≥ (m − 1)A coth(Ar) on BR . Arguing as in the proof of Theorem 6.16, this in turn implies that the first eigenvalue of the Dirichlet Laplacian on BR satisfies λΔ 1 (BR ) ≥

2

(m − 1) 2 A [coth(AR)]2 , 4

whence, using (6.125), Δ−c−1 m S(x)

λ1

(BR ) ≥

(m − 1)2 2 m(m − 2) 2 A [coth(AR)]2 − H >0 4 4

by condition (6.123).

Remark. The conclusion of the theorem in fact continues to hold even if equality holds in (6.123). This requires a different argument for the construction of the supersolution inside the ball BR . We refer to [RRV97] for the details. Note also that the upper bound (6.122) iii) for S(x) and the lower bound (6.78) for K(x) are used to construct the subsolution and determine a lower estimate for the solution u. As noted after the proof of Theorem 6.15, replacing (6.78) with K(x) ≥ −C1 r(x)β r(x) 1, with β < α allows us to produce a subsolution, and therefore a solution which satisfy the estimate u(x) ≥ Cr(x)γ , r(x) 1, with γ =

m−2 4 (β

+ α). In particular if β + α = 2, then u(x) ≥ Cr−

m−2 2

,

and we may guarantee that the conformal metric 4 , = u m−2 ,

is complete.

6.5. Appendix: the Monotone Iteration Scheme

6.5

191

Appendix: the Monotone Iteration Scheme

In this section we describe the monotone iteration scheme, due to H. Amann (see [Ama76]), that has been one of the main tools to obtain our existence results. Our presentation follows closely that of D. H. Sattinger, [Sat73]. Let Ω be an open domain, and let f (x, s) be a continuous function on Ω × R, and consider the differential equation Δu + f (x, u) = 0,

in Ω.

(6.127)

1,2 We say that a function u+ ∈ Wloc (Ω) ∩ L∞ loc (Ω) is a supersolution of (6.127) if it satisfies the differential inequality

Δu+ + f (x, u+ ) ≤ 0,

weakly in Ω,

that is, for every test compactly supported function ρ ≥ 0 in W 1,2 (Ω), − ∇u+ , ∇ρ + f (x, u+ )ρ ≤ 0. A function u− is a subsolution if it satisfies the reverse differential inequality. 1,2 Similarly, if Ω is bounded, and g is a continuous function on ∂Ω, u+ ∈ Wloc (Ω) ∩ C(Ω) is a supersolution of the boundary value problem Δu + f (x, u) = 0 in Ω, (6.128) u=g on ∂Ω, if

Δu+ + f (x, u+ ) ≤ 0 in Ω, u+ ≥ g on ∂Ω.

The definition of a subsolution is obtained reversing the inequalities. Theorem 6.19. Let Ω be a relatively compact open domain with smooth boundary older function such that s → f (x, s) ∂Ω and let f : Ω × R → R be a locally H¨ is locally Lipschitz with respect to s uniformly with respect to x. Suppose that ϕ and ψ ∈ C 0,1 (Ω) are respectively a subsolution and a supersolution of (6.128) satisfying ϕ ≤ ψ on Ω and ϕ ≤ g ≤ ψ on ∂Ω. Suppose moreover that g extends to a C 2,α (Ω) function, which we denote with the same letter. Then the boundary value problem (6.128) has a solution u ∈ C 2,α (Ω) satisfying ϕ ≤ u ≤ ψ. Proof. Since f (x, s) is locally Lipschitz in s uniformly with respect to x in Ω, there exists λ > 0 such that s → f (x, s) + λs = F (x, s)

192


is monotone increasing for every x in Ω and every s satisfying minΩ ϕ ≤ s ≤ maxΩ ψ. For every function w ∈ C α (Ω) we let v = T w ∈ C 2,α (Ω) be the solution of the boundary value problem (Δ − λ)v = −[f (x, w) + λw] in Ω, (6.129) v=g on ∂Ω, which exists and is unique by standard elliptic theory, (see, e.g., [GT01] Theorem 6.14). We claim that the operator T is monotone, that is, if w1 ≤ w2 on Ω, then T w1 ≤ T w2 . Indeed, by the monotonicity of the function F (x, s) the difference v = T w2 − T w1 satisfies (Δ − λ) v = − F (x, w1 ) − F (x, w2 )] ≤ 0 in Ω, v = 0 on ∂Ω, and therefore by the minimum principle, v ≥ 0 in Ω, i.e., v2 = T w2 ≥ v1 = T w1 , as claimed. + − − Now we set u− 1 = T ϕ, u1 = T ψ and for every k ≥ 1, uk+1 = T uk and + + uk+1 = T uk . Recalling that ϕ and ψ are respectively sub- and supersolutions of the boundary value problem (6.128) and arguing as above, the maximum principle and an induction argument show that − − + + + ϕ ≤ u− 1 ≤ u2 ≤ · · · ≤ uk ≤ · · · ≤ uk ≤ u2 ≤ u1 ≤ ψ. + − + Thus, as k tends to infinity u− k → u and uk → u . − + 2,α We claim that u and u are in C (Ω) and solve the boundary value (6.128). Indeed, the solution v of (6.129) satisfies Lp Sobolev estimates up to the boundary of the form ||v||W 2,p ≤ C ||g||W 2,p + ||F (x, w)||Lp ,

(see, e.g., [ADN59], Theorem 15.2) and it follows easily that for every p, the 2,p sequences {u± to a solution (u± ) of the differential equation k } converge in W Δu = f (x, u). By Morrey’s embedding lemma, see [GT01], Theorem 7.26, for every p > n the space W 1,p embeds continuously into C β (Ω) and therefore, by choosing p large enough, the above convergence takes place in C α (Ω). Finally, by [ADN59], Theorem 7.3, the solution v of (6.129) satisfies Schauder’s estimates up to the boundary ||v||C 2,α ≤ C( ||g||C 2,α + ||F (x, w)||C α , 2,α and therefore the sequences u± (Ω) to solutions u± of (6.128), as k converge in C required.


193

We conclude this section with the following result that extends to the situation at hand the well-known fact that the maximum of two subharmonic functions is subharmonic. Theorem 6.20. Let f be a continuous function on Ω × R and suppose that ui ∈ 1,2 L∞ loc (Ω) ∩ Wloc (Ω), i = 1, 2 are subsolutions of (6.127) on Ω. Then = max{u1 , u2 } is also a subsolution. Similarly if vi are supersolutions of (6.127), then so is v = min{v1 , v2 }. Proof. The proof is adapted from [Le98]. We only show that the maximum of subsolutions is a subsolution. The argument in the case of supersolutions is similar. 1,2 Set u = max{u1 , u2 } and note that u ∈ L∞ loc (Ω) ∩ Wloc (Ω) and ∇u =

∇u1 ∇u2

if u1 ≥ u2 , if u1 ≤ u2 .

By assumption, for every 0 ≤ ϕ ∈ C0∞ (Ω) we have ∇ui , ∇ϕ − f (x, ui ) ≤ 0, i = 1, 2. Let D ⊂ Ω be a relatively compact open set with smooth boundary such that supp ϕ ⊂ D, and set D1 = {x ∈ D : u1 (x) > u2 (x)}, Do = {x ∈ D : u1 (x) = u2 (x)} and D2 = {x ∈ D : u1 (x) < u2 (x)}. Let also γ : R → R be such that γ ≥ 0, γ(t) = 0 if t ≤ 0 and = 1 is t ≥ 1, and let γn (t) = γ(nt), so that γn (t) = 0 if t ≤ 0, = 1 if t ≥ 1/n and 0 ≤ γn (t) ≤ nM 1,2 with M = sup γ . Let w = u2 − u1 . Since w ∈ Wloc (Ω) there exists wn ∈ C0∞ (Ω) 1,2 such that wn → w in W (D). By passing to a subsequence, if necessary, we may also assume that wn → w a.e. in D and ||wn − w||W 1,2 (D) ≤ n12 . Finally, set ϕ1 = (1−γn (wn ))ϕ and ϕ2 = γn (wn )ϕ, so that 0 ≤ ϕi ∈ C0∞ (D). We claim that for a.e. x ∈ D2 there exists n0 = n0 (x) such that wn (x) > 1/n for every n ≥ n0 . Indeed, since w > 0 on D2 , there exists n1 = n1 (x) such that w(x) > 2/n1 . Moreover, wn → w a.e., for a.e. x there exists n0 = n0 (x) ≥ n1 such that for every n ≥ n0 |wn (x) − w(x)| ≤ 1/n1 , and therefore, for n ≥ n0 ≥ n1 , wn (x) ≥ w(x) − |wn (x) − w(x)| ≥

2 1 1 1 − = ≥ . n1 n1 n1 n

Thus, for a.e. x and every n ≥ n0 we have γn (wn (x)) = 1 and we conclude that γn (wn ) → 1 a.e. in D2 . Similarly, since w(x) < 0 in D1 , we see that γn (wn (x)) → 0 a.e. in D1 . Since 0 ≤ ϕi ∈ C0∞ (Ω), and ui are subsolutions we have ∇ui , ∇ϕi − f (x, ui ) ≤ 0,

194


whence, adding the two inequalities, and using the definition of ϕi , a computation yields

∇u1 , (1 − γn (wn ))∇ϕ − γn (wn )ϕ∇wn + ∇u2 , γn (wn )∇ϕ + γn (wn )ϕ∇wn − f (x, u1 )(1 − γn (wn )∇ϕ + f (x, u2 )γn (wn )∇ϕ = ∇u2 − ∇u1 , γn (wn )∇ϕ + ∇u2 − ∇u1 , γn (wn )ϕ∇wn + [f (xu1 ) − f (x, u2 )]γn (wn )ϕ + ∇u1 , ∇ϕ − f (x, u1 )ϕ

0≥

(6.130)

= I1 (n) + I2 (n) + I3 (n) + I4 . Observe now that, since ∇ui ∈ L2 (D), ∇ϕ ∈ L∞ (D), ∇u1 = ∇u2 on Do , 0 ≤ γn (wn ) ≤ 1, and γn (wn )tends to 1 a.e. on D2 and to 0 a.e. on D1 , , letting n → ∞, by dominated convergence we have I1 (n) =

D1 ∪D2

∇u2 − ∇u1 , γn (wn ))∇ϕ →

∇u2 − ∇u1 , ∇ϕ. D2

Similarly, since f (x, ui ) is bounded on D, f (x, u1 ) = f (x, u2 ) on Do and ϕ is compactly supported I4 (n) =

[−f (x, u2 )+f (x, u1 )]γn (wn ))ϕ →

D1 ∪D2

[−f (x, u2 )+f (x, u1 )]γn (wn ))ϕ. D2

To deal with I2 note that I2 = ≥

∇u2 −

∇u1 , ∇wn γn (wn )ϕ

=

∇w, ∇wn − ∇wγn (wn )ϕ ≥ −

∇w, ∇w + (∇wn − ∇w)γn (wn )ϕ |∇w||∇wn − ∇w|γn (wn )ϕ.

Since γn (wn ) ≤ M n,

and

||wn − w||W 1,2 (D) ≤

1 , n2

the integral on the rightmost side is bounded above by M n||ϕ||∞ ||∇ϕ||L2 (D) n12 = C n and we deduce that lim inf I2 (n) ≥ 0. n


195

Thus, letting n → +∞ in (6.130) we obtain 0 ≥ lim I1 (n) + lim inf I2 (n) + lim I3 (n) + I4 n n n ≥ ∇u2 − ∇u1 , ∇ϕ + [−f (x, u2 ) + f (x, u1 )ϕ] + ∇u1 , ∇ϕ − f (x, u1 )ϕ D2 D ∇u2 , ∇ϕ − f (x, u2 )ϕ + ∇u1 , ∇ϕ − f (x, u1 )ϕ. = D2

D\D2

Recalling that u = max{u1 , u2 } =

u2 u1

on D2 , and that ∇u = on D \ D2 ,

∇u2 ∇u1

on D2 , on D \ D2 ,

the last inequality amounts to 0≥

∇u, ∇ϕ − f (x, u)ϕ,

as required to show that u is a subsolution of (6.127).

Chapter 7

Some special cases In this final chapter we present some special cases where one can use particular techniques to relax assumptions. Typically this happens in Euclidean and hyperbolic spaces, and more generally in the case of models, or manifolds with some special symmetry. For the models, the main advantage is to have a precise expression for Δr which is available, on general manifolds, only in the form of an upper or a lower bound under certain assumptions on the curvature and the cut locus. We provide here refined techniques and a few results that, as a side product, show the degree of sharpness of the general theory and of the methods that we have developed dealing with generic complete Riemannian manifolds. As a final remark we underline that, here and there, in this chapter we explicitly link our geometric point of view to more familiar Euclidean procedures and tools (for instance, the Rellich-Pohozaev formula), as shown below in detail.

7.1 A nonexistence result We begin by showing that under appropriate assumptions on a(x) and b(x) there are no positive solutions u on M of Δu + a(x)u − b(x)uσ ≥ 0. To prove this fact we shall use a technique that does not require the determination of a priori estimates on u. The idea is to perform a sort of radialization and then apply ODE methods. However, as we will explain later, this has some geometrical heavy side effects. We need a collection of preliminary technical results; the next lemma is motivated by the work of K.S. Cheng and J. T. Lin, [CL87]. Lemma 7.1. Let 0 < R0 ≤ R1 ≤ R2 ≤ +∞, σ > 1 and assume that the functions 0 l, ψ, b ∈ C ([R0 , +∞)) satisfy i) l(t) > 0 on [R0 , +∞); ii) ψ(t) > 0 on [R0 , +∞); (7.1) iii) tb(t) ∈ L1 (+∞); iv) tb(t) > 0 and nonincreasing on [R1 , +∞).


197

198

Chapter 7. Some special cases

Suppose that lim inf t→+∞

l(t)ψ(t)σ > 0. tb(t)

(7.2)

If α is a positive solution of &

s

α(s) ≥ ψ(s) c1 + c2 Ro

' t σ l(t)α(t) dt 1− s

(7.3)

on [R0 , R2 ) for some constant c1 , c2 > 0, then R2 < +∞. Proof. We argue by contradiction and let R2 = +∞. We set β(s) =

α(s) ψ(s)

so that (7.3) implies that β satisfies ' & s t σ σ l(t)ψ(t) β(t) dt 1− ∀ s ≥ R0 . β(s) ≥ c1 + c2 s Ro Because of (7.2), there exist c3 > 0 and R3 ≥ R0 such that s t tb(t)β(t)σ dt ∀ s ≥ R ≥ R3 . β(s) ≥ c1 + c3 1− s R

(7.4)

We choose R4 > max {R1 , R3 } and we define

t

η(t) =

vb(v) dv + R42 b(R4 ).

(7.5)

R4

We observe that η(t) ≥ 0 and

i) η (t) = t b(t) > 0;

ii) η(t) → +∞ as t → +∞

(7.6)

because of assumptions (7.1) iii), iv). We let R ≥ R4 and we consider η : [R, s] → [ν = η(R), ξ = η(s)]. We indicate with t(η) its inverse. We set γ(η) = β(t(η)) and we perform a change of variable in (7.4) to obtain

ξ

γ(ξ) ≥ c1 + c3

! 1−

ν

" t(η) γ(η)σ dη. t(ξ)

(7.7)

7.1. A nonexistence result

199

Next, we observe that for η sufficiently large, say η ≥ η0 , η −1 t(η) is nondecreasing. Indeed, it is enough to show that ηt (η) − t(η) ≥ 0,

η ≥ η0 .

This, in turn, is equivalent to showing that

t

vb(v) dv + R42 b(R4 ) − t2 b(t) ≥ 0

(7.8)

R4

for t sufficiently large. But (7.1) iv) and our choice of R4 imply (7.8) for t ≥ R4 . From (7.7) we deduce

ξ

γ(ξ) ≥ c1 + c3 ν

η γ(η)σ dη 1− ξ

(7.9)

for all ξ > ν ≥ η(R4 ). With the aid of (7.6) ii) and (7.1) we then have: ∀ ν > η(R4 ) and ∀ ξ ∈ [ν, 2ν], ξ 1 γ(ξ) ≥ c1 + c3 (ξ − η)γ(η)σ dη. (7.10) 2ν ν To complete the proof we define g(ξ) = c1 + c3

1 2ν

ξ

(ξ − η)γ(η)σ dη

(7.11)

ν

and observe that g satisfies i) g(ν) = c1 , ii) g (ν) = 0 #ξ c3 γ(η)σ dη, iv) g (ξ) = iii) g (ξ) = 2ν ν

c3 σ 2ν γ(ξ)

≥

c3 σ 2ν g(ξ)

(7.12)

where in (7.12) iv) we have been using (7.10) and (7.11). From (7.12) we get

[g (ξ)]

2

= 2g (ξ)g (ξ) ≥

c3 c3 g(ξ)σ+1 . g (ξ)g(ξ)σ = ν (σ + 1)ν

(7.13)

Therefore, integrating (7.13) over [ν, ξ], with the aid of (7.12) ii) we deduce !

c3 g (ξ) ≥ (σ + 1)ν

" 12

g(ξ)σ+1 − g(ν)σ+1

12

.

(7.14)

Integrating (7.14) again over [ν, ξ] we have

ξ ν

!

g (x) dx [g(x)σ+1 − g(ν)σ+1 ]

1/2

c3 ≥ (σ + 1)ν

"1/2 (ξ − ν).

(7.15)

200


Next, we perform the change of variable u=

g(x) g(ν)

and we rewrite (7.15) in the form

g(ξ) c1

1

9 √

du uσ+1 − 1

≥

cσ−1 c3 1 (ξ − ν). (σ + 1)ν

(7.16)

− 1 But σ > 1; hence uσ+1 − 1 2 ∈ L1 ([1, +∞)). On the other hand, choosing ξ = 2ν in (7.16), we immediately obtain a contradiction by taking ν sufficiently large. To state the next result we introduce some notation. Given σ > 1 and a nonnegative f ∈ C 0 (M ), we set 1 ¯ f (r) = f, r ∈ [0, +∞); (7.17) vol(∂Br ) ∂Br '1−σ & 1 1 f 1−σ , r ∈ [0, +∞) (7.18) f¯σ (r) = vol(∂Br ) ∂Br respectively for the spherical mean (as in the previous chapters) and the weighted spherical mean of f , with the convention that f¯σ (r) = 0 in case the integral in (7.18) is infinite. Furthermore Br = Br (o) for some fixed origin o ∈ M as before. We prove Lemma 7.2. Let (M, , ) be a complete manifold of dimension m ≥ 2 and assume Δr ≥ (m − 1)z(r)

(7.19)

for some z(r) ∈ C 0 ([0, +∞)) in the weak sense on M . Let a(x), b(x) ∈ C 0 (M ) satisfy a(x) ≤ p(r(x)) on M \{o} (7.20) for some p(t) ∈ C 0 ((0, +∞)), b(x) ≥ 0

on M.

(7.21)

Let u ∈ C 2 (M ) be a positive solution of the differential inequality Δu + a(x)u − b(x)uσ ≥ 0, σ > 1,

on M.

(7.22)

Assume that, for any fixed 0 < δ < s, there exists a function ϕ ∈ C 2 ([δ, s]) with the following properties: ϕ + (m − 1)z(t)ϕ + p(t)ϕ ≥ 0 on [δ, s],

(7.23)


201

i) ϕ (s) =

1 ; vol(∂Bs )

−

ii) ϕ (t) ≥ 0 on [δ, s]

(7.24)

ϕ(s) = 0

(7.25)

ϕ(δ) ≤C ϕ (δ)

(7.26)

for some constant C > 0 independent of δ and s. Then, up to choosing δ sufficiently small (and consequently ϕ), there exists a constant C1 > 0, independent of δ and s, such that for all R ∈ [δ, s) u ¯(s) satisfies s u ¯(s) ≥ C1 ϕ (δ)δ m−1 − ϕ(t) vol(∂Bt ) ¯bσ (t) u ¯(t)σ dt. (7.27) R

Proof. We fix 0 < δ < s, ϕ and R ∈ [δ, s]. We consider the function ϕ(r(x)) on M ; because of (7.19), (7.20) and (7.24) ii) we have Δϕ(r)(x) ≥ −a(x)ϕ(r(x))

(7.28)

¯s \Bδ . The second Green’s identity, (7.24) i), (7.25) together in the weak sense on B with the definition (7.17) of u ¯ give uΔϕ − ϕΔu + uΔϕ − ϕΔu + u ∇ϕ, ∇r − ϕ ∇u, ∇r . u ¯(s) = Bs \BR

BR \Bδ

∂Bδ

(7.29) Observe that (7.22), (7.28), (7.21) and positivity of u yield uΔϕ − ϕΔu ≥ 0.

(7.30)

BR \Bδ

Similarly

Bs \BR

uΔϕ − ϕΔu ≥

ϕ(r)b(x)uσ .

(7.31)

Bs \BR

Next, a straightforward application of H¨ older’s inequality shows b(x)uσ ≥ ¯bσ (t)¯ u(t)σ vol(∂Bt )

(7.32)

∂Bt

where ¯bσ (t) has been defined in (7.18). Putting together (7.31), (7.32) and using the co-area formula (1.87), we finally get s uΔϕ − ϕΔu ≥ − ϕ(t)¯bσ (t)¯ u(t)σ vol(∂Bt ) dt. (7.33) Bs \BR

R

We now take care of the boundary term in (7.29). On the one hand, u ∇ϕ, ∇r − ϕ ∇u, ∇r = ϕ (δ)¯ u(δ) vol(∂bt ) − ϕ(δ) Δu. ∂Bδ

Bδ

(7.34)

202


On the other hand, vol(∂Bδ ) # δ m−1 ; vol(Bδ ) # δ m as δ ↓ 0+ .

(7.35)

Using positivity of u, (7.35), and (7.34) it is easy to see that, up to choosing 0 < δ ≤ δ0 sufficiently small, there exists a constant C1 > 0 independent of δ and s such that u ∇ϕ, ∇r − ϕ ∇u, ∇r ≥ C1 ϕ (δ)δ m−1 . (7.36) ∂Bδ

Estimate (7.27) now follows from (7.29), (7.30), (7.33) and (7.36). Remark. In case u is a radial solution of u + (m − 1)z(r)u + a(r)u − b(r)uσ ≥ 0, σ > 1 u (0) = 0, u(0) = u0 > 0 with z(r) =

h (r) h(r) ,

on [0, +∞),

(M, h) a model (see Chapter 4), (7.27) becomes

u(s) ≥ C1 ϕ (δ)δ m−1 −

s

ϕ(t)h(t)m−1 b(t)u(t)σ dt.

(7.37)

R

We apply the above results to produce a nonexistence theorem for complete Riemannian manifolds with sectional curvature bounded above by a negative constant −B 2 , B > 0. The case B = 0 can be dealt with similarly and later, in the special case of Euclidean space, we shall apply this technique to differential inequalities of a certain type. Theorem 7.3. Let (M, , ) be a complete manifold of dimension m ≥ 2 with a pole o ∈ M and suppose Riem ≤ −B 2 (7.38) for some constant B > 0. Let a(x), b(x) ∈ C 0 (M ) satisfy a(x) ≤ AB 2 coth(Br(x)) on M, with A ≤ b(x) ≥ 0

(m − 1)2 , 4

on M.

Suppose that for some constant σ > 1, ⎧ b(t) vol(∂Bt )σ−1 ⎪ ⎪ C ⎪ ⎨ tσ−1 e m−1 2 B(σ−1)t ¯bσ (t) ≥ tb(t) vol(∂Bt )σ−1 ⎪ ⎪ C ⎪ (m−1)2 ⎩ B m−1 −A∗ (σ−1)t 2 + 4 e

(7.39) (7.40)

in case A =

(m−1)2 , 4

in case A
0 and with A∗ = max {0, A},

(7.42)


203

where i) tb(t) ∈ L1 (+∞),

ii) tb(t) > 0 and nonincreasing at infinity.

(7.43)

Then the differential inequality Δu + a(x)u − b(x)uσ ≥ 0

(7.44)

has no positive C 2 -solution on M . Remark. In the proof of Theorem 7.3 we shall make use of the Hessian comparison theorem (see Chapter 1) which in the above assumptions yields Δr ≥ (m − 1)B coth Br

(7.45)

pointwise on M \{o}. Remark. We would like to comment on the case of hyperbolic space Hm −B 2 and Yamabe equation m+2

cm Δu + m(m − 1)B 2 u + k(x)u m−2 = 0, where cm = 4 m−1 m−2 . Here, vol(∂Bt ) ∼ C1 e(m−1)Bt

as t → +∞

for some C1 > 0; furthermore, a(x) =

m(m − 2) 2 B , 4

b(x) = −

1 k(x). cm

Thus, a(x) ≤ AB 2 coth(Br(x)),

with A =

m(m − 2) (m − 1)2 < . 4 4

It follows that the requirement (7.41) on ¯bσ (t) becomes ¯bσ (t) ≥ Ctb(t)e2Bt , t 1, C > 0. Now admissible choices of b(t) are b(t) =

1 , t2

b(t) =

t2

1 , log t

b(t) =

t2

1 , log t(log log t)

Thus, for instance, the above becomes 2Bt ¯bσ (t) ≥ e , t

t 1, C > 0.

. . . , t 1.

204


This is implied by b(x) ≥ C

e2Br(x) r(x)

for r(x) 1, C > 0.

(7.46)

This is expressed, in terms of k(x), as k(x) ≤ −C

e2Br(x) r(x)

for r(x) 1, C > 0.

(7.47)

Inequality (7.47) relaxes the second of (4.86) in the case of m-dimensional hyperbolic space Hm −B 2 . However, assumption (7.38) is definitely a stronger requirement than (4.84) of Theorem 4.8. Note that (7.47) is independent of the dimension of Hm −B 2 . Proof of Theorem 7.3. We reason by contradiction and we assume that (7.44) has a positive C 2 -solution u on M . Note that, by the first remark after the statement of the theorem, (7.45) holds on M . Next, we begin by considering the case A = (m−1)2 . We fix 0 < δ ≤ R < s and define 4 ϕ(t) =

m−1 t−s e 2 B(s−t) vol(∂Bs )

on [δ, s].

Then ϕ satisfies (7.23) of Lemma 7.2 with z(t) = B coth Bt. (7.24), (7.25) and (7.26) are verified up to choosing in (7.26) C = m−1 2 B. From Lemma 7.2 it follows that u ¯(s) satisfies (7.27), that is, for some constant C1 > 0 independent of δ and s, s u ¯(s) ≥ C1 ϕ (s)δ m−1 − ϕ(t) vol(∂Bt )¯bσ (t)¯ u(t)σ dt (7.48) R

for all R ∈ [δ, s). On the other hand, ϕ (δ) =

e

m−1 2 B(s−δ)

vol(∂Bs )

m−1 B 2

1 + (s − δ)

m−1

≥ C2

s e 2 Bs vol(∂Bs )

with C2 > 0 independent of s and δ ∈ (0, δ0 ] for some δ0 > 0 fixed. Furthermore, −ϕ(t) =

m−1 t − m−1 Bt s e 2 Bs 1− e 2 . vol(∂Bs ) s

Therefore we choose m−1

s e 2 Bs ψ(s) = , vol(∂Bs ) to obtain

l(t) = e−

& u ¯(s) ≥ ψ(s) C3 + C4

s R

m−1 2 Bt

vol(∂Bt )¯bσ (t)

t 1 − l(t)¯ u(t)σ dt s

'


205

that is, (7.3) of Lemma 7.1. Next we observe that ψ(t) > 0, l(t) > 0 for t sufficiently large (so that we shall choose R sufficiently large in the above). Furthermore assumption (7.41) implies that (7.2) is satisfied. From Lemma 7.1 it follows that u ¯(s) is defined on a finite interval; contradiction. 2 The case A < (m−1) is treated similarly choosing on [δ, s], 0 < δ < s, the 4 function ϕ as follows: eαB(s−t) 1 − e−αB(s−t) ϕ(t) = − α vol(∂Bs ) with m−1 α= + 2

:

(m − 1)2 − A∗ . 4

Remark. The proof above shows how (7.41) is sensitive to the coefficient A of (7.39). One can also explore the dependence of (7.41) on the possible decay of the right-hand side of (7.39) at infinity. This is a very delicate question 0 to which 1 we 2

shall return later. Note that, given (7.39), (7.41) is sharp for A ∈ 0, (m−1) . 4

We shall now apply the above result in a special geometrical setting. Consider the complete manifold Sm−p−1 × Hp+1 with the product metric (recall that Sm is the standard sphere of dimension m and that Hm is the standard hyperbolic space of dimension m: see also Chapter 2); a computation shows that its scalar curvature S(x) is S(x) = (m − 1)(m − 2 − 2p).

(7.49)

Therefore, S(x) is positive for zero for negative for

m−2 , 2 m−2 p= , 2 m−2 < p ≤ m − 2. 2

0≤p
0 on Sm−p−1 × Hp+1 . We restrict our attention to the special case k(θ, x) = k(x). Given w = w(θ, x) define 1 w(x) 6 = w(θ, x) dθ, (7.54) ωm−p−1 Sm−p−1

206


where dθ is the volume element on Sm−p−1 , can and ωm−p−1 = vol Sm−p−1 . Note that, for all σ > 1, using Hölder’s inequality we have w(x) 6 σ≤

1 ;σ (x). w ωm−p−1

(7.55)

The following result is straightforward. Lemma 7.4. Let u be a positive solution of (7.53) on Sm−p−1 × Hp+1 If k(x) ≤ 0, then u 6 satisfies m+2

cm ΔHp+1 u 6 − (m − 1)(m − 2 − 2p)6 u + ωm−p−1 k(x)6 u m−2 ≥ 0.

(7.56)

Proof. We observe, as one can easily check, that Δ = ΔSm−p−1 + ΔHp+1 . Therefore, using Stokes’ theorem we have ; = ΔS m−p−1 u + Δ Δu Hp+1 u =

1 ωm−p−1

Sm−p−1

ΔSm−p−1 u(θ, x) dθ

6. + Δ Hp+1 u = ΔHp+1 u(θ, x) = ΔHp+1 u Hence, m+2 ; = (m − 1)(m − 2 − 2p)6 u − k(x)u m−2 cm ΔHp+1 u 6 = cm Δu m+2 = (m − 1)(m − 2 − 2p)6 u − k(x)u m−2

and using k(x) ≤ 0 and (7.55) we obtain (7.56).

We are now ready to prove the following Theorem 7.5. Consider the complete manifold Sm−p−1 × Hp+1 with the product metric for m−2 < p ≤ m − 2. Then its metric cannot be conformally deformed to 2 a metric of scalar curvature k(θ, x) = k(x) ∈ C ∞ (Hp+1 ) satisfying k(x) ≤ 0 and k(x) ≤ −C

e2

2p−m+2 m−2 r(x)

r(x)

for r(x) 1,

(7.57)

where r(x) is the hyperbolic distance from a fixed origin in Hp+1 and C > 0 constant. Proof. Simply apply Theorem 7.3 to the differential inequality (7.56) reasoning by contradiction. Note the special case p =

m−2 2

k(x) ≤ −

in which (7.57) reduces to C r(x)

for r(x) 1.


7.1.1

207

A Rellich-Pohozaev formula

The aim of this section is to provide a Pohozaev formula, a tool well-known to analysts in Euclidean space, which is used in order to obtain nonexistence results. The formula can be readily generalized from Rm to a general Riemannian manifold M , but its effectiveness in this wider context depends on the possibility of choosing a conformal vector field X on M appropriately related to the problem at hand. For Euclidean space, one has the canonical simple choice X = r∇r, where r(x) is the distance function from the origin o. Similar choices are naturally available for spaces of constant sectional curvature or for models in the sense of Greene and Wu, [GW79]. However, for general Riemannian manifolds the mere existence of a globally defined conformal vector field X implies subtle geometrical restrictions, and the behavior of X and its divergence (or, alternatively, the Lie derivative of the metric in the direction of X) is difficult to understand. This is the reason why in this section we shall limit ourselves to special cases, and, in the next section, to hyperbolic space, and more specifically to its Poincaré disc B1 (0) realization. What follows can be considered as a geometric motivation of the Pohozaev formula in the particular case of the Yamabe equation in Rm : indeed, this special case relates the formula to the Kazdan-Warner obstruction obtained by integration of (2.17) (see Chapter 2, Theorem 2.3; see also the discussion in Section 5 of [PRS03a]). Let (Rm , , ) be the Euclidean space with its flat canonical metric. For the sake of simplicity let us suppose m ≥ 3. We consider the conformally related metric 4 , = u m−2 , and, as in the previous chapters, we denote by T and T the trace-free Ricci tensors of , and , respectively. Then clearly T ≡ 0 and equation (2.26) yields ! ! " " 1 1 2m −2 −1 2 T = 2u du ⊗ du − |∇u| , . (7.58) Δu , − Hess(u) + u m m−2 m Let now X = r∇r. Since Lr∇r , = 2 , , X is conformal with respect to , , and therefore also with respect to , . Recalling (2.17) we have +W + = m − 2 X(S). div 2m , to the Applying the divergence theorem on Br with respect to the metric above equation, and expressing the result in terms of the original metric we obtain 2m m−2 r ∇r, ∇S u m−2 dvolm = ru2 T(∇r, ∇r)dvolm−1 , (7.59) 2m Br ∂Br where we have used the fact that the outward unit normal to ∂Br with respect to , is u−2/(m−2) ∇r, that the m- and (m − 1)-dimensional volume elements are

208


+ m = u2m/(m−2) dvolm and dvol + m−1 = u2(m−1)/(m−2) dvolm−1 respecgiven by dvol + is defined by + , Y = T(X, Y ). Now, tively, and that, according to (2.14), W W from the expression of T and X = r∇r, setting ∂u for ∇r, ∇u we obtain ∂r

ru T(∇r, ∇r) = 2ru 2

!

. 5 " 2 1 1 2m ∂u 2 − |∇u| . Δu − Hess(u)(∇r, ∇r) + r m m−2 ∂r m

We insert this formula into (7.59) and observe that, using the expression for Δ in spherical coordinates, Hess(u)(∇r, ∇r) =

∂2u m − 1 ∂u = Δu − − Δ∂Br u. ∂r2 r ∂r

Now we apply the divergence theorem on ∂Br and use |∇u|2 =

∂u ∂r

2

+ |∇∂Br u|2∂Br ,

to deal with the term Δ∂Br u, and finally use m+2 m−2 cm Δu = −Su ,

cm = 4

m−1 m−2

to obtain, after some simplifications,

2m 2m m−2 m−2 r∇r, ∇Su dvolm = rSu dvolm−1 Br ∂Br 5 %. $ 2 ∂u 1 ∂u 2 u dvolm−1 . + mcm − |∇u| + r 2 ∂r m−2 ∂r ∂Br

(7.60)

The above formula is in the typical form of a Rellich-Pohozaev identity. We are going to describe a simpler and more general way to obtain it. Let u ∈ C 2 (M ) be a solution of the equation Δu = h(x)f (u) (7.61) on a complete Riemannian manifold (M, , ) with h ∈ C 0 (M ) and f ∈ C 0 (R). Let u F (u) = f (s) ds, 0

and define the vector field 1 W = h(x)F (u) + |∇u|2 X − X, ∇u∇u − αu∇u, 2

(7.62)


209

where X is a vector field on M and α ∈ R. Using the identities ∇|∇u|2 , X = 2 Hess(u)(∇u, X), 1 ∇u∇u, X = Hess(u)(∇u, X) + LX , (∇u, ∇u), 2 we compute div W = h(x)(F (u) div X − αuf (u)) 1 1 + div X − α |∇u|2 + F (u)∇h, X − LX , (∇u, ∇u). 2 2 If we assume that X is a conformal vector field, the above inequality becomes div W = h(x) F (u) div X − αuf (u) m−2 + div X − α |∇u|2 + F (u)∇h, X. 2m Integrating over the annulus Br \ Bro , 0 ≤ r0 < r, and using the divergence theorem we obtain r 1 ∂u ∂u 2 h(x)F (u) + |∇u| X, ∇r − X, ∇u div W = − αu . 2 ∂r ∂r Br ∂Bt ro

In particular, if (M, , ) is R with its canonical metric, and we choose X = r∇r, m+2 m−2 f (u) = u m−2 , h(x) = c−1 m S(x) and α = 2 , the above formula yields m

m−2 2m cm

∇ru m−2 r∇S, 2m

Br

0m−2

= ∂Br

2m cm

rSu

2m m−2

∂u 2 m − 2 ∂u 1r r − |∇u|2 + r + u , 2 ∂r 2 ∂r

(7.63)

ro

and, letting ro = 0, we recover (7.60). As an application of this identity, we consider the prescribed scalar curvature equation on Sm , m+2

cm ΔSm u − m(m − 1)u + K(x)u m−2 = 0. We use (the inverse of) stereographic projection to pull back the standard metric of Sm onto Rm . The pull-back metric is conformal to Euclidean metric with conformality factor 4(1 + |x|2 )−2 . Thus, the original conformal change of metric on Sm may be viewed as a conformal change of metric on Rm and we are led to investigating the equation m+2

cm Δu + b(x)u m−2 = 0

(7.64)

210


on Rm , where b(x) is bounded and admits a finite limit at infinity, and to look for positive solution u ∈ C ∞ (Rm ) such that u(x) ∼ |x|2−m

as |x| → +∞.

(7.65)

We claim that if b is radial, somewhere positive and monotone increasing where positive, then there are no positive radial solutions of (7.65) satisfying (7.65). Indeed, assume by contradiction that such a solution exists, and identify radial functions on Rm with functions on [0, +∞). Since b(x) is bounded, and u(r) ∼ r2−m as r → +∞, it follows from (7.64) that u (r) = O(r1−m ) as r → +∞. To prove this, note that since u is a radial solution of (7.64) it satisfies the ode m−1 m+2 m−1 u = −c−1 b(r)u m−2 r m r so that, integrating between ε and r, we obtain r m+2 m−1 m−1 −1 u (r) = ε u (ε) − cm b(t)tm−1 u(t) m−2 dt. r ε

Since b(t) is bounded, and u(t) ∼ t

2−m

, we have

m+2 b(t)tm−1 u(t) m−2 = O t−3 ,

so that the integral on the right-hand side is absolutely convergent at infinity, and we deduce that r

m+2 |u (r)| ≤ r1−m εm−1 |u (ε)|+c−1 |b(t)|tm−1 u(t) m−2 dt ≤ Er1−m as r → +∞ m ε

as required. Next, by the assumption on b, there exists ro ≥ 0 such that b(r) ≤ 0 for 0 ≤ r ≤ ro and b(r) > 0, b ≥ 0, b ≡ 0 for r > ro .

(7.66)

We apply equation (7.63) with b(r) instead of S and with the value of ro specified above, and let r → +∞. Since b ≥ 0, by monotone convergence, the integral over Br \Bro converges to the integral over Rm \Bro , while, by the asymptotic behavior of u and u the boundary integral over ∂Br tends to zero, and we obtain 2m m−2 m−2 ro 2 rb (r)u m−2 . (7.67) u (ro ) + u(ro )u (ro ) = − 2 2 2m c m m R \Bro Bro Now, if ro = 0, the left-hand side of the above equation vanishes, while the righthand side does not, and we immediately deduce the required contradiction. Otherwise, b ≤ 0 on Bro and it follows from (7.64) that Δu ≥ 0

on Bro ,

so that, by the maximum principle maxBro u = u(ro ) and u (ro ) ≥ 0, and, again, (7.67) yields a contradiction. We have therefore proved the following result due to W. Chen and C. Li ([CL95], Theorem 3)


211

Theorem 7.6. Let K be a nonconstant rotationally symmetric function on Sm , m ≥ 3,, and assume that K is somewhere positive, and monotone increasing in the region where it is positive. Then the equation m+2

cm Δu − m(m − 1)u + Ku m−2 = 0 has no positive smooth rotationally symmetric solutions on Sm .

7.1.2

A nonexistence result for hyperbolic space

As already mentioned in the introduction, in this section we will work in the Poincaré model for hyperbolic space. The main reason is that the Rellich-Pohozaev formula has a particularly nice expression in this setting, and allows us to give a more transparent proof of our main nonexistence result. We leave it to the interested reader to translate the result in the standard model Hm = Rm with metric in polar coordinates on Hm \ {0} = (0, +∞) × Sm−1 given by , Hm = dr2 + sinh2 dθ2 , which has been used throughout the book. We recall that the Poincaré model of Hm is (Bm , gH ), where Bm is the unit ball of Rm and gH is the metric conformally related to the canonical Euclidean metric , given by gH =

4 , , (1 − |x|2 )2

(7.68)

where, using standard notation, |x| = distRm (x, 0). We state the following problem: Given K ∈ C ∞ (Bm ), m ≥ 3, does there exist a positive function v on Bm such that 4

(i) gv = v m−2 gH has scalar curvature K and (ii) gv is a complete Riemannian metric on Bm ? Setting

5 u(x) = v(x)

. m−2 4

4 (1 − |x|2 )

2

,

we have that the above problem admits an affirmative answer if and only if the equation m+2 cm Δu + K(x)u m−2 = 0, (7.69)

212


where cm = 4 m−1 m−2 (as in the previous chapters) and Δ is the Euclidean Laplacian, admits a positive solution u such that the conformal metric 4

gu = u m−2 , is a complete Riemannian metric on Bm . We note that the requirement of completeness implies that u must diverge at the boundary of Bm , while equation (7.69) of course means that K is the scalar curvature of gu . We also recall that, letting r = |x| , then the Riemannian distance ρ from the origin of Bm with respect to the hyperbolic metric (7.68) is given by (see [RRV94b]) 1+r , (7.70) ρ = log 1−r while

ρ

eρ − 1 . (7.71) 2 eρ + 1 For the developments below, we will use the Rellich-Pohozaev-type formula which is obtained in the following way. Fix xo ∈ Bm , let X = x − xo and define a vector field by the formula & ' 2m cm m−2 2 m−2 W = (x−xo )−cm x−xo , ∇u∇u+cm αu∇u. (7.72) |∇u| − Ku 2 2m r = tanh

=

Using the fact that Lx−xo , = , , a computation similar to the one carried out in the previous section shows that 0 1 2m m−2 m−2 m−2 K− div W = cm − α |∇u|2 + α − x − xo , ∇K u m−2 , 2 2 m (7.73) and we obtain a Rellich-Pohozaev formula integrating and using the divergence theorem. We are now ready to state (see [RRV94b], Theorem 1.2) Theorem 7.7. Let K ∈ C ∞ (Bm ) and suppose that lim sup(1 − |x|2 )K(x) < 0. |x|→1−

Assume also that either (i) K ≤ 0 in Bm , or and xo ∈ Bm such that (ii) there exists α ≤ m−2 2 m−2 m−2 K(x) − α− x − xo , ∇K(x) ≥ 0 on Bm , 2 2m with strict inequality in at least one point.

(7.74)


213

Then equation (7.69) does not admit any positive solutions on Bm . Note that if we assume that u is nonnegative and satisfies (7.69), then, by the strong maximum principle (see, [GT01], p. 35), either u > 0 or u ≡ 0. We split the proof into several steps. We note that steps 1 to 3 are devoted to showing that the solution u(x) must tend to zero in a specific way as the point x tends to the boundary of Bm . This fact could be obtained as an application of Theorem 4.4 in Chapter 4. However we follow the alternative method presented below to describe a technique of common use in Analysis. Proof. We assume that (7.69) admits a positive solution u and derive a contradiction. Step 1. Assumption (7.74) implies that u is bounded. To prove this fact we first establish the following Proposition 7.8. Let BR ⊂ Rm be the ball of radius R centered at 0, let σ > 1, and let f be a positive function in C 0 (B R ). Then there exists a constant C = C(σ, m) > 0 such that, if u ∈ C 2 (BR ) satisfies the differential inequality Δu ≥ f (x)uσ on BR , then

1

u(o) ≤ C(R2 min f )− σ−1 .

(7.75) (7.76)

BR

Proof. Set A = minB R f , so that u satisfies Δu ≥ Auσ . For 0 < R1 < R define where

λ=

(7.77) 2

v(x) = λ(R12 − |x|2 )− σ−1 , B 2 R A 1

1 σ−1

B = max

) 2m σ+1 * . ,4 σ − 1 (σ − 1)2

A straightforward computation shows that Δv ≤ Av σ on BR1 ,

(7.78)

and v(x) → +∞ as |x| → R1 . Thus the function w = u − v attains an absolute maximum at a point xo ∈ BR1 , and satisfies the inequality Δw ≥ A(uσ − v σ ) = h(x)w, where h ∈ C 0 (BR1 ) is the nonnegative function defined by # u(x) σ−1 σA t dt if u(x) = v(x), v(x) h(x) = u(x)−v(x) if u(x) = v(x). σAu(x)σ−1

(7.79)

214


If w(xo ) > 0, then h(xo ) > 0, and Δw(xo ) ≥ h(xo )w(xo ) > 0, which is impossible since w attains a maximum at xo . Thus for every x ∈ BR1 , w(x) ≤ w(xo ) ≤ 0, that is u(x) ≤ v(x) in BR1 , and in particular, u(0) ≤ v(0) =

B A

1 σ−1

2 − σ−1

R1

− 1 1 = B σ−1 AR12 σ−1 ,

and letting R1 → R we obtain the desired estimate.

To prove the claim in Step 1, let u be a positive solution of (7.69). We fix γ in (0, 1) and applying Proposition 7.8 in the ball B γ (x) = {y ∈ Rm : |y − x| ≤ γ(1 − |x|)}, we deduce that − m−2 & 4 (−K(z)) . u(x) ≤ C (1 − |x|)2 max γ B (x)

Using (7.74) we deduce that u is bounded above as |x| → 1− and, it is therefore bounded. Step 2. We show that lim|x|→1− u(x) = 0. By (7.74) there exists constants 0 ≤ η < 1 and c > 0 such that c K(x) ≤ − (1 − |x|)−2 2

(7.80)

on the set Annη,1 = B1 \ B η = {x ∈ Bm : η < |x| < 1}. Let δ ∈ (η, 1) and let vδ be the maximal solution of the problem m+2 cm Δv = 2c (1 − |x|)−2 v m−2 on Annη,δ , lim|x|→η v(x) = lim|x|→δ v(x) = +∞, which exists by Lemma 6.4 in Chapter 6, and is radially symmetric since it must coincide with the maximal radial solution to (7.1.2). Since vδ (x) diverges as x tends to the boundary of Annη,δ , by usual comparison arguments, u < vδ on this set. Moreover, as δ → 1− , vδ decreases and converges to a radially symmetric function v which satisfies m+2 c cm Δv = (1 − |x|)−2 v m−2 on Annη,1 , 2


215

and 0 ≤ u(x) ≤ v(|x|). Therefore, it suffices to show that lim v(|x|) = 0.

|x|→1−

To this end, we first note that, arguing as in Step 1, we can show that v is bounded in a neighborhood of ∂Bm . Next, since v is radial, it satisfies the ode cm (rm−1 v ) =

m+2 c m−1 (1 − r)−2 v m−2 , r 2

for η < r < 1. It follows that the function rm−1 v (r) is increasing in η < r < 1, hence it has constant sign in a left neighborhood of 1. Thus v itself tends to a nonnegative limit β as r → 1− . Assume by contradiction that β > 0. Then there exists η < τ < 1 and a constant θ > 0 such that (rm−1 v ) ≥ θrm−1 (1 − r)−2 , and integrating twice we deduce that v(r) → +∞ as r → 1− contradicting the fact that v is bounded. The contradiction shows that β = 0 and therefore limr→1− u(x) = 0. Step 3. There exists a constant M > 0 such that u(x) ≤ M (1 − |x|).

(7.81)

Indeed, the function ψ(x) = M (1 − |x|) is a supersolution of (7.69) in Annη,1 and, by taking M large enough we can arrange that u(x) ≤ ψ(x) on ∂Annη,1 . Thus, the usual comparison principle shows that u ≤ ψ on Annη,1 , and by increasing M if necessary, the inequality holds on Bm . Step 4. Suppose that K ≤ 0 on Bm . Then u is a nonnegative subharmonic function which vanishes on ∂Bm , and by the maximum principle u ≡ 0. Step 5. Using the Pohozaev-type formula (7.73), we prove that, if K satisfies condition (ii) in the statement of the theorem, then (7.69) has no positive solution on Bm .

216

Chapter 7. Some special cases m

Suppose first that u ∈ C 1 (B ). We fix ε > 0, integrate (7.73) over B1−ε and let ε → 1− . Using the fact that u = 0 on ∂Bm−1 , so that ∇u = ∂u ∂r ∇r, we obtain cm m−2 2 − |∇u, ∇r| x − xo , ∇u = cm − α |∇u|2 2 ∂Bm 2 Bm " ! 2m m−2 m−2 K− α− + x − xo , ∇K u m−2 . 2 2m Bm Since the left-hand side of the equality is nonpositive, and α ≤ m−2 we conclude 2 that " ! 2m m−2 m−2 K− α− (7.82) x − xo , ∇K u m−2 ≤ 0. 2 2m m B On the other hand, condition (ii) in the statement of the theorem and u > 0 imply that the left-hand side is strictly positive, and the required contradiction follows. However, in general we do not know that u is C 1 up to the boundary, and we need to justify that one may take the limit as ε → 0 to obtain (7.82). This can be done by the following approximation technique. Write K = K + − K − , and, for a fixed δ > 0, let 1 Kδ− (x) = min{ , K − (x)}, δ and let v = vδ be the solution of the problem m+2 m+2 cm Δv − Kδ− v m−2 = −K + u m−2 v = 0 on ∂Bm .

(7.83)

in Bm ,

(7.84)

m+2

By the asymptotic behavior of K, the function K + u m−2 is Lipschitz and compactly supported in Bm , and since Kδ− is Lipschitz and bounded on Bm , the existence of v is guaranteed by a minimization argument, and by standard regularity, v is C 2 m in B . Moreover, if δ1 > δ2 , then Kδ−1 ≤ Kδ−2 , and by comparison, v δ2 ≤ v δ1

in Bm .

(7.85)

Recalling the equations satisfied by vδ and u and noting that Kδ− ≤ K − , we deduce that the function w = u − v satisfies m+2

m+2

cm Δw ≥ Kδ− (u m−2 − v m−2 ) = h(x)Kδ− w, m+2 . Since Kδ− ≥ 0 in Bm , and u = v = 0 with h(x) ≥ 0 defined by (7.79) and σ = m−2 m on ∂B , by the maximum principle w ≥ 0 and therefore u ≤ vδ for every δ > 0. Thus as δ → 0, vδ converges to a function vo which vanishes on ∂Bm and satisfies vo ≥ u. Now, by Step 3, vδ is a solution of (7.84) and is uniformly bounded on Bm . Moreover, if r < 1, Kδ− = K on Br for δ small enough, so that, by standard


217

elliptic estimates (see, e,g., [Aub98], Theorem 4.40), we deduce that vδ is uniformly bounded in C 1 (Br ) and therefore it converges to vo uniformly on Br . On the other hand, again by Step 3, vδ satisfies the estimate (7.81) and therefore {|∇r, ∇vδ |} is bounded and decreasing, and it follows that vδ tends to zero as x → ∂Bm uniformly in δ. Combining these two facts we conclude easily that vδ → vo uniformly on Bm . Integrating equation (7.74) against a test function and letting δ → 0 shows that vo belongs to W 1,2 (Bm ) ∩ C 0 (Bm ) and it is a weak solution of m+2

m+2

cm Δvo = K − vom−2 + K + u m−2 . Thus w = vo − u is a nonnegative weak solution of m+2 m+2 m−2 − m−2 = h(x)w, −u Δw = K vo where h ≥ 0 is defined in (7.79). Since w = 0 on ∂Bm , by comparison w ≤ 0 and therefore w ≡ 0, that is vo = u. Next we fix δ > 0 and we consider the vector field ! "' & 2m m+2 m−2 cm − m−2 2 + m−2 Kδ v δ (x − xo ) Wδ = −K u vδ |∇vδ | + 2 2m − cm x − xo , ∇vδ ∇vδ + cm αvδ ∇vδ . Writing v instead of vδ , a computation similar to that leading to (7.73) shows that 2m m+2 m−2 m−2 − α |∇v|2 + x − xo , ∇Kδ− vδm−2 − ∇K + u m−2 vδ divWδ = cm 2 2m 2m m+2 m−2 Kδ− vδm−2 − K + u m−2 vδ − L(v), (7.86) + α− 2 where L(v) = L(vδ ) =

m+2 4 m + 2 + m−2 m + 2 + m−2 K vu K vu + x − xo , ∇u 2 m−2 m+2 m + 2 m−2 + x − xo , ∇K + . vu 2m

Integrating (7.86) over Bm , noting that v is C 1 on Bm and vanishes on ∂Bm , and using the divergence theorem we obtain cm − x − xo , ∇r∇v, ∇r2 2 ∂Bm m+2 2m m−2 m−2 = cm − α |∇v|2 + x − xo , ∇Kδ− v m−2 − ∇K + u m−2 v 2 2m Bm

m+2 2m m−2 ∇Kδ− v m−2 − ∇K + u m−2 v − L(v). −α + 2 Bm Bm

218


Since K + has compact support in Bm , as δ → 0, the integral L(vδ ) converges to L(u). Moreover, L(u) =

m+2 2m m + 2 + m−2 m + 2 + m−2 + x − xo , ∇u K u K u 2 m−2 2m 2m m + 2 m−2 m+2 + x − xo , ∇K + = u div K + u m−2 (x − xo ) , 2m 2m

so that L(u) = 0 and L(vδ ) = o(δ) as δ → 0. Summing up, recalling that α ≤ m−2 2 , we have obtained that 0 m+2 2m m − 2 + m−2 α− − Kδ− v m−2 o(δ) + K vu 2 Bm 1 m+2 2m m−2 − x − xo , vu m−2 ∇K + − v m−2 ∇Kδ− , < 0 2m

(7.87)

with o(δ) → 0 as δ → 0. Next, we rewrite the integrand in the above formula as 0 1 2m m−2 m−2 (α − )(K + − Kδ− ) − x − xo , ∇(K + − Kδ− ) v m−2 2 2m 0 1 m+2 2m m−2 + m−2 + (α − )K − x − xo , ∇K + vu m−2 − v m−2 , 2 2m and note that since K + is compactly supported in Bm and vδ → u uniformly, the integral of the second summand in tends to zero as δ → 0. Hence, (7.87) yields 0 1 2m m−2 m−2 lim sup (α − )(K + − Kδ− ) − x − xo , ∇(K + − Kδ− ) v m−2 ≤ 0. 2 2m δ→0 Bm (7.88) On the other hand, let Gδ = {x : K(x) < −1/δ}, so that Gδ is open and K = −Kδ− = −1/δ on Gδ . It follows that ∇K = −∇Kδ− = 0 on Gδ , and, using again − − α ≤ m−2 2 , the fact that Kδ = K in the set where K ≥ −1/δ and using condition (ii) in the statement we deduce that

α−

m − 2 + m−2 (K − Kδ− )(x) − x − xo , ∇(K + − Kδ− )(x) ≥ 0 2 2m

(7.89)

almost everywhere on Bm , so that the limit in (7.88) exists and is equal to zero. Finally, since the integrand in (7.88) is nonnegative, and, as δ → 0, tends to 1 2m m − 2 + m−2 α− (K − K − )(x) − x − xo , ∇(K + − K − )(x) u m−2 2 2m almost everywhere in Bm , we conclude from Fatou’s lemma that " ! 2m m − 2 m−2 α− K− x − xo , ∇K u m−2 ≤ 0, 2 2m m B and, using (ii) and u > 0 a contradiction is reached as above.

(7.90)


7.1.3

219

An integral obstruction

The aim of this subsection is to give a necessary condition for the existence of a metric conformally related to the Poincaré metric (7.68) on Bm with assigned scalar curvature K(x); we recall that, as observed before, this problem consists in finding a positive solution of equation (7.69). The nature of the obstruction that we are going to describe is manifestly different from that of the Kazdan-Warner condition (see Chapter 2) which, basically, succesfully applies only in the compact setting. However, its generalization to a generic complete manifold is still missing. We have the following (see [RRV94b], Proposition 1.1) Proposition 7.9. Suppose that K ∈ L1 (Bm ). If equation (7.69) admits a positive solution u such that lim u(x) = +∞, (7.91) − |x|→1

then the following inequalities hold: B $

m

1

r1−m

K(x) < 0; %

(7.92)

K(x) dr < 0

(7.93)

|x| ε > 0 we consider a Proof. Step 1. First we prove (7.92). Let γ = m−2 family of nonnegative, nondecreasing functions ρε (r), r ∈ [0, 1), such that 0 if 0 ≤ r ≤ 2ε , (7.94) ρε (r) = r − 2ε if ε ≤ r < 1.

We multiply equation (7.69) by ρε cm Δu ρε

1 uγ

1 uγ

to obtain

+ Kuγ ρε

1 uγ

= 0.

(7.95)

The function ρε u1γ (x) is compactly supported in Bm , because of (7.91); thus we can apply the first Green’s formula in (7.95) to obtain - , 1 1 γ ∇u, ∇ ρε γ + (7.96) Ku ρε γ = 0. −cm u u Bm Bm An easy computation shows that the first integral in (7.96) is equal to 1 2 u−γ−1 ρε γ |∇u| . cm γ u Bm

220


Since ρε > 0 if u < ε1γ , we deduce that, for ε > 0 small enough, 1 1 2 Kuγ ρε γ = −γ u−γ−1 ρε γ |∇u| < 0. u u Bm Bm

(7.97)

We let ε → 0; since 0 < ρεr(r) < 1, by Lebesgue’s theorem the left-hand side term in (7.97) admits a limit, which is K(x) ≤ 0. (7.98) Bm

To be more precise, we can arrange to have that the function ε → ρε (r) is nondecreasing for all r ≥ 0; in this way, we conclude that the right-hand side in (7.97) decreases to its limit as ε tends to 0. Thus we find 2 K(x) = −γ u−α−1 |∇u| < 0. (7.99) Bm

Bm

Step 2. We prove now (7.93). First we observe that 2 div u−α ∇u = u−α Δu − αu−α−1 |∇u| .

(7.100)

Next, we rewrite equation (7.69) as −cm u−α Δu = K.

(7.101)

We use (7.100) in (7.101) and integrate over |x| < r to obtain 1 2 K(x) = −γ u−γ−1 |∇u| − u−α ∇u, ν , cm |x| 0 is small enough.

Using Lemma 7.18 we now proceed similarly to what we did in the proof of Proposition 7.16 to establish Proposition 7.19. Let a(t) ∈ C 0 ([0, +∞)), a(t) ≥ 0 and satisfying, for m ≥ 3, a(t) ≤

i)

A2 t2

ii) min

)

(m−2)2 A2 4t2 , t2+ε

*

for 0 ≤ A ≤

m−2 2 ,

for some ε > 0, A > 0.

(7.165)

Then the equation Δu + a(|x|)u = 0 has a positive radial solution u ∈ C 2 (Rm ) satisfying, for some constant C > 0 and |x| 1, in case (7.165) i),

C −1 (log |x|)|x|−

m−2 2

≤ u(x) ≤ C(log |x|)|x|−

m−2 2

C −1 |x|−γ ≤ u(x) ≤ C|x|−γ where γ =

1 2

m−2−

if A = if 0 ≤

m−2 2 , A < m−2 2

(7.166)

( (m − 2)2 − 4A2 ,

while in case (7.165) ii), C −1 ≤ u(x) ≤ C. We now go back to Lemma 7.18 to provide a

(7.167)

7.3. The case of Euclidean space and further results

235

Proof of the claim. We divide the argument into three steps. Step 1. We begin by considering the differential equation ϕ +

m−1 A2 ϕ + 2+ε = 0. t t

(7.168)

is not an integer, the general Assuming, without loss of generality, that ν = m−2 ε solution is given in terms of the Bessel function of the first kind by m−2 m−2 2A − ε 2A − ε ϕ(t) = C1 t− 2 Jν t 2 + C2 t− 2 J−ν t 2 , ε ε see [Wat66] page 97. Given s > 0 we define ϕ1,s (t) =

Γ(ν)Γ(−ν) − m−2 − m−2 (7.169) s 2 t 2 " ! ε 2A − ε 2A − ε 2A − ε 2A − ε 2 2 2 2 Jν + Jν J−ν × −J−ν s t s t ε ε ε ε

so that ϕ1,s is a solution of (7.168) satisfying ϕ1,s (s) = 0.

4

2 λ Since Jλ (x) behaves like πx cos x − λπ 2 − 4 as x → +∞, the function ϕ1,s is not of definite sign in (0, s); however, we claim that there exists p1 independent of s such that, for all s > p1 , ϕ1,s (t) < 0 in [p1 , s). Indeed, from the power series representation of Bessel functions Jλ (x) =

+∞ x λ

2

0

k

(−1)k (x/2) , Γ(k + 1)Γ(k + λ + 1)

it follows easily that Jν (x) (resp. Γ(1 − ν)J−ν (x)) is positive and increasing (resp. decreasing) in an interval (0, x+ ) (resp. (0, x− )). By rewriting the right-hand side of (7.169) as Γ(ν) 2A − ε 2A − ε 2 2 − m−2 m−2 Jν Γ(1 − ν)J−ν t s ε ε εt 2 s 2 5 2A − ε 2A − ε . Jν ε s 2 Γ(1 − ν)J−ν ε t 2 , × 1 − 2A ε − 2ε Jν ε t− 2 Γ(1 − ν)J−ν 2A ε s we may therefore conclude that there exists p1 > 0, which depends only upon x+ and x− , such that ϕ1,s (t) is negative for p1 ≤ t < s. We will also need to know the asymptotic behavior of ϕ1,s (t) and ϕ1,s (t) for s and t large. Using the formula x λ Jλ (x) =

2

Γ(1 + λ)

+ o(xλ )

as x → 0,

236


which can be read off from the power series representation above, and the recurrence formulas for the derivative of Jλ ,

xλ Jλ (x)

= xλ Jλ−1 (x),

x−λ Jλ (x)

= x−λ Jλ+1 (x),

a lengthy but straightforward computation shows that s2−m − t2−m 2−m + s2−m (ηt + ηs ), + t m−2 εA2 −(m−1) ϕ1,s (t) = t + s2−m t−ε−1 + t1−m (ηs + ηt ) (m − 2)(m − 2 − ε) ϕ1,s =

+ s2−m t−ε−1 (ηs + ηt ), where limt→+∞ ηt = 0 and lims→+∞ ηs = 0. This immediately gives lim sm−1 ϕ1,s (s) = 1

s→+∞

which implies that there exists p2 ≥ p1 , independent of s, such that |ϕ1,s (t)| ≤

2m − 1 2−m t , 2(m − 2)2

4m − 9 1−m ≤ ϕ1,s (t) ≤ 2t1−m t 4(m − 2)

(7.170)

for all t, p2 ≤ t ≤ s. Step 2. Fix p ∈ (p2 , s) and let ϕ2 (t) = ϕ2,p,s (t) be the solution of the differential equation m − 1 (m − 2)2 ϕ + ϕ = 0 on (0, +∞) ϕ + t 4t2 satisfying ϕ2 (p) = ϕ1,s (p), ϕ2 (p) = ϕ1,s (p). It is easy to verify that ϕ2 is given by the formula ! p " m−2 2−m 2−m m−2 ϕ2 (t) = p 2 ϕ1,s (p)t 2 − ϕ1,s (p) + pϕ1,s (p)t 2 log . 2 t According to (7.170) above, we have m − 3 2−m m−2 , ϕ1,s (p) + pϕ1,s (p) > p 2 2(m − 2) which implies that ϕ2 (t) < 0 A2 t2+ε

for 0 < t ≤ p.

Step 3. Let t0 be such that a(t) ≤ for t ≥ t0 and let p = max {t0 , p2 }. Define ϕ2,p,s (t) for 0 < t ≤ p, ϕs (t) = ϕ1,s (t) for t > p.

7.3. The case of Euclidean space and further results

237

By steps 1 and 2 above, ϕs is a piecewise C 2 -solution of m−1 ϕ (t) + a(t)ϕ(t) ≥ 0 t

ϕ (t) +

which is negative in (0, s], vanishes for t = s and satisfies lim sm−1 ϕs (s) = 1.

s→+∞

To conclude the proof of our claim, it remains to be shown that ϕs satisfies condition (iii), that is, there are constants c1 and c2 independent of s such that |ϕs (δ)| ≤ c1

1 log δ

− m−2 2 and

ϕs (δ)

≥ c2

1 −m δ 2 log δ

for δ small enough. Indeed, if δ < p the definition of ϕ2 and (7.170) imply |ϕs (δ)| ≤ p

m−2 2 +1

ϕ1,s (p)δ

m−2 2

log

2−m p ≤ 2p 2 (log p + 1) log δ

1 2−m δ 2 . δ

On the other hand, a simple computation gives ! m−2 m−2 m−2 ϕs (δ) = p 2 − ϕ1,s (p) + ϕ1,s (p) + pϕ1,s (p) 2 2 p " m m−2 δ− 2 × 1+ log 2 δ p m m − 2 m−2 ≥ p 2 mϕ1,s (p) + pϕ1,s (p) log δ− 2 , 2 δ whence again by (7.170) ϕs (δ) ≥

p m m − 3 2−m p 2 log δ− 2 4 δ

and the proof of the claim is completed.

As a first consequence of Proposition 7.19 we give the next uniqueness result for ground states (see Chapter 3 for the definition) of a Yamabe type equation in Rm . This is exactly the subtle case for uniqueness and in what follows we shall apply Corollary 5.8. Theorem 7.20. Let m ≥ 3 and a(x), b(x) ∈ C 0 (Rm ), b(x) ≥ 0, b(x) ≡ 0, with ⎧ ⎨i) A22 for 0 ≤ A ≤ m−2 |x| 2 , * ) a(x) ≤ (7.171) 2 (m−2)2 ⎩ii) min 4|x|2 , |x|A2+ε for some ε > 0, A > 0.

238


Let u(x) and v(x) be positive solutions of Δu + a(x)u − b(x)uσ = 0

on Rm .

Assume, as |x| → +∞, in case (7.171) i), ⎧

m−2 ⎪ ⎨u(x) − v(x) = o (log |x|)|x|− 2

√ − 1 m−2− (m−2)2 −4A2 ⎪ ⎩u(x) − v(x) = o |x| 2

if A =

m−2 2 ,

if 0 ≤ A
0} = ∅ and, in case (7.171) i), ⎧& '−1 0 1 β+1 ⎪ H (2−β) ⎪ − m−2 2(β+1) ⎨ # 2 (log |x|)|x| u ∈ L1 (+∞) ∂Br ) * −1 ⎪ # β+1 ⎪ ⎩ |x|−γ H (2−β) u2(β+1) ∈ L1 (+∞) ∂Br

( with γ = 12 m − 2 − (m − 2)2 − 4A2 , while in case (7.171) ii) & u

2(β+1)

'−1

∂Br

for some β > 1 and max 0, A ≤ β ≤ H − 1.

∈ L1 (+∞)

if A =

m−2 2 ,

if 0 ≤ A

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