Blow-up theory for elliptic PDEs in Riemannian geometry

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Author: Olivier Druet | Emmanuel Hebey | Frederic Robert

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Blow-up Theory for Elliptic PDEs in Riemannian Geometry

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Blow-up Theory for Elliptic PDEs in Riemannian Geometry

Olivier Druet Emmanuel Hebey ´ eric ´ Robert Fred

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

c 2004 by Princeton University Press Copyright Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 3 Market Place, Woodstock, Oxfordshire OX20 1SY All Rights Reserved The publisher would like to acknowledge the authors of this volume for providing the camera-ready copy from which this book was printed. Library of Congress Cataloging-in-Publication Data Druet, Olivier, 1976– Blow-up theory for elliptic PDEs in Riemannian geometry / Olivier Druet, Emmanuel Hebey, Frédéric Robert. p. cm Includes bibliographical references. ISBN: 0-691-11953-8 (pbk.: alk paper) 1. Calculus of Variations. 2. Differential Equations, Nonlinear. 3. Geometry, Riemannian. I. Hebey, Emmanuel, 1964– II. Robert, Frédéric, 1974– III. Title. QA315.D78 515.’353–dc22

2004 2003064801

British Library Cataloging-in-Publication Data is available Printed on acid-free paper. pup.princeton.edu Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

Contents

Preface Chapter 1. Background Material 1.1 1.2

Riemannian Geometry Basics in Nonlinear Analysis

Chapter 2. The Model Equations 2.1 2.2 2.3 2.4

Palais-Smale Sequences Strong Solutions of Minimal Energy Strong Solutions of High Energies The Case of the Sphere

Chapter 3. Blow-up Theory in Sobolev Spaces 3.1 3.2 3.3

The H12 -Decomposition for Palais-Smale Sequences Subtracting a Bubble and Nonnegative Solutions The De Giorgi–Nash–Moser Iterative Scheme for Strong Solutions

Chapter 4. Exhaustion and Weak Pointwise Estimates 4.1 4.2

Weak Pointwise Estimates Exhaustion of Blow-up Points

Chapter 5. Asymptotics When the Energy Is of Minimal Type 5.1 5.2

Strong Convergence and Blow-up Sharp Pointwise Estimates

Chapter 6. Asymptotics When the Energy Is Arbitrary 6.1 6.2 6.3

A Fundamental Estimate: 1 A Fundamental Estimate: 2 Asymptotic Behavior

vii 1 1 7 13 14 17 19 23 25 26 32 45 51 52 54 67 68 72 83 88 143 182

Appendix A. The Green’s Function on Compact Manifolds

201

Appendix B. Coercivity Is a Necessary Condition

209

Bibliography

213


Preface Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3. We let (hα ) be a sequence of smooth functions on M and consider equations like ∆g u + hα u = u2

−1

(Eα )

where ∆g = −divg ∇ is the Laplace-Beltrami operator, 2 = 2n/(n − 2) is the critical Sobolev exponent for the embedding of the Sobolev space H12 (M ) into Lebesgue’s spaces, and u is required to be positive. Such equations have been the target of investigation for decades. For instance, if hα = ((n − 2)Sg ) / (4(n − 1)), where Sg is the scalar curvature of g, then (Eα ) is the Yamabe equation. We let (uα ) be a bounded sequence in H12 (M ) of solutions of (Eα ) in the sense that, for any α, 2 −1 ∆g uα + hα uα = uα

and uα H12 ≤ Λ where Λ > 0 is independent of α. We know from important work developed in the 1980s how to describe the H12 -asymptotic behavior of the uα ’s as α → +∞. Let us assume, for instance, that the hα ’s are uniformly bounded and that they converge L2 to some function h∞ . Up to a subsequence, the asymptotic description with respect to the H12 -Sobolev space then gives that u α = u0 + Bαi + Rα (I1 ) where u0 is a solution of the limit equation ∆g u + h∞ u = u2

−1

,

(E∞ )

the sum in the right hand side of (I1 ) is a finite sum over i, Bαi is a bubble, obtained by rescaling fundamental positive solutions of the Euclidean equation ∆u = u2 −1 , and the Rα ’s are lower order terms which converge strongly to 0 in H12 (M ). This asymptotic description provides a very nice H12 -theory for the asymptotic behavior of solutions of equations like (Eα ). Let us assume now that the hα ’s converge C 0,θ to h∞ for some 0 < θ < 1. An important issue in the study of equations like (Eα ) is to get a theory in which the above asymptotic description holds also in the C 0 -space, where pointwise estimates are involved. Developing such a C 0 -theory is the purpose of these notes. We wanted these notes to be as self-contained as possible. They should be accessible to graduate students and researchers in other fields. For the sake of clearness, we decided to present our theory in the specific case of equations like (Eα ). However, the material we present in these notes is applicable to more general equations.

viii

PREFACE

In some aspects, this includes variants of equations (Eα ) involving the p-Laplacian. It also includes variants of equations (Eα ) involving fourth order operators like the square of the Laplacian. Chapter 1 is devoted to background material in Riemannian geometry and nonlinear analysis on manifolds. The existence of Palais-Smale sequences for equations (Eα ), and of strong solutions of minimal or arbitrary energies to equations (Eα ), is briefly discussed in Chapter 2. We present the H12 -decomposition theorem in Chapter 3. Another construction, providing weak pointwise estimates, is presented in Chapter 4. Chapter 5 describes the C 0 -theory we mentioned above when the solutions are of minimal type. The C 0 -theory in the general case of arbitrary energies is presented in Chapter 6. The authors thank Ellen Foos, Vickie Kearn, Alison Kalett, Jennifer Slater and the Princeton University Press for their constant support, their efficiency, and the wonderful job they did in the preparation of the manuscript.

The Authors Paris, July 2003

Chapter One Background Material We recall in this chapter basic facts concerning Riemannian geometry and nonlinear analysis on manifolds. For reasons of length, we are obliged to be succinct and partial. Possible references are Chavel [20], do Carmo [22], Gallot-Hulin-Lafontaine [36], Hebey [43], Jost [50], Kobayashi-Nomizu [53], Sakai [65], and Spivak [72]. As a general remark, we mention that Einstein’s summation convention is adopted: an index occurring twice in a product is to be summed. This also holds for the rest of this book. 1.1 RIEMANNIAN GEOMETRY We start with a few notions in differential geometry. Let M be a Hausdorff topological space. We say that M is a topological manifold of dimension n if each point of M possesses an open neighborhood that is homeomorphic to some open subset of the Euclidean space Rn . A chart of M is then a couple (Ω, ϕ) where Ω is an open subset of M , and ϕ is a homeomorphism of Ω onto some open subset of Rn . For y ∈ Ω, the coordinates of ϕ(y) in Rn are said to be the coordinates of y in(Ω, ϕ). An atlas of M is a collection of charts (Ωi , ϕi ), i ∈ I, such that M = i∈I Ωi . Given an atlas (Ωi , ϕi )i∈I , the transition functions are ϕj ◦ ϕ−1 : ϕi (Ωi ∩ Ωj ) → ϕj (Ωi ∩ Ωj ) i with the obvious convention that we consider ϕj ◦ ϕ−1 i if and only if Ωi ∩ Ωj = ∅. The atlas is then said to be of class C k if the transition functions are of class C k , and it is said to be C k -complete if it is not contained in a (strictly) larger atlas of class C k . As one can easily check, every atlas of class C k is contained in a unique C k -complete atlas. For our purpose, we will always assume in what follows that k = +∞ and that M is connected. One then gets the following definition of a smooth manifold: A smooth manifold M of dimension n is a connected topological manifold M of dimension n together with a C ∞ -complete atlas. Classical examples of smooth manifolds are the Euclidean space Rn itself, the torus T n , the unit sphere S n of Rn+1 , and the real projective space Pn (R). Given two smooth manifolds, M and N , and a smooth map f : M → N from M to N , we say that f is differentiable (or of class C k ) if for any charts (Ω, ϕ) and ˜ ϕ) ˜ the map (Ω, ˜ of M and N such that f (Ω) ⊂ Ω, −1 ˜ ϕ˜ ◦ f ◦ ϕ : ϕ(Ω) → ϕ( ˜ Ω) k is differentiable (or of class C ). In particular, this allows us to define the notion of diffeomorphism and the notion of diffeomorphic manifolds.

2

CHAPTER 1

We refer to the above definition of a manifold as the abstract definition of a smooth manifold. As a surface gives the idea of a two-dimensional manifold, a more concrete approach would have been to define manifolds as submanifolds of Euclidean space. According to a well-known result of Whitney, any paracompact (abstract) manifold of dimension n can be seen as a submanifold of some Euclidean space. Let us now say some words about the tangent space of a manifold. Given M a smooth manifold and x ∈ M , let Fx be the vector space of functions f : M → R which are differentiable at x. For f ∈ Fx , we say that f is flat at x if for some chart (Ω, ϕ) of M at x, D f ◦ ϕ−1 ϕ(x) = 0. Let Nx be the vector space of such functions. A linear form X on Fx is then said to be a tangent vector of M at x if Given Nx ⊂ KerX. We let Tx (M ) be the vector space of such tangent ∂vectors. ∈ T (Ω, ϕ) some chart at x, of associated coordinates xi , we define ∂x x (M ) i x by, for any f ∈ Fx , ∂ · (f ) = Di f ◦ ϕ−1 ϕ(x) . ∂xi x ∂ As a simple remark, one gets that the ∂x ’s form a basis of Tx (M ). Now, one i x defines the tangent bundle of M as the disjoint union of the Tx (M )’s, x ∈ M . If M is n-dimensional, one can show that T (M ) possesses a natural structure of a 2n-dimensional smooth manifold. Given a chart (Ω, ϕ) of M , Tx (M ), Φ x∈Ω

is a chart of T (M ), where for X ∈ Tx (M ), x ∈ Ω, Φ(X) = ϕ1 (x), . . . , ϕn (x), X(ϕ1 ), . . . , X(ϕn ) [the coordinates of x in (Ω, ϕ) and the components of X in (Ω, ϕ), that is, the coordinates of X in the basis of Tx (M ) associated with (Ω, ϕ) by the process described above]. By definition, a vector field on M is a map X : M → T (M ) such that for any x ∈ M , X(x) ∈ Tx (M ). Since M and T (M ) are smooth manifolds, the notion of a vector field of class C k makes sense. A manifold M of dimension n is said to be parallelizable if there exist n smooth vector fields Xi , i = 1, . . . , n, such that for any x ∈ M , the Xi (x)’s, i = 1, . . . , n, define a basis of Tx (M ). Given two smooth manifolds, M and N , a point x in M , and a differentiable map f : M → N at x, the tangent linear map of f at x (or the differential map of f at x), denoted by f (x), is the linear map from Tx (M ) to Tf (x) (N ) defined, for X ∈ Tx (M ) and g : N → R differentiable at f (x), by f (x) · (X) · (g) = X(g ◦ f ) . By extension, if f is differentiable on M , one gets the tangent linear map of f , denoted by f . That is the map f : T (M ) → T (N ) defined, for X ∈ Tx (M ), by f (X) = f (x).(X). As one can easily check, f is C k−1 if f is C k . Similar

3

BACKGROUND MATERIAL

to the construction of the tangent bundle, one can define the cotangent bundle of a smooth manifold M as the disjoint union of the Tx (M ) ’s, x ∈ M. In a more general way, one can define Tpq (M ) as the disjoint union of the Tpq Tx (M ) ’s, where Tpq (Tx (M )) is the space of (p, q)-tensors on Tx (M ). Then Tpq (M ) possesses a natural structure of a smooth manifold of dimension n 1 + np+q−1 . A map T : M → Tpq (M ) is then said to be a (p, q)-tensor field on M if for any x ∈ M , T (x) ∈ Tpq Tx (M ) . It is said to be of class C k if it is of class C k from the manifold M to the manifold Tpq (M ). Given two manifolds M and N , a map f : M → N of class C k+1 , and a (p, 0)-tensor field T of class C k on N , one can define the pullback f T of T by f , that is, the (p, 0)-tensor field of class C k on M defined for x ∈ M and X1 , . . . , Xp ∈ Tx (M ), by f T )(x) · X1 , . . . , Xp = T f (x) · f (x) · X1 , . . . , f (x) · Xp . We now define the notion of a linear connection. Denote by Γ(M ) the space of differentiable vector fields on M . A linear connection D on M is a map D : T (M ) × Γ(M ) → T (M ) which satisfies a certain number of propositions. In local coordinates, given a chart (Ω, ϕ), this is equivalent to having n3 smooth functions Γkij : Ω → R, that we refer to as the Christoffel symbols of the connection in (Ω, ϕ). They characterize the connection in the sense that for X ∈ Tx (M ), x ∈ Ω, and Y ∈ Γ(M ), ∂Y j ∂ j i i α + Γiα (x)Y (x) DX (Y ) = X (∇i Y )(x) = X ∂xi x ∂xj x where the X i ’s and Y i ’s denote the components of X and Y in the chart (Ω, ϕ), and for f : M → R differentiable at x, ∂f = Di f ◦ ϕ−1 ϕ(x) . ∂xi x As one can easily check, the Γkij ’s are not the components of a (2, 1)-tensor field. An important remark is that linear connections have natural extensions to differentiable tensor fields. Given a differentiable (p, q)-tensor field, T , a point x in M , X ∈ Tx (M ), and a chart (Ω, ϕ) of M at x, DX (T ) is the (p, q)-tensor on Tx (M ) defined by DX (T ) = X i (∇i T )(x), where j1 ...jq p ∂Ti1 ...ip j ...j j1 ...jq − Γα ∇i T (x)i11...ipq = iik (x)T (x)i1 ...ik−1 αik+1 ...ip ∂xi x k=1

+

q

j ...j

Γjiαk (x)T (x)i11...ipk−1

αjk+1 ...jq

.

k=1

The covariant derivative commutes with the contraction in the sense that DX Ckk12 T = Ckk12 DX (T ) where Ckk12 T stands for the contraction of T of order (k1 , k2 ). Given a (p, q)-tensor field of class C k+1 , T , we let ∇T be the (p + 1, q)-tensor field of class C k whose components in a chart are given by j1 ...jq j1 ...jq ∇T i ...i = ∇i1 T i2 ...ip+1 . 1 p+1

4

CHAPTER 1

By extension, one can then define ∇2 T , ∇3 T , and so on. For f : M → R a smooth function, one has that ∇f = df and, in any chart (Ω, ϕ) of M , 2 2 ∂ f ∂f − Γkij (x) ∇ f (x)ij = ∂xi ∂xj x ∂xk x where

∂2f ∂xi ∂xj

x

2 = Dij f ◦ ϕ−1 ϕ(x) .

2

In the Riemannian context, ∇ f is called the Hessian of f and is sometimes denoted by Hess(f ). The torsion T of a linear connection D can be seen as the smooth (2, 1)-tensor field on M whose components in any chart are given by the relation Tijk = Γkij −Γkji . One says that the connection is torsion-free if T ≡ 0. The curvature R of D can be seen as the smooth (3, 1)-tensor field on M whose components in any chart are given by the relation l = Rijk

∂Γlji ∂Γlki l α − + Γljα Γα ki − Γkα Γji . ∂xj ∂xk

l l As one can easily check, Rijk = −Rikj . Moreover, when the connection is torsionfree, one has that l l l Rijk + Rkij + Rjki = 0 and

(∇i R)lmjk + (∇k R)lmij + (∇j R)lmki = 0 . Such relations are referred to as the first Bianchi and second Bianchi identities. We now discuss Riemannian geometry. Let M be a smooth manifold. A Riemannian metric g on M is a smooth (2, 0)-tensor field on M such that for any x ∈ M , g(x) is a scalar product on Tx (M ). A smooth Riemannian manifold is a pair (M, g) where M is a smooth manifold and g a Riemannian metric on M . According to Whitney, for any paracompact smooth n-manifold there exists a smooth embedding f : M → R2n+1 . One then gets that any smooth paracompact manifold possesses a Riemannian metric. Just think of g = f ξ, where ξ is the Euclidean metric. Two Riemannian manifolds (M1 , g1 ) and (M2 , g2 ) are said to be isometric if there exists a diffeomorphism f : M1 → M2 such that f g2 = g1 . Given a smooth Riemannian manifold (M, g), and γ : [a, b] → M a curve of class C 1 , the length of γ is

b dγ dγ L(γ) = dt g(γ(t)) · , dt t dt t a dγ where ( dγ dt )t ∈ Tγ(t) (M ) is such that ( dt )t · f = f ◦ γ (t) for any f : M → R differentiable at γ(t). If γ is piecewise C 1 , the length of γ may be defined as the sum of the lengths of its C 1 pieces. For x and y in M , let Cxy be the space of piecewise C 1 curves γ : [a, b] → M such that γ(a) = x and γ(b) = y. Then dg (x, y) = inf L(γ) γ∈Cxy

BACKGROUND MATERIAL

5

defines a distance on M whose topology coincides with the original one of M . In particular, by Stone’s theorem, a smooth Riemannian manifold is paracompact. By definition, dg is the distance associated with g. Let (M, g) be a smooth Riemannian manifold. There exists a unique torsion-free connection on M having the property that ∇g = 0. Such a connection is the LeviCivita connection of g. In any chart (Ω, ϕ) of M , of associated coordinates xi , and for any x ∈ Ω, its Christoffel symbols are given by the relations ∂gmj 1 ∂gmi ∂gij k g(x)mk Γij (x) = + − 2 ∂xi x ∂xj x ∂xm x where the g ij ’s are such that gim g mj = δij . Let R be the curvature of the LeviCivita connection as introduced above. One defines 1. the Riemann curvature Rmg of g as the smooth (4, 0)-tensor field on M α , whose components in a chart are Rijkl = giα Rjkl 2. the Ricci curvature Rcg of g as the smooth (2, 0)-tensor field on M whose components in a chart are Rij = Rαiβj g αβ , and 3. the scalar curvature Sg of g as the smooth real-valued function on M whose expression in a chart is Sg = Rij g ij . As one can check, in any chart, Rijkl = −Rjikl = −Rijlk = Rklij , and the two Bianchi identities are Rijkl + Riljk + Riklj = 0 , ∇i Rmg jklm + ∇m Rmg jkil + ∇l Rmg jkmi = 0 . In particular, the Ricci curvature is symmetric, so that in any chart Rij = Rji . Given a smooth Riemannian manifold (M, g), and its Levi-Civita connection D, a smooth curve γ : [a, b] → M is said to be a geodesic, if for all t, dγ D dγ = 0. dt dt t This means again that in any chart, and for all k, k (t) + Γkij γ(t) γ i (t) γ j (t) = 0 . γ For any x ∈ M , and any X ∈ Tx (M ), there exists a unique geodesic γ : [0, ] → M such that γ(0) = x and ( dγ dt )0 = X. Let γx,X be this geodesic. For λ > 0 real, γx,λX (t) = γx,X (λt). Hence, for X small, where · stands for the norm in Tx (M ) associated with g(x), one has that γx,X is defined on [0, 1]. The exponential map at x is the map from a neighborhood of 0 in Tx (M ), with values in M , defined of by expx (X) = γx,X (1). If M is n-dimensional and up to the assimilation Tx (M ) to Rn via the choice of an orthonormal basis, one gets a chart Ω, exp−1 x of M at x. This chart is normal at x in the sense that the components gij of g in this

6

CHAPTER 1

chart are such that gij (x) = δij , with the additional property that the Christoffel symbols Γkij of the Levi-Civita connection in this chart are such that Γkij (x) = 0. The coordinates associated with this chart are referred to as geodesic normal coordinates. Given a smooth Riemannian n-manifold (M, g), one can define a natural positive Radon measure on M . In particular, can be ap the theory of the Lebesgue integral plied. For some atlas of M , Ωi , ϕi i∈I , we shall say that a family Ωj , ϕj , αj j∈J is a partition of unity subordinate to Ωi , ϕi i∈I if the following holds: (i) (α j )j is a smooth partition of unity subordinate to the covering (Ωi )i , (ii) Ωj , ϕj j is an atlas of M , and (iii) for any j, suppαj ⊂ Ωj . As one can easily check, for any atlas Ωi , ϕi i∈I of M , there exists a partition of unity Ωj , ϕj , αj j∈J subordinate to Ωi , ϕi i∈I . One can then define the Riemannian measure as follows: Given a continuous map f : M → R with compact support, and an atlas Ωi , ϕi i∈I of M ,

αj |g|f ◦ ϕ−1 f dv(g) = j dx M

j∈J

ϕj (Ωj )

where Ωj , ϕj , αj j∈J is a partition of unity subordinate to Ωi , ϕi i∈I , |g| stands of g in for the determinant of the matrix whose elements are the components Ωj , ϕj , and dx stands for the Lebesgue volume element of Rn .One can prove that such a construction does not depend on the choice of the atlas Ωi , ϕi i∈I and the partition of unity Ωj , ϕj , αj j∈J . The Laplacian acting on functions of a smooth Riemannian manifold (M, g) is the operator ∆g whose expression in a local chart of associated coordinates xi is 2 ∂u ∂ u . − Γkij ∆g u = −g ij ∂xi ∂xj ∂xk For u and v of class C 2 on M , one then has the following formula for integration by parts:

(∇u∇v) dv(g) = u ∆g v dv(g) ∆g u vdv(g) = M

M

M

where (·, ·) is the scalar product associated with g for 1-forms. Coming back to geodesics, one can define the injectivity radius of (M, g) at some point x, denoted by ig (x), as the largest positive real number r for which any geodesic starting from x and of length less than r is minimizing. One can then define the (global) injectivity radius by ig = inf ig (x) . x∈M

One has that ig > 0 for a compact manifold, but it may be zero for a complete noncompact manifold. In a similar way, one can define the cut locus Cx of x, where Cx is a subset of M , and prove that Cx has measure zero, that ig (x) = dg x, Cx ,

7

BACKGROUND MATERIAL

and that expx is a diffeomorphism from some star-shaped domain of Tx (M ) at 0 onto M \Cx . In particular, one gets that the distance function r to a given point is differentiable almost everywhere, with the additional property that |∇r| = 1 almost everywhere. As is well known, curvature assumptions may give topological and diffeomorphic information on the manifold. A striking example of the relationship that exists between curvature and topology is given by the Gauss-Bonnet theorem, whose present form is actually due to the works of Allendoerfer [2], Allendoerfer-Weil [3], Chern [21], and Fenchel [35]. One has here that the Euler-Poincaré characteristic χ(M ) of a compact manifold can be expressed as the integral of a universal polynomial in the curvature. For instance, when the dimension of M is 2,

1 χ(M ) = Sg dv(g) , 4π M and when the dimension of M is 4, as shown by Avez [8],

1 1 1 2 2 2 χ(M ) = |Wg | + Sg − |Eg | dv(g) , 16π 2 M 2 12 where | · | stands for the norm associated with g for tensors, and where Wg and Eg are, respectively, the Weyl tensor of g and the traceless Ricci tensor of g. In a local chart, the components of Wg are 1 Rik gjl + Rjl gik − Ril gjk − Rjk gil n−2 Sg gik gjl − gil gjk + (n − 1)(n − 2)

Wijkl = Rijkl −

where n stands for the dimension of the manifold. As another striking example of the relationship that exists between curvature and topology, one can refer to Hamilton’s theorem [39]: any three-dimensional, compact, simply connected Riemannian manifold of positive Ricci curvature must be diffeomorphic to the unit sphere S 3 . Conversely, by recent results of Lohkamp [59], negative sign assumptions on the Ricci curvature have no effect on the topology, since any compact manifold possesses a Riemannian metric of negative Ricci curvature. As another example of the relationship that exists between curvature and topology, we refer to the well-known sphere theorem of Berger [10], Klingenberg [51, 52], Rauch [61], and Tsukamoto [75]. 1.2 BASICS IN NONLINEAR ANALYSIS Given a smooth compact n-dimensional Riemannian manifold (M, g), one easily defines the Sobolev spaces Hkp (M ), following what is done in the more traditional Euclidean context. For instance, when k = 1 and p > 1, one may define the Sobolev space H1p (M ) as follows: for u ∈ C ∞ (M ), we let uH1p = up + ∇up

8

CHAPTER 1

where .p is the Lp -norm with respect to the Riemannian measure dvg . We then define H1p (M ) as the completion of C ∞ (M ) with respect to .H1p . A similar definition holds for Hkp (M ), with uHkp =

k

∇i up .

i=0

H1p

are that Lipschitz functions on M do belong to the Very useful properties of p,and that if u ∈ H1p (M ) for some p, then we have Sobolev spaces H1p (M ) for all p that |u| ∈ H1 (M ) and ∇|u| = ∇u almost everywhere. As for bounded open subsets of the Euclidean space, the Sobolev embedding theorem (continuous embeddings), and the Rellich-Kondrakov theorem (compact embeddings), do hold. In order to fix ideas, we let k = 1 and p = 2. Let 2 =

2n n−2

be the critical Sobolev exponent. Then, for any q ∈ [1, 2 ], H12 (M ) ⊂ Lq (M ) and this embedding is continuous, with the property that the embedding is also compact if q < 2 . The Sobolev inequality corresponding to the continuous embedding H12 (M ) ⊂ L2 (M ) is as follows: For any u ∈ H12 (M ), u2 ≤ A∇u2 + Bu2 where A and B are positive constants independent of u, but that may depend on the manifold. Another very useful inequality is the so-called Poincaré inequality. When dealing with H12 , it reads as the existence of a positive constant A such that, for any u ∈ H12 (M ), u − u2 ≤ A∇u2 where u=

1 Vg

udvg M

is the average of u, and Vg the volume of M with respect to g. In this particular case, thanks to the Rayleigh characterization of the first nonzero eigenvalue of the Laplacian, A may be taken to be the inverse of the square root of this eigenvalue. Combining the Sobolev inequality and the Poincaré inequality, one gets the socalled Sobolev-Poincaré inequality: for any u ∈ H12 (M ), u − u2 ≤ A∇u2 , where A is a positive constant, independent of u as usual. A very useful notion concerning Sobolev embeddings, which appeared to be crucial in many problems like the Yamabe problem, is that of best constants. In the particular case k = 1 and p = 2, the Sobolev inequality in the Euclidean space reads as

12 21 2 2 ≤A |∇u| dx |u| dx . Rn

Rn

9

BACKGROUND MATERIAL

The best constant A in this inequality, which we denote by Kn , is 4 Kn = 2/n n(n − 2)ωn where ωn is the volume of the unit n-sphere. Taking A = Kn in the above inequality, we get what we refer to as the sharp Euclidean Sobolev inequality. Its extremal functions are known. They are expressed as 1− n2 uλ,a,µ (x) = µ λ + |x − a|2 where λ > 0, µ ∈ R, and a ∈ Rn are arbitrary. For compact manifolds, we consider the Sobolev inequality u22 ≤ A∇u22 + Bu22 . As is easily checked, any such constant A must satisfy A ≥ Kn2 . Conversely, we refer to Hebey-Vaugon [47, 48], it can be proved that there exists a positive constant B such that for any u ∈ H12 (M ), u22 ≤ Kn2 ∇u22 + Bu22 . More developments on Sobolev spaces, Sobolev inequalities, and the notion of best constants are in Druet-Hebey [29] and Hebey [44]. Let (M, g) be a smooth compact Riemannian manifold. The equations we will be interested in are basically of the form ∆g u + au = f where a, f are given functions on M . A function u ∈ H12 (M ) is said to be a weak solution of this equation if, for all ϕ ∈ H12 (M ),

∇u, ∇ϕg dvg + a(x)uϕdvg = f (x)ϕdvg . M

M

M

Regularity results for this equation do hold. They are similar to the more traditional ones expressed in the Euclidean context (regularity is a local notion, so this is not very surprising). The regularity result we will mostly use is the following: if a is smooth, and f ∈ Hkp (M ) for some k ∈ N and p > 1, then a weak solution u to p (M ). In particular, it follows from this result and the the above equation is in Hk+2 Sobolev embedding theorem, that when f is smooth u is also smooth. Needless to say, the “bible” for such topics is the exhaustive Gilbarg-Trudinger [37]. A simpler, but very nice reference is the lecture notes [41] by Han and Lin. In parallel with regularity, the very useful maximum principles hold for the Laplacian on Riemannian manifolds. A currently used form is as follows: if a nonnegative u ∈ C 2 (M ) is such that, for any x ∈ M , ∆g u(x) ≥ u(x)f (x, u(x)) for some continuous functions f : M × R → R, then either u is everywhere positive, or u is the zero function. This easily follows from the Hopf maximum principle, as usually stated.

10

CHAPTER 1

In order to end this section, we give a basic example of a possible use of the above results. We discuss here the existence (and uniqueness) of a solution u to the Laplace equation ∆g u = f on a compact Riemannian manifold (M, g). Although not necessary, we assume here for convenience that f : M → R is smooth. Integrating the Laplace equation, one sees that a necessary condition for the existence of a solution is that

f dvg = 0 . M

The elementary result we wish to briefly discuss here is that the Laplace equation possesses a smooth solution if and only if M f dvg = 0. Moreover, the solution is unique, up to the addition of a constant. In order to prove this claim, we proceed as follows. As already mentioned, the condition that f is of null average is a necessary condition. We prove now that it is also a sufficient condition. Let

2 udvg = 0 and f udvg = 1 H = u ∈ H1 (M ) s.t. M

and

M

µ = inf

u∈H

|∇u|2 dvg .

M

Clearly, H = ∅. Consider a minimizing sequence (ui ) ∈ H for µ so that ui ∈ H for all i, and

|∇ui |2 dvg = µ . lim i→+∞

M

By the Poincaré inequality we discussed above, there exists A > 0 such that, if u ∈ H12 (M ) is of null average, then

u2 dvg ≤ A |∇u|2 dvg . M

M

It easily follows that the ui ’s are bounded in H12 (M ). Since H12 (M ) is a reflexive space, and the embedding H12 (M ) ⊂ L2 (M ) is compact by the Rellich-Kondrakov theorem (even when n = 2, noting that H12 ⊂ H1q for q < 2), there exists a function u ∈ H12 (M ) and a subsequence (ui ) of (ui ), such that (1) (ui ) converges weakly to u in H12 (M ) and (2) (ui ) converges to u in L2 (M ). By (2), u ∈ H. By (1), and a basic property of the weak limit (the norm of a weak limit is less than or equal to the infimum limit of the norms of the sequence), we get that

|∇u|2 dvg ≤ µ . M

Hence,

M

|∇u|2 dvg = µ

11

BACKGROUND MATERIAL

and µ is attained. By a well-known theorem of Lagrange, this gives the existence of two constants α and β, the Lagrange multipliers, such that, for all ϕ ∈ H12 (M ),

∇u, ∇ϕg dvg = α ϕdvg + β f ϕdvg . M

M

M

Taking ϕ = 1, one gets that α = 0. Taking ϕ = u, one gets that β = µ. Hence, u is a weak solution of the Laplace equation. By standard regularity results, u is smooth. The function µ−1 u is then the solution we were looking for. The proof of uniqueness is then very simple. If u and v are two solutions of the Laplace equation, then ∆g (v − u) = 0. Multiplying this relation by v − u and integrating over M gives that

|∇(v − u)|2 dvg = 0 . M

Hence, v − u is constant, and this ends the proof of the above claim. As is easily checked, everything in this proof comes from the compactness of the embedding H12 ⊂ L2 . The equations that we discuss below involve critical (noncompact) Sobolev embeddings.


Chapter Two The Model Equations Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3. We let (hα ) be a sequence of smooth functions on M , and consider equations like ∆g u + hα u = u2

−1

(Eα )

where ∆g = −divg ∇ is the Laplace-Beltrami operator, 2 = 2n/(n − 2) is the critical Sobolev exponent for the embedding of the Sobolev space H12 (M ) into Lebesgue’s spaces Lp (M ), and u is required to be positive. By standard regularity theory, as developed by Gilbarg-Trudinger [37], and thanks to the maximum principle, if u ∈ H12 (M ) is a nonnegative solution of (Eα ), then u is smooth and either u > 0 everywhere or u ≡ 0. As already mentioned, thanks to the Sobolev embedding theorem, H12 (M ) ⊂ Lp (M ) for p ≤ 2 , this embedding being compact if p < 2 and not compact if p = 2 . Equations like (Eα ) arise naturally in several problems. This is the case for the Yamabe problem (see Schoen [66]) but also for sharp constant problems in Sobolev inequalities. Possible survey articles on the Yamabe problem are Hebey [42], Lee and Parker [54], and Schoen [69]. Possible monographs on sharp constants problems are Druet and Hebey [29], and Hebey [44]. We refer also to Druet [27], Hebey [45], Hebey and Vaugon [49], and Schoen [67, 70] for extensions of these problems where understanding equations like (Eα ) is necessary. The reference Hebey [45] discusses the notion of the energy function. The reference Hebey-Vaugon [49] discusses the notion of critical functions. Among other topics, the references Druet [27] and Schoen [67, 70] discuss compactness results. We are concerned in these notes with the blow-up behavior of sequences of solutions of equations like (Eα ). A very nice H12 -theory for the blow-up behavior of such sequences was developed in the 1980s. Among other possible references, we refer to Brézis-Coron [12, 13], Lions [58], Sacks-Uhlenbeck [64], Schoen [67], Struwe [73], and Wente [76]. A partial survey of the subject is given by EkelandGhoussoub [34]. The H12 -theory as in Struwe [73], dealing with Palais-Smale sequences, is presented in Chapter 3 below. Atkinson-Peletier [5], with arguments from ordinary differential equation theory, and Brézis-Peletier [16], with arguments from partial differential equation theory, have been concerned with the description of the pointwise behavior of sequences of solutions of equations like (Eα ), dealing with radially symmetrical solutions uε of the semi-critical equations ∆u = u2 −1−ε on the unit ball of the Euclidean space. We refer also to AdimurthiPacella-Yadava [1] and Robert [62, 63] for more recent developments. Roughly speaking, a bubble is the rescaling of a fundamental positive solution of the critical

14

CHAPTER 2

Euclidean equation ∆u = u2 −1 . An estimate like the one we present in Chapter 5, stating that solutions of minimal energy of equations like (Eα ) are controlled from above by a standard bubble, appeared in Han [40], dealing with solutions uε of the equations ∆u = u2 −1−ε on bounded open subsets of the Euclidean space, in Hebey-Vaugon [47, 48], dealing with (Eα ) and arbitrary Riemannian manifolds, and in Li [55, 56], dealing with equations like (Eα ) on the unit sphere. Other possible references are Chang-Yang [18], Chang-Gursky-Yang [19], Druet [24, 25], Druet-Hebey [30], Druet-Robert [33], and Li-Zhu [57]. Weak estimates, like the ones we present in Chapter 4, are reminiscent of the definition by Schoen [68] of isolated blow-up points. We refer also to Schoen-Zhang [71]. Such estimates have been developed by Druet [23, 26] when studying sharp constant problems.

We briefly discuss in this very basic introductory chapter the existence of PalaisSmale sequences for (Eα ), and the existence of strong solutions for (Eα ) of minimal or arbitrarily high energies. More sophisticated examples can be found in Druet-Hebey [31]. 2.1 PALAIS-SMALE SEQUENCES Let h be a smooth function on M . Following standard terminology, we say that an operator like Lg = ∆g + h is coercive if there exists C > 0 such that

(Lg u)udvg (2.1.1) u2H 2 ≤ C 1

M

H12 (M ),

where the right hand side of this equation has to be understood for all u ∈ in the distributional sense, so that

(Lg u)udvg = |∇u|2 + hu2 dvg . M

M

Equation (2.1.1) just says that the H12 -norm of a function u is controlled by the energy of u with respect to Lg . It is easily checked that, if h > 0 everywhere, then Lg is coercive. As another remark, the existence of a uα > 0 solution of (Eα ) implies that the operator Lα g = ∆g + hα in the left hand side of equation (Eα ) is coercive. From now on, we let Igα be the functional defined on the Sobolev space H12 (M ) by

1 1 1 |∇u|2 dvg + hα u2 dvg − |u|2 dvg . Igα (u) = 2 M 2 M 2 M Let (uα ) be a sequence of functions in H12 (M ). Following standard terminology, we say that (uα ) is a Palais-Smale sequence for equation (Eα ), or that (uα ) is a Palais-Smale sequence for Igα , if the two following propositions hold: the sequence (Igα (uα )) is bounded and DIgα (uα ) → 0 strongly in H12 (M ) as α → +∞ .

(2.1.2)

As an elementary remark, it should be noted that if (uα ) is a bounded sequence in H12 (M ) of solutions of (Eα ), then (uα ) is a Palais-Smale sequence for (Eα ).

15

THE MODEL EQUATIONS

We assume in what follows that the operators Lα g are uniformly coercive. More precisely, we assume that the following holds: there exists h smooth, s.t. ∆g + h is coercive, and hα ≥ h for all α .

(2.1.3)

Taking inspiration from Brézis-Nirenberg [15], thanks to the mountain pass lemma of Ambrosetti and Rabinowitz [4], we prove the following proposition in this section. P ROPOSITION 2.1 Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, and (hα ) be a sequence of smooth functions on M . We assume that (2.1.3) holds. Then there exist Palais-Smale sequences of smooth positive functions for (Eα ). Proof. We let (εα ) be a sequence of positive real numbers such that εα → 0 as α → +∞, and let q ∈ (1, 2 − 1). We fix α, and let Φα be the functional defined on H12 (M ) by

1 1 2 |∇u| dvg + hα u2 dvg Φα (u) = 2 M 2 M

1 εα − (u+ )2 dvg − (u+ )q+1 dvg 2 M q+1 M where u+ = max(u, 0). It is easily seen that the mountain pass lemma of Ambrosetti and Rabinowitz [4], as stated in Brézis and Nirenberg [15], can be applied to Φα . It follows that there exists cα > 0 and a sequence (ϕj ) in H12 (M ) such that Φα (ϕj ) = cα + o(1) and DΦα (ϕj )(H12 ) = o(1)

(2.1.4)

where o(1) → 0 as j → +∞. Moreover, given u0 ∈ H12 (M ), u0 ≥ 0, and u0 ≡ 0, we can choose cα such that cα ≤ sup Φα (tu0 ) . t≥0

We claim that by taking for u0 local test functions as in Aubin [6], cα can be chosen such that cα < n1 Kn−n . More precisely, given δ > 0 small, x0 in M , and θ > 0, we let uθ be the function 1− n2 uθ (x) = η(x) θ2 + dg (x0 , x)2 where dg is the distance with respect to g, and η is a smooth cutoff function such that η = 1 in Bx0 (δ) and η = 0 in M \Bx0 (2δ), the geodesic balls of center x0 and radii δ and 2δ. Easy computations give that for any q > 1 if n ≥ 4, and any q > 3 if n = 3, uθ 1 n− n−2 n− n−2 2 (q+1) + o 2 (q+1) = − Aε θ θ sup Φα t α uθ 2 nKnn t≥0 where A = A(q, n) is a positive constant depending only on q and n. Assuming that q > 3 if n = 3, and letting u0 = uθ with θ > 0 sufficiently small, this proves

16

CHAPTER 2

the above claim. Following (2.1.4),

1 |∇ϕj |2 + hα ϕ2j dvg 2 M

1 εα 2 = (ϕ+ ) dv + (ϕ+ )q+1 dvg + cα + o(1) g 2 M j q+1 M j and

|∇ϕj |2 + hα ϕ2j dvg M

2 q+1 = (ϕ+ ) dv + ε (ϕ+ dvg + ϕj H12 o(1) . g α j j )

(2.1.5)

M

(2.1.6)

M

1 q+1 (2.1.6),

we get that Writing (2.1.5) −

1 1 − |∇ϕj |2 + hα ϕ2j dvg ≤ cα + o(1) + ϕj H12 o(1) . 2 q+1 M Because of (2.1.3), and since cα < nKn−n , we have proved that there exists C > 0, independent of α, such that for any j, ϕj H12 ≤ C. Up to a subsequence we may therefore assume that ϕj uα weakly in H12 (M ), that ϕj → uα strongly in Lp (M ) when p < 2 , and that ϕj → uα almost everywhere as j → +∞. It follows that 2 ∆g uα + hα uα = (u+ α)

−1

q + εα (u+ α) .

Regularity results and the maximum principle then give that uα is smooth and that either uα ≡ 0 or uα > 0 everywhere. Up to another subsequence, we can assume that ∇ϕj 22 → θ as j → +∞. If uα ≡ 0 we then get with (2.1.5) and (2.1.6) 1/2 as j → +∞ and that ncα = θ. On the other hand, thanks to that ϕ+ j 2 → θ Hebey and Vaugon [47, 48], there exists B > 0 such that + 2 2 2 2 ϕ+ j 2 ≤ Kn ∇ϕj 2 + Bϕj 2

so that, passing to the limit as j → +∞, we get that θ2/2 ≤ Kn2 θ. Noting that this is in contradiction with the equations ncα = θ and cα < n1 Kn−n , we have proved that uα > 0 everywhere. Hence, (uα ) is a sequence of smooth positive functions, solutions of the equations

2 −1 + εα uqα . ∆g uα + hα uα = uα

Moreover, since ϕj H12 ≤ C and ϕj uα weakly in H12 (M ), we also have that for any α, uα H12 ≤ C. Then

1 εα 1 α 2 − uα dvg + uq+1 dvg Ig (uα ) = 2 2 2 M α M is uniformly bounded in α. Similarly, for any ϕ ∈ H12 (M ),

DIgα (uα ).ϕ = εα uqα ϕdvg , M

17

THE MODEL EQUATIONS

and since q < 2 − 1,

DIgα (uα ).ϕ ≤ εα

u2α q/(2 −1) dvg

(2 −1)/2 ϕ2

M

≤ Cεα ϕH12 where C > 0 is independent of α. In particular, the conditions in (2.1.2) are satisfied. This proves the proposition. 2 A more classical construction of Palais-Smale sequences involves bubbles as defined in the following chapter. 2.2 STRONG SOLUTIONS OF MINIMAL ENERGY We assume in what follows that, for any α, the operator Lα g is coercive. We discuss the existence of strong solutions of minimal energy. Given u ∈ H12 (M ), we define the energy E(u) of u by E(u) = u2 . If u is a strong solution of (Eα ), Igα (u) =

1 E(u)2 . n

−(n−2)/2

where Kn is the sharp constant in the Euclidean We let also Λmin = Kn Sobolev inequality ϕ2 ≤ Kn ∇ϕ2 , ϕ ∈ C0∞ (Rn ). A very standard result is the following. P ROPOSITION 2.2 Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3 and (hα ) be a sequence of smooth functions on M . If, for any α, Lα g = ∆g + hα is coercive, and |∇u|2 dvg + M hα u2 dvg 1 M inf < 2 2/2 K 2 u∈H12 (M )\{0} n |u| dvg M

then (Eα ) possesses a sequence (uα ) of smooth positive solutions such that for any α, E(uα ) < Λmin . Proof. The proof of such an assertion is by now very classical. We fix α, and let I be the functional defined on H12 (M ) by

|∇u|2 dvg + hα u2 dvg . I(u) = M

M

We let also H be the subset of H12 (M ) defined by

2 |u|2 dvg = 1 H = u ∈ H1 (M ) s.t. M

and µ = inf I(u) . u∈H

is coercive, µ ≥ 0. By definition, a sequence (ui ) in H12 (M ) is a miniSince mizing sequence for µ if Lα g

18

CHAPTER 2

(i) ui ∈ H for all i and (ii) I(ui ) → µ as i → +∞. We let (ui ) be such a minimizing sequence, and assume that the ui ’s are nonnegative. Clearly, (ui ) is bounded in H12 (M ). After passing to a subsequence, we may thus assume that there exists u ∈ H12 (M ) such that ui u weakly in H12 (M ), ui → u strongly in L2 (M ), and ui → u almost everywhere as i → +∞. In particular, u is nonnegative. Independently, it easily follows from the weak convergence that ∇ui 22 = ∇(ui − u)22 + ∇u22 + o(1) for all i, where o(1) → 0 as i → +∞. We also have (see, for instance, Brézis-Lieb [14]) that ui 22 = ui − u22 + u22 + o(1)

for all i, where, as above, o(1) → 0 as i → +∞. As shown by Hebey and Vaugon [47, 48], there exists B > 0 such that, for any i, ui − u22 ≤ Kn2 ∇(ui − u)22 + Bui − u22 . Since ui ∈ H, it follows that 2/2 ≤ Kn2 ∇ui 22 − ∇u22 + o(1) . 1 − u22 Since I(ui ) → µ, and since ui → u in L2 (M ), we also have that

2 2 2 2 2 2 2 Kn ∇ui 2 − ∇u2 = Kn µ − Kn |∇u| dvg + hα u dvg + o(1) M

M

≤ Kn2 µ − Kn2 µu22 + o(1) . Hence, 2/2 1 − u22 ≤ Kn2 µ 1 − u22 . We assumed that µKn2 < 1. Noting that 2/2 , 1 − u22 ≤ 1 − u22 this implies that u2 = 1. Then, ∇ui 2 → ∇u2 as i → +∞, and since ∇ui 22 = ∇(ui − u)22 + ∇u22 + o(1) , we get that ui → u strongly in H12 (M ) as i → +∞. In particular, u is a minimizer for µ, µ > 0, and u is a weak nonnegative solution of the equation ∆g u + hα u = µu2

−1

.

By standard regularity results, as developed by Trudinger [74], and using the maximum principle, u is smooth and positive. Noting that uα = µ(n−2)/4 u is a solution 2 of (Eα ), this proves the proposition.

19

THE MODEL EQUATIONS

Let Sg be the scalar curvature of g. Local arguments as in Aubin [6] give that if n ≥ 4, and if there exists x ∈ M such that hα (x)
0 such that for any u ∈ H12 (M ), u22 ≤ Kn2 ∇u22 + Bu22 . Let B0 (g) be the smallest B in this inequality. Taking u = 1, it is easily seen that −2/n B0 (g) ≥ Vg , where Vg is the volume of M with respect to g. We also have that, for any u ∈ H12 (M ), u22 ≤ Kn2 ∇u22 + B0 (g)u22 . By the definition of B0 (g), if hα : M → R is such that hα < Kn−2 B0 (g), then the inequality on the infimum in Proposition 2.2 is satisfied. On the other hand, it was proved by Hebey and Vaugon [49] that there are compact manifolds for which the above sharp Sobolev inequality does not possess (nonzero) extremal functions. For such manifolds, if (hα ) is a sequence of smooth positive functions satisfying that hα < Kn−2 B0 (g) for all α, and that hα → Kn−2 B0 (g) in C 0 (M ) as α → +∞, then equation (Eα ) with respect to g possesses a sequence (uα ) of smooth positive solutions such that for any α, E(uα ) < Λmin , such that E(uα ) → Λmin as the parameter α → +∞, and such that uα 2 → 0 as α → +∞. In particular, the uα ’s necessarily blow up as α → +∞. We refer to Chapter 3 for the definition of blowing-up sequences of solutions. 2.3 STRONG SOLUTIONS OF HIGH ENERGIES We discuss the existence of strong positive solutions of arbitrary high energies in some specific cases. We claim that the following result holds. P ROPOSITION 2.3 There are smooth compact Riemannian manifolds with the following property: there exists Λ0 > 0 such that for any Λ1 > Λ0 there exist positive real numbers K1 (Λ1 ) < K2 (Λ1 ) such that, if (hα ) is any sequence of real numbers in (K1 (Λ1 ), K2 (Λ1 )), then (Eα ) possesses a sequence (uα ) of smooth nonconstant positive solutions such that for any α, Λ1 ≤ E(uα ) ≤ Λ1 + Λmin . Typical examples of such manifolds are the unit sphere in odd dimension and the product of a circle with any compact Riemannian manifold. We prove the above proposition in the second case.

20

CHAPTER 2

Proof. We let (M, g) be any smooth compact Riemannian manifold of dimension n − 1, and let (S 1 (t), ht ) be the circle in R2 of center 0 and radius t together with its standard metric. We let Mt = S 1 (t) × M and gt = ht + g be the product metric on Mt . Given m integer, we let Gm be the subgroup of O(2) generated by z→e

2iπ m

z

and regard Gm as acting on Mt by (x, y) → (σ(x), y). Then Mt /Gm = Mt/m . We let u be a smooth nonconstant function on M , and let ut be the function it induces on Mt by ut (x, y) = u(y). Then, ut ◦ σ = ut for all σ ∈ Gm . We know from Hebey and Vaugon [47, 48] that there exists B > 0 such that, for any u ∈ H12 (Mt ), u22 ≤ Kn2 ∇u22 + Bu22

where Kn is the sharp Euclidean constant, and 2 the critical Sobolev exponent. We let B0 (gt ) be the smallest constant B in the above inequality. Then, for any u ∈ H12 (Mt ), u22 ≤ Kn2 ∇u22 + B0 (gt )u22 . −2/n

Taking u = 1, it is easily seen that B0 (gt ) ≥ Vgt where Vgt is the volume of Mt −2/n with respect to gt . We claim now that for m sufficiently large B0 (g1/m ) > Vg1/m . If this is not the case, then, up to a subsequence, u1/m 22 . u1/m 22 ≤ Kn2 ∇u1/m 22 + Vg−2/n 1/m Noting that for all p,

M1/m

|∇u1/m |2 dvg1/m =

M1/m

|u1/m |p dvg1/m

1 m

1 = m

M1

M1

|∇u1 |2 dvg1 and

|u1 |p dvg1 ,

we get that for any m, Kn2 ∇u1 22 + Vg−2/n u1 22 . 1 m2/n This in turn implies that, for any m, u1 22 ≤

(2π)2/n Kn2 ∇u22 + Vg−2/n u22 m2/n where Vg is the volume of M with respect to g. Letting m → +∞, we get that u22 ≤

u22 ≤ Vg−2/n u22 , and this is impossible since u is nonconstant. The above claim is proved. In partic−2/n ular, there exists Λ0 > 0 such that, for any Λ1 > Λ0 , B0 (g1/m ) > Vg1/m where −2 2 2 greatest integer not exceeding m = E(Λ−2 min Λ1 ) + 1, and E(Λmin Λ1 ) is the −2 2 2 Λmin Λ1 . Given Λ1 > Λ0 , we let m = E(Λ−2 Λ 1 ) + 1. For real B satisfying min −2/n the condition Vg1/m < B < B0 (g1/m ), we let |∇u|2 + Kn−2 Bu2 dvg1/m M1/m λB = inf . 2/2 u∈H12 (M1/m )\{0} 2 |u| dvg1/m M1/m

21

THE MODEL EQUATIONS

Since B < B0 (g1/m ), λB < Kn−2 . Moreover, λB is nondecreasing in B, and it can be proved with the kind of arguments developed in the preceding section that lim

B→B0 (g1/m )

λB = Kn−2 .

2 Noting that Λmin = (Kn−2 )(n−2)/4 , and that m > Λ−2 min Λ1 , we let

Vg−2/n < T1 (Λ1 ) < B0 (g1/m ) 1/m be such that, for any T1 (Λ1 ) < B < B0 (g1/m ), 1/2 (n−2)/4 2 Λ1 ≤ E(Λ−2 λB . min Λ1 ) + 1 We set T2 (Λ1 ) = B0 (g1/m ), and let (hα ) be any sequence of real numbers in (K1 , K2 ) where K1 = Kn−2 T1 and K2 = Kn−2 T2 . We write that hα = Kn−2 βα , so that T1 < βα < T2 . We set λα = λβα . Since λα < Kn−2 , we get with the kind of arguments developed in the preceding section that there exists a minimizer uα for λα . Moreover, this minimizer can be chosen smooth and positive. Clearly, uα is not constant. If this is not the case, βα Vg2/n = λα Kn2 , 1/m 2/n

and this is impossible since λα Kn2 < 1 and βα Vg1/m > 1. Independently, we can choose uα such that it is a solution of the equation 2 −1 ∆g1/m uα + hα uα = λα uα

(n−2)/4

˜α = λα and such that uα 2 = 1. If we let u

˜ α + hα u ˜α = ∆g1/m u (n−2)/4

and ˜ uα 2 = λα Then

uα , then

2 −1 u ˜α

. We let u ˆα be the function on M1 such that u ˆα /Gm = u ˜α . 2 −1 ˆ α + hα u ˆα = u ˆα ∆ g1 u

(n−2)/4 1/2 . Noting that Λ21 + Λ2min ≤ Λ1 + and E(ˆ uα ) = ˆ uα 2 = m1/2 λα Λmin , it is easily seen from the above construction that

uα ) ≤ Λ1 + Λmin . Λ1 ≤ E(ˆ This proves the proposition.

2

As a remark, the same proof gives that for any sequence (hα ) of smooth functions on M , if for any α, K1 (Λ1 ) < hα < K2 (Λ1 ), then (Eα ) on M1 = S 1 × M possesses a sequence (uα ) of smooth nonconstant positive solutions such that, for any α, Λ1 ≤ E(uα ) ≤ Λ1 + Λmin . Another possible construction, mixing both the ideas of the preceding section and of this section, is as follows. Given m ≥ 1 integer, we let (hα ) be a sequence 1 ) × M for which equation (Eα ) on M1/m of smooth functions on M1/m = S 1 ( m possesses a sequence (uα ) of smooth positive solutions. Let Π : M1 → M1/m

22

CHAPTER 2

ˆ α and u ˆ α = hα ◦ Π and be the canonical projection, and let h ˆα be such that h u ˆα = uα ◦ Π. Then, for any α, 2 −1 ˆ αu ˆα + h ˆα = u ˆα ∆ g1 u

on M1 , and E(ˆ uα ) = m1/2 E(uα ). Let us now suppose that the dimension n of M is such that n ≥ 3, and that g is conformally flat. Then, g1/m is also conformally flat on M1/m . By “conformally flat” we mean that, up to conformal changes of the metric, we get local isometries with the Euclidean space. A conformal metric to some Riemannian metric g is expressed as f g where f is a smooth and positive function. When n ≥ 4, g is conformally flat if and only if Wg ≡ 0, where Wg is the Weyl curvature tensor of g. Following Hebey-Vaugon [49], there exists g˜m , a conformal metric to g1/m on M1/m , such that the sharp Sobolev inequality

gm )u22 u22 ≤ Kn2 ∇u22 + B0 (˜ with respect to g˜m does not possess (nonzero) extremal functions, where B0 (˜ gm ) is defined as in the proof of Proposition 2.3 (see also the remark at the end of section 2.2). We refer to Hebey-Vaugon [49] for a more general statement. Let (hα ) be gm ), and a sequence of positive real numbers such that for any α, hα < Kn−2 B0 (˜ such that hα → Kn−2 B0 (˜ gm ) as α → +∞. By the definition of B0 (˜ gm ), the kind of arguments of the preceding section give that, for any α, there exists uα smooth and positive on M1/m , and λα ∈ (0, Kn−2 ), such that 2 −1 ∆g˜m uα + hα uα = λα uα

(n−2)/4

and E(uα ) = 1. We let u ˜α be given by u ˜ α = λα uα . By the definition of gm ), and since the above sharp Sobolev inequality does not possess extremal B0 (˜ ˆα be the smooth functions, λα → Kn−2 and uα 2 → 0 as α → +∞. Let u ˆα = uα ◦ Π, where Π is as above, and let gˆm be positive function on M1 given by u the conformal metric to g1 on M1 given by gˆm = Π g˜m . Then, for any α, 2 −1 ∆gˆm u ˆ α + hα u ˆα = u ˆα .

(n−2)/4

, so that E(ˆ uα ) → m1/2 Λmin as α → +∞, Moreover, E(ˆ uα ) = m1/2 λα and ˆ uα 2 → 0 as α → +∞, where the energy and the L2 -norm in these equations are given with respect to gˆm . Summarizing, we have proved that, if (M, g) is a conformally flat Riemannian manifold of dimension n ≥ 3, then, for any m ≥ 1 integer, there exists a conformal metric gˆm to the standard product metric on M1 = S 1 × M , there exists a bounded sequence (hα ) of positive real numbers, and there exists a sequence (uα ) of smooth positive functions on M1 , solutions of equations uα ) → m1/2 Λmin as α → +∞, (Eα ) on M1 with respect to gˆm , and such that E(ˆ and ˆ uα 2 → 0 as α → +∞, where the energy and the L2 -norm in these equations ˆα ’s necessarily blow up as α → +∞. are with respect to gˆm . In particular, the u Here again, we refer to Chapter 3 for the definition of blowing-up sequences of solutions. Independently, note that the same construction can be carried out with ˆ → M of order m of S n , n odd, in place of S 1 × M , or with any covering Π = M a compact Riemannian manifold of dimension n ≥ 4 whose Weyl curvature tensor vanishes on some open subset of M .

23

THE MODEL EQUATIONS

2.4 THE CASE OF THE SPHERE Let (S n , h) be the unit n-sphere, n ≥ 3. Let Sh be the scalar curvature of h. Then Sh is constant and Sh = n(n − 1). We consider the equation n(n − 2) ∆h u + u = u2 −1 4 n−2 so that hα = 4(n−1) Sh . The positive solutions of this equation are known. Let x0 ∈ S n and r be the distance to x0 . Then, for any β > 1, n−2 4 1− n2 n(n − 2) 2 (β − 1) β − cos r uβ = 4 is a solution of the above equation. The energy of uβ is E(uβ ) = Λmin . A possible reference in book form is Hebey [44]. It is easily seen that uβ → 0 in 0 (S n \{x0 }) as β → 1. On the other hand, uβ (x0 ) → +∞ as β → 1. Thus the Cloc uβ ’s develop a singularity at x0 (they blow up at x0 ) as β → 1. Let xβ and µβ be given by 1− n 2

uβ (xβ ) = maxn uβ (x) = µβ x∈S

Then xβ = x0 and

µβ =

.

4(β − 1) . n(n − 2)(β + 1)

In particular, µβ → 0 as β → 1. Let Bβ be the function given by ⎛ ⎞ n−2 2 µ β ⎝ ⎠ Bβ (x) = . d (xβ ,x)2 µ2β + hn(n−2) We refer to Bβ (see the following chapters) as the standard bubble with respect to xβ and µβ . It is easily seen that there exists C > 1 such that 1 Bβ (x) ≤ uβ (x) ≤ CBβ (x) (2.4.1) C for all x ∈ S n and all β > 1. Similarly, for any ε > 0, there exists δε > 0 such that 1 Bβ (x) ≤ uβ (x) ≤ (1 + ε)Bβ (x) (2.4.2) 1+ε for all x ∈ Bx0 (δε ) when β > 1 is close to 1. Let Rβ be the function given by uβ = Bβ + Rβ . We claim that

Rβ → 0 in H12 (S n )

as β → 1. Noting that S n Bβ2 dvh → 0 as β → 1, we get with (2.4.1) that uβ → 0 in L2 (S n ) as β → 1, and then that Rβ → 0 in L2 (S n ) as β → 1. Independently (see, for instance, Chapter 3)

|∇Bβ |2 dvh → Λ2min and Bβ2 dvh → Λ2min Sn

Sn

24

CHAPTER 2

as β → 1. Moreover, since E(uβ ) = Λmin and uβ → 0 in L2 (S n ), and thanks to the equation satisfied by uβ ,

|∇uβ |2 dvh → Λ2min Sn

as β → 1. Writing that

2 2 |∇Rβ | dvh = |∇uβ | dvh + Sn

Sn

2

|∇Bβ | dvh − 2

Sn

we then get that

|∇Rβ |2 dvh = 2Λ2min − 2 Sn

(∇uβ ∇Bβ ) dvh , Sn

(∇uβ ∇Bβ ) dvh + o(1)

(2.4.3)

Sn

where o(1) → 0 as β → 1. Because of the equation satisfied by uβ , and since uβ → 0 and Bβ → 0 in L2 (S n ),

(∇uβ ∇Bβ ) dvh = uβ2 −1 Bβ dvh + o(1) . Sn

Sn

n Given δ > 0, uβ → 0 and Bβ → 0 in S \Bx0 (δ) as β → 1. Writing that

uβ2 −1 Bβ dvh = uβ2 −1 Bβ dvh + uβ2 −1 Bβ dvh 0 Cloc

Bx0 (δ)

Sn

S n \Bx0 (δ)

we then get with (2.4.2) that

uβ2 −1 Bβ dvh = Λ2min + o(1) . Sn

Hence,

Sn

(∇uβ ∇Bβ ) dvh = Λ2min + o(1) ,

and coming back to (2.4.3) we get that

|∇Rβ |2 dvh → 0 Sn

as β → 1. In particular, Rβ → 0 in H12 (S n ) as β → 1. The above claim is proved. On the other hand, in general, Rβ → 0 in C 0 (S n ) as β → 1, and the Rβ ’s are not even bounded in L∞ (S n ). Let us assume, for instance, that n ≥ 7, and let (yβ ) be a sequence of points in S n such that dh (x0 , yβ )2 = (β − 1)1+ε where ε > 0 is such that ε < n−6 4 . Then, Rβ (yβ ) → +∞ as β → 1. More complete information is as follows. When n ≥ 7, the Rβ ’s are not bounded in L∞ (S n ). When n = 6, the Rβ ’s are bounded in L∞ (S n ), but they do not converge to 0 in C 0 (S n ) as β → 1. When n = 3, 4, 5, the Rβ ’s converge to 0 in C 0 (S n ) as β → 1. Refinements of these constructions are in Druet-Hebey [31].

Chapter Three Blow-up Theory in Sobolev Spaces An important result of the 1980s describes the asymptotic behavior of Palais-Smale sequences associated with equations like ∆u = u2 −1 , (3.0.1) where ∆ is the Euclidean Laplacian, and u is required to vanish on the boundary of a smooth bounded open subset Ω of the Euclidean space Rn . This result was proved by Struwe [73]. Related references are Brézis [11], Brézis-Coron [12, 13], Lions [58], Sacks-Uhlenbeck [64], Schoen [67], and Wente [76]. Let D12 (Rn ) be the homogeneous Sobolev space defined as the completion of C0∞ (Rn ), the space of smooth functions with compact support in Rn , with respect to the norm

uD12 = |∇u|2 dx . Rn

Nonnegative solutions in D12 (Rn ) of (3.0.1) have been classified by Caffarelli, Gidas, and Spruck [17]. We refer also to Obata [60]. The result we use is as follows: if u ∈ D12 (Rn ), u nonnegative and nontrivial, is a solution of (3.0.1), then u = uλa for some a ∈ Rn and some λ > 0, where n−2 2 λ λ . (3.0.2) ua (x) = λ2 1 + n(n−2) |x − a|2 These functions are extremal functions for the sharp Euclidean Sobolev inequality 22

2 2 |u| dx ≤ Kn |∇u|2 dx . (3.0.3) Rn

Rn

Up to a nonzero multiplicative constant, they are the only nontrivial extremal functions for (3.0.3). Their energy is precisely the minimum energy in the sense that E(uλa ) = uλa 2 = Λmin . The free energy Ef , defined for u ∈ D12 (Rn ), is given by

1 1 |∇u|2 dx − |u|2 dx . (3.0.4) Ef (u) = 2 Rn 2 Rn It is easily checked that if u = uλa is a nonnegative nontrivial solution of (3.0.1), then Ef (u) = n1 Kn−n . We present in this chapter the Struwe result [73] for smooth compact Riemannian manifolds and equations like (Eα ) ∆g u + hα u = u2 −1 as considered in the preceding chapter. For the sake of clarity, we assume that the hα ’s are uniformly bounded and that they converge L2 to some limiting function.

26

CHAPTER 3

3.1 THE H12 -DECOMPOSITION FOR PALAIS-SMALE SEQUENCES Given (M, g) smooth, compact, of dimension n ≥ 3, we let (hα ) be a sequence of smooth functions on M . We assume that there exists C > 0 and a smooth (or only continuous) function h∞ on M such that the following holds: for any α and any x, |hα (x)| ≤ C and (3.1.1) hα → h∞ in L2 (M ) as α → +∞ . It clearly follows from (3.1.1) that, for any p ≥ 1, hα → h∞ in Lp (M ) as the parameter α → +∞. Let (uα ) be a Palais-Smale sequence of nonnegative functions for (Eα ). If

1 1 1 |∇u|2 dvg + hα u2 dvg − |u|2 dvg , Igα (u) = 2 M 2 M 2 M the uα ’s are characterized by the following: the sequence (Igα (uα )) is bounded and (3.1.2) DIgα (uα ) → 0 strongly in H12 (M ) as α → +∞ .

We let Ig∞ be the functional defined on H12 (M ) by

1 1 1 ∞ 2 2 |∇u| dvg + h∞ u dvg − |u|2 dvg Ig (u) = 2 M 2 M 2 M and (E∞ ) be the equation ∆g u + h∞ u = u2 −1 (E∞ ) where h∞ is as above. In the following, given δ > 0, ηδ denotes a smooth cutoff function in Rn such that ηδ = 1 in B0 (δ) and ηδ = 0 in Rn \B0 (2δ). For x ∈ M , where (M, g) is a smooth compact Riemannian manifold, and δ < ig /2, where ig is the injectivity radius, we let ηδ,x be the smooth cutoff function in M given by ηδ,x (y) = ηδ exp−1 x (y) where expx is the exponential map at x. Here and in what follows, we regard expx as defined in Rn . An intrinsic definition is possible if M is parallelizable. ˜ i , i = 1, . . . , N , be open subsets of M such that for any If not we let Ωi and Ω ˜ i , and such that M = ∪Ωi . The canonical ˜ i, Ωi is parallelizable and Ωi ⊂ Ω exponential map gives N maps expx defined in Ωi × Rn , and expx is, depending on the situation, one of these maps. A property of expx that holds for any x ∈ M should then be regarded as a property that holds for any i and any x ∈ Ωi .

Given a converging sequence (xα ) in M , and a sequence (Rα ) of positive real numbers such that Rα → +∞ as α → +∞, we define a bubble Bα as a sequence of functions on M such that n−2 Bα (x) = ηα (x)Rα 2 uλa Rα exp−1 xα (x) where ηα = ηδ,xα , δ < ig /2, and uλa is a nontrivial nonnegative solution of (3.0.1), thus given by (3.0.2). Roughly speaking, the Struwe result [73] we prove in this section is that, if (uα ) is a Palais-Smale sequence for (Eα ), then, in the H12 -sense, uα = solution of the limit equation + sum of bubbles with the additional property that the energies split also. An equivalent definition of bubbles is in Chapter 6. A more precise statement is as follows:

27

BLOW-UP THEORY IN SOBOLEV SPACES

T HEOREM 3.1 Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, (hα ) be a sequence of smooth functions on M such that (3.1.1) is satisfied and (uα ) be a Palais-Smale sequence of nonnegative functions for (Eα ). j j j ), Rα > 0, and Rα → +∞ as α → ∞, conThere exist m ∈ N, sequences (Rα j verging sequences (xα ) in M , a nonnegative solution u0 ∈ H12 (M ) of (E∞ ), and λ nontrivial nonnegative solutions uj = uajj ∈ D12 (Rn ) of (3.0.1), j = 1, . . . , m, such that, up to a subsequence, uα = u0 +

m

ηαj ujα + o(1)

(3.1.3)

j=1

where

j n−2 j 2 ujα (x) = Rα uj Rα exp−1 , j (x) x α

ηαj = ηδ,xjα , δ < ig /2, and o(1)H12 → 0 as α → +∞. Moreover, Igα (uα ) = Ig∞ (u0 ) +

m −n K + o(1) n n

(3.1.4)

where o(1) → 0 as α → ∞. The proof of Theorem 3.1 proceeds in several steps. We follow the proof of Struwe [73]. Uniqueness of the decomposition (3.1.3) is, for instance, discussed by Hebey [46]. We use the terminology of a Palais-Smale sequence for (Igα ) instead of a Palais-Smale sequence for (Eα ). First we claim that the following result holds: S TEP 1. Palais-Smale sequences for (Igα ) are bounded in H12 (M ). Proof of step 1. We let (uα ) be a Palais-Smale sequence for (Igα ). Then,

|∇uα |2 dvg + hα u2α dvg − |uα |2 dvg DIgα (uα ).uα = M M M = o uα H12 . Hence, Igα (uα )

1 = n

M

|uα |2 dvg + o uα H12

and, since Igα (uα ) ≤ C for some C > 0 independent of α, we get that

|uα |2 dvg ≤ C + o uα H12 . M

Thanks to Hölder’s inequality, this implies in turn that

2/2 u2α dvg ≤ C + o uα H 2 M

1

where C > 0, like all the constants below, is independent of α. Writing that

2 |∇uα |2 dvg + hα u2α dvg = 2Igα (uα ) + |uα |2 dvg , 2 M M M

28

CHAPTER 3

we also get that

|∇uα |2 dvg +

M

M

hα u2α dvg ≤ C + o uα H12 .

Noting that, thanks to (3.1.1),

|∇uα |2 + hα u2α dvg + Cuα 22 , uα 2H 2 ≤ 1

M

it follows from the above equations that 2/2 uα 2H 2 ≤ C + o uα H12 + o uα H 2 . 1

1

This clearly implies that the uα ’s are bounded in

H12 (M ).

Step 1 is proved.

2

A second step in the proof is as follows. We let Ig be the functional defined on H12 (M ) by

1 1 |∇u|2 dvg − |u|2 dvg Ig (u) = 2 M 2 M so that Ig = Igα when hα ≡ 0. S TEP 2. Let (uα ) be a Palais-Smale sequence of nonnegative functions for Igα , and u0 ∈ H12 (M ) a nonnegative function such that uα u0 weakly in H12 (M ), uα → u0 strongly in L2 (M ), and uα → u0 almost everywhere as α → +∞. Let uα ) is a Palais-Smale sequence for Ig and u ˆα = uα − u0 . Then (ˆ Ig (ˆ uα ) = Igα (uα ) − Ig∞ (u0 ) + o(1) where o(1) → 0 as α → +∞. Moreover, u0 is a solution of (E∞ ). Proof of step 2. We first observe that for any ϕ ∈ C ∞ (M ),

α 2 −1 (∇uα ∇ϕ) dvg + hα uα ϕdvg − uα ϕdvg DIg (uα ).ϕ = M

M

M

(3.1.5)

= o(1) where o(1) → 0 as α → +∞, and (∇uα ∇ϕ) is the scalar product with respect to g of ∇uα and ∇ϕ. By (3.1.1), hα → h∞ in Lp (M ) for any p ≥ 1 as α → +∞. Writing that

hα uα ϕdvg = (hα − h∞ )uα ϕdvg + h∞ uα ϕdvg (3.1.6) M

M

M

and noting that, from Hölder’s inequalities,

(hα − h∞ )uα ϕdvg ≤ uα 2 hα − h∞ n

2

1/2

|ϕ| dvg

M

M

we easily get by passing to the limit in (3.1.5) that

0 h∞ u0 ϕdvg = (u0 )2 −1 ϕdvg . ∇u ∇ϕ dvg + M

M

M

,

(3.1.7)

29


ˆα . We know by Hence, u0 is a solution of (E∞ ). Now we compute the energy of u step 1 that the uα ’s are bounded in H12 (M ). From the Sobolev embedding theorem and (3.1.6), (3.1.7) with ϕ = uα , we can thus write that

hα u2α dvg = h∞ (u0 )2 dvg + o(1) . M

M

It follows that Igα (uα )

=

Ig∞ (u0 )

1 + Ig (ˆ uα ) − 2

Φα dvg + o(1)

(3.1.8)

M

uα +u0 |2 −|ˆ uα |2 −|u0 |2 . Clearly, there exists C > 0, independent where Φα = |ˆ of α, such that

|Φα | dvg ≤ C |ˆ uα |2 −1 |u0 |dvg + C |u0 |2 −1 |ˆ uα |dvg

M

M

M

while basic integration theory gives that

|ˆ uα |2 −1 |u0 |dvg = o(1) and M

Therefore,

M

|u0 |2

−1

|ˆ uα |dvg = o(1) .

M

Φα dvg = o(1), and coming back to (3.1.8), we get that Ig (ˆ uα ) = Igα (uα ) − Ig∞ (u0 ) + o(1) .

Summarizing, we are left in the proof of step 2 with the proof that (ˆ uα ) is a PalaisSmale sequence for Ig . Let ϕ ∈ C ∞ (M ). From (3.1.6), (3.1.7), and the Sobolev embedding theorem,

hα uα ϕdvg = h∞ u0 ϕdvg + o ϕH12 . M

M

It follows that

DIgα (uα ).ϕ

= DIg (ˆ uα ).ϕ − M

Ψα ϕdvg + o ϕH12

(3.1.9)

where uα + u0 |2 Ψα = |ˆ

−2

(ˆ uα + u0 ) − |ˆ uα |2

−2

u ˆα − |u0 |2

−2 0

u .

Here again, it is easily checked that there exists C > 0, independent of α, such that

|Ψα ϕ|dvg ≤ C |ˆ uα |2 −2 |u0 ||ϕ|dvg + C |u0 |2 −2 |ˆ uα ||ϕ|dvg . M

M

M

From Hölder’s inequalities we then get that

|Ψα ϕ|dvg M 0 2 −2 ≤ C |ˆ uα |2 −2 u0 + |ˆ u | u ˆα 2 /(2 −1)

2 /(2 −1)

while basic integration theory gives that 0 2 −2 = o(1) and |ˆ u | u ˆα uα |2 −2 u0 |ˆ 2 /(2 −1)

2 /(2 −1)

ϕ2

= o(1) .

30

CHAPTER 3

Coming back to (3.1.9), and from the Sobolev embedding theorem, we then get that DIgα (uα ).ϕ = DIg (ˆ uα ).ϕ + o ϕH12 . This implies that (ˆ uα ) is a Palais-Smale sequence for Ig . Step 2 is proved.

2

A third step in the proof of Theorem 3.1 is the following. We let β = be the value of the free energy Ef when considered on the uλa ’s of (3.0.2).

1 −n n Kn

uα ) be a Palais-Smale sequence for Ig such that u ˆα 0 weakly S TEP 3. Let (ˆ uα ) → β as α → +∞, where β < β . in H12 (M ) as α → +∞, and such that Ig (ˆ Then u ˆα → 0 strongly in H12 (M ). Proof of step 3. By step 1, the u ˆα ’s are bounded in H12 (M ). Independently,

2 uα ).ˆ uα = |∇ˆ uα | dvg − |ˆ uα |2 dvg DIg (ˆ M M = o ˆ uα H12 . Hence, 1 1 ˆ uα 22 + o(1) = ∇ˆ uα 22 + o(1) = β + o(1) . (3.1.10) n n This already implies that β ≥ 0. Following Hebey-Vaugon [47, 48], there exists B > 0, independent of α, such that

Ig (ˆ uα ) =

ˆ uα 22 ≤ Kn2 ∇ˆ uα 22 + Bˆ uα 22 . Moreover, since the embedding H12 (M ) ⊂ L2 (M ) is compact, u ˆα → 0 strongly in L2 (M ) as α → +∞. Letting α → +∞ in the above sharp Sobolev inequality, it follows from (3.1.10) that (nβ)2/2 ≤ Kn2 nβ .

Since β < β , this implies that β = 0. By (3.1.10) we then get that u ˆα → 0 2 strongly in H12 (M ). This proves step 3. It follows from steps 1–3 that if (uα ) is a Palais-Smale sequence for Igα , and → β as α → +∞, where β < β , then, up to a subsequence, (uα ) converges strongly to some u0 in H12 (M ). In other words, compactness holds for Palais-Smale sequences when the energy is below the minimum energy. Another illustration of this fact is given by Proposition 2.2.

Igα (uα )

The following step is the main ingredient in the proof of Theorem 3.1. We postpone its proof to section 3.2. S TEP 4. Let (ˆ uα ) be a Palais-Smale sequence for Ig such that u ˆα 0 weakly in H12 (M ) but not strongly. Then there exists a sequence (Rα ) of positive real numbers, Rα → +∞ as α → ∞, a converging sequence (xα ) in M , and a nontrivial solution u ∈ D12 (Rn ) of the Euclidean equation ∆u = |u|2

−2

u

(3.1.11)

31


such that, up to a subsequence, the following holds: if vˆα = u ˆα − ηα Bα where

n−2 Bα (x) = Rα 2 u Rα exp−1 xα (x)

and ηα = ηδ,xα , δ < ig /2, then (ˆ vα ) is also a Palais-Smale sequence for Ig , vα ) = Ig (ˆ uα ) − Ef (u) + o(1), vˆα 0 weakly in H12 (M ) as α → +∞, and Ig (ˆ where o(1) → 0 as α → +∞. Following steps 1–4, we are now in a position to prove Theorem 3.1. The proof proceeds as follows. Proof of Theorem 3.1. A preliminary claim is that nontrivial solutions of the Euclidean equation (3.1.11) have a free energy Ef bounded from below by β . Indeed, if u ∈ D12 (Rn ) is a nontrivial solution of (3.1.11), we get with the sharp Euclidean Sobolev inequality (3.0.3) that

2 /2

|∇u|2 dx = |u|2 dx ≤ Kn2 |∇u|2 dx . Hence,

Rn

Rn

Rn

2

|∇u| dx ≥

Rn

Kn−n

and

1 1 − |∇u|2 dx ≥ β . Ef (u) = 2 2 Rn

This proves the above claim. In order to prove the theorem, we let (uα ) be a PalaisSmale sequence of nonnegative functions for Igα . According to step 1, (uα ) is bounded in H12 (M ). Up to a subsequence, we may therefore assume that for some u0 ∈ H12 (M ), uα u0 weakly in H12 (M ), uα → u0 strongly in L2 (M ), and uα → u0 almost everywhere as α → +∞. We may also assume that Igα (uα ) → c as α → +∞. By step 2, u0 is a nonnegative solution of (3.0.1) and u ˆ α = uα − u 0 is a Palais-Smale sequence for Ig such that Ig (ˆ uα ) = Igα (uα ) − Ig∞ (u0 ) + o(1) . If u ˆα → 0 strongly in H12 (M ), then uα = u0 + o(1), and the theorem is proved. If not, according to the claim at the beginning of this proof, we apply step 4 to get a new Palais-Smale sequence (ˆ u1α ) of energy Ig (ˆ u1α ) ≤ Ig (ˆ uα ) − β + o(1) . Here again, either u ˆ1α → 0 strongly in H12 (M ), in which case the theorem is proved, 1 or u ˆα 0 weakly but not strongly in H12 (M ), in which case we again apply step 4. By induction, at some point, either we do have compactness or the PalaisSmale sequence (ˆ um α ) we get with this process has an energy that converges to 2 ˆm some β < β . Then, by step 3, u α → 0 strongly in H1 (M ). Applying step 4 j in this process, we got some u ’s, j = 1, . . . , m. Assuming that the uj ’s are also nonnegative (see below) this ends the proof of Theorem 3.1. 2 The fact that the uj ’s we get in the above process are nonnegative is proved in the next section. We refer to Lemma 3.3.

32

CHAPTER 3

3.2 SUBTRACTING A BUBBLE AND NONNEGATIVE SOLUTIONS In this section, we prove step 4 and we prove that the uj ’s in Theorem 3.1 are nonnegative. We start with the proof of step 4. The notations are those of section 3.1. L EMMA 3.2 Let (ˆ uα ) be a Palais-Smale sequence for Ig such that u ˆα 0 weakly in H12 (M ) but not strongly. Then there exist a sequence (Rα ) of positive real numbers, Rα → +∞ as α → ∞, a converging sequence (xα ) in M , and a nontrivial solution u ∈ D12 (Rn ) of the Euclidean equation ∆u = |u|2

−2

u

such that, up to a subsequence, the following holds: if vˆα = u ˆα − ηα Bα where

n−2 Bα (x) = Rα 2 u Rα exp−1 xα (x)

and ηα = ηδ,xα , δ < ig /2, then (ˆ vα ) is also a Palais-Smale sequence for Ig , vα ) = Ig (ˆ uα ) − Ef (u) + o(1), vˆα 0 weakly in H12 (M ) as α → +∞, and Ig (ˆ where o(1) → 0 as α → +∞. uα ) → β as α → +∞. We Proof. Up to a subsequence, we may assume that Ig (ˆ may also assume that u ˆα is smooth, since if not there always exists uα smooth and ˆα H12 → 0. Then, (uα ) is a Palais-Smale sequence for Ig such such that uα − u that uα 0 weakly in H12 (M ) but not strongly, and, as is easily checked, if the uα ). Since DIg (ˆ uα ) → 0, we get as in claim holds for (uα ), then it holds also for (ˆ step 3 of section 3.1 that

|∇ˆ uα |2 dvg = nβ + o(1) (3.2.1) M

and that nβ ≥ Kn−n . For t > 0, we let µα (t) = max x∈M

Bx (t)

|∇ˆ uα |2 dvg .

Given t0 > 0 small, it follows from (3.2.1) that there exist x0 ∈ M and λ0 > 0 such that, up to a subsequence,

|∇ˆ uα |2 dvg ≥ λ0 Bx0 (t0 )

for all α. Then, since t → µα (t) is continuous, we get that for any λ ∈ (0, λ0 ), there exists tα ∈ (0, t0 ) such that µα (tα ) = λ. Clearly, there also exists xα ∈ M such that

µα (tα ) =

Bxα (tα )

|∇ˆ uα |2 dvg .

Up to a subsequence, (xα ) converges. We let r0 ∈ (0, ig /2) be such that for all x ∈ M and all y, z ∈ Rn , if |y| ≤ r0 and |z| ≤ r0 , then dg (expx (y), expx (z)) ≤ C0 |z − y|

33


for some C0 ∈ [1, 2] independent of x, y, and z. Given Rα ≥ 1 and x ∈ Rn such that |x| < ig Rα , we let − n−2 −1 ˆα expxα (Rα x) , u ˜α (x) = Rα 2 u −1 x) . g˜α (x) = expxα g (Rα Then,

−1 n x) = Rα |∇˜ uα |2 (x) , |∇ˆ uα |2 expxα (Rα

where the norm in the left hand side of this equation is with respect to g, and the norm in the right hand side is with respect to g˜α . It follows that, if |z| + r < ig Rα , then

2 |∇˜ uα | dvg˜α = (3.2.2) |∇ˆ uα |2 dvg . −1 Bz (r) expxα (Rα Bz (r)) When |z| + r < r0 Rα ,

−1 Bz (r) ⊂ Bexpx expxα Rα

α

while

−1 (Rα z)

−1 C0 rRα

−1 −1 expxα Rα B0 (C0 r) = Bxα C0 rRα .

(3.2.3) (3.2.4)

C0 rt−1 0

≥ 1. Then, for any λ ∈ (0, λ0 ), to Given r ∈ (0, r0 ), we fix t0 such that −1 = tα . By (3.2.2)–(3.2.4), be fixed later on, we let Rα ≥ 1 be such that C0 rRα n for any z ∈ R such that |z| < r0 Rα − r,

|∇˜ uα |2 dvg˜α ≤ λ and Bz (r)

(3.2.5) |∇˜ uα |2 dvg˜α = λ . B0 (C0 r)

We let δ ∈ (0, ig ) and C1 > 1 be such that for any x ∈ M , and any R ≥ 1, if g˜x,R (y) = expx g(R−1 y), then

1 2 2 |∇u| dx ≤ |∇u| dvg˜x,R ≤ C1 |∇u|2 dx (3.2.6) C1 Rn Rn Rn for all u ∈ D12 (Rn ) such that suppu ⊂ B0 (δR). Without loss of generality, we also assume that

1 |u|dx ≤ |u|dvg˜x,R ≤ C1 |u|dx (3.2.7) C1 Rn Rn Rn for all u ∈ L1 (Rn ) such that suppu ⊂ B0 (δR). We let η˜ ∈ C0∞ (Rn ) be a cutoff function such that η˜ ≤ 1, η˜ = 1 in B0 (1/4), and η˜ = 0 in Rn \B0 (3/4). We −10 ≤ −1 set η˜α (x) = η˜ δ Rα x , where δ is as above. Then,

|∇(˜ ηα u ˜α )|2 dvg˜α = O(1) Rn

˜α ) is bounded in D12 (Rn ). In and it follows from (3.2.6) that the sequence (˜ ηα u ˜α u particular, up to a subsequence, there exists u ∈ D12 (Rn ) such that η˜α u

34

CHAPTER 3

weakly in D12 (Rn ). Now we divide the proof of Lemma 3.2 into several steps. As a first step, we claim that the following holds. S TEP 1. For r and λ sufficientl small, η˜α u ˜α → u strongly in H12 (B0 (C0 r))

(3.2.8)

as α → +∞. Proof of step 1. We let x0 ∈ Rn , and for ρ > 0, we let hρ be the standard metric on ∂Bx0 (ρ). By Fatou’s lemma,

2r

Nξ (˜ ηα u ˜α )dvhρ dρ ≤ lim inf Nξ (˜ ηα u ˜α )dx ≤ C lim inf r

α→+∞

α→+∞

∂Bx0 (ρ) 2

Bx0 (2r)

2

where Nh (u) = |∇u| + u , the norm in Nh is with respect to h, and ξ is the Euclidean metric. It follows that there exists ρ ∈ [r, 2r] such that, up to a subsequence, and for all α,

Nξ (˜ ηα u ˜α )dvhρ ≤ C . ∂Bx0 (ρ)

As an easy consequence, we get that ˜ ηα u ˜α H 2 (∂Bx

)≤C where C > 0 is independent of α. The embedding 1

0 (ρ)

2 H12 (∂Bx0 (ρ)) ⊂ H1/2 (∂Bx0 (ρ))

is compact, and the trace operator u → u|∂B is continuous. It follows that, up to a subsequence, 2 η˜α u ˜α → u in H1/2 (∂Bx0 (ρ))

as α → +∞. Let A be the annulus A = Bx0 (3r)\Bx0 (ρ). Let also ϕα ∈ D12 (Rn ) ˜α − u in Bx0 (ρ + ε), and ϕα = 0 in Rn \Bx0 (3r − ε), ε > 0 be such that ϕα = η˜α u small. Then ˜ ηα u ˜α − uH 2 (∂Bx (ρ)) = ϕα H 2 (∂Bx (ρ)) 0 0 1/2 1/2 while there exists ϕ0α ∈ D12 (A), the closure of C0∞ (A) in H12 (A), such that 2 (∂A) . ϕα + ϕ0α H12 (A) ≤ Cϕα H1/2

Minimization arguments give that there exists zα ∈ H12 (A) such that ∆zα = 0 in A , zα − ϕα − ϕ0α ∈ D12 (A) , and zα H12 (A) ≤ Cϕα + ϕ0α H12 (A) . Hence, zα → 0 strongly in H12 (A) as α → +∞. We let ψα ∈ D12 (Rn ) be such that ψα = η˜α u ˜α − u in B x0 (ρ) , ψα = zα in B x0 (3r)\Bx0 (ρ) ,

35


and ψα = 0 otherwise. We let r be such that r < min(ig /6, δ/24), and let ψ˜α be such that n−2 ψ˜α (x) = Rα 2 ψα Rα exp−1 xα (x) if dg (xα , x) < 6r, and ψ˜α = 0 otherwise. Clearly, η˜ δ −1 exp−1 xα (x) = 1 if dg (xα , x) < 6r. If in addition |x0 | < 3r, then DIg (ˆ uα ).ψ˜α = DIg (ˆ ηα u ˆα ).ψ˜α

= (∇(˜ ηα u ˜α )∇ψα ) dvg˜α − Bx0 (3r)

Bx0 (3r)

|˜ ηα u ˜α |2

−2

(˜ ηα u ˜α )ψα dvg˜α

˜ where ηˆα (x) = η˜ δ −1 exp−1 xα (x) . We have that ψα H12 (M ) ≤ Cψα D12 (Rn ) . In uα ).ψ˜α = o(1). particular, the ψ˜α ’s are bounded in H12 (M ). It follows that DIg (ˆ 2 Noting that ψα → 0 strongly in H1 (A), and ψα 0 weakly in D12 (Rn ),

(∇(˜ ηα u ˜α )∇ψα ) dvg˜α Bx0 (3r)

=

Bx0 (ρ)

=

Rn

(∇(ψα + u)∇ψα ) dvg˜α + o(1)

|∇ψα |2 dvg˜α + o(1) .

Similarly, one easily gets that

|˜ ηα u ˜α |2 −2 (˜ ηα u ˜α )ψα dvg˜α = Bx0 (3r)

|ψα |2 dvg˜α + o(1) ,

Rn

and since DIg (ˆ uα ).ψ˜α = o(1), we have proved that

2 |∇ψα | dvg˜α − |ψα |2 dvg˜α = o(1) . Rn

Rn

(3.2.9)

By the strong convergence ψα → 0 in H12 (A), and the weak convergence ψα 0 in D12 (Rn ),

|∇ψα |2 dvg˜α = |∇(˜ ηα u ˜α − u)|2 dvg˜α + o(1) Rn

Bx0 (ρ)

= so that

2

Bx0 (ρ)

Rn

|∇(˜ ηα u ˜α )| dvg˜α −

2

|∇ψα | dvg˜α ≤

Bx0 (ρ)

Bx0 (ρ)

|∇u|2 dvg˜α + o(1)

|∇(˜ ηα u ˜α )|2 dvg˜α + o(1) .

Let N be an integer such that B0 (2) is covered by N balls of radius 1 and centered in B0 (2). Then there exist N points xi ∈ Bx0 (2r), i = 1, . . . , N , such that Bx0 (ρ) ⊂ Bx0 (2r) ⊂

N i=1

Bxi (r)

36

CHAPTER 3

and we get with (3.2.5) that for x0 and r such that |x0 | + 3r < r0 ,

|∇ψα |2 dvg˜α ≤ N λ + o(1) . Rn

(3.2.10)

Independently, from the Sobolev inequality, for C1 as in (3.2.6) and (3.2.7), and x0 and r such that |x0 | + 3r < δ, we also have that

2/2

2/2 2/2 2 2 |ψα | dvg˜α ≤ C1 |ψα | dx Rn Rn

2/2 ≤ C1 Kn2 |∇ψα |2 dx Rn

1+(2/2 ) 2 ≤ C1 Kn |∇ψα |2 dvg˜α . Rn

Following (3.2.9) and (3.2.10), we can then write that

|∇ψα |2 dvg˜α ≤ K |∇ψα |2 dvg˜α + o(1) Rn

Rn

where 1+(2 /2)

K = C1 Let λ > 0 be such that

1+(2 /2) 2 Kn (N λ)(2 /2)−1 C1

so that ψα → 0 strongly in

(2 /2)−1

Kn2 (N λ + o(1))

.

< 1. Then,

|∇ψα |2 dvg˜α = o(1)

Rn D12 (Rn )

as α → +∞. Since r ≤ ρ, it follows that

˜α → u strongly in H12 (Bx0 (r)) η˜α u

(3.2.11)

1+(2 /2) 2 Kn (N λ)(2 /2)−1 C1

< 1, |x0 | < 3r, and the convergence holds as soon as |x0 | + 3r < r0 , |x0 | + 3r < δ, and r < min(ig /6, δ/24). We fix λ > 0 sufficiently 1+(2 /2) 2 Kn (N λ)(2 /2)−1 < 1, and let r > 0 be sufficiently small such that C1 small such that r < min(ig /6, δ/24, r0 /6). Then (3.2.11) holds for any x0 such that |x0 | < 2r. Since C0 ≤ 2, B0 (C0 r) is covered by N balls of radius r and ˜α → u strongly in H12 (B0 (C0 r)). This centered in B0 (2r). It follows that η˜α u proves (3.2.8). 2 From (3.2.5) and (3.2.8), we can write that

|∇˜ uα |2 dvg˜α λ= B0 (C0 r)

= |∇(˜ ηα u ˜α )|2 dvg˜α B0 (C0 r)

≤ C1 |∇u|2 dx + o(1) . B0 (C0 r)

It follows that u ≡ 0. Let us assume that Rα → R as α → +∞, R ≥ 1. If ˆα 0 weakly in H12 (M ). R < +∞, then u ˜α 0 weakly in H12 (B0 (C0 r)) since u From (3.2.8), and since u ≡ 0, we get that lim Rα = +∞ .

α→+∞

(3.2.12)

37


We claim now that the following holds. S TEP 2. For any R > 0, u ˜α → u strongly in H12 (B0 (R)) as α → +∞, and u is a solution of the Euclidean equation ∆u = |u|2

(3.2.13)

−2

u.

Proof of step 2. We let R ≥ 1 be given. By (3.2.12), Rα ≥ R for α large, and (3.2.5) holds for z such that |z| < r0 R − r. Then, as is easily checked from the proof of step 1, (3.2.11) holds if |x0 | < 3r(2R − 1), |x0 | + 3r < r0 R, and ˜α → u |x0 | + 3r < δR. In particular, (3.2.11) holds if |x0 | < 2rR. Hence, η˜α u strongly in H12 (B0 (2rR)). Noting that for x in a compact subset of Rn , η˜α (x) = 1 for α large, and that R ≥ 1 is arbitrary, we easily get that (3.2.13) holds. Now we prove that u is a solution of the critical Euclidean equation ∆u = |u|2 −2 u. Let ϕ ∈ C0∞ (Rn ) and R0 > 0 be such that suppϕ ⊂ B0 (R0 ). Let also ϕˆα be given by n−2

ϕˆα (x) = Rα 2 ϕ(Rα x) . −1 R0 ). For α large, we let ϕα be the smooth function on M Then suppϕˆα ⊂ B0 (Rα given by ϕˆα = ϕα ◦ expxα . For α large,

(∇ˆ uα ∇ϕα ) dvg = (∇(˜ ηα u ˜α )∇ϕ) dvg˜α Rn

M

and

|ˆ uα |2

−2

u ˆα ϕα dvg =

M

Rn

|˜ ηα u ˜α |2

−2

(˜ ηα u ˜α )ϕdvg˜α .

From (3.2.12), g˜α → ξ in C 1 (B0 (R)) for any R > 0. Moreover, (ϕα ) is bounded in H12 (M ). Since (ˆ uα ) is a Palais-Smale sequence for Ig , and η˜α u ˜α → u in D12 (Rn ), we get by passing to the limit as α → +∞ in the above equations that

(∇u∇ϕ) dx = |u|2 −2 uϕdx . Rn

In other words, u ∈

Rn

D12 (Rn )

is such that ∆u = |u|2

−2

u. This proves step 2.

For x ∈ M and δˆ ∈ (0, δ/8), we let Vα be given by n−2 Vα (x) = ηα (x)Rα 2 u Rα exp−1 xα (x)

2

(3.2.14)

ˆα − Vα and claim that the following holds. where ηα = ηδ,x ˆ α . We let wα = u S TEP 3. The following relations hold. On the one hand, wα 0 weakly in H12 (M )

(3.2.15)

as α → +∞. On the other hand, DIg (Vα ) → 0 and DIg (wα ) → 0

(3.2.16)

strongly as α → +∞. Finally, uα ) − Ef (u) + o(1) Ig (wα ) = Ig (ˆ

(3.2.17)

38

CHAPTER 3

where o(1) → 0 as α → +∞ Proof of step 3. We start with the proof of (3.2.15). It suffices to prove that Vα 0 −1 R). For ϕ a smooth weakly in H12 (M ). Given R > 0, we let Ωα (R) = Bxα (Rα function on M and α large,

n−2 Vα ϕdvg = Rα 2 ηδˆ(x)u(Rα x)ϕ(expxα (x))dvgα −1 B0 (Rα R)

Ωα (R)

where gα = expxα g. It follows that for C > 0 such that dvgα ≤ Cdx,

− n+2 |u|dx . Vα ϕdvg ≤ Cϕ∞ Rα 2 Ωα (R) B0 (R) Similarly, by Hölder’s inequality,

− n+2 2 Vα ϕdvg ≤ Cϕ∞ Rα |u|dx M \Ωα (R) B0 (δRα )\B0 (R) 1/2

≤ Cϕ∞

|u|2 dx

B0 (δRα )\B0 (R)

.

Taking R > 0 sufficiently large and using (3.2.12), we get that M Vα ϕdvg → 0 as α → +∞. Similar arguments give that M (∇Vα ∇ϕ)dvg → 0 as α → +∞. This proves (3.2.15). Now we prove (3.2.16). Here again we let ϕ be a smooth function on M . Then,

DIg (Vα ).ϕ = (∇Vα ∇ϕ)dvg − |Vα |2 −2 Vα ϕdvg . M

Given R > 0, we write that

(∇Vα ∇ϕ)dvg =

Ωα (R)

M

M

(∇Vα ∇ϕ)dvg +

Easy computations give

Bxα (δ)\Ωα (R)

Bxα (δ)\Ωα (R)

(∇Vα ∇ϕ)dvg .

(∇Vα ∇ϕ)dvg = O ϕH12 εR

where εR → 0 as R → +∞. Let ϕα be the function of D12 (Rn ) given by −1 − n−2 ϕα (x) = Rα 2 ηα,δˆ(x) ϕ ◦ expxα (Rα x) −1 x). Then, for α large, where ηα,δˆ(x) = ηδˆ(Rα

(∇Vα ∇ϕ)dvg = (∇u∇ϕα )dvg˜α . Ωα (R)

B0 (R)

Noting that g˜α → ξ in C 1 (B0 (R )), R > R, and that

2 |∇ϕ| dvg = |∇ϕα |2 dvg˜α , Ωα (R)

B0 (R)

39


we get that

B0 (R)

(∇u∇ϕα )dvg˜α =

We also have that

B0 (R)

B0 (R)

(∇u∇ϕα )dx =

Rn

(∇u∇ϕα )dx + o ϕH12 .

(∇u∇ϕα )dx + O ϕH12 εR

where εR is as above. Therefore,

(∇Vα ∇ϕ)dvg = (∇u∇ϕα )dx+o ϕH12 +O ϕH12 εR . (3.2.18) Rn

M

In a similar way, we can prove that

2 −2 |Vα | Vα ϕdvg = |u|2 −2 uϕα dx n M R + o ϕH12 + O ϕH12 εR . Since u is a solution of ∆u = |u|2

(3.2.19)

−2

u, it follows from (3.2.18) and (3.2.19) that DIg (Vα ).ϕ = o ϕH12 + O ϕH12 εR .

Since R > 0 is arbitrary, we get that DIg (Vα ) → 0 strongly as α → +∞. This proves the first claim of (3.2.16). Now we write DIg (wα ).ϕ = DIg (ˆ uα ).ϕ − DIg (Vα ).ϕ − A(α).ϕ , where

(3.2.20)

A(α).ϕ =

Φα ϕdvg = M

ˆ Bxα (2δ)

Φα ϕdvg

and Φα = |wα |2

−2

wα − |ˆ uα |2

−2

u ˆα + |Vα |2

−2

Vα .

By the Hölder and Sobolev inequalities, |A(α).ϕ| ≤ Φα 2 /(2 −1) ϕH12 . ˆ Given R > 0, we let Ωα (R)c = Bxα (2δ)\Ω α (R), where Ωα (R) is as above. Then, for α large, Φα 2 /(2 −1) ≤ Φα L2 /(2 −1) (Ωα (R)) + Φα L2 /(2 −1) (Ωα (R)c ) . As in the proof of step 2 of section 3.1, we can write Φα L2 /(2 −1) (Ωα (R)c ) ≤ C Φ1α L2 /(2 −1) (Ωα (R)c ) + Φ2α L2 /(2 −1) (Ωα (R)c ) uα |2 −2 Vα and Φ2α = |Vα |2 −2 u ˆα . We have that where Φ1α = |ˆ

2 ˜ α | 22−1 dvg˜ |Φα | 2 −1 dvg = |Φ α

Ωα (R)

B0 (R)

40

CHAPTER 3

where ˜ α = |˜ uα − u|2 −2 (˜ uα − u) − |˜ uα |2 −2 u ˜α + |u|2 −2 u . Φ

Using (3.2.13), we then get that

2

Ωα (R)

|Φα | 2 −1 dvg = o(1) .

Independently,

2 |Φ1α | 2 −1 dvg =

ˆ α )\B0 (R) B0 (2δR

Ωα (R)c

|˜ ηα u ˜α |

≤C where

Rn \B0 (R)

|˜ ηα u ˜α |

2 (2 −2) 2 −1

2 (2 −2) 2 −1

2

2 −1

|u| 2 −1 ηˆα2

dvg˜α

2

|u| 2 −1 dx ,

−1 ηˆα (x) = ηδ,x ˆ α expxα (Rα x)

and C > 0 is such that dvg˜α ≤ Cdx. Without loss of generality, we may assume ˜α → u almost everywhere in Rn . We let that η˜α u ηα u ˜α | fα = |˜

2 (2 −2) 2 −1

and f = |u|

2 (2 −2) 2 −1

.

2 −1 2 −2

Then (fα ) is bounded in L (Rn ) and (fα ) converges almost everywhere to f . From standard integration theory, it follows that 2 −1

fα f weakly in L 2 −2 (Rn ) as α → +∞. Hence,

lim α→+∞

Rn \B

0 (R)

|˜ ηα u ˜α |

2 (2 −2) 2 −1

and we get that

|u|

lim lim sup

R→+∞ α→+∞

Similarly, we can prove that lim lim sup

R→+∞ α→+∞

Ωα (R)c

Ωα (R)c

2 2 −1

dx =

Rn \B

|u|2 dx ,

0 (R)

2

|Φ1α | 2 −1 dvg = 0 .

2

|Φ2α | 2 −1 dvg = 0 .

Coming back to (3.2.20), and since R > 0 is arbitrary, we get that DIg (wα ) → 0 strongly as α → +∞. This proves (3.2.16). Now we are left with the proof of (3.2.17). We have that

1 1 Ig (wα ) = |∇wα |2 dvg − |wα |2 dvg . (3.2.21) 2 M 2 M Concerning the first term in (3.2.21), we write

2 2 |∇wα | dvg = |∇wα | dvg + M

ˆ Bxα (2δ)

ˆ M \Bxα (2δ)

|∇ˆ uα |2 dvg .

41


Then, given R > 0,

|∇wα |2 dvg = ˆ Bxα (2δ)

Ωα (R)

|∇wα |2 dvg +

Ωα (R)c

|∇wα |2 dvg

where Ωα (R) and Ωα (R)c are as above. We have

2 |∇wα |2 dvg = |∇(˜ uα − u)| dvg˜α Ωα (R)

B0 (R)

so that, from (3.2.13),

Ωα (R)

|∇wα |2 dvg = o(1) .

Independently, it follows from rough estimates that

lim lim sup |∇Vα |2 dvg = 0 . R→+∞ α→+∞

Ωα (R)c

Since wα = u ˆα − Vα , and (ˆ uα ) is bounded in H12 (M ), we have that

|∇wα |2 dvg = |∇ˆ uα |2 dvg + BR (α) Ωα (R)c

Ωα (R)c

where lim lim sup BR (α) = 0 .

R→+∞ α→+∞

Therefore,

|∇wα |2 dvg = M

|∇ˆ uα |2 dvg −

Ωα (R)

M

where BR (α) satisfies (3.2.22). Noting that

2 |∇ˆ uα | dvg = Ωα (R)

(3.2.22)

|∇ˆ uα |2 dvg + BR (α) + o(1)

B0 (R)

|∇˜ uα |2 dvg˜α

and that g˜α → ξ in C 1 (B0 (R)), we get with (3.2.13) that

|∇ˆ uα |2 dvg = |∇u|2 dx + o(1) Ωα (R) B0 (R)

= |∇u|2 dx + εR + o(1) Rn

where εR → 0 as R → +∞. Hence,

2 2 |∇wα | dvg = |∇ˆ uα | dvg − M

M

Rn

|∇u|2 dx + BR (α) + o(1)

where BR (α) satisfies (3.2.22). Similarly, we can prove that

|wα |2 dvg = |ˆ uα |2 dvg − |u|2 dx + BR (α) + o(1) M

M

Rn

(3.2.23)

(3.2.24)

where BR (α) satisfies (3.2.22). Combining (3.2.21), (3.2.23), and (3.2.24), we then get that Ig (wα ) = Ig (ˆ uα ) − Ef (u) + BR (α) + o(1) ,

42

CHAPTER 3

and since R > 0 is arbitrary, it follows that uα ) − Ef (u) + o(1) . Ig (wα ) = Ig (ˆ This proves (3.2.17) and step 3.

2

From steps 1–3, Lemma 3.2 holds for some δ ∈ (0, ig /2) small. Given δ1 < δ2 in (0, ig /2), it is easily seen that (ηδ2 ,xα − ηδ1 ,xα )Bα H 2 = o(1) . 1

It follows that Lemma 3.2 holds for any δ ∈ (0, ig /2). This ends the proof of Lemma 3.2. 2 Another important ingredient we used in the proof of Theorem 3.1 is the following result. L EMMA 3.3 If (uα ) is a Palais-Smale sequence of nonnegative functions for (Igα ), λ

the uj ’s of Theorem 3.1 are also nonnegative. Therefore, when j ≥ 1, uj = uajj for some aj ∈ Rn and λj > 0, where uλa is as in (3.0.2). ˆα = uα − u0 and let Proof. That u0 is nonnegative is straightforward. We let u i i µα = 1/Rα . First we prove the following: for any N integer in [1, m], and for any s integer in [0, N − 1], there exist an integer p and sequences (yαj ) and (λjα ), j N j = 1, . . . , p, yαj ∈ M , and λjα > 0, such that for any j, dg (xN α , yα )/µα is bounded j N and λα /µα → 0, and such that for any R, R > 0, 2

s (3.2.25) uiα − uN ˆα − dvg = o(1) + ε(R ) u α p j N ˜ Ωα (R)\∪j=1 Ωα (R ) i=1

j ˜j j (R λ ), ε(R ) → 0 as R → +∞, where ΩN (RµN α (R) = BxN α ), Ωα (R ) = Byα α α and the (uiα )’s and (xiα )’s are the ordered sequences in i we get in the proof of Theorem 3.1. We proceed here by inverse induction on s. If s = N − 1, then, by (3.2.13), 2

N −1 uiα − uN ˆα − u α dvg = o(1) ΩN α (R) i=1

so that (3.2.25) holds with p = 0. Now we suppose that (3.2.25) holds for some s, s ≤ N − 1. If the dg (xsα , xN α )’s do not converge to 0, then, up to a subsequence, s ˜ ˜ > 0. As a consequence, (R) ∩ Ω ( R) = ∅ for R ΩN α α

|usα |2 dvg ≤ |usα |2 dvg p ˜j ΩN α (R)\∪j=1 Ωα (R )

˜ M \Ωsα (R)

and it follows (we refer to the proof of Lemma 3.2) that

|usα |2 dvg ≤ C p ˜j ΩN α (R)\∪j=1 Ωα (R )

Rn \B

˜ 0 (R)

|us |2 dx .

˜ > 0 is arbitrary and us ∈ L2 (Rn ), we get that Since R

|usα |2 dvg = o(1) p ˜j ΩN α (R)\∪j=1 Ωα (R )

43


and thus

2 s−1 uiα − uN ˆα − u dvg = o(1) + ε(R ) . α p j N ˜ Ωα (R)\∪j=1 Ωα (R )

i=1

In particular, (3.2.25) holds for s − 1. Let us now assume that dg (xsα , xN α ) → 0 as α → +∞. We let r0 and C ≥ 1 be such that for all x ∈ M , and all y, z ∈ Rn , if |y| ≤ r0 and |z| ≤ r0 , then 1 |z − y| ≤ dg (expx (y), expx (z)) ≤ C|z − y| . C If x ˜sα and y˜αj are such that xsα = expxN (µN ˜sα ) and yαj = expxN (µN ˜αj ), then αx αy α α j R λα λjα 1 −1 ˜ j j Ω exp (R ) ⊂ B C By˜αj ⊂ R (3.2.26) α y˜α xN α C µN µN µN α α α and

Bx˜sα

R µsα C µN α

⊂

1 exp−1 (Ωsα (R )) ⊂ Bx˜sα xN α µN α

R C

µsα µN α

.

(3.2.27)

˜ > 0, we have by (3.2.13) that Given R 2

s i uα dvg = o(1) . ˆα − u ˜ Ωsα (R) i=1

Hence, by (3.2.25),

2 |uN α | dvg = o(1) + ε(R ) ,

(

p ˜j ΩN α (R)\∪j=1 Ωα (R )

)

˜ ∩Ωsα (R)

and it follows from (3.2.26) and (3.2.27) that

|uN |2 dx = o(1) + ε(R ) . j µs B0 (R)\∪p j=1 B

j (R y ˜α

C λα µN α

˜

) ∩Bx (R ˜s α C

α µN α

)

(3.2.28)

Now we distinguish two cases. In the first case we assume that, as α → +∞, N s N s dg (xsα , xN α )/µα → +∞. Then we also have that dg (xα , xα )/µα → +∞, since if s N ˜ large enough that µ /µ → 0, while not, we get by (3.2.28) with R α α dg (xsα , xN dg (xsα , xN µN α) α) = × αs . s N µα µα µα s ˜ ˜ Then it follows that ΩN α (R) ∩ Ωα (R) = ∅ for R > 0, and we may proceed as in s N the case where the dg (xα , xα )’s do not converge to 0 to get that (3.2.25) holds for N s − 1. In the second case, we assume that, as α → +∞, the dg (xsα , xN α )/µα ’s s N p+1 s converge. By (3.2.28), we must have that µα /µα → 0. We set yα = xα and λp+1 = µsα . Then, α 2

s i N uα − uα dvg = o(1) + ε(R ) ˆα − u p+1 ˜ j ΩN α (R)\∪j=1 Ωα (R ) i=1

44

CHAPTER 3

and

|usα |2 dvg ≤

p+1 ˜ j ΩN α (R)\∪j=1 Ωα (R )

|usα |2 dvg

M \Ωsα (R )

≤ ε(R ) . It follows that

2 s−1 uiα − uN ˆα − u dvg = o(1) + ε(R ) α p+1 ˜ j N Ωα (R)\∪j=1 Ωα (R ) i=1

and (3.2.25) holds for s − 1. In particular, (3.2.25) is always true. Now we prove the claim that if the uα ’s in Theorem 3.1 are nonnegative, then u0 and the ui ’s of Theorem 3.1 are also nonnegative. As already mentioned, it is clear that u0 is nonnegative. We let u Ñ α be given by N n−2 N 2 u u Ñ ˆα expxN (µ x) . α (x) = (µα ) α α We apply (3.2.25) with s = 0. Then,

2 u dvg = o(1) + ε(R ) , ˆ α − uN α p ˜j ΩN α (R)\∪j=1 Ωα (R )

and it follows that

j λα B0 (R)\∪p j (R C N j=1 By ˜α µα

)

N 2 u ˜α − uN dx = o(1) + ε(R )

(3.2.29)

where the y˜αj are as above. The y˜αj ’s are bounded. Up to a subsequence we may assume that y˜αj → y j as α → +∞. Then we get from (3.2.29) that N ˜ u Ñ in L2loc B0 (R)\X α →u ˜ consists of the y˜j ’s, j = 1, . . . , p. Therefore, we may as α → +∞, where X N N assume that u ˜α → u almost everywhere in Rn as α → +∞. Let N n−2 2 u0 u ˜0,N (µN expxN α (x) = (µα ) α x) . α Then,

where g˜α (x) =

|u0 |2 dvg =

ΩN α (R) expxN g (µN α x). α

as α → +∞, and thus that vαN ’s given by

u ˜0,N α u ˜0,N α

2 |˜ u0,N ˜α α | dvg

B0 (R)

It follows that

→ 0 in L2 (B0 (R)) → 0 almost everywhere in Rn . Summarizing, the

vαN (x) = (µN α)

n−2 2

N uα expxN (µ x) α α

converge almost everywhere to uN . In particular, uN is nonnegative, and this proves Lemma 3.3. 2 As a remark, it follows from the above proof that for any i = j, j Rα Ri i j + α + Rα Rα dg (xiα , xjα )2 → +∞ j i Rα Rα as α → +∞.

45


3.3 THE DE GIORGI–NASH–MOSER ITERATIVE SCHEME FOR STRONG SOLUTIONS Given (M, g) smooth, compact, of dimension n ≥ 3, we let (hα ) be a sequence of smooth functions on M . We also let (uα ) be a bounded sequence in H12 (M ) of nonnegative solutions of 2 −1 . ∆g uα + hα uα = uα

(Eα )

As already mentioned, it is easily seen that (uα ) is a Palais-Smale sequence for (Eα ). If (3.1.1) holds we can therefore apply Theorem 3.1 to (uα ). The purpose of this section is to illustrate the idea that, because of (Eα ), we easily get complementary information by applying, for instance, the De Giorgi–Nash–Moser iterative scheme to the uα ’s. We also propose a very simple construction of an elementary blow-up. First, for the sake of completeness, we recall the content of the De Giorgi–Nash–Moser iterative scheme. Following standard arguments, as in Han and Lin [41], we can prove the following. L EMMA 3.4 Let B0 (3δ), δ > 0, be the ball in Rn of center 0 and radius 3δ, and let g be a Riemannian metric on B0 (3δ). Let A > 0 be such that for any smooth function ϕ with compact support in B0 (3δ), ϕL2 (B0 (2δ)) ≤ A∇ϕL2 (B0 (2δ)) where the norms are taken with respect to g, and u ∈ C 1 (B0 (2δ)), u > 0, be such that where

∆g u ≤ f u

B0 (2δ)

|f | dvg ≤ K for some r > n/2. Then, for all p > 0, r

sup u ≤ CuLp (B0 (2δ))

B0 (δ)

where the above norm is taken with respect to g, and C = C(n, A, K, p, r, δ) depends only on n, A, K, p, r, and δ. Another useful version (we refer, for instance, to Han and Lin [41]) is the fol lowing. With respect to the notations in Han and Lin [41], we let c = h − u2 −2 and f = 0. We also apply a Trudinger-type argument [74] in order to prove first that u ∈ Lk (M ) for some k > 2 . L EMMA 3.5 Let (M, g) be a smooth compact Riemannian n-manifold, n ≥ 3, h be a smooth function on M , and u ∈ H12 (M ), u ≥ 0, be such that for any nonnegative ϕ ∈ H12 (M ),

∇u, ∇ϕdvg + huϕdvg ≤ u2 −1 ϕdvg . M

M

M

Then u ∈ L∞ (M ). Moreover, for any x in M , any Λ > 0, any p > 0, any q > 2 , and any δ > 0, if u is some nonnegative function of H12 (M ) satisfying the above equation and

uq dvg ≤ Λ , Bx (2δ)

46

CHAPTER 3

then sup u(y) ≤ C

y∈Bx (δ)

Bx (2δ)

up dvg

1/p

where C > 0 does not depend on u. As a remark, one passes from small δ’s (given by Han-Lin [41]) to arbitrary δ’s by writing that Bx (δ) ⊂ By (δy ) where δy < δ. Hence Bx (δ) ⊂ equations

y∈Bx (δ) i=1,...,N

Byi (δyi ) for some yi ∈ Bx (δ). The

sup u ≤ Ci uLp (Bx (2δyi ))

Byi (δyi )

then give that sup u ≤ CuLp (Bx (2δ))

Bx (δ)

since Byi (2δyi ) ⊂ Bx (2δ). Unless otherwise stated, we do not assume here that (3.1.1) holds, but assume that the operators Lα g = ∆g + hα are uniformly coercive in the sense of (2.1.3). In other words, we assume that there exists h smooth, s.t. ∆g + h is coercive, and hα ≥ h for all α .

(3.3.1)

For the sake of simplicity, we also assume that uα 0 weakly but not strongly in H12 (M ) as α → +∞. Clearly, uα 2 → 0 as α → +∞. We let λα = u ˜α = uα −1 2 uα . Then 2 −1 ˜ α + hα u ˜α = λα u ˜α ∆g u

4/(n−2) uα 2

(3.3.2) and set ˜α ) (E

and ˜ uα 2 = 1. Moreover, the λα ’s are bounded, and it easily follows from (3.3.1) and (3.3.2) that there exists c > 0, independent of α, such that for any α, uα ) is bounded in H12 (M ), and that (˜ uα ) also λα ≥ c. It follows that the sequence (˜ satisfies (3.3.2). Following standard terminology, we say that x is a (geometrical) blow-up point for (˜ uα ), or for (uα ), if for any δ > 0,

u ˜2α dvg > 0 lim sup α→+∞

Bx (δ)

where Bx (δ) is the geodesic ball of center x and radius δ. We let S be the set consisting of the geometrical blow-up points for (˜ uα ). Clearly, S = ∅. We claim that, up to a subsequence, the following holds: S is a finite set and 0 uα → 0 in Cloc (M \S) .

(3.3.3)

47


We prove (3.3.3). Let x be some point in M , δ > 0, and 0 ≤ η ≤ 1 a smooth ˜α ) by function such that η = 1 in Bx (δ/2) and η = 0 in M \Bx (δ). Multiplying (E 2 θ ˜α , where 1 ≤ θ ≤ 2 − 1, we get that η u 2 +θ−1 ˜θα ∆g u ˜ α + hα η 2 u ˜θ+1 = λα η 2 u ˜α . η2 u α

(3.3.4)

Integrating by parts,

θ+1 4θ 2 θ 2 η u ˜ α ∆g u ˜α dvg = |∇(η˜ u )|2 dvg α (θ + 1)2 M M

2(θ − 1) 2 θ+1 − η(∆ η)˜ u dv − |∇η|2 u ˜θ+1 g g α α dvg (θ + 1)2 M θ+1 M while by Hölder’s inequalities,

η2 u ˜2α +θ−1 dvg ≤

M

θ+1 2

(η˜ uα )2 dvg

22

Bx (δ)

M

u ˜2α dvg

22−2 .

Moreover, one gets by the Sobolev inequalities that there exist A, B > 0, independent of α and θ, such that 22

θ+1 θ+1 2 2 2 2 (η˜ uα ) dvg ≤A |∇(η˜ uα )| dvg + B η2 u ˜θ+1 α dvg . (3.3.5) M

M

M

From (3.3.1), there also exists C > 0, independent of α and θ, such that

θ+1 4θ −1 |∇(η˜ uα 2 )|2 dvg C+ (θ + 1)2 M

θ+1 4θ 2 2 |∇(η˜ uα )| dvg + ≤ hα η 2 u ˜θ+1 α dvg (θ + 1)2 M M so that, for any θ ≥ 1 sufficiently close to 1,

θ+1 θ+1 C 4θ 2 2 |∇(η˜ u |∇(η˜ uα 2 )|2 dvg ≤ )| dv + hα η 2 u ˜θ+1 α g α dvg . 2 M (θ + 1)2 M M Integrating (3.3.4) then gives that

θ+1 |∇(η˜ uα 2 )|2 dvg ≤ C1 (α) M

θ+1

|∇(η˜ uα 2 )|2 dvg + C2 (θ, α) ,

M

uα 2 = 1, where, since λα ≤ λ for some λ > 0, and ˜ 2λA C1 (α) = C and

Bx (δ)

u ˜2α dvg

22−2

4(θ − 1) η(∆g η)˜ uθ+1 α dvg (θ + 1)2 C M

4 2λB + |∇η|2 u ˜θ+1 dv + η2 u ˜θ+1 g α α dvg . (θ + 1)C M C M

C2 (θ, α) =

(3.3.6)

48

CHAPTER 3

Clearly, there exists at least one geometrical blow-up point for the sequence (˜ uα ). Let x0 be such a point. We claim that for all δ > 0, 22−2

C 2 . (3.3.7) u ˜α dvg ≥ lim sup 2λA α→+∞ Bx (δ) 0 If this is not the case, then there exists C1 ∈ (0, 1) such that C1 (α) ≤ C1 for all α. By (3.3.6), and since (˜ uα ) is bounded in L2 (M ), it follows that there exists C2 > 0 such that for any α,

θ+1 |∇(η˜ uα 2 )|2 dvg ≤ C2 . (3.3.8) M

By Hölder’s inequalities,

u ˜2α dvg Bx0 (δ/2)

≤

2 (θ+1) 2

Bx0 (δ/2)

u ˜α

n+2 n(θ+1)

dvg

n(θ+1) nθ−2

Bx0 (δ/2)

u ˜α

nθ−2 n(θ+1)

dvg

,

so that, with (3.3.8) and according to the Sobolev inequality (3.3.5), nθ−2

n(θ+1)

u ˜2α dvg ≤ C3

n(θ+1)

Bx0 (δ/2)

Bx0 (δ/2)

u ˜αnθ−2 dvg

where C3 > 0 is independent of α. Since x0 is assumed to be a blow-up point for the sequence (˜ uα ), we then get the existence of C4 > 0 such that for some subsequence

n(θ+1) (3.3.9) u ˜αnθ−2 dvg ≥ C4 . Bx0 (δ/2)

Fix θ > 1 close to 1, and let θ1 = n(θ + 1)/(nθ − 2). Then 1 < θ1 < 2 . Since uα 1 → 0 as α → +∞. Then, by ˜ uα 2 → 0 as α → +∞, we also have that ˜ uα θ1 → 0 as α → +∞. Hölder’s inequalities, and since ˜ uα 2 is bounded, ˜ This contradicts (3.3.9), so that (3.3.7) is proved. Then, as an immediate consequence of (3.3.7) and the relation ˜ uα 2 = 1, we get that, up to a subsequence, (˜ uα ) has a finite number of geometrical blow-up points. Let S be the set of blow-up points of this subsequence, and let x now be a point in M \S. There exists δ > 0 such that u ˜α → 0 in L2 Bx (δ) . Coming back to (3.3.6), there exists θ > 1 such (θ+1)/2 that η˜ uα is bounded in H12 (M ). Thanks to the Sobolev embedding theorem, it follows that u ˜α is bounded in Lq Bx (δ/2) for some q > 2 . Noting that 2 −1 ∆g u ˜α + h˜ uα ≤ λα u ˜α

where h is as in (3.3.1), and applying the De Giorgi–Nash–Moser iterative scheme ˜α → 0 as α → +∞ in C 0 Bx (δ/4) . to the u ˜α ’s (see Lemma 3.5) we get that u This proves (3.3.3). As a final remark, let us assume that we also have (3.1.1), in addition to (3.3.1) and (3.3.2). Then Theorem 3.1 applies to the uα ’s with, in this context, u0 ≡ 0.


49

xl , l ≤ m, be the limits as α → +∞ of the xjα ’s of Theorem 3.1, We let x ˜1 ,. . . ,˜ j = 1, . . . , m. Let also S˜ be the set consisting of the x ˜i ’s, i = 1, . . . , l. We claim that S˜ = S, where S is as above. As is easily checked, it follows from the proof of Lemma 3.3 that, for any j = 1, . . . , m,

u2α dvg > 0 . lim inf α→+∞

B

j j (1/Rα ) xα

Hence, for any δ > 0, and any i = 1, . . . , l,

u2α dvg > 0 . lim inf α→+∞

Bx ˜i (δ)

˜ there exists This proves that S˜ ⊂ S. Conversely, (3.1.3) gives that for any x ∈ S, 2 ˜ δ > 0 such that uα → 0 in L Bx (δ) . Hence, x ∈ S, and S ⊂ S. This proves the above claim.


Chapter Four Exhaustion and Weak Pointwise Estimates Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3. We let (hα ) be a sequence of smooth functions on M , and consider equations like ∆g u + hα u = u2

−1

(Eα )

where u is required to be positive. We let (uα ) be a sequence of solutions to (Eα ), so that uα > 0 and 2 −1 ∆g uα + hα uα = uα

for all α. We describe in this chapter an exhaustion method for the blow-up behavior of the uα ’s. This method provides a constructive approach to Theorem 3.1, and weak (with respect to the material in Chapters 5 and 6) pointwise estimates on the uα ’s. We assume in what follows that there exist 0 < θ < 1 and a smooth (or only C 0,θ ) function h∞ on M such that the operator ∆g + h∞ is coercive and

(4.0.1)

hα → h∞ in C 0,θ (M ) as α → +∞ .

We also assume that there exists Λ > 0 such that E(uα ) ≤ Λ for all α, where E(u) = u2 is as in Chapter 2. Assuming (4.0.1) and that E(uα ) ≤ Λ for all α, it is easily seen that the uα ’s are bounded in H12 (M ). Indeed, we clearly have that hα n/2 ≤ C for all α, where C > 0 is independent of α. Multiplying (Eα ) by uα and integrating over M , we then get that

u2α dvg ≤ Λ2 . |∇uα |2 dvg + hα u2α dvg = M

M

From Hölder’s inequality,

2 hα uα dvg ≤ M

M

2/n

|hα |

n/2

u2α dvg

dvg

M

2/2 .

M

It follows that there exists C > 0 such that ∇uα 2 ≤ C for all α, and then that the uα ’s are bounded in H12 (M ) since they are also bounded in L2 (M ) and 2 < 2 . Up to a subsequence, we may therefore assume that for some u0 ∈ H12 (M ), uα u0 weakly in H12 (M ) 0

as α → +∞. It is easily seen that u is a solution of ∆g u0 + h∞ u0 = (u0 )2

−1

.

(4.0.2)

52

CHAPTER 4

Another easy remark is that (4.0.1) implies the uniform coercivity condition (3.3.1) for α sufficiently large. Indeed, if ∆g +h∞ is coercive, then there exists ε > 0 such that ∆g +(h∞ −ε) is still coercive. Since hα → h∞ in C 0,θ (M ), hα ≥ h∞ −ε for α sufficiently large. In particular, there exists h such that hα ≥ h for α sufficiently large, and such that ∆g + h is coercive. Without loss of generality, we can assume in the following that lim max uα = +∞

α→+∞ M

(4.0.3)

so that blow-up occurs. If this is not the case, up to a subsequence, we easily get from standard regularity theory that uα → u0 strongly in C 2 (M ) as α → +∞. The main result of this chapter, Theorem 4.1, is stated in section 4.1. Section 4.2 is concerned with the proof of this result. 4.1 WEAK POINTWISE ESTIMATES We set up some notation. Given N ∈ N , let (xi,α ), i = 1, . . . , N , be N converging sequences of points in M , and (µi,α ), i = 1, . . . , N , be N sequences of positive real numbers converging to 0. We set (4.1.1) S= lim xi,α , i = 1, . . . , N α→+∞

and when N ≥ 2 we set 1 lim exp−1 (x ) , x ∈ B (δ) , j = i Si = j,α j,α xi,α xi,α α→+∞ µi,α

(4.1.2)

for i = 1, . . . , N , where 0 < δ ≤ ig /2, ig is the injectivity radius of (M, g), and the limits, up to a subsequence, are assumed to exist. As in Chapter 3, we regard expx as defined in Rn . An intrinsic definition is possible if M is parallelizable. ˜ j , j = 1, . . . , k, be open subsets of M such that for any If not we let Ωj and Ω ˜ j , and such that M = ∪Ωj . The canonical ˜ j, Ωj is parallelizable and Ωj ⊂ Ω exponential map gives k maps expx defined in Ωj × Rn , and expx is, depending on the situation, one of these maps. A property of expx that holds for any x ∈ M should then be regarded as a property that holds for any j and any x ∈ Ωj . We also let ui,α be the function defined in B0 δµ−1 i,α , the Euclidean ball of center 0 and −1 radius δµi,α , by n 2 −1 ui,α (x) = µi,α uα expxi,α (µi,α x) . (4.1.3) Conversely, if u is a smooth function defined in Rn , we let vi,α = vi,α (u) be the function given by 1 1− n −1 2 expxi,α (x) if x ∈ Bxi,α (δ) and vi,α (x) = µi,α u µi,α (4.1.4) vi,α (x) = 0 otherwise .

EXHAUSTION AND WEAK POINTWISE ESTIMATES

53

Finally, we let u = u10 , where u10 is as in Chapter 3, be the function defined in Rn by 1− n2 |x|2 . (4.1.5) u(x) = 1 + n(n − 2) Then, u is a positive solution, the only one satisfying u(0) = 1 = maxRn u, of the equation ∆u = u2 −1 where ∆ is the Euclidean Laplacian. Its energy E(u) = u2 is E(u) = Λmin , −(n−2)/2 where Λmin = Kn is as in Chapter 2. In what follows, we let η be a smooth cutoff function in Rn such that η ≡ 1 in B0 (δ/2) and η ≡ 0 in Rn \B0 (δ). Then we define ηi,α by ηi,α (x) = η(µi,α x) for i = 1, . . . , N . When N ≥ 2, we consider the following statement: dg (xi,α , xj,α ) → +∞ (4.1.6) min (µi,α , µj,α ) N for all i = j, as α → +∞. We let also Rα be the function given by N Rα (x) = min dg (xi,α , x) . i=1,...,N

(4.1.7)

The main result of this chapter is the following theorem. T HEOREM 4.1 Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, (hα ) be a sequence of smooth functions on M such that (4.0.1) is satisfied and (uα ) be a sequence of positive solutions to (Eα ) such that E(uα ) ≤ Λ for some Λ > 0 and all α. We assume that (4.0.2) and (4.0.3) hold. Then there exist N ∈ N , converging sequences (xi,α ) in M , and sequences (µi,α ) of positive real numbers converging to 0, i = 1, . . . , N , such that (4.1.6) holds, and such that, up to a subsequence, the following propositions hold: 0 (M \S) as α → +∞, where u0 is given by (4.0.2), (P1) uα → u0 strongly in Cloc and there exist ε0 > 0 and xi ∈ Rn such that for any i = 1, . . . , N , lim ηi,α (x)ui,α (x) = u(x − xi ) α→+∞

2 weakly in D12 (Rn ), strongly in Cloc (Rn \Si ), and strongly in C 2 (B0 (ε0 )), where S is as in (4.1.1), Si is as in (4.1.2), ui,α is as in (4.1.3), and u is as in (4.1.5). n N (x) 2 −1 uα (x) ≤ C, where C > 0 is (P2) For any x ∈ M , and any α, Rα independent of α. Moreover, n N lim sup Rα (x) 2 −1 uα (x) − u0 (x) = 0 lim R→+∞ α→+∞ x∈M \Ωα (R)

N where u0 is given by (4.0.2), Rα is given by (4.1.7), and Ωα (R) is given by N Ωα (R) = ∪i=1 Bxi,α (Rµi,α ). (P3) The energy E, given by E(u) = u2 , satisfie N 2 lim E uα − u0 − vi,α ≤ lim E(uα )2 − u0 22 − N Λ2min α→+∞

i=1

α→+∞

where u0 is given by (4.0.2), vi,α = vi,α (ui ) is as in (4.1.4), and ui (x) = u(x−xi ) where u is as in (4.1.5) and xi is as in (P1).

54

CHAPTER 4

As a remark, because of (P3), we can saturate Theorem 4.1 in order to get some maximal N by adding sequences (xj,α ) and (µj,α ) such that (P1)–(P3) continue to hold. On the other hand, we clearly have that (4.0.1) implies (3.1.1). We let S be the set of geometrical blow-up points given by (4.1.1) that we get from Theorem 4.1. We also let S˜ be the set consisting of the limits of the (xik )’s we get from Theorem 3.1. An easy remark is that S and S˜ coincide. 4.2 EXHAUSTION OF BLOW-UP POINTS We prove Theorem 4.1 in this section. We proceed in several steps. The claims in what follows are up to a subsequence. First we claim that the following holds. S TEP 1. There exist a converging sequence (x1,α ) of points in M and a sequence (µ1,α ) of positive real numbers converging to 0 such that, up to a subsequence, the following propositions hold : 2 (A1) u1,α → u in Cloc (Rn ) as α → +∞,

(A2) lim uα − u0 − v1,α 22 ≤ lim E(uα )2 − u0 22 − Λ2min ,

α→+∞

α→+∞

where u1,α is as in (4.1.3), u is given by (4.1.5), and v1,α = v1,α (u) is as in (4.1.4). Proof of step 1. We let x1,α be a point in M where uα achieves its maximum and set 1− n

uα (x1,α ) = µ1,α 2 = max uα . M

From (4.0.3), lim µ1,α = 0 . and for x ∈ B0 δµ−1 1,α , the Euclidean ball of center 0 and α→+∞

i

We let 0 < δ < 2g radius δµ−1 1,α , we set

n 2 −1 uα expx1,α (µ1,α x) , u1,α (x) = µ1,α g1,α (x) = expx1,α g (µ1,α x) , h1,α (x) = hα expx1,α (µ1,α x) .

Since µ1,α → 0 as α → +∞, it is clear that 2 lim g1,α = ξ in Cloc (Rn )

α→+∞

where ξ is the Euclidean metric. Note that we also have that g1,α is controlled on both sides by ξ in the sense of bilinear forms. It is easily checked that u1,α (0) = u1,α L∞ (B0 (δµ−1 )) = 1

(4.2.1)

2 −1 ∆g1,α u1,α + h1,α µ21,α u1,α = u1,α

(4.2.2)

1,α

and that

55


−1

in B0 δµ1,α . We claim now that 2 (Rn ) lim u1,α = u1 in Cloc

(4.2.3)

α→+∞

where u1 ∈ D12 (Rn ), u1 ≡ 0, satisfies ∆u1 = u12

−1

(4.2.4)

where ∆ is the Euclidean Laplacian. We also claim that η1,α u1,α u1 in D12 (Rn ) as α → +∞, where η1,α is as in the introduction of this section. In order to prove these claims, we first note that the η1,α u1,α ’s are bounded in D12 (Rn ). Indeed, it is clear that there exists C > 0, independent of α, such that |∇η1,α | ≤ Cµ1,α . Letting s = 2 /(2 − 2), so that 2s = n, we then write that

|∇(η1,α u1,α )|2 dvg1,α Rn

2 ≤2 |∇η1,α |2 u21,α dvg1,α + 2 η1,α |∇u1,α |2 dvg1,α Rn

Rn

≤2

Rn

|∇η1,α |2s dvg1,α

1/s

2/2 u21,α dvg1,α

B0 (δ/µ1,α )

+ O(1)

≤ O(1) . This proves that the η1,α u1,α ’s are bounded in D12 (Rn ). We may therefore assume that η1,α u1,α u1 in D12 (Rn ) as α → +∞ for some u1 ∈ D12 (Rn ). From (4.0.1), 0 the coefficients in equation (4.2.2) are bounded and h1,α µ21,α → 0 in Cloc (Rn ) as α → +∞. By standard elliptic theory, using (4.2.1), this gives (4.2.3) and (4.2.4). It is clear from (4.2.1) and (4.2.3) that u1 (0) = u1 ∞ = 1. From CaffarelliGidas-Spruck [17], we necessarily have that u1 = u where u is as in (4.1.5). In particular, (A1) holds. Let us now prove that (A2) holds also. We write that

0 2 |uα − u0 − v1,α |2 dvg |uα − u − v1,α | dvg = M \Bx1,α (Rµ1,α )

M

+

|uα − u0 − v1,α |2 dvg

Bx1,α (Rµ1,α )

where R > 0 is arbitrary. From (A1),

lim |uα − u0 − v1,α |2 dvg = 0 . α→+∞

Bx1,α (Rµ1,α )

We write now that for any ε > 0 there exists Cε > 0 such that

|uα − u0 − v1,α |2 dvg M \Bx1,α (Rµ1,α )

≤ (1 + ε)

+Cε

M \Bx1,α (Rµ1,α )


|uα − u0 |2 dvg

2 v1,α dvg .

56

CHAPTER 4

Here, Cε does not depend on R and α, and Cε is defined by the property that for any x, y ∈ R, |x + y|2 ≤ (1 + ε) |x|2 + Cε |y|2 .

Direct computations give that lim

lim

R→+∞ α→+∞

2 v1,α dvg = 0 .


Combining these relations,

|uα − u0 − v1,α |2 dvg lim α→+∞ M

≤ lim |uα − u0 |2 dvg α→+∞ M

− lim lim |uα − u0 |2 dvg . R→+∞ α→+∞

From (A1),

lim

lim

R→+∞ α→+∞

Bx1,α (Rµ1,α )

|uα − u0 |2 dvg = lim

Bx1,α (Rµ1,α )

R→+∞

u2 dx

B0 (R)

= Λ2min .

Moreover, since uα u0 weakly in H12 (M ), and uα → u0 almost everywhere,

|uα − u0 |2 dvg = lim uα 22 − u0 22 . lim α→+∞

α→+∞

M

2

Thus (A2) is proved thanks to these relations. This ends the proof of step 1.

Let k ∈ N, k ≥ 1, and for i = 1, . . . , k, let (xi,α ) be k converging sequences of points in M and (µi,α ) be k sequences of positive real numbers converging to 0. We consider the following assertions: (B1) when k ≥ 2, (xi,α ) and (µi,α ) satisfy that +∞ for any i = j,

dg (xi,α ,xj,α ) µi,α

→ +∞ as α →

(B2) for any i ∈ {1, . . . , k}, ηi,α ui,α → u weakly in D12 (Rn ) and strongly in as α → +∞,

2 (Rn ) Cloc

(B3) lim uα − u0 − α→+∞

k i=1

vi,α 22 ≤ lim E(uα )2 − u0 22 − kΛ2min

α→+∞

where ui,α is as in (4.1.3), u is as in (4.1.5), and vi,α = vi,α (u) is as in (4.1.4). We let (Hk1 ) be this sequence of assertions and say that (Hk1 ) holds if there exist k converging sequences (xi,α ) of points in M and k sequences (µi,α ) of positive real numbers converging to 0 such that, up to a subsequence, (B1)–(B3) hold. By step 1, we know that (H11 ) holds. A second step in the proof of the theorem is the following:

57

EXHAUSTION AND WEAK POINTWISE ESTIMATES 1 ) holds or S TEP 2. Assume that (Hk1 ) holds. Then either (Hk+1 k Rα (x) 2 −1 uα (x) ≤ C n

k is given by (4.1.7) and C > 0 is independent for all x ∈ M and all α, where Rα of x and α.

Proof of step 2. Let us assume that k (4.2.5) (x) 2 −1 uα (x) → +∞ max Rα x∈M 1 holds. We let yα ∈ M be such that as α → +∞. We need to prove that Hk+1 n

k k max Rα (x) 2 −1 uα (x) = Rα (yα ) 2 −1 uα (yα ) n

n

x∈M

(4.2.6)

and we set 1− n

2 . xk+1,α = yα and uα (xk+1,α ) = µk+1,α

Since M is compact, (4.2.5) implies that µk+1,α → 0 as α → +∞. It also follows from (B2) of (Hk1 ) and (4.2.5) that for any i ∈ {1, . . . , k}, dg (xi,α , xk+1,α ) → +∞ µi,α i as α → +∞. Let 0 < δ < 2g . For x ∈ B0 δµ−1 k+1,α , the Euclidean ball of center 0 and radius δµ−1 k+1,α , we set n 2 −1 uα expxk+1,α (µk+1,α x) , uk+1,α (x) = µk+1,α gk+1,α (x) = expxk+1,α g (µk+1,α x) , hk+1,α (x) = hα expxk+1,α (µk+1,α x) . Since µk+1,α → 0 as α → +∞, lim gk+1,α = ξ

α→+∞

2 in Cloc (Rn ), where ξ is the Euclidean metric. We also have that gk+1,α is on both sides controled by ξ in the sense of bilinear forms. It is easily checked that

uk+1,α (0) = 1 and that 2 −1 ∆gk+1,α uk+1,α + hk+1,α µ2k+1,α uk+1,α = uk+1,α n in B0 δµ−1 k+1,α . Let x ∈ R . For i ∈ {1, . . . , k},

dg (xk+1,α , xi,α ) − µk+1,α |x| ≤ dg xi,α , expxk+1,α (µk+1,α x) ≤ dg (xk+1,α , xi,α ) + µk+1,α |x| . By (4.2.5), for any i ∈ {1, . . . , k}, dg (xi,α , xk+1,α ) → +∞ µk+1,α

58

CHAPTER 4

as α → +∞, so that lim

dg xi,α , expxk+1,α (µk+1,α x) dg (xi,α , xk+1,α )

α→+∞

=1

for all i ∈ {1, . . . , k}. From (4.2.6), it follows that lim sup uk+1,α (x) ≤ 1 α→+∞

for all x ∈ R . As in step 1, it is easily seen that, up to a subsequence, n

2 (Rn ) lim uk+1,α = uk+1 in Cloc

α→+∞

where uk+1 ≡ 0 is such that 2 −1 ∆uk+1 = uk+1

and where ∆ is the Euclidean Laplacian. We also get that ηk+1,α uk+1,α uk+1 weakly in D12 (Rn ) as α → +∞. From Caffarelli-Gidas-Spruck [17], uk+1 = u 1 . Independently, we have where u is as in (4.1.5). This proves (B2) of Hk+1 1 already seen that (B1) of Hk+1 is true. Thus, it remains to prove that (B3) of 1 Hk+1 is also true. Let R > 0 be given. From the expression for u, there exists C > 0, independent of α, such that for any i ∈ {1, . . . , k},

−2n 2 vi,α dvg ≤ Cµni,α dg (xi,α , x) dvg . Bxk+1,α (Rµk+1,α )

Bxk+1,α (Rµk+1,α )

1 Using (B1) of Hk+1 , it is easily checked that for α large enough and for any x ∈ Bxk+1,α (Rµk+1,α ), −2n

dg (xi,α , x) Thus,

−2n

≤ 2dg (xi,α , xk+1,α )

.

2 vi,α dvg


−2n n ≤ 2Cdg (xi,α , xk+1,α ) µi,α Volg Bxk+1,α (Rµk+1,α ) µnk+1,α µni,α ≤ C(R) n n dg (xi,α , xk+1,α ) dg (xi,α , xk+1,α )

where C(R) is independent of α, and for Ω ⊂ M , Volg (Ω) stands for the volume 1 , it follows that of Ω with respect to g. Using again (B1) of the sequence Hk+1 for any i ∈ {1, . . . , k} and any R > 0,

2 lim vi,α dvg = 0 . (4.2.7) α→+∞


1 From (B2) of Hk+1 , we also have that

lim lim R→+∞ α→+∞


u2α dvg = Λ2min .

(4.2.8)

59


We write now that

2 k 0 vi,α dvg uα − u − M \Bxk+1,α (Rµk+1,α ) i=1

k 2 0 = vi,α dvg u α − u − M i=1 2

k 0 − vi,α dvg . uα − u − Bx (Rµk+1,α ) i=1

k+1,α

Since

2 k 0 − vi,α dvg uα − u − Bxk+1,α (Rµk+1,α ) i=1

≤− u2α dvg + 2


+2


k

i=1

2 −1 0 uα u dvg

2 −1 uα vi,α dvg ,


it follows from (B3) of Hk1 , (4.2.7), and (4.2.8) that 2

k lim vi,α dvg lim uα − u0 − R→+∞ α→+∞ M \B (Rµk+1,α ) x

(4.2.9)

i=1

k+1,α

2

u0 22

1)Λ2min .

− (k + ≤ lim E(uα ) − α→+∞ 1 Using (B2) of Hk+1 , we also have that

|uα − vk+1,α |2 dvg = 0 lim

(4.2.10)

and direct computations give that

lim lim

(4.2.11)

α→+∞


R→+∞ α→+∞

Writing that ⎛

⎝ Bxk+1,α

+

M \Bxk+1,α (Rµk+1,α )

i=1

|uα − vk+1,α |2 dvg


k i=1

⎞ 21 2 k+1 vi,α dvg ⎠ u α − u 0 − (Rµk+1,α )

≤

2 vk+1,α dvg = 0 .


2 vi,α dvg

21

21 +


0 2

(u ) dvg

21 ,

60

CHAPTER 4

we get with (4.2.7) and (4.2.10) that 2

k+1 lim vi,α dvg = 0 . uα − u0 − α→+∞ B (Rµ ) x k+1,α

(4.2.12)

i=1

k+1,α

Given ε > 0 and R > 0, we write that 2

k+1 0 vi,α dvg uα − u − M i=1 2

k+1 ≤ vi,α dvg uα − u0 − Bxk+1,α (Rµk+1,α ) i=1 2

k 0 + (1 + ε) vi,α dvg uα − u − M \Bxk+1,α (Rµk+1,α ) i=1

2 +Cε vk+1,α dvg M \Bxk+1,α (Rµk+1,α )

where Cε does not depend on R and α. Passing to the limit in this expression as α → +∞, then as R → +∞ and finally as ε → 0, we get from (4.2.9), (4.2.11), and (4.2.12), that 2

k+1 vi,α dvg ≤ lim E(uα )2 −u0 22 −(k+1)Λ2min . lim uα − u0 − α→+∞ M α→+∞ i=1 1 Hence, (B3) of Hk+1 is true, and this ends the proof of Step 2. 2 Following steps 1 and 2, by induction, we get that there exists k ≥ 1, with Λ2min k ≤ Λ2 − u0 22 , that there exist k converging sequences (xi,α ) of points in M and that there exist k sequences (µi,α ) of positive real numbers converging to 0, such that (B1)–(B3) hold, and such that for any x ∈ M and any α, k Rα (x) 2 −1 uα (x) ≤ C n

(4.2.13)

where C > 0 is independent of α and x. By (4.2.13), uα L∞ (Ω) ≤ CΩ for any Ω ⊂⊂ M \S, where S, given by (4.1.1), consists of the limits of the xi,α ’s, and CΩ depends only on Ω. An easy claim then is that 0 lim uα = u0 in Cloc (M \S) .

α→+∞

This easily follows from standard elliptic theory. Let k ∈ N, k ≥ 1, and for i ∈ {1, . . . , k}, let (xi,α ) be k converging sequences of points in M , and (µi,α ) be k sequences of positive real numbers converging to 0. We consider the following assertions: (C1) when k ≥ 2, (xi,α ) and (µi,α ) satisfy that i = j,

dg (xi,α ,xj,α ) min{µi,α ,µj,α }

0 (C2) lim uα = u0 in Cloc (M \S) where S is as in (4.1.1), α→+∞

→ +∞ for any

61


(C3) there exists ε0 > 0 and there exists xi ∈ Rn , i = 1, . . . , k, such that for any 2 (Rn \Si ), i = 1, . . . , k, ηi,α ui,α → u ( · − xi ) weakly in D12 (Rn ), strongly in Cloc 2 and strongly in C (B0 (ε0 )) as α → +∞, where ui,α is as in (4.1.3), Si is as in (4.1.2), and u is given by (4.1.5), k (C4) for any x ∈ M and any α, Rα (x) 2 −1 uα (x) ≤ C where C > 0 is independent of x and α, k (C5) lim uα − u0 − i=1 vi,α 22 ≤ lim E(uα )2 − u0 22 − kΛ2min n

α→+∞

α→+∞

where vi,α = vi,α (ui ) is as in (4.1.4), and ui (x) = u(x − xi ), where u is as in (4.1.5) and xi is as in (C4). We let (Hk2 ) be this sequence of assertions and say that (Hk2 ) holds if there exist k converging sequences (xi,α ) of points in M and k sequences (µi,α ) of positive real numbers converging to 0 such that, up to a subsequence, (C1)–(C5) hold. According to what we just said, as a consequence of steps 1 and 2, there exists k ≥ 1 such that (Hk2 ) holds. A third step in the proof of the theorem is the following: 2 ) holds or S TEP 3. Assume that (Hk2 ) holds. Then either (Hk+1 n k −1 lim lim sup Rα (x) 2 uα (x) − u0 (x) = 0 R→+∞ α→+∞ x∈M \Ωα (R)

(4.2.14)

where Ωα (R) = ∪ki=1 Bxi,α (Rµi,α ). 2 ) Proof of step 3. We assume that (4.2.14) is false. We need to prove that (Hk+1 holds. Since (4.2.14) is assumed to be false, there exists a sequence (xk+1,α ) of points in M such that for any i = 1, . . . , k,

dg (xi,α , xk+1,α ) → +∞ µi,α

(4.2.15)

as α → +∞, and such that k (xk+1,α ) 2 −1 |uα (xk+1,α ) − u0 (xk+1,α ) | ≥ (4δ0 ) Rα n

n−2 2

(4.2.16)

for some δ0 > 0. We let 1− n

2 uα (xk+1,α ) = µk+1,α

2 and claim that Hk+1 holds when adding (xk+1,α ) and (µk+1,α ) to the (xi,α )’s 2 is a consequence and (µi,α )’s, i = 1, . . . , k. An easy remark is that (C1) of Hk+1 of (4.2.15). We also have that k Rα (xk+1,α ) → 0 2 as α → +∞. If not, (C2) of Hk implies that

(4.2.17)

|uα (xk+1,α ) − u0 (xk+1,α ) | → 0 as α → +∞, contradicting (4.2.16) since M is compact. Using (4.2.17), we can rewrite (4.2.16) as k (xk+1,α ) Rα ≥ 2δ0 µk+1,α

(4.2.18)

62

CHAPTER 4

for α large enough. This clearly implies that µk+1,α → 0 as α → +∞ . We let 0 < δ < ig /2, and, for x ∈ B0 δµ−1 k+1,α , the Euclidean ball of center 0 and radius δµ−1 k+1,α , we set n 2 −1 uk+1,α (x) = µk+1,α uα expxk+1,α (µk+1,α x) , gk+1,α (x) = expxk+1,α g (µk+1,α x) , hk+1,α (x) = hα expxk+1,α (µk+1,α x) . Since µk+1,α → 0 as α → +∞, 2 lim gk+1,α = ξ in Cloc (Rn ) .

(4.2.19)

α→+∞

We also have that gk+1,α is controlled on both sides by ξ in the sense of bilinear forms. It is easily checked that uk+1,α (0) = 1 and that 2 −1 ∆gk+1,α uk+1,α + hk+1,α µ2k+1,α uk+1,α = uk+1,α (4.2.20) in B0 δµ−1 k+1,α . We let 1 lim exp−1 (x ) , x ∈ B (δ) , 1 ≤ i ≤ k Sk+1 = i,α i,α xk+1,α xk+1,α α→+∞ µk+1,α

where, up to a subsequence, the limits are assumed to exist. From (4.2.18), 3 Sk+1 ∩ B0 δ0 = ∅ . (4.2.21) 2 Let R > 0 and let (xα ) be a sequence of points in B0 (R) such that dξ (xα , Sk+1 ) ≥

1 R

where dξ is the Euclidean distance. From (4.2.19), µ k+1,α k expxk+1,α (µk+1,α xα ) ≥ Rα 2R for α sufficiently large. Letting yα = expxk+1,α (µk+1,α xα ), it follows from (C4) of Hk2 that n

uk+1,α (xα ) ≤ (2R) 2

−1

n

k Rα (yα ) 2

−1

n

uα (yα ) ≤ (2R) 2

−1

C.

Hence, for any K ⊂⊂ R \Sk+1 , there exists CK > 0, independent of α, such that n

uk+1,α L∞ (K) ≤ CK .

(4.2.22)

63


Using (4.2.21) and (4.2.22), we can apply the De Giorgi–Nash–Moser iterative scheme as stated in Lemma 3.4 to equation (4.2.20). This gives the existence of some C > 0 independent of α, such that

12 1 = uk+1,α (0) ≤ C

B0 (δ0 )

u2k+1,α dvgk+1,α

.

As in steps 1 and 2, it is easily seen that, up to a subsequence, 2 (Rn \Sk+1 ) lim uk+1,α = uk+1 in Cloc

α→+∞

where uk+1 ≡ 0 is such that 2 −1 ∆uk+1 = uk+1

and where ∆ is the Euclidean Laplacian. We also get that ηk+1,α uk+1,α uk+1 weakly in D12 (Rn ) as α → +∞. Thanks to Caffarelli-Gidas-Spruck [17], (n−2)/2

uk+1 (x) = λk+1

u (λk+1 (x − xk+1 ))

where u is as in (4.1.5), λk+1 > 0, and xk+1 ∈ Rn are such that λk+1 = 1 +

λ2k+1 |xk+1 |2 . n(n − 2)

2 , up to Since Si with respect to Hk2 is a subset of Si with respect to Hk+1 2 −1 changing µk+1,α into λk+1 µk+1,α and xk+1 into λk+1 xk+1 , (C3) of Hk+1 is proved. By (4.2.17), the limit as α → +∞ of xk+1,α belongs to2 S, where S is the ’s, i = 1, . . . , k. Hence, (C2) of H set of the limits of the x i,α k implies (C2) of 2 Hk+1 . Noting that min

i=1,...,k+1

dg (xi,α , x) ≤ min dg (xi,α , x) ,

Hk2

i=1,...,k

2 implies (C4) of Hk+1 . As already mentioned, we also have that (C4) of 2 (C1) of Hk+1 follows from (4.2.15). We are thus left with the proof that (C5) of 2 Hk+1 holds. Let R > 0. For any ε > 0, there exists Cε independent of R and α such that 2

k+1 vi,α dvg uα − u0 − M \Bxk+1,α (Rµk+1,α ) i=1 2

k 0 ≤ (1 + ε) vi,α dvg uα − u − M \Bxk+1,α (Rµk+1,α ) i=1

2 +Cε vk+1,α dvg .


Noting that

lim

lim

R→+∞ α→+∞


2 vk+1,α dvg = 0 ,

64

CHAPTER 4

we easily get that

2 k+1 vi,α dvg uα − u0 − M \Bxk+1,α (Rµk+1,α ) i=1 2

k 0 ≤ vi,α dvg + εR (α) uα − u − M \Bx (Rµk+1,α )

(4.2.23)

i=1

k+1,α

where lim

lim εR (α) = 0 .

R→+∞ α→+∞

Letting

2 k+1 vi,α dvg IR (α) = uα − u0 − Bxk+1,α (Rµk+1,α ) i=1 2

k 0 − vi,α dvg , uα − u − Bx (Rµk+1,α )

(4.2.24)

i=1

k+1,α

it follows from (4.2.23) that 2 2

k+1 k vi,α dvg ≤ vi,α dvg uα − u0 − uα − u0 − M M i=1

i=1

(4.2.25)

+ IR (α) + εR (α) . Let

2 2 k+1 k Fα = uk+1,α − u0α − v˜i,α , v˜i,α − uk+1,α − u0α − i=1

i=1

where

n 2 −1 v˜i,α (x) = µk+1,α vi,α expxk+1,α (µk+1,α x)

and

n 2 −1 u0α (x) = µk+1,α u0 expxk+1,α (µk+1,α x) .

It is easily checked that

IR (α) =

Let

B0 (R)

VR =

Fα dvgk+1,α .

1 x ∈ R s.t. dξ (x, Sk+1 ) ≤ R

n

where dξ is the Euclidean distance. Similar arguments to the ones we used to prove (4.2.23) give that

Fα dvgk+1,α ≤ εR (α) . B0 (R)∩VR


Hence

65

IR (α) ≤

B0 (R)\VR

Fα dvgk+1,α + εR (α) .

(4.2.26)

Given i ∈ {1, . . . , k}, either dg (xi,α , xk+1,α ) ≤ Cµk+1,α for some C > 0 independent of α, or dg (xi,α , xk+1,α ) → +∞ µk+1,α as α → +∞. In the first case, we get by (4.2.15) that µk+1,α → +∞ µi,α as α → +∞. Hence, given i ∈ {1, . . . , k}, either dg (xi,α , xk+1,α ) → +∞ µk+1,α

(4.2.27)

µk+1,α → +∞ µi,α

(4.2.28)

as α → +∞, or

as α → +∞. If i ∈ {1, . . . , k} is such that (4.2.27) holds, we get as in step 2 that

2 vi,α dvg → 0 Bxk+1,α (Rµk+1,α )

as α → +∞. In particular, for any R > 0,

2 v˜i,α dvgk+1,α → 0 B0 (R)\VR

(4.2.29)

as α → +∞. Let now i ∈ {1, . . . , k} be such that (4.2.28) holds and (4.2.27) does not hold. Let also x ∈ B0 (R) \VR . For α sufficiently large, 1 µk+1,α , dg xi,α , expxk+1,α (µk+1,α x) ≥ 2R and it follows from the expression of v˜i,α that n2 −1 µi,α sup v˜i,α ≤ CR µk+1,α B0 (R)\VR for some CR > 0 independent of α. In particular, using (4.2.28), we get that for any R > 0, sup

B0 (R)\VR

v˜i,α → 0

2 , we also have that for any R > 0, as α → +∞. From (C3) of Hk+1

|uk+1,α − v˜k+1,α |2 dvgk+1,α → 0 B0 (R)\VR

(4.2.30)

(4.2.31)

66

CHAPTER 4

as α → +∞. Combining (4.2.26) and (4.2.29)–(4.2.31), we then get, as in step 2, that IR (α) ≤ −Λ2min + εR (α)

(4.2.32)

where lim

lim εR (α) = 0 .

R→+∞ α→+∞

From (4.2.24), (4.2.25), (4.2.32), and (C5) of Hk2 , it follows that 2

k+1 lim vi,α dvg ≤ lim E(uα )2 − u0 22 − (k + 1)Λ2min uα − u0 − α→+∞ M α→+∞ i=1 2 and (C5) of Hk+1 holds. This ends the proof of step 3. 2 We are now in position to prove Theorem 4.1. As already mentioned, it follows from steps 1 and 2 that there exists k ∈ N, Λ2 − u0 22 1≤k≤ , Λ2min

and that there exist k converging sequences (xi,α ) of points in M and k sequences (µi,α ) of positive real numbers converging to 0, such that Hk2 holds. From step 3, we then get by induction that there exists N ∈ N, Λ2 − u0 22 1≤k≤N ≤ , Λ2min

2 holds and such that such that HN lim

lim

sup

R→+∞ α→+∞ x∈M \Ωα (R)

n N Rα (x) 2 −1 uα (x) − u0 (x) = 0

where Ωα (R) = ∪N i=1 Bxi,α (Rµi,α ). This proves Theorem 4.1.

Chapter Five Asymptotics When the Energy Is of Minimal Type Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3. We let (hα ) be a sequence of smooth functions on M , and consider equations like ∆g u + hα u = u2

−1

(Eα )


for all α. For the reader’s convenience, we describe in this chapter the C 0 -theory for the blow-up behavior of the uα ’s when the energy of the uα ’s is of minimal type. The general situation of arbitrary energies is treated in the next chapter. As in Chapter 4, we assume in what follows that there exist 0 < θ < 1 and a smooth (or only C 0,θ ) function h∞ on M such that the operator ∆g + h∞ is coercive and (5.0.1) hα → h∞ in C 0,θ (M ) as α → +∞ . We also assume in this section that the energy of the uα ’s is of minimal type in the sense that E(uα ) ≤ Λmin

(5.0.2)

−(n−2)/2 Kn

for all α, where E(u) = u2 and Λmin = are as in Chapter 2. It is easily seen that the uα ’s are bounded in H12 (M ). Assuming (5.0.1), we clearly have that hα n/2 ≤ C for all α, where C > 0 is independent of α. Multiplying (Eα ) by uα and integrating over M , we then get that

2 2 |∇uα | dvg + hα uα dvg = u2α dvg ≤ Λ2min . M

M

From Hölder’s inequality,

2 hα uα dvg ≤ M

M

M

2/n

|hα |

n/2

dvg

u2α dvg

2/2 .

M

It follows that there exists C > 0 such that ∇uα 2 ≤ C for all α, and then that the uα ’s are bounded in H12 (M ) since they are also bounded in L2 (M ) and we have 2 < 2 . Up to a subsequence, we may therefore assume that for some u0 ∈ H12 (M ), uα u0 in H12 (M ) as α → +∞. Assuming (5.0.1) and (5.0.2), one can prove that, up to a subsequence, either u0 ≡ 0, and uα → u0 strongly in C 2 (M ) as α → +∞, or u0 ≡ 0 and uα 0 weakly and not strongly in H12 (M ) as α → +∞. This is the subject of Proposition 5.1 of section 5.1. Sharp estimates on the uα ’s are then proved in section 5.2.

68

CHAPTER 5

5.1 STRONG CONVERGENCE AND BLOW-UP We assume (5.0.1) and (5.0.2). As already mentioned, up to a subsequence, we have that uα u0 in H12 (M ) for some u0 ∈ H12 (M ) as α → +∞. We claim that the following proposition holds. P ROPOSITION 5.1 Under the above conditions (5.0.1) and (5.0.2), up to a subsequence, either u0 ≡ 0, and uα → u0 strongly in C 2 (M ) as α → +∞, or u0 ≡ 0, and uα 0 weakly and not strongly in H12 (M ) as α → +∞. Proof. An easy remark (we refer to Chapter 4) is that (5.0.1) implies the uniform coercivity condition (3.3.1) for α sufficiently large. In particular, up to a subsequence, there exists h such that hα ≥ h for all α, and such that ∆g + h is coercive. Then, from the Sobolev embedding theorem,

u2α dvg = |∇uα |2 dvg + hα u2α dvg M

M

M 2 ≥ |∇uα | dvg + hu2α dvg M M

≥ C1 |∇uα |2 dvg + u2α dvg M

u2α dvg

≥ C2

2/2

M

M

where C1 , C2 > 0 are independent of α, so that there exists c > 0 such that uα 2 ≥ c for all α. This proves that, if u0 ≡ 0, then uα 0 weakly and not strongly in H12 (M ) as α → +∞. In the following we assume that u0 ≡ 0. A possible proof that the convergence uα → u0 holds in C 2 (M ) is as follows. Thanks to regularity theory, it is easily seen that we need to prove only that the uα ’s are bounded in C 0 (M ). We proceed by contradiction and assume that, up to 4/(n−2) a subsequence, maxM uα → +∞ as α → +∞. We let λα = uα 2 and −1 u ˜α = uα 2 uα . Then 2 −1 ∆g u ˜ α + hα u ˜α = λα u ˜α

and

u ˜2α dvg = 1 .

M

As already mentioned, there exists c > 0 such that uα 2 ≥ c. We also have that λα ≤ Kn−2 for all α, and still up to a subsequence, we can assume that λα → λ as ˜ weakly in H12 (M ) as α → +∞, where u ˜ is given by α → +∞, and that u ˜α u −(n−2)/4 0 u . We let xα ∈ M and µα > 0 be such that u ˜=λ 1− n 2

max u ˜α = u ˜α (xα ) = µα M

.

Then µα → 0 as α → +∞. We use below only the fact that there exists r0 > n/2 − such that h− α r0 ≤ C for all α, where C > 0 is independent of α, and hα is the

69

ASYMPTOTICS WHEN THE ENERGY IS OF MINIMAL TYPE

negative part of hα . This easily follows from (5.0.1). Given 0 < δ < ig , where ig is the injectivity radius of (M, g), and x ∈ Rn such that µα |x| ≤ δ, we define n−2 ˜α expxα (µα x) u ˆα (x) = µα 2 u where expxα is the exponential map at xα . Then 2 −1 ˆ αu ∆gˆα u ˆα + h ˆα = λα u ˆα ˆ α (x) = µ2 hα exp (µα x) and gˆα (x) = (exp ) g(µα x). For any where h xα xα α R > 0, and α sufficiently large,

2 |∇ˆ uα | dvgˆα ≤ |∇˜ uα |2 dvg ≤ C and B0 (R) M

u ˆ2α dvgˆα ≤ u ˜2α dvg ≤ C

B0 (R)

M

where C > 0 is independent of α and R, and B0 (R) is the Euclidean ball of center 0 and radius R. Let η be a smooth cutoff function, 0 ≤ η ≤ 1, such that η = 1 in B0 (δ/2) and η = 0 in Rn \B0 (3δ/4). We let ηα (x) = η(µα x). Then, |∇ηα | ≤ Cµα for all α. Let s = 2 /(2 − 2). Then 2s = n, and we can write that

|∇(ηα u ˆα )|2 dvgˆα ≤ 2 |∇ηα |2 u ˆ2α dvgˆα + 2 ηα2 |∇ˆ uα |2 dvgˆα Rn

≤2

Rn

Rn

2s

1/s

|∇ηα | dvgˆα

B0 (3δ/4µα )

Rn

2/2

u ˆ2α dvgˆα

+ O(1) ≤ O(1) .

ˆα ’s are bounded in D12 (Rn ), where D12 (Rn ) is the homogeIt follows that the ηα u neous Sobolev space defined as the completion of C0∞ (Rn ) with respect to the norm uD12 = ∇u2 . Up to a subsequence, we can therefore assume that ˆα u in D12 (Rn ) as α → +∞, u ∈ D12 (Rn ), u ≥ 0. In particular, ηα u ˆα u ηα u in L2 (Rn ) as α → +∞, and thus ˆα 2 ≤ 1 . u2 ≤ lim inf ηα u α→+∞

(5.1.1)

We claim now that u ≡ 0. We can write that 2 −2 ˆ −| u ∆gˆα u ˆα . ˆ α ≤ λα u ˆα + |h α − r0 ˆ Moreover, u ˆα ≤ 1, and B0 (1) h dvgˆα ≤ C for all α, where C > 0 is indepenα dent of α, and r0 > n/2 is as above. Hence, thanks to the De Giorgi–Nash–Moser iterative scheme, as stated in Lemma 3.4, ˆα ≤ Cˆ uα L2 (B0 (1)) . 1 = max u B0 (1/2)

It clearly follows that u ≡ 0. This proves the above claim. We let ϕ ∈ C0∞ (Rn ), ϕ ≥ 0, and R > 0 such that suppϕ ⊂ B0 (R). For α large,

− 2 −1 ˆ u ˆα ϕdvgˆα (5.1.2) (∇ˆ uα ∇ϕ) dvgˆα + ˆα ϕdvgˆα ≤ λα hα u B0 (R)

B0 (R)

B0 (R)

70

CHAPTER 5

and we can write that

B0 (R)

2− rn

ˆ − |ˆ |h ˆα ≤ Cµα α uα ϕdvg

0

h− α r0 = o(1)

where C > 0 is independent of α, and r0 > n/2 is as above. Hence, passing to the limit as α → +∞ in (5.1.2), we get that, for all ϕ ∈ C0∞ (Rn ),

1 (∇u∇ϕ) dx ≤ 2 u2 −1 ϕdx . (5.1.3) K n Rn Rn By density, (5.1.3) is true for all ϕ ∈ D12 (Rn ), ϕ ≥ 0. Taking ϕ = u, it follows that

1 2 |∇u| dx ≤ 2 u2 dx . K n n R n R Noting that u ≡ 0, using the sharp Euclidean Sobolev inequality and (5.1.1), we can write that 1− 22

|∇u|2 dx 1 1 1 2 Rn ≤ ≤ u dx ≤ 2. 2/2 2 Kn2 K K n 2 R n n dx n u R

2

Hence, Rn u dx = 1, and u is an extremal function for the sharp Euclidean Sobolev inequality. In particular, ˆ α 2 u2 ≥ lim sup ηα u α→+∞

and, since L2 is uniformly convex, we get that ηα u ˆα → u strongly in L2 (Rn ) as α → +∞. Let Dα ⊂ M be a measurable set. Noting that there exists C > 0, such that for any x, y ∈ R, |x + y|2 − |x|2 − |y|2 ≤ C |x|2 −1 |y| + |y|2 −1 |x|

and that

|˜ uα − u ˜ |2

lim

α→+∞

we get that

−1

α→+∞

M

|˜ uα − u ˜|2 dvg =

Dα

|˜ u|2

|˜ u|dvg = lim

Dα

|˜ uα − u ˜|dvg = 0 ,

u ˜2 dvg + o(1) .

Dα

Taking Dα = M , it follows that

|˜ uα − u ˜|2 dvg = 1 − M

−1

M

u ˜2α dvg −

u ˜2 dvg + o(1) .

M

Taking Dα = Bxα (Rµα ), R > 0, it follows that

2 u ˜α dvg = |˜ uα − u ˜|2 dvg + o(1) . Bxα (Rµα )

Bxα (Rµα )


Now we can write that

B0 (R)

2

u dx ≤

|ηα u ˆα |2 dvgˆα + o(1)

B0 (R)

=

=

71

u ˜2α dvg + o(1)

Bxα (Rµα )

|˜ uα − u ˜|2 dvg + o(1)

Bxα (Rµα )

u ˜2 dvg + o(1) .

≤1 − M

Passing to the limit as α → +∞, and then as R → +∞, it follows that

1≤1− u ˜2 dvg , M

and we get the contradiction we were looking for. This ends the proof that the u ˜α ’s, and thus the uα ’s, are bounded in C 0 (M ). From regularity theory, and up to a subsequence, we then get that uα → u0 strongly in C 2 (M ) as α → +∞. This proves Proposition 5.1. 2 From now on we assume that uα 0 weakly and not strongly in H12 (M ) as α → +∞. As above, we assume in addition that (5.0.1) and (5.0.2) hold. Since (5.0.1) implies the uniform coercivity condition (3.3.1) for α large, there exists 4/(n−2) c > 0, independent of α, such that uα 2 ≥ c for all α. We let λα = uα 2 −1 and u ˜α = uα 2 uα . Then 2 −1 ˜ α + hα u ˜α = λα u ˜α ∆g u

and

˜α ) (E

u ˜2α dvg = 1 .

M

Kn−2

for all α, while u ˜α 0 weakly and not strongly in H12 (M ) as Clearly, λα ≤ α → +∞. An easy claim is as follows: up to a subsequence, there exists one point x0 in M such that for any δ > 0,

lim inf (5.1.4) u ˜2α dvg = 1 α→+∞

Bx0 (δ)

and such that 0 (M \{x0 }) u ˜α → 0 in Cloc

(5.1.5)

as α → +∞. In other words, up to a subsequence, the u ˜α ’s have one and only one geometrical blow-up point. In order to prove this claim, we proceed as follows. As already mentioned, (5.0.1) implies the uniform coercivity condition (3.3.1) for α large. From subsection 3.3 we then get that if S is the set of the geometrical ˜α → 0 in blow-up points of (˜ uα ), then, up to a subsequence, S is finite and u 0 (M \S). On the other hand, we clearly have that maxM u ˜α → +∞ as the Cloc parameter α → +∞. The above argument, used to prove Proposition 5.1, then

72

CHAPTER 5

shows that, up to a subsequence, there exists εR > 0, with the property that εR → 0 as R → +∞, such that for any R > 0,

1 − εR ≤ u ˜2α dvg + o(1) (5.1.6) Bxα (Rµα )

where o(1) → 0 as α → +∞, and xα , µα are as above. Still up to a subsequence, we can assume that xα → x0 as α → +∞. Given δ > 0, and passing to the limit as α → +∞ in (5.1.6), we then get that for any R > 0,

u ˜2α dvg 1 − εR ≤ lim inf α→+∞

Bx0 (δ)

≤ lim sup α→+∞

u ˜2α dvg ≤ 1 .

Bx0 (δ)

Passing to the limit as R → +∞, it follows that for any δ > 0,

lim u ˜2α dvg = 1 . α→+∞

Bx0 (δ)

In particular, S = {x0 }, and (5.1.4)–(5.1.5) are proved. 5.2 SHARP POINTWISE ESTIMATES We let (uα ) be a sequence of positive solutions to (Eα ). We assume that (5.0.1) and (5.0.2) hold. We also assume (see the discussion in section 5.1) that uα 0 weakly in H12 (M ) as α → +∞ .

(5.2.1)

We define xα ∈ M and µα > 0 by the relations 1− n 2

uα (xα ) = max uα = µα M

.

Clearly, xα → x0 and µα → 0 as α → +∞, where x0 is as in (5.1.4), (5.1.5). We know from Chapter 4 that there exists C > 0, independent of α, such that for any α and any x ∈ M , dg (xα , x)

n−2 2

uα (x) ≤ C

(5.2.2)

and that lim

lim

sup

R→+∞ α→+∞ M \Bx (Rµα ) α

dg (xα , x)

n−2 2

uα (x) = 0 ,

(5.2.3)

where dg is the distance with respect to g. We let u = u10 , where u10 is as in Chapter 3, be the function defined in Rn by 1− n2 |x|2 . u(x) = 1 + n(n − 2) Then, u is a positive solution, the only one satisfying u(0) = 1 = maxRn u, of the equation ∆u = u2

−1

73


where ∆ is the Euclidean Laplacian. Its energy E(u) = u2 is E(u) = Λmin , − n−2 2

−(n−2)/2

is as in Chapter 2. The standard bubble µα where Λmin = Kn is then given by the expression ⎞ n−2 ⎛ 2 n−2 x µα − 2 ⎠ ⎝ µα u . = 2 µα µ2α + |x|

u(µ−1 α x)

n(n−2)

We claim here that the following sharp estimates hold. T HEOREM 5.2 Under the above assumptions (5.0.1), (5.0.2), and (5.2.1), up to a subsequence, there exists C > 1, independent of α, such that for any α and any x ∈ M, ⎛ ⎛ ⎞ n−2 ⎞ n−2 2 2 µα µα 1 ⎝ ⎝ ⎠ ⎠ ≤ uα (x) ≤ C d (x ,x)2 C µ2α + dg (xα ,x)2 µ2α + g α n(n−2)

n(n−2)

where xα and µα are as above. Moreover, for any ε > 0, there exists δε > 0 such that, up to a subsequence, for any α and any x ∈ Bx0 (δε ), ⎛ ⎛ ⎞ n−2 ⎞ n−2 2 2 µα µ 1 ⎝ α ⎝ ⎠ ⎠ ≤ uα (x) ≤ (1 + ε) . d (x ,x)2 1 + ε µ2α + dg (xα ,x)2 µ2α + g α n(n−2)

n(n−2)

0

In particular, the uα ’s are C controlled, on both sides, by the standard bubble. Proof. We prove this theorem by coming back to the u ˜α ’s. As in the preceding 4/(n−2) and u ˜α = uα −1 u . Then, section, we let λα = uα 2 α 2 2 −1 ˜ α + hα u ˜α = λα u ˜α ∆g u

and

u ˜2α dvg = 1

˜α ) (E (5.2.4)

M

for all α. We know (see the proof of Proposition 5.1) that there exists c > 0, ˜α 0 weakly in independent of α, such that uα 2 ≥ c for all α. Clearly, u H12 (M ) as α → +∞, and (5.1.4), (5.1.5) hold. An easy claim is that 1 lim λα = 2 . (5.2.5) α→+∞ Kn On the one hand, we know that λα ≤ Kn−2 for all α. On the other hand, since the ˜α → 0 strongly embedding of H12 (M ) in L2 (M ) is compact, we can assume that u in L2 (M ). Thanks to the sharp Sobolev inequality, as proved by Hebey-Vaugon [47, 48], there exists B > 0, independent of α, such that ˜ uα 22 ≤ Kn2 ∇˜ uα 22 + B˜ uα 22 .

˜α ) and (5.2.4) we can then write that Using (E

hα u ˜2α dvg ≤ λα Kn2 ∇˜ uα 22 + B˜ uα 22 . ∇˜ uα 22 + M

74

CHAPTER 5

It follows that

uα 22 uα 22 ≤ C˜ 1 − λα Kn2 ∇˜

for all α, where C > 0 is independent of α. Clearly, there exists c > 0 such that ˜α ’s converge strongly to ∇˜ uα 2 ≥ c. If this is not the case, a subsequence of the u 0 in H12 (M ), contradicting (5.2.4). Passing to the limit as α → +∞ in the above equation, this proves (5.2.5). We define µ ˜α > 0 by the relations 1− n 2

˜α = µ ˜α u ˜α (xα ) = max u M

2/(n−2)

where xα is as above. Clearly, µ ˜α = uα 2 µα and µ ˜α → 0 as α → +∞. By (5.2.2), (5.2.3), and (5.2.5), we have that that there exists C > 0, independent of α, such that for any α and any x ∈ M , dg (xα , x)

n−2 2

u ˜α (x) ≤ C

and that lim

lim

sup

R→+∞ α→+∞ M \Bx (R˜ µα ) α

dg (xα , x)

n−2 2

u ˜α (x) = 0

(5.2.6)

where dg is the distance with respect to g. It is easily checked that the equations in Theorem 5.2 reduce to the existence of C > 1, independent of α, such that for any α and any x, ⎛ ⎞ n−2 ⎞ n−2 ⎛ 2 2 1 ⎝ µ ˜α µ ˜ α ⎠ ⎠ ⎝ ≤u ˜α (x) ≤ C 2/n 2/n ω d (x ,x)2 ω d (x ,x)2 C µ ˜2 + n g α µ ˜2 + n g α α

4

α

4

(5.2.7) and, for any ε > 0, to the existence of δε > 0, independent of α, such that for any α and any x ∈ Bx0 (δε ), ⎛ ⎞ n−2 ⎛ ⎞ n−2 2 2 µ ˜α µ ˜α 1 ⎝ ⎠ ⎝ ⎠ ≤ u ˜ (x) ≤ C α ε 2/n 2/n ω d (x ,x)2 ω d (x ,x)2 Cε µ ˜2α + n g4 α µ ˜2α + n g4 α (5.2.8) where Cε = 1 + ε. We prove (5.2.7) and (5.2.8) in what follows. The proof proceeds in several steps. We claim first that there exists C > 0 such that, up to a subsequence, n−2 2 µ ˜α u ˜α (x) ≤ C 2/n µ ˜2α + ωn4 dg (xα , x)2 for all α and all x ∈ M . It is easily checked that the proof of this claim reduces to the proof that there exists C > 0 such that for any α and any x ∈ M , ˜α (xα )˜ uα (x) ≤ C . dg (xα , x)n−2 u

(5.2.9)


75

As a first step we prove that for any ε > 0 there exists Cε > 0 such that for any α and any x ∈ M , − n−2 2 +ε

dg (xα , x)n−2−ε µ ˜α

u ˜α (x) ≤ Cε .

(5.2.10)

In step 2 we prove that we can take ε = 0 in (5.2.10), so that (5.2.9) is true. Step 3 is concerned with the exact asymptotic profile of the u ˜α ’s. The proof of (5.2.7) and (5.2.8), and thus of Theorem 5.2, follows from this exact asymptotic profile. S TEP 1. We prove that (5.2.10) is true. An easy claim is that it suffices to prove (5.2.10) for ε > 0 small. Indeed, let ε1 < ε2 be positive. It is easily seen that ˜α . On the other hand, if (5.2.10) with respect to ε = ε1 , ε2 is true if dg (xα , x) ≤ µ ˜α , then (5.2.10) is true with respect to ε1 , and dg (xα , x) ≥ µ − n−2 +ε

2 ˜α 2 u ˜α (x) dg (xα , x)n−2−ε2 µ ε −ε − n−2 +ε1 2 1 ≤ dg (xα , x)−1 µ ˜α dg (xα , x)n−2−ε1 µ ˜α 2 u ˜α (x)

so that (5.2.10) with respect to ε2 is also true. This proves the above claim. As already mentioned, (5.0.1) implies the uniform coercivity condition (3.3.1). We let h ∈ C 0,θ (M ) be such that, up to a subsequence, hα ≥ h for all α, and ∆g + h 0 is coercive. Clearly, there exists ε0 > 0 such that for ε > 0 small, ∆g + h−ε 1−ε is coercive. We fix ε > 0 small, and let Gε be the Green’s function of this operator. By the maximum principle, Gε is positive. We let Lα be the operator 2 −2 ˜α u. Lα u = ∆g u + hα u − λα u

˜α = 0 with u ˜α > 0 on M , we get from Berestycki-Nirenberg-Varadhan Since Lα u [9] that the maximum principle holds for Lα . It is easily seen that in M \{xα }, (xα , x) Lα G1−ε |∇Gε |2 (xα , x) ε 2 −2 = ε + h (x) − h(x) − λ u ˜ (x) + ε(1 − ε) 0 α α α G2ε (xα , x) G1−ε (xα , x) ε so that (xα , x) Lα G1−ε |∇Gε |2 (xα , x) ε 2 −2 ≥ ε . − λ u ˜ (x) + ε(1 − ε) 0 α α G2ε (xα , x) G1−ε (xα , x) ε

(5.2.11)

A standard property of the Green’s function we use in the following (we refer to Appendix A for more details) is that there exist C > 0 and ρ > 0 such that, for any α and any x ∈ Bxα (ρ)\{xα }, |∇Gε |(xα , x) C ≥ . Gε (xα , x) dg (xα , x)

(5.2.12)

Let R > 0, to be fixed later on. Combining (5.1.5) and (5.2.11), we get that for α sufficiently large (xα , x) ≥ 0 Lα G1−ε ε µα ), then, from (5.2.6), in M \Bxα (ρ). On the other hand, if x ∈ Bxα (ρ)\Bxα (R˜ 2 −2 ˜α (x) ≤ εR dg (xα , x)2 u

76

CHAPTER 5

where εR → 0 as R → +∞. Together with (5.2.11), (5.2.12), this gives that (xα , x) Lα G1−ε Cε(1 − ε) − λα εR ε ≥ ε0 + . 1−ε dg (xα , x)2 Gε (xα , x) We choose R > 0 sufficiently large such that Cε(1 − ε) − λα εR ≥ 0. Then (xα , x) ≥ 0 in Bxα (ρ)\Bxα (R˜ µα ), and we have proved that for α suffiLα G1−ε ε (x , x) ≥ 0 in M \B µα ). By another standard property ciently large, Lα G1−ε α xα (R˜ ε of the Green’s function (we refer here again to Appendix A for more details) there µα ), exists C > 0 such that for any α, and any x ∈ ∂Bxα (R˜ (xα , x) ≥ C µ ˜−(1−ε)(n−2) . G1−ε ε α (1−ε)(n−2)− n−2 2

˜α If we let Cα = C −1 µ

, we then get that

(xα , x) ≥ u ˜α (x) Cα G1−ε ε µα ). From the maximum principle, this implies that for all α and all x ∈ ∂Bxα (R˜ (xα , x) ≥ u ˜α (x) Cα G1−ε ε µα ). Noting that there exists C > 0 such that for for all α, and all x ∈ M \Bxα (R˜ any α, and any x ∈ M \{xα }, dg (xα , x)n−2 Gε (xα , x) ≤ C , it follows that for any ε > 0, any α, and any x ∈ M \{xα }, − n−2 ε 2 +ˆ

˜α dg (xα , x)n−2−ˆε µ

u ˜α (x) ≤ Cε

where εˆ = (n − 2)ε, and Cε > 0 is independent of α. In particular, (5.2.10) is true µα ). Noting that (5.2.10) is obviously satisfied in Bxα (R˜ µα ), this in M \Bxα (R˜ proves (5.2.10). S TEP 2. We prove that we can take ε = 0 in (5.2.10) so that (5.2.9) is true. ˜α (xα )˜ uα (x) is maximum. We need to We let yα be a point where dg (xα , x)n−2 u ˜α (xα )˜ uα (yα ) is bounded as α → +∞. We let G be the prove that dg (xα , yα )n−2 u Green’s function of the operator ∆g + h, where h is as above. Since G is positive,

G(yα , x) (∆g u ˜α + h˜ uα ) (x)dvg (x) u ˜α (yα ) =

M ≤ G(yα , x) (∆g u ˜ α + hα u ˜α ) (x)dvg (x) M

2 −1 = λα G(yα , x)˜ uα (x)dvg (x) . M

Given δ > 0 small, we write that

2 −1 G(yα , x)˜ uα (x)dvg (x) M

2 −1 = G(yα , x)˜ uα (x)dvg (x) + Bxα (δ)

M \Bxα (δ)

2 −1 G(yα , x)˜ uα (x)dvg (x) .

77


Using (5.2.10), we can write that

2 −1 G(yα , x)˜ uα (x)dvg (x) M \Bxα (δ)

( n−2 −ε)(2 −1) µ ˜α 2

=O

n+2 2 −(2 −1)ε

M \Bxα (δ)

G(yα , x)dvg (x)

G(yα , x)dvg (x) =O µ ˜α M n+2 −(2 −1)ε =O µ ˜α 2 . Therefore,

u ˜α (yα ) ≤ C

Bxα (δ)

n+2 −(2 −1)ε 2 −1 (5.2.13) G(yα , x)˜ uα (x)dvg (x) + O µ ˜α 2

where C > 0 is independent of α. Up to a subsequence, we can assume that yα → y0 as α → +∞. Let us assume first that y0 = x0 . We fix δ > 0 such that dg (x0 , y0 ) > 3δ. Then G(yα , .) is bounded in Bxα (δ) so that

2 −1 2 −1 G(yα , x)˜ uα (x)dvg (x) ≤ C u ˜α dvg . Bxα (δ)

Bxα (δ)

Independently, we can write that

− n+2 2 −1 2 u ˜α dvg ≤ µ ˜α Bxα (˜ µα )

n−2

Bxα (˜ µα )

dvg ≤ C µ ˜α 2 .

From (5.2.10), where we choose ε < 2/(2 − 1), we then get that

2 −1 2 −1 G(yα , x)˜ uα (x)dvg (x) ≤ C u ˜α dvg Bxα (δ)

n+2 2 −ε(2 −1)

Bxα (˜ µα )

+C µ ˜α

Bxα (δ)\Bxα (˜ µα )

It follows that

dg (xα , yα )

n−2 Bxα (δ)

dg (xα , x)(ε+2−n)(2

−1)

dvg (x) .

n−2

2 −1 G(yα , x)˜ uα (x)dvg (x) ≤ C µ ˜α 2 .

Coming back to (5.2.13), we then get that dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα ) ≤ C since ε < 2/(2 − 1). Now we assume that, up to a subsequence, yα → x0 as ˜α for some C > 0 independent of α. Then it α → +∞, and that dg (xα , yα ) ≤ C µ is clear that dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα ) ≤ C u ˜α (xα )−1 u ˜α (yα ) ≤ C .

78

CHAPTER 5

Finally, we assume that, up to a subsequence, yα → x0 as α → +∞ and that dg (xα ,yα ) → +∞ as α → +∞. Coming back to (5.2.13), we write that µ ˜α dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα )

n−2 − ≤ Cµ ˜α 2 dg (xα , yα )n−2

1 Bα

− n−2 2

+C µ ˜α

2 −1 G(yα , x)˜ uα (x)dvg (x)

dg (xα , yα )n−2

2 Bα

2 −1 G(yα , x)˜ uα (x)dvg (x)

−1)ε n−2 +O µ ˜2−(2 d (x , y ) g α α α 1 where Bα = x ∈ Bxα (δ) s.t. dg (yα , x) ≥ 12 dg (xα , yα ) , and Bα2 is given by the equation Bα2 = Bxα (δ)\Bα1 . By a standard property of the Green’s function (we refer to Appendix A) there exists C > 0 such that G(x, y) ≤ Cdg (x, y)2−n for all (x, y) ∈ M × M , x = y. We once again fix ε > 0 such that ε < 2/(2 − 1). Then,

− n−2 2 −1 dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα ) ≤ C µ ˜α 2 u ˜α dvg − n−2 2

+C µ ˜α

1 Bα


2 −1 dg (yα , x)2−n u ˜α (x)dvg (x) + o(1) .

2 Bα

As above, using (5.2.10), we easily get that

− n−2 2 −1 2 µ ˜α u ˜α dvg ≤ C 1 Bα

for some C > 0 independent of α. Independently, still using (5.2.10), we can write that

− n−2 2

µ ˜α


≤C

µ ˜α dg (xα , yα )

2 −1 dg (yα , x)2−n u ˜α (x)dvg (x)

2 Bα

2−(2 −1)ε

1 dg (xα , yα )2

2 Bα

dg (yα , x)2−n dvg (x)

2−(2 −1)ε µ ˜α = o(1) dg (xα , yα ) where C > 0 is independent of α. Hence, ˜α (xα )˜ uα (yα ) ≤ C dg (xα , yα )n−2 u where C > 0 is independent of α. Summarizing these cases, we have proved that, up to a subsequence, the sequence dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα ) is bounded. In particular, (5.2.9) is true. ≤C

From now on, we let G be the Green’s function of the operator ∆g + h∞ , where h∞ is given by (5.0.1). Let also y0 ∈ M , and a sequence (yα ) in M such that yα → y0 as α → +∞. S TEP 3. We claim that the two following asymptotics hold: if y0 = x0 , then, up to a subsequence, n−2 2 2/n ωn − n−2 2 2 dg (xα , yα ) µ ˜α + µ ˜α 2 u ˜α (yα ) → 1 (5.2.14) 4

79


as α → +∞, and if y0 = x0 , then, up to a subsequence, n−2 2 2/n ωn − n−2 2 2 µ ˜α + dg (xα , yα ) µ ˜α 2 u ˜α (yα ) 4 → (n − 2)ωn−1 dg (x0 , y0 )

n−2

(5.2.15)

G(x0 , y0 )

as α → +∞. In particular, from standard elliptic theory, n−2 2 u ˜α (xα )˜ uα (x) → 4ωn−2/n (n − 2)ωn−1 G(x0 , x) 2 in Cloc (M \{x0 }) as α → +∞. We prove (5.2.14) and (5.2.15). For α large, thanks to (5.0.1), ∆g + hα is coercive. We let Gα be the Green’s function of the operator ∆g + hα . Given δ > 0 and small, we write that

Gα (yα , x) (∆g u ˜ α + hα u ˜α ) dvg (x) u ˜α (yα ) = M

2 −1 = λα Gα (yα , x)˜ uα (x)dvg (x)

M 2 −1 = λα Gα (yα , x)˜ uα (x)dvg (x) Bxα (δ)

+λα From step 2,

2 −1 Gα (yα , x)˜ uα (x)dvg (x) .

M \Bxα (δ)

2 −1 Gα (yα , x)˜ uα (x)dvg (x)

M \Bxα (δ)

n+2 2

=O µ ˜α

M \Bxα (δ)

Gα (yα , x)dvg (x)

n+2 =O µ ˜α 2 . Therefore, u ˜α (yα ) = λα

Bxα (δ)

n+2 2 −1 . Gα (yα , x)˜ uα (x)dvg (x) + O µ ˜α 2

(5.2.16)

We again distinguish three cases. First we assume that y0 = x0 . We choose δ > 0 ˜−1 and small such that dg (x0 , y0 ) ≥ 3δ. For x ∈ B0 (δ µ α ), the Euclidean ball of −1 center 0 and radius δ µ ˜α , we let n−2 ˜α 2 u ˜α expxα (˜ µα x) u ˆα (x) = µ where expxα is the exponential map at xα . Then, 2 −1 ˆ αu ∆gˆα u ˆα + h ˆα = λα u ˆα ˆ α (x) = µ where h ˜2α hα expxα (˜ µα x) and gˆα (x) = (expxα ) g(˜ µα x). The u ˆα ’s are ˆ α → 0 uniformly. It follows from standard clearly bounded, and, from (5.0.1), h

80

CHAPTER 5 0,γ Cloc (Rn )

elliptic theory that the u ˆα ’s are bounded in for some γ > 0, and then, from Ascoli’s theorem, that, up to a subsequence, we also have that u ˆα → u in 0 n (R ) as α → +∞. Moreover, it is easily checked that u(0) = 1 = maxRn u, Cloc that Rn u2 dx = 1, and that ∆u =

1 2 −1 u Kn2

where ∆ is the Euclidean Laplacian. Hence, u(x) =

2/n

ωn |x|2 1+ 4

− n−2 2

where ωn is the volume of the unit n-sphere. Coming back to (5.2.16) we write that uα (yα ) u ˜α (xα )˜

− n−2 = λα µ ˜α 2

= λα

Bxα (δ)

B0 (δ µ ˜ −1 α )

2 2 −1 Gα (yα , x)˜ uα (x)dvg (x) + O µ ˜α

2 −1 2 Gα yα , expxα (˜ µα x) u ˆα (x)dvgˆα (x) + O µ ˜α .

An equivalent formulation of the claim in step 2 is that u ˆα ≤ Cu for some C > 0 independent of α, where u is as above. The Lebesgue-dominated convergence theorem and this estimate then give that

−2 lim u ˜α (xα )˜ uα (yα ) = Kn G(y0 , x0 ) u2 −1 dx α→+∞

Rn

where G is the Green’s function of the operator ∆g + h∞ . It follows that u ˜α (xα )˜ uα (yα ) →

4 2/n

ωn

n−2 2

(n − 2)ωn−1 G(x0 , y0 )

as α → +∞. This proves (5.2.15). Let us now assume that y0 = x0 and that dg (xα ,yα ) → R as α → +∞. We let zα be such that yα = expxα (˜ µα zα ). Then µ ˜α lim dg (xα , yα )

n−2

α→+∞

u ˜α (xα )˜ uα (yα ) = lim

α→+∞

dg (xα , yα ) µ ˜α

n−2 u ˆα (zα )

ˆα → u where u ˆα is as above. Clearly, |zα | → R as α → +∞, and since, as above, u dg (xα ,yα ) 0 n in Cloc (R ) as α → +∞, we get that if y0 = x0 and → R as α → +∞, µ ˜α then n−2 2 2 R ˜α (xα )˜ uα (yα ) → (5.2.17) dg (xα , yα )n−2 u 2/n 1 + ωn4 R2

81


as α → +∞. Finally, we assume that y0 = x0 and that α → +∞. Coming back to (5.2.16), we can write that dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα )

n−2 − = λα µ ˜α 2 dg (xα , yα )n−2 − n−2 2

2 −1 (x)dvg (x) Gα (yα , x)˜ uα


→ +∞ as

1 Bα

+λα µ ˜α

dg (xα ,yα ) µ ˜α

2 −1 Gα (yα , x)˜ uα (x)dvg (x)

2 Bα

˜2α +O dg (xα , yα )n−2 µ where Bα1 = x ∈ Bxα (δ) s.t. dg (yα , x) ≥ 12 dg (xα , yα ) , and Bα2 is given by the equation Bα2 = Bxα (δ)\Bα1 . Clearly,

− n−2 2 −1 Gα (yα , x)˜ uα (x)dvg (x) µ ˜α 2

=

Ωα

1 Bα

2 −1 Gα yα , expxα (˜ µα x) u ˆα (x)dvgˆα (x)

−1 1 where Ωα = µ ˜−1 µα x). For x ∈ Ωα , the ratio α expxα (Bα ). Let zα = expxα (˜ dg (yα ,zα ) → +∞ as α → +∞. Writing that µ ˜α

dg (yα , zα ) − dg (xα , zα ) ≤ dg (xα , yα ) ≤ dg (yα , zα ) + dg (xα , zα ) and noting that dg (xα , zα ) = µ ˜α |x|, we get that lim

α→+∞

dg (xα , yα ) = 1. dg (yα , zα )

By a standard property of the Green’s function (we refer to Appendix A for more details) this implies that 1 lim dg (xα , yα )n−2 Gα yα , expxα (˜ µα x) = α→+∞ (n − 2)ωn−1 where ωn−1 is the volume of the unit (n − 1)-sphere. Because of the Lebesguedominated convergence theorem and step 2, we then get that

− n−2 2 −1 ˜α 2 dg (xα , yα )n−2 Gα (yα , x)˜ uα (x)dvg (x) lim λα µ α→+∞

1 = (n − 2)ωn−1 Kn2

1 Bα

u

2 −1

dx =

4

n−2 2

2/n

ωn where u is as above. On the other hand, from step 2 [see (5.2.9)] we can write that

− n−2 2 −1 µ ˜α 2 dg (xα , yα )n−2 Gα (yα , x)˜ uα (x)dvg (x) ≤C ≤C

µ ˜α dg (xα , yα ) µ ˜α dg (xα , yα )

Rn

2 2

2 Bα

1 dg (xα , yα )2 = o(1)

2 Bα

dg (yα , x)2−n dvg (x)

82

CHAPTER 5 dg (xα ,yα ) µ ˜α

where C > 0 is independent of α. Therefore, if y0 = x0 and α → +∞, then n−2 2 4 dg (xα , yα )n−2 u ˜α (xα )˜ uα (yα ) → 2/n ωn

→ +∞ as

(5.2.18)

as α → +∞. Combining (5.2.17) and (5.2.18), and since u(0) = 1, this proves (5.2.14). Now we prove (5.2.7) and (5.2.8), and thus Theorem 5.2. We let vα be the function defined by − n−2 2 µ ˜α u ˜α (x) vα (x) = 2/n µ ˜2α + ωn4 dg (xα , x)2 and let yα be such that vα (yα ) = minM vα . Then, up to a subsequence, it follows from (5.2.14) and (5.2.15) that vα (yα ) ≥ c for some c > 0 independent of α. Together with step 2, this proves (5.2.7). Independently, lim dg (x0 , y)n−2 G(x0 , y) =

y→x0

1 . (n − 2)ωn−1

Hence, given ε > 0, there exists δε > 0 such that if y ∈ Bx0 (δε ), then 1 ≤ (n − 2)ωn−1 dg (x0 , y)n−2 G(x0 , y) ≤ 1 + ε . 1+ε If we let yα and y˜α be such that yα ) = max vα , vα (yα ) = min vα and vα (˜ Bx0 (δε )

Bx0 (δε )

we then get with (5.2.14) and (5.2.15) that, up to a subsequence, 1 ≤ vα (yα ) ≤ vα (˜ yα ) ≤ 1 + ε . 1+ε This proves (5.2.8) and thus Theorem 5.2.

2

Chapter Six Asymptotics When the Energy Is Arbitrary Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3. We let (hα ) be a sequence of smooth functions on M , and consider equations like ∆g u + hα u = u2

−1

(Eα )


for all α. We describe in this chapter the C 0 -theory for the blow-up behavior of the uα ’s when the energy of the uα ’s is arbitrary. As in Chapters 4 and 5, we assume in what follows that there exist 0 < θ < 1 and a smooth (or only C 0,θ ) function h∞ on M such that the operator ∆g + h∞ is coercive and

(6.0.1)

hα → h∞ in C 0,θ (M ) as α → +∞ .

We also assume that there exists Λ > 0 such that E(uα ) ≤ Λ for all α, where E(u) = u2 is as in Chapter 2. The existence of uα implies the coercivity of the operator ∆g + hα . An estimate like the one in Theorem 6.1 below implies the coercivity of the operator ∆g + h∞ (see Appendix B). Independently, as in Chapter 4, multiplying (Eα ) by uα and integrating over M , we get that

2 2 |∇uα | dvg + hα uα dvg = u2α dvg ≤ Λ2 . M

M

M

Noting that by Hölder’s inequalities,

2/n

n/2 2 h u dv |h | dv ≤ g α g α α M

and that by (6.0.1),

M

u2α dvg

2/2

M

|hα |n/2 dvg ≤ C M

for all α, where C > 0 is independent of α, we get that the uα ’s are bounded in H12 (M ). Up to a subsequence, we may therefore assume that for some function u0 ∈ H12 (M ), uα u0 weakly in H12 (M ) 0

as α → +∞. It is easily seen that u is a solution of ∆g u0 + h∞ u0 = (u0 )2

−1

.

(6.0.2)

84

CHAPTER 6

By the maximum principle and regularity theory, u0 ∈ C 2,θ (M ), and either u0 ≡ 0 or u0 > 0 everywhere in M . We assume in what follows that lim max uα = +∞

(6.0.3)

α→+∞ M

so that blow-up occurs. If this is not the case, up to a subsequence, we easily get from standard regularity theory that uα → u0 strongly in C 2 (M ) as α → +∞, and this provides a complete description of the behavior of the uα ’s. We prove in this chapter that the following theorem holds. The theorem was announced in Druet-Hebey-Robert [32]. Applications of the theorem are in Druet [27]. T HEOREM 6.1 Let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, (hα ) be a sequence of smooth functions on M such that (6.0.1) is satisfied and (uα ) be a sequence of positive solutions to (Eα ) such that E(uα ) ≤ Λ for some Λ > 0 and all α. We assume that (6.0.3) holds. Then there exist N ∈ N , converging sequences (xi,α ) in M , and sequences (µi,α ) of positive real numbers converging to 0, i = 1, . . . , N , such that, up to a subsequence, ⎛ ⎞ n−2 2 N µ 1 i,α 0 ⎝ ⎠ (1 − εα ) u (x) + C i=1 µ2 + dg (xi,α ,x)2 i,α

≤ uα (x) ≤ (1 + εα ) u0 (x) + C

n(n−2)

N

⎛ ⎝

i=1

µi,α µ2i,α

+

dg (xi,α ,x)2 n(n−2)

⎞ n−2 2 ⎠

0

for all x ∈ M and all α, where u is given by (6.0.2), C > 1 is independent of α and x, and (εα ), independent of x, is a sequence of positive real numbers converging to 0 as α → +∞. In particular, the uα ’s are C 0 -controlled, on both sides, by u0 and standard bubbles. A complement to Theorem 6.1 is that C can be chosen as close as we want to 1 if we restrict the equation in Theorem 6.1 to small neighborhoods of the geometrical blow-up points. For instance, if u0 ≡ 0, or if the uα ’s just have one geometrical blow-up point, then for any ε > 0, there exists δε > 0 such that, up to a subsequence, ⎛ ⎞ n−2 2 N 1 µ i,α 0 ⎝ ⎠ (1 − εα ) u (x) + 1 + ε i=1 µ2 + dg (xi,α ,x)2 i,α

n(n−2)

≤ uα (x) ≤ (1 + εα ) u0 (x) + (1 + ε)

N i=1

⎛ ⎝

µi,α µ2i,α +


⎞ n−2 2 ⎠

for all α, all x0 ∈ S, and all x ∈ Bx0 (δε ), where S is the set consisting of the limits of the xi,α ’s as α → +∞. Outside the Bx (δε )’s, x ∈ S, the uα ’s converge C 2,θ to u0 . The estimate then extends to M in the particular case where u0 ≡ 0. We refer to section 6.3 [see, in particular, (6.3.24), (6.3.25), and (6.3.37)] for more details and a refined statement.

ASYMPTOTICS WHEN THE ENERGY IS ARBITRARY

85

Another complement to Theorem 6.1 is that the bubbles in this theorem satisfy Theorem 3.1. More precisely, we also have that for any x ∈ M and any α, ⎛ ⎞ n−2 2 N µi,α 0 ⎝ ⎠ uα (x) = u (x) + + Rα (x) d (xi,α ,x)2 µ2i,α + gn(n−2) i=1 where the Rα ’s, Rα ∈ H12 (M ) for all α, are such that Rα → 0 strongly in H12 (M ) as α → +∞. Moreover, if

1 1 |u|2 dvg |∇u|2 + hα u2 dvg − Igα (u) = 2 M 2 M and

1 1 Ig∞ (u) = |u|2 dvg |∇u|2 + h∞ u2 dvg − 2 M 2 M are as in Chapter 3, then N Igα (uα ) = Ig∞ (u0 ) + Kn−n + rα n where the rα ’s, rα ∈ R for all α, are such that rα → 0 as α → +∞. We refer to the end of section 6.3 for the proof of these claims. Sections 6.1–6.3 below are devoted to the proof of Theorem 6.1. We set up some notation in the rest of this introduction. We let the xi,α ’s and µi,α ’s be as in Theorem 4.1, i = 1, . . . , N , and choose N to be maximal. As a remark, the xi,α ’s in Theorem 6.1 will be slightly distinct from those in Theorem 4.1, but the µi,α ’s are the same. Up to a subsequence, we may assume that for all i, j, k in {1, . . . , N }, the µ−1 k,α dg (xi,α , xj,α )’s have a limit in [0, +∞] as α → +∞. We let Sα = {xi,α , i = 1, . . . , N }

(6.0.4)

and rearrange the xi,α ’s in the following way. First, we let x1,α be such that µ1,α satisfies the condition µ1,α ≥ µi,α for all i = 1, . . . , N . We let N1 = 1 and Sα1 = {xN1 +1,α , . . . , xN2 −1,α } be the subset of Sα consisting of the xi,α ’s, i > N1 , which are such that lim

α→+∞

dg (xN1 ,α , xi,α ) < +∞ . µN1 ,α

We let then xN2 ,α be such that µN2 ,α ≥ µi,α for all i = N2 , . . . , N , and let Sα2 = {xN2 +1,α , . . . , xN3 −1,α } be the subset of Sα consisting of the xi,α ’s, i > N2 , which are such that lim

α→+∞

dg (xN2 ,α , xi,α ) < +∞ . µN2 ,α

We repeat the process up to the exhaustion of Sα . We then get the existence of some integer p such that Sα consists of the (ordered) sequence x1,α = xN1 ,α , x2,α , . . . , xN2 ,α , xN2 +1,α , . . . , xN3 ,α , . . . , xNp ,α , . . . , xN,α = xNp+1 −1,α .

86

CHAPTER 6

Given i = 1, . . . , p, we let Sα0 = {xNi ,α , i = 1, . . . , p} , Sαi = xNi +1,α , . . . , xNi+1 −1,α , Sˆα = ∪p S i .

(6.0.5)

i=1 α

The

Sαj ’s,

j = 0, . . . , p, are disjoint, and we have that Sα = Sα0 ∪ Sˆα .

As a remark, it might be, in this process of rearranging the xi,α ’s, that Ni+1 = Ni + 1 or, equivalently, that Sαi = ∅. In this case, xNi+1 ,α is chosen such that µNi+1 ,α ≥ µi,α for all i = Ni + 1, . . . , N . Given i = 1, . . . , p, we let yi,α = xNi ,α and νi,α = µNi ,α .

(6.0.6)

The above construction gives that ν1,α ≥ · · · ≥ νp,α

(6.0.7)

dg (yi,α , yj,α ) → +∞ νi,α

(6.0.8)

and that for i = j,

as α → +∞. We let R0 > 0 be such that for any j = 1, . . . , p and for any i = Nj , . . . , Nj+1 − 1, lim

α→+∞

dg (yj,α , xi,α ) R0 . ≤ νj,α 2

(6.0.9)

Independently, we let Rα be the function given by Rα (x) = min dg (yj,α , x) . j=1,...,p

(6.0.10)

Up to a subsequence, we may assume that there exists C > 0 such that, for any α, sup

and that lim

lim

sup

R→+∞ α→+∞ M \Ω0 (R)

Rα (x) 2 −1 uα (x) ≤ C , n

M \Ω0α (R0 )

n Rα (x) 2 −1 uα (x) − u0 (x) = 0

(6.0.11)

(6.0.12)

α

where Ω0α (R) = ∪pj=1 Byj,α (Rνj,α ). We prove (6.0.11) and (6.0.12) in what follows. Let (yα ) be a sequence of points in M \Ω0α (R0 ). We distinguish two cases. In the first case we assume that there exists j ∈ {1, . . . , p} such that dg (yj,α , yα ) →R νj,α as α → +∞, where R ≥ R0 . By the above construction, Rα (yα ) = dg (yj,α , yα )


87

for α large. We write that yα = expyj,α (νj,α zα ), where zα ∈ Rn is such that |zα | → R as α → +∞. From Theorem 4.1, proposition (P1), and (6.0.8), (6.0.9), there exists z0 ∈ Rn such that, if zα → z as α → +∞, then n n lim Rα (yα ) 2 −1 uα (yα ) = R 2 −1 u (z − z0 ) α→+∞

where u is given by (4.1.5). In particular, n lim sup Rα (yα ) 2 −1 uα (yα ) < +∞ . α→+∞

In the second case we assume that, for any j ∈ {1, . . . , p}, dg (yj,α , yα ) → +∞ νj,α N be as in Theorem 4.1. After passing to a subsequence, as α → +∞. We let Rα there exists k ∈ {1, . . . , N } such that N Rα (yα ) = dg (xk,α , yα ) . We let j ∈ {1, . . . , p} be such that Nj ≤ k ≤ Nj+1 − 1. By (6.0.9), and for α sufficiently large, N Rα (yα ) ≥ dg (yα , yj,α ) − dg (xk,α , yj,α ) ≥ dg (yα , yj,α ) − R0 νj,α 1 ≥ dg (yα , yj,α ) 2 1 ≥ Rα (yα ) . 2 By the first claim of proposition (P2) of Theorem 4.1, it follows that, here again, n lim sup Rα (yα ) 2 −1 uα (yα ) < +∞ . α→+∞

In particular, (6.0.11) is proved. Independently, it is easily checked that, for any i ∈ {1, . . . , p} and for any l ∈ {Ni , . . . , Ni+1 − 1}, dg (xl,α , yα ) νi,α dg (yi,α , yα ) ≥ − R0 µl,α µl,α νi,α so that dg (xl,α , yα ) → +∞ µl,α as α → +∞. In particular, for any R > 0, yα ∈ M \Ωα (R) for α large, where Ωα (R) is as in Theorem 4.1. By the second claim of proposition (P2) of Theorem 4.1, and since 1 N Rα (yα ) ≥ Rα (yα ) , 2 we get that n Rα (yα ) 2 −1 uα (yα ) − u0 (yα ) → 0 as α → +∞. This proves (6.0.12). The rest of this chapter is divided as follows. Section 6.1 is devoted to the proof of an upper estimate on the uα ’s, similar to the one in Theorem 6.1 when we replace Sα by Sα0 . In section 6.2 we prove the upper estimate on the uα ’s with respect to the full set Sα . Section 6.3 is devoted to the proof of the estimate from below in Theorem 6.1, and to the asymptotics of uα .

88

CHAPTER 6

6.1 A FUNDAMENTAL ESTIMATE: 1 This section is devoted to the proof of a fundamental pointwise estimate with respect to Sα0 , where Sα0 is as in (6.0.5). We borrow ideas from Druet-Robert [33] and Hebey-Vaugon [47, 48] where this kind of analysis was carried out in the case when the energy of the uα ’s is of minimal type (see Chapter 5). We divide this section into two subsections. In subsection 6.1.1 we consider the case where u0 ≡ 0. In subsection 6.1.2 we consider the case where u0 ≡ 0. All that follows is up to a subsequence. 6.1.1 The case u0 ≡ 0. We assume in this subsection that u0 ≡ 0. Let R0 be as in (6.0.9). We want to prove that there exists C > 0 such that for any y ∈ M \ ∪pj=1 Byj,α (R0 νj,α ), uα (y) ≤ C

p

n

−1

2−n

2 νj,α dg (yj,α , y)

(6.1.1)

j=1

where the yj,α ’s and νj,α ’s are as in (6.0.6). We split the proof of (6.1.1) into several steps that we refer to as claims. The proof of (6.1.1) is by induction. First we introduce some notation. We let h0 ∈ C 0,θ (M ) be such that h0 < h∞ and ∆g + h0 is coercive. We let G : M × M \ {(x, x) , x ∈ M } → R+ , be the Green’s function of the operator ∆g + h0 . Then G is the only symmetric positive function satisfying in the sense of distributions ∆g,y G (x, y) + h0 (y)G (x, y) = δx for all x ∈ M , where δx is the Dirac mass at x. We know (see Appendix A) that there exist ρ > 0, C1 > 0 and C2 > 0 such that, for any x = y in M , dg (x, y) ≤ ρ =⇒

C1 |∇y G(x, y)| ≥ G(x, y) dg (x, y)

(6.1.2)

and 1 n−2 ≤ dg (x, y) G (x, y) ≤ C2 . C2

(6.1.3)

Our first claim is the following: C LAIM 6.1.1. Assume that u0 ≡ 0, and let 0 < ε < 12 . If ε is sufficientl small, then there exist R(ε) > 0 and C (ε) > 0 such that for any α > 0 and any y in M \ ∪pj=1 Byj,α (R (ε) νj,α ), (1− n )(1−2ε) (n−2)(1−ε) ν1,α 2 Rα (y) uα (y) ≤ C (ε) where Rα is as in (6.0.10), the νj,α ’s are as in (6.0.6), and the yj,α ’s are as in (6.0.6). Proof of claim 6.1.1. Let Lα be the linear operator given by 2 −2 u. Lα (u) = ∆g u + hα u − uα

89


Since Lα uα = 0 and uα > 0, Lα satisfies the maximum principle on M (see for instance [9]). We let 0 < ε < 12 be such that h∞ > (1 − ε) h0 and set Hα (y) =

p

1−ε

Gj,α (y)

j=1

where Gj,α (y) = G (yj,α , y) and G is the Green’s function of ∆g + h0 . Easy computations give that for any y ∈ M \ {y1,α , . . . , yp,α }, Lα Hα (y) =

p

1−ε

Gj,α (y)

Aα (y)

(6.1.4)

j=1

where 2 −2

Aα (y) = hα (y) − (1 − ε) h0 (y) − uα (y)

|∇Gj,α (y)|2

+ ε (1 − ε)

2

Gj,α (y)

.

We claim that Lα Hα ≥ 0 in M \ ∪pj=1 Byj,α (Rνj,α ) for some R > 0 to be fixed later on, and for α large enough. In order to prove this claim, we let (yα ) be a sequence of points in M \ ∪pj=1 Byj,α (Rνj,α ). Up to a subsequence, yα → y0 as α → +∞. We distinguish two cases. Case 1. We assume that dg (y0 , S) ≥ ρ2 , where S is given by Theorem 4.1 and 0 (M \S). Coming ρ is as in (6.1.2). By point (P1) of Theorem 4.1, uα → 0 in Cloc back to (6.1.4), we can write in this case that, for α sufficiently large, Lα Hα (yα ) ≥

p

1−ε

Gj,α (yα )

(hα (yα ) − (1 − ε) h0 (yα ) + o (1))

j=1

≥0 since we fixed ε such that h∞ > (1 − ε) h0 . Case 2. We assume that dg (y0 , S) ≤ ρ2 , where S is given by Theorem 4.1 and ρ is as in (6.1.2). We let i ∈ {1, . . . , p} be such that, up to a subsequence, Rα (yα ) = dg (yi,α , yα ). We write that 1−ε

Lα Hα (yα ) ≥ ε (1 − ε) Gi,α (yα ) 2 −2

−uα (yα )

p

|∇Gi,α (yα )|2 2

Gi,α (yα ) 1−ε

Gj,α (yα )

.

j=1

Then, from (6.1.2) and (6.1.3), C12 (2−n)(1−ε)−2 Rα (yα ) C21−ε p 2 −2 1−ε (2−n)(1−ε) −uα (yα ) C2 dg (yj,α , yα ) ,

Lα Hα (yα ) ≥ ε (1 − ε)

j=1

90

CHAPTER 6

and we get that C12 (2−n)(1−ε)−2 Rα (yα ) C21−ε

Lα Hα (yα ) ≥ ε (1 − ε)

2 −2

−uα (yα ) (2−n)(1−ε)−2

(2−n)(1−ε)

C21−ε pRα (yα ) C12 2 2 −2 1−ε ε (1 − ε) 1−ε − pC2 Rα (yα ) uα (yα ) C2

≥ Rα (yα )

for α sufficiently large. We fix R > R0 , depending only on ε, such that, for α large, 2

2 −2

Rα (yα ) uα (yα )

≤

ε (1 − ε) C12 . 2p C22−2ε

Since u0 ≡ 0, this is always possible from (6.0.12). Then Lα Hα (yα ) ≥ 0. In particular, it follows from cases 1 and 2 that there exists R (ε) ≥ R0 such that, for α large, Lα Hα ≥ 0 in M \ ∪pj=1 Byj,α (R (ε) νj,α ) .

(6.1.5)

Since R (ε) ≥ R0 , we can write that, for α large, 1− n

uα (y) ≤ 2νj,α 2

(6.1.6)

for any j ∈ {1, . . . , p} and any y ∈ ∂Byj,α (R (ε) νj,α ). This follows from point (P1) of Theorem 4.1 and the choice of R0 we made in (6.0.9). Independently, (6.1.3) implies that, for any j ∈ {1, . . . , p}, (2−n)(1−ε)

Hα ≥ C2ε−1 (R (ε) νj,α )

(6.1.7)

on ∂Byj,α (R (ε) νj,α ). Using (6.1.6) and (6.1.7), and since ν1,α ≥ νj,α for all j ∈ {1, . . . , p}, we get that for any j ∈ {1, . . . , p} and any y ∈ ∂Byj,α (R (ε) νj,α ), 1− n

uα (y) ≤ 2νj,α 2

( n −1)(1−2ε) (2−n)(1−ε) ≤ 2νj,α2 νj,α (n−2)(1−ε)

≤ 2C21−ε R (ε)

( n2 −1)(1−2ε) ν1,α Hα (y) .

In particular, ( n2 −1)(1−2ε) Hα (y) uα (y) ≤ C(ε)ν1,α for all j ∈ {1, . . . , p} and all y ∈ ∂Byj,α R (ε) νj,α . Noting that Lα uα = 0, it follows from (6.1.5) that ( n2 −1)(1−2ε) Hα − u α ≥ 0 Lα C(ε)ν1,α in M \∪pj=1 Byj,α R (ε) νj,α . From the maximum principle, we then get that there exists C(ε) > 0, depending only on ε, such that ( n2 −1)(1−2ε) Hα (y) uα (y) ≤ C (ε) ν1,α

91


in M \

∪pj=1

Byj,α (R (ε) νj,α ). Together with (6.1.3), this proves claim 6.1.1. 2

The content of our second claim is that the inequality in claim 6.1.1 holds also with ε = 0. Claim 6.1.2 can be stated as follows: C LAIM 6.1.2. Assume that u0 ≡ 0. There exists C > 0 such that for any α > 0 and any y in M \ ∪pj=1 Byj,α (R0 νj,α ), −1

n

2−n

2 uα (y) ≤ Cν1,α Rα (y)

where Rα is as in (6.0.10), the νj,α ’s are as in (6.0.6), and the yj,α ’s are as in (6.0.6). Proof of claim 6.1.2. For j ∈ {1, . . . , p}, we let Σj,α = {y ∈ M s.t. Rα (y) = dg (yj,α , y)} . We also let G be the Green’s function of ∆g + h0 , where h0 ∈ C 0,θ (M ) is such that ∆g + h0 is coercive and h∞ > h0 . Let (yα ) be a sequence in M \ ∪pj=1 Byj,α (R0 νj,α ). We have to prove that 1− n

n−2

lim sup ν1,α 2 Rα (yα ) α→+∞

uα (yα ) < +∞ .

(6.1.8)

By the Green’s representation formula, we get, using (Eα ) and h∞ > h0 , that

2 −1 uα (yα ) ≤ G (yα , y) uα (y) dvg (6.1.9) M

for α large. We distinguish three cases. Case 1. We assume that, up to a subsequence, Rα (yα ) → 2δ as α → +∞ for some δ > 0. We then write that, for any j ∈ {1, . . . , p},

2 −1 G (yα , y) uα (y) dvg Σj,α

=

2 −1

Σj,α ∩Byj,α (δ)

G (yα , y) uα (y)

+

dvg

2 −1

Σj,α \Byj,α (δ)

G (yα , y) uα (y)

dvg .

By (6.1.3), G (yα , y) stays bounded in Byj,α (δ) since dg (yj,α , yα ) ≥ large. Thus we can write that

2 −1 G (yα , y) uα (y) dvg Σj,α ∩Byj,α (δ)

≤C

Σj,α ∩Byj,α (δ)

2 −1

uα (y)

(6.1.10)

3 2δ

for α

(6.1.11) dvg

92

CHAPTER 6

where C, as in the following, is a positive constant independent of α. By claim 6.1.1, for any ε > 0, there exist C (ε) and R (ε) such that

2 −1 uα (y) dvg Σj,α ∩Byj,α (δ)

≤

2 −1

Σj,α ∩Byj,α (R(ε)ν1,α ) (n+2)(1−2ε) 2

uα (y)

dvg

+C (ε) ν1,α

Σj,α ∩Byj,α (δ)\Byj,α (R(ε)ν1,α )

dg (yj,α , y)

−(n+2)(1−ε)

dvg .

Here we used that Σj,α ∩ Byi,α (R (ε) νi,α ) = ∅ for α large enough as soon as 2 , and using Hölder’s inequality, i = j. This follows from (6.0.8). Taking ε < n+2 we get that

2 −1 uα (y) dvg Σj,α ∩Byj,α (δ)

1 ≤ Volg Byj,α (R (ε) ν1,α ) 2 uα 22 −1

(n+2)(1−2ε) −(n+2)(1−ε) 2 +C (ε) ν1,α dg (yj,α , y) dvg

n−2 2

M \Byj,α (R(ε)ν1,α )

(n+2)(1−2ε) 2

= O ν1,α + O ν1,α n−2 = O ν1,α2 .

n−(n+2)(1−ε)

× ν1,α

Coming back to (6.1.11), we have obtained that

n−2 2 −1 . G (yα , y) uα (y) dvg = O ν1,α2 Σj,α ∩Byj,α (δ)

(6.1.12)

On the other hand, using claim 6.1.1 and (6.1.3), we can write that, for any ε > 0,

2 −1 G (yα , y) uα (y) dvg Σj,α \Byj,α (δ)

( n+2 2 )(1−2ε)

≤ Cν1,α

( n+2 2 )(1−2ε)

= O ν1,α Taking ε <

2 n+2 ,

Σj,α \Byj,α (δ)

G (yα , y) dvg (y)

.

this implies that 2 −1

Σj,α \Byj,α (δ)

G (yα , y) uα (y)

n 2 −1 . dvg = O ν1,α

(6.1.13)

Coming back to (6.1.10), we get with (6.1.12) and (6.1.13) that for any j, with j ∈ {1, . . . , p},

n 2 −1 2 −1 . G (yα , y) uα (y) dvg = O ν1,α Σj,α

93


Noting that M = that (6.1.8) holds.

∪pj=1 Σj,α ,

and using (6.1.9), we have proved that (yα ) is such

Case 2. We assume that there exists j ∈ {1, . . . , p} such that, up to a subsequence, dg (yα , yj,α ) →R νj,α as α → +∞, where R ≥ R0 . In this case, it easily follows from point (P1) of Theorem 4.1 and the definition (6.0.9) of R0 that (yα ) is such that (6.1.8) holds. Just note that ν1,α ≥ νj,α for any j ∈ {1, . . . , p}. Case 3. We assume that Rα (yα ) → 0 as α → +∞ and that for any j with j ∈ {1, . . . , p}, up to a subsequence, dg (yj,α , yα ) → +∞ νj,α as α → +∞. Given k ∈ {1, . . . , p}, we let 1 1 = y ∈ Σk,α ∩ Byk,α (δ) s.t. dg (yα , y) ≥ Rα (yα ) , Bk,α 2 1 2 = y ∈ Σk,α ∩ Byk,α (δ) s.t. dg (yα , y) < Rα (yα ) . Bk,α 2 We write that

2 −1 G (yα , y) uα (y) dvg Σk,α

=

2 −1

1 Bk,α

G (yα , y) uα (y)

+

2 Bk,α

dvg

2 −1

G (yα , y) uα (y)

(6.1.14) dvg

+

2 −1

Σk,α \Byk,α (δ)

G (yα , y) uα (y)

dvg .

As in case 1 [see the proof of (6.1.13)] we can write that

n 2 −1 2 −1 . G (yα , y) uα (y) dvg = O ν1,α

(6.1.15)

Σk,α \Byk,α (δ)

Independently, from (6.1.3), 2−n

G (yα , y) ≤ C2 dg (yα , y)

2−n

≤ 2n−2 C2 Rα (yα )

1 in Bk,α , so that

2 −1 2−n G (yα , y) uα (y) dvg ≤ CRα (yα ) 1 Bk,α

1 Bk,α

We proved in case 1 that

Σk,α ∩Byk,α (δ)

2 −1

uα (y)

2 −1

uα (y)

n 2 −1 . dvg = O ν1,α

dvg .

94

CHAPTER 6

It follows that

n 2 −1 2−n 2 −1 . G (yα , y) uα (y) dvg = O ν1,α Rα (yα )

(6.1.16)

1 Bk,α

2 , we have that In Bk,α

dg (yk,α , y) ≥ dg (yk,α , yα ) − dg (yα , y) 1 ≥ Rα (yα ) − Rα (yα ) 2 1 ≥ Rα (yα ) . 2 Let ε > 0, to be fixed later on. Since dg (yk,α , yα ) → +∞ νk,α 2 as α → +∞, we have that dg (yk,α , y) ≥ R (ε) νk,α for all y ∈ Bk,α and α large. Noting that for α large and i = k,

Σk,α ∩ Byi,α (R (ε) νi,α ) = ∅ , 2 we can apply claim 6.1.1. It follows that, for any y ∈ Bk,α ,

( n−2 )(1−2ε) −(n−2)(1−ε) dg (yk,α , y) . uα (y) ≤ Cν1,α2 Together with (6.1.3) we then get that

2 −1 G (yα , y) uα (y) dvg 2 Bk,α

( n+2 )(1−2ε) −(n+2)(1−ε) ≤ Cν1,α2 Rα (yα ) ( n+2 2 )(1−2ε)

≤ Cν1,α

−(n+2)(1−ε)

Rα (yα )

2 Bk,α

G (yα , y) dvg

Byα (

2−n

Rα (yα ) 2

)

dg (yα , y)

dvg .

This easily leads to

( n+2 )(1−2ε) 2 −1 2−(n+2)(1−ε) Rα (yα ) G (yα , y) uα (y) dvg ≤ Cν1,α2 2 Bk,α

n 2 −1

2−n

≤ Cν1,α Rα (yα )

Rα (yα ) ν1,α

ε(n+2)−2 .

−1 Let us assume that ν1,α Rα (yα ) → +∞ as α → +∞. We let ε > 0 be such that 2 ε < n+2 . Then,

n 2 −1 2−n 2 −1 . (6.1.17) G (yα , y) uα (y) dvg = O ν1,α Rα (yα ) 2 Bk,α

Since M = ∪pk=1 Σk,α , we get, by combining (6.1.9) with (6.1.14)–(6.1.17), that −1 Rα (yα ) = O (1), we get as (yα ) is such that (6.1.8) holds. If, on the contrary, ν1,α a straightforward consequence of (6.0.11) that (yα ) is such that (6.1.8) holds also.

95


From cases 1–3, if (yα ) is a sequence of points in 1− n 2

n−2

lim sup ν1,α Rα (yα ) α→+∞

M \ ∪pj=1 Byj,α

(R0 νj,α ), then

uα (yα ) < +∞ . 2

This ends the proof of claim 6.1.2. Given j ∈ {1, . . . , p} and ε > 0, we let Rj,α (y) = min dg (yk,α , y) and k=j,...,p

( n −1)(1−2ε) (2−n)(1−ε) Φεj,α (y) = νj,α2 dg (yj,α , y) .

(6.1.18)

Then we say that (6.1.19)j holds if there exists C > 0, independent of α, such that for any α > 0 and any y ∈ M \ ∪pi=1 Byi,α (R0 νi,α ), j−1 n −1 n 2−n 2−n 2 2 −1 + νj,α Rj,α (y) . (6.1.19)j νi,α dg (yi,α , y) uα (y) ≤ C i=1

By claim 6.1.2, (6.1.19)j holds for j = 1. We prove the following: C LAIM 6.1.3. Assume that u0 ≡ 0. Let 0 < ε < 12 , j ∈ {2, . . . , p}, and assume that (6.1.19)j−1 holds. If ε is sufficientl small, then there exist R (ε) > R0 and C (ε) > 0, such that for any α > 0 and any y ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ), j−1 ( n2 −1)(1−2ε) (2−n)(1−ε) ε Φi,α (y) + νj,α Rj,α (y) uα (y) ≤ C (ε) i=1

where the are as in (6.1.18), the Rj,α ’s are as in (6.1.18), the νj,α ’s are as in (6.0.6), and the yj,α ’s are as in (6.0.6). Φεi,α ’s

Proof of claim 6.1.3. We let 0 < ε < hε∞

1 2

sufficiently small such that

= h∞ − (1 − ε) h0 > 0

(6.1.20)

where h0 is as above. Given k ∈ {1, . . . , j − 1}, we let Ωεk,α = y ∈ M s.t. Φεk,α (y) ≥ Φεi,α (y) for all i = 1, . . . , j − 1 . (6.1.21) ε It is easily checked that ∪j−1 i=1 Ωi,α = M . We also let ε (1 − ε) C12 1 ; ρ2 inf hε∞ D (ε) = min 2(1−ε) 2 M 2p C

(6.1.22)

2

where C1 , C2 , and ρ are given by (6.1.2) and (6.1.3), and where hε∞ is given by (6.1.20). As a starting point we claim that there exists R (ε) > R0 such that for any y ∈ M \ ∪pi=1 Byi,α R (ε) νi,α and any k ∈ {1, . . . , j − 1}, 2

2 −2

y ∈ Ωεk,α =⇒ min {Rj,α (y) , dg (yk,α , y)} uα (y)

≤ D (ε)

(6.1.23)

for α large. We prove this claim using (6.1.19) j−1 and (6.0.12). Let (yα ) be a sequence of points in M \ ∪pi=1 Byi,α Rνi,α , R > R0 to be chosen later on, such

96

CHAPTER 6

that yα ∈ Ωεk,α for some k ∈ {1, . . . , j − 1}. We need to prove that, up to choosing R sufficiently large, 2

2 −2

min {Rj,α (yα ) , dg (yk,α , yα )} uα (yα )

≤ D (ε)

(6.1.24)

for α large. From (6.1.19)j−1 , and since n

−1

2−n

2 νi,α dg (yi,α , yα )

we have that uα (yα ) ≤ C

j−2

(n−2)ε

= νi,α

(2−n)ε

dg (yi,α , yα )

Φεi,α ,

n 2 −1

2−n

νi,α dg (yi,α , yα )

n 2 −1

2−n

+ νj−1,α Rj−1,α (yα )

i=1

≤C

j−2

R

(2−n)ε

Φεi,α

n 2 −1

2−n

(yα ) + νj−1,α Rj−1,α (yα )

.

i=1

Noting that Rj−1,α (yα ) = min {Rj,α (yα ) , dg (yj−1,α , yα )} , we easily get that 2−n

Rj−1,α (yα )

2−n

≤ Rj,α (yα )

2−n

+ dg (yj−1,α , yα )

.

Since yα ∈ Ωεk,α , it follows that j−1 n 2−n (2−n)ε ε 2 −1 R Φi,α (yα ) + νj−1,α Rj,α (yα ) uα (yα ) ≤ C !

i=1 n

−1

2−n

2 Rj,α (yα ) ≤ C (j − 1) R(2−n)ε Φεk,α (yα ) + νj−1,α

Therefore n −1 min {Rj,α (yα ) , dg (yk,α , yα )} 2 uα (yα ) ≤ C (j − 1) R

(2−n)ε

dg (yk,α , yα )

≤ C (j − 1) R

1− n 2

+

Rj,α (yα ) νj−1,α

n 2 −1

Φεk,α

(yα ) +

1− n2

Rj,α (yα ) νj−1,α

" .

1− n2

.

Assuming that Rj,α (yα ) → +∞ νj−1,α as α → +∞, and choosing R sufficiently large, it easily follows from the above equation that (6.1.24) holds. Let us assume now that Rj,α (yα ) = O (νj−1,α ) .

(6.1.25)

We claim that in such a case Rj,α (yα ) = Rα (yα ). Then (6.1.24) follows from (6.0.12) by choosing R sufficiently large. To see that Rj,α (yα ) = Rα (yα ), we let

97


l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yα , yl,α ). Given i ∈ {1, . . . , j − 1}, from (6.1.25), we have that dg (yα , yi,α ) ≥ dg (yi,α , yl,α ) − dg (yα , yl,α ) ≥ dg (yi,α , yl,α ) − O (νj−1,α ) dg (yi,α , yl,α ) ≥ νj−1,α − O (1) νj−1,α dg (yi,α , yl,α ) νi,α ≥ νj−1,α − O (1) . νi,α νj−1,α −1 → +∞ as α → +∞ by (6.0.8) and νj−1,α ≤ νi,α by Since dg (yi,α , yl,α ) νi,α (6.0.7), we get that −1 → +∞ dg (yα , yi,α ) νj−1,α

as α → +∞. From (6.1.25) we then get that, for α sufficiently large, dg (yα , yi,α ) ≥ Rj,α (yα ) for all i ∈ {1, . . . , j − 1}. In particular, Rj,α (yα ) = Rα (yα ). As already mentioned, this proves (6.1.24) and thus (6.1.23). We now let Aj,α be given by ( n2 −1)(1−2ε) Aj,α = where

sup

sup

i=1,...,j−1 y∈Γεi,α

Φεi,α (y) ( n −1)(1−2ε) (2−n)(1−ε) νj,α2 Rj,α (y)

Γεi,α = y ∈ Ωεi,α \ ∪pk=1 Byk,α (R (ε) νk,α ) 2

2 −2

s.t. dg (yi,α , y) uα (y)

> D (ε) .

(6.1.26)

(6.1.27)

By convention, Aj,α = −∞ if the Γεi,α ’s, i = 1, . . . , j − 1, are empty. We claim that, for R (ε) sufficiently large, lim sup Aj,α α→+∞

νj,α ≤ 1. νj−1,α

(6.1.28)

We prove (6.1.28) in what follows. We assume that Aj,α = −∞. Then, up to a subsequence, there exist k ∈ {1, . . . , j − 1} and yα ∈ Γεk,α such that Φεk,α (yα ) = (Aj,α νj,α )( 2

n

−1)(1−2ε)

(2−n)(1−ε)

Rj,α (yα )

.

We proceed by contradiction. Assuming that (6.1.28) is false, we have that ( n2 −1)(1−2ε) (2−n)(1−ε) Φεk,α (yα ) ≥ νj−1,α Rj,α (yα ) .

(6.1.29)

98

CHAPTER 6

From (6.1.19)j−1 , and since yα ∈ Γεk,α , this implies that D (ε)

n−2 4

n

≤ dg (yk,α , yα ) 2

−1

uα (yα ) j−1 ! n −1 −ε(n−2) ≤ Cdg (yk,α , yα ) 2 Φεi,α (yα ) R (ε) n 2 −1

2−n

"

i=1

+νj−1,α Rj,α (yα ) ! n −1 −ε(n−2) ε Φk,α (yα ) (j − 1) R (ε) ≤ Cdg (yk,α , yα ) 2 (n−2)ε " νj−1,α ( n2 −1)(1−2ε) (2−n)(1−ε) Rj,α (yα ) +νj−1,α Rj,α (yα ) ! n −ε(n−2) ε 2 −1 (j − 1) R (ε) Φk,α (yα ) ≤ Cdg (yk,α , yα ) (n−2)ε " νj−1,α +Φεk,α (yα ) Rj,α (yα ) (n−2)ε " ! νj−1,α 1− n 1− n ( 2 2 )(1−2ε) . + R (ε) ≤ C (j − 1) R (ε) Rj,α (yα ) Taking R (ε) sufficiently large, this implies in turn that Rj,α (yα ) = O (νj−1,α ) .

(6.1.30)

Coming back to (6.1.29), we get with (6.1.30) that n −1 (1−2ε) νk,α ( 2 ) (n−2)(1−ε) (n−2)(1−ε) dg (yk,α , yα ) ≤ Rj,α (yα ) νj−1,α n ( 2 −1)(1−2ε) n2 −1 = O νk,α νj−1,α . Since k ≤ j − 1, νj−1,α ≤ νk,α . Thus, dg (yk,α , yα ) = O (νk,α ) .

(6.1.31)

We let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ). From (6.1.30) and (6.1.31) we can write that dg (yk,α , yl,α ) ≤ dg (yk,α , yα ) + dg (yl,α , yα ) = O (νk,α ) + O (νj−1,α ) = O (νk,α ) , a contradiction with (6.0.8). This proves (6.1.28). Let G be the Green’s function of ∆g + h0 . We let θj,α = νj,α max {Aj,α ; 1}

(6.1.32)

and Hα (y) =

j−1 p ( n −1)(1−2ε) ( n −1)(1−2ε) 1−ε 1−ε νi,α2 Gi,α (y) + θj,α2 Gi,α (y) i=1

i=j

99


where Gi,α (y) = G (yi,α , y). It is clear from point (P1) of Theorem 4.1, (6.0.7), (6.0.9), (6.1.3), and (6.1.32) that there exists C (ε) > 0 such that uα ≤ C (ε) Hα on ∂Byi,α R (ε) νi,α (6.1.33) for all i ∈ {1, . . . , p}. We let Lα be the linear operator given by 2 −2 Lα (u) = ∆g u + hα u − uα u.

Straightforward computations give that j−1 ! ( n −1)(1−2ε) 1−ε hα (y) − (1 − ε) h0 (y) Lα Hα (y) = νi,α2 Gi,α (y) i=1 2 −2

− uα (y)

+ ε (1 − ε)

|∇Gi,α (y) |2 " 2

Gi,α (y) p ! n ( −1)(1−2ε) 1−ε + θj,α2 hα (y) − (1 − ε) h0 (y) Gi,α (y) i=j 2 −2

− uα (y)

+ ε (1 − ε)

(6.1.34)

|∇Gi,α (y) |2 " 2

Gi,α (y)

for all y ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ). We claim that, for α sufficiently large, Lα Hα ≥ 0 in M \ ∪pi=1 Byi,α (R (ε) νi,α ) . In order to prove this claim, we let (yα ) be a sequence of points yα ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ) . We assume that yα ∈ Ωεk,α for some k ∈ {1, . . . , j − 1} and distinguish four cases. Case 1. We assume that 2 −2

uα (yα )

D (ε) ρ2

≤

where D (ε) is as in (6.1.22) and ρ is given by (6.1.2). Then 2 −2

hα (yα ) − (1 − ε) h0 (yα ) − uα (yα )

>0

for α large. Thus Lα Hα (yα ) ≥ 0 for α large. Case 2. We assume that 2 −2

uα (yα )

≥

D (ε) , ρ2

that yα ∈ Γεk,α , and that dg (yk,α , yα ) ≥ Rj,α (yα ). Since yα ∈ Γεk,α , 2

dg (yk,α , yα ) 2 2 −2 D (ε) ≤ dg (yk,α , yα ) uα (yα ) ≤ D (ε) . 2 ρ Hence, Rj,α (yα ) ≤ dg (yk,α , yα ) ≤ ρ .

(6.1.35)

100

CHAPTER 6

We let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ). By (6.1.35), using (6.1.2) and (6.1.3), we get with (6.1.34) that, for α large, |∇Gk,α (yα ) |2 ( n2 −1)(1−2ε) 1−ε Lα Hα (yα ) ≥ νk,α Gk,α (yα ) ε (1 − ε) 2 Gk,α (yα ) j−1 ( n −1)(1−2ε) 1−ε 2 −2 uα (yα ) − νi,α2 Gi,α (yα ) i=1

( n2 −1)(1−2ε)

+θj,α

1−ε

Gl,α (yα )

⎛ p ( n2 −1)(1−2ε) ⎝

−θj,α

ε (1 − ε) ⎞ 1−ε ⎠

Gi,α (yα )

|∇Gl,α (yα ) |2 2

Gl,α (yα )

2 −2

uα (yα )

i=j

C12 ( n2 −1)(1−2ε) (2−n)(1−ε)−2 ν dg (yk,α , yα ) C21−ε k,α j−1 ( n −1)(1−2ε) (2−n)(1−ε) 2 −2 2 1−ε uα (yα ) −C2 νi,α dg (yi,α , yα )

≥ ε (1 − ε)

i=1

C12 ( n2 −1)(1−2ε) (2−n)(1−ε)−2 θ dg (yα , yl,α ) C21−ε j,α ⎛ ⎞ p n −1 (1−2ε) ) ( (2−n)(1−ε) ⎠ 2 −2 ⎝ −C21−ε θj,α2 dg (yi,α , yα ) uα (yα ) +ε (1 − ε)

i=j

C12 ε −2 ≥ ε (1 − ε) 1−ε Φk,α (yα ) dg (yk,α , yα ) C2 j−1 2 −2 −C21−ε Φεi,α (yα ) uα (yα ) i=1

! C12 ( n −1)(1−2ε) (2−n)(1−ε)−2 ε (1 − ε) 1−ε +θj,α2 Rj,α (yα ) C2 " 2 2 −2 . −C21−ε (p − j + 1) Rj,α (yα ) uα (yα )

Since yα ∈ Ωεk,α , Φεi,α (yα ) ≤ Φεk,α (yα ) for all i ∈ {1, . . . , j − 1}. Therefore, −2

Lα Hα (yα ) ≥ Φεk,α (yα ) dg (yk,α , yα ) ( n2 −1)(1−2ε)

+θj,α

Aα,ε (2−n)(1−ε)−2

Rj,α (yα )

Bα,ε

where Aα,ε = ε (1 − ε)

C12 2 2 −2 1−ε (j − 1) dg (yk,α , yα ) uα (yα ) 1−ε − C2 C2

and Bα,ε = ε (1 − ε)

C12 2 2 −2 1−ε (p − j + 1) Rj,α (yα ) uα (yα ) . 1−ε − C2 C2

101


Since yα ∈ Γεk,α and using (6.1.35), we have that 2

2 −2

Rj,α (yα ) uα (yα )

2

2 −2

≤ dg (yk,α , yα ) uα (yα )

≤ D (ε) .

In particular, from the definition (6.1.22) of D (ε), Lα Hα (yα ) ≥ 0 for α large. Case 3. We assume that

D (ε) , ρ2 Γεk,α , and that dg (yk,α , yα ) ≤ Rj,α (yα ). Since yα ∈ Γεk,α , here again that yα ∈ 2 −2

uα (yα )

≥

2

dg (yk,α , yα ) 2 2 −2 D (ε) ≤ dg (yk,α , yα ) uα (yα ) ≤ D (ε) ρ2 so that dg (yk,α , yα ) ≤ ρ. From (6.1.2), (6.1.3), and (6.1.34), |∇Gk,α (yα ) |2 ( n2 −1)(1−2ε) 1−ε Gk,α (yα ) ε (1 − ε) Lα Hα (yα ) ≥ νk,α 2 Gk,α (yα ) j−1 ( n −1)(1−2ε) 1−ε 2 −2 − νi,α2 Gi,α (yα ) uα (yα ) i=1

⎛ p ( n2 −1)(1−2ε) ⎝

−θj,α

⎞ 1−ε ⎠

Gi,α (yα )

2 −2

uα (yα )

i=j

C12 ε −2 ≥ ε (1 − ε) 1−ε Φk,α (yα ) dg (yk,α , yα ) C2 j−1 2 −2 −C21−ε Φεi,α (yα ) uα (yα ) i=1 ( n −1)(1−2ε) (2−n)(1−ε) 2 −2 − (p − j + 1) C21−ε θj,α2 Rj,α (yα ) uα (yα ) for α large. Since yα ∈ Ωεk,α , Φεi,α (yα ) ≤ Φεk,α (yα ) for all i ∈ {1, . . . , j − 1}. Thus, −2

Lα Hα (yα ) ≥ Φεk,α (yα ) dg (yk,α , yα ) ( − (p − j + 1) C21−ε θj,α

n 2 −1

)(1−2ε)

Aα,ε (2−n)(1−ε)

Rj,α (yα )

(6.1.36)

2 −2

uα (yα )

where Aα,ε is as in case 2. Since Rj,α (yα ) ≥ dg (yα , yk,α ), and using the definition (6.1.32) of θj,α , ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (yα ) θj,α2 ( n −1)(1−2ε) (2−n)(1−ε) ≤ θj,α2 dg (yk,α , yα ) ≤ (νj,α max {Aj,α ; 1})( 2

n

−1)(1−2ε)

(2−n)(1−ε)

dg (yk,α , yα )

.

By (6.1.28), Aj,α νj,α ≤ νj−1,α (1 + o (1)). Since j − 1 ≤ j, we also have by (6.0.7) that νj,α ≤ νj−1,α . Hence, νj,α max {Aj,α ; 1} ≤ νj−1,α (1 + o (1)) ≤ νk,α (1 + o (1))

102

CHAPTER 6

since k ≤ j − 1. It follows that ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (yα ) ≤ Φεk,α (yα ) (1 + o (1)) . θj,α2 Coming back to (6.1.36), we then get that −2

Lα Hα (yα ) ≥ Φεk,α (yα ) dg (yk,α , yα )

Cα,ε

where Cα,ε = ε (1 − ε)

C12 2 2 −2 1−ε (1 + o (1)) dg (yk,α , yα ) uα (yα ) . 1−ε − pC2 C2

Since 2

2 −2

dg (yk,α , yα ) uα (yα )

≤ D (ε) ,

and using the definition (6.1.22) of D (ε), we have that Cα,ε ≥ 0 for α large. Hence, Lα Hα (yα ) ≥ 0 for α large. Case 4. We assume that 2 −2

uα (yα )

≥

D (ε) ρ2

and that yα ∈ Γεk,α . Since yα ∈ Γεk,α , 2

2 −2

dg (yk,α , yα ) uα (yα )

> D (ε) .

Then (6.1.23) implies that 2

2 −2


≤ D (ε) .

It follows that Rj,α (yα ) ≤ ρ and that Rj,α (yα ) ≤ dg (yk,α , yα ). We let l with l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ). From (6.1.34), j−1 ( n −1)(1−2ε) 1−ε 2 −2 2 Lα Hα (yα ) ≥ − νi,α Gi,α (yα ) uα (yα ) i=1 2 ( n −1)(1−2ε) 1−ε |∇Gl,α (yα ) | +ε (1 − ε) θj,α2 Gl,α (yα ) 2 Gl,α (yα ) ⎛ ⎞ p ( n −1)(1−2ε) ⎝ 1−ε 2 −2 −θj,α2 Gi,α (yα ) ⎠ uα (yα ) . i=j

We can then use (6.1.2) and (6.1.3) to write that j−1 2 −2 1−ε ε Φi,α (yα ) uα (yα ) Lα Hα (yα ) ≥ −C2 i=1

C12 ( n2 −1)(1−2ε) (2−n)(1−ε)−2 dg (yα , yl,α ) 1−ε θj,α C2 ⎛ ⎞ p n ( −1)(1−2ε) ⎝ (2−n)(1−ε) ⎠ 2 −2 −C21−ε θj,α2 dg (yi,α , yα ) uα (yα ) +ε (1 − ε)

i=j

103


and then that Lα Hα (yα ) ≥ −C21−ε

j−1

2 −2

Φεi,α (yα ) uα (yα )

i=1

( n2 −1)(1−2ε)

+θj,α

(2−n)(1−ε)−2

Rj,α (yα )

Bα,ε

where Bα,ε = ε (1 − ε)

C12 2 2 −2 1−ε (p − j + 1) Rj,α (yα ) uα (yα ) 1−ε − C2 C2

is as in case 2. Since yα ∈ Ωεk,α , Φεi,α (yα ) ≤ Φεk,α (yα ) for all i ∈ {1, . . . , j − 1}. Moreover, since yα ∈ Γεk,α , it follows from the definition of Aj,α that Φεk,α (yα ) ≤ (Aj,α νj,α )( 2

n

−1)(1−2ε)

(2−n)(1−ε)

Rj,α (yα )

.

By the definition (6.1.32) of θj,α , this leads to ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (yα ) . Φεk,α (yα ) ≤ θj,α2 Coming back to the above estimate on Lα Hα (yα ), we get that ( n −1)(1−2ε) (2−n)(1−ε)−2 Rj,α (yα ) Lα Hα (yα ) ≥ θj,α2 # $ C12 2 2 −2 1−ε × ε (1 − ε) 1−ε − pC2 Rj,α (yα ) uα (yα ) . C2 Since Rj,α (yα ) ≤ dg (yk,α , yα ), (6.1.23) and the definition (6.1.22) of D(ε) give that Lα Hα (yα ) ≥ 0 for α large. It clearly follows from the above four cases that, for α sufficiently large, Lα Hα ≥ 0 in M \ ∪pi=1 Byi,α (R (ε) νi,α ) .

(6.1.37)

Combining (6.1.37) with (6.1.33) and the fact that Lα uα = 0 in M , the maximum principle gives the existence of some C (ε) > 0, independent of α, such that for any y ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ), uα (y) ≤ C (ε) Hα (y). From (6.1.3) it follows that there exists C (ε) > 0, independent of α, such that j−1 ( n2 −1)(1−2ε) (2−n)(1−ε) ε (6.1.38) Φi,α (y) + θj,α Rj,α (yα ) uα (y) ≤ C (ε) i=1

for all y ∈ M \

∪pi=1

Byi,α (R (ε) νi,α ).

In order to end the proof of claim 6.1.3 (see the definition (6.1.32) of θj,α ) it remains to prove that Aj,α is bounded from above. We proceed by contradiction and assume on the contrary that Aj,α → +∞

(6.1.39)

as α → +∞. By the definition (6.1.26) of Aj,α , there exist k ∈ {1, . . . , j − 1} and yα ∈ Γεk,α such that ( n2 −1)(1−2ε) = Aj,α

Φεk,α (yα ) ( n −1)(1−2ε) (2−n)(1−ε) νj,α2 Rj,α (yα )

.

104

CHAPTER 6

Since Aj,α → +∞ as α → +∞, we have that θj,α = Aj,α νj,α . Thus ( n −1)(1−2ε) (2−n)(1−ε) Φεk,α (yα ) = θj,α2 Rj,α (yα ) .

(6.1.40)

Combining (6.1.38) and (6.1.40) we then get that j−1 ε ε uα (yα ) ≤ C (ε) Φi,α (yα ) + Φk,α (yα ) . i=1

Since yα ∈

Ωεk,α ,

Φεi,α

(yα ) ≤

Φεk,α

(yα ) for all i ∈ {1, . . . , j − 1}. Hence,

uα (yα ) ≤ C (ε) jΦεk,α (yα ) . Independently, since yα ∈ Γεk,α , n

dg (yk,α , yα ) 2

−1

uα (yα ) ≥ D (ε)

n−2 4

.

It follows that n−2

n D (ε) 4 −1 ≤ dg (yk,α , yα ) 2 Φεk,α (yα ) C (ε) j ( n2 −1)(1−2ε) νk,α ≤ dg (yk,α , yα )

so that dg (yk,α , yα ) = O (νk,α ) .

(6.1.41)

Coming back to (6.1.40), we then get that n −1 (1−2ε) θj,α ( 2 ) (n−2)(1−ε) (n−2)(1−ε) = dg (yk,α , yα ) Rj,α (yα ) νk,α n ( 2 −1)(1−2ε) n2 −1 . = O θj,α νk,α Since θj,α = Aj,α νj,α = O (νj−1,α ) by (6.1.28), and from (6.0.7), it follows that Rj,α (yα ) = O (νk,α ) .

(6.1.42)

Let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ). By (6.1.41) and (6.1.42) we can write that dg (yl,α , yk,α ) ≤ dg (yk,α , yα ) + dg (yl,α , yα ) = O (νk,α ) , a contradiction with (6.0.8). Therefore, (6.1.39) is false, and θj,α = O (νj,α ). Then (6.1.38) reads as j−1 ( n2 −1)(1−2ε) (2−n)(1−ε) ε uα (y) ≤ C (ε) Φi,α (y) + νj,α Rj,α (yα ) i=1

and this ends the proof of claim 6.1.3. The last claim of subsection 6.1.1 is as follows.

2

105


C LAIM 6.1.4. Assume that u0 ≡ 0. If (6.1.19)j−1 holds for some j = 2, . . . , p, then (6.1.19)j holds also. Proof of claim 6.1.4. Let (yα ) be a sequence in M \ ∪pi=1 Byi,α (R0 νi,α ), where R0 is as in (6.0.9). We have to prove that lim sup j−1 α→+∞

uα (yα )

0 i=1 Φi,α

n

−1

2−n

2 (yα ) + νj,α Rj,α (yα )

< +∞

(6.1.43)

.

(6.1.44)

where n

−1

2−n

2 Φ0i,α (y) = νi,α dg (yi,α , y)

2 small. Given i = 1, . . . , j − 1, we let We fix 0 < ε < n+2 ( n2 −1)(1−2ε) (2−n)(1−ε) ε ε ε ˜ Rj,α (y) Ωi,α = y ∈ Ωi,α , Φi,α (y) ≥ νj,α

(6.1.45)

where Ωεi,α is as (6.1.21), and Φεi,α is as in (6.1.18). We also let ˜ εi,α . ˜ εj,α = M \ ∪j−1 Ω Ω i=1 Let k ∈ {1, . . . , j − 1}, and R (ε) be as in claim 6.1.3. We claim that ˜ε By (R (ε) νk,α ) ⊂ Ω k,α

k,α

(6.1.46) (6.1.47)

and that for any i = k, ˜ε = ∅ Byi,α (R (ε) νi,α ) ∩ Ω k,α

(6.1.48)

provided that α is large. First we prove (6.1.47). Let (zα ) be a sequence of points in Byk,α (R (ε) νk,α ). We need to prove that Φεk,α (zα ) ≥ Φεi,α (zα ) for all i such that i ∈ {1, . . . , j − 1}, and that ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (zα ) . Φεk,α (zα ) ≥ νj,α2 Given 1 ≤ i = k ≤ j − 1, we assume by contradiction that Φεi,α (zα ) ≥ Φεk,α (zα ). Since zα ∈ Byk,α R (ε) νk,α , n ( 2 −1)(1−2ε) n2 −1 (n−2)(1−ε) = O νi,α νk,α dg (yi,α , zα ) (n−2)(1−ε) . = O max {νi,α , νk,α } Thus dg (yi,α , zα ) = O (max {νi,α , νk,α }), and dg (yk,α , yi,α ) ≤ dg (zα , yk,α ) + dg (yi,α , zα ) = O (νk,α ) + O (max {νi,α , νk,α }) = O (max {νi,α , νk,α }) . Such an equation is in contradiction with (6.0.8). Hence Φεi,α (zα ) < Φεk,α (zα ) for all i ∈ {1, . . . , j − 1}, i = k, and zα ∈ Ωεk,α . Now we assume by contradiction that ( n −1)(1−2ε) (2−n)(1−ε) Φεk,α (zα ) ≤ νj,α2 Rj,α (zα ) .

106

CHAPTER 6

Since zα ∈ Byk,α R (ε) νk,α and j ≥ k, (n−2)(1−ε)

Rj,α (zα )

n ( 2 −1)(1−2ε) n2 −1 = O νj,α νk,α (n−2)(1−ε) . = O νk,α

Hence, Rj,α (zα ) = O (νk,α ). Let l ∈ {j, . . . , p} be such that Rj,α (zα ) is given by Rj,α (zα ) = dg (yl,α , zα ). Then dg (yk,α , yl,α ) ≤ dg (yl,α , zα ) + dg (yk,α , zα ) = O (νk,α ) , a contradiction with (6.0.8). This proves (6.1.47). Now we prove (6.1.48). Let i ∈ {1, . . . , p}, i = k, and assume by contradiction that there exists a sequence ˜ ε . Then dg (zα , yi,α ) = O (νi,α ), and, (zα ) of points in Byi,α (R (ε) νi,α ) ∩ Ω k,α from (6.0.8), Rα (zα ) = dg (yi,α , zα ). It easily follows that ( n −1)(1−2ε) (2−n)(1−ε) Φεk,α (zα ) ≥ νi,α2 dg (yi,α , zα ) and then that (n−2)(1−ε)

dg (yk,α , zα )

( n2 −1)(1−2ε)

n 2 −1

= O νk,α νi,α (n−2)(1−ε) . = O max {νk,α , νi,α }

As above, this is in contradiction with (6.0.8), so that (6.1.48) is proved. Now we claim that, for any k ∈ {j, . . . , p}, ˜ε Byk,α (R (ε) νj,α ) ⊂ Ω j,α

(6.1.49)

and that, for any k ∈ {1, . . . , j − 1}, ˜ε = ∅ Byk,α (R (ε) νk,α ) ∩ Ω j,α

(6.1.50)

provided that α is large. First we prove (6.1.49). Let k ∈ {j, . . . , p} and let (zα ) be a sequence of points in Byk,α (R (ε) νj,α ). Assume by contradiction that ˜ ε . Then, from the definition (6.1.46) of Ω ˜ ε , we get that there exists i zα ∈ Ω j,α j,α with i ∈ {1, . . . , j − 1} such that ( n −1)(1−2ε) (2−n)(1−ε) Φεi,α (zα ) ≥ νj,α2 Rj,α (zα ) . Since Rj,α (zα ) ≤ dg (yk,α , zα ) = O (νj,α ) and since i ≤ j, we can write, from (6.0.7), that n −1 (1−2ε) νi,α ( 2 ) (n−2)(1−ε) (n−2)(1−ε) ≤ Rj,α (zα ) dg (yi,α , zα ) νj,α n ( −1)(1−2ε) n2 −1 = O νi,α2 νj,α (n−2)(1−ε) . = O νi,α


107

Thus dg (yi,α , zα ) = O (νi,α ), and dg (yi,α , yk,α ) ≤ dg (yi,α , zα ) + dg (zα , yk,α ) = O (νi,α ) + O (νj,α ) = O (νi,α ) . Noting that this is in contradiction with (6.0.8), we get that (6.1.49) is true. Independently, it is easily seen that (6.1.50) is a direct consequence of (6.1.47). By claim 6.1.3, (6.1.47), and (6.1.48), we get that, for any k ∈ {1, . . . , j − 1} ˜ ε \By (R (ε) νk,α ), and any y ∈ Ω k,α k,α uα (y) ≤ CΦεk,α (y)

(6.1.51)

where C > 0 is independent of α. Independently, it follows from claim 6.1.3 and ˜ ε \ ∪p By (R (ε) νi,α ), (6.1.50) that, for any y ∈ Ω i,α j,α i=j ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (y) uα (y) ≤ Cνj,α2

(6.1.52)

where C > 0 is independent of α. Now we prove (6.1.43). Let us assume first that there exists k ∈ {1, . . . , j − 1} such that dg (yk,α , yα ) →R νk,α as α → +∞ for some R ≥ R0 . Then, from (6.0.8), (6.0.9) and point (P1) of Theorem 4.1, n 2−n 2 −1 . dg (yk,α , yα ) uα (yα ) = O νk,α In particular, it easily follows that (6.1.43) is true. As a consequence, we can assume that, for any k ∈ {1, . . . , j − 1}, dg (yk,α , yα ) → +∞ νk,α

(6.1.53)

as α → +∞. Let us assume in addition that Rj,α (yα ) = O (νj,α ). We claim that Rα (yα ) = Rj,α (yα ). In order to prove this claim, we let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yα , yl,α ). Then we can write that, for any k ∈ {1, . . . , j − 1}, dg (yk,α , yα ) dg (yk,α , yl,α ) dg (yl,α , yα ) ≥ − νj,α νj,α νj,α dg (yk,α , yl,α ) νk,α ≥ − O (1) . νk,α νj,α In particular, using (6.0.7) and (6.0.8), we get that dg (yk,α , yα ) → +∞ νj,α

108

CHAPTER 6

as α → +∞. Hence, dg (yk,α , yα ) > dg (yl,α , yα ) for α large, and, as an immediate consequence, we get that Rα (yα ) = Rj,α (yα ). The above claim is proved. By (6.0.11) we can then write that n −1 n n Rj,α (yα ) 2 n−2 1− 2 −1 νj,α uα (yα ) = Rα (yα ) 2 uα (yα ) Rj,α (yα ) νj,α = O (1) , and we have proved that, if Rj,α (yα ) = O (νj,α ), then (6.1.43) holds. Conversely, we assume, in addition to (6.1.53), that Rj,α (yα ) → +∞ (6.1.54) νj,α as α → +∞. Let G be the Green’s function of ∆g + h0 , where h0 is as above; namely, h0 ∈ C 0,θ (M ) is such that ∆g + h0 is coercive and h∞ > h0 . From the Green’s representation formula and equation (Eα ),

2 −1 uα (yα ) ≤ G (yα , y) uα (y) dvg (6.1.55) M

for α large. Let k ∈ {1, . . . , j − 1}, and 1 1 ε ˜ Bk,α = y ∈ Ωk,α , dg (yα , y) ≥ dg (yk,α , yα ) , 2 1 2 ε ˜ Bk,α = y ∈ Ωk,α , dg (yα , y) < dg (yk,α , yα ) . 2 ε 1 2 ˜ = B ∪ B . Independently, (6.1.47) and (6.1.53) give that Clearly, Ω k,α

k,α

k,α

1 Byk,α (R (ε) νk,α ) ⊂ Bk,α

for α large. It follows that

2 −1 G (yα , y) uα (y) dvg ˜ε Ω k,α

=

2 −1

Byk,α (R(ε)νk,α )

G (yα , y) uα (y)

+

2 −1

1 \B Bk,α yk,α (R(ε)νk,α )

+

2 Bk,α

dvg

G (yα , y) uα (y) 2 −1

G (yα , y) uα (y)

(6.1.56) dvg

dvg .

1 , From (6.1.3), for any y ∈ Bk,α 2−n

G (yα , y) ≤ C2 dg (yα , y) Hence,

2−n

≤ C2 2n−2 dg (yk,α , yα ) 2 −1


G (yα , y) uα (y) 2−n

≤ Cdg (yk,α , yα )

2−n


.

dvg


2 −1

uα (y)

dvg

1 Volg Byk,α (R (ε) νk,α ) 2 uα 22 −1

(6.1.57)

109


and, since uα 2 ≤ Λ, we get that

2 −1 G (yα , y) uα (y) dvg = O Φ0k,α (yα ) .

(6.1.58)


Using (6.1.51) and (6.1.57), we can also write that

2 −1 G (yα , y) uα (y) dvg 1 \B Bk,α yk,α (R(ε)νk,α )

2−n

≤ Cdg (yk,α , yα ) Since ε
0 is independent of α. Since uα 2 ≤ Λ, we get that

2 −1 G (yα , y) uα (y) dvg ∪p i=j Byi,α (R(ε)νj,α )

n 2 −1

2−n

= O νj,α Rj,α (yα )

(6.1.63)

.

1 , from (6.1.3), In Bk,α 2−n

G (yα , y) ≤ C2 dg (yα , y)

2−n

≤ C2 2n−2 Rj,α (yα )

.

Independently, using (6.1.52) and the definition of Σk,α , 2 −1

uα (y)

( n +1)(1−2ε) −(n+2)(1−ε) ≤ Cνj,α2 dg (yk,α , y) .

2 , As a consequence, since ε < n+2

2 −1 G (yα , y) uα (y) dvg 1 Bk,α

2−n

≤ CRj,α (yα )

2−n

≤ CRj,α (yα )

2−n

≤ CRj,α (yα )

( n +1)(1−2ε) νj,α2 ( n +1)(1−2ε) νj,α2

1 Bk,α

−(n+2)(1−ε)

dg (yk,α , y)

M \Byk,α (R(ε)νj,α )

dvg −(n+2)(1−ε)

dg (yk,α , y)

dvg

( n +1)(1−2ε)−(n+2)(1−ε)+n νj,α2 .

It follows that

n 2 −1 2−n 2 −1 . G (yα , y) uα (y) dvg = O νj,α Rj,α (yα ) 1 Bk,α

2 In Bk,α , we have that

dg (yk,α , y) ≥ dg (yk,α , yα ) − dg (yα , y) 1 ≥ Rj,α (yα ) − Rj,α (yα ) 2 1 ≥ Rj,α (yα ) 2

(6.1.64)

112

CHAPTER 6

so that, by (6.1.52) and the definition of Σk,α , 2 −1

uα (y)

( n +1)(1−2ε) −(n+2)(1−ε) ≤ Cνj,α2 Rj,α (yα ) .

Using (6.1.3) again, we can write that

( n +1)(1−2ε) 2 −1 −(n+2)(1−ε) G (yα , y) uα (y) dvg ≤ Cνj,α2 Rj,α (yα ) 2 Bk,α

×

2 Bk,α

G (yα , y) dvg

( n +1)(1−2ε) −(n+2)(1−ε) ≤ Cνj,α2 Rj,α (yα )

2−n × dg (yα , y) dvg 2 Bk,α

( n +1)(1−2ε) −(n+2)(1−ε) ≤ Cνj,α2 Rj,α (yα )

2−n × dvg dg (yα , y) Byα ( 12 Rj,α (yα )) ( n +1)(1−2ε) 2−(n+2)(1−ε) ≤ Cνj,α2 Rj,α (yα ) 2−ε(n+2) n νj,α 2−n 2 −1 ≤ Cνj,α Rj,α (yα ) . Rj,α (yα ) 2 , we then get that By (6.1.54), and since ε < n+2

n 2 −1 2−n 2 −1 . G (yα , y) uα (y) dvg = o νj,α Rj,α (yα )

(6.1.65)

Substituting (6.1.63)–(6.1.65) into (6.1.62), it follows that

n 2 −1 2−n 2 −1 Rj,α (yα ) G (yα , y) uα (y) dvg = O νj,α .

(6.1.66)

2 Bk,α

˜ε Ω j,α

˜ ε = M , we get with (6.1.55), (6.1.61), and (6.1.66) that Noting that ∪ji=1 Ω i,α (6.1.43) is true. This proves claim 6.1.4. 2 By claim 6.1.2, (6.1.19)j is true for j = 1. By induction, thanks to claim 6.1.4, we then get that (6.1.19)j is true for all j = 1, . . . , p. In particular, (6.1.19)p is true. Noting that equations (6.1.1) and (6.1.19)p are the same, (6.1.1) is proved. 6.1.2 The case u0 ≡ 0. We assume in this subsection that u0 ≡ 0. Let R0 be as in (6.0.9). We want to prove that there exists C > 0 and a sequence (εα ), εα → 0 as α → +∞, such that for any y ∈ M \ ∪pj=1 Byj,α (R0 νj,α ), uα (y) − u0 (y) ≤ C

p

n

−1

2−n

2 νj,α dg (yj,α , y)

+ εα

(6.1.67)

j=1

where the yj,α ’s and νj,α ’s are as in (6.0.6). As in subsection 6.1.1, we split the proof of (6.1.67) into several steps that we refer to as claims. Here again we proceed by induction. As in subsection 6.1.1, we let h0 ∈ C 0,θ (M ) be such that

113


h0 < h∞ and ∆g + h0 is coercive. We let G be the Green’s function of the operator ∆g + h0 . Our first claim is the following: C LAIM 6.1.5. Assume that u0 ≡ 0, and let 0 < ε < 21 . If ε is sufficientl small, then there exist R(ε) > 0 and C (ε) > 0 such that for any α > 0 and any y in M \ ∪pj=1 Byj,α (R (ε) νj,α ), n ( 2 −1)(1−2ε) (2−n)(1−ε) (2−n)ε Rα (y) + Rα (y) uα (y) ≤ C (ε) ν1,α where Rα is as in (6.0.10), the νj,α ’s are as in (6.0.6), and the yj,α ’s are as in (6.0.6). Proof of claim 6.1.5. We fix 0 < ε < 12 and x0 ∈ S, where S is as in Theorem 4.1. We let ε (1 − ε) C12 ε (1 − ε) C12 ; (6.1.68) D (ε) = min 2(1−ε) 4p 4p C22ε C 2

where C1 and C2 are given by (6.1.2) and (6.1.3). For ρ as in (6.1.2), we let 0 < δ (ε) < ρ2 and R (ε) ≥ R0 be such that, for α sufficiently large, and any y ∈ Bx0 (δ (ε)) \ ∪pj=1 Byj,α (R (ε) νj,α ), 2 2 −2 Rα (y) uα (y) + εh0 (y) − hα (y) ≤ D (ε) and (6.1.69) 2 2 −2 + (1 − ε) h0 (y) − hα (y) ≤ D (ε) . Rα (y) uα (y) The existence of δ (ε) and R (ε) easily follows from (6.0.12). Taking δ(ε) sufficiently small, we may also assume that dg (x0 , S\ {x0 }) ≥ 2δ (ε) .

(6.1.70)

Let Lα be the linear operator given by 2 −2 Lα (u) = ∆g u + hα u − uα u.

From [9], Lα satisfies the maximum principle. We let p p 1−ε ε 1 2 Hα (y) = Gj,α (y) and Hα (y) = Gj,α (y) j=1

j=1

where Gj,α (y) = G (yj,α , y). Straightforward computations give that p 1−ε 2 −2 1 Lα Hα (y) = Gj,α (y) hα (y) − (1 − ε) h0 (y) − uα (y) j=1

+ ε (1 − ε)

|∇Gj,α (y)|2

(y) =

p

(6.1.71)

2

Gj,α (y)

and Lα Hα2

ε

Gj,α (y)

2 −2

hα (y) − εh0 (y) − uα (y)

j=1

+ ε (1 − ε)

|∇Gj,α (y)|2 2

Gj,α (y)

(6.1.72)

114

CHAPTER 6

in M \ {y1,α , . . . , yp,α }. We claim that, for α sufficiently large, Lα Hα1 ≥ 0 and Lα Hα2 ≥ 0 in Bx0 (δ (ε)) \ ∪pj=1 Byj,α (R (ε) νj,α ). In order to prove this claim, we let (yα ) be a sequence in Bx0 (δ (ε)) \ ∪pj=1 Byj,α (R (ε) νj,α ), and let k ∈ {1, . . . , p} be such that dg (yk,α , yα ) = Rα (yα ). It is clear that Rα (yα ) ≤ 2δ (ε) < ρ for α large so that, by (6.1.2), |∇Gk,α (y)|2

≥

2

Gk,α (y)

C12

C12

=

2

dg (yα , yk,α )

.

2

Rα (yα )

Then, by (6.1.69), C12

1−ε

Lα Hα1 (yα ) ≥ ε (1 − ε) Gk,α (yα ) −2

−D (ε) Rα (yα )

2

Rα (yα )

p

1−ε

Gj,α (yα )

j=1

and Lα Hα2 (yα ) ≥ ε (1 − ε) Gk,α (yα )

C12

ε

−2

−D (ε) Rα (yα )

2

Rα (yα )

p

ε

Gj,α (yα )

j=1

for α large. From (6.1.3), 1 2−n dg (yk,α , yα ) C2 1 2−n Rα (yα ) ≥ C2

Gk,α (yα ) ≥

and, for any j ∈ {1, . . . , p}, 2−n

Gj,α (yα ) ≤ C2 dg (yj,α , yα ) 2−n

≤ C2 Rα (yα ) Hence, Lα Hα1

(2−n)(1−ε)−2

C12 − pD (ε) C21−ε ε (1 − ε) 1−ε C2

(yα ) ≥ Rα (yα )

and (2−n)ε−2

Lα Hα2 (yα ) ≥ Rα (yα )

.

ε (1 − ε)

C12 − pD (ε) C2ε C2ε

.

Coming back to the definition (6.1.68) of D (ε), it follows that Lα Hα1 (yα ) ≥ 0 and Lα Hα2 (yα ) ≥ 0 for α large enough. Summarizing, we proved that Lα Hα1 ≥ 0 , Lα Hα2 ≥ 0 in Bx0 (δ (ε)) \ ∪pj=1 Byj,α (R (ε) νj,α )

(6.1.73)

115


for α large. It easily follows from point (P1) of Theorem 4.1, (6.0.7), (6.0.9), and (6.1.3), that there exists C (ε) > 0, independent of α, such that for any j with j ∈ {1, . . . , p}, ( n2 −1)(1−2ε) 1−ε Gj,α (y) on ∂Byj,α (R (ε) νj,α ) . uα (y) ≤ C (ε) ν1,α

(6.1.74)

Independently, it easily follows from (6.1.70) and point (P1) of Theorem 4.1 that uα (y) ≤ C (ε) Hα2 (y) on ∂Bx0 (δ (ε))

(6.1.75)

for some C (ε) > 0 independent of α. Combining (6.1.73)–(6.1.75), we get with the maximum principle that # n $ ( 2 −1)(1−2ε) 1 uα (y) ≤ C (ε) ν1,α Hα (y) + Hα2 (y) in Bx0 (δ (ε)) \ ∪pj=1 Byj,α (R (ε) νj,α ). Since x0 ∈ S is arbitrary, and considering (6.1.3), this proves that the estimate of claim 6.1.5 holds when we replace M by a δ(ε)-neighborhood of S. As a straightforward consequence of point (P1) of Theorem 4.1, we also have that such an estimate holds outside this neighborhood of S. This ends the proof of claim 6.1.5. 2 Our next claim is the following: C LAIM 6.1.6. Assume that u0 ≡ 0. There exist C > 0 and a sequence (εα ), εα → 0 as α → +∞, such that for any α > 0 and any y in M \ ∪pj=1 Byj,α (R0 νj,α ), n

−1

2−n

2 uα (y) − u0 (y) ≤ Cν1,α Rα (y)

+ εα

where Rα is as in (6.0.10), the νj,α ’s are as in (6.0.6), and the yj,α ’s are as in (6.0.6). Proof of claim 6.1.6. For j = 1, . . . , p, we let yj = xNj be the point of Rn given by point (P1) of Theorem 4.1. We let D0 = 2 sup

sup |z|n−2 u (z − yj )

j=1,...,p z∈Rn

(6.1.76)

where u is given by (4.1.5). It easily follows from point (P1) of Theorem 4.1 that, up to a subsequence, sup

M \∪p j=1 Byj,α (δα )

|uα − u0 | → 0 as α → +∞

(6.1.77)

where (δα ) is a sequence of positive real numbers such that δα → 0 as α → +∞. We claim that there exists C > 0 such that for any sequence (yα ) of points in M \ ∪pj=1 Byj,α (R0 νj,α ), n 2−n 2 −1 lim sup uα (yα ) − u0 (yα ) − Cν1,α Rα (yα ) ≤ 0. (6.1.78) α→+∞

In order to prove (6.1.78), we distinguish four cases. Case 1. We assume that Rα (yα ) → δ as α → +∞ for some δ > 0. Then dg (yα , S) → δ as α → +∞ and point (P1) of Theorem 4.1 gives that (6.1.78) holds with C > 0.

116

CHAPTER 6

Case 2. We assume that there exists k ∈ {1, . . . , p} such that for some R ≥ R0 , dg (yk,α , yα ) →R νk,α as α → +∞. By (6.0.9) and point (P1) of Theorem 4.1, we have in that case that n −1 lim ν 2 uα α→+∞ k,α

(yα ) = u (y0 − yk )

where, up to a subsequence, y0 = lim

α→+∞

Independently, n

−1

2 lim νk,α

n

1 exp−1 yk,α (yα ) . νk,α

−1

2−n

2 νk,α dg (yk,α , yα )

α→+∞

= |y0 |2−n

so that lim

α→+∞

uα (yα ) n 2 −1

2−n

νk,α dg (yk,α , yα )

= |y0 |n−2 u (y0 − yk ) .

Using (6.0.7), we then get that uα (yα )

lim

α→+∞

n 2 −1

2−n

ν1,α Rα (yα )

≤

1 D0 2

where D0 is as in (6.1.76). Since Rα (yα ) ≤ dg (yk,α , yα ) ≤ 2Rνk,α , we get with (6.0.7) that n

−1

2−n

2 Rα (yα ) ν1,α

→ +∞

as α → +∞. It follows that (6.1.78) holds with C ≥ D0 . Case 3. We assume that, for any j ∈ {1, . . . , p}, dg (yj,α , yα ) → +∞ νj,α as α → +∞. We also assume that Rα (yα ) = O (ν1,α ). Using (6.0.12) we then get that 1− n

|uα (yα ) − u0 (yα ) | ν1,α 2 Rα (yα ) n2 −1 n Rα (yα ) −1 = Rα (yα ) 2 |uα (yα ) − u0 (yα ) | ν1,α n −1 = O Rα (yα ) 2 |uα (yα ) − u0 (yα ) | n−2

= o (1) . It follows that (6.1.78) holds with C > 0.

117


Case 4. We assume that

Rα (yα ) → +∞ ν1,α as α → +∞. We let Gα be the Green’s function of ∆g + hα . We write with the Green’s representation formula that

2 −1 2 −1 0 uα (yα ) − u (yα ) = Gα (yα , y) uα (y) − u0 (y) dvg M

(6.1.79) + Gα (yα , y) (h∞ (y) − hα (y)) u0 (y) dvg . Rα (yα ) → 0 and

M

We let C2 > 0 (we refer to Appendix A) be such that, for any α > 0 and any x, y ∈ M , x = y, 1 2−n 2−n dg (x, y) ≤ Gα (x, y) ≤ C2 dg (x, y) . (6.1.80) C2 We clearly have that

Gα (yα , y) (h∞ (y) − hα (y)) u0 (y) dvg = o (1) (6.1.81) M

and that

M \∪p i=1 Byi,α (δα )

2 −1 2 −1 dvg = o (1) Gα (yα , y) uα (y) − u0 (y)

where δα is as in (6.1.77). Coming back to (6.1.79), using (6.1.81) and the above equation, we get that

2 −1 0 uα (yα ) − u (yα ) ≤ Gα (yα , y) uα (y) dvg + o (1) . (6.1.82) ∪p i=1 Byi,α (δα )

Given j = 1, . . . , p, we now let Σj,α = {y ∈ M s.t. Rα (y) = dg (yj,α , y)} and let 1 Bj,α 2 Bj,α

1 = y ∈ Σj,α ∩ Byj,α (δα ) s.t. dg (yα , y) ≥ Rα (yα ) , 2 1 = y ∈ Σj,α ∩ Byj,α (δα ) s.t. dg (yα , y) < Rα (yα ) . 2

1 We fix 0 < ε < n+2 small. Let j ∈ {1, . . . , p}. Up to increasing δα so that ν1,α = o(δα ), we can write that

2 −1 Gα (yα , y) uα (y) dvg Σj,α ∩Byj,α (δα )

=

2 −1

Σj,α ∩Byj,α (R(ε)ν1,α )

Gα (yα , y) uα (y)

+

1 \B Bj,α yj,α (R(ε)ν1,α )

+


dvg

2 −1


2 −1


(6.1.83) dvg dvg

118

CHAPTER 6

−1 where R (ε) is given by claim 6.1.5. Since Rα (yα ) ν1,α → +∞ as α → +∞, we can write that, for α large and any y ∈ Σj,α ∩ Byj,α (R (ε) ν1,α ),

dg (yα , y) ≥ dg (yα , yj,α ) − dg (yj,α , y) ≥ Rα (yα ) − R (ε) ν1,α 1 ≥ Rα (yα ) . 2 Since uα 2 ≤ Λ, using (6.1.80) and Hölder’s inequality, we have that, for α large,

2 −1 Gα (yα , y) uα (y) dvg Σj,α ∩Byj,α (R(ε)ν1,α )

2−n

≤ C2 2n−2 Rα (yα )

2 −1


uα (y)

dvg

1 uα 22 −1 Volg Byj,α (R (ε) ν1,α ) 2 21 2−n 2 −1 ωn−1 n n R (ε) ν1,α ≤ 2n−1 C2 Rα (yα ) Λ n 1 n n −1 2−n n−1 2 −1 ωn−1 2 2 −1 ν1,α ≤ 2 C2 Λ R (ε) 2 Rα (yα ) . n 2−n

≤ C2 2n−2 Rα (yα )

Thus, for α large,

2 −1



n 2 −1

dvg (6.1.84)

2−n

≤ D1 ν1,α Rα (yα )

,

where D1 = 2n−1 C2 Λ2

−1

ω

n−1

n

21

n

R (ε) 2

−1

.

(6.1.85)

1 Given y ∈ Bj,α , we get from (6.1.80) that 2−n

Gα (yα , y) ≤ C2 2n−2 Rα (yα )

1 and, if y ∈ Bj,α \Byj,α (R (ε) ν1,α ), we get from claim 6.1.5 that 2 −1

uα (y)

≤ 22

!

( n2 +1)(1−2ε) −(n+2)(1−ε) dg (yj,α , y) ν1,α " −(n+2)ε +dg (yj,α , y) .

−2

2 −1

C (ε)

Using (6.0.8) we indeed have that Byi,α (R (ε) νi,α ) ∩ Σj,α = ∅

119


when α is large and i = j. Thus we can write that

2 −1 Gα (yα , y) uα (y) dvg 1 \B Bj,α yj,α (R(ε)ν1,α )

≤ 22

−2

2 −1

C (ε)

−(n+2)ε


2 −1

+22 +n−4 C (ε)

×

2−n

C2 Rα (yα )

1 n+2 ,

dg (yj,α , y)

dvg .

and since δα → 0 as α → +∞, it is easily checked with −(n+2)ε


Gα (yα , y) dg (yj,α , y)

≤

dvg

( n2 +1)(1−2ε) ν1,α

−(n+2)(1−ε)


Noting that ε < (6.1.80) that


Gα (yα , y)

n n−1

n−1 n dvg

= O δα1−(n+2)ε = o (1) . Similarly, it is easily checked that


n1 −n(n+2)ε

Byj,α (δα )

M

dvg

dg (yj,α , y)

−(n+2)(1−ε)

dg (yj,α , y)

dvg

dvg

2ωn−1 n−(n+2)(1−ε) (R (ε) ν1,α ) (n + 2) (1 − ε) − n for α large. Therefore,

2 −1 Gα (yα , y) uα (y) dvg ≤

1 \B Bj,α yj,α (R(ε)ν1,α ) n 2 −1

(6.1.86) 2−n

≤ D2 ν1,α Rα (yα ) where D2 = 22 +n−3

+ o (1)

ωn−1 2 −1 n−(n+2)(1−ε) C (ε) C2 R (ε) . (6.1.87) (n + 2) (1 − ε) − n 2 , we have that In Bj,α dg (yj,α , y) ≥ dg (yj,α , yα ) − dg (y, yα ) 1 ≥ Rα (yα ) − Rα (yα ) 2 1 ≥ Rα (yα ) 2 so that, applying claim 6.1.5, we get that ! 2 −1 2 −1 −(n+2)ε dg (yj,α , y) ≤ 22 −2 C (ε) uα (y) " ( n2 +1)(1−2ε) −(n+2)(1−ε) . +2(n+2)(1−ε) ν1,α Rα (yα )

120

CHAPTER 6

Once again, we used the fact that Byi,α (R (ε) νi,α ) ∩ Σj,α = ∅ for α large and any i = j. Using (6.1.80) we then get that


≤ 22

−2+(n+2)(1−ε)

×


2 −2

+2 ≤ 22

2 −1

C (ε)

2 −1

2 −1

C (ε)

Byα (

−2

2 −1

dg (yα , y)

)

Gα (yα , y) n−1 dvg M

= O ν1,α

−1

−n(n+2)ε

dvg

2−(n+2)(1−ε)

Rα (yα ) 2−n

2 Rα (yα ) = O ν1,α

n−1 n

n1

dg (yj,α , y)

( n2 +1)(1−2ε)

dvg

dvg n

C (ε)

Byj,α (δα )

n


( n2 +1)(1−2ε) −(n+2)(1−ε) ν1,α Rα (yα ) 2−n

Rα (yα ) 2

×

−(n+2)ε


+22

Gα (yα , y) dvg

C (ε)

−2+(n+2)(1−ε)

×C2

( n2 +1)(1−2ε) −(n+2)(1−ε) ν1,α Rα (yα )

ν1,α Rα (yα )

+ o (1)

2−(n+2)ε

+ o (1) .

−1 → +∞ as α → +∞, it follows that Since Rα (yα ) ν1,α


n 2 −1

2−n

= o ν1,α Rα (yα )

(6.1.88) + o (1) .

Substituting (6.1.84), (6.1.86), and (6.1.88) into (6.1.83), we then get that, for any j ∈ {1, . . . , p},

2 −1 Gα (yα , y) uα (y) dvg Σj,α ∩Byj,α (δα ) (6.1.89) n

−1

2−n

2 ≤ (D1 + D2 + o (1)) ν1,α Rα (yα )

+ o (1)

where D1 and D2 are given by (6.1.85) and (6.1.87). Since M = ∪pj=1 Σj,α and Byi,α (δα ) ∩ Σj,α ⊂ Byj,α (δα ) ∩ Σj,α , we get with (6.1.82) and (6.1.89) that n

−1

2−n

2 uα (yα ) − u0 (yα ) ≤ p (D1 + D2 + o (1)) ν1,α Rα (yα )

+ o (1) .


121

It follows that (6.1.78) holds with C > p (D1 + D2 ), where D1 is as in (6.1.85), and D2 is as in (6.1.87). From cases 1–4, for any sequence (yα ) in M \ ∪pj=1 Byj,α (R0 νj,α ), n 2−n 2 −1 ≤0 Rα (yα ) lim sup uα (yα ) − u0 (yα ) − Cν1,α α→+∞

if C > 0 is chosen such that C ≥ D0 + p (D1 + D2 ), where D0 is as in (6.1.76), D1 is as in (6.1.85), and D2 is as in (6.1.87). Clearly, this ends the proof of claim 6.1.6. 2 For j ∈ {1, . . . , p}, we say that (6.1.90)j holds if there exist C > 0, independent of α, and a sequence (εα ), εα → 0 as α → +∞, such that for any α > 0 and any y ∈ M \ ∪pi=1 Byi,α (R0 νi,α ), uα (y) − u0 (y) j−1 n −1 n −1 2−n 2−n 2 2 ≤C νi,α dg (yi,α , y) + νj,α Rj,α (y) + εα

(6.1.90)j

i=1

where Rj,α is defined by (6.1.18). By claim 6.1.6, (6.1.90)j holds for j = 1. The subject of claims 6.1.7 and 6.1.8 below is to prove that, for any j ∈ {2, . . . , p}, (6.1.90)j−1 holds ⇒ (6.1.90)j holds.

(6.1.91)

First we set up notation. We let j ∈ {2, . . . , p} and we assume that (6.1.90)j−1 1 holds. We fix 0 < ε < n+2 such that h∞ > (1 − ε) h0 . For any x ∈ S, S as in Theorem 4.1, we let Aj (x) = {i ∈ {1, . . . , j − 1} s.t. yi,α → x as α → +∞} , Bj (x) = {i ∈ {j, . . . , p} s.t. yi,α → x as α → +∞} .

(6.1.92)

For k = 1, . . . , j − 1, we let (6.1.93) Ωεk,α = y ∈ M s.t. Φεk,α (y) ≥ Φεi,α (y) for all i ∈ {1, . . . , j − 1} where ( n −1)(1−2ε) (2−n)(1−ε) dg (yi,α , y) . Φεi,α (y) = νi,α2 We also let

D (ε) = min

ε (1 − ε) C12 ε (1 − ε) C12 ; 2(1−ε) 4p 4p C22ε C

(6.1.94)

(6.1.95)

2

where C1 and C2 are given by (6.1.2) and (6.1.3). Assuming (6.1.90)j−1 , we claim now that there exist R (ε) ≥ R0 and δ (ε) > 0 such that the following assertions hold: (A1) for any k ∈ {1, . . . , j − 1}, and any y ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ), if y ∈ Ωεk,α , then 1 n D (ε) 2 −2 0 2 −1 (6.1.96) uα (y) − u (y) < min {Rj,α (y) , dg (yk,α , y)} 100

122

CHAPTER 6

when α is sufficiently large; (A2) for any x ∈ S, 2

2 −2

δ (ε) u0 (y)

≤

D (ε) 100

(6.1.97)

when y ∈ Bx (δ (ε)); (A3) for any x ∈ S, and any y ∈ Bx (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ), ! " 2 2 −2 − hα (y) + εh0 (y) ≤ D (ε) (6.1.98) Rα (y) uα (y) when α is sufficiently large. The proof of (A1) is based on (6.1.90)j−1 and (6.0.12). Let (yα ) be a sequence of points in M \ ∪pi=1 Byi,α (Rνi,α ), R > R0 to be fixed later on, such that yα ∈ Ωεk,α for some k ∈ {1, . . . , j − 1}. In order to prove (A1) we have to prove that, choosing R sufficiently large, n −1 uα (yα ) − u0 (yα ) min {Rj,α (yα ) , dg (yk,α , yα )} 2 1 (6.1.99) D (ε) 2 −2 < 100 for α large. From (6.1.90)j−1 , uα (yα ) − u0 (yα ) j−2 n −1 n 2−n 2−n 2 2 −1 ≤C νi,α dg (yi,α , yα ) + νj−1,α Rj−1,α (yα ) + o (1) . i=1

Since Rj−1,α (yα ) = min {Rj,α (yα ) ; dg (yj−1,α , yα )} , it is clear that 2−n

Rj−1,α (yα )

2−n

≤ Rj,α (yα )

2−n

+ dg (yj−1,α , yα )

.

Since yα ∈ Ωεk,α , dg (yi,α , yα ) ≥ Rνi,α , and n

−1

2−n

2 νi,α dg (yi,α , yα )

(n−2)ε

= Φεi,α (yα ) νi,α

(2−n)ε

dg (yi,α , yα )

,

it follows that uα (yα ) − u0 (yα ) j−1 n −1 n 2−n 2−n 2 2 −1 ≤C + νj−1,α Rj,α (yα ) + o (1) νi,α dg (yi,α , yα )

i=1

≤ C R(2−n)ε

j−1 i=1

n 2 −1

2−n

Φεi,α (yα ) + νj−1,α Rj,α (yα )

+ o (1)

" ! n 2−n 2 −1 Rj,α (yα ) + o (1) . ≤ C (j − 1) R(2−n)ε Φεk,α (yα ) + νj−1,α

123


Therefore,

uα (yα ) − u0 (yα ) n2 −1 n νj−1,α (2−n)ε ε 2 −1 dg (yk,α , yα ) Φk,α (yα ) + + o (1) ≤ C (j − 1) R Rj,α (yα ) n2 −1 n ν j−1,α + o (1) . ≤ C (j − 1) R1− 2 + Rj,α (yα ) n

min {Rj,α (yα ) , dg (yk,α , yα )} 2

−1

−1 → +∞ as α → +∞, then (6.1.99) holds with R sufficiently If Rj,α (yα ) νj−1,α large. Without loss of generality, we may thus assume that

Rj,α (yα ) = O (νj−1,α ) .

(6.1.100)

We claim that this implies that Rj,α (yα ) = Rα (yα ). In order to prove this claim, we let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yα , yl,α ). For any i ∈ {1, . . . , j − 1}, from (6.0.7) and (6.1.100), dg (yα , yi,α ) dg (yl,α , yi,α ) dg (yα , yl,α ) ≥ − νj−1,α νj−1,α νj−1,α dg (yl,α , yi,α ) νi,α Rj,α (yα ) ≥ − νi,α νj−1,α νj−1,α dg (yl,α , yi,α ) ≥ − O (1) . νi,α By (6.0.8), we then get that dg (yα , yi,α ) → +∞ νj−1,α as α → +∞, and it follows from (6.1.100) that, for any i ∈ {1, . . . , j − 1}, we have that dg (yi,α , yα ) > Rj,α (yα ) when α is large. This proves that Rj,α (yα ) is such that Rj,α (yα ) = Rα (yα ) for α large. Using (6.0.12), we then get that n

Rj,α (yα ) 2

−1

|uα (yα ) − u0 (yα ) | ≤ εR + o (1)

where εR → 0 as R → +∞. In particular, choosing R sufficiently large, (6.1.99) holds. This proves (A1). An elementary remark is that one can choose δ (ε) > 0 small such that (6.1.97) holds. This proves (A2). Another simple remark is that (6.1.98) follows from (6.0.12). This proves (A3). We now let Aj,α be defined by ( n2 −1)(1−2ε) Aj,α =

sup

sup

i=1,...,j−1 y∈Γεi,α

Φεi,α (y) ( n −1)(1−2ε) (2−n)(1−ε) νj,α2 Rj,α (y)

(6.1.101)

where

Γεi,α = y ∈ Ωεi,α \ ∪pk=1 Byk,α (R (ε) νk,α ) s.t. 1 n D (ε) 2 −2 0 2 −1 . uα (y) − u (y) ≥ dg (yi,α , y) 100

(6.1.102)

124

CHAPTER 6

By convention, Aj,α = −∞ if the Γεi,α ’s, i = 1, . . . , j − 1, are empty. We claim that for R (ε) sufficiently large, νj,α ≤ 1. (6.1.103) lim sup Aj,α ν α→+∞ j−1,α We prove (6.1.103) in what follows. We assume that Aj,α = −∞. Then, up to a subsequence, there exist k ∈ {1, . . . , j − 1} and yα ∈ Γεk,α such that Φεk,α (yα ) = (Aj,α νj,α )( 2

n

−1)(1−2ε)

(2−n)(1−ε)

Rj,α (yα )

.

We proceed by contradiction. Assuming that (6.1.103) is false, we have that, up to a subsequence, ( n2 −1)(1−2ε) (2−n)(1−ε) Rj,α (yα ) . Φεk,α (yα ) ≥ νj−1,α

(6.1.104)

From (6.1.90)j−1 , and since yα ∈ Γεk,α , this implies that 1 n D (ε) 2 −2 −1 ≤ dg (yk,α , yα ) 2 uα (yα ) − u0 (yα ) 100 j−1 n −1 n 2−n 2 2 −1 ≤ Cdg (yk,α , yα ) νi,α dg (yi,α , yα ) i=1

n 2 −1

2−n

+νj−1,α Rj,α (yα )

+ o (1)


n 2 −1

R (ε)

(2−n)ε

n

−1

2−n

Φεi,α (yα )

i=1

2 Rj,α (yα ) +νj−1,α

j−1

+ o (1)


n 2 −1

( n2 −1)(1−2ε)

+νj−1,α

(2−n)ε

(j − 1) R (ε)

(2−n)(1−ε)

Rj,α (yα )

≤ Cdg (yk,α , yα ) +Φεk,α

(yα )

≤ CR (ε)

1− n 2

n 2 −1

νj−1,α Rj,α (yα )

(2−n)ε

(j − 1) R (ε)


Φεk,α (yα ) (n−2)ε + o (1)

Φεk,α (yα )

(n−2)ε

(j − 1) + R (ε)

+ o (1) (n−2)ε


(n−2)ε + o (1) .

Taking R (ε) sufficiently large, this implies in turn that Rj,α (yα ) = O (νj−1,α ) .

(6.1.105)

125


Coming back to (6.1.104), we get with (6.1.105) that (n−2)(1−ε)

dg (yk,α , yα )

n −1 (1−2ε) νk,α ( 2 ) (n−2)(1−ε) ≤ Rj,α (yα ) νj−1,α n ( 2 −1)(1−2ε) n2 −1 = O νk,α νj−1,α .

Since k ≤ j − 1, νj−1,α ≤ νk,α . Thus, dg (yk,α , yα ) = O (νk,α ) .

(6.1.106)

We let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ). Using (6.1.105) and (6.1.106) we can write that dg (yk,α , yl,α ) ≤ dg (yk,α , yα ) + dg (yl,α , yα ) = O (νk,α ) + O (νj−1,α ) = O (νk,α ) , a contradiction with (6.0.8). This proves (6.1.103). We now let θj,α = νj,α max {Aj,α ; 1} .

(6.1.107)

Claim 6.1.7 can then be stated as follows: 1 C LAIM 6.1.7. Assume that u0 ≡ 0. We let 0 < ε < n+2 sufficientl small such that h∞ > (1 − ε) h0 , and assume that (6.1.90)j−1 holds for some j ∈ {2, . . . , p}. Let R (ε) > R0 be such that (6.1.96)–(6.1.98) and (6.1.103) hold. Then there exists C (ε) > 0 such that for any y ∈ M \ ∪pi=1 Byi,α (R (ε) νi,α ),

uα (y) ≤ C (ε)

j−1

Φεi,α

( n2 −1)(1−2ε)

(y) + θj,α

(2−n)(1−ε)

Rj,α (y)

(2−n)ε

+ Rα (y)

i=1

where Rα is as in (6.0.10), Rj,α is as in (6.1.18), the νj,α ’s are as in (6.0.6), the yj,α ’s are as in (6.0.6), Φεi,α is as in (6.1.94), and θj,α is as in (6.1.107). Proof of claim 6.1.7. Let Lα be the linear operator given by 2 −2 Lα (u) = ∆g u + hα u − uα u,

and Hα (y) =

j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (y) i=1

( n −1)(1−2ε) 1−ε +θj,α2 Gi,α (y) p

i=j

+

p i=1

ε

Gi,α (y)

126

CHAPTER 6

where Gi,α (y) = G (yi,α , y), and G is the Green’s function of ∆g + h0 . We fix x0 ∈ S, where S is as in Theorem 4.1. We let ρ 0 < δ (ε) < , 2 1 δ (ε) < dg (x0 , S\ {x0 }) 2 be such that (6.1.97) and (6.1.98) hold, where ρ is as in (6.1.2), (6.1.3). Since h∞ > (1 − ε) h0 , direct computations give that, for any y ∈ M \ {y1,α , . . . , yp,α } and for α large, j−1 ( n −1)(1−2ε) 1−ε Lα Hα (y) ≥ νi,α2 Gi,α (y) i=1

|∇Gi,α (y) |2

× ε (1 − ε)

2

Gi,α (y)

2 −2

− uα (y)

( n −1)(1−2ε) 1−ε Gi,α (y) + θj,α2 p

i=j

|∇Gi,α (y) |2

× ε (1 − ε) +

p

2

Gi,α (y)

ε

Gi,α (y)

(6.1.108)

2 −2

− uα (y)

2 −2

hα (y) − εh0 (y) − uα (y)

i=1

+ ε (1 − ε)

|∇Gi,α (y) |2

. 2 Gi,α (y) We claim that, for α sufficiently large, Lα Hα ≥ 0 in Bx0 (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ) . Let (yα ) be a sequence of points in Bx0 (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ). We let also l ∈ {1, . . . , p} be such that Rα (yα ) = dg (yl,α , yα ). Since δ (ε) < ρ2 , it is clear that Rα (yα ) ≤ ρ for α large. Thus we can apply (6.1.2) and (6.1.3). We get that p |∇Gi,α (yα ) |2 ε 2 −2 Gi,α (yα ) hα (yα ) − εh0 (yα ) − uα (yα ) + ε (1 − ε) 2 Gi,α (yα ) i=1 ≥

1 C12 (2−n)ε dg (yl,α , yα ) ε (1 − ε) 2 ε C2 dg (yl,α , yα ) p " ! ε 2 −2 − Gi,α (yα ) uα (yα ) − hα (yα ) + εh0 (yα ) . i=1

From (6.1.98),

p

ε

Gi,α (yα )

!

2 −2

uα (yα )

"

− hα (yα ) + εh0 (yα )

i=1 −2

≤ D (ε) Rα (yα )

p i=1

ε

Gi,α (yα )

127


so that, by (6.1.3), p

ε

Gi,α (yα )

!

2 −2

uα (yα )

"

− hα (yα ) + εh0 (yα )

i=1 −2

≤ D (ε) Rα (yα ) ≤

D (ε) pC2ε Rα

C2ε

p

(2−n)ε

dg (yα , yi,α )

i=1 −2+(2−n)ε

(yα )

.

Therefore, coming back to the definition (6.1.95) of D (ε), p ε 2 −2 Gi,α (yα ) hα (yα ) − εh0 (yα ) − uα (yα ) i=1

|∇Gi,α (yα ) |2

+ ε (1 − ε) ≥

(6.1.109)

2

Gi,α (yα )

3pD (ε) C2ε Rα

−2+(2−n)ε

(yα )

for α large. Now we estimate the other terms in the right-hand side of (6.1.108). Up to a subsequence, we may assume that yα ∈ Ωεk,α for some k ∈ {1, . . . , j − 1}. We distinguish four cases. Case 1. We assume that min {Rj,α (yα ) ; dg (yk,α , yα )} → δ as α → +∞ for some δ > 0. Then, by (6.1.96), uα (yα ) = O (1). From (6.1.3), and since yα ∈ Ωεk,α , this gives in turn that j−1 j−1 ( n2 −1)(1−2ε) 2 −2 1−ε ε uα (yα ) νi,α Gi,α (yα ) =O Φi,α (yα ) i=1

=O

i=1 Φεk,α

(yα ) .

In particular, we can write that 2 −2

uα (yα )

j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (yα ) = o (1)

(6.1.110)

i=1

since dg (yk,α , yα ) ≥ that

δ 2

for α large, and νk,α → 0 as α → +∞. Similarly, one has ( n −1)(1−2ε) 1−ε θj,α2 Gi,α (yα ) p

2 −2

uα (yα )

i=j

n ( −1)(1−2ε) (2−n)(1−ε) = O θj,α2 Rj,α (yα ) n ( −1)(1−2ε) . = O θj,α2 By (6.1.103) [see also (6.1.107)] θj,α → 0 as α → +∞. Hence, ( n −1)(1−2ε) 1−ε θj,α2 Gi,α (yα ) = o (1) . p

2 −2

uα (yα )

i=j

(6.1.111)

128

CHAPTER 6

Coming back to (6.1.108), we get with (6.1.109)–(6.1.111) that for α sufficiently large, Lα Hα (yα ) ≥ 0. Case 2. We assume that yα ∈ Γεk,α , that dg (yk,α , yα ) ≥ Rj,α (yα ), and that Rj,α (yα ) → 0 as α → +∞. With these assumptions, from (6.1.2), (6.1.3), and (6.1.96), it is not difficult to check that, for α large, p ( n −1)(1−2ε) 1−ε θj,α2 Gi,α (yα )

i=j

× ε (1 − ε)

|∇Gi,α (yα ) |2

(6.1.112)

2 −2

− uα (yα )

2

Gi,α (yα )

( n −1)(1−2ε) −2+(2−n)(1−ε) ≥ (3p + j − 1) D (ε) C21−ε θj,α2 Rj,α (yα ) . Since yα ∈ Γεk,α , 1 n D (ε) 2 −2 0 2 −1 uα (yα ) − u (yα ) ≤ dg (yk,α , yα ) 100 so that 2 2 −2 dg (yk,α , yα ) uα (yα ) D (ε) (6.1.113) 2 0 2 −2 2 −3 + dg (yk,α , yα ) u (yα ) . ≤ max 1; 2 100 If k ∈ Aj (x0 ), Aj (x0 ) as in (6.1.92), then, since δ (ε) < 12 dg (x0 , S\ {x0 }), we get that dg (yk,α , yα ) → δ as α → +∞ for some δ > 0. In particular, we get with (6.1.113) that uα (yα ) = O (1). As in the first case, we can then prove that j−1 ( n −1)(1−2ε) 2 −2 1−ε νi,α2 Gi,α (yα ) = o (1) (6.1.114) uα (yα ) i=1

and (6.1.114) holds under the assumption that k ∈ Aj (x0 ). Let us now assume that k ∈ Aj (x0 ). Then dg (yk,α , yα ) ≤ 2δ (ε) for α large so that, combining (6.1.97) and (6.1.113), 2

2 −2

≤ D (ε) dg (yk,α , yα ) uα (yα ) for α large. From (6.1.2) and (6.1.3), this implies that j−1 |∇Gi,α (yα ) |2 ( n2 −1)(1−2ε) 1−ε 2 −2 νi,α Gi,α (yα ) − uα (yα ) ε (1 − ε) 2 Gi,α (yα ) i=1 ( n2 −1)(1−2ε) ≥ νk,α

1

dg C21−ε

(2−n)(1−ε)

(yk,α , yα ) −2

−C21−ε D (ε) dg (yk,α , yα )

ε (1 − ε)

i=1

−2

2

dg (yk,α , yα )

j−1 ( n −1)(1−2ε) (2−n)(1−ε) νi,α2 dg (yi,α , yα )

C12 −2 ≥ ε (1 − ε) 1−ε dg (yk,α , yα ) Φεk,α (yα ) C2 −C21−ε D (ε) dg (yk,α , yα )

C12

j−1 i=1

Φεi,α (yα ) .

129


Since yα ∈ Ωεk,α , we then get that, when k ∈ Aj (x0 ), j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (yα ) i=1

× ε (1 − ε)

|∇Gi,α (yα ) |2 2

Gi,α (yα )

2 −2

− uα (yα )

(6.1.115)

−2

≥ (4p − (j − 1)) C21−ε D (ε) dg (yk,α , yα ) Φεk,α (yα ) . Coming back to (6.1.108), it follows from (6.1.109), (6.1.112), (6.1.114), and (6.1.115), that for α sufficiently large Lα Hα (yα ) ≥ 0. Case 3. We assume that yα ∈ Γεk,α , that dg (yk,α , yα ) ≤ Rj,α (yα ), and that dg (yk,α , yα ) → 0 as α → +∞. Using (6.1.2), (6.1.3), and (6.1.96), we can then write that j−1 |∇Gi,α (yα ) |2 ( n2 −1)(1−2ε) 1−ε 2 −2 νi,α Gi,α (yα ) − uα (yα ) ε (1 − ε) 2 Gi,α (yα ) i=1 ≥ ε (1 − ε)

C12 −2 dg (yk,α , yα ) Φεk,α (yα ) C21−ε −2

−D (ε) C21−ε dg (yk,α , yα )

j−1

Φεi,α (yα ) .

i=1

Since yα ∈ Ωεk,α , this implies that j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (yα ) i=1

× ε (1 − ε)

|∇Gi,α (yα ) |2 2

Gi,α (yα )

2 −2

− uα (yα )

(6.1.116)

−2

≥ (4p − (j − 1)) D (ε) C21−ε dg (yk,α , yα ) Φεk,α (yα ) . Similarly, since Rj,α (yα ) ≥ dg (yk,α , yα ), we can write using (6.1.3) and (6.1.96) that p ( n −1)(1−2ε) 2 −2 1−ε θj,α2 uα (yα ) Gi,α (yα ) i=j

( n −1)(1−2ε) −2 (2−n)(1−ε) ≤ (p − j + 1) D (ε) C21−ε θj,α2 dg (yk,α , yα ) Rj,α (yα ) ( n −1)(1−2ε) −2+(2−n)(1−ε) ≤ (p − j + 1) D (ε) C21−ε θj,α2 dg (yk,α , yα ) . By (6.0.7) and (6.1.103), since k ≤ j − 1 ≤ j, θj,α ≤ νk,α (1 + o (1)) . Therefore, p ( n −1)(1−2ε) 2 −2 1−ε uα (yα ) Gi,α (yα ) θj,α2 (6.1.117) i=j −2

≤ (p − j + 1) D (ε) C21−ε dg (yk,α , yα )

Φεk,α (yα ) (1 + o (1)) .

130

CHAPTER 6

Coming back to (6.1.108), we get with (6.1.109), (6.1.116), and (6.1.117) that, for α sufficiently large, Lα Hα (yα ) ≥ 0. Case 4. We assume that yα ∈ Γεk,α and that min {Rj,α (yα ) ; dg (yk,α , yα )} → 0 as α → +∞. Since yα ∈ Γεk,α , n

dg (yk,α , yα ) 2

uα (yα ) − u0 (yα ) ≥

−1

D (ε) 100

21−2 .

Using (6.1.96) we then get that Rj,α (yα ) ≤ dg (yk,α , yα ) and, since Rj,α (yα ) → 0, we have that 2

2 −2


≤

D (ε) + o (1) . 100

From (6.1.2) and (6.1.3), this implies that p |∇Gi,α (yα ) |2 ( n2 −1)(1−2ε) 1−ε 2 −2 θj,α ε (1 − ε) Gi,α (yα ) − uα (yα ) 2 Gi,α (yα ) i=j ( n −1)(1−2ε) (2−n)(1−ε)−2 ≥ (3p + j − 1) C21−ε D (ε) θj,α2 Rj,α (yα ) (6.1.118) for α large. On the other hand, still using (6.1.3), and since yα ∈ 2 −2

uα (yα )

Ωεk,α ,

j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (yα ) i=1 −2

≤ D (ε) Rj,α (yα )

C21−ε

j−1

Φεi,α (yα )

i=1 −2 1−ε ε C2 Φk,α

≤ (j − 1) D (ε) Rj,α (yα )

(yα ) .

We assumed that yα belongs to Γεk,α . By the definition (6.1.101) of Aj,α and the definition (6.1.107) of θj,α , this implies that Φεk,α (yα ) ≤ (Aj,α νj,α )( 2

n

−1)(1−2ε)

(2−n)(1−ε)

Rj,α (yα )

( n −1)(1−2ε) (2−n)(1−ε) ≤ θj,α2 Rj,α (yα ) . Hence, for α large, 2 −2

uα (yα )

j−1 ( n −1)(1−2ε) 1−ε νi,α2 Gi,α (yα )

(6.1.119)

i=1

( ≤ (j − 1) D (ε) C21−ε θj,α

n 2 −1

)(1−2ε)

−2+(2−n)(1−ε)

Rj,α (yα )

.

Coming back to (6.1.108), we get with (6.1.109), (6.1.118), and (6.1.119) that, for α sufficiently large, Lα Hα (yα ) ≥ 0.

131


It clearly follows from cases 1–4 that, for α sufficiently large, Lα Hα ≥ 0 in Bx0 (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ) .

(6.1.120)

Independently, using point (P1) of Theorem 4.1 and (6.1.3), we easily get the existence of some C (ε) > 0 such that uα ≤ C (ε) Hα on ∂ Bx0 (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ) . (6.1.121) Since Lα uα = 0, noting that Lα satisfies the maximum principle (see [9]), it follows from (6.1.120) and (6.1.121) that uα ≤ C (ε) Hα in Bx0 (δ (ε)) \ ∪pi=1 Byi,α (R (ε) νi,α ) . From (6.1.3), this proves that the estimate of claim 6.1.7 holds when we replace M by a δ(ε)-neighborhood of S. As a straightforward consequence of point (P1) of Theorem 4.1, we also have that such an estimate holds outside this neighborhood of S. This ends the proof of claim 6.1.7. 2 The last claim of subsection 6.1.2 is as follows. C LAIM 6.1.8. Assume that u0 ≡ 0. If (6.1.90)j−1 holds for some j = 2, . . . , p; then (6.1.90)j holds also. Proof of claim 6.1.8. The first part of the proof is devoted to proving the following estimate: there exists C > 0 such that for any sequence (yα ) of points in M \ ∪pi=1 Byi,α (R0 νi,α ), uα (yα ) − u0 (yα ) j−1 n 2−n 0 2 −1 ≤C Φi,α (yα ) + θj,α Rj,α (yα ) + o (1)

(6.1.122)

i=1

where n

−1

2−n

2 dg (yi,α , y) Φ0i,α (y) = νi,α

(6.1.123)

and θj,α is as in (6.1.107). The definition of θj,α depends on the choice of some 1 . We fix ε such that claim 6.1.7 applies, and consider that Aj,α and 0 < ε < n+2 θj,α , as defined in the proof of claim 6.1.7, are with respect to such an ε. Given i = 1, . . . , j − 1, we let n ˜ ε = y ∈ Ωε s.t. Φε (y) ≥ θ( 2 −1)(1−2ε) Rj,α (y)(2−n)(1−ε) (6.1.124) Ω i,α i,α i,α j,α where Ωεi,α is as in (6.1.93), Φεi,α is as in (6.1.94), and θj,α is as in (6.1.107). We also let ˜ εi,α . ˜ εj,α = M \ ∪j−1 Ω Ω i=1

(6.1.125)

By point (P1) of Theorem 4.1, up to a subsequence, there exists δα > 0, δα → 0 as α → +∞, such that sup

M \∪p i=1 Byi,α (δα )

|uα − u0 | → 0

(6.1.126)

132

CHAPTER 6

as α → +∞. Let k ∈ {1, . . . , j − 1}, and let R (ε) be such that (6.1.96)–(6.1.98) and (6.1.103) hold. We claim that, for α large, ˜ε (6.1.127) By (R (ε) νk,α ) ⊂ Ω k,α

k,α

and ˜ε = ∅ Byi,α (R (ε) νi,α ) ∩ Ω k,α

(6.1.128)

for all i = k. First we prove (6.1.127). Let (zα ) be a sequence of points in Byk,α (R (ε) νk,α ). We need to prove that, for α large, Φεk,α (zα ) ≥ Φεi,α (zα ) for all i ∈ {1, . . . , j − 1} and ( n −1)(1−2ε) (2−n)(1−ε) Rj,α (zα ) . Φεk,α (zα ) ≥ θj,α2 We assume by contradiction that Φεi,α (zα ) ≥ Φεk,α (zα ) for some i = 1, . . . , j − 1, i = k. Since zα ∈ Byk,α (R (ε) νk,α ), this gives that n ( 2 −1)(1−2ε) n2 −1 (n−2)(1−ε) = O νi,α νk,α dg (yi,α , zα ) (n−2)(1−ε) . = O max {νi,α ; νk,α } Thus dg (yi,α , zα ) = O (max {νi,α ; νk,α }), and we get that dg (yi,α , yk,α ) ≤ dg (yi,α , zα ) + dg (zα , yk,α ) = O (max {νi,α ; νk,α }) + O (νk,α ) = O (max {νi,α ; νk,α }) . Noting that this is in contradiction with (6.0.8), we have proved that zα ∈ Ωεk,α . Assume now by contradiction that ( n −1)(1−2ε) (2−n)(1−ε) Φεk,α (zα ) ≤ θj,α2 Rj,α (zα ) . By (6.1.103), θj,α = O (νj−1,α ). Since k ≤ j − 1 and zα ∈ Byk,α (R (ε) νk,α ), the above equation gives that n ( 2 −1)(1−2ε) n2 −1 (n−2)(1−ε) Rj,α (zα ) = O νj−1,α νk,α (n−2)(1−ε) . = O νk,α Hence, Rj,α (zα ) = O (νk,α ). Let l ∈ {j, . . . , p} be such that, up to a subsequence, Rj,α (zα ) = dg (yl,α , zα ). Then, dg (yl,α , yk,α ) ≤ dg (yl,α , zα ) + dg (zα , yk,α ) = O (νk,α ) , ˜ ε , and (6.1.127) is proved. and this is in contradiction with (6.0.8). Thus zα ∈ Ω k,α Now we prove (6.1.128). We let i ∈ {1, . . . , p}, i = k, and assume by contradic˜ ε . Then tion that there exists a sequence (zα ) of points in Byi,α (R (ε) νi,α ) ∩ Ω k,α dg (zα , yi,α ) = O (νi,α ), and ( n −1)(1−2ε) (2−n)(1−ε) dg (yi,α , zα ) . Φεk,α (zα ) ≥ νi,α2

133


If i ≤ j − 1, such an equation follows from the fact that zα ∈ Ωεk,α . If i ≥ j, we write that θj,α ≥ νj,α ≥ νi,α and that Rj,α (zα ) ≤ dg (yi,α , zα ). With such an equation we get that n ( 2 −1)(1−2ε) n2 −1 (n−2)(1−ε) = O νk,α νi,α dg (yk,α , zα ) (n−2)(1−ε) , = O max {νi,α ; νk,α } and this leads to a contradiction as above. In particular (6.1.128) is proved. We claim now that, for any k ∈ {j, . . . , p}, ˜ε Byk,α (R (ε) θj,α ) ⊂ Ω j,α

(6.1.129)

and that, for any k ∈ {1, . . . , j − 1}, ˜ε = ∅ Byk,α (R (ε) νk,α ) ∩ Ω j,α

(6.1.130)

when α is large. We prove (6.1.129). Let k ∈ {j, . . . , p}, and let (zα ) be a sequence of points in Byk,α (R (ε) θj,α ). We assume by contradiction that zα is ˜ ε . Hence we assume that there exists i ∈ {1, . . . , j − 1} such such that zα ∈ Ω j,α that ( n −1)(1−2ε) (2−n)(1−ε) Φεi,α (zα ) ≥ θj,α2 Rj,α (zα ) . Since Rj,α (zα ) ≤ dg (yk,α , zα ) = O (θj,α ), n −1 (1−2ε) νi,α ( 2 ) (n−2)(1−ε) (n−2)(1−ε) dg (yi,α , zα ) ≤ Rj,α (zα ) θj,α n ( 2 −1)(1−2ε) n2 −1 = O νi,α θj,α . By (6.1.103), θj,α = O (νj−1,α ), and since i ≤ j − 1, νj−1,α ≤ νi,α . As a consequence, dg (yi,α , zα ) = O (νi,α ), and we can write that dg (yi,α , yk,α ) ≤ dg (yi,α , zα ) + dg (zα , yk,α ) = O (νi,α ) + O (θj,α ) = O (νi,α ) + O (νj−1,α ) = O (νi,α ) . Such an equation is is in contradiction with (6.0.8). This proves (6.1.129). An easy remark is that (6.1.130) is a direct consequence of (6.1.127). It follows from claim 6.1.7 and (6.1.128) that, for any k ∈ {1, . . . , j − 1}, and ˜ ε \By (R (ε) νk,α ), any y ∈ Ω k,α k,α 2 −1 2 −1 −(n+2)ε ≤ C0 Φεk,α (y) + Rα (y) (6.1.131) uα (y) where C0 > 0 is independent of α. Similarly, it follows from claim 6.1.7 and ˜ ε \ ∪p By (R (ε) θj,α ), (6.1.130) that, for any y ∈ Ω i,α j,α i=j 2 −1

uα (y)

( n2 +1)(1−2ε)

≤ C0 θj,α

−(n+2)(1−ε)

Rj,α (y)

−(n+2)ε

+ Rα (y)

(6.1.132)

134

CHAPTER 6

where C0 > 0 is as above. Now we prove that (6.1.122) holds. For that purpose, we let (yα ) be a sequence of points in M \ ∪pi=1 Byi,α (R0 νi,α ), and distinguish four cases. Case 1. We assume that Rα (yα ) → δ as α → +∞ for some δ > 0. Then, for S as in Theorem 4.1, dg (yα , S) → δ as α → +∞, and point (P1) of Theorem 4.1 gives that (6.1.122) holds with C > 0. Case 2. We assume that there exists k ∈ {1, . . . , p}, and R ≥ R0 , such that dg (yk,α , yα ) →R νk,α as α → +∞. By point (P1) of Theorem 4.1, we then get that n

−1

2 uα (yα ) = u (y0 − yk ) lim νk,α

α→+∞

where yk = xNk is given by Theorem 4.1, u is given by (4.1.5), and, up to a subsequence, 1 y0 = lim exp−1 yk,α (yα ) . α→+∞ νk,α Clearly, |y0 | ≥ R0 . Independently, n n 2−n 2 −1 2 −1 lim νk,α νk,α = |y0 |2−n . dg (yk,α , yα ) α→+∞

It follows that lim

α→+∞

uα (yα ) = |y0 |n−2 u (y0 − yk ) . Φ0k,α (yα )

(6.1.133)

Letting D0 = 2 sup

sup |z|n−2 u (z − yk ) ,

(6.1.134)

k=1,...,p z∈Rn

it is easily checked that (6.1.133) gives that lim sup j−1 α→+∞

uα (yα ) − u0 (yα )

0 i=1 Φi,α

n 2 −1

2−n

(yα ) + θj,α Rj,α (yα )

≤

1 D0 . 2

In particular, (6.1.122) holds with C ≥ D0 . Case 3. We assume that, for any k ∈ {1, . . . , p}, dg (yk,α , yα ) → +∞ νk,α as α → +∞, and that Rj,α (yα ) = O (θj,α ). First we prove that Rα (yα ) = Rj,α (yα ). Let l ∈ {j, . . . , p} be such that Rj,α (yα ) = dg (yl,α , yα ) = O (θj,α ). We proceed by contradiction and assume that Rα (yα ) = dg (yk,α , yα ) for some k ∈ {1, . . . , j − 1}. Then dg (yk,α , yα ) ≤ dg (yl,α , yα ), so that dg (yk,α , yl,α ) ≤ dg (yk,α , yα ) + dg (yα , yl,α ) ≤ 2dg (yα , yl,α ) = O (θj,α ) = O (νj−1,α ) .

135


Since j − 1 ≥ k, it follows from the above equation that dg (yk,α , yl,α ) = O (νk,α ), and this is in contradiction with (6.0.8). Therefore, Rα (yα ) = Rj,α (yα ) = O (θj,α ) . −1 dg (yk,α , yα ) νk,α

Since from (6.0.12) that

Hence,

→ +∞ as α → +∞ for all k ∈ {1, . . . , p}, it follows

1− n uα (yα ) − u0 (yα ) = o Rα (yα ) 2 . uα (yα ) − u0 (yα ) 1− n n−2 = θj,α 2 Rα (yα ) uα (yα ) − u0 (yα ) n −1 Rα (yα ) 2 =o θj,α

1− n

n−2

θj,α 2 Rj,α (yα )

= o (1) . In particular, (6.1.122) holds with C > 0. Case 4. We assume that Rα (yα ) → 0 as α → +∞, that for any k ∈ {1, . . . , p}, dg (yk,α , yα ) → +∞ νk,α as α → +∞, and that Rj,α (yα ) → +∞ θj,α as α → +∞. We let Gα be the Green’s function of ∆g +hα . Thanks to the Green’s representation formula,

2 −1 2 −1 dvg Gα (yα , y) uα (y) − u0 (y) uα (yα ) − u0 (yα ) = M

(6.1.135) + Gα (yα , y) (h∞ (y) − hα (y)) u0 (y) dvg . M

From (6.0.1) and (6.1.80),

Gα (yα , y) (h∞ (y) − hα (y)) u0 (y) dvg = o (1) .

(6.1.136)

M

Independently, (6.1.80) and (6.1.126) give that

2 −1 2 −1 dvg = o (1) . Gα (yα , y) uα (y) − u0 (y) M \∪p i=1 Byi,α (δα )

Inserting this equation and (6.1.136) into (6.1.135), we get that uα (yα ) − u0 (yα )

≤

∪p i=1 Byi,α (δα )

2 −1


dvg + o (1) .

(6.1.137)

136

CHAPTER 6

Let k ∈ {1, . . . , j − 1}. We set 1 ˜ ε ∩ ∪p By (δα ) s.t. dg (yα , y) ≥ 1 dg (yk,α , yα ) , Bk,α = y∈Ω k,α i,α i=1 2 2 ˜ ε ∩ ∪p By (δα ) s.t. dg (yα , y) < 1 dg (yk,α , yα ) , = y∈Ω Bk,α k,α i,α i=1 2 where δα is as in (6.1.126). Then

˜ ε ∩(∪p By (δα )) Ω i=1 k,α i,α

2 −1


≤

2 −1

1 Bk,α


+

dvg

2 −1

2 Bk,α


dvg (6.1.138)

dvg .

−1 → +∞ as α → +∞, for any y ∈ Byk,α (R (ε) νk,α ), and Since dg (yk,α , yα ) νk,α for α large, 1 dg (yα , y) ≥ dg (yα , yk,α ) . 2 Using (6.1.127), we then get that 1 Byk,α (R (ε) νk,α ) ∩ ∪pi=1 Byi,α (δα ) ⊂ Bk,α (6.1.139)

for α large. From (6.1.80), 2−n

Gα (yα , y) ≤ C2 2n−2 dg (yk,α , yα )

1 for all y ∈ Bk,α . It follows that, for α large,

2 −1 dvg Gα (yα , y) uα (y) p Byk,α (R(ε)νk,α )∩(∪i=1 Byi,α (δα ))

2−n 2 −1 n−2 ≤ C2 2 dg (yk,α , yα ) uα (y) dvg Byk,α (R(ε)νk,α )

2−n

≤ C2 2n−2 dg (yk,α , yα )

1 uα 22 −1 Volg Byk,α (R (ε) νk,α ) 2 .

Hence, for α large,

2 −1

Byk,α (R(ε)νk,α )∩(∪p i=1 Byi,α (δα ))

≤

D1 Φ0k,α


dvg (6.1.140)

(yα )


−1

ω

n−1

n

21

n

R (ε) 2

−1

.

(6.1.141)

Independently, from (6.1.131), n ( 2 +1)(1−2ε) 2 −1 −(n+2)(1−ε) −(n+2)ε uα (y) ≤ C0 νk,α dg (yk,α , y) + Rα (y)

137

ASYMPTOTICS WHEN THE ENERGY IS ARBITRARY 1 \Byk,α (R (ε) νk,α ). Thus we can write with (6.1.80) that for all y ∈ Bk,α

2 −1 Gα (yα , y) uα (y) dvg 1 \B Bk,α yk,α (R(ε)νk,α )

≤ C0

−(n+2)ε

1 \B Bk,α yk,α (R(ε)νk,α )

Gα (yα , y) Rα (y)

dvg

n 2−n ( 2 +1)(1−2ε) +C0 C2 2n−2 dg (yk,α , yα ) νk,α

−(n+2)(1−ε) × dg (yk,α , y) dvg . 1 \B Bk,α yk,α (R(ε)νk,α )

Noting that ε
0 and a sequence (εα ) of positive real numbers converging to 0 as α → +∞ such that N uα (x) ≤ (1 + εα ) u0 (x) + C ϕi,α (x) (6.2.1) i=1

144

CHAPTER 6

for all x ∈ M and all α, where

⎛

ϕi,α (x) = ⎝

µi,α µ2i,α +


⎞ n−2 2 ⎠

.

It follows from section 6.1 [we refer to (6.1.1) and (6.1.67)] that (6.2.1) holds for all x ∈ M \ ∪pj=1 Byj,α (R0 νj,α ). It remains to prove that (6.2.1) holds also in Byj,α (R0 νj,α ) for all j ∈ {1, . . . , p}. We fix j0 ∈ {1, . . . , p}, and let ˜q,α = xNj0 +1 −1,α , x ˜1,α = xNj0 +1,α , . . . , x ˜q,α = µNj0 +1 −1,α . µ ˜1,α = µNj0 +1,α , . . . , µ It easily follows from point (P1) of Theorem 4.1 that (6.2.1) holds in the ball Byj0 ,α (R0 νj0 ,α ) if there exists C > 0 such that q n −1 1− n 2−n 2 2 µ ˜i,α dg (˜ xi,α , x) + νj0 ,α uα (x) ≤ C (6.2.2) i=1

x1,α , . . . , x ˜q,α } and all α. Without loss of generalfor all x ∈ Byj0 ,α (R0 νj0 ,α ) \ {˜ ity, we assume in what follows that the set {˜ xi,α }i=1,...,q is not empty. Otherwise, (6.2.2) is a direct consequence of Theorem 4.1. We rearrange the x ˜i,α ’s by a process analogous to that used in the introduction of this chapter. We let x ˜1,α be such that µ ˜1,α ≥ µ ˜i,α for all i ∈ {1, . . . , q}. We ˜q2 −1 the x ˜i,α ’s, i = 2, . . . , q, which are such that, up to denote then by x ˜2,α , . . . , x a subsequence, lim

α→+∞

dg (˜ x1,α , x ˜i,α ) < +∞ . µ ˜1,α

˜q2 ,α be such that µ ˜q2 ,α ≥ µ ˜i,α for all i ∈ {q2 , . . . , q}. We We let q1 = 1, and let x denote then by x ˜q2 +1,α , . . . , x ˜q3 −1,α the x ˜i,α ’s, q2 + 1 ≤ i ≤ q, which are such that, up to a subsequence, lim

α→+∞

dg (˜ xq2 ,α , x ˜i,α ) < +∞ . µ ˜q2 ,α

Going on with this process, we obtain a sequence of points ˜q1 ,α , x ˜2,α , . . . , x ˜q2 ,α , x ˜q2 +1,α , . . . , x ˜q3 ,α , . . . , x ˜q,α = x ˜qr+1 −1,α . x ˜1,α = x For i ∈ {1, . . . , r}, we let ˜qi ,α y˜i,α = x

and

ν˜i,α = µ ˜qi ,α .

The above construction gives that ν˜1,α ≥ · · · ≥ ν˜r,α

(6.2.3)

and that for any i, j ∈ {1, . . . , r}, i = j, yi,α , y˜j,α ) dg (˜ → +∞ ν˜i,α

(6.2.4)

145


˜ 0 > 0 be such that for any j ∈ {1, . . . , r} and for any as α → +∞. We let R i ∈ {qj + 1, . . . , qj+1 − 1}, ˜0 dg (˜ yj,α , x ˜i,α ) R . ≤ ν˜j,α 4

lim sup α→+∞

We have that

SNj0 =

lim

1

α→+∞

νj0 ,α

exp−1 yj0 ,α

(6.2.5)

(˜ yj,α ) j = 1, . . . , r

where SNj0 is as in (4.1.2). We claim now that there exists C > 0 such that ⎛ ⎞ r n n −1 1− 2−n 2 ν˜j,α dg (˜ yj,α , y) + νj0 ,α2 ⎠ uα (y) ≤ C ⎝ (6.2.6) j=1

˜ 0 ν˜j,α ) and all α. We prove this claim for all y ∈ Byj0 ,α (R0 νj0 ,α )\ ∪rj=1 By˜j,α (R in what follows. We divide the proof into several steps, following what we did when proving (6.1.67). Noting that the arguments developed below do not use any specific property of yj0 ,α , but only the properties satisfied by any concentration point and the above construction, we easily get by finite induction that an equation like (6.2.6) implies (6.2.2). In particular, the proof of (6.2.1) reduces to the proof of (6.2.6). The first step in the proof of (6.2.6) is the following: νj,α ), C LAIM 6.2.1. Given R > 0, for any y ∈ Byj0 ,α (R0 νj0 ,α ) \∪rj=1 By˜j,α (R˜ and for any α, ˜ α (y) 2 −1 |uα (y) − vN ,α (y) | ≤ εR,α R j0 n

where ˜ α (y) = min dg (˜ yj,α , y) , R j=1,...,r

where lim

lim εR,α = 0, and vNj0 ,α is as in Theorem 4.1.

R→+∞ α→+∞

Proof of claim 6.2.1. We proceed by contradiction and assume that claim 6.2.1 does not hold. Then there exists a sequence (xN +1,α ) in Byj0 ,α (R0 νj0 ,α ) such that, for any j ∈ {1, . . . , r}, yj,α , xN +1,α ) dg (˜ → +∞ ν˜j,α

(6.2.7)

as α → +∞, and such that ˜ α (xN +1,α ) 2 −1 |uα (xN +1,α ) − vN ,α (xN +1,α ) | ≥ (4δ0 ) 2 −1 R j0 n

n

(6.2.8)

for some δ0 > 0. We let 1− n

2 uα (xN +1,α ) = µN +1,α

(6.2.9)

and prove that Theorem 4.1 continues to hold if we add xN +1,α and µN +1,α to the xi,α ’s and µi,α ’s, i = 1, . . . N . Noting that this contradicts the maximality of N , as

146

CHAPTER 6

assumed in the introduction of this chapter, claim 6.2.1 will be proved. We follow the lines of step 3 of the proof of Theorem 4.1. We claim that lim ν −1 α→+∞ j0 ,α

exp−1 yj0 ,α (xN +1,α ) ∈ SNj0

(6.2.10)

where SNj0 is defined by (4.1.2). If (6.2.10) is false, point (P1) of Theorem 4.1 gives that 1− n |uα (xN +1,α ) − vNj0 ,α (xN +1,α ) | = o νj0 ,α2 . ˜ α (xN +1,α ) = O (νj ,α ), we get that Noting that this contradicts (6.2.8) since R 0 (6.2.10) holds. From (6.2.10), ˜ α (xN +1,α ) = o (νj ,α ) . R (6.2.11) 0 1− n2 Since vNj0 ,α (xN +1,α ) = O νj0 ,α , we get by combining (6.2.8), (6.2.9) and (6.2.11) that ˜ α (xN +1,α ) R ≥ 2δ0 µN +1,α

(6.2.12)

µN +1,α = o (νj0 ,α ) .

(6.2.13)

for α large, and that In particular, µN +1,α → 0 as α → +∞. Now we consider the following assertions: (D1) For any i = 1, . . . , N , dg (xi,α , xN +1,α ) → +∞ min {µi,α , µN +1,α } as α → +∞. (D2) There exists δ0 > 0 and xN +1 ∈ Rn such that lim ηN +1,α (x)uN +1,α (x) = u (x − xN +1 )

α→+∞

2 (Rn \SN +1 ), and strongly in C 2 (B0 (δ0 )), weakly in D12 (Rn ), strongly in Cloc where ηN +1,α is as in Chapter 4, SN +1 is as in (4.1.2), uN +1,α is as in (4.1.3), and u is as in (4.1.5).

(D3) The energy E, given by E(u) = u2 , satisfies 2 N +1 vi,α lim E uα − u0 − α→+∞

i=1 2

≤ lim E (uα ) α→+∞

− E(u0 )2 − (N + 1) Λ2min

where vi,α = vi,α (ui ) is as in (4.1.4), ui (x) = u (x − xi ), u is as in (4.1.5), and xi is as in (P2) and (D2). As is easily checked, the proof of claim 6.2.1 reduces to the proof of assertions (D1)–(D3). Indeed, (D1) implies that (4.1.6) is satisfied by the xi,α ’s and µi,α ’s, i = 1, . . . , N + 1. Then, point (P1) of Theorem 4.1 with respect to the xi,α ’s and

147


µi,α ’s, i = 1, . . . , N + 1, is a consequence of (D2). One may note here that the set S is unchanged when we add the sequence (xN +1,α ). Independently, point (P2) of Theorem 4.1 obviously continues to hold when we add (sequences of) blow-up points. Finally, (D3) implies that point (P3) of Theorem 4.1 holds for the xi,α ’s and µi,α ’s, i = 1, . . . , N + 1. Hence, if (D1)–(D3) hold, then Theorem 4.1 continues to hold if we add xN +1,α and µN +1,α to the xi,α ’s and µi,α ’s, i = 1, . . . N . This contradicts the maximality of N and proves claim 6.2.1. We prove (D1). First we consider the case where i ∈ {Nj0 , . . . , Nj0 +1 − 1}. Then, by construction of the yj,α ’s, dg (xi,α , yj0 ,α ) → +∞ νj0 ,α as α → +∞. Since dg (xN +1,α , yj0 ,α ) = O (νj0 ,α ) and using (6.2.13), this leads to dg (xi,α , xN +1,α ) → +∞ µN +1,α as α → +∞. Now we consider the case i = Nj0 . It is easily checked (see the proof of Theorem 4.1) that SNj0 ∩ B0 ε20 is empty for some ε0 > 0. From (6.2.10), we can then write that dg (yj0 ,α , xN +1,α ) ≥ ε40 νj0 ,α for α large, and it follows from (6.2.13) that dg (yj0 ,α , xN +1,α ) → +∞ µN +1,α as α → +∞. Finally, we consider the case where i ∈ {Nj0 + 1, . . . , Nj0 +1 − 1}. ˜k,α for some k ∈ {1, . . . , q}. We let j ∈ {1, . . . , r} be such that Then xi,α = x k ∈ {qj , . . . , qj+1 − 1}. By (6.2.5), we can write that dg (˜ xk,α , xN +1,α ) ≥ dg (xN +1,α , y˜j,α ) − dg (˜ xk,α , y˜j,α ) ˜0 R ≥ dg (xN +1,α , y˜j,α ) − ν˜j,α 2 d (˜ x

,x

)

for α large. Then (6.2.7) implies that g k,αν˜j,αN +1,α → +∞ as α → +∞. In particular, since µ ˜k,α ≤ ν˜j,α by construction, dg (˜ xk,α , xN +1,α ) → +∞ µ ˜k,α as α → +∞. Summarizing, from the above three cases, (D1) is proved. Now we prove (D2). Given x ∈ B0 (δµ−1 N +1,α ), the Euclidean ball of center 0 and radius , we let δµ−1 N +1,α n −1 uN +1,α (x) = µN2 +1,α uα expxN +1,α (µN +1,α x) , gN +1,α (x) = expxN +1,α g (µN +1,α x) , hN +1,α (x) = hα expxN +1,α (µN +1,α x) .

(6.2.14)

148

CHAPTER 6

Since µN +1,α → 0 as α → +∞, 2 lim gN +1,α = ξ in Cloc (Rn ) .

(6.2.15)

α→+∞

We also have that gN +1,α is controlled on both sides by ξ in the sense of bilinear forms. It is easily checked that uN +1,α (0) = 1

(6.2.16)

and that 2 −1 ∆gN +1,α uN +1,α + µ2N +1,α hN +1,α uN +1,α = uN +1,α

(6.2.17)

in B0 (δµ−1 N +1,α ). We let exp−1 xN +1,α (xi,α ) SN +1 = lim , xi,α ∈ BxN +1,α (δ) , 1 ≤ i ≤ N α→+∞ µN +1,α where, up to a subsequence, the limits are assumed to exist. By (6.2.12) [see also the proof of (D1)] 3 SN +1 ∩ B0 δ0 = ∅ . (6.2.18) 2 Let R > 0 and let (xα ) be a sequence of points in B0 (R) such that 1 R where dξ is the Euclidean distance. We can write with (6.2.15) that µ N +1,α min dg xi,α , expxN +1,α (µN +1,α xα ) ≥ i=1,...,N 2R dξ (xα , SN +1 ) ≥

(6.2.19)

for α large. Let yα = expxN +1,α (µN +1,α xα ). It follows from point (P2) of Theorem 4.1 that uN +1,α (xα ) ≤ (2R) 2

−1

n

−1

n

≤ (2R) 2

n

min dg (xi,α , yα ) 2

i=1,...,N

−1

uα (yα )

C

for some C > 0 independent of α. Thus (uN +1,α ) is locally uniformly bounded in Rn \SN +1 . By standard elliptic theory, and (6.0.1), (6.2.14), (6.2.15), and (6.2.17), we then get that, up to a subsequence, lim ηN +1,α uN +1,α = uN +1

α→+∞

(6.2.20)

2 (Rn \SN +1 ). We also have that ηN +1,α uN +1,α uN +1 weakly in the in Cloc space D12 (Rn ) as α → +∞. Moreover, uN +1 verifies that 2 −1 n ∆ξ uN +1 = uN +1 in R

and uN +1 (0) = 1 thanks to (6.2.16). By Caffarelli-Gidas-Spruck [17], n−2 2 λN +1 uN +1 (x) = 1 1 + n(n−2) λ2N +1 |x − xN +1 |2

149


for some λN +1 > 0 and some xN +1 ∈ Rn such that λN +1 = 1 +

1 λ2 |xN +1 |2 . n(n − 2) N +1

Up to changing µN +1,α into λ−1 N +1 µN +1,α and xN +1 into λN +1 xN +1 , (D2) is proved. Now, it remains to prove (D3). Let R > 0. For any ε > 0, there exists Cε independent of R and α such that 2

N +1 vi,α dvg uα − u0 − M \BxN +1,α (RµN +1,α ) i=1 2

N ≤ (1 + ε) vi,α dvg uα − u0 − M \BxN +1,α (RµN +1,α ) i=1

2 +Cε vN +1,α dvg . M \BxN +1,α (RµN +1,α )

Noting that

lim

lim

R→+∞ α→+∞

2 vN +1,α dvg = 0 ,

M \BxN +1,α (RµN +1,α )

we get that 2 N +1 0 vi,α dvg uα − u − M \BxN +1,α (RµN +1,α ) i=1 2

N vi,α dvg + εR,α ≤ uα − u0 − M \Bx (RµN +1,α )

i=1

N +1,α

where lim

(6.2.21)

lim εR,α = 0. Hence,

R→+∞ α→+∞

2 N +1 0 vi,α dvg uα − u − M i=1 2

N ≤ vi,α dvg + IR (α) + εR,α uα − u0 − M

(6.2.22)

i=1

where 2 N +1 vi,α dvg IR (α) = uα − u0 − BxN +1,α (RµN +1,α ) i=1 2

N − vi,α dvg . uα − u0 − Bx (RµN +1,α )

N +1,α

i=1

(6.2.23)

150

CHAPTER 6

We have that

2 N +1 v˜i,α dvgN +1,α IR (α) = uN +1,α − u0α − B0 (R) i=1 2

N − v˜i,α dvgN +1,α uN +1,α − u0α − B0 (R)

i=1

where

n −1 v˜i,α (x) = µN2 +1,α vi,α expxN +1,α (µN +1,α x)

and

n −1 u0α (x) = µN2 +1,α u0 expxN +1,α (µN +1,α x) .

We let VR be given by VR =

x ∈ Rn s.t. dξ (x, SN +1 ) ≤

Noting that

lim

lim

R→+∞ α→+∞

VR

1 R

.

2 vÑ +1,α dvgN +1,α = 0 ,

similar arguments to those used to prove (6.2.21) give that 2

N +1 0 v˜i,α dvgN +1,α IR (α) ≤ uN +1,α − uα − B0 (R)\VR i=1 (6.2.24) 2

N − v˜i,α dvgN +1,α + εR,α . uN +1,α − u0α − B0 (R)\VR i=1

Given i = 1, . . . , N , either dg (xi,α , xN +1,α ) = O (µN +1,α ) or dg (xi,α , xN +1,α ) → +∞ µN +1,α as α → +∞. Since (4.1.6) is satisfied by the xi,α ’s and µi,α ’s, i = 1, . . . , N + 1, µ +1,α → +∞ as α → +∞. Hence, given i = we get in the first case that Nµi,α 1, . . . , N , either dg (xi,α , xN +1,α ) → +∞ (6.2.25) µN +1,α as α → +∞, or µN +1,α → +∞ (6.2.26) µi,α as α → +∞. If i ∈ {1, . . . , N } is such that (6.2.25) holds, then

−2n 2 vi,α dvg = O µni,α dg (xi,α , xN +1,α ) BxN +1,α (RµN +1,α )

×Volg BxN +1,α (RµN +1,α ) −2n = O µni,α µnN +1,α dg (xi,α , xN +1,α ) n µi,α =o . dg (xi,α , xN +1,α )

151


yj,α ), From (6.2.7) and the construction of (yj,α ) and (˜ µi,α = O (dg (xi,α , xN +1,α )) for all i = 1, . . . , N . Thus,

if i ∈ {1, . . . , N } is such that (6.2.25) holds, then 2 vi,α dvg = o (1)

BxN +1,α (RµN +1,α )

and we get that

2 v˜i,α dvgN +1,α → 0

B0 (R)\VR

as α → +∞. Let now i ∈ {1, . . . , N } be such that (6.2.26) holds and (6.2.25) does not hold. Let also x ∈ B0 (R) \VR . Then, for α sufficiently large, 1 dg xi,α , expxN +1,α (µN +1,α x) ≥ µN +1,α 2R and it follows from the expression of v˜i,α that n2 −1 µi,α ˜ vi,α L∞ (B0 (R)\VR ) ≤ CR µN +1,α where CR > 0 is independent of α. In particular,

2 v˜i,α dvgN +1,α → 0

B0 (R)\VR

as α → +∞. Summarizing,

we proved that for any i = 1, . . . , N , 2 v˜i,α dvgN +1,α → 0 B0 (R)\VR

(6.2.27)

as α → +∞. Since point (P1) of Theorem 4.1 holds for xN +1,α and µN +1,α ,

|uN +1,α − vÑ +1,α |2 dvgN +1,α → 0 (6.2.28) as α → +∞, and lim

B0 (R)\VR

lim

R→+∞ α→+∞

u2N +1,α dvgN +1,α = Λ2min .

B0 (R)\VR

From (6.2.27) and (6.2.28), 2

N 0 v˜i,α dvgN +1,α → 0 uN +1,α − vÑ +1,α − uα − B0 (R)\VR i=1

as α → +∞. Independently, from (6.2.27) and (6.2.29), 2

N v˜i,α dvgN +1,α uN +1,α − u0α − B0 (R)\VR i=1

≥ u2N +1,α dvgN +1,α B0 (R)\VR

2 −1 0 −2 uN +1,α uα dvgN +1,α B0 (R)\VR

−2

N

i=1

≥

Λ2min

B0 (R)\VR

+ εR,α

2 −1 uN ˜i,α dvgN +1,α +1,α v

(6.2.29)

152

CHAPTER 6

where lim

lim εR,α = 0. Coming back to (6.2.24), we then get that

R→+∞ α→+∞

IR (α) ≤ −Λ2min + εR,α

and using (6.2.22), since R is arbitrary, we get that (D3) is proved. Therefore, (D1)–(D3) hold. As already mentioned, this is in contradiction with the maximality of N in Theorem 4.1. In particular, claim 6.2.1 is proved. 2 Going on with the proof of (6.2.6), mimicking what was done in section 6.1, we now prove the following: ˜ 0 and C (ε) > 0 such C LAIM 6.2.2. For any 0 < ε < 12 , there exist R (ε) ≥ R that n ( 2 −1)(1−2ε) ˜ (1− n )(1−2ε) ˜ (2−n)(1−ε) (2−n)ε uα (y) ≤ C (ε) ν˜1,α + νj0 ,α 2 Rα (y) Rα (y) ˜ α is as in for all y ∈ Byj0 ,α (R0 νj0 ,α ) \ ∪rj=1 By˜j,α (R (ε) ν˜j,α ) and all α, where R claim 6.2.1. Proof of claim 6.2.2. Let 0 < ε < 12 . Also let x0 ∈ SNj0 , where SNj0 is as in Theorem 4.1, and xα = expyj0 ,α (νj0 ,α x0 ). We set D (ε) =

ε (1 − ε) C12 4r C22ε

(6.2.30)

and (6.1.3). First we claim that there exist where C1 and C2 are given by (6.1.2) ˜ 0 and 0 < δ (ε) ≤ 1 dξ x0 , SN \ {x0 } such that R (ε) ≥ R j0 2 ˜ α (y)2 uα (y)2 R

−2

≤ min {D (ε) ; D (1 − ε)}

(6.2.31)

for all y ∈ Bxα (δ (ε) νj0 ,α ) \ By˜j,α (R (ε) ν˜j,α ) and α large. In order to prove (6.2.31), we let (yα ) be a sequence of points such that ∪rj=1

νj,α ) yα ∈ Bxα (δνj0 ,α ) \ ∪rj=1 By˜j,α (R˜ for all α. Using claim 6.2.1, we can write that n n ˜ α (yα ) 2 −1 uα (yα ) ≤ εR,α + R ˜ α (yα ) 2 −1 vN R

j0 ,α

where lim

(yα )

lim εR,α = 0. Thus,

R→+∞ α→+∞

˜ α (yα ) 2 −1 uα (yα ) ≤ εR,α + (2δνj ,α ) 2 −1 vN ,α (yα ) R 0 j0 n

n

n

−1

. ≤ εR,α + (2δ) 2 ˜ 0 sufficiently large, this proves Choosing δ > 0 sufficiently small and R ≥ R (6.2.31). Now we consider the linear operator Lα given by 2 −2 Lα (u) = ∆g u + hα u − uα u.

This operator satisfies the maximum principle by [9]. We let h0 ∈ C 0,θ (M ) be as in the introduction of section 6.1, and G be the Green’s function of ∆g + h0 . We set r r 1−ε ε Gj,α (y) and Hα2 (y) = Gj,α (y) Hα1 (y) = j=1

j=1

153


yj,α , y). Direct computations lead to where Gj,α (y) = G (˜ r 1−ε 2 −2 1 Gj,α (y) hα (y) − (1 − ε) h0 (y) − uα (y) Lα Hα (y) = j=1

+ε (1 − ε)

|∇Gj,α (y)|2 2

Gj,α (y)

and to Lα Hα2 (y) =

r

2 −2

ε

Gj,α (y)

hα (y) − εh0 (y) − uα (y)

j=1

+ε (1 − ε)

|∇Gj,α (y)|2

2

Gj,α (y)

in M \ {˜ y1,α , . . . , y˜r,α }. Let yα ∈ Bxα (δ (ε) νj0 ,α ) \ ∪rj=1 By˜j,α (R (ε) ν˜j,α ) . By (6.1.2), |∇Gj,α (yα )|2 2

Gj,α (yα )

≥

C12

2

dg (˜ yj,α , yα )

for α large. By (6.2.31), we also have that 2 −2

uα (yα )

˜ α (yα )−2 ≤ min {D (ε) ; D (1 − ε)} R

for α large. Let 1 ≤ l ≤ r be such that, up to a subsequence, we have that ˜ α (yα ). Thanks to (6.1.3) and the two equations above, we can yl,α , yα ) = R dg (˜ write that, for α large, Lα Hα1 (yα ) r 1−ε ˜ α (yα )−2 Gj,α (yα ) hα (yα ) − (1 − ε) h0 (yα ) − D (1 − ε) R ≥ j=1

+ε (1 − ε)

C12

2

dg (˜ yj,α , yα )

C12 (2−n)(1−ε)−2 dg (˜ yl,α , yα ) C21−ε r r 1−ε 1−ε ˜ α (yα )−2 −D (1 − ε) R Gj,α (yα ) −O Gj,α (yα )

≥ ε (1 − ε)

j=1

≥

j=1

C12 ˜ (2−n)(1−ε)−2 ε (1 − ε) 1−ε Rα (yα ) C2 ˜ α (yα )(2−n)(1−ε)−2 −rD (1 − ε) C21−ε R ˜ α (yα )(2−n)(1−ε) −O R

˜ α (yα ) ≥ 2rD (1 − ε) C21−ε R

(2−n)(1−ε)−2

.

154

CHAPTER 6

Similarly, we also have that ˜ α (y) Lα Hα2 (yα ) ≥ 2rD (ε) C2ε R

(2−n)ε−2

for α large. Hence, when α is large, Lα Hα1 ≥ 0 and Lα Hα2 ≥ 0 in Bxα (δ (ε) νj0 ,α ) \ since

∪rj=1

(6.2.32)

By˜j,α (R (ε) ν˜j,α ). From point (P1) of Theorem 4.1,

δ (ε) ≤

1 dξ x0 , SNj0 \ {x0 } , 2

we can write that 1− n

uα ≤ 2νj0 ,α2 on ∂Bxα (δ (ε) νj0 ,α ) ˜ 0 , we also have that for α large. Since R (ε) ≥ R 1− n

uα ≤ 2˜ νj,α 2 on ∂By˜j,α (R (ε) ν˜j,α ) for all j ∈ {1, . . . , r} and α large. By (6.1.3) and (6.2.3) we then get the existence of C (ε) > 0 such that n ( 2 −1)(1−2ε) 1 (1− n2 )(1−2ε) 2 uα ≤ C (ε) ν˜1,α Hα + νj0 ,α Hα on ∂ Bxα (δ (ε) νj0 ,α ) \ ∪rj=1 By˜j,α (R (ε) ν˜j,α ) . From (6.2.32), the above equation, and the fact that Lα uα = 0 in M , we can apply the maximum principle. It follows that n ( 2 −1)(1−2ε) 1 (1− n )(1−2ε) 2 Hα + νj0 ,α 2 Hα uα ≤ C (ε) ν˜1,α in Bxα (δ (ε) νj0 ,α ) \ ∪rj=1 By˜j,α (R (ε) ν˜j,α ). In particular, using (6.1.3), this proves that the estimate of claim 6.2.2 holds in Bxα (δ (ε) νj0 ,α ) \ ∪rj=1 By˜j,α (R (ε) ν˜j,α ) . x0 ∈SNj

0

Noting that, with a possibly larger C (ε), point (P1) of Theorem 4.1 gives that this estimate also holds in Byj0 ,α (R0 νj0 ,α ) \ ∪x0 ∈SNj Bxα (δ (ε) νj0 ,α ) , 0

2

this ends the proof of claim 6.2.2. Our next claim is the following:

C LAIM 6.2.3. There exist C > 0 and a sequence (εα ) of positive real numbers converging to 0 as α → +∞ such that n

−1

1− n

2 ˜ α (y) + εα νj0 ,α2 uα (y) − vNj0 ,α (y) ≤ C ν˜1,α R ˜ 0 ν˜j,α and all α, where vN ,α is as for all y ∈ Byj0 ,α (R0 νj0 ,α ) \ ∪rj=1 By˜j,α R j0 ˜ in Theorem 4.1, and Rα is as in claim 6.2.1.

2−n


155

Proof of claim 6.2.3. We need to prove that there exists C > 0 such that for any ˜ 0 ν˜j,α , sequence (yα ) of points in Byj0 ,α (R0 νj0 ,α ) \ ∪rj=1 By˜j,α R n n −1 2−n 2 −1 ˜ ≤ 0 . (6.2.33) Rα (yα ) lim sup νj20 ,α uα (yα ) − vNj0 ,α (yα ) − C ν˜1,α α→+∞

We study the quantity involved in (6.2.33) and distinguish four cases. Case 1. We assume that there exists δ > 0 such that, up to a subsequence, → δ as α → +∞. Then it follows from point (P1) of Theorem 4.1 that 1− n uα (yα ) − vNj0 ,α (yα ) = o νj0 ,α2 .

˜ α (yα ) R νj0 ,α

Hence, (6.2.33) holds with C ≥ 0. Case 2. We assume that there exists k ∈ {1, . . . , r} such that, up to a subsed (˜ y ,yα ) ˜ 0 . By (6.2.5) and point quence, g ν˜k,α → R as α → +∞ for some R ≥ R k,α (P1) of Theorem 4.1, we then get that, up to a subsequence, n −1 lim ν˜ 2 uα α→+∞ k,α

(yα ) = u (y0 − yk )

where, up to a subsequence, 1

exp−1 y˜k,α (yα ) ν˜k,α is given by Theorem 4.1. Noting that dg (˜ yk,α , yα ) lim = |y0 | , α→+∞ ν˜k,α y0 = lim

α→+∞

and yk = xNj0 +qk

we obtain that lim

α→+∞

uα (yα ) n 2 −1

2−n

ν˜k,α dg (˜ yk,α , yα )

= |y0 |n−2 u (y0 − yk ) .

From (6.2.3), it follows that lim sup α→+∞

uα (yα ) ≤ D0 ˜ α (yα )2−n R

n 2 −1

ν˜1,α

where D0 =

sup

sup |z|n−2 u (z − yj )

j=1,...,r z∈Rn

(6.2.34)

and yj = xNj0 +qj is as in Theorem 4.1. This gives that (6.2.33) holds with C > D0 . d (˜ y

,y )

α Case 3. We assume that g ν˜j,α → +∞ as α → +∞ for all j ∈ {1, . . . , r} j,α ˜ α (yα ) = O (˜ ν1,α ). From claim 6.2.1, and that R

1− ˜ n−2 ν˜1,α 2 R |uα (yα ) − vNj0 ,α (yα ) | α (yα ) n2 −1 ˜ α (yα ) n R ˜ α (yα ) 2 −1 |uα (yα ) − vN ,α (yα ) | R = j0 ν˜1,α n ˜ α (yα ) 2 −1 |uα (yα ) − vN ,α (yα ) | =O R j0 n

= o (1) .

156

CHAPTER 6

Hence, (6.2.33) holds with C > 0. ˜ α (yα ) = o (νj ,α ) and that R ˜ α (yα ) ν˜−1 → +∞ as Case 4. We assume that R 0 1,α α → +∞. We let (ηα ) be a sequence of smooth functions with compact support in Byj0 ,α (2R0 νj0 ,α ) such that ηα ≡ 1 in Byj0 ,α (R0 νj0 ,α ) and ∇ηα ∞ = O νj−1 , ∆g ηα ∞ = O νj−2 . (6.2.35) 0 ,α 0 ,α Let Gα be the Green’s function of ∆g + hα . We write with the Green’s representation formula that uα (yα ) − vNj0 ,α (yα )

= Gα (yα , y) ∆g ηα uα − vNj0 ,α (y) dvg Byj ,α (2R0 νj0 ,α )

0 + Gα (yα , y) hα (y) ηα (y) uα (y) − vNj0 ,α (y) dvg . Byj ,α (2R0 νj0 ,α ) 0 (6.2.36) From (6.0.9), (6.2.35), and point (P1) of Theorem 4.1, −1− n ∆g ηα uα − vNj0 ,α (y) = o νj0 ,α 2 in Byj0 ,α (2R0 νj0 ,α ) \Byj0 ,α (R0 νj0 ,α ). Thus we can write that

Gα (yα , y) ∆g ηα uα − vNj0 ,α (y) dvg Byj ,α (2R0 νj0 ,α )\Byj ,α (R0 νj0 ,α ) 0 0

n −1− 2 = o νj0 ,α Gα (yα , y) dvg . Byj ,α (2R0 νj0 ,α ) 0 By (6.1.80), we have that

2−n dvg Gα (yα , y) dvg = O dg (yα , y) Byj ,α (2R0 νj0 ,α ) Byj ,α (2R0 νj0 ,α ) 0 0 =O

Byα (3R0 νj0 ,α )

dg (yα , y)

2−n

dvg

= O νj20 ,α so that

Gα (yα , y) ∆g ηα uα − vNj0 ,α (y) dvg Ωα 1− n = o νj0 ,α2

(6.2.37)

where Ωα = Byj0 ,α (2R0 νj0 ,α ) \Byj0 ,α (R0 νj0 ,α ) . Similarly, it is easily checked that

Gα (yα , y) hα (y) ηα (y) uα (y) − vNj0 ,α (y) dvg Ωα 1− n = o νj0 ,α2 .

(6.2.38)

157


Independently, from (6.0.1), (6.1.80), and the definition of vNj0 ,α ,

1− n Gα (yα , y) hα (y) vNj0 ,α (y) dvg = o νj0 ,α2 . Byj ,α (R0 νj0 ,α ) 0

(6.2.39)

Coming back to (6.2.36), using (6.2.37)–(6.2.39), we then get that uα (yα ) − vNj0 ,α (yα )

= Gα (yα , y) (∆g uα (y) + hα (y) uα (y)) dvg Byj ,α (R0 νj0 ,α )

0 1− n − Gα (yα , y) ∆g vNj0 ,α (y) dvg + o νj0 ,α2 Byj ,α (R0 νj0 ,α ) 0

2 −1 = dvg Gα (yα , y) uα (y) Byj ,α (R0 νj0 ,α ) 0

1− n − Gα (yα , y) ∆g vNj0 ,α (y) dvg + o νj0 ,α2 . Byj ,α (R0 νj0 ,α ) 0 Passing through geodesic normal coordinates at yj0 ,α , it is easily checked that 1− n 2 −1 ∆g vNj0 ,α (y) = vNj0 ,α (y) + O νj0 ,α2 in Byj0 ,α (R0 νj0 ,α ). Thus, as above,

Gα (yα , y) ∆g vNj0 ,α (y) dvg Byj ,α (R0 νj0 ,α ) 0

2 −1 = dvg Gα (yα , y) vNj0 ,α (y) Byj ,α (R0 νj0 ,α ) 0

1− n 2 +O νj0 ,α Gα (yα , y) dvg Byj ,α (R0 νj0 ,α ) 0

1− n 2 −1 = dvg + o νj0 ,α2 Gα (yα , y) vNj0 ,α (y) Byj ,α (R0 νj0 ,α ) 0 and it follows that uα (yα ) − vNj0 ,α (yα )

2 −1 2 −1 = dvg − vNj0 ,α (y) Gα (yα , y) uα (y) Byj ,α (R0 νj0 ,α ) 0 1− n + o νj0 ,α2 .

(6.2.40)

From point (P1) of Theorem 4.1, up to a subsequence, there exists δα → 0 as α → +∞ such that (

)

sup

Byj ,α R0 νj0 ,α \∪rj=1 By˜j,α 0

n

(δα νj0 ,α )

−1

νj20 ,α |uα − vNj0 ,α | → 0

(6.2.41)

158

CHAPTER 6

as α → +∞. Using (6.1.80), as above, one then gets that

2 −1 2 −1 dvg − vNj0 ,α (y) Gˆα (y) uα (y) Byj ,α (R0 νj0 ,α )\∪rj=1 By˜j,α (δα νj0 ,α ) 0

−1− n 2

2−n

= o νj0 ,α

1− n = o νj0 ,α2 ,

Byj

(R0 νj0 ,α ) 0 ,α

dg (yα , y)

dvg

where Gˆα (y) = Gα (yα , y). Hence, (6.2.40) can be rewritten as uα (yα ) − vNj0 ,α (yα )

1− n 2 −1 ≤ dvg + o νj0 ,α2 . Gα (yα , y) uα (y) ∪rj=1 By˜j,α (δα νj0 ,α )

(6.2.42)

Given j = 1, . . . , r, we define ˜ α (y) = dg (˜ yj,α , y) , Σj,α = y ∈ M s.t. R 1˜ 1 = y ∈ Σj,α ∩ By˜j,α (δα νj0 ,α ) s.t. dg (yα , y) ≥ R (y ) , Bj,α α α 2 1˜ 2 = y ∈ Σj,α ∩ By˜j,α (δα νj0 ,α ) s.t. dg (yα , y) < R (y ) . Bj,α α α 2 1 n+2 .

We fix 0 < ε <

Let j ∈ {1, . . . , r}. We write that 2 −1

Σj,α ∩By˜j,α (δα νj0 ,α )


=

dvg 2 −1

Σj,α ∩By˜j,α (R(ε)˜ ν1,α )


+

2 −1

1 \B Bj,α ν1,α ) y ˜j,α (R(ε)˜


+

dvg

2 −1

2 \B Bj,α ν1,α ) y ˜j,α (R(ε)˜


(6.2.43) dvg dvg

where R (ε) is as in claim 6.2.2. We assume in (6.2.43) that ν˜1,α = o (δα νj0 ,α ). This is always possible up to changing δα . Since

˜ α (yα ) R ν ˜1,α

→ +∞ as α → +∞,

dg (yα , y) ≥ dg (yα , y˜j,α ) − dg (˜ yj,α , y) ˜ ≥ Rα (yα ) − R (ε) ν˜1,α 1˜ ≥ R α (yα ) 2

for α large and all y ∈ Σj,α ∩ By˜j,α (R (ε) ν˜1,α ). Using (6.1.80) we then get that

159


for α large

2 −1



˜ α (yα ) ≤ C2 2n−2 R

2−n

dvg

2 −1 uα dvg


1 uα 22 −1 Volg By˜j,α R (ε) ν˜1,α 2 21 n ˜ α (yα )2−n Λ2 −1 × 2 ωn−1 R (ε)n ν˜1,α ≤ C2 2n−2 R . n Hence, for α large,

2 −1 Gα (yα , y) uα (y) dvg ≤ C2 2

n−2

˜ α (yα ) R

2−n

Σj,α ∩By˜j,α (R(ε)˜ ν1,α ) n 2 −1

(6.2.44)

˜ α (yα ) ≤ D1 ν˜1,α R

2−n


−1

ω

n−1

21

n 1 , we can write using (6.1.80) that Given y ∈ Bj,α

n

R (ε) 2

˜ α (yα ) Gα (yα , y) ≤ C2 2n−2 R

2−n

−1

(6.2.45)

.

.

Independently, it comes from (6.2.4) that By˜i,α (R (ε) ν˜i,α ) ⊂ Σi,α for α large and all i ∈ {1, . . . , r}. Applying claim 6.2.2, we get that ! n +1 (1−2ε) (2 ) 2 −1 2 −1 −(n+2)(1−ε) uα (y) ν˜1,α ≤ 22 −2 C (ε) dg (˜ yj,α , y) " −( n +1)(1−2ε) −(n+2)ε +νj0 ,α2 dg (˜ yj,α , y) 1 \By˜j,α (R (ε) ν˜1,α ). Hence, for all y ∈ Bj,α

2 −1 Gα (yα , y) uα (y) dvg 1 \B Bj,α ν1,α ) y ˜j,α (R(ε)˜

≤ 22

C (ε)

× +22

1 n+2 ,

−( n +1)(1−2ε) νj0 ,α2 −(n+2)ε

1 \B Bj,α ν1,α ) y ˜j,α (R(ε)˜

× Since ε
0 such that j−1 ( n −1)(1−2ε) ˜ (2−n)(1−ε) Φεi,α (y) + θ 2 uα (y) ≤ C (ε) Rj,α (yα ) j,α

i=1

(1− n )(1−2ε) ˜ (2−n)ε +νj0 ,α 2 Rα (y)

˜ α (y) is for all y ∈ Byj0 ,α (R0 νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) and all α, where R ˜ as in claim 6.2.1, Rj,α (y) is as in (6.2.50), and θj,α is as in (6.2.63). Proof of claim 6.2.4. We let x0 ∈ SNj0 , where SNj0 is as in Theorem 4.1, and we let xα = expyj0 ,α (νj0 ,α x0 ). It is a direct consequence of (6.2.55) that there exists 0 < δ (ε) ≤ 12 dξ x0 , SNj0 \ {x0 } such that ˜ α (y)2 uα (y)2 R

−2

≤ D (ε)

(6.2.64)

for all y ∈ Bxα (δ (ε) νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) and α large. We consider the linear operator Lα given by 2 −2 u. Lα (u) = ∆g u + hα u − uα

We let Hα (y) =

j−1 ( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (y) i=1

( n −1)(1−2ε) 1−ε +θj,α2 Gi,α (y) r

(1− n )(1−2ε) +νj0 ,α 2

i=j r

ε

Gi,α (y)

i=1

where yi,α , y) , Gi,α (y) = G (˜ and G is the Green’s function of ∆g +h0 , where h0 is as in the beginning of section 6.1. We have already seen when proving claim 6.2.2 that r ε ˜ α (y)(2−n)ε−2 Lα Gi,α (y) ≥ 2rD (ε) C2ε R (6.2.65) i=1

for all y ∈ Bxα (δ (ε) νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) and α large. We just need to

167


replace (6.2.31) by (6.2.64). Then, easy computations give that Lα Hα (y) ≥

j−1 ( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (y) i=1

× ε (1 − ε)

|∇Gi,α (y) |2 2

Gi,α (y)

2 −2

− uα (y)

+ hα (y) − (1 − ε) h0 (y)

( n −1)(1−2ε) 1−ε + θj,α2 Gi,α (y) r

i=j

× ε (1 − ε)

|∇Gi,α (y) |2 2

Gi,α (y)

2 −2

− uα (y)

+ hα (y) − (1 − ε) h0 (y)

(1− n )(1−2ε) ˜ (2−n)ε−2 + 2rD (ε) C2ε νj0 ,α 2 Rα (y) (6.2.66) ∪ri=1

By˜i,α (R (ε) ν˜i,α ) and α large. Let (yα ) be a for all y ∈ Bxα (δ (ε) νj0 ,α ) \ sequence of points in Bxα (δ (ε) νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ). We claim that Lα Hα (yα ) ≥ 0 for α large. By (6.1.2), we have that, for α large, |∇Gi,α (yα ) |2 2

Gi,α (yα )

≥

C12

2

dg (˜ yi,α , yα )

for all i ∈ {1, . . . , r}. Thus, thanks to (6.2.66), we can write that, for α large, Lα Hα (yα ) ≥

j−1 ( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (yα ) i=1

× C12

ε (1 − ε) 2


2 −2

− uα (yα )

+ O (1)

( n −1)(1−2ε) 1−ε + θj,α2 Gi,α (yα ) r

i=j

×

(6.2.67)

C12

ε (1 − ε) 2


2 −2

− uα (yα )

+ O (1)

(1− n )(1−2ε) ˜ (2−n)ε−2 + 2rD (ε) C2ε νj0 ,α 2 . Rα (yα ) After passing to a subsequence, we assume that yα ∈ Ωεk,α for some 1 ≤ k ≤ j −1. We distinguish four cases. Case 1. We assume that

˜ j,α (yα ) , dg (˜ yk,α , yα ) min R νj0 ,α

→δ

168

CHAPTER 6

as α → +∞, for some δ > 0. Then, by (6.2.55), 2 −2 = O νj−2 . uα (yα ) 0 ,α Using (6.1.3) we then get that

2 −2

uα (yα )

+ O (1)

νj−2 0 ,α

j−1

Φεi,α

i=1 −2 ε O νj0 ,α Φk,α

=

( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (yα )

i=1

=O

j−1

(yα )

(yα )

since yα ∈ Ωεk,α . By (4.1.6) and the construction of the yj,α ’s, ν˜k,α = o (νj0 ,α ). We also have here that for α large, dg (˜ yk,α , yα ) ≥ 2δ νj0 ,α . Hence,

2 −2

uα (yα ) =o

+ O (1)

−1− n νj0 ,α 2

j−1

( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (yα )

i=1

(6.2.68)

.

Similarly, thanks in particular to (6.2.59), we have that

2 −2

uα (yα ) =o

r n −1 (1−2ε) ) ( 1−ε + O (1) θj,α2 Gi,α (yα )

−1− n νj0 ,α 2

(6.2.69)

i=j

.

Combining (6.2.67) with (6.2.68) and (6.2.69), we then get that, for α large, (1− n )(1−2ε) ˜ −1− n (2−n)ε−2 Lα Hα (yα ) ≥ 2rD (ε) C2ε νj0 ,α 2 + o νj0 ,α 2 . Rα (yα ) ˜ α (yα ) ≤ 2R0 νj ,α , it follows from this equation that Lα Hα (yα ) ≥ 0 Since R 0 when α is large. ˜ j,α (yα ), and that yk,α , yα ) ≥ R Case 2. We assume that yα ∈ Γεk,α , that dg (˜ ˜ j,α (yα ) = o (νj ,α ). By (6.2.55), we then get that R 0 ˜ j,α (yα )2 uα (yα )2 R

−2

≤ D (1 − ε)

for α large, and it follows [see, for instance, the proof of (6.2.32)] that ( n −1)(1−2ε) 1−ε Gi,α (yα ) θj,α2 r

i=j

×

C12

ε (1 − ε) 2


2 −2

− uα (yα )

+ O (1)

( n −1)(1−2ε) ˜ (2−n)(1−ε)−2 ≥ 2rD (1 − ε) C21−ε θj,α2 Rj,α (yα )

(6.2.70)

169


for α large. Since yα ∈ Γεk,α , n

yk,α , yα ) 2 dg (˜

−1

uα (yα ) − vNj0 ,α (yα ) ≤

D (1 − ε) 100

21−2

so that 2

2 −2

dg (˜ yk,α , yα ) uα (yα ) ≤ max 1, 22 −3 D (1 − ε) 2 2 −2 + dg (˜ . yk,α , yα ) vNj0 ,α (yα ) × 100

(6.2.71)

1− n2 d (˜ y ,yα ) If g νk,α → δ as α → +∞ for some δ > 0, then u (y ) = O νj0 ,α . It α α j0 ,α follows, as in case 1, that j−1 ( n −1)(1−2ε) 2 −2 1−ε + O (1) ν˜i,α2 Gi,α (yα ) uα (yα ) (6.2.72) i=1 −1− n 2 = o νj0 ,α . yk,α , yα ) = o (νj0 ,α ), we get with (6.2.71) that If dg (˜ 2

2 −2

dg (˜ yk,α , yα ) uα (yα )

≤ D (1 − ε)

for α large. Then, from (6.1.3),

j−1 ( n2 −1)(1−2ε) 1−ε Gi,α (yα ) ν˜i,α C12 i=1

ε (1 − ε)

2 −2

− uα (yα )

2


+ O (1)

ε (1 − ε) ( n2 −1)(1−2ε) C12 (2−n)(1−ε) ≥ ν˜k,α dg (˜ yk,α , yα ) 2 C21−ε dg (˜ yk,α , yα ) −2 −C21−ε D (1 − ε) dg (˜ yk,α , yα ) + O (1) ×

j−1 ( n −1)(1−2ε) (2−n)(1−ε) dg (˜ yi,α , yα ) ν˜i,α2 i=1

≥ ε (1 − ε)

C12 −2 dg (˜ yk,α , yα ) Φεk,α (yα ) C21−ε

j−1 −2 yk,α , yα ) + O (1) Φεi,α (yα ) . −C21−ε D (1 − ε) dg (˜ i=1

Since yα ∈ Ωεk,α , and from the definition (6.2.30) of D (1 − ε), this gives that j−1 ( n −1)(1−2ε) 1−ε ν˜i,α2 Gi,α (yα ) i=1

×

C12

ε (1 − ε) 2


2 −2

− uα (yα )

(6.2.73)

+ O (1) −2

yk,α , yα ) ≥ (4r − (j − 1) + o (1)) C21−ε D (1 − ε) dg (˜

Φεk,α (yα ) .

170

CHAPTER 6

Combining (6.2.67) with (6.2.70), (6.2.72), and (6.2.73), we then get, as in case 1, that Lα Hα (yα ) ≥ 0 for α large. ˜ j,α (yα ) and that Case 3. We assume that yα ∈ Γεk,α , that dg (˜ yk,α , yα ) ≤ R dg (˜ yk,α , yα ) = o (νj0 ,α ). We then get with (6.2.55) that 2

2 −2

yk,α , yα ) uα (yα ) dg (˜

≤ D (1 − ε)

for α large. As when proving (6.2.73), it follows that j−1 ( n −1)(1−2ε) 1−ε Gi,α (yα ) ν˜i,α2 i=1

×

C12

ε (1 − ε) 2


2 −2

− uα (yα )

(6.2.74)

+ O (1) −2

yk,α , yα ) ≥ (4r − (j − 1) + o (1)) C21−ε D (1 − ε) dg (˜

Φεk,α (yα ) .

˜ j,α (yα ) ≥ dg (˜ yk,α , yα ), and since Since R 2 −2

uα (yα )

−2

≤ D (1 − ε) dg (˜ yk,α , yα )

,

we can write with (6.1.3) that r ( n −1)(1−2ε) 2 −2 1−ε θj,α2 + O (1) Gi,α (yα ) uα (yα ) i=j

−2

≤ (r − j + 1) D (1 − ε) dg (˜ yk,α , yα )

+ O (1) C21−ε

( n −1)(1−2ε) ˜ (2−n)(1−ε) ×θj,α2 Rj,α (yα ) ( n −1)(1−2ε) −2+(2−n)(1−ε) ≤ (r − j + 1 + o (1)) D (1 − ε) C21−ε θj,α2 dg (˜ yk,α , yα ) . By (6.2.3), (6.2.59), and (6.2.63), θj,α ≤ ν˜k,α (1 + o (1)) since k ≤ j − 1. It follows that r ( n −1)(1−2ε) 2 −2 1−ε θj,α2 + O (1) Gi,α (yα ) uα (yα )

(6.2.75)

i=j

≤ (r − j + 1 + o (1)) D (1 −

ε) C21−ε dg

−2

(˜ yk,α , yα )

Φεk,α

(yα ) .

Combining (6.2.67) with (6.2.74) and (6.2.75), we then get that Lα Hα (yα ) ≥ 0 for α large. Case 4. We assume that yα ∈ Γεk,α and that ˜ j,α (yα ) ; dg (˜ yk,α , yα ) = o (νj0 ,α ) . min R Since yα ∈ Γεk,α , dg (˜ yk,α , yα )

n 2 −1

uα (yα ) − vNj0 ,α (yα ) ≥

D (1 − ε) 100

21−2 .

171


By (6.2.55) we then get that ˜ j,α (yα ) ≤ dg (˜ yk,α , yα ) , R ˜ j,α (yα ) = o (νj ,α ), we have that and since R 0

˜ j,α (yα )2 uα (yα )2 R

−2

≤ D (1 − ε)

for α large. As when proving (6.2.32), it follows that r ( n −1)(1−2ε) 1−ε θj,α2 Gi,α (yα ) i=j

×

C12

ε (1 − ε) 2


2 −2

− uα (yα )

+ O (1)

(6.2.76)

≥ (3r + j − 1 + o (1)) D (1 − ε) C21−ε ( n −1)(1−2ε) ˜ (2−n)(1−ε)−2 . × θj,α2 Rj,α (yα ) On the other hand, from (6.1.3) and the above equations, j−1 ( n −1)(1−2ε) 2 −2 1−ε uα (yα ) + O (1) ν˜i,α2 Gi,α (yα ) i=1

j−1 ˜ j,α (yα )−2 + O (1) C 1−ε ≤ D (1 − ε) R Φεi,α (yα ) 2 i=1

≤ (j − 1 +

o (1)) C21−ε D (1

˜ j,α (yα )−2 Φε (yα ) − ε) R k,α

since yα ∈ Ωεk,α . Noting that yα also belongs to Γεk,α , we get from the definitions (6.2.57) of Aj,α and (6.2.63) of θj,α that n −1 (1−2ε) ˜ (2−n)(1−ε) Φε (y ) ≤ (A ν˜ )( 2 ) R (y ) k,α

α

j,α j,α

( ≤ θj,α

n 2 −1

j,α

α

)(1−2ε) ˜ (2−n)(1−ε) . Rj,α (yα )

This leads to j−1 ( n −1)(1−2ε) 2 −2 1−ε + O (1) ν˜i,α2 Gi,α (yα ) uα (yα ) i=1

≤ (j − 1 + o (1)) C21−ε D (1 − ε)

(6.2.77)

( n −1)(1−2ε) ˜ (2−n)(1−ε)−2 × θj,α2 . Rj,α (yα ) Combining (6.2.67) with (6.2.76) and (6.2.77), we then get that Lα Hα (yα ) ≥ 0 for α large. The study of these four cases clearly proves that, for α large, L α Hα ≥ 0

(6.2.78)

∪ri=1

in Bxα (δ (ε) νj0 ,α ) \ By˜i,α (R (ε) ν˜i,α ). Using point 4.1, (P1) of Theorem (6.1.3), (6.2.3), (6.2.63), and the fact that δ (ε) ≤ 12 dξ x0 , SNj0 \ {x0 } , we also have that there exists C (ε) > 0 such that uα ≤ C (ε) Hα

(6.2.79)

172

CHAPTER 6

on ∂ Bxα (δ (ε) νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) . By [9], Lα satisfies the maximum principle. Since Lα uα = 0, we then get with (6.2.78), (6.2.79), and (6.1.3) that the equation of claim 6.2.4 holds in ∪x0 ∈SNj Bxα (δ (ε) νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) . 0

By point (P1) of Theorem 4.1, the equation extends to Byj0 ,α (R0 νj0 ,α ) \ ∪ri=1 By˜i,α (R (ε) ν˜i,α ) . This proves claim 6.2.4.

2

Now that claim 6.2.4 is proved, we prove that (6.2.52) holds. This is the subject of claim 6.2.5. Claim 6.2.5 is as follows: C LAIM 6.2.5. If (6.2.51)j−1 holds for some j = 2, . . . , r, then (6.2.51)j holds also. Proof of claim 6.2.5. We claim that there exists C > 0 such that for any sequence r ˜ (yα ) of points in Byj0 ,α (R0 νj0 ,α ) \ ∪i=1 By˜i,α R0 ν˜i,α , uα (yα ) − vNj0 ,α (yα ) j−1 n 1− n 2−n 0 2 −1 ˜ ≤C + o νj0 ,α2 Φi,α (yα ) + θj,α Rj,α (yα )

(6.2.80)

i=1

where n

−1

2−n

2 dg (˜ yi,α , y) Φ0i,α (y) = ν˜i,α

(6.2.81)

1 in the definition (6.2.57) and θj,α is as in (6.2.63). We fix ε such that 0 < ε < n+2 of Aj,α . Before proving (6.2.80), we introduce some notation and discuss some consequences of claim 6.2.4. Given i = 1, . . . , j − 1, we let ( n2 −1)(1−2ε) ˜ (2−n)(1−ε) ε ε ε ˜ (6.2.82) Rj,α (y) Ωi,α = y ∈ Ωi,α s.t. Φi,α (y) ≥ θj,α

where Ωεi,α is as in (6.2.53), and Φεi,α is as in (6.2.54). We let also ˜ ε = By ˜ε . Ω (R0 νj ,α ) \ ∪j−1 Ω j,α

j0 ,α

0

i=1

i,α

(6.2.83)

As in (6.1.127) and (6.1.129), for α large, ˜ ε , and if k ∈ {1, . . . , j − 1} then By˜k,α (R (ε) ν˜k,α ) ⊂ Ω k,α ε ˜ if k ∈ {j, . . . , r} then By˜ (R (ε) θj,α ) ⊂ Ω . k,α

(6.2.84)

j,α

By claim 6.2.4, (6.2.3), (6.2.63), and (6.2.84) [see also the proof of (6.1.128)] we ˜ ε \By˜ (R (ε) ν˜k,α ), get that for any k ∈ {1, . . . , j − 1}, and any y ∈ Ω k,α k,α $ # n −( +1)(1−2ε) 2 −1 2 −1 ˜ α (y)−(n+2)ε uα (y) (6.2.85) ≤ C0 Φεk,α (y) + νj0 ,α2 R when α is large, and where C0 does not depend on α. By claim 6.2.4, (6.2.3), ˜ ε \∪r By˜ (R (ε) θj,α ), (6.2.63), and (6.2.84), we also have that, for any y ∈ Ω i,α j,α i=j ! n ( 2 +1)(1−2ε) ˜ 2 −1 −(n+2)(1−ε) ≤ C0 θj,α uα (y) Rj,α (y) (6.2.86) " n −( 2 +1)(1−2ε) −(n+2)ε ˜ + νj0 ,α Rα (y)


173

when α is large, and where C0 is as above. We let (δα ) be a sequence of positive real numbers converging to 0 as α → +∞ such that (

)

n

sup

Byj ,α R0 νj0 ,α \∪ri=1 By˜i,α 0

(δα νj0 ,α )

−1

νj20 ,α |uα − vNj0 ,α | → 0

(6.2.87)

as α → +∞. The existence of (δα ) easily follows from point (P1) of Theorem 4.1. Now we prove that (6.2.80) holds. For that purpose, we let (yα ) be such that ˜ 0 ν˜i,α yα ∈ Byj0 ,α (R0 νj0 ,α ) \ ∪ri=1 By˜i,α R for all α, and distinguish four cases. Case 1. We assume that point (P1) of Theorem 4.1,

˜ α (yα ) R νj0 ,α

→ δ as α → +∞ for some δ > 0. Then, from

1− n |uα (yα ) − vNj0 ,α (yα ) | = o νj0 ,α2 .

In particular, (6.2.80) holds with C > 0. ˜ 0 such that Case 2. We assume that there exists k ∈ {1, . . . , r} and R ≥ R (˜ yk,α , yα ) → R as α → +∞. Then, as in case 2 of the proof of claim 6.2.4,

−1 dg ν˜k,α

lim sup α→+∞

uα (yα ) ≤ D0 Φ0k,α (yα )

where D0 = sup

sup |z|n−2 u (z − yk )

i=1,...,r z∈Rn

(6.2.88)

and yk = xNj0 +qk is as in Theorem 4.1. By (6.2.3) and (6.2.63) we then get that (6.2.80) holds with C > D0 . −1 dg (˜ yk,α , yα ) → +∞ as α → +∞ for all k such Case 3. We assume that ν˜k,α ˜ j,α (yα ) = O (θj,α ). Then, as in case 3 of the proof that k ∈ {1, . . . , r}, and that R ˜ α (yα ). Since ˜ j,α (yα ) = R of claim 6.1.8, we get that for α large, R −1 dg (˜ yk,α , yα ) → +∞ ν˜k,α

as α → +∞ for all k ∈ {1, . . . , r}, claim 6.2.1 gives that ˜ α (yα ) 2 −1 |uα (yα ) − vN ,α (yα ) | = o (1) . R j0 n

Hence,

1− n ˜ n−2 θj,α 2 R uα (yα ) − vNj0 ,α (yα ) j,α (yα ) n −1 ˜ j,α (yα ) 2 n R ˜ α (yα ) 2 −1 uα (yα ) − vN ,α (yα ) R = j0 θj,α ⎛ n2 −1 ⎞ ˜ Rj,α (yα ) ⎠ = o⎝ θj,α = o (1) .

174

CHAPTER 6

In particular, (6.2.80) holds with C > 0. ˜ α (yα ) = o (νj ,α ), that ν˜−1 dg (˜ yk,α , yα ) → +∞ as Case 4. We assume that R 0 k,α −1 ˜ α → +∞ for all k ∈ {1, . . . , r}, and that θj,α Rj,α (yα ) → +∞ as the parameter α → +∞. We let (ηα ) be a sequence of smooth functions with compact support in Byj0 ,α (2R0 νj0 ,α ) such that ηα ≡ 1 in Byj0 ,α (R0 νj0 ,α ) and , ∆g ηα ∞ = O νj−2 . ∇ηα ∞ = O νj−1 0 ,α 0 ,α We let Gα be the Green’s function of ∆g +hα . Thanks to the Green’s representation formula, uα (yα ) − vNj0 ,α (yα )

= Gα (yα , y) ∆g ηα uα − vNj0 ,α (y) dvg Byj ,α (2R0 νj0 ,α )

0 + Gα (yα , y) hα (y) ηα (y) uα (y) − vNj0 ,α (y) dvg . Byj ,α (2R0 νj0 ,α ) 0 As in case 4 in the proof of claim 6.2.3, this equation reduces to 1− n uα (yα ) − vNj0 ,α (yα ) = o νj0 ,α2

2 −1 2 −1 + dvg . − vNj0 ,α (y) Gα (yα , y) uα (y) Byj ,α (R0 νj0 ,α ) 0 Using (6.1.80) and (6.2.87), we then get that uα (yα ) − vNj0 ,α (yα )

1− n 2 −1 ≤ dvg + o νj0 ,α2 Gα (yα , y) uα (y) ∪ri=1 By˜i,α (δα νj0 ,α )

(6.2.89)

where δα → 0 is as in (6.2.87). Given k ∈ {1, . . . , j − 1}, we let r 1 1 ε ˜ yk,α , yα ) , Bk,α = y ∈ Ωk,α ∩ ∪i=1 By˜i,α (δα νj0 ,α ) s.t. dg (yα , y) ≥ dg (˜ 2 r 1 2 ε ˜ yk,α , yα ) . Bk,α = y ∈ Ωk,α ∩ ∪i=1 By˜i,α (δα νj0 ,α ) s.t. dg (yα , y) < dg (˜ 2 Then,

2 −1

(

˜ε ∩ Ω k,α

=

∪ri=1 By˜i,α

2 −1

1 Bk,α


+ Since

dg (˜ yk,α ,yα ) ν ˜k,α

(δα νj0 ,α ))


2 Bk,α

dvg

2 −1


dvg (6.2.90)

dvg .

→ +∞ as α → +∞, we may assume that 1 By˜k,α (R (ε) ν˜k,α ) ⊂ Bk,α

(6.2.91)

175

ASYMPTOTICS WHEN THE ENERGY IS ARBITRARY 1 , it follows from (6.1.80) that for α large, up to changing δα . Given y ∈ Bk,α 2−n

Gα (yα , y) ≤ C2 2n−2 dg (˜ yk,α , yα )

.

Thus we can write that

2 −1 Gα (yα , y) uα (y) dvg By˜k,α (R(ε)˜ νk,α )

≤ C2 2

n−2

2−n

dg (˜ yk,α , yα )

2 −1

By˜k,α (R(ε)˜ νk,α )

2−n

≤ C2 2n−2 dg (˜ yk,α , yα )

2 −1


where D1 = 2n−2 C2 Λ2 From (6.2.85), 2 −1

uα (y)

# ≤ C0

Φεk,α

dvg

1 uα 22 −1 Volg By˜k,α (R (ε) ν˜k,α ) 2 .

This gives that

By˜k,α (R(ε)˜ νk,α )

uα (y)

−1

dvg ≤ (D1 + o (1)) Φ0k,α (yα ) (6.2.92)

ω

n−1

21

n

2 −1

(y)

+

n

R (ε) 2

−1

−( n +1)(1−2ε) ˜α νj0 ,α2 R

(6.2.93)

.

−(n+2)ε

$

(y)

1 for all y ∈ Bk,α \By˜k,α (R (ε) ν˜k,α ). Noting that n+2

2 ν˜k,α

(1−2ε)

2−n

dg (˜ yk,α , yα )

2−ε(n+2)

= Φ0k,α (yα ) ν˜k,α

,

we can then write with (6.1.80) that

2 −1 Gα (yα , y) uα (y) dvg 1 \B Bk,α νk,α ) y ˜k,α (R(ε)˜

−( n +1)(1−2ε) ≤ C0 νj0 ,α2

1 \B Bk,α νk,α ) y ˜k,α (R(ε)˜

˜ α (y)−(n+2)ε dvg Gα (yα , y) R

2−ε(n+2)

+C0 C2 2n−2 Φ0k,α (yα ) ν˜k,α

−(n+2)(1−ε) × dg (˜ yk,α , y) dvg . 1 \B Bk,α νk,α ) y ˜k,α (R(ε)˜

Since ε
r (D1 + D2 ). From cases 1–4, equation (6.2.80) holds for any sequence (yα ) of points in Byj0 ,α (R0 νj0 ,α ) \ ∪ri=1 By˜i,α (R0 ν˜i,α ) when C > D0 + r (D1 + D2 ), where D0 is as in (6.2.88), D1 is as in (6.2.93), and D2 is as in (6.2.95). In order to prove that (6.2.51)j holds, and thus to end the proof of claim 6.2.5, it remains to prove that θj,α = O (˜ νj,α ) .

(6.2.103)

By (6.2.63), this reduces to proving that Aj,α = O (1) where Aj,α is as in (6.2.57). We proceed by contradiction and assume that Aj,α → +∞

(6.2.104)

as α → +∞. By the definition of Aj,α , there exists k ∈ {1, . . . , j − 1} and yα ∈ Γεk,α such that, up to a subsequence, (Aj,α ν˜j,α )( 2

n

−1)(1−2ε)

˜ j,α (yα )(2−n)(1−ε) = Φεk,α (yα ) (1 + o (1)) . R

It follows from (6.2.104) that θj,α = Aj,α ν˜j,α . Hence, ( n −1)(1−2ε) ˜ (2−n)(1−ε) (1 + o (1)) . Φεk,α (yα ) = θj,α2 Rj,α (yα ) Since yα ∈

Γεk,α ,

Φεk,α

D (1 − ε) 100

(yα ) ≥

21−2

Φεi,α

(6.2.105)

(yα ) for all i ∈ {1, . . . , j − 1}, and n

≤ dg (yα , y˜k,α ) 2

−1

uα (yα ) − vNj0 ,α (yα ) .

181


Using (6.2.80) and (6.2.105) we then get that 1 D (1 − ε) 2 −2 100 j−1 n n −1 −1 2−n 0 2 ˜ j,α (yα ) ≤ Cdg (˜ yk,α , yα ) 2 Φi,α (yα ) + θj,α R +o

i=1

dg (˜ yk,α , yα ) νj0 ,α

≤ Cdg (˜ yk,α , yα ) +

n2 −1

˜ j,α (yα ) R θj,α

n 2 −1

!

−ε(n−2)

R (ε)

−ε(n−2)

j−1

Φεi,α (yα )

i=1

" ( n −1)(1−2ε) ˜ (2−n)(1−ε) θj,α2 + o (1) Rj,α (yα )

! n −1 −ε(n−2) yk,α , yα ) 2 Φεk,α (yα ) (j − 1) R (ε) ≤ Cdg (˜ −ε(n−2) " ˜ j,α (yα ) R + o (1) . + θj,α In particular,

1 D (1 − ε) 2 −2 100 1− n (1−2ε) dg (˜ yk,α , yα ) ( 2 ) ≤C Kα + o (1) ν˜k,α

where

Kα = d (˜ y

˜ j,α (yα ) R θj,α

(6.2.106)

−ε(n−2) + O (1) .

,y )

α → +∞ as α → +∞. Let us assume by contradicNow we claim that g ν˜k,α k,α tion that dg (˜ yk,α , yα ) = O (˜ νk,α ). Then, coming back to (6.2.105), n −1 (1−2ε) θj,α ( 2 ) (n−2)(1−ε) (n−2)(1−ε) ˜ = dg (˜ yk,α , yα ) Rj,α (yα ) ν˜k,α n ( 2 −1)(1−2ε) n2 −1 . ν˜k,α = O θj,α

νj−1,α ). By (6.2.3), ν˜j−1,α ≤ ν˜k,α since k ≤ j − 1. Thus, By (6.2.59), θj,α = O (˜ ˜ j,α (yα ) = dg (˜ ˜ j,α (yα ) = O (˜ νk,α ). Let l ∈ {j, . . . , r} be such that R yl,α , yα ). R Then, yk,α , y˜l,α ) ≤ dg (˜ yk,α , yα ) + dg (yα , y˜l,α ) dg (˜ ≤ C ν˜k,α

182

CHAPTER 6

and this is in contradiction with (6.2.4). In particular, the above claim is proved and dg (˜ yk,α ,yα ) ˜ j,α (yα ) = o (θj,α ), → +∞ as α → +∞. Then, thanks to (6.2.106), R ν ˜k,α and coming back to (6.2.105) we get that n −1 (1−2ε) ν˜k,α ( 2 ) (n−2)(1−ε) ˜ j,α (yα )(n−2)(1−ε) dg (˜ yk,α , yα ) = R θj,α n ( 2 −1)(1−2ε) n2 −1 = o ν˜k,α θj,α n ( 2 −1)(1−2ε) n2 −1 ν˜j−1,α = o ν˜k,α (n−2)(1−ε) . = o ν˜k,α Since dg (˜ yk,α , yα ) → +∞ ν˜k,α as α → +∞, this is the contradiction we were looking for. In particular, (6.2.104) is absurd, and (6.2.103) is proved. As already mentioned, this ends the proof of claim 6.2.5. 2 By claim 6.2.3, (6.2.51)j is true for j = 1. By induction, thanks to claim 6.2.5, we then get that (6.1.51)j is true for all j = 1, . . . , r. In particular, (6.2.51)r is true. Noting that (6.2.51)r implies (6.2.6), we have proved that (6.2.6) holds. As already mentioned, by finite induction, repeating the above arguments, we in fact proved that (6.2.2) is true. Then, (6.2.1) is also true, and the upper estimate on the uα ’s of Theorem 6.1 holds. 6.3 ASYMPTOTIC BEHAVIOR We prove the estimate from below in Theorem 6.1. For that purpose, we let (xα ) be a sequence of points in M and compute an exact asymptotic expansion of uα (xα ). We split this section into two subsections. The first subsection is concerned with the case u0 ≡ 0. The second subsection is concerned with the case u0 ≡ 0. First we introduce some notations. After passing to a subsequence, we can assume that (xα ) converges and we let x∞ = lim xα .

(6.3.1)

x ˜k = lim xk,α .

(6.3.2)

α→+∞

For any k ∈ {1, . . . , N }, we let α→+∞

Once again, we pass to a subsequence so that these limits do exist. We clearly have that S = {˜ xk }k=1,...,N where S is as in Theorem 4.1. We let also for any k ∈ {1, . . . , N } x ˜k,α = expxk,α (µk,α xk )

(6.3.3)


183

where xk ∈ Rn is given by Theorem 4.1. We obviously have that lim x ˜k,α = x ˜k

α→+∞

for all k ∈ {1, . . . , N }. We let ψk,α be the bubble given by ⎛ ⎞ n2 −1 µk,α ⎠ ψk,α (x) = ⎝ , d (˜ xk,α ,x)2 2 µk,α + gn(n−2)

(6.3.4)

where x ∈ M and k ∈ {1, . . . , N }. It is easily checked that for any R > 0, ψk,α →0 (6.3.5) vk,α − 1 ∞ L (Bxk,α (Rµk,α )) as α → +∞, where vk,α is as in Theorem 4.1. Note also that there exists C > 1 independent of α such that 1 ϕk,α ≤ ψk,α ≤ Cϕk,α in M (6.3.6) C where ϕk,α is as in section 6.2, relation (6.2.1). In particular, the upper estimate of ˜i,α ’s, and leave the µi,α ’s Theorem 6.1 is also true if we replace the xi,α ’s by the x unchanged. Finally, we let S∞ ∈ C 0 (M × M ) be the function defined by ⎧ if x = y ⎨ 1 S∞ (x, y) = (6.3.7) ⎩ n−2 G∞ (x, y) if x = y (n − 2) ωn−1 dg (x, y) where (x, y) ∈ M × M , and G∞ is the Green’s function of ∆g + h∞ . From the property (P3) of Appendix A, S∞ is indeed continuous on M × M . We assume in the following that the xi,α ’s are ordered such that µ1,α ≥ µ2,α ≥ · · · ≥ µN,α . We define the sets Ωi,α , i = 1, . . . , N , in the following way: Ω1,α = x ∈ M s.t. ψ1,α (x) ≥ ψi,α (x) , i = 1, . . . , N and (6.3.8) Ωj,α = x ∈ M \ ∪j−1 i=1 Ωi,α s.t. ψj,α (x) ≥ ψi,α (x) , i = 1, . . . , N when j = 2, . . . , N . It is easily seen that M = ∪N i=1 Ωi,α and that the Ωi,α ’s are disjoint. From (6.2.1) and (6.3.6), for any k ∈ {1, . . . , N } and any x ∈ Ωk,α , uα (x) ≤ u0 (x) (1 + εα ) + Cψk,α (x)

(6.3.9)

where C > 0 is independent of α, and (εα ), independent of x, is a sequence of positive real numbers converging to 0 as α → +∞. 6.3.1 The case u0 ≡ 0. Since u0 ≡ 0, we get with (6.3.9) that for any k such that k ∈ {1, . . . , N }, and any x ∈ Ωk,α , uα (x) ≤ Cψk,α (x) .

(6.3.10)

184

CHAPTER 6

We write with the Green’s representation formula that

2 −1 uα (xα ) = Gα (xα , x) uα (x) dvg M

=

N

k=1

Ωk,α

Gα (xα , x) uα (x)

2 −1

(6.3.11) dvg

where Gα is the Green’s function of ∆g + hα . We fix k ∈ {1, . . . , N }, and let 1 1 = x ∈ Ωk,α s.t. dg (xα , x) ≥ dg (˜ xk,α , xα ) , Bk,α 2 1 2 = x ∈ Ωk,α s.t. dg (xα , x) < dg (˜ xk,α , xα ) . Bk,α 2 1 2 1 2 Clearly, Ωk,α = Bk,α ∪ Bk,α , and Bk,α ∩ Bk,α = ∅. Given R > 0, we write that

2 −1 Gα (xα , x) uα (x) dvg = IR,α + IIR,α + IIIR,α (6.3.12) Ωk,α

where

IR,α =

IIR,α =

IIIR,α =

Ωk,α ∩Bx ˜k,α (Rµk,α )

1 \B Bk,α x ˜k,α (Rµk,α )



2 −1


2 −1


2 −1

dvg , dvg , dvg .

Now we estimate these three integrals. We distinguish two cases. Case 1. We assume that, up to a subsequence, xk,α , xα ) dg (˜ → +∞ µk,α as α → +∞. We start by estimating IIIR,α . From (6.3.10) and the definition 2 , noting that dg (˜ xk,α , x) ≥ dg (˜ xk,α , xα ) − dg (xα , x), there exists C > 0 of Bk,α 2 independent of α such that, for any x ∈ Bk,α , n 2 −1− 2 n (˜ x , x) d 1 −1− g k,α 2 −1 ≤ Cµk,α 2 1 + uα (x) n (n − 2) µ2k,α n 2 −1− 2 (˜ x , x ) d 1 −1− n g k,α α ≤ Cµk,α 2 1 + 4n (n − 2) µ2k,α n

+1

−(n+2)

2 ≤ Cµk,α dg (˜ xk,α , xα )

In particular,

n 2 +1

−(n+2)

IIIR,α = O µk,α dg (˜ xk,α , xα )

.

2 Bk,α

Gα (xα , x) dvg

.

185


From (6.1.80),

Gα (xα , x) dvg = O 2 Bk,α

2−n

⎛

2 Bk,α

dg (xα , x)

dvg ⎞

⎜ =O⎝

Bxα

(

2−n

dg x ˜k,α ,xα 2

2

xk,α , xα ) = O dg (˜

) dg (xα , x)

⎟ dvg ⎠

.

Since dg (xα , x ˜k,α ) µ−1 k,α → +∞ as α → +∞, we then get that n 2−n 2 −1 . dg (˜ xk,α , xα ) IIIR,α = o µk,α Coming back to the expression for ψk,α (xα ), it easily follows that, for any R > 0, IIIR,α = o ψk,α (xα ) . (6.3.13) 1 Now we estimate IIR,α . From (6.1.80) and the definition of Bk,α , 2−n

Gα (xα , x) ≤ C2 2n−2 dg (˜ xk,α , xα ) 1 for all x ∈ Bk,α . Together with (6.3.10) this gives that

2−n xk,α , xα ) IIR,α ≤ Cdg (˜


2 −1 ψk,α dvg

where C > 0 is independent of α and R. It is easily checked that

n 2 −1 2 −1 ψk,α dvg = εR,α µk,α M \Bx ˜k,α (Rµk,α )

where limR→+∞ limα→+∞ εR,α = 0. It follows that IIR,α = εR,α ψk,α (xα )

(6.3.14)

where εR,α is such that limR→+∞ limα→+∞ εR,α = 0. Finally, we estimate IR,α . xk,α , xα ) µ−1 Since dg (˜ k,α → +∞ as α → +∞, we easily get (see Appendix A) that ˜k ) S∞ (x∞ , x n−2 dg (˜ →0 xk,α , xα ) Gα (xα , .) − (n − 2) ωn−1 L∞ (Bx˜ (Rµk,α )) k,α as α → +∞. It follows that ˜k ) S∞ (x∞ , x 2−n IR,α = + o (1) dg (˜ xk,α , xα ) (n − 2) ωn−1

2 −1 × uα dvg .

(6.3.15)


Let i ∈ {1, . . . , N }, i = k, and assume that there exists a sequence of points (zα ) in Bx˜k,α (Rµk,α ) such that ψi,α (zα ) ≥ ψk,α (zα ). Then, 2 xi,α , zα ) dg (˜ 1 R2 µi,α 1 + ≤ µ 1 + k,α n (n − 2) µ2i,α n (n − 2)

186

CHAPTER 6

so that 2

dg (˜ xi,α , zα ) ≤ n (n − 2) µ2i,α

#

µk,α µi,α

1+

R2 n (n − 2)

$ −1 .

xi,α , zα ) = O (µk,α ), and This is possible if and only if µi,α = O (µk,α ). Then dg (˜ xk,α , zα ) = O (µk,α ), we have that dg (˜ xi,α , x ˜k,α ) = O (µk,α ). Using since dg (˜ (4.1.6) and (6.3.3), we then obtain that µi,α = o (µk,α ), so that 2 xi,α , zα ) = O (µi,α µk,α ) = o µ2k,α . dg (˜ This clearly proves that, for any R > 0, Volg Bx˜k,α (Rµk,α ) \Ωk,α = o µnk,α . Writing that

2 −1 uα dvg =


(6.3.16)

2 −1 uα dvg

Bx ˜k,α (Rµk,α )

−

2 −1 uα dvg

Bx ˜k,α (Rµk,α )\Ωk,α

we get with (6.3.16), and from Hölder’s inequalities that

2 −1 2 −1 uα dvg = uα dvg Ωk,α ∩Bx ˜k,α (Rµk,α )


+O

=

1 Volg Bx˜k,α (Rµk,α ) \Ωk,α 2 n 2 −1 2 −1 . uα dvg + o µk,α


By (6.3.3), for R large, 1 Bxk,α ( Rµk,α ) ⊂ Bx˜k,α (Rµk,α ) ⊂ Bxk,α (2Rµk,α ) . 2 From point (P1) of Theorem 4.1, we then get that

1− n 2 −1 lim µk,α2 uα dvg = u2 −1 dx lim R→+∞ α→+∞


Rn

where u is as in (4.1.5). Since

n ωn−1 (n (n − 2)) 2 u2 −1 dx = n n R this leads to


2 −1 uα dvg =

ω

n−1

n

n n 2 −1 (n (n − 2)) 2 + εR,α µk,α

where limR→+∞ limα→+∞ εR,α = 0. Coming back to (6.3.15), we then get that n n −1 2−n 2 −1 xk , x∞ ) + εR,α µk,α dg (˜ xk,α , xα ) . IR,α = (n (n − 2)) 2 S∞ (˜

187


xk,α , xα ) µ−1 Since dg (˜ k,α → +∞ as α → +∞, we also have that lim

ψk,α (xα )

α→+∞

n 2 −1

n

2−n

µk,α dg (˜ xk,α , xα )

= (n (n − 2)) 2

−1

.

Hence, xk , x∞ ) + εR,α ) ψk,α (xα ) . IR,α = (S∞ (˜

(6.3.17)

Combining (6.3.12) with (6.3.13), (6.3.14), and (6.3.17), it follows that

2 −1 Gα (xα , x) uα (x) dvg = S∞ (˜ xk , x∞ ) + o (1) ψk,α (xα ) . (6.3.18) Ωk,α

This ends case 1. Case 2. We assume that, up to a subsequence, dg (˜ xk,α , xα ) → Rk µk,α as α → +∞ for some Rk ≥ 0. We start by estimating IIIR,α . By the definition of 2 2 , for any x ∈ Bk,α , Bk,α xk,α , x) ≤ dg (˜ xk,α , xα ) + dg (xα , x) dg (˜ 3 xk,α , xα ) . ≤ dg (˜ 2 Hence, for α large, 2 ⊂ Bx˜k,α (2Rk µk,α ) . Bk,α

It follows that 2 \Bx˜k,α (Rµk,α ) = ∅ Bk,α

when α is large and R is sufficiently large. In particular, lim

lim IIIR,α = 0 .

R→+∞ α→+∞

1 Now we estimate IIR,α . For any x ∈ Bk,α \Bx˜k,α (Rµk,α ), we have that

xk,α , x) − dg (xα , x ˜k,α ) dg (xα , x) ≥ dg (˜ xk,α , x) − (Rk + o (1)) µk,α ≥ dg (˜ ≥ (R − Rk + o (1)) µk,α . 1 \Bx˜k,α (Rµk,α ), Thus, for α large, for R large, and for any x ∈ Bk,α

1 dg (˜ xk,α , x) . 2 Then we can write, using (6.1.80) and (6.3.10), that

2−n 2 −1 dg (˜ xk,α , x) ψk,α (x) dvg IIR,α ≤ C dg (xα , x) ≥


(6.3.19)

188

CHAPTER 6

where C > 0 is independent of α and R. Direct computations give that

n 2−n 2 −1 2 −1 lim µk,α dg (˜ xk,α , x) ψk,α (x) dvg = 0 lim R→+∞ α→+∞

M \Bx ˜k,α (Rµk,α )

so that 1− n

IIR,α = εR,α µk,α2

where limR→+∞ limα→+∞ εR,α = 0. Since dg (˜ xk,α , xα ) µ−1 k,α → Rk as the parameter α → +∞, we also have that 1− n2 2 R 1− n k + o (1) . 1+ ψk,α (xα ) = µk,α2 n (n − 2) In particular, it follows that IIR,α = εR,α ψk,α (xα )

(6.3.20)

where limR→+∞ limα→+∞ εR,α = 0. Finally, we estimate IR,α . We let 1 exp−1 (6.3.21) xk,α (xα ) . µk,α Given 0 < δ < ig /2, and for x ∈ B0 δµ−1 k,α , the Euclidean ball of center 0 and zα =

radius δµ−1 k,α , we also let

n 2 −1 uα expxk,α (µk,α x) , uk,α (x) = µk,α

gk,α (x) = expxk,α g (µk,α x) . Then we can write that n

−1

2 IR,α = µk,α

where

Vk,α (R)

2 Gˆα (x)uk,α (x)

−1

dvgk,α

Gˆα (x) = Gα expxk,α (µk,α zα ) , expxk,α (µk,α x)

and Vk,α (R) =

1 µk,α

exp−1 ˜k,α (Rµk,α ) . xk,α Ωk,α ∩ Bx

Using the property (P3) of Appendix A, for any x ∈ Vk,α (R), Gα expxk,α (µk,α zα ) , expxk,α (µk,α x) 2−n 1 + o (1) = dg expxk,α (µk,α zα ) , expxk,α (µk,α x) (n − 2) ωn−1 1 + o (1) 2−n = µ2−n dg (zα , x) . (n − 2) ωn−1 k,α k,α Hence,

1 + o (1) 1− n 2−n 2 −1 µk,α2 1Vk,α (R) dgk,α (zα , x) uk,α (x) dvgk,α IR,α = (n − 2) ωn−1 Rn

189


where 1Vk,α (R) is the characteristic function of Vk,α (R). It is easily seen that 2 (Rn ) as α → +∞, where ξ is the Euclidean metric. Using (6.3.3) gk,α → ξ in Cloc and (6.3.16), we have that 1Vk,α (R) → 1Bxk (R) a.e. as α → +∞. From point (P1) of Theorem 4.1, we have that uk,α → u (. − xk ) a.e. as α → +∞. Finally, up to a subsequence, dgk,α (zα , .)

2−n

→ |z∞ − .|2−n a.e.

as α → +∞, where z∞ = limα→+∞ zα . By (6.3.10), 1Vk,α (R) uk,α is uniformly bounded in Rn . Using the Lebesgue-dominated convergence theorem we then easily get that

1 1− n 2 −1 IR,α = µk,α2 |z∞ − x|2−n u (x − xk ) dx + εR,α . (n − 2) ωn−1 Rn Since ∆u = u2

−1

, we have that

2 −1 |z∞ − x|2−n u (x − xk ) dx = u (z∞ − xk ) .

1 (n − 2) ωn−1

Rn

Independently, from (6.3.5), n

−1

2 µk,α ψk,α (xα ) → u (z∞ − xk )

as α → +∞. Hence, we have proved that IR,α = (1 + εR,α ) ψk,α (xα ) .

(6.3.22)

Combining (6.3.12) with (6.3.19), (6.3.20), and (6.3.22), we then get that

2 −1 Gα (xα , x) uα (x) dvg = (1 + o (1)) ψk,α (xα ) . (6.3.23) Ωk,α

This ends case 2. From cases 1 and 2, and since S∞ (x, x) = 1,

2 −1 Gα (xα , x) uα (x) dvg = S∞ (˜ xk , x∞ ) + o (1) ψk,α (xα ) Ωk,α

for all k ∈ {1, . . . , N }. By (6.3.11), we then get that uα (xα ) =

N

xk , x∞ ) + o (1) ψk,α (xα ) . (S∞ (˜

(6.3.24)

k=1

We already know from the preceding section and (6.3.6) that there exists C > 1 such that, for any x ∈ M , uα (x) ≤ C

N k=1

ψk,α (x) .

190

CHAPTER 6

Let C > 1 be sufficiently large such that S∞ (x, y) ≥ 2/C for all x, y ∈ M . Since S∞ is positive and continuous, such a C exists. Then, using (6.3.24), we get that, for α large, uα (x) ≥

N 1 ψk,α C k=1

and this proves Theorem 6.1. As a remark, the xi,α ’s of Theorem 6.1 are not those of Theorem 4.1, but the x ˜i,α ’s defined in (6.3.3). Summarizing, Theorem 6.1 is proved in the special case u0 ≡ 0. From standard elliptic theory, it easily follows from (6.3.24) that, up to a subsequence, n −1 N µi,α 2 1− n lim µ1,α 2 uα = Cn G∞ (˜ xi , .) (6.3.25) lim α→+∞ α→+∞ µ1,α i=1 2 (M \S), where in Cloc n

Cn = (n (n − 2)) 2

−1

(n − 2) ωn−1 ,

and S is as in Theorem 4.1. 6.3.2 The case u0 ≡ 0. Following (6.3.9), we have that for any k ∈ {1, . . . , N } and any x ∈ Ωk,α , 2 −1 2 −1 uα (x) ≤ C ψk,α (x) +1 (6.3.26) where C > 0 is independent of α. We write with the Green’s representation formula that

2 −1 2 −1 0 uα (xα ) − u (xα ) = dvg Gα (xα , x) uα (x) − u0 (x) M

+ Gα (xα , x) (h∞ (x) − hα (x)) u0 (x) dvg . M

By (6.0.1) and (6.1.80), we get that

Gα (xα , x) (h∞ (x) − hα (x)) u0 (x) dvg = o (1) M

so that uα (xα ) − u0 (xα ) =

2 −1 2 −1 dvg + o (1) . Gα (xα , x) uα (x) − u0 (x)

M

By point (P1) of Theorem 4.1, after passing to a subsequence, there exists δα → 0 as α → +∞ such that sup

M \∪N ˜i,α (δα ) i=1 Bx

|uα − u0 | → 0

as α → +∞. Then it follows from (6.1.80) that

2 −1 2 −1 Gα (xα , x) uα (x) − u0 (x) dvg = o (1) M \∪N ˜i,α (δα ) i=1 Bx

191


and that

∪N ˜i,α (δα ) i=1 Bx

Gα (xα , x) u0 (x)

2 −1

dvg = o (1) .

Hence, uα (xα ) − u0 (xα ) N

(6.3.27) 2 −1 = dvg + o (1) . Gα (xα , x) uα (x) N ˜i,α (δα )) k=1 Ωk,α ∩(∪i=1 Bx We fix k ∈ {1, . . . , N }, and let N 1 1 Bk,α = x ∈ Ωk,α ∩ ∪i=1 Bx˜i,α (δα ) s.t. dg (xα , x) ≥ dg (˜ xk,α , xα ) , 2 N 1 2 xk,α , xα ) . Bk,α = x ∈ Ωk,α ∩ ∪i=1 Bx˜i,α (δα ) s.t. dg (xα , x) < dg (˜ 2 1 2 1 2 ∪ Bk,α and Bk,α ∩ Bk,α = ∅. Up to replacing δα by Clearly, Ωk,α = Bk,α δˆα = max δα , ( max µi,α )1/2 , i=1,...,N

we can assume that for any R > 0, and any k ∈ {1, . . . , N }, Bx˜k,α (Rµk,α ) ⊂ Bx˜k,α (δα ) . Given R > 0, we can then write that for α large,

2 −1 dvg Gα (xα , x) uα (x) Ωk,α ∩(∪N ˜i,α (δα )) i=1 Bx

(6.3.28)

= IR,α + IIR,α + IIIR,α , where

IR,α =



IIR,α =



2 −1


2 −1

IIIR,α =


2 −1

dvg , dvg , dvg .

Now we estimate these three integrals. Here again we distinguish two cases. Case 1. We assume that, up to a subsequence, ˜k,α ) dg (xα , x → +∞ µk,α as α → +∞. First we estimate IIIR,α . From (6.3.26) and the definition of 2 , noting that dg (˜ xk,α , x) ≥ dg (˜ xk,α , xα ) − dg (xα , x), there exists C > 0 Bk,α 2 independent of α such that for any x ∈ Bk,α , ⎡ ⎤ n 2 −1− 2 n (˜ x , x ) d 1 −1− g k,α α 2 −1 ≤ C ⎣µk,α 2 1 + + 1⎦ uα (x) 4n (n − 2) µ2k,α n −(n+2) 2 +1 dg (˜ xk,α , xα ) +1 . ≤ C µk,α

192

CHAPTER 6

It follows that

n 2 +1

−(n+2)

xk,α , xα ) IIIR,α = O µk,α dg (˜

+O

Gα (xα , x) dvg

By (6.1.80),

2 Bk,α

2 Bk,α

2 Bk,α

2−n

2 Bk,α

⎛

Gα (xα , x) dvg

.

Gα (xα , x) dvg = O

dg (xα , x)

dvg ⎞

⎜ =O⎝

Bxα

(

2−n

dg x ˜k,α ,xα

2

2

xk,α , xα ) = O dg (˜

) dg (xα , x)

⎟ dvg ⎠

.

2 ⊂ ∪N Since δα → 0 and Bk,α i=1 Bxi,α (δα ), it also follows from (6.1.80) that

Gα (xα , x) dvg = o (1) . 2 Bk,α

Since dg (xα , x ˜k,α ) µ−1 k,α → +∞ as α → +∞, this gives that n 2−n 2 −1 dg (˜ xk,α , xα ) + o (1) . IIIR,α = o µk,α Coming back to the definition of ψk,α , it easily follows that IIIR,α = o ψk,α (xα ) + o (1)

(6.3.29)

1 , for all R > 0. Now we estimate IIR,α . From (6.1.80) and the definition of Bk,α 2−n

Gα (xα , x) ≤ C2 2n−2 dg (˜ xk,α , xα ) 1 for all x ∈ Bk,α . We can then write with (6.3.26) that

2−n IIR,α ≤ Cdg (˜ xk,α , xα )


+C


2 −1 ψk,α dvg

Gα (xα , x) dvg

where C > 0 is independent of α and R. It is easily checked that

n 2 −1 2 −1 ψk,α dvg = εR,α µk,α M \Bx ˜k,α (Rµk,α )

where limR→+∞ limα→+∞ εR,α = 0. As above, since δα → 0 as α → +∞ and 1 ⊂ ∪N Bk,α ˜i,α (δα ), we also get with (6.1.80) that i=1 Bx

Gα (xα , x) dvg = o (1) . 1 Bk,α


Hence, coming back to the definition of ψk,α , IIR,α = εR,α ψk,α (xα ) + 1 .

193

(6.3.30)

Finally, we estimate IR,α . It is easily checked that the arguments used to prove (6.3.17) can be used here. Following the proof of (6.3.17) we then get that xk , x∞ ) + εR,α ψk,α (xα ) . (6.3.31) IR,α = S∞ (˜ Combining (6.3.28) and (6.3.29)–(6.3.31), it follows that

2 −1 dvg Gα (xα , x) uα (x) Ωk,α ∩(∪N ˜i,α (δα )) i=1 Bx = S∞ (˜ xk , x∞ ) + o (1) ψk,α (xα ) + o (1) .

(6.3.32)

This ends case 1. Case 2. We assume that, up to a subsequence, xk,α , xα ) dg (˜ → Rk µk,α as α → +∞ for some Rk ≥ 0. First we estimate IIIR,α . By the definition of 2 2 , for any x ∈ Bk,α , Bk,α xk,α , x) ≤ dg (˜ xk,α , xα ) + dg (xα , x) dg (˜ 3 xk,α , xα ) . ≤ dg (˜ 2 Hence, for α large, 2 ⊂ Bx˜k,α (2Rk µk,α ) . Bk,α

It follows that 2 \Bx˜k,α (Rµk,α ) = ∅ Bk,α

when α is large and R is sufficiently large. In particular, lim

lim IIIR,α = 0 .

R→+∞ α→+∞

1 Now we estimate IIR,α . For any x ∈ Bk,α \Bx˜k,α (Rµk,α ), we have that

xk,α , x) − dg (xα , x ˜k,α ) dg (xα , x) ≥ dg (˜ xk,α , x) − (Rk + o (1)) µk,α ≥ dg (˜ ≥ (R − Rk + o (1)) µk,α . 1 \Bx˜k,α (Rµk,α ), Thus, for α large, for R large, and for any x ∈ Bk,α

1 dg (˜ xk,α , x) . 2 Then we can write, using (6.1.80) and (6.3.26), that

2−n 2 −1 dg (˜ xk,α , x) ψk,α (x) dvg IIR,α ≤ C dg (xα , x) ≥


+C

1 Bk,α

Gα (xα , x) dvg

(6.3.33)

194

CHAPTER 6

where C > 0 is independent of α and R. Direct computations give that

n 2−n 2 −1 2 −1 lim µk,α dg (˜ xk,α , x) ψk,α (x) dvg = 0 . lim R→+∞ α→+∞

M \Bx ˜k,α (Rµk,α )

Moreover, since dg (˜ xk,α , xα ) µ−1 k,α → Rk as α → +∞, we also have that 1− n2 2 R 1− n k + o (1) . ψk,α (xα ) = µk,α2 1+ n (n − 2) 1 Independently, since δα → 0 and Bk,α ⊂ ∪N ˜i,α (δα ), it also follows from i=1 Bx (6.1.80) that

Gα (xα , x) dvg = o (1) . 1 Bk,α

In particular, it follows that

IIR,α = εR,α ψk,α (xα ) + 1 .

(6.3.34)

Finally, we estimate IR,α . It is easily checked that the arguments used to prove (6.3.22) can be used here. Following the proof of (6.3.22) we then get that IR,α = 1 + εR,α ψk,α (xα ) . (6.3.35) Combining (6.3.28) with (6.3.33)–(6.3.35), it follows that

2 −1 dvg Gα (xα , x) uα (x) Ωk,α ∩(∪N ˜i,α (δα )) i=1 Bx

(6.3.36)

= (1 + o (1)) ψk,α (xα ) + o (1) . This ends case 2. From cases 1 and 2, and since S∞ (x, x) = 1,

2 −1 dvg Gα (xα , x) uα (x) Ωk,α ∩(∪N ˜i,α (δα )) i=1 Bx = (S∞ (˜ xk , x∞ ) + o (1)) ψk,α (xα ) + o (1) for all k ∈ {1, . . . , N }. By (6.3.27), we then get that uα (xα ) = u0 (xα ) (1 + o (1)) +

N

xk , x∞ ) + o (1) ψk,α (xα ) . (S∞ (˜

(6.3.37)

k=1

We already know form the preceding section and (6.3.6) that there exists C > 1 such that, for any x ∈ M , uα (x) ≤ u0 (x) (1 + o (1)) + C

N k=1

ψk,α (x) .

195


Let C > 1 be sufficiently large such that S∞ (x, y) ≥ 2/C for all x, y ∈ M . Since S∞ is positive and continuous, such a C exists. Then, thanks to (6.3.37), we get that, for α large, N 1 ψk,α (x) uα (x) ≥ u (x) (1 + o (1)) + C 0

k=1

and this proves Theorem 6.1. As a remark, the xi,α ’s of Theorem 6.1 are not those of Theorem 4.1, but the x ˜i,α ’s defined in (6.3.3). Summarizing, Theorem 6.1 is proved in the special case u0 ≡ 0. Together with subsection 6.3.1, Theorem 6.1 is proved. The remark following Theorem 6.1 easily follows from (6.3.37) and the fact that S∞ (x, x) = 1 for all x. We prove now that the bubbles of Theorem 6.1 satisfy Theorem 3.1. More precisely, we prove that, up to a subsequence, uα = u0 +

N

ψi,α + o (1)

(6.3.38)

i=1

where o (1) H12 (M ) → 0 as α → +∞, and that N Igα (uα ) = Ig∞ u0 + Kn−n + o (1) n where Igα (u) =

1 2

and Ig∞ (u) =

1 2

1 |∇u|2 + hα u2 dvg − 2

M

M

1 |∇u|2 + h∞ u2 dvg − 2

(6.3.39)

|u|2 dvg

M

|u|2 dvg .

M

Since 2 −1 ∆g uα + hα uα = uα

in M , we have that Igα

1 (uα ) = n

u2α dvg .

M

Since ∆g u0 + h∞ u0 = (u0 )2 in M , we also have that 1 Ig∞ u0 = n

−1

(u0 )2 dvg .

M

It follows that the proof of (6.3.39) reduces to the proof that

u2α dvg = (u0 )2 dvg + N Kn−n . lim α→+∞

M

M

(6.3.40)

196

CHAPTER 6

Using the asymptotic estimates (6.3.24) and (6.3.37) of this section,

lim u2α dvg α→+∞

M

0

= lim

u (x) +

α→+∞

2

N

M

ψi,α (x) S∞ (˜ xi , x)

(6.3.41) dvg .

i=1

Since S∞ is continuous on M × M ,

u0 (x) +

M

N

2 ψi,α (x) S∞ (˜ xi , x)

i=1

=

(u0 )2 dvg +

M

N

M

(u0 )2

+O M

+O

u

0

−1

M

ψi,α (x) S∞ (˜ xi , x) dvg

2 −1 ψi,α

dvg (6.3.42)

ψi,α

i=1

N

2

i=1

N

dvg

dvg

.

i=1

Direct computations give that

ψi,α dvg → 0 and M

2 −1 ψi,α dvg → 0

M

as α → +∞, for all i ∈ {1, . . . , N }. By (6.3.42) we then get that

u0 (x) +

M

N

2 ψi,α (x) S∞ (˜ xi , x)

i=1

=

(u0 )2 dvg +

N

M

M

dvg (6.3.43)

2 ψi,α (x) S∞ (˜ xi , x)

dvg + o (1) .

i=1

We write now that N

M

=

2 ψi,α (x) S∞ (˜ xi , x)

dvg

i=1

N

i=1

ψi,α (x)

2

2

S∞ (˜ xi , x)

M

⎛

+O⎝

i=j

M

dvg

⎞ 2 −1 ψi,α ψj,α dvg ⎠ .

(6.3.44)

197


Let i ∈ {1, . . . , N }. Given R > 0, we write that

2 2 ψi,α (x) S∞ (˜ xi , x) dvg M

2 2 = ψi,α (x) S∞ (˜ xi , x) dvg Bx ˜i,α (Rµi,α )

+

M \Bx ˜i,α (Rµi,α )

ψi,α (x)

2

2

S∞ (˜ xi , x)

= (1 + o (1))

Bx ˜i,α (Rµi,α )

+


ψi,α (x)

ψi,α (x)

2

2

dvg

dvg 2

S∞ (˜ xi , x)

dvg .

Here we used that S∞ is continuous on M × M and that S∞ (x, x) = 1 for all x ∈ M . Direct computations give that

2 lim ψi,α (x) dvg = Kn−n lim R→+∞ α→+∞

and that


lim

lim

R→+∞ α→+∞

Thus,


lim

α→+∞

ψi,α (x)

2

ψi,α (x)

2

S∞ (˜ xi , x)

2

dvg = 0 .

dvg = Kn−n

(6.3.45)

M

for all i ∈ {1, . . . , N }. Let i, j ∈ {1, . . . , N }, i = j, and R > 0. Assume for instance that µi,α ≥ µj,α . We write that

2 −1 2 −1 ψi,α ψj,α dvg + ψj,α ψi,α M

2 −1 2 −1 ψi,α ψj,α dvg ≤ + ψj,α ψi,α Bx ˜j,α (Rµj,α )

+

≤

M \Bx ˜j,α (Rµj,α )

2 ψj,α dvg

Bx ˜j,α (Rµj,α )

+



2 ψi,α dvg

+ M

22−1

2 ψi,α dvg

+

2 −1 2 −1 + ψj,α ψi,α ψi,α ψj,α dvg


22−1


2 ψj,α dvg

2 −1 2

2 ψi,α dvg

22−1

21

2 ψj,α dvg

2 ψi,α dvg

21

M


2 ψj,α dvg

21 .

21

198

CHAPTER 6

As above,


where lim

2 ψj,α dvg = εR,α

lim εR,α = 0. Since (ψi,α 2 ) and (ψj,α 2 ) are bounded, we

R→+∞ α→+∞

get that

2 −1 2 −1 ψi,α ψj,α + ψj,α ψi,α

⎛

dvg = O ⎝


M

21 ⎞ 2 ⎠ + εR,α ψi,α dvg

⎛ +O ⎝ If

dg (xi,α ,xj,α ) µi,α

⎞ 22−1 2 ⎠. ψi,α dvg


→ +∞ as α → +∞, we have that


−2n n 2 ψi,α dvg = O µni,α dg (xi,α , xj,α ) µj,α = o (1) .

If dg (xi,α , xj,α ) = O (µi,α ), we then have by (4.1.6) that µj,α = o (µi,α ) and thus that

2 n ψi,α dvg = O µ−n i,α µj,α Bx ˜j,α (Rµj,α )

= o (1) . Therefore,

2 −1 ψi,α ψj,α dvg → 0

(6.3.46)

M

as α → +∞, for all i, j ∈ {1, . . . , N }, i = j. Combining (6.3.41) and (6.3.43)– (6.3.46), we then get that (6.3.40) holds. As already mentioned, this proves that (6.3.39) holds. We now let R α = uα − u 0 −

N

ψi,α .

i=1

The proof of (6.3.38) reduces to the proof that, up to a subsequence,

2 |∇Rα |2 dvg → 0 and Rα dvg → 0 M

(6.3.47)

M

as α → +∞. Up to a subsequence, we may assume that uα → u0 in L2 (M ) as 2 dvg → 0 as α → +∞ for all i, we then get that α → +∞. Noting that M ψi,α

2 dvg → 0 (6.3.48) Rα M

199


as α → +∞. Independently, we have that

|∇Rα |2 dvg M

|∇uα |2 dvg +

=

M

+

M

i=j

−2

N

i=1

M

|∇ψi,α |2 dvg

M

∇uα ∇u0 dvg

(∇ψi,α ∇ψj,α ) dvg − 2

N

i=1

|∇u0 |2 dvg +

(6.3.49)

M

(∇uα ∇ψi,α ) dvg + 2

M

N

M

∇u0 ∇ψi,α dvg .

M

i=1

It easily follows from (6.3.40) that

2 |∇uα | dvg →

|∇u0 |2 dvg + N Kn−n

(6.3.50)

M

as α → +∞, while it is clear that

|∇u0 |2 dvg , ∇uα ∇u0 dvg → M M

0 ∇u ∇ψi,α dvg → 0 ,

M |∇ψi,α |2 dvg → Kn−n

(6.3.51)

M

as α → +∞, for all i. From (6.3.46) we can also prove that

(∇ψi,α ∇ψj,α ) dvg → 0

(6.3.52)

M

for all i = j. Independently,

(∇uα ∇ψi,α ) dvg = (∆g uα )ψi,α dvg M

M

so that, from the equations satisfied by uα ,

2 −1 ψi,α dvg + o(1) . (∇uα ∇ψi,α ) dvg = uα M

M

Given R > 0, we write that

2 −1 uα ψi,α dvg M

2 −1 = uα ψi,α dvg + Bx ˜i,α (Rµi,α )

Since


lim

lim

R→+∞ α→+∞


2 −1 ψi,α dvg . uα

ψi,α (x)

2

dvg = 0 ,

200

CHAPTER 6

it follows that

2 −1 uα ψi,α dvg

=

M

where

lim


2 −1 uα ψi,α dvg + εR,α

lim εR,α = 0. Independently, we easily get with (6.3.46) and

R→+∞ α→+∞

using the asymptotics (6.3.24) and (6.3.37), that

2 −1 uα ψi,α dvg = Kn−n + εR,α Bx ˜i,α (Rµi,α )

where εR,α is as above. Hence,

(∇uα ∇ψi,α ) dvg → Kn−n

(6.3.53)

M

as α → +∞. Combining (6.3.49)–(6.3.53), it follows that

|∇Rα |2 dvg → 0

(6.3.54)

M

as α → +∞. Using (6.3.48) and (6.3.54), we get that (6.3.47) holds. As already mentioned, this proves that (6.3.38) also holds.

Appendix A The Green’s Function on Compact Manifolds We discuss in this appendix existence and properties of Green’s functions for operators like ∆g + h. We let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, and let h ∈ C 0,θ (M ), 0 < θ < 1. We assume that ∆g + h is coercive, and let λ > 0 be such that for any u ∈ H12 (M ),

2 2 (A1) |∇u| + hu dvg ≥ λ |∇u|2 + u2 dvg . M

M

Given K > 0 and λ > 0, we let H(K, λ) be the set of the functions h in C 0,θ (M ) which are such that (A1) is satisfied and |h(x)| ≤ K for all x , |h(y) − h(x)| ≤ Kdg (x, y)θ for all x, y .

(A2)

Let DM be the subset of M ×M consisting of the (x, x), x ∈ M . Let h ∈ H(K, λ) for some K, λ > 0. We claim in this appendix that there exists a continuous function G : M × M \DM → R such that for any x ∈ M , Gx = G(x, .) is in L1 (M ), such that for any u ∈ C 2 (M ) and any x ∈ M ,

G(x, y) (∆g u(y) + h(y)u(y)) dvg (y) , (A3) u(x) = M

and such that the following holds: (P1) For any x, y ∈ M , x = y, G(x, y) = G(y, x), and G(x, y) > 0. Moreover, for any x ∈ M , the function Gx : M \{x} → R given by Gx (y) = G(x, y), is in 2,θ (M \{x}). Cloc (P2) For any x, y ∈ M , x = y, dg (x, y)n−2 G(x, y) ≤ C, where C > 0 depends only on (M, g), K and λ. (P3) There exists δ > 0, depending only on (M, g), such that for any x, y ∈ M , x = y, if dg (x, y) < δ, then 1 ≤ Cdg (x, y) , and dg (x, y)n−2 G(x, y) − (n − 2)ωn−1 dg (x, y)n−1 |∇y G(x, y)| − 1 ≤ Cdg (x, y) ωn−1 where ωn−1 is the volume of the unit (n − 1)-sphere, and C > 0 depends only on (M, g), K and λ. We prove the existence of G and (P1)–(P3) in what follows. We let δ > 0 be such that δ < ig /3, where ig is the injectivity radius of (M, g). We also let

202

APPENDIX A

η ∈ C ∞ (M × M ), 0 ≤ η ≤ 1, be such that η(x, y) = 1 if dg (x, y) ≤ δ, and η(x, y) = 0 if dg (x, y) ≥ 2δ. For x = y, we set H(x, y) =

ηx (y) (n − 2)ωn−1 dg (x, y)n−2

where ηx (y) = η(x, y), and ωn−1 is the volume of the unit (n − 1)-sphere. Following Aubin [7], it is easily seen that there exists C > 0 such that for any x, y, x = y, C , dg (x, y)n−2 C |∆g,y H(x, y)| ≤ dg (x, y)n−2

|H(x, y)| ≤

(A4)

and such that ∆g,y,dist. H(x, y) = δx + ∆g,y H(x, y) in the sense of distributions, where δx is the Dirac measure at x. Letting f (x, y) = ∆g,y H(x, y) , we thus have that for any ϕ ∈ C 2 (M ), and any x ∈ M ,

H(x, y)∆g ϕ(y)dvg (y) − f (x, y)ϕ(y)dvg (y) . ϕ(x) = M

(A5)

M

Let X, Y : M × M \DM → R be continuous functions such that for any x = y, |X(x, y)| ≤ CX dg (x, y)α−n and |Y (x, y)| ≤ CY dg (x, y)β−n where CX , CY > 0, and 0 < α, β < n are independent of x and y. For x = y, we let

Z(x, y) = X(x, z)Y (z, y)dvg (z) . M

From a result by Giraud [38] (see also Aubin [7]) Z : M × M \DM → R is continuous and there exists C(α, β, M, g) > 0, depending only on α, β, and (M, g), such that |Z(x, y)| ≤ CX CY C(α, β, M, g)dg (x, y)α+β−n if α + β < n , |Z(x, y)| ≤ CX CY C(α, β, M, g) (1 + | ln dg (x, y)|) if α + β = n , and (A6) |Z(x, y)| ≤ CX CY C(α, β, M, g) if α + β > n . Moreover, when α + β > n, Z is continuous on M × M . We let Γ1 be the function from M × M \DM to R given by Γ1 (x, y) = −∆g,y H(x, y) − h(y)H(x, y)

(A7)

and then, by induction, we define Γi : M × M \DM → R by

Γi (x, z)Γ1 (z, y)dvg (z) . Γi+1 (x, y) =

(A8)

M

THE GREEN’S FUNCTION ON COMPACT MANIFOLDS

We also let

203

n

+1 2 where E 2 is the greatest integer not exceeding n2 . It is easily seen from (A4) and (A6) that for h ∈ H(K, λ), any i ∈ 1, . . . , k + 1, and any x = y, k=E

n

|Γi (x, y)| ≤

C if 2i < n , dg (x, y)n−2i

|Γi (x, y)| ≤ C (1 + | ln dg (x, y)|) if 2i = n , and |Γi (x, y)| ≤ C if 2i > n ,

(A9)

where C > 0 depends only on (M, g) and K. We also have that Γk and Γk+1 are continuous on M × M . Given x ∈ M , we let ux be given by ∆g ux + hux = Γk+1,x

(A10)

where Γk+1,x (y) = Γk+1 (x, y). Since ∆g + h is coercive, ux exists. We then let G : M × M \DM → R be given by k

G(x, y) = H(x, y) + Γi (x, z)H(z, y)dvg (z) + u(x, y) (A11) i=1

M

where u(x, y) = ux (y). It is easily checked that for any x ∈ M , Gx = G(x, .) is ˜ ∈ C 2 (M ) and any x ∈ M , in L1 (M ), and that for any u

u ˜(x) = G(x, y) (∆g u ˜(y) + h(y)˜ u(y)) dvg (y) . M

In particular, (A3) is satisfied. Now we claim that there exists C > 0, depending only on (M, g) and K, such that for any x, y, y ∈ M , |Γk+1 (x, y ) − Γk+1 (x, y)| ≤ Cdg (y, y )θ .

(A12)

In order to prove this claim, we proceed as follows. It is easily checked from (A7) and (A9) that |Γk+1 (x, y ) − Γk+1 (x, y)|

θ ≤ Cdg (y, y ) + C |f (z, y ) − f (z, y)| dvg (z) M

1 1 +C dg (z, y)n−2 − dg (z, y )n−2 dvg (z) M

(A13)

where C > 0 depends only on (M, g) and K. We let δ > 0 to be chosen later on, and assume that dg (y, y ) < δ /2. We can write that

1 1 dvg (z) − dg (z, y)n−2 n−2 dg (z, y ) M

1 1 dvg (z) = − (A14) n−2 n−2 dg (z, y ) By (δ ) dg (z, y)

1 1 dvg (z) . + − n−2 dg (z, y )n−2 M \By (δ ) dg (z, y)

204

APPENDIX A

If dg (z, y) ≥ δ , then dg (z, y ) ≥ δ /2. It easily follows that

1 1 dvg (z) ≤ Cdg (y, y )θ − n−2 dg (z, y )n−2 M \By (δ ) dg (z, y)

(A15)

where C(δ ) > 0 depends only on (M, g), θ, and δ . Similarly, we can write that n−3 1 1 ≤ (n − 2) (dg (z, y) + dg (z, y )) − dg (y, y ) dg (z, y)n−2 dg (z, y )n−2 dg (z, y)n−2 dg (z, y )n−2 so that

1 1 dvg (z) − n−2 n−2 dg (z, y ) By (δ ) dg (z, y)

(A16)

≤ (n − 2)I(y, y )dg (y, y ) where I(y, y ) =

(dg (z, y) + dg (z, y )) dvg (z) . dg (z, y)n−2 dg (z, y )n−2 n−3

By (δ )

We choose δ > 0 sufficiently small such that for any x ∈ M , and any s, t ∈ Rn , if |s| ≤ δ and |t| ≤ δ , then 1 |t − s| ≤ dg (expx (s), expx (t)) ≤ C(δ )|t − s| C(δ ) where C(δ ) > 1 does not depend on x, s, and t. We let z = expy (t) and let y = expy (t0 ), |t0 | < δ /2. Then easy computations give that

n−3 (|t| + |t − t0 |) dt ≤ C(δ ) (A17) I(y, y ) ≤ C(δ ) n−2 |t − t |n−2 |t| 0 B0 (δ ) where C(δ ) > 0 depends only on (M, g) and δ . Combining (A14)–(A17), we proved that there exists C > 0, depending only on (M, g), such that for any points y, y ∈ M , if dg (y, y ) < δ /2, then

1 1 dvg (z) ≤ Cdg (y, y )θ . − (A18) dg (z, y)n−2 n−2 d (z, y ) g M Clearly, (A18) extends to all y, y ∈ M , so that, from (A13) and (A18), |Γk+1 (x, y ) − Γk+1 (x, y)|

θ ≤ Cdg (y, y ) + C |f (z, y ) − f (z, y)| dvg (z)

(A19)

M

for all y, y ∈ M , where C > 0 depends only on (M, g) and K. We can write (see, for instance, Aubin [7]) that f (x, y) =

Q(x, y) dg (x, y)n−1

where, in geodesic normal coordinates at x,

1 Qx = ∂r ln |g| ωn−1


205

and Qx (y) = Q(x, y). In particular, there exists C > 0, depending only on (M, g), such that for any x, y, y , |Q(x, y)| ≤ Cdg (x, y) and |Q(x, y ) − Q(x, y)| ≤ Cdg (y, y ) . It follows that

|f (z, y ) − f (z, y)| dvg (z) ≤ Cdg (y, y )

1 1 dvg (z) . +C dg (z, y) − n−1 )n−1 d (z, y) d (z, y g g M

M

Writing that n−2 1 1 ≤ (n − 1) (dg (z, y) + dg (z, y )) dg (y, y ) − dg (z, y)n−1 dg (z, y )n−1 dg (z, y)n−1 dg (z, y )n−1 we get, as when proving (A18), that for any y, y ∈ M ,

1 1 dvg (z) ≤ Cdg (y, y )θ dg (z, y) − dg (z, y)n−1 dg (z, y )n−1

(A20)

M

where C > 0 depends only on (M, g). Combining (A19) and (A20), it follows that there exists C > 0, depending only on (M, g) and K, such that for any x, y, y in M, |Γk+1 (x, y ) − Γk+1 (x, y)| ≤ Cdg (y, y )θ . This proves (A12). Now that we have (A12), we return to the definition (A10) of ux . Clearly, ux ∈ C 2,θ (M ). Thanks to standard elliptic estimates as in GilbargTrudinger [37], see for instance Theorems 6.2 and 8.17, we have that ux C 2,θ ≤ C ux C 0 + Γk+1,x C 0,θ and

ux C 0 ≤ C ux 2 + Γk+1,x C 0

where C > 0 depends only on (M, g) and K. Since h ∈ H(K, λ), C Γk+1,x C 0 λ and we get, using (A9), (A12), and the above equations, that for any x ∈ M , ux 2 ≤

ux C 2,θ ≤ C

(A21)

where C > 0 depends only on (M, g), K, and λ. Coming back to the definition (A11) of G, and combining (A4), (A6), (A9), and (A21), we get that there exists C > 0, depending only on (M, g), K, and λ, such that for any x, y ∈ M , dg (x, y)n−2 G(x, y) ≤ C . This proves (P2). Noting that ∆g (ux − ux ) + h(ux − ux ) = Γk+1,x − Γk+1,x we get, thanks to standard elliptic estimates as above, that ux − ux C 0 ≤ CΓk+1,x − Γk+1,x C 0

(A22)

206

APPENDIX A

where C > 0 depends only on (M, g), K, and λ. Writing that |u(x , y ) − u(x, y)| ≤ |ux (y) − ux (y)| + |ux (y) − ux (y )| ≤ ux − ux C 0 + ∇ux C 0 dg (y, y ) we then get with (A21)–(A22) that, for any x, x , y, y ∈ M , |u(x , y ) − u(x, y)| ≤ C (Γk+1,x − Γk+1,x C 0 + dg (y, y )) where C > 0 depends only on (M, g), K, and λ. In particular, u is continuous on M × M . As an easy consequence, coming back to the definition (A11) of G, it follows that G is continuous on M × M \DM . Noting that ∆g Gx + hGx = 0

(A23)

2,θ (M \{x}). Given x, y ∈ M such that in M \{x}, we also have that Gx is in Cloc 0 < dg (x, y) < 1, we get with (A4), (A6), (A9), and (A21), that

|G(x, y) − H(x, y)| ≤ Cdg (x, y)3−n

(A24)

where C > 0 depends only on (M, g), K, and λ. In particular, if dg (x, y) ≤ δ, then 1 2−n − Cdg (x, y) . Gx (y) ≥ dg (x, y) (n − 2)ωn−1 ˆ It follows that there exists δˆ > 0 such that for any x, y ∈ M , if 0 < dg (x, y) < δ, − then Gx (y) ≥ 0. Let Gx be the nonpositive part of Gx . According to what we − ˆ just said, G− x has its support in M \Bx (δ). Moreover, Gx is Lipschitz so that − 2 Gx ∈ H1 (M ). Using (A23) we can then write

0= G− x (∆g Gx + hGx ) dvg M

2 − 2 = |∇G− dvg . x | + h(Gx ) M

Since ∆g + h is coercive, it follows that G− x ≡ 0, and we have that Gx ≥ 0 in M \{x}. The Hopf maximum principle then gives that Gx is positive in M \{x}. Independently, by standard arguments, we get that for any ϕ ∈ C ∞ (M ),

(G(x, y) − G(y, x)) ϕ(y)dvg (y) = 0 . M

It follows that G(x, y) = G(y, x) for all x = y, and (P1) is proved. We are thus left with the proof of (P3). The first equation in (P3) is an easy consequence of (A24). It remains to prove the second equation in (P3). We fix x, y ∈ M such that 0 < dg (x, y) < δ, and let y˜ ∈ Rn be such that y = expx (˜ y ). Given X ∈ S n−1 , n the unit sphere of R , and t ∈ R, we set y + tX) . yt = expx (˜ Then, d (˜ y , X) H(x, yt )t=0 = − dt ωn−1 dg (x, y)n−1 |˜ y|

(A25)


where (., .) is the Euclidean scalar product, and we also have that d C H(z, yt ) ≤ dt t=0 dg (z, y)n−1

207

(A26)

where C > 0 depends only on (M, g). For any r > 0, and any i = 1, . . . , k, from (A26),

d Γi (x, z)H(z, yt )dvg (z) dt t=0 M \By (r) (A27)

d H(z, yt ) dvg (z) . = Γi (x, z) dt t=0 M \By (r) Independently, for r > 0 small, r < dg (x, y)/2, and t > 0 small, we can write that

H(z, yt ) − H(z, y) dvg (z) Γi (x, z) By (r) t

dg (z, yt )2−n − dg (z, y)2−n C dvg (z) ≤ dg (x, y)n−2 By (r) t

n−3 (dg (z, yt ) + dg (z, y)) C dvg (z) ≤ dg (x, y)n−2 By (r) dg (z, yt )n−2 dg (z, y)n−2 where C > 0 depends only on (M, g). As when proving (A18), it follows that

H(z, yt ) − H(z, y) dvg (z) ≤ C(r) Γi (x, z) (A28) By (r) t where C(r) → 0 as r → 0. Combining (A27) and (A28) we then get that for any i = 1, . . . , k,

d Γi (x, z)H(z, yt )dvg (z) dt t=0 M (A29)

d H(z, yt ) = Γi (x, z) dvg (z) . dt t=0 M Using (A25) and (A29), coming back to the definition (A11) of G, we can write that d G(x, yt )t=0 dt (˜ y , X) =− (A30) ωn−1 dg (x, y)n−1 |˜ y|

k d d + H(z, yt )t=0 dvg (z) + ux (yt )t=0 . Γi (x, z) dt dt i=1 M By (A21),

d ux (yt ) ≤ C dt t=0

(A31)

208

APPENDIX A

where C > 0 depends only on (M, g), K, and λ. Independently, as an easy consequence of (A6), (A9), and (A26), for any i = 1, . . . , k,

C d H(z, yt ) t=0 dvg (z) ≤ (A32) Γi (x, z) dt d (x, y)n−2 g M where C > 0 depends only on (M, g) and K. In geodesic normal coordinates at x, we refer for instance to Lemma 8 in Hebey-Vaugon [47], |g ij (y)−δ ij | ≤ Cdg (x, y) where C > 0 can be chosen such that it depends only on (M, g). Combining (A30)–(A32), we then get that ωn−1 dg (x, y)n−1 |∇y G(x, y)| − 1 ≤ Cdg (x, y) where C > 0 depends only on (M, g), K, and λ. This ends the proof of (P3).

Appendix B Coercivity Is a Necessary Condition We prove that the operator ∆g + h∞ has to be coercive if the equation in Theorem 6.1 is true, and thus that the coercivity of the operator ∆g + h∞ in Theorem 6.1 is a necessary condition. We let (M, g) be a smooth compact Riemannian manifold of dimension n ≥ 3, and for h ∈ C 0,θ (M ), θ ∈ (0, 1), and u ∈ H12 (M ), we let

|∇u|2 + hu2 dvg . Ih (u) = M

Then we let µh = inf u Ih (u) where the infimum is with respect to the u’s in H12 (M ) such that M u2 dvg = 1. The operator ∆g + h is said to be coercive [see (2.1.1)] if there exists C > 0 such that Ih (u) ≥ Cu2H 2 for all u ∈ H12 (M ). 1 Easy claims are as follows: C LAIM 1. The operator ∆g + h is coercive if and only if µh > 0. uh ∈ C 2,θ (M ), uh > 0 everywhere, such C LAIM 2. Whatever h is, there exists 2 that ∆g uh + huh = µh uh and M uh dvg = 1. C LAIM 3. If there exists u ∈ C 2,θ (M ) such that u > 0 and ∆g u + hu > 0 everywhere, then ∆g + h is coercive. We prove claim 1 by noting that, if µh > 0, then

|∇u|2 dvg + hu2 dvg M M

2 ≥ε |∇u| dvg + hu2 dvg + (1 − ε)µh u2 dvg M M M

≥ε |∇u|2 dvg + u2 dvg M

M

where ε > 0 is chosen such that (1 − ε)µh + εh ≥ ε. Claim 2 easily follows from standard minimization techniques, the maximum principle, and standard bootstrap arguments for regularity. Finally, we prove claim 3 using claim 2 by writing that

0< uh (∆g u + hu) dvg M

= µh uuh dvg M

so that µh > 0, and we are back to claim 1.

210

APPENDIX B

Now we let (hα ) be a sequence of functions in C 0,θ (M ), θ ∈ (0, 1), we let h∞ ∈ C 0,θ (M ), and let (uα ), uα > 0, be a sequence in C 2,θ (M ) such that for any α, 2 −1 ∆g uα + hα uα = uα

and such that hα → h∞ in C (1 − εα )u0 + where

1 C

0,θ

(B1)

(M ) as α → +∞. We assume that

N

Bαi ≤ uα ≤ (1 + εα )u0 + C

i=1

Bαi

(B2)

i=1

⎞ n−2 2

⎛ Bαi (x) = ⎝

N

µi,α µ2i,α

+

⎠


,

the xi,α ’s are converging sequences of points in M , the µi,α ’s are positive real numbers such that µi,α → 0 as α → +∞, C is a positive constant independent of α and x, N ∈ N , and u0 ∈ H12 (M ) is a nonnegative function such that uα u0 in H12 (M ). Then we claim that, necessarily, the operator ∆g + h∞ is coercive. We prove the claim in what follows. We let µ∞ = µh∞ and u∞ = uh∞ . By the maximum principle and regularity theory, u0 ∈ C 2,θ (M ), ∆g u0 + h∞ u0 = (u0 )2

−1

and either u0 ≡ 0 or u0 > 0 everywhere. If u0 > 0 everywhere, then ∆g + h∞ is coercive from claim 3. We may therefore assume that u0 ≡ 0. As an easy remark, we necessarily have that µ∞ ≥ 0, since for any u ∈ H12 (M ), Ih∞ (u) = lim Ihα (u) . α→+∞

We proceed by contradiction and assume that ∆g + h∞ is not coercive. Then, from claim 1, this means that µ∞ = 0. Using claim 2 we then get that u∞ ∈ C 2,θ (M ), u∞ > 0, is such that ∆ g u ∞ + h∞ u ∞ = 0 . We let µ1,α be such that µ1,α ≥ µi,α for all i. From (B1) we can write that

2 −1 (∆g uα + hα uα ) u∞ dvg = uα u∞ dvg . M

(B3)

M

Since µ∞ = 0,

(∆g uα + hα uα ) u∞ dvg M

= uα (∆g u∞ + h∞ u∞ ) uα dvg + (hα − h∞ ) uα u∞ dvg M

M = (hα − h∞ ) uα u∞ dvg M

and since hα → h∞ in C 0,θ (M ), we get that

(∆g uα + hα uα ) u∞ dvg = o M

M

uα dvg

.

(B4)

211

COERCIVITY IS A NECESSARY CONDITION

Then, using (B2), we have that

N

uα dvg ≤ C M

Bαi dvg

M

i=1

and we can write that

=

n 2 −1 Bαi dvg + O µi,α

Bxi,α (δ)

M

for all i, where δ > 0 is small. We have that

n 2 −1 Bαi dvg ≤ Cµi,α Bxi,α (δ)

δ

+1

2 ≤ Cµi,α

rn−1

µ2i,α + r2

0

n

Bαi dvg

δ µi,α

dr n−2 2

rn−1 n−2

(1 + r2 ) 2

µδ n i,α 2 +1 ≤ Cµi,α rdr C+ 0

dr

1

so that

Bxi,α (δ)

n 2 −1 . Bαi dvg = O µi,α

Then, by the definition of µ1,α and from (B4),

n 2 −1 . (∆g uα + hα uα ) u∞ dvg = o µ1,α

(B5)

M

Still, using (B2), and since u∞ > 0 everywhere, we can write that

2 −1 2 −1 uα u∞ dvg ≥ C uα dvg Bx1,α (µ1,α )

M

≥C

Bx1,α (µ1,α )

Bα1

2 −1

dvg .

Noting that

Bx1,α (µ1,α )

1 2 −1

Bα

dvg ≥ C

µ1,α

0 n 2 −1

≥ Cµ1,α it follows that

M

0

µ1,α 2 µ1,α + r2

1

rn−1 dr

rn−1 (1 + r2 ) n

−1

2 −1 2 uα u∞ dvg ≥ Cµ1,α

n+2 2

n+2 2

dr

(B6)

for some C > 0 independent of α. Inserting (B5) and (B6) into (B3), we get the desired contradiction. This proves the above claim.


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