Danilo Gregorin Afonso - University of São Paulo

Normalized solutions for aSchrödinger-Bopp-Podolsky system

Danilo Gregorin Afonso

Dissertation presented

to the

Institute of Mathematics and Statistics

of the

University of São Paulo

to

obtain the title

of

Master of Sciences

Program: Mathematics

Advisor: Prof. Dr. Gaetano Siciliano

During the development of this work the author was nancially supported by CNPq.

São Paulo, January 2020

Normalized solutions for aSchrödinger-Bopp-Podolsky system

This version of the dissertation contains the corrections and modications suggested

by the Judging Committee during the defense of the original version of the work,

which took place on 14/02/2020. A copy of the original version is available at the

Institute of Mathematics and Statistics of the University of São Paulo.

Judging Committee:

• Prof. Dr. Gaetano Siciliano (advisor) - IME-USP

• Prof. Dr. Antoine Laurain - IME-USP

• Prof. Dr. Francisco Odair Vieira de Paiva - UFSCar

Acknowledgements

To Gaetano, for all he has taught me and for the patience.

To my parents, Ana Laura and Waldemar, for the constant support.

To my mom, again, for her patience.

To Hanna, for her companionship.

To Elvis, Sereno, Bia, Vitor, Carol, Garapa, Felipe, Giovanni and Lissa, for being there.

To all my friends at the Institute, for all the coees and teas and discussions.

To the Wikipedia and MathStackExchange communities, for obvious reasons.

To CNPq for the nancial support.

i

ii

Resumo

GREGORIN AFONSO, D. Soluções normalizadas para um sistema de Schrödinger-

Bopp-Podolsky. 2020. Dissertação (Mestrado) - Instituto de Matemática e Estatística,

Universidade de São Paulo, São Paulo, 2020.

Objetivamos estudar um sistema de tipo Schrödinger-Bopp-Podolsky, que consiste de

duas equações diferenciais parciais não lineares. Apresentamos um resultado original de ex-

istência e multiplicidade de soluções fracas para o problema, ou seja, de existência de pontos

críticos de um funcional restrito a uma subvariedade de um espaço de Hilbert. É desenvoldida

a teoria do cálculo em espaços de Banach. Discutimos a teoria do gênero de Krasnoselskii e

apresentamos o Lema de Deformação e noções correlatas. Discutimos subvariedades em um

espaço de Banach e multiplicadores de Lagrange. Provamos a existência e multiplicidade de

soluções fracas para o problema proposto.

Palavras-chave: sistema de Schrödinger-Bopp-Podolsky, gênero de Krasnoselskii, Lema de

Deformação, multiplicadores de Lagrange, soluções fracas.

iii

iv

Abstract

GREGORIN AFONSO, D. Normalized solutions for a Schrödinger-Bopp-Podolsky

system. 2020. Dissertation (Masters) - Instituto de Matemática e Estatística, Universidade

de São Paulo, São Paulo, 2020.

The aim is to study a Schrödinger-Bopp-Podolsky system of partial dierential equations.

We present an original result for the existence and multiplicity of weak solutions to the prob-

lem, which consists in the determination of critical points for a functional constrained to a

submanifold of a Hilbert space. The calculus in Banach spaces is developed. Krasnoselskii's

genus theory is discussed, after which the Deformation Lemma and some related notions

are presented. Submanifolds of Banach spaces and Lagrange multipliers are discussed. The

existence and multiplicity of solutions to the proposed problem is proved.

Keywords: Schrödinger-Bopp-Podolsky system, Krasnoselskii genus, Deformation Lemma,

Lagrange multipliers, weak solutions.

v

vi

Contents

List of Figures ix

1 Introduction 1

2 Calculus in Banach spaces 3

2.1 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.2 Higher order derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1.3 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Local Inversion Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.1 The Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . 13

2.2.2 The Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . 18

3 Genus Theory 21

3.1 Denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Properties of the genus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Deformations 25

4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.2 Preliminary denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 The Palais-Smale condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.4 Tangent pseudo-gradient vector elds . . . . . . . . . . . . . . . . . . . . . . 28

4.5 The Deformation Lemma and its consequences . . . . . . . . . . . . . . . . . 32

4.6 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 Submersions, Manifolds, Lagrange Multipliers 37

5.1 Submersions and submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.2 The Theorem of Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . 40

6 Proof of the main result 43

6.1 An auxiliary problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.2 The manifold M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.3 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

vii

viii CONTENTS

6.4 Multiplicity of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

A Useful theorems 61

A.1 Neumann boundary value problems . . . . . . . . . . . . . . . . . . . . . . . 61

A.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

A.3 The theorem of Borsuk-Ulam . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Bibliography 63

Index 65

List of Figures

4.1 Graph of J(x) = x3 − 3x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2 Graph of J(x, y) = (x2 + y2)2 − 2(x2 + y2) . . . . . . . . . . . . . . . . . . . 26

4.3 Graph of J(x) = xe1−x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

ix

x LIST OF FIGURES

Chapter 1

Introduction

The Schrödinger-Poisson equation consists of a nonlinear coupling of the Schrödingerequation with a gravitational potential of newtonian form, representing the interaction of aparticle with its own gravitational eld.

In 1998, Benci and Fortunato (1998) treated a similar problem, where the coupling waswith Maxwell's equations and represented the interaction of the particle with its own elec-tromagnetic eld. This gives rise to a coupled system of PDEs. In their paper the authorsstudied the problem with Dirichlet boundary conditions on the unknowns u and φ andemployed variational methods and critical point theory to develop a procedure that wouldbecome standard to treat other similar problems.

Later, Pisani and Siciliano (2013) treated a Schrödinger-Poisson system with Neumannboundary conditions on the scalar eld φ and considering the case in which the interactionfactor responsible for the coupling of the equations (the q below) is non-constant. This givesrise to important and interesting considerations regarding the geometry of the manifold ofpossible solutions.

In this dissertation we treat a modication of the problem dealt with by Pisani and Sicil-iano consisting in the addition of a biharmonic term in the second equation and impositionof appropriate boundary conditions. This can be interpreted as a coupling of Schrödingerequation with Bopp-Podolsky electrodynamics (for more information on this matter, seed'Avenia and Siciliano (2019) and the references therein), although here we focus on themathematical aspects of the problem.

The aim is to study the following system of partial dierential equations in a connected,bounded, smooth open set Ω ⊂ R3:

−∆u+ qφu− κ|u|p−2u = ωu in Ω (1.1)

∆2φ−∆φ = qu2 in Ω (1.2)

where ω ∈ R. We assume the following boundary conditions:

u = 0 on ∂Ω (1.3)∂φ

∂n= h1 on ∂Ω (1.4)

∂∆φ

∂n= h2 on ∂Ω (1.5)

with h1, h2 continuous. The symbol n denotes the unit vector normal to ∂Ω pointing out-wards. Since u represents the amplitude of the wave function of a particle conned in Ω, we

1

2 INTRODUCTION 1.0

assume the following normalizing condition:∫Ω

u2 = 1. (1.6)

We also assume that the coupling factor q is continuous on Ω:

q ∈ C(Ω). (1.7)

Our main theorem is the following:

Theorem 1.0.1. Let

α :=

∫∂Ω

h2 ds−∫∂Ω

h1 ds.

Assume that infΩ q < α < supΩ q and that |q−1(α)| = 0. Then there exists a solution(u, ω, φ) ∈ H1

0 (Ω)×R×H2 (Ω) such that u ≥ 0. Moreover, there exists innitely manysolutions (un, ωn, φn) ∈ H1

0 (Ω)×R×H2 (Ω) to the problem (1.1) - (1.6), with∫Ω

|∇un|2 dx→ +∞.

The motivation for the denition of α and the importance of the hypothesis will be madeclear in Chapter 6.

In our computations, nothing important is lost if we drop the term κ|u|p−2u, so we makethis simplication. One only would have to take account of the appropriate values of p. SeeRemark 6.4.3.

It is now time to x some notation. In what follows, X, Y and eventually Z will denoteBanach spaces. Usually, U ⊂ X will be an open set. The symbol L(X, Y ) denotes the set ofcontinuous linear operators T : X −→ Y . The value of T at x ∈ X will be denoted by T [x].The letter c denotes a constant whose value can change from line to line.

Chapter 2

Calculus in Banach spaces

2.1 Derivatives

2.1.1 Basic notions

Recall from calculus in nite dimension the meaning of derivative of a function at somegiven point. The idea was to obtain for the given function a linear approximation for thefunction in some neighborhood of the point. In one dimension we had the angular coecientof the tangent line; in N dimensions we had the matrix of the linear transformation. Theconcept extends immediately to innitely many dimensions:

Denition 2.1.1. Let X, Y be Banach spaces and U ⊂ X be an open set. We say thatf : U −→ Y is Fréchet dierentiable(dierentiable for short) at the point u ∈ U if thereexists Lu ∈ L(X, Y ) such that

f(u+ h)− f(u) = Lu[h] +R(h),

with

lim||h||→0

||R(h)||||h||

= 0,

that is, R(h) = o(h)1. The map Lu is the dierential of f at u.

Remark 2.1.2. The denition of Fréchet dierentiablity depends only on the topology ofX and Y and not on the specic norms these spaces are endowed with.

This denition was rst introduced by Fréchet 2 in 1911.Of course, all generalizations must reduce to the original cases. It is easy to see that the

denition above reduces to the nite dimensional case if the spaces X and Y are taken tobe R or RN .

We remark that the dierential is unique. This is an immediate consequence of thefollowing

Lemma 2.1.3. Suppose L1, L2 ∈ L(X, Y ) are such that L1[h]−L2[h] = o(h). Then L1 = L2.

Proof. By hypothesis, we have that

||L1[h]− L2[h]||||h||

||h||→0−−−−→ 0.

1This is known as little-o notation.2Maurice Fréchet .

3

https://goo.gl/7wdZXM

4 CALCULUS IN BANACH SPACES 2.1

Now, suppose the Lemma to be false. Then there is some h∗ such that L1[h∗]− L2[h∗] 6= 0.Now, letting h = th∗ we have

limt→0

||L1[th∗]− L2[th∗]||||th∗||

= limt→0

t||L1[h∗]− L2[h∗]||t||h∗||

6= 0,

a contradiction.

The uniqueness of the dierential compels us to adopt the following

Notation. Let f : X −→ Y be a map dierentiable at u ∈ X. Then the dierential of f atu is the continuous linear transformation denoted by

f ′(u) or sometimes by df(u).

Remark 2.1.4. If f is dierentiable at u ∈ U then it is continuous at u. Indeed,

f(u+ h)− f(u) = f ′(u)[h] + o(h)h→0−−→ 0.

Denition 2.1.5. If a function f is dierentiable at all u ∈ U , we say that it is dierentiablein U . The map

f ′ : U −→ L(X, Y )u 7−→ f ′(u)

is the derivative of f . If f ′ is continuous, we say that f is of class C1.

Remark 2.1.6. Note the dierence between the concepts of dierential and derivative. Thedierential is a linear approximation for the function in the neighborhood of a given point,while the derivative is a map that associates to each point a linear map.

We now see some examples.

Example 2.1.7. f is constant =⇒ f ′(u) = 0 for all u ∈ U .

Example 2.1.8. If f ∈ L(X, Y ) then f ′(u) = f for all u ∈ U . Indeed, in this case

f(u+ h)− f(u) = f(h),

from which the claim immediately follows.

Example 2.1.9. Let b : X×Y −→ Z be a continuous bilinear form. Then b is dierentiableat any (u, v) ∈ X × Y and db(u, v) is the linear map

(h, k) 7→ b(u, k) + b(h, v)

Indeed, note that

b(u+ h, v + k)− b(u, v) = b(u, k) + b(h, v) + b(h, k)

but by continuity we have (see Brezis (2010), pp. 138)

||b(h, k)|| ≤ c||h|| ||k||

and of course

lim(h,k)→0

c||h|| ||k||||h||+ ||k||

= 0,

2.1 DERIVATIVES 5

from which the claim follows.

Example 2.1.10. Let b : X ×X −→ R be a continuous bilinear form. Dene

f : X −→ Ru 7−→ b(u, u)

Thenf ′(u) = b(u, · ) + b( · , u).

Of course, if b is symmetric then f ′(u) = 2b(u, · ). What if X is a Hilbert space andb(u, v) = (u|v), the scalar product? Then f(u) = ||u||2 and f ′(u) = 2(u| · ).

Example 2.1.11. Let Ω be a smooth open set in RN , A = A(x) be an N × N symmetricmatrix with aij ∈ L∞ (Ω) and let q ∈ L∞ (Ω). Then

J(u) =1

2

∫Ω

(A∇u) · ∇u dx+

∫Ω

qu2 dx

denes a functional on the Sobolev space H1 (Ω)3 whose Fréchet derivative is given by

J ′(u)[v] =

∫Ω

(A∇u) · ∇v dx+ 2

∫Ω

quv dx.

Indeed, note that J is the functional induced by the following bilinear form onH10 (Ω)×H1

0 (Ω):

b(u, v) =1

2

∫Ω

(A∇u) · ∇v dx+ 2

∫Ω

quv dx.

This completes the proof.

The usual rules of calculus remain true for the Fréchet derivative.

Proposition 2.1.12. Let f, g be dierentiable functions at u ∈ X and let a, b ∈ R. Thenaf + bg is dierentiable at u and

(af + bg)′(u) = af ′(u) + bg′(u).

Proof. Note that

(af + bg)(u+ h)− (af + bg)(u)− af ′(u)[h]− bg′(u)[h] = o(h),

from which the claim immediately follows.

Proposition 2.1.13 (Chain Rule). Let X, Y and Z be Banach spaces, u ∈ X and f : X −→Y , g : Y −→ Z such that f is dierentiable at u and g is dierentiable at v = f(u). Then

g f : X −→ Z

is dierentiable at u and(g f)′(u) = g′(f(u))f ′(u).

3See Brezis (2010).


Proof. We have

f(u+ h) = f(u) + f ′(u)[h] + o(h),

g(v + k) = g(v) + g′(v)[k] + o(k).

Then

g(f(u+ h))− g(f(u)) = g(f(u) + f ′(u)[h] + o(h))− g(f(u))

= g′(f(u))[f ′(u)[h] + o(h)

]+ o(f ′(u)[h] + o(h))

= g′(f(u))[f ′(u)[h]

]+ g′(f(u))[o(h)] + o(f ′(u)[h] + o(h)).

We must show that the two last terms are o(h). That

g′(f(u))[o(h)]

||h||h→0−−→ 0

is clear by linearity and continuity of g′(f(u)). Now, note that

o(f ′(u)[h] + o(h))

||h||=o(f ′(u)[h] + o(h))

||h||||o(f ′(u)[h] + o(h))||||o(f ′(u)[h] + o(h))||

.

But

||o(f ′(u)[h] + o(h))||||h||

=

∥∥∥∥∥f ′(u)

[h

||h||

]+o(h)

||h||

∥∥∥∥∥≤ ||f ′(u)||+ o(h)

||h||≤ C

near h = 0. Since f ′(u)[h] + o(h)h→0−−→ 0, we have that

||o(f ′(u)[h] + o(h))||||h||

h→0−−→ 0,

which completes the proof.

Just like in the case of RN , there is a concept of directional derivative. It was introducedby Gateaux4 in 1913.

Denition 2.1.14. Let f : X −→ Y . Then f is said to be Gateaux-dierentiable (G-dierentiable for short) at u ∈ X if there exists A ∈ L(X, Y ) such that for all h ∈ X thereexists

limt→0+

f(u+ th)− f(u)

t:= A[h].

The map A is uniquely determined (this follows by the argument in Lemma 2.1.3) andwill be denoted by f ′G(u).

It is immediate that Fréchet dierentiability implies Gateaux dierentiability. However,the converse is not true: Gateaux dierentiability doesn't even imply continuity. Consider

4René Gateaux .

2.1 DERIVATIVES 7

the following example in R2:

F (x, y) =

(x2y

x4 + y2

)2

, y 6= 0,

F (x, 0) = 0.

Notice that F ((√y, y)) = 1/2 but F (0, 0) = 0.

We now prove the so-called Mean-Value Inequality. It plays a fundamental role in whatfollows.

Notation. Given u, v ∈ U the segment between them is denoted by

[u, v] :=tu+ (1− t)v : t ∈ [0, 1]

.

Theorem 2.1.15 (Mean Value Inequality). Let f : U −→ Y be G-dierentiable at any pointof the open set U ⊂ X. Given u, v such that [u, v] ⊂ U it holds that

||f(u)− f(v)|| ≤ supw∈[u,v]

||f ′G(w)|| ||u− v||. (2.1)

Proof. If f(u) = f(v) the claim is trivial. Suppose then f(u) 6= f(v). By a corollary of theHahn-Banach theorem5 there exists ψ ∈ Y ∗ such that

||ψ|| = 1

andψ(f(u)− f(v)

)= ||f(u)− f(v)||.

For t ∈ [0, 1] letγ(t) = tu+ (1− t)v

and consider the map h : [0, 1] −→ R dened by

h(t) = ψ(f(γ(t)

).

Notice thatγ(t+ τ) = γ(t) + τ(u− v).

From this it follows that

h(t+ τ)− h(t)

τ= ψ

(f(γ(t) + τ(u− v))− f(γ(t))

τ

)Passing to the limit as τ → 0 we have

h′(t) = ψ(f ′G(γ(t))[u− v]

).

Now, by the classic Mean-Value Theorem we have that h(1)−h(0) = h′(θ) for some θ ∈ (0, 1).

5See Theorem 12.2 in Bachman and Narici (2000)


Now, note that

||f(u)− f(v)|| = h(1)− h(0)

= h′(θ)

= ψ(f ′G(θu+ (1− θv))[u− v]

)≤ ||ψ|| ||f ′G(θu+ (1− θ)v)|| ||u− v||

But recall that ||ψ|| = 1 and that θu+ (1− θ)v ∈ [u, v]. This completes the proof.

There is a well-known result in classic Calculus that says that if a function has continuouspartial derivatives at a point then it is dierentiable at the given point. We have an analogousrelation between the Gateaux and Fréchet derivatives.

Proposition 2.1.16. Let f : U −→ Y be a G-dierentiable function in U and suppose f ′Gto be continuous at u ∈ U . Then f is dierentiable at u and f ′(u) = f ′G(u).

Proof. The proposition gives us a hint to the proof: it suces to show that

R(h) = f(u+ h)− f(u)− f ′G(u)[h]

is o(h).Note that R(h) is G-dierentiable for h small enough:

R(h+ tk)−R(h)

t=

1

t

(f(u+ h+ tk)− f ′G(u)[h+ tk]− f(u+ h) + f ′G(u)[h]

)=f(u+ h+ tk)− f(u+ h)

t− f ′G[k]

→ f ′G(u+ h)[k]− f ′G(u)[k].

(h must be small enough so that u+ h ∈ U).Note that R(0) = 0. We can then apply the Mean Value Inequality on the segment [0, h]

to obtain||R(h)|| ≤ sup

0≤t≤1||R′G(th)|| ||h||.

But, as we have seen, R′G(h) = f ′G(u+ h)− f ′G(u). Thus

||R(h)||||h||

≤ sup0≤t≤1

||f ′G(u+ th)− f ′G(u)|| h→0−−→ 0,

since f ′G is continuous. This completes the proof.

Remark 2.1.17. Computing Fréchet derivatives may be dicult, but Gateaux derivativesare more treatable. In view of this, Proposition 2.1.16 provides a method to nd Fréchetderivatives: one computes the Gateaux derivative and then show that it is continuous.

The next example will be useful afterwards.

Example 2.1.18. Let Ω be a smooth, bounded open set and q ∈ C(Ω). The map

u ∈ L6 (Ω) 7→ qu2 ∈ L6/5 (Ω)

2.1 DERIVATIVES 9

is of class C1. Indeed:

limt→0+

f(u+ th)− f(u)

t= lim

t→0+uh+ th2 = uh.

Hence f ′G(u)[h] = uh. Let us show that f ′G is continuous with respect to u. Let un → u inL6 (Ω). Then

||f ′G(u)− f ′G(un)|| = sup||h||6=1

||(u− un)h||6/5 ≤ ||u− un||6/4 → 0

by Hölder's inequality, which completes the proof.

We close this section with an example that will be important in Section 2.1.3.

Example 2.1.19. Let f ∈ C([a, b], Y ) and let6

F (t) =

∫ t

a

f(ξ) dξ.

Then F is dierentiable and F ′(t) = f(t). Indeed, we have

F (t+ h)− F (t)

h=

1

h

∫ t+h

t

f(ξ) dξ.

But ∫ t+h

t

f(t) dξ = hf(t).

Then we have

F (t+ h)− F (t)

h− f(t) =

1

h

[∫ t+h

t

f(ξ) dξ −∫ t+h

t

f(t) dξ

]

=1

h

∫ t+h

t

(f(ξ)− f(t)) dξ.

Hence, ∥∥∥∥F (t+ h)− F (t)

h− f(t)

∥∥∥∥ ≤ 1

|h||h| sup

[t,t+h]

||f(ξ)− f(t)|| → 0.


2.1.2 Higher order derivatives

Denition 2.1.20. Let f ∈ C(U, Y ) (where, as always, U ⊂ X is an open subset) be adierentiable function. If the map

f ′ : U −→ L(X, Y )

is dierentiable at u ∈ U then f is said to be twice Fréchet dierentiable at u. The seconddierential of f at u is

f ′′(u) = (f ′)′(u).

6The theory of integration works just like for functions into RN . See Lang (1997) for details.


Higher order dierentials are dened in an analogous manner:

f (k)(u) = (f (k−1))′(u).

Note thatf ′′(u) ∈ L(X,L(X, Y )).

Indeed, f ′ takes values in L(X, Y ).It is useful, however, to see f ′′(u) as a bilinear map. Recall that we can identify L(X,L(X, Y ))

with L2(X, Y ), the space of continuous bilinear forms from X × X to Y . Indeed, givenA ∈ L(X,L(X, Y )) let ΦA be the bilinear form

ΦA(u1, u2) = [A(u1)](u2).

Of course ΦA is continuous, since both A and A(u1) are continuous. Conversely, any Φ ∈L2(X, Y ) denes a map

φ : X −→ L(X, Y )h 7−→ Φ(h, · )

.

Note that we have just dened an isomorphism between L(X,L(X, Y )) and L2(X, Y ). More-over,

||φ||L(X,L(X,Y )) = sup||h||≤1

||Φ(h)||L(X,Y )

= sup||h||≤1

sup||k||≤1

||Φ(h, k)||

= ||Φ||L2(X,Y )

and thus we actually have an isometry.From now on, f ′′(u) will be viewed as the bilinear form just dened. Its value at a pair

(h, k) will be denotedf ′′(u)[h, k].

Denition 2.1.21. If f is twice dierentiable at every u ∈ U we say that f ′′ : U −→L2(X, Y ) is the second derivative of f . Higher order derivatives are dened in an analogousmanner.

Proposition 2.1.22. Let f ∈ C2(U, Y ). Then f ′′(u) is symmetric for all u ∈ U .

Proof. Let ϕ ∈ Y ∗. Dene

g : (t, s) ∈ R2 7→ ϕ(f(u+ tv + sw)) ∈ R .

By the Chain Rule we have that

∂g

∂t(t, s) = ϕ(f ′(u+ tv + sw)[v]),

∂

∂s

(∂g

∂t

)(t, s) = ϕ(f ′′(u+ tv + sw)[v, w]).

Analogously,∂

∂t

(∂g

∂s

)(t, s) = ϕ(f ′′(u+ tv + sw)[w, v]).

2.1 DERIVATIVES 11

Now, since g is of class C2 (it is a composition of C2 maps), it follows that

ϕ(f ′′(u+ tv + sw)[v, w]) = ϕ(f ′′(u+ tv + sw)[w, v]).

The equality holds for all ϕ ∈ Y ∗, thus

f ′′(u+ tv + sw)[v, w] = f ′′(u+ tv + sw)[w, v]

for all t, s. In particular, for t = s = 0 we get

f ′′(u)[v, w] = f ′′(u)[w, v],

and the proof is complete.

An analogous theorem holds for higher order derivatives. See Lang (1997) for details.Just like in classic Calculus, second derivatives are very useful for the study of extrema

of functionals. For an extensive discussion, see Blanchard and Brüning (1992).If f (k) : U −→ Y is a continuous map we say that f ∈ Ck(U, Y ).We can now generalize Taylor's formula.

Theorem 2.1.23 (Taylor's formula). Let f : U −→ Y be a map of class Cn dened in anopen, convex neighborhood of u. Then

f(u+ h) = f(u) +n−1∑j=1

1

j!f (j)(u)[h]j +Rn (2.2)

with

Rn =

∫ 1

0

(1− t)n−1

(n− 1)!f (n)(u+ th)hn dt.

Here, [h]j = [h, . . . , h] j times.

Proof. For t ∈ [0, 1] and u, h ∈ U let γ(t) = u+ th and

φ : [0, 1] −→ Yt 7−→ f(γ(t))

.

Then φ is of class Cn and

φ′(t) = f ′(u+ th)[h],

φ′′(t) = f ′′(u+ th)[h]2,

...

φ(n)(t) = f (n)(u+ th)[h]n.

But then

φ(1) = φ(0) + φ′(0) + . . .+1

(n− 1)!φ(n)(0) +

1

(n− 1)!

∫ 1

0

(1− t)n−1φ(n)(t) dt,

and hence the formula follows.

For more details on higher order derivatives, see Lang (1997) and Ambrosetti and Prodi(1993).


2.1.3 Partial derivatives

The formulation of partial derivatives is as simple as in classic Calculus.Consider a product of Banach spaces:

X = X1 × . . .×Xn

and let Ui ⊂ Xi be open subset. Denote U = U1 × . . .× Un and let

f : U −→ Y

be a map. We can write elements u ∈ U in terms of their coordinates: u = (u1, . . . , un).

Denition 2.1.24. Let u1, . . . , ui−1, ui+1, . . . un be xed. If the map

ui 7→ f(u1, . . . , un)

is dierentiable in U then its derivative is the partial derivative of f in the i-th coordinate,and its value at u ∈ U is denoted by Dif(u) or f ′ui(u).

Note that for each u ∈ U we have

Dif(u) ∈ L(Xi, Y ).

Just as in Calculus in nite dimension, smoothness is equivalent to smoothness of thepartial derivatives:

Proposition 2.1.25. Let Ui ⊂ Xi be open sets and let f : U = ΠN1 Ui −→ Y . Then f is of

class Ck if and only if each partial derivative

Dif : U −→ L(Xi, Y )

is of class Ck−1. If this is the case, then

f ′(u)[h] =n∑1

Dif(u)[hi]

for h = (h1, . . . , hn) ∈ Πn1Xi

Proof. It suces to write a proof for n = 2. For greater n, the argument is exactly the same.First, suppose the partial derivatives exist and are continuous. Let (u, v) ∈ U and h =

(h1, h2). Then

f(u+ h1, v + h2)− f(u, v)

= f(u+ h1, v + h2)− f(u+ h1, v) + f(u+ h1, v)− f(u, v)

=

∫ 1

0

D2f(u+ h1, v + th2)[h2] dt+

∫ 1

0

D1f(u+ th1, v)[h1] dt.

Letψ(h1, th2) := D2f(u+ h1, v + th2)−D2f(u, v)

2.2 LOCAL INVERSION THEOREMS 13

The rst integral can be written as∫ 1

0

D2f(u+ h1, v + th2)[h2] dt =

∫ 1

0

D2f(u, v)[h2] dt+

∫ 1

0

ψ(h1, th2)[h2] dt

= D2f(u, v)[h2] +

∫ 1

0

ψ(h1, th2)[h2] dt.

Now, note that ∣∣∣∣∣∫ 1

0

ψ(h1, th2)[h2] dt

∣∣∣∣∣ ≤ supt∈[0,1]

||ψ(h1, th2)|| ||h2||

≤ ||h|| supt∈[0,1]

||ψ(h1, th2)||

= o(h)

since, by continuity,

ψ(h1, th2) = D2f(u+ h1, v + th2)−D2f(u, v)h→0−−→ 0.

Now letφ(th1, h2) = D1f(u+ th1, v)−D1f(u, v).

We have ∫ 1

0

D1f(u+ th1, v)[h1] dt =

∫ 1

0

D1f(u, v)[h1] dt+

∫ 1

0

φ(th1, h2) dt

= D1f(u, v)[h1] +

∫φ(th1, h2) dt.

Estimating the error as we have just done we conclude that

f(u+ h1, v + h2)− f(u, v) = D1f(u, v)[h1] +D2f(u, v)[h2] + o(h),

hence f is dierentiable and the derivative is given as above.To prove the converse statement one must only evaluate the derivative along the direc-

tions (h1, 0) and (0, h2).

2.2 Local Inversion Theorems

One of the most important theorems in classic Calculus is the Inverse Function Theorem,together with its cousin Implicit Function Theorem. In this chapter we will see how theygeneralize to innite-dimensional spaces.

We follow Ambrosetti and Prodi (1993) and Lang (1995).

2.2.1 The Inverse Function Theorem

Denition 2.2.1. Let X and Y be Banach spaces. A map A ∈ L(X, Y ) is said to beinvertible if there exists A−1 ∈ L(Y,X) such that

A−1 A = IX , A A−1 = IY .


We setInv(X, Y ) :=

A ∈ L(X, Y ) : A is invertible

.

In order to prove our next proposition we need a preliminary lemma:

Lemma 2.2.2. Let A ∈ L(X,X) such that ||A|| < 1. Then I−A is invertible and its inverseis given by the convergent series

(I − A)−1 =∞∑0

An

Proof. Note that ||AA|| ≤ ||A|| ||A||, thus ||An|| ≤ ||A||n. Then

||I + A+ . . .+ An|| ≤ 1 + ||A||+ . . .+ ||A||n

which converges, since ||A|| < 1 (it is the geometric series). Thus the series really converges.Now, note that

(I − A)(I + A+ . . .+ An) = I − An+1 = (I + A+ . . .+ An)(I − A)

and that An → 0. Taking the limit completes the proof.

The above lemma shows that there is a ball of radius one around the identity of invertibleoperators. Indeed, if ||I − A|| < 1 then A = I − (I − A) is invertible.

Proposition 2.2.3. If A ∈ Inv(X,X) then every T ∈ L(X,X) such that

||T − A|| < 1

||A−1||

is invertible. Hence, Inv(X,X) is a open subset of L(X,X).

Proof. Note that

||TA−1 − I|| = ||(T − A)A−1|| ≤ ||A−1|| ||T − A|| ≤ 1.

Thus TA−1 is invertible. Hence, T = TA−1A is invertible.

Proposition 2.2.4. The map

ϕ : A ∈ Inv(X,X) 7→ A−1 ∈ L(X,X)

is innitely dierentiable, with

ϕ′(A)[B] = −A−1 B A−1.

Proof. For B suciently small we have that

(A+B)−1 − A−1 = (A(I + A−1B))−1 − A−1

= [(I + A−1B)−1 − I]A−1.

By the above lemma, if B is so small that ||A−1B|| < 1 then

(I + A−1B)−1 = I − A−1B + (A−1B)2g(B)


where g is a convergent power series. Hence

(A+B)−1 − A−1 = −A−1BA−1 + (A−1B)2g(B)A−1.

Now, notice that the series is bounded, because it is convergent. Then the second term abovecan be estimated:

||(A−1B)2g(B)A−1|| ≤ C||B||2.

Thusϕ′(A)[B] = −A−1 B A−1.

That ϕ is innitely dierentiable follows from the fact that ϕ′ is a composition of inversesand continuous bilinear maps (the composition of linear operators).

Remark 2.2.5. By Proposition 2.2.3 the set of invertible linear maps is open, and thus thederivative above really makes sense.

Above we dened the set of invertible linear maps. Let us now treat general maps.

Denition 2.2.6. Let f ∈ C(X, Y ) be a continuous map and let u ∈ X. Then f is said tobe locally invertible at u if there exist neighborhoods U of u and V of f(u) and a mapf−1 : V −→ U such that

f f−1 = IV , f−1 f = IU .

We say that f ∈ Hom(U, V ).

Proposition 2.2.7. The following properties hold:

(i) (Transitivity) Let f ∈ C(X, Y ) be locally invertible at u ∈ X and g ∈ C(Y, Z) be locallyinvertible at v = f(u). Then g f is locally invertible at u.

(ii) (Stability) Let f be locally invertible at u. Then f is locally invertible at any point insome neighborhood of u.

Proof. (i) We have f ∈ Hom(U, V ′) and g ∈ Hom(V ′′,W ). Take V = V ′ ∩ V ′′. Then

g f : f−1(V ) −→ g(V )

is well dened and is clearly a bijection, thus is an invertible map.

(ii) We have f ∈ Hom(U, V ). Let u′ ∈ U and take some ball B around u′ contained in U .Then f |B : B −→ f(B) is a bijection, thus f is locally invertible at u′.

Let f be a locally invertible map at u0 ∈ U and suppose that f is dierentiable at u0

and that f−1 is dierentiable at v0 = f(u0). We have that

f−1 f(u) = u ∀u ∈ U, f f−1(v) = v ∀v ∈ V.

Dierentiating and applying the chain rule we get

(f−1)′(v0) f ′(u0) = IX , f ′(u0) (f−1)′(v0) = IY .

Hence, f ′(u0) ∈ Inv(X, Y ) and (f ′(u0))−1 = (f−1)′(v0).We have just proved that invertibility of maps implies the invertibility of derivatives.

Now, derivatives are linear maps, for which it is much easier to study invertibility. It would


then be interesting to try and obtain information about the invertibility of maps from theinvertibility of derivatives. This is our next theorem.

Theorem 2.2.8 (Inverse Function Theorem). Let f ∈ C1(X, Y ) and suppose f ′(u0) ∈Inv(X, Y ). Then f is locally invertible at u0 with C1 inverse. In other words, there existneighborhoods U of u0 and V of v0 = f(u0) such that

(i) f ∈ Hom(U, V ).

(ii) f−1 ∈ C1(V,X) and(f−1)′(v0) = f ′(u0)−1

(iii) If f ∈ Ck(X, Y ) then f−1 ∈ Ck(V,X).

Proof. We begin by making some simplications to our problem.If we prove the theorem for Af , where A is any invertible linear map, then the theorem

is proved for f . Indeed, we know that A−1 is locally invertible. Then, by the transitivityproperty, the map

A−1 [A f ] = f

is locally invertible. Choosing A = [f ′(0)]−1 we reduce the problem to proving the theoremfor F : X −→ X, F (u) = [f ′(0)]−1 f(u).

We consider the case where u0 = 0 and v0 = F (0) = 0. The general case follows imme-diately from this one by considering compositions with translations (which are, of course,locally invertible).

Let g(u) = u−F (u). Note that g ∈ C1(X,X) and that g′(0) = 0. By continuity, there issome ball of radius r > 0 such that for p ∈ Br(0) we have

||g′(p)|| < 1

2.

Now, by the Mean Value Inequality (Theorem 2.1.15) we have that

||g(p)− g(q)|| ≤ supw∈[p,q]

||g′(w)|| ||p− q||

≤ 1

2||p− q||

for all p, q ∈ Br(0). Hence, g is a contraction on Br(0). In particular,

||g(p)|| ≤ 1

2||p|| ∀p ∈ Br(0).

For v ∈ X, deneφv(u) = v − g(u).

Note that φv is a contraction in Br(0). Moreover, if for u ∈ Br(0) and v ∈ Br/2(0) we havethat

||φv(u)|| ≤ ||v||+ ||g(u)|| ≤ r

Then, if we choose v ∈ Br/2(0) then φv is a contraction that maps Br(0) into itself. Then,by the Banach Fixed Point Theorem it has a unique xed point u ∈ Br(0) such that

u = v − g(u).


Thereof,v = F (u).

Hence F is bijective from Br(0) into Br/2(0) and thus we can dene an inverse

F−1 : Br/2(0) −→ Br(0).

To show that F−1 is continuous let u = F−1(v) and w = F−1(z), that is,

u+ g(u) = v

w + g(w) = z

Then

||u− w|| ≤ ||v − z||+ ||g(u)− g(w)||

≤ ||v − z||+ 1

2||u− w||

and thus||F−1(v)− F−1(z)|| ≤ 2||v − z||.

We have thus proved thatF |U ∈ Hom(U, V )

where V = Br/2(0) and U = Br(0) ∩ F−1(V ). Hence (i) is proved.For (ii), suppose r small enough such that F ′(u) is invertible. Since F is dierentiable

at u, we may write

F (u)− F (w) = F ′(u)(u− w) + ||u− w||ψ(u− w)

where ψ is some function such that

limw→u

ψ(u− w) = 0.

Now, note that

||F−1(v)− F−1(z)− F ′(u)−1(v − z)|| = ||u− w − F ′(u)−1(F (u)− F (w))||= ||u− w − u+ w + F ′(u)−1||u− w||ψ(u− w)||≤ C||v − z|| ||ψ(F−1(v)− F−1(z))||→ 0.

Hence, F−1 is dierentiable and

(F−1)′(v) = F ′(u)−1.

To see that (F−1)′ : v 7→ F ′(F−1(v))−1 is continuous, note that it is a composition ofcontinuous maps:

vF−1

−−→ uF ′−→ F ′(u)

ϕ−→ (F ′(u))−1.

Item (iii) follows by induction, using the same arguments presented above and notingthat ϕ ∈ C∞.

Remark 2.2.9. It can be show that the continuity of the derivative cannot be dropped in


the theorem. Indeed, consider the following elementary counterexample:

f(x) =

x+ 2x2 sin(1/x), x 6= 0

0, x = 0

We have that f ′(x) = 1 + 4x sin(x) − 2 cos(1/x), so the limit limx→0 f′(x) does not exist,

although f ′(0) = 1. Hence it is discontinuous at x = 0. Moreover, it oscillates up and down,so it is not invertible around the origin.

2.2.2 The Implicit Function Theorem

Suppose we know a given point (u0, v0) is a solution to the equation

f(u, v) = 0.

It may be of interest to nd a path of solutions near the given point, that is, to nd amap u 7→ v(u) such that v(u0) = v0 and f(u, v(u)) = 0. In other words, we want to nd afunction v = v(u) implicitly dened by f . But does such a function always exist? If so, whatsmoothness properties does it have? Does it depend on f? How so?

The Implicit Function Theorem answers all these questions.

Theorem 2.2.10 (Implicit Function Theorem). Let U ⊂ X × Y be an open neighborhoodof (u0, v0) and f : U −→ Z, where X, Y, Z are Banach spaces. Suppose that

f(u0, v0) = 0.

Also, assume that fv(u0, v0) exists and is bijective and that both f and fv are continuous at(u0, v0). Then

(i) There are r0, r > 0 such that for every u ∈ Br0(u0) there is exactly one v(u) ∈ Br(v0) ⊂Y and f(u, v(u)) = 0.

(ii) The sequencevn+1 = vn − f ′v(u0, v0)−1f(u, vn(u))

converges to v(u) for u ∈ Br0(u0).

(iii) If f is continuous in a neighborhood of (u0, v0) then v( · ) is continuous in a neighbor-hood of u0.

(iv) If f is of class Ck then v( · ) is of class Ck.

Proof. Without loss of generality, we can suppose (u0, v0) = 0 (it suces to compose withtranslations).

Setg(u, v) := f ′v(0, 0)[v]− f(u, v).

Then the equation f(u, v) = 0 is equivalent to

v = f ′v(0, 0)−1[g(u, v)] =: Tu(v)

Here u is an index to the function T , not a partial derivative. We must show that thereexists a unique solution to this equation.


Let ||u||, ||v||, ||w|| ≤ r. Note that

g′v(u, v) = f ′v(0, 0)− f ′v(u, v),

hence g′v is continuous at (0, 0) (because f ′v is) and g′v(0, 0) = 0. Then Taylor's Theoremimplies that

||g(u, v)− g(u,w)|| ≤ sup0<t<1

||g′v(u, t(v − w)|| ||v − w|| r→0−−→ 0.

Moreover, since g is continuous at (0, 0) and g(0, 0) = 0,

||g(u, v)|| r→0−−→ 0.

Let M =v ∈ Y : ||v|| ≤ r

. Then if r, r0 > 0 are small enough we have that

||Tuv|| ≤ r, ||Tu(v)− Tu(w)|| ≤ 1

2||v − w||

for all v, w ∈M and xed u ∈ X such that ||u|| ≤ r0.It then follows that Tu : M −→ M is a contraction. Thus for each xed u there is a

unique xed point, v = v(u), that is the solution to the equation f(u, v) = 0. This proves(i).

Item (ii) follows from the construction of the xed point in the proof of the BanachFixed Point Theorem.

The continuity of v follows from the continuity of the xed point on a parameter (seeZeidler (1986), Proposition 1.2). This proves (iii).

It remains to prove (iv). First, let k = 1. We will compute the Gateaux derivative of vand show that it is continuous. From the existence and continuity of v(u) we have that

f(u+ th, v(u+ th)) = f(u, v).

But since f is dierentiable we have

0 = f(u+ th, v(u+ th))− f(u, v(u))

= f ′u(u, v(u))[th] + f ′v(u, v(u))[v(u+ th)− v(u)] + o(t).

Thenv(u+ th)− v(u) = −tf ′v(u, v(u))−1f ′u(u, v(u))[h] + o(t)

and hence the Gateaux derivative of v exists:

v′G(u)[h] = −f ′v(u, v(u))−1f ′u(u, v(u))[h]

Note that f ′v(u, v(u)) is invertible in a neighborhood of the origin by Proposition 2.2.3. Sincef is of class C1 then both f ′u and f ′v are continuous, then vG! is continuous. Thus followsthat v is of class C1.

Let k = 2. Note that

G(u) = f ′v(u, v(u))v′(u)[h] + f ′u(u, v(u))[h] = 0

for all x in a neighborhood of the origin. Therefore G(u+ tk) = 0 for small t. Repeating theargument above we can show that the Gateaux derivative of v exists and is continuous.


For higher k, the argument is analogous. The key fact in the argument is the existenceof fv(u, u(v))−1.

The proof is now complete.

Remark 2.2.11. Note that the proof above gives us an expression for v′(u):

v′(u) = −f ′v(u, v(u))−1f ′u(u, v(u)). (2.3)

Chapter 3

Genus Theory

In this chapter we study a topological tool that is of fundamental importance in theproof of our main theorem: the Krasnoselskii genus.

It was rst introduced, as the reader may guess, by Krasnoselskii in the study of nonlinearintegral equations. The original denition can be found, for example, in Krasnoselskii (1964).We use an equivalent denition by Coman (1969). The proof of the equivalence can be foundin Rabinowitz (1973). For the properties of the genus we follow Kavian (1993). Anotherinteresting reference is Ambrosetti and Malchiodi (2007), which presents some properties ofthe genus and a general framework for the application of the theory to the existence andmultiplicity of critical points of even unbounded functionals.

3.1 Denitions

We begin by dening the domain of genus theory.

Denition 3.1.1. Let X be a Banach space. Then s(X) denotes the family of closed,symmetric subsets of X that do not contain the origin:

s(X) :=A ⊂ X : A = A,A = −A, 0 /∈ A

We now dene the genus of a set in s(X).

Denition 3.1.2. For A ∈ s(X), the genus of A is the least integer n such that there exista continuous, odd function between A and Rn \0:

γ(A) = infn ≥ 1 : ∃ϕ : A −→ Rn \0 continuous and odd

.

For convenience, we set γ(∅) = 0. If no such function exists, we set γ(A) =∞.

Example 3.1.3. Let x0 ∈ X andR < ||x0||. Set A = BR(x0)∪BR(−x0). Of course A ∈ s(X).Then we can compute γ(A) = 1. Indeed, it suces to take ϕ : A −→ R with ϕ(x) = 1 ifx ∈ BR(x0) and ϕ(x) = −1 otherwise.

We now present an important class of sets of genus N .

Example 3.1.4. Let X = RN and let Ω be an open, symmetric set containing the origin.Then γ(∂Ω) = N .

It is immediately seen that ∂Ω ∈ s(RN). Of course we have that γ(∂Ω) ≤ N , sincethe identity is a trivial odd, continuous function from ∂Ω onto RN \0. Now, suppose

21

22 GENUS THEORY 3.2

γ(∂Ω) ≤ N − 1. Then there exists and odd, continuous function ϕ : ∂Ω −→ RN−1 \0.Then, by the theorem of Borsuk-Ulam (see Kavian (1993), or the Appendix) there is somex ∈ ∂Ω such that ϕ(x) = 0, a contradiction. Hence γ(∂Ω) = N .

3.2 Properties of the genus

We now present some of the most useful properties of the genus.

Theorem 3.2.1. Let X be a Banach space and let A,B ∈ s(X).

(i) If there exist and odd, continuous function f : A −→ B then γ(A) ≤ γ(B).

(ii) If A ⊂ B then γ(A) ≤ γ(B).

(iii) If there is and odd homeomorphism f : A −→ B then γ(A) = γ(B).

(iv) γ(A ∪B) ≤ γ(A) + γ(B).

(v) If A is compact then γ(A) <∞.

(vi) If A is compact, it has a closed neighborhood of the same genus. More precisely, thereexists an ε > 0 such that if Aε =

x ∈ X : d(x,A) ≤ ε

then γ(Aε) = γ(A).

(vii) If γ(B) <∞ thenγ(A \B) ≥ γ(A)− γ(B).

Proof. (i) If γ(B) = ∞ there is nothing to show. Assume then that γ(B) = n < ∞. Thenthere exists and odd, continuous function ϕ : B −→ Rn \0. Now, the composition

ϕ f : A −→ Rn \0

is an odd, continuous function. Hence γ(A) ≤ n.

(ii) The canonical inclusion is an odd, continuous function from A to B, hence the resultfollows from (i).

(iii) By (i) we have that γ(A) ≤ γ(B) and also the converse inequality, hence γ(A) = γ(B).

(iv) If the genus of either A or B is ∞, then there is nothing to prove. Suppose then thatγ(A) = m <∞, γ(B) = n <∞. Then there exist two odd, continuous functions

ϕ : A −→ Rm \0

andψ : B −→ Rn \0.

By Tietze's theorem we can extend both functions to odd, continuous functions denedin X:

ϕ : X −→ Rm odd, continuous such that ϕ|A = ϕ

andψ : X −→ Rn odd, continuous such that ψ|B = ψ.

Now, setf : A ∪B −→ Rm+n \0

x 7−→ (ϕ(x), ψ(x)).

3.2 PROPERTIES OF THE GENUS 23

Note that f is an odd, continuous function and that indeed f(x) 6= 0 for all x ∈ A∪B.Hence γ(A ∪B) ≤ m+ n.

(v) Let x ∈ A. Take R(x) = ||x||/2 and set

ω(x) = BR(x)(x) ∪BR(x)(−x),

A(x) = ω(x). We know from Example 3.1.3 that γ(A(x)) = 1. Now, of course thefamily

ω(x)

x∈A is an open cover of A, hence, by compactness, there is some nite

set of points x1, . . . , xn such that

A ⊂⋃

1≤i≤n

ω(xi)

and henceA ⊂

⋃1≤i≤n

A(xi).

By the subaditivity property of the genus ((iv)) we have that γ(A) ≤ n.

(vi) By (v) we know that γ(A) = n < ∞, so there exists an odd, continuous functionϕ : A −→ Rn \0. By Tietze's theorem, such a function admits an odd, continuousextension ϕ : X −→ Rn.

By the compactness of A, there exists an ε > 0 such that 0 /∈ ϕ(Aε). Indeed, supposethere is no such ε. Then there are sequences (xk) ⊂ X, (yk) ⊂ A such that

ϕ(xk) = 0, d(xk, yk) = ||xk − yk|| ≤1

k.

By the compactness of A, the sequence (yk) is convergent (up to a subsequence),yk → y ∈ A. Consequently, xk → y:

||xk − y|| ≤ ||xk − yk||+ ||yk − y|| → 0.

But then, since ϕ is continuous, ϕ(y) = ϕ(y) = 0, a contradiction. Consequently, forsome ε > 0 we have that ϕ(Aε) ⊂ Rn \0. Hence

n = γ(A) ≤ γ(Aε) ≤ n.

(vii) Of course we have that A ⊂ B ∪ A \B. Then

γ(A) ≤ γ(B) + γ(A \B).

Since γ(B) <∞,γ(A \B) ≥ γ(A)− γ(B),


24 GENUS THEORY 3.2

Chapter 4

Deformations

The present chapter is devoted to the Deformation Lemma. It basically says that thetopological structure of the sublevels of a functional change when we pass through a criticalvalue, or rather that such structure remains the same in a neighborhood of a regular value.

Such a change in topology is useful to characterize the critical levels, and thus providesus with a tool for nding critical points.

We present here a fairly general version, for functionals constrained on a manifold of codi-mension 1 and such that the map dening the constraint is of class C1,1

loc . It can be extendedin a number of ways, for example to the case of C1 manifolds of any nite codimension, seeBonnet (1993), or to the case of complete C1,1-Finsler manifolds, see Palais (1970).

After some motivations to look at the topology of sublevels when dealing with criticalpoints, we present the Palais-Smale condition, the construction of tangent pseudo-gradientvector elds, state and prove the Deformation Lemma and present some of its applications.We also present, briey, some generalizations that will be useful in Chapter 6.

Notation. In what follows, let X be a Banach Space and J : X −→ R a C1 functional. TheFréchet derivative of J is denoted by J ′. Given b ∈ R, J b denotes the sublevel x ∈ X; J(x) ≤b. The sets Jb and so on are dened analogously. Kb := u ∈ X; J(u) = b, J ′(u) = 0denotes the set of critical points of J at level b.

4.1 Motivation

The topology of the sublevels of a functional is intrinsically related to the existence ofcritical points for the given functional.

Example 4.1.1. Consider the real functional J(x) = x3 − 3x. It is easily seen that J ′(x) =3x2 − 3 and so the critical points are x = ±1, with critical values c1 := J(1) = −2 andc2 := J(1) = 2. Now, note that

• for a1 < c1 we have Ja1 = [−∞, J−1(a1)] (one connected component).

• for c1 < a2 < c2 we have Ja2 = [−∞, α] ∪ [β, γ] for some α < β < γ (two connectedcomponents).

• for a3 > c2 we have Ja3 = [−∞, J−1(a3)] (again, one connected component).

Note that for each ai there is some ε > 0 small enough so that Jai and Jai+ε have thesame topological structure, i.e. are homeomorphic, and so can be continuously deformedinto one another. For the values ci, however, it is impossible to nd such ε. Indeed, for small

25

26 DEFORMATIONS 4.1

Figure 4.1: Graph of J(x) = x3 − 3x

values of ε > 0 the sets J ci−ε and J ci+ε have a dierent number of connected componentsand thus are not homeomorphic.

However, a change in the number of connected components is not enough as a criterionfor the existence of critical points.

Example 4.1.2. Consider in R2 the functional J(x, y) = (x2 + y2)2 − 2(x2 + y2).

Figure 4.2: Graph of J(x, y) = (x2 + y2)2 − 2(x2 + y2)

It is clear that J has critical values c1 := −1 and c2 := 0. Now, note that

• for a1 < c1 we have Ja1 = ∅

• for c1 < a2 < c2 the sublevel is a ring, with the inner radius shrinking as a2 goes to c2

• for a3 > c2 the sublevel is some ball BR(0).

Now, in the last two cases the presence of a critical level did not cause a change in thenumber of connected sets of the sublevels. Nonetheless, there was a (more subtle) change oftopology: while Ja3 is simply connected, Ja2 is not.

Now, a change in the topology of sublevels, alone, is not enough to assert the existenceof critical points

4.2 PRELIMINARY DEFINITIONS 27

Figure 4.3: Graph of J(x) = xe1−x

Example 4.1.3. Consider the real functional J(x) = xe1−x (see Figure 4.3). We have thatJ ′(x) = (1 − x)e1−x, so J has a (maximum) critical point at x = 1, with critical valuec = 1. We also have that limx→+∞ J(x) = 0. Then, although 0 is not a critical level, for any0 < a < c the sublevel Ja has two connected components, while for a < 0 the sublevel Ja isan interval.

In the previous example, it is intuitive that some compactness is missing. To overcomesuch an issue, the notion of Palais-Smale condition is introduced.

4.2 Preliminary denitions

The formal denition of a constraint is as follows:

Denition 4.2.1. Let X be a Banach space. A constraint is a set

M = v ∈ X; G(v) = 0

where G ∈ C1(X,R) is such that G′(v) 6= 0 for all v ∈M .

It can be shown (see Ambrosetti and Malchiodi (2007)) that M is a submanifold ofcodimension 1 of X.

Denition 4.2.2. LetM be a constraint dened by the map G and let p ∈M . The tangentspace to M at the point p is the set

TpM := v ∈ X; G′(p)[v] = 0.

Note that the derivative of the function that denes the restriction is "orthogonal" tothe restriction at each point. Indeed, X = TpM ⊕ 〈G′(p)〉.

Denition 4.2.3. Let J ∈ C1(X,R) be a functional and let M be a constraint dened byG. A constrained critical point is a point u ∈M such that

J ′(u)[v] = 0 ∀v ∈ TuM.

Equivalently, u is such thatJ ′(u) = λG′(u).

28 DEFORMATIONS 4.4

The real number λ is called Lagrange multiplier.

We provide more details on Lagrange Multipliers in the next chapter.Note that when dealing with functional restricted to manifolds the critical points are

those such that the derivative in the direction of the manifold (that is, in the tangent space)is identically zero.

4.3 The Palais-Smale condition

As we have seen in Example 4.1.3, some notion of compactness is necessary if we wish todeal with critical points via topology. The Palais-Smale condition is a notion that emulatesthe necessary compactness and allows us to construct the tangent pseudo-gradient vectoreld later on.

Denition 4.3.1. Let X be a Banach space, J ∈ C1(X,R) and M a constraint dened bythe map G. Then J |M satises the Palais-Smale condition if every sequence (un) such that

(J(un)) is bounded, J |′M(un)→ 0

has a converging subsequence. If J(un)→ b we say that J satises the Palais-Smale conditionat level b.

Remark 4.3.2. Let (un) be a sequence satisfying the Palais-Smale condition. Then thereis a sequence (λn) such that λn → λ, the Lagrange multiplier.

Remark 4.3.3. It is not strictly necessary that the functional be dened in all of X. Onemay ask for a functional dened in a neighborhood of M as well as in M only, if taking careof the denition of the derivative on manifolds.

Remark 4.3.4. With the notation as above, if J |M satises the Palais-Smale condition thenthe following sets are compact:

(u, λ) ∈M × R; J(u) = b, J ′(u) = λG′(u),Kb := u ∈M ; J(u) = b and ∃λ ∈ R such that J ′(u) = λG′(u),∪a≤b≤c Kb

for all a, c ∈ R.

4.4 Tangent pseudo-gradient vector elds

The idea of the proof of the Deformation Lemma in nite dimensions (or in Hilbertspaces in general) is to construct the deformations as ows in the direction of the gradient.Since in general Banach spaces there is no notion of gradient at our disposal, we emulatethe necessary properties with the notion of a pseudo-gradient vector eld. In the case ofconstrained functionals, we talk about tangent pseudo-gradient vector elds.

Notation. Let X be a Banach space and let M be a constraint dened by G. For J ∈C1(X,R) and u ∈M denote

||J ′(u)||∗ := sup|J ′(u)[y]|; y ∈ X, ||y|| = 1, G′(u)[y] = 0

.

4.4 TANGENT PSEUDO-GRADIENT VECTOR FIELDS 29

Note that ||J ′(u)||∗ is the norm of the projection of J ′(u) onto the hyperplane tangentto M at u. Also note that ||J ′(u)||∗ = 0 means precisely that J ′(u) = λG′(u), i.e., u is acritical points of J |M .

Notation. With the notation as above, denote by M the set of regular points of J |M , thatis,

M := u ∈M ; ∀λ ∈ R J ′(u)− λG′(u) 6= 0.

We now dene tangent pseudo-gradient vectors and vector elds.

Denition 4.4.1. Let X be a Banach space, J ∈ C1(X,R) and M a constraint dened byG. Then v is a tangent pseudo-gradient vector of J with respect to S at u if

||v(|| ≤ 2||J ′(u)||∗;J ′(u)[v] ≥ ||J ′(u)||2∗ if G′(u)[v] = 0.

A map V : M −→ X is a tangent pseudo-gradient vector eld of J if V is locally Lipschitzin M and V (u) is a tangent pseudo-gradient vector for every u ∈ M .

Remark 4.4.2. Let u ∈ S. It is immediately seen that the the set of tangent pseudo-gradient vectors at u is convex. It then follows that convex combinations of two or moretangent pseudo-gradient vector elds is also a tangent pseudo-gradient vector eld.

Tangent pseudo-gradient vector elds do not always exist, but under some conditions onthe constraint we can prove they exist. We require a little more than C1 regularity.

Notation. The set of C1 functionals such that the derivative is locally Lipschitz from Xonto X∗ is denoted by C1,1

loc (X,R).

Proposition 4.4.3. Let X be a Banach space, J ∈ C1(X,R), G ∈ C1,1loc (X,R) and M a

constraint dened by G. Suppose that J is not constant on M . Then there exists a map Vthat is locally Lipschitz in an open neighborhood NM of M and whose restriction to M is atangent pseudo-gradient vector eld. Moreover, if J and G are even then NM can be chosensymmetric with respect to the origin and V is odd.

Proof. Let u0 ∈ M . Then, by the denition of ||J(u0)||∗ there exists y0 ∈ X such that

||y0|| = 1, G′(u0)[y0] = 0, J ′(u0)[y0] ≥ 2

3||J ′(u0)||∗.

Set

v0 =5

3||J ′(u0)||∗y0.

Then we have ||v0|| =

5

3||J ′(u0)||∗

J ′(u0)[v0] ≥ 10

9||J ′(u0)||2∗, G′(u0)[v0] = 0.

Now, since u0 ∈M , then G′(u0) 6= 0 and so there exists z0 ∈ X such that ||z0|| = 1 and

G′(u0)[z0] ≥ 2

3||G′(u0)||.

Settingx0 :=

z0

G′(u0)[z0]

30 DEFORMATIONS 4.4

we have that

G′(u0)[x0] = 1, ||x0|| ||G′(u0)|| ≥ 3

2.

Now, follows form the continuity of G′ that there exists R1 > 0 such that for all u ∈ BR1(u0)we have

G′(u)[x0] ≥ 1

2, ||x0|| ||G′(u)|| ≤ 2.

Then, taking R < R1 small enough we have that

v := v(u, u0) = v0 −G′(u)[v0]

G′(u)[x0]x0

is a tangent pseudo-gradient vector of J at u for all u ∈ BR(u0)∩ M . Indeed, rst note that

‖v‖ ≤‖v0‖+ 2

∣∣G′(u)[v0]∣∣

G′(u0)[z0].

By continuity, there is R such that for some 0 < α < 2 we have

||v0|| ≤ α||J ′(u)||∗

and

|G′(u)[v0]| ≤ 2− α2

G′(u0)[z0]||J ′(u)||∗

from what follows that||v|| ≤ 2||J ′(u)||∗

Now, note that we can nd an R > 0 small enough such that for all u ∈ BR(u0) ∩ M wehave

J(u)[v] = J ′(u)[v0]− G′(u)[v0]

G′(u)[x0]J ′(u)[x0]

= J ′(u)[v0]− G′(u)[v0]

G′(u)[z0]J ′(u)[z0]

≥(

10

9− ε)||J ′(u)||2∗ −

|G′(u)[v0]||G′(u)[z0]|

||J ′(u)||∗

≥(

10

9− ε)||J ′(u)||2∗ −

(3

2||G′(u0)||+ ε′

)|G′(u)[v0]| ||J ′(u)||∗

≥(

10

9− ε)||J ′(u)||2∗ −

(3

2||G′(u0)||+ ε′

)|G′(u)[v0]| sup

BR(u0)

||J ′(u)||∗

≥ ||J ′(u)||2∗

Various R where considered, it suces to take the smallest one.Let us show now that

u 7→ v(u, u0)

is a locally Lipschitz map in BR(u0). Let K be some compact subset. Then, since G′ is locally

4.4 TANGENT PSEUDO-GRADIENT VECTOR FIELDS 31

Lipschitz, for u,w ∈ K we have:

∥∥v(u, u0)− v(w, u0)∥∥ =

∥∥∥∥G′(u)[v0]

G′(u)[x0]x0 −

G′(w)[v0]

G′(w)[x0]x0

∥∥∥∥≤ 1∣∣G′(u0)[z0]

∣∣∥∥∥∥G′(u)[v0]

G′(u)[x0]− G′(w)[v0]

G′(w)[x0]

∥∥∥∥= c

∥∥∥∥∥G′(u)[v0]

G′(u)[x0]− G′(w)[v0]

G′(u)[x0]−(G′(w)[v0]

G′(w)[x0]− G′(w)[v0]

G′(u)[x0]

)∥∥∥∥∥≤ c∥∥G′(u)−G′(w)

∥∥‖v0‖+

∥∥∥∥G′(w)[v0]

G′(w)[x0]− G′(w)[v0]

G′(u)[x0]

∥∥∥∥= c‖u− w‖+

∥∥∥∥G′(w)[v0]

G′(w)[x0]− G′(w)[v0]

G′(u)[x0]

∥∥∥∥Now, ∥∥∥∥G′(w)[v0]

G′(w)[x0]− G′(w)[v0]

G′(u)[x0]

∥∥∥∥ ≤ supK

∣∣G′(w)[v0]∣∣∥∥∥∥ 1

G′(w)[x0]− 1

G′(u)[x0]

∥∥∥∥≤ 4 sup

K

∣∣G′(w)[v0]∣∣∥∥G′(u)[x0]−G′(w)[x0]

∥∥≤ c‖u− v‖

and thus u 7→ v(u, u0) is locally Lipschitz.Since

NM :=⋃M

BR(u)

is paracompact (because it is a subset of a metric space) there is a locally nite renement(ωj)j∈I and a partition of unity (θj)j∈I subordinated to (ωj)j∈I . For each j ∈ I there is someu0j such that ωj ⊂ BR(u0j). Then

V (u) =∑j

= θj(u)v(u, u0j)

is a locally Lipschitz tangent pseudo-gradient vector eld for J over M .Suppose now that G and J are even. Then, taking R(u) as minR(u), R(−u) we have

that ⋃M

BR(u)

is symmetric with respect to the origin. Then

V1(u) =1

2(V (u)− V (−u))

is an odd tangent pseudo-gradient vector eld. It suces to show that V− := −V (−u) is atangent pseudo-gradient vector for J at u ∈ M . Note that, by the symmetry of G and Jthe construction of y0, z0 and x0 is the same for both u0 and −u0. Thus, if u ∈ BR(u0) and−u ∈ BR(−u0),

v(u, u0) = v(−u,−u0).

It then follows that V (−u) is a convex combination of tangent pseudo-gradient vectors at

32 DEFORMATIONS 4.5

u, and thus the claim follows.

Remark 4.4.4. Note that for all u ∈ BR(u0) we have

G′(u)[v] = G′(u)[v0]− G′(u)[v0]

G′(u)[x0]G′(u)[x0] = 0.

This implies that G′(u)[V (u)] = 0 for all u ∈ NM .

4.5 The Deformation Lemma and its consequences

We are now in position to prove the Deformation Lemma for constrained functionals.We remark that it would suce to consider a functional dened in a neighborhood of themanifold M .

Theorem 4.5.1 (Deformation Lemma). Let X be a Banach space, G ∈ C1,1loc (X,R) dening

the constraint M , J ∈ C1(X,R) such that J |M satises the Palais-Smale condition, J |M isnot constant and b ∈ R is not a critical value for J |M . Then there exists ε0 > 0 such thatfor all 0 < ε < ε0 there exists a map η ∈ C(R×M,M) such that

(i) η(0, u) = u for all u ∈M ;

(ii) If u /∈ J |Mb+ε0b−ε0 then η(t, u) = u for all t ∈ R;

(iii) For each t ∈ R η(t, · ) is a homeomorphism from M onto M ;

(iv) For all u ∈M the function t 7→ J |M(η(t, u)

)is decreasing;

(v) η(1, J |b+εM ) ⊂ J |b−εM ;

(vi) If J |M and G are even then η(t, · ) odd.

Proof. The idea is to build a ow as the solution of a convenient ordinary dierential equationand then to check that it veries conditions (i) to (vi).

We know by Proposition 4.4.3 that there exists a map V dened in an open neighborhoodNM of M , locally Lipschitz over NM such that V |M is a tangent pseudo-gradient vector eld.If J |M and G are even then NM is symmetric with respect to the origin.

Let us note that there exists ε1 > 0 and δ > 0 such that

||J |′M(u)||∗ ≥ δ ∀u ∈ J |Mb+ε1b−ε1 .

It is convenient to choose δ ≤ 1. Indeed, suppose that the assertion is not true. Then thereare sequences εn → 0, δn → 0 such that for all n ∈ N

∃un ∈ J |Mb−εnb−εn such that ||J |′M(un)|| < δn.

But then (un) is a Palais-Smale sequence at level b for J . Hence there is a convergingsubsequence, which implies, by continuity, that b is a critical level, a contradiction.

Dene

ε0 := min

ε1,

δ2

8

.

4.5 THE DEFORMATION LEMMA AND ITS CONSEQUENCES 33

For each v ∈ M set Rv = 12d(v,X \NM

). Dene

Ω :=⋃M

BRv(v).

Now, for 0 < ε < ε0 dene

A := J |Mb−ε0 ∪ J |Mb+ε0 ∪(X \ Ω

), B := J |Mb+ε

b−ε

and

α(u) =d(u,A)

d(u,A) + d(u,B).

Of course, α(B) = 1 and α(A) = 0. We also know that α is locally Lipschitz in X. Incase J |M and G are even then A and B are symmetric with respect to the origin and α iseven.

Let

W (u) :=

α(u) min

1, 1||V (u)||

V (u), u /∈ A

0 otherwise

Then W is locally Lipschitz and ||W (u)|| ≤ 1 for all u ∈ X. It then follows that givenu ∈ X there exists a unique solution to the problem

(∗)

d

dtη(t, u) = −W (η(t, u))

η(0, u) = u

dened on the whole real line. We know that η( · , u) ∈ C1(R, X), that η(t, η(s, u)) =η(t + s, u) and that η(t, · ) is an homeomorphism of X onto X for all t ∈ R. It remains toprove that η satises properties (i) to (vi).

Property (i) is trivial, by the initial condition of the problem (∗).Property (ii) is trivial, since for u /∈ J |Mb+ε0

b−ε0 the right hand side of the equation in (∗)is zero and thus the solution is constant.

To prove property (iii) it suces to show that if u ∈ M then η(t, u) ∈ M for all t ∈ R.Indeed, the inverse is given by η(−t, u) and continuity of both follows from continuity ofthe solution to (∗) with respect to initial conditions. The claim is trivial if u /∈ J |Mb+ε0

b−ε0 . Letu ∈ J |Mb+ε0

b−ε0 . Then, u ∈ M . Since t 7→ η(t, u) is continuous we have that η(t, u) ∈ NM for tsmall enough. But then we have that

d

dtG(η(t, u)) = G′(η(t, u))

[d

dtη(t, u)

]= −α(η(t, u)) min

1,

1

||V (η(t, u))||

G′(η(t, u))

[V (η(t, u))

]= 0.

Thus t 7→ G(η(t, u)) is constant, which implies that η(t, u) ∈M for all t ∈ R.

34 DEFORMATIONS 4.5

To prove (iv), note that

d

dtJ |M(η(t, u)) = J |′M(η(t, u))

[d

dtη(t, u)

]= −α(η(t, u)) min

1,

1

||V (η(t, u))||

J |′M(η(t, u))

[V (η(t, u))

]≤ −α(η(t, u)) min

1,

1

||V (η(t, u))||

||J |′M(η(t, u))||2∗

≤ 0.

For (v), since we have (iv) it suces to prove that J |M(η(1, u)) ≤ b− ε for u ∈ J |Mb+εb−ε.

First, note that||J |′M(u)||2∗ ≤ J |′M(u)[V (u)] ≤ ||J |′M(u)||∗ ||V (u)||,

thus ||J |′M(u)||∗ ≤ ||V (u)|| ≤ 2||J |′M(u)||∗. Then we have that

d

dtJ |M(η(t, u)) ≤ −α(η(t, u)) min

1,

1

||V (η(t, u))||

||J |′M(η(t, u))||2∗

≤ −1

4min

1,

1

||V (η(t, u))||

||V (η(t, u))||2

≤

−1

4if ||V (η(t, u))|| ≥ 1

−δ2

4if ||V (η(t, u))|| < 1

Integrating, we conclude that

J |M(η(1, u)) ≤ −δ2

4+ J |M(η(0, u)) ≤ −δ

2

4+ b+ ε ≤ b− ε.

At last, note that if J |M and G are even, then V , and consequently W , can be taken tobe odd. In this case, it is easy to see that −η(t, u) is a solution to the Cauchy problem (∗)with initial condition −η(0, u) = −u. By the uniqueness of the solution, η(t, · ) is odd forall t ∈ R.

Remark 4.5.2. There are plenty versions of the Deformation Lemma. One of them doesnot require that b be a regular value for J |M . In this case, the deformation is takes thecomplement of a neighborhood of the critical points to a lower sublevel. The proof is verysimilar, and the key ideas are the same. See Struwe (2008) for details.

We now analyze some important consequences of the Deformation Lemma.

Theorem 4.5.3 (Morse Theorem). Let X be a Banach space, M be a constraint dened byG ∈ C1,1

loc (X,R), J ∈ C1(X,R) such that J |M is not constant and b a regular value. Thenthere exists ε0 > 0 such that for all 0 < ε < ε0 the sublevel J |b−εM is a deformation retract ofJ |b+εM .

Proof. With the notation as in the proof of Theorem 4.5.1, dene

ϕ(t, u) = η

(4t

δ2(J |M(u)− b+ ε)+, u

)

4.5 THE DEFORMATION LEMMA AND ITS CONSEQUENCES 35

We show that ϕ denes a deformation of J b+ε into J b−ε. That is, ϕ ∈ C(

[0, 1]× J |b+εM , J |b−εM

)is such that

ϕ(0, · ) = Id, ϕ (t, · ) = Id ∀t ∈ [0, 1], ϕ(1, J |c+εM

)= J |c−εM .

Of course, ϕ(0, · ) = η(0, · ) = Id. If u ∈ J |b−εM then J |M(u) − b + ε ≤ −2ε < 0 soϕ(t, u) = η(0, u) = u as well.

Now, let u ∈ J |Mb+εb−ε and τ > 0 such that η(t, u) ∈ J |Mb+ε

b−ε for 0 ≤ t ≤ τ . Recall from theproof of Theorem 4.5.1 that

b− ε < J |M(η(τ, u)) ≤ J |M(u)− δ2

4τ,

that is,

τ <4

δ2(J |M(u)− b+ ε) =: t0.

Thenϕ(1, u) = η(t0, u) ∈ J |b−εM .

Indeed, note that

J |M(η(t0, u)) ≤ J |M(u)− δ2

4t0 = b− ε,


Theorem 4.5.4 (Minimax Principle). Let X be a Banach space, M be a constraint denedby G ∈ C1,1

loc (X,R), J ∈ C1(X,R) such that J |M is not constant and satises the Palais-Smale condition, and let B a nonempty family of nonempty subsets of M . Suppose that foreach b ∈ R and ε > 0 the ow η constructed in the proof of the Deformation Lemma is suchthat

η(1, A) ∈ B ∀A ∈ B.

Letb := inf

A∈Bsupv∈A

J |M(v).

Then if b ∈ R we have that b is a critical value of J |M .

Proof. Suppose b ∈ R and is not a critical point of J |M . By the denition of b, there is anε > 0 small enough such that for some A ∈ B we have

c ≤ supv∈A

J |M(v) ≤ b+ ε.

But then η(1, A) ∈ B ⊂ J |b−εM , a contradiction.

The families B of sets used in the Minimax Principle are dened with the help of theLjusternik-Schnirelman category (see Ambrosetti and Malchiodi (2007)) or of the genus. Forexample, we have the following theorem:

Theorem 4.5.5. Let X be a Banach space, G ∈ C1,1loc (X,R), M a constraint dened by G

and J ∈ C1(X,R). Suppose that G and J |M are even, that J |M is not constant and satisesthe Palais-Smale condition on M and that 0 /∈M . For k ∈ N, set

Bk =A ∈ s(X) : A ⊂M,γ(A) ≥ k

36 DEFORMATIONS 4.6

andbk := inf

A∈Bksupv∈A

J |M(v).

Then

(i) For all k ∈ N such that Bk 6= ∅ and bk ∈ R, bk is a critical value of J |M . Moreover,

bk ≤ bk+1

and if for some j ∈ N we have Bj+k 6= ∅ and bk = bj+k ∈ R then γ(Kbk) ≥ j + 1.

(ii) If for all k ∈ N we have Bk 6= ∅ and bk ∈ R, then

limn→∞

bk = +∞

4.6 Generalizations

As already mentioned, the theorems of this chapter can be modied to more generalsituations.

Of remarkable interest is the following theorem, which gives a Minimax Principle formanifolds that are only C1:

Theorem 4.6.1. Let M be a closed symmetric submanifold of a Banach space X such that0 /∈M . Let J ∈ C1(M,R) be even and bounded from below. Dene

bk = infA∈Bk

supv∈A

J(v)

where Bk =A ∈ s(X) : A ⊂M,γ(A) ≥ k,A is compact

. If Bk 6= ∅ for some k and if J

satises the Palais-Smale condition then J has at least k distinct pairs of critical points.

The proof relies heavily on the category of Ljusternik-Schnirelman and on the Finslerstructure of the manifold. See Szulkin (1988).

There is also a generalization of the category and of the genus and for the results presentedhere to variational problems that admit some kind of index theory. See Struwe (2008) andthe references therein.

Chapter 5

Submersions, Manifolds, Lagrange

Multipliers

In the proof of Theorem 1.0.1, we will consider the restriction of a functional to a subsetof our Hilbert space. The fact that this subset is a manifold has a fundamental role in thetheorems we will apply.

In this chapter, we recall some denitions and prove two important theorems: Ljusternik'stheorem on the construction of manifolds via submersions and the Theorem of LagrangeMultipliers.

5.1 Submersions and submanifolds

Denition 5.1.1. Let X be a Banach space. A subset A of X is a (smooth) submanifoldof X if for every u ∈ A there exists a neighborhood Uu of u in X and a (smooth) mapξu : Uu −→ Vu, where Vu is an open subset of X, and a closed complemented subspace S ofX such that

ξu(Uu ∩ A) = Vu ∩ S.

Denition 5.1.2. Let X, Y be two Banach spaces. A map G : X −→ Y is said to be asubmersion at u0 ∈ X if

(i) G is continuously dierentiable in a neighborhood of u0;

(ii) G′(u0) is surjective;

(iii) there exists a continuous projection operator P : X −→ kerG′(u0). Hence,

X = kerG′(u0)⊕ (I − P )(X).

Remark 5.1.3. If X is a Hilbert space and Y = Rk then condition (iii) is immediatelysatised and the mapG = (G1, . . . , Gk) is a submersion at u0 if and only ifG′1(u0), . . . , G′k(u0)are linearly independent.

Denition 5.1.4. Let M ⊂ X and u0 ∈M be xed.

(i) An admissible curve in M through u0 is a map u : t ∈ (−ε, ε) 7→ u(t) ∈ M such thatu(0) = u0 and the derivative u′(0) exists.

(ii) h is called a tangent vector toM at u0 if and only if there is an admissible curve suchthat h = u′(0).

37

38 SUBMERSIONS, MANIFOLDS, LAGRANGE MULTIPLIERS 5.1

(iii) If the set of tangent vectors form a vector space then this space is denoted by Tu0Mand called the tangent space at u0.

We have the following theorem for the construction of manifolds by means of submersions:

Theorem 5.1.5 (Ljusternik). Let X, Y be Banach spaces and let G : X −→ Y be a sub-mersion at u0 with G(u0) = 0. Let M = G−1(0). Then

(i) Tu0M = kerG′(u0).

(ii) There exists a homeomorphism ϕ : V ⊂ Tu0M −→ M of a neighborhood of 0 onto aneighborhood of u0 in M . Moreover, ϕ is dierentiable and ϕ′(0) = I.

(iii) If the domain D(G) of G is open and G is a submersion for all u ∈ M then M is aC1-manifold. If G is of class Ck them M is a Ck-manifold.

Proof. The proof relies heavily on the Implicit Function Theorem.Let

X = X1 ⊕X2

where X1 = kerG′(u0) = PX and X2 = (I − P )X. Thus

u = u1 + u2, u1 ∈ X1, u2 ∈ X2.

We assume X1 6= 0.We want to solve the equation

F (u1, u2) = G(u0 + u1 + u2) = 0

in terms of a function u2 = ψ(u1), as given by the Implicit Function Theorem, in a neigh-borhood of the origin in X1. Let us verify that the hypothesis of the theorem are satised.Suppose F dened on a neighborhood U of (0, 0).

First, note that F (0, 0) = G(u0) = 0.According to the Chain Rule, we have

Fui(u1, u2)[h] = G′(u0 + u1 + u2)[h], h ∈ Xi

on U . So both partial derivatives exist and are continuous, thus F ∈ C1(U).Now, note that Fu2(0, 0) is the restriction of G′(u0) to X2. It is a continuous linear

transformation, as a derivative. Also, it is bijective. Indeed, by hypothesis the range ofG′(u0) is Y , since it is a submersion. On the other hand, h ∈ X2 and G′(u0)[h] = 0 if andonly if h = 0, by the way we dened X2.

By the Implicit Function Theorem there are δ > 0 and a neighborhood W of (0, 0) inX1×X2 such that for each u1 ∈ Bδ(0) ⊂ X1 there is one ψ(u1) ∈ X2 such that (u1, ψ(u1)) ∈W and

F (u1, ψ(u1)) = 0

and ψ ∈ C1. Dierentiating with respect to u1 we have

Fu1(0, 0) + Fu2(0, 0)ψ′(0) = 0.

But Fu1(0, 0) = G′(u0)|X1 = 0, thus ψ′(0) = 0 = ψ(0). Then

ψ(u1) = o(u1).

5.2 SUBMERSIONS AND SUBMANIFOLDS 39

Now setϕ(u1) := u0 + u1 + ψ(u1)

This is an homeomorphism of a suciently small neighborhood of zero in X1 onto a neigh-borhood of u0 in M . Indeed:

• Injectivity follows from the injectivity of ψ.

• For each u in a neighborhood of u0 we have u = ϕ(P [u]), thus ϕ is surjective.

• Continuity follows from ψ(u1) = o(u1).

• The inverse is given by projection onto X1, hence is continuous.

We remark that the suciently small neighborhood refers to the fact that we need ψto exist, and it happens on a suciently small neighborhood of zero.

We are now in position to prove each assertion.

(i) Let h be a tangent vector. Then there is a curve u(t) inM with u(0) = u0 and u′(0) = h.Now, since u(t) lies in M we have that

G(u(t)) = 0 =⇒ G′(u(0))[u′(0)] = G′(u0)[h] = 0

=⇒ h ∈ kerG′(u0).

Conversely, if G′(u0)[h] = 0 then th is in the domain of ϕ and u(t) = ϕ(th) is anadmissible curve. Indeed, G(u(t)) = 0, u(0) = 0 and

u′(0) = ϕ′(0)[h] = I[h] = h.

We have thus proved that the tangent space at u0 is indeed a vector space, preciselythe kernel of G′(u0). Note that it is closed.

(ii) The construction is essentialy done. It remains to note that if G is of class Ck then Fis of class Ck and consequently ψ and ϕ are of class Ck.

(iii) Suppose G is a submersion at every point in M . Then, by what we have constructed,the tangent space exists and is closed, and every point has a homeomorphism betweena neighborhood of zero in the tangent space and a neighborhood of itself. Now, let u′0be another point in M with corresponding map ϕ′. We must show that the coordinatechanges are as regular as G. Let u ∈M . Then it has two representations:

u = ϕ(h), h ∈ Tu0M,

u = ϕ′(h′), h ∈ Tu′0M.

Now, by the denition of ϕ we have that

h = P (u− u0) = P (ϕ′(h′)− u0) = ϕ−1 ϕ′(h′).

Now, if G is of class Ck then ψ is of class Ck, by the inverse function theorem. Thenfollows that ϕ′ is of class Ck. Since P is linear, it is innitely dierentiable. Henceϕ−1 ϕ′ is of class Ck.

This completes the proof.


5.2 The Theorem of Lagrange Multipliers

Let J : X −→ R be a C1 functional.

Denition 5.2.1. The mapping

J ′(u)|TuM : h ∈ TuM 7→ J ′(u)[h] ∈ R

is the restriction of J ′(u) to the tangent space TuM . The point u ∈M is said to be a criticalpoint of J |M if

J |′M = J ′(u)|TuM = 0

Theorem 5.2.2. Let G : Ω ⊂ X −→ Y be a submersion at u0 ∈ M = G−1(0) and letf : X −→ Z be dierentiable at u0, where Z is another Banach space. Then the followingassertions are equivalent

(i) u0 is a critical point of f |M .

(ii) There exists Λu0 ∈ L(Y, Z) such that u0 is a critical point of

f − Λu0 G : X −→ Z,

that is,f ′(u0)[v]− Λu0(G′(u0)[v]) = 0 ∀v ∈ X.

Moreover, if one of the above holds true then the function Λu0 is unique, and it is called theLagrange Multiplier.

Proof. First we show that (i) =⇒ (ii). We have

X = kerG′(u0)⊕ (I − P )(X).

To simplify the notation, we denote

Q := I − P, L := (I − P )(X) = ImQ.

Note thatHu0

:= G′(u0)|L : L −→ Y

is a linear and continuous isomorphism. Dene

Ku0: v ∈ L = ImQ 7→ f ′(u0)[u] ∈ Z such that v = Qu

We remark that Ku0 is well dened: let v = Qu1 = Qu2. Then

u1 − u2 ∈ kerQ = kerG′(u0) ⊂ ker f ′(u0)

and thus f ′(u0)[u1] = f ′(u0)[u2].If v ∈ L we have that

v = Qu for some u ∈ X, u = Pu+Qu = Pu+ v.

5.2 THE THEOREM OF LAGRANGE MULTIPLIERS 41

Therefore,

Ku0(v) = f ′(u0)[Pu+ v]

= f ′(u0)[v]

and we conclude thatKu0 = f ′(u0)|L

is linear and continuous.Dene

Λu0 = Ku0 Hu0: Y −→ Z.

It is linear and continuous, as a composition of linear and continuous maps. Moreover, foreach v ∈ X we have that

Λu0(G′(u0)[v]) = Λu0(G′(u0)[Pu+Qu])

= Λu0(G′(u0)[Qu])

= Λu0(Hu0(Qu))

= Ku0(Qu)

= f ′(u0)[Qu]

= f ′(u0)[u].

We now prove that (ii) =⇒ (i). It suces to note that if h ∈ Tu0M = kerG′(u0) then

f ′(u0)[h] = Λu0(G′(u0)[h]) = 0.

Finally, we show that if (i) or (ii) holds then Λu0 is unique. Suppose that given u0 thereare Λu0 and Λu0 . Given y ∈ Y , there is u ∈ X such that

y = G′(u0)[u]

(since G′(u0) is surjective). Then

Λu0(y) = Λu0(G′(u0)[u])

= Λu0(G′(u0)[Qu])

= f ′(u0)[Qu]

= Λu0(y),


The next example will be useful afterwards.

Example 5.2.3. Suppose Y = R2 and Z = R. Then

G′(u0)[v] =

[G′1(u0)[v]G′2(u0)[v]

]

andΛu0 =

[λ1 λ2

],


hence

Λu0(G′(u0)[v]) =[λ1 λ2

] [G′1(u0)[v]G′2(u0)[v]

]= λ1G

′1(u0)[v] + λ2G

′2(u0)[v],

which completes the example.

Chapter 6

Proof of the main result

We are now in position to prove Theorem 1.0.1.

6.1 An auxiliary problem

Our aim is to build a functional whose critical points will be the weak solutions to theproblem. We want to deal with homogeneous boundary conditions. This will help us buildthe functional in a more convenient way.

Let

α =

∫∂Ω

h2 ds−∫∂Ω

h1 ds.

Consider the following auxiliary problem:

∆2χ−∆χ = α/|Ω| in Ω, (6.1)∂χ

∂n= h1 on ∂Ω, (6.2)

∂∆χ

∂n= h2 on ∂Ω. (6.3)∫

Ω

χ dx = 0. (6.4)

We make the following change of variables:

θ = ∆χ.

By Theorem A.1.1 we know that there is a unique weak solution to

∆θ − θ = α/|Ω| in Ω (6.5)∂θ

∂n= h2 on ∂Ω (6.6)

and the solution satises ∫Ω

θ dx =

∫∂Ω

h1 ds (6.7)

43

44 PROOF OF THE MAIN RESULT 6.1

by the way we have dened α. Then, by Theorem A.1.2 there is a unique weak solution to

∆χ = θ, in Ω (6.8)∂χ

∂n= h1 in ∂Ω (6.9)∫

Ω

χ dx = 0 (6.10)

and such solution is the solution to the auxiliary problem. Note that the solution is unique.Indeed, let χ1, χ2 be two solutions to (6.8)-(6.10). Then w = χ1 − χ2 satises

∆w = 0 in Ω,∂w

∂n= 0 on ∂Ω.

Then multiplying by w and integrating we have∫Ω

w∆w dx = 0 =⇒∫

Ω

|∇w|2 dx+

∫∂Ω

w∂w

∂nds = 0

hence ∇w = 0 and thus w is a constant. Hence, there is uniqueness up to a constant. Thenthe hypothesis

∫Ωχ dx = 0 implies uniqueness.

Setϕ = φ− χ− µ,

where

µ =1

|Ω|

∫Ω

φ dx.

With the new variables (u, ω, ϕ, µ) our problem becomes

−∆u+ q(χ+ ϕ)u = ωu− µqu in Ω, (6.11)

∆2ϕ−∆ϕ = qu2 − α/|Ω| in Ω, (6.12)

u = 0 on ∂Ω, (6.13)∫Ω

u2 dx = 1, (6.14)

∂ϕ

∂n= 0 on ∂Ω, (6.15)

∂∆ϕ

∂n= 0 on ∂Ω, (6.16)∫

Ω

ϕ dx = 0. (6.17)

Notice that the compatibility condition between (6.12) and (6.15) and (6.16) now readsas ∫

Ω

qu2 dx = α.

6.2 THE MANIFOLD M 45

Let

S :=

u ∈ H1

0 (Ω) :

∫Ω

u2 dx = 1

, (6.18)

N :=

u ∈ H1

0 (Ω) :

∫Ω

qu2 dx = α

, (6.19)

M := S ∩N. (6.20)

Notice that α depends on both the boundary conditions to the original problem.If the problem has a solution, then M 6= ∅. Hence,

qmin ≤ α ≤ qmax (6.21)

whereqmin = inf

Ωq and qmax = sup

Ωq.

Indeed, suppose α < qmin. Then

α =

∫Ω

qu2 dx ≥ qmin > α,

a contradiction. The case α > qmax is analogous.From (6.21) we deduce that q−1(α) is not empty, with its measure playing a major role.

Suppose α = qmin and |q−1(α)| = 0. Then∫Ω

qu2 dx =

∫Ω

χq−1(α)αu2 dx+

∫Ω

(1− χq−1(α))qu2 dx > α.

If α = qmax and |q−1(α)| = 0 we proceed in an analogous manner to conclude that M isempty and so the problem has no solutions. Therefore, we arrive at the following necessarycondition for the existence of solutions: either

qmin < α < qmax (6.22)

or|q−1(α)| 6= 0. (6.23)

6.2 The manifold M

We rst note that M is symmetric with respect to the origin: if u ∈ M then −u ∈ M .This follows trivially from the denition of M .

We also note thatM is weakly closed in H10 (Ω). Indeed, let (un)n ⊂M such that un u

in H10 (Ω). Since H1

0 (Ω) has a compact embedding into L2 (Ω) it follows that un convergesstrongly in L2 (Ω). Then by ||u − un||2 ≥

∣∣||u||2 − ||un||2∣∣ and since ||u − un|| → 0 we knowthat ∣∣||u||2 − ||un||2∣∣→ 0


and thus ||u||2 = 1 and u ∈ S. Now, note that∫Ω

qu2 dx−∫

Ω

qu2n dx =

∫Ω

q(u2 − u2n) dx→ 0

by the Dominated Convergence Theorem (note that u2 − u2n → 0 a.e. and that |u2 − u2

n| ≤u2 + u2

n ∈ L1 (Ω)).Now, we want to show that under condition (6.22) the set M is not empty. For this, we

introduce the following notation.Let A ⊂ Ω be an open subset,

SA :=

u ∈ H1

0 (A) :

∫A

u2 dx = 1

and

gA : u ∈ SA 7→∫A

qu2 dx ∈ R .

It is immediately seen that.

gA(SA) ⊂ [infAq, sup

Aq]. (6.24)

Lemma 6.2.1. The following inclusion holds:

(infAq, sup

Aq) ⊂ gA(SA). (6.25)

Proof. The idea is to take a number in (infA q, supA q) and show that it is the limit of asequence (gA(un)).

Let d ∈ (infA q, supA q) and y, z ∈ A such that

q(y) < d < q(z).

Since intervals on the real line are convex, there exists a λ ∈ (0, 1) such that

d = (1− λ)q(y) + λq(z).

From the existence of cuto functions we have that there exist two smooth functions v, wsupported on disjoint balls centered in y and z, respectively, and contained in A, such that∫

A

v2 dx = 1− λ,∫A

w2 dx = λ.

For n ∈ N denevn(x) = n3/2v(n(x− y) + y)

andwn(x) = n3/2w(n(x− z) + z).

Note that supp v1 = supp v and analogously for w,w1, and that the supports shrink as nincreases. Thus for every n ∈ N the functions vn, wn have disjoint supports. Also note that∫

A

v2n dx = 1− λ,

∫A

w2n dx = λ.


Deneun := vn + wn.

By the above, we have that un ∈ SA for all n ∈ N.Now, we note that

limn→∞

gA(vn) = limn→∞

∫A

q(x)n3v2(n(x− y) + y) dx

= limn→∞

∫A

q

(t− yn

+ y

)v2(t) dt

= (1− λ)q(y)

by the Dominated Convergence Theorem (note that all functions are bounded, as well asthe domain of integration). Analogously,

limn→∞

∫A

gA(wn) dx = λq(z).

It follows at once thatlimn→∞

gA(un) = d,

and so the proof is complete.

We can conclude:

Proposition 6.2.2. Let A ⊂ Ω be an open subset. If α ∈ (infA q, supA q) then there existsu ∈ H1

0 (A) such that ∫A

u2 dx = 1 and

∫A

qu2 dx = α.

Proof. From inclusions (6.24) and (6.25) we have that[infAq, sup

Aq

]= gA(SA).

Note that SA is connected. Indeed, consider the following map:

F : H10 (A) \ 0 −→ SA

u 7−→ u

||u||2.

Of course F is continuous and takes the connected set L2(A) \ 0 onto SA. Since the imageof a connected set via a continuous function is connected, the claim follows.

Moreover, since gA is continuous we have that

(infAq, sup

Aq) ⊂ gA(SA).

The claim is now immediate.

Theorem 6.2.3. Assume that infΩ q < α < supΩ q. Then M is not empty.

Proof. Take A = Ω and use the above proposition.

Note that M has subsets of arbitrarily large genus:


Theorem 6.2.4. Let u1, . . . , uk ∈M be functions with disjoint supports and let

Vk = 〈u1, . . . , uk〉

be the space spanned by u1, . . . , uk. Then M ∩ Vk is the (k − 1)-dimensional sphere, henceγ(M ∩ Vk) = k.

Proof. Since the supports are disjoint the functions u1, . . . , uk are all linearly independentand hence the space Vk has dimension k. Notice that S is the L2-sphere in H1

0 (Ω), so S ∩Vkis just the L2-sphere in Vk. Hence we need to show that

M ∩ Vk = S ∩ Vk.

Now, of courseM ∩ Vk ⊂ S ∩ Vk.

On the other hand, let u =∑k

1 αiui ∈ S ∩ Vk. Then

1 = ||u||22 =k∑1

α2i .

Therefore ∫Ω

qu2 dx =

∫Ω

qk∑1

α2iu

2i dx =

k∑1

α2i

∫Ω

qu2i dx = α

k∑1

α2i = α,

hence u ∈M ∩ Vk.

Now, it is natural if one raises the question of whether there exists such functions withdisjoint supports for arbitrary k. The answer is positive:

Theorem 6.2.5. If (6.22) holds then for every k ≥ 2 there exist k functions u1, . . . , uk ∈Mwith disjoint supports. Hence γ(M) = +∞.

Proof. By hypothesis, the sets

Ω+ :=x ∈ Ω : q(x) > α

, Ω− :=

x ∈ Ω : q(x) < α

are open and non-empty. There exists 2k disjoint balls Y1, . . . , Yk ⊂ Ω− and Z1, . . . , Zk ⊂ Ω+.Let

Ai = Yi ∪ Zi, i = 1, . . . , k.

Then, by construction,infAi

q < α < supAi

q.

Then, by Proposition 6.2.2 for each i there exists ui ∈ H10 (Ai) such that∫

Ω

u2i dx =

∫Ai

u2i dx = 1 and

∫Ω

qu2i dx =

∫Ai

qu2i dx = α.

All the functions ui have disjoint supports and belong to M .That γ(M) = +∞ follows immediately by the subaditivity property of the genus.


Let

G1 : u ∈ H10 (Ω) 7→

∫Ω

u2 dx− 1 ∈ R, (6.26)

G2 : u ∈ H10 (Ω) 7→

∫Ω

qu2 dx− α ∈ R (6.27)

andG = (G1, G2).

ThenM =

u ∈ H1

0 (Ω) : G1(u) = G2(u) = 0

= G−1(0).

We note that G is of class C1, for

G′1(u)[v] = 2

∫Ω

uv dx, v ∈ H10 (Ω) .

Note that G′1 is linear. Let us show that G′1 is continuous. Let (un) ⊂ H10 (Ω) be such that

un → u ∈ H10 (Ω). Then (un) also converges in L2 (Ω). Hence, for all v ∈ H1

0 (Ω) we have

||G′1(un)−G′1(u)|| = sup||v||

H10 (Ω)

=1

|G′1(un − u)[v]|

= sup||v||

H10 (Ω)

=1

∣∣∣∣∫Ω

(un − u)v dx

∣∣∣∣≤ sup||v||

H10 (Ω)

=1

||un − u||2||v||2

≤ c||un − u||2 → 0

By an analogous argument we see that G′2 is continuous as well. Thus, G1 and G2 are ofclass C1,1. Now, G′ is linear and continuous, because its components are continuous. HenceG is C1,1 as well.

If we show that G′1(u) and G′2(u) are linearly independent, then G will be a submersionand M a submanifold (of codimension 2).

Proposition 6.2.6. Assume M is not empty. The dierentials G′1(u) and G′2(u) are linearlyindependent for every u ∈M if and only if

|q−1(α)| = 0. (6.28)

Proof. First, assume (6.28). We will show that G′1(u) and G′2(u) are linearly independent,for all u ∈M . Suppose that there are a, b ∈ R such that

aG′1(u) + bG′2(u) = 0 in H−1 (Ω)

for some u ∈M . Evaluating this expression in u, we nd that a+ bα = 0. Then

aG′1(u)[v] + bG′2(u)[v] = b

(−α∫

Ω

uv dx+

∫Ω

quv dx

)= 0 ∀v ∈ H1

0 (Ω),

that is,

b

∫Ω

(q − α)uv dx = 0 ∀v ∈ H10 (Ω) .


If b 6= 0 then we would have (q − α)u = 0 a.e., and hence, in view of (6.28), u = 0, acontradiction. Thus G′1(u) and G′2(u) are linearly independent for all u ∈M .

Now, suppose (6.28) is not satised. Then q−1(α) has not empty interior, hence there issome test function u with support in q−1(α) such that ||u||2 = 1. It is immediately seen thatu ∈M because qu = αu and hence

G′2(u) = αG′1(u),


6.3 Existence of solutions

We now proceed to study the variational framework of the problem. Our aim is to con-struct a functional whose critical points will be the weak solutions of the problem.

Following Gazzola et al. (2010), let

V =

ξ ∈ H2 (Ω) :

∂ξ

∂n= 0 on ∂Ω

.

We remark that V is a closed subspace of H2 (Ω). Indeed, let (vn) ⊂ V such that vn → vin V. Then 0 = γ1(vn) → γ1(v) and hence γ1(v) = 0, where γ1 denotes the trace operatorwhich for smooth functions gives the directional derivative on the boundary. Being a closedsubspace, V inherits the Hilbert space structure of H2 (Ω).

Recall thatϕ = φ− χ− µ

where

µ =1

|Ω|

∫Ω

φ dx.

In this way, we have ϕ = 0. Consider then the following decomposition of V :

V = V ⊕ R (6.29)

whereV = η ∈ V : η = 0 .

On V we have the equivalent norm

||η||V =(||∇η||22 + ||∆η||22

)1/2.

Consider the functional F : H10 (Ω)×H2 (Ω) dened below:

F (u, ϕ) =1

2

∫Ω

|∇u|2 dx+1

2

∫Ω

q(ϕ+ χ)u2 dx− 1

4

∫Ω

(∆ϕ)2 dx

− 1

4

∫Ω

|∇ϕ|2 dx− α

2|Ω|

∫Ω

ϕ dx.

It is easy to see that this functional is of class C1 and that for every u ∈ H10 (Ω) and

6.3 EXISTENCE OF SOLUTIONS 51

ϕ ∈ H2 (Ω) we have

F ′u(u, ϕ)[v] =

∫Ω

∇u · ∇v dx+

∫Ω

q(ϕ+ χ)uv dx

F ′ϕ(u, ϕ)[ξ] =1

2

∫Ω

qξu2 dx− 1

2

∫Ω

∆ϕ∆ξ dx− 1

2

∫Ω

∇ϕ · ∇ξ dx− α

2|Ω|

∫Ω

ξ dx

for every v ∈ H10 (Ω) and ξ ∈ H2 (Ω).

Theorem 6.3.1. Let (u, ϕ) ∈ H10 (Ω)×H2 (Ω). Then there exist ω, µ ∈ R such that (u, ϕ, ω, µ)

is a solution to (6.11)-(6.17) if and only if (u, ϕ) is a critical point of F constrained onM×V ,in which case the real constants ω, µ are the two Lagrange multipliers with respect to F ′u.

Proof. Note that (u, ϕ, ω, µ) is a weak solution to (6.11)-(6.17) if (u, ϕ) ∈M × V and

∀v ∈ H10 (Ω) F ′u(u, ϕ)[v] = ω

∫Ω

uv dx− µ∫

Ω

quv dx,

∀ξ ∈ V F ′ϕ(u, ϕ)[ξ] = 0.

Now, (u, ϕ) is a critical point of F constrained on M × V if

∀v ∈ TuM F ′u(u, ϕ)[v] = 0,

∀ξ ∈ V F ′ϕ(u, ϕ)[ξ] = 0.

Note that the tangent space to V at ϕ is V itself.Recall that

TuM =v ∈ H1

0 (Ω) : G′(u)[v] = 0.

But

G′(u)[v] =

(2

∫Ω

uv dx, 2

∫Ω

quv dx

)= 0⇐⇒

∫Ω

uv dx =

∫Ω

quv dx = 0

Now, it is easy to see that a weak solution is a constrained critical point. Suppose that(u, ϕ) is a constrained critical point. Then, by the theorem of Lagrange multipliers we havethat

F ′u(u, ϕ)[v] = ω

∫Ω

uv dx− µ∫

Ω

quv dx.

It remains to prove that F ′ϕ(u, ϕ)[ξ] = 0 ∀ξ ∈ V . But this follows from decomposition (6.29).Note that F ′ϕ(u, ϕ)[r] = 0 for every constant r ∈ R.

The functional F constrained on M × V is unbounded from above and from below.Indeed, for ϕ = 0 we have

F (u, 0) =1

2

∫Ω

|∇u|2 dx+1

2

∫Ω

qχu2 dx

which is unbounded from above as ||∇u||2 → +∞. On the other hand, let u, ϕ be xed.Note that for all t > 0 we have tϕ ∈ V . Then

F (u, tϕ) = ||∇u||2 + t1

2

∫Ω

qϕu2 dx+

∫Ω

qχu2 dx− t2 1

4

∫Ω

(∆ϕ)2 dx− t2 1

4

∫Ω

|∇ϕ|2 dx

= c1 + tc2 − t2c3t→+∞−−−−→ −∞.


Note that c3 > 0.The unboundedness is a serious diculty. This issue has been addressed by Benci and Fortunato

(1998) and in many subsequent papers. Their standard method goes as follows:

(i) For every xed u there exists a unique Φ(u) such that F ′ϕ(u,Φ(u)) = 0.

(ii) The map u 7→ Φ(u) is of class C1.

(iii) The graph of Φ is a manifold, and we are reduced to study the functional J(u) =F (u,Φ(u), possibly constrained.

The method sketched above fails in our situation, for two reasons. First, we see thatF ′ϕ(u, ϕ) = 0 with ϕ ∈ V is just

∆2ϕ−∆ϕ− qu2 + α/|Ω| = 0 in Ω,

∂ϕ

∂n= 0 on ∂Ω,

∂∆ϕ

∂n= 0 on ∂Ω,∫

Ω

ϕ dx = 0.

The problem above has a unique solution if and only if u ∈ N (one can see this by integratingthe equations and using Green's formula). Also, since N is not a manifold (unless α 6= 0) wecannot require the map Φ : u 7→ Φ(u) to be of class C1. We shall then extend such a map Φ.

Proposition 6.3.2. For every w ∈ L6/5 (Ω) there exists a unique L(w) ∈ V solution of

∆2ϕ−∆ϕ− w + w = 0 in Ω,

∂ϕ

∂n= 0 on ∂Ω,

∂∆ϕ

∂n= 0 on ∂Ω,∫

Ω

ϕ dx = 0.

The map L : L6/5 (Ω) −→ V is linear and continuous, hence of class C∞.

Proof. We begin by noting that V is a Hilbert space. Indeed, let (un) ⊂ V be a Cauchysequence. Then it is a Cauchy sequence in H2 (Ω), which is a Hilbert space, and thus un →u ∈ H2 (Ω). But

0 =

∫Ω

un dx→∫

Ω

u dx

and hence u ∈ V .Now, weak solutions to the problem are functions ϕ ∈ V such that∫

Ω

∆ϕ∆v dx+

∫Ω

∇ϕ · ∇v dx−∫

Ω

wv dx = 0 ∀v ∈ V .

Consider then the bilinear form b : V × V −→ R given by

b(ϕ, v) =

∫Ω

∆ϕ∆v dx+

∫Ω

∇ϕ · ∇v dx.


This bilinear form actually gives the scalar product in V so it suces to apply the RieszTheorem.

Proposition 6.3.3. The map

u ∈ L6 (Ω) 7→ qu2 ∈ L6/5 (Ω)

is of class C1.

Proof. See Example 2.1.18.

As a consequence of the previous propositions, we can dene the following map:

Φ : u ∈ H10 (Ω) 7→ L(qu2) ∈ V .

It is clear thatΦ(u) = Φ(−u) = Φ(|u|).

Moreover, for every (u, ϕ) ∈ H10 (Ω)×V we have that ϕ = Φ(u) if and only if for every η ∈ V∫

Ω

∆ϕ∆η dx+

∫Ω

∇ϕ∇η dx =

∫Ω

qu2η dx. (6.30)

Taking η = Φ(u) we see that∫Ω

(∆Φ(u))2 dx+

∫Ω

|∇Φ(u)|2 dx =

∫Ω

qu2Φ(u) dx. (6.31)

from which we have

||Φ(u)||2V≤ ||q||∞

∫Ω

u2Φ(u) dx

≤ c||u2||2||Φ(u)||2

= c

(∫Ω

u4 dx

)1/2

||Φ(u)||2

= c||u||24||Φ(u)||2≤ c||u||22||∇Φ(u)||2≤ c||∇u||22||Φ(u)||V

and hence||Φ(u)||V ≤ c||∇u||22, (6.32)

that is, Φ is bounded on bounded sets. We actually have more:

Lemma 6.3.4. The map Φ is compact. Moreover, if un u in H10 (Ω) then∫

Ω

qu2nΦ(un) dx→

∫Ω

qu2Φ(u) dx.

Proof. Let un u in H10 (Ω) and dene Bn, B : V −→ R by

Bn(η) :=

∫Ω

qu2nη dx, B(η) :=

∫Ω

qu2η dx.


Such operators are continuous due to Hölder's inequality. For example:∣∣∣∣∫Ω

qu2η dx

∣∣∣∣ ≤ ||qu2||6/5||η||6 ≤ ||q||∞||u||24||η||2 ≤ c||∇η||2 ≤ c||η||V .

Recall that H10 (Ω) → Lr (Ω) for r ∈ [1, 6], with compact embedding if r < 6. Hence

un u in Lr (Ω) for all r ∈ [1, 6], hence (un), and consequently (un + u) are bounded. Sou2n → u2 in L6/5 (Ω). Indeed:∫

Ω

(u2n − u2)6/5 dx =

∫Ω

(un − u)6/5(un + u)6/5 dx

≤ ||(un − u)6/5||2||(un + u)6/5||2

≤ c

∫Ω

(un − u)12/5 dx→ 0

From the above it follows that

|Bn(η)−B(η)| ≤ ||q||∞ ||u2n − u2||6/5 ||η||6

≤ c||q||∞ ||u2n − u2||6/5 ||η||V .

Hence

||Bn −B|| ≤ supη 6=0

c||u2n − u2||6/5||η||V||η||V

→ 0

and thus Bn → B in L(V ).On the other hand, we have that Φ(un) Φ(u) in V . Indeed, let g ∈ V ∗. Then there is

some vg ∈ V such that

g(Φ(un)) =

∫Ω

∇Φ(un) · ∇vg dx+

∫Ω

∆Φ(u)∆vg dx =

∫Ω

qu2nvg dx.

But then

g(φ(un))− g(Φ(u)) =

∫Ω

q(u2n − u2)vg dx

≤ ||q||∞||u2n − u2||2||vg||2 → 0

since u2n → u2 in L2 (Ω) as well.

We then conclude that ∫Ω

qu2nΦ(un) dx→

∫Ω

qu2Φ(u) dx

and hence Φ(un)→ Φ(u) in V by (6.31).

Note that for every u ∈ N we have that F ′ϕ(u,Φ(u)) = 0. Indeed, Φ(u) is the uniquesolution to the problem in Proposition 6.3.2 with w = qu2.

We now dene the reduced functional of a single variable:

J : H10 (Ω) −→ R

u 7−→ F (u,Φ(u))


With the notation ϕu := Φ(u) the functionl J is explicitly given by

J(u) =1

2

∫Ω

|∇u|2 dx+1

2

∫Ω

qϕuu2 dx+

1

2

∫Ω

qχu2 dx− 1

4

∫Ω

(∆ϕu)2 dx

− 1

4

∫Ω

|∇ϕu|2 dx−α

2|Ω|

∫Ω

ϕu dx

=1

2

∫Ω

|∇u|2 dx+1

4

∫Ω

(∆ϕu)2 dx+

1

4

∫Ω

|∇ϕu|2 dx+

∫Ω

qχu2 dx.

We note that J is of class C1 on H10 (Ω) and even. Moreover, for every u ∈ M we have

thatJ ′(u)[v] = F ′u(u, ϕu)[v] + F ′ϕ(u, ϕu)[Φ

′(u)[v]] = F ′u(u, ϕu)[v] ∀v ∈ H10 (Ω)

and hence we have the following

Theorem 6.3.5. The pair (u, ϕ) ∈M × V is a critical point of F constrained on M × V ifand only if u is a critical point of J |M and ϕ = Φ(u).

The following lemma gives the existence of solutions to our modied problem.

Lemma 6.3.6. The functional J on M is weakly lower semicontinuous and coercive. Inparticular, it has a minimum u ∈M .

Proof. Recall that

J(u) =1

2

∫Ω

|∇u|2 dx+1

4

∫Ω

(∆ϕu)2 dx+

1

4

∫Ω

|∇ϕu|2 +

∫Ω

qχu2 dx

The two terms depending on ϕu are positive. Also,∫Ω

qχu2 dx ≥ − supΩ|q| sup

Ω|χ|.

Hence,

J(u) ≥∫

Ω

|∇u|2 dx− c

and thus J is coercive and bounded from below on M .Now, let (un) ⊂M such that u u. Since M is weakly closed, u ∈M . By Lemma 6.3.4

we know that

1

4

∫Ω

(∆ϕun)2 dx+1

4

∫Ω

|∇ϕun|2 dx→1

4

∫Ω

(∆ϕu)2 dx+

1

4

∫Ω

|∇ϕu|2 dx.

We also know that u2n → u2 in L6/5 (Ω) so∫

Ω

qχ(u2n − u2) dx ≤ c

∫Ω

u2n − u2 dx

≤ c||un − u||6/5 → 0.

Finally, the rst term is the norm of u in H10 (Ω), so it is convex and continuous and hence

lower semicontinuous. Adding all terms, we have that

J(u) ≤ lim inf J(un)


and thus J is weakly lower semicontinuous. The existence of the minimum is just given bythe Weierstrass theorem.

Remark 6.3.7. Note that J(u) = J(|u|) so the minimum may be assumed to be positive.This completes the rst part of Theorem 1.0.1.

6.4 Multiplicity of solutions

We will use a deformation argument to show that there are innitely many solutions. Acrucial point is that the functional satises the Palais-Smale condition.

Proposition 6.4.1. The functional J satises the Palais-Smale condition on M .

Proof. Let (un) ⊂M be a Palais-Smale sequence for J . We want to prove that there is someu0 ∈M such that un → u0 in H1

0 (Ω). We know that

J(un) is bounded

and thatJ |′M(un)→ 0. (6.33)

But (6.33) means precisely that there exists two sequences of real numbers (αn), (βn) and asequence (vn) ⊂ H−1 such that vn → 0 and

−∆un + q(ϕn + χ)un = αnun + βnqun + vn (6.34)

where ϕn := ϕun .Since J is coercive and J(un) is bounded, we know that (un) is bounded in H1

0 (Ω).Hence there exists u ∈ H1

0 (Ω) such that un u, up to a subsequence. By Lemma 6.3.4 weknow that ϕn → ϕ0. Also, since M is weakly closed, we know that u0 ∈M . It only remainsto show that un → u0 in H1

0 (Ω).By (6.34) we have that∫

Ω

|∇un|2 dx+

∫Ω

q(ϕn + χ)u2n dx− 〈vn, un〉 = αn + αβn. (6.35)

We note that

||∇un||2 is bounded,∫

Ω

q(ϕn + χ)u2n dx→

∫Ω

q(ϕ0 + χ)u20 dx, 〈vn, un〉 → 0.

Indeed,

• The boundedness of (un) means precisely that ||∇un||2 is bounded;

• Splitting the integral, we have that the rst term converges, by Lemma 6.3.4:∫Ω

qϕnu2n dx→

∫Ω

qϕ0u20 dx.

Also, ∫Ω

qχu2n dx→

∫Ω

qχu20 dx

6.4 MULTIPLICITY OF SOLUTIONS 57

because, as in the proof of Lemma 6.3.4, H10 (Ω) → L6/5 with compact embedding and

so u2n → u2

0 in L6/5 (Ω) and hence∫Ω

qχ(u2n − u2

0) dx ≤ ||qχ||6||u2n − u2

0||6/5 → 0;

• At last,|〈vn, un〉| ≤ ||vn||H−1 ||un||H1

0 (Ω) ≤ c||vn||H−1 → 0

We can then deduce that the right-hand side of (6.35) is bounded. Since bounded sequencesof real numbers converge up to subsequences, we can assume that

αn + αβn = λ+ o(1)

with λ ∈ R and o(1) means that this term goes to zero as n goes to innity. Then (6.34)becomes

−∆un + q(ϕn + χ)un − vn = (λ+ o(1))un − βn(q − α)un. (6.36)

Now, since u0 ∈ M we know that ||u0||2 = 1. This, together with the assumption|q−1(α)| = 0 implies that (q − α)u0 is not identically zero. Then there exists a test functionw0 ∈ C∞

c(Ω) such that ∫

Ω

(q − α)u0w0 dx 6= 0.

Evaluating (6.36) on w0 we get∫Ω

∇un · ∇w0 dx+

∫Ω

q(ϕn + χ)unw0 dx− 〈vn, w0〉 − (λ+ o(1))

∫Ω

unw0 dx

= βn

∫Ω

(q − α)unw0 dx.

Note that every term in the left-hand side converges. In particular, we note that∫Ω

qϕnunw0 dx−∫

Ω

qϕ0u0w0 dx ≤ c

∫Ω

ϕnun − ϕ0u0 dx

= c

∫Ω

ϕnun − ϕ0un dx+ c

∫Ω

ϕ0un − ϕ0u0 dx

≤ c||ϕn − ϕ0|| + c

∫Ω

ϕ0un − ϕ0u0 dx

→ 0.

The other terms are continuous linear functionals on un and thus converge.Also, by the weak convergence of (un),∫

Ω

(q − α)unw0 dx→∫

Ω

(q − α)u0w0 dx.

This implies that (βn) is bounded, which in turn implies that (αn) is bounded.


Applying (6.36) to un − u0 we get∫Ω

∇un∇(un − u0) dx+

∫Ω

q(ϕn + χ)un(un − u0) dx− 〈vn, un − u0〉

= (λ+ o(1))

∫Ω

un(un − u0) dx+ βn

∫Ω

(q − α)un(un − u0) dx.

Since ∫Ω

q(ϕn + χ)un(un − u0) dx→ 0, 〈vn, un − u0〉 → 0,

(λ+ o(1))

∫Ω

un(un − u0) dx→ 0, βn

∫Ω

(q − α)un(un − u0) dx→ 0

we conclude that ||∇un||2 → ||∇u0||2 and so un → u0 in H10 (Ω).

Let us show the above convergences. We begin by splitting the rst integral. Then, it isimmediately seen that∫

Ω

qχun(un − u0) dx ≤ c

∫Ω

un(un − u0) dx

≤ c||un||2||un − u0||2 → 0,

since un → u in L2 (Ω). We need to show that∫Ω

qϕnun(un − u0) dx→ 0.

Note that ∫Ω

|qϕnun(un − u0)| dx ≤ ||q||∞∫

Ω

|ϕnun(un − u0)| dx

≤ c||ϕn||2||un(un − u0)||2= c||ϕn||2||un||4||un − u0||4≤ c||ϕV ||un||4||un − u0||4 → 0.

As to the second convergence, note that

|〈vn, un − u0〉| ≤ ||vn|| ||un − u0||H10 (Ω) → 0

since (un − u0) is bounded in H10 (Ω).

Regarding the third term, note that∣∣∣∣(λ+ o(1))

∫Ω

un(un − u0) dx

∣∣∣∣ ≤ c||un||2||un − u0||2 → 0

by the compact Sobolev embedding.The last convergence follows by an argument similar to the one above.

By Theorem 6.2.4 M has compact, symmetric subsets of genus k for every k ∈ N.Hence, by Theorem 4.6.1 J has innitely many critical points. The critical points give riseto Lagrange multipliers ωn, µn and then, by the decomposition ϕ = φ − χ − µ to solutions(un, ωn, φn) ∈ H1

0 (Ω)×R×H2 (Ω) of the original problem.

6.4 MULTIPLICITY OF SOLUTIONS 59

Finally, we show that ||∇un||2 → +∞. We begin by noting that, by Theorem 4.5.5 thecritical levels are divergent, that is,

bn = J(un) =1

2

∫Ω

|∇un|2 dx+1

4

∫Ω

(∆ϕn)2 dx+1

4

∫Ω

|∇ϕn|2 dx−1

2

∫Ω

qχu2n dx→ +∞.

We proceed, nonetheless, with a direct proof. Since the functional satises the Palais-Smale condition and is bounded from below the sublevels have nite genus:

Lemma 6.4.2. For any b ∈ R the sublevel

J b =u ∈M : J(u) ≤ b

has nite genus.

Proof. We argue by contradiction. Suppose that

D =b ∈ R : γ(J b) =∞

6= ∅.

Since J |M is bounded from below, then D is bounded from below. Then

−∞ < b = inf D <∞.

Moreover, since J |M satises the Palais-Smale condition, the set

Z =u ∈M : J(u) = b, J |′M(u) = 0

is compact. Hence there exists a closed symmetric neighborhood UZ of Z such that γ(UZ) <

∞. By the Deformation Lemma, there exists an ε > 0 such that J b−ε includes a deformationretract of J b+ε \ UZ . Then, by the properties of the genus,

γ(J b+ε) ≤ γ(J b+ε \ UZ) + γ(UZ) ≤ γ(J b−ε) + γ(UZ) <∞,

a contradiction.

Let k ∈ N. By Lemma 6.4.2 there exists some n ∈ N depending on k such that

γ(Jk) = n.

LetAn+1 =

A ⊂M : A = −A,A = A, γ(A) = n+ 1

By Theorem 6.2.4 we know that An+1 is not empty. By the monotonicity property of thegenus, any A ∈ An+1 is not contained in Jk, then supA J > k and therefore

ck = infA∈An+1

supu∈A

J(u) ≥ k

Then ck is a critical value of J |M . In conclusion: for any k ∈ N there is some critical pointuk ∈M such that J(uk) ≥ k. Thus follows the claim about the divergence of critical levels.

Now we note that ∫Ω

qχu2n dx ≤ ||qχ||∞

60 PROOF OF THE MAIN RESULT

so the last term is bounded. Also, from (6.32) we have that

||ϕn||V =

∫Ω

(∆ϕn)2 dx+

∫Ω

|∇ϕn|2 dx ≤ c||∇u||22

and hence ||∇un||2 → +∞.

Remark 6.4.3. We would still get the same results if we perturbed the rst equation witha nonlinearity of the form κ|u|p−2u, for some values of p. The appropriate range for p willdepend on the sign of the constant κ. See Pisani and Siciliano (2007) for details.

Appendix A

Useful theorems

In this appendix we collect some theorems that are useful in solving the problem butwhich we felt did not belong to the main body of the text. They are stated here for theconvenience of the reader, and are presented without proof but with due references.

A.1 Neumann boundary value problems

For the proofs of the following two theorems, see Taylor (1996)

Theorem A.1.1. For k = 0, 1, . . ., given f ∈ Hk (Ω) and g ∈ Hk+1/2(∂Ω) there is a uniquesolution u ∈ Hk+2 (Ω) to the problem

−∆u+ u = f in Ω

∂u

∂n= g on ∂Ω.

Theorem A.1.2. For k = 0, 1, . . ., f ∈ Hk (Ω) and g ∈ Hk+1/2(∂Ω) the problem

∆u = f in Ω

∂u

∂n= g on ∂Ω

has a solution u ∈ Hk+2 (Ω) if and only if∫Ω

f dx =

∫∂Ω

g ds.

A.2 Sobolev spaces

We present now two important theorems concerning embeddings of Sobolev spaces.

Theorem A.2.1. Let Ω ⊂ RN be a bounded domain with Lipschitz boundary and let k ≥1, 1 ≤ p ≤ ∞.

(i) If kp < N then W k,p (Ω) → Lq (Ω) for all 1 ≤ q ≤ Np/(N − kp). The embedding iscompact provided that q < Np/(N − kp).

(ii) If kp = n then W k,p (Ω) → Lq (Ω) for all 1 ≤ q <∞ and the embedding is compact.

61

62 APPENDIX A

(iii) If kp > n then W k,p (Ω) → C0,α (Ω) where

α =

k − n/p if k − n/p < 1

1 if k − n/p > 1.

If k − n/p = 1 then the embedding holds for all α ∈ [0, 1).

Our version is as in Ambrosetti and Malchiodi (2007). For a proof of a fairly more com-plete version, see Adams and Fournier (2008).

Theorem A.2.2. Let Ω ⊂ RN be a connected open set having the segment property andsatisfying the cone condition, and let k1− k2 > N/2. Then the elements from Hk1 (Ω) are ofclass Ck2 and the embedding

Hk1 (Ω) → Ck2(Ω)

is continuous.

For a proof and for a discussion of the segment and cone properties, see Wloka (1987).For us, it suces to know that a smooth bounded open set satises both conditions.

Another important concept is that of the trace. See Evans (2010) or Kesavan (1989) foran introduction or Wloka (1987) for an extensive discussion.

Theorem A.2.3. Let Ω ⊂ RN be a bounded open set of class Cm+1. Then there existscontinuous linear trace operator from Hm (Ω) into (L2(∂Ω)m such that

(i) If v ∈ C∞(Ω) then

γ0(v) = v|∂Ω, γ1(v) =∂v

∂n

∣∣∣∣∣∂Ω

, . . . , γm−1(v) =∂mv

∂ nm

∣∣∣∣∣∂Ω

.

(ii) The range of γ is the spacem−1∏j=0

Hm−j−1/2(∂Ω)

(iii) The kernel of γ is Hm0 (Ω).

A.3 The theorem of Borsuk-Ulam

Theorem A.3.1. Let Ω ⊂ RN be an open, bounded and symmetric set containing the originand ϕ ∈ C(∂Ω,RN) and odd function such that ϕ(∂Ω) ⊂ E, where E is an ane subspaceof dimension less than or equal to N − 1. Then there exists x ∈ ∂Ω such that ϕ(x) = 0.

The proof depends on the theory of topological degree. See Kavian (1993) or Deimling(1985).

Bibliography

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Index

Chain Rule, 5constraint, 27

Deformation Lemma, 32derivative, 4

Fréchet dierentiable function, 3

Gateaux dierentiable function, 6

Implicit Function Theorem, 18Inverse Function Theorem, 16

locally invertible map, 15

Mean Value Inequality, 7Minimax Principle, 35

Palais-Smale condition, 28partial derivative, 12

second derivative, 10sublevel, 25

tangent pseudo-gradient vector eld, 29Taylor's formula, 11Theorem

Morse, 34

65

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Danilo Gregorin Afonso - University of São Paulo