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Tesis de Jinggang Tan

A mis padres, Chong-Ying Tan y Chun-Ying Deng,Con mucho Amor...

AGRADECIMIENTOS

Deseo expresar mi mas sincero agradecimiento a los profesores PatricioFelmer y Xavier Cabre, mis directores, por la motivacion transmitida, confianzaen mis estudios y apoyo en Santiago y Barcelona. Gracias a su profesionalidady dedicacion ha sido posible este trabajo.

Mi agradecimiento al Centro de Modelamiento Matematico (CMM) enChile, a la Comision Nacional de Investigacion Cientıfica y Tecnologica (CON-ICYT), y al Centre de Recerca Matematica (CRM) en Barcelona por financiareste trabajo.

Me complace agradecer el apoyo y los medios recibidos en los distintosdepartamentos donde he desarrollado parte de mi doctorado: el Departamentode Ingenierıa Matematica (DIM) de la Universidad de Chile y el Departamentde Matematica Aplicada 1 (MA1) de la Universitat Politecnica de Catalunya.

Quiero agradecer a los profesores P. Dartnell, S. Martınez, M. Kowal-czyk(UChile), J. Sola-Morales(UPC), J.A. Carrillo(UAB): muchısimas graciaspor su ayuda. Tambien agradezco a los profesores G. Li, J.F. Yang, H.S. Zhou,H.J. Zhao(WIPM), W.M. Ni, Y.Y. Li, Y. Lou, C. Gui, D.M. Cao, J. Wei, K.Tanaka, M. Tang por sus consejos.

Quiero agradecer a los miembros de la comision de mi tesis.

A todos mis companeros de estudio con los cuales he compartido oficina ymi vida. En particular a J. Mayorga, S. Alarcon y Pancho, Erika, a quienesagradezco sus consejos y apoyos en los momentos difıciles. Gracias por suamistad.

Finalmente, un profundo agradecimiento a toda mi familia, muy especial aAnnie (y su hijo Yaoyao), AQing (y su hijo Bo Li), ASheng, Lifan.

RESUMEN DE LA TESIS

El primer problema abordado en esta tesis es la demostracion de existenciade soluciones periodicas para un sistema de ecuaciones en derivadas parcialesque modela el movimiento de un retıculo elastico dos dimensional. Mas pre-cisamente, el estado de cada punto l = 1, 2, · · · , N del retıculo se representadapor ul(x, t). Este sistema con condiciones periodicas de Dirichlet posee un

Hamiltoniano con energıa cinetica∑N

l=1

∫ π

0

∫ 2π

0(|∂tul|2−|∂xul|2) dxdt y energıa

potencial∑N

l=1

∫ π

0

∫ 2π

0|ul+1−ul|p+1

p+1dxdt, donde uN+1 = u1. Puesto que el retıculo

elastico involucra al operador de ondas ∂tt−∂xx, el funcional correspondiente esfuertemente indefinido. En el caso autonomo, aplicamos el teorema del enlacede Benci y Rabinowitz a este funcional definido en un espacio Hilbert, lo cualconduce a la existencia de infinitas soluciones periodicas. Para tratar el casode un retıculo forzado, debemos utilizar metodos globales si las fuerzas exter-nas no son pequenas. Nuestro estudio se basa tambien en metodos clasicos enecuaciones en derivadas parciales del calculo variacional que son inspirados porel caso autonomo. La demostracion de infinitas soluciones en el caso forzadose basa en el metodo de perturbacion de simetrıa, que fue desarrollado porBahri, Berestycki, Struwe, Rabinowitz y Tanaka. Combinando las estimacionesdel ındice de Morse y el analisis del espectro del operador de ondas multidi-mensional, y tambien usando un teorema de puntos crıticos de Rabinowitz,establecemos la existencia de un numero infinito de soluciones periodicas.

La segunda parte de esta tesis esta consagrada al estudio de problemasno lineales que involucran un operador positivo no local: la raız cuadrada delLaplaciano −∆ en un dominio acotado Ω de Rn con condicion de Dirichlet nulaen la frontera. Designamos a este operador por A1/2 y estudiamos problemas nolineales A1/2u = f(u) en Ω y u = 0 sobre ∂Ω con metodos de calculo variacionalen ecuaciones en derivadas parciales. Una herramienta importante en nuestroanalisis es realizar este problema no local a traves de un problema local en elsemi cilindro Ω × (0,∞) con condiciones no lineales de Neumann en la parteΩ × 0 de la frontera del semi-cilindro y con condicion nula de Dirichlet enla parte ∂Ω × [0,∞) del borde. Demostramos una formula de tipo Pohozaevpara conseguir un resultado de no existencia en los casos crıtico y super-crıticocuando Ω es estrellado: f(u) = up, para p ≥ n+1

n−1. Establecemos la existencia de

soluciones positivas para el caso subcrıtico 1 0). Demostramos la regularidad y

una estimacion L∞ de soluciones debiles. Tambien obtenemos un resultado desimetrıa de tipo Gidas-Ni-Nirenberg usando el metodo de los planos moviles.

v

vi

Finalmente demostramos las estimaciones a priori de tipo Gidas-Spruck en elcaso de linealidades subcrıticas f(u) = up, 1 < p < n+1

n−1. Para ello, los teoremas

de Liouville para A1/2 en el espacio total y en el semi-espacio son ingredientesimportantes. El segundo caso no era conocido y lo demostramos usando latransformacion del Kelvin, el metodo de los planos moviles y una identidadhamiltoniana.

ABSTRACT

In the first part of this thesis, we investigate the existence of periodic so-lutions for a system of partial differential equations which describes the peri-odic motion of a two dimensional elastic lattice coupled with nearest neigh-bor interaction potentials of power-type functions. More precisely, the stateof each point l = 1, 2, · · · , N of the lattice is represented by ul(x, t). Thetwo dimensional elastic lattice with periodic-Dirichlet conditions possesses aHamiltonian with kinetic energy

∑Nl=1

∫ π

0

∫ 2π

0(|∂tul|2 − |∂xul|2) dxdt and po-

tential energy∑N

l=1

∫ π

0

∫ 2π

0|ul+1−ul|p+1

p+1dxdt, where uN+1 = u1. Since it involves

the wave operator ∂2t − ∂2

x, the corresponding functional is strongly indefinite.In the autonomous case, we apply the Benci-Rabinowitz linking theorem tothis functional in a Hilbert space to establish the existence of infinitely manyperiodic solutions. To treat a forced lattice, we must use global methods ifthe external forces are not small. Our approach also uses variational methodswhich are suggested by the autonomous case. Following a perturbation argu-ment from symmetry, which had been developed by Bahri, Berestycki, Struwe,Rabinowitz and Tanaka, we combine the use of the Morse index estimates andthe spectral analysis of the wave operator to establish the existence of infinitelymany periodic solutions by using a minimax theorem of Rabinowitz.

The second part of the thesis is devoted to the study of nonlinear problemsinvolving a nonlocal positive operator: the square root of the Laplacian −∆ ina bounded domain Ω of Rn with zero Dirichlet boundary condition. We denotethis nonlocal operator by A1/2, and study nonlinear problems A1/2u = f(u) inΩ and u = 0 on ∂Ω by using variational methods. An important tool in ouranalysis is to realize our nonlocal problem as a nonlinear local problem in ahalf cylinder C = Ω × (0,∞) with a nonlinear Neumann boundary conditionon the part Ω × 0 of the boundary and zero Dirichlet condition on theboundary part ∂Ω × [0,∞). We show a Pohozaev type formula and then weobtain a non-existence result for the critical and supercritical cases f(u) = up,for p ≥ n+1

n−1when Ω is star-shaped. We establish the existence of positive

solutions for the subcritical case 1 0), where

we follow the procedure of Brezis-Nirenberg. We show the regularity and anL∞ estimate of weak solutions. We also obtain a symmetry result of Gidas-Ni-Nirenberg type. Finally by using an analogue of the moving planes method weestablish a priori estimates of Gidas-Spruck type for problems with subcriticalnonlinearities f(u) = up and 1 < p < n+1

n−1. For this, important ingredients are

Liouville theorems for A1/2 in the whole space and in the half space. The half

vii

viii

space case was unknown, and we prove it by use of the Kelvin transform, themoving planes method and a Hamiltonian identity.

Table of Contents

1 Introduction and summary of results 11.1 An elastic lattice . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The square root of the Laplacian . . . . . . . . . . . . . . . . . 51.3 Summary of results . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Un retıculo elastico . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 La raız cuadrada del Laplaciano . . . . . . . . . . . . . . . . . . 171.6 Resumen de los resultados . . . . . . . . . . . . . . . . . . . . . 21

2 A two dimensional lattice 252.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3 An autonomous lattice . . . . . . . . . . . . . . . . . . . . . . . 362.4 A modified functional . . . . . . . . . . . . . . . . . . . . . . . . 402.5 Minimax methods . . . . . . . . . . . . . . . . . . . . . . . . . . 492.6 Morse index and spectral analysis . . . . . . . . . . . . . . . . . 592.7 Proof of theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3 The square root of the Laplacian 673.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.3 Regularity of solutions . . . . . . . . . . . . . . . . . . . . . . . 843.4 Maximum principles . . . . . . . . . . . . . . . . . . . . . . . . 883.5 The subcritical case and a Pohozaev type identity . . . . . . . . 913.6 Nonlinear problems in the critical case . . . . . . . . . . . . . . 963.7 Symmetry of solutions . . . . . . . . . . . . . . . . . . . . . . . 1053.8 A priori estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 107

ix

Chapter 1

Introduction and summary ofresults (Introduccion y resumende los resultados)

This thesis is concerned with the existence and non-existence of periodic mo-tion for a two dimensional elastic lattice involving the wave operator and fornonlinear problems involving the square root of the Laplacian operator. Westudy the solutions of these nonlinear problems by using methods of nonlinearanalysis.

The first part of the thesis is devoted to the periodic motion of a twodimensional elastic lattice coupled with nearest neighbor interaction potentialsof power-type functions. We show the existence of infinitely many periodicsolutions for the autonomous case by using Benci-Rabinowitz linking theoremand the existence of infinitely many periodic solutions for the forced lattice. Forthis we apply a minimax theorem of Rabinowitz to the perturbed functional,combined with the use of estimates of the Morse index and the spectral analysisof the wave operator. In some cases, there is no solution for the forced lattice.

The second part of the thesis is concerned with nonlinear problems involv-ing a nonlocal positive operator: the square root of the Laplacian in a boundeddomain with zero Dirichlet boundary condition. We establish the existenceof positive solutions for problems with power nonlinearities in the subcriticalcase, Brezis-Nirenberg type existence results for the critical problems undera small perturbation, non-existence of positive solutions in some supercriticalproblems, the regularity and an L∞ estimate of weak solutions, a symmetryresult of Gidas-Ni-Nirenberg type, and a priori estimates of Gidas-Spruck type.

In the next sections we expose briefly the problems treated in this thesisand make a summary of our main results about the above problems. All ofthese are written both in English and in Spanish.

1

2 Chapter 1. Introduction and summary of results

1.1 An elastic lattice

In Chapter 2, we consider the periodic motion of a two dimensional elastic lat-tice coupled with nearest neighbor interaction potentials of power-type func-tions:

ul = |ul+1 − ul|p−1(ul+1 − ul)− |ul − ul−1|p−1(ul − ul−1) + gl(x, t) (1.1.1)

for x ∈ (0, π), t ∈ R, l ∈ N, satisfying the periodic-Dirichlet conditions

ul(0, t) = 0 = ul(π, t), ∀t ∈ R, l ∈ N, (1.1.2)

ul(x, t) = ul(x, t+ 2π), ∀x ∈ (0, π), t ∈ R, l ∈ N, (1.1.3)

where ul = (ultt − ulxx), N = 1, 2, . . . , N, N ∈ N, uN+1 = u1, u0 = uN ,1 < p < ∞ and gl(x, t + 2π) = gl(x, t) for all l ∈ N. We are looking forsolutions which are periodic in time.

In last decades a considerable effort has been devoted to the mathematicalstudy of periodic motion of elastic lattices constituted by coupled flexible orelastic elements as strings, beams, membranes or plates. They can be viewedas systems of partial differential equations on networks or graphs. These sys-tems are known as multi-link or multi-body structures. They may generatenew, unexpected phenomena and could have a huge application in the future.However, the mathematical models describing the motion of elastic lattices arequite complex. For the control problems of linear systems, we can find wideinformation in [8].

On the other hand, a one dimensional Toda lattice takes the form:

d2ul

dt2= exp(ul+1 − ul)− exp(ul − ul−1).

Its complete integrability and existence of periodic solutions are well estab-lished in [1],[9],[10] and [22]. In particular, the classical Toda lattice was shownto be completely integrable, with explicit periodic and soliton solutions in[22]. It is well known by the KAM (Kolmogorov-Arnold-Moser) theory [1] thatperiodic and quasi-periodic solutions on the Toda lattice persist under smallperturbations. A surprising result of the existence of soliton was obtained by[9], for the interactions of power-type functions as follows:

d2ul

dt2= |ul+1 − ul|p−1(ul+1 − ul)− |ul − ul−1|p−1(ul − ul−1).

In [10] travelling wave solutions (periodic oscillations and heteroclinic solu-tions) were constructed for the lattice, corresponding to mass particles inter-acting nonlinearly with their nearest neighbor (the Fermi-Pasta-Ulam model).

1.1. An elastic lattice 3

These models have wide application in many physical systems and biologymodels.

Here we only consider a simplified version of these models of lattices coupledwith nearest neighbor interactions on one line. This may give us the way tolater address more complex situations. To the best of our knowledge, periodicmotion of a nonlinear 2-dimensional Toda lattice:

∂ttul − ∂xxu

l = exp(ul+1 − ul)− exp(ul − ul−1)

was first introduced by Mikhailov in [13]. He showed its integrability by usinginverse problem method. Then the reduced problem about the two-dimensionalgeneralized Toda lattice was considered in [14], all involving exponential po-tentials. The existence of periodic motion of the 2-dimensional Toda latticehas been explored in [12] by nonlinear analysis methods. Here we will use min-imax methods to find the critical points which correspond to the solutions of(1.1.1)-(1.1.3), but with the interactions of power-type functions.

First let us denote u(x, t) = (u1(x, t), . . . , uN(x, t)) for (x, t) ∈ Ω = (0, π)×(0, 2π). For p ∈ [1,∞) set by Lp(Ω) the vector of 2π-time periodic functions oft whose p-th powers are integrable, i.e.,

‖u‖p = (

∫Ω

N∑l=1

|ul(x, t)|p dxdt)1/p <∞.

The vector of smooth functions satisfying periodic-Dirichlet conditions (1.1.2)-(1.1.3) has a Fourier expansion of the form

u = (aljk) =N∑l=1

∞∑j=1

∞∑k=−∞

aljk sin jxeiktel, alj,−k = aljk, (1.1.4)

where e1, · · · , eN is the usual orthogonal basis in RN .We consider the functional for the autonomous case g ≡ 0:

G0(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

F0(u) dxdt,

where F0(u) = 1p+1

∑Nl=1 |ul+1−ul|p+1, uN+1 = u1. Applying variational meth-

ods to study existence and multiplicity of periodic solutions for (1.1.1)-(1.1.3),we consider the strongly indefinite functional G0(u), whose stable and unsta-ble manifolds at a critical point are of infinite dimension. So we shall applythe linking theorem of Benci and Rabinowtz [5] to obtain the existence of pe-riodic solutions. This technique was useful for Hamiltonian systems or waveequations.

The forced case, g 6≡ 0, will be considered with the aid of the standardargument for a perturbation from Z2 symmetry. We will study the functional:

G1(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

Fg(u) dxdt, (1.1.5)

where Fg(u) = 1p+1

∑Nl=1 |ul+1 − ul|p+1 − (g, u), uN+1 = u1.

In several papers, similar questions have been studied for the problems ofelliptic equations, Hamiltonian systems and wave equations. Bahri & Beresty-cki, Struwe and Rabinowitz considered the following elliptic problem:

−∆u = |u|p−1u+ f(x) x ∈ D,u = 0 x ∈ ∂D,

(1.1.6)

where D ⊂ Rn is a bounded domain with a smooth boundary ∂D and f ∈L2(D), they showed the existence of infinitely many solutions of (1.1.6) undera certain condition of p (see [18] and [21] for the details). They were concernedwith the following functional

Φ(u) =1

2

∫D

|∇u|2 dx− 1

p+ 1

∫D

|u|p+1 dx−∫D

fu dx

in H10 (D) and critical points of Φ(u) were found by Lusternik-Schnirelman

theory and energy estimates for Φ(u). Then Bahri & Lions [3] improved theresults and showed that the existence of infinitely many solutions under thecondition 1 < p < n/(n − 2). To get this existence result, they applied ageneral result giving lower bounded estimates of the Morse index as criticalpoints, which obtained through dual minimax theorems together with suitableestimates for the eigenvalues in the bounded domain D with zero Dirichletboundary conditions. By the way, Bolle, Ghoussoub and Tehrani [7] testedanother approach on several problems, including (1.1.6), proving the existenceof infinitely many solutions for a larger range, namely 1 < p < (n+1)/(n−1).

The existence of T -periodic solutions of the following forced Hamiltoniansystem of ordinary differential equations was considered by Bahri & Berestyski[4]:

dpdt

= −∂H(p,q)∂q

+ f1(t),dqdt

= ∂H(p,q)∂p

+ f2(t),(1.1.7)

where (p(t), q(t)), (f1(t), f2(t)) ∈ R2n, p(0) = p(T ), q(0) = q(T ) and H(p, q)is a given Hamiltonian. Assuming that H is superquadratic, they proved that(1.1.7) has infinitely many solutions for all (f1, f2) by using the perturbationargument from an S1-symmetric functional.

1.2. The square root of the Laplacian 5

Tanaka [21] considered the nonlinear vibrating string equation:∂ttu− ∂xxu+ |u|p−1u = f(x, t) (x, t) ∈ (0, π)× R,u(0, t) = u(π, t) = 0 t ∈ R,u(t+ 2π, x) = u(t, x) (x, t) ∈ (0, π)× R,

(1.1.8)

where p > 1 and f(x, t + 2π) = f(x, t). An unbounded sequence of weaksolutions was obtained in this paper with the aid of a standard perturbationfrom Z2 symmetry argument, combined with the estimates for the Morse indexof the functional.

In Chapter 2, we will concentrate in the functional for lattice (1.1.5). Ourapproach for the existence of infinitely many solutions is carried out by theperturbation argument, which is based on the Morse index estimates and aminimax theorem of Rabinowitz. It is very much influenced by the work ofTanaka [21].

1.2 The square root of the Laplacian

In Chapter 3 we are concerned with positive solutions to nonlinear problems in-volving a nonlocal positive operator: the square root of the Laplacian operatorin a bounded domain with zero Dirichlet boundary condition. We establish theexistence and nonexistence of positive solutions for problems with power-typenonlinearities, the regularity and an L∞-estimate of weak solutions, a symme-try result of Gidas-Ni-Nirenberg type, and a priori estimates of Gidas-Sprucktype. Particularly, we are looking for a function u satisfying the nonlinearproblem involving the square root of the Laplacian:

A1/2u = f(u) in Ω,u = 0 on ∂Ω,u > 0 in Ω,

(1.2.1)

where Ω is a smooth bounded domain in Rn and A1/2 stands for the squareroot of the Laplacian operator −∆ in Ω with zero Dirichlet boundary value on∂Ω.

The fractional powers of the Laplacian, which are called fractional Lapla-cians and correspond to Levy stable processes, appear in anomalous diffusionphenomena in physics, biology as well as other areas. They occur in flamepropagation, chemical reaction in liquids, population dynamics. Levy diffusionprocesses have discontinuous sample paths and heavy tails, while Brownianmotion has continuous sample paths and exponential decaying tails. So these


processes have been applied to American options in mathematical finance formodelling the jump processes of the financial derivatives such as futures, for-wards, options, and swaps, see [26] and references therein. Moreover, fractionalLaplacians play an important role in the study of the quasi-geostrophic equa-tions in geophysical fluid dynamics. Recently the fractional Laplacians attractmuch interest in nonlinear analysis. Caffarelli and Silvestre [35] gave a new for-mulation of the fractional Laplacians through Dirichlet-Neumann maps. Theregularity of the obstacle problems for the fractional powers of the Laplacianoperator was proved by Silvestre [51]. Moreover, Caffarelli et al [36],[34] stud-ied a free boundary problem: the Signorini problem involving the fractionalLaplacians as well as random homogenization of fractional obstacle problems.

One of the purposes of this thesis is to look for functions satisfying nonlinearproblems involving the square root of the Laplacian A1/2 as in (1.2.1). Notethat A1/2 is a nonlocal operator in Ω, but we will realize it through a localproblem in Ω× (0,∞). We mention that the half Laplacian in the whole spaceis a well studied operator. Let u be a smooth function u ∈ C∞

0 (Rn). There is aunique harmonic extension v ∈ C∞(Rn+1

+ ) of u in the half space such that itsderivatives Dkv(x, y) → 0 as |(x, y)| → ∞, for all k ≥ 0 and v(x, 0) = u(x). Itis the solution of the following Laplacian problem:

∆v = 0 in Rn+1+ ,

v = u on Rn = ∂Rn+1+ .

Consider the operator T : u 7→ −∂yv(·, 0). Since ∂yv is still a harmonic function,if we apply the operator twice, we obtain

T Tu = ∂yyv |y=0= −∆xv |y=0= −∆xu in Rn.

Thus the operator T that maps the Dirichlet-type data u to the Neumann-typedata −∂yv(x, 0) is actually the half Laplacian.

In Chapter 2, we introduce a new analogue operator in a bounded domainΩ ⊂ Rn. Consider the harmonic extension v of u in the half-cylinder Ω×(0,∞)vanishing on the lateral boundary ∂Ω. Then, since ∂yv is harmonic and alsovanishes on the lateral boundary as before, the Dirichlet-Neumann map of theharmonic extension v on the bottom of the half cylinder is the square root ofthe Laplacian A1/2 = B−1

1/2. That is, we have the properties:

A1/2 A1/2 = −∆ and B1/2 B1/2 = (−∆)−1,

where (−∆)−1 is the inverse Laplacian in Ω with zero Dirichlet boundary valueon ∂Ω. In this way we can study problem (1.2.1) by variational methods for a

1.2. The square root of the Laplacian 7

local problem. More precisely, we need to study the following mixed boundaryvalue problem in the half cylinder:

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= f(v) on Ω× 0,v > 0 in C,

(1.2.2)

where ν is the unit outer normal to Ω × 0. If v satisfies (1.2.2), then thetrace u on Ω × 0 of the function v will be a solution of (1.2.1). Moreover,we will obtain that the operator A1/2 is self-adjoint and positive definite andthat A1/2 has a spectral representation in terms of the eigenvalues and theeigenfunctions of −∆ in Ω with zero Dirichlet boundary conditions.

By studying the problem (1.2.2), we establish existence and non-existenceresults for problem (1.2.1) with power-type nonlinearities, the regularity andan L∞ estimate of weak solutions, a symmetry result of Gidas-Ni-Nirenbergtype, and a priori estimates of Gidas-Spruck type.

The analogue problem to (1.2.1) for the Laplacian operator has been inves-tigated widely in the last decades. This is the following problem

−∆u = f(u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω,

(1.2.3)

where Ω is a smooth bounded domain in Rn. If f(u) = up in problem (1.2.3),then there is a sharp contrast between the subcritical case p < n+2

n−2, for which

the problem admits a solution, and the critical case p = n+2n−2

, for which theSobolev embedding is not compact. Pohozaev discovered that there is no pos-itive weak solution for the critical or supercritical problem (1.2.3)

f(u) = up and p ≥ n+ 2

n− 2,

when Ω is a star-shaped domain.In the case of the square root of Laplacian A1/2, since the Sobolev trace

embedding is compact in the subcritical case in bounded domains, we get apositive solution by studying a corresponding minimization problem for (1.2.2).We also build a Pohozaev type identity which leads to the nonexistence ofpositive weak solution for problem (1.2.1) with critical or supercritical powernonlinearities in star-shaped domains.

In the case of f(u) = un+2n−2 + µu and µ > 0, the existence of positive

solutions of problem (1.2.3) was studied by Brezis and Nirenberg [31]. For this


they considered the constrained variational problem

min∫

Ω

(|∇u|2 − µ|u|2) dx | u ∈ H10 (Ω), ‖u‖

L2n

n−2 (Ω)= 1.

The critical points of this problem correspond to solutions of (1.2.3) for the

nonlinearity f(u) = un+2n−2 + µu, (µ > 0). But this energy functional may lose

compactness, since the nonlinearity involves the critical exponent. While thefunctional does not satisfy the Palais-Smale condition globally, some compact-ness will still hold in the range determined by the best constant of the Sobolevinequality. The steps of the proof need a careful analysis introduced by Brezisand Nirenberg about the energy level computed on cut-off functions of theextremal functions for the best constant in the Sobolev inequality. Their tech-nique has been extended to many other situations.

There is another approach to Brezis-Nirenberg result, based on a carefulstudy of the compactness properties for the Palais-Smale sequences of thefunctional Φ(u) defined as follows:

Φ(u) =1

2

∫Ω

|∇u|2 dx− n− 2

2n

∫Ω

|u|2n

n−2 dx− µ

2

∫Ω

|u|2 dx.

Both approaches are completely equivalent. However, the second approachbrings out the peculiarities of the limiting case more clearly, which gives theenergy estimates of the Palais-Smale sequences. We will prove the existence ofpositive solutions of (1.2.1) involving the square root of the Laplacian by usingboth procedures.

The fundamental estimates for elliptic equations (1.2.3) include huge mate-rials: the W 2,p-estimates proved by Calderon-Zygmund inequality, the Holderestimates developed by De Giorgi and Moser, the L∞ estimate due to Brezis-Kato based on Moser’s iteration technique. Here we will show some analogueregularity results for (1.2.1) and also an L∞ estimate for weak solutions of(1.2.1) by following the procedure of Brezis-Kato.

Gidas, Ni and Nirenberg [42] obtained symmetry properties of solutions forproblem (1.2.3) when f is Lipschitz continuous and Ω has certain symmetries.The proof of these symmetry results uses the maximum principle and a methodof Alexandroff, developed in the framework of partial differential equations bySerrin, called the moving planes method. The improved version by Berestycki-Nirenberg [29] replaces the use of Hopf’s lemma by the maximum principlein domains of small measure. In Chapter 3, we obtain a symmetry result ofGidas-Ni-Nirenberg type via the moving planes method. We also develop amaximum principle in domains of small measure for problem (1.2.1). We need

1.3. Summary of results 9

to point out that Chipot, Chlebık, Fila and Shafrir [37] have studied a differentbut related problem:

−∆v = g(v) in B+R = z ∈ Rn+1 | |z| ≤ R, zn+1 > 0,

v = 0 on ∂B+R ∩ zn+1 > 0,

∂v∂ν

= f(v) on ∂B+R ∩ zn+1 = 0,

v > 0 in B+R ,

(1.2.4)

where f, g ∈ C1(R) and ν is the unit outer normal. They proved the existence,non-exsitence and axial symmetry results of solutions for (1.2.4).

Gidas and Spruck [43] established a priori estimates for positive solutionsof problem (1.2.3) when f(u) = up and 1 < p < n+2

n−2. The proof involves the

method of blow-up combined with important ingredients: nonlinear Liouvilletype results in the whole space and in the half space. The moving planesmethod or the moving spheres method are strong tools to prove nonlinearLiouville theorems. The Kelvin transform is also useful to prove nonlinearLiouville theorems for A1/2. Yanyan Li & L. Zhang [46] and Ou [50] provedthat there is no positive solution for problem

(−∆)1/2u = up in Rn,

when 1 < p < n+1n−1

. Regarding the Liouville theorem in a half space for A1/2

-previously unknown- we establish it using an idea introduced by Cabre andSola-Morales [33]. They studied layer solutions which are monotone with re-spect to one variable of

(−∆)1/2u = f(u) in Rn,

where f is balanced bistable type. That is, if G(u) = −∫ u

0f(s) ds, then G has

two, and only two, absolute minima of the same height. They developed a non-local Modica type estimate, as well as a conserved Hamiltonian quantity forevery layer solution. These tools are useful to prove a nonlinear Liouville theo-rem involving A1/2 in the half space. The proof uses the Kelvin transform, themoving planes method and the Hamiltonian identity. Then by these nonlinearLiouville theorems, we will present a priori estimates for positive solutions ofproblem (1.2.1) with nonlinearity f(u) = up and 1 < p < n+1

n−1.

1.3 Summary of results

An elastic lattice

We study a strongly indefinite functional in a Hilbert space and applyBenci-Rabinowitz linking theorem to this functional, then we obatin a peri-odic solution for the two dimensional elastic lattice in autonomous case. Themultiplicity follows from the simple argument of Nirenberg [15].

Theorem 1.3.1. If gl ≡ 0 for all l ∈ N = 1, 2, . . . , N, then lattice (1.1.1)-(1.1.3) has infinitely many nontrivial periodic solutions.

The second result is the existence of periodic motion of the two dimensionalforced lattice. The proof follows a perturbation argument from a Z2-symmetricfunctional. It depends on the estimates of the Morse index and the spectralanalysis of the wave operator developed by Tanaka [21].

Theorem 1.3.2. Assume that gl ∈ Lα/(α−1)(Ω) satisfies∫

Ωglζ dxdt = 0 for

all ζ ∈ L∞(Ω)∩ ker(∂tt − ∂xx), where Ω = (0, π)× (0, 2π), p > 1, max2, p <α < p + 1, l ∈ N = 1, 2, . . . , N. Then lattice (1.1.1)-(1.1.3) has infinitelymany nontrivial periodic solutions.

Moreover, if gl ∈ L∞(Ω) ∩ ker(∂tt − ∂xx) satisfies∑

l gl 6= 0, then problem

(1.1.1)-(1.1.3) has no periodic solution.

The square root of the Laplacian

By the minimizing procedure and a Pohozaev type identity we can de-rive, respectively, the following existence and non-existence results of positivesolutions to problem (1.2.1):

Theorem 1.3.3. Let n ≥ 1 be an integer and 2] = 2nn−1

when n ≥ 2. Supposethat Ω is a smooth bounded domain in Rn and f(u) = up. Then,(i) Problem (1.2.1) has at least one C2(Ω) solution whenever 1 < p < 2]− 1 =n+1n−1

if n ≥ 2, or 1 < p <∞ if n = 1.

(ii) Let p ≥ 2]− 1 = n+1n−1

and n ≥ 2. If Ω is star-shaped with respect to a pointof Rn, then there exists no weak bounded solution of (1.2.1).

We apply the Brezis-Nirenberg technique to problem (1.2.2) and obtain ananalogue result to theirs but now for problem (1.2.1) involving the square rootof the Laplacian A1/2:

Theorem 1.3.4. Let n ≥ 2. Suppose that Ω is a smooth bounded domain of

Rn, f(u) = un+1n−1 + µu, and that µ1 =

√λ1 is the first eigenvalue of A1/2,

where λ1 is the first eigenvalue of the Laplacian −∆ in Ω with zero Dirichletboundary condition.

Then, for every µ ∈ (0, µ1), there exists at least one C2(Ω) solution of(1.2.1). Furthermore, there exists no weak solution of (1.2.1) for µ ≥ µ1.

1.3. Summary of results 11

We also show a symmetry result of Gidas-Ni-Nirenberg type for problem(1.2.1) in symmetric domains. For this we use the moving planes method.

Theorem 1.3.5. Let Ω be a bounded domain of Rn which is convex in thex1 direction and symmetric with respect to the hyperplane x1 = 0. Let f beLipschitz continuous and u be a C2(Ω) solution of (1.2.1).

Then u is symmetric with respect to x1, i.e., u(−x1, x′) = u(x1, x

′) for all(x1, x

′) ∈ Ω. In addition, ∂u∂x1

< 0 for x1 > 0.

As a consequence of Theorem 1.3.5, let v be a solution of (1.2.2) under theassumption: Ω is a ball and f is a Lipschitz continuous function. Then we canconclude here that v is axially symmetric with respect to the y-axis.

We finally establish the following a priori estimates of Gidas-Spruck typefor positive solutions of problem (1.2.1) in the subcritical case. For this, we usethe blow up technique.

Theorem 1.3.6. Let n ≥ 2 and 2] = 2nn−1

. Assume that Ω ⊂ Rn is a smooth

bounded domain and f(u) = up, 1 < p < 2] − 1.Then there exists a constant C(p,Ω), dependent only on p and Ω, such that

every weak solution u of (1.2.1) satisfies

‖u‖L∞(Ω) ≤ C(p,Ω).

The proof of the previous result uses two important ingredients: nonlinearLiouville theorems for the half Laplacian in Rn and in Rn

+. The one for all spacewas proved by Y.Y. Li and L. Zhang [46] by using the moving spheres methodand Ou [50] by using the moving planes method. The corresponding Liouvilletheorem in a half space was not available and we establish it in Chapter 3for bounded solutions. Here, in contrast with the whole space Rn case, we caninclude the critical nonlinearity.


. Denote

Rn+1++ = z = (x1, x2, · · · , xn, y) | xn > 0, y > 0.

If 1 0, y = 0,v = 0 on xn = 0, y > 0,v > 0 in Rn+1

++ ,

(1.3.1)

where ν is the unit outer normal to xn > 0, y = 0.The proof of this result combines the use of the Kelvin transform, the

moving planes method, as well as a Hamiltonian identity in spirit of [33].

Introduccion general y resumen de losresultados

En esta tesis se aborda el estudio de soluciones periodicas para un retıculoelastico que involucra el operador de ondas y para soluciones positivas de prob-lemas no lineales que involucran la raız cuadrada del operador Laplaciano.Estudiamos estos problemas no lineales usando metodos del analisis no lineal.

La primera parte de la tesis esta dedicada al estudio del movimiento de unretıculo elastico: un sistema de ecuaciones en derivadas parciales, con fuerzainterna entre los vecinos mas cercanos y fuerza externa para cada punto delretıculo. La existencia de un numero infinito de soluciones periodicas es de-mostrada para el caso autonomo usando el teorema de enlace de Benci y Rabi-nowitz. La demostracion de infinitas soluciones del caso forzado se basa en elmetodo de perturbacion de simetrıa combinado con las estimaciones del ındicede Morse y un teorema de los puntos crıticos de Rabinowitz.

En segundo lugar, nos concentramos en problemas no lineales que involu-cran un operador positivo no local: la raız cuadrada del Laplaciano en dominiosacotados con la condicion nula en la frontera de tipo Dirichlet. Establecemosla existencia y no existencia de soluciones positivas para el caso de no lineal-idades dadas por funciones potencias, la regularidad y una estimacion L∞ desoluciones debiles, un resultado de simetrıa de tipo Gidas-Ni-Nirenberg, y lasestimaciones a priori de tipo Gidas-Spruck.

En las siguientes secciones exponemos brevemente los problemas que mo-tivan esta tesis y presentamos un resumen de nuestros principales resultadosobtenidos.

1.4 Un retıculo elastico

En el Capıtulo 2 consideramos un sistema de ecuaciones en derivadas parcialesque modelan el movimiento de un retıculo elastico dos dimensional, con fuerzainterna entre los vecinos mas cercanos y fuerza externa para cada punto, de laforma:


para x ∈ (0, π), t ∈ R, l ∈ N, con condiciones de borde periodicas-Dirichlet

ul(0, t) = 0 = ul(π, t), ∀t ∈ R, l ∈ N, (1.4.2)


donde ul = (ultt − ulxx), N = 1, 2, . . . , N, N ∈ N, 1 < p < ∞, uN+1 =u1, u0 = uN y gl(x, t+ 2π) = gl(x, t) para l ∈ N.


En los ultimos anos mucho esfuerzo ha sido dedicado al estudio matematicodel movimiento de los retıculos elasticos, constituido por elementos flexibles oelasticos ligados, como secuencias de vigas, membranas o placas. Pueden servistos como sistemas de ecuaciones en derivadas parciales en redes o grafos.Estos sistemas se conocen como estructuras de multi-acoplamiento o multi-cuerpo, que pueden dar origen a fenomenos nuevos e inesperados de enormeimportancia practica. Sin embargo, los modelos matematicos que describenel movimiento de los retıculos elasticos pueden ser tremendamente complejos.Para los problemas del control de los sistemas lineales, informacion muy ampliase puede encontrar en [8].

Por otro lado, un retıculo elastico de Toda unidimensional se escribe como

d2ul

dt2= exp(ul+1 − ul)− exp(ul − ul−1).

La variable ul(t) para cada l = 1, · · ·N o l ∈ Z, representa la informacion(usualmente posicion) de las partıculas del sistema fısico, quımico o biologico.Su integrabilidad total y la existencia de soluciones periodicas fueron estable-cidas en [1], [9], [10] y [22]. Particularmente, en [22] se demostro que el retıculoclasico de Toda es totalmente integrable, con soluciones periodicas y de tiposoliton explıcitas. Es bien sabido por la teorıa KAM, (Kolmogorov-Arnold-Moser) [1], que las soluciones periodicas y cuasi-periodicas del retıculo de Todapersisten bajo perturbaciones pequenas. Un resultado asombroso fue la exis-tencia de una solucion de tipo soliton que fue conseguida en [9] para el tipo deinteracciones siguiente

d2ul

dt2= |ul+1 − ul|p−1(ul+1 − ul)− |ul − ul−1|p−1(ul − ul−1).

En [10] se construyen soluciones de ondas viajeras (oscilaciones periodicas ysoluciones heteroclınicas) para un retıculo que corresponde al modelo de tipoFermi-Pasta-Ulam, que tiene muchas aplicaciones en sistemas fısicos y modelosde la biologıa.

En esta tesis consideramos solamente una version sencilla de los modelosde ecuaciones en derivadas parciales sobre un retıculo con interacciones entrelos vecinos mas cercanos, desplegados periodicamente. Esto permitira la posi-bilidad de tratar situaciones mas complejas en el futuro. El problema no linealdos dimensional sobre un retıculo de Toda se puede escribir como

∂ttul − ∂xxu

l = exp(ul+1 − ul)− exp(ul − ul−1),

fue introducido por Mikhailov en [13] y su integrabilidad fue demostrada por elmetodo del problema inverso. Mas tarde, el problema de la reduccion sobre un

1.4. Un retıculo elastico 15

retıculo general dos dimensional fue considerados en [14], con todas las fuerzasinternas modeladas por funciones exponenciales. La existencia de solucionesperiodicas para un retıculo dos dimensional de Toda se ha explorado en [12] pormetodos de analisis no lineal. Aquı utilizaremos el metodo de puntos crıticospara encontrar soluciones de (1.4.1)-(1.4.3), pero donde las fuerzas internas sonfunciones potencias.

Denotaremos u(x, t) = (u1(x, t), . . . , uN(x, t)) para (x, t) ∈ Ω = (0, π) ×(0, 2π). Sea Lp el espacio vectorial de las funciones 2π-periodicas en t, que sonp-integrable para todo p ∈ [1,∞), es decir,

‖u‖p = (

∫Ω

N∑l=1


Un vector de funciones suaves que satisfaga (1.4.2),(1.4.3) tiene la forma:

u = (aljk) =N∑l=1

∞∑j=1

∞∑k=−∞


donde e1, · · · , eN es la base canonica de RN .Para encontrar soluciones de (1.4.1)-(1.4.3) vamos a estudiar el funcional

en caso que g ≡ 0:

G0(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

F0(u) dxdt,

donde F0(u) = 1p+1

∑Nl=1 |ul+1 − ul|p+1, uN+1 = u1. Al utilizar los metodos

variacionales para buscar resultados de existencia y multiplicidad de solucionesperiodicas para (1.4.1)-(1.4.3), tenemos que tomar en cuenta que el funcionalG0(u) es fuertemente indefinido, es decir, la variedad estable y la variedad in-estable tienen dimension infinita en cada punto crıtico. Aplicaremos el teoremadel enlace de Benci y Rabinowitz de [5] para obtener nuestros resultados, tec-nica bien conocida para los sistemas Hamiltonianos o las ecuaciones de ondas.

El caso g 6≡ 0 sera analizado usando una tecnica de perturbacion de unfuncional simetrico en Z2. Estudiaremos el funcional:

G1(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

F0,g(u) dxdt, (1.4.5)

donde F0,g(u) = 1p+1

∑Nl=1 |ul+1 − ul|p+1 − (g, u), uN+1 = u1.

En varios artıculos, se han estudiado problemas similares para ecuacioneselıpticas, sistemas Hamiltonianos y ecuaciones de ondas. Bahri & Berestycki,

Struwe y Rabinowitz consideraron el siguiente problema elıptico:−∆u = |u|p−1u+ f(x) x ∈ D,u = 0 x ∈ ∂D,

(1.4.6)

donde D ⊂ Rn es un dominio acotado con frontera suave ∂D y f ∈ L2(D).Demostraron la existencia de un numero infinito de soluciones para (1.4.6)bajo cierta condicion sobre p (ver [18] y [21]). Ellos consideraron el siguientefuncional:

Φ(u) =1

2

∫D

|∇u|2 dx− 1

p+ 1

∫D

|u|p+1 dx−∫D

fu dx

en H10 (D) y los puntos crıticos de Φ(u) son obtenidos usando el esquema de

Lusternik & Schnirelman y estimaciones de energıa para Φ(u). Luego Bahri &Lions en [3] mejoraron los resultados y demostraron que la existencia de unnumero infinito de soluciones vale para 1 < p < n/(n− 2). Para conseguir esteresultado de existencia, ellos aplicaron un resultado general que da una cotapor abajo para el ındice de Morse en puntos crıticos, y usaron teoremas dualesde los puntos crıticos junto a una estimacion conveniente de los valores propiosdel problema con las condiciones nulas de tipo Dirichlet en la frontera de D.Por otro lado, Bolle, Ghoussoub y Tehrani [7] utilizaron otro enfoque paraproblemas incluyendo (1.4.6), obteniendo la existencia de infinitas solucionespara el rango mas grande 1 < p < (n+ 1)/(n− 1).

El metodo para demostrar la existencia de soluciones periodicas fue uti-lizado para los siguiente sistemas Hamiltonianios de ecuaciones diferencialesordinarias en [4]:

dpdt

= −∂H(p,q)∂q

+ f1(t),dqdt

= ∂H(p,q)∂p

+ f2(t),(1.4.7)

donde (p(t), q(t)), (f1(t), f2(t)) ∈ R2n, p(0) = p(T ), q(0) = q(T ) y H(p, q) es unHamiltoniano dado. Bajo una condicion sobre el crecimiento y suponiendo queH es super cuadratica, mostraron que (1.4.7) tiene infinitas soluciones paratodo (f1, f2) por un metodo de perturbacion del funcional S1-simetrico.

Tanaka considero el problema de la cuerda vibrante no lineal con una fuerzaexterna de la siguiente forma:

∂ttu− ∂xxu+ |u|p−1u = f(x, t) (x, t) ∈ (0, π)× R,u(0, t) = u(π, t) = 0 t ∈ R,u(t+ 2π, x) = u(t, x) (x, t) ∈ (0, π)× R,

(1.4.8)

donde p > 1 y f(x, t + 2π) = f(x, t). Tanaka obtuvo una sucesion no aco-tada de soluciones aplicando la perturbacion del funcional correspondiente,

1.5. La raız cuadrada del Laplaciano 17

con simetrıa de Z2 y combinando con estimaciones del ındice de Morse depuntos crıticos de un funcional asociado.

En el Capıtulo 2 nos concentraremos en el funcional (1.4.5) para el casodel retıculo elastico. La demostracion de existencia de infinitas soluciones estadada usando el metodo de perturbacion, que esta basado en el teorema delpunto de silla y estimaciones del ındice de Morse. La prueba esta influenciadapor el trabajo de Tanaka en [21].

1.5 La raız cuadrada del Laplaciano

En el Capıtulo 3, estamos interesados en soluciones positivas para ciertos prob-lemas no lineales con un operador positivo no local: la raız cuadrada del oper-ador Laplaciano. Establecemos la existencia y no existencia para el caso de nolinealidades de tipo potencia, la regularidad y una estimacion L∞ de solucionesdebiles, un resultado de simetrıa de tipo Gidas-Ni-Nirenberg, y estimaciones apriori de tipo Gidas-Spruck. Mas precisamente, intentaremos buscar una fun-cion u que satisfaga el siguiente problema con la raız cuadrada del Laplaciano:

A1/2u = f(u) en Ω,u = 0 sobre ∂Ω,u > 0 en Ω,

(1.5.1)

donde A1/2 denota la raız cuadrada del operador Laplaciano −∆ con la condi-cion nula de tipo Dirichlet y Ω es un dominio acotado suave en Rn.

Los Laplacians fraccionarios: (−∆)α para 0 < α < 1 corresponden a losprocesos estables de Levy que aparecen en fenomenos de difusion anomalos enfısica, biologıa ası como otras areas. Ellos ocurren en propagacion de llamas,reacciones quımicas en lıquidos, y dinamica demografica. Los procesos de di-fusion de Levy tienen trayectorias discontinuas y colas pesadas, mientras queel movimiento Browniano tiene trayectorias continuas y colas con decaimientoexponencial. Los procesos de Levy se han aplicado extensamente a opcionesamericanas en finanzas matematicas para modelar los procesos de salto de losderivados financieros como futuros, adelante, opciones, y cambios, ver [26] ylas referencias que allı aparecen. Ademas, este operador es muy importante enel estudio de las ecuaciones cuasi-geostroficas en la dinamica fluida geofısica.Recientemente los Laplacianos fraccionarios atraen mucho interes en analisisno lineal. Caffarelli y Silvestre [35] dieron una nueva formulacion de Lapla-cianos fraccionarios a traves de mapas de Dirichlet-Neumann y la regularidaddel problemas del obstaculo con los operadores Laplacianos fraccionarios esmostrada por Silvestre [51]. Ademas, Caffarelli et al [36] [34] estudiaron el


problema de frontera libre de tipo Signorini y la homogeneizacion de los prob-lemas del obstaculo que involucran Laplacianos fraccionarios.

El proposito de este tercer capıtulo es estudiar las funciones u que satisfacenlos problemas no lineales que involvcran la raız cuadrada del Laplaciano A1/2

como en (1.5.1). Observamos que aunque A1/2 es un operador no local, (1.5.1)sera transformado en un problema local. Es bien conocido el operador raızcuadrada del Laplaciano en todo el espacio Rn. Sea una funcion suave u ∈C∞

0 (Rn), entonces existe una funcion armonica v ∈ C∞(Rn+1+ ) extension de u

tal que para todo k ≥ 0, Dkv(x, y) → 0 cuando |(x, y)| → ∞, la cual es lasolucion unica del problema

∆v = 0 in Rn+1+ ,

v = u on Rn = ∂Rn+1+ .

Consideramos el operador T : u 7→ −∂yv(·, 0). Puesto que ∂yv es tambien unafuncion armonica, si aplicamos el operador dos veces obtenemos que

T Tu = ∂yyv |y=0= −∆xv |y=0= −∆u in Rn.

Ası el operador T de Dirichlet-Neumann es realmente el medio Laplaciano.De la misma manera, haremos la extension armonica v de u hacia el semicilindro, y entonces el operador de Dirichlet-Neumann de v en la parte inferiordel cilindro actua como el operador A1/2 = B−1

1/2, lo que significa

A1/2 A1/2 = −∆ and B1/2 B1/2 = (−∆)−1,

donde (−∆)−1 es el inverso del operador Laplaciano en Ω con condiciones deDirichlet nulas en ∂Ω. De esta manera se reduce el problema (1.5.1) a un prob-lema local, lo cual sera tratado por metodos variacionales. Mas precisamente,estudiaremos el siguiente problema local en un semi cilindro:

−∆v = 0 en C = Ω× (0,∞),v = 0 sobre ∂LC = ∂Ω× [0,∞),∂v∂ν

= f(v) sobre Ω× 0,v > 0 en C,

(1.5.2)

donde ν es la normal exterior unitaria hacia Ω×0. La funcion v(x, 0) que sat-isface (1.5.2) sera la solucion de (1.5.1). Ademas, observamos que el operadorA1/2 es autoadjunto y definido positivo y se puede representar como combi-nacion lineal de las funciones propias ponderadas en terminos de los valorespropios de −∆ en Ω, con condiciones nulas en la frontera de tipo Dirichlet.

Por lo tanto, analizando el problema (1.5.2), estableceremos los resultadosde existencia y no existencia para (1.5.1), la regularidad y una estimacion L∞,

1.5. La raız cuadrada del Laplaciano 19

un resultado de simetrıa de tipo Gidas-Ni-Nirenberg, y estimaciones a prioride tipo Gidas-Spruck.

El problema analogo para el operador Laplaciano se ha investigado exten-samente en los ultimos anos. Se trata del problema

−∆u = f(u) en Ω,

u = 0 sobre ∂Ω,(1.5.3)

donde Ω es un dominio acotado y suave en Rn. Cuando f(u) = up en elproblema (1.5.3), hay un furte contraste entre el caso subcrıtico 1 < p < n+2

n−2,

para el cual se tiene soluciones, y el caso crıtico p = n+2n−2

, para el cual lainyeccion de Sobolev no es compacta, y no existe solucion positiva para el dichoproblema si Ω es dominio acotado de tipo estrellado por la celebrada identidadde Pohozaev. Puesto que la inclusion es compacta en el caso subcrıtico endominios acotados, estudiando un problema correspondiente de minimizacionpara (1.5.2), obtenemos una solucion positiva para (1.5.1). Al mismo tiempo,construimos una identidad de tipo Pohozaev que implica que no hay solucionpositiva para el problema crıtico (1.5.1) en dominios de tipo estrellado.

En el caso crıtico f(u) = un+2n−2 + µu, µ > 0, la existencia de soluciones

positivas del problema (1.5.3) ha sido estudiada por Brezis y Nirenberg [31].Para probar este resultado, estudiaron el problema utilizando el metodo deminimizacion con restriccion:

min∫

Ω

(|∇u|2 − µ|u|2) dx | u ∈ H10 (Ω), ‖u‖

L2n

n−2= 1.

Los puntos crıticos de este funcional son las soluciones debiles de (1.5.3). Peroaquı el correspondiente funcional puede perder compacidad, puesto que la nolinealidad esta en el caso del exponente crıtico. Si bien el funcional no satisfacela condicion de Palais-Smale globalmente, una cierta compacticidad se man-tendra en la zona determinada por la mejor constante de la desigualdad deSobolev. Se necesita analizar cuidadosamente el nivel de energıa de la funcionextremal para la mejor constante de la desigualdad de Sobolev, idea que fueintroducida por Brezis y Nirenberg. Sus tecnicas se han extendido a muchassituaciones.

Hay otro enfoque para el resultado de tipo Brezis-Nirenberg que se basa enun estudio cuidadoso de las caracterısticas de compacidad para las sucesionesde Palais-Smale del siguiente funcional:

Φ(u) =1

2

∫Ω

|∇u|2 dx− n− 2

2n

∫Ω

|u|2n

n−2 dx− µ

2

∫Ω

|u|2 dx.

Ambos caminos son totalmente equivalentes. Sin embargo, el segundo nosda la estimacion de energıa de las sucesiones de Palais-Smale en el caso lımitemas claramente. Estudiaremos la existencia de soluciones para (1.5.2) por am-bos metodos.

Las estimaciones fundamentales para ecuaciones elıpticas (1.5.3) incluyenmuchos estudios: las estimaciones W 2,p provadas por desigualdades de tipoCalderon-Zygmund, las estimaciones de Holder desarrolladas por De Giorgi yMoser, la estimacion debida a Brezis-Kato que se basa en la tecnica de iteracionde Moser. En analogıa con algunos de estos resultados, nosotros mostraremosla regularidad de soluciones debiles de (1.5.1) y una estimacion L∞ para solu-ciones debiles de (1.5.1).

A continuacion, el resultado de simetrıa de soluciones positivas del prob-lema (1.5.3) fue demostrado por Gidas, Ni and Nirenberg en [42]. La tecnicade demostracion depende del principio del maximo y de un metodo de Alexan-droff, desarrollado en el caso de ecuaciones en derivadas parciales por Serrin,llamado el metodo de los planos moviles. En el metodo mejorado, se substituyeel uso del lema de Hopf por el principio del maximo para dominios con me-dida pequena [29]. Conseguiremos un resultado de simetrıa similar para A1/2

usando el metodo de los planos moviles. Tambien estableceremos el principiodel maximo en dominios con medida pequena para el problema (1.5.1). Necesi-tamos precisar que Chipot, Chlebık, Fila y Shafrir [37] trataron otro problemade tipo mixto en la frontera:

−∆v = g(v) en B+R = z ∈ Rn+1 | |z| ≤ R, zn+1 > 0,

v = 0 sobre ∂B+R ∩ zn+1 > 0,

∂v∂ν

= f(v) sobre ∂B+R ∩ zn+1 = 0,

v > 0 en B+R ,

(1.5.4)

donde f, g ∈ C1(R) y ν es la normal exterior unitaria. Hallaron existencia ypropiedades de simetrıa axial para las soluciones de (1.5.4).

Las estimaciones a priori para soluciones positivas del problema (1.5.3),cuando f(u) = up y 1 < p < n+2

n−2, fueron establecidas por Gidas & Spruck

en [43]. La demostracion usa el metodo de blow-up combinado con dos in-gredientes importantes: teoremas de Liouville en todo el espacio y en el semiespacio. El metodo de los planos moviles o el metodo de las esferas movilesproveen formas simples para demostrar estos teoremas de Liouville no lineales.La transformacion de Kelvin es tambien util en la demostracion para el casoA1/2. El caso del todo espacio fue considerado en el trabajo de Y.Y. Li & L.Zhang [46] y Ou [50], donde se demuestra que no existe solucion positiva para

(−∆)1/2u = up in Rn,

1.6. Resumen de los resultados 21

si 1 < p < n+1n−1

. Pero necesitamos demostrar tambien el resultado analogoen el caso del semi espacio, lo cual llevamos a cabo en esta tesis. Para ello,usamos una idea fue introducida por Cabre y Sola-Morales [33] para estudiarsoluciones “layer” de

(−∆)1/2u = f(u) in Rn,

donde f verifica que si G(u) = −∫ u

0f(s) ds, G tiene dos, y solamente dos,

mınimos absolutos. Ellos desarrollaron una estimacion no local y una cantidadconservada o Hamiltoniana, que es satisfecha por cada solucion “layer”. Laidea antedicha es util para probar el teorema de tipo Liouville en nuestro caso.Nuestra prueba usa tambien la transformacion de Kelvin, el metodo de losplanos moviles y una identidad Hamiltoniana. Usando estos teoremas de tipoLiouville, presentaremos las estimaciones a priori de tipo Gidas-Spruck parasoluciones positivas del problema (1.5.1) con nolinealidad subcrıticas.

1.6 Resumen de los resultados

Un retıculo elastico

Estudiamos el funcional fuertemente indefinido en el espacio de Hilbertcorrespondiente y utilizamos el teorema de enlace de Benci y Rabinowitz paraeste funcional y un metodo de Nirenberg [15]. Obtenemos soluciones para elmovimiento del sistema elastico en un retıculo en el caso autonomo.

Teorema 1.6.1. Supongamos que gl ≡ 0 para todo l ∈ N = 1, 2, . . . , N. En-tonces existen infinitas soluciones periodicas no triviales para (1.4.1)-(1.4.3).

El segundo resultado es la existencia de movimientos en el caso forzado. Lademostracion se sigue por un proceso de perturbacion de un funcional simetricode Z2. Depende de las estimaciones del ındice de Morse de ciertos puntoscrıticos y del analisis de espectro del operador de ondas, como fue desarrolladopor Tanaka [21].

Teorema 1.6.2. Sea Ω = (0, π)×(0, 2π) y p > 1. Supongamos que max2, p <α < p + 1 y gl ∈ Lα/(α−1)(Ω) satisface

∫Ωglζ dxdt = 0 para todo ζ ∈ L∞(Ω) ∩

ker (∂tt − ∂xx), l ∈ N = 1, 2, . . . , N.Entonces el problema (1.4.1)-(1.4.3) tiene infinitas soluciones periodicas no

triviales. Ademas si gl ∈ L∞(Ω) ∩ ker (∂tt − ∂xx) satisface∑

l gl 6= 0, entonces

el problema (1.4.1)-(1.4.3) no tiene solucion periodica.

La raız cuadrada del Laplaciano

Por procedimientos de minimizacion y una identidad de tipo Pohozaevdemostramos, respectivamente, la existencia y no existencia de soluciones pos-itivas del problema (1.5.1):

Teorema 1.6.3. Sea n ≥ 1 un entero y 2] = 2nn−1

para n ≥ 2. Supongamos queΩ es un dominio acotado suave en Rn y f(u) = up. Entonces,(i) El problema (1.5.1) tiene al menos una solucion de clase C2(Ω) cuando1 < p < 2] − 1 = n+1

n−1si n ≥ 2, o 1 < p <∞ si n = 1.

(ii) Sea p ≥ 2] − 1 = n+1n−1

y n ≥ 2. Si Ω es un dominio de tipo estrellado,entonces (1.5.1) no tiene solucion debil en L∞(Ω).

En el caso crıtico, obtenemos resultados mas precisos con la tecnica deBrezis-Nirenberg usada aquı para el problema (1.5.1) con la raız cuadrada delLaplaciano A1/2:

Teorema 1.6.4. Sea n ≥ 2. Supongamos que Ω es un dominio acotado suave

en Rn y f(u) = un+1n−1 +µu. Denotamos por µ1 =

√λ1 al primer valor propio de

A1/2, donde λ1 es el primer valor propio del Laplaciano −∆ en Ω con condicionnula de tipo Dirichlet.

Entonces, para todo µ ∈ (0, µ1), existe una solucion de clase C2(Ω) de(1.5.1). Ademas, no existe solucion debil de (1.5.1) de µ ≥ µ1.

Tambien demostramos un resultado de simetrıa de tipo Gidas-Ni-Nirenbergpara el problema (1.5.1) por el metodo de los planos moviles:

Teorema 1.6.5. Sea Ω un dominio acotado en Rn que es convexo en la direc-cion x1 y es simetrico con respecto al hiperplano x1 = 0. Supongamos que ues una solucion de clase C2(Ω) de (1.5.1) y f es una funcion Lipschitz.

Entonces u es simetrica con respecto a x1, i.e. u(−x1, x′) = u(x1, x

′) paratodo (x1, x

′) ∈ Ω. Ademas, ∂u∂x1

< 0 para x1 > 0.

Como consecuencia del Teorema 1.6.5, sea v una solucion del problema(1.5.2), suponiendo que Ω es una bola y f es Lipschitz. Entonces concluimosque v es axialmente simetrica con respecto al eje y.

Finalmente damos estimaciones a priori de tipo Gidas-Spruck para solu-ciones positivas de (1.5.1) en el caso subcrıtico por el metodo de blow up:

Teorema 1.6.6. Sea n ≥ 2. Supongamos que Ω ⊂ Rn es un dominio acotadosuave y f(u) = up, 1 < p < n+1

n−1.

Entonces existe un constante C, dependiente solo de p y Ω, tal que todaslas soluciones debiles u de (1.5.1) satisfacen ‖u‖∞ ≤ C(p,Ω).

1.6. Resumen de los resultados 23

Para demostrar este teorema necesitamos resultados no lineales de tipoLiouville para A1/2. En el caso de espacio Rn este resultado fue probado en[46],[50]. Tambien necesitamos este resultado en el semi espacio Rn

+. Este noera conocido, y lo demostramos aquı:

Teorema 1.6.7. Sea Rn+1++ = z = (x1, x2, · · · , xn, y) | xn > 0, y > 0 y

sea 1 0, y = 0,v = 0 sobre xn = 0, y > 0v > 0 en Rn+1

++ ,

(1.6.1)

donde ν es la normal exterior unitaria de xn > 0, y = 0.

Para la demostracion de este resultado se combina la transformacion deKelvin, el metodo de los planos moviles, y una identidad Hamiltoniana comoen [33].

Chapter 2

The existence of infinitely manyperiodic solutions for a twodimensional lattice

2.1 Introduction

In this chapter we consider the periodic motion of a two dimensional elas-tic lattice coupled with nearest neighbor interaction potentials of power-typefunctions:


for x ∈ (0, π), t ∈ R, l ∈ N, satisfying the periodic-Dirichlet conditions

ul(0, t) = 0 = ul(π, t), ∀t ∈ R, l ∈ N, (2.1.2)


where ul = (ultt − ulxx), N = 1, 2, . . . , N, N ∈ N, uN+1 = u1, u0 = uN ,1 < p <∞ and gl(x, t+ 2π) = gl(x, t) for l ∈ N. We are looking for solutionswhich are periodic in time.

In last decades a considerable effort has been devoted to the mathematicalstudy of periodic motion of elastic lattices constituted by coupled flexible orelastic elements as strings, beams, membranes or plates. They can be viewedas systems of partial differential equations on networks or graphs. These sys-tems are known as multi-link or multi-body structures. They may generatenew, unexpected phenomena and could have a huge application in the future.However, the mathematical models describing the motion of elastic lattices arequite complex. For the control problems of this kind of systems, we can findwide information in [8].

25

26 Chapter 2. A two dimensional lattice

On the other hand, a 1-dimensional lattice takes the form as (2.1.1)-(2.1.3)by replacing the operator with d2/dt2. Its complete integrability and exis-tence of periodic solutions are well established in [1],[9],[10] and [22]. In par-ticular, the classical Toda lattice was shown to be completely integrable, withexplicit periodic and soliton solutions in [22]. It is well known by the KAM(Kolmogorov-Arnold-Moser) theory [1] that periodic and quasi-periodic solu-tions of the Toda lattice persist under small perturbations. A surprising result,the existence of soliton was obtained for the power-type interactions in [9]. In[10] travelling wave solutions (periodic oscillations and heteroclinic solutions)were constructed on the lattice, corresponding to mass particles interactingnonlinearly with their nearest neighbor (the Fermi-Pasta-Ulam model). Thesemodels have wide application in many physical systems and biology models.

For simplicity, here we only consider a simplified version of those modelsof lattices coupled with nearest neighbor interactions on one line. This maygive us the way to later address more complex situations. To the best of ourknowledge, the nonlinear problem about motion of a 2-dimensional Toda latticelike (2.1.1)-(2.1.3) was first introduced by Mikhailov in [13], which showedits integrability by using inverse problem method. Then the reduced problemabout the two-dimensional generalized Toda lattice was considered in [14],all involving exponential potentials. The existence of periodic motion of the2-dimensional Toda lattice has been explored in [12] by nonlinear analysismethods. Here we will use minimax methods to find the critical points whichcorrespond to the solutions of (2.1.1)-(2.1.3) but with interactions of power-type functions. Our main goals are to prove the following two theorems.

Theorem 2.1.1. If gl ≡ 0 for all l ∈ N = 1, 2, . . . , N, then lattice (2.1.1)-(2.1.3) has infinitely many nontrivial periodic solutions.

Theorem 2.1.2. Assume that gl ∈ Lα/(α−1)(Ω) satisfies∫

Ωglζ dxdt = 0 for all

ζ ∈ L∞(Ω)∩ ker(∂tt− ∂xx), where Ω = (0, π)× (0, 2π), p > 1 and max2, p <α < p + 1, l ∈ N = 1, 2, . . . , N. Then lattice (2.1.1)-(2.1.3) has infinitelymany nontrivial periodic solutions.

Moreover, if gl ∈ L∞(Ω) ∩ ker(∂tt − ∂xx) satisfies∑

l gl 6= 0, then problem

(2.1.1)-(2.1.3) has no periodic solution.

Consider the functional for the case autonomous case g ≡ 0. Applyingvariational methods to the problem of the existence and multiplicity of periodicsolutions for (2.1.1)-(2.1.3), we have the functional

G0(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

F0(u) dxdt,

2.1. Introduction 27

where

F0(u) =1

p+ 1

N∑l=1

|ul+1 − ul|p+1, uN+1 = u1.

Note that 12

∫Ω(|∂tu|2 − |∂xu|2) dxdt corresponds to the kinetic energy and∫

ΩF0(u) dxdt presents the potential energy of the system.Denote u(x, t) = (u1(x, t), . . . , uN(x, t)) for (x, t) ∈ Ω = (0, π) × (0, 2π).

For p ∈ [1,∞) set by Lp(Ω) or Lp the vector of 2π-time periodic functions oft whose p-th powers are integrable, i.e.,

‖u‖p = (

∫Ω

N∑l=1


A vector of smooth functions satisfying periodic-Dirichlet conditions (2.1.2)-(2.1.3) has a Fourier expansion of the form

u = (aljk) =N∑l=1

∞∑j=1

∞∑k=−∞

aljk sin jxeiktel, alj,−k = aljk,

where e1, · · · , eN is the usual orthogonal basis in RN . From here observethat the space of functions spansin jxeiktel, j = |k|, l ∈ N is contained inthe kernel of the wave operator vector (∂ltt − ∂lxx)

Nl=1. It will be convenient to

use the Lp+1 norm on this space. On the complementary space of functionsspansin jxeiktel, j 6= |k|, l ∈ N, where the wave operator does not vanishesfor non-zero vectors, we define an inner product and corresponding norm asfollows: for u and v, we let

〈u, v〉 =1

4|Ω|

N∑l=1

∞∑j=1

∞∑k=−∞,|k|6=j

|k2 − j2|aljkbljk, ‖u‖2 = 〈u, u〉.

On the other hand, it is easy to see that if u satisfies equations (2.1.1),then

∑Nl=1 ul =

∑Nl=1 g

l, so that in the case autonomous case we have that∑Nl=1 ul = 0. So it is natural to consider our problem in the space E ⊕ E0

where

E0 := span |Lp+1 (sin jxeiktel), for j = |k|, l ∈ N ∩ u |N∑l=1

ul = 0

and

E := span(sin jxeiktel), for j 6= |k|, l ∈ N ∩ u |N∑l=1

ul = 0.


In E the closure is considered with respect to the norm just defined above.

Moreover, we will see in Proposition 2.2.3, that still in the non-autonomouscase we may consider the problem just in E0 ⊕ E.

We denote by E+ the subspace of E where the kinetic energy part of G0 ispositive and by E− the subspace of E where it is negative. We observe thatE+, E− and E0 are all infinite dimensional.

So we are led to consider the strongly indefinite functional G0(u) in E⊕E0.Furthermore, since the kernel of the wave operator is infinite dimensional andthe embedding operator from vector space of p-integrable functions to thiskernel is not compact, we turn to study the auxiliary functional I(ϕ) as Tanaka[21]:

I(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ),

where

Q0(ϕ) := minψ∈E0

∫Ω

F0(ϕ+ ψ) dxdt

for ϕ = ϕ+ + ϕ− ∈ E := E+ ⊕ E−. As a consequence of Lemma 2.2.2, we seethat Q0 is a strictly convex functional in E. Then we apply linking theorem ofBenci and Rabinowtz in [5] to obtain a critical point ϕ of I in E. Let ϕ ∈ Ebe the critical point for I and ψ ∈ E0 be the minimizer for Q0. Thus the sumϕ+ ψ is the nontrivial solution for the autonomous lattice we are looking for.The multiplicity follows by the simple argument of Nirenberg [15]. These aregiven in Section 2.3.

Theorem 2.1.2 is established from Section 2.4 to Section 2.7. It will beproved with the aid of a standard argument for functionals perturbed from Z2

symmetry. Here our steps follow Tanaka’s framework in [21]. We introduce afunctional I associated to problem (2.1.1)-(2.1.3),

I(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q(ϕ),

where

Q(ϕ) := minψ∈E0

∫Ω

Fg(ϕ+ ψ) dxdt

and

Fg(u) =1

p+ 1

N∑l=1

|ul+1 − ul|p+1 − (g, u),

for ϕ = ϕ+ +ϕ− ∈ E := E+⊕E−. Notice that Q is a strictly convex functionalin E. Since there is the force term g 6≡ 0, the functional I is no longer Z2


symmetric, hence the method need to be adapted. We change I by making asimple modification in Section 2.4:

J(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ)− χ(ϕ)(Q(ϕ)−Q0(ϕ)),

for an appropriate function χ(ϕ). Then we can apply the methods of Rabi-nowitz [17] to J in Section 2.5 and obtain the existence of infinitely manysolutions of (2.1.1)-(2.1.3) under a crucial assumption. For q ∈ N, we definethe critical values for a certain set Dq of E and a family of maps from Dq toE:

bq = infγ∈Γq

supϕ∈Dq

J(γ(ϕ)).

Under the crucial assumption

bqj ≥ C1q(p+1)/[(p−1)(1−ε)]j − C2 (2.1.4)

for a subsequence qj ⊂ N, using a comparison argument deviced by Rabi-nowitz [17], we will prove that the functional J has large critical values thatin turn correspond to critical values of I.

To prove (2.1.4), we follow Tanaka’s steps, which were introduced by Bahri-Berestycki, to find a comparing functional in E+ defined as

K(ϕ) =1

2‖ϕ‖2 − a0

p+ 1‖ϕ‖p+1

p+1,

where a0 is a constant. Then, for q, n ∈ N and n > q, we define the criticalvales for K

σnq = supσ∈An

q

miny∈Sn−q

K(σ(y)),

where Sn−q is the unit sphere and Anq is the family of symmetric maps fromSn−q to a finite dimensional subspace E+

n of E. Rather than the number ofσnq we consider σq = limn→∞ σnq for comparison, proving in Section 2.5 thatσq ≤ bq + C2 for all q. Then we follow the arguments by Tanaka [21] by usingthe Morse index estimates and the spectral analysis in Section 2.6 and Section2.7 to prove that for large qj ∈ N,

σqj ≥ C1q(p+1)/[(p−1)(1−ε)]j . (2.1.5)

Consequently, we obtain (2.1.4) and finally we complete the proof of Theorem2.1.2 at the end of this chapter.


2.2 Preliminaries

For convenience, we use the following standard notation. Let E be a Hilbertspace, 〈·, ·〉 be the inner product in the Hilbert space E or the dual bracketbetween the Hilbert space E and its dual space E∗. We denote by (·, ·) theusual inner product in RN .

Let Ω = (0, π) × (0, 2π). We recall basic properties of the wave operator∂tt − ∂xx acting on integrable functions which are 2π−periodic and satisfyDirichlet boundary conditions. It is well known that (∂tt − ∂xx)ζ = 0 (in theweak sense) if and only if

ζ(x, t) = γ(x+ t)− γ(x− t) (2.2.1)

for some γ ∈ L1loc(R), 2π-periodic and such that

∫ 2π

0γ(s) ds = 0. Also, let h be

an integrable function satisfying∫Ω

hζ dxdt = 0 for all ζ ∈ L∞(Ω) ∩ ker(∂tt − ∂xx).

Then there is a unique 2π-periodic continuous function ϑ satisfying

ϑtt − ϑxx = h (2.2.2)

and the boundary condition ϑ(0, t) = ϑ(π, t) = 0 for all t ∈ (0, 2π). From this,it is clear that if 0 6= h ∈ ker(∂tt − ∂xx), then there is no solution for problem(2.2.2).

Denote u(x, t) = (u1(x, t), . . . , uN(x, t)) for (x, t) ∈ Ω. For p ∈ [1,∞) wedenote by Lp(Ω) the vector space of 2π-time periodic functions of u(x, t) whosep-th powers are integrable, i.e.,

‖u‖p = (

∫Ω

N∑l=1


A vector of smooth functions satisfying (2.1.2),(2.1.3) has a Fourier expansionof the form

u =N∑l=1

∞∑j=1

∞∑k=−∞


where e1, · · · , eN is the usual orthogonal basis in RN . We define

〈u, v〉 =1

4|Ω|

N∑l=1

∞∑j=1

∞∑k=−∞,|k|6=j

|k2 − j2|aljkbljk, ‖u‖ := 〈u, u〉,

2.2. Preliminaries 31

for u =∑

l,j,k aljk sin jxeiktel and v =

∑l,j,k b

ljk sin jxeiktel. We observe that

‖ · ‖ is a norm on the set u| aljk = 0 if j = |k|. Let

W+ = span(sin jxeiktel)| j < |k|, l = 1, 2, · · · , N,W− = span(sin jxeiktel)| j > |k|, l = 1, 2, · · · , N,W = W+ ⊕W−,

where the closures are taken under the norm ‖ · ‖. Note that (W, 〈·, ·〉) is aHilbert space. Further set

W 0 = span |Lp+1 (sin jxe±ijtel)| j ∈ N, l = 1, 2, · · · , N.

Then W+,W− and W 0 are complementary subspaces of the vector space offunctions satisfying (2.1.2),(2.1.3). Moreover, we will prove in this section thatthe space W has the following properties:

Lemma 2.2.1. There is a generic constant Cτ such that

‖u‖τ ≤ Cτ‖u‖ for all u ∈ W and τ ∈ [1,∞),

the embedding W → Lτ is compact for all τ ∈ [1,∞).

However, the embedding W 0 → Lτ is not compact for all τ ∈ [1,∞).

Since∑N

l=1 ul =∑N

l=1 gl, we will study lattice (2.1.1)-(2.1.3) in a subspace

E of W , which will be our working space. Let 11 := (1, 1, · · · , 1) ∈ RN and let11⊥ be its orthogonal space. Given ξ ∈ RN , we define [ξ] = maxξ1−ξN , ξ2−ξ1, · · · , ξN − ξN−1. It is known from [12] that

[ξ] ≥ 2

N − 1|ξ|∞ ≥ 2√

N(N − 1)|ξ| for ξ ∈ 11⊥.

Moreover, we have

Lemma 2.2.2. Let

F0(ξ) =1

p+ 1

N∑l=1

|ξl+1 − ξl|p+1,

where ξN+1 = ξ1. Then we have the qualitative properties for F0:(i) F0 ∈ C2(RN ∩ 11⊥,R) is strictly convex and coercive.(ii) For every α ∈ (0, p + 1], we have (ξ,∇F0(ξ)) ≥ αF0(ξ) > 0 for all ξ ∈RN ∩ 11⊥ and ξ 6= 0.(iii) There are generic positive constants C1, C2 > 0 such that for ξ ∈ RN ∩11⊥,

F0(ξ) ≥ C1|ξ|α + C2. (2.2.4)

Let w(x, t)11 := (w(x, t), w(x, t), · · · , w(x, t)) ∈ W ⊕ W 0. Denote X1 =w11 ∈ W ⊕W 0 and by X2 = X⊥

1 its orthogonal space with respect to theinner product in RN . It is clear that X2 = (u0, · · · , uN)|

∑Nl=1 u

l = 0. Definethe function spaces

E+ = W+ ∩X2, E− = W− ∩X2,

E = E+ ⊕ E−, E0 = W 0 ∩X2,(2.2.5)

endowed with the norm of W . Note that E has the same compact embeddingproperties of the space W and that W ⊕W 0 = [(W ⊕W 0)∩X1]⊕ (E ⊕E0).

On the other hand, it is clear that the solutions of problem (2.1.1) satisfyingconditions (2.1.2),(2.1.3) correspond to critical points of the functional foru ∈ W ⊕W 0

G1(u) :=1

2

∫Ω

(|∂tu|2 − |∂xu|2) dxdt−∫

Ω

Fg(u) dxdt, (2.2.6)

where

Fg(u) =1

p+ 1

N∑l=1

|ul+1 − ul|p+1 − (g, u), uN+1 = u1.

If g ∈ W ⊕ W 0, then we have the decomposition g = h11 + g such thath11 ∈ (W ⊕W 0)∩X1 and g ∈ E⊕E0 = (W ⊕W 0)∩X2. It is easy to see thatproblem (2.1.1)-(2.1.3) in (W ⊕W 0) ∩X1 becomes a scalar equation

vtt − vxx = h, (2.2.7)

and v is a 2π periodic function satisfying Dirichlet condition. Furthermore,we know from (2.2.2) that there is a unique solution v if h ∈ L1(Ω) andh 6∈ ker(∂tt − ∂xx). Suppose that for max2, p < α < p + 1, gl ∈ Lα/(α−1)(Ω)satisfies ∫

Ω

glζ dxdt = 0 for all ζ ∈ L∞(Ω) ∩ ker(∂tt − ∂xx), l ∈ N,

and that there is a critical point u of G1 in E, which is the main fact we needto prove in this chapter. Then u + v11 is a solution of (2.1.1)-(2.1.3), wherev is the solution of (2.2.7). To see this, we have that for ζ ∈ E ⊕ E0 andξ ∈ (W ⊕W 0) ∩X1,

〈(G1)′(u+ v11), (ζ + ξ)〉 =

∫Ω

N∑l=0

[ultζ

lt + vtξ

lt − ulxζ

lx − vxξ

lx

− (F0(u), ζ) + (h11, ξ) + glζ l]dxdt = 〈(G1|E⊕E0)′(u), ζ〉,

where g = h11 + g is such that h11 ∈ (W ⊕W 0) ∩X1 and g ∈ E ⊕ E0.Summarizing the above facts, we have:


Proposition 2.2.3. Set the spaces of functions as in (2.2.5). Let g = h11 + gas above. Then there is a solution v of (2.2.7) for lattice (2.1.1)-(2.1.3), whichwe call the trivial solution, if h ∈ L1(Ω) and h 6∈ ker(∂tt − ∂xx). Moreover, letu be a critical point of G1 in E⊕E0. Then u+v11 is the nontrivial solution forlattice (2.1.1)-(2.1.3). In particular, If g ≡ 0, then there is always the trivialsolution as in (2.2.1) for lattice (2.1.1)-(2.1.3). Otherwise, h ∈ ker(∂tt − ∂xx),then there is no solution for problem (2.1.1)-(2.1.3).

For this reason, we may assume∑N

l=1 gl = 0 without loss of generality.

Thus, the critical points of G1 in E ⊕ E0 gives the existence of solutions forlattice (2.1.1)-(2.1.3) that we are looking for in this chapter.

To find the critical points of G1, we will introduce an auxiliary functional totreat the loss of compactness of the embedding E0 → Lp+1. Letting u = ϕ+ψ,where ϕ = ϕ++ϕ− ∈ E and ψ ∈ E0, we observe that G1 ∈ C2(E+⊕E−⊕E0,R)and

G1(u) = G1(ϕ+ + ϕ− + ψ)

=1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −

∫Ω

Fg(ϕ+ + ϕ− + ψ) dxdt.

The kinetic energy part corresponding to the wave operator is positive definite,negative definite and null on E+, E− and E0 respectively. Observe also that forfixed ϕ = ϕ+ + ϕ−, the functional ψ 7→ G1(ϕ

+ + ϕ− + ψ) is a strictly concavefunction of ψ ∈ E0. Since F0 is strictly convex in 11⊥, there is a one to onecorrespondence between critical points of G1 in E ⊕ E0 and those of I in E,where I : E → R is the auxiliary functional for ϕ = ϕ+ + ϕ− ∈ E, defined as

I(ϕ) = maxψ∈E0

G1(ϕ+ ψ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q(ϕ), (2.2.8)

where

Q(ϕ) := minψ∈E0

∫Ω

Fg(ϕ+ ψ) dxdt for ϕ ∈ E. (2.2.9)

We treat problem (2.1.1)-(2.1.3) by finding the critical points of the functionalI in E.

Note thatQ(ϕ) can be also defined for all ϕ ∈ Lp+1 by (2.2.9). Now we provesome properties that we mentioned above about the space E, the function F0

and treat Q as a functional in Lp+1(Ω).

Proof of Lemma 2.2.1. The proof of compactness of the embedding is stan-dard, see [20] or [11]. Clearly, W is compactly embedded in L2(Ω), hence inLτ (Ω) for τ ∈ [1, 2] by Holder’s inequality. We assume τ > 2. So we observe

that u ∈ W has the form:

u =N∑l=1

∞∑j=1

∑|k|6=j

aljk sin jxeiktel, alj,−k = aljk.

Denote ajk = (a1jk, · · · , aNjk) by |ajk| its modulus in RN . We first note that for

p > 1, ∑j 6=|k|

|j2 − k2|−p ≤ 4∑j<|k|

(k2 − j2)−p +∑j≥1

j−2p

=∑j,l≥1

l−p(2j + l)−p +∑j≥1

j−2p ≤ C∑j,l≥1

l−pj−p +∑j≥1

j−2p

≤ C∑j≥1

j−2p ≤ 3(p

p− 1)2.

Then by F. Riesz’s Theorem in [23], Holder’s inequality, letting p = τ/(τ − 2)in the above inequality, we have that for τ > 2,

‖u‖τ ≤ π2/τ( ∑j 6=|k|

|ajk|τ/(τ−1))(τ−1)/τ

≤ π2/τ( ∑j 6=|k|

|j2 − k2||ajk|2)1/2( ∑

j 6=|k|

|j2 − k2|−τ/(τ−2))(τ−2)/2τ

≤ Cτ‖u‖.

Hence, we see that W is continuously embedded in Lτ (Ω) for τ > 2. Then bythe interpolation inequality, the compactness follows. 2

Proof of Lemma 2.2.2. (i) Since

d2F0(ξ + sζ)

ds2|s=0 =

N∑l=1

p|ξl − ξl−1|p−1(ζ l − ζ l−1)2,

this gives that F0 is strictly convex in 11⊥.(ii) It is clear that for α ≤ p+ 1,

〈∇F0(ξ), ξ〉 − αF0(ξ) = (1− α

p+ 1)[

N∑l=1

|ξl − ξl−1|p+1] ≥ 0.

Hence we have that, for α ∈ (0, p+ 1), and for all ξ ∈ 11⊥, ξ 6= 0,

〈∇F0(ξ), ξ〉 ≥ αF0(ξ) > 0.


(iii) Let s = |ξ| and η = ξ|ξ| . So ξ = sη and then

〈∇F0(sη), sη〉 ≥ αF0(sη).

It directly gives that for s ≥ 1

F0(sη) ≥ C1sα,

from where we can get (iii). 2

Lemma 2.2.4. (i) For all ϕ ∈ Lp+1(Ω), there exists a unique ψ(ϕ) ∈ E0 suchthat

Q(ϕ) =

∫Ω

Fg(ϕ+ ψ(ϕ)) dxdt. (2.2.10)

(ii) ψ : Lp+1(Ω) → E0 is continuous.(iii) Q is of class C1 on E and

〈Q′(ϕ), h〉 =

∫Ω

(∇Fg(ϕ+ ψ(ϕ)), h) dxdt for all ϕ, h ∈ E. (2.2.11)

In particular, Q′ : E → E∗ is compact.

Proof. The above lemma is a slight generalization from Tanaka [21].(i) From the property (i) of F0 in Lemma 2.2.2, we see that

ψ →∫

Ω

Fg(ϕ+ ψ) dxdt (2.2.12)

is a strictly convex and coercive functional on E0. Then we have the firstassertion.(ii) Suppose that ϕj → ϕ in Lp+1(Ω). We show that ψ(ϕj) → ψ(ϕ) strongly inE0. By the definition of ψ(ϕj), we have∫

Ω

Fg(ϕj + ψ(ϕ)) dxdt ≥∫

Ω

Fg(ϕj + ψ(ϕj)) dxdt, (2.2.13)

from where we find that ψ(ϕj) is bounded in E0 (i.e., in Lp+1(Ω)). We extracta subsequence, still denoted by ψ(ϕj), that converges point-wise a.e in Ω andconverges weakly to ψ in E0. Letting j → ∞ in (2.2.13), by Fatou’s lemmaand weak continuity, we obtain∫

Ω

Fg(ϕ+ ψ(ϕ)) dxdt ≥ lim supj→∞

∫Ω

Fg(ϕj + ψ(ϕj)) dxdt

≥∫

Ω

Fg(ϕ+ ψ) dxdt.


By the uniqueness of ψ(ϕ), we observe that ψ = ψ(ϕ) and lim supFg(ϕj +ψ(ϕj)) = Fg(ϕ+ ψ(ϕ)). Thus we obtain ψ(ϕj) → ψ(ϕ) in E0.(iii) By the convexity of (2.2.12), we find that for w ∈ E0,

w = ψ(ϕ) iff

∫Ω

(∇Fg(ϕ+ w), ζ) dxdt = 0 for all ζ ∈ E0. (2.2.14)

From the convexity of the function Fg and minimality property of ψ(ϕ), wehave that, for all ϕ, h ∈ E and s > 0,

Q(ϕ+sh)−Q(ϕ) ≥∫

Ω

(∇Fg(ϕ+ψ(ϕ)), sh+ψ(ϕ+sh)−ψ(ϕ)) dxdt. (2.2.15)

Since ψ(ϕ+ sh)− ψ(ϕ) ∈ E0, by (2.2.14) we get

Q(ϕ+ sh)−Q(ϕ) ≥∫

Ω

(∇Fg(ϕ+ ψ(ϕ)), sh) dxdt. (2.2.16)

Similarly we have

Q(ϕ+ sh)−Q(ϕ) ≤∫

Ω

(∇Fg(ϕ+ sh+ ψ(ϕ+ sh)), sh) dxdt. (2.2.17)

Dividing by s and letting s → 0 in (2.2.16) and (2.2.17), we obtain (2.2.11).Thus Q ∈ C1(E,R). Moreover from Lemma 2.2.1 and the continuity of ψ :Lp+1(Ω) → E0, we deduce that Q′ : E → E∗ is compact. 2

2.3 Periodic solutions for an autonomous lat-

tice

In this section we prove the existence of infinitely many solutions for a au-tonomous lattice: g ≡ 0. Consider the functional

I(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ) for ϕ = ϕ+ + ϕ− ∈ E.

The critical points of I are the solutions of our problem. Let Br be an openball with radius r and ai, ri, i ∈ N denote nonnegative constants. Recall thefollowing infinite dimensional linking theorem built by Benci and Rabinowitz.

Lemma 2.3.1. [5] Let E be a real Hilbert space, E1 a closed subspace of E,and E2 = E⊥

1 . Denote I(u) = Φ(u) + b(u). Suppose that I ∈ C1(E,R) andsatisfies(I1) Φ(u) = 1

2〈Lu, u〉 where u = u1 + u2 ∈ E1 ⊕ E2, Lu = L1u1 + L2u2 and

2.3. An autonomous lattice 37

Li : Ei → Ei, i = 1, 2 is a (bounded) linear self adjoint mapping.(I2) b is weakly continuous and uniformly differentiable on bounded subsets ofE. Here the weak continuity of b means that b(um) converges to b(u), wheneverum converges weakly to u. The uniformly differentiability of b on bounded setsmeans that for every R, ε > 0, there exists a δ > 0, depending only on R and ε,such that if u, u+ v are contained in BR := u ∈ E | ‖u‖ ≤ R and ‖v‖ ≤ δ,then

|b(u+ v)− b(u)− 〈b′(u), v〉| ≤ ε‖v‖,

where b′(u) denotes the Frechet derivative of b at u.(I3) If for a sequence um∞m=1, I(um) is bounded from above and I ′(um) → 0as m→∞, then um is bounded.(I4) There are constants r1, r2, α, ρ, ω with r1 > ρ, α > ω and r1, r2, ρ > 0 andthere is an e ∈ ∂B1 ∩E1 such that (i) I ≥ α on S := ∂Bρ ∩E1, (ii) I ≤ ω on∂Σ where Σ := re| 0 ≤ r ≤ r1 ⊕ (Br2 ∩ E2).

Then I possesses a critical value c ≥ α.

Proof of Theorem 2.1.1. The results follows directly by using Lemma 2.3.1.Let E1 = E+, E2 = E− and

〈Liϕi, ζi〉 =

∫Ω

(∂tϕi∂tζi − ∂xϕi∂xζi) dxdt,

for all ζi ∈ Ei, i = 1, 2. Then I satisfies (I1) with Liϕi be defined for ϕi ∈ Ei,for i = 1, 2. Hence for ϕ ∈ E,

Φ(ϕ) =1

2〈Lϕ, ϕ〉 =

1

2

∫Ω

(|∂tϕ|2 − |∂xϕ|2) dxdt

and

b(ϕ) = −Q0(ϕ) = −∫

Ω

F0(ϕ+ ψ(ϕ)) dxdt.

By Lemma 2.2.4 (ii) we get I ∈ C1(E,R). It remains to prove that I(ϕ) satisfies(I2)-(I4).

Step 1. We claim that b is weakly continuous. To prove this, assume thatϕj ⊂ E and ϕj converges weakly to ϕ in E. Then by Lemma 2.2.1, ϕj →ϕ in Lβ for all β ∈ [1,∞). Since

F0(ϕ) ≤ C|ϕ|p+1, (2.3.1)

choosing β = p+ 1, we see that ϕj → ϕ in Lp+1 and b(ϕj) → b(ϕ) as j →∞.Step 2. By Lemma 2.2.4 and Lemma 2.2.1, we see that b(ϕ) = −Q0(ϕ)

has uniform differentiability on bounded subsets of E as [16]. Thus we verifycondition (I2).

2.3. An autonomous lattice 39

Then, by (2.3.4) we obtain

|〈Q′0(ϕj), ϕ

+j 〉| ≤ C(Q0(ϕj)

p/α + 1)‖ϕ+j ‖.

Also by (2.3.5) we get

|〈Q′0(ϕj), ϕ

−j 〉| ≤ C(Q0(ϕj)

p/α + 1)‖ϕ−j ‖.

Therefore, by (2.3.3) and (2.3.2), we see

‖ϕj‖2 ≤ C(Q0(ϕj)p/α‖ϕj‖+ ‖ϕj‖)

≤ C(‖ϕj‖p/α + 1)‖ϕj‖.

Recall that by hypothesis we have max2, p < α < p+ 1. So ‖ϕj‖ is boundedbecause of α > p.

Step 4. We verify that I satisfies (I4) (i). The form of F0 and the definitionQ imply

Q0(ϕ) ≤∫

Ω

F0(ϕ) dxdt

≤ C

∫Ω

|ϕ|p+1 dxdt ≤ C‖ϕ‖p+1.

Thus for ϕ ∈ E1,

I(ϕ) ≥ 1

2‖ϕ‖2 − C‖ϕ‖p+1

= ‖ϕ‖2(1

2− C‖ϕ‖p−1).

Choosing ρ so that Cρp−1 < 12

gives I(ϕ) ≥ 12ρ2 for ϕ ∈ S = ∂Bρ ∩ E1.

Step 5. We show that there exists a set Σ with r1 and r2 such that I satisfies(I4)(ii). In fact, let ϕ = ϕ− ∈ Br2 ∩ E2 where r2 > 0 and consider

I(ϕ+ re) =r2

2− ‖ϕ−‖2

2−

∫Ω

F0(ϕ− + re+ ψ(ϕ− + re)) dxdt, (2.3.6)

where e ∈ ∂B1 ∩ E1. Note that for r = 0, I(ϕ) ≤ 0, since F0 ≥ 0. By (2.2.4),


we have ∫Ω

F0(ϕ+ re+ ψ(ϕ+ re)) dxdt

≥C1

∫Ω

|ϕ+ re+ ψ(ϕ+ re)|α dxdt+ C2

≥C1(

∫Ω

|ϕ+ re+ ψ(ϕ+ re)|2 dxdt)α2 + C2

=C1(

∫Ω

|ϕ|2 + |re|2 + |ψ(ϕ+ re)|2 dxdt)α2 + C2

≥C1r2α

( ∫Ω

|e|2 dxdt)α

2+ C2. (2.3.7)

Hence by (2.3.6) and (2.3.7),

I(ϕ+ re) ≤ r2

2− ‖ϕ−‖2

2− C1r

α − C2. (2.3.8)

Choosing r1 such that

h(r) :=r2

2− C1 − C2r

α ≤ 0 (2.3.9)

for r ≥ r1, it then shows from (2.3.8)-(2.3.9) and F0 ≥ 0 that I ≤ 0 =: ω on∂Σ.

Step 6. By Lemma 2.3.1 we have at least one solution u for the autonomouslattice. To prove the existence of infinitely many periodic solutions one relies onthe following simple argument of [15]: Assume that u is the solutions found bythe previous argument. First, we prove that u = (u1, · · · , uN) depends on thetime t. Suppose that u is independent of t, then by multiplying the equationu+∇F0(u) = 0 with u, we have

∫Ωu2x dxdt+ (p+ 1)

∫ΩF0(u) dxdt = 0. This

yields a contradiction. Therefore, u has a minimal period≤ 2π, say 2π/j for j ∈N. Consider the same equations, but on the space H1((0, π)× (0, 2π/j)). Thisyields another solution with minimal period ≤ 2π/j. Repeating this procedure,we can find infinitely many distinct periodic solutions. 2

2.4 A modified functional

In the following sections, we treat the non-autonomous case. The functionalI will be replaced by a modified functional J . In this section, we study therelationship between functionals J and I and prove that J satisfies the Palais-Smale condition.

2.4. A modified functional 41

Let χ ∈ C∞(R) be such that χ(τ) = 1 for τ ≤ 1, χ(τ) = 0 for τ ≥ 2

and χ′(τ) ≤ 0, 0 ≤ χ ≤ 1 for τ ∈ R. Denote χ(ϕ) = χ( Q0(ϕ)

a[I(ϕ)2+1]1/2 ). For

ϕ = ϕ+ + ϕ− ∈ E+ ⊕ E− = E, we define the functional J :

J(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ)− χ(ϕ)(Q(ϕ)−Q0(ϕ)), (2.4.1)

where a = max1, 12/(p−1). Recall thatQ(ϕ) = minψ∈E0

∫ΩFg(ϕ+ψ(ϕ)) dxdt

and Q0(ϕ) = minψ∈E0

∫ΩF0(ϕ+ ψ(ϕ)) dxdt for ϕ ∈ E.

Proposition 2.4.1. The functional J is of class C1 and satisfies:(i) There is a constant c1 > 0 such that for ϕ ∈ E

|J(ϕ)− J(−ϕ)| ≤ c1(|J(ϕ)|1/α + 1). (2.4.2)

(ii) Assume that there is a constant M0 > 0 such that J(ϕ) ≥ M0 and‖J ′(ϕ)‖E∗ ≤ 1. Then we have J(ϕ) = I(ϕ).

To prove this proposition, we need some lemmas that we prove in whatfollows.

Lemma 2.4.2. There is a constant C > 0 such that for ϕ ∈ E,

|Q(ϕ)| ≤ C(Q0(ϕ) + 1), (2.4.3)

|Q(ϕ)−Q0(ϕ)| ≤ C(Q0(ϕ)1/α + 1). (2.4.4)

Proof. By the definition Q0 and Lemma 2.2.2 (iii), we have

Q(ϕ)−Q0(ϕ) = minψ∈E0

∫Ω

Fg(ϕ+ ψ) dxdt−∫

Ω

F0(ϕ+ ψ0(ϕ)) dxdt

≤ |〈g, ϕ+ ψ0(ϕ)〉| ≤ ‖g‖α/(α−1)‖ϕ+ ψ0(ϕ)‖α≤ C1(Q0(ϕ) + C2)

1/α ≤ C(Q0(ϕ)1/α + 1). (2.4.5)

By using Young’s inequality, we obtain (2.4.3). Similarly we can prove

Q0(ϕ)−Q(ϕ) ≤ C(|Q0(ϕ)|1/α + 1).

Thus we get (2.4.4) from the above inequality and (2.4.5). 2

Lemma 2.4.3. Assume that there is a constant M1 > 0 such that J(ϕ) ≥M1

and ϕ ∈ suppχ. Then we have I(ϕ) ≥ 13J(ϕ).


Proof. Form the definition of J(ϕ) and Lemma 2.4.2,

J(ϕ) = I(ϕ) + (1− χ(ϕ))(Q(ϕ)−Q0(ϕ))

≤ I(ϕ) + C(Q0(ϕ)1/α + 1).

By definition of χ, we get for ϕ ∈ suppχ that Q0(ϕ) ≤ a(I2(ϕ)2 + 1)1/2, then

J(ϕ) ≤ I(ϕ) + C(|I(ϕ)|1/α + 1)

≤ I(ϕ) +1

2|I(ϕ)|+ C1.

Choosing M1 = 2C1, we get the desired result. 2

Lemma 2.4.4. For all ϕ = ϕ+ + ϕ− ∈ E = E+ ⊕ E− and h ∈ E∗,

〈J ′(ϕ), h〉 = (1 + T1(ϕ))〈ϕ+ − ϕ−, h〉 − (1 + T2(ϕ))〈Q′0(ϕ), h〉

−(χ(ϕ) + T1(ϕ))〈Q′(ϕ)−Q′0(ϕ), h〉,

(2.4.6)

where T1, T2 ∈ C(E,R) are functionals satisfying

sup|Ti(ϕ)||ϕ ∈ E, J(ϕ) ≥M2, i = 1, 2 → 0 as M2 →∞. (2.4.7)

Proof. For all ϕ = ϕ+ + ϕ− ∈ E, we have

〈J ′(ϕ), h〉 = 〈ϕ+ − ϕ−, h〉 − 〈Q′0(ϕ), h〉

− 〈χ′(ϕ), h〉(Q(ϕ)−Q0(ϕ))− χ(ϕ)〈Q′(ϕ)−Q′0(ϕ), h〉,

(2.4.8)

where

〈χ′(ϕ), h〉 = χ′( Q0(ϕ)

a[I(ϕ)2 + 1]1/2

)·[− Q0(ϕ)I(ϕ)〈I ′(ϕ), h〉

a(I(ϕ)2 + 1)3/2

+〈Q′

0(ϕ), h〉a(I(ϕ)2 + 1)1/2

]and

〈I ′(ϕ), h〉 = 〈ϕ+ − ϕ−, h〉 − 〈Q′0(ϕ), h〉 − 〈Q′(ϕ)−Q′

0(ϕ), h〉.By regrouping terms, we obtain (2.4.6) with

T1(ϕ) := a−1χ′(·)(I2(ϕ) + 1)−3/2I(ϕ)Q0(ϕ)(Q(ϕ)−Q0(ϕ)), (2.4.9)

T2(ϕ) := T1(ϕ) + a−1χ′(·)(I2(ϕ) + 1)−1/2(Q(ϕ)−Q0(ϕ)), (2.4.10)

which converge to zero. We prove (2.4.7). Suppose that ϕ ∈ E satisfies J(ϕ) ≥M2. Using (2.4.4) we get

|T1(ϕ)| ≤ C|χ′(·)|(I2(ϕ) + 1)−1Q0(ϕ)(Q0(ϕ)1/α + 1).


If ϕ 6∈ suppχ, then T1(ϕ) = 0. Otherwise, by the definition of χ, we have

Q0(ϕ) ≤ a(I2(ϕ) + 1)1/2.

On the other hand, from Lemma 2.4.3 we see

a(I2(ϕ) + 1)1/2 ≥ I(ϕ)

≥ 1

3J(ϕ) ≥ 1

3M2.

Hence we obtain that for M2 →∞,

|T1(ϕ)| ≤ C(a(I2(ϕ) + 1)1/2)−(α−1)/α)

≤ CM−(α−1)/α)2 → 0.

Similarly, we can deduce that T2(ϕ) → 0 as M2 →∞. Thus we get (2.4.7). 2

Lemma 2.4.5. For all ϕ ∈ E, there exists a constant C such that

(i) |〈Q′(ϕ), ϕ〉 − (p+ 1)Q(ϕ)| ≤ C(|Q(ϕ)|1/α + 1), (2.4.11)

(ii) |〈Q′0(ϕ), ϕ+ − ϕ−〉| ≤ C(|Q0(ϕ)|p/α + 1)‖ϕ‖, (2.4.12)

(iii) |〈Q′(ϕ), ϕ+ − ϕ−〉| ≤ C(|Q(ϕ)|p/α + 1)‖ϕ‖, (2.4.13)

(iv) |〈Q′(ϕ)−Q′0(ϕ), ϕ〉| ≤ C(|Q0(ϕ)|p/α + 1)‖ϕ‖. (2.4.14)

Proof. (i) By (2.2.10), (2.2.11) and (2.2.14), we obtain that for ϕ ∈ E,

|(p+ 1)Q(ϕ)− 〈Q′(ϕ), ϕ〉| ≤ p|∫

Ω

(g, ϕ+ ψ(ϕ)) dxdt|

≤ C‖g‖α/(α−1)‖ϕ+ ψ(ϕ)‖α.

On the other hand, by Lemma 2.2.2, we see∫Ω

|ϕ+ ψ(ϕ)|α dxdt

≤C1

∫Ω

Fg(ϕ+ ψ(ϕ)) dxdt+ C1

∫Ω

(g, ϕ+ ψ(ϕ)) dxdt+ C2

≤C1|Q(ϕ)|+ C1‖g‖α/(α−1)‖ϕ+ ψ(ϕ)‖α + C2.

By Young’s inequality, we have∫Ω

|ϕ+ ψ(ϕ)|α dxdt

≤C1|Q(ϕ)|+ ε‖ϕ+ ψ(ϕ)‖αα + Cε(C1‖g‖α/(α−1))α/(α−1) + C2.


So we can prove (i) by the following fact

‖ϕ+ ψ(ϕ)‖α ≤ C(|Q(ϕ)|1/α + 1).

(ii) Note that for ϕ ∈ E we have

〈Q′0(ϕ), ϕ+〉 ≤ C

∫Ω

N∑l=1

|ϕl + ψl0(ϕ)|p|ϕ+| dxdt

≤C(

∫Ω

|ϕ+ ψ0(ϕ)|α dxdt)p/α(∫

Ω

|ϕ+|α

α−p dxdt)(α−p)/α

≤C(

∫Ω

|ϕ+ ψ0(ϕ)|α dxdt)p/α‖ϕ+‖.

Similarly, we get

〈Q′0(ϕ), ϕ−〉 ≤ C(

∫Ω

|ϕ+ ψ0(ϕ)|α dxdt)p/α‖ϕ−‖.

Since ∫Ω

|ϕ+ ψ0(ϕ)|α dxdt ≤ C1

∫Ω

F0(ϕ+ ψ0(ϕ)) dxdt+ C2,

we get |〈Q′0(ϕ), ϕ+ − ϕ−〉| ≤ C(Q0(ϕ)p/α + 1)‖ϕ‖.

(iii) We can prove (iii) in the similar way,

〈Q′(ϕ), ϕ+〉

≤C∫

Ω

N∑l=1

|ϕl + ψl(ϕ)|p|ϕ+| dxdt+

∫Ω

(g, ϕ+ ψ(ϕ)) dxdt

≤C(

∫Ω

|ϕ+ ψ(ϕ)|α dxdt)p/α(∫

Ω

|ϕ+|α

α−p dxdt)(α−p)/α + ‖g‖α/(α−1)‖ϕ+‖

≤C[(

∫Ω

|ϕ+ ψ(ϕ)|α dxdt)p/α + 1]‖ϕ+‖.

Similarly, we have

〈Q′(ϕ), ϕ−〉 ≤ C[(

∫Ω

|ϕ+ ψ(ϕ)|α dxdt)p/α + 1]‖ϕ−‖.

Lemma 2.2.2 and Young’s inequality imply that∫Ω

|ϕ+ ψ(ϕ)|α dxdt ≤ C1

∫Ω

F0(ϕ+ ψ(ϕ)) dxdt+ C2

≤C1|∫

Ω

Fg(ϕ+ ψ(ϕ)) dxdt|+ |∫

Ω

(g, ϕ+ ψ(ϕ)) dxdt|+ C2

≤C1|Q(ϕ)|+∫

Ω

(ε|ϕ+ ψ(ϕ)|α + Cε|g|α/(α−1)) dxdt+ C2.


Therefore, ∫Ω

|ϕ+ ψ(ϕ)|α dxdt ≤ C(|Q(ϕ)|+ 1).

So we get |〈Q′(ϕ), ϕ+ − ϕ−〉| ≤ C(|Q(ϕ)|p/α + 1)‖ϕ‖.(iv) The proof is similar to (ii) and (iii). It is easy to verify

〈Q′(ϕ), ϕ〉 ≤ C(Q0(ϕ)p/α + 1)‖ϕ‖,〈Q′

0(ϕ), ϕ〉 ≤ C(Q0(ϕ)p/α + 1)‖ϕ‖.

Those imply (iv). 2

Proof of Proposition 2.4.1 (i) From the definition of J(ϕ), we have

|J(ϕ)− J(−ϕ)| ≤ χ(ϕ)|Q(ϕ)−Q0(ϕ)|+ χ(−ϕ)|Q(−ϕ)−Q0(−ϕ)|.

Suppose that −ϕ ∈ suppχ, so that Q0(ϕ) ≤ 2a(I(−ϕ)2 + 1)1/2. From thedefinition of J(ϕ), we see

I(−ϕ) = J(ϕ) +Q0(ϕ)−Q(−ϕ) + χ(ϕ)(Q(ϕ)−Q0(ϕ)).

By Lemma 2.4.2, we get

|I(−ϕ)| ≤ |J(ϕ)|+ C(Q0(ϕ)1/α + 1) ≤ |J(ϕ)|+ C(I(−ϕ)2 + 1)1/2α.

By using Young’s inequality, we deduce

|I(−ϕ)| ≤ 2|J(ϕ)|+ C.

Hence we obtain that for −ϕ ∈ suppχ,

Q0(−ϕ) ≤ a(I(−ϕ)2 + 1)1/2 ≤ C|J(ϕ)|+ C.

Similarly, we have that for ϕ ∈ suppχ,

Q0(ϕ) ≤ C|J(ϕ)|+ C.

Combining the above inequalities and Lemma 2.4.2, we see that for ϕ ∈ E,

|J(ϕ)− J(−ϕ)| ≤ C(χ(ϕ) + χ(−ϕ))(Q0(ϕ)1/α).

Thus, the desired result is justified.(ii) It is sufficient to show that χ = 1, that is, by the definition of χ,

Q0(ϕ) ≤ a(I(ϕ)2 + 1)1/2, (2.4.15)


for ϕ ∈ E such that J(ϕ) ≥ M0 and ‖J ′(ϕ)‖ ≤ 1. For a sufficiently largeM0 > 0 and we can assume by (2.4.7) that |T1(ϕ)| ≤ 1

2, |T2(ϕ)| ≤ 1 and

(p+ 1)(1 + T2(ϕ))

2(1 + T1(ϕ))− 1 >

p− 1

4= b.

Notice that 3/b ≤ a. From (2.4.6), we obtain

I(ϕ)− 1

2(1 + T1(ϕ))〈J ′(ϕ), ϕ〉

=−Q(ϕ) +1 + T2(ϕ)

2(1 + T1(ϕ))〈Q′

0(ϕ), ϕ〉

+χ(ϕ) + T1(ϕ)

2(1 + T1(ϕ))〈Q′(ϕ)−Q′

0(ϕ), ϕ〉

=L1(ϕ) + L2(ϕ) + L3(ϕ),

where

L1(ϕ) :=((p+ 1)(1 + T2(ϕ))

2(1 + T1(ϕ))− 1

)Q0(ϕ),

L2(ϕ) := −(Q(ϕ)−Q0(ϕ))

and

L3(ϕ) :=χ(ϕ) + T1(ϕ)

2(1 + T1(ϕ))〈Q′(ϕ)−Q′

0(ϕ), ϕ〉.

By (2.4.4) we know

|L2(ϕ)| ≤ C(Q0(ϕ)1/α + 1). (2.4.16)

By using (2.4.3) and (2.4.4), we see

|L3(ϕ)| ≤ C|〈Q′(ϕ)−Q′0(ϕ), ϕ〉|

≤ |(p+ 1)Q(ϕ)− 〈Q′(ϕ), ϕ〉|+ (p+ 1)|Q(ϕ)−Q0(ϕ)|≤ C(Q0(ϕ)1/α + 1). (2.4.17)

On the other hand, letting h = ϕ+ − ϕ− in (2.4.6) we get

〈J ′(ϕ), ϕ+ − ϕ−〉=(1 + T1(ϕ))‖ϕ‖2 − (1 + T2(ϕ))〈Q′

0(ϕ), ϕ+ − ϕ−〉· (χ(ϕ) + T1(ϕ))〈Q′(ϕ)−Q′

0(ϕ), ϕ+ − ϕ−〉.(2.4.18)


Therefore, from (2.4.18) and Lemma 2.4.4, T1(ϕ) ≤ 12

and the assumption:‖J ′(ϕ)‖E∗ ≤ 1, we deduce

1

2‖ϕ‖2 ≤ ‖J ′(ϕ)‖E∗‖ϕ‖+ C(Q0(ϕ)p/α + 1)‖ϕ‖

≤ C(Q0(ϕ)p/α + 1)‖ϕ‖.

That is,

‖ϕ‖ ≤ C(Q0(ϕ)p/α + 1). (2.4.19)

By (2.4.16) and (2.4.17), we obtain

I(ϕ)− 1

2(1 + T1(ϕ))〈J ′(ϕ), ϕ〉 ≥ L1(ϕ)− C(Q0(ϕ)1/(p+1) + 1).

It follows by (2.4.19) and ‖J ′(ϕ)‖E∗ ≤ 1 that

I(ϕ) ≥ −C‖J ′(ϕ)‖E∗‖ϕ‖+ L1(ϕ)− C(Q0(ϕ)1/(p+1) + 1)

≥ bQ0(ϕ)/2− C0. (2.4.20)

We claim that as M →∞,

infQ0(ϕ)| ‖J ′(ϕ)‖E∗ ≤ 1 and J(ϕ) ≥M → ∞.

This follows from (2.4.19). In fact, J(ϕ) → ∞ implies ‖ϕ‖ → ∞. So it givesthat Q0(ϕ) →∞, by (2.4.19).

By the properties T1(ϕ), T2(ϕ) in (2.4.7), we see that J(ϕ) ≥ M0. Thisimplies bQ0(ϕ)/6 − C0 ≥ 0, so that I(ϕ) ≥ bQ0(ϕ)/3, by (2.4.20). Then wehave

Q0(ϕ) ≤ aI(ϕ) ≤ a(I(ϕ)2 + 1)1/2.

2

Proposition 2.4.6. J ∈ C1(E,R) satisfies the following Palais-Smale com-pactness condition (P.S.): Whenever a sequence ϕj∞j=1 in E satisfies for alarge M2 and some M3 > 0,

M2 ≤ J(ϕj) ≤M3 for all j,

J ′(ϕj) → 0 in E∗ as j →∞,

there is a subsequence of ϕj which is convergent in E.


Proof. Letting h = ϕj and h = ϕ+j − ϕ−j in (2.4.6), we have∣∣∣(1 + T1(ϕj))

(‖ϕ+

j ‖2 − ‖ϕ−j ‖2)− (1 + T2(ϕj))〈Q′

0(ϕj), ϕj〉

−(χ(ϕj) + T1(ϕj)

)〈Q′(ϕj)−Q′

0(ϕj), ϕj〉∣∣∣ ≤ m‖ϕj‖, (2.4.21)∣∣∣(1 + T1(ϕj))‖ϕj‖2 − (1 + T2(ϕj))〈Q′

0(ϕj), ϕ+j − ϕ−j 〉

−(χ(ϕj) + T1(ϕj)

)〈Q′(ϕj)−Q′

0(ϕj), ϕ+j − ϕ−j 〉

∣∣∣ ≤ m‖ϕj‖, (2.4.22)

where m = sup ‖J ′(ϕj)‖E∗ . Since

J(ϕj) =1

2‖ϕ+

j ‖2 − 1

2‖ϕ−j ‖2

−Q0(ϕj)− χ(ϕj)(Q(ϕj)−Q0(ϕj))

≤M3,

from (2.4.21) we see

1 + T2(ϕj)

2(1 + T1(ϕj))〈Q′

0(ϕj), ϕj〉 −Q0(ϕj)−χ(ϕj) + T1(ϕj)

2(1 + T1(ϕj))〈Q′(ϕj)−Q′

0(ϕj), ϕj〉

− χ(ϕj)(Q(ϕj)−Q0(ϕj)) ≤M3 +m‖ϕj‖.Hence, by Lemma 2.4.4, we obtain that for large M2, T1(ϕj), T2(ϕj) are small,there is a constant C0 such that

1 + T2(ϕj)

2(1 + T1(ϕj))〈Q′

0(ϕj), ϕj〉 −Q0(ϕj) ≤ C0Q0(ϕj).

By Lemma 2.4.2 and 2.4.5, we know

C0Q0(ϕj)− C1(|Q0(ϕj)|+ 1)p/α ≤M3 +m‖ϕj‖.Hence by Young’s inequality,

|Q0(ϕj)| ≤ C1(1 + ‖ϕj‖) for all j. (2.4.23)

Then, by (2.4.22) we obtain that for large j,

‖ϕj‖2 ≤∣∣∣1 + T2(ϕj)

1 + T1(ϕj)〈Q′

0(ϕj), ϕ+j − ϕ−j 〉

− χ(ϕj) + T1(ϕj)

1 + T1(ϕj)〈Q′(ϕj)−Q′

0(ϕj), ϕ+j − ϕ−j 〉

∣∣∣ +m‖ϕj‖

≤C2[(|Q(ϕj)|p/α + ‖g‖α/(α−1)‖ϕj‖α)‖ϕj‖+ ‖ϕj‖]≤C2(‖ϕj‖p/α + 1)‖ϕj‖.

So ‖ϕj‖ is bounded. Observe that J ′(ϕj) = ϕ+i −ϕ−j +P (ϕj) where P : E → E∗

is compact operator and J ′(ϕj) → 0 as j →∞. Hence, ϕ+j −ϕ−j is precompact

in E and then ϕj is precompact in E. Thus the proof is completed. 2

2.5. Minimax methods 49

2.5 Minimax methods

In this section, we construct critical points of J via minimax methods. We startwith the definition of the usual lexicographical order for 2-tuples (k, i) ∈ D asfollows, where D = N× 1, 2, · · · , N,

(j,m) = (k, i), if j = k and m = i,

(j,m) < (k, i), if j < k or j = k and m < i.

Moreover, we write (k, i) := (k + [ iN

], i− [ iN

]) for every i ∈ N, where [a] is theinteger part of a, (k, 0) ≡ (k − 1, N) for k ∈ N.

We observe that the positive eigenvalues of the wave operator under periodic-Dirichlet conditions are j2 − k2| j ∈ N, k ∈ Z and corresponding eigen-functions are sin jx cos kt and sin jx sin kt. We arrange the eigenvalues in thefollowing order, denoted by 0 < µ1 ≤ µ2 ≤ · · · ≤ µn ≤ · · · with repetitionsaccording to the multiplicity of each eigenvalue and denote by vn the eigen-functions which correspond to µn. We assume 〈vi, vj〉 = δij for i, j ∈ N. Wedenote by e1, · · · , eN the usual orthogonal basis in RN . Define vjk = vjek forj ∈ N and 1 ≤ k ≤ N . Let

E+q = E+

mi = spanvjk| (1, 1) ≤ (j, k) ≤ (m, i),

where 1 ≤ i ≤ N and q = mN + i.

Lemma 2.5.1. For all θ ∈ (0, 1/α), there is a constant C > 0, independentof m ∈ N, such that

‖ϕ‖α ≤ Cµ−θm ‖ϕ‖ for ϕ ∈ (E+q )⊥,

where (E+q )⊥ is the orthogonal space of E+

q in E+ and m = [q/N ] is the integerpart of q/N .

Proof. By the definition of µm, we have that, for all l ∈ N,∫Ω

|ϕl|2 dxdt ≤ µ−1m

∫Ω

(|ϕlt|2 − |ϕlx|2) dxdt for ϕ ∈ (E+q )⊥.

Summing the inequalities from 1 to N , we get

‖ϕ‖2 ≤ µ−1/2m ‖ϕ‖ for ϕ ∈ (E+

q )⊥.

On the other hand, by Lemma 2.2.1,

‖ϕ‖s ≤ Cs‖ϕ‖ for all ϕ ∈ E+ and s ∈ [1,∞).

Using Holder’s inequality, we obtain that, for s ∈ (α,∞),

‖ϕ‖α ≤ ‖ϕ‖τ2‖ϕ‖1−τs for ϕ ∈ E+,

where

τ =2(s− α)

α(s− 2)∈ (0,

2

α).

Combining the above inequalities, we have

‖ϕ‖α ≤ C1−τs µ−τ/2m ‖ϕ‖ for ϕ ∈ (E+

q )⊥.

2

Note that

‖ϕ‖ ≤ µ1/2m ‖ϕ‖2 for ϕ ∈ E+

q ,m = [q/N ].

For ϕ = ϕ+ + ϕ− ∈ E+q ⊕ E−, by Lemma 2.2.2 and 2.5.1, we have

J(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ)− χ(ϕ)(Q(ϕ)−Q0(ϕ))

≤ 1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ) + C(Q0(ϕ)1/α + 1)

≤ 1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 − 1

2Q0(ϕ) + C

≤ 1

2‖ϕ+‖2 − 1

2

∫Ω

F0(ϕ+ − ϕ− + ψ) dxdt− 1

2‖ϕ−‖2 + C

≤ 1

2‖ϕ+‖2 − c‖ϕ+ + ϕ− + ψ‖αα −

1

2‖ϕ−‖2 + C

≤ 1

2‖ϕ+‖2 − c‖ϕ+ + ϕ− + ψ‖α2 −

1

2‖ϕ−‖2 + C

≤ 1

2‖ϕ+‖2 − c‖ϕ+‖α2 −

1

2‖ϕ−‖2 + C

≤ 1

2‖ϕ+‖2 − cµ−α/2m ‖ϕ+‖α − 1

2‖ϕ−‖2 + C.

Hence, for every q ∈ N there is Rq such that

J(ϕ) ≤ 0, for all ϕ ∈ E+q ⊕ E− =: Eq with ‖ϕ‖ ≥ Rq. (2.5.1)

We may assume that Rq < Rq+1 for all q ∈ N. Let

BR = ϕ ∈ E| ‖ϕ‖ ≤ R for R ≥ 0,

Dq = BRq ∩ Eq,Γq = γ ∈ C(Dq, E)| γ satisfies (γ1)− (γ3),


where(γ1) γ is odd, i.e., γ(−ϕ) = −γ(ϕ) for all ϕ ∈ Dq,(γ2) γ(ϕ) = ϕ for all ϕ ∈ ∂Dq,(γ3) for ϕ ∈ Dq, γ(ϕ) = α(ϕ)ϕ + κ(ϕ) where α ∈ C(Dq, [1, α]) is a evenfunctional (α depends on γ) and κ is a compact operator such that α(ϕ) = 1and κ = 0 on ∂Dq.Moreover, let

Uq = ϕ = w + τvq+1|w ∈ BRq+1 ∩ Eq, τ ∈ [0, Rq+1], ‖ϕ‖ ≤ Rq+1

andΛq = λ ∈ C(Uq, E)|λ satisfies (λ1)− (λ3),

where(λ1) λ|Dq ∈ Γq,(λ2) λ(ϕ) = ϕ on ∂Uq \Dq,(λ3) for ϕ ∈ Uq, λ(ϕ) = α(ϕ)ϕ + κ(ϕ) where α ∈ C(Uq, [1, α]) is a evenfunctional (α depends on λ) and κ is a compact operator such that α(ϕ) = 1and κ = 0 on ∂Uq \Dq.

Define for q ∈ N,

bq = infγ∈Γq

supϕ∈Dq

J(γ(ϕ)), cq = infλ∈Λq

supϕ∈Uq

J(λ(ϕ)). (2.5.2)

Notice that cq ≥ bq for all q ∈ N.

Lemma 2.5.2. Suppose that c > M0 is a regular value of J , that is, J ′(ϕ) 6= 0when J(ϕ) = c, where M0 is given in Proposition 2.4.1. Then for every ε thereexists an ε ∈ (0, ε) and η ∈ C([0, 1]× E,E) such that(i) η(t, ·) is odd, for t ∈ [0, 1] if g(x, t), the force term, identically vanishes;(ii) η(t, ·) is homeomorphism of E onto E for all t;(iii) η(0, ϕ) = ϕ for all ϕ ∈ E;(iv) η(t, ϕ) = ϕ if J(ϕ) 6∈ [c− ε, c+ ε];(v) J(η(1, ϕ)) ≤ c− ε if J(ϕ) ≤ c+ ε;(vi) for ϕ = ϕ+ +ϕ− ∈ E+⊕E−, η(1, ϕ) = α+(ϕ)ϕ+ +α−(ϕ)ϕ− +κ(ϕ) whereα+ ∈ C(E, [0, 1]), α− ∈ C(E, [1, α]) is an even functional (α ≥ 1 is constant)and κ is a compact operator.

Proof. Since J(ϕ) ∈ C1(E,R) and satisfies (P.S.) condition by Proposition2.4.6, the assertions are standard. By Proposition A.18 in [18], we know thatthere is an η satisfying the initial value problem

dη

dt= −ω(η)V (η), η(0, ϕ) = ϕ,


where ω ∈ C0,1(E,R) satisfying 0 ≤ ω ≤ 1 and V is a pseudogradient vectorfield for J ′ on E∗. By Lemma 2.4.4, we have

dηdt

= −ω(η)[(1 + T1(η))(η+ − η−) + P(η)],

η(0) = ϕ,

where P(η) is compact. It yieldsdη±

dt= −ω(η)[±(1 + T1(η))η

± + P±P(η)],η±(0) = ϕ±,

where P± : E → E± are the orthogonal projection. Integrating these shows

η±(t, ϕ) = eA(t)[ϕ± −∫ t

0

e−A(τ)ω(η(s, ϕ))P±P(η(τ, ϕ)) dτ ],

where A(t) = ∓∫ t

0ω(η(s, ϕ))(1 + T1(η(s, ϕ)))) ds. Thus η is of the asserted

form. 2

Proposition 2.5.3. Suppose that cq > bq ≥M0, where M0 is given in Propo-sition 2.4.1. Let d ∈ (0, cq − bq) and

Λq(d) = λ ∈ Λq| J(λ) ≤ bq + d on Dq.

Definecq(d) = inf

λ∈Λq(d)supϕ∈Uq

J(λ(ϕ)), (≥ cq). (2.5.3)

Then cq(d) is a critical value of J .

Proof. By Proposition 2.4.1 and 2.4.6, we obtain that if ϕ ∈ E satisfies J ′(ϕ) =0 and J(ϕ) ≥M0, then I(ϕ) = J(ϕ) and I ′(ϕ) = 0, and that J(ϕ) satisfies thePalais-Smale condition on AM0 = ϕ ∈ E| J(ϕ) ≥M0. Note that

J ′(ϕ) = (1 + T1(ϕ))(ϕ+ − ϕ−) + (compact),

where |T1(ϕ)| ≤ 12, on ϕ ∈ E |J(ϕ) ≥M0, see Lemma 2.4.4. So we can show

that cq(d) is a critical value of J as in [17]. In fact, note that by the definitionof bq and Λq, Λq(d) 6= ∅. Choose ε = 1

2(cq− bq− d) > 0. If cq(d) is not a critical

value of J , there are ε > 0 and η as in Lemma 2.5.2. Choose λ ∈ Λq(d) suchthat

sup J(λ(ϕ)) ≤ cq(d) + ε. (2.5.4)

Consider η(1, λ(ϕ)) ∈ C(Uq, E). Note that if ‖ϕ‖ = Rq+1 or ϕ ∈ (BRq+1\BRq)∩Eq, J(λ(ϕ)) = J(ϕ) ≤ 0, so η(1, λ(ϕ)) = ϕ by Lemma 2.5.2. So η(1, λ) ∈ Λq.Moreover on Dq, J(λ(ϕ)) ≤ bq + d ≤ cq − ε ≤ cq(d) − ε from our choice of d


and ε. Then η(1, λ) = λ, J(η(1, λ)) ≤ bq + d on Dq, again by Lemma 2.5.2.Thus η(1, λ) ∈ Λq(d) and by (2.5.4) and Lemma 2.5.2,

sup J(λ(ϕ)) ≤ cq(d)− ε,

contrary to the definition of cq(d). Hence cq(d) is a critical value of J . 2

Proposition 2.5.4. If cq = bq for all q ≥ q0, then there is a constant C > 0such that

bq ≤ Cq(p+1)/p for all q ∈ N.Proof. Let q ≥ q0, ε > 0 and γ ∈ Λq such that

supγ∈Uq

J(γ(ϕ)) ≤ bq + ε.

Let γ(ϕ) = γ(ϕ) for ϕ ∈ Γq and γ(−ϕ) = −γ(ϕ) for ϕ ∈ −Γq. Note thatUq ∪ (−Uq) = Dq+1. Moreover, since γ ∈ Λq implies γ ∈ Γq+1. Therefore,

bq+1 ≤ bq + C(|bq|1/α + 1).

Hence for all q ∈ N,

bq ≤ Cqα

α−1 ≤ Cqp+1

p .

2

Therefore, the existence of a subsequence of cq which satisfy cq > bq ≥M0

guarantees the existence of critical values for functional I. In other words, weshould show the existence of subsequence qj such that

cqj > bqj ≥M0, for large qj ∈ N, and

bqj →∞ as qj →∞.

To show the above properties, we will prove the existence of a subsequenceqj such that for every ε > 0, there is a Cε > 0 satisfying

bqj ≥ Cεq(p+1)/[(p−1)(1+ε)]j for large qj ∈ N. (2.5.5)

Choosing ε ∈ (0, 2p−1

) and using p > 1, we make sure that the case in Proposi-

tion 2.5.4 does not happen. The rest of this chapter is devoted to prove (2.5.5).

We now look for a comparison functional for J . By using Lemma 2.2.4 andLemma 2.4.2, we have

J(ϕ) =1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 −Q0(ϕ)− χ(ϕ)(Q(ϕ)−Q0(ϕ))

≥ 1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 − 2Q0(ϕ)− a1

≥ 1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 − 2

∫Ω

F0(ϕ+ + ϕ−) dxdt− a1

≥ 1

2‖ϕ+‖2 − 1

2‖ϕ−‖2 − a0

p+ 1

‖ϕ+‖p+1

p+1 + ‖ϕ−‖p+1p+1

− a1, (2.5.6)


where a0, a1 > 0 are constants independent of ϕ.Define

K(ϕ+) :=1

2‖ϕ+‖2 − a0

p+ 1‖ϕ+‖p+1

p+1 ∈ C2(E+,R). (2.5.7)

Here we recall the definitions of (P.S.)∗ and (P.S.)n conditions for the func-tional K in E+.

K satisfies (P.S.)∗ condition if ϕn ⊂ E+ has the properties such thatϕn ∈ E+

n , K(ϕn) ≤ C and (K|E+n)′(ϕn) converges to 0 in (E+

n )∗ as n → ∞,then ϕn is relatively compact in E+.

K satisfies (P.S.)n condition if ϕm ⊂ E+n has the properties such that

K(ϕm) ≤ C and (K|E+n)′(ϕm) converges to 0 in (E+

n )∗ as m→∞, then ϕmis relatively compact in E+

n .

Then we have the following properties for functional K.

Lemma 2.5.5. (i) J(ϕ+) ≥ K(ϕ+)− a1 for all ϕ+ ∈ E+.(ii) K satisfies the (P.S.), (P.S.)∗ and (P.S.)n conditions on E+.

Proof. The arguments to show (P.S.), (P.S.)∗ and (P.S.)n are very similar,so we just give the proof of (P.S.)∗. Let ϕn ⊂ E+ be a sequence such thatϕn ∈ E+

n , K(ϕn) ≤ C for all n ∈ N and

limn→∞

‖(K|E+n)′(ϕn)‖(E+

n )∗ = 0.

That is, for all h ∈ E+, as n→∞,

〈ϕn, h〉E+ − a0

∫Ω

|ϕn|p−1ϕnh dxdt = εn → 0.

It leads to ‖ϕn‖E+ = ‖ϕn‖E+n≤ C. Hence, there is a subsequence, denoted

again by ϕn, such that ϕn → ϕ weakly in E+, strongly in Lp+1(Ω), asn→∞. Consequently, we have that, as n→∞,∫

Ω

|ϕn|p+1 dxdt→∫

Ω

|ϕ|p+1 dxdt

and ∫Ω

|ϕn|p−1ϕnϕdxdt→∫

Ω

|ϕ|p+1 dxdt.

Thus,

‖ϕn‖2 = a0

∫Ω

|ϕn|p+1 dxdt+ εn → a0

∫Ω

|ϕ|p+1 dxdt = ‖ϕ‖2.

It implies that ϕn → ϕ in E+, as n→∞. 2

Now we are concerned with functional K in E+ and state index propertiesfor Bahri-Berestycki’s max-min value σq. For n > q, n, q ∈ N. We define

Anq = σ ∈ C(Sn−q, E+n )| σ(−y) = −σ(y) for all y, (2.5.8)

andσnq = sup

σ∈Anq

miny∈Sn−q

K(σ(y)). (2.5.9)

Lemma 2.5.6. ([21]) Let a, b ∈ N. Suppose that h1 ∈ C(Sa,Ra+b) and h2 ∈C(Rb,Ra+b) are continuous such that

h1(−y) = −h1(y) for all y ∈ Sa, h2(−y) = −h2(y) for all y ∈ Rb,

and there is a r0 > 0 such that h2(y) = y for |y| ≥ r0. Then h1(Sa)∩h2(Rb) 6= ∅.

Lemma 2.5.7. For all σ ∈ Anq ,

σ(Sn−q) ∩ E+q 6= ∅. (2.5.10)

Proof. Apply Lemma 2.5.6 to h1 = σ : Sn−q → E+n and h2 = id : E+

q → E+n .

Then we get the result. 2

Proposition 2.5.8. (i) 0 ≤ σnq ≤ σnq+1 for all q, n ∈ N;(ii) For all n ∈ N there exist ν(q) and ν(q) such that

0 ≤ ν(q) ≤ σnq ≤ ν(q) <∞ for all n ≥ q + 1; (2.5.11)

(iii) Moreover, ν(q) →∞ as q →∞.

Proof. (i) For every σ ∈ Anq , it is clear that there is a σ ∈ Anq+1 withσ(Sn−q−1) ⊂ σ(Sn−q). Hence we have σnq ≤ σnq+1.(ii) We prove the existence of ν(q). By Lemma 2.5.7 we have for all σ ∈ Anq ,

miny∈Sn−q

K(σ(y)) ≤ supϕ∈E+

q

K(ϕ). (2.5.12)

For all ϕ ∈ E+q and m = [q/N ], we have

K(ϕ) ≤ 1

2‖ϕ‖2 − C‖ϕ‖p+1

p+1 ≤1

2‖ϕ‖2 − C‖ϕ‖p+1

2

≤ 1

2‖ϕ‖2 − Cµ−(p+1)

m ‖ϕ‖p+1.(2.5.13)

Thus, the right-hand of (2.5.13) is finite and independent of σ and n. Set

ν(q) = supϕ∈E+

q

K(ϕ) <∞,


then we obtainσnq = sup

σ∈Anq

miny∈Sn−q

K(σ(y)) ≤ ν(q).

(iii) To prove the existence of ν(q), we construct a special σ ∈ Anq . Write

Sn−q = y = (yq, . . . , yn) ∈ Rn−q+1|n∑i=q

y2i = 1.

Let vi be the eigenfunction corresponding to µi given at the beginning of Sec-tion 2.5. We define

w(y) =∑

(m,i)≤(l,j)≤(m,i)

ylN+jvlej,

where q = mN + i and n = mN + i. Define σ : Sn−q → E+q \ 0 by

σ(y) = a−1/(p−1)0 ‖w(y)‖−(p+1)/(p−1)

p+1 w(y).

Obviously we have σ ∈ Anq . Since ‖w(y)‖ = 1 on Sn−q, we obtain

K(σ) ≥ (1

2− 1

p+ 1)a−2/(p−1)0 ‖w‖−2(p+1)/(p−1)

p+1 .

Sincew(y) ∈ (E+

q−1)⊥, ‖w(y)‖ = 1 for all y ∈ Sn−q,

we get that‖w(y)‖p+1 ≤ Cθµ

−θq−1,

where θ ∈ (0, 1/(p+1)) and Cθ is a constant independent of n and y. Therefore,

K(ϕ) ≥ C ′θµ

θ(p+1)/(p−1)q−1 := ν(q)

and then we have

σnq ≥ miny∈Sn−q

K(σ(y)) ≥ ν(q) for n > q.

Since µq−1 →∞ as q →∞, we obtain ν(q) →∞ as q →∞. 2

To continue the proof, we define

σq := lim supn→∞

σnq .

Proposition 2.5.9. Suppose that ν(q) > 0. Then σnq is a critical value of therestriction of K to E+

n . Furthermore, σq is a critical value of K ∈ C2(E+,R),σq ≤ σq+1 for all q ∈ N and σq →∞ as q →∞.

Proof. Since K satisfies (P.S.), (P.S.)∗ and (P.S.)n by Lemma 2.5.5, we haveσnq is a critical value of KE+

n∈ C2(E+

n ,R). By Proposition 2.5.8, we can choosea sequence nj such that nj →∞ as j →∞,

σq = limj→∞

σnjq exist for all q ∈ N. (2.5.14)

Using the (P.S.)∗ condition, we can extract a convergent subsequence such thatϕnjq → ϕq, and then we observe that K(ϕq) = limnj→∞ σ

njq and K ′(ϕq) = 0.

Therefore σq is a critical value of K ∈ C2(E+,R), the other properties followdirectly from Proposition 2.5.8. 2

Next we state the relation between bq and σq.

Proposition 2.5.10. For all q ∈ N,

bq ≥ σq − a1, (2.5.15)

where a1 is the constant appeared in (2.5.6).

To prove this proposition, we need the following lemma:

Lemma 2.5.11. For all γ ∈ Γq and σ ∈ Anq , n > q, we have(Pn γ(Dq) ∪ ϕ ∈ E+

q ⊕ E−| ‖ϕ‖ ≥ Rq)∩ σ(Sn−q) 6= ∅,

where Pn : E = E+ ⊕ E− → E+n ⊕ E− is the orthogonal projection.

Proof. We extend γ to γ ∈ C(E+q ⊕ E−, E) by

γ(ϕ) = γ(ϕ) if ‖ϕ‖ ≤ Rq γ(ϕ) = ϕ if ‖ϕ‖ ≥ Rq.

Obviously, γ(ϕ) is well defined and odd in E+q ⊕ E−. We observe

Pn γ(E+q ⊕ E−) = Pn γ(Dq) ∪ ϕ ∈ Eq ⊕ E−| ‖ϕ‖E ≥ Rq.

So it is sufficient to prove Pn γ(E+q ⊕E−)∩σ(Sn−q) 6= ∅. Let · · · ≤ νn ≤ · · · ≤

ν2 ≤ ν1 < 0 be the negative eigenvalues of the wave operator, with repetitionsaccording to multiplicity of each eigenvalue and denote wn the eigenfunctionthat corresponds to νn. We define wjk = wjek, where e1, e2, · · · , eN is theusual basis of RN . Now we consider the spaces

E−s = spanwjk| (1, 1) ≤ (k, j) ≤ (m, i),

where 1 ≤ i ≤ N and s = mN + 1. We let Pn,s : E = E+ ⊕ E− → E+n ⊕ E−

s

be the orthogonal projection. Consider the operators

σ : Sn−q → E+n ⊂ E+

n ⊕ E−s , Pn,s γ : E+

q ⊕ E−s → E+

n ⊕ E−s .


Applying Lemma 2.5.6 to h1 = σ and h2 = Pn,s γ, we get that, for someys ∈ Sn−q and ϕs ∈ E+

q ⊕ E−s ,

σ(ys) = Pn,s γ(ϕs).

Since Sn−q is compact, there is a subsequence ysisuch that, as si →∞,

ysi→ y in Sn−q, (2.5.16)

σ(ysi) → σ(y) in E+

n . (2.5.17)

On the other hand, by (γ3), we see that

Pn,s γ(ϕs) = Pn,s [α(ϕs)ϕs + κ(ϕs)] = α(ϕs)ϕs + Pn,s κ(ϕs),

where α(ϕ) ≥ 1 on E+q ⊕ E− and κ(E+

q ⊕ E−) = κ(Dq) is compact. Hence wehave

ϕs =1

α(ϕs)Pn,s [γ(ϕs)− κ(ϕs)] =

1

α(ϕs)Pn,s [σ(ys)− κ(ϕs))].

By (2.5.17), we know that there is a convergent subsequence ϕsi of ϕs

such that, as si →∞,ϕsi

→ ϕ in E+q ⊕ E−.

Passing to the limit, we obtain

Pn γ(ϕ) = σ(y), i.e., Pn γ(E+q ⊕ E−) ∩ σ(Sn−q) 6= ∅.

2

Proof of Proposition 2.5.10. Since J(ϕ) ≤ 0 on ϕ ∈ E+q ⊕E−| ‖ϕ‖ ≥ Rq,

we have from Lemma 2.5.11

miny∈Sn−q

J(σ(y)) ≤ supϕ∈Dq

J(Pn γ(ϕ))

for all γ ∈ Γq and σ ∈ Anq . By Lemma 2.5.5,

miny∈Sn−q

K(σ(y))− a1 ≤ supϕ∈Dq

J(Pn γ(ϕ)).

Hence we obtain

supσ∈An

q

miny∈Sn−q

K(σ(y))− a1 ≤ infγ∈Γq

supϕ∈Dq

J(Pn γ(ϕ)) =: bnq .

Then, taking lim sup at both sides, we see that

σq − a1 ≤ lim supn→∞

bnq .

Thus it suffices to show the following lemma.

2.6. Morse index and spectral analysis 59

Lemma 2.5.12. For q ∈ N, bq = lim supn→∞ bnq .

Proof. Since PnΓq = Pn γ| γ ∈ Γq ⊂ Γq, it is clear that bq ≤ bnq for n > q.We prove bq ≥ lim supn→∞ bnq for q ∈ N. From the definition of bq, for everyε > 0, there is γ ∈ Γq such that

supϕ∈Dq

J(γ(ϕ)) ≤ bq + ε.

By (γ3), we see that γ(ϕ) takes a form γ(ϕ) = α(ϕ)ϕ + κ(ϕ) where α(ϕ) ∈C(Dq, [1, α]) and κ(Dq) is compact. Since Pn κ(ϕ) → κ(ϕ) as n → ∞ uni-formly in Dq, we have

Pn γ(ϕ) = α(ϕ)ϕ+ Pn κ(ϕ) → α(ϕ)ϕ+ κ(ϕ) = γ(ϕ)

uniformly in Dq. Hence

supϕ∈Dq

J(Pn γ(ϕ)) → supϕ∈Dq

J(γ(ϕ)),

as n→∞. By the above inequality, we obtain

lim supn→∞

bnq ≤ lim supn→∞

supϕ∈Dq

J(Pn γ(ϕ))

= supϕ∈Dq

J(γ(ϕ)) ≤ bq + ε.

Since ε is arbitrary, we get the desired result. 2

2.6 Morse index and spectral analysis

In this section, we prove the lower and upper bounds for the Morse index ofK ′′. For ϕ ∈ E+, define the Morse index of K ′′(ϕ) by

indexK ′′(ϕ) = the number of eigenvalues of K ′′(ϕ)

which are non-positive.

That is,

indexK ′′(ϕ) = maxdimH|H is subspace such that

〈K ′′(ϕ)h, h〉 ≤ 0 for all h ∈ H.

First we have lower bounds of the Morse index as given in the following twopropositions:

Proposition 2.6.1. Assume that σnq < σnq+1, n > q + 1. Then there exists aϕnq ∈ E+

n such that

K(ϕnq ) ≤ σnq , (2.6.1)

(K|E+n)′(ϕnq ) = 0, (2.6.2)

index (K|E+n)′′(ϕnq ) ≥ q. (2.6.3)

Proposition 2.6.2. Assume that σq < σq+1. Then there exists a ϕq ∈ E+

such that

K(ϕq) ≤ σq, (2.6.4)

K ′(ϕq) = 0, (2.6.5)

indexK ′′(ϕq) ≥ q. (2.6.6)

To prove those propositions, we need several lemmas. We start with anapproximation lemma:

Lemma 2.6.3. ([4][21]) Let U be a C2 open subset of some Hilbert space Hand let φ ∈ C2(U,R) satisfy the (P.S.) condition. Assume that φ′′ is a Fredholmoperator (of null index) on the critical set Z(φ) = x ∈ U |φ′(x) = 0. and thatZ(φ) is compact. Then for every ε > 0, there exists ψ ∈ C2(U,R) satisfying(P.S.) with the following properties:(i) ψ = φ(x) if dist(x, Z(φ)) ≥ ε;(ii) |φ(x)− ψ(x)|, ‖φ′(x)− ψ′(x)‖, ‖φ′′(x)− ψ′′(x)‖ ≤ ε for all x ∈ U ;(iii) The critical points of ψ are finite in number and non-degenerate.

Notice that K|E+n∈ C2(E+

n ,R) satisfies (P.S.) condition and that all criticalvalues of K|E+

nare nonnegative. To see this fact, we suppose that ϕ ∈ E+

n is acritical point of K|E+

n, then we have

K(ϕ) = K(ϕ)− 1

2〈(K|E+

n)′(ϕ), ϕ〉 = (

1

2− 1

p+ 1)a0‖ϕ‖p+1

p+1 ≥ 0.

On the other hand, there is a constant Rn such that K(ϕ) < 0 for ϕ ∈ E+n with

‖ϕ‖ ≥ Rn. Thus Z(K|E+n) is compact. By using Lemma 2.6.3, we see that, for

all ε > 0, there exists a ψε ∈ C2(E+n ,R) satisfying (P.S.) with the following

properties:

|ψε(ϕ)−K(ϕ)|, ‖ψ′ε(ϕ)− (K|E+n)′(ϕ)‖, ‖ψ′′ε (ϕ)− (K|E+

n)′′(ϕ)‖ ≤ ε (2.6.7)

for all ϕ ∈ E+n ; the critical points of ψε are finite in number and nondegenerate.

We set, for n > q and ε > 0,

σnq (ε) = supσ∈An

q

miny∈Sn−q

ψε(σ(y)).

By (2.6.7),σnq − ε ≤ σnq (ε) ≤ σnq + ε.

Moreover we have the following lemmas as Tanaka [21], on the critical groupsof the functional ψε.

Lemma 2.6.4. [21] Suppose that aε ∈ R satisfies

σnq (ε) < aε − 2ε < aε < σnq+1(ε).

Thenπn−q−1([ψε ≥ aε]n, ζ) 6= 0 for some ζ ∈ [ψε ≥ aε]n, (2.6.8)

where [ψε ≥ aε]n = ϕ ∈ E+n |ψε(ϕ) ≥ aε and πs(·, ζ) is the s-th homotopy

groups at the point ζ, for s ∈ N.

Lemma 2.6.5. [21] For a regular value a ∈ R of ψε, let

L(ε; a) = maxindexψ′′ε (x)|ψε(x) ≤ a, ψ′ε(x) = 0.

Then

πs([ψε(x) ≥ a]n, ζ) = 0 for all ζ ∈ [ψε ≥ a]n and s ≤ n− L(ε; a)− 2.

Proof of Proposition 2.6.1. Since σnq < σnq+1 and the critical points of ψε arefinite in the number and nondegenerate, we observe that for every 0 < ε ≤ ε0there is a sequence aε ∈ R such that

aε is a regular value of ψε, (2.6.9)

σnq (ε) < aε − 2ε < aε < σnq+1(ε), (2.6.10)

aε → σnq as ε→ 0. (2.6.11)

Using Lemma 2.6.4 and Lemma 2.6.5, we observe

L(ε; a) ≥ q for 0 < ε < ε0.

Therefore, there is a ϕε ∈ E+n such that

ψε(ϕε) ≤ aε, (2.6.12)

ψ′ε(ϕε) = 0, (2.6.13)

indexψ′′ε (ϕε) ≥ q. (2.6.14)

It follows from (2.6.7) that K(ϕε) is bounded and (K|E+n)′(ϕε) → 0 as ε →

0. Since K|E+n

satisfies (P.S.) condition on E+n , we can choose a convergent

subsequence such that ϕεj → ϕnq as εj → 0. 2

Proof of Proposition 2.6.2. Since σq < σq+1, we have σnjq < σ

nj

q+1 for suffi-ciently large j. Hence there is a ϕ

njq ∈ E+

njsatisfying (2.6.1)-(2.6.3) by Propo-

sition 2.6.1. Since K ∈ C2(E+,R) satisfies (P.S.)∗ condition, ϕnjq has a con-

vergent subsequence ϕnj′q . Let ϕq = limj′→∞ ϕ

nj′q . Then (2.6.4) and (2.6.5)

follow from (2.6.1) and (2.6.2) easily. Let us prove (2.6.6).First we have for all n ∈ N,

indexK ′′(ϕnq ) ≥ index (K|E+n)′′(ϕnq ).

On the other hand, we observe that K ′′(ϕq) is an operator of type id +(compact). Hence there is an ε > 0 such that for all h ∈ E+,

〈K ′′(ϕq)h, h〉 ≤ 0 iff 〈K ′′(ϕq)h, h〉 ≤ ε‖h‖2.

That is,

indexK ′′(ϕq) = index (K ′′(ϕq)− ε).

Since K ∈ C2(E+,R), there is j′0 such that, for j′ > j′0,

‖K ′′(ϕnj′q )−K ′′(ϕq)‖ ≤ ε.

Thus for j′ ≥ j′0 and h ∈ E+, we have

〈K ′′(ϕq)h, h〉 − ε‖h‖2 ≤ 〈K ′′(ϕnj′q )h, h〉.

That is,

indexK ′′(ϕnj′q ) ≤ index (K ′′(ϕq)− ε).

Therefore, by the above inequality, we complete the proof. 2

Now we prove the upper bounds for the Morse index.

Proposition 2.6.6. For every ε > 0, there is a constant Cε > 0 such that forϕ ∈ E+,

indexK ′′(ϕ) ≤ Cε‖ϕ‖(p−1)(1+ε)(p−1)(1+ε). (2.6.15)

Note that for ϕ, h ∈ E+,

〈K ′′(ϕ)h, h〉 = ‖h‖2 − pa0〈|ϕ|p−1h, h〉. (2.6.16)

From the definition of index K ′′(ϕ), we observe that

indexK ′′(ϕ) = maxdimH|H ⊂ E+ subspace such that

‖h‖2 ≤ pa0〈|ϕ|p−1h, h〉, for all h ∈ H.(2.6.17)

We define an operator D : L2 → E+, for v =∑

l,j,k anjksinjxe

iktel, by

Dv =∑j<|k|

(k2 − j2)−1/2∑l

aljksinjxeiktel. (2.6.18)

Define the space L2+ = spanL2 sinjxeiktel| j < |k| and let (L2

+)⊥ be theorthogonal space of L2

+ in L2. It is clear that D is an isometry from L2+ to E+

and D = 0 on (L2+)⊥.

Setting h = Dv as in (2.6.18), we get

indexK ′′(ϕ) = maxdimH|H ⊂ L2 such that

‖v‖22 ≤ 〈pa0|ϕ|p−1Dv,Dv〉, for v ∈ L2,

(2.6.19)

which means that

indexK ′′(ϕ) is the number of the eigenvalues of

D∗(pa0|ϕ|p−1)D that are greater than or equal to 1.

For the above reason, we are concerned with an operator TV,θ in L2 definedby

TV,θ = V (x, t)∑j,k

θjk∑l

aljksinjxeiktel, (2.6.20)

for v =∑

l,j,k aljksinjxe

iktel ∈ L2, where V (x, t) is a function on Ω and θ =θjk is a sequence for (j, k) ∈ N× Z. If we set

V (x, t) =√pa0|ϕ|(p−1)/2, (2.6.21)

θij =

(k2 − j2)−1/2 if j < |k|,

0 if j ≥ |k|, (2.6.22)

then letting θ = θij, we have

D∗(pa0|ϕ|(p−1))D = T ∗V ,θTV ,θ. (2.6.23)

To analyze the operator TV,θ, we recall the definition of the singular valuesof a compact operator. Let A : L2 → L2 be a compact operator. The singularvalues of A, sn(A) are the eigenvalues of |A| =

√A∗A listed according to

s1(A) ≥ s2(A) ≥ . . . . For 1 ≤ τ < ∞, A is said to lie in trace ideal Iτ if andonly if

‖A‖Iq = (∞∑n=1

sn(A)τ )1/τ <∞ for 1 ≤ τ <∞.

For τ = ∞, let

I∞ = linear bounded operators : L2 → L2

and‖A‖I∞ = sup‖Av‖2| ‖v‖2 ≤ 1 <∞.

The following properties of trace ideals are known :(i) I2 is the Hilbert-Schmidt class on L2.(ii) Let B denote the family of orthogonal sequences in L2, then

‖A‖Iτ = supun,vn∈B

(∞∑n=1

|〈un, Avn〉|τ )1/τ .

When τ = 2, for every complete orthogonal sequence vn in L2,

‖A‖I2 = (∞∑n=1

‖Avn‖22)

1/2. (2.6.24)

(iii) For τ ≥ 2, A ∈ Iτ if and only if A∗A ∈ Iτ/2 and ‖A‖2Iτ

= ‖A∗A‖Iτ/2.

Denote by lτ = lτ (N× Z) the space of sequences θ = (θjk) which satisfy

‖θ‖lτ = (∑jk

|θjk|τ )1/τ <∞ for τ ∈ [1,∞),

‖θ‖l∞ = supjk|θjk| <∞.

Lemma 2.6.7. ([21]) Suppose that V ∈ Lτ and θ = (θjk) ∈ lτ for τ ∈ [2,∞].Then TV,θ ∈ Lτ and there is a constant Cτ > 0, independent of V and θ, suchthat

‖TV,θ‖Iτ ≤ CτN‖V ‖τ‖θ‖lτ for all V and θ. (2.6.25)

Proof. The proof follows directly as [21]. First we prove the case τ = 2.Expanding the function v by using the basis 1

πsin jxeiktel in (2.6.24), we

have

‖TV,θ‖I2 =∑j,k,l

1

π2‖TV,θ(sin jxeiktel)‖2

2

=∑j,k,l

1

π2‖V (x, t)θjk sin jxeiktel‖2

2

≤∑j,k,l

1

π2‖V (x, t)‖2

2|θjk|2 =N

π2‖θ‖2

l2‖V ‖22.

Next we deal with the case τ = ∞. For v =∑

l,j,k alj,k sin jxeiktel, we get

‖TV,θv‖22 = ‖V

∑l,j,k

θjkalj,k sin jxeiktel‖2

2 ≤ ‖V ‖2∞‖θ‖2

l∞‖v‖22.

That is,‖TV,θ‖I∞ = sup

‖v‖2=1

‖TV,θv‖2 ≤ ‖V ‖∞‖θ‖l∞ .

Lastly, for 2 ≤ τ ≤ ∞, fixing un, vn ∈ B, we consider the operator Lτ ×lτ → lτ defined by (V, θ) → 〈un, TV,θvn〉n∈N. By the case τ = 2,∞, we obtain

‖〈un, TV,θvn〉‖l2 ≤ ‖TV,θ‖I2 ≤1

π2‖θ‖l2‖V ‖2,

‖〈un, TV,θvn〉‖l∞ ≤ ‖TV,θ‖I∞ ≤ ‖θ‖l∞‖V ‖∞.

By using the complex interpolation [6], we see that, for τ ∈ (2,∞),

‖〈un, TV,θvn〉‖lτ ≤ Cτ‖θ‖lτ‖V ‖τ

where Cτ is a constant independent of un, vn ∈ B. By the definition of‖TV,θ‖Iτ , we complete the proof. 2

Proof of Proposition 2.6.6. Since T ∗V,θTV,θ is a positive self-adjoint operator,we see that, for τ ≥ 2,

‖T ∗V,θTV,θ‖Iτ/2= (

∑n

sτ/2n )2/τ

where sn are the eigenvalues of T ∗V,θTV,θ. Hence we have, from the definition ofIτ and (2.6.19), that

indexK ′′(ϕ) ≤ ‖T ∗V,θTV,θ‖τ/2Iτ/2

≤ ‖TV,θ‖τIτ for τ ≥ 2.

Let V and θ as in (2.6.21) and (2.6.22). Then we have from (2.6.23) that

indexK ′′(ϕ) ≤ ‖TV ,θ‖τIτ for τ ∈ (2,∞].

Note that for every τ ∈ (2,∞] as in [21],

‖θ‖τlτ =∑j<|k|

(k2 − j2)−τ/2 = 2∑j,s∈N

((j + s)2 − j2)−τ/2

= 2∑j,s

[s(2j + s)])−τ/2 ≤∑j,s

s−τ/2j−τ/2 <∞.

Then, from Lemma 2.6.7 we deduce

indexK ′′(ϕ) ≤ N‖TV ,θ‖τIτ ≤ CτN‖θ‖τlτ‖V ‖ττ ≤ CN‖ϕ‖(p−1)τ/2

(p−1)τ/2.

So taking τ = 2(1 + ε), we get the desired result. 2

2.7 Proof of Theorem 2.1.2

As we already mentioned, by Proposition 2.5.3 and Proposition 2.5.4, we knowthat

bqj ≥ Cεq(p+1)/[(p−1)(1+ε)]j for large qj ∈ N,

ensures the existence of an unbounded sequence ϕj of critical points of J .Then by Proposition 2.4.1, we obtained that the unbounded critical pointsϕj of J satisfy I(ϕj) = J(ϕj) and I ′(ϕj) = 0.

By Proposition 2.5.10, it is sufficient to show the following property: forevery ε > 0, there is a Cε > 0 such that

σqj ≥ Cεq(p+1)/[(p−1)(1+ε)]j for large qj ∈ N.

Since σq → ∞ as q → ∞, there is a sequence qj such that σqj < σqj+1. ByProposition 2.6.2, there is a sequence ϕj ∈ E+ such that

K(ϕj) ≤ σqj , (2.7.1)

K ′(ϕj) = 0, (2.7.2)

indexK ′′(ϕj) ≥ qj, for large j ∈ N. (2.7.3)

Next by using Proposition 2.6.6, we get

Cε‖ϕj‖(p−1)(1+ε)(p−1)(1+ε) ≥ qj.

Choosing ε ∈ (0, 2/(p− 1)), we obtain that, for j ∈ N,

‖ϕj‖p+1p+1 ≥ C‖ϕj‖p+1

(p−1)(1+ε) ≥ C ′εq

(p+1)/[(p−1)(1+ε)]j . (2.7.4)

On the other hand, (2.7.2) implies

〈K ′(ϕj), ϕj〉 = ‖ϕj‖2 − a0‖ϕj‖p+1p+1 = 0.

Then, by (2.7.1) we have

σqj ≥ K(ϕj) =1

2‖ϕj‖2 − a0

p+ 1‖ϕj‖p+1

p+1 = (1

2− a0

p+ 1)‖ϕj‖p+1

p+1.

By (2.7.4), we conclude that there exists a unbounded sequence qj, as j →∞,such that

σqj ≥ Cεq(p+1)/[(p−1)(1+ε)]j .

Therefore we complete the proof by Proposition 2.2.3. 2

Chapter 3

Positive solutions of nonlinearproblems involving the squareroot of the Laplacian

3.1 Introduction

In this chapter we are concerned with positive solutions to nonlinear prob-lems involving a nonlocal positive operator: the square root of the Laplacianoperator in a bounded domain with zero Dirichlet boundary condition. We es-tablish the existence and nonexistence of positive solutions for problems withpower-type nonlinearities, the regularity and an L∞-estimate of some weak so-lutions, a symmetry result of Gidas-Ni-Nirenberg type, and a priori estimatesof Gidas-Spruck type. Particularly, we are looking for a function u satisfyingthe nonlinear problem involving the square root of the Laplacian:

A1/2u = f(u) in Ω,u = 0 on ∂Ω,u > 0 in Ω,

(3.1.1)

where Ω is a smooth bounded domain in Rn and A1/2 stands for the squareroot of the Laplacian operator −∆ in Ω with zero Dirichlet boundary value on∂Ω.

The fractional powers of the Laplacian, which are called fractional Lapla-cians and correspond to Levy stable processes, appear in anomalous diffusionphenomena in physics, biology as well as other areas. They occur in flamepropagation, chemical reaction in liquids, population dynamics. Levy diffusionprocesses have discontinuous sample paths and heavy tails, while Brownianmotion has continuous sample paths and exponential decaying tails. These

67

68 Chapter 3. The square root of the Laplacian

processes have been applied to American options in mathematical finance formodelling the jump processes of the financial derivatives such as futures, for-wards, options, and swaps, see [26] and references therein. Moreover, they playan important role in the study of the quasi-geostrophic equations in geophys-ical fluid dynamics. Recently the fractional Laplacians attract much interestin nonlinear analysis. Caffarelli and Silvestre [35] gave a new formulation ofthe fractional Laplacians through Dirichlet-Neumann maps. The regularity ofthe obstacle problem for the fractional powers of the Laplacian operator wasproved by Silvestre [51]. Moreover, Caffarelli et al [36],[34] studied a free bound-ary problem: the Signorini problem involving fractional Laplacians as well asrandom homogenization of fractional obstacle problems.

The purpose of this chapter is to look for functions satisfying nonlinearproblems involving the square root of the Laplacian A1/2 as in (3.1.1). Notethat A1/2 is a nonlocal operator in Ω, but we will realize it through a localproblem in Ω× (0,∞). We mention that the half Laplacian in the whole spaceis a well studied operator. Let u be a smooth function u ∈ C∞

0 (Rn). Thereis a unique harmonic extension v ∈ C∞(Rn+1

+ ) of u in a half space such thatDkv(x, y) → 0 as |(x, y)| → ∞, for all k ≥ 0 and v(x, 0) = u(x). It is thesolution of the following Laplacian problem:

∆v = 0 in Rn+1+ ,

v = u on Rn = ∂Rn+1+ .

Consider the operator T : u 7→ −∂yv(·, 0). Since ∂yv is still a harmonic function,if we apply the operator twice, we obtain

T Tu = ∂yyv |y=0= −∆xv |y=0= −∆u in Rn.

Thus the operator T that maps the Dirichlet-type data u to the Neumann-typedata −∂yv(x, 0) is actually the half Laplacian.

In this chapter, we introduce a new analogue operator but now in a boundeddomain Ω ⊂ Rn. Consider the harmonic extension v of u in the half-cylinderΩ × (0,∞) vanishing on the lateral boundary ∂Ω × [0,∞). Then since ∂yv isharmonic and also vanishes on the lateral boundary, as before the Dirichlet-Neumann map of the harmonic extension v on the bottom of the half cylinder isthe square root of the Laplacian: A1/2 = B−1

1/2. That is, we have the properties:

A1/2 A1/2 = −∆ and B1/2 B1/2 = (−∆)−1,

where −∆ is the Laplacian in Ω with zero Dirichlet boundary value on ∂Ω.In this way we can study problem (3.1.1) by variational methods for a local

problem. More precisely, we will study the following mixed value boundaryproblem in a half cylinder:

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= f(v) on Ω× 0,v > 0 in C,

(3.1.2)

where ν is the unit outer normal to Ω×0. If v satisfies (3.1.2), then the traceu on Ω×0 of the function v will be a solution of problem (3.1.1). Moveover,we will have that the operator A1/2 is self-adjoint and positive definite andthat A1/2 has a spectral representation in terms of the eigenvalues and theeigenfunctions of −∆ in Ω with zero Dirichlet boundary conditions.

By studying problem (3.1.2), we establish existence and non-existence re-sults for problem (3.1.1) with power-type nonlinearities, the regularity and anL∞ estimate of weak solutions, a symmetry result of Gidas-Ni-Nirenberg typeand a priori estimates of Gidas-Spruck type.

The analogue problem to (3.1.1) for the Laplacian operator has been inves-tigated widely in the last decades. This is the following problem

−∆u = f(u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω,

(3.1.3)

where Ω is a smooth bounded domain in Rn. If f(u) = up in problem (3.1.3),then there is a sharp contrast between the subcritical case p < n+2

n−2, for which

the problem admits a solution, and the critical case p = n+2n−2

, for which theSobolev embedding is not compact. Pohozaev discovered that there is no pos-itive solution for the critical or supercritical problem

f(u) = up and p ≥ n+ 2

n− 2,

when Ω is a star-shaped domain.

In the case of the square root of the Laplacian A1/2, since the Sobolevtrace embedding is compact in the subcritical case in bounded domains, bystudying a corresponding minimization problem for (3.1.2), we get a positivesolution for it. We also build a Pohozaev type identity which implies that thereis no positive solution for the critical problem or supercritical in star-shapeddomains. More precisely, we derive the following existence and non-existenceresults for positive solutions to problem (3.1.1):

Theorem 3.1.1. Let n ≥ 1 be an integer and 2] = 2nn−1

when n ≥ 2. Supposethat Ω is a smooth bounded domain in Rn and f(u) = up. Then,(i) Problem (3.1.1) has at least one C2,α(Ω) solution when 1 < p < 2]−1 = n+1

n−1

if n ≥ 2, or 1 0, the existence of positive solutions

of problem (3.1.3) was studied in a famous paper by Brezis and Nirenberg [31].For this they studied the minimizing problem:

min

∫Ω

(|∇u|2 − µ|u|2) dx | u ∈ H10 (Ω), ‖u‖

L2n

n−2 (Ω)= 1

.

The critical points of this constrained functional correspond to weak solutionsof problem (3.1.3). But this energy functional may lose compactness, sincethe nonlinearity involves the critical exponent. While the functional does notsatisfy the Palais-Smale condition globally, some compactness will hold in therange determined by the best constant of the Sobolev inequality. The steps ofthe proof need a careful analysis introduced by Brezis and Nirenberg aboutthe energy level computed on cut-off functions of the extremal functions forthe best constant in the Sobolev inequality. Their technique has been extendedto many other situations. In the following result we use the Brezis-Nirenbergtechnique to build an analogue result for problem (3.1.1) related to the squareroot of the Laplacian A1/2:

Theorem 3.1.2. Let n ≥ 2 and 2]. Suppose that Ω is a smooth bounded

domain in Rn and f(u) = u2]−1+ µu = u

n+1n−1 + µu, and that µ1 =

√λ1 is the

first eigenvalue of A1/2, where λ1 is the first eigenvalue of the Laplacian −∆in Ω with zero Dirichlet boundary condition.

Then, for every µ ∈ (0, µ1), there exists at least one C2,α(Ω) solution of(3.1.1). Furthermore, there exists no weak solution of (3.1.1) for µ ≥ µ1.

There is another approach to the Brezis-Nirenberg result, based on a care-ful study of the compactness properties for Palais-Smale sequences of the func-tional Φ(u) associated to problem (3.1.3):

Φ(u) =1

2

∫Ω

|∇u|2 dx− n− 2

2n

∫Ω

|u|2n

n−2 dx− µ

2

∫Ω

|u|2 dx.

Both approaches are completely equivalent. However, the second approachbrings out the peculiarities of the limiting case more clearly, which gives theenergy estimate of Palais-Smale sequences. We will prove the existence of pos-itive solutions of (3.1.1) via both procedures. The steps of our proofs need a

careful analysis about the energy level computed on cut-off functions of theextremal functions for the best constant in the Sobolev trace inequality.

Gidas, Ni and Nirenberg [42] obtained symmetry properties of solutionsfor problem (3.1.3) when f is Lipschitz continuous and Ω has certain symme-tries. The proof of these symmetry results uses the maximum principle and amethod of Alexandroff, developed in the framework of partial differential equa-tions by Serrin, called the moving planes method. In the improved method ofBerestycki-Nirenberg [29], it replaces the use of Hopf’s lemma by the maximumprinciple in domains of small measure.

In this chapter, we obtain a symmetry result of Gidas-Ni-Nirenberg typefor problem (3.1.1) in symmetric domains. This symmetry result is proved byusing the moving planes method.

Theorem 3.1.3. Let Ω be a bounded domain of Rn which is convex in thex1 direction and symmetric with respect to the hyperplane x1 = 0. Let f beLipschitz continuous and u be a C2(Ω) solution of (3.1.1).

Then u is symmetric with respect to x1, i.e., u(−x1, x′) = u(x1, x

′) for all(x1, x

′) ∈ Ω. In addition, ∂u∂x1

< 0 for x1 > 0.

For the proof we also establish maximum principle in domains of smallmeasure for problem (3.1.1). As a consequence of Theorem 3.1.3, we have

Corollary 3.1.4. Assume that f(u) = up for 1 < p ≤ n+1n−1

, Ω is a ball in Rn

and that v is a weak solution of (3.1.2). Then we have that u(x) = v(x, 0) =u(|x|) for all x in the ball and that v is axially symmetric with respect to they-axis.

We point out that Chipot, Chlebık, Fila and Shafrir [37] studied the prob-lem:

−∆v = g(v) in B+R = z ∈ Rn+1 | |z| ≤ R, zn+1 > 0,

v = 0 on ∂B+R ∩ zn+1 > 0,

∂v∂ν

= f(v) on ∂B+R ∩ zn+1 = 0,

v > 0 in B+R ,

(3.1.4)

where f, g ∈ C1(R) and ν is the unit outer normal. They proved existence,non-existence and axial symmetry results for solutions of (3.1.4).

Gidas and Spruck [43] established a priori estimates for positive solutions ofproblem (3.1.3) when f(u) = up and p < n+2

n−2. The proof involves the method of

blow-up combined with some important ingredients: nonlinear Liouville typeresults in all space and in a half space. The moving planes method or themoving spheres method are strong tools to prove such Liouville theorems. TheKelvin transform is also useful in the proof.

In this chapter, we also establish the following Gidas-Spruck type a prioriestimates for positive solutions of problem (3.1.1). For this, we use the blowup technique.


. Assume that Ω ⊂ Rn is a smooth

bounded domain and f(u) = up, 1 < p < 2] − 1. Then there exists a constantC(p,Ω), dependent only on p and Ω, such that every weak solution of (3.1.1)satisfies

‖u‖L∞(Ω) ≤ C(p,Ω).

The proof of the previous results uses two important ingredients: nonlinearLiouville theorems for the half Laplacian in Rn and in Rn

+. The one for thewhole space is proved by Y.Y. Li and L. Zhang [46] by using the movingspheres method and Ou [50] by using the moving planes method. It states thatthere is no positive solution for problem

(−∆)1/2u = up in Rn,

where 1 0, y > 0.

If 1 0, y = 0,v = 0 on xn = 0, y > 0,v > 0 in Rn+1

++ ,

(3.1.5)

where ν is the unit outer normal to xn > 0, y = 0.

The proof of this Liouville theorem in the half space for A1/2 uses sometools developed by Cabre and Sola-Morales [33]. They studied layer solutions(solutions which are monotone with respect to one variable) of

(−∆)1/2u = f(u) in Rn,

where f is of balanced bistable type. That is, if G(u) = −∫ u

0f(s) ds, then G

has two, and only two, absolute minima at the same height. They developedsome new ingredients, a nonlocal Modica type estimate, as well as a conserved


Hamiltonian quantity for every layer solution. These tools are useful to provenonlinear Liouville theorem in the half space. The proof of our result combinesthe Kelvin transform, the moving planes method, as well as the Hamiltonianidentity in spirit of [33]. The result is still open without the assumption ofboundedness of the solutions. We recall that then we use this Liouville theoremand the method of blow up, to obtain a priori estimates of positive solutionsto problem (3.1.1) as stated in Theorem 3.1.5.

The chapter is organized as follows. In Section 3.2, we give the definitionof the square root of the Laplacian operator and of the appropriate functionspaces. Regularity results and maximum principles can be found in Sections3.3 and 3.4. The proofs of Theorems 3.1.1 and 3.1.2 are given in Sections 3.5and 3.6. We prove Theorem 3.1.3 in Section 3.7, and Theorems 3.1.5 and 3.1.6in Section 3.8.

3.2 Preliminaries: function spaces and the op-

erator A1/2

In this section, we collect preliminary facts for future reference. First of all, letus write the standard notations which we will use in this chapter.

Rn+1+ = z = (x, y) = (x1, · · · , xn, y) ∈ Rn+1 | y > 0.

Denote by Hs(U) = W s, 2(U) the Sobolev space in a domain U of Rn or Rn+1+ .

For example, letting U ⊂ Rn and s > 0, Hs(U) is a Sobolev space with itsnorm

‖u‖Hs(U) =∫

U

∫U

|u(x)− u(x)|2

|x− x|n+2sdxdx+

∫U

|u(x)|2 dx1/2

.

Let Ω be a bounded smooth domain in Rn. Denote the half cylinder withbase Ω by

C = Ω× (0,∞)

and its lateral boundary by

∂LC = ∂Ω× [0,∞).

To treat the nonlocal problem (3.1.1), we will study a corresponding exten-sion problem in one more dimension, which allows us to investigate problem(3.1.1) by studying a local problem via classical nonlinear variational methods.We define a Sobolev space of functions whose traces vanish on ∂LC:

H10, L(C) = v ∈ H1(C) | v = 0 a.e. on ∂LC , (3.2.1)

equipped with the norm

‖v‖ = (

∫C|∇v|2 dxdy)1/2. (3.2.2)

We denote by trΩ the trace operator on Ω× 0 for functions in H10, L(C):

trΩv := v(x, 0), for v ∈ H10, L(C).

We have that trΩv ∈ H1/2(Ω), since it is well known that traces of H1 functionsare H1/2 functions on the boundary.

Recall the well known spectral theory of the Laplacian −∆ in a smoothbounded domain Ω with zero Dirichlet boundary value. We repeat each eigen-value of −∆ in Ω with zero Dirichlet boundary condition according to its(finite) multiplicity:

0 < λ1 < λ2 ≤ · · · ≤ λk ≤ · · · → ∞, as k →∞

and we denote by ϕk ∈ H10 (Ω) an eigenfunction corresponding to λk for k =

1, 2, · · · . Namely, −∆ϕk = λkϕk in Ω,ϕk = 0 on Ω.

(3.2.3)

We can take them to form an orthonormal basis ϕk of L2(Ω), in particular,∫Ω

ϕ2k dx = 1,

and to belong to C0(Ω) ∩ C2(Ω) by regularity theory.Now we can state the main results which we are going to prove at the end

of this section.

Proposition 3.2.1. Let V0(Ω) be the space of all traces on Ω×0 of functionsin H1

0, L(C). Then we have the following properties:

V0(Ω) := u = trΩv | v ∈ H10, L(C)

= u ∈ H1/2(Ω) |∫

Ω

u2(x)

d(x)dx < +∞

= u ∈ L2(Ω) | u =∞∑k=1

bkϕk satisfying∑k

b2kλ1/2k < +∞,

where d(x) = dist(x, ∂Ω), and λk, ϕk is the spectral decomposition of −∆ in Ωas above, with ϕk an orthonormal basis in L2(Ω).

Proposition 3.2.2. If u ∈ V0(Ω), then there exists a unique harmonic exten-sion v in C of u such that v ∈ H1

0, L(C). In particular, if the expansion of u iswritten by u(x) =

∑∞k=1 bkϕk(x) ∈ V0(Ω), then

v(x, y) =∞∑k=1

bkϕk(x) exp(−λ1/2k y) ∈ H1

0, L(C),

where λk, ϕk is the spectral decomposition of −∆ in Ω as above, with ϕk anorthonormal basis in L2(Ω). Let us define the operator A1/2 : V0(Ω) → V∗0 (Ω)by

A1/2u :=∂v

∂ν|Ω×0,

where V∗0 (Ω) is the dual space of V0(Ω). Then

A1/2u =∞∑k=1

bkλ1/2k ϕk,

and A21/2 is equal to −∆ in Ω with zero Dirichlet boundary value on ∂Ω. More

precisely, the inverse B1/2 := A−11/2 is the unique square root of the inverse

Laplacian (−∆)−1 in Ω with zero Dirichlet boundary value on ∂Ω.

The proofs of these two propositions need the development of several tools.First let us give some properties of the space H1

0, L(C). Denote by D1,2(Rn+1+ )

the closure of the set of smooth functions compactly supported in Rn+1+ with

respect to the norm of ‖w‖D1,2(Rn+1+ ) =

( ∫Rn+1

+|∇w|2 dxdy

)1/2

. The well known

Sobolev trace inequality [48] states that for w ∈ D1, 2(Rn+1+ ),( ∫

Rn

|w(x, 0)|2n/(n−1)dx)(n−1)/2n

≤ C( ∫

Rn+1+

|∇w(x, y)|2dxdy)1/2

, (3.2.4)

where C depends only on n.Denote for n ≥ 2,

2] =2n

n− 1and 2] − 1 =

n+ 1

n− 1.

We say that p is subcritical if 1 2] − 1 = n+1n−1

for n ≥ 2.Consider

S0 = inf

∫Rn+1

+|∇w(x, y)|2dxdy

(∫

Rn |w(x, 0)|2]dx)2/2] | w ∈ D1,2(Rn+1+ )

. (3.2.5)

It is known [40] that S0 is achieved by the extremal functions

Uε(x, y) =ε(n−1)/2

|(x, y + ε)|n−1, (3.2.6)

where ε > 0 is arbitrary.For v ∈ H1

0, L(C), its extension by zero in Rn+1+ \ C can be approximated

by functions compactly supported in Rn+1+ . Thus the Sobolev trace inequality

(3.2.4) leads to:

Lemma 3.2.3. Let n ≥ 2 and 2] = 2nn−1

. There exists a constant C, dependingonly on n, such that, for all v ∈ H1

0, L(C),

( ∫Ω

|v(x, 0)|2]

dx)1/2]

≤ C( ∫

C|∇v(x, y)|2dxdy

)1/2

. (3.2.7)

Therefore by Holder’s inequality, since Ω is bounded, the above lemma leadsto:

Lemma 3.2.4. Let 1 ≤ q ≤ 2] for n ≥ 2. Then we have that for all v ∈H1

0, L(C), ( ∫Ω

|v(x, 0)|qdx)1/q

≤ C( ∫

C|∇v(x, y)|2dxdy

)1/2

, (3.2.8)

where C depends only on n, q and the measure of Ω. Moreover, (3.2.8) holdsif 1 ≤ q <∞ for n = 1.

This lemma states that trΩ(H10, L(C)) ⊂ Lq(Ω), where 1 ≤ q ≤ 2] for n ≥ 2

and 1 ≤ q <∞ for n = 1. We also have the following compact embedding.

Lemma 3.2.5. Let 1 ≤ q < 2] = 2nn−1

for n ≥ 2 and 1 ≤ q < ∞ for n = 1.Then trΩ(H1

0, L(C)) is compactly embedded in Lq(Ω).

Proof. It is well known that trΩ(H10, L(C)) ⊂ H1/2(Ω) and that H1/2(Ω) ⊂⊂

Lq(Ω) when 1 ≤ q < 2] = 2nn−1

for n ≥ 2 and 1 ≤ q < ∞ for n = 1. Here⊂⊂ denotes the compact embedding. This completes the proof of the lemma.However, here we would like to give another proof without using the abovefractional Sobolev space.

Considering the restriction of functions in C to Ω×(0, 1), it suffices to showthat the embedding is compact with C replaced by Ω× (0, 1). To prove this, letvm ∈ H1

0, L(Ω× (0, 1)) := v ∈ H1(Ω× (0, 1)) | v = 0 a.e. on ∂Ω× (0, 1) suchthat vm 0 weakly in H1

0, L(Ω × (0, 1)), as m → ∞. We may assume by the

classical Rellich’s theorem in Ω× (0, 1) that vm → 0 strongly in L2(Ω× (0, 1)),as m→∞. We introduce the function wm = vm(1− y). It is clear that

wm|Ω×0 = vm, wm|y=1 = 0.

By direct computations we have∫Ω

|vm(x, 0)|2 dx =

∫Ω

|wm(x, 0)|2 dx = −∫ 1

0

∫Ω

∂y(w2m(x, y)) dxdy

≤ 2( ∫ 1

0

∫Ω

w2m(x, y) dxdy

)1/2( ∫ 1

0

∫Ω

|∇wm(x, y)|2 dxdy)1/2

.

Therefore, since |wm| ≤ |vm|, we find that, as m→∞,

vm(x, 0) → 0 strongly in L2(Ω) and hence also in L1(Ω). (3.2.9)

On the other hand, since q is subcritical, using the following interpolationinequality, we have

‖vm(x, 0)‖Lq(Ω) ≤ ‖vm(x, 0)‖θL1(Ω)‖vm(x, 0)‖1−θL2]

(Ω),

where 0 < θ < 1. This completes the proof since we have established that vmstrongly converges to zero in L1(Ω). 2

We also need to establish a trace boundary Hardy inequality:

Lemma 3.2.6. We have

trΩ(H10, L(C)) ⊂ H1/2(Ω).

In addition, for every v ∈ H10, L(C),∫

Ω

|v(x, 0)|2

d(x)dx ≤ C

∫C|∇v(x, y)|2 dxdy,

where d(x) = dist(x, ∂Ω) and the constant C depends only on Ω.

Proof. The first statement is clear since the traces of H1(C) functions belongto H1/2(∂C). Regarding the second statement is proved in two steps. (1) LetΩ = (0, 1). Then (x, y) ∈ (0, 1) × (0,∞). For 0 < x0 < 1/2, consider thesegment from (0, x0) to (x0, 0) in C = (0, 1)× (0,∞). We have

v(x0, 0) = v(t, x0 − t) |x0t=0=

∫ x0

0

(∂xv − ∂yv)(t, x0 − t) dt.

Then

|v(x0, 0)|2 ≤ x0

∫ x0

0

2|∇v(t, x0 − t)|2 dt.

Integrating in x0 over (0, 1/2), and making the change of variables x = t,y = x0 − t, we deduce∫ 1/2

0

|v(x0, 0)|2

x0

dx0 ≤ 2

∫ 1/2

0

∫ x0

0

|∇v|2 dxdy ≤ 2

∫C|∇v|2 dxdy.

(2) In the general case, after straightening a piece of the boundary ∂Ω andrescaling the new variables, we can consider the inequality in a domain D =x = (x′, xn) | |x′| < 1, 0 < xn < 1/2 and assume that v = 0 on xn =0, |x′| < 1 × (0,∞), since the flatting procedure possesses equivalent norms.By the argument in (1) above, we have∫ 1/2

0

|v(x, 0)|2

xndxn ≤ C

∫ 1/2

0

∫ ∞

0

|∇v|2 dxndy,

for all x′ with |x′| < 1. From this, integrating in x′ we see∫D

|v(x, 0)|2

xndx =

∫D

∫ 1/2

0

|v(x, 0)|2

xndx′dxn

≤C∫D×(0,∞)

|∇v|2 dxdy.

Since xn is comparable to d(x) = dist(x, ∂Ω), this is the desired inequality. 2

Recall that the fractional Sobolev space H1/2(Ω) is a Banach space withthe norm

‖u‖2H1/2(Ω) =

∫Ω

∫Ω

|u(x)− u(x)|2

|x− x|n+1dxdx+

∫Ω

|u(x)|2 dx. (3.2.10)

Note that the closure H1/20 (Ω) of smooth functions with compact support

C∞c (Ω) in H1/2(Ω) is all the space H1/2(Ω), by Theorem 11.1 in [47]; that is,

C∞c (Ω) is dense in H1/2(Ω). Recall that in Proposition 3.2.1 we have denoted

by V0(Ω) the space of traces on Ω× 0 of functions in H10, L(C):

V0(Ω) := u = trΩv | v ∈ H10, L(C) ⊂ H1/2(Ω), (3.2.11)

endowed with the norm of H1/2(Ω). The dual space of V0(Ω) is denoted byV∗0 (Ω), equipped with the norm

‖g‖V∗0 = sup〈u, g〉 | u ∈ V0(Ω), ‖u‖H1/2(Ω) ≤ 1.

Next we give the first characterization of the space V0(Ω):

Lemma 3.2.7. Let V0(Ω) be the space of traces on Ω × 0 of functions inH1

0, L(C) defined as (3.2.11). Then we have

V0(Ω) =

u ∈ H1/2(Ω) |

∫Ω

u2(x)

d(x)dx < +∞

.

Proof. The inclusion⊂ follows Lemma 3.2.6. Next we show the other inclusion.Let u be the extension of u in Rn by zero in Rn \ Ω. The quantity ‖u‖H1/2(Rn)

given by expression (3.2.10) with u and Ω replaced u and Rn, respectively, can

be bounded –using u ≡ 0 in Rn \ Ω– by‖u‖2

H1/2(Ω)+

∫Ωu2(x)d(x)

dx1/2

, where

d(x) = dist(x, ∂Ω), that we assume to be finite. Hence, u ∈ H1/2(Rn) and thusu is the trace in Rn = ∂Rn+1

+ of a function w ∈ H1(Rn+1+ ).

Next, we claim that there is a bi-Lipschitz map from Rn+1+ onto Ω×[0,∞) =

C which is the identity on Ω×0 and which maps Rn \Ω = (∂Rn+1+ ) \Ω onto

∂Ω×[0,∞). By composing this map with the function w, we obtain an H10, L(C)

function with trace on Ω× 0 as desired.Finally we prove the claim on the existence of the bi-Lipschitz map. First,

consider the one dimensional case: Ω = (0,∞). Then simply take the bi-Lipschitz map

L :(x, y) ∈ (0,∞)× (0,∞) = Ω× (0,∞)

7→ (x2 − y2√x2 + y2

,2xy√x2 + y2

) ∈ R× (0,∞),

whose Jacobian can be checked to be identically 2. In the general case, we canflatten the boundary ∂Ω and use the previous map. 2

Now we consider, for a function u ∈ V0(Ω) on Ω ⊂ Rn, the minimizingproblem:

J(v) = inf∫C|∇v|2 dxdy, | v ∈ H1

0, L(C), v = u on Ω.

Note that the set of functions v where we minimize is non empty by the def-inition of V0(Ω) and the fact that u ∈ V0(Ω). By lower weak semi-continuityand by Lemma 3.2.5, we will see next that there is a minimizer of J .

We call v a weak solution of the problem−∆v = 0 in C,v = 0 on ∂LC,v = u on Ω× 0.

(3.2.12)

The existence of the minimizer is proved as follows:


Lemma 3.2.8. If u ∈ V0(Ω), then there exists a unique minimizer v ∈ H10, L(C)

of J(v). The function v is the harmonic extension of u (in the weak sense) toC and vanishing on ∂LC.

Proof. By the definition of V0(Ω), we have that, for every u ∈ V0(Ω), thereexists at least one w ∈ H1

0, L(C) such that trΩ(w) = u. Then the standardminimization argument gives (using lower semi-continuity and Lemma 3.2.5)the existence of a minimizer. The uniqueness follows automatically from theidentity of the parallelogram used for two possible minimizers v1 and v2:

0 ≤ J(v1 − v2

2) =

1

2J(v1) +

1

2J(v2)− J(

v1 + v2

2) ≤ 0.

2

By Lemma 3.2.8, there exists a function v ∈ H10, L(C), which is the harmonic

extension of u in C vanishing on ∂LC, denoted by

v := h-ext(u).

It is easy to see that for every ξ ∈ C∞ and ξ ≡ 0 on ∂LC,∫C∇v∇ξ dxdy =

∫Ω

∂v

∂νξ dx. (3.2.13)

By the trace theorem from H1(C) onto H1/2(∂C), we know that, for everyu ∈ V0(Ω),

‖u‖H1/2(Ω) ≤ C‖h-ext(u)‖H10, L(C),

for some constant C depending on Ω. Since the h-ext operator is bijective fromV0(Ω) to H1

0, L(C), we also have the reverse inequality:

‖h-ext(u)‖H10, L(C) ≤ ‖u‖H1/2(Ω) = C‖u‖V0(Ω),

for all u ∈ V0(Ω). From this we deduce the following. Given ξ ∈ V0(Ω), con-

sider the h-ext(ξ) and call it still ξ. Now use (3.2.13), to obtain∣∣∣ ∫

Ω∂v∂νξ dx

∣∣∣ ≤C‖u‖H1/2(Ω)‖ξ‖H1/2(Ω). That is, ∂v

∂ν|Ω∈ V∗0 (Ω) and there is a bound for its

norm. Hence we have:

Definition 3.2.9. Define the operator A1/2 : V0(Ω) → V∗0 (Ω) by

A1/2u :=∂v

∂ν|Ω×0, (3.2.14)

where v = h-ext(u) ∈ H10, L(C). It is clear that A1/2 is linear and bounded from

V0(Ω) to V∗0 (Ω).

We now give the spectral representation of A1/2 and the correspondingstructure of the space V0(Ω).

Lemma 3.2.10. (i) Let ϕk be an orthonormal basis of L2(Ω) giving anspectral decomposition of −∆ in Ω with Dirichlet boundary conditions as in(3.2.3). Then we have

V0(Ω) = u =∞∑k=1

bkϕk ∈ L2(Ω) |∞∑k=1

b2kλ1/2k < +∞.

(ii) Let u ∈ V0(Ω). Then we have, for u =∑∞

k=1 bkϕk,

A1/2u =∞∑k=1

bkλ1/2k ϕk.

Proof. Let u ∈ V0(Ω), which is contained in L2(Ω). Then its expansion iswritten by u(x) =

∑∞k=1 bkϕk(x). Consider the smooth function for y > 0,

v(x, y) =∞∑k=1

bkϕk(x) exp(−√λky). (3.2.15)

Observe that v(x, 0) = u(x) in Ω and, for y > 0,

∆v(x, y) =∞∑k=1

bk−λkϕk exp(−√λky) + λkϕk exp(−

√λky) = 0.

So v is a harmonic extension. We will see that v = h-ext(u) by uniqueness oncewe find the condition on bk for v to belong to H1

0, L(C). But such condition issimple: using (3.2.15) and that ϕk are eigenfunctions of −∆ and orthonormalin L2(Ω). We have∫ ∞

0

∫Ω

|∇v|2 dxdy =

∫ ∞

0

∫Ω

|∇xv|2 + |∂yv|2 dxdy

= 2∞∑k=1

b2kλk

∫ ∞

0

exp(−2λ1/2k y) dy

= 2∞∑k=1

b2kλk1

2λ1/2k

=∞∑k=1

b2kλ1/2k .

This means that v ∈ H10, L(C) if and only if

∑∞k=1 b

2kλ

1/2k < ∞. Therefore, this

condition on bk is equivalent to u ∈ V0(Ω).Assertion (ii) follows from the direct computation of−∂v

∂y|y=0 using (3.2.15).

2


Definition 3.2.11. Define the operator B1/2 : V∗0 (Ω) → V0(Ω), by g 7→ trΩv,where v is found by solving the following problem:

−∆v = 0 in C,v = 0 on ∂LC,∂v∂ν

= g(x) on Ω× 0,(3.2.16)

as we indicate next.

We say that v is a weak solution of (3.2.16), whenever v ∈ H10L(C) and∫

C∇v∇ξ dxdy = 〈g(x), ξ(·, 0)〉 (3.2.17)

for all ξ ∈ H10, L(C). We see that there is a unique weak solution of (3.2.16)

by the Lax-Milgram theorem, via studying the corresponding functional inH1

0, L(C):

I(v) =1

2

∫C|∇v|2 dxdy − 〈g(x), v(·, 0)〉,

where g ∈ V∗0 (Ω) is given. Observe that the operator B1/2 is clearly the inverseof the operator A1/2.

On the other hand, let us compute now B1/2 B1/2. Let ϕ ∈ H10 (Ω)∩H2(Ω)

be the solution of Poisson’s problem for the Laplacian for a given g ∈ L2(Ω),−∆ϕ = g in Ω,ϕ = 0 on ∂Ω.

Since H10 (Ω) ⊂ V0(Ω) (for instance, by Lemma 3.2.10) we have that the har-

monic extension ψ ∈ H10, L(C) of ϕ in C is the solution of

−∆ψ = 0 in C,ψ = 0 on ∂LC,ψ = ϕ on Ω× 0.

Moreover, ψ := ψ(x, y)− ϕ(x) solves−∆ψ = ∆ϕ = −g in C,ψ = 0 on ∂LC,ψ = 0 on Ω× 0.

Considering the odd reflections ψod and god of ψ and g with respect to Ω×0,we have

−∆ψod = −god in Ω× (−∞,∞),

ψod = 0 on ∂Ω× (−∞,∞).


Therefore, since god ∈ L2(Ω× (−1, 1)), we obtain that ψod ∈ H2(Ω× (−1, 1))and hence ψ ∈ H2(Ω× (0, 1)). We deduce, by the exponential decay in y of theharmonic function ψ, that ψ ∈ H1

0, L(C)∩H2(C). It follows that−∂yψ ∈ H10, L(C)

solves −∆(−∂yψ) = 0 in C,−∂yψ = 0 on ∂LC,∂∂ν

(−∂yψ) = −∆xψ = −∆ϕ = g on Ω× 0.Since V0(Ω) ⊂ L2(Ω), L2(Ω) ∼= L2(Ω)∗ ⊂ V∗0 (Ω), we deduce that the solutionv of (3.2.16) is v = −∂yψ, because of the uniqueness of H1

0, L(Ω) solution of(3.2.16). In particular, B1/2g = v(·, 0) = −∂yψ(·, 0). On the other hand, sinceψ ∈ H1

0, L(C) solves−∆ψ = 0 in C,ψ = 0 on ∂LC,∂ψ∂ν≡ −∂yψ(·, 0) = v(·, 0) on Ω× 0,

we conclude that

B1/2 B1/2g = B1/2v(·, 0) = ψ(·, 0) = ϕ = (−∆)−1g.

Summarizing the above argument, we have:

Proposition 3.2.12. B1/2 B1/2 |L2(Ω)= (−∆Ω)−1 : L2(Ω) → L2(Ω), where(−∆Ω)−1 is the inverse Laplacian in Ω with the zero Dirichlet boundary con-dition.

Note that B1/2 : L2(Ω) → L2(Ω) is a self-adjoint operator. In fact, since forv1, v2 ∈ H1

0,L(C),∫C(v2∆v1 − v1∆v2) dxdy =

∫Ω

(v2∂v1

∂ν− v1

∂v2

∂ν) dx,

we see ∫Ω

B1/2g2 · g1 =

∫Ω

B1/2g1 · g2

and ∫Ω

v2(x, 0)A1/2v1(x, 0) dx =

∫Ω

v1(x, 0)A1/2v2(x, 0) dx.

On the other hand, by using (3.2.17) with ξ = v and Lemma 3.2.5, we obtainthat B1/2 is a positive compact operator in L2(Ω). Hence by the operatortheory of compact, self adjoint operators we have that all the eigenvalues ofB1/2 are real, positive, and there are corresponding eigenfunctions which makeup an orthonormal basis of L2(Ω). Furthermore, such basis and eigenvalues areexplicit in terms of those of the Laplacian with Dirichlet boundary conditions,since we have the expression of A1/2 given in Lemma 3.2.10 (ii). Summarizing:

Proposition 3.2.13. Let ϕk be an orthonormal basis of L2(Ω) giving an spec-tral decomposition of −∆ in Ω with Dirichlet boundary conditions as in (3.2.3).Then for all k ≥ 1,

A1/2ϕk = λ1/2k ϕk in Ω,

ϕk = 0 on ∂Ω.(3.2.18)

In particular, ϕk is also a basis of eigenfunctions of A1/2, with eigenvalues

λ1/2k .

Proof of Proposition 3.2.1 It follows from Lemma 3.2.7 and Lemma 3.2.10.2

Proof of Proposition 3.2.2 It follows from Lemma 3.2.8, Lemma 3.2.10 andPropositions 3.2.12 and 3.2.13. 2

3.3 Regularity of solutions

Now we study the regularity of weak solutions for linear and nonlinear problemsinvolving A1/2. First we consider the linear problem

A1/2u = g(x) in Ω,u = 0 on ∂Ω,

(3.3.1)

where Ω is a smooth bounded domain in Rn. By the construction of the previoussection, the precise meaning of (3.3.1) is that u = trΩv, where the functionv ∈ H1

0, L(C) with v(·, 0) = u ∈ V0(Ω) satisfies−∆v = 0 in C,v = 0 on ∂LC,∂v∂ν

= g(x) on Ω× 0.(3.3.2)

We will say that v is a weak solution of (3.3.2) and u is a weak solution of(3.3.1). Then we have the following regularity result:

Proposition 3.3.1. Let α ∈ (0, 1), Ω be a C2, α bounded domain of Rn, v ∈H1

0, L(C) be the weak solution of (3.3.2) and u = trΩv. Then,(i) If g ∈ L∞(Ω), then v ∈ W 1,q(Ω × (0, R)), for all R > 0 and 1 < q < ∞.Moreover, v ∈ Cα(C) and u ∈ Cα(Ω).(ii) If g ∈ Cα(Ω) and g|∂Ω ≡ 0, then v ∈ C1, α(C) and u ∈ C1, α(Ω).(iii) If g ∈ C1,α(Ω) and g|∂Ω ≡ 0, then v ∈ C2, α(C) and u ∈ C2, α(Ω).

As a consequence, we deduce the regularity of weak solutions for the non-linear problem:

A1/2u = f(u) in Ω,u = 0 on ∂Ω,

(3.3.3)

3.3. Regularity of solutions 85

where Ω is a smooth bounded domain in Rn. As before, the precise meaningfor (3.3.3) is that v ∈ H1

0, L(C), v(x, 0) = u, and v is a weak solution of−∆v = 0 in C,v = 0 on ∂LC,∂v∂ν

= f(v(·, 0)) on Ω× 0.(3.3.4)

Proposition 3.3.2. Let α ∈ (0, 1), Ω be a C2, α bounded domain of Rn, fbe a C1,α function such that f(0) = 0. If u is a bounded weak solution of(3.3.3), and thus v ∈ H1

0, L(C) ∩ L∞(C) is a weak solution of (3.3.4), then

u ∈ C2,α(Ω) ∩ C0(Ω). In addition, v ∈ C2,α(C).

Proof of Proposition 3.3.2. By (i) of Proposition 3.3.1 we have u ∈ Cα(Ω).Next, by (ii) of Proposition 3.3.1 and since on ∂Ω × 0, f(v) = f(0) = 0,we have u ∈ C1,α(Ω). Finally, v ∈ C2,α(C) and u ∈ C2,α(Ω) from (iii) ofProposition 3.3.1 since f is of class C1,α and f(u) = f(0) = 0 on ∂Ω. 2

Proof of Proposition 3.3.1. (i) Let v be a weak solution of (3.3.2). Weproceed with a very useful method introduced by Cabre and Sola Morales [33]by using the auxiliary function:

w(x, y) =

∫ y

0

v(x, t) dt in C. (3.3.5)

Since (∆w)y = 0, we have that ∆w is independent of y. Hence we can computeit on y = 0. On y = 0, since w ≡ 0, we have ∆w = wyy = vy. Thus w is asolution of the Dirichlet problem

−∆w(x, y) = g(x) in C,w = 0 on (Ω× 0) ∪ ∂LC.

(3.3.6)

We extend w to the whole cylinder Ω× R by odd reflection:

wod(x, y) =

w(x, y) for y ≥ 0,−w(x,−y) for y ≤ 0.

Moreover, we put

god(x, y) =

g(x) for y > 0,−g(x) for y < 0.

Then we obtain −∆wod = god in Ω× R,wod = 0 on ∂Ω× R.

(3.3.7)

Since god ∈ Lq(Ω × (−R,R)) for all R > 0 and 1 < q < ∞, the regularity forthe Dirichlet problem (3.3.7) gives wod ∈ W 2, q(Ω× (−R,R)) for all R > 0 and1 < q <∞. In particular, w ∈ C1,α(C) for all α ∈ (0, 1). Therefore, v ∈ Cα(C)and u ∈ Cα(Ω).(ii) Choose a smooth domain H such that Ω ⊂ H, and let

gH =

g in Ω,0 in H \ Ω.

We have that gH ∈ Cα(H), since g |∂Ω= 0. Consider the weak solution vH of−∆vH = 0 in H × (0,∞),vH = 0 on ∂H × (0,∞),∂vH

∂ν= gH(x) on H × 0.

Consider also the auxiliary function

wH(x, y) =

∫ y

0

vH(x, t) dt in H × (0,∞),

which solves problem (3.3.6) with Ω and g replaced by H and gH . Using bound-ary (but away from the corners ofH×[0,∞)) regularity theory for this Dirichletproblem we see that wH is C2,α(H × [0,∞)) (again here we do not claim reg-ularity at the corners of ∂H × 0) and thus wH ∈ C2,α(C) (here instead weinclude the corners ∂Ω× 0 of C).

Consider the difference ϕ = wH − w in C, where w is defined by (3.3.5). Itis clear that

−∆ϕ = 0 in C,ϕ = wH on ∂LC,ϕ = 0 on Ω× 0.

We extend ϕ to the whole cylinder Ω× R by odd reflection:

ϕod(x, y) =

ϕ(x, y) for y ≥ 0,−ϕ(x,−y) for y ≤ 0.

Moreover, we put

wH,od(x, y) =

wH(x, y) for y > 0,−wH(x,−y) for y ≤ 0.

Then we have −∆ϕod = 0 in Ω× R,ϕod = wH,od on ∂Ω× R.

3.3. Regularity of solutions 87

Since wH ∈ C2,α(C), wH ≡ 0 on ∂Ω×0, and ∂yywH = ∂yvH = −gH = −g onΩ×0 and ∂yywH = −g = 0 on ∂Ω×0, we deduce that wH,od ∈ C2,α(∂Ω×R).It follows that ϕod ∈ C2,α(Ω × R). Thus, ϕ ∈ C2,α(C), w ∈ C2,α(C) andv = ∂yw ∈ C1,α(C).(iii) Choose a smooth bounded domain B such that B ⊂ Ω. B could be thesame as H in (ii), for instance a ball, but we change its name for notationclarity. Since g ∈ C1,α(Ω) there is an extension gB ∈ C1,α(B), see [44]. Considerthe solution vB of

−∆vB = 0 in B × (0,∞),vB = 0 on ∂B × (0,∞),∂vB

∂ν= gB on B × 0.

Consider the auxiliary function

wB(x, y) =

∫ y

0

vB(x, t) dt in B × (0,∞).

As before, interior boundary regularity gives that wB ∈ C3,α(B× [0,∞)) sincegB ∈ C1,α(B) (away from the corners ∂B×0). Thus, vB ∈ C2,α(B× [0,∞)).Thus, vB ∈ C2,α(C). Consider the difference between vB and v in C, where v isa weak solution of (3.3.2). So we have that ψ = vB − v satisfies

−∆ψ = 0 in C,ψ = vB on ∂LC,∂ψ∂ν

= 0 on Ω× 0.

We extend ψ to the whole cylinder Ω× R by even reflection:

ψev(x, y) =

ψ(x, y) for y ≥ 0,ψ(x,−y) for y ≤ 0.

Moreover, we put

vB,ev(x, y) =

vB(x, y) for y > 0,vB(x,−y) for y ≤ 0.

Then, since ∂ψ∂ν

= 0 on Ω× 0, we have−∆ψev = 0 in Ω× R,ψev = vB,ev on ∂Ω× R.

Since vB ∈ C2,α(C), −∂yvB = gB = g on Ω × 0, and g|∂Ω = 0, we deducethat vB,ev ∈ C2,α(∂Ω×R). Therefore, it follows that ψev ∈ C2,α(Ω×R). Thus,ψ ∈ C2,α(C), and v ∈ C2,α(C). 2

3.4 Maximum principles

Now we give some maximum principles for A1/2.

Lemma 3.4.1. (maximum principle) Assume that u ∈ C0(Ω)∩C2(Ω) sat-isfies

A1/2u+ c(x)u ≥ 0 in Ω,u = 0 on ∂Ω,

where Ω is a smooth bounded domain in Rn and c(x) ≥ 0 in Ω. Then u ≥ 0 inΩ.

Proof. Consider the extension v = h-ext(u). If we prove that v ≥ 0 in C, thenu ≥ 0 in Ω. Suppose by contradiction that v is negative somewhere in C. Thenthe fact ∆v = 0 implies that v is negative somewhere in Ω× 0 and that theinfC v < 0 will be achieved at some point (x0, 0) ∈ Ω× 0. Thus, we have

infCv = v(x0, 0) < 0.

By Hopf’s lemma,vy(x0, 0) > 0.

It follows∂v

∂ν= −vy(x0, 0) = A1/2v(x0, 0) < 0.

Therefore, since c ≥ 0,

A1/2v(x0, 0) + c(x0)v(x0, 0) < 0.

This is a contradiction with the hypothesis A1/2u+ c(x)u ≥ 0. 2

Lemma 3.4.2. (strong maximum principle) Assume that u ∈ C0(Ω) ∩C2(Ω) satisfies

A1/2u+ c(x)u ≥ 0 in Ω,u ≥ 0 in Ω,u = 0 on ∂Ω,

where Ω is a smooth bounded domain in Rn and c ∈ L∞(Ω). Then, either u > 0in Ω, or u ≡ 0 in Ω.

Proof. The proof is similar to that of Lemma 3.4.1. Consider v = h-ext(u).We observe that v ≥ 0 in C. Suppose that v 6≡ 0 but v = 0 somewhere inΩ× [0,∞). Then there exists a minimum point x0 ∈ Ω such that v(x0, 0) = 0.

Then by Hopf’s lemma we see that ∂v(x0,0)∂y

> 0. This implies that A1/2u(x0) +

c(x0)u(x0) < 0, because of v(x0, 0) = u(x0) = 0. 2

3.4. Maximum principles 89

Lemma 3.4.3. (“Hopf’s lemma”) Assume that 0 6≡ u ∈ C0(Ω) ∩ C2(Ω)satisfies

A1/2u+ c(x)u ≥ 0 in Ω,u ≥ 0 in Ω,u = 0 on ∂Ω,

where Ω is a smooth bounded domain in Rn and c ∈ L∞(Ω). Then ∂u∂ν0

< 0 on∂Ω, where ν0 is the unit outer normal to ∂Ω.

Proof. We follows the proof given in [37].(1) We shall first prove the lemma in the case c ≡ 0. Set v = h-ext(u). Withoutloss of generality we may assume that P1 = (b1, 0, · · · , 0) ∈ ∂Ω × 0, b1 > 0and ν0 = (1, 0, · · · , 0). Hence we need to prove

∂v(P1)

∂x1

< 0.

Since Ω is smooth, we find two concentric half-balls included in the domain C,such that P1 is the only point in the closed balls belonging also to ∂LC. Chooseb2 with 0 < b2 < b1, and set P2 = (b2, 0, · · · , 0) ∈ Ω. Denote

B+1 (P2) := z = (x, y) | |z − P2| < |P1 − P2|, y > 0 ⊂ C,

B+1/2(P2) := z = (x, y) | |z − P2| < |P1 − P2|/2, y > 0,

A = B+1 (P2) \B+

1/2(P2).

Consider the function on A:

ϕ = exp(−λ|z − P2|2)− exp(−λ|P1 − P2|2)

with λ > 0 to be determined later. Note that

∆ϕ = exp(−λ|z − P2|2)n+1∑i=1

(4λ2|zi − P2|2 − 2λ).

So we can choose λ > 0 large enough such that ∆ϕ ≥ 0 in A. On the otherhand, by Lemma 3.4.2, we see that v > 0 in A \ P1. Hence, since ϕ ≡ 0 on∂B+

1 (P2) ∩ y > 0, we can take an ε > 0 such that

v − εϕ ≥ 0 on ∂A ∩ y > 0.

Since −∆(v−εϕ) ≥ 0 in A, and on ∂A∩y = 0 we have ∂y(v−εϕ) = −∂yv =A1/2u ≥ 0, by the maximum principle as in Lemma 3.4.1 we obtain

v − εϕ ≥ 0 in A.

Thus, from v − εϕ = 0 at P1 we see that ∂(v−εϕ)(P1)∂x1

≤ 0. So ∂v(P1)∂x1

≤ ε∂ϕ(P1)∂x1

=

−2λ1(b1 − b2)e−λ|P1−P2|2 < 0, and then we have the desired result.

(2) In the case c 6≡ 0, we define the function w = v exp(−βy) for some β > 0to be determined. From a direct calculation, we see that for β large enough,

−∆w − 2β∂yw = β2w ≥ 0 in Cand

−∂yw ≥ [β − c(x)]w ≥ 0 on Ω× 0,choosing β ≥ ‖c‖L∞(Ω). Then we can apply the first approach for w with ∆replaced by ∆ + 2β∂y, and then we obtain the assertion. 2

Lemma 3.4.4. (maximum principle in “small” domains) (i) Assumethat u ∈ C0(Ω) ∩ C2(Ω) satisfies

A1/2u+ c(x)u ≥ 0 in Ω,u = 0 on ∂Ω,

where Ω is a smooth bounded domain in Rn and c ∈ L∞(Ω). Then there existsδ > 0 depending only on n and ‖c−‖L∞(Ω), such that if |Ω| ≤ δ, then u ≥ 0 inΩ.(ii) Assume that Ω is a boundary (not necessary smooth) domain of Rn andc ∈ L∞(Ω). Let v ∈ C2(C) ∩ L∞(C), where C = Ω× (0,∞), satisfy

−∆v = 0 in C,v ≥ 0 on ∂LC,∂v∂ν

+ c(x)v ≥ 0 on Ω× 0.Then, there exists δ > 0 depending only on n and ‖c−‖L∞(Ω), such that if wehave |Ω ∩ v(·, 0) < 0| ≤ δ then v ≥ 0 in C.Proof. For part (i) of the theorem, consider v = h-ext(u). We see that vsatisfies the assumptions on part (ii) of the theorem. Hence, if it is enough toprove part (ii). For this, let v− = max0,−v ≥ 0. Since v− = 0 on ∂Ω×[0,∞),we see

0 =

∫Cv−∆v dxdy =

∫Ω×0

v−∂v

∂ν+

∫C|∇v−|2 dxdy.

Then, ∫C|∇v−|2 dxdy = −

∫Ω×0

v−∂v

∂νdx

≤∫

Ω×0v−cv dx =

∫Ω

−c(v−)2 dx

≤∫

Ω∩v−(·,0)>0c−(v−(·, 0))2 dx

≤|Ω ∩ v−(·, 0) > 0|1/n‖c−‖L∞(Ω)‖v−(·, 0)‖2L2n/(n−1)(Ω).

3.5. The subcritical case and a Pohozaev type identity 91

Thus, extending v− by 0 outside C we obtain an H1(Rn+1+ ) function and thus

we have

0 < S0 ≤

∫Rn+1

+|∇v−|2 dxdy

‖v−(·, 0)‖2L2n/(n−1)(Rn)

=

∫C |∇v

−|2 dxdy‖v−(·, 0)‖2

L2n/(n−1)(Ω)

≤|Ω ∩ v−(·, 0) > 0|1/n‖c−‖L∞(Ω),

where S0 is the best constant of the Sobolev trace inequality in Rn+1+ . If |Ω ∩

v−(·, 0) > 0| is small enough, we arrive at a contradiction. 2

3.5 Nonlinear problem in the subcritical case

and a Pohozaev identity

In this section, we study the nonlinear problem (3.1.1) in the subcritical case.Namely we look for a function v(x, y), for x ∈ Ω and y ∈ R+, satisfying theproblem

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= vp on Ω× 0,v > 0 in C,

(3.5.1)

where ν is the unit outer normal to Ω × 0 and 1 < p < 2] − 1 if n ≥ 2or 1 < p < ∞ if n = 1. If v is a solution of (3.5.1), then v(x, 0) = u(x) is asolution of problem (3.1.1) with the nonlinearity f(u) = up.

In order to find a solution for problem (3.5.1) whenever 1 < p < 2] − 1 ifn ≥ 2 or 1 < p < ∞ if n = 1, as stated in Theorem 3.1.1 (i) which we willprove in this section, we consider the following minimization problem:

I0 = inf

∫C|∇v(x, y)|2dxdy | v ∈ H1

0, L(C),

∫Ω

|v(x, 0)|p+1dx = 1

.

We show that I0 is achieved.

Proposition 3.5.1. Assume that 1 < p < 2] − 1 if n ≥ 2 or 1 < p < ∞ ifn = 1. Then I0 is achieved in H1

0, L(C) by a nonnegative function v.

Proof. First, there is a function v ∈ H10, L(C) such that∫

C|∇v(x, y)|2dxdy <∞ and

∫Ω

|v(x, 0)|p+1dx = 1.

In fact, it suffices to take any C∞ function with compact support in Ω× [0,∞)and not identically zero on Ω×0, and multiply it by an appropriate constant.Next we complete the proof by weak lower semi-continuity of the Dirichletintegral and by the compact embedding property in Proposition 3.2.5. Finally,|v| ≥ 0 is a nonnegative minimizer if v is a minimizer. 2

Now we give an L∞-estimate by the technique of bootstrap for subcriticalor critical nonlinear problems. Let g0 be a Caratheodory function in Ω × Rsatisfying the growth condition

|g0(·, s)| ≤ C(1 + |s|p), (3.5.2)

where Ω is a smooth domain in Rn, 1 ≤ p ≤ n+1n−1

if n ≥ 2, or 1 ≤ p < ∞ ifn = 1. We consider the following problem

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= g0(·, v) on Ω× 0.(3.5.3)

Proposition 3.5.2. Let v ∈ H10, L(C) be a weak solution of (3.5.3) and assume

the growth condition (3.5.2) for g0, where g0 satisfies condition (3.5.2) and1 ≤ p ≤ n+1

n−1if n ≥ 2, or 1 ≤ p <∞ if n = 1. Then v(·, 0) ∈ L∞(Ω).

Proof. The proof is essentially due to Brezis-Kato. First of all, let us rewritethe condition on g0 as

|g0(x, v)| ≤ a(x)(1 + |v(x, 0)|)

with a function

a(x) :=|g0(x, v(x, 0))|1 + |v(x, 0)|

which satisfies0 ≤ a ≤ C(1 + |v(·, 0)|p−1) ∈ Ln(Ω)

since p− 1 ≤ 2n−1

and since v ∈ H10, L(C), v(·, 0) ∈ L

2nn−1 (Ω).

DenoteB+r = (x, y) | |(x, y)| < r and y > 0.

For β ≥ 0 and T > 1, let ϕβ, T = vv2βT ∈ H1

0, L(C) with vT = min|v|, T.Denote

DT = (x, y) ∈ C | |v| ≤ T.By direct computation, we see∫

C|∇(vvβT )|2 dxdy =

∫Cv2βT |∇v|

2 dxdy

+

∫DT

(2β + β2)|v|2β|∇v|2 dxdy.


Multiplying (3.5.3) by ϕβ, T and integrating by parts, we obtain∫Cv2βT |∇v|

2 dxdy + 2β

∫DT

|v|2β|∇v|2 dxdy

=

∫C∇v∇(vv2β

T ) dxdy =

∫Ω×0

g0(x, v)vv2βT dx

≤∫

Ω×0a(x)(1 + |v|2)v2β

T dx.

Combining these facts, we have∫C|∇(vvβT )|2 dxdy ≤C(β + 1)

∫Ω×0

a(x)(1 + |v|2)v2βT dx,

where C denotes different constants independent of T and of β. By Lemma3.2.4, we deduce( ∫

Ω×0|vvβT |

2]

dx)2/2]

≤C(β + 1)

∫Ω×0

a(x)(1 + |v|2)v2βT dx. (3.5.4)

Assume that |v(·, 0)|β+1 ∈ L2(Ω) for some β ≥ 0. Then∫

Ω×0 |v|2v2βT dx is

bounded uniformly in T and also∫

Ω×0 v2βT dx. In what follows, let C denote

constants independent of T–but that depend on β and ‖v(·, 0)β+1‖L2(Ω). GivenM0 > 0, we have∫

Ω×0a|v|2v2β

T dx ≤M0

∫Ω×0

|v|2v2βT dx+

∫a≥M0

a|v|2v2βT dx

≤ CM0 +( ∫

a≥M0an dx

)1/n( ∫Ω×0

|vvβT |2]

dx)2/2]

≤ CM0 + ε(M0)( ∫

Ω×0|vvβT |

2]

dx)2/2]

,

where ε(M0) = (∫a≥M0 a

n dx)1/n → 0, as M0 → ∞. Note that we can deal

with∫

Ω×0 v2βT dx in the analogue procedure. Therefore, we deduce from the

last inequalities and (3.5.4), taking Cε(M0) = 12, that( ∫

Ω×0|vvβT |

2]

dx)2/2]

≤ C(1 +M0). (3.5.5)

Thus letting T → ∞, since we said that C is independent of T , we ob-tain |v|β+1(·, 0) ∈ L2]

(Ω). This conclusion followed simply from assuming|v(·, 0)|β+1 ∈ L2.

Hence, by iterating β0 = 0, βi + 1 = (βi−1 + 1) nn−1

if i ≥ 1 in (3.5.5), weconclude that v(·, 0) ∈ Lq(Ω) for all q < ∞. Finally, the proof of part (i) inProposition 3.3.1 –which only use g ∈ Lq(Ω) for all q < ∞ and g ∈ L∞(Ω)–used with g(x) = g0(x, v(x, 0))), which satisfies |g| ≤ C(1 + |v(·, 0)|p) ∈ Lq(Ω)for all q <∞, leads to v(·, 0) ∈ Cα(Ω) ⊂ L∞(Ω). 2

Next we prove a Pohozaev type formula for the problem−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= f(v) on Ω× 0.(3.5.6)

Lemma 3.5.3. Assume that f is a C1 function with primitive F (s) =∫ s

0f(t) dt

and that v is a weak solution of (3.5.6) in H10, L(C) ∩ L∞(C). Then v satisfies

a Pohozaev type identity:

1

2

∫∂LC

|∇v|2(x, ν) dσ −∫

Ω×0nF (v) dx+

n− 1

2

∫Ω×0

vf(v) dx = 0. (3.5.7)

Proof. Let v be a weak bounded solution of (3.5.6) and z = (x, y). Then weknow by Proposition 3.3.2 that v ∈ C2(C). The following identity is known:

div

(z,∇v)∇v − z

|∇v|2

2

+ (

n+ 1

2− 1)|∇v|2 = (z,∇v)∆v.

Thus, by (3.5.6), we know that in C,

div

(z,∇v)∇v − z

|∇v|2

2

+ (

n+ 1

2− 1)|∇v|2 = 0.

Integrating the above equation over Ω× (0, R), by the divergence theorem andusing that v ≡ 0 on ∂Ω× [0,∞), we see

1

2

∫∂Ω×(0,R)

|∇v|2(x, ν) dσ +

∫Ω×y=0

(x,∇xv)(∇v, ν) dx

+ (n+ 1

2− 1)

∫Ω×(0,R)

|∇v|2 dxdy

+

∫Ω×y=R

((x,∇xv) +R∂yv)∂yv −R|∇v|2

2 dx = 0.

(3.5.8)

We have ∫Ω×0

(x,∇xv)(∇v, ν) dx =

∫Ω×0

(x,∇xv)f(v) dx

=

∫Ω×0

(x,∇xF (v)) dx = −∫

Ω×0nF (v) dx,


since F (0) = 0 and v ≡ 0 on ∂Ω × 0. Next we claim that there exists asequence Rm →∞ such that

limm→∞

∫Ω×y=Rm

((x,∇xv) +Rm∂yv)∂yv −Rm|∇v|2

2 dx = 0.

To prove this, note first that

|∫

Ω×y=R((x,∇xv) +R∂yv)∂yv −R

|∇v|2

2 dx|

≤C(diam(Ω) + 1)

∫Ω×y=R

R|∇v|2 dx.

If

lim infR→∞

∫Ω×y=R

R|∇v|2 dx = c0 > 0,

then there exists R0 such that, for all R1 ≥ R0,∫ R1

R0

∫Ω×y=R

|∇v|2 dR ≥ c02

∫ R1

R0

1

RdR =

c02

logR1

R0

.

Letting R1 ↑ ∞, this contradicts that v ∈ H10, L(C). The only remanding term in

our equality (3.5.8) is n−12

∫Ω×(0,R)

|∇v|2 dxdy. Integrating by parts, the integral

is equal to∫

Ω×0 vf(v) dx +∫

Ω×R v∂yv dx. The second integral goes to zero

for a sequence Rm by the previous argument and since v ∈ L∞.Thus, we obtain by taking R = Rm →∞

1

2

∫∂LC

|∇v|2(x, ν) dσ −∫

Ω×0nF (v) dx+

n− 1

2

∫Ω×0

vf(v) dx = 0.

2

Proposition 3.5.4. Suppose that f(v) = |v|p−1v in (3.5.6). If p ≥ 2] − 1 andΩ is star-shaped with respect to some point in Rn, then (3.5.1) has no weakbounded solution v ∈ H1

0, L(C).

Proof. Since f(v) = |v|p−1v has primitive F (v) = 1p+1

|v|p+1, from Lemma3.5.3, we have

1

2

∫∂LC

|∇v|2(x, ν) dσ = (n

p+ 1− n− 1

2)

∫Ω×0

|v|p+1 dx.

If p ≥ n+1n−1

, then the right hand side is nonpositive, but the left hand side ispositive if after a translation in Rn we take Ω star-shaped with respect to theorigin. This gives a contradiction. 2


Proof of Theorem 3.1.1. Regarding part (i), Proposition 3.5.1 gives the exis-tence of a weak solution to (3.5.1) after multiplying the nonnegative minimizerof I0 by a constant to take are of the Lagrange multiplier. Then, Proposition3.5.2 and Proposition 3.3.2 give that u ∈ C2,α(Ω), since f(s) = |s|p is a C1,α

function for α ∈ (0, 1) small. Finally the strong maximum principle Lemma3.4.2 gives that u > 0 in Ω. Lastly part (ii) follows from Proposition 3.5.4 . 2

3.6 Nonlinear problems in the critical case

Now, we establish the Brezis-Nirenberg type result of Theorem 3.1.2 concerningthe nonlinearity f(u) = u2]−1+µu, where µ > 0 and n ≥ 2. Namely we look forpositive solutions of problem (3.1.1). Equivalently, we consider the followingproblem:

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= v2]−1 + µv on Ω× 0,v > 0 in C.

(3.6.1)

Let

µ1 = inf∫C|∇v(x, y)|2dxdy | v ∈ H1

0, L(C),

∫Ω

|v(x, 0)|2dx = 1 .

Note that, by Proposition 3.2.13, µ1 = λ1/21 is the first eigenvalue of A1/2

and a minimizer φ1 > 0 of the above minimization problem is an H10, L(C)

harmonic function with ∂φ1

∂ν= µ1φ1 on Ω × 0. Its trace ϕ1 = φ1(·, 0) is the

first eigenfunction of A1/2 in Proposition 3.2.13.

Lemma 3.6.1. If µ1 ≤ µ, then (3.6.1) does not admit (a positive) solution.

Proof. Assume that the problem admits a positive solution v. Since

0 =

∫C[v∆φ1 − φ1∆v] dxdy =

∫Ω×0

[v∂φ1

∂ν− φ1

∂v

∂ν] dx,

we have∫Ω×0

µ1vφ1 dx =

∫Ω×0

φ1∂v

∂νdx

=

∫Ω×0

[v2]−1φ1 + µvφ1

]dx >

∫Ω×0

µvφ1 dx.

3.6. Nonlinear problems in the critical case 97

Thenµ1 > µ.

2

In order to prove Theorem 3.1.2, we need to estimate the functional:

Qµ(v) =

∫C |∇v|

2 dxdy − µ∫

Ω×0 |v|2 dx

(∫

Ω×0 |v|2] dx)2/2] .

Proposition 3.6.2. Let µ ∈ (0, µ1) and denote by S0 the best constant for theSobolev trace inequality defined by (3.2.5). Then we have

Sµ = Sµ(Ω) := infQµ(v) | v ∈ H10, L(C) < S0.

Proof. Let Uε be the minimizers for S0, which are given by expression (3.2.6).We see that ∫

Rn×0|Uε|2

]

dx =

∫Rn

εn

(ε2 + |x|2)ndx

= ωn

∫ ∞

0

rn−1

(1 + r2)ndr =: K1,

where ωn is the surface area of unit sphere in Rn. Denote

B+ρ = (x, y) | |(x, y)| < ρ, and y > 0.

Let η ∈ C∞(C), 0 ≤ η ≤ 1 and for small fixed ρ,

η(x, y) =

1 (x, y) ∈ B+

ρ/2,

0 (x, y) 6∈ B+ρ ,

|∇η| ≤ Const./ρ. (3.6.2)

We take ρ small enough so that B+ρ ⊂ C ∪ (Ω × 0). Thus the function

ηUε ∈ H10, L(C) and we will use it as test function v in the expression for Qµ

above. In what follows, O(εa) will have usual meaning, with constants thatmay depend on ρ, which is fixed.

We have ∫Ω×0

|ηUε|2]

dx =

∫Rn

εnη2](x, 0)

(ε2 + |x|2)ndx

= K1 +

∫Rn

εn(η2](x, 0)− 1)

(ε2 + |x|2)ndx

= K1 + εn∫

Rn\Bρ/2

η2](x, 0)− 1

(ε2 + |x|2)ndx

= K1 +O(εn).

Hence we see ( ∫Rn

|ηUε|2]

dx)2/2]

= K2/2]

1 +O(εn).

Let now

K2 :=

∫Rn+1

+

|∇Uε|2 dxdy.

Since Uε are minimizers for S0, we have that

K2

K2/2]

1

= S0. (3.6.3)

We see ∫C|∇(ηUε)|2 dxdy =

∫Cη2|∇Uε|2 dxdy +O(εn−1),

since in the second term all integrals are computed in B+ρ \ B+

ρ/2. Thus, by

uniform integrability of ε1−n|∇Uε|2 in Rn+1+ \B+

ρ/2, we deduce∫C|∇(ηUε)|2 dxdy =

∫Rn+1

+

|∇Uε|2 dxdy +O(εn−1)

= K2 +O(εn−1).

On the other hand, we have for all ε < ρ/2∫Ω×0

|ηUε|2 dx =

∫Ω×0

εn−1η2(x, 0)

(|x|2 + ε2)n−1dx

≥∫|x|<ρ/2

εn−1

(|x|2 + ε2)n−1dx

≥∫|x|<ε

εn−1

(2ε2)n−1dx+

∫ε<|x|<ρ/2

εn−1

(2|x|2)n−1dx

= c1ε+ c2εn−1

∫ ρ

ε

r1−n dr

=

c3ε+O(εn−1) for n ≥ 3,c4ε ln(1/ε) +O(ε) for n = 2,

where c1, c2, c3 and c4 are positive constants.We compute, using the above,

Qµ(ηUε) =

∫C |∇(ηUε)|2 dxdy − µ

∫Ω×0 |ηUε|

2 dx

(∫

Ω×0 |ηUε|2] dx)2/2] .

In the case n ≥ 3, we have, recalling (3.6.3),

Qµ(ηUε) =K2 − µc3ε+O(εn−1)

K2/2]

1 +O(εn)

=S0 − µc3K

−2/2]

1 ε+O(εn+1)

1 +O(εn).

Then

Qµ(ηUε) = S0 − µc3ε

K2/2∗

1

+O(εn−1) < S0,

if we take ε > 0 small enough.In the case n = 2, we see

Qµ(ηUε) =K2 − µc4ε ln(1/ε) +O(ε)

K2/2]

1 +O(ε2)

=S0 − µc4K

−2/2]

1 ε ln(1/ε) +O(ε)

1 +O(ε2)

=S0 − µc4ε ln(1/ε)

K2/2∗

1

+O(ε) < S0,

for ε small enough. 2

We will prove that infQµ(v) | v ∈ H10, L(C) is achieved. Recall the Brezis-

Lieb Lemma:

Lemma 3.6.3. [30] Let U be an open subset of Rn and let vm ⊂ Lq(U),2 ≤ q < ∞. Assume that (i) vm weakly converges to v in Lq(U); (ii) vm → va.e. (almost everywhere) in U , as m→∞. Then we have

limm→∞

(‖vm‖qLq(U) − ‖vm − v‖qLq(U)) = ‖v‖qLq(U).

Proposition 3.6.4. For µ ∈ (0, µ1), we have that

Sµ = Sµ(Ω) := infQµ(v) | v ∈ H10, L(C)

is achieved.

Proof. Note that first that Qµ ≥ 0 since µ ≤ µ1. Let vm ⊂ H10, L(C)

be a minimizing sequence for Sµ. Normalize it to satisfy ‖vm‖L2](Ω×0) = 1.

Replacing vm by |vm|, we may assume vm ≥ 0. Since ‖vm‖L2](Ω×0) is bounded,

the minimizing property leads to∫C |∇vm|

2 dxdy ≤ C. Then, by Lemma 3.2.5,we extract a subsequence, still denoted by vm, such that, as m→∞,

vm v weakly in H10, L(C),

vm(·, 0) → v(·, 0) strongly in Lq(Ω), 2 ≤ q < 2],

vm(·, 0) → v(·, 0) a.e. in Ω.

Developing the square and by weak convergence, we have∫C|∇(vm − v)|2 dxdy =

∫C|∇vm|2 dxdy −

∫C|∇v|2 dxdy + o(1).

On the other hand, by Lemma 3.6.3, we have

‖vm(·, 0)− v(·, 0)‖2]

L2](Ω)

= ‖vm(·, 0)‖2]

L2](Ω)

− ‖v(·, 0)‖2]

L2](Ω)

+ o(1).

Therefore, we see

Qµ(vm) =

∫C|∇(vm − v)|2 dxdy +

∫C|∇v|2 dxdy

− µ

∫Ω

|v(x, 0)|2 dx+ o(1)

≥S0‖vm − v‖2

L2](Ω×0) + Sµ‖v‖2

L2](Ω×0) + o(1)

≥S0‖vm − v‖2]

L2](Ω×0) + Sµ‖v‖2]

L2](Ω×0) + o(1)

=(S0 − Sµ)‖vm − v‖2]

L2](Ω×0) + Sµ‖vm‖2]

L2](Ω×0) + o(1).

Hence we have

Sµ ≥ (S0 − Sµ)‖vm − v‖2]

L2](Ω×0) + Sµ + o(1).

This implies, since S0 − Sµ > 0 by Proposition 3.6.2, that

vm(·, 0) → v(·, 0) in L2]

(Ω).

Hence, by lower semi-continuity, we see that v ≥ 0 is a minimizer for Qµ. 2

Proof of Theorem 3.1.2. It follows from Proposition 3.6.4. Indeed, let v ≥ 0be the minimizer for Qµ of Proposition 3.6.4. Since µ < µ1, we deduce thatSµ > 0. Now compute the first variation of Qµ, we see that a positive multipleof v a solution of (3.6.1). The C2,α(Ω) regularity of the solution follows fromProposition 3.3.2. Finally Lemma 3.6.1 gives the nonexistence of solution forµ ≥ µ1. 2

We now also give the second approach for Theorem 3.1.2, based on a carefulstudy of the compactness properties for Palais-Smale sequences of the freefunctional Iµ(v) defined as follows:

Iµ(v) =1

2

∫C|∇v|2 dxdy − µ

2

∫Ω×0

|v|2 dx− 1

2]

∫Ω×0

|v|2]

dx. (3.6.4)

Both approaches are completely equivalent. However, the second approach willbring out the peculiarities of the limiting case more clearly. We begin from thefollowing lemma which gives the energy estimates of Palais-Smale sequences.

Lemma 3.6.5. Assume that the functional Iµ(v) is defined as in (3.6.4) andthat Ω is a smooth bounded domain in Rn, n ≥ 2. Then for every µ ∈ R, everysequence vm in H1

0, L(C) such that, as m→∞,

Iµ(vm) → β <1

2nSn0 , I ′µ(vm) → 0, (3.6.5)

is relatively compact in H10, L(C).

Proof. It is easy to check that vm is bounded in H10, L(C). In fact,

1

2nSn0 + o(1)(1 + ‖vm‖) ≥ Iµ(vm)− 1

2〈I ′µ(vm), vm〉

=(1

2− 1

2])

∫Ω×0

|vm|2]

dx ≥ C1(

∫Ω×0

|vm|2 dx)2]/2.

So

‖vm‖2 = 2Iµ(vm) + µ

∫Ω×0

|vm|2 dx+2

2]

∫Ω×0

|vm|2]

dx

≤ C2 + o(1)‖vm‖,

where o(1) → 0 as m→∞. Thus it follows that vm is bounded in H10, L(C).

By Lemma 3.2.5, we can extract a subsequence, still denoted by vm, suchthat, as m→∞,

vm v weakly in H10, L(C),

vm(·, 0) → v(·, 0) strongly in Lq(Ω), 2 ≤ q < 2],

vm(·, 0) → v(·, 0) a.e. in Ω.

In particular, for every ϕ ∈ H10, L(C), we obtain that, as m→∞,

〈I ′µ(vm), ϕ〉 =

∫C∇vm∇ϕdxdy −

∫Ω×0

[µvmϕ+ |vm|2]−2vmϕ] dx

→∫C∇v∇ϕdxdy −

∫Ω×0

[µvϕ+ |v|2]−2vϕ] dx = 〈I ′µ(v), ϕ〉.


Since by hypothesis I ′(vm) → 0, we deduce 〈I ′µ(v), ϕ〉 = 0. Thus, v ∈ H10, L(C)

solves the three first equations of problem (3.6.1). We have by choosing ϕ = v,

〈I ′µ(v), v〉 =

∫C|∇v|2 dxdy −

∫Ω×0

[µv2 + |v|2]

] dx = 0.

Hence, we see

Iµ(v) = (1

2− 1

2])

∫Ω×0

|v|2]

dx =1

2n

∫Ω×0

|v|2]

dx ≥ 0.

Moreover, Lemma 3.6.3 leads to∫Ω×0

|vm|2]

dx =

∫Ω×0

|vm − v|2]

dx+

∫Ω×0

|v|2]

dx+ o(1).

Notice that, developing the square and by weak convergence,∫C|∇vm|2 dxdy =

∫C|∇vm −∇v|2 dxdy +

∫C|∇v|2 dxdy + o(1).

So we haveIµ(vm) = Iµ(v) + I0(vm − v) + o(1). (3.6.6)

Furthermore,∫Ω×0

(|vm|2]−1vm − |v|2]−2v)(vm − v) dx

=

∫Ω×0

(|vm|2] − |vm|2

]−2vmv) dx+ o(1)

=

∫Ω×0

(|vm|2] − |v|2]

) dx+ o(1) =

∫Ω×0

(|vm − v|2]

) dx+ o(1).

It gives

o(1) = 〈I ′µ(vm), vm − v〉=〈I ′µ(vm)− I ′µ(v), vm − v〉

=

∫C|∇(vm − v)|2 dxdy −

∫Ω×0

|vm − v|2]

dx+ o(1).

Then we obtain

I0(vm − v) =1

2n

∫C|∇(vm − v)|2 dxdy + o(1).

On the other hand, by (3.6.6) and since Iµ(v) ≥ 0, we see that there is alarge m0 > 0 such that, for m ≥ m0,

I0(vm − v) = Iµ(vm)− Iµ(v) + o(1)

≤Iµ(vm) + o(1) <1

2nSn0 .

Therefore, we have the following inequality

‖vm − v‖2 < Sn0 .

This implies∫C|∇(vm − v)|2 dxdy < Sn0 ≤

( ∫C |∇(vm − v)|2 dxdy

(∫

Ω×0 |vm − v|2] dx)(n−1)/n

)n.

It gives

1 > c5 ≥

∫Ω×0 |vm − v|2]

dx∫C |∇(vm − v)|2 dxdy

,

for all m ≥ m0. Then we obtain that, as m→∞,

(1− c5)‖vm − v‖2 ≤ ‖vm − v‖2(1−

∫Ω×0 |vm − v|2]

dx∫C |∇(vm − v)|2 dxdy

)

≤∫C|∇(vm − v)|2 dxdy −

∫Ω×0

|vm − v|2]

dx = o(1),

establishing that vm → v strongly in H10, L(C). 2

Before giving the second proof of Theorem 3.1.2, let us recall the MountainPass Lemma which was developed in [25].

Lemma 3.6.6. (Mountain Pass Lemma): Let E be a real Banach space withits dual space E∗ and suppose that I ∈ C1(E,R) satisfies the condition

maxI(0), I(u1) ≤ α < β ≤ inf‖u1‖=ρ

I(u),

for some α < β, ρ > 0 and u1 ∈ E with ‖u1‖ > ρ. Let c ≥ β be characterizedby

c = infγ∈Γ

max0≤τ≤1

I(γ(τ)),

where Γ = γ ∈ C([0, 1], E) | γ(0) = 0, γ(1) = u1 is the set of continuouspaths joining 0 and u1. Then, there exists a sequence um ⊂ E such that, asm→∞,

I(um) → c ≥ β and I ′(um) |E∗→ 0.

LetΣ = v ∈ H1

0, L(C) \ 0 | 〈I ′µ(v), v〉 = 0.We define critical values for the functionals as follows:

c∗ = infv∈Σ

Iµ(v),

c = infγ∈Γ

supt∈[0,1]

Iµ(γ(t)),

c∗∗ = infv∈H1

0, L(C)\0supt≥0

Iµ(tv),

where Γ := γ ∈ C([0, 1], H10, L(C)) | γ(0) = 0, Iµ(γ(1)) < 0. We have the

following relations, whose proofs are standard.

Lemma 3.6.7.c = c∗ = c∗∗.

Proof of Theorem 3.1.2. It is sufficient to prove that Iµ satisfies the condi-tion (3.6.5). Considering the functional Iµ, for every v ∈ H1

0, L(C) and t ≥ 0,we have

Iµ(tv) =A1t

2

2− A2t

2]

2],

where

A1 =

∫C|∇v|2 dxdy − µ

∫Ω×0

|v|2 dx

and

A2 =

∫Ω×0

|v|2]

dx.

We see that Iµ(tv) has its maximum at t0 = (A1

A2)1/(2]−2) = (A1

A2)(n−1)/2. Hence,

we obtain

supt≥0

Iµ(tv) = maxt≥0

Iµ(tv) = Iµ(tv)|t=t0 =1

2n

( A1

A2/2]

2

)n.

This implies

inf0 6=v∈H1

0, L(C)supt≥0

Iµ(tv) ≤1

2n

(inf

0 6=v∈H10, L(C)

Qµ(v))n

<1

2nSn0 ,

by Proposition 3.6.2. Then by using Lemma 3.6.6, 3.6.7 and 3.6.5, we obtain

c∗∗ = inf0 6=v∈H1

0, L(C)supt≥0

Iµ(tv)

3.7. Symmetry of solutions 105

is a critical value of Iµ. Finally we complete the proof of regularity of thesolution by Proposition 3.3.2. 2

Remark 3.6.8. By minimizing the functional Iµ(v) on the Nehari manifoldΣ = v ∈ H1

0, L(C) | 〈I ′µ(v), v〉 = 0, one can get another proof of Theorem3.1.2.

3.7 Symmetry of solutions

The goal of this section is to prove a symmetry result of Gidas-Ni-Nirenbergtype for positive solutions of nonlinear problems involving the operator A1/2,as stated in Theorem 3.1.3, by using the moving planes method.

Proof of Theorem 3.1.3. Let x = (x1, x′) ∈ Ω and λ > 0. Consider the sets

Σλ = (x1, x′) ∈ Ω | x1 > λ and Tλ = (x1, x

′) ∈ Ω | x1 = λ.

For x ∈ Σλ, define xλ = (2λ− x1, x′). By the hypotheses on the domain Ω we

seexλ | x ∈ Σλ ⊂ Ω.

Let uλ(x1, x′) = u(2λ − x1, x

′) = u(xλ). Consider the harmonic extensionv = h-ext(u) of u to C. Recall that v = h-ext(u) satisfies the following problem

−∆v = 0 in C = Ω× (0,∞),v = 0 on ∂LC = ∂Ω× [0,∞),∂v∂ν

= f(v) on Ω× 0,v > 0 in C.

For (x, y) ∈ Σλ × [0,∞), let us define vλ(x, y) = v(xλ, y) = v(2λ − x1, x′, y)

and wλ(x1, y) = (vλ − v)(x, y). Note that vλ satisfies−∆vλ = 0 in Σλ × (0,∞),vλ ≥ 0 on (∂Ω ∩ Σλ)× (0,∞),∂vλ

∂ν= f(vλ) on Σλ × (0,∞).

Thus, since ∂Σλ = (∂Ω∩Σλ)∪Tλ and wλ ≡ 0 on Tλ, we have that wλ satisfies−∆wλ = 0 in Σλ × (0,∞),wλ ≥ 0 on (∂Σλ)× (0,∞),∂wλ

∂ν+ cλ(x)wλ = 0 on Σλ × (0,∞)

where cλ = −f(uλ)−f(u)uλ−u

∈ L∞(Σλ).

Let λ∗ = supλ | Σλ 6= ∅ and let ε > 0 be a small number. If λ∗ > x1 >λ∗ − ε, then Σλ has small measure and we have by part (ii) of Lemma 3.4.4that

wλ ≥ 0 in Σλ × (0,∞).

Note here that Σλ is not a smooth domain but that Lemma 3.4.4 (2) doesnot require smoothness of domains. By the strong maximum principle (or anadaptation of the proof Lemma 3.4.2) we see that wλ is identically equal tozero or strictly positive. Since wλ > 0 in (∂Ω∩∂Σλ)× (0,∞), we conclude thatwλ > 0 in Σλ × (0,∞).

Let λ0 = infλ > 0 | wλ ≥ 0 in Σλ × (0,∞). We are going to prove thatλ0 = 0. Suppose that λ0 > 0 by contradiction. First, by continuity, we havewλ0 ≥ 0 in Σλ0 × (0,∞). Then as before we deduce wλ0 > 0 in Σλ0 × (0,∞).Next, let δ > 0 be a constant and K ⊂ Σλ0 be a compact set such that|Σλ0 \K| ≤ δ/2. We have wλ0(·, 0) ≥ η > 0 in K for some constant η, since Kis compact. So we obtain that wλ0−ε(·, 0) > 0 in K and that |Σλ0−ε \K| ≤ δfor ε small enough.

Now we apply again part (ii) of Lemma 3.4.4 in the domain Σλ0−ε×(0,∞) tothe function wλ0−ε. We know that wλ0−ε(·, 0) ≥ 0 in K, and hence (wλ0−ε)

− >0 ⊂ Σλ0−ε \K, which has measure at most δ, the constant of part (ii) Lemma3.4.4. We deduce that

wλ0−ε ≥ 0 in Σλ0−ε × (0,∞).

This is a contradiction to the definition of λ0. Thus, λ0 = 0.

So we have proved, letting λ ↓ 0 that

v(−x1, x′, 0) ≥ v(x1, x

′, 0) in (Ω ∩ x1 > 0)× 0

and, since wλ = 0 on Tλ,

ux1 = −1

2

∂wλ∂x1

< 0 for x1 > 0,

by Hopf’s lemma. Finally replacing x1 by −x1 we deduce the desired symmetryv(−x1, x

′, 0) = v(x1, x′, 0).

2

Proof of Corollary 3.1.4. By Proposition 3.5.2 u is bounded and then byProposition 3.3.2, we have u ∈ C2(Ω) ∩ C0(Ω). Then the conclusion followsfrom Theorem 3.1.3.

2

3.8. A priori estimates 107

3.8 A priori estimates of positive solutions

In this section we prove Theorem 3.1.5. Namely, we will get a priori estimatesof Gidas-Spruck type for weak solutions of problem

∆v = 0 in C = Ω× (0,∞) ⊂ Rn+1+ ,

∂v∂ν

= vp on Ω ⊂ Rn,

v = 0 on ∂LC = ∂Ω× (0,∞),

v > 0 in C,

(3.8.1)

where Ω ⊂ Rn(n ≥ 2) is a bounded smooth domain and 1 < p < n+1n−1

.

For this, we need two nonlinear Liouville theorems for problems involvingA1/2: one in the whole space, another in the half space. The first one is asfollows:

Proposition 3.8.1. [46][50] If 1 0 in Rn+1+ .

(3.8.2)

As usual in low dimensions (here in dimension n+ 1 = 2), a much stronger(but quite simple to prove) Liouville theorem can be proved for supersolutionsof the linear homogeneous problem. Indeed, we have the following result fromProposition 6.1 of [37]. Its proof essentially compares in an appropriate waythe solution v with log(| · |).

Proposition 3.8.2. [37] Suppose that v is a weak solution of problem−∆v ≥ 0 in R2

+,∂v∂ν≥ 0 on ∂R2

+,

v ≥ 0 in R2+.

(3.8.3)

Then v is a constant.

We need to prove a new nonlinear Liouville type results involving A1/2

in the half space. As we will see, this nonlinear Liouville theorem for A1/2

in any dimension will be reduced to the one dimensional case for A1/2 byusing the moving planes method. Then we prove that there exists no positivesolution for the nonlinear Neumann boundary problem in the quarter R2

++,which corresponds to the nonlinear Liouville theorem involving A1/2 in thehalf line.

Proposition 3.8.3. Let Rn+1++ = z = (x1, x2, · · · , xn, y) | xn > 0, y > 0.

Assume that v is a solution of problem−∆v = 0 in Rn+1

++ ,∂v∂ν

= vp on xn > 0, y = 0,v = 0 on xn = 0, y ≥ 0,v > 0 in Rn+1

++ ,

(3.8.4)

for 1 ≤ p ≤ n+1n−1

. Then v depends only on xn and y.

Proof. We shall follows the steps as in [43]. Let en = (0, · · · , 0, 1, 0) andN = n+ 1. Consider the conformal transformation

z = T (z) =z + en|z + en|2

and the Kelvin transformation w of v

w(z) = |z + en|N−2v(z) = |z|2−Nv(z).

Denote B+1/2(

en

2) = z = (x, y) | |z − 1

2en| < 1

2, y > 0, S+

1/2(en

2) :=

∂B+1/2(

en

2) ∩ y > 0, Γ0,1/2 := ∂B+

1/2(en

2) ∩ y = 0.

Note that Rn+1++ = x > 0, y > 0 goes into the half ball B+

1/2(en

2), the

boundary xn > 0, y = 0 becomes the half ball Γ0,1/2, xn = 0, y ≥ 0 goes tothe half sphere S+

1/2(en

2), and the infinity goes to z = 0. We see that w satisfies

−∆w = 0 in B+1/2(

en

2),

w = 0 on S+1/2(

en

2),

∂w(z)∂ν

= |z|p(N−2)−Nwp(z) on Γ0,1/2,

w > 0 in B+1/2(

en

2).

Since |z|p(N−2)−N is nonincreasing in the zi direction for i = 1, · · · , n−1 (in fact,in any direction orthogonal to the zn-axis), the moving planes method used asin [37] gives that w is symmetric about all the zi-axis for i = 1, · · · , n − 1.This leads to w = w(|z′|, zn, y), where z′ = (z1, · · · , zn−1) and hence v =v(|x′|, xn, y). Now, since we may perform the Kelvin’s transform with respectto any point (x′0, 0, 0) on xn = 0, y = 0 we conclude that v = v(xn, y) asclaimed. 2

Proposition 3.8.4. Assume that f is a continuous function such that f ≥ 0in [0,∞), f > 0 in (0,∞) and f(0) = 0. Let C be a positive constant. Thenthere is no bounded solution of problem:

−∆v = 0 in R2++ = x > 0, y > 0,

∂v∂ν

= f(v) on x > 0, y = 0,v = 0 on x = 0, y ≥ 0,0 < v ≤ C in R2

++.

(3.8.5)

Proof. First we claim that v(x, 0) → 0 as x → ∞. Suppose by contradictionthat there exists a sequence am → ∞ (m → ∞) such that v(am, 0) → α > 0.Let vm(x, y) := v(x+ am, y). It is clear that vm is a solution of problem (3.8.5)in Um := (x, y) | x > −am, y > 0. Moreover, vm(0, 0) = v(am, 0) → α.Therefore there exists a subsequence, still denoted by vm, such that vm → v inC2loc(R2

+), m→∞, and v is a solution of the following problem:∂xxv + ∂yyv = 0 in (x, y) | y > 0 ⊂ R2,∂v∂ν

= f(v) ≥ 0 on y = 0,0 ≤ v ≤ C in y > 0.

(3.8.6)

Notice that

v(0, 0) = α > 0.

On the other hand, by Proposition 3.8.2 we know that v ≡ Const., which isimpossible due to the nonlinear Neumann condition because of f > 0 in (0,∞)and f(v(0, 0)) = f(α) > 0. Then we conclude the claim that v(x, 0) → 0 asx→∞.

Note that we can reflect the function v with respect to x = 0, y > 0:v(x, y) = −v(−x, y) for x < 0, and obtain a bounded harmonic function v inall R2

+ = y > 0, since v ≡ 0 on x = 0, y > 0. Applying interior gradientestimates to the harmonic function v in the ball Bt(x, t) ⊂ R2

+, we obtain

|∇v(x, t)| ≤ C‖v‖∞t

≤ C

t, for all t >

1

2, x > 0.

On the other hand, by Proposition 3.3.2 since f(0) = 0, we have that |∇v| and

|D2v| are bounded in R2++ ∩ 0 ≤ y ≤ 1. Then |∇v| and |D2v| are bounded

in R2+. In addition, we can deduce

|∇v(x, t)| ≤ C

t+ 1, for all t > 0, x > 0.


We have that, by using the same argument to the partial derivatives of vinstead of v,

|D2v(x, t)| ≤ C

t2 + 1, for all t > 0, x > 0.

Moreover, we see∣∣∣ ∂∂x

|∂xv(x, t)|2 − |∂yv(x, t)|2

2

∣∣∣ ≤ C

t3 + 1.

By these facts, we know that function

Φ(x) :=

∫ ∞

0

|∂xv(x, t)|2 − |∂yv(x, t)|2

2dt

is well defined and dΦdx

is also.Using the limy→∞ |∇v(x, y)| = 0, we obtain

d

dx[Φ(x) + F (v(x, 0))] =

∫ ∞

0

[∂xxv∂xv − ∂yv∂xyv](x, t) dt+ [f(v)∂xv](x, 0)

=[∂yv∂xv + f(v)∂xv](x, 0) = 0,

thanks to the Neumann boundary condition where F (v) =∫ v

0f(s) ds. This

implies

Φ(x) + F (v(x, 0)) ≡ Const., for all x > 0.

Furthermore, using limx→∞ v(x, 0) = 0 (and thus also that limx→∞ v(x, y) = 0uniformly in compact sets in y) together with the above bounds for |∇v(x, y)|for y large, we deduce

limx→∞

Φ(x) = 0.

From all these we obtain

Φ(x) + F (v(x, 0)) ≡ 0, for x > 0.

Since v has zero value and ∂yv = 0 along the y-axis, we see by the definitionof Φ(0) that

0 = Φ(0) =1

2

∫ ∞

0

|∂xv|2(0, t) dt.

This implies that ∂xv = 0 on x = 0, y > 0, which contradicts to Hopf’slemma. Thus the contradiction means that there is no positive solution of theproblem. 2

Before proving Theorems 3.1.6 and 3.1.5, we mention that:


Remark 3.8.5. Theorem 3.1.6 is still open without the boundedness assump-tion on v.

In this respect, let us give some examples of functions in the quarter planeR2

++. There is an unbounded harmonic function in the quarter plane R2++. It

is easy to see that v = x is a solution of problem:−∆v = 0, v ≥ 0 in R2

++,∂v∂ν

= 0 on x > 0, y = 0,v = 0 on x = 0, y > 0.

This tells us that the result as in Proposition 3.8.2 (which did not requireboundedness of the solution in the half plane) but in a quarter plane is nottrue.

On the other hand, it is clear that v(x, y) = π2

arctan xy+1

satisfies ∆v = 0

and −∂yv |y=0=πx

2(1+x2)≥ 0. Hence there exists a bounded harmonic function

in the quarter plane R2++ such that−∆v = 0, v ≥ 0 in R2

++,∂v∂ν≥ 0 on x > 0, y = 0,

v = 0 on x = 0, y > 0.

Thus the nonlinear condition ∂v∂ν

= vp on y = 0 is important in Theorem3.1.6.

Proof of Theorem 3.1.6. It follows from Proposition 3.8.3 and Proposition3.8.4. 2

Proof of Theorem 3.1.5. We know by Proposition 3.5.2 and Proposition3.3.2 that all weak solutions u ∈ C2(Ω)∩C0(Ω). Assume by contradiction thatthe theorem is not true and that there is a sequence um of solutions with

Km = ‖um‖L∞(Ω) →∞.

Since vm = h-ext(um) is a positive harmonic function in C, vanishing on ∂LC,we may assume that vm attains its maximum Km at the point (xm, 0) ∈ Ω.Let Ωm = Kp−1

m (Ω− xm) and, define

vm(x, y) = v(xm +K1−pm x,K1−p

m y), x ∈ Ωm, y > 0.

Then we see that ‖vm‖L∞(Ωm×(0,∞)) ≤ 1 and−∆vm = 0, vm > 0 in Cm := Ωm × (0,∞),∂vm

∂ν= vpm on Ωm × 0,

vm = 0 on ∂Ωm × (0,∞).

(3.8.7)

Notice that vm(0, 0) = 1. Let

dm = dist(xm, ∂Ω).

Two cases may occur as m→∞: either case (a)

Kp−1m dm →∞

for a subsequence still denoted as before, or case (b):

Kp−1m dm is bounded.

If case (a) occurs, since Kp−1m Bdm(0) = BKp−1

m dm(0) ⊂ Ωm by local compact-

ness (Arzela-Ascoli) of bounded solutions (recall ‖vm‖L∞(Ωm) ≤ 1) we obtaina solution v of problem (3.8.2) in all of Rn+1

+ = Rn × (0,∞) with v(0, 0) = 1.This is a contradiction to Proposition 3.8.1.

Assume now that case (b), Kp−1m dm is bounded, occurs. Note first that by

Proposition 3.3.1 (i) and (ii), we have C1 estimates for problem (3.8.1) for vmin Ω. Since the nonlinearity |vm|p = vpm ≤ Kp

m, we deduce that ‖∇vm‖L∞(Ω) ≤CKp

m for a constant C independent of m. Now, since vm |∂Ω×0≡ 0, we getKm = vm(xm, 0) ≤ ‖∇vm‖L∞dist(xm, ∂Ω) ≤ CKp

mdm. We deduce that

0 < c ≤ Kp−1m dm

for some positive constant c. Thus, in this case (b), we may assume that, upto a subsequence,

Kp−1m dm → a ∈ (0,∞)

for some constant a > 0.We deduce that, up to a certain rotation of Rn for each index m (since

Kp−1m → ∞, dm → 0, and BKp−1

m dm(0) is tangent to ∂Ω), the domains Ωm

converge to the half space Rn+ = xn > −a. Thus, through a subsequence of

vm, we obtain a solution v of problem (3.8.4) in Rn+1++ = xn > −a, y > 0 with

v bounded by 1 and v > 0 (since vm(0, 0) = 1 for all m). This is a contradictionwith Propositions 3.8.3 and 3.8.4. 2

Remark 3.8.6. From Theorem 3.1.5 we have a priori bounds for eventualsolutions of problem (3.8.1) with 1 < p < n+1

n−1. Then by topological methods:

Leray-Schauder degree theory, we can obtain the existence of positive solutionsto problem (3.8.1) with 1 < p < n+1

n−1.

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