UNIVERSIDADE FEDERAL DE PERNAMBUCO CENTRO DE CIÊNCIAS EXATAS E DA NATUREZA PROGRAMA DE PÓS-GRADUAÇÃO EM MATEMÁTICA

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1 UNIVERSIDADE FEDERAL DE PERNAMBUCO CENTRO DE CIÊNCIAS EXATAS E DA NATUREZA PROGRAMA DE PÓS-GRADUAÇÃO EM MATEMÁTICA EUDES MENDES BARBOZA Hénon type equations with nonlinearities in the critical growth range RECIFE 2017

2 EUDES MENDES BARBOZA Hénon type equations with nonlinearities in the critical growth range Tese apresentada ao Programa de Pós-graduação em Matemática da Universidade Federal de Pernambuco como requisito parcial para obtenção do título de Doutor em Matemática. Orientador: Prof. Dr. JOÃO MARCOS BEZERRA DO Ó RECIFE 2017

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4 EUDES MENDES BARBOZA Hénon type equations with nonlinearities in the critical growth range Tese apresentada ao Programa de Pós-graduação em Matemática da Universidade Federal de Pernambuco como requisito parcial para obtenção do título de Doutor em Matemática. Aprovado em: 30/05/2017. BANCA EXAMINADORA - Prof. Doutor João Marcos Bezerra do Ó (Orientador) Universidade Federal de Pernambuco - Prof. Doutor Miguel Fidencio Loayza Lozano Universidade Federal de Pernambuco - Prof. Doutor Bruno Henrique Carvalho Ribeiro Universidade Federal da Paraíba - Prof. Doutor Pedro Ubilla López Universidad de Santiago de Chile - Prof. Doutor Ederson Moreira dos Santos Universidade de São Paulo - São Carlos

5 À família, a meus professores, alunos e amigos.

6 Agradecimentos Gostaria de agradecer ao professor Dr. João Marcos Bezerra do Ó por ter me guiado no estudo da Análise e das Equações Diferenciais. Agradeço também ao professor Dr. Bruno Ribeiro por todos os momentos de discussão que muito me ajudaram nesta fase final do Doutoramento. E ao CNPq pelo apoio financeiro desde a graduação. Muito obrigado.

7 Abstract In this work, using variational methods we have investigated the existence of solutions for some Hénon type equations, which are characterized by the presence of the weight x α in the nonlinearity with α > 0. When we are working in the radial context, this characteristic modifies the critical growth of the nonlinearities in some senses. This fact allows us to study some well-known problems under new perspectives. For this purpose, we have considered three different classes of problems with critical nonlinearity which presents the weight of Hénon. Firstly, we have studied the class of problem with a Trudinger- Moser nonlinearity in critical range in R 2. In the subcritical case, there was no difference if we have looked for weak solutions in H0( 1 ) or in H0,rad 1 (). Nevertheless, in the critical case we have needed to adapt some hypotheses when we have changed the space where we were seeking the solutions. For the second problem, we have kept working with exponential nonlinearity in R 2, but we were treating an Ambrosseti-Prodi problem for which we have searched two weak solutions. In the subcritical case, analogously to first problem, the radially symmetric solutions were obtained as the solutions in H0( 1 ), what have not happened in the critical case. Thus, again some assumptions have had to depend on the context where we were searching for the solutions. Lastly, we have studied a natural version of the second problem with the nonlinearity involving critical Sobolev growth in R N (N 3). In this last problem, we have searched the existence of solutions only in the radial critical case because the others cases were almost identical to problems with nonlinearities without the weight of Hénon. Keywords: Partial differential equations. Hénon type equations. Nonlinearity in critical growth range. Variational methods.

8 Resumo Neste trabalho, utilizando métodos variacionais investigamos a existência de soluções para algumas equações do tipo Hénon, que são caracterizadas pela presença do peso x α na não-linearidade com α > 0. Quando estamos trabalhando no contexto radial, essa característica modifica o crescimento crítico das não-linearidades em alguns sentidos. Este fato nos permite estudar problemas bem conhecidos sob novas perspectivas. Com este propósito, consideramos três classes diferentes de problemas com uma não-linearidade que apresenta o peso de Hénon. Em primeiro lugar, estudamos a classe de problema envolvendo uma não-linearidade do tipo Trudinger-Moser com imagem crítica em R 2. No caso subcrítico, não houve diferença se procuramos soluções fracas em H0( 1 ) ou em H0,rad 1 (). No entanto, no caso crítico, precisamos adaptar algumas hipóteses quando mudamos o espaço onde buscávamos as soluções. Para o segundo problema, continuamos trabalhando com uma não-linearidade exponencial em R 2, mas desta vez tratando de um problema do tipo Ambrosseti-Prodi para o qual buscamos duas soluções fracas. No caso subcrítico, analogamente ao primeiro problema, as soluções radialmente simétricas foram obtidas do mesmo modo das soluções em H0( 1 ), o que não aconteceu no caso crítico. Assim, algumas hipótese novamente tiveram que depender do contexto em que buscávamos as soluções. Por fim, estudamos uma versão natural do segundo problema com a não-linearidade envolvendo o crescimento crítico do tipo Sobolev em R N (N 3). Neste último problema, pesquisamos a existência apenas de soluções radiais no caso crítico porque os outros casos eram quase idênticos a problemas com não-linearidades sem o peso de Hénon. Palavras-chave: Equações diferenciais parciais. Equações tipo Hénon. Não-linearidades com imagem crítica. Métodos variacionais.

9 ILLUSTRATIONS LIST Figure Q := {v + sz n : v H k, v R and 0 s R}. 63 Figure Q := {v + sw n : v H k, v R and 0 s R}. 64 Figure Support of z r n. 92 Figure Support of v H r k is a subset of \ B r Figure Supports of ψ r and f r. 115 Figure Support of u r ε. 124

10 C, C 1, C 2,... Positive constants. SYMBOL LIST R N N-dimensional Euclidian space, where x = (x 1,..., x n ), x i R, i = 1,..., n. x x x 2 N. B R (x) = B(x, R) Open ball of Euclidian space R N centered at x with radius R > 0, i.e., B R (x) = {y R n ; x y < R}. B R Open ball of Euclidian space R N centered at origin. B R (x) = B(x, R) Closet ball of Euclidian space R N centered at x with radius R > 0, i.e., B R (x) = {y R n ; x y R}. B R Closet ball of Euclidian space R N centered at origin. B R (x) Circumference of R n centered at x with radius R > 0, i.e., B R (x) = {y R n ; y x = R}. u + (x) max{u(x), 0}. suppu The closure of {x in domain of u; u(x) 0}. u x i = u xi i th partial derivative of function u. u The vector ( u x 1,..., u x N ). u N i=1 2 u. x 2 i

11 u(x) = o(g(x)) u(x) = O(g(x)) u(x) lim x x0 g(x) = 0. u(x) lim x x0 g(x) C for C > 0. C c (Ω) The set of C -function with compact support. L q (Ω) {u : Ω R; u q < }. Ω L 1 loc (Ω) The set of functions which are integrables on every compact subset of its domain of definition. L (Ω) {u : Ω R; supess u < }. C 0 (Ω) The set of continuous functions in Ω. C k (Ω) The set of functions in Ω which has continuos partial derivative in Ω of order less than or equal to k. C (Ω) The set of functions in Ω which has continuos partial derivative in Ω of all orders k N. W k,p (Ω) Space of functiosn weakly k-differenciable and p-integrable. W k,p 0 The closure of C c (Ω) in W k,p (Ω). H 1 0(Ω) W 1,2 0 (Ω). H 1 0,rad (Ω) The set of radially symmetric functions which belong to H1 0(Ω).

12 H The orthogonal subspace to space H. H H Direct sum of subspaces H and H. dim E Dimention of vectorial espace E. u n u The sequence (u n ) converges to u. u n u The sequence (u n ) converges weakly to u. e t = exp(t) Exponential function. O(N) Space of rotations in R N.

13 Contents 1 Introduction 15 2 Preliminaries Sobolev spaces Elliptic Differential Partial Equation Fredholm alternative The Principle of Symmetric Criticality of Palais Spectrum of a operator Spectrum of (, H0( 1 )) Spectrum of (, H0,rad 1 ()) Critical Point Theorems Other results Hénon elliptic equations in R 2 with subcritical and critical exponential growth Hypotheses Statement of main results Outline The variational formulation The boundedness of a (P S) c sequence The (P S) c condition Proof of Theorems and 1.2.2: Subcritical Case Geometric conditions Proof of Theorems and completed Proof of Theorems and 1.2.4: Critical Case in H0( 1 )

14 1.5.1 Geometric conditions Estimate of minimax levels Proof of Theorems and completed Proof of Theorems and 1.2.6: Critical case in H0,rad 1 () The geometric conditions Estimate of minimax level Proof of Theorems and completed Hénon type equations with one-sided exponential growth Hypotheses Statement of main results Outline The Linear Problem The variational formulation Palais-Smale condition Proof of Theorem 2.2.2: The subcritical case The geometric condition Proof of Theorem completed Proof of Theorem 2.2.3: Critical case in H0( 1 ) The geometric conditions Estimate of minimax levels Proof of Theorem completed Proof of Theorem 2.2.4: Critical case in H0,rad 1 () The geometric conditions Estimate of minimax levels Proof of Theorem completed Hénon type equation with jumping nonlinearity in critical growth range Hypotheses Statement of main results Outline The Linear Problem The variational formulation

15 3.4.1 Preliminary properties Palais-Smale sequences Proof of Theorem The geometric conditions Estimates of minimax levels Proof of Theorem completed References 143

16 1 Introduction This work has as main theme the study of Hénon type Problems with nonlinearity in critical range. The Hénon equations arose in 1973, when M. Hénon [33], using the concentric shell model in order to investigate numerically the stability of spherical stellar system steady state with respect to spherical disturbances, studied the following problem u = x α u p 1, and u > 0 in ; (1) u = 0 on, where is the unit ball centered at the origin in R N (N 3), p > 1 and the power α is positive (HÉNON, 1973). Since H 1 0( ) is compactly embedded in L q ( ) for 1 q < 2 = 2N/(N 2), we can use standard arguments of Nonlinear Analysis in the search of solution for (1). However it is known that the presence of the weight x α modifies, in some sense, the global homogeneity properties of the equation. Indeed, for instance, we notice that the usual arguments based on the Pohozaev identity yield that Problem (1) has no solution if p 2 α = 2(N + α)/(n 2) (see [27, 30](FIGUEIREDO; LION, 1982; FIGUEIREDO et al, 2008)). Still on these topics, W. Ni [41] used the power-like decay away from the origin of radially symmetric functions of H 1 0( ) (Strauss Lemma [3, 54] AMBROSETTI; MALCHIODI, 2007; STRAUSS, 1977) together with the weight x α in order to prove the compactness of the embedding H 1 0,rad () in L q (, x α ) for any 1 q < 2 α. Due to this fact, he obtained via the Mountain Pass argument that (1) admits at least one radially symmetric solution for every 2 < p < 2 α (NI, 1982). Although the initial study of this kind of equation is only numerical, the paper of W. Ni motivated subsequent researches which showed that the above problem exhibits very rich features from the functional-analytic point of view, see for example [7, 8, 13, 17, 18, 19, 30, 35, 36, 37, 43, 50, 51, 53, 56](BADIALE; SERRA, 2004; 15

17 BARUELLO; SECCHI, 2008; BONHEURE et al, 2008; CALANCHI; TERRANEO, 2005; CARRIÃO et al, 2009; CHEN, 2000; FIGUEIREDO et al, 2008; LONG; YANG, 2012; LONG; YANG, 2010; LI; PENG, 2009; PENG, 2006; SECCHI; SERRA, 2006; SERRA, 2005; SMETS et al, 2002; TARSI; 2008). In particular, various questions that arise quite naturally concerning existence, multiplicity and qualitative properties of solutions have given the Hénon type equation the role of a very interesting item in nonlinear analysis and critical point theory. Considering, more specifically, the existence of solution for Hénon equations, we can emphasize that D. Smets, J. Su and M. Willem [53] proved that for any p (2, 2 ) if N 3 and for all p (2, ) if N = 2, there exists at least a non-radial solution of the Hénon problem provided that α is large enough. From this fact together with the result of W. Ni mentioned above, we can conclude that (1) admits the coexistence of radial and non-radial positive solutions for N 3 and larger α (SMETS et al, 2002). In [51], E. Serra showed that (1) also possesses a non-radial solution for N 3 and p critical, that is, p = 2 (SERRA, 2006). In this direction, considering the critical Sobolev exponent, W. Long and J. Yang [36] showed the existence of a positive solution for the following Hénon type equation involving Sobolev critical growth u = λu + x α u 2 1 in Ω; for Ω R N u = 0 on Ω, is a smooth bounded domain, N 4, λ < λ 1 and α > 0 is small enough (LONG; YANG, 2010). In [35], for the same nonlinearity they proved the identical result for λ > λ 1 and N 7 (LONG; YANG, 2012). If we focus our attention to the plane, due to the well-known Trudinger-Moser inequality, we can say, with relevant differences, that the growth e 4πt2 to the critical growth t 2 (2) in R 2 corresponds in R N for N 3. This can be made precise by introducing the class of Orlicz spaces, see [40, 57] (MOSER, 1971; TRUDINGER, 1967). This inequality can be summarized as follows sup e βu2 u H0 1() u 2 =1 < + if β 4π; dx = + if β > 4π. The embedding in the sense of the Trudinger-Moser inequality for the Sobolev space of radial functions H 1 0,rad () was considered by (CALANCHI; TERRANEO, 2005) [17] (see (3) 16

18 also [13] BONHEURE et al, 2008). In this case the behavior of the weight x α at zero produces a higher embedding than the usual case (3), precisely, < + if β 2π(2 + α); sup x α e βu2 dx u H0,rad 1 () = + if β > 2π(2 + α). u 2 =1 Based on the fact, M. Calanchi and E. Terraneo, proved the existence of a non-radial solution, analogously to [53] (SMETS et al, 2002), for the following Hénon type equation in R 2 with exponential growth nonlinearity u = x α (e p u γ 1 p u γ ) in (0) R 2 ; u = 0 on (0), where α > 0 and p > 0, 1 < γ < 2 or 0 < p < 4π, γ = 2. After this work, some attention has been given to the study of this type of equation with Trudinger-Moser nonlinearity, as we can see in [13, 56] (BONHEURE et al, 2008; TARSI, 2008). Motivated by the results above, we questioned: How can the presence of the weight x α influence a search for solutions to relevant problems of Nonlinear Analysis? Can we improve some assumptions of classical problems if we seek radial solutions when we add this weight on the nonlinearity? In order to answer some points of these questions, we studied three problems with nonlinearity in the critical range with the weight x α : the first two problems with nonlinearities in the critical Trudinger-Moser growth range and the last problem with critical Sobolev growth. For the matter, we use these well-known critical point results (the Mountain Pass Theorem or the Linking Theorem). Thus, we need to prove some geometric conditions satisfied by the functionals associated with each problems. addition, for the critical case, we have to estimate the minimax levels which have to lie below some appropriate constants. In order to present these studies, we organized this work in four chapters. Firstly, in Chapter, we present some basic facts and previous results used along the text and, in the others, we trate our problems. In Chapter 1, inspired by studies of the Hénon type equations with the Trudinger- Moser growth nonlinearity in R 2, we focus our attention on the natural two-dimensional extension for problem (2). More specifically, we study the Dirichlet problem in the unit (4) In 17

19 ball of R 2 for the Hénon type equation of the form u = λu + x α f(u) in, u = 0 on, where f(t) is a C 1 -function in the critical growth range motivated by the Trudinger-Moser inequality. In this chapter, we search for a weak solutions as critical points of functional associated to problem in two spaces: H 1 0( ) and H 1 0,rad (). In the subcritical case, our arguments in the proof of boundedness of minimax levels are the same for the both spaces. However, in the critical case, we must consider the Trudinger-Moser inequality as (3), when we work in H 1 0( ) (see [57] TRUDINGER, 1967), or as (4), when we look for a solution in H 1 0,rad () (see [13] BONHEURE, 2008). Then, in the last case, the minimax levels increase its boundedness from above. Because of this fact, to prove the existence of radially symmetric solutions, we modify in some points our arguments to work in Sobolev space H 1 0,rad (). In Chapters 2 and 3, we intend to establish a link between Ambrosetti-Prodi problems and Hénon type equations. Ambrosetti-Prodi problems have been studied, explored and extended by an enormous variety of authors. The resulting literature is nowadays rather rich (see, for instance, [24, 29, 46] FIGUEIREDO, 1980; FIGUEIREDO; YANG, 1999; RAMOS, 1988 and references therein). nonhomogeneous problems such that u = g(x, u) + f(x) in Ω, In this direction, one searches for solution of u = 0 on Ω, where g jumps eigenvalues, meaning that the limits lim s g(s)/s < lim s + g(s)/s form an interval that contains at least one eigenvalue of (, H 1 0( )). (5) This class of problems are also called problems with jumping nonlinearities in a terminology introduced by S. Fu cik (see [31] FU CIK; 1980). Talking specifically about multiplicity of solutions for this kind of problem, we note that, in almost all Ambrosetti-Prodi problems since the work of J. Kazdan and F. Warner in [34] (KAZDAN; WARNER, 1975), it is extensively considered the heavy dependence of an usual hypothesis regarding a suitable parametrization of the forcing term f. More precisely, we assume that f satisfies f(x) = h(x)tφ 1 (x), 18

20 where h L µ ( ), φ 1 is the eigenfunction associated to the first eigenvalue of the Laplacian operator in, µ > N and hφ 1 dx = 0. Then, one should have that g(x, s) C s p 1 + C with 2 < p 2. The subcritical case (p < 2 ) of (5) was studied by B. Ruf and P. N. Srikanth [49] (RUF; SRIKANTH, 1986) considered g(x, s) = s p + + λs and λ > λ 1 (see also [23] FIGUEIREDO, 1988). The first paper that addressed Ambrosetti-Prodi problem (5) involving critical nonlinearity of the type g(s) = s K(s) was the work of D. YinBin, [58](YINBIN; 1991), where K(s) is a lower perturbation of the expression with the critical exponent. Thus, we can say that the author considered a function that was superlinear both in + and and, so, asymptotically jumping all the eigenvalues of (, H 1 0(Ω)). In [29], D. Figueiredo and J. Yang dealt with one-sided critical growth, which allowed to explore more jumping possibilities. Due to natural limitations on the techniques they used, the existence of multiple solutions was investigated in dimensions N 7 only (FIGUEIREDO; YANG, 1999). In [15], M. Calanchi and B. Ruf extended their results and proved that the same problem has at least two solutions provided N 6. They also added a lower order growth term to the nonlinearity, which guaranteed the existence of solutions in lower dimensions as well (CALANCHI; RUF, 2002). If we turn our attention to problems on domais in R 2, where the critical growth is wellknown to be exponential, we find the paper of M. Calanchi et al. [16], which considered the following problem: u = λu + g(u + ) + f(x) in Ω, u = 0 on Ω, where Ω is bounded and smooth in R 2 and g has a subcritical or critical Trudinger-Moser growth. They proved the existence of two solutions for some forcing terms f, depending on the usual parametrization f(x) = h(x) + tφ 1 (x) (CALANCHI et al, 2004). (6) This result was a natural extension to the ones found in [15, 29] (CALANCHI; RUF, 2002; FIGUEIREDO; YANG, 1999). 19

21 In Chapter 2, we consider a similar problem to (6) with the weight x α nonlinearity. Thus, we deal with the following class of problems u = λu + x α g(u + ) + f(x) in, u = 0 on, in the (7) where is the unit ball in R 2, g is a C 1 function in [0, + ) which is assumed to be in the subcritical or critical growth range of Trudinger-Moser type and f L µ ( ) for some µ > 2. Under suitable hypotheses on the constant λ, we prove existence of at least two solutions for this problem using variational methods. If we consider that f is radially symmetric, the two solutions will be radially symmetric as well. We remark that existence of a first solution for Ambrosseti-Prodi problems does not require any growth of the nonlinearity. Only when we find out a second solution, we need to consider the growth of the nonlinearity at infinite. For this, we get a first solution ψ using the Fredholm Alternative in a related linear equation. A second solution is obtained via the Mountain Pass Theorem, if λ < λ 1, or the Linking Theorem, if λ > λ 1. In Chapter 3, we studied a natural extension for problem (7) for N 3. But we concentrated on the radial critical case, once the critical case in H 1 0( ) is very near to the result of M. Calanch and B.Ruf in [15](CALANCHI; RUF, 2002). Thus, we search the existence of radially symmetric solutions for problems with jumping nonlinearities in the presence of a typical weight of a Hénon equation. More specifically, our goal is to study the following class of Hénon type problems u = λu + x α k(u + ) + f(x) in, u = 0 on, where is the unit ball in R N, k(t) is a C 1 function in [0, + ) which is assumed to be in the critical level with subcritical perturbation, f is radially symmetric and belongs to L µ ( ) for suitable µ depending on N 3. Under appropriate hypotheses on the constant λ, we prove the existence of at least two radial solutions. Analogously to Chapter 3, we get a first solution ψ using the Fredholm Alternative in a related linear equation. A second solution is obtained via Critical Points Theorems, consequently we need to show the geometric condition and estimate the minimax level. For this matter, we consider k(t) = t 2 α +g(t), where g(t) is an appropriated subcritical perturbation, which can vanish in high dimensions. However, the assumptions on g(t) are crucial in order to obtain the 20

22 estimate of minimax level in lower dimensions. Because of this, we add this lower order growth term to the nonlinearity for all N 3 to guarantee a second radial solution of this last problem. Throughout this work, we realize that a nonlinearity with this new characteristic, the weight x α, allows us to deal with a well-studied problem under new perspectives. For instance, if we work with a critical growth nonlinearity, we need to assume suitable hypotheses when we are seeking radial solutions for the three problems. Due to this fact, we must adapt some technique to obtain radial versions for classic problems with Hénon weight. It is important we emphasize that there are two technical differences between Chapter 1 and Chapters 2 and 3, when we search for a solution in the critical cases using Critical Points Theorems. In Chapter 2, we show that the associated functional of the problem satisfies the (P S) c condition for levels c in an interval bounded from above for a suitable constant. In Chapters 2 and 3, we only prove the boundedness of the minimax levels by appropriate constants, consequently, we also need to show that the solution obtained this away is not trivial. In these last chapters, we use a technique based on the paper of B. Ruf and F. Gazzola [32] (RUF; GAZZOLA, 1997), which is not used in Chapter 1. Precisely, in Chapters 2 and 3 when we work in Linking case, we separate the supports of the functions used to prove the estimates of the minimax levels (functions of Moser and functions of Talenti, respectly) and the supports of functions spanned by the first k th - eigenfunctions. Thus, we make some estimates easier to handle, due to lack of interference between these modified functions. 21

23 2 Preliminaries This chapter is devoted to present some important results used along the text. Throughout this chapter, Ω is an open set in R N. We start introducing the notions of Sobolev spaces. 0.1 Sobolev spaces Before we define Sobolev space, we need to know what is a weak derivative. Definition Assume u, v L 1 loc (Ω) and σ = (σ 1,..., σ N ) is a multiindex of order σ = σ σ k. We say that v is the σ th -weak partial derivative of u, written D σ u = v provided for all test functions φ C c ud σ φdx = ( 1) σ vφdx Ω Ω (Ω). Definition Fix 1 p and let k a nonnegative integer. The Sobolev space W k,p (Ω) consists of all locally summable functions u : Ω R such that for each multiindex σ with σ k, D σ u exists in the weak sense and belongs to L p (Ω). Definition If u W k,p (Ω), we define its norm to be u W k,p (Ω) := σ k σ k Ω D σ u p dx 1/p if 1 p < ; ess sup D σ u if p =. Ω 22

24 Preliminaries 23 Example If p = 2, we usually write H k (Ω) = W k,2 (Ω) (k = 0, 1,...). We can prove that H k (Ω) with the norm above is a Hilbert space. We still notice that H 0 (Ω) = L 2 (Ω). Definition We denote by the closure of Cc (Ω) in W k,p (Ω). It is customary to write W k,p 0 (Ω) H k 0 (Ω) = W k,2 0 (Ω). 0.2 Elliptic Differential Partial Equation Initially, we consider the following problem Lu = f in Ω u = 0 on Ω, (8) where u : Ω R is the unknown, u = u(x), f : U R is given, and L denotes a second-order partial differential operator having the form n n Lu = (a ij (x)u xi ) xj + b i (x)u xi + c(x)u i,j=1 i=1 for given coefficient functions a ij = a ji, b j, c(i, j = 1,..., n). Definition The partial differential operator L is elliptic if there exists a constant ϑ > 0 such that n a ij (x)ξ i ξ j ϑ ξ 2 i,j=1 for almost every x Ω and all ξ R N. Thus, we can say that ellipticity means that for each point x Ω, the symmetric N N matrix (a ij ) is positive definite, with smallest eigenvalue greater than or equal to ϑ. Example Let Ω R N and a C 2 (Ω)-function. The Laplace operator of u, denoted by u is given by n u = u xi x i = divdu. j=1 The Laplace operator is a second-order elliptic operator.

25 Preliminaries 24 Definition The bilinear form B[, ] associated with the elliptic operator L is N N B[u, v] := a ij u xi v xj + b j u xi v + cuvdx Ω i,j=1 i=1 for u, v H0(Ω). 1 Definition We say that u H0(Ω) 1 is a weak solution of the boundary-value problem (8) if B[u, v] = (f, v) for all v H0(Ω), 1 where (, ) denotes the inner product in L 2 (Ω). This identity is called thevariational form of problem (8). 0.3 Fredholm alternative The Fredholm alternative will be used with more emphasis in last two chapters, but this important result is applied to the Laplace Operator. For this purpose we present its short version as following. Theorem (Fredholm alternative). Precisely one of the following statements holds: either (a) for each f L 2 (Ω) there exists a unique weak solution u of the problem Lu = f in Ω u = 0 on Ω, or else (b) there exists a weak solution u 0 of the homogeneous problem Lu = 0 in Ω u = 0 on Ω. See [9, 21]. 0.4 The Principle of Symmetric Criticality of Palais Another important result which will be used in last chapters is the Principle of Symmetric Criticality of Palais. This celebrated result is also used to study some characteristics of eigenfunctions when we are working in radial contexts.

26 Preliminaries 25 Definition We consider an action of a group G on a normed vectorial space H. (i) The invariant space under G is the closet subspace of H defined by Fix(G) = {u H; g(u) = u for all g G}. (ii) A set A H is called G-invariant if g(a) = A for all g G. (iii) A functional J : H R is called G-invariant if J g = J for all g G. (iv) A function f : H H is called G-invariant if f g = g f for all g G. Theorem (The Principle of Symmetric Criticality of Palais). Consider an isometric action of a group G under a Hilbert Space (H,, ). If the functional J : H R is G- invariant and u is a critical point of J on Fix(G), then u is a critical point of J in H. See [42]. 0.5 Spectrum of a operator This section is based on results and definitions that can be found in [9, 21]. We will use them some times throughout this work. Theorem (i) There exists an at most countable set R such that the problem Lu = λu + f in Ω (9) u = 0 on Ω has a unique weak solution for each f L 2 (Ω) if and only if λ. (ii) If is infinite, then = {λ k } k=1, the values of a nondecreasing sequence with λ k +. Definition We call the spectrum of the operator L. We notice that Problem (9) with f 0 has a nontrivial solution φ if and only if λ, in which case λ is called eigenvalue of L and φ a corresponding eigenfunction. From now on, we will focus our attention to (, ), where is the unity ball of R N for N 2. This is because we search solutions for problems with this specific operator.

27 Preliminaries Spectrum of (, H 1 0( )) We denote by 0 < λ 1 < λ 2 λ 3... the eigenvalues of (, H 1 0( )). The next result gives us a basis for H 1 0( ) associated to these eigenvalues. Theorem There exists an orthonormal basis {φ k } k=1 of L2 ( ) and H 1 0( ), where φ k is an eigenfunctions corresponding to λ k : φ k = λ k φ k in ; for k = 1, 2,... φ k = 0 on Since is smooth, from regularity theory, we can conclude that φ k C ( ) for k = 1, 2,... Definition We call λ 1 > 0 the principal eigenvalue of (, H 1 0( )). Theorem (Variational principle of the principal eigenvalue). We have λ 1 = inf u 2 u 2 2 =1 2 for all u H0( 1 ) \ {0}. The number λ 1 is attained for a function φ 1, which is radially symmetric and positive within. We can write this characterization as λ 1 = inf u H 1 0 ()\{0} u 2 2. u 2 2 Definition If k 2 we say that λ k is a high order eigenvalue. We define H k as the finite dimensional subspace of H 1 0( ) spanned by φ 1, φ 2,...φ k. Notice that we can write H 1 0( ) = H k H k = H j, which are called orthonormal decompositions of H 1 0( ). j=1 by Similar to λ 1, we have a variational characterization to λ k and λ k+1, respectively, given λ k = u 2 2 sup u H k \{0} u 2 2 u 2 2 and λ k+1 = inf. u Hk \{0} u 2 2

28 Preliminaries Spectrum of (, H 1 0,rad ()) Since we also seek radially symmetric solutions for our problems, in these moments, we need to consider a suitable subspace of H0( 1 ). More specifically, some times we work in H0,rad( 1 ) := {u H0( 1 ); u(sx) = u(x) for all S O(N)}, where O(N) is space of rotations in R N. First, we observe that H0,rad 1 () is an infinite dimensional subspace of H0( 1 ). Indeed, let us take {e n x 2 e n with n N} H0,rad 1 (), which is linearly independent. Furthermore, H0,rad 1 () is a Hilbert space with the norm of H0( 1 ). And analogously to H0( 1 ), we can write H0,rad 1 () as an infinite union of finite dimensional orthogonal subspaces. From the Principle of Symmetric Criticality of Palais, we can see that φ 1 is radially symmetric. So we have H 1 = H 1 H0,rad 1 (). Now, we set Hk = H k H0,rad( 1 ) for all k N and we have H 1 0,rad( ) = Hk. (10) Moreover, it is straightforward to prove that the spectrum of (, H 1 0,rad ()) is a k=1 subsequence of (λ k ) that we will denote by λ 1 = λ 1 < λ 2 λ 3... λ k... where λ j λ j for all j = 1, 2, 3... This fact guarantees that the variational characterizations of eigenvalues in H 1 0,rad () do not change. 0.6 Critical Point Theorems In this section, we introduce the theorems which are used to guarantee our existence results. We need the following concept. Definition We consider real C 1 functional Φ defined on a Banach space E. When looking for critical points of Φ it has become standard to assume the following compactness condition (P S) c : any sequence (u j ) in E such that Φ(u j ) c and Φ (u j ) 0 in E (11) has a convergence subsequence. A sequence that satisfies (11) is called a (P S) sequence (at level c).

29 Preliminaries 28 We shall use two well-known critical-point theorems, namely, the Mountain Pass and the Linking Theorems, to λ < λ 1 and λ j < λ < λ j+1, respectively. Let us state them here, for the sake of completeness. Initially, we have the Mountain Pass and the Linking Theorems in the versions which will be used in the Chapter 1. For the proofs we refer the reader to [5, 12, 44, 52]. Theorem A (Mountain Pass). Let Φ be a C 1 functional on a Banach space E satisfying (Φ 0 ) There exist ϱ > 0 such that Φ satisfies (P S) condition for levels in the interval (0, ϱ). (Φ 1 ) There exist constants ρ, δ > 0 such that Φ(u) δ for all u E such that u = ρ, (Φ 2 ) Φ(0) < δ and Φ(v) < δ for some v E \ {0} and v ρ. Considere Γ := {η C([0, 1], E) : η(0) = 0 and η(1) = v} and set c = inf max Φ(η(t)). δ. η Γ t [0,1] Then c (0, ϱ) and it is a critical value of Φ. Moreover, if f(t) is a odd function, one has many infinitely solutions. Theorem B (Linking Theorem). Let Φ be a C 1 functional on a Banach space E = E 1 E 2 such that dime 1 is finite. If Φ satisfies (Φ 0 ) and (Φ 3 ) There exist constants δ, ρ > 0 such that Φ(u) δ for all u E 2 such that u = ρ, (Φ 4 ) There exist an z E 1 with z = 1 and R > ρ such that if Q := {v + sz : v E 1, v R and 0 s R}, then Φ(u) 0 for all u Q. Consider Γ := {η C( Q, E) : η(u) = u if u Q} and set c = inf max η Γ u Q Φ(η(u)) δ. Then c (0, ϱ) and it is a critical value of Φ. Moreover, if f(t) is a odd function, one has many infinitely solutions.

30 Preliminaries 29 In Chapters 3 and 4, we also shall use these well-known critical-point theorems, but now both without the (P S) condition. For the proofs we refer the reader to [11, 12, 25, 39, 44]. Theorem C (Mountain-Pass without the (P S) condition). Let Φ be a C 1 functional on a Banach space E and satisfying (Φ1 ) There exist constants ρ, δ > 0 such that Φ(u) δ if u E and u = ρ, (Φ2 ) Φ(0) < δ and Φ(v) < δ for some v E if u > ρ. Consider Γ := {η C([0, 1], E) : η(0) = 0 and η(1) = v} and set c = inf max Φ(η(t)) δ. η Γ t [0,1] Then there exists a sequence (u j ) in E satisfying (11). Moreover, if Φ satisfies (P S) c, then c is a critical value for Φ. Theorem D (Linking Theorem without the (P S) condition). Let Φ be a C 1 functional on a Banach space E = E 1 E 2 such that dime 1 is finite and satisfying (Φ3 ) There exist constants δ, ρ > 0 such that Φ(u) δ if u E 2 and u = ρ, (Φ4 ) There exists z E 2 with z = 1 and there exists R > ρ such that Φ(u) 0 for all u Q, where Q := {v + sz : v E 1, v R and 0 s R}. Consider Γ := {η C( Q, E) : η(u) = u if u Q} and set c = inf max η Γ u Q Φ(η(u)) δ. Then there exists a sequence (u j ) in E satisfying (11). Moreover, if Φ satisfies (P S) c, then c is a critical value for Φ.

31 Preliminaries Other results Here, we will introduce results used in this work, which are proved in our references. Lemma Let (u n ) be a sequence of functions in L 1 (Ω) converging to u in L 1 (Ω). Assume that f(x, u n (x)) and f(x, u(x)) are also L 1 functions. If f(x, u n (x))u n (x) dx C 1 then f(x, u n ) converges in L 1 to f(x, u). See [28, Lemma 2.1]. Ω Now we consider the so-called Moser functions (log n) 1/2, 0 x 1 ; n z n (x) = 1 2π log 1 x (log n) 1/2 1 n x 1; 0 x > 1. (12) It is known that z n H 1 0( ), where R 2 and z n = 1 for all n N and z n 2 = O(1/ log n) 1/2, for details see [40]. Now, we have the following lemma, which gives us some estimates used when we work with the orthogonal projection on (H k ). Lemma Let T k : H 1 0( ) (H k ) be the orthogonal projection. We consider (12) and define The following estimates hold where A k, B k > 0 are such that w n (x) = T k z n (x). (13) 1 A k log n w n 2 1. (14) B k (log n) for all x ; 1/2 w n (x) (15) (log n) 1/2 2π B k (log n) 1/2 for all x /n, u A k u 2 and u (B k /B) u 2 for all u H k, and B satisfies z n 2 B/(log n) 1/2 for all n N.

32 Preliminaries 31 See [48, Lemma 2]. Theorem (Dominated convergence theorem). Let {f n } be a sequence of measurable functions on a measure space (Ω, Σ, µ). Suppose that the sequence converges pointwise to a function f and is dominated by some integrable function g in the sense that f n (x) g(x) for all numbers n in the index set of the sequence and all points x Ω. integrable and See [22]. lim f n dµ = f Ω Ω dµ. Then f is Theorem Let (u n ) W 1,N 0 satisfy: u n N 1. Without loss of generality we may assume that u n u, u n 2 ν weakly. Then either ν = δ x0 (the Dirac measure of mass 1 concentrated at some x 0 Ω ) for some x 0 Ω and u n 0, e β N u n N/N 1 cδ x0 for some c 0, or there exists β > 0 such that e β N u n N/N 1 is bounded in L 1 (Ω) and thus In particular this is the case if u 0. See [38, Theorem I.6]. e β N u n N/N 1 e β N u N/N 1 in L 1 (Ω). Now we consider the following sequence s y n (s) = n (1 δ n) 1/2, 0 s n; 1/2 ( ) 1 An + 1 log + (n(1 δ (n(1 δ n )) 1/2 A n + e (s n) n ) 1/2, n s, where δ n = 2 log n/n and A n = 1/(en) + O(1/n 4 ). (16) Theorem Let F (t) be a C 1 (R)-function with Trudinger-Moser critical growth, then lim F (u n ) dx [0, e N 1 B1 (0) ] (0) for any normalized concentrating sequence {u n }. In particular, there exists an explicit normalized concentrating sequence {y n } with lim F (y n ) dx = e N 1 B1 (0). See [26, Theorem 1.4]. (0)

33 Chapter 1 Hénon elliptic equations in R 2 with subcritical and critical exponential growth In this chapter, using variational methods, we prove the existence of a nontrivial solution for the following class of Hénon type equations u = λu + x α f(u) in, u = 0 on, (P1) where λ > 0, α 0 (for Hénon type equations we should have α > 0, but in this work we also consider α as possibly equal to 0) and is a unity ball centred at origin of R 2, which is a natural two dimensional extension of (2). For this purpose, we assume that f(t) has the maximum growth which allows us to study (P1) variationally in Sobolev Spaces, motivated by the well-known Trudinger- Moser inequality, given by (3), when we seek a solution in H 1 0( ), and by (4), if we search a radial solution. Therefore, from (3)-(4), we have naturally associated notions of subcriticality and criticality: we say that (SG) f(t) has subcritical growth at + if for all β one has f(t) lim = 0; t + e βt2 (CG) f(t) has critical growth at + if there exists β 0 > 0 such that f(t) f(t) lim = 0, β > β 0 ; lim = +, β < β 0. t + e βt2 t + e βt2 32

34 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 33 Similarly we define subcritical and critical growth at. In this chapter we will consider problems where the nonlinearities have critical or subcritical growth. 1.1 Hypotheses Before stating our main results, we shall introduce the following assumptions on the nonlinearity f(t): (H 1 ) The function f(t) is continuous and f(0) = 0. (H 2 ) There exist t 0 and M > 0 such that 0 < F (t) =: t (H 3 ) 0 < 2F (t) f(t)t for all t R \ {0}. 0 f(s) ds M f(t) for all t > t 0. Remark (a) We can see that from conditions (H 1 ) and (H 2 ), it follows that there are constants C, t 0 > 0 such that F (t) > Ce t /M for all t > t 0. (1.1) Hence we can see that F (t) and f(t) have exponential growth for t large enough. (b) Also by (H 1 ) and (H 2 ), we notice that f(t) satisfies the well-known Ambrosetti Rabinowitz condition, in exact terms, there exist ϑ > 2 and t ϑ > 0 such that ϑf (t) f(t)t for all t t ϑ. (1.2) The Ambrosetti Rabinowitz condition has been used in most of the studies for superlinear problems and plays an important role in studying the existence of nontrivial solutions of many nonlinear elliptic boundary value problems. Here, we use (1.2) in order to ensure that the functional associated with our problem has the geometric condition of the Mountain Pass Theorem if λ < λ 1 or the Linking Theorem if λ k < λ < λ k+1 and also guarantee the boundedness of the Palais Smale sequence. Thus, we obtain the weak convergence of the Palais Smale sequence to the solution. Then condition (H 2 ) guarantees the nontriviality of our weak solutions. (c) In the subcritical case, we can replace condition (H 3 ) by the following weak hypothesis

35 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 34 (H 3) F (t) > 0 and f(t)t > 0 for all t R \ {0}. (d) We mention that we do not use condition (H 3 ) to prove the existence of solutions attaining the minimax level in the subcritical case. For the critical case, we can show the existence of weak solution of (P1) without condition (H 3 ). Nevertheless, we use (H 3 ), in our arguments, to prove the existence of a critical point achieving the minimax level. In some cases, these solutions will be ground state solutions. 1.2 Statement of main results Here, we divide our results in six theorems, which are considered depending on the growth conditions of the nonlinearity and also on the interval where λ lies. In the critical case, since the weight x α has an important role on the estimate of the minimax levels, the variational setting and methods used in H 1 0( ) and H 1 0,rad () are different and, therefore, are given in separate theorems. Another aspect, which we should consider in this case, is an asymptotic hypothesis on f(t)t/e β 0t 2 asymptotic behavior for f(t)t/e β 0t 2 at infinity. We shall suppose a certain kind of at infinity, if λ < λ 1 and α 0, in order to find a solution in H 1 0( ) or in H 1 0,rad (). If λ > λ 1 and α > 0, an analogous hypothesis can be assumed, ensuring, thus, the existence of a solution in H 1 0( ). However, if we search a solution for (P1) in H 1 0,rad (), when we are supposing λ > λ 1 and α 0, we demand a stronger hypothesis on the behavior of f(t)t/e β 0t 2. Then, we have the following results. Theorem (The subcritical case, local minimum at 0). Assume (H 1 ) (H 3 ) and that f(t) has subcritical growth (SG) at both + and. Furthermore, suppose that 0 < λ < λ 1 and 2F (t) (H 4 ) lim sup λ t 0 t 2 1 λ. Then, problem (P1) has a nontrivial (radially symmetric) solution. Moreover, if f(t) is an odd function, then (P1) has infinitely many solutions. Theorem (The subcritical case, saddle point at 0). Assume (H 1 ) (H 3 ) and that f(t) has subcritical growth (SG) at both + and. Furthermore, suppose that λ k < λ < λ k+1 and (H 5 ) There exist δ > 0 and 0 < µ < λ k+1 λ, such that F (t) µ 2 t2, for all t δ.

36 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 35 Then, problem (P1) has a nontrivial (radially symmetric) solution. Moreover, if we assume that f(t) is an odd function, then (P1) has infinitely many solutions. For the next two theorems, we will use the following notations: for 0 ε < 1 let us consider M ε = lim n 1 0 ne n((1 ε)t2 t) dt and M = lim ε 0 M ε. We use the notation M 0 for M ε with ε = 0, and using calculus, one can see that M 0 = 2 and M = 1. We also consider constants 0 < r, d < 1 such that there exists a ball B r so that x > d for all x B r. Theorem (The critical case, local minimum at 0). Assume (H 1 ) (H 4 ) and that f(t) has critical growth (CG) at both + and. Furthermore, suppose that 0 < λ < λ 1 and f(t)t (H 6 ) lim ξ, ξ > t + e β 0t 2 4 with d > 0. β 0 M 0 d α r2 Then, problem (P1) has a nontrivial solution. Theorem (The critical case, saddle point at 0 with α > 0). Assume (H 1 ) (H 3 ), (H 5 ) and that f(t) has critical growth (CG) at both + and. Furthermore, suppose that λ k < λ < λ k+1, α > 0 and f(t)t (H 7 ) lim ξ, with ξ > t + e β 0t 2 where 0 < r d < 1 are such that 4 β 0 r 2+α [( ) α d 2 ] 1, r 2 + α ( ) α d > 2 r 2 + α. Then problem (P1) has a nontrivial solution. Theorem (The radial critical case, local minimum at 0). Assume (H 1 ) (H 4 ) and that f(t) has critical growth (CG) at both + and. Furthermore, suppose that 0 < λ < λ 1 and f(t)t (H 6,rad ) lim ξ, ξ > t + e β 0t 2 8 (2 + α)β 0 e. Then, problem (P1) has a nontrivial radially symmetric solution. Theorem (The radial critical case, saddle point at 0). Assume (H 1 ) (H 3 ), (H 5 ) and that f(t) has critical growth (CG) at both + and. Furthermore, suppose that λ k < λ < λ k+1 and for all γ 0 there exists c γ 0 such that

37 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 36 (H 8 ) f(t)t e β 0t 2 lim inf t + γh(t) for all t > c γ, where h : R R + is a Carathéodory function satisfying log(h(t)) t > 0. Then problem (P1) has a nontrivial radially symmetric solution. Remark (a) Note that an example for nonlinearity with subcritical growth which satisfies (H 1 ) (H 3 ) is given by f(t) = e t q t p+q 2 t + p q t p 2 te t q + ct, where 1 q < 2 < p. Moreover, if λ < λ 1 and 0 < c λ 1 λ, assumption (H 4 ) is satisfied. Thus, applying Theorem 1.2.1, we obtain a nontrivial (radial) solution of (P1). Similarly, in view of Theorem 1.2.2, we can obtain a nontrivial (radial) solution of (P1) provided that λ k < λ < λ k+1. (b) Our Theorem is a version of Theorem 1.1 of [28], for α = 0 and the autonomous nonlinearity and Theorem in this chapter is analogous to Theorem 1.2 in [28] (FIGUEIREDO et al, 1995). (c) For examples of nonlinearities with critical growth satisfying the assumptions of Theorems 1.2.3, or 1.2.5, one can consider f(t) = e β 0t 2 t p t + p 2β 0 t p 2 te β 0t2 + ct, where p > 1 and c can be chosen so that f(t) satisfies (H 4 ) or (H 5 ). (d) We also realize that Theorem is a version of Theorem 1.3 in [28] for the autonomous case. Moreover, Theorem is a radial version of Theorem and, consequently, of Theorem 1.3 in [28] with autonomous linearity when α = 0. We also mention that for the radial case we obtain an improvement in assumption of behavior of f(t)/t at infinity. Indeed, in radial context, the estimate from bellow in (H 6,rad ) is smaller than the one in (H 6 ) (FIGUEIREDO et al, 1995). (e) Our Theorem is not a similar version of Theorem 1.4 in [28], because here we are demanding that α > 0, which was crucial to guarantee that the minimax level is in the range of compactness of the associated functional (FIGUEIREDO et al,

38 Hénon elliptic equations in R 2 with subcritical and critical exponential growth ). In order to obtain a similar result to Theorem with α equal to 0 we need to replace (H 7 ) with (H 8 ), which would make this result weaker. (f) For p > 1 and appropriated choice of constants c > 0, we can see that the following nonlinearity f(t) = e β 0t 2 +t t p t + satisfies the hypotheses of Theorem p 2β 0 t p 2 te β 0t2 +t + ct (g) In Theorem 1.2.6, although α can be taken equal to 0, we do not obtain a similar result to the autonomous case for Theorem 1.4 in [28], because we need to assume (H 8 ) instead of (H 7 ) (FIGUEIREDO et al, 1995). (h) Following a similar argument as the one due to P. H. Rabinowitz [45, Proposition 3.11] (RABINOWITZ, 1992), we can prove that the Mountain Pass solution obtained in Theorems 1.2.1, and are indeed ground state solution provided that f(t) = o( t ) at 0 and f(t)/t is increasing. Since we prove the existence of a solution for the Hénon type equations using variational methods, we considere the nonlinearity with subcritical or critical exponential growth. We do so in the same spirit of D. Figueiredo, O. Miyagaki and B. Ruf [28] (FIGUEIREDO et al, 1995), following the Brezis-Nirenberg argument [11] (BREZIS; NIRENBERG, 1983). As we can see in the proofs of Theorems 1.2.3, 1.2.4, and 1.2.6, the minimax levels of the functional depend heavily on the asymptotic behavior of f(t)t/e β 0t 2 at infinity. Moreover, the parameters α, λ and the Sobolev spaces (H 1 0( ) or H 1 0,rad ()) exert a significant influence on the choice of our assumptions (H 6 ), (H 6,rad ), (H 7 ) and (H 8 ). For example, in H 1 0( ), we were inspired by Adimurthi, in [2] (ADIMURTHI, 1990), and D. Figueiredo, O. Miyagaki and B. Ruf, in [28] (FIGUEIREDO, 1995), who showed the existence of solution for similar problems to (P1), in a bounded domain Ω R 2, provided that f satisfies (among other conditions) the asymptotic hypothesis (h 0 ) lim t f(x, t)t e β 0t 2 > C(r), where r is the radius of the largest open ball in the domain. To study the existence of solutions for (P1) in H 1 0( ) for nonlinearity with the vanishing weight at origin of the type f(x, t) = x α f(t), we assume that an analogous condition to

39 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 38 (h 0 ) holds uniformly on a small ball B(x 0, δ) \ {0} in order to avoid the influence of the weight x α. Nevertheless, we can prove the existence of solution for (P1) in H 1 0( ) in the critical case supposing λ k < λ < λ k+1 and (H 7 ) for α > 0. For the saddle point at 0 case, we cannot see how it is possible to find a solution for (P1) in H 1 0( ) with α = 0 or in H 1 0,rad () for α 0, using a hypothesis similar to (h 0 ). Hence we have to assume a stronger hypothesis than (H 7 ), which will be used in Theorem In Theorem 1.2.5, although we search for a radial solution for (P1), since λ < λ 1, we can use a technique introduced by D. Figueiredo, J. do Ó and B. Ruf, in [26] (FIGUEIREDO et al, 2002), and assume a hypothesis for f(t) as in (h 0 ). Moreover, this technique together with the weight x α in H 1 0,rad () minimize the constant which bounded from below the asymptotic behavior of f(t). Consequently, we achieve better estimatives for the last case. Indeed, in [28] (FIGUEIREDO, 1995), it is used the hypothesis as (h 0 ) with C 1 (r) = 4 M 0 β 0 r 2 in order to prove a similar result to Theorem 1.2.5, in which we use the analogous hypothesis (H 6,rad ) with C 2 (r) = 8 (2 + α)β 0 er 2. Once C 1 (r) > C 2 (r), we can notice that, even with α equal to zero, we obtain a significant improvement in this kind of estimate Outline This Chapter is organized as follows. In Sect. 1.3, we introduce the variational framework and prove the boundedness of Palais-Smale sequences of the functional associated to Problem (P1). After that, we consider the subcritical case and prove that the associated functional of the problem satisfies the (P S) c condition for all c > 0. For critical case, we show a similar result for levels c in an interval bounded from above for a suitable constant, which depends only on β 0 in H0( 1 ) and on β 0 and α in H0,rad 1 (). In Sect. 1.4, we consider a nonlinearity with subcritical growth and we obtain the geometric conditions for the functional in order to prove the existence of a solution to both Sobolev Spaces, H0( 1 ) and H0,rad 1 (). In Sect. 1.5 and 1.6, we study Problem (P1) in critical growth range and guarantee the existence of a solution in H0( 1 ) and H0,rad 1 (), respectively. In

40 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 39 these cases, we also prove the geometric conditions and the boundedness of the minimax levels. 1.3 The variational formulation ) 1/2 We can notice that by (SC) or (CG) there exist positive constants C and β such that f(t) Ce βt2 for all t R. (1.3) We denote H = H0,rad 1 () in the radial case or H = H0( 1 ) in the another case, with the norm ( u = u 2 dx for all u H. Recalling that our efforts are searching for a nontrivial weak solution to Problem (P1), we define the functional Φ λ : H R by Φ λ (u) = 1 u 2 dx λ u 2 dx x α F (u) dx, 2 2 where α 0 (only in Theorem 1.2.4, we should have α strictly positive) and λ > 0. By (H 1 ) (H 2 ) and (1.3), we see that the functional Φ λ is C 1 and its derivative is given by Φ λ(u), v = u v dx λ uv dx x α f(u)v dx for all v H and the critical points of Φ λ are a (weak) solution to (P1). Hence to find a solution of (P1) one can look for critical points of the functional Φ λ with critical value c > The boundedness of a (P S) c sequence We will use (1.2) to prove the following lemma that shows that a (P S) c sequence of the functional Φ λ is bounded. Let us notice that the proof of the Palais-Smale condition for the functional associated to the problem follows traditional methods. Then we will present a brief proof for this condition. Lemma Assume (H 1 ) (H 3 ). Let (u n ) H be a (P S) c sequence of Φ λ. Then (u n ) is bounded in H. Proof: Let (u n ) H be a (P S) c sequence, i.e. 1 2 u n 2 λ 2 u n 2 2 x α F (u n ) dx c 0 (1.4)

41 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 40 and u n v dx λ u n v dx x α f(u n )v dx ε n v for all v H, (1.5) where ε n 0. It follows from (1.2) that, for ϑ = 4, we obtain t 4 > 0 such that F (t) 1 4 f(t)t for all t t 4. Using (1.4) and (1.5), we have c + ε n u n 1 x α f(u n )u n x α F (u n ) 2 2 = 1 x α f(u n )u n 2 {u n t 4 } x α F (u n ) {u n t 4 } x α F (u n ) 1 2 x α f(u n )u n 1 4 {u n t 4 } x α f(u n )u n C 0 So we have and, consequently, 1 4 Indeed, given M 0 > 0, we have x α f(u n ) = x α f(u n )u n C 0. x α f(u n )u n C + ε n u n with C > 0, (1.6) x α f(u n ) dx C + ε n u n. (1.7) {u n M 0 } C 1 + x α f(u n ) + {u n M 0 } {u n M 0 } x α f(u n ) u n u n x α f(u n ) C M 0 {u n M 0 } x α f(u n )u n C M 0 x α f(u n )u n C + ε n u n.

42 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 41 Firstly, we consider 0 < λ < λ 1. characterization of λ 1, we have ε n u n Therefore, (u n ) is a bounded sequence. Using (1.5) and (1.6) and the variational (1 λλ1 ) u n 2 (C + ε n u n ). Now we consider λ k < λ < λ k+1. It is convenient to decompose H into appropriate subspaces: where H k is finite dimensional and defined by H = H k H k, (1.8) H k = [φ 1,..., φ k ], (1.9) where φ j is the eigenfunction associated to eigenvalue λ k. This notation is standard when dealing with this framework of high order eigenvalues, and we will use it throughout this chapter. and For all u in H, let us take u = u k + u, where u k H k and u Hk. We notice that u u k dx λ uu k dx = u k 2 λ u k 2 2 (1.10) u u dx λ uu dx = u 2 λ u 2 2. (1.11) By (1.5), (1.10) and the variational characterization of λ k, we can see that ε n u k n u n u k n dx λ u n u k n dx x α f(u n )u k n dx ) (1 λλk u k n 2 x α f(u n )u k n dx. Therefore, C u k n 2 ε n u k n x α f(u n )u k n dx (1.12) ε n u k n + u k n x α f(u n ).

43 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 42 Then, since H k is a finite dimensional subspace, by (1.7) and (1.12) C u k n 2 ε n u k n + u k n x α f(u n ) ε n u k n + C u k n (C + ε n u n ) (1.13) C + C u n + Cε n u n 2. Now we notice that, by (1.4), (1.11) and the characterization of λ k+1, we obtain ε n u n u n u n + λ u n u n + x α f(u n )u n ( ) λ 1 u n 2 + x α f(u n )u n. λ k+1 Therefore, C u n 2 ε n u n + x α f(u n )u n. (1.14) By (1.7), (1.6) and (1.14), we get C u n 2 ε n u n + x α f(u n )u n + u k n (C + ε n u n ) ε n u n + C + ε n u n + C u n (C + ε n u n ) (1.15) C + C u n + Cε n u n 2. By summing the inequalities in (1.13) and (1.15), we can see that u n 2 C + C u n + Cε n u n 2, (1.16) proving the boundedness of the sequence (u n ) as desired. Remark (a) In the proof of Lemma there is no difference between assuming subcritical or critical growth or considering the radial case or not. Then we can conclude that even in the case of critical growth, every Palais-Smale sequence is bounded. (b) Lemma still implies that u n C, u n 2 C, f(u n )u n dx C, F (u n ) dx C (1.17) and if we assume (H 3 ), then the two integrals above are non-negative.

44 Hénon elliptic equations in R 2 with subcritical and critical exponential growth The (P S) c condition Now we show that (P S) c condition is satisfied by functional Φ λ. However, we can observe that there is an important difference in assuming subcritical or critical growth of nonlinearity in order to guarantee the (P S) c condition. Indeed, if we consider subcritical case, we shall obtain the (P S) c condition for all levels in R. In critical case, we only obtain this condition for certain levels, which are bounded from above by suitable constants. Working in H0( 1 ), we achieve a constant that depends on β 0, given in (CG). In H0,rad 1 (), this constant also depends on α. First, we show this condition for subcritical case. Lemma Assume (H 1 ) (H 2 ) and (SG). Then the functional Φ λ satisfies the (P S) c condition for all c R. Proof: We consider (u n ) a (P S) c sequence. By Lemma 1.3.1, we have a subsequence denoted again by (u n ) such that, for some u H, we have u n u in H0( 1 ), u n u in L q ( ) or on L q rad ( x α, ) for all q 1, u n (x) u(x) almost everywhere in (0). Notice that there is nothing else to prove in the case when u n 0. Thus, one may suppose that u n k > 0 for n sufficiently large. It follows from (H 2 ) and [28, Lemma 2.1] that x α F (u n ) dx x α F (u) dx. We will prove that x α f(u n )u n dx x α f(u)u dx. (1.18) In fact, we have x α f(u n )u n dx x α f(u)u dx [f(u n ) f(u)]u dx + f(u n )(u n u) dx B. 1

45 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 44 Initially, let us focus on the second integral in the left side of this last estimate. By (SG), we get x α f(u n )(u n u) dx C e βu2 n un u dx for all β > 0. Using Höder inequality, we obtain e βu2 n un u dx ( ) 1 e qβ( un ) 2 u n 2 dx q un u q, where 1/q + 1/q = 1. We take q > 1 and by (SG) and Lemma 1.3.1, we can choose β sufficiently small such that qβ u n 2 4π. Thus by the Trudinger-Moser inequality, we have e βu2 n un u dx C 1 u n u q. (1.19) Since u n u strongly in L q, one has x α f(u n )(u n u) dx 0. It remains to show that x α [f(u n ) f(u)]u dx 0 as well. Indeed, let ε > 0 be given. By similarly arguments used to prove (1.19), there exists C 2 such that f(u n ) f(u) 2, x α C 2. Consider ξ C0 ( ) such that ξ u 2 < ε/2c 2. Now, since x α f(u n ) f(u) dx 0, for this ε, there exists n ε such that x α f(u n ) f(u) dx < ε 2 ξ for all n n ε. Therefore x α [f(u n ) f(u)]u dx B 1 x α f(u n ) f(u) ξ dx + x α f(u n ) f(u) ξ u dx < ε

46 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 45 for all n n ε, as desired. Consequently, we conclude that (1.18) holds. Now taking v = u and n in (1.5), we have u 2 = λ u x α f(u)u dx. On the other hand, if n in (1.5) with v = u n, since (1.18) is true, we have u n 2 λ u x α f(u)u dx. Hence, u n u and so u n u in H. Remark Notice that (1.19) holds because, since we are supposing (SG), we can choose β small enough. This fact is not true if we assume (CG). Thus, only in the subcritical case, we can conclude the (P S) c condition is satisfied for all c R. Now we consider the critical case, which will dramatically change the arguments in order to show the (P S) c condition for Φ λ. Initially, we are working in H0( 1 ), so we will prove the (P S) c condition for c below an appropriate constant depending on β 0. Lemma Assume (H 1 ) (H 3 ), (CG) and H = H0( 1 ). Then the functional Φ λ satisfies the (P S) c condition for < 0 < 2π/β 0. Proof: First, we consider (u n ) a (P S) c sequence. By Lemma 1.3.1, we obtain a subsequence denoted again by (u n ) such that, for some u H, we have u n u in H0( 1 ), u n u in L q ( ) or on L q rad ( x α, ) for all q 1, u n (x) u(x) almost everywhere in (0). It follows from (H 2 ) and [28, Lemma 2.1], that x α F (u n ) x α F (u) in L 1 ( (0)). Using (1.4), we obtain and, consequently, by (1.5) it follows lim u n 2 = 2c + λ u x α F (u) dx (1.20) lim x α f(u n )u n dx = 2c + 2 x α F (u) dx. (1.21)

47 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 46 By (H 3 ) and (1.21), we conclude that c 0. Then any (P S) c sequence approaches a non-negative level. It follows from [28, Lemma 2.1] and (1.5) that u ψ dx = λ uψ dx + x α f(u)ψ dx, for all ψ C0 ( (0)). Using elliptic regularity one can see that u is a classical solution for (P1). We still have that u 2 λ u 2 2 = x α f(u)u dx 2 x α F (u) dx. Consequently, we can see that Φ λ (u) 0. Now we need to show that u n u H, in order to have this convergence, we separate the proof into three cases. Case 1 c = 0. In this case, using (1.20) and the convergence of u n in L q for all q > 1, we have 0 Φ λ (u) lim inf Φ λ (u n ) = lim inf 1 2 u n 2 Then u n u and u n u in H, as we desire. ( ) λ 2 u x α F (u) dx = 0. Case 2 c 0 and u = 0. We show that this case cannot happen for a (P S) c sequence. For some q > 1 and a (P S) c sequence (u n ) H, such that u n 0, we claim that f(u n ) q dx C 0. (1.22) Since u = 0, it follows from (1.20) that, given ε > 0, for n large enough we have u n 2c + ε. (1.23) In order to estimate the integral (1.22) using (CG), we notice that f(u n ) q dx C qβ0u2 e n dx C e qβ 0 (u n) 2 ( (un) ) 2 dx. (1.24) By the Trudinger-Moser inequality, the integral in (1.24) is bounded, independently from n, if qβ 0 u n 2 4π. Because of (1.23), this will be indeed the case c 2π/β 0, if we choose q > 1 sufficiently close to 1 and ε is sufficiently small. Now using (1.5) with v = u n, we have u n 2 λ u n 2 2 x α f(u n )u n dx ε n u n (c + ε)ε n. (1.25)

48 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 47 We estimate the last integral above using the Hölder inequality. Initially, we consider q such that (1/q) + (1/q ) = 1. Then by (1.24), we have x α f(u n )u n dx ( C u n q ) 1/q dx. Then, from (1.22), since u n 0 in L q, we conclude that u n 0. This contradicts (1.20), which says, u n 2 2c 0. Case 3 c 0 and u 0. We shall prove that Φ λ (u) = c. Assume by contraction that Φ λ < c. Thus, ( ) u 2 < 2 c + λ/2 u x α F (u) dx. (1.26) Let v n = u n / u n and v = u 2(c + λ/2 u x α F (u) dx) Since v n v, v n = 1, and v < 1, it follows by [38, Theorem I.6] that 1 sup e 4πpv2 n dx <, for all p < 1 v. 2 Using (CG), we can obtain the following estimate of f(u n ) in L q f(u n ) q dx C e qβ 0u n qβ0u2 dx C e n dx C e qβ 0 u n 2v2 n dx and this will be bounded if q and p can be chosen such that qβ 0 u n 2 4πp 4π c + λ/2 u x α F (u) dx c Φ λ (u) = 4π 1 v 2. For large n, this will happen if β 0 2π < 1 c Φ λ (u), which is actually so, since Φ λ (u) 0 and c < 2π/β 0. Then, from (1.22), since u n u in L q and f(u n ) is bounded in L q for some q, we conclude that u n u. Consequently, u n u in H. This is impossible in view of (1.20) and (1.26) and we conclude that Φ λ (u) = c. Thus, the proof of Lemma is complete.

49 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 48 If we consider H = H0,rad 1 () and f(t) having a critical growth, we should also prove that the (P S) c condition is satisfied by minimax level, which are bounded from above by an appropriate constant. But in this context, this constant, in addition to relying on β 0, it also depends on α, which allows an increasing in the range of the minimax levels. Lemma Assume (H 1 ) (H 3 ), (CG) and H = H0,rad 1 (). Then the functional Φ λ satisfies the (P S) c condition for c < (2 + α)π/β 0. Proof: Since we work in H = H0,rad 1 () and the wieght x α is in the nonlinearity, we can change (3) into (4), so the proof follows exactly the same steps of Lemma Proof of Theorems and 1.2.2: Subcritical Case This section is devoted to the proof of Theorems and Here, we consider H = H0( 1 ), although, in the case that H = H0,rad 1 (), the proof uses the same arguments. We begin proving the geometric conditions of the Mountain Pass Theorem and the Linking Theorem are satisfied for associated functional Φ λ in Theorems and 1.2.2, respectively Geometric conditions Initially, we will show that Φ λ (u) 0, when u belongs to a finite dimensional subspace of H and norm of u is large enough. This fact helps us to prove the geometric condition of the Mountain Pass Theorem to λ < λ 1 and of the Linking Theorem to λ k < λ < λ k+1. Proposition Assume (H 1 ) and (H 2 ). Let V be a finite dimensional subspace of H. Then Φ λ is bounded from above in V, and moreover, given M > 0 there is an R > 0 such that Φ λ (u) M for all u R and u V. Proof: Given u 0 V, let us define η(t) = Φ λ (tu 0 ) = t2 2 u 0 2 λ 2 u x α F (tu 0 ) dx for all t R.

50 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 49 It follows from (H 2 ) (see (1.1)) that for σ > 2 there is a constant C > 0 such that F (t) C t σ C for all t R. (1.27) Thus, η(t) t2 2 u 0 2 t2 λ 2 u C x α t σ u 0 σ dx + C for all t. By the equivalence of norms in V, we obtain η(t) t2 2 u 0 2 C 2π 2 + α t σ u 0 σ + C for all t, which implies that η(t) as t. The result follows by compactness. Proposition Assume 0 < λ < λ 1, (H 1 ), (H 2 ), (H 4 ) and (1.3). Then there exists a > 0 and ρ > 0 such that Φ λ (u) a if u = ρ. Proof: By (H 4 ) we know that there exist µ < λ 1 λ and δ > 0 such that F (t) µ 2 t2, if t δ. On the other hand, from (1.3) we obtain F (t) Ce βt2 t 3 for all t R for β large enough. Hence we obtain F (t) µ 2 t2 + Ce βt2 t 3 for all t R, which implies Φ λ (u) 1 2 u 2 1 ( ) µ + λ u 2 2 C 2 λ 1 1 [ ( )] ( µ + λ 1 u 2 C 2 Using the Trudinger-Moser inequality we have that e 2βu2 λ 1 e βu2 u 3 dx ) 1/2 dx u 3 6. e 2βu2 dx = e 2β u 2 ( u ) u 2 dx < C 0 if u 2 τ and 2βτ < 4π if u H0( 1 ). So Φ λ (u) 1 [ ( )] µ + λ 1 u 2 C u 3. 2 λ 1

51 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 50 Now we choose ρ > 0 as the point where the function g(s) = 1 2 assumes its maximum and take a = g(ρ). The result follows. [ 1 ( µ + λ λ 1 )] s 2 Cs 3 Here, we consider φ j to be the eigenfunctions of (, H 1 0( )) corresponding to the eigenvalues λ j for all j N and let us denote the subspace H k as in (1.9). Proposition Assume λ k < λ < λ k+1, (H 1 ), (H 2 ), (1.3) and (H 5 ). Then, there exist a, ρ > 0 such that Φ λ (u) a if u = ρ and u H k. Using the variational characterization of λ k+1, the proof follows the same way of Proposition See below. From (H 5 ), we notice that there exist µ < λ k+1 λ and δ > 0 such that Moreover, from (1.3) we get Consequently, we obtain F (t) µ 2 t2, if t δ. F (t) Ce βt2 t 3 for all t R. F (t) µ 2 t2 + Ce βt2 t 3 for all t R, which implies, together with the variational characterization of λ k+1, that Φ λ (u) 1 2 u 2 1 ( ) µ + λ u 2 2 C e βu2 u 3 dx 2 λ λk+1 1 [ ( )] ( ) 1/2 µ + λ 1 u 2 C dx u λ k+1 e 2βu2 By the Trudinger-Moser inequality, we know that dx = e 2β u 2 ( u ) u 2 dx < C 0 e 2βu2 if u δ 1/2, where βδ < 2π. So Φ λ (u) C u 2 C u 3 > 0. for all u < min{1, δ 1/2 }. Remark Propositions 1.4.1, and are available for both cases (subcritical and critical).

52 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 51 Proposition Suppose that (H 1 ) (H 3 ), (1.3) and (H 5 ) hold and λ k < λ < λ k+1. Let Q be as Q := {v + sφ k+1 : v H k, v R and 0 s R}, where φ k+1 is eigenfunctions corresponding to eigenvalue λ k+1. Then Φ λ (u) 0 for all u Q. Proof: We set R 0 > ρ and R R 0. Now we consider a usual split Q = Q 1 Q 2 Q 3, where Q 1 = {v H k : v R}; Q 2 = {v + sφ k+1 : v = R and 0 s R}; Q 3 = {v + Rφ k+1 : v R}. Let v be on Q 1, then by (H 3 ) and the variational characterization of λ k, it follows that Φ λ (v) 1 2 u 2 λ 2 u ) (1 λλk v 2 0, 2 (1.28) independently of R > 0. On Q 2, we have v + sφ k+1 2 = v + s 2 φ k+1 2 v 2 = R 2 and on Q 3, we can see that v + Rφ k+1 2 = R 2 φ k v 2 R 2. Let us take R R 0 large enough, such that by Proposition one has Φ λ (u) 0 for u Q. Remark Propositions and are proved exactly the same way when we work in H0,rad 1 (). However, in order to show that Propositions and are also available in H0,rad 1 (), we need to split it into two orthogonal subspaces as we usually do with H0( 1 ). Let us recall the notation introduced in (1.9) and that H0,rad( 1 ) = Hk. (1.29) Then, in order to prove Lemma 1.3.1, when one considers the radial case, we should use the decomposition given by (1.29). k=1

53 Hénon elliptic equations in R 2 with subcritical and critical exponential growth Proof of Theorems and completed We know that from the hypotheses in both theorems that Φ λ satisfies (P S) c for all c R. Using the Propositions and and applying the Mountain Pass Theorem, we prove Theorem From Propositions and 1.4.4, Theorem follows, applying the Linking Theorem. If we consider H = H 0,rad ( ) in the proofs of these last Theorems, we have that solutions are radially symmetric. 1.5 Proof of Theorems and 1.2.4: Critical Case in H0( 1 ) This section is dedicated to the proof of Theorems and Since we cannot obtain a Palais-Smale condition for the functional Φ λ as we have done in the subcritical case, we need to show that the minimax levels are below an appropriate constant in order to recover a similar compactness property for the functional. Before that, we show the geometric conditions of the Mountain Pass Theorem when λ < λ 1 and in the Linking Theorem if λ k < λ < λ k Geometric conditions Initially, we need to introduce the so-called Moser functions (log n) 1/2, 0 x 1 z n (x) = 1 ; n log 1 x 1 x 1; 2π (log n) 1/2 n 0 x > 1. (1.30) This kind of sequence is usually used to guarantee that minimax levels lies under an appropriate constant, which will allow us to recover the compactness properties for Φ λ that are lost when dealing with critical growth ranges. Let us consider a suitable translation of Moser s functions in a region of far from the origin where the presence of x α can be neglected. First, let us set x 0 and 0 < r d < 1 such that B r (x 0 ) and x > d in B r (x 0 ) and define the functions z n (x) = z n ( x x0 r which are in H 1 0( ), with z n = 1 and suppz n B r (x 0 ). ), (1.31)

54 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 53 Remark If α = 0 and 0 < λ < λ 1, we do not need to have x > d, thus it is sufficient to consider x 0 equal to origin and r = 1. Fist of all, we will see that for large n we still have the same geometric condition proven in the subcritical case. For 0 < λ < λ 1, we see that Φ λ has a local minimum at 0. Indeed, by Proposition 1.4.2, we have that there exist a, ρ > 0 such that Φ λ (u) a if u = ρ for all u H and from Proposition 1.4.1, we see that, for n large enough, there exist R n > ρ such that Φ λ (R n z n ) 0. For λ k < λ < λ k+1, the proof of Theorem is accomplished by the use of the Linking Theorem. As in Proposition 1.4.3, we have that there exist a, ρ > 0 Φ λ (u) a if u = ρ for all u Hk. Now, we prove the following proposition in order to conclude that geometric conditions of the saddle point case are satisfied. Proposition Suppose that λ k < λ < λ k+1, (H 1 ) (H 3 ), (CG) and (H 5 ) hold. We consider z H k. Let Q be as follows Q := {v + sz : v H k, v R and 0 s R}, Proof: First, we split Q as usual Φ λ (v) 0 for all v Q. where Q = Q 1 Q 2 Q 3 Q 1 = {v H k : v R}; Q 2 = {v + sz : v = R, 0 s R}; Q 3 = {v + Rz : v R}. If v Q 1, using (H 3 ) and the variational characterization of λ k, we have Φ λ (v) 1 2 v 2 λ 2 v ) (1 λλk v 2 0, 2

55 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 54 independently of R. On Q 2, let us take R 1 > ρ, with v = R 1 and 0 s R 1, such that there exists R 2 as in Proposition satisfying v + sz > R 2 > ρ, and Φ λ (v + sz) 0. Analogously, on Q 3 let us take R 3 > ρ, so there exists R 4 such that v + R 3 z R 4 > ρ, and we also obtain by Proposition Φ λ (v + R 3 z) 0. For R max{r 1, R 3, R 4 }, the result follows. Remark Notice that z n, given in (1.31)is not in H k for all n N, so we will use Proposition with z = z n Estimate of minimax levels For the Mountain Pass case, we define the minimax level of Φ λ by c = c(n) = inf max Φ λ(w) (1.32) υ Γ w υ([0,1]) where Γ = {υ C([0, 1], H) : υ(0) = 0 and υ(1) = R n z n }, R n being such that Φ λ (R n z n ) 0 and z n is given in (1.31). Proposition Let c(n) be given in (1.32). Then there exists n large enough such that c(n) < 2π β 0. Proof: We claim that there exists n such that max t 0 Φ λ(tz n ) < 2π β 0.

56 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 55 Suppose by contradiction that is not case. So, for all n, this maximum is larger than or equal to 2π/β 0 (it is indeed a maximum, in view of Proposition 1.4.1). Let t n > 0 be such that Then and, consequently, Φ λ (t n z n ) = max t 0 Φ λ(tz n ). (1.33) Φ λ (t n z n ) 2π β 0 for all n N, Let us prove that t 2 n 4π/β 0. From (1.33) we obtain d dt (Φ λ(tz n )) = 0. t=tn t 2 n 4π β 0 for all n N. (1.34) Multiplying this last equation by t n and noticing that z n = 1, λ, f(t) 0, we have, for n large enough, that t 2 n B r(x 0 ) x α f(t n z n )t n z n dx, By (1.31), one has x d, for all x B r (x 0 ). Thus, t 2 n d α f(t n z n )t n z n dx. (1.35) B r(x 0 ) By (H 6 ), it follows that given ε > 0 there is t ε > 0 such that f(t)t (ξ ε)e β 0t 2, for all t > t ε. Then from (1.35), for n large enough, we obtain t 2 n (ξ ε)d α B r(x 0 ) = (ξ ε)πd α r 2 e 2 log n e β 0t 2 n z2 n dx ( ) β 0 t 2 n 4π 1, (1.36) which implies that t n is bounded. And moreover, (1.34) together with (1.36) gives us that t 2 n 4π/β 0. Let us estimate (1.35) more precisely. Let A n = {x B r (x 0 ) : t n z n (x) t ε }, B n = B r (x 0 ) \ A n, and split the integral in (1.35) into a sum of the integrals over A n and B n. Using (H 6 ), we have t 2 n d [(ξ α ε) B r(x 0 ) ] e β 0t 2 nzn 2 dx + f(t n z n )t n z n dx (ξ ε) e B β0t2 nz n dx. (1.37) n B n

57 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 56 Since t n z n < t ε for x B n, one has χ Bn 1 almost everywhere in B r (x 0 ) as n. Thus, by Lebesgue Dominated Convergence Theorem, we conclude that f(t n z n )t n z n dx 0 B n and e β0t2 nz n dx r 2 π. B n Passing to the limit in (1.37), we obtain [ 4π d α (ξ ε) lim β 0 B r(x 0 ) ] e 4π2z2 n dx r 2 π. (1.38) The last integral in (1.38), denoted by J n, is evaluated as follows { } J n = r 2 e 4π z2 n dx = r 2 π 1 1 ( n 2 e4π 2π log n + 2π e 4π 1 log r) 1 2 2π log n r dr. (1.39) Changing variables in the integral in (1.39), s = log( 1 )/ log n, we have r 1 } J n = r {π 2 + 2π log n e s2 log n 2s log n ds. Thus from (1.38), we obtain 0 1/n 4π β 0 (ξ ε)d α r 2 πm 0 for all ε > 0, which implies a contradiction to (H 6 ). ξ 4 β 0 M 0 d α r 2, In the Linking case with α > 0, we define the minimax level of Φ λ by ĉ = ĉ(n) = inf sup υ Γ w υ(q) Φ λ (w) (1.40) where Γ = {υ C(Q, H) : υ(ω) = ω if ω Q}. First, in order to prove Theorem 1.2.4, let us prove the following proposition. Proposition Let ĉ(n) be given in (1.40). Assume α > 0 and (H 7 ), then there exists n large enough such that ĉ(n) < 2π β 0.

58 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 57 Proof: In the same spirit of [2], we demand to show that there is n > ρ such that max{φ λ (v + sz n ) : v R n, 0 s R n } < 2π β 0 (1.41) where we set z n as in (1.31). First, we need to take d and r, given in (1.31), according to hypothesis (H 7 ). Since, α > 0, we can choose 0 < r d < 1, previously fixed, satisfying ( ) α d 1 > 2 r 2 + α. (1.42) Now, we assume, by contradiction, that (1.41) is not true. Then for all n max{φ λ (v + sz n ) : v R n, 0 s R n } 2π β 0. Let u n = v n + t n z n be the point where this maximum is achieved. So Φ λ (u n ) 2π β 0 (1.43) and using the fact that the derivative of Φ λ, restricted to H k Rz n, is zero at u n, we obtain u n 2 λ u n 2 2 x α f(u n )u n dx = 0. (1.44) Initially, we accept the following claims, which will be proved later. Claim 1. (v n ), (t n ) are bounded sequences. We can assume that the subsequences also denoted by (v n ) and (t n ) such that v n v 0 and t n t 0. Claim 2. v 0 = 0 and t 2 0 = 4π β 0. From (1.44), we notice that lim u n 2 lim and consider B r(x 0 ) x α f(u n ) dx (1.45) Ã n = {x B r (x 0 ) : u n t ε }, Bn = B r (x 0 ) \ Ān, and split the integral in (1.45) into a sum of the integrals over Ãn and B n. Using (H 7 ) we can see that x α f(u n )u n dx (ξ ε) x α e β 0u 2 n dx B r(x 0 ) (1.46) + x α f(u n )u n dx (ξ ε) x α e β 0u 2n dx. B n Bn

59 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 58 Since χ Bn 1, from Claim 2 and Dominated Convergence Theorem, we have lim x α f(u n )u n dx = 0 and lim x α e β 2πr 0u 2+α 2n dx = B n Bn 2 + α. Now we notice that, for n large enough, given ε > 0, there is C ε such that u 2 n (1 ε)t 2 nz 2 n C ε v 2 n. Since t 2 n 4π/β 0, we obtain for n large enough u 2 n (1 ε) 2 4π β 0 z 2 n C ε v. (1.47) Thus, using (1.42) and (1.47), the first integral on the right side in (1.46) is estimated from below by B r(x 0 ) x α e β 0u 2 n dx d α e Cε v B r(x 0 ) e (1 ε)2 4π z 2 n dx. (1.48) We denote the integral over B r (x 0 ) in (1.48) by J n and we can evaluate it as follows J n = r 2 (1 ε)24πz2 e n dx { 1 } (1.49) = r 2 π e (1 ε)2 2 log n + 2π e (1 ε)2 (log 1 n 2 r )2 (log n) 1 r dr. Changing variables in the integral, s = log ( 1 r ) / log n, we can see that the second integral in (1.49) converges to π M. Hence, J n r 2 π M. Hence we obtain ( ) [( ) u n 2 (ξ ε)πr 2 d α 2r M α α d = (ξ ε)πr 2+α 2 ]. (1.50) 2 + α r 2 + α On the hand, since v n 0, for any ε, we have lim u n 2 = lim t 2 n z n 2 + lim 2t n Then it follows from (1.48) and (1.50) that (1 + ε) 4π β 0 (ξ ε)πr 2+α 1/n z n v n dx + lim v n 2 (1 + ε) lim t 2 n = (1 + ε) 4π β 0. [( ) α d 2 ], for all ε > 0, r 2 + α which implies a contradiction to (H 7 ). ξ 4 β 0 r 2+α [( ) α d 2 ] 1, r 2 + α

60 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 59 Remark We notice that the last contradiction only happens because α > 0 and r and d satisfy (1.42). If α = 0, we should assume (H 8 ), so we can obtain a result as Proposition 1.63 in the next section. Proof of Claim 1: Given (t n ) and (v n ) as above, one of the following two possibilities has to hold: (i) either there exists a constant C 0 > 0 such that t n / v n C 0, or (ii) there are subsequences, denoted also by (t n ) and (v n ), such that t n / v n 0. If (i) holds, then there is a constant C > 0 such that which applied to (1.44) gives C 2 t 2 n u n 2 u n v n + t n Ct n, B r(x 0 ) x α f(u n )u n dx d α (ξ ε) e β0u2 n dx, B r/n (1.51) where (H 7 ) was used. We can write ( u n (x) = t n (log n) 1/2 v n (x) 2π t n ) 2π (log n) + 1 1/2 for all x B r/n (x 0 ). Hence, given ε (0, 1), we have u n (x) (1 ε)t n (log n) 1/2 / 2π for large n and x B r/n (x 0 ). Together with (1.51), we obtain which can be written as C 2 t 2 n d α (ξ ε) πr2 n 2 eβ 0/2πt2 n (1 ε)2 log n, 1 d α πr 2 (ξ ε) C 2 t 2 nn 2 eβ 0/2πt2 n (1 ε)2 log n (1.52) It follows that t n is bounded. Consequently, v n is also bounded in this case. Now, we assume that (ii) is true. Then, for n sufficiently large, we have t n v n, which implies u n 2 v n. We suppose, by contradiction, that v n. As before, let s ε > 0 be such that f(t)t (ξ ε)e β 0s 2, for all s s ε. Thus by (1.44), it follows that 1 d α f(u n )u n u n s ε u n 2 dx (ξ ε) dα 4 un sε e β 0u 2 n dx. (1.53) v n 2

61 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 60 On the other hand, we notice that Hence, we can see that u n v n χ u n s ε = v n v n + t n v n z n u n v n χ u n<s ε. u n (x) v n (x) χ u n s ε (x) v a. e. in H 1 0( ), where v H k, with v n / v n v and v = 1. Then using Fatou s Lemma in (1.53) and since we have assumed that v n, we come to a contradiction. Thus v n is bounded and, consequently, t n is also bounded. Proof of Claim 2: First, we will show that Once z n converges weakly to zero in H, we have t 2 0 4π/β 0. (1.54) u n 2 t v 0 2. On the other hand, we notice that u n v 0 in L 1 ( ) and by (1.44) we have x α f(u n )u n dx u n 2 C and using (H 2 ), we ensure that the other hypotheses of [28, Lemma 2.1] are satisfied. Therefore, we obtain x α F (u n ) dx x α F (v 0 ) dx (1.55) Using this information in (1.43), we can see that 2π 1 β 0 2 (t2 0 + v 0 2 λ v 0 2 2) x α F (v 0 ) dx. From the variational characterization of λ k and since λ > λ k, we obtain 2π 1 ) β 0 2 t2 0 + (1 λλk v t2 0, which implies that t 2 0 4π/β 0. We will work again with the alternative set in the proof of Claim 1. We first observe that by (1.54) the alternative (ii) cannot hold. Hence let us assume (i). We conclude from (1.52) that t 2 0 4π β 0 (1 ε) 2 t2 0 4π β 0.

62 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 61 This fact together with (1.54) implies t 2 0 = 4π/β 0. Now we must have v 0 = 0. First, observe that v n v 0, t n t 0, z n = 1 and z n 2 2 0, from (1.43) and (1.55), we can see that Φ λ (v 0 ) + t π β 0. Then we conclude that Φ λ (v 0 ) 0. But, we have that Φ λ (v 0 ) 1 ( vn 2 λ v 0 2) 1 ) (1 λλk v Hence, we obtain Φ λ (v 0 ) = 0. Now we must show that if v 0 H k and Φ λ (v 0 ) = 0, then v 0 = 0 and we finish the proof of Claim 2. Consider v 0 H k, then 0 = Φ λ (v 0 ) = 1 2 v 0 2 λ 2 v x α F (v 0 ) dx 1 2 ) (1 λλk v 0 2 x α F (v 0 ) dx x α F (v 0 ) dx. Since F (t) 0, we can see that x α F (v 0 ) dx = 0. Thus Thus we can see that which give us v 0 = 0. 0 = Φ λ (v 0 ) = 1 2 v 0 2 λ 2 v (1 λλk ) v 0 2. ) 1 (1 λλk v 0 2 0, 2

63 Hénon elliptic equations in R 2 with subcritical and critical exponential growth Proof of Theorems and completed Proof: We know that from the hypotheses in these three theorems that Φ λ satisfies (P S) c for c < 2π/β 0. From the Propositions 1.4.1, and and applying the Mountain Pass Theorem, we prove Theorem By Propositions 1.4.3, and 1.5.3, Theorem satisfies the hypotheses of the Linking Theorem. 1.6 Proof of Theorems and 1.2.6: Critical case in H0,rad 1 () In this section, we treat the radial case with f(t) having critical growth. In this case, a solution u for Problem (P1) should be H = H0,rad 1 (), which will force us to change some calculations The geometric conditions Analogously to Subsection 1.5.1, we need to introduce sequences of functions, which helps us to guarantee the minimax levels lie under an appropriate constant, in order to achieve the compactness properties for Φ λ as before. If 0 < λ < λ 1, we consider the following sequence (y n ), given by s y n (s) = n (1 δ n) 1/2, 0 s n; 1/2 ( ) 1 An + 1 log + (n(1 δ (n(1 δ n )) 1/2 A n + e (s n) n )) 1/2, n s, (1.56) where δ n = 2 log n/n and A n = 1/(en) + O(1/n 4 ). Since f(t) satisfies (CG), by [26, Theorem 1.4], we have 0 y n(t) 2 dt = 1 and lim F (y n ) dt = πe. (1.57) n Using this sequence together with Proposition and Proposition 1.4.2, we see that Φ λ has a local minimum at 0. If λ k < λ < λ k+1, we have to consider the split H = H l ( (H l ) H 0,rad ( ) ), as in (1.29) and similarly to Proposition 1.4.3, we have that there exist constants a, ρ > 0 such that Φ λ (u) a for all u ( (H l ) H 0,rad ( ) ) with u = ρ. Taking z n as in (1.30), we

64 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 63 set T l : H0,rad 1 () ( (Hl ) H 0,rad ( ) ) as the orthogonal projection. More explicitly, we have w n (x) = T l z n (x). (1.58) Since Hl H k, we have that w n satisfies (14) and (15) for all n N. By the same away of Proposition 1.4.4, we conclude that, if n is large enough, there exists R n = R(n) > ρ such that for Q := {v + s w n : v Hl ; v R n and 0 s R n }, we have Φ λ (u) 0 for all u Q. Remark Sequences (y n ), given in (1.56), and ( w n ), given in (1.58), will be used in the proofs of Theorems and 1.2.6, respectively. Remark By definition, w n is not in H k for all n. From (15), we realize that w n 0, given in (1.58), for all n N. Since w n is the projection of z n on H 0,rad ( )), we have that z n, given in (1.31), is also not in H k for all n N, so we will use Proposition with z = z n and with z = w n in order to prove Theorems and 1.2.6, respectively. Geometrically, the next figures illustrate this fact. Figure 1.1: Q := {v + sz n : v H k, v R and 0 s R}.

65 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 64 Figure 1.2: Q := {v + sw n : v H k, v R and 0 s R} Estimate of minimax level First, we consider λ < λ 1. From Lemma 1.3.4, we know that Φ λ satisfies (P S) c for c < (2 + α)π/β 0. To conclude via the Mountain Pass Theorem that (P1) has a radial solution, we need to show that there is a z H 1 0,rad () with z = 1 such that max{φ λ (tz) : t 0} < (2 + α)π/β 0. It is common to use the Moser sequence, given in (1.30), in order to show this kind of result, but here we will use sequence (y n ) given in (1.56) and, thus, we can obtain better estimates. Initially, we notice that, since we are working in the radial case, we can rewrite the functional J λ in radial coordinates: 1 ( 1 2π 2 u r 2 λ ) 2 u 2 r α F (u) r dr, 0 where u r denotes the derivative of u in relation to the variable r. Cancelling the factor 2π, we see that 1 0 ( 1 2 u r 2 λ ) 2 u 2 r α F (u) r dr satisfies (P S) c for c < (2 + α)/2β 0. Next, we perform a change of variables to transform the interval (0, 1) into the interval (0, + ). Let and hence we obtain r = e s/2 ; dr = 1 2 e s/2 ds u s = u r dr ds = 1 2 u re s/2, 0 ( 1 2 2u se s/2 2 λ 2 u 2 e sα/2 F (u) ) 1 2 e s ds.

66 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 65 Multiplying by 2, we see that the functional (2 u s 2 λ2 ) u 2 e s F (u)e s(1+α/2) ds 0 satisfies (P S) c for (2 + α)/β 0. Finally, substitute y = 2 πu and multiply by π to achieve ( 1 Φ λ (y) = 2 y s 2 λ ( ) ) 1 8 y 2 e s πf 2 π y e s(1+α/2) ds, (1.59) 0 which again satisfies (P S) c for c < (2 + α)π/β 0. Now we can define the minimax level ċ = ċ(n) = inf max Φ λ(ν(w)) (1.60) ν Γ w ν([0,1]) where Γ = {ν C([0, 1], H); ν(0) = 0 and ν(1) = R n y n }. Proposition Let ċ(n) be given as in (1.60). Then there exists n large enough such that ċ(n) < Proof: We claim that there exists n such that max t 0 Φ λ(ty n ) < (2 + α)π β 0. (2 + α)π β 0. Suppose by contradiction that this is not the case. So, for all n, this maximum is larger than or equal to (2 + α)π/β 0 (it is indeed a maximum, in view of Proposition 1.4.2). Then Let t n > 0 be such that and, consequently, Φ λ (t n y n ) = max t 0 Φ λ(ty n ). (1.61) Φ λ (t n y n ) t 2 n (2 + α)π β 0 for all n N, 2(2 + α)π β 0 for all n N. In order to prove that t 2 n 2(2 + α)π/β 0, first, we observe that, from (1.61), we have d dt (Φ λ(ty n )) = 0. t=tn So, using condition (H 6,rad ), we conclude that ( t 2 tn n π f 2 π y n 0 ) tn 2 π y ne s(1+ 2 ) ds α (ξ ε)π n e β 0t 2 n/4π(n 2 log n) s(1+ α 2 ) ds.

67 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 66 We suppose that lim n t2 n > 2(2 + α)π β 0. Consequently, there exists a subsequence of t n with t 2 n δ + 2(2 + α)π/β 0 for some δ > 0. Then we have t 2 n (ξ ε)π n ( ) 2(2+α)π β0 +δ β e 0 ) β0 = (ξ ε)πe δ β 0 4π n ( 2(2+α)π β 0 +δ 4π (n 2 log n) t(1+ α 2 ) ds 2π log n. Thus, we conclude that t 2 n is bounded and converges to 2(2 + α)π/β 0. In order to estimate more precisely, we set A > 0 and set [0, b n ) = {s [0, ) : t n y n (s) < A}. Since y n (s) = s((1 δ n )/n) 1/2 0 for every fixed s 0, we can see that b n. Then we have t 2 n (ξ ε)π (ξ ε)π 0 bn 0 bn e β 0/4πt 2 n y2 n s(1+ α 2 ) ds + π f ( ) tn 2 π y tn n 2 π y ne s 2 ds 0 (1.62) e β 0/4πt 2 n y2 n s(1+ α 2 ) ds. α The last integral in (1.62) goes to 1. In fact, we have bn 0 bn bε bn e s(1+ α 2 ) ds e s2 n y2 n t(1+ α 2 ) ds = e s2 n y2 n t(1+ α 2 ) ds + 0 where we choose for given ε > 0 the number b ε > 0 such that bn b ε bn e β0/4πt2 nyn s(1+ 2 α 2 ) ds e β 0 /4πA 2 0 b ε e s(1+ α 2 ) ε ds 2 b ε e s2 n y2 n t(1+ α 2 ) ds, for all n. Now, using that y n (s) ε n 0 on [0, b ε ], we choose N ε sufficiently large such that bε 0 e β 0/4πt 2 ny 2 n t(1+ α 2 ) ds e β 0 /4πε 2 n bε 0 e t(1+ α 2 ) ds 1 + ε 2 for all n > N ε. The second integral in (1.62) is positive, and in fact goes to zero (as can be seen using a similar argument). Hence we have in the limit ( ) lim n t2 n = (ξ ε)π lim e β 0/4πt 2 nyn s(1+ 2 α 2 ) ds 1 n ( 0 ) = (ξ ε)π e (1+ α 2 )yn 2 s(1+ α 2 ) ds 1 lim n 0

68 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 67 First consider the change of variable s = (1 + α/2)t. (1 + α/2) 1 x n (s). From (1.57), it follows that 2(2 + α) ( π (ξ ε) 1 + α ) ( 2 π lim β 0 2 n for ε small enough, which implies ( = (ξ ε) 1 + α ) 2 πe, 2 ξ 8 (2 + α)β 0 e, 0 After, changing y n (s) into ) e x2n t dt 1 a contradiction to (H 6,rad ). Now, we consider the Linking case and define the minimax level where Γ = {ν C(Q, H); ν(u) = u if u Q}. c = c(n) = inf max Φ λ(h(w)) (1.63) ν Γ w ν( Q) Proposition Let c(n) be given as in (1.63). Then there exists n large enough such that c(n) < (2 + α)π β 0. Proof: Suppose by contradiction that for all n we have c(n) (2 + α)π/β 0. Thus we obtain c(n) max{φ λ (v + t w n ) : v H l, v R n and t 0}. It follows that for each n there exist v n H l and t n > 0 such that Φ λ (v n + t n w n ) = max{φ λ (v + t w n ); v H l B Rn, t 0} (1.64) and we have Φ λ (v n + t n w n ) So, since w n (H l ), from (1.28) we obtain (2 + α)π β 0 for all n N. (1.65) t 2 n 2(2 + α)π β 0 for all n N. (1.66) Analogously to Proposition 1.5.2, let us assume the following claims, whose proofs we will give later.

69 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 68 Claim 3. (v n ) and (t n ) are bounded sequences. Claim 4. t 2 n 2(2 + α)π β 0 in R and v n 0 in H. Since v n H l, in view of Claim 4, we also obtain v 0. However, we have v n + t n w n uniformly in /n. We can suppose, without loss of generality, that K 0 log(h(t)) t C 0 (1.67) for all t large enough for some C 0, K 0 > 0. Indeed, note that if h(s) satisfies (H 8 ), there exists h(t) such that h(t) h(t) for all t large enough and 0 < lim inf t + log( h(t)) t lim sup t + log( h(t)) t < +. Finally, consider γ such that γ > (2 + α)2 β 0 ( ) 4(2 + α)πbk exp + C2 0. (1.68) 2π 4β 0 From (1.64), using the fact that derivative of Φ λ, restricted to H k Rw n is zero at v n +t n w n, we obtain v n + t n w n 2 λ v n + t n w n 2 2 x α f(v n + t n w n )(v n + t n w n ) dx = 0 (1.69) and we can see, for n large enough, we have t 2 n x α f(v n + t n w n )(v n + t n w n ) dx. /n Since t n 2(2 + α)π/β 0 > 0 and v n 0, by (15) we can take n large enough that t n w n v n c λ, where c λ is given in (H 8 ). Thus, we obtain t 2 n γ x α h (t n w n v n ) exp ( β 0 (t n w n v n ) 2) dx. /n We have t 2 n [ ( ) 2 γ x α log[h(tn w n v n )] exp /n 2 β 0 (t n w n v n ) ( +β 0 t n w n v n + log[h(t ) ] 2 n w n v n )] dx. 2β 0 (t n w n v n )

70 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 69 But if n is large, (1.67) shows that ( ) 2 log[h(tn w n v n )] 2 C2 0 β 0 (t n w n v n ) 4β 0 and log[h(t n w n v n )] 2β 0 (t n w n v n ) K 0 2β 0, from which we obtain ( ) ( t 2 n γ exp C2 0 x α exp (β 0 t n w n v n + K ) ) 2 0 4β 0 /n 2β 0 ( ) γ exp C2 0 x α exp(β 0 t 2 n w 4 n) 2 dx. /n Consequently, from (15) and (1.66) we can see that ( ) t 2 n γ exp C2 0 x α exp[2(2 + α)π w 4 n] 2 dx /n ( ) = γ exp C2 0 4 = γe [ ( 2π (log n) 1/2 exp 2(2 + α)π (2 + α)n2+α 2π ( ) C2 0 2π 4 (2+α)n 2+α +[(2+α) log n]+( 4(2+α)πB k/ 2π)+[2(2+α)πBk 2/ log n]. Now, letting n in the inequality above, one gets γ (2 + α)2 β 0 exp ( ) 4(2 + α)πbk + C2 0, 2π 4β 0 dx ) 2 ] B k (log n) 1/2 which is contrary to the choice of γ in (1.68). This contradiction follows the assumption č(n) (2 + α)π/β 0 for all N, which concludes the proof. Although Claims 3 and 4 are similar to Claims 1 and 2, respectively, we must use different arguments in order to prove them, because we assume (H 8 ) instead of (H 7 ). Now we will prove them briefly. Proof of Claim 3: It is sufficient to prove that all subsequence of (t n ) and (v n ) have bounded subsequences. Let us suppose that this is not true. So, we can find subsequences, which by convenience we still denote by (t n ) and (v n ), respectively, such that all of their subsequences are unbounded. That means we can assume the following t nk + v nk for all subsequences (n k ). (1.70) Therefore, one of the following two possibilities has that hold:

71 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 70 (i) either there exists a constant C 0 > 0 such that t n / v n C 0, or (ii) there exist subsequences such that t n / v n 0. Assume that (i) holds and using (1.70), we have that t n. Now we can see from (H 8 ) and (1.64) that t 2 n x α f(v n + t n w n )(v n + t n w n ) dx B r/n (x 0 ) γ x α h(v n + t n w n ) exp ( β 0 (v n + t n w n ) 2) dx B r/n (x 0 ) for n large enough from (H 8 ) we have when s is large. Thus we have t 2 n γ C B r/n (x 0 ) h(s) C (1.71) x α exp ( β 0 (v n + t n w n ) 2) dx. (1.72) We notice that since H k has finite dimension, we have that v n /t n is bounded for all x B r/n (x 0 ). This fact, together with (15), give us ( v n (x) + t n w n (x) = t n w n (x) 1 + v ) n(x) 1 t n w n t n 2π ((log n) 1/2 2πBk (log n) 1/2 ) ( 1 C ( (log n) 1/2 2π ) 1 ) B k (log n) 1/2 t n 2 1 2π ((log n) 1/2 2πBk (log n) 1/2 and taking n such that (log n) 1/2 2πB k /(log n) 1/2 (1/2)(log n) 1/2, we obtain ) and by (1.72) it follows that t 2 n γ C v n (x) + t n w n (x) B r/n (x 0 ) x α exp ( t n 4 (log n)1/2 2π β 0 ( tn 4 2π (log n)1/2 ) 2 ) dx = γ (( r2+α Cπ 2 + α exp β 0 ) ) t 2 n 32π 2 log n.

72 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 71 Consequently, t n must be bounded in case (i), which contradicts t n. Then (ii) occurs. Since lim n t n / v n = 0, by (1.70) we conclude that v n. From (1.69), we obtain u n 2 x α f(t n w n + v n )(t n w n + v n ) dx. (1.73) Using (H 8 ) and (1.71), for n large enough, we have u n 2 γ C x α e β 0(t n w n+v n) 2 dx. {u n c γ} Since we are supposing (ii), it follows that 1 Cγ x α eβ 0(t n w n+v n) 2 dx u n C γ 2 2 {u n c γ} {u n c γ} x α eβ 0(t n w n+v n) 2 v n 2 dx (1.74) and just like with Claim 2, using Fatou s Lemma, we come to a contradiction. Then v n is bounded and, consequently, t n is also bounded. Proof of Claim 3: First, we notice for some appropriated subsequences we have v n v 0 in H and t n t 0. Since z n 0, we have w n 0 and w n 0 for all x. Then, it follows that Moreover, in view of (1.69) we see that v n + t n w n v 0 almost everywhere in. (1.75) x α f(v n + t n w n )(v n + t n w n ) dx v n + t n w n 2 C. (1.76) However, using [28, Lemma 2.1], once (H 1 ), (1.75) and (1.76) hold we have x α F (v n + t n w n ) dx x α F (v 0 ) dx. From (1.65) and (1.76), we can see that Φ λ (v 0 ) + t2 0 2 and since v 0 H k, in view of Φ λ (v 0 ) 0, we have t 2 0 (2 + α)π β 0. (1.77) 2(2 + α)π β 0. Let us prove that t 2 0 = 2(2 + α)π/β 0. Then, suppose that it is not true. We have t 2 0 > 2(2 + α)π/β 0. Thus we can take small enough ε > 0 so that t 2 n > (1 + ε) 2(2 + α)π β 0

73 Hénon elliptic equations in R 2 with subcritical and critical exponential growth 72 for all large n. We consider ε n = sup /n v n (x) t n w n, clearly we see that ε n 0, which yields C γ C x α exp[β 0 (v n + t n w n ) 2 ] dx γ C B r/n (x 0 ) B r/n (x 0 ) x α exp[β 0 ( ε n t n w n + t n w n ) 2 ] dx. Now, using (H 8 ) and (15) and large n, we see that ( [ C γ C πr 2+α (log n) exp β (2 + α)n2+α 0 (1 ε n ) 2 t 2 1/2 n 2π ] 2 ) B k (log n) 1/2 πr 2+α = γ C (2 + α)e (2+α) log n exp ( [ B β 0 (1 ε n ) 2 t 2 2 k n log n 2 B ]) ( ) k (1 ε n ) 2 t 2 n log n exp β 0. 2π 2π Because t 2 n 2(2 + α)π/β 0 and ε n 0, we notice that ( [ B exp β 0 (1 ε n ) 2 t 2 2 k n log n 2 B ]) k > C 1 2π for n large enough and some C 1 > 0. Using t 2 n (1 + ε)4π/β 0, we have ( ) r2+α C C 1 γ Cπ 2 + α exp β 0 (1 ε n ) 2 t 2 log n n (2 + α) 2π C 1 γ Cπ r2+α 2 + α exp ( (2 + α) log n[(1 ε n ) 2 (1 + ε) 1] ), which is a contradiction. Consequently, we must have t 2 0 = 2(2 + α)π/β 0 as desired. Thus, by (1.77), we obtain Φ λ (v 0 ) 0. But we know that v 0 H k, so by (1.28), we have Φ λ (v 0 ) = 0. As with Claim 2, we can conclude that if v 0 H k and Φ λ (v 0 ) = 0, then v 0 = 0. This finishes the proof of Claim Proof of Theorems and completed Proof: We know that from the hypotheses in both Theorems that Φ λ satisfies (P S) c for c < (2 + α)π/β 0. For λ < λ 1, we see that Φ λ has a local minimum point at 0, thus, applying the Mountain Pass Theorem, we prove Theorem If λ k < λ < λ k+1, we have a saddle point at 0 and Theorem follows from the Linking Theorem.

74 Chapter 2 Hénon type equations with one-sided exponential growth In this chapter, we establish a link between Ambrosetti Prodi problems and Hénon equations. First of all, we remember that the study these problem starts with the celebrate paper by A. Ambrosetti and G. Prodi [4] in Ever since, this kind of problems has been studied, explored and extended by an enormous variety of authors. For a brief review, we refer the reader to [24, 29, 46]. In short, it deals with non-homogeneous problems such as u = g(x, u) + f(x) in Ω, u = 0 on Ω, where at least one eigenvalue of (, H0( 1 )) is in the interval, whose the extremes are the limits lim s g(s)/s < lim s + g(s)/s. Moreover, the existence of multiple solutions depends heavily on an usual hypothesis regarding a suitable parametrization of the forcing term f, which is extensively considered in almost all Ambrosetti-Prodi problems since the work of J. Kazdan and F. Warner in [34]. Since our work deals with nonlinearities in critical growth range, let us focus our discussion to more specifically related studies. As far as we know, the first paper that addressed an Ambrosetti-Prodi problem involving critical nonlinearities was the work of D. YinBin, [58]. There, the author considered a function that was superlinear both in + and and, so, asymptotically jumping all the eigenvalues of (, H0(Ω)). 1 In [29], D. Figueiredo and J. Yang explored more jumping possibilities with one-sided critical growth equation. However, their techniques have natural limitations, which allowed to investigate 73

75 Hénon type equations with one-sided exponential growth 74 the existence of multiple solutions in dimensions N 7 only. In [15], M. Calanchi and B. Ruf improved these results, once they proved that the same problem has at least two solutions provided N 6. They also guarantee the existence of solutions in lower dimensions, when they added a lower order growth term to the nonlinearity. Considering similar problems on domains in R 2 with the exponential growth, we have the work of CALANCHI et al, 2004 [16], which studied the following problem: u = λu + g(u + ) + f(x) in Ω, u = 0 on Ω, where Ω is bounded and smooth in R 2 and g has a subcritical or critical Trudinger-Moser growth. They proved the existence of two solutions for some forcing terms f, depending on the usual parametrization f(x) = h(x) + tφ 1 (x). Here we consider a similar problem with the weight x α on the nonlinearity, which is proper of Hénon equations. More specifically, we we study the solvability of problems of the type u = λu + x α g(u + ) + f(x) in, u = 0 on, (P2) where λ, α 0 and = {x R 2 : x < 1}. We assume that g has the maximum growth which allows us to treat problem (P2) variationally in suitable Sobolev Spaces, due to the well-known Trudinger-Moser inequality (see [40, 57]), which is given by (3) and (4) (this last in radial context with the typical Hénon weight). Motivated by these inequalities, similarity to we did in previous chapter, we say that g has subcritical growth at + if g(t) lim = 0, β (2.1) t + e βt2 and g has critical growth at + if there exists β 0 > 0 such that g(t) g(t) lim = 0, β > β 0 ; lim = +, β < β 0. (2.2) t + e βt2 t + e βt2 2.1 Hypotheses Before stating our main results, we shall introduce the following assumptions on the nonlinearity g: (g 0 ) g C(R, R + ), g(s) = 0 for all s 0.

76 Hénon type equations with one-sided exponential growth 75 (g 1 ) There exist s 0 and M > 0 such that 0 < G(s) = (g 2 ) g(s) = o( s ) when s 0. s 0 g(t) dt Mg(s) for all s > s 0. We recall that we consider λ 1 < λ 2 λ 3... λ j... the sequence of eigenvalues of (, H 1 0( )) and φ j is a j th eigenfunction of (, H 1 0( )). We observe that, using assumption (g 0 ), one can see that ψ is a nonpositive solution of (P2) if and only if it is a nonpositive solution for the linear problem ψ = λψ + f(x) in, ψ = 0 on. In order to get such solutions for (2.3), let us assume that (f 1 ) f(x) = h(x) + tφ 1 (x), where h L µ ( ), µ > 2 and hφ 1 dx = 0. (2.3) For that matter, the parameter t plays a crucial role. We shall use this hypothesis in the first Theorem of this chapter. 2.2 Statement of main results We divide our results in four theorems. The first one deals with the first solution of the problem, which is nonpositive and is obtained by a simple remark about a linear problem related to our equation. The other theorems concern the second solution and are considered depending on the growth conditions of the nonlinearity. In the critical case, since the weight x α has an important role on the estimate of the minimax levels, the variational setting and methods used in H0( 1 ) and H0,rad 1 () are different and, therefore, are given in separate theorems. Theorem (The linear Problem). Assume that λ λ j for all j N and (f 1 ) holds, then there exists a constant T = T (h) > 0 such that: (i) If λ < λ 1, there exists ψ t < 0, a solution for (2.3) and, consequently, for (P2), for all t < T.

77 Hénon type equations with one-sided exponential growth 76 (ii) If λ > λ 1, there exists ψ t < 0, a solution for (2.3) and, consequently, for (P2), for all t > T. Furthermore, if f is radially symmetric, then ψ t is radially symmetric as well. Theorem (The subcritical case). Let f L µ ( ) with µ > 2 such that there exists a nonpositive solution ψ for (P2). Assume (g 0 ) (g 2 ), (2.1) and λ λ j for all j N. Then, problem (P2) has a second solution. Furthermore, if ψ is radially symmetric, the second solution is also radially symmetric. Theorem (The critical case). Let f L µ ( ) with µ > 2 such that there exists a nonpositive solution ψ for (P2). Assume (g 0 ) (g 2 ), (2.2), λ λ j for all j N. Furthermore suppose that for all γ 0 there exists c γ 0 such that g(s)s γe (β 0s 2) h(s) for all s > c γ, (2.4) where h : R R + is a continuous function satisfying lim inf s + log(h(s)) s > 0. (2.5) Then, problem (P2) has a second solution. Theorem (The radial critical case). Let f L µ ( ) with µ > 2 such that there exists a radial and nonpositive solution ψ for (P2). Assume (g 0 ) (g 2 ), (2.2), (2.4) with h : R R + being a continuous function satisfying lim inf s + log(h(s)) s 4β 0 ψ, (2.6) where β 0 is given in (2.2). If λ λ j for all j N, problem (P2) has a second solution which is radially symmetric. Example For examples of non-linearities with critical growth satisfying the assumptions of Theorem 2.2.3, one can consider e β 0s 2 +K 0 s (2β 0 s 3 + K 0 s 2 + 2s) if s 0 g(t) = 0 if s < 0 with h(s) = e K 0s (2β 0 s 3 + K 0 s 2 + 2s) and K 0 > 0. If K 0 4β 0 ψ, then g(s) satisfies the hypotheses of Theorem ,

78 Hénon type equations with one-sided exponential growth 77 Remark We notice that if λ < λ 1, we shall use the Mountain Pass Theorem in the proofs of Theorems 2.2.2, and On the other hand, if λ > λ 1, we need to use the Linking Theorem. Remark We also must point out an important fact: the hypothesis (2.5) and (2.6) are definitely stricter than the one used in [16], since they do not require this additional function h. What happens is that, as far as we are concerned, it is not proven, nor we were able to prove that the estimate given in Lemma 3.5 of that article is uniform in m = m(n). It concerned the minimax level c(k) and, although it is not clear there, this level actually depends also on the parameter n, which surely changes with the considered terms of the Moser functions z k,r(n). So, an uniform estimate in this level must be proven in order to conclude that the weak limit of the associated (P S) sequence is actually nontrivial, as established on Proposition 3.4 of that paper. In Theorems 2.2.2, and 2.2.4, we assume that f is such that (P2) admits a nonpositive solution ψ. Then, a second solution will be given by u + ψ, where u is a nontrivial solution of the following problem: u = λu + x α g(u + ψ) + in ; u = 0 on. (2.7) This means that we will focus our attention in looking for a nontrivial solution for Problem (2.7), considering each of the cases given by the Theorems above. This translation has the advantage of working with a homogeneous problem, without a forcing term that could hinder the desired estimates on the minimax levels. On the other hand, the function ψ plays this role in complicating the estimates, but we could handle them by performing some delicate arguments involving the additional hypothesis (2.4), with h satisfying (2.5) or (2.6). Under certain condition, we can see that the first solution we obtained is radially symmetric, due to a celebrated result of R. Palais, namely: the Principle of Symmetric Criticality (see [42]). as well. After that, we get a second solution that is radially symmetric In the subcritical case, we obtain this second solution with almost the same arguments regardless the space we consider, H 1 0( ) or H 1 0,rad (). However, in the critical case, we must consider the Trudinger-Moser inequality as given in (3), when we work in H 1 0( ), or as in (4), in H 1 0,rad (), which will dramatically change the arguments. In the

79 Hénon type equations with one-sided exponential growth 78 last case, the minimax levels have a higher upper boundedness, that should indicate an easier task, but, since we can only work with radial functions, it is impossible to follow the steps we give in the H 1 0( ) framework Outline This chapter is organized as follows. In Sect. 2.3, we study a linear problem related to our equation. This is an important step to guarantee a first solution for the problem, which will be denoted by ψ and is nonpositive. This case is almost identical to the previous results in [15, 16, 29, 47], but we give a little remark concerning existence of a radially symmetric solution when the forcing term is also radial. In Sect. 2.4, we introduce the variational framework and prove the boundedness of Palais-Smale sequences of the functional associated to Problem (2.7). We also show that this functional satisfies the (P S) condition, in the subcritical case. In Sect. 2.5, we obtain the geometric conditions for the functional in order to prove the existence of a second solution to the problem, considering the subcritical growth both in H0( 1 ) and H0,rad 1 (). In Sect. 2.6 and 2.7, we consider the problem in the critical growth range and guarantee the existence of a second solution in H0( 1 ) and H0,rad 1 (), respectively. In these cases, we also show that the geometric conditions are satisfied and we prove the boundedness of the minimax levels by appropriate constants depending on β 0 when we consider the functional defined in H0( 1 ) and on β 0 and α when the functional is considered in H0,rad 1 (). 2.3 The Linear Problem In this section, we prove Theorem 2.2.1, for that we consider the linear problem ψ = λψ + f(x) in, ψ = 0 on. (2.8) It is easy to see that if f is such that this linear problem admits a nonpositive solution, then it will also be a solution for Problem (P2). Considering f decomposed as in (f 1 ), we will see that the sign of the (unique) solution of (2.8) can be established depending on t. Moreover, we give an idea on how to obtain radial solutions as well. Proof of Theorem 1: Up to the point where we discuss the radial case, this proof follows exactly the same arguments found in [15, 16, 29, 47], but we bring it here for the

80 Hénon type equations with one-sided exponential growth 79 sake of completeness. Since λ λ j for all j N, we obtain, by the Fredholm Alternative, a unique solution ψ of (2.8) in H 1 0( ). Using (f 1 ), we can write ψ = ψ t = ω + s t φ 1, with ωφ 1 dx = 0 and s t = t λ 1 λ. We recall that f L µ ( ), with µ > 2. Thus, by elliptic regularity, we have ω C 1,ν for some 0 < ν < 1. Then λ 1 λ C ψ φ 1 = t 1 λ 1 λ t ω C. 1 Let ε > 0 be such that v > 0 for all v C 1,ν such that v φ 1 C 1 ψ < 0, we must have < ε. Since we want λ λ 1 t > 0 and λ 1 λ t ω C 1 < ε So there exists T > 0 such that for λ < λ 1 and t < T or for λ > λ 1 and t > T, we have ψ < 0. Now we notice that ψ is a solution for (2.8) if only if it is a critical point of the following functional I : H0( 1 ) R, given by I(ψ) := 1 ψ 2 dx λ 1 ψ 2 dx f(x)ψ dx. 2 2 When we restrict I to H0,rad 1 (), we also obtain a critical point of this functional on this subspace. If f is radially symmetric, by the Principle of Symmetric Criticality of Palais (see [42]), we can see that all critical points on H0,rad 1 () are also critical points on H0( 1 ). So, due to the fact that I admits only one critical point in the whole space, we get that ψ is also radially symmetric. This completes the proof of Theorem The variational formulation Again, we will denote H = H0( 1 ) or H = H0,rad 1 (), depending on which theorem we are considering, with the Dirichlet norm ( ) 1/2 u = u 2 dx for all u H.

81 Hénon type equations with one-sided exponential growth 80 Recalling that our efforts are searching for a nontrivial weak solution to Problem (2.7), we define the functional J λ : H R as J λ (u) = 1 u 2 dx λ u 2 dx x α G(u + ψ) + dx, (2.9) 2 2 where α, λ 0 and λ λ j for all j N. By (g 0 ), (g 1 ) and (2.1) or (2.2), we have that the functional J λ is C 1 and we can see that its derivative is given by J λ(u), v = u v dx λ uv dx x α g(u + ψ) + v dx for all v H (2.10) and the critical points of J λ are (weak) solution of (2.7). We observe the u = 0 satisfies J λ (0) = 0, which corresponds to the negative solution ψ for (2.3). To find a second solution of (P2) we shall look for critical points of the functional J λ with critical values c > Palais-Smale condition Initially, from (g 1 ), we can see that for every σ > 0 there exists s σ > 0 such that 0 < G(s) 1 σ g(s)s for all s s σ. (2.11) The proof of Palais Smale condition for the subcritical case is essentially standard. For the reader s convenience, we sketch the proof in the next lemma. We must point out that it is not necessary to suppose that g(s) is O(s 2 ) at s = 0 as required in [16]. The well-known techniques we used here can also be handled in case α = 0 and for that it is sufficient to admit that g(s) = o(s) at s = 0, following usual assumptions. Lemma Suppose (g 0 ) (g 2 ). Let (u n ) H be a (P S) c sequence of J λ. Then (u n ) is bounded in H. Proof: Let (u n ) H be a (P S) c sequence of J λ, that is, 1 2 u n 2 λ 2 u n 2 2 x α G(u n + ψ) + dx c 0 (2.12) and u n v dx λ u n v dx x α g(u n + ψ) + v dx ε n v for all v H, (2.13)

82 Hénon type equations with one-sided exponential growth 81 where ε n 0. By (2.11), let us take s 0 > 0 such that G(s) 1 4 g(s)s for all s s 0. (2.14) Using (2.12), (2.13) and (2.14), we get c + ε n u n 1 x α g(u n + ψ) + (u n ) x α G(u n + ψ) = 1 x α g(u n + ψ) + (u n ) 2 x α G(u n + ψ) + {(u n+ψ) + s 0 } {(u n+ψ) + s 0 } x α G(u n + ψ) x α g(u n + ψ) + (u n ) 1 4 {(u n+ψ) + s 0 } x α g(u n + ψ) + (u n + ψ) + C x α g(u n + ψ) + (u n + ψ) C 0. We notice that the last integral above is obtained because ψ 0. Thus, we have x α g(u n + ψ) + (u n + ψ) + dx C + ε n u n with C > 0 (2.15) and, similarly to (1.7) in previous chapter, we obtain x α g(u n + ψ) + dx C + ε n u n. (2.16) Initially, we consider the case: 0 λ < λ 1. Using (2.13) and (2.15), we have ) ε n u n (1 λλ1 u n 2 (C + ε n u n ). Thus, (u n ) is a bounded sequence. Now we consider the case: λ k appropriate subspaces: < λ < λ k+1. It is convenient to decompose H into H = H k H k, (2.17) where H k is finite dimensional and defined by H k = [φ 1,..., φ k ]. (2.18) This notation is standard when dealing with this framework of high order eigenvalues, and we will use it throughout this work.

83 Hénon type equations with one-sided exponential growth 82 For all u in H, let us take u = u k + u, where u k H k and u Hk. We notice that u u k dx λ uu k dx = u k 2 λ u k 2 2 (2.19) and u u dx λ uu dx = u 2 λ u 2 2. By (2.13), (2.19) and the characterization of λ k, we can see that ε n u k n u n u k n dx λ u n u k n dx x α g(u n + ψ) + u k n dx ) (1 λλk u k n 2 x α g(u n + ψ) + u k n dx. Therefore, C u k n 2 ε n u k n x α g(u n + ψ) + u k n dx with C > 0. (2.20) Similarly, we get C u n 2 ε n u n + x α g(u n + ψ) + u n dx. (2.21) Then, since H k is a finite dimensional subspace, by (2.16) and (2.20), we obtain C u k n 2 ε n u k n + u k n x α g(u n + ψ) + dx ε n u k n + C u k n (C + ε n u n ) (2.22) C + C u n + Cε n u n 2. Using (2.15), (2.16) and (2.21), we have C u n 2 ε n u n + x α g(u n + ψ) + u n dx + u k n (C + ε n u n ) ε n u n + x α g(u n + ψ) + (u n + ψ) dx + ψ x B α g(u n + ψ) + dx + u k n (C + ε n u n ) 1 ε n u n + C + ε n u n + ψ g(u n + ψ) + dx + C u n (C + ε n u n ) C + C u n + Cε n u n 2. (2.23)

84 Hénon type equations with one-sided exponential growth 83 By summing the inequalities in (2.23) and (2.22), we reach u n 2 C + C u n + Cε n u n 2, proving the boundedness of the sequence (u n ) as desired. Remark In the proof of Lemma there is no difference between assuming subcritical or critical growth or considering the radial case or not. So we can conclude that even in the case of critical growth, every Palais-Smale sequence is bounded. In case of subcritical growth, we can obtain the (P S) c condition for all levels in R. Lemma Assume (g 0 ) (g 2 ) and (2.1). Then the functional J λ satisfies the (P S) c condition for all c R. Proof: Let (u n ) be a (P S) c sequence. By Lemma 2.4.1, we know that (u n ) is bounded. So we consider a subsequence denoted again by (u n ) such that, for some u H, we have u n u weakly in H, u n (x) u(x) almost everywhere in, u n u strongly in L q ( ) or L q rad ( x α ) for all q 1. Notice that there is nothing else to prove if u n 0. Thus, one may suppose that u n k > 0 for n sufficiently large. It follows from (g 1 ) and [28, Lemma 2.1] that x α G(u n + ψ) + dx x α G(u + ψ) + dx. We will prove that x α g(u n + ψ) + u n dx x α g(u + ψ) + u dx. (2.24) In fact, we have x α g(u n + ψ) + u n dx x α g(u + ψ) + u dx x α [g(u n + ψ) + g(u + ψ) + ]u dx + x B α g(u n + ψ) + (u n u) dx. 1

85 Hénon type equations with one-sided exponential growth 84 we get First, let us focus on the second integral in the left side of this last estimate. By (2.1), x α g(u n + ψ) + (u n u) dx C e βu2 n un u dx for all β > 0. Using the Höder inequality, we obtain e βu2 n un u dx ( ) 1 e qβ( un ) 2 u n 2 dx q un u q, where 1/q + 1/q = 1. We take q > 1 and by (2.1) and Lemma 2.4.1, we can choose β sufficiently small such that qβ u n 2 4π. Thus by the Trudinger-Moser inequality, we have e βu2 n un u dx C 1 u n u q. (2.25) Since u n u strongly in L q, one has x α g(u n + ψ) + (u n u) dx 0. It remains to prove that x α [g(u n + ψ) + g(u + ψ) + ]u dx 0 as well. Indeed. Let ε > 0 be given. By analogous arguments used to prove (2.25), there exists C 2 such that g(u n + ψ) + g(u + ψ) + 2, x α C 2. Consider ξ C 0 ( ) such that ξ u 2 < ε/2c 2. Now, since x α g(u n + ψ) + g(u + ψ) + dx 0, for this ε, there exists n ε such that x α g(u n + ψ) + g(u + ψ) + dx < ε 2 ξ for all n n ε. Therefore x α [g(u n + ψ) + g(u + ψ) + ]u dx B 1 x α g(u n + ψ) + g(u + ψ) + ξ dx + x α g(u n + ψ) + g(u + ψ) + ξ u dx < ε

86 Hénon type equations with one-sided exponential growth 85 for all n n ε, as desired. Consequently, we conclude that (2.24) holds. Now taking v = u and n in (2.13), we have u 2 = λ u x α g(u + ψ) + u dx. On the other hand, if n in (2.13) with v = u n, u n 2 λ u x α g(u + ψ) + u dx again by (2.24). Consequently, u n u and so u n u in H. 2.5 Proof of Theorem 2.2.2: The subcritical case This section is devoted to the proof of Theorem Here, we consider ψ radially symmetric and H = H0,rad 1 (). In the case that ψ is not necessarily radial and H = H0( 1 ), the proof uses the same arguments The geometric condition First of all, we will prove the following lemma, which will be used in the proofs of geometric conditions for the associated functionals. Lemma Assume (g 0 ) (g 2 ). Then, for all ε > 0 and β > 0 if g satisfies (2.1) or β > β 0 if g satisfies (2.2), there exists K ε such that G(u) εu 2 + K ε u q e βu2, for all q 1. (2.26) Proof: Given ε, by (g 2 ) there exists C ε > 0 such that if u C ε, we have u u ( ) G(u) G(0) = g(t) dt ε C 1 t dt εc 2 u2. Since G(0) = 0, we see that 0 G(u) εcu 2 if u C ε. By (2.1) or (2.2), there exists C 1 such that if u > C 1, then for ssuitable β > 0. g(u)u εce βu2 u εce βu2 u q 0

87 Hénon type equations with one-sided exponential growth 86 From (2.11) we obtain G(u) εce βu2 u q for u > C 1. Once the set where C ε u C 1 is compact, using (g 0 ) we complete the proof. Now, we consider λ < λ 1. We will show that the hypotheses of the Mountain Pass Theorem hold for the functional J λ. Proposition Suppose that λ < λ 1, (g 0 ) (g 2 ) and (2.1). Then there exist ρ, a > 0 such that J λ (u) a if u = ρ. Proof: Since λ < λ 1, the variational characterization of the eigenvalues gives us ) J λ (u) (1 λλ1 u 2 x α G(u + ψ) + dx and the Lemma 2.26 with q = 3 give us J λ (u) C u 2 ε x α (u + ψ) 2 + dx K ε x α e β(u+ψ)2 + (u + ψ) 3 + dx. From the monotonicity and Hölder s inequality, we have x α (u + ψ) 2 + dx ( x 2α dx ) 1/2 ( (u + ψ) 4 + dx ) 1/2 and x α e β(u+ψ)2 + (u + ψ) 3 + dx ( ) 1/2 1 u 4 dx = (2α + 2) 1/2 1 (2α + 2) 1/2 u 2 C α u 2. ( e 2β(u+ψ)2 + dx ) 1/2 ( x 4α dx 1 (2α + 2) 1/2 u 2 4 ) 1/4 ( ( ) 1/2 ) 1/4 ( e 2βu2 + dx x (B 4α dx u 12 dx 1 (u + ψ) 12 + dx ) 1/4 ) 1/4 ( 1/2 C α dx) u 3. e 2βu2 Since we are supposing (2.1), β can be chosen such that 2β 4π. So by Trudinger-Moser inequality, we obtain J λ (u) C u 2 C u 3 > 0

88 Hénon type equations with one-sided exponential growth 87 for all u H with u < min{1, δ 1/2 }. The second constant comes from the Trudinger- Moser inequality. Therefore, take ρ = u and a = Cρ 2 Cρ 3. Proposition Suppose (g 0 ) (g 2 ) and (2.1) hold. Then, there exists R > ρ such that J λ (Rφ 1 ) 0, where φ 1 is a first eigenfunction of (, H) (with φ 1 > 0 and φ 1 = 1) and ρ is given in Proposition Proof: We fix R 0 > ρ and 0 < r < 1 such that We observe that (2.11) gives us φ 1 (x) 2 ψ R 0 almost everywhere in B r. for σ > 2, and C σ, D σ 0. Thus, we obtain J λ (Rφ 1 ) R2 2 Rσ C σ x α B r G(t) C σ t σ D σ (2.27) + ( φ 1 + ψ ) σ R + dx + πr 2 D σ. Let R R 0. We estimate the last integral ( x α φ 1 + ψ ) σ ( ) σ ψ 2r 2+α π dx B r R R α = τ > 0. It follows that Since σ > 2, we can choose R > ρ such that J λ (Rφ 1 ) R2 2 C σr σ τ + πr 2 D σ. J λ (Rφ 1 ) 0, which is the desired conclusion. Next, we consider λ k < λ < λ k+1. Before the proof of the geometric conditions of the Linking Theorem for J λ, we need to split H0,rad 1 () into two orthogonal subspaces as we have done to H0( 1 ) in (2.17). Recalling the notation Hk = H k H0,rad( 1 ) for all k N. We have that H0,rad( 1 ) = Hk. (2.28) k=1

89 Hénon type equations with one-sided exponential growth 88 Remark In the proof of Lemma 2.4.1, when we consider the radial case we use the decomposition given by (2.28). For λ k < λ, we consider the corresponding subspace H k and we write H 1 0,rad () as H = H 1 0,rad( ) = H l ( (H l ) H 1 0,rad( ) ) where Hl H k with l = max{j : Hj H k }. Since Hl H k, we notice that ( (H l ) H0,rad 1 () ) Hk. This decomposition will allow us to use the same variational inequalities that characterize λ k and λ k+1 in the H 1 0( ) environment. Proposition Suppose that λ k < λ < λ k+1, (g 0 ) (g 2 ) and (2.1) hold. Then there exist ρ, a > 0 such that J λ (u) a for all u ( (H l ) H 1 0,rad () ) with u = ρ. Proof: Since λ < λ k+1, the variational characterization of the eigenvalues gives us ( J λ (u) 1 λ ) u 2 x α G((u + ψ) + ) dx λ k+1 and by Lemma with q = 3, we have J λ (u) C u 2 + C ε x B α (u + ψ) 2 + dx C x α e β(u+ψ)2 + (u + ψ) 3 + dx. 1 and The Hölder s inequality and the monotonicity give us x α (u + ψ) 2 + dx C α u 2. x α e β(u+ψ)2 + (u + ψ) 3 + dx C α ( e 2βu2 dx) 1/2 u 3. Choosing β 2π, by Trudinger-Moser inequality, we obtain J λ (u) C u 2 C u 3 > 0, for all u ( (H l ) H 1 0,rad () ) with u < min{1, δ 1/2 }. The second constant comes from the Trudinger-Moser inequality. Therefore, take ρ = u and a = Cρ 2 Cρ 3. Proposition Suppose that λ k < λ < λ k+1, (g 0 ) (g 2 ) and (2.1) hold. There exists z ( (H l ) H 1 0,rad () ) H k and R > 0 such that R z > ρ and J λ (u) 0 for all u Q. where Q := {v H l : v R} {sz : 0 s R}.

90 Hénon type equations with one-sided exponential growth 89 Proof: We fix R 0 > ρ and choose z ( (H l ) H 1 0,rad () ) H k and x 0 such that (a) z 2 < λ λ k 1; (b) B r (x 0 ) and z(x) ( K + 2 ψ ) R 0 almost everywhere in B r (x 0 ) with K > 0 satisfying v K v for all v H k. This choice is possible because ( (H l ) H 1 0,rad () ) possesses unbounded functions and H k has finite dimension. We consider a usual split Q = Q 1 Q 2 Q 3, where Q 1 = {v H l : v R}, Q 2 = {v + sz : v Hl, v = R and 0 s R}, Q 3 = {v + Rz : v H l and v R}. Let u be on Q 1, by (g 2 ) and characterization of λ k, it follows independently of R > 0. For Q 2, we get J λ (u) 1 2 u 2 λ 2 u (1 λλk ) u 2 0 J λ (v + sz) 1 2 v s2 z 2 λ 2 v 2 2 independently of R > R R2 z 2 λ 2λ k R R2 ( 1 λ λ k + z 2 ) < 0, Now for Q 3, using (2.27) and obtain J λ (v + Rz) R2 2 z 2 x α G((v + Rz + ψ) + ) dx ( R2 2 z 2 R σ C σ x α z + ψ + v ) σ R + dx D σ.

91 Hénon type equations with one-sided exponential growth 90 Let R R 0 be, we estimate the last integral ( x α z + ψ + v ) σ dx R + B r(x 0 ) ( x α z ψ ) σ + v dx R ( ) σ ( ) σ ψ x α πr2+α ψ = τ > 0. R B r(x 0 ) 2 + α R Thus, we obtain Since σ > 2, we finish the proof. J λ (v + Rz) R2 2 z 2 C σ R σ τ D σ Proof of Theorem completed We have proved that J λ satisfies the geometric and the compactness conditions required in the Mountain Pass Theorem when λ < λ 1 and in the Linking Theorem if λ k < λ < λ k+1. Thus there exists a nontrivial critical point for J λ and thus a solution of (2.7). 2.6 Proof of Theorem 2.2.3: Critical case in H 1 0( ) It is well known that for non-linear elliptic problems involving critical growth some concentration phenomena may occur, due to the action of the non-compact group of dilations. For problems (3) and (4) there are loss of compactness at the limiting exponent β = 4π and β = 2π(2 + α) respectively. Thus, the energy functional J λ fails to satisfy the (P S) c condition for certain levels c. Such a failure makes it difficulty to apply standard variational approach to this class of problems. Our proofs here relies on a Brezis-Nirenberg type argument: We begin by proving the geometric condition of the Mountain Pass Theorem when λ < λ 1 and the geometric condition in the Linking Theorem if λ k < λ < λ k+1. In the second step we show that the minimax levels belong to the intervals where the (P S) condition holds for the functional J λ The geometric conditions Initially we consider λ < λ 1. We will prove the geometric condition of the Mountain Pass Theorem.

92 Hénon type equations with one-sided exponential growth 91 Proposition Assume (g 0 ) (g 2 ), λ < λ 1 and (2.2). Then, there exist a, ρ > 0 such that J λ (u) a if u = ρ. Proof: We can now proceed analogously to the proof of Proposition Now, let us recall the so-called Moser sequence. That is, for each n, we define (log n) 1/2, 0 x 1 z n (x) = 1 ; n 2π log 1 x 1 x 1. (log n) 1/2 n (2.29) It is known that z n H, z n = 1 for all n and z n 2 = O ( 1/(log n) 1/2). For details see [40]. In order to apply our techniques, we shall consider a suitable translation of Moser s functions in a region of far from the origin where the presence of x α can be neglected. We begin by noticing that, from (2.5), we obtain ε 0 and s 0 such that log(h(s)) s ε 0, (2.30) for all s > s 0. Since ψ 0 on, we can fix r > 0, small enough, and x 0, sufficiently close to, such that with B r (x 0 ). ψ,r := ψ Br 2 (x 0 ) ε 0 2β 0 and x 1 2 in B r 2(x 0), (2.31) Let us take the following family of functions: (log n) 1/2, 0 x x 0 r2 ; n z r n(x) = 1 2π log r 2 x x 0 (log n) 1/2, r 2 n x x 0 r 2 ; (2.32) 0, x x 0 r 2. We notice that suppz r n = B r 2(x 0 ), as we can see in Figure 3.1, and, for all n N, one has zn r 2 = 2 ( ) z r 2 n 1 dx = 2π x i 2π log n i=1 r 2 r 2 /n 1 rd r = 1, (2.33) r 2

93 Hénon type equations with one-sided exponential growth 92 where r = x x 0. Figure 2.1: Support of z r n. This sequence will be used to guarantee the existence of a minimax level lying under an appropriate constant which will allow us to recover the compactness properties for J λ that are lost when dealing with critical growth ranges. Before that, we see in the next proposition that for large n we still have the same geometric condition proved in Proposition with z r n taking the place of φ 1. Proposition Suppose that (g 0 ) (g 2 ) and the condition (2.2). Then, there exists R n = R(n) > ρ such that J λ (R n z r n) 0. Proof: We fix R 0 > ρ and N N, such that for all n > N, we have z r n(x) 2 ψ R 0 almost everywhere in B r 2 /n(x 0 ). Using (2.27), let us take R > R 0. We get J λ (Rzn) r R2 2 B1 x α G((Rz rn + ψ) + ) dx R2 2 Rϑ C σ x α Let R R 0 be, we estimate the last integral ( x α zn r + ψ ) σ ( dx x α zn r ψ ) σ dx R R + B r 2 /n (x 0 ) ( zn r + ψ ) σ D σ dx. R + ( ) σ ( ) σ ψ x α ψ 2πr 2(2+α) dx = = τ = τ(n, α). R B r 2 /n (x 0 ) R (2 + α)n2+α Thus we obtain

94 Hénon type equations with one-sided exponential growth 93 J λ (Rz r n) R2 2 C σr σ τ + r4 n 2 πd σ. Since σ > 2, we can choose R n R 0 such that J λ (R n z r n) 0 and the proof is complete. Now we consider λ k < λ < λ k+1 and H = H0( 1 ). Before proving the geometry of linking we want split the supports of the Moser sequence and the k th eigenfunctions. Consequently, we can make the proof easier to be carried on. Let us take x 0 and B r (x 0 ) as in (2.31) and set ζ r : R by 0 in B r 2(x 0 ); ζ r (x) = x x 0 r r 2 r r r r 2 r in B r (x 0 ) \ B r 2(x 0 ); 1 in \ B r (x 0 ). Define φ r j = ζ r φ j and consider the following finite-dimensional subspace H r k = [φ r 1, φ r 2,..., φ r k]. We observe that if v H r k, then suppv \ B r 2, as we can see in the next figure.

95 Hénon type equations with one-sided exponential growth 94 Figure 2.2: Support of v H r k is a subset of \ B r 2. Now we will prove the next result, which is a two-dimensional version of [32, Lemma 2]. Lemma If r 0, then φ r j φ j in H for all j = 1,..., k. Moreover, for each r small enough, we have that there exists c k such that v 2 (λ k + ϱ r c k ) v 2 2 for all v Hk, r where lim r 0 ϱ r = 0. Proof: For j {1,..., k} fixed, we have φ j φ r j 2 = (φ j φ r j) 2 dx = φ j (1 ζ r ) φ j ζ r 2 dx φ j 2 1 ζ r 2 dx + 2 B r + φ j 2 ζ r 2 dx B r\b r 2 r 2+2 r B r\b r 2 φ j (1 η r ) φ j ζ r dx r+1/2 C 1 φ j 2 (r r r 2 r ) + C r φ 2 j φ j (r r r 2 r ) 2 r 2+2 r 1 +C 3 φ j 2 (r r r 2 r ) 2 r C (1 r r ). 2 (2.34) It is straightforward to prove that Consequently, φ j φ r j 0. lim r 0 r (1 r r ) 2 = 0.

96 Hénon type equations with one-sided exponential growth 95 Now, let us take v r H r k such that v r 2 = 1. Notice that v r = k j=1 c jζ r φ j = ζ r v where v = k j=1 c jφ j H k and v r v = o(1) when r 0. Thus, v r 2 = ( v r 2 v 2 ) + v 2 C k ϱ r + λ k v 2 2 C k ϱ r + λ k [( v 2 2 v r 2 2) + v r 2 2] = c k ϱ r + λ k, as desired. As usual, we must continue by choosing an appropriate decomposition for H. Notice that for r small enough, we can split the space H = H r k H k. Fix r small enough such that we obtain and δ > 0 such that λ λ k + c k ϱ r 1 > 0 (2.35) δ 2 < λ λ k + c k ϱ r 1. (2.36) The next two propositions regard the geometric conditions of the Linking Theorem using this non-orthogonal direct sum. Proposition Suppose that λ k < λ < λ k+1, (g 0 ) (g 2 ) and (2.2) hold. Then there exist ρ, a > 0 such that J λ (u) a if u H k with u = ρ. Proof: We can now proceed analogously to the proof of Proposition For the following proposition, let us remark that, since suppz r n suppv = B r 2(x 0 ) = 0 for all v H r k. (2.37) This fact is easily seen if we observe Figures 3.1 and 3.2. Thus, one has z r n (H r k ). Proposition Suppose that λ k < λ < λ k+1, (g 0 ) (g 2 ) and (2.2) hold. For each n large enough, there exists R n = R(n) > 0 such that J λ (u) 0 for all u Q δ r, where Q δ r := {v + sδz r n : v H r k, v R n and 0 s R n }. Proof: Fix R 0 > ρ and N N such that for all n > N, we have z r n(x) 2 ψ,r δr in B r 2 /n(x 0 ) for all R > R 0. (2.38)

97 Hénon type equations with one-sided exponential growth 96 Let us take R > R 0 and as in Proposition 2.5.4, we split Q δ r as follows Q 1 = {v Hk r : v R}; Q 2 = {v + sδzn r : v Hk r, v = R and 0 s R}; Q 3 = {v + Rδzn r : v Hk r, v R}. If v Q 1, from Lemma 2.6.1, we obtain J λ (v) 1 ( ) λ 1 v 2 for all v H 2 λ k + c k ϱ k. r r Since λ > λ k and due to the choice of r in (2.35), we can take C 1 > 0 such that J λ (v) C 1 v 2 0 (2.39) for all v Hk r independently of R > 0. Using (2.37), we observe that J λ (v + sδzn) r = J λ (v) + J λ (sδzn) r for all v Hk. r (2.40) For Q 2, using (2.33), and the choice of δ in (2.36), we get J λ (v + sδzn) r = J λ (v) + J λ (Rδzn) r 1 ) λ (1 2 R2 + δ 2 < 0, λ k + c k ϱ r independently of R > 0. For Q 3, using (2.27), (2.39) and (2.40), we obtain J λ (v + Rδ z 2 ) J λ (v) + J λ (sδ z n ) C 1 v 2 + R2 2 δ2 x α G(R z r n + ψ) + dx δ2 R 2 2 R σ C σ B r 2 /n (x 0 ) ( x α δzn r ψ ) σ,n R + dx D σ. Let us choose n sufficiently large that δ z n = δ 2π log 1/2 n 2 ψ,n R 0 for all x B r 2 /n(x 0 ).

98 Hénon type equations with one-sided exponential growth 97 Taking R R 0, we can estimate the last integral as it follows ( x α δzn r + ψ ) σ ( dx x α δzn r ψ ) σ,n dx R B r 2 /n (x 0 ) R + ( ψ,n R ) σ x α dx B r 2 /n (x 0 ) ( ψ,n R ) σ 2πr 2(2+α) = τ(n) > 0. (2 + α)n2+α Thus, we obtain J λ (v + δ z) R2 δ 2 2 Since σ > 2, this complete the proof. C σ R σ τ D σ Estimate of minimax levels For the Mountain Pass case, we define the minimax level of J λ by c = c(n) = inf max J λ(w) (2.41) υ Γ w υ([0,1]) where Γ = {υ C([0, 1], H) : υ(0) = 0 and υ(1) = R n z r n}, R n being such that J λ (R n z r n) 0 as in Proposition Proposition Let c(n) be given as in (2.41). Then there exists n large enough such that c(n) < 2π β 0. Proof: We claim that there exists n such that max t 0 J λ(tz r n) < 2π β 0. (2.42) Let us fix some constants that we shall use in this proof. We can assume, without loss of generality, that there exist C 0 > 0 ε 0 log(h(s)) s C 0 (2.43) for all s large enough, where ε 0 is given in (2.30).

99 Hénon type equations with one-sided exponential growth 98 Indeed, note that if h(s) satisfies (2.5), there exists h(s) such that h(s) h(s) for all s large enough and 0 < lim inf s + log( h(s)) s lim sup s + log( h(s)) s < +. By (2.31), we have ψ,r ε 0 2β 0. (2.44) Finally, we consider γ given by (2.4) such that ( ) γ > 22+α C 2 exp 0. (2.45) r 4 β 0 4β 0 Now suppose by contradiction that (2.42) is not true. So, for all n, this maximum is larger than or equal to 2π/β 0 (it is indeed a maximum, in view of Proposition 2.6.2). Let t n > 0 be such that Then and, consequently, from (2.33), Let us prove that t 2 n 4π/β 0. From (2.46) we get d dt (J λ(tzn)) r = 0. t=tn So we have t 2 n B r 2 /n (x 0 ) J λ (t n z r n) = max t 0 J λ(tz r n). (2.46) J λ (t n z r n) 2π β 0 for all n N, (2.47) x α g(t n zn r + ψ) + t n zn r dx t 2 n 4π β 0 for all n N. (2.48) B r 2 /n (x 0 ) x α g(t n z r n + ψ) + (t n z r n + ψ) + dx. Then, (2.4) and (2.31) imply that there exists s 0 large enough such that t 2 n 1 ( ) ( ( ) ) 2 2 γ tn h log 1/2 tn n + ψ exp β α 0 log 1/2 n + ψ B r 2 /n (x 0 ) 2π 2π dx, where we have taken n sufficiently large such that ( ) ( ) tn log 1/2 tn n + ψ log 1/2 n ψ,r s 0 2π 2π in B r 2 /n(x 0 ).

100 Hénon type equations with one-sided exponential growth 99 We still have t 2 n 1 ( 2 γ log[h((t n / ) 2 2π) log exp 1/2 n + ψ)] α B r 2 /n (x 0 ) 2 β 0 ((t n / + 2π) log 1/2 n + ψ) ( +β 0 (t n / 2π) log 1/2 n + ψ + log[h((t n/ ) 2 2π) log 1/2 n + ψ)] 2 β 0 ((t n / dx 2π) log 1/2 n + ψ) and taking an even larger n, (2.43) shows that t 2 n 1 ( ) ( ( 2 γ πr4 α n exp C2 0 tn exp β 2 0 log 1/2 n ψ,r + ε ) ) β 0 2π 2β 0 By (2.44), we see that t 2 n 1 ( ) ( ) 2 exp C2 0 γ πr4 α 4β 0 n exp t 2 n β 2 0 2π log n = 1 ( ) (( ) ) 2 exp C2 0 γπr 4 t 2 n exp β α 0 4β 0 2π 2 log n, (2.49) which implies that t n is bounded. Moreover, (2.49) together with (2.48) give t 2 n 4π/β 0. Letting n in (2.49), one gets ( γ 22+α C 2 exp 0 r 4 β 0 which is contrary to the choice of γ in (2.45). This contradiction happens because we are supposing c(n) 2π/β 0, so we conclude this proof. 4β 0 ), Now we can define the minimax level for the Linking case where ĉ = ĉ(n) = inf max ν Γ w Q δ r Γ = {ν C(Q δ r; H) : ν(w) = w if w Q}. J λ (ν(w)) (2.50) Proposition Let ĉ(n) be given as in (2.50). Then there exists n large enough such that ĉ(n) < 2π β 0. Proof: Since we split the support of the functions in H r k of the support of zr n, we have ĉ(n) max{j λ (v + tz r n) : v H r k, v R n and t 0} = max{j λ (v) + J λ (tz r n) : v H r k, v R n and t 0} max{j λ (v) : v H r k and v R n} + max{j λ (tz r n); t 0}

101 Hénon type equations with one-sided exponential growth 100 By (2.39), we see that J λ (v) 0 for all v Hk r. It follows that ĉ(n) max{j λ (tz r n) : t 0}. From now on, we can proceed analogously to the proof of Proposition Proof of Theorem completed Let us take n such that c(n) < 2π/β 0, where c(n) = c(n), if λ < λ 1 or c(n) = ĉ(n), if λ > λ 1. Let us consider (u m ) a (P S) sequence at level c(n). Since it is bounded by Lemma 2.4.1, Then, up to a subsequence, we may assume that u m u weakly in H, u m u strongly in L p ( ) for all p 1 and almost everywhere in. Therefore, we notice that u is a solution for (2.7). Indeed, for each v Cc ( ) we have 0 J λ(u m ), v = u m v dx λ u m v dx x α g(u m + ψ) + v dx. To see this fact we use that u m v dx u v dx, u m v dx uv dx and x α g(u m + ψ) + v dx x α g(u + ψ) + v dx, (this last one because of (2.15), and using Lemma 2.1 in [28] and (g 1 )). Thus, we can easily conclude that J λ (u), v = 0 for all v C c for (2.7). We still need to ensure that u 0. ( ). Consequently u is a weak solution Suppose, instead, that u 0. So, we must have u m 2 0 and, again using (2.15), (g 1 ) and Lemma 2.1 in [28], we have x α G(u m + ψ) + dx x α G(u + ψ) + dx = 0. Moreover, since c(n) = lim m J λ(u m ) = 1 2 lim m u m 2, and c(n) < 2π/β 0, one can find δ > 0 and m 0 such that u m 2 4π β 0 δ for all m m 0.

102 Hénon type equations with one-sided exponential growth 101 Consider a small ε > 0 and p > 1 (sufficiently close to 1) in order to have p(β 0 + ε)(4π/β 0 δ) 4π. By (2.2), one can take C > 0 sufficiently large such that g(s) p e p(β 0+ε)s 2 + C for all s 0. (2.51) From the fact that J λ (u m), u m = ε m 0, we can see that u m 2 λ u m x α g(u m + ψ) + u m dx + ε m. We need to estimate the integral on the right of this last inequality. Hölder s inequality and (2.51) give x α g(u m + ψ) + u m dx ( (g(u m + ψ) + ) p dx ) 1 p ( B1 ) 1 u m p p dx [ ( exp ( ) ] 1 ) p(β 0 + ε)u 2 p 1 m dx + Cπ p u m p and then we have ( ( ( u m 2 λ u m ε m + exp p(β 0 + ε) 4 π ) ( ) ) )1 2 p um δ dx β 0 u m ( ( λ u m ε m + exp 4π ( ) ) 2 um u m dx ) 1 p + Cπ 1 p u m p. + Cπ 1 p u m p Since the last integral in the estimates above is bounded (because of Trudinger-Moser inequality), we get u m 0. Hence u m 0 in H and then J λ (u m ) 0 = c(n). This is impossible since c(n) a > 0 for all n. Thus u 0 is the desired solution. 2.7 Proof of Theorem 2.2.4: Critical case in H 1 0,rad () In this section, we treat the radial case with g having critical growth. In this case, a solution u for Problem (2.7) is in H = H0,rad 1 (), which will force us to change some calculations.

103 Hénon type equations with one-sided exponential growth The geometric conditions If λ < λ 1, we observe that the geometric conditions follow from Proposition and Proposition 2.6.2, replacing zn, r given in (2.32), with z n, given in (2.29). If λ k < λ < λ k+1, we use the same arguments developed in Proposition in order to prove that 0 is a local minimum of J λ in ( (Hl ) H0,rad 1 () ). However, we cannot take the Moser sequence with disjoint support from the eigenfunctions, as we did in Proposition Indeed, since now we work in an environment of radial functions, we need to use the sequence set in (2.29), instead of the one given in (2.32). This replacement brings some difficulties because we lose the advantage of being close to the boundary, where the interference of ψ could be considered negligible. Initially, let l = max{j : Hj H k } and we set T l : H0,rad 1 () ( (Hl ) H0,rad 1 () ) as the orthogonal projection. We consider (2.29) and define w n (x) = T l z n (x). Since ( (Hl ) H0,rad 1 () ) Hk, we have that estimates of [48, Lemma 2] hold (see Lemma 0.7.2). Proposition Suppose that λ k < λ < λ k+1, (g 0 ) (g 2 ) and (2.2) hold. For each n large enough, there exists R n = R(n) > 0 such that if Q := {v Hl : v R n } {sδw n : 0 s R n }, with δ 2 λ λ k 1, (2.52) then J λ (u) 0 for all u Q; moreover, R n when n. Proof: Fix R 0 > ρ and let us take R > R 0. As it is usual, we split Q as follows Q 1 = {v H l : v R}; Q 2 = {v + sδw n : v Hl, v = R and 0 s R}; Q 3 = {v + Rδw n : v Hl, v R}. Recalling that G(s) 0 for all s R and considering v Q 1 we get ) J λ (v) (1 λλk v 2 0, (2.53)

104 Hénon type equations with one-sided exponential growth 103 independently of R. For Q 2, let us take δ satisfying (2.52). By (14), we obtain ) J λ (v + sδw n ) (1 λλk R + δ , independently of R > 0. For Q 3, we take δ > 0 given in (2.52) and v + Rδw n with v R and v H l. Using (2.27), we obtain J λ (v + Rδw n ) 1 + ( ) σ δ2 R 2 R σ x α ( v + ψ ) + δw n dx + D σ. 2 R + Since H l H k has finite dimension, it follows that v C k R for some C k > 0. We can suppose that R 0, previously fixed, satisfies ψ R 0 C k. Then, by (15) and considering R > R 0, we have where C n = J λ (v + Rδw n ) 1 + δ2 R 2 2 C n R σ + πd σ, ( ( (log n) 1/2 2C k + δ 2π B k )) σ (log n) 1/2 + 2π > 0. (2 + α)n2+α Since σ > 2, the result is then proved by taking R n large enough so that δ w n R n > ρ and, thus, we obtain J λ (v + R n δw n ) 1 + δ2 R n 2 C n Rn σ + πd σ 0 and since C n 0 when n, we can see that R n. This concludes the proof Estimate of minimax levels Here we can notice another important difference between the cases studied in H0( 1 ) and in H0,rad 1 (). Since, in the radial case, we need to work with radially symmetric functions whose supports have the center at 0 R 2, we cannot neglect the weight x α as we did in the critical case. Actually, although the estimates are harder to obtain, it turns out to be an advantage because this environment of symmetric functions changes the boundedness of the minimax levels: in the critical case, we had that they were bounded by the constant

105 Hénon type equations with one-sided exponential growth 104 2π/β 0. On the other hand, working in H0,rad 1 (), this boundedness can be obtained by a greater value that depends on β 0 and α. For the Mountain Pass problem, we define the minimax level of J λ by c = c(n) = inf max J λ(w), (2.54) υ Γ w υ([0,1]) where Γ = {υ C([0, 1], H) : υ(0) = 0 and υ(1) = R n z n }, R n being such that J λ (R n z n ) 0. Proposition Let c(n) be given in (2.54). Then there exists n large enough such that c(n) < Proof: We claim that exists n such that max t 0 J λ(tz n ) < (2 + α)π β 0. (2.55) (2 + α)π β 0. (2.56) First of all, let us fix some constants that we shall use in this proof. Analogously as we have done in Proposition 2.6.5, we can assume, without loss of generality, that for all s large enough. K 0 log(h(s)) s By (2.6), K 0 can be taken large enough in order to have C 0 (2.57) ψ K 0 2β 0. (2.58) Finally, consider γ such that γ > (2 + α)2 β 0 ( ) C 2 exp 0. (2.59) 4β 0 We suppose by contradiction that (2.56) is not true. So, for all n, this maximum is large or equal to (2 + α)π/β 0. Let t n > 0 be such that J λ (t n z n ) = max t 0 J λ(tz n ). (2.60) Then J λ (t n z n ) (2 + α)π β 0 for all n N,

106 Hénon type equations with one-sided exponential growth 105 and, consequently, Thus, t 2 n Let us prove that t 2 n 2(2 + α)π/β 0. From (2.60) we get d dt (J λ(tz n )) = 0. t=tn 2(α + 2)π β 0 for all n N. (2.61) t n z n 2 λt z n 2 2 x α g((t n z n + ψ) + )z n dx = 0. Multiplying this last equation by t n and since z n = 1, λ 0, ψ < 0 and g 0, for n large enough, we have t 2 n x α g((t n z n + ψ) + )t n z n dx /n x α g((t n z n + ψ) + )(t n z n + ψ) + dx. /n Since t n 2(2 + α)π/β 0 > 0, we can take n large enough that (t n ( 2π) 1 log 1/2 n ψ ) c λ, where c λ is given in (2.4). Since in /n and so (2.4) implies that (t n z n + ψ) + = (t n z n + ψ) t n ( 2π) 1 log 1/2 n ψ ( ) t n γ x α tn h log 1/2 n ψ exp /n 2π ( β 0 ( tn 2π log 1/2 n ψ ) 2 ) We have ( t 2 n γ x α log[h((t n / ) 2 2π) log exp 1/2 n ψ )] /n 2 β 0 ((t n / 2π) log 1/2 n ψ ) ( +β 0 (t n / 2π) log 1/2 n ψ + log[h((t n/ ) 2 2π) log 1/2 n ψ )] 2β 0 ((t m / dx. 2π) log 1/2 n ψ ) But if n is large, (2.57) shows that ( log[h((t n / ) 2 2π) log 1/2 n ψ )] 2 β 0 ((t n / C2 0 2π) log 1/2 n ψ ) 4β 0 and log[h((t n / 2π) log 1/2 n ψ )] 2β 0 ((t m / 2π) log 1/2 n ψ ) K 0 2β 0, from which we obtain ( ) ( ( t 2 2π n γ exp C2 0 tn exp β (2 + α)n2+α 0 log 1/2 n ψ + K ) ) β 0 2π 2β 0 dx.

107 Hénon type equations with one-sided exponential growth 106 By (2.58), we see that ( ) t 2 n exp C2 0 γ 4β 0 ( ) = exp C2 0 γ 4β 0 2π (2 + α)n 2π (2 + α) exp ( exp 2+α β 0 t n 2π log n ) ( ) t n β 0 (2 + α) log n, 2π (2.62) which implies that t n is bounded. Moreover, (2.62) together with (2.61) gives that t 2 n 2(2 + α)π/β 0. Now, letting n in (2.62), one gets γ (2 + α)2 β 0 ( ) C 2 exp 0. 4β 0 It contradicts (2.59), then we have that (2.55) holds. Remark This last proposition explains why we had to assume (2.6) instead of (2.5). In the H0( 1 ) environment, we could control the L norm of ψ by moving the supports of our Moser sequence far away from 0 and close enough to so that ψ would be sufficiently small and then it would not interfere in the estimates. Since this could not be done in the radial case, the interference was avoided by using (2.58). Remark We also point out the different choices we have made for the constant γ in (2.45) and (2.59), which explains, in part, the role of x α in the radial case. Notice that γ in (2.45) must be greater than the one given in (2.59). Now, for the linking problem, we can define the minimax level where Γ = {ν C(Q, H) : ν(w) = w if w Q}. č = č(n) = inf max J λ(ν(w)) (2.63) ν Γ w ν(q) We remark that since we did not separate the supports of the eigenfunctions from the Moser sequence, the estimates done in Proposition will not work in this radial case. Therefore, we need do handle more delicate arguments in order to achieve analogous results. This is done in the following proposition. Proposition Let č(n) be given as in (2.63). Then there exists n large enough such that č(n) < (2 + α)π β 0.

108 Hénon type equations with one-sided exponential growth 107 Proof: Suppose by contradiction that for all n we have č(n) (2 + α)π/β 0. We notice that č(n) max{j λ (v + tw n ) : v H l with v R n, t 0} and it follows that for each n there exist v n H l and t n > 0 such that J λ (v n + t n w n ) = max{j λ (v + tw n ) : v H l with v R n, t 0}. (2.64) Therefore, we have J λ (v n + t n w n ) So, since w n ( (H l ) H 1 0,rad () ), from (2.53) we obtain (2 + α)π β 0 for all n N. (2.65) t 2 n 2(2 + α)π β 0 for all n N. (2.66) Let us assume the following claims whose proofs we give later: Claim 5. (v n ) and (t n ) are bounded sequences. Claim 6. t 2 n 2(2 + α)π β 0 in R and v n 0 in H. Since v n Hl, in view of Claim 6, we also get v n 0. However, we have v n + t n w n uniformly in /n. As we also have done in Proposition 2.7.2, let us observe that we can assume, without loss of generality, that K 0 log(h(s)) s C 0 (2.67) for all s large enough. From (2.6), K 0 can be taken large enough in order to have and for n large enough. Finally, consider γ such that γ > (2 + α)2 β 0 ψ K 0 4β 0 v n K 0 4β 0 ( ) 4(2 + α)πbk exp + C2 0. (2.68) 2π 4β 0

109 Hénon type equations with one-sided exponential growth 108 From (2.64), using the fact that derivative of J λ, restricted to Hl Rw n is zero at v n +t n w n, we obtain v n + t n w n 2 λ v n + t n w n 2 2 x α g(v n + t n w n + ψ) + (v n + t n w n ) dx = 0, (2.69) and we can see, for n large enough, we have t 2 n x α g((v n + t n w n + ψ) + )(v n + t n w n + ψ) + dx /n Since t n 2(2 + α)π/β 0 > 0 and v n 0, by (15) we can take n large enough that (t n w n ( v + ψ ) c λ, where c λ is given in (2.4). Thus, we obtain t 2 n γ x α h (t n w n ( v + ψ ) exp ( β 0 (t n w n ( v + ψ ) 2) dx, /n consequently, [ t 2 n γ x α exp /n ( ) 2 log[h(tn w n ( v + ψ )] 2 β 0 (t n w n ( v + ψ ) +β 0 ( t n w n ( v + ψ + log[h(t nw n ( v + ψ )] 2β 0 (t n w n ( v + ψ ) ) 2 ] dx. If n is large, (2.67) shows that and ( ) 2 log[h(tn w n ( v + ψ )] 2 C2 0 β 0 (t n w n ( v + ψ ) 4β 0 log[h(t n w n ( v + ψ )] 2β 0 (t n w n ( v + ψ ) K 0 2β 0, from which we obtain ( ) ( t 2 n γ exp C2 0 x α exp (β 0 t n w n ( v n + ψ ) + K ) ) 2 0 4β 0 /n 2β 0 γ exp ( C2 0 4 ) /n x α exp(β 0 t 2 nw 2 n) dx. dx

110 Hénon type equations with one-sided exponential growth 109 Consequently, from (15) and (2.66) we can see that ( ) t 2 n γ exp C2 0 x α exp[2(2 + α)πw 4 n] 2 dx /n ( ) = γ exp C2 0 4 = γe [ ( 2π (log n) 1/2 ) exp 2(2 + α)π (2 + α)n2+α 2π ( ) C2 0 2π 4 (2+α)n 2+α )+[(2+α) log n]+( 4(2+α)πB k/ 2π)+[2(2+α)πBk 2/ log n]. Now, letting n in the inequality above, we can conclude that γ (2 + α)2 β 0 which is contrary to the choice of γ in (2.68). ( ) 4(2 + α)πbk exp + C2 0, 2π 4β 0 ) 2 ] B k (log n) 1/2 This contradiction follows from the assumption č(n) (2 + α)π/β 0 for all n N, which concludes the proof. Proof of Claim 5: It is sufficient to prove that all subsequences of (t n ) and (v n ) have bounded subsequences. Let us suppose that this is not true. So, we can find subsequences, which by convenience we still denote by (t n ) and (v n ), respectively, such that all of their subsequences are unbounded. That means, we can assume that t nk + v nk for all subsequences (n k ). (2.70) Therefore, one of the following two possibilities has to hold: (i) either there exists a constant C 0 > 0 such that t n / v n C 0, or (ii) there are subsequences such that t n / v n 0. Assume that (i) holds and using (2.70), we have that t n. Now we can see from (2.4) that t 2 n x α g(v n + t n w n + ψ) + (v n + t n w n + ψ) + dx /n γ x α h(v n + t n w n + ψ) exp ( β 0 (v n + t n w n + ψ) 2) dx /n for n large enough and we can see from (2.6) that h(s) C (2.71)

111 Hénon type equations with one-sided exponential growth 110 for all s large enough. So we have We notice that since H l t 2 n γ C x α exp ( β 0 (v n + t n w n + ψ) 2) dx. /n has finite dimension, we have that v n /t n is bounded for all x /n. We also know that ψ /t n is bounded as well. These facts together with (15), give us v n (x) + t n w n (x) + ψ(x) ( = t n w n (x) 1 + v ) n(x) + ψ(x) 1 t n w n t n 2π ((log n) 1/2 2πBk (log n) 1/2 ) ( 1 C ( (log n) 1/2 2π ) 1 ) B k (log n) 1/2 t n 2 1 2π ((log n) 1/2 2πBk (log n) 1/2 and taking n such that (log n) 1/2 2πB k /(log n) 1/2 (1/2)(log n) 1/2, we obtain ) v n (x) + t n w n (x) + ψ(x) t n 4 (log n)1/2 2π and by (2.71) it follows that ( ( t 2 n γ C x α tn exp β 0 /n 4 2π (log n)1/2 ) 2 ) dx = γ (( 2π C 2 + α exp β 0 t 2 n 32π ) ) (2 + α) log n. Consequently, t n must be bounded in case (i), which contradicts t n in case (i). So (ii) occurs. Since lim n t n / v n = 0, by (2.70) we conclude that v n. By (2.69), we get t n w n + v n 2 x α g(t n w n + v n + ψ) + (t n w n + v n ) + dx. Using (2.4) and (2.71), for n large enough, we have t n w n + v n 2 γ C x α e β 0(t nw n+v n+ψ) 2 dx. {t nw n+v n+ψ c γ}

112 Hénon type equations with one-sided exponential growth 111 Since we are supposing (ii), it follows that 1 Cγ C γ 2 {t nw n+v n+ψ c γ} {t nw n+v n+ψ c γ} x α eβ 0(t nw n+v n+ψ) 2 t n w n + v n 2 x α eβ 0(t nw n+v n+ψ) 2 v n 2 dx dx. (2.72) On the other hand, we notice that t n w n + v n + ψ χ {tnw v n n+v n+ψ c γ} = v n v n + t n Hence, we can see that v n w n t nw n + v n χ {tnw v n n+v n+ψ c γ} + ψ v n χ {t nw n+v n+ψ c γ}. t n w n + v n (x) + ψ χ {tnw v n (x) n+v n+ψ c γ}(x) v a. e. in H0( 1 ), where v H k, with v n / v n v and v = 1. So using Fatou s Lemma in (2.72) and since we have assumed that v n, we reach a contradiction. So v n is bounded and, consequently, t n is also bounded. Proof of Claim 6: First, we notice for some appropriated subsequences we have v n v 0 in H and t n t 0 and since z n 0 we get w n 0 and w n 0 for all x. Then it follows Moreover, in view of (2.69) we see that v n + t n w n v 0 almost everywhere in. (2.73) x α g(v n + t n w n ψ) + (v n + t n w n ψ) dx v n + t n w n 2 C. (2.74) However, using [28, Lemma 2.1] and recalling (g 1 ), (2.73) and (2.74), we have x α G(v n + t n w n + ψ) + dx x α G(v 0 + ψ) + dx. (2.75) From (2.65) and (2.75) we can see that J λ (v 0 ) + t2 0 2 (2 + α)π β 0. (2.76) and since v 0 H l, in view of J λ(v 0 ) 0 we have t 2 0 2(2 + α)π β 0.

113 Hénon type equations with one-sided exponential growth 112 Now we prove that t 2 0 = (2(2 + α)π)/β 0. Let us suppose that this is not true. We have t 2 0 > (2(2 + α)π)/β 0. Thus we can take small enough ε > 0 so that for all large n. We consider t 2 2(2 + α)π n > (1 + ε) β 0 v n (x) + ψ(x) ε n = sup, /n t n w n (x) clearly we see that ε n 0, which, together with (2.71), yields C γ C x α exp[β 0 (v n + ψ + t n w n ) 2 ] dx /n γ C x α exp[β 0 ( ε n t n w n + t n w n ) 2 ] dx. /n Using (2.4) and (2.74) and n large enough, we see that ( [ ] C γ C 2π (log n) exp β (2 + α)n2+α 0 (1 ε n ) 2 t 2 1/2 2 ) B n k 2π (log n) 1/2 = γ C 2π (2 + α)e We notice that (2+α) log n exp ( [ B β 0 (1 ε n ) 2 t 2 2 k n log n 2 B ]) ( ) k (1 ε n ) 2 t 2 n log n exp β 0. 2π 2π ( [ B exp β 0 (1 ε n ) 2 t 2 2 k n log n 2 B ]) k > C 1 2π for n large enough and some C 1 > 0, due to the facts t 2 n > 2π(2 + α)/β 0 and ε n 0. Thus, using t 2 n > (1 + ε)2π(2 + α)/β 0, we have ( ) 2π C C 1 γ C (2 + α) exp β 0 (1 ε n ) 2 t 2 log n n (2 + α) log n 2π C 1 γ C 2π (2 + α) exp ( (2 + α) log n[(1 ε n ) 2 (1 + ε) 1] ), which is a contradiction. Consequently, we must have t 2 0 = 2π(2 + α)/β 0 as desired. So by (2.76) we get J λ (v 0 ) 0. But we know that v 0 H l, so by (2.53), we have J λ(v 0 ) = 0. Now we must show that if v 0 H l proof of Claim 6. and J λ (v 0 ) = 0, then v 0 = 0 and we finish the

114 Hénon type equations with one-sided exponential growth 113 Consider v 0 Hl, then 0 = J λ (v 0 ) = 1 2 v 0 2 λ 2 v x α G(v 0 + ψ) + dx 1 2 ) (1 λλk v 0 2 x α G(v 0 + ψ) + dx x α G(v 0 + ψ) + dx. Since G 0, we can see that x α G(v 0 + ψ) + dx = 0. Thus 0 = J λ (v 0 ) = 1 2 v 0 2 λ 2 v (1 λλk ) v 0 2. So we can see which gives us v 0 = 0. ) 1 (1 λλk v Proof of Theorem completed Let us take n such that the minimax level is below (2 + α)π/β 0. Consider that (u m ) is a (P S) c sequence that it is bounded by Lemma 2.4.1, so there exists a subsequence of (u m ) and u H such that u m u weakly in H, u m u in L p rad () for all p 1 and almost everywhere in. In the same way of the Theorem 2.2.3, we conclude that u is a solution for (2.7) and u 0.

115 Chapter 3 Hénon type equation with jumping nonlinearity in critical growth range In this Chapter, we search for two nontrivial radially symmetric solutions of the Dirichlet problem involving a Hénon-type equation of the form u = λu + x α k(u + ) + f(x) in, u = 0 on, (P3) where λ > 0, α 0, is a unity ball centered at the origin of R N (N 3), k(s) = s 2 α 1 + g(s) with 2 α = 2(N + α)/(n 2) and g(s) is a C 1 function in [0, + ) which is assumed to be in the subcritical growth range. The proofs are based on variational methods and to ensure that the considered minimax levels lie in a suitable range, special classes of approximating functions which have disjoint support with Talenti functions (Hénon version) are constructed. We notice that Problem (P3) is naturally a Sobolev version of Problem (P2). However, here we only seek radially symmetric solutions because for obtaining solutions in H 1 0( ), we would apply arguments very close to those used by CALANCH and RUF in [15]. As in Chapter 3, we begin studying a linear problem related to our equation. This is an important step to guarantee a first solution for the problem, which will be denoted by ψ and is nonpositive. This case is almost identical to the previous results in [15, 29], but we also give a little remark concerning existence of a radially symmetric solution when the forcing term is also radial. After that, we notice that to find a second solution of 114

116 Hénon type equation with jumping nonlinearity in critical growth range 115 (P3), we could set u = v + ψ, then v is a solution for v = λv + x α (v + ψ) 2 α x α g(v + ψ) + in ; v = 0 on. (3.1) Obviously, v = 0 is a solution of (3.1) and this gives back the negative solution ψ for (P3). In [29], the authors searched for critical points of the associated functional of a similar problem to (3.1). This translation has the advantage of working with a homogeneous problem, but the interference of ψ in the estimates of minimax level was a great setback. Here we use a different technique. In the same spirit of [32], we separate the supports of the negative solution and the function given in (3.7) in order to make some estimates easier to handle. This is done by cutting a small hole into the function ψ. Initially, we consider B r as a ball of radius r and center in origin. Let us take η r C 1 ( ), given by Define the functions 0 in B r 2; x r r η r (x) = 2 r r r in B r 2 r r \ B r 2; 1 in \ B r. ψ r (x) = η r (x)ψ(x) in (3.2) and set f r (x) = ψ r (x) λψ r (x). (3.3) We can notice that the supports of ψ r and f r are \ B 2 r. See the Figure 4.1. Figure 3.1: Supports of ψ r and f r. Using as before u = v+ψ r in (P3), we see that v = ψ ψ r solves the following equation v = λv + x α (v + ψ r ) 2 α x α g(v + ψ r ) + + (f f r ) in ; (3.4) v = 0 on.